Q&A: practice questions with answers

Concept: In a chase, the gap closes at the relative speed (difference of speeds).

1. Relative speed $=13-11=2\text{km/h}$; gap $=0.3$ km.
2. Time $=\frac{0.3}{2}=0.15$ h.
3. Policeman's distance $=13\times 0.15=1.95$ km.

Trick: Distance by chaser $=\frac{\text{gap}\times \text{chaser speed}}{\text{relative speed}}=\frac{0.3\times 13}{2}$. (1.65 km is the thief's run.)

Answer: (D) 1.95

Concept: $A=P{(1+\frac{R}{100})}^n$, CI $=A-P$.

1. $(1.07{)}^2=1.1449$, $(1.07{)}^4=1.31079601$.
2. $(1.07{)}^5=1.40255173$.
3. A $=100000\times 1.40255\approx 140255$ ⇒ CI $\approx 40255$.

Trick: SI would be 35,000; CI must be a little above it – only 40,255 fits the exact value.

Answer: (D) ₹40,255

Concept: Curved surface area of a cone $=\pi rl$.

1. CSA $=\frac{22}{7}\times 7\times 10$.
2. $=22\times 10=220{\text{cm}}^2$.

Trick: When r = 7, $\pi r=22$ instantly.

Answer: (D) $220{\text{cm}}^2$

Concept: Arc length $=\frac{\theta }{{360}^{\circ }}\times 2\pi r$.

1. $\frac{45}{360}\times 2\times \frac{22}{7}\times r=22$.
2. $\frac{1}{8}\times \frac{44}{7}r=22$ ⇒ $\frac{11r}{14}=22$.
3. $r=28$ cm.

Trick: ${45}^{\circ }$ is $\frac{1}{8}$ of the circle ⇒ circumference $=8\times 22=176=2\pi r$ ⇒ $r=28$.

Answer: (C) 28 cm

Concept: LCM = product of the highest powers of all prime factors.

1. $210=2\times 3\times 5\times 7$
2. $336=2^4\times 3\times 7$, $504=2^3\times 3^2\times 7$
3. LCM $=2^4\times 3^2\times 5\times 7=5040$.

Trick: The LCM must be divisible by 504 and 5; among options only 5040 is.

Answer: (A) 5040

Concept: $\text{LCM of fractions}=\frac{\text{LCM of numerators}}{\text{HCF of denominators}}$.

1. LCM(1, 3, 4, 9) $=36$.
2. HCF(3, 5, 7, 16) $=1$.
3. LCM $=\frac{36}{1}=36$.

Yaad rakhein: LCM → LCM/HCF; HCF → HCF/LCM (numerators/denominators).

Answer: (A) 36

Concept: $\text{A as \% of B}=\frac{A}{B}\times 100$.

1. Chennai: MBA $=1342$, M.A. $=1666$.
2. $\frac{1342}{1666}\times 100=80.55\%$ (approx.).

Trick: $1666\times 0.8=1332.8$; extra $9.2$ is about $0.55\%$ of 1666 ⇒ $80.55\%$.

Answer: (B) $80.55\%$

Concept: $\cot ({90}^{\circ }-A)=\tan A$; get the third side from Pythagoras.

1. $\cos A=\frac{15}{17}$ ⇒ base 15, hypotenuse 17.
2. Perpendicular $=\sqrt{{17}^2-{15}^2}=\sqrt{64}=8$.
3. $\tan A=\frac{8}{15}$.

Yaad rakhein: Triplet $(8,15,17)$.

Answer: (A) $\frac{8}{15}$

Concept: Pick the right cells, add, then subtract.

1. Q + S in September $=189+154=343$.
2. P + R in November $=168+168=336$.
3. Difference $=343-336=7$.

Trick: Compare unit digits first: $9+4\to 3$, $8+8\to 6$ ⇒ difference ends in 7.

Answer: (A) 7

Concept: Volume of cone $=\frac{1}{3}\pi r^{2}h$.

1. Diameter 14 ⇒ $r=7$ cm.
2. $\frac{1}{3}\pi \times 49\times h=196\pi$.
3. $h=\frac{196\times 3}{49}=12$ cm.

Trick: $\pi$ cancels; $h=\frac{3V}{\pi r^2}$.

Answer: (B) 12 cm

Concept: $\text{LCM of fractions}=\frac{\text{LCM of numerators}}{\text{HCF of denominators}}$.

1. LCM(3, 81, 9) $=81$.
2. HCF(2, 16, 8) $=2$.
3. LCM $=\frac{81}{2}$.

Trick: The LCM must be a multiple of the largest fraction $\frac{81}{16}$: $\frac{81}{2}=8\times \frac{81}{16}$ ✓.

Answer: (B) $\frac{81}{2}$

Concept: Express $\frac{L}{A}$ and $\frac{C}{N}$ using $\frac{B}{M}$ from the two equations.

1. $\frac{M}{B}=1-\frac{A}{L}=\frac{L-A}{L}$ ⇒ $\frac{B}{M}=\frac{L}{L-A}$.
2. $\frac{N}{C}=1-\frac{B}{M}=\frac{-A}{L-A}$ ⇒ $\frac{C}{N}=\frac{A-L}{A}=1-\frac{L}{A}$.
3. $\frac{L}{A}+\frac{C}{N}=\frac{L}{A}+1-\frac{L}{A}=1$.

Trick: Put values: A = 1, L = 2 ⇒ $\frac{M}{B}=\frac{1}{2}$, $\frac{B}{M}=2$ ⇒ $\frac{N}{C}=-1$; then $2+(-1)=1$.

Answer: (C) $1$

Concept: Perimeter of sector $=2r+l$; area $=\frac{1}{2}lr$.

1. $2\times 3+l=24$ ⇒ arc $l=18$ cm.
2. Area $=\frac{1}{2}\times 18\times 3=27{\text{cm}}^2$.

Trick: Area of sector $=\frac{1}{2}\times \text{arc}\times \text{radius}$ – no angle or $\pi$ needed.

Answer: (C) 27

Concept: A continuous decrease means every year's value is less than the previous year's.

1. R: 21 → 17 → 16 → 15 → 13 → 8 – falls every year.
2. S rises every year; Q rises in 1991; T rises in 1991.

Trick: Any single rise in a row rules that car out.

Answer: (C) R

Concept: If $A+B={90}^{\circ }$, then $\cos A=\sin B$ and $\sin A=\cos B$.

1. ${65}^{\circ }+{25}^{\circ }={90}^{\circ }$ ⇒ $\frac{\cos {65}^{\circ }}{\sin {25}^{\circ }}=1$.
2. ${19}^{\circ }+{71}^{\circ }={90}^{\circ }$ ⇒ $\frac{5\sin {19}^{\circ }}{\cos {71}^{\circ }}=5$.
3. ${28}^{\circ }+{62}^{\circ }={90}^{\circ }$ ⇒ $\frac{3\cos {28}^{\circ }}{\sin {62}^{\circ }}=3$.
4. Value $=1+5-3=3$.

Trick: Each ratio of complementary sin/cos pairs equals 1, so just use the coefficients: $1+5-3$.

Answer: (D) 3

Concept: Use the shopping share to find the total expenditure, then subtract the known items.

1. $11.25\%$ of expenditure $=7200$ ⇒ expenditure $=\frac{7200\times 100}{11.25}=64000$.
2. Food $=64000-(12800+32000+7200)=64000-52000=12000$.
3. Check: income $=\frac{64000}{0.8}=80000$; $15\%$ of it $=12000$ ✓.

Trick: $11.25\%=\frac{9}{80}$ ⇒ expenditure $=7200\times \frac{80}{9}=64000$.

Answer: (C) 12000

Concept: Group into perfect squares and use $x^2-y^2=(x-y)(x+y)$.

1. Numerator $=a^2-(b+c{)}^2=(a-b-c)(a+b+c)$.
2. Denominator $=(a+b{)}^2-c^2=(a+b-c)(a+b+c)$.
3. Cancel $(a+b+c)$: $\frac{a-b-c}{a+b-c}$.

Trick: Put $a=3,b=1,c=1$: expression $=\frac{9-1-2-1}{9+1+6-1}=\frac{5}{15}=\frac{1}{3}$ and option (B) $=\frac{1}{3}$ ✓.

Answer: (B) $\frac{a-b-c}{a+b-c}$

Concept: Take CP = 100 and apply mark-up and discount in turn.

1. CP $=100$ ⇒ MP $=175$.
2. SP $=175\times \frac{60}{100}=105$.
3. Profit $=5\%$.

Trick: Net $=75-40-\frac{75\times 40}{100}=75-40-30=5\%$.

Answer: (D) $5\%$

Concept: Alligation: cheaper : dearer $=(\text{dearer}-\text{mean}):(\text{mean}-\text{cheaper})$.

1. Cheaper ₹55, dearer ₹70, mean ₹65.
2. Ratio $=(70-65):(65-55)=5:10$.
3. $=1:2$.

Trick: The mean (65) is nearer to 70, so more of the ₹70 rice is needed – ratio less than 1.

Answer: (A) 1 : 2

Concept: Add fractions using the LCM of denominators.

1. LCM(12, 16, 8) $=48$.
2. $\frac{4}{48}+\frac{3}{48}+\frac{6}{48}=\frac{13}{48}$.

Trick: The sum is clearly less than $\frac{1}{3}$; $\frac{13}{48}\approx 0.27$ fits.

Answer: (A) $\frac{13}{48}$