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August 1, 2023
This page is all about Tips, Tricks and shortcuts of HCF Questions. You will also get to know about its definition and methods to solve it. Suppose there are two integers a and b. There is another Suppose there are two integers a and b. There is another integer c which divides each of the integers i.e. a and b. Then, integer c is the Highest Common Factor.
One of the simplest methods to calculate the HCF of a number. The following steps help to calculate HCF of a number.a) Expand the given number as the product of their prime factors.
b) Check for common prime
Example :
Find out HCF of 3/5, 9/15, 18/25
HCF = HCF of Numerators / LCM of Denominators
HCF of 3, 9 and 18 = 3LCM of 5, 15 and 25 = 75
HCF of fraction = 3/75
Step 1 : Make the same number of decimal places in all the given numbers by suffixing zero(s) in required numbers as needed.
Step 2 : Now find the LCM/HCF of these numbers without decimal.
Step 3 : Put the decimal point in the result obtained in step 2 leaving as many digits on its right as there are in each of the numbers
Question 1 .The H.C.F. of two numbers is 40 and the other two factors of their L.C.M. are 15 and 20. Find the larger number?
Options
A. 600
B. 460
C. 800
D. 1400
Solution: The numbers are 40 * 15 and 40 * 20
40 * 15 = 600
40 * 20 = 800
Correct option: C
Question 2. Find the HCF of \frac{3}{10}, \frac{4}{7}, \frac{6}{5}, and \frac{2}{9}
A. \frac{1}{630}
B. \frac{3}{630}
C. \frac{2}{543}
D. 630
Solution We know that
HCF = HCF of Numerator/LCM of Denominators
HCF = \frac{HCF(3,4,6,2)}{LCM(10,7,6,9)}
HCF = \frac{1}{630}
Correct Option : A
Question 3 : Find the HCF of .63, 1.05, 2.1
Options:
A. .32
B. .21
C. .12
D. .65
Solution : the numbers can be written as .63, 1.05, 2.10
Now find the HCF of these number without decimal.
HCF of 63, 105 and 210 = 21
we need to put decimal point in the result obtained in step 2 leaving two digits on its right.
HCF (.63, 1.05, 2.1) = .21
Question 4 : For any integer n, what is HCF (22n + 7, 33n + 10) equal to?
A. n
B. 1
C. 11
D. None of these
Solution :
HCF of (22n + 7, 33n + 10) is always 1.On placing different values in the given equation; n = 1, 2, 3, ……
We get, For n = 1, HCF (22n + 7, 33n + 10) = (29, 43) ⇒ HCF = 1 For n = 2, HCF (22n + 7, 33n + 10) = (51, 76) ⇒ HCF = 1 For n = 3, HCF (22n + 7, 33n + 10) = (73, 109) ⇒ HCF = 1.
Question 5 : The greatest number which can divide 1356, 1868 and 2764 leaving the same remainder 12 in each case is
A. 64
B. 124
C.128
D. 132
Required number = HCF of (1356 – 12) , (1868 – 12 ), (2764 – 12) HCF of 1344, 1856 and 2752 = 64.
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