Q&A: practice questions with answers

Concept: Let the numbers be $x,5x,6x$ (1, 5, 6 have no common factor, so HCF $=x$).

1. LCM $=x\times \text{LCM}(1,5,6)=30x$.
2. $30x=5250$ ⇒ $x=175$.
3. HCF $=175$ (numbers 175, 875, 1050).

Trick: HCF $=\frac{\text{LCM}}{\text{LCM of ratio terms}}=\frac{5250}{30}$.

Answer: (B) 175

Concept: Surface area $=4\pi r^2$, volume $=\frac{4}{3}\pi r^3$.

1. $4\pi r^2=4\pi$ ⇒ $r^2=1$ ⇒ $r=1$ cm.
2. $V=\frac{4}{3}\pi \times 1^3=\frac{4\pi }{3}{\text{cm}}^3$.

Trick: $V=\frac{S\times r}{3}=\frac{4\pi \times 1}{3}$.

Answer: (D) $\frac{4\pi }{3}{\text{cm}}^3$

Concept: Find profit per article, then divide the required profit by it.

1. CP per bag $=\frac{48884}{44}=$ ₹1111.
2. SP per bag $=\frac{35252}{28}=$ ₹1259.
3. Profit per bag $=1259-1111=$ ₹148.
4. Bags needed $=\frac{6956}{148}=47$.

Trick: Check the last digit: $148\times 47$ ends in 6, matching 6956.

Answer: (C) 47

Concept: $\text{SP}=\text{CP}+\text{Profit}$; cost price stays the same when SP changes.

1. Let initial profit $=P$, so initial $\text{SP}=C+P$.
2. New SP $=3(C+P)$ and new profit $=4P$, so $3(C+P)=C+4P$.
3. $3C+3P=C+4P\Rightarrow P=2C$.
4. Initial profit $\%=\frac{2C}{C}\times 100=200\%$.

Trick: Write both SPs as $C+\text{profit}$ and equate – the cost cancels to a simple ratio.

Answer: (C) 200

Concept: A number is divisible by 9 if the sum of its digits is divisible by 9.

1. Digit sum $=5+1+8+8+3+p+4+7=36+p$.
2. 36 is already a multiple of 9, so $p$ must be 0 or 9.
3. Maximum value of $p=9$.

Trick: If the known digits already sum to a multiple of 9, the blank can be only 0 or 9.

Answer: (D) 9

Concept: Total $=\text{average}\times \text{count}$; then divide the total in the given ratio.

1. Total price $=15015\times 3=45045$.
2. Sum of ratio terms $=3+5+7=15$.
3. Costliest item $=\frac{7}{15}\times 45045=7\times 3003=21021$.

Trick: Average of the ratio is $\frac{15}{3}=5$ parts $=15015$, so 1 part $=3003$ and 7 parts $=21021$.

Answer: (C) ₹21,021

Concept: If $a\%$ of $x=b$, then $x=\frac{b\times 100}{a}$.

1. $\frac{2}{100}\times x=360$.
2. $x=\frac{360\times 100}{2}=18000$.

Trick: $2\%=360\Rightarrow 1\%=180\Rightarrow 100\%=18000$.

Answer: (B) 18000

Concept: Write everything as powers of small primes, then cancel.

1. Numerator: $2^2\times 9^3=4\times 729=2916$.
2. Denominator: $\sqrt{4}\times \sqrt{6561}=2\times 81=162$.
3. Value $=\frac{2916}{162}=18$.

Trick: $9^3=3^6$ and $\sqrt{6561}=3^4$, so $\frac{4\times 3^6}{2\times 3^4}=2\times 9=18$.

Answer: (B) 18

Concept: Convert each statement into an equation and add the two final equations.

1. $2P+J=48$ … (i)
2. $S=2J$, $M=P+4$ and $S+M=46\Rightarrow 2J+P+4=46\Rightarrow P+2J=42$ … (ii)
3. Add (i) and (ii): $3P+3J=90\Rightarrow P+J=30$.

Trick: When coefficients are swapped ($2P+J$ and $P+2J$), just add and divide by 3.

Answer: (C) 30

Concept: Translate the sentence directly into an equation.

1. $1200\%$ of $25=12\times 25=300$.
2. $25\%$ of $k=300-10=290$.
3. $\frac{k}{4}=290\Rightarrow k=1160$.

Trick: $25\%=\frac{1}{4}$, so multiply by 4: $290\times 4=1160$.

Answer: (D) 1160

Concept: Take tank capacity = LCM of times; work = rate × time for each pipe.

1. Capacity $=\text{LCM}(21,42)=42$ units. Rate of A $=2$ units/min, B $=1$ unit/min.
2. A runs all 16 minutes: $16\times 2=32$ units.
3. Remaining $42-32=10$ units are filled by B ⇒ B runs $10$ minutes.

Trick: Fill-time of the pipe kept open for the full period first; the shortfall gives the other pipe's time.

Answer: (B) 10 minutes

Concept: For $a{x}^2+bx+c=0$, $D=b^2-4ac$; $D<0$ ⇒ roots imaginary, $D=0$ ⇒ real and equal, $D>0$ ⇒ real and distinct.

1. $a=3$, $b=-4$, $c=2$.
2. $D=(-4{)}^2-4\times 3\times 2=16-24=-8$.
3. $D<0$ ⇒ the roots are imaginary (not real).

Trick: A negative discriminant can never give real roots – options with $-8$ and "real" are impossible.

Answer: (A) Discriminant $=-8$, roots are imaginary

Concept: Third proportional of $a$ and $b$ is $c$ where $a:b=b:c$, i.e. $c=\frac{b^2}{a}$.

1. $a=10x^2$, $b=30xy$.
2. $c=\frac{(30xy{)}^2}{10x^2}=\frac{900x^2{y}^2}{10x^2}$.
3. $c=90y^2$.

Trick: Numbers alone: $\frac{{30}^2}{10}=90$; variables: $\frac{x^2{y}^2}{x^2}=y^2$.

Answer: (D) $90y^2$

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

1. $\frac{8}{6}\text{ of }\frac{6}{6}=\frac{8}{6}\times 1=\frac{8}{6}$.
2. $3\times \frac{8}{6}=4$.
3. $50÷4=\frac{25}{2}$.

Trick: $\frac{6}{6}=1$, so the bracket is just $3\times \frac{4}{3}=4$.

Answer: (B) $\frac{25}{2}$

Concept: Parallel lines cut any two transversals proportionally: $\frac{AB}{CD}=\frac{EF}{GH}$.

1. $\frac{5}{3}=\frac{15}{GH}$.
2. $GH=\frac{15\times 3}{5}=9\text{cm}$.

Trick: T's intercepts are $\frac{15}{5}=3$ times S's, so $GH=3\times 3=9$.

Answer: (A) 9

Concept: Same direction ⇒ relative speed = difference; opposite ⇒ sum. Distance = total length.

1. Overtaking the man: relative speed $=\frac{120}{18}=\frac{20}{3}\text{m/s}=24\text{km/h}$ ⇒ train speed $=24+9=33\text{km/h}$.
2. Opposite direction: relative speed $=33+27=60\text{km/h}=\frac{50}{3}\text{m/s}$.
3. Distance $=120+230=350\text{m}$.
4. Time $=350÷\frac{50}{3}=21\text{s}$.

Trick: $1\text{m/s}=\frac{18}{5}\text{km/h}$: $\frac{20}{3}\times \frac{18}{5}=24$.

Answer: (C) 21

Concept: In an isosceles triangle the two base angles are equal and all angles sum to ${180}^{\circ }$.

1. Base angles together $={180}^{\circ }-{14}^{\circ }={166}^{\circ }$.
2. Each base angle $=\frac{{166}^{\circ }}{2}={83}^{\circ }$.

Trick: Each base angle $={90}^{\circ }-\frac{\text{vertex}}{2}={90}^{\circ }-7^{\circ }={83}^{\circ }$.

Answer: (C) ${83}^{\circ }$ and ${83}^{\circ }$

Concept: Check the differences between consecutive terms.

1. $982-980=2$, $980-976=4$, $976-970=6$, $970-962=8$.
2. Differences are even numbers increasing by 2, so the next difference is $10$.
3. $962-10=952$.

Trick: Subtract consecutive even numbers: $-2,-4,-6,-8,-10$.

Answer: (A) 952

Concept: Apply the rule digit by digit.

1. 596432: odd digits $5,9,3$ become $2,6,0$; even digits $6,4,2$ become $8,6,4$.
2. New number: 2 6 8 6 0 4.
3. Greatest digit $=8$, smallest digit $=0$, sum $=8+0=8$.

Trick: Only the extreme digits matter: largest even digit $6\to 8$ and odd digit $3\to 0$.

Answer: (A) 8

Concept: Replace the letters by signs, then use BODMAS.

1. Expression: $43-15+7\times 9-(28÷2)$.
2. Bracket: $28÷2=14$; multiplication: $7\times 9=63$.
3. $43-15+63-14=77$.

Trick: Write the sign-converted expression first, then solve bracket → ×/÷ → +/−.

Answer: (D) 77