Algebraic Equations (Class-7, Experience- 12)

[Exercise- Page 240]

E.12.1. Raka's friend has 10 tk more than Raka. The product of their amount is 119. Determine how much they have each.

Solution:

Let's represent Raka's amount as x Tk. 

Since Raka's friend has 10 Tk more, their amount would be (x + 10) Tk.

Given that the product of their amounts is 119, we can write the equation: 

      x ✕ (x + 10) = 119

or, x² + 10x = 119 

or, x² + 10x - 119 = 0 

Now, we can solve this quadratic equation to find the value of x, which represents Raka's amount.

Using the quadratic formula: 

x = -b ± √ (b² - 4ac)           2a

Plugging in the values: 

  x = -10 ± √ 10² - 4 ✕ 1 ✕ (-119)               2 ✕ 1  [where a = 1, b = 10, and c = -119]

or, x = -10 ± √ 100 + 476              2 

or, x = -10 ± √ 576         2 

or, x = -10 ± 24      2

Now, we have two possible solutions:

1. x = (-10 + 24)        2 =  14   2 = 7 Tk

2. x = (-10 - 24)        2 = - 34 2 = -17 Tk

Since the amount cannot be negative, the valid solution is x = 7 Tk.

So, Raka has 7 Tk, and 

Raka's friend has 7 + 10 = 17 Tk. (Answer)

E.12.2. The sum of the two digits of a two-digit number is 12 and their product 27. What is the number?

Solution:

Let's represent the two-digit number as (10a + b), where 'a' represents the tens digit and 'b' represents the units digit.

Given that the sum of the two digits is 12, we have the equation:

a + b = 12 .....(1)

And given that their product is 27, we have the equation: 

ab = 27 .....(2)

We can use substitution or elimination to solve this system of equations. Let's use substitution. From equation (1), we can express 'a' as:

a = 12 - b 

Substituting this expression for 'a' into equation (2):

ab = 27

or, (12 - b)b = 27

or, 12b - b² = 27 

or, b² - 12b + 27 = 0  [Rearranging to form a quadratic equation]

Now, we can solve this quadratic equation to find the value(s) of 'b'. Using the quadratic formula:

x = -b ± √ (b² - 4ac)           2a

Plugging in the values: 

      b = 12 ± √ 12² - 4 ✕ 1 ✕ 27                (2 ✕ 1)

or, b = 12 ± √ 144 - 108             2

or, b = 12 ± √ 36       2

or, b = 12 ± 6    2

Now, we have two possible solutions for 'b':

b = 12 + 6    2 

b =  18   2 

b = 9 

or,

b = (12 - 6)     2 

b =   6    2 

b = 3

Now, we can find the corresponding values of 'a' using equation (1):

If b = 9, then a = 12 - 9 = 3

If b = 3, then a = 12 - 3 = 9

So, the two possible two-digit numbers are 39 and 93.

E.12.3. The area of a floor of rectangular room is 132 sq.m. If the length is reduced by 6 m. and the width is made twice, the area of the floor remains unchanged. Determine the length and width of the floor.

Solution:

Let's denote, 

the original length of the rectangular room as L meters 

and the original width as W meters.

Given that the area of the floor is 132 sq.m, we have the equation:

L ✕ W = 132 .....(1)

After reducing the length by 6 meters and doubling the width, 

the new length becomes (L - 6) meters 

and the new width becomes 2W meters 

and area of the floor remains unchanged. 

So, we have the equation:

(L - 6) ✕ (2W) = L ✕ W .....(2)

Expanding equation (2):

     (L - 6) ✕ (2W) = L ✕ W

or, 2W(L - 6) = LW

or, 2WL - 12W = LW

or, 2WL - LW = 12W

or, L(2W - W) = 12W

or, L ✕ W = 12W  [From equation (1), we know that L ✕ W = 132]

or, 132 = 12W

or, W = 132 12

∴ W = 11

Now that we have found the value of W, we can substitute it back into equation (1) to find the value of L:

L ✕ 11 = 132

or. L = 132 11

∴ L = 12

So, the original length of the floor is 12 meters and the original width is 11 meters. (Answer)

E.12.4. The length of the hypotenuse of a right angled triangle is 13 meter and the difference of the length of other two sides is 7 meter. What is the area and circumference of the triangle?

Solution:

Answer will be posted soon. Please wait...

E.12.5. The height of a triangle is 2 cm less than thrice its base. If the area of the triangle is 20 sq. cm, then determine the its height.

Solution:

Given, 

The area of a triangle is 20 cm²,

And the height of the triangle is 2 cm less than thrice its base, we can write the equation:

h = 3b - 2

The formula for the area of a triangle is:

Area =  1  2 ✕ base ✕ height

or, 20 =  1  2 ✕ b ✕ (3b - 2) [substituting the given values]

or, 20 =  1  2 ✕ 3b² - 2b

or, 40 = 3b² - 2b

or, 3b² - 2b - 40 = 0

Now, we can solve this quadratic equation for b. We can either factor it or use the quadratic formula. Let's use the quadratic formula:

x = -b ± √ (b² - 4ac)       2a

Plugging in the values: 

b = -(-2) ± √ (-2)² - 4 ✕ 3 ✕ (-40)              (2 ✕ 3)

or, b = 2 ± √ 4 + 480         6

or, b = 2 ± √ 484       6

or, b = 2 ± 22   6

Since b cannot be negative, we take the positive solution:

    b = 2 + 22   6 

or, b =   24     6 

∴ b = 4

Now, we can find the height of the triangle using the equation 

h = 3b - 2

or, h = 3 ✕ 4 - 2 

or, h = 12 - 2 

∴ h = 10

So, the height of the triangle is 10 cm.

E.12.6. Cutting papers, solve the quadratic equation x² + 6x - 7 = 0.

Solution:

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