The product of 1/10 of a number and 200 less than a number is twice the number. Which of the following could be the unknown number?
A: 440
B: 210
C: 220
D: 180

Answer:

C: 220

Step-by-step explanation:

hey there,

< First let’s put the equation out in numbers. Product of (something) and (something) is the first part. The first “something” is 1/10 of a number. We don’t know what the number is so we can just put the variable “n”. “of” always means multiplication, so we can just put the two together. The other “something” is 200 less than a number. That means it would be (n-200) since we took away 200 from that number

[tex]\frac{1}{10}n[/tex] × [tex](n-200)[/tex]

That’s the first part. Now it says this is equal to something else. Whenever there is an “is”, that usually always means “equals” in mathematical terms. So, here, it says that it equals twice the number. Twice means multiplied by two. So it would be 2n.

Answer:C: 220

Step-by-step explanation:hey there,< First let’s put the equation out in numbers. Product of (something) and (something) is the first part. The first “something” is 1/10 of a number. We don’t know what the number is so we can just put the variable “n”. “of” always means multiplication, so we can just put the two together. The other “something” is 200 less than a number. That means it would be (n-200) since we took away 200 from that number

[tex]\frac{1}{10}n[/tex] × [tex](n-200)[/tex]

That’s the first part. Now it says this is equal to something else. Whenever there is an “is”, that usually always means “equals” in mathematical terms. So, here, it says that it equals twice the number. Twice means multiplied by two. So it would be 2n.

Now, let’s create the full equation:

[tex]\frac{1}{10}n[/tex] × [tex](n-200)[/tex] = [tex]2n[/tex]

Solve the problem for the unknown variable which is “n”.

Move all numbers to the left since they include the variable and let the equation be equal to 0.

n = 0, 220

In our question, the answer choice does not include 0 but it does include 220 so

C.is your final answer. >Hope this helped! Feel free to ask anything else.