NCERT Solutions for Class 7 Mathematics Chapter 4 - Simple Equations

Answer:

In the following we find the value of expression (10y – 20) by taking 5 different values of y.
When y = 1; 10y – 20 = 10 × 1 – 20 = 10 – 20 = – 10
When y = 2; 10y – 20 = 10 × 2 – 20 = 20 – 20 = 0
When y = 3; 10y – 20 = 10 × 3 – 20 = 30 – 20 = 10
When y = 4; 10y – 20 = 10 × 4 – 20 = 40 – 20 = 20
When y = 5; 10y – 20 = 10 × 5 – 20 = 50 – 20 = 30

From the above table; We observe that for each value of variable y there corresponds to a unique value of expression (10y – 20).
Thus, it is verified that the value of expression (10y – 20) depends on the value of y.
For 5 different values of y, we could not find the solution to the equation 10y – 20 = 50.
Let us try for more values to y.
When y = 6; 10y – 20 = 10 × 6 – 20 = 60 – 20 = 40
When y = 7; 10y – 20 = 10 × 7 – 20 = 70 – 20 = 50
Since equation 10y – 20 = 50 is satisfied at y = 7. Hence by trial and error method y = 7 is the solution to the equation 10y – 20 = 50.

Answer:

Statement forms for equation 5p = 20 are as follows:
1. If you multiply a number p by 5; you get 20.
2. A number p multiplied by 5 gives 20.

Answer:

Statement forms for equation 3n + 7 = 1 are as follows:
1. The sum three times n and 7 is 1.
2. Three times a number n plus 7 gives 1.
3. If you add 7 to a number n multiplied by 3; you get 1.

Answer:

Statement forms for equation m/5 – 2 = 6 are as follows:
1. If you subtract 23 from a number m divided by 5; you get 6.
2. One fifth of m is greater by 6 than 2.
3. The difference between one fifth of a number m and 2 is 6.

Answer:

No.
Reason: For x = 3; x + 3 = 3 + 3 = 6
[Put x = 3; on the LHS of the equation x + 3 = 0]
But result on the RHS of the equation x + 3 = 0 is 0.
Since, 6 $\ne$ 0.
Therefore, the given equation is not satisified for x = 3.

Answer:

No.
Reason: For x = 0; x + 3 = 0 + 3 = 3
[Put x = 0; on the LHS of the equation x + 3 = 0]
But result on the RHS of the equation x + 3 = 0 is 0.
Since, 3 $\ne$ 0.
i.e. LHS $\ne$ RHS
Therefore, equation is not satisfied for x = 0.

Answer:

Yes.
Reason: For x = –3; x + 3 = – 3 + 3 = 0
[Put x = – 3; on the LHS of the equation x + 3 = 0]
Result on the RHS of the equation x + 3 = 0 is also 0.
Since, 0 = 0.
Therefore, given equation is satisfied for x = –3.

Answer:

No.
Reason: For x = 7; x – 7 = 7 – 7 = 0
[Put x = 7; on the LHS of the equation x – 7 = 1]
But result on the RHS of the equation x – 7 = 0 is 1.
Since, 0 $\ne$ 1.
i.e. LHS $\ne$ RHS
Therefore, the given equation is not satisfied for x = 7.

Answer:

Yes.
Reason: For x = 8; x – 7 = 8 – 7 = 1
[Put x = 8; on the LHS of the equation x – 7 = 1]
Result on the RHS of the equation x – 7 = 1 is also 1.
Since, 1 = 1.
Therefore, the given equation is satisfied for x = 8.

Answer:

No.
Reason: For x = 0; 5x = 5 × 0 = 0
[Put x = 0; on the LHS of the equation 5x = 25]
But result on the RHS of the equation 5x = 25 is 25.
Since, 0 $\ne$ 25
Therefore, the given equation is not satisfied for x = 0.

Answer:

Yes.
Reason: For x = 5; 5x = 5 × 5 = 25
[Put x = 5; on the LHS of the equation 5x = 25]
Result on the RHS of the equation is also 25.
Since, 25 = 25.
Therefore, the given equation is satisfied for x = 5.

Answer:

No.
Reason: For x = –5; 5x = 5 × (–5) = –25
[Put x = –5; on the LHS of the equation 5x = 25]
But result on the RHS of the equation 5x = 25 is 25.
Since, –25 $\ne$ 25.
i.e. LHS $\ne$ RHS
Therefore the given equation is not satisfied for x = –5.

Answer:

No.
Reason: For m = –6; m/3 = − 6/3 = – 2
[Put m = –6; on the LHS of the equation m/3 = 2]
But result on the RHS of the equation m/3 = 2 is 2.
Since, –2 $\ne$ 2.
Therefore, the given equation is not satisfied for m = –6.

Answer:

No.
Reason: For m = 0; m/3 = 0/3 = 0
[Put m = 0; on the LHS of the equation m/3 = 2]
But result on the RHS of the equation m/3 = 2 is 2.
Since, 0 $\ne$ 2.
Therefore, the given equation is not satisfied for m = 0.

Answer:

Yes.
Reason: For m = 6; m/3 = 6/3 = 2
[Put m = 6; on the LHS of the equation m/3 = 2]
Result on the RHS of the equation m/3 = 2 is also 2.
Since, 2 = 2.
i.e. LHS = RHS
Therefore, the given equation is satisfied for m = 6.

Answer:

Equation n + 5 = 19 is a condition on variable n.
It states that the value of the expression n + 5 is 19.
When n = 1; n + 5 = 1 + 5 = 1 + 5 = 6
[Put n = 1 in LHS of equation 7n + 5 = 19]
Since, 12 $\ne$ 6.
i.e. LHS $\ne$ RHS
It implies that equation n + 15 = 19 is not satisfied when n = 1.
Hence, n = 1 is not a solution to the given equation.

Answer:

Equation 7n + 5 = 19 is a condition on variable n.
It states that the value of the expression 7n + 5 is 19.
When n = –2; 7n + 5 = 7 (–2) + 5 = –14 + 5 = –9
[Put n = –2 in LHS of equation n + 5 = 19]
Since, –9 $\ne$ 19
i.e. LHS $\ne$ RHS
It implies that equation 7n + 15 = 19 is not satisified when n = –2.
Hence, n = – 2 is not a solution to the given equation.

Answer:

Equation 7n + 5 = 19 is a condition on variable n.
It states that the value of the expression 7n + 5 is 19.
When n = 2; 7n + 5 = 7 × 2 + 5 = 14 + 5 = 19
[Put n = 2 in LHS of equation 7n + 5 = 19]
Since, 19 = 19
It implies that the equation 7n + 15 = 19 is satisfied when n = 2.
Hence n = 2 is a solution to the given equation.

Answer:

Equation 4p – 3 = 13 is a condition on variable p.
It states that the value of the expression 4p – 3 is 13.
When p = 1; 4p – 3 = 4 × 1 – 3 = 4 – 3 = 1
[Put p = 1 in LHS of equation 4p – 3 = 13]
Since, 1 $\ne$ 13.
i.e. LHS $\ne$ RHS
It implies that equation 4p + 3 = 13 is not satisfied when p = 1.
Hence, p = 1 is not a solution to the given equation.

Answer:

Equation 4p – 3 = 13 is a condition on variable p.
It states that the value of the expression 4p – 3 is 13.
When p = –4; 4p – 3 = 4 (–4) – 3 = –16 – 3 = –19
[Put p = –4 in LHS of equation 4p – 3 = 13]
Since, –19 $\ne$ 13.
i.e. LHS $\ne$ RHS
It implies that equation 4p + 3 = 13 is not satisfied when p = – 4.
Hence, and p = – 4 is not a solution to the given equation.

Answer:

Equation 4p – 3 = 13 is a condition on variable p.
It states that the value of the expression 4p – 3 is 13.
When p = 0; 4p – 3 = 4 × 0 – 3 = 0 – 3 = –3
[Put p = 0 in LHS of equation 4p – 3 = 13]
Since, –3 $\ne$ 13.
It implies that equation 4p – 3 = 13 is not satisfied when p = 0.
Hence, and p = 0 is not a solution to the given equation.

Answer:

To find the solution of equation 5p + 2 = 17 by trial and error method; we give different values to variable p.
These values could be anything 1, 2, 3, 4, . . .
Then we find different values of expression 5p + 2 corresponding to the assumed different values of variable p.
These are given in the following table as follows:

We observe that equation 5p + 2 = 17 is satisfied at p = 3.
Hence, p = 3 is a solution of the given equation.

Answer:

To find the solution of equation 3m – 14 = 4 by trial and error method; we give different values to variable m.
These values could be anything 1, 2, 3, 4, . . .
Then we find different values of expression 3m – 14 corresponding to the assumed different values of variable m.
These are given in the following table as follows:

We observe that equation 3m – 14 = 4 is satisfied at m = 6.
Hence, m = 6 is a solution of the given equation.

Answer:

Sum of x and 4 is x + 4.
The sum is 9.
Therefore, the required equation is x + 4 = 9.

Answer:

Difference of y and 2 is y – 2.
The difference is 8.
Therefore, the required equation is y – 2 = 8.

Answer:

Ten times a is 10a.
It is 70.
Therefore, the required equation is 10a = 70.

Answer:

The number b divided by 5 is b/5.
It is 6
Therefore, the required equation is b/5 = 6.

Answer:

Three fourth of t is 3/4 t
It is 15
Therefore, the required equation is 3/4 t = 15.

Answer:

m plus 7 means we add 7 to m.
One gets m + 7
7 times m plus 7 is 7 (m + 7)
The result is 77.
Therefore; the required equation is 7 (m + 7) = 77.

Answer:

Take a number to be n.
One fourth of n is n/4
One fourth of the number minus 4 is n/4 – 4
The result is 4.
Therefore; the required equation is n/4 – 4 = 4.

Answer:

6 times y is 6y.
To take away 6 from 6 times y means subtracting 6 from 6y.
One gets 6y – 6
The result is 68.
Therefore, the required equation is 6y – 6 = 60.

Answer:

One third of z is n/4 z.
Addition of 3 to one third of z is 1/3 z + 3.
Therefore, the result is 30.
Therefore; the required equation is 1/3 z + 3 = 30.

Answer:

The sum of numbers p and 4 is 15.

Answer:

Taking away 7 from m gives 3.

Answer:

Twice a number m is 7.

Answer:

One-fifth of a number m is 3.

Answer:

Three fifth of a number m is 6.

Answer:

Three times p when added to 4 gives you 25.

Answer:

2 subtracted from four times a number p is 18.

Answer:

Add 2 to half of a number p to get 8.

Answer:

Take m to be the number of Parmit’s marble.
Five times m is 5m
7 more than five times m = 5m + 7
But 7 more than five times m = Number of Irfan’s marbles
Therefore, 5m + 7 = 37, which is an equation in variable m.

Answer:

Take y years to be the Laxmi’s age.
Three times Laxmi’s age = 3y
4 years older than three times Laxmi’s age = 3y + 4
But 4 years older than three times Laxmi’s age = age of Laxmi’s father
Therefore 3y + 4 = 49, which is an equation in variable y.

Answer:

Take the lowest marks to be l.
Twice the lowest marks = 2l
Twice the lowest marks plus 7 = 2l + 7
But twice the lowest marks plus 7 = Highest marks obtained by a student.
Therefore, 2l + 7 = 87, which is an equation in variable l.

Answer:

In an isosceles triangle, let b be either base angle in degree.
$\therefore$ Vertical angle = Twice the base angle = 2b
Sum of the angles of triangle = 180$°$
Therefore, b + b + 2b = 180$°$
or, 4b = 180$°$, which is an equation in variable b.

Answer:

x – 1 = 0 is an equation in variable x.
The LHS is x – 1.
In the first step, we shall add 1 to it to get x.
Therefore, adding 1 to both sides of given equation, we get:
x – 1 + 1 = 0 + 1      (Step 1)
or, x = 1; which is the required solution.

Answer:

x + 1 = 0 is an equation in variable x.
The LHS is x + 1.
In the first step; we shall subtract 1 from it to get x.
Therefore, subtracting 1 from both sides of given equation, we get:
x + 1 – 1 = 0 – 1      (Step 1)
or, x = – 1; which is the required solution.

Answer:

x – 1 = 5 is an equation in variable x.
The LHS is x – 1.
In the first step; we shall add 1 to it to get x.
Therefore, adding 1 to both sides of given equation, we get:
x – 1 + 1 = 5 + 1     (Step 1)
or, x = 6; which is the required solution.

Answer:

x + 6 = 2 is an equation in variable x.
The LHS is x + 6.
In the first step; we shall subtract 6 from it to get x.
Therefore, subtracting 6 from both sides of given equation, we get:
x + 6 – 6 = 2 – 6       (Step 1)
or, x = –4; which is the solution.

Answer:

y – 4 = – 7 is an equation in variable y.
The LHS is y – 4.
In the first step; we shall add 4 to it to get y.
Therefore, adding 4 to both sides of the given equation, we get:
y – 4 + 4 = – 7 + 4       (Step 1) or, y = –3; which is the solution.

Answer:

y – 4 = 4 is an equation in variable.
The LHS is y – 4.
In the first step; we shall add 4 to it to get y.
Therefore, adding 4 to both sides of given equation, we get:
y – 4 + 4 = 4 + 4       (Step 1) or, y = 8; which is the solution.

Answer:

y + 4 = 4 is an equation in variable y.
The LHS is y + 4.
In the first step; we shall subtract 4 from it to get y.
Therefore; subtracting 4 from both sides of given equation, we get:
y + 4 – 4 = 4 – 4       (Step 1)
or, y = 0; which is the solution.

Answer:

y + 4 = –4 is an equation is variable y.
The LHS is y + 4.
In the first step; we shall subtract 4 from it to get y.
Therefore, subtracting 4 from both sides of given equation, we get:
y + 4 – 4 = – 4 – 4
or, y = –8; which is the solution.

Answer:

3l = 42 is an equation in variable l.
The LHS is 3l.
In the first step; we shall divide it by 3 to get l.
Therefore, divide both sides by 3; this will separate l.

or l = 14; which is the solution.

Answer:

Answer:

Answer:

4x = 25 is an equation in variable x.
The LHS is 4x. In first step; we divide it by 4 to get x.
Therefore, divide both sides by 4; this will separate x.

Answer:

8y = 36 is an equation in variable y.
The LHS is 8y. In the first step; we divide it by 8 to get y.
Therefore, divide both sides by 8; this will sepatate y.

Answer:

Answer:

Answer:

20t = –10 is an equation in variable t.
The LHS is 20t. In the first step; we divide it by 20 to get t.
Therefore, divide both sides by 20; this will separate t.

Answer:

3n – 2 = 46 is an equation in variable n.
Add 2 to both sides, we get:
3n – 2 + 2 = 46 + 2       (Step 1)
or       3n = 48
Divide both sides by 3; this will separate n.

Answer:

5m + 7 = 17 is an equation in variable m. We go stepwise to separate the variable m on the LHS of the equation. The LHS is 5m + 7. We shall first subtract 7 from it so that we get 5m. From this; in the next step we shall divide by 5 to get m.
Now we do the same operations on both sides of the equation.
Therefore; subtracting 7 from both sides, we get:
5m + 7 – 7 = 17 – 7       (Step 1)
or,       5m = 10
Now; divide both sides by 5, we get:

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Answer:

3s = –9 is an equation in variable s. Divide both sides by 3; this will separate variable s.

or s = –3; which is the solution.

Answer:

3s + 12 = 0 is an equation in variable s.
Subtract 12 from both sides, we get:
3s + 12 – 12 = 0 –12
or       3s = – 12
Divide both sides by 3 this will separate s.

or      s = – 4; which is the solution.

Answer:

3s = 0 is an equation in variable s.
Divide both sides by 3; this will separate s.

or      s = 0; which is the solution.

Answer:

2q = 6 is an equation in variable q.
Divide both sides by 2; this will separate q.

or q = 3; which is the solution.

Answer:

2q – 6 = 0 is an equation in variable q.
Add 6 to both sides, we get:
2q – 6 + 6 = 0 + 6
or      2q = 6
Divide both sides by 2; this will separate q.

or q = 3; which is the solution.

Answer:

2q + 6 = 0 is an equation in variable q.
Subtract 6 from both sides, we get:
2q + 6 – 6 = 0 – 6
or      2q = –6
Divide both sides by 2; this will separate q.

or q = –3; which is the solution.

Answer:

2q + 6 = 12 is an equation in variable q.
Subtract 6 from both sides, we get:
2q + 6 – 6 = 12 – 6
or      2q = 6
Divide both sides by 2; this will separate q.

or q = 3; which is the solution.

Answer:

Let us follow the path to make a number puzzle in one variable with the solution 11.

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Answer:

2(x + 4) = 12 is an equation in variable x. Divide both the sides by 2. This will remove the brackets in the LHS, we get:

Answer:

3(n – 5) = 21 is an equation in variable n. Divide both the sides by 3. This will remove the brackets in the LHS, we get:

or n – 5 = 7
or n = 7 + 5 (Transposing – 5 to RHS)
or n = 12; which is the solution.
Check:
Putting n = 12; in the LHS of equation
3 (n – 5) = 21.
LHS = 3 (n – 5)
= 3 (12 – 5)
= 3 × 7
= 21 = RHS as required.

Answer:

3(n – 5) = – 21 is an equation in variable n.
Divide both the sides by 3. This will remove the brackets in the LHS, we get:

Answer:

– 4 (2 + x) = 8 is an equation in variable x.
Dividing both sides by –4, we get:

Answer:

4 (2 – x) = 8 is an equation in variable x.
Divide both the sides by 4. This will remove the bracket in the LHS; we get:

or 2 – x = 2
or – x = 2 – 2 (Transposing 2 to RHS)
or – x = 0
or x = 0; which is the solution.
Check:
Putting x = 0; in the LHS of the equation
4 (2 – x) = 8
LHS = 4(2 – 0)
= 4(2)
= 8 = RHS as required.

Answer:

4 = 5(p – 2), is an equation in variable p
or 4 = 5p – 10
– 5p = – 10 – 4
[Transposing 5p to LHS and 4 to RHS]
or – 5p = – 14
Divide both the sides by – 5; this will separate variable p.

Answer:

–4 = 5(p – 2), is an equation in variable p.
Divide both the sides by 5; this will remove the brackets in the RHS, we get:

Answer:

16 = 4 + 3(t + 2) is an equation in variable t.
or 16 – 4 = 3(t + 2) (Transposing 4 to LHS)
or 12 = 3(t + 2)
Dividing both side by 3, we get:

Answer:

4 + 5(p – 1) = 34 is an equation in variable p.
or 5(p – 1) = 34 – 4 (Transposing 4 to RHS)
or 5(p – 1) = 30
Dividing both side by 5, we get:

Answer:

0 = 16 + 4 (m – 6) is an equation in variable m.
Transpose expression 4 (m – 6) to LHS, we get:
0 – 4 (m – 6) = 16
or – 4 (m – 6) = 16
Divide both the sides by – 4. This will remove the brackets on the LHS, we get:

or m – 6 = – 4
or m = – 4 + 6 (Transpose – 6 to RHS)
or m = 2; which is the solution.
Check:
Putting m = 2; in the RHS of given
equation,
RHS = 16 + 4 (2 – 6)
= 16 + 4 ( – 4)
= 16 – 16
= 0 = LHS as required.

Answer:

To build up the first equation;
Start with x = 2
Multiply both 3x = 6
sides by 3;
Add 7 to both sides, 3x + 7 = 13
This has resulted in an equation.
To build up the second equation;
Start with x = 2
Subtract 3 from both
the sides, x – 3 = – 1
Multiply both the sides
by 4, 4 (x – 3) = – 4
Add 11 to both the 4 (x – 3) + 11 = 7
sides,
This has resulted in an equation.
To build up the third equation;
Start with x = 2
Multiply both
the sides by –1, x (– 1) = 2 (– 1)
– x = – 2
Add 9 to both sides, – x + 9 = – 2 + 9
or 9 – x = 7
This has resulted in equation.

Answer:

To build up the first equation;
Start with x = –2
Add 8 to the both
sides, x + 8 = 6
Multiply both the
sides by 5, 5 (x + 8) = 30
This has resulted in an equation.
To build up the second equation;
Start with x = – 2
Subtract 6
from the both sides, x – 6 = – 2 – 6
or x – 6 = – 8
Divide both ½(x – 6) = –4
the sides by 2,
This has resulted in an equation.
To build up the third equation;
Start with x = – 2
Multiply both 9x = – 18
sides by 9,
Divide both 9/2 x = – 9
sides by 2,
Add 5 to both 9/2 x + 5 = – 4
the sides,
This has resulted in an equation.

Answer:

Let, required number be x
Multiplying number with 6, we get 6x
By subtracting 5 from the product, we get 7
So, equation becomes
6x – 5 = 7 ... (1)
On transposing 5, we get
6x = 7 + 5
or 6x = 12
On dividing by 6, we get
or x = 12/6 = 2
$\therefore$ Required number be 2.
Check:
Put x = 2 in LHS of (1), we get
LHS = 6 (2) – 5
= 12 – 5 = 7
= RHS
Hence, x = 2 is verified.

Answer:

Let required number be x
One third of number added to 5 gives 8
So, equation becomes
x/3 + 5 = 8 ... (1)
On transposing 5, we get
x/3 = 8 – 5
or x/3 = 3
Multiplying by 3, we get
or x/3 × 3 = 3 × 3
or x = 9
$\therefore$ Required number be 9.
Check:
Put x = 9 in LHS of (1), we get
LHS = 9/3 + 5
= 3 + 5 = 8
= RHS
Hence, x = 9 is verified.

Answer:

Let, number of mangoes is smaller type box be x Number of mangoes in larger type box are 4 more than the number of mangoes contained in 8 boxes of the smaller type. So, equation becomes
8x + 4 = 100 ... (1)
On transposing 4, we get
8x = 100 – 4
or 8x = 96
Dividing both sides by 8, we get

Answer:

Let required number be x.
Add 4 to eight times the number, we get 60.
$\therefore$ 8x + 4 = 60 ... (1)
On transposing 4, we get
8x = 60 – 4
8x = 56
On dividing both sides by 8, we get

or x = 7
Hence, required number be 7.
Check:
Put x = 7 is LHS of (1), we get
LHS = 8 (7) + 4
= 56 + 4
= 60 = RHS
Hence, x = 7 is verified.

Answer:

Let required number be x.
One-fifth of the number minus 4 gives 3,
So, equation becomes

Answer:

Let required number be x.
On taking three-fourth of the number and
count up 3 more, we get 21.
So, equation becomes

Answer:

Let required number be x.
On subtracting 11 from twice the number,
we get 15
So, equation becomes
2x – 11 = 15 ... (1)
On transposing 11, we get
2x = 15 + 11
or 2x = 26
On dividing both sides by 2, we get

Answer:

Let required number be x.
On subtracting 11 from twice the number,
we get 15
So, equation becomes
2x – 11 = 15 ... (1)
On transposing 11, we get
2x = 15 + 11
or 2x = 26
On dividing both sides by 2, we get

or x = 13
Hence, required number be 13.
Check:
Put x = 13 in LHS of (1), we get
LHS = 2 (13) – 11
= 26 – 11
= 15 = RHS
Hence, x = 13 is verified.

Answer:

Let number of note books Munna has x.
On subtracting thrice the number of note
books from 50, he gets 8.
So, equation becomes
50 – 3x = 8 ... (1)
On transposing 50, we get
– 3x = 8 – 50
or – 3x = – 42
On dividing both sides by – 3, we get

or x = 14
Hence, required number of note books are 14.
Check:
Put x = 14 in LHS of (1), we get
LHS = 50 – 3 (14)
= 50 – 42
= 8 = RHS
Hence, x = 14 is verified.

Answer:

Let required number be x.
If we add 19 to it and divide the sum by 5, we get 8
So, equation becomes

Answer:

Answer:

Let the lowest marks in the class be x.
If we twice the lowest marks plus 7 we score 87 marks
So, equation becomes
2x + 7 = 87 ... (1)
On transposing 7, we get
2x = 87 – 7
2x = 80
Dividing both sides by 2, we get

or x = 40
Hence, lowest marks in the class is 40.
Check:
Put x = 40 in LHS of (1), we get
LHS = 2 (40) + 7
= 80 + 7
= 87 = RHS
Hence, x = 40 is verified.

Answer:

Let ABC be an isosceles triangle whose base angles are equal and of measure x$°$.
Also, vertex angle is 40$°$.
Since, sum of three angles of a triangle is 180$°$.
$\angle$A + $\angle$B + $\angle$C = 180$°$
or 40$°$ + x$°$ + x$°$ = 180$°$
or 40$°$ + 2x$°$ = 180$°$ ... (1)

Hence, base angles are of measure 70$°$.
Check:
Put x = 70$°$ in LHS of (1), we get
LHS = 40$°$ + 2 (70$°$)
= 40$°$ + 140$°$
= 180$°$ = RHS
Hence, x = 70$°$ is verified.

Answer:

Let runs scored by Rahul be x and runs scored by Sachin be 2x.
Since, the sum of their runs be two short of a double century.
So, equation becomes
x + 2x + 2 = 200
3x + 2 = 200 ... (1)
On trasnsposing 2, we get
3x = 200 – 2
or 3x = 198
Dividing both sides by 3, we get

or x = 66
Hence, runs scored by Rahul be 66 and by Sachin be 132.
Check:
Put x = 66 in LHS of (1), we get
LHS = 3 (66) + 2
= 198 + 2
= 200 = RHS
Hence, x = 66 is verified.

Answer:

Let, number of marbles Parmit has x.
If five times number of marbles of Parmit is added to 7, we get 37 marbles.
So, equation becomes
5x + 7 = 37 ... (1)
On transposing 7, we get
5x = 37 – 7
or 5x = 30
On dividing both sides by 5, we get

or x = 6
Hence, 6 marbles have Parmit.
Check:
Put x = 6 in LHS of (1), we get
LHS = 5 (6) + 7
= 30 + 7
= 37 = RHS
Hence, x = 6 is verified.

Answer:

Let Laxmi’s age be x years.
Laxmi’s father is 49 years old is 4 years older than three times Laxmi’s age.
So, equation becomes
3x + 4 = 49 ... (1)
On transposing 4, we get
3x = 49 – 4
3x = 45
On dividing both sides by 3, we get

or x = 15
Hence, Laxmi’s age be 15 years.
Check:
Put x = 15 in LHS of (1), we get
LHS = 3 (15) + 4
= 45 + 4
= 49 = RHS
Hence, x = 15 is verified.

Answer:

Let number of fruit trees planted be x.
Sum of fruit trees and three times the number of fruit trees are two less from 102.
So, equation becomes
x + (3x + 2) = 102
4x + 2 = 102 ... (1)
On transposing 2, we get
4x = 102 – 2
or 4x = 100
On dividing both sides by 4, we get

Answer:

Let required number be x.
(i) Seven time of this number be 7x
(ii) Then add a fifity to it so it becomes
7x + 50
(iii) To reach a triple century, we need forty,
So, equation becomes
7x + 50 + 40 = 300
7x + 90 = 300 ... (1)
On transposing 90, we get
7x = 300 – 90
or 7x = 210
On dividing both sides by 7, we get

or x = 30
Hence, required number be 30.
Check:
Put x = 30 in LHS of (1), we get
LHS = 7 (30) + 90 = 210 + 90 = 300
= RHS
Hence, x = 30 is verified.