OP Malhotra Class 9 ICSE Maths Solutions Chapter 5 Simultaneous Linear Equations in Two Variables
Welcome to NCTB Solutions. Here with this post we are going to help 9th class students for the Solutions of OP Malhotra Class 9 ICSE Math Book, Chapter 5, Simultaneous Linear Equations in Two Variables. Here students can easily find step by step solutions of all the problems for Simultaneous Linear Equations in Two Variables, Exercise 5a and 5b Also here our mathematics teacher’s are solved all the problems with easily understandable methods with proper guidance so that all the students can understand easily. Here in this post students will get chapter 5 solutions. Here in this post all the solutions are based on ICSE latest Syllabus.
Simultaneous Linear Equations in Two Variables Exercise 5(a) Solution :
Question no – (1)
Solution :
Here, 2x + y – 7 = 0
= 2 × 5 + k – 7 = 0
= 3 + k = 0
= k = -3
Therefore, the value of k will be -3
Question no – (2)
Solution :
Given, x + y = 5, x – y = 3,
= x + y = 5 – 0
x – y = 3
= x = 3 + y
Substitute the value of x in (i)
3 + y + y = 5
= 2y = 5 – 3
= y = 2/2 = 1
x = 3 + y
= 3 + 1 = 4
Question no – (3)
Solution :
y = 2x – 6 …..(i)
y = 0 ….. (ii)
From (i) and (ii)
2x – 6 = 0
2x = 6
x = 6/2 = 3
∴ y = 2 × 3 – 6
= 0
∴ x = 3, y = 0
Question no – (4)
Solution :
In the question we get,
p = 2q – 1 …………(i)
q = 5 – 3p …………..(ii)
Substitute the value of p in (ii)
q = 5 – 3 (2q – 1)
= 5 – 6q + 3
= q = 8 – 6q
= q + 6q = 8
= 7q = 8
= q = 8/7
Now, p = 2 × (8/7) – 1
= 16/7 – 1
= 16 – 7/7
= 9/7
∴ p = 9/7, q = 8/7
Question no – (5)
Solution :
Given in the question,
9x + 4y = 5 …………(i)
4x – 5y = 9 …………((ii)
(i) No × by 4 and (ii) No × 9
36x + 16y = 20
36x – 45y = 81
——————————-
61 y = -61
= y = -61/61
= -1
Putting value in (i)
= 9x + 4y = 5
= 9x + 4(1) = 5
= 9x = 5 + 4
= x = 9/9
∴ x = 1
Question no – (6)
Solution :
From the question we get,
x + 3y = 5 ………….(i) × 3
3x – y = 5 ………….(ii) × 1
Now,
3x + 9y = 15
3x – y = 5
——————————-
10y = 10
= y = 10/10
= y = 1
Putting value in (ii) 3x – y = 5
3x – 1 = 5
3x = 5 + 1
= x = 6/3
∴ x = 2
Question no – (7)
Solution :
According to the question,
3x – 7y + 10 = 0 …………(i) × 1
y – 2x – 3 = 0 …………(ii) × 7
Now,
3x – 7y = -10
-14x + 7y = 21
——————————-
-11x = 11
= x = -1
Putting value in (ii)
y – 2x = 3
= y – 2 × (-1) = 3
= y = 3 – 2
∴ y = 1
Question no – (8)
Solution :
According to the question,
20u – 30v = 13 ………(i) × 1
-10u + 10v = -5 ………(ii) × 2
Now,
20u – 30v = 13
-20u + 20v = -10
——————————-
-10v = 3
= v = -3/10
Putting value in (i)
20u – 30 × (-3/10) = 13
= 20u = 13 – 9
= u = 4/20
∴ u = 1/5
Question no – (9)
Solution :
2x – 3y = 1.3 …… (i) × 1
= y – x = 0.5
= -x + y = 0.5 ……(ii) × 2
Now, 2x – 3y = 1.3
-2x + 2y = 1.0
——————————-
-y = 2.3
= y = -2.3
Putting value in (ii)
-x + y = 0.5
= -x – 2.3 = 0.5
= -x = 0.5 + 2.3
= x = -2.8
∴ x = -2.8 and y = -2.3
Question no – (10)
Solution :
11x + 15y + 23 = 0
= 11x + 15y = -23 …… (i) × 2
7x – 2y – 20 = 0
= 7x – 2y = 20 …… (ii) × 15
Now,
22x + 30y = -46
105x – 30y = 300
——————————-
127x = 254
= x = 254/154
= x = 2
Putting value x = 2 in …(ii)
11x + 15y = -23
= 11 × 2 + 15y = -23
= 15y = -23 -22
= y = -45/15
= y = -3
∴ x = 2, and y = -3
Question no – (12)
Solution :
= 2x + y/4 = 6
= 2x + y = 24 ………..(i)
and, x/5 – y/2 = 0
= 2x – 5y/10 = 0
= 2x – 5y = 0 ………..(ii)
Now, (i) – (ii),
2x + y = 24
2x – 5y = 0
——————————-
6y = 24
= y = 24/6
∴ y = 4
Question no – (14)
Solution :
(i) Given in the question,
65x – 33y = 97 …….. (i)
33x – 65y = 1 …….. (ii)
First, (i) + (ii),
98x – 98y = 98
= x – y = 1 ……..(iii)
Second, (i) – (ii),
32x + 32y = 96
= x + y = 2 …..(iv)
Now, (iii) – (iv),
2x = 3
= x = 3/2
Putting value in …(4)
x + y = 2
= y = 2 – x
= 2 – 3/2
= 4 – 3/2
= 1/2
(ii) 23x + 31y = 77 – (i)
= 31x + 23y = 85 – (ii)
(i) + (ii)
54x + 54y = 162
= x + y = 3 – (iii)
Now, (ii) – (i)
= 8x – 8y = 8
= x – y = 1 – (iv)
= Now, (iii) + (iv)
= 2x = 4
= x = 4/2
= x = 2
Putting value 9n ….(i)
= 23x × 31y = 77
= 31y = 77 – 56
= 3 – 1/31
= 1
Question no – (15)
Solution :
(i) From the given question,
4x + 6/y = 15 – (i) – × 3
6x – 8/y = 14 – (ii) × 2
Now,
12x + 18/y = 45
= 12x – 16/y = 28
(-) (+) (-)
——————————-
34/y = 17
= y = 34/17
= 2
Now, value in (i)
= 4x + 6/2 = 15
= 4x = 15 – 3
= x = 12/4 = 3
∴ x = 38 y = 2
Now, y = px – 2
= 2 = 3p – 2
= 3p = 2 + 2
= p = 4/3
Therefore, p = 4/3
(ii) Given, 4/x + 5y = 7, 3/x + 4y = 5 ; x ≠ 0
= 4/x + 5t = 7 ……. (i) × 4
= 3/x + 4y = 5 ……. (ii) × 5
So, Now,
16/x + 20y = 28
15/x + 2-0y = 25
(-) (-) (-)
——————————-
= 1/x = 3
= x = 1/3
Putting value in (i)
4/x + 5y = 7
= 4/13 + 5y = 7
= 12 + 5y = 7
= 5y = 7 – 12
= y = – 5/5 = – 1
∴ x = 1/3, y = – 1
Question no – (17)
Solution :
(i) Given, 2 (3u – v) = 5uv, 2 (u + 3v) = 5uv
Now, divide by uv
= 6/v – 2/u = 5 – (i) × 1
and, 2 (u + 3v) = 5uv
= 2u + 6v = 5uv
Divided by ‘uv’
2/V + 6/U = 5 (ii) × 3
Now,
6/v – 2/u = 5
6/v + 18/u = 15
(-) (-) (-)
——————————-
– 20/u = – 10
= u = 20/10 = 2
Putting value in (ii)
= 2/v + 6/2 = 5
= 2/v + 6/2 = 5
= 2/v = 5 – 3
= v = 1
= 2
(ii) Give, 3 (a + 3b) = 11ab, 3 (2a + b) = 7ab
= 3a + 9b = 11ab
Divide by ab
3/b + 9/a = 11 (i) × 2
And, 3 (2a + b) = 7ab
= 6a + 3b = 7ab
Divide by ab
6/b + 3/a = 7 (ii)
6/b + 18/a = 22
6/b + 3/a = 7
(-) (-) (-)
——————————-
= 15/a = 15
= a = 15/15 = 1
= a = 15/15 = 1
Putting value in (i)
3/b + 9/1 = 11
= 3/b = 11 – 9
= 2
= b = 3/2
Question no – (18)
Solution :
According to the question,
= 3x – 6 + 3 – 5y – 4/2 = 5y/2
= 3x – 6 + 12 – 16y + 8/4 = 5y/2
= (3x – 10y + 14) = 10y
= 3x – 10y – 10y + 14 = 0
= 3x – 20y = – 14 ……(i) |
And, y – x/4 + x/8 – 7x – 5y/3 = y – 2x
= 6y – 6x + 3x – 56x + 40y/24 = y – 2x
= 6y – 6x + 3x – 56x + 40y – 24y + 48x = 0
= – 11x + 22y = 0
= x = -22/11 = + 2
= x = 2
From 0,
3x – 20y = – 14
= 3 × 2y – 20y = – 14
= 14y = – 14
∴ y = 1
Question no – (19)
Solution :
According to the question,
x – y = 0.9
= 9/10 (i) |
= 11/2(x + 2y) = 1
= 2x + 2y = 11
= x + y = 11/2 – (ii)
(i) + (ii)
x – y = – 9/10
= x + y = 11/2
——————————-
2x = (9/10 + 11/2)
= 9 + 55/10
= 64/10
= x = 64/10 × 2
= x = 3.2
Now, Putting value in (ii)
= x + y = 11/2
= y = 11/2 – x
= y = 11/2 – 3.2
∴ y = 2.3
Question no – (20)
Solution :
According to the question,
= x/2 + y = 0.8
= x/2 + y = 8/10
= 5x + 1y = 8 (i) × 2
and, 7/x + y/2 = 10
= 7 = 10 (x + y/2)
= 10x + 5y = 7 – (ii) × 1
10x + 20y = 16
10x + 5y = 7
(-) (-) (-)
——————————-
15y = 9
= y = 9/15 = 3/5
Putting value of y in (i)
= 5x + 10y = 8
= 5x + 10 × 3/5 = 8
5x = 8 – 6
∴ x = 2/5
Question no – (21)
Solution :
2x + y = 235 (i) × 4
3x + 4y = 65 – (ii) × 1
So, 8x + 4y = 140
3x + 4y = 65
——————————-
5x = 75
= x = 75/5 = 15
Putting value in (i)
2 × 15 + y = 35
= y = 35 – 30
= 5
∴ x/y = 15/5
= 3
Thus, the value of x/y is 3
Question no – (22)
Solution :
= 3/x – 4y = 7 ……. (i) – 2 ……. × 2
= 2/x + 7y = 5 ……. (ii) – × 3
= 5x + 4/y = 9 ……. (iii) So,
6/x + 8y = 14
– 6/x = 21y = 15
= 29y = 29
= y = 29/29
= 1
Putting in value (i) 3/x + 4 × 1 = 7
= 3/x = 7 – 4
= x = 3/3 = 1
Substitute the value x and y in (ii)
= 5x + 4/y = 9
= 5 × 1 + 4/1 = 9
= 5 + 4 = 9
= 9 = 9
Therefore, it is True.
Simultaneous Linear Equations in Two Variables Exercise 5(b) Solution :
Question no – (1)
Solution :
Let, the no are = x and y
x + y = 16 ……(i)
and, x = 3y
= x – 3y = 0 …….(ii)
From (i) 3y + y = 16
= 4y = 16
= y = 4
∴ x = 3y = 3 × 4 = 12
Thus, the required numbers are 4, 8, and 12
Question no – (2)
Solution :
Let, 1st no x 2nd no y.
x + y = 6 ……(i)
x – y = 4 ……(ii)
(i) + (ii)
= 2x = 10
= x = 5
Putting value in (i)
= 5 + y = 6
= y = 6 – 5
= 1
Therefore, the other numbers are 5 and 1.
Question no – (3)
Solution :
Let, one’s digit x
ten’s digit y
∴ Number vis = x + 10y
After reversing, no is 10x + y
∴ x + y = 13 – (i)
and, (y + 10x) – (x + 10y) = 45
= y + 10x – x – 10y = 45
= 9x – 9y = 45
= x – y 45/9 = 5 ……(ii)
Now, (i) + (ii)
x + y = 13
x – y = 5
——————————-
= 2x = 18
= x 18/2 = 9
Putting in (i)
9 + y = 13
= y = 13 – 9
= 4
∴ The required number,
= x + 10y
= 9 + 10 × 4
= 49
Therefore, the number will be 49
Question no – (4)
Solution :
Let, unit digit = x
ten’s digit= y
∴ Number is = x + 10y
After reversing, no is = 10x + y
Now, x + y = 9
and, y + 10x – (x + 10y) – (i)
y + 10x – (x + 10y) = 9
= y + 10x – x – 10y = 9
= – 5x + 9x = 9
= x – y = 1
Now, (i) + (ii)
x + y = 9
x – y = 1
——————
2x = 10
= x = 15
Putting value in …..(i)
y = 9 – x
= 9 – 5
= 4
∴ Required number is,
= x + 10y
= 5 + 104
= 45
Therefore, 45 will be the required number.
Question no – (5)
Solution :
Let, unit digit is = x
Tan’s digit is = y
∴ number is = x + 10y
After revering, no is = 10x + y
So, x – y = 3 (i)
and, x + 10y + y + 10x = 121
= 11x + 11y = 121
= x + y = 11 – (ii)
(i) + (v)
= 2x = 14
= x = 14/2 = 7
So, y = 11- x
= 11 – 7
= 4
∴ Number will be,
= x + 10 × 4
= 47
and, new number,
= 10x + 4
= 10 × 7 + 4
= 74
Thus, 74 is the required number.
Question no – (6)
Solution :
Let, one’s digit = x tens digit = y
∴ Number is = x + 10y
By reversing 10x + y is no
Now, x + y = 3 – (i)
and, 7 (x + 10y) = 4 (y + 10x)
= 7x + 70y = 4y + 40x
= 70y – 4y = 40x – 7x
= 66y = 33x = 0
= x – 2y = 0 – (ii)
(i) – (ii)
3y = 3
= y = 3/3 = 1
Now, x = 2y
= x = 2.1
= 2
∴ Required number is,
= x + 10y
= 2 + 10 × 1
= 12
Therefore, the required number will be 12
Question no – (7)
Solution :
Let, largest no is x
Smallest no is y
From questions 3x = 4y + 3
= 3x – 4y = 3 (i) × 7
and, 7y = x × 5 + 1
= 5x – 7y = – 1 (ii) × 4
So, 21x – 28y = 21
20x – 28y = – 4
(-) (+) (+)
—————————-
x = 25
Putting value in (i) 3x – 4y = 3
= 3 × 25 – 4y = 3
= – 4y = 3 – 75
= – 72
= y = 72/4 = 18
Thus, the Largest number will be 25 and smallest will be 18.
Question no – (8)
Solution :
Let, no is (result) = x
1st no = x – 2
2nd no = x + 2
3rd no = x/2
4th no = 2x
According to condition,
= x – 2 + x + 2 + x/2 + 2x = 36
= 4x + x/2 = 36
= 8x + x/2 = 36
= 9x = 72
= x = 72/9 = 8
∴ 1st number will be,
= x – 2 = 8 – 2
= 6
∴ 2nd number will be,
= x + 2 = 8 + 2
= 10
∴ 3rd number will be,
= x/2 = 8/2
= 4
∴ 4th number will be,
= 2 × 4
= 8
Question no – (9)
Solution :
Let, Numerator of fractions is = x
Denominator of fractions is = y
∴ x + 2/y + 1 = 5/8
= 8x + 16 = 5y + 5
= 8x – 5y = 5 – 16
= – 11 – (i) × 1
and, x + 1/y + 1 = 1/2
= 2x + 2 = y + 1
= 2x – y = 1 – 2
= 2x – y = 1 – 2
= – 1 (ii) × 4
Now,
8x – 5y = – 11
8x – 4y = – 4
(-) (+) (+)
—————————-
– y = – 7
Putting value y = 7 in (i)
= 8x – 5 × 7 = – 11
= 8x = – 11 + 35
= 8x = 24
= x = 24/8 = 3
Therefore, the required Fraction will be 3/7
Question no – (10)
Solution :
Let, fractions = x/y
From question,
x/y + 1= 1/2
= 2x = y + 1
= 2x – y = 1 – (i) × 1
= 2x – y = 1
and, x + 1/y = 1
= x + 1 = y
= x – y = – 1 (ii) × 2
Now, 2x – y = 1
2x – 2y = – 2
(-) (+) (+)
—————————-
y = 3
Putting value in (2)
x – y = – 1
= x – 3 = – 1
= x = – 1 + 3
= 2
Therefore, the required Fraction will be 2/3
Question no – (11)
Solution :
Let, fraction x/4 [N = x, D = y]
From question = x + 4/y = x/y + 2/3
= x + 4/y = x/y = 2/3
= x + 4/y = 2/3
= 4/y = 2/3
= 2y = 12
= y = 12/2
Hence, the denominator will be 6
Question no – (12)
Solution :
Let, age of father 8 son x and y years.
From questions x = 6y – (i)
4 year since Age of father will be = x + 4
4 year since age of son will be = y + 4
∴ (x + 4) = 4 (y + 4)
= x + 4 = 4y + 16
= x – 4y = 16 – 4
= 12
= x = 4y + 12 – (ii)
From (i) and (ii)
= 6y = 4y + 12
= 6y – 4y = 12
= 2y = 12
= y = 12/2 = 6
Therefore, x = 6y
= 6 × 6
= 36.
∴ Present age of father = 36 years.
∴ Present age of son = 6 years.
Question no – (13)
Solution :
Let, age of son = x years
And, age of father = y rears
From questions y = 3x + 3 – (i)
After 3 years age of son will be = (x + 3) years
age of father will be = (y + 3)
∴ y + 3 = 2 (x + 3) + 10
= y + 3 = 2x + 6 + 10
= y= 2x + 16 – 3
= 2x + 13 (ii)
From (i) and (ii)
∴ 3x + 3 = 2x + 13
= 3x – 2x = 13 – 3
= x = 10
Putting value in (i)
y = 3 × 10 + 3
= 33
∴ Present age of father = 33 years.
∴ Present age of son = 10 years.
Question no – (14)
Solution :
Let, of age of man = x years
age of children = y years
x = 3y …….(i)
5 years hence, age of man = (x + 5) years
and, the sum of ages two children = (y + 10) years
= x + 5 = 2 )(t + 10)
= x + 5 = 2y + 20
= x – 2y = 20 – 5
= 15
= x = 2y + 15 ……. (ii)
From (i) and (ii)
3y = 2y + 15
= 3y – 2y = 15
= y = 15
So, x = 3y
= 3 × 15
= 45 years
Therefore, his present age will be 45 years.
Question no – (15)
Solution :
Let, age of Ram = x years
Age of is father = y years
∴ y = 4m (i)
= 5 years ago,, age of Ram = (x – 5) years
∴ Age of father = (y – 5) years
Therefore, y – 5 = 9 (x – 5)
= y – 5 = 9x – 45
= y – 9x = – 45 + 5
= – 40
= y = 9x + 40 (ii)
From (i) and (ii)
= 9x + 40 = 4x
= 9x – 4x = 40
= 5x = 40
= x = 40/5
= 8
Putting value in (i)
y = 4x = 4 × 8
= 32 years
Age of father 32 years.
∴ Age of Son (Ram) = 4x = y
= 4x = 32
= x = 32/4
= 8 years
Hence, the present age of Ram will be 8 years and present age of his father will be 32 years.
Question no – (16)
Solution :
At the time marriage,
Let, age of man = x years
Age of wife = y years
∴ x = y + 6 – (i)
After 12 years age of man = (x + 12) years
in wife = (y + 12) 6/5
∴ x + 12 = (y + 12) 6/5
= 5x + 60 = 6y + 72
= 5x – 6y = 72 – 60
= 12 – (ii)
From, (i) and (ii)
= 5 (y + 6) – 6y = 12
= 5y + 30 – 6y = 12
= – y = 12 – 30
= – 18
∴ y = 18
Putting value in (i)
x = 18 + 6
= 24
∴ Age of man = 24 years.
∴ Age of wife = 18 years.
Question no – (17)
Solution :
Let, age of Ram = x years
Age of Shyam = y years
∴ x/y = 5/6
= 6x = 5y
= x = 5y/6 ……….. (i)
And, 5 years ago, age of Ram = (x – 5) years
Age of Shyam = (y – 5) years
∴ x – 5/y – 5 = 4/5
= 5x – 25 = 4y – 20
= 5x- 4y = – 20 + 25
= 5 ……….(ii)
From, (i) and (ii)
= 5 × (5y/6) – 4y = 5
= 25y/6 – 4y = 5
= 25y – 24y/6 = 5
= y = 30
∴ Putting value y = 30, in (ii)
x = 5y/6
= 5 × 306
= 25
∴ Present age of Ram will be = 25 years
∴ Present age of Shyam will be = 30 years.
Question no – (18)
Solution :
Let, cost of pencil = xp
Cost of erasers = yp
So, 5x + 2y = 1.80 – (i)
3x + 2y = 9.20 – (ii) × 3
∴ 5x + 6y = 180
= 9x + 6y = 276
(-) (-) (-)
—————————-
= – 4x = – 96
= x = 96/4 = 24p
Putting value in (i)
= 5 × 24 + 6y = 1,80
= 6y = 180 – 120
= 60
= y = 60/6 = 10
∴ Cost of 1 pencil = 24p
∴ Cost of 1 eraser = 10p
Therefore, the cost of pencil will be 24 paise and eraser will be 10p.
Question no – (19)
Solution :
Let, Cost of one chair = x Rs
Cost of one table = y Rs
9x + 5y = 90 – (i) × 4
= 5x + 4y = 61 – (ii) × 5
Now,
36x + 20y = 360
25x + 20y = 305
(-) (-) (-)
—————————-
11x = 55
= x = 55/11 = 5
Putting value in (i)
9x + 5y = 90
= 9 × 5 + 5y = 90
= 5y = 90 – 45
= y = 45/5
= 9
∴ Cost of one chair = 5 Rs
∴ Cost of table = 9 Rs
∴ Total cost of 6 chairs and 3 tables,
= (6 × 5 + 9 × 3)
= (30 + 27)
= 57 Rs
Hence, the price of 6 chairs and 3 tables will be 57 Rs.
Question no – (20)
Solution :
Let, Cost of 1 horse = x Rs
Cost of 1 cow = y Rs
From questions, x + 2y = 680
and, x = y + 80
y + 80 + 2y = 680
= 3y = 680 – 80
= 600
= y = 600/3
= 200
and, putting value in (i)
x = y + 80
= 200 + 80
= 280
Therefore, the cost of horse is Rs 280 and cow is Rs 200.
Question no – (21)
Solution :
Let, cost 1 kg sweets = x Rs
Cost of 1 kg apples = y Rs
= x + 3y = 33 – (i) × – (2)
= 2x + y = 31 – (ii) × – (1)
∴ 2x + 6y = 66
2x + y = 31
(-) (-) (-)
—————————-
= 5y = 35
= y = 35/5 = 7
Putting value in (i)
= x + 3 × 7 = 33
= x = 33 – 21
= 12
∴ Price of 1 kg sweets = 12 Rs
∴ Price of 1 kg apples = 7 Rs
Therefore, cost of per kg sweets will be 12 Rs and apples will be 7 Rs.
Question no – (22)
Solution :
Let, Cost of 1 pen 8 pencil x and y Rs,
∴ x = y + 3.5 – (i)
and, 3x + 2y = 13 – (i)
From (i) and (ii) substitute,
3 (y + 3.5) + 3y = 13
= 2y + 10.5 + 2y = 13
= 5y = 13 – 10.5
= y = 2.5/5
= 0.5 Rs
Now, putting value,
= x = y + 3.5
= 0.5 + 3.5
= 4.0 Rs
∴ Cost of pencil = 50 p
∴ Cost of Pen = 4 Rs
Question no – (23)
Solution :
Let, no of stamps if 3p = x
Number of stamps = 5p = y
From questions, x + y = 22 – (i)
and, x × 3/100 + y + 5/100 = 1
= 3x + 5y = 100 – (ii)
Substitute value,
3 (222 – y) + 5y = 100
= 2y = 100 – 66
= y = 34/2
= 17
Putting value in (i)
x + 17 = 22
= x = 22 – 17
= 5
Therefore, the number of stamps of 3p will be 5
Question no – (24)
Solution :
Let, number of children = x
Number of adults = y
From, 25x + 75y = 1503 Rs – (I)
= 150300p
And, x + y = 2500
= y = 2500 – x – (ii)
Now, substitute the value,
25x + 75 = (2500 – x) = 150300
= 25x + 75 × 2500 – 75x = 150300
= – 50x = 150300 – 187500
= – 37200
= x = 37200/50
= 744 children
Putting value in (ii)
= y = 2500 – 744
= 1756 adults
Therefore, total 744 children’s saw the fair.
Question no – (25)
Solution :
Let, 1st angle is = x°
2nd angle is = y°
and third angle = x + y
∴ x + y + x + y = 180
= 2x + 2y = 180
= x + y = 90° – (i)
and, x – y = 50° – (ii)
(i) + (ii)
2x = (90 + 50)
= 140°
= x = 70°
So, x + y = 90°
= y = 90 – x
= 90 – 70
= 20°
Hence, the required angles are 20°, 70°, and 90°
Question no – (26)
Solution :
Let, 1st angle = x°
∴ x + y = 180° – (i)
and, x = 2/5 y, – (ii)
(i) and (ii) substitute
= 2/5y + y = 180°
= 2y + 5y/5 = 190
= 7y = 5 × 180
= y = 900°/7
∴ x = 2/5 × y
= 2/5 × 900/7
= 360°/7
Therefore, opposite angles are 360°/7, 900°/7
Question no – (27)
Solution :
Let, length of rectangle is x m
breadth = y m
∴ Area = xy m2
From questions,
Now, length = (x + 1)
breath = (y + 1)
∴ (x + 1) (y + 1)
= (xy + x + y + 1)
= xy + x + y + 1
= xy + 21
= x + y
= 21 – 1
= 20 – (i)
and, 2nd Area now, (x + 1) (y – 1)
= xy – x + y – 1
= xy – (x – y + 1)
= xy – 5
= – 5 + 1
= – 4 (ii)
From, (i) and (ii)
2x = 24
= x = 24/2 = 12
Subtract (ii) From (i)
2y = 16
= y = 8
∴ Length = 12m
∴ Breadth = 8 m
Now, Perimeter = 2 (l + b)
= 2 (12 + 8)
= 2 × 20
= 40 m
Therefore, the perimeter of the room will be 40 m.
Question no – (28)
Solution :
As per the question,
In downstream time taken = 40/60 hr
In downstream time taken = 60 min = 1 hr
Let, speed of sailor in still water = x km/hr
Speed of current y km/hr,
From, 8/x + y = 40/60
= 8/x + y = 3/2
= 2x + 2y = 24
= x + y = 24/2
= 12 – (i)
and, 8/x – y = 1
= x – y = 8 – (ii)
(i) + (ii)
= 2x + 12 + 8
= 20
= x = 20/2
= 10
Putting value in (i)
= 10 + y = 12
= y = 12 – 10
= 2
Therefore, Speed of sailor in still water will be 10 km/hr.
Question no – (31)
Solution :
Let, the speed of walking = x km/hr
∴ Speed of cycling = y km/hr
∴ y – x = 7
= y = x + 7 – (i)
Now, the distance covered by walking = x × 6 km
the distance covered cycling = y × 2 km
∴ 6x = 2y
= x = 2/6 y
= y = 6/2 x
= 3x – (ii)
From (i) and (ii)
3x = x + 7
= 3x – x = 7
= 2x = 7
= x = 7/2
= 3.5
Putting values of x,
= y = 3x
= 3 × 3.5
= 10.5
∴ Average speed of man by walking = 3.5 km/hr
∴ Average speed of man by cycling = 10.5 km/hr
Question no – (32)
Solution :
Let, speed of train x km/hr
speed of car y km/hr
∴ Distance travelled by car
= 600 – 200
= 400 km
∴ 400/x + 200/y = 6 1/2 hr
= 13/2 hr – (i) × 2
and, 200/x + 400/y = 7 – (ii) × 1
800/x + 400/y = 13
200/x + 400/y = 7
(-) (-) (-)
—————————-
= 600/x = 6
= x = 600/6 = 100
∴ Putting in value (i)
∴ 400/100 + 200/y = 13/2
= 200/y = 13/2 – 4
= 13 – 8/2
= 200/x = 5/2
= 15y = 2 × 200
= y = 200 × 2/5
= 80
∴ Speed of train is = 100km/hr.
∴ Speed of car is = 80km/hr.
Next Chapter Solution :
👉 Chapter 6 👈