Q 1 – 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.
Ans.
Let the length of the third side be x cm.
Each equal side = 2x cm.
As per the condition of the question, we have
Perimeter = x + 2x + 2x = 30
⇒ 5x = 30
⇒ x = 6
Thus, the third side of the triangle = 6 cm
and other two equal sides are 2 × 6 = 12 cm each
Q 2 – Write the following statement in the form of an equation:
The sum of three times x and 10 is 13.
(a) 3x + 10 = 13
(b) 3x – 10 = 13
(c) 3x + 13 = 10
(d) none of these
Ans. (a) 3x + 10 = 13
Q 3 – Write the following statement in the form of an equation:
Taking away 5 from x gives 10
(a) x – 5 = 10
(b) x + 5 = 10
(c) x – 10 – 5
(d) none of these
Ans. (a) x – 5 = 10
Q 4 – Write the following statement in the form of an equation:
Add 1 to three times n to get 7
(a) 3n + 1 = 7
(b) 3n – 1 = 7
(c) 3n + 7 = 1
(d) none of these
Ans. (a) 3n + 1 = 7
Q 5 – Convert the following equations in statement form:
(a) 5x = 20
(b) 3y + 7 = 1
Ans. (a) Five times a number x gives 20.
(b) Add 7 to three times a number y gives 1
Q 6 – Solve the following equations:
3(y – 2) = 2(y – 1) – 3
Ans.
3(y – 2) = 2(y – 1) – 3
⇒ 3y – 6 = 2y – 2 – 3 (Removing the brackets)
⇒ 3y – 6 = 2y – 5
⇒ 3y – 2y = 6 – 5 (Transposing 6 to RHS and 2y to LHS)
⇒ y = 1
Thus y = 1
Q 7 – The solution of the equation x – 6 = 1 is
(a) 1
(b) 6
(c) -7
(b) 7
Ans. (d) 7
Q 8 – The solution of the equation 7n + 5 = 12 is
(a) 0
(b) – 1
(c) 1
(d) 5
Ans. (c) 1
Q 9 – If 5 is added to twice a number, the result is 29. Find the number.
Ans. Let the required number be x.
Step I: 2x + 5
Step II: 2x + 5 = 29
Solving the equation, we get
2x + 5 = 29
⇒ 2x = 29 – 5 (Transposing 5 to RHS)
⇒ 2x = 24
⇒ x = 12 (Dividing both sides by 2)
⇒ x = 12
Thus the required number is 12.
Q 10 – The length of a rectangle is twice its breadth. If its perimeter is 60 cm, find the length and the breadth of the rectangle.
Ans. Let the breadth of the rectangle be x cm.
its length = 2x
Perimeter = 2 (length + breadth) = 2(2x + x) = 2 × 3x = 6x
As per the condition of the question, we have
6x = 60 ⇒ x = 10
Thus the required breadth = 10 cm
and the length = 10 × 2 = 20 cm.
Q 11 – The sum of three consecutive multiples of 2 is 18. Find the numbers.
Ans. Let the three consecutive multiples of 2 be 2x, 2x + 2 and 2x + 4.
As per the conditions of the question, we have
2x + (2x + 2) + (2x + 4) = 18
⇒ 2x + 2x + 2 + 2x + 4 = 18
⇒ 6x + 6 = 18
⇒ 6x = 18 – 6 (Transposing 6 to RHS)
⇒ 6x = 12
⇒ x = 2
Thus, the required multiples are
2 × 2 = 4, 4 + 2 = 6, 6 + 2 = 8 i.e., 4, 6 and 8.
Q 12 – The solution of the equation 4p – 3 = 9 is
(a) 1
(b) 2
(c) 3
(d) 4
Ans. (c) 3
Q 13 – The solution of the equation p + 4 = 4 is
(a) 0
(b) 4
(c) -4
(d) 8
Ans. (a) 0
Q 14 – The solution of the equation 2m = 4 is
(a) 1
(b) 2
(c) -1
(d) -2
Ans. (b) 2
Q 15 – The solution of the equation 3m/ 7= 6 is
(a) 6
(b) 7
(c) 14
(d) 3
Ans. (c) 14
Q 16 – The solution of the equation 3 m + 7=16 is
(a) 1
(b) 2
(c) 3
(d) 4
Ans. (c) 3
Q 17 – The solution of the equation 4x + 5 = 9 is
(a) 1
(b) 2
(c) 3
(d) 4
Ans. (a) 1
Q 18 – The solution of the equation 10 t = – 20 is
(a) 1
(b) -1
(c) 2
(d) -2
Ans. (d) -2
Q 19 – The solution of the equation 3s + 6 = 0 is
(a) 1
(b) -1
(c) 2
(d) -2
Ans. (d) -2
Q 20 – The solution of the equation 12p – 11 = 13 is
(a) 1
(b) 2
(c) 3
(d) 4
Ans. (b) 2