Ratio And Proportion Exercise – 12.1
Question 1. There are 20 girls and 1 5 boys in a class.
- What is the ratio of number of girls to the number of boys?
- What is the ratio of number of girls to the total number of students in the class?
Solution:
- The ratio of the number of girls to that of boys \(=\frac{20}{15}=\frac{4}{3}=4: 3\)
- 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
- Several students like football to several students like tennis.
- 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
- The ratio of the number of students liking football to that of tennis = \(\frac{6}{12}=\frac{1}{2}=1: 2\)
- 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
- Number of triangles to the number of circles inside the rectangle.
- Number of squares to all the figures inside the rectangle.
- Number of circles to all the figures inside the rectangle.
Solution:
- The ratio of the number of triangles to that of circles \(=\frac{3}{2}=3: 2\)
- The ratio of the number of squares to that of all figures \(=\frac{2}{7}=2: 7\)
- 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: 
[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 \(\frac{25}{30}.\)
Yes, these are equivalent ratios.
Question 6. Find the ratio of the following :
- 81 to 108
- 98 to 63
- 33 km to 121 km
- 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:
- 30 minutes to 1.5 hours
- 40 cm to 1.5 m
- 55 paise to? 1
- 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
⇒ \(500 \mathrm{ml}: 2000 \mathrm{ml}=\frac{500}{2000}=\frac{1}{4}=1: 4\)
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
- Number of girls to the total number of students.
- Number of boys to the number of girls.
- 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
- The number of students who opted for basketball to the number of students who opted for table tennis.
- Several students opted for cricket and several students opted for basketball.
- 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.
- The ratio of the number of students who opted for basketball to that who opted for table tennis = \(\frac{750}{250}=\frac{3}{1}=3: 1\)
- The ratio of the number of students who opted for cricket to that who opted for basketball\(=\frac{800}{750}=\frac{16}{15}=16: 15\)
- 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
- Present age of father to the present age of son.
- Age of the father to the age of the son, when the son was 12 years old.
- Age of father after 1 0 years to the age of son after 1 0 years.
- 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.
- 15,45,40,120
- 33,121,9,96
- 24,28,36,48
- 32,48,70,210
- 4,6,8,12
- 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 :
- 16: 24:: 20: 30
- 21: 6:: 35: 10
- 12: 18:: 28: 12
- 8 : 9:: 24: 27
- 5 : 2: 3.9:: 3:4
- 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?
- 40 persons : 200 persons = ?15:? 75
- 7.5 litres: 1 5 litres = 5 kg: 1 0 kg
- 9 kg : 45 kg = ?44: ? 20
- 32 m :64 m = 6 sec: 12 sec
- 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.
- 25cm:1 m and? 40:? 160
- 39 litres: 65 litres and 6 bottles: 1 0 bottles
- 2 kg: 8 kg and 25 g: 625 g
- 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.
- What will be the cost of 8 kg of wheat?
- 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.