Q&A: practice questions with answers

Concept: $\text{Area}=\frac{1}{2}\times \text{base}\times \text{height}$.

1. Area $=\frac{1}{2}\times 12\times 8$
2. $=6\times 8=48{\text{cm}}^2$

Yaad rakhein: Height is the perpendicular from the opposite vertex to the base.

Answer: (C) $48{\text{cm}}^2$

Concept: Form equations from the age conditions; the age gap never changes.

1. Let present ages be father $F$, son $S$. Three years ago: $F-3=2(S-3)+24$ → $F=2S+21$.
2. After $x$ years: $F+x=2(S+x)$.
3. $2S+21+x=2S+2x$ → $x=21$.

Trick: If $F=2S+k$ now, father becomes twice the son after exactly $k$ years.

Answer: (D) 21

Concept: Form two linear equations and solve by elimination.

1. Let pen $=p$, pencil $=q$: $12p+7q=285$.
2. $11(p-6)+3(q+5)=118$ → $11p+3q=169$.
3. $3\times$(i) $-7\times$(ii): $36p-77p=855-1183$ → $p=8$; then $q=\frac{285-96}{7}=27$.
4. $2p+2q=2(8+27)=70$.

Trick: Verify: $12\times 8+7\times 27=96+189=285$.

Answer: (D) ₹70

Concept: $\text{SI}=\frac{P\times R\times T}{100}$, with $R$ and $T$ in the same time unit.

1. Rate is 4% per month and time is 9 months.
2. $\text{SI}=\frac{1200\times 4\times 9}{100}=432$.
3. (If the rate were 4% per annum, SI would be $\frac{1200\times 4\times 9}{100\times 12}=36$.)

Yaad rakhein: Always match the unit of rate and time before using the formula.

Answer: (B) ₹432

Concept: $\text{Profit}\%=\frac{\text{SP}-\text{CP}}{\text{CP}}\times 100$.

1. CP of 1 pencil $=\frac{180}{12}=15$.
2. SP $=17$ → profit $=2$.
3. $\text{Profit}\%=\frac{2}{15}\times 100=13.33\%$

Yaad rakhein: $\frac{1}{15}=6.67\%$, so $\frac{2}{15}=13.33\%$.

Answer: (D) 13.33

Concept: Sum = average × count.

1. Sum of first two $=2\times 80=160$; third $=109$.
2. Sum of last two $=655-160-109=386$.
3. Average $=\frac{386}{2}=193$.

Yaad rakhein: Convert every average to a total first.

Answer: (C) 193

Concept: Each share = total × (its part ÷ sum of parts).

1. Sum of parts $=18+12+2=32$; 1 part $=\frac{658}{32}$.
2. Difference (largest − smallest) $=18-2=16$ parts.
3. $16\times \frac{658}{32}=\frac{658}{2}=329$.

Trick: 16 parts is exactly half of 32 → half of ₹658.

Answer: (A) 329

Concept: Rule out options by quick tests, then verify the remaining one by division.

1. Last digit 9 → not divisible by 10; odd → not by 12.
2. Digit sum $=4+4+4+3+4+8+2+6+9=44$ → not divisible by 9.
3. Check 19: $444348269÷19=23386751$ exactly.

Trick: Elimination: three options fail simple rules, so 19 must be the answer.

Answer: (B) 19

Concept: Successive changes multiply: $P\times (1+\frac{r_1}{100})(1+\frac{r_2}{100})(1-\frac{r_3}{100})$.

1. $14680\times 1.25\times 1.70\times 0.80$
2. $1.25\times 0.80=1$, so it is $14680\times 1.7$.
3. $=24956$.

Trick: +25% and −20% cancel each other ($\frac{5}{4}\times \frac{4}{5}=1$).

Answer: (A) 24,956

Concept: Combined average = total runs ÷ total matches.

1. Runs in 23 matches $=23\times 42=966$.
2. Runs in 10 matches $=10\times 15=150$.
3. Average $=\frac{966+150}{33}=\frac{1116}{33}=33.82$.

Yaad rakhein: Never average the two averages directly when counts differ.

Answer: (C) 33.82

Concept: Rearrange the digits and compare position by position.

1. Original: 7 1 6 4 2 9 3 8
2. Ascending: 1 2 3 4 6 7 8 9
3. Only the 4th position (digit 4) is the same.

Trick: Write both rows one under the other and tick matches.

Answer: (C) One

Concept: Facing the centre, right = anticlockwise and left = clockwise.

1. Fix E; B is 3rd to the right of E; F is immediate left of B.
2. C sits next to B (other side from F) and A next to C.
3. Two seats remain beside E; G cannot be next to F, so D is between E and F and G is between A and E.
4. Going left from G: A, C, B, F, then D → 4 people between.

Trick: Place the most fixed clue (B from E) first.

Answer: (A) Four

Concept: Swap the signs as told, then apply BODMAS.

1. Swap + ↔ × and − ↔ ÷: $38-7\times 5+20÷4$.
2. $=38-35+5$
3. $=8$

Trick: Rewrite the whole expression first, then calculate.

Answer: (C) 8

Concept: Decode each symbol and build a family tree.

1. A + E: A is the daughter of E. E − F: E is the wife of F.
2. F × N: F is the father of N. N ÷ S: N is the brother of S.
3. A, N and S are all children of E and F; A is female → A is the sister of S.

Trick: Mark gender as soon as a relation reveals it (daughter → female).

Answer: (D) Sister

Concept: Here 2 is added to every digit (even and odd).

1. 7145362 → 9 3 6 7 5 8 4.
2. First two: $9+3=12$; last two: $8+4=12$.
3. Sum $=12+12=24$.

Trick: Only the four end digits matter: $(7+1)+(6+2)+4\times 2=24$.

Answer: (B) 24

Concept: Draw the least-overlap Venn diagram.

1. All physicists lie inside chemists.
2. Chemists and biologists do not touch, so physicists cannot touch biologists.
3. I: No biologist is a physicist – true. II: Some biologists are physicists – false.

Yaad rakhein: All A are B + No B is C → No A is C (and No C is A).

Answer: (C) Only conclusion (I) follows.

Concept: Swap the signs as told ($+\leftrightarrow ÷$, $-\leftrightarrow \times$), then simplify using BODMAS (÷ and × before + and −).

1. The equation printed in the paper is $21÷57-3\times 8+4=?$.
2. After interchanging: $21+57\times 3-8÷4$.
3. $=21+171-2$.
4. $=190$.
5. The value $190$ is not among the options (56, 23, 45, 34), so no option is correct.

Note: Always apply the interchange first, then BODMAS. 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: Common letters in two words share the common codes.

1. MADE → {4,1,7,2}; HAME → {2,4,7,5}.
2. Common letters M, A, E ↔ common codes 4, 7, 2.
3. Left over in HAME: H ↔ 5 (and D ↔ 1 in MADE).

Trick: Unique letter ↔ unique code.

Answer: (C) 5

Concept: Look at the differences.

1. Differences: $8,16,24,32$ – multiples of 8.
2. Next difference $=40$.
3. $107+40=147$.

Trick: Difference series $8n$.

Answer: (B) 147

Concept: The letter common to both words gets the code common to both.

1. ALPS → {3,2,5,7}; PUNE → {1,3,4,6}.
2. Only common letter: P. Only common digit: 3.
3. So P = 3.

Trick: One common letter ↔ one common digit.

Answer: (D) 3