Real Numbers most important questions

Real Numbers most important questions

sms (smart studies ) February 04, 2021

 Real Numbers

 




 

 

(A). Terminating / Non-Terminating(2^n 5^m type)  2009,2010, 2017

(B). Find HCF 2017, 2018

(C). HCF,LCM formula’s  2010, 2018, 2020.

 (D). Prove Irrational 2009, 2010, 2020

Q1. Write whether the rational number 7/75 will have a terminating decimal expansion or a non-terminating repeating decimal expansion. (CBSE 2017-1mark).

 

Que2: Has the rational 441/(2^2 X 5^5 X 7^2 )  a terminating or non terminating decimal expansion? (CBSE 2010-1mark)

 

Que3: Has the rational number 51/1500  a terminating or non terminating decimal expansion? ? (CBSE 2009-1mark).

 

[B]

 

Ques4: What is the HCF of the smallest prime number and the smallest composite number? (CBSE 2018 – 2mark).

 

Ques5: If two positive integers p and q are written  p = a^2 b^3 and q = = a^3 b ; a, b are prime numbers, then verify LCM (p, q) x HCF (p, q) = pq .( CBSE 2017 – 2mark).

 

[C]

Qus6: Find HCF and LCM of 404 and 96 and verify that HCF x LCM = product of the given  numbers. (CBSE 2018 – 3mark).

 

Ques7:Find HCF and LCM of 306 and 54 and verify that HCF x LCM = product of the two given numbers.(CBSE 2010 -3mark).

 

Ques8:The HCF and LCM of 12,21,15 respectively are?

Or

The HCF and LCM of 12,21,15 respectively are
 

(a)3,140 (b)12,420
(C)3,420 (D)420,3

 

Ques9: Prove (2√3)/5   that is an irrational number. ( CBSE 2010 – 3mark).

 

Ques10: Prove 7+3√2 that is not rational number. ( CBSE 2009 – 3mark).

 

Ques11: Prove 2-3√5 that is an irrational number. (CBSE 2010 – 3 mark)

 

Ques12: Identify the rational number : 5-√3, 5+√3, 4+√2, 5+√9 (CBSE 2010 – 1mark.

 

 

Q. Write whether the rational number 7/75 will have a terminating decimal expansion or a non-terminating repeating decimal expansion. (CBSE 2017-1mark)

Ans: The rational number 7/75

The denominator = 75 = 3 x 25 = 3 x 2^0 x 5^2

The denominator cannot be written in form (2^n 5^m type). So it is non-terminating decimal expansion.

 

Que: Has the rational 441/(2^2 X 5^5 "x" 7^2 )  a terminating or non terminating decimal expansion? (CBSE 2010-1mark)

Ans: we have

441/(2^2 X5^5 "x" 7^2 ) = 63/(2^2 X 5^5 "x" 7) = 9/2^2 X 5^5 

The condition for terminating decimal representation is that denominator should of form 2^n 5^m in its lowest form.

so since lowest form of this fraction is  9/((2^2 ) ̅"x" 5^5 )   so this will have a terminating decimal expansion.

 

Que: Has the rational number 51/1500  a terminating or non terminating decimal expansion? ? (CBSE 2009-1mark)

 Solution :

simplest form of the given fraction=> 51/1500  = 17/500

since denominator of the fraction has 2 and 5 only as its factors so it will have terminating decimal expansion.

the decimal expansion will be 0.034.

 

 

Ques: What is the HCF of the smallest prime number and the smallest composite number? (CBSE 2018 – 2mark)

Ans:

Smallest prime number is 2 and smallest composite number is 4.

Clearly, 2 is a factor of 4, so their H.C.F is 2

Thus the H.C.F of the smallest prime number and the smallest composite number is 2.

  

Ques: If two positive integers p and q are written  p = a^2 b^3 and q = = a^3 b ; a, b are prime numbers, then verify LCM (p, q) x HCF (p, q) = pq .( CBSE 2017 – 2mark)

Answer

p = a^2 b^3

q = = a^3 b

P = a  x a  x  b x b x b

Q= a  x a  x  a x b

H.C.F.(p,q)= a^2 b 

P = a  x a  x  b x b x b

Q= a  x a  x a x  b

L.C.M.(p,q)= a^3 b^3(highest power)

L.C.M.(p,q)×H.C.F.(p,q)= a^5 b^4

a^2 b x a^3 b^3 = a^(2+3) b^(1+3) = a^5 b^4

pq= a^5 b^4

Therefore, L.C.M.(p,q)×H.C.F.(p,q)=pq

 

Qus: Find HCF and LCM of 404 and 96 and verify that HCF x LCM = product of the given  numbers. (CBSE 2018 – 3mark)

Ans:

Expressing the number as a product of prime number

404 = 2 x 2 x 101

96 = 2 x 2 x 2 x 2 x 2 x 3

LCM of 404 and 96 is 2 x 2 x 2 x 2 x 2 x 101 x 3 = 9696

HCF of 404 and 96 is 2 x 2 = 4

Now HCF x LCM = 4 x 9696 = 38784

Product of two given number 404 x 96 = 38784

Hence HCM x LCM = product of two given number.

Find HCF and LCM of 306 and 54 and verify that HCF x LCM = product of the two given numbers.(CBSE 2010 -3mark).

Step-by-step explanation:

Given :

Two numbers 306 and 54

Prime factorization of numbers are as :

306 = 2 × 3 × 3 × 17

54 = 2 × 3 × 3 × 3

Now H.C.F = common factors :

H.C.F = 2 × 3 × 3 = > 18

L.C.M. = 18 × 3 × 17 = 918

Verification :

Products of No. = H.C.F × L.C.M

306 × 54 = 18 × 918

16,524 = 16,524

L.H.S. = R.H.S

Hence verified .

 

The HCF and LCM of 12,21,15 respectively are?

Or

The HCF and LCM of 12,21,15 respectively are
(a)3,140 (b)12,420
(C)3,420 (D)420,3

Step-by-step explanation:

given numbers :

12  , 21, 15

factors of

12 = 2×2×3

21 = 3×7

15=3×5

then HCF is = 3

LCM = 3×7×5×2×2

LCM = 420

hence  HCF and LCM of 12,21,15 are

option (c)

3,420

 

Prove (2√3)/5   that is an irrational number. ( CBSE 2010 – 3mark)

Sol: let it be rational number
therefore it can be written in form of a and b where a and b are co-prime numbers.
2√3=5a/b
5a/b is rational number as it is of the form p/q which is a rational number.
but we know that √3 is irrational number so our assumption is wrong.
(2√3)/5  is irrational.

 

Prove 7+3√2 that is not rational number. ( CBSE 2009 – 3mark)

Sol: Step-by-step explanation :

let 7 + 3√2 be an rational number where

7+3√2 = a/b [ a and b are coprime and b is not equal to zero]

3√2= a/b - 7

3√2 =  ((a-7b))/b

√2 = ((a-7b))/3b.....(i)


Now ,from equation (i) ,we get that √2 is rational but we know that √2 is irrational. so actually 7 + 3√2 is irrational not rational. thus our assumption is wrong. The number is irrational.

 

Prove 2-3√5 that is an irrational number. (CBSE 2010 – 3 mark)

Ans:

Assume 2-3√5 be a rational no
therefore 2-3√5= a/b
hence √5= (a-2b)/3b
therefore  (a-2b)/3b  is a rational no
therefore √5 is a rational no
but we know that √5 is a irrational no
therefore our assumption was wrong that 2-3√5 is a rational no
therefore 2-3√5 is a irrational no.

 

Identify the rational number : 5-√3, 5+√3, 4+√2, 5+√9 (CBSE 2010 – 1mark

Answer:

Step-by-step explanation: we know that,

If q is Rational and s is irrational then q+s , q-s, qs and q/s (s≠0) are irrational .

Here ,

i) q = 5, s = √3

5-√3 is irrational

ii) q = 5 , s = √3

5+√3 is irrational,

iii ) q = 4, s = √2,.

4+√2 is irrational

iv ) q = 5 (rational)

√9 = 3 (rational)

Therefore,.

5-√9 = 5-3 = 2 (Rational).

 

 

 

 

 


 




 

 

 

 

 

 

 

 

 

 

 

 

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