Q&A: practice questions with answers

Concept: Evaluate powers and roots first, then add/subtract left to right; $\sqrt{a^2}=a$ for $a>0$.

1. $2^2=4$, $\sqrt{8^2}=8$, $2\sqrt{4}=2\times 2=4$
2. Value $=4+8-4-11$
3. $=12-15=-3$

Trick: $\sqrt{x^2}$ just removes the square: $\sqrt{8^2}=8$.

Answer: (D) $-3$

Concept: If the numbers are $n$ and $n+1$, then $n(n+1)=272$; look for two neighbours around $\sqrt{272}$.

1. $\sqrt{272}\approx 16.5$, so try 16 and 17.
2. $16\times 17=272$ ✓
3. Greater number $=17$.

Trick: Product of consecutive numbers ≈ square: $\sqrt{272}$ lies between 16 and 17.

Answer: (C) 17

Concept: BODMAS: innermost bracket first, then multiplication, then addition, then the outer division.

1. $(2-7)=-5$
2. $5\times (-5)=-25$, so $\{27-25\}=2$
3. $290÷2=145$

Trick: Always solve ( ) → { } → [ ].

Answer: (A) 145

Concept: Least number to add $=\text{divisor}-\text{remainder}$.

1. $73\times 1118=81614$
2. Remainder $=81661-81614=47$
3. Number to add $=73-47=26$; check: $81687=73\times 1119$ ✓

Trick: Add (divisor − remainder); subtract the remainder itself.

Answer: (C) 26

Concept: Angle sum of a triangle $={180}^{\circ }$; with $BC\parallel AD$, alternate angles $\angle BCA=\angle CAD$; in an isosceles triangle the angles opposite equal sides are equal.

1. In $△ABC$: $\angle BCA={180}^{\circ }-{113}^{\circ }-{26}^{\circ }={41}^{\circ }$.
2. $BC\parallel AD$ and $AC$ is a transversal, so $\angle CAD=\angle BCA={41}^{\circ }$.
3. $AC=CD$, so in $△ACD$: $\angle CDA=\angle CAD={41}^{\circ }$.
4. $\angle ACD={180}^{\circ }-{41}^{\circ }-{41}^{\circ }={98}^{\circ }$.
5. ${98}^{\circ }$ is not among the options (168°, 165°, 41°, 166°); 41° is $\angle CAD$, not $\angle ACD$. No option is correct.

Note: Mark each angle on the figure with its vertex letters before matching it to the options. In the exam, every candidate gets full marks (bonus) for such a question.

Answer: Bonus – no option is correct as printed; full marks to all.

Concept: $\text{Discount}=\text{Marked price}\times \frac{\text{Discount}\%}{100}$

1. Discount $=32000\times \frac{9}{100}$
2. $=320\times 9=2880$

Trick: 1% of ₹32,000 is ₹320, so 9% $=9\times 320$.

Answer: (B) ₹2,880

Concept: Take tank capacity = LCM of the times; each pipe’s hourly work = capacity ÷ its time.

1. LCM$(14,24,17)=2856$ units
2. A $=\frac{2856}{14}=204$, B $=\frac{2856}{24}=119$, C $=\frac{2856}{17}=168$ units/h
3. Together $=204+119+168=491$ units/h
4. Time $=\frac{2856}{491}=5\frac{401}{491}$ hours

Trick: Rough check: $\frac{1}{14}+\frac{1}{24}+\frac{1}{17}\approx 0.17$, so time ≈ 5.8 h – only the option starting with 5 fits.

Answer: (A) $5\frac{401}{491}$ hours

Concept: No overall profit or loss ⇒ loss on the first frock = gain on the second.

1. Let CP of first $=x$, second $=1200-x$.
2. $10\%$ of $x=15\%$ of $(1200-x)$ ⇒ $10x=18000-15x$
3. $25x=18000$ ⇒ $x=720$
4. Check: loss $=72$, gain $=15\%$ of $480=72$ ✓

Trick: Loss% : Gain% $=10:15$ ⇒ CPs are in ratio $15:10=3:2$ ⇒ first $=\frac{3}{5}\times 1200=720$.

Answer: (D) 720

Concept: Take distance MN $=D$; speeds are $\frac{D}{\text{time}}$; trains moving towards each other close the gap at the sum of speeds.

1. A takes 7 h (8:05 a.m.–3:05 p.m.), speed $\frac{D}{7}$; B takes 5 h (10:05 a.m.–3:05 p.m.), speed $\frac{D}{5}$.
2. By 10:05 a.m., A covers $\frac{2D}{7}$; gap $=\frac{5D}{7}$.
3. Relative speed $=\frac{D}{7}+\frac{D}{5}=\frac{12D}{35}$
4. Time $=\frac{5D}{7}÷\frac{12D}{35}=\frac{25}{12}\text{h}=2\text{h}5\text{min}$ ⇒ 10:05 + 2:05 = 12:10 p.m.

Trick: Take $D=35$ (LCM of 7 and 5): A $=5$, B $=7$ per h; gap $35-10=25$ at 12 per h $=\frac{25}{12}$ h.

Answer: (A) 12:10 p.m.

Concept: Total surface area of a cuboid $=2(lb+bh+hl)$.

1. $lb=8\times 5=40$, $bh=5\times 3=15$, $hl=3\times 8=24$
2. Sum $=40+15+24=79$
3. TSA $=2\times 79=158{\text{cm}}^2$

Trick: Lateral area $=2h(l+b)=78$; add top and bottom $2lb=80$ ⇒ 158.

Answer: (C) $158{\text{cm}}^2$

Concept: Take total work = LCM of the given days; then one-day work (efficiency) of each person can be added or subtracted.

1. Total work $=\text{LCM}(22,11,8)=88$ units.
2. Priya $=\frac{88}{22}=4$; Priya + Varsha $=\frac{88}{11}=8$, so Varsha $=4$.
3. All three $=\frac{88}{8}=11$, so Nisha $=11-8=3$ units/day.
4. Priya + Nisha $=4+3=7$ units/day → time $=\frac{88}{7}$ days.

Trick: Nisha alone $=\frac{1}{8}-\frac{1}{11}=\frac{3}{88}$; add Priya $\frac{4}{88}$ → $\frac{7}{88}$, so $\frac{88}{7}$ days.

Answer: (D) $\frac{88}{7}$

Concept: Open the brackets; the $x^2$ terms cancel, leaving a linear equation.

1. $25x^2-35-25x^2-10x+20=15$
2. $-10x-15=15$
3. $-10x=30\Rightarrow x=-3$

Trick: Check options: $x=-3$ → $5(45-7)-5(45-6-4)=190-175=15$ ✓.

Answer: (C) $-3$

Concept: $A=P(1+\frac{RT}{100})$ for simple interest.

1. Total interest rate $=4\times 5=20\%$.
2. $P\times 1.2=2280$
3. $P=\frac{2280}{1.2}=1900$

Trick: Amount $=120\%$ of P → $\frac{2280}{120}\times 100=1900$.

Answer: (B) ₹1,900

Concept: Rule out options by quick tests, then divide.

1. 726179 is odd → not divisible by 28 or 20.
2. $726179÷23=31573$ exactly ($23\times 31573=726179$).
3. $726179÷29$ leaves remainder 9 → not divisible.

Trick: An odd number can never be divisible by an even number – strike 28 and 20 instantly.

Answer: (C) 23

Concept: Third proportional of a and b is $x=\frac{b^2}{a}$ (since $a:b=b:x$).

1. $54:18=18:x$
2. $x=\frac{{18}^2}{54}=\frac{324}{54}=6$

Trick: $54\to 18$ is $÷3$; continue: $18÷3=6$.

Answer: (B) 6

Concept: Convert averages into totals; the common person is counted twice in the pair sums.

1. $V+Mo+Mm=3\times 46=138$
2. $V+Mo=2\times 36=72$, $Mo+Mm=2\times 43=86$
3. $Mo=72+86-138=20$ kg

Trick: Common person = (sum of the two pair totals) − (total of all three).

Answer: (B) 20

Concept: If the overall literacy rate equals the male literacy rate, the female rate must be the same (weighted average).

1. Literates $=63\%\times 364000=229320$; literate males $=63\%\times 178000=112140$.
2. Females $=186000$; literate females $=229320-112140=117180$.
3. $\frac{117180}{186000}\times 100=63\%$

Trick: Average of two groups = 63% and one group = 63% → the other group is also 63%.

Answer: (C) 63%

Concept: The car is treated as a point, so distance covered = length of the bridge $=\text{speed}\times \text{time}$.

1. Time $=1.4$ min $=\frac{1.4}{60}$ h.
2. Length $=75\times \frac{1.4}{60}=\frac{105}{60}=1.75$ km

Trick: 75 km/h $=1.25$ km per minute → $1.25\times 1.4=1.75$ km.

Answer: (B) 1.75 km

Concept: Form two linear equations in mother’s age M and daughter’s age D.

1. $2M=9D+27$ and $6D=M-9\Rightarrow M=6D+9$.
2. $2(6D+9)=9D+27\Rightarrow 12D+18=9D+27\Rightarrow D=3$
3. $M=6\times 3+9=27$; difference $=27-3=24$ years.

Trick: Substitute $M=6D+9$ straight into the first equation – one step to D.

Answer: (C) 24

Concept: Count days excluding the day of deposit and including the day of withdrawal; 2024 is a leap year (366 days).

1. Days: February $29-23=6$, March 31, April 24 → total 61 days.
2. $SI=\frac{4000\times 6.75\times 61}{100\times 366}$
3. $=270\times \frac{61}{366}=270\times \frac{1}{6}=45$

Trick: $\frac{61}{366}=\frac{1}{6}$ exactly – the numbers are chosen for it.

Answer: (C) ₹45