Q&A: practice questions with answers

Concept: If 75% goes to rent, the other 25% covers all remaining expenses and savings.

1. Remaining amount $=653+6229+446=7328$ rupees.
2. $25\%$ of salary $=7328$.
3. Salary $=7328\times 4=29312$.

Trick: $25\%=\frac{1}{4}$, so just multiply the leftover by 4.

Answer: (A) ₹29,312

Concept: Volume is conserved on melting; number of cubes needed must be a whole number (round up).

1. Volume of one cube $={3.6}^3=46.656{\text{cm}}^3$.
2. Two cones $=\frac{1}{3}\pi r^2(h_1+h_2)=\frac{1}{3}\times 3.14\times 29.16\times 84=2563.7472{\text{cm}}^3$.
3. Cubes needed $=\frac{2563.7472}{46.656}\approx 54.95$, so 55 cubes are used.
4. Surplus $=731-55=676$.

Trick: Joined cones with a common base: add the heights, $\frac{1}{3}\pi r^2(h_1+h_2)$.

Answer: (C) 676

Concept: Angle of depression from the top equals the angle of elevation at that point; use $\tan \theta =\frac{\text{opposite}}{\text{adjacent}}$.

1. Bottom of post (60°): $d=\frac{60}{\tan {60}^{\circ }}=\frac{60}{\sqrt{3}}=20\sqrt{3}\text{m}$.
2. Top of post (30°): $60-h=d\tan {30}^{\circ }=20\sqrt{3}\times \frac{1}{\sqrt{3}}=20$.
3. $h=60-20=40\text{m}$.

Trick: With depressions 30° and 60°, the post is $\frac{2}{3}$ of the building: $\frac{2}{3}\times 60=40$.

Answer: (A) $40\text{m}$

Concept: If profit % = loss % for two selling prices, CP is their average.

1. $\text{CP}=\frac{3471+2577}{2}=\frac{6048}{2}=3024$.
2. For 25% profit: $\text{SP}=3024\times \frac{125}{100}=3780$.

Trick: Equal profit and loss % → CP = mean of the two SPs.

Answer: (B) 3,780

Concept: In $a:b::b:x$, $b$ is the mean proportional, so $b^2=a\times x$.

1. ${40.6}^2=2.5\times x$.
2. ${40.6}^2=1648.36$.
3. $x=\frac{1648.36}{2.5}=659.344$.

Trick: Dividing by 2.5 = multiplying by 0.4: $1648.36\times 0.4=659.344$.

Answer: (D) 659.344

Concept: $\text{SI}=\frac{P\times R\times T}{100}$.

1. $3500=\frac{14000\times x\times 4}{100}=560x$.
2. $x=\frac{3500}{560}=6.25$.

Trick: Total interest is $\frac{3500}{14000}=25\%$ in 4 years → $\frac{25}{4}=6.25\%$ p.a.

Answer: (B) 6.25

Concept: Supplementary angles add up to ${180}^{\circ }$.

1. Let the smaller angle be $x$; larger $=x+{62}^{\circ }$.
2. $x+x+{62}^{\circ }={180}^{\circ }\Rightarrow 2x={118}^{\circ }$.
3. $x={59}^{\circ }$ (larger $={121}^{\circ }$).

Trick: Smaller $=\frac{{180}^{\circ }-{62}^{\circ }}{2}={59}^{\circ }$.

Answer: (B) ${59}^{\circ }$

Concept: Take CP $=100$: MP and SP become simple percentages.

1. MP $=134$ (34% above CP); SP $=110$ (10% profit).
2. Discount $=134-110=24$.
3. Discount $\%=\frac{24}{134}\times 100\approx 17.91\%$.
4. (Check: CP $=\frac{578}{1.34}\approx 431.34$, SP $\approx 474.48$, discount $\approx 103.52$ on 578.)

Trick: Discount $\%=\frac{M-P}{100+M}\times 100=\frac{34-10}{134}\times 100$.

Answer: (B) 17.91%

Concept: If $M\propto \frac{1}{N}$, then $M\times N$ is constant.

1. $M\times N=18\times 10=180$.
2. When $N=9$: $M=\frac{180}{9}=20$.

Trick: Inverse proportion: product stays the same.

Answer: (D) 20

Concept: Express every age in terms of Urvashi's age $U$.

1. Ayushi $=2U-6$; Sanju $=$ Ayushi $+8=2U+2$; Mohan $=4U$.
2. $4U+(2U+2)=50\Rightarrow 6U=48$.
3. $U=8$ years.

Trick: Check options: $U=8$ → Mohan 32, Sanju 18, total 50 ✓.

Answer: (A) 8

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

1. RIPE → 5, 9, 8, 7 and FIRE → 8, 7, 5, 1.
2. Common letters: R, I, E; common digits: 5, 8, 7.
3. Leftover letter in FIRE is F and leftover digit is 1, so F = 1.

Trick: Cancel common letters with common digits – the odd ones pair up.

Answer: (D) 1

Concept: Check the difference between consecutive terms.

1. $29-23=6,35-29=6,41-35=6,47-41=6$
2. Constant difference +6.
3. Next term $=47+6=53$

Trick: Equal gaps → arithmetic series; just add the gap.

Answer: (A) 53

Concept: Change each digit by the rule, then pick the required positions.

1. 3154872: odd digits +1 (3→4, 1→2, 5→6, 7→8); even digits −2 (4→2, 8→6, 2→0).
2. New number: 4 2 6 2 6 8 0
3. Second from left = 2, second from right = 8; difference $=8-2=6$

Trick: Write the new digits directly under the old ones to avoid mistakes.

Answer: (D) 6

Concept: Fix the definite clues first, then place blocks of people.

1. S = Tuesday; 3 people between S and V → V = Saturday.
2. T, P, U must be consecutive (T–P–U); the only 3 free consecutive days are Wed, Thu, Fri.
3. One person between T (Wed) and R → R = Monday (Friday is taken by U); Q = Sunday.

Trick: Place fixed persons, then fit the longest block into the only gap left.

Answer: (B) R

Concept: Find the shift from the given right pair, then apply the same to #.

1. Try # = POL: P−3 = M, O−3 = L, L−3 = I → MLI ✓
2. Apply −3 to LKH: L−3 = I, K−3 = H, H−3 = E → IHE
3. So # = POL and % = IHE.

Trick: Test each option quickly with the same letter-shift on both sides.

Answer: (C) # = POL, % = IHE

Concept: Facing the centre: right = anticlockwise, left = clockwise.

1. Fix G; A is third to the right of G.
2. C is second to the right of D and C is not next to G → the only fit is D second to the right of G and C second to the right of D (order G, _, D, A, C).
3. E is immediately left of B → B and E take the remaining adjacent seats; F sits between G and D.
4. From D moving left (clockwise): F, G, then B → 2 people between.

Trick: Draw the circle and fix the person with most clues first.

Answer: (B) Two

Concept: Identify the highest and lowest numbers, then pick the asked digits.

1. Numbers: 194, 932, 446, 225, 493
2. Highest = 932 → first digit 9; lowest = 194 → second digit 9
3. Sum $=9+9=18$

Trick: Compare only first digits to spot highest/lowest fast.

Answer: (C) 18

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

1. HI – CD, QR – LM, PQ – KL: each cluster is two consecutive letters in forward order, and the second cluster is 5 letters back (H−5 = C, Q−5 = L, P−5 = K).
2. LK – KJ: letters are in reverse order and the shift is only 1 (L−1 = K).
3. So LK – KJ is the odd one.

Trick: First check forward/backward order inside clusters – often enough to find the odd one.

Answer: (A) LK - KJ

Concept: Track the net east–west and north–south displacement.

1. North–south: $+9-10+1=0$, so P is on the same level as A.
2. East–west: $+7$ east then $8$ west → net $1\text{km}$ west of A.
3. To return, Mukesh must drive 1 km towards the east.

Trick: Add north, subtract south; add east, subtract west – net values give the answer.

Answer: (B) 1 km to the east

Concept: Four left turns starting north go: North → West → South → East → North.

1. North–south: $+8-12+4=0$ → P is level with A.
2. East–west: $9$ west then $11$ east → net $2\text{km}$ east of A.
3. So Radhe must drive 2 km towards the west to reach A.

Trick: Net east 2 km means the return trip is 2 km west.

Answer: (B) 2 km to the west