Questions are made tough when instead of a precise value, a range of value is given. And instead of a unique answer, either the maximum or minimum value is asked. Try this question as an example.

Q.

In a group, there are 30 people whose age is 50 years or above. The average age of all the people in the group is 32 years. If the number of people in the group is not more than 75, find the maximum average age of all those who are less than 50 year old.

TITA type i.e. Type In The Answer type

There are a lot of unknowns here. We would need to know how do the different unknowns affect the average age of those below 50 years. But first lets express the data in math terms …

There are two groups, one aged less than 50 years; another aged 50 and above. And we also have average of both the groups combined i.e. of all people. So the setting of Weighted Average should be evident.

Let group 1 be those aged less than 50 years. If their average age is A1, we need to maximise A1. Also the number of people in this group will be relevant, say n1. Now as per the given conditions, 1 ≤ n1 ≤ 45 (to find average age of this group, there needs to be atleast one person)

The other group, aged 50 years or above … let their average age be A2. And we know that n2 = 30.

The average of the two groups combined is given as 32 years.

Points to think on …

1. How does A1 react to changes in n1? As n1 increases from 1 to 45, will A1 increase or decrease?

2. How does A1 react to changes in A2? If A2 increases, will A1 increase or decrease?

And if the answers to the above are not very forthcoming, then go the mechanical way. Select any reasonable values for n1 and A2 and find A1. Change one of n1 or A2 and re-calculate A1. Now find the answers to the above two questions.

For exam purposes, if you are smart enough to guess that the correct answer will lie at extreme values of n1 and A2, then try the extreme possibilities and find which of them maximises A1. The extreme values of n1 are obvious … 1 & 45. And for A2, we just know that A2 ≥ 50. So try 50 once and try some random large value, say 100, to check the effect of increasing A2.

Do the above, if you are serious. Only then watch the video given in the explanation.

20 years

Learn the relevant theory and the question being solved here:

Do study the entire theory as explained in the video explanation to above question. And only once you have learnt it, try the following question.

Q.

The average marks of all students who scored more than 70% marks is 74%. The average of all students who scored more than or equal to 60% is 68%. Of the students scoring more than or equal to 60%, at least what percent of students have scored between 60% and 70%?

A. 25%

B. 34.25%

C. 37.77%

D. 42.84%

D. 42.84%

Consider the two groups as …. those scoring between 60% and 70%; and those scoring more than 70%.

Those scoring more than 60% are the two groups combined.

A1 is unknown, A2 is 74% and Ar is 68%.

And we need to minimise w1 : w2

i.e. increase the weight of 2nd group, relatively

i.e. Ar has to be closer to A2 than A1. But Ar and A2 are fixed. So they can be bought closer, relatively, by placing A1 as far away from Ar as possible

i.e. w1 : w2 will be minimised when A1 is least

i.e. when A1 = 60 (all in the group score between 60 & 70)

Doing Alligation with: 60 … 68 … 74, we get the ratio w1 : w2 as 3 : 4

Thus, the number of students in first group is atleast 3/7the of the total i.e. 42.84% of the total.

Q.

38 students took a test that had 35 as the passing marks and 100 as the maximum marks. If the average marks of all the 38 students is 78, find the maximum number of students who could have failed.

TITA type i.e. Type In The Answer type

12

Consider two groups of students …. group 1 as those who failed i.e. scored less than 35; and group 2 as those who passed i.e. scored 35 or more.

We are given that the weighted average of both the groups combined is 78. And we want to maximise the number of students in group 1 i.e. maximise the weight of group 1, relative to group 2

i.e. we need A1 to be as close to 78 as possible and A2 to be as far away

The closest that A1 can be to 78 is when the students who failed, just failed i.e got 34.99999 marks. As a limiting case, A1 = 35

And the furthest that A2 can be from 78 is when all those who passed scored the maximum possible i.e. A2 = 100

Doing Alligation, 35 … 78 … 100, we get w1 : w2 as 22 : 43.

Thus, the maximum who could fail is \(\frac{22}{65} \times 38\) i.e. 12.86

Since number of failing students has to be a whole number, the maximum who can fail is 12.