Let the possible average marks of all the students combined be a.

We know that 32 < a < 60.

If the overall average is ‘a’, the ratio of the number of students in class A to the number of students in class B is (60 – a) : (a – 32).

Also when there are equal numbers of students in class A and class B, the value of a is simply 60+32/2=46

. If there are more students in class B than class A, the value of a is greater than 46 and if there are fewer students in class B than class A, the value of a is less than 46.

Since there are 10 more students in class B than class A, 46 < a < 60.

The extreme values of the number of students depend on the extreme values of a.

When a = 47, the ratio of students in class A to class B is 13 : 15

15x – 13x = 10

2x = 10

x = 5

Total number of students in class A = 13x = 65.

When a = 56, the ratio of students in class A to class B is 1 : 6

6x – x = 10

5x = 10

x = 2

Total number of students in class A = x = 2.

The difference between the maximum and minimum possible number of students in section A is 65 – 2 = 63

 We know the sum of the first pair of numbers is 14 * 2 = 28

Sum of the last pair of numbers is 28 * 2 = 56

To maximize the average of all the 6 numbers, we must try to maximize the two numbers in between. This is possible when the last pair of numbers are 27 and 29.

The maximum average case is

a, b, 25, 26, 27, 29

Where a + b = 28

The average of these 6 numbers is 22.5

On average, Manu targets saving ₹550 per month.

In the first 9 months, Manu earns ₹4000 per month and spends ₹3500.

So, he ends up saving only ₹450 per month. In other words, he misses his target by ₹50 per month.

So he saves ₹50 ×

 9 = ₹450 less than his target.

In the next 3 months of the year, his expenses are ₹3700 per month. To save ₹550 per month in these three months, his income should be ₹3700 + ₹550 = ₹4250 per month.

But also, he should earn some more to compensate for the ₹450 he was short of in the first 9 months. This ₹450 is earned over a span of 3 months. Or he should earn ₹150 extra each month.

So Manu’s income in the last 3 months = ₹4250 + ₹150 = ₹4400

 Let the four integers be I1,I2,I3

 and I4

.

We know that the average of I1,I2,I3

is 13.

If I4

is also 13, the average will be unchanged. The new average is also 13.

If I4

 is 13 + 4, the average will increase by 1. The new average is 14.

If I4

 is 13 + 4(2), the average will increase by 2. The new average is 15.

If I4

 is 13 + 4(x), the average will increase by x. The new average is 13 + x.

If I4

 is 13 – 4, the average will decrease by 1. The new average is 12.

If I4

 is 13 – 4(2), the average will decrease by 2. The new average is 11.

If I4

 is 13 – 4(x), the average will decrease by x. The new average is 13 – x.

We observe that for the average to remain odd and an integer, I4

 should be 13±8x

 where x is a whole number. Since we intend to find the smallest value of I4

, the question transpires to be What is the maximum value of x such that 13 – 8x is a natural number?

The maximum value of x is 1. Therefore, the minimum value of 13 – 8x is 5.

That minimum value of I4

 such that it satisfies all the given conditions is 5.

 Let the weight of Silver in the initial Alloy be Ag kgs

and the weight of Copper in the initial Alloy be Cu kgs

So, the total weight of the initial Alloy = (Ag + Cu) kgs

This Alloy is mixed with 3kgs of Pure Silver to form a new mixture.

Total weight of the mixture = (Ag + Cu) + 3 kgs

Weight of Silver in the mixture = (Ag + 3) kgs

Since this mixture contains 90% Silver,

 Weight of Silver  Total weight of the Mixture =90100

Ag+3/Ag+Cu+3=90/100

 — (1)

The initial Alloy is mixed with 2kgs of another alloy having 90% Silver by weight.

Total weight of the mixture = (Ag + Cu) + 2 kgs

Weight of Silver in the mixture = (Ag + 90% of 2) kgs = (Ag + 1.8) kgs

Since this mixture contains 84% Silver,

 Weight of Silver  /Total weight of the Mixture =84/100

Ag+1.8/Ag+Cu+2=84/100

 — (2)

Simplifying (1)

Ag+3/Ag+Cu+3=90/100

Ag+3/Ag+Cu+3=9/10

10(Ag + 3) = 9(Ag + Cu + 3)

Ag + 3 = 9 Cu

Ag = 9 Cu – 3

Simplifying (1)

Ag+1.8/Ag+Cu+2=84/100

Ag+1.8/Ag+Cu+2=21/25

25(Ag + 1.8) = 21(Ag + Cu + 2)

4 Ag + 3 = 21 Cu

Ag = 21Cu−3/4

9 Cu – 3 = 21Cu−3/4

4(9 Cu – 3) = 21 Cu – 3

36 Cu – 12 = 21 Cu – 3

15 Cu = 9

Cu = 9/15= 3/5

= 0.6

Ag = 9 Cu – 3 = 9(0.6) – 3 = 5.4 – 3 = 2.4

The weight of the initial Alloy = Ag + Cu = 2.4 + 0.6 = 3 kgs

 The arithmetic mean of the scores of 25 students = 50

Sum of scores of these students = 25 × 50 = 1250

For the scores of the top 5 students to be as high as possible, the score of the bottom 20 students should be as low as possible.

The minimum score is 30, and the scores of the bottom 20 students are distinct integers.

In order for the bottom 20 scores to be as low as possible, they must be,

30, 31, 32, … 49

Sum of the bottom 20 scores

= 30 + 31 + 32 + … + 49

= (30 + 0) + (30 + 1) + (30 + 2) + … + (30 + 19)

= 20 × 30 + (0 + 1 + 2 + 3 + … + 19)

= 600 + 19×20/2

= 600 + 190

= 790

Therefore, the maximum sum of the scores of the top 5 students = 1250 – 790 = 460

The maximum score of the five toppers, since they all scored the same marks = 460/5

= 92.

 Let the number of matches already played be x, then over the next ten matches if one more goal is scored, the average goals become 0.15. Further if the player scores two goals the average becomes 0.2.

From this we can see that one extra goal is increasing the average by 0.05 (0.2 – 0.15). Hence it can be said that the total number of matches played till the end is 20.

Therefore, x + 10 = 20

Hence, x = 10

 Given that onion is sold for 5 consecutive months at the rate of Rs 10, 20, 25, 25, and 50 per kg respectively.

Let’s consider the amount spend on onions in these five months to be 100, 100, 100, 50, 50.

Hence, the number of kgs in these five months should be 10, 5, 4, 2, 1 respectively.

The average expense on onions per kg over these five months is given by

=  total expense on onions in these five months/  total number of kgs of onions 

= 400/22

 = 18.181818

Therefore, the closest average would be 18. 

 Let’s consider Tom’s age to be ‘x’ years.

Dick is thrice as old as Tom,

Dick’s age = 3 × x = 3x

Harry is twice as old as Dick,

Harry’s age = 2 × 3x = 6x

Dick’s age is 1 year less than the average age of all three,

3x = Average(x, 3x, 6x) – 1

3x = x+3x+6x/3

 – 1

3x = 10x/3

 – 1

1 = 10x/3 – 3x

1 = 10x−9x/3

1 = x/3

3 = x

x = 3

Harry’s age = 6x = 6 × 3 = 18 years.

 The average score of the last 2 innings = 38+15/2

 = 26.5

The average of the first n innings = 30

The average of all the n+2 innings = 29

29(n + 2) = n(30) + 2(26.5)

n = 5 (You may choos eto use the allegation method too)

The average of the first 5 innings = 30 and in none of the innings did the batsman score 38 or above.

Let’s assume the scores of the 5 innings of the batsman in ascending order to be 30, 30, 30, 30, 30. (This complies with the avberage being 30)

The maximum score that he can score is 37 runs, Let’s distribute the score of the first innings among the rest 4, as 7 per innings so that the score of the first innings is minimized.

Now the scores will be 2, 37, 37, 37, 37

Now we can’t distribute the score of the first innings anymore, because, that makes the score of atleast one of the remainin 4 overs 38 or greater.

Hence the least score of any inning is 2.

 Here a2 to a9 is common to both the terms

So, a1 + (a2 to a9) = 42 × 9

a10 + (a2 to a9) = 47 × 9

Solving these two a10 – a1 = 45

a1, a2, a3,………………………, a9, (a1 + 45)

One instance is every number is 42

42, 42, …………………, 42, 42+45 (a1 to a9 are equal) ———–(1)

Another instance is every number is 47

47 – 45, 47, …………………, 47, 47 (a2 to a10 are equal) ———(2)

Mean of (1) = 46.5

Mean of (2) = 42.5

(1) – (2) = 4

 a + b+c/2 = 5 —> 2a + b + c = 10

b + a+c/2 = 7 —> 2b + a + c = 14

Solving these two, we get b – a = 4

b = a + 4

Substituting this in the first equation

2a + a + 4 + c = 10

3a + c = 6

Given all three as positive integers, maximum value we can get for a = 1

Ex: When substituting a = 2, c = 0 (Not possible)

So a = 1, then b = 5

Sum of a + b = 6

We are told that exactly 20 of the 30 integers do not exceed 5…

That means exactly 10 of the 30 integers do exceed 5.

In order to keep the average of the 20 integers as high as possible, we need to keep the average of the 10 integers above 5 as low as possible. Since we are dealing with integers, the least value that the 10 integers above 5 can take is 6.

So, the sum of the 10 integers = 10 * 6 = 60

So the sum of the remainng 20 integers = Total sum – 60 = 5 * 50 – 60 = 90

Hence the average of the remaining 20 is 90/20 = 4.5

 It is given that, a1, a2, a3,….. Ramesh, Gautham, …., a22 writes an examination

Given that, Average score of the 21 students other than Gautham = 62

So, Total Score – Gautham’s score = 62 x 21

Ramesh scored 82.5

It is given that, when Ramesh leaves, the average score drops down by 1 mark.

Which means that Ramesh scored more than the Overall class average.

Since his departure has resulted in the decrease of the overall class average by 1, his score is 21 more than the average.

Overall Class average = 82.5 – 21 = 61.5 marks

Total Score – Gautham’s score = 62 x 21

61.5 x 22 – 62 x 21 = Gautham’s score

1353 – 1302 = Gautham’s score

Gautham’s score = 51 marks

 Let there be ‘n’ tests and the Overall Average score be k

Average (n) = k => Total marks = nk

So, when we ignore the first 10 questions,

Average (n-10) = k+1

Similarly, if we ignore the last 10 questions,

Average (first (n-10)) = k-1

It is given that when the first 10 tests are not considered, the overall average increases by 1 (Each question carries 20 marks)

Total marks – 20×10 = (k+1) (n-10)

kn – 200 = (k+1) (n-10) —(1)

Similarly, If the last 10 tests are not considered, the overall average decreases by 1 (Each question carries 30 marks)

kn – 300 = (k-1) (n-10) —(2)

Solving (1) and (2), we get

kn – 200 = (k+1) (n-10)

(-) kn – 300 = (k-1) (n-10)

————————————-

100 = 2(n-10)

50 = n-10

n = 60 questions

 Given that the Average age of all the residents in the complex = 38 years

Number of people having ages 51 and above = 30

Number of people having ages below 51 ≤ 39

Since we need to maximize the average age of people below 51 and to maintain the combined average as 38 years, we need to consider the ages of people above 51 as 51 (Any more than that would lower the other value)

Also, We need to take the Number of people having ages below as 39 (Again, for maximizing the value)

Averages – Finding the largest average age

So, 13 : 38-x = 39 : 30 = 13 : 10

38 – x = 10

x = 28

Average age of people whose ages are below 51 = 28 years

 From the given data let us assume the average height of 22 toddlers to be a

If two of them leaves this group then average height of 20 toddlers increases by 2 so a + 2

If the average height of two toddlers who leaves the group is one third the average height of the original 22

No.of toddlers × average = Total

Hence we can write the equation in such a way that,

22a = 20(a + 2) + 2a/3

2a – 2a/3

 = 40

⟹ 4a/3

 = 40

a = 30

We need to find the average of remaining 20 toddlers.

⟹ a + 2 = 30 + 2 = 32

  Given that an elevator has a weight limit of 630 kg.

It is carrying a group of people of whom the heaviest weighs 57 kg and the lightest weighs 53 kg.

We have to find the maximum possible number of people in the group.

We can take one person to be 57 kg and since they have asked for the maximum possible number of people it can accommodate in the lift we can take more no. of people with lightest weight.

So, at least one guy with 57 kg.

⟹ 630 – 57 = 573

⟹ 573/53

 = 10.8 or at most there can 10 people.

Hence, 10 + 1 = 11 people

The maximum possible number of people in the group is 11.