# Races and games - Quantitative Aptitude (MCQ) questions

Dear Readers, Welcome to Quantitative Aptitude Races and games questions and answers with explanation. These Races and games solved examples with shortcuts and tricks will help you learn and practice for your Placement Test and competitive exams like Bank PO, IBPS PO, SBI PO, RRB PO, RBI Assistant, LIC,SSC, MBA - MAT, XAT, CAT, NMAT, UPSC, NET etc.

After practicing these tricky Races and games multiple choice questions, you will be exam ready to deal with any objective type questions.

1)   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?
- Published on 06 Jul 15

a. 48
b. 60
c. 54
d. 90
 Answer  Explanation ANSWER: 60 Explanation: When P scores 80, Q scores 64.When R scores 80, P scores 60Hence, when R scores 150, Q scores (60 * 64 * 150) / (80 * 80) = 90Therefore, in a game of 150, R can give 60 points to Q.

2)   If X can run 48m and Y 42m, then in a race of 1km, X beats Y by:
- Published on 06 Jul 15

a. 140m
b. 125m
c. 100m
d. 110m
 Answer  Explanation ANSWER: 125m Explanation: When X runs 48m, Y runs 42m.Hence, when X runs 1000m, Y runs (1000 * 42) / 48 = 875mX beats Y by 125m.

3)    If in a race of 80m, A covers the distance in 20 seconds and B in 25 seconds, then A beats B by:
- Published on 06 Jul 15

a. 20m
b. 16m
c. 11m
d. 10m
 Answer  Explanation ANSWER: 16m Explanation: The difference in the timing of A and B is 5 seconds. Hence, A beats B by 5 seconds.The distance covered by B in 5 seconds = (80 * 5) / 25 = 16mHence, A beats B by 16m.

4)   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?
- Published on 15 Jun 15

a. 225m
b. 120m
c. 220m
d. 160m
 Answer  Explanation ANSWER: 225m Explanation: Let X be the length of the courseaccording to the question, A beats B by 15 metres and C by 29 metresApplying this, If A runs X metres, B runs (X-15) and C runs ( X-29) metres.B and C together run, B runs X metres and C runs (X-15)=> If B runs 1 metre, C runs X-15/X=> B runs X-15 m, C runs (X-15/x) * (X-15)APPLYING THIS, X-29 = (X-15)(X-15) / XSolving,X2 -29X= X2 – 30 X + 225X = 225Thus, length of the course is 225 metres.

5)   For a race a distance of 224 meters can be covered by P in 28 seconds and Q in 32 seconds. By what distance does P defeat Q eventually?
- Published on 19 Oct 15

a. 26m
b. 32m
c. 24m
d. 28m
 Answer  Explanation ANSWER: 32m Explanation: This is a simple speed time problem. Given conditions: =>Speed of P= 224/28 = 8m/s=>Speed of Q=224/32 = 7m/s=>Difference in time taken = 4 secondsTherefore, Distance covered by P in that time= 8m/s x 4 seconds = 32 metres

1