- Average = Sum of quantities/ Number of quantities
- Sum of quantities = Average × Number of quantities
- The average of first n natural numbers is (n +1) / 2
- The average of the squares of first n natural numbers is (n +1)(2n+1 ) / 6
- The average of cubes of first n natural numbers is n(n+1)
^{2}/ 4 - The average of first n odd numbers is given by (last odd number +1) / 2
- The average of first n even numbers is given by (last even number + 2) / 2
- The average of squares of first n consecutive even numbers is 2(n+1)(2n+1) / 3
- The average of squares of consecutive even numbers till n is (n+1)(n+2) / 3
- The average of squares of squares of consecutive odd numbers till n is n(n+2) / 3
- If the average of n consecutive numbers is m, then the difference between the smallest and the largest number is 2(m-1)
- If the number of quantities in two groups be n
_{1}and n_{2}and their average is x and y respectively, the combined average is (n_{1}x+n_{2}y) / (n_{1}+ n_{2}) - The average of n quantities is equal to x. When a quantity is removed, the average becomes y. The value of the removed quantity is n(x-y) + y
- The average of n quantities is equal to x. When a quantity is added, the average becomes y. The value of the new quantity is n(y-x) + y

A. 34

B. 35

C. 36

D. 37

x = 34 + 3 = 37

A. 5

B. 5.5

C. 6

D. 6.5

Average = 55/10 = 5.5

A. 9

B. 12

C. 15

D. 18

x = 3

3x = 9

A. 6

B. 7

C. 8

D. 9

1 to 5 = 5 × 10 = 50

5 to 9 = 5 × 8 = 40

5th = 50 + 40 = 90 – 81 = 9

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A. 625

B. 632

C. 656

D. 665

A. 17

B. 18

C. 19

D. 20

So, the average after the 15th inning will be (P-3)

Hence, 15(P-30) + 64 = 16P => P = 19

A. 23

B. 25

C. 27

D. 29

If each number is increased by 4, the total increase = 4 × 10 = 40

The new sum = 230 + 40 = 270 The new average = 270/10 = 27

A. 162

B. 174

C. 186

D. 198

A. 5.75

B. 6.25

C. 6.75

D. 7.25

A. 75 kgs

B. 80 kgs

C. 85 kgs

D. Data inadequate

Weight of new person = (65 + 20) kg = 85 kgs

A. 15 kgs

B. 16 kgs

C. 18 kgs

D. 19 kgs

Given, Rony / Titu = 15 / 8

Hence, the weigth of Titu = 16 kgs.

A. Rs. 7314/-

B. Rs. 7324/-

C. Rs. 7334/-

D. Rs. 7344/-

When the number of students is increased by 36 then the total expenditure becomes = 459x + 81

Given, average = 459x + 81 / 495 = x - 1

So, x = 16

Hence, the original expenditure = 16 × 459 = Rs. 7344/-

A. Rs. 10,800/-

B. Rs. 11,600/-

C. Rs. 12,400/-

D. Rs. 13,600/-

As Redmi is 70% higher than Realme, the cost of Redmi = x + 0.7x = 1.7x

Excluding Realme and Redmi, there are 30 mobiles with an average cost of Rs. 8880/-

Total cost of these 30 mobiles = 30 × 8880 = Rs. 266400/-

Total cost of all 32 mobiles = 32 × 9000 = Rs. 288000/-

Cost of Realme + Cost of Redmi = Total cost of all mobiles - Total cost of 30 mobiles

x + 1.7x = 288000 - 266400

2.7x = 21600

x = 8000 (Cost of Realme)

Cost of Redmi = 1.7x = 1.7 × 8000 = Rs. 13,600/-

A. 160

B. 165

C. 170

D. 175

Total runs scored by Dhoni in n innings = 8000.

a = 8000 / n

Average runs scored by Dhoni in (n+5) innings = 8085 / (n+5)

Now, we know that the average is reduced by 1 run. So, 8085 / (n+5) = a - 1

Now, after solving, we will get a quadratic equation => n² + 90n – 40000 = 0

(n + 250) (n - 160) = 0

n = 160

Total innings = n + 5 = 165.

A. 30 kgs

B. 31 kgs

C. 32 kgs

D. 33 kgs

Given, average weight of 38 students except the heaviest and the lightest ones = 32 − 1 = 31 kgs.

Total weight = 38 × 31 = 1178 kgs.

Sum of weights of the heaviest and the lightest = 1280 − 1178 = 102 kgs.

Average weight of 39 students except the heaviest one = 31 kgs.

Total weight = 39 × 31 = 1209 kgs.

Weight of the heaviest = 1280 − 1209 = 71 kgs.

Weight of the lightest = 102 − 71 = 31 kgs.

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