Ratio and Proportion

(i) $\frac{1}{6} : \frac{1}{9}$

By Division Method,

$$\begin{array}{l|r} 2 & 6, 9 \\ \hline 3 & 3, 9 \\ \hline 3 & 1, 3 \\ \hline & 1, 1 \end{array}$$

L.C.M. of 6 and 9 = 2 x 3 x 3 = 18. Multiply both terms by 18:

$\frac{1}{6} \times 18 : \frac{1}{9} \times 18 = 3 : 2$

Hence, the answer is 3 : 2.

(ii) $4\frac{1}{2} : 1\frac{1}{8}$

$4\frac{1}{2} = \frac{9}{2}$ and $1\frac{1}{8} = \frac{9}{8}$

So the ratio is $\frac{9}{2} : \frac{9}{8}$

By Division Method,

$$\begin{array}{l|r} 2 & 2, 8 \\ \hline 2 & 1, 4 \\ \hline 2 & 1, 2 \\ \hline & 1, 1 \end{array}$$

L.C.M. of 2 and 8 = 2 x 2 x 2 = 8. Multiply both terms by 8:

$\frac{9}{2} \times 8 : \frac{9}{8} \times 8 = 36 : 9 = 4 : 1$

Hence, the answer is 4 : 1.

(iii) $\frac{1}{5} : \frac{1}{10} : \frac{1}{15}$

By Division Method,

$$\begin{array}{l|r} 2 & 5, 10, 15 \\ \hline 3 & 5, 5, 15 \\ \hline 5 & 5, 5, 5 \\ \hline & 1, 1 \end{array}$$

L.C.M. of 5, 10 and 15 = 2 x 3 x 5 = 30. Multiply each term by 30:

$\frac{1}{5} \times 30 : \frac{1}{10} \times 30 : \frac{1}{15} \times 30 = 6 : 3 : 2$

Hence, the answer is 6 : 3 : 2.

(i) ₹5 to 50 paise

Convert rupees into paise: ₹5 = 5 × 100 paise = 500 paise.

Ratio = 500 : 50 = 10 : 1

Hence, the answer is 10 : 1.

(ii) 3 km to 300 m

Convert km into m: 3 km = 3 × 1000 m = 3000 m.

Ratio = 3000 : 300 = 10 : 1

Hence, the answer is 10 : 1.

(iii) 9 m to 27 cm

Convert m into cm: 9 m = 9 × 100 cm = 900 cm.

Ratio = 900 : 27

By Prime Factorization,

$$\begin{array}{l|r} 2 & 900 \\ \hline 2 & 450 \\ \hline 3 & 225 \\ \hline 3 & 75 \\ \hline 5 & 25 \\ \hline 5 & 5 \\ \hline & 1 \end{array} \quad \text{ and } \quad \begin{array}{l|r} 3 & 27 \\ \hline 3 & 9 \\ \hline 3 & 3 \\ \hline & 1 \end{array}$$

900 = 2 x 2 x 3 x 3 x 5 x 5

27 = 3 x 3 x 3

H.C.F. of 900 and 27 = 3 x 3 = 9.

$\frac{900 \div 9}{27 \div 9} = \frac{100}{3} = 100 : 3$

Hence, the answer is 100 : 3.

(iv) 15 kg to 210 g

Convert kg into g: 15 kg = 15 × 1000 g = 15000 g.

Ratio = 15000 : 210

By Prime Factorization,

$$\begin{array}{l|r} 2 & 15000 \\ \hline 2 & 7500 \\ \hline 2 & 3750 \\ \hline 3 & 1875 \\ \hline 5 & 625 \\ \hline 5 & 125 \\ \hline 5 & 25 \\ \hline 5 & 5 \\ \hline & 1 \end{array} \quad \text{ and } \quad \begin{array}{l|r} 2 & 210 \\ \hline 3 & 105 \\ \hline 5 & 35 \\ \hline 7 & 7 \\ \hline & 1 \end{array}$$

15000 = 2 x 2 x 2 x 3 x 5 x 5 x 5 x 5

210 = 2 x 3 x 5 x 7

H.C.F. of 15000 and 210 = 2 x 3 x 5 = 30.

$\frac{15000 \div 30}{210 \div 30} = \frac{500}{7} = 500 : 7$

Hence, the answer is 500 : 7.

(v) 25 minutes to 1.5 hours

Convert hours into minutes: 1.5 hours = 1.5 × 60 minutes = 90 minutes.

Ratio = 25 : 90

By Prime Factorization,

$$\begin{array}{l|r} 5 & 25 \\ \hline 5 & 5 \\ \hline & 1 \end{array} \quad \text{ and } \quad \begin{array}{l|r} 2 & 90 \\ \hline 3 & 45 \\ \hline 3 & 15 \\ \hline 5 & 5 \\ \hline & 1 \end{array}$$

25 = 5 x 5

90 = 2 x 3 x 3 x 5

H.C.F. of 25 and 90 = 5.

$\frac{25 \div 5}{90 \div 5} = \frac{5}{18} = 5 : 18$

Hence, the answer is 5 : 18.

(vi) 30 days to 36 hours

Convert days into hours: 30 days = 30 × 24 hours = 720 hours.

Ratio = 720 : 36 = 20 : 1

Hence, the answer is 20 : 1.

A : B = 3 : 4

B : C = 8 : 9

To combine, make the value of B the same in both ratios. B is 4 in the first and 8 in the second. L.C.M. of 4 and 8 is 8.

Multiply the first ratio by 2: A : B = (3 × 2) : (4 × 2) = 6 : 8

Now, A : B = 6 : 8 and B : C = 8 : 9

So, A : B : C = 6 : 8 : 9

∴ A : C = 6 : 9 = 2 : 3

Hence, A : C = 2 : 3.

A : B = 5 : 8

B : C = 18 : 25

To combine, make the value of B the same in both ratios. B is 8 in the first and 18 in the second. L.C.M. of 8 and 18 is 72.

Multiply the first ratio by 9: A : B = (5 × 9) : (8 × 9) = 45 : 72

Multiply the second ratio by 4: B : C = (18 × 4) : (25 × 4) = 72 : 100

Now, A : B = 45 : 72 and B : C = 72 : 100

Hence, A : B : C = 45 : 72 : 100.

Let 3A = 2B = 5C = k

Then A = $\frac{k}{3}$, B = $\frac{k}{2}$ and C = $\frac{k}{5}$

So, A : B : C = $\frac{k}{3} : \frac{k}{2} : \frac{k}{5}$

L.C.M. of 3, 2 and 5 is 30. Multiply each term by 30:

$\frac{k}{3} \times 30 : \frac{k}{2} \times 30 : \frac{k}{5} \times 30 = 10k : 15k : 6k = 10 : 15 : 6$

Hence, A : B : C = 10 : 15 : 6.

Given:

Daily income = ₹600

Spending = ₹450

Saving = Income − Spending = ₹600 − ₹450 = ₹150

(i) Spending to income = 450 : 600

By Prime Factorization,

$$\begin{array}{l|r} 2 & 450 \\ \hline 3 & 225 \\ \hline 3 & 75 \\ \hline 5 & 25 \\ \hline 5 & 5 \\ \hline & 1 \end{array} \quad \text{ and } \quad \begin{array}{l|r} 2 & 600 \\ \hline 2 & 300 \\ \hline 2 & 150 \\ \hline 3 & 75 \\ \hline 5 & 25 \\ \hline 5 & 5 \\ \hline & 1 \end{array}$$

450 = 2 x 3 x 3 x 5 x 5

600 = 2 x 2 x 2 x 3 x 5 x 5

H.C.F. of 450 and 600 = 2 x 3 x 5 x 5 = 150.

$\frac{450 \div 150}{600 \div 150} = \frac{3}{4} = 3 : 4$

Hence, the ratio of spending to income is 3 : 4.

(ii) Saving to income = 150 : 600

By Prime Factorization,

$$\begin{array}{l|r} 2 & 150 \\ \hline 3 & 75 \\ \hline 5 & 25 \\ \hline 5 & 5 \\ \hline & 1 \end{array} \quad \text{ and } \quad \begin{array}{l|r} 2 & 600 \\ \hline 2 & 300 \\ \hline 2 & 150 \\ \hline 3 & 75 \\ \hline 5 & 25 \\ \hline 5 & 5 \\ \hline & 1 \end{array}$$

150 = 2 x 3 x 5 x 5

600 = 2 x 2 x 2 x 3 x 5 x 5

H.C.F. of 150 and 600 = 2 x 3 x 5 x 5 = 150.

$\frac{150 \div 150}{600 \div 150} = \frac{1}{4} = 1 : 4$

Hence, the ratio of saving to income is 1 : 4.

(iii) Saving to spending = 150 : 450

By Prime Factorization,

$$\begin{array}{l|r} 2 & 150 \\ \hline 3 & 75 \\ \hline 5 & 25 \\ \hline 5 & 5 \\ \hline & 1 \end{array} \quad \text{ and } \quad \begin{array}{l|r} 2 & 450 \\ \hline 3 & 225 \\ \hline 3 & 75 \\ \hline 5 & 25 \\ \hline 5 & 5 \\ \hline & 1 \end{array}$$

150 = 2 x 3 x 5 x 5

450 = 2 x 3 x 3 x 5 x 5

H.C.F. of 150 and 450 = 2 x 3 x 5 x 5 = 150.

$\frac{150 \div 150}{450 \div 150} = \frac{1}{3} = 1 : 3$

Hence, the ratio of saving to spending is 1 : 3.

Given:

Total weight of alloy = 5 grams

Weight of copper = $3\frac{3}{4}$ grams = $\frac{15}{4}$ grams

Weight of nickel = Total − Copper = $5 - \frac{15}{4} = \frac{20}{4} - \frac{15}{4} = \frac{5}{4}$ grams

Ratio of nickel to copper = $\frac{5}{4} : \frac{15}{4}$

Multiply both terms by 4:

$\frac{5}{4} \times 4 : \frac{15}{4} \times 4 = 5 : 15 = 1 : 3$

Hence, the ratio by weight of nickel to copper is 1 : 3.

Let the length of the second piece = x metres.

Then the length of the first piece = $\frac{1}{2}$ x metres.

Total length of the pole = 3 metres.

$\frac{1}{2}x + x = 3$

$\frac{3x}{2} = 3$

$x = \frac{3 \times 2}{3} = 2$

So, second piece = 2 m and first piece = $\frac{1}{2} \times 2 = 1$ m.

Hence, the lengths of the two pieces are 1 m and 2 m.

Given:

Height of Anshul = 1.04 m = 1.04 × 100 cm = 104 cm

Height of Dhruv = 78 cm

Ratio of their heights = 104 : 78

By Prime Factorization,

$$\begin{array}{l|r} 2 & 104 \\ \hline 2 & 52 \\ \hline 2 & 26 \\ \hline 13 & 13 \\ \hline & 1 \end{array} \quad \text{ and } \quad \begin{array}{l|r} 2 & 78 \\ \hline 3 & 39 \\ \hline 13 & 13 \\ \hline & 1 \end{array}$$

104 = 2 x 2 x 2 x 13

78 = 2 x 3 x 13

H.C.F. of 104 and 78 = 2 x 13 = 26.

$\frac{104 \div 26}{78 \div 26} = \frac{4}{3} = 4 : 3$

Total number of parts = 4 + 3 = 7

Value of 1 part = $\frac{35}{7}$ = 5 sweets

Anshul's share = 4 × 5 = 20 sweets

Dhruv's share = 3 × 5 = 15 sweets

Hence, Anshul gets 20 sweets and Dhruv gets 15 sweets.

The given ratio is $\frac{1}{3} : \frac{1}{4} : \frac{1}{6}$

By Division Method,

$$\begin{array}{l|r} 2 & 3, 4, 6 \\ \hline 2 & 3, 2, 3 \\ \hline 3 & 3, 1, 3 \\ \hline & 1, 1, 1 \end{array}$$

L.C.M. of 3, 4 and 6 = 2 x 2 x 3 = 12. Multiply each term by 12:

$\frac{1}{3} \times 12 : \frac{1}{4} \times 12 : \frac{1}{6} \times 12 = 4 : 3 : 2$

Total number of parts = 4 + 3 + 2 = 9

Value of 1 part = $\frac{180}{9}$ = ₹20

First child's share = 4 × 20 = ₹80

Second child's share = 3 × 20 = ₹60

Third child's share = 2 × 20 = ₹40

Hence, the shares of the three children are ₹80, ₹60 and ₹40.

Let the two parts be 7x and 11x.

Difference of the two parts = 11x − 7x = 4x

According to the question,

4x = 20

$x = \frac{20}{4} = 5$

First part = 7 × 5 = 35

Second part = 11 × 5 = 55

The number = 35 + 55 = 90

Hence, the number is 90 and the two parts are 35 and 55.

Let the two parts be 9x and 13x.

Second part = 13x = ₹260

$x = \frac{260}{13} = 20$

First part = 9 × 20 = ₹180

Total amount = ₹180 + ₹260 = ₹440

Hence, the total amount is ₹440.

Let the three parts be 5x, 7x and 8x.

First part = 5x = ₹225

$x = \frac{225}{5} = 45$

Second part = 7 × 45 = ₹315

Third part = 8 × 45 = ₹360

Total amount = ₹225 + ₹315 + ₹360 = ₹900

Hence, the total amount is ₹900 and the other two parts are ₹315 and ₹360.

Let the second part = x.

First part = $\frac{2}{3}$ x

Second part : Third part = 4 : 7

$\therefore \frac{\text{Second part (x)}}{\text{Third part}} = \frac{4}{7}$

Third part = $\frac{7}{4}x$

Total = ₹1312

$\frac{2}{3}x + x + \frac{7}{4}x = 1312$

By Division Method,

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

L.C.M. of 3 and 4 = 2 x 2 x 3 = 12.

$$\begin{aligned}\frac{8x}{12} + \frac{12x}{12} + \frac{21x}{12} = 1312\\[1em] \frac{8x + 12x + 21x}{12} = 1312\\[1em] \frac{41x}{12} = 1312\\[1em] x = \frac{1312 \times 12}{41} = 32 \times 12 = 384\end{aligned}$$

Second part = ₹384

First part = $\frac{2}{3} \times 384$ = ₹256

Third part = $\frac{7}{4} \times 384$ = ₹672

Hence, the three parts are ₹256, ₹384 and ₹672.

Let the present ages of Saanvi and Navya be 2x years and 3x years.

Five years hence:

Saanvi's age = (2x + 5) years

Navya's age = (3x + 5) years

According to the question,

$\frac{2x + 5}{3x + 5} = \frac{3}{4}$

By cross multiplication,

4(2x + 5) = 3(3x + 5)

8x + 20 = 9x + 15

20 − 15 = 9x − 8x

x = 5

Saanvi's present age = 2 × 5 = 10 years

Navya's present age = 3 × 5 = 15 years

Hence, Saanvi is 10 years old and Navya is 15 years old.

Let the present ages of A and B be 5x years and 6x years.

Three years ago:

A's age = (5x − 3) years

B's age = (6x − 3) years

According to the question,

$\frac{5x - 3}{6x - 3} = \frac{4}{5}$

By cross multiplication,

5(5x − 3) = 4(6x − 3)

25x − 15 = 24x − 12

25x − 24x = −12 + 15

x = 3

A's present age = 5 × 3 = 15 years

B's present age = 6 × 3 = 18 years

Hence, A is 15 years old and B is 18 years old.

Let the two numbers be 5x and 6x.

According to the question,

$\frac{5x + 2}{6x + 3} = \frac{4}{5}$

By cross multiplication,

5(5x + 2) = 4(6x + 3)

25x + 10 = 24x + 12

25x − 24x = 12 − 10

x = 2

First number = 5 × 2 = 10

Second number = 6 × 2 = 12

Hence, the two numbers are 10 and 12.

Total number of students = 1430

Ratio of boys to girls = 7 : 6

Total number of parts = 7 + 6 = 13

Value of 1 part = $\frac{1430}{13}$ = 110

Number of boys = 7 × 110 = 770

Number of girls = 6 × 110 = 660

After admitting 26 new girls, number of girls = 660 + 26 = 686

Let the number of new boys admitted = y.

Then number of boys = (770 + y)

According to the question,

$\frac{770 + y}{686} = \frac{8}{7}$

By cross multiplication,

7(770 + y) = 8 × 686

5390 + 7y = 5488

7y = 5488 − 5390

7y = 98

$y = \frac{98}{7} = 14$

Hence, 14 new boys should be admitted.

(i) 5 : 6 or 6 : 7

Write the ratios as fractions: $\frac{5}{6}$ and $\frac{6}{7}$

By cross multiplication, compare 5 × 7 and 6 × 6:

5 × 7 = 35 and 6 × 6 = 36

Since 35 < 36, we have $\frac{5}{6} \lt \frac{6}{7}$

Hence, 6 : 7 is the greater ratio.

(ii) 13 : 24 or 17 : 32

Write the ratios as fractions: $\frac{13}{24}$ and $\frac{17}{32}$

By cross multiplication, compare 13 × 32 and 17 × 24:

13 × 32 = 416 and 17 × 24 = 408

Since 416 > 408, we have $\frac{13}{24} \gt \frac{17}{32}$

Hence, 13 : 24 is the greater ratio.

(i) Increase the number 150 in ratio 5 : 7

To increase a number in the ratio 5 : 7, multiply it by $\frac{7}{5}$.

New number = $150 \times \frac{7}{5} = 30 \times 7 = 210$

Hence, the new number is 210.

(ii) A man earns ₹18,000 per month. His income is increased in the ratio 12 : 13.

To increase in the ratio 12 : 13, multiply by $\frac{13}{12}$.

New income = $18000 \times \frac{13}{12} = 1500 \times 13 = ₹19500$

Hence, his new monthly income is ₹19,500.

(iii) Savita weighs 55 kg. She reduced her weight in the ratio 11 : 9.

To reduce in the ratio 11 : 9, multiply by $\frac{9}{11}$.

New weight = $55 \times \frac{9}{11} = 5 \times 9 = 45$ kg

Hence, her new weight is 45 kg.

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

(i) 2.5 : 1.5 :: 7.0 : 4.2

Product of extremes = 2.5 × 4.2 = 10.5

Product of means = 1.5 × 7.0 = 10.5

Since product of extremes = product of means, the statement is true.

Hence, the statement is True.

(ii) $\frac{1}{2} : \frac{1}{3} = \frac{1}{3} : \frac{1}{4}$

Product of extremes = $\frac{1}{2} \times \frac{1}{4} = \frac{1}{8}$

Product of means = $\frac{1}{3} \times \frac{1}{3} = \frac{1}{9}$

Since $\frac{1}{8} \neq \frac{1}{9}$, the statement is false.

Hence, the statement is False.

(iii) 24 men : 16 men = 33 horses : 22 horses

$24 : 16 = \frac{24}{16} = \frac{3}{2}$

$33 : 22 = \frac{33}{22} = \frac{3}{2}$

Both ratios are equal, so the statement is true.

Hence, the statement is True.

Therefore, statements (i) and (iii) are true.

Four numbers a, b, c, d are in proportion if product of extremes = product of means, i.e. a × d = b × c.

(i) 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.

(ii) 3, $3\frac{1}{2}$, 4, $4\frac{1}{2}$

Product of extremes = $3 \times 4\frac{1}{2} = 3 \times \frac{9}{2} = \frac{27}{2} = 13.5$

Product of means = $3\frac{1}{2} \times 4 = \frac{7}{2} \times 4 = 14$

Since 13.5 ≠ 14, the numbers are not in proportion.

Hence, 3, $3\frac{1}{2}$, 4, $4\frac{1}{2}$ are not in proportion.

(iii) 0.1, 0.2, 0.3, 0.6

Product of extremes = 0.1 × 0.6 = 0.06

Product of means = 0.2 × 0.3 = 0.06

Since 0.06 = 0.06, the numbers are in proportion.

Hence, 0.1, 0.2, 0.3, 0.6 are in proportion.

Therefore, the numbers in (i) and (iii) are in proportion.

(i) x : 4 = 9 : 12

Product of extremes = product of means.

x × 12 = 4 × 9

12x = 36

$x = \frac{36}{12} = 3$

Hence, x = 3.

(ii) $\frac{1}{13} : x :: \frac{1}{2} : \frac{1}{5}$

Product of extremes = product of means.

$\frac{1}{13} \times \frac{1}{5} = x \times \frac{1}{2}$

$\frac{1}{65} = \frac{x}{2}$

$x = \frac{2}{65}$

Hence, x = $\frac{2}{65}$.

(iii) 3.6 : 0.4 = x : 0.5

Product of extremes = product of means.

3.6 × 0.5 = 0.4 × x

1.8 = 0.4x

$x = \frac{1.8}{0.4} = 4.5$

Hence, x = 4.5.

If a, b, c, x are in proportion, then x is the fourth proportional and a × x = b × c.

(i) 42, 12, 7

42 : 12 :: 7 : x

42 × x = 12 × 7

42x = 84

$x = \frac{84}{42} = 2$

Hence, the fourth proportional is 2.

(ii) $\frac{1}{3}, \frac{1}{4}, \frac{1}{5}$

$\frac{1}{3} : \frac{1}{4} :: \frac{1}{5} : x$

$\frac{1}{3} \times x = \frac{1}{4} \times \frac{1}{5}$

$\frac{x}{3} = \frac{1}{20}$

$x = \frac{3}{20}$

Hence, the fourth proportional is $\frac{3}{20}$.

(iii) 3 kg, 12 kg, 15 kg

3 : 12 :: 15 : x

3 × x = 12 × 15

3x = 180

$x = \frac{180}{3} = 60$ kg

Hence, the fourth proportional is 60 kg.

Three numbers a, b, c are in continued proportion if a : b :: b : c, i.e. a × c = b × b.

Here a = 7, b = 49, c = 343.

Product of extremes = 7 × 343 = 2401

Product of means = 49 × 49 = 2401

Since 2401 = 2401, the numbers are in continued proportion.

Hence, yes, 7, 49, 343 are in continued proportion.

If a, b, x are in continued proportion, then x is the third proportional and a × x = b × b.

(i) 36, 18

36 : 18 :: 18 : x

36 × x = 18 × 18

36x = 324

$x = \frac{324}{36} = 9$

Hence, the third proportional is 9.

(ii) $5\frac{1}{4}$, 7

$$\begin{aligned}5\frac{1}{4} = \frac{21}{4}\\[1em] \frac{21}{4} : 7 :: 7 : x\\[1em] \frac{21}{4} \times x = 7 \times 7\\[1em] \frac{21x}{4} = 49\\[1em]\end{aligned}$$

$x = \frac{49 \times 4}{21} = \frac{196}{21} = \frac{28}{3} = 9\frac{1}{3}$

Hence, the third proportional is $9\frac{1}{3}$.

(iii) 3.2, 0.8

3.2 : 0.8 :: 0.8 : x

3.2 × x = 0.8 × 0.8

3.2x = 0.64

$x = \frac{0.64}{3.2} = 0.2$

Hence, the third proportional is 0.2.

Given:

Length : Width = 7 : 5

Width = 20.5 cm

Let the length = x cm.

$\frac{x}{20.5} = \frac{7}{5}$

By cross multiplication,

5 × x = 7 × 20.5

5x = 143.5

$x = \frac{143.5}{5} = 28.7$ cm

Hence, the length of the sheet is 28.7 cm.

Given:

Advik : Anaya = 4 : 5

Advik's age = 4 years 8 months = (4 × 12 + 8) months = 56 months

Let Anaya's age = x months.

$\frac{56}{x} = \frac{4}{5}$

By cross multiplication,

4 × x = 56 × 5

4x = 280

$x = \frac{280}{4} = 70$ months

70 months = 5 years 10 months

Hence, the age of Anaya is 5 years 10 months.

Given:

Cost of 6 bowls = ₹90

Cost of 1 bowl = $\frac{90}{6}$ = ₹15

Cost of 10 bowls = 10 × ₹15 = ₹150

Hence, the cost of 10 bowls is ₹150.

Given:

Cost of 10 pencils = ₹15

Cost of 1 pencil = $\frac{15}{10}$ = ₹1.50

Number of pencils that can be bought with ₹72 = $\frac{72}{1.50}$ = 48

Hence, 48 pencils can be bought with ₹72.

Given:

Cost of 400 g cake = ₹800

Cost of 1 g cake = $\frac{800}{400}$ = ₹2

1.5 kg = 1.5 × 1000 g = 1500 g

Cost of 1500 g cake = 1500 × ₹2 = ₹3000

Hence, a 1.5 kg cake would cost ₹3000.

Given:

Earning in 3 months = ₹18000

Earning in 1 month = $\frac{18000}{3}$ = ₹6000

(i) Time taken to earn ₹30000 = $\frac{30000}{6000}$ = 5 months

Hence, he would take 5 months to earn ₹30000.

(ii) Money earned in 7 months = 7 × ₹6000 = ₹42000

Hence, he will earn ₹42000 in 7 months.

Given:

Weight of 12 mangoes = 2.4 kg

Weight of 1 mango = $\frac{2.4}{12}$ = 0.2 kg

Weight of 8 mangoes = 8 × 0.2 kg = 1.6 kg

Hence, the weight of 8 mangoes is 1.6 kg.

Given:

Weight of 12 sheets = 40 grams

$2\frac{1}{2}$ kg = $\frac{5}{2}$ × 1000 g = 2500 g

Number of sheets weighing 40 g = 12

Number of sheets weighing 1 g = $\frac{12}{40}$

Number of sheets weighing 2500 g = $2500 \times \frac{12}{40} = \frac{30000}{40} = 750$

Hence, 750 sheets would weigh $2\frac{1}{2}$ kilograms.

Given:

Diesel needed for 90 km = 25 litres

Diesel needed for 1 km = $\frac{25}{90}$ litre = $\frac{5}{18}$ litre

Diesel needed for 288 km = $288 \times \frac{5}{18} = 16 \times 5 = 80$ litres

Hence, 80 litres of diesel is needed to cover 288 kilometres.

Given:

Cost of $\frac{4}{5}$ metre cloth = ₹36

Cost of 1 metre cloth = $36 \div \frac{4}{5} = 36 \times \frac{5}{4} = ₹45$

$2\frac{1}{5}$ metres = $\frac{11}{5}$ metres

Cost of $\frac{11}{5}$ metres = $\frac{11}{5} \times 45 = 11 \times 9 = ₹99$

Hence, the cost of $2\frac{1}{5}$ metres of cloth is ₹99.

Given:

Number of men to pack 540 parcels = 15

Number of parcels packed by 1 man = $\frac{540}{15}$ = 36

Number of men needed to pack 396 parcels = $\frac{396}{36}$ = 11

Hence, 11 men are needed to pack 396 parcels per day.

For 12 kg potatoes:

Cost of 12 kg = ₹132

Cost of 1 kg = $\frac{132}{12}$ = ₹11

For 16 kg potatoes:

Cost of 16 kg = ₹168

Cost of 1 kg = $\frac{168}{16}$ = ₹10.50

Since ₹10.50 < ₹11, the second option is cheaper per kilogram.

Hence, 16 kg potatoes for ₹168 is the better buy.

To convert km/h into m/sec, multiply by $\frac{5}{18}$.

(i) 72 km/h

$= 72 \times \frac{5}{18} = 4 \times 5 = 20$ m/sec

Hence, 72 km/h = 20 m/sec.

(ii) 9 km/h

$= 9 \times \frac{5}{18} = \frac{45}{18} = 2.5$ m/sec

Hence, 9 km/h = 2.5 m/sec.

(iii) 1.2 km/minute

1.2 km = 1.2 × 1000 m = 1200 m and 1 minute = 60 seconds.

$= \frac{1200}{60} = 20$ m/sec

Hence, 1.2 km/minute = 20 m/sec.

(iv) 600 m/hour

1 hour = 3600 seconds.

$= \frac{600}{3600} = \frac{1}{6}$ m/sec

Hence, 600 m/hour = $\frac{1}{6}$ m/sec.

To convert m/sec into km/h, multiply by $\frac{18}{5}$.

(i) 15 m/sec

$= 15 \times \frac{18}{5} = 3 \times 18 = 54$ km/h

Hence, 15 m/sec = 54 km/h.

(ii) 1.5 m/sec

$= 1.5 \times \frac{18}{5} = \frac{27}{5} = 5.4$ km/h

Hence, 1.5 m/sec = 5.4 km/h.

Let us convert 30 m/sec into km/h by multiplying by $\frac{18}{5}$.

$30 \text{ m/sec} = 30 \times \frac{18}{5} = 6 \times 18 = 108$ km/h

Now compare 108 km/h and 30 km/h.

Since 108 km/h > 30 km/h, a speed of 30 m/sec is greater.

Hence, a speed of 30 m/sec is greater.

Given:

Speed of aeroplane = 720 km/h

(i) Time = $\frac{\text{Distance}}{\text{Speed}} = \frac{1800}{720} = \frac{5}{2} = 2\frac{1}{2}$ hours

Hence, the aeroplane will take $2\frac{1}{2}$ hours to cross the two cities.

(ii) 40 minutes = $\frac{40}{60}$ hour = $\frac{2}{3}$ hour

Distance = Speed × Time = $720 \times \frac{2}{3} = 240 \times 2 = 480$ km

Hence, the aeroplane covers 480 km in 40 minutes.

(iii) Convert the speed into m/sec: $720 \times \frac{5}{18} = 40 \times 5 = 200$ m/sec

Distance covered in 15 seconds = 200 × 15 = 3000 m = 3 km

Hence, the aeroplane will fly 3 km in 15 seconds.

Given:

Speed of dog = 6 km/h

(i) 5 minutes = $\frac{5}{60}$ hour = $\frac{1}{12}$ hour

Distance = Speed × Time = $6 \times \frac{1}{12} = \frac{1}{2}$ km = 500 metres.

Hence, the dog covers 500 metres in 5 minutes.

(ii) Speed = 6 km/h = 6 × 1000 m/h = 6000 m/h

Time = $\frac{\text{Distance}}{\text{Speed}} = \frac{200}{6000}$ hour = $\frac{1}{30}$ hour

$\frac{1}{30}$ hour = $\frac{1}{30} \times 60$ minutes = 2 minutes

Hence, the dog would take 2 minutes to cover 200 metres.

Given:

Length of pool = 50 metres

Total distance covered = 50 × 2 = 100 metres = $\frac{100}{1000}$ km = 0.1 km

Time = 5 minutes = $\frac{5}{60}$ hour = $\frac{1}{12}$ hour

Speed = $\frac{\text{Distance}}{\text{Time}} = \frac{0.1}{\frac{1}{12}} = 0.1 \times 12 = 1.2$ km/h

Hence, his swimming speed is 1.2 km/h.

Given:

Speed = 50 km/h, Time = 48 minutes = $\frac{48}{60}$ hour = $\frac{4}{5}$ hour

Distance = Speed × Time = $50 \times \frac{4}{5} = 10 \times 4 = 40$ km

Now, at a speed of 30 km/h:

Time = $\frac{\text{Distance}}{\text{Speed}} = \frac{40}{30} = \frac{4}{3}$ hours

$\frac{4}{3}$ hours = $\frac{4}{3} \times 60$ minutes = 80 minutes = 1 hour 20 minutes

Hence, it will take 1 hour 20 minutes to cover the same distance at 30 km/h.

(i) The simplest form of the ratio $\frac{1}{6} : \frac{1}{4}$ is 2 : 3.

(Reason: L.C.M. of 6 and 4 is 12. So, $\frac{1}{6} \times 12 : \frac{1}{4} \times 12 = 2 : 3$.)

(ii) 75 cm : 1.25 m = 3 : 5.

(Reason: 1.25 m = 125 cm. So, 75 : 125 = 3 : 5.)

(iii) If two ratios are equivalent, then the four quantities are said to be in proportion.

(iv) If 8, x, 48 and 18 are in proportion then the value of x is 3.

(Reason: 8 × 18 = x × 48 ⇒ 144 = 48x ⇒ x = 3.)

(v) If the cost of 10 pencils is ₹15, then the cost of 6 pencils is ₹9.

(Reason: Cost of 1 pencil = $\frac{15}{10}$ = ₹1.50. Cost of 6 pencils = 6 × 1.50 = ₹9.)

(vi) If a cyclist is travelling at a speed of 15 km/h, then the distance covered by him in 20 minutes is 5 km.

(Reason: 20 minutes = $\frac{20}{60}$ hour = $\frac{1}{3}$ hour. Distance = $15 \times \frac{1}{3}$ = 5 km.)

(i) False. A ratio can be less than 1 also. For example, 2 : 3 = $\frac{2}{3}$, which is less than 1.

(ii) False. Half an hour = 30 minutes = 30 × 60 = 1800 seconds. So the ratio = 1800 : 20 = 90 : 1, not 30 : 20.

(iii) False. $\frac{5}{7}$ and $\frac{5}{6}$: cross multiplying, 5 × 6 = 30 and 5 × 7 = 35. Since 30 < 35, $\frac{5}{7} \lt \frac{5}{6}$. So 5 : 7 is smaller than 5 : 6.

(iv) True. 3 × x = 5 × 12 ⇒ 3x = 60 ⇒ x = 20.

(v) False. $\frac{3}{2} = 1.5$ and $\frac{4}{5} = 0.8$, which are not equal.

6 : 10 = 3 : 5 (dividing both terms by 2).

Checking option 2: 18 : 30 = 3 : 5 (dividing both terms by 6).

So, 18 : 30 is equivalent to 6 : 10.

Hence, option 2 is the correct option.

$\frac{2}{3} : \frac{3}{4}$

By Division Method,

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

L.C.M. of 3 and 4 = 2 x 2 x 3 = 12. Multiply both terms by 12:

$\frac{2}{3} \times 12 : \frac{3}{4} \times 12 = 8 : 9$

Hence, option 2 is the correct option.

3 litres = 3 × 1000 mL = 3000 mL.

Ratio = 75 : 3000

By Prime Factorization,

$$\begin{array}{l|r} 3 & 75 \\ \hline 5 & 25 \\ \hline 5 & 5 \\ \hline & 1 \end{array} \quad \text{ and } \quad \begin{array}{l|r} 2 & 3000 \\ \hline 2 & 1500 \\ \hline 2 & 750 \\ \hline 3 & 375 \\ \hline 5 & 125 \\ \hline 5 & 25 \\ \hline 5 & 5 \\ \hline & 1 \end{array}$$

75 = 3 x 5 x 5

3000 = 2 x 2 x 2 x 3 x 5 x 5 x 5

H.C.F. of 75 and 3000 = 3 x 5 x 5 = 75.

$\frac{75 \div 75}{3000 \div 75} = \frac{1}{40} = 1 : 40$

Hence, option 3 is the correct option.

Number of sides of a rectangle = 4

Number of edges of a cuboid = 12

Ratio = 4 : 12 = 1 : 3

Hence, option 2 is the correct option.

Number of girls = 20

Number of boys = 35 − 20 = 15

Ratio of boys to girls = 15 : 20 = 3 : 4

Hence, option 1 is the correct option.

7 parts represent boys, so 7 parts = 21.

1 part = $\frac{21}{7}$ = 3

Number of girls = 6 parts = 6 × 3 = 18

Hence, option 3 is the correct option.

Total number of parts = 3 + 5 = 8

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

Larger number = 5 × 18 = 90

Hence, option 4 is the correct option.

x : 12 :: 8 : 32

Product of extremes = product of means.

x × 32 = 12 × 8

32x = 96

$x = \frac{96}{32} = 3$

Hence, option 3 is the correct option.

3, 12, x are in continued proportion, so 3 : 12 :: 12 : x.

Product of extremes = product of means.

3 × x = 12 × 12

3x = 144

$x = \frac{144}{3} = 48$

Hence, option 4 is the correct option.

Weight of 5 bags = 27 kg

Weight of 1 bag = $\frac{27}{5}$ = 5.4 kg

Hence, option 3 is the correct option.

One dozen = 12 notebooks

Cost of 12 notebooks = ₹66

Cost of 1 notebook = $\frac{66}{12}$ = ₹5.50

Hence, option 4 is the correct option.

To convert km/h into m/sec, multiply by $\frac{5}{18}$.

$90 \times \frac{5}{18} = 5 \times 5 = 25$ m/sec

Hence, option 3 is the correct option.

Statement I: Total parts = 4 + 5 = 9. Value of 1 part = $\frac{27}{9}$ = 3. Smaller number = 4 × 3 = 12. So, Statement I is true.

Statement II: Multiplying or dividing both terms of a ratio by the same non-zero number gives an equivalent ratio. So, Statement II is true.

Hence, option 3 is the correct option.

Statement I: Cost of 1 banana = $\frac{49}{7}$ = ₹7. Cost of a dozen (12) bananas = 12 × 7 = ₹84, not ₹70. So, Statement I is false.

Statement II: Cost of 1 pen = $\frac{77}{7}$ = ₹11, so a dozen pens cost 12 × 11 = ₹132. Cost of 1 notebook = $\frac{60}{5}$ = ₹12, so a dozen notebooks cost 12 × 12 = ₹144. Since ₹132 < ₹144, a dozen pens cost less than a dozen notebooks. So, Statement II is false.

Hence, option 4 is the correct option.

Statement I: Convert the speed of car A into m/sec: $57 \times \frac{5}{18} = \frac{285}{18} = 15.8\overline{3}$ m/sec.

Since 17.5 m/sec > 15.83 m/sec, car B is faster than car A.

So, Statement I is true.

Statement II: Speed = $\frac{\text{Distance}}{\text{Time}}$ is the correct formula.

So, Statement II is true.

Hence, option 3 is the correct option.

Given:

Length = 120 m

Breadth = 75 m

(i) Ratio of breadth to length = 75 : 120

By Prime Factorization,

$$\begin{array}{l|r} 3 & 75 \\ \hline 5 & 25 \\ \hline 5 & 5 \\ \hline & 1 \end{array} \quad \text{ and } \quad \begin{array}{l|r} 2 & 120 \\ \hline 2 & 60 \\ \hline 2 & 30 \\ \hline 3 & 15 \\ \hline 5 & 5 \\ \hline & 1 \end{array}$$

75 = 3 x 5 x 5

120 = 2 x 2 x 2 x 3 x 5

H.C.F. of 75 and 120 = 3 x 5 = 15.

$\frac{75 \div 15}{120 \div 15} = \frac{5}{8} = 5 : 8$

Hence, the ratio of breadth to length is 5 : 8.

(ii) Perimeter of the park = 2 × (length + breadth) = 2 × (120 + 75) = 2 × 195 = 390 m

Ratio of length to perimeter = 120 : 390

By Prime Factorization,

$$\begin{array}{l|r} 2 & 120 \\ \hline 2 & 60 \\ \hline 2 & 30 \\ \hline 3 & 15 \\ \hline 5 & 5 \\ \hline & 1 \end{array} \quad \text{ and } \quad \begin{array}{l|r} 2 & 390 \\ \hline 3 & 195 \\ \hline 5 & 65 \\ \hline 13 & 13 \\ \hline & 1 \end{array}$$

120 = 2 x 2 x 2 x 3 x 5

390 = 2 x 3 x 5 x 13

H.C.F. of 120 and 390 = 2 x 3 x 5 = 30.

$\frac{120 \div 30}{390 \div 30} = \frac{4}{13} = 4 : 13$

Hence, the ratio of length to perimeter is 4 : 13.

The sum of the angles of a triangle is 180°.

Total number of parts = 2 + 3 + 4 = 9

Value of 1 part = $\frac{180°}{9}$ = 20°

First angle = 2 × 20° = 40°

Second angle = 3 × 20° = 60°

Third angle = 4 × 20° = 80°

Hence, the three angles are 40°, 60° and 80°.

Given:

Height of Anshul = 1.04 m = 104 cm

Height of Ankita = 1.30 m = 130 cm

Height of Dhruv = 91 cm

Ratio of their heights = 104 : 130 : 91

By Prime Factorization,

$$\begin{array}{l|r} 2 & 104 \\ \hline 2 & 52 \\ \hline 2 & 26 \\ \hline 13 & 13 \\ \hline & 1 \end{array} \quad \text{ and } \quad \begin{array}{l|r} 2 & 130 \\ \hline 5 & 65 \\ \hline 13 & 13 \\ \hline & 1 \end{array} \quad \text{ and } \quad \begin{array}{l|r} 7 & 91 \\ \hline 13 & 13 \\ \hline & 1 \end{array}$$

104 = 2 x 2 x 2 x 13

130 = 2 x 5 x 13

91 = 7 x 13

H.C.F. of 104, 130 and 91 is 13.

$\frac{104}{13} : \frac{130}{13} : \frac{91}{13} = 8 : 10 : 7$

Total number of parts = 8 + 10 + 7 = 25

Value of 1 part = $\frac{100}{25}$ = 4 sweets

Anshul's share = 8 × 4 = 32 sweets

Ankita's share = 10 × 4 = 40 sweets

Dhruv's share = 7 × 4 = 28 sweets

Hence, Anshul gets 32 sweets, Ankita gets 40 sweets and Dhruv gets 28 sweets.

Given:

Divya : Himanshu = 5 : 7

Himanshu's weight = 28 kg

7 parts represent Himanshu's weight, so 7 parts = 28 kg.

1 part = $\frac{28}{7}$ = 4 kg

Divya's weight = 5 parts = 5 × 4 = 20 kg

Hence, the weight of Divya is 20 kg.

Let the areas of flat A, flat B and flat C be 5x, 6x and 8x square metres respectively.

Difference of areas of flat C and flat A = 8x − 5x = 3x

According to the question,

3x = 180

$x = \frac{180}{3} = 60$

Area of flat B = 6x = 6 × 60 = 360 square metres

Hence, the area of flat B is 360 square metres.

Let the original income be 7x and the new income be 8x.

Increase in income = 8x − 7x = x

According to the question,

x = ₹4500

New income = 8x = 8 × 4500 = ₹36000

Hence, his new income is ₹36000 per month.

3A = 5B

$\frac{A}{B} = \frac{5}{3}$, so A : B = 5 : 3

4B = 6C

$\frac{B}{C} = \frac{6}{4} = \frac{3}{2}$, so B : C = 3 : 2

Since the value of B is 3 in both ratios,

A : B : C = 5 : 3 : 2

∴ A : C = 5 : 2

Hence, A : C = 5 : 2.

Write the ratios as fractions: $\frac{9}{13}$ and $\frac{7}{11}$

By cross multiplication, compare 9 × 11 and 7 × 13:

9 × 11 = 99 and 7 × 13 = 91

Since 99 > 91, we have $\frac{9}{13} \gt \frac{7}{11}$

Hence, 7 : 11 is the smaller ratio.

(i) 4, 7, 20

4 : 7 :: 20 : x

Product of extremes = product of means.

4 × x = 7 × 20

4x = 140

$x = \frac{140}{4} = 35$

Hence, the fourth proportional is 35.

(ii) $2\frac{1}{2}$, $1\frac{1}{4}$, 2.2

$2\frac{1}{2} = \frac{5}{2}$ and $1\frac{1}{4} = \frac{5}{4}$

$\frac{5}{2} : \frac{5}{4} :: 2.2 : x$

Product of extremes = product of means.

$\frac{5}{2} \times x = \frac{5}{4} \times 2.2$

$\frac{5x}{2} = \frac{5 \times 2.2}{4} = \frac{11}{4}$

$x = \frac{11}{4} \times \frac{2}{5} = \frac{22}{20} = 1.1$

Hence, the fourth proportional is 1.1.

Given:

3 hours 30 minutes = (3 × 60 + 30) minutes = 210 minutes

Time to type 70 pages = 210 minutes

Time to type 1 page = $\frac{210}{70}$ = 3 minutes

Time to type 300 pages = 300 × 3 = 900 minutes

900 minutes = $\frac{900}{60}$ = 15 hours

Hence, she will take 15 hours to type 300 pages.

Given:

Cloth woven by 12 looms = 210 m

Cloth woven by 1 loom = $\frac{210}{12}$ = 17.5 m

Cloth woven by 8 looms = 8 × 17.5 = 140 m

Hence, 8 looms will weave 140 m of cloth per day.

Given:

4 hours 30 minutes = $4\frac{1}{2}$ hours = $\frac{9}{2}$ hours

Distance = Speed × Time = $60 \times \frac{9}{2} = 30 \times 9 = 270$ km

Convert 15 m/sec into km/h: $15 \times \frac{18}{5} = 3 \times 18 = 54$ km/h

Time taken at 54 km/h = $\frac{\text{Distance}}{\text{Speed}} = \frac{270}{54} = 5$ hours

Hence, the same journey will take 5 hours at a speed of 15 m/sec.

Let the present ages of Rohit and Mayank be 11x years and 8x years.

8 years hence:

Rohit's age = (11x + 8) years

Mayank's age = (8x + 8) years

According to the question,

$\frac{11x + 8}{8x + 8} = \frac{5}{4}$

By cross multiplication,

4(11x + 8) = 5(8x + 8)

44x + 32 = 40x + 40

44x − 40x = 40 − 32

4x = 8

x = 2

Rohit's present age = 11 × 2 = 22 years

Mayank's present age = 8 × 2 = 16 years

Hence, Rohit is 22 years old and Mayank is 16 years old.

Let the length and breadth of the rectangle be 3x m and 2x m.

According to the question, length is 5 m more than the breadth.

3x − 2x = 5

x = 5

Length = 3 × 5 = 15 m

Breadth = 2 × 5 = 10 m

Perimeter = 2 × (length + breadth) = 2 × (15 + 10) = 2 × 25 = 50 m

Hence, the perimeter of the rectangle is 50 m.