Q 1) Say whether the equation is satisfied.
Equation: x + 3 = 0, Value: x = 3
Answer: x + 3 = 0
Substituting x = 3 in above equation
L.H.S. = x + 3
= 3 + 3
= 6
R.H.S. = 0
L.H.S. ≠ R.H.S.
Therefore, equation is not satisfied.
Q 2) Check whether the value given in the brackets is a solution to the given equation or not.
(i) n + 5 = 19 (n = 1)
Answer: n + 5 = 19
Substituting n = 1 in above equation
L.H.S. = n + 5
= 1 + 5
= 6
R.H.S. = 19
L.H.S. ≠ R.H.S.
Therefore, n = 1 is not a solution to the given equation.
(ii) 7n + 5 = 19 (n = – 2)
Answer: 7n + 5 = 19
Substituting n = -2 in above equation
L.H.S. = 7n + 5
= 7 x (-2) + 5
= (-14) + 5
= (-9)
R.H.S. = 19
L.H.S. ≠ R.H.S.
Therefore, n = -2 is not a solution to the given equation.
Q 3) Solve the following equation by trial and error method.
5p + 2 = 17
Answer:
5p + 2 = 17
Substituting p = 1 in above equation
L.H.S. = 5p + 2
= (5 x 1) + 2
= 5 + 2
= 7
R.H.S. = 17
L.H.S. ≠ R.H.S.
Therefore, p = 1 is not a solution to the given equation.
Substituting p = 2 in above equation
L.H.S. = 5p + 2
= (5 x 2) + 2
= 10 + 2
= 12
R.H.S. = 17
L.H.S. ≠ R.H.S.
Therefore, p = 2 is not a solution to the given equation.
Substituting p = 3 in above equation
L.H.S. = 5p + 2
= (5 x 3) + 2
= 15 + 2
= 17
R.H.S. = 17
L.H.S. = R.H.S.
Therefore, p = 3 is a solution to the given equation.
Q 4) Write equations for the following statements.
(i) The sum of numbers x and 4 is 9.
Answer: x + 4 = 9
(ii) The number b divided by 5 gives 6.
Answer: b/5 = 6
(iii) Seven times m plus 7 gets you 77.
Answer: 7m + 7 = 77
(iv) One-fourth of a number x minus 4 gives 4.
Answer: (x/4) – 4 = 4
Q 5) Write the following equations in statement forms.
(i) p + 4 = 15
Answer: The sum of numbers p and 4 is 15.
(ii) 3p + 4 = 25
Answer: Three times p plus 4 gives you 25.
(iii) (p/2) + 2 = 8
Answer: If you add half of a number p to 2, you get 8.
(iv) 4p – 2 = 18
Answer: Four times p minus 2 gives you 18.
Q 6) Set up an equation in the following cases.
(i) Irfan says that he has 7 marbles, more than five times the marbles Parmit has. Irfan has 37 marbles (Take m to be the number of Parmit’s marbles).
Answer:
Given
Irfan’s marbles = 37
To Find
Parmit’s marbles = ?
Condition
Irfan has 7 marbles more than five times Parmit’s marbles
Solution
Let Parmit’s marbles be m
Therefore, Irfan’s marbles = 5m + 7
But it is given that Irfan’s marbles = 37
Therefore, 5m + 7 = 37
(ii) In an isosceles triangle, the vertex angle is twice either base angle (Let the base angle be b in degrees. Remember that the sum of angles of a triangle is 180 degrees).
Answer:
Given
Sum of angles of a triangle = 180o
To Find
Measure of the base angles = ?
Condition
In an isosceles triangle, the vertex angle is twice either base angle
Solution
Let base angle be b degrees
Therefore, Vertex angle = 2b
As sum of angles of a triangle = 180o
2b + b + b = 180o
Therefore, 4b = 180o
Q 7) Give first the step you will use to separate the variable and then solve the equation.
(i) x – 1 = 0
Answer: x – 1 = 0
Adding 1 to both sides of the equation
x – 1 + 1 = 0 + 1
x = 1
(ii) y – 4 = -7
Answer: y – 4 = -7
Adding 4 to both sides of the equation
y – 4 + 4 = -7 + 4
y = -3
(iii) b/2 = 6
Answer: = 6
Multiplying both sides of the equation by 2
(b/2) x 2 = 6 x 2
b = 12
Q 8) Solve the following equations.
(i) 2q + 6 = 12
Answer: 2q + 6 = 12
Transpose 6 from LHS to RHS
2q = 12 – 6
2q = 6
Dividing both sides by 2
q = 6/2
q = 3
(ii) 2q – 6 = 0
Answer: 2q – 6 = 0
Transpose (-6) from LHS to RHS
2q = 0 + 6
2q = 6
Dividing both sides by 2
q = 6/2
q = 3
(iii) 4 + 5(p – 1) =34
Answer: 4 + 5(p – 1) =34
4 + (5 x p) – (5 x 1) = 34
4 + 5p – 5 = 34
5p – 1 = 34
Transpose (-1) from LHS to RHS
5p = 34 + 1
5p = 35
Dividing both sides by 5
p = 35/5
p = 7
Q 9) Solve the following.
(i) The teacher tells the class that the highest marks obtained by a student in her class are twice the lowest marks plus 7. The highest score is 87. What is the lowest score?
Answer:
Given
Highest score in class = 87
To Find
Lowest score = ?
Condition
Highest score is twice the lowest score plus 7
Solution
Let lowest score be x
Therefore, highest score = 2x + 7
But it is given that highest score = 87
Therefore, 2x + 7 = 87
Transpose 7 from LHS to RHS
2x = 87 – 7
2x = 80
Dividing both sides by 2
x = 80/2
x = 40
Therefore, lowest score in class is 40 marks.
(ii) People of Sundargram planted trees in the village garden. Some of the trees were fruit trees. The number of non-fruit trees was two more than three times the number of fruit trees. What was the number of fruit trees planted if the number of non-fruit trees planted was 77?
Answer:
Given
Non-fruit trees = 77
To Find
Fruit trees = ?
Condition
Number of non-fruit trees was two more than three times the number of fruit trees
Solution
Let fruit trees be x
Therefore, non-fruit trees = 3x + 2
But it is given that non-fruit trees = 77
Therefore, 3x + 2 = 77
Transpose 2 from LHS to RHS
3x = 77 – 2
3x = 75
Dividing both sides by 3
x = 75/3
x = 25
Therefore, number of fruit trees is 25.
Practice Questions
- If one-third of a number exceeds its one-fourth by 1, find the number.
- The length of a rectangle is twice its breadth. If its perimeter is 60 cm, find the length and the breadth of the rectangle.
- Seven times a number is 12 less than thirteen times the same number. Find the number.
- The present age of a son is half the present age of his father. Ten years ago, the father was thrice as old as his son. What are their present age?
- The sum of three consecutive multiples of 2 is 18. Find the three multiples.
- Each of the 2 equal sides of an isosceles triangle is twice as large as the third side. If the perimeter of the triangle is 30 cm, find the length of each side of the triangle.