Q&A: practice questions with answers

Concept: Evaluate the root, then do $\times$ and $÷$ from left to right, then solve for $y$.

1. $\sqrt{3025}=55$.
2. $26\times 5\times 715÷55=130\times 13=1690$ (since $715÷55=13$).
3. $y=1690-913=777$.

Trick: Divide first where it cancels cleanly: $715=55\times 13$ saves a big multiplication.

Answer: (A) 777

Concept: In an isosceles triangle the altitude from the apex bisects the base; use Pythagoras to get the height.

1. $BD=DC=\frac{12}{2}=6$ cm.
2. $AD=\sqrt{{10}^2-6^2}=\sqrt{64}=8$ cm.
3. Area $=\frac{1}{2}\times 12\times 8=48{\text{cm}}^2$.

Trick: 6-8-10 is a 3-4-5 triple – spot it and skip the square root.

Answer: (C) 48 cm²

Concept: Overall average $=\frac{\text{total amount}}{\text{total members}}$ (weighted mean).

1. Total $=85\times 407+23\times 365+42\times 280=34595+8395+11760=54750$.
2. Members $=85+23+42=150$.
3. Average $=\frac{54750}{150}=365$.

Trick: Take 365 as base: deviations $85\times 42=3570$ and $42\times (-85)=-3570$ cancel exactly, B is at the base → answer 365 without division.

Answer: (D) 365

Concept: For $n$ sorted values the median is the $\frac{n+1}{2}$-th value (odd $n$) or the mean of the two middle values (even $n$).

1. 13 values (already sorted): median = 7th value = 28.
2. Remove 55 → 12 values: median $=\frac{6\text{th}+7\text{th}}{2}=\frac{25+28}{2}=26.5$.
3. Difference $=28-26.5=1.5$ years.

Trick: Removing the largest value shifts the median down by half the gap between the two middle values – here $\frac{28-25}{2}=1.5$.

Answer: (B) 1.5 years

Concept: Step down from $x^4+\frac{1}{x^4}$ to $x-\frac{1}{x}$ using ${(x^2+\frac{1}{x^2})}^2=x^4+\frac{1}{x^4}+2$ and ${(x-\frac{1}{x})}^2=x^2+\frac{1}{x^2}-2$.

1. $x^4+\frac{1}{x^4}=322$ → $x^2+\frac{1}{x^2}=\sqrt{324}=18$.
2. ${(x-\frac{1}{x})}^2=18-2=16$ → $x-\frac{1}{x}=4$ (positive as $x\ge 1$).
3. $x^3-\frac{1}{x^3}={(x-\frac{1}{x})}^3+3(x-\frac{1}{x})=64+12=76$.

Trick: Remember the chain $322\to 18\to 4$: each step is "add or subtract 2, then square-root".

Answer: (A) 76

Concept: The ladder, wall and ground form a right triangle; the side opposite the angle at the top (between ladder and wall) is the foot-to-wall distance, so use $\sin$.

1. Angle with the wall $={30}^{\circ }$, hypotenuse (ladder) $=24$ m.
2. Distance from wall $=24\sin {30}^{\circ }=24\times \frac{1}{2}=12$ m.

Trick: Angle with the wall → $\sin$ gives the ground distance; angle with the ground → $\cos$ gives it. Here it is simply half the ladder.

Answer: (A) 12

Concept: CSA of a cone $=\pi rl$; when $r$ and $l$ change by $a\%$ and $b\%$, net change $=a+b+\frac{ab}{100}\%$.

1. $a=+8$, $b=-15$: net $=8-15+\frac{8\times (-15)}{100}=-7-1.2=-8.2\%$.
2. Check: $1.08\times 0.85=0.918$ → a $8.2\%$ decrease.

Trick: For a product of two quantities, use the successive-percentage formula; the sign of $\frac{ab}{100}$ is negative when one rises and the other falls.

Answer: (A) decreases by 8.2%

Concept: Let CP of each article be ₹1 and MP be ₹$m$. The customer pays for 120 articles at the discounted price but receives 123.

1. Cost to trader $=123\times 1=123$; revenue $=120\times 0.8m=96m$.
2. Profit 28% → $96m=1.28\times 123=157.44$ → $m=1.64$.
3. MP exceeds CP by $64\%$.

Trick: $\frac{\text{MP}}{\text{CP}}=\frac{(1+\text{profit})}{(1-\text{discount})}\times \frac{\text{articles given}}{\text{articles paid}}=\frac{1.28}{0.8}\times \frac{123}{120}=1.6\times 1.025=1.64$.

Answer: (A) 64%

Concept: Treat the letters and the number separately; each follows its own fixed jump.

1. First letters: X(24), T(20), P(16), L(12) → each $-4$, next is H(8).
2. Inside each cluster the 2nd letter is first letter $-2$ and the 3rd is first letter $-1$: XVW, TRS, PNO, LJK → HFG.
3. Numbers: 26, 22, 18, 14 → $-4$ each, next $10$. (Also number = first letter position $+2$.)

Trick: In such series the number is usually tied to the first letter – here $26=24+2$, so for H it must be $8+2=10$.

Answer: (D) HFG 10

Concept: Identify the required numbers first, then pick the digits by position.

1. Numbers: 235, 879, 584, 635, 159. Highest $=879$, lowest $=159$.
2. Third digit of 879 $=9$; second digit of 159 $=5$.
3. Sum $=9+5=14$.

Trick: Underline the max and min numbers before reading the digit instruction – it prevents mixing up "second" and "third" digits.

Answer: (D) 14

Concept: Remove every symbol (@, ©, £, %) and count from the left in the remaining series.

1. Original: H 2 @ Q I 3 L © £ R S 4 4 U © 9 1 W Y % 6 £.
2. After dropping symbols: H 2 Q I 3 L R S 4 4 U 9 1 W Y 6.
3. Fifth from the left $=3$.

Trick: Only the first few elements matter for "fifth from left" – scan until you have counted 5 non-symbols: H, 2, Q, I, 3.

Answer: (A) 3

Concept: Apply the rule digit by digit, then read off the largest and smallest new digits.

1. 53764821: odd digits $5,3,7,1$ → $6,4,8,2$; even digits $6,4,8,2$ → $5,3,7,1$.
2. New number: 6 4 8 5 3 7 1 2 → digits $\{6,4,8,5,3,7,1,2\}$.
3. Greatest $=8$, smallest $=1$; product $=8\times 1=8$.

Trick: Odd $+1$ gives even, even $-1$ gives odd – so a 9 becomes 10 only if present (not here), and a 0 would appear only from a 1 under subtraction (not applicable).

Answer: (C) 8

Concept: Look at the differences between consecutive terms – they themselves form a pattern.

1. $80\to 55\to 39\to 30\to 26$: differences $25,16,9,4$.
2. These are $5^2,4^2,3^2,2^2$; next difference $=1^2=1$.
3. $26-1=25$.

Trick: Decreasing differences that are perfect squares are a favourite – write $5^2,4^2,3^2,\dots$ as soon as you see 25, 16, 9.

Answer: (D) 25

Concept: Place the pair with a fixed gap first, test each position against the odd-floor condition, then fit the remaining pair.

1. F is immediately above G; E is 3 floors away from G (two people between) and on an odd floor. Try G = 4, F = 5 → E = 1 (odd ✓). (G = 2 gives E = 5 but then Y–W cannot be consecutive in 1, 4, 6; other G values fail.)
2. Remaining floors 2, 3, 6 for W, X, Y with Y immediately below W → Y = 2, W = 3, X = 6.
3. Between X (6) and W (3) are floors 5 and 4 → two people (F, G).

Trick: With 6 floors, "two people between" the pair forces (1,4), (2,5) or (3,6); check the odd-floor clue against these three only.

Answer: (B) Two

Concept: Check each statement alone; combine only if neither alone pins down the shortest tree.

1. (I): N > V > T and M > T. L is not placed, so the shortest could be T or L – not sufficient.
2. (II): L is taller than exactly one tree → L is 2nd shortest, but who is below L is unknown – not sufficient.
3. Together: L is 2nd shortest and T is below V, N and M, so the only tree below L is T → T is shortest. Sufficient.

Trick: In "who is shortest" questions, a statement is sufficient only if every other member is shown to be taller than one candidate.

Answer: (A) Both statements together are sufficient

Concept: Compare the position jump of each letter from the first cluster to the second.

1. HJ-KM: H→K $+3$, J→M $+3$. MO-PR: $+3,+3$. CE-FH: $+3,+3$.
2. GI-JM: G→J $+3$ but I→M $+4$ – breaks the pattern.

Trick: Write positions quickly (G7 I9 J10 M13): differences 3 and 4 jump out immediately.

Answer: (A) GI-JM

Concept: Each pair is two consecutive letters mapped to two consecutive letters; check the size of the jump.

1. FG-IJ: F→I $+3$, G→J $+3$. CD-FG: $+3,+3$. PQ-ST: $+3,+3$.
2. MN-QR: M→Q $+4$, N→R $+4$ – a different jump.

Trick: Count the letters skipped between the clusters: three pairs skip two letters (H, I / D, E / Q, R), MN-QR skips three (N, O, P).

Answer: (A) MN-QR

Concept: Draw the sets; a conclusion follows only if it is true in every possible diagram.

1. All grapes are lemons; no grape is a melon. Lemons that are not grapes may still be melons → "No lemon is a melon" need not be true. (I) does not follow.
2. No melon is a strawberry says nothing about grapes; a strawberry can be a grape → (II) does not follow.

Trick: Two negative statements (No A is B, No B is C) never give a conclusion about A and C.

Answer: (C) Neither conclusion (I) nor (II) follows

Concept: Facing the centre, a person's left is clockwise and right is anticlockwise. Fix one person and place the rest.

1. Put G at the top; U is immediately right (anticlockwise) of G.
2. Three people between U and E counted from U's left (clockwise): U → G, _, _, E – so E is 4th clockwise from U. Three between G and T from G's left → T is next to E (clockwise).
3. Remaining seats: two between G and E, one between T and U. D immediately right of S → S and D occupy the two seats between G and E (D nearer G, S nearer E); F takes the seat between T and U.
4. From S's left (clockwise): E, T, then F → two people between.

Trick: "Counted from the left of X" fixes the direction – always count clockwise for people facing the centre.

Answer: (B) Two

Concept: Apply the rule digit by digit: odd digit $+1$, even digit $-2$, then read the new number.

1. 6587134 → 6 is even → 4; 5 odd → 6; 8 even → 6; 7 odd → 8; 1 odd → 2; 3 odd → 4; 4 even → 2.
2. New number = 4668242.
3. Second digit from the left = 6.

Trick: Write the original digits in a row and the changed digits directly below them – no digit gets skipped or mixed up.

Answer: (C) 6