Q&A: practice questions with answers

Concept: Empirical relation: $\text{Mode}=3\text{Median}-2\text{Mean}$, so $\text{Mode}-\text{Median}=2(\text{Median}-\text{Mean})$.

1. $\text{Mode}-\text{Median}=71.24$
2. $2(\text{Median}-\text{Mean})=71.24$
3. $\text{Median}-\text{Mean}=35.62$

Trick: Median sits between mean and mode, twice as far from the mode: just halve the gap.

Answer: (B) 35.62

Concept: Take total work = LCM of days and work in units per day.

1. Total work $=\text{LCM}(12,18)=36$ units; A $=3$, B $=2$ units/day.
2. In 2 days together: $2\times 5=10$ units.
3. Remaining $26$ units by B: $\frac{26}{2}=13$ days.
4. Total $=2+13=15$ days

Trick: LCM method avoids fractions completely.

Answer: (C) 15

Concept: If two chords AB and CD meet externally at P, then $PA\times PB=PC\times PD$ (secant–secant theorem).

1. Let $PB=x$, so $PA=x+7$; $PD=5$, $PC=5+1=6$.
2. $x(x+7)=5\times 6=30$
3. $x^2+7x-30=0\Rightarrow (x+10)(x-3)=0$
4. $x=3$ cm

Trick: Outside part × whole = outside part × whole. Try options: $3\times 10=30$ ✓.

Answer: (A) 3

Concept: To cross a platform, a train covers (train length + platform length).

1. Speed $=36\times \frac{5}{18}=10\text{m/s}$; time $=1.7\times 60=102\text{s}$
2. Distance $=10\times 102=1020\text{m}$
3. Platform $=1020-591=429\text{m}$
4. Difference $=591-429=162\text{m}$

Trick: 36 km/h = 10 m/s, so distance in metres = 10 × seconds.

Answer: (A) 162 m

Concept: $\sin ({90}^{\circ }-\theta )=\cos \theta$, $\cos ({90}^{\circ }-\theta )=\sin \theta$, $\csc ({90}^{\circ }-\theta )=\sec \theta$.

1. $\sin {55}^{\circ }=\cos {35}^{\circ }\Rightarrow \frac{\cos {35}^{\circ }}{\sin {55}^{\circ }}=1$
2. $\cos {79}^{\circ }=\sin {11}^{\circ }\Rightarrow \frac{\sin {11}^{\circ }}{\cos {79}^{\circ }}=1$
3. $\csc {63}^{\circ }=\sec {27}^{\circ }\Rightarrow 2\cos {27}^{\circ }\csc {63}^{\circ }=2$
4. Value $=1+1-2=0$

Trick: Angles adding to ${90}^{\circ }$ in a ratio always give 1.

Answer: (A) 0

Concept: Compare each ratio with $\frac{1}{2}$ or convert to decimals.

1. $\frac{19}{37}\approx 0.51,\frac{10}{15}\approx 0.67$
2. $\frac{15}{33}\approx 0.45,\frac{18}{37}\approx 0.49$
3. Greatest is $\frac{10}{15}$.

Trick: Only $\frac{10}{15}=\frac{2}{3}$ is well above $\frac{1}{2}$; others are near $\frac{1}{2}$.

Answer: (B) 10 : 15

Concept: Factorise; every prime must have an even power in a perfect square.

1. $6760=676\times 10={26}^2\times 2\times 5$
2. The unpaired primes are 2 and 5.
3. Multiply by $2\times 5=10$: $67600={260}^2$

Trick: Spot $676={26}^2$ quickly; leftover $10$ is the multiplier.

Answer: (D) 10

Concept: Find maximum marks from Kalpana’s score, then Sudha’s percentage.

1. Max marks $=\frac{696}{87}\times 100=800$
2. Sudha $=\frac{584}{800}\times 100=73\%$

Trick: Direct ratio: $\frac{584}{696}\times 87=73$ (since $696=8\times 87$, $584=8\times 73$).

Answer: (A) 73%

Concept: Use the ratio unit: numbers are $5k,16k,21k$.

1. $20\%$ of $5k=k=26$
2. Difference of third and second $=21k-16k=5k=130$
3. $40\%$ of $130=52$

Trick: $40\%$ of $5k=2k=2\times 26=52$ directly.

Answer: (B) 52

Concept: Cost price stays the same; profit = SP − CP.

1. Let CP $=C$, SP $=S$. New SP $=2S$, new profit $=6(S-C)$.
2. $2S-C=6S-6C\Rightarrow 5C=4S\Rightarrow S=\frac{5}{4}C$
3. Initial profit $=\frac{S-C}{C}\times 100=\frac{1}{4}\times 100=25\%$

Trick: Test option: CP 100, SP 125, profit 25; SP 250 → profit 150 = 6 × 25 ✓.

Answer: (D) 25

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

1. Let diameters be $D$ (larger) and $d$ (smaller); $D-d=2$.
2. $\pi D^2-\pi d^2=528\Rightarrow \frac{22}{7}(D-d)(D+d)=528$.
3. $(D+d)=\frac{528\times 7}{22\times 2}=84$.
4. $D=\frac{84+2}{2}=43\text{cm}$.

Trick: Use $\pi (D^2-d^2)$ directly with diameters – avoids halving radii.

Answer: (A) $43$

Concept: Average speed $=\frac{\text{total distance}}{\text{total time}}$.

1. Forward: $d$ at 84 and $2d$ at 28; time $=\frac{d}{84}+\frac{2d}{28}=\frac{7d}{84}=\frac{d}{12}$.
2. Return distance $=3d$, so total distance $=6d$; total time $=\frac{6d}{24}=\frac{d}{4}$.
3. Return time $=\frac{d}{4}-\frac{d}{12}=\frac{d}{6}$.
4. Return speed $=\frac{3d}{d/6}=18\text{km/h}$.

Trick: Take $d=84$: forward time $=1+6=7$ h, total time $=\frac{504}{24}=21$ h, return $=14$ h for 252 km → 18.

Answer: (C) $18$

Concept: A quadratic with roots $p,q$ is $x^2-(p+q)x+pq=0$.

1. For $x^2-6x-5=0$: $\alpha +\beta =6,\alpha \beta =-5$.
2. New sum $=3\alpha +3\beta =18$.
3. New product $=9\alpha \beta =-45$.
4. Equation: $x^2-18x-45=0$.

Trick: Roots multiplied by $k$: replace $x$ by $\frac{x}{k}$ → $x^2-6(3)x-5(9)=0$.

Answer: (B) $x^2-18x-45=0$

Concept: $\text{Increase}\%=\frac{\text{increase}}{\text{original}}\times 100$.

1. Increase $=193900-170014=23886$.
2. $\%=\frac{23886}{170014}\times 100\approx 14.05\%$.
3. Rounded: $14\%$.

Trick: 10% of 170014 ≈ 17000, 15% ≈ 25500; 23886 lies just below 15% → 14%.

Answer: (C) $14\%$

Concept: A pure recurring decimal $0.\bar{a}=\frac{a}{9}$.

1. $6.555\dots =6+0.\bar{5}$.
2. $0.\bar{5}=\frac{5}{9}$.
3. $6+\frac{5}{9}=\frac{54+5}{9}=\frac{59}{9}$.

Trick: Check: $59÷9=6.555\dots$ ✓

Answer: (A) $\frac{59}{9}$

Concept: For $n$ (odd) observations in order, median $={\left(\frac{n+1}{2}\right)}^{\text{th}}$ value.

1. Ascending order: 17, 30, 38, 48, 59, 61, 67, 80, 80, 88, 91.
2. $n=11$, so median is the $\frac{11+1}{2}=6^{\text{th}}$ value.
3. 6th value $=61$.

Trick: Always sort first – the unsorted middle value (88) is a trap.

Answer: (A) $61$

Concept: For similar triangles, ratio of areas $=$ (ratio of corresponding sides)${}^2$.

1. ${\left(\frac{AB}{PQ}\right)}^2=\frac{25}{49}$.
2. $\frac{AB}{PQ}=\sqrt{\frac{25}{49}}=\frac{5}{7}$.
3. $AB:PQ=5:7$.

Trick: Sides = square root of area ratio; area = square of side ratio.

Answer: (C) $5:7$

Concept: Solve the innermost bracket first, then $÷$ and $\times$ from left to right.

1. $15-\frac{11}{4}=\frac{60-11}{4}=\frac{49}{4}$.
2. $8÷4\times \frac{49}{4}=2\times \frac{49}{4}=\frac{49}{2}$.
3. $22\times 6=132$.
4. $132\times \frac{49}{2}=66\times 49=3234$.

Trick: $66\times 49=66\times 50-66=3300-66=3234$.

Answer: (B) $3234$

Concept: Same work: (workers × days) must be equal.

1. $(13m+3w)\times 7=(5m+6w)\times 18$.
2. $91m+21w=90m+108w$.
3. $m=87w$.
4. So 1 man = 87 women.

Trick: Cross-multiply efficiencies with days, collect $m$ on one side, $w$ on the other.

Answer: (A) $87$

Concept: On melting and recasting, volume stays the same. Hemisphere $=\frac{2}{3}\pi r^3$, cylinder $=\pi r^{2}h$.

1. $\pi r^{2}h=\frac{2}{3}\pi r^3$.
2. $h=\frac{2}{3}r=\frac{2}{3}\times 6$.
3. $h=4\text{cm}$.

Trick: Same radius: cylinder height $=\frac{2}{3}$ of radius (for hemisphere).

Answer: (D) $4\text{cm}$