Download ww-8-Cubes-and-Cube-Root

ID : ww-8-Cubes-and-Cube-Root [1]
Grade 8
Cubes and Cube Root
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Answer t he quest ions
(1)
T ake a number x, and multiply it with 4. T ake the cube of the resulting number. What is the ratio
of this number to the cube of the original number?
(2)
Solve the f ollowing :
(3)
What is the value of 413?
(4) If the volume of a cube is 35937 m3, then what is the surf ace area of the cube?
(5)
If you have a container in the shape of a cube that has a volume of 50653 m3, then what is the
area of each of the f aces of the cube?
(6) If a cube has a surf ace area of 726 m2, then what is the volume of the cube?
(7)
What is the value of (
10
)3
11
(8)
A cubical box has all sides of 6 m. What is its volume?
(9) If you subtract a number x f rom 20 times that number, and then take the cube of the dif f erence,
what will be the result?
Choose correct answer(s) f rom given choice
(10) If you add a number x with another number that is 3 times the value of x, and then take the cube
of the sum, what will be the result?
a. 64x3
b. 9x3
c. 179x3
d. -59x3
(11) Which of the f ollowing choices is the cube root of 512
a. 8
b. 6
c. 14
d. 15
(12) Find the value of A if
a. 21
b. 3
c. 4
d. 7
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ID : ww-8-Cubes-and-Cube-Root [2]
(13) T here are three numbers a, b and c.
b is 2 times a.
c is 0.2 times b.
T he sums of their cubes if 72.512.
Find the value of b
a. 8
b. 6
c. 0
d. 4
(14) Which of the f ollowing choices is the value of
.
a. -17
b. -6
c. -2
d. -8
Fill in t he blanks
(15) If the surf ace areas of two cubes are in the ratio 1:4, and the volume of the f irst cube is 512
m3, then the volume of the second cube is
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m3.
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ID : ww-8-Cubes-and-Cube-Root [3]
Answers
(1)
64:1
Step 1
If we multiply the number x with 4, the resulting number is: 4x.
Step 2
T he cube of the resulting number 4x = (4x)3
= 64x3
Step 3
T he original number is x.
Cube of the original number = x3.
Step 4
T he ratio of the new number to the original number =
New number
Original number
=
64x3
x3
=
64
1
Step 5
T heref ore, the ratio of the new number to the cube of the original number is 64:1.
(2)
15625
= {√(49 + 576)}3
= {√(625)}3
= {√(252)}3
= (25)3
= 15625
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ID : ww-8-Cubes-and-Cube-Root [4]
(3)
68921
Step 1
According to the question, we have to f ind the value of 413.
Step 2
Now,
413 = 41 × 41 × 41
= 1681 × 41
= 68921
Step 3
T heref ore, the value of 413 is 68921.
(4) 6534 m2
Step 1
We know that the volume of a cube with side a = a3
T he given volume of the cube = 35937 m3
Let us write these two f acts as an equation and f ind the value of a:
a3 = 35937 m3
⇒ a3 = 333 m3
⇒ a = 33 m
Step 2
Now we know that side of the cube, a = 33 m. Let us f ind the surf ace area which is equal to
6a2:
Surf ace area = 6a2
= 6(33)2
= 6(1089)
= 6534 m2
Step 3
T heref ore, the surf ace area of the cube is 6534 m2.
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ID : ww-8-Cubes-and-Cube-Root [5]
(5)
1369 m2
Step 1
Let us assume the side of a cube shape container is a.
Step 2
Volume of cube = 50653 m3
also, volume of the cube = a3
compare both, a3 = 50653
⇒ a = (50653)
1
3
⇒ a = 37
Step 3
Area of side of the cube = a2
= 37 2
= 1369
Step 4
T heref ore, the area of the f ace of cube shape container is 1369 m2.
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ID : ww-8-Cubes-and-Cube-Root [6]
(6) 1331 m3
Step 1
We need to f ind the volume of the cube. T he volume of a cube with side a is a3. T his means
if we can f ind the value of a, we can easily f ind the value of the volume of the cube.
Step 2
Let us f ind the value of the side a of the cube. We have been told that the surf ace area of
the cube is 726 m2. Let us use this f act to write an equation and f ind the value of a.
Step 3
T he surf ace area of the cube with side a = 6a2.
T he given surf ace area of the cube = 726 m2.
T his means, 726 = 6a2
⇒ a2 =
726
6
⇒ a2 = 121
⇒ a = √121
⇒ a = 11
Step 4
Now that we know that the value of the side of the cube is 11, let us f ind the volume of the
cube, which is = a3
= (11)3
= 1331 m3
Step 5
T heref ore, the volume of the cube is 1331 m3.
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ID : ww-8-Cubes-and-Cube-Root [7]
(7)
1000
1331
Step 1
We have been asked to f ind the value of (
10
)3.
11
Step 2
Now,
(
10
)3=
11
=
103
113
=
10 × 10 × 10
11 × 11 × 11
1000
1331
Step 3
T heref ore, the value of (
10
11
(8)
)3 is
1000
.
1331
216 m3
Step 1
We know that the volume of a cube (or a cubical box) is calculated by taking the cube of its
side.
Step 2
We have been told that the side of the cube in question is 6 m. T heref ore, its volume = (6)3
m3
= 6 × 6 × 6 m3
= 216 m3
Step 3
T heref ore, the volume of the cubical box is 216 m3.
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ID : ww-8-Cubes-and-Cube-Root [8]
(9) 6859x3
Step 1
According to the question,
the f irst number is x and
the second number is 20 x.
Step 2
Dif f erence of the numbers = (20 x - x)
Step 3
Cube of the dif f erence = (20 x - x)3
= (19 x)3
= 6859 x3
Step 4
T heref ore, the result will be 6859x3.
(10) a. 64x3
Step 1
According to the question,
the f irst number is x and the second number is 3 x.
Step 2
Sum of the two numbers = (x + 3 x)
Step 3
Cube of the sum = (x + 3 x)3
= (4 x)3
= 64 x3
Step 4
T heref ore, the result will be 64x3.
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ID : ww-8-Cubes-and-Cube-Root [9]
(11) a. 8
Step 1
We have been asked to f ind the cube root of 512.
Step 2
Prime f actors of 512 are,
2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2.
Step 3
Now, make group of three prime f actors.
(2 × 2 × 2) × (2 × 2 × 2) × (2 × 2 × 2)
Step 4
T here is no prime f actor lef t which is not part of any group. Cube root is given by the
multiplication of a element f rom each group of prime f actors. i.e.
2×2×2
=8
Step 5
T heref ore, the cube root of 512 is 8.
(12) b. 3
Step 1
We have been asked to f ind the value of A f rom the f ollowing equation,
³√(3087A) = 21.
Step 2
³√(3087A) = 21
T aking cube both side,
⇒ (³√(3087A))3 = (21)3
⇒ 3087A = 21 × 21 × 21
⇒A=
7 ×7 ×7 ×3×3×3
3087
⇒A=
7 ×7 ×7 ×3×3×3
7 ×7 ×7 ×3×3
⇒A=
7 ×7 ×7 ×3×3×3
7 ×7 ×7 ×3×3
⇒A= 3
Step 3
T heref ore, the value of A is 3.
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ID : ww-8-Cubes-and-Cube-Root [10]
(13) d. 4
Step 1
According to the question, there are three numbers a, b and c.
b is 2 times a, that is, b = 2a.
c is 0.2 times b, that is, c = 0.2b.
or c = 0.2 × 2a = 0.4a
Step 2
It is also given that sum of their cubes is 72.512,
a3 + b3 + c3 = 72.512
Step 3
Put the value of b and c in above equation.
a3 + (2a)3 + (0.4a)3 = 72.512
⇒ a3(1 + 8 + 0.064) = 72.512
72.512
⇒ a3 =
9.064
⇒ a3 = 8
⇒ a3 = (2)3
⇒a=2
Step 4
Now put value of a in b=2a
⇒b=2×2
⇒b=4
Step 5
T heref ore, the value of b is 4.
(14) b. -6
Step 1
Let's f irst f ind all prime f actors of 216.
216 = 2 × 2 × 2 × 3 × 3 × 3
Step 2
Now,
= -1 × 3√(2 × 2 × 2 × 3 × 3 × 3)
= -1 × 3√(23 × 33)
= -1 × 2 × 3
= -6
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ID : ww-8-Cubes-and-Cube-Root [11]
(15)
4096
Step 1
Let us assume the side of the f irst cube is a .
Let us assume the side of the second cube is b.
Surf ace area of f irst cube = 6a2
Surf ace area of second cube = 6b2
Ratio of surf ace area of two cubes = 1:4
Step 2
Volume of f irst cube, a3 = 512 m3
⇒ a3 = 83
⇒a=8
Step 3
Now the ratio of surf ace area of cubes,
6a2
=
4
6b2
⇒
1
82
1
=
4
b2
⇒ 64 × 4 = 1 × b2
⇒
256
= b2
1
⇒ 256 = b2
⇒ 162 = b2
⇒ b = 16
Step 4
Volume of second cube, b3 = (16)3
= 16 × 16 × 16
= 4096
Step 5
T heref ore, the volume of the second cube is 4096 m3.
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