Exercise 2.1

 

Solve the following equations.

Q.1 x−2=7
Sol. Given, x−2=7
Adding 2 on both the sides, we get,
x−2+2=7+2
x=7+2
x=9

Q.2y+3=10
Sol. Given, y+3=10
Subtracting 3 on both the sides, we get,
y+3−3=10−3
y=10−3
y=7

Q.3 6=z+2
Sol. Given, 6=z+2
On rearranging the terms, we get,
z+2=6
Subtracting 2 on both the sides, we get,
z+2−2=6−2
z=4

Q.4 37+x=177
Sol. Given, 37+x=177
Subtracting 37on both the sides, we get,
37+x−37=177−37
x=17−37
x=147
x=2

Q.5 6x=12
Sol. Given, 6x=12
Dividing by 6 on both the sides, we get,
6x6=126
x=126
x=2

Q.6 t5=10
Sol. Given, t5=10
Multiplying by 5 on both the sides, we get,
t5×5=10×5
t=10×5
t=50

Q.7 2x3=18
Sol. Given, 2x3=18
Multiplying by 3 on both the sides, we get,
2x3×3=18×3
2x=54
Dividing by 2 on both the sides, we get,
2x2=542
x=27

Q.8 1.6=y1.5
Sol. Given, 1.6=y1.5
Multiplying 1.5 on both the sides, we get,
1.6×1.5=y1.5×1.5
1.6×1.5=y
2.40=y
y=2.4

Q.9 7x−9=16
Sol. Given, 7x−9=16
Adding 9 on both the sides, we get,
7x−9+9=16+9
7x=25
Dividing by 7 on both the sides, we get,
7x7=257
x=257

Q.10 14y−8=13
Sol. Given, 14y−8=13
Adding 8 on both the sides, we get,
14y−8+8=13+8
14y=21
Dividing by 14 on both the sides, we get,
14y14=2114
y=32

Q.11 17 + 6p = 9
Sol. Given, 17 + 6p = 9
Subtracting 17 on both the sides, we get,
17 + 6p – 17 = 9 – 17
6p = – 8
Dividing by 6 on both the sides, we get,
6p6=−86
p=−43

Q.12 x3+1=715
Sol. Given, x3+1=715
Transposing 1 from LHS to RHS, we get,
x3=715−1
x3=7−1515
x3=−815
Multiplying by 3 on both the sides, we get,
x3×3=−815×3
x=−85

 

Exercise 2.2

Q.1 If you subtract12from a number and multiply the result by 12, you get 18. What is the number?
Sol. Let the number be x.
Given, (x−12)×12=18
Dividing both the sides by 12, we get,
(x−12)×12÷12=18÷12
(x−12)×12×21=18×21
(x−12)=28
(x−12)=14
Transposing −12to RHS, we get,
x=14+12
x=1+24
x=34

Q.2 The perimeter of a rectangular swimming pool is 154 m. Its length is 2 m more than twice its breadth. What are the length and the breadth of the pool?
Sol. Let the breadth be x m. Therefore, the length will be (2x + 2) m.
Now, given perimeter of swimming pool = 2(length + breadth) = 154 m.
2 (2x + 2 + x) = 154
2 (3x + 2) = 154
Dividing both the sides by 2, we get,
2(3x+2)2=1542
(3x + 2) = 77
Transposing 2 from LHS to RHS, we get,
3x = 77 – 2
3x = 75
Dividing by 3 on both the sides, we get,
3x3=753
x = 25
Therefore, (2x + 2) = (2×25 + 2) = 52
Hence, breadth and length of the pool are 25 m and 52 m respectively.

Q.3 The base of an isosceles triangle is cm. The perimeter of the triangle is 4 cm. What is the length of either of the remaining equal sides?
Sol. Let the length of equal sides be x cm.
Now, perimeter of triangle = x + x + base = 4215
2x+43=6215
Transposing 43from LHS to RHS, we get,
2x=6215−43
2x=62−4×515
2x=62−2015
2x=4215
Dividing both the sides by 2, we get,
2x2=4215×12
x=75=125
Therefore, the length of equal sides is 125 cm.

Q.4 Sum of two numbers is 95. If one exceeds the other by 15, find the numbers.
Sol. Let one number be x. Thus, the other number will be x + 15.
Given, x + x + 15 = 95
2x + 15 = 95
Transposing 15 from LHS to RHS, we get,
2x = 95 – 15
2x = 80
Dividing both the sides by 2, we get,
2x2=802
x = 40
x + 15 = 40 + 15 = 55
Thus, the numbers are 40 and 55.

Q.5 Two numbers are in the ratio 5:3. If they differ by 18, what are the numbers?
Sol. Let the first number be 5x and second number be 3x.
Given, 5x 3x = 18
2x = 18
Dividing by 2 on both the sides, we get,
2x2=182
x = 9
Since, first number = 5x = 5 × 9 = 45
Second number = 3x = 3 × 9 = 27
Therefore, the numbers are 45 and 27.

Q.6 Three consecutive integers add up to 51. What are these integers?
Sol. Let the three consecutive integers be x, x + 1 and x + 2.
Given, x + x + 1 + x + 2 = 51
3x + 3 = 51
Transposing 3 from LHS to RHS, we get,
3x = 51 – 3
3x = 48
Dividing by 3 on both the sides, we get,
3x3=483
x = 16
Therefore, x + 1 = 17 and x + 2 = 18.
Hence, three consecutive integers are 16, 17 and 18.

Q.7 The sum of three consecutive multiples of 8 is 888. Find the multiples.
Sol. Let the three consecutive multiples of 8 be 8a, 8(a + 1) and 8(a + 2).
Given, 8a + 8(a + 1) + 8(a + 2) = 888
8(a + a + 1 + a + 2) = 888
8(3a + 3) = 888
Dividing by 8 on both the sides, we get,
8(3a+3)8=8888
3a + 3 = 111
Transposing 3 from LHS to RHS, we get,
3a = 111 – 3
3a = 108
Dividing by 3 on both the sides, we get,
3a3=1083
a = 36
Therefore, 8a = 8 × 36 = 288,
8(a + 1) = 8(36 + 1) = 8 × 37 = 296
8(a + 2) = 8(36 + 2) = 8 × 38 = 304
Thus, the required numbers are 288, 296 and 304.

Q.8 Three consecutive integers are such that when they are taken in increasing order and multiplied by 2, 3 and 4 respectively, they add up to 74. Find these numbers.
Sol. Let the three consecutive integers be a, a + 1 and a + 2.
Given, 2a + 3(a + 1) + 4(a + 2) = 74
2a + 3a + 3 + 4a + 8 = 74
9a + 11 = 74
Transposing 11 from LHS to RHS, we get,
9a = 74 – 11
9a = 63
Dividing by 9 on both the sides, we get,
9a9=633
a = 7
Therefore, a + 1 = 7 + 1 = 8, a + 2 = 7 + 2 = 9
Therefore, the required numbers are 7, 8 and 9.

Q.9 The ages of Rahul and Haroon are in the ratio 5:7. Four years later the sum of their ages will be 56 years. What are their present ages?
Sol. Let the age of Rahul be 5x and the age of Haroon be 7x.
After 4 years, the age of Rahul will be (5x + 4) and that of Haroon will be (7x + 4)
Given, (5x + 4) + (7x + 4) = 56
12x + 8 = 56
Transposing 8 from LHS to RHS, we get,
12x = 56 – 8
12x = 48
Dividing by 12 on both the sides, we get,
12x12=4812
x = 4
Therefore, age of Rahul = 5x = 5 × 4 = 20 years
Age of Haroon = 7x = 7 × 4 = 28 years

Q.10 The number of boys and girls in a class are in the ratio 7:5. The number of boys is 8 more than the number of girls. What is the total class strength?
Sol. Let the number of boys be 7x and number of girls be 5x.
Given, 7x – 8 = 5x
Transposing 7x from LHS to RHS, we get,
– 8 = 5x – 7x
– 8 = – 2x
8 = 2x
Dividing both the sides by 2, we get,
82=2x2
4 = x
x = 4
Therefore, number of boys = 7x = 7 × 4 = 28
Number of boys = 5x = 5 × 4 = 20
Hence, total class strength = 28 + 20 = 48 students.

Q.11 Baichung’s father is 26 years younger than Baichung’s grandfather and 29 years older than Baichung. The sum of the ages of all the three is 135 years. What is the age of each one of them?
Sol. Let Baichung’s father’s age be a years.
Hence, age of Baichung’s father and Baichung’s grandfather will be (a – 29) and (a + 26) respectively.
Given, a + (a – 29) + (a + 26) = 135
3a – 3 = 135
Transposing 3 from LHS to RHS, we get,
3a = 135 + 3
3a = 138
Dividing by 3 on both the sides, we get,
3a3=1383
a = 46
Thus, age of Baichung’s father = a years = 46 years
Age of Baichung = (a – 29) = (46 – 29) = 17 years
Age of Baichung’s Grandfather = (a + 26) = (46 + 26) = 72 years.

Q.13 A rational number is such that when you multiply it by 52and add 23 to the product, you get−712. What is the number?
Sol. Let the number be a.
Given, 52a+23=−712
Transposing from LHS to RHS, we get,
52a=−712−23
52a=−7−(2×4)12
52a=−1512
Multiplying by on both the sides, we get,
a=−1512×25
a=−12
Thus, the required rational number is −12.

Q.14 Lakshmi is a cashier in a bank. She has currency notes of denominations Rs 100, Rs 50 and Rs 10, respectively. The ratio of the number of these notes is 2:3:5. The total cash with Lakshmi is Rs 4,00,000. How many notes of each denomination does she have?
Sol. Let the number of Rs 100 notes, Rs 50 notes and Rs 10 notes be 2x, 3x, and 5x respectively.
Therefore, amount of Rs 100 notes = Rs (100 × 2x) = Rs 200x
Amount of Rs 100 notes = Rs (50 × 3x) = Rs 150x
Amount of Rs 100 notes = Rs (10 × 5x) = Rs 50x
Given, 200x + 150x + 50x = 400000
400x = 400000
Dividing by 400 on both the sides, we get,
x = 1000
Thus, number of Rs 100 notes = 2x = 2 × 1000 = 2000
Number of Rs 50 notes = 3x = 3 × 1000 = 3000
Number of Rs 10 notes = 5x = 5 × 1000 = 5000

Q.15 I have a total of Rs 300 in coins of denomination Re 1, Rs 2 and Rs 5. The number of Rs 2 coins is 3 times the number of Rs 5 coins. The total number of coins is 160. How many coins of each denomination are with me?
Sol. Let the number of Rs 5 coins be a.
Number of Rs 2 coins = 3 x number of Rs 5 coins = 3a
Number of Rs 1 coins = 160 – (number of coins of Rs 5 and of Rs 2) = 160 – (3a + a) = 160 – 4a.
Now, amount of Rs 1 coins = Rs [1 x (160 – 4a)] = Rs (160 – 4a)
Amount of Rs 2 coins = Rs (2 x 3a) = Rs 6a
Amount of Rs 5 coins = Rs (5 x a) = Rs 5a
Given, 160 – 4a + 6a + 5a = 300
160 + 7a = 300
Transposing 160 from LHS to RHS, we get,
7a = 300 – 160
7a = 140
Dividing by 7 on both the sides, we get,
7a7=1407
a = 20
Thus, number of Rs 5 coins = a = 20
Number of Rs 2 coins = 3a = 3 x 20 = 60
Number of Rs 1 coins = 160 – 4a = 160 – 4 x 20 = 160 – 80 = 80

Q.16 The organisers of an essay competition decide that a winner in the competition gets a prize of Rs 100 and a participant who does not win gets a prize of Rs 25. The total prize money distributed is Rs 3,000. Find the number of winners, if the total number of participants is 63.
Sol. Let the number of winners be y.
Hence, the number of participants who did not win will be 63 – y.
Amount given to the winners = Rs (100 x y) = Rs 100y
Amount given to the participants who did not win = Rs [25 (63 – y)] = Rs (1575 – 25y)
Given, 100y + 1575 – 25y = 3000
Transposing 1575 from LHS to RHS, we get,
75y = 3000 – 1575
75y = 1425
Dividing by 75 on both the sides, we get,
75y75=142575
y = 19.
Thus, number of winners are 19.

 

Exercise 2.3

 

Solve the following equations and check your results.

Q.1 3x = 2x + 18
Sol. Given, 3x = 2x + 18
Transposing 2x from RHS to LHS, we get,
3x – 2x = 18
x = 18
Verification:
LHS = 3x = 3 x 18 = 54
RHS = 2x + 18 = 2 x 18 + 18 = 36 + 18 = 54
Here, LHS = RHS
Thus, the result obtained is correct.

Q.2 5t – 3 = 3t – 5
Sol. Given, 5t – 3 = 3t – 5
Transposing 3t from RHS to LHS and – 3 from LHS to RHS, we get,
5t – 3t = -5 – (- 3)
2t = – 2
Dividing both the sides by 2, we get,
t = -1
Verification:
LHS = 5t – 3 = 5 x (-1) – 3 = – 8
RHS = 3t – 5 = 3 x (-1) – 5 = -3 – 18 = -8
Here, LHS = RHS
Thus, the result obtained is correct.

Q.3 5x + 9 = 5 + 3x
Sol. Given, 5x + 9 = 5 + 3x
Transposing 3x from RHS to LHS and 9 from LHS to RHS
5x – 3x = 5 – 9
2x = -4
Dividing both the sides by 2, we get,
x = -2
Verification:
LHS = 5x + 9 = 5 x (-2) + 9 = – 1
RHS = 5 + 3x = 5 + 3 x (-2) = 5 – 6 = -1
Here, LHS = RHS
Thus, the result obtained is correct.

Q.4 4z + 3 = 6 + 2z
Sol. Given, 4z + 3 = 6 + 2z
Transposing 2z from RHS to LHS and 3 from LHS to RHS, we get,
4z – 2z = 6 – 3
2z = 3
Dividing by 2 on both the sides, we get,
z=32
Verification:
LHS = 4z + 3 = 4 x (32)+ 3 = 6 + 3 = 9
RHS = 6 + 2z = 6 + 2 x (32)= 6 + 3 = 9
Here, LHS = RHS
Thus, the result obtained is correct.

Q.5 2x – 1 = 14 – x
Sol. Given, 2x – 1 = 14 – x
Transposing x from RHs to LHS and 1 from LHS to RHS, we get,
2x + x = 14 + 1
3x = 15
Dividing by 3 on both the sides, we get,
x = 5
Verification:
LHS = 2x – 1 = 2 x (5) – 1 = 10 – 1= 9
RHS = 14 – x = 14 – 5 = 9
Here, LHS = RHS
Thus, the result obtained is correct.

Q.6 8x + 4 = 3 (x – 1) + 7
Sol. Given, 8x + 4 = 3 (x – 1) + 7
8x + 4 = 3x – 3 + 7
Transposing 3x from RHS to LHS and 4 from LHS to RHS, we get,
8x – 3x = – 3 + 7 – 4
5x = – 7 + 7
x = 0
Verification:
LHS = 8x + 4 = 8 x (0) + 4 = 4
RHS = 3(x – 1) + 7 = 3(0 – 1) + 7 = – 3 + 7 = 4
Here, LHS = RHS
Thus, the result obtained is correct.

Q.7 x=45(x+10)
Sol. Given, x=45(x+10)
Multiplying by 5 on both the sides, we get,
5x = 4(x + 10)
5x = 4x + 40
Transposing 4x from RHS to LHS, we get,
5x – 4x = 40
x = 40
Verification:
LHS = x = 40
RHS =45(x+10)= 45(40+10)= 45×50= 40
Here, LHS = RHS
Thus, the result obtained is correct.

Q.8 2x3+1=7x15+3
Sol. Given, 2x3+1=7x15+3
Transposing 7x15 from RHS to LHS and 1 from LHS to RHS, we get,
2x3−7x15=3−1
5×2x−7x15=2
3x15=2
x5=2
Multiplying by 5 on both the sides, we get,
x = 10
Verification:
LHS = 2x3+1= 2×103+1=2×10+1×33=223
RHS = 7x15+3= 7×215+3= 1415+3=
14+3×1515=233=
Here, LHS = RHS
Thus, the result obtained is correct.

Q.9 2y+53=263−y
Sol. Given, 2y+53=263−y
2y+y=263−53
3y=213=7
Dividing by 3 on both the sides, we get,
y=73
Verification:
LHS = 2y+53= 2×73+53 = 143+53 = 193
RHS = 263−y= 263−73= 193
Here, LHS = RHS
Thus, the result obtained is correct.

Q.10 3m=5m−85
Sol. Given, 3m=5m−85
Transposing 5m from RHS to LHS, we get,
3m−5m=−85
−2m=−85
Dividing by -2 on both the sides, we get,
m=45
Verification:
LHS = 3m=3×45=125
RHS = 5m−85=5×45−85=125
Here, LHS = RHS
Thus, the result obtained is correct.

 

Exercise 2.4

 

Q.1 Amina thinks of a number and subtracts from it. She multiplies the result by 8. The result now obtained is 3 times the same number she thought of. What is the number?
Sol. Let the number be a.
Given, 8(a−52)=3a
8a – 20 = 3a
Transposing 3a from RHS to LHS and – 20 from LHS to RHS, we get,
8a – 3a = 20
5a = 20
Dividing by 5 on both the sides, we get,
a = 4
Therefore, the number is 4.

Q.2 A positive number is 5 times another number. If 21 is added to both the numbers, then one of the new numbers becomes twice the other new number. What are the numbers?
Sol. Let the numbers be a and 5a.
Given, 21 + 5a = 2(a + 21)
21 + 5a = 2a + 42
Transposing 2a from RHS to LHS and 21 from LHS to RHS, we get,
5a – 2a = 42 – 21
3a = 21
Dividing both the sides by 3, we get,
a = 7
Hence, 5a = 5 x 7 = 35
Thus, the numbers are 7 and 35 respectively.

Q.3 Sum of the digits of a two-digit number is 9. When we interchange the digits, it is found that the resulting new number is greater than the original number by 27. What is the two-digit number?
Sol. Let the digit at ten’s place and one’s place be s and (9 – s) respectively.
Hence, original number = 10s + (9 – s) = 9s + 9
Now, new number on interchanging the digits = 10(9 – s) + s = 90 – s
Given, 90 – 9s = 9s + 9 + 27
90 – 9s = 9s + 36
Transposing 9s from LHS to RHS and 36 from RHS to LHS, we get,
90 – 36 = 18s
54 = 18s
Dividing by 18 on both the sides, we get,
s = 3
Therefore, 9 – s = 9 – 3 = 6
Thus, the two digit number is 9s + 9 = 9 × 3 + 9 = 36.

Q.4 One of the two digits of a two digit number is three times the other digit. If you interchange the digits of this two-digit number and add the resulting number to the original number, you get 88. What is the original number?
Sol. Let the digits at ten’s place and one’s place be s and 3s respectively.
Hence, original number = 10s + 3s = 13s
Now, number after interchanging = 10 x 3s + s = 30s + s = 31s
Given, 13s + 31s = 88
44s = 88
Dividing by 44 on both the sides, we get,
s = 2
Therefore, original number = 13s = 13 x 2 = 26
And number after interchanging = 31 x 2 = 62
Hence, the two-digit number might be 26 or 62.

Q.5 Shobo’s mother’s present age is six times Shobo’s present age. Shobo’s age five years from now will be one third of his mother’s present age. What are their present ages?
Sol. Let the age of Shobo be a years. Hence, Shobo’s mother’s age will be 6a years.
Given, a+5=6a3
a + 5 = 2a
Transposing a from LHS to RHS, we get,
5 = 2a – a
5 = a
a = 5
Therefore, 6a = 6 x 5 = 30
Thus, the present ages of Shobo and his mother are 5 years and 30 years respectively.

Q.6 There is a narrow rectangular plot, reserved for a school, in Mahuli village. The length and breadth of the plot are in the ratio 11:4.At the rate Rs100 per metre it will cost the village panchayat Rs 75000 to fence the plot. What are the dimensions of the plot?
Sol. Let the common ratio between the length and breadth of the plot be s.
Therefore, the length and the breadth of the rectangular plot will be 11s and 4s respectively.
Perimeter of the plot = 2(length + breadth) = 2(11s + 4s) = 30s
Given, 100 x Perimeter = 75000
100 x 30s = 75000
3000s = 75000
Dividing by 3000 on both the sides, we get,
s = 25
Therefore, length = 11s = (11 x 25) = 275
Breadth = 4s = (4 x 25) = 100
Thus, the dimensions of the plot are 275 m and 100 m respectively.

Q.7 Hasan buys two kinds of cloth materials for school uniforms, shirt material that costs him Rs 50 per metre and trouser material that costs him Rs 90 per metre. For every 2 meters of the trouser material he buys 3 metres of the shirt material. He sells the materials at 12% and 10% profit respectively. His total sale is Rs 36,660. How much trouser material did he buy?
Sol. Let trouser material and shirt material bought be 2a and 3a respectively.
Selling price of trouser material per metre = Rs (90+90×12100) = Rs 100.80
Selling price of shirt material per metre = Rs (50+50×10100) = Rs 55
Given, 100.80 x 2a + 55 x 3a = 36660
201.60a + 165a = 36660
366.60a = 36660
Dividing by 366.60 on both the sides, we get,
a = 100
Therefore, trouser material = 2a = 2 x 100 = 200
Shirt material = 3a = 3 x 100 = 300
Thus, Hasan should buy 200 m of trouser material.

Q.8 Half of a herd of deer are grazing in the field and three fourths of the remaining are playing nearby. The rest 9 are drinking water from the pond. Find the number of deer in the herd.
Sol. Let the number of deer be a.
So, number of deer grazing in the field = a2
Number of deer playing nearby =34 x Number of remaining deer
=34×(a−a2) = 34×(a2) = 3a8
Given, a−(a2+3a8)=9
a−(4a+3a8)=9
a−(7a8)=9
a8=9
Multiplying by 8 on both the sides, we get,
a = 72
Thus, the total number of deer is 72.

Q.9 A grandfather is ten times older than his granddaughter. He is also 54 years older than her. Find their present ages.
Sol. Let the age of granddaughter be s years. Hence, the age of grandfather will be 10s years.
Given, 10s = s + 54
Transposing s from RHS to LHS, we get,
10s – s = 54
9s = 54
s = 6
Therefore, age of grandfather = 10s = 10 x 6 = 60 years.
Thus, the present ages of granddaughter and grandfather are 6 and 60 respectively.

Q.10 Aman’s age is three times his son’s age. Ten years ago he was five times his son’s age. Find their present ages.
Sol. Let the age of Aman’s son be a years. Hence, age of Aman will be 3a years.
Before ten years, age of Aman and his son will be (a – 10) and (3a – 10) respectively.
Given, (3a – 10) = 5(a – 10)
3a – 10 = 5a – 50
Transposing 3a from LHS to RHS and 50 RHS to LHS, we get,
50 – 10 = 5a – 3a
40 = 2a
Dividing by 2 on both the sides, we get,
20 = a
Therefore, age of Aman’s son = a = 20
Age of Aman = 3a = 3 x 20 = 60
Hence, the present ages of Aman and his son are 60 years and 20 years respectively.

 

Exercise 2.5

 

Solve the following linear equations.

Q.1 x2−15=x3+14
Sol. Given, x2−15=x3+14
Transposing x3 from RHS to LHS and 15 from LHS to RHS, we get,
x2−x3=14+15
3x−2x6=5+420
x6=920
x=9×620
x=2710

Q.2 n2−3n4+5n6=21
Sol. Given, n2−3n4+5n6=21
6n−9n+10n12=21
7n12=21
n=21×127
n=36

Q.3x+7−8x3=176−5x2
Sol. Given, x+7−8x3=176−5x2
Transposing 5x2 from RHS to LHS and 7 from LHS to RHS, we get,
x+5x2−8x3=176−7
6x−16x+15x6=17−426
5x6=−256
x=−25×66×5
x=−5

Q.4 x−53=x−35
Sol. Given, x−53=x−35
5 (x – 5) = 3 (x – 3)
5x – 25 = 3x – 9
Transposing 3x from RHS to LHS and 25 from LHS to RHS, we get,
5x – 3x = -9 + 25
2x = 16
x = 8

Q.5 3t−24−2t+33=23−t
Sol. Given, 3t−24−2t+33=23−t
Transposing from RHS to LHS, we get,
3t−24−2t+33+t=23
3(3t−2)−4(2t+3)+12t4=23
9t−6−8t−12+12t12=23
3t−1812=23
3×(13t−18)=2×12
39t−54=24
39t=24+54
39t=78
t=7839
t=2

Q.6 m−m−12+m−23=1
Sol. Given, m−m−12+m−23=1
Transposing m−23from RHS to LHS, we get,
m−m−12+m−23=1
6m−3(m−1)+2(m−2)2=1
6m−3m+2+2m−46=1
5m−16=1
5m−1=6
5m=7
m=75

Simplify and solve the following linear equations.

Q.7 3(t-3) = 5(2t+1)
Sol. Given, 3(t-3) = 5(2t+1)
3t – 9 = 10t + 5
3t-10t = 5 + 9
-7t = 14
t=14−7
t = -2

Q.8 15 (y-4) -2 (y-9) + 5 (y+6) = 0
Sol. Given,  15 (y-4) -2 (y-9) + 5 (y+6) = 0
15y – 60 – 2y + 18 + 5y + 30 = 0
18y – 12 = 0
18y = 12
y=1218
y=23

Q.9 3(5z – 7) – 2 (9z – 11) = 4 (8z  – 13) -17
Sol. Given, 3(5z – 7) – 2 (9z – 11) = 4 (8z  – 13) -17
15z – 21 -18z + 22 = 32z – 52 – 17
-3z + 1 = 32z – 69
-3z -32 z = – 69 – 1
-35 z = -70
z=−70−35
z = 2

Q.10 0.25 (4f – 3) = 0.05 (10f – 9)
Sol. Given, 0.25 (4f – 3) = 0.05 (10f – 9)
1.00 f -0.75 = 0.50 f – 0.45
1.00 f -0.50 f = -0.45 +0.75
0.50 f = 0.3
f=0.30.50
f = 0.6

 

Exercise 2.6

 

Solve the following linear equations.

Q.1 8x−33x=2
Sol. Given, 8x−33x=2
Multiplying by 3x on both the sides, we get,
(8x – 3) = 2 X 3x
(8x – 3) = 6x
8x – 6x = 3
2x = 3
x=32

Q.2 9x7−6x=15
Sol. Given, 9x7−6x=15
Multiplying by (7 – 6x) on both the sides, we get,
9x = 15 (7 – 6x)
9x =105 -90x
9x +90x = 105
99x =105
x=10599
x=3533

Q.3 zz+15=49
Sol. Given, zz+15=49
Multiplying by 9(z + 15) on both the sides, we get,
9z = 4 (z + 15)
9z = 4z +60
9z – 4z = 60
5z = 60
z=605
z=15

Q.4 3y+42−6y=−25
Sol. Given, 3y+42−6y=−25
Multiplying by 5(2 – 6y) on both the sides, we get,
5(3y + 4) = -2 (2 – 6y)
15y + 20 = -4 + 12y
15y – 12y = -4 -20
3y = -24
y=−243
y = -8

Q.5 7y+4y+2=−43
Sol. Given, 7y+4y+2=−43
Multiplying by 3(y + 2) on both the sides, we get,
3 (7y + 4) = -4 (y + 2)
21y + 12 = -4y – 8
21y + 4y = -8-12
25y = -20
y=−45

Q.6 The ages of Hari and Harry are in the ratio 5:7. Four years from now the ratio of their ages will be 3:4. Find their present ages.
Sol. Let the ages of Hari and Harry be 5x and 7x respectively.
After 4 years, the ages of Hari and Harry will be (5x + 4) and (7x + 4) respectively.
Given, 5x+47x+4=34
4(5x + 4) = 3(7x + 4)
20x +16 = 21x + 12
20 – 21x =12 – 16
-x = -4
x = 4
Therefore, age of Hari = 5x = 5 x 4 = 20
Age of Harry = 7x = 7 x 4 = 28
Thus, the present ages of Hari and Harry are 20 and 28 years respectively.

Q.7 The denominator of a rational number is greater than its numerator by 8. If the numerator is increased by 17 and the denominator is decreased by 1, the number obtained is . Find the rational number.
Sol. Let the numerator of a rational number be a. Thus, denominator will be a + 8.
Therefore, rational number will be aa+8
Given, a+17a+8−1=32
2 (a + 17) = 3 (a + 7)
2a + 34 = 3a + 21
2a – 3a = 21 – 34
-a = -13
a = 13
Thus, the required rational is aa+8=1313+8=1321