Aptitude - Average - Discussion

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.

Very easy and best trick, understand the average as equal distribution of data.

As average of A,B,C = 45.
means all is 45.

a+b is 40 , but as both should be 45, means someone took 5 from both to make them 40.
means 10 is taken so keep value -10 in mind.

Similarly for a+c = 43 , 2 is less so someone taken 2 from.both means -4 is been taken from both.

So.from both case 10 and 4.

-14 is been taken.
.
And as b is the question lets make b as the robbery and if the average should be 45 and as b taken 14 from them.

means b is 45-14 = 31.

Let consider 3 Equation are there:

a + b + c = 45
a + b = 40
b + c = 43

Let multiply the average of the equation by the number of persons.

a + b + c = 45 * 3 = 135 -----> 1
a + b = 40 * 2 = 80 ----->2
b + c = 43 * 2 = 86 ---->3
Now put equation 2 in 1.
80 + c = 135.
c = 135 - 80 = 55.

Put the value of c in Equation 3.
b + 55 = 86,
b = 86 -55.
b = 31 hence the average of B is 31 i.e option D.

Hello Rsm:

Given data:

avg. of A,B,C is 45,

i.e.(A+B+C)/3=45,

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

avg. of A,B is 40,

i.e. (A+B)/2=40,

=>A+B=80.......................(2)

avg. of B,C is 43,

i.e (B+C)/2=43,

=>B+C=86.......................(3)

from (1) & (2), we get C=55,

from (1) & (3), we get A=49,

subst. A & C in (1).

we get B=31.

Most simply answer:

We know that,

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

A+B/2 = 40 ........(a).
B+C/2 = 43 ........(b).

Add equation 'a' and 'b'.

=> (A+B/2)+(B+C/2) = 86.
=> (A+2B+C)/2 = 86.
=> A+B+C = 166 ........(ii).

Subtract equation (ii) to (i) we get value of 'B'.

B = 166 - 135 = 31.

B = 31(answer).

A + B + C = 45 * 3 = 135-----> eq(1)
A + B = 40 * 2 = 80---------> eq(2)
B + C = 43 * 2 = 86----------> eq(3)
Using eq(1) & eq(3)
(A) + (B+C) = 135
A + 86 = 135
A = 135 - 86
= 49.

By using eq(2)
A + B = 80.
49 + B = 80.
B= 80-49
Ans =31.

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.

so simple brother....


A+B+C= 45*3=135......1

A+B= 40*2= 80..........2

B+C= 43*2=86.......3

BY SUBTRACTING 1 FROM 2

WE GET 135-80= C=55,

THEN PUT VALUE OF C IN 3....

B=86-55= 31.

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=135......(1)
B+C=86.....(2)
SUBTRACT EQ. 2 FROM 1, WE GET.
(A+B+C)-(B+C)=135-86=49.
A=49.
A+B=80.....(3).

PUT THE VALUE OF A IN ABOVE EQUATION.
49+B=80,
B=80-49=31.