Q&A: practice questions with answers

Concept: If a number is decreased by $r\%$, the result is $(100-r)\%$ of the number.

1. After a $7\%$ decrease, $93\%$ of the number $=3720$.
2. Number $=\frac{3720\times 100}{93}$.
3. $\frac{3720}{93}=40$, so number $=40\times 100=4000$.

Trick: Check options: $7\%$ of $4000=280$ and $4000-280=3720$ ✔.

Answer: (C) $4000$

Concept: Open brackets by multiplying each term, then combine like terms.

1. $3(\frac{7}{3}{x}^2-22x+19)=7x^2-66x+57$.
2. $-7(x^2+9x-14)=-7x^2-63x+98$.
3. Add: $x^2$ terms: $7x^2-7x^2=0$; $x$ terms: $-66x-63x=-129x$; constants: $57+98=155$.
4. Result: $-129x+155$.

Trick: Put $x=0$: expression $=3(19)-7(-14)=57+98=155$, so the constant must be $+155$ → only (B) fits with $-129x$.

Answer: (B) $-129x+155$

Concept: Sum $=$ average $\times$ number of items.

1. Sum of first two numbers $=2\times 80=160$.
2. Third number $=102$.
3. Sum of remaining two $=655-160-102=393$.
4. Average of remaining two $=\frac{393}{2}=196.5$.

Trick: Odd sum $393$ ⇒ average ends in $\mathrm{.5}$ → only $196.5$ or $197.5$; $\frac{393}{2}=196.5$.

Answer: (A) $196.5$

Concept: For successive discounts $a\%$ and $b\%$: $\text{SP}=\text{MP}\times \frac{100-a}{100}\times \frac{100-b}{100}$.

1. After $15\%$: $20000\times \frac{85}{100}=17000$.
2. After $10\%$ more: $17000\times \frac{90}{100}=15300$.
3. Final SP $=$ ₹15,300.

Trick: Net discount $=a+b-\frac{ab}{100}=15+10-1.5=23.5\%$; $23.5\%$ of $20000=4700$; $20000-4700=15300$.

Answer: (C) ₹15,300

Concept: $22=2\times 11$, so the number must be even AND divisible by $11$ (difference of sums of alternate digits $=0$ or a multiple of $11$).

1. 6136901 is odd → out.
2. 7467614: digits from right $4,1,6,7,6,4,7$. Odd places: $4+6+6+7=23$; even places: $1+7+4=12$; difference $=11$ ✔.
3. 6580350: $0+3+8+6=17$, $5+0+5=10$, difference $7$ ✘. 8132950: $0+9+3+8=20$, $5+2+1=8$, difference $12$ ✘.

Trick: First drop odd numbers, then apply the $11$-test only to the even ones.

Answer: (A) $7467614$

Concept: Total cost price includes all expenses; $\text{Loss}\%=\frac{\text{Loss}}{\text{CP}}\times 100$.

1. CP $=940+10=950$; SP $=659$. Since SP $<$ CP, there is a loss.
2. Loss $=950-659=291$.
3. Loss% $=\frac{291}{950}\times 100=\frac{29100}{950}=\frac{582}{19}\%$ $\approx 30.63\%$.

Trick: SP is below CP, so reject both "profit" options at once; then $\frac{291}{950}\times 100=\frac{582}{19}$.

Answer: (B) $\frac{582}{19}\%$ loss

Concept: Let Ram's age $=R$ and Ramya's age $=r$; translate each sentence into an equation.

1. "2 times Ram is 14 more than 6 times Ramya": $2R=6r+14$ ⇒ $R=3r+7$.
2. "4 times Ramya is 4 less than Ram": $4r=R-4$ ⇒ $R=4r+4$.
3. $3r+7=4r+4$ ⇒ $r=3$, $R=16$.
4. Difference $=16-3=13$ years.

Trick: Equate the two expressions of $R$ directly: $3r+7=4r+4$.

Answer: (C) $13$

Concept: Third proportional of $a$ and $b$ is $x$ such that $a:b=b:x$, i.e. $x=\frac{b^2}{a}$.

1. $4:22=22:x$.
2. $x=\frac{{22}^2}{4}=\frac{484}{4}=121$.

Trick: Third proportional $=\frac{(\text{second}{)}^2}{\text{first}}$.

Answer: (B) $121$

Concept: For a decimal, take the square root of the digits and halve the number of decimal places.

1. $\sqrt{121}=11$.
2. $\sqrt{0.0121}=0.11$ (4 decimal places → 2).
3. $\sqrt{7.84}=2.8$ since ${28}^2=784$.
4. $11+0.11-2.8=8.31$.

Trick: Decimal places in the square root $=$ half of those under the root.

Answer: (B) $8.31$

Concept: A number is divisible by $11$ if (sum of digits at odd places) $-$ (sum at even places), counted from the right, is $0$ or a multiple of $11$.

1. N01M22 from right: odd places $2,M,0$ → $M+2$; even places $2,1,N$ → $N+3$.
2. Condition: $(M+2)-(N+3)=M-N-1$ must be $0$ or $\pm 11$.
3. Since $M,N$ are digits, $M-N-1=0$ ⇒ $M-N=1$ (e.g. $M=3,N=2$: $201322=11\times 18302$).
4. $M-N=5$ gives $4$, $M=N$ gives $-1$; $M+N=-1$ is impossible for digits.

Trick: Mark alternate digits $+,-$ from the right and set the total to $0$.

Answer: (A) $M-N=1$

Concept: At every vertex of a polygon, interior angle + exterior angle $={180}^{\circ }$; for a regular polygon $n=\frac{{360}^{\circ }}{\text{exterior angle}}$.

1. $I+E={180}^{\circ }$ and $I-E={140}^{\circ }$.
2. Subtracting: $2E={40}^{\circ }\Rightarrow E={20}^{\circ }$ (and $I={160}^{\circ }$).
3. $n=\frac{{360}^{\circ }}{{20}^{\circ }}=18$.

Trick: Exterior angle $=\frac{{180}^{\circ }-\text{difference}}{2}=\frac{{180}^{\circ }-{140}^{\circ }}{2}={20}^{\circ }$, then divide 360 by it.

Answer: (C) 18

Concept: $SI=\frac{P\times R\times T}{100}$, with R and T in the same time unit (here per month and months).

1. $P=1200$, $R=4\%$ per month, $T=1$ month.
2. $SI=\frac{1200\times 4\times 1}{100}=48$.

Trick: 4% of 1200: $12\times 4=48$ — 1% of a hundreds-number is just that number ÷ 100.

Answer: (D) ₹48

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

1. $r^2={6.5}^2=42.25$.
2. $4\times \frac{22}{7}\times 42.25=\frac{88\times 42.25}{7}=\frac{3718}{7}$.
3. $\frac{3718}{7}\approx 531.14\approx 531{\text{cm}}^2$.

Trick: $4\pi r^2=\frac{88}{7}{r}^2$ — multiply $r^2$ by 88 and divide by 7.

Answer: (C) $531{\text{cm}}^2$

Concept: Distance = Speed × Time; convert minutes to hours first. (A car's own length is negligible.)

1. Time $=3$ min $=\frac{3}{60}=\frac{1}{20}$ h.
2. Length of bridge $=67\times \frac{1}{20}=3.35\text{km}$.

Trick: In 3 minutes a vehicle covers $\frac{1}{20}$ of its km/h speed: $67÷20=3.35$.

Answer: (B) 3.35 km

Concept: Product of two numbers = HCF × LCM.

1. $88\times x=11\times 792$.
2. $x=\frac{11\times 792}{88}=\frac{792}{8}=99$.

Trick: Check: $99=11\times 9$ and $88=11\times 8$; LCM $=11\times 8\times 9=792$ ✓.

Answer: (A) 99

Concept: Two successive discounts $a\%$ and $b\%$ equal a single discount of $a+b-\frac{ab}{100}$ %. The scheme with the LEAST discount is best for the company.

1. First: $25+10-\frac{250}{100}=32.5\%$.
2. Second: $20+15-\frac{300}{100}=32\%$.
3. Third: $30+5-\frac{150}{100}=33.5\%$.
4. Least discount $=32\%$ → Second scheme.

Trick: Best for the buyer = biggest discount; best for the seller/company = smallest discount.

Answer: (D) Second scheme

Concept: Lateral surface area of a cuboid $=2(l+b)h$.

1. Side of each cube: $a^3=3375\Rightarrow a=15\text{cm}$.
2. Five cubes in a row: $l=75$, $b=15$, $h=15$.
3. LSA $=2(75+15)\times 15=2\times 90\times 15=2700{\text{cm}}^2$.

Trick: For $n$ cubes of side $a$ joined in a line, LSA $=2(n+1)a^2=2\times 6\times 225=2700$.

Answer: (A) 2700

Concept: One-day work adds up: if a group does a job in $d$ days, its one-day work is $\frac{1}{d}$.

1. Anjali $=\frac{1}{21}$; Ayushi $=\frac{1}{12}-\frac{1}{21}=\frac{7-4}{84}=\frac{1}{28}$.
2. Ankita $=\frac{1}{8}-\frac{1}{12}=\frac{3-2}{24}=\frac{1}{24}$.
3. Anjali + Ankita $=\frac{1}{21}+\frac{1}{24}=\frac{8+7}{168}=\frac{15}{168}=\frac{5}{56}$.
4. Days $=\frac{56}{5}$.

Trick: Take total work = LCM(21, 12, 8) = 168 units: Anjali 8/day, all three 21/day, Anjali+Ayushi 14/day → Ankita 7/day → $\frac{168}{8+7}=\frac{56}{5}$.

Answer: (D) $\frac{56}{5}$ days

Concept: Rationalise: $\sqrt{\frac{1+\cos A}{1-\cos A}}=\frac{1+\cos A}{\sin A}$, using $1-{\cos }^{2}A={\sin }^{2}A$.

1. Multiply inside by $\frac{1+\cos A}{1+\cos A}$: $\sqrt{\frac{(1+\cos A{)}^2}{{\sin }^{2}A}}=\frac{1+\cos A}{\sin A}$ (positive for acute A).
2. $\frac{1+\cos A}{\sin A}=\text{cosec}A+\cot A$.
3. $M=(\text{cosec}A+\cot A)-\text{cosec}A-\cot A=0$.

Trick: Put $A={60}^{\circ }$: $\sqrt{\frac{1.5}{0.5}}=\sqrt{3}$ and $\text{cosec}{60}^{\circ }+\cot {60}^{\circ }=\frac{2}{\sqrt{3}}+\frac{1}{\sqrt{3}}=\sqrt{3}$, so $M=0$.

Answer: (B) 0

Concept: When the sum and LCM are given, test options: the pair must add to 52 and have LCM 627.

1. $627=3\times 11\times 19=19\times 33$.
2. 19 and 33 are co-prime, so LCM $=19\times 33=627$ ✓ and $19+33=52$ ✓.
3. Others fail: LCM(12, 40) = 120, LCM(17, 35) = 595, LCM(24, 28) = 168.

Trick: Factorise the LCM and look for two factors whose sum is 52.

Answer: (A) 19, 33