Q&A: practice questions with answers

Concept: Savings = 100% minus total spending.

1. Spending $=30+25+15+16=86\%$.
2. Savings $=14\%=42000$.
3. Salary $=\frac{42000}{14}\times 100=300000$, i.e. ₹3,00,000.

Trick: $1\%=3000$, so $100\%=3,00,000$.

Answer: (B) 3,00,000

Concept: Write all terms with the same base: ${\left(\frac{10}{11}\right)}^n={\left(\frac{11}{10}\right)}^{-n}$, then add powers.

1. LHS $={\left(\frac{11}{10}\right)}^{7-10+9}={\left(\frac{11}{10}\right)}^6={\left(\frac{10}{11}\right)}^{-6}$.
2. So $3m+17=-6$.
3. $3m=-23$ ⇒ $m=-\frac{23}{3}$.

Trick: Flipping the fraction just changes the sign of the power.

Answer: (B) $-\frac{23}{3}$

Concept: $\sin 3\theta =3\sin \theta -4{\sin }^3\theta$. From $\cot \theta =\frac{3}{4}$ build a 3-4-5 right triangle.

1. Base $=3$, perpendicular $=4$, hypotenuse $=5$ ⇒ $\sin \theta =\frac{4}{5}$.
2. $\sin 3\theta =3\times \frac{4}{5}-4\times \frac{64}{125}$.
3. $=\frac{300}{125}-\frac{256}{125}=\frac{44}{125}$.

Trick: 3-4-5 triple: $\cot =\frac{\text{base}}{\text{perp}}$, so perp $=4$.

Answer: (A) $\frac{44}{125}$

Concept: Product of two numbers $=\text{LCM}\times \text{HCF}$. With ratio $17:7$ the numbers are $17k$ and $7k$, HCF $=k$.

1. $17k\times 7k=119$ ⇒ $k^2=1$ ⇒ $k=1$.
2. HCF $=1$, LCM $=119$.
3. $\frac{1}{119}+\frac{1}{1}=\frac{120}{119}$.

Trick: Product equals $17\times 7$ itself, so the numbers are 17 and 7 (co-prime).

Answer: (A) $\frac{120}{119}$

Concept: Average $=\frac{\text{sum}}{\text{count}}$.

1. Sum $=308+127+90+45+125+103=798$.
2. Average $=\frac{798}{6}=133$.

Trick: Check unit digits: $8+7+0+5+5+3=28$ → sum ends in 8; $133\times 6=798$ ✓.

Answer: (D) 133

Concept: $\sigma =\sqrt{\frac{\sum (x-\overline{x}{)}^2}{n}}$.

1. Mean $\overline{x}=\frac{225}{6}=37.5$.
2. Deviations $\pm 2.5,\pm 7.5,\pm 12.5$; squares $6.25,56.25,156.25$ (each twice).
3. $\sum (x-\overline{x}{)}^2=2\times 218.75=437.5$.
4. $\sigma =\sqrt{\frac{437.5}{6}}=\sqrt{72.92}\approx 8.54$.

Trick: Equally spaced data with gap $d$: $\sigma =d\sqrt{\frac{n^2-1}{12}}=5\sqrt{\frac{35}{12}}\approx 8.54$.

Answer: (C) 8.54

Concept: $a^3+b^3+c^3-3abc=(a+b+c)(a^2+b^2+c^2-ab-bc-ca)$.

1. Take $a=2x,b=3y,c=-4z$.
2. Numerator: $8x^3+27y^3-64z^3-3(2x)(3y)(-4z)=8x^3+27y^3-64z^3+72xyz$ ✓.
3. Denominator: $4x^2+9y^2+16z^2-6xy+12yz+8zx$ ✓.
4. Quotient $=a+b+c=2x+3y-4z$.

Trick: The sign of $z$ in $-64z^3$ fixes $c=-4z$; answer must contain $-4z$ with $+3y$.

Answer: (B) $(2x+3y-4z)$

Concept: Read 'sells at 75% of the profit' as a selling price fixed at 75% profit, then reduced by the 14% discount; the second pen is sold at 48% profit.

1. Pen 1: $444\times 1.75=777$; after discount $777\times 0.86=668.22$; profit $=224.22$.
2. Pen 2: profit $=356\times 0.48=170.88$.
3. Total profit $=224.22+170.88=395.10$.

Trick: Pen 1 multiplier $1.75\times 0.86=1.505$ → profit $=444\times 0.505=224.22$.

Answer: (D) ₹395.10

Concept: Replace each sign by its meaning, then apply BODMAS.

1. Given: $9÷12\times 450+8208-18$
2. After interchange: $9\times 12+450-8208÷18$
3. $8208÷18=456$, $9\times 12=108$
4. $108+450-456=102$

Trick: Write the new expression first, then do ÷ and × before + and −.

Answer: (C) $102$

Concept: Find the shift applied to each letter.

1. J→M, A→D, Y→B: each letter $+3$ (Y $+3$ wraps to B)
2. P→S, A→D, N→Q: each $+3$
3. UPC: U→X, P→S, C→F = XSF ✓; HEN would give KHQ, RIC gives ULF, KUT gives NXW

Trick: Constant $+3$ on all three letters – check only the last letter of each option first.

Answer: (D) UPC : XSF

Concept: Trace the path on axes (East $+x$, North $+y$), starting at A $(0,0)$.

1. 13 km west → $(-13,0)$; right (north) 2 km → $(-13,2)$
2. Right (east) 7 km → $(-6,2)$; left (north) 5 km → $(-6,7)$
3. Right (east) 6 km → P $(0,7)$
4. P is 7 km north of A, so he must drive 7 km south.

Trick: East–west: $-13+7+6=0$; North: $2+5=7$ → straight 7 km back south.

Answer: (C) 7 km; South

Concept: Facing the centre, a person's left is clockwise and right is anticlockwise.

1. Fix G. I is immediate right of G; H is 3rd to the left of G.
2. I is 2nd to the left of D → D is 2nd to the right of I, i.e. 3rd to the right of G.
3. D is 3rd to the left of A (two people between) → A sits immediate left of G.
4. Seats left for B and C: the one next to H must be C (B is not H's neighbour); B sits between D and I.
5. From C, going right (anticlockwise): A, G, I → I is 3rd to the right of C.

Trick: Draw G at the bottom first; every clue then fixes one seat.

Answer: (C) I

Concept: Check the gap inside each cluster and between the two clusters.

1. NR – MQ: N $+4$ = R, M $+4$ = Q; first letters N → M ($-1$)
2. EI – DH and RV – QU follow the same: $+4$ inside, $-1$ between
3. XZ – TL: X $+2$ = Z and T → L is $-8$; breaks the pattern

Trick: Pattern = (letter, letter $+4$) – (letter $-1$, letter $+3$); XZ fails at the first check.

Answer: (D) XZ - TL

Concept: Look for number–number–number triplets.

1. Series: 4 @ 7 4 £ 9 9 # @ 1 4 $ % 4 3 8 2 © © 1 5 #
2. Triplets with a number on both sides: 4 3 8 and 3 8 2.
3. Others fail: e.g. 7 4 £ (followed by £), 9 9 # (followed by #), 1 4 $, 1 5 #.

Trick: Scan only runs of 3 or more digits in a row: here only '4 3 8 2'.

Answer: (C) $2$

Concept: Find the rule using the complete pair from an option, then apply it to MRV.

1. Try # = WFN: W $+4$ = A, F $+3$ = I, N $+2$ = P → AIP ✓
2. Apply $+4,+3,+2$ to MRV: M → Q, R → U, V → X = QUX
3. So # = WFN and % = QUX.

Trick: % must be MRV shifted forward: only QUX starts with Q (M $+4$).

Answer: (B) # = WFN, % = QUX

Concept: Position from left $=\text{Total}-\text{position from right}+1$.

1. Sima from left $=81-12+1=70$
2. Manju is 9 places to the left: $70-9=61$
3. Manju is ${61}^{\text{st}}$ from the left end.

Trick: Facing north, 'left' means towards the left end, so subtract.

Answer: (C) ${61}^{\text{st}}$

Concept: Each position of the three-letter term moves forward by one letter.

1. 1st letters: K, L, M, N → O
2. 2nd letters: I, J, K, L → M
3. 3rd letters: O, P, Q, R → S

Trick: Every letter $+1$: NLR → OMS.

Answer: (B) OMS

Concept: Trace the path on axes (East $+x$, North $+y$), starting at A $(0,0)$.

1. 12 km south → $(0,-12)$; right (west) 7 km → $(-7,-12)$
2. Right (north) 14 km → $(-7,2)$; right (east) 10 km → $(3,2)$
3. Right (south) 2 km → P $(3,0)$
4. P is 3 km east of A, so he must drive 3 km towards the west.

Trick: Net east $=-7+10=3$; net north $=-12+14-2=0$.

Answer: (D) 3 km to the west

Concept: Find the rule using the complete pair from an option, then apply it to RVN.

1. Try # = DHL: D $+1$ = E, H $+3$ = K, L $+2$ = N → EKN ✓
2. Apply $+1,+3,+2$ to RVN: R → S, V → Y, N → P = SYP
3. So # = DHL and % = SYP.

Trick: % must start with R $+1$ = S and end with N $+2$ = P → SYP.

Answer: (A) # = DHL, % = SYP

Concept: Facing the centre, a person's left is clockwise and right is anticlockwise.

1. Fix T. R is immediate left of T; V is 3rd to the left of T.
2. S is a neighbour of V. The seat between R and V cannot be S (Q also needs to be next to S), so S is on V's other side and Q next to S.
3. Remaining seats: between R and V, and between Q and T. U is not next to V → U sits between Q and T; P between R and V.
4. From S going right (anticlockwise): V, P, R, T, then U → 4 people between.

Trick: Place T first, then the 'third to the left' person; the neighbour clues finish the circle.

Answer: (C) $4$