Q&A: practice questions with answers

Concept: Apply BODMAS: division and multiplication first (left to right), then addition and subtraction. Keep the sign rules for negatives.

1. $(-56)÷(-14)=4$ and $(-2)\times 6=-12$.
2. Expression $=(-9)-4+(-12)=-9-4-12=-25$.

Trick: Negative ÷ negative = positive; negative × positive = negative. Write the simplified terms before adding to avoid sign slips.

Answer: (C) -25

Concept: Empirical relation: $\text{Mode}=3\text{Median}-2\text{Mean}$, which rearranges to $\text{Mode}-\text{Median}=\frac{2}{3}(\text{Mode}-\text{Mean})$.

1. Given $\text{Mode}-\text{Mean}=33.6$.
2. $\text{Mode}-\text{Median}=\frac{2}{3}\times 33.6=22.4$.
3. Check: then $\text{Mean}-\text{Median}$ would be $-11.2$, i.e. Median exceeds Mean by $11.2$, and $33.6=3\times 11.2$ ✓

Trick: Mean – Median – Mode are in the ratio of gaps $1:2$: Mode−Median is twice Median−Mean, so Mode−Mean splits as $\frac{2}{3}$ and $\frac{1}{3}$.

Answer: (A) 22.4

Concept: Evaluate the root first, then follow BODMAS: division before multiplication/addition in left-to-right order.

1. $\sqrt{3136}=56$ (since ${56}^2=3136$).
2. $560÷56=10$, so LHS $=18\times 7\times 10=1260$.
3. $1260=y+703\Rightarrow y=557$.

Trick: Recognise squares ending in 6: ${56}^2=3136$ (and ${44}^2=1936$, ${54}^2=2916$). Note $560=10\times 56$ for an instant division.

Answer: (A) 557

Concept: A chord of the outer circle that touches the inner circle is perpendicular to the radius of the inner circle at the point of contact, and that radius bisects the chord.

1. Let O be the common centre, AB the chord touching the smaller circle at M. Then $OM=15$, $OB=17$, $OM⟂AB$.
2. In right $△OMB$: $MB=\sqrt{{17}^2-{15}^2}=\sqrt{289-225}=\sqrt{64}=8\text{cm}$.
3. Chord $AB=2\times MB=16\text{cm}$.

Trick: Chord length $=2\sqrt{R^2-r^2}$. Spot the Pythagorean triple 8-15-17 and double the 8.

Answer: (D) 16 cm

Concept: Sum of observations $=\text{average}\times \text{count}$.

1. Sum of the three numbers $=20\times 3=60$.
2. Third number $=60-(13+22)=60-35=25$.

Trick: Deviation method: $13$ is $7$ below $20$, $22$ is $2$ above → net $-5$, so the third must be $5$ above: $25$.

Answer: (C) 25

Concept: A pure recurring part of $n$ digits equals that block divided by $n$ nines: $0.\overline{12}=\frac{12}{99}$.

1. $22.121212\dots =22+0.\overline{12}=22+\frac{12}{99}=22+\frac{4}{33}$.
2. $=\frac{22\times 33+4}{33}=\frac{726+4}{33}=\frac{730}{33}$.
3. Check: $\frac{730}{33}=22.1212\dots$ ✓

Trick: Direct formula: $\frac{2212-22}{99}=\frac{2190}{99}=\frac{730}{33}$ (whole number with repeating block minus whole number, over 99).

Answer: (C) $\frac{730}{33}$

Concept: If $a,b,c$ are in continued proportion then $\frac{a}{b}=\frac{b}{c}$, i.e. $b^2=ac$ ($b$ is the mean proportional).

1. $x^2=8\times 18=144$.
2. $x=\sqrt{144}=12$ (taking the positive value).
3. Check: $8:12=12:18=2:3$ ✓

Trick: Mean proportional $=\sqrt{ac}$. Here $8\times 18=144$, a perfect square – answer is $12$ at sight.

Answer: (C) 12

Concept: Change every digit by the rule, then read the new number.

1. Number 6134527: odd digits $+1$, even digits $+2$.
2. $6\to 8,1\to 2,3\to 4,4\to 6,5\to 6,2\to 4,7\to 8$ → new number 8246648.
3. First digit $8$, last digit $8$ → sum $=8+8=16$.

Trick: Only the first and last digits are asked – just convert those two: $6\to 8$ and $7\to 8$.

Answer: (D) 16

Concept: Write the alphabet positions and compare the gaps.

1. YCX: $25\to 3$ is $+4$ (wrapping past Z), $3\to 24$ is $-5$.
2. GKF: $7\to 11(+4),11\to 6(-5)$; NRM: $14\to 18(+4),18\to 13(-5)$.
3. TYS: $20\to 25(+5),25\to 19(-6)$ – different pattern, so TYS is the odd one.

Trick: Three clusters follow $+4,-5$; the third letter is one place before the first letter (F before G, M before N, X before Y) – but S is not just before T.

Answer: (A) TYS

Concept: Common letters in two words get common codes; the leftover letter takes the leftover code.

1. ROAD → 8657, DEAR → 4567. Common letters R, A, D ↔ common codes 5, 6, 7.
2. In ROAD the extra letter O gets the extra code 8.
3. In DEAR the extra letter E gets the extra code 4.

Trick: Don't match letter positions – just cancel the common letters and common digits.

Answer: (D) 4

Concept: Cancel the common letters and common digits of the two words.

1. REST → 5729, SEAT → 7625. Common letters S, E, T ↔ common codes 5, 7, 2.
2. REST's extra letter R → extra code 9.
3. SEAT's extra letter A → extra code 6.

Trick: Whatever digit appears in only one word belongs to the letter that appears in only that word.

Answer: (D) 6

Concept: Check each statement alone, then together.

1. Statement I: Ritu, Seema $<$ Tanya $<$ Usha. Varsha is unknown – she could be above Usha. Not sufficient.
2. Statement II: Tanya $<$ Varsha $<$ Usha. Ritu and Seema unknown. Not sufficient.
3. Together: Usha is above Tanya, Varsha, Ritu and Seema → Usha scored the highest. Both needed.

Trick: For "who is highest", every person must be placed below someone – any unplaced name kills sufficiency.

Answer: (A) Both statements together are sufficient

Concept: Convert to positions; check the gaps inside and between the pairs.

1. SU–VX: $19,21|22,24$ → gaps $+2,+1,+2$.
2. BD–EG: $2,4|5,7$ and IK–LN: $9,11|12,14$ → same $+2,+1,+2$.
3. LM–NP: $12,13|14,16$ → $+1,+1,+2$ – breaks the pattern.

Trick: In each valid pair the two letters skip one letter (S_U, B_D, I_K); LM are consecutive – spot it instantly.

Answer: (D) LM – NP

Concept: Facing the centre, a person's left is clockwise and right is anticlockwise. Start from the clue that fixes two people.

1. P is immediate right of Q; L is third to the left of Q.
2. P is second to the left of R → R sits two places anticlockwise from P. N: two people between R and N from N's left → N is between Q and K's side (see figure).
3. Remaining seats: K and M; M is not next to L, so M goes beside P and K beside N. Order (clockwise): Q, N, K, L, R, M, P.
4. Third to the right of K (anticlockwise 3): N → Q → P.

Trick: Place the "immediate" clue first (Q–P), then count only clockwise for "left" to avoid confusion.

Answer: (D) P

Concept: Arrange the digits, then pick the positions asked.

1. Digits of 749216358 in descending order: 9 8 7 6 5 4 3 2 1.
2. Second from left $=8$; third from right $=3$.
3. Product $=8\times 3=24$.

Trick: All nine digits 1–9 appear once, so the arrangement is simply 987654321 – no sorting needed.

Answer: (B) 24

Concept: Find the operation between consecutive numbers and test the options.

1. $100\to 50(÷2),50\to 60(+10),60\to 30(÷2)$; also $300\to 150\to 160\to 80$ – same.
2. Option D: $80÷2=40,40+10=50,50÷2=25$ ✓.
3. A fails at $90\to 180$, B at $120\to 30$, C at $50\to 40$.

Trick: Pattern "$÷2,+10,÷2$" – check the last step first; only D halves at the end.

Answer: (D) 80 – 40 – 50 – 25

Concept: Check each letter: is a symbol on both sides?

1. Series: = N * R ^ S V L ^ W O P Y $ @ H A F ? E R
2. N: between = and * ✓. R: between * and ^ ✓.
3. S (^ S V), W (^ W O), H (@ H A), F (A F ?), E (? E R) – each has a letter on one side ✗.

Trick: Scan only around symbols; a valid letter must be sandwiched symbol–letter–symbol.

Answer: (A) Two

Concept: Look at the differences.

1. $38-34=4,46-38=8,58-46=12,74-58=16$.
2. Differences grow by 4 → next difference $20$ → $74+20=94$.

Trick: Differences are multiples of 4: $4,8,12,16,20$.

Answer: (A) 94

Concept: Draw the basic Venn diagram and test each conclusion against it.

1. "Some tables are boards" converts to "Some boards are tables" → conclusion I follows.
2. "No chair is a desk" says nothing about the rest of the tables – a desk can be a table outside the chairs circle. So "No table is a desk" is not certain → II does not follow.

Trick: Some A are B ⇔ Some B are A (always). A negative about a subset (chairs) never transfers to the superset (tables).

Answer: (A) Only conclusion (I) follows

Concept: Fix the definite clue first, then place the others.

1. R = Thursday. Two people between F and R → F = Monday or Sunday. Four between S and F → (F Mon, S Sat) or (F Sun, S Tue).
2. W is after F, so F cannot be Sunday → F = Mon, S = Sat.
3. Left: Tue, Wed, Fri, Sun for W, E, O, L. E is immediately before O → Tue–Wed. W before L → W = Fri, L = Sun.

Trick: "X is after Y" eliminates the end-of-week position for Y – use it to kill one of the two cases.

Answer: (C) W