Average Questions Advanced with Solution

Solution:

If there are an odd number of terms, say, 2n + 1, then the median is the middle term. And if average is lesser than the middle term, there will at least be n + 1 terms greater than the average. So, there will be more terms above the average than below it. 

However, this need not be the case when there are an even number of terms.  When there are 2n distinct terms, n of them will be greater than the median and n will be lesser than the median. The average of these two terms can be  such that there are n terms above the average and n below it. 

For instance, if the numbers are 0, 1, 7, 7.5. The median is 4, average is  3.875.  Average is less than the median. And there are more 2 numbers above the average and 2 below the average. 

the answer is "No, it doesn't mean that". 

Solution:

Let's say that the students are named a, b, c, d, and e, in increasing order of weights. The average of a, b, c, and d is 40 kg, whereas the average of b, c, d, and e is 45 kg. 

The sum of a, b, c, and d is 160 kg, and the sum of b, c, d and e is 180 kg. 

What is the total weight of all the students? 

There are two ways of looking at this. 

160 + e; 180 + a; Or, e is 20 more than a. 

The total weight is 160 + e. So, the highest value of e will correspond to the highest possible average. The highest possible value of e occurs when it is  20 higher than the highest possible value for a, which is 40 (all the first 4  scores are equal to 40). 

So, the highest possible average is (160 + 60) / 5 = 44 

This will be the case when the weights are 40 kgs, 40kgs, 40 kgs, 40 kgs, and  60 kgs. Conversely, the least possible value for the average occurs when a is the least. This happens when e is the least too (since a is 20 less than e). The least possible value for e is 45 = 180 / 4 

So, the least possible value for a would be 25. 

The least possible average = 180 + 25 / 5 = 41 

This will be the case when the weights are 25 kgs, 45 kgs, 45 kgs, 45 kgs, and  45 kgs. So, the difference between maximum possible and minimum possible average = 3 kgs. 

Solution:

Let the free luggage allowance be ‘f’ kg. Let the weight of the luggage carried by Ravind be ‘r’ kg and the weight of the luggage carried by Pranas be ‘p’ kg. Thus, the excess luggage weights carried by Ravind and Pranas respectively are (r – f) kg and (p – f) kg. 

Thus, the total luggage charge for both would be (r – f)k + (p – f)k if k is the charge per kg. 

Thus, (r – f) k + (p – f) k = 1100. 

(r + p – 2f) k = 1100 .......................(1) 

If Ravind carried twice the luggage weight he actually did, i.e., if he carried  2r kg, then the excess luggage weight he carried would have been 2r – f and the corresponding charge would have been (2r – f) k. 

Therefore, (2r – f) k = 2000 ................(2) 

Likewise, If Pranas carried twice the luggage he actually did i.e., if he carried 2p kg, then the excess luggage he carried would have been 2p –f and the corresponding charge would have been (2p – f) k. 

Therefore, (2p – f) k = 1000 ................(3) 

Adding (2) and (3) and simplifying, we get, 

(r + p – f) k = 1500 ........................(4) 

Dividing (4) by (1) and simplifying, we get, 

19f = 4r + 4p .............................(5) 

Dividing (2) by (3) and simplifying, we get, 

–f = 2r – 4p .............................(6) 

Solving (5) and (6) for r, we get, 

r = 3f ...................................(7) 

Subtracting (1) from (4) and simplifying, we get, 

fk = 400. 

Ravind’s luggage charge = (r – f) k. 

But, according to equation (7), r = 3f. Therefore, Ravind’s luggage charge =  2fk 

But, fk = 400. Therefore, Ravind’s luggage charge = Rs. 800.

Solution:

Yes. We can have 24 zeroes and 1500 as the 25 numbers. In this case, there are 24 numbers below the average. 

No. In a sequence of 25 numbers, 12 will be greater than or equal to the median and 12 will be lesser than or equal to the median. We cannot have  20 terms in between the average and median 

Solution:

All the terms appear the same number of times when we select them 4  at a time. So, the average of averages will be equal to the overall average.

Solution:

Let us first rewrite the numbers a touch – 40% of students in Class A are boys, and 37.5% of boys in Class B are boys. The overall percentage of boys should lie between these two numbers. 

Now, class A has at least twice as many students as class B. So, the overall weighted average should definitely be closer to the percentage of boys in class A, or closer to 40%. 

Now, the number of students in class A can be much higher than the number 

in class B, in which case the overall percentage would practically be 40%, This is the maximum percentage that can be there. 

For the minimum percentage, we need to consider the other extreme-where class A has exactly twice as many students as class B. 

The weighted average would be (2 ∗ 40 + 1 ∗ 37.5) / 3 = 39.17% 

Let average in class C = x. Average of class B = x + 16. Average of class A  = y 

y = 2 + 20y + 30 ( x +16 ) + 50x / 100 → y - x = 8.5

Solution:

Let us try to construct a scenario for getting the smallest value of the average. We need to have the minimum value for each of the 5 medians.

Now, what is the lowest value the median of any of the 5 groups can take? 

The median is the middle term among the 5 terms. So, it cannot be 1 or 2,  but it can be 3. If the set were 1, 2, 3, 4, 5, the median would be 3. If we choose a sub-group such as this, the smallest median fro the next group will be 8. The next group could be 6, 7, 8, 9, 10. The idea is to create sub-groups in such a way so that the second group can have a smaller median than 8. 

Now, even a sub-group that has the numbers 1, 2, 3, 24, 25 would have the  median as 3. So, if we want to keep the medians as small as possible, we should choose sub-groups that have small numbers till the median, and then very large numbers. Because these large numbers do not affect the median,  we will still have small numbers to choose from for the next group. So,  choose sub-groups such that the first 3 numbers are small, the final two are as large as possible. 

1, 2, 3, 24, 25 

4, 5, 6, 22, 23 

7, 8, 9, 20, 21 

10, 11, 12, 18, 19 

13, 14, 15, 16, 17 

would be an ideal set of groups. The medians would be 3, 6, 9, 12, 15, and the average of the medians would be 9. 

Similar set of groups can be found to find the highest value of the average.  The medians would be 23, 20, 17, 14, 11 and the average would be 17.

Solution:

We can assume a, b,c d are in ascending order (with the caveat that numbers  can be equal to each other) 

a + b + c = 90 

b + c + d = 120 

We need to find the maximum and minimum value of a + b + c + d. a + b + c + d = 120 + a. So, this will be minimum when a is minimum.  Given a + b + c = 90. a is minimum when b + c is maximum. If b + c is maximum, d should be the minimum. 

Given that b + c + d = 120, the minimum value d can take is 40 as d cannot  be less than b or c. 

The highest value b + c can take is 80, when b = c = d = 40. When b = c = d  = 40, a = 10. 

a + b + c + d = 130. Average = 32.5

Similarly, a + b + c + d = 90 + d. So, this will be maximum when d is maximum. 

Given b + c + d = 120. d is maximum when b + c is minimum. If b + c is minimum, a should be maximum. 

Given that a + b + c = 90, the maximum value a can take is 30 as a cannot be greater than b or c. The lowest value b + c can take is 60, when a = b = c  = 30. When a = b = c = 30, d = 60. 

a + b + c + d = 150. Average = 37.5 

So, the average has to range from 32.5 to 37.5

Solution:

The table shows the composition of the two alloys. The total percentage of tin is the total weight of the tin as a percentage of the total weight of the two alloys put together. 

Percentage of Tin overall = (25 + 3x / 5 ) * ( 100 / 100  + x) 

This number should be between 40% and 50% 

40 ≤ (25 + 3x / 5) * ( 100/100 + x) ≤ 50 

Solving the inequality, x has to lie between 75 and 250 kg 

Alternatively, we can find the range by thinking of it as two mixtures problems for the two boundary conditions. 

(i) 100 kgs of 25% Tin mixed with x kgs of 60% Tin to give 40% Tin →  This gives x = 75 kgs. 

(ii) 100 kgs of 25% Tin mixed with x kgs of 60% Tin to give 50% Tin →  This gives x = 250 kgs. 

Hence, the answer is "75 kgs ≤ x ≤ 250 kgs". 

Let the numbers be a, b, c, d, e in ascending order. a + b + c + d + e = 165.  Average of the three largest numbers is 39, so c + d + e = 117, or a + b = 48. 

We need to find the maximum and minimum possible values of c. 

For minimum value, a and b have to be minimum. a + b = 48. Let us assume  a = 23, b = 25, ca can be as low as 26. 

23, 25, 26, 45, 46 is a possible sequence that satisfies the conditions. For maximum value, we need d + e to be minimum as c + d + e = 117. we 

can have c = 38, d = 39 and e = 40. Or, the maximum value c can take = 38. 

23, 25, 38, 39, 40 is a possible sequence that satisfies the conditions specified. 

Hence, the minimum value of the median is 26 and maximum value of the  median is 38

Solution:

Average of a, b, c, d, and e is k, Average of b, c, d, and e is also k, this implies that a = k. 

If we remove b, the average drops, this implies that b is higher than the average. 

e is less than k, or e is less than a. c is less than d. 

e < a < b, c < d 

Average of a, b, c, and e is greater than average a, c, d, and e. This tells us  that d < b 

From this, we get that b is the largest number 

b - c = d - e 

b + e = c + d 

a + b + c + d + e = 5a 

Or, b + c + d + e = 4a 

b + e = 2a, c + d = 2a. Or, e, a, b is an AP, c, a, d is an AP. 

e is the smallest number (as b is the largest number). 

Or, the ascending order should be e c a d b

Solution:

Let a, b, c, d, e, f be the six numbers in ascending order. 

e = 43 ; a + b + c + d + e + f = 198 

Median of 6 integers = c + d / 2 

c+d/2 is maximum when c and d are maximum since e = 43, dmax = 42 and  cmax = 41 

and is minimum when c and d are minimum amin = 1, bmin = 2, cmin = 3,  dmin = 4 

Therefore, = c + d / 2 min = 3.5 

Max – min = 41.5 – 3.5 = 38 

Let us assume class A has 2x boys and 3x girls, class B has 4y boys and 5y  girls. 

2x+4y/3x+5y = 34 

8x + 16y = 9x + 15y → x = y 

Required ratio = 3x / 5y (since x = y) = 3/5