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:
137 comments Page 7 of 14.
Suman said:
10 years ago
Run = (no of runs took in one over)/(total possible no of runs),
So, instead of 3.2 = x/10, it should be 3.2 = x/(10 * 6),
Since, 1 over has 6 runs?
So, instead of 3.2 = x/10, it should be 3.2 = x/(10 * 6),
Since, 1 over has 6 runs?
Bhargavi said:
9 years ago
A batsman makes a score of 87runs in the 17th inning & thus increases his avg by 3. Find his avg after 17th inning? Can anyone solve this problem?
Karthi said:
1 year ago
First 10 - run rate 3.2
3.2 = sum of runs/10
Sum = 3.2×10 = 32
40 over total runs 282.
So, 282 - 32 = 250 remaining runs.
Run rate = 250/40 = 6.25.
3.2 = sum of runs/10
Sum = 3.2×10 = 32
40 over total runs 282.
So, 282 - 32 = 250 remaining runs.
Run rate = 250/40 = 6.25.
(52)
Siva said:
9 years ago
@Chahat.
First, 15 overs team scored 120 runs at rate 6.2 and 183 runs need to be scored to reach 303 in 35 overs so req rate is 183/35 = 5.22.
First, 15 overs team scored 120 runs at rate 6.2 and 183 runs need to be scored to reach 303 in 35 overs so req rate is 183/35 = 5.22.
Tintu mathew said:
1 decade ago
The simple method is
total runs takes in 10 over=3.2*10=32
we know dat target is 282
so
32+40*x=282
from this we get required run rate.
total runs takes in 10 over=3.2*10=32
we know dat target is 282
so
32+40*x=282
from this we get required run rate.
GAUTAM said:
1 decade ago
THE AVERAGE OF THREE NO.IS
X+Y+Z/3=8
THE SUM OF FIRST TWO NO. IS 11,SO
(X+Y)+Z/3=8
11+Z/3=8
BY TRANSPOSITION
11+Z=3*8
11+Z=24
Z=24-11
Z=13
Z
X+Y+Z/3=8
THE SUM OF FIRST TWO NO. IS 11,SO
(X+Y)+Z/3=8
11+Z/3=8
BY TRANSPOSITION
11+Z=3*8
11+Z=24
Z=24-11
Z=13
Z
Rashmi said:
1 decade ago
Is there any short cut for this?
If we solve this way like long method, rest of the questions will be pending. Any one suggest me?
If we solve this way like long method, rest of the questions will be pending. Any one suggest me?
Balaji said:
5 years ago
First 10 overs run rate 3.2=3.2 * 10 = 32.
Next, 40 overs is 282 = 282 - 32 = 250.
250 ÷ 40 overs = 6.25.
So, the answer is 6.25.
Next, 40 overs is 282 = 282 - 32 = 250.
250 ÷ 40 overs = 6.25.
So, the answer is 6.25.
(6)
Shruti godi said:
8 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)
Muhundhan said:
2 decades ago
So we calculate the runs in the first 10 overs
first 10 overs they got 32 runs
remaining 40 overs they got 250
250 / 40 =6.25
first 10 overs they got 32 runs
remaining 40 overs they got 250
250 / 40 =6.25
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