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ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 7 Ratio and Proportion Ex 7.2

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 7 Ratio and Proportion Ex 7.2

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 value of x in the following proportions :
(i) 10 : 35 = x : 42
(ii) 3 : x = 24 : 2
(iii) 2.5 : 1.5 = x : 3
(iv) x : 50 :: 3 : 2
Solution:
(i) 10 : 35 = x : 42
⇒ 35 × x = 10 × 42

Question 2.
Find the fourth proportional to
(i) 3, 12, 15
(ii) \(\frac { 1 }{ 3 } ,\frac { 1 }{ 4 } ,\frac { 1 }{ 5 } \)
(iii) 1.5, 2.5, 4.5
(iv) 9.6 kg, 7.2 kg, 28.8 kg
Solution:
(i) Let fourth proportional to
3, 12, 15 be x.
then 3 : 12 :: 15 : x

Question 3.
Find the third proportional to
(i) 5, 10
(ii) 0.24, 0.6
(iii) Rs. 3, Rs. 12
(iv) \(5 \frac { 1 }{ 4 } \) and 7.
Solution:
(i) Let x be the third proportional to 5, 10,
then 5 : 10 :: 10 : x

Question 4.
Find the mean proportion of:
(i) 5 and 80
(ii) \(\\ \frac { 1 }{ 12 } \) and \(\\ \frac { 1 }{ 75 } \)
(iii) 8.1 and 2.5
(iv) (a – b) and (a³ – a²b), a> b
Solution:
(i) Let x be the mean proportion of 5 and 80 ,
then 5 : x : : x : 80
x² = 5 x 80
⇒ x = \(\sqrt { 5\times 80 } =\sqrt { 400 } \) = 20
x = 20

Question 5.
If a, 12, 16 and b are in continued proportion find a and b.
Solution:
∵ a, 12, 16, b are in continued proportion, then

Question 6.
What number must be added to each of the numbers 5, 11, 19 and 37 so that they are in proportion ? (2009)
Solution:
Let x be added to 5, 11, 19 and 37 to make them in proportion.
5 + x : 11 + x : : 19 + x : 37 + x

Question 7.
What number should be subtracted from each of the numbers 23, 30, 57 and 78 so that the remainders are in proportion ? (2004)
Solution:
Let x be subtracted from each term, then
23 – x, 30 – x, 57 – x and 78 – x are proportional
23 – x : 30 – x : : 57 – x : 78 – x

Question 8.
If 2x – 1, 5x – 6, 6x + 2 and 15x – 9 are in proportion, find the value of x.
Solution:
∵ 2x – 1, 5x – 6, 6x + 2 and 15x – 9 are in proportion.
then (2x – 1) (15x – 9) = (5x – 6) (6x + 2)

Question 9.
If x + 5 is the mean proportion between x + 2 and x + 9, find the value of x.
Solution:
∵ x + 5 is the mean proportion between x + 2 and x + 9, then
(x + 5)² = (x + 2) (x + 9)
⇒ x² + 10x + 25 = x² + 11x + 18
⇒ x² + 10x – x² – 11x = 18 – 25
⇒ – x = – 7
∵ x = 7 Ans.

Question 10.
What number must be added to each of the numbers 16, 26 and 40 so that the resulting numbers may be in continued proportion?
Solution:
Let x be added to each number then
16 + x, 26 + x and 40 + x
are in continued proportion.

Question 11.
Find two numbers such that the mean proportional between them is 28 and the third proportional to them is 224.
Solution:
Let the two numbers are a and b.
∵ 28 is the mean proportional
∵ a : 28 : : 28 : b

Question 12.
If b is the mean proportional between a and c, prove that a, c, a² + b², and b² + c² are proportional.
Solution:
∵ b is the mean proportional between a and c, then,
b² = a × c ⇒ b² = ac …(i)

Question 13.
If b is the mean proportional between a and c, prove that (ab + bc) is the mean proportional between (a² + b²) and (b² + c²).
Solution:
b is the mean proportional between a and c then
b² = ac …(i)
Now if (ab + bc) is the mean proportional

Question 14.
If y is mean proportional between x and z, prove that
xyz (x + y + z)³ = (xy + yz + zx)³.
Solution:
∵ y is the mean proportional between
x and z, then
y² = xz …(i)

Question 15.
If a + c = mb and \(\frac { 1 }{ b } +\frac { 1 }{ d } =\frac { m }{ c } \), prove that a, b, c and d are in proportion.
Solution:
a + c = mb and \(\frac { 1 }{ b } +\frac { 1 }{ d } =\frac { m }{ c } \)
a + c = mb

Question 16.
If \(\frac { x }{ a } =\frac { y }{ b } =\frac { z }{ c } \), prove that
(i)\(\frac { { x }^{ 3 } }{ { a }^{ 2 } } +\frac { { y }^{ 3 } }{ { b }^{ 2 } } +\frac { { z }^{ 3 } }{ { c }^{ 2 } } =\frac { { \left( x+y+z \right) }^{ 3 } }{ { \left( a+b+c \right) }^{ 2 } } \)

Solution:
\(\frac { x }{ a } =\frac { y }{ b } =\frac { z }{ c } \)
∴ x = ak, y = bk, z = ck

Question 17.
If \(\frac { a }{ b } =\frac { c }{ d } =\frac { e }{ f } \) prove that :
(i) (b² + d² + f²) (a² + c² + e²) = (ab + cd + ef)²

Solution:
\(\frac { a }{ b } =\frac { c }{ d } =\frac { e }{ f } \) = k(say)
∴ a = bk, c = dk, e =fk

Question 18.
If ax = by = cz; prove that
\(\frac { { x }^{ 2 } }{ yz } +\frac { { y }^{ 2 } }{ zx } +\frac { { z }^{ 2 } }{ xy } \) = \(\frac { bc }{ { a }^{ 2 } } +\frac { ca }{ { b }^{ 2 } } +\frac { ab }{ { c }^{ 2 } } \)
Solution:
Let ax = by = cz = k

Question 19.
If a, b, c and d are in proportion, prove that:
(i) (5a + 7b) (2c – 3d) = (5c + 7d) (2a – 3b)
(ii) (ma + nb) : b = (mc + nd) : d
(iii) (a⁴ + c⁴) : (b⁴ + d⁴) = a² c² : b² d².

Solution:
∵ a, b, c, d are in proportion
\(\\ \frac { a }{ b } \) = \(\\ \frac { c }{ d } \) = k(say)

Question 20.
If x, y, z are in continued proportion, prove that:\(\frac { { \left( x+y \right) }^{ 2 } }{ { \left( y+z \right) }^{ 2 } } =\frac { x }{ z } \). (2010)
Solution:
x, y, z are in continued proportion
Let \(\\ \frac { x }{ y } \) = \(\\ \frac { y }{ z } \) = k

Question 21.
If a, b, c are in continued proportion, prove that:
\(\frac { { pa }^{ 2 }+qab+{ rb }^{ 2 } }{ { pb }^{ 2 }+qbc+{ rc }^{ 2 } } =\frac { a }{ c } \)
Solution:
Given a, b, c are in continued proportion
\(\frac { { pa }^{ 2 }+qab+{ rb }^{ 2 } }{ { pb }^{ 2 }+qbc+{ rc }^{ 2 } } =\frac { a }{ c } \)
Let \(\\ \frac { a }{ b } \) = \(\\ \frac { b }{ c } \) = k

Question 22.
If a, b, c are in continued proportion, prove that:
(i) \(\frac { a+b }{ b+c } =\frac { { a }^{ 2 }(b-c) }{ { b }^{ 2 }(a-b) } \)

Solution:
As a, b, c, are in continued proportion
Let \(\\ \frac { a }{ b } \) = \(\\ \frac { b }{ c } \) = k

Question 23.
If a, b, c, d are in continued proportion, prove that:
(i) \(\frac { { a }^{ 3 }+{ b }^{ 3 }+{ c }^{ 3 } }{ { b }^{ 3 }+{ c }^{ 3 }+{ d }^{ 3 } } =\frac { a }{ d } \)

Solution:
a, b, c, d are in continued proportion
∴ \(\frac { a }{ b } =\frac { b }{ c } =\frac { c }{ d } =k(say)\)

Hope given ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 7 Ratio and Proportion Ex 7.2 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 7.1

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 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.

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