Q&A: practice questions with answers

Concept: With the same base, multiply by adding powers; and ${\left(\frac{a}{b}\right)}^n={\left(\frac{b}{a}\right)}^{-n}$.

1. Write ${\left(\frac{4}{29}\right)}^2={\left(\frac{29}{4}\right)}^{-2}$.
2. LHS $={\left(\frac{29}{4}\right)}^{3-2+11}={\left(\frac{29}{4}\right)}^{12}={\left(\frac{4}{29}\right)}^{-12}$.
3. So $9m+7=-12\Rightarrow 9m=-19\Rightarrow m=-\frac{19}{9}$.

Trick: Flipping the base flips the sign of the power – convert everything to one base first.

Answer: (A) $-\frac{19}{9}$

Concept: For $a{x}^2+bx+c=0$, discriminant $D=b^2-4ac$. $D>0$: two real roots; $D=0$: equal roots; $D<0$: no real roots.

1. Here $a=1$, $b=-2$, $c=13$.
2. $D=(-2{)}^2-4\times 1\times 13=4-52=-48$.
3. $D<0$, so there are no real roots.

Trick: If $c$ is positive and $b^2<4ac$ (here $4<52$), the roots are imaginary at once.

Answer: (B) $-48$, No real roots

Concept: Weighted average $=\frac{\sum (\text{share}\times \text{average})}{\sum \text{share}}$.

1. Children $=100\%-20\%-50\%=30\%$.
2. Average $=0.2\times 45+0.5\times 60+0.3\times 20$.
3. $=9+30+6=45\text{kg}$.

Trick: Take 10 people: 2 men, 5 women, 3 children → total $90+300+60=450$, average $=45$.

Answer: (B) 45

Concept: $a:b::c:d\Rightarrow ad=bc$; and if $p:y::y:q$ then $y^2=pq$.

1. $(26+x)(34+x)=(40+x)(22+x)$.
2. $884+60x+x^2=880+62x+x^2\Rightarrow 2x=4\Rightarrow x=2$.
3. $4x=8$, $7x-6=8$; so $y^2=8\times 8=64$.
4. $y>0\Rightarrow y=8$.

Trick: Check options: $x=2$ gives $28:42=24:36$, both $=2:3$ ✓.

Answer: (B) 8

Concept: In the right triangle formed by the tower and the ground, $\tan \theta =\frac{\text{height}}{\text{base}}$.

1. $\tan {30}^{\circ }=\frac{h}{48}$.
2. $\frac{1}{\sqrt{3}}=\frac{h}{48}\Rightarrow h=\frac{48}{\sqrt{3}}$.
3. $h=\frac{48\sqrt{3}}{3}=16\sqrt{3}\text{m}$.

Trick: For ${30}^{\circ }$: height $=\frac{\text{base}}{\sqrt{3}}$; dividing by $\sqrt{3}$ is the same as $\frac{\text{base}}{3}\times \sqrt{3}$, i.e. $16\sqrt{3}$.

Answer: (D) $16\sqrt{3}\text{m}$

Concept: Angles on a straight line add to ${180}^{\circ }$, and the angles of a triangle add to ${180}^{\circ }$.

1. $\angle PEB$ and $\angle BEF$ are a linear pair: $\angle BEF={180}^{\circ }-{56}^{\circ }={124}^{\circ }$.
2. G lies inside $\angle BEF$: $\angle GEF={124}^{\circ }-{32}^{\circ }={92}^{\circ }$.
3. In $△EGF$: $\angle EGF={180}^{\circ }-{92}^{\circ }-{47}^{\circ }={41}^{\circ }$.

Trick: Shortcut: $\angle EGF={180}^{\circ }-({180}^{\circ }-{56}^{\circ }-{32}^{\circ })-{47}^{\circ }={56}^{\circ }+{32}^{\circ }-{47}^{\circ }={41}^{\circ }$.

Answer: (D) ${41}^{\circ }$

Concept: $\text{CP}=\frac{\text{SP}}{1+\text{gain}\%}$; number bought for ₹1 $=\frac{1}{\text{CP of one}}$.

1. SP of 1 chocolate $=₹\frac{1}{5}$.
2. CP of 1 chocolate $=\frac{1}{5}÷1.4=\frac{1}{7}$ rupee.
3. So for ₹1 he bought 7 chocolates.

Trick: With gain, he buys fewer per rupee than he sells: $5\times 1.4=7$.

Answer: (D) 7

Concept: A sum doubles when $\text{SI}=P$, so $R\times T=100$.

1. $\text{SI}=\frac{P\times R\times T}{100}=P\Rightarrow R\times 12=100$.
2. $R=\frac{100}{12}=8.33\%$ (approx.).

Trick: To become $n$ times in SI: $R=\frac{(n-1)\times 100}{T}$; here $\frac{100}{12}=8\frac{1}{3}\%$.

Answer: (B) $8.33\%$

Concept: Look at the differences between consecutive terms.

1. Differences: +18, +11, +18, +11, +18, …
2. Next difference = +11: $113+11=124$.

Trick: Alternating add pattern (+18, +11) – check differences first.

Answer: (D) 124

Concept: A letter common to two words gets the digit common to their codes.

1. GIVE = {G,I,V,E} → {4,6,2,8}; VOLT = {V,O,L,T} → {3,5,6,7}.
2. Common letter: V; common digit: 6.
3. So V = 6.

Trick: Common letter ↔ common digit.

Answer: (C) 6

Concept: Find the letter shifts from first cluster to second on one side and apply them on the other side.

1. Try (A): BPJ → FLN: B+4=F, P−4=L, J+4=N → pattern (+4, −4, +4).
2. TXB → T+4=X, X−4=T, B+4=F = XTF ✓.

Trick: Test each option on the left pair first; reuse the same shifts on the right.

Answer: (A) # = BPJ, % = XTF

Concept: Compare letters at the same position in the two clusters.

1. KL − FG: K−5=F, L−5=G; XY − ST: −5, −5; OP − JK: −5, −5.
2. HK − LP: H+4=L, K+5=P – different rule (also H, K are not consecutive).

Trick: Three pairs = "consecutive letters, each moved back 5".

Answer: (A) HK - LP

Concept: Replace letters by signs, then apply BODMAS.

1. $1+45÷2\times 2-4$
2. $45÷2\times 2=45$
3. $1+45-4=42$

Trick: ÷ and × are done left to right before + and −.

Answer: (B) 42

Concept: Facing the centre: right = anticlockwise, left = clockwise.

1. U fixed; Q is immediate right of U; P is 3rd to the left of Q (Q 3rd right of P); V is 3rd to the right of U.
2. Remaining 3 seats for R, S, T. R is 2nd to the left of T; S not next to U → T next to U, then R, then S.
3. Order (clockwise): U, T, P, R, V, S, Q.
4. From S going right: V, R, P, then T → 3 people.

Trick: Fix the person with most links (U) first.

Answer: (C) 3

Concept: Change each digit separately: even + 2, odd − 1.

1. 6→8, 1→0, 5→4, 4→6, 9→8, 3→2, 2→4
2. New number: 8046824
3. 8 appears twice, 4 appears twice; 0, 6, 2 once → two digits.

Trick: Write the new number fully, then tally.

Answer: (C) Two

Concept: Find the shift from # to LPR and apply the same shift to PSV.

1. (A) HNT → LPR: H+4=L, N+2=P, T−2=R → (+4, +2, −2).
2. PSV → P+4=T, S+2=U, V−2=T = TUT ✓.

Trick: Check the left pair for each option; keep the one whose shifts also give a listed %.

Answer: (A) # = HNT, % = TUT

Concept: Find the shift from # to AND and apply it to RWX.

1. (C) XQB → AND: X+3=A (wrap-around), Q−3=N, B+2=D → (+3, −3, +2).
2. RWX → R+3=U, W−3=T, X+2=Z = UTZ ✓.

Trick: Remember wrap-around: X → Y → Z → A.

Answer: (C) # = XQB, % = UTZ

Concept: Check each letter position separately.

1. EL-GN-IK: E→G→I (+2, +2); L→N (+2), N→K (−3).
2. GN-IP-KM: same rule.
3. IP-KR-MO: I→K→M; P→R (+2), R→O (−3) ✓.

Trick: Option (B) gives MQ – the second letter must be R−3 = O, not Q.

Answer: (C) IP-KR-MO

Concept: Vowel → next letter; consonant → previous letter.

1. ACT → B B S (no vowel)
2. BUS → A V R (A is a vowel)
3. CAR → B B Q (no vowel); DAD → C B C (no vowel)
4. So 3 clusters have no vowel.

Trick: A consonant B becomes vowel A – watch for B → A.

Answer: (D) Three

Concept: Track every turn on a sketch (A at origin).

1. 3 km S; right → West 2 km; left → South 5 km; right → West 3 km; right → North 8 km.
2. Vertical: $-3-5+8=0$ → P is level with A.
3. Horizontal: $2+3=5$ km West of A → go 5 km East.

Trick: Net N–S = 0 means the answer is purely East/West.

Answer: (D) 5 km; East