Free Cubes and Cube Roots 01 Practice Test - 8th Grade
Question 1
Which of the following is a perfect cube?
532
729
827
943
SOLUTION
Solution : B
By prime factorization, we get 729 = 3×3×3×3×3×3=36
So cube root of 729=(729)13=363=32=9
Since, the cube root of 729 is a natural number, 729 is a perfect cube.
Question 2
The volume of a closed cubical tank is 1331 cubic metres. Find the cost of painting its complete surface area from outside at the rate of ₹ 10 per square meter.
7260
7240
7000
7280
SOLUTION
Solution : A
Let the side of the given cubical tank be 'a' metres .
Volume of the given cube is a×a×a
By prime factorization, we get cube root of 1331 as 11
Surface area of each side of a cube is a×a
Total surface area of the cube is 6×a×a
Cost incurred is then 10×6×11×11 = ₹ 7260
Question 3
a3 - b3 is always divisible by:
a - b
a × b
a + b
2a + b
SOLUTION
Solution : A
Take a=2 and b=1 and check.
Alternatively, we know the algebraic identity,a3−b3=(a−b)(a2+2ab+b2)
Hence , we can see that (a−b) is a factor of a3 - b3
Thus, a3 - b3 is divisible by (a−b)
Question 4
What is the minimum number that should be multiplied with 360 in order to make it a perfect cube ?
75
70
65
80
SOLUTION
Solution : A
For a number to be a perfect cube, its prime factors should be in the form of triplets i.e. every prime factor should be raised to the power in multiples of 3
Prime factorizing the given number, we get 360=2×2×2×3×3×5
Hence, we can see that the given number must be multiplied by 3×5×5 i.e.75
Question 5
The number of students in class 20 minutes ago was 729 . If the number of students joining the class triples every 10 minutes, what is the cube root of the number of students joining after 40 minutes from now?
81
27
64
48
SOLUTION
Solution : A
For every 10 mins passage of time, the number of students joining the class gets tripled i.e. multiplied by 3
The number of students joining after 40 minutes from now on is 729×3×3×3×3×3×3 ( from 20 mins back to 40 mins forward)
The cube root of 729 is 9.
Cube root of given number is 9×3×3=81
Question 6
What is the sum of the cubes of the first 7 natural numbers?
784
774
441
788
SOLUTION
Solution : A
Sum of cubes of first n natural numbers is (n(n+1))24
Putting n = 7,
(7(7+1))24
(7×8)24= 784.
Question 7
The units digit in the expansion of 163 is
SOLUTION
Solution :The units digit of the cube of any number with 6 in its unit's place is always 6. Alternatively, we can see that 6 when multiplied with itself three times yields 216, whose last digit is 6.
Question 8
If the value of (a3−b3a3+b3) when a = 7 and b = 2 is 335p, then p is 321.
True
SOLUTION
Solution : B
a3- b3 = (a−b)( a2+ab+ b2)
a3+ b3 =(a+b)( a2-ab+ b2)
Puting the values of a and b, we get the value of p as 351.
Alternatively, we can observe that since a and b are positive, so denominator has to be greater than numerator. Thus, a3+ b3 will be greater than a3- b3
Question 9
The difference of cubes of two consecutive numbers is always divisible by 3.
True
False
SOLUTION
Solution : B
Take two numbers, for example, 2 and 1 and check.
The difference of cubes of 2 and 1 is 7, which is not divisible by 3.
Question 10
m3 ≥ m2 . To satisfy this condition, m can be
a positive integer
SOLUTION
Solution : A and D
The cube of a positive number is positive but the cube of a negative number is negative.
e.g: 23>22and (−2)3<(−2)2
The cube of any number between 0 and 1 is less than its square. Any number greater than 1 will have its cube greater than its square.
e.g 0.23<0.22
as 0.23=0.008 0.22=0.04