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Now let us look at Example 2 where the numbers are not comfortable.

Example 2:

Average of 75 students is 82, out of which average of 42 students is 79.
What is the average of the remaining 33 students?

Average of 75 members = 82.

Two groups of 42 and 33 and we want each group to have an average of 82.

But the first group i.e. 42 students they had an average of 79.
We fell short by 3 in the average.
So the loss in the sum= 3x42.

So the average of the second group i.e. 33 students must be 82 + (3x42)/33 = 82+ 42/11 = 85.82

I am sure this is much faster than what we did earlier.

THIS QUESTION IS DISCUSSED HERE.
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Problem 5:

The average of batsmen up to certain number of innings was 45. In the next inning he was out for a duck and his average reduced to 40.5. Find the total number of innings played by him including the latest inning.

Sol:

Again, if the batsmen had scored 45 in his latest inning his average would remain in tact i.e. it would have been 45. But he scored 0 runs.

Since he did not score 45, he lost an average of 4.5 upon the total number of innings, which includes the latest innings as well.

So 45 runs upon x innings will give an average of 4.5

Hence 45/x = 4.5 i.e. x = 10. (Remember, x includes the latest inning as well)

THIS QUESTION IS DISCUSSED HERE.


Problem 6:

The average of 8 scores is 87. Of these, the highest is 2 more than the one next in magnitude. If these two scores are eliminated the average of the remaining score is 85. What was the highest score?

Sol:

If the sum of the top two scores taken out is 87+87 i.e. 174, the average of the remaining 6 scores will remain 87.
But, since the average is reduced by 2 upon 6 scores, the sum of the top two scores taken out is 174+12 = 186.
So the top two scores are 94 and 92.

THIS QUESTION IS DISCUSSED HERE.
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Problem 7:

The average age of a committee of 8 members is 40 years. A member, aged 55 years, retired and he was replaced by a member aged 39 years. The average age of the present committee is?

Sol:

If the age of the new person who is replacing the retired person is also 55 years, the average would remain same i.e. 40 years. But since the age of the new person is 16 years less than that of the retired person, the loss in the sum is 16 years.

This loss of 16 years in the sum will result in a loss of 16/8 = 2 yrs in the average (remember that there is no change in the total strength).

So the present average = 40-2 = 38 yrs.

THIS QUESTION IS DISCUSSED HERE.


Problem 8:

The average of marks obtained by 120 candidates in a certain examination is 35. If the average of passed candidates is 39 and that of the failed candidates is 15. The number of candidates who passed the examination is?

Sol:

Folks, look at the relative calculation here.

If we consider that all the 120 candidates passed the examination, the average of the entire class must be 39. But it is given that average of the entire class is 35.

So we are getting an extra of 4 in the average i.e. an extra of 4x120=480 in the total sum.

We got extra total sum because some students who failed were also considered to be passed. For every one student considered as passed who actually failed we get 39-15 = 24 marks extra.

Since we got 480 marks extra, total number of failed students who were considered as passed = 480/24 = 20.

Hence the total number of students who passed = 100

You can solve this by assuming that all the 120 failed the examination but this would take a bit extra time.

THIS QUESTION IS DISCUSSED HERE.


Folks, most of the questions solved here can also be solved quickly by using a technique called Alligation. I will soon come up with a tutorial on this topic and I will discuss these questions in that tutorial besides some other questions as well. So stay tuned....
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cicerone
Problem 5:

The average of batsmen up to certain number of innings was 45. In the next inning he was out for a duck and his average reduced to 40.5. Find the total number of innings played by him including the latest inning.

Sol:

Again, if the batsmen had scored 45 in his latest inning his average would remain in tact i.e. it would have been 45. But he scored 0 runs.

Since he did not score 45, he lost an average of 4.5 upon the total number of innings, which includes the latest innings as well.

So 45 runs upon x innings will give an average of 4.5

Hence 45/x = 4.5 i.e. x = 10. (Remember, x includes the latest inning as well)

Problem 6:

The average of 8 scores is 87. Of these, the highest is 2 more than the one next in magnitude. If these two scores are eliminated the average of the remaining score is 85. What was the highest score?

Sol:

If the sum of the top two scores taken out is 87+87 i.e. 174, the average of the remaining 6 scores will remain 87.
But, since the average is reduced by 2 upon 6 scores, the sum of the top two scores taken out is 174+12 = 186.
So the top two scores are 94 and 92.

Can you please explain how you arrived at 94 and 92
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cicerone
Problem 7:

The average age of a committee of 8 members is 40 years. A member, aged 55 years, retired and he was replaced by a member aged 39 years. The average age of the present committee is?

Sol:

If the age of the new person who is replacing the retired person is also 55 years, the average would remain same i.e. 40 years. But since the age of the new person is 16 years less than that of the retired person, the loss in the sum is 16 years.

This loss of 16 years in the sum will result in a loss of 16/8 = 2 yrs in the average (remember that there is no change in the total strength).

So the present average = 40-2 = 38 yrs.

Problem 8:

The average of marks obtained by 120 candidates in a certain examination is 35. If the average of passed candidates is 39 and that of the failed candidates is 15. The number of candidates who passed the examination is?

Sol:

Folks, look at the relative calculation here.

If we consider that all the 120 candidates passed the examination, the average of the entire class must be 39. But it is given that average of the entire class is 35.

So we are getting an extra of 4 in the average i.e. an extra of 4x120=480 in the total sum.

We got extra total sum because some students who failed were also considered to be passed. For every one student considered as passed who actually failed we get 39-15 = 24 marks extra.

Since we got 480 marks extra, total number of failed students who were considered as passed = 480/24 = 20.

Hence the total number of students who passed = 100

You can solve this by assuming that all the 120 failed the examination but this would take a bit extra time.

Folks, most of the questions solved here can also be solved quickly by using a technique called Alligation. I will soon come up with a tutorial on this topic and I will discuss these questions in that tutorial besides some other questions as well. So stay tuned....

I guess there is one shorter way as you have already explained in one of the above questions:

Let n=number of passed students, then, 120-n= number of failed students
If Gain in average=Loss of average
=> 4n=(120-n)20
=> n=100

Am I correct?
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cicerone
Problem 5:

The average of batsmen up to certain number of innings was 45. In the next inning he was out for a duck and his average reduced to 40.5. Find the total number of innings played by him including the latest inning.

Sol:

Again, if the batsmen had scored 45 in his latest inning his average would remain in tact i.e. it would have been 45. But he scored 0 runs.

Since he did not score 45, he lost an average of 4.5 upon the total number of innings, which includes the latest innings as well.

So 45 runs upon x innings will give an average of 4.5


Hence 45/x = 4.5 i.e. x = 10. (Remember, x includes the latest inning as well)

rahulms
is there another way to solve the batsman problem? i'm quite confused with this method.

Here you go.

Let us assume that total no. of innings played as x

45(x-1) + 0 = 40.5 x --> 45 is average of (x-1) innings, 40.5 is new average.

4.5x = 45 --> x =10

Hope this helps.
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Using the above method, how would you apply it to the following problem ?

At a certain fruit stand, the price of each apple is 40 cents and the price of each orange is 60 cents. Mary selects a total of 10 apples and oranges from the fruit stand, and the average (arithmetic mean) price of the 10 pieces of fruit is 56 cents. How many oranges must Mary put back so that the average price of the pieces of fruit that she keeps is 52 cents?

(A) 1
(B) 2
(C) 3
(D) 4
(E) 5
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cicerone
Problem 5:

The average of batsmen up to certain number of innings was 45. In the next inning he was out for a duck and his average reduced to 40.5. Find the total number of innings played by him including the latest inning.

Sol:

Again, if the batsmen had scored 45 in his latest inning his average would remain in tact i.e. it would have been 45. But he scored 0 runs.

Since he did not score 45, he lost an average of 4.5 upon the total number of innings, which includes the latest innings as well.

So 45 runs upon x innings will give an average of 4.5

Hence 45/x = 4.5 i.e. x = 10. (Remember, x includes the latest inning as well)

Problem 6:

The average of 8 scores is 87. Of these, the highest is 2 more than the one next in magnitude. If these two scores are eliminated the average of the remaining score is 85. What was the highest score?

Sol:

If the sum of the top two scores taken out is 87+87 i.e. 174, the average of the remaining 6 scores will remain 87.
But, since the average is reduced by 2 upon 6 scores, the sum of the top two scores taken out is 174+12 = 186.
So the top two scores are 94 and 92.

Can you please explain how you arrived at 94 and 92
So if you got the part till the average of the top two scores being 186, the next bit is simple.

Assume the top two scores were equal. Then each would receive 186/2=93.

But the question clearly tells us that the highest score is two more than the one next to it. Which Implies highest score is >93.
Borrow 1 from the second highest number i.e (93-1=92) and add it to the highest number i.e (93+1 =94).

So the final answer is 94.

Regards,
Shreya

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Using the above method, how would you apply it to the following problem ?

At a certain fruit stand, the price of each apple is 40 cents and the price of each orange is 60 cents. Mary selects a total of 10 apples and oranges from the fruit stand, and the average (arithmetic mean) price of the 10 pieces of fruit is 56 cents. How many oranges must Mary put back so that the average price of the pieces of fruit that she keeps is 52 cents?

(A) 1
(B) 2
(C) 3
(D) 4
(E) 5
Given, 10 fruits - > 56

Now consider Mary puts x oranges back. This implies
(10-x) fruits - > 52
x - > 60 (because the number of oranges will be an integer value so the average cost will be just the price of one orange).

56-52=4.

Using the above method,
56+[4(10-x)]/x = 60.
Solving this we get x =5 which is the most meet of oranges Mary has to put back.

So the answer is E.

Hope this helps.

Regards,
Shreya


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