LCM and HCF Test - 2

Let the three men be A, B and C. A takes steps in multiples of 68; B takes steps in multiples of 51; and C takes steps in multiples of 85. They start in sync. They will again be in step when they are at a distance which is an integral multiple of 68, 51 and 85, which is the LCM, i.e. 2 x 2 x 3 x 5 x 17 = 1020 cm = 10.2 m.

LCM of 9, 12, 16 and 18 can be calculated as:
9 = 3 × 3
12 = 2 × 3 × 2
16 = 2 × 2 × 2 × 2
18 = 2 × 3 × 3

LCM = 3 × 3 × 2 × 2 × 2 × 2 = 144 (Taking the highest power out of the factorisation of all the numbers)

So, the smallest weight that can be divided into the exact number of parcels weighing 9 kg, 12 kg, 16 kg and 18 kg is 144 kg.
OR
Among options, only 144 is exactly divisible by 9, 12, 16 and 18.

Let b the number be n
As the HF is 13, so n = 13 a
Now, 455 = 5 x 7 x 13
So, n = 13 , or 5 x 13 or 7 x 13 or 5 x 7 x 13
ie. n = 13, or 65, or 91 or 455
As n lies between 75 and 125, n = 91

HCF of 408 and 312 = 24
The number of boys or girls that can be placed in a section = 24.

Total sections formed = 408/24 + 312/24 = 30

Numbers without remainder are (1625 – 8), (1281 – 4) and (4218 – 5)= (1617), (1277), (4213). These are exactly divisible by 11. Hence, 11 is the required answer.

Largest number of boys and girls that can be placed in a group = HCF of 408 and 312 = 24.

LCM = a + b; HCF = k (a - b);
Let the number to be found out be n.
Now, product of numbers = HCF × LCM
So, nk = k(a + b)(a - b) = k(a² - b²)
Or n = (a² - b²)

LCM = 120 seconds = 2 minutes
So, in 30 minutes, they toll together = 30/2 = 15 times
The bells also toll in the starting that is at 0 minute.
Therefore, they will toll together 16 times.

21 = 3 × 7 ; 36 = 2² × 3² 66 = 2 × 3 × 11
LCM of 21, 36, 66 = 2 × 2 × 3 × 3 × 7 × 11 = 2772

Multiples of 2772 less than 10000 are : 2772, 5544, 8316.

Greatest possible number of fruits in a heap = H.C.F. of 1323, 2457 and 3024 = 189