Average
- There are two groups A and B of a class, consisting of 42 and 28 students respectively. If the average weight of group A is 25 kg and that of group B is 40 kg, find the average weight of the whole class.
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Required average weight = 42 × 25 + 28 × 40 42 + 28 Required average weight = 1050 + 1120 70 Required average weight = 2170 70
Required average weight = 31 kg
Second method to find the required average ,
Here, n1 = 42, a1 = 25
n2 = 28, a2 = 40Average = n1a1 + n2a2 n1 + n2 Correct Option: B
Required average weight = 42 × 25 + 28 × 40 42 + 28 Required average weight = 1050 + 1120 70 Required average weight = 2170 70
Required average weight = 31 kg
Second method to find the required average ,
Here, n1 = 42, a1 = 25
n2 = 28, a2 = 40Average = n1a1 + n2a2 n1 + n2 Average weight = 42 × 25 + 28 × 40 42 + 28 Average weight = 1050 + 1120 70 Average weight = 2170 = 31 kg 70
- Find the average of 1.11, 0.01, 0.101, 0.001, 0.11
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We know that ,
Required average = Sum of given numbers Total number of terms Required average = 1.11 + 0.01 + 0.101 + 0.001 + 0.11 5
Correct Option: A
We know that ,
Required average = Sum of given numbers Total number of terms Required average = 1.11 + 0.01 + 0.101 + 0.001 + 0.11 5 Required average = 1.332 = 0.2664 5
- Out of 20 boys, 6 are each of 1 m 15 cm height, 8 are of 1 m 10 cm and rest of 1 m 12 cm. The average height of all of them is
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We can find the required average height with the help of given formula ,
Here, n1 = 6 , a1 = 1.15 m ,
n2 = 8 , a2 = 1.10 m , n2 = 6 , a2 = 1.12 m∴ Average height = n1a1 + n2a2 + n3a3 n1 + n2 + n3 Average height = 6 × 1.15 + 8 × 1.1 + 6 × 1.12 20
Correct Option: A
We can find the required average height with the help of given formula ,
Here, n1 = 6 , a1 = 1.15 m ,
n2 = 8 , a2 = 1.10 m , n2 = 6 , a2 = 1.12 m∴ Average height = n1a1 + n2a2 + n3a3 n1 + n2 + n3 Average height = 6 × 1.15 + 8 × 1.1 + 6 × 1.12 20 Average height = 6.9 + 8.8 + 6.72 20 Average height = 22.42 20
Average height = 1 meter 12.1 m
- The average of the largest and smallest 3 digit numbers formed by 0 , 2 and 4 would be
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Largest 3–digit number formed by 0, 2 and 4 = 420
Smallest number of three digits = 204∴ Required average = 420 + 204 = 624 2 2 Correct Option: A
Largest 3–digit number formed by 0, 2 and 4 = 420
Smallest number of three digits = 204∴ Required average = 420 + 204 = 624 = 312 2 2
- The mean of 9 observations is 16. One more observation is included and the new mean becomes 17. The 10th observation is
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Tenth observation = Mean of Ten observations – Mean of nine observations
Tenth observation = 10 × 17 – 16 × 9
Tenth observation = 170 – 144 = 26
Second method to find the required average ,
Here, N = 9, T = 16 n = 1, t = 110th observation = T+ 
N + 1 
t n
Correct Option: C
Tenth observation = Mean of Ten observations – Mean of nine observations
Tenth observation = 10 × 17 – 16 × 9
Tenth observation = 170 – 144 = 26
Second method to find the required average ,
Here, N = 9, T = 16 n = 1, t = 110th observation = T+ N + 1 t n 10th observation = 16 + 9 + 1 × 1 1
10th observation = 16 + 10 = 26
