Two numbers are such that the square of one is 224 less than 8 times the square of the other. If the numbers are in the ratio of 3 : 4, then their values are [SSC (CGL) 2012] |
A) 12, 16
B) 6, 8
C) 9, 12
D) 12, 9
Correct Answer: B
Solution :
Let the numbers be 3x and 4x, respectively. |
Then, according to the question |
\[{{(4x)}^{2}}=8\times {{(3x)}^{2}}-224\] |
\[\Rightarrow \]\[16{{x}^{2}}=72{{x}^{2}}-224\]\[\Rightarrow \]\[72{{x}^{2}}-16{{x}^{2}}=224\] |
\[\Rightarrow \]\[56{{x}^{2}}=224\]\[\Rightarrow \]\[{{x}^{2}}=\frac{224}{56}=4\] |
\[\Rightarrow \]\[x=\sqrt{4}=2\] |
\[\therefore \]Numbers = 6 and 8 |
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