| Latest Maths NCERT Books Solution | ||||||
|---|---|---|---|---|---|---|
| 6th | 7th | 8th | 9th | 10th | 11th | 12th |
| Content On This Page | ||
|---|---|---|
| Example 1 to 4 (Before Exercise 4.1) | Exercise 4.1 | Example 5 (Before Exercise 4.2) |
| Exercise 4.2 | Example 6 to 10 (Before Exercise 4.3) | Exercise 4.3 |
Chapter 4 Simple Equations
Welcome to the solutions guide for Chapter 4, "Simple Equations," a foundational chapter in algebra presented in the latest Class 7 NCERT mathematics textbook for the academic session 2024-25. This chapter marks a significant step into the world of algebra, teaching students how to represent relationships mathematically and solve for unknown values. Mastering simple equations is not just about finding a numerical answer; it's about developing logical reasoning and problem-solving skills that are applicable across various fields. These solutions provide invaluable assistance, offering systematic and clear methods for tackling the exercises.
The core focus of this chapter is on understanding and solving linear equations in one variable. These are equations where the highest power of the variable (like $x$, $y$, or any other letter representing an unknown value) is one, typically taking forms such as $ax + b = c$ or involving the variable on both sides, like $ax + b = cx + d$. The solutions meticulously guide students through the primary techniques used to find the value of the unknown variable that makes the equation true:
- The Balancing Method: This intuitive approach involves treating the equation like a balance scale. To maintain equality, whatever mathematical operation (addition, subtraction, multiplication, division) is performed on one side of the equation must also be performed on the other side. For example, to solve $y + 3 = 10$, we subtract 3 from both sides: $y + 3 - 3 = 10 - 3$, which simplifies to $y = 7$.
- The Transposition Method: This is often seen as a shortcut derived from the balancing method. It involves moving a term from one side of the equation to the other, changing its sign or operation in the process. For instance, in $2x - 5 = 11$, transposing $-5$ to the right side changes it to $+5$, giving $2x = 11 + 5$, or $2x = 16$. Then, transposing the multiplication by 2 becomes division by 2 on the other side: $x = \frac{16}{2} = 8$.
A crucial skill emphasized throughout the chapter and detailed in the solutions is the ability to set up equations from practical word problems or verbal statements. This involves translating real-world scenarios into the symbolic language of algebra. The solutions meticulously break down this process: identifying the unknown quantity, assigning a variable to it, representing the relationships described in the problem as an equation, solving the formulated equation using the methods described above, and finally, interpreting the solution back into the context of the original problem. Common types of word problems encountered include those related to ages, numbers, geometric figures (e.g., perimeter problems), and everyday situations potentially involving costs (perhaps in $\textsf{₹}$).
Furthermore, the solutions often incorporate the important step of checking the solution. This involves substituting the value found for the variable back into the original equation to verify that it makes the left-hand side (LHS) equal to the right-hand side (RHS), ensuring the answer's correctness. For example, if we found $x=8$ for $2x-5=11$, checking involves calculating $2(8)-5 = 16-5 = 11$, which matches the RHS, confirming the solution.
Regarding the rationalized syllabus for 2024-25, Chapter 4, "Simple Equations," in the Class 7 NCERT textbook has retained its fundamental focus on solving linear equations in one variable and setting them up from word problems. The rationalization primarily involved adjusting the complexity and number of word problems and equation types, ensuring a strong grasp of core algebraic techniques without introducing overly complicated scenarios at this stage. The essential methods of solving (balancing, transposition) and the application to contextual problems remain central. By utilizing these detailed solutions, students can develop logical and systematic approaches to solving equations, enhance their algebraic manipulation skills, gain confidence in translating verbal descriptions into mathematical expressions, and build a critical foundation for future, more advanced algebraic studies.
Example 1 to 4 (Before Exercise 4.1)
Example 1. Write the following statements in the form of equations:
(i) The sum of three times x and 11 is 32.
(ii) If you subtract 5 from 6 times a number, you get 7.
(iii) One fourth of m is 3 more than 7.
(iv) One third of a number plus 5 is 8.
Answer:
Solution:
Let's translate each statement into a mathematical equation:
(i) The sum of three times x and 11 is 32.
Three times x can be written as $3x$.
The sum of $3x$ and 11 is $3x + 11$.
The statement says this sum "is 32".
So, the equation is:
$3x + 11 = 32$
(ii) If you subtract 5 from 6 times a number, you get 7.
Let the number be represented by a variable, say $y$.
Six times a number is $6y$.
Subtracting 5 from $6y$ gives $6y - 5$.
The statement says you "get 7".
So, the equation is:
$6y - 5 = 7$
(iii) One fourth of m is 3 more than 7.
One fourth of m can be written as $\frac{1}{4}m$ or $\frac{m}{4}$.
3 more than 7 is $7 + 3$, which simplifies to 10.
The statement says "One fourth of m is 3 more than 7".
So, the equation is:
$\frac{m}{4} = 7 + 3$
Simplifying the right side, we get:
$\frac{m}{4} = 10$
(iv) One third of a number plus 5 is 8.
Let the number be represented by a variable, say $n$.
One third of a number is $\frac{1}{3}n$ or $\frac{n}{3}$.
One third of a number plus 5 is $\frac{n}{3} + 5$.
The statement says this "is 8".
So, the equation is:
$\frac{n}{3} + 5 = 8$
Example 2. Convert the following equations in statement form:
(i) x – 5 = 9
(ii) 5p = 20
(iii) 3n + 7 = 1
(iv) $\frac{m}{5} - 2 = 6$
Answer:
Solution:
Let's convert each equation into its statement form:
(i) $x – 5 = 9$
This equation means that 5 is subtracted from a number $x$, and the result is 9.
Statement: 5 subtracted from x is 9.
(ii) $5p = 20$
This equation means that 5 times a number $p$ is equal to 20.
Statement: 5 times p is 20.
(iii) $3n + 7 = 1$
This equation means that 3 times a number $n$ is added to 7, and the result is 1.
Statement: 3 times n plus 7 is 1.
(iv) $\frac{m}{5} - 2 = 6$
This equation means that a number $m$ is divided by 5, and then 2 is subtracted from the result, which gives 6.
Statement: 2 subtracted from one-fifth of m is 6.
Example 3. Consider the following situation:
Raju’s father’s age is 5 years more than three times Raju’s age. Raju’s father is 44 years old. Set up an equation to find Raju’s age.
Answer:
Solution:
Let Raju's age be represented by the variable $r$ years.
Three times Raju's age is $3r$ years.
Raju’s father’s age is 5 years more than three times Raju’s age, which can be expressed as $3r + 5$ years.
We are also given that Raju's father is 44 years old.
To set up an equation, we equate the two expressions for Raju's father's age:
$\text{Age of Raju's father} = 3r + 5$
And we know,
$\text{Age of Raju's father} = 44$ years
(Given)
Therefore, the equation to find Raju's age is:
$3r + 5 = 44$
Example 4. A shopkeeper sells mangoes in two types of boxes, one small and one large. A large box contains as many as 8 small boxes plus 4 loose mangoes. Set up an equation which gives the number of mangoes in each small box. The number of mangoes in a large box is given to be 100.
Answer:
Solution:
Given:
A large box contains mangoes equivalent to 8 small boxes plus 4 loose mangoes.
The total number of mangoes in a large box is 100.
To Find:
Set up an equation to find the number of mangoes in each small box.
Let:
Let the number of mangoes in one small box be represented by the variable $x$.
Formulating the Equation:
Number of mangoes in 8 small boxes $= 8 \times x = 8x$.
A large box contains 8 small boxes plus 4 loose mangoes. So, the number of mangoes in a large box can be expressed as:
Number of mangoes in a large box $= 8x + 4$
We are given that the number of mangoes in a large box is 100.
Therefore, we can equate the two expressions for the number of mangoes in a large box:
$8x + 4 = 100$
This is the required equation that gives the number of mangoes in each small box ($x$).
Exercise 4.1
Question 1. Complete the last column of the table.
| S.No | Equation | Value | Say, whether the Equation is Satisfied. (Yes/ No) |
|---|---|---|---|
| (i) | x + 3 = 0 | x = 3 | |
| (ii) | x + 3 = 0 | x = 0 | |
| (iii) | x + 3 = 0 | x = -3 | |
| (iv) | x - 7 = 1 | x = 7 | |
| (v) | x - 7 = 1 | x = 8 | |
| (vi) | 5x = 25 | x = 0 | |
| (vii) | 5x = 25 | x = 5 | |
| (viii) | 5x = 25 | x = -5 | |
| (ix) | $\frac{m}{3} = 2$ | m = -6 | |
| (x) | $\frac{m}{3} = 2$ | m = 0 | |
| (xi) | $\frac{m}{3} = 2$ | m = 6 |
Answer:
Solution:
To check if the equation is satisfied by the given value, we substitute the value into the equation and see if the Left Hand Side (LHS) is equal to the Right Hand Side (RHS).
Here is the completed table:
| S.No | Equation | Value | Say, whether the Equation is Satisfied. (Yes/ No) |
| (i) | $x + 3 = 0$ | $x = 3$ | No ($3 + 3 = 6 \neq 0$) |
| (ii) | $x + 3 = 0$ | $x = 0$ | No ($0 + 3 = 3 \neq 0$) |
| (iii) | $x + 3 = 0$ | $x = -3$ | Yes ($-3 + 3 = 0$) |
| (iv) | $x - 7 = 1$ | $x = 7$ | No ($7 - 7 = 0 \neq 1$) |
| (v) | $x - 7 = 1$ | $x = 8$ | Yes ($8 - 7 = 1$) |
| (vi) | $5x = 25$ | $x = 0$ | No ($5 \times 0 = 0 \neq 25$) |
| (vii) | $5x = 25$ | $x = 5$ | Yes ($5 \times 5 = 25$) |
| (viii) | $5x = 25$ | $x = -5$ | No ($5 \times (-5) = -25 \neq 25$) |
| (ix) | $\frac{m}{3} = 2$ | $m = -6$ | No ($\frac{-6}{3} = -2 \neq 2$) |
| (x) | $\frac{m}{3} = 2$ | $m = 0$ | No ($\frac{0}{3} = 0 \neq 2$) |
| (xi) | $\frac{m}{3} = 2$ | $m = 6$ | Yes ($\frac{6}{3} = 2$) |
Question 2. Check whether the value given in the brackets is a solution to the given equation or not:
(a) n + 5 = 19 (n = 1)
(b) 7n + 5 = 19 (n = – 2)
(c) 7n + 5 = 19 (n = 2)
(d) 4p – 3 = 13 (p = 1)
(e) 4p – 3 = 13 (p = – 4)
(f) 4p – 3 = 13 (p = 0)
Answer:
Solution:
To check if the given value is a solution, we substitute the value into the equation and see if the Left Hand Side (LHS) equals the Right Hand Side (RHS).
(a) $n + 5 = 19$ ($n = 1$)
LHS $= n + 5$
Substitute $n = 1$:
LHS $= 1 + 5 = 6$
RHS $= 19$
Since LHS $\neq$ RHS ($6 \neq 19$), the equation is not satisfied.
Therefore, $n=1$ is not a solution.
(b) $7n + 5 = 19$ ($n = – 2$)
LHS $= 7n + 5$
Substitute $n = -2$:
LHS $= 7(-2) + 5 = -14 + 5 = -9$
RHS $= 19$
Since LHS $\neq$ RHS ($-9 \neq 19$), the equation is not satisfied.
Therefore, $n=-2$ is not a solution.
(c) $7n + 5 = 19$ ($n = 2$)
LHS $= 7n + 5$
Substitute $n = 2$:
LHS $= 7(2) + 5 = 14 + 5 = 19$
RHS $= 19$
Since LHS $=$ RHS ($19 = 19$), the equation is satisfied.
Therefore, $n=2$ is a solution.
(d) $4p – 3 = 13$ ($p = 1$)
LHS $= 4p – 3$
Substitute $p = 1$:
LHS $= 4(1) – 3 = 4 – 3 = 1$
RHS $= 13$
Since LHS $\neq$ RHS ($1 \neq 13$), the equation is not satisfied.
Therefore, $p=1$ is not a solution.
(e) $4p – 3 = 13$ ($p = – 4$)
LHS $= 4p – 3$
Substitute $p = -4$:
LHS $= 4(-4) – 3 = -16 – 3 = -19$
RHS $= 13$
Since LHS $\neq$ RHS ($-19 \neq 13$), the equation is not satisfied.
Therefore, $p=-4$ is not a solution.
(f) $4p – 3 = 13$ ($p = 0$)
LHS $= 4p – 3$
Substitute $p = 0$:
LHS $= 4(0) – 3 = 0 – 3 = -3$
RHS $= 13$
Since LHS $\neq$ RHS ($-3 \neq 13$), the equation is not satisfied.
Therefore, $p=0$ is not a solution.
Question 3. Solve the following equations by trial and error method:
(i) 5p + 2 = 17
(ii) 3m – 14 = 4
Answer:
Solution:
We will solve the equations by substituting different values for the variable until the LHS becomes equal to the RHS.
(i) $5p + 2 = 17$
Let's try different values for $p$:
If $p = 0$, LHS $= 5(0) + 2 = 0 + 2 = 2$. RHS $= 17$. LHS $\neq$ RHS.
If $p = 1$, LHS $= 5(1) + 2 = 5 + 2 = 7$. RHS $= 17$. LHS $\neq$ RHS.
If $p = 2$, LHS $= 5(2) + 2 = 10 + 2 = 12$. RHS $= 17$. LHS $\neq$ RHS.
If $p = 3$, LHS $= 5(3) + 2 = 15 + 2 = 17$. RHS $= 17$. LHS $=$ RHS.
Since LHS = RHS when $p = 3$, the equation is satisfied.
Thus, the solution is $\mathbf{p=3}$.
(ii) $3m – 14 = 4$
Let's try different values for $m$:
If $m = 0$, LHS $= 3(0) – 14 = 0 – 14 = -14$. RHS $= 4$. LHS $\neq$ RHS.
If $m = 1$, LHS $= 3(1) – 14 = 3 – 14 = -11$. RHS $= 4$. LHS $\neq$ RHS.
If $m = 2$, LHS $= 3(2) – 14 = 6 – 14 = -8$. RHS $= 4$. LHS $\neq$ RHS.
If $m = 3$, LHS $= 3(3) – 14 = 9 – 14 = -5$. RHS $= 4$. LHS $\neq$ RHS.
If $m = 4$, LHS $= 3(4) – 14 = 12 – 14 = -2$. RHS $= 4$. LHS $\neq$ RHS.
If $m = 5$, LHS $= 3(5) – 14 = 15 – 14 = 1$. RHS $= 4$. LHS $\neq$ RHS.
If $m = 6$, LHS $= 3(6) – 14 = 18 – 14 = 4$. RHS $= 4$. LHS $=$ RHS.
Since LHS = RHS when $m = 6$, the equation is satisfied.
Thus, the solution is $\mathbf{m=6}$.
Question 4. Write equations for the following statements:
(i) The sum of numbers x and 4 is 9.
(ii) 2 subtracted from y is 8.
(iii) Ten times a is 70.
(iv) The number b divided by 5 gives 6.
(v) Three-fourth of t is 15.
(vi) Seven times m plus 7 gets you 77.
(vii) One-fourth of a number x minus 4 gives 4.
(viii) If you take away 6 from 6 times y, you get 60.
(ix) If you add 3 to one-third of z, you get 30.
Answer:
Solution: We need to write the given statements as mathematical equations.
(i) The sum of numbers x and 4 is 9.
"Sum of numbers x and 4" means adding x and 4, which is $x + 4$.
"is 9" means it is equal to 9.
So, the equation is: $x + 4 = 9$.
(ii) 2 subtracted from y is 8.
"2 subtracted from y" means we start with y and take away 2, which is $y - 2$.
"is 8" means it is equal to 8.
So, the equation is: $y - 2 = 8$.
(iii) Ten times a is 70.
"Ten times a" means multiplying a by 10, which is $10 \times a$ or simply $10a$.
"is 70" means it is equal to 70.
So, the equation is: $10a = 70$.
(iv) The number b divided by 5 gives 6.
"The number b divided by 5" can be written as $b \div 5$ or $\frac{b}{5}$.
"gives 6" means it is equal to 6.
So, the equation is: $\frac{b}{5} = 6$.
(v) Three-fourth of t is 15.
"Three-fourth of t" means $\frac{3}{4} \times t$ or $\frac{3}{4}t$.
"is 15" means it is equal to 15.
So, the equation is: $\frac{3}{4}t = 15$.
(vi) Seven times m plus 7 gets you 77.
"Seven times m" is $7m$.
"plus 7" means adding 7 to $7m$, so $7m + 7$.
"gets you 77" means it is equal to 77.
So, the equation is: $7m + 7 = 77$.
(vii) One-fourth of a number x minus 4 gives 4.
"One-fourth of a number x" is $\frac{1}{4} \times x$ or $\frac{1}{4}x$ or $\frac{x}{4}$.
"minus 4" means subtracting 4 from $\frac{1}{4}x$, so $\frac{1}{4}x - 4$.
"gives 4" means it is equal to 4.
So, the equation is: $\frac{1}{4}x - 4 = 4$.
(viii) If you take away 6 from 6 times y, you get 60.
"6 times y" is $6y$.
"If you take away 6 from 6 times y" means $6y - 6$.
"you get 60" means it is equal to 60.
So, the equation is: $6y - 6 = 60$.
(ix) If you add 3 to one-third of z, you get 30.
"One-third of z" is $\frac{1}{3} \times z$ or $\frac{1}{3}z$ or $\frac{z}{3}$.
"If you add 3 to one-third of z" means $\frac{1}{3}z + 3$.
"you get 30" means it is equal to 30.
So, the equation is: $\frac{1}{3}z + 3 = 30$.
Question 5. Write the following equations in statement forms:
(i) p + 4 = 15
(ii) m – 7 = 3
(iii) 2m = 7
(iv) $\frac{m}{5} = 3$
(v) $\frac{3m}{5} = 6$
(vi) 3p + 4 = 25
(vii) 4p – 2 = 18
(viii) $\frac{p}{2} + 2 = 8$
Answer:
Solution: We need to write the given equations as statements in words.
Here are the statement forms for the given equations:
(i) $p + 4 = 15$
The equation shows that a number 'p' is added to 4, and the result is 15.
Statement: The sum of a number p and 4 is 15.
(ii) $m – 7 = 3$
The equation shows that 7 is subtracted from a number 'm', and the result is 3.
Statement: 7 subtracted from a number m is 3.
(iii) $2m = 7$
The equation shows that a number 'm' is multiplied by 2 (which is "twice m" or "2 times m"), and the result is 7.
Statement: Twice a number m is 7. (or Ten times m is 7)
(iv) $\frac{m}{5} = 3$
The equation shows that a number 'm' is divided by 5 (which is "m divided by 5" or "one-fifth of m"), and the result is 3.
Statement: The number m divided by 5 is 3. (or One-fifth of a number m is 3)
(v) $\frac{3m}{5} = 6$
The equation shows that a number 'm' is multiplied by 3 and then divided by 5 (which is "three-fifths of m"). The result is 6.
Statement: Three-fifths of a number m is 6.
(vi) $3p + 4 = 25$
The equation shows that a number 'p' is multiplied by 3 (which is "3 times p"), and then 4 is added to this result. The final result is 25.
Statement: Three times a number p plus 4 is 25.
(vii) $4p – 2 = 18$
The equation shows that a number 'p' is multiplied by 4 (which is "4 times p"), and then 2 is subtracted from this result. The final result is 18.
Statement: Four times a number p minus 2 is 18. (or If you take away 2 from four times p, you get 18)
(viii) $\frac{p}{2} + 2 = 8$
The equation shows that a number 'p' is divided by 2 (which is "half of p"), and then 2 is added to this result. The final result is 8.
Statement: Half of a number p plus 2 is 8.
Question 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.)
(ii) Laxmi’s father is 49 years old. He is 4 years older than three times Laxmi’s age. (Take Laxmi’s age to be y years.)
(iii) The teacher tells the class that the highest marks obtained by a student in her class is twice the lowest marks plus 7. The highest score is 87. (Take the lowest score to be l.)
(iv) 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:
Solution:
Let's set up an equation for each case based on the given information.
(i) Irfan’s marbles:
Let the number of Parmit's marbles be $m$.
Five times the marbles Parmit has is $5m$.
Irfan has 7 marbles more than five times Parmit's marbles, so Irfan's marbles $= 5m + 7$.
We are given that Irfan has 37 marbles.
Equating the two expressions for Irfan's marbles:
$5m + 7 = 37$
(ii) Laxmi’s father’s age:
Let Laxmi's age be $y$ years.
Three times Laxmi’s age is $3y$ years.
Laxmi’s father is 4 years older than three times Laxmi’s age, so father's age $= 3y + 4$ years.
We are given that Laxmi's father is 49 years old.
Equating the two expressions for the father's age:
$3y + 4 = 49$
(iii) Highest score:
Let the lowest score be $l$.
Twice the lowest marks is $2l$.
The highest marks are twice the lowest marks plus 7, so highest score $= 2l + 7$.
We are given that the highest score is 87.
Equating the two expressions for the highest score:
$2l + 7 = 87$
(iv) Angles of an isosceles triangle:
Let the base angle be $b$ degrees.
In an isosceles triangle, the two base angles are equal. So, the two base angles are $b$ and $b$ degrees.
The vertex angle is twice either base angle, so the vertex angle $= 2b$ degrees.
The sum of angles in a triangle is $180^\circ$.
Sum of angles $= \text{base angle} + \text{base angle} + \text{vertex angle} = b + b + 2b$.
So, the equation is:
$b + b + 2b = 180$
Combining the terms on the left side:
$4b = 180$
Example 5 (Before Exercise 4.2)
Example 5. Solve:
(a) 3n + 7 = 25
(b) 2p – 1 = 23
Answer:
Solution:
We will solve the equations by performing inverse operations on both sides to isolate the variable.
(a) $3n + 7 = 25$
The variable term is $3n$. To isolate it, we need to remove the $+7$. We do this by subtracting 7 from both sides of the equation.
$3n + 7 - 7 = 25 - 7$
$3n = 18$
Now, the variable $n$ is multiplied by 3. To isolate $n$, we divide both sides by 3.
$\frac{3n}{3} = \frac{18}{3}$
$n = 6$
The solution is $\mathbf{n=6}$.
(b) $2p – 1 = 23$
The variable term is $2p$. To isolate it, we need to remove the $-1$. We do this by adding 1 to both sides of the equation.
$2p – 1 + 1 = 23 + 1$
$2p = 24$
Now, the variable $p$ is multiplied by 2. To isolate $p$, we divide both sides by 2.
$\frac{2p}{2} = \frac{24}{2}$
$p = 12$
The solution is $\mathbf{p=12}$.
Exercise 4.2
Question 1. Give first the step you will use to separate the variable and then solve the equation:
(a) x – 1 = 0
(b) x + 1 = 0
(c) x – 1 = 5
(d) x + 6 = 2
(e) y – 4 = – 7
(f) y – 4 = 4
(g) y + 4 = 4
(h) y + 4 = – 4
Answer:
Solution:
(a) $x – 1 = 0$
The variable is $x$. To isolate $x$, we need to remove the $-1$ from the left side.
First step: Add 1 to both sides of the equation.
$x - 1 + 1 = 0 + 1$
$x = 1$
The solution is $\mathbf{x=1}$.
(b) $x + 1 = 0$
The variable is $x$. To isolate $x$, we need to remove the $+1$ from the left side.
First step: Subtract 1 from both sides of the equation.
$x + 1 - 1 = 0 - 1$
$x = -1$
The solution is $\mathbf{x=-1}$.
(c) $x – 1 = 5$
The variable is $x$. To isolate $x$, we need to remove the $-1$ from the left side.
First step: Add 1 to both sides of the equation.
$x - 1 + 1 = 5 + 1$
$x = 6$
The solution is $\mathbf{x=6}$.
(d) $x + 6 = 2$
The variable is $x$. To isolate $x$, we need to remove the $+6$ from the left side.
First step: Subtract 6 from both sides of the equation.
$x + 6 - 6 = 2 - 6$
$x = -4$
The solution is $\mathbf{x=-4}$.
(e) $y – 4 = – 7$
The variable is $y$. To isolate $y$, we need to remove the $-4$ from the left side.
First step: Add 4 to both sides of the equation.
$y - 4 + 4 = -7 + 4$
$y = -3$
The solution is $\mathbf{y=-3}$.
(f) $y – 4 = 4$
The variable is $y$. To isolate $y$, we need to remove the $-4$ from the left side.
First step: Add 4 to both sides of the equation.
$y - 4 + 4 = 4 + 4$
$y = 8$
The solution is $\mathbf{y=8}$.
(g) $y + 4 = 4$
The variable is $y$. To isolate $y$, we need to remove the $+4$ from the left side.
First step: Subtract 4 from both sides of the equation.
$y + 4 - 4 = 4 - 4$
$y = 0$
The solution is $\mathbf{y=0}$.
(h) $y + 4 = – 4$
The variable is $y$. To isolate $y$, we need to remove the $+4$ from the left side.
First step: Subtract 4 from both sides of the equation.
$y + 4 - 4 = -4 - 4$
$y = -8$
The solution is $\mathbf{y=-8}$.
Question 2. Give first the step you will use to separate the variable and then solve the equation:
(a) 3l = 42
(b) $\frac{b}{2} = 6$
(c) $\frac{p}{7} = 4$
(d) 4x = 25
(e) 8y = 36
(f) $\frac{z}{3} = \frac{5}{4}$
(g) $\frac{a}{5} = \frac{7}{15}$
(h) 20t = – 10
Answer:
Solution:
(a) $3l = 42$
The variable $l$ is multiplied by 3. To separate the variable, we need to perform the inverse operation of multiplication by 3, which is division by 3.
First step: Divide both sides of the equation by 3.
$\frac{3l}{3} = \frac{42}{3}$
$l = 14$
The solution is $\mathbf{l=14}$.
(b) $\frac{b}{2} = 6$
The variable $b$ is divided by 2. To separate the variable, we need to perform the inverse operation of division by 2, which is multiplication by 2.
First step: Multiply both sides of the equation by 2.
$\frac{b}{2} \times 2 = 6 \times 2$
$b = 12$
The solution is $\mathbf{b=12}$.
(c) $\frac{p}{7} = 4$
The variable $p$ is divided by 7. To separate the variable, we need to perform the inverse operation of division by 7, which is multiplication by 7.
First step: Multiply both sides of the equation by 7.
$\frac{p}{7} \times 7 = 4 \times 7$
$p = 28$
The solution is $\mathbf{p=28}$.
(d) $4x = 25$
The variable $x$ is multiplied by 4. To separate the variable, we need to perform the inverse operation of multiplication by 4, which is division by 4.
First step: Divide both sides of the equation by 4.
$\frac{4x}{4} = \frac{25}{4}$
$x = \frac{25}{4}$
The solution is $\mathbf{x=\frac{25}{4}}$.
(e) $8y = 36$
The variable $y$ is multiplied by 8. To separate the variable, we need to perform the inverse operation of multiplication by 8, which is division by 8.
First step: Divide both sides of the equation by 8.
$\frac{8y}{8} = \frac{36}{8}$
$\frac{\cancel{36}^{9}}{\cancel{8}_{2}}$
$y = \frac{9}{2}$
The solution is $\mathbf{y=\frac{9}{2}}$.
(f) $\frac{z}{3} = \frac{5}{4}$
The variable $z$ is divided by 3. To separate the variable, we need to perform the inverse operation of division by 3, which is multiplication by 3.
First step: Multiply both sides of the equation by 3.
$\frac{z}{3} \times 3 = \frac{5}{4} \times 3$
$z = \frac{15}{4}$
The solution is $\mathbf{z=\frac{15}{4}}$.
(g) $\frac{a}{5} = \frac{7}{15}$
The variable $a$ is divided by 5. To separate the variable, we need to perform the inverse operation of division by 5, which is multiplication by 5.
First step: Multiply both sides of the equation by 5.
$\frac{a}{5} \times 5 = \frac{7}{15} \times 5$
$a = \frac{7}{\cancel{15}_{3}} \times \cancel{5}^{1}$
$a = \frac{7}{3}$
The solution is $\mathbf{a=\frac{7}{3}}$.
(h) $20t = – 10$
The variable $t$ is multiplied by 20. To separate the variable, we need to perform the inverse operation of multiplication by 20, which is division by 20.
First step: Divide both sides of the equation by 20.
$\frac{20t}{20} = \frac{-10}{20}$
$t = -\frac{\cancel{10}^{1}}{\cancel{20}_{2}}$
$t = -\frac{1}{2}$
The solution is $\mathbf{t=-\frac{1}{2}}$.
Question 3. Give the steps you will use to separate the variable and then solve the equation:
(a) 3n – 2 = 46
(b) 5m + 7 = 17
(c) $\frac{20p}{3} = 40$
(d) $\frac{3p}{10} = 6$
Answer:
Solution:
We will solve each equation by applying inverse operations to isolate the variable.
(a) $3n – 2 = 46$
The term $-2$ is subtracted from $3n$. To remove $-2$, we perform the inverse operation, which is adding 2.
First step: Add 2 to both sides of the equation.
$3n - 2 + 2 = 46 + 2$
$3n = 48$
Now, the variable $n$ is multiplied by 3. To remove the multiplication by 3, we perform the inverse operation, which is dividing by 3.
Second step: Divide both sides of the equation by 3.
$\frac{3n}{3} = \frac{48}{3}$
$n = 16$
The solution is $\mathbf{n=16}$.
(b) $5m + 7 = 17$
The term $+7$ is added to $5m$. To remove $+7$, we perform the inverse operation, which is subtracting 7.
First step: Subtract 7 from both sides of the equation.
$5m + 7 - 7 = 17 - 7$
$5m = 10$
Now, the variable $m$ is multiplied by 5. To remove the multiplication by 5, we perform the inverse operation, which is dividing by 5.
Second step: Divide both sides of the equation by 5.
$\frac{5m}{5} = \frac{10}{5}$
$m = 2$
The solution is $\mathbf{m=2}$.
(c) $\frac{20p}{3} = 40$
The variable $p$ is multiplied by 20 and divided by 3. To separate the variable, we first remove the division by 3 by multiplying by 3.
First step: Multiply both sides of the equation by 3.
$\frac{20p}{3} \times 3 = 40 \times 3$
$20p = 120$
Now, the variable $p$ is multiplied by 20. To remove the multiplication by 20, we divide by 20.
Second step: Divide both sides of the equation by 20.
$\frac{20p}{20} = \frac{120}{20}$
$\frac{\cancel{120}^{6}}{\cancel{20}_{1}}$
$p = 6$
The solution is $\mathbf{p=6}$.
(d) $\frac{3p}{10} = 6$
The variable $p$ is multiplied by 3 and divided by 10. To separate the variable, we first remove the division by 10 by multiplying by 10.
First step: Multiply both sides of the equation by 10.
$\frac{3p}{10} \times 10 = 6 \times 10$
$3p = 60$
Now, the variable $p$ is multiplied by 3. To remove the multiplication by 3, we divide by 3.
Second step: Divide both sides of the equation by 3.
$\frac{3p}{3} = \frac{60}{3}$
$\frac{\cancel{60}^{20}}{\cancel{3}_{1}}$
$p = 20$
The solution is $\mathbf{p=20}$.
Question 4. Solve the following equations:
(a) 10p = 100
(b) 10p + 10 = 100
(c) $\frac{p}{4} = 5$
(d) $\frac{-p}{3} = 5$
(e) $\frac{3p}{4} = 6$
(f) 3s = –9
(g) 3s + 12 = 0
(h) 3s = 0
(i) 2q = 6
(j) 2q – 6 = 0
(k) 2q + 6 = 0
(l) 2q + 6 = 12
Answer:
Solution:
(a) $10p = 100$
Divide both sides by 10:
$\frac{10p}{10} = \frac{100}{10}$
$p = 10$
(b) $10p + 10 = 100$
Subtract 10 from both sides:
$10p + 10 - 10 = 100 - 10$
$10p = 90$
Divide both sides by 10:
$\frac{10p}{10} = \frac{90}{10}$
$p = 9$
(c) $\frac{p}{4} = 5$
Multiply both sides by 4:
$\frac{p}{4} \times 4 = 5 \times 4$
$p = 20$
(d) $\frac{-p}{3} = 5$
Multiply both sides by 3:
$\frac{-p}{3} \times 3 = 5 \times 3$
$-p = 15$
Multiply both sides by -1:
$-p \times (-1) = 15 \times (-1)$
$p = -15$
(e) $\frac{3p}{4} = 6$
Multiply both sides by 4:
$\frac{3p}{4} \times 4 = 6 \times 4$
$3p = 24$
Divide both sides by 3:
$\frac{3p}{3} = \frac{24}{3}$
$p = 8$
(f) $3s = –9$
Divide both sides by 3:
$\frac{3s}{3} = \frac{-9}{3}$
$s = -3$
(g) $3s + 12 = 0$
Subtract 12 from both sides:
$3s + 12 - 12 = 0 - 12$
$3s = -12$
Divide both sides by 3:
$\frac{3s}{3} = \frac{-12}{3}$
$s = -4$
(h) $3s = 0$
Divide both sides by 3:
$\frac{3s}{3} = \frac{0}{3}$
$s = 0$
(i) $2q = 6$
Divide both sides by 2:
$\frac{2q}{2} = \frac{6}{2}$
$q = 3$
(j) $2q – 6 = 0$
Add 6 to both sides:
$2q - 6 + 6 = 0 + 6$
$2q = 6$
Divide both sides by 2:
$\frac{2q}{2} = \frac{6}{2}$
$q = 3$
(k) $2q + 6 = 0$
Subtract 6 from both sides:
$2q + 6 - 6 = 0 - 6$
$2q = -6$
Divide both sides by 2:
$\frac{2q}{2} = \frac{-6}{2}$
$q = -3$
(l) $2q + 6 = 12$
Subtract 6 from both sides:
$2q + 6 - 6 = 12 - 6$
$2q = 6$
Divide both sides by 2:
$\frac{2q}{2} = \frac{6}{2}$
$q = 3$
Example 6 to 10 (Before Exercise 4.3)
Example 6. Solve: 12p – 5 = 25
Answer:
Solution:
We are given the equation $12p – 5 = 25$.
To solve for $p$, we first isolate the term with the variable, $12p$. We do this by removing the constant term, $-5$.
Add 5 to both sides of the equation:
$12p – 5 + 5 = 25 + 5$
$12p = 30$
Now, the variable $p$ is multiplied by 12. To isolate $p$, we divide both sides of the equation by 12.
$\frac{12p}{12} = \frac{30}{12}$
$p = \frac{\cancel{30}^{5}}{\cancel{12}_{2}}$
$p = \frac{5}{2}$
The solution is $\mathbf{p=\frac{5}{2}}$.
Example 7. Solve
(a) 4(m + 3) = 18
(b) – 2(x + 3) = 8
Answer:
Solution:
(a) $4(m + 3) = 18$
To isolate the term $(m+3)$, divide both sides of the equation by 4:
$\frac{4(m + 3)}{4} = \frac{18}{4}$
$m + 3 = \frac{\cancel{18}^{9}}{\cancel{4}_{2}}$
$m + 3 = \frac{9}{2}$
To isolate $m$, subtract 3 from both sides of the equation:
$m + 3 - 3 = \frac{9}{2} - 3$
$m = \frac{9}{2} - \frac{6}{2}$
$m = \frac{9 - 6}{2}$
$m = \frac{3}{2}$
The solution is $\mathbf{m = \frac{3}{2}}$.
(b) $-2(x + 3) = 8$
To isolate the term $(x+3)$, divide both sides of the equation by -2:
$\frac{-2(x + 3)}{-2} = \frac{8}{-2}$
$x + 3 = -4$
To isolate $x$, subtract 3 from both sides of the equation:
$x + 3 - 3 = -4 - 3$
$x = -7$
The solution is $\mathbf{x = -7}$.
Example 8. The sum of three times a number and 11 is 32. Find the number.
Answer:
Solution:
Let the unknown number be represented by the variable $n$.
According to the statement, "three times a number" is $3n$.
The "sum of three times a number and 11" is $3n + 11$.
The statement says this sum "is 32".
So, we can write the equation:
$3n + 11 = 32$
Now, we solve this equation for $n$.
Subtract 11 from both sides of the equation to isolate the term with $n$:
$3n + 11 - 11 = 32 - 11$
$3n = 21$
Divide both sides by 3 to isolate $n$:
$\frac{3n}{3} = \frac{21}{3}$
$n = 7$
Thus, the required number is 7.
Example 9. Find a number, such that one-fourth of the number is 3 more than 7.
Answer:
Solution:
To Find:
The number.
Let:
Let the required number be $y$.
Formulating the Equation:
According to the given statement:
"one-fourth of the number" is $\frac{1}{4}y$ or $\frac{y}{4}$.
"3 more than 7" is $7 + 3 = 10$.
The statement says "one-fourth of the number is 3 more than 7".
So, the equation is:
$\frac{y}{4} = 10$
Solving the Equation:
We have the equation $\frac{y}{4} = 10$.
To isolate $y$, we perform the inverse operation of division by 4, which is multiplication by 4.
Multiply both sides of the equation by 4:
$\frac{y}{4} \times 4 = 10 \times 4$
$y = 40$
The required number is 40.
Example 10. Raju’s father’s age is 5 years more than three times Raju’s age. Find Raju’s age, if his father is 44 years old.
Answer:
Solution:
Given:
Raju’s father’s age is 5 years more than three times Raju’s age.
Raju’s father is 44 years old.
To Find:
Raju's age.
Let:
Let Raju’s age be $r$ years.
Formulating the Equation:
According to the given statement, Raju’s father’s age is 5 years more than three times Raju’s age.
Three times Raju's age $= 3 \times r = 3r$ years.
5 years more than three times Raju's age $= 3r + 5$ years.
So, Raju's father's age can be expressed as $3r + 5$ years.
We are given that Raju's father is 44 years old.
Equating the two expressions for the father's age, we get the equation:
$3r + 5 = 44$
Solving the Equation:
We need to solve the equation $3r + 5 = 44$ for $r$.
First, subtract 5 from both sides of the equation to isolate the term $3r$:
$3r + 5 - 5 = 44 - 5$
$3r = 39$
Next, divide both sides of the equation by 3 to isolate $r$:
$\frac{3r}{3} = \frac{39}{3}$
$r = 13$
Thus, Raju's age is 13 years.
Exercise 4.3
Question 1. Set up equations and solve them to find the unknown numbers in the following cases:
(a) Add 4 to eight times a number; you get 60.
(b) One-fifth of a number minus 4 gives 3.
(c) If I take three-fourths of a number and add 3 to it, I get 21.
(d) When I subtracted 11 from twice a number, the result was 15.
(e) Munna subtracts thrice the number of notebooks he has from 50, he finds the result to be 8.
(f) Ibenhal thinks of a number. If she adds 19 to it and divides the sum by 5, she will get 8.
(g) Anwar thinks of a number. If he takes away 7 from $\frac{5}{2}$ of the number, the result is 23
Answer:
Solution:
(a) Add 4 to eight times a number; you get 60.
Let:
Let the unknown number be $x$.
Equation:
Eight times the number is $8x$.
Adding 4 to $8x$ gives $8x + 4$.
This result is 60.
$8x + 4 = 60$
Solve:
Subtract 4 from both sides:
$8x + 4 - 4 = 60 - 4$
$8x = 56$
Divide both sides by 8:
$\frac{8x}{8} = \frac{56}{8}$
$x = 7$
Answer:
The unknown number is 7.
(b) One-fifth of a number minus 4 gives 3.
Let:
Let the unknown number be $y$.
Equation:
One-fifth of the number is $\frac{y}{5}$.
Subtracting 4 from $\frac{y}{5}$ gives $\frac{y}{5} - 4$.
This result is 3.
$\frac{y}{5} - 4 = 3$
Solve:
Add 4 to both sides:
$\frac{y}{5} - 4 + 4 = 3 + 4$
$\frac{y}{5} = 7$
Multiply both sides by 5:
$\frac{y}{5} \times 5 = 7 \times 5$
$y = 35$
Answer:
The unknown number is 35.
(c) If I take three-fourths of a number and add 3 to it, I get 21.
Let:
Let the unknown number be $t$.
Equation:
Three-fourths of a number is $\frac{3}{4}t$.
Adding 3 to $\frac{3}{4}t$ gives $\frac{3}{4}t + 3$.
This result is 21.
$\frac{3}{4}t + 3 = 21$
Solve:
Subtract 3 from both sides:
$\frac{3}{4}t + 3 - 3 = 21 - 3$
$\frac{3}{4}t = 18$
Multiply both sides by $\frac{4}{3}$ (or multiply by 4 and then divide by 3):
$\frac{3}{4}t \times \frac{4}{3} = 18 \times \frac{4}{3}$
$t = \frac{\cancel{18}^{6} \times 4}{\cancel{3}_{1}}$
$t = 6 \times 4$
$t = 24$
Answer:
The unknown number is 24.
(d) When I subtracted 11 from twice a number, the result was 15.
Let:
Let the unknown number be $n$.
Equation:
Twice a number is $2n$.
Subtracting 11 from $2n$ gives $2n - 11$.
This result was 15.
$2n - 11 = 15$
Solve:
Add 11 to both sides:
$2n - 11 + 11 = 15 + 11$
$2n = 26$
Divide both sides by 2:
$\frac{2n}{2} = \frac{26}{2}$
$n = 13$
Answer:
The unknown number is 13.
(e) Munna subtracts thrice the number of notebooks he has from 50, he finds the result to be 8.
Let:
Let the number of notebooks Munna has be $m$.
Equation:
Thrice the number of notebooks is $3m$.
Subtracting $3m$ from 50 gives $50 - 3m$.
The result is 8.
$50 - 3m = 8$
Solve:
Subtract 50 from both sides:
$50 - 3m - 50 = 8 - 50$
$-3m = -42$
Divide both sides by -3:
$\frac{-3m}{-3} = \frac{-42}{-3}$
$m = 14$
Answer:
The number of notebooks Munna has is 14.
(f) Ibenhal thinks of a number. If she adds 19 to it and divides the sum by 5, she will get 8.
Let:
Let the number Ibenhal thinks of be $k$.
Equation:
She adds 19 to the number: $k + 19$.
She divides the sum by 5: $\frac{k + 19}{5}$.
She will get 8.
$\frac{k + 19}{5} = 8$
Solve:
Multiply both sides by 5:
$\frac{k + 19}{5} \times 5 = 8 \times 5$
$k + 19 = 40$
Subtract 19 from both sides:
$k + 19 - 19 = 40 - 19$
$k = 21$
Answer:
The number Ibenhal thinks of is 21.
(g) Anwar thinks of a number. If he takes away 7 from $\frac{5}{2}$ of the number, the result is 23.
Let:
Let the number Anwar thinks of be $z$.
Equation:
$\frac{5}{2}$ of the number is $\frac{5}{2}z$.
Taking away 7 from $\frac{5}{2}z$ gives $\frac{5}{2}z - 7$.
The result is 23.
$\frac{5}{2}z - 7 = 23$
Solve:
Add 7 to both sides:
$\frac{5}{2}z - 7 + 7 = 23 + 7$
$\frac{5}{2}z = 30$
Multiply both sides by $\frac{2}{5}$ (or multiply by 2 and then divide by 5):
$\frac{5}{2}z \times \frac{2}{5} = 30 \times \frac{2}{5}$
$z = \frac{\cancel{30}^{6} \times 2}{\cancel{5}_{1}}$
$z = 6 \times 2$
$z = 12$
Answer:
The number Anwar thinks of is 12.
Question 2. Solve the following:
(a) The teacher tells the class that the highest marks obtained by a student in her class is twice the lowest marks plus 7. The highest score is 87. What is the lowest score?
(b) In an isosceles triangle, the base angles are equal. The vertex angle is 40°. What are the base angles of the triangle? (Remember, the sum of three angles of a triangle is 180°).
(c) Sachin scored twice as many runs as Rahul. Together, their runs fell two short of a double century. How many runs did each one score?
Answer:
Solution:
(a) Finding the lowest score:
Given:
Highest score = 87
Highest score = twice the lowest marks + 7
To Find:
The lowest score.
Let:
Let the lowest score be $l$.
Equation:
According to the statement, twice the lowest marks is $2l$.
Twice the lowest marks plus 7 is $2l + 7$.
This is equal to the highest score, which is 87.
So, the equation is:
$2l + 7 = 87$
Solve:
Subtract 7 from both sides:
$2l + 7 - 7 = 87 - 7$
$2l = 80$
Divide both sides by 2:
$\frac{2l}{2} = \frac{80}{2}$
$l = 40$
Answer:
The lowest score is 40.
(b) Finding the base angles of an isosceles triangle:
Given:
The triangle is isosceles, so base angles are equal.
Vertex angle = $40^\circ$.
Sum of angles in a triangle = $180^\circ$.
To Find:
The measure of each base angle.
Let:
Let each base angle be $b$ degrees.
Equation:
The angles of the triangle are $b^\circ$, $b^\circ$, and $40^\circ$.
The sum of the angles is $180^\circ$.
So, the equation is:
$b + b + 40 = 180$
Combining like terms:
$2b + 40 = 180$
Solve:
Subtract 40 from both sides:
$2b + 40 - 40 = 180 - 40$
$2b = 140$
Divide both sides by 2:
$\frac{2b}{2} = \frac{140}{2}$
$b = 70$
Answer:
Each base angle of the isosceles triangle is $\mathbf{70^\circ}$.
(c) Finding runs scored by Sachin and Rahul:
Given:
Sachin scored twice as many runs as Rahul.
Their combined score was two short of a double century (200 runs).
To Find:
Runs scored by Rahul and Sachin.
Let:
Let the number of runs scored by Rahul be $r$.
Since Sachin scored twice as many runs as Rahul, Sachin's runs $= 2r$.
Equation:
The total runs scored by them together is $r + 2r$.
A double century is 200 runs. Two short of a double century is $200 - 2 = 198$ runs.
So, the total runs are 198.
The equation is:
$r + 2r = 198$
Combining like terms on the left side:
$3r = 198$
Solve:
Divide both sides by 3:
$\frac{3r}{3} = \frac{198}{3}$
$r = 66$
So, Rahul scored 66 runs.
Sachin's runs $= 2r = 2 \times 66 = 132$.
Answer:
Rahul scored 66 runs and Sachin scored 132 runs.
Question 3. Solve the following:
(i) Irfan says that he has 7 marbles more than five times the marbles Parmit has. Irfan has 37 marbles. How many marbles does Parmit have?
(ii) Laxmi’s father is 49 years old. He is 4 years older than three times Laxmi’s age. What is Laxmi's age?
(iii) People of Sundargram planted trees in the village garden. Some of the trees were fruit trees. The number of non-fruit trees were 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:
Solution:
(i) Finding the number of marbles Parmit has:
Given:
Irfan has 7 marbles more than five times the marbles Parmit has.
Irfan has 37 marbles.
To Find:
The number of marbles Parmit has.
Let:
Let the number of marbles Parmit has be $m$.
Equation:
Five times the marbles Parmit has is $5m$.
7 marbles more than five times Parmit's marbles is $5m + 7$.
This is equal to the number of marbles Irfan has, which is 37.
So, the equation is:
$5m + 7 = 37$
Solve:
Subtract 7 from both sides of the equation:
$5m + 7 - 7 = 37 - 7$
$5m = 30$
Divide both sides by 5:
$\frac{5m}{5} = \frac{30}{5}$
$m = 6$
Answer:
Parmit has 6 marbles.
(ii) Finding Laxmi's age:
Given:
Laxmi’s father is 49 years old.
He is 4 years older than three times Laxmi’s age.
To Find:
Laxmi's age.
Let:
Let Laxmi's age be $y$ years.
Equation:
Three times Laxmi's age is $3y$ years.
4 years older than three times Laxmi's age is $3y + 4$ years.
This is equal to Laxmi's father's age, which is 49 years.
So, the equation is:
$3y + 4 = 49$
Solve:
Subtract 4 from both sides of the equation:
$3y + 4 - 4 = 49 - 4$
$3y = 45$
Divide both sides by 3:
$\frac{3y}{3} = \frac{45}{3}$
$y = 15$
Answer:
Laxmi's age is 15 years.
(iii) Finding the number of fruit trees:
Given:
Number of non-fruit trees = 77.
Number of non-fruit trees = two more than three times the number of fruit trees.
To Find:
The number of fruit trees planted.
Let:
Let the number of fruit trees be $f$.
Equation:
Three times the number of fruit trees is $3f$.
Two more than three times the number of fruit trees is $3f + 2$.
This is equal to the number of non-fruit trees, which is 77.
So, the equation is:
$3f + 2 = 77$
Solve:
Subtract 2 from both sides of the equation:
$3f + 2 - 2 = 77 - 2$
$3f = 75$
Divide both sides by 3:
$\frac{3f}{3} = \frac{75}{3}$
$f = 25$
Answer:
The number of fruit trees planted was 25.
Question 4. Solve the following riddle:
I am a number,
Tell my identity!
Take me seven times over
And add a fifty!
To reach a triple century
You still need forty!
Answer:
Solution:
Let the number be represented by the variable $N$.
According to the riddle:
"Take me seven times over" translates to $7N$.
"And add a fifty!" translates to $7N + 50$.
"To reach a triple century You still need forty!" means that the current value ($7N + 50$) is 40 less than 300 (a triple century).
So, the equation is:
$7N + 50 = 300 - 40$
$7N + 50 = 260$
Now, we solve the equation $7N + 50 = 260$ for $N$.
Subtract 50 from both sides:
$7N + 50 - 50 = 260 - 50$
$7N = 210$
Divide both sides by 7:
$\frac{7N}{7} = \frac{210}{7}$
$N = 30$
The number is 30.