EXERCISE 12.1

1. There are 20 girls and 15 boys in a class. (a) What is the ratio of number of girls to the number of boys? (b) What is the ratio of number of girls to the total number of students in the class?

Solution:

Given,

Number of girls = 20 girls

Number of boys = 15 boys

Total number of students = 20 + 15 = 35

Ratio of number of girls to number of boys = 20/15

= 4/3

2. Out of 30 students in a class, 6 like football, 12 like cricket and remaining like tennis. Find the ratio of

(a) Number of students liking football to number of students liking tennis.

(b) Number of students liking cricket to total number of students.

Solution:

Number of students liking tennis =30−(6+12)=30−18=12

a) Ratio of number of students liking football to number of students liking tennis
=6:12=1:2

b) Ratio of number of students liking cricket to total number of students
=12:30=2:5

Hence,
Ratio of number of students liking football to number of students liking tennis =1:2
Ratio of number of students liking cricket to total number of students =2:5

3. See the figure and find the ratio of

(a) Number of triangles to the number of circles inside the rectangle.

(b) Number of squares to all the figures inside the rectangle.

(c) Number of circles to all the figures inside the rectangle.

Solution:

From the given figure we get to know that

Number of triangles = 3

Number of circles = 2

Number of squares = 2

Total number of figures = 7

(a) Ratio of number of triangles to the number of circles inside the rectangle

= 3 / 2

(b) Ratio of number of squares to all the figures inside the rectangle

= 2 / 7

(c) Ratio of number of circles to all the figures inside the rectangle

= 2 / 7

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

Solution:

We know that, Speed = Distance Time

Speed of Hamid = 9 km/1 h = 9 km/h

Speed of Akhtar = 12 km/1 h = 12 km/h

Ratios are used to compare quantities which have same units. In this case, units of speed of Hamid and Akhtar is same i.e., km/h. Hence, we can find the required ratio.

∴ Ratio of speed of Hamid to the speed of Akhtar

= Speed of Hamid /Speed of Akhtar

912 = 34 = 3:4 [Dividing numerator and denominator by HCF of 9 and 12 i.e., 3]

5. Fill in the following blanks:

   [Are these equivalent ratios?]

Solution:

Given: 15/18= [ ]/ 6=10/[ ] = [ ] /30

15/18=5×3/6×3

⇒15/18=5/6 ….[1]

5/6=5×2/6×2 (By multiplying both numerator and denominator by 2 )

⇒5/6=10/12 … [2]

5/6=5×5/6×5 (By multiplying both numerator and denominator by 5 )

⇒5/6=25/30 …[3]

By comparing [1], [2] and [3], we can say that, 15/18=5/6=10/12=25/30

Hence, 15/18=5/6=10/12=25/30.

Yes, they are equivalent ratios.

6. Find the ratio of the following:

(a) 81 to 108

(b) 98 to 63

(c) 33 km to 121 km

(d) 30 minutes to 45 minutes

Solution:

• 81 to 108

= Ratio = 81/108

= (81/ 108) * (27/27)

= 3/4

= 3:4 is ratio.

• 98 to 63

= Ratio = 98/63

= (98/63) * (7/7)

= 14/9

= 14:9 is ratio.

• 33 km to 121 km

= Ratio = 33/121 km

= (33/121) * (11/11)

= 3/11

= 3:11 km is ratio.

• 30 minutes to 45 minutes

= ratio = 30/45 min

= (30/45) * (15/15)

= 2/3

= 2:3 min is ratio.

7. Find the ratio of the following:

(a) 30 minutes to 1.5 hours

(b) 40 cm to 1.5 m

(c) 55 paise to Rs. 1

(d) 500 mL to 2 litres

Solution:

• 30 minutes to 1.5 hours

= 30 min to 90 min ( 1 hours = 60 min)

= (30/90) * (30/30)

= 1/3

= 1:3 min is ratio.

• 40 cm to 1.5 m

= 40 cm to 150 cm (1m = 100 cm)

= 40/150

= 4/15

= 4:15 cm is ratio.

• 55 paise to Rs. 1

= 55 paise to 100 paise (1Rs. = 100 paise)

= (55/100) * (5/5) = 11/20

= 11:20 paise is ratio.

• 500 ml to 2 litres

= 500 ml to 2000 ml (1 litres = 1000 ml)

= (500/2000) * (500/500)

= 1/4

= 1:4 ml is ratio.

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

(a) Money that Seema earns to the money she saves.

(b) Money that she saves to the money she spends.

Solution:

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:

Number of teachers in the school = 102

Number of students in the school = 3300

The ratio of the number of teachers to the number of students = Number of teachers in the school / Number of students in the school

= 102 / 3300

Here,

The factors of 102 are 2 × 3 × 17 and the factors of 3300 is 2 × 2 × 3 × 5 × 5 × 11 

Now, the HCF of 102 and 3300 is 2 × 3 = 6.

= (6 × 17) / (6 × 550)

= 17 / 550

Therefore, 17:550 is the resultant ratio of the number of teachers to the number of students.

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

(a) Number of girls to the total number of students.

(b) Number of boys to the number of girls.

(c) Number of boys to the total number of students.

Solution:

(a) We have,

Total number of students=4320

Number of girls=2300

Therefore, ratio of number of girls to the total number of students=Number of girls Total number of students=2300/4320

Dividing numerator and denominator by their HCF i.e., 20=115/216=115:216

Hence, required ratio is 115:216

(b) Total number of students in school = 4320

Number of girls = 2300

Therefore, number of boys = Total number of students – Number of girls

= 4320 – 2300 = 2020

Ratio of number of boys to that of girls

= Number of boys Number of girls

= 20202300 = 101115 = 101:115 [Dividing numerator and denominator by HCF of 2020 and 2300 i.e. 20]

(c) Total number of students in school = 4320

Number of girls = 2300

Therefore, number of boys = Total number of students – Number of girls

= 4320 – 2300 = 2020

Ratio of number of boys to total number of students

= Number of boys/Total number of students

= 2020/4320 = 101/216 = 101:216 [Dividing numerator and denominator by HCF of 2020 and 4300 i.e.,

20]

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

(a) Number of students who opted basketball to the number of students who opted table tennis.

(b) Number of students who opted cricket to the number of students opting basketball.

(c) Number of students who opted basketball to the total number of students.

Solution:

(a) Required ratio = 750/250 = 3

(b) Required ratio =800/750 = 16/15

(c) Required ratio =750/1800 = 25/60 = 5/12

12. Cost of a dozen pens is ` 180 and 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) = Rs 180

∴ Cost of 1 pen = 180/12 = Rs 15

Cost of 8 ball pens = Rs 56

∴ Cost of 1 ball pen = 56/8 = Rs 7

Ratio of cost of one pen to that of one ball pen

= Cost of 1 pen/Cost of 1 ball pen

= 157 = 15:7

13. Consider the statement: 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.

Solution:

Two ratios are called equivalent ratios when their simplest forms (or standard forms) are equal.

Ratio of breadth to length = 2 : 5 = 25

1025=2×55×5

∴ Other equivalent ratios are = 25×1010=2050, 25×2020=40100

Thus,

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

Now,

Let the total number of pens be (3 + 2) x = 5x 

5x = 20

x = 4

The total number of pens with Sheela = 3x

3 × 4 = 12

The total number of pens with Sangeeta = 2x

2 × 4 = 8

Therefore, Sheela gets 12 pens and Sangeeta gets 8 pens.

15. Mother wants to divide ` 36 between her daughters Shreya and Bhoomika in the ratio of their ages. If age of Shreya is 15 years and age of Bhoomika is 12 years, find how much Shreya and Bhoomika will get.

Solution:

Ratio of ages =15/12 = 5/4

Therefore, mother wants to divide Rs 36 in a ratio of 5: 4.

Terms of 5: 4 are 5 and 4.

Sum of these terms = 5 + 4 = 9

Shreya will get 5/9 of the total money and Bhoomika will get 4/9 of it.

Amount that Shreya will get = (5/9) * 36 = 20

Amount that Bhoomika will get = (4/9) * 36=16

Therefore, Shreya and Bhoomika will get Rs 20 and Rs 16 respectively.

16. Present age of father is 42 years and that of his son is 14 years. Find the ratio of

(a) Present age of father to the present age of son.

(b) Age of the father to the age of son, when son was 12 years old.

(c) Age of father after 10 years to the age of son after 10 years.

(d) Age of father to the age of son when father was 30 years old.

Solution:

The correct option is D

10:3

When the son was 12 years, i.e., 2 years ago, father’s age would have been (42 – 2) = 40 years.
Therefore, the ratio of their ages = 4012​ = 103​ = 10:3.

EXERCISE 12.2

1. Determine if the following are in proportion.

(a) 15, 45, 40, 120

(b) 33, 121, 9,96

(c) 24, 28, 36, 48

(d) 32, 48, 70, 210

(e) 4, 6, 8, 12

(f) 33, 44, 75, 100

Solution:

(a) First will combine two terms and find the ratio of those two terms 

15, 45, 40, 120

15 / 45 = 1 / 3

40 / 120 = 1 / 3

Hence, 15: 45 = 40:120

∴ The given set of numbers are in a proportion

(b) First will combine two terms and find the ratio of those two terms 

33, 121, 9, 96

33 / 121 = 3 / 11

9 / 96 = 3 / 32

Hence 33:121 ≠ 9: 96

∴ The given set of numbers are not in a proportion

(c)  24, 28, 36, 48

24 / 28 = 6 / 7

36 / 48 = 3 / 4

Hence, 24: 28 ≠ 36:48

∴ The given set of numbers are not in a proportion

(d)  32, 48, 70, 210

32 / 48 = 2 / 3

70 / 210 = 1 / 3

Hence, 32: 48 ≠ 70: 210

∴ The given set of numbers are not in a proportion

(e) First will combine two terms and find the ratio of those two terms

4, 6, 8, 12

4 / 6 = 2 / 3

8 / 12 = 2 / 3

Hence 4: 6 = 8: 12

∴ The given set of numbers are in a proportion

(f) First will combine two terms and find the ratio of those two terms

33, 44, 75, 100

33/ 44 = 3/ 4

75 / 100 = 3 / 4

Hence, 33:44 = 75: 100

∴ The given set of numbers are in a proportion

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

(a) 16: 24: :20 : 30

(b) 21: 6 :: 35 : 10

(c) 12 : 18 :: 28 : 12

(d) 8 : 9 :: 24 : 27

(e) 5.2 : 3.9 :: 3 : 4

(f) 0.9 : 0.36 :: 10 : 4

Solution:

(a) The given statement is True

Explanation

16: 24 :: 20: 30

16 / 24 = 2 / 3

20 / 30 = 2 / 3

Hence, 16: 24 = 20: 30

Hence, True

(b)  The given statement is True

Explanation

21: 6:: 35: 10

21 / 6 = 7 / 2

35 / 10 = 7 / 2

Hence, 21: 6 = 35: 10

Hence, True

(c) The given statement is False

Explanation

12: 18 :: 28: 12

12 / 18 = 2 / 3

28 / 12 = 7 / 3

Hence, 12: 18 ≠ 28:12

Hence, False

(d) The given statement is True

Explanation

8: 9:: 24: 27

We know that = 24 / 27 = (3 × 8) / (3 × 9)

= 8 / 9

Hence, 8: 9 = 24: 27

Hence, True

(e) The given statement is False

Explanation

5.2: 3.9:: 3: 4

As 5.2 / 3.9 = 3 / 4

Hence, 5.2: 3.9 ≠ 3: 4

Hence, False

(f) The given statement is True

Explanation

0.9: 0.36:: 10: 4

0.9 / 0.36 = 90 / 36

= 10 / 4

Hence, 0.9: 0.36 = 10: 4

Hence, True

3. Are the following statements true?

(a) 40 persons: 200 persons = ` 15 : ` 75

(b) 7.5 litres : 15 litres = 5 kg : 10 kg

(c) 99 kg : 45 kg = ` 44 : ` 20

(d) 32 m : 64 m = 6 sec : 12 sec

(e) 45 km : 60 km = 12 hours : 15 hours

Solution:

(a) 40 persons: 200 persons = Rs 15: Rs 75

True

(b) 7.5 l: 15 l = 5 kg: 10 kg

True

(c) 99 kg: 45 kg = Rs 44: Rs 20

True

(d) 32 m: 64 m = 6 sec: 12 sec

True

(e) 45 km: 60 km = 12 hrs: 15 hrs

False

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

(a) 25 cm : 1 m and Rs. 40 : Rs. 160

(b)39 litres : 65 litres and 6 bottles : 10 bottles

(c) 2 kg : 80 kg and 25 g : 625 g

(d) 200 mL : 2.5 litre and Rs. 4 : Rs. 50

Solution:

(a) 25cm: 1 m and Rs 40: Rs 160

25 cm = = 0.25 m

Yes. These are in proportion.

Middle terms are 1m, Rs 40.

Extreme terms are 25 cm, Rs 160.

(b) 39 l: 65 l and 6 bottles: 10 bottles

Yes. These are in proportion.

Middle terms are 65 l, 6 bottles.

Extreme terms are 39 l, 10 bottles.

(c) 2 kg: 80 kg and 25g: 625 g

No. These are not in proportion.

(d) 200 mL: 2.5 l and Rs 4: Rs 50

1 l = 1000 mL

2.5 l = 2500 mL

Yes. These are in proportion.

Middle terms are 2.5 l, Rs 4.

Extreme terms are 200 mL, Rs 50.

EXERCISE 12.3

1. If the cost of 7 m of cloth is Rs. 1470, find the cost of 5 m of cloth.

Solution:

Cost of 7 m cloth = Rs. 1470

Cost of 1 m cloth = 1470/7

= Rs. 210

Therefore, Cost of 5 cloth = 210 * 5

= Rs. 1050

2. Ekta earns Rs. 3000 in 10 days. How much will she earn in 30 days?

Solution:

Given the salary of Ekta in 10days is rs.30008

Ekta earns in a day = 3000/10 = 300

therefore, Ekta will earn in 30days=300*30=9000

hence, Ekta will earn rs.9000 in 30days.

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:

the measure of rain in 3 days is 276 mm.

The measure of rain in one day = 276 / 3 = 92 mm

Therefore, the measure of rain in one week that is 7 days = 92 × 7 = 644 mm

To change from mm to cm, we need to divide by 644 by 10 since, 1 cm = 10 mm

= 644 / 10

= 64.4 cm

Thus, the measure of rain in one week that is 7 days, will be 64.4 cm.

4. Cost of 5 kg of wheat is Rs. 91.50.

(a) What will be the cost of 8 kg of wheat?

(b) What quantity of wheat can be purchased in Rs. 183?

Solution:

Given that,

Cost of 5 kg of wheat =₹ 91.50

Quantity of wheat purchased in ₹ 1 =5/91.50 kg

Quantity of wheat purchased in ₹ 183

=5/91.50×183 kg=10 kg

Hence, the quantity of wheat can be purchased in ₹183 is 10 kg.

• The temperature dropped 15 degrees Celsius in the last 30 days. If the rate of temperature drop remains the same, how many degrees will the temperature drop in the next ten days?

Solution:

It is given in the problem that in the last 30 days the temperature dropped by 15 degrees Celsius.

So, the rate of drop in temperature is 15/30=1/2 degree Celsius per day.

Now, as per the given condition, the rate of drop in temperature remains constant.

Therefore, in the next 10 days at this rate of drop, the temperature will drop by (1/2) × 10= 5 degree Celsius.

• Shaina pays Rs. 15000 as rent for 3 months. How much does she have to pay for a whole year, if the rent per month remains same?

Solution:

Rent paid by Shaina in 3 months = ₹ 15000

Rent for 1 month will be 15000 / 3 = ₹ 5000

To calculate the rent for 12 months, we will multiply 5000 with 12

Therefore, rent for 12 months (1 year) = 5000 × 12 = ₹ 60,000

Thus, rent paid by Shaina in 1 year is ₹ 60,000.

7. Cost of 4 dozen bananas is Rs. 180. How many bananas can be purchased for Rs. 90?

Solution:

Number of bananas bought in₹ 180 = 4 dozens

= 4 × 12

= 48 bananas

Number of bananas bought in ₹ 1 = 48 / 180

So, number of bananas bought in ₹ 90 = (48 / 180) × 90

= 24 bananas

∴ 24 bananas can be purchased in ₹ 90

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

Solution:

Weight of 72 books = 9 kg

Weight of 1 book = 9 / 72

= 1 / 8 kg

So, weight of 40 books = (1 / 8) × 40

= 5 kg

∴ Weight of 40 books is 5 kg

9. A truck requires 108 litres of diesel for covering a distance of 594 km. How much diesel will be required by the truck to cover a distance of 1650 km?

Solutions:

Diesel required for 594 km = 108 litres

Diesel required for 1 km = 108 / 594

= 2 / 11 litre

So, diesel required for 1650 km = (2 / 11) × 1650

= 300 litres

∴ Diesel required by the truck to cover a distance of 1650 km is 300 litres

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

Solutions:

Pens purchased by Raju in ₹ 150 = 10 pens

Cost of 1 pen = 150 / 10

= ₹ 15

Pens purchased by Manish in ₹ 84 = 7 pens

Cost of 1 pen = 84 / 7

= ₹ 12

∴ Pens purchased by Manish are cheaper than Raju

11. Anish made 42 runs in 6 overs and Anup made 63 runs in 7 overs. Who made more runs per over?

Solutions:

Runs made by Anish in 6 overs = 42

Runs made by Anish in 1 over = 42 / 6

= 7

Runs made by Anup in 7 overs = 63

Runs made by Anup in 1 over = 63 / 7

= 9 ∴ Anup scored more runs than Anish