A maths question.Please answer!?

Question

Here is whole question:

In the year 65BC Marcus was half as old as Jules.

Thirty-six years later, the age of Marcus was three-quarters

the age of Jules. In what year was Marcus born?

Answer 1

Let Marcus be x years old and so Jules is 2x.

x + 36 = 3/4 (2x + 36)

4x + 144 = 6x + 108

36 = 2x

x = 18

So, Marcus was born in 83 BC (you need to add the age of 18 to 65 BC).

Answer 2

This is simple.....

Assume age of Marcus in the year 65 BC to be "M" and that of Jules be "J".

Now, as stated in year 65 BC, M = 1/2(J) ..........(1)

Also, after 36 years , Marcus would be (M + 36) and Jules would be (J+36).

Again given - (M+36) = 3/4(J+36)

substituting J = 2M from (1) in this

M + 36 = 3/4( 2M + 36)

Solving this, M = 18

So in year 65BC, Marcus was 18 years old. He was born 18 years ago = (-65 - 18) = 83 BC.

Thanks.

Answer 3

 Let’s assume:
Marcus’ age = m
Jules’ age = j


 To find the year, we have to find their ages first.

Make the equation:

In the year 65BC, Marcus was half as old as Jules.
m = ½ j ........................... eq (1)

Thirty-six years later, the age of Marcus was three-quarters the age of Jules.
m + 36 = ¾ j ....................... eq (2)


Substitute (1) to (2)
(1/2 j) + 36 = ¾ (j + 36)
½ j + 36 = ¾ j + 27
36 – 27 = ¾ j – 1/2j
9 = 1/4j
j = 9/(1/4) = 36

m = ½ j = ½(36) = 18

So in the year 65 BC, Marcus was 18 years old and Jules was 36 years old.


 Year Marcus was born is:
65 + 18 = 83 BC

Answer 4

two equations can be formed here

in 65 BC
Marcus is x years old
Jules is y years old
and 2x = y

36 years later
Marcus is x+36
Jules is y+36
And
x+36= 3/4 (y+36) Multiply both sides by 4 to get:
4x + 144 = 3y + 108
4x + 36 = 3(2x)
36 = 2x
x=18
y=36
Marcus is 36 years old in 65 BC so he was born in
65+18= 83 OR 83 BC

Answer 5

83 B.C.

At 65BC, 2X = Y
At 29BC, 4X = 3Y
Therefore 36 = 2x =2y
Or X = 18 years.

so 65+18 = 83BC

check
At 65BC, Marcus is 18, Jules is 36, ratio 0.5
At 29BC, Marcus is 54. jules is 72, ration 0.75.

Answer 6

M = J/2 . . . . . . . . equation 1 . . . substitute to equation 2
M + 36 = 3/4 (J + 36 ) . . . . equation 2
J/2 + 36 = 3/4 (J + 36 )
( J + 72 ) /2 = (3J + 108 ) / 4
J + 72 = (3J + 108 ) / 2
2 J + 144 = 3 J + 108
J = 36
M = 18
Marcus was born in year 83BC