My First Experiment

1.      The proportion between the length of two similar sides is 2:3. If the diagonal length of the small rectangle is 30 cm, so the length of the diagonal of the bigger rectangle is … .

a.       20 cm

b.      45 cm

c.       50 cm

d.      55 cm

Solution:

Known, there are two rectangle A (small) and B (bigger). Proportion between the length of two similar sides is . It shows that length A similar with length B. So, . Diagonal of small rectangle is 30 cm. Because the rectangle A and B are similar, so, diagonal A is similar to  diagonal B. That is state  .

To find the diagonal B, we can calculate with equation  =  ,  = .

Then dB = . Three multiplied by 30 then divided by 2 equal to 45.

We can conclude, that the length of the diagonal of the bigger rectangle is 45 cm.

The answer is b.

Source: Mathematics for Junior High School 3, Bab 1, Exercise 1 question A.11, page 48

2.      Mother buys a half of watermelon. The perimeter of surrounding watermelon is 62.8 cm. (the water melon is assumed as a sphere). The volume of the watermelon is … .

a.       628 cm³

b.      1,256 cm³

c.       2,093.33 cm³

d.      4186.67 cm³

Solution:

Known the surrounding a half of watermelon is 62.8 cm. We will calculate the volume of that watermelon. So, we need radius of the watermelon. To find the radius, we can use surrounding formula of circle that is S=2r. The formula can be changed r =  . that we use is 3.14 to make easy when we calculate it.

2 multiplied by 3.14 equal to 6,28 then 62,8 by 6.28 equal to10 cm.

After we find the radius, we can calculate the volume of a half watermelon with volume formula of sphere divided by 2, V=  /2. V=  x 3.14 x 1000 /2.  multiplied by  multiplied by 1000 equal to 4,186.67. Then 4,186.67 divided by 2 equal to 2,093.33.

So, the volume of a half watermelon is 2,0933.33 cm³. The answer is c.

Source: Mathematics for Junior High School 3, Bab 2, Exercise 2 question A.15, page 80

3.      The mean of the art test score from a certain group of student consist of 5 students is 75. After a new member join the group, the average turns into 73. The art test score of the new member is … .

a.       73

b.      70

c.       63

d.      60

Solution:

Known, the first mean from a certain group of student consist of 5 students is 75. Total score of 5 students can be calculated by mean multiplied by total student, 75 multiplied by 5 equal to 375. Then known that a new student join the group, so, the total member is 5 students plus 1 students equal to 6 students. Then mean change be 73. Total score of 6 students can be calculated by 73 multiplied by 6 equal to 438.

The score of the new member can be calculated by total second score minus total first score, 438 minus 375 equal to 63.

So, we can conclude that the art test score of the new member is 63. The answer is c.

Source: Mathematics for Junior High School 3, Bab 3, Exercise 3 question A.5, page 132

4.      There are 2 boxes. Each box contains 5 balls. The ball in every box are labeled 1 to 5. Then, from each box 1 ball is taken randomly all at once.

a.       What is the probability of obtaining that balls of each box has the same label?

b.      What is the probability of obtaining an odd numbered ball from the first box?

Solution:

S₁ is sample space of balls in the first box and member space of the first box is S₁={1,2,3,4,5}. So, the total member of sample space S₁ is n(S₁)=5.

S₂ is sample space of balls in the first box and member space of the first box is S₂={1,2,3,4,5}. So, the total member of sample space S₂ is n(S₂)=5.

a.       We will calculate probability of obtaining that the balls of each box has the same label. So, the event of the balls from the fist and second box is same. The event from the first box is A={1,2,3,4,5} and total member of event A is n(A)=5. The event from the first box is B={1,2,3,4,5} and total member of event B is n(B)=5.

Event A and event B is called independent events. To find the probability, it is used P(AB)=P(A)xP(B). P(A) is probability of event A. It is can be found by member of event A divided by member space of sample S₁ then the formula is  .

P(B) is probability of event B. It is can be found by member of event B divided by member space of sample S₂ then the formula is 

Then the steps are P(AB)=P(A)xP(B)

We can conclude that the probability of obtaining that the balls of each box has the same label is 1.

b.      If A is an event of obtaining an odd numbered ball from the first box, so, the member of event A is A={1,3,5} and the total member is n(A)=3. Because the event B in the second box is not change, so, the member is B={1,2,3,4,5} and total member is n(B)=5

Then                      P(AB)=P(A)xP(B)

We can conclude that the probability of obtaining an odd numbered ball from the first is .

Source: Mathematics for Junior High School 3, Bab 3, Exercise 3 question B.4, page 136

5.      The first term of arithmetic sequences is 6, and the 5^th term is 22. Find the value of the divergent of the arithmetic sequences.

Solution:

From the question we have first term, a=6 and 5^th term, U₅=22. So, we can calculate its divergent by using formulas U_n=a+(n-1)b.

Then U_n=a+(n-1)b

U₅=6+(5-1)b

22=6+4b

4b=16

  b=4

So, we can conclude that the value of the divergent of the arithmetic sequences is 4.

Source: Mathematics for Junior High School 3, Bab 5, Exercise number 2, page 195