Q&A: practice questions with answers

Concept: A 50% loss means SP $=\frac{1}{2}$ of CP.

1. SP $=25\%$ of $x=\frac{x}{4}$.
2. $\frac{x}{4}=\frac{1}{2}\text{CP}\Rightarrow \text{CP}=\frac{x}{2}$.
3. Selling at $x$: profit $=x-\frac{x}{2}=\frac{x}{2}$.
4. Profit% $=\frac{x/2}{x/2}\times 100=100\%$.

Trick: SP $=x$ is double CP $=\frac{x}{2}$ ⇒ 100% profit.

Answer: (A) 100%

Concept: Cube has 12 equal edges; cuboid edge sum $=4(l+b+h)$.

1. Cube volume $512=8^3$ ⇒ edge 8 cm; edge sum $=12\times 8=96$.
2. Cuboid: $4(4k+3k+k)=32k=96\Rightarrow k=3$.
3. Dimensions: 12, 9, 3 cm.
4. Volume $=12\times 9\times 3=324{\text{cm}}^3$.

Trick: Volume $=12k^3=12\times 27=324$.

Answer: (C) 324

Concept: Find each value, then subtract.

1. $60\%\text{ of }40=\frac{60}{100}\times 40=24$.
2. $\frac{1}{5}\text{ of }15=3$.
3. Difference $=24-3=21$.

Trick: $60\%=\frac{3}{5}$, so $\frac{3}{5}\times 40=24$ instantly.

Answer: (B) 21

Concept: Empirical relation: $\text{Mode}=3\text{Median}-2\text{Mean}$.

1. $\text{Mode}=3\times 78-2\times 43$.
2. $=234-86=148$.

Yaad rakhein: Remember "3 Median minus 2 Mean".

Answer: (C) 148

Concept: Downstream speed $=B+S$, upstream speed $=B-S$, $B=\frac{D+U}{2}$.

1. Downstream speed $=\frac{66}{6}=11\text{km/h}$.
2. Upstream speed $=\frac{42}{6}=7\text{km/h}$.
3. Speed in still water $=\frac{11+7}{2}=9\text{km/h}$.
4. Stream speed $=\frac{11-7}{2}=2\text{km/h}$.

Trick: Still water = average of the two speeds.

Answer: (B) 9

Concept: Solve innermost brackets first, then ÷ and × from left to right.

1. $26÷\frac{2}{2}=26÷1=26$.
2. $6÷2\times \frac{12-2}{2}=3\times 5=15$.
3. $5\times 6÷3=10$.
4. $26\times 15÷10=390÷10=39$.

Yaad rakhein: Answer must be a multiple of 13 (from 26): 39, 52, 65, 78 – compute to pick 39.

Answer: (A) 39

Concept: Fourth proportional to $a,b,c$ is $\frac{bc}{a}$; third proportional to $a,b$ is $\frac{b^2}{a}$.

1. Fourth proportional to 3, 9, 4 $=\frac{9\times 4}{3}=12$.
2. Third proportional to $A$ and 12 $=\frac{144}{A}$.
3. $\frac{144}{A}=12\Rightarrow A=12$.

Trick: If $\frac{b^2}{a}=b$, then $a=b$.

Answer: (A) 12

Concept: Net filling time $=\frac{xy}{y-x}$ when the filler takes $x$ h and the emptier $y$ h ($y>x$).

1. Net work per hour $=\frac{1}{12}-\frac{1}{14}=\frac{7-6}{84}=\frac{1}{84}$.
2. Time $=84$ hours.

Trick: LCM method: tank = 84 units; filler 7, emptier 6 ⇒ net 1 unit/h ⇒ 84 h.

Answer: (D) 84

Concept: Going back $n$ years: $P_{\text{past}}=\frac{P}{(1+r/100{)}^n}$. With $25\%=\frac{1}{4}$, each year back multiply by $\frac{4}{5}$.

1. 2 years ago: $140000\times \frac{4}{5}\times \frac{4}{5}=89600$
2. 3 years ago: $89600\times \frac{4}{5}=71680$
3. Difference $=89600-71680=17920$

Trick: Difference = population 2 years ago $\times \frac{1}{5}$ $=\frac{89600}{5}=17920$.

Answer: (D) 17,920

Concept: In the division method the last divisor is the HCF. Rebuild the numbers backwards: dividend $=$ divisor $\times$ quotient $+$ remainder.

1. HCF $=29$; last step: $29\times 6=174$ (remainder 0)
2. Previous step: $174\times 4+29=725$
3. First step: $725\times 1+174=899$ → numbers are 899 and 725
4. LCM $=\frac{899\times 725}{29}=31\times 725=22475$

Trick: Numbers $=29\times 31$ and $29\times 25$ → LCM $=29\times 31\times 25=22475$.

Answer: (A) 22475

Concept: $(a+b{)}^2-(a-b{)}^2=4ab$.

1. $(a+b{)}^2={46}^2=2116$
2. $4ab=2116-612=1504$
3. $ab=\frac{1504}{4}=376$

Trick: Direct: $ab=\frac{(a+b{)}^2-(a-b{)}^2}{4}$.

Answer: (B) 376

Concept: Monthly savings $=$ salary $-$ expenses; in the birthday month savings are 0.

1. Expenses $=4000+1500=₹5500$
2. Monthly savings $=13000-5500=₹7500$
3. Saving months $=11$ → $11\times 7500=₹82500$

Trick: Yearly $=12\times 7500-7500=82500$.

Answer: (C) 82,500

Concept: Median of $n$ (odd) observations $=$ value at position $\frac{n+1}{2}$ after sorting.

1. Sorted: 16, 16, 27, 32, 37, 43, 64, 65, 67, 77, 85
2. $n=11$ → position $\frac{11+1}{2}=6$
3. 6th value $=43$

Trick: Always sort first; repeated values (16, 16) are both counted.

Answer: (C) 43

Concept: $\text{MP}\times \frac{100-d}{100}=\text{CP}\times \frac{100+g}{100}$.

1. SP $=56000\times 1.71=₹95760$
2. MP $\times 0.19=95760$
3. MP $=\frac{95760}{0.19}=₹504000$

Trick: $\text{MP}=56000\times \frac{171}{19}=56000\times 9=504000$.

Answer: (D) ₹5,04,000

Concept: Converse tests for parallel lines: equal corresponding angles, equal alternate interior angles, or supplementary co-interior angles.

1. A and B are true (equal corresponding / alternate angles ⇒ parallel).
2. D is true: co-interior angles summing to ${180}^{\circ }$ ⇒ parallel.
3. C is false: co-interior angles must be supplementary (${180}^{\circ }$), not complementary (${90}^{\circ }$).

Trick: Co-interior = "C-shape" angles, sum ${180}^{\circ }$.

Answer: (D) Only C

Concept: A product stays the same if the total shift of decimal points is zero.

1. Option B: $24.184$ is $÷100$; $7958$ is $\times 10$; $3.735$ is $\times 10$
2. Net factor $=\frac{10\times 10}{100}=1$
3. Others: A $=÷10$, C $=\times 100$, D $=÷100$

Trick: Count decimal places: original total = 1+1+4 = 6; option B = 3+0+3 = 6 ✓ (same digits).

Answer: (B) $24.184\times 7958\times 3.735$

Concept: Combined average $=\frac{\text{total of all ages}}{\text{total people}}$.

1. Group 1 total $=4\times 52=208$
2. Group 2 total $=6\times 58=348$
3. Average $=\frac{208+348}{10}=\frac{556}{10}=55.6$ years

Trick: Weighted: $52+\frac{6}{10}\times 6=55.6$.

Answer: (B) 55.6 years

Concept: If SP of $x$ items = CP of $y$ items, gain $\%=\frac{y-x}{x}\times 100$.

1. Let CP of 1 book $=₹1$; SP of 30 books $=₹36$
2. Gain on 30 books $=36-30=6$
3. Gain $\%=\frac{6}{30}\times 100=20\%$

Trick: $y>x$ → gain; divide by the items sold (30), not 36.

Answer: (D) $20\%$ gain

Concept: Make Q common in both ratios, then divide the profit.

1. $Q:R=3:12=1:4$; $P:Q=9:7$ → $P:Q:R=9:7:28$
2. Sum $=44$
3. R $=\frac{28}{44}\times 25300=28\times 575=₹16100$

Trick: $25300÷44=575$ – one part.

Answer: (B) 16,100

Concept: Expand the new areas and cancel $lb$ to get two linear equations.

1. $(l+6)(b-5)=lb+7$ → $-5l+6b=37$
2. $(l-6)(b+7)=lb+31$ → $7l-6b=73$
3. Add: $2l=110$ → $l=55$; then $6b=37+275=312$ → $b=52$
4. Perimeter $=2(55+52)=214$ m

Trick: Adding the two equations removes $b$ at once ($+6b$ and $-6b$).

Answer: (D) 214