__Average Problems with Solutions for Bank exams 2024-:__

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__PROBLEMS ON AVERAGE Bankersway__

**1). **There are 45 students in a class. The no. of boys is 5 more than girls. The average weight of boys is 5 more than average weight of girls. The total weight of boys is 100 more than girls. Calculate the average weight of the class?

(a) 50.1kg (b) 51.1 kg (c) 49.1kg (d) 50kg (e)52kg

**2). **If there are 10 people in a trip then the average height of all is 160 cm, but if 8 people go in the trip then the average height reduces by 1 cm. find the average height of remaining two?

(a) 160 cm (b) 164 cm (c) 155 cm (d) 170 cm (e) 164.5 cm

**3). **In a family of 8 of three generation, there are two fathers, two mothers, two brothers, and two male children. The total age of the female member of the family is 5 less than total age of male member of the family. What is the difference in the average age of male and female member of the family?

(a) 5 (b) 10 (c) 15 (d) 12 (e) Data inadequate

**4). **The average of three numbers is 59. When six more numbers are included the average becomes one-third more of the previous average, what is difference between the averages of the two set of numbers?

(a) 19.4 (b) 19.3 (c) 19.5 (d) 9.7 (e) 20.17

**5). **Two trains moving in opposite direction have an average speed of 56km/hr both. If both runs in same direction than the average speed of both train will be?

(a) 56km/hr (b) 50 km/hr (c) 56.7 km/hr (d) cannot be determined (e) None

**6). **There are 35 students in a lecture of physics, in which the difference between the girls and boys is 5. The average of marks got by girls is 5 more than boys and the total marks got by boys are 50 less than girls. What is the average mark of total students?

(a) 160/7 (b) 150/9 (c) 130/9 (d) 190/9 (e)190/7

**7). **Four boxes full of apples are placed one by one in a big box. The average height and length of the small boxes are in ratio of 5:7. The average of breadth of all boxes is 4 more than average of height. Calculate the approximate average volume of all four boxes if total height of all boxes is 60 cm.

(a) 48857 (b) 48867 (c) 47885 (d) 48897 (e) None of these

**8). **The average age of Arun and his wife was 25 at the time of their marriage and it increases by 3 at the birth of their child. What is the average age of the three at present at the 5th birthday of their son?

(a) 53/3 (b) 65/3 (c) 71/3 (d) 25 (e)23

**9). **The difference between the averages of two set of numbers is 6, if total numbers are 10 and the total sum

of second set is 56. What is the average of sum of both set?

(a) 50 (b) 65 (c) 60 (d) cannot be determined

(e) None of these

**10). **The average score of a cricketer is 100 in 15 matches, if his average score for 7 matches is 10 more than

average score of 8 matches than what is difference in the averages of first 7 matches and last 8 matches if

the cricketer scores 480 runs in last 8 matches?

(a) 5 (b) 4 (c) 6 (d) 12 (e) 10

**11). **The average runs of a cricketer in a tournament, in which he played 14 matches, was 47. His average runs in the first seven matches are 57 and that in the last five matches are 44. If the runs made by him in 8th match are 15, how many runs did he make in 9th match?

(a) 24 (b) 25 (c) 20 (d) 16 (e) 30

**12). **The average weight of 70 students in a class calculated as 40 kg. Later it was found that the weight of one of the student was calculated as 64kg instead of 32kg. What is the actual average weight of the students in the class?

(a) 34.5 (b) 40.5 (c) 39 (d) 39.5 (e) 40.1

**13). **There are 45 students in a hostel. Due to the admission of 15 new students the expenses of the mess were increased by Rs. 60 per day while the average expenditure per head diminished by Re. 1. What was the original expenditure of the mess?

(a) 675 (b) 670 (c) 360 (d) 480 (e) 390

**14). **In an examination, a student‘s average marks were 65 per paper. If he had obtained 20 more marks for his English paper and 5 more marks for his science paper, his average per paper would have been 70. How many papers were there in the examination?

(a) 5 (b) 10 (c) 6 (d) 11 (e) NOT

**15). **A motorist travels to a place 200 km away at an average speed of 40 km/hr and returns at50 km/hr. his average speed for the journey is how much less than its total speed?

(a) 300/7 (b) 400/9 (c) 200/9 (d) 350/9 (e) NOT

16). 16). The arithmetic mean of the scores of a group of students in a test of mathematics was 48. The brightest

20% of them secured a mean score of 75 and dullest 25% a mean score of 30. The mean score of the

remaining 55% is:

(a) 560/11 (b) 56 (c) 510/11 (d) 670/11 (e) NOT

**17). **Of the 5 numbers the average of first 4 numbers is greater than average of last 4 numbers by 6. What is the average of 5 numbers if the sum of middle 3 numbers is 59 and also last number is less than the average by 2?

(a) 27 (b) 29 (c) 26 (d) cannot be determined (e) 3

**18). **A company produces on an average of 500 items per month for the first 4 months. How many items it must produce on an average per month over the next 8 month, to make average of 1000 items per month?

(a) 1250 (b) 1350 (c) 1150 (d) 1550 (e) 1300

**19). **The average of 5 numbers is a and the average of three of these is b. if the average of last two is c, then

which of the following is correct relation between a, b and c?

(a) b=a+c (b) 2a=3b+2c (c) 5a=3b+2c (d) a=b+c

(e) a=b/c

**20). **In a coaching center the fees of girls is less than boys as some percent concession is given to girls to encourage them towards higher education. The average fee taken by 50 girls is Rs 2500 per year and the number of boys joining the center is 200. What is the average fee of all students if fees of each boy are 50% more than that of girls?

(a) 3000 (b) 2700 (c) 3500 (d) cannot be determined (e) 4500

__SOLUTION __

**1.(b)**

Total student =45

Let no of boys be x then girls will be 45-x.

As given x = 45-x +5.

Then x = 25.

Let total weight of boys and girls be A and B respectively.

Then A/25 + 5 = B/20, also A = 100 + B.

On solving we get A+ B =1900 dividing by 45 we get 51.1 as answer.

**2.(b)**

Total height of 10 people is 1600 cm (10*160)

When people = 8, total height is 1272 cm (8*159)

Average of remaining two is (1600-1272)/ 2 = 164 cm.

**3.(e)**

Female are 3 and male are 5

Total age of male – total age of female = 5.

No other information is given so data inadequate

**4.(e)**

a+b+c= (3*59)

Let six new numbers be y then

a+b+c +y= (59 +1/3) * 9

On solving we get difference as 20.17

**5.(d)**

Let speed of both the train be x and y, then 2xy/(x+y) = 56 km/hr.

If both run in same direction than average speed will be 2xy/(x-y).

Since there is no data for x and y so answer is (d).

**6.(e)**

Case 1: when boys> girls

Total student =35

Let no of boys be x then girls will be 35-x.

As given x = 35-x +5.

Then x = 20. Let total weight of boys and girls be A and B respectively.

Then A/15 + 5 = B/20, also A = 50 + B.

On solving we get a negative value which is incorrect, so this assumption is wrong.

Case 2: when boys< girls

Total student =35

Let no of girls be x then boys will be 35-x.

As given x = 35-x +5.

Then x = 20. Let total weight of boys and girls be A and B respectively.

Then A/20 – 5 = B/15, also B = 50 + A.

On solving we get average marks as 190/9

**(a)**

The ratio between height and length is 5:7 and also average breath is 4 more than average height. Total height is 60 cm.

Then we get total breath as 19*4=68 then total length will be 300/7.

Then average volume (total L* total B * total H) will be total volume/4. i.e.48857.

**8.(c)**

Arun +wife = 50

Arun + wife = 56 at the birth of their son.

After 5 years average age =71/3

**9.(d)**

Let the two set sum be A and B and total numbers in each set be n and m respectively.

A/n –B/m =6

Also m+n =10, B=56

On solving inadequate data problem take place so answer is (d).

**10.(e)**

Total score = 1500 runs.

Also total runs in 7 matches/7 = 10 + total runs in last 8 matches/8

Total runs in 8 matches are 480. On solving we get average of 7 matches as 70,

Difference is 10.

**11.(a)**

Correct equation is

(47*14)- (57*7)-(44*5)-15= 9th match run

**12.(d)**

(70*40-32)/70 = 39.5

**13.(c)**

Let original average expenditure be y then

(y-1) – 45y = 60 The y= 8

The original expenditure is 8*45= 360.

**14.(e)**

Let no. of papers be x Then 65x + 20 +5 = 70x x=5

**15.(b)**

2*40*50/ (40+50) =400/9

**16.(c)**

Let the mean score be x. 20*75 + 25* 30 + 55*x =4800 On solving x= 510/11

**17.(b)**

Let the five numbers be a, b, c, d and e and their average be x.

Then a+b+c+d/4- b+c+d+e/4 =6 and a+b+c+d +e = 5 *y

Also b+c+d = 59. And e=y +2

59 +a+e=5y => 61 + a = 4y

Also a-e =24 => putting e = y+2, a-y = 26

On solving both equation we get y = 29.

**18.(a)**

Let items in 8 month be x, then 2000+8x= 12000

On solving x=1250

**19.(c)**

Sum of 5/5 =a, sum of 3/3 =b, sum of remaining 2/2=c.

So a= (3b+2c)/5

**20.(c)**

Total fees of girls = 125000 in a year.

Total fees of boys (boys=200) will be 150% of 2500 = 3750 *200.

Average fees = (3750*200 +125000)/250= 3500

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