Aptitude - Average - Discussion

A+B+C=45*3=>A+B+C=135----(1)
A+B=40*2=>A+B=80----(2)
B+C=43*2=>B+C=86----(3)
Sub 3 in 1
A+86=135
A=41
Sub A value in 2 we get
41+B=80
B=31
Is this correct..
Pls anyone Reply me..

A+B+C = 45*3.
A+B = 80.
B+C = 86.

We want 'B', So A+B+C-(A+B)-(B+C).
=> A+B+C-A-B-B-C = 135-80-86.
=> -B = -31.
=> B = 31.

@Shree hema when we subtract 135-86,
Then result will be 49.

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 of B is:

Let A, B, C represent their respective weights. Then, we have:

A + B + C = (45 x 3) = 135 .... (i).

A + B = (40 x 2) = 80 .... (ii).

B + C = (43 x 2) = 86 ....(iii).

Adding (ii) and (iii), we get: A + B+ B + C = 80+86 .... (iv).
i.e B+A+B+C =166.

But A + B + C = 135 so B+(A+B+C)=166 i.e B+135=166 then B=166-135.
Subtracting (i) from (iv), we get : B = 31.

B's weight = 31 kg.

A+B+C = 135---- (1).

A+B = 80---- (2).
B+C = 86---- (3).

Substitute (3) in (1).

A+86 = 135.
A = 49.

Substitute A in (2).

49+B = 80.

B = 80-49.
B = 31.

Explained answer is perfect by using elimination method.

If we add equation 1 and 3, we get 135 plus 86 = 221. So how it can be 166?

When average is 45 all three so why multiply by three.

a + b + c/3 = 45.
a + b + c = 135------>1.

We know b + c = 86.
substitute in eqn 1.

We get a + 86 = 135.
a = 49.
a + b/2 = 40.
49 + b = 80.
b = 31 //. Answer.

A + B + C = 45 * 3 = 135.
A + B = 40 * 2 = 80.
B + C = 43 * 2 = 86.

Now, (A + B + C ) - (A + B) = C i.e. 55,
Vice versa for the value of A i.e. 49.

Substitute the value of A in A + B = 80.
B will be 31.