Word problems on quadratic equations play a major role quantitative aptitude test. There is no competitive exam without the questions from this topic. We have already learned this topic in our lower classes.Even though we have been already taught this topic in our lower classes, we need to learn some more short cuts which are being used to solve the problems in the above topic.

The only thing we have to do is, we need to apply the appropriate short cut and solve the problems in a limited time. This limited time will be one minute or less than one minute in most of the competitive exams.

Students who are preparing to improve their aptitude skills and those who are preparing for this type of competitive test must prepare this topic in order to have better score. Because, today there is no competitive exam without questions from the topic quadratic equations problems. Whether a person is going to write placement exam to get placed or a students is going to write a competitive exam in order to get admission in university, they must be prepared to solve word problems on equations. This is the reason for why people must study this topic.

As we mentioned in the above paragraph, a person who wants to get placed in a company and a students who wants to get admission in university for higher studies must write competitive exams like placement test and entrance exam. To meet the above requirement, it is very important to score more marks in the above mentioned competitive exams. To score more marks, they have to prepare the topics like this . Preparing this topic would definitely improve their marks in the above exams. Preparing this topic is not difficult task. We are just going to remember the stuff that we have already learned in our lower classes

Students have to learn few basic operations in this topic and some additional tricks. Already we are much clear with the four basic operations which we often use in math. They are addition, subtraction, multiplication and division. Even though we are much clear with these four basic operations, we have to be knowing some more stuff to do the problems which are being asked from this topic in competitive exams. The stuff which I have mentioned above is nothing but the tricks and shortcuts which need to solve the problems in a very short time.

**Here,
we are going to have some word problems on quadratic equations. You
can check your answer online and see step by step solution.**

1. Difference between a number and its positive square root is 12. Find the number.

Let "x" be the required number. Its positive square root is √x

Difference = 12

x - √x = 12

x - 12 = √x

(x-12)^{2} = x

x^{2} - 24x + 144 = x

x^{2} - 25x + 144 = 0

(x-9)(x-16) = 0

x = 9 or x = 16

x = 9 does not satisfy the condition given in the question.

Hence, the required number is 16.

Difference = 12

x - √x = 12

x - 12 = √x

(x-12)

x

x

(x-9)(x-16) = 0

x = 9 or x = 16

x = 9 does not satisfy the condition given in the question.

Hence, the required number is 16.

2. A piece of iron rod costs $60. If the rod was 2 meter shorter and each meter costs $1 more, the cost would remain unchanged. What is the length of the rod?

Let "x" be the length of the given rod.

Then the length of the rod 2 meter shorter is (x-2) and the total cost of both the rods is $60 (because cost would remain unchanged)

Cost of one meter of the given rod = 60/x

Cost of one meter of the rod which is 2 meter shorter = 60/(x-2)

From the question, 60/(x-2) = 60/x + 1

60/(x-2) = (60+x)/x

60x = (x-2)(60+x)

60x = x^{2} + 58x - 120

x^{2} -2x -120 = 0

(x+10)(x-12) = 0

x = -10 or x = 12

x = -10 can not be accepted. Because length can not be negative.

Hence, the length of the given rod is 12 m

Then the length of the rod 2 meter shorter is (x-2) and the total cost of both the rods is $60 (because cost would remain unchanged)

Cost of one meter of the given rod = 60/x

Cost of one meter of the rod which is 2 meter shorter = 60/(x-2)

From the question, 60/(x-2) = 60/x + 1

60/(x-2) = (60+x)/x

60x = (x-2)(60+x)

60x = x

x

(x+10)(x-12) = 0

x = -10 or x = 12

x = -10 can not be accepted. Because length can not be negative.

Hence, the length of the given rod is 12 m

3. Divide 25 in two parts so that sum of their reciprocals is 1/6

Let "x" be one of the parts of 25. Then the other part is 25-x.

Sum of the reciprocals of the parts = 1/6

1/x + 1/(25-x) = 1/6

(25-x+x)/x(25-x) = 1/6

25/(25x-x^{2}) = 1/6

6x25 = 25x-x^{2}

150 = 25x-x^{2}

x^{2}- 25x + 150 = 0

(x-15)(x-10) = 0

x= 15 or x = 10

Hence, the two parts of the 25 are 10 and 15.

Sum of the reciprocals of the parts = 1/6

1/x + 1/(25-x) = 1/6

(25-x+x)/x(25-x) = 1/6

25/(25x-x

6x25 = 25x-x

150 = 25x-x

x

(x-15)(x-10) = 0

x= 15 or x = 10

Hence, the two parts of the 25 are 10 and 15.

4. The hypotenuse of a right angled triangle is 20 cm. The difference between its other two sides is 4 cm. The sides are

Let "x" and "x+4" be the lengths of other two sides.

Using Pythagorean theorem, (x+4)^{2} + x^{2} = 20^{2}

x^{2} + 8x + 16 + x^{2} - 400 = 0

2x^{2} + 8x - 384 = 0

x^{2} + 4x - 192 = 0

(x+16)(x-12) = 0

x = -16 or x = 12

x = -16 can not be accepted. Because length can not be negative.

If x = 12, x+4 = 12+4 = 16

Hence, the other two sides of the triangle are 12 cm and 16 cm

Using Pythagorean theorem, (x+4)

x

2x

x

(x+16)(x-12) = 0

x = -16 or x = 12

x = -16 can not be accepted. Because length can not be negative.

If x = 12, x+4 = 12+4 = 16

Hence, the other two sides of the triangle are 12 cm and 16 cm

5. The sides of an equilateral triangle are shortened by 12 units, 13 units and 14 units respectively and a right angle triangle is formed. The side of the equilateral triangle is

Let "x" be the length of each side of the equilateral triangle.

Then, the sides of the right angle triangle are (x-12), (x-13) and (x-14)

In the above three sides, the side represented by x-12 is hypotenuse (because that is the longest side).

Using Pythagorean theorem, (x-12)^{2} = (x-13)^{2} + (x-14)^{2}

x^{2}-24x+144 = x^{2}-26x+169 + x^{2}-28x+196

x^{2} -30x +221 = 0

(x-13)(x-17) = 0

x = 13 or x = 17.

x = 13 can not be accepted. Because, if x=13, one of the sides of the right angle triangle would be negative.

Hence, the side of the equilateral triangle is 17 units

Then, the sides of the right angle triangle are (x-12), (x-13) and (x-14)

In the above three sides, the side represented by x-12 is hypotenuse (because that is the longest side).

Using Pythagorean theorem, (x-12)

x

x

(x-13)(x-17) = 0

x = 13 or x = 17.

x = 13 can not be accepted. Because, if x=13, one of the sides of the right angle triangle would be negative.

Hence, the side of the equilateral triangle is 17 units

6. The area of a rectangular field is 2000 sq.m and its perimeter is 180 m. The length and breadth are

Let "x" and "y" be the length and breadth of the rectangle respectively.

Perimeter = 180 m (given)

2x + 2y = 180 ---> x + y = 90

So, y = 90 - x

Area = 2000 sq.m

xy = 2000

x(90-x) = 2000

90x - x^{2} = 2000

x^{2} - 90x + 2000 = 0

(x-50)(x-40) = 0

x = 50 or x = 40

If x = 50, y = 90-50 = 40

If x = 40, y = 90-40 = 50

Hence, the length and breadth of the rectangle are 40 m and 50 m respectively

Perimeter = 180 m (given)

2x + 2y = 180 ---> x + y = 90

So, y = 90 - x

Area = 2000 sq.m

xy = 2000

x(90-x) = 2000

90x - x

x

(x-50)(x-40) = 0

x = 50 or x = 40

If x = 50, y = 90-50 = 40

If x = 40, y = 90-40 = 50

Hence, the length and breadth of the rectangle are 40 m and 50 m respectively

7. A distributor of apple juice has 5000 bottles in the store that it wishes to distribute in a month. From experience, it is known that demand D (in number of bottles) is given by D = -2000p² + 2000p + 17000. The price per bottle that will result zero inventory is

Stock in the store is 5000 bottles. If the inventory be zero, the demand has to be 5000 bottles.

To know the price for zero inventory, we have to plug D = 5000 in the quadratic equation D = -2000p^{2} + 2000p + 17000

5000 = -2000p^{2} + 2000p + 17000

2000p^{2} - 2000p - 12000 = 0

p^{2} - p - 6 = 0

(p+3)(p-2) = 0

p = -3 or p = 2

p = -3 can not be accepted. Because, price can not be negative.

Hence, price per bottle that will result zero inventory is $2

To know the price for zero inventory, we have to plug D = 5000 in the quadratic equation D = -2000p

5000 = -2000p

2000p

p

(p+3)(p-2) = 0

p = -3 or p = 2

p = -3 can not be accepted. Because, price can not be negative.

Hence, price per bottle that will result zero inventory is $2

8. The sum of two irrational numbers multiplied by the larger one is 70 and their difference is multiplied by the smaller one is 12. The two numbers are.

The sum of two irrational numbers multiplied by the larger one is 70 which is rational.

When we check the options (A) and (B), we have unlike irrational numbers.

If the sum of two unlike irrational numbers multiplied by the larger one, the result would be irrational. But, we have to get the answer 70 which is rational.

Clearly, option (A) and (B) are incorrect.

In option (C), we have like irrational numbers.

Let us check the given conditions in option (C)

(5√2+2√2)5√2 = 7√2x5√2 = 70 ---(1)

(5√2-2√2)2√2 = 3√2x2√2 = 12 ---(2)

From (1)&(2), option(C) satisfies the given conditions.

Hence, option (C)2√2, 5√2 is correct

When we check the options (A) and (B), we have unlike irrational numbers.

If the sum of two unlike irrational numbers multiplied by the larger one, the result would be irrational. But, we have to get the answer 70 which is rational.

Clearly, option (A) and (B) are incorrect.

In option (C), we have like irrational numbers.

Let us check the given conditions in option (C)

(5√2+2√2)5√2 = 7√2x5√2 = 70 ---(1)

(5√2-2√2)2√2 = 3√2x2√2 = 12 ---(2)

From (1)&(2), option(C) satisfies the given conditions.

Hence, option (C)2√2, 5√2 is correct

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