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?
6.25
6.5
6.75
7
Answer: Option
Explanation:

Required run rate = 282 - (3.2 x 10) = 250    = 6.25
40 40

Video Explanation: https://youtu.be/GhK9d8tcqvA

Discussion:
135 comments Page 13 of 14.

RIDDHI said:   8 years ago
282 ie the target run.
10 over + 40 over = 50 over ie in 50 overs 282 should be made in order to win.
So run rate in 50 over needed is 282/50 = 5.64.

USING METHOD OF DEVATION.
3.2--5.64= --2.44*10 = --24.4......deviation for 10 overs (equ 1).
(Let X be avg for next 40 overs).
(X--5.64)*40 = 40X--225.6.......deviation for 40 overs.
(Equ 2)
Solving equ 1 and 2 we get,
40X--255.6--24.4 = 40X--250.
X = 250/40 = 6.25 run rate required in 40 overs.

Kaviya said:   8 years ago
Sum(h+w+c)three years ago:

(x-3)+(y-3)+(y-3)=27*3,
x+y+z=90....................eqn(1),
sum(w+c)five years ago:
(y-5)+(z-5)=20*2.
y+z=50.........................eqn(2).
simplify 1&2.
x=40.............................(ans).

Sh g said:   8 years ago
Thanks @Pratik.

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:   8 years ago
Why .6.25?

BISWAJIT said:   8 years ago
Not understanding this, please explain.


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