Arithmetic: previous-year questions

Answer: (B) 36

24, 72 and 168 are all multiples of 24 ($24 \times 1$, $24 \times 3$, $24 \times 7$), but 36 is not ($36 \div 24 = 1.5$), so 36 is the odd one out. A second test gives the same answer: 36 is the only perfect square ($6^2$) in the list. In odd-number-out questions, check in this order: divisibility by a common factor, squares and cubes, prime or composite, and sum of digits, then pick the option that fails a property shared by the other three.

Answer: (B) 30

The terms are products of consecutive numbers, $n(n+1)$: $1 \times 2 = 2$, $2 \times 3 = 6$, $3 \times 4 = 12$, $4 \times 5 = 20$, so the next is $5 \times 6 = 30$. The differences confirm it: 4, 6, 8, and therefore 10 next ($20 + 10 = 30$). Such numbers ($n^2 + n$) are called pronic or oblong numbers, and the series continues 42, 56, 72. In any number series, first write down the differences; if they form a simple pattern (here, consecutive even numbers) the answer follows at once.

Answer: (C) She suffers a loss of 1%

When two articles are sold at the same selling price, one at a gain of $x\%$ and the other at a loss of $x\%$, there is always an overall loss of $\frac{x^2}{100}\%$. Here $\frac{10^2}{100} = 1\%$ loss. Check with numbers: if each goat sells for Rs 99, the cost prices are $\frac{99}{1.1} = 90$ and $\frac{99}{0.9} = 110$, total cost 200 against total sale 198, a loss of 2 on 200, i.e. $1\%$. The trap is option A: 'no profit, no loss' is true only when the two cost prices (not the selling prices) are equal.

Answer: (B) 2

This is a derangement problem: arrangements in which no object occupies its own place. For three letters and three envelopes there are $3! = 6$ arrangements in all, and only two of them, (2, 3, 1) and (3, 1, 2), put every letter in a wrong envelope. The general formula is $D_n = n!\left(1 - \frac{1}{1!} + \frac{1}{2!} - \frac{1}{3!} + \dots \pm \frac{1}{n!}\right)$, which gives $D_2 = 1$, $D_3 = 2$, $D_4 = 9$ and $D_5 = 44$, values worth memorising. Option D (6) counts all arrangements, and option A (1) is the number in which every letter is correctly placed.

Answer: (C) 80%

Use alligation on the averages: boys 52, girls 42, combined 50. Boys : girls $= (50 - 42) : (52 - 50) = 8 : 2 = 4 : 1$, so boys are $\frac{4}{5} = 80\%$ of the class. Algebra gives the same result: if $b$ is the fraction of boys, $52b + 42(1 - b) = 50$, so $b = 0.8$. Option D is the trap, since the actual numbers are not needed; a ratio is enough to give a percentage. Common-sense check: the combined average (50) lies much closer to the boys' average, so boys must be the large majority.

Answer: (B) 24 m/s

When equal distances are covered at speeds $v_1$ and $v_2$, the average speed is the harmonic mean $\frac{2v_1v_2}{v_1 + v_2} = \frac{2 \times 20 \times 30}{20 + 30} = 24$ m/s. Option C (25 m/s) is the simple average, which would be correct only if equal times, not equal distances, were spent at each speed; the true value is lower because the body spends longer on the slower half. Option A confuses average speed with average velocity in a round trip. Check with a distance of 120 m: time $= \frac{60}{20} + \frac{60}{30} = 5$ s, so speed $= \frac{120}{5} = 24$ m/s.

Answer: (C) 2:1

Let cows $= x$ and crows $= y$. Heads: $x + y = 30$. Legs: $4x + 2y = 100$. Subtracting twice the first equation from the second gives $2x = 40$, so $x = 20$ cows and $y = 10$ crows, a ratio of $20 : 10 = 2 : 1$. Shortcut: if all 30 were crows there would be 60 legs; the 40 extra legs come from cows at 2 extra legs each, so there are 20 cows. The trap is option A (1 : 2), which is the ratio of crows to cows; always check the order asked in the question.

Answer: (A) is 16 m/s.

For two equal distances the average speed is the harmonic mean: $\frac{2 \times 10 \times 40}{10 + 40} = \frac{800}{50} = 16$ m/s. It does not depend on the total distance (option D), because the distance cancels out: for a total distance $2d$, time $= \frac{d}{10} + \frac{d}{40} = \frac{d}{8}$, and speed $= \frac{2d}{d/8} = 16$. Option C (25 m/s) is the arithmetic mean, valid only when equal times are spent at each speed. Notice how strongly the slow half pulls the average down: the answer is much nearer 10 than 40.

Answer: (C) 115

With $c$ correct, $u$ unattempted and the rest wrong ($c + u \le 30$), the score is $4c + u$. Scores of 113 ($c = 28$, $u = 1$), 114 ($c = 28$, $u = 2$) and 117 ($c = 29$, $u = 1$) are all possible. For 115 we would need $c = 28$ with $u = 3$ (31 questions) or $c = 29$ with $u = -1$, both impossible, so 115 cannot be scored. In fact, near the maximum of 120 the unattainable scores are 119, 118 and 115. Method for such questions: fix the number of correct answers at the highest value that does not overshoot, then see whether the remaining marks fit into the remaining questions.

Answer: (C) 30

Since the installments are equal and interest-free, 18 installments correspond to 60% of the loan, so the total number is $\frac{18}{0.60} = 30$. Each installment is therefore $\frac{100}{30} = 3\frac{1}{3}\%$ of the loan, and 12 installments (40%) remain. Unitary check: 60% is 18 installments, so 10% is 3 installments and 100% is 30. The other options fail the same check, since 60% of 22, 24 and 33 is 13.2, 14.4 and 19.8, not 18.

Answer: (B) decrease by 4%.

New area $= (0.8L) \times (1.2B) = 0.96\,LB$, i.e. 96% of the original, a decrease of 4%. General rule: when one dimension increases by $x\%$ and the other decreases by $x\%$, the area always falls by $\frac{x^2}{100}\%$; here $\frac{20^2}{100} = 4\%$. For unequal changes of $a\%$ and $b\%$ the net change is $a + b + \frac{ab}{100}$, which gives $20 - 20 - \frac{400}{100} = -4\%$. Option A is the usual trap of assuming that equal percentage increase and decrease cancel out. The same $\frac{x^2}{100}$ rule gives the net loss when two articles are sold at one price with equal profit and loss percentages.

Answer: (C) 13

This is the pigeonhole principle: with 12 months (pigeonholes), 12 members could all have different birth months, but a 13th member must share a month with someone. So the minimum that guarantees a match is $12 + 1 = 13$. Option B (12) is the largest group that can still avoid a match, and option D (24) is a distractor based on 'two per month'. Extension often asked: to guarantee that at least three members share a birth month you need $2 \times 12 + 1 = 25$; to guarantee two people with the same birthday you need 367 people.

Answer: (B) $x = 6$, $y = 8$

From $\frac{x}{y} = \frac{3}{4}$, $x = \frac{3y}{4}$. Substituting in $2x - 3y = -12$: $\frac{3y}{2} - 3y = -12$, so $-\frac{3y}{2} = -12$, giving $y = 8$ and $x = 6$. The quickest exam method is to test the options: only $x = 6$, $y = 8$ satisfies both conditions ($12 - 24 = -12$ and $\frac{6}{8} = \frac{3}{4}$). Option A fails the ratio ($\frac{6}{9} = \frac{2}{3}$), and options C and D give a ratio greater than 1.

Answer: (A) 10:10 am

The cars move towards each other, so their relative speed is $50 + 30 = 80$ km/h. Time to meet $= \frac{160}{80} = 2$ hours, and 2 hours after 8:10 am is 10:10 am. They meet $50 \times 2 = 100$ km from A and $30 \times 2 = 60$ km from B. Rule: for bodies moving in opposite directions add the speeds; for the same direction subtract them. If the two cars were travelling in the same direction, the faster one would close a 160 km gap in $\frac{160}{20} = 8$ hours.

Answer: (B) 500 m

60 km/h is 1 km per minute, so in 2 minutes the train travels 2 km. To pass completely through a tunnel a train must cover the length of the tunnel plus its own length, so its length is $2 - 1.5 = 0.5$ km $= 500$ m. The same rule applies to bridges and platforms: distance covered equals the length of the object plus the length of the train, whereas for passing a pole or a standing man it is the train's length alone. Useful conversion: 1 km/h $= \frac{5}{18}$ m/s, so 60 km/h $= 16\frac{2}{3}$ m/s.

Answer: (D) 6

The rows form an arithmetic progression with common difference 3. If the front (largest) row has $a$ children and there are $n$ rows, the total is $na - \frac{3n(n-1)}{2} = 630$, so $a = \frac{630}{n} + \frac{3(n-1)}{2}$, which must be a whole number. For $n = 3$, $a = 213$; for $n = 4$, $a = 162$; for $n = 5$, $a = 132$; but for $n = 6$, $a = 105 + 7.5 = 112.5$, which is impossible. Check for $n = 5$: $132 + 129 + 126 + 123 + 120 = 630$. Shortcut: with an odd number of rows the middle row equals the average ($\frac{630}{n}$), so $n$ need only divide 630; with an even number of rows the average must end in .5, which fails for 6 ($\frac{630}{6} = 105$).

Answer: (A) 4%

Candidates who failed in at least one subject = $55+65-24=96\%$ (inclusion–exclusion, since the 24% who failed in both would otherwise be counted twice). Therefore those who passed in both subjects = $100-96=4\%$. As a check: failed in English only = $55-24=31\%$, failed in Mathematics only = $65-24=41\%$, failed in both = 24%, and $31+41+24=96\%$. A Venn diagram of two overlapping circles makes such questions quick.

Answer: (B) 21-08-2002

The gaps between successive dates grow by one day each time: 18 July to 25 July is 7 days; 25 July to 2 August is 8 days (July has 31 days); 2 August to 11 August is 9 days. The next gap must be 10 days, and 11 August + 10 days = 21-08-2002. The usual slip is to forget that July has 31 days when counting across the month-end. Months with 31 days: January, March, May, July, August, October and December.

Answer: (B) 12

Let $x$ be the number of ₹200 notes and $y$ the number of ₹100 notes. Then $x+y=48$ and $200x+100y=6000$, i.e. $2x+y=60$. Subtracting the first equation from the second gives $x=12$, so $y=36$. Check: $12\times200+36\times100=2400+3600=6000$. Option (C) 36 is the number of ₹100 notes, placed there to catch candidates who do not read which denomination is asked. Shortcut: if all 48 notes were ₹100 the amount would be ₹4,800; the shortfall of ₹1,200 is made up at ₹100 per ₹200 note, giving 12 notes.

Answer: (A) 20

Work from the innermost brackets. $7*2$: since $7>2$, it is $7-2=5$. Then $5*5$: the numbers are equal, so it is $2\times5=10$. Next $5*7$: since $5<7$, it is $5+7=12$. Then $12*2$: since $12>2$, it is $12-2=10$. Adding the two parts, $10+10=20$. The trap lies in the middle step: after the first bracket becomes 5, the operation $5*5$ must use the "equal" rule ($2x$), not addition or subtraction.