The average weight of the students in four sections A, B, C and D is 60 kg. The average weight of the students in sections A, B, C and D individually are 45 kg, 50 kg, 72 kg and 80 kg respectively. If the average weight of the students of section A and B together is 48 kg and that of B and C together is 60 kg, what is the ratio of the number of students in section A and D?
The average weight of the students in four sections A, B, C and D is 60 kg. The average weight of the students in sections A, B, C and D individually are 45 kg, 50 kg, 72 kg and 80 kg respectively. If the average weight of the students of section A and B together is 48 kg and that of B and C together is 60 kg, what is the ratio of the number of students in section A and D?
1). 5 : 7
2). 1 : 4
3). 3 : 2
4). 4 : 7
1 answers
We know that, formula,
$(\Rightarrow \left\{ {{A_E} = \frac{{{S_E}}}{{{n_E}}}\;or\;{S_E} = {A_E} \times {n_E}} \right\})$
Where,
SE = sum of entities,
nE = number of entities,
AE = Average of entities.
Let number of students in the sections A, B, C and D be a, b, c and d respectively.
Then, total weight of students of section
A =45a
Total weight of students of section B = 50b
Total weight of students of section C =72c
Total weight of students of section D = 80d
According to the question,
Average weight of students of sections A and B = 48 kg
$(\therefore \frac{{45a + 50b}}{{a + b}} = 48)$
⇒ 45a + 50b = 48a + 48b
⇒ 3a = 2b
⇒ 15a = 10b
And average weight of students of sections B and C = 60kg.
⇒ 50b + 72c = 60(b + c)
⇒ 10b = 12c
Now, average weight of students of A, B, C and D = 60 kg
∴ 45a + 50b + 72c + 80d
= 60(a + b + c + d)
⇒ 15a + 10b – 12c -20d = 0
⇒ 15a = 20d
⇒ a : d = 4 : 3
Hence, the required ratio is 4 : 3.