Average
- The average weight of A, B and C is 45 kg. If the average weight of A and B be 40 kg and that of B and C be 43 kg, then the weight (in kg) of B is
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Weight of B = (A + B)’s weight + (B + C)’s weight – (A + B + C)’s weight
Weight of B = 40 × 2 + 43 × 2 – 45 × 3
Weight of B = 80 + 86 – 135Correct Option: C
Weight of B = (A + B)’s weight + (B + C)’s weight – (A + B + C)’s weight
Weight of B = 40 × 2 + 43 × 2 – 45 × 3
Weight of B = 80 + 86 – 135
Weight of B = 166 – 135 = 31 kg.
- There are 100 students in 3 sections A, B and C of a class. The average marks of all the 3 sections was 84. The average of B and C was 87.5 and the average marks of A is 70. The number of students in A was
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Number of students in section A = n
∴ Number of students in sections B and C = (100 – n)
∴ n × 70 + (100 – n) × 87.5 = 84 × 100
⇒ 70n + 87.5 × 100 – 87.5n = 8400
⇒ 8750 – 17.5n = 8400
⇒ 17.5n = 8750 – 8400 = 350Correct Option: C
Number of students in section A = n
∴ Number of students in sections B and C = (100 – n)
∴ n × 70 + (100 – n) × 87.5 = 84 × 100
⇒ 70n + 87.5 × 100 – 87.5n = 8400
⇒ 8750 – 17.5n = 8400
⇒ 17.5n = 8750 – 8400 = 350⇒ n = 350 = 20 17.5
- The arithmetic mean of the following numbers
1, 2, 2, 3, 3, 3, 4, 4, 4, 4, 5, 5, 5, 5, 5, 6, 6, 6, 6, 6, 6 and 7, 7, 7, 7, 7, 7, 7 is
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Required mean = 1 × 1 + 2 × 2 + 3 × 3 + 4 × 4 + 5 × 5 + 6 × 6 + 7 × 7 1 + 2 + 3 + 4 + 5 + 6 + 7 Required mean = 1 + 4 + 9 + 16 + 25 + 36 + 49 1 + 2 + 3 + 4 + 5 + 6 + 7
Correct Option: B
Required mean = 1 × 1 + 2 × 2 + 3 × 3 + 4 × 4 + 5 × 5 + 6 × 6 + 7 × 7 1 + 2 + 3 + 4 + 5 + 6 + 7 Required mean = 1 + 4 + 9 + 16 + 25 + 36 + 49 1 + 2 + 3 + 4 + 5 + 6 + 7 Required mean = 140 28
Required mean = 5
- The average of 30 numbers is 15. The average of the first 18 numbers is 10 and that of the next 11 numbers is 20. The last number is
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The average of 30 numbers = 15
The average of the first 18 numbers = 10
and the average of next 11 numbers = 20
Let the last number be n.
According to the question,
18 × 10 + 11 × 20 + n = 30 × 15Correct Option: D
The average of 30 numbers = 15
The average of the first 18 numbers = 10
and the average of next 11 numbers = 20
Let the last number be n.
According to the question,
18 × 10 + 11 × 20 + n = 30 × 15
⇒ 180 + 220 + n = 450
⇒ 400 + n = 450
⇒ n = 450 – 400 = 50
- The average of 100 numbers is 44. The average of these 100 numbers and 4 other new numbers is 50. The average of the four new numbers will be
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Sum of 4 new numbers = 50 ×104 – 100 × 44
Sum of 4 new numbers = 5200 – 4400 = 800∴ Average = 800 = 200 4
Second method to find the required average ,
Here, N = 100, T = 44 n = 4, y = 50 – 44 = 6∴ Average of new numbers = T+ 
N + 1 
t n
Correct Option: B
Sum of 4 new numbers = 50 ×104 – 100 × 44
Sum of 4 new numbers = 5200 – 4400 = 800∴ Average = 800 = 200 4
Second method to find the required average ,
Here, N = 100, T = 44 n = 4, y = 50 – 44 = 6∴ Average of new numbers = T+ N + 1 t n Average of new numbers = 44 + 100 + 1 × 6 4
Average of new numbers = 44 + 26 × 6 = 44 + 156 = 200
