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1. Final Exam Revision
Teacher :Saeed YSaeed Grade:6
Date: May 31 , 2023 Subject : Math

2. Question 1:
a) b)
c) d)
A) make net to represent the following .
B) Find the surface area of the following .

3. ANSWER:
a) b)
c) d)
A)

4. ANSWER:
a) Area =front + back + top + bottom + side + side
Front=back , top=bottom , side=side .
𝐴𝑟𝑒𝑎𝑓𝑟𝑜𝑛𝑡 = 𝐿 𝑋 𝑊 = 10 X 8 = 80
𝐴𝑟𝑒𝑎𝑇𝑂𝑃 = 𝐿 𝑋 𝑊 = 10 X 6 = 60
𝐴𝑟𝑒𝑎𝑆𝑖𝑑𝑒 = 𝐿 𝑋 𝑊 = 6 X 8 = 48
Front=back=80 , top=bottom=60 , side=side=48
Area =front + back + top + bottom + side + side
= 80 + 80 + 60 + 60 + 48 + 48
= 376
b)Area = Base + base + top + bottom + side
𝐴𝑟𝑒𝑎𝑡𝑟𝑎𝑖𝑛𝑔𝑢𝑙𝑎𝑟 𝑏𝑎𝑠𝑒 =
1
2
ℎ 𝑋 𝑏 =
1
2
𝑥 12 𝑥 5 =30
A) 𝐴𝑟𝑒𝑎𝑡𝑟𝑎𝑖𝑛𝑔𝑢𝑙𝑎𝑟 𝑏𝑎𝑠𝑒 =
1
2
ℎ 𝑋 𝑏 =
1
2
𝑥 12 𝑥 5 =30
𝐴𝑟𝑒𝑎𝑡𝑟𝑎𝑖𝑛𝑔𝑢𝑙𝑎𝑟 𝑏𝑎𝑠𝑒 =
1
2
ℎ 𝑋 𝑏 =
1
2
𝑥 12 𝑥 5 =30
𝐴𝑟𝑒𝑎𝑡𝑜𝑝 =L X W =13 𝑥 15
=195
𝐴𝑟𝑒𝑎𝐵𝑂𝑇𝑇𝑂𝑀 =L X W =12 𝑥 15
=180
𝐴𝑟𝑒𝑎𝑆𝑖𝑑𝑒 =L X W =15 𝑥 5
=75
Area = Base + base + top + bottom + side
= 30 + 30 + 195 + 180 + 75

5. ANSWER:
c) Area =square base +1st triangle + 2nd triangle + 3rd triangle + 4th triangle
1st triangle = 2nd triangle = 3rd triangle = 4th triangle.
𝐴𝑟𝑒𝑎𝑠𝑞𝑢𝑎𝑟𝑒 𝑏𝑎𝑠𝑒 = 𝑆2= S X S =4 X 4 = 16
𝐴𝑟𝑒𝑎1𝑠𝑡 𝑡𝑟𝑖𝑎𝑛𝑔𝑙𝑒 =
1
2
ℎ 𝑋 𝑏 =
1
2
X 4 X 7.23 =14.46
1st triangle = 2nd triangle = 3rd triangle = 4th triangle =14.46.
Area =square base + 1st triangle + 2nd triangle + 3rd triangle + 4th triangle
= 16 + 14.46 + 14.46 + 14.46 + 14.46 =73.84
D) Area =Triangular base +1st triangle + 2nd triangle + 3rd triangle
1st triangle = 2nd triangle = 3rd triangle
𝐴𝑟𝑒𝑎𝑡𝑟𝑖𝑎𝑛𝑔𝑢𝑙𝑎𝑟 𝑏𝑎𝑠𝑒 =
1
2
ℎ 𝑋 𝑏 =
1
2
x 9 x 7.8 = 35.1
𝐴𝑟𝑒𝑎1𝑠𝑡 𝑡𝑟𝑖𝑎𝑛𝑔𝑙𝑒 =
1
2
ℎ 𝑋 𝑏 =
1
2
X 9 X 7.8 = 35.1
1st triangle = 2nd triangle = 3rd triangle= 35.1
Area =triangular base + 1st triangle + 2nd triangle + 3rd triangle
= 35.1 + 35.1 + 35.1 + 35.1 = 140.4

6. Question 2:
A
B

7. A)
SOLUOTION :
B)

8. Question 3 :
A)

9. ANSWER :
A)
B) Mean=8.094 , also
Mean =
𝒔𝒖𝒎 𝒐𝒇 𝒕𝒉𝒆 𝒅𝒂𝒕𝒂 𝒗𝒂𝒍𝒖𝒆𝒔
𝒏𝒖𝒎𝒃𝒆𝒓 𝒐𝒇 𝒕𝒉𝒆 𝒅𝒂𝒕𝒂 𝒗𝒂𝒍𝒖𝒆𝒔
=
𝟏𝟎.𝟗𝟐 + 𝟗.𝟐𝟐 + 𝟕.𝟒𝟔 + 𝟓.𝟔𝟑 + 𝒅
𝟓
=
𝟑𝟑.𝟐𝟑 + 𝒅
𝟓
.
Mean = Mean
8.094 =
𝟑𝟑.𝟐𝟑 + 𝒅
𝟓
( multiply by 5 both sides )
40.47 = 33.23 + d ( subtract 33.23 both sides )
7.24 = d
So d = 7.24 km

10. A )
B )
Question 1:

11. ANSWER:
Step 1 : ordered the values from least to greatest
2 , 3 , 4 , 6 , 7 , 10 , 12
Step 2 : remove one from left and one from right tell you
reach the middle
2 , 3 , 4 , 6 , 7 , 10 , 12
2 , 3 , 4 , 6 , 7 , 10 , 12
2 , 3 , 4 , 6 , 7 , 10 , 12
So the median = 6
A ) B )
Step 1 : ordered the values from least to greatest
1.65 , 1.97 ,2.35 ,2.45 , 2.49 , 2.87 ,2.87 ,3.75
Step 2 : remove one from left and one from right tell
you reach the middle
1.65 , 1.97 ,2.35 ,2.45 , 2.49 , 2.87 ,2.87 ,3.75
because there is an even number of data values find
the two values closest to the middle
1.65 , 1.97 ,2.35 ,2.45 , 2.49 , 2.87 ,2.87 ,3.75
1.65 , 1.97 ,2.35 ,2.45 , 2.49 , 2.87 ,2.87 ,3.75
1.65 , 1.97 ,2.35 ,2.45 , 2.49 , 2.87 ,2.87 ,3.75
Step 3 : find the median of the two middle data : 2.45 , 2.49
Mean =
𝟐.𝟒𝟓+𝟐.𝟒𝟗
𝟐
=
𝟒.𝟗𝟒
𝟐
= 2.47
So the median of the data is 2.47

12. A ) B )
Question 1:

13. ANSWER:
Step 1 :find the mean Mean =
𝒔𝒖𝒎 𝒐𝒇 𝒕𝒉𝒆 𝒅𝒂𝒕𝒂 𝒗𝒂𝒍𝒖𝒆𝒔
𝒏𝒖𝒎𝒃𝒆𝒓 𝒐𝒇 𝒕𝒉𝒆 𝒅𝒂𝒕𝒂 𝒗𝒂𝒍𝒖𝒆𝒔
=
𝟏𝟓+𝟐𝟕+𝟏𝟎+𝟏𝟗+𝟐𝟒+𝟐𝟏+𝟐𝟖+𝟏𝟔
𝟖
=
𝟏𝟔𝟎
𝟖
=20
Step 2 :find the distance between each data value and the mean
l 15 –20l= 5
l 27 – 20 l= 7
l 10 – 20 l= 10
l 19 – 20 l= 1
l 24 – 20 l= 4
l 21 – 20 l= 1
l 28 – 20 l= 8
l 16 – 20 l= 4
A )
Step 3 :find the mean of the distances :
Mean =
𝒔𝒖𝒎 𝒐𝒇 𝒕𝒉𝒆 𝒅𝒂𝒕𝒂 𝒗𝒂𝒍𝒖𝒆𝒔
𝒏𝒖𝒎𝒃𝒆𝒓 𝒐𝒇 𝒕𝒉𝒆 𝒅𝒂𝒕𝒂 𝒗𝒂𝒍𝒖𝒆𝒔
=
𝟓+𝟕+𝟏𝟎+𝟏+𝟒+𝟏+𝟖+𝟒
𝟖
=
𝟒𝟎
𝟖
= 5
The average distance for each value from the mean is 5
days .

14. ANSWER:
Step 1 :find the mean of school A
Mean =
𝒔𝒖𝒎 𝒐𝒇 𝒕𝒉𝒆 𝒅𝒂𝒕𝒂 𝒗𝒂𝒍𝒖𝒆𝒔
𝒏𝒖𝒎𝒃𝒆𝒓 𝒐𝒇 𝒕𝒉𝒆 𝒅𝒂𝒕𝒂 𝒗𝒂𝒍𝒖𝒆𝒔
=
𝟕𝟎 + 𝟕𝟗 + 𝟖𝟎 + 𝟖𝟐 + 𝟗𝟓
𝟓
=
𝟒𝟎𝟔
𝟓
=81.2
Step 2 :find the distance between each data
value and the mean
l 70 –81.2l= 11.2
l 79 –81.2l= 2.2
l 80 –81.2l= 1.2
l 82 –81.2l= 0.8
l 95 –81.2l= 13.8
Step 3 :find the mean of the distances :
Mean =
𝒔𝒖𝒎 𝒐𝒇 𝒕𝒉𝒆 𝒅𝒂𝒕𝒂 𝒗𝒂𝒍𝒖𝒆𝒔
𝒏𝒖𝒎𝒃𝒆𝒓 𝒐𝒇 𝒕𝒉𝒆 𝒅𝒂𝒕𝒂 𝒗𝒂𝒍𝒖𝒆𝒔
=
𝟏𝟏.𝟐+𝟐.𝟐+𝟏.𝟐+𝟎.𝟖+𝟏𝟑.𝟖
𝟓
=
𝟐𝟗.𝟐
𝟓
= 5.84
B )
Step 1 :find the mean of school B
Mean =
𝒔𝒖𝒎 𝒐𝒇 𝒕𝒉𝒆 𝒅𝒂𝒕𝒂 𝒗𝒂𝒍𝒖𝒆𝒔
𝒏𝒖𝒎𝒃𝒆𝒓 𝒐𝒇 𝒕𝒉𝒆 𝒅𝒂𝒕𝒂 𝒗𝒂𝒍𝒖𝒆𝒔
=
𝟕𝟕 +𝟖𝟑 + 𝟖𝟑 + 𝟖𝟏 +𝟖𝟐
𝟓
=
𝟒𝟎𝟔
𝟓
=81.2
Step 2 :find the distance between each data
value and the mean
l 77–81.2l= 4.2
l 83 –81.2l= 1.8
l 83 –81.2l= 1.8
l 81 –81.2l= 0.2
l 82 –81.2l= 0.8
Step 3 :find the mean of the distances :
Mean =
𝒔𝒖𝒎 𝒐𝒇 𝒕𝒉𝒆 𝒅𝒂𝒕𝒂 𝒗𝒂𝒍𝒖𝒆𝒔
𝒏𝒖𝒎𝒃𝒆𝒓 𝒐𝒇 𝒕𝒉𝒆 𝒅𝒂𝒕𝒂 𝒗𝒂𝒍𝒖𝒆𝒔
=
𝟒.𝟐+𝟏.𝟖+𝟏.𝟖+𝟎.𝟐+𝟎.𝟖
𝟓
=
𝟖.𝟖
𝟓
= 1.76
COMPARE SCHOOL A
and B :
The mean absolute
deviation for school A is
greater than that for
school B . This means
the scores for school B
are closer together and
clustered around the
mean . The scores for
school A are more
spread out and not as
clustered around the
mean.