A) 10
B) 50
C) 25
D) 20
Correct Answer: B
Solution :
(b): Let the two parts be x and \[\left( 60-x \right)\] Then \[\frac{1}{x}+\frac{1}{60-x}=\frac{3}{25}\] \[\Rightarrow \frac{60-x+x}{x\left( 60-x \right)}=\frac{3}{25}\] \[\Rightarrow {{x}^{2}}-60x+500=0\] \[\Rightarrow {{x}^{2}}-10x-50x+500=0\] \[\Rightarrow x(x-10)-50(x-10)=0\] \[x=10,x=50\] Hence largest number be 50You need to login to perform this action.
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