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 60 Hence, when R scores 150, Q scores (60 * 64 * 150) / (80 * 80) = 90 Therefore, 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 = 875m X 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 = 16m Hence, 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 course
according to the question, A beats B by 15 metres and C by 29 metres
Applying this, If A runs X metres, B runs (X15) and C runs ( X29) metres.
B and C together run, B runs X metres and C runs (X15)
=> If B runs 1 metre, C runs X15/X
=> B runs X15 m, C runs (X15/x) * (X15)
APPLYING THIS, X29 = (X15)(X15) / X
Solving,
X^{2} 29X= X^{2} – 30 X + 225 X = 225
Thus, 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 seconds
Therefore, Distance covered by P in that time= 8m/s x 4 seconds = 32 metres

