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, repsectively.

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

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