Average Questions Moderate with Solution

Solution:

The average of 13 papers is 40, so the sum = 13 × 40 = 520 

The average of first 7 papers is 42, so the sum will be = 7 × 42 = 294 The average of last 7 papers is 35, so the sum will be = 7 × 35 = 245 So, the marks obtained in the 7th paper will be = 539 – 520 = 19

Solution:

The salary expectation for the whole year is 5000. So, the total earning will  be 60000.

Earning for 11 months is 4200 so the total earnings are 46200. Now for the last month he has to earn 60000 – 46200 = 13800. 

Sum of x, y and z will be 135. As the average is 45 and Also given that , X - 45 = 45 - Y, X + Y = 90, therefore, 90 + Z=135  ; Z = 45

Solution:

Here, as sum of n terms will 0, so we can have at the most n-1 terms  negative and the last term whose magnitude will be equal to the sum of all  the negative terms.

Solution:

The sum of salary of officers will be = 30 × 120 = 3600 

Let the number of labourers = X. 

ATQ, 3600 + 40X = 50(30+X) 

2100 = 10X 

X= 210 

Solution:

Average marks for 40 students is equal to 52.15, the marks were taken as 49  instead of 85 so there will be an increase of 36 which is now to be distributed equally amongst 40 students, so 36/40 = 0.9 which is to be distributed amongst all. 

So, new average stands out to be 52.15 + 0.9 = 53.05

Solution:

Let initial run rate be x so, the final run rate would have be X + 8 ATQ, 9 × X + 100 = 10 × (X + 8) 

Solving, we get X = 20 

So, the new average will be 28.

Solution:

The average of 5 consecutive terms is n, implies that the 3rd term is n. Now  as the next 2 terms are included implies that the new average for 7 terms  would be the 4th term. So, the 4th term would be n+1.

Solution:

Runs scored for 1st 20 overs = 100 

Total required = 300 

So, in the next 20 overs, the team has to score 200 runs. 

So, run rate required = 200 / 20 = 10

If the weight of the man would have been 30, then the average weight would  have been the same. So, the extra 100 kg that the obese man brings with him  would be distributed equally amongst all of them, i.e. 100 / 40 = 2.5 So, the average becomes 30 + 2.5 = 32.5

Solution:

If the overall average weight has to increase after the new people are added, the average weight of the new entrants has to be higher than 40. So, n > 40 

Consequently, m has to be < 10 (as n + m = 50) . we know that the total  additional weight added by m students would be (n - 40) each, above the  already existing average of 40. m ( n - 40) is the total extra additional weight  added, which is shared amongst 40 + m students. 

So, m * (n − 40) / (m + 40) has to be maximum for the overall average to be  maximum. At this point, use the trial and error approach (or else, go with the  answer options) to arrive at the answer. The maximum average occurs when  m = 5, and n = 45. And the average is 40 + (45 – 40) * 5/45 = 40.56 kgs

Solution:

10 students have scored 600 marks amongst them, and no one is allowed to score lesser than 40 or higher than 100. The idea now is to maximize what the highest scorer gets. 

The 5 least scores have an average of 55, which means that they have scored  55 x 5 = 275 marks amongst them. This leaves 325 marks to be shared amongst the top 5 students. Lets call them a, b, c, d, and e. Now, in order to  maximize what the top scorerall the others have to get the least possible  scores (and at the same time, they should also get distinct integers.) 

The least possible score of the top 5 should be at least equal to the highest of the bottom 5. Now we want to make sure that the highest of the bottom 5 is the least possible. This can be done by making all scores equal to 55. If some scores are less than 55, some other scores have to be higher than 55 to compensate and make the average 55. Thus the highest score is the least only when the range is 0. 

So now, we have the lowest value that the top 5 can score, which is 55. The others have to get distinct integer scores, and as few marks as possible, so that gets the maximum. 

So, 55 + 56 + 57 + 58 + e = 325 

e = 99 marks.

Solution:

Step 1: A vs B: 

Since A has a higher percentage of alcohol than B, A has to give B 200 ml of wine. Considering that A has a 60% strength mixture, B will end up having  800 ml of wine that has 60/4 

Step 2: B vs C: 

B has a higher percentage of alcohol than C, so B has to give 200 ml of alcohol to C now. C will end up having wine that has 51/4 = 50.25% Step 3: A gives C 200 ml of wine. We see that computing percentages is going to cumbersome. So, let us just forget the numbers and think about how the game pans out. 

Step 4: A would give B 200 ml of wine. We note straight away that A will always give the wine. A had the highest percentage alcohol to start with.  And no mixture can reach 60% alcohol levels. So, A will give all his wine away by this round. 

Post this, we notice that, between B and C whoever has higher percentage alcohol will keep giving wine to the other. Eventually, all the wine will be with just one person. At the rate of 3 minutes per round, the game, the game will definitely be over before 1 hour. 

So, the concentration of alcohol in the wine D drinks is 60/3 = 52.67%.

Solution:

Let's employ the idea of a total of 25 students (all of the same weight) sitting on a see-saw, which has numbers from 51 to 100 marked on it. At least as many students are sitting on 76 (or to its right), as there are sitting to the left of 75. Now this means that you can have only one person sitting to the left of 75 and all the rest sitting beyond 76. But you can't do that, as you have other constraints as well.

First of all, you have to seat 3 students from 51 to 60, and 8 students from  61 to 70. Secondly, you also have to make sure that the average is the least.  This means that the see-saw should be tilting as much to the left as possible,  which in turn means that the number of people sitting to the left of 75 should be the highest possible. 

This makes it 12 students to the left of 75, and the remaining 13 students on  76 or to its right. 

Next, how do you ensure that the average is least, i.e. how do you ensure that the balance tilts as much as possible to the left? Make each student score as little as possible given the constraints. 

So, the first 3 students only score 51 each. The next 8 students score only 61  each. 11 Students are now fixed. The 12th student has to be below 75, so seat him on 71. The remaining 6 students (who are in the 71 to 80 range)  have to score 76. The next 4 score 81 and the next 3 score 91. 

This would give you the least average. 

The lowest possible average would be: 

((3 ∗ 51) + (8 ∗ 61) + 71 + (6 ∗ 76) + (4 ∗ 81) + (3 ∗ 91) 25 (3 ∗ 51) + (8 ∗ 61) + 71 + (6 ∗ 76) + (4 ∗ 81) + (3 ∗ 91)) / 25 

→70.6