Question

2. A painter paid R400 for paint at Rx per litre. Two days later, she returned to the
shop and found that the price had been increased by R10 per litre.
She again paid R400, but this time left the shop with 2 litres less paint than on
the previous occasion. Determine the original price per litre and the number of
litres she bought on the first occasion.

​

Answer 1

Answer:

Step-by-step explanation:

Compare 2x² - √5x + 1 = 0 with

ax²+bx+c=0 , we get

a = 2 , b = -√5 , c = 1

Discreminant ( D ) = b² - 4ac

= ( -√5 )² - 4 × 2 × 1

= 5 - 8

= -3

D < 0

No real roots .

Therefore ,

Quadratic equation has no real roots

Answer 2

Let :

On first occasion she bought y litre of paint

Given :

On first occasion

• Rate of paint( original price) = Rs x per litre
• Number of litre of paint bought = y
• Amount paid = Rs 400

On second occasion

• Rate of paint = Rs 10 more than original price
• Number of litre of paint bought = 2 less than original
• Amount paid = Rs 400

To find :

• Original price
• Number of litre she bought on first occasion

Solution :

On first occasion

Amount = Number of litres bought × Rate

➝ Rs 400 = y litre × Rs x/litre

➝ 400 = xy

➝ x = (400/y) equation 1

________________________________

On second occasion

• Rate = Rs ( x+10 )/litre
• Number of litre of paint bought = ( y-2 )

➝ Amount = (y-2) × (x+10)

On putting value of x from equation 1 into above equation, we get ;

$\sf \implies \: 400 \: = \: (y - 2) \times ( \dfrac{400}{y} + 10)$

$\sf \implies \: 400 \: = \: (y - 2) \times \bigg( \dfrac{(400 \times 1) + (10 \times y)}{y} \bigg)$

$\sf \implies \: 400 \: = \: (y - 2) \times \bigg( \dfrac{400 + 10y}{y} \bigg)$

$\sf \implies \: 400y \: = \: (y - 2) \times (400 + 10y)$

➝ 400y = 400y + 10y² - 800 - 20y

➝ 10y² - 20y + 400y - 400y - 800 = 0

➝ 10y² - 20y - 800 = 0

➝ 10(y² - 2y - 80) = 0

➝ y² - 2y - 80 = 0

Using quadratic formula

$\sf \implies \: y \: = \: \dfrac{ - ( - 2) \: \pm \: \sqrt{( { - 2)}^{2} - 4(1)( - 80)} }{2(1)}$

$\sf \implies \: y \: = \: \dfrac{ 2\: \pm \: \sqrt{4 + 320} }{2}$

$\sf \implies \: y \: = \: \dfrac{ 2\: \pm \: \sqrt{ 324} }{2}$

$\sf \implies \: y \: = \: \dfrac{ 2\: \pm \: 18 }{2}$

$\sf \implies \: y \: = \: \dfrac{ 2\: + 18}{2} \: \: \: \bold{or} \: \: \: y \: = \: \dfrac{ 2\: - 18}{2}$

$\sf \implies \: y \: = \: \dfrac{ 20}{2} \: \: \: \bold{or} \: \: \: y \: = \: \dfrac{ - 16}{2}$

$\sf \implies \: y \: = \: 10 \: \: \: \: \: \: \bold{or} \: \: \: \: \: \: y \: = \: \: - 8$

As y denote number of litre of paint, it can't be negative ,

therefore

➝ y = 10

________________________________

On putting value of y in equation 1 we get,

➝ x = 400/10

➝ x = 40

________________________________

ANSWER :

• Original price ( x ) = Rs 40 per litre
• Number of litre of paint bought on first occasion = 10