Aptitude - Problems on Numbers - Discussion
Discussion Forum : Problems on Numbers - General Questions (Q.No. 7)
7.
The sum of the squares of three numbers is 138, while the sum of their products taken two at a time is 131. Their sum is:
Answer: Option
Explanation:
Let the numbers be a, b and c.
Then, a2 + b2 + c2 = 138 and (ab + bc + ca) = 131.
(a + b + c)2 = a2 + b2 + c2 + 2(ab + bc + ca) = 138 + 2 x 131 = 400.
(a + b + c) = 400 = 20.
Video Explanation: https://youtu.be/qmJ-0X8j_xQ
Discussion:
10 comments Page 1 of 1.
Akil said:
6 years ago
8 7 5.
8*7 + 7*5 + 5*8 = 131.
8*7 + 7*5 + 5*8 = 131.
(1)
Niya said:
7 years ago
What is about 131?
Shiv said:
7 years ago
And the following three numbers are 8, 7, 5 which is 64+49+25 = 138.
And also 8+7+5 = 20.
And also 8+7+5 = 20.
(3)
Johrge kutty said:
9 years ago
@KAILASH.
(A+B+C)^2 400
And they asked only sum.
ie (A + B + C) = sqrt(400) = 20.
(A+B+C)^2 400
And they asked only sum.
ie (A + B + C) = sqrt(400) = 20.
(1)
Najar said:
9 years ago
Please tell me the answer for this qestion in detail.
Geetha said:
1 decade ago
They ask just sum why you take (a+b+c)^2?
Bhupathi said:
1 decade ago
Any more easiest method?
Yashable said:
1 decade ago
And the following three numbers are 8, 7, 5 which is 64+49+25 = 138.
And also 8+7+5 = 20.
And also 8+7+5 = 20.
(2)
SAI KUMAR said:
1 decade ago
How to find a, b, c values? If asked in the question?
(1)
Kailash said:
1 decade ago
Why not it is (a + b + c)^2??
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