Real Numbers Test

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Real Numbers Test - 31

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Weekly Quiz Competition

Question 1
1 / -0

The H. C. F. of $$252$$, $$324$$ and $$594$$ is ____________.

A
$$36$$

B
$$18$$

C
$$12$$

D
$$6$$

Solution

H.C.F is the greatest number which divides each of them individually.
All the numbers are written in form of factors of prime numbers.
$$252$$ $$\rightarrow$$ $$2$$$$\times$$$$2$$$$\times$$$$3$$$$\times$$$$3$$$$\times$$$$7$$
$$324$$ $$\rightarrow$$ $$2$$$$\times$$$$2$$$$\times$$$$3$$$$\times$$$$3$$$$\times$$$$3$$$$\times$$$$3$$
$$594$$ $$\rightarrow$$ $$2$$$$\times$$$$3$$$$\times$$$$3$$$$\times$$$$3$$$$\times$$$$11$$
From this greatest factor is $$2$$$$\times$$$$3$$$$\times$$$$3$$ $$\rightarrow$$ $$18$$
Hence, option B is correct.
Question 2
1 / -0

The product of two irrational numbers is

A
Always irrational

B
Always rational

C
Can be both rational or irrational

D
always an integer

Solution

According to the properties of irrational numbers, the product of two irrational numbers can be both rational or irrational.
For example, let $$p=\sqrt 3$$ and $$q=\sqrt 3$$ be two irrational numbers
$$\therefore \ pq=\sqrt 3\times \sqrt 3=3$$
Here, $$3$$ is a rational number.

Now, let $$p=\sqrt 3$$ and $$q=\sqrt 2$$
$$\therefore \ pq=\sqrt 3\times \sqrt 2=\sqrt 6$$
Here, $$\sqrt6$$ is an irrational number.

Hence, the product can be both rational or irrational.

So, option C is correct.
Question 3
1 / -0

$$\frac{2}{3}$$ is a rational number whereas $$\frac{\sqrt{2}}{\sqrt{3}}$$ is?

A
Also a rational number

B
An irrational number

C
Not a number

D
A natural periodic number

Solution
Question 4
1 / -0

Without actually performing the long division, state whether the following rational number will have terminating decimal expansion or a non-terminating repeating decimal expansion. Also, find the numbers of places of decimals after which the decimal expansion terminates.
$$\dfrac { 13 }{ 3125 } $$

A
$$3$$

B
$$4$$

C
$$5$$

D
$$6$$

Solution

The given value is $$\dfrac{13}{3125}$$ the denominator is 3125 which can be written as:

$$3125=2^0 \times 5^5$$ it is in the form of $$2^m \times 5^n$$

$$max(m,n)=5$$

$$\therefore$$ the expansion is terminating decimal it terminates after

$$max(m,n)=5$$ places from the decimal [since $$ m=0,n=5$$]
Question 5
1 / -0

If HCF of numbers $$408$$ and $$1032$$ can be expressed in the form of $$1032x -408 \times 5$$, then find the value of $$x$$.

A
$$0$$

B
$$1$$

C
$$2$$

D
$$3$$

Solution

$$408=2\times2\times2\times3\times17$$

$$1032=2\times2\times2\times3\times43$$

Hence, $$HCF=2\times2\times2\times3=24$$

Now, $$1032x-408\times5=24\Rightarrow 1032x=2064\Rightarrow x=2$$
Question 6
1 / -0

Find the $$HCF$$ of integer pairs $$963$$ & $$654$$.

A
$$3$$

B
$$5$$

C
$$9$$

D
$$1$$

Solution
Question 7
1 / -0

GCF of $$99$$ and $$100$$ is __________

A
$$1$$

B
$$2$$

C
$$3$$

D
$$4$$

Solution

when we divide $$100$$ with $$99$$ we will get remainder $$1$$ and hence next we divide 99 with $$1$$ which gives us remainder $$0$$.

therefore $$1$$ is the HCF or GCF of $$99 $$ and $$100$$
Question 8
1 / -0

The decimal expansion of the rational number $$\dfrac{43}{2^4\times 5^3}$$ will terminate after:

A
$$3$$

B
$$4$$

C
$$2$$

D
$$1$$

Solution
Question 9
1 / -0

A rectangular veranda is of dimension $$18$$ m $$72$$ cm $$\times 13$$ m $$20$$ cm. Square tiles of the same dimensions are used to cover it. Find the least number of such tiles.

A
$$4290$$

B
$$4540$$

C
$$4620$$

D
$$4230$$

Solution

The edge of rectangular veranda are $$18\ m\ 72\ cm=1872\ cm$$ and $$13\ m\ 20\ cm=1320\ cm$$.

On taking $$HCF$$ of $$1872$$ and $$1320$$ using prime factorisation, we get

$$HCF=24$$

Therefore,
No. of tiles required $$=$$ $$\dfrac{Area\ of\ Veranda}{Area\ of\ tiles}$$

$$=\dfrac{1872\times 1320}{24\times 24}$$

$$=4290$$

Hence, this is the answer.
Question 10
1 / -0

$$n^2 -1$$ is divisible by $$8$$ , if n is

A
an integer

B
a natural number

C
an odd integer

D
an even integer

Solution

Given,

Any odd positive integer $$n$$ can be written in form of $$4q + 1$$ or $$4q + 3$$.

If $$n = 4q + 1$$, when $$n^2 - 1 = (4q + 1)^2 - 1 = 16q^2 + 8q + 1 - 1 = 8q(2q + 1)$$ which is divisible by $$8.$$

If $$n = 4q + 3$$, when $$n^2 - 1 = (4q + 3)^2 - 1 = 16q^2 + 24q + 9 - 1 = 8(2q^2 + 3q + 1)$$ which is divisible by $$8$$.

So, it is clear that $$n^2 - 1$$ is divisible by $$8$$, if $$n$$ is an odd positive integer.

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Re-Attempt Weekly Quiz Competition

Real Numbers Test - 31

Result

Real Numbers Test - 31

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Rank

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SHARING IS CARING

Question 1

$$36$$

$$18$$

$$12$$

$$6$$

Solution

Question 2

Always irrational

Always rational

Can be both rational or irrational

always an integer

Solution

Question 3

Also a rational number

An irrational number

Not a number

A natural periodic number

Solution

Question 4

$$3$$

$$4$$

$$5$$

$$6$$

Solution

Question 5

$$0$$

$$1$$

$$2$$

$$3$$

Solution

Question 6

$$3$$

$$5$$

$$9$$

$$1$$

Solution

Question 7

$$1$$

$$2$$

$$3$$

$$4$$

Solution

Question 8

$$3$$

$$4$$

$$2$$

$$1$$

Solution

Question 9

$$4290$$

$$4540$$

$$4620$$

$$4230$$

Solution

Question 10

an integer

a natural number

an odd integer

an even integer

Solution

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