# WORD PROBLEMS ON QUADRATIC EQUATIONS WITH SOLUTIONS

Problem 1 :

A cottage industry produces a certain number of toys in a day. The cost of production of each toy (in rupees) was found to be 55 minus the number of toys produced in a day, the total cost of production was \$750. Find the number of toys produced on that day.

Solution :

Let 'x' be the number of toys produced.

Cost of production of one toy  =  55 - x

Total cost  =  Number of toys produced on that day x cost of one toy

Total cost  =  x (55 - x)

750  =  55 x - x2

x2 - 55x + 750  =  0

x2 - 30x - 25x + 750  =  0

x(x - 30) - 25(x - 30)  =  0

(x - 30)(x - 25)  =  0

x - 30  =  0  or  x - 25  =  0

x  =  30  or  x  =  25

So, the number of toys produced on that particular day is 30 or 25.

Problem 2 :

Find two numbers whose sum is 27 and product is 182.

Solution :

Let 'x' be one of the two numbers.

Then, the other number is (27 - x).

Product of those numbers  =  182

x(27 - x)  =  182

27x - x2  =  182

x2 - 27x + 182  =  0

x2 - 14x - 13x + 182  =  0

x(x - 14) - 13(x - 14)  =  0

(x - 14)(x - 13)  =  0

x - 14  =  0  or  x - 13  =  0

x  =  14  or  x  =  13

If x  =  14,

27 - x  =  27 - 14  =  13

If x  =  13,

27 - x  =  27 - 13  =  14

So, the numbers are 13 and 14.

Problem 3 :

Find two consecutive positive integers, sum of whose squares is 365.

Solution :

Let 'x' and 'x + 1' be the two consecutive integers.

Sum of squares  =  365

x2 + (x + 1)2  =  365

x2 + x+ 2x + 1  =  365

2x2 + 2x + 1 - 365  =  0

2x2 + 2x - 364  =  0

Divide the whole equation by 2.

x2 + x - 182  =  0

x2 + 13x - 12x - 182  =  0

x(x + 13) - 12(x + 13)  =  0

(x - 12)(x + 13)  =  0

x - 12  =  0  or  x + 13  =  0

x  =  12  or  x  =  -13

Because x is positive integer, it can not take a negative value.

Then,

x  =  12

x + 1  =  13

Therefore, the two consecutive integers are 12 and 13. Apart from the stuff given in this sectionif you need any other stuff in math, please use our google custom search here.

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