it is basically trial and error.. trying to match the "start" ratio and "end" ratio with the changes ie + 3 children and -2 adults.Sruthi wrote:There were 3/5 as many children as adults in the bus. At the next bus stop, 3 children boarded and 2 adults alighted from the bus. Then, there were 5/6 as many children as adults on the bus. How many children were there on the bus at first?

The method from the teacher not able to understand .

How did we get 12:20 .

Can anyone explain please ?

before ratio-> 3: 5 = 6:10 = 12:20

after ratio-> 5:6=10:12 = 15:18

12 children (before) --> 15 children (after) .. + 3 children

20 adults (before) --> 18 adults (after) ... - 2 adults

so,

at first, number of children : adults = 12: 20

in the end, number of children : adults = 15 : 18

a more mathematical way is to examine the start ratio and end ratio.

Start ratio, children : adults = 3:5

end ratio, children : adults = 5:6

we see that the number of children at first is a multiple of 3, after 3 more children boarded, the final number is a multiple of 5.

Adding 3 to a multiple of 3 will still yield a multiple of 3, therefore the final number of children must be a multiple of both 3 & 5 --> LCM = 15

so, in the end, possible number of children : adults

= 15: 18 = 30:36 = 45:54.....

we can now reverse the changes (-3 children and + 2 adults) and list (listing heuristics) possible number of children and adults at first, and check which will yield the ratio of 3:5

ratio of children : adults

at end = 15:18 = 30: 36 = 45:54 ..

at start = 12:20 = 27: 38 = 42:56...

since 12:20 = 3:5,

therefore the number of children and adults at start are 12 and 20 respectively, and

the number of children and adults in the end are 15 and 18 respectively.

by the way, this question can be solved using model method.

cheers.