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Three consecutive whole numbers are such that if they be divided by 5,3 and 4 respectively, the sum of the quotient is 40. What are the numbers?

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2 Answers

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;
Three consecutive whole numbers are such that if they be divided by 5,3 and 4 respectively, the sum of the quotient is 40.
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