Q&A: practice questions with answers

Concept: With a false weight, the shopkeeper charges for 1000 g but gives less; his real gain depends on the goods actually given.

1. Let cost be ₹1 per gram. He "sells 1 kg" at a 60% loss, i.e. collects ₹$0.40\times 1000=400$.
2. Real gain is 90%, so the cost of goods actually given $=\frac{400}{1.9}=210.53$ → he gives about $210.5$ g.
3. Check: cost ₹210.5, sold for ₹400 → gain $\frac{189.5}{210.5}\approx 90\%$ ✓.

Trick: $\text{Weight given}=1000\times \frac{100-\text{claimed loss}\%}{100+\text{actual gain}\%}=1000\times \frac{40}{190}$.

Answer: (A) 210.5 gm

Concept: Evaluate roots and "of" first, then multiplication, then subtraction.

1. $15\%\text{ of }80=12$; $\sqrt[3]{216}=6$; $\sqrt{36}=6$.
2. $\frac{3}{6}\times 6\times 6=18$.
3. $y=12-18=-6$.

Trick: "of" means multiply and ranks with brackets – do it before the minus sign.

Answer: (A) $-6$

Concept: In a cyclic quadrilateral the exterior angle at a vertex equals the interior opposite angle.

1. PQ is produced to T, so $\angle RQT$ is the exterior angle at Q.
2. Interior opposite angle of Q is $\angle PSR$ (at S).
3. $\angle PSR=\angle RQT={58}^{\circ }$.
4. Check: $\angle PQR={180}^{\circ }-{58}^{\circ }={122}^{\circ }$ and ${122}^{\circ }+{58}^{\circ }={180}^{\circ }$ ✓.

Trick: Exterior angle = opposite interior angle (because opposite angles sum to ${180}^{\circ }$). Option A (${122}^{\circ }$) is the trap – that is $\angle PQR$.

Answer: (D) ${58}^{\circ }$

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

1. Ratio $=1050:1510:1280=105:151:128$ (divide by 10).
2. Total $=105+151+128=384$.
3. Pooja's share $=\frac{128}{384}\times 1650=\frac{1}{3}\times 1650=550$.

Trick: $128\times 3=384$, so Pooja gets exactly one-third of the profit – no long division needed.

Answer: (A) ₹550

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

1. Other number $=\frac{1116\times 186}{372}$.
2. $\frac{1116}{372}=3$, so the number $=3\times 186=558$.
3. Check: HCF(372, 558) = 186 and LCM = 1116 ✓.

Trick: Cancel first: $372=2\times 186$ and $1116=3\times 372$ → answer $=3\times 186$.

Answer: (C) 558

Concept: If $a:b::b:c$, then $b$ is the mean proportional: $b^2=ac$.

1. $b^2=ac$ → $36=a\times 4a=4a^2$ → $a^2=9$ → $a=3$ (positive).
2. $c=4a=12$.
3. Check: $3:6::6:12$ ✓.

Trick: Mean proportional: $b=\sqrt{ac}$. Here $\sqrt{a·4a}=2a=6$ → $a=3$ instantly.

Answer: (D) 12

Concept: Trains crossing in opposite directions: relative speed = sum of speeds; distance covered = sum of lengths.

1. Relative speed $=110+120=230\text{km/h}=230\times \frac{5}{18}=\frac{575}{9}\text{m/s}$.
2. Total length $=\frac{575}{9}\times 12=\frac{2300}{3}=766.67\text{m}$.
3. Other train $=766.67-450=316.67\text{m}$.

Trick: Opposite directions → add speeds; same direction → subtract. Convert km/h to m/s with $\times \frac{5}{18}$.

Answer: (A) 316.67

Concept: Take total work = LCM of days; work in 2-day cycles, then finish the remainder day by day.

1. Work $=36$ units. Reema $=3$ units/day, Juhi $=1$ unit/day.
2. One cycle (Reema day + Juhi day) $=4$ units in 2 days. $8$ cycles $=32$ units in $16$ days.
3. Day 17 (Reema): $32+3=35$. Day 18 (Juhi): $35+1=36$ – done.
4. Total $=18$ days.

Trick: Never divide $36÷4=9$ cycles blindly – the last cycle may end mid-way; always check the final days.

Answer: (C) 18

Concept: Curved surface area of a cone $=\pi rl$, where slant height $l=\sqrt{r^2+h^2}$.

1. $l=\sqrt{7^2+{24}^2}=\sqrt{49+576}=\sqrt{625}=25\text{m}$.
2. CSA $=\frac{22}{7}\times 7\times 25=550{\text{m}}^2$.
3. Cost $=550\times 40=$ ₹22,000.

Trick: 7–24–25 is a Pythagorean triplet; with $r=7$ and $\pi =\frac{22}{7}$ the 7 cancels → CSA $=22\times l$.

Answer: (B) ₹22,000

Concept: On a straight line adjacent angles sum to ${180}^{\circ }$; for parallel lines, corresponding angles are equal.

1. At M, $\angle PMB$ and $\angle BMN$ form a linear pair: $5x+3x+2y=180$ → $8x+2y=180$ … (i)
2. $\angle BMN$ and $\angle DNQ$ are corresponding angles (AB ∥ CD): $3x+2y=x+3y$ → $y=2x$ … (ii)
3. From (i): $8x+4x=180$ → $x=15$, $y=30$.
4. $\angle BMN=3(15)+2(30)=45+60={105}^{\circ }$.

Trick: Check: $\angle PMB={75}^{\circ }$ and ${75}^{\circ }+{105}^{\circ }={180}^{\circ }$ ✓; $\angle DNQ=15+90={105}^{\circ }$ ✓.

Answer: (D) ${105}^{\circ }$

Concept: Sum = Average × Number. The 6^th result is counted in both the first six and the last six, so it appears twice when the two sums are added.

1. Total of 11 results $=11\times 61=671$.
2. First 6: $6\times 64=384$; last 6: $6\times 61=366$; together $=750$.
3. 6^th result $=750-671=79$.

Trick: Middle term = (sum of first group + sum of last group) − total sum.

Answer: (D) 79

Concept: For grouped data, $\text{Mode}=l+\frac{f_1-f_0}{2f_1-f_0-f_2}\times h$, where $l$ = lower limit of the modal class (highest frequency), $f_1$ its frequency, $f_0$ and $f_2$ the frequencies just before and after, $h$ = class width.

1. Highest frequency 38 → modal class 70–80, so $l=70$, $f_1=38$, $f_0=18$, $f_2=26$, $h=10$.
2. $\text{Mode}=70+\frac{38-18}{76-18-26}\times 10=70+\frac{20}{32}\times 10$.
3. $=70+6.25=76.25$.

Trick: The denominator is simply $(f_1-f_0)+(f_1-f_2)=20+12=32$.

Answer: (D) 76.25

Concept: Compare cost and selling price of the same unit (one dozen) – the number of dozens does not change the percentage.

1. CP of one dozen = ₹520.
2. SP of one dozen $=12\times 52=$ ₹624.
3. Profit $=624-520=104$; Profit% $=\frac{104}{520}\times 100=20\%$.

Trick: $\frac{624}{520}=1.2$ → 20% profit at a glance; "59 dozens" is just a distraction.

Answer: (D) 20%

Concept: Express $x$ from the simplest equation, substitute, and reduce to two equations in $y$ and $z$ – or simply test the options.

1. From $x-3z=-5$: $x=3z-5$.
2. Put in the 2nd: $2(3z-5)-y+2z=16\Rightarrow 8z-y=26$. Put in the 3rd: $7(3z-5)-3y-5z=19\Rightarrow 16z-3y=54$.
3. From $y=8z-26$: $16z-24z+78=54\Rightarrow z=3$, $y=-2$, $x=4$.
4. Check: $4-9=-5$ ✓, $8+2+6=16$ ✓, $28+6-15=19$ ✓.

Trick: In the exam, plug each option into $x-3z=-5$ first – only (B) satisfies it ($4-9=-5$), so no further work is needed.

Answer: (B) $x=4,y=-2,z=3$

Concept: Successive changes multiply: final $=\text{initial}\times (1\pm r_1)(1\pm r_2)\dots$

1. Let original price $=100$.
2. After three decreases: $100\times 0.6\times 0.4\times 0.4=9.6$.
3. After 70% increase: $9.6\times 1.7=16.32$.
4. Net change $=16.32-100=-83.68$ → decrease of 83.68%.

Trick: Only option (B) and (D) are decreases; $0.6\times 0.4\times 0.4\times 1.7$ ends in …32, so 83.68.

Answer: (D) Decrease by 83.68%

Concept: The first $n$ even natural numbers are $2,4,\dots ,2n$ – an AP, so the average is the mean of the first and last terms.

1. Sum $=2+4+\cdots +2n=n(n+1)$, so average $=\frac{n(n+1)}{n}=n+1$.
2. For $n=62$: average $=62+1=63$.

Trick: First $n$ even numbers → average $n+1$; first $n$ odd numbers → average $n$.

Answer: (B) 63

Concept: Convert each ratio to a decimal (or compare with a benchmark like $\frac{1}{2}$) and pick the largest.

1. $\frac{5}{12}=0.417$, $\frac{14}{31}=0.452$, $\frac{8}{33}=0.242$, $\frac{17}{39}=0.436$.
2. Largest value is $0.452$ → 14 : 31.

Trick: Compare with $\frac{1}{2}$: 14 is short of 15.5 by 1.5 (least gap relative to 31), so 14 : 31 is closest to half and the biggest.

Answer: (B) 14 : 31

Concept: Total provision is fixed: $\text{men}\times \text{days}$ = constant (inverse proportion).

1. Provision $=100\times 26=2600$ man-days.
2. Remaining men $=100-50=50$.
3. Days $=\frac{2600}{50}=52$.

Trick: Half the men → double the days: $26\times 2=52$.

Answer: (C) 52

Concept: SP = MP × (1 − d₁)(1 − d₂); both articles have the same marked price, so find MP from X and use it for Y.

1. SP of X $=800+523=1323$. Also SP of X $=MP\times 0.55\times 0.27=0.1485MP$.
2. SP of Y $=MP\times 0.11$. So $\frac{{\text{SP}}_Y}{{\text{SP}}_X}=\frac{0.11}{0.1485}=\frac{20}{27}$.
3. SP of Y $=1323\times \frac{20}{27}=49\times 20=980$.
4. Profit% on Y $=\frac{980-800}{800}\times 100=22.5\%$.

Trick: Never compute MP (an ugly number); take the ratio of the two selling fractions instead – $1323=27\times 49$ cancels neatly.

Answer: (D) 22.5%

Concept: $\sqrt[3]{2744}=14$ (since ${14}^3=2744$) and $\sqrt[4]{256}=4$ (since $4^4=256$).

1. Expression $=\frac{14}{4}\times \frac{2}{42}\times 72$.
2. $\frac{14}{42}=\frac{1}{3}$, so it becomes $\frac{1}{3}\times \frac{2}{4}\times 72=\frac{1}{3}\times 36=12$.

Trick: Memorise cubes up to 20 (${14}^3=2744$) and powers of 2 ($2^8=256$).

Answer: (C) 12