Ratio, Proportion and Unitary Method Exercise 10.1 |

Ratio, Proportion and Unitary Method Exercise 10.2 |

Q1: Find each of the following ratios in the simplest form:

(i) 24 to 56

(ii) 84 paise to Rs. 3

(iii) 4 kg to 750 g

(iv) 1.8 kg to 6 kg

(v) 48 minutes to 1 hour

(vi) 2.4 km to 900 m

Sol:

(i) 24:56 = \(\frac{24}{56} = \frac{24 \div 8}{56 \div 8} = \frac{3}{7}\)

As the H.C.F of 3 and 7 is 1, the simplest form of 24 : 56 = 3 : 7

(ii) 84 paise to Rs. 3 = Rs. 0.84 to Rs. 3

\(\frac{0.84}{3} = \frac{0.84 \div 3}{3 \div 3} = \frac{0.28}{1} = \frac{28}{100} = \frac{28 \div 4}{100 \div 4} = \frac{7}{25}\)

As the H.C.F. of 7 and 25 is 1, the simplest form of 0.84 : 3 is 7 : 25.

(iii) 4 kg : 750 g = 4000g : 750 g = \(\frac{4000 \div 250}{750 \div 250} = \frac{16}{3}\)

As the H.C.F of 16 and 3 is 1, the simplest form of 4000:750 is 16 : 3.

(iv) 1.8 kg : 6 kg = \(\frac{1.8}{6} = \frac{18}{60} = \frac{18 \div 6}{60 \div 6} = \frac{3}{10} \)

As the H.C.F. of 3 and 10 is 1, the simplest form of 1.8 : 6 = 3 : 10.

(v) 48 minutes to 1 hour = 48 minutes to 60 minutes = 48 : 60 = \(\frac{48 \div 12}{60 \div 12} = \frac{4}{5}\)

As the H.C.F. of 4 and 5 is 1, the simplest form of 48 : 60 = 4 : 5.

(vi) 2.4 km to 900 m = 2400 m to 900 m = \(\frac{2400}{900} = \frac{24}{9} = \frac{24 \div 3}{9 \div 3} = \frac{8}{3}\)

As the H.C.F. of 8 and 3 is 1, the simplest form of 2400 : 900 = 8 : 3.

Q2: Express each of the following ratios in the simplest form:

(i) 36 : 90

(ii) 324 : 144

(iii) 85 : 561

(iv) 480 : 384

(v) 186 : 403

(vi) 777 : 1147

Sol:

(i) 36 : 90 = \(\frac{36}{90} = \frac{36 \div 18}{90 \div 18} = \frac{2}{5}\) (As the H.C.F. of 36 and 90 is 18)

Since the H.C.F. of 2 and 5 is 1, the simplest form of 36 : 90 is 2 : 5.

(ii) 324 : 144 = \(\frac{324}{144} = \frac{324 \div 36}{144 \div 36} = \frac{9}{4}\) (As the H.C.F. of 324 and 144 is 36)

Since the H.C.F. of 9 and 4 is 1, the simplest form of 324 : 144 is 9 : 4.

(iii) 85 : 561 = \(\frac{85}{561} = \frac{85 \div 17}{561 \div 17} = \frac{5}{33}\) (As the H.C.F. of 85 and 561 is 17)

Since the H.C.F. of 5 and 33 is 1, the simplest form of 85 : 561 is 5 : 33.

(iv) 480 : 384 = \(\frac{480}{384} = \frac{480 \div 96}{384 \div 96} = \frac{5}{4}\) (As the H.C.F. of 480 and 384 is 96)

Since the H.C.F. of 5 and 4 is 1, the simplest form of 480 : 384 is 5 : 4

(v) 186 : 403 = \(\frac{186}{403} = \frac{186 \div 31}{403 \div 31} = \frac{6}{13}\) (As the H.C.F. of 186 and 403 is 31)

Since the H.C.F. of 6 and 13 is 1, the simplest form of 186 : 403 is 6 : 13

(vi) 777 : 1147 = \(\frac{777}{1147} = \frac{777 \div 37}{1147 \div 37} = \frac{21}{31}\) (As the H.C.F. of 777 and 1147 is 37)

Since the H.C.F. of 21 and 31 is 1, the simplest form of 777 : 1147 is 21 : 31.

Q3: Write each of the following ratios in the simplest form:

(i) Rs. 6.30 to Rs. 16.80

(ii) 3 weeks : 30 days

(iii) 3m 5 cm : 35 cm

(iv) 48 min : 2 hours 40 min

(v) 1 L 35 mL : 270 mL

(vi) 4 kg : 2kg 500g

Sol:

(i) Rs. 6.30 to Rs. 16.80

\(\frac{6.30}{16.80} = \frac{63}{168} = \frac{63 \div 21}{168 \div 21} = \frac{3}{8}\) (H.C.F. of 63 and 168 is 21.)

Ratio = 3 : 8

(ii) 3 weeks : 30 days = 21 days : 30 days

\(\frac{21}{30} = \frac{21 \div 3}{30 \div 3} = \frac{7}{10}\) (H.C.F. of 21 and 30 is 3.)

Ratio = 7 : 10.

(iii) 3m 5 cm : 35 cm = 305 cm : 25 cm (as 1 m = 100 cm)

\(\frac{305}{35} = \frac{305 \div 5}{35 \div 5} = \frac{61}{7}\) (H.C.F. of 305 and 35 is 5.)

Ratio = 61 : 7.

(iv) 48 min : 2 hours 40 min = 48 min : 160 min (as 1 hour = 60 min)

\(\frac{48}{160} = \frac{48 \div 16}{160 \div 16} = \frac{3}{10}\) (H.C.F. of 48 and 160 is 16.)

Ratio = 3 : 10.

(v) 1 L 35 mL : 270 mL = 1035 mL : 270 mL ( as 1 L = 1000 mL)

\(\frac{1035}{270} = \frac{1035 \div 45}{270 \div 45} = \frac{23}{6}\) (H.C.F. of 1035 and 270 is 45.)

Ratio = 23 : 6.

(vi) 4 kg : 2kg 500g = 4000 g : 2500 g (as 1 kg = 1000 g)

\(\frac{4000}{2500} = \frac{4000 \div 500}{2500 \div 500} = \frac{8}{5}\) (H.C.F. of 4000 and 2500 is 500.)

Ratio = 8 : 5.

Q4: Mr Sahai and his wife are both school teachers and earn Rs. 16800 and Rs. 10500 per months respectively. Find the ratio of

(i) Mr. Sahil’s income to his wife’s income.

(ii) Mrs. Sahil’s income to her husband’s income.

(iii) Mr. Sahil’s income to the total income of the two.

Sol:

Mr Sahai’s earning = Rs. 16800

Mrs. Sahai’s earning = Rs. 10500

(i) Ratio = 16800 : 10500 = 168 : 105 = \(\frac{168 \div 21}{105 \div 21} = \frac{8}{5}\) (H.C.F. of 168 and 105 is 21.)

Mr Sahai’s income : Mrs. Sahai’s income = 8 : 5

(ii) Ratio = 10500 : 16800 = 105 : 168 = \(\frac{105 \div 21}{168 \div 21} = \frac{5}{8}\) (H.C.F. of 168 and 105 is 21.)

Mrs. Sahai’s income : Mr Sahai’s income = 5 : 8

(iii) Total income = 16800 + 10500 = Rs. 27300

Ratio = 16800 : 27300 = 168 : 273 = \(\frac{168 \div 21}{273 \div 21} = \frac{8}{13}\) (H.C.F. of 168 and 273 is 21.)

Mrs. Sahai’s income : Total income : 8 : 13

Q5: Rohit earns Rs. 15300 and saves Rs. 1224 per month. Find the ratio of

(i) his income and savings

(ii) his income and expenditure

(iii) his expenditure and saving

Sol:

Rohit’s income = Rs. 15300

Rohit’s saving = Rs. 1224

(i) Income : Savings = 15300 : 1224 = \(\frac{15300 \div 612}{1224 \div 612} = \frac{25}{2}\) (H.C.F. of 15300 and 1224 is 612.)

Income : Savings = 25 : 2

(ii) Monthly expenditure = Rs. (15300 – 1224) = Rs. 14076

Income : Expenditure = 14076 : 1224 = \(\frac{15300 \div 612}{14076 \div 612} = \frac{25}{23}\) (H.C.F. of 15300 and 14076 is 612.)

Income : Expenditure = 25 : 23

(iii) Expenditure : Savings = 14076 : 1224 = \(\frac{14076 \div 612}{1224 \div 612} = \frac{23}{2}\) (H.C.F. of 14076 and 1224 is 612.)

Expenditure : Savings = 23 : 2

Q6: The ratio of the number of males and females workers in a textile mill is 5 : 3. If there are 115 male worker, what is the number of female workers in the mill?

Sol:

Number of males : Number of females = 5 : 3

Let the number be x.

Number of males = 5x

Number of females = 3x

Number of male workers = 115

Now 5x = 115

\(\Rightarrow\) x = 23

Number of female workers in the mill = 3x = 3 \(\times\) 23 = 69

Q7: The boys and the girls in a school are in the ratio 9 : 5. If the total strength of the school is 448, find the number of girls.

Sol:

Boys : Girls = 9 : 5

Let the number of boys = 9x

Let the number of girls = 5x

Total strength of the school = 448

According to given condition, we have:

9x + 5x = 448

\(\Rightarrow\) 14x = 448

\(\Rightarrow\) x = 35

Number of boys = 9x = 9 \(\times\) 35 = 288

Number of girls = 5x = 5 \(\times\) 35 = 160

Q8: Divide Rs. 1575 between Kamal and Madhu in the ratio 7 : 2.

Sol:

Kamal : Madhu = 7 : 2

Sum of the ratio terms = 7 + 2 = 9

Kamal’s share = \(\frac{7}{9} \times 1575 = \frac{11025}{9}\) = Rs. 1225

Madhu’s share = \(\frac{2}{9} \times 1575 = \frac{3150}{9}\) = Rs. 350

Q9: Divide Rs. 3450 among A, B and C in the ratio 3 : 5 : 7.

Sol:

A: B : C = 3 : 5 : 7

Sum of the ratio terms = 3 + 5 + 7 = 15

A’s share = \(\frac{3}{15} \times 3450 = \frac{10350}{15}\) = Rs. 690

B’s share = \(\frac{5}{15} \times 3450 = \frac{17250}{15}\) = Rs. 1150

C’s share = \(\frac{7}{15} \times 3450 = \frac{24150}{15}\) = Rs. 1610

Q10: Two numbers are in the ratio 11 : 12 and their sum is 460. Find the numbers.

Sol:

Two numbers are in the ratio 11 : 12

Let the numbers be 11x and 12x.

Given: 11x + 12x = 460

\(\Rightarrow\) 23x = 460

\(\Rightarrow\) x = 20

First number = 11x = 11 \(\times\) 20 = 220

Second number = 12x = 12 \(\times\) 20 = 240

Hence, the numbers are 220 and 240.

Q11: A 35 cm line segment is divided into two parts in the ratio 4 : 3. Find the length of each part.

Sol:

Ratio of the two parts of line segment = 4 : 3

Sum of the ratio terms = 4 + 3 = 7

First part = \(\frac{4}{7} \times 35 = 4 \times 5\) = 20 cm

Second part = \(\frac{3}{7} \times 35 = 3 \times 5\) = 15 cm

Q12: A factory produces electric bulbs. If 1 out of every 10 bulbs is defective and the factory produces 630 bulbs per day, find the number of defective bulbs produced each day.

Sol:

Number of bulbs produced each day = 630

Out of 10 bulbs, 1 is defective.

Number of defective bulbs = \(\frac{630}{10} = 63\)

Therefore, Number of defective bulbs produced each day = 63

Q13: Find the ratio of the prices of a pencil to that of the ball pen if pencil cost Rs. 96 per score and ball pens cost Rs. 50.40 per dozen.

Sol:

Price of pencil = Rs. 96 per score

Price of ball pen = Rs. 50.40 per dozen

Price per unit of pencil = \(\frac{96}{20} = 4.8\)

Price per unit of ball pen = \(\frac{50.40}{12} = 4.2\)

Ratio = \(\frac{4.8}{4.2} = \frac{48}{42} = \frac{48 \div 6}{42 \div 6} = \frac{8}{7} \)

Price of pencil : Price of ball pen = 8 : 7

Q14: The ratio of the length of a field to its width is 5 : 3. Find its length if the width is 42 metres.

Sol:

Length : Width = 5 : 3

Let the length and the width of the field be 5x and 3x m respectively.

Width = 42 m

3x = 42

x = \(\frac{42}{3} = 14\)

Therefore,5x = 5 \(\times\) 14 = 70 m

Q15: The ratio of income to savings of a family is 11 : 2. Find the expenditure if the savings is Rs. 1520.

Sol:

Income : Savings = 11 : 2

Let the income and the savings be Rs. 11x and Rs. 2x respectively.

Savings = Rs. 1520

2x = 1520

x = \(\frac{1520}{2} = 760\)

Therefore,Income = Rs 11x = Rs (11 \(\times\) 760) = Rs. 8360

Expenditure = Income – Saving

= Rs (8360 – 1520)

Rs. 6840

Q16: The ratio of income to expenditure of a family is 7 : 6. Find the savings if the income is Rs. 14000.

Sol:

Income : Expenditure = 7 : 6

Let the income and the expenditure be Rs. 7x and 6x, respectively.

Income = Rs 14000

7x = 14000

X = 2000

Expenditure = Rs 6x = Rs 6 \(\times\) 2000 = Rs. 12000

Therefore,Saving = Income – Expenditure

= Rs. (14000 – 12000)

= Rs. 2000

Q17: The ratio of zinc and copper in an alloy is is 7 : 9. If the weight of copper in the alloy is 11.7 kg, find the weight of zinc in it.

Sol:

Let the weight of zinc be x kg.

Ratio of zinc and copper = 7.9

Weight of copper in the alloy = 11.7 kg

\(\frac{7}{9} = \frac{x}{11.7}\)

\(\Rightarrow x = \frac{7 \times 11.7}{9} = \frac{81.9}{9} = 9.1\)

Weight of zinc = 9.1 kg

Q18: A bus covers 128 km in 2 hours and a train covers 240 km in 3 hours. Find the ratio of their speeds.

Sol:

A bus covers 128 km in 2 hours.

Speed of the bus = \(\frac{Distance}{time} = \frac{128}{2} = 64\) km/h

A train covers 240 km in 3 hours = \(\frac{Distance}{time} = \frac{240}{3} = 80\) km/h

Ratio of their speeds = 64 : 80 = \(\frac{64}{80} = \frac{64 \div 16}{80 \div 16} = \frac{4}{5} \)

Therefore,Ratio of the speeds of the bus and the train = 4 : 5

Q19: From each of the given pairs, find which ratio is larger:

(i) (3 : 4) or (9 : 16)

(ii) (5 : 12 ) or (17 : 30)

(iii) (3 : 7) or (4 : 9)

(iv) (1 : 2) or (13 : 27)

Sol:

(i) (3 : 4) or (9 : 16)

Making the denominator equal

\(\frac{3 \times 4}{4 \times 4} = \frac{12}{16}\) and

\(\frac{12}{16} > \frac{9}{16}\)

Therefore, 3 : 4 > 9 : 16

(ii) (5 : 12 ) or (17 : 30)

Making the denominator equal

\(\frac{5 \times 5}{12 \times 5} = \frac{25}{60}\) and

\(\frac{17 \times 2}{30 \times 2} = \frac{34}{60}\)

\(\Rightarrow \frac{25}{60} < \frac{34}{60}\)

Therefore, 5 : 12 < 17 : 30

(iii) (3 : 7) or (4 : 9)

Making the denominator equal

\(\frac{3 \times 9}{7 \times 9} = \frac{27}{54}\) and

\(\frac{4 \times 7}{9 \times 7} = \frac{28}{63}\)

\(\Rightarrow \frac{27}{63} < \frac{28}{63}\)

Therefore, 3 : 7 < 4 : 9

(iv) (1 : 2) or (13 : 27)

Making the denominator equal

\(\frac{1 \times 27}{2 \times 27} = \frac{27}{54}\) and

\(\frac{13 \times 2}{27 \times 2} = \frac{26}{54}\)

\(\Rightarrow \frac{27}{54} > \frac{26}{54}\)

Therefore, 1 : 2 > 13 : 27

Q20: Fill in the blank places:

(i) \(\frac{24}{40} = \frac{}{5} = \frac{12}{}\)

(ii) \(\frac{36}{63} = \frac{4}{} = \frac{}{21}\)

(iii) \(\frac{5}{7} = \frac{}{28} = \frac{35}{}\)

Sol:

(i) \(\frac{24}{40} = \frac{24 \div 8}{40 \div 8} = \frac{3}{5} = \frac{3 \times 4}{5 \times 5} = \frac{12}{20} \)

(ii) \(\frac{36}{63} = \frac{36 \div 9}{63 \div 9} = \frac{4}{7} = \frac{4 \times 3}{7 \times 3} = \frac{12}{21} \)

(iii) \(\frac{5}{7} = \frac{5 \times 4}{7 \times 4} = \frac{20}{28} = \frac{5 \times 7}{7 \times 7} = \frac{35}{49} \)