# 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:

126 comments Page 1 of 13.
Shreya Dakhale said:
4 months ago

Total Runs in 10 overs/number of overs = run rate.

Total Runs in 10 overs/10 = 3.2.

Total Runs in 10 overs = 32.

Scores to achieve - Total Runs in 10 overs.

282 - 32 = 250.

250/40 = run rate.

So, run rate = 6.25.

Total Runs in 10 overs/10 = 3.2.

Total Runs in 10 overs = 32.

Scores to achieve - Total Runs in 10 overs.

282 - 32 = 250.

250/40 = run rate.

So, run rate = 6.25.

(28)

Ahmed Raafat said:
6 months ago

Run rate = number of runs/number of overs.

(9)

ANVITHA said:
3 years ago

Runs scored= run rate*overs.

= 3.2*10,

= 32.

In remaining 40 overs they score 250 runs.

Runs scored=run rate * overs.

250 = run rate * 40.

Run rate = 250/40.

= 6.25.

= 3.2*10,

= 32.

In remaining 40 overs they score 250 runs.

Runs scored=run rate * overs.

250 = run rate * 40.

Run rate = 250/40.

= 6.25.

(92)

Jyothis said:
3 years ago

Assume run rate as average,

Then, Avg of first 10 overs = 3.2.

Avg(10) = total runs for 10overs/no.of overs.

3.2=total runs/10.

Total runs for 10 overs = 3.2 * 10 = 32,

Remaining runs=282-32 = 250,

Avg runs remaining 40 overs = 250/40 = 6.25.

Then, Avg of first 10 overs = 3.2.

Avg(10) = total runs for 10overs/no.of overs.

3.2=total runs/10.

Total runs for 10 overs = 3.2 * 10 = 32,

Remaining runs=282-32 = 250,

Avg runs remaining 40 overs = 250/40 = 6.25.

(33)

Manalan said:
3 years ago

First 10 over,

3.2 = 10 * 3.2 = 32runs.

40 overs is 282 = 282-32.

= 250.

250 Ã· 40 = 6.25,

6.25.

3.2 = 10 * 3.2 = 32runs.

40 overs is 282 = 282-32.

= 250.

250 Ã· 40 = 6.25,

6.25.

(8)

Balaji said:
3 years ago

First 10 overs run rate 3.2=3.2 * 10 = 32.

Next, 40 overs is 282 = 282 - 32 = 250.

250 Ã· 40 overs = 6.25.

So, the answer is 6.25.

Next, 40 overs is 282 = 282 - 32 = 250.

250 Ã· 40 overs = 6.25.

So, the answer is 6.25.

(5)

Masthan said:
4 years ago

10 * 3.2 = 32

40 * X = 282

=> 282 - 32 = 250.

40X = 250.

X = 250/40.

X = 6.25.

40 * X = 282

=> 282 - 32 = 250.

40X = 250.

X = 250/40.

X = 6.25.

(9)

Suchi said:
4 years ago

Use the Average formula :

Average = sum of observations/number of observations.

For first 10 overs :

3.2 = Sum/10 --> Sum = 32--> Run in first 10 overs.

The total score is 282. First 10 overs, we scored 32 runs. So, remaining (282-32) runs=250 runs in 40 overs.

Applying the formula:

Average = sum of observations/number of observations.

Average= 250/40 = 6.25.

Average = sum of observations/number of observations.

For first 10 overs :

3.2 = Sum/10 --> Sum = 32--> Run in first 10 overs.

The total score is 282. First 10 overs, we scored 32 runs. So, remaining (282-32) runs=250 runs in 40 overs.

Applying the formula:

Average = sum of observations/number of observations.

Average= 250/40 = 6.25.

(13)

Shreya said:
4 years ago

In a cricket match, Ashwin scored 40perct. Of total run scored by his team & Arijit scored 140 run which is 200/9perct less than Ashwin's score, find total run scored by Ashwin's team, Anyone please answer it.

(3)

Rahul Vishwakarma said:
4 years ago

Left over = 40-10 = 30overs.

So, required run rate = (target /left over ) - current run rate.

= (282/30)- 3.2.

= 6.2.

So, required run rate = (target /left over ) - current run rate.

= (282/30)- 3.2.

= 6.2.

(2)

Post your comments here:

Quick links

Quantitative Aptitude

Verbal (English)

Reasoning

Programming

Interview

Placement Papers