Q&A: practice questions with answers

Concept: No profit, no loss ⇒ loss on one = gain on the other.

1. Let CP of first $=x$, second $=1200-x$.
2. $\frac{7}{100}x=\frac{28}{100}(1200-x)\Rightarrow x=4(1200-x)$
3. $5x=4800\Rightarrow x=960$
4. Check: loss $=67.2$, gain $=0.28\times 240=67.2$ ✓

Trick: Costs are in inverse ratio of the percentages: $x:(1200-x)=28:7=4:1$, so $x=\frac{4}{5}\times 1200=960$.

Answer: (B) 960

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

1. $72:12:36=6:1:3$ (divide by 12); total $=10$ parts.
2. 1 part $=3560÷10=356$
3. $C-B=(3-1)\times 356=712$

Trick: Difference = (difference of ratio terms ÷ sum) × total $=\frac{2}{10}\times 3560$.

Answer: (A) ₹712

Concept: Sum of the first $n$ odd numbers $=n^2$.

1. Sum $={148}^2$
2. Average $=\frac{{148}^2}{148}=148$
3. Or: average $=\frac{1+295}{2}=148$ (last odd number $=2\times 148-1=295$).

Trick: Average of first $n$ odd numbers $=n$; of first $n$ even numbers $=n+1$.

Answer: (D) 148

Concept: If $4A=6B=7C$, then $A:B:C=\frac{1}{4}:\frac{1}{6}:\frac{1}{7}$.

1. LCM of 4, 6, 7 $=84$; multiply by 168: $A:B:C=42:28:24$.
2. Sum $=94$ parts $=37600\Rightarrow$ 1 part $=400$.
3. $A=42\times 400=16800$

Trick: For $aA=bB=cC$: $A:B:C=bc:ca:ab=42:28:24$.

Answer: (D) ₹16,800

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

1. SP $=850\times \frac{85}{100}=722.5$
2. CP $=722.5\times \frac{100}{125}=578$

Trick: One line: CP $=850\times \frac{85}{100}\times \frac{4}{5}=578$.

Answer: (A) ₹578

Concept: Area of triangle $=\frac{1}{2}\times b\times h$ (area is in square units).

1. Let height $=h$, base $=h+2$.
2. $\frac{1}{2}(h+2)h=220\Rightarrow h^2+2h-440=0$
3. $(h+22)(h-20)=0\Rightarrow h=20$
4. Base $=20+2=22\text{cm}$

Trick: $b\times h=440$: two numbers differing by 2 → $22\times 20$.

Answer: (A) 22 cm

Concept: Make decimal places equal, find LCM of the whole numbers, then put back the decimal.

1. $4.9=\frac{4900}{1000}$, $0.008=\frac{8}{1000}$
2. $\text{LCM}(4900,8)$: $4900=2^2\times 5^2\times 7^2$, $8=2^3$ → LCM $=2^3\times 5^2\times 7^2=9800$
3. LCM $=\frac{9800}{1000}=9.8$

Trick: LCM must be a multiple of 4.9: $9.8=2\times 4.9$ and $9.8÷0.008=1225$ ✓.

Answer: (C) 9.8

Concept: Expand each bracket, then combine like terms.

1. $3(\frac{7}{3}{x}^2-25x+12)=7x^2-75x+36$
2. $-7(x^2+8x-16)=-7x^2-56x+112$
3. Add: $7x^2-7x^2-75x-56x+36+112=-131x+148$

Trick: $x^2$ terms cancel; just track signs: $x$-term $=-75-56$, constant $=36+112$.

Answer: (D) $-131x+148$

Concept: Write in $\sin$, $\cos$: $cosec\theta +\cot \theta =\frac{1+\cos \theta }{\sin \theta }$; and $1-{\cos }^2\theta ={\sin }^2\theta$.

1. $(1-\cos \theta )·\frac{1+\cos \theta }{\sin \theta }=\frac{1-{\cos }^2\theta }{\sin \theta }=\frac{{\sin }^2\theta }{\sin \theta }=\sin \theta$
2. Expression $=cosec\theta \times \sin \theta =1$

Trick: Put $\theta ={90}^{\circ }$: $1\times (1-0)\times (1+0)=1$.

Answer: (D) $1$

Concept: To make a number divisible by $13$, add $13-\text{remainder}$.

1. $63130÷13$: $13\times 4856=63128$, remainder $=2$.
2. Least number to add $=13-2=11$.
3. Check: $63130+11=63141=13\times 4857$.

Trick: Find the remainder first; the answer is (divisor − remainder), not the remainder itself (tempting option is $13$).

Answer: (A) $11$

Concept: $SI=\frac{P\times R\times T}{100}$; amount $=P+SI$.

1. $4\frac{2}{9}=\frac{38}{9}$ years: $P\times \frac{4.5}{100}\times \frac{38}{9}=0.19P=8778$, so $P=46200$ (₹).
2. $2\frac{5}{7}=\frac{19}{7}$ years: $SI=46200\times \frac{5.5}{100}\times \frac{19}{7}=6600\times 1.045=6897$.
3. Amount $=46200+6897=53097$.

Trick: $4.5\times \frac{38}{9}=19\%$ in total – so $P=\frac{8778}{0.19}$ at once.

Answer: (B) ₹53,097

Concept: Angle sum of a triangle is ${180}^{\circ }$; with parallel lines, alternate angles are equal; equal sides face equal angles.

1. In $△ABC$: $\angle ACB={180}^{\circ }-{38}^{\circ }-{65}^{\circ }={77}^{\circ }$.
2. $BC\parallel AD$ ⇒ $\angle CAD=\angle ACB={77}^{\circ }$ (alternate angles).
3. $AC=CD$ ⇒ $\angle CDA=\angle CAD={77}^{\circ }$.
4. $\angle ACD={180}^{\circ }-{77}^{\circ }-{77}^{\circ }={26}^{\circ }$.

Trick: Isosceles triangle: apex angle $={180}^{\circ }-2\times \text{base angle}$.

Answer: (B) $26$

Concept: For decimals, $\sqrt{}$ halves the number of decimal places: $\sqrt{0.0324}=0.18$.

1. $\sqrt{289}=17$.
2. $\sqrt{0.0324}=\sqrt{\frac{324}{10000}}=\frac{18}{100}=0.18$.
3. $\sqrt{4.41}=2.1$.
4. $17+0.18-2.1=15.08$.

Trick: Only option $15.08$ has two decimal places ending in $8$ – matches $0.18$.

Answer: (A) $15.08$

Concept: A $100\%$ increase means the number becomes $2$ times itself.

1. Number $\times (1+1)=2810$.
2. Number $=\frac{2810}{2}=1405$.

Trick: $+100\%$ = double; $+50\%$ = $1.5$ times.

Answer: (B) $1405$

Concept: Amount $=P(1+\frac{RT}{100})$ under simple interest.

1. $RT=5\times 5=25\%$, so amount $=1.25P$.
2. $1.25P=2280$ ⇒ $P=\frac{2280}{1.25}=1824$.

Trick: $25\%=\frac{1}{4}$: amount $=\frac{5}{4}P$, so $P=\frac{4}{5}\times 2280$.

Answer: (D) ₹1,824

Concept: Successive discounts multiply: $SP=MP\times (1-d_1)(1-d_2)$.

1. After $10\%$: $50000\times 0.9=45000$.
2. After $8\%$ more: $45000\times 0.92=41400$.

Trick: Net discount $=10+8-\frac{10\times 8}{100}=17.2\%$; $50000\times 0.828=41400$.

Answer: (C) ₹41,400

Concept: Rates add: part filled per hour $=\frac{1}{a}+\frac{1}{b}+\frac{1}{c}$.

1. $\frac{1}{13}+\frac{1}{26}=\frac{3}{26}$.
2. $\frac{3}{26}+\frac{1}{17}=\frac{51+26}{442}=\frac{77}{442}$.
3. Time $=\frac{442}{77}=5\frac{57}{77}$ hours.

Trick: All options share $\frac{57}{77}$; just check the whole part: $77\times 5=385<442<462$.

Answer: (C) $5\frac{57}{77}$ hours

Concept: Volume of cuboid $=\text{base area}\times \text{height}$.

1. Height $=\frac{440}{88}=5\text{cm}$.

Trick: $88\times 5=440$ – check by multiplying back.

Answer: (C) $5\text{cm}$

Concept: Selling at a fraction of a price changes SP proportionally; compare with CP.

1. Let actual SP $=S$, CP $=C$. $\frac{4}{10}S=0.72C$ (loss $28\%$).
2. $S=\frac{0.72C}{0.4}=1.8C$.
3. At $82\%$: $0.82\times 1.8C=1.476C$ ⇒ profit $=47.6\%$.

Trick: Take $C=100$: $S=180$, $0.82\times 180=147.6$ ⇒ $47.6\%$.

Answer: (B) $47.6\%$

Concept: Take distance $MN=D$; speed $=\frac{D}{\text{journey time}}$; after both start, gap closes at the sum of speeds.

1. A takes $9$ h (speed $\frac{D}{9}$), B takes $6$ h (speed $\frac{D}{6}$).
2. By 8:35 AM, A has run $2$ h: covered $\frac{2D}{9}$, gap $=\frac{7D}{9}$.
3. Relative speed $=\frac{D}{9}+\frac{D}{6}=\frac{5D}{18}$; time $=\frac{7}{9}÷\frac{5}{18}=2.8$ h $=2$ h $48$ min.
4. Meeting time $=8:35+2:48=11:23$ AM.

Trick: Distance cancels – always use $D$ or LCM of times ($18$ units).

Answer: (D) 11:23 AM