Quantitative Aptitude: Average Questions Set 5

Word Problems on Average for SBI PO, IBPS PO/Clerk, IBPS RRB, NIACL, NICL, RBI, OICL, UIICL, BoB and other competitive exams.

1. There are, 20 balls; some of them are green and the others are yellow. The average cost of all balls is Rs. 38, average cost of green balls Rs. 35 and that of yellow balls is Rs. 40, the number of yellow balls is:
   A) 8
   B) 12
   C) 14
   D) 10
   E) None
   
   View Answer

   Option B
   Solution:
   green ball price …………….yellow ball price
   …..35 ……………………………………..40
   .                               38
   2                                                           3
   Ratio 2:3
   Then 5 20(total ball)
   3 ? ==> 12 balls.
2. The average marks obtained by 50 students of a class is 92. If the 5 highest marks are removed, the average reduces by two mark. The average marks of the top 5 students is
   A) 95
   B) 102
   C) 110
   D) 105
   E) None
   
   View Answer

   Option C
   Solution:
   Average mark of 50 student is 92
   Their sum is 50*92=4600
   5 highest marks are removed.
   Average mark of 45 student is 90
   Their sum is 45*90=4050
   Sum of the removed 5 students is 4600-4050=550
   Their average 550/5=110.
3. If the average mark of 1/4th class is 85% and that of 1/3 rd of class is 70% and the average mark of rest of the class is 56%.then the average of the whole class is
   A) 67.91
   B) 70.32
   C) 68.5
   D) 69.25
   E) None
   
   View Answer

   Option A
   Solution:
   Let x be the total no of students in whole class.
   Then (x/4*85/100 + x/3 *70/100+ 5x/12 * 56/100 )*100
   =67.91.
4. The average marks of Naveen decreased by one, when he replaced the subject in which he has scored 40 marks by the other two subjects in which he has scored 23 and 25 marks respectively. Later he has also included 57 marks of Maths, then the average marks increased by two. How many subjects were there initially?
   A) 16
   B) 12
   C) 14
   D) 15
   E) None
   
   View Answer

   Option D
   Solution:
   Total subjects = x ; Average Marks = y
   (x + 1)(y – 1) = (xy – 40) + (23 + 25)
   y – x = 9 –(1)
   (x + 2)(y + 1) = (xy – 40) + (23 + 25)+ 57
   xy + 2y + x + 2 = xy + 65
   2y + x = 63 –(2)
   From Equation (1) and (2) => x = 15.
5. Out of the three annual examinations, each with a total of 500 marks, a student secured average marks of 56% and 64% in the first and second annual examinations. To have an overall average of 64%, how many marks does the student need to secure in the third annual examination ?
   A) 360
   B) 340
   C) 320
   D) 270
   E) None
   
   View Answer

   Option A
   Solution:
   Total marks for three examinations = 3x 500 = 1500
   Total required marks in three examinations = 64% of 1500 = 960
   Marks secured in first examination = 56 % of 500 =280
   Marks secured in third examination = 64 % of 500 =320
   Thus, the required marks in third examination
   =960 – ( 280 + 320 )
   =960 – 600
   = 360.
6. The average weight of 18 students in a class is 52.5 kg and that of the remaining 6 students is 43.25 kg. Find the average weights of all the students in the class.
   A) 49kg
   B) 47.65kg
   C) 50.18kg
   D) 51.2kg
   E) None
   
   View Answer

   Option C
   Solution:
   The average weights of all the students in the class
   =[(18*52.5) +(6*43.25)]/(18+6)
   =945+259.5 /24
   =50.18kg.
7. When I was married 10 years ago my wife is the 6th member of the family. Today my father died and a baby born to me.The average age of my family during my marriage is same as today. What is the age of Father when he died ?
   A) 56yrs
   B) 55yrs
   C) 58yrs
   D) 60yrs
   E) None
   
   View Answer

   Option D
   Solution:
   Let the Father be x years when he died
   Average Age 10 years ago be A
   Total Age 10 years ago = 6*A
   Total Age after 10 years (Just before father’s Death) = 6A + 6*10 = 6A + 60
   Father Died and Baby was born => the Total number of people in the family is Same (6)
   (6A +60 – x)/6 = 6A/6
   = A + 10 -(x/6) = A
   = x/6 = 10
   = x = 60yrs.
8. The average age of the group having 3 members is 84. One more person joins the group and now the average becomes 80. Now a fifth person comes whose age is 3 years more than that of fourth person replaces the first person. After this the average age of the group becomes 79. What is the weight of the first person?
   A) 75
   B) 64
   C) 72
   D) 78
   E) None
   
   View Answer

   Option A
   Solution:
   Let the group of members are A, B, C, D, E
   A+B+C = 84*3 = 252
   A+B+C+D = 80*4 = 320
   Then D = 320-252 = 68 and E = 68+3 = 71
   Now B+C+D+E = 79*4 = 316
   (A+B+C+D) – (B+C+D+E) = 320- 316
   A-E = 4
   Then A = 71+4=75.
9. A train cover a certain distance at a speed of 80 km/hr however it will halt for fixed time interval in each hour its average speed reduced 60 km/hr. What is the time interval for which the train halt in each hour?it will halt for fixed time interval in each hour its average speed reduced 50 km/hr. What is the time interval for
   A) 12min
   B) 15min
   C) 10min
   D) 14min
   E) None
   
   View Answer

   Option B
   Solution:
   Train Halt time,
   = 1 – (slower speed/faster speed)
   = 1 – 60/80
   = 1/4 hours = 15 min.
10. The average age of all the 100 employees in an office is 29 years, where 2/5 employees are ladies and the ratio of average age of men to women is 5 : 7. The average age of female employees is
    A) 26yrs
    B) 28yrs
    C) 35yrs
    D) 32yrs
    E) None
    
    View Answer

    Option C
    Solution:
    Let the average of men and women be 5x and 7x resp.
    (3*5x/5)*100 + (2*7x/5)*100=29*100
    300x+280x = 2900
    X=5.
    Then average age of females 7x=7*5=35yrs.

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