Q&A: practice questions with answers

Concept: Equal amounts at different times: $M(1.1{)}^{16}=A(1.1{)}^{17}$.

1. Cancel $(1.1{)}^{16}$: $M=1.1A$, i.e. Manoj : Anand $=11:10$.
2. Total $=21$ parts $=1407\Rightarrow 1$ part $=67$.
3. Anand's share $=10\times 67=670$.

Trick: If amounts are equal after $n$ and $n+1$ years, the shares are in the ratio $(100+R):100$ – here $110:100=11:10$.

Answer: (C) 670

Concept: If $a=7+4\sqrt{3}$, then $\frac{1}{a}=7-4\sqrt{3}$ (rationalising, since $(7+4\sqrt{3})(7-4\sqrt{3})=49-48=1$), and $a^3+\frac{1}{a^3}={(a+\frac{1}{a})}^3-3(a+\frac{1}{a})$.

1. $a+\frac{1}{a}=(7+4\sqrt{3})+(7-4\sqrt{3})=14$.
2. $a^3+\frac{1}{a^3}={14}^3-3\times 14=2744-42=2702$.

Trick: Whenever $a=p+q\sqrt{r}$ with $p^2-q^{2}r=1$, $a+\frac{1}{a}=2p$ instantly – here $2\times 7=14$.

Answer: (D) $2702$

Concept: Take CP = 100; then MP = 240 and required SP = 168; $\text{Discount}\%=\frac{\text{MP}-\text{SP}}{\text{MP}}\times 100$.

1. MP $=100+140=240$; SP for 68% gain $=168$.
2. Discount $=240-168=72$; $\frac{72}{240}\times 100=30\%$.

Trick: $\frac{\text{SP}}{\text{MP}}=\frac{168}{240}=0.7$ → 70% of MP is kept, so discount = 30%.

Answer: (C) $30\%$

Concept: For two numbers, $\text{HCF}\times \text{LCM}=\text{product of the numbers}$.

1. Other number $=\frac{6\times 72}{18}$.
2. $=\frac{432}{18}=24$.

Trick: Check: HCF(18, 24) = 6 and LCM(18, 24) = 72 ✓ – always verify with the pair.

Answer: (A) 24

Concept: Margin = difference of the two candidates' shares of valid votes.

1. Let registered voters $=N$. Polled $=0.8N$; valid $=0.8\times 0.8N=0.64N$.
2. Winner 55%, loser 45% → margin $=10\%$ of valid $=0.064N$.
3. $0.064N=368\Rightarrow N=\frac{368}{0.064}=5750$.

Trick: Margin % of registered $=80\%\times 80\%\times 10\%=6.4\%$; $368÷0.064=5750$.

Answer: (C) 5750

Concept: $\text{Mean}=\frac{\sum f_i{x}_i}{\sum f_i}$.

1. $\sum f_i{x}_i=30\times 31+40X+55\times 50+95\times 17=930+40X+2750+1615=5295+40X$.
2. $\sum f_i=31+X+50+17=98+X$.
3. $\frac{5295+40X}{98+X}=41\Rightarrow 5295+40X=4018+41X\Rightarrow X=1277$.

Trick: Use deviations from 41: $31(-11)+X(-1)+50(14)+17(54)=0\Rightarrow -341-X+700+918=0\Rightarrow X=1277$.

Answer: (C) 1277

Concept: For $A{x}^2+Bx+C$ with $A<0$ the parabola opens downward, so the vertex gives the maximum: at $x=-\frac{B}{2A}$, value $=\frac{4AC-B^2}{4A}$.

1. Complete the square: $A{x}^2+Bx+C=A{(x+\frac{B}{2A})}^2+C-\frac{B^2}{4A}$.
2. Since $A<0$, the squared term is $\le 0$; the maximum is $C-\frac{B^2}{4A}=\frac{4AC-B^2}{4A}$.

Trick: Extreme value of a quadratic $=-\frac{D}{4A}$ where $D=B^2-4AC$ – maximum if $A<0$, minimum if $A>0$.

Answer: (B) $\frac{4AC-B^2}{4A}$

Concept: Place the fixed person first, then use the 'gap' clue and the consecutive block.

1. C → Friday. Three people between C and F → F is on Monday (Tuesday–Thursday are the three days between).
2. E, D, B are on consecutive days in that order (E, then D, then B). The only three consecutive free days are Tue, Wed, Thu → E = Tuesday, D = Wednesday, B = Thursday.
3. A takes the remaining day, Saturday.

Trick: Fix the definite clue (C = Friday), then slot the three-in-a-row block E–D–B; the answer appears without trial and error.

Answer: (C) Tuesday

Concept: When two words share exactly one letter, the common digit in their codes is the code of that letter.

1. BUST → 3, 5, 1, 9 and LIDS → 6, 8, 9, 4.
2. The only common letter is S; the only common digit is 9.
3. So S = 9.

Trick: Match the common letters with the common digits – no need to decode the whole word.

Answer: (B) 9

Concept: Look at the differences between consecutive terms.

1. $145-141=4$, $141-133=8$, $133-121=12$, $121-105=16$.
2. The differences 4, 8, 12, 16 increase by 4, so the next difference is 20.
3. $105-20=85$.

Trick: When the series falls slowly then faster, check for differences in an arithmetic pattern (multiples of a number).

Answer: (D) 85

Concept: Treat the letters and the number as two separate series.

1. Letters: TRZ → RPX → PNV → NLT: each letter moves 2 steps back (T→R→P→N, R→P→N→L, Z→X→V→T). Next: N−2 = L, L−2 = J, T−2 = R → LJR.
2. Numbers: 2, 5, 8, 11 → +3 each → next 14.
3. Required cluster: LJR 14.

Trick: Note that the first letter of each cluster is the second letter of the previous one (T R Z → R P X → P N V…) – a quick confirmation that letters shift by −2.

Answer: (D) LJR 14

Concept: Letters common to both words get the digits common to both codes; the leftover letter gets the leftover digit.

1. BITE → 4, 5, 6, 8 and TIME → 4, 8, 2, 6.
2. Common letters: I, T, E → common digits 4, 6, 8.
3. The letter only in TIME is M; the digit only in its code is 2. So M = 2.

Trick: Elimination: strike out the shared digits (4, 6, 8) from TIME's code – what remains (2) is M.

Answer: (C) 2

Concept: Build the family tree generation by generation, then read the relation from E's side.

1. B is the mother of C and D (C is D's brother). A is B's sister → A is the maternal aunt of D.
2. E is D's husband → D is E's wife. A is the sister of E's wife's mother.
3. So A is E's wife's mother's sister.

Trick: Write 'E's wife = D' first; whatever A is to D, prefix it with 'wife's'. A = D's mother's sister → wife's mother's sister.

Answer: (B) Wife's mother's sister

Concept: Fix the couple first: Kc's mother is Ef and Kc's father is Cw, so Ef and Cw are husband and wife.

1. Qk is Ef's brother and Lu is Qk's father → Lu is also Ef's father.
2. Ef is Cw's wife, so Lu is Cw's wife's father (father-in-law).

Trick: Whenever two persons are mother and father of the same child, link them as a couple at once – the chain shortens immediately.

Answer: (A) Wife's father

Concept: Scan each symbol and check both neighbours: letter on the left AND letter on the right.

1. Series: K { J % I + H = G * F / D : - ! B C ? A #.
2. Qualifying symbols: { (K…J), % (J…I), + (I…H), = (H…G), * (G…F), / (F…D), ? (C…A) – 7 symbols.
3. Rejected: ':' and '!' are next to '-', '-' is between two symbols, and '#' has nothing on its right.

Trick: Underline every symbol, then tick only those sandwiched between two letters; symbols adjacent to another symbol or at the end are out.

Answer: (A) SEVEN

Concept: Start from the fixed floor and use the 'only n between' clues; six floors, 1 at the bottom.

1. D = 1. Only one person between D and C → C = 3. Only two between O and C → O = 6 (floor 0 does not exist).
2. Q is below C with two people between Q and P: Q = 2 and P = 5 (Q = 1 is taken by D).
3. The last person N takes floor 4. Above N are P (5) and O (6) → two people.

Trick: Build the tower bottom-up: 1-D, 2-Q, 3-C, 4-N, 5-P, 6-O. 'Above N' = floors higher than 4.

Answer: (A) Two

Concept: Find one operation on the whole number that works for both given pairs.

1. $63\times 3=189$, $189-30=159$ ✓
2. $44\times 3=132$, $132-30=102$ ✓ – rule: $3n-30$.
3. $57\times 3-30=171-30=141$.

Trick: Check the difference of the pairs: $159-102=57$ and $63-44=19$; $57=3\times 19$, so the multiplier is 3, then adjust the constant.

Answer: (C) 141

Concept: Find the shift from the right-hand pair where both terms are partly known, then apply it (in reverse) on the left.

1. Right side XHP : %. Testing option letters: X(24)+6 = 30 → D(4), H(8)+6 = N(14), P(16)+6 = V(22) → % = DNV (each letter +6, cyclic).
2. So the rule is 'first + 6 = second'. On the left, # + 6 = KUC → # = K−6 = E, U−6 = O, C−6 = W (3−6 = −3 → 23 = W).
3. # = EOW, % = DNV.

Trick: Only option (D) has both EOW and DNV; options with DNO or FOW break the uniform +6 shift.

Answer: (D) # = EOW % = DNV

Concept: Compare letter positions in the given pair to find the position-wise shift.

1. RILA → SKOE: R+1 = S, I+2 = K, L+3 = O, A+4 = E. Pattern +1, +2, +3, +4.
2. Check: KEPO → K+1 = L, E+2 = G, P+3 = S, O+4 = S → LGSS ✓
3. MJRT → M+1 = N, J+2 = L, R+3 = U, T+4 = X → NLUX.

Trick: When the first letters differ by 1 and the last by 4, suspect the +1, +2, +3, +4 staircase.

Answer: (B) NLUX

Concept: Facing the centre: 'left' means clockwise and 'right' means anticlockwise. Start with the clue linking two fixed people.

1. F is to the immediate right of C. C is third to the left of G, so G is third to the right of C. F is third to the right of E → E is third to the left of F.
2. Going clockwise from C: C, _, E, _, G, _, F (E is 3 places left of F, G is 3 right of C, F just right of C).
3. Only one person between D and B counted from the left of B → D is 2nd to the left of B. The empty seats are the 1st, 3rd and 5th clockwise from C; A is not next to C, so A is at the 5th seat (between G and F), B at the 1st and D at the 3rd.
4. From A's left (clockwise): F, then C → one person between A and C.

Trick: Draw 7 dots, mark C, and put F anticlockwise of C at once; the rest of the clues fill in without any trial.

Answer: (B) 1