# Free Simple Equations Subjective Test 02 Practice Test - 7th grade

### Question 1

Is 4x+7 an equation? [1 MARK]

#### SOLUTION

Solution :No, because every equation must be equated with some quantities by including "=”. So, the given problem is an expression.

### Question 2

When you subtract 10 from thrice a number, the result is 17. Find the unknown number. [2 MARKS]

#### SOLUTION

Solution :Framing the equation: 1 Mark

Result: 1 Mark

Let the number be y

Then, 3y−10=17

⇒3y=17+10

⇒3y=27

⇒y=9

### Question 3

If 2x+13x−2=910, then the value of 'x' is: [2 MARKS]

#### SOLUTION

Solution :Steps: 1 Mark

Result: 1 Mark

2x+13x−2=910

⇒10(2x+1)=9(3x−2)

⇒20x+10=27x−18

⇒10+18=27x−20x

⇒28=7x

⇒7x=28

⇒x=287=4

The value of 'x' is 4.

### Question 4

Find x, when 3x+57=167. [2 MARKS]

#### SOLUTION

Solution :Steps: 1 Mark

Result: 1 Mark

3x+57=167

Or, (3x+5)×7=16×7

Or, 21x+35=112

Or, 21x=112−35

Or, 21x=77

Or, x=7721=113.

### Question 5

What is the solution of the equation 4x+25=7x+10 ? [2 MARKS]

#### SOLUTION

Solution :Steps: 1 Mark

Result: 1 Mark

4x+25=7x+10

Now, subtracting 10 on both sides, we get

4x+25−10=7x+10−10

4x+15=7x

Now, subtracting 4x on both sides, we get

4x−4x+15=7x−4x

15=3x

Dividing both sides by 3, we get:

153=3x3

x=5

### Question 6

Answer the following: [3 MARKS]

a) Make two equations where the solution is 6.

b) Make two equations where the solution is -8.

#### SOLUTION

Solution :Each part: 1.5 Mark

a) There are many numbers of equations that satisfy y=6

First equation: 4y−8=16

Second equation: 7y3−4=10;

b) There are number of equations that satisfy y=−8

First equation: 4y−8=−40

Second equation: 7y4+24=10

### Question 7

In thirteen years Gaurav will be twenty-four years old. Write the expression for Gaurav's age for 20 years from now. If after 20 years his father's age is 3 years less than 6 times Gaurav's present age. What is Gaurav's father's present age? [3 MARKS]

#### SOLUTION

Solution :Framing the equation: 1 Mark

Steps: 1 Mark

Result: 1 Mark

Let the present age of Gaurav be x.

According to question, In 13 years, Gaurav' age =24

⇒x+13=24

⇒x=11

The expression for Gaurav's age for 20 years from now = x+20

After 20 years, Gaurav's age will be:

x+20

11+20=31 years

Since Gaurav's present age is 11 years

∴ After 20 years father's age will be = 6(11)−3=66−3=63

Hence, father's present age = 63−20=43 years

### Question 8

Sum of two numbers is 38 and their difference is 16. Find the numbers? [3 MARKS]

#### SOLUTION

Solution :Framing of the equation: 1 Mark

Steps: 1 Mark

Answer: 1 Mark

Let the number be ′x′.

Then, the other number is (38−x)

The difference between the numbers is 16.

⇒(38−x)−x=16

⇒38−2x=16

⇒38−16=2x

⇒2x=22

⇒x=11

Other number

=38−x=38−11=27

### Question 9

The sum of four consecutive multiples of 9 is 378. Find these multiples? [3 MARKS]

#### SOLUTION

Solution :Framing the equation: 1 Mark

Steps: 1 Mark

Result: 1 Mark

Let the first number be 9x

Then, second consecutive number =9(x+1)

Third consecutive number =9(x+2),

And, fourth consecutive number =9(x+3)

According to question:

9x+9(x+1)+9(x+2)+9(x+3)=378

⇒9x+9x+9+9x+18+9x+27=378

⇒36x+54=378

⇒36x=378−54

⇒36x=324

⇒x=32436=9

The numbers are 81, 90, 99 and 108.

### Question 10

(a) Write the statement of the following equations. [4 MARKS]

i) 7x+4=46

ii) 5x+46=13

(b) 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.

#### SOLUTION

Solution :(a) Correct statement of equation: 1 Mark each

(b) Framing of equation: 1 Mark

Solution: 1 Mark

(a) i) When a number is multiplied with 7 and 4 is added to their product, we get 46 as the final result.

ii) When a number is multiplied with 5, 4 is added to their product and their final sum is divided by 6, the result which we get is 13.

(b) Let Raju's age be x.

Raju's father's age = 3x+5=44

⇒3x+5=44

⇒3x=44−5=39

⇒x=393

⇒x=13 years

Raju's age is 13 years.

### Question 11

Find the number in the following cases: [4 MARKS]

a) When a number is multiplied by 6 and 4 is added to the product, it becomes 54.

b) Adding 6 to 4 times a number gives 88.

#### SOLUTION

Solution :Each part: 2 Marks

a) Let the number be y

According to question, (y×6)+4=54

⇒6y=54−4

⇒6y=50

⇒y=506

b) Let the number is y;

According to question, 6+4×y=88.

⇒6+4y=88

⇒4y=88−6

⇒4y=82

⇒y=824

### Question 12

A fraction is given in which numerator is less than denominator by three. If we add number seven in the numerator and subtract one from the denominator, 32 is the new fraction which we get. Find the original fraction? [4 MARKS]

#### SOLUTION

Solution :Framing the equation: 1 Mark

Steps: 2 Marks

Result: 1 Mark

Let the numerator of a rational number =x

Then the denominator of a rational number =x+3

When numerator is increased by 7, then new numerator =x+7

When denominator is decreased by 1, then new denominator =x+3−1

The new number formed =32

According to the question,

x+7x+3−1=32

⇒x+7x+2=32

⇒2(x+7)=3(x+2)

⇒2x+14=3x+6

⇒3x−2x=14−6

⇒x=8

Numerator = 8, then denominator = 8+3 = 11

Original fraction =811

### Question 13

Ethan is given the responsibility of buying a week's supply of food and medication for the dogs and cats at a local shelter. The food and medication for each dog cost twice as much as the supplies for a cat. He needs to feed 164 cats and 24 dogs. His budget is Rs. 4240. How much can Ethan spend on each dog for food and medication? [4 MARKS]

#### SOLUTION

Solution :Framing the equation 2 Marks

Steps: 1 Mark

Result: 1 Mark

Let the expenditure on a cat be y

The expenditure on a dog =2y.

Total expenditure on cats =164×y=164y

Total expenditure of dogs =24×2y=48y

According to question:

164y+24×2y=4240

⇒164y+48y=4240

⇒212y=4240

⇒y=4240212=20

Hence, expenditure of a dog =2y=40

So, for 24 dogs the total expenditure is

=24×40=Rs.960

### Question 14

(a) Aditya has a certain number of pencils. The cost of pencils is 10 greater than 4 times the number of pencils which Aditya has. The total cost of the pencils is Rs 162. Frame the equation and find the number of pencils.

(b) People of Sundargram planted a total of 102 trees in the village garden. Some of the trees were fruit trees. The number of non-fruit trees was two more than three times the number of fruit trees. What was the number of fruit trees? [4 MARKS]

#### SOLUTION

Solution :(a) Steps: 1 Mark

Result: 1 Mark

(b) Steps: 1 Mark

Result: 1 Mark

(a) Let Aditya have x pencils

Now 4 times the number of pencils which Aditya has is 4×x=4x

10 greater than 4x=4x+10

Now, given that the total cost is Rs 162, we get

(4x+10)=162

⇒4x=162−10

⇒4x=152

⇒x=1524

⇒x=38

So, Aditya has 38 pencils.

(b) Total trees planted = 102

Let the number of fruit trees planted be x

∴ The number of non-fruit trees planted is 3x+2

No. of Non-fruit trees + No. of fruit trees = Total no. of trees

3x+2 + x = 102

4x+2=102

2x+1=51

2x=50

x=25

Therefore no. of fruit trees was 25.

### Question 15

After 7 years Sarosh's father's age will be 3 times that of Sarosh's age which will be equal to 6 times of Sarosh's present age. What is the sum of their present ages? [4 MARKS]

#### SOLUTION

Solution :Framing of equation: 1 Mark

Steps: 2 Marks

Result: 1 Mark

Let Sarosh's present age be x

After 7 years, Sarosh's age will be x+7

Father's age after 7 years is 3(x+7)

This is equal to 6 times Sarosh's present age =6x

So, 6x=3(x+7)=3x+21

⇒6x=3x+21

⇒6x−3x=21

⇒3x=21

⇒x=213=7

Sarosh's present age is 7 years.

Father's age after 7 years is 6×7=42 years

His present age is 42−7=35 years

Sum of their present ages =35+7=42 years