Practice Quantitative Aptitude questions on Races and Games, covering race distances, speed comparisons, head starts, winning margins, relative speeds, and problems involving points, matches, and game outcomes. Takes about 20–25 minutes and is useful for placement tests, TCS NQT, SSC, banking and MBA entrance exams.
Question 1
If in a race of 80m, A covers the distance in 20 seconds and B in 25 seconds, then A beats B by:
Explanation
B's speed = 8025 m/s. In the 20 seconds A takes to finish, B covers 8025 × 20 = 64 m. A beats B by 80 – 64 = 16 m.
Question 2
If X can run 48m and Y 42m, then in a race of 1km, X beats Y by:
Explanation
When X covers 1000 m, Y covers 1000 × 4248 = 875 m. X beats Y by 1000 – 875 = 125 m.
Question 3
In a race of 800m, if the ratio of the speeds of two participants P and R is 3:4, and P has a start of 120m, then R beats P by:
Explanation
Since P has a 120m start, P only needs to cover 800 – 120 = 680 m. When R covers 800 m, P (at 34 R's speed) covers 800 × 34 = 600 m, which is 680 – 600 = 80 m short of finishing. R beats P by 80 m.
Question 4
If in a race of 120m X can beat Y by 20m and Z by 35m, then Y can beat Z by:
Explanation
When X runs 120 m, Y runs 100 m and Z runs 85 m (at the same instant). Scaling Y up to 120 m: Z = 120 × 85100 = 102 m. Y beats Z by 120 – 102 = 18 m.
Question 5
If in a game of 80, P can give 16 points to Q and R can give 20 points to P, then in a game of 150, how many points can R give to Q?
Explanation
Speed ratios: P:Q = 80:64 = 5:4, and R:P = 80:60 = 4:3. Combining, R:P:Q = 20:15:12, so R:Q = 5:3. When R = 150, Q = 150 × 35 = 90. Points given by R to Q = 150 – 90 = 60.
Question 6
A 100 m race is organised. X covers the distance in 36 seconds while Y covers it in 45 seconds. By how many metres does X beat Y?
Explanation
When X finishes (at 36 sec), Y has covered (10045)x36 = 80 m. X beats Y by 100-80=20 m.
Question 7
A can run 22.5 m while B runs 25 m. In a race of 1 km, by how many meters can B beat A?
Explanation
Speed ratio A:B = 22.5:25 = 9:10. When B covers 1000m, A covers (910)x1000=900m. B beats A by 100m.
Question 8
A and B run a 1560 m race ending in a dead heat. At first, the speeds of A and B are in the ratio of 7:5. B then quickens his pace and for the remainder of the race runs 40 percent faster than A. How many metres did A run before B quickened his pace?
Explanation
Let A run d metres before B speeds up. Up to then the speeds are 7 : 5, so B has run 5d7 m.
After that B runs 40% faster than A: 1.4 × 7 = 9.8 against A's 7.
Both finish together: d7 + 1560 − 5d/79.8 = 15607, so 9.8d + 10920 − 5d = 15288, 4.8d = 4368 and d = 910 m.
After that B runs 40% faster than A: 1.4 × 7 = 9.8 against A's 7.
Both finish together: d7 + 1560 − 5d/79.8 = 15607, so 9.8d + 10920 − 5d = 15288, 4.8d = 4368 and d = 910 m.
Question 9
A and B started running simultaneously from a point P in opposite directions along a circular track of radius 7 m. After every meeting, they exchange their speeds as well as directions. B started running with a speed thrice that of A. They meet at point Q for the first time, at point R for the second time, at point S for the third time, and so on. At how many different points on the track do they meet?
Explanation
Swapping both speed and direction at each meeting is equivalent (from an outside observer's view) to the two runners simply passing through each other unchanged. With speed ratio 1:3, the two velocity profiles going around the circle coincide at exactly 4 distinct points before repeating.
Question 10
In a race, A beats B by 15 metres and C by 29 metres. If B and C run over the course together, B wins by 15 metres. What is the length of the course?
Explanation
Let the length of the course be L metres. When A finishes, B has run L - 15 and C has run L - 29, so the speeds of B and C are in the ratio (L - 15) : (L - 29).
When B and C run the course together, B wins by 15 m, so when B runs L, C runs L - 15: L - 15L = L - 29L - 15.
(L - 15)² = L(L - 29), so L² - 30L + 225 = L² - 29L and L = 225 m.
When B and C run the course together, B wins by 15 m, so when B runs L, C runs L - 15: L - 15L = L - 29L - 15.
(L - 15)² = L(L - 29), so L² - 30L + 225 = L² - 29L and L = 225 m.
Question 11
Sam, Suresh and Sanket run a race of 1400 meters. Sam runs twice as fast as Suresh while Suresh runs thrice as fast as Sanket. If Sanket finishes the race by clocking 15 minutes, how far was Suresh from the finish line when Sam must have finished the race?
Explanation
Let Sanket's speed be v. Suresh runs at 3v and Sam at 6v. Sanket takes 15 minutes, so v = 140015 m per minute.
Sam takes 156 = 2.5 minutes. In that time Suresh covers 3 × 140015 × 2.5 = 700 m.
Suresh is 1400 - 700 = 700 m from the finish line.
Sam takes 156 = 2.5 minutes. In that time Suresh covers 3 × 140015 × 2.5 = 700 m.
Suresh is 1400 - 700 = 700 m from the finish line.
