Real Numbers Test

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Real Numbers Test - 30

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Question 1
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Using the theory that any positive odd integers are of the form $$4 q + 1$$ or $$4 q + 3$$ where $$q$$ is a positive integer, If the quotient is $$4$$, the dividend is $$19$$ then what will be the remainder?

A
$$17$$

B
$$2$$

C
$$1$$

D
$$3$$

Solution

Let $$a$$ be any positive integer.
We know by Euclid's algorithm, if $$a$$ and $$b$$ are two positive integers, there exist unique integers $$q$$ and $$r$$ satisfying, $$a= bq + r$$ where $$0 \leq r < b$$.
Take $$b= 4$$ we get,
$$ a = 4q + r, 0 \leq r < b$$.
Since $$0 \leq r < 4$$, therefore the possible remainders are $$0, 1, 2$$ and $$3$$.
That is, $$a$$ can be $$4q, 4q+1$$, or $$4q+2$$ or $$4q+3$$, where $$q$$ is the quotient.
Since $$a$$ is odd, $$a$$ cannot be $$4q$$ or $$4q + 2$$ as they are both divisible by $$2$$.
Therefore, any odd integer is of the form $$4q + 1$$ or $$4q + 3$$.
Given the quotient $$q=4$$ we get,
$$4q+1= 4 \times 4 +1 = 17$$, or $$4q + 3= 4 \times 4 + 3 = 19$$
Since dividend is $$19$$ the remainder is $$3.$$
Therefore$$,$$ $$D$$ is the correct answer.
Question 2
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The HCF of two consecutive even numbers is

A
$$6$$

B
$$3$$

C
$$4$$

D
$$2$$

Solution

$$HCF$$ of two consecutive even numbers is always $$2$$. For example:

$$HCF$$ of $$2$$ and $$4$$ is $$2$$ and similarly,

$$HCF$$ of $$22$$ and $$24$$ is $$2$$ and we can do so on..
Question 3
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There are five odd numbers $$1, 3, 5, 7, 9$$. What is the HCF of these odd numbers?

A
$$2$$

B
$$1$$

C
$$6$$

D
$$5$$

Solution

Let us calculate and check what is the highest common factor for these numbers.
First finding the factors of individual numbers:

Factors of $$1=1$$
Factors of $$3=1, 3$$
Factors of $$5=1, 5$$
Factors of $$7=1, 7$$
Factors of $$9=1, 3, 9$$

HCF is $$1$$ for the five odd numbers.
Therefore, $$B$$ is the correct answer.
Question 4
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Find HCF of $$70$$ and $$245$$ using Fundamental Theorem of Arithmetic.

A
$$35$$

B
$$30$$

C
$$15$$

D
$$25$$

Solution

Using Fundamental Theorem of Arithmetic we can write the given integers as a product of their prime factors,
$$70 =2\times 5\times 7$$
$$245=5\times 7 \times 7 $$

Hence, common factors of $$70$$ and $$245$$ are $$5$$ and $$7.$$
$$HCF = 5\times 7$$
$$=35$$

Hence, option $$A$$ is correct.
Question 5
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Three ropes are $$7\ m, 12\ m\ 95\ cm$$ and $$3\ m\ 85\ cm$$ long. What is the greatest possible length that can be used to measure these ropes?

A
$$35\ cm$$

B
$$55\ cm$$

C
$$1\ m$$

D
$$65\ cm$$

Solution

The given three ropes are $$7$$m, $$12$$ m $$95$$cm and $$3$$m$$85$$cm long. We know that $$1$$m=$$100$$cm, therefore,

The length of the respective ropes will be:

1st rope $$=7\times 100=700$$cm
2nd rope $$=(12\times 100)+95=1200+95=1295$$cm
3rd rope $$=(3\times 100)+85=300+85=385$$cm

Now, let us factorize the length of the ropes as follows:

$$700=2\times 2\times 5\times 5\times 7\\ 1295=5\times 7\times 37\\ 385=5\times 7\times 11$$

The highest common factor (HCF) is $$5\times 7=35$$

Hence, the greatest possible length that can be used to measure these ropes is $$35$$cm.
Question 6
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Identify a non-terminating repeating decimal.

A
$$\dfrac{24}{1600}$$

B
$$\dfrac{171}{800}$$

C
$$\dfrac{123}{2^2 \times 5^3}$$

D
$$\dfrac{145}{2^3 \times 5^2 \times 7^2}$$

Solution

We know that, a fraction will have terminating decimal expansion only if its denominator is in the form of $${2}^{m}\times{{5}^{n}}$$.
Let's check the denominator of each option:
$$(i)\ \dfrac{24}{1600}\\$$
$$1600=2\times 2\times 2\times 2\times 2\times 2\times 5\times 5=2^6\times 5^2$$

Hence, the denominator is $$2^6\times 5^2$$

$$(ii)\ \dfrac{171}{800}\\$$
$$800=2\times 2\times 2\times 2\times 2\times 5\times 5=2^5\times 5^2$$

Hence, the denominator is $$2^5\times 5^2$$

$$(iii)\ \dfrac{123}{2^3\times 5^3}\\$$
Hence, the denominator is $$2^3\times 5^3$$

$$(iv)\ \dfrac{145}{2^3\times 5^2\times 7^2}\\$$
Hence, the denominator is $$2^3\times 5^2\times 7^2$$

Only option D does not have a denominator of the form $$2^m\times 5^n$$.
Hence, it will have a non-terminating decimal expansion. Further, the expansion will be repeating.

Hence, option D is correct.
Question 7
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A rational number can be expressed as a terminating decimal if the denominator has factors _________.

A
$$2$$ or $$5$$

B
$$2$$, $$3$$ or $$5$$

C
$$3$$ or $$5$$

D
Only $$2$$ and $$3$$

Solution

We know, any rational number with its denominator is in the form of $${ 2 }^{ m }\times { 5 }^{ n }$$, where $$m,n$$ are positive integers are terminating decimals.
So, option $$A$$ is correct.
Question 8
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Which one of the following is not true?

A
$$\sqrt{2}$$ is an irrational number

B
If a is a rational number and $$\sqrt{b}$$ is an irrational number then $$a\sqrt{b}$$ is irrational number

C
Every surd is an irrational number

D
The square root of every positive integer is always irrational

Solution

(a) All numbers that are not rational are considered irrational. An irrational number can be written as a decimal, but not as a fraction. An irrational number has endless non-repeating digits to the right of the decimal point. Here are some irrational numbers:

$$π = 3.141592…$$
$$\sqrt {2} = 1.414213…$$

Therefore, $$\sqrt {2}$$ is an irrational number.

(b) Let us take a rational number $$a=\dfrac {2}{1}$$ and an irrational number $$b=\sqrt {2}$$, then their product can be determined as:

$$a\times b=2\times \sqrt { 2 } =2\sqrt { 2 }$$ which is also an irrational number.

Therefore, if $$a$$ is a rational number and $$\sqrt {b}$$ is an irrational number than $$a\sqrt {b}$$ is an irrational number.

(c) A surd is an irrational root of a rational number. So we know that surds are always irrational and they are always roots.

For example, $$\sqrt {2}$$ is a surd since $$2$$ is rational and $$\sqrt {2}$$ is irrational.

Surds are numbers left in root form $$\sqrt {\ \ }$$ to express its exact value. It has an infinite number of non-recurring decimals.

Therefore, every surd is an irrational number.

(d) Let us take a positive integer $$4$$, now square root of $$4$$ will be:

$$\sqrt {4}=2$$ which is not an irrational number

Hence, the square root of every positive integer is not always irrational.
Question 9
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A real number $$\displaystyle \frac{2^2 \times 3^2 \times 7^2}{2^5 \times 5^3 \times 3^2 \times 7}$$ will have _________.

A
Terminating decimal

B
Non-terminating decimal

C
Non-terminating and non-repeating decimal

D
Terminating repeating decimal

Solution

Using the law of exponent,
$$\dfrac{2^{2}\times3^{2}\times7^{2}}{2^{5}\times5^{3}\times3^{2}\times7} = 2^{2-5}\times3^{2-2}\times5^{-3}\times7^{2-1}$$
$$=2^{-3}\times3^{0}\times5^{-3}\times7^{1}$$

$$= \dfrac{7}{2^{3}\times5^{3}}$$

$$= \dfrac{7}{1000}$$ $$= 0.007$$.

$$\therefore 0.007$$ is a terminating and non-repeating decimal.
That is, option $$A$$ is correct.
Question 10
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The greatest integer that divides $$358,\ 376$$ and $$ 232$$. The same remainder in each case is

A
$$6$$

B
$$7$$

C
$$8$$

D
$$9$$

Solution

If the remainder is the same in each case and the remainder is not given, HCF of the differences of the numbers is the required greatest number

Given: The greatest number that will divide 358,376 and 232 leaving the same remainder in each case.

To find: The number?

Solution :

First, we find the difference between these numbers.

The required numbers are

376 - 232 = 144

376-358= 18

358 - 232 =126

Now, We find the HCF of 18, 126, and 144

$$18 =2\times3^2$$

$$126= 2\times3^2\times7$$

$$144= 2^4\times3^2$$

$$HCF(18,126,144)= 2\times3\times3$$

Therefore, The required largest number is 18.

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Real Numbers Test - 30

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Real Numbers Test - 30

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Question 1

$$17$$

$$2$$

$$1$$

$$3$$

Solution

Question 2

$$6$$

$$3$$

$$4$$

$$2$$

Solution

Question 3

$$2$$

$$1$$

$$6$$

$$5$$

Solution

Question 4

$$35$$

$$30$$

$$15$$

$$25$$

Solution

Question 5

$$35\ cm$$

$$55\ cm$$

$$1\ m$$

$$65\ cm$$

Solution

Question 6

$$\dfrac{24}{1600}$$

$$\dfrac{171}{800}$$

$$\dfrac{123}{2^2 \times 5^3}$$

$$\dfrac{145}{2^3 \times 5^2 \times 7^2}$$

Solution

Question 7

$$2$$ or $$5$$

$$2$$, $$3$$ or $$5$$

$$3$$ or $$5$$

Only $$2$$ and $$3$$

Solution

Question 8

$$\sqrt{2}$$ is an irrational number

If a is a rational number and $$\sqrt{b}$$ is an irrational number then $$a\sqrt{b}$$ is irrational number

Every surd is an irrational number

The square root of every positive integer is always irrational

Solution

Question 9

Terminating decimal

Non-terminating decimal

Non-terminating and non-repeating decimal

Terminating repeating decimal

Solution

Question 10

$$6$$

$$7$$

$$8$$

$$9$$

Solution

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