Two numbers are in the ratio 3:5. If 9 is subtracted from the numbers, the ratio becomes 12:23. The numbers are?

Let the original numbers be 3k and 5k, respectively.

Write an equation for the ratio after subtracting 9 from each number:

(3k-9)/(5k-9) = 12/23

Cross multiply:

23(3k-9) = 12(5k-9)

69k - 207 = 60k - 108

Group like terms:

69k - 60k = 207 - 108

9k = 99

k = 11

3k → 3(11) = 33

5k → 5(11) = 55

Answer:

The original numbers are 33 and 55.

Let x, y be the numbers, then

x/y=3/5=>x=3y/5------(1)

(x-9)/(y-9)=12/23------(2)

Combining (1) & (2), get

23(3y/5-9)=12(y-9)

=>

69y-1035=60y-540

=>

9y=495

=>

y=55

x=33 from (1)

Two numbers are in the ratio 3:5 → a/b = 3/5 → a = 3b/5

If 9 is subtracted from the numbers, the ratio becomes 12:23

(a - 9)/(b - 9) = 12/23

23.(a - 9) = 12.(b - 9)

23a - 207 = 12b - 108

23a - 12b = 99 → recall: a = 3b/5

23.(3b/5) - 12b = 99

(69b/5) - 12b = 99 → you multiply by 5 both sides

69b - 60b = 495

9b = 495

→ b = 55

Recall: a = 3b/5

a = (3 * 55)/5

→ a = 33

Two numbers are in the ratio 3 : 5. 

If 9 is subtracted from the numbers, 

the ratio becomes 12 : 23.  

Let the number be 3x and 5x

then according to the question

3x − 9/5x − 9 = 12/23

​⇒ 23(3x − 9) = 12(5x − 9)

⇒ 69x − 60x = 207 − 108

⇒ 9x = 99

⇒ x = 11

The numbers are 33 and 55.

We have the ratio of two numbers equal to 3:5, so:

x/y = 3/5

If 9 is subtracted from each, the ratio is 12:23, so:

(x - 9) / (y - 9) = 12/23

We now have a system of two equations and two unknowns that can be solved:

x/y = 3/5 and (x - 9) / (y - 9) = 12/23

Let's solve the first equation for x in terms of y and the second I'll multiply both sides by y - 9:

x = (3/5)y and x - 9 = (12/23)(y - 9)

If we susbtitute the expression in terms of y, for x, in the second equation:

x - 9 = (12/23)(y - 9)

(3/5)y - 9 = (12/23)(y - 9)

Now, we can solve for y.  I'll multiply both sides by the LCD of 115 to get rid of the fractions:

115[(3/5)y - 9] = 115(12/23)(y - 9)

69y - 1035 = 60(y - 9)

69y - 1035 = 60y - 540

9y = 495

y = 55

Now we can solve for x:

x = (3/5)y

x = (3/5)(55)

x = 3(11)

x = 33

The two original numbers are 33 and 55