H.C.F and L.C.M- Aptitude Questions and Answers - in Hindi

The first wheel spins at 60 rpm which means that it completes 60/60 = 1 revolution every second.
The second wheel spins at 36 rpm. So it completes 36/60 = 0.6 of a revolution every second
The third wheel spins at 24 rpm. So it completes 24/60 = 0.4 of a revolution every second.

So we have the values: 1, 0.6, 0.4
If we wait 10 seconds, then we multiply each value by 10 to get: 10, 6, 4
Now we can divide by the GCF 2 to get 5, 3, 2

This means that it will take 5 seconds for the spots to line up again (since the first number corresponds to the amount of time that has gone by in seconds)

The least number which is exactly divisible by 7, 10, 15, 21 and 25 is their L.C.M, i.e.
L.C.M of 7, 10, 15, 21 and 25 =1050

The Largest number of five digits is 99999.
Required number must be divisible by L.C.M of 7, 10, 15, 21 and 25 i.e. 1050
On dividing 99999 by 1050 get 249 as remainder.
So required number is = (99999-249) = 99750.

H.C.F of numbers is 30 So required numbers should be multiple of 30.
lets the numbers are 30a and 30b ,there the product of the numbers = 5400
Then,
30a * 30b = 5400 <=> ab = 6

the co-primes or possible pairs of product 6 either (1, 6) and (2,3), or (6,1) and (3, 2)
that means
a*b=6
30a*30b = 5400

30*180 = (30,180) = 5400 (there (1, 6) a=1 , b=6)
60*90 = (60,90) = 5400 (there (2, 3) a=2 , b=3)

thus,two numbers of pairs are possible whose product is 5400 and H.C.F. is 30.

lets other number is Z
as per rule, Product of two numbers =Product of their H.C.F. and L.C.M
So Z*24=8*48
Z=384/24 => 16

Least common multiples (LCM) of Interval of bells toll:
= (36 to 40 to 48)/2
= (18 to 20 to 24)/2
= (9 to 10 to 12)/2
= (9 to 5 to 6)/3
= (3 to 5 to 2)

So,They will start ringing together again after: = (2 * 2 * 2 * 3 * 3 * 5 * 2) sec = 720 sec
Thus , next toll together after 720 sec i.e. 12 mins

HCF is the product of all common prime factors using the least power of each common prime factor
HCF = 23 x 32 x 11 = 792

The least number which is exactly divisible by is their LCM of (4, 12, 15, and 18)
=4,12,15,18|2
=2,6,15,9|3
=2,2,5,3|2
=1,1,5,3
LCM=2*3*2*1*5*3=180

Product of two numbers = Product of their HCF and LCM
77 x a = 11 x 693
Other number a = 11 x 693 / 77 =99.

The largest number is the H.C.F. of 210, 315, 147, and 168, which is 21.

Product of two numbers = Product of their HCF and LCM.

First number = 2 x 44 = 88
Second number = (LCM x HCF ) / First number
So, Second number = 44 x 264 / 88 = 132.

2² * 3² =>2*2*3*3
2 * 3⁴ * 7=>2*3*3*3*3*7

HCF is the product of all common prime factors using the least power of each common prime factor.
So we got 2*3*3*7 = 126

Let’s solve this question by factorization method.
54 =2 * 3³
288 = 2⁵ * 3²
360 = 2³ * 3² * 5
So HCF is the product of all common prime factors using the least power of each common prime factor in all three, i.e. 2 * 3² =18

The required number, which is exactly divisible by each one of the given numbers, should be divisible by LCM of (6, 9, 12, 17) i.e. 612
The Largest number of four digits is 9999.by dividing 9999 to 612 get 207 as remainder.
So, the greatest number of 4 digits divisible by any numbers from 6, 9, 12 or 17 = (9999-207) = 9792.

But we should get the remainder as 2,so we need to add 2 to 9792.So the number PQRS is 9794.

Thus, required value (P + R)/(Q + S) = (9+9)/(7+4)
=>18/11

L.C.M. of 6, 5, 7, 10 and 12 is 420.

So, the bells will toll together after every 420 seconds i.e. 7 minutes.

Now, 7 x 8 = 56 and 7 x 9 = 63.

Thus, in 1-hour (or 60 minutes), the bells will toll together 8 times, excluding the one at the start.

As given,
LCM + HCF = 84 ----eq1
LCM - HCF = 60 ----eq2

When add eq1 and 2 then
2*LCM = 144
LCM => 72
after implement it in eq1, then
72 + HCF = 84
HCF = 84-72 =>12

LCM × HCF = Ist number ×IInd number Or Product of numbers = HCF × LCM
LCM = 72, HCF = 12
One number x = 24 ,Let other number be y
So, x*y = LCM * HCF
24 * y = 72 * 12
y = (72 * 12)/24 = 36
other number will be 36

Suppose two numbers are 3x and 5x.
Then, 3x * 5x = HCF * LCM
15x² = 16 * 240
x² = 256
x = 16

First number= 66 * 2 = 132
132 * second number = HCF * LCM
132 * x = 66 * 384
x = 192