CBSE Class 7 Maths Important Questions Chapter 4 - Simple Equations

Write equations for the following

1. Sum of the numbers \[\mathrm{P}\] and \[\mathrm{5}\] is \[\mathrm{11}\].

Ans: \[\text{P+5=11}\]

2. Two third of a number \[\mathrm{Z}\] subtracted with \[\mathrm{3}\] gives \[\mathrm{5}\].

Ans: \[\frac{2}{3}\text{Z}-3=5\]

3. Write statements for the following:

\[\mathrm{3q+8=15}\]

Ans: Eight is added to three times the number \[\text{q}\] gives \[\text{15}\].

4. \[\frac{\mathrm{3}}{\mathrm{4}}\mathrm{m=15}\]

Ans: Three fourth of a number \[\text{m}\]gives \[\text{15}\].

5. \[\mathrm{n-17=18}\]

Ans: \[\text{17}\] is subtracted from a number \[\text{n}\] gives \[\text{18}\].

6. Check whether the value given in brackets is Solution to the given equation or not

a. \[\mathrm{n-9=19}\]\[\left( \mathrm{n=10} \right)\]

Ans:

Putting \[\text{n}=10\] in LHS

\[\begin{align} & \text{n}-9=10-9=1 \\ & 1\ne 9 \\ \end{align}\]

\[\text{n}=10\] is not a solution for the given equation \[\text{n}-9=19\].

b. \[\mathrm{17n+5=25}\]\[\left( \mathrm{n=2} \right)\]

Ans:

Putting \[\text{n}=2\] in LHS

\[\begin{align} & \text{17n}+5=34+5=39 \\ & 39\ne 25 \\ \end{align}\]

\[\text{n}=2\] is not a solution for the given equation \[\text{17n}+5=25\].

7. Give the first step to solve the following equations and then solve the equation.

a. \[\mathrm{P+9=3}\]

Ans:

Subtracting \[9\] from both sides, we obtain

\[\begin{align} & \text{P}+9=3 \\ & \text{P}+9-9=3-9 \\ & \text{P}=-6 \\ \end{align}\]

b. \[\mathrm{a-12=15}\]

Ans:

Adding \[12\] both sides, we obtain

\[\begin{align} & \text{a}-12+12=15+12 \\ & \text{a}=27 \\ \end{align}\]

8. Solve the equation \[\mathrm{3z+}\frac{\mathrm{1}}{\mathrm{3}}\mathrm{=}\frac{\mathrm{17}}{\mathrm{27}}\] .

Ans:

\[\begin{align} & \text{3z}+\frac{1}{3}=\frac{17}{27} \\ & \text{3z}=\frac{17}{27}-\frac{1}{3} \\ & 3\text{z}=\frac{17-9}{27} \\ & \text{z}=\frac{\frac{8}{27}}{3} \\ & \text{z}=\frac{8}{27}\times \frac{1}{3} \\ & \text{z}=\frac{8}{81} \\ \end{align}\]

9. Solve \[\mathrm{2}\left( \mathrm{P+9} \right)\mathrm{=-4}\]

Ans:

\[\begin{align} & \text{2p}+18=-4 \\ & \text{2p}=-4-18 \\ & \text{2p}=-22 \\ & \text{p}=-\frac{22}{2} \\ & \text{p}=-11 \\ \end{align}\]

10. Write an equation and solve the digit \[\mathrm{5}\] is added to \[\mathrm{9}\] times a number: gives you \[\mathrm{77}\].

Ans:

Let the number \[\text{x}\]

\[\text{9}\] times the number \[=9\text{x}\]

\[\begin{align} & \text{9x}+5=77 \\ & \text{9x}=77-5 \\ & \text{9x}=72 \\ & \text{x}=\frac{72}{9} \\ & \text{x}=8 \\ \end{align}\]

11. In an isosceles triangle base angles are equal if the vertex angle is \[\mathrm{5}{{\mathrm{0}}^{\mathrm{0}}}\]. What are base angles?

Ans:
Let us assume the base angles of the isosceles  triangle as \[\text{x}\].

We know that

\[\begin{align} & \text{x}+\text{x}+{{50}^{0}}={{180}^{0}} \\ & 2\text{x}+{{50}^{0}}={{180}^{0}} \\ & 2\text{x}={{130}^{0}} \\ & \text{x}=\frac{{{130}^{0}}}{2} \\ & \text{x}={{65}^{0}} \\ \end{align}\]

12. If Manvita’s age is year less than the thirteen times of her father’s age. Find her age if her father’s age is \[\mathrm{32}\]years.

Ans:

Let’s Manvita’s age \[\mathrm{=}\text{x}\]years.

\[\begin{align} & 13\text{x}-\frac{1}{2}=32 \\ & 13\text{x}=32+\frac{1}{2} \\ & 13\text{x}=\frac{64+1}{2} \\ & \text{x}=\frac{65}{2}\times \frac{1}{13} \\ & \text{x}=\frac{5}{2} \\ & \text{x}=2.5\text{ years} \\ \end{align}\]

13. Solve \[\frac{\mathrm{a}}{\mathrm{6}}\mathrm{=}\frac{\mathrm{8}}{\mathrm{18}}\].

Ans:

\[\begin{align} & \frac{\text{a}}{6}=\frac{8}{18} \\ & \text{a}=\frac{8}{18}\times 6 \\ & \text{a}=\frac{48}{18} \\ & \text{a}=\frac{8}{3} \\ & \text{a=2}\frac{2}{3} \\ \end{align}\]

14. Solve \[\mathrm{0=18+3}\left( \mathrm{m+12} \right)\].

Ans:

\[\begin{align} & 0=18+3\left( \text{m}+12 \right) \\ & 3\text{m}=-54 \\ & \text{m}=-\frac{54}{3} \\ & \text{m}=-18 \\ \end{align}\]

15. Solve \[\mathrm{5+6}\left( \mathrm{q-1} \right)\mathrm{=36}\].

Ans:

\[\begin{align} & 5+6\left( \text{q}-1 \right)=36 \\ & 5+6\text{q}-6=36 \\ & 6\text{q}=36+1 \\ & 6\text{q}=37 \\ & \text{q}=\frac{37}{6} \\ & \text{q}=6\frac{1}{6} \\ \end{align}\]

16. Construct \[\mathrm{3}\] equations starting with \[\mathrm{x=5}\].

Ans:

\[\text{X=5}\]

Multiplying both sides by \[\text{5}\].

\[\text{5x}=25..........\left( i \right)\]

Subtract \[5\] on both sides

\[5\text{x}-5=25-5\]

\[5\text{x}-5=20.................\left( ii \right)\]

Divide both sides by \[2\]

\[\frac{5\text{x}-2}{2}=10\]

\[\frac{5\text{x}}{2}-\frac{5}{2}=10..............\left( iii \right)\]

Here \[\left( i \right),\left( ii \right)\] and \[\left( iii \right)\] are three equations.

17. Solve \[\mathrm{3m-18=6}\] by trial and error method.

Ans:

Put \[\text{m}=1\]

\[\begin{align} & 3\left( 1 \right)-18=6 \\ & 3-18=6 \\ & -15\ne 6 \\ \end{align}\]

Put \[\text{m}=2\]

\[\begin{align} & 3\left( 2 \right)-18=6 \\ & 6-18=6 \\ & -12\ne 6 \\ \end{align}\]

Put \[\text{m}=6\]

\[\begin{align} & 3\left( 6 \right)-18=6 \\ & 18-18=6 \\ & 0\ne 6 \\ \end{align}\]

Put \[\text{m}=8\]

\[\begin{align} & 3\left( 8 \right)-18=6 \\ & 24-18=6 \\ & 6=6 \\ \end{align}\] \[\text{m}=8\] is the solution of the given equation.

18. Write equation and solve

a. Add \[\mathrm{4}\] to one fourth of a number gives \[\mathrm{20}\].

Ans:

Let the number be \[\text{x}\].

\[\frac{1}{4}\text{x}+4=20\]

Transferring \[\text{4}\] from LHS to RHS.

\[\frac{1}{4}\text{x}=\text{20}-4\]

\[\frac{1}{4}\text{x}=16\]

Cross multiplying with \[\text{4}\] on both sides, we get

\[\text{x}=64\]

b. If you take away \[\mathrm{5}\] from \[\mathrm{5}\] times a number you get \[\mathrm{50}\].

Ans:

Let the number be \[\text{y}\].

\[\text{5y}-5=50\]

Transferring \[\text{5}\] from LHS to RHS.

\[5\text{y}=50+5\]

\[5\text{y}=55\]

Dividing by \[\text{5}\] on both sides.

\[\text{y}=\frac{55}{5}\]

\[\text{y}=11\]

19. Solve

a. \[\frac{\mathrm{5p}}{\mathrm{6}}\mathrm{=5}\]

Ans:

\[\frac{\text{5p}}{6}=30\]
Multiplying with  \[\text{6}\] on both sides.

\[\text{5p}=30\]

Dividing by \[\text{5}\]on both sides

\[\text{p}=\frac{30}{5}\]

\[\text{p}=6\]

b. \[\mathrm{2x+12=28}\]

Ans:

\[2\text{x}+12=28\]

Transferring \[12\] from LHS to RHS.

\[2\text{x}=28-12\]

\[\text{2x=16}\]

Dividing by \[2\]on both sides

\[\text{x=8}\]

20. Sum of \[\mathrm{5}\] times a number and \[\mathrm{12}\] is \[\mathrm{47}\]. Find the number.

Ans:

Let the number be \[\text{x}\].

\[\text{5x}+12=47\]

Transferring \[12\] from LHS to RHS.

\[5\text{x}=47-12\]

\[\text{5x}=35\]

Dividing by \[\text{5}\] on both sides

\[\text{x}=\frac{35}{5}\]

\[\text{x}=7\]

CBSE Class 7 Maths Chapter - 4 Important Questions - Free PDF Download

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Important Questions For Class 7 Maths Chapter 4

1. Which of the Following is Not Allowed in a Given Equation?

(a) Adding the same number to both sides of the equation.

(b) Subtracting the same number from both sides of the equation.

(c) Multiplying both sides of the equation by the same non-zero number.

(d) Dividing both sides of the equation by the same number.

2. The Solution of Which of the Following Equations is Neither a Fraction Nor an Integer?

(a) 2x + 6 = 0

(b) 3x – 5 = 0

(c) 5x – 8 = x + 4

(d) 4x + 7 = x + 2

3. If 7x + 4 = 25, then x is Equal to

(a) 29/7 (b) 100/7 (c) 2 (d) 3

4. If k + 7 = 16, then the Value of 8k – 72 is

(a) 0 (b) 1 (c) 112 (d) 56

5. The Equation Having 5 as a Solution is:

(a) 4x + 1 = 2 (b) 3 – x = 8 (c) x – 5 = 3 (d) 3 + x = 8

5 Important Points for Class 7 Chapter 4 Simple Equations You Shouldn’t Miss!

Understanding simple equations is essential for solving various mathematical problems. Here are five important points from Chapter 4 that every Class 7 student should know:

1. General Form of a Linear Equation:

A simple equation can be expressed in the form:

$ax + b = c$

where $a$, $b$, and $c$ are constants, and $x$ is the variable to be solved.

2. Isolating the Variable:

To solve for $x$, rearrange the equation to isolate the variable on one side:

$x = \dfrac{c - b}{a}$

3. Substitution Method:

If you have two equations, you can substitute one equation into the other to find the value of the variable:

Given $y = mx + b$ and $ax + by = c$, substitute $y$ into the second equation to find $x$.

4. Balancing Method:

To maintain equality, whatever you do to one side of the equation, do the same to the other side. For example, if you add a number to one side:

$ax + b + d = c + d$

5. Verification of Solutions:

Always check your solution by substituting the value back into the original equation to ensure both sides are equal. If $ax + b = c$ holds true, then $x$ is the correct solution.

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Conclusion

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Important Study Materials for Class 7 Maths Chapter 4

CBSE Class 7 Maths Important Questions for All Chapters

Class 7 Maths Important Questions and Answers cover key topics, aiding in thorough preparation and making revision simpler.

Important Study Materials for Class 7 Maths