Following quiz provides Multiple Choice Questions (MCQs) related to H.C.F & L.C.M.. You will have to read all the given answers and click over the correct answer. If you are not sure about the answer then you can check the answer using Show Answer button. You can use Next Quiz button to check new set of questions in the quiz.
Making the same number of decimal places, the given numbers are 0.63, 1.05 and 2.10.
without decimal places, these numbers are 63, 105, and 210
Now, H.C.F of 63, 105 and 210 is 21 so H.C.F of 0.63, 1.05 and 2.1 is 0.21
L.C.M of 63, 105 and 210 is 630 so L.C.M of 0.63, 1.05 and 2.1 is 6.30
Q 2 - The greatest number of four digits which is exactly divisible by 15, 25, 40, and 75 is?
Greatest number of 4 digits is 9999. L.C.M of 15, 25, 400 and 75 is 600.
On dividing 9999 by 600, the remainder is 399.
Required number = (9999 - 399) = 9600.
Q 3 - The least number, which when divided by 48, 60, 72, 108 and 140 leaves 38, 50, 62, 98 and 130 as remainders respectively, is?
Q 4 - If the sum of two numbers is 55 and the H.C.F and L.C.M of these numbers are 5 and 120 respectively then the sum of the reciprocals of the numbers is equal to?
Product of numbers = 11 x 385.
let the numbers be 11a and 11b. then 11a x 11b = (4235) = ab = 35
Now co-primes with product 35 are (1, 35) (5, 7).
So the numbers are (11 x 1, 11 x 35) and (11 x 5, 11 x 7).
Since one number lies between 75 and 125 the suitable pair is (55, 77).
Required number = 77.
Let the numbers be X & Y
HCF*LCM=Product of two numbers =XY=12x72=864
XY=864 -------- (1)
Given X-Y=24 -------- (2)
On solving 1 & 2 we get X=12 Y=36
Their sum = 12+36=48
Q 8 - Having H.C.F and L.C.M of two numbers as 21 and 4641 respectively. If one of the numbers is in between 200 and 300, then the two numbers are :
Let the numbers be 21 a and 21 b, where a and b are co-primes.
Then ,21 a* 21 b= (21* 4641)⇒ab= 221.
Two co-primes with product 221 are 13 and 17.
∴ Numbers are (21*13, 21*17) , i.e (273,357)
Q 9 - The sum of two numbers is 528 and their H.C.F is 33. The numbers of such pairs possible are:
let the numbers be 33a and 33b , where a and b are co-primes.
Then, 33a + 33b = 528 ⇒ 33(a+b) = 528 ⇒ a+b = 16
∴ (a=1, b= 15), (a =3, b=13), (a=5, b=11), (a=7 , b=9)
Numbers are (33*1, 33*15 ), (33*3 ,33*13), (33*5 ,33 *11) , (33*7, 33*9)
Possible numbers of pairs = 4
Q 10 - Find the largest fraction among the following:
