Average Quantitative Aptitude Questions Solutions Set 3

Advanced 467 Reasoning Puzzles Book SBI PO 2018
Average - An average, or more accurately an arithmetic mean is, in crude terms, the sum of n different data divided by n.

The two mostly used formula in this chapter are:
  1. Average = Sum of quantities/Number of quantities.
  2. Suppose a man covers a certain distance at x kmph and an equal distance at y kmph, then the average speed during the whole journey is (2xy/x+y) kmph.
Nowadays, almost in all competitive exams  Average Quantitative Aptitude Questions Solutions is must ask and especially in Bank Recruitment Exams.

Following are some  Average Quantitative Aptitude Questions Solutions which are likely to be asked/ already asked in competitive exams. Following  Average Quantitative Aptitude Questions Solutions will help you securing good marks in upcoming SBI PO, SBI Clerk, IBPS PO­, IBPS Clerk­, IBPS RRB Exams in 2017 2018.

Average Quantitative Aptitude Questions Solutions

1. There are two sections A and B of a class, consisting of 36 and 44 students respectively. If the average weight of section A is 40kg and that of section B is 35kg, find the average weight of the whole class.
  • a) 30 kg
  • b)35 kg
  • c) 42.5 kg
  • d) 37.25 kg
Answer: d - Total weight of (36+44) Students=(36 x 40 + 44 x 35) kg = 2980 kg
Therefore average weight of the whole class= (2980/80) kg Therefore average weight = 37.25 kg.

2. Distance between two stations A and B is 778 km. A train covers the journey from A to B at 84 k m per hour and returns back to A with a uniform speed of 56 km per hour. Find the average speed of train during the whole journey.
  • a) 60 km/hr
  • b)30 km/hr
    C) 67 km/hr
  • d) 67.2 km/hr
Answer: d - required average speed = (2xy/x+y) km/hr = (2 x 84 x 56/84 + 56) km/hr
= (2 x 84 x 56/140) km/hr = 67.2 km/hr

3. A Batsman makes a score of 87 runs in the 17th inning and thus increases his average by 3. Find his average after 17th inning.
  • a) 40
  • b) 39
  • c) 52
  • d) 55
Answer: b - let the average after 17th inning = x
Then average after 16th inning = (x - 3)
Therefore 16(x - 3) + 87 = 17x
Therefore x = 39

4. There were 35 students in a hostel. Due to the admission of 7 new students the expenses of the mess were increased by Rs.42 per day while the average expenditure per head diminished by Re 1. What was the original expenditure of the mess?
  • a) Rs.450
  • b) Rs.320
  • c) Rs.550
  • d) Rs.420
Answer: d - let the original average expenditure be Rs.x. then,
42(x - 1) - 35x = 42
=>7x = 84
=>x = 12
Therefore original expenditure =Rs (35 x 12)= Rs.420.

5. Nine persons went to a hotel for taking their meals. Eight of them spent Rs.12 each on their meals and the ninth spent Rs.8 more then the average expenditure of all the nine. What was the total money spent by them.
  • a)Rs.115
  • b) Rs. 116
  • c) Rs. 117
  • d) Rs. 118
Answer: b - Let the average expenditure of all the nine be RS.X
Then, 12 x 8 + (X + 8) = 9X
Therefore X = 13.
Total money spent = 9X = Rs (9 x 13)= Rs 117.

6. David obtained 76, 65, 82, 67 and 85 marks (out of 100) in English, mathematics, physics, chemistry and biology. What are his average marks?
  • a) 65
  • b) 69
  • c) 75
  • d) none of above
Answer: c - Average = (76 + 65 + 82 + 67 + 85)/5 = 375/5 = 75 Hence average = 75.

7. Find the average of all numbers between 6 and 34 which are divisible by 5.
  • a) 18
  • b) 20
  • c) 24
  • d) 30
Answer: b -  Average = (10 + 15 + 20 + 25 + 30)/5 = 100/5 = 20 Hence average = 20.

8. A student was asked to find the arithmetic mean of the numbers 3,11,7,9,15,13,8,19,17,21,14 and X. He found the mean to be 12. What should be the number in place of X?
  • a) 3
  • b) 7
  • c) 17
  • d) 31
Answer: b - clearly we have (3 + 11 + 7 + 9 + 15 + 13 + 8 + 19 + 17 + 21 + 14 + X)/12 = 12
Therefore 137 + X =144
Therefore X = 144 - 137 = 7

9. The average of five numbers is 27. If one number is excluded, the average becomes 25. The excluded number is?
  • a) 25
  • b) 27
  • c) 30
  • d) 35
Answer: c - excluded number = (27 x 5) - (25 x 4) = 135 - 100 = 35.

10. The average of runs of a cricket player of 10 innings was 32. How many runes must be make in his next innings so as to increase his average of runs by 4?
  • a) 2
  • b)4
  • b) 70
  • d)76
Answer: d - Average after 11 innings = 36
Required number of runs = (36 x 11) - (32 x 10) = 396 - 320 = 76.