Ratio & Proportion Questions and Answers

Answer: B

The value of y is 1.5.

If x varies inversely as y, their product is constant:

\[ \begin{aligned} xy &= 3.5 \times 2.4 = 8.4 \\ y &= \frac{8.4}{5.6} = 1.5 \end{aligned} \]

As x increases, y must decrease, which rules out 2.8. 1.4 and 2.1 would come from treating the variation as direct or from arithmetic slips.

Answer: B

The value of \(\sqrt{12x + 9}\) is 9.

For the numbers to be in proportion, (14 + x) : (12 + x) = (34 + x) : (30 + x). Cross-multiplying:

\[ \begin{aligned} (14 + x)(30 + x) &= (12 + x)(34 + x) \\ 420 + 44x + x^{2} &= 408 + 46x + x^{2} \\ 12 &= 2x, \quad x = 6 \end{aligned} \]

Check: 20 : 18 = 40 : 36, both equal to 10 : 9. Then

\[ \sqrt{12(6) + 9} = \sqrt{81} = 9 \]

8, 11 and 13 are the square roots of 64, 121 and 169, which would need x = 4.6, 9.3 or 13.3, none of which makes the four numbers proportional.

Answer: D

The true statements are (a), (b) and (c).

Write \(x = kp\) and \(y = kq\) for some non-zero k. Then:

\[ \begin{aligned} \tfrac{x + y}{y} &= \tfrac{k(p + q)}{kq} = \tfrac{p + q}{q} \\ \tfrac{x - y}{y} &= \tfrac{p - q}{q}, \quad \tfrac{x}{p} = k = \tfrac{y}{q} \end{aligned} \]

These are componendo (a), dividendo (b) and alternendo (c). Statement (d) is wrong: componendo and dividendo give \(x + y : x - y = p + q : p - q\), not \(p - q : p + q\), so any option containing (d) is ruled out.

Answer: C

The value of x is 8. Since x varies directly with y when z is constant and directly with z when y is constant, x = kyz for some constant k. From y = 1.5, z = 4 and x = 2.5: \[2.5 = k \times 1.5 \times 4\] so \(k = \frac{2.5}{6} = \frac{5}{12}\). For y = 8 and z = 2.4: \[x = \frac{5}{12} \times 8 \times 2.4\] \[x = \frac{5 \times 19.2}{12} = 8\] The options 5, 6 and 10 do not satisfy this relation.

Answer: C

The youngest employee is 8 years old. The total of the three ages is \(3\times 16 = 48\) years. The ages are in the ratio 2 : 4 : 6, which makes 12 parts, so one part is \(48 \div 12 = 4\) years. The youngest has 2 parts: \[ 2\times 4 = 8\text{ years} \] The ages are 8, 16 and 24, which add to 48. The 18 years and 10 years answers do not match any share, and 5 years would need a part of 2.5, which gives a total of 30, not 48.

As a check, 8 : 16 : 24 reduces to 1 : 2 : 3, the same as 2 : 4 : 6, and the mean of 8, 16 and 24 is 16.

Answer: C

The ratio of the increased seats is 70 : 90 : 128. The seats in the ratio 5 : 6 : 8 are increased by 40%, 50% and 60%, so they become \[ 5\times 1.4 = 7,\quad 6\times 1.5 = 9 \] \[ 8\times 1.6 = 12.8 \] The new ratio is 7 : 9 : 12.8, and multiplying by 10 to remove the decimal gives 70 : 90 : 128. The other options do not equal 7 : 9 : 12.8 in any scale, so they cannot be the ratio of the increased seats.

Dividing 70 : 90 : 128 by 10 gives 7 : 9 : 12.8, so the ratio and the option are the same; the 10 only clears the decimal.

Answer: D

The ratio of the two numbers is 14 : 15. Let the third number be 100. Then the numbers that are 12% and 20% more than it are 112 and 120. The ratio is \[ 112 : 120 = 14 : 15 \] on dividing both by 8. The other ratios, 22 : 15, 12 : 17 and 15 : 18, are not equal to 112 : 120 in lowest terms, which is 14 : 15.

Answer: C

Let the boys be 3k and the girls 4k. After 200 girls leave, the girls are 4k - 200 and the ratio of boys to girls is 6 : 7.

\[ \frac{3k}{4k - 200} = \frac{6}{7} \] \[ 21k = 24k - 1200 \Rightarrow k = 400 \]

The number of boys is \(3k = 1200\). Check: the girls were 1600, then 1400, and 1200 : 1400 = 6 : 7. The value 400 is k itself, 300 and 900 do not fit. The answer is 1200, option C.

Answer: B

Four numbers a, b, c and d are in proportion when \(a : b = c : d\), that is, \(a \times d = b \times c\). Let x be added to each of 8, 12, 16 and 22.

\[ (8 + x)(22 + x) = (12 + x)(16 + x) \]

\[ 176 + 30x + x^2 = 192 + 28x + x^2 \] \[ 2x = 16 \Rightarrow x = 8 \]

Check: 16, 20, 24 and 30 have 16 : 20 = 24 : 30 = 4 : 5. The values 7, 9 and 10 do not make the products equal. The answer is 8, option B.

Answer: A

The value of p is 2. In a proportion the product of the means equals the product of the extremes. \[\frac{3}{2} \times \frac{1}{6} = \frac{1}{p} \times \frac{1}{p}\\\] \[\frac{1}{4} = \frac{1}{p^2}\] so \(p^2 = 4\), and since p is positive, p = 2. Check: 3/2 : 1/2 = 3 : 1 and 1/2 : 1/6 = 3 : 1, so the two ratios are equal. The other options 3, 3.5 and 4 give squares of 9, 12.25 and 16 instead of 4.