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.
Kumar said:
1 decade ago
The question makes no sense yaar. What does the question maker mean by saying 'sum of their sqaures'. A sense English means x^2+y^2 , but the solution shows (x+y)^2 [20+3] when the given question should have been like 20^2+3^2. The question should have been 'square of their sum' and not 'sum of their squares'.
(1)
Abhishek Patel said:
1 decade ago
Can anybody make me understand, why in the given solution we are adding 240 in 289 whereas actually it should be deducted from 289.
(1)
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).
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)
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)
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)
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)
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)
Tirth said:
7 years ago
120*2 = 240.
240+289 = 529.
Then;
√529 = 23.
240+289 = 529.
Then;
√529 = 23.
(6)
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)
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