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.
LCM of (80, 85, 90) can be found by prime factorizing them.
80 ? 2 × 2 × 2 × 2 × 5
85 ? 17 × 5
90 ? 2 × 3 × 3 × 5
L.C.M of (80,85,90) = 2 × 2 x 2 × 2 × 3 × 3 × 5 × 17
= 16 x 9 x 85
= 144 x 85
= 12240
L.C.M of (80,85,90) = 12240.
To get the least number of coconuts :
LCM = 30 => 30 + 1 = 31
60 is not a multiple of 8
HCF of 210 and 55 is 5
Now, 210x5 + 55P = 5
=> 1050 + 55P = 5
=> 55P = -1045
=> P = -1045/55
=> P = -19.
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 =
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