Assume x is the average expenditure of 25 people. Then, 25*x = 10*84 + 15 (x + 4)  10*x = 900; x = 90. Thus the total expenditure is 90*25 = 2250.

Total marks of 28 students is 28*50 = 1400. After leaving of 8 students total marks is 20*55 = 1100. Decrease in total marks is 1400-1100 = 300. This decrement is the total marks of 8 students who left the class. Thus average is 300/8 = 37.5

The decrease in the average of three persons is 90 – 85 = 5 kg. Total decrease for three persons is 5 × 3 = 15 kg. The weight of Bumi must be New average – Decrease i.e. 85 – 15 = 70 kg. Marko is weight is 3 kg more i.e. 73 kg replaces Kora and thus the average becomes 81. Now if you compare the four persons with Marko the average decreases by 4, thus the total decrease is 4 × 4 = 16 kg. The new included person Marko must be 16 kg lighter than the person who is excluded i.e. Kora. Thus weight of A is 73 + 16 = 89 kg.

Total 3 years ago was 30 × 3 = 90, the present total must be 90 + 9 = 99 years. The total of Kaya and Daznza 7 years ago was 20 × 2 = 40, the present total age of Kaya and Dazna must be 40 + 14 = 54 years. The present age of Katara is 45 years

  On the other hand, the students who left in the October would have been of age 19 years and 6 months (December – March) on 1^st April. This means extra 6 months would decrease the average of all. Now these six month time means 180 days, which would be equally reducing the average of all the 60 students by 180/60 = 3 days.  The real average of the class will be 19 years – 3 days = 18 years + 11 months + 27 days

Average speed = (Total Distance)/(Time Taken); D/(3D/(4*200) + D/(4*240)) = 208.6

Let the 4 consecutive odd numbers be x - 3, x - 1, x + 1 and x + 3. Sum of these numbers is 4x. √4x/10 or 2√x/10 is a perfect square. Now take different values of x. At x = 10, it becomes a perfect square. So, the four numbers are 7, 9, 11 and 13.

(Hinata + Saske)/2 = 25 (7 years ago) Their total age 7 years ago = 50 years Hence, their total present age = 50 + 7 + 7 = 64 years Total age of their family (Hinata + Saske + Child) = 22 * 3 = 66 years Age of child = 66 – 64 = 2 years. Hence, child was born 7 – 2 = 5 years after marriage.

Whole numbers start from 0; hence n whole numbers means n-1 natural numbers. Sum first n-1 natural numbers is {(n-1)(n-1+1)}/2 ; Average is ((n-1)*n)/(2*n) , Average is (n-1)/2.

Worth of rice which is damaged = 60/100*134650 = 80790. Its selling price is 80790*.8 = 64632. Total selling price should be 134650*1.2 = 161580. Remaining rice is sold at 161850 - 64632 = Rs. 97218 to get overall 20% profit. So, profit is (96948 - 53860)/53860*100 = 80% (approximately).

Let the total number of laborers be x. (120*30 + 40*x)/(30 + x) = 50; x = 210. Hence, total number of employees is 210 + 30 = 240.

(Daffny + Kirito + Kakashi)/3 = 55. Daffny – 55 = 55 – Kirito, Daffny + Kirito = 110. Their total age is 165 and hence, Kakashi’s age is 55.

Average of consecutive integers is always middle term. So, the average of first 129 terms is (129 + 1)/2= 65^th term, which is n. Average of 137 terms is (137 + 1)/2 = 69^th term. So, 69^th term is 65^th term + 2*4 = n + 8.

11 years ago, age of 6 members = 205  3 years later, total age of 6 members = 205 + 6*3 - 55 = 168 After another 3 years, total age of 6 members = 168 + 6*3 - 55 = 131 Current total age of 6 members = 131 + 5*6 = 161 Thus, the average is 161/6 = 27 (approximately)

 By taking 75 instead of 39, total increment is 36. These 36 marks are equally distributed in all 50 students. Hence. marks of each student are increased by 0.72. Hence, average becomes 64.15 + 0.72 = 64.87.