2 Marks Questions

Question 512 Marks

Write down the decimal expansions of the following rational numbers by writing their denominators in the form $2^m \times 5^n$, where, m, n are non-negative integers.
$\frac{13}{125}$

Answer

The given number is $\frac{13}{125}.$
$\frac{13}{125} = \frac{13}{5^{3}} = \frac{13}{5^{3} \times 2^0}$
$= \frac{13\times 2^{3}}{5^{3}\times 2^{3}} = \frac{13\times 8}{(10)^3} = \frac{104}{1000} = 0.104$
Thus, the decimal expansion of $\frac{13}{125}$ is 0.014.

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Question 522 Marks

Without actually performing the long division, state whether state whether the following rational numbers will have a terminating decimal expansion or a non terminating repeating decimal expansion.
$\frac{129}{2^2\times5^7\times7^{17}}$

Answer

Since the denominator is not of the form $2^m 5^n,$ and it also has $7$ as its factor, the decimal expansion of $\frac{129}{2^2\times5^77^{17}}$ is non-terminating repeating.

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Question 532 Marks

Write 98 as product of its prime factors.

Answer

Using factors tree for prime factorisation of 98.

prime factoris of 98 = 2 × 7 × 7

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Question 542 Marks

Define HCF of two positive integers and find the HCF of the following pairs of numbers:
240 and 6552

Answer

By applying Euclid’s division lemma
6552 = 240 × 27 + 72
Since remainder ≠ 0, apply division lemma on divisor of 240 and remainder 72.
240 = 72 × 3 + 24
Since remainder ≠ 0, apply division lemma on divisor of 72 and remainder 24.
72 = 24 × 3 + 0
$\therefore$ H.C.F of 6552 and 240 is = 24.

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Question 552 Marks

Find the HCF of the following pairs of integers and express it as a linear combination of them.
1288 and 575

Answer

By applying Euclid’s division lemma
1288 = 575 × 2 + 138 …..(i)
Since remainder ≠ 0, apply division lemma on division 575 and remainder 138.
575 = 138 × 4 + 23 …..(ii)
Since remainder ≠ 0, apply division lemma on division 138 and remainder 23 …..(iii)
HCF = 23
Now, 23 = 575 - 138 × 4 [from (ii)]
= 575 - [1288 - 575 × 2] × 4 [from (i)]
= 575 - 1288 × 4 + 575 × 8
= 575 × 9 - 1288 × 4

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Question 562 Marks

Use Euclid's division algorithm to find the HCF of:
136, 170 and 255

Answer

Given integers are 136, 170 and 255.
Let us first find the HCF of 136, 170 by Euclid lemma.
Clearly, 170 > 136. So, we will apply Euclid’s division lemma to 136 and 170.
170 = 136 × 1 + 34
Remainder is 34 which is a non-zero number.
Now, apply Euclid’s division lemma to 136 and 34.
136 = 34 × 4 + 0
The remainder at this stage is zero.
Therefore, 34 is the HCF of 136 and 170.
Now, again use Euclid’s division lemma to find the HCF of 34 and 255.
255 = 34 × 7 + 17
Remainder is 17 which is a non-zero number.
Now, apply Euclid’s division lemma to 34 and 17.
34 = 17 × 2 + 0
The remainder at this stage is zero.
Therefore, 17 is the HCF of 136, 170 and 255.

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Question 572 Marks

What can you say about the prime factorisations of the 43 of the following rationals:
0.120120012000120000...

Answer

We have,
0.120120012000120000...
We can see that it is a non-terminating repeating decimal expersion.
So, its denominator has factors others them 2 or 5.

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Question 582 Marks

Define $HCF$ of two positive integers and find the $HCF$ of the following pairs of numbers:
$75$ and $243$

Answer

Prime factors of 75 and 243 are
$75 = 3 \times 5 \times 5 = 3 \times 5^2$
$243 = 3 \times 3 \times 3 \times 3 \times 3 = 3^5$
$HCF (75, 243) =$ Product of the smallest power of each common prime factor in the number
$= 3$

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Maths 2 Marks Questions