Q&A: practice questions with answers

Concept: If $a+b+c=0$, then $a^3+b^3+c^3=3abc$.

1. $22+(-6)+(-16)=0$
2. So value $=3\times 22\times (-6)\times (-16)$
3. $=3\times 2112=6336$

Trick: Always check whether the bases add to zero before cubing.

Answer: (C) 6336

Concept: Fee income for the year = fees received − outstanding at start (belongs to last year) + outstanding at end.

1. Fees for the year $=7643-1105+916=7454$
2. Total income $=7454+2198=9652$
3. Expenditure $=5287+3120=8407$
4. Surplus $=9652-8407=1245$

Trick: Opening outstanding → subtract; closing outstanding → add.

Answer: (D) ₹1,245 surplus

Concept: Bill = fixed part $F$ + $v\times$(number of students).

1. 15 extra students add $25000-17500=7500$, so $v=500$.
2. $F=17500-25\times 500=5000$
3. For 60: $5000+60\times 500=35000$

Trick: From 40 to 60 is 20 more students: $25000+20\times 500=35000$.

Answer: (B) ₹35,000

Concept: 3 and 4 are co-prime, so a number divisible by both is divisible by $3\times 4=12$.

1. Divisible by 4 and 3 → divisible by 12.
2. Hence remainder on division by 12 is 0.
3. Check: $A=2$ gives 75492 (last two digits 92, digit sum 27) $=12\times 6291$.

Trick: Divisible by co-prime factors ⇒ divisible by their product.

Answer: (D) 0

Concept: Overall average = total sales ÷ total employees.

1. Total $=40\times 120000+30\times 180000+30\times 150000$
2. $=4800000+5400000+4500000=14700000$
3. Employees $=100$, average $=\frac{14700000}{100}=147000$

Trick: Work in thousands: $\frac{40\times 120+30\times 180+30\times 150}{100}=147$ thousand.

Answer: (D) ₹1,47,000

Concept: $\text{Time}=\frac{\text{Distance}}{\text{Speed}}$ for each leg.

1. A → B: $\frac{78}{13}=6$ h
2. B → A: $\frac{78}{26}=3$ h
3. Total $=6+3=9$ h

Trick: Do not average the speeds; add the times.

Answer: (B) 9

Concept: Mean proportional of $a$ and $c$ is $b=\sqrt{ac}$ (since $a:b=b:c$).

1. $b=\sqrt{25\times 100}$
2. $=\sqrt{2500}=50$

Trick: Check: $25:50=50:100=1:2$.

Answer: (B) 50

Concept: Compare the price paid per toy as a % of marked price; the highest % is the least beneficial.

1. A: $0.94\times 0.61=0.5734$ → $57.34\%$
2. B: $0.52\times 0.84\times 0.89\approx 0.3888$ → $38.88\%$
3. C: $\frac{2\times 63+2\times 71}{4}=67\%$
4. D: 6 toys for the price of 4 → $\frac{4}{6}\approx 66.67\%$
5. Highest cost per toy is C.

Trick: Turn every offer into "% paid per toy" and compare.

Answer: (C) C

Concept: Single equivalent of $a\%$ and $b\%$ is $a+b-\frac{ab}{100}$.

1. Single offer: $30\%$ of 800 $=240$
2. Successive: $25+5-\frac{125}{100}=28.75\%$ → $0.2875\times 800=230$
3. Difference $=240-230=10$

Trick: Difference $=1.25\%$ of 800 $=10$.

Answer: (B) ₹10

Concept: Form two equations from the present sum and the past ratio.

1. $A+R=34$
2. $A-5=2(R-5)$ ⇒ $A=2R-5$
3. $3R-5=34$ ⇒ $R=13$, $A=21$

Trick: Option check: $A=21$, $R=13$; 5 years ago $16=2\times 8$ ✓.

Answer: (C) 21

Concept: Rates add: draining pipes $+$, filling pipe $-$ (taking tank = 1).

1. A + B drain $\frac{1}{10}+\frac{1}{15}=\frac{1}{6}$ per h; in 2 h → $\frac{1}{3}$ drained, $\frac{2}{3}$ left.
2. With C: net $\frac{1}{6}-\frac{1}{12}=\frac{1}{12}$ per h.
3. Time $=\frac{2}{3}÷\frac{1}{12}=8$ h; total from start $=2+8=10$ h.

Trick: LCM 60 units: A = 6, B = 4, C = −5; first 2 h drain 20, then 40 at 5 per h = 8 h.

Answer: (A) 10 hours

Concept: BODMAS: brackets first, and "of" is done before $÷$.

1. $\frac{12}{27}÷\frac{12}{2}=\frac{12}{27}\times \frac{2}{12}=\frac{2}{27}$
2. $\frac{1}{11}\times \frac{22}{3}+\frac{5}{9}=\frac{2}{3}+\frac{5}{9}=\frac{11}{9}$
3. $\frac{2}{27}÷\frac{11}{9}=\frac{2}{33}$
4. $\frac{33}{22}\text{ of }\frac{22}{9}=\frac{33}{9}$, so $\frac{2}{9}÷\frac{33}{9}=\frac{2}{33}$
5. Sum $=\frac{2}{33}+\frac{2}{33}=\frac{4}{33}$

Trick: "of" means multiply and has higher priority than $÷$.

Answer: (B) $\frac{4}{33}$

Concept: Total cost = sum of (number × average cost) of each item.

1. Notebooks: $20\times 50=1000$; pencils: $50\times 10=500$
2. Pens: $3500-1000-500=2000$
3. Average per pen $=\frac{2000}{100}=20$

Trick: Subtract the known totals first, then divide.

Answer: (A) ₹20

Concept: Take total work = LCM of days; one-day work = units per day.

1. Work $=24$ units; A $=2$, B $=1$ unit/day.
2. In 4 days together: $4\times 3=12$ units.
3. Remaining $12$ units by B: $\frac{12}{1}=12$ days.

Trick: Together they finish half in 4 days; B alone does half in $\frac{24}{2}=12$ days.

Answer: (D) 12

Concept: Find time for each half separately: $\text{Time}=\frac{\text{Distance}}{\text{Speed}}$.

1. Each half $=60$ km.
2. $\frac{60}{30}=2$ h and $\frac{60}{60}=1$ h
3. Total $=3$ h

Trick: Average speed $=\frac{2\times 30\times 60}{90}=40$ km/h → $\frac{120}{40}=3$ h.

Answer: (A) 3 hours

Concept: $5^2+{12}^2={13}^2$, so the triangle is right-angled. Altitude to the hypotenuse $=\frac{\text{product of legs}}{\text{hypotenuse}}$.

1. Area $=\frac{1}{2}\times 5\times 12=30$
2. Also area $=\frac{1}{2}\times 13\times h$
3. $h=\frac{60}{13}\approx 4.615\approx 4.62$ cm

Trick: Altitude on hypotenuse $=\frac{ab}{c}$; it is always less than the shorter leg (5).

Answer: (D) 4.62 cm

Concept: A regular hexagon = 6 equilateral triangles: Area $=\frac{3\sqrt{3}}{2}{a}^2$.

1. $a=10$, $a^2=100$
2. Area $=\frac{3\times 1.732}{2}\times 100=2.598\times 100$
3. $\approx 259.8$ m²

Trick: 173.2 is the area of just 4 such triangles — a common trap.

Answer: (B) 259.8 square meters

Concept: $\text{Profit}\%=\frac{\text{Profit}}{\text{CP}}\times 100$.

1. Profit $=575-500=75$
2. $\frac{75}{500}\times 100=15\%$

Trick: Profit % is always on cost price, not selling price.

Answer: (B) 15%

Concept: $SI=\frac{P\times R\times T}{100}$.

1. $SI=\frac{3200\times 12\times 5}{100}$
2. $=32\times 60=1920$

Trick: Total rate $=12\times 5=60\%$; $60\%$ of 3200 $=1920$.

Answer: (C) ₹1920

Concept: $68=4\times 17$; a number must be divisible by both 4 and 17.

1. Test by 4 (last two digits): 17350 → 50 ✗; 35350 → 50 ✗.
2. 23800 $=68\times 350$ ✓; 28560 $=68\times 420$ ✓.
3. Not divisible: (i) and (iv).

Trick: The quick test by 4 eliminates (i) and (iv) at once.

Answer: (C) ( i ) and ( iv )