Q&A: practice questions with answers

Concept: Inverse proportion ⇒ $xy=$ constant.

1. $k=xy=28\times 30=840$.
2. When $y=20$: $x=\frac{840}{20}=42$.

Trick: $y$ falls from 30 to 20 (ratio $3:2$), so $x$ rises in ratio $2:3$: $28\times \frac{3}{2}=42$.

Answer: (B) 42

Concept: $a^3-b^3=(a-b)(a^2+ab+b^2)$ with $a=2x$, $b=3y$.

1. $8x^3-27y^3=(2x-3y)(4x^2+6xy+9y^2)$.
2. Match with $(Ax+By)(C{x}^2-D{y}^2+Exy)$: $A=2$, $B=-3$, $C=4$, $-D=9\Rightarrow D=-9$, $E=6$.
3. $A+B+C+D+E=2-3+4-9+6=0$.

Trick: Watch the sign in front of $D$ – the identity has $+9y^2$, so $D=-9$, not $9$.

Answer: (C) 0

Concept: Same marked price $M$ for both; use X to find $M$, then apply the single discount to Y.

1. X: SP $=M\times 0.85\times 0.70=0.595M=1000+337=1337\Rightarrow M=\frac{1337}{0.595}\approx 2247.06$.
2. Y: SP $=0.51M=0.51\times 2247.06\approx 1146$.
3. Profit on Y $=1146-1000=146\Rightarrow 14.6\%$.

Trick: ${\text{SP}}_Y=1337\times \frac{0.51}{0.595}=1337\times \frac{6}{7}=1146$ – skip finding $M$.

Answer: (A) 14.6%

Concept: $a\times b=\text{HCF}\times \text{LCM}$ and $\frac{1}{a}-\frac{1}{b}=\frac{b-a}{ab}$.

1. $ab=3\times 1350=4050$. Let $a=3p$, $b=3q$ (co-prime): $pq=450$, $p+q=43\Rightarrow p=18,q=25$.
2. So $a=54$, $b=75$ ($54+75=129$ ✓), $b-a=21$.
3. $\frac{1}{a}-\frac{1}{b}=\frac{21}{4050}=\frac{7}{1350}$.

Trick: $(b-a{)}^2=(a+b{)}^2-4ab=16641-16200=441\Rightarrow b-a=21$ – no need to find $a,b$ separately.

Answer: (A) $\frac{7}{1350}$

Concept: Area of a regular hexagon $=\frac{3\sqrt{3}}{2}{a}^2$; equal perimeters link the square and hexagon.

1. Square side $=\sqrt{3969}=63$ cm, perimeter $=4\times 63=252$ cm.
2. Hexagon side $a=\frac{252}{6}=42$ cm.
3. Area $=\frac{3\sqrt{3}}{2}\times {42}^2=\frac{3\sqrt{3}}{2}\times 1764=2646\sqrt{3}{\text{cm}}^2$.

Trick: ${63}^2=3969$ (since $63=9\times 7$); hexagon side $=\frac{2}{3}$ of square side when perimeters are equal.

Answer: (D) $2646\sqrt{3}$

Concept: Words are directly proportional to the number of lines.

1. 4 lines → 36 words ⇒ 1 line → $9$ words.
2. 93 lines → $93\times 9=837$ words.

Trick: $93\times 9=93\times 10-93=930-93=837$.

Answer: (C) 837

Concept: For equal distances, average speed $=\frac{2uv}{u+v}$.

1. $\frac{2\times 144\times v}{144+v}=126$.
2. $288v=126\times 144+126v\Rightarrow 162v=18144$.
3. $v=112\text{km/h}$.

Trick: $\frac{1}{v}=\frac{2}{126}-\frac{1}{144}=\frac{1}{63}-\frac{1}{144}=\frac{144-63}{9072}=\frac{81}{9072}=\frac{1}{112}$.

Answer: (D) 112 km/h

Concept: Decode the expression pair by pair and build the family tree.

1. W – O: W is the brother of O. O x R: O is the wife of R. R + K: R is the son of K. K ÷ G: K is the father of G.
2. So R and G are children of K, i.e. R is G's brother.
3. O is R's wife $=$ G's brother's wife; W is O's brother.
4. Hence W is G's brother's wife's brother.

Trick: Always start from the person asked about at the end (G) and walk back along the chain.

Answer: (A) Brother's wife's brother

Concept: Find one rule that links both given pairs and apply it to the new number.

1. $10\to 120$: $10\times 12=10\times (10+2)=120$.
2. $12\to 168$: $12\times 14=12\times (12+2)=168$. Rule: $n\times (n+2)$.
3. $18\to 18\times 20=360$.

Trick: Equivalent form: $n^2+2n$. Check: $100+20=120$ ✓, $144+24=168$ ✓, $324+36=360$.

Answer: (C) 360

Concept: Fix the most restricted person first (X), test each case and discard contradictions.

1. X is on an even floor other than 4 $\Rightarrow$ X $=2$ or $6$. Two people between X and T $\Rightarrow$ (X, T) $=(2,5)$ or $(6,3)$.
2. S is odd but not 1 $\Rightarrow$ S $=3$ or $5$; two between S and Y $\Rightarrow$ (S, Y) $=(3,6)$ or $(5,2)$; Z is just below S.
3. If X $=2$, T $=5$: S $=3$ makes Z $=2$ (clash with X) and S $=5$ clashes with T. Rejected.
4. So X $=6$, T $=3$, S $=5$, Y $=2$, Z $=4$ and U gets floor 1. Between U (1) and Z (4): floors 2 and 3 – two people.

Trick: "Immediately below" clues are the fastest contradiction-finders – test them first in each case.

Answer: (A) Two

Concept: In number-coding of words, a letter common to two words gets the digit common to both codes.

1. WINS $\to$ 8 2 4 6 and NORM $\to$ 3 8 5 1.
2. Only letter common to both words: N. Only digit common to both codes: 8.
3. So N $=8$.

Trick: Compare letters and digits side by side – the single overlap gives the answer in one look.

Answer: (C) 8

Concept: Apply the rule digit by digit, then read the required positions in the new number.

1. 8 3 7 1 4 5 6: odd digits $+1$, even digits $-2$.
2. $8\to 6,3\to 4,7\to 8,1\to 2,4\to 2,5\to 6,6\to 4$ $\Rightarrow$ new number 6 4 8 2 2 6 4.
3. Second from left $=4$, second from right $=6$; sum $=10$.

Trick: Write the new digits directly below the old ones – positions never get mixed up.

Answer: (D) 10

Concept: Treat the letters and the numbers as two separate series.

1. Letters: Y$\to$S$\to$M$\to$G (each $-6$), so G$-6=$ A. Q$\to$K$\to$E$\to$Y (also $-6$), so Y$-6=$ S. M$\to$G$\to$A$\to$U, so U$-6=$ O. Cluster: ASO.
2. Numbers: 2, 6, 12, 20 – differences 4, 6, 8, so next difference 10 $\Rightarrow 30$.

Trick: 2, 6, 12, 20, 30 is $n(n+1)$: $1\times 2,2\times 3,3\times 4,4\times 5,5\times 6$.

Answer: (B) ASO 30

Concept: Check the gaps between consecutive letters in each cluster.

1. EHJ: E$\stackrel{+3}{\to }$H$\stackrel{+2}{\to }$J. NQS: $+3,+2$. BEG: $+3,+2$.
2. DGH: D$\stackrel{+3}{\to }$G$\stackrel{+1}{\to }$H – the second gap is $+1$, not $+2$.

Trick: Convert to positions (E5 H8 J10, D4 G7 H8) – the differing gap jumps out.

Answer: (D) DGH

Concept: The same letter-shift must turn # into LHN and PLR into %. Test the pattern suggested by the options.

1. Options give # $=$ MIO (mostly). MIO $\to$ LHN: M$-1=$L, I$-1=$H, O$-1=$N – each letter $-1$.
2. Apply the same to PLR: P$-1=$O, L$-1=$K, R$-1=$Q $\Rightarrow$ % $=$ OKQ.
3. Only option B has both # $=$ MIO and % $=$ OKQ.

Trick: When the left term is unknown, work backwards from the second pair and eliminate options with a single mismatched letter.

Answer: (B) # = MIO , % = OKQ

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

1. MINT $\to$ 6 1 3 4; TIME $\to$ 4 3 6 5.
2. Common letters M, I, T $\leftrightarrow$ common digits 6, 3, 4.
3. The extra letter in TIME is E and the extra digit is 5, so E $=5$.

Trick: Cross out the common letters and common digits – whatever is left is the pair you need.

Answer: (B) 5

Concept: Facing the centre, "left" means clockwise and "right" means anticlockwise. Place P first and count carefully.

1. From P's right (anticlockwise) two people, then U. Two between U and V and three between P and T fix V next to P (clockwise) and T between V-side and U, as in the figure.
2. Remaining seats: one between V and T, and two between U and P. S is to the immediate left (clockwise) of R, so R and S take the two adjacent seats; Q takes the single seat between V and T.
3. From the right (anticlockwise) of Q: V, then P. Only one person (V) sits between P and Q.

Trick: Draw 7 dots, mark "right $=$ anticlockwise" once at the top, and never re-think direction mid-question.

Answer: (A) 1

Concept: Plot the path on axes (East $=+x$, North $=+y$); a left turn rotates the heading anticlockwise: W $\to$ S $\to$ E $\to$ N $\to$ W.

1. A(0,0) $\stackrel{W42}{\to }$ $(-42,0)$ $\stackrel{S40}{\to }$ $(-42,-40)$ $\stackrel{E49}{\to }$ $(7,-40)$.
2. $\stackrel{N46}{\to }$ $(7,6)$ $\stackrel{W7}{\to }$ P$(0,6)$.
3. P is directly 6 km north of A, so he must drive 6 km South to reach A.

Trick: Net E–W: $-42+49-7=0$; net N–S: $-40+46=+6$ (north). Just net the opposite legs.

Answer: (D) 6 km to the South

Concept: Pin the fixed days first (B on Monday, N on Sunday), then use the "between" clues.

1. B $=$ Monday. Three people between B and L $\Rightarrow$ L $=$ Friday.
2. One person between L and A $\Rightarrow$ A $=$ Wednesday or Sunday; Sunday is N, so A $=$ Wednesday.
3. Remaining days Tue, Thu, Sat for I, S, G with I before S before G $\Rightarrow$ I $=$ Tue, S $=$ Thu, G $=$ Sat.
4. Thursday: S.

Trick: "Only three between" from Monday always lands on Friday (Mon $+4$).

Answer: (D) S

Concept: Test each statement alone, then both together; stop as soon as the shortest tree is uniquely fixed.

1. (I) alone: N is shorter than only three $\Rightarrow$ N is 4th tallest, i.e. exactly one tree is shorter than N – but we don't know which. Not sufficient.
2. (II) alone: M $<$ N $<$ T. L and V are unplaced, so the shortest could be M, L or V. Not sufficient.
3. Together: only one tree is below N, and (II) says M is below N $\Rightarrow$ M is the shortest. Sufficient.

Trick: "Shorter than only three" $=$ 4th position in a group of 5 – convert such phrases to ranks immediately.

Answer: (D) Both statements I and II together (and not statement I alone or statement II alone) are sufficient