Ratio & Proportion Questions and Answers

Answer: C

The incomes are Rs. 18,000 and Rs. 14,000.

Let the incomes be 9x and 7x and the expenditures 4y and 3y. Each saves Rs. 2,000:

\[ \begin{aligned} 9x - 4y &= 2000 \\ 7x - 3y &= 2000 \end{aligned} \]

Multiply the first by 3 and the second by 4 and subtract:

\[ \begin{aligned} 27x - 28x &= 6000 - 8000 \\ x &= 2000 \end{aligned} \]

So the incomes are 9 × 2000 = 18,000 and 7 × 2000 = 14,000 (expenditures 16,000 and 12,000, ratio 4 : 3). The other pairs have the right 9 : 7 ratio but give different savings.

Answer: D

Anil gets ₹7,500.

The contributions are 5 : 3 : 2, which add up to 10 parts. Anil put in 3 of the 10 parts, so his share of the prize is

\[ \frac{3}{10} \times 25000 = 7500 \]

Aman receives ₹12,500 and Atul ₹5,000, and the three shares add to ₹25,000 as a check. ₹5,000 is Atul's share, not Anil's; ₹4,000 and ₹6,500 do not correspond to any of the three fractions.

Answer: C

Raj has ₹24,000.

Let Rakesh have x. Then Raj has 1.5x and Ravi has 1.5 × 1.5x = 2.25x. Together:

\[ \begin{aligned} x + 1.5x + 2.25x &= 4.75x = 76000 \\ x &= 16000 \end{aligned} \]

So Rakesh has ₹16,000, Raj ₹24,000 and Ravi ₹36,000. The options ₹16,000 and ₹36,000 are the other two people's amounts, which is the trap; ₹12,000 matches nobody.

Answer: C

The largest part is 88.

Let the common value be k. Then half the first part is k, a third of the second is k and a quarter of the third is k, so the parts are 2k, 3k and 4k:

\[ \begin{aligned} 2k + 3k + 4k &= 198 \\ 9k &= 198 \Rightarrow k = 22 \end{aligned} \]

The parts are 44, 66 and 88, and the largest is 88. Check: 44/2 = 66/3 = 88/4 = 22. 68, 78 and 98 do not give equal fractions.

Answer: D

The white part is 25 m long.

The ratio 5 : 6 : 7 has \(5 + 6 + 7 = 18\) parts in all, so

\[ \text{white} = \frac{5}{18} \times 90 = 25\ \text{m} \]

The black part is 30 m and the blue 35 m, which together with 25 m make 90 m. 30 m is the black part, the usual slip, and 15 m and 20 m do not follow from the ratio.

Answer: A

The ratio is 2:1.

Take \(a = 3k\), \(b = 4k\), \(c = 7k\). Then

\[ \begin{aligned} (a + b + c) : c &= (3 + 4 + 7)k : 7k \\ &= 14 : 7 = 2 : 1 \end{aligned} \]

3:1 would need the total to be 21 parts, and 2:3 and 1:3 are smaller than 1, which is impossible, since \(a + b + c\) is always greater than \(c\).

Answer: B

The number of one rupee coins is 78.

Let there be \(x\) one rupee coins and \(y\) fifty paise coins. Their values are Rs \(x\) and Rs \(0.5y\), in the ratio 13:11:

\[ \begin{aligned} \frac{x}{0.5y} &= \frac{13}{11} \\ y &= \frac{22x}{13} \\ x + \frac{22x}{13} &= 210 \\ \frac{35x}{13} &= 210, \quad x = 78 \end{aligned} \]

Check: 78 one rupee coins and 132 fifty paise coins make 210 coins, worth Rs 78 and Rs 66, and \(78 : 66 = 13 : 11\). The other options fail this check; the ratio is of values, not of numbers of coins.

Answer: B

The total cost is 165 paise/km.

Coal use varies as the square of the speed, so halving the speed to 30 km/hour cuts it to a quarter:

\[ \begin{aligned} \text{coal} &= 1000 \times (30/60)^2 = 250 \text{ kg/h} \\ \text{coal cost} &= 250 \times 15 = 3750 \text{ paise/h} \\ \text{total} &= 3750 + 1200 = 4950 \text{ paise/h} \end{aligned} \]

\[ \text{per km} = 4950 / 30 = 165 \text{ paise/km} \]

49·5 is the hourly cost in rupees, not a cost per kilometre. 270 paise/km is the cost at the original 60 km/hour: (15000 + 1200) / 60. 'None of the above' is wrong because 165 is offered.

Answer: B

\(\frac{X}{Y} = \) \(\frac{33}{32}\).

Chain the given ratios through W:

\[ \begin{aligned} \frac{X}{Y} &= \frac{X}{W} \times \frac{W}{Y} \\ &= \frac{11}{6} \times \frac{9}{16} = \frac{99}{96} = \frac{33}{32} \end{aligned} \]

The ratio \(\frac{Y}{Z}\) is not needed. \(\frac{32}{33}\) is \(\frac{Y}{X}\), the reciprocal, and \(\frac{16}{11}\) or \(\frac{11}{16}\) come from combining the wrong pair of ratios.

Answer: B

The number to add is 11.

For \((94 + x), (24 + x), (100 + x), (26 + x)\) to be in proportion, the product of the extremes equals the product of the means:

\[ \begin{aligned} (94 + x)(26 + x) &= (24 + x)(100 + x) \\ 2444 + 120x &= 2400 + 124x \\ x &= 11 \end{aligned} \]

Check: 105 : 35 = 3 : 1 and 111 : 37 = 3 : 1. The other values do not make the two ratios equal.

A quick check from the options: adding 11 makes each pair a 3 : 1 ratio, while adding 10 gives 104 : 34 and 110 : 36, which are unequal.