Aptitude - Average - Discussion

First 10 overs they scored 32 runs====3.2*10.

Then we required to score 250 runs to get the score of 282 in 40 overs=======(28232 = 250 in 40 overs).

Therefore: 250/40 = 6.25.

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?

10*3.2 = 32.
40*? = ?
Required = 282.

In 10 overs run scored is 250, and in remaining 40 overs to get the total target of 282, remaining runs are 250 as 32 has been scored. Divide 250 by 40 that will be the required run-rate.

Sorry @Palani your point of view to check answer is wrong!

GAUTAM AND AYUSH you BOTH are RIGHT.

Gautam takes the sum of first two number is 11, and Ayush takes the average first two number is 11.

So both answer is correct.

Why we Divided 40 over because remaining over 30 and you also given the 10 over run ate?

@Aayush - Sorry to say but your solution is wrong.
@Aymen - The solution for your problem stated.

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.

i.e) X+Y=11.
Z=13.

HENCE THE AVG X+Y+Z/3=8.

SO, Gautam IS RIGHT YOU'RE WRONG Ayush.

The total target for 50 overs is 282 out of that first 10 overs they scored at The run rate of 3.2, which means they scored 32,
For remaining 40 overs the target is 282-32=250,

For 40 overs the target is 250.

Then for 1 over it is 250/40 = 6.25.

@Gautam - Sorry to say but your solution is wrong.
@Aymen - The solution for your problem stated.

"The average of three numbers is 8, that of the first two is 11. Find the third number?please help me." is:

Let three numbers be x,y,z.
Average of first three numbers is 8.
Hence, x+y+z/3 = 8.
=> x+y+z = 24.....eq(1).

Average of first two numbers is = 11.
Hence, x+y/2 = 11.
=> x+y = 22....eq(2).

Now substitute eq(2) value in eq(1).
Thus, 22+z = 24.
=> z = 24-22 = 2.
Answer is z = 2.
:).

What is the formula for this sum?