Two numbers are such that their difference,th | vedclass

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

Question

(C) Let the two numbers be $x$ and $y$.Given the ratio of their difference,sum,and product is $1: 7: 24$.Let the common multiplier be $a$. Then:$x - y

Vedclass · Accepted Answer

$48$

Vedclass · Answer

(C) Let the two numbers be $x$ and $y$.Given the ratio of their difference,sum,and product is $1: 7: 24$.Let the common multiplier be $a$. Then:$x - y = 1a$$x + y = 7a$$xy = 24a$Adding the first two equations: $(x - y) + (x + y) = 1a + 7a \Rightarrow 2x = 8a \Rightarrow x = 4a$.Subtracting the first from the second: $(x + y) - (x - y) = 7a - 1a \Rightarrow 2y = 6a \Rightarrow y = 3a$.Now,substitute $x$ and $y$ into the product equation:$xy = (4a)(3a) = 12a^2$.Since $xy = 24a$,we have $12a^2 = 24a$.Dividing by $12a$ (assuming $a 
eq 0$),we get $a = 2$.The product of the numbers is $xy = 24a = 24 	imes 2 = 48$.

Two numbers are such that their difference,their sum,and their product are in the ratio of $1: 7: 24$. The product of the numbers is:

Similar Questions

$\sqrt[3]{1331} + \sqrt[3]{1728} = ?$

Which fraction among $\frac{3}{7}, \frac{5}{11}$,and $\frac{6}{13}$ is the largest?

The sum of squares of two numbers is $90$ and the square of their difference is $46$. The product of the two numbers is

Three times the first of three consecutive odd integers is $3$ more than twice the third. The third integer is

$(2+\sqrt{5})^{2}=?+4 \sqrt{5}$

Vedclass Test Series

Exam Paper Generator

Online Exam Module