Aptitude - Average - Discussion
Discussion Forum : Average - General Questions (Q.No. 11)
11.
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:
Answer: Option
Explanation:
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 + 2B + C = 166 .... (iv)
Subtracting (i) from (iv), we get : B = 31.
B's weight = 31 kg.
Discussion:
40 comments Page 4 of 4.
Ankur said:
1 decade ago
Simplified.
Sum of weights = 45*3 = 135.
A+B+C = (45*3).
A+B = (40*2)----1.
B+C = (43*2)----2.
From 1, 80+C=135 => c=55.
from 2, B+55=86=> B=31.
Sum of weights = 45*3 = 135.
A+B+C = (45*3).
A+B = (40*2)----1.
B+C = (43*2)----2.
From 1, 80+C=135 => c=55.
from 2, B+55=86=> B=31.
Rabindra said:
1 decade ago
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).
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).
Kaustubh l said:
1 decade ago
A+B+C = 45*3 = 135---(1).
A+B = 80.
Hence (1) becomes:
80+C = 135.
C = 55.
B+C = 43*2 = 86.
B = 86-55 = 31.
A+B = 80.
Hence (1) becomes:
80+C = 135.
C = 55.
B+C = 43*2 = 86.
B = 86-55 = 31.
Pradeep said:
1 decade ago
@shree hema.
You said simple method. Thanks.
You said simple method. Thanks.
MANDEEP SHARMA said:
1 decade ago
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=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.
M.V.KRISHNA/PALVONCHA said:
1 decade ago
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.
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.
Rsm said:
1 decade ago
Anyone explain the logic behind it. Thanks in advance.
ANIL said:
1 decade ago
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)
S0 80+86-135=31
B=31
A+B=40*2=>A+B=80----(2)
B+C=43*2=>B+C=86----(3)
S0 80+86-135=31
B=31
Kantha said:
1 decade ago
Ya. Its 100% right. There will be many ways to get answers for aptitude questions. No worries.
Shree Hema said:
1 decade ago
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=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..
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