Q&A: practice questions with answers

Concept: BODMAS: brackets first, then "of", then division/multiplication, then addition.

1. $\frac{20}{36}÷\frac{20}{5}=\frac{20}{36}\times \frac{5}{20}=\frac{5}{36}$.
2. $\frac{5}{11}\times \frac{22}{15}+\frac{3}{12}=\frac{2}{3}+\frac{1}{4}=\frac{11}{12}$; so $\frac{5}{36}÷\frac{11}{12}=\frac{5}{33}$.
3. "of" first: $\frac{33}{21}\times \frac{21}{5}=\frac{33}{5}$; then $\frac{5}{5}÷\frac{33}{5}=\frac{5}{33}$.
4. Total $=\frac{5}{33}+\frac{5}{33}=\frac{10}{33}$.

Trick: "of" is solved before $÷$ — the commonest trap in such questions.

Answer: (D) $\frac{10}{33}$

Concept: Net change for $-a\%$ then $+b\%$: $-a+b-\frac{ab}{100}$.

1. $-30+50-\frac{30\times 50}{100}=20-15=5$.
2. Positive, so it is a $5\%$ increase.
3. Check: $100\to 70\to 105$.

Trick: Multiply factors: $0.7\times 1.5=1.05\Rightarrow +5\%$.

Answer: (D) 5% increase

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

1. $655\times \frac{80}{100}=655\times 0.8$.
2. $=524$.

Trick: $20\%$ of 655 $=131$; $655-131=524$.

Answer: (D) ₹524

Concept: Half-yearly: rate $=\frac{8}{2}=4\%$ per period, periods $=2$; $A=P{(1+\frac{r}{100})}^n$.

1. $10816=P(1.04{)}^2=1.0816P$.
2. $P=\frac{10816}{1.0816}=10000$.

Trick: $(1.04{)}^2=1.0816$ — the amount\'s digits "10816" give away $P=10,000$.

Answer: (C) ₹10,000

Concept: $(a-b{)}^2=(a+b{)}^2-4ab$.

1. $(a-b{)}^2={14}^2-4\times 45=196-180=16$.
2. $a-b=4$ (numbers are 9 and 5).

Trick: Find two numbers with sum 14, product 45: $9$ and $5$, difference $4$.

Answer: (B) 4

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

1. Work $=\text{LCM}(12,6)=12$ units; A = 1 unit/day, B = 2 units/day.
2. Together for 2 days: $2\times 3=6$ units.
3. Remaining $6$ units by B: $\frac{6}{2}=3$ days.
4. Total $=2+3=5$ days.

Trick: B alone takes 6 days; in 2 days A did $\frac{2}{12}=\frac{1}{6}$ of work = 1 day of B, so B needs $6-1=5$ days in all.

Answer: (D) 5

Concept: LCM of fractions $=\frac{\text{LCM of numerators}}{\text{HCF of denominators}}$.

1. Write $15=\frac{15}{1},10=\frac{10}{1},0.4=\frac{2}{5}$.
2. LCM of numerators $(15,10,2)=30$; HCF of denominators $(1,1,5)=1$.
3. LCM $=\frac{30}{1}=30$.

Trick: Check: $30÷0.4=75$, $30÷15=2$, $30÷10=3$ — all whole numbers.

Answer: (D) 30

Concept: A number is divisible by 9 if its digit sum is divisible by 9.

1. Digit sum of 95061 $=9+5+0+6+1=21$; remainder on $÷9$ is $3$.
2. Below: $95061-3=95058$ (digit sum 27 ✔). Above: $95061+6=95067$.
3. 95058 is nearer (3 away) and is among the options.

Trick: Subtract the remainder of the digit sum: $21\to$ remainder 3, so $95061-3$.

Answer: (A) 95058

Concept: Take present ages as multiples of the ratio and use the future ratio.

1. Father $=12x$, son $=7x$.
2. $\frac{12x+4}{7x+4}=\frac{5}{3}\Rightarrow 36x+12=35x+20\Rightarrow x=8$.
3. Sum $=19x=19\times 8=152$ years.

Trick: Sum must be a multiple of $12+7=19$: only $152=19\times 8$ fits.

Answer: (B) 152 years

Concept: $\tan \theta =\frac{\text{height}}{\text{shadow}}$, where $\theta$ is the sun\'s angle of elevation.

1. $\tan \theta =\frac{\sqrt{3}}{1}=\sqrt{3}$.
2. $\tan {60}^{\circ }=\sqrt{3}\Rightarrow \theta ={60}^{\circ }$.

Trick: Height $>$ shadow means angle $>{45}^{\circ }$; $\sqrt{3}:1$ is the ${60}^{\circ }$ pair.

Answer: (C) ${60}^{\circ }$

Concept: Q\'s selling price is the price R pays; compare it with the 25% profit price.

1. Let CP $=x$. Q buys at $1.12x$ and sells to R at $1.12x+26$.
2. Direct sale at $25\%$ profit: $1.25x$.
3. $1.25x=1.12x+26\Rightarrow 0.13x=26\Rightarrow x=200$.

Trick: $25\%-12\%=13\%$ of CP equals ₹26, so CP $=\frac{26}{13}\times 100=200$.

Answer: (D) 200

Concept: Time saved occurs only on the 98 km part: $\frac{98}{v}-\frac{98}{v+7}=\frac{42}{60}$ h.

1. $98·\frac{7}{v(v+7)}=0.7$.
2. $v(v+7)=\frac{686}{0.7}=980$.
3. $28\times 35=980\Rightarrow v=28\text{km/h}$.

Trick: Test options in $v(v+7)=980$: $28\times 35=980$ ✔ directly.

Answer: (B) 28 km/hr

Concept: CSA $=2\pi rh$, Volume $=\pi r^{2}h$.

1. $2\times \frac{22}{7}\times 3.5\times h=616\Rightarrow 22h=616\Rightarrow h=28\text{cm}$.
2. $V=\frac{22}{7}\times 3.5\times 3.5\times 28=38.5\times 28$.
3. $V=1078{\text{cm}}^3$.

Trick: $V=\text{CSA}\times \frac{r}{2}=616\times \frac{3.5}{2}=1078$.

Answer: (D) 1078 cm³

Concept: Average $=\frac{\text{sum}}{\text{count}}$.

1. Primes between 5 and 30: 7, 11, 13, 17, 19, 23, 29 (7 numbers).
2. Sum $=119$.
3. Average $=\frac{119}{7}=17$.

Trick: "Between 5 and 30" excludes 5; including it would wrongly give $\frac{124}{8}=15.5$, not an option.

Answer: (A) 17

Concept: Compare the actual cost of the goods given with the money received.

1. For "1 kg" he gives 900 g and gets ₹150.
2. Cost of 900 g $=120\times 0.9=108$.
3. Profit $\%=\frac{150-108}{108}\times 100=\frac{42}{108}\times 100=38\frac{8}{9}\%$.

Trick: Combined: $\frac{150}{120}\times \frac{1000}{900}=\frac{25}{18}\Rightarrow \frac{7}{18}\times 100=38\frac{8}{9}\%$.

Answer: (A) $38\frac{8}{9}\%$

Concept: Remove the deductions first; the rest is split in the given ratio.

1. Total deducted $=25+52+18=95$; remaining $=1472-95=1377$.
2. $15k+19k+17k=51k=1377\Rightarrow k=27$.
3. Vikas after deduction $=17\times 27=459$.

Trick: Remaining amount must be divisible by $51$: $1377÷51=27$.

Answer: (D) ₹459

Concept: Selling price after successive discounts $=MP\times (1-d_1)(1-d_2)\dots$

1. I: $0.11\times 0.32=0.0352$ of MP.
2. II: $0.50\times 0.49=0.245$; III: $0.75\times 0.89=0.6675$.
3. IV: $0.06\times 0.51\times 0.79\approx 0.0242$ of MP — the smallest.

Trick: Compare the "remaining" fractions, not the discount totals; IV leaves only about $2.4\%$ of MP.

Answer: (B) IV

Concept: Third proportional of $a$ and $b$: $a:b=b:x\Rightarrow x=\frac{b^2}{a}$.

1. $6:12=12:x$.
2. $x=\frac{{12}^2}{6}=\frac{144}{6}=24$.

Trick: $6\to 12$ doubles, so $12\to 24$.

Answer: (A) 24

Concept: Write everything as powers of 3: $(a^m{)}^n=a^{mn}$, $a^m\times a^n=a^{m+n}$.

1. Numerator: $(3^2\times 3^2{)}^3=(3^4{)}^3=3^{12}$.
2. Denominator: ${27}^2\times 3^4=3^6\times 3^4=3^{10}$.
3. $\frac{3^{12}}{3^{10}}=3^2=9$.

Trick: Convert 9 and 27 to base 3 first — then just subtract exponents.

Answer: (C) 9

Concept: If every number decreases by $k$, the average also decreases by $k$.

1. New average $=43-7=36$.

Yaad rakhein: Adding/subtracting the same value to all shifts the average by that value; multiplying scales it.

Answer: (B) 36