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3 years ago

Sam is 2 years and his mother is 28 years old. In how many years will Sam's mother be 3 times as old as he will be?

Mathematics

2 answers:

xz_007 [3.2K]3 years ago

5 0

Answer:

$x=13$

Step-by-step explanation:

Let:

$A_s=Age\hspace{3}of\hspace{3}Sam\\A_m=Age\hspace{3}of\hspace{3}Sam's\hspace{3}mother\\x=Years\hspace{3}required\hspace{3}for\hspace{3}Sam's\hspace{3}mother\hspace{3}be\hspace{3}3\hspace{3}times\hspace{3}as\hspace{3}old\hspace{3}as\hspace{3}sam\hspace{3}will\hspace{3}be$

Now, let's write the problem in form of an algebraic equation. Basically, the future age will be the current age plus x quantity of years. So mathematically we can calculate in how many years will Sam's mother be 3 times as old as he will be as:

$A_m+x=3(A_s+x)$

Solving for x:

$A_m+x=3A_s+3x\\3x-x=A_m-3A_s\\2x=A_m-3A_s\\x=\frac{A_m-3A_s}{2}$

Replacing the values of $A_m$ and $A_s$

$x=\frac{28-3(2)}{2}=\frac{28-6}{2}=\frac{22}{2}=11$

Hence, the years required for Sam's mother be 3 times as old as sam will be is 11

stepan [7]3 years ago

4 0

When Sam is 13 his mother will be 39 which is 3 times as old as he is. So Sam's mother will be 3 times as old as Sam will be in 11 years. (Your answer is 11 years.)

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3 years ago

Two lighthouses are located 75 miles from one another on a north-south line. If a boat is spotted S 40o E from the northern ligh

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Answer:

The northern lighthouse is approximately $24.4\; \rm mi$ closer to the boat than the southern lighthouse.

Step-by-step explanation:

Refer to the diagram attached. Denote the northern lighthouse as $\rm N$ , the southern lighthouse as $\rm S$ , and the boat as $\rm B$ . These three points would form a triangle.

It is given that two of the angles of this triangle measure $40^{\circ}$ (northern lighthouse, $\angle {\rm N}$ ) and $21^{\circ}$ (southern lighthouse $\angle {\rm S}$ ), respectively. The three angles of any triangle add up to $180^{\circ}$ . Therefore, the third angle of this triangle would measure $180^{\circ} - (40^{\circ} + 21^{\circ}) = 119^{\circ}$ (boat $\angle {\rm B}$ .)

It is also given that the length between the two lighthouses (length of $\rm NS$ ) is $75\; \rm mi$ .

By the law of sine, the length of a side in a given triangle would be proportional to the angle opposite to that side. For example, in the triangle in this question, $\angle {\rm B}$ is opposite to side $\rm NS$ , whereas $\angle {\rm S}$ is opposite to side ${\rm NB}$ . Therefore:

$\begin{aligned} \frac{\text{length of NS}}{\sin(\angle {\rm B})} = \frac{\text{length of NB}}{\sin(\angle {\rm S})} \end{aligned}$ .

Substitute in the known measurements:

$\begin{aligned} \frac{75\; \rm mi}{\sin(119^{\circ})} = \frac{\text{length of NB}}{\sin(21^{\circ})} \end{aligned}$ .

Rearrange and solve for the length of $\rm NB$ :

$\begin{aligned} & \text{length of NB} \\ =\; & (75\; \rm mi) \times \frac{\sin(21^{\circ})}{\sin(119^{\circ})} \\ \approx\; & 30.73\; \rm mi\end{aligned}$ .

(Round to at least one more decimal places than the values in the choices.)

Likewise, with $\angle {\rm N}$ is opposite to side ${\rm SB}$ , the following would also hold:

$\begin{aligned} \frac{\text{length of NS}}{\sin(\angle {\rm B})} = \frac{\text{length of SB}}{\sin(\angle {\rm N})} \end{aligned}$ .

$\begin{aligned} \frac{75\; \rm mi}{\sin(119^{\circ})} = \frac{\text{length of SB}}{\sin(40^{\circ})} \end{aligned}$ .

$\begin{aligned} & \text{length of SB} \\ =\; & (75\; \rm mi) \times \frac{\sin(40^{\circ})}{\sin(119^{\circ})} \\ \approx\; & 55.12\; \rm mi\end{aligned}$ .

In other words, the distance between the northern lighthouse and the boat is approximately $30.73\; \rm mi$ , whereas the distance between the southern lighthouse and the boat is approximately $55.12\; \rm mi$ . Hence the conclusion.

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