Question

54. A person carries 500 and wants to
buy apples and oranges out of it. If the
cost of one apple is 5 and the cost of
one orange is *7, then what is the
number of ways in which a person can
buy both apples and oranges using
total amount ?
(a) 10
ob 14
(c) 15
how to solve​

Answer 1

Answer:

i think the Question is wrong

Step-by-step explanation:

Answer 2

Answer:

Step-by-step explanation:

Lets Suppose he bought 50 oranges that will cost = (7x50) = $ 350 and the remaining money will be = (500-350) = $ 150 , in which he can buy 30 Apples (Since Cost of one Apple is $ 5).

If he buys one orange it will cost $ 7 and remaining money will be = (500-7)= $493. If he buy apples with The remaining money $ 493, the number of apples is going to be in decimal number, Since he can't buy half of apple or quarter of apple and so on. Therefor the cost of Oranges must be divisible by 5 or multiple of 5.

In the multiplication of 7, the number ending (Unit Digit) with 5 or 0(Zero) will be divisible by 5.(Example. 7x5 = 35, 7x10 = 70, 7x15 = 105 etc. will be divisible by 5). That is the number of multiplication of 7 which are divisible by 5 is going to be the number of possibilities or ways that he can buy both apples and oranges.

Since he carries $ 500, the limit is 500. therefor the nearest of 500 in 7 multiplication is 490, which is 7x70(Since 7x75 = 525 which exceeded 500)

Therefor the possibly ways are :

7x5=35 => 5 oranges and 93 Apples, (1st Possibility) or number of 5 = 1

7x10=70 => 10 oranges and 86 Apples,(2nd Possibility)or numbers of 5's = 2

7x20,.. until  7x70=490 => 70 Oranges and 2 Apples, (14th Possibility) or the number of 5's in 70 = (70/5) = 14.

That is 14 possible ways are there.