The difference between two numbers is 18 and their HCF and LCM are 6 and 168, respectively. What is the sum of squares of the two numbers? [RBI (Assistant) 2015] |
A) 2280
B) 2260
C) 2420
D) 2340
E) 2380
Correct Answer: D
Solution :
Let one number be \[x.\] |
Then, another number is \[x+18.\] |
Now, product of two numbers \[=HCF\times LCM\] |
\[\Rightarrow \] \[x\times (x+18)=6\times 168\] |
\[\Rightarrow \] \[{{x}^{2}}+18x=1008\] |
\[\Rightarrow \] \[{{x}^{2}}+18x-1008=0\] |
\[\Rightarrow \]\[{{x}^{2}}-24x+42x-1008=0\] |
\[\Rightarrow \]\[x\,\,(x-24)+42x\,\,(x-24)=0\] |
\[\Rightarrow \] \[(x+42)(x-24)=0\] |
\[\Rightarrow \] \[x=24,\]\[x\ne -\,\,42\] |
\[\therefore \]Another number is 42. |
Required sum of squares \[={{(24)}^{2}}+{{(42)}^{2}}\] |
\[=576+1764=2340\] |
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