3 Mark Question

Question 13 Marks

Solve the following equation by trial and error method:
2x - 3 = 9

Answer

The given equation is 2x - 3 = 9
We guess and try several values of x to find L.H.S. and R.H.S. and stop when L.H.S. = R.H.S.

x	L.H.S.	R.H.S.
3	2 × 3 - 3 = 3	9
4	2 × 4 - 3 = 5	9
5	2 × 5 - 3 = 7	9
6	2 × 6 - 3 = 9	9

When x = 6, we have L.H.S. = R.H.S
So, x = 6 is the solution of the given equation.

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Question 23 Marks

Solve: $\frac{\text{2x}}{5}-\frac{\text{x}}{2}=\frac{5}{2}$

Answer

$\frac{\text{2x}}{5}-\frac{\text{x}}{2}=\frac{5}{2}$
Or, $\frac{\text{4x-5x}}{10}=\frac{5}{2}$ [Taking L.C.M. as 10]
$\Rightarrow\frac{\text{x}}{10}=\frac{5}{2}$
$\Rightarrow\frac{\text{x}}{10}\times(-10)=\frac{5}{2}\times(-10)$ [Multiplying both the sides by (-10)]
$\text{x}=-25$

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Question 33 Marks

Solve the following equation and verify the answer:
x - 7 = 6

Answer

x - 7 = 6
Adding 7 on both the sides:
⇒ x - 7 + 7 = 6 + 7
⇒ x = 13
Verification:
Substituting x = 13 in the L.H.S.
⇒ 13 - 7 = 6 = R.H.S.
L.H.S. = R.H.S.
Hence, verified.

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Question 43 Marks

When Raju multiplies a certain number by 17 and adds 4 to the product, he gets 225. Find that number.

Answer

Let the required number is xThen, 17x + 4 = 225
Or, 17x = 225 - 4
$\Rightarrow\text{x}=\frac{221}{17}$
x = 13
So, the number is 13

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Question 53 Marks

Solve the following equation and verify the answer:
x + 5 = 12

Answer

x + 5 = 12
Subtracting 5 from both the sides:
⇒ x + 5 - 5 = 12 - 5
⇒ x = 7
Verification:
Substituting x = 7 in the L.H.S.
⇒ 7 + 5 = 12 = R.H.S.
L.H.S. = R.H.S.
Hence, verified.

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Question 63 Marks

If a number is tripled and the result is increased by 5, we get 50. Find the number.

Answer

Let the required numbers is xThen, 3x + 5 = 50
Or, 3x = 50 - 5
$\Rightarrow\text{x}=\frac{45}{3}$
x = 15
So, the number is 15

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Question 73 Marks

Solve the following equation and verify the answer:
4x + 7 = 15

Answer

4x + 7 = 15
⇒ 4x + 7 - 7 = 15 - 7 [Subtracting 7 from both the sides]
⇒ 4x = 8
$\Rightarrow\frac{\text{4x}}{4}=\frac{8}{4}$ [Dividing both the sides by 4]
⇒ x = 2
Verification:
Substituting x = 2 in the L.H.S.
⇒ 4 × 2 + 7 = 8 + 7 = 15 = R.H.S.
L.H.S. = R.H.S.
Hence, verified.

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Question 83 Marks

Solve the following equation by trial and error method:
x - 7 = 10

Answer

The given equation is x - 1 = 10
We guess and try several values of x to find L.H.S. and R.H.S. and stop when L.H.S. = R.H.S.

x	L.H.S.	R.H.S.
14	14 - 7 = 7	10
15	15 - 7 = 8	10
16	16 - 7 = 9	10
17	17 - 7 = 10	10

When x = 17, we have L.H.S. = R.H.S
So x = 17 is the solution of the given equation.

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Question 93 Marks

The length of a rectangular park is thrice its breadth. If the perimeter of the park is 168 metres, fund its dimensions.

Answer

Let the Breadth of the park is x m. and length is 3x m. Then, 2(x + 3x) = 168 Or, 2x + 6x = 168 ⇒ 8x = 168 $\Rightarrow\text{x}=\frac{168}{8}$x = 21
So, the breadth is 21m and length is (3 × 21) = 63m

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Question 103 Marks

The sum of three consecutive natural numbers is 51. Find the numbers.

Answer

Let the three consecutive natural numbers be x, (x + 1) and (x + 2)
$\therefore$ x + (x + 1) + (x + 2) = 51
⇒ 3x + 3 = 51
⇒ 3x + 3 - 3 = 51 - 3 [Subtracting 3 from both the sides]
⇒ 3x = 48
$\Rightarrow\frac{\text{3x}}{3}=\frac{48}{3}$ [Dividing both the sides by 3]
x = 16
Thus, the three natural numbers are x = 16, x + 1 = 17 and x + 2 = 18

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Question 113 Marks

A man is thrice as old as his son. Five years ago the man was four times as old as his son. Find their present ages.

Answer

Let the age of son is x and and age of the man is 3x
Before 5 years their age are (x - 5) and (3x - 5)
Then, 3x - 5 = 4(x - 5)
Or, 3x - 5 = 4x - 20
⇒ 3x - 4x = -20 + 5
⇒ -x = -15
x = 15
So, present age of the son is 15 years and present age of the man is (3 × 15) = 45 years

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Question 123 Marks

The number of boys in a school is 334 more than the number of girls. If the total strength of the school is 572, find the number of girls in the school.

Answer

Let the number of girls is x Then, x + 334 + x = 572 Or, 2x = 572 - 334 $\Rightarrow\text{x}=\frac{238}{2}$x = 119
So, the number of girls is 119

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Question 133 Marks

Lalit earns Rs x per day and spends Rs y per day. How much does he save in 30 days?

Answer

Lalit's earning per day = Rs. x
$\therefore$ Lalit's earning in 30 days = Rs. 30 × x
= Rs. 30x
Similarly, Lalit's expenditure per day = Rs. y
$\therefore$ Lalit's expenditure in 30 days = Rs. 30 × y
= Rs. 30 y
$\therefore$ In 30 days, Lalit saves = (Total earnings - Total expenditure)
= Rs. (30x - 30y)
= Rs. 30(x - y)

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Question 143 Marks

Solve the following equation and verify the answer:
3(x + 6) = 24

Answer

3(x + 6) = 24
$\Rightarrow\frac{3(\text{x}+6)}{3}=\frac{24}{3}$
(Dividing both sides by 3)
x + 6 = 8
⇒ x = 8 - 6
(Transposing 6 to R.H.S.)
⇒ x = 2
x = 2 is a solution of the given equation.
Check: Substituting the value of x = 2
in the given equation, we get
L.H.S. = 3(2 + 6 ) = 3 × 8 = 24
and RH.S. = 24
$\therefore$ When x = 2, we have L.H.S. = R.H.S.

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Question 153 Marks

Solve the following equation by trial and error method:
2y + 4 = 3y

Answer

The given equation is 2y + 4 = 3y
We guess and try several values of y to find L.H.S. and R.H.S. and stop when L.H.S. = R.H.S.

x	L.H.S.	R.H.S.
1	2 × 1 + 4 = 6	3 × 1 = 3
2	2 × 2 + 4 = 8	3 × 2 = 6
3	2 × 3 + 4 = 10	3 × 3 = 9
4	2 × 4 + 4 = 12	3 × 4 = 12

When y = 4, we have L.H.S. = R.H.S
So, y = 4 is the solution of the given equation.

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Question 163 Marks

Solve the following equation by trial and error method:
y + 9 = 13

Answer

The given equation is y + 9 = 13
We try several values of y and find L.H.S. and the R.H.S. and stop when L.H.S. = R.H.S.

y	L.H.S.	R.H.S.
2	2 + 9 = 11	13
3	3 + 9 = 12	13
4	4 + 9 = 13	13

When y = 4, we have L.H.S. = R.H.S.
So y = 4 is the solution of the given equation.

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Question 173 Marks

The length of a rectangular hall is 5 metres more than its breadth. If the perimeter of the hall is 74 metres, find its length and breadth.

Answer

Let the breadth of the hall is x m. and length is (x + 5) m. Then, 2(x + x + 5) = 74 Or, 4x + 10 = 74 ⇒ 4x = 74 - 10 $\Rightarrow\text{x}=\frac{64}{4}$x = 16
So, the breadth is 16m and the length is (16 + 5) = 21m

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Question 183 Marks

After 32 years, Rahim will be 5 times as old as he was 8 years ago. How old is Rahim today?

Answer

Let the age of Rahim be x years After 32 years his age is x + 32 years. 8 years ago his age was (x - 8) Then, 5(x - 8) = x + 32 Or, 5x - 40 = x + 32 ⇒ 5x - x = 32 + 40 ⇒ 4x = 72 $\Rightarrow\text{x}=\frac{72}{4}$ x = 18So, Rahim is 18 years old today

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Question 193 Marks

Solve the following equation by trial and error method:
11 + x = 19

Answer

The given equation is 11 + x = 19
We try several values of x and find L.H.S. and the R.H.S. and stop when L.H.S. = R.H.S.

x	L.H.S.	R.H.S.
3	11 + 3 = 14	19
4	11 + 4 = 15	19
6	11 + 6 = 17	19
7	11 + 7 = 18	19
8	11 + 8 = 9	19

When x = 8, we have L.H.S. = R.H.S.
So x = 8 is the solution of the given equation.

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Question 203 Marks

The cost of 1 pen is Rs. 16 and the cost of 1 pencil is Rs. 5. What is the total cost of x pens and y pencils.

Answer

Cost of 1 pen = Rs. 16
$\therefore$ Cost of x pens = Rs. 16 × x
= Rs. 16x
Similarly, cost of 1 pencil = Rs. 5
$\therefore$ Cost of y pencils = Rs. 5 × y
= Rs. 5y
$\therefore$ Total cost of x pens and y pencils = Rs. (16x + 5y)

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Question 213 Marks

Three times a number added to 8 gives 20. Find the number.

Answer

Let the required number be x
Three times this number is 3x
On adding 8, the number becomes 3x + 8
3x + 8 = 20
Or, 3x + 8 - 8 = 20 - 8 [Subtracting 8 from both the sides]
⇒ 3x = 12
$\Rightarrow\frac{\text{3x}}{3}=\frac{12}{3}$
[Dividing both the sides by 3]
⇒ x = 4
$\therefore$ Required number = 4

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Question 223 Marks

Solve the following equation by trial and error method:
z - 3 = 2z - 5

Answer

The given equation is z - 3 = 2z - 5
We guess and try several values of z to find L.H.S. and R.H.S. and stop when L.H.S. = R.H.S.

x	L.H.S.	R.H.S.
1	1 - 3 = -2	2 × 1 - 5 = -3
2	2 - 3 = -1	2 × 2 - 5 = -1

When z = 2, we have L.H.S. = R.H.S
So, z = 2 is the solution of the given equation.

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Question 233 Marks

After 16 years, Fatima will be three times as old as she is now. Find her present age.

Answer

Then, 3x = x + 16
Or, 3x - x = 16
⇒ 2x = 16
$\Rightarrow\text{x}=\frac{16}{2}$
⇒ x = 8
So, the present age of fatima is 8 years

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Question 243 Marks

Solve the following equation and verify the answer:
x - 2 = -5

Answer

x - 2 = -5
Adding 2 on both the sides:
⇒ x - 2 + 2 = -5 + 2
⇒ x = -3
Verification:
Substituting x = -3 in the L.H.S.
⇒ -3 - 2 = -5 = R.H.S.
L.H.S. = R.H.S.
Hence, verified.

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Question 253 Marks

Solve the following equation and verify the answer:
$\frac{\text{x}}{5}=12$

Answer

$\frac{\text{x}}{5}=12$ $\Rightarrow\frac{\text{x}}{5}\times5=12\times5$ [Multiplying both the sides by 5] $\Rightarrow \text{x}=60$ Verification: Substituting x = 60 in the L.H.S.$\Rightarrow\frac{60}{5}=12=\text{R.H.S.}$
$\Rightarrow\text{L.H.S.}=\text{R.H.S.}$
Hence, verified.

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Question 263 Marks

A man is 4 times as old as his son. After 16 years he will be only twice as old as his son. Find their present ages.

Answer

Let the age of son is x and his father's age is 4x After 16 years the age of son is x + 16 and his father age is 4x + 16 Then, 4x + 16 = 2(x + 16) Or, 4x + 16 = 2x + 32 ⇒ 4x - 2x = 32 - 16 ⇒ 2x = 16$\Rightarrow\text{x}=\frac{16}{2}$
x = 8 So, the present age of son is 8 years and his father's age is (4 × 8) = 32 years

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Question 273 Marks

A bag contains 25-paisa and 50-paisa coins whose total value is Rs. 30. If the number of 25-paisa coins is four times that of 50-paisa coins, find the number of each type of coins.

Answer

Let the amount of 50 paise is x and 25 paise is 4 x50x + (25 × 4x) = 50x + 100x = 150x
And, Rs. 30 = 3000 paise
Then, 150x = 3000
Or, $\text{x}=\frac{3000}{150}$
x = 20
So, the required number of 50 paise is 20 and 25 paise is (4 × 20) = 80

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Question 283 Marks

Solve the following equation by trial and error method:
$\frac{1}{2}\text{x}+7=11$

Answer

The given equation is $\frac{1}{2}\text{x}+7=11$ We guess and try several values of x to find L.H.S. and R.H.S. and stop when L.H.S. = R.H.S.

x	L.H.S.	R.H.S.
2	$\frac{1}{2}\times2+7=8$	11
4	$\frac{1}{2}\times4+7=9$	11
6	$\frac{1}{2}\times6+7=10$	11
8	$\frac{1}{2}\times8+7=11$	11

When x = 8, we have L.H.S. = R.H.S So, x = 8 is the solution of the given equation.

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Question 293 Marks

Solve the following equation and verify the answer:
x + 3 = -2

Answer

x + 3 = -2
Subtracting 3 from both the sides:
⇒ x + 3 - 3 = -2 - 3
⇒ x = -5
Verification:
Substituting x = -5 in the L.H.S.
⇒ -5 + 3 = -2 = R.H.S.
L.H.S. = R.H.S.
Hence, verified.

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Question 303 Marks

Solve: 3(x + 2) - 2(x - 1) = 7

Answer

3(x + 2) - 2(x - 1) = 7.
Or, 3 × x + 3 × 2 - 2 × x - 2 × (-1) = 7 [On expanding the brackets]
⇒ 3x + 6 - 2x + 2 = 7
⇒ x + 8 = 7
⇒ x + 8 - 8 = 7 - 8 [Subtracting 8 from both the sides]
⇒ x = −1

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Question 313 Marks

Solve the following equation and verify the answer:
6x + 5 = 2x + 17

Answer

6x + 5 = 2x + 17 ⇒ 6x - 2x = 17 - 5 (Transposing 2x to L.H.S. and 5 to R.H.S.) ⇒ 4x = 12$\Rightarrow\frac{\text{4x}}{4}=\frac{12}{4}$
(Dividing both sides by 4) ⇒ x = 3 x = 3 is a solution of the given equation. Check : Substituting x = 3 in the given equation, we get L.H.S. = 6 × 3 + 5 = 18 + 5 = 23 R.H.S. = 2 × 3 + 17 = 6 + 17 = 23 $\therefore$ When x = 3, we have L.H.S. = R.H.S.

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Question 323 Marks

The sum of two consecutive even numbers is 74. Find the numbers.

Answer

Let the required numbers are x and x + 2
Then, x + x + 2 = 74
Or, 2x = 74 - 2
$\Rightarrow\text{x}=\frac{72}{2}$
x = 36
So, the numbers are 36 and (36 + 2) = 38

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Question 333 Marks

Solve the following equation and verify the answer:
3x - 5 = 13

Answer

3x - 5 = 13⇒ 3x - 5 + 5 = 13 + 5 [Adding 5 on both the sides]
⇒ 3x = 18
$\Rightarrow\frac{\text{3x}}{3}=\frac{18}{3}$ [Dividing both the sides by 3]
⇒ x = 6
Verification:
Substituting x = 6 in the L.H.S.
⇒ 3 × 6 - 5 = 18 - 5 = 13 = R.H.S.
L.H.S. = R.H.S.
Hence, verified.

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Question 343 Marks

If $x = 1, y = 2$ and $z = 3$, find the value of $x^2 + y^2 + 2xyz.$

Answer

Given:
$x =1$
$y = 2$
$z = 3$
Substituting $x = 1, y = 2$ and $z = 3$ in the given equation:
$(x^2 + y^2 + 2xyz)$
$\Rightarrow (1)^2 + ( 2)^2 + 2(1)(2)(3)$
$\Rightarrow 1 + 4 + 12 = 17$

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Question 353 Marks

One out of two numbers is thrice the other. If their sum is 124, find the numbers.

Answer

Let the required number are x and 3xThen, x + 3x = 124
Or, 4x = 124
$\Rightarrow\text{x}=\frac{124}{4}$
x = 31
So, the numbers are 31 and (3 × 31) = 93

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Question 363 Marks

Solve: 4x + 9 = 17

Answer

4x + 9 = 17
Or, 4x + 9 - 9 = 17 - 9 [Subtracting 9 from both the sides]
⇒ 4x = 8
$\Rightarrow\frac{\text{4x}}{4}=\frac{8}{4}$ [Dividing both the sides with 4]
x = 2

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Question 373 Marks

Solve the following equation by trial and error method:
3y = 36

Answer

The given equation is 3y = 36
We guess and try several values of y to find L.H.S. and R.H.S. and stop when L.H.S. = R.H.S.

y	L.H.S.	R.H.S.
7	3 × 7 = 21	36
9	3 × 9 = 27	36
10	3 × 10 = 30	36
12	3 × 12 = 36	36

When y = 12, we have L.H.S. = R.H.S
So y = 12 is the solution of the given equation.

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Question 383 Marks

After 16 years, Seema will be three times as old as she is now. Find her present age.

Answer

Let the present age of Seema be x years.
After 16 years:
Seema's age = x + 16
After 16 years, her age becomes thrice of her age now
$\therefore$ x + 16 = 3x
Or, 16 = 3x - x [Transposing x to the R.H.S.]
⇒ 2x = 16
$\Rightarrow\frac{\text{2x}}{2}=\frac{16}{2}$ [Dividing both the sides by 2]
x = 8 years

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Question 393 Marks

Find two numbers such that one of them is five times the other and their difference is 132.

Answer

Let the required numbers are x and 5x
Then, 5x - x - = 132
Or, 4x = 132
$\Rightarrow\text{x}=\frac{132}{4}$
x = 33
So, the numbers are 33 and (33 × 5) = 165

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Question 403 Marks

Five times the price of a pen is Rs. 17 more than three times its price. Find the price of the pen.

Answer

Let the price of each pen is xThen, 5x = 3x + 17
Or, 5x - 3x = 17
⇒ 2x = 17
$\Rightarrow\text{x}=\frac{17}{2}$
x = 8.50
So, the price of the pen is 8.50 rupees

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Question 413 Marks

Find two numbers such that one of them exceeds the other by 18 and their sum is 92.

Answer

Let the required number are x and x + 18Then, x + x + 18 = 92
Or, 2x = 92 - 18
$\Rightarrow\text{x}=\frac{74}{2}$
x = 37
So, the numbers are 37 and (37 + 18) = 55

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Question 423 Marks

Solve the following equation and verify the answer:
$\frac{\text{3x}}{5}=15$

Answer

$\frac{\text{3x}}{5}=15$ $\Rightarrow\frac{\text{3x}}{5}\times5=15\times5$ [Multiplying both the sides by 5] $\Rightarrow \text{3x}=75$ $\Rightarrow\frac{\text{3x}}{3}=\frac{75}{3}$ [Dividing both sides by 3] $\Rightarrow\text{x}=25$ Verification: Substituting x = 25 in the L.H.S.$\Rightarrow\frac{3\times25}{3}=15=\text{R.H.S.}$
$\Rightarrow\text{L.H.S.}=\text{R.H.S.}$
Hence, verified.

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Question 433 Marks

Mrs. Goel is 27 years older than her daughter Rekha. After 8 years she will be twice as old as Rekha. Find their present ages.

Answer

Let Rekha's age is x years and Mrs. Goel's age is x + 27 years after 8 years the age of Rekha is x + 8 and Mrs. Goel is x + 27 + 8 Then, x + 27 + 8 = 2(x + 8) Or, x + 35 = 2x + 16⇒ 2x - x = 35 - 16
x = 19 So, the age of Rekha is 19 years and Mrs. Goel is 19 + 27 = 46

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Question 443 Marks

Solve the following equation and verify the answer:
$\frac{\text{x}}{4}-8=1$

Answer

$\frac{\text{x}}{4}-8=1$
$\Rightarrow\frac{\text{x}}{4}=1+8$
(Transporting -8 to R.H.S.)
$\Rightarrow\frac{\text{x}}{4}=9\Rightarrow\frac{\text{x}}{4}\times4=9\times4$
(Multiplying both sides by 4)
⇒ x = 36
$\therefore\text{x}=36$ is a solution of the given equation.
Substituting x = 36 in the given equation, we get
$\text{L.H.S.}=\frac{36}{4}-8=9-8=1$ and
$\text{R.H.S.}=1$
$\therefore$ When x = 36, we have L.H.S. = R.H.S.

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Question 453 Marks

Solve the following equation by trial and error method:
4x = 28

Answer

The given equation is 4x = 28
We guess and try several values of x to find L.H.S. and R.H.S. and stop when L.H.S. = R.H.S.

x	L.H.S.	R.H.S.
2	4 × 2 = 8	28
4	4 × 4 = 16	28
5	4 × 5 = 20	28
6	4 × 6 = 24	28
7	4 × 7 = 28	28

When x = 7, we have L.H.S. = R.H.S
So x = 7 is the solution of the given equation.

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Question 463 Marks

Solve the following equation and verify the answer:
5x - 3 = x + 17

Answer

5x - 3 = x + 17
⇒ 5x - x = 17 + 3
[Transporting x to L.H.S. -3 to R.H.S.]
$\Rightarrow4\text{x}=20\Rightarrow\frac{\text{4x}}{4}=\frac{20}{4}$
(Dividing both sides by 4)
⇒ x = 5
So. x = 5 is a solution of the given equation.
Check: Substituting x = 5 in the given equation, we get
L.H.S. = 5 × 5 - 3 = 25 - 3 = 22
R.H.S. 5 + 17 = 22
$\therefore$ When x = 5, we have L.H.S. = R.H.S.

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Question 473 Marks

The sum of three consecutive odd numbers is 21. Find the numbers.

Answer

Let the required numbers are x, x + 2 and x + 4
Then, x + x + 2 + x + 4 = 21
Or, 3x = 21 - 6
$\Rightarrow\text{x}=\frac{15}{3}$
x = 5
So, the numbers are 5, (5 + 2 ) = 7 and (5 + 4) = 9

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Question 483 Marks

Solve the following equation by trial and error method:
$\frac{\text{x}}{3}=4$

Answer

The given equation is $\frac{\text{x}}{3}=4$ We guess and try several values of x to find L.H.S. and R.H.S. and stop when L.H.S. = R.H.S.

x	L.H.S.	R.H.S.
3	$\frac{3}{3}=1$	4
6	$\frac{6}{3}=2$	4
9	$\frac{9}{3}=3$	4
12	$\frac{12}{3}=4$	4

When x = 12, we have L.H.S. = R.H.S So x = 12 is the solution of the given equation.

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Question 493 Marks

A wire of length 86cm is bent in the form of a rectangle such that its length is 7cm more than its breadth. Find the length and the breadth of the rectangle so formed.

Answer

Let breadth is x cm. and length is (x + 7) cmThen, 2(x + x +7) = 86
Or, 4x + 14 = 86
⇒ 4x = 86 - 14
$\Rightarrow\text{x}=\frac{72}{4}$
x = 18
So, the breadth of rectangle formed is 18cm and the length is (18 + 7) = 25cm

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