Percentage Questions and Answers

Answer: C

The class has 75 students.

If 44% are females, the males are \(100 - 44 = 56\%\) of the class, and they number 42.

\[ \begin{aligned} 0.56 \times N &= 42 \\ N &= \frac{42}{0.56} = 75 \end{aligned} \]

Check: 44% of 75 is 33 females, and \(33 + 42 = 75\).

Taking 42 as 44% of the class (the female share) gives about 95, which is not offered. 70 would leave 28 females, 40% of the class, and 65 would make the males alone almost 65% of it. 45 would leave only 3 females, under 7% of the class.

Answer: C

The price before tax is ₹ 3,000.

The price including tax is 110% of the price before tax. Let the price before tax be \(P\):

\[ \begin{aligned} 1.10P &= 3300 \\ P &= \frac{3300}{1.10} = 3000 \end{aligned} \]

Check: 10% of 3,000 is 300, and 3,000 + 300 = 3,300. The trap is ₹ 2,970, which is 3,300 minus 10% of 3,300: the tax is 10% of the pre-tax price, not of the total. ₹ 3,030 and ₹ 2,700 come from adding or subtracting the tax amount the wrong way.

Answer: B

The keyed answer is 5%.

With five people, each pays 1/5 = 20% of the rent. When one withdraws, each of the remaining four pays 1/4 = 25% of the rent. So each person's share of the total rent rises from 20% to 25%, that is, by 5 percentage points of the rent, which is the reading the key takes.

Measured against each person's own earlier payment, the increase is (25 - 20)/20 = 25%, which is not among the options; 4%, 6% and 7% fit neither reading.

Answer: B

His monthly income is ₹ 1500.

If Ram spends 65% of his salary, he saves the remaining 35%, and that is ₹ 525:

\[ \begin{aligned} 0.35 \times S &= 525 \\ S &= \frac{525}{0.35} = 1500 \end{aligned} \]

Check: 65% of 1500 is 975 and 1500 - 975 = 525. The other options give savings of ₹ 420, ₹ 630 and ₹ 735, so none matches. A common slip is dividing 525 by 0.65 (giving about ₹ 808), which treats the savings as the spent share.

Answer: C

The total discount is 44%.

Successive discounts multiply rather than add. After 30% off, 70% of the price remains; after a further 20% off, 80% of that remains:

\[ \begin{aligned} 0.70 \times 0.80 &= 0.56 \\ 1 - 0.56 &= 0.44 = 44\% \end{aligned} \]

The shortcut for successive discounts is a + b - ab/100 = 30 + 20 - 6 = 44. 50% is the mistake of simply adding the two discounts, and 56% is the share of the price that is still paid, not the discount.

Answer: C

About 266.6 kg of ore is needed.

Only the alloy carries iron, and the alloy is 25% of the ore, of which 90% is iron. So the iron fraction of the ore is

\[ \begin{aligned} 0.25 \times 0.90 &= 0.225 \\ \text{ore} &= \frac{60}{0.225} \\ &= 266.67\ \text{kg} \end{aligned} \]

which the option gives as 266.6 kg. 300 kg would be right if the ore were 20% iron, and 270.6 kg and 250.67 kg do not correspond to the 22.5% iron content.

Answer: A

The area decreases by 28%.

The new length is 80% and the new breadth 90% of the originals, so the new area is

\[ \begin{aligned} 0.80 \times 0.90 &= 0.72 \\ \text{decrease} &= 1 - 0.72 = 0.28 = 28\% \end{aligned} \]

The shortcut for two successive decreases of \(a\%\) and \(b\%\) is \(a + b - \frac{ab}{100} = 20 + 10 - 2 = 28\).

Adding the losses gives 30%, which is not offered; 20% and 10% take one dimension only, and 22% has no basis in the data.

Answer: D

10 % of the students failed in both subjects.

Find the percentage who passed in at least one subject, using the rule for the union of two sets:

\[ \begin{aligned} P(E \cup M) &= 85 + 65 - 60 = 90\% \\ \text{failed both} &= 100 - 90 = 10\% \end{aligned} \]

15% is the share who failed English (100 − 85) and 35% the share who failed Mathematics (100 − 65); neither is the share who failed both. 40% is the share who did not pass both subjects (100 − 60), a larger group that includes those who passed one of them.

Answer: B

The present population is \((XY)^{1/2}\), the geometric mean of X and Y.

Let the present population be P and the yearly growth factor be r. Then last year's population was \(X = P/r\) and next year's will be \(Y = Pr\). Multiplying:

\[ \begin{aligned} XY &= \frac{P}{r} \cdot Pr = P^{2} \\ \Rightarrow P &= \sqrt{XY} \end{aligned} \]

Growth at a constant rate is multiplicative, so the middle term of three is the geometric mean, not the arithmetic mean \((X+Y)/2\), which would apply to growth by a constant number each year. \(2XY/(X+Y)\) is the harmonic mean, and \(XY/2\) has the wrong units altogether.

Answer: A

The percentage rise is 12 %.

Rs. 140 and 49 paise is Rs. 140.49. Three equal rises of r% multiply the salary by \((1 + r/100)^{3}\):

\[ \begin{aligned} 100\left(1 + \tfrac{r}{100}\right)^{3} &= 140.49 \\ 1.12^{3} &= 1.404928 \approx 1.4049 \end{aligned} \]

So r = 12. A check of the neighbours: \(1.11^{3} \approx 1.3676\), giving Rs. 136.76, and \(1.13^{3} \approx 1.4429\), giving Rs. 144.29. Dividing the 40.49 rise by 3 to get about 13.5% ignores compounding.