Ratio and Proportion

Express the following ratios in simplest form:

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

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

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

Answer

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

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:

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

Hence, the answer is 3 : 2.

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

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

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

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:

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

Hence, the answer is 4 : 1.

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

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:

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

Hence, the answer is 6 : 3 : 2.

Find the ratio of each of the following in simplest form:

(i) ₹5 to 50 paise

(ii) 3 km to 300 m

(iii) 9 m to 27 cm

(iv) 15 kg to 210 g

(v) 25 minutes to 1.5 hours

(vi) 30 days to 36 hours

Answer

(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.

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

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.

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

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.

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

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.

If A : B = 3 : 4 and B : C = 8 : 9, then find A : C.

Answer

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.

If A : B = 5 : 8 and B : C = 18 : 25, then find A : B : C.

Answer

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.

If 3A = 2B = 5C, then find A : B : C.

Answer

Let 3A = 2B = 5C = k

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

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

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

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

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

Out of daily income of ₹600, a worker spends ₹450 on food and shelter and saves the rest. Find the ratio of his

(i) spending to income

(ii) saving to income

(iii) saving to spending

Answer

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.

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

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.

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

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.

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

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

5 grams of an alloy contains $3\dfrac{3}{4}$ grams copper and the rest is nickel. Find the ratio by weight of nickel to copper.

Answer

Given:

Total weight of alloy = 5 grams

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

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

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

Multiply both terms by 4:

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

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

A pole of height 3 metres is struck by a speeding car and breaks into two pieces such that the first piece is $\dfrac{1}{2}$ of the second. Find the length of both pieces.

Answer

Let the length of the second piece = x metres.

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

Total length of the pole = 3 metres.

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

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

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

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

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

Heights of Anshul and Dhruv are 1.04 m and 78 cm respectively. Divide 35 sweets between them in the ratio of their heights.

Answer

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.

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

Total number of parts = 4 + 3 = 7

Value of 1 part = $\dfrac{35}{7}$ = 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.

₹180 are to be divided among three children in the ratio $\dfrac{1}{3} : \dfrac{1}{4} : \dfrac{1}{6}$. Find the share of each child.

Answer

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

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:

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

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

Value of 1 part = $\dfrac{180}{9}$ = ₹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.

A natural number has been divided into two parts in the ratio 7 : 11. If the difference of two parts is 20, find the number and the two parts.

Answer

Let the two parts be 7x and 11x.

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

According to the question,

4x = 20

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

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.

A certain sum of money has been divided into two parts in the ratio 9 : 13. If the second part is ₹260, find the total amount.

Answer

Let the two parts be 9x and 13x.

Second part = 13x = ₹260

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

First part = 9 × 20 = ₹180

Total amount = ₹180 + ₹260 = ₹440

Hence, the total amount is ₹440.

A certain sum of money is divided into three parts in the ratio 5 : 7 : 8. If the first part is ₹225, find the total amount and the other two parts.

Answer

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

First part = 5x = ₹225

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

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.

Divide ₹1312 into three parts such that first part is $\dfrac{2}{3}$ of the second and the ratio between second and third parts is 4 : 7.

Answer

Let the second part = x.

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

Second part : Third part = 4 : 7

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

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

Total = ₹1312

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

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.

$\dfrac{8x}{12} + \dfrac{12x}{12} + \dfrac{21x}{12} = 1312\\[1em] \dfrac{8x + 12x + 21x}{12} = 1312\\[1em] \dfrac{41x}{12} = 1312\\[1em] x = \dfrac{1312 \times 12}{41} = 32 \times 12 = 384$

Second part = ₹384

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

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

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

The ratio of the present ages of Saanvi and Navya is 2 : 3. Five years hence, the ratio of their ages will be 3 : 4. Find their present ages.

Answer

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,

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

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.

The present ages of A and B are in the ratio 5 : 6. Three years ago, their ages were in the ratio 4 : 5. Find their present ages.

Answer

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,

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

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.

Two numbers are in the ratio 5 : 6. When 2 is added to first and 3 is added to second, they are in the ratio 4 : 5. Find the numbers.

Answer

Let the two numbers be 5x and 6x.

According to the question,

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

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.

The ratio of number of boys to the number of girls in a school of 1430 students is 7 : 6. If 26 new girls are admitted in the school, find how many new boys should be admitted so that the ratio of number of boys to the number of girls changes to 8 : 7.

Answer

Total number of students = 1430

Ratio of boys to girls = 7 : 6

Total number of parts = 7 + 6 = 13

Value of 1 part = $\dfrac{1430}{13}$ = 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,

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

By cross multiplication,

7(770 + y) = 8 × 686

5390 + 7y = 5488

7y = 5488 − 5390

7y = 98

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

Hence, 14 new boys should be admitted.

Which ratio is greater:

(i) 5 : 6 or 6 : 7

(ii) 13 : 24 or 17 : 32

Answer

(i) 5 : 6 or 6 : 7

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

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

5 × 7 = 35 and 6 × 6 = 36

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

Hence, 6 : 7 is the greater ratio.

(ii) 13 : 24 or 17 : 32

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

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

13 × 32 = 416 and 17 × 24 = 408

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

Hence, 13 : 24 is the greater ratio.

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

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

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

Answer

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

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

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

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 $\dfrac{13}{12}$.

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

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 $\dfrac{9}{11}$.

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

Hence, her new weight is 45 kg.

Which of the following statements are true?

(i) 2.5 : 1.5 : : 7.0 : 4.2

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

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

Answer

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) $\dfrac{1}{2} : \dfrac{1}{3} = \dfrac{1}{3} : \dfrac{1}{4}$

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

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

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

Hence, the statement is False.

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

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

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

Both ratios are equal, so the statement is true.

Hence, the statement is True.

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

Check whether the following numbers are in proportion or not:

(i) 18, 10, 9, 5

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

(iii) 0.1, 0.2, 0.3, 0.6

Answer

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, $3\dfrac{1}{2}$, 4, $4\dfrac{1}{2}$

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

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

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

Hence, 3, $3\dfrac{1}{2}$, 4, $4\dfrac{1}{2}$ 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.

Find x in the following proportions:

(i) x : 4 = 9 : 12

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

(iii) 3.6 : 0.4 = x : 0.5

Answer

(i) x : 4 = 9 : 12

Product of extremes = product of means.

x × 12 = 4 × 9

12x = 36

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

Hence, x = 3.

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

Product of extremes = product of means.

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

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

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

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

(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

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

Hence, x = 4.5.

Find the fourth proportional to

(i) 42, 12, 7

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

(iii) 3 kg, 12 kg, 15 kg

Answer

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

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

Hence, the fourth proportional is 2.

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

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

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

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

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

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

(iii) 3 kg, 12 kg, 15 kg

3 : 12 :: 15 : x

3 × x = 12 × 15

3x = 180

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

Hence, the fourth proportional is 60 kg.

Check whether 7, 49, 343 are in continued proportion or not.

Answer

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.

Find the third proportional to

(i) 36, 18

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

(iii) 3.2, 0.8

Answer

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

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

Hence, the third proportional is 9.

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

$5\dfrac{1}{4} = \dfrac{21}{4}\\[1em] \dfrac{21}{4} : 7 :: 7 : x\\[1em] \dfrac{21}{4} \times x = 7 \times 7\\[1em] \dfrac{21x}{4} = 49\\[1em]$

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

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

(iii) 3.2, 0.8

3.2 : 0.8 :: 0.8 : x

3.2 × x = 0.8 × 0.8

3.2x = 0.64

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

Hence, the third proportional is 0.2.

The ratio between the length and width of a rectangular sheet of paper is 7 : 5. If the width of the sheet is 20.5 cm, find its length.

Answer

Given:

Length : Width = 7 : 5

Width = 20.5 cm

Let the length = x cm.

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

By cross multiplication,

5 × x = 7 × 20.5

5x = 143.5

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

Hence, the length of the sheet is 28.7 cm.

The ages of Advik and Anaya are in the ratio 4 : 5. If Advik is 4 years 8 months old, find the age of Anaya.

Answer

Given:

Advik : Anaya = 4 : 5

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

Let Anaya's age = x months.

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

By cross multiplication,

4 × x = 56 × 5

4x = 280

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

70 months = 5 years 10 months

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

6 bowls cost ₹90. What would be cost of 10 such bowls?

Answer

Given:

Cost of 6 bowls = ₹90

Cost of 1 bowl = $\dfrac{90}{6}$ = ₹15

Cost of 10 bowls = 10 × ₹15 = ₹150

Hence, the cost of 10 bowls is ₹150.

Ten pencils cost ₹15. How many pencils can be bought with ₹72?

Answer

Given:

Cost of 10 pencils = ₹15

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

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

Hence, 48 pencils can be bought with ₹72.

400 grams cake costs 800 rupees. How much would a 1.5 kg cake cost?

Answer

Given:

Cost of 400 g cake = ₹800

Cost of 1 g cake = $\dfrac{800}{400}$ = ₹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.

A man earns ₹18000 in 3 months.

(i) How much time would he take to earn ₹30000?

(ii) How much money will he earn in 7 months?

Answer

Given:

Earning in 3 months = ₹18000

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

(i) Time taken to earn ₹30000 = $\dfrac{30000}{6000}$ = 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.

12 mangoes weigh 2.4 kg. What is the weight of 8 mangoes?

Answer

Given:

Weight of 12 mangoes = 2.4 kg

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

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

Hence, the weight of 8 mangoes is 1.6 kg.

If the weight of 12 sheets of thick paper is 40 grams, how many sheets of the same paper would weigh $2\dfrac{1}{2}$ kilograms?

Answer

Given:

Weight of 12 sheets = 40 grams

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

Number of sheets weighing 40 g = 12

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

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

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

A bus consumes 25 litres of diesel in covering a distance of 90 kilometres. How much diesel is needed to cover 288 kilometres?

Answer

Given:

Diesel needed for 90 km = 25 litres

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

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

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

If $\dfrac{4}{5}$ metre cloth costs ₹36, find the cost of $2\dfrac{1}{5}$ metres of cloth.

Answer

Given:

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

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

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

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

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

If 15 men can pack 540 parcels per day, how many men are needed to pack 396 parcels per day?

Answer

Given:

Number of men to pack 540 parcels = 15

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

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

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

Which is a better buy : 12 kg potatoes for ₹132 or 16 kg potatoes for ₹168?

Answer

For 12 kg potatoes:

Cost of 12 kg = ₹132

Cost of 1 kg = $\dfrac{132}{12}$ = ₹11

For 16 kg potatoes:

Cost of 16 kg = ₹168

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

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

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

Convert the following speeds into m/sec :

(i) 72 km/h

(ii) 9 km/h

(iii) 1.2 km/minute

(iv) 600 m/hour

Answer

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

(i) 72 km/h

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

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

(ii) 9 km/h

$= 9 \times \dfrac{5}{18} = \dfrac{45}{18} = 2.5$ 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.

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

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

(iv) 600 m/hour

1 hour = 3600 seconds.

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

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

Convert the following speeds into km/h :

(i) 15 m/sec

(ii) 1.5 m/sec

Answer

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

(i) 15 m/sec

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

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

(ii) 1.5 m/sec

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

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

Which is greater — a speed of 30 m/sec or 30 km/h?

Answer

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

$30 \text{ m/sec} = 30 \times \dfrac{18}{5} = 6 \times 18 = 108$ 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.

An aeroplane is flying at a speed of 720 km/h

(i) If the aerial distance between two cities is 1800 km, how much time will the aeroplane take in crossing these cities?

(ii) How much distance does the aeroplane cover in 40 minutes?

(iii) How far will it fly in 15 seconds?

Answer

Given:

Speed of aeroplane = 720 km/h

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

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

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

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

Hence, the aeroplane covers 480 km in 40 minutes.

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

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

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

A dog is walking at a speed of 6 km/h.

(i) How much distance does it cover in 5 minutes?

(ii) How much time would it take to cover 200 metres?

Answer

Given:

Speed of dog = 6 km/h

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

Distance = Speed × Time = $6 \times \dfrac{1}{12} = \dfrac{1}{2}$ 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 = $\dfrac{\text{Distance}}{\text{Speed}} = \dfrac{200}{6000}$ hour = $\dfrac{1}{30}$ hour

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

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

A swimming pool is 50 metres long. A boy can swim across the length and then return to his starting position in 5 minutes. What is his swimming speed in km/h?

Answer

Given:

Length of pool = 50 metres

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

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

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

Hence, his swimming speed is 1.2 km/h.

A bus takes 48 minutes to cover a certain distance when travelling at a speed of 50 km/h. How much time will it take to cover the same distance when travelling at a speed of 30 km/h?

Answer

Given:

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

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

Now, at a speed of 30 km/h:

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

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

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

Fill in the blanks:

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

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

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

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

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

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

Answer

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

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

(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 = $\dfrac{15}{10}$ = ₹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 = $\dfrac{20}{60}$ hour = $\dfrac{1}{3}$ hour. Distance = $15 \times \dfrac{1}{3}$ = 5 km.)

State whether the following statements are true (T) or false (F):

(i) A ratio is always greater than 1.

(ii) Ratio of half an hour to 20 seconds is 30 : 20

(iii) The ratio 5 : 7 is greater than the ratio 5 : 6

(iv) If the numbers 3, 5, 12 and x are in proportion then the value of x is 20

(v) The ratios 3 : 2 and 4 : 5 are equivalent.

Answer

(i) False. A ratio can be less than 1 also. For example, 2 : 3 = $\dfrac{2}{3}$, 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. $\dfrac{5}{7}$ and $\dfrac{5}{6}$: cross multiplying, 5 × 6 = 30 and 5 × 7 = 35. Since 30 < 35, $\dfrac{5}{7} \lt \dfrac{5}{6}$. So 5 : 7 is smaller than 5 : 6.

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

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

A ratio equivalent to 6 : 10 is

1. 3 : 4

2. 18 : 30

3. 12 : 40

4. 5 : 3

Answer

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.

A ratio equivalent to the ratio $\dfrac{2}{3} : \dfrac{3}{4}$ is

1. 4 : 6

2. 8 : 9

3. 6 : 8

4. 9 : 8

Answer

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

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:

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

Hence, option 2 is the correct option.

The ratio of 75 mL to 3 litres is

1. 25 : 1

2. 40 : 1

3. 1 : 40

4. 3 : 200

Answer

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.

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

Hence, option 3 is the correct option.

The ratio of the number of sides of a rectangle to the number of edges of a cuboid is

1. 1 : 2

2. 1 : 3

3. 2 : 3

4. none of these

Answer

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.

In a class of 35 students, the number of girls is 20. The ratio of number of boys to the number of girls in the class is

1. 3 : 4

2. 4 : 3

3. 7 : 4

4. 7 : 3

Answer

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.

The ratio of number of girls to the number of boys in a class is 6 : 7. If there are 21 boys in the class, then the number of girls in the class is

1. 39

2. 24

3. 18

4. 13

Answer

7 parts represent boys, so 7 parts = 21.

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

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

Hence, option 3 is the correct option.

Two numbers are in the ratio 3 : 5. If the sum of the numbers is 144, then the larger number is

1. 48

2. 54

3. 72

4. 90

Answer

Total number of parts = 3 + 5 = 8

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

Larger number = 5 × 18 = 90

Hence, option 4 is the correct option.

If x, 12, 8 and 32 are in proportion, then x is

1. 6

2. 4

3. 3

4. 2

Answer

x : 12 :: 8 : 32

Product of extremes = product of means.

x × 32 = 12 × 8

32x = 96

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

Hence, option 3 is the correct option.

If 3, 12 and x are in continued proportion, then x is

1. 4

2. 6

3. 16

4. 48

Answer

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

Product of extremes = product of means.

3 × x = 12 × 12

3x = 144

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

Hence, option 4 is the correct option.

If the weight of 5 bags of sugar is 27 kg, then the weight of one bag of sugar is

1. 5.04 kg

2. 5.2 kg

3. 5.4 kg

4. 5.6 kg

Answer

Weight of 5 bags = 27 kg

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

Hence, option 3 is the correct option.

Sonali bought one dozen notebooks for ₹66. What did she pay for one notebook?

1. ₹6.50

2. ₹6.60

3. ₹5.60

4. ₹5.50

Answer

One dozen = 12 notebooks

Cost of 12 notebooks = ₹66

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

Hence, option 4 is the correct option.

The speed of 90 km/h is equal to

1. 10 m/sec

2. 18 m/sec

3. 25 m/sec

4. none of these

Answer

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

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

Hence, option 3 is the correct option.

Statement I: Two numbers are in the ratio 4 : 5. If the sum of the numbers is 27, then the smaller number is 12

Statement II: We can multiply or divide both the terms of a ratio by the same non-zero integer.

1. Statement I is true but Statement II is false.

2. Statement I is false but Statement II is true.

3. Both Statement I and Statement II are true.

4. Both Statement I and Statement II are false.

Answer

Statement I: Total parts = 4 + 5 = 9. Value of 1 part = $\dfrac{27}{9}$ = 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: If 7 bananas cost ₹49, then the cost of a dozen bananas is ₹70

Statement II: If 7 pens cost ₹77 and 5 notebooks cost ₹60, then the cost of one dozen pens is higher than the cost of one dozen notebooks.

1. Statement I is true but Statement II is false.

2. Statement I is false but Statement II is true.

3. Both Statement I and Statement II are true.

4. Both Statement I and Statement II are false.

Answer

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

Statement II: Cost of 1 pen = $\dfrac{77}{7}$ = ₹11, so a dozen pens cost 12 × 11 = ₹132. Cost of 1 notebook = $\dfrac{60}{5}$ = ₹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: If the speed of car A is 57 km/h and the speed of car B is 17.5 m/s, then car B is faster than car A.

Statement II: Speed = $\dfrac{\text {Distance}}{\text {Time}}$

1. Statement I is true but Statement II is false.

2. Statement I is false but Statement II is true.

3. Both Statement I and Statement II are true.

4. Both Statement I and Statement II are false.

Answer

Statement I: Convert the speed of car A into m/sec: $57 \times \dfrac{5}{18} = \dfrac{285}{18} = 15.8\overline{3}$ 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 = $\dfrac{\text{Distance}}{\text{Time}}$ is the correct formula.

So, Statement II is true.

Hence, option 3 is the correct option.

A rectangular park is 120 m long and 75 m wide. Find the ratio of:

(i) breadth to its length

(ii) length to its perimeter

Answer

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.

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

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.

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

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

Divide the angles of a triangle in the ratio 2 : 3 : 4

Answer

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

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

Value of 1 part = $\dfrac{180°}{9}$ = 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°.

Heights of Anshul, Ankita and Dhruv are 1.04 m, 1.30 m and 91 cm respectively. Divide 100 sweets among them in the same ratio as their heights.

Answer

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.

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

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

Value of 1 part = $\dfrac{100}{25}$ = 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.

The weights of Divya and Himanshu are in the ratio 5 : 7. If Himanshu weighs 28 kg, find the weight of Divya.

Answer

Given:

Divya : Himanshu = 5 : 7

Himanshu's weight = 28 kg

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

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

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

Hence, the weight of Divya is 20 kg.

The areas of three flats are in the ratio 5 : 6 : 8. If the differences in the areas of flat C and flat A is 180 square metres, find the area of the flat B.

Answer

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

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

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

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

The income of a man is increased in the ratio 7 : 8. If the increase in his income is ₹4500 per month, find his new income.

Answer

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.

If 3A = 5B and 4B = 6C, then find A : C.

Answer

3A = 5B

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

4B = 6C

$\dfrac{B}{C} = \dfrac{6}{4} = \dfrac{3}{2}$, 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.

Which ratio is smaller — 9 : 13 or 7 : 11?

Answer

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

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

9 × 11 = 99 and 7 × 13 = 91

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

Hence, 7 : 11 is the smaller ratio.

Find the fourth proportional to

(i) 4, 7, 20

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

Answer

(i) 4, 7, 20

4 : 7 :: 20 : x

Product of extremes = product of means.

4 × x = 7 × 20

4x = 140

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

Hence, the fourth proportional is 35.

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

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

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

Product of extremes = product of means.

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

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

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

Hence, the fourth proportional is 1.1.

A typist types 70 pages in 3 hours 30 minutes. How long will she take to type 300 pages?

Answer

Given:

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

Time to type 70 pages = 210 minutes

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

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

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

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

12 looms weave 210 m cloth per day. How many metres of cloth will 8 looms weave per day?

Answer

Given:

Cloth woven by 12 looms = 210 m

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

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

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

A journey takes 4 hours 30 minutes at a speed of 60 km/h. How long will the same journey take at a speed of 15 m/sec?

Answer

Given:

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

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

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

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

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

Present ages of Rohit and Mayank are in the ratio 11 : 8. 8 years hence ratio of their ages will be 5 : 4. Find their present ages.

Answer

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,

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

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.

Ratio of length and breadth of a rectangle is 3 : 2. If the length of rectangle is 5 m more than the breadth, find the perimeter of the rectangle.

Answer

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.