8th Grade > Mathematics
CUBES AND CUBE ROOTS MCQs
Total Questions : 57
| Page 2 of 6 pages
Answer: Option A. -> 7260
:
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
:
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
Answer: Option B. -> False
:
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.
:
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.
Answer: Option B. -> False
:
B
a3- b3= (a−b)( a2+ab+ b2)
a3+ b3=(a+b)( a2-ab+ b2)
Puting the values of a and b,weget 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+ b3will be greater thana3-b3
:
B
a3- b3= (a−b)( a2+ab+ b2)
a3+ b3=(a+b)( a2-ab+ b2)
Puting the values of a and b,weget 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+ b3will be greater thana3-b3
:
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.
Answer: Option A. -> 3
:
A
The units digit of the cube of 7 i.e.343 is 3.
Thus, the units digit of the cube of the number whose unit digit is 7 will be 3.
:
A
The units digit of the cube of 7 i.e.343 is 3.
Thus, the units digit of the cube of the number whose unit digit is 7 will be 3.
Answer: Option C. -> 5
:
C
For a numberto be a perfect cube, its prime factors should be expressed as triplets
By prime factorization,we get 1715 = 7×7×7×5
We have three 7’s but only one 5 so we need to divide the number by 5.
:
C
For a numberto be a perfect cube, its prime factors should be expressed as triplets
By prime factorization,we get 1715 = 7×7×7×5
We have three 7’s but only one 5 so we need to divide the number by 5.
Answer: Option A. -> True
:
A
100 = 102
= 13+ 23+ 33+ 43
= 1+8+27+64
:
A
100 = 102
= 13+ 23+ 33+ 43
= 1+8+27+64
:
Since the side of the smaller cube is 1 cm, its volume will be 1 cubic cm.
We know that the volume of cubes of sides 1,2,3,4,... are 1,8,27,64...cubic cm.
But Anshul and Balendu have only 50 small cubes, so they can make cubes of 3 sizes with the help of 1,8 and 27 smaller cubes.
It should be noted that (50-1-8-27) = 14 smaller cubes will be left unused.
Answer: Option C. -> 5
:
C
For a number to be a perfect cube, its prime factors should be in the form of triplets
By prime factorization, 43200 = 26× 33× 52
Only two 5’s are there so we need one more 5 to make a triplet.
So if 43200 is multiplied by 5, we get a perfect cube.
:
C
For a number to be a perfect cube, its prime factors should be in the form of triplets
By prime factorization, 43200 = 26× 33× 52
Only two 5’s are there so we need one more 5 to make a triplet.
So if 43200 is multiplied by 5, we get a perfect cube.
Answer: Option D. -> 81
:
D
By prime factorization, we have
343=7×7×71331=11×11×11512=8×8×881=3×3×3×3
For a number to be a perfect cube ,each of its factors must be repeated thrice.
343, 1331 and 512 are the perfect cubes of 7, 11 and 8 respectively, whereas 81 is not, as 3 is repeating 4 times which is not a triplet.
81 is not a perfect cube.
:
D
By prime factorization, we have
343=7×7×71331=11×11×11512=8×8×881=3×3×3×3
For a number to be a perfect cube ,each of its factors must be repeated thrice.
343, 1331 and 512 are the perfect cubes of 7, 11 and 8 respectively, whereas 81 is not, as 3 is repeating 4 times which is not a triplet.
81 is not a perfect cube.