Ratio & Proportion

(i) 20 : 35

The given ratio is 20 : 35.

Let us find H.C.F. of 20 and 35:

$$\begin{array}{r} 1 \\ 20 \overline{) 35} \\ \underline{-20} \\ 15 \overline{) 20} ( 1 \\ \underline{-15} \\ 5 \overline{) 15} ( 3 \\ \underline{-15} \\ 0 \end{array}$$

H.C.F. = 5

$∴ 20 : 35 = \frac{20 \div 5}{35 \div 5} = \frac{4}{7} = 4 : 7$

Hence, the answer is 4 : 7

(ii) 95 : 57

The given ratio is 95 : 57.

Let us find H.C.F. of 95 and 57:

$$\begin{array}{r} 1 \\ 57 \overline{) 95} \\ \underline{-57} \\ 38 \overline{) 57} ( 1 \\ \underline{-38} \\ 19 \overline{) 38} ( 2 \\ \underline{-38} \\ 0 \end{array}$$

H.C.F. = 19

$∴ 95 : 57 = \frac{95 \div 19}{57 \div 19} = \frac{5}{3} = 5 : 3$

Hence, the answer is 5 : 3

(iii) 2.1 : 1.2

First, multiply both by 10 to remove decimals:

(2.1 x 10 : 1.2 x 10) = 21 : 12

Let us find H.C.F. of 21 and 12:

$$\begin{array}{r} 1 \\ 12 \overline{) 21} \\ \underline{-12} \\ 9 \overline{) 12} ( 1 \\ \underline{-9} \\ 3 \overline{) 9} ( 3 \\ \underline{-9} \\ 0 \end{array}$$

H.C.F. = 3

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

Hence, the answer is 7 : 4

(iv) $3\frac{1}{4} : 6\frac{1}{2}$

Convert mixed fractions to improper fractions: $\frac{13}{4} : \frac{13}{2}$

Let us find L.C.M. of 4 and 2:

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

L.C.M. = 2 x 2 = 4

Multiply both terms by 4 to convert fractions to whole number:

$\Big(\frac{13}{4} \times 4\Big) : \Big(\frac{13}{2} \times 4\Big) = 13 : 26$

Let us find H.C.F of 13 and 26:

$$\begin{array}{r} 2 \\ 13 \overline{) 26} \\ \underline{-26} \\ 0 \end{array}$$

H.C.F = 13

$\frac{13 \div 13}{26 \div 13}$ = 1 : 2

Hence, the answer is 1 : 2

(v) 6a : 9b

Given ratio is 6a : 9b

Let us find H.C.F. of 6 and 9:

$$\begin{array}{r} 1 \\ 6 \overline{) 9} \\ \underline{-6} \\ 3 \overline{) 6} ( 2 \\ \underline{-6} \\ 0 \end{array}$$

H.C.F. = 3

$\frac{6a}{9b} = \frac{6a \div 3}{9b \div 3} = \frac{2a}{3b} = 2a : 3b$

Hence, the answer is 2a : 3b

(vi) 8ab² : 6a²b

Find the common factors in the coefficients and variables:

Let us find H.C.F. of 8 and 6:

$$\begin{array}{r} 1 \\ 6 \overline{) 8} \\ \underline{-6} \\ 2 \overline{) 6} ( 3 \\ \underline{-6} \\ 0 \end{array}$$

H.C.F. = 2

Common variables: ab.

Divide both terms by 2ab:

$\frac{8ab^2}{2ab} : \frac{6a^2b}{2ab} = 4b : 3a$

Hence, the answer is 4b : 3a

(i) $\frac{1}{8} : \frac{1}{12}$

Let us find L.C.M of 8 and 12:

$$\begin{array}{l|l} 2 & 8, 12 \\ \hline 2 & 4, 6 \\ \hline 2 & 2, 3 \\ \hline 3 & 1, 3 \\ \hline & 1, 1 \end{array}$$

L.C.M. = 2 x 2 x 2 x 3 = 24.

Multiplying both terms by 24 to convert fractions to whole number:

$\left(\frac{1}{8} \times 24\right) : \left(\frac{1}{12} \times 24\right) = 3 : 2$

Hence, the answer is 3 : 2

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

Let us find L.C.M of 9 and 6:

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

L.C.M. = 3 x 3 x 2 = 18.

Multiplying both terms by 18 to convert fractions to whole number:

$\left(\frac{1}{9} \times 18\right) : \left(\frac{1}{6} \times 18\right) = 2 : 3$

Hence, the answer is 2 : 3

(iii) $\frac{2}{3} : \frac{5}{6}$

Let us find L.C.M of 3 and 6:

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

L.C.M. = 3 x 2 = 6.

Multiplying both terms by 6 to convert fractions to whole number:

$\left(\frac{2}{3} \times 6\right) : \left(\frac{5}{6} \times 6\right) = 4 : 5$

Hence, the answer is 4 : 5

(iv) $\frac{1}{4} : \frac{1}{6} : \frac{1}{8}$

Let us find L.C.M of 4, 6 and 8:

$$\begin{array}{l|l} 2 & 4, 6, 8 \\ \hline 2 & 2, 3, 4 \\ \hline 2 & 1, 3, 2 \\ \hline 3 & 1, 3, 1 \\ \hline & 1, 1, 1 \end{array}$$

L.C.M. = 2 x 2 x 2 x 3 = 24.

Multiplying each term by 24 to convert fractions to whole number:

$\left(\frac{1}{4} \times 24\right) : \left(\frac{1}{6} \times 24\right) : \left(\frac{1}{8} \times 24\right) = 6 : 4 : 3$

Hence, the answer is 6 : 4 : 3

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

Convert to improper fractions: $\frac{5}{3} : \frac{5}{2} : \frac{7}{4}$.

Let us find L.C.M of 3, 2 and 4:

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

L.C.M. = 2 x 2 x 3 = 12.

Multiplying each term by 12 to convert fractions to whole number:

$\left(\frac{5}{3} \times 12\right) : \left(\frac{5}{2} \times 12\right) : \left(\frac{7}{4} \times 12\right) = 20 : 30 : 21$

Hence, the answer is 20 : 30 : 21

(vi) $2\frac{1}{3} : 3\frac{1}{4} : 1\frac{1}{6}$

Convert to improper fractions: $\frac{7}{3} : \frac{13}{4} : \frac{7}{6}$.

Let us find L.C.M of 3, 4 and 6:

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

L.C.M. = 2 x 2 x 3 = 12.

Multiplying each term by 12 to convert fractions to whole number:

$\left(\frac{7}{3} \times 12\right) : \left(\frac{13}{4} \times 12\right) : \left(\frac{7}{6} \times 12\right) = 28 : 39 : 14$

Hence, the answer is 28 : 39 : 14

(i) 35 paise : ₹ 1

Convert rupee into paise: ₹ 1 = 100 paise.

The ratio is 35 : 100.

Let us find H.C.F. of 35 and 100:

$$\begin{array}{r} 2 \\ 35 \overline{) 100} \\ \underline{-70} \\ 30 \overline{) 35} ( 1 \\ \underline{-30} \\ 5 \overline{) 30} ( 6 \\ \underline{-30} \\ 0 \end{array}$$

H.C.F. = 5

$\frac{35 \div 5}{100 \div 5} = \frac{7}{20} = 7 : 20$

Hence, the answer is 7 : 20

(ii) 1 kg : 375 g

Convert kg into g: 1 kg = 1000 g.

The ratio is 1000 : 375.

Let us find H.C.F. of 1000 and 375:

$$\begin{array}{r} 2 \\ 375 \overline{) 1000} \\ \underline{-750} \\ 250 \overline{) 375} ( 1 \\ \underline{-250} \\ 125 \overline{) 250} ( 2 \\ \underline{-250} \\ 0 \end{array}$$

H.C.F. = 125

$\frac{1000 \div 125}{375 \div 125} = \frac{8}{3} = 8 : 3$

Hence, the answer is 8 : 3

(iii) 60 cm : 1 m

Convert m into cm: 1 m = 100 cm.

The ratio is 60 : 100.

Let us find H.C.F. of 60 and 100:

$$\begin{array}{r} 1 \\ 60 \overline{) 100} \\ \underline{-60} \\ 40 \overline{) 60} ( 1 \\ \underline{-40} \\ 20 \overline{) 40} ( 2 \\ \underline{-40} \\ 0 \end{array}$$

H.C.F. = 20

$\frac{60 \div 20}{100 \div 20} = \frac{3}{5} = 3 : 5$

Hence, the answer is 3 : 5

(iv) 400 m : 1 km

Convert km into m: 1 km = 1000 m.

The ratio is 400 : 1000.

Let us find H.C.F. of 400 and 1000:

$$\begin{array}{r} 2 \\ 400 \overline{) 1000} \\ \underline{-800} \\ 200 \overline{) 400} ( 2 \\ \underline{-400} \\ 0 \end{array}$$

H.C.F. = 200

$\frac{400 \div 200}{1000 \div 200} = \frac{2}{5} = 2 : 5$

Hence, the answer is 2 : 5

(v) 8 months : $1\frac{2}{3}$ years

Convert $1\frac{2}{3}$ years to improper fraction: $\frac{5}{3}$ years.

Convert years into months: $\frac{5}{3} \times 12 = 20$ months.

The ratio is 8 : 20.

Let us find H.C.F. of 8 and 20:

$$\begin{array}{r} 2 \\ 8 \overline{) 20} \\ \underline{-16} \\ 4 \overline{) 8} ( 2 \\ \underline{-8} \\ 0 \end{array}$$

H.C.F. = 4

$\frac{8 \div 4}{20 \div 4} = \frac{2}{5} = 2 : 5$

Hence, the answer is 2 : 5

(vi) $6\frac{2}{3}$ hours : 1 day

Convert $6\frac{2}{3}$ hours to improper fraction: $\frac{20}{3}$ hours.

Convert 1 day into hours: 24 hours.

The ratio is $\frac{20}{3}$ : 24.

Multiply both sides by 3 to remove the denominator:

$\Big(\frac{20}{3} \times 3\Big) : 24 \times 3 = 20 : 72$

Let us find H.C.F. of 20 and 72:

$$\begin{array}{r} 3 \\ 20 \overline{) 72} \\ \underline{-60} \\ 12 \overline{) 20} ( 1 \\ \underline{-12} \\ 8 \overline{) 12} ( 1 \\ \underline{-8} \\ 4 \overline{) 8} ( 2 \\ \underline{-8} \\ 0 \end{array}$$

H.C.F. = 4

$\frac{20 \div 4}{72 \div 4} = \frac{5}{18} = 5 : 18$

Hence, the answer is 5 : 18

Given:

Ratio = 5 : 6 : 7.

The sum of all angles in any triangle is always 180°.

Find total parts: 5 + 6 + 7 = 18 parts.

Find the value of 1 part: 180° ÷ 18 = 10°

Now, calculate each angle by multiplying each part with 10°:

First angle: 5 x 10° = 50°

Second angle: 6 x 10° = 60°

Third angle: 7 x 10° = 70°

Angles are 50°, 60°, and 70°.

Given:

Length = 84 cm

Ratio = 4 : 3

Find total parts: 4 + 3 = 7 parts.

Find the value of 1 part: 84 ÷ 7 = 12 cm.

Now, calculate each part by multiplying with 12 cm:

First part = 4 x 12 cm = 48 cm

Second part = 3 x 12 cm = 36 cm

Lengths are 48 cm and 36 cm.

Given:

Total money = ₹ 3900

Ratio = 15 : 11

Find total parts: 15 + 11 = 26 parts.

Find the value of 1 part: Divide ₹ 3900 by 26.

3900 ÷ 26 = 150. So, 1 part = ₹ 150.

Now calculate shares by multiplying each part with ₹ 150:

Kunal's share: 15 x ₹ 150 = ₹ 2250

Kirti's share: 11 x ₹ 150 = ₹ 1650

Kunal's share = ₹2250 and Kirti's share = ₹1650.

Given:

Total Money: ₹ 16250

Ratio: 5 : 7 : 13

Find total parts: 5 + 7 + 13 = 25 parts.

Find the value of 1 part: 16250 ÷ 25 = 650. So, 1 part = ₹ 650.

Now calculate shares by multiplying each part with ₹ 650:

A gets: 5 x ₹ 650 = ₹ 3250

B gets: 7 x ₹ 650 = ₹ 4550

C gets: 13 x ₹ 650 = ₹ 8450

A's share = ₹3250, B's share = ₹4550, C's share = ₹8450

Given:

Total Sweets: 35

Ratio = $\frac{1}{2} : \frac{1}{3}$

Let us find L.C.M. of 2 and 3.

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

L.C.M. = 2 x 3 = 6

Multiply both by 6 to convert fractions to whole numbers:

$\Big(\frac{1}{2} \times 6\Big) : \Big(\frac{1}{3} \times 6\Big) = 3 : 2$

Total parts: 3 + 2 = 5 parts.

Value of 1 part: 35 ÷ 5 = 7 sweets.

Calculate sweets by multiplying each part with 7 sweets:

Sudha gets: 3 x 7 sweets = 21 sweets

Tanvy gets: 2 x 7 sweets = 14 sweets

Sudha gets 21 sweets and Tanvy gets 14 sweets.

Given:

Total Rice: 468 g

Ratio: $\frac{1}{6} : \frac{1}{8} : \frac{1}{12}$

Let us find L.C.M. of 6, 8, and 12:

$$\begin{array}{l|rrr} 2 & 6, & 8, & 12 \\ \hline 2 & 3, & 4, & 6 \\ \hline 2 & 3, & 2, & 3 \\ \hline 3 & 3, & 1, & 3 \\ \hline & 1, & 1, & 1 \end{array}$$

L.C.M. = 2 x 2 x 2 x 3 = 24

Multiply all three terms with 24 to convert fractions to whole numbers:

$\Big(\frac{1}{6} \times 24\Big) : \Big(\frac{1}{8} \times 24\Big) : \Big(\frac{1}{12} \times 24\Big)$

= 4 : 3 : 2

Total parts: 4 + 3 + 2 = 9 parts.

Value of 1 part: 468 g ÷ 9 = 52 g.

Calculate heaps by multiplying each part with 52 g:

First heap: 4 x 52 g = 208 g

Second heap: 3 x 52 g = 156 g

Third heap: 2 x 52 g = 104 g

The three heaps are 208 g, 156 g and 104 g.

Given:

Total Money: ₹ 715

B gets 3 times A

C gets half of B.

Let A's share = 1 part.

Then B's share = 3 times A = 3 x 1 = 3 parts.

Then C's share = Half of B = 3 ÷ 2 = $\frac{3}{2}$ parts.

Ratio (A : B : C) = 1 : 3 : $\frac{3}{2}$

Multiply by 2, to convert fraction into whole number = 2 : 6 : 3.

Total parts: 2 + 6 + 3 = 11 parts.

Value of 1 part: 715 ÷ 11 = ₹ 65.

Calculate shares by multiplying each part with ₹ 65:

A gets: 2 x ₹ 65 = ₹ 130

B gets: 6 x ₹ 65 = ₹ 390

C gets: 3 x ₹ 65 = ₹ 195

A's share = ₹130, B's share = ₹390, C's share = ₹195.

Given:

Total Money: ₹ 760

A = $\frac{5}{6}$ of B

B : C = 3 : 4.

Since B : C is 3 : 4, let's look at A.

A = $\frac{5}{6} \times 3 = \frac{5}{2}$ parts.

Ratio (A : B : C) = $\frac{5}{2}$ : 3 : 4.

Multiply by 2 to convert fraction into whole number = 5 : 6 : 8.

Total parts: 5 + 6 + 8 = 19 parts.

Value of 1 part: 760 ÷ 19 = ₹ 40.

Calculate shares by multiplying each part with ₹ 40:

A gets: 5 x ₹ 40 = ₹ 200

B gets: 6 x ₹ 40 = ₹ 240

C gets: 8 x ₹ 40 = ₹ 320

A's share = ₹200, B's share = ₹240, C's share = ₹320

Given:

Ratio (Boys : Girls) = 8 : 3

Total number of Girls = 405

Since 3 parts represent the girls, 3 parts = 405.

1 part = 405 ÷ 3 = 135.

Boys = 8 parts = 8 x 135 = 1080.

There are 1080 boys.

Given:

Ratio (Copper : Zinc) = 7 : 3

Copper = 12.6 g

Since 7 parts represent copper, 7 parts = 12.6 g

1 part = 12.6 ÷ 7 = 1.8 g

Zinc = 3 x 1.8 g = 5.4 g

Total weight = Copper + Zinc

Substituting the values in above:

Total weight = 12.6 g + 5.4 g = 18 g

The weight of the alloy is 18 g.

Given:

Original Weight = 63 kg

Reduction Ratio = Old weight : New weight = 9 : 8

9 parts = 63 kg

1 part = 63 ÷ 9 = 7 kg.

The new weight is represented by 8 parts.

New weight 8 x 7 kg = 56 kg.

Her new weight is 56 kg.

Given:

Ratio = 9 : 13

Total Sum = 176

Total parts = 9 + 13 = 22 parts.

Value of 1 part: 176 ÷ 22 = 8. (So, 1 part = 8).

Calculate the numbers by multiplying each part with 8:

First number = 9 x 8 = 72

Second number = 13 x 8 = 104

The numbers are 72 and 104.

Given:

Ratio = 5 : 8

Difference = 12

The larger part is 8 and the smaller is 5.

Difference in parts = 8 - 5 = 3 parts.

Difference given = 12

3 parts = 12

1 part = 12 ÷ 3 = 4

Calculate the numbers by multiplying each part with 4:

First number = 5 x 4 = 20

Second number = 8 x 4 = 32

The numbers are 20 and 32.

Given:

a : b = 3 : 2

b : c = 4 : 5

$$\begin{aligned}a : b = \frac{a}{b} = \frac{3}{2} \\[1em] b : c = \frac{b}{c} = \frac{4}{5}\end{aligned}$$

Now,

$$\begin{aligned}a : c = \frac{a}{c} = \Big( \frac{a}{b} \times \frac{b}{c} \Big) \\[1em] = \Big( \frac{3}{2} \times \frac{4}{5} \Big) \\[1em] = \Big( \frac{3}{1} \times \frac{2}{5} \Big) \\[1em] = \frac{6}{5} \\[1em] = 6 : 5\end{aligned}$$

∴ a : c = 6 : 5

Given:

a : b = 5 : 6

b : c = 2.8 : 3.5

Multiply by 10 to remove decimals in the second ratio.

b : c = (2.8 x 10) : (3.5 x 10) = 28 : 35

b : c = 4 : 5 $\quad$ [Dividing by 7]

$$\begin{aligned}a : b = \frac{a}{b} = \frac{5}{6} \\[1em] b : c = \frac{b}{c} = \frac{4}{5}\end{aligned}$$

Now,

$$\begin{aligned}a : c = \frac{a}{c} = \left( \frac{a}{b} \times \frac{b}{c} \right) \\[1em] = \left( \frac{5}{6} \times \frac{4}{5} \right) \\[1em] = \left( \frac{1}{6} \times \frac{4}{1} \right) \\[1em] = \frac{4}{6} \\[1em] = \frac{2}{3} \\[1em] = 2 : 3\end{aligned}$$

∴ a : c = 2 : 3

Given:

p : q = $1\frac{1}{3} : 1\frac{1}{2}$

q : r = $\frac{1}{2} : \frac{1}{3}$

Convert p : q to improper fractions:

p : q = $\frac{4}{3} : \frac{3}{2}$

L.C.M. of 3 and 2 is 6.

Multiply p : q and q : r with 6 to convert fractions to whole numbers:

$p : q = (\frac{4}{3} \times 6) : (\frac{3}{2} \times 6) = 8 : 9$

And

$q : r = (\frac{1}{2} \times 6) : (\frac{1}{3} \times 6) = 3 : 2$

$$\begin{aligned}\frac{p}{q} = \frac{8}{9} \\[1em] \frac{q}{r} = \frac{3}{2}\end{aligned}$$

Now,

$$\begin{aligned}p : r = \frac{p}{r} = \left( \frac{p}{q} \times \frac{q}{r} \right) \\[1em] = \left( \frac{8}{9} \times \frac{3}{2} \right) \\[1em] = \left( \frac{8}{3} \times \frac{1}{2} \right) \\[1em] = \frac{8}{6} \\[1em] = \frac{4}{3} \\[1em] = 4 : 3\end{aligned}$$

∴ p : r = 4 : 3

Given:

a : b = 3 : 4

b : c = 8 : 9

We make b, the same in both the ratios.

Let us find L.C.M. of 4 and 8:

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

L.C.M. = 2 x 2 x 2 = 8

So, we make b equal to 8 in each case.

Now, multiply the first ratio by 2:

$$\begin{aligned}a : b = 3 : 4 \\[1em] = \frac{3 \times 2}{4 \times 2} \\[1em] = \frac{6}{8} \\[1em] = 6 : 8 \\[1em]\end{aligned}$$

In second ratio b has 8 in denominator already, so keep it as it is.

b : c = 8 : 9

∴ a : b : c = 6 : 8 : 9

Given:

l : m = $2\frac{1}{2} : 1\frac{2}{3}$

m : n = $1\frac{1}{4} : 3\frac{1}{2}$

Convert mixed to improper fractions:

l : m = $\frac{5}{2} : \frac{5}{3}$

m : n = $\frac{5}{4} : \frac{7}{2}$

Let us simplify the ratios into whole numbers.

l : m = $\frac{5}{2} : \frac{5}{3}$

Let us find L.C.M. of 2 and 3:

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

L.C.M. = 2 x 3 = 6

l : m = $\Big(\frac{5}{2} \times 6\Big) : \Big(\frac{5}{3} \times 6\Big)$ = 15 : 10

l : m = 3 : 2 $\quad$[Dividing by 5]

m : n = $\frac{5}{4} : \frac{7}{2}$

Let us find L.C.M. of 4 and 2:

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

L.C.M. = 2 x 2 = 4

m : n = $\Big(\frac{5}{4} \times 4\Big) : \Big(\frac{7}{2} \times 4\Big)$

m : n = 5 : 14

We make m, the same in both the ratios.

Let us find L.C.M. of 2 and 5:

$$\begin{array}{l|rr} 2 & 2, & 5 \\ \hline 5 & 1, & 5 \\ \hline & 1, & 1 \end{array}$$

L.C.M. = 2 x 5 = 10

So, we make m equal to 10 in each case.

Now, multiply the first ratio by 5:

$$\begin{aligned}l : m = 3 : 2 \\[1em] = \frac{3 \times 5}{2 \times 5} \\[1em] = \frac{15}{10} \\[1em] = 15 : 10\end{aligned}$$

Multiply the second ratio by 2:

$$\begin{aligned}m : n = 5 : 14 \\[1em] = \frac{5 \times 2}{14 \times 2} \\[1em] = \frac{10}{28} \\[1em] = 10 : 28\end{aligned}$$

Now we have,

l : m = 15 : 10 and m : n = 10 : 28

∴ l : m : n = 15 : 10 : 28

(i)

The given ratios are 3 : 4 and 5 : 7, which are equivalent to the fractions $\frac{3}{4}$ and $\frac{5}{7}$ respectively.

Let us find the L.C.M. of 4 and 7:

$$\begin{array}{l|rr} 2 & 4, & 7 \\ \hline 2 & 2, & 7 \\ \hline 7 & 1, & 7 \\ \hline & 1, & 1 \end{array}$$

L.C.M. = 2 x 2 x 7 = 28.

$∴ 3 : 4 = \frac{3}{4} = \frac{3 \times 7}{4 \times 7} = \frac{21}{28}$

$5 : 7 = \frac{5}{7} = \frac{5 \times 4}{7 \times 4} = \frac{20}{28}$

Clearly, $\frac{21}{28} \gt \frac{20}{28}$.

Hence, (3 : 4) > (5 : 7).

(ii)

The given ratios are 11 : 21 and 19 : 28, which are equivalent to the fractions $\frac{11}{21}$ and $\frac{19}{28}$ respectively.

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

$$\begin{array}{l|l} 7 & 21, 28 \\ \hline 3 & 3, 4 \\ \hline 4 & 1, 4 \\ \hline & 1, 1 \end{array}$$

L.C.M. = 7 x 3 x 4 = 84.

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

$19 : 28 = \frac{19}{28} = \frac{19 \times 3}{28 \times 3} = \frac{57}{84}$

Clearly, $\frac{57}{84} \gt \frac{44}{84}$.

Hence, (19 : 28) > (11 : 21).

(i) False
Reason —

First Ratio (51 : 68):

Let us find H.C.F. of 51 and 68:

$$\begin{array}{r} 1 \\ 51 \overline{) 68} \\ \underline{-51} \\ 17 \overline{) 51} ( 3 \\ \underline{-51} \\ 0 \end{array}$$

H.C.F. is 17.

Simplest form: $\frac{51 \div 17}{68 \div 17} = {\frac{3}{4}}$

Second Ratio (85 : 102):

Let us find H.C.F. of 85 and 102:

$$\begin{array}{r} 1 \\ 85 \overline{) 102} \\ \underline{-85} \\ 17 \overline{) 85} ( 5 \\ \underline{-85} \\ 0 \end{array}$$

H.C.F. is 17.

Simplest form: $\frac{85 \div 17}{102 \div 17} = {\frac{5}{6}}$

Since $3:4 \neq 5:6$, the statement is False.

(ii) True
Reason —

First Ratio = 1.5 : 2.5

Multiply by 10 to remove decimals = (1.5 x 10 : 2.5 x 10) = 15 : 25

Let us find H.C.F. of 15 and 25:

$$\begin{array}{r} 1 \\ 15 \overline{) 25} \\ \underline{-15} \\ 10 \overline{) 15} ( 1 \\ \underline{-10} \\ 5 \overline{) 10} ( 2 \\ \underline{-10} \\ 0 \end{array}$$

H.C.F = 5

Simplest form: $\frac{15 \div 5}{25 \div 5} = {\frac{3}{5}}$

Second Ratio = 3.6 : 6

Multiply by 10 to remove decimals = (3.6 x 10 : 6 x 10) = 36 : 60

Let us find H.C.F. of 36 and 60:

$$\begin{array}{r} 1 \\ 36 \overline{) 60} \\ \underline{-36} \\ 24 \overline{) 36} ( 1 \\ \underline{-24} \\ 12 \overline{) 24} ( 2 \\ \underline{-24} \\ 0 \end{array}$$

H.C.F. = 12.

Simplest form: $\frac{36 \div 12}{60 \div 12} = {\frac{3}{5}}$

Since both simplify to 3:5, the statement is True.

(iii) True
Reason —

First Ratio = 30 : 18

Let us find H.C.F. of 30 and 18:

$$\begin{array}{r} 1 \\ 18 \overline{) 30} \\ \underline{-18} \\ 12 \overline{) 18} ( 1 \\ \underline{-12} \\ 6 \overline{) 12} ( 2 \\ \underline{-12} \\ 0 \end{array}$$

H.C.F. = 6.

Simplest form: $\frac{30 \div 6}{18 \div 6} = {\frac{5}{3}}$

Second Ratio = 450 : 270

First, cancel the zeros: 45 : 27.

Let us find H.C.F. of 45 and 27:

$$\begin{array}{r} 1 \\ 27 \overline{) 45} \\ \underline{-27} \\ 18 \overline{) 27} ( 1 \\ \underline{-18} \\ 9 \overline{) 18} ( 2 \\ \underline{-18} \\ 0 \end{array}$$

H.C.F. = 9.

Simplest form: $\frac{45 \div 9}{27 \div 9} = {\frac{5}{3}}$

Since both simplify to 5:3, the statement is True.

(iv) False
Reason —

First Ratio = 32 : 36

Let us find H.C.F. of 32 and 36:

$$\begin{array}{r} 1 \\ 32 \overline{) 36} \\ \underline{-32} \\ 4 \overline{) 32} ( 8 \\ \underline{-32} \\ 0 \end{array}$$

H.C.F. = 4.

Simplest form: $\frac{32 \div 4}{36 \div 4} = {\frac{8}{9}}$

Second Ratio = 8 : 9

This is already in its simplest form: ${\frac{8}{9}}$

So numerically both ratios are equal.

But a ratio should exist only between quantities of the same kind. Here the quantities are kg and ₹, which are different kinds. Therefore the statement is false.

(i) 30, 40, 45, 60

The given numbers are 30, 40, 45, 60.

We have:

$30 : 40 = \frac{30}{40} = \frac{3}{4}$

$45 : 60 = \frac{45}{60} = \frac{45 \div 15}{60 \div 15} = \frac{3}{4}$

∴ (30 : 40) = (45 : 60)

Hence, 30, 40, 45 and 60 are in proportion.

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

The given numbers are $2, 2\frac{1}{2}, 3, 3\frac{1}{2}$.

Convert mixed to improper fraction:

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

$3\frac{1}{2} = \frac{7}{2}$

We have:

$2 : \frac{5}{2} = \frac{2}{5/2} = 2 \times \frac{2}{5} = \frac{4}{5}$

$3 : \frac{7}{2} = \frac{3}{7/2} = 3 \times \frac{2}{7} = \frac{6}{7}$

∴ $(2 : 2\frac{1}{2})$ ≠ $(3 : 3\frac{1}{2})$.

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

(iii) 0.8, 3, 2.4, 9

The given numbers are 0.8, 3, 2.4, 9.

Multiply by 10 to convert decimals to whole numbers:

We have:

$0.8 : 3 = \frac{0.8 \times 10}{3 \times 10} = \frac{8}{30} = \frac{4}{15}$

$2.4 : 9 = \frac{2.4 \times 10}{9 \times 10} = \frac{24}{90} = \frac{24 \div 6}{90 \div 6} = \frac{4}{15}$

∴ (0.8 : 3) = (2.4 : 9)

Hence, 0.8, 3, 2.4 and 9 are in proportion.

(iv) $\frac{1}{5},\frac{1}{8},\frac{1}{4},\frac{1}{10}$

The given numbers are $\frac{1}{5},\frac{1}{8},\frac{1}{4},\frac{1}{10}$.

We have:

$\frac{1}{5} : \frac{1}{8} = \frac{1}{5} \times \frac{8}{1} = \frac{8}{5}$

$\frac{1}{4} : \frac{1}{10} = \frac{1}{4} \times \frac{10}{1} = \frac{10}{4} = \frac{5}{2}$

$∴ \frac{1}{5} : \frac{1}{8}$ ≠ $\frac{1}{4} : \frac{1}{10}$

Hence, $\frac{1}{5},\frac{1}{8},\frac{1}{4},\frac{1}{10}$ are not in proportion.

(v) $\frac{1}{2},\frac{1}{5},\frac{1}{6},\frac{1}{15}$

The given numbers are $\frac{1}{2},\frac{1}{5},\frac{1}{6},\frac{1}{15}$.

We have:

$\frac{1}{2} : \frac{1}{5} = \frac{1}{2} \times \frac{5}{1} = \frac{5}{2}$

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

$∴ \frac{1}{2} : \frac{1}{5} = \frac{1}{6} : \frac{1}{15}$.

Hence, $\frac{1}{2},\frac{1}{5},\frac{1}{6},\frac{1}{15}$ are in proportion.

(i) 42 : 12 : : 7 : x

In a proportion, we know that:

product of extremes = product of means

∴ 12 × 7 = 42 × x

⇒ 84 = 42x

⇒ x = $\frac{84}{42} = 2$

Hence, x = 2

(ii) 1.8 : x : : 2.4 : 6.0

In a proportion, we know that:

product of extremes = product of means

∴ x × 2.4 = 1.8 × 6.0

⇒ 2.4x = 10.8

⇒ x = $\frac{10.8}{2.4} = \frac{108}{24} = \frac{9}{2} = 4.5$

Hence, x = 4.5

(iii) 6 : 0.8 : : x : 10

In a proportion, we know that:

product of extremes = product of means

∴ 0.8 × x = 6 × 10

⇒ 0.8x = 60

⇒ x = $\frac{60}{0.8} = \frac{600}{8} = 75$.

Hence, x = 75

(iv) x : 1.6 : : 2.1 : 8.4

In a proportion, we know that:

product of extremes = product of means

∴ 1.6 × 2.1 = x × 8.4

⇒ 3.36 = 8.4x

⇒ x = $\frac{3.36}{8.4} = \frac{336}{840} = 0.4$

Hence, x = 0.4

(v) $\frac{1}{9}$ : x : : $\frac{1}{3} : \frac{1}{4}$

In a proportion, we know that:

product of extremes = product of means

$∴ x \times \frac{1}{3} = \frac{1}{9} \times \frac{1}{4}$

⇒ $\frac{x}{3} = \frac{1}{36}$

⇒ x = $\frac{3}{36} = \frac{1}{12}$.

Hence, x = $\frac{1}{12}$

(vi) 16 : x : : x : 25

In a proportion, we know that:

product of extremes = product of means

∴ x × x = 16 × 25

⇒ x² = 400

⇒ x = $\sqrt{400} = 20$

Hence, x = 20

(i) 4, 9, 32

Let the fourth proportional be x. Then 4, 9, 32, x are in proportion.

product of extremes = product of means

4 × x = 9 × 32

⇒ x = $\frac{9 \times 32}{4}$

⇒ x = 9 × 8 = 72

Hence, the fourth proportional is 72.

(ii) 15, 6, 7

Let the fourth proportional be x. Then 15, 6, 7, x are in proportion.

product of extremes = product of means

15 × x = 6 × 7

⇒ x = $\frac{42}{15} = \frac{14}{5} = 2.8$

Hence, the fourth proportional is 2.8

(iii) 0.6, 1.5, 3

Let the fourth proportional be x. Then 0.6, 1.5, 3, x are in proportion.

product of extremes = product of means

0.6 × x = 1.5 × 3

⇒ x = $\frac{4.5}{0.6} = \frac{4.5 \times 10}{0.6 \times 10} = \frac{45}{6} = 7.5$

Hence, the fourth proportional is 7.5

(iv) $\frac{1}{3}, \frac{2}{5}, 6$

Let the fourth proportional be x. Then $\frac{1}{3}, \frac{2}{5}, 6$, x are in proportion.

product of extremes = product of means

$$\begin{aligned}\frac{1}{3} \times x = \frac{2}{5} \times 6 \\[1em] \Rightarrow \frac{x}{3} = \frac{12}{5} \\[1em] \Rightarrow x = \frac{12 \times 3}{5} \\[1em] \Rightarrow x = \frac{36}{5} \\[1em] \Rightarrow x = 7\frac{1}{5}\end{aligned}$$

Hence, the fourth proportional is $7\frac{1}{5}$

(v) $2\frac{1}{2}, 2\frac{6}{7}, 3\frac{1}{2}$

Convert to improper fractions: $\frac{5}{2}, \frac{20}{7}, \frac{7}{2}$. Let the fourth proportional be x.

Then $\frac{5}{2}, \frac{20}{7}, \frac{7}{2}$, x are in proportion.

product of extremes = product of means

$$\begin{aligned}\frac{5}{2} \times x = \frac{20}{7} \times \frac{7}{2} \\[1em] \Rightarrow \frac{5x}{2} = 10 \\[1em] \Rightarrow 5x = 10 \times 2 \\[1em] \Rightarrow 5x = 20 \\[1em] \Rightarrow x = \frac{20}{5} \\[1em] \Rightarrow x = 4\end{aligned}$$

Hence, the fourth proportional is 4

(vi) 3 hrs 12 min, 24 min, 1 m 68 cm

First, convert to the same units for each ratio:

1 hour = 60 min,

∴ 3 hrs = 3 x 60 min = 180 min

3 hrs 12 min = 180 min + 12 min = 192 min.

1 m = 100 cm,

∴ 1 m 68 cm = 100 cm + 68 cm = 168 cm.

Let the fourth proportional be x (in cm). We have:

192 : 24 :: 168 : x

Then 192 min, 24 min, 168 cm, x are in proportion.

product of extremes = product of means

192 x x = 24 x 168

⇒ x = $\frac{24 \times 168}{192}$

⇒ x = $\frac{1 \times 168}{8}$

⇒ x = 21

Hence, the fourth proportional is 21 cm.

(i) 81 and 121

Mean proportion between 81 and 121

$$\begin{aligned}= \sqrt{81 \times 121} \\[1em] = \sqrt{81} \times \sqrt{121} \\[1em] = 9 \times 11 \\[1em] = 99 \\[1em]\end{aligned}$$

Hence, the answer is 99

(ii) 1.8 and 0.2

Mean proportion between 1.8 and 0.2

$$\begin{aligned}= \sqrt{1.8 \times 0.2} \\[1em] = \sqrt{0.36} \\[1em] = 0.6\end{aligned}$$

Hence, the answer is 0.6

(iii) $\frac{2}{3}$ and $\frac{8}{27}$

Mean proportion between $\frac{2}{3}$ and $\frac{8}{27}$

$$\begin{aligned}= \sqrt{\frac{2}{3} \times \frac{8}{27}} \\[1em] = \sqrt{\frac{16}{81}} \\[1em] = \frac{\sqrt{16}}{\sqrt{81}} \\[1em] = \frac{4}{9}\end{aligned}$$

Hence, the answer is $\frac{4}{9}$

(iv) 0.32 and 0.08

Mean proportion between 0.32 and 0.08

$$\begin{aligned}= \sqrt{0.32 \times 0.08} \\[1em] = \sqrt{0.0256} \\[1em] = 0.16\end{aligned}$$

Hence, the answer is 0.16

(v) $\frac{1}{25}$ and $\frac{1}{16}$

Mean proportion between $\frac{1}{25}$ and $\frac{1}{16}$

$$\begin{aligned}= \sqrt{\frac{1}{25} \times \frac{1}{16}} \\[1em] = \sqrt{\frac{1}{400}} \\[1em] = \frac{1}{\sqrt{400}} \\[1em] = \frac{1}{20}\end{aligned}$$

Hence, the answer is $\frac{1}{20}$

(i) 36, 12

Let the third proportional be x.

Then, 36 : 12 :: 12 : x.

product of extremes = product of means

36 × x = 12 × 12

⇒ x = $\frac{144}{36}$

⇒ x = 4

Hence, the third proportional is 4

(ii) 1.2, 0.6

Let the third proportional be x.

Then, 1.2 : 0.6 :: 0.6 : x.

product of extremes = product of means

1.2 × x = 0.6 × 0.6

⇒ x = $\frac{0.36}{1.2}$

⇒ x = $\frac{3.6}{12}$

⇒ x = 0.3

Hence, the third proportional is 0.3

(iii) $\frac{1}{9}, \frac{2}{3}$

Let the third proportional be x.

Then, $\frac{1}{9} : \frac{2}{3} :: \frac{2}{3} : x$.

product of extremes = product of means

$$\begin{aligned}\frac{1}{9} \times x = \frac{2}{3} \times \frac{2}{3} \\[1em] \Rightarrow \frac{x}{9} = \frac{4}{9} \\[1em] \Rightarrow x = \frac{4}{9} \times 9 \\[1em] \Rightarrow x = 4\end{aligned}$$

Hence, the third proportional is 4

(iv) 1 m 60 cm, 40 cm

First, convert to the same unit:

1 m = 100 cm

∴ 1 m 60 cm = 100 cm + 60 cm = 160 cm.

Let the third proportional be x.

Then, 160 : 40 :: 40 : x

product of extremes = product of means

160 × x = 40 × 40

⇒ x = $\frac{1600}{160}$

⇒ x = 10

Hence, the third proportional is 10 cm

(v) 1 kg 250 g, 500 g

First, convert to the same unit:

1 kg = 1000 g

∴ 1 kg 250 g = 1000 g + 250 g = 1250 g

Let the third proportional be x.

Then, 1250 : 500 :: 500 : x

product of extremes = product of means

1250 x x = 500 x 500

⇒ x = $\frac{250000}{1250}$

⇒ x = $\frac{25000}{125}$

⇒ x = 200

Hence, the third proportional is 200 g

(vi) ₹ 2.40, ₹ 4.80

Let the third proportional be x.

Then, 2.40 : 4.80 :: 4.80 : x.

product of extremes = product of means

2.40 × x = 4.80 × 4.80

⇒ x = $\frac{4.80 \times 4.80}{2.40}$

⇒ x = 2 x 4.80

⇒ x = 9.60

Hence, the third proportional is ₹ 9.60

If 6, 36, 216 are in continued proportion, then it should satisfy the condition b² = a x c

Here a = 6, b = 36, c = 216

∴ 36² = 6 x 216

1296 = 1296

Since it satisfies the condition b² = a x c, the numbers 6, 36, and 216 are in continued proportion.

∴ 6, 36, 216 are in continued proportion

Given:

Cost of 8 pens = ₹ 356

Cost of 14 pens = ?

Let the cost of 14 pens = ₹ x

Ratio of pens : Ratio of costs

8 : 14 :: 356 : x

By the rule: Product of Extremes = Product of Means:

8 × x = 14 × 356

⇒ x = $\frac{14 \times 356}{8}$

⇒ x = 14 x 44.5

⇒ x = 623

Hence, the cost of 14 pens is ₹ 623.

Given:

Weight of 7 m iron bar = 22.4 kg

Weight of 13 m iron bar = ?

Let the weight of 13 m iron bar = x kg

Ratio of lengths : Ratio of weights

7 : 13 :: 22.4 : x

By the rule: Product of Extremes = Product of Means:

7 × x = 13 × 22.4

⇒ x = $\frac{13 \times 22.4}{7}$

⇒ x = 13 x 3.2

⇒ x = 41.6

Hence, the bar of length 13 m weighs 41.6 kg.

Given:

Map distance 1.7 cm = Actual distance 68 km

Map distance 8.5 cm = ?

Let actual distance for 8.5 cm = x km

Ratio of map distances : Ratio of actual distances

1.7 : 8.5 :: 68 : x

By the rule: Product of Extremes = Product of Means:

1.7 x x = 8.5 x 68

⇒ x = $\frac{8.5 \times 68}{1.7}$

⇒ x = 5 x 68

⇒ x = 340

Hence, the distance represented is 340 km.

Given:

Distance covered in 6 hours = 435 km

Distance covered in 8 hours = ?

Let distance covered in 8 hours = x km

Ratio of times : Ratio of distances

6 : 8 :: 435 : x

By the rule: Product of Extremes = Product of Means:

6 × x = 8 × 435

⇒ x = $\frac{8 \times 435}{6}$

⇒ x = 8 x 72.5

⇒ x = 580

Hence, the bus will cover 580 km in 8 hours.

Given:

Men required for 35 m trench = 15 men

Men required for 84 m trench = ?

Let men required for 84 m trench = x men

Length of trench : Number of men

35 : 84 :: 15 : x

By the rule: Product of Extremes = Product of Means:

35 × x = 84 × 15

x = $\frac{84 \times 15}{35}$

x = $\frac{84 \times 3}{7}$

x = 12 x 3

x = 36

Hence, 36 men are required to dig the 84 m trench.

Given:

First quantity = 6 cm

Second quantity = 20 mm

First, convert to the same unit (1 cm = 10 mm).

6 cm = 6 x 10 = 60 mm.

Ratio = 60 : 20 = $\frac{60}{20} = \frac{3}{1}$ = 3 : 1

Hence, option 1 is the correct option.

Given:

Ratio: $\frac{1}{8} : \frac{1}{12}$

Let us find the L.C.M. of denominators 8 and 12:

$$\begin{array}{l|rr} 2 & 8, & 12 \\ \hline 2 & 4, & 6 \\ \hline 2 & 2, & 3 \\ \hline 3 & 1, & 3 \\ \hline & 1, & 1 \end{array}$$

L.C.M. = 2 x 2 x 2 x 3 = 24

Multiply both terms by 24:

$\Big(\frac{1}{8} \times 24\Big) : \Big(\frac{1}{12} \times 24\Big) = 3 : 2$

Hence, option 2 is the correct option.

Given:

A : B = 2 : 5

B : C = 4 : 5

To find C : A, first find A : C by multiplying the ratios:

$$\begin{aligned}\frac{A}{C} = \frac{A}{B} \times \frac{B}{C} \\[1em] = \frac{2}{5} \times \frac{4}{5} \\[1em] = \frac{8}{25}\end{aligned}$$

So, A : C = 8 : 25.

For C : A, we reverse the ratio = 25 : 8.

Hence, option 4 is the correct option.

Given:

Original number = 91

Increased ratio = 7 : 13

The original part is 7.

7 parts = 91

1 part = 91 ÷ 7 = 13

The increased value is 13 parts:

New value = 13 x 13 = 169

Hence, option 2 is the correct option.

Given:

Proportion: 12 : x : : 15 : 25

Product of Means = Product of Extremes

x × 15 = 12 × 25

15x = 300

⇒ x = $\frac{300}{15}$

⇒ x = 20

Hence, option 4 is the correct option.

Given:

Numbers = 9 and 18

Let the third proportional be x.

9 : 18 : : 18 : x

Product of Means = Product of Extremes

9 × x = 18 × 18

⇒ x = $\frac{18 \times 18}{9}$

⇒ x = 2 x 18

⇒ x = 36

Hence, option 4 is the correct option.

Given:

Numbers = 5 and 45

Mean Proportion = $\sqrt{a \times b}$

= $\sqrt{5 \times 45} = \sqrt{225}$

= 15

Hence, option 3 is the correct option.

Let us consider the given numbers as: a, b, c

Check each option using b² = a x c:

8² = 64, 4 × 12 = 48 (No)

15² = 225, 5 × 25 = 125 (No)

36² = 1296, 6 × 216 = 1296 (Yes)

12² = 144, 9 × 18 = 162 (No)

Hence, option 3 is the correct option.

(i) Ratio has no unit.

(ii) To convert a ratio a : b in its simplest form, we divide a and b by H.C.F. of a and b.

(iii) If a : b : : b : c, then a, b, c are said to be in continued proportion.

(iv) If a, b, c are in continued proportion, then c is called the third proportional to a and b.

(v) In a proportion, the first and fourth terms are called the extremes.

(i) False
Reason — For a, b, c, d to be in proportion (a : b :: c : d), the rule is Product of Extremes = Product of Means. This means a x d = b x c, or ad = bc. The statement says ac = bd, which is incorrect.

(ii) False
Reason — When four terms are in the form a : b :: c : d, they are simply said to be in proportion. There is no standard mathematical term called "absolute proportion" used in this context.

(iii) False
Reason — If a, b, c are in continued proportion, then a : b :: b : c. This means b² = ac or $b = \sqrt{ac}$. The formula $\frac{a+c}{2}$ is for the arithmetic mean, not the mean proportional.

(iv) True
Reason — By definition, if x is the third proportional to a and b, then a, b, and x are in continued proportion. In this sequence, b is the mean (repeated) term.

(v) False
Reason — To check if 1, 2, 3, 4 are in proportion, we test if 1 x 4 = 2 x 3:

Product of Extremes (1 x 4) = 4

Product of Means (2 x 3) = 6

Since 4 ≠ 6, they are not in proportion.

(1) Given:

Value of second property = ₹ 60,00,000

Ratio (Priya : Seema) = $\frac{1}{5} : \frac{1}{7}$

Let us find L.C.M. of 5 and 7:

$$\begin{array}{l|rr} 5 & 5, & 7 \\ \hline 7 & 1, & 7 \\ \hline & 1, & 1 \end{array}$$

L.C.M. = 5 x 7 = 35

Priya : Seema = $(\frac{1}{5} \times 35) : (\frac{1}{7} \times 35)$

Priya : Seema = 7 : 5

Total parts = 7 + 5 = 12

Value of 1 part = ₹ 60,00,000 ÷ 12 = ₹ 5,00,000

Priya's amount = 7 parts x ₹ 5,00,000 = ₹ 35,00,000

Hence, option 4 is the correct option.

(2) Given:

Value of first property = ₹ 38,00,000

Ratio (Priya : Seema) = 7 : 12

Total parts = 7 + 12 = 19

Value of 1 part = ₹ 38,00,000 ÷ 19 = ₹ 2,00,000

Seema's amount = 12 parts x ₹ 2,00,000 = ₹ 24,00,000

Hence, option 2 is the correct option.

(3)

Calculate total for Priya.

From 1st property:

Ratio (Priya : Seema) = 7 : 12 $\quad$Given

Value of 1 part = ₹ 2,00,000 $\quad$[From previous step]

∴ 7 x 2,00,000 = ₹ 14,00,000

From 2nd property:

Priya's amount = ₹ 35,00,000 $\quad$[From step 1]

Total = ₹ 14,00,000 + ₹ 35,00,000 = ₹ 49,00,000

Calculate total for Seema.

From 1st property:

Seema's amount = ₹ 24,00,000 $\quad$[From step 2]

From 2nd property:

Ratio (Priya : Seema) = $\frac{1}{5} : \frac{1}{7}$

L.C.M. of 5 and 7 is 35.

Priya : Seema = $(\frac{1}{5} \times 35) : (\frac{1}{7} \times 35)$

Priya : Seema = 7 : 5

Value of 1 part = ₹ 5,00,000

∴ 5 x ₹ 5,00,000 = ₹ 25,00,000

Total = ₹ 24,00,000 + ₹ 25,00,000 = ₹ 49,00,000

Difference = Priya - Seema

= ₹ 49,00,000 - ₹ 49,00,000 = ₹ 0

Hence, option 1 is the correct option.

(4)

Seema's 1st amount = ₹ 24,00,000 $\quad$[From step 2]

Seema's 2nd amount = ₹ 25,00,000 $\quad$[From previous step]

Ratio = 24,00,000 : 25,00,000

Ratio = 24 : 25

Hence, option 3 is the correct option.

(1) Given:

Elephant statue is made from variety 1.

Variety 1 ratio (Copper : Zinc) = 7 : 4

Total weight of statue = 176 g

Total parts = 7 + 4 = 11

Value of 1 part = 176 g ÷ 11 = 16 g

Copper = 7 parts x 16 g = 112 g

Hence, option 2 is the correct option.

(2) Given:

Horse statue is made from variety 2.

Variety 2 ratio (Copper : Zinc) = 5 : 3

Quantity of Zinc = 135 g

3 parts of Zinc = 135 g

Value of 1 part = 135 g ÷ 3 = 45 g

Copper = 5 parts x 45 g

Copper = 225 g

Hence, option 4 is the correct option.

(3) Given:

Zinc in Horse = 135 g

Zinc in Elephant:

Variety 1 ratio (Copper : Zinc) = 7 : 4

Value of 1 part = 16 g $\quad$[From step 1]

∴ 4 parts x 16 g = 64 g

Zinc in Elephant = 64 g

Total Zinc = 135 g + 64 g = 199 g

Hence, option 4 is the correct option.

(4)

Calculate Copper in Elephant:

Variety 1 ratio (Copper : Zinc) = 7 : 4

Value of 1 part = 16 g $\quad$[From step 1]

Copper in Elephant = 7 x 16 g = 112 g

Copper in Horse = 225 g $\quad$[From step 2]

Total Copper = Copper in Elephant + Copper in Horse

Total Copper = 112 g + 225 g = 337 g $\quad$[Substituting the values]

Total Copper = 337 g

Total Zinc = 199 g $\quad$[From previous step]

The ratio of the quantities of copper and zinc = 337 : 199

Hence, option 4 is the correct option.

Both Assertion (A) and Reason (R) are true but Reason (R) is not the correct explanation of Assertion (A).

Explanation

Given:

₹1250 is divided between Dinesh and Anmol in the ratio 3 : 7.

Total parts: 3 + 7 = 10

Value of one part: 1250 ÷ 10 = 125

Shares:

Dinesh = 3 × 125 = 375

Anmol = 7 × 125 = 875

Difference: 875 − 375 = 500

So the Assertion is true.

A ratio is indeed a comparison of two quantities of the same kind, so it is a fraction and has no units. The Reason is True.

This statement is true, but it does not explain why the difference is ₹500.

Hence, option 2 is the correct option.

Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of Assertion (A).

Explanation

Given:

Numbers 4, 8, 16 are in continued proportion.

Numbers to be in continued proportion, it should satisfy 4 : 8 = 8 : 16

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

Ratio 2 (8 : 16) = $\frac{8}{16} = \frac{1}{2}$

Since the ratios are equal, the Assertion is True.

The reason states that a, b, c are in continued proportion if a : b = b : c. This is the mathematical definition of continued proportion.

This statement is correct and explains the assertion.

Hence, option 1 is the correct option.

Given:

Distance covered by bus = 126 km

Time taken by bus = 3 hours

Distance covered by train = 315 km

Time taken by train = 5 hours

Step 1: Find the speed of the bus and the train

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

Speed of bus = $\frac{126}{3}$ = 42 km/h

Speed of train = $\frac{315}{5}$ = 63 km/h

Step 2: Find the ratio of their speeds

Ratio of speeds = Speed of bus : Speed of train

= 42 : 63

= $\frac{42}{63}$ $\hspace{2cm}$[Writing the ratio as a fraction]

= $\frac{42 \div 21}{63 \div 21}$

[Dividing numerator and denominator by H.C.F. of 42 and 63, which is 21]

= $\frac{2}{3}$

= 2 : 3

∴ The ratio of speeds of the bus and the train is 2 : 3.

Hence, option 2 is the correct option.

Given:

4, a, a, 36 are in proportion.

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

Here, extremes are 4 and 36, and means are a and a.

According to the question, the equation can be written as:

4 × 36 = a × a

⇒ a² = 144

⇒ a = $\sqrt{144}$

⇒ a = 12 $\hspace{2cm}$ [∵ 12 × 12 = 144]

∴ The value of a is 12.

Hence, option 2 is the correct option.

Given:

u : v = 4 : 7

v : w = 9 : 7

u = 72

Step 1: Find the value of v

u : v = 4 : 7

$\frac{u}{v}$ = $\frac{4}{7}$

$\frac{72}{v}$ = $\frac{4}{7}$ $\hspace{2cm}$[Substituting the value of u]

⇒ v = $\frac{72 \times 7}{4}$ $\hspace{2cm}$[Cross multiplying]

⇒ v = 18 × 7

⇒ v = 126

Step 2: Find the value of w

v : w = 9 : 7

$\frac{v}{w}$ = $\frac{9}{7}$

$\frac{126}{w}$ = $\frac{9}{7}$ $\hspace{2cm}$[Substituting the value of v]

⇒ w = $\frac{126 \times 7}{9}$ $\hspace{2cm}$[Cross multiplying]

⇒ w = 14 × 7

⇒ w = 98

∴ The value of w is 98.

Hence, option 1 is the correct option.

Given:

x : y = 2 : 3

y : z = 4 : 5

To find z : x, we make y the same in both the ratios.

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

So, we make y equal to 12 in each case.

Step 1: Make y = 12 in both ratios

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

y : z = 4 : 5 = $\frac{4 \times 3}{5 \times 3}$ = $\frac{12}{15}$ = 12 : 15

Step 2: Combine to get x : y : z

x : y : z = 8 : 12 : 15

Step 3: Find z : x

z : x = 15 : 8

∴ The ratio z : x is 15 : 8.

Hence, option 2 is the correct option.

Given:

Ratio of number of ₹2, ₹5 and ₹10 coins = 5 : 7 : 8

Total value of coins = ₹1250

Let the number of ₹2 coins = 5k, the number of ₹5 coins = 7k and the number of ₹10 coins = 8k.

Step 1: Find the total value in terms of k

Value of ₹2 coins = 2 × 5k = ₹10k

Value of ₹5 coins = 5 × 7k = ₹35k

Value of ₹10 coins = 10 × 8k = ₹80k

Total value = 10k + 35k + 80k = ₹125k

Step 2: Apply the total value condition

According to the question, the equation can be written as:

125k = 1250

⇒ k = $\frac{1250}{125}$

⇒ k = 10

Step 3: Find the number of ₹5 coins

Number of ₹5 coins = 7k = 7 × 10 = 70

∴ The number of ₹5 coins in the bag is 70.

Hence, option 1 is the correct option.

Given:

A : B = 7 : 8

B : C = 7 : 9

To find A : B : C, we make B the same in both the ratios.

L.C.M. of 8 and 7 = 56.

So, we make B equal to 56 in each case.

Step 1: Make B = 56 in both ratios

A : B = 7 : 8 = $\frac{7 \times 7}{8 \times 7}$ = $\frac{49}{56}$ = 49 : 56

B : C = 7 : 9 = $\frac{7 \times 8}{9 \times 8}$ = $\frac{56}{72}$ = 56 : 72

Step 2: Combine to get A : B : C

A : B : C = 49 : 56 : 72

∴ The ratio A : B : C is 49 : 56 : 72.

Hence, option 2 is the correct option.

Given:

Total number of coins = 220

Total amount = ₹160 = 16000 paise $\hspace{2cm}$[∵ ₹1 = 100 paise]

Number of 1 rupee coins = 3 × Number of 25 paise coins

Step 1: Express the number of each type of coin

Let the number of 25 paise coins = x.

Then, the number of 1 rupee coins = 3x.

Number of 50 paise coins = 220 − x − 3x

= 220 − 4x

Step 2: Set up the equation using the total amount

Value of 25 paise coins = 25x paise

Value of 50 paise coins = 50 × (220 − 4x) = (11000 − 200x) paise

Value of 1 rupee coins = 100 × 3x = 300x paise

According to the question, the equation can be written as:

25x + (11000 − 200x) + 300x = 16000

25x + 11000 − 200x + 300x = 16000

125x + 11000 = 16000

125x = 16000 − 11000

125x = 5000

⇒ x = $\frac{5000}{125}$

⇒ x = 40

Step 3: Find the number of 50 paise coins

Number of 50 paise coins = 220 − 4x

= 220 − 4 × 40

= 220 − 160

= 60

∴ The number of 50 paise coins is 60.

Hence, option 1 is the correct option.

Given:

Pages read by Deepa per day = 6

Pages read by Mahima per day = 9

Page number Deepa is on = 72

Since both started on the same day, they have been reading for the same number of days.

Step 1: Find the number of days Deepa has been reading

Number of days = $\frac{\text{Total pages read}}{\text{Pages per day}}$

Number of days = $\frac{72}{6}$ = 12 days

Step 2: Find the page Mahima is on

Pages read by Mahima in 12 days = 9 × 12 = 108

∴ Mahima will be on page 108 when Deepa is on page 72.

Hence, option 4 is the correct option.

Given:

Cost of 5 mangoes and 4 oranges = Cost of 3 mangoes and 7 oranges

Let the cost of 1 mango = ₹m and the cost of 1 orange = ₹o.

Step 1: Set up the equation

Cost of 5 mangoes and 4 oranges = ₹(5m + 4o)

Cost of 3 mangoes and 7 oranges = ₹(3m + 7o)

According to the question, the equation can be written as:

5m + 4o = 3m + 7o

Step 2: Solve for the ratio m : o

5m − 3m = 7o − 4o $\hspace{2cm}$[Rearranging the terms]

2m = 3o

$\frac{m}{o}$ = $\frac{3}{2}$ $\hspace{2cm}$[Dividing both sides by 2o]

m : o = 3 : 2

∴ The ratio of the cost of 1 mango to that of 1 orange is 3 : 2.

Hence, option 3 is the correct option.

Given:

In class 6, ratio of girls to boys = 3 : 5

In class 7, ratio of girls to boys = 9 : 13

A ratio only compares two quantities; it does not tell us their actual values.

The ratio 3 : 5 in class 6 means that for every 3 girls there are 5 boys, but the total number of students in class 6 is not given.

Similarly, the ratio 9 : 13 in class 7 means that for every 9 girls there are 13 boys, but the total number of students in class 7 is not given.

Without knowing the total number of students in each class, the actual number of girls in each class cannot be determined.

∴ We can't say in which class there are more girls.

Hence, option 4 is the correct option.