If the cost of bananas is increased by ` 10 per dozen, one can get 3 dozen less for ` 600. Find the
original cost of one dozen of banana
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40 rupee is cost for 1 dozen banana it is original price....
40*15 dozens =600
after 10 rupees increased per bannana
50*12 dozens=600
40*15 dozens =600
after 10 rupees increased per bannana
50*12 dozens=600
Answered by
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Heya
Let the actual cost on one dozen of bananas be x
Cost after increasing the price is x+10
Let the number of dozens bought(with the actual price) for 600 be y
After the hike in price, the number of dozens bought is y-3
(y-3)(x+10) = 600
we also know that before the increase in prices yx= 600
y=600/x
Substituting this in the first equation, we get
(600/x - 3)(x+10)= 600
(x+10)(600/x-3) = 600
(x+10)(600–3x)/x = 600
(10+x)(600 -3x) = 600x
6000 + 570x - 3x^2 = 600x
6000 - 30x - 3x^2 = 0
2000 - 10x - x^2 = 0
x^2 + 10x - 2000 = 0
(x+50)(x-40) = 0
We then get the values of x as -50 and 40
As the price cannot be a negative value, the price of one dozen of bananas is Rs 40
Let the actual cost on one dozen of bananas be x
Cost after increasing the price is x+10
Let the number of dozens bought(with the actual price) for 600 be y
After the hike in price, the number of dozens bought is y-3
(y-3)(x+10) = 600
we also know that before the increase in prices yx= 600
y=600/x
Substituting this in the first equation, we get
(600/x - 3)(x+10)= 600
(x+10)(600/x-3) = 600
(x+10)(600–3x)/x = 600
(10+x)(600 -3x) = 600x
6000 + 570x - 3x^2 = 600x
6000 - 30x - 3x^2 = 0
2000 - 10x - x^2 = 0
x^2 + 10x - 2000 = 0
(x+50)(x-40) = 0
We then get the values of x as -50 and 40
As the price cannot be a negative value, the price of one dozen of bananas is Rs 40
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