NCERT Exemplar Solutions: Algebra

# NCERT Exemplar Solutions: Algebra | Mathematics (Maths) Class 6 PDF Download

## Exercise Page: 107

In questions 1 to 23, out of the four given options, only one is correct. Write the correct answer.
Q.1. If each match box contains 50 matchsticks, the number of matchsticks required to fill n such boxes is
(a) 50 + n
(b) 50n
(c) 50 ÷ n
(d) 50 – n

From the question it is given that, number of match sticks in each match box = 50
Then, the number of matchsticks required to fill n such boxes = 50 × n
= 50n

Q.2. Amulya is x years of age now. 5 years ago her age was
(a) (5 – x) years
(b) (5 + x) years
(c) (x – 5) years
(d) (5 ÷ x) years

Current age of Amulya = x years
Then, 5 years ago her age was = (x – 5) years

Q.3. Which of the following represents 6 × x
(a) 6x
(b) x/6
(c) 6 + x
(d) 6 – x

Q.4. Which of the following is an equation?
(a) x + 1
(b) x – 1
(c) x – 1 = 0
(d) x + 1 > 0

An expression with a variable, constants and the sign of equality (=) is called an equation.

Q.5. If x takes the value 2, then the value of x + 10 is
(a) 20
(b) 12
(c) 5
(d) 8

From the question it is given that, value of x is 2.
Now substitute the value of x in x + 10
= 2 + 10
= 12

Q.6. If the perimeter of a regular hexagon is x metres, then the length of each of its sides is
(a) (x + 6) metres
(b) (x ÷ 6) metres
(c) (x – 6) metres
(d) (6 ÷ x) metres

We know that, perimeter of hexagon = number of sides × length of each side
Given, the perimeter of a regular hexagon is x metres
Then, the length of each of its side = (x/6) metres
= (x ÷ 6) metres

Q.7. Which of the following equations has x = 2 as a solution?
(a) x + 2 = 5
(b) x – 2 = 0
(c) 2x + 1 = 0
(d) x + 3 = 6

Transforming – 2 from left hand side to right hand side it becomes 2.
Then, x = 2

Q.8. For any two integers x and y, which of the following suggests that operation of addition is commutative ?
(a) x + y = y + x
(b) x + y > x
(c) x – y = y – x
(d) x × y = y × x

Let us assume a and b are the two integers,
Then, commutative law of addition = a + b = b + a
Where, a = x, b = y
Therefore, commutative law of addition = x + y = y + x

Q.9. Which of the following equations does not have a solution in integers?
(a) x + 1 = 1
(b) x – 1 = 3
(c) 2x + 1 = 6
(d) 1 – x = 5

Consider the equation, 2x + 1 = 6
Transforming 1 from left hand side to right hand side it becomes -1.
2x = 6 – 1
2x = 5
X = 5/2

Q.10. In algebra, a × b means ab, but in arithmetic 3 × 5 is
(a) 35
(b) 53
(c) 15
(d) 8

Q.11. In algebra, letters may stand for
(a) Known quantities
(b) Unknown quantities
(c) Fixed numbers
(d) None of these

Q.12. “Variable” means that it
(a) Can take different values
(b) Has a fixed value
(c) Can take only 2 values
(d) Can take only three values

The word ‘variable’ means something that can vary, i.e., change. The value of a variable is not fixed. We use a variable to represent a number and denote it by any letter such as l, m, n, p, x, y, z etc.

Q.13. 10 – x means
(a) 10 is subtracted x times
(b) x is subtracted 10 times
(c) x is subtracted from 10
(d) 10 is subtracted from x

Q.14. Savitri has a sum of Rs x. She spent Rs 1000 on grocery, Rs 500 on clothes and Rs 400 on education, and received Rs 200 as a gift. How much money (in Rs) is left with her?
(a) x – 1700
(b) x – 1900
(c) x + 200
(d) x – 2100

From the question it is given that,
Savitri has a sum of Rs x
She spent money on grocery = ₹ 1000
She spent money on clothes = ₹ 500
She spent money on education = ₹ 400
Total money spent by Savitri = 1000 + 500 + 400 = ₹ 1900
Then,
Total money left with her after deducting = ₹ (x – 1900)
Therefore, money left with her after adding gift money = (x – 1900) + 200
= x – 1700

Q.15. The perimeter of the triangle shown in Fig. 7.1 is

(a) 2x + y
(b) x + 2y
(c) x + y
(d) 2x – y

Given triangle is an isosceles triangle,
So, perimeter of an isosceles triangle = 2 × x + y
= 2x + y

Q.16. The area of a square having each side x is
(a) x × x
(b) 4x
(c) x + x
(d) 4 + x

We know that, area of square = side × side
Given, square having a side x.
So, area of a square = x × x
= x2

Q.17. The expression obtained when x is multiplied by 2 and then subtracted from 3 is
(a) 2x – 3
(b) 2x + 3
(c) 3 – 2x
(d) 3x – 2

From the question it is given that,
X is multiplied by 2 = x × 2 = 2x
Then, x is multiplied by 2 and then subtracted from 3 = 3 – 2x

Q.18. q/2 = 3 has a solution
(a) 6
(b) 8
(c) 3
(d) 2

Consider the given equation q/2 = 3
By cross multiplication we get, q = 6

Q.19. x – 4 = – 2 has a solution
(a) 6
(b) 2
(c) – 6
(d) – 2

Consider the given equation x – 4 = -2
Transform – 4 from left hand side to right hand side it becomes 4.
x = – 2 + 4
x = 2

Q.20. 4/2 = 2 denotes a
(a) numerical equation
(b) algebraic expression
(c) equation with a variable
(d) false statement

4/2 = 2
By cross multiplication we get,
4 = 4

Q.21. Kanta has p pencils in her box. She puts q more pencils in the box. The total number of pencils with her are
(a) p + q
(b) pq
(c) p – q
(d) p q

From the question it is given that,
Kanta has p pencils in her box
She puts q more pencils in the box
The total number of pencils with her are = p + q

Q.22. The equation 4x = 16 is satisfied by the following value of x
(a) 4
(b) 2
(c) 12
(d) –12

Consider the given equation 4x = 16
Then, value of x is,
x = 16/4 … [divide both numerator and denominator by 4]
x = 4

Q.23. I think of a number and on adding 13 to it, I get 27. The equation for this is
(a) x – 27 = 13
(b) x – 13 = 27
(c) x + 27 = 13
(d) x + 13 = 27

Let us assume the number be ‘x’,
Then, adding 13 to the number = x + 13
Therefore, x + 13 = 27

### In question 24 to 40, fill in the blanks to make the statements true:

Q.24. The distance (in km) travelled in h hours at a constant speed of 40km per hour is __________.

The distance (in km) travelled in h hours at a constant speed of 40km per hour is 40h.
From the question,
Time taken to travel a distance = h hours
Travel at a speed of 40 km/h
So, total distance travelled = 40 × h
= 40h

Q.25. p kg of potatoes are bought for Rs 70. Cost of 1kg of potatoes (in Rs) is __________.

p kg of potatoes are bought for Rs 70. Cost of 1kg of potatoes (in Rs) is 70/p.
Given, p kg of potatoes are bought for ₹ 70
Then, cost of 1 kg of potato = 70/p

Q.26. An auto rickshaw charges Rs 10 for the first kilometre then Rs 8 for each such subsequent kilometre. The total charge (in Rs) for d kilometres is __________.

An auto rickshaw charges Rs 10 for the first kilometre then Rs 8 for each such subsequent kilometre. The total charge (in Rs) for d kilometres is 2 + 8d.
From the question it is given that,
An auto rickshaw charges ₹ 10 for the first kilometre
Then ₹ 8 for each such subsequent kilometre.
So, the total charge (in Rs) for d kilometres is = 10 + (d – 1)8
= 10 + 8d – 8
= 2 + 8d

Q.27. If 7x + 4 = 25, then the value of x is __________.

If 7x + 4 = 25, then the value of x is 3.
Consider the equation, 7x + 4 = 25
Transposing 4 from left hand side to right hand side it becomes -4,
7x = 25 – 4
7x = 21
x = 21/7
x = 3

Q.28. The solution of the equation 3x + 7 = –20 is __________.

The solution of the equation 3x + 7 = –20 is -9.
Consider the equation, 3x + 7 = –20
Transposing 7 from left hand side to right hand side it becomes -7,
3x = – 20 – 7
3x = – 27
x = -27/3
x = – 9

Q.29. ‘x exceeds y by 7’ can be expressed as __________.

‘x exceeds y by 7’ can be expressed as x – y = 7.

Q.30. ‘8 more than three times the number x’ can be written as __________.

‘8 more than three times the number x’ can be written as 3x + 8.
As per the condition given in the question, three times the number x = 3x
So, 8 more than three times the number x = 3x + 8

Q.31. Number of pencils bought for Rs x at the rate of Rs 2 per pencil is __________.

Number of pencils bought for Rs x at the rate of Rs 2 per pencil is x/2.
From the question it is given that, cost of pencil = ₹ x
No. of pencils bought for ₹ 2 = 1
Therefore, number of pencil bought for ₹ x = ₹ x/2

Q.32. The number of days in w weeks is __________.

The number of days in w weeks is 7w.
We know that, there are 7 days in a week.
Therefore, number of days in w weeks is 7w.

Q.33. Annual salary at r rupees per month along with a festival bonus of Rs 2000 is __________.

Annual salary at r rupees per month along with a festival bonus of Rs 2000 is ₹12r + 2000.
From the question it is given that,
Salary per month is r rupees
A festival bonus of ₹ 2000
Therefore, Annual salary at r rupees per month along with a festival bonus of Rs 2000 is ₹ 12r + 2000

Q.34. The two digit number whose ten’s digit is ‘t’ and units’s digit is ‘u’ is __________.

The two digit number whose ten’s digit is ‘t’ and units’s digit is ‘u’ is 10t + u.
From the question,
Two digit number whose ten’s digit is ‘t’
Two digit number whose unit’s digit is ‘u’
Then, the number = 10 × t + 1 × u
= 10t + u

Q.35. The variable used in the equation 2p + 8 = 18 is __________.

The variable used in the equation 2p + 8 = 18 is p.
The word ‘variable’ means something that can vary, i.e., change. The value of a variable is not fixed. We use a variable to represent a number and denote it by any letter such as l, m, n, p, x, y, z etc

Q.36. x metres = __________ centimetres

x metres = x × 100 centimetres
We know that, 1 metre = 100 centimetre.
Therefore, x metres × 100 centimetres = 100x centimetres

Q.37. p litres = __________ millilitres

p litres = p× 1000 millilitres
we know that, 1 litre = 1000 millilitres
Therefore, p litres × 1000 millilitres = 1000p milliliters.

Q.38. r rupees = __________ paise

r rupees = 100r paise
We know that, 1 rupee = 100 paise
Therefore, r rupees = 100r paise

Q.39. If the present age of Ramandeep is n years, then her age after 7 years will be __________.

If the present age of Ramandeep is n years, then her age after 7 years will be n + 7 years.

Q.40. If I spend f rupees from 100 rupees, the money left with me is __________ rupees.

If I spend f rupees from 100 rupees, the money left with me is 100 – f rupees.

### In question 41 to 45, state whether the statements are true or false.

Q.41. 0 is a solution of the equation x + 1 = 0

False.
Consider the equation, x + 1 = 0
Then, x = -1

Q.42. The equations x + 1 = 0 and 2x + 2 = 0 have the same solution.

True.
Consider equations x + 1 = 0
So, x = -1
Consider the equation, 2x + 2 = 0
Divide both the side by 2,
Then we get, x + 1 = 0
Therefore, x = – 1

Q.43. If m is a whole number, then 2m denotes a multiple of 2.

True.

Q.44. The additive inverse of an integer x is 2x.

False.
Additive inverse of x is – x

Q.45. If x is a negative integer, – x is a positive integer.

True.

Q.46. 2x – 5 > 11 is an equation.

False.
An expression with a variable, constants and the sign of equality (=) is called an equation.

Q.47. In an equation, the LHS is equal to the RHS.

True

Q.48. In the equation 7k – 7 = 7, the variable is 7.

False.
In the equation 7k – 7 = 7, the variable is k.

Q.50. The distance between New Delhi and Bhopal is not a variable.

True

Distance between New Delhi and Bhopal is fixed.

Clearly, it is not a variable, so the statement is true.

Q.51. 1 minutes are equal to 601 seconds.

True
We know that, 1 minute = 60 seconds [transposing +2 to RHS]
minutes = 60 x f seconds = 601 seconds So, the statement is true.

Q.52. The difference between the ages of two sisters Leela and Yamini is a variable.

False

Difference between the ages of two sisters Leela and Yamini is not a variable because Leela’s and Yamini’s ages are fixed. But the value of a variable is not fixed.

So, the statement false.

Q.53. The number of lines that can be drawn through a point is a variable.

True

Because infinite number of lines can be drawn through a point.

In questions 56 to 74, choose a letter x, y, z, p etc.,

wherever necessary, for the unknown (variable) and write the corresponding expressions.

Q.54. One more than twice the number.

Let the number be x and twice the number x = 2x

Now, according to question,

The expression = 2x + 1

Hence, required expression is 2x + 1.

Q.55. 20°C less than the present temperature.

Let the present temperature be f°C.

∴ Required expression

= Present temperature – 20°C = (f – 20)°C

Q.56. The successor of an integer.

Let the integer be n.

Successor of n = n + 1

∴ Required expression = n + 1

Note: If 7 is added to a number, we get its successor.

Q.57. The perimeter of an equilateral triangle, if side of the triangle is m.

Given, side of triangle is m.

In equilateral triangle, all sides are equal.

∴ Perimeter of an equilateral triangle =Sum of all the sides

= m + m + m = 3m

Hence, the perimeter of an equilateral triangle is 3m.

Q.58. Area of the rectangle with length k units and breadth n units.

Given, length of rectangle = k units

Breadth of rectangle = n units

Now, area of rectangle = Length x Breadth = k x n = kn units

Hence, area of the rectangle is kn sq units.

Q.59. Omar helps his mother 1 hour more than his sister does.

Let sister’s helping hours = x years

Then, Omar’s helping hour = Sister’s helping hour +1 = (x + 1)years

∴ Required expression = (x + 1) years

Q.60. Two consecutive even integers.

Any even integer can be written as 2n, where n is an integer. So, next even integer will be 2n + 2.
Hence, two consecutive even integers are 2n and 2n + 2.

The document NCERT Exemplar Solutions: Algebra | Mathematics (Maths) Class 6 is a part of the Class 6 Course Mathematics (Maths) Class 6.
All you need of Class 6 at this link: Class 6

## Mathematics (Maths) Class 6

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## FAQs on NCERT Exemplar Solutions: Algebra - Mathematics (Maths) Class 6

 1. What is the importance of solving algebra problems?
Ans. Solving algebra problems is important as it helps in developing critical thinking skills, logical reasoning, and problem-solving abilities. It is a fundamental branch of mathematics that is widely used in various fields such as science, engineering, economics, and computer science.
 2. How can I improve my algebra skills?
Ans. To improve algebra skills, it is important to practice regularly. Start by understanding the basic concepts and principles of algebra. Solve a variety of problems, starting from simple to complex. Seek help from teachers, online resources, or join study groups to clarify doubts and gain additional practice.
 3. What are the common mistakes to avoid while solving algebraic equations?
Ans. Some common mistakes to avoid while solving algebraic equations include: - Forgetting to apply the distributive property - Misplacing negative signs - Not simplifying expressions before solving - Forgetting to multiply or divide both sides of the equation by the same value when simplifying
 4. Can algebra be used in real-life situations?
Ans. Yes, algebra can be used in various real-life situations. It helps in solving problems related to finance, budgeting, statistics, measurements, and many other fields. For example, algebra can be used to calculate compound interest, analyze data trends, or determine the amount of ingredients needed for a recipe.
 5. Are there any tips for understanding and retaining algebraic concepts?
Ans. Yes, here are some tips for understanding and retaining algebraic concepts: - Start by mastering the basics before moving on to more complex topics. - Break down problems into smaller steps to make them more manageable. - Practice regularly and review previous concepts to reinforce learning. - Seek help from teachers or online resources whenever needed. - Create a study schedule and allocate dedicated time for algebraic practice.

## Mathematics (Maths) Class 6

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