Q&A: practice questions with answers

Concept: If two numbers are in ratio $a:b$, write them as $ax$ and $bx$.

1. Numbers $=7x$ and $5x$; sum $12x=360\Rightarrow x=30$.
2. Difference $=7x-5x=2x=2\times 30=60$.

Trick: Difference $=\frac{\text{sum}\times (7-5)}{7+5}=\frac{360\times 2}{12}=60$.

Answer: (D) 60

Concept: Negative power flips the base: $a^{-n}={\left(\frac{1}{a}\right)}^n$; power $\frac{1}{2}$ means square root.

1. ${\left(\frac{1}{4}\right)}^{-1/2}=4^{1/2}$.
2. $4^{1/2}=\sqrt{4}=2$.

Trick: First flip the fraction to remove the minus sign, then take the root.

Answer: (B) 2

Concept: $x\%$ of $N=\frac{x}{100}\times N$.

1. $1.25\%=\frac{1.25}{100}=\frac{1}{80}$.
2. $\frac{1}{80}\times 1600=20$, i.e. ₹20.

Trick: $1\%$ of 1600 is 16 and $0.25\%$ is 4, so $1.25\%=16+4=20$.

Answer: (C) ₹20

Concept: Total marks obtained $=\text{average}\%\times \text{total maximum marks}$.

1. Maximum marks $=5\times 105=525$.
2. Total obtained $=80\%\times 525=420$.
3. Given four subjects: $89+92+98+81=360$.
4. Science $=420-360=60$.

Trick: Find total first, then subtract the known marks.

Answer: (C) 60

Concept: Part of a quantity in ratio $a:b$: $\frac{b}{a+b}\times \text{total}$.

1. Total parts $=3+2=5$.
2. Water $=\frac{2}{5}\times 10=4$ litres.

Trick: One part $=10÷5=2$ litres; water has 2 parts, so $2\times 2=4$ litres.

Answer: (D) 4 litres

Concept: Simplify signs first, then add using the LCM of denominators.

1. $\frac{9}{-16}=-\frac{9}{16}$ and $\frac{-5}{-12}=+\frac{5}{12}$.
2. LCM of 16 and 12 $=48$.
3. $-\frac{27}{48}+\frac{20}{48}=\frac{-7}{48}$.

Yaad rakhein: Two negatives give a positive: $\frac{-5}{-12}=\frac{5}{12}$.

Answer: (D) $\frac{-7}{48}$

Concept: When a transversal cuts two parallel lines, alternate interior angles are equal.

1. $\angle 1$ and $\angle 2$ lie on opposite sides of the transversal, between the parallel lines.
2. So $\angle 2=\angle 1={75}^{\circ }$.

Yaad rakhein: Alternate interior angles form a "Z" shape and are always equal; co-interior angles (a "C"/"U" shape) add to ${180}^{\circ }$.

Answer: (A) ${75}^{\circ }$

Concept: $\text{SI}=\frac{P\times R\times T}{100}\Rightarrow P=\frac{\text{SI}\times 100}{R\times T}$.

1. $P=\frac{1056\times 100}{16\times 3}=\frac{105600}{48}$.
2. $P=2200$, i.e. ₹2,200.

Trick: Interest for 3 years $=3\times 16\%=48\%$ of P; $48\%=1056\Rightarrow 1\%=22\Rightarrow 100\%=2200$.

Answer: (A) ₹2,200

Concept: Mode is the value that occurs most often.

1. Without $2k+1$: 6 occurs 3 times and 7 occurs 3 times (a tie).
2. For mode to be 6 alone, $2k+1=6$.
3. $2k=5\Rightarrow k=2.5$.

Trick: The unknown term must join the value given as the mode to break the tie.

Answer: (D) 2.5

Concept: Empirical formula: $\text{Mode}=3\text{Median}-2\text{Mean}$.

1. $3\text{Median}=\text{Mode}+2\text{Mean}=61.1+2\times 58.7$.
2. $3\text{Median}=61.1+117.4=178.5$.
3. $\text{Median}=\frac{178.5}{3}=59.5$.

Trick: Median always lies between mean and mode (here between 58.7 and 61.1) – only 59.5 fits.

Answer: (A) 59.5

Concept: Half-yearly: rate $=\frac{R}{2}$, time $=2T$; $A=P{(1+\frac{r}{100})}^n$.

1. Rate $=\frac{8}{2}=4\%$, periods $=2$.
2. $A=7500\times (1.04{)}^2=7500\times 1.0816$.
3. $A=8112$, i.e. ₹8,112.

Trick: Effective rate for $4\%$ twice $=4+4+\frac{4\times 4}{100}=8.16\%$; $8.16\%$ of 7500 $=612$, so $A=8112$.

Answer: (B) ₹8,112

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

1. Numerator $=(5.32+3.64)(5.32-3.64)=8.96\times 1.68$.
2. Divide by $1.68$: result $=8.96$.

Trick: When the denominator equals $a-b$, the answer is simply $a+b=5.32+3.64$.

Answer: (B) 8.96

Concept: $\text{CP}=\text{SP}-\text{Profit}$; price after $25\%$ discount $=75\%$ of MRP.

1. CP $=18600-600=18000$.
2. He paid $75\%$ of MRP: $\frac{3}{4}\times \text{MRP}=18000$.
3. MRP $=18000\times \frac{4}{3}=24000$, i.e. ₹24,000.

Trick: $25\%$ off means paying $\frac{3}{4}$; so MRP $=\frac{4}{3}\times$ price paid.

Answer: (B) ₹24,000

Concept: Translate the sentence into an equation.

1. Let Radhika’s age $=x$.
2. Teacher $=2x+20=60$.
3. $2x=40\Rightarrow x=20$ years.

Trick: Check options: $2\times 20+20=60$ ✓.

Answer: (C) 20 years

Concept: $1+{\cot }^2\theta ={\csc }^2\theta$, $1+{\tan }^2\theta ={\sec }^2\theta$, $(1+\cos \theta )(1-\cos \theta )={\sin }^2\theta$, $(1-\sin \theta )(1+\sin \theta )={\cos }^2\theta$.

1. First term $={\csc }^2\theta ·{\sin }^2\theta =1$.
2. Second term $={\cos }^2\theta ·{\sec }^2\theta =1$.
3. Value $=1-1=0$.

Trick: Put $\theta ={45}^{\circ }$: $(2)\left(\frac{1}{2}\right)-\left(\frac{1}{2}\right)(2)=0$.

Answer: (A) 0

Concept: BODMAS: solve the innermost bracket first.

1. $(7-3)=4$.
2. $\{10-5+4\}=9$.
3. $20+9=29$.

Yaad rakhein: Order of brackets: ( ) → { } → [ ].

Answer: (D) 29

Concept: Work per minute: filling pipe positive, emptying pipe negative.

1. A fills $\frac{1}{30}$, B empties $\frac{1}{45}$ per minute; net $=\frac{1}{30}-\frac{1}{45}=\frac{1}{90}$.
2. In 30 min: $30\times \frac{1}{90}=\frac{1}{3}$ filled.
3. Remaining $\frac{2}{3}$ by A alone: $\frac{2}{3}\times 30=20$ min.

Trick: Take capacity $=90$ units (LCM): A $=+3$, B $=-2$; 30 min → 30 units; remaining 60 at 3 per min → 20 min.

Answer: (B) 20 min

Concept: HCF of two numbers also divides their difference.

1. Difference $=2700-2410=290=2\times 5\times 29$.
2. 2410 and 2700 both end in 0, so divisible by 10; 29 does not divide 2700.
3. HCF $=10$.

Trick: HCF must be a factor of the difference (290) – check its factors from largest down.

Answer: (C) 10

Concept: Sum $S$ and difference $D$: greater $=\frac{S+D}{2}$, smaller $=\frac{S-D}{2}$.

1. Greater $=\frac{58+18}{2}=\frac{76}{2}=38$.
2. Smaller $=58-38=20$; check $38-20=18$ ✓.

Yaad rakhein: Greater number $=\frac{\text{sum}+\text{difference}}{2}$.

Answer: (D) 38

Concept: Count of multiples of $n$ up to $N=\lfloor \frac{N}{n}\rfloor$.

1. Multiples of 7 up to 90: $\lfloor \frac{90}{7}\rfloor =12$.
2. Multiples of 7 up to 11: $\lfloor \frac{11}{7}\rfloor =1$ (only 7).
3. Between 11 and 90: $12-1=11$ (14, 21, …, 84).

Trick: First $=14=7\times 2$, last $=84=7\times 12$ → $12-2+1=11$.

Answer: (D) 11