Total Cards = 52. Numbered Cards = (2, 3, 4, 5, 6, 7, 8, 9, 10) 9 from each suit 4 × 9 = 36
P (E) = 36/52 = 9/13

Or

If you count an ace as a number card there are 40 in a pack so p(get a number card) = 40/52 = 10/13.

If aces don’t count then there are 9 x 4 = 36 in a pack so p(get a number card) = 36/52 = 9/13.

If we want to pack the drinks in the least number of cans possible, then each can should contain the maximum numbers of liters possible.As each can contains the same number liters of a drink, the number of liters in each can is a comman factor for 80,144 and 368; and it is also the highest such factor, as we need to store the maximum number of liters in each can.

So, the number of liters in each can  = HCF of 80,144 and 368 = 16 liters.

Now, number of cans of Maaza = 80/16 = 5

Number of cans of Pepsi = 144/16 = 9

Number of cans of Sprite = 368/16 = 23

Thus, the total number of cans required = 5 + 9 + 23 = 37

Let us calculate both the length and width of the room in centimeters.

Length = 6 meters and 24 centimeters = 624 cm

width = 4 meters and 32 centimeters = 432 cm

As we want the least number of square tiles required, it means the length of each square tile should be as large as possible.Further,the length of each square tile should be a factor of both the length and width of the room.

Hence, the length of each square tile will be equal to the HCF of the length and width of the room = HCF of 624 and 432 = 48

Thus, the number of square tiles required = (624 x 432 ) / (48 x 48) = 13 x 9 = 117

Required Number = H.C.F  of  (91- 43), (183- 91) and (183-43)

= H.C.F of 48, 92, and 140 = 4

Required Length = H.C.F of 700 cm, 385 cm and 1295 c

                             = 35 cm.

60 is not a multiple of 8

Required number of students = H.C.F  of 1001 and 910  = 91

HCF = 11
LCM=7700
ONE NUMBER = 275
We are that,
One number× Other number = HCF×LCM
275× OTHER NUMBER = 11×7700
OTHER NUMBER = 11×7700/275=308

Required Number = (L.C.M  of 12, 16, 18,21,28)+7

= 1008 + 7

= 1015

H. C. F of two prime numbers is 1.  Product of numbers = 1 x 161 = 161.

Let the numbers be a and b . Then , ab= 161.

Now, co-primes with  product 161 are (1, 161) and (7, 23).

Since x and y are prime numbers and x >y , we have x=23 and y=7.

Therefore,  3y-x = (3 x 7)-23 = -2

Let the numbers be  3x and 4x . Then their H.C.F = x. So, x=4

Therefore, The numbers are 12 and 16

L.C.M of 12 and 16 = 48

L.C.M. of 6, 9, 15 and 18 is 90.

Let required number be 90k + 4, which is multiple of 7.

Least value of k for which (90k + 4) is divisible by 7 is k = 4.

=>Required number = (90 x 4) + 4   = 364.

H.C.F of co-prime numbers is 1. So, L.C.M = 117/1 =117

2X*3X = 48

6X = 48

X = 48/6

X = 8

so the numbers are 16 and 24 = 16+24=40