Find three no. Such that,the 2nd is twice the 1st,the 3rd is three times the first and their sum is 180.
Ans is 30,60,90
when u have the answer y r u asking the question?
The numerator of a fraction is less than its denominator is increased by 9 and the numerator by 2,the fraction bcms two-third.find d fractn.
Ans is 26/33
I want soltn bcz dis qustn didn't solve frm me
x+2x+3x=180
add them.
6x=180
divide
180/6= 30
M sorry i don't undrstnd dis soltn,any solve dis qustn plz and ans is 30,60,90
please give way answer
To find three numbers that satisfy the given conditions, we can use algebraic equations.
Let's assume that the first number is "x".
According to the conditions, the second number is twice the first number, so it can be written as "2x".
Similarly, the third number is three times the first number, so it can be written as "3x".
The sum of these three numbers is given as 180, so we can write the equation:
x + 2x + 3x = 180
Simplifying the equation, we have:
6x = 180
Dividing both sides by 6, we get:
x = 30
Now, we can substitute the value of x back into the equations to find the other two numbers.
The second number is twice the first number:
Second number = 2x = 2 * 30 = 60
The third number is three times the first number:
Third number = 3x = 3 * 30 = 90
Therefore, the three numbers that satisfy the given conditions are 30, 60, and 90.