CBSE Notes For Class 6 Social Science Chapterwise

Geography

  • Chapter 1 The Earth In The Solar System Notes
  • Chapter 2 Globe: Latitudes and Longitudes Notes
  • Chapter 3 Motions of The Earth Notes
  • Chapter 4 Maps
  • Chapter 5 Major Domains of the Earth Notes
  • Chapter 6 Our Country- India Notes

Civics

  • Chapter 1 Understanding Diversity Notes
  • Chapter 2 Diversity and Discrimination Notes
  • Chapter 3 What is Government ? Notes
  • Chapter 4 Panchayati Raj Notes
  • Chapter 5 Rural Administration Notes
  • Chapter 6 Urban Administration Notes
  • Chapter 7 Rural Livelihoods Notes
  • Chapter 8 Urban Livelihoods Notes

History

CBSE Notes For Class 6 Science Chapterwise

CBSE Solutions For Class 6 Maths Chapter 4 Basic Geometrical Ideas

Exercise – 4.1

1. Use the figure to name :

Use the figure to name

(1) Five points
(2) Aline
(3) Four rays
(4) Five line segments

Solution: (1) Five points : O, B, C, D, E

(2) A line: DB

(3) Four rays.: OD, OE,OC, OB

(4) Five line segments: DE, OE, OC, OB, OD

2. Name the line given in all possible (twelve) ways, choosing only two letters at a time from the four given.

Name the line

Solution: Possible lines are AB, AC, AD, BC, BD,CD, BA, CA, DA, CB, DB, DC

3. Use the figure to name:

. Use the figure to name

(1) Line containing point E.
(2) Line passing through A.
(3) Line on which O lies
(4) Two pairs of intersecting lines.

Solution: (1) A line containing point £ is AE.

(2) A line passing through A is AE.

(3) A line on which O lies is CO or OC

(4) Two pairs of intersecting lines are AD, CO and AE, FE

4. How many lines can pass through (1) one given point? (2) two given points?

Solution: (1) Infinite number of lines can pass through one given point.

Infinite number of lines can pass through one given point

(2) Only one line can pass through two given points.

Only one line can pass through two give points

5. Draw a rough figure and label suitably in each of the following cases:

(1) Point Plies on AB.
(2) XY and PQ intersect at M.
(3) Line/ contains E and F but not D.
(4) OP and OQ meet at O.

(1). Point Plies on AB

(2). XY and PQ intersect at M

(3).  Line contains E and F but not D

(4). OP and OQ meet at O

6. Consider the following figure of line IVIN. Say whether following statements are true or false in context of the given figure.

Consider the following figure of line MN

(1) Q, M, 0, N, P are points on the line MN.
(2) M, 0, N are points on a line segment MN.
(3) M and N are end points of line segment MN.
(4) O and N are end points of line segment OP.
(5) M is one of the end points of line segment QO.
(6) M is point on ray OP-
(7) Ray OP is different from ray QP.
(8) Ray OP is same as ray OM.
(9) Ray OM is not opposite to ray OP.
(10) 0 is not an initial point of OP.
(11) N is the initial point of NP and NM

Solution: (1) True

(2) True

(3) True

(4) False

(5) False

(6) False

(7) True

(8) False

(9) False

(10) False

(11) True

Exercise – 4.2

1. Classify the following curves as (2) Open or (3) Closed

Classify the following curves as

Solution: (1) Open curve

(2) Closed curve

(3) Open curve

(4) Closed curve

(5) Closed curve

2. Draw rough diagrams to illustrate the following :

(1) Open curve
(2) Closed curve.

Solution: (1) Open curves :

Open curves

(2) Closed curves :

Closed curves

3. Draw any polygon and shade its interior.

Solution:

Draw any polygon and shade its interior

ABCDEF is the required polygon.

4. Consider the given figure and answer the questions :

Consider the given figure

(1) Is it a curve?
(2) Is it closed?

Solution: (1) Yes, it is a curve.

(2) Yes, it is closed.

5. Illustrate, if possible, each one of the following with a rough diagram:

(1) A closed curve that is not a polygon.
(2) An open curve made up entirely of line segments.
(3) A polygon with two sides.

Solution:

(1) A closed curve that is not a polygon

(2) An open curve made up entirely of line segment

(3) Polygon with two sides cannot be drawn.

CBSE Solutions Class 6 Maths Chapter 4 Basic Geometrical Ideas

Exercise – 4.3

1. Name the angles in the given figure.

Name the angles

Solution: There are four angles in the given figure i.e, ∠ABC, ∠CDA, ∠DAB, ∠DCB

2. In the given diagram, name the point(s)

(1) In the interior of ∠DOE
(2) In the exterior of ∠EOF
(3) On /EOF

In the given diagram, name the point(s)

Solution: (1) Point in the interior of ∠DOE : A

(2) Points in the exterior of ∠EOF :C, A, D

(3) Points on ∠EOF : E, O, B, F

3. Draw rough diagrams of two angles such that they have

(1) One point in common.
(2) Two points in common.
(3) Three points in common.
(4) Four points in common.
(5) One ray in common.

Solution:  (1) One point in common

Here, two angles are AOd and BOC and point O is common.

(2)

Two points in common

Here, two angles are ∠AOB and ∠CDE and two points F and G are common.

(3)

Three points in common

Here, two angles are ∠AOB and ∠CDE and three points F, D and G are common.

(4)

four points in common

Here, two angles are ∠AOB and ∠CDE and four points F, G, H and I are common.

(5)

one ray in common

Here, two angles are ∠AOB and ∠AOC and ray OA is common.

CBSE Solutions For Class 6 Maths Chapter 12 Ratio And Proportion

Ratio And Proportion Exercise – 12.1

Question 1. There are 20 girls and 1 5 boys in a class.

  1. What is the ratio of number of girls to the number of boys?
  2. What is the ratio of number of girls to the total number of students in the class?

Solution:

  1. The ratio of the number of girls to that of boys \(=\frac{20}{15}=\frac{4}{3}=4: 3\)
  2. The ratio of the number of girls to the total number of students \(=\frac{20}{20+15}=\frac{20}{35}=\frac{4}{7}=4: 7\)

Question 2. Out of 30 students in a class, 6 like football, 1 2 like cricket and remaining like tennis. Find the ratio of

  1. Several students like football to several students like tennis.
  2. Number of students liking cricket to a total number of students.

Solution:

Total number of students = 30

Number of students who like football = 6

Number of students who like cricket = 12

Thus, the number of students like tennis = 30-6-12 = 12

  1. The ratio of the number of students liking football to that of tennis = \(\frac{6}{12}=\frac{1}{2}=1: 2\)
  2. The ratio of the number of students liking cricket to that of total students \(=\frac{12}{30}=\frac{2}{5}=2: 5\)

Question 3. See the figure and find the ratio of

  1. Number of triangles to the number of circles inside the rectangle.
  2. Number of squares to all the figures inside the rectangle.
  3. Number of circles to all the figures inside the rectangle.

CBSE Solutions For Class 6 Maths Chapter 12

Solution:

  1. The ratio of the number of triangles to that of circles \(=\frac{3}{2}=3: 2\)
  2. The ratio of the number of squares to that of all figures \(=\frac{2}{7}=2: 7\)
  3. The ratio of the number of circles to that of all figures \(=\frac{2}{7}=2: 7\)

Question 4. Distances travelled by Hamid and Akhtar in an hour are 9 km and 12 km. Find the ratio of the speed of Hamid to the speed of Akhtar.
Solution:

We know that, speed \(=\frac{\text { Distance }}{\text { Time }}\)

Speed of Hamid \(=\frac{9 \mathrm{~km}}{1 \mathrm{~h}}=9 \mathrm{~km} / \mathrm{h}\)

Speed Of Akhtar \(=\frac{12 \mathrm{~km}}{1 \mathrm{~h}}=12 \mathrm{~km} / \mathrm{h}\)

The ratio of the speed of Hamid to that of Akhtar \(\)

\(=\frac{9}{12}=\frac{3}{4}=3: 4\)

Question 5. Fill in the following blanks: Ratio and Proportion Number of circles to all the figures inside The Rectangles [Are these equivalent ratios?]
Solution:

To get the first missing number, we consider the fact that 18 = 3×6, i.e., we got 6 when we divided 18 by 3. This indicates that to get the missing number of the second ratio, 15 must also be divided by 3. When we divide, we get 15 + 3 = 5. Hence, the second ratio is \(\frac{5}{6}.\)

Similarly, to get the third ratio, we multiply both terms of the second ratio by 2. Hence, the third ratio is \(\frac{10}{12}.\)

To get the fourth ratio, we multiply both terms of the second ratio by 5. Hence, the fourth ratio is \(\frac{25}{30}.\)

Ratio and Proportion Number of These Are Equivalent Rate

Yes, these are equivalent ratios.

Question 6. Find the ratio of the following :

  1. 81 to 108
  2. 98 to 63
  3. 33 km to 121 km
  4. 30 minutes to 45 minutes

Solution:

1. The ratio of 81 to 108 \(=\frac{81}{108}=\frac{3}{4}=3: 4\)

2. The ratio of 98 to 63\(=\frac{98}{63}=\frac{14}{9}=14: 9\)

3. The ratio of 33 km to 121 km\(=\frac{33}{121}=\frac{3}{11}=3: 11\)

4. The ratio of 30 minutes to 45 minutes\(=\frac{30}{45}=\frac{2}{3}=2: 3\)

Question 7. Find the ratio of the following:

  1. 30 minutes to 1.5 hours
  2. 40 cm to 1.5 m
  3. 55 paise to? 1
  4. 500 ml to 2 litres

Solution:

1. 1.5 hours = 1.5 x 60 minutes = 90 minutes [ 1 hour = 60 minutes]

Now, the ratio of 30 minutes to 1.5 hours = 30 minutes: 1.5 hours

= 30 minutes: 90 minutes \(=\frac{30}{90}=\frac{1}{3}=1: 3\)

2. 1.5 m 1.5 x 100 cm 150 cm

[1 m = 100 cm]

Now, the ratio of 40 cm to 1.5 m = 40 cm: 1.5 m

= 40 cm :150 cm \(=\frac{40}{150}=\frac{4}{15}=4: 15\)

3. ₹ 1 = 100 paise

Now, the ratio of 55 paise to ₹ 1 = 55 paise : ₹ 1 = 55 paise : 100 paise \(=\frac{55}{100}=\frac{11}{20}=11: 20\)

2 litres = 2 x 1000 ml = 2000 ml [1 litre = 1000 ml]

Now, the ratio of 500 ml to 2 litres

=500 ml: 2 litres

⇒ \(500 \mathrm{ml}: 2000 \mathrm{ml}=\frac{500}{2000}=\frac{1}{4}=1: 4\)

CBSE Solutions Class 6 Maths Chapter 12 Ratio And Proportion

Question 8. In a year, Seema earns ₹ 1,50,000 and saves ₹ 50,000. Find the ratio of

The money that Seema earns is the money she saves.

Money that she saves to the money she spends.

Solution:

Total earnings of Seema = ? 1,50,000 and savings =? 50,000

Money spent by her = ? 1,50,000-? 50,000 = ? 1,00,000

The ratio of money earned to the money saved by Seema = \(\frac{150000}{50000}=\frac{3}{1}=3: 1\)

The ratio of money saved to the money spent by Seema = \(\frac{50000}{100000}=\frac{1}{2}=1: 2\)

Question 9. There are 102 teachers in a school of 3300 students. Find the ratio of the number of teachers to the number of students.
Solution:

The ratio of the number of teachers to that of students \(=\frac{102}{3300}=\frac{17}{550}=17: 550\)

Question 10. In a college, out of 4320 students, 2300 are girls. Find the ratio of

  1. Number of girls to the total number of students.
  2. Number of boys to the number of girls.
  3. Number of boys to the total number of students.

Solution:

Total number of students in the college = 4320

Number of girls = 2300

Therefore, number of boys = 4320- 2300

= 2020

1. The ratio of the number of girls to the total number of students = \(\frac{2300}{4320}=\frac{115}{216}=115: 216\)

2. The ratio of the number of boys to that of girls = \(=\frac{2020}{2300}=\frac{101}{115}=101: 115\)

3. The ratio of the number of boys to the total number of students =\(\frac{2020}{4320}=\frac{101}{216}=101: 216\)

Question 11. Out of 1800 students in a school, 750 opted for basketball, 800 opted for cricket and the remaining opted for table tennis. If a student can opt for only one game, find the ratio of

  1. The number of students who opted for basketball to the number of students who opted for table tennis.
  2. Several students opted for cricket and several students opted for basketball.
  3. Number of students who opted for basketball to the total number of students.

Solution:

Total number of students = 1800 Number of students who opted for basketball = 750

Number of students who opted for cricket = 800 Therefore, the number of students who opted for table tennis = 1800- (750 + 800) = 250.

  1. The ratio of the number of students who opted for basketball to that who opted for table tennis = \(\frac{750}{250}=\frac{3}{1}=3: 1\)
  2. The ratio of the number of students who opted for cricket to that who opted for basketball\(=\frac{800}{750}=\frac{16}{15}=16: 15\)
  3. The ratio of the number of students who opted for basketball to the total number of students =\(\frac{750}{1800}=\frac{5}{12}=5: 12\)

Question 12. The cost of a dozen pens Is ₹ 180 and the cost of 8 ball pens Is ₹ 56. Find the ratio of the cost of a pen to the cost of a ball pen.
Solution:

Cost of a dozen pens (12 pens) = ₹ 180

Cost of 1 pen = \(₹ \frac{180}{12}=₹ 15\)

Cost of 8 ball pens = ₹ 56

Cost of1 ball pen = \(₹ \frac{56}{8}= ₹ 7\)

Hence, the ratio of the cost of one pen to that of one ballpen = \(\frac{15}{7}=15: 7\)

Question 13. Consider the statement: The ratio of breadth and length of a hall is 2: 5. Complete the following table that shows some possible breadths and lengths of the hall.

table

Solution:

Ratio of breadth to length of the hall \(=2: 5=\frac{2}{5}\)

Other equivalent ratios are \(=\frac{2}{5} \times \frac{10}{10}=\frac{20}{50}, \frac{2}{5} \times \frac{20}{20}=\frac{40}{100}\)

thus

table

Question 14. Divide 20 pens between Sheela and Sangeeta in the ratio of 3:2.
Solution:

The ratio of dividing pens between Sheela and Sangeeta = 3:2.

The two parts are 3 and 2.

Sum of the parts = 3 + 2 = 5

Therefore, part of Sheela \(=\frac{3}{5} \text { of total pens }\)

⇒ \(=\frac{3}{5} \times 20=12 \text { pens }\)

And part of Sangeeta \(=\frac{2}{5} \text { of total pens }\)

⇒ \(=\frac{2}{5} \times 20=8 \text { pens }\)

Question 15. Mother wants to divide? 36 between her Exercise – 12.2 daughters Shreya and Bhoomika in the ratio of their ages. If the age of Shreya is 15 years and the age of Bhoomika is 12 years, find how much Shreya and Bhoomika will get.
Solution:

The ratio of the age of Shreya to that of Bhoomika = \(\frac{15}{12}=\frac{5}{4}=5: 4\)

Titus, ? 36 will be divided between Shreya and Bhoomika in the ratio of 5: 4.

Shreya will get \(=\frac{5}{9} \text { of } ₹ 36\)

⇒ \(₹ \frac{5}{9} \times 36=₹ 20\)

and Bhoomika will get =\(\frac{4}{9} \text { of ₹ } 36\)

⇒ \(₹ \frac{4}{9} \times 36=₹ 16\)

Question 16. The father’s present age is 42 years and his son’s is 1 4 years. Find the ratio of

  1. Present age of father to the present age of son.
  2. Age of the father to the age of the son, when the son was 12 years old.
  3. Age of father after 1 0 years to the age of son after 1 0 years.
  4. Age of father to the age of son when father was 30 years old

Solution:

The ratio of the father’s present age to that of a son \(=\frac{42}{14}=\frac{3}{1}=3: 1\)

When the son was 12 years old, i.e., 2 years ago, then the father was (42 – 2) = 40 years old.

Therefore, the required ratio of their ages \(=\frac{40}{12}=\frac{10}{3}=10: 3\)

Age of father after 10 years = (42 + 10) years = 52 years

Age of son after 10 year = (14 + 10) years = 24 years

Therefore, the required ratio of their ages \(=\frac{52}{24}=\frac{13}{6}=13: 6\)

When the father was 30 years old, i.e., 12 years ago, then the son was (14 – 12) = 2 years old.

Therefore, the required ratio of their ages \(=\frac{30}{2}=\frac{15}{1}=15: 1\)

Ratio and Proportion Exercise – 12.2

Question 1 . Determine if the following are in proportion.

  1. 15,45,40,120
  2. 33,121,9,96
  3. 24,28,36,48
  4. 32,48,70,210
  5. 4,6,8,12
  6. 33,44,75,100

Solution:

⇒ \(15: 45=\frac{15}{45}=\frac{1}{3}=1: 3\)

⇒ \(40: 120=\frac{40}{120}=\frac{1}{3}=1: 3\)

Since, 15: 45 = 40: 120

Therefore, 15, 45, 40, 120 and are in proportion.

⇒ \(33: 121=\frac{33}{121}=\frac{3}{11}=3: 11\)

⇒ \(9: 96=\frac{9}{96}=\frac{3}{32}=3: 32\)

Since, 33: 121 * 9: 96

Therefore, 33, 121, 9, and 96 are not in proportion.

⇒ \(24: 28=\frac{24}{28}=\frac{6}{7}=6: 7\)

⇒ \(36: 48=\frac{36}{48}=\frac{3}{4}=3: 4\)

Since, 24: 2 36: 48

Therefore, 24, 28, 36, 48 are not in proportion

⇒ \(32: 48=\frac{32}{48}=\frac{2}{3}=2: 3\)

⇒ \(70: 210=\frac{70}{210}=\frac{1}{3}=1: 3\)

Since, 32: 48 70: 210

Therefore, 32, 48, 70, 210 are not in proportion

⇒ \(4: 6=\frac{4}{6}=\frac{2}{3}=2: 3\)

⇒ \(8: 12=\frac{8}{12}=\frac{2}{3}=2: 3\)

Since, 4:6 = 8:12

Therefore, 4, 6, and  8, 12 are in proportion.

⇒ \(33: 44=\frac{33}{44}=\frac{3}{4}=3: 4\)

⇒ \(75: 100=\frac{75}{100}=\frac{3}{4}=3: 4\)

Since, 33 : 44 = 75 : 100

Therefore, 33, 44, 75, 100 are in proportion

Question 2. Write True ( T ) or False ( F ) against each of the following statements :

  1. 16: 24:: 20: 30
  2. 21: 6:: 35: 10
  3. 12: 18:: 28: 12
  4. 8 : 9:: 24: 27
  5. 5 : 2: 3.9:: 3:4
  6. 0.9: 0.36:: 10: 4

Solution:

1. True

Since

⇒ \(16: 24=\frac{16}{24}=\frac{2}{3}\) and \(20: 30=\frac{20}{30}=\frac{2}{3}\)

2. True

Since \(21: 6=\frac{21}{6}=\frac{7}{2}\) and \(35: 10=\frac{35}{10}=\frac{7}{2}\)

3. False

Since \(12: 18=\frac{12}{18}=\frac{2}{3}\) and \(28: 12=\frac{28}{12}=\frac{7}{3}\)

4. True

\(\begin{aligned}
& \text { Since } 8: 9=\frac{8}{9} \\
& \text { and } 24: 27=\frac{24}{27}=\frac{8}{9}
\end{aligned}\)

5. False

\(Since 5.2: 3.9=\frac{5.2}{3.9}=\frac{4}{3} and 3: 4=\frac{3}{4}$\)

6. True

\(Since 0.9: 0.36=\frac{0.9}{0.36}=\frac{5}{2} and 10: 4=\frac{10}{4}=\frac{5}{2}\)

Question 3. Are the following statements true?

  1. 40 persons : 200 persons = ?15:? 75
  2. 7.5 litres: 1 5 litres = 5 kg: 1 0 kg
  3. 9 kg : 45 kg = ?44: ? 20
  4. 32 m :64 m = 6 sec: 12 sec
  5. 45 km : 60 km = 12 hours : 15 hours

Solution:

1. 40 persons : 200 persons \(=\frac{40}{200}=\frac{1}{5}=1: 5\)

⇒ \(₹ 15: ₹ 75=\frac{15}{75}=\frac{1}{5}=1: 5\)

40 persons: 20 persons-? 1 : * 75 Hence, the statement is true.

2. 7.5 litres: 15 litres \(=\frac{7.5}{15}=\frac{75}{150}=\frac{1}{2}=1: 2\)

⇒ \(5 \mathrm{~kg}: 10 \mathrm{~kg}=\frac{5}{10}=\frac{1}{2}=1: 2\)

7.5 litres : 15 litres = 5 kg : 10 kg

Hence, the statement is true.

3. \(99 \mathrm{~kg}: 45 \mathrm{~kg}=\frac{99}{45}=\frac{11}{5}=11: 5\)

⇒ \(₹ 44: ₹ 20=\frac{44}{20}=\frac{11}{5}=11: 5\)

9 kg : 45 kg =  44 : ? 20

Hence, the statement is true

4. \(32 \mathrm{~m}: 64 \mathrm{~m}=\frac{32}{64}=\frac{1}{2}=1: 2\)

⇒ \(6 \mathrm{sec}: 12 \mathrm{sec}=\frac{6}{12}=\frac{1}{2}=1: 2\)

32 m : 64 m = 6 sec : 12 sec

Hence, the statement is true.

45 km : 60 km = \(\frac{45}{60}=\frac{3}{4}=3: 4\)

⇒ \(12 \text { hours : } 15 \text { hours }=\frac{12}{15}=\frac{4}{5}=4: 5\)

45 km: 6 km  12 hours: 1 hour

Hence, the statement is not true.

Question 4. Determine if the following ratios form a proportion. Also, write the middle and extreme terms where the ratios form a proportion.

  1. 25cm:1 m and? 40:? 160
  2. 39 litres: 65 litres and 6 bottles: 1 0 bottles
  3. 2 kg: 8 kg and 25 g: 625 g
  4. 200 ml : 2.5 litre and ? 4: ? 50

Solution:

1. 25 cm : 1 m = 25 cm : (1 X 100) cm

= 25 cm : 100 cm = \(\frac{25}{100}=\frac{1}{4}=1: 4\)

⇒ \(₹ 40: ₹ 160=\frac{40}{160}=\frac{1}{4}=1: 4\)

Since the ratios are equal, therefore these are in proportion. Middle terms are1 m and ? 40 and extreme terms are 25 cm and ? 160

2. 39 litres: 65 litres \(=\frac{39}{65}=\frac{3}{5}=3: 5\)

6 bottles : 10 bottles = \(\frac{6}{10}=\frac{3}{5}=3: 5\)

Since the ratios are equal, therefore these are in proportion.

Middle terms are 65 litres and 6 bottles and extreme terms are 39 litres and 10 bottles

3. 2kg:80kg =\(\frac{2}{80}=\frac{1}{40}=1: 40\)

25g : 625g = \(\frac{25}{625}=\frac{1}{25}=1: 25\)

Since the ratios are not equal, therefore these are not in proportion.

200 ml : 2.5 litres = 200 ml : (2.5 x 1000) ml

= 200 ml : 2500 ml =\(\frac{200}{2500}=\frac{2}{25}=2: 25\)

⇒ \(₹ 4: ₹ 50=\frac{4}{50}=\frac{2}{25}=2: 25\)

Since the ratios are equal, therefore these are in proportion.

Middle terms are 2.5 litres and ? 4 and extreme terms are 200 ml and ? 50.

Ratio And Proportion Exercise – 12.3

Question 1. If the cost of 7 m of cloth? 1470, find the cost of 5 m of cloth.
Solution:

Cost of 7 m of cloth =? 1470

Cost of lm of cloth = \(₹ \frac{1470}{7}=₹ 210\)

Cost of 5 m of cloth =? 210 x 5 = ? 1050

Thus, the cost of 5 m of cloth is? 150

Question 2. Ekta earns? 3000 in 10 days. How much will she earn in 30 days?
Solution: Earning of 10 das =? 3000

Earning of 1 day = \(₹ \frac{3000}{10}=₹ 300\)

Earning of 30 days =? 300 *30 =? 9000

Thus, the earnings for 30 days are 9000.

Question 3. If it has rained 276 mm in the last 3 days, how many cm of rain will fall in one full week (7 days)? Assume that the rain continues to fall at the same rate.
Solution:

Rainfall in 3 days = 276 mm

Rainfall in1 day = \(\frac{276}{3} \mathrm{~mm}=92 \mathrm{~mm}\)

Rainfall in 7 days = 92 x 7 mm = 644 mm

Thus, the rainfall in one full week is 64.4 cm.

Question 4. CostWhat cost5 kg oofheat.50.

  1. What will be the cost of 8 kg of wheat?
  2. What quantity of wheat can be purchased? 183?

Solution:

Cost of 5 kg of wheat =? 91.50

Cost of 1 kg of wheat = \(₹ \frac{91.50}{5}=₹ \frac{9150}{500}\)

= ? 18.30

Cost of 8 kg of wheat =? 18.30 x 8

= ? 146.40

From? 91.50, quantity of wheat can be purchased = 5 kg

From ? 1, quantity of wheat can be purchased = \(\frac{5}{91.50} \mathrm{~kg}\)

From? 83, quantity of wheat can be purchased \(\)

⇒ \(=\frac{5}{91.50} \times 183 \mathrm{~kg}=\frac{5}{9150} \times 18300 \mathrm{~kg}=10 \mathrm{~kg}\)

Question 5. The temperature is degrees Celsius in the last 30 days, and the rate of temperature remains the same, how many degrees will the temperature drop in the next 10 days?
Solution:

Temperature dropped in last 30 days = 15 degrees

Temperature dropped in 1 day \(=\frac{15}{30} \text { degree }=\frac{1}{2} \text { degree }\)

Temperature will drop in next 10 days \(=\frac{1}{2} \times 10 \text { degrees }=5 \text { degrees }\)

Thus, 5 degrees temperature will drop in the next 10 days.

Question 6. ShaiDoes na payspay5000 as rent for 3 months. How much does she have to pay for a whole year, if the rent per month remains the same?
Solution:

Rent paid for 3 months =? 15000

Rent paid for1 month \(=₹ \frac{15000}{3}\)

= ? 5000

Rent paid for 12 months =? 500 x 12

= ? 60,000

Thus, the total rent for the year is? 6,000.

Question 7. What cost of 4 dozen bananas? 180. How many bananas can be purchased? 90?
Solution:

Cost of 4 dozen bananas =? 180

Cost of 48 bananas =? 180

[4 dozen = 4×12 = 48]

From? 180, number of bananas can be purchased = 48

From ? 1, number of bananas can be purchased = \(\frac{48}{180}=\frac{4}{15}\)

From? 90, number of bananas can be purchased = \(\frac{4}{15} \times 90=4 \times 6=24\)

Thus, 24 bananas can be purchased for X 90.

Question 8. The weight of 72 books is 9 kg. What is the weight of 40 such books?
Solution:

The weight of 72 books = 9 kg

The weight of 1 book = \(\frac{9}{72} \mathrm{~kg}=\frac{1}{8} \mathrm{~kg}\)

The weight of 40 books = \(\frac{1}{8} \times 40 \mathrm{~kg}=5 \mathrm{~kg}\)

Thus, the weight of 40 books is 5 kg.

Question 9. A truck requires 08 litres of diesel to cover a distance of 594 km. How much diesel will be required by the truck to cover a distance of 1650 km?
Solution:

For covering 594 km, required diesel = 108 litres

For covering1 km, required diesel \(=\frac{108}{594} \text { litres }=\frac{2}{11} \text { litres }\)

For covering 1650 km, required diesel \(=\frac{2}{11} \times 1650 \text { litres }=300 \text { litres }\)

Thus, 300 litres of diesel will be required by the truck to cover a distance of 1650 km.

Question 10. Raju purchases 10 pens for X 150 and Manish buys 7 pens for X 84. Can you say who got the pens cheaper?
Solution:

Cost of 10 pens for Raju = X 150

Cost of1 pen for Raju = \(₹ \frac{150}{10}=₹ 15\)

Cost of 7 pens for Manish = X 84

Cost of1 pen for Manish = \(₹ \frac{84}{7}=₹ 12\)

Question 11. Anish made 42 runs in 6 overs and Anup made 63 runs in 7 overs. Who made more runs per over?
Solution: Runs made by Anish in 6 overs = 42

Runs made by Anish in1 over = \(\frac{42}{6}=7\)

Runs made by Anup in 7 overs = 63

Runs made by Anup in1 over \(=\frac{63}{7}=9\)

Thus, Anup made more runs per over.

CBSE Solutions For Class 6 Maths Chapter 11 Algebra

Algebra Exercise – 11.1

Question 1. Find the rule which gives the number of matchsticks required to make the following matchstick patterns. Use a variable to write the rule.

Algebra matchstick patterns

Solution:

1.  Number of matchsticks required to make oneAlgebra T =2

∴ Number of matchsticks required to make a pattern of letter T as Algebra T =2n

2.  Number of matchsticks required to make one Algebra Z = 3

∴ Number of matchsticks required to make a pattern of letter Z as Algebra Z = 3n

3.  Number of matchsticks required to make one Algebra U = 3

∴ The number of matchsticks required to make a pattern of the letter U is Algebra U = 3n

4.  Number of matchsticks required to make one Algebra V= 2

∴ Number of matchsticks required to make a pattern of letter V as Algebra V = 2n

5.  Number of matchsticks required to make one Algebra E = 5

∴ Number of matchsticks required to make a pattern of letter E as Algebra E = 5n

6.  Number of matchsticks required to make one Algebra S = 5

∴ The number of matchsticks required to make a pattern of the letter S is Algebra S = 5n

7. Number of matchsticks required to make oneAlgebra A = 6

∴ The number of matchsticks required to make a pattern of letter A is Algebra A= 6n

Question 2. We already know the rule for the pattern of letters L, C and F. Some of the letters from Q.1 (given above) give us the same rule as that given by L. Which are these? Why does this happen?
Solution:

Part (a) & (d) i.e., letters T and V have the same rule as that given by L. Because the number of matchsticks required in each of them is 2.

Question 3. Cadets are marching in a parade. There are 5 cadets in a row. What is the rule which gives the number of cadets, given the number of rows? (Use n for the number of rows.)
Solution:

Let the number of rows = n

Number of cadets in each row = 5

Therefore, the total number of cadets = 5n

Question 4. If there are 50 mangoes in a box, how will you write the total number of mangoes in terms of the number of boxes? (Use b for the number of boxes.)
Solution:

Let the number of boxes = b

Number of mangoes in each box = 50

Therefore, the total number of mangoes = 50b

Question 5. The teacher distributes 5 pencils per student. Can you tell how many pencils are needed, given the number of students? (Use s for the number of students.)
Solution:

Let the number of students = s

Number of pencils distributed to each Student = 5

Therefore, the total number of pencils needed = 5s

Question 6. A bird flies 1 kilometre in one minute. Can you express the distance covered by the bird in terms of its flying time in minutes? (Use t for flying time in minutes.)
Solution:

Let the flying time of the bird be two minutes.

Distance covered by the bird in minute =1 km

∴ Distance covered by the bird in t minutes =1 x t km = t km

Question 7. Radha is drawing a dot Rangoli (a beautiful pattern of lines joining dots) with chalk powder. She has 9 dots in a row. How many dots will her Rangoli have for r rows? How many dots are there if there are 8 rows? If there are 1 0 rows?
Solution:

The number of dots in each row = 9

Number of rows = r

Therefore, the total number of dots = 9r

When there are 8 rows, then the number of dots = 9 x 8 = 72

When there are 10 rows, then the number of dots = 9×10 = 90

CBSE Solutions Class 6 Maths Chapter 11 Algebra

Question 8. Leela is Radha’s younger sister. Leela is 4 years younger than Radha. Can you write Leela’s age in terms of Radha’s age? Take Radha’s age to be x years.
Solution:

Radha’s age = x years

Since Leela is 4 years younger than Radha.

Therefore, Leela’s age = (x- 4) years.

Question 9. Mother has made laddus. She gives some laddus to guests and family members; still, 5 laddus remain. If the number of laddus’s mother gave away is l, how many laddus did she make?
Solution:

Number of laddus gave away = l

Number of remaining laddus = 5

∴ Total number of laddus = (l + 5)

Question 10. Oranges are to be transferred from larger boxes into smaller boxes. When a large box is emptied, the oranges fill two smaller boxes, and 10 oranges remain outside. If the number of oranges in a small box is taken to be x, what is the number of oranges in the larger box?
Solution:

Number of oranges in a small box = x

Number of smaller boxes = 2

Therefore, the total number of oranges in smaller boxes = 2x

Number of remaining oranges = 10

Thus, the number of oranges in the larger box = 2x + 10

Question 11. 1. Look at the following matchstick pattern of squares. The squares are not separate. Two neighbouring squares have a common matchstick.

Observe the patterns and find the rule that gives the number of matchsticks in terms of the number of squares. (Hint: If you remove the vertical stick at the end, you will get a pattern ofCs.)

CBSE Solutions For Class 6 Maths Chapter 11

2. The given figure gives a matchstick pattern of triangles. As in Exercise 11 (a) above, find the general rule that shows the number of matchsticks in terms of the number of triangles.

Algebra The General Rule

Solution:

Algebra Make Multiples

If we remove 1 matchstick from each figure, then they make multiples of 3, i.e., 3, 6, 9, 12, 8,

So the required equation = 3x + 1, where x is number of squares

Algebra Make Multiple

If we remove 1 matchstick from each figure, then they make multiples of 2, i.e., 2, 4, 6,8…

So the required equation = 2x + 1, where x is several triangles.

CBSE Solutions For Class 6 Maths Chapter 7 Fractions

Fractions Exercise – 7.1

Question 1. Write the fraction representing the shaded portion.

Fraction repractaing

Solution:

  1. \(\frac{2}{4}\)
  2. \(\frac{8}{9}\)
  3. \(\frac{4}{8}\)
  4. \(\frac{1}{4}\)
  5. \(\frac{3}{7}\)
  6. \(\frac{3}{12}\)
  7. \(\frac{10}{10}\)
  8. \(\frac{4}{9}\)
  9. \(\frac{4}{8}\)
  10. \(\frac{1}{2}\)

Question 2. Color the part according to the given fraction.

CBSE Solutions For Class 6 Maths Chapter 7

Solution:

Fraction Of according Part

Question 3. Identify the error, if any.

Fraction Error

Solution:

Shaded parts do not represent the given fractions, because all the figures are not equally divided. For making fractions, it is necessary that the figure is to be divided into equal parts.

Question 4. What fraction of a day is 8 hours?
Solution: Since, 1 day = 24 hours

Therefore, the fraction of 8 hours = \(\frac{8}{24}\)

Question 5. What fraction of an hour is 40 minutes?
Solution: Since, 1 hour = 60 minutes.

Therefore, the fraction of 40 minutes = \(\frac{40}{60}\)

CBSE Solutions Class 6 Maths Chapter 7 Fractions

Question 6. Arya, Abhimanyu, and Vivek shared lunch. Arya has brought two sandwiches, one made of vegetables and one of jam. The other two boys forgot to bring their lunch. Arya agreed to share his sandwiches so that everyone would have an equal share of each sandwich.

  1. How can Arya divide his sandwiches so everyone has an equal share?
  2. What part of a sandwich will each boy receive?

Solution:

  1. Arya will divide each sandwich into three equal parts and give one part of each sandwich to each one of them.
  2. Each boy will get part \(\frac{1}{3}\) of a sandwich.

Question 7. Kanchan dyes dresses. She had to dye 30 dresses. She has so far finished 20 dresses. What fraction of dresses has she finished?

Solution: Total number of dresses = 30

Work finished = 20

Fraction of finished work \(=\frac{20}{30}=\frac{2}{3}\)

Question 8. Write the natural numbers from 2 to 1 2. What fraction of them are prime numbers?
Solution:

Natural numbers from 2 to 12 :

2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12

Prime numbers from 2 to 12 :

2,3,5,7,11

Hence, fraction of prime numbers \(=\frac{5}{11}\)

Question 9. Write the natural numbers from 102 to 113. What fraction of them are prime numbers?
Solution:

Natural numbers from 102 to 113 :

102, 103, 104, 105, 106, 107, 108, 109, 110, 111, 112, 113

Prime numbers from 102 to 113 :

103, 107, 109, 113

Hence, fraction of prime numbers \(=\frac{4}{12}\)

Question 10. What fraction of these circles have X’s in them?

Fraction Circle

Solution:

Total number of circles = 8

And the number of circles having ‘X’ = 4

Hence, the fraction \(=\frac{4}{8}\)

Question 11. Kristin received a CD player for her birthday. She bought 3 CDs and received 5 others as gifts. What fraction of her total CDs did she buy and what fraction did she receive as gifts?
Solution:

Total number of CDs = 3 + 5 = 8

Number of CDs purchased = 3

Fraction of CDs purchased \(=\frac{3}{8}\)

Fraction of CDs received as gifts \(=\frac{5}{8}\)

Fractions Exercise – 7.2

Question 1. Draw number lines and locate the points on them:

  1. \(\frac{1}{2}, \frac{1}{4}, \frac{3}{4}, \frac{4}{4}\)
  2. \(\frac{1}{8}, \frac{2}{8}, \frac{3}{8}, \frac{7}{8}\)
  3. \(\frac{2}{5}, \frac{3}{5}, \frac{8}{5}, \frac{4}{5}\)

Solution:

Fraction locate
Fraction locates

Question 2. Express the following as mixed fractions:

  1. \(\frac{20}{3}\)
  2. \(\frac{11}{5}\)
  3. \(\frac{17}{7}\)
  4. \(\frac{28}{5}\)
  5. \(\frac{19}{6}\)
  6. \(\frac{35}{9}\)

Solution:

Fraction Mixed Fraction

∴ \(\frac{20}{3}=6 \frac{2}{3}\)

Fraction Mixed Fraction

∴\(\frac{11}{5}=2 \frac{1}{5}\)

Fraction Mixed Fraction

∴\( \frac{17}{7}=2 \frac{3}{7}\)

Fraction Mixed Fraction

∴\( \frac{28}{5}=5 \frac{3}{5}\)

Fraction Mixed Fraction

∴\( \frac{19}{6}=3 \frac{1}{6}\)

Fraction Mixed Fraction

∴\( \frac{35}{9}=3 \frac{8}{9}\)

Question 3. Express the following as improper fractions:

  1. \(7 \frac{3}{4}\)
  2. \(5 \frac{6}{7}\)
  3. \(2 \frac{5}{6}\)
  4. \(10 \frac{3}{5}\)
  5. \(9 \frac{3}{7}\)
  6. \(8 \frac{4}{9}\)

Solution:

  1. \(7 \frac{3}{4}=\frac{(7 \times 4)+3}{4}=\frac{28+3}{4}=\frac{31}{4}\)
  2. \(5 \frac{6}{7}=\frac{(5 \times 7)+6}{7}=\frac{35+6}{7}=\frac{41}{7}\)
  3. \(2 \frac{5}{6}=\frac{(2 \times 6)+5}{6}=\frac{12+5}{6}=\frac{17}{6}\)
  4. \(10 \frac{3}{5}=\frac{(10 \times 5)+3}{5}=\frac{50+3}{5}=\frac{53}{5}\)
  5. \(9 \frac{3}{7}=\frac{(9 \times 7)+3}{7}=\frac{63+3}{7}=\frac{66}{7}\)
  6. \(8 \frac{4}{9}=\frac{(8 \times 9)+4}{9}=\frac{72+4}{9}=\frac{76}{9}\)

Fractions Exercise – 7.3

Question 1. Write the fractions. Are all these fractions equivalent?

Fractions Equivalent

Solution:

  1. \(\frac{1}{2}, \frac{2}{4}, \frac{3}{6}, \frac{4}{8}\)
    • Yes, all of these fractions are equivalent
  2. \(\frac{4}{12}, \frac{3}{9}, \frac{2}{6}, \frac{1}{3}, \frac{6}{15}\)
    • No, all these fractions are not equivalent.

Question 2. Write the fractions and pair up the equivalent fractions from each row.

Fraction Pair Of Eqivalent

Solution:

A. \(\frac{1}{2}\)

B. \(\frac{4}{6}\)

C. \(\frac{3}{9}\)

D. \(\frac{2}{8}\)

E. \(\frac{3}{4}\)

  1. \(\frac{6}{18}\)
  2. \(\frac{4}{8}\)
  3. \(\frac{12}{16}\)
  4. \(\frac{8}{12}\)
  5. \(\frac{4}{16}\)

Pairs of equivalent fractions are

(A), (2); (B), (4); (C), (1); (D), (5); (E), (3)

Question 3. Replace each of the following with the correct number:

1.

2. Fractions

3.  Fraction

4.  Fraction

5.  Fraction

Solution:

  1. \(\frac{2}{7}=\frac{2 \times 4}{7 \times 4}=\frac{8}{28}\)
  2. \(\frac{5}{8}=\frac{5 \times 2}{8 \times 2}=\frac{10}{16}\)
  3. \(\frac{3}{5}=\frac{3 \times 4}{5 \times 4}=\frac{12}{20}\)
  4. \(\frac{45}{60}=\frac{45+3}{60+3}=\frac{15}{20}\)
  5. \(\frac{18}{24}=\frac{18+6}{24+6}=\frac{3}{4}\)

Question 4. Find the equivalent fraction of \(\frac{3}{5}\) having

  1. Denominator 20
  2. Numerator 9
  3. Denominator 30
  4. Numerator 27

Solution:

  1. \(\frac{3}{5}=\frac{3 \times 4}{5 \times 4}=\frac{12}{20}\)
  2. \(\frac{3}{5}=\frac{3 \times 3}{5 \times 3}=\frac{9}{15}\)
  3. \(\frac{3}{5}=\frac{3 \times 6}{5 \times 6}=\frac{18}{30}\)
  4. \(\frac{3}{5}=\frac{3 \times 9}{5 \times 9}=\frac{27}{45}\)

Question 5. Find the equivalent fraction of \(\frac{36}{48}\) with

  1. Numerator 9
  2. Denominator 4

Solution:

  1. \(\frac{36}{48}=\frac{36/4}{48/4}=\frac{9}{12}\)
  2. \(\frac{36}{48}=\frac{36+12}{48+12}=\frac{3}{4}\)

Question 6. Check whether the given fractions are equivalent.

  1. \(\frac{5}{9}, \frac{30}{54}\)
  2. \(\frac{3}{10}, \frac{12}{50}\)
  3. \(\frac{7}{13}, \frac{5}{11}\)

Solution:

⇒ \(\frac{5}{9}, \frac{30}{54}\)

⇒ \(\frac{5}{9}=\frac{5 \times 6}{9 \times 6}=\frac{30}{54}\)

Therefore \(\frac{5}{9} \text { and } \frac{30}{54}\) are equivalent.

⇒ \(\frac{3}{10}, \frac{12}{50}\)

⇒ \(\frac{3}{10}=\frac{3 \times 5}{10 \times 5}=\frac{15}{50} \neq \frac{12}{50}\)

Therefore \(\frac{3}{10} \text { and } \frac{12}{50}\) are not equivalent.

⇒ \(\frac{7}{13}, \frac{5}{11}\)

⇒ \(\frac{7}{13}=\frac{7 \times 11}{13 \times 11}=\frac{77}{143}, \frac{5}{11}=\frac{5 \times 13}{11 \times 13}=\frac{65}{143}\)

∴ \( \frac{77}{143} \neq \frac{65}{143}\)

Therefore,\(\frac{7}{13} \text { and } \frac{5}{11}\) are not equivalent.

Question 7. Reduce the following fractions to simplest from:

  1. \(\frac{48}{60}\)
  2. \(\frac{150}{60}\)
  3. \(\frac{84}{98}\)
  4. \(\frac{12}{52}\)
  5. \(\frac{7}{28}\)

Solution:

  1. \(\frac{48}{60}=\frac{2 \times 2 \times 2 \times 2 \times 3}{2 \times 2 \times 3 \times 5}=\frac{4}{5}\)
  2. \(\frac{150}{60}=\frac{3 \times 5 \times 10}{2 \times 3 \times 10}=\frac{5}{2}\)
  3. \(\frac{84}{98}=\frac{2 \times 3 \times 14}{7 \times 14}=\frac{6}{7}\)
  4. \(\frac{12}{52}=\frac{2 \times 2 \times 3}{2 \times 2 \times 13}=\frac{3}{13}\)
  5. \(\frac{7}{28}=\frac{7}{2 \times 2 \times 7}=\frac{1}{4}\)

Question 8. Ramesh had 20 pencils, Sheelu had 50 pencils and Jamaal had 80 pencils. After 4 months, Ramesh used up 10 pencils, Sheelu used up 25 pencils and Jamaal used up 40 pencils. What fraction did each use up? Check if each has used up an equal fraction of her or his pencils.
Solution:

Ramesh: Total pencils = 20

Pencils used = 10

Fraction = \(\frac{10}{20}=\frac{1}{2}\)

Sheelu: Total pencils = 50

Pencils used = 25

Fraction = \(\frac{25}{50}=\frac{1}{2}\)

Jamaal : Total pencils = 80

Pencils used = 40

Fraction = \(\frac{40}{80}=\frac{1}{2}\)

Since all of them used half of their pencils, therefore, each has used up an equal fraction of pencils.

Question 9. Match the equivalent fractions and write two more for each.

  1. \(\frac{250}{400}\)
  2. \(\frac{180}{200}\)
  3. \(\frac{660}{990}\)
  4. \(\frac{180}{360}\)
  5. \(\frac{220}{550}\)
  6. \(\text { (a) } \frac{2}{3}\)
  7. \(\frac{2}{5}\)
  8. \(\frac{1}{2}\)
  9. \(\frac{5}{8}\)
  10. \(\frac{9}{10}\)

Solution:

(1) → (d); (2) → (e); (3) → (a); (4) →(c); (5)→(b);

1. \(\frac{250}{400}=\frac{250 \div 50}{400 \div 50}=\frac{5}{8}\)

Also, equivalent fractions of \(\frac{5}{8}\) are

⇒ \(\frac{5}{8}=\frac{5 \times 2}{8 \times 2}=\frac{10}{16}\)

∴ \(\frac{5}{8}=\frac{5 \times 3}{8 \times 3}=\frac{15}{24}\)

2. \(\frac{180}{200}=\frac{180 \div 20}{200 \div 20}=\frac{9}{10}\)

Also,equivalent fractions of \(\frac{9}{10}\) are

⇒ \(\frac{9}{10}=\frac{9 \times 2}{10 \times 2}=\frac{18}{20}\)

∴ \(\frac{9}{10}=\frac{9 \times 3}{10 \times 3}=\frac{27}{30}\)

3. \(\frac{660}{990}=\frac{660 \div 330}{990 \div 330}=\frac{2}{3}\)

Also, equivalent fractions of \(\frac{2}{3}\) are

∴ \(\frac{2 \times 2}{3 \times 2}=\frac{4}{6}, \frac{2 \times 3}{3 \times 3}=\frac{6}{9}\)

4. \(\frac{180}{360}=\frac{180 \div 180}{360 \div 180}=\frac{1}{2}\)

Also,equivalent fractions of \(\frac{1}{2}\) are

⇒ \(\frac{1}{2}=\frac{1 \times 2}{2 \times 2}=\frac{2}{4}\)

∴ \(\frac{1}{2}=\frac{1 \times 3}{2 \times 3}=\frac{3}{6}\)

5. \(\frac{220}{550}=\frac{220 \div 110}{550 \div 110}=\frac{2}{5}\)

Also, equivalent fractions of \(\frac{2}{5}\) are

⇒ \(\frac{2}{5}=\frac{2 \times 2}{5 \times 2}=\frac{4}{10}\)

∴ \(\frac{2}{5}=\frac{2 \times 3}{5 \times 3}=\frac{6}{15}\)

Fractions Exercise – 7.4

Question 1. Write the shaded portion as a fraction. Arrange them in ascending and descending order using the correct sign ‘<‘, ‘=’, ’>’ between the fractions:

Fraction shaded portion as fraction

(3) Show \(\frac{2}{6}, \frac{4}{6}, \frac{8}{6} \text { and } \frac{6}{6}\) on the number line. Put appropriate signs between the fractions given.

Fraction shaded portion as fraction

Solution:

1. \(\frac{3}{8}, \frac{6}{8}, \frac{4}{8}, \frac{1}{8}\)

Ascendung order: \(\frac{1}{8}<\frac{3}{8}<\frac{4}{8}<\frac{6}{8}\)

Descending order: \(\frac{6}{8}>\frac{4}{8}>\frac{3}{8}>\frac{1}{8}\)

2. \(\frac{8}{9}, \frac{4}{9}, \frac{3}{9}, \frac{6}{9}\)

Ascendung order: \(\frac{3}{9}<\frac{4}{9}<\frac{6}{9}<\frac{8}{9}\)

Descending order: \(\frac{8}{9}>\frac{6}{9}>\frac{4}{9}>\frac{3}{9}\)

Number line

Fraction Number Line

Fraction Number Lines

Question 2. Compare the fractions and put an appropriate sign.

1.  Fraction appropriate

2.  Fraction appropriate

3.  Fraction appropriate

4.  Fraction appropriate

Solution:

⇒ \(\frac{3}{6} \text { and } \frac{5}{6}\) are like fractions.

Also, numerator of \(\frac{5}{6}\) is greater than numerator of \(\frac{3}{6}\)

⇒ \(\frac{3}{6}<\frac{5}{6}\)

⇒ \(\frac{1}{7} \text { and } \frac{1}{4}\) are unlike fractions with same numerator. Also, denominator of \(\frac{1}{7}\) greater than denominator of \(\frac{1}{4}\)

⇒ \( \frac{1}{7}<\frac{1}{4}\)

⇒ \(\frac{4}{5} \text { and } \frac{5}{5}\) are unlike fractions with same numerator. Also, denominator of \(\frac{5}{5}\) greater than denominator of \(\frac{4}{5}\)

⇒ \( \frac{4}{5}<\frac{5}{5}\)

⇒ \(\frac{3}{5}\) and \( \frac{3}{7}\) are unlike fractions with same numerator.

Also, denominator of \(\frac{3}{7}\) greater than denominator of \(\frac{3}{5}\)

⇒ \(\frac{3}{5}>\frac{3}{7}\)

Question 3. Make five more such pairs and put appropriate signs.
Solution:

1.   put appropriate A

2.    put appropriate B

3.     put appropriate C

4.put appropriate D

5.put appropriate E

Question 4. Look at the figures and write ‘<‘ or ‘=’ between the given pairs of fractions.

Pairs Of Fraction

1.  Fraction appropriate

2.  Fraction appropriate

3.  Fraction appropriate

4.  Fraction appropriate

5.  Fraction appropriate

Make five more such problems and solve them with your friends. are unlike fractions with
Solution:

fraction five problem

fraction five problem

fraction five problem

fraction five problem

fraction five problem

Five more problems:

1.  Fraction

2.  Five more problem

3. Five more problems

4.  Five more problem

5. Five more problem

The above problems are solved as follows:

Fraction Problem

Question 5. How quickly can you do this? Fill appropriate sign. (‘<”=”>’)

Fraction Quickly

Solution:

⇒ \(\frac{1}{2} \) and \( \frac{1}{5}\) are unlike fractions with same numerator. Also, denominator of \(\frac{1}{5}\) is greater than denominator of \(\frac{1}{2}\)

⇒ \(\frac{1}{2}>\frac{1}{5}\)

⇒ \(\frac{2}{4}=\frac{2 \div 2}{4 \div 2}=\frac{1}{2}\)

⇒ \(\frac{3}{6}=\frac{3 \div 3}{6 \div 3}=\frac{1}{2}\)

⇒ \(\frac{2}{4}=\frac{3}{6}\)

⇒ \(\frac{3}{5} \) and \( \frac{2}{3}\) are unlike fractions with different numerator.

⇒ \(\frac{3}{5}=\frac{3 \times 3}{5 \times 3}=\frac{9}{15}\)

⇒ \(\frac{2}{3}=\frac{2 \times 5}{3 \times 5}=\frac{10}{15}\)

⇒ \(\frac{9}{15}<\frac{10}{15}\)

⇒ \(\frac{3}{5}<\frac{2}{3}\)

⇒ \(\frac{3}{4} \) and \( \frac{2}{8}\) are unlike fractions with different numerators.

⇒ \(\frac{3}{4}=\frac{3 \times 2}{4 \times 2}=\frac{6}{8}\)

⇒ \(\frac{2}{8}=\frac{2}{8}\)

⇒ \(\quad \frac{6}{8}>\frac{2}{8}\)

⇒ \(\frac{3}{4}>\frac{2}{8}\)

⇒ \(\frac{3}{5} \) and \( \frac{6}{5}\) are like fractions. Also,numerator of \(\frac{6}{5}\) is greater thannumerator of \(\frac{3}{5}\)

⇒ \( \frac{3}{5}<\frac{6}{5}\)

⇒ \(\frac{7}{9} \) and \( \frac{3}{9}\) arelike fractions. Also, numerator of \(\frac{7}{9}\) is greater thannumerator of \(\frac{3}{9}\)

⇒ \( \frac{7}{9}>\frac{3}{9}\)

⇒ \(\frac{1}{4} \) and \( \frac{2}{8}\) are unlike fractions with different numerators

⇒ \(\frac{1}{4}=\frac{1 \times 2}{4 \times 2}=\frac{2}{8}\)

⇒ \(\quad \frac{2}{8}=\frac{2}{8}\)

⇒ \(\frac{1}{4}=\frac{2}{8}\)

⇒ \(\frac{6}{10} \) and \( \frac{4}{5}\) are unlike fractions with different numerator

⇒ \(\frac{6}{10}=\frac{6 \div 2}{10 \div 2}=\frac{3}{5}\)

⇒ \(\quad \frac{3}{5}<\frac{4}{5}\)

⇒ \(\frac{6}{10}<\frac{4}{5}\)

⇒ \(\frac{3}{4} \) and \( \frac{7}{8}\) are unlike fractions with different numerators.

⇒ \(\frac{3}{4}=\frac{3 \times 2}{4 \times 2}=\frac{6}{8}\)

⇒ \( \quad \frac{6}{8}<\frac{7}{8}\)

⇒ \(\frac{3}{4}<\frac{7}{8}\)

⇒ \(\frac{6}{10} \) and \( \frac{3}{5}\) are unlike fractions with different numerators.

⇒ \(\frac{6}{10}=\frac{6 \div 2}{10 \div 2}=\frac{3}{5}\)

⇒ \(\quad \frac{3}{5}=\frac{3}{5}\)

⇒ \(\frac{6}{10}=\frac{3}{5}\)

⇒ \(\frac{5}{7} \) and \( \frac{15}{21}\) are unlike fractions with different numerators.

⇒ \(\frac{5}{7}=\frac{5 \times 3}{7 \times 3}=\frac{15}{21}\)

⇒ \(\quad \frac{15}{21}=\frac{15}{21}\)

∴ \(\frac{5}{7}=\frac{15}{21}\)

Question 6. The following fractions represent just three different numbers. Separate them into three groups of equivalent fractions, by changing each one to its simplest form.

  1. \(\frac{2}{12}\)
  2. \(\frac{3}{15}\)
  3. \(\frac{8}{50}\)
  4. \(\frac{16}{100}\)
  5. \(\frac{10}{60}\)
  6. \(\frac{15}{75}\)
  7. \(\frac{12}{60}\)
  8. \(\frac{16}{96}\)
  9. \(\frac{12}{75}\)
  10. \(\frac{12}{72}\)
  11. \(\frac{3}{18}\)
  12. \(\frac{4}{25}\)

Solution:

  1. \(\frac{2}{12}=\frac{2 \div 2}{12 \div 2}=\frac{1}{6}\)
  2. \(\frac{3}{15}=\frac{3 \div 3}{15 \div 3}=\frac{1}{5}\)
  3. \(\frac{8}{50}=\frac{8 \div 2}{50 \div 2}=\frac{4}{25}\)
  4. \(\frac{16}{100}=\frac{16 \div 4}{100 \div 4}=\frac{4}{25}\)
  5. \(\frac{10}{60}=\frac{10 \div 10}{60 \div 10}=\frac{1}{6}\)
  6. \(\frac{15}{75}=\frac{15 \div 15}{75 \div 15}=\frac{1}{5}\)
  7. \(\frac{12}{60}=\frac{12 \div 12}{60 \div 12}=\frac{1}{5}\)
  8. \(\frac{16}{96}=\frac{16 \div 16}{96 \div 16}=\frac{1}{6}\)
  9. \(\frac{12}{75}=\frac{12 \div 3}{75 \div 3}=\frac{4}{25}\)
  10. \(\frac{12}{72}=\frac{12 \div 12}{72 \div 12}=\frac{1}{6}\)
  11. \(\frac{3}{18}=\frac{3 \div 3}{18 \div 3}=\frac{1}{6}\)
  12. \(\frac{4}{25}=\frac{4}{25}\)

Equivalent groups :

1. group :\(\frac{1}{5}[(b),(f),(g)]\)

2. group:\(\frac{1}{6}[(\mathrm{a}),(\mathrm{e}),(\mathrm{h}),(\mathrm{j}),(\mathrm{k})]\)

3.group:\(\frac{4}{25}[(\mathrm{c}),(\mathrm{d}),(\mathrm{i}),(\mathrm{l})]\)

Question 7. Find answers to the following. Write and indicate how you solved them.

Is \(\frac{5}{9} \text { equal to } \frac{4}{5} \text { ? }\)

Is \(\frac{9}{16} \text { equal to } \frac{5}{9} ?\)

Is \(\frac{4}{5} \text { equal to } \frac{16}{20} ?\)

Is \(\frac{1}{15} \text { equal to } \frac{4}{30} \text { ? }\)

Solution:

⇒ \(\frac{5}{9} \text { and } \frac{4}{5} \Rightarrow \frac{5 \times 5}{9 \times 5}=\frac{25}{45} \text { and } \frac{4 \times 9}{5 \times 9}=\frac{36}{45}\)

[L.C.M. of 9 and 5 is 45]

Since, \(\frac{25}{45} \neq \frac{36}{45}\)

Therefore, \(\frac{5}{9} \neq \frac{4}{5}\)

⇒ \(\frac{9}{16} \text { and } \frac{5}{9}\)

⇒ \(\Rightarrow \frac{9 \times 9}{16 \times 9}=\frac{81}{144} \text { and } \frac{5 \times 16}{9 \times 16}=\frac{80}{144}\)

[L.C.M. of 16 and 9 is 144]

Since, \(\frac{81}{144} \neq \frac{80}{144}\)

Therefore, \(\frac{9}{16} \neq \frac{5}{9}\)

⇒ \(\frac{4}{5} \text { and } \frac{16}{20}\)

⇒ \(\frac{4 \times 4}{5 \times 4}=\frac{16}{20} \text { and } \frac{16 \times 1}{20 \times 1}=\frac{16}{20}\)

[L.C.M. of 5 and 20 is 20]

Since, \(\frac{16}{20}=\frac{16}{20}\)

Therefore, \(\frac{4}{5}=\frac{16}{20}\)

⇒ \(\frac{1}{15} \text { and } \frac{4}{30}\)

⇒ \(\frac{1 \times 2}{15 \times 2}=\frac{2}{30} \text { and } \frac{4 \times 1}{30 \times 1}=\frac{4}{30}\)

[L.C.M. of 15 and 30 is 30]

Since, \(\frac{2}{30} \neq \frac{4}{30}\)

∴ \(\frac{1}{15} \neq \frac{4}{30}\)

Question 8. Ila read 25 pages of a book containing 100 pages. Lalita read \(\frac{2}{5}\) of the same book. Who reads less?
Solution:

Ila read 25 pages out of 100 pages. Fraction of reading the pages

⇒ \(=\frac{25}{100}=\frac{1}{4}\) of book

⇒ \(\frac{1}{4} \text { and } \frac{2}{5}\frac{1 \times 5}{4 \times 5}=\frac{5}{20} \text { and } \frac{2 \times 4}{5 \times 4}=\frac{8}{20}\)

[L.C.M. of 5 and 4 is 20]

⇒ \(\frac{5}{20}<\frac{8}{20}\)

∴ \(\quad \frac{1}{4}<\frac{2}{5}\)

Therefore, Ila read less.

Question 9. Rafiq exercised for \(\frac{3}{6}\) of an hour, while Rohit exercised for \(\frac{3}{4}\) of an hour. Who exercised for a longer time?
Solution:

Rafiq exercised \(\frac{3}{6}\) of an hour.

\(\) Rohit exercised \(\frac{3}{4}\) of an hour.

Since, \(\frac{3}{4}>\frac{3}{6}\)

Therefore, Rohit exercised for a longer time.

Question 10. In a class A of 25 students, 20 passed with 60% or more marks; in another class B of 30 students, 24 passed with 60% or more marks. In which class was a greater fraction of students getting 60% or more marks?
Solution:

In class A, 20 passed with 60% or more marks out of 25

Fraction overpassed students \(=\frac{20}{25}=\frac{4}{5}\)

In class B, 24 passed with 60% or more marks out of 30

Fraction of passed students \(=\frac{24}{30}=\frac{4}{5}\)

Hence, both classes have the same fraction of students getting 60% or more marks

Fractions Exercise – 7.5

Question 1. Write these fractions appropriately as additions or subtractions:

Fraction Apporacimate

Solution:

\(+: \frac{1}{5}+\frac{2}{5}=\frac{1+2}{5}=\frac{3}{5}\) \(\text { (b) }-: \frac{5}{5}-\frac{3}{5}=\frac{5-3}{5}=\frac{2}{5}\) \(+: \frac{2}{6}+\frac{3}{6}=\frac{2+3}{6}=\frac{5}{6}\)

Question 2. Solve:

⇒ \(\frac{1}{18}+\frac{1}{18}\)

⇒ \(\frac{8}{15}+\frac{3}{15}\)

⇒ \(\frac{7}{7}-\frac{5}{7}\)

⇒ \(\frac{1}{22}+\frac{21}{22}\)

⇒ \(\frac{12}{15}-\frac{7}{15}\)

⇒ \(\frac{5}{8}+\frac{3}{8}\)

⇒ \(1-\frac{2}{3}\left(1=\frac{3}{3}\right)\)

⇒ \(\frac{1}{4}+\frac{0}{4}\)

∴ \(3-\frac{12}{5}\)

Solution:

⇒ \(\frac{1}{18}+\frac{1}{18}=\frac{1+1}{18}=\frac{2}{18}=\frac{1}{9}\)

⇒ \(\frac{8}{15}+\frac{3}{15}=\frac{8+3}{15}=\frac{11}{15}\)

⇒ \(\frac{7}{7}-\frac{5}{7}=\frac{7-5}{7}=\frac{2}{7}\)

⇒ \(\frac{1}{22}+\frac{21}{22}=\frac{1+21}{22}=\frac{22}{22}=1\)

⇒ \(\frac{12}{15}-\frac{7}{15}=\frac{12-7}{15}=\frac{5}{15}=\frac{1}{3}\)

⇒ \(\frac{5}{8}+\frac{3}{8}=\frac{5+3}{8}=\frac{8}{8}=1\)

⇒ \(1-\frac{2}{3}=\frac{3}{3}-\frac{2}{3}=\frac{3-2}{3}=\frac{1}{3}\)

⇒ \(\frac{1}{4}+\frac{0}{4}=\frac{1+0}{4}=\frac{1}{4}\)

∴ \(3-\frac{12}{5}=\frac{15}{5}-\frac{12}{5}=\frac{15-12}{5}=\frac{3}{5}\)

Question 3. Shubham painted \(\frac{2}{3}\) of the wall space in his room. His sister Madhavi helped and painted \(\frac{1}{3}\) of the wall space. How much did they paint together?
Solution:

Fraction of wall painted by Shubham = \(\frac{2}{3}\)

Fraction of wall painted by Madhavi = \(\frac{1}{3}\)

Total painting by both of them \(=\frac{2}{3}+\frac{1}{3}=\frac{2+1}{3}=\frac{3}{3}=1\)

Therefore, they painted a complete wall.

Question 4. Fill in the missing fractions.

Missing Fractions

Solution:

⇒ \(\frac{7}{10}-\frac{4}{10}=\frac{7-4}{10}=\frac{3}{10}\)

⇒ \(\frac{8}{21}-\frac{3}{21}=\frac{8-3}{21}=\frac{5}{21}\)

⇒ \(\frac{6}{6}-\frac{3}{6}=\frac{6-3}{6}=\frac{3}{6}\)

∴\(\frac{7}{27}+\frac{5}{27}=\frac{7+5}{27}=\frac{12}{27}\)

Question 5. Javed was given \(\frac{5}{7}\) of a basket of oranges. What fraction of oranges was left in the basket?
Solution:

Consider the total number of oranges to be the whole portion or 1.

Fraction of oranges left \(=1-\frac{5}{7}\)

⇒ \(=\frac{7}{7}-\frac{5}{7}=\frac{7-5}{7}=\frac{2}{7}\)

Thus, \(\frac{2}{7}\) oranges was left in the basket.

Fractions Exercise – 7.6

Question 1. Solve:

⇒ \(\frac{2}{3}+\frac{1}{7}\)

⇒ \(\frac{3}{10}+\frac{7}{15}\)

⇒ \(\frac{4}{9}+\frac{2}{7}\)

⇒ \(\frac{5}{7}+\frac{1}{3}\)

⇒ \(\frac{2}{5}+\frac{1}{6}\)

⇒ \(\frac{4}{5}+\frac{2}{3}\)

⇒ \(\frac{3}{4}-\frac{1}{3}\)

⇒ \(\frac{5}{6}-\frac{1}{3}\)

⇒ \(\frac{2}{3}+\frac{3}{4}+\frac{1}{2}\)

⇒ \(\frac{1}{2}+\frac{1}{3}+\frac{1}{6}\)

⇒ \(1 \frac{1}{3}+3 \frac{2}{3}\)

⇒ \(4 \frac{2}{3}+3 \frac{1}{4}\)

⇒ \(\frac{16}{5}-\frac{7}{5}\)

∴ \(\frac{4}{3}-\frac{1}{2}\)

Solution:

L.C.M. of 3 and 7 is 21

⇒ \(\frac{2}{3}+\frac{1}{7}=\frac{2 \times 7+1 \times 3}{21}=\frac{14+3}{21}=\frac{17}{21}\)

L.C.M. of 10 and 15 is 30

⇒ \(\frac{3}{10}+\frac{7}{15}=\frac{3 \times 3+7 \times 2}{30}=\frac{9+14}{30}=\frac{23}{30}\)

L.C.M. of 9 and 7 is 63

⇒ \(\frac{4}{9}+\frac{2}{7}=\frac{4 \times 7+2 \times 9}{63}=\frac{28+18}{63}=\frac{46}{63}\)

L.C.M. of 7 and 3 is 21

⇒ \(\frac{5}{7}+\frac{1}{3}=\frac{5 \times 3+7 \times 1}{21}=\frac{15+7}{21}=\frac{22}{21}=1 \frac{1}{21}\)

L.C.M. of 5 and 6 is 30

⇒ \(\frac{2}{5}+\frac{1}{6}=\frac{2 \times 6+5 \times 1}{30}=\frac{12+5}{30}=\frac{17}{30}\)

L.C.M. of 5 and 3 is 15

⇒ \(\frac{4}{5}+\frac{2}{3}=\frac{4 \times 3+2 \times 5}{15}=\frac{12+10}{15}=\frac{22}{15}=1 \frac{7}{15}\)

L.C.M. of 4 and 3 is 12

⇒ \(\frac{3}{4}-\frac{1}{3}=\frac{3 \times 3-4 \times 1}{12}=\frac{9-4}{12}=\frac{5}{12}\)

L.C.M. of 6 and 3 is 6

⇒ \(\frac{5}{6}-\frac{1}{3}=\frac{5 \times 1-2 \times 1}{6}=\frac{5-2}{6}=\frac{3}{6}=\frac{1}{2}\)

L.C.M. of 2, 3, and 6 is 6

⇒ \(\frac{2}{3}+\frac{3}{4}+\frac{1}{2}=\frac{2 \times 4+3 \times 3+1 \times 6}{12}\)

⇒ \(=\frac{8+9+6}{12}=\frac{23}{12}=1 \frac{11}{12}\)

L.C.M. of 3 and 3 is 3

⇒ \(\frac{1}{2}+\frac{1}{3}+\frac{1}{6}=\frac{1 \times 3+1 \times 2+1 \times 1}{6}\)

⇒ \(=\frac{3+2+1}{6}=\frac{6}{6}=1\)

L.C.M. of 3 and 4 is 12

⇒ \(\frac{14}{3}+\frac{13}{4}=\frac{14 \times 4+13 \times 3}{12}\)

⇒ \(=\frac{56+39}{12}=\frac{95}{12}=7 \frac{11}{12}\)

L.C.M. of 5 and 5 is 5

⇒ \(\frac{16}{5}-\frac{7}{5}=\frac{16-7}{5}=\frac{9}{5}=1 \frac{4}{5}\)

L.C.M. of 3 and 2 is 6

⇒ \(\frac{4}{3}-\frac{1}{2}=\frac{4 \times 2-1 \times 3}{6}=\frac{8-3}{6}=\frac{5}{6}\)

Question 2. Sarita bought \(\frac{2}{5}\) metre of ribbon and Lalita \(\frac{3}{4}\) metre ofribbon. What is the total length of the ribbon they bought?

Ribbon bought by Sarita = \(\frac{2}{5}\)m

And Ribbon bought by Lalita =\(\frac{3}{4}\)

TinTotal length, of ribbon = \(\frac{2}{5}+\frac{3}{4}=\frac{2 \times 4+5 \times 3}{20}\)

[L.C.M. of 5 and 4 is 20]

⇒ \(=\frac{8+15}{20}=\frac{23}{20}=1 \frac{3}{20} \mathrm{~m}\)

Therefore, they bought\(1 \frac{3}{20} \mathrm{~m}\) of ribbon.

Question 3. Naina was given \(1 \frac{1}{2}\) piece of cake and Najma was given \(1 \frac{1}{3}\) a piece of cake. Please find the total amount of cake that was given to both of them.
Solution:

Cake taken by Naina = \(1 \frac{1}{2}\) Piece.

And cake taken by Najma =\(1 \frac{1}{3}\) Piece.

Total cake taken = \(1 \frac{1}{2}+1 \frac{1}{3}=\frac{3}{2}+\frac{4}{3}\)

⇒ \(=\frac{3 \times 3+4 \times 2}{6}\)

[ L.C.M. of 2 and 3 is 6]

⇒ \(=\frac{9+8}{6}=\frac{17}{6}=2 \frac{5}{6}\)

Therefore, the total amount of cake given to both of them = \(2 \frac{5}{6}.\)

Question 4. Fill in the boxes:

Fraction Of Fill In The boxes

Solution:

⇒ \(\frac{1}{4}+\frac{5}{8}=\frac{2+5}{8}=\frac{7}{8}\)

⇒ \(\frac{1}{2}+\frac{1}{5}=\frac{5+2}{10}=\frac{7}{10}\)

⇒ \(\frac{1}{2}-\frac{1}{6}=\frac{3-1}{6}=\frac{2}{6}=\frac{1}{3}\)

Question 5. Complete the addition-subtraction box

Addition And Subtraction

Solution:

Addition And Subtraction .

Question 6. A piece of wire \(\frac{7}{8}\) meter long broke into two pieces. One piece was \(\frac{1}{4}\) metre long. How long is the other piece?
Solution:

Total length of wire = \(\frac{7}{8}\)

Length of first part = \(\frac{7}{8}\)

Remaining part = \(\frac{7}{8}-\frac{1}{4}=\frac{7 \times 1-2 \times 1}{8}\)

[L.C.M. of 8 and 4 is 8]

⇒ \(=\frac{7-2}{8}=\frac{5}{8} \text { metre }\)

Therefore, the length of the remaining part is \(\frac{5}{8}\) meter.

Question 7. Nandini’s house is \(\frac{9}{10}\) km from her school. She walked some distance and then took a bus for \(\frac{1}{2}\) km to reach the school. How far did she walk?
Solution:

The total distance between school and

house = \(\frac{9}{10}\)km

Distance covered by bus = \(\frac{1}{2} \mathrm{~km}\)

Remaining Distance = \(\frac{9}{10}-\frac{1}{2}=\frac{9 \times 1-1 \times 5}{10}\)

[ L.C.M. of 10 and 2 is 10]

⇒ \(=\frac{9-5}{10}=\frac{4}{10}=\frac{2}{5} \mathrm{~km}\)

Therefore, distance covered by walking is \(\frac{2}{5} \mathrm{~km}\)

Question 8. Asha and Samuel have bookshelves of the 5 same size partly filled with books. Asha’s shelf is \(\frac{5}{6}\)th full and Samuel’s shelf is \(\frac{2}{5}\)th full. Whose bookshelf is full? By what fraction?
Solution:

\(\frac{5}{6} \text { and } \frac{2}{5}\) \(\Rightarrow \frac{5}{6}=\frac{5}{6} \times \frac{5}{5}=\frac{25}{30} \text { and } \frac{2}{5}=\frac{2}{5} \times \frac{6}{6}=\frac{12}{30}\)

[ L.C.M. of 6 and 5 is 30]

⇒ \(\frac{25}{30}>\frac{12}{30} \Rightarrow \frac{5}{6}>\frac{2}{5}\)

Asha’s bookshelf is more covered than Samuel’s.

Difference = \(\frac{25}{30}-\frac{12}{30}=\frac{13}{30}\)

Question 9. Jaidev takes \(2 \frac{1}{5}\) minutes to walk across the school ground. Rahul takes \(\frac{7}{4}\) minutes to do the same. Who takes less time and by what fraction?
Solution:

Time taken by Jaidev = \(2 \frac{1}{5}\)minutes

=\(\frac{11}{5}\)minutes

Time taken by Rahul = \(\frac{7}{4} \text { minutes }\)

Difference = \(=\frac{11}{5}-\frac{7}{4}=\frac{11 \times 4-7 \times 5}{20}\)

[ L.C.M. of 5 and 4 is 20]

⇒ \(=\frac{44-35}{20}=\frac{9}{20} \text { minutes }\)

Thus, Rahul takes less time, which is \(\frac{9}{20}\) minutes.

CBSE Solutions For Class 6 Maths Chapter 10 Mensuration

Mensuration Exercise – 10.1

Question 1. Find the perimeter of each of the following figures:

CBSE Solutions For Class 6 Maths Chapter 10

Solution:

  1. Perimeter = Sum of all the sides = 4 cm + 2 cm +1 cm + 5 cm = 12 cm
  2. Perimeter = Sum of all the sides = 23 cm + 35 cm + 40 cm + 35 cm = 133 cm
  3. Perimeter = Sum of all the sides = 15 cm + 15 cm + 15 cm + 15 cm = 60 cm
  4. Perimeter = Sum of all the sides =4 cm+4 cm+4 cm+4 cm+ 4 cm=20 cm
  5. Perimeter = Sum of all the sides =1 cm + 4 cm + 0.5 cm + 2.5 cm + 2.5 cm + 0.5 cm + 4 cm = 15 cm
  6. Perimeter = Sum of all the sides = 4cm +lcm + 3cm + 2cm + 3cm + 4cm +lcm + 3cm + 2cm + 3cm + 4cm + lcm + 3cm + 2cm + 3cm + 4cm +lcm + 3cm + 2cm + 3cm = 52 cm

Question 2. The lid of a rectangular box of sides 40 cm by 10 cm is sealed all round with tape. What is the length of the tape required?
Solution:

Total length of tape required

= Perimeter of rectangle

= 2 (length + breadth)

= 2(40 +10) cm = 2 x 50 cm = 100 cm =1 m

Mensuration The lid of a rectangular

Thus, the total length of tape required is 100 cm or 1 m.

Question 3. A table-top measures 2 m 25 cm by 1 m 50 cm. What is the perimeter of the tabletop?
Solution:

Length of table-top = 2m 25 cm = 2.25 m

Breadth of table-top =1 m 50 cm = 1.50 m

Perimeter of table-top = 2 x (length + breadth)

= 2 x (2.25 + 1.50) m

= 2 x 3.75 m = 7.50 m

Thus, the perimeter of the table-top is 7.5 m

Question 4. What Is the length of the wooden strip required to frame a photograph of length and breadth of 32 cm and 21 cm respectively?
Solution:

Length of the wooden strip

= Perimeter of photograph

= 2 x (length + breadth)

= 2 (32 + 21) cm = 2 x 53 cm = 106 cm

Thus, the length of the wooden strip required is 106 cm

Question 5. A rectangular piece of land measures 0.7 km by 0.5 km. Each side is to be fenced with 4 rows of wires. What is the length of the wire needed?
Solution:

Since 4 rows of wires are needed.

Therefore, the total length of the wire is equal to 4 times the perimeter of land.

Perimeter of land = 2 x (length + breadth)

= 2 x (0.7 + 0.5) km = 2 x 1.2 km = 2.4 km

= 2.4 x 1000 m = 2400 m

Thus, the length of wire = 4 x 2400 m = 9600 m = 9.6 km

Question 6. Find the perimeter of each of the following shapes :

  1. A triangle of sides 3 cm, 4 cm and 5 cm.
  2. An equilateral triangle of side 9 cm.
  3. An isosceles triangle with equal sides 8 cm each and third side 6 cm.

Solution:

1. Perimeter of ΔABC

Mensuration A triangle of sides

=AB+BC+CA

= 3cm+5cm+4cm

= 12 cm

2. Perimeter of equilateral

ΔABC

Mensuration An isosceles triangle

= 3 x side

= 3 x 9 cm

= 27 cm

3. Perimeter of ΔABC

Mensuration An equilateral triangle

= AB + BC + CA

= 8cm + 6cm + 8cm

= 22 cm

CBSE Solutions Class 6 Maths Chapter 10 Mensuration

Question 7. Find the perimeter of a triangle with sides measuring 1 0 cm, 1 4 cm and 1 5 cm.
Solution:

Perimeter of triangle = Sum of all three sides

= 10 cm + 14 cm + 15 cm = 39 cm

Thus, the perimeter of the triangle is 39 cm.

Question 8. Find the perimeter of a regular hexagon with each side measuring 8 m.
Solution:

The perimeter of a regular hexagon

= 6 x length of one side

= 6 x 8m = 48m

Thus, the perimeter of a regular hexagon is 48 m.

Question 9. Find the side of the square whose perimeter is 20 m,
Solution:

The perimeter of the square = 4 x side

20 m = 4 x side ⇒ side = \(=\frac{20}{4} \mathrm{~m}=5 \mathrm{~m}\)

Thus, the side of the square is 5 m.

Question 10. The perimeter of a regular pentagon is 1 00 cm. How long is it on each side?
Solution:

The perimeter of a regular pentagon = 5 x side

100 cm = 5 x side ⇒ side = \(\frac{100}{5} \mathrm{~cm}=20 \mathrm{~cm}\)

Thus, the side of the regular pentagon is 20 cm.

Question 11. A piece of string is 30 cm long. What will be the length of each side if the string is used to form:

  1. A square?
  2. An equilateral triangle?
  3. A regular hexagon?

Solution:

Length of string= Perimeter of each shape

The perimeter of the square = 4 x side

30 cm = 4 x side ⇒ side = \(\frac{30}{4} \mathrm{~cm}=7.5 \mathrm{~cm}\)

Thus, the length of each side of the square will be 7.5 cm

The perimeter of equilateral triangle = 3 x side

30 cm = 3 x side ⇒ side =\(\frac{30}{3} \mathrm{~cm}=10 \mathrm{~cm}\)

Thus, each side of the equilateral triangle will be 10 cm long.

The perimeter of a regular hexagon = 6 x side

30 cm = 6 x side ⇒ side = \(\frac{30}{6} \mathrm{~cm}=5 \mathrm{~cm}\)

Thus, each side of a regular hexagon will be 5 cm long.

Question 12. Two sides of a triangle are 12 cm and 14 cm. The perimeter of the triangle is 36 cm. What is its third side?
Solution:

Let the length of the third title be x cm. The length of the other two sides are 12 cm and 14 cm

Now, the perimeter of the triangle = 36 cm

12 + 14 + x = 36

26 + x = 36

X = 36- 26 ⇒ x = 10

Thus, the length of the third side is 10 cm

Question 13. Find the cost of fencing a square park of side 250 m at the rat of? 20 per metre.
Solution:

Side of square park = 250 m

The perimeter of the square park = 4 x side

= 4 x 250 m = 1000 m

Sin ce, cost of fencing for metre = ₹ 20

Therefore, the cost of fencing for 1000 metres

= ₹ 20×1000 = ₹ 20,000

Question 14. Find the cost of fencing a rectangular park of length 175 m and breadth 125 m at the rate of ₹ 12 per metre
Solution:

Length of rectangular park = 175 m

The breadth of the rectangular park = 125 m

Perimeter of park = 2 x (length + breadth)

= 2 x (175 + 125) m

= 2 x 300 m = 600 m

Since, the cost of fencing a park for metre = ? 12

Therefore, the cost of fencing the park for 600 m

= ₹ 12 x 600 = ₹ 7,200

Question 15. Sweety runs around a square park of side 75 m. Bulbul runs around a rectangular park with a leaf of 60 m a and breadth of h 45 m. Who covers less distance?
Solution:

Distance covered by Sweety

= Perimeter of square park = 4 x side

= 4 x 75 m = 300 m

Thus, the distance covered by Sweety is 300 m.

Nthe ow, distance covered by Bulbul

= Perimeter of a rectangular park

= 2 x (length + breadth)

= 2 x (60 + 45) m

= 2 x 105 m = 210 m

Thus, Bulbul covers a distance of 210 m.

So, Bulbul covers less distance.

Question 16. What is the perimeter of each of the following figures? What do you infer from the answers?

Mensuration Perimeter

Solution:

Perimeter of square = 4 x side = 4 x 25 cm = 100 cm

Perimeter of rectangle = 2 x (length + breadth)

= 2 x (40 + 10) cm = 2 x 50 cm = 100 cm

Perimeter of rectangle = 2 x (length + breadth)

= 2 x (30 + 20) cm = 2 x 50 cm = 100 cm

Perimeter of triangle = Sum of all sides

= 30 cm + 30 cm + 40 cm

= 100 cm

Thus, all the figures have the same perimeter

Question 17. Avneet buys 9 square paving slabs, each with a side of \(\frac{1}{2}\) m. He lays them in the form of a square.

  1. What is the perimeter of his arrangement?
  2. Shari does not like his arrangement. She gets him to lay them out like a cross. What is the perimeter of her arrangement? ent.
  3. Which a has greater perimeter?
  4. Avneet wonders if there is a way of getting an even greater perimeter. Can you find a way of doing this? (The paving slabs must meet along complete edges i.e. they cannot be broken.)

Mensuration Greater Perimeter

Solution:

1. Side of one small square = \(\frac{1}{2}\) m

Side of given square ⇒ \(\frac{3}{2}\) perimetereter of square = 4 x side

⇒\(4 \times \frac{3}{2} \mathrm{~m}=6 \mathrm{~m}\)

2. The perimeter of the given figure

⇒ Sum of all sides = \(20 \times \frac{1}{2} \mathrm{~m}=10 \mathrm{~m}\)

3. The arrangement crosses a greater perimeter.

4. It is not possible to determine the arrangement with a perimeter greater than 10 m.

Mensuration Exercise – 10.2

Question 1. Find the areas of the following figures by countsquaresuare:

Mensuration The Counting square

Solution:

1. Number of filled squares = 9

∴ Area covered by filled squares

= (9 x 1) sq units = 9 sq units

2. Number of filled squares = 5

∴ Area covered by filled squares

= (5 x 1) sq units = 5 sq units

3. Number of fully-filled squares = 2

Number of half-filled squares = 4

∴ Area covered by fully-filled squares

= (2 x 1) sq units = 2 sq units

Area covered by half-filled squares

⇒ \(\left(4 \times \frac{1}{2}\right) \text { sq units }=2 \text { sq units }\)

Total area = (2 + 2) sq units = 4 sq units

4. Number of filled squares = 8

∴ Area covered filled squares

= (8 x 1) sq units = 8 sq units

5. Number of filled squares = 10

∴ Area covered by filled squares

= (10 x 1) sq units = 10 sq units

6. Number of fully-filled squares = 2

Number of half-filled squares = 4

∴ Area covered by fully-filled squares

= (2 x 1) sq units = 2 sq units

Area covered by half-filled squares

⇒ \(\left(4 \times \frac{1}{2}\right) \text { sq units }=2 \text { sq units }\)

Total area = (2 + 2) sq units = 4 sq units

7. Number of fully-filled squares = 4

Number of half-filled squares = 4

∴ Area covered by fully-filled squares

= (4 x 1) sq units = 4 sq units

Area covered by half-filled squares

⇒ \(\left(4 \times \frac{1}{2}\right) \text { sq units }=2 \text { sq units }\)

∴ Total area = (4 + 2) sq units = 6 sq units

8. Number of filled squares = 5

∴ Area covered by filled squares

= (5 x 1) sq units = 5 sq units

9. Number of filled squares = 9

∴ Area covered by filled squares

= (9 x 1) sq units = 9 sq units

10. Number of fully-filled squares = 2

Number of half-filled squares = 4

∴ Area covered by fully-filled squares

= (2 x 1) sq units = 2 sq units

Area covered by half-filled squares

⇒ \(\left(4 \times \frac{1}{2}\right) \text { sq units }=2 \text { sq units }\)

∴ Total area = (2 + 2) sq units = 4 sq units

11. Number of fully-filled squares = 4

Number of half-filled squares = 2

∴ Area covered by fully-filled squares

= (4 x 1) sq units = 4 sq units

Area covered by half-filled squares

⇒ \(\left(2 \times \frac{1}{2}\right) \text { sq units }=1 \text { sq units }\)

Total area = (4+1) sq units = 5 sq units

12. Number of fully-filled squares = 3,

Number of half-filled squares* 2,

Number of more than half-filled

squares = 4

and number of less than half-filled squares = 4.

Now, estimated area covered by fully-filled squares = 3 sq units, half-filled squares = \(\left(2 \times \frac{1}{2}\right) \text { sq units }\)

=1 sq unit

more than half-filled squares = 4 sq units and less than half-filled squares = 0 sq unit

∴ Total area- (3 +1 + 4 + 0) sq units = 8 sq units

13. Number of fully-filled squares = 7,

Number of more than half-filled

squares = 7

The number of less than half-filled

squares = 5

Estimated area covered by

fully-filled squares = 7 sq units,

more than half-filled squares = 7 sq units

and less than half-filled squares = 0 sq unit

∴ Total area = (7 + 7 + 0) sq units = 14 sq units

14. Number of fully-filled squares = 10,

Number of more than half-filled squares = 8

and number of less than half-filled squares = 5

Estimated area covered by

fully-filled squares = 10 sq units,

more than half-filled squares = 8 sq units

less than half-filled squares = 0 squint

∴ Total area= (10 + 8 + 0) sq units = 18 sq units

Mensuration Exercise – 10.3

Question 1. Find the areas of the rectangles whose sides are:

  1. 3 cm and 4 cm
  2. 12 m and 21 m
  3. 2 km and 3 km
  4. 2 m and 70 cm

Solution:

  1. Area of rectangle = length x breadth = 3 cm x 4 cm = 12 cm²
  2. Area of rectangle- length x breadth = 12 m x 21 m = 252 m²
  3. Area of rectangle = length x breadth = 2 km x 3km = 6 km²
  4. Area of rectangle = length x breadth = 2 m x 70 cm = 2 m x 0.7 m = 1.4 m²

Question 2. Find the areas of the squares whose sides are:

  1. 10 cm
  2. 14 cm
  3. 5 m

Solution:

  1. Area of square = side x side = 10 cm x 10 cm = 100 cm²
  2. Area of square = side x side = 14 cm x 14 cm = 196 cm²
  3. Area of square = side x side =5mx5m = 25m²

Question 3. The length and breadth  of the three rectangles are as given below:

  1. 9 m and 6 m
  2. 17 m and 3 m
  3. 4 m and 14 m

Which one has the largest area and which one has the smallest?
Solution:

  1. Area of rectangle = length x breadth =9mx6m= 54m²
  2. Area of rectangle = length x breadth = 17m x 3 m = 51 m²
  3. Area of rectangle = length x breadth = 4 m x 14 m = 56 m²

Thus, rectangle (c) has the largest area, therefore 56 m2 and rectangle (b) has the smallest area, therefore, 51 m²

Question 4. The area of a rectangular garden 50 m long is 300 sq m. Find the width of the garden.
Solution:

Length of rectangle = 50 m

Area of rectangle = 300 m²

Since, area other rectangle = length x breadth

Therefore, breadth  \(=\frac{\text { area of rectangle }}{\text { length }}\)

⇒ \(=\frac{300}{50} \mathrm{~m}=6 \mathrm{~m}\)

Thus, the breadth of the garden is 6 m

Question 5. What is the cost of tiling a rectangular plot of land 500 m long and 200 m wide at the rate of ₹ 8 per hundred sq m?
Solution:

Length of land = 500 m

Breadth of land = 200 m

Area of land = length x breadth = 500 m x 200 m = 1,00,000 sqm

Cost of tiling 100 sq m of land = ₹ 8

Cost of tiling 1,00,000 sq m of land

⇒ \(₹ \frac{8 \times 100000}{100}=₹ 8000\)

Question 6. A table-top measures 2 m by 1 m 50 cm. What is its area in square metres?
Solution:

Length of table-top = 2 m

Breadth of table-top =1 m 50 cm = 1.50m

Area of table-top = length x breadth = 2 m x 1.50 m = 3 m²

Question 7. A room is 4 m long and 3 m 50 cm wide. How many square metres of carpet is needed to cover the floor of the room?
Solution:

Length of room = 4 m

And breadth of room = 3 m 50 cm = 3.50 m

Area of carpet = length x breadth = 4 m x 3.50 m = 14 m²

Question 8. A floor is 5 m long and 4 m wide. A square carpet of sides 3 m is laid on the floor. Find the area of the floor that is not carpeted.
Solution:

Length of floor = 5 m

The d breadth of the floor = 4 m

Area of floor = length x breadth = 5mx4m = 20m²

Now, side the f square carpet = 3 m

Area of square carpet = side x side = 3mx3m = 9m²

Area of floor that is not carpeted = 20 m²- 9 m² = 11 m²

Question 9. Five square flower beds each of sides 1 m are dug on a piece of land 5 m long and 4 m wide. What is the area of the remaining part of the land?
Solution:

Side of square flower bed =1 m

Area of square flower bed = side x side =lm x lm = lm²

Area of 5 square flower beds = (1×5) m²

= 5 m²

Now, length of land = 5 m

The d breadth of land = 4 m

Area of land = length x breadth – 5 m x 4 m

=20 m²

Area of remaining part

= Area of land- Area of 5 flower beds

= 20 m²- 5 m² = 15 m²

Question 10. By splitting the following figures into rectangles, find their areas (The measures are given in centimetres).

Mensuration The splitting the Rectangle

Solution:

1. we have,

Mensuration The Rectangle

Area of square HKLM = 3×3 cm²= 9 cm²

Area of rectangle IJGH-1×2 cm² = 2 cm²

Area of square FEDG = 3×3 cm² = 9 cm²

Area of rectangle ABCD = 2×4 cm² = 8 cm²

Total area of the figure = (9 + 2 + 9 + 8) cm² = 28 cm²

2. we have,

Mensuration The Area Of The Rectangle

Area of rectangle ABCD =3×1 cm² = 3 cm²

Area of rectangle BJEF = 3×1 cm² = 3 cm²

Area of rectangle FGHI = 3×1 cm² = 3 cm²

Total area of the figure = (3 + 3 + 3) cm² = 9 cm²

Question 11. Split the following shapes into rectangles and find their areas. (The measures are given in centimetres).

Mensuration The Shapes Into Rectangles

Solution:

1.  We have,

Mensuration The Shapes Into Rectangles

Area of rectangle ABCD = 2 x 10 cm² = 20 cm²

Area of rectangle DEFG = 10 x 2 cm² = 20 cm²

Total area of the figure = (20 + 20) cm²

= 40 cm²

2.  We have,

Mensuration The Shapes Into Rectangle

There are 5 squares each of side 7 cm.

Area of one square = 7×7 cm² = 49 cm²

Area of 5 squares = 5 x 49 cm² = 245 cm²

3.  We have,

Mensuration The Shapes Into Rectangls

Area of rectangle ABCD = 5 x 1 = 5 cm²

Area of rectangle EFGH = 4 x 1 = 4 cm²

Total area of the figure = (5+4) cm²

= 9 cm²

Question 12. How many tiles whose length and breadth are 12 cm and 5 cm respectively will be needed to fit in a rectangular region whose length and breadth are respectively:

  1. 100 cm and 144 cm
  2. 70 cm and 36 cm.

Solution:

1.  The area of the rectangular region

=length x breadth= 100 cm x 144 cm= 14400 cm²

Area of one tile = 12 cm x 5 cm = 60 cm²

Number of tiles = \(=\frac{\begin{array}{r}
\text { Area of rectangular region }
\end{array}}{\text { Area of one tile }}\)

\(=\frac{14400}{60}=240\)

Thus, 240 tiles are required.

2.  The area of the rectangular region

= length x breadth = 70 cm x 36 cm = 2520 cm²

Area of one tile = 12 cm x 5 cm- 60 cm²

Number of tiles \(=\frac{\text { Area of rectangular region }}{\text { Area of one tile }}=\frac{2520}{60}=42\)

Thus, 42 tiles are required.

CBSE Solutions For Class 6 Maths Chapter 9 Data Handling

Data Handling

Data Handling Exercise – 9.1

Question 1. In a Mathematics test, the following marks were obtained by 40 students. Arrange these marks in a table using tally marks.

Data Handling using tally marks

  1. Find how many students obtained marks equal to or more than 7.
  2. How many students obtained marks below 4?

Solution:

CBSE Solutions For Class 6 Maths Chapter 9

5 + 4 + 3 = 12 students obtained marks equal to or more than 7.

2 + 3 + 3 = 8 students obtained marks below 4.

Question 2. Following is the choice of sweets for students of Class VI.

Ladoo, Barfi, Ladoo, Jalebi, Ladoo, Rasgulla, Jalebi, Ladoo, Barfi, Rasgulla, Ladoo, Jalebi, Jalebi, Rasgulla, Ladoo, Rasgulla, Jalebi, Ladoo, Rasgulla, Ladoo, Ladoo, Barfi, Rasgulla, Rasgulla, Jalebi, Rasgulla, Ladoo, Rasgulla, Jalebi, Ladoo.

  1. Arrange the names of sweets on a table using tally marks.
  2. Which sweet is preferred by most of the students?

Solution:

Data Handling sweet is preferred by most of the Student

Ladoo is preferred by most of the students.

Question 3. Catherine threw a die 40 times and noted the number appearing each time as shown below:

Data Handling Catherine threw a dice 40 times

Make a table and enter the data using tally marks. Find the number that appeared

  1. The minimum number of times.
  2. Tire maximum number of times.
  3. Find those numbers that appear an equal number of times.

Solution:

Data Handling Tire maximum number of times

  1. 4 appeared a minimum number of times.
  2. 5 appeared the maximum number of times
  3. 1 and 6 appeared an equal number of times.

CBSE Solutions Class 6 Maths Chapter 9 Data Handling

Question 4. The following pictograph shows the number of tractors in five villages.

The number of tractors in five villages

Observe the pictograph and answer the following questions.

  1. Which village has the minimum number of tractors?
  2. Which village has the maximum number of tractors?
  3. How many more tractors does village C have as compared to village B?
  4. What is the total number of tractors in all the five villages?

Solution:

  1. VillageD has the minimum number of tractors
  2. Village C has the maximum number of tractors.
  3. Village C has 8-5=3 more tractors than village B.
  4. Total number of tractors =6+5+8+3+6 = 28

Question 5. The number of female students in each class of a co-educational middle school is depicted by the pictograph :

Data Handling The number of girl students in each class of a co-educational middle school

  1. Observe this pictograph and answer the following questions:
  2. Which class has the minimum number of girl students?
  3. Is the number of girls in Class 6 less than the number of girls in Class 5?
  4. How many girls are there in Class 7?

Solution:

Data handling How many girls are there in Class 7

  1. Class 6 has the minimum number of female students.
  2. No, the number of girls in Class 6 is greater than the number of girls in Class 5.
  3. There are 12 girls in Class 7

Question 6. The sale of electric bulbs on different days of the week is shown below:

The sale of electric bulbs on different days of a week

Observe the pictograph and answer the following questions:

  1. How many bulbs were sold on Friday?
  2. On which day were the maximum number of bulbs sold?
  3. On which of the day’s same number of bulbs were sold?
  4. On which of the day is a minimum number of bulbs sold?
  5. If one big carton can hold 9 bulbs. How many cartons were needed in the given week?

Solution:

Data Handling big carton can hold 9 bulbs

  1. Number of bulbs sold on Friday is 14.
  2. A maximum number of bulbs were sold on Sunday.
  3. Same number of bulbs were sold on Wednesday and Saturday.
  4. A minimum number of bulbs were sold on Wednesday and Saturday.
  5. The long number of bulbs sold in the given week – 86

Number of cartons required for 9 bulbs- 1

∴ Number of cartons required for 86 bulbs = 86 + 9 = 9.55 = 10

Therefore, 10 cartons were needed in the given week

Question 7. In a village, six fruit merchants sold the following number of fruit baskets in a particular season:

In a village six fruit merchants

Observe this pictograph and answer the following questions:

  1. Which merchant sold the maximum number of baskets?
  2. How many fruit baskets were sold by Anwar?
  3. The merchants who have sold 600 or more baskets are planning to buy a godown for the next season. Can you name them?

Solution:

Data Handling Number of baskets are planning

  1. Martin sold the maximum number of baskets.
  2. 700 fruit baskets were sold by Anwar.
  3. Anwar, Martin, and Ranjit Singh have sold more than 600 baskets.

CBSE Solutions For Class 6 Maths Chapter 3 Playing With Numbers

Class 6 Maths Chapter 3 Playing With Numbers Exercises

1. Write all the factors of the following numbers:

(1) 24
(2) 15
(3) 21
(4) 27
(5) 12
(6) 20
(7) 18
(8)23
(9) 36

Solution: (1) 24 =1×24 = 2×12 = 3 x 8 = 4 x 6 = 6 x4

Factors of 24 are 1, 2, 3, 4, 6, 8, 12 and 24

(2) 15 =1×15 = 3×5 = 5×3

Factors of 15 are 1, 3, 5 and 15

(3) 21 = 1×21 = 3×7 = 7×3

Factors of 21 are 1, 3, 7 and 21

(4) 27 =1×27 = 3×9 = 9×3

Factors of 27 are 1, 3, 9 and 27

(5) 12 =1×12 = 2×6 =3×4 = 4×3 = 6×2

Factors of 12 are 1, 2, 3, 4, 6 and 12

(6) 20 =1×20 = 2×10 = 4×5 = 5×4

Factors of 20 are 1, 2, 4, 5, 10 and 20

(7) 18 = 1×18 = 2×9 = 3×6 = 6×3 = 9×2

Factors of 18 are 1, 2, 3, 6, 9 and 18

(8)23 = 1×23

Factors of 23 are1 and 23

(9) 36 =1×36 = 2×18 = 3×12 = 4×9 =6×6

Factors of 36 are 1, 2, 3, 4, 6, 9, 12, 18 and 36

2. Write first five multiples of:

(1) 5
(2) 8
(3) 9

Solution: (1) 5 x 1 = 5, 5 x 2 = 10, 5 x 3 = 15, 5 x 4 = 20, 5 x 5 = 25

First five multiples of 5 are 5, 10, 15, 20, 25.

(2) 8 x 1 = 8, 8 x 2 = 16, 8 x 3 = 24, 8 x 4 = 32, 8 x 5 = 40

First five multiples of 8 are 8, 16, 24, 32, 40.

(3) 9 x 1 = 9, 9 x 2 = 18, 9 x 3 = 27, 9 x 4 = 36, 9×5 = 45

First five multiples of 9 are 9, 18, 27, 36, 45

3. Match the items in column 1 with the items in column 2.

Column 1                                  Column 2

(i) 35                                    (1) Multiple of 8
(ii) 15                                   (2) Multiple of 7
(iii) 16                                  (3) Multiple of 70
(iv) 20                                  (4) Factor of 30
(v) 25                                   (5) Factor of 50
(6) Factor of 20

Solution:  (i) -> (2); (ii) -> (4); (iii) (1); (iv) -> (6); (v) -> (5)

(1) Multiples of 8 are 8, 16, 24, 32, 40,
(2) Multiples of 7 are 7, 14, 21, 28, 35,
(3) Multiples of 70 are 70, 140, 210,
(4) Factors of 30 are 1, 2, 3, 5, 6, 10, 15, 30.
(5) Factors of 50 are 1, 2, 5, 10, 25.
(6) Factors of 20 are 1, 2, 4, 5, 10, 20.

4. Find all the multiples of 9 upto 100.

Solution: Multiples of 9 upto 100 are 9, 18, 27, 36,45, 54, 63, 72, 81, 90, 99

Exercise – 3.2

1. What is the sum of any two
(1) Odd numbers?
(2) Even numbers?

Solution: (1) The sum of any two odd numbers is an even number.

As like, 1+3 = 4, 3 + 5 = 8

(2) The sum of any two even numbers is an even number.

As like, 2 + 4 = 6, 6 + 8 = 14

2. State whether the following statements are True or False:

(1) The sum of three odd numbers Is even.
(2) The sum of two odd numbers and one even number is even.
(3) The product of three odd numbers is odd.
(4) If an even number is divided by 2, the quotient is always odd.
(5) All prime numbers are odd.
(6) Prime numbers do not have any factors.
(7) Sum of two prime numbers is always even.
(8)2 is the only even prime number.
(9) All even numbers are composite numbers.
(10) The product of two even numbers is always even.

Solution: (1) False

Since, sum of two odd numbers is even and sum of one odd number and one

(2) True

Since, sum of two odd numbers is even and sum of two even numbers is always even.

(3) True

(4) False

If an even number is divided by 2, then the quotient is either odd or even.

(5) False
Since, prime number 2 is even.

(6) False

Factors of prime numbers are1 and the number itself.

(7) False

Sum of two prime numbers is either even or odd.

(8)True

(9) False

Since, even number 2 is prime i.e., not composite.

(10) True

3. The numbers 13 and 31 are prime numbers. Both these numbers have same digits 1 and 3. Find such pairs of prime numbers upto 100.

Solution: Pairs of prime numbers having same digits upto 100 are 17 and 71; 37 and 73; 79 and 97

4. Write down separately the prime and composite numbers less than 20.

Solution: Prime numbers less than 20 are 2, 3, 5,7, 11, 13, 17, 19

Composite numbers less than 20 are 4, 6, 8, 9,10, 12, 14, 15, 16, 18

5. What is the greatest prime number between 1 and 10?

Solution: The greatest prime number between 1 and 10 is 7.

CBSE Solutions Class 6 Maths Chapter 3 Playing With Numbers

6. Express the following as the sum of two odd primes.

(1) 44
(2) 36
(3) 24
(4) 18

Solution: (1) 44 = 3 + 41

(2) 36 = 5 + 31

(3) 24 = 7 + 17

(4) 18 = 7 +11

7. Give three pairs of prime numbers whose difference is 2. [Remark: Two prime numbers whose difference is 2 are called twin primes].

Solution: Three pairs of prime numbers whose difference is 2 are 3 and 5; 5 and 7; 11 and 13.

8. Which of the following numbers are prime?
(1) 23
(2) 51
(3) 37
(4) 26

Solution: 23 and 37 are prime numbers and 51 and 26 are composite numbers. Thus, numbers in option (1) and (3) are prime.

9. Write seven consecutive composite numbers less than 1 00 so that there is no prime number between them.

Solution: Seven consecutive composite numbers less than 100 are 90, 91, 92, 93, 94, 95, 96

10. Express each of the following numbers as the sum of three odd primes:

(1) 21
(2) 31
(3) 53
(4) 61

Solution: (1) 21 = 3 + 7 + 11

(2) 31 = 3 + 11 + 17

(3) 53 = 13 + 17 + 23

(4) 61 = 13 + 19 + 29

11. Write five pairs of prime numbers less than 20 whose sum is divisible by 5. (Hint: 3 + 7 = 10)

Solution: Since, 2 + 3 = 5; 7 + 13 = 20; 3 + 17 = 20; 2 + 13 = 15; 5 + 5 = 10 and 5, 10, 15, 20 all are divisible by 5.

So, five pairs of prime numbers less than 20 whose sum is divisible by 5 are 2, 3; 2, 13; 3,17; 7, 13; 5, 5.

12. Fill in the blanks:

(1) A number which has only two factors is called a____________.

Solution: Prime number

(2) A number which has more than two factors is called a__________.

Solution: Composite number

(3) 1 is neither__________ nor________.

Solution: Prime number, composite number

(4) The smallest prime number is________.

Solution: 2

(5) The smallest composite number is__________.

Solution: 4

(6) The smallest even number is________.

Solution: 2

Exercise – 3.3

1. Using divisibility tests, determine which of the following numbers are divisible by 2; by 3; by 4; by 5; by 6; by 8; by 9; by 1 0; by 11 (say, yes or no):

divisibility test

Solution:

divisibility test 1

 

2. Using divisibility tests, determine which of the following numbers are divisible by 4; by 8:

(1) 572
(2) 726352
(3) 5500
(4) 6000
(5) 12159
(6) 14560
(7) 21084
(8)31795072
(9) 1700
(10) 2150

Solution: (1) 572 is divisible by 4 as its last two digits are divisible by 4, butit is not divisible by 8 as its last three digits are not divisible by 8.

(2) 726352 is divisible by 4 as its last two digits are divisible by 4 and it is also divisible by 8 as its last three digits are divisible by 8.

(3) 5500 is divisibleby 4 asits last two digits are divisible by 4, but it is not divisible by 8 as its last three digits are not divisible by 8.

(4) 6000 is divisibleby 4 asits last two digits are divisible by 4 and it is also divisible by 8 as its last three digits are divisible by 8.

(5) 12159 is not divisible by 4 and 8 as it is an odd number.

(6) 14560 is divisible by 4 as its last two digits are divisible by 4 and it is also divisible by 8 as its last three digits are divisible by 8.

(7) 21084 is divisible by 4 as its last two digits are divisible by 4, but it is not divisible by 8 as its last three digits are not divisible by 8.

(8)31795072 is divisible by 4 as its last two digits are divisible by 4 and it is also divisible by 8 as its last three digits are divisible by 8.

(9) 1700 is divisibleby 4 as its last two digits are divisible by 4, but it is not divisible by 8 as its last three digits are not divisible by 8.

(10) 2150 is not divisible by 4 as its last two digits are not divisible by 4 and it is not divisible by 8 as its last three digits are not divisible by 8

3. Using divisibility tests, determine which of following numbers are divisible by 6:

(1) 297144
(2) 1258
(3) 4335
(4) 61233
(5) 901352
(6) 438750
(7) 1790184
(8)12583
(9) 639210
(10) 17852

Solution: (1) 297M4 is divisible by 2 as its ones place is aneven number and it is also divisible by 3 as sum of its digits (= 27) is divisible by 3.

Since, the number is divisible by both 2 and 3. Therefore, it is also divisible by 6.

(2) 1258 is divisible by 2 as its ones place is an even number, but it is not divisible by 3 as sum of its digits (= 16) is not divisible by 3.

Since, the number is not divisible by both 2 and 3. Therefore,itisnot divisible by 6.

(3) 4335 isnot divisibleby 2 as its ones place is not an even number, butit is divisible by 3 as sum ofits digits (=15) is divisible
by 3.

Since, the number is not divisible by both 2 and 3. Therefore,itisnot divisible by 6.

(4) 61233 is not divisible by 2 as its place is not an even number, but it is divisibleby 3 as sum ofits digits (= 15) is divisible by 3.

Since, the number is not divisible by both 2 and 3. Therefore,itisnot divisible by 6.

(5) 901352 is divisible by 2 as its ones place is an even number, butit is not divisible by 3 as sum of its digits (= 20) is not divisible by 3.

Since, the number is not divisible by both 2 and 3. Therefore,itisnot divisible by 6.

(6) 438750 is divisible by 2 as its ones place is an evennumber anditis also divisible by 3 as sum ofits digits (= 27) is divisible by 3.

Since, the number is divisible by both 2 and 3. Therefore, it is also divisible by 6.

(7) 1790184 is divisibleby 2 as its ones place is an even number anditis also divisible by 3 as sum ofits digits (= 30) is divisible
by 3.

Since, the number is divisible by both 2 and 3. Therefore, it is also divisible by 6.

(8)12583 is not divisible by 2 as its ones place is not an even number and it is alsonot divisibleby 3 as sum ofits digits
(= 19) is not divisible by 3.

Since, the number is not divisible by both 2 and 3. Therefore, it is not divisible by 6.

9. 639210 is divisible by 2 as its ones place is an even number and it is also divisible by 3 as sum ofits digits (= 21) is divisible
by 3.

Since, the number is divisible by both 2 and 3. Therefore, it is divisible by 6.

10. 17852 is divisible by 2 as its ones place is an evennumber, butit is not divisible by 3 as sum of its digits (= 23) is not
divisible by 3.

Since, the number is not divisible by both 2 and 3. Therefore,itisnot divisible by 6.

4. Using divisibility tests, determine which of the following numbers are divisible by 11:

(1) 5445
(2) 10824
(3) 7138965
(4) 70169308
(5) 10000001
(6) 901153

Solution: (1) In 5445, sum of the digits at odd places = 5 + 4 = 9

Sum of the digits at even places = 4 + 5=9

Difference ofboth sums =9-9 = 0

Since the difference is 0. Therefore, the number is divisible by 11.

(2) In 10824, sum of the digits at odd places = 4 + 8 +1 = 13

Sum of the digits at even places = 2 + 0=2

Difference of both sums = 13- 2 = 11

Since the difference is divisible by 11.

Therefore, the number is divisibleby 11.

(3) In 7138965, sum of the digits at odd places = 5 + 9 + 3 + 7 = 24

Sum of the digits at even places = 6 + 8 +1 = 15

Difference ofboth sums = 24- 15 = 9

Since the difference is neither 0 nor divisible by 11. Therefore, the number is not divisible by 11.

(4) In 70169308, sum of the digits at odd places = 8 + 3 + 6 + 0 = 17

Sum of the digits at even places =0+9+1+7=17

Difference ofboth sums = 17- 17 = 0

Since the difference is 0. Therefore, the number is divisible by 11.

(5) In 10000001, sum of the digits at odd places =l + 0 + 0 + 0 =l

Sum of the dibits at even places = 0 + 0 + 0+1=1

Difference of both sums =1-1=0

Since the difference is 0. Therefore, the number is divisible by 1 1.

(6) In 901 153, sum of the digits at odd places =3+1+ 0=4

Sum of the digits at even places = 5 + 1+9 = 15

Difference of both sums = 15- 4 = 11

Since the difference is 11. Therefore, the number is divisible by 11.

5. Write the smallest digit and the greatest digit in the blank space of each of the following numbers so that the number formed is divisible by 3:

(1) _6724
(2) 4765 _2

Solution: We know that a number is divisible by 3 if the sum of all digits is divisible by 3.

(1) The smallest digit will be 2.

Tire number formed is 26724 and 2 + 6 + 7 + 2 + 4 = 21, which is divisible by 3.

And the greatest digit will be 8.

The number formed is 86724 and 8 + 6 + 7 + 2 + 4 = 27, which is divisible by 3.

(2) The smallest digit will be 0.

The number formed is 476502 and 4 + 7 + 6 + 5 + 0 + 2 = 24, which is divisible by 3.

And the greatest digit will be 9.

The number formed is 476592 and 4 + 7+ 6 + 5 + 9 + 2 = 33, which is divisible by 3.

6. Write a digit in the blank space of each of the following numbers so that the number formed is divisible by 11 :

(1) 92 _ 389
(2) 8_ 9484

Solution: (1) We know that a number is divisible by 11 if the difference of the sum of the digits at odd places and that of even places is either 0 or divisible by 11.

The number formed is 928389

Sum of digits at odd places = 9 + 3 + 2 = 14

Sum of digits at even places = 8 + 8 + 9 = 25

Their difference = 25 – 14 = 11, which is divisible by 11.

(2) We know that a number is divisible by 11 if the difference of the sum of the digits at odd places and that of even places Is either 0
or divisible by II.

The number formed is 869484

Sum of digits at odd places – 4 + 4 + 6- 14

Sum of digits at even places – 8 + 9 + 8 = 25

Their difference – 25 – 14 – 11, which is divisible by 11.

Exercise – 3.4

1. Find the common factors of:

(1) 20 and 28
(2) 15 and 25
(4) 56 and 120
(3) 35 and 50

Solution: (1) Factors of 20 are 1, 2, 4, 5, 10 and 20

Factors of 28 are 1, 2, 4, 7, 14 and 28

Common factors of 20 and 28 are 1, 2 and 4

(2) Factors of 15 are 1, 3, 5 and 15

Factors of 25 are 1, 5 and 25

Common factors of 15 and 25 are1 and 5

(3) Factors of 35 are 1, 5, 7 and 35

Factors of 50 are 1, 2, 5, 10, 25 and 50

Common factors of 35 and 50 are 1 and 5

(4) Factors of 56 are 1, 2, 4, 7, 8, 14, 28 and 56

Factors of 120 are 1, 2, 3, 4, 5, 6, 8, 10, 12,15, 20, 24, 30, 40, 60 and 120

Common factors of 56 and 120 are 1, 2, 4 and 8

2. Find the common factors of: (1) 4, 8 and 12 (2) 5, 15 and 25

Solution: (1) Factors of 4 are 1, 2 and 4

Factors of 8 are 1, 2, 4 and 8

Factors of 12 are 1, 2, 3, 4, 6 and 12

Common factors of 4, 8 and 12 are 1, 2 and 4

(2) Factors of 5 are1 and 5

Factors of 15 are 1, 3, 5 and 15

Factors of 25 are 1, 5 and 25

Common factors of 5, 15 and 25 are 1 and 5

3. Find first three common multiples of: (1) 6 and 8

Solution: (1) Multiples of 6 are 6, 12, 18, 24, 30, 36, 42, 48, 54, 60, 66, 72,………..

Multiples of 8 are 8, 16, 24, 32, 40, 48, 56, 64, (2) 12 and 18 72,……….

First three common multiples of 6 and 8 are 24, 48 and 72

(2) Multiples of 12 are 12, 24, 36, 48, 60, 72,84, 96, 108, 120,……….

Multiples of 18 are 18, 36, 54, 72, 90, 108, 126,……….

First three common multiples of 12 and 18 are 36, 72 and 108

4. Write all the numbers less than 1 00 which are common multiples of 3 and 4.

Solution: Multiples of 3 are 3, 6, 9, 12, 15, 18, 21,24. 27, 30, 33, 36, 39, 42, 45, 48, 51, 54, 57, 60,63, 66, 69, 72, 75, 78, 81, 84, 87, 90, 93, 96, 99,…………

Multiples of 4 are 4, 8, 12, 16, 20, 24, 28, 32, 36,40, 44, 48, 52, 56, 60, 64, 68, 72, 76, 80, 84, 88,92, 96, 100,………..

Common multiples of 3 and 4 which are less than 100 are 12, 24, 36, 48, 60, 72, 84 and 96

5. Which of the following numbers are co-prime?

(1) 18 and 35
(2) 15 and 37
(4) 17 and 68
(6) 81 and 16
(3) 30 and 415
(5) 216 and 215

Solution: (1) Factors of 18 are 1, 2, 3, 6, 9 and 18

Factors of 35 are 1, 5, 7 and 35

Common factor of 18 and 35 is1

Since, both have only one common factor,i.e., 1. Therefore, 18 and 35 are co-prime numbers.

(2) Factors of 15 are 1, 3, 5 and 15

Factors of 37 are 1 and 37

Common factor of 15 and 37 is 1

Since, both have only one common factor, i.e., 1. Therefore, 15 and 37 are co-prime numbers.

(3) Factors of 30 are 1, 2, 3, 5, 6, 10, 15 and 30 ……, 83 and 415

Factors of 415 are 1, 5,

Common factors of 30 and 415 are 1 and 5 ?

Since, bothhave more than one common factor. Therefore, 30 and 415 are not co-prime numbers.

(4) Factors of 17 are 1 and 17

Factors of 68 are 1, 2, 4, 17, 34 and 68

Common factors of 17 and 68 are1 and 17

Since, both have more than one common factor. Therefore, 17 and 68 are not co-prime numbers.

(5) Factors of 216 are 1, 2, 3, 4, 6, 8, 9, 12, 18,24, 27, 36, 54, 72, 108 and 216

Factors of 215 are 1, 5, 43 and 215

Common factor of 216 and 215 is1

Since, both have only one common factor, i.c., I. Therefore, 216 and 215 are co-prime numbers.

(6) Factors of 81 are 1, 3, 9, 27 and 81

Factors of 16 are 1, 2, 4, 8 and 16

Common factor of 81 and 16 is 1

Since, both have only one common factor, i.e., 1. Therefore, 81 and 16 are co-prime numbers.

6. A number is divisible by both 5 and 12. By which other number will that number be always divisible?

Solution: Since 5 x 12 = 60. The number divisible by both 5 and 12, must also be divisible by 60.

7. A number is divisible by 12. By what other numbers will that number be divisible?

Solution: Factors of 12 are 1, 2, 3, 4, 6 and 12.

Therefore, the number divisible by 12, will also be divisible by 1, 2, 3, 4 and 6.

Exercise – 3.5

1. Here are two different factor trees for 60. Write the missing numbers.

Solution:

CBSE Solutions For Class 6 Maths Chapter 3

Factors of 60 are 1, 2, 3, 4, 5, 6, 10, 12, 15, 20, 30, 60.

(1) Since 6 = 2*3 and 10 = 5 * 2

The missing numbers are 3 and 2.

The missing numbers are 3 and 2

(2) Since, 60- 30 x 2 30- 10 x 3 and 10- 5 x 2

2 The missing numbers are 3 and 2

2. Which factors are not included in the prime factorisation of a composite number?

Solution:1 and the number it self are not included in the prime factorisation of a composite number.

3. Write the greatest 4-digit number and express it in terms of its prime factors.

Solution: The greatest four-digit number is 9999.

The greatest four digit number is 9999

9999 = 3x3x 11 x101.

4. Write the smallest 5-digit number and express it in the form ofits prime factors.

Solution: The smallest five digitnumber is 10000.

The smallest five digit number is 10000

10000 = 2x2x2x2x5x5x5x5.

5. Find all the prime factors of 1 729 and arrange them in ascending order. Now state the relation, if any; between two consecutive
prime factors.

Solution:

prime factors of 1729

1729 = 7 x 13 x 19.

The difference of two consecutive prime factors is 6. (v 13- 7 = 6 and 19- 13 = 6)

6. The product of three consecutive numbers is always divisible by 6. Verify this statement with the help of some examples.

Solution: Among the three consecutive numbers, there must be atleast one even number and one multiple of 3. Thus, the product must be divisible by 6.

For example: (1) 2 x 3 x 4 = 24
(2) 4x5x6 = 120, where both 24 and 120 are divisible by 6.

7. The sum of two consecutive odd numbers is divisible by 4. Verify this statement with the help of some examples.

Solution: The sum of two consecutive odd numbers is divisible by 4.

For example : 3 + 5 = 8 and 8 is divisible by 4.

5 + 7 = 12 and 12 is divisible by 4.

7 + 9 = 16 and 16 is divisible by 4.

9 + 11 = 20 and 20 is divisible by 4.

8. In which of the following expressions, prime factorisation has been done?

(1) 24 = 2x3x4
(2) 56 = 7x2x2x2
(3) 70 = 2x5x7
(4) 54 = 2x3x9

Solution: In expressions (2) and (3), prime factorisation has been done.

9. 18 is divisible byboth 2 and 3.It is also divisible by 2 x 3 = 6. Similarly, a number is divisible by both 4 and 6. Can we say that the number must also be divisible by 4 x 6 = 24? Ifnot, give an example to justify your answer.

Solution: No. The number 12 is divisible by both 6 and 4, but 12 is not divisible by 24.

A number divisible by both 4 and 6 may or may not be divisible by 4 x 6 = 24.

10. I am the smallest number, having four different prime factors. Can you find me?

Solution: Since, 2 x 3 x 5 x 7= 210

210 is the smallest number, having 4 different prime factors i.e, 2, 3, 5 and 7.

Exercise – 3.6

1. Find the HCF of the following numbers:

(1) 18,48
(2) 30,42
(3) 18,60
(4) 27,63
(5) 36,84
(6) 34,102
(7) 70,105,175
(8) 91,112,49
(9) 18,54,81
(10) 12,45,75

Solution: (1) The prime factorisation of 18 and

48 are; 18 = 2 x 3 x 3

48=2x2x2x2x3

HCF (18, 48) = 2×3 = 6

(2) The prime factorisation of 30 and 42 are; 30 = 2 x 3 x 5

42 = 2x3x7

HCF (30, 42) =2×3 = 6

(3) The prime factorisation of 18 and 60 are; 18 = 2 x 3 x 3

60 = 2x2x3x5

HCF (18, 60) = 2×3 = 6

(4) The prime factorisation of 27 and 63 are; 27 = 3 x 3 x 3

63 =3x3x7

HCF (27, 63) = 3×3 = 9

(5) The prime factorisation of 36 and 84 are; 36 = 2x2x3x3

84 = 2x2x3 x7

HCF (36, 84) = 2 x 2 x 3 = 12

(6) The prime factorisation of 34 and 102 are; 34 = 2 x 17

102 = 2x3x17

HCF (34, 102) = 2×17 = 34

(7) The prime factorisation of 70, 105 and 175 are; 70 = 2 x 5 x 7

105 =3x5x7

175 =5x5x7

HCF (70, 105, 175) = 5 x 7 = 35

(8)The prime factorisation of 91, 112 and

49 are; 91 = 7 x 13

112 = 2x2x2x2x7

49 = 7 x 7

HCF (91, 112, 49) = 7

(9) The prime factorisation of 18, 54 and 81 are; 18 = 2 x 3 x 3

54 = 2x3x3x3

81=3x3x3x3

HCF (18, 54, 81) = 3×3 = 9

(10) The prime factorisation of 12, 45 and 75 are; 12 = 2 *2 *3

45 = 3 x 3 x 5

75 = 3 x 5 x 5

HCF (12, 45, 75) = 3

2. What is the HCF of two consecutive

(1) numbers?
(2) even numbers?
(3) odd numbers?

Solution:(1) HCF of two consecutive numbers is 1.

(2) HCF of two consecutive even numbers is 2.

(3) HCF of two consecutive odd numbers is 1.

3. HCF of co-prime numbers 4 and 1 5 was found as follows by factorisation: 4 = 2×2 and 15 = 3×5 since there is no common prime factor, so HCF of 4 and 15 is 0. Is the answer correct? If not, what is the correct HCF?

Solution: No. The correct HCF of 4 and 15 is 1.

Exercise – 3.7

1. Renu purchases two bags of fertiliser of weights 75 kg and 69 kg. Find the maximum value of weight which can measure the weight of the fertiliser exact number of times.

Solution: For finding maximum weight, we have to find HCF of 75 and 69.

The prime factorisation of 75 and 69 are;

75 = 3x5x5

69 = 3×23

So, HCF of 75 and 69 = 3

Therefore, the required weight is 3 kg.

2. Three boys step off together from the same spot. Their steps measure 63 cm, 70 cm and 77 cm respectively. What is the minimum distance each should cover so that all can cover the distance in complete steps?

Solution: For finding minimum distance, we have to find LCM of 63, 70, 77.

minimum distance

LCM of 63, 70 and 77 = 2x3x3x 5 x7x 11 = 6930

Therefore, the minimum distance three boys should cover is 6930 cm.

3. The length, breadth and height of a room are S25 cm, 675 cm and 450 cm respectively. Find the longest tape which can measure the three dimensions of the room exactly.

Solution: The measurement of the longest tape = HCF of 825 cm, 675 cm and 450 cm.

The prime factorisation of 825, 675 and 450 are;

825 =3x5x5x 11

675 =3x3x3x5x5

450 = 2x3x3x5x5

Now, HCF of 825, 675 and 450 = 3 x 5 x 5 = 75

Therefore, the measurement of longest tape is 75 cm.

4. Determine the smallest 3-digit number which is exactly divisible by 6, 8 and 1 2.

Solution: The smallest 3-digit number = 100

smallest 3 digit number

LCM of 6, 8 and 12 =2x2x2x3 = 24

To find the number, we have to divide 100 by 24.

Required number

Therefore, the required number

= 100 + (24 -4) = 120.

5. Determine the greatest 3-digit number exactly divisible by 8, 1 0 and 1 2.

Solution: The greatest three digit number = 999.

The greatest three digit number

LCM of 8, 10 and 12=2x2x2x3x5=120

Now, to find the number, we have to divide 999 by 120.

the required The greatest three digit number

Therefore, the required number = 999- 39 = 960

6. The traffic lights at three different road crossings change after every 48 seconds, 72 seconds and 108 seconds respectively. If they change simultaneously at 7 a.m., at what time will they change simultaneously again?

Solution:

LCM of 48, 72 and 108

LCM of 48, 72 and 108 = 2 x 2 x 2 x 2 x 3 x 3 x 3 = 432

After 432 seconds, the traffic lights change simultaneously.

432 seconds = 7 minutes 12 seconds

Therefore, the required time = 7 a.m. + 7 minutes 12 seconds i.e., 7 minutes 12 seconds past 7 a.m.

7. Three tankers contain 403 litres, 434 litres and 465 litres of diesel respectively. Find the maximum capacity of a container that can measure the diesel of the three containers exact number of times.

Solution: The maximum capacity of a container = HCF (403, 434, 465)

The prime factorisation of 403, 434 and 465 are;

403 = 13 x 31

434 = 2x7x31

465 = 3x5x31

HCF of 403, 434,465 = 31

Therefore, a container of capacity 31 litres is required to measure the diesel of the three containers.

8. Find the least number which when divided by 6, 15 and 18 leave remainder 5 in each case

Solution:

lcm of 6, 15,18

LCM of 6, 15 and 18 = 2 x 3 x 3 x 5 = 90

Therefore, the required number = 90 + 5 = 95

9. Find the smallest 4-digit number which is divisible by 18, 24 and 32.

Solution: The smallest four digit number = 1000

lcm of 8,24 and 32

LCM of 18, 24 and 32 = 2x2x2x2x2x3x3 = 288

Now,

The Required number is 1152

Therefore, the required number is 1000 + (288- 136) = 1152.

10. Find the LCM of the following numbers:

(1) 9 and 4
(2) 12 and 5
(3) 6 and 5
(4) 15 and 4

Observe a common property in the obtained LCMs. Is LCM the product of two numbers in each case?

Solution: (1) We have,

lcm of 9 and 4

LCM of 9 and 4 = 2x2x3x3 = 36

(2) We have

lcm of 12 and 5

LCM of 12 and 5 = 2 x 2 x 3 x 5 = 60

(3) We have

lcm of 6 and 5

LCM of 6 and 5 = 2x3x5 = 30

(4) We have

lcm of 15 and 4

LCM of 15 and 4 = 2x2x3x5 = 60

Yes, the LCM is equal to the product of two numbers in each case and all LCMs are also the multiple of 3.

11. Find the LCM of the following numbers in which one number is the factor of the other.

(1) 5,20
(2) 6,18
(3) 12,48
(4) 9,45

What do you observe in the results obtained?

Solution:(1) We have

lcm of 5 and 20

LCM of 5 and 20 = 2 x 2 x 5 = 20

(2) We have,

lcm of 6 and 18

LCM of 6 and 18 = 2 x 3 x 3 = 18

(3) We have,

lcm of 12 and 48

LCM of 12 and 48 = 2 x 2 x 2 x 2 x 3 = 48

(4) We have,

lcm of 9 and 45

LCM of 9and 45 = 3x3x5 = 45

From above, we observe that the LCM of the given numbers in each case is the larger of the two numbers.