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Races And Games - Aptitude MCQ Questions and Solutions with Explanations

Races And Games MCQ Questions and answers with easy and logical explanations.Arithmetic Ability provides you all type of quantitative and competitive aptitude mcq questions on Races And Games with easy and logical explanations. Races And Games MCQ is important for exams like Banking exams,IBPS,SCC,CAT,XAT,MAT etc. Page-4 section-1

31
Amit and Beenu can cover a 400 m race in 44 seconds and 50 seconds, respectively. When Amit finished the race, then Beenu is at what distance from the finishing line?
Discuss
Answer & Solution
Answer: Option A
Solution:
Beenu,
50 sec → 400 m
1 sec → 8 m
44 sec → 44 × 8
Amit,
Races and Games mcq question image
32
In a 1000 m race, Arjun, Balaji and Charan are running. Arjun beats Balaji by 100 m, and Balaji beats Charan by 100 m. In the next 1000 m race (the speeds are the same as in the previous), Balaji gives Charan a head start of 100 m, and Arjun gives Balaji a head start of 100 m. Find the distance by which the winner is ahead of the person just behind him.
Discuss
Answer & Solution
Answer: Option A
No explanation is given for this question. Let's Discuss on Board
33
In a linear race of 1000 m, A beats B by 50 m or 5 seconds. What is the difference between the speeds (in m/s) of A and B?
Discuss
Answer & Solution
Answer: Option B
Solution:
Races and Games mcq question image
$$\eqalign{ & {\text{Speed of B}} = \frac{{50}}{5} = 10{\text{ m/sec}} \cr & {\text{Time same}} \cr & {\text{Then,}} \cr} $$
Races and Games mcq question image
$${\text{Difference in speed}} = \frac{{200}}{1} - \frac{{10}}{1} = \frac{{10}}{{19}}$$
34
Geeta runs $$\frac{5}{2}$$ times as fast as Babita. In a race, if Geeta gives a lead of 40 m to Babita, find the distance from the starting point where both of them will meet (correct up to two decimal places).
Discuss
Answer & Solution
Answer: Option A
Solution:
$$\eqalign{ & {\text{Babita's speed}} = 1 \cr & {\text{Geeta's speed}} = \frac{5}{2} \cr} $$
Races and Games mcq question image
$$\eqalign{ & \Rightarrow {\text{Relative speed in same direction}} \cr & = \frac{5}{2} - 1 = \frac{3}{2} \cr & \Rightarrow {\text{Time}} = \frac{{40}}{{\frac{3}{2}}} = \frac{{80}}{3} = 26.66 \cr & {\text{Distance travel by Babita}} \cr & = 1 \times 26.66\, - - - - - \,26.66 \cr & {\text{Total distance}} = 40 + 26.66 \cr & = 66.66 \cr & = 66.67 \cr} $$
35
In a race of 800 m, Ram can beat Shyam by 50 m, and in a race of 600 m, Shyam can beat Ghanshyam by 40 m. By how many metres will Ram beat Ghanshyam in a race of 400 m?
Discuss
Answer & Solution
Answer: Option B
No explanation is given for this question. Let's Discuss on Board
36
In a race, a cyclist covers a distance of 500 m in 625 sec in the first lap. He covers the second lap of the same length in 375 sec. What is the average speed (in m/sec) of the cyclist?
Discuss
Answer & Solution
Answer: Option B
Solution:
$$\eqalign{ & {\text{Average speed}} = \frac{{{\text{Total distance}}}}{{{\text{Total time}}}} \cr & = \frac{{500 + 500}}{{625 + 375}} \cr & = \frac{{1000}}{{1000}} \cr & = 1{\text{ m/s}} \cr} $$
37
In a 1500 m race, if vehicle P gives vehicle Q a start of 200 m, then vehicle P wins the race by 8 sec. Alternatively, if vehicle P gives vehicle Q a start of 400 m, the race ends in a dead heat. How long does vehicle P take to run 1500 m?
Discuss
Answer & Solution
Answer: Option A
Solution:
$$\eqalign{ & \frac{{1300}}{{\text{Q}}} - \frac{{1500}}{{\text{P}}} = 8{\text{ }}{\text{. }}{\text{. }}{\text{. }}{\text{. }}\left( {\text{i}} \right) \cr & {\text{and, }}\frac{{1100}}{{\text{Q}}} = \frac{{1500}}{{\text{P}}}{\text{ }}{\text{. }}{\text{. }}{\text{. }}{\text{. }}\left( {{\text{ii}}} \right) \cr & \frac{{\text{P}}}{{\text{Q}}} = \frac{{15}}{{11}} = \frac{{15x}}{{11x}} \cr & {\text{From equation }}\left( {\text{i}} \right) \cr & \frac{{1300}}{{11x}} - \frac{{1500}}{{15x}} = 8 \cr & 19500 - 16500 = 8 \times 11x \times 15x \cr & 3000 = 8 \times 11 \times 15x \cr & x = \frac{{25}}{{11}} \cr & {\text{Time of P to travel 1500m}} = \frac{{1500}}{{15x}} \cr & = \frac{{1500 \times 11}}{{15 \times 25}} \cr & = 44\sec \cr & \cr & {\bf{Alternate}}\,{\bf{solution:}} \cr} $$
Races and Games mcq question image
$$\eqalign{ & 11\mu \to 25 \cr & 1\mu \to \frac{{25}}{{11}} \cr & 15\mu \to \frac{{15 \times 25}}{{11}} \cr & {\text{Speed of Q}} = \frac{{200}}{8} = 25 \cr & {\text{Time of P}} = \frac{{1500 \times 11}}{{15 \times 25}} = 44\sec \cr} $$
38
In a 100 m race, A beats B by 20 m and B beats C by 20 m. By how much distance does A beat C.
Discuss
Answer & Solution
Answer: Option D
Solution:
Races and Games mcq question image
$${\text{A beat C by}} = \frac{{100}}{{25}} \times 9 = 36\,{\text{m}}$$
39
A gives B a head-start of 10 seconds in a 1500 m race and both finish the race at the same time. What is the time taken by A (in minutes) to finish the race if speed of B is 6 m/s?
Discuss
Answer & Solution
Answer: Option B
Solution:
B's speed = 6 m/s
B's distance in 10 second = 10 × 6 = 60 m
\[\begin{array}{*{20}{c}} {}&{\text{A}}&{\text{B}} \\ {{\text{Distance}} \to }&{1500}&{1440} \\ {{\text{Speed}} \to }&x&6 \\ {{\text{Time}} \to }&{240{\text{ second}}}&{4\min } \end{array}\]
∵ Time same
Time of B = Time of A
Time of A = 4 min
40
In a 1200 m race, bike A beats bike B by 100 m. Bike B beats bike C by 100 m in a 600 m race. If bike A beats bike C by 30 sec in a 720 m race, then what is the speed of bike C?
Discuss
Answer & Solution
Answer: Option A
Solution:
Races and Games mcq question image
\[\begin{array}{*{20}{c}} {\text{B}}&{\text{C}} \\ {600}&{500} \\ {600\,{\text{m}}}&{{\text{Race}}} \end{array}\]
$${\text{Speed}} = \frac{{170}}{{30}} = \frac{{17}}{3}{\text{m/s}}$$