ID : eu-8-Cubes-and-Cube-Root [1]
Grade 8
Cubes and Cube Root
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Answer t he quest ions
(1)
A cubical box has all sides of 7 m. What is its volume?
(2)
If the volume of a cube is 15625 m3, then what is the length of each side of the cube?
(3)
Which of the f ollowing statements is f alse:
T he sum of the cubes of the first n natural numbers is equal to the square of their sum.
For an integer, the cube root will always be greater than the square root.
If a number is divisible by a prime number p then the cube of that number is divisible by
p 3.
If a negative integer is a perfect cube, then the corresponding positive integer will also
be a perfect cube.
(4) Solve the f ollowing :
(5)
Some guavas are packed into a box in stacks. Each stack has the same number of guavas in
each row as the number of rows. T he number of stacks is also the same as the number of
rows. If there are 84 stacks of guavas in the box, what is the total number of guavas in the
box?
(6) Which of the f ollowing numbers is a perf ect cube?
343, 1005, 35939, 21949
(7) Calculate the cube root of 512
Choose correct answer(s) f rom given choice
(8)
If you subtract a number x f rom 21 times that number, and then take the cube of the dif f erence,
what will be the result?
a. 8000x3
b. 441x3
c. 455x2
d. 17x3
(9) What is the value of
a.
-7
.
b.
3
c.
-2
7
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-3
7
d.
-2
3
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ID : eu-8-Cubes-and-Cube-Root [2]
(10) Which of the f ollowing numbers is not a perf ect cube?
(11)
a. 216
b. 2197
c. 3000
d. 2744
What is the value of (
3
)3?
7
27
a.
b.
2401
27
27
c.
d.
343
(12)
343
27
7
=?
a. 19
b. 17
c. 7
d. 12
Fill in t he blanks
(13) T he number of people living in a small town is f ound to be a perf ect cube. We know that the
number of men in the town is 47417, and the number of women in the town is 55631.
By an odd coincidence, we count and f ind that the number of children in the town is also a
perf ect cube. If we give you a hint that this number is more than 21950, then the smallest
possible number of children in the town is
(14)
.
If a cube has a surf ace area of 100 m2, then the volume of the cube is
m3
(15) If you have a container in the shape of a cube that has a volume of 15625 m3, then the area of
each of the f aces of the cube is
m2.
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ID : eu-8-Cubes-and-Cube-Root [3]
Answers
(1)
343 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 7 m. T heref ore, its volume = (7)3
m3
= 7 × 7 × 7 m3
= 343 m3
Step 3
T heref ore, the volume of the cubical box is 343 m3.
(2)
25 m
Step 1
We know that the volume of a cube with side a = a3
T he given volume of the cube = 15625 m3
Step 2
Let us write these two f acts as an equation and f ind the value of a:
a3 = 15625 m3
⇒ a3 = 253 m3
⇒ a = 25 m
Step 3
T heref ore, the length of each side of the cube is 25 m.
(3)
For an integer, the cube root will always be greater than the square root
(4) 39304
= {√(256 + 900)}3
= {√(1156)}3
= {√(34 2)}3
= (34)3
= 39304
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ID : eu-8-Cubes-and-Cube-Root [4]
(5)
592704
Step 1
It is given that the total number of stack in the box is 84.
T otal number of rows in one stack = 84
number of columns in the stack equal to the number of rows = 84
Step 2
T otal number of guavas in the box = T otal number of stack × T otal number of rows in one
stack × T otal number of column on one stack
= 84 × 84 × 84
= 592704.
Step 3
T heref ore, the total number of guavas in the box are 592704.
(6) 343
Step 1
T he def inition of a perf ect cube is a number that is the result of multiplying an integer by
itself three times.Steps to f ind the perf ect cube are as f ollows.
i. Resolve the given number into prime f actors.
ii. Group the f actors in 3 in such a way that each number of the group is same.
iii. If there is nothing lef t with the group of prime numbers so the number is perf ect cube
otherwise not.
Step 2
Prime f actors of 343 = 7 x 7 x 7
T here is no prime f actor lef t with the group of prime f actors. So, 343 is a perf ect cube.
Step 3
Prime f actors of 35939 = 83 x 433
T here is a prime f actor lef t. So, 35939 is not a perf ect cube.
Step 4
Prime f actors of 21949 = 47 x 467
T here is a prime f actor lef t. So, 21949 is not a perf ect cube.
Step 5
Prime f actors of 1005 = 3 x 5 x 67
T here is a prime f actor lef t. So, 1005 is not a perf ect cube.
Step 6
T heref ore, 343 is the perf ect cube.
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ID : eu-8-Cubes-and-Cube-Root [5]
(7) 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.
(8)
a. 8000x3
Step 1
According to the question,
the f irst number is x and
the second number is 21 x.
Step 2
Dif f erence of the numbers = (21 x - x)
Step 3
Cube of the dif f erence = (21 x - x)3
= (20 x)3
= 8000 x3
Step 4
T heref ore, the result will be 8000x3.
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ID : eu-8-Cubes-and-Cube-Root [6]
(9)
-2
d.
3
Step 1
Let's f irst f ind prime f actors of 56,
56 = 2 × 2 × 2 × 7
Step 2
Similarly prime f actors of 189,
189 = 7 × 3 × 3 × 3
Step 3
Now,
=
=
=
-2
3
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ID : eu-8-Cubes-and-Cube-Root [7]
(10) c. 3000
Step 1
T he def inition of a perf ect cube is a number that is the result of multiplying an integer by
itself three times. Steps to f ind that a number is not the perf ect cube.
i. Resolve the given number into prime f actors.
ii. Create group of prime f actors such that each group has three equal f actors
(triplets).
iii. If there is something lef t af ter grouping of prime f actors, the number is not the
perf ect cube otherwise perf ect cube.
Step 2
Prime f actors of 216 = 2 x 2 x 2 x 3 x 3 x 3
T here is no prime f actor lef t af ter creating group of triplets of prime f actors. T heref ore,
216 is the perf ect cube.
Step 3
Prime f actors of 2197 = 13 x 13 x 13
T here is no prime f actor lef t af ter creating group of triplets of prime f actors. T heref ore,
2197 is the perf ect cube.
Step 4
Prime f actors of 3000 = 2 x 2 x 2 x 3 x 5 x 5 x 5
Af ter creating group of triplets, 3 will remain un-grouped. T heref ore, 3000 is not the perf ect
cube.
Step 5
Prime f actors of 2744 = 2 x 2 x 2 x 7 x 7 x 7
T here is no prime f actor lef t af ter creating group of triplets of prime f actors. T heref ore,
2744 is the perf ect cube.
Step 6
T heref ore, 3000 is not the perf ect cube.
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ID : eu-8-Cubes-and-Cube-Root [8]
(11)
27
c.
343
Step 1
We have been asked to f ind the value of (
3
)3.
7
Step 2
Now,
(
3
)3=
7
=
33
=
73
3×3×3
7 ×7 ×7
27
343
Step 3
T heref ore, the value of (
3
7
)3 is
27
.
343
(12) d. 12
Step 1
Let's f irst f ind prime f actors of 125 and 343
125 = 5 × 5 × 5 = 53
343 = 7 × 7 × 7 = 7 3
Step 2
Now,
³√125 + ³√343
= ³√(53) + ³√(7 3)
=5+7
= 12
Step 3
T heref ore, the value of ³√125 + ³√343 is 12.
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ID : eu-8-Cubes-and-Cube-Root [9]
(13)
21952
Step 1
T otal population of the town is a perf ect cube.
Number of men in the town = 47417
Number of women in the town = 55631
Step 2
It is given that the number of children in the village is also the perf ect cube, and is more
than 21950. Let us f irst f ind the values of perf ect cubes more than 21950.
Step 3
Cube root of 21950 = 27.999149634038 and the perf ect cubes more than 21950 will be the
cubes of all integers more than 27.999149634038, i.e. 283, 293, 303, etc.
Step 4
T he population of children will be the smallest value out of 283, 293, 303, etc, f or which the
total population of the village is a perf ect cube.
Step 5
Let us see if the population of the children can be 283 (= 21952). In this case, the total
population of the village will be = 47417 + 55631 + 283
= 47417 + 55631 + 21952
= 125000
Since the cube root of 125000 is 50, which is an integer, the total population of the village
in this case is indeed a perf ect cube.
Step 6
T heref ore, the smallest possible number of children in the village is 283 = 21952.
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ID : eu-8-Cubes-and-Cube-Root [10]
(14)
1000
Step 1
Let us assume the length of a side of the cube is a.
Step 2
T he area of one f ace of cube is 100 m2.
Area of cube of one f ace = a2
⇒ 100 = a2
⇒ 102 = a2
⇒ a = 10 m
Step 3
Volume of the cube,
V = a3
= (10)3
= 10 × 10 × 10
= 1000 m3
Step 4
T heref ore, the volume of the cube is 1000 m3.
(15)
625
Step 1
Let us assume the side of a cube shape container is a.
Step 2
Volume of cube = 15625 m3
also, volume of the cube = a3
compare both, a3 = 15625
⇒ a = (15625)
1
3
⇒ a = 25
Step 3
Area of side of the cube = a2
= 252
= 625
Step 4
T heref ore, the area of the f ace of cube shape container is 625 m2.
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