Q&A: practice questions with answers

Concept: Monthly saving = income − expenses; in the birthday month the saving becomes zero.

1. Monthly saving $=10000-6000-2500=1500$
2. Savings occur in 11 months only.
3. Yearly savings $=1500\times 11=16500$

Trick: One month lost → multiply by 11, not 12.

Answer: (B) ₹16,500

Concept: Net filling rate = filling rate − emptying rate.

1. Net per minute $=\frac{1}{10}-\frac{1}{20}=\frac{1}{20}$
2. Time = 20 minutes.

Trick: Time $=\frac{ab}{b-a}=\frac{10\times 20}{20-10}=20$.

Answer: (D) 20

Concept: Make the common term (y) equal to combine the ratios.

1. $x:y=2:5=8:20$, $y:z=4:7=20:35$ → $x:y:z=8:20:35$
2. Total parts = 63; one part $=rac1512063=240$
3. $x=1920,y=4800,z=8400$

Trick: Multiply crosswise: $x:y:z=2\times 4:5\times 4:5\times 7$.

Answer: (D) 1,920, 4,800 and 8,400

Concept: If 40% is spent, 60% is saved.

1. $60\%$ of income $=18000$
2. Income $=\frac{18000\times 100}{60}=30000$

Trick: $1\%=300$ → $100\%=30000$.

Answer: (B) ₹30,000

Concept: Covered $=\frac{4}{3}$ × remaining → covered : remaining = 4 : 3.

1. Covered $=\frac{4}{7}\times 37=\frac{148}{7}\text{km}$
2. Time $=1\text{h}5\text{min}=\frac{13}{12}\text{h}$
3. Speed $=\frac{148}{7}\times \frac{12}{13}=\frac{1776}{91}\approx 19.5\text{km/h}$

Trick: "a is $\frac{p}{q}$ of b" → split total in $p:q$.

Answer: (B) $19.5$

Concept: $a:b::c:x$ means $\frac{a}{b}=\frac{c}{x}$, so $x=\frac{b\times c}{a}$.

1. $x=\frac{17.4\times 14}{3.5}$
2. $\frac{14}{3.5}=4$ → $x=17.4\times 4=69.6$

Trick: 14 is 4 times 3.5, so x is 4 times 17.4.

Answer: (C) $69.6$

Concept: Range $=y-x$; sum of marks $=\text{mean}\times n$.

1. Sum $=64\times 12=768$; known 10 marks add to $648$ → $x+y=120$
2. $y-x=70$
3. $x=25,y=95$ (25 below 28 and 95 above 83 ✓)

Trick: Sum and difference of x, y → add and halve.

Answer: (C) $(25,95)$

Concept: "At 70% profit" means the price is marked 70% above cost; a discount is then taken on that price.

1. Pen 1: $630\times 1.70=1071$; after 9% discount $1071\times 0.91=974.61$ → profit $344.61$
2. Pen 2: profit $=722\times 0.71=512.62$
3. Total $=344.61+512.62=857.23$

Trick: Pen 1 profit directly: $630\times (1.7\times 0.91-1)=630\times 0.547$.

Answer: (B) ₹857.23

Concept: Use $a^2-b^2=(a+b)(a-b)$.

1. $a+b=3x+4$, $a-b=x+2$
2. $(3x+4)(x+2)=3x^2+6x+4x+8=3x^2+10x+8$

Trick: Check with $x=0$: $9-1=8$ ✓.

Answer: (A) $3x^2+10x+8$

Concept: Solve the fraction first, then add/subtract.

1. $\frac{72}{6}=12$
2. $70+12+29-30=81$

Trick: Group: $70-30=40$, $40+12+29=81$.

Answer: (B) 81

Concept: Curved surface area of a cone $=\pi rl$ and $l^2=r^2+h^2$.

1. $r=\frac{100}{2}=50$; $\pi \times 50\times l=6500\pi \Rightarrow l=130$
2. $h=\sqrt{{130}^2-{50}^2}=\sqrt{16900-2500}=\sqrt{14400}=120$

Trick: 50, 120, 130 = 10 × (5, 12, 13) triplet.

Answer: (B) 120

Concept: Find litres per hour of each pipe and add.

1. A: $\frac{3700}{9}$, B: $\frac{3700}{6}$ litres/h
2. Together $=3700(\frac{1}{9}+\frac{1}{6})=3700\times \frac{5}{18}=\frac{9250}{9}$ litres/h
3. Time $=\frac{6400\times 9}{9250}=\frac{57600}{9250}=\frac{1152}{185}$ hours

Trick: Together they fill 3700 L in $\frac{18}{5}$ h → 6400 L in $\frac{18}{5}\times \frac{6400}{3700}$ h.

Answer: (A) $\frac{1152}{185}$ hours

Concept: For 2 years: $\text{CI}-\text{SI}=P{\left(\frac{R}{100}\right)}^2$.

1. $47100\times {\left(\frac{9}{100}\right)}^2=47100\times 0.0081$
2. $=381.51$

Trick: 2-year difference = interest on the first year's interest.

Answer: (C) ₹381.51

Concept: New mean = old mean + $\frac{\text{total increase}}{n}$.

1. Total increase $=10\times 12\frac{1}{2}+10\times 15=125+150=275$
2. Increase in mean $=\frac{275}{25}=11$
3. New mean $=110+11=121$

Trick: Spread only the extra amount over all 25 numbers.

Answer: (B) 121

Concept: The largest container that divides both volumes exactly = HCF.

1. $156=2^2\times 3\times 13$
2. $256=2^8$
3. HCF $=2^2=4$ litres

Trick: "Maximum size that fits exactly" → HCF.

Answer: (B) 4

Concept: Follow BODMAS: innermost bracket first, then division, then addition.

1. Innermost bracket: $1-6+5=0$.
2. Square bracket: $6\times 0=0$.
3. Division: $0÷12=0$.
4. Addition: $194+0=194$.

Trick: Spot that $(1-6+5)=0$ at once – then the whole bracket term vanishes and the answer is just 194.

Answer: (B) 194

Concept: Curved surface area of a cone $=\pi rl$ and $l^2=r^2+h^2$.

1. Radius $r=\frac{144}{2}=72\text{cm}$.
2. $\pi \times 72\times l=5400\pi \Rightarrow l=\frac{5400}{72}=75\text{cm}$.
3. $h=\sqrt{{75}^2-{72}^2}=\sqrt{(75-72)(75+72)}=\sqrt{3\times 147}=\sqrt{441}=21\text{cm}$.

Trick: Use $a^2-b^2=(a-b)(a+b)$ to avoid squaring big numbers; $(72,75)=3\times (24,25)$ so $h=3\times 7=21$.

Answer: (B) 21

Concept: Mode $=l+\frac{f_1-f_0}{2f_1-f_0-f_2}\times h$, where $l$ is the lower limit of the modal class.

1. Mode 140 is the upper end of class 135–140, so take this as the modal class: $l=135$, $h=5$, $f_1=33$, $f_0=30$, $f_2=x$.
2. $140=135+\frac{33-30}{66-30-x}\times 5$.
3. $5=\frac{15}{36-x}\Rightarrow 36-x=3\Rightarrow x=33$.

Trick: Mode $=l+h$ happens only when $f_1-f_0=2f_1-f_0-f_2$, i.e. $f_2=f_1$, so $x=33$ directly.

Answer: (B) 33

Concept: Form one equation for the total cost using the total number of items.

1. Let sarees $=s$, so suits $=8-s$.
2. $1500s+1000(8-s)=10000$.
3. $500s+8000=10000\Rightarrow s=4$; suits $=8-4=4$.

Trick: If all 8 were suits: ₹8,000. Extra ₹2,000, and each saree costs ₹500 more, so sarees $=\frac{2000}{500}=4$.

Answer: (A) 4 sarees and 4 suits

Concept: The angle subtended by an arc (chord) at the centre is double the angle it subtends at any point on the remaining part of the circle.

1. Let the angle at the circumference be $\theta$.
2. Then the angle at the centre $=2\theta$.
3. Ratio $=2\theta :\theta =2:1$.

Trick: Centre angle is always double – e.g. $\angle ACB={40}^{\circ }\Rightarrow \angle AOB={80}^{\circ }$.

Answer: (B) $2:1$