Question

One number is 3 1/3 larger than the other. The tripled sum of these numbers is 133.What is the largest number?

Answer 1

Let denote the two numbers with X and Y. n represent one number

Y=X + 3 1/3

The sum of the two numbers is: X+Y=X + X + 3 1/3

The triple sum is: 3(X + X + 3 1/3)

3(X + X + 3 1/3) = 133

3X + 3X + 10 = 133            

6X + 10 = 133                    

6X = 123                            

X = 12 1/2                  , Y= X+ 3 1/3 =  12 1/2 = 3 1/3 is the largest number

Answer 2

Answer:

$23\frac{5}{6}$

Step-by-step explanation:

Let x be the smaller number,

Thus, according to the question,

Second number or larger number = $x+ 3\frac{1}{3}$

Now, sum of these number = $x+x+ 3\frac{1}{3}=2x+\frac{10}{3}$

Again according to the question,

$3\times ( 2x + \frac{10}{3})= 133$

$6x + 10 = 133$

$6x = 123$

$\implies x = \frac{123}{6}$

Thus, the larger number = $\frac{123}{6}+3\frac{1}{3}=\frac{123}{6}+\frac{10}{3}=\frac{123+20}{6}=\frac{143}{6}=23\frac{5}{6}$