Q&A: practice questions with answers

Concept: Write both sides with base 50 and compare powers: $(a^m{)}^n=a^{mn}$.

1. $x^z=({50}^{0.92}{)}^z={50}^{0.92z}$.
2. $y^3=({50}^{0.26}{)}^3={50}^{0.78}$.
3. $0.92z=0.78$ ⇒ $z=\frac{0.78}{0.92}\approx 0.848\approx 0.85$.

Trick: $z=\frac{3\times 0.26}{0.92}=\frac{78}{92}=\frac{39}{46}$, a little less than $0.85$.

Answer: (C) 0.85

Concept: A number is divisible by 5 only if its last digit is 0 or 5.

1. 7132956 ends in 6, 6175753 in 3, 6567359 in 9 → not divisible.
2. 7048030 ends in 0 → divisible by 5 (and by 10).

Yaad rakhein: For 5 just look at the unit digit: 0 or 5.

Answer: (C) 7048030

Concept: For the same time period, profit is shared in the ratio of investments.

1. Mohsin's share $=\frac{3800}{19000}=\frac{1}{5}$ of profit ⇒ Abdul : Mohsin $=4:1$.
2. Mohsin's investment $=\frac{1}{5}\times 76800=15360$; Abdul's $=61440$.
3. Difference $=61440-15360=46080$.

Trick: Difference $=\frac{4-1}{5}\times 76800=\frac{3}{5}\times 76800=46080$.

Answer: (D) ₹46,080

Concept: Convert each ratio to a decimal (or compare with $\frac{1}{2}$).

1. $\frac{28}{74}\approx 0.38$, $\frac{33}{72}\approx 0.46$ — both below $\frac{1}{2}$.
2. $\frac{45}{85}\approx 0.53$.
3. $\frac{40}{53}\approx 0.75$ — the greatest.

Trick: Only 40 : 53 has the first term close to the second; it is clearly above 0.7.

Answer: (D) 40 : 53

Concept: Sum of first n even numbers $=n(n+1)$, so their average $=n+1$.

1. Sum $=153\times 154$.
2. Average $=\frac{153\times 154}{153}=154$.

Yaad rakhein: Average of first n even numbers $=n+1$; of first n odd numbers $=n$; of first n natural numbers $=\frac{n+1}{2}$.

Answer: (B) 154

Concept: Expand the brackets and combine like terms.

1. $x(7x-5)=7x^2-5x$.
2. $7(x^2-2)=7x^2-14$.
3. Total $=7x^2-5x+7x^2-14+18=14x^2-5x+4$.

Trick: Put $x=0$: expression $=-14+18=4$; only options with $+4$ survive, and the $x^2$ coefficient is clearly positive.

Answer: (C) $14x^2-5x+4$

Concept: Convert to man-days and woman-days to find one person's daily work.

1. Work $=6\times 8=48$ man-days ⇒ 1 man does $\frac{1}{48}$ per day.
2. Work $=3\times 18=54$ woman-days ⇒ 1 woman does $\frac{1}{54}$ per day.
3. 9 women + 8 men per day $=\frac{9}{54}+\frac{8}{48}=\frac{1}{6}+\frac{1}{6}=\frac{1}{3}$.
4. Time $=3$ days.

Trick: 8 men alone need $\frac{48}{8}=6$ days, 9 women alone need $\frac{54}{9}=6$ days; together half of 6 = 3 days.

Answer: (C) 3 days

Concept: $\text{Discount}\%=\frac{\text{MP}-\text{SP}}{\text{MP}}\times 100$.

1. Discount $=3480-2958=522$.
2. Rate $=\frac{522}{3480}\times 100=15\%$.

Trick: 10% of 3480 = 348, 5% = 174; $348+174=522$ ⇒ 15%.

Answer: (A) 15%

Concept: Angle sum of a triangle is ${180}^{\circ }$; an exterior (linear pair) angle is ${180}^{\circ }$ minus the interior one.

1. In right △BDC: $\angle C={180}^{\circ }-{90}^{\circ }-{54}^{\circ }={36}^{\circ }$.
2. In △AEC: $\angle AEC={180}^{\circ }-{72}^{\circ }-{36}^{\circ }={72}^{\circ }$.
3. B, E, C are collinear ⇒ $\angle AEB={180}^{\circ }-{72}^{\circ }={108}^{\circ }$.

Trick: Exterior angle of △AEC at E $=\angle CAE+\angle C={72}^{\circ }+{36}^{\circ }={108}^{\circ }$.

Answer: (D) ${108}^{\circ }$

Concept: BODMAS: division and multiplication before addition and subtraction; watch the signs.

1. $(-60)÷(-6)=10$.
2. $(-4)\times 9=-36$.
3. $-9-10+(-36)=-55$.

Trick: Negative ÷ negative = positive, then the minus before it makes it $-10$.

Answer: (B) $-55$

Concept: Track each letter position of the clusters separately.

1. 1st letters: U, V, W, X, Y (+1).
2. 2nd letters: X, S, N, I, D (−5).
3. 3rd letters: X, U, R, O, L (−3).

Trick: First letter must be X — only XLV and XIO remain; N − 5 = I settles it.

Answer: (B) XIO

Concept: Check each number's left and right neighbour.

1. 8: preceded by #, followed by $ → both symbols ✔.
2. 1: preceded by $, followed by % ✔.
3. 5, 2, 4, 9 have a number on at least one side ✘; 7 is the last element (nothing after it) ✘.

Trick: Numbers inside a run of digits (5 2 4 9) can never qualify — skip them.

Answer: (D) Two

Concept: Letters common to both words share the common digits; the leftover letter takes the leftover digit.

1. LUMA → 5, 7, 9, 6; MALI → 8, 5, 6, 9.
2. Common letters L, M, A ↔ common digits 5, 6, 9.
3. LUMA's extra U = 7; MALI's extra I = 8.

Trick: Find the unmatched letter and unmatched digit — they pair up.

Answer: (A) 8

Concept: Number the letters of the first word 1-2-3-4 and track the order.

1. GULF (1234) → FLGU = 4-3-1-2 → GLUF = 1-3-2-4.
2. Check ALTS: STAL (4312), ATLS (1324) ✔.
3. SICK: 4312 = KCSI ✔; 1324 = SCIK ✔.

Trick: Third word = first word with middle two letters swapped — test this first.

Answer: (A) SICK – KCSI – SCIK

Concept: All A are B + All B are C ⇒ All A are C, so some C are A.

1. Peaches ⊂ Pineapples ⊂ Lemons.
2. All Peaches are Lemons ⇒ Some Lemons are Peaches (II) ✔.
3. No Lemons are Peaches (I) contradicts this ✘.

Yaad rakhein: 'All A are B' always gives 'Some B are A' (conversion).

Answer: (A) Only conclusion (II) follows.

Concept: Change only the even digits, then count repeated digits.

1. 36719542: even digits 6, 4, 2 → 7, 5, 3.
2. New number: 3 7 7 1 9 5 5 3.
3. Repeated: 3, 7, 5 → three digits.

Trick: Even + 1 always gives odd — the new number is all odd digits, so repeats are likely.

Answer: (C) Three

Concept: Find the gaps between letter positions (A = 1 … Z = 26).

1. TWA: T(20) +3 W(23) +4 A(27 ≡ 1).
2. JMQ: 10, 13, 17 → +3, +4. CFJ: 3, 6, 10 → +3, +4.
3. HLO: 8, 12, 15 → +4, +3 ✘ (reversed).

Trick: Write the gap pattern of each — one pattern is reversed.

Answer: (B) HLO

Concept: Find the shift applied to each letter position.

1. UFNE → YCRB: U+4, F−3, N+4, E−3.
2. Check CZVY → GWZV: C+4, Z−3, V+4, Y−3 ✔.
3. KTDS: K+4 = O, T−3 = Q, D+4 = H, S−3 = P → OQHP.

Trick: Pattern +4, −3, +4, −3; first letter O cuts options to OPGR and OQHP.

Answer: (D) OQHP

Concept: Alternate addition and multiplication with 1, 2, 3, 4, 5, 6.

1. $1+1=2$, $2\times 2=4$, $4+3=7$.
2. $7\times 4=28$, $28+5=33$.
3. $33\times 6=198$.

Trick: A sudden jump (7 → 28) signals multiplication — look for alternating +n, ×n.

Answer: (A) 198

Concept: Total $=$ position from left $+$ position from right $-1$.

1. Barun (15th) is exactly between Puja (13th) and Risabh ⇒ Risabh is 2 places further: 17th from left.
2. Risabh is 12th from right.
3. Total $=17+12-1=28$.

Trick: 'Exactly between' ⇒ equal gaps on both sides: $15+(15-13)=17$.

Answer: (C) 28