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:
135 comments Page 4 of 14.
Mounika said:
8 years ago
Am not understanding this, Please explain in simple way.
Ram said:
8 years ago
Thanks for the explanation @Deepa.
Seeramsettykavya said:
8 years ago
What is the logic behind this?
Shravya said:
8 years ago
Why do we minus 282-(3.2*10)?
Kavya said:
8 years ago
Super & excellent explanation @Muhundhan.
Sh g said:
8 years ago
Thanks @Pratik.
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).
(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).
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.
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.
Deepak said:
8 years ago
10 overs was completed with a run rate of 3.2 for over ....which means they scored 3.2 runs in single overs there for 3.2 x 10 = 32.
Target was 282 so subtract 282 by 32 therefore 282-32 = 250.
250 should be scored in 40 overs.
So, divide 250 by 40.
250/40 = 6.25.
Target was 282 so subtract 282 by 32 therefore 282-32 = 250.
250 should be scored in 40 overs.
So, divide 250 by 40.
250/40 = 6.25.
Malini said:
9 years ago
The average weight of 8 men is increased by 1.5 kg when one of the men who weighs of 9 mangoes increases by 20g, if one of them weighing 120g, is replaced by another. The weight of the new mango is?
Please help me to solve the problem.
Please help me to solve the problem.
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