The question paper will be designed to test the candidates ability to complete sequences making logical conclusion based on simple patter of numbers, statements, figures, letters etc as may be expected of a rational thinking person without any special study of the subject.
Territorial Army 2 2017 Exam Mock Test Series
5 Practice MCQs On Topic HCF LCM For TA 2 2017 Maths Exam
Tips on HCF LCM
The H.C.F. of two or more than two numbers is the greatest number that divided each of them exactly.
There are two methods of finding the H.C.F. of a given set of numbers:
- Factorization Method
- Division method
Factorization Method: Express the each one of the given numbers as the product of prime factors. The product of least powers of common prime factors gives H.C.F.
Division Method: Suppose we have to find the H.C.F. of two given numbers, divide the larger by the smaller one. Now, divide the divisor by the remainder. Repeat the process of dividing the preceding number by the remainder last obtained till zero is obtained as remainder. The last divisor is required H.C.F.
Least Common Multiple (L.C.M.)
The least number which is exactly divisible by each one of the given numbers is called their L.C.M.
There are two methods of finding the L.C.M. of a given set of numbers:
- Factorization Method,
- Division Method (Division Method is short cut method of LCM)
Factorization Method: Resolve each one of the given numbers into a product of prime factors. Then, L.C.M. is the product of highest powers of all the factors.
Division Method : Arrange the given numbers in a row in any order. Divide by a number which divided exactly at least two of the given numbers and carry forward the numbers which are not divisible. Repeat the above process till no two of the numbers are divisible by the same number except 1. The product of the divisors and the undivided numbers is the required L.C.M. of the given numbers
5 Practice MCQs On Topic HCF LCM
| Three number are in the ratio of 3 : 4 : 5 and their L.C.M. is 2400. Their H.C.F. is: | ||||||||
Let the numbers be 3x, 4x and 5x. Then, their L.C.M. = 60x. So, 60x = 2400 or x = 40. The numbers are (3 x 40), (4 x 40) and (5 x 40). Hence, required H.C.F. = 40. |
| The least number which should be added to 2497 so that the sum is exactly divisible by 5, 6, 4 and 3 is: | ||||||||
L.C.M. of 5, 6, 4 and 3 = 60. On dividing 2497 by 60, the remainder is 37. Number to be added = (60 – 37) = 23. |
| The smallest number which when diminished by 7, is divisible 12, 16, 18, 21 and 28 is: | ||||||||
Required number = (L.C.M. of 12,16, 18, 21, 28) + 7 = 1008 + 7 = 1015 |
| Find the highest common factor of 36 and 84. | ||||||||
36 = 22Â x 32 84 = 22Â x 3 x 7 H.C.F. = 22Â x 3 = 12. |
| Which of the following has the most number of divisors? | ||||||||
99 = 1 x 3 x 3 x 11 101 = 1 x 101 176 = 1 x 2 x 2 x 2 x 2 x 11 182 = 1 x 2 x 7 x 13 So, divisors of 99 are 1, 3, 9, 11, 33, .99 Divisors of 101 are 1 and 101 Divisors of 176 are 1, 2, 4, 8, 11, 16, 22, 44, 88 and 176 Divisors of 182 are 1, 2, 7, 13, 14, 26, 91 and 182. Hence, 176 has the most number of divisors. |
