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
The number of liters in each can = HCF of 80, 144 and 368 = 16 liters.
Number of cans of Maaza = 368/16 = 23
Number of cans of Pepsi = 80/16 = 5
Number of cans of Sprite = 144/16 = 9
The total number of cans required = 23 + 5 + 9 = 37 cans.
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
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
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