Q&A: practice questions with answers

Concept: Half-yearly compounding: rate per half-year $=\frac{7.8}{2}=3.9\%$; each deposit earns for the half-years it stays.

1. 1 January deposit stays 2 half-years: CI $=9920\times ({1.039}^2-1)=9920\times 0.079521=788.848$.
2. 1 July deposit stays 1 half-year: CI $=9920\times 0.039=386.88$.
3. Total interest $=788.848+386.88=1175.728\approx ₹1175.73$.

Trick: ${1.039}^2=1.079521$; interest on two deposits $=9920\times (0.079521+0.039)$.

Answer: (A) ₹1,175.73

Concept: Area of sector $=\frac{\theta }{{360}^{\circ }}\times \pi r^2$

1. $\theta ={150}^{\circ }$, $r=12\text{cm}$.
2. Area $=\frac{150}{360}\times \pi \times 144=\frac{5}{12}\times 144\pi$.
3. $=60\pi {\text{cm}}^2$.

Trick: $\frac{150}{360}=\frac{5}{12}$ and $144÷12=12$, so $5\times 12=60$.

Answer: (D) $60\pi {\text{cm}}^2$

Concept: The diameter divides the circle into two semicircles of ${180}^{\circ }$; a central angle equals the degree measure of its arc.

1. Arc PQ + arc QR = semicircle = ${180}^{\circ }$.
2. Arc PQ $=\frac{1}{2}$ arc QR ⇒ $x+2x={180}^{\circ }$ ⇒ $x={60}^{\circ }$.
3. So $\angle POQ$ = arc PQ $={60}^{\circ }$ (and $\angle QOR={120}^{\circ }$).

Trick: Split 180° in ratio 1 : 2 → 60° and 120°.

Answer: (D) ${60}^{\circ }$

Concept: For $a{x}^2+bx+c=0$, real and equal roots ⇔ $D=b^2-4ac=0$.

1. Here $a=2$, $b=-8$, $c=K$.
2. $D=64-8K=0$ ⇒ $K=8$.
3. $K^2-5=64-5=59$.

Trick: Equal roots: $c=\frac{b^2}{4a}=\frac{64}{8}=8$ directly.

Answer: (B) 59

Concept: Mode is the value that occurs the greatest number of times.

1. Count: 41 occurs 3 times.
2. 47, 48, 53, 54 occur twice each; all others once.
3. So mode = 41.

Trick: Sort the data quickly or tally – the highest tally wins.

Answer: (A) 41

Concept: Brackets first, then division/multiplication, then addition.

1. $(5-9)=-4$ and $\frac{123}{41}=3$.
2. $10\times (-4)=-40$.
3. $-40+3=-37$.

Trick: $41\times 3=123$ – check easy divisions first.

Answer: (D) −37

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

1. SP $=3930\times 0.6=₹2358$.
2. $1.2\text{CP}=2358$.
3. CP $=\frac{2358}{1.2}=₹1965$.

Trick: CP $=\text{MP}\times \frac{60}{120}=\frac{3930}{2}=1965$.

Answer: (B) ₹1,965

Concept: Net filling rate = filling rate − emptying rate.

1. Net per minute $=\frac{1}{18}-\frac{1}{19}=\frac{19-18}{342}=\frac{1}{342}$.
2. Full tank: 342 minutes.
3. One-third: $\frac{342}{3}=114$ minutes.

Trick: For $\frac{1}{a}-\frac{1}{b}$ with $b-a=1$, time $=a\times b$.

Answer: (A) 114

Concept: Required number = LCM of the divisors + the amount diminished.

1. $\text{LCM}(13,20,10,23)$: 10 divides 20, so $=13\times 20\times 23=5980$.
2. Number $=5980+9=5989$.

Trick: Subtract 9 from each option and test divisibility by 13 and 23 – only 5989 works.

Answer: (A) 5989

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

1. Times: $\frac{445}{89}=5$ h, $\frac{495}{55}=9$ h, $\frac{250}{50}=5$ h → total $19$ h.
2. Distance $=445+495+250=1190\text{km}$.
3. Average $=\frac{1190}{19}=62.63\text{km/h}$.

Trick: Never average the speeds directly; each leg's time here is a whole number.

Answer: (A) 62.63

Concept: CSA $=2\pi rh$, TSA $=2\pi r(r+h)$.

1. $2\pi rh=\frac{1}{2}\times 2\pi r(r+h)$.
2. $2h=r+h$ ⇒ $r=h$.
3. $r=102\text{cm}$.

Trick: CSA = half TSA ⇔ the two circular ends equal the CSA ⇔ r = h.

Answer: (D) 102

Concept: $\text{Average}=\frac{\text{Sum}}{\text{Number of terms}}$

1. Average $=\frac{300}{6}=50$.

Trick: Divide the total by the count.

Answer: (D) 50

Concept: Swap the signs first, then solve using BODMAS.

1. After swapping: $43\times 13+648÷8-52$.
2. $43\times 13=559$, $648÷8=81$.
3. $559+81-52=588$.

Trick: Rewrite the whole expression with new signs before calculating — never swap mid-way.

Answer: (A) 588

Concept: Apply the rule digit by digit, then pick the required positions.

1. Number: 8 5 7 4 1 6 2. Odd digits +1, even digits −2.
2. New digits: 6 6 8 2 2 4 0 → 6682240.
3. 2nd from left = 6, 2nd from right = 4; sum $=6+4=10$.

Trick: Only the two required positions need changing: 5→6 and 6→4.

Answer: (C) 10

Concept: Check each condition in order; apply the one that fits.

1. AUTEA: A=2, U=%, T=$, E=@, A=2 (direct codes).
2. Conditions 1 and 2 need the last letter to be a consonant — last is A (vowel), so no.
3. Condition 3: 4th letter E is a vowel — no.
4. Condition 4: 1st A and 2nd U are vowels → both coded Q: Q Q $ @ 2.

Trick: First check the last letter — it rules out conditions 1 and 2 at once.

Answer: (A) Q Q $ @ 2

Concept: Code = sum of alphabet positions of the letters $+4$.

1. GSTA: $7+19+20+1=47$, $47+4=51$ ✓.
2. XMBJ: $24+13+2+10=49$, $49+4=53$ ✓.
3. YDOM: $25+4+15+13=57$, $57+4=61$.

Trick: Learn positions by EJOTY: E=5, J=10, O=15, T=20, Y=25.

Answer: (D) 61

Concept: Replace the letters by signs and evaluate each option with BODMAS.

1. W = ÷, X = ×, Y = +, Z = −.
2. Option C: $48÷8-56+24\times 6$.
3. $=6-56+144=94$ ✓.

Trick: Look for an option where a big product (like $24\times 6=144$) is added — it can reach 94.

Answer: (C) 48 W 8 Z 56 Y 24 X 6

Concept: Change each digit by the rule, then find the highest and lowest digits.

1. 8629475: even +1 → 8→9, 6→7, 2→3, 4→5; odd −1 → 9→8, 7→6, 5→4.
2. New number: 9738564.
3. Highest 9, lowest 3; sum $=12$.

Trick: Smallest even digit +1 (2→3) vs smallest odd −1 (5→4): the lower one is the minimum.

Answer: (C) 12

Concept: Find the letter shifts and the number change.

1. PH14 → SD30: P+3 = S, H−4 = D, $14+16=30$.
2. JI18 → ME34: J+3 = M, I−4 = E, $18+16=34$ ✓.
3. MU6 → M+3 = P, U−4 = Q, $6+16=22$ → PQ22.

Trick: Number always +16 — only option with $6+16=22$ is PQ 22.

Answer: (B) PQ 22

Concept: Place the towns on a grid step by step.

1. C is west of D and E is east of C → E lies on line CD.
2. G is north of E and also north-west of D → E must be to the west of D.
3. So E lies between C and D, west of D.

Trick: If X is due north of E and NW of D, then E is due west of D.

Answer: (C) West