Elsa Winda Prastiana
09313244015
International Math Education ‘09
Exercise No.1 Page 95 Mathematics for Junior High School 2
1.Determine the line equation through point (0,0) and has gradient :
a.12
b.16
c.-3
Solution :
a.m=12
Since the line through point (0,0) then generally the line has equation y=mx. Given m=12, thus the line equation is y=12x.
b.m=16
Since the line through point (0,0) then generally the line has equation y=mx. Given m=16, thus the line equation is y=16x.
c.m=-3
Since the line through point (0,0) then generally the line has equation y=mx. Given m=-3, thus the line equation is y=-3x.
Exercise No.3 Page 95 Mathematics for Junior High School 2
2.Find the equation of a line through point (0,0) and is perpendicular to a line whose gradient of -1/6 .
Solution :
Since, general equation of line a is y =mx. Known the line a is perpendicular with another line with gradient of -1/6. Since the multiple gradient result of the two perpendicular line is -1, then the line gradient of a is 6. So, the line equation of a is y=6x.
Exercise No.2 Page 115 Mathematics for Junior High School 2
3.The price of 2kg of oranges and 3 kg mangos is Rp 38,000.00. Determine the price of 1 kg of orange if known that 1 kg of mango is Rp7,000.00
Solution :
Change the given information in the word problem into mathematical expression. For instance, let 1 kg orange in a and 1 kg mango in b. Accordingly, 2kg of oranges in 2a and then 3kg of mangos is 3b. Therefore we will get an LETV of 2a+3b=38.000. Fom the information we know that b=7.000. Subtitute b=7.000 to the equation of 2a+3b=38.000, thus resulting
↔2a+3(7.000)=38.000
↔2a+21.000=38.000
↔2a=38.000-21.000
↔2a=17.000
↔a=8.500
So, the price of 1 kg of orange (a) = Rp 8.500,00
Exercise No.1 Page 45 Mathematics for Junior High School 3
4.In the afternoon, the length of student’s shadow that has 150cm high is 50cm. If at the same time, the length of the tower’s shadow is 10cm, what is the height of the tower ?
Solution :
The student’s height is 150cm, the length of the student’s shadow is 50cm, and the length of tower’s shadow is 10m (1000cm). The height of the students corresponds to the height of the tower. The length of student’s shadow corresponds to the length of the tower’s shadow so that he proportion the corresponding sides is :
(the height of the tower)/(the height of the student)=(the length of the tower's shadow)/(the length of the student's shadow)
Let the height of tower is h cm, so by using the proportion in similarity, we get :
h/150=1000/50 ↔ 50h=1000×150
↔ 50h=15.000
↔ h=15.000/50
↔ h=300
So, the height of the tower is 300cm=3m.
Exercise No.2 Page 189 Mathematics for Junior High School 1
5.In a shelve, there are Mathematic books, Indonesian Language books, and Physics books with the ratio of 4∶2∶3. Determine the numbers of Physics books and Mathematical books if the Indonesian Language books is 6 books.
Solution :
The ratio among Mathematic books, Indonesian language books, and physics books is 4∶2∶3. The number of Indonesian Language books is 6 books.
The number of Indonesia Language books = (ratio of indonesia language books)/(total ratio)×total books
6 = 2/9 ×total books
6× 9/2 = total books
27 = total books
From that we know that the total number of books is 27 books.
So the number of Math books = (ratio of math books)/(total ratio)×total books
= 4/9×27 = 12 books
The number of Physics books = (ratio of physics books)/(total ratio)×total books
= 3/9×27 = 9 books
