Q&A: practice questions with answers

Concept: Both sat the same exam, so maximum marks are the same: $\text{Max}=\frac{\text{marks}}{\text{percent}}\times 100$.

1. $26\%$ of Max $=624\Rightarrow \text{Max}=\frac{624\times 100}{26}=2400$
2. Sudha: $\frac{720}{2400}\times 100=30\%$

Trick: Percent is proportional to marks: $26\times \frac{720}{624}=26\times \frac{15}{13}=30\%$.

Answer: (A) $30\%$

Concept: Half-yearly compounding: rate is halved and time is doubled.

1. Rate $=\frac{6}{2}=3\%$ per half-year, $n=2$
2. $\text{CI}=820000[(1.03{)}^2-1]=820000\times 0.0609$
3. $=49938$ (₹)

Trick: Effective rate for two periods of 3%: $3+3+\frac{3\times 3}{100}=6.09\%$ → $6.09\%$ of 820000.

Answer: (B) ₹$49,938$

Concept: Convert mixed fractions to improper fractions, then use BODMAS (multiply before adding).

1. $2\frac{2}{9}=\frac{20}{9}$, $2\frac{2}{3}=\frac{8}{3}$, $8\frac{3}{4}=8.75$
2. $\frac{20}{9}\times \frac{8}{3}\times 567=\frac{160}{27}\times 567=160\times 21=3360$
3. $3360+8.75=3368.75$

Trick: $567=27\times 21$ cancels the denominator 27; answer must end in .75 → only (C).

Answer: (C) $3368.75$

Concept: Work per hour: pump fills $\frac{1}{4}$; with the leak the net filling is $\frac{1}{9\frac{1}{3}}=\frac{3}{28}$.

1. Leak per hour $=\frac{1}{4}-\frac{3}{28}=\frac{7-3}{28}=\frac{4}{28}=\frac{1}{7}$
2. So the leak alone empties the full tank in $7$ hours.

Trick: Shortcut: $\text{Leak time}=\frac{a\times b}{b-a}=\frac{4\times \frac{28}{3}}{\frac{28}{3}-4}=\frac{112}{16}=7$.

Answer: (D) $7$ hours

Concept: Volume of a hollow sphere $=\frac{4}{3}\pi (R^3-r^3)$; mass = density × volume.

1. $R^3-r^3=125-8=117{\text{cm}}^3$
2. $V=\frac{4}{3}\times \frac{22}{7}\times 117=\frac{10296}{21}\approx 490.29{\text{cm}}^3$
3. Mass $=28\times \frac{10296}{21}=\frac{4}{3}\times 10296=13728\text{g}=13.728\text{kg}$

Trick: 28 and the 21 in the denominator cancel to $\frac{4}{3}$: mass $=\frac{4}{3}\times \frac{4}{3}\times 22\times 117$ g.

Answer: (A) $13.728$ kg

Concept: For $x^2-x-6=0$: $\alpha +\beta =1$, $\alpha \beta =-6$. Use ${\alpha }^4+{\beta }^4=({\alpha }^2+{\beta }^2{)}^2-2(\alpha \beta {)}^2$.

1. ${\alpha }^2+{\beta }^2=(\alpha +\beta {)}^2-2\alpha \beta =1+12=13$
2. ${\alpha }^4+{\beta }^4={13}^2-2\times 36=169-72=97$

Trick: Roots are $3$ and $-2$: $3^4+(-2{)}^4=81+16=97$.

Answer: (D) $97$

Concept: Mode is the observation that occurs most often.

1. Frequencies: 24 → 4 times; 25, 27, 34 → 2 times each; all others once.
2. Highest frequency is for 24.

Trick: Just tally the repeats – no calculation needed.

Answer: (C) $24$

Concept: Write the condition as an equation and form the ratio.

1. $\frac{56}{100}a=\frac{5}{9}b\Rightarrow \frac{a}{b}=\frac{5}{9}\times \frac{100}{56}$
2. $=\frac{500}{504}=\frac{125}{126}$
3. So $a:b=125:126$

Trick: $a\times 56\%=b\times \frac{5}{9}\Rightarrow a:b=\frac{5}{9}:\frac{56}{100}$ – cross-multiply.

Answer: (D) $125:126$

Concept: BODMAS: Brackets, then Multiplication, then Addition/Subtraction from left to right.

1. Bracket: $5+2=7$
2. Multiply: $4\times 7=28$
3. $81-28+54=107$

Trick: Never subtract before multiplying: $81-4$ first would give a wrong value.

Answer: (C) $107$

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

1. Total distance $=60+80+100=240\text{km}$
2. Total time $=3$ h
3. Average speed $=\frac{240}{3}=80\text{km/h}$

Trick: Equal time intervals → average speed = simple average of the speeds.

Answer: (D) $80\text{km/h}$

Concept: Find the value of one ratio part from the given difference.

1. C − B $=11-7=4$ parts $=196$ (₹)
2. 1 part $=\frac{196}{4}=49$ (₹)
3. A $=14\times 49=686$ (₹)

Trick: A is $\frac{14}{4}$ of the difference: $196\times 3.5=686$.

Answer: (B) ₹$686$

Concept: Convert a ratio of fractions into whole numbers by multiplying each term by the LCM of the denominators.

1. Ratio $=\frac{17}{2}:\frac{14}{6}:\frac{2}{5}=\frac{17}{2}:\frac{7}{3}:\frac{2}{5}$.
2. LCM of 2, 3, 5 is 30: $\frac{17}{2}\times 30=255,\frac{7}{3}\times 30=70,\frac{2}{5}\times 30=12$.
3. Sum $=255+70+12=337$ parts; one part $=\frac{674}{337}=2$.
4. Third monkey $=12\times 2=24$.

Trick: 674 is exactly $2\times 337$, so each part is 2 bananas.

Answer: (A) 24

Concept: If divisor − remainder is the same for all divisors, the number = k × LCM − (that common difference).

1. $32-4=28,41-13=28,48-20=28$, so $N+28$ is divisible by 32, 41 and 48.
2. $\text{LCM}(32,41,48)=96\times 41=3936$, so $N=3936k-28$.
3. $k=1$: $3908$ (not divisible by 53). $k=2$: $7872-28=7844=53\times 148$ ✓.
4. $N=7844$, which lies between 7834 and 7854.

Trick: Common 'shortfall' 28 → subtract it from multiples of the LCM, then test the extra condition.

Answer: (D) 7834 and 7854

Concept: Use $\sec A-\tan A=\frac{1-\sin A}{\cos A}$ and $1-{\sin }^{2}A={\cos }^{2}A$.

1. $\sec A(1+\sin A)(\sec A-\tan A)=\frac{1}{\cos A}\times (1+\sin A)\times \frac{1-\sin A}{\cos A}$.
2. $=\frac{1-{\sin }^{2}A}{{\cos }^{2}A}=\frac{{\cos }^{2}A}{{\cos }^{2}A}=1$.
3. Expression $=7\times 1-4=3$.

Trick: Put $A=0^{\circ }$: $7\times 1\times 1\times 1-4=3$.

Answer: (B) 3

Concept: Volume of cuboid $=l\times b\times h$.

1. $7524=33\times 4\times h=132h$.
2. $h=\frac{7524}{132}=57\text{cm}$.

Trick: Unit digit check: $132\times 57$ ends in 4 ($2\times 7=14$) ✓.

Answer: (B) 57

Concept: When a transversal cuts two parallel lines, corresponding angles are equal and angles on a straight line add up to ${180}^{\circ }$.

1. $\angle QND={50}^{\circ }$ (given).
2. $\angle QMB$ and $\angle QND$ are corresponding angles ($AB\parallel CD$), so $\angle QMB={50}^{\circ }$.
3. $\angle PMB$ and $\angle QMB$ form a linear pair: $\angle PMB={180}^{\circ }-{50}^{\circ }={130}^{\circ }$.

Trick: The angle at N is acute (${50}^{\circ }$); $\angle PMB$ opens on the other side of the transversal, so it is obtuse $={180}^{\circ }-{50}^{\circ }$.

Answer: (C) ${130}^{\circ }$

Concept: Successive selling: final price = original cost × product of the multipliers.

1. Loss $60\%$ → multiplier $0.4$; profit $25\%$ → multiplier $1.25$.
2. C's price $=x\times 0.4\times 1.25=0.5x$.
3. $0.5x=42206\Rightarrow x=84412$.

Trick: $0.4\times 1.25=\frac{1}{2}$, so A's cost is just double C's price.

Answer: (A) ₹84,412

Concept: Net filling rate = filling rate − emptying rate.

1. Net rate $=\frac{1}{27}-\frac{1}{28}=\frac{28-27}{756}=\frac{1}{756}$ tank per hour.
2. Full tank takes 756 hours.
3. One-fourth tank $=\frac{756}{4}=189$ hours.

Trick: For consecutive numbers $n,n+1$: net time $=n(n+1)$; here $27\times 28=756$.

Answer: (D) 189

Concept: Successive percentage increases are applied one after another as multipliers.

1. After year 1: $120000\times 1.10=132000$.
2. After year 2: $132000\times 1.30=171600$.

Trick: Net change $=10+30+\frac{10\times 30}{100}=43\%$; $120000\times 1.43=171600$.

Answer: (C) 1,71,600

Concept: If profit % at one SP equals loss % at another SP, the cost price is the average of the two SPs.

1. $\frac{863-CP}{CP}=\frac{CP-433}{CP}\Rightarrow CP=\frac{863+433}{2}=648$.
2. SP for $25\%$ profit $=648\times \frac{125}{100}=810$.

Trick: Equal profit% and loss% → $CP=\frac{S{P}_1+S{P}_2}{2}$.

Answer: (B) ₹810