A Kaya has 5 more red peppers than green peppers In all she

A. Kaya has 5 more red peppers than green peppers. In all, she has 17 peppers. How many red peppers does Kaya have?

B. The sum of two numbers is 62. Their difference is 6. What are the two numbers?

C. The quotient of two numbers is 4. Their difference is 21. What are the two numbers?

Solution

A. Let the number of green peppers with Kaya be x. Then the number of red peppers with Kaya is x+5 so that x +x+5 = 17 or, 2x+5 = 17     or, 2x = 17-5 = 12 so that x = 12/2 = 6. Then x+5 = 6+5 = 11. Thus, Kaya has 11 red peppers.

B. Let the 2 numbers be x and y. Then x+y = 62 …(1) and x-y = 6…(2). On adding the 2 equations, we get x+y+x-y= 62+6 or, 2x = 68 so that x = 34. Now, on substituting x =34 in the 2^nd equation, we get 34-y = 6 or, y = 34-6 = 28. Thus, the required numbers are 34 and 28.

C. Let the 2 numbers be x and y. We assume that x and y are positive integers. Then x-y = 21…(1) and x = 4y +c…(1), where c is the remainder when we divide x by y. Now, on substituting x = 4y+c in the 1^st equation, we get 4y+c-3y = 21 or, 3y = 21-c. Now, if x is completely divisible by y,i.e. if there is no remainder when we divide x by y, then c = 0 so that 3y = 21 or y = 7. Then , from the 2^nd equation, we have x = 7*4 +0 = 28. Then, in this case, the 2 numbers are 28 and 7.

Note: If c 0, we will have a different set of numbers.