1 One number is 3 times as large as another The sum of their

1) One number is 3 times as large as another. The sum of their reciprocals is 40/3. Find the two numbers.

2) If 11/13 is added to twice the reciprocal of a number, the result is 1. Find the number.

3) The numerator of a certain fraction is 3 more than the denominator. If 7/5 is added to the fraction, the result is 3. Find the fraction.

4) The sum of the reciprocals of two consecutive integers is 7/12. Find the two integers.

Solution

1) Let the numbers be x and y. and let x = 3y. Then 1/x + 1/y = 40/3 or 1/x + 3/x = 40/3 ( x = 3y so y = x/3 and 1/y = 3/x). or, 4/x = 40/3 or, 1/x = 10/3 so that x = 3/10. Then y = x/3 = 1/10.Thus, the required numbers are 3/10 and 1/10. ( since 3/10 = 3*1/10 and since 10/3 + 10 = 40/3, the result is correct).

2) Let the number be x. Then 2/x + 11/13 = 1 or, (2*13 +11x) /13x = 1 or, 26 + 11x = 13x or, 2x = 26 so that x = 13.Thus, the required number is 13.( since 1/13 + 11/13 = 13/13 = 1, the result is correct)

3) Let the denominator of the required number be x. Then this number is (x + 3)/x. Also, (x + 3)/x. + 7/5 = 3, so that [5(x + 3) +7x] / 5x = 3 or, 12x + 15 = 15x or, 3x = 15. Therefore, x = 15/3 = 5. Tyhus, the required number is (x+3)/x = 8/5.( since 8/5 + 7/5 = 15/5 = 3, the result is correct)

4) Let the two consecutive integers be x and x+ 1. Then 1/x + 1/(x + 1) = 7/12 or, (x + 1 + x)/ x(x + 1) = 7/12 or, 12(2x + 1) = 7x(x + 1) or, 24x + 12 = 7x² + 7x or, 7x² - 17x -12 = 0. or, 7x² -21x + 4x - 12 = 0 or, 7x( x -3) +4( x -3) = 0 or, (x -3)(7x + 4)= 0. Therefore, either x = 3 or x = -4/7. Since x is an integer, we have x = 3. Thus, the required numbers are 3 and 4.( The sum of 1/3 and 1/4 is 7/12. Hence, the result is correct)