Anne, Becky and Charlotte had sums of money in the ratio 7:6:5. One of them gave £9 to one of the others and this changed the ratio (in the same order of names) to 6:5:4. The total sum of money remained the same; what was it?

The total sum of money is £810

Step-by-step explanation:

The parameters given are;

The ratio of the sum Anne, Becky and Charlotte had = 7:6:5

One of them gave one of the others £9 and the new ratio becomes 6:5:4

The total sum of money remains the same = M

The initial fraction amount of the total sum of money each had is given as follows;

Anne = 7/(7 + 6 + 5)×M = 7/18·M = 0.389·M

Becky = 6/18·m = 1/3·M

Charlotte = 5/18·M = 0.278·M

After one of them gave £9 to one of the others, the fraction amount of the total sum of money each had became;

Anne = 6/(6 + 5 + 4)×M = 6/15·M = 0.4·M

Becky = 5/15·m = 1/3·M

Charlotte = 4/15·M = 0.267·M

Therefore, given that Anne's fraction of the total sum of money increased while Charlotte's fraction of the total sum of money decreased, it was Charlotte that gave Anne £9

Therefore, the fraction of the total sum of money equivalent to £9 is found s follows;

5/18·M - 4/15·M = 1/90·M = £9

Therefore;

M = 90 × £9 = £810

The total sum of money = £810.

∠cbe and ∠deb are same-side interior angles.

i took the test

answer:   c ?

step-by-step explanation: