Q&A: practice questions with answers

Concept: Evaluate roots and percentages first, then multiplication, then subtraction.

1. $16\%\text{ of }50=8$; $\sqrt[3]{729}=9$; $\sqrt{25}=5$.
2. $\frac{4}{5}\times 9\times 5=36$.
3. $y=8-36=-28$.

Trick: Remember cubes: $9^3=729$, $8^3=512$, $7^3=343$ – they appear again and again.

Answer: (C) $-28$

Concept: Downstream speed $=b+s$, upstream speed $=b-s$; time $=\frac{\text{distance}}{\text{speed}}$.

1. $\frac{36}{b+2}+\frac{24}{b-2}=6$.
2. Try $b=10$: $\frac{36}{12}+\frac{24}{8}=3+3=6$ ✓.
3. Downstream time $=3$ h, upstream time $=3$ h.

Trick: Check options instead of solving the quadratic – only option (A) gives equal times $3+3=6$ with speeds 12 and 8.

Answer: (A) (3.0 and 3.0)

Concept: Take total work = LCM of days; find per-day units and work in 2-day cycles.

1. Work $=\text{LCM}(12,15,18)=180$. A $=15$, B $=12$, C $=10$ units/day.
2. Day 1 (C+B) $=22$, Day 2 (A+B) $=27$ → 2-day cycle $=49$ units.
3. 3 cycles (6 days) $=147$; Day 7 (C+B) $=22$ → total $169$, left $=11$.
4. Day 8 (A+B at 27/day): $\frac{11}{27}$ day. Total $=7\frac{11}{27}$ days.

Trick: Never multiply cycles blindly – stop before the work overshoots and finish the last day by fraction.

Answer: (D) $7\frac{11}{27}$ days

Concept: Final price = original cost × (1 + profit%) × (1 − loss%).

1. ${\text{CP}}_P\times 1.53\times 0.32=3060$.
2. $1.53\times 0.32=0.4896$.
3. ${\text{CP}}_P=\frac{3060}{0.4896}=6250$.

Trick: Check: $6250\times 1.53=9562.5$; $9562.5\times 0.32=3060$ ✓.

Answer: (A) ₹6,250

Concept: $\text{cosec}A=\frac{\text{hyp}}{\text{perp}}$; build the right triangle and read the other ratios.

1. $\text{cosec}A=\frac{5}{4}$ → perp $=4$, hyp $=5$, base $=\sqrt{25-16}=3$.
2. $\cot A=\frac{3}{4}$, $\sec A=\frac{5}{3}$.
3. $4\cot A\sec A+3=4\times \frac{3}{4}\times \frac{5}{3}+3=5+3=8$.

Trick: $\cot A\sec A=\frac{\cos A}{\sin A}\times \frac{1}{\cos A}=\text{cosec}A$, so the expression is just $4\text{cosec}A+3=5+3$.

Answer: (B) 8

Concept: Centroid $G=(\frac{x_1+x_2+x_3}{3},\frac{y_1+y_2+y_3}{3})$.

1. x: $\frac{9+4+q}{3}=5\Rightarrow 13+q=15\Rightarrow q=2$.
2. y: $\frac{5+p+4}{3}=5\Rightarrow 9+p=15\Rightarrow p=6$.
3. So $p=6$, $q=2$.

Trick: Sum of x-coordinates $=3\times$ centroid x; same for y – one line each.

Answer: (C) 6 , 2

Concept: In $a:b::c:d$, $d=\frac{b\times c}{a}$.

1. $3:27::5.6:x\Rightarrow x=\frac{27\times 5.6}{3}=9\times 5.6=50.4$.
2. $5:35::6:y\Rightarrow y=\frac{35\times 6}{5}=42$.
3. $x:y=50.4:42=504:420=6:5$.

Trick: $27=3\times 9$ and $35=5\times 7$ – so $x=9\times 5.6$, $y=7\times 6$ directly.

Answer: (B) 6 : 5

Concept: Sum = average × number; each person ages 1 year per year.

1. 8 years ago A + B $=2\times 24=48$ → today A + B $=48+16=64$.
2. Today A + B + C $=3\times 32=96$ → C $=96-64=32$.
3. After 16 years C $=32+16=48$.

Trick: When moving a group's sum forward n years, add n for each member (here $2\times 8$).

Answer: (C) 48 years

Concept: Sum of first $n$ odd numbers $=n^2$; sum of an AP $=\frac{n}{2}(\text{first}+\text{last})$.

1. Odd numbers 1 to 100: 50 terms → sum $={50}^2=2500$.
2. Even numbers 150 to 200: $n=\frac{200-150}{2}+1=26$; sum $=\frac{26}{2}(150+200)=13\times 350=4550$.
3. Ratio $=2500:4550=50:91$.

Trick: All options start with 50 – so only compute $4550÷50=91$.

Answer: (C) 50:91

Concept: Expenditure = Income − Saving. Put incomes as $6x$ and $7x$ and use the expenditure ratio.

1. Expenditures: $6x-63000$ and $7x-63000$ in ratio $2:4=1:2$.
2. $\frac{6x-63000}{7x-63000}=\frac{1}{2}\Rightarrow 12x-126000=7x-63000$.
3. $5x=63000\Rightarrow x=12600$.
4. Ajay's income $=6x=6\times 12600=75600$.

Trick: When savings are equal, check options: 75,600 and 88,200 (ratio 6:7) minus 63,000 give 12,600 and 25,200 = 1 : 2 ✓.

Answer: (D) 75,600

Concept: $(x+y{)}^2=x^2+y^2+2xy$ with $x=2a$, $y=3b$.

1. $(2a+3b{)}^2=4a^2+9b^2+2(2a)(3b)=4a^2+9b^2+12ab$.
2. Substitute: $=49+12\times 6=49+72=121$.
3. Positive value: $2a+3b=\sqrt{121}=11$.

Trick: Whenever you see $4a^2+9b^2$ and $ab$, think of $(2a\pm 3b{)}^2$ – the cross term is $12ab$.

Answer: (C) 11

Concept: A number is divisible by 4 if the number formed by its last two digits is divisible by 4.

1. Last two digits of 65770A4 are "A4".
2. Two-digit numbers ending in 4 divisible by 4: 04, 24, 44, 64, 84 → A = 0, 2, 4, 6, 8.
3. Minimum value of A = 0 (the number 6577004 ends in 04, which is divisible by 4).

Trick: Units digit 4 ⇒ tens digit must be even for divisibility by 4; smallest even digit is 0.

Answer: (B) 0

Concept: BODMAS – do Division and Multiplication before Addition and Subtraction.

1. $20÷4=5$ and $3\times 3=9$.
2. Expression becomes $16+5-9$.
3. $21-9=12$.

Trick: Never work left to right blindly – circle the ÷ and × parts first.

Answer: (B) 12

Concept: Total surface area of a cone $=\pi rl+\pi r^2=\pi r(l+r)$.

1. $r=7$ cm, $l=25$ cm, so $l+r=32$.
2. TSA $=\frac{22}{7}\times 7\times 32$.
3. $=22\times 32=704{\text{cm}}^2$.

Trick: With $r=7$ and $\pi =\frac{22}{7}$, the 7 cancels – TSA is simply $22\times (r+l)$.

Answer: (B) $704{\text{cm}}^2$

Concept: For equal time, profit is shared in the ratio of investments.

1. $12:32=3:8$, total $=11$ parts.
2. 1 part $=\frac{2904}{11}=264$.
3. Difference $=(8-3)\times 264=5\times 264=1320$.

Trick: Difference of profits $=\frac{\text{difference of ratio terms}}{\text{sum}}\times \text{total}=\frac{5}{11}\times 2904$.

Answer: (A) ₹1,320

Concept: Melting preserves volume. Diagonal of a cube $=a\sqrt{3}$.

1. Total volume $=1^3+6^3+8^3=1+216+512=729{\text{cm}}^3$.
2. New edge $a=\sqrt[3]{729}=9$ cm.
3. Diagonal $=9\sqrt{3}$ cm.

Trick: $1,6,8\to 9$ is a famous cube triplet ($1^3+6^3+8^3=9^3$), like $3,4,5\to 6$.

Answer: (A) $9\sqrt{3}$

Concept: Add the per-hour work of filling taps and subtract the emptying tap.

1. Net part filled in 1 hour $=\frac{1}{4}+\frac{1}{26}-\frac{1}{26}=\frac{1}{4}$.
2. So the full cistern fills in 4 hours.
3. One-fourth of the cistern $=\frac{1}{4}\times 4=1$ hour.

Trick: The 26-hour filler and 26-hour emptier cancel exactly – only the 4-hour tap matters.

Answer: (B) 1

Concept: Factorise the LCM; A must supply every prime factor that 64 and 128 cannot.

1. $4480=128\times 35=2^7\times 5\times 7$; $64=2^6$, $128=2^7$.
2. Since 128 already gives $2^7$, A must contain $5\times 7=35$ and may have $2^k$ with $0\le k\le 7$.
3. Minimum A $=35$, maximum A $=35\times 2^7=4480$.
4. Difference $=4480-35=4445$.

Trick: Max A is always the LCM itself; Min A = LCM with the powers of 2 already covered by 128 removed.

Answer: (D) 4445

Concept: If the sum becomes $n$ times, SI $=(n-1)P$ and $R=\frac{(n-1)\times 100}{T}$.

1. Five times ⇒ SI $=4P$ in 9 years.
2. $R=\frac{4\times 100}{9}=\frac{400}{9}$.
3. $\frac{400}{9}=44\frac{4}{9}\%$ per annum.

Trick: "$n$ times in $T$ years" ⇒ $R\times T=(n-1)\times 100$.

Answer: (C) $44\frac{4}{9}\%$

Concept: $a{x}^2+bx+c=0$ has identical (equal) roots when the discriminant $b^2-4ac=0$.

1. Here $a=1$, $b=-(2+h)$, $c=h^2-4h+4$.
2. $(2+h{)}^2-4(h^2-4h+4)=0\Rightarrow 4+4h+h^2-4h^2+16h-16=0$.
3. $-3h^2+20h-12=0\Rightarrow 3h^2-20h+12=0\Rightarrow (3h-2)(h-6)=0$.
4. $h=\frac{2}{3}$ or $h=6$.

Trick: Check quickly: $h=6$ gives $x^2-8x+16=(x-4{)}^2$ ✓.

Answer: (C) $h=\frac{2}{3},h=6$