# Average- Aptitude Questions and Answers

 A man bought 13 items worth Rs 70, 15 items worth Rs 60 and 12 items worth 65 rupees. Average value per item is? A) 60.25 B) 64.75 C) 65.75 D) 63.25 Correct Answer :64.75 Explanation :Average value=(13*70)+(15*60)+(12*65) / (13+15+12) =2590/40 =64.75 If the average of 20-digit X1, X2, ... X20 is Y, then what is the average of X1-101, X2-101, X3-101 ..... X20-101? A) Y-20 B) Y-101 C) 20Y D) 101Y Correct Answer :Y-101 Explanation :Average = (X1+ X2+X3 +... +X20)/20 - (101*20)/20 =Y-101 The average of 27 numbers is 60. If instead of 28 one number has changed to 82, what will be the average? A) 56 B) 58 C) 62 D) 64 Correct Answer :62 Explanation :New average = (27*60)- 28 + 82 / 27 =1620-28+82 / 27 =1674/27 =62 A factory buys 8 machines.3 Machine A, 2 Machine B and rest Machine C. Prices of the machines are 100000, 80000 and 45000 respectively.Calculate the average cost of these machines? A) 74375 B) 75000 C) 75625 D) 72875 Correct Answer :74375 Explanation :Average cost of all the machines =(3*100000 + 2*80000 + 3*45000) / 8 = 595000/8 = 74375 The mean of 5 numbers is 15. If one more number is included, the mean of the 6 numbers becomes 17. What is the included number? A) 24 B) 25 C) 26 D) 27 Correct Answer :27 Explanation :Sum of these 5 numbers = Mean*Total Number = 15*5 = 75 Let the number included is x. As given 75+x / 6 = 17 75+x = 17*6 x = 102-75 x = 27 The mean marks obtained by 300 students in a subject are 60. The mean of top 100 students was found to be 80 and the mean of last 100 students was found to be 50. The mean marks of the remaining 100 students are? A) 70 B) 65 C) 60 D) 50 Correct Answer :50 Explanation :Total marks of remaining/middle 100 students = 300*60 - 100*80 - 100*50 = 18000 - 8000 - 5000 = 18000 - 13000 = 5000 Mean marks of remaining 100 students = 5000/100 = 50 In an asymmetrical distribution, of the mean and median of the distribution are 270 and 220 respectively, then the mode of the data is? A) 120 B) 220 C) 380 D) 370 Correct Answer :120 Explanation :Mode = 3 Median - 2 Mean = 3*220 - 2*270 = 660 - 540 = 120 Let a, b, c, d, e, f, g be consecutive even numbers and j, k, l, m, n be consecutive odd numbers. What is the average of all the numbers? A) 3(a + n) /2 B) (5l + 7d )/4 C) (a + b + m + n)/4 D) (7d + 5l) /12 Correct Answer :(7d + 5l) /12 Explanation :Let a = x (even) b = x + 2, c = x + 4, d = x + 6 e = x + 8, f = x + 10, g = x + 12 a + b + c + d + e + f + g = 7x + 42 = 7(x + 6) = 7d And j = y (odd), k = y + 2, l = y + 4, m = y + 6, n = y + 8 j + k + l + m+ n = 5y + 20 = 5( y + 4) = 5l Average of all the numbers = (7d + 5l) /12