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.