Q&A: practice questions with answers

Concept: Multiply by 111 using $111=100+10+1$.

1. $82\times 100=8200$, $82\times 10=820$, $82\times 1=82$.
2. $8200+820+82=9102$.

Trick: For a 2-digit $ab\times 111$: write $a,(a+b),(a+b),b$ with carry → 8, 10, 10, 2 → 9102.

Answer: (C) 9102

Concept: Make Q common in both ratios to combine them.

1. $P:Q=15:4=45:12$ and $Q:R=3:1=12:4$.
2. $P:Q:R=45:12:4$; total parts $=61$.
3. One part $=\frac{12200}{61}=200$; R's share $=4\times 200=800$.

Trick: Total 12,200 is divisible by 61 – a quick check that the ratio is right.

Answer: (B) 800

Concept: $\text{SI}=\frac{P\times R\times T}{100}$.

1. Amount $=3P$, so $\text{SI}=2P$.
2. $2P=\frac{P\times R\times 10}{100}\Rightarrow R=20\%$.

Trick: To become $n$ times in $T$ years: $R=\frac{(n-1)\times 100}{T}=\frac{2\times 100}{10}=20\%$.

Answer: (A) $20\%$

Concept: Write the condition as an equation and find $\frac{a}{b}$.

1. $\frac{20}{100}a=\frac{3}{4}b\Rightarrow \frac{a}{5}=\frac{3b}{4}$.
2. $\frac{a}{b}=\frac{3}{4}\times 5=\frac{15}{4}$.
3. Ratio $a:b=15:4$.

Trick: If $x\%$ of A = $y$ of B, then $A:B=y:\frac{x}{100}$ → $\frac{3}{4}:\frac{1}{5}=15:4$.

Answer: (A) 15 : 4

Concept: $(a+b{)}^2-(a-b{)}^2=4ab$.

1. $(a+b{)}^2={70}^2=4900$.
2. $4ab=4900-208=4692$.
3. $ab=\frac{4692}{4}=1173$.

Trick: $ab=\frac{(a+b{)}^2-(a-b{)}^2}{4}$ – one line.

Answer: (A) 1173

Concept: In an isosceles triangle the two base angles are equal, and the angles add up to ${180}^{\circ }$.

1. Let each base angle be $x$: ${44}^{\circ }+2x={180}^{\circ }$.
2. $2x={136}^{\circ }\Rightarrow x={68}^{\circ }$.

Trick: Base angle $={90}^{\circ }-\frac{\text{vertex}}{2}={90}^{\circ }-{22}^{\circ }={68}^{\circ }$.

Answer: (C) ${68}^{\circ }$ and ${68}^{\circ }$

Concept: Extra selling price corresponds to the extra gain percentage on the cost price.

1. Gain rises from $39\%$ to $64\%$ → difference $=25\%$ of CP.
2. $25\%$ of CP $=₹558$.
3. CP $=558\times 4=₹2232$.

Trick: Difference in gain % × CP = difference in SP.

Answer: (B) 2,232

Concept: Check every symbol: the element just before it must be an odd number and the element just after it must NOT be an even number.

1. & : 7 before, 9 after → ✔
2. ^ : 9 before, # after → ✔
3. * : 1 before, 5 after → ✔ ; ≺ : 5 before but 2 after → ✘
4. $ : 3 before, @ after → ✔ ; others (#, ≡, ∆, @) are not preceded by an odd number.
5. Total = 4, i.e. more than three.

Trick: First list symbols after odd numbers (&, ^, *, ≺, $), then strike out those followed by an even number (≺).

Answer: (D) More than three

Concept: Replace letters by signs (S = +, R = −, Q = ÷) and use BODMAS.

1. $81+28-39+(98÷7)$
2. $=81+28-39+14$
3. $=84$

Trick: Bracket first, then add/subtract left to right.

Answer: (C) $84$

Concept: Place the fixed pair J–L, then test where F (odd floor, two people between F and L) can fit, and finally fit the H–K pair.

1. L = 2, J = 3 → F = 5, but no two consecutive floors are left for H–K ✘. L = 3 or 5 → F would be on 6 or 2 (even) ✘. L = 1 → F = 4 (even) ✘.
2. L = 4, J = 5 → F = 1 (odd) ✔. Free floors 2, 3, 6 → H = 2, K = 3, G = 6.
3. Between G (6) and K (3): J and L → two people.

Trick: Start with the pair whose position is most restricted (J–L with F odd); only one case survives.

Answer: (B) Two

Concept: Write the direct codes, then check which condition applies to the first and last elements.

1. @&238: first @ (symbol), last 8 (even number) → condition (iii) applies; (i) and (ii) need a number first, so they fail.
2. Direct codes: @ = N, & = D, 2 = T, 3 = P, 8 = S.
3. By (iii) first and last both become the code for 3 = P → P D T P P.

Trick: Always check the type of first and last element (number/symbol, even/odd) before coding.

Answer: (C) PDTPP

Concept: Identify the highest and lowest numbers first, then pick the required digits.

1. Highest = 503 → second digit 0.
2. Lowest = 144 → first digit 1.
3. $0+1=1$

Trick: Arrange: 144 < 158 < 325 < 391 < 503.

Answer: (C) $1$

Concept: Find one operation that turns the first number into the second in both given pairs.

1. $72÷8=9$ and $24÷8=3$ → rule: divide by 8
2. $64÷8=8$

Trick: Both pairs share the same ratio 8 : 1 – check division first when the second number is much smaller.

Answer: (B) $8$

Concept: Change each digit as instructed, then add the even digits of the new number.

1. 6597423 → 6+2 = 8, 5−2 = 3, 9−2 = 7, 7−2 = 5, 4+2 = 6, 2+2 = 4, 3−2 = 1
2. New number: 8375641
3. Even digits 8, 6, 4 → sum = 18

Trick: Adding/subtracting 2 keeps parity: only the old even digits (6, 4, 2) stay even → 8 + 6 + 4.

Answer: (D) $18$

Concept: Look at the differences between consecutive terms.

1. 135 → 123 → 109 → 93 → 75: differences −12, −14, −16, −18
2. Next difference −20 → 75 − 20 = 55

Trick: Differences grow by 2 each time.

Answer: (B) $55$

Concept: Find the letter-wise shift in the right pair and apply the same shift on the left.

1. Option (A): KFI → PKN: K+5 = P, F+5 = K, I+5 = N.
2. TOR → YTW: T+5 = Y, O+5 = T, R+5 = W.
3. Both pairs follow +5 on every letter.

Trick: Use place values: K(11)→P(16), T(20)→Y(25) – same +5 shift.

Answer: (A) # = KFI, % = YTW

Concept: Write the alphabet positions and check the gaps between letters (wrap around after Z).

1. GLOS: 7, 12, 15, 19 → +5, +3, +4
2. ZEHL: 26, 5, 8, 12 → +5, +3, +4 ; YDGK: 25, 4, 7, 11 → +5, +3, +4
3. EJNQ: 5, 10, 14, 17 → +5, +4, +3 – different order

Trick: After Z continue A = 27, so Z→E is +5.

Answer: (B) EJNQ

Concept: Seven seats; five people between T and A puts them at the two ends.

1. N is immediately left of A, so A cannot be at the left end → A = 7, N = 6, T = 1.
2. One person between N and L → L = 4.
3. Seats 2, 3, 5 remain for B, M, S with B left of M left of S → B = 2, M = 3, S = 5.
4. Third from the left = M.

Trick: In a row of 7, '5 between' always means both are at the ends.

Answer: (C) M

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

1. 'please go down' → ta wk vb
2. 'down the drain' → cm dg wk
3. Common word 'down', common code 'wk'.

Trick: Common word ↔ common code.

Answer: (D) wk

Concept: Each letter is moved by a fixed number of places in the alphabet.

1. KORJ → DHKC: K(11)→D(4), O(15)→H(8), R(18)→K(11), J(10)→C(3); each letter $-7$.
2. Check IMPH → BFIA: I(9)→B(2), M→F, P→I, H→A; again $-7$.
3. NRUM: N(14)→G(7), R(18)→K(11), U(21)→N(14), M(13)→F(6).

Trick: Find the shift with the first letter only, verify with one more pair, then apply.

Answer: (B) GKNF