Q&A: practice questions with answers

Concept: Sum of first $n$ odd natural numbers $=n^2$, so their average $=\frac{n^2}{n}=n$.

1. Here $n=150$.
2. Sum $={150}^2=22500$.
3. Average $=\frac{22500}{150}=150$.

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

Answer: (C) $150$

Concept: With CI, the 2nd-year interest is the rate applied to (principal + 1st-year interest): $I_2=P\times \frac{r}{100}\times (1+\frac{r}{100})$.

1. Take $P=100$: 1st-year interest $=20$, amount $=120$.
2. 2nd-year interest $=20\%$ of $120=24$, i.e., $24\%$ of $P$.
3. $0.24P=3342\Rightarrow P=\frac{3342}{0.24}=13925$.

Trick: 2nd-year CI $=P\times \frac{r}{100}\times \frac{100+r}{100}$; at $20\%$ it is $\frac{6}{25}P$.

Answer: (B) ₹13,925

Concept: Form two linear equations with son's age $s$ and father's age $f$ and solve.

1. $2s+f=36$ … (i) and $s+2.5f=72$ … (ii).
2. Multiply (ii) by 2: $2s+5f=144$.
3. Subtract (i): $4f=108\Rightarrow f=27$.
4. Check: $s=\frac{36-27}{2}=4.5$ and $4.5+2.5\times 27=72$ ✓.

Trick: Make the coefficient of $s$ equal (×2) and subtract to remove $s$ in one step.

Answer: (A) $27$

Concept: If HCF $=h$, the numbers are $ha$ and $hb$ where $a,b$ are co-prime.

1. $9a\times 9b=8100\Rightarrow ab=\frac{8100}{81}=100$.
2. Factor pairs of $100$: $(1,100),(2,50),(4,25),(5,20),(10,10)$.
3. Co-prime pairs: only $(1,100)$ and $(4,25)$.
4. Numbers: $(9,900)$ and $(36,225)$ – so 2 pairs.

Trick: Co-prime pairs with product $p$ $=2^{k-1}$, where $k$ = number of distinct primes of $p$; $100=2^2\times 5^2$ gives $2^1=2$.

Answer: (C) $2$

Concept: Two consecutive numbers $n$ and $n+1$ lie close to $\sqrt{\text{product}}$.

1. $\sqrt{650}\approx 25.5$, so try $25$ and $26$.
2. $25\times 26=650$ ✓.
3. Greater number $=26$.

Trick: Unit digit of $650$ is $0$: consecutive numbers ending in $5$ and $6$ give unit digit $0$ ($5\times 6=30$).

Answer: (C) $26$

Concept: For the same distance, time is inversely proportional to speed: $S_1{T}_1=S_2{T}_2$.

1. Distance $=72\times 16=1152\text{km}$.
2. Nisha's time $=\frac{1152}{64}=18$ hours.
3. Check by ratio: $T_2=16\times \frac{72}{64}=16\times \frac{9}{8}=18$.

Trick: Speed ratio $72:64=9:8$ ⇒ time ratio $8:9$; $8\to 16$ so $9\to 18$.

Answer: (B) $18$ hours

Concept: Successive discounts are applied one after another: $\text{SP}=\text{MP}\times \frac{100-d_1}{100}\times \frac{100-d_2}{100}$.

1. After $20\%$: $1200\times 0.8=960$.
2. After $10\%$: $960\times 0.9=864$.
3. Equivalent single discount $=20+10-\frac{20\times 10}{100}=28\%$; $1200\times 0.72=864$.

Trick: Two discounts $a\%,b\%$ ⇒ single discount $a+b-\frac{ab}{100}$ (not $a+b$).

Answer: (A) ₹864

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, $T=7$ months.
2. $\text{SI}=\frac{1200\times 3\times 7}{100}=12\times 21=252$.

Trick: Rate per month × months directly – no need to convert to years ($3\%\times 7=21\%$ of $1200$).

Answer: (A) $252$

Concept: For two numbers: $\text{HCF}\times \text{LCM}=\text{product of the numbers}$.

1. $51\times 306=102\times x$.
2. $x=\frac{51\times 306}{102}=\frac{306}{2}=153$.
3. Check: $153=51\times 3$, $102=51\times 2$; HCF $=51$, LCM $=51\times 6=306$ ✓.

Trick: $102=2\times 51$, so $x=\frac{306}{2}$ instantly.

Answer: (D) $153$

Concept: Find cost price per item, then $\text{Profit}\%=\frac{\text{SP}-\text{CP}}{\text{CP}}\times 100$.

1. CP of one sharpener $=\frac{180}{12}=$ ₹15; SP = ₹20.
2. Profit $=20-15=$ ₹5.
3. $\text{Profit}\%=\frac{5}{15}\times 100=33.33\%$.

Trick: Profit is $\frac{1}{3}$ of CP ⇒ $33\frac{1}{3}\%$.

Answer: (A) $33.33$

Concept: One day's work of a person $=\frac{1}{\text{days}}$; work done together in $t$ days $=t(\frac{1}{a}+\frac{1}{b})$.

1. One day's work of A + B $=\frac{1}{17}+\frac{1}{19}=\frac{19+17}{323}=\frac{36}{323}$.
2. Work in 6 days $=6\times \frac{36}{323}=\frac{216}{323}$.
3. Work left $=1-\frac{216}{323}=\frac{107}{323}$.

Trick: Take total work $=17\times 19=323$ units; A does 19, B does 17 units/day → 36 units/day → $6\times 36=216$ done, $323-216=107$ left.

Answer: (A) $\frac{107}{323}$

Concept: Distance $=$ Speed $\times$ Time. Required speed $=\frac{\text{remaining distance}}{\text{remaining time}}$.

1. First part: $37\times 8=296\text{km}$.
2. Second part: 136 km at 34 km/h takes $\frac{136}{34}=4\text{h}$.
3. Covered $=296+136=432\text{km}$ in $8+4=12\text{h}$.
4. Remaining: $861-432=429\text{km}$ in $15-12=3\text{h}$ → speed $=\frac{429}{3}=143\text{km/h}$.

Trick: Keep a running total of distance and time; only the leftover part decides the new speed.

Answer: (B) 143 km/hr

Concept: BODMAS: Division and Multiplication are done before Addition and Subtraction.

1. $12÷6=2$ and $3\times 4=12$.
2. Expression becomes $16+2-12$.
3. $=18-12=6$.

Trick: Never go left to right blindly: first finish all $÷$ and $\times$, then $+$ and $-$.

Answer: (D) 6

Concept: Open the brackets; the $x^2$ terms cancel, leaving a linear equation in $x$.

1. $3(2x^2-5)=6x^2-15$ and $2(3x^2+4x-8)=6x^2+8x-16$.
2. $6x^2-15-6x^2-8x+16=20$ → $1-8x=20$.
3. $-8x=19$ → $x=-\frac{19}{8}$.

Trick: Check the sign when distributing the minus: $-2(\dots -8)$ gives $+16$.

Answer: (C) $-\frac{19}{8}$

Concept: BODMAS: brackets and powers first, then $\times$ / $÷$ from left to right, then $+$ / $-$.

1. $3^2=9$, so $3\times 9=27$.
2. $18\times 27=486$, then $486÷13=\frac{486}{13}$.
3. $66-74=-8=-\frac{104}{13}$.
4. Total $=\frac{486-104}{13}=\frac{382}{13}$.

Trick: All options have denominator 13, so convert the integer part to thirteenths: $8=\frac{104}{13}$.

Answer: (C) $\frac{382}{13}$

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

1. Digits of N65M04: odd places (1st, 3rd, 5th) $=N,5,0$; even places $=6,M,4$.
2. Difference $=(N+5)-(10+M)=N-M-5$.
3. With single digits, $N-M-5=0$ gives $N-M=5$, i.e. $M-N=-5$ (e.g. N = 7, M = 2 → 765204 $=11\times 69564$).

Trick: Alternate sum from the right: $4-0+M-5+6-N$ must be $0$ or $\pm 11$.

Answer: (C) M − N = −5

Concept: Write everything in $\sin A$ and $\cos A$ and use $a^2-b^2=(a-b)(a+b)$.

1. $\frac{1}{\sec A(1-\tan A)}=\frac{\cos A}{1-\frac{\sin A}{\cos A}}=\frac{{\cos }^{2}A}{\cos A-\sin A}$.
2. $\frac{1}{\csc A(1-\cot A)}=\frac{\sin A}{1-\frac{\cos A}{\sin A}}=\frac{{\sin }^{2}A}{\sin A-\cos A}=-\frac{{\sin }^{2}A}{\cos A-\sin A}$.
3. Sum $=\frac{{\cos }^{2}A-{\sin }^{2}A}{\cos A-\sin A}=\cos A+\sin A$.

Trick: Put $A=0^{\circ }$: expression $=\frac{1}{1\times 1}+0=1$; only $\cos A+\sin A=1$ matches.

Answer: (D) $\cos A+\sin A$

Concept: A number is divisible by 3 if the sum of its digits is divisible by 3.

1. Digit sum of 311916399 $=3+1+1+9+1+6+3+9+9=42$, and $42=3\times 14$ → divisible by 3.
2. Last digit 9 is odd → not divisible by 2.
3. Direct division leaves remainder 4 by 7 and remainder 2 by 13.

Trick: Always test 2, 3, 5 first – they need only the last digit or the digit sum.

Answer: (D) 3

Concept: Selling price $=$ Marked price $\times \frac{100-\text{discount}\%}{100}$.

1. Discount $=25\%$ of ₹800 $=\frac{25}{100}\times 800=200$ → ₹200.
2. SP $=800-200=600$ → ₹600.

Trick: 25% off means pay $\frac{3}{4}$: $\frac{3}{4}\times 800=600$.

Answer: (A) ₹600

Concept: Work in fractions of the registered voters $V$; the margin is (winner% $-$ loser%) of valid votes.

1. Votes polled $=0.8V$; valid $=80\%$ of $0.8V=0.64V$.
2. Winner $60\%$, loser $40\%$ of valid → margin $=20\%$ of $0.64V=0.128V$.
3. $0.128V=2144$ → $V=\frac{2144}{0.128}=16750$.

Trick: Margin fraction $=0.8\times 0.8\times 0.2=0.128$; $2144÷0.128=16750$.

Answer: (D) 16750