Question : If $x$ is subtracted from each of the numbers 20, 37, 54 and 105, then the numbers so obtained in this order are in proportion. What is the mean proportional between $(7 x-5)$ and $(x+1) $?
Option 1: 8
Option 2: 9
Option 3: 6
Option 4: 12
Correct Answer: 8
Solution : If $x$ is subtracted from each of the numbers 20, 37, 54 and 105, then the numbers so obtained in this order are in proportion. So, $\frac{(20-x)}{(37-x)} = \frac{(54-x)}{(105-x)}$ $⇒[(20-x)(105-x)]=[(54-x)(37-x)]$ $⇒34x=102$ $⇒x = 3$ The mean proportional between $(7x-5)$ and $(x+1)$ is the square root of their product. So, the mean proportional is $\sqrt{(7x-5)(x+1)}$. Substituting $x = 3$ into the above expression, we get $\sqrt{(7\times3-5)(3+1)} = \sqrt{(16)(4)} = 8$ Hence, the correct answer is 8.
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Question : If $x^2+8 y^2+12 y-4 x y+9=0$, then the value of $(7 x+8 y)$ is:
Option 1: –33
Option 3: 33
Option 4: –9
Question : The ratio between the two numbers is 3 : 4. If each number is increased by 6, the ratio becomes 4 : 5. What is the difference of the numbers?
Option 1: 1
Option 2: 3
Option 4: 8
Question : If $x^2+\frac{1}{x^2}=98$, then the value of $x+\frac{1}{x}$ is:
Option 1: 10
Option 2: 8
Option 3: 7
Option 4: 9
Question : If $x^2-3 x+1=0$, then the value of $\left(x^4+\frac{1}{x^2}\right) \div\left(x^2+1\right)$ is:
Option 1: 5
Option 2: 6
Option 3: 9
Option 4: 7
Question : If $x+y+z=13,x^2+y^2+z^2=133$ and $x^3+y^3+z^3=847$, then the value of $\sqrt[3]{x y z}$ is:
Option 1: $8$
Option 2: $7$
Option 3: $-9$
Option 4: $-6$
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