Q&A: practice questions with answers

Concept: Time $=\frac{\text{Distance}}{\text{Speed}}$; one-way distance $d$ is covered once downstream and once upstream.

1. $\frac{d}{22}+\frac{d}{5.5}=8\Rightarrow \frac{d}{22}+\frac{4d}{22}=8$.
2. $\frac{5d}{22}=8\Rightarrow d=\frac{176}{5}\text{km}$ (one way).
3. Total distance (to and fro) $=2d=\frac{352}{5}\text{km}$.

Trick: Speeds ratio $22:5.5=4:1$ → times $1:4$ → downstream takes $\frac{8}{5}$ h, so $d=22\times \frac{8}{5}=\frac{176}{5}$; double it.

Answer: (B) $\frac{352}{5}\text{km}$

Concept: $\text{SP}=\text{CP}\times \frac{100\pm \text{gain/loss}\%}{100}$.

1. At 20% loss: $112=\text{CP}\times 0.8\Rightarrow \text{CP}=$ ₹140.
2. For 50% gain: $\text{SP}=140\times 1.5=$ ₹210.

Trick: $80\%\to 112$, so $150\%\to \frac{112}{80}\times 150=210$.

Answer: (B) 210

Concept: In a right triangle, $\sin A=\frac{\text{perpendicular}}{\text{hypotenuse}}$ and $\cos A=\frac{\text{base}}{\text{hypotenuse}}$.

1. $\sin A=\frac{3}{4}$ → perpendicular $3$, hypotenuse $4$, base $=\sqrt{16-9}=\sqrt{7}$.
2. $\cos A=\frac{\sqrt{7}}{4}$.
3. $\frac{2\cos A}{\sqrt{7}}=\frac{2\times \frac{\sqrt{7}}{4}}{\sqrt{7}}=\frac{2}{4}=\frac{1}{2}$.

Trick: Use $\cos A=\sqrt{1-{\sin }^{2}A}=\sqrt{1-\frac{9}{16}}=\frac{\sqrt{7}}{4}$ – the $\sqrt{7}$ cancels.

Answer: (C) $\frac{1}{2}$

Concept: British Flag theorem: for any point P inside rectangle ABCD, $P{A}^2+P{C}^2=P{B}^2+P{D}^2$ (opposite corners pair up).

1. $P{A}^2+P{C}^2={69}^2+{38}^2=4761+1444=6205$.
2. $P{D}^2=6205-P{B}^2=6205-{27}^2=6205-729=5476$.
3. $PD=\sqrt{5476}=74\text{cm}$.

Trick: A and C are opposite, B and D are opposite – the sums of squares of distances to opposite corners are equal.

Answer: (D) 74

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

1. $P=1200$, $R=3\%$ per month, $T=1$ month.
2. $\text{SI}=\frac{1200\times 3\times 1}{100}=36$.

Trick: 1 month at 3% per month = simply 3% of principal: $12\times 3=36$.

Answer: (B) 36

Concept: $x^3+\frac{1}{x^3}={(x+\frac{1}{x})}^3-3(x+\frac{1}{x})$.

1. Put $x+\frac{1}{x}=4$.
2. $x^3+\frac{1}{x^3}=4^3-3\times 4=64-12=52$.

Trick: If $x+\frac{1}{x}=a$, then $x^3+\frac{1}{x^3}=a^3-3a$.

Answer: (A) 52

Concept: Income = Expenditure + Savings.

1. Expenditure $=33\%$ → savings $=67\%$ of income $=6298$.
2. $1\%=\frac{6298}{67}=94$.
3. Expenditure $=33\times 94=3102$.

Trick: Income $=9400$; go straight to $33\%$ of it.

Answer: (C) 3,102

Concept: Overall average $=\frac{n_1{a}_1+n_2{a}_2}{n_1+n_2}$.

1. Remaining boys $=50-15=35$.
2. Total height $=15\times 163+35\times 165=2445+5775=8220$.
3. Average $=\frac{8220}{50}=164.4\text{cm}$.

Trick: Start at 163: extra $=\frac{35\times 2}{50}=1.4$ → $164.4$.

Answer: (A) 164.4

Concept: Empirical relation: $\text{Mode}=3\text{Median}-2\text{Mean}$.

1. $3\text{Median}=\text{Mode}+2\text{Mean}=61.7+2\times 51.5=61.7+103=164.7$.
2. $\text{Median}=\frac{164.7}{3}=54.9$.

Trick: $\text{Median}=\text{Mean}+\frac{\text{Mode}-\text{Mean}}{3}=51.5+\frac{10.2}{3}=54.9$.

Answer: (D) 54.9

Concept: Profit is shared in the ratio of $\text{capital}\times \text{time}$.

1. Take capitals 1300, 2000, 1500. X: $1300\times 12=15600$.
2. Y keeps $67\%$ after 7 months: $2000\times 7+1340\times 5=14000+6700=20700$.
3. Z withdraws $24\%$ with 5 months left: $1500\times 7+1140\times 5=10500+5700=16200$.
4. Total $=52500$; Z's share $=\frac{16200}{52500}\times 16100=4968$.

Trick: Ratio $15600:20700:16200=52:69:54$ (sum 175); Z $=\frac{54}{175}\times 16100=54\times 92=4968$.

Answer: (D) 4,968

Concept: Total persons $=(\text{rank from top})+(\text{rank from bottom})-1$, because the person is counted twice.

1. Rank from top $=17$, rank from bottom $=44$.
2. Total $=17+44-1=60$.

Trick: Add both ranks and subtract 1.

Answer: (B) $60$

Concept: Facing the centre: a person's right is anticlockwise and left is clockwise. Fix O first, then place the others.

1. P is immediately right of O; P is third to the right of R, so R is third to the left of P.
2. O is third to the left of T, so T is third to the right of O.
3. Free seats remain on O's left side and the top. S is not next to O, and Q is second to the left of X, so X sits immediately left of O, Q two seats further and S in the last seat.
4. Counting from the right of S: T, Q, R, then X → three people in between.

Trick: Draw O at the bottom; right = anticlockwise for centre-facing people.

Answer: (C) Three

Concept: Interchange the signs first, then apply BODMAS (÷, × before +, −).

1. Interchanging '+'↔'−' and '×'↔'÷': $173\times 3+135÷5-17$.
2. $173\times 3=519$ and $135÷5=27$.
3. $519+27-17=529$.

Trick: Rewrite the whole expression with new signs before calculating.

Answer: (A) $529$

Concept: Identify the highest and lowest numbers, then pick the required digits.

1. Numbers: 611, 733, 279, 265, 370. Highest $=733$, lowest $=265$.
2. Second digit of 733 $=3$; first digit of 265 $=2$.
3. $3+2=5$.

Trick: Comparing only first digits quickly gives highest (7xx) and lowest (2xx; 265 is less than 279).

Answer: (B) $5$

Concept: Find the position change of each letter: here the pattern is $+4,-4,+4,-4$.

1. CSVG → GOZC: C$+4$=G, S$-4$=O, V$+4$=Z, G$-4$=C.
2. KKDY → OGHU: K$+4$=O, K$-4$=G, D$+4$=H, Y$-4$=U ✓.
3. SCLQ: S$+4$=W, C$-4$=Y (wrapping back from A to Z), L$+4$=P, Q$-4$=M → WYPM.

Trick: C $-4$: C(3) → $3-4+26=25$ = Y.

Answer: (B) WYPM

Concept: Check each condition on the given group; apply only those that hold, otherwise use the table codes.

1. A P U X R: 1st letter A is a vowel and last letter R is a consonant → condition 1 applies (interchange codes of A and R).
2. Condition 2: 2nd letter P is a consonant ✗. Condition 3: U is a vowel ✗. Condition 4: P is a consonant ✗.
3. Direct codes: A = 2, P = 9, U = %, X = 6, R = 4 (the option pattern 4 9 % 6 2 shows R's code is 4).
4. After interchanging the 1st and last codes: 4 9 % 6 2.

Trick: Test condition 1 first: vowel at start + consonant at end → just swap the end codes.

Answer: (D) 4 9 % 6 2

Concept: Find one operation that works for both given pairs.

1. $5\times 7=35$.
2. $2\times 7=14$ ✓ – the rule is 'multiply by 7'.
3. $9\times 7=63$.

Trick: Both outputs are multiples of 7 – divide and check.

Answer: (A) $63$

Concept: Number positions 1 (top) to 7 (bottom) and place the boxes with fixed positions first.

1. Three boxes above G → G is at 4. One box between F and G → F at 2 or 6.
2. Three boxes between F and B with B above G: F = 6 gives B = 2 ✓ (F = 2 gives B = 6, below G ✗). E just below B → E = 3.
3. Free places 1, 5, 7. C is not next to F (6) → C = 1. A above D → A = 5, D = 7.
4. Below A: F and D → 2 boxes.

Trick: Fix G first, then F–B pair; the rest falls in place.

Answer: (A) $2$

Concept: Identify the highest and lowest numbers, then pick the required digits.

1. Numbers: 388, 739, 783, 483, 785. Highest $=785$, lowest $=388$.
2. First digit of 785 $=7$; third digit of 388 $=8$.
3. $8-7=1$.

Trick: 783 vs 785: compare the last digit → 785 is highest.

Answer: (B) $1$

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

1. M – N: M is the brother of N.
2. N × O: N is the wife of O (N is female).
3. O ÷ P: O is the father of P, so N is P's mother.
4. P + Q: P is the mother of Q. So N is Q's mother's mother and M is her brother.

Trick: Read the chain backwards from Q: mother (P) → her mother (N) → N's brother (M).

Answer: (A) Mother's Mother's Brother