Understanding ICSE Mathematics Class 10 ML Aggarwal Solutions

ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 6 Factorization 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 6 Factorization Chapter Test

More Exercises

• ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 6 Factorization EX 6
• ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 6 Factorization MCQS
• ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 6 Factorization Chapter Test

Question 1.
Find the remainder when 2x³ – 3x² + 4x + 7 is divided by
(i) x – 2
(ii) x + 3
(iii) 2x + 1
Solution:
f(x) = 2x³ – 3x² + 4x + 7
(i) Let x – 2 = 0, then x = 2
Substituting value of x in f(x)
f(2) = 2 (2)³ – 3 (2)² + 4 (2) + 7
= 2 × 8 – 3 × 4 + 4 × 2 + 7
= 16 – 12 + 8 + 7 = 19
Remainder = 19
(ii) Let x + 3 = 0, then x = – 3
Substituting the value of x in f(x)

Question 2.
When 2x³ – 9x² + 10x – p is divided by (x + 1), the remainder is – 24.Find the value of p.
Solution:
Let x + 1 = 0 then x = -1
Substituting the value of x in f(x)
f(x) = 2x³ – 9x² + 10x – p

Question 3.
If (2x – 3) is a factor of 6x² + x + a, find the value of a. With this value of a, factorise the given expression.
Solution:
Let 2x – 3 = 0 then 2x = 3
⇒ x = \(\\ \frac { 3 }{ 2 } \)
Substituting the value of x in f(x)

Question 4.
When 3x² – 5x + p is divided by (x – 2), the remainder is 3. Find the value of p. Also factorise the polynomial 3x² – 5x + p – 3.
Solution:
f(x) = 3x² – 5x+ p
Let (x – 2) = 0, then x = 2
f(2) = 3 (2)² – 5(2) + p
= 3 x 4 – 10 + p
= 12 – 10 + p
= 2 + p

Question 5.
Prove that (5x + 4) is a factor of 5x³ + 4x² – 5x – 4. Hence factorize the given polynomial completely.
Solution:
f(x) = 5x³ + 4x² – 5x – 4
Let 5x + 4 = 0, then 5x = -4
⇒ x = \(\\ \frac { -4 }{ 2 } \)

Question 6.
Use factor theorem to factorise the following polynomials completely:
(i) 4x³ + 4x² – 9x – 9
(ii) x³ – 19x – 30
Solution:
(i) f(x) = 4x³ + 4x² – 9x – 9
Let x = -1, then
f(-1) = 4 (-1)³ + 4 (-1)² – 9 (-1) – 9

Question 7.
If x³ – 2x² + px + q has a factor (x + 2) and leaves a remainder 9, when divided by (x + 1), find the values of p and q. With these values of p and q, factorize the given polynomial completely.
Solution:
f(x) = x³ – 2x² + px + q
(x + 2) is a factor
f(-2) = (-2)³ – 2(-2)² + p (-2) + q

Question 8.
If (x + 3) and (x – 4) are factors of x³ + ax² – bx + 24, find the values of a and b: With these values of a and b, factorise the given expression.
Solution:
f(x) = x³ + ax² – bx + 24
Let x + 3 = 0, then x = -3
Substituting the value of x in f(x)

Question 9.
If 2x³ + ax² – 11x + b leaves remainder 0 and 42 when divided by (x – 2) and (x – 3) respectively, find the values of a and b. With these values of a and b, factorize the given expression.
Solution:
f(x) = 2x³ + ax² – 11 x + b
Let x – 2 = 0, then x = 2,
Substituting the vaue of x in f(x)

Question 10.
If (2x + 1) is a factor of both the expressions 2x² – 5x + p and 2x² + 5x + q, find the value of p and q. Hence find the other factors of both the polynomials.
Solution:
Let 2x + 1 = 0, then 2x = -1
x = \(– \frac { 1 }{ 2 } \)
Substituting the value of x in

Question 11.
When a polynomial f(x) is divided by (x – 1), the remainder is 5 and when it is,, divided by (x – 2), the remainder is 7. Find – the remainder when it is divided by (x – 1) (x – 2).
Solution:
When f(x) is divided by (x – 1),
Remainder = 5
Let x – 1 = 0 ⇒ x = 1

Hope given ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 6 Factorization 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 6 Factorization 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 6 Factorization MCQS

More Exercises

• ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 6 Factorization EX 6
• ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 6 Factorization MCQS
• ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 6 Factorization Chapter Test

Choose the correct answer from the given four options (1 to 5) :

Question 1.
When x³ – 3x² + 5x – 7 is divided by x – 2,then the remainder is
(a) 0
(b) 1
(c) 2
(d) – 1
Solution:
f(x) = x³ – 3x² + 5x – 7
g(x) = x – 2, if x – 2 = 0, then x = 2
Remainder will be

Question 2.
When 2x³ – x² – 3x + 5 is divided by 2x + 1, then the remainder is
(a) 6
(b) – 6
(c) – 3
(d) 0
Solution:
f(x) = 2x³ – x² – 3x + 5
g(x) = 2x + 1

Question 3.
If on dividing 4x² – 3kx + 5 by x + 2, the remainder is – 3 then the value of k is
(a) 4
(b) – 4
(c) 3
(d) – 3
Solution:
f(x) = 4x² – 3kx + 5
g(x) = x + 2
Remainder = – 3
Let x + 2 = 0, then x = – 2
Now remainder will be

Question 4.
If on dividing 2x³ + 6x² – (2k – 7)x + 5 by x + 3, the remainder is k – 1 then the value of k is
(a) 2
(b) – 2
(c) – 3
(d) 3
Solution:
f(x) = 2x³ + 6x² – (2k – 7)x + 5
g(x) = x + 3
Remainder = k – 1

Question 5.
If x + 1 is a factor of 3x³ + kx² + 7x + 4, then the value of k is
(a) – 1
(b) 0
(c) 6
(d) 10
Solution:
f(x) = 3x³ + kx² + 7x + 4
g(x) = x + 1
Remainder = 0

Hope given ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 6 Factorization 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 6 Factorization Ex 6

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 6 Factorization Ex 6

More Exercises

• ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 6 Factorization EX 6
• ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 6 Factorization MCQS
• ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 6 Factorization Chapter Test

Question 1.
Find the remainder (without divisions) on dividing f(x) by x – 2, where
(i) f(x) = 5x² – 1x + 4
(ii) f (x) = 2x³ – 7x² + 3
Solution:
Let x – 2 = 0, then x = 2
(i) Substituting value of x in f(x)
f(x) = 5x² – 7x + 4
⇒ f(2) = 5(2)² – 7(2) + 4

Question 2.
Using remainder theorem, find the remainder on dividing f(x) by (x + 3) where
(i) f(x) = 2x² – 5x + 1
(ii) f(x) = 3x³ + 7x² – 5x + 1
Solution:
Let x + 3 = 0
⇒ x = -3
Substituting the value of x in f(x),

Question 3.
Find the remainder (without division) on dividing f(x) by (2x + 1) where
(i) f(x) = 4x² + 5x + 3
(ii) f(x) = 3x³ – 7x² + 4x + 11
Solution:
Let 2x + 1 = 0, then x = \(– \frac { 1 }{ 2 } \)
Substituting the value of x in f(x):
(i) f(x) = 4x² + 5x + 3

Question 4.
(i) Find the remainder (without division) when 2x³ – 3x² + 7x – 8 is divided by x – 1 (2000)
(ii) Find the remainder (without division) on dividing 3x² + 5x – 9 by (3x + 2)
Solution:
(i) Let x – 1 = 0, then x = 1
Substituting value of x in f(x)

Question 5.
Using remainder theorem, find the value of k if on dividing 2x³ + 3x² – kx + 5 by x – 2, leaves a remainder 7. (2016)
Solution:
f(x) = 2x² + 3x² – kx + 5
g(x) = x – 2, if x – 2 = 0, then x = 2
Dividing f(x) by g(x) the remainder will be

Question 6.
Using remainder theorem, find the value of a if the division of x³ + 5x² – ax + 6 by (x – 1) leaves the remainder 2a.
Solution:
Let x – 1 = 0, then x = 1
Substituting the value of x in f(x)

Question 7.
(i) What number must be subtracted from 2x² – 5x so that the resulting polynomial leaves the remainder 2, when divided by 2x + 1 ?
(ii) What number must be added to 2x³ – 7x² + 2x so that the resulting polynomial leaves the remainder – 2 when divided by 2x – 3?
Solution:
(i) Let a be subtracted from 2x² – 5x,
Dividing 2x² – 5x by 2x + 1,

Question 8.
(i) When divided by x – 3 the polynomials x² – px² + x + 6 and 2x³ – x² – (p + 3) x – 6 leave the same remainder. Find the value of ‘p’
(ii) Find ‘a’ if the two polynomials ax³ + 3x² – 9 and 2x³ + 4x + a, leaves the same remainder when divided by x + 3.
Solution:
By dividing
x³ – px² + x + 6
and 2x³ – x² – (p + 3) x – 6

Question 9.
By factor theorem, show that (x + 3) and (2x – 1) are factors of 2x² + 5x – 3.
Solution:
Let x + 3 = 0 then x = – 3
Substituting the value of x in f(x)

Question 10.
Show that (x – 2) is a factor of 3x² – x – 10 Hence factorise 3x² – x – 10.
Solution:
Let x – 2 = 0, then x = 2
Substituting the value of x in f(x),

Question 11.
Show that (x – 1) is a factor of x³ – 5x² – x + 5 Hence factorise x³ – 5x² – x + 5.
Solution:
Let x – 1 = 0, then x = 1
Substituting the value of x in f(x),

Question 12.
Show that (x – 3) is a factor of x³ – 7x² + 15x – 9. Hence factorise x³ – 7x² + 15 x – 9
Solution:
Let x – 3 = 0, then x = 3,
Substituting the value of x in f(x),

Question 13.
Show that (2x + 1) is a factor of 4x³ + 12x² + 11 x + 3 .Hence factorise 4x³ + 12x² + 11x + 3.
Solution:
Let 2x + 1 = 0,
then x = \(– \frac { 1 }{ 2 } \)
Substituting the value of x in f(x),

Question 14.
Show that 2x + 7 is a factor of 2x³ + 5x² – 11x – 14. Hence factorise the given expression completely, using the factor theorem. (2006)
Solution:
Let 2x + 7 = 0, then 2x = -7
x = \(\\ \frac { -7 }{ 2 } \)
substituting the value of x in f(x),

Question 15.
Use factor theorem to factorise the following polynominals completely.
(i) x³ + 2x² – 5x – 6
(ii) x³ – 13x – 12.
Solution:
(i) Let f(x) = x³ + 2x² – 5x – 6

Question 16.
(i) Use the Remainder Theorem to factorise the following expression : 2x³ + x² – 13x + 6. (2010)
(ii) Using the Remainder Theorem, factorise completely the following polynomial: 3x² + 2x² – 19x + 6 (2012)
Solution:
(i) Let f(x) = 2x³ + x² – 13x + 6
Factors of 6 are ±1, ±2, ±3, ±6
Let x = 2, then

Question 17.
Using the Remainder and Factor Theorem, factorise the following polynomial: x³ + 10x² – 37x + 26.
Solution:
f(x) = x³ + 10x² – 37x + 26
f(1) = (1)³ + 10(1)² – 37(1) + 26
= 1 + 10 – 37 + 26 = 0
x = 1

Question 18.
If (2 x + 1) is a factor of 6x³ + 5x² + ax – 2 find the value of a
Solution:
Let 2x + 1 = 0, then x = \(– \frac { 1 }{ 2 } \)
Substituting the value of x in f(x),
f(x) = 6x³ + 5x² + ax – 2

Question 19.
If (3x – 2) is a factor of 3x³ – kx² + 21x – 10, find the value of k.
Solution:
Let 3x – 2 = 0, then 3x = 2
⇒ x = \(\\ \frac { 2 }{ 3 } \)
Substituting the value of x in f(x),
f(x) = 3x³ – kx² + 21x – 10

Question 20.
If (x – 2) is a factor of 2x³ – x² + px – 2, then
(i) find the value of p.
(ii) with this value of p, factorise the above expression completely
Solution:
(i) Let x – 2 = 0, then x = 2
Now f(x) = 2x³ – x² + px – 2

Question 21.
Find the value of ‘K’ for which x = 3 is a solution of the quadratic equation, (K + 2)x² – Kx + 6 = 0.
Also, find the other root of the equation.
Solution:
(K + 2)x² – Kx + 6 = 0 …(1)
Substitute x = 3 in equation (1)

Question 22.
What number should be subtracted from 2x³ – 5x² + 5x so that the resulting polynomial has 2x – 3 as a factor?
Solution:
Let the number to be subtracted be k and the resulting polynomial be f(x), then
f(x) = 2x³ – 5x² + 5x – k
Since, 2x – 3 is a factor of f(x),
Now, converting 2x – 3 to factor theorem

Question 23.
Find the value of the constants a and b, if (x – 2) and (x + 3) are both factors of the expression x³ + ax² + bx – 12.
Solution:
Let x – 2 = 0, then x = 0
Substituting value of x in f(x)
f(x) = x³ + ax² + bx – 12

Question 24.
If (x + 2) and (x – 3) are factors of x³ + ax + b, find the values of a and b. With these values of a and b, factorise the given expression.
Solution:
Let x + 2 = 0, then x = -2
Substituting the value of x in f(x),
f(x) = x³ + ax + b

Question 25.
(x – 2) is a factor of the expression x³ + ax² + bx + 6. When this expression is divided by (x – 3), it leaves the remainder 3. Find the values of a and b. (2005)
Solution:
As x – 2 is a factor of
f(x) = x³ + ax² + bx + 6

Question 26.
If (x – 2) is a factor of the expression 2x³ + ax² + bx – 14 and when the expression is divided by (x – 3), it leaves a remainder 52, find the values of a and b.
Solution:
f(x) = 2x³ + ax² + bx – 14
∴ (x – 2) is factor of f(x)
f(2) = 0

Question 27.
If ax³ + 3x² + bx – 3 has a factor (2x + 3) and leaves remainder – 3 when divided by (x + 2), find the values of a and 6. With these values of a and 6, factorise the given expression.
Solution:
Let 2x + 3 = 0 then 2x = -3
⇒ x = \(\\ \frac { -3 }{ 2 } \)
Substituting the value of x in f(x),
f(x) = ax³ + 3x² + 6x – 3

Question 28.
Given f(x) = ax² + bx + 2 and g(x) = bx² + ax + 1. If x – 2 is a factor of f(x) but leaves the remainder – 15 when it divides g(x), find the values of a and b. With these values of a and b, factorise the expression. f(x) + g(x) + 4x² + 7x.
Solution:
f(x) = ax² + bx + 2
g(x) = bx² + ax + 1
x – 2 is a factor of f(x)
Let x – 2 = 0
⇒ x = 2

Hope given ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 6 Factorization Ex 6 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 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.

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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.