Q&A: practice questions with answers

Concept: Net effect of successive changes $a\%$ and $b\%$: $a+b+\frac{ab}{100}$.

1. $a=-15,b=+10$
2. Net $=-15+10+\frac{(-15)(10)}{100}=-5-1.5=-6.5\%$
3. Check: $100\to 85\to 93.5$, i.e. $6.5\%$ lower.

Trick: Take salary = 100: $100\times 0.85\times 1.1=93.5$.

Answer: (C) $6.5\%$

Concept: If $m$ is the mean proportional between $a$ and $b$, then $m^2=ab$.

1. ${260}^2=x\times 338$
2. $x=\frac{67600}{338}=200$
3. Check: $200\times 338=67600={260}^2$ ✓

Trick: Factorise: $260=13\times 20$, $338=2\times {13}^2$ ⇒ $x=\frac{{13}^2\times 400}{2\times {13}^2}=200$.

Answer: (A) 200

Concept: Class average $=\frac{\text{total marks}}{\text{total students}}$.

1. Total of 12 students $=12\times 60=720$; rest $=18$ students, total $=18\times 62=1116$.
2. Class total $=720+1116=1836$
3. Average $=\frac{1836}{30}=61.2$

Trick: Deviation method: base 60; extra $=\frac{18\times 2}{30}=1.2$ ⇒ $61.2$.

Answer: (A) 61.2

Concept: Whole numbers start from 0, so the first 14 whole numbers are 0, 1, 2, …, 13.

1. Sum $=\frac{13\times 14}{2}=91$
2. Average $=\frac{91}{14}=6.5$

Trick: Average of consecutive numbers $=\frac{\text{first}+\text{last}}{2}=\frac{0+13}{2}=6.5$

Answer: (A) 6.5

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

1. $r=\frac{98}{2}=49\text{cm}$
2. $4\pi r^2=4\times \frac{22}{7}\times 49\times 49$
3. $=4\times 22\times 7\times 49=30184{\text{cm}}^2$

Trick: Using diameter: $\text{SA}=\pi d^2=\frac{22}{7}\times 98\times 98=30184$.

Answer: (C) $30184{\text{cm}}^2$

Concept: Form an equation from the "after 8 years" condition.

1. Let daughter $=d$, Rahul $=3d$.
2. After 8 years: $3d+8=2(d+8)+5$
3. $3d+8=2d+21\Rightarrow d=13$, Rahul $=3\times 13=39$ years.
4. Check: $39+8=47$ and $2\times 21+5=47$ ✓. (With 30: $38\ne 2\times 18+5=41$.)

Trick: Always verify the chosen option in both conditions.

Answer: (D) 39 years

Concept: Take total work = LCM of the given days and find each person's units per day.

1. 1 hall $=\text{LCM}(3,7)=21$ units. Varun + Ashish $=\frac{21}{3}=7$ units/day; Varun $=\frac{21}{7}=3$ units/day.
2. Ashish $=7-3=4$ units/day.
3. 4 halls $=4\times 21=84$ units ⇒ days $=\frac{84}{4}=21$.

Trick: Ashish alone for 1 hall $=\frac{3\times 7}{7-3}=\frac{21}{4}$ days; for 4 halls $=21$ days.

Answer: (D) 21

Concept: Speed of stream $=\frac{\text{downstream speed}-\text{upstream speed}}{2}$.

1. Upstream $=\frac{721}{42}=17.1\bar{6}\text{km/h}$
2. Downstream $=\frac{721}{15}=48.0\bar{6}\text{km/h}$
3. Stream $=\frac{48.07-17.17}{2}=\frac{30.9}{2}=15.45\text{km/h}$

Trick: Exact: $\frac{721}{2}(\frac{1}{15}-\frac{1}{42})=\frac{721}{2}\times \frac{27}{630}=15.45$.

Answer: (D) $15.45\text{km/hr}$

Concept: Savings $\%=100\%-\text{expenditure}\%$.

1. Savings $=100-45=55\%$ of income.
2. $55\%$ of income = ₹66000 ⇒ income $=\frac{66000\times 100}{55}$
3. $=1200\times 100=$ ₹120000

Trick: $\frac{66000}{55}=1200$ per 1% ⇒ ×100.

Answer: (D) 1,20,000

Concept: $\text{CP}=\frac{\text{SP}\times 100}{100\pm \text{profit/loss}\%}$; overall loss % is on total CP.

1. Dryer 1: $\text{CP}=\frac{63000\times 100}{105}=60000$
2. Dryer 2: $\text{CP}=\frac{63000\times 100}{70}=90000$
3. Total CP $=150000$, total SP $=126000$, loss $=24000$
4. Loss $\%=\frac{24000}{150000}\times 100=16\%$

Trick: With equal SP, never just average the percentages – find each CP.

Answer: (D) $16\%$

Concept: For 2 years: $\text{CI}-\text{SI}=P{\left(\frac{r}{100}\right)}^2$.

1. $797=P\times {\left(\frac{16}{100}\right)}^2=P\times 0.0256$
2. $P=\frac{797}{0.0256}=31132.8\dots$
3. Rounded ⇒ ₹31133

Trick: $P=\frac{\text{diff}\times 10000}{r^2}=\frac{7970000}{256}$.

Answer: (D) ₹31,133

Concept: Convert all fractions to the LCM of denominators (16) and add.

1. $\frac{3}{8}+\frac{1}{8}+\frac{2}{8}=\frac{6}{8}=\frac{12}{16}$
2. $\frac{12}{16}+\frac{5}{16}=\frac{17}{16}$
3. $\frac{17}{16}-2=\frac{17-32}{16}=-\frac{15}{16}$

Trick: Add the same-denominator fractions first, then convert once.

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

Concept: If $a:b::c:d$ then $ad=bc$; for mean proportion $y^2=ac$.

1. $(30+x)(11+x)=(15+x)(21+x)$ ⇒ $330+41x=315+36x$ ⇒ $x=-3$
2. Then $6x=-18$ and $3x-9=-18$.
3. $y^2=(-18)\times (-18)=324$ ⇒ $y=18$ (since $y>0$).

Trick: Check: $27,12,18,8$ give $\frac{27}{12}=\frac{18}{8}=2.25$ ✓.

Answer: (A) $18$

Concept: Diagonal AC splits ABCD into two triangles; both turn out to be right-angled (Pythagorean triplets).

1. $△ABC$: $8^2+{15}^2=289={17}^2$ → right angle at C, area $=\frac{1}{2}\times 8\times 15=60$
2. $△ACD$: $9^2+{12}^2=225={15}^2$ → right angle at D, area $=\frac{1}{2}\times 9\times 12=54$
3. Area of ABCD $=60+54=114{\text{cm}}^2$

Trick: Spot triplets 8-15-17 and 9-12-15 (3-4-5 × 3) – no Heron needed.

Answer: (B) $114$

Concept: $SI=\frac{P\times R\times T}{100}$

1. $SI=2280-1200=₹1080$
2. $1080=\frac{1200\times 4\times T}{100}=48T$
3. $T=\frac{1080}{48}=22.5$ years

Trick: Interest per year $=4\%$ of 1200 $=48$; divide total interest by 48.

Answer: (A) $22.5$

Concept: $SP=CP\times \frac{100+\text{gain}\%}{100}$

1. CP of 1 lemon $=₹\frac{1}{7}$
2. SP of 1 lemon $=\frac{1}{7}\times \frac{175}{100}=₹\frac{1}{4}$
3. Lemons sold for ₹1 $=4$

Trick: For ₹1 both ways: lemons sold $=\frac{7\times 100}{175}=4$.

Answer: (B) $4$

Concept: Supplementary angles add up to ${180}^{\circ }$.

1. Let smaller $=x$, larger $=x+{50}^{\circ }$
2. $2x+{50}^{\circ }={180}^{\circ }$ ⇒ $x={65}^{\circ }$
3. Larger $={115}^{\circ }$ (check: $65+115=180$)

Trick: Smaller $=\frac{180-50}{2}={65}^{\circ }$.

Answer: (A) ${65}^{\circ }$

Concept: In $a:b::c:d$, product of extremes = product of means: $ad=bc$.

1. $7.5\times x=19.5\times 16$
2. $x=\frac{312}{7.5}=41.6$

Trick: $\frac{19.5}{7.5}=2.6$, so $x=16\times 2.6=41.6$.

Answer: (C) $41.6$

Concept: BODMAS: bracket → multiplication → addition/subtraction.

1. $17+6=23$
2. $7\times 23=161$
3. $127-161+93=220-161=59$

Trick: Add the positive terms together first: 127 + 93 = 220.

Answer: (A) $59$

Concept: Work done = rate × time; an emptying pipe has a negative rate.

1. In 6 min A fills $\frac{6}{18}=\frac{1}{3}$; remaining $=\frac{2}{3}$
2. Net rate $=\frac{1}{18}-\frac{1}{20}=\frac{10-9}{180}=\frac{1}{180}$ per min
3. Time $=\frac{2}{3}\times 180=120$ min

Trick: LCM method: tank $=180$ units, A $=+10$, B $=-9$, net $=1$ unit/min; remaining $120$ units → $120$ min.

Answer: (A) $120$