Class- X-CBSE-Mathematics Quardatic Equations Practice more on Quardatic Equations Page - 1 www.embibe.com CBSE NCERT Solutions for Class 10 Mathematics Chapter 4 Back of Chapter Questions 1. Check whether the following are quadratic equations: (i) (+ 1) 2 = 2(− 3) (ii) 2 − 2=(−2)(3 −) (iii) (− 2)(+ 1) = (− 1)(+ 3) (iv) (− 3)(2+ 1) = (+ 5) (v) (2− 1)(− 3) = (+ 5)(− 1) (vi) 2 +3+1=(− 2) 2 (vii) (+ 2) 3 =2(2 − 1) (viii) 3 − 42 − +1=(− 2) 3 Solution: (i) We know that any equation of the form 2 + + =0 is called a quadratic equation, where , , are real numbers and ≠ 0. Given equation: (+ 1) 2 = 2(− 3) Using the formula (+ ) 2 = 2 +2+ 2 ⇒ 2 +2+1=2− 6 ⇒ 2 +7=0 Here, = 1, = 0 and = 7. Thus, the given equation is a quadratic equation as ≠ 0. (ii) We know that any equation of the form 2 + + =0 is called a quadratic equation, where , , are real numbers and ≠ 0. Given equation: 2 − 2=(−2)(3 −) ⇒ 2 − 2= −6+2⇒ 2 − 4+6=0 Here, = 1, = −4 and = 6. Thus, the given equation is a quadratic equation as ≠ 0. (iii) We know that any equation of the form 2 + + =0 is called a quadratic equation, where , , are real numbers and ≠ 0.
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Class- X-CBSE-Mathematics Quardatic Equations
Practice more on Quardatic Equations Page - 1 www.embibe.com
CBSE NCERT Solutions for Class 10 Mathematics Chapter 4 Back of Chapter Questions
1. Check whether the following are quadratic equations:
(i) (𝑥𝑥 + 1)2 = 2(𝑥𝑥 − 3)
(ii) 𝑥𝑥2 − 2𝑥𝑥 = (−2)(3 − 𝑥𝑥)
(iii) (𝑥𝑥 − 2)(𝑥𝑥 + 1) = (𝑥𝑥 − 1)(𝑥𝑥 + 3)
(iv) (𝑥𝑥 − 3)(2𝑥𝑥 + 1) = 𝑥𝑥(𝑥𝑥 + 5)
(v) (2𝑥𝑥 − 1)(𝑥𝑥 − 3) = (𝑥𝑥 + 5)(𝑥𝑥 − 1)
(vi) 𝑥𝑥2 + 3𝑥𝑥 + 1 = (𝑥𝑥 − 2)2
(vii) (𝑥𝑥 + 2)3 = 2𝑥𝑥(𝑥𝑥2 − 1)
(viii) 𝑥𝑥3 − 4𝑥𝑥2 − 𝑥𝑥 + 1 = (𝑥𝑥 − 2)3
Solution:
(i) We know that any equation of the form 𝑎𝑎𝑥𝑥2 + 𝑏𝑏𝑥𝑥 + 𝑐𝑐 = 0 is called aquadratic equation, where 𝑎𝑎, 𝑏𝑏, 𝑐𝑐 are real numbers and 𝑎𝑎 ≠ 0.
Given equation: (𝑥𝑥 + 1)2 = 2(𝑥𝑥 − 3)
Using the formula (𝑎𝑎 + 𝑏𝑏)2 = 𝑎𝑎2 + 2𝑎𝑎𝑏𝑏 + 𝑏𝑏2
⇒ 𝑥𝑥2 + 2𝑥𝑥 + 1 = 2𝑥𝑥 − 6
⇒ 𝑥𝑥2 + 7 = 0
Here, 𝑎𝑎 = 1, 𝑏𝑏 = 0 and 𝑐𝑐 = 7.
Thus, the given equation is a quadratic equation as 𝑎𝑎 ≠ 0.
(ii) We know that any equation of the form 𝑎𝑎𝑥𝑥2 + 𝑏𝑏𝑥𝑥 + 𝑐𝑐 = 0 is called aquadratic equation, where 𝑎𝑎, 𝑏𝑏, 𝑐𝑐 are real numbers and 𝑎𝑎 ≠ 0.
Given equation: 𝑥𝑥2 − 2𝑥𝑥 = (−2)(3 − 𝑥𝑥)
⇒ 𝑥𝑥2 − 2𝑥𝑥 = −6 + 2𝑥𝑥
⇒ 𝑥𝑥2 − 4𝑥𝑥 + 6 = 0
Here, 𝑎𝑎 = 1, 𝑏𝑏 = −4 and 𝑐𝑐 = 6.
Thus, the given equation is a quadratic equation as 𝑎𝑎 ≠ 0.
(iii) We know that any equation of the form 𝑎𝑎𝑥𝑥2 + 𝑏𝑏𝑥𝑥 + 𝑐𝑐 = 0 is called aquadratic equation, where 𝑎𝑎, 𝑏𝑏, 𝑐𝑐 are real numbers and 𝑎𝑎 ≠ 0.
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⇒ 𝑥𝑥3 + 8 + 6𝑥𝑥2 + 12𝑥𝑥 = 2𝑥𝑥3 − 2𝑥𝑥
⇒ 𝑥𝑥3 − 14𝑥𝑥 − 6𝑥𝑥2 − 8 = 0
This equation is not of the form 𝑎𝑎𝑥𝑥2 + 𝑏𝑏𝑥𝑥 + 𝑐𝑐 = 0
So, the given equation is not a quadratic equation.
(viii) We know that any equation of the form 𝑎𝑎𝑥𝑥2 + 𝑏𝑏𝑥𝑥 + 𝑐𝑐 = 0 is called a quadratic equation, where 𝑎𝑎, 𝑏𝑏, 𝑐𝑐 are real numbers and 𝑎𝑎 ≠ 0.
Given equation: 𝑥𝑥3 − 4𝑥𝑥2 − 𝑥𝑥 + 1 = (𝑥𝑥 − 2)3
Using the formula (𝑎𝑎 − 𝑏𝑏)3 = 𝑎𝑎3 − 3𝑎𝑎2𝑏𝑏 + 3𝑎𝑎𝑏𝑏2 − 𝑏𝑏3
⇒ 𝑥𝑥3 − 4𝑥𝑥2 − 𝑥𝑥 + 1 = 𝑥𝑥3 − 8 − 6𝑥𝑥2 + 12𝑥𝑥
⇒ 2𝑥𝑥2 − 13𝑥𝑥 + 9 = 0
Here, 𝑎𝑎 = 2, 𝑏𝑏 = −13 and 𝑐𝑐 = 9.
Thus, the given equation is a quadratic equation as 𝑎𝑎 ≠ 0.
2. Represent the following situations in the form of quadratic equations:
(i) The area of a rectangular plot is 528 m2. The length of the plot (in metres) is one more than twice its breadth. We need to find the length and breadth of the plot.
(ii) The product of two consecutive positive integers is 306. We need to find the integers.
(iii) Rohan’s mother is 26 years older than him. The product of their ages (in years) 3 years from now will be 360. We would like to find Rohan’s present age.
(iv) A train travels a distance of 480 km at a uniform speed. If the speed had been 8 km h⁄ less, then it would have taken 3 hours more to cover the same distance. We need to find the speed of the train.
Solution:
(i) Let the breadth of the plot be 𝑥𝑥 m.
Hence, the length of the plot is (2𝑥𝑥 + 1) m. (Since, given that length is one more than twice its breadth)
Therefore, area of a rectangle = length × breadth
Given: area of rectangle = 528 m2
∴ 528 = 𝑥𝑥(2𝑥𝑥 + 1)
⇒ 2𝑥𝑥2 + 𝑥𝑥 − 528 = 0 ………(i), which is of the form 𝑎𝑎𝑥𝑥2 + 𝑏𝑏𝑥𝑥 + 𝑐𝑐 = 0
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Also given that the train will take 3 more hours to cover the same distance.
Therefore, time take to travel 480 km = �480𝑥𝑥
+ 3� hrs
Speed × Time = Distance
(𝑥𝑥 − 8) �480𝑥𝑥
+ 3� = 480
⇒ 480 + 3𝑥𝑥 −3840𝑥𝑥
− 24 = 480
⇒ 3𝑥𝑥 −3840𝑥𝑥
= 24
⇒ 3𝑥𝑥2 − 24𝑥𝑥 − 3840 = 0
⇒ 𝑥𝑥2 − 8𝑥𝑥 − 1280 = 0…….. (i), which is of the form 𝑎𝑎𝑥𝑥2 + 𝑏𝑏𝑥𝑥 + 𝑐𝑐 = 0
Here 𝑎𝑎 = 1(≠ 0), 𝑏𝑏 = −8 and 𝑐𝑐 = −1280
Thus, quadratic equation (i) represents the situation given in the question and positive root of this equation will represent the speed of train.
EXERCISE 4.2
1. Find the roots of the following quadratic equations by factorisation:
(i) 𝑥𝑥2 − 3𝑥𝑥 − 10 = 0
(ii) 2𝑥𝑥2 + 𝑥𝑥 − 6 = 0
(iii) √2 𝑥𝑥2 + 7𝑥𝑥 + 5√2 = 0
(iv) 2𝑥𝑥2 − 𝑥𝑥 + 18
= 0
(v) 100𝑥𝑥2 − 20𝑥𝑥 + 1 = 0
Solution:
(i) To find the roots of given quadratic equation, lets first factorise the given quadratic expression 𝑥𝑥2 − 3𝑥𝑥 − 10. The given quadratic expression can be written as follows:
𝑥𝑥2 − 3𝑥𝑥 − 10
= 𝑥𝑥2 − 5𝑥𝑥 + 2𝑥𝑥 − 10 (we factorise by method of splitting the middle term)
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Now, the roots of this quadratic equation are the values of 𝑥𝑥 for which (𝑥𝑥 − 5)(𝑥𝑥 + 2) = 0
∴ 𝑥𝑥 − 5 = 0 or 𝑥𝑥 + 2 = 0
𝑖𝑖. 𝑒𝑒. , 𝑥𝑥 = 5 or 𝑥𝑥 = −2
Hence, the roots of this quadratic equation are 5 and−2.
(ii) To find the roots of given quadratic equation, lets first factorise the given quadratic expression 2𝑥𝑥2 + 𝑥𝑥 − 6 . The given quadratic expression can be written as follows:
2𝑥𝑥2 + 𝑥𝑥 − 6
= 2𝑥𝑥2 + 4𝑥𝑥 − 3𝑥𝑥 − 6 (we factorise by method of splitting the middle term)
= 2𝑥𝑥(𝑥𝑥 + 2) − 3(𝑥𝑥 + 2)
= (𝑥𝑥 + 2)(2𝑥𝑥 − 3)
Now, the roots of this quadratic equation are the values of 𝑥𝑥 for which (𝑥𝑥 + 2)(2𝑥𝑥 − 3) = 0
∴ 𝑥𝑥 + 2 = 0 or 2𝑥𝑥 − 3 = 0
𝑖𝑖. 𝑒𝑒. , 𝑥𝑥 = −2 or 𝑥𝑥 = 32
Hence, the roots of this quadratic equation are −2 and 32.
(iii) To find the roots of given quadratic equation, lets first factorise the given quadratic expression √2𝑥𝑥2 + 7𝑥𝑥 + 5√2 . The given quadratic expression can be written as follows:
√2𝑥𝑥2 + 7𝑥𝑥 + 5√2
= √2𝑥𝑥2 + 5𝑥𝑥 + 2𝑥𝑥 + 5√2 (we factorise by method of splitting the middle term)
= 𝑥𝑥�√2𝑥𝑥 + 5� + √2�√2𝑥𝑥 + 5�
= �√2𝑥𝑥 + 5��𝑥𝑥 + √2�
Now, the roots of this quadratic equation are the values of 𝑥𝑥 for which
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Hence, the roots of this quadratic equation are − 5√2
and −√2.
(iv) To find the roots of given quadratic equation, lets first factorise the given quadratic expression 2𝑥𝑥2 − 𝑥𝑥 + 1
8 . The given quadratic expression can be
written as follows:
2𝑥𝑥2 − 𝑥𝑥 +18
=18
(16𝑥𝑥2 − 8𝑥𝑥 + 1)
= 18
(16𝑥𝑥2 − 4𝑥𝑥 − 4𝑥𝑥 + 1) (we factorise by method of splitting the middle term)
=18�4𝑥𝑥(4𝑥𝑥 − 1) − 1(4𝑥𝑥 − 1)�
=18
(4𝑥𝑥 − 1)2
Now, the roots of this quadratic equation are the values of 𝑥𝑥 for which (4𝑥𝑥 − 1)2 = 0
Thus, (4𝑥𝑥 − 1) = 0 or (4𝑥𝑥 − 1) = 0
𝑖𝑖. 𝑒𝑒. , 𝑥𝑥 = 14 or 𝑥𝑥 = 1
4
Hence, the roots of this quadratic equation are 14 and 1
4.
(v) To find the roots of given quadratic equation, lets first factorise the given quadratic expression 100𝑥𝑥2 − 20𝑥𝑥 + 1. The given quadratic expression can be written as follows:
100𝑥𝑥2 − 20𝑥𝑥 + 1
= 100𝑥𝑥2 − 10𝑥𝑥 − 10𝑥𝑥 + 1 (we factorise by method of splitting the middle term)
= 10𝑥𝑥(10𝑥𝑥 − 1) − 1(10𝑥𝑥 − 1)
= (10𝑥𝑥 − 1)2
Now, the roots of this quadratic equation are the values of 𝑥𝑥 for which (10𝑥𝑥 − 1)2 = 0
Thus, (10𝑥𝑥 − 1) = 0 or (10𝑥𝑥 − 1) = 0
i. e., 𝑥𝑥 = 110
or 𝑥𝑥 = 110
Hence, the roots of this quadratic equation are 110
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2. Solve the problems given below.
Represent the following situations mathematically:
(i) John and Jivanti together have 45 marbles. Both of them lost 5 marbles each, and the product of the number of marbles they now have is 124. We would like to find out how many marbles they had to start with.
(ii) 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. On a particular day, the total cost of production was ₹ 750. We would like to find out the number of toys produced on that day.
Solution:
(i) Let the number of John's marbles be 𝑥𝑥.
So, the number of Jivanti's marbles = 45 − 𝑥𝑥
If both lost 5 marbles each,
Then number of marbles left with John = 𝑥𝑥 − 5
Then number of marbles left with Jivanti = 45 − 𝑥𝑥 − 5 = 40 − 𝑥𝑥
Given that the product of their marbles is 124.
∴ (𝑥𝑥 − 5)(40 − 𝑥𝑥) = 124
⇒ 𝑥𝑥2 − 45𝑥𝑥 + 324 = 0
⇒ 𝑥𝑥2 − 36𝑥𝑥 − 9𝑥𝑥 + 324 = 0
⇒ 𝑥𝑥(𝑥𝑥 − 36) − 9(𝑥𝑥 − 36) = 0
⇒ (𝑥𝑥 − 36)(𝑥𝑥 − 9) = 0
Either 𝑥𝑥 = 36 = 0 or 𝑥𝑥 − 9 = 0
𝑖𝑖. 𝑒𝑒. , 𝑥𝑥 = 36 or 𝑥𝑥 = 9
If the number of John's marbles = 36
Then, the number of Jivanti's marbles = 45 − 36 = 9
If the number of John's marbles = 9
Then, the number of Jivanti's marbles = 45 − 9 = 36.
(ii) Let the number of toys produced on that day be 𝑥𝑥.
∴ The cost of production of each toy that day = ₹ (55 − 𝑥𝑥)
So, the total cost of production that day = 𝑥𝑥(55 − 𝑥𝑥)
As per the question, the total cost of production of the toys = ₹ 750
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6. A cottage industry produces a certain number of pottery articles in a day. It was observed on a particular day that the cost of production of each article (in rupees) was 3 more than twice the number of articles produced on that day. If the total cost of production on that day was ₹. 90, find the number of articles produced and the cost of each article.
Solution:
Let the number of articles produced on that day be 𝑥𝑥.
So, the cost of production of each article = ₹(2𝑥𝑥 + 3)
According to the question, the total cost of production on that day was ₹ 90.
We know that
Total cost of production = Cost of each article × Number of articles produced
∴ 𝑥𝑥(2𝑥𝑥 + 3) = 90
⇒ 2𝑥𝑥2 + 3𝑥𝑥 − 90 = 0
⇒ 2𝑥𝑥2 + 15𝑥𝑥 − 12𝑥𝑥 − 90 = 0
⇒ 𝑥𝑥(2𝑥𝑥 + 15) − 6(2𝑥𝑥 + 15) = 0
⇒ (2𝑥𝑥 + 15)(𝑥𝑥 − 6) = 0
Either 2𝑥𝑥 + 15 = 0 or 𝑥𝑥 − 6 = 0, 𝑖𝑖. 𝑒𝑒. , 𝑥𝑥 = −152
or 𝑥𝑥 = 6
It’s clear that number of articles produced can only be a positive integer, so, 𝑥𝑥 can only be 6.
Therefore, number of articles produced on that day = 6
Cost of each article = (2 × 6) + 3 = ₹ 15
EXERCISE 4.3
1. Find the roots of the following quadratic equations, if they exist, by the method of completing the square:
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Let Rehman’s present age = 𝑥𝑥 years.
His age three years ago = (𝑥𝑥 − 3) years.
His age five years from now = (𝑥𝑥 + 5) years.
As per the question, the sum of the reciprocals of Rehman's ages 3 years ago and 5 years from now is 1
3.
∴1
𝑥𝑥 − 3+
1𝑥𝑥 + 5
=13
𝑥𝑥 + 5 + 𝑥𝑥 − 3(𝑥𝑥 − 3)(𝑥𝑥 + 5) =
13
2𝑥𝑥 + 2(𝑥𝑥 − 3)(𝑥𝑥 + 5) =
13
⇒ 3(2𝑥𝑥 + 2) = (𝑥𝑥 − 3)(𝑥𝑥 + 5)
⇒ 6𝑥𝑥 + 6 = 𝑥𝑥2 + 2𝑥𝑥 − 15
⇒ 𝑥𝑥2 − 4𝑥𝑥 − 21 = 0
⇒ 𝑥𝑥2 − 7𝑥𝑥 + 3𝑥𝑥 − 21 = 0
⇒ 𝑥𝑥(𝑥𝑥 − 7) + 3(𝑥𝑥 − 7) = 0
⇒ (𝑥𝑥 − 7)( 𝑥𝑥 + 3) = 0
⇒ 𝑥𝑥 = 7 or 𝑥𝑥 = −3
It’s clear that age is always positive.
Thus, Rehman's present age is 7 years.
5. In a class test, the sum of Shefali’s marks in Mathematics and English is 30. Had she got 2 marks more in Mathematics and 3 marks less in English, the product of their marks would have been 210. Find her marks in the two subjects.
Solution:
Let Shefali’s marks in Mathematics = 𝑥𝑥.
Then, her marks in English = 30 − 𝑥𝑥. (Given in question)
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⇒ 𝑥𝑥2 − 12𝑥𝑥 − 13𝑥𝑥 + 156 = 0
⇒ 𝑥𝑥(𝑥𝑥 + 2) − 13(𝑥𝑥 − 12) = 0
⇒ (𝑥𝑥 − 12)(𝑥𝑥 − 13) = 0
⇒ 𝑥𝑥 = 12 or 𝑥𝑥 = 13
If the marks in Mathematics is 12, then marks in English will be 30 − 12 = 18
If the marks in Mathematics is 13,then marks in English will be 30 − 13 = 17
6. The diagonal of a rectangular field is 60 metres more than the shorter side. If the longer side is 30 meters more than the shorter side, find the sides of the field.
Solution:
Let the shorter side of the rectangle be 𝑥𝑥 m.
Then, longer side of the rectangle = (𝑥𝑥 + 30)m
Diagonal of the rectangle = �𝑥𝑥2 + (𝑥𝑥 + 30)2 (By Pythagoras theorem)
But in question, it is given that the diagonal of the rectangular field is 60 𝑚𝑚 more than the shorter side.
∴ �𝑥𝑥2 + (𝑥𝑥 + 30)2 = 𝑥𝑥 + 60
⇒ 𝑥𝑥2 + (𝑥𝑥 + 30)2 = (𝑥𝑥 + 60)2 (By squaring on both the sides)
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But, side of a rectangle cannot be negative. So, the length of the shorter side is 90 m. Thus, the length of the longer side will be (90 + 30) m = 120 m
7. The difference of squares of two numbers is 180. The square of the smaller number is 8 times the larger number. Find the two numbers.
Solution:
Let the larger and smaller number be 𝑥𝑥 and 𝑦𝑦 respectively.
It is given in the question that,
𝑥𝑥2 − 𝑦𝑦2 = 180 and 𝑦𝑦2 = 8𝑥𝑥
⇒ 𝑥𝑥2 − 8𝑥𝑥 = 180
⇒ 𝑥𝑥2 − 8𝑥𝑥 − 180 = 0
⇒ 𝑥𝑥2 − 18𝑥𝑥 + 10𝑥𝑥 − 180 = 0
⇒ 𝑥𝑥(𝑥𝑥 − 18) + 10(𝑥𝑥 − 18) = 0
⇒ (𝑥𝑥 − 18)(𝑥𝑥 + 10) = 0
⇒ 𝑥𝑥 = 18,−10
If larger number, 𝑥𝑥 = −10
then smaller number, 𝑦𝑦 = ±√8𝑥𝑥
= ±�8(−10)
= ±√−80
Since we cannot have negative number in roots
𝑥𝑥 = −10 is not possible
Therefore, the larger number will be 18 only.
𝑥𝑥 = 18
∴ 𝑦𝑦2 = 8𝑥𝑥 = 8 × 18 = 144
⇒ 𝑦𝑦 = ±√144 = ±12
∴ Smaller number = ±12
Therefore, the numbers are 18 and 12 or 18 and −12.
8. A train travels 360 km at a uniform speed. If the speed had been 5 km h⁄ more, it would have taken 1 hour less for the same journey. Find the speed of the train.
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⇒ 150𝑥𝑥 − 750 = 8𝑥𝑥2 − 80𝑥𝑥
⇒ 8𝑥𝑥2 − 230𝑥𝑥 + 750 = 0
⇒ 8𝑥𝑥2 − 200𝑥𝑥 − 30𝑥𝑥 + 750 = 0
⇒ 8𝑥𝑥(𝑥𝑥 − 25) − 30(𝑥𝑥 − 25) = 0
⇒ (𝑥𝑥 − 25)(8𝑥𝑥 − 30) = 0
𝑖𝑖. 𝑒𝑒. , 𝑥𝑥 = 25 or 𝑥𝑥 = 308
= 154
Taking 𝑥𝑥 = 154
Time taken by smaller tap
= 𝑥𝑥 = 154
hrs
Time taken by larger tap = 𝑥𝑥 − 10
= 154− 10 = 15−40
4= −25
4
Since time is negative,
𝑥𝑥 = 154
is not the solution
Thus, the time taken separately by the smaller diameter tap and the larger diameter tap will be 25 hours and (25 − 10) = 15 hours respectively.
10. An express train takes 1 hour less than a passenger train to travel 132 km between Mysore and Bangalore (without taking into consideration the time they stop at intermediate stations). If the average speed of the express train is 11 km h⁄ more than that of the passenger train, find the average speed of the two trains.
Solution:
Let the average speed of passenger train be 𝑥𝑥 km/h.
Average speed of express train = (𝑥𝑥 + 11) km/h (Given in question)
According to the question, the time taken by the express train to cover 132 km is
1 hour less than a passenger train.
Therefore, time taken by passenger train – time taken by express train = 1 hour
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⇒132 × 11𝑥𝑥(𝑥𝑥 + 11)
= 1
⇒ 132 × 11 = 𝑥𝑥(𝑥𝑥 + 11)
⇒ 𝑥𝑥2 + 11𝑥𝑥 − 1452 = 0
⇒ 𝑥𝑥2 + 44𝑥𝑥 − 33𝑥𝑥 − 1452 = 0
⇒ 𝑥𝑥(𝑥𝑥 + 44) − 339𝑥𝑥 + 44) = 0
⇒ (𝑥𝑥 + 44)(𝑥𝑥 − 33) = 0
⇒ 𝑥𝑥 = −44, 33
Average speed of passenger train cannot be negative. Hence, the speed of the passenger train is 33km/h and thus, the speed of the express train will be 33 +11 = 44 km/h.
11. Sum of the areas of two squares is 468 m2. If the difference of their perimeters is 24 m, find the sides of the two squares.
Solution:
Let the sides of the two squares be 𝑥𝑥 m and 𝑦𝑦 m.
So, their perimeter will be 4𝑥𝑥 and 4𝑦𝑦 respectively and their areas will be 𝑥𝑥2and 𝑦𝑦2 respectively.