Aptitude - Problems on Numbers - Discussion
Discussion Forum : Problems on Numbers - General Questions (Q.No. 12)
12.
The product of two numbers is 120 and the sum of their squares is 289. The sum of the number is:
Answer: Option
Explanation:
Let the numbers be x and y.
Then, xy = 120 and x2 + y2 = 289.
(x + y)2 = x2 + y2 + 2xy = 289 + (2 x 120) = 529
x + y = 529 = 23.
Discussion:
12 comments Page 1 of 2.
Argghx said:
2 months ago
A^2+b^2=289 a*b=120apply (a+b) ^2 formula and replace the value.
If you are thinking square rooting 289 and getting a+b = 17 is right,
Here it doesn't satisfy thee first case of a*b=120.
If you are thinking square rooting 289 and getting a+b = 17 is right,
Here it doesn't satisfy thee first case of a*b=120.
Ugyen Jigme Chogyel said:
4 months ago
The product of two numbers is 120.
x + y = 120 and x^2 +y^2 = 289.
Lets take 120/15 = 8, or 15 * 8 = 120.
15^2 + 8^2 = 225 + 64.
= 289.
So, the sum of the number is:
15 + 8 = 23(the answer)
Hope it helped:)
x + y = 120 and x^2 +y^2 = 289.
Lets take 120/15 = 8, or 15 * 8 = 120.
15^2 + 8^2 = 225 + 64.
= 289.
So, the sum of the number is:
15 + 8 = 23(the answer)
Hope it helped:)
(1)
Sourav Yadav said:
6 years ago
It clearly says that "sum of squares ". It does not say" square of the sum".
So basically considering this case the question should be carried out by doing x^2 + y^2 and not (x+y) ^2.
So basically considering this case the question should be carried out by doing x^2 + y^2 and not (x+y) ^2.
(3)
Tirth said:
7 years ago
120*2 = 240.
240+289 = 529.
Then;
√529 = 23.
240+289 = 529.
Then;
√529 = 23.
(6)
Pooja P Kumar said:
8 years ago
You are absolutely wrong @Kumar.
The question is correct and the method used is perfect.
Just apply the formula (a+b)^2 = a^2+b^2+2ab, you will get the solution.
The question is correct and the method used is perfect.
Just apply the formula (a+b)^2 = a^2+b^2+2ab, you will get the solution.
(2)
Ram said:
9 years ago
x^2 + y^2 = 289 and we say it as x^2 + y^2 = 17^2 and take root at both sides.
It will give answer as 17. If this is correct or 23 is correct?
Please tell me.
It will give answer as 17. If this is correct or 23 is correct?
Please tell me.
(3)
Mohita said:
9 years ago
Shouldn't the solution be -
x^2 + y^2 + 2xy = 289 + 2xy
If we are opening the (x+y)^2 bracket on one side of the equation, a 2xy is being added to the LHS which must also be compensated for on the RHS of the equation. Please correct me if I'm wrong.
x^2 + y^2 + 2xy = 289 + 2xy
If we are opening the (x+y)^2 bracket on one side of the equation, a 2xy is being added to the LHS which must also be compensated for on the RHS of the equation. Please correct me if I'm wrong.
(2)
Jayesh said:
9 years ago
The two numbers are 15 and 8.
Where,
15 + 8 = 23.
15* 8 = 120.
225 + 64 = 289.
Where,
15 + 8 = 23.
15* 8 = 120.
225 + 64 = 289.
(3)
Mujahed said:
1 decade ago
Here ab = 120, a^2+b^2 = 289.
Very simple (a+b)^2 = a^2+b^2+2ab.
Here given a^2+b^2 = 289 & ab = 120.
= 289 + 2(120).
(a+b)^2 = 289 + 240.
(a+b)^2 = 529.
(a+b) = 23.
Very simple (a+b)^2 = a^2+b^2+2ab.
Here given a^2+b^2 = 289 & ab = 120.
= 289 + 2(120).
(a+b)^2 = 289 + 240.
(a+b)^2 = 529.
(a+b) = 23.
(2)
Pavan said:
1 decade ago
@Abishek.
In this question we already know their products and sum of the squares of their products only one we have to find out is their sum. So here we are using (a+b)whole square formulae:
Hence we are adding 289+2(120).
In this question we already know their products and sum of the squares of their products only one we have to find out is their sum. So here we are using (a+b)whole square formulae:
Hence we are adding 289+2(120).
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