Number Systems Test

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Number Systems Test - 33

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Question 1
1 / -0

Following are the steps to represent $$\sqrt5$$ on number line.
Arrange them in order.
1) Draw OC on line with $$l(OC)=l(OB)$$,
2) Draw $$AB \perp OA\ and\ l(AB) =1$$
3) Take $$l(OA)=2$$
4) $$l(OC)=\sqrt5$$, C is required point on real line.

A
$$1,2,3,4$$

B
$$2,4,1,3$$

C
$$3,2,4,1$$

D
$$3,2,1,4$$

Solution

The correct order of representing $$\sqrt { 5 } $$ on number line is

Step 1. Take $$l(OA) = 2$$

Step 2. Draw $$AB$$$$\perp $$$$OA\ and\ l(AB)$$ $$= 1$$

Step 3. Draw $$OC$$ on line with $$l(OC) = l(OB)$$

Step 4. $$l(OC) =$$ $$\sqrt { 5 } $$, $$C$$ is the required point on real line

Therefore, option(D) is correct.
Question 2
1 / -0

In mathematics, the number $$\phi$$ is an irrational number that is roughly equal to $$1.618$$... Which letter corresponds to the correct placement of $$\phi$$ on the number line below?

A
A

B
B

C
C

D
D

Solution

Since the part between $$1$$ and $$2$$ is divided into $$10$$ parts , so, each part measures $$0.1$$.
$$\phi=1.618$$ is to be located on the number line.
$$1.6<1.618<1.7$$
So, on analysis, we can say that point B is the correct point.
Question 3
1 / -0

Which of the following number is different from others?

A
$$\sqrt 7$$

B
$$\sqrt 6$$

C
$$\sqrt {25}$$

D
$$\sqrt{10}$$

Solution

$$\sqrt{7}$$ is an irrational number
$$\sqrt{6}$$ is an irrational number
$$\sqrt{10}$$ is an irrational number
$$\sqrt{25}=5$$ is different from others because others are irrational number but $$\sqrt{25}$$ is a rational number
Hence, option C is correct.
Question 4
1 / -0

If $$p$$ is prime, then $$\sqrt {p}$$ is:

A
Composite number

B
Rational number

C
Positive integer

D
Irrational number

Solution

SInce, we know that prime numbers are those which are never perfect square and not divisible by any other number except by itself.
which are $$2,3,5,7,...$$
Clearly, if $$p$$ is prime then $$\sqrt p $$ is irrational number.
Option $$D$$ is correct.
Question 5
1 / -0

Which one of the following has a terminating decimal expansion?

A
$$\displaystyle\frac{5}{32}$$

B
$$\displaystyle\frac{7}{9}$$

C
$$\displaystyle\frac{8}{15}$$

D
$$\displaystyle\frac{1}{12}$$

Solution

We know that a terminating decimal is a decimal that ends. It's a decimal with a finite number of digits. For example $$\dfrac {1}{4}=0.25$$, it has only two decimal digits.

Now consider the fraction $$\dfrac {5}{32}$$ whose decimal form will be:
$$\dfrac {5}{32}=0.15625$$
The resulting decimal number ends with five decimal digits and therefore, it is terminating decimal.
Hence, $$\dfrac {5}{32}$$ has a terminating decimal expansion.

While, $$\dfrac{7}{9}=0.7777... $$, non-terminating,
$$\dfrac{8}{15}=0.5333... $$, non-terminating
and $$\dfrac{7}{9}=0.08333... $$, non-terminating

Therefore, option $$A$$ is correct.
Question 6
1 / -0

Solve : $$(16)-(14\sqrt{2})$$

A
$$(2\sqrt{2})$$

B
$$2$$

C
$$(10\sqrt{2})$$

D
$$(16-14\sqrt{2})$$

Solution

We have $$(16)-(14\sqrt2)$$, where 16 is rational number and$$14\sqrt2$$ is irrational number.
We can not add rational and irrational numbers directly. So, we can not solve the given numbers directly.
$$(16-14\sqrt2)$$ is the answer.
Question 7
1 / -0

Simplify the following using law of exponents.
$$9^2 \times 9^{18}\times 9^{10}$$

A
$$9^{30}$$

B
$$9^{25}$$

C
$$9^{15}$$

D
$$9^{10}$$

Solution

We know that,

$$a^{m}\times a^{n}\times a^{p}=a^{m+n+p}$$

Therefore,

$$9^{2}\times 9^{18}\times 9^{10}=9^{2+18+10}$$

$$=9^{30}$$
Question 8
1 / -0

Simplify: $$(\sqrt{3}) + (5\sqrt{3}) - (8\sqrt{3}) + (10\sqrt{3})$$

A
$$24\sqrt{3}$$

B
$$8\sqrt{3}$$

C
$$12\sqrt{3}$$

D
$$16\sqrt{3}$$

Solution

In the given expression we can take $$\sqrt{3}$$ common
Now we have,
$$\sqrt3(1+5-8+10)=8\sqrt3$$
Question 9
1 / -0

Simplify: $$(9\sqrt{3}) + (5\sqrt{3}) - (10\sqrt{2}) - (2\sqrt{2})$$

A
$$(\sqrt{6})$$

B
$$(14\sqrt{3})-(8\sqrt{2})$$

C
$$(14\sqrt{3})-(12\sqrt{2})$$

D
$$(2\sqrt{6})$$

Solution

We can also write the given expression, After taking $$\sqrt3 $$ and $$-\sqrt2$$, as
$$=\sqrt3(9+5)-\sqrt2(10+2)\\=14\sqrt3-12\sqrt2$$
Question 10
1 / -0

Which of the following irrational number lies between $$2.5$$ and $$2.6$$?

A
$$2.512$$

B
$$2.77321$$

C
$$2.5333...$$

D
$$2.5135145...$$

Solution

$$2.512$$ and $$2.77321$$ are rational numbers as the decimal expansion terminates.
Option $$C$$ can be rewritten as $$2.5\bar{3}$$, which is also a rational number, as it is non-terminating recurring.
But option $$D$$ is non-terminating non-recurring and therefore is an irrational number and lies between $$2.5$$ and $$2.6$$.
So, option $$D$$ is the correct answer.

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Re-Attempt Weekly Quiz Competition

Number Systems Test - 33

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Number Systems Test - 33

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Question 1

$$1,2,3,4$$

$$2,4,1,3$$

$$3,2,4,1$$

$$3,2,1,4$$

Solution

Question 2

A

B

C

D

Solution

Question 3

$$\sqrt 7$$

$$\sqrt 6$$

$$\sqrt {25}$$

$$\sqrt{10}$$

Solution

Question 4

Composite number

Rational number

Positive integer

Irrational number

Solution

Question 5

$$\displaystyle\frac{5}{32}$$

$$\displaystyle\frac{7}{9}$$

$$\displaystyle\frac{8}{15}$$

$$\displaystyle\frac{1}{12}$$

Solution

Question 6

$$(2\sqrt{2})$$

$$2$$

$$(10\sqrt{2})$$

$$(16-14\sqrt{2})$$

Solution

Question 7

$$9^{30}$$

$$9^{25}$$

$$9^{15}$$

$$9^{10}$$

Solution

Question 8

$$24\sqrt{3}$$

$$8\sqrt{3}$$

$$12\sqrt{3}$$

$$16\sqrt{3}$$

Solution

Question 9

$$(\sqrt{6})$$

$$(14\sqrt{3})-(8\sqrt{2})$$

$$(14\sqrt{3})-(12\sqrt{2})$$

$$(2\sqrt{6})$$

Solution

Question 10

$$2.512$$

$$2.77321$$

$$2.5333...$$

$$2.5135145...$$

Solution

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