A question on ratio and proportion?

Question

Mr ram has five sons -a,b,c,e and g.He had some chocolates with him which he distributed in the following manner .Twice the no.of chocolates received by a,thrice the no.of chocolates received by c and four times the no .of chocolates received by e are equal.Five times the no .of chocolates received by b,six times the no.of chocolates received by e and 8 times the no.of chocolates received by g are equal.Find th e minimum no.of of chocolates that could have been distributed by MR.RAM.

Answer 1

umm.

2a = 3c = 4e

LCM is 12 so

ratio of a:c:e is

6 : 4 : 3

then 5b = 6e = 8g

lcm is 120

so b: e : g is

24 : 20 : 15

a:c:e : b :g

5:4:3

     20:24:15

the e in the first ratio is 3 and the e in the second one is 20... lcm is 60 so ans is

a    : c: e :  b : g

100:80:60:72 : 45

so... i guess since it can't be simplified... the minimum is 100 + 80 + 60 + 72 + 45 which is = 357

Answer 2

Let the number of chocolates got to a, b, c, d, e, g be as per their names.

2a = 3c = 4e

5b = 6e = 8g

consider 4e and 6e , LCM of 4 and 6 is 12

2a = 3c = 4e multiply by 3

6a = 9c = 12 e

5b = 6e = 8g multiply by 2

10 b = 12 e = 16 g

comparing

6a = 9c = 12 e = 10 b = 12 e = 16 g

a = 2e

b = 6/5 e

c = 4/3 e

g = 3/4 e

total = 2 e + 6/5 e + 4/3 e + e + 3/4 e

= (120 e + 72 e + 80e + e + 45e)/60

= 318e/60 = 53 e/10

to get full number e = 10

a = 20

b = 12

c = 40/3 ---------- odd number

g = 15/2 ------- odd number.

e = 10

to get full number of chocolates multiply be LCM of 2 and 3 = 6

a = 120

b = 72

c = 80

g = 45

e = 60

Total = 378 chocolates

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