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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 7.2

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:

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:

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:

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:

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

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:

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:

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

Question 9.
If x + 5 is the mean proportion between x + 2 and x + 9, find the value of x.
Solution:

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:

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

Question 12.
If b is the mean proportional between a and c, prove that a, c, a² + b², and b² + c² are proportional.
Solution:

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:

Question 14.
If y is mean proportional between x and z, prove that
xyz (x + y + z)³ = (xy + yz + zx)³.
Solution:

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:

Question 16.

Solution:

Question 17.

Solution:

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:

Question 19.

Solution:

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:

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:

Question 22.

Solution:

Question 23.

Solution:

ML Aggarwal Class 10 Solutions for ICSE Maths

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 7.1

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:

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:

Question 3.
Find the duplicate ratio of
(i) 2 : 3
(ii) √5 : 7
(iii) 5a : 6b
Solution:

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

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

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

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

Question 8.
Arrange the following ratios in ascending order of magnitude:
2 : 3, 17 : 21, 11 : 14 and 5 : 7
Solution:

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:

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:

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:

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:

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:

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:

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:

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:

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:

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:

Question 19.
(i) In a mixture of 45 liters, 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:

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 fewer girls, then the ratio would have been 2 : 1, How many students were there in the class?
Solution:

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:

ML Aggarwal Class 10 Solutions for ICSE Maths

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

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:

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

Question 3.
If (2x – 3) is a factor of 6x² + x + a, find the value of a. With this value of a, factorize the given expression.
Solution:

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:

Question 5.
Prove that (5x + 4) is a factor of 5x³ + 4x² – 5x – 4. Hence factorize the given polynomial completely.
Solution:

Question 6.
Use factor theorem to factorise the following polynomials completely:
(i) 4x³ + 4x² – 9x – 9
(ii) x³ – 19x – 30
Solution:

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:

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, factorize the given expression.
Solution:

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:

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:

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:

ML Aggarwal Class 10 Solutions for ICSE Maths

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

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

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:

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:

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:

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:

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:

ML Aggarwal Class 10 Solutions for ICSE Maths

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

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

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:

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:

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:

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:

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:

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:

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:

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:

Question 9.
By factor theorem, show that (x + 3) and (2x – 1) are factors of 2x² + 5x – 3.
Solution:

Question 10.
Show that (x – 2) is a factor of 3x² – x – 10. Hence factorise 3x² – x – 10.
Solution:

Question 11.
Show that (x – 1) is a factor of x³ – 5x² – x + 5 Hence factorise x³ – 5x² – x + 5.
Solution:

Question 12.
Show that (x – 3) is a factor of x³ – 7x² + 15x – 9. Hence factorise x³ – 7x² + 15 x – 9
Solution:

Question 13.
Show that (2x + 1) is a factor of 4x³ + 12x² + 11 x + 3. Hence factorise 4x³ + 12x² + 11x + 3.
Solution:

Question 14.
Show that 2x + 7 is a factor of 2x³ + 5x² – 11x – 14. Hence factorize the given expression completely, using the factor theorem. (2006)
Solution:

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

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:

Question 17.
Using the Remainder and Factor Theorem, factorize the following polynomial: x³ + 10x² – 37x + 26.
Solution:

Question 18.
If (2x + 1) is a factor of 6x³ + 5x² + ax – 2 find the value of a
Solution:

Question 19.
If (3x – 2) is a factor of 3x³ – kx² + 21x – 10, find the value of k.
Solution:

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, factorize the above expression completely
Solution:

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:

Question 22.
What number should be subtracted from 2x³ – 5x² + 5x so that the resulting polynomial has 2x – 3 as a factor?
Solution:

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:

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, factorize the given expression.
Solution:

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:

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:

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, factorize the given expression.
Solution:

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:

ML Aggarwal Class 10 Solutions for ICSE Maths

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

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:

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

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

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

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:

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:

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:

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

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:

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:

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:

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

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:

Question 14.
Find two natural numbers which differ by 3 and whose squares have the sum 117.
Solution:

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:

Question 16.
Two natural numbers are in the ratio 3 : 4. Find the numbers if the difference between their squares is 175.
Solution:

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:

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:

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

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:

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:

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:

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:

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:

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:

Question 26.
Forty years hence, Mr Pratap’s age will be the square of what it was 32 years ago. Find the present age.
Solution:

ML Aggarwal Class 10 Solutions for ICSE Maths