Free Playing with Numbers 03 Practice Test - 6th grade
Question 1
Which of the following represents the prime factorisation of 60?
2 x 2 x 3 x 5
2 x 2 x 5
2 x 2 x 3
2 x 2 x 3 x 5 x 5
SOLUTION
Solution : A
Prime factorisation is expressing a number as the product of its prime factors.
60 = 2 x 30
= 2 x 2 x 15
= 2 x 2 x 3 x 5
(2, 3 and 5 are prime numbers)
Hence, the prime factorisation of 60 is 2 x 2 x 3 x 5.
Question 2
The prime factors of a number is given as 3 × 3 × 11 × 101. Which of the following is correct about the number?
It is an even number.
It is the smallest 4 digit odd number.
The number is 10000.
It is the greatest 4 digit number.
SOLUTION
Solution : D
On simplification of the given expression, we get that
3 × 3 × 11 × 101 = 9999.
We can observe that it is the greatest 4 digit number.
Also, we can see that this is an odd number.
Question 3
H.C.F of 45, 81 and 27 is _____.
45
27
9
SOLUTION
Solution : C
Factors of 45 = 1, 3, 5, 9, 15, 45
Factors of 81 = 1, 3, 9, 27, 81
Factors of 27 = 1, 3, 9, 27
∴ The common factors of 45, 81 and 27 are 1, 3 and 9.So, the H.C.F of 45, 81 and 27 = 9.
Question 4
The H.C.F of two co-prime numbers is
4
1
2
3
SOLUTION
Solution : B
Numbers, which do not have any common factor between them other than one, are called co-prime numbers.
For example, 3 and 7 are co-prime numbers. They only have 1 as a common factor.
Question 5
L.C.M. of 1 and 90 is __.
90
360
45
1
SOLUTION
Solution : A
Multiples of 90 will be 90, 180, ...... and
Multiples of 1 will be 1, 2, 3...90 and so on.
From the above, we can say that, the least common multiple of 1 and 90 will be 90.
The L.C.M. of 1 and any other number, say 'x' is always 'x'.
Question 6
LCM of two prime numbers is ________________.
product of both numbers
sum of two numbers
difference of both the numbers
division of both the numbers
SOLUTION
Solution : A
Prime numbers have only two factors 1 and the number itself.
Hence, LCM of two prime numbers is always the product of the two numbers.For example, LCM of 3 and 5 = 3 × 5 = 15
Similarly, LCM of 7 and 11 = 7 × 11 = 77
Question 7
A farmer has 3 storage units for rice. Each storage unit has the same capacity. Rice is available to the farmer in bags of 80 kg, 85 kg and 90 kg. Find the minimum storage capacity of each of the storage unit considering that he has to take rice in bags of 80 kg, 85 kg and 90 kg only.
14220 kg
12240 kg
12440 kg
14422 kg
SOLUTION
Solution : B
The capacity of each storage unit should be same as well as minimum. The required minimum capacity should be the LCM of 80 kg, 85 kg & 90 kg.
2×2×2×2×5×9×17=12240LCM of 80,85,90=12240
Question 8
Find the least number which when divided by 32, 64 & 128 leaves the remainder of 8.
1024
1034
136
128
SOLUTION
Solution : C
The smallest number that is divisible by 32, 64, 128 is the LCM of 32, 64, 128.
LCM of 32, 64, 128 =
= 2×2×2×2×2×2×2
=128
When 128 is divided by 32, 64 & 128 there is no remainder. So, to get remainder 8 when divided by 32, 64 & 128, 8 is to be added to 128 i.e., 128+8 = 136
Question 9
The LCM of 18, 24 & 32 is _______.
288
32
24
432
SOLUTION
Solution : A
LCM of 18, 24 and 32 is
So, LCM = 2×2×2×2×2×3×3
= 288
Question 10
LCM of two numbers is 12 and HCF of same two numbers is 2. If one of the numbers is 6 then the other number is _____.
4
SOLUTION
Solution : B
It is given that LCM = 12; HCF=2
Product of 2 numbers = HCF × LCM of the respective numbers
⇒ 6 × the other number = 12 × 2
⇒ 6 × the other number = 24
⇒the other number = 246 = 4
Hence, the other number = 4