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Jimmy L. | Experienced Wall Street Analyast/CFAExperienced Wall Street Analyast/CFA

here is how I do it:

1) "Three consecutive whole numbers": n, n+1, n+2;

2) "if they be divided by 5,3 and 4 respectively": n/5, (n+1)/3, (n+2)/4

3) "the sum of the quotient is 40": 40= n/5 + (n+1)/3 + (n+2)/4

4) solve n: 40x5x4x3= nx3x4 + (n+1) x 5 x4 + (n+2) x5X3

--> n=50

Answer: these numbers are: 50, 51, 52

Double check:

50/5=10

51/3=17

52/4=13

10+17+13=40

Hi Abhishek;

n

n+1

n+2

n/5

(n+1)/3

(n+2)/4

40=(n/5)+[(n+1)/3]+[(n+2)/4]

Let's eliminate annoying fractions.

Multiply (5)(3)(4)=60

Let's multiply both sides by 60...

(60)(40)=60{(n/5)+[(n+1)/3]+[(n+2)/4]}

2400=12n+[20(n+1)]+[15(n+2)]

2400=12n+20n+20+15n+30

Combine like terms...

2400=47n+50

Subtract 50 from both sides...

2400-50=47n+50-50

2350=47n

Divide both sides by 47...

2350/47=(47n)/47

50=n

51=n+1

52=n+2

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