Question

Mr. Mehra sends his servant to the market to buy oranges worth Rs.15. The servant having eaten 3 oranges on the way, Mr. Mehra pays 25 paise per orange more than the market price. Find the no.of oranges which Mr. Mehra receives.
A) 9
B) 11
C) 10
D) 12

Answer 1

Hint: First find the total price given by Mr. Mehra. Then solve further, a quadratic equation will be formed. Now solve the equation by either factoring or by quadratic formula $\left( {x = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}} \right)$ to get the actual number of oranges received by Mr. Mehra.

Formula used: $x = $ $\dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}$

Complete step-by-step answer:
Given: -
Number of oranges eaten by servent = 3
The extra cost paid by Mr.Mehra to the servant= 25 paise
That is Rs. $\dfrac{{25}}{{100}}$ ( 1rupee=100 paise)
Let the number of oranges received by Mr. Mehra be n.
The cost per orange is Rs.$\dfrac{{15}}{n}$
The number of oranges bought by servant should be $n + 3$.
The actual cost of oranges should be $\dfrac{{15}}{{n + 3}}$.
So, the total price paid by Mr. Mehra is,
$\dfrac{{15}}{{n + 3}}$+$\dfrac{{25}}{{100}}$=$\dfrac{{15}}{n}$
Move the constant part to the right side and the variable to the left side of the equation,
$\dfrac{{15}}{{n + 3}}$$ - $$\dfrac{{15}}{n}$= $ - $ $\dfrac{{25}}{{100}}$
Take L.C.M. on the left side and cancel out the common factor from the right side,
$\dfrac{{15n - 15(n + 3)}}{{n(n + 3)}}$ = $ - \dfrac{1}{4}$
Open the brackets and multiply with 15,
$\dfrac{{15n - 15n - 45}}{{n(n + 3)}} = - \dfrac{1}{4}$
Add or subtract the like terms,
$\dfrac{{ - 45}}{{n(n + 3)}} = - \dfrac{1}{4}$
Cancel out the negative side from both sides of the equation,
$\dfrac{{45}}{{n(n + 3)}} = \dfrac{1}{4}$
Cross multiply the equation,
$45 \times 4 = n(n + 3)$
Open the bracket and multiply the terms both sides,
$180 = {n^2} + 3n$
Move all the terms to one side to factorize the equation,
${n^2} + 3n - 180 = 0$
Factorization can be done in 2 ways.
Method 1:
$x = $ $\dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}$
Put a= 1, b= 3 and c= -180,
$n = \dfrac{{ - 3 \pm \sqrt {{3^2} - 4 \times 1 \times ( - 180)} }}{{2 \times 1}}$
Multiply and square the terms and add them in the square root,
$n = \dfrac{{ - 3 \pm \sqrt {729} }}{2}$
Then,
$n = \dfrac{{ - 3 \pm 27}}{2}$
Then,
$n = - 15,12$
Since quantity cannot be negative. So,
$n = 12$
Hence, the number of oranges is 12.
Method 2:
${n^2} + 3n - 180 = 0$
We can write 3 as (15-12)
${n^2} + (15 - 12)n - 180 = 0$
Open the brackets and multiply the terms,
${n^2} + 15n - 12n - 180 = 0$
Take common factors from the equation,
$n(n + 15) - 12(n + 15) = 0$
Take common factors from the equation,
$(n - 12)(n + 15) = 0$
Then,
$n = - 15,12$
Since quantity cannot be negative. So,
$n = 12$

Hence, the option (D) is the correct answer.

Note: A quadratic equation is a polynomial equation of degree 2. A quadratic equation has two solutions. Either two distinct real solutions, one double real solution or two imaginary solutions.
There are several methods you can use to solve a quadratic equation:
Factoring
Completing the Square
Quadratic Formula
Graphing
All methods start with setting the equation equal to zero.