SOLUTION: One number is 3 times another. If the sum of their reciprocals is 4/21, find the two numbers.

Question 59953: One number is 3 times another. If the sum of their reciprocals is 4/21, find the two numbers.
Found 2 solutions by praseena, chitra:
Answer by praseena(37) (Show Source): You can put this solution on YOUR website!
suppose first number is x
Then the second number is given as 3 times the other
so second number can be taken as 3x

Now let us take the reciprocals of the numbers
Reciprocal of first number is 1/x
and reciprocal of the second number is 1/3x
Also it is given that sum of their reciprocals is 4/21
so we can write this condition as, 1/x + 1/3x = 4/21
Now multiply the whole expression by the lowest common multiple of the denominators, x,3x,21 which is 21x
21x*1/x + 21x * 1/3x = 21x * 4/21
21 + 7 = 4x
28 = 4x
Dvide both sides by 4
==>7=x
so solution is x=7
so the first number is 4 and the second number is 3*7=21
so the numbers are 7 and 21
Answer by chitra(359) (Show Source): You can put this solution on YOUR website!
Let "x" and "y" be two numbers.

Given that one is thrice the other.

Hence x = 3y.

Given that the sum of their reciprocals equals 4/21

Thus, the equation is:

1/(3y + y) = 4/21

1/4y = 4/21

16y = 21

y = 21/16

Therefore, x = 3y

x = 3(21/16)

x = 63/16

Thus the two numbers are x = 63/16 and y = 21/16

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