Q&A: practice questions with answers

Concept: Fourth proportional $x$ of $a,b,c$: $a:b=c:x\Rightarrow x=\frac{b\times c}{a}$.

1. $24:8=9:x$.
2. $x=\frac{8\times 9}{24}=3$.

Yaad rakhein: Product of extremes = product of means: $24\times x=8\times 9$.

Answer: (C) 3

Concept: $\text{SI}=\frac{P\times R\times T}{100}$, with rate and time in the same unit (here per month and months).

1. $P=800$, $R=7\%$ per month, $T=9$ months.
2. $\text{SI}=\frac{800\times 7\times 9}{100}=8\times 63=504$.

Trick: $1\%$ of 800 $=8$; total rate $=7\times 9=63\%$, so SI $=8\times 63=504$.

Answer: (D) ₹504

Concept: $\text{SP}=\text{CP}\times \frac{100+\text{gain}\%}{100}$ and $\text{SP}=\text{MP}\times \frac{100-\text{discount}\%}{100}$.

1. $\text{SP}=136\times \frac{120}{100}=163.2$.
2. $\text{MP}\times \frac{85}{100}=163.2\Rightarrow \text{MP}=\frac{163.2\times 100}{85}=192$.

Trick: $\text{MP}=136\times \frac{120}{85}=136\times \frac{24}{17}=8\times 24=192$.

Answer: (D) ₹192

Concept: Mode is the observation that occurs the maximum number of times.

1. Count: 6 appears 4 times; every other value appears only once.
2. So mode $=6$.

Yaad rakhein: Mode = most frequent value; mean = average; median = middle value.

Answer: (A) 6

Concept: Items of a product $=\frac{\text{its }\%}{100}\times \text{total items}$.

1. From the chart: Raincoats $=13\%$, Jackets $=8\%$.
2. Together $=21\%$ of 1000 $=210$.

Trick: Total is 1000, so $1\%=10$ items; $21\%=210$.

Answer: (D) 210

Concept: Sum of the four angles of a quadrilateral $={360}^{\circ }$.

1. $3x+5x+4x+6x={360}^{\circ }\Rightarrow 18x={360}^{\circ }\Rightarrow x={20}^{\circ }$.
2. $A={60}^{\circ }$, $B={100}^{\circ }$.
3. $3A+2B={180}^{\circ }+{200}^{\circ }={380}^{\circ }$.

Trick: $3A+2B=(3\times 3+2\times 5)x=19x=19\times {20}^{\circ }={380}^{\circ }$.

Answer: (C) ${380}^{\circ }$

Concept: Compare words in pairs: a letter common to two words gets the digit common to their codes.

1. PLATE = 25693, MASK = 3417: common letter A, common digit 3 $\Rightarrow$ A = 3.
2. MASK, KEPT: common K, common digit 4 $\Rightarrow$ K = 4.
3. PLATE, KEPT: common P, T, E $\leftrightarrow$ 6, 9, 5.
4. In PLATE the digits left are 2, 5, 6, 9, 3; removing A (3) and P, T, E (6, 9, 5) leaves L = 2.

Trick: Eliminate what is already decoded; the leftover digit belongs to the leftover letter.

Answer: (A) 2

Concept: Read each position separately across the terms.

1. 1st letters: T, S, R, Q $\to$ P (each $-1$).
2. 2nd letters: W, V, U, T $\to$ S ($-1$).
3. 3rd letters: P, O, N, M $\to$ L ($-1$).

Trick: Every letter moves one step back: QTM $\to$ PSL.

Answer: (B) PSL

Concept: Use alphabet positions and check the gaps inside each cluster.

1. LG-IM: L(12)$-5=$G(7); I(9)$+4=$M(13).
2. JE-GK: J(10)$-5=$E(5); G(7)$+4=$K(11). SN-PT: S(19)$-5=$N(14); P(16)$+4=$T(20).
3. RM-OT: R(18)$-5=$M(13) but O(15)$+5=$T(20) — the second gap is $+5$, not $+4$.

Trick: Also first letters go back by 3 (L$\to$I, J$\to$G, S$\to$P, R$\to$O) but second letters go $+6$ except M$\to$T ($+7$).

Answer: (A) RM-OT

Concept: Track each position of the three-letter groups.

1. 1st letters: F, H, J, L $\to$ N ($+2$).
2. 2nd letters: R, T, V, X $\to$ Z ($+2$).
3. 3rd letters: E, D, C, B $\to$ A ($-1$).

Yaad rakhein: Position values: F6, H8, J10, L12, N14 — EJOTY (5, 10, 15, 20, 25) helps count fast.

Answer: (A) NZA

Concept: Swap the signs first, then solve using BODMAS (÷, × before +, −).

1. After swapping: $10÷5+13-4\times 3$.
2. $=2+13-12$.
3. $=3$.

Yaad rakhein: Interchange first, then BODMAS — never solve the original expression.

Answer: (B) 3

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

1. Fix F. K is third to the right of F (two between).
2. B is immediately left of G, so B–G occupy two adjacent seats; three people (B, G, K) sit between F and D on F's right side, so D is third to the left of F.
3. Only two between K and C places C to the immediate left of F; E then sits between C and D, and G, B fill the seats between K and F.
4. Clockwise: F, C, E, D, K, G, B. Third to the left of E: D, K, G.

Trick: Draw K first (it is fixed from F), then test D's two possible seats — only one lets B, G sit together.

Answer: (A) G

Concept: Find the shift of each letter from the first cluster to the second.

1. IEMJ $\to$ GCKH: I$-2=$G, E$-2=$C, M$-2=$K, J$-2=$H.
2. MIQN $\to$ KGOL: again $-2$ each.
3. PLTQ $\to$ P$-2=$N, L$-2=$J, T$-2=$R, Q$-2=$O $=$ NJRO.

Trick: Same shift for every letter: just move each letter 2 steps back.

Answer: (A) NJRO

Concept: Number positions 1 (bottom) to 7 (top) and place the fixed boxes first.

1. F = 2. Three boxes between F and E $\Rightarrow$ E = 6. D below F $\Rightarrow$ D = 1.
2. Only A between E and C $\Rightarrow$ C = 4, A = 5. B above E $\Rightarrow$ B = 7; G = 3.
3. Between E (6) and D (1): A, C, G, F = 4 boxes.

Trick: Boxes between positions $p$ and $q$ $=|p-q|-1=6-1-1=4$.

Answer: (C) 4

Concept: Check the differences between consecutive terms.

1. Differences: $29,33,37,41$ — increasing by 4.
2. Next difference $=41+4=45$.
3. Next term $=491+45=536$.

Trick: When first differences grow steadily, the second difference (here 4) is constant.

Answer: (A) 536

Concept: Replace letters with signs (A $=+$, B $=\times$, C $=÷$, D $=-$), then apply BODMAS.

1. $14+2\times 6-20÷5$.
2. $=14+12-4$.
3. $=22$.

Yaad rakhein: Division and multiplication first, then addition and subtraction from left to right.

Answer: (D) 22

Concept: Test each option: the same rule must turn X into 65 and 15 into Y.

1. Option (C): X = 12, Y = 80.
2. $12\times 5+5=65$ ✓ and $15\times 5+5=80$ ✓.
3. Others fail, e.g. (A): $19\times 3+8=65$ but $15\times 3+8=53\ne 85$.

Trick: Equivalently the rule is $(n+1)\times 5$: $13\times 5=65$, $16\times 5=80$.

Answer: (C) X = 12, Y = 80

Concept: Count the position directly; from the right, start with the last digit.

1. 1st digit from the left $=7$.
2. From the right: 3, 0, 3, 6, 9, 4, 5 — 7th from the right $=5$.
3. Sum $=7+5=12$.

Trick: $k$th from right in a 30-digit series $=(30-k+1)$th from left $=24$th from left $=5$.

Answer: (C) 12

Concept: Count positions carefully from the required end.

1. From the left: 2, 9, 7, 2, 1, 5, 9, 6, 8, 6, 0, 1 — 12th $=1$.
2. From the right: 5, 1, 7, 0, 5, 2, 0 — 7th $=0$.
3. Sum $=1+0=1$.

Trick: Mark counted digits in groups of five to avoid miscounting.

Answer: (D) 1

Concept: A course of action follows if it is practical and directly tackles the problem stated.

1. Problem 1: no concrete action plan. Redefining top-level objectives to make a feasible plan (I) addresses this — follows.
2. Problem 2: a dozen regular programmes not run. A report on the factors stopping them (II) helps fix the cause — follows.

Yaad rakhein: Action must solve or reduce the problem without being extreme; both steps here are reasonable.

Answer: (A) Both I and II follow