Q&A: practice questions with answers

Concept: After 30% rent, the remaining 70% covers conveyance, grocery and savings.

1. $751+8481+715=9947$
2. $70\%\text{ of salary}=9947$
3. $\text{Salary}=\frac{9947\times 100}{70}=14210$

Trick: Check: $30\%$ of $14210=4263$; $4263+9947=14210$ ✓

Answer: (B) ₹14,210

Concept: Convert each ratio to a decimal (or compare with one half).

1. $\frac{18}{57}\approx 0.32,\frac{27}{57}\approx 0.47,\frac{27}{54}=0.5$
2. $\frac{30}{49}\approx 0.61$ – the largest

Trick: Only $\frac{30}{49}$ is more than half; all others are half or less.

Answer: (D) 30 : 49

Concept: Mode is the observation that occurs most often.

1. Frequencies: 4 → 4 times; 1, 5, 6 → 2 times each; others once.
2. Highest frequency belongs to 4.

Trick: Tally only the repeated numbers – quickest way to find mode.

Answer: (A) 4

Concept: Expenditure = Salary − Savings = (100 − 35)% of salary.

1. $65\%\text{ of }5140=\frac{65}{100}\times 5140$
2. $=3341$

Trick: $10\%=514$; $65\%=6.5\times 514=3341$.

Answer: (D) ₹3,341

Concept: Rolling along the width: the width becomes the circumference ($2\pi r$) and the length becomes the height. $V=\pi r^{2}h$

1. $2\pi r=22\Rightarrow r=\frac{11}{\pi }$
2. $h=17$
3. $V=\pi \times \frac{121}{{\pi }^2}\times 17=\frac{2057}{\pi }{\text{cm}}^3$

Trick: $V=\frac{(\text{circumference}{)}^2\times h}{4\pi }=\frac{484\times 17}{4\pi }$.

Answer: (C) $\frac{2057}{\pi }$

Concept: $\text{Area of parallelogram}=\text{base}\times \text{height}$

1. $97\times 43=97\times 40+97\times 3$
2. $=3880+291=4171{\text{cm}}^2$

Trick: $97\times 43=100\times 43-3\times 43=4300-129=4171$.

Answer: (C) $4171$

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

1. $4000=\frac{16000\times x\times 2}{100}=320x$
2. $x=\frac{4000}{320}=12.5$

Trick: Interest is $\frac{1}{4}$ of principal in 2 years → $25\%$ in 2 years → $12.5\%$ per year.

Answer: (C) 12.5

Concept: $\sin \theta =\cos ({90}^{\circ }-\theta )$, and $\cos \alpha =\cos \beta \Rightarrow \alpha =\pm \beta$ (within one turn).

1. $\sin 5A=\cos ({90}^{\circ }-5A)$, so $\cos ({90}^{\circ }-5A)=\cos (A-{42}^{\circ })$.
2. Case 1: ${90}^{\circ }-5A=A-{42}^{\circ }\Rightarrow 6A={132}^{\circ }\Rightarrow A={22}^{\circ }$; then $5A={110}^{\circ }$, which is not acute, so this case is rejected.
3. Case 2: ${90}^{\circ }-5A=-(A-{42}^{\circ })\Rightarrow 4A={48}^{\circ }\Rightarrow A={12}^{\circ }$; then $5A={60}^{\circ }$ (acute). Check: $\sin {60}^{\circ }=\frac{\sqrt{3}}{2}=\cos (-{30}^{\circ })$ ✓.
4. So $A={12}^{\circ }$, which is not among the options (21°, 22°, 20°, 23°). No option is correct.

Note: Always check the stated condition (here "5A is acute") after solving. In the exam, every candidate gets full marks (bonus) for such a question.

Answer: Bonus – no option is correct as printed; full marks to all.

Concept: $\text{Average speed}=\frac{\text{Total distance}}{\text{Total time}}$

1. Times: $\frac{819}{63}=13\text{h}$, $\frac{1053}{81}=13\text{h}$
2. Total distance $=819+1476+1053=3348\text{km}$; total time $=\frac{3348}{54}=62\text{h}$
3. Middle part time $=62-26=36\text{h}$
4. $x=\frac{1476}{36}=41\text{km/h}$

Trick: Find the total time first; the unknown part's time is what is left.

Answer: (B) 41

Concept: If SP of $a$ articles = CP of $b$ articles, $\text{gain}\%=\frac{b-a}{a}\times 100$.

1. Let CP of one book $=1$; SP of 10 books $=16$, CP of 10 books $=10$
2. $\text{Gain}\%=\frac{16-10}{10}\times 100=60\%$

Trick: SP side has fewer articles → gain; divide by the SP-side count.

Answer: (A) $60\%$ gain

Concept: Find one rule linking the three numbers that works for both given sets.

1. (41, 34, 24): $41-24=17$ and $17\times 2=34$.
2. (63, 80, 23): $63-23=40$ and $40\times 2=80$.
3. Rule: second number $=2\times (\text{first}-\text{third})$.
4. (43, 40, 23): $43-23=20$, $20\times 2=40$ ✓. Others fail, e.g. (46, 86, 24): $2\times 22=44\ne 86$.

Trick: Middle number is always even and here equals twice the gap between the outer numbers.

Answer: (B) (43, 40, 23)

Concept: Check the change in each part separately.

1. IL 17 → LN 12: I+3=L, L+2=N, $17-5=12$.
2. MN 12 → PP 7: M+3=P, N+2=P, $12-5=7$.
3. VR 4: V+3=Y, R+2=T, $4-5=-1$ → YT −1.

Trick: Rule: (+3, +2, −5).

Answer: (C) YT −1

Concept: Fix the definite clue first, then place chains (T < U < D and S < C < E).

1. F = Thursday. T is after F and U, D come after T in order → T = Fri, U = Sat, D = Sun.
2. Mon, Tue, Wed remain for S < C < E → S = Mon, C = Tue, E = Wed.
3. Between C (Tue) and U (Sat): Wed, Thu, Fri → E, F, T = 3 people.

Trick: A chain of 3 after Thursday must fill Fri–Sat–Sun exactly.

Answer: (B) Three

Concept: Replace P → $÷$, Q → $\times$, R → $+$, S → $-$ and use BODMAS.

1. (C) 24 P 4 S 28 R 12 Q 6 → $24÷4-28+12\times 6$.
2. $=6-28+72=50$ ✓.
3. Check (A): $24-4\times 28÷12+6=24-9.33+6\ne 50$, so it is not (A).

Trick: Test options one by one; stop when you get 50.

Answer: (C) 24 P 4 S 28 R 12 Q 6

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

1. M ÷ N: M is the father of N.
2. N + O: N is the sister of O, so O is also M’s child.
3. O × P: O is the wife of P, so O is female and P is M’s son-in-law.
4. P – Q: P is the brother of Q. So M is the father of the wife of Q’s brother.

Trick: Read the chain from Q’s side: Q → brother (P) → wife (O) → father (M).

Answer: (D) Brother’s Wife’s Father

Concept: Look at the differences between consecutive terms.

1. $942-2=940$, $940-4=936$, $936-6=930$, $930-8=922$.
2. The subtracted number goes up by 2 each time, so next is 10.
3. $922-10=912$.

Trick: Differences −2, −4, −6, −8, −10 (even numbers).

Answer: (D) 912

Concept: Track the first letter, second letter and the number separately.

1. First letters: B, E, H, K → +3 each → N.
2. Second letters: I, L, O, R → +3 each → U.
3. Numbers: 16, 24, 32, 40 → +8 each → 48.
4. Next term: NU48.

Trick: All parts move by a fixed step: (+3, +3, +8).

Answer: (B) NU48

Concept: Compare each letter of the first cluster with the matching letter of the second.

1. KO – LP: K+1=L, O+1=P.
2. QU – RV: Q+1=R, U+1=V.
3. NR – OS: N+1=O, R+1=S.
4. PI – KJ: P−5=K, I+1=J → pattern broken.

Trick: Three pairs are “+1, +1” (and each cluster has a gap of 4); PI–KJ is the odd one.

Answer: (A) PI - KJ

Concept: Change each digit as told, then read the required positions.

1. 3518642: odd digits +1 → 3→4, 5→6, 1→2; even digits −2 → 8→6, 6→4, 4→2, 2→0.
2. New number: 4 6 2 6 4 2 0.
3. Second from left = 6; second from right = 2.
4. Sum $=6+2=8$.

Trick: Only the 2nd and 6th digits matter: 5→6 and 4→2, so $6+2=8$.

Answer: (A) 8

Concept: A letter common to both words gets the digit common to both codes.

1. OWNS → 5391 and SIDE → 2458.
2. Common letter: S. Common digit: 5.
3. So S = 5.

Trick: Match common letters with common digits; order does not matter.

Answer: (B) 5