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# RD Chapter 1- Real Numbers Ex-1.4 Interview Questions Answers

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Question 1 :
Find the L.C.M. and H.C.F. of the following pairs of integers and verify that L.C.M. x H.C.F. = Product of the integers.
(i) 26 and 91
(ii) 510 and 92
(iii) 336 and 54

(i) 26 and 91
26 = 2 x 13
91 = 7 x 13
H.C.F. = 13
and L.C.M. = 2 x 7 x 13 = 182
Now, L.C.M. x H.C.F. = 182 x 13 = 2366
and 26 x 91 = 2366
L.C.M. x H.C.F. = Product of integers

Question 2 :
Find the L.C.M. and H.C.F. of the following integers by applying the prime factorisation method :
(i) 12, 15 and 21
(ii) 17, 23 and 29
(iii) 8, 9 and 25
(iv) 40, 36 and 126
(v) 84, 90 and 120
(vi) 24,15 and 36

Question 3 : Given that HCF (306, 657) = 9, Find LCM (306, 657). [NCERT]

Answer 3 : HCF of 306, 657 = 9

Question 4 : Can two numbers have 16 as their H.C.F. and 380 as their L.C.M. ? Give reason.

H.C.F. of two numbers = 16
and their L.C.M. = 380

We know the H.C.F. of two numbers is a factor of their L.C.M. but 16 is not a factor of 380 or 380 is not divisible by 16
It can not be possible.

Question 5 : The H.C.F. of two numbers is 145 and their L.C.M. is 2175. If one number is 725, find the other.

First number = 725
Let second number = x
Their H.C.F. = 145
and L.C.M. = 2175

Second number = 435

Question 6 : The H.C.F. of two numbers is 16 and their product is 3072. Find their L.C.M.

H.C.F. of two numbers = 16
and product of two numbers = 3072

Question 7 : The L.C.M. and H.C.F. of two numbers are 180 and 6 respectively. If one of the number is 30, find the other number.

First number = 30
Let x be the second number
Their L.C.M. = 180 and H.C.F. = 6
We know that
first number x second number = L.C.M. x H.C.F.
30 x x = 180 x 6
⇒ x =  = 36
Second number = 36

Question 8 : Find the smallest number which when increased by 17 is exactly divisible by both 520 and 468.

L.C.M. of 520 and 468

= 2 x 2 x 9 x 10 x 13 = 4680
The number which is increased = 17
Required number 4680 – 17 = 4663

Question 9 : Find the smallest number which leaves remainders 8 and 12 when divided by 28 and 32 respectively.

Dividing by 28 and 32, the remainders are 8 and 12 respectively
28 – 8 = 20
32 – 12 = 20
Common difference = 20
Now, L.C.M. of 28 and 32

= 2 x 2 x 7 x 8 = 224
Required smallest number = 224 – 20 = 204

Question 10 : What is the smallest number that, when divided by 35, 56 and 91 leaves remainders of 7 in each case ?

L.C.M. of 35, 56, 91

= 5 x 7 x 8 x 13 = 3640
Remainder in each case = 7
The required smallest number = 3640 + 7 = 3647

krishan