Q&A: practice questions with answers

Concept: Mean proportional of $a$ and $b$ is $\sqrt{ab}$.

1. $\sqrt{75\times 12}=\sqrt{900}=30$
2. $30-13=17$

Trick: $75\times 12=900$ is a perfect square, so the root is $30$ instantly.

Answer: (C) $17$

Concept: BODMAS: brackets first, then “of”, then $÷$ and $\times$ left to right, then $+$ and $-$.

1. $8\text{ of }3=24$, so $(24-16)=8$
2. $8÷4\times 20=2\times 20=40$
3. $[44-40]=4$
4. $64÷4=16$

Trick: Treat “of” as multiplication done before $÷$.

Answer: (B) $16$

Concept: Solve the innermost bracket first; convert mixed fractions to improper fractions with LCM $120$.

1. $\frac{3}{5}\times \frac{1}{10}÷\frac{4}{15}=\frac{3}{50}\times \frac{15}{4}=\frac{9}{40}$
2. $\frac{5}{6}-\frac{9}{40}=\frac{100-27}{120}=\frac{73}{120}$
3. $\frac{7}{2}-\frac{73}{120}=\frac{420-73}{120}=\frac{347}{120}$
4. $\frac{21}{4}-\frac{347}{120}=\frac{630-347}{120}=\frac{283}{120}=2\frac{43}{120}$

Trick: Use one common denominator ($120$) for all steps to avoid repeated LCMs.

Answer: (A) $2\frac{43}{120}$

Concept: The table is cumulative (“and above”). Students below a mark = total − students at or above it.

1. Aggregate “$40$ and above” $=78$ students
2. Total students $=100$
3. Below $40\%$ $=100-78=22$

Trick: “Less than $x$” $=$ total $-$ “$x$ and above” — read only the Aggregate row.

Answer: (D) $22$

Concept: Average speed $=\frac{\text{distance}}{\text{time}}$, with distance in km and time in hours.

1. Distance $=37\text{km}500\text{m}=37.5\text{km}$
2. Time $=90\text{min}=1.5\text{h}$
3. Speed $=\frac{37.5}{1.5}=25\text{km/h}$

Trick: $90$ min is $\frac{3}{2}$ h, so speed $=37.5\times \frac{2}{3}=25$.

Answer: (A) $25$

Concept: Angle sum of a triangle is ${180}^{\circ }$; at most one angle can be $\ge {90}^{\circ }$.

1. A: all angles $<{60}^{\circ }$ gives sum $<{180}^{\circ }$ — false.
2. B: one obtuse angle (e.g. ${110}^{\circ }$) with two small acute angles — true.
3. C: two right angles already make ${180}^{\circ }$, third angle $0^{\circ }$ — false.
4. D: every triangle has at least two acute angles — true.

Yaad rakhein: Equilateral triangle (${60}^{\circ }$ each) shows angles can never all be below ${60}^{\circ }$.

Answer: (B) B and D

Concept: Open brackets from inside out; product of two negatives is positive.

1. $-4(-a)=4a$, so $x[-5\{4a\}]=-20ax$
2. $-3(-a)=3a$, so $6[-3\{3a\}]=6(-9a)=-54a$
3. $-20ax-54a=6a\Rightarrow -20ax=60a$
4. $x=-3$

Trick: Count minus signs: even count gives $+$, odd gives $-$.

Answer: (A) $-3$

Concept: If $x+y+z=0$, then $x^3+y^3+z^3=3xyz$.

1. Let $x=a$, $y=2b$, $z=3c$; then $x+y+z=0$
2. $a^3+(2b{)}^3+(3c{)}^3=3·a·2b·3c$
3. $=18abc$

Trick: Multiply $3$ by all the coefficients: $3\times 1\times 2\times 3=18$.

Answer: (A) $18abc$

Concept: Required $\%=\frac{\text{F total}}{\text{C total}}\times 100$.

1. C: $95+110=205$ thousand
2. F: $70+80=150$ thousand
3. $\frac{150}{205}\times 100=73.17\%$

Trick: $\frac{150}{205}=\frac{30}{41}$; $\frac{30}{41}\approx 0.7317$.

Answer: (C) $73.17\%$

Concept: Add the pair rates: $2(A+B+C)$ = sum of pair rates. Then $D=(A+B+C+D)-(A+B+C)$.

1. $2(A+B+C)=\frac{1}{25}+\frac{1}{36}+\frac{1}{30}=\frac{36+25+30}{900}=\frac{91}{900}$
2. $A+B+C=\frac{91}{1800}$
3. $D=\frac{1}{18}-\frac{91}{1800}=\frac{100-91}{1800}=\frac{9}{1800}=\frac{1}{200}$
4. D alone takes $200$ days

Trick: Total work $=1800$ units: A+B+C do $91$ units/day, all four do $100$ units/day, so D does $9$ units/day $\Rightarrow \frac{1800}{9}=200$ days.

Answer: (C) $200$

Concept: Change in average $=\frac{\text{change in total}}{\text{number of items}}$.

1. Excess entered $=99-66=33$
2. Increase in average $=\frac{33}{55}=0.6$

Trick: Only the error matters, not the actual marks of others.

Answer: (C) $0.6$

Concept: Saving $=$ Income $-$ Expenditure. Take income $=100$.

1. Old: income $100$, expense $68$, saving $32$
2. New: income $115$, expense $68\times 1.1=74.8$, saving $40.2$
3. Increase $=\frac{40.2-32}{32}\times 100=\frac{8.2}{32}\times 100=25.625\%$
4. To 2 decimal digits: $25.62\%$

Trick: Assume income $=100$ so every figure is directly a percentage.

Answer: (D) $25.62\%$

Concept: Volume of a sphere $=\frac{4}{3}\pi r^3$; number of small spheres $=\frac{V_{\text{big}}}{V_{\text{small}}}={\left(\frac{R}{r}\right)}^3$.

1. $\frac{R}{r}=\frac{5}{2.5}=2$
2. Number $=2^3=8$

Trick: Halving the radius gives $8$ times as many spheres.

Answer: (A) $8$

Concept: Sum of angles of a triangle $={180}^{\circ }$.

1. Let smallest $=x$; others $2x$ and $3x$
2. $x+2x+3x=6x={180}^{\circ }$
3. $x={30}^{\circ }$

Yaad rakhein: Ratio $1:2:3$ always gives ${30}^{\circ },{60}^{\circ },{90}^{\circ }$.

Answer: (C) ${30}^{\circ }$

Concept: Surface area of sphere $=4\pi r^2$; rate $=\frac{\text{total cost}}{\text{area}}$.

1. $r=\frac{14}{2}=7\text{cm}$
2. Area $=4\times \frac{22}{7}\times 49=616{\text{cm}}^2$
3. Rate $=\frac{21560}{616}=35$

Trick: For $r=7$, $4\pi r^2=616$ — worth remembering.

Answer: (B) ₹$35$

Concept: Opposite directions: relative speed $=$ sum of speeds; distance $=$ sum of lengths.

1. Relative speed $=42+48=90\text{km/h}=90\times \frac{5}{18}=25\text{m/s}$
2. Distance $=230+270=500\text{m}$
3. Time $=\frac{500}{25}=20\text{s}$

Trick: Multiply km/h by $\frac{5}{18}$ to get m/s.

Answer: (B) $20$ seconds

Concept: $x^3+y^3+z^3-3xyz=\frac{1}{2}(x+y+z)[(x-y{)}^2+(y-z{)}^2+(z-x{)}^2]$.

1. $x+y+z=110+111+112=333$
2. $(x-y{)}^2+(y-z{)}^2+(z-x{)}^2=1+1+4=6$
3. Value $=\frac{1}{2}\times 333\times 6=999$

Trick: For three consecutive numbers the value is always $3\times$ their sum.

Answer: (A) $999$

Concept: MP $=$ CP $\times (1+\text{markup})$; SP $=$ MP $\times (1-\text{discount})$.

1. MP $=1600\times 2.3=3680$
2. SP $=3680\times 0.75=2760$

Trick: Net factor $=2.3\times 0.75=1.725$; $1600\times 1.725=2760$.

Answer: (A) $2760$

Concept: Add each row, then subtract.

1. Cricket $=100+150+200+300+200=950$
2. Tennis $=200+220+210+210+230=1070$
3. Difference $=1070-950=120$

Trick: Subtract column-wise: $100-20-10+90-30=120$ (Tennis − Cricket).

Answer: (A) $120$

Concept: Downstream $D=B+S$, upstream $U=B-S$, stream $S=\frac{D-U}{2}$.

1. $D=\frac{2}{10/60}=12\text{km/h}$
2. $U=\frac{12}{2}=6\text{km/h}$
3. $S=\frac{12-6}{2}=3\text{km/h}$

Trick: Stream speed is half the difference of downstream and upstream speeds.

Answer: (C) $3\text{km/h}$