Question

two numbers are respectively 20% and 10% more than a third number how much percent is the first numbers more than the second​

Answer 1

Answer:

Step-by-step explanation:

let x , y and z be the numbers.

x = 1.2z  and  y = 1.1z,  then

x / y = 1.2 / 1.1 ~ 1.09

thus the first number is approximately 9% more then the second.

Answer 2

The third number is 9.09 percent more than the second number.

Step-by-step explanation:

Given: Two numbers more than a third number by 20% and 10%

To find: The percent by which the first number is more than the second

Solution:

Let the three numbers be a, b, and c

Thus,

• a = c + 20% of c = $c + \frac{20}{100}c = c + 0.2 c = 1.2 c$
• b = c + 10% of c = $c + \frac{10}{100}c = c + 0.1c = 1.1 c$

So, a is 1.2 c and b is 1.1 c

• $a-b = 1.2c -1.1 c =0.1c$

• The percent by which a is greater than c

$= \frac{0.1c}{1.1c} \times 100$

$=\frac{100}{11}$

$=9.09\%$

Thus, the third number is 9.09 percent more than the second number.