**Average Questions for SBI PO 2020 PDF**

Download SBI PO Average Questions & Answers PDF for SBI PO Prelims and Mains exam. Top-15 Very Important SBI PO Questions with solutions for Banking Exams.

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**Question 1: **A person travels from P to Q at a speed of 40 kmph and returns to Q by increasing his speed by 50%. What is his average speed for both the trips?

a) 36 kmph

b) 45 kmph

c) 48 kmph

d) 50 kmph

e) None of these

**Question 2: **On combining two groups of students having 30 and 40 average marks respectively in an exam, the resultant group has an average score of 34. Find the ratio of the number of students in the first group to the number of students in the second group.

a) 2 : 1

b) 3 : 2

c) 3 : 1

d) 4 : 3

e) None of these

**Question 3: **The average of five numbers is 49 the average of the first and the second numbers is 48 and the average of the fourth and fifth number is 28. What is the third number ?

a) 92

b) 91

c) 95

d) Cannot be determined

e) None of these

**Question 4: **The average of five numbers is 57.8.The average of the first and the second numbers is 77.5 and the average of the fourth and fifth numbers is 46.What is the third number ?

a) 45

b) 43

c) 42

d) Cannot be determined

e) None of these

**Question 5: **The average speed of a train is 3 times the average speed of a car . The car covers a distance of 520 kms in 8 hours .how much distance will the train cover in 13 hours ?

a) 2553 kms

b) 2585 kms

c) 2355 kms

d) 2535 kms

e) None of these

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**Question 6: **The average score of a cricketer for 13 matches is 42 runs. If the average score for the first 5 matches is 54, then what is his average score (in runs) for last 8 matches?

a) 37

b) 39

c) 34.5

d) 33.5

e) 37.5

**Question 7: **The average weight of 21 boys was recorded as 64kg. If the weight of the teacher is added, the average increased by one kg. What was the teacher’s weight?

a) 86 kg

b) 64 kg

c) 72 kg

d) 98 kg

e) None of these

**Question 8: **Find the average of the following sets of sources.

253, 124, 255, 534, 836, 375, 101, 443, 760

a) 427

b) 413

c) 441

d) 490

e) None of these

**Question 9: **The average of four consecutive numbers A, B, C and D respectively is 49.5. What is the product of B and D?

a) 2499

b) 2352

c) 2450

d) 2550

e) None of these

**Question 10: **What will be the average of the following set of scores (rounded of to the nearest integer)?

62, 76, 42, 84, 21, 47, 28

a) 57

b) 54

c) 51

d) 62

e) 66

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**Question 11: **The total of the ages of a class of 75 girls is 1050 years the average age of 25 of them is 12 years and that of another 25 is 16 years Find the average age of the remaining girls

a) 12 years

b) 13 years

c) 14 years

d) 15 years

e) None of these

**Question 12: **What will be the average of the following set of scores ?

48, 47, 64, 91, 72, 68, 93

a) 65

b) 69

c) 72

d) 61

e) 75

**Question 13: **The average weight of 15 girls was recorded as 54 kg. If the weight of the teacher was added the average increased by 2 kg What was the teacher’s weight ?

a) 75 kg

b) 95 kg

c) 78 kg

d) 86 kg

e) None of these

**Question 14: **Find the average of the following set of scores

152, 635, 121, 423, 632, 744, 365, 253, 302

a) 403

b) 396

c) 428

d) 383

e) None of these

**Question 15: **The average of four consecutive numbers A,B,C and D respectively is 56.5 What is the product of A and C ?

a) 3363

b) 3306

c) 3192

d) 3080

e) None of these

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**Answers & Solutions:**

**1) Answer (C)**

Return speed = $\frac{40}{100}$ x 150 = 60 kmph

Average speed = $\frac{2 \times 40 \times 60}{(40 + 60)}$

$\frac{4800}{100}$ = 48kmph

**2) Answer (B)**

Required ratio = (40-34)/(34-30)=6/4=3/2

**3) Answer (E)**

Let the numbers be a,b,c,d,e.

Hence (a+b+c+d+e)/5=49

(a+b)/2=48, a+b=96.

(d+e)/2=28,d+e=56

c=245-96-56=93

**4) Answer (C)**

Let the numbers be a,b,c,d and e.

Now $\frac{a+b+c+d+e}{5}$=57.8*5

a+b+c+d+e=292

Now, $\frac{a+b}{2}$=77.5, a+b=155

Also, $\frac{d+e}{2}$=46, d+e=92.

Now, c=292-(a+b)-(d+e)=292-92-155=42.

Hence, the correct option is C.

**5) Answer (D)**

Let speed of train be x and speed of car be y.

Car covers 520km in 8 hrs, it’s speed= 520/8= 65km/hr

Speed of train = 3 * 65= 195km/hr

Distance covered by train in 13 hrs = 13*195= 2535km

**6) Answer (C)**

The average score of a cricketer for 13 matches is = 42 runs

Total score in 13 matches will be = 42*13 = 546

Total score of first 5 matches will be = 54*5 = 270

Hence, total score of last 8 matches = 546-270 = 276

So average of the last 8 matches will be = $\frac{276}{8} = 34.5$

**7) Answer (A)**

The average weight of 21 boys is 64 Kgs.

The average weight of 21 boys and the teacher is 65 Kgs.

So, total weight of 21 boys is 21*64 = 1344

The weight of 21 boys and the teacher = 22*65 = 1430

Hence, the weight of the teacher is 1430 – 1344 = 86 Kgs

**8) Answer (E)**

There are 9 numbers in the series:

Average =$\frac{253+124+255+534+836+375+101+443+760}{9}$ = 3681/9 = 409

**9) Answer (A)**

Since the ages of A, B, C, D are consecutive

Let the ages of A, B, C, D be n, n+1,n+2,n+3

$\frac{n+n+1+n+2+n+3}{4} = 49.5$

4n+6 = 49.5*4 = 198

4n = 192

n = 48

Ages of A, B, C, D = 48, 49, 50 ,51

Product of ages of B and D = 49*51 = (50-1)(50+1) =$50^2-1$ = 2500-1 = 2449.

**10) Answer (C)**

We can calculate the average as follows

= $\frac{62+76+42+84+21+47+28}{7} = 51.42$

**11) Answer (C)**

Let there be three groups of 25 girls each having averages a, b and c.

25a+25b+25c=1050

a+b+c=42

Now, a = 12 and b = 16

hence, c = 14

Therefore average age of remaining 25 girls = c = 14 years.

Hence, option C is correct option.

**12) Answer (B)**

Average of the given scores can be calculated as

$\frac{48+47+64+91+72+68+93}{7}$ = $\frac{483}{7}$ = 69

**13) Answer (D)**

The average weight of 15 girls is 54 kgs.

Therefore, total weight = 810

Now, after adding teachers weight the average increase to 56 kgs

Let teachers weight be x.

(810+x)/16 = 56

x= 86

The teachers weight is 86 kgs.

The correct option is option D.

**14) Answer (A)**

Average of all the scores = $\frac{Sum of all the scores}{Number of scores}$

=$\frac{152 + 635 + 121 + 423 + 632 + 744 + 365 + 253 + 302}{9}$

= 403

Option A is the correct answer.

**15) Answer (E)**

Let A = x

B= x+1

C= x+2

D=x+3

Now, average = 56.5

(A+B+C+D)/4 = 56.5

4x +6 = 56.5*4

4x +6 = 226

4x = 220

x = 55

Therefore A,B,C and D are 55,56,57 and 58 respectively.

Product of A and C = 55*57 = 3135

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