Q&A: practice questions with answers

Concept: Take total cost as 100 and add the profit (or loss) of each part.

1. 50% of goods at 16% profit: profit $=50\times 0.16=8$.
2. 40% of the remaining 50% = 20% of goods at 35% profit: $20\times 0.35=7$.
3. Remaining 30% at 25% loss: $30\times 0.25=7.5$ loss.
4. Net $=8+7-7.5=7.5$ on cost 100, so $x=7.5$.

Trick: Weighted average: $0.5(16)+0.2(35)+0.3(-25)=8+7-7.5=7.5\%$.

Answer: (A) 7.5

Concept: $\text{Discount}\%=\frac{MP-SP}{MP}\times 100$.

1. Let CP = 100. MP = 180% above CP = $280$.
2. For 12% gain, SP = $112$.
3. Discount $=280-112=168$; $\frac{168}{280}\times 100=60\%$.

Trick: $\frac{SP}{MP}=\frac{112}{280}=\frac{2}{5}=40\%$, so discount $=60\%$.

Answer: (A) 60%

Concept: Compare cost and selling price of one article, then $\text{Profit}\%=\frac{SP-CP}{CP}\times 100$.

1. CP of one article $=\frac{100}{12}=\frac{25}{3}$ rupees.
2. SP of one article $=75$; profit $=75-\frac{25}{3}=\frac{200}{3}$.
3. $\text{Profit}\%=\frac{200/3}{25/3}\times 100=8\times 100=800\%$.

Trick: $\frac{SP}{CP}=\frac{75\times 12}{100}=9$, so SP is 9 times CP → profit $=800\%$. The '62 dozens' is a distractor.

Answer: (A) 800%

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

1. Sum $=312+126+100+99=637$.
2. Average $=\frac{637}{4}=159.25$.

Trick: Take 100 as base: deviations $+212,+26,0,-1$ sum to $237$; $\frac{237}{4}=59.25$ → $159.25$.

Answer: (B) 159.25

Concept: $\text{Overall average}=\frac{\sum (\text{members}\times \text{average})}{\text{total members}}$.

1. Total spent $=72\times 404+12\times 316+87\times 424=29088+3792+36888=69768$.
2. Total members $=72+12+87=171$.
3. Average $=\frac{69768}{171}=408$.

Trick: Take 400 as base: deviations $72\times 4+12\times (-84)+87\times 24=288-1008+2088=1368$; $\frac{1368}{171}=8$ → 408.

Answer: (C) 408

Concept: Form the equation, find the ratio A : B, then compare with B as base.

1. $160\%\text{ of }75\%\text{ of }A=1.6\times 0.75A=1.2A$.
2. $2\times 150\%\text{ of }B=3B$. So $1.2A=3B\Rightarrow A=2.5B$.
3. A exceeds B by $1.5B$, i.e. $150\%$ more.

Trick: $A:B=3:1.2=5:2$; difference 3 on base 2 $=150\%$.

Answer: (B) 150%, more

Concept: An open box has 5 faces: base + 4 walls. Internal surface area $=lb+2h(l+b)$.

1. $l=8$, $b=17$, $h=22$ (cm).
2. Base $=8\times 17=136$.
3. Walls $=2\times 22\times (8+17)=44\times 25=1100$.
4. Total $=136+1100=1236{\text{cm}}^2$.

Trick: Total surface area $2(lb+bh+hl)$ minus one $lb$ gives the open-box area; here $2(136+374+176)-136=1236$.

Answer: (A) 1236

Concept: The LCM of two numbers carries the highest power of each prime present in either number.

1. $24=2^3\times 3$, $36=2^2\times 3^2$, $72=2^3\times 3^2$.
2. $2^3$ appears in LCM(A,B) and LCM(C,A) but not in LCM(B,C), so $2^3$ must be in A (not B, not C).
3. $3^2$ appears in LCM(B,C) and LCM(C,A) but not in LCM(A,B), so $3^2$ is in C.
4. Smallest: $A=8$, $C=9$; B needs a 3 (for LCM 24) and a 4 (for LCM 36) → $B=12$. Sum $=8+12+9=29$.

Trick: A prime power missing from exactly one LCM must sit in the number not involved in that LCM.

Answer: (C) 29

Concept: $\text{Median}=l+\frac{\frac{N}{2}-cf}{f}\times h$, where $l$ = lower limit of the median class.

1. $N=11+18+28+15+22+31+13=138$, so $\frac{N}{2}=69$.
2. Cumulative frequencies: 11, 29, 57, 72, … → 69 falls in class 50–60 ($cf=57$, $f=15$, $h=10$).
3. $\text{Median}=50+\frac{69-57}{15}\times 10=50+8=58$.

Trick: The median class is the first class whose cumulative frequency reaches $\frac{N}{2}$.

Answer: (C) 58

Concept: Let present ages be $F$ and $S$; write one equation for each time statement.

1. Six years hence: $F+6=3(S+6)\Rightarrow F=3S+12$.
2. Six years ago: $F-6=7(S-6)\Rightarrow F=7S-36$.
3. $3S+12=7S-36\Rightarrow 4S=48\Rightarrow S=12$, $F=48$.

Trick: Check with options: $48+6=54=3\times 18$ ✓ and $48-6=42=7\times 6$ ✓.

Answer: (D) 48 and 12

Concept: Evaluate percentages and roots first, then multiply, then subtract (BODMAS).

1. $30\%\text{ of }80=24$.
2. $\sqrt[3]{729}=9$, $\sqrt{49}=7$, so $\frac{4}{7}\times 9\times 7=36$.
3. $y=24-36=-12$.

Trick: The 7 in $\frac{4}{7}$ cancels with $\sqrt{49}=7$, leaving $4\times 9=36$ instantly.

Answer: (C) $-12$

Concept: Time is directly proportional to the number of pages.

1. Time per page $=\frac{40}{2}=20$ minutes.
2. For 91 pages: $91\times 20=1820$ minutes.

Trick: $91\times 2=182$, add a zero → 1820.

Answer: (C) 1820

Concept: The ladder, wall and ground form a right triangle. With the angle of elevation $\theta$ at the foot, $\cos \theta =\frac{\text{distance from wall}}{\text{ladder}}$.

1. Foot of ladder from wall $=11.5\text{m}$, $\theta ={60}^{\circ }$.
2. $\cos {60}^{\circ }=\frac{1}{2}=\frac{11.5}{L}$
3. $L=2\times 11.5=23\text{m}$.

Trick: At ${60}^{\circ }$ the hypotenuse is always double the side next to the angle ($1:\sqrt{3}:2$ triangle).

Answer: (A) 23 m

Concept: Evaluate roots and percentages first, then multiply, then subtract (BODMAS).

1. $45\%\text{ of }20=9$; $\sqrt[3]{729}=9$; $\sqrt{36}=6$.
2. $\frac{4}{6}\times 9\times 6=4\times 9=36$.
3. $y=9-36=-27$.

Trick: $\frac{4}{6}\times 6$ cancels at once to 4 – avoid decimals.

Answer: (A) $-27$

Concept: Check the differences between consecutive terms.

1. $40-33=7$, $54-40=14$, $75-54=21$, $103-75=28$.
2. Differences are multiples of 7: $7,14,21,28$, so next is $35$.
3. $103+35=138$.

Trick: Differences in a table of a number (here 7) is the commonest CBT 1 pattern.

Answer: (C) 138

Concept: From $a+\frac{1}{a}$ build higher powers: $a^2+\frac{1}{a^2}=(a+\frac{1}{a}{)}^2-2$, $a^3+\frac{1}{a^3}=(a+\frac{1}{a}{)}^3-3(a+\frac{1}{a})$, and $a^5+\frac{1}{a^5}=(a^2+\frac{1}{a^2})(a^3+\frac{1}{a^3})-(a+\frac{1}{a})$.

1. $a^2+\frac{1}{a^2}=16-2=14$.
2. $a^3+\frac{1}{a^3}=64-12=52$.
3. $a^5+\frac{1}{a^5}=14\times 52-4=728-4=724$.

Trick: Multiply the 2nd and 3rd power sums and subtract the 1st – works for any $a+\frac{1}{a}$.

Answer: (A) 724

Concept: In $a:b::c:d$, $d=\frac{b\times c}{a}$.

1. $x=\frac{28\times 4.1}{4}=7\times 4.1=28.7$.
2. $y=\frac{32\times 3}{4}=24$.
3. $x:y=28.7:24=287:240$.

Trick: $4:28$ means ×7 and $4:32$ means ×8 – just multiply $4.1\times 7$ and $3\times 8$.

Answer: (A) 287 : 240

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

1. Ratio $31:34$, total parts $=65$; difference of parts $=3$.
2. One part $=\frac{44850}{65}=690$.
3. Difference $=3\times 690=$ ₹$2,070$.

Trick: Directly compute $\frac{\text{difference of ratio}}{\text{sum of ratio}}\times \text{profit}=\frac{3}{65}\times 44850$.

Answer: (D) ₹2,070

Concept: The angle in a semicircle is ${90}^{\circ }$, and the angles of a triangle add to ${180}^{\circ }$.

1. In $△ACB$: $\angle ACB={90}^{\circ }$, $\angle CAB={48}^{\circ }$ → $\angle CBA={42}^{\circ }$.
2. In $△ADB$: $\angle ADB={90}^{\circ }$, $\angle ABD={55}^{\circ }$ → $\angle BAD={35}^{\circ }$.
3. C and D are on opposite sides of AB, so $\angle CBD={42}^{\circ }+{55}^{\circ }={97}^{\circ }$ and $\angle CAD={48}^{\circ }+{35}^{\circ }={83}^{\circ }$.
4. Difference $={97}^{\circ }-{83}^{\circ }={14}^{\circ }$.

Trick: $\angle CBD+\angle CAD={180}^{\circ }$ (opposite angles of cyclic quadrilateral ACBD); difference $=(42+55)-(48+35)$.

Answer: (A) ${14}^{\circ }$

Concept: Provisions are fixed, so $\text{men}\times \text{days}$ is constant.

1. Remaining men $=100-55=45$.
2. $100\times 27=45\times d$.
3. $d=\frac{2700}{45}=60$ days.

Trick: Fewer men → more days; $\frac{100}{45}=\frac{20}{9}$, so $27\times \frac{20}{9}=60$.

Answer: (A) 60