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標題:
比例和變數法的數學題
發問:
1.已知x+y隨x-y正變 ，求證x隨y正變2.三箱盛滿A,B及C三種茶,第一只盛A茶16kg,第二只盛B茶28kg,而第三只盛C茶 36kg. 若將各茶混合后分別放回各箱,求第三箱中各種茶多少kg.3.3種物質A,B,C之體積比為2:3:4,若體積相同他們的重量比為3:4:5,若這三種物 質之混合物重57/8 kg,問各物質之重量.4.兩種金屬A,B做成兩種合金,第一種A:B=5:3 ; 第二種A:B=4:7,如有A 253噸,B 161噸,且全部金屬皆用完,求各種金屬之重量.5.X,Y,Z 合作營商.最初X投資2500元,Y 3000元及Z 1500元. 若7個月后,Y取回 ... 顯示更多 1.已知x+y隨x-y正變 ，求證x隨y正變 2.三箱盛滿A,B及C三種茶,第一只盛A茶16kg,第二只盛B茶28kg,而第三只盛C茶 36kg. 若將各茶混合后分別放回各箱,求第三箱中各種茶多少kg. 3.3種物質A,B,C之體積比為2:3:4,若體積相同他們的重量比為3:4:5,若這三種物 質之混合物重57/8 kg,問各物質之重量. 4.兩種金屬A,B做成兩種合金,第一種A:B=5:3 ; 第二種A:B=4:7,如有A 253噸,B 161噸,且全部金屬皆用完,求各種金屬之重量. 5.X,Y,Z 合作營商.最初X投資2500元,Y 3000元及Z 1500元. 若7個月后,Y取回 1000元,而8個月后Z再投資2000，如一年之利潤5568元。問如何分配？
最佳解答:
Answers are: 1) Let ( x + y ) = k( x - y ), k being a non-zero constant. ( k + 1 )y = ( k - 1 )x y = [ ( k - 1 )/( k + 1 ) ] x, both ( k - 1 ) and ( k + 1 ) are constants Hence, y varies directly as x ======================= 2) Ratio of A : B : C in the mixture = 16 : 28 : 36 = 4 : 7 : 9 Hence, A occupies 4/20 = 20% of the total. B occupies 7/20 = 35% of the total. C occupies 9/20 = 45% of the total. After mixture and put 36 kg into container C, Weight of Tea A = 36 kg x 20% = 7.2 kg ================================= Weight of Tea B = 36 kg x 35% = 12.6 kg ================================== Weight of Tea C = 36 kg x 45% = 16.2 kg ================================== 3) Weight ratio of A : B : C = 2(3) : 3(4) : 4(5) = 6 : 12 : 20 = 3 : 6 : 10 Using Total weight is 57/8 kg and Ratio 3 : 6 : 10 Weight of A = ( 3/19 )( 57/8 kg ) = 9/8 kg =============================== Weight of B = ( 6/19 )( 57/8 kg ) = 9/4 kg =============================== Weight of C = ( 10/19 )( 57/8 kg ) = 15/4 kg ================================= 4) Let Weight of A and B in Alloy 1 be 5x kg and 3x kg resp. ( x non-zero ) Let Weight of A and B in Alloy 2 be 4y kg and 7y kg resp. ( y non-zero ) Since all A and B are used up, we can set up Equation 1: 5x + 4y = 253 [ Total weight of A ] Equation 2: 3x + 7y = 161 [ Total weight of B ] Solve them, we have ( x, y ) = ( 49, 2 ) Total weight of Alloy 1 = 8x kg = 392 kg ================================= Total weight of Alloy 2 = 11y kg = 22 kg ================================= 5) No concrete answer for we do not know how long is the profit obtained be compounded ( monthly or quarterly or else ). And please be reminded for the people who withdraws the money, it is possible his profit in the first several months would not be counted. But the problem has not clearly defined. Hope I can help you.
其他解答:
比例和變數法的數學題
發問:
1.已知x+y隨x-y正變 ，求證x隨y正變2.三箱盛滿A,B及C三種茶,第一只盛A茶16kg,第二只盛B茶28kg,而第三只盛C茶 36kg. 若將各茶混合后分別放回各箱,求第三箱中各種茶多少kg.3.3種物質A,B,C之體積比為2:3:4,若體積相同他們的重量比為3:4:5,若這三種物 質之混合物重57/8 kg,問各物質之重量.4.兩種金屬A,B做成兩種合金,第一種A:B=5:3 ; 第二種A:B=4:7,如有A 253噸,B 161噸,且全部金屬皆用完,求各種金屬之重量.5.X,Y,Z 合作營商.最初X投資2500元,Y 3000元及Z 1500元. 若7個月后,Y取回 ... 顯示更多 1.已知x+y隨x-y正變 ，求證x隨y正變 2.三箱盛滿A,B及C三種茶,第一只盛A茶16kg,第二只盛B茶28kg,而第三只盛C茶 36kg. 若將各茶混合后分別放回各箱,求第三箱中各種茶多少kg. 3.3種物質A,B,C之體積比為2:3:4,若體積相同他們的重量比為3:4:5,若這三種物 質之混合物重57/8 kg,問各物質之重量. 4.兩種金屬A,B做成兩種合金,第一種A:B=5:3 ; 第二種A:B=4:7,如有A 253噸,B 161噸,且全部金屬皆用完,求各種金屬之重量. 5.X,Y,Z 合作營商.最初X投資2500元,Y 3000元及Z 1500元. 若7個月后,Y取回 1000元,而8個月后Z再投資2000，如一年之利潤5568元。問如何分配？
最佳解答:
Answers are: 1) Let ( x + y ) = k( x - y ), k being a non-zero constant. ( k + 1 )y = ( k - 1 )x y = [ ( k - 1 )/( k + 1 ) ] x, both ( k - 1 ) and ( k + 1 ) are constants Hence, y varies directly as x ======================= 2) Ratio of A : B : C in the mixture = 16 : 28 : 36 = 4 : 7 : 9 Hence, A occupies 4/20 = 20% of the total. B occupies 7/20 = 35% of the total. C occupies 9/20 = 45% of the total. After mixture and put 36 kg into container C, Weight of Tea A = 36 kg x 20% = 7.2 kg ================================= Weight of Tea B = 36 kg x 35% = 12.6 kg ================================== Weight of Tea C = 36 kg x 45% = 16.2 kg ================================== 3) Weight ratio of A : B : C = 2(3) : 3(4) : 4(5) = 6 : 12 : 20 = 3 : 6 : 10 Using Total weight is 57/8 kg and Ratio 3 : 6 : 10 Weight of A = ( 3/19 )( 57/8 kg ) = 9/8 kg =============================== Weight of B = ( 6/19 )( 57/8 kg ) = 9/4 kg =============================== Weight of C = ( 10/19 )( 57/8 kg ) = 15/4 kg ================================= 4) Let Weight of A and B in Alloy 1 be 5x kg and 3x kg resp. ( x non-zero ) Let Weight of A and B in Alloy 2 be 4y kg and 7y kg resp. ( y non-zero ) Since all A and B are used up, we can set up Equation 1: 5x + 4y = 253 [ Total weight of A ] Equation 2: 3x + 7y = 161 [ Total weight of B ] Solve them, we have ( x, y ) = ( 49, 2 ) Total weight of Alloy 1 = 8x kg = 392 kg ================================= Total weight of Alloy 2 = 11y kg = 22 kg ================================= 5) No concrete answer for we do not know how long is the profit obtained be compounded ( monthly or quarterly or else ). And please be reminded for the people who withdraws the money, it is possible his profit in the first several months would not be counted. But the problem has not clearly defined. Hope I can help you.
其他解答:
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