Aptitude - Average - Discussion
Discussion Forum : Average - General Questions (Q.No. 1)
1.
In the first 10 overs of a cricket game, the run rate was only 3.2. What should be the run rate in the remaining 40 overs to reach the target of 282 runs?
Answer: Option
Explanation:
| Required run rate = |  |
282 - (3.2 x 10) |  |
= | 250 | = 6.25 |
| 40 | 40 |
Video Explanation: https://youtu.be/GhK9d8tcqvA
Discussion:
134 comments Page 11 of 14.
Kavya said:
8 years ago
Super & excellent explanation @Muhundhan.
Shravya said:
8 years ago
Why do we minus 282-(3.2*10)?
Seeramsettykavya said:
8 years ago
What is the logic behind this?
Ram said:
8 years ago
Thanks for the explanation @Deepa.
Mounika said:
8 years ago
Am not understanding this, Please explain in simple way.
Abhishek said:
7 years ago
Why .6.25?
BISWAJIT said:
7 years ago
Not understanding this, please explain.
Nigus said:
7 years ago
@Biswaji.
it means ; 10x3.2+40x(x)=282.
so,
32+40x=282,
40x=282-32,
40x=250,
x=250/40,
x=6.25.
it means ; 10x3.2+40x(x)=282.
so,
32+40x=282,
40x=282-32,
40x=250,
x=250/40,
x=6.25.
Shruti godi said:
7 years ago
First 10 overs ----> 3.2 runrate.
Remaining 40 overs ----> x runrate.
(10*3.2)+(40*x) = 282,
32+40x = 282,
X = 25/4,
X = 6.25.
Remaining 40 overs ----> x runrate.
(10*3.2)+(40*x) = 282,
32+40x = 282,
X = 25/4,
X = 6.25.
(2)
Yeri Tazu said:
7 years ago
For 1st 10 over:
3.2=(y/10).
Let y be the run.
y= 32 runs in 10 over.
For next 40 over:
x=(y+282)/50.
This match is talking about 1 day match which is of 50 overs. Also, 10+40=50.
Put value of 'y'
x=(32+282)/50,
x=6.28~6.25(ans).
Hope you understand.
3.2=(y/10).
Let y be the run.
y= 32 runs in 10 over.
For next 40 over:
x=(y+282)/50.
This match is talking about 1 day match which is of 50 overs. Also, 10+40=50.
Put value of 'y'
x=(32+282)/50,
x=6.28~6.25(ans).
Hope you understand.
Post your comments here:
Quick links
Quantitative Aptitude
Verbal (English)
Reasoning
Programming
Interview
Placement Papers