Q&A: practice questions with answers

Concept: Take total work = LCM of days, find each person's efficiency.

1. LCM$(11,33,48)=528$ units ⇒ A $=48$, B $=16$, C $=11$ units/day.
2. In 2 days together: $2\times (48+16+11)=150$ units.
3. Remaining $=528-150=378$; B + C do $27$ units/day ⇒ $\frac{378}{27}=14$ days.

Trick: LCM method avoids fractions completely.

Answer: (C) 14

Concept: Income = Savings + Expenditure.

1. Savings $=40\%$ ⇒ expenditure $=60\%$ of income.
2. $60\%$ of income $=360$
3. Income $=\frac{360\times 100}{60}=600$

Trick: $60\%\to 360$ ⇒ $10\%\to 60$ ⇒ $100\%\to 600$.

Answer: (A) 600

Concept: Let CP $=100$; find SP after mark-up and discount.

1. MP $=100+64=164$
2. SP $=164\times \frac{75}{100}=123$
3. Gain $=123-100=23$ ⇒ $23\%$ gain

Trick: Successive change: $64-25-\frac{64\times 25}{100}=39-16=23\%$.

Answer: (B) $23\%$ gain

Concept: $\text{Median}=L+\frac{\frac{N}{2}-cf}{f}\times h$

1. $N=16+15+31+24+38=124$, $\frac{N}{2}=62$.
2. Cumulative frequencies: 16, 31, 62, 86, 124 ⇒ median class 210-260 (cf reaches 62).
3. $L=210,cf=31,f=31,h=50$
4. Median $=210+\frac{62-31}{31}\times 50=210+50=260$

Trick: When cf equals exactly $\frac{N}{2}$ at a class end, the median is that upper boundary: 260.

Answer: (D) 260

Concept: For $n$ observations (odd), median = value at position $\frac{n+1}{2}$ after arranging in order.

1. Ascending order: 21, 26, 35, 36, 37, 38, 49, 66, 71, 73, 73
2. $n=11$ ⇒ position $\frac{11+1}{2}=6$
3. 6th value $=38$

Trick: Always sort first — the middle of the unsorted list (73) is a trap.

Answer: (B) 38

Concept: The largest side $x$ is the hypotenuse: $x^2=(x-18{)}^2+(x-25{)}^2$.

1. $x^2=2x^2-86x+949$ ⇒ $x^2-86x+949=0$
2. $(x-73)(x-13)=0$ ⇒ $x=73$ ($x=13$ makes sides negative).
3. Legs: $73-18=55$ and $73-25=48$.
4. Area $=\frac{1}{2}\times 55\times 48=1320{\text{cm}}^2$

Trick: Check Pythagorean triple: ${48}^2+{55}^2=2304+3025=5329={73}^2$ ✓.

Answer: (D) 1320

Concept: $\text{Average speed}=\frac{\text{Total distance}}{\text{Total time}}$

1. Time 1 $=\frac{80}{40}=2$ h; Time 2 $=\frac{44}{9}$ h.
2. Total time $=2+\frac{44}{9}=\frac{62}{9}$ h; total distance $=124$ km.
3. Average $=124\times \frac{9}{62}=18\text{km/h}$

Trick: Never average the speeds directly ($\frac{40+9}{2}$ is wrong) — use total distance ÷ total time.

Answer: (D) 18

Concept: Ratio of areas of similar triangles = square of the ratio of corresponding sides.

1. Sides ratio $=9:1$
2. Areas ratio $=9^2:1^2=81:1$

Yaad rakhein: Sides $a:b$ ⇒ areas $a^2:b^2$ ⇒ volumes (similar solids) $a^3:b^3$.

Answer: (C) $81:1$

Concept: Form two equations in pen ($p$) and pencil ($q$) prices and solve.

1. $4p+5q=193$ and $3p+6q=165$ ⇒ $p+2q=55$.
2. $p=55-2q$ ⇒ $4(55-2q)+5q=193$ ⇒ $220-3q=193$ ⇒ $q=9$, $p=37$.
3. $4p+6q=148+54=202$

Trick: Shortcut: $4p+6q=(4p+5q)+q=193+9=202$.

Answer: (A) ₹202

Concept: If numbers are $5k,4k,3k$ (5, 4, 3 co-prime), then HCF $=k$ and LCM $=\text{LCM}(5,4,3)\times k=60k$.

1. Numbers $=5k,4k,3k$
2. $60k=4980$
3. $k=\frac{4980}{60}=83$
4. HCF $=k=83$

Trick: HCF $=\frac{\text{LCM}}{\text{LCM of ratio terms}}=\frac{4980}{60}=83$.

Answer: (B) 83

Concept: Write the condition as an equation and find $\frac{a}{b}$.

1. $\frac{38}{100}a=\frac{5}{9}b$
2. $\frac{a}{b}=\frac{5}{9}\times \frac{100}{38}=\frac{500}{342}$
3. $=\frac{250}{171}$

Trick: Cross-multiply directly: $a:b=(5\times 100):(9\times 38)=500:342=250:171$.

Answer: (A) $250:171$

Concept: Use $(a-b)(a+b)=a^2-b^2$.

1. $691\times 709=(700-9)(700+9)$
2. $={700}^2-9^2=490000-81$
3. $=489919$

Trick: Numbers equidistant from a round number: (middle)${}^2$ − (gap)${}^2$. Unit digit check: $1\times 9=9$ — only (D) ends in 9.

Answer: (D) 489919

Concept: Let CP $=x$. Compare the two selling prices.

1. Actual SP $=0.89x$ (11% loss).
2. New CP $=0.95x$; new SP $=1.20\times 0.95x=1.14x$.
3. Difference: $1.14x-0.89x=0.25x=739$
4. $x=739\times 4=2956$

Trick: Difference of % on the same base: $114\%-89\%=25\%$ of CP = ₹739 ⇒ CP $=4\times 739$.

Answer: (C) ₹2,956

Concept: Area of a parallelogram $=\text{base}\times \text{height}$.

1. Area $=(3x-5y+4)(x-5y)$
2. $=3x^2-15xy-5xy+25y^2+4x-20y$
3. $=3x^2+25y^2-20xy+4x-20y$

Trick: Check signs: $(-5y)(-5y)=+25y^2$ and $xy$ terms $-15xy-5xy=-20xy$ — only (A) has both.

Answer: (A) $(3x^2+25y^2-20xy+4x-20y){\text{m}}^2$

Concept: $h=\sqrt{l^2-r^2}$ and $V=\frac{1}{3}\pi r^{2}h$.

1. $h=\sqrt{{25}^2-7^2}=\sqrt{576}=24\text{cm}$
2. $V=\frac{1}{3}\times \frac{22}{7}\times 7^2\times 24$
3. $=22\times 7\times 8=1232{\text{cm}}^3$

Trick: 7–24–25 is a Pythagorean triplet, so $h=24$ instantly.

Answer: (C) 1232

Concept: Simplify the fractions and convert to a whole-number ratio using the LCM of denominators.

1. $\frac{5}{20}:\frac{8}{16}:\frac{8}{14}=\frac{1}{4}:\frac{1}{2}:\frac{4}{7}$
2. Multiply by 28: $7:14:16$
3. Largest − smallest $=16k-7k=9k=54\Rightarrow k=6$
4. Largest $=16\times 6=96$

Trick: Compare decimals to spot the order: $0.25,0.5,0.57$ — so $\frac{8}{14}$ is largest.

Answer: (D) 96

Concept: Upstream speed $=b-s$, downstream speed $=b+s$.

1. Upstream speed $=\frac{32.2}{76/60}\approx 25.42\text{km/h}$
2. $b:s=9:8\Rightarrow$ upstream $=1$ part, downstream $=17$ parts
3. Downstream $=17\times 25.42\approx 432.2\text{km/h}$
4. Time $=\frac{44.9}{25.42}+\frac{57.8}{432.2}\approx 1.77+0.13\approx 1.9\text{h}$

Trick: With $b:s=9:8$, downstream speed is $\frac{9+8}{9-8}=17$ times the upstream speed.

Answer: (A) 1.9

Concept: In a perfect square every prime factor has an even power.

1. $2527=7\times 361=7\times {19}^2$
2. 19 already has an even power; 7 has power 1.
3. Multiply by 7: $2527\times 7=17689={133}^2$

Trick: Try dividing by small primes: $2527÷7=361={19}^2$ — leftover prime is the answer.

Answer: (C) 7

Concept: Take total work = LCM(10, 14, 21) = 210 units. Efficiency: Robert 21, Chris 15, Jeremy 10 units/day.

1. Let the work take $T$ days. Robert works 5 days, Chris $(T-2)$ days, Jeremy $T$ days.
2. $21\times 5+15(T-2)+10T=210$
3. $105-30+25T=210\Rightarrow 25T=135$
4. $T=5.4$ days

Trick: Write “days worked × efficiency” for each person and equate to total work.

Answer: (D) 5.4

Concept: Rotating a right triangle about one leg gives a cone: that leg = height, other leg = radius, hypotenuse = slant height.

1. $h=60$, $r=45$, $l=75\text{cm}$ (check: ${45}^2+{60}^2={75}^2$).
2. CSA $=\pi rl=\pi \times 45\times 75$
3. $=3375\pi {\text{cm}}^2$

Trick: The side you rotate about is always the height; CSA needs only $r$ and $l$.

Answer: (B) $3375\pi$