Q&A: practice questions with answers

Concept: Divide the figure into zones and count each.

1. Small top cap above the rectangle: 3.
2. Four corner pieces of the rectangle: 4.
3. Upper triangle on the middle horizontal (apex at top): 3.
4. Lower triangle on the same horizontal (apex at bottom): 3.
5. Left and right halves of the central kite: 2.

Trick: Total = 3 + 4 + 3 + 3 + 2 = 15. A triangle split by one median line gives 3.

Answer: (C) 15

Concept: Count part by part: left half, right square, then the lower triangle.

1. Left half (one diagonal): 2.
2. Right square with both diagonals: 8.
3. Long triangle across the top using the left diagonal and right diagonal: 1.
4. Lower triangle pieces (below the middle line): 4.
5. Whole lower triangle + the tall triangle from the top middle to the bottom right edge: 2.

Trick: Total = 2 + 8 + 1 + 4 + 2 = 17.

Answer: (C) 17

Concept: Build the fixed block first, then place it.

1. A is between D and F; F is between E and A → block E-F-A-D.
2. B is 2nd from right → seat 5; block fills seats 1-4, C gets seat 6.
3. D is not at an end → order E F A D B C (left to right).

Trick: Persons left of D = E, F, A = 3.

Answer: (D) 3

Concept: A conclusion must come only from the statement.

1. The email says class timings will change.
2. So 'timings of classes are changing' follows directly.
3. Closing, vacation, exam cancellation are not mentioned.

Trick: Reject any option that brings in a new idea not present in the statement.

Answer: (B) Timings of classes are changing

Concept: Right = 90° clockwise, Left = 90° anticlockwise.

1. Start: South.
2. (D) Left → East, Left → North, Right → East ✓
3. (A) and (B) end at West; (C) ends at South.

Trick: Two lefts and one right = net one left; left of South = East.

Answer: (D) left, left, right

Concept: Diagram: a middle class overlapping two classes that do not overlap each other.

1. Principal can be a man or a woman → overlaps both.
2. Man and Woman have nothing in common → separate.
3. (A) all separate; (D) Groom cannot be a Girl.

Trick: Spot the two mutually exclusive classes first (Man/Woman), then find what links both.

Answer: (C) Principal, Man, Woman

Concept: Track how each corner element moves from left to right figure.

1. In (A), (B), (D) every element jumps to the diagonally opposite corner – a half-turn (180°).
2. In (C) two elements stay fixed and only the other two swap – a different rule.

Trick: If an element stays in the same corner, it cannot be a 180° turn.

Answer: (C)

Concept: Track each element separately.

1. Black square moves cell by cell downward: next in row 3, left column.
2. Dot moves upward cell by cell: next in row 2, right column.
3. Circle alternates: left with top half shaded / right with right half shaded → next: left, top shaded.

Trick: Fix the circle first (pattern repeats every 2), then check the square and dot.

Answer: (A)

Concept: Code = sum of letter positions + half of first letter's value.

1. BOOK: $2+15+15+11=43$; $+\frac{2}{2}=1\to 44$ ✓
2. PAGE: $16+1+7+5=29$; $+\frac{16}{2}=8\to 37$ ✓
3. PAPER: $16+1+16+5+18=56$; + 8 → 64

Trick: PAPER = PAGE pattern with same first letter P: just add 8 to the letter sum.

Answer: (C) 64

Concept: CLEAR needs C, L, E, A, R.

1. CLEANER, CALENDAR, RECALL – all have these five letters.
2. CANDLE has no R → CLEAR cannot be made.

Trick: Check the rarest letter first – here R.

Answer: (D) CANDLE

Concept: Profit is divided in the ratio of $(\text{money}\times \text{months})$.

1. A = $6000\times 10=60000$, B = $4000\times 8=32000$, C = $5000\times 8=40000$
2. Ratio = 15 : 8 : 10 (÷ 4000); total 33 parts
3. 1 part = $330000÷33=$ ₹10,000
4. B = $8\times 10000=$ ₹80,000

Trick: ₹$3,30,000÷33=10,000$ is clean – a sign the ratio is right.

Answer: (B) ₹80,000

Concept: Use $(a+b{)}^2=a^2+b^2+2ab$ and $a^3+b^3=(a+b)(a^2+b^2-ab)$.

1. $225=117+2ab\to ab=54$
2. $a^3+b^3=15\times (117-54)$
3. $=15\times 63=945$

Trick: Guess numbers: $a+b=15,ab=54\to a=9,b=6$; $729+216=945$.

Answer: (D) 945

Concept: $\text{Total}=\text{students}\times \text{candies each}$: $R(R+6)=432$.

1. $R^2+6R-432=0$
2. $(R+24)(R-18)=0$
3. $R=18$ (students cannot be negative)

Trick: Check options: $18\times 24=432$ ✓ – no equation needed.

Answer: (D) 18

Concept: $\text{Volume of cone}=\frac{1}{3}\pi r^{2}h$.

1. $\frac{1}{3}\times \frac{22}{7}\times r^2\times 9=4158$
2. $\frac{66}{7}{r}^2=4158$
3. $r^2=4158\times \frac{7}{66}=441$
4. $r=21$ cm

Trick: Volume has $\frac{22}{7}$ → r is usually a multiple of 7; only 21 fits.

Answer: (C) 21 cm

Concept: $\text{Exterior angle}=\frac{{360}^{\circ }}{n}$ and $\text{interior}={180}^{\circ }-\text{exterior}$.

1. Smaller: $\text{exterior}={180}^{\circ }-{150}^{\circ }={30}^{\circ }\to n=\frac{360}{30}=12$
2. Larger: $3\times 12=36$ sides
3. $\text{Exterior}=\frac{360}{36}={10}^{\circ }\to \text{interior}={170}^{\circ }$

Trick: Perimeter and side ratio are distractors – angle depends only on number of sides.

Answer: (B) 170*

Concept: Show $△ABD\cong △ACE$, then equate matching sides.

1. $AD=AE\to \angle ADE=\angle AED\to$ their supplements $\angle ADB=\angle AEC$.
2. With $\angle BAD=\angle CAE$ and $AD=AE\to △ABD\cong △ACE$ (ASA).
3. So $BD=CE\to 3p=q$, and $AB=AC\to 2p+4=3q-8$.
4. $2p+4=9p-8\to p=\frac{12}{7},q=\frac{36}{7}\to p+q=\frac{48}{7}$.

Trick: Once you see $BD=CE$, write $q=3p$ and substitute – one-variable equation.

Answer: (B) $\frac{48}{7}$

Concept: A cuboid has 12 edges: total = $4(l+b+h)$.

1. $l+b+h=280÷4=70$
2. $3x+5x+6x=14x=70\to x=5$
3. $l=15,b=25,h=30$
4. Volume = $15\times 25\times 30=11250{\text{cm}}^3$

Trick: $\text{Volume}=90x^3=90\times 125=11250$.

Answer: (B) 11250 cubic cm

Concept: Volume stays the same after melting.

1. $n=\frac{4}{3}\pi (21{)}^3÷\frac{2}{3}\pi (7{)}^3=2\times 3^3=54$
2. $\text{CSA of one hemisphere}=2\pi r^2=2\pi \times 49=98\pi$
3. Total = $54\times 98\pi =5292\pi {\text{cm}}^2$

Trick: Radius ratio 3:1 → sphere = 27 small spheres = 54 hemispheres.

Answer: (B) $5292\pi {\mathrm{cm}}^2$

Concept: $\text{LSA}=\frac{1}{2}\times \text{perimeter of base}\times \text{slant height}$.

1. Base perimeter $=4\times 11.5=46$ m
2. $\text{LSA}=\frac{1}{2}\times 46\times 20=460{\text{m}}^2$

Trick: Half the perimeter (23) × slant height (20) = 460 – one-step mental calculation.

Answer: (C) $460m^2$

Concept: Break-even: profit amount = loss amount.

1. $20\%{\text{ of CP}}_1=30\%{\text{ of CP}}_2$
2. $\frac{{\text{CP}}_1}{{\text{CP}}_2}=\frac{30}{20}=\frac{3}{2}$

Trick: Ratio of CPs = inverse of the percentages: 20 : 30 → 3 : 2.

Answer: (C) $\mathrm{CP}=3:2$