Q&A: practice questions with answers

Concept: For consecutive numbers $n(n+1)=650$, $n$ is close to $\sqrt{650}$.

1. $\sqrt{650}\approx 25.5$, so try $25$ and $26$.
2. $25\times 26=650$ ✓
3. Greater number $=26$.

Trick: Product ending in 0 with numbers near $\sqrt{N}$: $25\times 26$ (5 × even → ends in 0).

Answer: (D) $26$

Concept: Total = average × number of persons.

1. $R+An+Am=3\times 45=135$
2. $R+An=2\times 44=88$, $An+Am=2\times 42=84$
3. Adding: $R+2An+Am=172$
4. $An=172-135=37\text{kg}$

Trick: Common person = (sum of pair totals) − (total of all) $=88+84-135=37$.

Answer: (A) $37$

Concept: $a^m÷a^n=a^{m-n}$ and $a^m\times a^n=a^{m+n}$ (left to right).

1. ${69}^{-4}÷{69}^{15}={69}^{-4-15}={69}^{-19}$
2. ${69}^{-19}\times {69}^{-9}={69}^{-19-9}={69}^{-28}$

Trick: Add the powers with their signs: $-4-15-9=-28$.

Answer: (D) ${69}^{-28}$

Concept: Selling price = marked price × $\frac{100-\text{discount}\%}{100}$.

1. $\text{SP}=46000\times \frac{84}{100}$
2. $=460\times 84=38640$

Trick: $16\%$ of $46000=7360$; $46000-7360=38640$.

Answer: (A) ₹38,640

Concept: Let Vinay take $t$ hours. Doubling the speed halves the time.

1. Amit's time $=t+8$; with double speed $\frac{t+8}{2}=t-9$
2. $t+8=2t-18\Rightarrow t=26$
3. Amit's time $=34\text{h}$
4. Amit's speed $=\frac{102}{34}=3\text{km/h}$

Trick: Check options: speed $3\Rightarrow 34\text{h}$ and $17\text{h}$; difference $34-17=17=8+9$ ✓.

Answer: (B) $3\text{km/h}$

Concept: Use $\sec A-\tan A=\frac{1-\sin A}{\cos A}$, $\text{cosec}A+\cot A=\frac{1+\cos A}{\sin A}$ and $1-{\sin }^{2}A={\cos }^{2}A$, $1-{\cos }^{2}A={\sin }^{2}A$.

1. Numerator: $(1+\sin A)\frac{1-\sin A}{\cos A}=\frac{{\cos }^{2}A}{\cos A}=\cos A$
2. Denominator: $\text{cosec}A(1-\cos A)\frac{1+\cos A}{\sin A}=\text{cosec}A·\frac{{\sin }^{2}A}{\sin A}=\text{cosec}A\sin A=1$
3. Value $=\frac{\cos A}{1}=\cos A$

Trick: Put $A={60}^{\circ }$: numerator $=\frac{1}{2}$, denominator $=1$ → $\frac{1}{2}=\cos {60}^{\circ }$.

Answer: (D) $\cos A$

Concept: Form two equations in mother's age $M$ and daughter's age $D$.

1. $2M=6D+26\Rightarrow M=3D+13$
2. $4D=M-4\Rightarrow M=4D+4$
3. $4D+4=3D+13\Rightarrow D=9$, $M=40$
4. Difference $=40-9=31$ years

Trick: Check: $2\times 40=80=6\times 9+26$ ✓ and $4\times 9=36=40-4$ ✓.

Answer: (C) $31$

Concept: $\text{SI}=\frac{P\times R\times T}{100}$, Amount $=P+\text{SI}$.

1. $2\frac{7}{8}=\frac{23}{8}$ years: $8671=\frac{P\times 2.9\times 23}{800}\Rightarrow P=\frac{8671\times 800}{66.7}=104000$
2. New SI $=\frac{104000\times 4.5\times 4.4}{100}=104000\times 0.198=20592$
3. Amount $=104000+20592=124592$

Trick: Rate × time $=4.5\times 4.4=19.8\%$ of $P$; $P\times 1.198=124592$.

Answer: (D) ₹1,24,592

Concept: Expand the brackets; the $x^2$ terms cancel, leaving a linear equation.

1. $9x^2-18-9x^2-15x+21=19$
2. $3-15x=19$
3. $-15x=16\Rightarrow x=-\frac{16}{15}$

Trick: Watch the sign: $-3\times (-7)=+21$.

Answer: (C) $-\frac{16}{15}$

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

1. Ratio $=1130:1030:1840=113:103:184$ (sum $400$)
2. Uday's share $=\frac{184}{400}\times 1800=184\times 4.5=828$

Trick: Uday's fraction $=\frac{1840}{4000}=46\%$; $46\%$ of $1800=828$.

Answer: (D) ₹828

Concept: Work on the part of salary left after each payment.

1. Let salary $=S$. After debt: $S-36\%S=0.64S$.
2. He spends $50\%$ of this, so savings $=0.5\times 0.64S=0.32S$.
3. $0.32S=704\Rightarrow S=\frac{704}{0.32}=2200$.

Trick: Savings fraction $=0.64\times 0.5=0.32$ of salary; divide savings by $0.32$ directly.

Answer: (C) ₹2,200

Concept: $\text{Distance}=\text{Speed}\times \text{Time}$, with time in hours.

1. Time $=18$ min $=\frac{18}{60}=0.3$ h.
2. Length of bridge $=21\times 0.3=6.3\text{km}$.

Trick: 18 min $=\frac{3}{10}$ h, so distance $=\frac{3}{10}$ of the speed: $\frac{3\times 21}{10}=6.3$.

Answer: (C) 6.3 km

Concept: Follow BODMAS: innermost bracket first, then multiplication, then addition, then division.

1. $(3-4)=-1$.
2. $7\times (-1)=-7$, so $\{13-7\}=6$.
3. $138÷6=23$.

Yaad rakhein: Solve brackets from inside out: ( ) → { } → [ ].

Answer: (A) 23

Concept: Take total work = LCM of the days; find the work done together and subtract.

1. Total work $=\text{LCM}(11,23)=253$ units.
2. A does $\frac{253}{11}=23$ units/day, B does $\frac{253}{23}=11$ units/day; together $34$ units/day.
3. In 2 days: $68$ units. Left $=253-68=185$ units $=\frac{185}{253}$.

Trick: Work left $=1-2(\frac{1}{11}+\frac{1}{23})=1-\frac{68}{253}$.

Answer: (C) $\frac{185}{253}$

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

1. Write with 3 decimal places: $2.900$ and $0.031$ → $2900$ and $31$.
2. $2900=2^2\times 5^2\times 29$ and $31$ is prime, so $\text{LCM}=2900\times 31=89900$.
3. Put back 3 decimal places: $89.900=89.9$.

Trick: LCM of decimals $=\frac{\text{LCM of numerators}}{\text{HCF of denominators}}$: $\frac{\text{LCM}(29,31)}{\text{HCF}(10,1000)}=\frac{899}{10}=89.9$.

Answer: (A) 89.9

Concept: Volume is conserved: number of balls $=\frac{\text{volume of big sphere}}{\text{volume of one ball}}$.

1. Big sphere radius $R=\frac{64}{2}=32$ cm, ball radius $r=4$ cm.
2. $n=\frac{\frac{4}{3}\pi R^3}{\frac{4}{3}\pi r^3}={\left(\frac{32}{4}\right)}^3=8^3=512$.

Trick: For spheres, $n={\left(\frac{R}{r}\right)}^3$ — no need to use $\pi$.

Answer: (D) 512

Concept: If $2A=4B=7C$, the shares are in the ratio $\frac{1}{2}:\frac{1}{4}:\frac{1}{7}$.

1. Multiply by LCM 28: $A:B:C=14:7:4$; total parts $=25$.
2. One part $=\frac{45800}{25}=1832$.
3. Share of A $=14\times 1832=25648$.

Trick: $xA=yB=zC\Rightarrow A:B:C=yz:xz:xy=28:14:8=14:7:4$.

Answer: (C) ₹25,648

Concept: Number to add $=\text{divisor}-\text{remainder}$.

1. $76\times 262=19912$, so $19964-19912=52$ is the remainder.
2. Number to add $=76-52=24$; check: $19988=76\times 263$.

Yaad rakhein: Least to add $=d-r$; least to subtract $=r$.

Answer: (C) 24

Concept: $\text{Sum}=\text{Average}\times \text{Count}$.

1. First two numbers: $80\times 2=160$; third $=106$.
2. Last two: $655-160-106=389$.
3. Average $=\frac{389}{2}=194.5$.

Trick: Remove known parts from the total, then divide by the count left.

Answer: (C) 194.5

Concept: Successive discounts are applied one after another on the reduced price.

1. After $20\%$: $1200\times 0.8=960$.
2. After $10\%$: $960\times 0.9=864$.

Trick: Net discount of $20\%$ and $10\%$ $=20+10-\frac{20\times 10}{100}=28\%$; $72\%$ of 1200 $=864$.

Answer: (D) ₹864