1. The cost $C of hiring a piano is the sum of the deposit $D plus the rental fee. It is known that the monthly rental fee is $R .a) If Miss Lee rents a piano for n months, use C,D,R and n to formulate a formula for the required fee.b) It is given that to hire a piano, the deposit is $3800 and the monthly rental fee is 250. If Miss Lee hires a piano for 2 years, find the required fee.
2.It is know that when the sum of x and 6 is devided by 4, the quotient obtained is less than -1.
a) Formulate an inequality to represent the above situation.
b) Write down three numbers that satisfy the inequality in (a)
3.In the following sequence, the 1st term and the 2nd term are 2 and 3 respectively. For any positive integer n, the (n+2)th term is the product of the nth term and the (n+1)th term. Find the 7th term of the sequence.
2, 3, 6, 18......
Solve the following equations（4，5）
4. (2n-3)/4 + (1-4n)/5 + 7/10 = 0
5. (r+2)/8 - 3(r-2)/6 = 10-r/4
6. If the sum of three consecutive odd number is 81, find the largest odd number.
7.40% of the students in S1A failed in physical test. They had to sit for another test in which 62.5% of the failed again.
a) Find the percentage of students who failed in both tests.
b) If 10 student failed in both tests, how many students are there in S1A?
8. A shopkeeper sold a book for $175, making a profit of $50, Find the precentage profit.
9. Miss Ko sells a chain for $1680, making a precentage profit of 50%. Find the profit gained by her.
10. Mr Chueng buys two watches for $500 and $740 respectively. Then he sells first one at a profit of 10% and sells the second one at loss of 8%. Find the overall percentage profit or less made by him, correct to the nearset at 0.01%
thz=】
S.1-S.2數學:)唔該20點
2011-08-30 6:32 am
回答 (2)
2011-09-09 7:52 am
1a)the formula for the required fee
=n(D+R)=Cb) the required fee
=$3800+$250x(2x12)
=$4050+$250x24
=$4050+$6000
=$100502a) Formulate an inequality to represent the above situation.
=(x+6)/4﹤-1 b) Write down three numbers that satisfy the inequality in (a)
∵(x+6)/4﹤-1 x+6﹤-4 x﹤-10
∴x﹤-10
∴Three numbers of x that can satisfy the inequality is -11,-12 and -13.
3. the 4th term of the sequence
=6x18
=108
the 5th term of the sequence
=108x18
=1944
the 6th term of the sequence
=1944x108
=209952
the 7th term of the sequence
=209952x1944
=408146688
4. (2n-3)/4 + (1-4n)/5 + 7/10 = 0
2n/4-3/4+1/5-4n/5+7/10=0
nx5/4x5-3x5/4x5+1x4/5x4+7x2/10x2=0
(5n-15+4+14)/20=0
(5n+3)/20=0
5n/20+3/20=0
5n/20=3/20
5n=3
n=3/5
5. (r+2)/8 - 3(r-2)/6 = 10-r/4 r/8+2/8-3r/6+6/6=10-r/4
r/8-3r/6+1/4+1=10-r/4
rx6/8x6-3rx8/6x8+5/4=10-rx12/4x12
6r/48-24r/48+5/4+12r/48=10
-6r/48=10-5/4
-r/8=34/4 -r=(34/4)x8
-r=68
r=68x(-1)
r=-68
6.the smallest odd number
=(81-2-4)/3
=75/3
=25
∴The three consecutive odd number is25.27 and29.
7a)the percentage of students who failed in both tests
=40%x62.5%
=25%
7b)there are S1A students
=10/25%
=10x4
=40
8)the precentage profit
= 50/(175- 50)x100%
=50/125x100%
=40%
9)the profit gained by her
=1680x2/1680x3x3x100%
=3360/15120x100%
=2/9x100%
=22 2/9
10)the overall percentage less made by him
=(740x8%-500x10%)/(500+740)x100%
=(59.2-50)/1240x100%
=9.2/1240x100%
=3/4%
11)the overall percentage profit
=5400x(20%-10%)/(5400x2)x100%
=5400x10%/(10800)x100%
=540x100%/10800
=54000%/10800
=5%
12)he percentage profit
=300x(100%-2%)x2.7-450/450
=300x98%x2.7-450/450
=294x2.7-450/450
=793.8-450/450
=343.8/450
=0.764
=76.4%
13)the selling price of the trousers
=228x(100%+140%)/(100+90%)
=228x240%/190%
=$288
14a)the cost price
=1850/40%
=$4625
b)the original cost price of the ring for the jeweler
=4625/(100%+85%)
=4625/185%
=$2500
=n(D+R)=Cb) the required fee
=$3800+$250x(2x12)
=$4050+$250x24
=$4050+$6000
=$100502a) Formulate an inequality to represent the above situation.
=(x+6)/4﹤-1 b) Write down three numbers that satisfy the inequality in (a)
∵(x+6)/4﹤-1 x+6﹤-4 x﹤-10
∴x﹤-10
∴Three numbers of x that can satisfy the inequality is -11,-12 and -13.
3. the 4th term of the sequence
=6x18
=108
the 5th term of the sequence
=108x18
=1944
the 6th term of the sequence
=1944x108
=209952
the 7th term of the sequence
=209952x1944
=408146688
4. (2n-3)/4 + (1-4n)/5 + 7/10 = 0
2n/4-3/4+1/5-4n/5+7/10=0
nx5/4x5-3x5/4x5+1x4/5x4+7x2/10x2=0
(5n-15+4+14)/20=0
(5n+3)/20=0
5n/20+3/20=0
5n/20=3/20
5n=3
n=3/5
5. (r+2)/8 - 3(r-2)/6 = 10-r/4 r/8+2/8-3r/6+6/6=10-r/4
r/8-3r/6+1/4+1=10-r/4
rx6/8x6-3rx8/6x8+5/4=10-rx12/4x12
6r/48-24r/48+5/4+12r/48=10
-6r/48=10-5/4
-r/8=34/4 -r=(34/4)x8
-r=68
r=68x(-1)
r=-68
6.the smallest odd number
=(81-2-4)/3
=75/3
=25
∴The three consecutive odd number is25.27 and29.
7a)the percentage of students who failed in both tests
=40%x62.5%
=25%
7b)there are S1A students
=10/25%
=10x4
=40
8)the precentage profit
= 50/(175- 50)x100%
=50/125x100%
=40%
9)the profit gained by her
=1680x2/1680x3x3x100%
=3360/15120x100%
=2/9x100%
=22 2/9
10)the overall percentage less made by him
=(740x8%-500x10%)/(500+740)x100%
=(59.2-50)/1240x100%
=9.2/1240x100%
=3/4%
11)the overall percentage profit
=5400x(20%-10%)/(5400x2)x100%
=5400x10%/(10800)x100%
=540x100%/10800
=54000%/10800
=5%
12)he percentage profit
=300x(100%-2%)x2.7-450/450
=300x98%x2.7-450/450
=294x2.7-450/450
=793.8-450/450
=343.8/450
=0.764
=76.4%
13)the selling price of the trousers
=228x(100%+140%)/(100+90%)
=228x240%/190%
=$288
14a)the cost price
=1850/40%
=$4625
b)the original cost price of the ring for the jeweler
=4625/(100%+85%)
=4625/185%
=$2500
2011-08-30 8:23 am
1a.) nR+D=C 1b.) the required fee= 2x12(250)+3800=$9800
2a.) (x+6)/4<-1 2b.) (x+6)/4<-1 x+6<-4 x<-10 the required numbers are -11,-12,-13
3.) 20 9952
4.) (2n-3)/4+(1-4n)/5+7/10=0 (10n-15+4-16n+14)/20=0 -6n+3=0 n=1/2
5.) (r+2)/8-3(r-2)/6=(10-r)/4 (3r+6-12r+24)/24=(60-r)/24 15r+30=60-r r=15/8
6.) let a be the required number =3a-3=81 a=28
7a.) The required percentage =40/100x62.5/100=25% 7b.) Let x be the total no. of student 25%x=10 x=40
8.) the required percentage =(50/175-50)x100%=40%
9.) let x be the original price (1680-x)/x x100%=50% x=$1120
the profit: 1680-1120=$560
10.) sum of selling price of 2 watches =500x(1+10%)+740x(1-8%) =$1230.8
overall loss =500+740-1230.8 =$9.2
9.2/1240 x100%=0.74% (cor. to the nearest 0.01%)
2011-08-30 00:31:43 補充：
11.) let $x be the original price of one of the camera and $y be another one
0.9x=5400 x=6000
1.2y=5400 y=4500
the sum= 6000+4500= $10500
overall selling price=5400x2=$10800
overall profit=$10800-10500=$300
overall profit percentage= 300/10500 x100%= 0.03% (cor. to the nearest 0.01%)
2011-08-30 00:36:24 補充：
12.) the percentage profit= [300(1-2%)x2.7]-450 /450 x100%=0.44%
13.) let $x be the original price= 1.9x=228 x=120
the required selling price= 120x(2.4) =$288
2a.) (x+6)/4<-1 2b.) (x+6)/4<-1 x+6<-4 x<-10 the required numbers are -11,-12,-13
3.) 20 9952
4.) (2n-3)/4+(1-4n)/5+7/10=0 (10n-15+4-16n+14)/20=0 -6n+3=0 n=1/2
5.) (r+2)/8-3(r-2)/6=(10-r)/4 (3r+6-12r+24)/24=(60-r)/24 15r+30=60-r r=15/8
6.) let a be the required number =3a-3=81 a=28
7a.) The required percentage =40/100x62.5/100=25% 7b.) Let x be the total no. of student 25%x=10 x=40
8.) the required percentage =(50/175-50)x100%=40%
9.) let x be the original price (1680-x)/x x100%=50% x=$1120
the profit: 1680-1120=$560
10.) sum of selling price of 2 watches =500x(1+10%)+740x(1-8%) =$1230.8
overall loss =500+740-1230.8 =$9.2
9.2/1240 x100%=0.74% (cor. to the nearest 0.01%)
2011-08-30 00:31:43 補充：
11.) let $x be the original price of one of the camera and $y be another one
0.9x=5400 x=6000
1.2y=5400 y=4500
the sum= 6000+4500= $10500
overall selling price=5400x2=$10800
overall profit=$10800-10500=$300
overall profit percentage= 300/10500 x100%= 0.03% (cor. to the nearest 0.01%)
2011-08-30 00:36:24 補充：
12.) the percentage profit= [300(1-2%)x2.7]-450 /450 x100%=0.44%
13.) let $x be the original price= 1.9x=228 x=120
the required selling price= 120x(2.4) =$288
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