Selina Concise Mathematics Class 10 ICSE Solutions Chapter 9 Matrices Ex 9B

These Solutions are part of Selina Concise Mathematics Class 10 ICSE Solutions. Here we have given Selina Concise Mathematics Class 10 ICSE Solutions Chapter 9 Matrices Ex 9B.

Other Exercises

Question 1.
Evaluate

Solution:

Question 2.
Find x and y if :

Solution:

Question 3.

Solution:

Question 4.

Solution:

Question 5.

Solution:

Question 6.

Solution:

Question 7.

Solution:

Question 8.

Solution:

Question 9.

Solution:

Question 10.
If I is the unit matrix of order 2 x 2; find the matrix M, such that:

Solution:

Question 11.

Solution:

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Selina Concise Mathematics Class 10 ICSE Solutions Chapter 9 Matrices Ex 9A

These Solutions are part of Selina Concise Mathematics Class 10 ICSE Solutions. Here we have given Selina Concise Mathematics Class 10 ICSE Solutions Chapter 9 Matrices Ex 9A.

Other Exercises

Question 1.
State, whether the following statements are true or false. If false, give a reason.
(i) If A and B are two matrices of orders 3 x 2 and 2 x 3 respectively; then their sum A + B is possible,
(ii) The matrices A 2 x3, and B 2 x 3, are conformable for subtraction,
(iii) Transpose of a 2 x 1 matrix is a 2 x 1 matrix,
(iv) Transpose of a square matrix is a square matrix,
(v) A column matrix has many columns and only one row.
Solution:
(i) False, because the orders of both the matrices are not same,
(ii) True,
(iii) False, because transpose of a 2 x 1 matrix is a 1 x 2 matrix,
(iv) True,
(v) False, because it has only one column and may have many rows.

Question 2.

Solution:
Comparing the elements in order, we get that :
x = 3
y + 2 = 1 ⇒ y = 1 – 2 = -1
z – 1 = 2 ⇒ z = 2 + 1 = 3
Hence x = 3, y = -1 , z = 3

Question 3.
Solve for a, b and c; if :

Solution:
(i) Comparing the elements in order, we get
-4 = b + 4 ⇒ – b = 4 + 4 = 8 ⇒ b = – 8
a + 5 = 2 ⇒ a = 2 – 5= -3
2 = c – 1 ⇒ c = 2 + 1 = 3
Hence a = -3, b = -8, c = 3
(ii) Comparing the elements in order, we get
a = 3
a – b = – 1 ⇒ 3 – b = -1 ⇒ -b = -1 – 3 ⇒ -b = -4 ⇒ b = 4
b + c = 2 ⇒ 4 + c = 2 ⇒ c = 2 – 4 = -2
Hence a = 3, b = 4, c = – 2

Question 4.

Solution:

Question 5.

Solution:

Question 6.
Wherever possible, write each of the following as a single matrix.

Solution:

This is not possible because both the matrices are not of the same order.

Question 7.
Find, x and y from the following equations:

(ii) [-8 x] + [y -2] = [-3 2]
Solution:

Comparing, the elements of two equal martrices:
3 – x = 7 ⇒ x = -7 + 3 = -4
and y + 2 = 2 ⇒ y = 2 – 2 = 0
Hence x = – 4, y = 0
(ii) [-8 x] + [y -2] = [-3 2]
⇒ [-8 + y x – 2] = [-3 2]
Comparing the elements, we get :
-8 + y = -3 ⇒ y = -3 + 8 = 5
x – 2 = 2 ⇒ x = 2 + 2 = 4
Hence x = 4, y = 5

Question 8.

Solution:

Question 9.
Write the additive inverse of matrices A, B and C

Solution:

Question 10.
Given A = [ 2 -3], B = [0 2] and C = [-1 4]; find the matrix X in each of the following :
(i) X + B = C – A
(ii) A – X = B + C
Solution:
A = [2 -3], B = [0 2], C = [-1 4]
(i) X + B = C – A
X + [0 2] = [-1 4] – [2 -3]
⇒ X + [0 2] = [- 1 – 2 4 – (-3)]
X + [0 2] = [-3 4 + 3] = [-3 7]
X = [-3 7] – [0 2] = [-3 – 0 7 – 2] = [-3 5]
(ii) A – X = B + C
⇒ -X = B + C – A
⇒ X = A – B – C = [2 -3] – [0 2] – [-1 4] = [2 – 0 – (-1) -3 – 2 – 4] = [ 2 + 1 – 9] = [3 – 9]

Question 11.

Solution:

Hope given Selina Concise Mathematics Class 10 ICSE Solutions Chapter 9 Matrices Ex 9A are helpful to complete your math homework.

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Selina Concise Mathematics Class 10 ICSE Solutions Chapter 8 Remainder and Factor Theorems Ex 8C

These Solutions are part of Selina Concise Mathematics Class 10 ICSE Solutions. Here we have given Selina Concise Mathematics Class 10 ICSE Solutions Chapter 8 Remainder and Factor Theorems Ex 8C.

Other Exercises

Question 1.
Show that (x – 1) is a factor of x3 – 7x2 + 14x – 8. Hence, completely factorise the given expression.
Solution:

Question 2.
Using Remainder Theorem, factorise : x3 + 10x2 – 37x + 26 completely. (2014)
Solution:
f(x) = x3 + 10x2 – 37x + 26
f(1) = (1)3 + 10(1)2 – 37(1) + 26 = 1 + 10 – 37 + 26 = 0
x = 1
x – 1 is factor of f(x)

Question 3.
When x3 + 3x2 – mx + 4 is divided by x – 2, the remainder is m + 3. Find the value of m.
Solution:
Let f(x) = x3 + 3x2 – mx + 4
and x – 2 = 0 then x = 2
f(2) = (2)3 + 3(2)2 – m(2) + 4 = 8 + 12 – 2m + 4 = 24 – 2m
Remainder = 24 – 2m
But, remainder is given m + 3
m + 3 = 24 – 2m
⇒ m + 2m = 24 – 3
⇒ 3m = 21
⇒ m = 7
Hence m = 7

Question 4.
What should be subtracted from 3x3 – 8x2 + 4x – 3, so that the resulting expression has x + 2 as a factor ?
Solution:
The number to be subtracted = Remainder obtained by dividing 3x3 – 8x2 + 4x – 3 by x + 2
Let f(x) = 3x3 – 8x2 + 4x – 3
and x + 2 = 0, then x = – 2
Remainder = f(-2) = 3 (-2)3 – 8 (-2)2 + 4 (-2) – 3 = -24 – 32 – 8 – 3 = -67
Hence the number to be subtracted = – 67

Question 5.
If (x + 1) and (x – 2) are factors of x3 + (a + 1) x2 – (b – 2) x – 6, find the values of a and 6. And then, factorise the given expression completely.
Solution:

Question 6.
If x – 2 is a factor of x2 – ax + b and a + b = 1, find the values of a and b.
Solution:
(x – 2) is a factor of x2 + ax + b
Let x – 2 = 0 ⇒ x = 2
Now x2 + ax + b = (2)2 + a x 2 + b = 4 + 2a + b = 2a + b + 4
x – 2 is the factor Remainder = 0 or 2a + b + 4 = 0
⇒ 2a + b = -4 …(i)
But a + b = 1 (given) …(ii)
Subtracting, we get : a = -5
Substituting the value of a in (ii)
-5 + b = 1 ⇒ b = 1 + 5 ⇒ b = 6
Hence a = -5, b = 6

Question 7.
Factorise x3 + 6x2 + 11x + 6 completely using factor theorem.
Solution:

Question 8.
Find the value of ‘m’ if mx3 + 2x2 – 3 and x2 – mx + 4 leave the same remainder when each is divided by x – 2.
Solution:
Let f(x) = mx3 + 2x2 – 3
g (x) = x2 – mx + 4
Let x – 2 = 0, then x = 2
f(2) = m (2)3 + 2 (2)2 – 3 = 8m + 8 – 3 = 8m + 5
g(2) = (2)2 – mx2 + 4 = 4 – 2m + 4 = 8 – 2m
In both cases the remainder is same
8m + 5 = 8 – 2m
⇒ 8m + 2m = 8 – 5
⇒ 10m = 3
⇒ m = $$\frac { 3 }{ 10 }$$

Question 9.
The polynomial px3 + 4x2 – 3x + q is completely divisible by x2 – 1; find the values of p and q. Also, for these values of p and q, factorize the given polynomial completely.
Solution:

Question 10.
Find the number which should be added to x2 + x + 3 so that the resulting polynomial is completely divisible by (x + 3).
Solution:
Let k be added to f(x), then f(x) = x2 + x + 3 + k
Let x + 3 = 0, then x = -3
f(-3) = (-3)2 + (-3) + 3 + k = 9 – 3 + 3 + k = 9 + k
f(x) is divisible by x + 3, then remainder will be 0.
9 + k = 0 ⇒ k = -9

Question 11.
When the polynomial x3 + 2x2 – 5ax – 7 is divided by (x – 1), the remainder is A and when the polynomial x3 + ax2 – 12x + 16 is divided by (x + 2), the remainder is B. Find the value of ‘a’ if 2A + B = 0.
Solution:
Let f(x) = x3 + 2x2 – 5ax – 1
and let x – 1 = 0, then x = 1
f(1) = (1)3 + 2(1)2 – 5a x 1 – 7 = 1 + 2 – 5a – 7 = -5a – 4
-5a – 4 = A ….(i)
Let g (x) = x3 + ax2 +12x + 16
and let x + 2 = 0, then x = -2
g (-2) = (-2)3 + a (-2)2 – 12 (-2) + 16 = -8 + 4a + 24 + 16 = 32 + 4a
32 + 4a = B ….(ii)
2A + B = 0
2 (-5a – 4) + 32 + 4a = 0
⇒ -10a – 8 + 32 + 4a = 0
⇒ -6a + 24 = 0
⇒ 6a = 24
⇒ a = 4
a = 4

Question 12.
(3x + 5) is a factor of the polynomial (a – 1) x3 + (a + 1) x2 – (2a + 1) x – 15. Find the value of ‘a’. For this value of ‘a’, factorise the given polynomial completely.
Solution:
Let f(x) = (a – 1) x3 + (a + 1) x2 – (2a + 1) x – 15

Question 13.
When divided by x – 3 the polynomials x3 – px2 + x + 6 and 2x3 – x2 – (p + 3) x – 6 leave the same remainder. Find the value of ‘p.’
Solution:
When (x – 3) divides x3 – px2 + x + 6,
then Remainder = p(3) = (3)3 – p(3)2 + (3) + 6 = 27 – 9p + 9 = 36 – 9p
When (x – 3) divides 2x3 – x2 – (p + 3) x – 6,
then Remainder = p(3) = 2(3)3 – (3)2 – (p + 3) (3) – 6
= 54 – 9 – 3p – 9 – 6 = 30 – 3p
A.T.Q. both remainders are equal
⇒ 36 – 9p = 30 – 3p
⇒ 36 – 30 = -3p + 9p
⇒ 6 = 6p
⇒ p = 1

Question 14.
Use the Remainder Theorem to factorise the following expression : 2x3 + x2 – 13x + 6
Solution:
(a) By hit and trial, putting x = 2, we have
2 (8) + 4 – 26 + 6 = 0
⇒ (x – 2) is the factor of 2x3 + x2 – 13x + 6

Question 15.
Using remainder theorem, find the value of k if on dividing 2x3 + 3x2 – kx + 5 by x – 2, leaves a remainder 7. (2016)
Solution:
Let f(x) = 2x3 + 3x2 – kx + 5 By the remainder theorem,
f(2) = 7
⇒ 2(2)3 + 3(2)2 – k(2) + 5 = 7
⇒ 2(8) + 3(4) – k(2) + 5 = 7
⇒ 16 + 12 – 2k + 5 = 7
⇒ 2k = 16 + 12 + 5 – 7
⇒ 2k = 26
⇒ k = 13
The value of k is 13.

Hope given Selina Concise Mathematics Class 10 ICSE Solutions Chapter 8 Remainder and Factor Theorems Ex 8C are helpful to complete your math homework.

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Selina Concise Mathematics Class 10 ICSE Solutions Chapter 8 Remainder and Factor Theorems Ex 8B

These Solutions are part of Selina Concise Mathematics Class 10 ICSE Solutions. Here we have given Selina Concise Mathematics Class 10 ICSE Solutions Chapter 8 Remainder and Factor Theorems Ex 8B.

Other Exercises

Question 1.
Using the Factor Theorem, show that:
(i) (x – 2) is a factor of x3 – 2x2 – 9x +18. Hence, factorise the expression x3 – 2x2 – 9x + 18 completely.
(ii) (x + 5) is a factor of 2x3 + 5x2 – 28x – 15. Hence, factorise the expression 2x3 + 5x2 – 28x – 15 completely.
(iii) (3x + 2) is a factor of 3x3 + 2x2 – 3x – 2. Hence, factorise the expression 3x3 + 2x2 – 3x – 2. completely.
(iv) 2x + 7 is a factor of 2x3 + 5x2 – 11x – 14. Hence, factorise the given expression completely.
Solution:

Question 2.
Using the Remainder Theorem, factorise each of the following completely:
(i) 3x3 + 2x2 – 19x + 6
(ii) 2x3 + x2 – 13x + 6
(iii) 3x3 + 2x2 – 23x – 30
(iv) 4x3 + 7x2 – 36x – 63
(v) x3 + x2 – 4x – 4. (2004)
Solution:

Question 3.
Using the Remainder Theorem factorise the expression 3x3 + 10x2 + x – 6. Hence, solve the equation. 3x3 + 10x2 + x – 6 = 0
Solution:

Question 4.
Factorise the expression f(x) = 2x3 – 7x2 – 3x + 18. Hence, find all possible values of x for which f(x) = o.
Solution:

Question 5.
Given that x – 2 and x +1 are factors of f(x) = x3 + 3x2 + ax + b; calculate the values of a and b. Hence, find all the factors of f(x).
Solution:

Question 6.
The expression 4x3 – bx2 + x – c leaves remainders 0 and 30 when divided by x + 1 and 2x – 3 respectively. Calculate the values of b and c. Hence, factorise the expression completely.
Solution:

Question 7.
If x + a is a common factor of expressions f(x) = x2 + px + q and g (x) = x2 + mx + n show that a = $$\frac { n – q }{ m – p }$$
Solution:

Question 8.
The polynomials ax3 + 3x2 – 3 and 2x3 – 5x + a, when divided by x – 4, leave the same remainder in each case. Find the value of a.
Solution:

Question 9.
Find the value of ‘a’ if (x – a) is a factor of x3 – ax2 + x + 2.
Solution:

Question 10.
Find the number that must be subtracted from the polynomial 3y3 + y2 – 22y + 15, so that the resulting polynomial is completely divisible by y + 3.
Solution:

⇒ 9 – k = 0 ⇒ k = 9
Hence 9 should be subtracted.

Hope given Selina Concise Mathematics Class 10 ICSE Solutions Chapter 8 Remainder and Factor Theorems Ex 8B are helpful to complete your math homework.

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Selina Concise Mathematics Class 10 ICSE Solutions Chapter 8 Remainder and Factor Theorems Ex 8A

These Solutions are part of Selina Concise Mathematics Class 10 ICSE Solutions. Here we have given Selina Concise Mathematics Class 10 ICSE Solutions Chapter 8 Remainder and Factor Theorems Ex 8A.

Other Exercises

Question 1.
Find in each case, the remainder when :
(i) x4 – 3x2 + 2x + 1 is divided by x – 1.
(ii) x3 + 3x2 – 12x + 4 is divided by x – 2.
(iii) x4 + 1 is divided by x + 1.
(iv) 4x3 – 3x2 + 2x – 4 is divided by 2x + 1.
(v) 4x3 + 4x2 – 21x + 16 is divided by 2x – 3.
(vi) 2x3 + 9x2 – x – 15 is divided by 2x + 3.
Solution:

Question 2.
Show that:
(i) x – 2 is a factor of 5x2 + 15x – 50.
(ii) 3x + 2 is a factor of 3x2 – x – 2.
(iii) x + 1 is a factor of x3 + 3x2 + 3x + 1.
Solution:

Question 3.
Use the Remainder Theorem to find which of the following is a factor of 2x3 + 3x2 – 5x – 6.
(i) x + 1
(ii) 2x – 1
(iii) x + 2
(iv) 3x – 2
(v) 2x – 3.
Solution:
(i) f(x) = 2×3 + 3×2 – 5x – 6

Question 4.
(i) If 2x + 1 is a factor of 2x2 + ax – 3, find the value of a.
(ii) Find the value of k, if 3x – 4 is a factor of expression 3x2 + 2x – k.
Solution:

Question 5.
Find the values of constants a and b when x – 2 and x + 3 both are the factors of expression, x3 + ax2 + bx – 12.
Solution:

(x + 3) is a factor of f(x)
f(-3) = 0
9a – 3b – 39 – 0
⇒ 3a – b – 13 = 0
⇒ 3a – b = 13
Adding (i) and (ii) we get:
5a = 15 ⇒ a = 3
Substituting the value of a in (i)
2(3) + b = 2
⇒ 6 + b = 2
⇒ b = 2 – 6 = – 4
Hence a = 3, b = – 4

Question 6.
Find the value of k, if 2x + 1 is a factor of (3k + 2) x3 + (k – 1).
Solution:

Question 7.
Find the value of a, if x – 2 is a factor of 2x5 – 6x4 – 2ax3 + 6ax2 + 4ax + 8.
Solution:

Question 8.
Find the values of m and n so that x – 1 and x + 2 both are factors of x3 + (3m + 1) x2 + nx – 18.
Solution:

Question 9.
When x3 + 2x2 – kx + 4 is divided by x – 2, the remainder is k. Find the value of constants.
Solution:
f(x) = x3 + 2x2 – kx + 4

Question 10.
Find the value of a, if the division of ax3 + 9x2 + 4x – 10 by x + 3 leaves a remainder 5.
Solution:

Question 11.
If x3 + ax2 + bx + 6 has x – 2 as a factor and leaves a remainder 3 when divided by x – 3, find the values of a and b. [2005]
Solution:

Question 12.
The expression 2x3 + ax2 + bx – 2 leaves remainder 7 and 0 when divided by 2x – 3 and x + 2 respectively. Calculate the values of a and b.
Solution:

Question 13.
What number should be added to 3x3 – 5x2 + 6x so that when resulting polynomial is divided by x – 3, the remainder is 8 ?
Solution:

Question 14.
What number should be subtracted from x3 + 3x2 – 8x + 14 so that on dividing it by x – 2, the remainder is 10.
Solution:

Question 15.
The polynomials 2x3 – 7x2 + ax – 6 and x3 – 8x2 + (2a + 1) x – 16 leave the same remainder when divided by x – 2. Find the value of ‘a’.
Solution:

Question 16.
If (x – 2) is a factor of the expression 7x3 + 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 17.
Find ‘a’ if the two polynomials ax3 + 3x2 – 9 and 2x3 + 4x + a, leave the same remainder when divided by x + 3. (2015)
Solution:

Hope given Selina Concise Mathematics Class 10 ICSE Solutions Chapter 8 Remainder and Factor Theorems Ex 8A are helpful to complete your math homework.

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Selina Concise Mathematics Class 10 ICSE Solutions Chapter 7 Ratio and Proportion (Including Properties and Uses) Ex 7D

These Solutions are part of Selina Concise Mathematics Class 10 ICSE Solutions. Here we have given Selina Concise Mathematics Class 10 ICSE Solutions Chapter 7 Ratio and Proportion Ex 7D.

Other Exercises

Question 1.
If a : b = 3 : 5, find : (10a + 3b) : (5a + 2b)
Solution:

Question 2.
If 5x + 6y : 8x + 5y = 8 : 9, find x : y
Solution:

Question 3.
If k (3x – 4y) : (2x – 3y) = (5x – 6y) : (4x – 5y), find x : y.
Solution:

Question 4.
Find the :
(i) duplicate ratio of 2√2 : 3√5
(ii) triplicate ratio of 2a : 3b,
(iii) sub-duplicate ratio of 9x2 a4 : 25y6 b2
(iv) sub-triplicate ratio of 216 : 343
(v) reciprocal ratio of 3 : 5
(vi) ratio compounded of the duplicate ratio of 5 : 6, the reciprocal ratio of 25 : 42 and the sub-duplicate ratio of 36 : 49.
Solution:

Question 5.
Find the value of x, if :
(i) (2x + 3) : (5x – 38) is the duplicate ratio of √5 : √6
(ii) (2x + 1) : (3x + 13) is the sub-duplicate ratio of 9 : 25.
(iii) (3x -7): (4x + 3) is the sub-triplicate ratio of 8 : 27.
Solution:

Question 6.
What quantity must be added to each term of the ratio x : y so that it may become equal to c : d ?
Solution:

Question 7.
A woman reduces her weight in the ratio 7 : 5. What does her weight become if originally it was 84 kg?
Solution:
Ratio in reduction of weight = 7 : 5
Originally weight of the woman = 84 kg
Reduced weight = $$\frac { 84 x 5 }{ 7 }$$ = 60 kg

Question 8.
If 15 (2x² – y²) = 7xy, find x : y, if x and y both are positive.
Solution:

Question 9.
Find the :
(i) fourth proportional to 2xy, x² and y².
(ii) third proportional to a² – b² and a + b.
(iii) mean proportion to (x – y) and (x3 – x²y)
Solution:
(i) Let a be the fourth proportional
Then 2xy : x² :: y² : a

Question 10.
Find two numbers such that the mean proportional between them is 14 and third proportional to them is 112.
Solution:
Let x and y be the two numbers.
Then 14 is the mean proportional between x and y
xy = 14² => xy = 196 ….(i)
and 112 is the third proportional to x, y
y² = 112 x ….(ii)

Question 11.
If x and y be unequal and x : y is the duplicate ratio of x + z and y + z, prove that z is mean proportional between x and y.
Solution:

Question 12.
If q is the mean proportional between p and r, prove that:

Solution:

Question 13.
If a, b and c are in continued proportion, prove that: a : c = (a² + b²) : (b² + c²).
Solution:

Question 14.

Solution:

Question 15.
If (4a + 9b) (4c – 9d) = (4a – 9b) (4c + 9d), prove that a : b = c : d.
Solution:

Question 16.

Solution:

Question 17.

Solution:

Question 18.
There are 36 members in a student council in a school and the ratio of the number of boys to the number of girls is 3 : 1. How many more girls should be added to the council so that the ratio of number of boys to the number of girls may be 9 : 5 ?
Solution:
Total number of members = 36
Ratio in boys and girls = 3 : 1

Question 19.
If 7x – 15y = 4x + y, find the value of x : y. Hence, use componendo and dividendo to find the values of:

Solution:

Question 20.

Solution:

Question 21.
If x, y, z arc in continued proportion, prove that

Solution:

Question 22.

Solution:

Question 23.

Solution:

Question 24.
Using componendo and dividendo, find the value

Solution:

Question 25.

Solution:

Question 26.

Solution:

Question 27.

Solution:

Hope given Selina Concise Mathematics Class 10 ICSE Solutions Chapter 7 Ratio and Proportion Ex 7D are helpful to complete your math homework.

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Selina Concise Mathematics Class 10 ICSE Solutions Chapter 7 Ratio and Proportion (Including Properties and Uses) Ex 7C

These Solutions are part of Selina Concise Mathematics Class 10 ICSE Solutions. Here we have given Selina Concise Mathematics Class 10 ICSE Solutions Chapter 7 Ratio and Proportion Ex 7C.

Other Exercises

Question 1.
If a : b = c : d, prove that:
(i) 5a + 7b : 5a – 7b = 5c + 7d : 5c – 7d.
(ii) (9a + 13b) (9c – 13d) = (9c + 13d) (9a – 13b).
(iii) xa + yb : xc + yd = b : d.
Solution:

Question 2.
If a : b = c : d, prove that: (6a + 7b) (3c – 4d) = (6c + 7d) (3a – 4b).
Solution:

Question 3.

Solution:

Question 4.

Solution:

Question 5.
If (7a + 8b) (7c – 8d) = (7a – 8b) (7c + 8d), prove that a : b = c : d.
Solution:

Question 6.

Solution:

Question 7.
If (a + b + c + d) (a – b – c + d) = (a + b – c – d) (a – b + c – d), prove that a : b = c : d.
Solution:

Question 8.

Solution:

Question 9.

Solution:

Question 10.
If a, b and c are in continued proportion, prove that

Solution:

Question 11.
Using properties of proportion, solve for x:

Solution:

But x = – 1 does not satisfy it
x = 1

Question 12.

Solution:

Question 13.
Using the properties of proportion, solve for x, given

Solution:

Question 14.

Solution:

Question 15.

Solution:

Hope given Selina Concise Mathematics Class 10 ICSE Solutions Chapter 7 Ratio and Proportion Ex 7C are helpful to complete your math homework.

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Selina Concise Mathematics Class 10 ICSE Solutions Chapter 22 Heights and Distances Ex 22B

These Solutions are part of Selina Concise Mathematics Class 10 ICSE Solutions. Here we have given Selina Concise Mathematics Class 10 ICSE Solutions Chapter 22 Heights and Distances Ex 22B

Other Exercises

Question 1.
In the figure given below, it is given that AB is perpendicular to BD and is of length X metres. DC = 30 m. ∠ADB = 30° and ∠ACB = 45°. Without using tables, find X.
Solution:

Question 2.
Find the height of a tree when it is found that on walking away from it 20m, in a horizontal line through its base, the elevation of its top Changes from 60° to 30°.
Solution:
Let AB be the tree and its height be x DC = 20 m.

Question 3.
Find the height of a building, when it is found that on walking towards it 40 m in a horizontal line through its base the angular elevation of its top changes from 30° to 45°.
Solution:
Let AB be the building and its
height be x and DC =40m

Question 4.
From the top of a light house 100 m high, the angles of depression of two ships are observed as 48° and 36° respectively. Find the distance between the two ships (in the nearest metre) if :
(i) the ships are on the same side of the light house,
(ii) the ships are on the opposite sides of the light house.
Solution:
In right angled ∆ABD

Question 5.
Two pillars of equal heights stand 011 either side of a roadway, which is 150m wide. At a point in the roadway between the pillars the elevations of the tops of the pillars are 60° and 30”; find the height of the pillars and the position of the point.
Solution:
Let AB and CD be the two pillars which stand on either side of a road BD, then BD = 150m
Let AB = CD = h
Let P be the point on the road such that the angles of elevation from P to the top of the pillars are 60° and 30° respectively.
Let BP = x then PD = 150 – x

Question 6.
From the figure given below, calculate the length of CD.

Solution:

Question 7.
The angle of elevation of the top of a tower is observed to be 60°. At a point 30 m vertically above the first point of observation, the elevation is found to be 45°. Find :
(i) the height of the tower,
(ii) its horizontal distance from the points of observation.
Solution:
Let height of tower = x i.e. AB = x

Question 8.
From the top of a cliff. 60 metre high, the angles of depression of the top and bottom of a tower are observed to be 30° and 60°. Find the height of the tower.
Solution:
Height of the cliff AB = 60m
Let the height of tower = x
Draw TR $$\parallel$$ SB. meeting AB in R

Question 9.
A man on a cliff observes a boat, at an angle of depression 30°, which is sailing towards the shore to the point immediately beneath him. Three minutes later the angle of depression of the boat is found to be 60°. Assuming that the boat sails at a uniform speed, determine :
(i) how much more time it will take to reach the shore.
(ii) the speed of the boat in metre per second, if the height of the cliff is 500 m.
Solution:
Height of cliff = 500 m.
In right ∆ACD,

Question 10.
A man in a boat rowing away from a light house 150m high, takes 2 minutes to change the angle of elevation of the top of the lighthouse from 60° to 45°. Find the speed of the boat.
Solution:
Height of the lighthouse AB = 150m
Now is right ∆ACB,

Question 11.
A person standing on the bank of a river observes that the angle of elevation of the top of a tree standing on the opposite bank is 60°. When he moves 40 m away from the bank, he finds the angle of elevation to be 30°. Find :
(i) the height of the tree, correct to 2 decimal places,
(ii) the width of the river.
Solution:
TR is the height of tree and RQ is the width of the river.
Let TR = x
Now in right ∆TQR

Question 12.
The horizontal distance between two towers is 75 m and the angular depression of the top of the first tower as seen from the top of the second, which is 160m high, is 45°. Find the height of the first tower.
Solution:

Let the height of first tower CD = x
and height of second tower AB = 160 m
Distance between them DB = 75 m.
In right ∆ACE,

Question 13.
The length of the shadow of a tower standing on level plane is found to be 2y metres longer when the sun’s altitude is 30″ than when it was
45°. Prove that the height of the tower is y ($$\sqrt { 3 }$$ +1) metres.
Solution:
Let AB be the tower and AB = x
Distance CD = 2y

Question 14.
An aeroplane flying horizontally 1 km above the ground and going away from the observer is observed at an elevation of 60°. After 10 seconds, its elevation is observed to be 30°, find the uniform speed of the aeroplane in km per hour.
Solution:
Height of aeroplane = 1 km = 1000 m.
In right ∆ACB,

Question 15.
From the top of a hill, the angles of depression of two consecutive kilometre stones, due east, are found to be 30° and 45° respectively. Find the distance of the two stones from the foot of the hill. (2007)
Solution:

Hope given Selina Concise Mathematics Class 10 ICSE Solutions Chapter 22 Heights and Distances Ex 22B are helpful to complete your math homework.

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Selina Concise Mathematics Class 10 ICSE Solutions Chapter 7 Ratio and Proportion (Including Properties and Uses) Ex 7B

These Solutions are part of Selina Concise Mathematics Class 10 ICSE Solutions. Here we have given Selina Concise Mathematics Class 10 ICSE Solutions Chapter 7 Ratio and Proportion Ex 7B.

Other Exercises

Question 1.
Find the fourth proportional to :
(i) 1.5, 4.5 and 3.5
(ii) 3a, 6a² and 2ab²
Solution:
(i) Let x be the fourth proportional to 1.5, 4.5

Question 2.
Find the third proportional to :
(i) 2$$\frac { 2 }{ 3 }$$ and 4
(ii) a – b and a² – b²
Solution:

Question 3.
Find the mean proportional between :
(i) 6 + 3√3 and 8 – 4√3
(ii) a – b and a3 – a²b.
Solution:

Question 4.
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
(x + 5)² = (x + 2) (x + 9) {b² = ac}
⇒ x² + 10x + 25 = x² + 11x + 18
⇒ x² + 10x – x² – 11x = 18 – 25
⇒ -x = -7
⇒ x = 7
Hence x = 7

Question 5.
If x², 4 and 9 are in continued proportion, find x.
Solution:
x², 4 and 9 are in continued proportion

Question 6.
What least number must be added to each of the numbers 6, 15, 20 and 43 to make them proportional. (2005, 2013)
Solution:
Let x to be added to each number then 6 + x, 15 + x, 20 + x and 43 + x are in proportion.

Question 7.

Solution:

Question 8.
What least number must be subtracted from each of the numbers 7,17 and 47 so that the remainders are in continued pro-portion ?
Solution:
Let x be subtracted from each of the numbers 7, 17 and 47.
Then 7 – x, 17 – x and 47 – x are in continued proportion.
7 – x : 17 – x : : 17 – x : 47 – x
⇒ (7 – x) (47 – x) = (17 – x) (17 – x)
⇒ 329 – 7x – 47x + x² = 289 – 17x – 17x + x²
⇒ -7x – 47x + x² + 17x + 17x – x² = 289 – 329
⇒ -54x + 34x = – 40
⇒ -20x = -40
⇒ x = 2
2 is to be the subtracted

Question 9.
If y is the mean proportional between x and z; show that xy + yz is the mean proportional between x² + y² and y² + z².
Solution:
y is the mean proportional between x and z.
y² = xz
Now, we have to prove that
xy + yz is the mean proportional between x² + y² and y² + z²
i.e., (xy + yz)² = (x² + y²) (y² + z²)
L.H.S. (xy + yz)² = [y(x + z)]² = y² (x + z)² = xz (x + z)²
R.H.S. (x² + y²) (y² + z²) = (x² + xz) (xz + z²) = x (x + z) z (x + z) = xz (x + z)²
L.H.S. = R.H.S.
Hence proved.

Question 10.
If q is the mean proportional between p and r, show that: pqr (p + q + r)3 = (pq + qr + pr)3.
Solution:
q is the mean proportional between p and r,
q² = pr
Now L.H.S. = pqr (p + q + r)3
= qq² (p + q + r)3
= q3 (p + q + r)3
= [q(p + q + r)]3
= (pq + q² + qr)3
= (pq + pr + qr)3
= (pq + qr + pr)3
= R.H.S.

Question 11.
If three quantities are in continued proportion; show that the ratio of the first to the third is the duplicate ratio of the first to the second.
Solution:
Let x, y and z are three quantities which are
in continued proportion
Then x : y : y : z ⇒ y² = zx
Now, we have to prove that
x : z = x² : y² or xy² = zx²
L.H.S. = xy² = x x zx (y² = zx)
= x² z = R.H.S.
Hence Proved.

Question 12.
If y is the mean proportional between x and z, prove that:

Solution:

Question 13.
Given four quantities a, b, c and d are in proportion. Show that:
(a – c) b² : (b – d) cd = (a² – b² – ab) : (c² – d² – cd)
Solution:
a, b, c and d are in proportion Then a : b :: c : d

Question 14.
Find two numbers such that the mean proportional between them is 12 and the third proportional to them is 96.
Solution:
Let a and b be the two numbers, whose mean proportional is 12

Question 15.

Solution:

Question 16.
If p : q = r : s ; then show that: mp + nq : q = mr + ns : s.
Solution:

Question 17.

Solution:

Question 18.

Solution:

Question 19.

Solution:

Question 20.

Solution:

Hope given Selina Concise Mathematics Class 10 ICSE Solutions Chapter 7 Ratio and Proportion Ex 7B are helpful to complete your math homework.

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Selina Concise Mathematics Class 10 ICSE Solutions Chapter 22 Heights and Distances Ex 22A

These Solutions are part of Selina Concise Mathematics Class 10 ICSE Solutions. Here we have given Selina Concise Mathematics Class 10 ICSE Solutions Chapter 22 Heights and Distances Ex 22A

Other Exercises

Question 1.
The height of a tree is $$\sqrt { 3 }$$ times the length of its shadow. Find the angle of elevation of the sun.
Solution:
Let AB be the tree and BC be its shadow.

∴ θ = 60°
∵ Angle of elevation of the sun = 60°

Question 2.
The angle of elevation of the top of a tower, from a point on the ground and at a distance of 160 m from its foot, is found to be 60°. Find the height of the tower.
Solution:
Let AB be the tower and C is the point which is 160 m away from the foot of the tower,
i.e. CB = 160 m
Let height of the tower be x

Question 3.
A ladder is placed along a wall such that its upper end is resting against a vertical wall. The foot of the ladder is 2.4 m from the wall and the ladder is making an angle of 68“ with the ground. Find the height, upto which the ladder reaches.
Solution:
Let AB be the wall and

AC be the ladder, which is placed against the wall. If foot is 2.4 m away from the wall i.e. CB = 2.4m1.
Let AB =x m.
In right ∆ ABC,
tan θ = $$\frac { AB }{ BC }$$ ⇒ tan 68° = $$\frac { X }{ 2.4 }$$
∴ x = 2.4 x tan 68° = 2.4 x 2.4751
= 5.94 m

Question 4.
Two persons are standing on the opposite sides of a tower. They observe the angles of elevation of the top of the tower to be 30° and 38° respectively. Find the distance between them, if the height of the tower is 50 m.
Solution:

Two persons A and B are standing on the opposite side of the tower TR and height of tower TR = 50 m and angles of elevation with A and B are 30° and 38° respectively. Let AR = x and RB = y
Now in right ∆TAR,

Question 5.
A kite is attached to a string. Find the length of the string, when the height of the kite is 60 m. and the string makes an angle 30° with the ground.
Solution:
Let KT be the height of kite and PK is the string which makes an angle of 30° with the ground.
∴ KT = 60 m
Let KP = xm.
Now in right ∆PKT,

Question 6.
A boy 1.6m tall, is 20 m away from a tower and observes the angle of elevation of the top of the tower to be (i) 45° (ii) 60°. Find the height of the tower in each case.
Solution:
(i) Let AB be the tower and MN be the boy who is 20m away from the foot of the tower.
Let AB = x and angle of elevation = 45°

Question 7.
The upper part of a tree, broken over by the wind, makes an angle of 45° with the ground; and the distance from the root to the point where the top of the tree touches the ground, is 15m. What was the height of the tree before it was broken ?
Solution:
Let AB be the tree which was broken at the point C which makes an angle of elevation of 45°, with the ground at a distance of 15m.
BD = 15m
AC = CD

Question 8.
The angle of elevation of the top of an unfinished tower at a point distance 80 m from its base is 30°. How much higher must the tower be raised so that its angle of elevation at the same point may be 60° ?
Solution:
Distance of a point from the tower = 80 m
Angle of elevation = 30°
In second case the elevation of lower = 60°
In first case,

Question 9.
At a particular time, when the sun’s altitude is 30°, the length of the shadow of’C vertical tower is 45 m. Calculate :
(i) height of the tower.
(ii) the length of the shadow of the same tower, when the sun’s altitude is (a) 45° (b) 60°.
Solution:
Shadow of the tower = 45 m and angle of elevation = 30°
Let AB be the lower and BC is its shadow.
∴ CB = 45 m.
Now in right ∆ABC,

(ii) In second case,
(a) Angle of elevation = 45°
and height of tower = 25.98 m or 15$$\sqrt { 3 }$$ m

(b) Angle of elevation = 60°
and height of tower = 25.98 m or 15$$\sqrt { 3 }$$ m.
Let shadow of the tower DB = xm

Question 10.
Two vertical poles are on either side of a road. A 30 m long ladder is placed between the two poles. When the ladder rests against one pole, it makes angle 32°24′ with the pole and when it is turned to rest against another pole, it makes angle 32°24′ with the road. Calculate the width of the road.
Solution:
Two poles AB and CD which are at the either end of a road BD. A ladder 30 m long subtends an angle of 32° 24′ with the first pole AB and 32°24′ with the road when it is turned to rest against the second pole CD.
Now in right ∆ABE.

Question 11.
Two climbers are at points A and Bona vertical cliff face. To an observer C, 40 m from the foot of the cliff on the level ground, A is at an elevation of 48° and B of 57°. What is the distance between the climbers ?
Solution:
A and B are two climbers on the cliff and ob-server is at C, 40 m from the foot of the cliff while the angles of elevations of each climber is 48° and 57° respectively.
In right ∆ACD,

Question 12.
A man stands 9m away from a flag-pole. He observes that angle of elevation of the top of the pole is 28° and the angle of depression of the bottom of the pole is 13°. Calculate the height of the pole.
Solution:
Let PL is the pole and MN is the man The angle of elevation of the top of the pole = 28°
arid the angle of depression of the bottom of the pole =13°
Man is 9 m away from the pole,
i.e. MQ = 9 m
Now in right ∆PMQ,

Question 13.
From the top of a cliff 92 m high, the angle of depression of a buoy is 20°. Calculate to the nearest metre the distance of the buoy from the foot of the cliff.
Solution:
Let CD be the cliff and CD = 92m, B is the buoy,
then from C ,
the angle of depression is 20°

Hope given Selina Concise Mathematics Class 10 ICSE Solutions Chapter 22 Heights and Distances Ex 22A 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.

Selina Concise Mathematics Class 10 ICSE Solutions Chapter 11 Geometric Progression Ex 11D

These Solutions are part of Selina Concise Mathematics Class 10 ICSE Solutions. Here we have given Selina Concise Mathematics Class 10 ICSE Solutions Chapter 11 Geometric Progression Ex 11D

Other Exercises

Question 1.
Find the sum of G.P. :
(i) 1 + 3 + 9 + 27 +….to 12 terms.
(ii) 0.3 + 0.03 + 0.003 + 0.0003 +….to 8 terms.
(iii) $$1-\frac { 1 }{ 2 } +\frac { 1 }{ 4 } -\frac { 1 }{ 8 } ….to\quad 9\quad terms$$
(iv) $$1-\frac { 1 }{ 3 } +\frac { 1 }{ { 3 }^{ 2 } } -\frac { 1 }{ { 3 }^{ 3 } } ….to\quad n\quad terms$$
(v) $$\frac { x+y }{ x-y } +1+\frac { x-y }{ x+y } +….upto\quad n\quad terms$$
(vi) $$\sqrt { 3 } +\frac { 1 }{ \sqrt { 3 } } +\frac { 1 }{ 3\sqrt { 3 } } +….to\quad n\quad terms$$
Solution:
(i) 1 + 3 + 9 + 27 +….to 12 terms.
Here a = 1, r = 3 and n = 12

Question 2.
How many terms of the geometric progression 1 + 4 + 16 + 64 +…. must be added to get sum equal to 5461 ?
Solution:
Sn = 5461 and G.P. is
1 + 4 + 16 + 64 +…..
Here, a = 1, r = 4 (r > 1)

Question 3.
The first term of a G.P. is 27 and its 8th term is $$\\ \frac { 1 }{ 81 }$$. Find the sum of its first 10 terms.
Solution:
First term of a G.P (a) = 27
T8 = $$\\ \frac { 1 }{ 81 }$$, n = 10
a = 27

Question 4.
A boy spends Rs 10 on first day, Rs 20 on second day, Rs 40 on third day and so on. Find how much, in all, will he spend in 12 days?
Solution:
A boy spends Rs 10 on first day,
Rs 20 on second day
Rs 40 on third day and so on
G.P. is 10 + 20 + 40 +…. 12 terms
Here a = 10, r = 2 and n = 12 (r > 1)

Question 5.
The 4th and the 7th terms of a G.P. are $$\\ \frac { 1 }{ 27 }$$ and $$\\ \frac { 1 }{ 729 }$$ respectively. Find the sum of n terms of this G.P.
Solution:
In a G.P.
T4 = $$\\ \frac { 1 }{ 27 }$$

Question 6.
A geometric progression has common ratio = 3 and last term = 486. If the sum of its terms is 728 ; find its first term.
Solution:
In a G.P.
Common ratio (r) = 3
Last term (l) = 486
Sum of its terms (Sn) = 728
Let a be the first term, then

Question 7.
Find the sum of G.P. : 3, 6, 12, …… 1536.
Solution:
G.P. is 3, 6, 12,….1536
Here a = 3, r = $$\\ \frac { 6 }{ 3 }$$ = 2

Question 8.
How many terms of the series 2 + 6 + 18 +…. must be taken to make the sum equal to 728 ?
Solution:
G.P. is 2 + 6 + 18 +….
Here a = 2, r = $$\\ \frac { 6 }{ 2 }$$ = 3, Sn = 728

Question 9.
In a G.P., the ratio between the sum of first three terms and that of the first six terms is 125 : 152. Find its common ratio.
Solution:
In a G.P.
Sum of first 3 terms : Sum of 6 terms = 125 : 152

Question 10.
Find how many terms of G.P.$$\frac { 2 }{ 9 } -\frac { 1 }{ 3 } +\frac { 1 }{ 2 }$$… must be added to get the sum equal to $$\\ \frac { 55 }{ 72 }$$ ?
Solution:
$$\frac { 2 }{ 9 } -\frac { 1 }{ 3 } +\frac { 1 }{ 2 }$$…
Now,Sn = $$\\ \frac { 55 }{ 72 }$$

Question 11.
If the sum of 1 + 2 + 22 +…..+ 2n – 1 is 255, find the value of n.
Solution:
1 + 2 + 22 +…..+ 2n – 1 = 255

Question 12.
Find the geometric mean between :
(i) $$\\ \frac { 4 }{ 9 }$$ and $$\\ \frac { 9 }{ 4 }$$
(ii) 14 and $$\\ \frac { 7 }{ 32 }$$
(iii) 2a and 8a3
Solution:
(i) G.M between $$\\ \frac { 4 }{ 9 }$$ and $$\\ \frac { 9 }{ 4 }$$

Question 13.
The sum of three numbers in G.P. is $$\\ \frac { 39 }{ 10 }$$ and their product is 1. Find the numbers.
Solution:
Sum of three numbers in G.P. = $$\\ \frac { 39 }{ 10 }$$
and their product = 1
Let number be $$\\ \frac { a }{ r }$$, a, ar, then

Question 14.
The first term of a G.P. is – 3 and the square of the second term is equal to its 4th term. Find its 7th term.
Solution:
In G.P.
T1 = – 3

Question 15.
Find the 5th term of the G.P. $$\\ \frac { 5 }{ 2 }$$, 1,…..
Solution:
Given G.P is $$\\ \frac { 5 }{ 2 }$$, 1,…..

Question 16.
The first two terms of a G.P. are 125 and 25 respectively. Find the 5th and the 6th terms of the G.P.
Solution:
Given, First term = a = 125….(i)
and Second term = ar = 25…..(ii)
Now, Divide eq. (ii) by eq (i), we get

Question 17.
Find the sum of the sequence $$– \frac { 1 }{ 3 }$$, 1, – 3, 9,….upto 8 terms.
Solution:
Here, First Term, a = $$– \frac { 1 }{ 3 }$$…(i)
and Second Term, ar = 1 …(ii)
Dividing eq. (i) by eq. (ii), we get

Question 18.
The first term of a G.P. in 27. If the 8th term be $$\\ \frac { 1 }{ 81 }$$, what will be the sum of 10 terms ?
Solution:
Given, First term (a) = 27, n = 10

Question 19.
Find a G.P. for which the sum of first two terms is – 4 and the fifth term is 4 times the third term.
Solution:
Let a be the first term and r be the common ratio of the G.P.
According to the given conditions,

Hope given Selina Concise Mathematics Class 10 ICSE Solutions Chapter 11 Geometric Progression Ex 11D 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.