# Free Simple Equations 02 Practice Test - 7th grade

### Question 1

The sum of a number and its half is five less than twice the number. Which of the following equations represents given statement mathematically?

x+x2 = (2x−5)

x+a2 = (5−2x)

x+a2 = (5−2a)

a+x2 = 2x−5

#### SOLUTION

Solution :A

Let the number be 'x'.

Half of the number is x2.

Similarly, twice of the number =2x.

Sum of the number and its half =x+x2

5 less than twice the number =2x−5

Given: Sum of the number and its half is five less than twice of the number.

∴The required equation is x+x2=2x−5

### Question 2

If a=x then which number satisfies the equation: x+a2=(2x−5)?

10

5

2

7

#### SOLUTION

Solution :A

The equation in the above question is x+a2=(2x−5)

We need to find the value of x.

Substituting the value 'x=a' and solving.⇒x+x2=(2x−5)

⇒3x2=(2x−5)

⇒2x−3x2=5

⇒x2=5

⇒x=5×2

⇒x=10

### Question 3

After buying a toy from the shop, I am left with ₹15 from my 100 rupees of pocket money. Then, the cost of toy (in rupees) is:

15

100

85

75

#### SOLUTION

Solution :C

Let the cost of the toy be x as it is the unknown.

x + 15 = 100

x = 100 - 15 = ₹85

So, the cost of the toy= ₹85

### Question 4

A shopkeeper has an unknown weight in his shop. To find out the weight, he first weighs 1 kg of rice, and then weighs 1 kg of rice and the unknown weight together. The combined weight of the rice and the unknown weight is 1.5 kg. Find the unknown weight (in grams).

#### SOLUTION

Solution :Let the unknown weight be x kg

Given that,

weight of rice = 1 kg = 1000 g

weight of rice+unknown weight=1.5 kg =1500 g

⇒x+1000 g=1500 g

⇒x=(1500−1000)g=500 g

So, unknown weight = 500 g

### Question 5

Karan spends ₹** **5000 on food which is 58th of his income. Karan's income is

#### SOLUTION

Solution :Let Karan' s income be x

58 of x = ₹ 5000

x=₹ 5000×85

x = ₹ (1000 x 8) = ₹ 8000

### Question 6

Pavani has a certain number of chocolates; she gave one-third of the chocolates to her brother Arjun, one-sixth to her friend Preeti. She was then left with 30 chocolates. What is the sum of the number of chocolates did she give to Arjun and Preeti?

#### SOLUTION

Solution :Let the number of chocolates Pavani has be x.

x3 are given to Arjun. So now she is left with x−x3=2x3 chocolates

She gave x6 to Preeti. So, she is left with 2x3 - x6 = 4x6 - x6 = 3x6 = x2

So, x2=30

x=30×2=60

Initially, Pavani had 60 chocolates. She gave 13rd i.e. 603 =20 to Arjun.

16th i.e. 10 chocolates to Preeti.

Sum = 10 + 20 = 30

### Question 7

What is the value of x if 4x−7=3x+5 ?

1

12

3

5

#### SOLUTION

Solution :B

The given equation is,

4x−7=3x+5

On transposing,

4x−3x=7+5 x=12

### Question 8

In the equation x=10y−24, sum of the coefficients of dependent and independent variables is

#### SOLUTION

Solution :The value of x depends on the value of y. So the dependent variable is x and the independent variable is y.

Sum of the coefficients of dependent and independent variable =(10+1)=11

### Question 9

The sum of three consecutive numbers is 36. Find the numbers.

12,13,14

10,11,12

11,12,13

13,12,11

#### SOLUTION

Solution :C

Let the first number be a.

So the three consecutive numbers will be a, a+1, a+2.

a +a +1 +a +2 = 36

⇒3a +3 = 36

⇒ 3a = 36 -3 = 33

⇒a=333=11

So, the three consecutive numbers will be 11,12,13.

### Question 10

3x+15=10 is a simple equation. If 5 is added to the left-hand side of the equation, the equation remains the same as the given equation.

True

False

#### SOLUTION

Solution :B

If 5 is added to the left-hand side of the equation 3x+15=10, we get 3x+15+5=10.

The value on the left-hand side is increased but it remains same on the right-hand side. So, equality doesn't hold good now. 3x+15+5≠10.

Equality sign remains unchanged only when same quantity is added on both sides, i.e., 3x+15+5=10+5.

Thus, the equation does not remain same as the given equation.