Ratio & Proportion Questions and Answers

Answer: C

The class needs 5 more boys.

With 64 students in the ratio 5 : 3, there are \(64 \times \frac{5}{8} = 40\) boys and 24 girls. After 6 more girls join, there are 30 girls. For the ratio of boys to girls to be 3 : 2:

\[ \text{boys} = 30 \times \frac{3}{2} = 45 \]

So \(45 - 40 = 5\) more boys must be admitted. Check: 45 : 30 = 3 : 2. 6, 8 and 12 more boys would give 46 : 30, 48 : 30 and 52 : 30, none of which is 3 : 2.

Answer: C

The ratio is 4 : 1.

From the table, the most common blood group is O with 12 students, and the rarest is AB with 3 students. The ratio of the most common to the rarest is:

\[ 12 : 3 = 4 : 1 \]

The order matters: the question asks for most common to rarest, so 1 : 4 reverses it. 3 : 1 compares A (9) with AB (3), and 1 : 3 is that ratio reversed.

Answer: B

The product of their corresponding values remains constant.

Two quantities vary inversely when one increases in the same proportion as the other decreases:

\[ y \propto \frac{1}{x} \quad\Rightarrow\quad xy = k \]

For example, if 6 workers take 10 days, 12 workers take 5 days: 6 × 10 = 12 × 5 = 60.

A constant ratio defines direct variation, which is the common confusion. A constant sum or difference is not a form of variation at all.

Answer: B

The number to subtract is 17.

If x is subtracted, (50 − x) : (61 − x) = (92 − x) : (117 − x). The product of the extremes equals the product of the means:

\[ \begin{aligned} (50 - x)(117 - x) &= (61 - x)(92 - x) \\ 5850 - 167x &= 5612 - 153x \\ 238 &= 14x \\ x &= 17 \end{aligned} \]

Check: 33, 44, 75, 100, and 33 : 44 = 3 : 4 = 75 : 100. Subtracting 14, 19 or 23 does not give equal ratios (for example 14 gives 36 : 47 and 78 : 103).

Answer: B

The difference between A and C is ₹ 450.

Let 4A = 6B = 9C = k. Then A = k/4, B = k/6 and C = k/9, so A : B : C = 9 : 6 : 4 (multiplying through by 36).

\[ \begin{aligned} 9 + 6 + 4 &= 19\ \text{parts} = 1710 \\ 1\ \text{part} &= 90 \\ A &= 810,\ C = 360 \\ A - C &= 450 \end{aligned} \]

₹ 360 is C's share itself, not the difference. The usual trap is writing the ratio as 4 : 6 : 9 instead of its reciprocal; that gives A = 360 and C = 810, with the same difference, so check the shares make 4A = 6B = 9C (4 × 810 = 3240 = 9 × 360).

Answer: B

The total number of fruits is 32.

Let A have 7t fruits and B have 9t. After moving six from A to B:

\[ \begin{aligned} \frac{7t - 6}{9t + 6} &= \frac{1}{3} \\ 21t - 18 &= 9t + 6 \\ 12t &= 24,\ t = 2 \end{aligned} \]

So A has 14 and B has 18, a total of 16t = 32. Check: 8 : 24 = 1 : 3. The total does not change when fruits move between baskets, so it must be a multiple of 16; 28, 36 and 40 are not.

Answer: B

The actual distance is about 11.57 km.

Each centimetre on the map stands for 845 m, so multiply:

\[ \begin{aligned} 13.7 \times 845 &= 11576.5\ \text{m} \\ &\approx 11.58\ \text{km} \end{aligned} \]

11.57 km is the option this rounds to (11.5765 km cut to two decimal places). 12.90 km, 12.50 km and 11.02 km would correspond to map distances of about 15.3 cm, 14.8 cm and 13.0 cm.

Answer: B

The value of \((3x + 1)\) is 13.

Four numbers are in proportion when the product of the extremes equals the product of the means.

\[ \begin{aligned} (4 + x)(24 + x) &= (10 + x)(12 + x) \\ 96 + 28x + x^{2} &= 120 + 22x + x^{2} \\ 6x &= 24 \\ x &= 4 \end{aligned} \]

Check: the numbers become 8, 14, 16, 28, and \(8 : 14 = 16 : 28 = 4 : 7\). So \(3x + 1 = 13\). The value 16 is \(4x\), and 10 and 19 would need \(x = 3\) and \(x = 6\), which do not make the numbers proportional.

Answer: C

The sum of the two numbers is 48.

Let the numbers be \(3x\) and \(5x\). Adding 12 to each gives the new ratio 5:7.

\[ \begin{aligned} \frac{3x + 12}{5x + 12} &= \frac{5}{7} \\ 21x + 84 &= 25x + 60 \\ 4x &= 24, \quad x = 6 \end{aligned} \]

The numbers are 18 and 30, and their sum is \(8x = 48\). Check: \(18 + 12 = 30\) and \(30 + 12 = 42\), and \(30 : 42 = 5 : 7\).

The other sums (32, 40, 56) are multiples of 8 too, but only \(x = 6\) satisfies the changed ratio.

Answer: B

The sum of the three numbers is 63.

Let the numbers be 2k, 3k and 4k. Then:

\[ \begin{aligned} (2k)^{3} + (3k)^{3} + (4k)^{3} &= 99k^{3} = 33957 \\ k^{3} &= 343 \Rightarrow k = 7 \\ 2k + 3k + 4k &= 9k = 63 \end{aligned} \]

The numbers are 14, 21 and 28 (check: 2744 + 9261 + 21952 = 33957). 54, 72 and 81 are 9k for k = 6, 8 and 9, whose cubes do not give 33957.