Q&A: practice questions with answers

Concept: $\text{Area of rhombus}=\frac{1}{2}\times d_1\times d_2$.

1. $d_1=15$ cm, $d_2=24$ cm.
2. $\text{Area}=\frac{1}{2}\times 15\times 24=15\times 12=180$ cm².

Trick: Halve the even diagonal first: $24÷2=12$, then $12\times 15=180$ – no fractions.

Answer: (B) $180{\mathrm{cm}}^2$

Concept: Take total work = LCM of days; find work done in one full cycle, then the leftover.

1. $\text{LCM}(6,8,12,16)=48$ units. Daily work: Avi 8, Avni 6, Aarti 4, Reeta 3.
2. In $\frac{1}{4}$ day each: $2+1.5+1+0.75=5.25$ units per cycle (1 day).
3. 9 cycles = $9\times 5.25=47.25$ units; left = 0.75 unit.
4. Next turn is Avi's; he does 2 units in his quarter, so he finishes the 0.75 unit.

Trick: Count full cycles that fit ($48÷5.25\approx 9.1$). The remainder is tiny and the next cycle starts with Avi.

Answer: (C) Avi

Concept: If $\frac{A}{192}=\frac{48}{A}$, then $A^2=192\times 48$ (A is the mean proportional).

1. $A^2=192\times 48=9216$.
2. $A=\sqrt{9216}=96$.

Trick: $192=4\times 48$, so $A^2=4\times {48}^2$ → $A=2\times 48=96$.

Answer: (D) 96

Concept: $\text{Female literates}=\text{total literates}-\text{male literates}$; then divide by female population.

1. Work in thousands: females = $385-168=217$ thousand.
2. Literate people = $0.52\times 385=200.2$ thousand; literate men = $0.21\times 168=35.28$ thousand.
3. Literate women = $200.2-35.28=164.92$ thousand.
4. Female literacy = $\frac{164.92}{217}\times 100=76\%$.

Trick: Work in thousands: $385\times 0.52=200.2$, $168\times 0.21=35.28$ → $164.92÷217=0.76$.

Answer: (C) 76%

Concept: For a fixed distance, $\text{time}\propto \frac{1}{\text{speed}}$.

1. Speed becomes $\frac{1}{8}$ → time becomes 8 times: usual t, new 8t.
2. Delay = $8t-t=7t=21$ min.
3. t = 3 minutes.

Trick: At $\frac{1}{n}$ of speed, $\text{delay}=(n-1)\times \text{usual time}$ → usual time = $21÷7=3$.

Answer: (B) 3

Concept: For two numbers: $\text{first}\times \text{second}=\text{HCF}\times \text{LCM}$.

1. $99\times x=9\times 198$.
2. $x=9\times 198÷99=9\times 2=18$.

Trick: $198÷99=2$ at once, so $x=9\times 2$. Also check: answer must be a multiple of HCF 9 → only 18.

Answer: (C) 18

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

1. $\text{CP}=\frac{222\times 100}{120}$.
2. $\text{CP}=222\times \frac{5}{6}$ = ₹185.

Trick: 20% gain = $\frac{1}{5}$ → SP is 6 parts, CP 5 parts: $222÷6=37,\times 5=185$.

Answer: (C) 185

Concept: $\text{Average speed}=\frac{\text{total distance}}{\text{total time}}$ (not the mean of speeds).

1. Distance = $4\times 2=8$ km.
2. Time = $\frac{2}{10}+\frac{2}{20}+\frac{2}{30}+\frac{2}{40}=\frac{12+6+4+3}{60}=\frac{25}{60}=\frac{5}{12}$ h.
3. Average speed = $8÷\frac{5}{12}=\frac{96}{5}=19.2$ km/h.

Trick: For equal distances, average = harmonic mean, always less than simple mean (25) – reject 21.2 quickly by calculation.

Answer: (C) 19.2

Concept: $\text{New member's value}=\text{new total}-\text{old total}$.

1. Students' total = $48\times 10=480$.
2. With teacher: $49\times 11=539$.
3. Teacher = $539-480=59$ years.

Trick: Teacher = new average + (old count × rise) = $11+48\times 1=59$.

Answer: (C) 59

Concept: $\text{Weighted average}=\frac{\Sigma (\text{quantity}\times \text{value})}{\Sigma \text{quantity}}$.

1. Ratio of barrels 200 : 100 : 100 = 2 : 1 : 1.
2. $\frac{2\times 85+90+95}{4}=\frac{355}{4}=88.75\%$.

Trick: Use the simple ratio 2 : 1 : 1 instead of the big barrel numbers.

Answer: (A) 88.75%

Concept: Third proportional of a and b = $\frac{b^2}{a}$.

1. $c=\frac{{30}^2}{18}=\frac{900}{18}$.
2. $c=50$.

Trick: Check: 18 : 30 = 3 : 5 and 30 : 50 = 3 : 5 ✓.

Answer: (D) 50

Concept: If amounts are equal, the share kept for one more year is smaller by the factor $(1+r)$.

1. Mandar's money grows 10 years, Ketan's 9 years.
2. $M\times {\left(\frac{21}{20}\right)}^{10}=K\times {\left(\frac{21}{20}\right)}^9$ → $K=M\times \frac{21}{20}$ → M : K = 20 : 21.
3. Ketan = $24600\times \frac{21}{41}=600\times 21$ = ₹12600.

Trick: 1-year gap at 5% → shares 20 : 21 (younger gets 20). Total 41 parts.

Answer: (C) 12600

Concept: If $p\%\text{ of }x=y$, then $x=\frac{y\times 100}{p}$.

1. $x=\frac{36\times 100}{4}$.
2. $x=900$.

Trick: $4\%=\frac{1}{25}$ → $x=36\times 25=900$.

Answer: (A) 900

Concept: At the same rate and time, difference in SI comes only from difference in principal.

1. Atul has ₹4000 more ($9400-5400$).
2. Extra interest per year = $720÷4$ = ₹180.
3. $\text{Rate}=\frac{180}{4000}\times 100=4.5\%$.

Trick: Option check: 1% of 4000 for 4 years = 160; $720÷160=4.5$.

Answer: (B) 4.5

Concept: Smallest number divisible by all = LCM (highest power of each prime).

1. $12=2^2\times 3,28=2^2\times 7,36=2^2\times 3^2,27=3^3$.
2. $\text{LCM}=2^2\times 3^3\times 7=4\times 27\times 7=756$.

Trick: Option test: must be divisible by 7 and 27 → $756=27\times 28$ ✓; others fail.

Answer: (C) 756

Concept: Apply discounts one after another on the reduced price.

1. After 10%: $60000\times 0.9=54000$.
2. After 15%: $54000\times 0.85=45900$.
3. After 20%: $45900\times 0.8$ = ₹36720.

Trick: Multiply factors: $0.9\times 0.85\times 0.8=0.612$ → $60000\times 0.612=36720$.

Answer: (A) ₹36,720

Concept: For 2 years, $\text{CI}-\text{SI}$ = interest on one year's SI.

1. One year's SI = $40÷2$ = ₹20.
2. $\text{CI}-\text{SI}=5=20\times \frac{R}{100}$ → $R=25\%$.
3. $P=\frac{20\times 100}{25}$ = ₹80.

Trick: $P=\frac{{\left(\frac{\text{SI}}{2}\right)}^2}{\text{CI}-\text{SI}}=\frac{{20}^2}{5}=80$.

Answer: (A) 80

Concept: Solve innermost brackets first, then ×/÷, then +/−.

1. $(8-5)=3$ → $\frac{28}{3}\times 3=28$.
2. $\frac{63}{7}+28=9+28=37$.
3. $90÷3=30$ → $30\times 37=1110$.

Trick: Answer = 30 × something → must end in 0. Only 1110.

Answer: (D) 1110

Concept: $\text{Profit}\%=\frac{\text{error}}{\text{true weight given}}\times 100$.

1. Marked 1000 g, actually gives 680 g (32% less).
2. Gain = 320 g on 680 g.
3. Profit% = $\frac{320}{680}\times 100=\frac{8}{17}\times 100=\frac{800}{17}=47\frac{1}{17}\%$.

Trick: $\text{Profit}\%=\frac{x}{100-x}\times 100=\frac{32}{68}\times 100$. Denominator 17 → divide 800 by 17.

Answer: (C) $47\frac{1}{17}\%$

Concept: $\text{HCF}\times \text{LCM}=\text{product of the two numbers}$.

1. Numbers x and 3x → $x\times 3x=1200$.
2. $3x^2=1200$ → $x^2=400$ → $x=20$.
3. Numbers: 20 and 60.

Trick: Option check: product must be 1200 → $20\times 60=1200$ ✓.

Answer: (A) 20, 60