Q&A: practice questions with answers

Concept: Brackets, then ÷/×, then +/− from left to right.

1. $(12+18)÷2=15$; $16\times 4=64$; $13(-18+15)=-39$.
2. $15-11+64-28-39=1$.

Yaad rakhein: Treat $13(\dots )$ as multiplication.

Answer: (C) 1

Concept: Equivalent discount $=a+b-\frac{ab}{100}$.

1. $40+20-\frac{40\times 20}{100}=60-8=52\%$.

Trick: Factor: $0.6\times 0.8=0.48$ ⇒ $52\%$ discount.

Answer: (D) 52%

Concept: Sides $3k,4k,5k$ form a right triangle (Pythagorean triple).

1. Area $=\frac{1}{2}\times 3k\times 4k=6k^2=486\Rightarrow k^2=81\Rightarrow k=9$.
2. Perimeter $=12k=108\text{cm}$.

Trick: 3 : 4 : 5 ⇒ area $=6k^2$, perimeter $=12k$.

Answer: (A) 108 cm

Concept: Split the original drink by ratio, add the new quantities, equate to the new ratio.

1. 160 L in $9:7$: water $=90$, juice $=70$.
2. $\frac{90+x}{70+2x}=\frac{13}{15}\Rightarrow 1350+15x=910+26x\Rightarrow x=40$.
3. Drink $=160+x+2x=160+120=280$ L.

Trick: Check: water 130, juice 150 ⇒ $13:15$ ✓.

Answer: (B) 280

Concept: Non-voters = registered × (100 − voted)%. Then % more $=\frac{\text{difference}}{\text{base}}\times 100$.

1. B: $30\%$ of $5400=1620$.
2. C: $15\%$ of $6500=975$.
3. $\frac{1620-975}{975}\times 100=\frac{645}{975}\times 100=66\frac{2}{13}\%$.

Trick: $\frac{645}{975}=\frac{43}{65}$; $\frac{4300}{65}=66\frac{10}{65}=66\frac{2}{13}$.

Answer: (C) $66\frac{2}{13}\%$

Concept: $\frac{\tan A+\tan B}{1-\tan A\tan B}=\tan (A+B)$.

1. $A+B={71}^{\circ }+{19}^{\circ }={90}^{\circ }$.
2. $\tan {71}^{\circ }\tan {19}^{\circ }=\tan {71}^{\circ }\cot {71}^{\circ }=1$, so the denominator is $0$.
3. $\tan {90}^{\circ }$ is not defined.

Yaad rakhein: Complementary angles: $\tan A\tan ({90}^{\circ }-A)=1$.

Answer: (C) Not defined

Concept: Add each row over all years, then take the ratio.

1. Educated females $=600+820+950+980+1000=4350$.
2. Females $=900+1000+1020+1100+1200=5220$.
3. $\frac{4350}{5220}=\frac{5}{6}=0.8333$, i.e. $83.33\%$.

Trick: $4350:5220$ – divide both by 870 ⇒ $5:6$.

Answer: (C) 0.8333

Concept: Remove brackets from the innermost: ( ), { }, [ ].

1. $(9-7+1)=3$.
2. $\{17-3\}=14$; $[15+14]=29$.
3. $63-29=34$.

Yaad rakhein: Order: small ( ) → curly { } → square [ ].

Answer: (C) 34

Concept: Equal distances at speeds $x$ and $y$: average speed $=\frac{2xy}{x+y}$.

1. $\frac{2\times 90\times 65}{90+65}=\frac{11700}{155}$.
2. $=75.48\text{km/h}$.

Yaad rakhein: Average speed is not $\frac{90+65}{2}=77.5$; distance 800 km does not matter.

Answer: (A) 75.48

Concept: Write everything in $\sin \theta$, $\cos \theta$ and use $(a+b)(a-b)=a^2-b^2$.

1. First $=\frac{\cos \theta +\sin \theta +1}{\cos \theta }$, second $=\frac{\sin \theta +\cos \theta -1}{\sin \theta }$.
2. Numerator $=(\sin \theta +\cos \theta {)}^2-1=2\sin \theta \cos \theta$.
3. Value $=\frac{2\sin \theta \cos \theta }{\sin \theta \cos \theta }=2$.

Trick: Put $\theta ={45}^{\circ }$: $(2+\sqrt{2})(2-\sqrt{2})=2$.

Answer: (C) 2

Concept: Value of a sector $=\frac{\text{angle}}{{360}^{\circ }}\times \text{total}$.

1. $1^{\circ }=\frac{7200}{360}=$ ₹20 crore.
2. Infrastructure $=84.7\times 20=$ ₹1,694 crore.

Trick: Find the value of $1^{\circ }$ first, then multiply.

Answer: (D) ₹1,694 crores

Concept: Compare $\frac{\text{High}}{\text{Low}}$ for Jmbuja cement in each exchange.

1. Mumbai $\frac{150}{155}<1$, Uttarakhand $\frac{120}{125}<1$, UP $\frac{145}{170}<1$.
2. Rajasthan $\frac{160}{135}\approx 1.19>1$ – the only one above 1.

Trick: Only where High > Low can the ratio exceed 1.

Answer: (A) Rajasthan

Concept: Sum and difference given ⇒ larger $=\frac{\text{sum}+\text{diff}}{2}$.

1. $g+b=18$, $g-b=8$.
2. $g=13$, $b=5$.
3. Product $=13\times 5=65$.

Trick: $\frac{{18}^2-8^2}{4}=\frac{260}{4}=65$.

Answer: (D) 65

Concept: $a^3+b^3+c^3-3abc=(a+b+c)[(a+b+c{)}^2-3(ab+bc+ca)]$.

1. $(a+b+c{)}^2=169$; $3(ab+bc+ca)=66$.
2. Value $=13\times (169-66)=13\times 103=1339$.

Yaad rakhein: $a^2+b^2+c^2-ab-bc-ca=(a+b+c{)}^2-3(ab+bc+ca)$.

Answer: (C) 1339

Concept: Food left (in student-days) stays the same: $M_1{D}_1=M_2{D}_2$.

1. After 20 days, food left for 300 students for 10 days $=3000$ student-days.
2. Remaining students $=250$: $\frac{3000}{250}=12$ days.

Yaad rakhein: Use only the remaining days (30 − 20).

Answer: (B) 12 days

Concept: Equal SP ⇒ CPs are in the inverse ratio of their SP factors.

1. $0.85C_1=1.19C_2\Rightarrow C_1:C_2=119:85=7:5$.
2. $C_1=\frac{7}{12}\times 11520=$ ₹6,720.

Trick: $119=7\times 17$, $85=5\times 17$ ⇒ $7:5$.

Answer: (B) ₹6,720

Concept: $SI=\frac{P\times R\times T}{100}$ with $T$ in years ($\text{months}÷12$).

1. $S{I}_1=P\times \frac{19.2}{100}\times \frac{20}{12}=0.32P$.
2. $S{I}_2=P\times \frac{11.2}{100}\times \frac{15}{12}=0.14P$.
3. $0.18P=2920\Rightarrow P=16222\approx$ ₹16,200 (nearest hundred).

Trick: Rate × time: $32\%$ vs $14\%$, difference $18\%$.

Answer: (D) 16,200

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

1. $lb=77$, $bh=28$, $hl=44$.
2. TSA $=2(77+28+44)=2\times 149=298{\text{m}}^2$.

Yaad rakhein: Three pairs of opposite faces ⇒ factor 2.

Answer: (C) 298

Concept: A is what % of B $=\frac{A}{B}\times 100$.

1. 2017 $=15$, 2019 $=25$.
2. $\frac{15}{25}\times 100=60\%$.

Trick: $\frac{15}{25}=\frac{3}{5}=60\%$.

Answer: (B) 60

Concept: Volume of a hemisphere $=\frac{2}{3}\pi r^3$.

1. $r^3={18}^3=5832$; $\frac{2}{3}\times 5832=3888$.
2. $V=\frac{22}{7}\times 3888=\frac{85536}{7}=12219.43{\text{cm}}^3$.

Yaad rakhein: Hemisphere = half of $\frac{4}{3}\pi r^3$.

Answer: (B) 12219.43