Q&A: practice questions with answers

Concept: A number is divisible by 11 if (sum of digits at odd places) − (sum of digits at even places) is $0$ or a multiple of $11$.

1. Odd places of N29M11: $N+9+1=N+10$.
2. Even places: $2+M+1=M+3$.
3. Difference $=(N+10)-(M+3)=N-M+7$.
4. Setting it $=0$ gives $M-N=7$ (e.g. $N=1,M=8$: $129811=11\times 11801$).

Trick: Write $N-M+7=0$ or $11$; only $M-N=7$ is in the options.

Answer: (A) $M-N=7$

Concept: $V=\pi r^{2}h$, so $V\propto r^{2}h$.

1. Take $r=10$, $h=10$: $r^{2}h=1000$.
2. New $r=8$ ($20\%$ less), new $h=14$ ($40\%$ more): $r^{2}h=64\times 14=896$.
3. Change $=\frac{896-1000}{1000}\times 100=-10.4\%$, i.e. a decrease.

Trick: Multiply factors: ${0.8}^2\times 1.4=0.896$ ⇒ $10.4\%$ decrease.

Answer: (A) $10.4\%$ decrease

Concept: Sum of the first $n$ even numbers $=n(n+1)$, so their average $=n+1$.

1. Here $n=154$.
2. Sum $=154\times 155$.
3. Average $=\frac{154\times 155}{154}=155$.

Trick: Average of first $n$ even numbers $=n+1$; of first $n$ odd numbers $=n$.

Answer: (C) $155$

Concept: Two successive discounts $a\%$ and $b\%$ equal a single discount of $a+b-\frac{ab}{100}\%$.

1. Net discount $=20+10-\frac{20\times 10}{100}=28\%$.
2. $28\%$ of MP = ₹23,800.
3. MP $=\frac{23800\times 100}{28}=85000$, i.e. ₹85,000.

Trick: Take MP $=100$: $100\to 80\to 72$, discount $28$ units.

Answer: (D) ₹85,000

Concept: If the bases are equal, the exponents can be equated: $(a^m{)}^n=a^{mn}$.

1. $x^z=({103}^{0.2}{)}^z={103}^{0.2z}$.
2. $y^6=({103}^{0.43}{)}^6={103}^{2.58}$.
3. So $0.2z=2.58$ ⇒ $z=\frac{2.58}{0.2}=12.9$.

Trick: $z=\frac{6\times 0.43}{0.2}=12.9$ directly.

Answer: (A) $12.9$

Concept: Write present ages as $F$ (father) and $S$ (son) and convert each statement into an equation.

1. 4 years ago: $F-4=2(S-4)+37$ ⇒ $F=2S+33$.
2. After $x$ years: $F+x=2(S+x)$.
3. $2S+33+x=2S+2x$ ⇒ $x=33$.

Trick: The extra '$+33$' over twice the son's age vanishes after $33$ years.

Answer: (B) $33$

Concept: In a trapezium, the segment joining the midpoints of the diagonals is parallel to the bases and equals half their difference: $EF=\frac{DC-AB}{2}$.

1. Let the midline through E and F be drawn parallel to AB and DC.
2. In $△ADC$ midpoint segment = $\frac{DC}{2}$; in $△ADB$ midpoint segment = $\frac{AB}{2}$; on the same line their difference is $EF$.
3. $EF=\frac{53-45}{2}=\frac{8}{2}=4\text{cm}$. (BC and AD are not needed.)

Trick: Midpoints of diagonals → half the difference; median → half the sum.

Answer: (A) $4$

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

1. Mohan + Nishant = 4280 and Mohan − Nishant = 1070 ⇒ Mohan = 2675, Nishant = 1605; ratio $2675:1605=5:3$.
2. Let Nishant's capital be $x$; Mohan's $=x+89000$.
3. $\frac{3(x+89000)}{2x}=\frac{5}{3}$ ⇒ $9x+801000=10x$ ⇒ $x=801000$.
4. Mohan's capital $=801000+89000=890000$, i.e. ₹8,90,000.

Trick: First turn the profit sum and difference into a ratio ($5:3$), then use capital × time.

Answer: (C) ₹8,90,000

Concept: Increasing a number by $100\%$ doubles it: new value $=x\times \frac{200}{100}=2x$.

1. $2x=2800$.
2. $x=\frac{2800}{2}=1400$.
3. Check: $1400+100\%\text{ of }1400=2800$.

Trick: $+100\%$ = double, so halve the result.

Answer: (A) $1400$

Concept: Expand the brackets and then combine like terms.

1. $x(7x-4)=7x^2-4x$.
2. $2(x^2-2)=2x^2-4$.
3. Sum $+11$: $7x^2+2x^2-4x-4+11=9x^2-4x+7$.

Trick: Put $x=1$: expression $=3+(-2)+11=12$, and $9-4+7=12$ ✓.

Answer: (A) $9x^2-4x+7$

Concept: Use ${\sin }^2\alpha +{\cos }^2\alpha =1$, i.e. $1-{\sin }^2\alpha ={\cos }^2\alpha$.

1. Take LCM $(1-\sin \alpha )\cos \alpha$: $\frac{{\cos }^2\alpha -(1+\sin \alpha )(1-\sin \alpha )}{(1-\sin \alpha )\cos \alpha }$
2. $(1+\sin \alpha )(1-\sin \alpha )=1-{\sin }^2\alpha ={\cos }^2\alpha$
3. Numerator $={\cos }^2\alpha -{\cos }^2\alpha =0$, so the value is $0$.

Trick: Put $\alpha =0^{\circ }$: $\frac{1}{1}-\frac{1}{1}=0$. Only option (D) matches.

Answer: (D) $0$

Concept: $SI=\frac{P\times R\times T}{100}$, with T in years = days/366 (2024 is a leap year). The deposit day is not counted, the withdrawal day is.

1. Days: Feb $(29-12)=17$, March $31$, April $13$ → $17+31+13=61$ days
2. $T=\frac{61}{366}=\frac{1}{6}$ year
3. $SI=\frac{4000\times 7.5}{100}\times \frac{1}{6}=300\times \frac{1}{6}=50$

Trick: Count one of the two end dates only. $61$ days of a leap year $=\frac{1}{6}$ year exactly.

Answer: (D) ₹50

Concept: $(a+b{)}^2=a^2+2ab+b^2$.

1. ${182}^2=(180+2{)}^2$
2. $=32400+2\times 180\times 2+4=32400+720+4$
3. $=33124$

Trick: Use ${180}^2=32400$ and add $4\times 180+4=724$ to get 33124 directly.

Answer: (D) 33124

Concept: Total literates = male literates + female literates.

1. Total literates $=39\%\times 356000=138840$
2. Male literates $=59\%\times 178000=105020$
3. Females $=356000-178000=178000$; female literates $=138840-105020=33820$
4. $\frac{33820}{178000}\times 100=19\%$

Trick: Males = females (half each), so average: $\frac{59+x}{2}=39\Rightarrow x=19$.

Answer: (D) 19%

Concept: Share $=\frac{\text{ratio part}}{\text{sum of parts}}\times \text{total}$.

1. Sum of parts $=5+17+7=29$; one part $=\frac{638}{29}=22$
2. Largest − smallest $=17-5=12$ parts
3. $12\times 22=264$

Trick: Find value of 1 part first, then multiply by the difference of ratio terms.

Answer: (B) ₹264

Concept: A number $n$ is prime if no prime $\le \sqrt{n}$ divides it ($\sqrt{401}\approx 20$, test 2, 3, 5, 7, 11, 13, 17, 19).

1. Even numbers (394, 396, 398, 400) and 395 (by 5) are out
2. 393 $=3\times 131$, 399 $=3\times 133$ → not prime
3. 397 and 401 are not divisible by 7, 11, 13, 17, 19 → prime

Trick: Between consecutive odd numbers here, strike out multiples of 3 and 5 first; only 397, 401 survive.

Answer: (D) 2

Concept: Divisible by 3 if digit sum is divisible by 3; by 2, 4 or 12 the number must be even.

1. Digit sum $=9+5+6+1+6+9=36$, divisible by 3 ✓
2. Last digit 9 is odd → not divisible by 2, 4 or 12

Trick: An odd number can never be divisible by any even number.

Answer: (A) 3

Concept: Two consecutive even numbers: $n$ and $n+2$.

1. $n^2+(n+2{)}^2=1460\Rightarrow 2n^2+4n-1456=0\Rightarrow n^2+2n-728=0$
2. $(n+28)(n-26)=0\Rightarrow n=26$
3. Numbers 26 and 28: $676+784=1460$ ✓; sum $=54$

Trick: $\frac{1460}{2}=730\approx {27}^2$, so the numbers sit around 27: 26 and 28.

Answer: (C) 54

Concept: If $pA=qB$, then $A:B=q:p$.

1. $\frac{40}{100}A=\frac{2}{7}B\Rightarrow \frac{2}{5}A=\frac{2}{7}B$
2. $\frac{A}{B}=\frac{2}{7}\times \frac{5}{2}=\frac{5}{7}$

Trick: Cancel the common 2 in $\frac{2}{5}$ and $\frac{2}{7}$: ratio is $5:7$ directly.

Answer: (A) $5:7$

Concept: Final price $=MP\times (1-\frac{d_1}{100})(1-\frac{d_2}{100})$.

1. After 25%: $4000\times \frac{3}{4}=3000$
2. After 20%: $3000\times \frac{4}{5}=2400$

Trick: Net discount $=25+20-\frac{25\times 20}{100}=40\%$; $60\%$ of 4000 $=2400$.

Answer: (B) ₹2,400