Download Grade 8 Cubes and Cube Root

ID : pk-8-Cubes-and-Cube-Root [1]
Grade 8
Cubes and Cube Root
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Answer t he quest ions
(1)
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 117670, and the number of women in the town is 166011.
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 59310, then what is the smallest
possible number of children in the town?
(2)
Which of the f ollowing numbers is a perf ect cube?
13822, 21952, 518, 42879
(3)
If one f ace of a cube has an area of 49 m2, then what is the volume of the cube?
(4) 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 9.064.
Find the value of b
(5)
T here are two numbers such that sum of the numbers is 27 and their dif f erence is 7. Find the
dif f erence of their cubes.
(6) Find the cube root of 60 correct to one place of decimal.
(7) What is the value of 403?
(8)
What is the value of
(9) Find the value of
Choose correct answer(s) f rom given choice
(10) Solve the f ollowing :
a. 125
b. 534
c. -590
d. 845
(11) What is the value of
a.
12
.
b.
21
c.
23
12
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12
23
d.
13
23
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ID : pk-8-Cubes-and-Cube-Root [2]
(12) Solve the f ollowing:
a. 41
b. 48
c. -48
d. -41
(13) Some pears are packed into a box in stacks. Each stack has the same number of pears 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 82 stacks of pears in the box, what is the total number of pears in the box?
a. 552572
b. 551379
c. 551055
d. 551368
(14) Which of the f ollowing statements is f alse:
a. 512 is a perf ect cube
b. Cubes of even numbers are always even
c. A number that has the last digit as 6 will
have 6 as the last digit of its cube
d. T he sum of the cubes of f our consecutive
natural numbers will always be odd
Fill in t he blanks
(15)
If the volume of a cube is 29791 m3, then surf ace area of the cube is
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m2
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generated at www.edugain.com
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ID : pk-8-Cubes-and-Cube-Root [3]
Answers
(1)
59319
Step 1
T otal population of the town is a perf ect cube.
Number of men in the town = 117670
Number of women in the town = 166011
Step 2
It is given that the number of children in the village is also the perf ect cube, and is more
than 59310. Let us f irst f ind the values of perf ect cubes more than 59310.
Step 3
Cube root of 59310 = 38.998027513652 and the perf ect cubes more than 59310 will be the
cubes of all integers more than 38.998027513652, i.e. 393, 403, 413, etc.
Step 4
T he population of children will be the smallest value out of 393, 403, 413, 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 393 (= 59319). In this case, the total
population of the village will be = 117670 + 166011 + 393
= 117670 + 166011 + 59319
= 343000
Since the cube root of 343000 is 70, 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 39 3 = 59319.
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ID : pk-8-Cubes-and-Cube-Root [4]
(2)
21952
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 21952 = 2 x 2 x 2 x 2 x 2 x 2 x 7 x 7 x 7
T here is no prime f actor lef t with the group of prime f actors. So, 21952 is a perf ect cube.
Step 3
Prime f actors of 42879 = 3 x 14293
T here is a prime f actor lef t. So, 42879 is not a perf ect cube.
Step 4
Prime f actors of 13822 = 2 x 6911
T here is a prime f actor lef t. So, 13822 is not a perf ect cube.
Step 5
Prime f actors of 518 = 2 x 7 x 37
T here is a prime f actor lef t. So, 518 is not a perf ect cube.
Step 6
T heref ore, 21952 is the perf ect cube.
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ID : pk-8-Cubes-and-Cube-Root [5]
(3)
343 m3
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 49 m2.
Area of cube of one f ace = a2
⇒ 49 = a2
⇒ 7 2 = a2
⇒a=7m
Step 3
Volume of the cube,
V = a3
= (7)3
=7 ×7 ×7
= 343 m3
Step 4
T heref ore, the volume of the cube is 343 m3.
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ID : pk-8-Cubes-and-Cube-Root [6]
(4) 2
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 9.064,
a3 + b3 + c3 = 9.064
Step 3
Put the value of b and c in above equation.
a3 + (2a)3 + (0.4a)3 = 9.064
⇒ a3(1 + 8 + 0.064) = 9.064
9.064
⇒ a3 =
9.064
⇒ a3 = 1
⇒ a3 = (1)3
⇒a=1
Step 4
Now put value of a in b=2a
⇒b=2×1
⇒b=2
Step 5
T heref ore, the value of b is 2.
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ID : pk-8-Cubes-and-Cube-Root [7]
(5)
3913
Step 1
Let us assume that x and y are the two numbers.
Step 2
According to the question, the sum of two numbers is 27, that is,
x + y = 27 ........... (1)
Step 3
And the dif f erence of two numbers is 7, that is,
x - y = 7 ........... (2)
Step 4
On adding equations (1) and (2),
x + y + x - y = 27 + 7
⇒ 2x = 34
⇒ x = 17
Step 5
Now using equation (1),
x + y = 27
⇒ 17 + y = 27
⇒ y = 27 - 17 = 10
Step 6
Now, dif f erence of their cubes = x3 - y3
= (17)3 - (10)3
= 3913
Step 7
T heref ore, the value of the dif f erence of their cubes is 3913.
(6) 3.9
(7) 64000
Step 1
According to the question, we have to f ind the value of 403.
Step 2
Now,
403 = 40 × 40 × 40
= 1600 × 40
= 64000
Step 3
T heref ore, the value of 403 is 64000.
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ID : pk-8-Cubes-and-Cube-Root [8]
(8)
-5
7
Step 1
Let's f irst f ind prime f actors of 375,
375 = 5 × 5 × 5 × 3
Step 2
Similarly prime f actors of 1029,
1029 = 3 × 7 × 7 × 7
Step 3
Now,
=
=
=
-5
7
(9) -9
Step 1
Let's f irst f ind all prime f actors of 729.
729 = 3 × 3 × 3 × 3 × 3 × 3
Step 2
Now,
= -1 × 3√(3 × 3 × 3 × 3 × 3 × 3)
= -1 × 3√(33 × 33)
= -1 × 3 × 3
= -9
(10) a. 125
= {√(9 + 16)}3
= {√(25)}3
= {√(52)}3
= (5)3
= 125
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ID : pk-8-Cubes-and-Cube-Root [9]
(11)
12
b.
23
Step 1
We have been asked to f ind the value of
.
Step 2
Now,
=(
(12 × 12 × 12)
)
(23 × 23 × 23)
(12)3
=(
= ((
3
12
1
)3)
3
1
(23)3
23
=
)
1
3
12
23
Step 3
T heref ore, the value of
is
12
.
23
(12) c. -48
Step 1
Let's f irst f ind prime f actors of -288,
-288 = -1 × 2 × 2 × 2 × 3 × 3 × 4
Step 2
Similarly f ind prime f actors of 384,
384 = 2 × 2 × 2 × 3 × 4 × 4
Step 3
Now,
³√(-288) ³√(384)
= ³√(-1 × 2 × 2 × 2 × 3 × 3 × 4) ³√(2 × 2 × 2 × 3 × 4 × 4)
= ³√(-1 × 2 × 2 × 2 × 3 × 3 × 4 × 2 × 2 × 2 × 3 × 4 × 4)
= ³√[-1 × (2 × 2 × 2) × (2 × 2 × 2) × (3 × 3 × 3) × (4 × 4 × 4)]
= ³√[-13 × 23 × 23 × 33 × 4 3]
= (-1 × 2 × 2 × 3 × 4)
= -48
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ID : pk-8-Cubes-and-Cube-Root [10]
(13) d. 551368
Step 1
It is given that the total number of stack in the box is 82.
T otal number of rows in one stack = 82
number of columns in the stack equal to the number of rows = 82
Step 2
T otal number of pears in the box = T otal number of stack × T otal number of rows in one
stack × T otal number of column on one stack
= 82 × 82 × 82
= 551368.
Step 3
T heref ore, the total number of pears in the box are 551368.
(14) d. T he sum of the cubes of f our consecutive natural numbers will always be odd
(15)
5766
Step 1
We know that the volume of a cube with side a = a3
T he given volume of the cube = 29791 m3
Let us write these two f acts as an equation and f ind the value of a:
a3 = 29791 m3
⇒ a3 = 313 m3
⇒ a = 31 m
Step 2
Now we know that side of the cube, a = 31 m. Let us f ind the surf ace area which is equal to
6a2:
Surf ace area = 6a2
= 6(31)2
= 6(961)
= 5766 m2
Step 3
T heref ore, the surf ace area of the cube is 5766 m2.
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