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:
134 comments Page 8 of 14.

Padmashri said:   1 decade ago
RR runs Target-282.

1-10 overs 3.2 32.

11-40 overs ? 250.

RR = Runs/Overs.

RR = 250/40.

Answer: 6.25.

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.

Manalan said:   4 years ago
First 10 over,
3.2 = 10 * 3.2 = 32runs.
40 overs is 282 = 282-32.
= 250.

250 ÷ 40 = 6.25,
6.25.
(9)

Colbie said:   1 decade ago
Mr.A average on 4 test is 80. what does he need on his fifth test to raise his average to 84?

Ravi said:   1 decade ago
We need to calculate remaining average so we have to deduct score whick is already scored.

Pradeep said:   1 decade ago
Total run in 10 over=10*3.2=32
Required run in 40 over=282-32=250
Hence avg=250/40 =6.25

Manish said:   1 decade ago
Why we Divided 40 over because remaining over 30 and you also given the 10 over run ate?

Vasavi said:   1 decade ago
@Apple.

You give example test match in that problem 80*4 = 320 but you tack it as 360?

Rahul said:   9 years ago
x = req run rate.
(10 * 3.2) + (40 * x) = 282,
32 + 40x = 282,
40x = 250,
x = 6.25.

Srijan said:   9 months ago
I understand the solution with the help of the given explanation. Thanks, everyone.
(24)


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