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11. If 19 and 1140 are the respective HCF and LCM of two numbers, which are greater than 19 then what will be the possible number of such pair?
Answer: Option C
Explanation:Product of HCF and LCM = product of the numbers
Then, product of the numbers = 19 x 1140
Let 19a and 19b be the numbers.
19a x 19b = 19 x 1140
ab = 19 x 1140 / 19 x 19 = 60
If ab = 60 then (a,b) = (1,60), (2,30), (3,20), (4,15), (5,12) and (6,10).
Since a and b are co-primes then (a,b) = (1,60), (4,15) and (5,12)
Hence the number of such pairs = 3a
Workspace
Tags: IBM
12. The ratio of two numbers is 3:4 and their HCF is 4.Their LCM is:
Answer: Option D
Explanation:Given Two No, ratio = 3:4 and their HCF = 4
So, No. = 3*4 =12 and 4*4=16
LCM of 12,16 = 48
Workspace
Tags: Accenture
13. Let 13 and 273 are the HCF and LCM of two numbers respectively, and if one of them is less than 140 and greater than 60 then what will be that number?
Answer: Option B
Explanation:Let two numbers be ah and bh.
As h is 13, we get the numbers as 13a, 13b.
LCM = 13ab.
So 13ab = 273
=> ab = 21.
So a = 7 or 3.
One of this number is 39 or 91. Given that the number is greater than 60, we take 91 as the required number.
Workspace
Tags: Accenture
14. find the numbers between 100 to 400 which is divisible by either 2,3,5,7..
15. A teacher can divide her class into groups into groups of 5,13 and 17. What is the smallest possible strength of the class?
Answer: Option C
Explanation:For smallest possible class strength, we consider LCM of the given numbers. So LCM (5, 13, 17) = 1105.
Workspace
Tags: Accenture
16. If an integer k is divisible by 2,5 and 13, what is the next largest number that is divisible by all the three given numbers ?
17. What is the least amount that a person can have, such that when he distributes it into groups of Rs.16 or Rs. 18 or Rs. 20 or Rs. 25, he is always left with Rs. 4?
Answer: Option D
Explanation:Find the lcm of 16 18 20 and 25 i.e 3600.
so the amount is (3600 + 4) = Rs. 3604
Workspace
Tags: Cognizant
18. HCF of 2472,1284 and a third number 'n'is 12.If their LCM is 8*9*5*103*107.then the number 'n'is..
Answer: Option A
Explanation:2472 = 2^3×3×103
1284 = 2^2×3×107
HCF = 2^2×3
LCM = 2^3×3^2×5×103×107
HCF of the numbers is the highest number which divides all the numbers. So N should be a multiple of 2^2×3
LCM is the largest number that is divided by the given numbers. As LCM contains 32×5 these two are from N.
So N = 2^2×3^2×5^1
Workspace
Tags: TCS
19. In a circular racetrack of length 100 m, three persons A, B and C start together. A and B start in the same direction at speeds of 10 m/s and 8 m/s respectively. While C runs in the opposite at 15 m/s. When will all the three meet for the first time after the start?
Answer: Option A
Explanation:Since the track is a circular track A and B will meet every 50 seconds. i.e. 100 / (10-8).
Since it is a multiple of 50 they will be meeting at the starting point every 50 Seconds.
If you multiply 15 x 50 you will get 750 and after the second 50 it will be 1500.
All of them will meet at the starting point after 100
Workspace
Tags: TCS
20. The least number which when divided by 5, 6 , 7 and 8 leaves a remainder 3, but when divided by 9 leaves no remainder, is:
Answer: Option B
Explanation:L.C.M. of 5, 6, 7, 8 = 840.
Required number is of the form 840k + 3
Least value of k for which (840k + 3) is divisible by 9 is k = 2.
Required number = (840 * 2 + 3) = 1683.
Workspace
Tags: Syntel Inc.