Q&A: practice questions with answers

Concept: The mean proportional between $a$ and $b$ is $\sqrt{ab}$ (since $a:x::x:b\Rightarrow x^2=ab$).

1. $x=\sqrt{9\times 25}=\sqrt{225}=15$.
2. Check: $9:15=15:25$ (both $=3:5$).

Trick: $\sqrt{9}\times \sqrt{25}=3\times 5=15$ – multiply the square roots directly.

Answer: (D) 15

Concept: For grouped data, Median $=l+\frac{\frac{N}{2}-cf}{f}\times h$, where the median class is the first class whose cumulative frequency reaches $\frac{N}{2}$.

1. Cumulative frequencies: 5, 14, 28, 50, 80, 100. $N=100$, $\frac{N}{2}=50$.
2. The cf first reaches 50 in class 30–40, so $l=30$, $cf=28$ (previous), $f=22$, $h=10$.
3. Median $=30+\frac{50-28}{22}\times 10=30+10=40$.

Trick: When $\frac{N}{2}$ equals a cumulative frequency exactly, the median is the upper limit of that class – here 40 at once.

Answer: (D) 40.0

Concept: Overall average $=\frac{\text{total amount}}{\text{total members}}$ – a weighted mean, not the mean of the three averages.

1. Total $=32\times 266+52\times 470+73\times 168=8512+24440+12264=45216$.
2. Members $=32+52+73=157$.
3. Average $=\frac{45216}{157}=288$ (since $157\times 288=45216$).

Trick: Take 288 as a guess: deviations $-22\times 32$, $+182\times 52$, $-120\times 73$ $=-704+9464-8760=0$ ✓.

Answer: (A) 288

Concept: Mean $=\frac{\sum f_i{x}_i}{\sum f_i}$ with $x_i$ the class marks.

1. Class marks: 15, 25, 35, 45, 55.
2. $\sum f_i{x}_i=42\times 15+19\times 25+23\times 35+26\times 45+55X=630+475+805+1170+55X=3080+55X$.
3. $\sum f_i=110+X$. Mean $=40\Rightarrow 3080+55X=40(110+X)=4400+40X$.
4. $15X=1320\Rightarrow X=88$.

Trick: Deviation method with $A=40$: $42(-25)+19(-15)+23(-5)+26(5)+X(15)=0\Rightarrow -1050-285-115+130+15X=0\Rightarrow X=88$.

Answer: (B) 88

Concept: Write $\cot \theta =\frac{\cos \theta }{\sin \theta }$ and factor out $\cos \theta$.

1. $\cot \theta -\cos \theta =\cos \theta (\frac{1}{\sin \theta }-1)=\cos \theta (cosec\theta -1)$.
2. $\cot \theta +\cos \theta =\cos \theta (cosec\theta +1)$.
3. Ratio $=\frac{cosec\theta -1}{cosec\theta +1}$.

Trick: Put $\theta ={30}^{\circ }$: $\frac{\sqrt{3}-\frac{\sqrt{3}}{2}}{\sqrt{3}+\frac{\sqrt{3}}{2}}=\frac{1}{3}$; option B gives $\frac{2-1}{2+1}=\frac{1}{3}$ ✓ (A gives $\frac{2/\sqrt{3}-1}{2/\sqrt{3}+1}\ne \frac{1}{3}$).

Answer: (B) $\frac{cosec\theta -1}{cosec\theta +1}$

Concept: Translate the sentence into an equation: $20\%$ of $N$ $=20\%$ of $620+190$.

1. $20\%$ of $620=124$.
2. $0.2N=124+190=314$.
3. $N=\frac{314}{0.2}=1570$.

Trick: $20\%$ of the difference is 190, so $N-620=190\times 5=950\Rightarrow N=1570$.

Answer: (A) 1570

Concept: Do division and multiplication before addition and subtraction, keeping track of signs.

1. $(-56)÷(-14)=+4$ and $(-3)\times 6=-18$.
2. Expression $=(-9)-4+(-18)=-9-4-18$.
3. $=-31$.

Trick: Negative $÷$ negative $=$ positive; then "$-(+4)$" is a subtraction – the common slip is writing $+4$.

Answer: (B) $-31$

Concept: Average speed $=\frac{\text{total distance}}{\text{total time}}$, never the plain average of speeds.

1. Distance $=20\times 90+70\times 60=1800+4200=6000\text{m}$.
2. Time $=90+60=150\text{min}$.
3. Average speed $=\frac{6000}{150}=40\text{m/min}$.

Trick: Time-weighted mean: $\frac{20\times 3+70\times 2}{3+2}=\frac{200}{5}=40$ (times in ratio $3:2$).

Answer: (C) 40

Concept: Compare cost and selling price per article; $\text{Profit}\%=\frac{\text{SP}-\text{CP}}{\text{CP}}\times 100$.

1. CP per article $=\frac{150}{12}=$ ₹12.50; SP per article $=$ ₹78.
2. Profit per article $=78-12.5=$ ₹65.50.
3. Profit$\%=\frac{65.5}{12.5}\times 100=524\%$.

Trick: Per dozen: CP 150, SP $78\times 12=936$; $\frac{936-150}{150}=\frac{786}{150}=5.24\Rightarrow 524\%$. The "21 dozens" is never needed.

Answer: (B) 524%

Concept: Express all ages in one variable; subtract 3 years per person for "3 years ago".

1. Son $=s$, daughter $=s-3$, Vanraj $=3s$. Present sum $=5s-3$.
2. 3 years ago each is 3 less: $5s-3-9=5s-12=93\Rightarrow 5s=105\Rightarrow s=21$.
3. Vanraj $=3\times 21=63$ years.

Trick: Present sum $=93+3\times 3=102$; then $5s-3=102$. Only option C (63) is a multiple of 3 with a sensible son's age.

Answer: (C) 63

Concept: Volume of a cone $=\frac{1}{3}\pi r^{2}h$. The length 66 cm becomes the base circumference and the breadth 30 cm is taken as the height (largest possible).

1. $2\pi r=66\Rightarrow r=\frac{66\times 7}{44}=10.5\text{cm}$, $h=30\text{cm}$.
2. Full cone: $V=\frac{1}{3}\times \frac{22}{7}\times {10.5}^2\times 30=\frac{1}{3}\times \frac{22}{7}\times 110.25\times 30=3465{\text{cm}}^3$.
3. Stapled 2 cm from the vertex: the tiny cone removed is similar to the big one, $r^{\prime }=10.5\times \frac{2}{30}=0.7$, volume $=\frac{1}{3}\times \frac{22}{7}\times 0.49\times 2\approx 1.03{\text{cm}}^3$.
4. Usable volume $\approx 3465-1.03\approx 3464{\text{cm}}^3$.

Trick: A cone cut parallel to the base at $\frac{1}{k}$ of its height has volume $\frac{1}{k^3}$ of the whole – here $\frac{3465}{{15}^3}\approx 1$.

Answer: (C) 3464

Concept: Each of the four properties is a standard test for a parallelogram.

1. Opposite sides equal ⇒ parallelogram; each diagonal dividing it into congruent triangles ⇒ parallelogram.
2. Opposite angles equal ⇒ parallelogram; diagonals bisecting each other ⇒ parallelogram.
3. A rectangle also satisfies these but is a special case – nothing here forces a right angle, so the figure "has to be" only a parallelogram.

Yaad rakhein: Rectangle needs one right angle (or equal diagonals); kite/trapezium fail "opposite sides equal".

Answer: (C) parallelogram

Concept: Divisible by 6 ⇔ divisible by both 2 and 3.

1. The number 56339A8 ends in 8, so it is even for every A.
2. Digit sum $=5+6+3+3+9+A+8=34+A$ must be a multiple of 3.
3. $34+2=36$ → smallest $A=2$ (next would be 5 or 8).

Trick: For "minimum A", find the first multiple of 3 at or above the fixed digit sum.

Answer: (C) 2

Concept: Simplify each equation; it is quadratic only if, after simplification, the highest power of $x$ is exactly 2.

1. (A) $x^3-3x^2+3x-1=(x-1{)}^3$ is an identity (true for all $x$) – not an equation to solve.
2. (B) $\frac{3x}{x-1}=3\Rightarrow 3x=3x-3$ – linear, no solution. (D) $\frac{(x-1)(x+1)}{x+1}=x-2\Rightarrow x-1=x-2$ – linear.
3. (C) $(x^2+1{)}^2-x^4=2x+1\Rightarrow 2x^2+1=2x+1\Rightarrow 2x^2-2x=0$ – a quadratic.

Trick: Expand and cancel first; the degree you see before simplifying can be misleading.

Answer: (C) $(x^2+1{)}^2-x^4=2x+1$

Concept: With a false weight, actual gain $=\frac{\text{claimed SP}}{\text{cost of goods actually given}}$. Let the true weight used be $w$ g per "1 kg".

1. Selling "1 kg" at 60% loss: he receives $0.4C$, where $C$ = cost of 1000 g.
2. He actually gives $w$ g costing $\frac{w}{1000}C$. Gain 70% ⇒ $\frac{0.4C}{\frac{w}{1000}C}=1.7$.
3. $w=\frac{400}{1.7}=235.29\approx 235.3$ g.

Trick: $w=1000\times \frac{100-\text{loss}\%}{100+\text{gain}\%}=1000\times \frac{40}{170}$.

Answer: (A) 235.3 gm

Concept: Time is directly proportional to the number of pages.

1. 2 pages → 40 min, so 1 page → $20$ min.
2. 87 pages → $87\times 20=1740$ min.

Trick: $\frac{40}{2}\times 87$ – reduce to one unit first, then multiply.

Answer: (C) 1740

Concept: Margin = (winner's % − loser's %) of valid votes. Let registered voters $=N$.

1. Votes polled $=0.8N$; valid $=85\%$ of polled $=0.68N$.
2. Winner 75%, loser 25% ⇒ margin $=50\%$ of valid $=0.34N$.
3. $0.34N=289\Rightarrow N=850$.

Trick: $N=\frac{289}{0.8\times 0.85\times 0.5}=\frac{289}{0.34}=850$.

Answer: (C) 850

Concept: Area of a sector $=\frac{\theta }{{360}^{\circ }}\times \pi r^2$.

1. $\pi r^2=\frac{22}{7}\times 21\times 21=1386{\text{cm}}^2$.
2. $\frac{\theta }{{360}^{\circ }}\times 1386=385\Rightarrow \frac{\theta }{{360}^{\circ }}=\frac{385}{1386}=\frac{5}{18}$.
3. $\theta =\frac{5}{18}\times {360}^{\circ }={100}^{\circ }$.

Trick: Sector area ÷ circle area = angle ÷ ${360}^{\circ }$; $\frac{385}{1386}$ cancels by 77 to $\frac{5}{18}$.

Answer: (D) ${100}^{\circ }$

Concept: Under compound interest, if money doubles in $n$ years it becomes $2^k$ times in $kn$ years.

1. Doubles in 7 years ⇒ $\times 2$ every 7 years.
2. $16=2^4$, so four doublings are needed.
3. Time $=4\times 7=28$ years.

Trick: CI multiplies, it does not add – 2 → 4 → 8 → 16 takes 7 years each step.

Answer: (B) 28

Concept: Cube both expressions directly: $x^3=31+14\sqrt{3}$, $y^3=31-14\sqrt{3}$.

1. $x^3+y^3=62$.
2. $x^3{y}^3=(31+14\sqrt{3})(31-14\sqrt{3})=961-588=373$.
3. $x^3+y^3+3x^3{y}^3=62+3\times 373=62+1119=1181$.

Trick: Conjugate surds multiply to $a^2-b^{2}c$ – no need to simplify the cube roots.

Answer: (A) 1181