Q&A: practice questions with answers

Concept: $\text{Mean}=\frac{\text{Sum of observations}}{\text{Number of observations}}$.

1. Sum $=28+31+40+63+57+37+34+70+99=459$.
2. Number of observations = 9.
3. Mean $=\frac{459}{9}=51$.

Trick: Pair to round numbers: (31 + 99) + (63 + 37) + (70 + 40) + (28 + 57 + 34) = 130 + 100 + 110 + 119 = 459.

Answer: (D) 51

Concept: If $a:b::c:d$ then $ad=bc$; mean proportional $y$ of $p$ and $q$ satisfies $y^2=pq$.

1. $\frac{22+x}{26+x}=\frac{19+x}{21+x}\Rightarrow (22+x)(21+x)=(19+x)(26+x)$.
2. $462+43x=494+45x\Rightarrow 2x=-32\Rightarrow x=-16$ (check: 6, 10, 3, 5 are in proportion).
3. $2x=-32$, $4x-8=-72$; $y^2=(-32)(-72)=2304$.
4. $y>0\Rightarrow y=48$.

Trick: $y=\sqrt{32\times 72}=\sqrt{2304}=48$.

Answer: (A) 48

Concept: $x^3-\frac{1}{x^3}={(x-\frac{1}{x})}^3+3(x-\frac{1}{x})$.

1. Here $x-\frac{1}{x}=4$.
2. $x^3-\frac{1}{x^3}=4^3+3\times 4=64+12$.
3. $=76$.

Trick: If $x-\frac{1}{x}=k$, then $x^3-\frac{1}{x^3}=k^3+3k$.

Answer: (A) 76

Concept: The largest container that exactly measures both quantities = HCF of the quantities.

1. $200=2^3\times 5^2$.
2. $274=2\times 137$ (137 is prime).
3. HCF $=2$ → maximum volume of each container = 2 litres.

Trick: $274-200=74=2\times 37$; HCF must divide 74 and 200 → only 2 works.

Answer: (A) 2

Concept: Simplify, then factorise by splitting the middle term.

1. $4m^2+6m+2=0\Rightarrow 2m^2+3m+1=0$.
2. $2m^2+2m+m+1=0\Rightarrow (2m+1)(m+1)=0$.
3. $m=-\frac{1}{2}$ or $m=-1$.

Trick: All coefficients positive → both roots negative; sum $=-\frac{6}{4}=-\frac{3}{2}$ and product $=\frac{2}{4}=\frac{1}{2}$ fit $-\frac{1}{2},-1$.

Answer: (C) $-\frac{1}{2}$ and $-1$

Concept: $\text{Savings}\%=100\%-\text{Expenditure}\%$.

1. Spends 40% → saves 60% of income.
2. $60\%$ of income = ₹24,000.
3. Income $=\frac{24000\times 100}{60}$ = ₹40,000.

Trick: 60% = 24,000 → 10% = 4,000 → 100% = 40,000.

Answer: (B) 40,000

Concept: Cost price stays the same; $\text{Profit}=\text{SP}-\text{CP}$.

1. Let CP $=C$, initial profit $=p$, so SP $=C+p$.
2. New SP $=2(C+p)$, new profit $=11p$: $2(C+p)-C=11p$.
3. $C+2p=11p\Rightarrow C=9p$.
4. Initial profit $\%=\frac{p}{9p}\times 100=\frac{100}{9}=11\frac{1}{9}\%$.

Trick: Doubling SP adds one more SP $=(C+p)$ to profit: $C+p=10p\Rightarrow C=9p$.

Answer: (C) $11\frac{1}{9}\%$

Concept: $\text{CP}=\frac{\text{MP}}{1+\text{markup}}$, $\text{SP}=\text{MP}\times (1-\text{discount})$.

1. CP $=\frac{3100\times 100}{125}$ = ₹2480.
2. SP $=3100\times \frac{88}{100}$ = ₹2728.
3. Profit $=2728-2480$ = ₹248; Profit $\%=\frac{248}{2480}\times 100=10\%$.

Trick: Net effect $=25-12-\frac{25\times 12}{100}=10\%$ – no need for actual prices.

Answer: (D) 10%

Concept: Check the alphabet gaps between consecutive letters.

1. SUP: S +2 U, U −5 P
2. NPK: N +2 P, P −5 K
3. GID: G +2 I, I −5 D
4. PRV: P +2 R, R +4 V — breaks the pattern

Trick: Write position numbers: S19 U21 P16 → +2, −5.

Answer: (C) PRV

Concept: Fix the certain positions first (top = 1, bottom = 7).

1. Only A above D → A = 1, D = 2; four between D and F → F = 7.
2. C and G (two between) fill 3 & 6; E–B pair takes 4–5.
3. If G = 3, G is immediately above E ✘ → C = 3, E = 4, B = 5, G = 6.
4. Below E: B, G, F = 3 boxes.

Trick: Always place the "only X above" clue first — it fixes the top.

Answer: (D) 3

Concept: Apply the same change to each part.

1. WJ15 → UN11: W −2 = U, J +4 = N, 15 −4 = 11
2. QH8 → OL4: Q −2 = O, H +4 = L, 8 −4 = 4
3. RI3 → R −2 = P, I +4 = M, 3 −4 = −1 → PM −1

Trick: Rule: (−2, +4, −4).

Answer: (D) PM −1

Concept: A word common to two sentences gets the code common to both.

1. ‘poet’ appears in both: joy lab poet / poet win union.
2. Common code in ‘to dl dh’ and ‘dl fo kb’ is dl.

Trick: Common word ↔ common code.

Answer: (A) dl

Concept: Arrange the digits in ascending order and count from the left.

1. Digits of 8142769: 8, 1, 4, 2, 7, 6, 9
2. Ascending: 1 2 4 6 7 8 9
3. Fourth from left = 6

Trick: With 7 digits, 4th from left is the middle (median) digit.

Answer: (A) 6

Concept: Track net east–west and north–south movement (left/right turns change facing).

1. West 12 → left (S) 6 → left (E) 4 → right (S) 2 → right (W) 3 → left (S) 3 → left (E) 11.
2. East–West: $-12+4-3+11=0$
3. North–South: $-6-2-3=-11$ (11 km south of A)
4. So P is 11 km south of A → drive 11 km north.

Trick: Horizontal moves cancel exactly, so the answer is just the total southward distance.

Answer: (C) 11 km north

Concept: Compare each letter of the first cluster with the matching letter of the second.

1. OP → KL: −4, −4
2. IJ → EF: −4, −4
3. MN → IJ: −4, −4
4. LD → KH: −1, +4 — different

Trick: In the group, the first pair is also consecutive letters; LD is not.

Answer: (A) LD - KH

Concept: Find the shift for each letter (cyclic alphabet).

1. DREAM → ZNAWI: every letter −4 (D→Z, A→W wraps around).
2. DANCE → ZWJYA: again −4 each.
3. DOUBT: D→Z, O→K, U→Q, B→X, T→P = ZKQXP

Trick: −4 = move 4 steps back; after A comes Z.

Answer: (D) ZKQXP

Concept: Interchange the signs, then use BODMAS.

1. $56-12\times 4÷9+5$ becomes $56+12÷4\times 9-5$
2. $12÷4=3$, $3\times 9=27$
3. $56+27-5=78$

Trick: Rewrite the whole expression first, then solve — never swap mentally.

Answer: (C) 78

Concept: Track each letter position separately.

1. 1st letters: V T R P N (−2)
2. 2nd letters: Q O M K I (−2)
3. 3rd letters: L J H F D (−2)

Trick: All three positions move −2 → NID.

Answer: (A) NID

Concept: Apply the rule to each digit of 1437586.

1. Odd +1: 1→2, 3→4, 7→8, 5→6
2. Even −2: 4→2, 8→6, 6→4
3. New number: 2 2 4 8 6 6 4
4. Highest − lowest $=8-2=6$

Trick: Only the extreme digits matter: 7 becomes 8 (max), 1 and 4 become 2 (min).

Answer: (A) 6

Concept: Look at the differences.

1. 901 → 900 → 897 → 892 → 885: differences −1, −3, −5, −7
2. Next difference −9
3. $885-9=876$

Trick: Subtracting consecutive odd numbers: $901-n^2$ pattern ($901-25=876$).

Answer: (B) 876