Q&A: practice questions with answers

Concept: $\tan \theta =\frac{\text{height}}{\text{base}}$.

1. $\tan {60}^{\circ }=\frac{18}{d}\Rightarrow \sqrt{3}=\frac{18}{d}$
2. $d=\frac{18}{\sqrt{3}}=\frac{18\sqrt{3}}{3}=6\sqrt{3}\text{m}$

Trick: At ${60}^{\circ }$, base $=\frac{\text{height}}{\sqrt{3}}$; at ${30}^{\circ }$, base $=\text{height}\times \sqrt{3}$.

Answer: (A) $6\sqrt{3}$

Concept: Surface area of a sphere $=4\pi r^2$.

1. Sum $=4\pi (2^2+3^2+6^2)=4\pi \times 49$
2. $=4\times \frac{22}{7}\times 49=616{\text{cm}}^2$

Trick: $r^2$ sum $=49$ cancels the 7 in $\pi =\frac{22}{7}$ neatly.

Answer: (A) $616{\text{cm}}^2$

Concept: Convert each scheme into an effective % discount on 5 toys; least beneficial = smallest discount.

1. (A) $1-0.76\times 0.67=1-0.5092\Rightarrow 49.08\%$
2. (B) $1-0.61\times 0.63\times 0.77=1-0.2959\Rightarrow 70.4\%$
3. (C) $\frac{2\times 24+3\times 12}{5}=\frac{84}{5}=16.8\%$
4. (D) pays for 5, gets 7 → $\frac{2}{7}\approx 28.6\%$
5. Smallest discount is scheme C

Trick: Successive discounts multiply the remaining fractions, they never simply add.

Answer: (B) C

Concept: Selling price after successive discounts $=\text{MP}\times (1-d_1)(1-d_2)$.

1. $3000\times 0.9=2700$
2. $2700\times 0.8=2160$

Trick: Single equivalent discount $=10+20-\frac{10\times 20}{100}=28\%$; $72\%$ of $3000=2160$.

Answer: (D) ₹2,160

Concept: $\text{Mode}=l+\frac{f_1-f_0}{2f_1-f_0-f_2}\times h$, using the class with the highest frequency.

1. Modal class $50-55$: $l=50$, $f_1=13$, $f_0=4$, $f_2=5$, $h=5$
2. $\text{Mode}=50+\frac{13-4}{26-4-5}\times 5=50+\frac{45}{17}$
3. $=50+2.647=52.65$

Trick: Mode must lie inside the modal class (50–55) — only 51.34 and 52.65 qualify; formula picks 52.65.

Answer: (C) 52.65

Concept: Position from left $=$ Total $-$ Position from right $+1$. Facing north, left of a person is towards the left end.

1. Keerthi from left $=38-13+1=26$th.
2. Tanvi is 7 places to the left of Keerthi: $26-7=19$th from the left.

Trick: Moving left reduces the position from the left end – just subtract.

Answer: (A) 19^th

Concept: Compare each letter with the first letter using alphabet positions.

1. GHE: G(7), H(8) $=+1$, E(5) $=-2$.
2. LMJ: L(12), M $=+1$, J(10) $=-2$. STQ: S(19), T $=+1$, Q(17) $=-2$.
3. CDB: C(3), D $=+1$, but B(2) $=-1$ – pattern breaks.

Trick: Write the pattern as “$+1$, $-2$” from the first letter; the option that does not fit is odd.

Answer: (A) CDB

Concept: Treat the 1st, 2nd and 3rd letters of the clusters as three separate series.

1. 1st letters: C, D, E, ?, G → $+1$ → F.
2. 2nd letters: D, F, H, ?, L → $+2$ → J.
3. 3rd letters: X, U, R, ?, L → $-3$ → O.

Trick: Check only the letter that differs among options: here the 2nd letter J alone decides.

Answer: (D) FJO

Concept: Count the letters from the given end, then count letters between them in the alphabet.

1. The series has 25 letters. 8th from the left: H U N V Y M L N → N.
2. 5th from the right = $25-5+1=21$st from left: …N Z W M J I N F Q N → I.
3. Between I(9) and N(14) in the alphabet: J, K, L, M → $14-9-1=4$ letters.

Trick: Letters between two letters $=$ difference of positions $-1$.

Answer: (D) Four

Concept: Number positions 1 (bottom) to 7 (top) and fix the definite clues first.

1. Only J above S → J = 7, S = 6. Only three boxes below I → I = 4.
2. Two boxes between I and R → R = 1 (7 is taken).
3. G, H, Q fill 2, 3, 5 with G above H above Q → Q = 2, H = 3, G = 5.
4. Third above Q (2) is position 5 = G.

Trick: “Only X above Y” instantly fixes the top two places.

Answer: (D) G

Concept: Check each letter position and the number separately.

1. 1st letter: R, W, B, G → $+5$ → L.
2. 2nd letter: H, M, R, W → $+5$ → B.
3. 3rd letter: E, C, A, Y → $-2$ → W.
4. Number: 53, 61, 69, 77 → $+8$ → 85.

Trick: Numbers end in 85 for two options; the first letter L settles it.

Answer: (C) LBW 85

Concept: Look at the differences between consecutive terms.

1. Differences: $146-135=11$, $168-146=22$, $201-168=33$, $245-201=44$.
2. Next difference $=55$: $245+55=300$.

Trick: Differences are multiples of 11 ($11\times 1,11\times 2,\dots$).

Answer: (A) 300

Concept: Take A as origin (0, 0); east and north positive. Right turns go W → N → E → S → W.

1. West 16 → $(-16,0)$; right (north) 8 → $(-16,8)$.
2. Right (east) 19 → $(3,8)$; right (south) 14 → $(3,-6)$.
3. Right (west) 3 → $P=(0,-6)$.
4. P is 6 km directly south of A, so he must drive 6 km towards the north.

Trick: Net E–W: $-16+19-3=0$; net N–S: $8-14=-6$ → straight back 6 km north.

Answer: (B) 6 km to the north

Concept: Write the original and rearranged numbers one below the other and compare place by place.

1. Original: 7 6 3 2 1 9 8.
2. Ascending: 1 2 3 6 7 8 9.
3. Compare: 7–1, 6–2, 3–3, 2–6, 1–7, 9–8, 8–9 → only the 3rd digit (3) is unchanged.

Trick: In ascending order a digit stays only if exactly (its rank $-1$) digits are smaller and all of them lie to its left.

Answer: (A) ONE

Concept: Find the relation between the letters of each pair using alphabet positions.

1. KJ-PZ: K→J $-1$, K(11)→P(16) $+5$, J(10)→Z(26) $+16$.
2. FE-KU: $-1$, $+5$, E(5)→U(21) $+16$. PO-UE: $-1$, P→U $+5$, O(15)→E $+16$ (looping past Z).
3. HG-NW: $-1$, but H(8)→N(14) $=+6$ – pattern breaks.

Trick: Check the first letters of the two clusters: only H→N is $+6$, all others $+5$.

Answer: (B) HG – NW

Concept: Relate the 2nd and 3rd letters to the 1st letter.

1. ROS: R(18) → O(15) $=-3$, S(19) $=+1$.
2. MJN: M(13) → J $-3$, N $+1$. LIM: L(12) → I $-3$, M $+1$.
3. JKO: J(10) → K $+1$, O $+5$ – does not follow.

Trick: Pattern “$-3$, $+1$” from the first letter; JKO breaks it.

Answer: (A) JKO

Concept: Split each cluster into 1st, 2nd and 3rd letters.

1. 1st: P, Q, R, S → $+1$ → T.
2. 2nd: C, G, K, O → $+4$ → S.
3. 3rd: O, L, I, F → $-3$ → C.

Trick: The first letter must be T – only one option starts with T.

Answer: (D) TSC

Concept: Decode each symbol from left to right and draw the family tree.

1. V − D: V is the daughter of D. D + Q: D is the sister of Q (D is female).
2. Q × T: Q is the son of T. T ÷ R: T is the husband of R, so R is the mother of Q and D.
3. V is the daughter of D, and D is R’s daughter → V is R’s daughter’s daughter.

Trick: A sister of someone shares the same parents – put D beside Q under T–R.

Answer: (D) Daughter’s daughter

Concept: Find the change in each letter and in the number separately.

1. DI 12 → JE 3: D(4)→J(10) $+6$, I(9)→E(5) $-4$, $12-9=3$.
2. EW 11 → KS 2: E→K $+6$, W(23)→S(19) $-4$, $11-9=2$ ✓.
3. KN 19: K(11)→Q(17), N(14)→J(10), $19-9=10$ → QJ 10.

Trick: Apply “$+6$, $-4$, $-9$” directly; the number 10 alone matches only one option.

Answer: (C) QJ 10

Concept: Find even digits with a perfect square (1, 4 or 9) just before and an odd digit just after.

1. Scan for patterns (1, 4 or 9) – even – odd.
2. Found: 4 8 7, 4 6 3, 9 8 7, 9 8 7, 1 2 5, 9 6 3, 9 8 7, 1 4 5.
3. Total $=8$.

Trick: Search around every 1, 4 and 9 only – each is a possible “preceding” digit.

Answer: (C) 8