Rational And Irrational Numbers - Real Numbers, Class 10 Mathematics - Class 10 MCQ

According to Euclid’s division lemma,

a = 3q + r, where 0  r < 3

As the number is divided by 3.So the remainder cannot be greater than divisor 3 also r is an integer. Therefore, the values of r can be 0, 1 or 2.

p = a × b × b
q = a × a × a × b
Since L.C.M is the product of the greatest power of each prime factor involved in the numbers.

Therefore, L.C.M of p and q = a³b²

Explanation: After simplification,

 

As the denominator has factor 5³ × 2² and which is of the type 5^m × 2ⁿ, So this is a terminating decimal expansion.

Since 5 and 8 are the remainders of 70 and 125, respectively. Thus after subtracting these remainders from the numbers, we have the numbers

65 = (70 − 5), 117 = (125 − 8) which is divisible by the required number.

Now required number = H.C.F of (65,117)

117 = 65 x 1 + 52

65 = 52 x 1 + 13
52 = 13 x 4 + 0

H.C.F of (65, 117) = 13

We have, 

 

On dividing 1056 by 23, we got 21 as remainder.

⇒ If we add 23 – 21 = 2 to the dividend 1056, we will get a number completely divisible by 23.

∴ Required number = (23 – 21) = 2

Product of a non-zero rational and an irrational number is always irrational. 

i.e., 

Since a = p × p × p × q × q,

b = p × q × q × q

Therefore H.C.F of a and b = pq²

Suppose the number be x. Since it divides 542 and 128 leave a remainder 2, using the Euclid’s division algorithm for 542 and x, 542 = ax + 2, where a is an integer. So, ax = 542 – 2 = 540, i.e. x is a factor of 540. Similarly x is a factor of (128 – 2).

The HCF of 542 – 2 = 540, 128 – 2 = 126:
Step 1: 540 = 126 × 4 + 36
Step 2: 126 = 36 × 3 + 18
Step 3: 36 = 18 × 2 + 0

HCF is 18
3, 6, 9 are factors of 18. 12 is the only number which isn’t a factor and hence doesn’t satisfy the above property.

Finding the biggest number that will divide both 324 and 144 is same as finding the HCF of both.
Prime factorizing the two numbers we get,

324 = 3 × 3 × 3 × 3 × 2 × 2

144 = 3 × 3 × 2 × 2 × 2 × 2
Now, taking the common factors between them gives us the HCF.
Therefore, HCF = 2 × 2 × 3 × 3 = 36

Maximum number of columns = HCF of 616 and 32 
616 = 2³ × 7 × 11
32 = 2⁵
∴ HCF of 616 and 32 =  8

Consider first two numbers 1848 and 3058, where 3058 > 1848.
3058 = 1848 × 1 + 1210
1848 = 1210 × 1 + 638
1210 = 638 × 1 + 572
638 = 572 × 1 + 66
572 = 66 × 8 + 44
66 = 44 × 1 + 22
44 = 22 × 2 + 0
∴ HCF of 1848 and 3058 is 22.
Let us find the HCF of the numbers 1331 and 22.
1331 = 22 × 60 + 11
22 = 11 × 2 + 0
HCF of 1331 and 22 is 11
∴ HCF of the three given numbers 1848, 3058 and 1331 is 11.
 

As the number 2m will always be even, so if we add 1 to it then, the number will always be odd.

It is rational because decimal expansion is terminating. Therefore, it can be expressed in p/q form where factors of q are of the form. and n and m are non-negative integers.

If the denominator of a rational number is in the form of., where m and n are non-negative integers, then the rational number has terminating decimal expansion.

The given number is 23 / 8

We know a rational number is expressed in simplest form p/q and if q can be expressed as ×, then it is a terminating decimal. Clearly, 2 and 5 are not the factors of 23.
8 can be expressed in terms of its primes as 2 × 2 × 2

Or, 8 =., where m = 3 and n = 0

Since q can be expressed in the form of., we can say 23/8 is a terminating decimal.

To get the denominator in powers of 10, multiply both numerator and denominator by 2⁴

‘s’ is called irrational if it cannot be written in the form of p/q, where p and q are integers and q ≠ 0

Non-terminating and non-repeating decimals are irrational.

If the denominator of a rational number is in the form of  where n and m are non-negative integers, then the rational number is terminating.

0.2 = 2 / 10 = 2 / (2 x 5)

∴ 0.2 has terminating decimal and is a rational number.

3.1415926535… is non-terminating and non-repeating.

∴ 3.1415926535… is irrational.

1 / 0.2 = 10 / 2

∴ 1 / 0.2 has terminating decimal and is a rational number.


 

(0.2)² has terminating decimal and is a rational number.

If the number is divisible by another number, it will be divisible by its factors too.
The factors of 80 = 2 × 5 × 8
For the number to be divisible by both 2 and 5, the last digit should be 0
So, y = 0
So we can rewrite the number as 653 × 0
Now For the number to be divisible by 8; last 3 digits should be divisible by 8
So 3×0 should be divisible by 8
So, x can be either 2 or 6 since 320 and 360 are divisible by 8
If x = 2; then the number becomes 65320 which is not divisible by 80
If x = 6; then the number becomes 65360 which is divisible by 80
Hence the value of x = 6
So, x + y = 6 + 0 = 6
 

For a number to be divisible by 8, the last three digits must be divisible by 8.

Here, 6y2 must be divisible by 8.

For y = 3, 632 is divisible by 8.

We know that the square of a natural number never ends in 2, 3, 7 and 8.

∴ 42437 can’t be the square of a natural number.

If p = d × q + r, (p > q) where p, q, d, r are integers and for a given (p, d), there exist a unique (q, r), then HCF (p, d) = HCF (d, r). Because this relation holds true, the Euclid’s Division Algorithm exists in a step by step manner. So, to find the HCF (1008, 20), we use Euclid’s division lemma at every step.
 

Step 1: 1008 = 20 × 50 + 8 ⇒ HCF(1008, 20) = HCF(20, 8) ⇒ a could be 8

Step 2: 20 = 8 × 2 + 4 ⇒ HCF(20, 8) = HCF(8, 4) ⇒ b could be 4

Step 3: 8 = 4 × 2 + 0

HCF = 4

Since 1008 = 20 × q + a where q and a are positive integers satisfy Euclid’s Division Lemma, we must have 0 ≤ a < 20. So a is surely 8 and b is 4.

There exist integers m and n such that a = 7m + p and b = 6n + q such that

0 < p < 7

⇒ Maximum value of p will be 6.

0 < q < 6 

⇒ Maximum value of p will be 5.

Therefore, the maximum value of p + q will be 11.