The sum of the square of the first number and | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(C) Let the first number be $x$ and the second number be $y$.According to the problem,the square of the second number is $15$ less than the square of

Vedclass · Accepted Answer

$9$

Vedclass · Answer

(C) Let the first number be $x$ and the second number be $y$.According to the problem,the square of the second number is $15$ less than the square of $8$:$8^2 - y^2 = 15$$64 - y^2 = 15$$y^2 = 64 - 15 = 49$Since the number is positive,$y = \sqrt{49} = 7$.Next,the sum of the square of the first number and the cube of the second number is $568$:$x^2 + y^3 = 568$Substituting $y = 7$:$x^2 + 7^3 = 568$$x^2 + 343 = 568$$x^2 = 568 - 343 = 225$Since the number is positive,$x = \sqrt{225} = 15$.Finally,we need to find $\frac{3}{5}$ of the first number:$\frac{3}{5} 	imes 15 = 3 	imes 3 = 9$.

The sum of the square of the first number and the cube of the second number is $568$. Also,the square of the second number is $15$ less than the square of $8$. What is the value of $\frac{3}{5}$ of the first number? (Assuming both numbers are positive)

Similar Questions

The smallest number by which $3600$ must be multiplied to make it a perfect cube is

$\sqrt{\sqrt[3]{\sqrt[3]{\sqrt[3]{\sqrt[3]{\sqrt[3]{3}}}}}} = ?$

$(x)^{3} = 729$,find the value of $x$.

The least perfect cube which is completely divisible by $21, 24$ and $27$ is

$(4 \times 4)^{3} \div (512 \div 8)^{4} \times (32 \times 8)^{4} = (2 \times 2)^{?} + ?^{4}$

Vedclass Test Series

Exam Paper Generator

Online Exam Module