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# 英文數學題(20點,急!)

1)Find two fractions with different denominators such that their sum is 19/24.

2)A piece of string is cut into four parts.Two parts are equal in length,and the difference between the lengths of the two remaining parts is 2cm.Suggest the original length of the piece of string.

3)Write down two numbers with H.C.F. as a multiple of 4 and L.C.M. as 240.

4)Suggest two fractions with different denominators but the same numerators such that their difference is 5/12.

5)Suggest three different mixed numbers such that their product is an integer.

謝了！

no missing

### 1 Answer

- 比分快!開心哂Lv 61 decade agoFavorite Answer
Q2. any missing parts?

2008-08-26 10:22:23 補充：

1. let the fractions be y/Y and x/X, such that y<Y and x<X, then y/Y +x/X= 19/24 = (Xy+Yx)/XY, therefore Xy+Yx=19, and XY=24

for XY =24, combinations may be 1x24, 2x12, 3x8 and 4x6 only

for 1x24, if either X or Y is 1, their sum will be greater than 1, so 1x24 is ignored.

for 2x12, both even X and Y multiple with y and x will give out even nominnator, so 2x12 and 4x6 also be neglected.

only left 3x8, i.e 3y+8x=19 or(8y+3x=19, is same), since y<Y=8 and x<X=3, by plotting inequality diagram, only the point (2,1) match these requirements, therefore the two fractions are

1/8 and 2/3

2. should be some missing information, pls clarity

3. factorise 240, give 1x240, 2x120, 3x80, 4x60, 5x48, 6x40, 8x30, 10x24, 12x20

for HCF is multiple of 4, only 4x60, 5x48, 6x40 and 12x20 are possible for HCF = 4, 20 and 48 is the combination

for HCF = 8, 40 and 48 is the combination

for HCF = 12, 60 and 48 is the combination

4. let the fractions be y/Y and y/X, such that y<Y and y<X and Y>1 and X>1, then y/Y-y/X =5/12, therefore Xy-Yy=5 and XY=12

For XY =12, combinations may be 1x12, 2x6, 3x4, only 3x4 combination will output odd nominator,

Therefore 4y-3y=5, y=5, therefore Y=3 and X=4, the two fractions are 5/3 and 5/4

5. A(X/Y)*B(Y/Z)*C(W/X), where A, B, C can be any integer and Z>Y>X>W, where W=Z-X

for C(W/X) can be written as C[(Z-X)/X],

one combination for Z,Y,X and W is 9,8,7,2 i.e. A(7/8)*B(8/9)*C(2/7) =A(7/8)*B(8/9)*(C)[(9-7)/7]

= A(7/8)*B(8/9)*(C-1)(9/7) =A*B*(C-1) which must be an integer

for C=1, product will became A*B instead of zero.

2008-08-26 13:16:12 補充：

sorry misunderstood the question

5 should be

One possible example of the three mixed numbers are 1(1/2), 1(1/3), 2(1/2) --> (3/2)(4/3)(5/2) = 5

2008-08-27 13:29:13 補充：

Q2. any missing information or missing numbers?

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