Time Speed Distance – MBA CET questions with solutions

Approach: Chase problem: head start ÷ relative speed.

Solution:
In the first 2 s the thief covers 5 × 2 = 10 m.
Relative speed = 10 − 5 = 5 m/s.
Time to close 10 m = 10 / 5 = 2 s (from when the policeman starts).

Cetking shortcut: Policeman is twice as fast, so he catches up in the same time as the head-start time (2 s).

Answer: B

Approach: Race starts: convert to distances run in the same time.

Solution:
When A runs 1000 m, B runs 950 m and C runs 931 m.
So B : C speed = 950 : 931.
When B runs 1000 m, C runs 1000 × 931/950 = 980 m.
B can give C a start of 1000 − 980 = 20 m.

Why the other options are wrong:
B (19 m) is the naive 69 − 50 subtraction; A, D, E do not fit the ratio.

Cetking shortcut: 931/950 = 0.98 exactly (950 × 0.98 = 931) → 20 m.

Answer: C

Approach: Equal distances → average speed is the harmonic mean.

Solution:
Average speed = 3 / (1/30 + 1/40 + 1/60).
LCM 120: 1/30 + 1/40 + 1/60 = (4 + 3 + 2)/120 = 9/120.
Average speed = 3 × 120/9 = 40 km/h.

Why the other options are wrong:

• The arithmetic mean would be ≈ 43.3, which is not an option; B (42), C (45), D (48), E (50) are not the harmonic mean.

Cetking shortcut: Take distance = LCM 120 km each leg: times 4 + 3 + 2 = 9 h for 360 km → 40 km/h.

Answer: A

Approach: Head start, then relative speed for trains moving towards each other.

Solution:
From 9:30 to 10:15 = 45 min = 3/4 h; train A covers 72 × 3/4 = 54 km alone.
Gap at 10:15 = 480 − 54 = 426 km.
Relative speed (towards each other) = 72 + 48 = 120 km/h.
Time = 426/120 = 3.55 h = 3 h 33 min.
Meeting time = 10:15 + 3:33 = 1:48 p.m.

Why the other options are wrong:

• A (1:30), B (1:45): ignore part of the time — 3.55 h is 3 h 33 min, not 3 h 30 min.
• D (2:00) would need 3.75 h after 10:15; E (2:15) comes from 480/120 = 4 h after 10:15, ignoring A's 54 km head start.

Cetking shortcut: 0.55 h × 60 = 33 min — convert decimals of an hour carefully; 3.5 h would be 3 h 30 min.

Answer: C

Approach: Speed = (hops per unit time) × (length of one hop).

Solution:
4 dog hops = 5 rabbit hops → rabbit hop : dog hop = 4 : 5. Let rabbit hop = 4, dog hop = 5 units.
In the same time: rabbit 6 hops = 6 × 4 = 24 units; dog 5 hops = 5 × 5 = 25 units.
Speed ratio rabbit : dog = 24 : 25.

Why the other options are wrong:

• A (5 : 4), D (25 : 24): reverse the hop-length ratio.
• B (6 : 5): uses only hop counts.
• E (5 : 6): inverse of hop counts.

Cetking shortcut: Speed ratio = (hop-rate ratio) × (hop-length ratio) = (6/5) × (4/5) = 24/25.

Answer: C

Approach: Race with head start — B covers less distance but takes 15 s more.

Solution:
A's speed = 8 km/h = 8 × 5/18 = 20/9 m/s.
A's time for 100 m = 100 ÷ 20/9 = 45 s.
B starts 4 m ahead → runs 96 m, and finishes 15 s after A → 45 + 15 = 60 s.
B's speed = 96/60 = 1.6 m/s = 1.6 × 18/5 = 5.76 km/h ≈ 5.8 km/h.

Why the other options are wrong:

• A (4.5), B (5.2), D (6.4), E (7): none equals 96 m in 60 s (5.76 km/h); they come from using a wrong distance or time for B.

Cetking shortcut: Convert once: 96 m in 60 s = 96 × 60 m/h = 5.76 km/h.

Answer: C

Approach: Races with a head start — compare the times to finish.

Solution:
A runs 500 − 140 = 360 m at speed 3 ⇒ time = 120 units.
In 120 units B runs 4 × 120 = 480 m.
When A finishes, B is 500 − 480 = 20 m short ⇒ A wins by 20 m.

Why the other options are wrong:
A (10), C (30), D (40), E (60) – wrong time base (e.g. using 500 m for A).

Cetking shortcut: Distance ratio = speed ratio: B covers 360 × 4/3 = 480 m while A covers 360 m.

Answer: B

Approach: Constant speeds ⇒ speed ratios chain: (A/B) × (B/C) = (A/C).

Solution:
Let the race length = L. When A finishes, B has run L − 12 and C has run L − 18.
When B finishes, C has run L − 8 ⇒ C/B = (L − 8)/L.
Also C/B = (L − 18)/(L − 12).
(L − 8)/L = (L − 18)/(L − 12) ⇒ L² − 20L + 96 = L² − 18L ⇒ 2L = 96 ⇒ L = 48 m.
Check: A : B : C = 48 : 36 : 30; B/C = 36/30 = 48/40 ✓.

Why the other options are wrong:
A (36), B (24), C (60), D (72) – fail the chained ratio check.

Cetking shortcut: B beats C by 6 m over the last L − 12 m and by 8 m over L m: 6/(L − 12) = 8/L ⇒ L = 48.

Answer: E

Approach: Average speed = total distance / total time (not the mean of speeds).

Solution:
Take total distance = 6 units (LCM of 2 and 3): parts 3, 2, 1.
Time = 3/30 + 2/40 + 1/20 = 0.1 + 0.05 + 0.05 = 0.2.
Average speed = 6 / 0.2 = 30 km/h.

Cetking shortcut: Better: take distance = 120 km → 60/30 + 40/40 + 20/20 = 2 + 1 + 1 = 4 h → 120/4 = 30.

Answer: D

Approach: Harmonic mean for equal distances.

Solution: Average speed = 2 × 50 × 30 / (50 + 30) = 3000/80 = 37.5 km/hr.
Check: time = 150/50 + 150/30 = 3 + 5 = 8 h; 300/8 = 37.5.

Why the other options are wrong:
E (40): simple average — the classic trap.
A, B, D: calculation slips.

Answer: C

Approach: Time is the same in both cases.

Solution: Extra distance = (14 − 10) × t = 20 ⇒ t = 5 hours.
Actual distance = 10 × 5 = 50 km.

Why the other options are wrong:
C (70 km): distance at 14 km/hr.
B, D, E: arithmetic slips.

Answer: A

Approach: Relative speed with distance as 1 unit.

Solution: X takes 60 min, Y takes 90 min (4:00 to 5:30).
Let distance = 180 units. Speed X = 3 units/min, Y = 2 units/min.
Meeting time = 180 ÷ (3 + 2) = 36 min after 4 pm.
They cross at 4:36 pm.

Why the other options are wrong:
B–E: come from wrong times (e.g. averaging) — the formula gives exactly 36 min.

Cetking shortcut: T = (60 × 90)/(60 + 90) = 5400/150 = 36 min.

Answer: A

Approach: Speed ratio = √(time of the other after meeting).

Solution: Howrah train: 9 h after meeting; Patna train: 16 h.
S1 : S2 = √16 : √9 = 4 : 3.

Why the other options are wrong:
D (9 : 16): takes the time ratio directly.
A (2 : 3): no valid working.
C, E: no valid working.

Answer: B

Approach: Distance B covers in A's time.

Solution: B's speed = 100/45 m/s.
In 36 s B runs 36 × 100/45 = 80 m.
A beats B by 100 − 80 = 20 m.

Why the other options are wrong:
D (9 m): takes 45 − 36 seconds as metres.
B (25 m): uses A's speed × 9 s.
C, E: arithmetic slips.

Answer: A

Approach: Use the beat-margin to find B's speed.

Solution: B runs 35 m in 7 s ⇒ speed = 5 m/s.
B's time for 200 m = 200 ÷ 5 = 40 s.
A's time = 40 − 7 = 33 s.

Why the other options are wrong:
A (40 s): B's time.
B (47 s): adds 7 instead of subtracting.
D, E: no valid working.

Answer: C

Approach: Relative movement on steps — count step positions after each move.

Solution: Shyam starts at the bottom (step 0) and goes up 1 step per move; Ram starts at the top (step 23) and comes down 2 steps per move. The gap of 23 steps closes by 3 steps per move.
After 7 moves: Shyam is on step 7, Ram on step 23 − 14 = 9.
The only step between them now is step 8; in the next move both reach/cross it, so they meet at step 8 from the bottom.

Cetking shortcut: Gap ÷ combined steps = 23 ÷ 3 = 7.67 → they meet during the 8th move, i.e. on the 8th step.

Answer: A

Approach: Trains meeting rule: S_P/S_Q = √(t_Q/t_P), where t = time taken after meeting.

Solution:
Time of P after meeting t_P = 25 h; time of Q after meeting t_Q = 16 h.
S_P/S_Q = √(16/25) = 4/5
100/S_Q = 4/5 ⇒ S_Q = 100 × 5/4 = 125 km/h

Cetking shortcut: The train that takes LESS time after meeting is FASTER — so Q > 100; only 120, 125, 140, 115 qualify, and √(25/16) = 5/4 gives 125.

Answer: B

Approach: Distance = Speed × Time; same distance both ways.

Solution:
Let one-way distance = D km.
Time going = D/12, time returning = D/6.
D/12 + D/6 = 6 ⇒ (D + 2D)/12 = 6 ⇒ 3D = 72 ⇒ D = 24 km.

Cetking shortcut: Average speed for equal distances = 2×12×6/(12+6) = 8 km/h; total distance = 8 × 6 = 48 km, one way = 24 km.

Answer: C

Approach: Gap between two runners in the same direction = (difference of speeds) × time; max gap is between fastest and slowest.

Solution:
Speeds in m/s: A = 15; B = 67 × 5/18 ≈ 18.61; C = 1020/60 = 17.
Fastest = B, slowest = A ⇒ speed difference = 18.61 − 15 = 3.61 m/s.
Time = 8 h 30 min 39 s = 30600 + 39 = 30639 s.
Maximum gap = 3.61 × 30639 ≈ 110606.8 m ≈ 110.606 km (rounding the speed to 3.61 m/s; the unrounded 3.6111 gives 110.64 km, still closest to E).

Why the other options are wrong:
Ignoring the 39 seconds gives ≈ 110.47 km — not offered.
A, B, C, D: other pairs/slips; B−C and C−A gaps are far smaller (≈ 49 km and ≈ 61 km).

Cetking shortcut: Only the fastest-minus-slowest pair matters: 3.61 m/s × 30639 s.

Answer: E

Approach: Uphill one way is downhill the other way; add the two journeys so hill parts combine.

Solution:
Let uphill (A→B) = x, plain = y, downhill (A→B) = z; x + y + z = 11.5
A→B: x/3 + y/4 + z/5 = 2.9 h (2 h 54 min)
B→A: x/5 + y/4 + z/3 = 3.1 h (3 h 6 min)
Add: (1/3 + 1/5)(x + z) + y/2 = 6 → (8/15)(11.5 − y) + y/2 = 6
× 30: 16(11.5 − y) + 15y = 180 → 184 − y = 180 → y = 4 km
(Also x = 3, z = 4.5: 1 + 1 + 0.9 = 2.9 ✓)

Cetking shortcut: Adding the two trips removes the direction issue — every hill km costs 8/15 h round trip, every plain km 1/2 h. If all 11.5 km were hill: 11.5 × 8/15 = 6.133 h; actual 6 h; each plain km saves 1/30 h → 0.133 × 30 = 4 km.

Answer: E

Approach: Distances covered in the same time are in the ratio of speeds.

Solution:
When A runs 100, B runs 90 and C runs 87
So B : C = 90 : 87 = 30 : 29
When B runs 180, C runs 180 × 29/30 = 174
B beats C by 180 − 174 = 6 m

Why the other options are wrong:
A (5.4 m): uses 3 m per 100 m instead of 3 m per 90 m.
B, D, E: arithmetic slips; only 6 m satisfies the 30 : 29 speed ratio.

Cetking shortcut: B beats C by 3 m in 90 m → 1 m per 30 m → 6 m in 180 m.

Answer: C

Approach: Net inflow = leak rate − pump rate; ship must reach shore before 68 t accumulates.

Solution:
Leak: (7/3) t in (13/2) min → 14/39 t/min → 840/39 = 280/13 t/h
Net inflow = 280/13 − 15 = 85/13 t/h
Time to sink = 68 ÷ 85/13 = 68 × 13/85 = 10.4 h
Speed = 156/10.4 = 15 km/h

Why the other options are wrong:
A/B/D/E: rounding the leak rate early or forgetting the pump gives wrong times (e.g. ignoring pump → 3.16 h).

Cetking shortcut: Keep fractions: 68 × 13/85 = 4 × 13/5 = 10.4 h exactly → 156/10.4 = 15.

Answer: C

Approach: Fixed distance — speed down, time up.

Solution:
Let the usual speed = s km/hr; scheduled time = 600/s.
600/(s − 200) − 600/s = 1/2.
600 × 200 = ½ × s(s − 200) ⇒ s² − 200s − 240000 = 0 ⇒ (s − 600)(s + 400) = 0 ⇒ s = 600 km/hr.
Scheduled duration of the flight = 600/600 = 1 hr (the slowed flight took 1 hr + 30 min = 1.5 hr).
The duration asked (scheduled flight time) = 1 hr.

Why the other options are wrong:
B (2 hrs): s = 300 ⇒ 600/100 − 2 = 4 hr extra, not 0.5.
C, D, E: likewise give extra time ≠ 30 min.

Cetking shortcut: Test options as the scheduled time t: 600/t − 600/(t + 0.5) must be 200 ⇒ t = 1: 600 − 400 = 200 ✓.

Answer: A

Approach: Races — head start and time margin.

Solution:
A's speed = 5 km/hr = 25/18 m/s ⇒ A's time for 100 m = 100 × 18/25 = 72 s.
B starts 8 m ahead ⇒ runs 92 m, and finishes 8 s after A ⇒ B's time = 80 s.
B's speed = 92/80 = 1.15 m/s = 1.15 × 18/5 = 4.14 km/hr.

Why the other options are wrong:
B, C, D, E: use 100 m or 72/80 s wrongly.

Cetking shortcut: B's speed = 5 × (92/100) × (72/80) = 5 × 0.92 × 0.9 = 4.14 km/hr.

Answer: A

Approach: Sum the distances of the three segments.

Solution: Segment times: 15 min = 1/4 h (speed x), 20 min = 1/3 h (speed 2x), 10 min = 1/6 h (speed x).
Distance = x/4 + 2x/3 + x/6 = (3x + 8x + 2x)/12 = 13x/12.
13x/12 = 39 → x = 36 km/h.

Cetking shortcut: Effective 'x-minutes' = 15 + 2×20 + 10 = 65 min at speed x → 39 km in 65 min → x = 39 × 60/65 = 36.

Answer: B

Approach: Distance = speed × time for each part; times in hours.

Solution:
45 min = 15 min at x + 20 min at 2x + 10 min at x.
Distance = x(15/60) + 2x(20/60) + x(10/60) = x/4 + 2x/3 + x/6 = 13x/12.
13x/12 = 39 → x = 39 × 12/13 = 36 km/h.

Cetking shortcut: Time at x is equivalent to 15 + 40 + 10 = 65 min at speed x → x = 39 × 60/65 = 36.

Answer: C

Approach: Use the motorbike's distance to A to find its speed, then add both distances.

Solution:
Motorbike: 1 pm → 3:40 pm = 8/3 h; car: 2 pm → 3:40 pm = 5/3 h.
Motorbike speed x: (8/3)x = 168 → x = 63 km/h; car = 126 km/h.
Car distance = 126 × 5/3 = 210 km.
AB = 168 + 210 = 378 km.

Cetking shortcut: Car covers 2 × (5/8) = 5/4 of the motorbike's distance: 168 × 5/4 = 210.

Answer: E

Approach: Relative speed for the catch time; a chaser at the thief's own speed never closes the gap.

Solution:
Head start of thief in 15 min = 60 × 0.25 = 15 km.
X closes at 65 − 60 = 5 km/h → catch time = 15/5 = 3 h after 12:15.
In those 3 h, X runs 195 km; the second policeman (60 km/h, same start time as X) runs 180 km.
Gap at the moment of catching = 195 − 180 = 15 km.

Cetking shortcut: The second policeman matches the thief's speed, so he stays exactly as far behind as the thief's head start: 15 km.

Answer: E

Approach: Convert both animals to the same distance unit (deer steps) and per-time rates.

Solution:
2 deer steps = 3 tiger steps → 1 tiger step = 2/3 deer step.
In the time the deer takes 5 steps (5 deer-steps of distance), the tiger takes 6 steps = 6 × 2/3 = 4 deer-steps.
Gain = 1 deer-step per 5 deer steps.
Gap = 40 deer-steps → deer needs 40 × 5 = 200 steps.

Cetking shortcut: Distance ratio per unit time = 5 : 4 → deer gains 1 part in 5; 40 × 5 = 200.

Answer: A

Approach: On a circular track, runners meet for the first time when the faster gains one full lap.

Solution:
Let slowest = x m/min, fastest = 2x m/min.
First meeting after 5 min: gain of one lap → (2x − x) × 5 = 1000 → x = 200 m/min.
Fastest = 400 m/min → 4000 m race takes 4000/400 = 10 minutes.

Cetking shortcut: In 5 min the fastest runs exactly 2 laps (the slow one 1), so 2000 m per 5 min → 4000 m in 10 min.

Answer: C

Approach: Use the meeting-point distance from A to find the bike's speed, then add the car's distance.

Solution: Let bike speed = x km/h; car speed = 2x.
Bike time (1:00 → 3:40 pm) = 2 h 40 min = 8/3 h. Meeting point is 168 km from A, so (8/3)x = 168 → x = 63 km/h.
Car time (2:00 → 3:40 pm) = 1 h 40 min = 5/3 h. Car distance = 126 × 5/3 = 210 km.
AB = 168 + 210 = 378 km.

Why the other options are wrong:
A, B, C, D: none matches 168 + 210; D (375) is a near-miss trap.

Cetking shortcut: Distances ∝ speed × time = 1 × 8 : 2 × 5 = 8 : 10. Bike's 8 parts = 168 → 1 part = 21 → total 18 parts = 378.

Answer: E

Approach: For a fixed distance, time is inversely proportional to speed.

Solution:
Time ratio = 1/5 : 1/4 : 1/6.
Multiply by LCM 60: 12 : 15 : 10.
So times are in the ratio 12 : 15 : 10.

Why the other options are wrong:
5 : 4 : 6: this is the speed ratio itself.
15 : 12 : 11, 12 : 10 : 13, 10 : 20 : 30: not the reciprocal ratio.

Cetking shortcut: Inverse of a : b : c = bc : ac : ab → 4×6 : 5×6 : 5×4 = 24 : 30 : 20 = 12 : 15 : 10.

Answer: E

Approach: Same-direction chase: gap closes at relative speed (difference of speeds).

Solution:
Relative speed = 20 − 15 = 5 km/h.
Gap 200 m = 0.2 km → time = 0.2 / 5 = 0.04 h.
Distance walked by the friend = 15 × 0.04 = 0.6 km = 600 m.

Why the other options are wrong:
900 m: no working gives this (Lala walks 800 m, the friend 600 m).
6 km / 0.6 m / 6 m: unit slips.

Cetking shortcut: Distance ratio = speed ratio: friend : Lala = 15 : 20 = 3 : 4, and the difference (1 part) = 200 m → friend = 3 × 200 = 600 m.

Answer: A

Approach: Find time for each leg, then total distance / total time.

Solution:
Leg 1: 50 km in 40 min
Leg 2: 50 km at 2 km/min = 25 min
Leg 3: 30 km at 1 km/min = 30 min
Total: 130 km in 95 min
Average speed = 130 × 60/95 = 7800/95 = 82.1 km/h

Cetking shortcut: 130/95 ≈ 1.368 km/min × 60 ≈ 82.1 — just above 82, which rules out 81.2 and 80.8.

Answer: E