Q&A: practice questions with answers

Concept: In a partnership for the same time, profit is shared in the ratio of investments.

1. Ratio $=1540:1760:1170=154:176:117$ (divide by 10).
2. Total parts $=154+176+117=447$.
3. Rishi’s share $=\frac{117}{447}\times 1490$. Since $1490=447\times \frac{10}{3}$, share $=117\times \frac{10}{3}=390$.

Trick: $447\times 3=1341$ and $1490\times 3=4470$ → each part = $\frac{10}{3}$, so Rishi = $\frac{1170}{3}=390$.

Answer: (B) ₹390

Concept: Amount $A=P(1+\frac{R\times T}{100})$ for simple interest.

1. Total interest rate $=5\%\times 4=20\%$.
2. So $A=P\times \frac{120}{100}=2280$.
3. $P=2280\times \frac{100}{120}=1900$.

Trick: $120\%\to 2280$, so $1\%\to 19$ and $100\%\to 1900$.

Answer: (A) ₹1,900

Concept: Apply BODMAS: innermost bracket first, then multiplication, then addition, then division.

1. Innermost: $(4-8)=-4$.
2. $2\times (-4)=-8$, so $\{29+(-8)\}=21$.
3. $147÷21=7$.

Trick: Solve brackets from inside out: ( ) → { } → [ ].

Answer: (D) 7

Concept: For equal distances at speeds $x$ and $y$, average speed $=\frac{2xy}{x+y}$ (harmonic mean, not simple average).

1. Here $x=6$, $y=12$ km/h.
2. Average speed $=\frac{2\times 6\times 12}{6+12}=\frac{144}{18}=8$ km/h.
3. Check: take distance $12\text{km}$ → times $2\text{h}$ and $1\text{h}$ → $\frac{24}{3}=8$.

Trick: Simple average 9 is wrong – the slower leg takes more time, so the answer is less than 9.

Answer: (D) 8

Concept: Use $\frac{1}{{\tan }^2\theta }={\cot }^2\theta$, $\tan \theta =\frac{\sin \theta }{\cos \theta }$ and $1+{\cot }^2\theta ={\csc }^2\theta$.

1. First term: $\frac{1}{{\tan }^2\theta }={\cot }^2\theta$.
2. Second term: $\frac{{\tan }^2\theta }{{\sin }^2\theta }=\frac{1}{{\cos }^2\theta }$, so $\frac{1}{{\cos }^2\theta }\times {\cos }^2\theta =1$.
3. Sum $={\cot }^2\theta +1={\csc }^2\theta$.

Trick: Put $\theta ={45}^{\circ }$: $1+\frac{1}{0.5}\times 0.5=2$, and ${\csc }^2{45}^{\circ }=2$.

Answer: (C) ${\csc }^2\theta$

Concept: In $△ABC$ the angle sum is ${180}^{\circ }$; with $BC\parallel AD$, alternate angles are equal; in an isosceles triangle base angles are equal.

1. In $△ABC$: $\angle BCA={180}^{\circ }-{45}^{\circ }-{104}^{\circ }={31}^{\circ }$.
2. $BC\parallel AD$, so $\angle CAD=\angle BCA={31}^{\circ }$ (alternate angles).
3. $AC=CD$, so $\angle CDA=\angle CAD={31}^{\circ }$.
4. $\angle ACD={180}^{\circ }-{31}^{\circ }-{31}^{\circ }={118}^{\circ }$.

Trick: Answer $={180}^{\circ }-2\times ({180}^{\circ }-{45}^{\circ }-{104}^{\circ })={180}^{\circ }-{62}^{\circ }$.

Answer: (D) ${118}^{\circ }$

Concept: For the same base: $a^m÷a^n=a^{m-n}$ and $a^m\times a^n=a^{m+n}$.

1. ${69}^{-4}÷{69}^{10}={69}^{-4-10}={69}^{-14}$.
2. ${69}^{-14}\times {69}^{-17}={69}^{-14-17}={69}^{-31}$.

Trick: Add powers with signs: $-4-10-17=-31$ (dividing flips the sign of 10).

Answer: (B) ${69}^{-31}$

Concept: LCM of numbers in power form = product of each base with its highest power (the bases 5, 9, 14, 19 are pairwise co-prime).

1. Highest power of 5: $5^3$; of 9: $9^3$.
2. Highest power of 14: ${14}^2$; of 19: ${19}^2$.
3. LCM $=5^3\times 9^3\times {14}^2\times {19}^2$.

Yaad rakhein: LCM → highest powers; HCF → lowest powers of common bases.

Answer: (A) $5^3\times 9^3\times {14}^2\times {19}^2$

Concept: Let a pen cost $p$ and a pencil cost $c$ (in ₹); form two equations and solve.

1. $13p+3c=152$ … (i)
2. New prices: $18(p-1)+3(c+7)=165\Rightarrow 18p+3c=165+18-21=162$ … (ii)
3. (ii) − (i): $5p=10\Rightarrow p=2$; then $3c=152-26=126\Rightarrow c=42$.
4. $12p+11c=24+462=486$.

Trick: Both equations have $3c$ – subtract directly to get $p$.

Answer: (B) ₹486

Concept: To test a number $n$ for primality, check division by primes up to $\sqrt{n}$ (here up to 19, since $\sqrt{409}\approx 20.2$).

1. Even numbers 402, 404, 406, 408 and 405 (ends in 5) are not prime.
2. $403=13\times 31$ and $407=11\times 37$ – not prime.
3. 409 is not divisible by 2, 3, 5, 7, 11, 13, 17 or 19 – prime. So only 1 prime.

Trick: Odd numbers not ending in 5 are the only candidates: 403, 407, 409.

Answer: (A) 1

Concept: Total runs = average $\times$ number of matches; combined average = total runs $÷$ total matches.

1. Runs in 23 matches $=23\times 40=920$
2. Runs in next 10 matches $=10\times 15=150$
3. Total $=920+150=1070$ in 33 matches
4. Average $=\frac{1070}{33}=32.42$

Trick: Deviation method: $40-\frac{10\times (40-15)}{33}=40-\frac{250}{33}=40-7.58=32.42$

Answer: (B) $32.42$

Concept: If two numbers are $a$ and $b$, both their sum and their LCM must be checked; the quickest way is to test the options.

1. $351=3^3\times 13=27\times 13$
2. Option (A): $13+27=40$ ✓ and $\text{LCM}(13,27)=351$ ✓ (they are co-prime)
3. (B) $\text{LCM}(18,22)=198$, (C) $\text{LCM}(11,29)=319$, (D) $\text{LCM}(6,34)=102$ — all wrong

Trick: Factorise the LCM first: $351=27\times 13$, and $27+13=40$ instantly.

Answer: (A) 13, 27

Concept: Savings = Income $-$ Expenditure. Increase income and expenditure separately, then compare savings.

1. Savings $=24\%$ of 41400 $=9936$; Expenditure $=41400-9936=31464$
2. New income $=41400\times 1.22=50508$
3. New expenditure $=31464\times 1.25=39330$
4. New savings $=50508-39330=11178$; change $=11178-9936=+1242$

Trick: Change in savings $=22\%$ of $41400-25\%$ of $31464=9108-7866=1242$ (increase).

Answer: (A) Increase of ₹1,242

Concept: $\text{Average speed}=\frac{\text{Total distance}}{\text{Total time}}$ (never the average of the speeds).

1. Times: $\frac{234}{78}=3$ h, $\frac{441}{49}=9$ h, $\frac{488}{61}=8$ h
2. Total time $=3+9+8=20$ h
3. Total distance $=234+441+488=1163\text{km}$
4. Average speed $=\frac{1163}{20}=58.15\text{km/h}$

Yaad rakhein: Each distance here is an exact multiple of its speed — compute times first.

Answer: (C) $58.15$

Concept: Volume of a sphere $V=\frac{4}{3}\pi r^3$.

1. $r^3=7^3=343$
2. $V=\frac{4}{3}\times \frac{22}{7}\times 343=\frac{4\times 22\times 49}{3}$
3. $=\frac{4312}{3}=1437.33\approx 1437{\text{cm}}^3$

Trick: With $r=7$: $\frac{4}{3}\times 22\times 49$ — cancel 7 first, then divide by 3 at the end.

Answer: (B) $1437{\text{cm}}^3$

Concept: Write each ratio as a fraction (decimal) and compare.

1. $\frac{37}{56}\approx 0.661$
2. $\frac{38}{75}\approx 0.507$
3. $\frac{33}{54}\approx 0.611$
4. $\frac{35}{80}=0.4375$ → largest is $37:56$

Trick: Only $37:56$ and $33:54$ are above $\frac{1}{2}$ comfortably; cross-multiply: $37\times 54=1998>33\times 56=1848$.

Answer: (A) $37:56$

Concept: Increasing a number by $80\%$ multiplies it by $1.8$.

1. $x\times \frac{180}{100}=2430$
2. $x=\frac{2430\times 100}{180}=1350$

Trick: $180\%\to 2430$, so $10\%\to 135$ and $100\%\to 1350$.

Answer: (D) 1350

Concept: Form two linear equations from the two conditions and solve.

1. Now: $8P=Q+12$
2. After 8 years: $3(Q+8)=8(P+8)-28\Rightarrow 3Q+24=8P+36$
3. Put $8P=Q+12$: $3Q+24=Q+48\Rightarrow 2Q=24$
4. $Q=12$ years (and $P=3$)

Trick: Check option (A): $Q=12\Rightarrow P=3$; later $3\times 20=60$ and $8\times 11-28=60$ ✓

Answer: (A) 12

Concept: Take total work = LCM of the days; find each person’s daily units.

1. Total $=\text{LCM}(5,15)=15$ units; Abhay $=3$/day, Bharat $=1$/day
2. In 2 days together: $2\times (3+1)=8$ units
3. Remaining $=15-8=7$ units
4. Abhay alone: $\frac{7}{3}=2\frac{1}{3}$ days

Trick: Work done together $=2(\frac{1}{5}+\frac{1}{15})=\frac{8}{15}$; rest $\frac{7}{15}\times 5=\frac{7}{3}$ days.

Answer: (B) $2\frac{1}{3}$

Concept: Apply the divisibility tests for 5, 4, 9 and 12 (= 3 and 4).

1. Last digit is 5 → divisible by 5 ✓
2. Last two digits 05 → not divisible by 4, hence not by 12
3. Digit sum $7+4+0+7+0+5=23$ → not divisible by 9 (or 3)

Yaad rakhein: An odd number can never be divisible by 4 or 12 — check the last digit first.

Answer: (D) 5