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In the first 10 overs of a cricket game, the run rate was only 3.2. Wh

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In the first 10 overs of a cricket game, the run rate was only 3.2. Wh  [#permalink]

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New post 21 Feb 2019, 00:42
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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?

A. 6
B. 6.25
C. 6.5
D. 7
E. 7.25

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Re: In the first 10 overs of a cricket game, the run rate was only 3.2. Wh  [#permalink]

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New post 21 Feb 2019, 03:50
Bunuel wrote:
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?

A. 6
B. 6.25
C. 6.5
D. 7
E. 7.25


In 10 overs total run = Runrate * number of overs = 3.2*10 = 32

Remaining runs to score = 282 - 32 = 250

Run Rate for remaining 40 overs = 250/40 = 6.25

Answer: Option B
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Re: In the first 10 overs of a cricket game, the run rate was only 3.2. Wh  [#permalink]

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New post 21 Feb 2019, 09:25
Bunuel wrote:
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?

A. 6
B. 6.25
C. 6.5
D. 7
E. 7.25



run scored in 1-10 overs ; 32
and balance score to reach target = 282-32 ; 250
and overs 40 ; avg reqd ; 250/40 ; 6.25 IMO B
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Re: In the first 10 overs of a cricket game, the run rate was only 3.2. Wh  [#permalink]

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New post 24 Feb 2019, 06:45
Bunuel wrote:
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?

A. 6
B. 6.25
C. 6.5
D. 7
E. 7.25


Letting n = run rate of the remaining 40 overs, we can create the equation:

10(3.2) + 40n = 282

32 + 40n = 282

40n = 250

n = 250/40 = 25/4 = 6.25

Answer: B
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Re: In the first 10 overs of a cricket game, the run rate was only 3.2. Wh   [#permalink] 24 Feb 2019, 06:45
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