Category: ICSE Class 10

ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 7 Ratio and Proportion Ex 7.1

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 7 Ratio and Proportion Ex 7.1

More Exercises

• ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 7 Ratio and Proportion Ex 7.1
• ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 7 Ratio and Proportion Ex 7.2
• ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 7 Ratio and Proportion Ex 7.3
• ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 7 Ratio and Proportion MCQS
• ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 7 Ratio and Proportion Chapter Test

Question 1.
An alloy consists of \(27 \frac { 1 }{ 2 } \) kg of copper and \(2 \frac { 3 }{ 4 } \) kg of tin. Find the ratio by weight of tin to the alloy
Solution:
Copper = \(27 \frac { 1 }{ 2 } \) kg = \(\\ \frac { 55 }{ 2 } \) kg,
Tin = \(2 \frac { 3 }{ 4 } \) kg = \(\\ \frac { 11 }{ 4 } \) kg

Question 2.
Find the compounded ratio of:
(i) 2 : 3 and 4 : 9
(ii) 4 : 5, 5 : 7 and 9 : 11
(iii) (a – b) : (a + b), (a + b)² : (a² + b²) and (a⁴ – b⁴) : (a² – b²)²
Solution:
(i) 2 : 3 and 4 : 9
Compound ratio = \(\\ \frac { 2 }{ 3 } \) x \(\\ \frac { 4 }{ 9 } \)
= \(\\ \frac { 8 }{ 27 } \) or 8 : 27
(ii) 4 : 5, 5 : 7 and 9 : 11

Question 3.
Find the duplicate ratio of
(i) 2 : 3
(ii) √5 : 7
(iii) 5a : 6b
Solution:
(i) Duplicate ratio of 2 : 3 = (2)² : (3)² = 4 : 9
(ii) Duplicate ratio of √5 : 7 = (√5)² : (7)² = 5 : 49
(iii) Duplicate ratio of 5a : 6b = (5a)² : (6b)² = 25a² : 36b²

Question 4.
Find the triplicate ratio of
(i) 3 : 4
(ii) \(\\ \frac { 1 }{ 2 } \) : \(\\ \frac { 1 }{ 3 } \)
(iii) 1³ : 2³
Solution:
(i) Triplicate ratio of 3 : 4
= (3)³ : (4)³
= 27 : 64

Question 5.
Find the sub-duplicate ratio of
(i) 9 : 16
(ii) \(\\ \frac { 1 }{ 4 } \) : \(\\ \frac { 1 }{ 9 } \),
(iii) 9a² : 49b²
Solution:
(i) Sub-duplicate ratio of 9 : 16
= √9 : √16
= 3 : 4

Question 6.
Find the sub-triplicate ratio of
(i) 1 : 216
(ii) \(\\ \frac { 1 }{ 8 } \) : \(\\ \frac { 1 }{ 125 } \)
(iii) 27a³ : 64b³
Solution:
(i) Sub-triplicate ratio of 1 : 216
= \(\sqrt [ 3 ]{ 1 } :\sqrt [ 3 ]{ 216 } \)

Question 7.
Find the reciprocal ratio of
(i) 4 : 7
(ii) 3² : 4²
(iii) \(\frac { 1 }{ 9 } :2 \)
Solution:
(i) Reciprocal ratio of 4 : 7 = 7 : 4
(ii) Reciprocal ratio of 3² : 4² = 4² : 3² = 16 : 9
(iii) Reciprocal ratio of \(\frac { 1 }{ 9 } :2 \) = \(2:\frac { 1 }{ 9 } \) = 18 : 1

Question 8.
Arrange the following ratios in ascending order of magnitude:
2 : 3, 17 : 21, 11 : 14 and 5 : 7
Solution:
Writing the given ratios in fraction
\(\frac { 2 }{ 3 } ,\frac { 17 }{ 21 } ,\frac { 11 }{ 14 } ,\frac { 5 }{ 7 } \)

Question 9.
(i) If A : B = 2 : 3, B : C = 4 : 5 and C : D = 6 : 7, find A : D
(ii) If x : y = 2 : 3, and y : z = 4 : 7, find x : y : z
Solution:
Let A : B = 2 : 3, B : C = 4 : 5 and C : D = 6 : 7
\(\frac { A }{ B } =\frac { 2 }{ 3 } ,\frac { B }{ C } =\frac { 4 }{ 5 } ,\frac { C }{ D } =\frac { 6 }{ 7 } \)

Question 10.
(i) If A: B = \(\frac { 1 }{ 4 } :\frac { 1 }{ 5 } \) and B : C = \(\frac { 1 }{ 7 } :\frac { 1 }{ 6 } \), find A : B : C.
(ii) If 3A = 4B = 6C, find A : B : C
Solution:
A : B = \(\frac { 1 }{ 4 } \times \frac { 5 }{ 1 } =\frac { 5 }{ 4 } \)
B : C = \(\frac { 1 }{ 7 } \times \frac { 6 }{ 1 } =\frac { 6 }{ 7 } \)

Question 11.
(i) If \(\frac { 3x+5y }{ 3x-5y } =\frac { 7 }{ 3 } \) , Find x : y
(ii) ) If a : b = 3 : 11, find (15a – 3b) : (9a + 5b). a
Solution:
(i) \(\frac { 3x+5y }{ 3x-5y } =\frac { 7 }{ 3 } \)
⇒ 9x + 15y = 21x – 35y [By cross multiplication]
⇒ 21x – 9x = 15y + 35y

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Question 12.
(i) If (4x² + xy) : (3xy – y²) = 12 : 5, find (x + 2y) : (2x + y).
(ii) If y (3x – y) : x (4x + y) = 5 : 12. Find (x² + y²) : (x + y)².
Solution:
(4x² + xy) : (3xy – y²) = 12 : 5
⇒ \(\frac { { 4x }^{ 2 }+xy }{ 3xy-{ y }^{ 2 } } =\frac { 12 }{ 5 } \)
⇒ 20x² + 5xy = 36xy – 12y²

Question 13.
(i) If (x – 9) : (3x + 6) is the duplicate ratio of 4 : 9, find the value of x.
(ii) If (3x + 1) : (5x + 3) is the triplicate ratio of 3 : 4, find the value of x.
(iii) If (x + 2y) : (2x – y) is equal to the duplicate ratio of 3 : 2, find x : y.
Solution:
(i) \(\frac { x-9 }{ 3x+6 } ={ \left( \frac { 4 }{ 9 } \right) }^{ 2 }\)
⇒ \(\frac { x-9 }{ 3x+6 } =\frac { 16 }{ 81 } \)

Question 14.
(i) Find two numbers in the ratio of 8 : 7 such that when each is decreased by \(12 \frac { 1 }{ 2 } \), they are in the ratio 11 : 9.
(ii) The income of a man is increased in the ratio of 10 : 11. If the increase in his income is Rs 600 per month, find his new income.
Solution:
(i) The ratio is 8 : 7
Let the numbers be 8x and 7x,
According to condition,

Question 15.
(i) A woman reduces her weight in the ratio 7 : 5. What does her weight become if originally it was 91 kg.
(ii) A school collected Rs 2100 for charity. It was decided to divide the money between an orphanage and a blind school in the ratio of 3 : 4. How much money did each receive?
Solution:
(i) Ratio between the original weight and reduced weight = 7 : 5
Let original weight = 7x
then reduced weight = 5x
If original weight = 91 kg.

Question 16.
(i) The sides of a triangle are in the ratio 7 : 5 : 3 and its perimeter is 30 cm. Find the lengths of sides.
(ii) If the angles of a triangle are in the ratio 2 : 3 : 4, find the angles.
Solution:
(i) Perimeter of a triangle = 30 cm.
Ratio among sides = 7 : 5 : 3
Sum of ratios 7 + 5 + 3 = 15

Question 17.
Three numbers are in the ratio \(\frac { 1 }{ 2 } :\frac { 1 }{ 3 } :\frac { 1 }{ 4 } \) If the sum of their squares is 244, find the numbers.
Solution:
The ratio of three numbers \(\frac { 1 }{ 2 } :\frac { 1 }{ 3 } :\frac { 1 }{ 4 } \)
= \(\frac { 6:4:3 }{ 12 } \)
= 6 : 4 : 3
Let first number 6x, second 4x and third 3x
.’. According to the condition
(6x)² + (4x)² + (3x)² = 244
⇒ 36x² + 16x² + 9x² = 244

Question 18.
(i) A certain sum was divided among A, B and C in the ratio 7 : 5 : 4. If B got Rs 500 more than C, find the total sum divided.
(ii) In a business, A invests Rs 50000 for 6 months, B Rs 60000 for 4 months and C, Rs 80000 for 5 months. If they together earn Rs 18800 find the share of each.
Solution:
(i) Ratio between A, B and C = 7 : 5 : 4
Let A’s share = 7x
B’s share = 5x
and C’s share = 4x
Total sum = 7x + 5x + 4x = 16x

Question 19.
(i) In a mixture of 45 litres, the ratio of milk to water is 13 : 2. How much water must be added to this mixture to make the ratio of milk to water as 3 : 1 ?
(ii) The ratio of the number of boys to the number of girls in a school of 560 pupils is 5 : 3. If 10 new boys are admitted, find how many new girls may be admitted so that the ratio of the number of boys to the number of girls may change to 3 : 2.
Solution:
(i) Mixture of milk and water = 45 litres
Ratio of milk and water =13 : 2
Sum of ratio = 13 + 2 = 15

Question 20.
(i) The monthly pocket money of Ravi and Sanjeev are in the ratio 5 : 7. Their expenditures are in the ratio 3 : 5. If each saves Rs 80 every month, find their monthly pocket money.
(ii) In class X of a school, the ratio of the number of boys to that of the girls is 4 : 3. If there were 20 more boys and 12 less girls, then the ratio would have been 2 : 1, How many students were there in the class?
Solution:
(i) Let the monthly pocket money of Ravi and Sanjeev be 5x and 7x respectively.
Also, let their expenditure be 3y and 5y respectively.

Question 21.
In an examination, the ratio of passes to failures was 4 : 1. If 30 less had appeared and 20 less passed, the ratio of passes to failures would have been 5 : 1. How many students appeared for the examination
Solution:
Let the number of passes = 4x
and number of failures = x
The total number of students appeared = 4x + x = 5x
In the second case, the number of students appeared = 5x – 30
and number of passes = 4x – 20

Hope given ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 7 Ratio and Proportion Ex 7.1 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 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 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 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 7 Ratio and Proportion 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 7 Ratio and Proportion Chapter Test

More Exercises

• ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 7 Ratio and Proportion Ex 7.1
• ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 7 Ratio and Proportion Ex 7.2
• ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 7 Ratio and Proportion Ex 7.3
• ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 7 Ratio and Proportion MCQS
• ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 7 Ratio and Proportion Chapter Test

Question 1.
Find the compound ratio of:
(a + b)² : (a – b )² ,
(a² – b²) : (a² + b²),
(a⁴ – b⁴) : (a + b)⁴
Solution:
(a + b)² : (a – b )² ,
(a² – b²) : (a2 + b²),
(a⁴ – b⁴) : (a + b)⁴

Question 2.
If (7 p + 3 q) : (3 p – 2 q) = 43 : 2 find p : q
Solution:
(7p + 3q) : (3p – 2q) = 43 : 2
⇒ \(\frac { 7p+3q }{ 3p-2q } =\frac { 43 }{ 2 } \)

Question 3.
If a : b = 3 : 5, find (3a + 5b): (7a – 2b).
Solution:
a : b = 3 : 5
⇒ \(\frac { a }{ b } =\frac { 3 }{ 5 } \)
⇒ 3a + 5n : 7a – 2b
Dividing each term by b

Question 4.
The ratio of the shorter sides of a right angled triangle is 5 : 12. If the perimeter of the triangle is 360 cm, find the length of the longest side.
Solution:
Let the two shorter sides of a right-angled triangle be 5x and 12x.
Third (longest side)

Question 5.
The ratio of the pocket money saved by Lokesh and his sister is 5 : 6. If the sister saves Rs 30 more, how much more the brother should save in order to keep the ratio of their savings unchanged?
Solution:
Let the savings of Lokesh and his sister are 5x and 6x.
and the Lokesh should save Rs y more Now, according to the problem,
⇒ \(\frac { 5x+y }{ 6x+30 } =\frac { 5 }{ 6 } \)
⇒ 30x + 6y = 30x + 150

Question 6.
In an examination, the number of those who passed and the number of those who failed were in the ratio of 3 : 1. Had 8 more appeared, and 6 less passed, the ratio of passed to failures would have been 2 : 1. Find the number of candidates who appeared.
Solution:
Let number of passed = 3 x
and failed = x
Total candidates appeared = 3x + x = 4x.
In second case
No. of candidates appeared = 4 x + 8
and No. of passed = 3 x – 6
and failed = 4x + 8 – 3x + 6 = x + 14
then ratio will be = 2 : 1
Now according to the condition

Question 7.
What number must be added to each of the numbers 15, 17, 34 and 38 to make them proportional ?
Solution:
Let x be added to each number, then numbers will be
15 + x, 17 + x, 34 + x, and 38 + x.
Now according to the condition

Question 8.
If (a + 2 b + c), (a – c) and (a – 2 b + c) are in continued proportion, prove that b is the mean proportional between a and c.
Solution:
(a + 2 b + c), (a – c) and (a – 2 b + c) are in continued proportion
⇒ \(\frac { a+2b+c }{ a-c } =\frac { a-c }{ a-2b+c } \)

Question 9.
If 2, 6, p, 54 and q are in continued proportion, find the values of p and q.
Solution:
2, 6, p, 54 and q are in continued proportional then
⇒ \(\frac { 2 }{ 6 } =\frac { 6 }{ p } =\frac { p }{ 54 } =\frac { 54 }{ q } \)

Question 10.
If a, b, c, d, e are in continued proportion, prove that: a : e = a⁴ : b⁴.
Solution:
a, b, c, d, e are in continued proportion
⇒ \(\frac { a }{ b } =\frac { b }{ c } =\frac { c }{ d } =\frac { d }{ e } \) = k (say)

Question 11.
Find two numbers whose mean proportional is 16 and the third proportional is 128.
Solution:
Let x and y be two numbers
Their mean proportion = 16
and third proportion = 128

Question 12.
If q is the mean proportional between p and r, prove that:
\({ p }^{ 2 }-{ 3q }^{ 2 }+{ r }^{ 2 }={ q }^{ 4 }\left( \frac { 1 }{ { p }^{ 2 } } -\frac { 3 }{ { q }^{ 2 } } +\frac { 1 }{ { r }^{ 2 } } \right) \)
Solution:
q is mean proportional between p and r
q² = pr

Question 13.
If \(\frac { a }{ b } = \frac { c }{ d } = \frac { e }{ f } \), prove that each ratio is
(i) \(\sqrt { \frac { { 3a }^{ 2 }-{ 5c }^{ 2 }+{ 7e }^{ 2 } }{ { 3b }^{ 2 }-{ 5d }^{ 2 }+{ 7f }^{ 2 } } } \)
(ii) \({ \left[ \frac { { 2a }^{ 3 }+{ 5c }^{ 3 }+{ 7e }^{ 3 } }{ { 2b }^{ 3 }+{ 5d }^{ 3 }+{ 7f }^{ 3 } } \right] }^{ \frac { 1 }{ 3 } } \)
Solution:
\(\frac { a }{ b } = \frac { c }{ d } = \frac { e }{ f } \) = k(say)
∴ a = k, c = dk, e = fk

Question 14.
If \(\frac { x }{ a } = \frac { y }{ b } = \frac { z }{ c } \), prove that
\(\frac { { 3x }^{ 3 }-{ 5y }^{ 3 }+{ 4z }^{ 3 } }{ { 3a }^{ 3 }-{ 5b }^{ 3 }+{ 4c }^{ 3 } } ={ \left( \frac { 3x-5y+4z }{ 3a-5b+4c } \right) }^{ 3 }\)
Solution:
\(\frac { x }{ a } = \frac { y }{ b } = \frac { z }{ c } \) = k (say)
x = ak, y = bk, z = ck

Question 15.
If x : a = y : b, prove that
\(\frac { { x }^{ 4 }+{ a }^{ 4 } }{ { x }^{ 3 }+{ a }^{ 3 } } +\frac { { y }^{ 4 }+{ b }^{ 4 } }{ { y }^{ 3 }+{ b }^{ 3 } } =\frac { { \left( x+y \right) }^{ 4 }+{ \left( a+b \right) }^{ 4 } }{ { \left( x+y \right) }^{ 3 }+{ \left( a+b \right) }^{ 3 } } \)
Solution:
\(\frac { x }{ a } = \frac { y }{ b } \) = k (say)
x = ak, y = bk

Question 16.
If \(\frac { x }{ b+c-a } =\frac { y }{ c+a-b } =\frac { z }{ a+b-c } \) prove that each ratio’s equal to :
\(\frac { x+y+z }{ a+b+c } \)
Solution:
\(\frac { x }{ b+c-a } =\frac { y }{ c+a-b } =\frac { z }{ a+b-c } \) = k(say)
x = k(b + c – a),
y = k(c + a – b),
z = k(a + b – c)

Question 17.
If a : b = 9 : 10, find the value of
(i) \(\frac { 5a+3b }{ 5a-3b } \)
(ii) \(\frac { { 2a }^{ 2 }-{ 3b }^{ 2 } }{ { 2a }^{ 2 }+{ 3b }^{ 2 } } \)
Solution:
a : b = 9 : 10
⇒ \(\frac { a }{ b } = \frac { 9 }{ 10 }\)

Question 18.
If (3x² + 2y²) : (3x² – 2y²) = 11 : 9, find the value of \(\frac { { 3x }^{ 4 }+{ 25y }^{ 4 } }{ { 3x }^{ 4 }-{ 25y }^{ 4 } } \) ;
Solution:
\(\frac { { 3x }^{ 4 }+{ 25y }^{ 4 } }{ { 3x }^{ 4 }-{ 25y }^{ 4 } } =\frac { 11 }{ 9 } \)
Applying componendo and dividendo

Question 19.
If \(x=\frac { 2mab }{ a+b } \) , find the value of
\(\frac { x+ma }{ x-ma } +\frac { x+mb }{ x-mb } \)
Solution:
\(x=\frac { 2mab }{ a+b } \)
⇒ \(\frac { x }{ ma } +\frac { 2b }{ a+b } \)
Applying componendo and dividendo

Question 20.
If \(x=\frac { pab }{ a+b } \) ,prove that \(\frac { x+pa }{ x-pa } -\frac { x+pb }{ x-pb } =\frac { 2\left( { a }^{ 2 }-{ b }^{ 2 } \right) }{ ab } \)
Solution:
\(x=\frac { pab }{ a+b } \)
⇒ \(\frac { x }{ pa } +\frac { b }{ a+b } \)
Applying componendo and dividendo

Question 21.
Find x from the equation \(\frac { a+x+\sqrt { { a }^{ 2 }-{ x }^{ 2 } } }{ a+x-\sqrt { { a }^{ 2 }-{ x }^{ 2 } } } =\frac { b }{ x } \)
Solution:
\(\frac { a+x+\sqrt { { a }^{ 2 }-{ x }^{ 2 } } }{ a+x-\sqrt { { a }^{ 2 }-{ x }^{ 2 } } } =\frac { b }{ x } \)
Applying componendo and dividendo,

Question 22.
If \(x=\frac { \sqrt [ 3 ]{ a+1 } +\sqrt [ 3 ]{ a-1 } }{ \sqrt [ 3 ]{ a+1 } -\sqrt [ 3 ]{ a-1 } } \), prove that :
x³ – 3ax² + 3x – a = 0
Solution:
\(x=\frac { \sqrt [ 3 ]{ a+1 } +\sqrt [ 3 ]{ a-1 } }{ \sqrt [ 3 ]{ a+1 } -\sqrt [ 3 ]{ a-1 } } \)
Applying componendo and dividendo,

Question 23.
If \(\frac { by+cz }{ b^{ 2 }+{ c }^{ 2 } } =\frac { cz+ax }{ { c }^{ 2 }+{ a }^{ 2 } } =\frac { ax+by }{ { a }^{ 2 }+{ b }^{ 2 } } \), prove that each of these ratio is equal to \(\frac { x }{ a } =\frac { y }{ b } =\frac { z }{ c } \)
Solution:
\(\frac { by+cz }{ b^{ 2 }+{ c }^{ 2 } } =\frac { cz+ax }{ { c }^{ 2 }+{ a }^{ 2 } } =\frac { ax+by }{ { a }^{ 2 }+{ b }^{ 2 } } \)

Hope given ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 7 Ratio and Proportion 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 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.

ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 7 Ratio and Proportion 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 7 Ratio and Proportion MCQS

More Exercises

• ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 7 Ratio and Proportion Ex 7.1
• ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 7 Ratio and Proportion Ex 7.2
• ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 7 Ratio and Proportion Ex 7.3
• ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 7 Ratio and Proportion MCQS
• ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 7 Ratio and Proportion Chapter Test

Choose the correct answer from the given options (1 to 10):

Question 1.
The ratio of 45 minutes to \(5 \frac { 3 }{ 4 } \) hours is
(a) 180:23
(b) 3:23
(c) 23:3
(d) 6:23
Solution:
ratio of 45 minutes to \(5 \frac { 3 }{ 4 } \) hours is
45 minutes to : \(5 \frac { 3 }{ 4 } \) hours

Question 2.
The ratio of 4 litres to 900 mL is
(a) 4 : 9
(b) 40 : 9
(c) 9 : 40
(d) 20 : 9
Solution:
4l : 900 ml
= 4000 ml : 900 ml
= 4000 : 900
= 40 : 9 (b)

Question 3.
When the number 210 is increased in the ratio 5 : 7, the the new number is
(a) 150
(b) 180
(c) 294
(d) 420
Solution:
210 is increased in the ratio 5 : 7, then
New increased number will be
= 210 × \(\\ \frac { 7 }{ 5 } \)
= 294 (c)

Question 4.
Two numbers are in the ratio 7 : 9. If the sum of the numbers is 288, then the smaller number is
(a) 126
(b) 162
(c) 112
(d) 144
Solution:
Ratio in two number = 7 : 9
Sum of numbers = 288
Sum of ratios = 7 + 9
= 16
Smaller number = \(\\ \frac { 288\times 7 }{ 16 } \)
= 126 (a)

Question 5.
A ratio equivalent to the ratio \(\\ \frac { 2 }{ 3 } \) : \(\\ \frac { 5 }{ 7 } \) is
(a) 4:6
(b) 5:7
(c) 15:14
(d) 14:15
Solution:
\(\\ \frac { 2 }{ 3 } \) : \(\\ \frac { 5 }{ 7 } \)
Multiply and divide \(\\ \frac { 2 }{ 3 } \) by 7 and
Multiply and divide \(\\ \frac { 5 }{ 7 } \) by 3

Question 6.
The ratio of number of edges of a cube to the number of its faces is
(a) 2 : 1
(b) 1 : 2
(c) 3 : 8
(d) 8 : 3
Solution:
No. of edges of the cube = 12
No. of faces = 6
Ratio in edges a cube to the number of faces = 12 : 6
= 2 : 1 (a)

Question 7.
If x, 12, 8 and 32 are in proportion, then the value of x is
(a) 6
(b) 4
(c) 3
(d) 2
Solution:
x, 12, 8, 32 are in proportion, then
x × 32 = 12 × 8 (∵ ad = bc)
⇒ x = \(\\ \frac { 12\times 8 }{ 32 } \) = 3
x = 3 (c)

Question 8.
The fourth proportional to 3, 4, 5 is
(a) 6
(b) \(\\ \frac { 20 }{ 3 } \)
(c) \(\\ \frac { 15 }{ 4 } \)
(d) \(\\ \frac { 12 }{ 5 } \)
Solution:
The fourth proportion to 3, 4, 5 will be
= \(\\ \frac { 4\times 5 }{ 3 } \)
= \(\\ \frac { 20 }{ 3 } \) (b)

Question 9.
The third proportional to \(6 \frac { 1 }{ 4 } \) and 5 is
(a) 4
(b) \(8 \frac { 1 }{ 2 } \)
(c) 3
(d) none of these
Solution:
The third proportional to \(6 \frac { 1 }{ 4 } \) and 5 is
⇒ \(6 \frac { 1 }{ 4 } \) : 5 :: 5 : x
⇒ \(\\ \frac { 25 }{ 4 } \) : 5 :: 5 : x
⇒ x = \(\\ \frac { 5\times 5 }{ 25 } \) × 4
⇒ 4 (a)

Question 10.
The mean proportional between \(\\ \frac { 1 }{ 2 } \) and 128 is
(a) 64
(b) 32
(c) 16
(d) 8
Solution:
The mean proportional between \(\\ \frac { 1 }{ 2 } \) and 128 is
= \(\sqrt { \frac { 1 }{ 2 } \times 128 } \)
= √64
= 8 (d)

Hope given ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 7 Ratio and Proportion 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 Ex 5.2

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 5.2

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 24) by factorization

Question 1.
(i) 4x² = 3x
(ii) \(\frac { { x }^{ 2 }-5x }{ 2 } =0\)
Solution:
(i) 4x² = 3x
x(4x – 3) = 0
Either x = 0,

Question 2.
(i) (x – 3) (2x + 5) = 0
(ii) x (2x + 1) = 6
Solution:
(i) (x – 3) (2x + 5) = 0
Either x – 3 = 0,
Then x = 3

Question 3.
(i) x² – 3x – 10 = 0
(ii) x(2x + 5) = 3
Solution:
(i) x² – 3x – 10 = 0
⇒ x² – 5x + 2x – 10 = 0
⇒ x(x – 5) + 2(x – 5) = 0

Question 4.
(i) 3x² – 5x – 12 = 0
(ii) 21x² – 8x – 4 = 0
Solution:
(i) 3x² – 5x – 12 = 0
⇒ 3x² – 9x + 4x – 12 = 0
⇒ 3x (x – 3) + 4(x – 3) = 0

Question 5.
(i) 3x² = x + 4
(ii) x(6x – 1) = 35
Solution:
(i) 3x² = x + 4
⇒ 3x² – x – 4 = 0
⇒ 3x² – 4x + 3x – 4 = 0

Question 6.
(i) 6p² + 11p – 10 = 0
(ii) \(\frac { 2 }{ 3 } { x }^{ 2 }-\frac { 1 }{ 3 } x=1 \)
Solution:
(i) 6p² + 11p – 10 = 0
⇒ 6p² + 15p – 4p – 10 = 0
⇒ 3p(2p + 5) – 2(2p + 5) = 0

Question 7.
(i) (x – 4)² + 5² = 13²
(ii) 3(x – 2)² = 147
Solution:
(i) (x – 4)² + 5² = 13²
x² – 8x + 16 + 25 = 169

Question 8.
(i) \(\\ \frac { 1 }{ 7 } \)(3x – 5)² = 28
(ii) 3(y² – 6) = y(y + 7) – 3
Solution:
(i) \(\\ \frac { 1 }{ 7 } \)(3x – 5)² = 28
(3x – 5)² = 28 × 7
⇒ 9x² – 30x + 25 = 196

Question 9.
x² – 4x – 12 = 0,when x∈N
Solution:
x² – 4x – 12 = 0
⇒ x² – 6x + 2x – 12 = 0
⇒ x (x – 6) + 2 (x – 6) = 0
⇒ (x – 6) (x + 2) = 0
Either x – 6 = 0, then x = 6
or x + 2 = 0, then x = -2
But -2 is not a natural number
∴ x = 6

Question 10.
2x² – 8x – 24 = 0 when x∈I
Solution:
2x² – 8x – 24 = 0
⇒ x² – 4x – 12 = 0 (Dividing by 2)
⇒ x² – 6x + 2x – 12 = 0
⇒ x (x – 6) + 2 (x – 6) = 0
⇒ (x – 6) (x + 2) = 0
Either x – 6 = 0, then, x = 6
or x + 2 = 0, then x = – 2
Hence x = 6, – 2

Question 11.
5x² – 8x – 4 = 0 when x∈Q
Solution:
5x² – 8x – 4 = 0
∵ 5 × ( – 4) = – 20
-20 = – 10 + 2
-8 = – 10 + 2

Question 12.
2x² – 9x + 10 = 0,when
(i)x∈N
(ii)x∈Q
Solution:
2x² – 9x + 10 = 0
⇒ 2x² – 4x – 5x + 10 = 0
⇒ 2x(x – 2) – 5(x – 2) = 0

Question 13.
(i) a²x² + 2ax + 1 = 0, a≠0
(ii) x² – (p + q)x + pq = 0
Solution:
(i) a²x² + 2ax + 1 = 0
⇒ a²x² + ax + ax + 1 = 0

Question 14.
a²x² + (a² + b²)x + b² = 0, a≠0
Solution:
a²x² + (a² + b²)x + b² = 0
⇒ a²x(x + 1) + b²(x + 1) = 0

Question 15.
(i) √3x² + 10x + 7√3 = 0
(ii) 4√3x² + 5x – 2√3 = 0
Solution:
(i) √3x² + 10x + 7√3 = 0
[ ∵ √3 x 7√3 = 7 x 3 = 21]
⇒ √3x(x + √3) + 7(x + √3) = 0

Question 16.
(i) x² – (1 + √2)x + √2 = 0
(ii) \(x+ \frac { 1 }{ x } \) = \(2 \frac { 1 }{ 20 } \)
Solution:
(i) x² – (1 + √2)x + √2 = 0
⇒ x² – x – √2x + √2 = 0

Question 17.
(i) \(\frac { 2 }{ { x }^{ 2 } } -\frac { 5 }{ x } +2=0,x\neq 0 \)
(ii)\(\frac { { x }^{ 2 } }{ 15 } -\frac { x }{ 3 } -10=0 \)
Solution:
(i) \(\frac { 2 }{ { x }^{ 2 } } -\frac { 5 }{ x } +2=0,x\neq 0 \)
⇒ 2 – 5x + 2x² = 0

Question 18.
(i) \(3x-\frac { 8 }{ x } =2 \)
(ii) \(\frac { x+2 }{ x+3 } =\frac { 2x-3 }{ 3x-7 } \)
Solution:
(i) \(3x-\frac { 8 }{ x } =2 \)
\(\frac { { 3x }^{ 2 }-8 }{ x } =2 \)

Question 19.
(i) \(\frac { 8 }{ x+3 } -\frac { 3 }{ 2-x } =2 \)
(ii) \(\frac { x }{ x-1 } +\frac { x-1 }{ x } =2\frac { 1 }{ 2 } \)
Solution:
(i) \(\frac { 8 }{ x+3 } -\frac { 3 }{ 2-x } =2 \)
\(\frac { 16-8x-3x-9 }{ (x+3)(2-x) } =2 \)

Question 20.
(i) \(\frac { x }{ x+1 } +\frac { x+1 }{ x } =\frac { 34 }{ 15 } \)
(ii) \(\frac { x+1 }{ x-1 } +\frac { x-2 }{ x+2 } =3 \)
Solution:
(i) \(\frac { x }{ x+1 } +\frac { x+1 }{ x } =\frac { 34 }{ 15 } \)
\(\frac { { x }^{ 2 }+{ x }^{ 2 }+2x+1 }{ x(x+1) } =\frac { 34 }{ 15 } \)

Question 21.
(i) \(\frac { 1 }{ x-3 } -\frac { 1 }{ x+5 } =\frac { 1 }{ 6 } \)
(ii) \(\frac { x-3 }{ x+3 } +\frac { x+3 }{ x-3 } =2\frac { 1 }{ 2 } \)
Solution:
(i) \(\frac { 1 }{ x-3 } -\frac { 1 }{ x+5 } =\frac { 1 }{ 6 } \)

Question 22.
(i) \(\frac { a }{ ax-1 } +\frac { b }{ bx-1 } =a+b,a+b\neq 0,ab\neq 0\)
(ii) \(\frac { 1 }{ 2a+b+2x } =\frac { 1 }{ 2a } +\frac { 1 }{ b } +\frac { 1 }{ 2x } \)
Solution:
(i) \(\frac { a }{ ax-1 } +\frac { b }{ bx-1 } =a+b\)
⇒ \(\left( \frac { a }{ ax-1 } -b \right) +\left( \frac { b }{ bx-1 } -a \right) =0\)

Question 23.
\(\frac { 1 }{ x+6 } +\frac { 1 }{ x-10 } =\frac { 3 }{ x-4 } \)
Solution:
\(\frac { 1 }{ x+6 } +\frac { 1 }{ x-10 } =\frac { 3 }{ x-4 } \)
⇒ \(\frac { x-10+x+6 }{ (x+6)(x-10) } =\frac { 3 }{ x-4 } \)

Question 24.
(i) \(\sqrt { 3x+4 } =x\)
(ii) \(\sqrt { x(x-7) } =3\sqrt { 2 } \)
Solution:
(i) \(\sqrt { 3x+4 } =x\)
Squaring on both sides

Question 25.
Use the substitution y = 3x + 1 to solve for x : 5(3x + 1 )² + 6(3x + 1) – 8 = 0
Solution:
y = 3x + 1
Now, 5(3x + 1)² + 6(3x + 1) – 8 = 0

Question 26.
Find the values of x if p + 1 =0 and x² + px – 6 = 0
Solution:
p + 1 = 0, then p = – 1
Substituting the value of p in the given quadratic equation
x² + ( – 1)x – 6 = 0
⇒ x² – x – 6 = 0
⇒ x² – 3x + 2x – 6 = 0
⇒ x (x – 3) + 2 (x – 3) = 0
⇒ (x – 3) (x + 2) = 0
Either x – 3 = 0, then x = 3
or x + 2 = 0, then x = – 2
Hence x = 3, -2

Question 27.
Find the values of x if p + 7 = 0, q – 12 = 0 and x² + px + q = 0,
Solution:
p + 7 = 0, then p = – 7
and q – 12 = 0, then q = 12
Substituting the values of p and q in the given quadratic equation,
x² – 7x + 12 = 0
⇒ x² – 3x – 4x + 12 = 0
⇒ x (x – 3) – 4 (x – 3) = 0
⇒ (x – 3) (x – 4) = 0
Either x – 3 = 0, then x = 3
or x – 4 = 0, then x = 4
Hence x = 3, 4

Question 28.
If x = p is a solution of the equation x(2x + 5) = 3, then find the value of p.
Solution:
Given, x = p and x(2x + 5) = 3
Substituting the value of p, we get
p(2p + 5) = 3
⇒ 2p² + 5p – 3 = 0
⇒ 2p² + 6p – p – 3 = 0

Hope given ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 5 Quadratic Equations in One Variable Ex 5.2 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 7 Ratio and Proportion Ex 7.3

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 7 Ratio and Proportion Ex 7.3

More Exercises

• ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 7 Ratio and Proportion Ex 7.1
• ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 7 Ratio and Proportion Ex 7.2
• ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 7 Ratio and Proportion Ex 7.3
• ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 7 Ratio and Proportion MCQS
• ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 7 Ratio and Proportion Chapter Test

Question 1.
If a : b : : c : d, prove that
(i) \(\frac { 2a+5b }{ 2a-5b } =\frac { 2c+5d }{ 2c-5d } \)
(ii) \(\frac { 5a+11b }{ 5c+11d } =\frac { 5a-11b }{ 5c-11d } \)
(iii) (2a + 3b)(2c – 3d) = (2a – 3b)(2c + 3d)
(iv) (la + mb) : (lc + mb) :: (la – mb) : (lc – mb)
Solution:
(i) a : b : : c : d
then \(\frac { a }{ b } =\frac { c }{ d } \)
⇒ \(\frac { 2a }{ 5b } =\frac { 2c }{ 5d } \) (multiply by \(\\ \frac { 2 }{ 5 } \) )

Question 2.
(i) If \(\frac { 5x+7y }{ 5u+7v } =\frac { 5x-7y }{ 5u-7v } \) , Show that \(\frac { x }{ y } =\frac { u }{ v } \)
(ii) \(\frac { 8a-5b }{ 8c-5d } =\frac { 8a+5b }{ 8c+5d } \) , prove that \(\frac { a }{ b } =\frac { c }{ d } \)
Solution:
(i) \(\frac { 5x+7y }{ 5u+7v } =\frac { 5x-7y }{ 5u-7v } \)
Applying alternendo \(\frac { 5x+7y }{ 5u+7v } =\frac { 5x-7y }{ 5u-7v } \)
Applying componendo and dividendo

Question 3.
If (4a + 5b) (4c – 5d) = (4a – 5d) (4c + 5d), prove that a, b, c, d are in proporton.
Solution:
(4a + 5b) (4c – 5d) = (4a – 5d) (4c + 5d)
⇒ \(\frac { 4a+5b }{ 4a-5b } =\frac { 4c+5d }{ 4c-5d } \)
Applying componendo and dividendo

Question 4.
If (pa + qb) : (pc + qd) :: (pa – qb) : (pc – qd) prove that a : b : : c : d
Solution:
(pa + qb) : (pc + qd) :: (pa – qb) : (pc – qd)
⇒ \(\frac { pa+qb }{ pc+qd } =\frac { pq-qb }{ pc-qd } \)
⇒ \(\frac { pa+qb }{ pc-qd } =\frac { pq+qb }{ pc-qd } \)
Applying componendo and dividendo
⇒ \(\frac { pa+qb+pa-qb }{ pa+qb-pa+qb } =\frac { pc+qd+pc-qd }{ pc-qd-pc+qd } \)

Question 5.
If (ma + nb): b :: (mc + nd) : d, prove that a, b, c, d are in proportion.
Solution:
(ma + nb): b :: (mc + nd) : d
⇒ \(\frac { ma+nb }{ b } =\frac { mc+nd }{ d } \)
⇒ mad + nbd = mbc + nbd
⇒ mad = mbc
⇒ ad = bc
⇒ \(\frac { a }{ b } =\frac { c }{ d } \)
Hence a : b :: c : d.

Question 6.
If (11a² + 13b²) (11c² – 13d²) = (11a² – 13b²)(11c² + 13d²), prove that a : b :: c : d.
Solution:
(11a² + 13b²) (11c² – 13d²) = (11a² – 13b²)(11c² + 13d²)
⇒ \(\frac { 11a+{ 13b }^{ 2 } }{ { 11a }^{ 2 }-{ 13b }^{ 2 } } =\frac { { 11c }^{ 2 }+{ 13d }^{ 2 } }{ { 11c }^{ 2 }-{ 13d }^{ 2 } } \)
Applying componendo and dividendo

Question 7.
If (a + 3b + 2c + 6d) (a – 3b – 2c + 6d) = (a + 3b – 2c – 6d) (a – 3b + 2c – 6d), prove that a : b :: c : d.
Solution:
\(\frac { a + 3b + 2c + 6d }{ a – 3b + 2c – 6d } =\frac { a + 3b – 2c – 6d }{ a – 3b – 2c + 6d } \)
⇒ \(\frac { a + 3b + 2c + 6d }{ a + 3b – 2c – 6d } =\frac { a – 3b + 2c – 6d }{ a – 3b – 2c + 6d } \) (by altenendo)
Applying componendo and dividendo

Question 8.
If \(x=\frac { 2ab }{ a+b } \) find the value of \(\frac { x+a }{ x-a } +\frac { x+b }{ x-b } \)
Solution:
\(x=\frac { 2ab }{ a+b } \)
⇒ \(\frac { x }{ a } =\frac { 2b }{ a+b } \)
Applying componendo and dividendo,

Question 9.
If \(x=\frac { 8ab }{ a+b } \) find the value of \(\frac { x+4a }{ x-4a } +\frac { x+4b }{ x-4b } \)
Solution:
\(x=\frac { 8ab }{ a+b } \)
⇒ \(\frac { x }{ 4a } =\frac { 2b }{ a+b } \)
Applying componendo and dividendo,

Question 10.
If \(x=\frac { 4\sqrt { 6 } }{ \sqrt { 2 } +\sqrt { 3 } } \) find the value of \(\frac { x+2\sqrt { 2 } }{ x-2\sqrt { 2 } } +\frac { x+2\sqrt { 3 } }{ x-2\sqrt { 3 } } \)
Solution:
\(x=\frac { 4\sqrt { 6 } }{ \sqrt { 2 } +\sqrt { 3 } } \)
⇒ \(\frac { 4\sqrt { 2 } \times \sqrt { 3 } }{ \sqrt { 2 } +\sqrt { 3 } } \)

Question 11.
Solve \(x:\frac { \sqrt { 36x+1 } +6\sqrt { x } }{ \sqrt { 36x+1 } -6\sqrt { x } } =9 \)
Solution:
\(\frac { \sqrt { 36x+1 } +6\sqrt { x } }{ \sqrt { 36x+1 } -6\sqrt { x } } =\frac { 9 }{ 1 } \)
Applying componendo and dividendo,

Question 12.
Find x from the following equations :
(i) \(\frac { \sqrt { 2-x } +\sqrt { 2+x } }{ \sqrt { 2-x } -\sqrt { 2+x } } =3 \)
(ii) \(\frac { \sqrt { x+4 } +\sqrt { x-10 } }{ \sqrt { x+4 } -\sqrt { x-10 } } =\frac { 5 }{ 2 } \)
(iii) \(\frac { \sqrt { 1+x } +\sqrt { 1-x } }{ \sqrt { 1+x } -\sqrt { 1-x } } =\frac { a }{ b } \)
(iv) \(\frac { \sqrt { 12x+1 } +\sqrt { 2x-3 } }{ \sqrt { 12x+1 } -\sqrt { 2x-3 } } =\frac { 3 }{ 2 } \)
(v) \(\frac { 3x+\sqrt { { 9x }^{ 2 }-5 } }{ 3x-\sqrt { { 9x }^{ 2 }-5 } } =5 \)
(vi) \(\frac { \sqrt { a+x } +\sqrt { a-x } }{ \sqrt { a+x } -\sqrt { a-x } } =\frac { c }{ d } \)
Solution:
(i) \(\frac { \sqrt { 2-x } +\sqrt { 2+x } }{ \sqrt { 2-x } -\sqrt { 2+x } } =3 \)
Applying componendo and dividendo,

Question 13.
Solve \(\frac { 1+x+{ x }^{ 2 } }{ 1-x+{ x }^{ 2 } } =\frac { 62\left( 1+x \right) }{ 63\left( 1-x \right) } \)
Solution:
\(\frac { 1+x+{ x }^{ 2 } }{ 1-x+{ x }^{ 2 } } =\frac { 62\left( 1+x \right) }{ 63\left( 1-x \right) } \)
⇒ \(\frac { \left( 1-x \right) \left( 1+x+{ x }^{ 2 } \right) }{ \left( 1+x \right) \left( 1-x+{ x }^{ 2 } \right) } =\frac { 62 }{ 63 } \)
⇒ \(\frac { \left( 1+x \right) \left( 1-x+{ x }^{ 2 } \right) }{ \left( 1-x \right) \left( 1+x+{ x }^{ 2 } \right) } =\frac { 63 }{ 62 } \)

Question 14.
Solve for \(x:16{ \left( \frac { a-x }{ a+x } \right) }^{ 3 }=\frac { a+x }{ a-x } \)
Solution:
\(x:16{ \left( \frac { a-x }{ a+x } \right) }^{ 3 }=\frac { a+x }{ a-x } \)
⇒ \(\left( \frac { a+x }{ a-x } \right) \times { \left( \frac { a+x }{ a-x } \right) }^{ 3 }=16 \)

Question 15.
If \(x=\frac { \sqrt { a+x } +\sqrt { a-1 } }{ \sqrt { a+1 } -\sqrt { a-1 } } \) , using properties of proportion , show that x² – 2ax + 1 = 0
Solution:
We have \(x=\frac { \sqrt { a+x } +\sqrt { a-1 } }{ \sqrt { a+1 } -\sqrt { a-1 } } \)
⇒ \(\frac { x+1 }{ x-1 } =\frac { 2\sqrt { a+1 } }{ 2\sqrt { a-1 } } \)

Question 16.
Given \(x=\frac { \sqrt { { a }^{ 2 }+{ b }^{ 2 } } +\sqrt { { a }^{ 2 }-{ b }^{ 2 } } }{ \sqrt { { a }^{ 2 }+{ b }^{ 2 } } -\sqrt { { a }^{ 2 }-{ b }^{ 2 } } } \) Use componendo and dividendo to prove that \({ b }^{ 2 }=\frac { { 2a }^{ 2 }x }{ { x }^{ 2 }+1 } \)
Solution:
If \(\frac { x }{ 1 } =\frac { \sqrt { { a }^{ 2 }+{ b }^{ 2 } } +\sqrt { { a }^{ 2 }-{ b }^{ 2 } } }{ \sqrt { { a }^{ 2 }+{ b }^{ 2 } } -\sqrt { { a }^{ 2 }-{ b }^{ 2 } } } \)
Applying componendo and dividendo both sides

Question 17.
Given that \(\frac { { a }^{ 3 }+3{ ab }^{ 2 } }{ { b }^{ 3 }+{ 3a }^{ 2 }b } =\frac { 63 }{ 62 } \). Using componendo and dividendo find a: b. (2009)
Solution:
Given that \(\frac { { a }^{ 3 }+3{ ab }^{ 2 } }{ { b }^{ 3 }+{ 3a }^{ 2 }b } =\frac { 63 }{ 62 } \)
By componendo and dividendo

a : b = 3 : 2

Question 18.
Give \(\frac { { x }^{ 3 }+12x }{ { 6x }^{ 2 }+8 } =\frac { { y }^{ 3 }+27y }{ { 9y }^{ 2 }+27 } \) Using componendo and dividendo find x : y.
Solution:
Give \(\frac { { x }^{ 3 }+12x }{ { 6x }^{ 2 }+8 } =\frac { { y }^{ 3 }+27y }{ { 9y }^{ 2 }+27 } \)
Using componendo-dividendo, we have

Question 19.
Using the properties of proportion, solve the following equation for x; given
\(\frac { x^{ 3 }+3x }{ { 3x }^{ 2 }+1 } =\frac { 341 }{ 91 } \)
Solution:
\(\frac { x^{ 3 }+3x }{ { 3x }^{ 2 }+1 } =\frac { 341 }{ 91 } \)
Applying componendo and dividendo

Question 20.
If \(\frac { x+y }{ ax+by } =\frac { y+z }{ ay+bz } =\frac { z+x }{ az+bx } \) , prove that each of these ratio is equal to \(\\ \frac { 2 }{ a+b } \) unless x + y + z = 0
Solution:
\(\frac { x+y }{ ax+by } =\frac { y+z }{ ay+bz } =\frac { z+x }{ az+bx } \)
= \(\frac { x+y+y+z+z+x }{ ax+by+ay+bz+az+bx } \)

Hope given ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 7 Ratio and Proportion Ex 7.3 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.

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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