Q&A: practice questions with answers

Concept: Successive increases multiply: $\text{New}=\text{Old}\times (1+\frac{a}{100})(1+\frac{b}{100})$.

1. $x\times \frac{104}{100}\times \frac{124}{100}=161200$
2. $x\times 1.2896=161200$
3. $x=\frac{161200}{1.2896}=125000$

Trick: Check option (C): $125000\times 1.04=130000$, $130000\times 1.24=161200$ ✓.

Answer: (C) ₹1,25,000

Concept: Apply BODMAS: division and multiplication first, then subtraction/addition; mind the signs.

1. $(-60)÷(-12)=5$
2. $(-2)\times 8=-16$
3. $(-9)-5+(-16)=-9-5-16=-30$

Trick: Negative ÷ negative = positive; then just add all the negatives.

Answer: (A) $-30$

Concept: Take total work = LCM of the days, then find each person's units per day.

1. Total work $=\text{LCM}(30,16,15)=240$ units.
2. Anjani $=\frac{240}{30}=8$; A + K $=\frac{240}{16}=15$; A + K + S $=\frac{240}{15}=16$ units/day.
3. Sushmita $=16-15=1$ unit/day.
4. Anjani + Sushmita $=8+1=9$ units/day → time $=\frac{240}{9}=\frac{80}{3}$ days.

Trick: S alone = (A+K+S) − (A+K); no need to find Khushbu.

Answer: (B) $\frac{80}{3}$

Concept: Required speed = remaining distance ÷ remaining time.

1. First part: $53\times 8=424\text{km}$ in $8$ h.
2. Next part: $150\text{km}$ at $50\text{km/h}$ → $3$ h.
3. Covered $=574\text{km}$ in $11$ h; remaining $=886-574=312\text{km}$ in $17-11=6$ h.
4. Speed $=\frac{312}{6}=52\text{km/h}$

Trick: Draw the journey as a line, write distance and time under each part.

Answer: (A) $52\text{km/h}$

Concept: Under compound interest, if $P$ becomes $kP$ in $t$ years, it becomes $k^{n}P$ in $nt$ years.

1. $3P$ in $5$ years.
2. $9P=3^{2}P$ → time $=2\times 5=10$ years.

Trick: Powers of the multiple ↔ multiples of the time: $3\to 5$, $9\to 10$, $27\to 15$.

Answer: (C) $10$ years

Concept: Let CP $=100$; find SP after mark-up and discount.

1. Marked price $=136$; SP $=136\times \frac{75}{100}=102$.
2. Profit $=102-100=2$ units $=$ ₹171 → 1 unit $=$ ₹85.5.
3. List price $=136\times 85.5=11628$

Trick: Net effect $=36-25-\frac{36\times 25}{100}=+2\%$ profit.

Answer: (C) ₹11,628

Concept: For 7 and 13: make 3-digit groups from the right and take the alternating sum; the number is divisible if this sum is.

1. Groups: $788|973|289$ → $289-973+788=104$.
2. $104=8\times 13$ → divisible by $13$; $104÷7$ leaves remainder $6$ → not by $7$.
3. Last two digits $89$ are not divisible by $4$ → not by $12$; the number is odd → not by $18$.

Trick: $1001=7\times 11\times 13$, so one alternating 3-digit sum tests 7, 11 and 13 together.

Answer: (D) $13$

Concept: If $7A=4B=2C=k$, then $A:B:C=\frac{1}{7}:\frac{1}{4}:\frac{1}{2}$.

1. Multiply by LCM $28$: $A:B:C=4:7:14$.
2. Total parts $=4+7+14=25$ → 1 part $=\frac{15500}{25}=620$.
3. Share of A $=4\times 620=2480$

Trick: Coefficients 7, 4, 2 → shares in ratio of LCM ÷ coefficient: $\frac{28}{7}:\frac{28}{4}:\frac{28}{2}$.

Answer: (C) ₹2,480

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

1. $4^2=16$, so $19\times 16=304$.
2. $9\times 304=2736$ and $2736÷14=\frac{1368}{7}$.
3. $13-86=-73=-\frac{511}{7}$.
4. $\frac{1368}{7}-\frac{511}{7}=\frac{857}{7}$.

Trick: Since $14=2\times 7$, the answer must have denominator 7; only the numerator needs care: $1368-511=857$.

Answer: (D) $\frac{857}{7}$

Concept: $\text{SP}=\text{MP}\times \frac{100-\text{Discount}\%}{100}$.

1. After 20% discount, SP $=80\%$ of MP.
2. $0.8\times \text{MP}=1800$.
3. $\text{MP}=\frac{1800}{0.8}=2250$.

Trick: $80\%\to 1800$, so $10\%\to 225$ and $100\%\to 2250$.

Answer: (B) ₹2,250

Concept: Use divisibility tests: 2 (last digit even), 3 (digit sum divisible by 3), 4 (last two digits divisible by 4).

1. Last digit is 5 (odd) → not divisible by 2 or 4.
2. Digit sum $=3+4+2+1+6+5=21$, divisible by 3 → 342165 is divisible by 3.
3. $7\times 48880=342160$, remainder 5 → not divisible by 7.

Trick: An odd number can never be divisible by 2 or 4 – strike those out first.

Answer: (B) 3

Concept: Make equal decimal places, find LCM of the whole numbers, then put back the decimal places.

1. $3.8=3.800$ and $0.052$ → both have 3 decimal places: 3800 and 52.
2. $3800=2^3\times 5^2\times 19$, $52=2^2\times 13$.
3. LCM $=2^3\times 5^2\times 19\times 13=49400$.
4. Put back 3 decimal places: $49.400=49.4$.

Trick: The LCM must be a multiple of 3.8: $3.8\times 13=49.4$ ✓ and $49.4÷0.052=950$ ✓.

Answer: (A) 49.4

Concept: $\text{SI}=\frac{P\times R\times T}{100}$; equal interest for the same time means $P_1{R}_1=P_2{R}_2$.

1. Let the sum at 10% be $x$; the rest $20300-x$ is at 4%.
2. $\frac{x\times 10\times 4}{100}=\frac{(20300-x)\times 4\times 4}{100}$ → $10x=4(20300-x)$.
3. $14x=81200$ → $x=5800$.

Trick: Sums are in inverse ratio of rates: $10:4$ → sums $=4:10=2:5$; $\frac{2}{7}\times 20300=5800$.

Answer: (B) 5,800

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

1. $\sqrt{98\times 18}=\sqrt{1764}$.
2. $98\times 18=(2\times 49)\times (2\times 9)=4\times 49\times 9$.
3. $\sqrt{4\times 49\times 9}=2\times 7\times 3=42$.

Trick: Pair the factors: $\sqrt{2^2\times 7^2\times 3^2}=42$ – no long multiplication needed.

Answer: (C) 42

Concept: In 'buy $x$ get $y$ free', the customer receives $x+y$ items but pays for only $x$; discount = value of free items.

1. Customer gets $5+2=7$ items, pays for 5.
2. Free items $=2$, each worth ₹250.
3. Discount $=2\times 250=500$ rupees.

Yaad rakhein: Discount % in such a scheme $=\frac{y}{x+y}\times 100=\frac{2}{7}\times 100\approx 28.6\%$.

Answer: (A) ₹500

Concept: Take total work = LCM of the days; each person's efficiency = work ÷ days.

1. Total work $=\text{LCM}(21,15,2)=210$ units.
2. A $=\frac{210}{21}=10$, B $=\frac{210}{15}=14$, A+B+C $=\frac{210}{2}=105$ units/day.
3. C $=105-10-14=81$ units/day.
4. C alone $=\frac{210}{81}=\frac{70}{27}=2\frac{16}{27}$ days.

Trick: $\frac{1}{C}=\frac{1}{2}-\frac{1}{21}-\frac{1}{15}=\frac{27}{70}$; C must be faster than both, so the answer is below 15 – only $2\frac{16}{27}$ fits.

Answer: (B) $2\frac{16}{27}$ days

Concept: Circumference $=2\pi r$; volume of cylinder $=\pi r^{2}h$.

1. $2\times \frac{22}{7}\times r=44$ → $r=7\text{cm}$.
2. $V=\frac{22}{7}\times 7^2\times 20$.
3. $V=22\times 7\times 20=3080{\text{cm}}^3$.

Trick: Circumference 44 always gives $r=7$ with $\pi =\frac{22}{7}$; base area $=154$, so $V=154\times 20=3080$.

Answer: (A) 3080 cu. cm

Concept: If $x+\frac{1}{x}=2$ then $x=1$; here $x=\sin \theta$ since $\text{cosec}\theta =\frac{1}{\sin \theta }$.

1. $\sin \theta +\frac{1}{\sin \theta }=2$ → $(\sin \theta -1{)}^2=0$ → $\sin \theta =1$.
2. Then $\text{cosec}\theta =1$.
3. ${\sin }^{120}\theta +{\text{cosec}}^{120}\theta =1^{120}+1^{120}=2$.

Trick: $x+\frac{1}{x}=2\Rightarrow x^n+\frac{1}{x^n}=2$ for every power $n$.

Answer: (A) 2

Concept: Form an equation from the past condition, then use it for the future condition.

1. Let present ages be F and S. $F-9=2(S-9)+25$ → $F=2S+16$.
2. After $x$ years: $F+x=2(S+x)$ → $x=F-2S$.
3. $x=(2S+16)-2S=16$ years.

Trick: 'Twice' condition: the gap $F-2S$ shrinks by 1 each year. 9 years ago $F-2S$ was 25; now it is $25-9=16$, so after 16 years it becomes 0.

Answer: (D) 16

Concept: Total surface area of a cube $=6a^2$.

1. $a^2={74}^2=5476$.
2. $6\times 5476=32856{\text{cm}}^2$.

Trick: Unit digit: $4^2$ ends in 6, $6\times 6$ ends in 6 → options ending in 6: 32856, 32896, 32846; ${74}^2\approx 5500$, $6\times 5476=32856$.

Answer: (A) $32856{\text{cm}}^2$