write ten simple equations questions with answer
Answers
Answer:
Question 1.
Write the following statements in the form of equations.
(a) The sum of four times a number and 5 gives a number five times of it.
(b) One-fourth of a number is 2 more than 5.
Solution:
(a) Let the number be x.
Sum of 4x and 5 = 4x + 5
The sum is 5x.
The equation is 4x + 5 = 5x as required.
(b) Let the number be x.
\frac { 1 }{ 4 }x = 5 + 2
⇒ \frac { 1 }{ 4 }x = 7 as required.
Question 2.
Convert the following equations in statement form:
(a) 5x = 20
(b) 3y + 7 = 1
Solution:
(a) Five times a number x gives 20.
(b) Add 7 to three times a number y gives 1
Answer:
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Step-by-step explanation:
Question 1.
Write the following statements in the form of equations.
(a) The sum of four times a number and 5 gives a number five times of it.
(b) One-fourth of a number is 2 more than 5.
Solution:
(a) Let the number be x.
Sum of 4x and 5 = 4x + 5
The sum is 5x.
The equation is 4x + 5 = 5x as required.
(b) Let the number be x.
14x = 5 + 2
⇒ 14x = 7 as required.
Question 2.
Convert the following equations in statement form:
(a) 5x = 20
(b) 3y + 7 = 1
Solution:
(a) Five times a number x gives 20.
(b) Add 7 to three times a number y gives 1.
Question 3.
If k + 7 = 10, find the value of 9k – 50.
Solution:
k + 7 = 10
⇒ k = 10 – 7 = 3
Put k = 3 in 9k – 50, we get
9 × 3 – 50 = 27 – 50 = -23
Thus the value of k = -23
Question 4
Solve the following equations:
3(y – 2) = 2(y – 1) – 3
Solution:
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
Question 5.
Solve the following equations:
3(y – 2) = 2(y – 1) – 3
Solution:
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
Question 6.
If 5 is added to twice a number, the result is 29. Find the number.
Solution:
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.