Quadratic Equations Exercise 9.11 – Questions
- Find two consecutive positive odd numbers such that the sum of their squares is equal to 130.
- Find the whole number such that four times the number subtracted from three times the square of the number makes 15.
- The sum of two natural numbers is 8. Determine the numbers, if the sum of their reciprocals is 8/15
- A two digit number is such that the product of the digits is 12. When 36 is added to this number the digits interchange their places. Determine the number
- Find three consecutive positive integers such that the sum of the square of the first and the product of other two is 154.
- The ages of Kavya and Karthik are 11 years and 14 years. In how many years times will the product of their ages be 304?
- The age of the mother is twice the square of the age of her son. Eight years hence, the age of the mother will be 4 years more than three times the age of her son. Find their present age.
- The area of a rectangle is 56 cm2. If the measure of its base is represented by (x + 5) and the measure of its height by (x – 5), find the dimensions of the rectangle.
- The altitude of a triangle is 6 cm greater than its base. If its area is 108cm2 . Find its base and height.
- In rhombus ABCD, the diagonals AC and BD intersect at E. If AE = x, BE = (x + 7) and AB = (x + 8), find the lengths of the diagonals AC and BD.
- In an isosceles triangle ABC, AB = BC and BD is the altitude to base AC.IF DC = x , BD = 2x – 1 and BC = 2x + 1, fibd the lengths of all three sides of the triangle.
- A motor boat whose speed is 15km/hr in still water 30 km downstream and comes back in a total of 4 hours 30 minutes. Determine the speed of the stream.
- A dealer sells an article for Rs. 24 and gains as much percent as the cost price of the article. Find the cost price of the article.
- Shwetha takes 6 ways less than the number of days taken by Ankitha to complete a piece of work. If both Shwetha and Ankitha together can complete the same work in 4 days, in how many days will Ankitha alone complete the work?
Quadratic Equations Exercise 9.11 – Solutions:
- Find two consecutive positive odd numbers such that the sum of their squares is equal to 130.
Solution:
Let one of the odd positive number be x.
The other odd positive number be (x + 2).
The sum of the numbers = x2 + (x + 2)2 = 130
x2 + x2 + 2x + 4 = 130
2x2 + 2x + 4 = 130
2x2 + 2x – 126 = 0
x2 + x – 63 = 0
x2 + 9x – 7x – 63 = 0
x(x + 9) -7(x + 9) = 0
(x – 7)(x + 9) = 0
x = 7 or x = -9
Therefore, one of the odd positive number, x = 7 and the other odd positive number is x + 2 = 7 + 2 = 9
Therefore, two consecutive positive odd numbers 7 and 9.
- Find the whole number such that four times the number subtracted from three times the square of the number makes 15.
Solution:
Let the whole number be x.
Four times the number be 4x and three times the number be 3x2
Therefore,
4x – 3x2 + 15 = 0
-3x2 + 4x + 15 = 0
-3x2 + 9x – 5x + 15 = 0
3x(x – 3)-5(x – 3) = 0
(x – 3)(3x – 5) = 0
x = 3 or 3x = 5
x = 3 or x = 5/3
Therefore the whole number is 3.
- The sum of two natural numbers is 8. Determine the numbers, if the sum of their reciprocals is 8/15
Solution:
The sum of two natural numbers is 8. i.e., x + y = 8
x = 8 – y
The sum of their reciprocals is 8/15 i.e., 1/x + 1/y = 8/15
1/x + 1/y = 8/15
y + x/xy = 8/15
(x + y)15 = 8xy
15 x 8 = 8 xy
xy = 15
(8 – y)y = 15
8y – y2 = 15
-y2 + 8y – 15 = 0
-y2 + 3y + 5y – 15 = 0
-y(y – 3)+5(y – 3) = 0
(y – 3)(-y + 5) = 0
y = 3 or –y + 5 = 0
y = 3 or y = 5
Therefore the numbers be 3, 5 .
- A two digit number is such that the product of the digits is 12. When 36 is added to this number the digits interchange their places. Determine the number
Solution:
Let ten’s digit of the number = x and its unit digit = y. Then two digit number is 10x + y
xy = 12 and 10x + y + 36 = 10y + x
y = 12/x and 9x + 36 = 9y i.e., x + 4 = y
Substituting y = 12/x in x + 4 = 4; we get,
x + 4 = 12/x
x2 + 4x – 12 = 0
x2 +6x – 2x – 12 = 0
x(x + 6)-2(x + 6) = 0
(x + 6)(x – 2) = 0
x = -6 and x = 2
Since x is a digit, therefore, x = 2 and y = 12/x = 12/2 = 6
Therefore the required number is 10x + y = 10×2 + 6 = 26
- Find three consecutive positive integers such that the sum of the square of the first and the product of other two is 154.
Solution:
Let three consecutive numbers be x, x + 1, x + 2
x2 +(x+1)(x+2) = 154
x2 + (x2 + 2x + x + 2) = 154
2x2 + 3x + 2 = 154
2x2 + 3x – 152 = 0
2x2 + 19x – 16x – 152 = 0
x(2x + 19) – 8(2x +19) = 0
(2x + 19)(x – 8) = 0
2x + 19 = 0 or x – 8 = 0
2x = -19 or x = 8
x = –19/2 or x = 8
Therefore, three consecutive numbers be 8, 9 and 10.
- The ages of Kavya and Karthik are 11 years and 14 years. In how many years time will the product of their ages be 304?
Solution:
The age of Kavya be 11 years and the age of the Karthik be 14 years.
In x years the product of their ages will be 304. i.e., (x + 11)(x + 14) = 304
x2 + 14x + 11x + 154 = 304
x2 + 25x + 154 – 304 = 0
x2 + 25x + 150 = 0
x2 + 30x – 5x + 150 = 0
x(x + 30) – 5(x + 30) = 0
(x + 30)(x – 5) = 0
x + 30 = 0 or x – 5 = 0
x = -30 or x = 5
Therefore, in 5 years the product of their ages will be 304.
- The age of the mother is twice the square of the age of her son. Eight years hence, the age of the mother will be 4 years more than three times the age of her son. Find their present age.
Solution:
Let the age of the mother be x.
Hence the present age of son = 2x2
Given the age of the mother is twice the square of the age of her son i.e., x = 2x2
Age of son after 8 years = (x + 8)years
Age of mother after 8 years = (2x2 + 8) years
Given that, (2x2 + 8) = 3(x + 8) + 4
2x2 + 8 = 3x + 24 + 4
2x2 – 3x + 8 – 28 = 0
2x2 – 3x – 20 =0
2x2 – 8x + 5x – 20 = 0
2x(x – 4) +5(x – 4) = 0
(2x + 5)(x – 4) = 0
x – 4 = 0 or 2x + 5 = 0
x = 4 or 2x = -5
x = 4 or x = –5/2
Thus, x = 4
Hence present age of son is 4 years and present age of mother is 2x2 = 2(4)2 = 32.
- The area of a rectangle is 56 cm2. If the measure of its base is represented by (x + 5) and the measure of its height by (x – 5), find the dimensions of the rectangle.
Solution:
A = 56 m2
56 = (x + 5)(x – 5)
56 = x2 – 5x + 5x – 25
56 = x2 – 25
56 + 25 = x2
81 = x2
x = √81 = 9
Therefore, base of the rectangle = x + 5 = 9 + 5 = 14
height of the rectangle = x – 5 = 9 – 5 = 4
- The altitude of a triangle is 6 cm greater than its base. If its area is 108cm2. Find its base and height.
Solution:
Let the base of the triangle be x.
Then altitude of the triangle = x + 6
Area of the triangle = 108 cm2
Formula to find the area of triangle = 1/2 x base x height
108 = 1/2 x(x + 6)
216 = x(x + 6)
216 = x2 + 6x
x2 + 6x – 216 = 0
x2 + 18x – 12x – 216 = 0
x(x + 18) -12(x + 18) = 0
(x – 12)(x + 18) = 0
x – 12 = 0 or x + 18 = 0
x = 12
Let the base of the triangle be 12cm
Then height of the triangle 18cm
- In rhombus ABCD, the diagonals AC and BD intersect at E. If AE = x, BE = (x + 7) and AB = (x + 8), find the lengths of the diagonals AC and BD.
Solution:
. In rhombus ABCD, the diagonals AC and BD intersect at E. Therefore, angles at E are right angles. Hence, all the triangles formed by the diagonals AC and BD are right angled triangle.
In right angled triangle ABE, by Pythagoras theorem. We have
AB2 = AE2 + EB2
(x + 8)2 = x2 + (x + 7)2
x2 + 16x + 64 = x2 + x2 + 14x + 49
x2 + 16x + 64 – x2 – x2 – 14x – 49 = 0
-x2 + 2x + 15 = 0
-x2 + 5x – 3x + 15 = 0
x(-x + 5) +3(-x + 5) = 0
(-x + 5)(x + 3) = 0
x + 3 = 0 or –x +5 = 0
x = -3 or x = 5
The diagonal AC = AE + EC = 2(AE) = 2x = 10 cm
The diagonal BD = BE + ED = 2(BE)= 2(x+7) = 2(5 +7) = 2(12) = 24 cm
- In an isosceles triangle ABC, AB = BC and BD is the altitude to base AC.IF DC = x , BD = 2x – 1 and BC = 2x + 1, find the lengths of all three sides of the triangle.
Solution:
Given, AB = BC = 2x + 1
In an isosceles triangle ABC, AB = BC and BD is the altitude to base AC.
Therefore triangle BCD is a right angled triangle.
By Pythagoras theorem, BC2 = BD2 + CD2
(2x + 1)2 = (2x – 1)2 + x2
4x2 + 4x + 1 = 4x2 – 4x + 1 + x2
4x2 + 4x + 1 – 4x2 + 4x – 1 – x2 = 0
-x2 + 8x = 0
x(-x + 8) = 0
-x + 8 = 0
x = 8
We know, AB = BC = 2x + 1 = 2(8) + 1 = 17cm
AC = AD + DC = 2(DC) = 2x = 2(8) = 16 cm
- A motor boat whose speed is 15km/hr in still water 30 km downstream and comes back in a total of 4 hours 30 minutes. Determine the speed of the stream.
Solution:
Let the speed of the stream be x km/hr. Then speed downstream = (15 + x) kn/hr
Speed upstream = (15 –x )km/hr
given, it took 4 hours 30 minutes to travel back to same place.
So, we have,
30/15 + x – 30/15 – x = 41/2
30*30/(15+x)(15-x) = 41/2
900/225 – x^2 = 9/2
900 *2 = (225 – x2)*9
1800 = 2025 – 9x2
9x2 = – 1800 + 2025
9x2 = 225
x2 = 225/9
x2 = 25
x = 5km/hr
- A dealer sells an article for Rs. 24 and gains as much percent as the cost price of the article. Find the cost price of the article.
Solution:
Let x be the cost price of the article.
The selling price of the article Rs. 24
x = 24 – x/x * 100
x2 = (24 – x)100
x2 = 2400 – 100 x
x2 + 100x – 2400 = 0
x2 + 120x – 20x – 2400 = 0
x(x + 120) -20(x + 120) = 0
(x + 120)(x – 20) = 0
x = -120 or x = 20
Therefore, cost price of the article is 20 Rs.
- Shwetha takes 6 ways less than the number of days taken by Ankitha to complete a piece of work. If both Shwetha and Ankitha together can complete the same work in 4 days, in how many days will Ankitha alone complete the work?
Solution:
Let Shwetha take x – 6 days and Ankitha takes x days.
Shwetha one day work = 1/(x – 6) and Ankitha one day work = 1/x
Given, (A + B) can finish work in 4 day.
1/x + 1/(x – 6) = 1/4
(x-6)+x/x(x-6) = 1/4
2x -6/x(x – 6) = 1/4
4(2x – 6) = x(x – 6)
8x – 24 = x2 – 6x
x2 – 6x – 8x + 24 = 0
x2 – 14x + 24 = 0
x2 – 12x – 2x + 24 = 0
x(x – 12)-2(x – 12) = 0
(x – 12)(x – 2) = 0
x = 12 days taken by Ankitha alone.