Average- 2 - Free MCQ Practice Test with solutions, CLAT Quant Techniques

SUm of 5 odd numbers = 255
Average = 255/5 = 51
47…..49…..51……53……55
Average = 51
Sum of largest and small number = 47+55 = 102

Average = sum/n
Average of 9 students = x
n = 9
​
Seven of them give Rs. 5 = 7 x 5 = 35
35+(5+x)+(9+x)
∴ other two give Rs. 5 and Rs. 9 more than the average contribution
​
9x = 49 + 2x
9x - 2x = 49
7x = 49
x = 7

Let the original number be (10b+a)

  ​​ ​​​​ Because of interchange, the mistaken number is (10a+b)

  ​​ ​​​​ Given that, 10b+a = 10a+b +(2.7*10) => 9b-9a = 27 => b-a = 3

So, the correct answer is A. 

The problem can be solved using the concept of weighted average.

Let's say the quantity of Rs. 60/kg rice required is 'x' kg.

Then, the total cost of the 100 kg Rs. 40/kg rice is 100*40 = Rs. 4000
And, the total cost of 'x' kg Rs. 60/kg rice is 60*x = Rs. 60x

The total quantity of the rice mix is 100 + x kg and the total cost of the mix is Rs. 4000 + 60x.

But we know that the cost of the rice mix is Rs. 50/kg, so the total cost of the mix is also 50*(100 + x) = Rs. 5000 + 50x

Setting these two expressions for the total cost equal to each other gives:

4000 + 60x = 5000 + 50x

Solving this equation gives:

10x = 1000

So, x = 100 kg

Therefore, 100 kg of Rs. 60/kg rice should be mixed with the 100 kg Rs. 40/kg rice to get a Rs. 50/kg rice mix. 

Option (a) 49 is correct

Explanation:- 

  total run scored in 10 innings :-

=> 21.5 * 10 = 215

Total run he must score after 11 innings=  24*11 = 264

He must score in 11 th innings:-

( 264- 215) = 49

Let the average age of board of consultants before replacement by a old consultant be a years. The sum of the ages of the consultants on the board = 10*a

The average age of the board of consultants 2 years age = a-2

The sum of the ages of the consultants on the board 2 years ago = 10(a-2)​​ 

After the older consultant was replaced with a younger man, the average remains the same as it was 2 yrs ago i.e., 10a hence, the younger man was 20 years younger to the old consultant.​​

So, the correct option is B.  

Average of 5 numbers = 15

  ​​ ​​​​ Sum of these 5 numbers = 15*5 = 75

  ​​ ​​​​ Average of 3 of those numbers = 11

  ​​ ​​​​ Sum of these 3 numbers = 11*3 = 33

  ​​ ​​​​ Sum of remaining two numbers = 75-33 = 42

  ​​ ​​​​ Average of remaining two numbers = 42/2 = 21

So, the correct answer is B 

Total weight of boys in IX = 40 × 52 = 2080
Total weight of girls in IX = 30 × 48 = 1440
Total students = 70

Average = (2080 + 1440) ÷ 70 = 3520 ÷ 70 = 50.29 ≈ 50.4 kg

Total girls = 30 + 45 = 75

Weight IX girls = 30 × 48 = 1440
Weight X girls = 45 × 50 = 2250

Total = 3690

Average = 3690 ÷ 75 = 49.2 kg

Total boys = 40 + 35 = 75

Weight IX boys = 2080
Weight X boys = 1925

Total = 4005

Average = 4005 ÷ 75 ≈ 53.4 kg

Boys’ weight = 35 × 55 = 1925
Girls’ weight = 45 × 50 = 2250
Total weight = 4175
Total students = 80

Average = 4175 ÷ 80 = 52.19 ≈ 52.2 kg

Total weight IX = 3520
Total weight X = 4175

Grand total = 7695
Total students = 150

Average = 7695 ÷ 150 ≈ 51.3 kg

Old weight = 1925
Added weight = 5 × 60 = 300

New total = 2225
New number = 40

Average = 2225 ÷ 40 = 55.625 ≈ 55.6 kg

New girls’ average = 50

Girls’ weight = 30 × 50 = 1500
Boys’ weight = 2080

Total = 3580
Students = 70

Average = 3580 ÷ 70 = 51.14 ≈ 51.1 kg

Boys’ average = 53.4 kg
Girls’ average = 49.2 kg

Difference = 53.4 − 49.2 = 4.2 kg

Original weight = 2250
Weight of leaving girls = 10 × 45 = 450

New total = 1800
Remaining girls = 35

Average = 1800 ÷ 35 ≈ 51.43 ≈ 51.4 kg

Therefore, Class X has higher average weight.

Old average ≈ 51.3 kg
New average = 52.3 kg

Total students = 150

New total weight = 150 × 52.3 = 7845 kg