Friday, 5 February 2016

MATHS (SOLVED IN STEPS) Problems on Average - (Page -1)

MATHS (SOLVED IN STEPS)
Problems on Average - (Page -1)
1. The average monthly income of A and B is Rs. 5050. The average monthly income of B and C is Rs. 6250 and the average monthly income of A and C is Rs.5200. What is the monthly income of A?
A. 2000             B. 3000             C. 4000         D. 5000
Answer : Option C
Explanation :
Let the monthly income of A = a
monthly income of B = b
monthly income of C = a
a +b=2×5050------------(Equation1)
b + c=2×6250------------(Equation2)
a +c=2×5200------------(Equation3)
(Equation 1) + (Equation 3) - (Equation 2)
=>  ( a +b +a +c)   “  (b +c)
=(2×5050)+(2×5200)”(2×6250)
=> 2a = 2(5050 + 5200 - 6250)
=> a = 4000
=> Monthly income of A = 4000

2. In the first 10 overs of a cricket game, the run rate was only 3.2. What should be the run rate in the remaining 40 overs to reach the target of 282 runs?
A. 6.25,        B. 5.5,        C. 7.4 ,             D. 5
Answer : Option A
Explanation :
Runs scored in the first 10 overs = 10×3.2=32
Total runs = 282
remaining runs to be scored = 282 - 32 = 250
remaining overs = 40
Run rate needed = 250/40=6.25

3. A grocer has a sale of Rs. 6435, Rs. 6927, Rs. 6855, Rs. 7230 and Rs. 6562 for 5 consecutive months. How much sale must he have in the sixth month so that he gets an average sale of Rs.6500?
A. 4800      B. 4991        C. 5004         D. 5000
Answer : Option B
Explanation : Let the sale in the sixth month = x
Then( 6435+6927+6855+7230+6562+x)/6=6500
= > 6435 + 6927 + 6855+ 7230 + 6562 + x
= 6 × 6500 = 39000
= > 34009 + x = 39000
= > x = 39000 - 34009 = 4991

4. The average of 20 numbers is zero. Of them, How many of them may be greater than zero, at the most?
A. 1           B. 20     C. 0    D. 19
Answer : Option D
Explanation : Average of 20 numbers = 0
=>Sum of 20 numbers/20=0
=> Sum of 20 numbers = 0
Hence at the most, there can be 19 positive numbers.
(Such that if the sum of these 19 positive numbers is x, 20th number will be -x)

5. The captain of a cricket team of 11 members is 26 years old and the wicket keeper is 3 years older. If the ages of these two are excluded, the average age of the remaining players is one year less than the average age of the whole team. Find out the average age of the team.
A. 23 years    B. 20 years   C. 24 years        D. 21 years
Answer : Option A
Explanation :Number of members in the team = 11
Let the average age of the team = x
=>(Sum of the ages of all the 11 members of the team)/11=x
=> Sum of the ages of all the 11 members of the team = 11x
Age of the captain = 26
Age of the wicket keeper = 26 + 3 = 29
Sum of the ages of 9 members of the team excluding captain and wicket keeper = 11x - 26 - 29 = 11x - 55
Average age of 9 members of the team excluding captain and wicket keeper =(11x”55)/9
Given that( 11x”55)/9=(x”1)
=> 11x - 55 = 9(x - 1)
=> 11x - 55 = 9x - 9
=> 2x = 46
=>x=46/2=23 years

6. In Kiran’s opinion, his weight is greater than 65 kg but less than 72 kg. His brother does not agree with Kiran and he thinks that Kiran’s weight is greater than 60 kg but less than 70 kg. His mother’s view is that his weight cannot be greater than 68 kg. If all are them are correct in their estimation, what is the average of different probable weights of Kiran?
 A. 70 kg                 B. 69 kg              C. 61 kg              D. 67 kg
Answer : Option D
Explanation : Let Kiran’s weight = x. Then
According to Kiran, 65 < x < 72 ——————(equation 1)
According to brother, 60 < x < 70 ——————(equation 2)
According to mother, xd”68 ——————(equation3) 
Given that equation 1,equation 2 and equation 3 are correct. By combining these equations, we can write as 65<xd”68
That is x = 66 or 67 or 68
average of different probable weights of Kiran = (66+67+68)/3=67

7. The average weight of 16 boys in a class is 50.25 kg and that of the remaining 8 boys is 45.15 kg. Find the average weights of all the boys in the class.
 A. 48.55                B. 42.25              C. 50               D. 51.25
Answer : Option A
Explanation :
Average weight of 16 boys = 50.25
Total Weight of 16 boys = 50.25×16
Average weight of remaining 8 boys = 45.15
Total Weight of remaining 8 boys = 45.15×8
Total weight of all boys in the class = (50.25×16)+(45.15×8)
Total boys = 16+8=24
Average weight of all the boys = (50.25×16)+(45.15×8)/24
= (50.25×2)+(45.15×1)/3
= (16.75×2)  +  15.05
=33.5+15.05=48.55

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