Q&A: practice questions with answers

Concept: Equal amounts after different periods: $P_1(1+r{)}^{16}=P_2(1+r{)}^{17}\Rightarrow \frac{P_1}{P_2}=(1+r{)}^{17-16}$.

1. $\frac{\text{Akshay}}{\text{Arjun}}=1.05=\frac{21}{20}$.
2. Total parts $=21+20=41$; one part $=\frac{1394}{41}=34$.
3. Arjun $=20\times 34=$ ₹680 (Akshay $=21\times 34=$ ₹714).

Trick: Difference of years $=1$ → shares are in ratio $(100+r):100=21:20$; the one invested longer (Arjun) gets the smaller share.

Answer: (A) 680

Concept: Compare cost and selling price of one article; $\text{Profit}\%=\frac{SP-CP}{CP}\times 100$.

1. CP per article $=\frac{300}{12}=$ ₹25.
2. SP per article $=$ ₹90 → profit $=90-25=$ ₹65.
3. Profit% $=\frac{65}{25}\times 100=260\%$.

Trick: The "48 dozens" is a distraction – percentage profit is the same for one article as for all.

Answer: (A) 260%

Concept: Assumed-mean method: $\overline{x}=a+\frac{\sum f_i{d}_i}{\sum f_i}$ with $d_i=x_i-a$. If $\overline{x}=a$, then $\sum f_i{d}_i=0$.

1. Mid-points: 15, 25, 35, 45, 55. Take $a=35$ (= given mean), so $d_i=-20,-10,0,10,20$.
2. $\sum f_i{d}_i=15(-20)+21(-10)+43(0)+41(10)+X(20)=-300-210+410+20X=20X-100$.
3. Mean $=35$ means $20X-100=0\Rightarrow X=5$.

Trick: When the given mean equals a mid-point, deviations about it must sum to zero – no division needed.

Answer: (D) 5

Concept: For two numbers, $\text{HCF}\times \text{LCM}=a\times b$.

1. $12\times 840=84\times b$.
2. $b=\frac{12\times 840}{84}=12\times 10=120$.
3. Check: HCF(84, 120) = 12, LCM = 840 ✓.

Trick: $\frac{840}{84}=10$, so the other number is simply $\text{HCF}\times 10=120$.

Answer: (C) 120

Concept: If A is $x\%$ more than B, then B is $\frac{x}{100+x}\times 100\%$ less than A.

1. Let B $=100$, so A $=213$.
2. B is less than A by $213-100=113$ on a base of 213.
3. $\frac{113}{213}\times 100=53.05\%$.

Trick: "More" → base is the smaller; "less" → base is the bigger. Never reuse 113% as the answer.

Answer: (B) 53.05%

Concept: The first $n$ multiples of $k$ form an AP; average $=k\times \frac{n+1}{2}$ (middle term).

1. Multiples: 7, 14, …, $7\times 350$.
2. Average $=\frac{7+2450}{2}=\frac{2457}{2}=1228.5$.
3. Or $7\times \frac{351}{2}=7\times 175.5=1228.5$.

Trick: Average of first $n$ multiples of $k$ = $k\times \frac{n+1}{2}$; for even $n$ it always ends in .5 × k.

Answer: (B) 1228.5

Concept: Same distance both ways: $(b+s)\times t_1=(b-s)\times t_2$.

1. $(b+4)\times 6=(b-4)\times 8$.
2. $6b+24=8b-32\Rightarrow 2b=56\Rightarrow b=28$.
3. Check: downstream $32\times 6=192$ km, upstream $24\times 8=192$ km ✓.

Trick: Time ratio $6:8=3:4$ → speed ratio down:up $=4:3$. Difference (1 part) $=2s=8$ → part $=8$; boat $=\frac{4+3}{2}\times 8=28$ km/h.

Answer: (B) 28 km/hr

Concept: Expenditure = Income − Savings; set up the ratio of expenditures.

1. Incomes $6x$ and $7x$; savings ₹66,000 each.
2. $\frac{6x-66000}{7x-66000}=\frac{1}{3}\Rightarrow 18x-198000=7x-66000$.
3. $11x=132000\Rightarrow x=12000$.
4. Vipul's income $=6x=$ ₹72,000.

Trick: Check: expenditures $72000-66000=6000$ and $84000-66000=18000$ → $1:3$ ✓.

Answer: (B) 72,000

Concept: $\overline{x}=\frac{\sum fx}{\sum f}$, ${\sigma }^2=\frac{\sum f{x}^2}{\sum f}-{\overline{x}}^2$, $\sigma =\sqrt{{\sigma }^2}$; SD is always smaller than variance when variance $>1$.

1. $\sum fx=84+104+108+110=406$; mean $=\frac{406}{48}\approx 8.46$ – the only value near 8.5 is item 1, so i → 1.
2. $\sum f{x}^2=588+832+972+1100=3492$; $\frac{3492}{48}=72.75$; variance $=72.75-{8.458}^2\approx 72.75-71.54=1.207$ → iii → 2.
3. SD $=\sqrt{1.207}\approx 1.099$ → ii → 3.

Trick: Of the two small numbers, variance is the larger (1.207) and SD its square root (1.099) – that alone fixes ii and iii.

Answer: (C) i - 1, ii - 3 and iii - 2

Concept: Angle sum of any polygon $=(n-2)\times {180}^{\circ }$; a trapezium has $n=4$.

1. A diagonal splits the trapezium into two triangles.
2. Each triangle has angle sum ${180}^{\circ }$ → total $=2\times {180}^{\circ }={360}^{\circ }$.
3. Equivalently $(4-2)\times {180}^{\circ }={360}^{\circ }$.

Trick: Every quadrilateral – square, kite, trapezium, any shape – has interior angles summing to ${360}^{\circ }$.

Answer: (C) ${360}^{\circ }$

Concept: Decode each symbol step by step and build the family tree, starting from the end (E) to reach D.

1. D × N → D is the wife of N. N % P → N is the father of P, so D is P's mother.
2. P + K → P is the mother of K. K # E → K is the daughter of E, so E is K's father and P's husband.
3. D is the mother of P, and P is E's wife → D is E's wife's mother.

Trick: Whenever one person is "mother of X" and another is "father of X", they are husband and wife – that single link solves most coded-relation questions.

Answer: (C) Wife's mother

Concept: Write the place values (A=1 … Z=26) and compare the gaps between consecutive letters.

1. LPU: 12, 16, 21 → +4, +5. WAF: 23, 1, 6 → +4 (W→A cyclic), +5. DHM: 4, 8, 13 → +4, +5.
2. JNT: 10, 14, 20 → +4, +6 – the second gap breaks the pattern.

Trick: Check the first gap of all four; if all match, the odd one hides in the second gap. Remember that after Z the alphabet wraps to A.

Answer: (A) JNT

Concept: Scan each symbol and check the element just before it (must be a letter) and just after it (must be a number).

1. Symbols in the series: & (G, R), % (K, 7), @ (7, 2), & (S, 6), # (T, 3), * (Y, D).
2. Letter before and number after: K % 7, S & 6, T # 3 → three.

Trick: Underline only the symbols first (there are just six), then test neighbours – far faster than reading the whole series.

Answer: (C) Three

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

1. Letters: each position moves +5. T→Y→D→I→N; W→B→G→L→Q; U→Z→E→J→O.
2. Numbers: 89, 81, 73, 65 → each −8 → 57.
3. Next term: NQO 57.

Trick: Decide the number first (57 → options C or D), then check just the first letter: I+5 = N in both, so check the second letter L+5 = Q → NQO.

Answer: (D) NQO 57

Concept: Compare the two words: letters common to both must share the common digits; the leftover letter gets the leftover digit.

1. SEMI → 3615, TIME → 1523. Common letters: E, M, I. Common digits: 1, 5, 3.
2. So in SEMI the uncommon letter S ↔ uncommon digit 6; in TIME the uncommon letter T ↔ uncommon digit 2.

Trick: In such questions the codes are jumbled – never match by position, always by "common vs uncommon" elimination.

Answer: (D) 2

Concept: Find one rule connecting the three numbers of each given set, then test it on the options.

1. (19, 24, 7): $2\times (19-7)=24$ ✓. (12, 14, 5): $2\times (12-5)=14$ ✓. Rule: second = 2 × (first − third).
2. (3, 4, 1): $2\times (3-1)=4$ ✓. Others fail: (7,5,3) → 8; (12,5,3) → 18; (12,5,4) → 16.

Trick: When the middle number is the largest, try "middle = (first ± third) × k" – here $2\times (a-c)$ fits both sets at once.

Answer: (C) (3, 4, 1)

Concept: Find the position-wise shift from the first word to the second and apply the same shift.

1. DMKZ → EKNV: D+1=E, M−2=K, K+3=N, Z−4=V. Pattern: +1, −2, +3, −4.
2. Check LOJT → MMMP: L+1=M, O−2=M, J+3=M, T−4=P ✓.
3. SRLQ: S+1=T, R−2=P, L+3=O, Q−4=M → TPOM.

Trick: Alternating signs with increasing magnitude (+1, −2, +3, −4) is a very common RRB pattern – verify with the second pair, then apply.

Answer: (C) TPOM

Concept: Change each digit as instructed, then pick the greatest and smallest digits of the new number.

1. 48572613 → 4−1=3, 8−1=7, 5+1=6, 7+1=8, 2−1=1, 6−1=5, 1+1=2, 3+1=4 → new number 37681524.
2. Greatest digit 8, smallest digit 1 → product $8\times 1=8$.

Trick: Odd +1 and even −1 just swaps each digit with its neighbour pair (1↔2, 3↔4, 5↔6, 7↔8); only 9 would become 10 and 0 would become −1 – neither occurs here.

Answer: (A) 8

Concept: Start with the definite clue (top floor), then fix the rest using "only two between" constraints.

1. No one above D → D on floor 6. Only two between D and C → C on floor 3.
2. Q below C but above N → Q = 2, N = 1. Only two between O and Q → O = 5. Remaining P = 4.
3. Between P (4) and N (1) are floors 3 and 2 → two people (C and Q).

Trick: Draw the 6 boxes first; "only two between X and Y" means X and Y are exactly 3 floors apart.

Answer: (D) Two

Concept: Decode each symbol and draw the family tree; the gender of D is given by "brother".

1. D − F → D is the brother of F. F + G → F is the daughter of G, so D is also G's son.
2. G ÷ H → G is the father of H, so D, F and H are siblings. H × J → H is the wife of J.
3. D is the brother of J's wife → wife's brother.

Trick: Siblings of a parent's other children are your own siblings – once D and H share father G, D is H's brother, hence J's brother-in-law.

Answer: (A) Wife's brother