maths(20)

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馬小姐的上班時間是9點正，公司規定員工遲到會扣減工資，每過5分鐘扣減10元，如此類推。馬小姐在新界居住，每次上班必須先乘坐巴士45分鐘，再乘專線小巴18分鐘，最後步行15分鐘。若巴士的平均行車速率為每小時65公里，專線小巴行車速率為每小時50公里，平均步行速率為每小時4公里。
(a)問馬小姐的住處與公司相距多遠?
(b)馬小姐最遲要在什麼時候由住處出發才可準時回到公司?
(c)若以每小時30公里的平均速率走畢全程，馬小姐需要提早多少分鐘由住處出發，才可準時回到公司。
(d)馬小姐今天8時正由住處出發上班，途中發生了交通意外，巴士路段只能以車速每小時40公里行走，中途轉車耽誤了20分鐘，小巴路段時速亦比平時慢了10公里。若其他因素不變，問馬小姐於什麼時候才能返抵公司?被扣減了多少工資?

(a)∵距離d=速率s x 時間t(hour)
∴d=45/60 x 65 + 18/60 x 50 + 15/60 x4
=64.75km

(b)假設沒有任何影響因素
0900-(45mins+18mins+15mins)
=0742

(c)時間=距離/速率
=64.75/30
=129.5mins
=2hours & 9.5mins
∴佢大約要早0742-[0900-(2hours+ 9.5mins)]
=0742-0650(必需把9.5當成10)
=52mins

(d)∵巴士要行走的距離是48.75km
∴共需時(48.75/40)=73.125mins
∵小巴要行走的距離是15km
∴共需時[15/(50-10)]=22.5mins
加上中途轉車耽誤了20分鐘和15分鐘的步行
整個行程需時73.125+22.5+20+15=2hours & 10.625mins
∴大約0800+(2ours & 10.625mins)=1011到達公司
∴要扣減工資(60/5)x10+(10/5)x10=$140

1.it is given that a=30 and b=50.if a and b are decreased by 5% and 8% respectively,
find the percentage decrease in their sum.

2.A square has side 15 cm.if each side is increased by 20%,find
(a) the percentage increase in perimeter.
(b)the percentage increase in area.

3.given that a=20 and b=15,
(a)by what percentage is a greater than b?
(b)by what percentage is b less than a?

4.the number of members in a Mathematics is increase by 10% to become 110.what is the original number of members?

5.the price of a computer is decreased by 20% to become $6000.what is the original price of the computer?

6.55% of F.1 students in a school pass an English test.if 110 students pass the test,find the total number of students in F.1

1. (new value-original value)/original value x 100%
original value is given as 30+50=80
the new value is 30(1-5%)+50(1-8%)=74.5
hence (74.5-80)/80 x 100%=-6.875%
∴The percentage decreased in their sum is 6.875%

2(a). The original side is 15
hence the perimeter is 15x4=60
The changed side is 15(1+20%)=18
hence the perimeter is 18x4=72
∵The percentage changes=(new value-original value)/original value x 100%
∴The percentage increase in perimeter is (72-60)/60 x 100%=20%
(b)The area of 15cm side is 225
The new area of 18cm side is 324
Therefore The percentage increase in area is (324-225)/225 x 100%=44%

3. I do not know what is the question about

4. The original number of members is 110÷(1+10%)=100

5. The original price of the computer is $6000÷(1－20%)=$7500

6. According to the formula
(Sample/Population) x 100%=The percentage of sample in population
let P be the unknown as population
thus (110/P) x 100%=55%
P=200
∴The total number of students in F.1 is 200

1) Given that (x ^ 2) + 6 x identically equal to [ ( x + p ) ^ 2 ] + q,
find the values of the constant p and q.

2) If [ ( 2 x + 1 ) ^ 2 ] - p x identically equal to ( q x ^ 2 ) + 1,
where p and q are constant, find the values of p and q.

1. x^2+6x≡(x+p)^2+q
x^2+6x≡x^2+2xp+p^2+q
hence 6x=2xp
p=3
∵p^2+q=0
∴3^2+q=0
q= -9

2. (2x+1)^2-px≡qx^2+1
4x^2+(4-p)x+1≡qx^2+1
we can see that q=4
and 4-p=0
thus p=4

23+64-7=80

5-1=4

84+6=90

150x50%=150x50/100=150x1/2=75

300x80%=300x80/100=300x4/5=240

500x15%=500x15/100=500x3/20=75

希望幫到您=]

計算下例各題:

1+1=2

2+2=4

3+3=6