Q&A: practice questions with answers

Concept: Equation with roots $\alpha ,\beta$: $x^2-(\alpha +\beta )x+\alpha \beta =0$.

1. Sum $=3+(-2)=1$, product $=3\times (-2)=-6$.
2. Equation: $x^2-x-6=0$.

Trick: Check: $(x-3)(x+2)=x^2-x-6$.

Answer: (B) $x^2-x-6=0$

Concept: Downstream speed $=u+v$, upstream speed $=u-v$ ($u$ = still-water speed, $v$ = stream speed).

1. Boat A downstream: $(10+5)\times 3=45\text{km}$ = distance PQ.
2. Boat B upstream (Q→P): $\frac{45}{5}=9\text{km/h}=u-5$ → $u=14\text{km/h}$.
3. Boat B downstream (P→Q): $14+5=19\text{km/h}$ → time $=\frac{45}{19}=2\frac{7}{19}$ h.

Trick: Downstream − upstream $=2v=10$ → B's downstream speed = $9+10=19$ directly.

Answer: (C) $2\frac{7}{19}$

Concept: Area of regular hexagon $=\frac{3\sqrt{3}}{2}{a}^2$ (six equilateral triangles).

1. Square side $=\sqrt{5184}=72\text{cm}$; perimeter $=288\text{cm}$.
2. Hexagon side $a=\frac{288}{6}=48\text{cm}$.
3. Area $=\frac{3\sqrt{3}}{2}\times {48}^2=\frac{3\sqrt{3}}{2}\times 2304=3456\sqrt{3}{\text{cm}}^2$.

Trick: ${72}^2=5184$ is worth memorising; hexagon side = $\frac{2}{3}$ of square side.

Answer: (D) $3456\sqrt{3}$

Concept: A square is both a rectangle and a rhombus, so it inherits the diagonal properties of both.

1. Rectangle property: diagonals are equal.
2. Parallelogram/rhombus property: diagonals bisect each other (and in a square also at ${90}^{\circ }$).

Yaad rakhein: Square diagonals – equal, bisect each other, perpendicular, and bisect the vertex angles.

Answer: (A) The diagonals of a square are equal and bisect each other.

Concept: For even $n$, median = average of the $\frac{n}{2}$th and $(\frac{n}{2}+1)$th values after sorting.

1. Sorted: 15, 16, 20, 23, 24, 25, 25, 27, 30, 32, 40, 45.
2. $n=12$ → 6th and 7th values are 25 and 25 → median $=\frac{25+25}{2}=25$.

Trick: Always sort first – the raw order is a trap.

Answer: (B) 25

Concept: Profit % $=\frac{\text{SP}-\text{CP}}{\text{CP}}\times 100$; compare per article.

1. CP per article $=\frac{300}{12}=25$ rupees; SP per article = ₹34.
2. Profit $=34-25=9$ → $\frac{9}{25}\times 100=36\%$.

Trick: The "67 dozens" is irrelevant – percentage profit is the same per article.

Answer: (D) $36\%$

Concept: Mean $=\frac{\sum f_i{x}_i}{\sum f_i}$, with $x_i$ = class mid-point.

1. Mid-points 25, 35, 45, 55, 65; frequencies 49, 19, 25, 21, X.
2. $\sum f_i{x}_i=1225+665+1125+1155+65X=4170+65X$; $\sum f_i=114+X$.
3. $\frac{4170+65X}{114+X}=45$ → $4170+65X=5130+45X$ → $20X=960$ → $X=48$.

Trick: Use deviations from assumed mean 45: $49(-20)+19(-10)+0+21(10)+20X=0$ → $-980-190+210+20X=0$ → $X=48$.

Answer: (B) 48

Concept: $\tan ({90}^{\circ }-\theta )=\cot \theta$, $\cot ({90}^{\circ }-\theta )=\tan \theta$; identities ${\csc }^2\theta -{\cot }^2\theta =1$, ${\sec }^2\theta -{\tan }^2\theta =1$.

1. ${\tan }^2{27}^{\circ }={\cot }^2{63}^{\circ }$ → ${\csc }^2{63}^{\circ }-{\tan }^2{27}^{\circ }={\csc }^2{63}^{\circ }-{\cot }^2{63}^{\circ }=1$.
2. ${\cot }^2{34}^{\circ }={\tan }^2{56}^{\circ }$ → $3{\sec }^2{56}^{\circ }-3{\cot }^2{34}^{\circ }=3({\sec }^2{56}^{\circ }-{\tan }^2{56}^{\circ })=3$.
3. Total $=1+3=4$.

Trick: Pair angles adding to ${90}^{\circ }$ ($63+27$, $56+34$) and convert to one angle.

Answer: (A) 4

Concept: When a transversal cuts parallel lines, co-interior (same-side interior) angles are supplementary.

1. The angle $x^{\circ }$ marked above P equals its vertically opposite angle below P (on the opposite side of the transversal).
2. That angle and $(48+x{)}^{\circ }$ at Q lie between the parallels on the same side → co-interior: $x+48+x=180$.
3. $2x=132$ → $x=66$.

Trick: One acute + one obtuse angle at parallel lines always add to ${180}^{\circ }$.

Answer: (A) 66

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

1. $720=\frac{9000\times 4\times T}{100}=360T$.
2. $T=\frac{720}{360}=2$ years.

Trick: $4\%$ of 9000 = 360 per year → 720 needs 2 years.

Answer: (D) 2

Concept: Net work per hour = sum of filling rates − emptying rate.

1. Rates: $\frac{1}{4}+\frac{1}{29}-\frac{1}{29}=\frac{1}{4}$ of the cistern per hour.
2. The second filling tap and the emptying tap cancel each other exactly.
3. One-fourth of the cistern at $\frac{1}{4}$ per hour $\Rightarrow$ $1$ hour.

Trick: Whenever a filling and an emptying pipe have the same time, strike both out – only the remaining pipe matters.

Answer: (B) 1

Concept: Successive discounts multiply: $\text{SP}=\text{MP}\times (1-d_1)(1-d_2)$. Same marked price for both articles.

1. SP of X $=3500+889=4389$.
2. $\text{MP}\times 0.55\times 0.40=0.22\text{MP}=4389\Rightarrow \text{MP}=19950$.
3. SP of Y $=19950\times 0.31=6184.5$; profit $=6184.5-3500=2684.5$.
4. Profit% $=\frac{2684.5}{3500}\times 100=76.7\%$.

Trick: $45\%$ and $60\%$ off together leave $0.55\times 0.40=22\%$ of MP – never add discounts.

Answer: (A) 76.7%

Concept: Make the common term (third group) equal in both ratios, then combine.

1. Second : Third $=3:7$ and First : Third $=2:7$ – third is already 7 in both.
2. First : Second : Third $=2:3:7$, total $12$ parts.
3. 1 part $=\frac{300}{12}=25$; second group $=3\times 25=75$.

Trick: When the shared quantity already has the same number in both ratios, just write the three numbers side by side.

Answer: (C) 75

Concept: Running in opposite directions, first meeting time $=\frac{\text{track length}}{\text{sum of speeds}}$.

1. Relative speed $=22+15=37\text{km/h}=37\times \frac{5}{18}=\frac{185}{18}\text{m/s}$.
2. Time $=\frac{1850}{185/18}=1850\times \frac{18}{185}=10\times 18=180$ seconds.

Trick: $1850=185\times 10$ – spot the $185$ and the $\frac{5}{18}$ conversion cancels neatly.

Answer: (C) 180 seconds

Concept: Let cost price be ₹1 per gram (₹1000 per kg). He sells a "kg" at a 35% loss on the claimed cost but actually gives only $w$ grams.

1. SP of the "kg" $=0.65\times 1000=650$.
2. Actual cost $=w$; gain 75% $\Rightarrow 1.75w=650$.
3. $w=\frac{650}{1.75}=371.43\approx 371.4$ g.

Trick: $\text{weight used}=\frac{100-\text{loss}\%}{100+\text{gain}\%}\times 1000=\frac{65}{175}\times 1000$.

Answer: (A) 371.4 gm

Concept: Put $t=p+4$ and reduce the condition to $t+\frac{1}{t}=k$; then $t^3+\frac{1}{t^3}=k^3-3k$.

1. $p=t-4$: $(t-4{)}^2+3(t-4)=3\Rightarrow t^2-5t+1=0$.
2. Divide by $t$: $t+\frac{1}{t}=5$.
3. $t^3+\frac{1}{t^3}=5^3-3\times 5=125-15=110$.

Trick: $x+\frac{1}{x}=a\Rightarrow x^3+\frac{1}{x^3}=a^3-3a$ – memorise; the whole question is just finding $a$.

Answer: (A) 110

Concept: If every observation is multiplied by $k$, the average is also multiplied by $k$.

1. Old average $=56$, each mark doubled $\Rightarrow k=2$.
2. New average $=2\times 56=112$ (the number of students, 16, does not matter).

Trick: Add $x$ to all $\Rightarrow$ average $+x$; multiply all by $k$ $\Rightarrow$ average $\times k$.

Answer: (C) 112

Concept: Take total work = LCM of the days; find the work done in one 2-day cycle.

1. Total work $=\text{LCM}(20,30)=60$ units; P does $3$/day, Q does $2$/day.
2. One cycle (P day + Q day) $=5$ units in 2 days.
3. $12$ cycles $=60$ units exactly $\Rightarrow 12\times 2=24$ days, finishing on Q's day with no remainder.

Trick: Always check whether the total divides exactly by the cycle work – here $60÷5=12$, so no fractional last day.

Answer: (A) 24 days

Concept: Monthly saving = salary − expenses; count only the months in which the saving is actually kept.

1. Saving per month $=18000-5500-1500=11000$.
2. In the birthday month the whole saving is spent $\Rightarrow$ only $11$ months of saving.
3. Yearly saving $=11\times 11000=121000$.

Trick: Option B ($132000=12\times 11000$) is the trap for those who forget the birthday month.

Answer: (D) 121000

Concept: With options given, substitute; or eliminate one variable.

1. Add $2\times$(eq.1) to (eq.2): $8x-5z=1$. Subtract (eq.3) from $5\times$(eq.1): $7x-13z=-25$.
2. From $8x-5z=1$, $x=\frac{1+5z}{8}$; substituting gives $7+35z-104z=-200\Rightarrow z=3$, $x=2$.
3. Eq.1: $4+y-9=-4\Rightarrow y=1$. Check eq.2: $8-2+3=9$ ✓.

Trick: In the exam just test options in the first equation: only $(2,1,3)$ and $(2,-1,-3)$ survive; eq.2 then kills $(2,-1,-3)$ ($8+2-3\ne 9$).

Answer: (B) x = 2, y = 1, z = 3