Q&A: practice questions with answers

Concept: Upstream speed $=B-S$, downstream speed $=B+S$.

1. $B:S=8:7$ ⇒ upstream $=1$ part, downstream $=15$ parts.
2. Upstream speed $=\frac{40.6}{27/60}=\frac{812}{9}\text{km/h}$; downstream $=15\times \frac{812}{9}=1353.3\text{km/h}$.
3. Time up $=\frac{77.7\times 9}{812}=0.861$ h; time down $=\frac{52.5}{1353.3}=0.039$ h.
4. Total $\approx 0.9$ h.

Trick: Downstream is 15 times upstream speed here, so its time is tiny – total stays just above the upstream time.

Answer: (B) $0.9$

Concept: Remove decimals by pairing digits: $\sqrt{0.012321}=\frac{\sqrt{12321}}{1000}$.

1. $\sqrt{12321}=111\Rightarrow \sqrt{0.012321}=0.111$
2. $\sqrt{1369}=37\Rightarrow \sqrt{0.001369}=0.037$
3. $\frac{0.111}{0.037}=3$

Trick: Same number of decimal places (6 each), so compare $\sqrt{\frac{12321}{1369}}=\sqrt{9}=3$ directly.

Answer: (D) $3$

Concept: Successive discounts multiply: $SP=MP\times (1-d_1)(1-d_2)(1-d_3)$.

1. $SP=640\times 0.70\times 0.75\times 0.75$
2. $640\times 0.70=448$; $448\times 0.75=336$
3. $336\times 0.75=252$

Trick: $0.7\times 0.75\times 0.75=0.39375$; $640\times 0.39375=252$.

Answer: (B) ₹$252$

Concept: $\text{Sum}=\text{Average}\times \text{Count}$

1. Sum of four numbers $=40\times 4=160$
2. Sum of the three given numbers $=40+42+55=137$
3. Fourth number $=160-137=23$

Trick: Deviation method: from 40, the deviations are $0,+2,+15$ → total $+17$, so the fourth is $40-17=23$.

Answer: (B) $23$

Concept: $\text{Still-water speed}=\frac{\text{Downstream}+\text{Upstream}}{2}$

1. Downstream $=40\text{km/h}$, upstream $=26\text{km/h}$
2. Speed in still water $=\frac{40+26}{2}=\frac{66}{2}=33\text{km/h}$
3. (Stream speed $=\frac{40-26}{2}=7\text{km/h}$)

Trick: Still-water speed is the average of the two speeds – it must lie between 26 and 40.

Answer: (A) $33\text{km/h}$

Concept: In the right triangle, $\tan \theta =\frac{\text{height}}{\text{base}}$.

1. Base $=30\text{m}$, $\theta ={30}^{\circ }$
2. $\tan {30}^{\circ }=\frac{h}{30}\Rightarrow \frac{1}{\sqrt{3}}=\frac{h}{30}$
3. $h=\frac{30}{\sqrt{3}}=\frac{30\sqrt{3}}{3}=10\sqrt{3}\text{m}$

Trick: At ${30}^{\circ }$, height $=\frac{\text{base}}{\sqrt{3}}$; at ${60}^{\circ }$, height $=\text{base}\times \sqrt{3}$.

Answer: (A) $10\sqrt{3}$ metres

Concept: Multiply each ratio term by its own growth factor.

1. A: $3\times 1.10=3.3$
2. B: $10\times 1.61=16.1$
3. C: $8\times 1.20=9.6$
4. New ratio $=3.3:16.1:9.6=33:161:96$

Trick: Only option (D) has A’s term $=33$ ($3\times 11$); the answer is fixed at a glance.

Answer: (D) $33:161:96$

Concept: When a solid is melted and recast, its volume stays the same.

1. $\frac{4}{3}\pi r^3=\frac{1}{3}\pi r^{2}h$ (same radius $r$)
2. Cancel $\frac{1}{3}\pi r^2$: $4r=h$
3. $h=4\times 53=212\text{cm}$

Trick: Sphere → cone of same radius: $h=4r$. Sphere → cylinder of same radius: $h=\frac{4}{3}r$.

Answer: (D) $212$

Concept: Let a pen cost ₹$p$ and a pencil ₹$c$; form two equations.

1. $6p+5c=259$ … (i)
2. $19(p-9)+2(c+4)=90\Rightarrow 19p+2c=253$ … (ii)
3. (i)$\times 2$: $12p+10c=518$; (ii)$\times 5$: $95p+10c=1265$ → $83p=747$, $p=9$
4. $5c=259-54=205\Rightarrow c=41$
5. $13p+2c=117+82=199$

Trick: Eliminate the variable whose coefficients are easy to equalise ($5c$ and $2c$ → $10c$).

Answer: (B) ₹$199$

Concept: A root of an equation satisfies it: substitute the common root.

1. Check: ${11}^2-9(11)-22=121-99-22=0$ ✓
2. Put $x=11$ in $2x^2-3x+c=0$: $2(121)-33+c=0$
3. $242-33+c=0\Rightarrow c=-209$

Trick: Only the equation containing c needs the substitution; the second one only confirms 11 is a root.

Answer: (D) $-209$

Concept: Solve brackets from inside out: ( ) → { } → [ ].

1. $(27÷9)=3$
2. $\{63-3\}=60$, then $60÷6=10$
3. $[7+10]=17$
4. $49+17=66$

Trick: Innermost bracket first, then division before addition.

Answer: (D) $66$

Concept: $\text{Sum}=\text{Average}\times \text{Count}$

1. First two numbers: $77\times 2=154$; third $=129$
2. Remaining two $=655-154-129=372$
3. Their average $=\frac{372}{2}=186$

Trick: Subtract known parts from the total, then divide by how many are left.

Answer: (B) $186$

Concept: Mode is the value that occurs most often.

1. Count: 9 → 4 times; 2 → 3 times; 4 → 3 times; 5 → 3 times
2. 1, 6, 8 → once each
3. Highest frequency belongs to 9.

Trick: Tally only the repeated values – the one with the longest tally is the mode.

Answer: (D) $9$

Concept: Convert all fractions to the common denominator 16.

1. $\frac{2}{8}=\frac{4}{16}$, so $\frac{4}{16}+\frac{4}{16}+\frac{1}{16}+\frac{4}{16}=\frac{13}{16}$
2. $1=\frac{16}{16}$
3. $\frac{13}{16}-\frac{16}{16}=-\frac{3}{16}$

Trick: $3\times \frac{2}{8}=\frac{3}{4}$; $\frac{3}{4}+\frac{1}{16}$ is less than 1, so the answer must be negative.

Answer: (C) $-\frac{3}{16}$

Concept: $\text{Mode}=l+\frac{f_1-f_0}{2f_1-f_0-f_2}\times h$

1. Mode 165 lies at the end of 160–165, so take modal class 160–165: $l=160$, $h=5$, $f_1=37$, $f_0=26$, $f_2=x$
2. $165=160+\frac{37-26}{74-26-x}\times 5$
3. $5=\frac{55}{48-x}\Rightarrow 48-x=11$
4. $x=37$

Trick: Plug the options in: only $x=37$ makes $\frac{11}{48-x}\times 5=5$.

Answer: (C) $37$

Concept: Sum of exterior angles of any polygon $={360}^{\circ }$; sum of interior angles $=(n-2)\times {180}^{\circ }$.

1. Interior sum $={360}^{\circ }+100\%\text{ of }{360}^{\circ }={720}^{\circ }$
2. $(n-2)\times 180=720\Rightarrow n=6$
3. Each interior angle $=\frac{720}{6}={120}^{\circ }$

Trick: Interior sum $=k\times {360}^{\circ }$ → $n=2k+2$. Here $k=2$, so $n=6$.

Answer: (D) $120$

Concept: Successive discounts multiply the remaining fractions: $\text{Remaining}=(1-d_1)(1-d_2)(1-d_3)$.

1. $0.94\times 0.72=0.6768$
2. $0.6768\times 0.85=0.57528$
3. Single discount $=1-0.57528=0.42472=42.47\%$

Trick: Two at a time: $6\%,28\%\to 6+28-\frac{6\times 28}{100}=32.32\%$; then with $15\%$: $32.32+15-\frac{32.32\times 15}{100}=42.47\%$.

Answer: (D) $42.47\%$

Concept: Convert each ratio to a decimal (or compare by cross-multiplication).

1. $\frac{16}{57}\approx 0.281$, $\frac{17}{53}\approx 0.321$
2. $\frac{24}{54}=\frac{4}{9}\approx 0.444$
3. $\frac{28}{60}=\frac{7}{15}\approx 0.467$
4. Check: $28\times 54=1512>24\times 60=1440$, so $28:60$ is the greatest.

Trick: Simplify first: $\frac{4}{9}$ vs $\frac{7}{15}$ → cross-multiply $60$ vs $63$.

Answer: (D) $28:60$

Concept: A conclusion follows only if it is true in every possible Venn diagram of the statements.

1. "All bikes are roads" puts Bikes inside Roads – it never makes all roads bikes, so (I) does not follow.
2. "Some roads are TVs" means only a part of Roads meets TVs; roads outside TVs are possible, so (II) does not follow.
3. Both conclusions are "All" type, but the statements give no such link.

Trick: "All A are B" never gives "All B are A" – the reverse of an All statement is only "Some B are A".

Answer: (C) Neither conclusion (I) nor (II) follows

Concept: Find the shift of the first letters and of the second letters separately.

1. OL-QN-SU: O→Q→S is $+2,+2$; L→N→U is $+2,+7$.
2. LI-NK-PR: L→N→P is $+2,+2$; I→K→R is $+2,+7$. Same rule.
3. RO-TQ-VX: R→T→V is $+2,+2$; O→Q→X is $+2,+7$ ✔. In (C) Q→Y is $+8$, in (A)/(D) O→T or O→R breaks $+2$.

Trick: Check only the last letter first: second-letter jump must be $+7$ (Q→X).

Answer: (B) RO-TQ-VX