Get the simplified Class 7 Maths Extra Questions and Class 7 Maths Part 2 Chapter 7 Finding the Unknown Extra Questions and Answers with complete explanation.
Class 7 Finding the Unknown Extra Questions
Finding the Unknown Extra Questions Class 7
Finding the Unknown Class 7 Very Short Question Answer
Question 1.
The sum of a number and 17 is 45. Find the number.
Solution :
Let the number be x. then,
x + 17 = 45
→ x = 45-17
→ x = 28
So, the number is 28.
Question 2.
Three consecutive integers add up to 84.
Find the integers.
Solution :
Let the first integer be x.
Then the next two are (x + 1) and (x + 2).
According to question,
x + (x + 1) + (x + 2) = 84
→ 3 x + 3 = 84
→ 3 x = 81
→ x = 27
Hence, the integers are 27, 28 and 29.
Question 3.
Solve the following equations
2(y-6) = 3(y + 5)
Solution :
2(y-6) = 3(y + 5)
2 y-12 = 3 y + 15
3 y-2 y = -12-15
y = -27
Question 4.
The perimeter of a rectangle is 48 cm. If its length is 3 cm more than twice its breadth, find its length and breadth.
Solution :
Let breadth (b) = x cm
then length (l) = 2 x + 3
Perimeter = 2(l + b)
48 = 2(2 x + 3 + x)
3x + 3 = 24
3x = 21
x = 7 cm
∴ l = 2x + 3 = 2(7) + 3 = 17 cm
So, length = 17 cm and breadth = 7 cm.
Question 5.
Solve for x :
5 x + 7 = 32
Solution :
5 x + 7 = 32
5 x = 32-7
5 x = 25
x = 5
Question 6.
The denominator of a fraction is 4 more then its numerator. If the fraction equals \(\frac{3}{7}\), find the fraction.
Solution :
Let the numerator be x.
Then denominator = x + 4
So,
\(\frac{x}{x + 4}\) = \(\frac{3}{7}\)
7 x = 3(x + 4)
7 x = 3 x + 12
4 x = 12
x = 3
Numerator = 3; Denominator = 3 + 4 = 7
∴ The fraction is \(\frac{3}{7}\).
Question 7.
Can you give a real life situation that can be modelled as 15 x + 20 = 80.
Solution :
Ramesh pays ₹ 80 to the delivery boy including delivery charges for a certain quantity of onions. If the delivery charges are ₹ 20 and rate of onions is ₹ 15 per kg, what is the quantity of onions bought?
Question 8.
Solve : 28(x + 4) + 300 = 1000
Solution :
28(x + 4) + 300 = 1000
28 x + 112 + 300 = 1000
28 x + 412 = 1000
28 x = 588
x = 21
Finding the Unknown Class 7 Short Question Answer
Question 1.
Express the given statement as an equation.
“If 1 is subtracted from a number and the difference is multiplied by \(\frac{1}{2}\), the result is 7.”
Solution :
Let the number be x.
According to question,
(x-1) × [/latex]\frac{1}{2}[/latex] = 7
Thus, \(\frac{1}{2}\) (x-1) = 7 is the required equation.
Question 2.
A number exceeds the other number by 12. If their sum is 72, find the numbers.
Solution :
Let the number be x.
Then the other number be (x + 12).
According to the question,
x + (x + 12) = 72
2x = 72-12
x = \(\frac{60}{2}\) = 30
Hence, (x + 12) = 30 + 12 = 42
Thus, the numbers are 30 and 42.
Question 3.
Subramaniam and Naidu donate some money in a Relief Fund. The amount paid by Naidu is ₹ 125 more than that of Subramaniam. If the total money paid by them is ₹ 975, find the amount of money donated by Subramaniam.
Solution :
Let amount of money donated by Subramaniam be ₹ x.
Then the amount of money donated by Naidu be ₹ (x + 125).
The required equation is :
x + (x + 125) = 975
2 x = 975-125
x = [/latex]\frac{850}{2}[/latex]
x = 425
Thus, the amount of money donated by Subramaniam is ₹ 425.
Question 4.
My younger sister’s present age is three times what her age will be 3 years from now, minus three times what her age was 3 years ago. Find her present age.
Solution : Let her present age be x years. Then after 3 years, her age be (x + 3) years and 3 years ago her age was (x-3) years. The linear equation formed is :
x = 3 [(x + 3)-(x-3)
x = 3 [x + 3-x + 3]
x = 3 [6]
x = 18
Thus, her present age is 18 years.
Question 5.
A man travelled two-fifth of his journey by train, one-third by bus, onefourth by car and the remaining 3 km on foot. What is the length of his total journey?
Solution :
Let the length of his total journey be x km}.
Then his journey by train be \(\frac{2 x}{5}\), by bus be \(\frac{x}{3}\) and by car be \(\frac{x}{4}\).
The linear equation we get is :
x = \(\frac{2 x}{5}\) + \(\frac{x}{3}\) + \(\frac{x}{4}\) + 3
→ x-\(\frac{2 x}{5}\)–\(\frac{x}{3}-\frac{x}{4}\) = 3
\(\frac{x}{60}\) = 3
→ x = 180
Thus, the length of his total journey is 180 km.
Question 6.
If one side of a square is represented by 18 x-20 and the adjacent side is represented by 42-13 x, find the length of the side of the square.
Solution : We know that :
All sides of a square are equal.
18 x-20 = 42-13 x
By further calculation
18 x + 13 x = 42 + 20
31 x = 62
Dividing both sides by 31
x = 2
Substituting the value of x
Side of square = 18 x-20 = 18 × 2-20
= 36-20 = 16
Therefore, the length of the side of the square is 16 units.
Question 7.
If marks scored by Chandni are doubled, it becomes 42 more than marks obtained by Sanjana. If Sanjana scored 210 marks, then how many marks were scored by Chandni?
Solution :
Let Chandni’s marks be x.
According to the question, the required equation is :
2 x-42 = 210
2 x = 210 + 42 = 252
x = 126
So, Chandni scored 126 marks in examination.
Question 8.
Construct three equations for the solution :
(i) x = 4
(ii) a = -5
Solution :
(i) Equations for x = 4
(a) x + 3 = 7
(b) 2 x = 8
(c) 15-x = 11
(ii) Equations for a = -5
(a) a + 7 = 2
(b) 3 a = -15
(c) 10-a = 15