Problem 1 :
Difference between a number and its positive square root is 12. Find the number.
Solution :
Let x be the required number.
Its positive square root is √x
Given : Difference between x and √x = 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.
So, the required number is 16.
Problem 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?
Solution :
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 is
= 60/x
Cost of one meter of the rod which is 2 meter shorter is
= 60/(x - 2)
Given : If the rod was 2 meter shorter and each meter costs $1 more.
That is, 60/(x-2) is $1 more than 60/x.
[60/(x - 2)] - [60/x] = 1
[60x - 60(x - 2)]/[x(x - 2)] = 1
[60x - 60x + 120]/[x^{2} - 2x] = 1
120/(x^{2} - 2x) = 1
120 = x^{2} - 2x
0 = x^{2} + 2x - 120
x^{2} + 2x - 120 = 0
(x + 10)(x - 12) = 0
x = - 10 or x = 12
Because length can not be a negative number, we can ignore - 10.
So, the length of the given rod is 12 m.
Problem 3 :
Divide 25 in two parts so that sum of their reciprocals is 1/6.
Solution :
Let x be one of the parts of 25. Then the other part is (25 - x).
Given : Sum of the reciprocals of the parts is 1/6.
1/x + 1/(25 - x) = 1/6
(25 - x + x)/x(25 - x) = 1/6
25/(25x - x^{2}) = 1/6
6(25) = 25x - x^{2}
150 = 25x - x^{2}
x^{2} - 25x + 150 = 0
(x - 15)(x - 10) = 0
x = 15 or x = 10
When x = 15,
25 - x = 25 - 15
25 - x = 10
When x = 10,
25 - x = 25 - 10
25 - x = 15
So, the two parts of the 25 are 10 and 15.
Problem 4 :
The hypotenuse of a right angled triangle is 20 cm. The difference between its other two sides is 4 cm. Find the length of the sides.
Solution :
Let x and x + 4 be the lengths of other two sides.
Using Pythagorean theorem,
(x + 4)x^{2} + x^{2} = 20x^{2}
x^{2} + 8x + 16 + x^{2} = 400
2x^{2} + 8x + 16 = 400
Subtract 400 from both sides.
2x^{2} + 8x - 384 = 0
Divide both sides by 2.
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
So, the other two sides of the triangle are 12 cm and 16 cm.
Problem 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. Find the length of each side of the equilateral triangle.
Solution :
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, the side represented by (x - 14) will be negative.
So, the side of the equilateral triangle is 17 units.
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