Q&A: practice questions with answers

Concept: Apply BODMAS: 'of' first, then division/multiplication, then addition/subtraction.

1. $\frac{5}{8}$ of $184=\frac{5\times 184}{8}=115$
2. $115\times 15÷5=115\times 3=345$
3. $7\frac{5}{8}+345-(?)=0\Rightarrow (?)=352\frac{5}{8}$

Trick: 'of' is just multiplication done first; $184÷8=23$ makes it quick: $23\times 5=115$.

Answer: (D) $352\frac{5}{8}$

Concept: Read the 2015 column and pick the largest value.

1. 2015 earnings: P = 56, Q = 55, R = 68, S = 60 (₹ crore).
2. The largest of these is 68, which belongs to firm R.

Trick: Look only at the asked column; ignore the other years.

Answer: (C) R

Concept: $\text{SP}=\text{MP}\times \frac{100-d}{100}$ and $\text{CP}=\frac{\text{SP}\times 100}{100+p}$.

1. $\text{SP}=7200\times \frac{85}{100}=6120$
2. $\text{CP}=\frac{6120\times 100}{120}=5100$

Trick: Directly: $\text{CP}=7200\times \frac{85}{120}=60\times 85=5100$.

Answer: (D) ₹5,100

Concept: $\frac{{\sec }^{2}A}{{\csc }^{2}A}=\frac{1/{\cos }^{2}A}{1/{\sin }^{2}A}=\frac{{\sin }^{2}A}{{\cos }^{2}A}={\tan }^{2}A$.

1. Given $\tan A=\frac{2}{5}$.
2. $\frac{{\sec }^{2}A}{{\csc }^{2}A}={\tan }^{2}A={\left(\frac{2}{5}\right)}^2=\frac{4}{25}$

Trick: $\sec /\csc$ always simplifies to $\tan$; just square the given value.

Answer: (B) $\frac{4}{25}$

Concept: $\text{CI}=P[{(1+\frac{r}{100})}^n-1]$.

1. $(1.12{)}^3=1.404928$
2. $\text{CI}=25000\times 0.404928=10123.20$

Trick: Effective rate for 3 years at $12\%$: $12+12+12+3\times 1.44+0.1728=40.4928\%$; $40.4928\%$ of 25000 = 10123.2.

Answer: (D) ₹10,123.20

Concept: A number divisible by 2, 4, 6 and 8 must be a multiple of their LCM.

1. $\text{LCM}(2,4,6,8)=24$
2. $1351=24\times 56+7$, remainder 7.
3. Number to add $=24-7=17$ (check: $1368=24\times 57$).

Trick: Least number to ADD = LCM − remainder; least number to SUBTRACT = remainder.

Answer: (D) 17

Concept: SAS rule: two sides and the angle included between them are equal ⇒ triangles are congruent.

1. Given: AC = PR (side) and ∠A = ∠P (angle).
2. ∠A lies between sides AB and AC; ∠P lies between PQ and PR.
3. So if AB = PQ, the equal angle is the included angle ⇒ congruent by SAS.

Trick: SSS needs all 3 sides; ASS and AAA are NOT congruence rules.

Answer: (B) AB = PQ by SAS

Concept: Amount on an item = its percentage × total income.

1. Education of children $=12\%$.
2. $12\%$ of ₹1,00,000 $=\frac{12}{100}\times 100000=12000$

Trick: With total ₹1,00,000, every $1\%$ is ₹1,000 — just read the percentage.

Answer: (B) 12,000

Concept: Cost price = price after both discounts + overhead (transport).

1. Price paid $=850\times 0.8\times 0.9=612$
2. Total CP $=612+55=667$; profit $=980-667=313$
3. $\text{Profit}\%=\frac{313}{667}\times 100\approx 46.93\%$

Trick: Successive discounts 20% and 10% = single discount $20+10-\frac{20\times 10}{100}=28\%$.

Answer: (D) $46.93\%$

Concept: Product of two numbers = $\text{HCF}\times \text{LCM}$.

1. Let numbers be $4k$ and $3k$: $12k^2=2700\Rightarrow k^2=225\Rightarrow k=15$
2. Numbers: 60 and 45.
3. Difference $=60-45=15$ (= $k$).

Trick: For ratio $a:b$ with $a-b=1$, the difference equals $k$ itself.

Answer: (A) 15

Concept: With simple interest at different rates for different periods, add up rate × time.

1. Total rate-years $=5\times 4+8\times 6+12\times 2=20+48+24=92\%$
2. $92\%$ of P $=9016\Rightarrow P=\frac{9016\times 100}{92}=9800$

Trick: Last part runs only $12-10=2$ years — the usual trap.

Answer: (B) ₹9,800

Concept: Area of an equilateral triangle $=\frac{\sqrt{3}}{4}{a}^2$.

1. $\frac{\sqrt{3}}{4}{a}^2=4\sqrt{3}$
2. $a^2=16\Rightarrow a=4\text{cm}$

Trick: Cancel $\sqrt{3}$ on both sides first: $\frac{a^2}{4}=4$.

Answer: (B) 4

Concept: All colleges have 4 years, so the least average = the least total.

1. A: 68 + 72 + 78 + 80 = 298
2. B: 78 + 86 + 91 + 87 = 342; C: 56 + 78 + 82 + 84 = 300
3. D: 78 + 82 + 85 + 87 = 332. Least total = 298 → College A.

Trick: Compare totals instead of averages when the count is the same; C (300) is the close trap.

Answer: (D) College A

Concept: New person = person replaced + (number of people × increase in average).

1. Total increase $=15\times 3.2=48\text{kg}$
2. New person $=52+48=100\text{kg}$

Trick: Increase in average → add; decrease → subtract.

Answer: (C) 100 kg

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

1. From the graph: boys in section B = 30, girls in section E = 55.
2. $\frac{30}{55}\times 100=54.54\%\approx 55\%$

Trick: $\frac{6}{11}\approx 54.5\%$ — remember $\frac{1}{11}=9.09\%$.

Answer: (B) $55\%$

Concept: If $x-\frac{1}{x}=k$, then $x^2+\frac{1}{x^2}=k^2+2$.

1. $x(5-\frac{2}{x})=\frac{5}{x}\Rightarrow 5x-2=\frac{5}{x}$
2. $5x-\frac{5}{x}=2\Rightarrow x-\frac{1}{x}=\frac{2}{5}$
3. $x^2+\frac{1}{x^2}=\frac{4}{25}+2=\frac{54}{25}$

Trick: Minus form → add 2; plus form $(x+\frac{1}{x})$ → subtract 2.

Answer: (A) $\frac{54}{25}$

Concept: The largest square tile that fits both length and breadth exactly has side = HCF of the two.

1. $126=2\times 3^2\times 7$, $90=2\times 3^2\times 5$
2. $\text{HCF}=2\times 9=18\text{ft}$
3. Area of each tile $={18}^2=324\text{sq ft}$

Trick: Largest tile → HCF; least number of tiles = (area of floor) ÷ (tile area).

Answer: (C) 324

Concept: Check each column: every next value must be larger than the previous one.

1. Wheat: 62 → 81 → 116 → 124 → 137 (rises every time).
2. Rice falls 197 → 187; coarse grains keep falling; pulses fall 64 → 46 → 44.

Trick: One single fall is enough to reject an option.

Answer: (D) Wheat

Concept: LCM of two distinct primes = their product.

1. $533=13\times 41$, so $x=41$, $y=13$ ($x>y$).
2. $4y-x=52-41=11$

Trick: To factor 533, test primes up to $\sqrt{533}\approx 23$: $533÷13=41$.

Answer: (B) 11

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

1. $\frac{\cot A}{\cot B}=\frac{\cot A}{\tan A}={\cot }^{2}A$
2. ${\cos }^{2}A+{\cos }^{2}B={\cos }^{2}A+{\sin }^{2}A=1$
3. Expression $=1+{\cot }^{2}A={\csc }^{2}A$

Trick: Check with $A=B={45}^{\circ }$: $1+\frac{1}{2}+\frac{1}{2}=2={\csc }^2{45}^{\circ }$ — only (B) and (D) give 2; (B) matches in general.

Answer: (B) ${\csc }^{2}A$