Q&A: practice questions with answers

Concept: Compute the fraction of marked price actually paid for 10 toys; the scheme with the highest payment is least beneficial.

1. (A) $0.56\times 0.88=0.4928\to$ pays $49.3\%$
2. (B) $0.82\times 0.57\times 0.77\approx 0.36\to$ pays $36\%$
3. (C) $\frac{4\times 0.92+6\times 0.62}{10}=\frac{3.68+3.72}{10}=0.74\to$ pays $74\%$
4. (D) pays for 7 of 10 $\to 70\%$

Trick: Only (C) and (D) are close; $74\%>70\%$, so (C) costs the most.

Answer: (C) C

Concept: Decrease in average $=\frac{\text{total decrease}}{k}$.

1. Total decrease $=48\times 2=96$
2. New average $=x-12\Rightarrow \frac{96}{k}=12$
3. $k=8$

Trick: Average falls by 12 while sum falls by 96 $\Rightarrow 96÷12=8$ numbers.

Answer: (A) 8

Concept: In equal time, distance $\propto$ speed.

1. Faster car: $140$ km corresponds to ratio part 7, so one part $=20$ km
2. Slower car: $5\times 20=100$ km

Trick: Distances in the same time are in the ratio $5:7$ directly.

Answer: (C) 100 km

Concept: Take tank capacity $=$ LCM of times as units; add inlet rates, subtract the leak.

1. Capacity $=\text{LCM}(18,9,27)=54$ units
2. M $=3$/h, N $=6$/h, leak $=-2$/h; net $=3+6-2=7$/h
3. Time $=\frac{54}{7}=7\frac{5}{7}$ hours

Trick: Without the leak it takes 6 h; the leak slows it, so the answer must exceed 6 h – rules out nothing but confirms direction.

Answer: (B) $7\frac{5}{7}$ hours

Concept: Diagonals of a rhombus bisect each other at right angles; side $=\sqrt{{\left(\frac{d_1}{2}\right)}^2+{\left(\frac{d_2}{2}\right)}^2}$.

1. Half-diagonals $=112$ m and $210$ m
2. Side $=\sqrt{{112}^2+{210}^2}=\sqrt{12544+44100}=\sqrt{56644}=238$ m
3. Perimeter $=4\times 238=952$ m

Trick: $112:210=8:15$, a Pythagorean triplet $(8,15,17)$ scaled by 14 $\Rightarrow$ side $=17\times 14=238$.

Answer: (A) 952

Concept: In a three-circle Venn diagram, "Teachers who are also Dentists but not Surgeons" is the part of the Teacher–Dentist overlap that lies outside the Surgeons circle.

1. The Teachers and Dentists circles overlap in two regions: one inside Surgeons (23) and one outside Surgeons (29).
2. We need only the region outside the Surgeons circle → 29.
3. 14 = only Teachers, 16 = Teachers and Surgeons only, 5 = only Surgeons – none of these fit the condition.

Trick: "A and B but not C" → take the A∩B lens and drop the central (all-three) number.

Answer: (C) 29

Concept: Write the alphabet positions and check how each letter of a cluster moves to the next cluster.

1. 1st letters: M(13) → J(10) → G(7) → D(4) → ? : each time $-3$, so next = A(1).
2. 2nd letters: F(6) → C(3) → Z(26) → W(23) → ? : $-3$ each, so next = T(20).
3. 3rd letters: J(10) → G(7) → D(4) → A(1) → ? : $-3$, so next = X(24).
4. Required cluster = ATX.

Trick: All three letters fall by 3 – check only one letter of each option: the first letter must be A and the last must be X.

Answer: (C) ATX

Concept: Rearrange the digits in increasing order and then count positions from both ends.

1. Digits of 52178436 in ascending order: 1 2 3 4 5 6 7 8.
2. Third from the left = 3; third from the right = 6.
3. Sum $=3+6=9$.

Trick: The eight digits here are exactly 1–8, so the ascending number is simply 12345678 – no sorting effort needed.

Answer: (C) 9

Concept: Sort the digits in ascending order, then read the required positions.

1. Digits of 36857219 in ascending order: 1 2 3 5 6 7 8 9 (note: no 4).
2. Second from the left = 2; third from the right = 7.
3. Sum $=2+7=9$.

Trick: Write the sorted digits in a line and count from the right by pointing with a finger – most errors come from miscounting "third from the right".

Answer: (B) 9

Concept: Check every D in the series: the letter just before it must be a vowel (A, E, I, O, U) and the letter just after it must be a consonant.

1. Series: K D D U R V U D E C Q D V A D R D G D E D Z D J.
2. D's with a vowel before them: UDE (after it is E, a vowel – rejected), ADR (R is a consonant – accepted), EDZ (Z is a consonant – accepted).
3. All other D's have a consonant before them (K, D, Q, R, G, Z) – rejected.
4. Count = 2.

Trick: First underline the vowels (U, U, E, A, E) – only the D's right after a vowel need checking.

Answer: (B) Two

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

1. DIP → GLS: D+3 = G, I+3 = L, P+3 = S. BST → EVW: B+3 = E, S+3 = V, T+3 = W. Rule: every letter $+3$.
2. HKD → KNG: H+3 = K, K+3 = N, D+3 = G ✓.
3. TEL → VHP is (+2, +3, +4); CIG → FKJ is (+3, +2, +3); EHL → IFP is (+4, −2, +4) – none follow the rule.

Trick: Test only the first letter of each option for "+3": T→V (+2 ✗), H→K ✓, C→F ✓, E→I ✗; then check the second letter of the two survivors.

Answer: (B) HKD : KNG

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

1. S – D: S is the brother of D. D + F: D is the son of F. So S is also a son of F.
2. F ÷ H: F is the father of H. H × K: H is the wife of K, so H is female and K is her husband.
3. S and H are children of the same father F, so S is H's brother; H is K's wife.
4. Hence S is K's wife's brother.

Trick: Draw one generation line for all children of F (S, D, H); the person at the far end (K) is only connected by marriage, so the answer must contain "wife's".

Answer: (C) Wife's brother

Concept: Find the differences between consecutive terms; they often form a simple pattern.

1. $70-67=3$, $97-85=12$, $112-97=15$ → differences are multiples of 3: 3, 6, 9, 12, 15.
2. So ? $=70+6=76$, and check: $76+9=85$ ✓.

Trick: When the last two differences are 12 and 15, go backwards in steps of 3 (9, 6, 3) to fill the gap.

Answer: (C) 76

Concept: Apply the rule letter by letter: vowel → next letter, consonant → previous letter.

1. C U R R E N T → C−1 = B, U+1 = V, R−1 = Q, R−1 = Q, E+1 = F, N−1 = M, T−1 = S.
2. New group: B V Q Q F M S.
3. Third from the right: S (1st), M (2nd), F (3rd).

Trick: Only the vowels U and E move forward; every other letter steps back one – write the new letters directly under the word.

Answer: (C) F

Concept: Find one operation that links the three numbers of each given set.

1. (2, 7, 14): $2\times 7=14$. (3, 16, 48): $3\times 16=48$. Rule: third number = product of the first two.
2. (5, 3, 15): $5\times 3=15$ ✓.
3. (4, 6, 20): $24\ne 20$; (8, 6, 56): $48\ne 56$; (6, 7, 48): $42\ne 48$.

Trick: When the third number is clearly bigger, test multiplication of the first two before anything else.

Answer: (B) (5, 3, 15)

Concept: In a row of $n$ people, position from the right $=n-(\text{position from left})+1$.

1. Sunil is 17th from the left. Deepak is 3rd to his right → Deepak is $17+3=20$th from the left.
2. Position from the right $=22-20+1=3$rd.

Trick: Left position + Right position $=n+1$; so $20+R=23\Rightarrow R=3$.

Answer: (C) 3rd

Concept: Convert every position to "from the left" first, then move along the row.

1. Gagan is 13th from the right in a row of 54 → from the left he is $54-13+1=42$nd.
2. Nitin is 17th to the left of Gagan → $42-17=25$th from the left.

Trick: "To the left" means subtract from the left-position; "to the right" means add.

Answer: (B) 25th

Concept: Perfect-square digits are 1, 4 and 9. We need an even digit whose left neighbour is 1, 4 or 9 and whose right neighbour is odd.

1. Series: 4 6 2 9 1 4 9 7 1 7 6 5 3 6 4 1 4 3 8 4 2 2 3 9 9 7 6.
2. Even digits with 1, 4 or 9 before them: 4-6-2 (next 2 is even ✗), 1-4-9 (9 odd ✓), 1-4-3 (3 odd ✓), 4-2-2 (next 2 even ✗).
3. Count = 2.

Trick: Circle every 1, 4, 9 first; only the digit immediately after each circle needs to be tested.

Answer: (B) Two

Concept: Compare the positions of corresponding letters in the two given pairs to find the shift.

1. JDHK → OIMP: J(10)+5 = O(15), D(4)+5 = I(9), H(8)+5 = M(13), K(11)+5 = P(16). Rule: $+5$ to every letter.
2. MGKN → RLPS: M+5 = R, G+5 = L, K+5 = P, N+5 = S ✓ (same rule).
3. PJNQ → P+5 = U, J+5 = O, N+5 = S, Q+5 = V → UOSV.

Trick: Options differ only in the order of the same letters – keep the letters in sequence (U, O, S, V) and don't let the look-alike options confuse you.

Answer: (C) UOSV

Concept: Find the position relation inside each pair and see which pair breaks it.

1. LO–QJ: L+3 = O, O+2 = Q, and J = L−2. OR–TM: O+3 = R, R+2 = T, M = O−2. HK–MF: H+3 = K, K+2 = M, F = H−2.
2. Pattern: 2nd = 1st+3, 3rd = 2nd+2, 4th = 1st−2.
3. EH–JB: E+3 = H ✓, H+2 = J ✓, but B = E−3 (should be C = E−2) ✗.

Trick: When the first three letters match the pattern in all options, check only the last letter: L→J, O→M, H→F are all "−2", E→B is "−3".

Answer: (A) EH–JB