1.Solve the equation: x – 2 = 7.

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We have: x – 2 = 7

⇒x – 2 + 2 = 7 + 2 (adding 2 on both sides)

⇒ x = 9

**Check:**

Keep x=9 in the given equation:

L.H.S= x-2=9-2=7=R.H.S

Hence we have x=9 as the solution.

2.Solve the equation: y + 3 = 10.

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We have: y + 3 = 10

⇒y + 3 - 3 = 10 - 3 (subtracting 3 on both sides)

⇒ y = 7

**Check:**

Keep y=7 in the given equation:

L.H.S= y+3=7+3=10=R.H.S

Hence we have y=7 as the solution.

3.Solve the equation: 6 = z + 2

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We have:6 = z + 2

⇒6 - 2=z +2 - 2 (subtracting 2 on both sides)

⇒ z=4

**Check:**

Keep z=4 in the given equation:

L.H.S= z+2=4+2=6=R.H.S

Hence we have z=4 as the solution.

4.Solve the equations: \(\frac { 3 }{ 7 } + x = \frac { 17 }{ 7 }\)

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We have:\(\frac { 3 }{ 7 } + x = \frac { 17 }{ 7 }\)

⇒\(\frac { 3 }{ 7 } -\frac{3}7 + x = \frac { 17 }{ 7 }-\frac{3}7 \quad \) (subtracting \(\frac{3}7\) on both sides)

⇒ x=\(\frac{17-3}7\)

⇒ x=\(\frac{14}7=2\)

⇒ x=2

**Check:**

Keep x=2 in the given equation:

L.H.S= x+\(\frac{3}7\)=\(\frac{3}7+2\)=\(\frac{3+2×7}{7}\)= \(\frac{17}7 \)=R.H.S

Hence we have x=2 as the solution.

5.Solve the equation 6x = 12.

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Here,we have 6x = 12

\(\Rightarrow\) 6x ÷ 6 = 12 ÷ 6 (dividing each side by 6)

\(\Rightarrow\) x = 2

Thus, x = 2 is the required solution.

**Check:**

Keep the value of x obtained in the given equation:

L.H.S=6x=6×2=12=R.H.S

Hence we have x=2 as the solution of the equation

6.Solve the equation \(\frac { t }{ 5 } = 10\).

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Here we have \(\frac { t }{ 5 } = 10\)

\(\Rightarrow\) \(\frac { t }{ 5 } × 5 = 10 × 5\) (multiplying both sides by 5)

\(\Rightarrow\) t = 50

Thus, t = 50 is the required solution.

**Check:**

Keep t=50 in the given equation:

L.H.S= \(\frac{t}5=\frac{50}5=10\)=R.H.S

So, t=50 is the required solution

7.Solve the equation \(\frac { 2x }{ 3 } = 18\).

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Here,we have \(\frac{2x}3 = 18\)

\(\Rightarrow\) \(\frac{2x}3 × 3 = 18 ×3 \quad \)(multiplying each side by 3)

\(\Rightarrow\) \(2x = 54\)

\(\Rightarrow\)2x ÷ 2 = 54 ÷ 2 (dividing both sides by 2)

\(\Rightarrow\) x = 27

Thus, x=27 is the required solution.

**Check:**

Substitute x=27 in the given equation

L.H.S=\(\frac{2x}3=\frac{2×27}{3}=2×9=18\)=R.H.S

Hence, x=27 is the required solution.

8.Solve the equation \(1.6 = \frac { y }{ 1.5 }\)

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We have: \(1.6 = \frac { y }{ 1.5 }\)

\(\Rightarrow 1.6 × 1.5 = \frac { y }{ 1.5 } × 1.5 \quad\) (multiplying both sides by 1.5)

\(\Rightarrow\) 2.40 = y

Thus, y = 2.40 is the required solution.

**Check:**

Keep y=2.40 in the given solution

R.H.S=\(\frac{y}{1.5}=\frac{2.4}{1.5}=1.6\)=L.H.S

Hence y=2.40 is the required solution.

9.Solve the equation 7x – 9 = 16.

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Here we have, 7x – 9 = 16

\(\Rightarrow\) 7x – 9 + 9 = 16 + 9 (adding 9 to both the sides)

\(\Rightarrow\) 7x = 25

\(\Rightarrow\) 7x ÷ 7 = 25 ÷ 7 (dividing both the sides by 7)

\(\Rightarrow\) x = \(\frac { 25 }{ 7 }\)

Thus, x = \(\frac { 25 }{ 7 }\) is the required solution.

**Check:**

Keep x = \(\frac { 25 }{ 7 }\) in the given solution

L.H.S=\(7x-9=7×\frac{25}{7}-9=25-9=16\)=R.H.S

Hence x = \(\frac { 25 }{ 7 }\) is the required solution.

10.Solve the equation 14y – 8 = 13.

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Here,we have 14y – 8 = 13

\(\Rightarrow\) 14y – 8 + 8 = 13 + 8 (adding 8 to both sides)

\(\Rightarrow\) 14y = 21

\(\Rightarrow\) 14y ÷ 14 = 21 ÷ 14 (dividing both sides by 14)

\(\Rightarrow\) y =\(\frac { 21 }{ 14 }\)

\(\Rightarrow\)y = \(\frac { 3 }{ 2 }\)

Thus, y = \(\frac { 3 }{ 2 }\) is the required solution.

**Check:**

Keepy = \(\frac { 3 }{ 2 }\) in the given solution

L.H.S=\(14y-8=14×\frac{3}{2}-8=7×3-8=21-8=13\)=R.H.S

Hence y = \(\frac { 3 }{ 2 }\) is the required solution.

11.Solve the equation 17 + 6p = 9.

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Here,we have, 17 + 6p = 9

\(\Rightarrow\) 17 – 17 + 6p = 9 – 17 (subtracting 17 from the both sides)

\(\Rightarrow\) 6p = -8

\(\Rightarrow\) 6p ÷ 6 = -8 ÷ 6 (dividing both the sides by 6)

\(\Rightarrow\) p = \(\frac { -8 }{ 6 }\)

\(\Rightarrow\) p = \(\frac { -4 }{ 3 }\)

Thus, p = \(\frac { -4 }{ 3 }\) is the required solution.

**Check:**

Keep p = \(\frac { -4 }{ 3 }\) in the given solution

L.H.S=\(17+6p=17+6×\frac{-4}{3}=17-8=9\)=L.H.S

Hencep = \(\frac { -4 }{ 3 }\) is the required solution.

12.Solve the equation \(\frac { x }{ 3 } + 1 = \frac { 7 }{ 15 }\)

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Here we have \(\frac{x}3+1=\frac{7}{15}\)

\(\Rightarrow \frac{x}3+1-1=\frac{7}{15}-1 \quad \)(Subtracting 1 from both the sides)

\(\Rightarrow \frac{x}3=\frac{7}{15}-1 \)

\(\Rightarrow \frac{x}3=\frac{7-15}{15}\)

\(\Rightarrow \frac{x}3=\frac{-8}{15}\)

\(\Rightarrow \frac{x}3×3=\frac{-8}{15}×3\)(multiplying both the sides by 3)

\(\Rightarrow x=\frac{-8}{5}\)

So, we have \(x=\frac{-8}{5}\) as the required solution.

**Check:**

Keep \(x=\frac{-8}{5}\) in the given solution

L.H.S= \(\frac { x }{ 3 } + 1 =\frac{-8}{5}×\frac{1}{3}+1=\frac{-8+15}{15}=\frac{7}{15}=R.H.S\)

Hence \(x=\frac{-8}{5}\) is the required solution.

1.If you subtract \(\frac { 1 }{ 2 }\) from a number and multiply the result by \(\frac { 1 }{ 2 }\), you get \(\frac { 1 }{ 8 }\). What is the number?

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Let the number be x

Condition 1: \(x-\frac{1}2\)

Condition 2: \((x-\frac{1}2) ×\frac{1}2\)

Condition 3: \(x-\frac{1}2) ×\frac{1}2 =\frac{1}8\)

\(\Rightarrow (x-\frac{1}2) ×\frac{1}2 × 8=\frac{1}8×8\quad \)[multiplying both the sides by 8]

\(\Rightarrow (x-\frac{1}2) ×4=1\)

\(\Rightarrow 4x-\frac{1}2 ×4=1\quad \)[opening the brackets]

\(\Rightarrow 4x-2=1\)

\(\Rightarrow 4x-2+2=1+2 \quad \)[adding 2 to both the sided]

\(\Rightarrow 4x=3\)

\(\Rightarrow 4x÷4= 3÷4\quad \)[dividing both the sides by 4]

\(\Rightarrow x=\frac{3}4\)

Thus, the required number is \(\frac{3}4\)

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?

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Let the breadth of the rectangular swimming pool be x

Condition I: Length of the pool = (2x + 2) m.
Condition II: Perimeter of the pool =2×(length+breadth)

\(\Rightarrow\)2×(2x+2+x)= 154 m.

\(\Rightarrow\)2 × [3x + 2] = 154

\(\Rightarrow\) 6x + 4 = 154 [solving the bracket]

\(\Rightarrow\) 6x+4-4 = 154 – 4 [subtracting 4 from both the sides]

\(\Rightarrow\) 6x = 150

\(\Rightarrow\) 6x÷6 = 150 ÷ 6 [dividing 6 on both the sides]

\(\Rightarrow\) x = 25

Thus, the required breadth x= 25 m

and the length = 2×x + 2 = 2×25+2=50 + 2 = 52 m.

3.The base of an isosceles triangle is \(\frac{4}3\) cm. The perimeter of the triangle is \(4\frac{2}{15}\) cm. What is the length of either of the remaining equal sides?

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We have length of base as \(\frac{4}3\)

Let the length of the equal sides by each of x.

Perimeter of the isosceles triangle=sum of the base +sum of two equal sides=\(4\frac{2}{15}\)

\(\qquad = (\frac{4}3+x+x)cm =4\frac{2}{15}\)

∴ \(2x+\frac{4}3=4\frac{2}{15}\)

\(\Rightarrow 2x+\frac{4}3=\frac{62}{15}\)

\(\Rightarrow 2x=\frac{62}{15}-\frac{4}3\quad \)[Transposing \(\frac{4}3\) from (+) to (-)]

\(\Rightarrow 2x=\frac{62-20}{15}\)

\(\Rightarrow 2x=\frac{42}{15}\)

\(\Rightarrow x=\frac{42}{15}÷2\quad \)[Transposing 2 from (x) to ÷]

\(\Rightarrow x=\frac{42}{15}×\frac{1}2\)

\(\Rightarrow x=\frac{7}{5}=1\frac{2}5\)

Hence, the length of each equal side is =\(1\frac{2}5\)

4.Sum of two numbers be 95. If one exceeds the other by 15, find the numbers.

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Let one number be x and the other be y.

As per given we have y = x + 15

Also we are given that x+y=95

\(\Rightarrow\)x + (x + 15) = 95 [Substituting the value of y]

\(\Rightarrow\) x + x + 15 = 95

\(\Rightarrow\) 2x + 15 = 95

\(\Rightarrow\) 2x = 95 – 15 [transposing 15 from (+) to (-)]

\(\Rightarrow\) 2x = 80

\(\Rightarrow\) x =\(\frac { 80 }{ 2 }\quad\) [transposing 2 from (×) to (÷)]

\(\Rightarrow\) x = 40

∴Other number y= x + 15 = 40+15 = 55

Thus, the required numbers are 40 and 55.

5.Two numbers are in the ratio 5 : 3. If they differ by 18, what are the numbers?

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Let the two numbers be 5x and 3x.

As per the given question, we get

5x – 3x = 18

\(\Rightarrow\)2x = 18

\(\Rightarrow\) x = 18 ÷ 2 [Transposing 2 from (×) to (÷)]

\(\Rightarrow\) x = 9.

Thus, 5x =5×9= 45 and 3x=3×9 =27

Hence, we have the numbers as 45 and 27.

6.Three consecutive integers add up to 51. What are these integers?

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Let the three consecutive integers be x, x + 1 and x + 2.

As per the condition given, we have

x + (x + 1) + (x + 2) = 51

\(\Rightarrow\)x + x + 1 + x + 2 = 51

\(\Rightarrow\) 3x + 3 = 51

\(\Rightarrow\) 3x +3 -3= 51 – 3 [subtracting 3 from both the sides]

\(\Rightarrow\)3x = 48

\(\Rightarrow\) x ÷3= 48 ÷ 3 [dividing 3 to both the sides]

\(\Rightarrow\)x = 16

Thus, the required integers are 16, 16 + 1 = 17 and 16 + 2 = 18, i.e., 16, 17 and 18.

7.The sum of three consecutive multiples of 8 is 888. Find the multiples.

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Let the three consecutive multiples of 8 be 8x, 8x + 8 and 8x + 16.

As per the given conditions in the question we have,

8x+8x+8+8x+16=888,

\(\Rightarrow\)24x+24=888

\(\Rightarrow\)24x+24-24=888-24[subtracting 24 from both the sides]

\(\Rightarrow\)24x=864

\(\Rightarrow\)24x÷24=864÷24[dividing both the sides by 24 ]

\(\Rightarrow\) x= 36

So we have the consecutive numbers as :

8x=8×36=288

8x+8=8×36+8=294

8x+16=8×36+16=304

Thus, we have the required numbers as 288,294 and 304.

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.

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Let the three consecutive integers be x, x + 1 and x + 2.

As per the condition given, we have

2x + 3(x + 1) + 4(x + 2) = 74

\(\Rightarrow\) 2x + 3x + 3 + 4x + 8 = 74

\(\Rightarrow\) 9x + 11 = 74

\(\Rightarrow\) 9x = 74 – 11 (transposing 11 to RHS)

\(\Rightarrow\) 9x = 63

\(\Rightarrow\) x = 63 ÷ 9 (transposing 9 to RHS)

\(\Rightarrow\) x = 7

Thus, the required numbers are 7, 7 + 1 = 8 and 7 + 2 = 9, i.e., 7, 8 and 9.

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

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Let the present ages of Rahul and Haroon he 5x years and 7x years respectively.

As per the conditions given we have

4 years later, the age of Rahul will be (5x + 4) years.

4 years later, the age of Haroon will be (7x + 4) years.

As per the conditions, we get

(5x + 4) + (7x + 4) = 56

\(\Rightarrow\) 5x + 4 + 7x + 4 = 56

\(\Rightarrow\) 12x + 8 = 56

\(\Rightarrow\) 12x = 56 – 8 (transposing 8 to RHS)

\(\Rightarrow\) 12x = 48

\(\Rightarrow\) x = 48 ÷ 12 = 4 (transposing 12 to RHS)

Hence, the required age of Rahul = 5 × 4 = 20 years.

and the required age of Haroon = 7 × 4 = 28 years.

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 numbers of girls. What is the total class strength?

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Let the number of boys be 7x

and the number of girls be 5x

As per the conditions given, we get

7x – 5x = 8

\(\Rightarrow\) 2x = 8

\(\Rightarrow\)x = 8 ÷ 2 = 4 (transposing 2 to RHS)

Hence we got the required number of boys = 7× 4 = 28

and the number of girls = 5 × 4 = 20

Hence, total class strength = 28 + 20 = 48

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?

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Let the age of Baichung be x years.

As given,the age of his father = x + 29 years,

and the age of his grandfather = x + 29 + 26 = (x + 55) years.

As per the conditions given, we get

x + x + 29 + x + 55 = 135

\(\Rightarrow\) 3x + 84 = 135

\(\Rightarrow\) 3x = 135 – 84 (transposing 84 to RHS)

\(\Rightarrow\) 3x = 51

\(\Rightarrow\) x = 51 ÷ 3 (transposing 3 to RHS)

\(\Rightarrow\)x = 17

Hence Baichung’s age = 17 years

Baichung’s father’s age = 17 + 29 = 46 years,

and grand father’s age = 46 + 26 = 72 years.

12.Fifteen years from now Ravi’s age will be four times his present age. What is Ravi’s present age?

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Let the present age of Ravi be x years.

After 15 years, his age will be = (x + 15) years

As per the conditions given, we have

\(\Rightarrow\) x + 15 = 4x

\(\Rightarrow\) 15 = 4x – x (transposing x to RHS)

\(\Rightarrow\) 15 = 3x

\(\Rightarrow\)15 ÷ 3 = x (transposing 3 to LHS)

\(\Rightarrow\) x = 5

Hence, the present age of Ravi = 5 years.

13.A rational number is such that when you multiply it by \(\frac { 5 }{ 2 }\) and add \(\frac { 2 }{ 3 }\) to the product, you get \(\frac { -7 }{ 12 }\). What is the number?

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Let the number be x

As per given, we have

\(x× \frac{5}2 +\frac{2}3 = \frac{-7}{12}\)

\(\Rightarrow \frac{5x}2 = \frac{-7}{12}-\frac{2}3 \quad\)[ subtracting \(\frac{2}3\) from both the sides]

\(\Rightarrow \frac{5x}2 = \frac{-7-8}{12}\)

\(\Rightarrow \frac{5x}2 = \frac{-15}{12}\)

\(\Rightarrow x = \frac{-15}{12}÷\frac{5}2\quad \)[transposing \(\frac{5}2\) to RHS]

\(\Rightarrow X= \frac{-30}{60}=\frac{-1}2\)

Hence, the required rational number is \(\frac{-1}2\)

14.Lakshmi is a cashier in a bank. She has currency notes of denominations ₹ 100, ₹ 50 and ₹ 10, respectively. The ratio of these notes is 2 : 3 : 5. The total cash with Lakshmi is ₹ 4,00,000. How many notes of each denomination does she have?

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Let the number of ₹ 100, ₹ 50 and ₹10 notes be 2x, 3x and 5x respectively as per the given ratio.

Converting all the denominations into rupees, we have

2x ×100, 3x × 50 and 5x × 10 i.e. 200x, 150x and 50x

As per the conditions given in the question, we have

200x + 150x + 50x = 4,00,000

\(\Rightarrow\) 400x = 4,00,000

\(\Rightarrow\) x = 4,00,000 ÷ 400 (transposing 400 to RHS)

\(\Rightarrow\) x = 1,000

Hence, the required number of notes of

₹ 100 notes = 2×1000 = 2000

₹ 50 notes = 3 × 1000 = 3000

and ₹ 10 notes = 5 ×1000 = 5000

15.I have a total of ₹300 in coins of denomination ₹1, ₹2 and ₹5. The number of ₹2 coins is 3 times the number of ₹5 coins. The total number of coins is 160. How many coins of each denomination are with me?

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As per the ratio given,

Let the number of ₹5 coins be x.

Number of ₹2 coins = 3x

Total number of coins = 160

Number of ₹1 coin = 160 – (x + 3x) = 160 – 4x

Converting the number of coins into rupees, we have

x coins of ₹5 amount = ₹5x

3x coins of ₹2 amount = ₹ 3x × 2 = ₹6x

and (160 – 4x) coins of ₹ 1 amount =₹1 × (160 – 4x) =₹(160 – 4x)

As per the conditions given, we got

5x + 6x + 160 – 4x = 300

\(\Rightarrow\) 7x + 160 = 300

\(\Rightarrow\) 7x = 300 – 160 (transposing 160 to RHS)

\(\Rightarrow\) 7x = 140

\(\Rightarrow\) x = 140 ÷ 7 (transposing 7 to RHS)

\(\Rightarrow\) x = 20

Thus, number of ? 5 coins = 20

Number of ₹2 coins = 3 × 20 = 60

and Number of ₹ 1 coins = 160 – 4 × 20 = 160 – 80 = 80

16.The organizers of an essay competition decide that a winner in the competition gets a prize of ₹100 and a participant who does not win gets a prize of ₹25. The total prize money distributed is ₹3,000. Find the number of winners, if the total number of participants is 63.

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Let the number of winners = x

Number of participants who do not win the prize = 63 – x

Amount won by winners = ₹ 100 × x = ₹ 100x

Amount won by losers = ₹ (63 – x) × 25 = ₹ (1575 – 25x)

As per the conditions given, we get

100x + 1575 – 25x = 3000

\(\Rightarrow\) 75x + 1575 = 3000

\(\Rightarrow\) 75x = 3000 – 1575 (transposing 1575 to RHS)

\(\Rightarrow\) 75x = 1425

\(\Rightarrow\) x = 1425 ÷ 75 (Transposing 75 to RHS)

\(\Rightarrow\) x = 19

Thus, the number of winners = 19

1. 3x = 2x + 18

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Given, 3x = 2x + 18

\(\Rightarrow\) 3x – 2x = 18 (Transposing 2x to LHS)

\(\Rightarrow\)x = 18

Hence, x = 18 is the required solution.

**Checking:**

3x = 2x + 18

Substitute x = 18 in the given equation, we have

LHS = 3 × 18 = 54

RHS = 2 × 18 + 18 = 36 + 18 = 54

LHS = RHS

Hence x=18 is the required solution.

2. 5t – 3 = 3t – 5

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Given 5t – 3 = 3t – 5

\(\Rightarrow\) 5t – 3t – 3 = -5 (Transposing 3t to LHS)

\(\Rightarrow\)2t = -5 + 3 (Transposing -3 to RHS)

\(\Rightarrow\)2t = -2

\(\Rightarrow\) t = -2 ÷ 2

\(\Rightarrow\) t = -1

Hence, t = -1 is the required solution.

**Checking: **

5t – 3 = 3t – 5

Substitute t = -1 in the given equation, we have

LHS = 5t – 3 = 5 × (-1)-3 = -5 – 3 = -8

RHS = 3t – 5 = 3 × (-1) – 5 = -3 – 5 = -8

LHS = RHS

Hence, t = -1 is the required solution.

3. 5x + 9 = 5 + 3x

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Given that 5x + 9 = 5 + 3x

\(\Rightarrow\) 5x – 3x + 9 = 5 (Transposing 3x to LHS) => 2x + 9 = 5

\(\Rightarrow\) 2x = 5 – 9 (Transposing 9 to RHS)

\(\Rightarrow\) 2x = -4

\(\Rightarrow\) x = -4 ÷ 2 = -2

Hence x = -2 is the required solution.

**Checking: **

5x + 9 = 5 + 3x

Substituting x = -2 in the given equation, we have

LHS = 5 × (-2) + 9 = -10 + 9 = -1

RHS = 5 + 3 × (-2) = 5 – 6 = -1

LHS = RHS

Hence x = -2 is the required solution.

4. 4z + 3 = 6 + 2z

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Given, 4z + 3 = 6 + 2z

\(\Rightarrow\) 4z – 2z + 3 = 6 (Transposing 2z to LHS)

\(\Rightarrow\)2z + 3 = 6

\(\Rightarrow\)2z = 6 – 3 (Transposing 3 to RHS)

\(\Rightarrow\)2z = 3

\(\Rightarrow\) z = \(\frac { 3 }{ 2 }\)

Hence z = \(\frac { 3 }{ 2 }\) is the required solution.

**Checking: **

4z + 3 = 6 + 2z

Substituting z = \(\frac { 3 }{ 2 }\) in the given equation, we have

LHS = 4z + 3 = 4 × \(\frac { 3 }{ 2 }\) + 3 = 6 + 3 = 9

RHS = 6 + 2z = 6 + 2 × \(\frac { 3 }{ 2 }\) = 6 + 3 = 9

LHS = RHS

Hence z = \(\frac { 3 }{ 2 }\) is the required solution.

5. 2x – 1 = 14 – x

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Given 2x – 1 = 14 – x

\(\Rightarrow\) 2x + x = 14 + 1 (Transposing x to LHS and 1 to RHS)

\(\Rightarrow\) 3x = 15

\(\Rightarrow\)x = 15 ÷ 3 = 5

Hence x = 5 is the required solution.

**Checking:**

2x – 1 = 14 – x

Substituting x = 5 in the given equation,we have,

LHS= 2x – 1 = 2 × 5 – 1 = 10 – 1 = 9

RHS = 14 – x = 14 – 5 = 9

LHS = RHS

Hence x = 5 is the required solution.

6. 8x + 4 = 3(x – 1) + 7

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Given, 8x + 4 = 3(x – 1) + 7

\(\Rightarrow\) 8x + 4 = 3x – 3 + 7 (Solving the bracket)

\(\Rightarrow\)8x + 4 = 3x + 4

\(\Rightarrow\) 8x – 3x = 4 – 4 [Transposing 3x to LHS and 4 to RHS]

\(\Rightarrow\) 5x = 0

\(\Rightarrow\) x = 0 ÷ 5 [Transposing 5 to RHS]

or x = 0

Hence, x = 0 is the required solution.

**Checking: **

8x + 4 = 3(x – 1) + 7

Substituting x = 0 in the given equation, we have

8 × 0 + 4 = 3(0 – 1) + 7

\(\Rightarrow\) 0 + 4 = -3 + 7

\(\Rightarrow\) 4 = 4

LHS = RHS

Hence, x = 0 is the required solution.

7. x =\( \frac { 4 }{ 5 } (x + 10)\)

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Given x = \(\frac { 4 }{ 5 } (x + 10)\)

\(\Rightarrow\) 5 × x = 4(x + 10) (Transposing 5 to LHS)

\(\Rightarrow\) 5x = 4x + 40 (Solving the bracket)

\(\Rightarrow\) 5x – 4x = 40 (Transposing 4x to LHS)

\(\Rightarrow\) x = 40

Thus x = 40 is the required solution.

**Checking:**

x = \(\frac { 4 }{ 5 }\) (x + 10)

Substituting x = 40 in the given equation, we have

40 = \(\frac { 4 }{ 5 } \)(40 + 10)

\(\Rightarrow\) 40 =\( \frac { 4 }{ 5 }\) × 50

\(\Rightarrow\) 40 = 4× 10

\(\Rightarrow\) 40 = 40

LHS = RHS

Hence x = 40 is the required solution.

8.\(\frac { 2x }{ 3 } + 1 = \frac { 7x }{ 15 } + 3\)

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Given that:\(\frac { 2x }{ 3 } + 1 = \frac { 7x }{ 15 } + 3\)

15\((\frac { 2x }{ 3 } + 1)\) = 15\((\frac { 7x }{ 15 } + 3)\quad \)[multiplying both the sides by 15]

\(\Rightarrow \frac { 2x }{ 3 } × 15 + 1 ×15 = \frac { 7x }{ 15 } ×15 + 3 ×15 \)

\(\Rightarrow \) 2x × 5 + 15 = 7x + 45

\(\Rightarrow \) 10x + 15 = 7x + 45

\(\Rightarrow \) 10x – 7x = 45 – 15 [Transposing 7x to LHS and 15 to RHS]

\(\Rightarrow \)3x = 30

\(\Rightarrow \) x = 30 ÷ 3 = 10 [Transposing 3 to RHS]

Thus, the required solution is x = 10.

**Checking**

Substitute x=10 in the given equation, we have

\(\frac { 2×10 }{ 3 } + 1 = \frac { 7×10 }{ 15 } + 3\)

\(\Rightarrow \frac { 20 }{ 3 } + 1 = \frac { 70 }{ 15 } + 3\)

\(\Rightarrow \frac { 20 + 3}{ 3 } = \frac { 70 + 45}{ 15 } \)

\(\Rightarrow \frac { 23}{ 3 } = \frac { 115}{ 15 } \)

\(\Rightarrow \) LHS= RHS

Hence,the required solution is x = 10.

9. \(2y + \frac { 5 }{ 3 } =\frac { 26 }{ 3 } – y\).

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Given that, \(2y + \frac { 5 }{ 3 } = \frac { 26 }{ 3 } – y\)

\(\Rightarrow 2y + y = \frac { 26 }{ 3 } – \frac { 5 }{ 3 }\quad \)[Transposing -y to LHS and \(\frac{5}3\) to RHS]

\(\Rightarrow 3y = \frac { 21 }{ 3 }\)

\(\Rightarrow 3y =7\)

\(\Rightarrow \frac{3y}{3} = \frac { 7 }{ 3 }\quad \)[dividing both the sides by 3]

\(\Rightarrow y = \frac {7 }{ 3 }\)

Hence, \(y=\frac {7 }{ 3 }\) is the required value

**Checking **

Substitute \(y=\frac {7 }{ 3 }\) in the given equation, we have

LHS=\(2×\frac{7}3+\frac{5}3=\frac{14}3+\frac{5}3=\frac{19}3\)

RHS=\(\frac{26}3-\frac{7}3=\frac{19}3\)

LHS=RHS

Hence,\(y=\frac {7 }{ 3 }\) is the required value.

10.\(3m=5m-\frac { 8 }{ 5 } \)

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Given that, \(3m=5m-\frac { 8 }{ 5 } \)

\(\Rightarrow 3m-5m=-\frac {8 }{ 5 } \quad \)[Transposing 5m to LHS]

\(\Rightarrow -2m=-\frac {8 }{ 5 } \)

\(\Rightarrow \frac{-2m}{-2}=\frac{-\frac {8 }{ 5 }}{-2}\quad \)[Dividing both the sides by -2]

\(\Rightarrow m=\frac {4}{ 5 }\)

Thus,\( m=\frac {4}{ 5 }\) is the required solution

**Checking: **

Substituting \(m=\frac {4}{ 5 }\) in the given equation, we have

LHS=\(3×\frac{4}5 =\frac{12}5\)

RHS=\(4×\frac{4}5 - \frac{8}5=\frac{20}5-\frac{8}5 =\frac{12}5\)

RHS=LHS

Hence,\( m=\frac {4}{ 5 }\) is the required solution

1. Amina thinks of a number and subtracts \(\frac { 5 }{ 2 }\) 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?

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Let the number be x.

Condition I: x – \(\frac { 5 }{ 2 }\)

Condition II: 8 × (x – \(\frac { 5 }{ 2 }\))

Condition III: 8 × (x – \(\frac { 5 }{ 2 }\)) = 3x

\(\Rightarrow\) 8x – \(\frac { 5 }{ 2 }\) × 8 = 3x [Solving the bracket]

\(\Rightarrow\)8x – 20 = 3x

\(\Rightarrow\) 8x – 3x = 20 [Transposing 3x to LHS and 20 to RHS]

\(\Rightarrow\) 5x = 20

\(\Rightarrow\) x = 20 ÷ 5 = 4 [Transposing 5 to RHS]

Hence, x = 4 is the required number.

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 number. What are the numbers?

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Let the number be x.

And the other number be y

As given y= 5x and also,

Condition I: x + 21 and 5x + 21

Let the number y be twice of x

Condition II: 5x + 21 = 2 (x + 21)

\(\Rightarrow\) 5x + 21 = 2x + 42 (Solving the bracket)

\(\Rightarrow\) 5x – 2x = 42 – 21 (Transposing 2x to LHS and 21 to RHS)

\(\Rightarrow\) 3x = 21

\(\Rightarrow\) x = 21 ÷ 3 = 7 (Transposing 3 to RHS)

Hence, the required numbers are 7 and 7 × 5 = 35.

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?

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Let unit place digit be x.

Ten’s place digit y= 9 – x[Sum of both digits: x+y=9]

Original number = x + 10(9 – x)

Condition I: 10x + (9 – x) (Interchanging the digits)

Condition II: New number = original number + 27

\(\Rightarrow\) 10x + (9 – x) = x + 10(9 – x) + 27

\(\Rightarrow\) 10x + 9 – x = x + 90 – 10x + 27 (solving the brackets)

\(\Rightarrow\) 9x + 9 = -9x + 117 (Transposing 9x to LHS and 9 to RHS)

\(\Rightarrow\) 9x + 9x = 117 – 9

\(\Rightarrow\) 18x = 108

\(\Rightarrow\) x = 108 ÷ 18 (Transposing 18 to RHS)

\(\Rightarrow\) x = 6

Hence we have,

Unit place digit = 6

Tens place digit = 9 – 6 = 3

Thus, the required number = 6 + 3 × 10 = 6 + 30 = 36

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?

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Let unit place digit be x.

And the ten’s place digit be y= 3x

Original number = x + 3x × 10 = x + 30x = 31x

Condition I: 10x + 3x = 13x [interchanging the digits]

Condition II: New number + original number = 88

13x + 31x = 88

\(\Rightarrow\) 44x = 88

\(\Rightarrow\) x = 88 ÷ 44 [Transposing 44 to RHS]

\(\Rightarrow\) x = 2

Thus, we have,

The original number = 31x = 31 × 2 = 62

Hence the required number = 62

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?

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Let unit place digit be x.

And the ten’s place digit be y= 3x

Original number = x + 3x × 10 = x + 30x = 31x

Condition I: 10x + 3x = 13x [interchanging the digits]

Condition II: New number + original number = 88

13x + 31x = 88

\(\Rightarrow\) 44x = 88

\(\Rightarrow\) x = 88 ÷ 44 [Transposing 44 to RHS]

\(\Rightarrow\) x = 2

Thus, we have,

The original number = 31x = 31 × 2 = 62

Hence the required number = 62

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?

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Let Shobo’s present age be x years.

And Shobo’s mother’s age = 6x years.

After 5 years Shobo’s age=(x + 5) years.

As per the condition given, we have

x + 5 =\( \frac { 1 }{ 3 }\) × 6x

\(\Rightarrow\) x + 5 = 2x

\(\Rightarrow\) 5 = 2x – x (Transposing x to RHS)

\(\Rightarrow\) 5 = x

Hence, Shobo’s present age = 5 years

and Shobo’s mother’s present age 6x = 6 × 5 = 30 years.

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 of ₹ 100 per metre, it will cost the village panchayat ₹ 75000 to fence the plot. What are the dimensions of the plot?

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Let the length and breadth of the plot be 11x m and 4x m respectively.

Fencing all around = perimeter of the rectangular plot

Perimeter of the plot = \(\frac { 75000 }{ 100 }\) = 750 m

2(length + breadth) = 750

\(\Rightarrow\)2(11x + 4x) = 750

\(\Rightarrow\)2(15x) = 750

\(\Rightarrow\) 30x = 750

\(\Rightarrow\) x = 750 ÷ 30 = 25

Hence we have,

length =11x= 11 × 25 m = 275 m

and breadth = 4x=4 × 25 m = 100 m

7.Hasan buys two kinds of cloth materials for school uniforms, shirt material that costs him ₹ 50 per metre and trousers material that costs him ₹ 90 per metre. For every 3 metres of the shirt material, he buys 2 metres of the trouser material. He sells the materials at 12% and 10% profit respectively. His total sale is ₹ 36,600. How much trouser material did he buy?

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Given that for every 3 metres of the shirt material, he buys 2 metres so the ratio of shirt material brought to the trouser material bought = 3 : 2

Let the shirt material bought = 3x m

and trouser material bought = 2x m

Cost of shirt material = 50 × 3x = ₹ 150x

Cost of trouser material = 90 × 2x = ₹ 180x

Also, profit in case of shirt material=12%

∴ It's selling price=\(₹(150x×\frac{112}{100}\))=₹168x

Also, profit in case of trouser material=10%

∴ It's selling price=\(₹(180x×\frac{110}{100}\))=₹198x

∴ Total selling price=₹(168x+198x)=₹366x

But we have the total selling price as ₹36660

∴ 366x=36660

\(\Rightarrow\ x=\frac{36660}{366}=100.16\)

Thus, the trouser material purchased =2x=2×100.16m≈200 meters.

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.

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Let the number of deer = x.

As per the condition given, we have

\(\frac { x }{ 2 }\) deer are grazing in the field.

Remaining deers=\(x-\frac{x}2=\frac{x}2\)

∴ we have \(\frac{x}2×\frac{3}4=\frac{3x}8 \) deers playing near by.

Number of the deers drinking water=9

∴ we have, \(\frac{x}2+\frac{4x}2+9=x\)

\(\Rightarrow 8×\frac{x}2+8×\frac{3x}8+8×9=8×x\)

\(\Rightarrow 4x+3x+72=8x\)

\(\Rightarrow 7x-8x=-72\)

\(\Rightarrow -x=-72\)

\(\Rightarrow x=72\)

Hence the number of deer in the herd is 72.

9. A grandfather is ten times older than his granddaughter. He is also 54 years older than her. Find their present ages.

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Let the present age of granddaughter be x years.

and the present age of grandfather = 10x years.

As per the conditions, we have

10x – x = 54

\(\Rightarrow\) 9x = 54

\(\Rightarrow\) x = 54 ÷ 9 = 6 [Transposing 9 to RHS]

Hence,we have

the present age of the granddaughter = 6 years

and the present age of grandfather = 6 × 10 = 60 years.

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.

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Let the present age of the son be x years.

Present age of Aman = 3x years

Also,10 years ago, the son’s age was = (x – 10) years

10 years ago, the father’s age was = (3x – 10) years

As per the conditions, we have

5(x – 10) = 3x – 10

\(\Rightarrow\)5x – 50 = 3x – 10

\(\Rightarrow\) 5x – 3x = 50 – 10[Transposing 3x to LHS and 50 to RHS]

\(\Rightarrow\) 2x = 40

\(\Rightarrow\)x = 40 ÷ 2 = 20

Hence,we have,

The son’s age = 20 years.

and the age of Aman = 20 × 3 = 60 years.

1.\(\frac{x}2-\frac{1}5=\frac{x}3+\frac{1}4\)

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We have, \(\frac{x}2-\frac{1}5=\frac{x}3+\frac{1}4\)

LCM of denominators 2,5,3 and 4 is 60

So we have \(\frac{x}2×60-\frac{1}5×60=\frac{x}3×60+\frac{1}4×60\quad \)[Multiplying both the sides by 60 ]

\(\Rightarrow 30x -12 =20x +15\)

\(\Rightarrow 30x -20x =15+12 \)[Transposing 20x to LHS and 12 to RHS]

\(\Rightarrow 10x =27\)

\(\Rightarrow x =\frac{27}{10}\)[Dividing by 10 on both the sides]

2.\(\frac{n}2-\frac{3n}4+\frac{5n}6=21\)

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Given that,\(\frac{n}2-\frac{3n}4+\frac{5n}6=21\)

LCM of denominators 2,4 and 6=12

So, \(\frac{n}2×12-\frac{3n}4×12+\frac{5n}6×12=21×12\quad \)[Multiplying both the sides by 12]

\(\Rightarrow 6n-9n +10n = 252\)

\(\Rightarrow 7n = 252\)

\(\Rightarrow n = 252÷7\quad \)[Dividing both the sides by 7]

\(\Rightarrow n = 36\)

3.\(x+7-\frac{8x}3=\frac{17}6-\frac{5x}2\)

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Given that,\(x+7-\frac{8x}3=\frac{17}6-\frac{5x}2\)
LCM of denominators 3,6 and 2 =6

∴\(6×x+7-6×\frac{8x}3=6×\frac{17}6-6×\frac{5x}2\quad \)[Multiplying both the sides by 6]

\(\Rightarrow 6x+42-16x=17-15x\)

\(\Rightarrow -10x + 42 = 17 – 15x \)

\(\Rightarrow-10x + 15x = 17 – 42 \)[Transposing 15x to LHS and 42 to RHS]

\(\Rightarrow 5x = -25 \)

\(\Rightarrow x = -25 ÷ 5 \) [Transposing 5 to RHS]

\(\Rightarrow x = -5 \)

4.\(\frac{x-5}3=\frac{x-3}5\)

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Given that,\(\frac{x-5}3=\frac{x-3}5\)

LCM of denominators 3 and 5=15

So, \(\frac{x-5}3×15=\frac{x-3}5×15\quad \)[Multiplying both the sides by 15]

\(\Rightarrow (x-5)×5=(x-3)×3\)

\(\Rightarrow 5x-25 = 3x-9\quad \)[Solving the brackets]

\(\Rightarrow 5x-3x = 25-9\quad \)[Transposing 3x to LHS and 25 to RHS]

\(\Rightarrow 2x = 16\)

\(\Rightarrow x = 16÷2\)[Dividing both the sides by 2]

\(\Rightarrow x = 8\)

5.\(\frac{3t-2}4-\frac{2t+3}3=\frac{2}3-t\)

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Given that,\(\frac{3t-2}4-\frac{2t+3}3=\frac{2}3-t\)

LCM of denominators 3 and 4=12

So, \(\frac{3t-2}4×12-\frac{2t+3}3×12=\frac{2}3×12-t×12 \quad \)[Multiplying both the sides by 12]

\(\Rightarrow (3t – 2) × 3 – (2t + 3) × 4 = 2 × 4 – 12t\)

\(\Rightarrow 9t – 6 – 8t – 12 = 8 – 12t \quad \)[Solving the brackets]

\(\Rightarrow t – 18 = 8 – 12t \quad \)

\(\Rightarrow t + 12t = 8 + 18 \)[Transposing 12t to LHS and 18 to RHS]

\(\Rightarrow 13t = 26\)

\(\Rightarrow t = 2 \)[Transposing 13 to RHS]

Hence t = 2 is the required solution.

6.\(m-\frac{m-1}2=1-\frac{m-2}3\)

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6.\(m-\frac{m-1}2=1-\frac{m-2}3\)
Given that,\(m-\frac{m-1}2=1-\frac{m-2}3\)

LCM of denominators 3 and 4=6

So, \(m×6-\frac{m-1}2×6=1×6-\frac{m-2}3×6 \quad \)[Multiplying both the sides by 6]

\(\Rightarrow 6m – (m – 1) × 3 = 6 – (m – 2) × 2\)

\(\Rightarrow 6m – 3m + 3 = 6 – 2m + 4 \quad \)[Solving the brackets]

\(\Rightarrow 3m + 3 = 10 – 2m \)

\(\Rightarrow 3m + 2m = 10 – 3 \)[Transposing 2m to LHS and 3 to RHS]

\(\Rightarrow 5m = 7\)

\(\Rightarrow m = \frac { 7 }{ 5 } \)[Transposing 5 to RHS]

Hence \(m = \frac { 7 }{ 5 } \) is the required solution.

Given that,

3(t – 3) = 5(2t + 1)

\(\Rightarrow\) 3t – 9 = 10t + 5 [Solving the brackets]

\(\Rightarrow\) 3t – 10t = 9 + 5 [Transposing 10t to LHS and 9 to RHS]

\(\Rightarrow\) -7t = 14

\(\Rightarrow\) t = -2[Transposing -7 to RHS]

Thus, t = -2 is the required solution.

8. 15(y – 4) – 2(y – 9) + 5(y + 6) = 0

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Given that, 15(y – 4) – 2(y – 9) + 5(y + 6) = 0

\(\Rightarrow\)15y – 60 – 2y + 18 + 5y + 30 = 0 [Solving the brackets]

\(\Rightarrow\) 8y – 12 = 0

\(\Rightarrow\) 8y = 12 [Transposing 12 to RHS]

\(\Rightarrow\) y =\( \frac { 2 }{ 3 }\)

Hence, y = \(\frac { 2 }{ 3 }\) is the required solution.

9. 3(5z – 7) – 2(9z – 11) = 4(8z – 13) – 17

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Given that, 3(5z – 7) – 2(9z – 11) = 4(8z – 13) – 17

\(\Rightarrow\)15z – 21 – 18z + 22 = 32z – 52 – 17 [Solving the bracket]

\(\Rightarrow\) -3z + 1 = 32z – 69

\(\Rightarrow\) -3z – 32z = – 69 – 1 [Transposing 32z to LHS and 1 to RHS]

\(\Rightarrow\) -35z = -70

\(\Rightarrow\) z = 2

Thus, z = 2 is the required solution.

10. 0.25(4f – 3) = 0.05(10f – 9)

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Given that, 0.25(4f – 3) = 0.05(10f – 9)

\(\Rightarrow\) 0.25 × 4f – 3 × 0.25 = 0.05 × 10f – 9 × 0.05 [Solving the brackets]

\(\Rightarrow\) 1.00f – 0.75 = 0.5f – 0.45

\(\Rightarrow\) f – 0.5f = -0.45 + 0.75 [Transposing 0.5 to LHS and 0.75 to RHS]

\(\Rightarrow\) 0.5f = 0.30

\(\Rightarrow\) f = 0.6[Dividing both the sides by 0.5]

Hence, f = 0.6 is the required solution

1. \(\frac { 8x-3 }{ 3x } =2\)

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Given that, \(\frac { 8x-3 }{ 3x } =2\)

\(\Rightarrow \frac { 8x-3 }{ 3x } = \frac { 2 }{ 1 }\)

\(\Rightarrow 8x – 3 = 2 × 3x \quad \)[Cross-multiplication]

\(\Rightarrow 8x – 3 = 6x\)

\(\Rightarrow 8x – 6x = 3\) [Transposing 6x to LHS and 3 to RHS]

\(\Rightarrow 2x = 3\quad \)[Dividing both the sides by 2]

\(\Rightarrow x = \frac { 3 }{ 2 }\)

2. \(\frac { 9x }{ 7-6x } = 15\)

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Given that, \(\frac { 9x }{ 7-6x } = 15\)

\(\Rightarrow \frac { 9x }{ 7-6x } = \frac { 15 }{ 1 }\)

\(\Rightarrow 9x = 15(7 – 6x)\quad \) [Cross-multiplication]

\(\Rightarrow 9x = 105 – 90x\quad \) [Solving the bracket]

\(\Rightarrow 9x + 90x = 105 \quad \)[Transposing 90x to LHS]

\(\Rightarrow 99x = 105\)

\(\Rightarrow x = \frac { 105 }{ 99 }\)

\(\Rightarrow x = \frac { 35 }{ 33 }\)

3.\(\frac { z }{ z+15 } =\frac { 4 }{ 9 }\)

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Given that, \(\frac { z }{ z+15 } =\frac { 4 }{ 9 }\)

\(\Rightarrow 9z = 4 (z + 15)\quad \) [Cross-multiplication]

\(\Rightarrow 9z = 4z + 60 \quad \)[Solving the bracket]

\(\Rightarrow 9z – 42 = 60 \)

\(\Rightarrow 5z = 60\)

\(\Rightarrow z = 12\)

4. \(\frac { 3y+4 }{ 2-6y } =\frac { -2 }{ 5 }\)

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Given that, \(\frac { 3y+4 }{ 2-6y } =\frac { -2 }{ 5 }\)

\(\Rightarrow 5(3y + 4) = -2(2 – 6y)\quad \) [Cross-multiplication]

\(\Rightarrow 15y + 20 = -4 + 12y \quad \)[Solving the bracket]

\(\Rightarrow 15y – 12y = -4 – 20\quad \) [Transposing 12y to LHS and 20 to RHS]

\(\Rightarrow 3y = -24\quad \) [Transposing 3 to RHS]

\(\Rightarrow y = -8\)

5. \(\frac { 7y+4 }{ y+2 } =\frac { -4 }{ 3 }\)

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Given that, \(\frac { 7y+4 }{ y+2 } =\frac { -4 }{ 3 }\)

\(\Rightarrow 3(7y + 4) = -4 (y + 2)\quad \) [Cross-multiplication]

\(\Rightarrow 21y + 12 = -4y – 8\quad \) [Solving the bracket]

\(\Rightarrow 21y + 4y = -12 – 8\quad \) [Transposing 4y to LHS and 12 to RHS]

\(\Rightarrow 25y = -20\quad\) [Transposing 25 to RHS]

\(\Rightarrow y = \frac { -4 }{ 5 }\)

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.

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Let the present ages of Hari=5x and Harry=7x years.

After 4 years Hari’s age will be (5x + 4) years and Harry’s age will be (7x + 4) years.

As per the conditions given, we have

\(\frac { 5x+4 }{ 7x+4 } =\frac { 3 }{ 4 }\)

\(\Rightarrow 4(5x + 4) = 3(7x + 4)\quad\) [Cross-multiplication]

\(\Rightarrow 20x + 16 = 21x + 12\quad \) [Solving the bracket]

\(\Rightarrow 20x – 21x = 12 – 16 \quad \)[Transposing 21x to LHS and 16 to RHS]

\(\Rightarrow -x = -4\)

\(\Rightarrow x = 4\)

Hence, the present ages of Hari and Harry are 5 × 4 = 20years and 7 × 4 = 28years respectively.

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 \frac { 3 }{ 2 }. Find the rational number.

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Let the numerator be x.

and denominator be y

As given y= (x + 8)

Also as per the conditions given, we have

\(\frac{x+17}{x+8-1}=\frac{3}2\)

\( \Rightarrow \frac{x+17}{x+7}=\frac{3}2\)

\(\Rightarrow 2(x + 17) = 3(x + 7) \quad\)[Cross-multiplication]

\(\Rightarrow 2x + 34 = 3x + 21\quad \)[Solving the bracket]

\(\Rightarrow 2x – 3x = 21 – 34 \quad \)[Transposing 3x to LHS and 34 to RHS]

\(\Rightarrow -x = -13\)

\(\Rightarrow x = 13\)

Hence, numerator = 13

and denominator = 13 + 8 = 21

Thus, the rational number is \(\frac { 13 }{ 21 }\).