Aptitude Discussion

Q. |
A man purchased 40 fruits; apples and oranges for Rs 17. Had he purchased as many as oranges as apples and as many apples as oranges, he would have paid Rs 15. Find the cost of one pair of an apple and an orange. |

✖ A. |
70 paise |

✖ B. |
60 paise |

✔ C. |
80 paise |

✖ D. |
1 rupee |

**Solution:**

Option(**C**) is correct

Man buys $x$ apples at $m$ price and $y$ oranges at $n$ price,then:

\(x+y=40\)-------- (i)

\(mx+ny=17\)-------- (iI)

On solving both the equations we get:

\((m+n)(x+y)=17+15\)

\(m+n=\dfrac{32}{40}\)

= Rs 0.80

= **80 paise**

**JonSnow**

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Actually I got it. (m+n)(x+y) = mx+my+nx+ny which is addition of

n apples & m oranges as well as m apples & n oranges

Got it :)

**Gawthem**

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Sorry I misunderstood the question

**Gawthem**

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When he buy equal no it costs rs15 i.e

40/2 = 20

20m+20n=15

m+n=15/20 = 75paise

Isn't that right?

**Kaushal Chutiya**

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please someone explain this.

Can someone please tell me the solving of equations.

How did 17+15 came??