Real Numbers Test

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Real Numbers Test - 28

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Weekly Quiz Competition

Question 1
1 / -0

If the H.C.F. of $$A$$ and $$B$$ is $$24$$ and that of $$C$$ and $$D$$ is $$56,$$ then the H.C.F. of $$A, B, C$$ and $$D$$ is

A
$$4$$

B
$$12$$

C
$$8$$

D
$$3$$

Solution

Given the H.C.F. of $$A$$ and $$B$$ is $$24$$ and that of $$C$$ and $$D$$ is $$56.$$
Then the H.C.F. of $$A, B, C$$ and $$D$$ is the HCF of $$24$$ and $$56$$ which is $$8.$$
Question 2
1 / -0

Find the HCF of $$867$$ and $$255$$ using Euclid theorem.

A
$$50$$

B
$$51$$

C
$$52$$

D
$$53$$

Solution

The given numbers are $$867$$ and $$255$$
According to the definition of Euclid's theorem,
$$a = b \times q + r$$ where $$0 \leq r < b .$$
$$867=255\times 3+102$$
$$255=102\times 2+51$$
$$102=51\times 2+0$$

The HCF of $$867$$ and $$255$$ is $$51$$.
So, option $$B$$ is correct.
Question 3
1 / -0

The HCF of 75 and 180 is equal to

A
$$11$$

B
$$12$$

C
$$13$$

D
$$15$$

Solution

The given numbers are $$75$$ and $$180$$.
To find out: The HCF of the given numbers.

We can represent the given numbers as the product of their prime factors as:
$$75=3\times 5\times 5$$
$$180=2\times 2\times 3\times 3\times 5$$

We know that, the HCF of two numbers is the product of the factors common between them.

Here, $$3$$ and $$5$$ are the factors common between $$75$$ and $$180$$.

So, HCF $$=3\times 5$$
$$=15$$

Hence, the HCF of $$75$$ and $$180$$ is $$15$$.
Question 4
1 / -0

If we apply Euclid"s division algorithm for $$20$$ using $$ 8$$ then the correct answer will be:

A
$$20=8\times 2+4$$

B
$$20=8\times 4+0$$

C
$$20=8\times 3+3$$

D
none of the above

Solution

Here, Euclid"s division algorithm is used to represent $$20$$ using $$8.$$
The only option in which L.H.S $$=$$ R.H.S is option A.
i.e. $$ 20 = 2 \times 8 + 4$$
$$=16+4=20$$
So, option A is correct.
Question 5
1 / -0

The HCF of $$196$$ and $$38416$$ using Euclid algorithm is

A
$$191$$

B
$$192$$

C
$$193$$

D
$$196$$

Solution

Let us first state Euclid’s division algorithm:

To obtain the HCF of two positive integers, say $$c$$ and $$d$$, with $$c > d$$, follow the steps below:

Step 1 : Apply Euclid’s division lemma, to $$c$$ and $$d$$. So, we find whole numbers, $$q$$ and $$r$$ such that $$c = dq + r, 0 ≤ r < d$$.

Step 2 : If $$r = 0$$, $$d$$ is the HCF of $$c$$ and $$d$$. If $$r ≠ 0$$, apply the division lemma to $$d$$ and $$r$$.

Step 3 : Continue the process till the remainder is zero. The divisor at this stage will be the required HCF.

Now, let us apply the Euclid’s division algorithm to the given numbers $$196$$ and $$38416$$ as follows:

$$38416=196\times 196+0$$

Since the remainder is $$0$$.

Hence, the HCF of $$196$$ and $$38416$$ is $$196$$.
Question 6
1 / -0

The HCF of $$455$$ and $$42$$ using Euclid algorithm is

A
$$7$$

B
$$6$$

C
$$5$$

D
$$4$$

Solution

According to the definition of Euclid's theorem,
$$a = b \times q + r$$ where $$0 \leq r < b .$$
$$455=42\times 10+35$$
$$42=35\times 1+7$$
$$35=7\times 5+0$$
The HCF of $$455$$ and $$42$$ is $$7$$.
So, option A is correct.
Question 7
1 / -0

If $$\displaystyle d=\frac { 1 }{ { 2 }^{ 3 }\times { 5 }^{ 7 } } $$ is expressed as a terminating decimal, how many non zero digits will $$d$$ have?

A
One

B
Two

C
Three

D
Seven

E
Ten

Solution

Given, $$d=\cfrac{1}{2^3\times5^7 } $$.
Multiply and divide by $${2^4}$$,
$$\Rightarrow d=\cfrac{2^4}{2^3 \times 5^7 \times 2^4}$$
= $$\cfrac{16}{2^7 \times 5^7 }$$
= $$\cfrac{16}{10^7 }$$
=$$0.0000016 $$
Hence, $$d$$ will have two non-zero digits, $$1$$ and $$6$$, when expressed as a decimal.
Therefore, option $$B$$ is the correct answer.
Question 8
1 / -0

Apply Euclid"s theorem for $$17, 5$$.

A
$$17=5\times 3+2$$

B
$$17=5\times 2+7$$

C
$$17=5\times 4-3$$

D
None of the above

Solution

According to the definition of Euclid's theorem,
$$a = b \times q + r$$ where $$0 \leq r < b .$$
so,
$$17=5\times 3+2$$
Question 9
1 / -0

The HCF of $$135$$ and $$225$$ using Euclid algorithm is

A
$$41$$

B
$$42$$

C
$$43$$

D
$$45$$

Solution

According to the definition of Euclid's theorem,
$$a = b \times q + r$$ where $$0 \leq r < b .$$
$$225=135\times 1+90$$ [by Euclid"s algorithm]
$$135=90\times 1+45$$
$$90=45\times 2+0$$
$$\therefore$$ HCF of $$135$$ and $$225$$ is $$45$$.
So, option $$D$$ is correct.
Question 10
1 / -0

The the HCF of $$248$$ and $$492$$ is equal to

A
$$2$$

B
$$3$$

C
$$4$$

D
$$5$$

Solution

$$248=2^3 \times 31$$
$$492=2^2 \times 3 \times 41$$
HCF is $$4.$$

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Real Numbers Test - 28

Result

Real Numbers Test - 28

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SHARING IS CARING

Question 1

$$4$$

$$12$$

$$8$$

$$3$$

Solution

Question 2

$$50$$

$$51$$

$$52$$

$$53$$

Solution

Question 3

$$11$$

$$12$$

$$13$$

$$15$$

Solution

Question 4

$$20=8\times 2+4$$

$$20=8\times 4+0$$

$$20=8\times 3+3$$

none of the above

Solution

Question 5

$$191$$

$$192$$

$$193$$

$$196$$

Solution

Question 6

$$7$$

$$6$$

$$5$$

$$4$$

Solution

Question 7

One

Two

Three

Seven

Ten

Solution

Question 8

$$17=5\times 3+2$$

$$17=5\times 2+7$$

$$17=5\times 4-3$$

None of the above

Solution

Question 9

$$41$$

$$42$$

$$43$$

$$45$$

Solution

Question 10

$$2$$

$$3$$

$$4$$

$$5$$

Solution

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