Real Numbers Test

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Real Numbers Test - 25

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Weekly Quiz Competition

Question 1
1 / -0

H.C.F. of $$6, 72$$ and $$120$$ is:

A
$$6$$

B
$$2$$

C
$$3$$

D
$$1$$

Solution

Factors of 6 are $$1,2,3,6$$
Factors of 72 are $$1,2,3,4,6,8,9,12,18,24,36,72$$
Factors of 120 are $$1,2,3,4,5,6,8,10,12,15,20,24,30,40,60,120$$
The highest factor in all the three is $$6$$ .
Question 2
1 / -0

Write whether every positive integer can be of the form $$4q + 2$$, where $$q$$ is an integer.

A
Yes

B
No

C
Ambiguous

D
Data Insufficient

Solution

No, all positive integers cannot be written in the form of $$ 4q+2$$
Because, $$4q+2=2\left( 2q+1 \right) $$
Therefore, $$4q+2$$ is an even number.
So, we can't write odd numbers in the form $$4q+2$$ (where $$q$$ is an integer)
Also $$2q+1$$ is an odd number, hence the maximum power of $$2$$ that divides $$4q+1$$ is $$1$$. Therefore, we can't represent the numbers divisible by $$4$$ this form.
Question 3
1 / -0

Euclids division lemma states that if a and b are any two $$+$$ ve integers, then there exists unique integers q and r such that :

A
$$a=bq+r, 0 < r < b$$

B
$$a=bq+r, 0 \leq r \leq b$$

C
$$a=bq+r, 0 \leq r < b$$

D
$$a=bq+r, 0 < b < r$$

Solution

Division Algorithm: Euclids division lemme states that if a and b are any two positive integers then there exist unique integers q and r such that, $$ a = q b + r \, \, 0 \leq r < b $$
$$ \therefore $$ option C is correct.
Proof: We begin by proving that the set s = {a - xb | x is an integer: a-xb $$\geq$$ 0 }
is non-empty. To do this, if suffices to exhibit a value of x making a-xb nonnegative a value of x making a-xb nonnegative. Because the integer b $$\geq$$1 , we have |a| b $$\geq$$ |a| and so -
$$ a - (-|a|) b = a + |a| b \geq a + |a| \geq 0 $$
For the choice x = -|a|, then a - xb lies in s. This paves the way for an application of well ordering principle, from which we infer that the set S contains a smallest integer : call it r. By the definition S, $$\exists$$ an integer q satisfying , r = a - qb , $$0\leq r$$
we argue that r < b. If this were not this case, then r $$\geq$$ 0 and a - ( q + 1 ) b = ( a - q b ) - b = r - b $$\geq$$ 0
This implication is that the integer a - ( q + 1 ) b has the proper form to belong to the set S. But a - ( q + 1 ) b = r - b < r , leading to a contradiction of the choice of r as the smallest member of S. Hence r < b .
Next we turn to the task of showing the uniqueness of q and r. Suppose that a has two representation of the desired form say,
$$ a = q b + r = q^{'}b + r^{'} $$
Where $$ 0 \leq r < b, \, 0 \leq r^{'} < b . $$ Then $$r^{'}- r = b (q - q^{'}) $$ and, owing to the fact that
$$ | r^{'} - r | = b | q - q^{'} | $$ upon adding two inequalities $$-b < -r \leq 0 $$ and $$ 0 \leq r^{'} < b $$ we obtain $$ -b < r^{'}-r < b \Rightarrow |r^{'} - r | < b $$
Thus, $$b|q-q^{'}| < b $$ which yields $$ 0 \leq |q-q^{'} | < 1 $$
As, $$ |q-q^{'} | $$ is positive, the only possibility is that $$ |q-q^{'} | = 0 $$ where $$q = q^{'} : $$ this is turns give $$ r = r^{'} $$ and ends the proof.
Question 4
1 / -0

H.C.F. of $$26$$ and $$91$$ is:

A
$$13$$

B
$$2366$$

C
$$91$$

D
$$182$$

Solution

Prime factors of $$26=2\times 13$$
Prime factors of $$91=7\times 13$$
As $$13$$ is the only common factor.
Therefore,
$$HCF(26,91)=13$$
Question 5
1 / -0

Use Euclid's division lemma to find the HCF of the following

16 and 176

A
$$16$$

B
$$14$$

C
$$10$$

D
$$12$$
Question 6
1 / -0

Euclid's division lemma states that for two positive integers a and b, there exist unique integers q and r such that $$a = bq + r$$, where r must satisfy

A
1 < r < b

B
$$0 < r \leq b$$

C
$$0\leq r < b$$

D
0 < r < b

Solution

If $$r$$ must satisfy$$ 0 \le r < b$$
Proof,
$$..,a-3b,a-2b,a-b,a,a+b,a+2b,a+3b,..$$
clearly it is an arithmetic progression with common difference $$b$$ and it extends infinitely in both directions.
Let $$r$$ be the smallest non-negative term of this arithmetic progression.Then,there exists a non-negative integer $$q$$ such that,
$$a-bq=r$$
$$=>a=bq+r$$
As,$$r$$ is the smallest non-negative integer satisfying the result.Therefore, $$0\quad \le \quad r\quad \le \quad b$$
Thus, we have
$$a=bq_1+r_1$$, $$0 \le r_1 \le b$$
Question 7
1 / -0

If HCF of $$210$$ and $$55$$ is of the form $$(210) (5) + 55 y$$, then the value of $$y$$ is :

A
$$-19$$

B
$$-18$$

C
$$5$$

D
$$55$$

Solution

HCF of 210 and 55 is 5
Given HCF = $$(210)(5)+55y$$
$$5 =(210)(5)+55y$$
$$55y$$=$$-1045$$
On solving this equation we get $$y=-19$$
So, correct answer will be option A
Question 8
1 / -0

Use Euclid's division lemma to find the HCF of $$40$$ and $$248$$.

A
$$4$$

B
$$10$$

C
$$6$$

D
$$8$$

Solution

Here, $$a=248$$ and $$b=40$$
by using Euclid's division lemma,
$$a=bq+r$$ ; $$0\le r<b$$
$$248=40\times 6 +8$$
$$40=8\times 5+0$$
$$\therefore H.C.F (40,248)=8$$
hence option $$D$$ is correct.
Question 9
1 / -0

The decimal expansion of the rational number $$\displaystyle\frac{23}{2^{3}5^{2}}$$, will terminate after how many places of decimal?

A
1

B
3

C
4

D
5

Solution

$$\displaystyle\frac{23}{2^{3}5^{2}}$$

= $$\displaystyle\frac{23}{2(2)^{2}(5)^{2}}$$

= $$\displaystyle\frac{23}{2(2\times5)^{2}}$$

= $$\displaystyle\frac{23}{2(10)^{2}}$$

= $$\displaystyle\frac{11.5}{100} = 0.115$$

Therefore, the decimal expansion of the rational number $$\displaystyle\frac{23}{2^{3}5^{2}}$$
will terminate after three places of decimal.
Question 10
1 / -0

Use Euclid's division lemma to find the HCF of the following

27727 and 53124

A
$$233$$

B
$$200$$

C
$$154$$

D
$$212$$

Solution

Here, $$a=53124$$ and $$b=27727$$
by using Euclid's division lemma,
$$a=bq+r$$ ; $$0<r<b$$
$$53124=(27727)(1)+25397$$
$$27727=(25397)(1)+2330$$
$$25397=(2330)(10)+2097$$
$$2330=(2097)(1)+233$$
$$2097=(233)(9)+0$$
H.C.F is $$233$$

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Re-Attempt Weekly Quiz Competition

Real Numbers Test - 25

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Real Numbers Test - 25

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SHARING IS CARING

Question 1

$$6$$

$$2$$

$$3$$

$$1$$

Solution

Question 2

Yes

No

Ambiguous

Data Insufficient

Solution

Question 3

$$a=bq+r, 0 < r < b$$

$$a=bq+r, 0 \leq r \leq b$$

$$a=bq+r, 0 \leq r < b$$

$$a=bq+r, 0 < b < r$$

Solution

Question 4

$$13$$

$$2366$$

$$91$$

$$182$$

Solution

Question 5

$$16$$

$$14$$

$$10$$

$$12$$

Question 6

1 < r < b

$$0 < r \leq b$$

$$0\leq r < b$$

0 < r < b

Solution

Question 7

$$-19$$

$$-18$$

$$5$$

$$55$$

Solution

Question 8

$$4$$

$$10$$

$$6$$

$$8$$

Solution

Question 9

1

3

4

5

Solution

Question 10

$$233$$

$$200$$

$$154$$

$$212$$

Solution

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