Question

A and B are whole numbers, and A/11 + B/3 = 31/33. find the value of A and the value of B.

Answer 1

The value of A is 3 and the value of B is 2.

Whole Numbers

They are the numbers represented by positive real numbers where the fractions and decimal numbers are not included.

The question gives:

A and B are whole numbers;

• $\frac{A}{11}+\frac{B}{3} =\frac{31}{33}$

Thus, you need to find A and B for these conditions.

When we calculated the  Least Common Multiple ( LCM ) between 3 and 11, we can rewrite the equation as:

                                            $\frac{A}{11}+\frac{B}{3} =\frac{31}{33}\\ \\ \frac{3A+11B}{33}=\frac{31}{33}$

Thus, we have:

                                          $3A+11B=31\\ \\ 3A=31-11B\\ \\ A=\frac{31-11B}{3}$ (1)

From this condition, for that A is a whole number, then 31-11B >0.

Then,

                                                    $31-11B > 0\\ \\ -11B > -31\\ \\ 11B < 31\\ \\ B < 2.8$

In this case, for that B should be a whole number, B can be 0,1 and 2.

Now, you should replace probable numbers for B in equation (1). The value of B will be the number from that you will find the whole number for A. See below.

For B=0, you have $A=\frac{31-11*0}{3}=\frac{31}{3}=10.33$ . In this case, A is not a whole number. Then, B can't be 0 (zero).

For B=1, you have $A=\frac{31-11*1}{3}=\frac{31-11}{3}=\frac{20}{3}=6.67$ . In this case, A is not a whole number. Then, B can't be 1 (one).

For B=2, you have $A=\frac{31-11*2}{3}=\frac{31-22}{3}=\frac{9}{3}=3$ . In this case, A is a whole number (3). Then, B can be 2 (two).

Let's replace this information in the given equation in your question.

                                                    $\frac{A}{11}+\frac{B}{3} =\frac{31}{33}\\ \\ \frac{3}{11}+\frac{2}{3} =\frac{31}{33}\\ \\ \frac{9+22}{33}= \frac{31}{33}\\ \\ \frac{31}{33}=\frac{31}{33}$  

Hence, A=3 and B=2.

Read more about the whole number here:

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