痛过的人

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jug x and y contain different amounts of water at first.50% of the water in jug x was poured into jug y.then 40% of the water in jug y was poured into jug x.the final ratio of the amount of water in jug x to the amount of water in jug y was 7:6
a)what was the ratio of the amount of water in jug x to the amount of water in jug y at first?
b)if there were 10 litres of water in jug x at first,how much water was there in jug x at the end?

痛过的人

回复 静雯 的帖子

alina and adeline had some stickers in the ratio 3:5.when alina bought 42 more stickers and adeline bought 7 more stickers,the ratiobecame 6:7.find the niumber of stickers alina had at first.

frekiwang

Please help with the following question:

A. B and C have some marbles.
The ratio of the marbles will become 5:3:4 if both B and C give 6 marbles each to A;
The ratio of the marbles will become 5:6:10 if both A and B give 3 marbles each to C.
Find the number of marbles each person has originally.

静雯

回复 痛过的人 的帖子

Solution: (Using working backwards)
a)  After the second pouring, the final ratio = 7 : 6

(The second pouring, 40% of the water in jug y was poured into jug x.  
The ratio of the amount of water poured out to the amount of water left in jug y = 40% : 60% = 4 : 6.  
That means 4 units water poured into jug x, so)
After the first pouring, the ratio of the amount of water in jug x to that of y =(7-4):6+4)=3:10
(The first  pouring, 50% of the water in jug x was poured into jug y.  There were 3 units (50%) water left in jug x. That means the original water in jug x is (3+3) units.
The original water in jug y = (10-3) units.)
The original water in jug x : the original water in jug y = (3+3) : (10-3) = 6: 7.
The ratio of the amount of water in jug x to the amount of water in jug y at first is 6:7.
b) 6 units = 10 litres  ==>   1 units = 10/6=5/3 litres  ==>  7 units = 5/3*7=35/3 litres.
There was 35/3 litres water in jug x at the end.

Note: It's a good habit to check your answer.

静雯

回复 痛过的人 的帖子

When you ask your question next time, please let me know whether the problem comes from secondary book or primary book.  Because we use different methods at different period. Thanks.

静雯

回复 痛过的人 的帖子

If it is a primary problem, you can do like the following:
Solution:
(The following is a table)
                              Alina        Adeline
before change:       3       :        5       =   6 :  10   ( 6 units :  10 units)
   their change:      +42           +7
after   change:        6       :        7      = 6 UNITS :  7 UNITS
(According to the above table, you can get:   )
6 units + 42  = 6 UNITS       ==>      1 unit +7 =  1 UINT      (From Alina's changed)
                                             ==>  7 UINTS = 7 units +49
10 units +7 = 7 units +49   ==>  3 units = 42   ==> 1 unit =14
6 units = 14 * 6 =84

Alina had 84 stickers at first.

静雯

回复 frekiwang 的帖子

The key to find the solution of the question is the total amount of the marbles they had unchanged.

Sorry.  Could you check your data in your problem?  Or scan your problem?

frekiwang

Sorry, I got the answer already because there was a typo erron in the question.

In case 1, A has 5/12 of the total while in case 2 A has 5/21 of the total. The difference in the number of marbles that A has(receiving 6x2 and giving 3) is 6x2+3=15, which is equivalent to 5/12-5/21=5/28.
15 ->5/28
3 -> 1/28
total=3x28=84
A has 84 x 5/12 - 12 = 23
B has 84 x 3/12 + 6 = 27
C has 84 x 4/12 + 6 =34

静雯

回复 frekiwang 的帖子

Well done!  

Your working is very neat!

静雯

回复 frekiwang 的帖子

Could you correct your type error here or type your problem again?

Thanks.