Ratio & Proportion

(i) 20 : 40

The given ratio is 20 : 40.

Let us find H.C.F. of 20 and 40:

$$\begin{array}{l} 20\overline{\smash{\big)}40\smash{\big(}} 2 \\ \phantom{)}\phantom{)}\underline{-40} \\ \phantom{++}0 \end{array}$$

H.C.F. = 20

$\therefore 20 : 40 = \frac{20 \div 20}{40 \div 20} = \frac{1}{2} = 1 : 2$.

Hence, the answer is 1 : 2

(ii) 40 : 20

The given ratio is 40 : 20.

Let us find H.C.F. of 40 and 20:

$$\begin{array}{r} 2 \\ 20 \overline{) 40} \\ \underline{-40} \\ 0 \end{array}$$

H.C.F. = 20

$\therefore 40 : 20 = \frac{40 \div 20}{20 \div 20} = \frac{2}{1} = 2 : 1$

Hence, the answer is 2 : 1

(iii) 81 : 108

The given ratio is 81 : 108.

Let us find H.C.F. of 81 and 108:

$$\begin{array}{l} 81\overline{\smash{\big)}108\smash{\big(}} 1 \\ \phantom{ac}\phantom{)}\underline{-81} \\ \phantom{{x^2 =}} 27 \overline{\smash{\big)}81\smash{\big(}} 3 \\ \phantom{ac = }\phantom{))}\underline{-81} \\ \phantom{++++}0 \end{array}$$

H.C.F. = 27

$\therefore 81 : 108 = \frac{81 \div 27}{108 \div 27} = \frac{3}{4} = 3 : 4$

Hence, the answer is 3 : 4

(iv) 98 : 63

The given ratio is 98 : 63.

Let us find H.C.F. of 98 and 63:

$$\begin{array}{l} 63\overline{\smash{\big)}98\smash{\big(}} 1 \\ \phantom{)}\phantom{)}\underline{-63} \\ \phantom{{x^2 ()}} 35 \overline{\smash{\big)}63\smash{\big(}} 1 \\ \phantom{ac())}\phantom{)}\underline{-35} \\ \phantom{{x^2 )} + 2)}28\overline{\smash{\big)}35 \smash{\big(}} 1 \\ \phantom{ac = sc ()}\phantom{)}\underline{-28} \\ \phantom{{x^2 )} + 2x = a}7\overline{\smash{\big)}28\smash{\big(}}4 \\ \phantom{ac = sc + sa}\phantom{)}\underline{-28} \\ \phantom{+++++++}0 \end{array}$$

H.C.F. = 7

$\therefore 98 : 63 = \frac{98 \div 7}{63 \div 7} = \frac{14}{9} = 14 : 9$

Hence, the answer is 14 : 9

(i) $\frac{14}{21} = \frac{...}{3} = \frac{6}{...}$

Let,

$\frac{14}{21} = \frac{...}{3} = \frac{6}{...}$ = $\frac{14}{21} = \frac{x}{3} = \frac{6}{y}$

First, reduce $\frac{14}{21}$ to its simplest form:

$\frac{14 \div 7}{21 \div 7} = \frac{2}{3}$

$\Rightarrow$ x = 2

Now, $\frac{2}{3} = \frac{6}{y}$

Cross multiplication,

2y = 18

$\Rightarrow$ y = 9

Hence, the missing numbers are 2 and 9.

(ii) $\frac{15}{18} = \frac{...}{6} = \frac{10}{...} = \frac{...}{30}$

First, reduce $\frac{15}{18}$ to its simplest form:

$\frac{15}{18} = \frac{15 \div 3}{18 \div 3} = \frac{5}{6}$

Now compare each fraction with $\frac{5}{6}$:

$\frac{...}{6} = \frac{5}{6} \Rightarrow \frac{5}{6}$

$\frac{10}{...} = \frac{5}{6} \Rightarrow \frac{5 \times 2}{6 \times 2} = \frac{10}{12}$

$\frac{...}{30} = \frac{5}{6} \Rightarrow \frac{5 \times 5}{6 \times 5} = \frac{25}{30}$

Thus,

$\frac{15}{18} = \frac{5}{6} = \frac{10}{12} = \frac{25}{30}$.

Hence, the missing numbers are 5, 12 and 25 respectively.

(i) 2.1 m to 1.2 m

Multiply both terms by 10 to remove decimals:

2.1 m : 1.2 m = 21 : 12

H.C.F. of 21 and 12 is 3.

$\frac{21 \div 3}{12 \div 3} = \frac{7}{4} = 7 : 4$

Hence, the answer is 7 : 4.

(ii) 91 cm to 1.04 m

Convert m into cm: 1.04 m = 1.04 × 100 cm = 104 cm.

Ratio = 91 : 104

H.C.F. of 91 and 104 is 13.

$\frac{91 \div 13}{104 \div 13} = \frac{7}{8} = 7 : 8$

Hence, the answer is 7 : 8.

(iii) 3.5 kg to 250 g

Convert kg into g: 3.5 kg = 3.5 × 1000 g = 3500 g.

Ratio = 3500 : 250

H.C.F. of 3500 and 250 is 250.

$\frac{3500 \div 250}{250 \div 250} = \frac{14}{1} = 14 : 1$

Hence, the answer is 14 : 1.

(iv) 60 paise to 4 rupees

Convert rupees into paise: 4 rupees = 4 × 100 paise = 400 paise.

Ratio = 60 : 400

H.C.F. of 60 and 400 is 20.

$\frac{60 \div 20}{400 \div 20} = \frac{3}{20} = 3 : 20$

Hence, the answer is 3 : 20.

(v) 1 minute to 15 seconds

Convert minute into seconds: 1 minute = 60 seconds.

Ratio = 60 : 15

H.C.F. of 60 and 15 is 15.

$\frac{60 \div 15}{15 \div 15} = \frac{4}{1} = 4 : 1$

Hence, the answer is 4 : 1.

(vi) 15 mm to 2 cm

Convert cm into mm: 2 cm = 2 × 10 mm = 20 mm.

Ratio = 15 : 20

H.C.F. of 15 and 20 is 5.

$\frac{15 \div 5}{20 \div 5} = \frac{3}{4} = 3 : 4$

Hence, the answer is 3 : 4.

Given:

Length = 125 m

Breadth = 60 m

Ratio of length to breadth = 125 : 60

H.C.F. of 125 and 60 is 5.

$\frac{125 \div 5}{60 \div 5} = \frac{25}{12} = 25 : 12$

Hence, the ratio of length to breadth is 25 : 12.

Given:

Total population = 4800

Number of females = 2160

Number of males = Total population − Number of females

Number of males = 4800 − 2160 = 2640

Ratio of males to females = 2640 : 2160

H.C.F. of 2640 and 2160 is 240.

$\frac{2640 \div 240}{2160 \div 240} = \frac{11}{9} = 11 : 9$

Hence, the ratio of males to females is 11 : 9.

Given:

Number of boys = 30

Number of girls = 25

Total number of students = 30 + 25 = 55

(i) Ratio of boys to girls = 30 : 25

H.C.F. of 30 and 25 is 5.

$\frac{30 \div 5}{25 \div 5} = \frac{6}{5} = 6 : 5$

Hence, the ratio of boys to girls is 6 : 5.

(ii) Ratio of girls to total = 25 : 55

H.C.F. of 25 and 55 is 5.

$\frac{25 \div 5}{55 \div 5} = \frac{5}{11} = 5 : 11$

Hence, the ratio of girls to total number of students is 5 : 11.

(iii) Ratio of boys to total = 30 : 55

H.C.F. of 30 and 55 is 5.

$\frac{30 \div 5}{55 \div 5} = \frac{6}{11} = 6 : 11$

Hence, the ratio of boys to total number of students is 6 : 11.

Given:

Money Reena earns = ₹1,50,000

Money Reena saves = ₹50,000

Money Reena spends = Money earned − Money saved

Money Reena spends = ₹1,50,000 − ₹50,000 = ₹1,00,000

(i) Ratio of money earned to money saved = 1,50,000 : 50,000 = 150 : 50 = 3 : 1

Hence, the ratio of money earned to money saved is 3 : 1.

(ii) Ratio of money saved to money spent = 50,000 : 1,00,000 = 50 : 100 = 1 : 2

Hence, the ratio of money saved to money spent is 1 : 2.

Given:

Original expenses = ₹350

Increased expenses = ₹500

Increase in expenses = ₹500 − ₹350 = ₹150

(i) Ratio of increase in expenses to original expenses = 150 : 350

H.C.F. of 150 and 350 is 50.

$\frac{150 \div 50}{350 \div 50} = \frac{3}{7} = 3 : 7$

Hence, the required ratio is 3 : 7.

(ii) Ratio of original expenses to increased expenses = 350 : 500

H.C.F. of 350 and 500 is 50.

$\frac{350 \div 50}{500 \div 50} = \frac{7}{10} = 7 : 10$

Hence, the required ratio is 7 : 10.

(iii) Ratio of increased expenses to increase in expenses = 500 : 150

H.C.F. of 500 and 150 is 50.

$\frac{500 \div 50}{150 \div 50} = \frac{10}{3} = 10 : 3$

Hence, the required ratio is 10 : 3.

Given:

Total students = 30

Students liking football = 6

Students liking cricket = 12

Students liking tennis = 30 − 6 − 12 = 12

(i) Ratio of football lovers to tennis lovers = 6 : 12

H.C.F. of 6 and 12 is 6.

$\frac{6 \div 6}{12 \div 6} = \frac{1}{2} = 1 : 2$

Hence, the required ratio is 1 : 2.

(ii) Ratio of cricket lovers to total students = 12 : 30

H.C.F. of 12 and 30 is 6.

$\frac{12 \div 6}{30 \div 6} = \frac{2}{5} = 2 : 5$

Hence, the required ratio is 2 : 5.

(i) 3 : 4 or 2 : 3

The given ratios are 3 : 4 and 2 : 3, which are equivalent to the fractions $\frac{3}{4}$ and $\frac{2}{3}$ respectively.

Let us find the L.C.M. of 4 and 3:

$$\begin{array}{l|rr} 2 & 4, & 3 \\ \hline 2 & 2, & 3 \\ \hline 3 & 1, & 3 \\ \hline & 1, & 1 \end{array}$$

L.C.M. = 2 × 2 × 3 = 12.

$3 : 4 = \frac{3}{4} = \frac{3 \times 3}{4 \times 3} = \frac{9}{12}$

$2 : 3 = \frac{2}{3} = \frac{2 \times 4}{3 \times 4} = \frac{8}{12}$

Clearly, $\frac{9}{12} \gt \frac{8}{12}$.

Hence, (3 : 4) > (2 : 3).

(ii) 11 : 21 or 15 : 28

The given ratios are 11 : 21 and 15 : 28, which are equivalent to the fractions $\frac{11}{21}$ and $\frac{15}{28}$ respectively.

Let us find the L.C.M. of 21 and 28:

$$\begin{array}{l|rr} 2 & 21, & 28 \\ \hline 2 & 21, & 14 \\ \hline 3 & 21, & 7 \\ \hline 7 & 7, & 7 \\ \hline & 1, & 1 \end{array}$$

L.C.M. = 2 x 2 × 3 × 7 = 84.

$11 : 21 = \frac{11}{21} = \frac{11 \times 4}{21 \times 4} = \frac{44}{84}$

$15 : 28 = \frac{15}{28} = \frac{15 \times 3}{28 \times 3} = \frac{45}{84}$

Clearly, $\frac{45}{84} \gt \frac{44}{84}$.

Hence, (15 : 28) > (11 : 21).

Given:

Total money = ₹560

Ratio = 3 : 2

Find total parts: 3 + 2 = 5 parts.

Find the value of 1 part: 560 ÷ 5 = ₹112. So, 1 part = ₹112.

Now calculate shares by multiplying each part with ₹112:

Ramu's share: 3 × ₹112 = ₹336

Munni's share: 2 × ₹112 = ₹224

Hence, Ramu gets ₹336 and Munni gets ₹224.

Given:

First investment = ₹15,000

Second investment = ₹25,000

Ratio of investments = 15000 : 25000 = 15 : 25 = 3 : 5

Total profit = ₹12,000

Find total parts: 3 + 5 = 8 parts.

Find the value of 1 part: 12000 ÷ 8 = ₹1,500. So, 1 part = ₹1,500.

Now calculate shares by multiplying each part with ₹1500:

First person's share: 3 × ₹1,500 = ₹4,500

Second person's share: 5 × ₹1,500 = ₹7,500

Hence, the first person gets ₹4,500 and the second person gets ₹7,500.

Given:

Ratio (Ankur : Roma) = 9 : 11

Ankur's money = ₹540

Since 9 parts represent Ankur's money, 9 parts = ₹540.

1 part = 540 ÷ 9 = ₹60.

Roma's money = 11 parts = 11 × ₹60 = ₹660.

Hence, Roma has ₹660.

Given:

Ratio (Tin : Zinc) = 2 : 5

Total weight of alloy = 31.5 g

Find total parts: 2 + 5 = 7 parts.

Find the value of 1 part: 31.5 ÷ 7 = 4.5 g. So, 1 part = 4.5 g.

Weight of zinc = 5 parts = 5 × 4.5 g = 22.5 g.

Hence, the weight of zinc in the alloy is 22.5 g.

In a proportion, product of extremes = product of means.

(i) 4 : 6 and 12 : 18

Product of extremes = 4 × 18 = 72

Product of means = 6 × 12 = 72

Since product of extremes = product of means, the ratios are in proportion.

Hence, 4 : 6 and 12 : 18 form a proportion.

(ii) 15 : 45 and 40 : 120

Product of extremes = 15 × 120 = 1800

Product of means = 45 × 40 = 1800

Since product of extremes = product of means, the ratios are in proportion.

Hence, 15 : 45 and 40 : 120 form a proportion.

(iii) 14 : 4 and 18 : 6

Product of extremes = 14 × 6 = 84

Product of means = 4 × 18 = 72

Since product of extremes ≠ product of means, the ratios are not in proportion.

Hence, 14 : 4 and 18 : 6 do not form a proportion.

(iv) 12 : 18 and 28 : 12

Product of extremes = 12 × 12 = 144

Product of means = 18 × 28 = 504

Since product of extremes ≠ product of means, the ratios are not in proportion.

Hence, 12 : 18 and 28 : 12 do not form a proportion.

(i) 16 : 24 = 20 : 30

Product of extremes = 16 × 30 = 480

Product of means = 24 × 20 = 480

Product of extremes = product of means.

Hence, the statement is True.

(ii) 16 : 24 = 30 : 20

Product of extremes = 16 × 20 = 320

Product of means = 24 × 30 = 720

Product of extremes ≠ product of means.

Hence, the statement is False.

(iii) 21 : 6 :: 35 : 10

Product of extremes = 21 × 10 = 210

Product of means = 6 × 35 = 210

Product of extremes = product of means.

Hence, the statement is True.

(iv) 5.2 : 3.9 :: 3 : 4

Product of extremes = 5.2 × 4 = 20.8

Product of means = 3.9 × 3 = 11.7

Product of extremes ≠ product of means.

Hence, the statement is False.

Four numbers a, b, c, d are in proportion if and only if a × d = b × c.

(i) 12, 16, 6, 8

Product of extremes = 12 × 8 = 96

Product of means = 16 × 6 = 96

Since 96 = 96, the numbers are in proportion.

Hence, 12, 16, 6, 8 are in proportion.

(ii) 2, 3, 4, 5

Product of extremes = 2 × 5 = 10

Product of means = 3 × 4 = 12

Since 10 ≠ 12, the numbers are not in proportion.

Hence, 2, 3, 4, 5 are not in proportion.

(iii) 18, 10, 9, 5

Product of extremes = 18 × 5 = 90

Product of means = 10 × 9 = 90

Since 90 = 90, the numbers are in proportion.

Hence, 18, 10, 9, 5 are in proportion.

(iv) 18, 9, 10, 5

Product of extremes = 18 × 5 = 90

Product of means = 9 × 10 = 90

Since product of extremes = product of means, the numbers are in proportion.

Hence, 18, 9, 10, 5 are in proportion.

(i) 39 kg : 36 kg = 26 men : 24 men

$39 : 36 = \frac{39}{36} = \frac{13}{12}$

$26 : 24 = \frac{26}{24} = \frac{13}{12}$

Both ratios are equal.

Hence, the statement is True.

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

$45 : 60 = \frac{45}{60} = \frac{3}{4}$

$12 : 15 = \frac{12}{15} = \frac{4}{5}$

$\frac{3}{4} \neq \frac{4}{5}$. The ratios are not equal.

Hence, the statement is False.

(iii) 40 people : 200 people = ₹1000 : ₹5000

$40 : 200 = \frac{40}{200} = \frac{1}{5}$

$1000 : 5000 = \frac{1000}{5000} = \frac{1}{5}$

Both ratios are equal.

Hence, the statement is True.

(iv) 7.5 litres : 15 litres = 15 children : 30 children

$7.5 : 15 = \frac{7.5}{15} = \frac{1}{2}$

$15 : 30 = \frac{15}{30} = \frac{1}{2}$

Both ratios are equal.

Hence, the statement is True.

(i) 25 cm : 1 m and ₹40 : ₹160

Convert m into cm: 1 m = 100 cm.

First ratio = 25 : 100 = 1 : 4

Second ratio = 40 : 160 = 1 : 4

Both ratios are equal, so they form a proportion.

Proportion: 25 : 100 :: 40 : 160

Hence, extreme terms are 25 cm and ₹160; middle terms are 100 cm and ₹40.

(ii) 39 litre : 65 litre and 6 bottles : 10 bottles

First ratio = 39 : 65 = 3 : 5

Second ratio = 6 : 10 = 3 : 5

Both ratios are equal, so they form a proportion.

Proportion: 39 : 65 :: 6 : 10

Hence, extreme terms are 39 litre and 10 bottles; middle terms are 65 litre and 6 bottles.

(iii) 2 kg : 80 kg and 30 sec : 5 minutes

Convert 5 minutes into seconds: 5 min = 5 × 60 sec = 300 sec.

First ratio = 2 : 80 = 1 : 40

Second ratio = 30 : 300 = 1 : 10

Since 1 : 40 ≠ 1 : 10, the ratios are not equal.

Hence, the ratios do not form a proportion.

(iv) 200 g : 2.5 kg and ₹4 : ₹50

Convert kg into g: 2.5 kg = 2.5 × 1000 g = 2500 g.

First ratio = 200 : 2500 = 2 : 25

Second ratio = 4 : 50 = 2 : 25

Both ratios are equal, so they form a proportion.

Proportion: 200 : 2500 :: 4 : 50

Hence, extreme terms are 200 g and ₹50; middle terms are 2500 g and ₹4.

Given:

Cost of 9 m cloth = ₹378

Cost of 1 m cloth = $\frac{378}{9}$ = ₹42

Cost of 4 m cloth = 4 × ₹42 = ₹168

Hence, the cost of 4 m cloth is ₹168.

Given:

Weight of 36 books = 12 kg

Weight of 1 book = $\frac{12}{36}$ kg = $\frac{1}{3}$ kg

Weight of 75 books = 75 × $\frac{1}{3}$ kg = 25 kg

Hence, the weight of 75 books is 25 kg.

Given:

Cost of 5 pens = ₹115

Cost of 1 pen = $\frac{115}{5}$ = ₹23

Number of pens that can be bought in ₹207 = $\frac{207}{23}$ = 9

Hence, 9 pens can be bought in ₹207.

Given:

Distance covered in 8 litres of petrol = 100 km

Distance covered in 1 litre of petrol = $\frac{100}{8}$ km = 12.5 km

Distance covered in 26 litres of petrol = 26 × 12.5 km = 325 km

Hence, the car will travel 325 km in 26 litres of petrol.

Given:

Diesel needed for 594 km = 108 litres

Diesel needed for 1 km = $\frac{108}{594}$ litre = $\frac{2}{11}$ litre

Diesel needed for 1650 km = $1650 \times \frac{2}{11}$ = 150 × 2 = 300 litres

Hence, 300 litres of diesel will be required to cover 1650 km.

Given:

Charge for carrying 80 quintals = ₹5400

Charge for carrying 1 quintal = $\frac{5400}{80}$ = ₹67.50

Charge for carrying 126 quintals = 126 × ₹67.50 = ₹8505

Hence, the transport company will charge ₹8505 to carry 126 quintals.

Given:

Cloth required for 20 shirts = 42 m

Cloth required for 1 shirt = $\frac{42}{20}$ m = 2.1 m

Cloth required for 36 shirts = 36 × 2.1 m = 75.6 m

Hence, 75.6 m of cloth will be required to make 36 shirts.

Given:

Cost of 5 kg of rice = ₹107.50

Cost of 1 kg of rice = $\frac{107.50}{5}$ = ₹21.50

(i) Cost of 8 kg of rice = 8 × ₹21.50 = ₹172

Hence, the cost of 8 kg of rice is ₹172.

(ii) Quantity of rice that can be bought in ₹64.5 = $\frac{64.5}{21.5}$ kg = 3 kg

Hence, 3 kg of rice can be purchased in ₹64.5.

Given:

20 tons of iron = 20 × 1000 kg = 20000 kg

Cost of 20000 kg of iron = ₹6,00,000

Cost of 1 kg of iron = $\frac{6{,}00{,}000}{20000}$ = ₹30

Cost of 560 kg of iron = 560 × ₹30 = ₹16800

Hence, the cost of 560 kg of iron is ₹16,800.

Given:

Number of bananas in 4 dozen = 4 × 12 = 48 bananas

Cost of 48 bananas = ₹180

Cost of 1 banana = $\frac{180}{48}$ = ₹3.75

Number of bananas that can be bought in ₹37.50 = $\frac{37.50}{3.75}$ = 10

Hence, 10 bananas can be purchased for ₹37.50.

For Aman:

Cost of 12 pens = ₹156

Cost of 1 pen = $\frac{156}{12}$ = ₹13

For Payush:

Cost of 9 pens = ₹108

Cost of 1 pen = $\frac{108}{9}$ = ₹12

Since ₹12 < ₹13, Payush got the pens at a cheaper rate.

Hence, Payush got the pens cheaper.

For Rohit:

Runs in 6 overs = 42

Runs per over = $\frac{42}{6}$ = 7 runs

For Virat:

Runs in 7 overs = 63

Runs per over = $\frac{63}{7}$ = 9 runs

Since 9 > 7, Virat made more runs per over.

Hence, Virat made more runs per over.

(i) 18% of ₹450

$= \frac{18}{100} \times 450 = \frac{18 \times 450}{100} = \frac{8100}{100} = ₹81$

Hence, 18% of ₹450 = ₹81.

(ii) 14% of $16\frac{2}{3}$ kg

$16\frac{2}{3} = \frac{50}{3}$

$= \frac{14}{100} \times \frac{50}{3} = \frac{14 \times 50}{100 \times 3} = \frac{700}{300} = \frac{7}{3} = 2\frac{1}{3}$ kg

Hence, 14% of $16\frac{2}{3} \text kg= 2\frac{1}{3}$ kg.

(iii) $27\frac{3}{4}$% of ₹1200

$27\frac{3}{4}$% = $\frac{111}{4}$%

$= \frac{111}{4 \times 100} \times 1200 = \frac{111 \times 1200}{400} = ₹333$

Hence, $27\frac{3}{4}$% of ₹1200 = ₹333.

(iv) $\frac{5}{8}$% of 600 m

$= \frac{5}{8 \times 100} \times 600 = \frac{5 \times 600}{800} = \frac{3000}{800} = 3.75$ m

Hence, $\frac{5}{8}$% of 600 m = 3.75 m.

(v) $6\frac{1}{4}$% of 1 hour 20 minutes

$6\frac{1}{4}$% = $\frac{25}{4}$%

1 hour 20 minutes = 60 + 20 = 80 minutes

$= \frac{25}{4 \times 100} \times 80 = \frac{25 \times 80}{400} = \frac{2000}{400} = 5$ minutes

Hence, $6\frac{1}{4}$% of 1 hour 20 minutes = 5 minutes.

(vi) 0.6% of 5 km

5 km = 5 × 1000 m = 5000 m

$= \frac{0.6}{100} \times 5000 = \frac{0.6 \times 5000}{100} = \frac{3000}{100} = 30$ m

Hence, 0.6% of 5 km = 30 m.

Given:

Total students = 60

Percentage of girls = 45%

Number of girls = 45% of 60 = $\frac{45}{100} \times 60 = \frac{2700}{100} = 27$

Number of boys = Total students − Number of girls = 60 − 27 = 33

Hence, there are 33 boys in the class.

Given:

Monthly salary = ₹12750

Savings = 22% of salary

Savings = $\frac{22}{100} \times 12750 = \frac{22 \times 12750}{100} = \frac{280500}{100} = ₹2805$

Expenditure = Salary − Savings = ₹12750 − ₹2805 = ₹9945

Hence, Mr. Malkani's monthly expenditure is ₹9945.

Given:

Percentage of students present = 94%

Percentage of students absent = 100% − 94% = 6%

Number of students absent = 174

Let the total strength of the school = x.

Then, 6% of x = 174

$\frac{6}{100} \times x = 174$

$x = \frac{174 \times 100}{6} = \frac{17400}{6} = 2900$

Hence, the total strength of the school is 2900.

Given:

Speed of car = $105\frac{1}{5}$ km/h = $\frac{526}{5}$ km/h

Time = $3\frac{3}{5}$ hours = $\frac{18}{5}$ hours

Distance = Speed × Time

$= \frac{526}{5} \times \frac{18}{5} = \frac{526 \times 18}{25} = \frac{9468}{25} = 378\frac{18}{25}$ km

Hence, the car covers $378\frac{18}{25}$ km in $3\frac{3}{5}$ hours.

Given:

Speed of car = 50.4 km/h

Time = 3.6 hours

Distance = Speed × Time = 50.4 × 3.6

$50.4 \times 3.6 = \frac{504 \times 36}{100} = \frac{18144}{100} = 181.44$ km

Hence, the car covers 181.44 km in 3.6 hours.

Given:

Distance = 201.25 km

Time = 3.5 hours

Speed = $\frac{\text{Distance}}{\text{Time}} = \frac{201.25}{3.5}$

$= \frac{201.25 \times 10}{3.5 \times 10} = \frac{2012.5}{35} = 57.5$ km/h

Hence, the speed of the car is 57.5 km/h.

Given:

Distance travelled by bus in 4 hours = 160 km

Distance travelled by bus in 1 hour = $\frac{160}{4}$ = 40 km

Distance travelled by train in 5 hours = 320 km

Distance travelled by train in 1 hour = $\frac{320}{5}$ = 64 km

Ratio of distances travelled in one hour (bus : train) = 40 : 64

H.C.F. of 40 and 64 is 8.

$\frac{40 \div 8}{64 \div 8} = \frac{5}{8} = 5 : 8$

Hence, the ratio of the distances travelled by them in one hour is 5 : 8.

(i) In the ratio 3 : 5, the first term is 3 and second term is 5.

(ii) In a ratio, the first term is also called antecedent and second term is also called consequent.

(iii) If two terms of a ratio have no common factor (except 1), then the ratio is said to be in simplest form (or lowest terms).

(iv) To simplify a ratio, we divide the two terms by their H.C.F. (Highest Common Factor).

(v) The simplest form of the ratio 8 : 12 is 2 : 3.

(Reason: H.C.F. of 8 and 12 is 4. So, 8 : 12 = 2 : 3.)

(vi) 90 cm : 1.5 m = 3 : 5.

(Reason: 1.5 m = 150 cm. So, 90 : 150 = 3 : 5.)

(vii) When two ratios are equal, they are said to be in proportion.

(viii) 4.5% of ₹40 is equal to ₹1.80.

(Reason: $\frac{4.5}{100} \times 40 = \frac{180}{100} = ₹1.80$.)

(i) True. A ratio is a comparison of two quantities of the same kind in the same units.

(ii) True. A ratio is a pure number and has no units.

(iii) False. The order in a ratio matters. In general, a : b ≠ b : a (unless a = b).

(iv) True. Multiplying or dividing both terms by the same non-zero number gives equivalent ratios.

(v) True. A ratio a : b is in simplest form when H.C.F. of a and b is 1.

To find an equivalent ratio, multiply both terms of 5 : 7 by the same non-zero number.

5 : 7 = (5 × 4) : (7 × 4) = 20 : 28

Hence, option 3 is the correct option.

H.C.F. of 384 and 480 is 96.

$\frac{384 \div 96}{480 \div 96} = \frac{4}{5} = 4 : 5$

Hence, option 4 is the correct option.

1 hour = 60 minutes.

Ratio = 20 : 60 = 1 : 3.

Hence, option 2 is the correct option.

2 kg = 2000 g.

Ratio = 150 : 2000

H.C.F. of 150 and 2000 is 50.

$\frac{150 \div 50}{2000 \div 50} = \frac{3}{40} = 3 : 40$

Hence, option 3 is the correct option.

Number of students playing cricket = 25

Number of students playing tennis = 40 − 25 = 15

Ratio = 25 : 15 = 5 : 3

Hence, option 2 is the correct option.

Total parts = 3 + 5 = 8

Value of 1 part = $\frac{144}{8}$ = 18

Smaller number = 3 × 18 = 54

Hence, option 1 is the correct option.

5 parts represent girls, so 5 parts = 25.

1 part = $\frac{25}{5}$ = 5

Number of boys = 4 parts = 4 × 5 = 20

Hence, option 2 is the correct option.

Number of sides of a square = 4

Number of edges of a cube = 12

Ratio = 4 : 12 = 1 : 3

Hence, option 2 is the correct option.

2 parts represent green-cover books, so 2 parts = 18.

1 part = $\frac{18}{2}$ = 9

Number of brown-cover books = 3 parts = 3 × 9 = 27

Hence, option 3 is the correct option.

Total parts = 4 + 7 = 11

The total number of marbles must be a multiple of 11.

Among the given options, 22 = 2 × 11 is a multiple of 11.

Hence, option 3 is the correct option.

If a, b, c, d are in proportion, then a : b :: c : d, which means:

Product of extremes = Product of means

⇒ a × d = b × c

Hence, option 2 is the correct option.

Weight of 1 bag = $\frac{272}{5}$ = 54.4 kg

Hence, option 2 is the correct option.

Cost of 1 pencil = $\frac{35}{7}$ = ₹5

Cost of 1 dozen (12) pencils = 12 × ₹5 = ₹60

Hence, option 1 is the correct option.

$2 : 3 = \frac{2}{3} = \frac{2}{3} \times 100% = \frac{200}{3}% = 66\frac{2}{3}%$

Hence, option 3 is the correct option.

$0.025 = 0.025 \times 100% = 2.5%$

Hence, option 4 is the correct option.

Let the total number of students be x.

Then, 45% of x = 18.

$\frac{45}{100} \times x = 18$

$x = \frac{18 \times 100}{45} = \frac{1800}{45} = 40$

Hence, option 2 is the correct option.

Statement I: A ratio is a comparison of two quantities of the same kind (in the same units) by division — not of different kinds. So, Statement I is false.

Statement II: 6 hours : 24 hours = $\frac{6}{24} = \frac{1}{4}$ = 1 : 4. So, Statement II is true.

Hence, option 2 is the correct option.

Statement I: Dividing both terms of a ratio by the same non-zero number gives an equivalent ratio. So, Statement I is true.

Statement II: By definition, in a ratio a : b, the first term a is called the antecedent. So, Statement II is true.

Hence, option 3 is the correct option.

Statement I: Ratio of price of sugar to price of tea = 40 : 300 = $\frac{40}{300} = \frac{2}{15}$ = 2 : 15. So, Statement I is true.

Statement II: The order matters in a ratio because a : b is not the same as b : a in general. So, Statement II is true.

Hence, option 3 is the correct option.

Statement I: By definition, an equality of two ratios is called a proportion. So, Statement I is true.

Statement II: If a, b, c, d are in proportion, then a × d = b × c, NOT ab = cd. So, Statement II is false.

Hence, option 1 is the correct option.

Statement I: Ratio of Maths period to recess = 40 : 30 = $\frac{40}{30} = \frac{4}{3}$ = 4 : 3. So, Statement I is true.

Statement II: A ratio is a pure number and has no units. So, Statement II is true.

Hence, option 3 is the correct option.

Statement I: Total marks obtained = 80 + 70 + 90 = 240. Maximum marks = 3 × 100 = 300.

Percentage = $\frac{240}{300} \times 100% = 80%$. So, Statement I is true.

Statement II: $\frac{2}{5} = \frac{2}{5} \times 100% = 40%$. So, Statement II is true.

Hence, option 3 is the correct option.

From the figure, on counting we have:

Number of triangles = 3

Number of circles = 2

Number of squares = 2

Total number of figures inside the rectangle = 3 + 2 + 2 = 7

(i) Ratio of number of triangles to number of circles = 3 : 2

Hence, the required ratio is 3 : 2.

(ii) Ratio of number of squares to total number of figures = 2 : 7

Hence, the required ratio is 2 : 7.

(iii) Number of remaining figures (excluding circles) = 3 + 2 = 5

Ratio of number of circles to number of remaining figures = 2 : 5

Hence, the required ratio is 2 : 5.

Given:

Length of pencil = 16 cm

Diameter of pencil = 6 mm

Convert cm into mm: 16 cm = 16 × 10 mm = 160 mm.

Ratio of diameter to length = 6 : 160

H.C.F. of 6 and 160 is 2.

$\frac{6 \div 2}{160 \div 2} = \frac{3}{80} = 3 : 80$

Hence, the ratio of the diameter to the length of the pencil is 3 : 80.

Given:

Length of steel tape = 10 m

Breadth of steel tape = 2.4 cm

Convert m into cm: 10 m = 10 × 100 cm = 1000 cm.

Ratio of length to breadth = 1000 : 2.4

Multiply both terms by 10 to remove the decimal:

= (1000 × 10) : (2.4 × 10) = 10000 : 24

H.C.F. of 10000 and 24 is 8.

$\frac{10000 \div 8}{24 \div 8} = \frac{1250}{3} = 1250 : 3$

Hence, the ratio of the length to the breadth is 1250 : 3.

Given:

Total members = 100

Tennis players = 25

Badminton players = 28

Chess players = 12

Members who do not play any game = 100 − 25 − 28 − 12 = 35

(i) Ratio of badminton players to chess players = 28 : 12

H.C.F. of 28 and 12 is 4.

$\frac{28 \div 4}{12 \div 4} = \frac{7}{3} = 7 : 3$

Hence, the required ratio is 7 : 3.

(ii) Ratio of badminton players to non-players = 28 : 35

H.C.F. of 28 and 35 is 7.

$\frac{28 \div 7}{35 \div 7} = \frac{4}{5} = 4 : 5$

Hence, the required ratio is 4 : 5.

(iii) Ratio of tennis players to non-players = 25 : 35

H.C.F. of 25 and 35 is 5.

$\frac{25 \div 5}{35 \div 5} = \frac{5}{7} = 5 : 7$

Hence, the required ratio is 5 : 7.

(iv) Players of badminton or chess = 28 + 12 = 40

Ratio of tennis players to (badminton or chess) players = 25 : 40

H.C.F. of 25 and 40 is 5.

$\frac{25 \div 5}{40 \div 5} = \frac{5}{8} = 5 : 8$

Hence, the required ratio is 5 : 8.

Given:

First ratio: 15 cm to 3 m

Convert m into cm: 3 m = 3 × 100 cm = 300 cm.

First ratio = 15 : 300 = 1 : 20

Second ratio: 25 seconds to 3 minutes

Convert minutes into seconds: 3 min = 3 × 60 = 180 seconds.

Second ratio = 25 : 180 = 5 : 36

Compare: 1 : 20 and 5 : 36.

Product of extremes = 1 × 36 = 36

Product of means = 20 × 5 = 100

Since 36 ≠ 100, the ratios are not equal.

Hence, the given ratios do not form a proportion.

Given:

Total money = ₹500

Ratio = 3 : 7

Find total parts: 3 + 7 = 10 parts.

Find the value of 1 part: 500 ÷ 10 = ₹50. So, 1 part = ₹50.

Now calculate shares by multiplying each part with ₹50:

Suresh's share: 3 × ₹50 = ₹150

Awanti's share: 7 × ₹50 = ₹350

Hence, Suresh gets ₹150 and Awanti gets ₹350.

Given:

Ratio (Girls : Boys) = 9 : 11

Number of boys = 2035

Since 11 parts represent the boys, 11 parts = 2035.

1 part = $\frac{2035}{11}$ = 185.

(i) Number of girls = 9 parts = 9 × 185 = 1665.

Hence, the number of girls is 1665.

(ii) Total students = Girls + Boys = 1665 + 2035 = 3700.

Hence, the total number of students in the school is 3700.

Given:

Ratio (Income : Expenditure) = 7 : 6

Income = ₹42,000

Since 7 parts represent the income, 7 parts = ₹42,000.

1 part = $\frac{42000}{7}$ = ₹6,000.

Expenditure = 6 parts = 6 × ₹6,000 = ₹36,000.

Savings = Income − Expenditure = ₹42,000 − ₹36,000 = ₹6,000.

Hence, the savings of the family is ₹6,000.

Given:

Earnings in 3 months = ₹72,000

Earnings in 1 month = $\frac{72000}{3}$ = ₹24000.

(i) Earnings in 7 months = 7 × ₹24000 = ₹1,68,000.

Hence, the employee earns ₹1,68,000 in 7 months.

(ii) Number of months to earn ₹3,60,000 = $\frac{3{,}60{,}000}{24000}$ = 15 months.

Hence, the employee will earn ₹3,60,000 in 15 months.

Given:

Distance travelled by train = 110 km in 2 hours

Speed of train = $\frac{110}{2}$ = 55 km/h

Distance travelled by car = 245 km in $3\frac{1}{2}$ hours = $\frac{7}{2}$ hours

Speed of car = $\frac{245}{\frac{7}{2}} = 245 \times \frac{2}{7} = \frac{490}{7} = 70$ km/h

Ratio (Speed of train : Speed of car) = 55 : 70

H.C.F. of 55 and 70 is 5.

$\frac{55 \div 5}{70 \div 5} = \frac{11}{14} = 11 : 14$

Hence, the ratio of the speed of the train to that of the car is 11 : 14.

Prime numbers between 1 and 45 are:

2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43

Number of prime numbers = 14.

The number 1 is neither prime nor composite.

Total numbers from 1 to 45 = 45.

Composite numbers = 45 − (number of primes) − 1 (for the number 1)

Composite numbers = 45 − 14 − 1 = 30.

Ratio (Primes : Composites) = 14 : 30

H.C.F. of 14 and 30 is 2.

$\frac{14 \div 2}{30 \div 2} = \frac{7}{15} = 7 : 15$

Hence, the required ratio is 7 : 15.

Given:

Total money = ₹6000

Ratio = 3 : 5 : 7

Find total parts: 3 + 5 + 7 = 15 parts.

Find the value of 1 part: 6000 ÷ 15 = ₹400. So, 1 part = ₹400.

Now calculate shares by multiplying each part with ₹400:

Irfan's share: 3 × ₹400 = ₹1200

Nagma's share: 5 × ₹400 = ₹2000

Ishan's share: 7 × ₹400 = ₹2800

Hence, Irfan gets ₹1200, Nagma gets ₹2000 and Ishan gets ₹2800.

Given:

Sapna's weight on earth = 54 kg

Sapna's weight on moon = 9 kg

Ratio of earth weight to moon weight = 54 : 9 = 6 : 1

So, weight on earth = 6 × weight on moon.

Monkey's weight on moon = 3.5 kg

Monkey's weight on earth = 6 × 3.5 = 21 kg.

Hence, the monkey will weigh 21 kg on the earth.

Given:

5 men complete the work in 14 days.

This is an inverse proportion problem — more men means fewer days.

Total work = 5 men × 14 days = 70 man-days.

Let 7 men take x days to complete the same work.

Then, 7 × x = 70

⇒ x = $\frac{70}{7}$ = 10 days.

Hence, 7 men will take 10 days to complete the construction work.