Q&A: practice questions with answers

Concept: For mutually exclusive events $P(E\cup F)=P(E)+P(F)$, and 'not E and not F' is the complement of $E\cup F$.

1. $P(E\cup F)=\frac{1}{6}+\frac{1}{2}=\frac{1}{6}+\frac{3}{6}=\frac{4}{6}=\frac{2}{3}$
2. $P(\text{not E and not F})=1-P(E\cup F)=1-\frac{2}{3}=\frac{1}{3}$

Trick: De Morgan: $P(E^{\prime }\cap F^{\prime })=1-P(E\cup F)$.

Answer: (D) $\frac{1}{3}$

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

1. $x=\frac{-16+(-2)+2}{3}=\frac{-16}{3}$
2. $y=\frac{13+(-9)+13}{3}=\frac{17}{3}$
3. $G=(-\frac{16}{3},\frac{17}{3})$

Trick: Centroid = average of the three vertices; do x and y sums separately.

Answer: (D) $(-\frac{16}{3},\frac{17}{3})$

Concept: Solve inside each bracket first, doing × and ÷ before +.

1. $2.5\times 0.48÷0.6=1.2÷0.6=2$
2. $3.5+1.6÷0.8=3.5+2=5.5$
3. $4.8\times 0.25÷0.8=1.2÷0.8=1.5$
4. $2+5.5-1.5=6$

Trick: Convert to fractions: $0.48÷0.6=\frac{48}{60}=0.8$, $0.25÷0.8=\frac{25}{80}$ – fewer decimal slips.

Answer: (B) 6

Concept: If HCF $=h$, the numbers are $ha$ and $hb$ with $a,b$ coprime and $ab=\frac{\text{LCM}}{h}$.

1. $ab=\frac{760}{19}=40$; coprime pairs: $(1,40)$ and $(5,8)$
2. $(1,40)\Rightarrow 19,760$ – but 19 is a two-digit number, rejected.
3. $(5,8)\Rightarrow 95,152$ – both three-digit ✓
4. Number of such pairs $=1$

Trick: Pairs with $ab=40$: $(1,40),(5,8)$ only ($(2,20),(4,10)$ are not coprime); then apply the digit condition.

Answer: (D) 1

Concept: Volume of hemisphere $=\frac{2}{3}\pi R^3$, volume of sphere $=\frac{4}{3}\pi r^3$. After the ball sinks, water + ball exactly fill the bowl.

1. Ball volume $=\frac{4}{3}\pi (2{)}^3=\frac{32\pi }{3}{\text{cm}}^3$.
2. Bowl volume $=\frac{992\pi }{3}+\frac{32\pi }{3}=\frac{1024\pi }{3}$.
3. $\frac{2}{3}\pi R^3=\frac{1024\pi }{3}\Rightarrow R^3=512\Rightarrow R=8\text{cm}$.

Trick: Keep $\pi$ and $\frac{1}{3}$ as they are; only $2R^3=1024$ needs solving. $512=8^3$.

Answer: (A) $8\text{cm}$ (the ball's radius is 2 cm, not the bowl's).

Concept: The three angles of a triangle always add up to ${180}^{\circ }$.

1. $(2x+5)+x+(3x-17)=180$.
2. $6x-12=180\Rightarrow 6x=192$.
3. $x=32$. Check: ${69}^{\circ }+{32}^{\circ }+{79}^{\circ }={180}^{\circ }$.

Trick: Add the $x$ terms and the constants separately: $6x$ and $-12$.

Answer: (A) $x=32$

Concept: In similar triangles, corresponding altitudes, medians and sides are all in the same ratio $k$, and areas are in the ratio $k^2$.

1. $k=\frac{AM}{PS}=\frac{3}{5}$, so $\frac{BC}{QR}=\frac{3}{5}\Rightarrow QR=\frac{6\times 5}{3}=10\text{cm}$.
2. $\frac{\text{ar}(ABC)}{\text{ar}(PQR)}=\frac{9}{25}\Rightarrow \text{ar}(ABC)=150\times \frac{9}{25}=54{\text{cm}}^2$.
3. $\frac{1}{2}\times BC\times AM=54\Rightarrow 3AM=54\Rightarrow AM=18\text{cm}$.
4. $AM+QR=18+10=28\text{cm}$.

Trick: Any linear ratio (side, altitude, median) $=k$; area ratio $=k^2$. The median data is extra information.

Answer: (B) $28\text{cm}$

Concept: Curved surface area of a cylinder $=2\pi rh$. Convert cm to m first.

1. $r=\frac{28}{2}=14\text{cm}=0.14\text{m}$, $h=25\text{m}$.
2. CSA $=2\times \frac{22}{7}\times 0.14\times 25=22{\text{m}}^2$.
3. Cost $=22\times 50=$ ₹1100.

Trick: $\frac{22}{7}\times 0.14=0.44$; then $2\times 0.44\times 25=22$.

Answer: (A) ₹1100

Concept: $(a^m{)}^n=a^{mn}$. Write both sides as powers of 112 and equate the exponents.

1. $x^z={112}^{0.17z}$ and $y^6={112}^{0.94\times 6}={112}^{5.64}$.
2. $0.17z=5.64\Rightarrow z=\frac{5.64}{0.17}=\frac{564}{17}$.
3. $z\approx 33.18$.

Trick: Same base ⇒ equate exponents. $17\times 33=561$, remainder 3 ⇒ $33.18$.

Answer: (C) $33.18$

Concept: Standard deviation $\sigma =\sqrt{\frac{\sum (x-\overline{x}{)}^2}{n}}$. Compute each subset separately.

1. All six (48, 50, 52, 55, 60, 65): $\overline{x}=55$, $\sum (x-\overline{x}{)}^2=49+25+9+0+25+100=208$, $\sigma =\sqrt{34.67}\approx 5.88$ → I.
2. Middle four (50, 52, 55, 60): $\overline{x}=54.25$, $\sum =56.75$, $\sigma =\sqrt{14.19}\approx 3.77$ → II.
3. First three (48, 50, 52): $\overline{x}=50$, $\sum =8$, $\sigma =\sqrt{2.67}\approx 1.63$ → IV.
4. Last three (55, 60, 65): $\overline{x}=60$, $\sum =50$, $\sigma =\sqrt{16.67}\approx 4.08$ → III.

Trick: Equally spaced triples: gap 2 gives $\sigma =2\sqrt{\frac{2}{3}}=1.63$; gap 5 gives $5\sqrt{\frac{2}{3}}=4.08$. The wider spread (D) must have the larger $\sigma$, so C–IV and D–III.

Answer: (A) A − I, B − II, C − IV, D − III

Concept: For $a{x}^2+bx+c=0$, discriminant $D=b^2-4ac$. Here $a\to a^2$, $b\to a^{3/2}$, $c\to 3$.

1. $D=(a^{3/2}{)}^2-4·a^2·3=a^3-12a^2$.
2. Given $D=28a$: $a^3-12a^2-28a=0\Rightarrow a(a^2-12a-28)=0$.
3. $a\ne 0$ (else it is not quadratic), so $a^2-12a-28=0$; product of roots $=\frac{c}{a}=-28$.

Trick: You need not solve for $a$: product of roots of $a^2-12a-28=0$ is simply $-28$.

Answer: (D) $-28$

Concept: $T_n=a+(n-1)d$. Form two equations and solve for $a$ and $d$.

1. $T_8+T_{12}=(a+7d)+(a+11d)=2a+18d=100\Rightarrow a+9d=50$.
2. $T_{20}=a+19d=95$. Subtract: $10d=45\Rightarrow d=4.5$, $a=50-40.5=9.5$.
3. $T_7=a+6d=9.5+27=36.5$.

Trick: $T_8+T_{12}=2T_{10}$, so $T_{10}=50$; then $T_{20}-T_{10}=10d=45$.

Answer: (D) $36.5$

Concept: The string is the hypotenuse; height is the side opposite the angle, so $\sin \theta =\frac{\text{height}}{\text{string}}$.

1. $\sin {30}^{\circ }=\frac{h}{40}$.
2. $h=40\times \frac{1}{2}=20\text{m}$.

Trick: In a ${30}^{\circ }$ right triangle the side opposite ${30}^{\circ }$ is half the hypotenuse.

Answer: (A) $20\text{m}$

Concept: $S_n=\frac{n}{2}[2a+(n-1)d]$.

1. $S_6=\frac{6}{2}[2\times 18+5\times 30]$.
2. $=3[36+150]=3\times 186=558$.

Trick: For $n=6$, $S_6=3(2a+5d)$ — one line.

Answer: (B) $558$

Concept: Split on the first question: $P(\text{2nd correct})=P(Q_1\text{ correct})·P(Q_2|Q_1)+P(Q_1\text{ wrong})·P(Q_2|\overline{Q_1})$.

1. $P(\text{2nd correct})=\frac{1}{5}\times \frac{5}{6}+\frac{4}{5}\times \frac{1}{4}=\frac{1}{6}+\frac{1}{5}=\frac{11}{30}$.
2. $P(\text{2nd incorrect})=1-\frac{11}{30}=\frac{19}{30}$.

Trick: Find the probability of 'correct' first (easier), then subtract from 1.

Answer: (A) $\frac{19}{30}$

Concept: A point on the x-axis is $P(x,0)$. Equidistant ⇒ $P{A}^2=P{B}^2$.

1. $(x-2{)}^2+3^2=(x+4{)}^2+5^2$.
2. $x^2-4x+13=x^2+8x+41\Rightarrow -12x=28$.
3. $x=-\frac{7}{3}$, so $P(-\frac{7}{3},0)$. Check (3,0): $P{A}^2=10$, $P{B}^2=74$ — not equal.

Trick: Square the distances; the $x^2$ terms cancel and a linear equation remains.

Answer: (C) $(-\frac{7}{3},0)$

Concept: $n(X\cup Y)=n(X)+n(Y)-n(X\cap Y)$.

1. $56=45+33-n(X\cap Y)$.
2. $n(X\cap Y)=78-56=22$.

Trick: Overlap $=$ (sum of the two sets) $-$ (union).

Answer: (B) $22$

Concept: For $x^2-5x+6=0$: $\alpha +\beta =5$, $\alpha \beta =6$; ${\alpha }^2+{\beta }^2=(\alpha +\beta {)}^2-2\alpha \beta$.

1. ${\alpha }^2+{\beta }^2=5^2-2\times 6=25-12=13$.
2. Check: roots are 2 and 3, $4+9=13$.

Trick: $x^2-5x+6$ factors as $(x-2)(x-3)$ — roots 2 and 3 instantly.

Answer: (A) $13$

Concept: A set with $n$ elements has $2^n$ subsets, so $n(P(S))=2^n$.

1. $n(S)=8$.
2. $n(P(S))=2^8=256$.

Trick: $2^8=256$ — remember $2^{10}=1024$ and work down.

Answer: (B) $256$

Concept: Stability factor $S=\frac{d{I}_C}{d{I}_{CO}}$ tells how much collector current changes with leakage current. A smaller $S$ means better thermal stability.

1. $S_1=101$ is typical of fixed bias (worst case, $S=1+\beta$).
2. $S_2=15$ is much smaller, so $I_C$ drifts less with temperature.
3. Hence circuit 2 is thermally more stable.

Trick: Lower S = better stability; ideal $S=1$. Voltage-divider bias gives small S.

Answer: (D) Circuit 2