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Published on: 16/09/2019
Linear Equations in One Variable
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1.
The length and breadth of a rectangular field are in the ratio 6: 4. Find the length and breadth if the cost of fencing the field at the rate of Rs 80 per metre is Rs 16,000.
2.
The sum of the digits of a two-digit number is 13. If the digits are interchanged, and the resulting number is added to the original number, then we get 143. What is the original number?
3.
The present age of Prabha's mother is three times the present age of Prabha. After 5 years their ages will add to 66 years. Find their present ages.
4.
The perimeter of a rectangle is 13 cm. If its width is \(2{3\over 4}\) cm, then find its length.
5.
The sum of two numbers is 74. If one of the numbers is 10 more than the other, then what are the numbers?
6.
Identify the linear equations in one variable from the following.x2 + 1 = 4, y+y2 = 3,z+4 = z2 +z3 5x= 1,\(\frac{4}{3}\) = y, 4z-12 = \(\frac{5}{7}\)
7.
Solve 0.44 t - 1.05 = 2 (0.71 t -0.01) + 1.11
8.
What should be subtracted from thrice the rational number \(\frac { -13 }{ 4 } \) to get \(\frac { 5 }{ 8 } \) ?
9.
Solve \(\frac { 13 }{ 5 } -5x=13\)
10.
Find the solution of \(\frac { y-(4-3y) }{ 2y-(3+4y) } =\frac { 1 }{ 5 } \).
11.
If the length and breadth of a rectangular field are in the ratio 6:4. Find the length and breadth, if cost of fencing the field at the rate of Rs. 80 per metre is Rs. 16000.
12.
Solve the following equations and check your result. \(3m=5m-\frac { 8 }{ 5 } \)
13.
Solve the following equations and check your result. 5x + 9 = 5 + 3x.
14.
Find the root the following equations \(2p+9=\frac { -17 }{ 7 } \)
15.
Find the root the following equations \(3x-\frac { 1 }{ 3 } =\frac { 2 }{ 3 } \)
1.
Length: 60 m; Breadth: 40 m
2.
85 (or 58)
3.
Let Sahil’s present age be x years.
| Sahil | Mother | Sum | |
| Present age | x | 3x | |
| Age 5 years later | x + 5 | 3x + 5 | 4x + 10 |
It is given that this sum is 66 years
Therefore, 4x + 10 = 66
This equation determines Sahil’s present age which is x years. To solve the equation,
we transpose 10 to RHS,
4x = 66 – 10
or 4x = 56
or \(x=\frac{56}{4}=14\)
Thus, Sahil’s present age is 14 years and his mother’s age is 42 years. (You may easily check that 5 years from now the sum of their ages will be 66 years.)
4.
Assume the length of the rectangle to be x cm.
The perimeter of the rectangle = 2 x (length + width)
\(
=2 \times\left(x+2 \frac{3}{4}\right) \)
\(=2\left(x+\frac{11}{4}\right)
\)
The perimeter is given to be 13 cm. Therefore,
\(2\left(x+\frac{11}{4}\right)=13\)
\(
x+\frac{11}{4} =\frac{13}{2}
\)
\(x =\frac{13}{2}-\frac{11}{4}
\)
\(=\frac{26}{4}-\frac{11}{4}=\frac{15}{4}=3 \frac{3}{4}\)
5.
32 and 42
6.
5x = 1,\(\frac{4}{3}\)= y, 4z-12 = \(\frac{5}{7}\) are linear equations in one variable and x2 + 1= 4, y + y2 = 3,z + 4 = Z2 + z3 are not linear equations in one va'riable, because in all of these equations highest power of variable is greater than 1.
7.
We have, 0.44t-1.05 = 2 x 0.71t- 2 x 0.01 + 1.11
0.44t-1.05 = 1.42 t- 0.02+ 1.11
0.44t-1.42t = 1.05-0.02+ 1.11
-0.98t = 2.14
\(t=\frac { 214\times 100 }{ -0.98\times 100 } \)
\(t=\frac { 214 }{ -98 } =\frac { -107 }{ 49 } \)
8.
Let x be the requires number
\(\therefore\) \(3x\left( \frac { -13 }{ 4 } \right) -x=\frac { 5 }{ 8 } \)
\(\Rightarrow\) \(\frac { -39 }{ 4 } -x=\frac { 5 }{ 8 } \Rightarrow \frac { -39 }{ 4 } -\frac { 5 }{ 8 } =x\)
\(\Rightarrow\) \(\frac { -78-5 }{ 8 } =5x\Rightarrow x=-\frac { 83 }{ 8 } \)
9.
We have \(\frac { 15 }{ 3 } -5x=13\)
\(\Rightarrow\) \(\frac { 13 }{ 6 } -13=5x\Rightarrow \frac { 13-65 }{ 5 } =5x\)
\(\Rightarrow\) \(\frac { -52 }{ 5 } =5x\Rightarrow x=-\frac { 52 }{ 25 } \)
10.
\(y=\frac { 17 }{ 22 } \)
11.
Length = 60 m, breadth = 40 m.
12.
m = \(\frac { 4 }{ 5 } \)
13.
We have, 5x + 9 = 5 + 3x
\(\Rightarrow\) 5x -3x = 5-9 [rransposing Sx to LHS and 9 to RHS]
\(2x=-4\)
\(\Rightarrow\) x =\(\frac { -4 }{ 2 } \) [dividing both sides by 2]
\(\therefore\) x = -2 which is the required solution
14.
\(p=-5\frac { 5 }{ 7 } \)
15.
\(x=\frac { 1 }{ 3 } \)
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