Johnny has apples bananas and carrots He has three times as

Johnny has apples, bananas, and carrots. He has three times as many fruits (apples and bananas combined) as vegetables (carrots). To even it up. he throws away 5 bananas and still has 2 more banana than apples. Then a monkey eats half of the apples and an additional 4 bananas. After these events he is left with 22 pieces of food. How many of each food did he have originally?

Solution

Let us presume that Johnny had x apples, y bananas and z carrots to begin with. Since the apples and bananas together are 3 times the carrots, we have x + y = 3z ...(1)

After throwing 5 banana, he has y - 5 bananas left with him. Since, the bananas are still 2 more than apples, we have y - 5 = x + 2... or, y = x + 7...(2)

Now, a monkey eats x/2 apples and 4 bananas so that the number of apples and bananas left with Jhonny are x - x/2 = x/2 and y - 5 - 4 = y - 9. Now, since Jhonny is left with 22 apples, bananas and carrots together, we have x/2 + y - 9 + z = 22... or, x + 2y - 18 + 2z = 44 ( on multiplying both the sides by 2) or, x + 2y + 2z = 44 +18 = 62 ...(3)

On substituting the value of y ( = x + 7) from the 2nd equation in the 1st equation, we get x + x + 7 = 3z or, 2x + 7 = 3z or,   2x - 3z = - 7 ...(4) Similarly , on substituting y = x + 7 in the 3rd equation, we get x + 2 ( x + 7) + 2z = 62 or, 3x + 14 + 2z = 62 or, 3x + 2z = 62 -14 = 48...(5) Now, on multiplying the 4th equation by 3 and the 5th eqyation by 2, we get 6x - 9z = - 21...(6) and 6x + 4z = 96...(7) Then, on subtracting the 6th equation from the 7th equation, we get 6x + 4z - 6x + 9z = 96 + 21 0r, 13z = 117. Therefore z = 117/13 = 9 Then from the 4th equation, we have 2x - 3*9 = -7 or 2x -27 = -7 or 2x = 27- 7 = 20. Therefore, x = 10 . Now, y = x + 7 = 10 + 7 = 17. Thus Jhonny had 10 apples, 17 bananas and 9 carrots originally.