Exercise 5: Simple Equations
Solve the following equations (1 – 12):
1. $x + 4 = 13$
Solution: $x + 4 = 13$
or, $x = 13 – 4$ [by transposition]
or, $x = 9$
$\therefore$Required solution:$x = 9$
2. $x + 5 = 9$
Solution: $x + 5 = 9$
or, $x = 9 – 5$ [by transposition]
or, $x = 4$
$\therefore$Required solution:$x = 4$
3. $y + 1 = 10$
Solution: $y + 1 = 10$
or, $y = 10 – 1$ [by transposition]
or, $y = 9$
$\therefore$Required solution:$y = 9$
4. $y – 5 = 11$
Solution: $y – 5 = 11$
or, $y = 11 + 5$ [by transposition]
or, $y = 16$
$\therefore$Required solution:$y = 16$
5. $z + 3 = 15$
Solution: $z + 3 = 15$
or, $z = 15 – 3$ [by transposition]
or, $z = 12$
$\therefore$Required solution:$z = 12$
6. $3x = 12$
Solution: $3x = 12$
or, $x = \frac{12}{3}$ [dividing both sides by $3$]
or, $x = 4$
$\therefore$Required solution:$x = 4$
7. $2x + 1 = 9$
Solution: $2x + 1 = 9$
or, $2x = 9 – 1$ [by transposition]
or, $2x = 8$
or, $x = \frac{8}{2}$ [dividing both sides by $2$]
or, $x = 4$
$\therefore$Required solution:$x = 4$
8. $4x – 5 = 11$
Solution: $4x – 5 = 11$
or, $4x = 11 + 5$ [by transposition]
or, $4x = 16$
or, $x = \frac{16}{4}$ [dividing both sides by $4$]
or, $x = 4$
$\therefore$Required solution:$x = 4$
9. $3x – 5 = 17$
Solution: $3x – 5 = 17$
or, $3x = 17 + 5$ [by transposition]
or, $3x = 22$
or, $x = \frac{22}{3}$ [dividing both sides by $3$]
or, $x = 7\frac{1}{3}$
$\therefore$Required solution:$x = \frac{22}{3}$ (or $7\frac{1}{3}$)
10. $7x – 2 = x + 16$
Solution: $7x – 2 = x + 16$
or, $7x – x = 16 + 2$ [by transposition]
or, $6x = 18$
or, $x = \frac{18}{6}$ [dividing both sides by $6$]
or, $x = 3$
$\therefore$Required solution:$x = 3$
11. $3 – x = 14$
Solution: $3 – x = 14$
or, $-x = 14 – 3$ [by transposition]
or, $-x = 11$
or, $x = -11$ [multiplying both sides by $-1$]
$\therefore$Required solution:$x = -11$
12. $2x + 9 = 3$
Solution: $2x + 9 = 3$
or, $2x = 3 – 9$ [by transposition]
or, $2x = -6$
or, $x = \frac{-6}{2}$ [dividing both sides by $2$]
or, $x = -3$
$\therefore$Required solution:$x = -3$
Solve by form equation (13 – 18):
13. If $6$ is added to twice of a number, the sum will be $14$. What is the number?
Solution: Let, the number $= x$
$\therefore$ Double of the number $= 2x$
According to the question,
$2x + 6 = 14$
or, $2x = 14 – 6$ [by transposition]
or, $2x = 8$
or, $x = \frac{8}{2}$ [dividing both sides by $2$]
or, $x = 4$
$\therefore$Required number:$4$
14. If $5$ is subtracted from a number, the difference will be $11$. What is the number?
Solution: Let, the number $= x$
According to the question,
$x – 5 = 11$
or, $x = 11 + 5$ [by transposition]
or, $x = 16$
$\therefore$Required number:$16$
15. What is the number whose $7$ times will be equal to $21$. ?
Solution: Let, the number $= x$
$\therefore 7$ times the number $= 7x$
According to the question,
$7x = 21$
or, $x = \frac{21}{7}$ [dividing both sides by $7$]
or, $x = 3$
$\therefore$Required number:$3$
16. If $3$ is added to $4$ times a number, the sum will be $23$. What is the number?
Solution: Let, the number $= x$
$\therefore 4$ times the number $= 4x$
According to the question,
$4x + 3 = 23$
or, $4x = 23 – 3$ [by transposition]
or, $4x = 20$
or, $x = \frac{20}{4}$ [dividing both sides by $4$]
or, $x = 5$
$\therefore$Required number:$5$
17. If the price of a pen is less than its specified price by $2$ taka, the price would be $10$ taka. What is the price of the pen?
Solution:
Let the price of the pen $= x$ Taka
According to the question,
$x – 2 = 10$
or, $x = 10 + 2$ [by transposition]
or, $x = 12$
$\therefore$Price of the pen:$12$ Taka
18. Monika has 4 times more chocolates than Kanika. They have $25$ chocolates together. How many chocolates does Kanika have?
Solution: Let the number of chocolates Konika has $= x$
$\therefore$ The number of chocolates Monika has $= 4x$
According to the question,
$x + 4x = 25$
or, $5x = 25$
or, $x = \frac{25}{5}$ [dividing both sides by $5$]
or, $x = 5$
$\therefore$Konika has $5$ chocolates.
Sample Questions (Multiple Choice Questions)
1. What is the perimeter of a rectangular garden having length of $x$ metre and breadth of $y$ metre?
Answer:
(d) $2(x+y)$
Explanation: We know, Perimeter of a rectangle $= 2 \times (\text{Length} + \text{Width}) = 2(x+y)$ metre.
2. Find out the value of $x$ when $3$ is added to twice $x$ that gives a sum of $9$.
Answer:
(a) $3$
Explanation:
According to the question, $2x + 3 = 9$
or, $2x = 9 – 3$
or, $2x = 6$
or, $x = \frac{6}{2}$
or, $x = 3$
3. If $a, b, c$ are any numbers and $a = b$, then—
(i) $ac = bc$
(ii) $a + c = b + c$
(iii) $a – c = b – c$
Which one is correct?
Answer:
(d) i, ii, and iii
Answer question 4-5 using the information given below:
The subtraction of two numbers is $30$ and the greater number is 4 times the smaller number.
4. What is the ratio of greater and smaller number?
Answer:
(d) $4:1$
Explanation: The larger number is $4$ times the smaller number.
That is, Larger number : Smaller number $= 4 : 1$.
5. What is the smaller number?
Answer:
(b) $10$
Explanation: Let, the smaller number $= x$
Therefore, the larger number $= 4x$
According to the question,
$4x – x = 30$
or, $3x = 30$
or, $x = \frac{30}{3}$
or, $x = 10$
Therefore, the smaller number is $10$.
Creative Questions
Scenario-1: The sum of three consecutive natural numbers is $33$.
Scenario-2: Rony goes to the market and buys $5$ apples and $1$ dozen bananas for $216$ Taka, where the price of one apple is equal to the price of three bananas.
a) If $7$ times a number is $35$, then what is $3$ times that number.
Solution: Let the number $= x$
According to the condition,
$7x = 35$
or, $x = \frac{35}{7}$
or, $x = 5$
Therefore, the number is $5$.
Now, $3$ times the number $= 5 \times 3 = 15$
Answer:$15$
b) According to Scenario 1, find the largest of the three natural numbers.
Solution:
Let, The first consecutive natural number $= x$
The second consecutive natural number $= x + 1$
The third (largest) consecutive natural number $= x + 2$
According to the question,
$x + (x + 1) + (x + 2) = 33$
or, $3x + 3 = 33$
or, $3x = 33 – 3$
or, $3x = 30$
or, $x = \frac{30}{3}$
or, $x = 10$
Therefore, the largest natural number $= x + 2 = 10 + 2 = 12$
Answer:$12$
c) According to Scenario 2, determine the price of one apple.
Solution:
We know, $1$ dozen $= 12$ pieces.
According to the stimulus,
Price of $1$ apple $=$ Price of $3$ bananas
Therefore,
Price of $5$ apples $=$ Price of $(5 \times 3)$ bananas
$=$ Price of $15$ bananas.
Roni bought $5$ apples and $12$ bananas.
That is, Total purchase $=$ Price of $15$ bananas $+$ Price of $12$ bananas
$=$ Price of $27$ bananas.
According to the question,
Price of $27$ bananas $= 216$ Taka
or, Price of $1$ banana $= \frac{216}{27}$ Taka $= 8$ Taka
Therefore,
Price of $1$ apple $= 3 \times$ Price of $1$ banana
$= 3 \times 8$ Taka
$= 24$ Taka
Answer:$24$ Taka
Short Answer Questions
1. Solve: $8x – 3 = 3x + 17$
Solution: $8x – 3 = 3x + 17$
or, $8x – 3x = 17 + 3$ [by transposition]
or, $5x = 20$
or, $x = \frac{20}{5}$ [dividing both sides by $5$]
or, $x = 4$
$\therefore$Required solution:$x = 4$
2. If three times a number plus $7$ equals $34$, find the number.
Solution: Let the number $= x$
According to the question,
$3x + 7 = 34$
or, $3x = 34 – 7$ [by transposition]
or, $3x = 27$
or, $x = \frac{27}{3}$ [dividing both sides by $3$]
or, $x = 9$
$\therefore$The number is$9$
3. If $3$ times number is added to $5$ times that number, the sum will be $32$. Find the number.
Solution: Let the number $= x$
According to the question,
$5x + 3x = 32$
or, $8x = 32$
or, $x = \frac{32}{8}$ [dividing both sides by $8$]
or, $x = 4$
$\therefore$Required number:$4$
4. If twice of a number is subtracted from four times the number, the difference will be $24$. Find is the number.
Solution: Let the number $= x$
According to the question,
$4x – 2x = 24$
or, $2x = 24$
or, $x = \frac{24}{2}$ [dividing both sides by $2$]
or, $x = 12$
$\therefore$Required number:$12$
5. The sum of two consecutive even natural numbers is $30$, find the numbers.
Solution: Let,
First consecutive even natural number $= x$
Second consecutive even natural number $= x + 2$
According to the question,
$x + (x + 2) = 30$
or, $2x + 2 = 30$
or, $2x = 30 – 2$ [by transposition]
or, $2x = 28$
or, $x = \frac{28}{2}$ [dividing both sides by $2$]
or, $x = 14$
First number $= 14$
Second number $= 14 + 2 = 16$
$\therefore$The two numbers are $14$ and $16$ respectively.
6. If the sum of three consecutive odd natural numbers is $27$, find the numbers.
Solution: Let,
First consecutive odd natural number $= x$
Second consecutive odd natural number $= x + 2$
Third consecutive odd natural number $= x + 4$
According to the question,
$x + (x + 2) + (x + 4) = 27$
or, $3x + 6 = 27$
or, $3x = 27 – 6$ [by transposition]
or, $3x = 21$
or, $x = \frac{21}{3}$ [dividing both sides by $3$]
or, $x = 7$
First number $= 7$
Second number $= 7 + 2 = 9$
Third number $= 7 + 4 = 11$
$\therefore$The three numbers are $7, 9,$ and $11$ respectively.