Since their HCFs are 7, numbers are divisible by 7 and are of the form 7x and 7y
Difference = 14
=> 7x - 7y = 14
=> x - y = 2
product of numbers = product of their hcf and lcm
=> 7x * 7y = 441 * 7
=> x * y = 63
Now, we have
x * y = 63 , x - y = 2
=> x = 9 , y = 7
The numbers are 7x and 7y
=> 63 and 49
HCF of 210 and 55 is 5
Now, 210x5 + 55P = 5
=> 1050 + 55P = 5
=> 55P = -1045
=> P = -1045/55
=> P = -19.
Required number of students = H.C.F of 1001 and 910 = 91
Let the numbers be a and b . Then, a+b =55 and ab = 5 x 120 = 600.
Therefore, Required sum =
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 the numbers be x and 4x. Then,
Hence Larger Number = 4x = 84
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
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