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ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 5 Quadratic Equations in One Variable Chapter Test

These Solutions are part of ML Aggarwal Class 10 Solutions for ICSE Maths. Here we have given ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 5 Quadratic Equations in One Variable Chapter Test

More Exercises

• ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 5 Quadratic Equations in One Variable Ex 5.1
• ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 5 Quadratic Equations in One Variable Ex 5.2
• ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 5 Quadratic Equations in One Variable Ex 5.3
• ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 5 Quadratic Equations in One Variable Ex 5.4
• ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 5 Quadratic Equations in One Variable Ex 5.5
• ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 5 Quadratic Equations in One Variable MCQS
• ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 5 Quadratic Equations in One Variable Chapter Test

Solve the following equations (1 to 4) by factorisation :

Question 1.
(i) x² + 6x – 16 = 0
(ii) 3x² + 11x + 10 = 0
Solution:
x² + 6x – 16 = 0
⇒ x² + 8x – 2x – 16 = 0
⇒ x (x + 8) – 2 (x + 8) = 0

Question 2.
(i) 2x² + ax – a² = 0
(ii) √3x² + 10x + 7√3 = 0
Solution:
(i) 2x² + ax – a² = 0
⇒ 2x² + 2ax – ax – a² = 0

Question 3.
(i) x(x + 1) + (x + 2)(x + 3) = 42
(ii) \(\frac { 6 }{ x } -\frac { 2 }{ x-1 } =\frac { 1 }{ x-2 } \)
Solution:
(i) x(x + 1) + (x + 2)(x + 3) = 42
⇒ 2x² + 6x + 6 – 42 = 0

Question 4.
(i)\(\sqrt { x+15 } =x+3 \)
(ii)\(\sqrt { { 3x }^{ 2 }-2x-1 } =2x-2\)
Solution:
(i) \(\sqrt { x+15 } =x+3 \)
Squaring on both sides
x + 15 = (x + 3)²

Solve the following equations (5 to 8) by using formula :

Question 5.
(i) 2x² – 3x – 1 = 0
(ii) \(x\left( 3x+\frac { 1 }{ 2 } \right) =6\)
Solution:
(i) 2x² – 3x – 1 = 0
Here a = 2, b = -3, c = -1

Question 6.
(i) \(\frac { 2x+5 }{ 3x+4 } =\frac { x+1 }{ x+3 } \)
(ii) \(\frac { 2 }{ x+2 } -\frac { 1 }{ x+1 } =\frac { 4 }{ x+4 } -\frac { 3 }{ x+3 } \)
Solution:
(i) \(\frac { 2x+5 }{ 3x+4 } =\frac { x+1 }{ x+3 } \)
(2x + 5)(x + 3) = (x + 1)(3x + 4)

Question 7.
(i) \(\frac { 3x-4 }{ 7 } +\frac { 7 }{ 3x-4 } =\frac { 5 }{ 2 } ,x\neq \frac { 4 }{ 3 } \)
(ii) \(\frac { 4 }{ x } -3=\frac { 5 }{ 2x+3 } ,x\neq 0,-\frac { 3 }{ 2 } \)
Solution:
(i) \(\frac { 3x-4 }{ 7 } +\frac { 7 }{ 3x-4 } =\frac { 5 }{ 2 } ,x\neq \frac { 4 }{ 3 } \)
let \(\frac { 3x-4 }{ 7 } \) = y,then

Question 8.
(i)x² + (4 – 3a)x – 12a = 0
(ii)10ax² – 6x + 15ax – 9 = 0,a≠0
Solution:
(i)x² + (4 – 3a)x – 12a = 0
Here a = 1, b = 4 – 3a, c = -12a

Question 9.
Solve for x using the quadratic formula. Write your answer correct to two significant figures: (x – 1)² – 3x + 4 = 0. (2014)
Solution:
(x – 1)² – 3x + 4 = 0
x² + 1 – 2x – 3x + 4 = 0

Question 10.
Discuss the nature of the roots of the following equations:
(i) 3x² – 7x + 8 = 0
(ii) x² – \(\\ \frac { 1 }{ 2 } x\) – 4 = 0
(iii) 5x² – 6√5x + 9 = 0
(iv) √3x² – 2x – √3 = 0
Solution:
(i) 3x² – 7x + 8 = 0
Here a = 3, b = -7, c = 8

Question 11.
Find the values of k so that the quadratic equation (4 – k) x² + 2 (k + 2) x + (8k + 1) = 0 has equal roots.
Solution:
(4 – k) x² + 2 (k + 2) x + (8k + 1) = 0
Here a = (4 – k), b = 2 (k + 2), c = 8k + 1

or k – 3 = 0, then k= 3
k = 0, 3 Ans.

Question 12.
Find the values of m so that the quadratic equation 3x² – 5x – 2m = 0 has two distinct real roots.
Solution:
3x² – 5x – 2m = 0
Here a = 3, b = -5, c = -2m

Question 13.
Find the value(s) of k for which each of the following quadratic equation has equal roots:
(i)3kx² = 4(kx – 1)
(ii)(k + 4)x² + (k + 1)x + 1 =0
Also, find the roots for that value (s) of k in each case.
Solution:
(i)3kx² = 4(kx – 1)
⇒ 3kx² = 4kx – 4
⇒ 3kx² – 4kx + 4 = 0

Question 14.
Find two natural numbers which differ by 3 and whose squares have the sum 117.
Solution:
Let first natural number = x
then second natural number = x + 3
According to the condition :
x² + (x + 3)² = 117

Question 15.
Divide 16 into two parts such that the twice the square of the larger part exceeds the square of the smaller part by 164.
Solution:
Let larger part = x
then smaller part = 16 – x
(∵ sum = 16)
According to the condition

Question 16.
Two natural numbers are in the ratio 3 : 4. Find the numbers if the difference between their squares is 175.
Solution:
Ratio in two natural numbers = 3 : 4
Let the numbers be 3x and 4x
According to the condition,

Question 17.
Two squares have sides A cm and (x + 4) cm. The sum of their areas is 656 sq. cm.Express this as an algebraic equation and solve it to find the sides of the squares.
Solution:
Side of first square = x cm .
and side of second square = (x + 4) cm
Now according to the condition,

or x – 16 = 0 then x = 16
Side of first square = 16 cm
and side of second square = 16 + 4 – 4 = 20 cm

Question 18.
The length of a rectangular garden is 12 m more than its breadth. The numerical value of its area is equal to 4 times the numerical value of its perimeter. Find the dimensions of the garden.
Solution:
Let breadth = x m
then length = (x + 12) m
Area = l × b = x (x + 12) m²
and perimeter = 2 (l + b) = 2(x + 12 + x) = 2 (2x + 12) m
According to the condition.

Question 19.
A farmer wishes to grow a 100 m² rectangular vegetable garden. Since he has with him only 30 m barbed wire, he fences three sides of the rectangular garden letting compound wall of his house act as the fourth side fence. Find the dimensions of his garden.
Solution:
Area of rectangular garden = 100 cm²
Length of barbed wire = 30 m
Let the length of the side opposite to wall = x

Question 20.
The hypotenuse of a right-angled triangle is 1 m less than twice the shortest side. If the third side is 1 m more than the shortest side, find the sides of the triangle.
Solution:
Let the length of shortest side = x m
Length of hypotenuse = 2x – 1
and third side = x + 1
Now according to the condition,

Question 21.
A wire ; 112 cm long is bent to form a right angled triangle. If the hypotenuse is 50 cm long, find the area of the triangle.
Solution:
Perimeter of a right angled triangle = 112 cm
Hypotenuse = 50 cm
∴ Sum of other two sides = 112 – 50 = 62 cm
Let the length of first side = x
and length of other side = 62 – x

Question 22.
Car A travels x km for every litre of petrol, while car B travels (x + 5) km for every litre of petrol.
(i) Write down the number of litres of petrol used by car A and car B in covering a distance of 400 km.
(ii) If car A uses 4 litres of petrol more than car B in covering 400 km. write down an equation, in A and solve it to determine the number of litres of petrol used by car B for the journey.
Solution:
Distance travelled by car A in one litre = x km
and distance travelled by car B in one litre = (x + 5) km
(i) Consumption of car A in covering 400 km

Question 23.
The speed of a boat in still water is 11 km/ hr. It can go 12 km up-stream and return downstream to the original point in 2 hours 45 minutes. Find the speed of the stream
Solution:
Speed of a boat in still water = 11 km/hr
Let the speed of stream = x km/hr.
Distance covered = 12 km.
Time taken = 2 hours 45 minutes

Question 24.
By selling an article for Rs. 21, a trader loses as much per cent as the cost price of the article. Find the cost price.
Solution:
S.P. of an article = Rs. 21
Let cost price = Rs. x
Then loss = x%

Question 25.
A man spent Rs. 2800 on buying a number of plants priced at Rs x each. Because of the number involved, the supplier reduced the price of each plant by Rupee 1.The man finally paid Rs. 2730 and received 10 more plants. Find x.
Solution:
Amount spent = Rs. 2800
Price of each plant = Rs. x
Reduced price = Rs. (x – 1)

Question 26.
Forty years hence, Mr. Pratap’s age will be the square of what it was 32 years ago. Find his present age.
Solution:
Let Partap’s present age = x years
40 years hence his age = x + 40
and 32 years ago his age = x – 32
According to the condition

Hope given ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 5 Quadratic Equations in One Variable Chapter Test are helpful to complete your math homework.

If you have any doubts, please comment below. Learn Insta try to provide online math tutoring for you.

ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 5 Quadratic Equations in One Variable MCQS

These Solutions are part of ML Aggarwal Class 10 Solutions for ICSE Maths. Here we have given ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 5 Quadratic Equations in One Variable MCQS

More Exercises

• ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 5 Quadratic Equations in One Variable Ex 5.1
• ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 5 Quadratic Equations in One Variable Ex 5.2
• ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 5 Quadratic Equations in One Variable Ex 5.3
• ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 5 Quadratic Equations in One Variable Ex 5.4
• ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 5 Quadratic Equations in One Variable Ex 5.5
• ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 5 Quadratic Equations in One Variable MCQS
• ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 5 Quadratic Equations in One Variable Chapter Test

Choose the correct answer from the given four options (1 to 15) :

Question 1.
Which of the following is not a quadratic equation ?
(a) (x + 2)² = 2(x + 3)
(b) x² + 3x = ( – 1) (1 – 3x)
(c) (x + 2) (x – 1) = x² – 2x – 3
(d) x³ – x² + 2x + 1 = (x + 1)³
Solution:
(a) (x + 2)² = 2(x + 3)
⇒ x² + 4x + 4 = 2x + 6
⇒ x² + 4x – 2x + 4 – 6 = 0
⇒ x² + 2x – 2
It is a quadratic equation.

Question 2.
Which of the following is a quadratic equation ?
(a) (x – 2) (x + 1) = (x – 1) (x – 3)
(b) (x + 2)³ = 2x(x² – 1)
(c) x² + 3x + 1 = (x – 2)²
(d) 8(x – 2)³ = (2x – 1)³ + 3
Solution:
(a) (x – 2) (x + 1) = (x – 1) (x – 3)
⇒ x² + x – 2x – 2 = x² – 3x – x + 3
⇒ 3x + x – 2x + x = 3 + 2
⇒ 3x = 5
It is not a quadratic equation.

Question 3.
Which of the following equations has 2 as a root ?
(a) x² – 4x + 5 = 0
(b) x² + 3x – 12 = 0
(c) 2x² – 7x + 6 = 0
(d) 3x² – 6x – 2 = 0
Solution:
(a) x² – 4x + 5 = 0
⇒ (2)² – 4x² + 5 = 0
⇒ 4 – 8 + 5 = 0
⇒ 9 – 8 ≠ 0
2 is not its root.

Question 4.
If \(\\ \frac { 1 }{ 2 } \) is a root of the equation x² + kx – \(\\ \frac { 5 }{ 4 } \) = 0, then the value of k is
(a) 2
(b) – 2
(c) \(\\ \frac { 1 }{ 4 } \)
(d) \(\\ \frac { 1 }{ 2 } \)
Solution:
\(\\ \frac { 1 }{ 2 } \) is a root of the equation
x² + kx – \(\\ \frac { 5 }{ 4 } \) = 0
Substituting the value of x = \(\\ \frac { 1 }{ 2 } \) in the

Question 5.
If \(\\ \frac { 1 }{ 2 } \) is a root of the quadratic equation 4x² – 4kx + k + 5 = 0, then the value of k is
(a) – 6
(b) – 3
(c) 3
(d) 6
Solution:
\(\\ \frac { 1 }{ 2 } \) is a root of the equation
4x² – 4kx + k + 5 = 0

Question 6.
The roots of the equation x² – 3x – 10 = 0 are
(a) 2,- 5
(b) – 2, 5
(c) 2, 5
(d) – 2, – 5
Solution:
x² – 3x – 10 = 0

x = 5, – 2 or – 2, 5 (b)

Question 7.
If one root of a quadratic equation with rational coefficients is \(\frac { 3-\sqrt { 5 } }{ 2 } \), then the other
(a)\(\frac { -3-\sqrt { 5 } }{ 2 } \)
(b)\(\frac { -3+\sqrt { 5 } }{ 2 } \)
(c)\(\frac { 3+\sqrt { 5 } }{ 2 } \)
(d)\(\frac { \sqrt { 3 } +5 }{ 2 } \)
Solution:
One root of a quadratic equation is \(\frac { 3-\sqrt { 5 } }{ 2 } \)
then other root will be \(\frac { 3+\sqrt { 5 } }{ 2 } \) (c)

Question 8.
If the equation 2x² – 5x + (k + 3) = 0 has equal roots then the value of k is
(a)\(\\ \frac { 9 }{ 8 } \)
(b)\(– \frac { 9 }{ 8 } \)
(c)\(\\ \frac { 1 }{ 8 } \)
(d)\(– \frac { 1 }{ 8 } \)
Solution:
2x² – 5x + (k + 3) = 0
a = 2, b = -5, c = k + 3

Question 9.
The value(s) of k for which the quadratic equation 2x² – kx + k = 0 has equal roots is (are)
(a) 0 only
(b) 4
(c) 8 only
(d) 0, 8
Solution:
2x² – kx + k = 0
a = 2, b = -k, c = k

Question 10.
If the equation 3x² – kx + 2k =0 roots, then the the value(s) of k is (are)
(a) 6
(b) 0 Only
(c) 24 only
(d) 0
Solution:
3x² – kx + 2k = 0
Here, a = 3, b = -k, c = 2k

Question 11.
If the equation {k + 1)x² – 2(k – 1)x + 1 = 0 has equal roots, then the values of k are
(a) 1, 3
(b) 0, 3
(c) 0, 1
(d) 0, 1
Solution:
(k + 1)x² – 2(k – 1)x + 1 = 0
Here, a = k + 1, b = -2(k – 1), c = 1

k = 0, 3 (b)

Question 12.
If the equation 2x² – 6x + p = 0 has real and different roots, then the values ofp are given by
(a)p < \(\\ \frac { 9 }{ 2 } \)
(b)p ≤ \(\\ \frac { 9 }{ 2 } \)
(c)p > \(\\ \frac { 9 }{ 2 } \)
(d)p ≥ \(\\ \frac { 9 }{ 2 } \)
Solution:
2x² – 6x + p = 0
Here, a = 2, b = -6, c = p

Question 13.
The quadratic equation 2x² – √5x + 1 = 0 has
(a) two distinct real roots
(b) two equal real roots
(c) no real roots
(d) more than two real roots
Solution:
2x² – √5x + 1 = 0
Here, a = 2, b = -√5, c = 1

Question 14.
Which of the following equations has two distinct real roots ?
(a) 2x² – 3√2x + \(\\ \frac { 9 }{ 4 } \) = 0
(b) x² + x – 5 = 0
(c) x² + 3x + 2√2 = 0
(d) 5x² – 3x + 1 = 0
Solution:
(a) 2x² – 3√2x + \(\\ \frac { 9 }{ 4 } \) = 0
b² – 4ac = ( -3√2)² – 4 x 2 x \(\\ \frac { 9 }{ 4 } \) = 18 – 18 = 0
.’. Roots are real and equal.

Question 15.
Which of the following equations has no real roots ?
(a) x² – 4x + 3√2 = 0
(b) x² + 4x – 3√2 = 0
(c) x² – 4x – 3√2 = 0
(d) 3x² + 4√3x + 4 = 0
Solution:
(a) x² – 4x + 3√2 = 0
b² – 4ac = ( -4)² – 4 × 1 × 3√2
= 16 – 12√2
= 16 – 12(1.4)
= 16 – 16.8
= -0.8
b² – 4ac < 0
Roots are not real. (a)

Hope given ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 5 Quadratic Equations in One Variable MCQS are helpful to complete your math homework.

If you have any doubts, please comment below. Learn Insta try to provide online math tutoring for you.

ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 5 Quadratic Equations in One Variable Ex 5.5

These Solutions are part of ML Aggarwal Class 10 Solutions for ICSE Maths. Here we have given ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 5 Quadratic Equations in One Variable Ex 5.5

More Exercises

• ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 5 Quadratic Equations in One Variable Ex 5.1
• ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 5 Quadratic Equations in One Variable Ex 5.2
• ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 5 Quadratic Equations in One Variable Ex 5.3
• ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 5 Quadratic Equations in One Variable Ex 5.4
• ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 5 Quadratic Equations in One Variable Ex 5.5
• ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 5 Quadratic Equations in One Variable MCQS
• ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 5 Quadratic Equations in One Variable Chapter Test

Question 1.
(i) Find two consecutive natural numbers such that the sum of their squares is 61.
(ii) Find two consecutive integers such that the sum of their squares is 61.
Solution:
Let the first natural number = x
then second natural number = x + 1
According to the condition, (x)² + (x + 1)² = 61

Question 2.
(i) If the product of two positive consecutive even integers is 288, find the integers.
(ii) If the product of two consecutive even integers is 224, find the integers.
(iii) Find two consecutive even natural numbers such that the sum of their squares is 340.
(iv) Find two consecutive odd integers such that the sum of their squares is 394.
Solution:
(i) Let first positive even integer = 2x
then second even integer = 2x + 2
According to the condition,
2x × (2x + 2) = 288
⇒ 4x² + 4x – 288 = 0
⇒ x² + x – 72 = 0 (Dividing by 4)

Question 3.
The sum of two numbers is 9 and the sum of their squares is 41. Taking one number as x, form ail equation in x and solve it to find the numbers.
Solution:
Sum of two numbers = 9
Let first number = x
then second number = 9 – x
Now according to the condition,

Question 4.
Five times a certain whole number is equal to three less than twice the square of the number. Find the number.
Solution:
Let number = x
Now according to the condition,
5x = 2x² – 3

Question 5.
Sum of two natural numbers is 8 and the difference of their reciprocal is 2/15. Find the numbers.
Solution:
Let x and y be two numbers
Given that, x + y = 8 ……(i)

Question 6.
The difference between the squares of two numbers is 45. The square of the smaller number is 4 times the larger number. Determine the numbers.
Solution:
Let the larger number = x
then smaller number = y
Now according to the condition,

Question 7.
There are three consecutive positive integers such that the sum of the square of the first and the product of other two is 154. What are the integers?
Solution:
Let the first integer = x
then second integer = x + 1
and third integer = x + 2
Now according to the condition,

Question 8.
(i) Find three successive even natural numbers, the sum of whose squares is 308.
(ii) Find three consecutive odd integers, the sum of whose squares is 83.
Solution:
(i) Let first even number = 2x
second even number = 2x + 2
third even number = 2x + 4
Now according to the condition,

Question 9.
In a certain positive fraction, the denominator is greater than the numerator by 3. If 1 is subtracted from both the numerator and denominator, the fraction is decreased by \(\\ \frac { 1 }{ 14 } \). Find the fraction.
Solution:
Let the numerator of a fraction = x
then denominator = x + 3
then fraction = \(\\ \frac { x }{ x+3 } \)
Now according to the condition,

Question 10.
The sum of the numerator and denominator of a certain positive fraction is 8. If 2 is added to both the numerator and denominator, the fraction is increased by \(\\ \frac { 4 }{ 35 } \). Find the fraction.
Solution:
Let the denominator of a positive fraction = x
then numerator = 8 – x

Question 11.
A two digit number contains the bigger at ten’s place. The product of the digits is 27 and the difference between two digits is 6. Find the number.
Solution:
Let unit’s digit = x
then tens digit = x + 6
Number = x + 10(x + 6)
= x + 10x + 60
= 11x + 60

Question 12.
A two digit positive number is such that the product of its digits is 6. If 9 is added to the number, the digits interchange their places. Find the number. (2014)
Solution:
Let 2-digit number = xy = 10x + y
Reversed digits = yx = 10y + x
According to question,

Question 13.
A rectangle of area 105 cm² has its length equal to x cm. Write down its breadth in terms of x. Given that the perimeter is 44 cm, write down an equation in x and solve it to determine the dimensions of the rectangle.
Solution:
Perimeter of rectangle = 44 cm
length + breadth = \(\\ \frac { 44 }{ 2 } \) = 22 cm
Let length = x
then breadth = 22 – x
According to the condition,

Question 14.
A rectangular garden 10 m by 16 m is to be surrounded by a concrete walk of uniform width. Given that the area of the walk is 120 square metres, assuming the width of the walk to be x, form an equation in x and solve it to find the value of x. (1992)
Solution:
Length of garden = 16 m
and width = 10 m
Let the width of walk = x m
Outer length = 16 + 2x
and outer width = 10 + 2x
Now according to the condition,

Question 15.
(i) Harish made a rectangular garden, with its length 5 metres more than its width. The next year, he increased the length by 3 metres and decreased the width by 2 metres. If the area of the second garden was 119 sq m, was the second garden larger or smaller ?
(ii) The length of a rectangle exceeds its breadth by 5 m. If the breadth were doubled and the length reduced by 9 m, the area of the rectangle would have increased by 140 m². Find its dimensions.
Solution:
In first case,
Let length of the garden = x m
then width = (x – 5) m
Area = l x b = x(x – 5) sq. m
In second case,

Question 16.
The perimeter of a rectangular plot is 180 m and its area is 1800 m². Take the length of the plot as x m. Use the perimeter 180 m to write the value of the breadth in terms of x. Use the values of length, breadth and the area to write an equation in x. Solve the equation to calculate the length and breadth of the plot. (1993)
Solution:
The perimeter of a rectangular field = 180 m
and area = 1800 m²
Let length = x m

Question 17.
The lengths of the parallel sides of a trapezium are (x + 9) cm and (2x – 3) cm and the distance between them is (x + 4) cm. If its area is 540 cm², find x.
Solution:
Area of a trapezium = \(\\ \frac { 1 }{ 2 } \)
(sum of parallel sides) x height
Lengths of parallel sides are (x + 9) and (2x – 3)
and height = (x + 4)
According to the condition,

Question 18.
If the perimeter of a rectangular plot is 68 m and the length of its diagonal is 26 m, find its area.
Solution:
Perimeter = 68 m and diagonal = 26 m
Length + breadth = \(\\ \frac { 68 }{ 2 } \) = 34 m
Let length = x m
then breadth = (34 – x) m

Question 19.
If the sum of two smaller sides of a right – angled triangle is 17cm and the perimeter is 30cm, then find the area of the triangle.
Solution:
The perimeter of the triangle = 30 cm.
Let one of the two small sides = x
then, other side = 17 – x

Question 20.
The hypotenuse of grassy land in the shape of a right triangle is 1 metre more than twice the shortest side. If the third side is 7 metres more than the shortest side, find the sides of the grassy land.
Solution:
Let the shortest side = x
Hypotenuse = 2x + 1
and third side = x + 7
According to the condition,

Question 21.
Mohini wishes to fit three rods together in the shape of a right triangle. If the hypotenuse is 2 cm longer than the base and 4 cm longer than the shortest side, find the lengths of the rods.
Solution:
Let the length of hypotenuse = x cm
then base = (x – 2) cm
and shortest side = x – 4
According to the condition,

Question 22.
In a P.T. display, 480 students are arranged in rows and columns. If there are 4 more students in each row than the number of rows, find the number of students in each row.
Solution:
Total number of students = 480
Let the number of students in each row = x

Question 23.
In an auditorium, the number of rows are equal to the number of seats in each row.If the number of rows is doubled and number of seats in each row is reduced by 5, then the total number of seats is increased by 375. How many rows were there?
Solution:
Let the number of rows = x
then no. of seats in each row = x
and total number of seats = x × x = x²
According to the condition,

Question 24.
At an annual function of a school, each student gives the gift to every other student. If the number of gifts is 1980, find the number of students.
Solution:
Let the number of students = x
then the number of gifts given = x – 1
Total number of gifts = x (x – 1)
According to the condition,
x (x – 1) = 1980

Question 25.
A bus covers a distance of 240 km at a uniform speed. Due to heavy rain, its speed gets reduced by 10 km/h and as such it takes two hours longer to cover the total distance. Assuming the uniform speed to be ‘x’ km/h, form an equation and solve it to evaluate x. (2016)
Solution:
Distance = 240 km
Let speed of a bus = x km/hr

Question 26.
The speed of an express train is x km/hr and the speed of an ordinary train is 12 km/hr less than that of the express train. If the ordinary train takes one hour longer than the express train to cover a distance of 240 km, find the speed of the express train.
Solution:
Let the speed of express train = x km
Then speed of the ordinary train = (x – 12) km
Time is taken to cover 240 km by the express

Question 27.
A car covers a distance of 400 km at a certain speed. Had the speed been 12 km/h more, the time taken for the journey would have been 1 hour 40 minutes less. Find the original speed of the car. (1996)
Solution:
Let the original speed of the car = x km/h.
Distance covered = 400 km

Question 28.
An aeroplane travelled a distance of 400 km at an average speed of x km/hr. On the return journey, the speed was increased by 40 km/hr. Write down an expression for the time taken for
(i)the onward journey,
(ii) the return journey.
If the return journey took 30 minutes less than the onward journey, write down an equation in x and find its value. (2002)
Solution:
Distance = 400 km
Speed of aeroplane = x km/hr

Question 29.
The distance by road between two towns A and B, is 216 km, and by rail it is 208 km. A car travels at a speed of x km/hr, and the train travels at a speed which is 16 km/hr faster than the car. Calculate :
(i) The time taken by the car, to reach town B from A, in terms of x ;
(ii) The time taken by the train, to reach town B from A, in terms of x ;
(iii) If the train takes 2 hours less than the car, to reach town B, obtain an equation in x and solve it.
(iv) Hence find the speed of the train. (1998)
Solution:
The distance by road between A and B = 216 km
and the distance by rail = 208 km
speed of car = x km/hr

Question 30.
An aeroplane flying with a wind of 30 km/hr takes 40 minutes less to fly 3600 km, than what it would have taken to fly against the same wind. Find the planes speed of flying in still air.
Solution:
Let the speed of the plane in still air = x km/hr
Speed of wind = 30 km/hr
Distance = 3600 km

Question 31.
A school bus transported an excursion party to a picnic spot 150 km away. While returning, it was raining and the bus had to reduce its speed by 5 km/hr, and it took one hour longer to make the return trip. Find the time taken to return.
Solution:
Distance = 150 km
Let the speed of bus = x km/hr

Question 32.
A boat can cover 10 km up the stream and 5 km down the stream in 6 hours. If the speed of the stream is 1.5 km/hr. find the speed of the boat in still water.
Solution:
Distance up stream = 10 km
and down stream = 5 km
Total time is taken = 6 hours
Speed of stream = 1.5 km/hr
Let the speed of a boat in still water = x km/hr
According to the condition,

Question 33.
Two pipes running together can fill a tank in \({ 11 }^{ 1/9 }\) minutes. If one pipe takes 5 minutes more than the other to fill the tank, find the time in which each pipe would/fill the tank.
Solution:
Let the time taken by one pipe = x minutes
Then time taken by second pipe = (x + 5) minutes
Time taken by both pipes = \({ 11 }^{ 1/9 }\) minutes
Now according to the condition.

Question 34.
(i) Rs. 480 is divided equally among ‘x’ children. If the number of children was 20 more then each would have got Rs. 12 less. Find ‘x’.
(ii) Rs. 6500 is divided equally among a certain number of persons. Had there been 15 more persons, each would have got Rs. 30 less. Find the original number of persons.
Solution:
(i) Share of each child = Rs \(\\ \frac { 480 }{ x } \)
According to the question

Question 35.
2x articles cost Rs. (5x + 54) and (x + 2) similar articles cost Rs. (10x – 4), find x.
Solution:
Cost of 2x articles = 5x + 54
Cost of 1 article = \(\\ \frac { 5x+54 }{ 2x } \) ….(i)
Again cost of x + 2 articles = 10x – 4

Question 36.
A trader buys x articles for a total cost of Rs. 600.
(i) Write down the cost of one article in terms of x. If the cost per article were Rs. 5 more, the number of articles that can be bought for Rs. 600 would be four less.
(ii) Write down the equation in x for the above situation and solve it to find x. (1999)
Solution:
Total cost = Rs. 600,
No. of articles = x

Question 37.
A shopkeeper buys a certain number of books for Rs 960. If the cost per book was Rs 8 less, the number of books that could be bought for Rs 960 would be 4 more. Taking the original cost of each book to be Rs x, write an equation in x and solve it to find the original cost of each book.
Solution:
Let original cost = Rs x
No. of books bought = \(\\ \frac { 960 }{ x } \)
New cost of books = Rs (x – 8)

Question 38.
A piece of cloth costs Rs. 300. If the piece was 5 metres longer and each metre of cloth costs Rs. 2 less, the cost of the piece would have remained unchanged. How long is the original piece of cloth and what is the rate per metre?
Solution:
The total cost of cloth piece = Rs. 300
Let the length of the piece of cloth in the beginning = x m

Question 39.
The hotel bill for a number of people for an overnight stay is Rs. 4800. If there were 4 more, the bill each person had to pay would have reduced by Rs. 200. Find the number of people staying overnight. (2000)
Solution:
Let the number of people = x
Amount of bill = Rs. 4800

Question 40.
A person was given Rs. 3000 for a tour. If he extends his tour programme by 5 days, he must cut down his daily expenses by Rs. 20. Find the number of days of his tour programme.
Solution:
Let the number of days of tour programme = x
Amount = Rs. 3000

Question 41.
Ritu bought a saree for Rs. 60 x and sold it for Rs. (500 + 4x) at a loss of x%. Find the cost price.
Solution:
The cost price of saree = Rs. 60x
and selling price = Rs. (500 + 4x)
Loss = x%
Now according to the condition

Question 42.
(i) The sum of the ages of Vivek and his younger brother Amit is 47 years. The product of their ages in years is 550. Find their ages. (2017)
(ii) Paul is x years old and his father’s age is twice the square of Paul’s age. Ten years hence, the father’s age will be four times Paul’s age. Find their present ages.
Solution:
(i) Let Vivek’s present age be x years.
His brother’s age = (47 – x) years
According to question,
x(47 – x) = 550
⇒ 47x – x² = 550
⇒ x² – 47x + 550 = 0
⇒ x² – 25x – 22x + 550 = 0
⇒ x(x – 25) – 22(x – 25) – 0

Question 43.
The age of a man is twice the square of the age of his son. Eight years hence, the age of the man will be 4 years more than three times the age of his son. Find the present age.
Solution:
Let the present age of the son = x years
then, the present age of the man = 2x² years.
8 years hence,
The age of son will be = (x + 8) years and the
age of man = (2x² + 8) years
According to the problem,

Question 44.
Two years ago, a man’s age was three times the square of his daughter’s age. Three years hence, his age will be four times his daughter’s age. Find their present ages.
Solution:
2 years ago,
Let the age of daughter = x
age of man = 3x²
then present age of daughter = x + 2
and mean = 3x² + 2

Question 45.
The length (in cm) of the hypotenuse of a right-angled triangle exceeds the length of one side by 2 cm and exceeds twice the length of another side by 1 cm. Find the length of each side. Also, find the perimeter and the area of the triangle.
Solution:
Let the length of one side = x cm
and other side = y cm.
then hypotenues = x + 2, and 2y + 1

Question 46.
If twice the area of a smaller square is subtracted from the area of a larger square, the result is 14 cm². However, if twice the area of the larger square is added to three times the area of the smaller square, the result is 203 cm². Determine the sides of the two squares.
Solution:
Let the side of smaller square = x cm
and side of bigger square = y cm
According to the condition,

Hope given ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 5 Quadratic Equations in One Variable Ex 5.5 are helpful to complete your math homework.

If you have any doubts, please comment below. Learn Insta try to provide online math tutoring for you.

ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 5 Quadratic Equations in One Variable Ex 5.4

These Solutions are part of ML Aggarwal Class 10 Solutions for ICSE Maths. Here we have given ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 5 Quadratic Equations in One Variable Ex 5.4

More Exercises

• ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 5 Quadratic Equations in One Variable Ex 5.1
• ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 5 Quadratic Equations in One Variable Ex 5.2
• ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 5 Quadratic Equations in One Variable Ex 5.3
• ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 5 Quadratic Equations in One Variable Ex 5.4
• ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 5 Quadratic Equations in One Variable Ex 5.5
• ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 5 Quadratic Equations in One Variable MCQS
• ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 5 Quadratic Equations in One Variable Chapter Test

Question 1.
Find the discriminant of the following equations and hence find the nature of roots:
(i) 3x² – 5x – 2 = 0
(ii) 2x² – 3x + 5 = 0
(iii) 7x² + 8x + 2 = 0
(iv) 3x² + 2x – 1 = 0
(v) 16x² – 40x + 25 = 0
(vi) 2x² + 15x + 30 = 0.
Solution:
(i) 3x² – 5x – 2 = 0
Here a = 3, b = -5, c = -2

Question 2.
Discuss the nature of the roots of the following quadratic equations :
(i) x² – 4x – 1 = 0
(ii) 3x² – 2x + \(\\ \frac { 1 }{ 3 } \) = 0
(iii) 3x² – 4√3x + 4 = 0
(iv) x² – \(\\ \frac { 1 }{ 2 } x\) + 4 = 0
(v) – 2x² + x + 1 = 0
(vi) 2√3x² – 5x + √3 = 0
Solution:
(i) x² – 4x – 1 = 0
Here a = 1, b = -4, c = -1

Question 3.
Find the nature of the roots of the following quadratic equations:
(i) x² – \(\\ \frac { 1 }{ 2 } x\) – \(\\ \frac { 1 }{ 2 } \) = 0
(ii) x² – 2√3x – 1 = 0 If real roots exist, find them.
Solution:
(i) x² – \(\\ \frac { 1 }{ 2 } x\) – \(\\ \frac { 1 }{ 2 } \) = 0
Here a = 1, b = \(– \frac { 1 }{ 2 } \), c = \(– \frac { 1 }{ 2 } \)

Question 4.
Without solving the following quadratic equation, find the value of ‘p’ for which the given equations have real and equal roots:
(i) px² – 4x + 3 = 0
(ii) x² + (p – 2)x + p = 0.
Solution:
(i) px² – 4x + 3 = 0
Here a = p, b = -4, c = 3

Question 5.
Find the value (s) of k for which each of the following quadratic equation has equal roots :
(i) kx² – 4x – 5 = 0
(ii) (k – 4) x² + 2(k – 4) x + 4 = 0
Solution:
(i) kx² – 4x – 5 = 0
Here a = k, b = -4, c = 5

Question 6.
Find the value(s) of m for which each of the following quadratic equation has real and equal roots:
(i) (3m + 1)x² + 2(m + 1)x + m = 0
(ii) x² + 2(m – 1) x + (m + 5) = 0
Solution:
(i) (3m + 1)x² + 2(m + 1)x + m = 0
Here a = 3m + 1, b = 2(m + 1), c = m

Question 7.
Find the values of k for which each of the following quadratic equation has equal roots:
(i) 9x² + kx + 1 = 0
(ii) x² – 2kx + 7k – 12 = 0
Also, find the roots for those values of k in each case.
Solution:
(i) 9x² + kx + 1 = 0
Here a = 9, b = k, c = 1

Question 8.
Find the value(s) of p for which the quadratic equation (2p + 1)x² – (7p + 2)x + (7p – 3) = 0 has equal roots. Also find these roots.
Solution:
The quadratic equation given is (2p + 1)x² – (7p + 2)x + (7p – 3) = 0
Comparing with ax² + bx + c = 0, we have

Question 9.
If – 5 is a root of the quadratic equation 2x² + px – 15 = 0 and the quadratic equation p(x² + x) + k = 0 has equal roots, find the value of k.
Solution:
-5 is a root of the quadratic equation
2x² + px – 15 = 0, then
⇒ 2(5)² – p( -5) – 15 = 0
⇒ 50 – 5p – 15 = 0
⇒ 35 – 5p = 0

Question 10.
Find the value(s) of p for which the equation 2x² + 3x + p = 0 has real roots.
Solution:
2x² + 3x + p = 0
Here, a = 2, b = 3, c = p

Question 11.
Find the least positive value of k for which the equation x² + kx + 4 = 0 has real roots.
Solution:
x² + kx + 4 = 0
Here, a = 1, b = k, c = 4

Question 12.
Find the values of p for which the equation 3x² – px + 5 = 0 has real roots.
Solution:
3x² – px + 5 = 0
Here, a = 3, b = -p, c = 5

Hope given ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 5 Quadratic Equations in One Variable Ex 5.4 are helpful to complete your math homework.

If you have any doubts, please comment below. Learn Insta try to provide online math tutoring for you.

ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 5 Quadratic Equations in One Variable Ex 5.3

These Solutions are part of ML Aggarwal Class 10 Solutions for ICSE Maths. Here we have given ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 5 Quadratic Equations in One Variable Ex 5.3

More Exercises

• ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 5 Quadratic Equations in One Variable Ex 5.1
• ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 5 Quadratic Equations in One Variable Ex 5.2
• ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 5 Quadratic Equations in One Variable Ex 5.3
• ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 5 Quadratic Equations in One Variable Ex 5.4
• ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 5 Quadratic Equations in One Variable Ex 5.5
• ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 5 Quadratic Equations in One Variable MCQS
• ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 5 Quadratic Equations in One Variable Chapter Test

Solve the following (1 to 8) equations by using formula :

Question 1.
(i) 2x² – 7x + 6 = 0
(ii) 2x² – 6x + 3 = 0
Solution:
(i) 2x² – 7x + 6 = 0
Here a = 2, b = -7, c = 6

Question 2.
(i) x² + 7x – 7 = 0
(ii) (2x + 3)(3x – 2) + 2 = 0
Solution:
(i) x² + 7x – 7 = 0
Here a = 1, b = 7, c = -7

Question 3.
(i)256x² – 32x + 1 = 0
(ii) 25x² + 30x + 7 = 0
Solution:
(i) 256x² – 32x + 1 = 0
Here a = 256, b = -32, c = 1

Question 4.
(i) 2x² + √5x – 5 = 0
(ii) √3x² + 10x – 8√3 = 0
Solution:
(i) 2x² + √5x – 5 = 0
Here a = 2, b = √5, c = -5

Question 5.
(i) \(\frac { x-2 }{ x+2 } +\frac { x+2 }{ x-2 } =4\)
(ii) \(\frac { x+1 }{ x+3 } =\frac { 3x+2 }{ 2x+3 } \)
Solution:
(i) \(\frac { x-2 }{ x+2 } +\frac { x+2 }{ x-2 } =4\)
⇒ \(\frac { { (x-2) }^{ 2 }+{ (x+2) }^{ 2 } }{ (x+2)(x-2) } =4\)

Question 6.
(i) a (x² + 1) = (a² + 1) x , a ≠ 0
(ii) 4x² – 4ax + (a² – b²) = 0
Solution:
(i) a (x² + 1) = (a² + 1) x
ax² – (a² + 1)x + a = 0

Question 7.
(i)\(x-\frac { 1 }{ x } =3,x\neq 0\)
(ii)\(\frac { 1 }{ x } +\frac { 1 }{ x-2 } =3,x\neq 0,2\)
Solution:
(i)\(x-\frac { 1 }{ x } =3\)
x² – 1 = 3x
⇒ x² – 3x – 1 = 0

Question 8.
\(\frac { 1 }{ x-2 } +\frac { 1 }{ x-3 } +\frac { 1 }{ x-4 } =0\)
Solution:
\(\frac { 1 }{ x-2 } +\frac { 1 }{ x-3 } +\frac { 1 }{ x-4 } =0\)
⇒ \(\frac { 1 }{ x-2 } +\frac { 1 }{ x-3 } =-\frac { 1 }{ x-4 } \)

Question 9.
Solve for \(x:2\left( \frac { 2x-1 }{ x+3 } \right) -3\left( \frac { x+3 }{ 2x-1 } \right) =5,x\neq -3,\frac { 1 }{ 2 } \)
Solution:
\(x:2\left( \frac { 2x-1 }{ x+3 } \right) -3\left( \frac { x+3 }{ 2x-1 } \right) =5 \)
Let \(\frac { 2x-1 }{ x+3 } =y\) then \(\frac { x+3 }{ 2x-1 } =\frac { 1 }{ y } \)

Question 10.
Solve the following equation by using quadratic equations for x and give your
(i) x² – 5x – 10 = 0
(ii) 5x(x + 2) = 3
Solution:
(i) x² – 5x – 10 = 0
On comparing with, ax² + bx + c = 0

Question 11.
Solve the following equations by using quadratic formula and give your answer correct to 2 decimal places :
(i) 4x² – 5x – 3 = 0
(ii) 2x – \(\\ \frac { 1 }{ x } \) = 1
Solution:
(i) Given equation 4x² – 5x – 3 = 0
Comparing with ax² + bx + c = 0, we have

Question 12.
Solve the following equation: \(x-\frac { 18 }{ x } =6\). Give your answer correct to two x significant figures. (2011)
Solution:
\(x-\frac { 18 }{ x } =6\)
⇒ x² – 6x – 18 = 0
a = 1, b = -6, c = -18

Question 13.
Solve the equation 5x² – 3x – 4 = 0 and give your answer correct to 3 significant figures:
Solution:
We have 5x² – 3x – 4 = 0
Here a = 5, b = – 3, c = – 4

Hope given ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 5 Quadratic Equations in One Variable Ex 5.3 are helpful to complete your math homework.

If you have any doubts, please comment below. Learn Insta try to provide online math tutoring for you.