Q&A: practice questions with answers

Concept: Expand each bracket and combine like terms.

1. $3(\frac{4}{3}{x}^2-21x+12)=4x^2-63x+36$
2. $-4(x^2+4x-14)=-4x^2-16x+56$
3. Sum: $(4-4)x^2+(-63-16)x+(36+56)=-79x+92$

Trick: $x^2$ cancels, so look only at $x$ and constant: $-63-16=-79$, $36+56=92$.

Answer: (C) $-79x+92$

Concept: In a right triangle the largest side $(x+4)$ is the hypotenuse; use Pythagoras.

1. $x^2+(x+2{)}^2=(x+4{)}^2\Rightarrow x^2-4x-12=0$
2. $(x-6)(x+2)=0\Rightarrow x=6$
3. Sides 6, 8, 10; area $=\frac{1}{2}\times 6\times 8=24$ ✓
4. Perimeter $=6+8+10=24\text{cm}$

Trick: Sides in steps of 2 → think of the triplet 6-8-10.

Answer: (B) 24

Concept: Use the triangle angle sum, alternate angles (BC ∥ AD) and the isosceles triangle ACD.

1. In △ABC: $\angle BCA={180}^{\circ }-{70}^{\circ }-{62}^{\circ }={48}^{\circ }$.
2. BC ∥ AD, AC is a transversal → $\angle CAD=\angle BCA={48}^{\circ }$ (alternate angles).
3. AC = CD → $\angle CDA=\angle CAD={48}^{\circ }$.
4. $\angle ACD={180}^{\circ }-{48}^{\circ }-{48}^{\circ }={84}^{\circ }$

Trick: Answer $={180}^{\circ }-2({180}^{\circ }-{70}^{\circ }-{62}^{\circ })$.

Answer: (D) 84

Concept: Evaluate powers and roots first (BODMAS).

1. $3^2=9$, $\sqrt{4^2}=4$, $5\sqrt{81}=5\times 9=45$
2. $9+4-45-4=-36$

Trick: $\sqrt{a^2}=a$ for positive $a$.

Answer: (D) $-36$

Concept: Mode is the value that occurs most often.

1. Frequencies: 20 → 4 times; 19, 14, 18, 12, 11, 17 → once each.
2. Highest frequency is for 20.

Yaad rakhein: Just count repeats — the most repeated value is the mode.

Answer: (A) 20

Concept: If SP of $a$ articles = CP of $b$ articles and $a>b$, there is a loss: $\text{Loss}\%=\frac{a-b}{a}\times 100$.

1. Let CP of 1 book = ₹1, so SP of 40 books = CP of 30 books = ₹30.
2. CP of 40 books = ₹40, SP = ₹30 → loss = ₹10.
3. $\text{Loss}\%=\frac{10}{40}\times 100=25\%$ (note $\frac{100}{10}\%=10\%$, which is wrong).

Trick: Goods count more than money count → loss; divide difference by the larger count.

Answer: (A) 25% loss

Concept: $SI=\frac{P\times R\times T}{100}$, with $T$ in years; count days excluding the deposit day (2024 is a leap year, 366 days).

1. Days: February $29-24=5$, March 31, April 25 → $5+31+25=61$ days.
2. $T=\frac{61}{366}=\frac{1}{6}$ year.
3. $SI=\frac{4000\times 6.75\times 1}{100\times 6}=\frac{270}{6}=45$.

Trick: 61 days of a 366-day year is exactly $\frac{1}{6}$ year — the numbers are chosen to cancel.

Answer: (D) ₹45

Concept: $(a^m{)}^n=a^{mn}$; equal bases ⇒ equal exponents.

1. $x^z=({95}^{0.53}{)}^z={95}^{0.53z}$.
2. $y^8=({95}^{0.83}{)}^8={95}^{6.64}$.
3. $0.53z=6.64\Rightarrow z=\frac{6.64}{0.53}\approx 12.53$.

Trick: Just compare powers of 95: $z=\frac{8\times 0.83}{0.53}$.

Answer: (A) 12.53

Concept: Literate females = total literate − literate males; then express as a percentage of total females.

1. Females $=342000-228000=114000$.
2. Total literate $=44\%\times 342000=150480$; literate males $=59\%\times 228000=134520$.
3. Literate females $=150480-134520=15960$.
4. $\frac{15960}{114000}\times 100=14\%$.

Trick: Weighted average: males are $\frac{2}{3}$ and females $\frac{1}{3}$ of population, so $44=\frac{2\times 59+f}{3}\Rightarrow f=14$.

Answer: (C) 14%

Concept: Take distance $MN=D$. Speed = $\frac{D}{\text{time}}$; after both trains are moving, time to meet = remaining gap ÷ sum of speeds.

1. A takes 7 h → speed $\frac{D}{7}$; B takes 5 h → speed $\frac{D}{5}$.
2. By 9:55 a.m., A has covered $2\times \frac{D}{7}=\frac{2D}{7}$; gap $=\frac{5D}{7}$.
3. Time $=\frac{5D}{7}÷(\frac{D}{7}+\frac{D}{5})=\frac{5}{7}\times \frac{35}{12}=\frac{25}{12}$ h = 2 h 5 min.
4. 9:55 a.m. + 2 h 5 min = 12:00 p.m.

Trick: Take $D=35$ (LCM of 7 and 5): speeds 5 and 7, gap at 9:55 is 25, time $=\frac{25}{12}$ h.

Answer: (D) 12:00 p.m.

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

1. $\frac{79}{100}a=\frac{4}{9}b$.
2. $\frac{a}{b}=\frac{4}{9}\times \frac{100}{79}=\frac{400}{711}$.
3. So $a:b=400:711$.

Trick: Cross-multiply directly: $a:b=(4\times 100):(9\times 79)$.

Answer: (A) 400 : 711

Concept: $SP=MP\times \frac{100-d}{100}$, so $MP=\frac{SP\times 100}{100-d}$.

1. $MP=\frac{34340\times 100}{85}$.
2. $\frac{34340}{85}=404$, so $MP=40400$.
3. Check: 15% of 40400 = 6060; $40400-6060=34340$ ✓.

Trick: SP is 85% of MP — divide by 0.85.

Answer: (B) ₹40,400

Concept: Test the options: the pair must add to 52 and have LCM 651.

1. $651=3\times 7\times 31$.
2. 31 and 21: sum $=52$, and since they are co-prime, $\text{LCM}=31\times 21=651$ ✓.
3. Other pairs: (24, 28) LCM 168; (29, 23) LCM 667; (36, 16) LCM 144.

Trick: Factorise 651 and look for a pair of its factors adding to 52: $31+21$.

Answer: (D) 31, 21

Concept: Simplify the bracket first, then take LCM.

1. $1+\frac{5}{7}=\frac{12}{7}$, so $\frac{1}{12/7}=\frac{7}{12}$.
2. Expression $=\frac{5}{4}+\frac{7}{12}-\frac{3}{5}$.
3. LCM 60: $\frac{75+35-36}{60}=\frac{74}{60}=\frac{37}{30}$.

Trick: Reciprocal of $1+\frac{5}{7}$ is $\frac{7}{12}$ — do this before anything else.

Answer: (D) $\frac{37}{30}$

Concept: Surface area of a sphere $=4\pi r^2$.

1. $r=\frac{64}{2}=32$ cm.
2. $4\pi r^2=4\pi \times 1024=4096\pi {\text{cm}}^2$.

Trick: With diameter $d$: area $=\pi d^2=\pi \times {64}^2=4096\pi$.

Answer: (A) $4096\pi {\text{cm}}^2$

Concept: $1+{\cot }^2\theta ={\csc }^2\theta$ and $1+{\tan }^2\theta ={\sec }^2\theta$.

1. $\frac{1+{\cot }^2\theta }{1+{\tan }^2\theta }=\frac{{\csc }^2\theta }{{\sec }^2\theta }$.
2. $=\frac{{\cos }^2\theta }{{\sin }^2\theta }={\cot }^2\theta$.
3. ${\cot }^2\theta ={\csc }^2\theta -1$.

Trick: Put $\theta ={45}^{\circ }$: expression $=1$; only ${\csc }^2\theta -1=2-1=1$ fits.

Answer: (A) ${\csc }^2\theta -1$

Concept: Successive change: net $\%=a-b-\frac{ab}{100}$ for $+a\%$ then $-b\%$.

1. $24-13-\frac{24\times 13}{100}=11-3.12=7.88\%$.
2. Check: $100\to 124\to 124\times 0.87=107.88$.

Trick: Take CP = 100; MP = 124; SP = $124-16.12=107.88$.

Answer: (C) 7.88%

Concept: For equal distances at speeds $a$ and $b$: average speed $=\frac{2ab}{a+b}$.

1. $\frac{2\times 9\times 18}{9+18}=\frac{324}{27}=12$.

Trick: Take distance 18 km: time $2+1=3$ h, total 36 km → $\frac{36}{3}=12$.

Answer: (B) 12

Concept: Form two linear equations in mother's age $M$ and daughter's age $D$.

1. $2M=5D+25$ and $3D=M-9\Rightarrow M=3D+9$.
2. $2(3D+9)=5D+25\Rightarrow D=7$.
3. $M=30$; difference $=30-7=23$.

Trick: Check options quickly using $M=3D+9$: difference $=2D+9$ must be odd.

Answer: (D) 23

Concept: A number is divisible by 9 if its digit sum is divisible by 9.

1. Digit sum $=4+8+8+3+0+9+1+3+9=45$, divisible by 9 ✓.
2. Not by 10 (last digit 9).
3. By 11: alternating sum $=(9+1+0+8+4)-(3+9+3+8)=22-23=-1$, not divisible.

Trick: Digit sum 45 → divisible by both 3 and 9.

Answer: (A) 9