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Find the answer to the fourth grade math problem in primary school! High score! ! !
A, B, C * * has 103 yuan, of which A is 2 times less than B, 3 yuan, C is 3 times more than A, and 7 yuan. How many are there in A, B and C?

Analysis: A is twice as 3 yuan as B. If 3 yuan is given to A, then A is exactly twice as as B; C is three times more than 7 yuan of A. If you take away 7 yuan, then C is exactly three times that of A, and then return 3×3=9 yuan, which is 2×3 times that of B. That is to say, the total amount of money plus 3+3× 3 yuan to be given, minus the 7 yuan taken away, is exactly 2+ 1+2× 3 of B's money.

Answer: b [103+(3+3× 3)-7] ÷ 2+1+2× 3.

= 108÷9

= 12 (yuan) a112× 2-3 = 21(yuan) C 2 1× 3+7 = 70 (yuan)

Party A, Party B and Party C * * have 100 yuan, of which Party A is 2 times less than Party B, and 3 yuan and Party C are 3 times more than Party B, so how much are Party A, Party B and Party C respectively?

Analysis: A is twice as 3 yuan as B. If 3 yuan is given to A, then A is exactly twice as as B; C is three times more than B, 7 yuan. If you take away seven dollars, then C is exactly three times that of B. That is to say, adding three dollars to A's money and subtracting seven dollars from it is exactly 2+ 1+3 of B's money.

Answer: b (100+3-7) ÷ (2+1+3)

=96÷6

= 16 (yuan) A 16× 2-3 = 29 (yuan) C 16× 3+7 = 52 (yuan)

Note: This question is simpler than the last one.