16. A sum of money was supposed to be distributed
between M and N in the ratio 4: 5 but was instead
distributed among M, N and P in the ratio 3 : 2 : 1.
If M gained 5,000 due to this, what was the total
amount distributed?

Answer :

Answer:

Let's denote the original amount of money intended for distribution between M and N as

9

9x (since the ratio of distribution was 4:5, and 4 + 5 = 9 parts).

According to the problem, due to incorrect distribution, the money was distributed among M, N, and P in the ratio 3:2:1. Hence, the total parts in this distribution ratio are

3

+

2

+

1

=

6

3+2+1=6 parts.

Let the actual amounts received by M, N, and P be

3

3y,

2

2y, and

y respectively. Then we have:

3

+

2

+

=

9

3y+2y+y=9x

6

=

9

6y=9x

=

9

6

=

3

2

y=

6

9x

=

2

3x

From the problem, M gained 5,000 due to this incorrect distribution. Therefore, M's gain is

3

4

=

5000

3y−4x=5000.

Substituting

=

3

2

y=

2

3x

into the equation gives:

3

(

3

2

)

4

=

5000

3(

2

3x

)−4x=5000

9

2

4

=

5000

2

9x

−4x=5000

9

8

2

=

5000

2

9x−8x

=5000

2

=

5000

2

x

=5000

=

2

×

5000

=

10000

x=2×5000=10000

Now, we can find the total amount distributed originally:

9

=

9

×

10000

=

90000

9x=9×10000=90000

Therefore, the total amount distributed was

90000

90000

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