Real Numbers Test

Result

Real Numbers Test - 9

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Weekly Quiz Competition

Question 1
1 / -0

The least number n so that 5ⁿ is divisible by 3, where n is:

A
a whole number

B
no natural number

C
a real number

D
a natural number

Solution

Since 5 is a prime number so it is not divisible by 3.

Therefore there is no natural number n

such that 5ⁿ is divisible by 3.
Question 2
1 / -0

For every positive integer ‘n’, n² - n is divisible by

A
6

B
8

C
4

D
2

Solution

n² - n = n(n - 1) . Since n and (n - 1) are consecutive integers. Therefore, one of them must be divisible by 2.
Question 3
1 / -0

If two positive integers ‘a’ and ‘b’ are written as a = p q² and b = p³q², where ‘p’ and ‘q’ are prime numbers, then LCM(a, b) =

A
p²q³

B
p³q²

C
p²q²

D
pq

Solution

We know that LCM = product of the highest powers of all the prime factors of the numbers pq², p³q² is p³q².
Question 4
1 / -0

The LCM of 2³× 3² and 2²× 3³ is

A
2×3²

B
2³×3³

C
2²×3

D
2²×3²

Solution

L.C.M. of 2³×3² and 2²×3³ is the product of all prime numbers with the greatest power of every given number = 2³×3³
Question 5
1 / -0

The least positive integer divisible by 20 and 24 is

A
480

B
240

C
120

D
360

Solution

Least positive integer divisible by 20 and 24 is

LCM of (20, 24).

20 = 2²× 5

24 = 2³× 3

∴∴ LCM (20, 24) = 2³. 3 . 5 = 120

Thus 120 is divisible by 20 and 24.

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Re-Attempt Weekly Quiz Competition

Real Numbers Test - 9

Result

Real Numbers Test - 9

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SHARING IS CARING

Question 1

a whole number

no natural number

a real number

a natural number

Solution

Question 2

6

8

4

2

Solution

Question 3

p2q3

p3q2

p2q2

pq

Solution

Question 4

2×32

23×33

22×3

22×32

Solution

Question 5

480

240

120

360

Solution

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Question Analysis

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p²q³

p³q²

p²q²

2×3²

2³×3³

2²×3

2²×3²