Q&A: practice questions with answers

Concept: Percentage $=\frac{\text{marks}}{\text{maximum}}\times 100$ — subjects have different maximum marks.

1. 1985: History $\frac{90}{200}=45\%$, Geography $\frac{35}{100}=35\%$, Maths $\frac{88}{200}=44\%$.
2. Science $\frac{120}{200}=60\%$, English $\frac{90}{150}=60\%$.
3. Science and English are equal.

Trick: Never compare raw marks when the totals differ.

Answer: (C) Science, English

Concept: Three equal diameters fill the big diameter, so $R=3r$. Area of semicircle $=\frac{1}{2}\pi r^2$.

1. Red $=3\times \frac{1}{2}\pi r^2=\frac{3}{2}\pi r^2$.
2. Big semicircle $=\frac{1}{2}\pi (3r{)}^2=\frac{9}{2}\pi r^2$; blue $=\frac{9}{2}\pi r^2-\frac{3}{2}\pi r^2=3\pi r^2$.
3. Red : Blue $=\frac{3}{2}:3=1:2$.

Trick: Areas scale as $r^2$: big $=9$ units, red $=3$, blue $=6$ → $1:2$.

Answer: (B) $1:2$

Concept: Average speed $=\frac{\text{total distance}}{\text{total time}}$ (not the average of speeds).

1. Time $=\frac{100}{50}+\frac{140}{70}=2+2=4$ h.
2. Average $=\frac{240}{4}=60\text{km/h}$.

Trick: Equal times (2 h each) → simple average $\frac{50+70}{2}=60$.

Answer: (A) $60\text{km/h}$

Concept: Longest pole in a room = body diagonal $=\sqrt{l^2+b^2+h^2}$.

1. $\sqrt{{60}^2+{30}^2+{20}^2}=\sqrt{3600+900+400}$.
2. $=\sqrt{4900}=70$ ft.

Trick: $60,30,20=10\times (6,3,2)$ and $6^2+3^2+2^2=49=7^2$.

Answer: (B) $70$ feet

Concept: In a bar graph the tallest bar shows the maximum value.

1. Bars for Jan, Mar, Apr, Jun are around or below 6000; Feb is a little higher.
2. May's bar is the tallest (about 7000+).

Yaad rakhein: Just compare the bar heights — no calculation needed.

Answer: (C) May

Concept: On melting, volume is conserved: $n\times V_{\text{sphere}}=V_{\text{cylinder}}$.

1. Sphere ($r=3$): $\frac{4}{3}\pi (3{)}^3=36\pi$.
2. Cylinder ($r=4$, $h=18$): $\pi (4{)}^2(18)=288\pi$.
3. $n=\frac{288\pi }{36\pi }=8$.

Trick: $n=\frac{3R^{2}H}{4r^3}=\frac{3\times 16\times 18}{4\times 27}=8$.

Answer: (C) $8$

Concept: Euclid's division method: HCF$(a,b)=$ HCF$(b,amodb)$.

1. $832=3\times 238+118$.
2. $238=2\times 118+2$.
3. $118=59\times 2+0$ → HCF $=2$.

Trick: $238=2\times 7\times 17$, $832=2^6\times 13$ — only 2 is common.

Answer: (C) $2$

Concept: Volume of a cuboid $=l\times b\times h$.

1. Let breadth $=b$; height $=6b$; length $=\frac{1}{2}\times 6b=3b$.
2. $3b\times b\times 6b=18b^3=23958\Rightarrow b^3=1331$.
3. $b=\sqrt[3]{1331}=11$ cm.

Trick: Check: $18\times {11}^3=18\times 1331=23958$ ✓ (with 21 cm the volume would be $166698$).

Answer: (C) $11$ cm

Concept: Fix a convenient base (Sohan $=100$) and apply the changes step by step.

1. Sohan $=100$ → Shyam $=100-30\%=70$.
2. Ram $=70+70\%$ of $70=70\times 1.7=119$.
3. Ram : Sohan $=119:100$.

Trick: $0.7\times 1.7=1.19$.

Answer: (A) $119:100$

Concept: Divisible by 16 → last four digits divisible by 16; by 11 → (sum of odd-place digits − sum of even-place digits) is $0$ or a multiple of 11.

1. Last four digits $30M6$: $3056=16\times 191$ ✓; 3016, 3036, 3076, 3096 are not multiples of 16 — so $M=5$.
2. Number $34R0503056$: odd places $3+R+5+3+5=16+R$; even places $4+0+0+0+6=10$.
3. $16+R-10=6+R$ must be $11$ → $R=5$.

Trick: Only option (C) has $M=5$, so it is chosen right after the 16-test.

Answer: (C) $5$ and $5$

Concept: A number is divisible by 9 if its digit sum is a multiple of 9.

1. Digit sum $=x+5+8+y=13+(x+y)$.
2. Next multiple of 9 after 13 is 18 → $x+y=5$.
3. ($x+y=-4$ is impossible), so least value $=5$.

Trick: $13\to 18$ needs $5$.

Answer: (B) $5$

Concept: $2{\cos }^2\theta -1=\cos 2\theta =1-2{\sin }^2\theta$.

1. Numerator $=\cos \theta (2{\cos }^2\theta -1)=\cos \theta \cos 2\theta$.
2. Denominator $=\sin \theta (1-2{\sin }^2\theta )=\sin \theta \cos 2\theta$.
3. Ratio $=\frac{\cos \theta }{\sin \theta }=\cot \theta$.

Trick: Put $\theta ={30}^{\circ }$: $\frac{2·\frac{3\sqrt{3}}{8}-\frac{\sqrt{3}}{2}}{\frac{1}{2}-\frac{1}{4}}=\sqrt{3}=\cot {30}^{\circ }$.

Answer: (C) $\cot \theta$

Concept: Surface area of a sphere $=4\pi r^2$.

1. $4\pi (5{)}^2=4\pi \times 25$.
2. $=100\pi {\text{cm}}^2$.

Yaad rakhein: Sphere SA is 4 times the area of its great circle.

Answer: (A) $100\pi {\text{cm}}^2$

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

1. Other number $=\frac{126\times 9}{18}$.
2. $=\frac{1134}{18}=63$.

Trick: Use the product rule directly: $\frac{\text{HCF}\times \text{LCM}}{\text{given number}}$.

Answer: (A) $63$

Concept: Find each average separately, then average them.

1. $J=\frac{1500+1460+1105}{3}=\frac{4065}{3}=1355$.
2. $K=\frac{1320+1180+1170}{3}=\frac{3670}{3}=1223.33$.
3. $\frac{J+K}{2}=\frac{2578.33}{2}=1289.17$.

Trick: $\frac{J+K}{2}=\frac{4065+3670}{6}=\frac{7735}{6}=1289.17$.

Answer: (A) $1289.17$

Concept: Below 36 years = groups up to 15, 16–25 and 26–35; use proportion.

1. Below 36: $15+20.25+16.75=52\%$ ↔ $30$ million.
2. Age 56–65: $5.5\%$.
3. $\frac{5.5}{52}\times 30=\frac{165}{52}\approx 3.17$ million.

Trick: Unitary: $1\%=\frac{30}{52}$ million.

Answer: (B) $3.17$ million

Concept: $(x-y{)}^2=(x+y{)}^2-4xy$.

1. $xy=24$; $(x-y{)}^2=100-96=4$, so $x-y=2$ (as $x>y$).
2. $x=6$, $y=4$.
3. $2x-y=12-4=8$.

Trick: Two numbers with sum 10 and product 24: 6 and 4.

Answer: (B) $8$

Concept: After both start, the gap closes at the sum of speeds.

1. By 9 a.m. train A covers $20$ km; gap $=120-20=100$ km.
2. Relative speed $=20+30=50\text{km/h}$; time $=\frac{100}{50}=2$ h.
3. They meet at $9+2=11$ a.m.

Trick: Head start first, then (remaining gap ÷ sum of speeds).

Answer: (C) 11:00 a.m.

Concept: Greatest length that measures all exactly = HCF of the lengths.

1. $62=2\times 31$, $186=6\times 31$, $279=9\times 31$.
2. HCF$(2,6,9)=1$, so HCF $=31$ m.

Trick: 62 is not a factor of 279, and 93 not of 62 — so the answer must be smaller: 31.

Answer: (C) $31$ m

Concept: In BODMAS, 'of' is done before division.

1. $2000$ of $(0.2{)}^2=2000\times 0.04=80$.
2. $(0.2{)}^3\times 400=0.008\times 400=3.2$.
3. $3.2÷80=0.04=\frac{1}{25}$.

Trick: $\frac{(0.2{)}^3\times 400}{2000\times (0.2{)}^2}=\frac{0.2\times 400}{2000}=\frac{1}{25}$.

Answer: (A) $\frac{1}{25}$