標題:
F.1 maths
發問:
1) 3 consecutive numbers, the smallest number is 19/20 of the largest one,find the middle nunber 2) 2 consecutive even numbers such that 2 times the larger one plus 1/4 of the smaller one equals to 49 let _____ be y neet step 更新: 3) 16 years later,she will be twice as old as she was 12 years old. find her present age. let _____ be y neet step
最佳解答:
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1) let the smallest number is y, then the middle would be y+1, the largest would be y+2 because the smallest number is 19/20 of the largest one, so y = (y+2)*19/20 y = (19y+38)/20 20y = 19y+38 20y-19y = 38 y = 38 so the smallest number is 38, and the middle number is 38+1=39 2) Let the smaller number is y, then the larger number is y+2 because 2 times the larger one plus 1/4 of the smaller one equals to 49, so 2*(y+2) + 1/4(y) = 49 2y + 4 + y/4 = 49 2y + y/4 = 45 (8y + y)/ 4 = 45 9y = 180 y = 20 so the smaller number is 20, while the larger number is 22
其他解答:
3) let present age be y y+16 = 2 x 12 y+16 = 24 y = 8 present age is 8 years old|||||1.let the largest number be y ans: 19/20y, (19/20y+y)2 ,y 明嗎?? 2.有d唔明問題 係咪最大個數 是 最細個數既1/4 若係既話 let the smallest one be y y+1/4y+1/4y = 49 第二條唔明...因此唔肯定..sor...
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