RD Sharma Solutions Class 6 Maths Chapter 8 Introduction to Algebra

Read RD Sharma Solutions Class 6 Maths Chapter 8 Introduction to Algebra below, students should study RD Sharma class 6 Mathematics available on Studiestoday.com with solved questions and answers. These chapter wise answers for class 6 Mathematics have been prepared by teacher of Grade 6. These RD Sharma class 6 Solutions have been designed as per the latest NCERT syllabus for class 6 and if practiced thoroughly can help you to score good marks in standard 6 Mathematics class tests and examinations

Exercise 8.1

 

Question 1:   Write the following using numbers, literals and signs of basic operations. State what each letter represents:

(i) The diameter of a circle is twice its radius.

(ii) The area of a rectangle is the product of its length and breadth.

(iii) The selling price equals the sum of the cost price and the profit.

(iv) The total amount equals the sum of the principal and the interest.

(v) The perimeter of a rectangle is two times the sum of its length and breadth.

(vi) The perimeter of a square is four times its side. 

Solution 1:

(i) Let the diameter of a Circle be = d

and radius of a circle be = r

It is given that diameter of a circle is twice its radius.

Thus, d = 2r.

 

(ii) Let the area of a rectangle be = A

Length of a rectangle be = l

Breadth of a rectangle be = b

It is given that area of a rectangle is the product of its length and breadth.

Thus, A = l × b.

 

(iii) Let the selling price be = S.P.

Cost price be = C.P.

Profit be = P

It is given that the selling price equals the sum of the cost price and the profit.

Thus, S.P = C.P + P

 

(iv) Let the amount be = A

Principal be = P

Interest be = I

It is given that the total amount equals the sum of the principal and the interest.

Thus, we get A = P + I

 

(v) Let perimeter of a rectangle be = P

Length of a rectangle be = l

Breadth of a rectangle be = b

It is given that the perimeter of a rectangle is two times the sum of its length and breadth.

Thus, P = 2 (l + b)

 

(vi) Let perimeter of a Square be = P

Side of a square be = s

It is given that the perimeter of a square is four times its side.

Thus, P = 4a

 

Question 2:  Write the following using numbers, literals and signs of basic operations:

(i) The sum of 6 and x.

(ii) 3 more than a number y.

(iii) One-third of a number x.

(iv) One-half of the sum of number x and y.

(v) Number y less than a number 7.

(vi) 7 taken away from x.

(vii) 2 less than the quotient of x and y.

(viii) 4 times x taken away from one-third of y.

(ix) Quotient of x by 3 is multiplied by y.

Solution 2:

RD Sharma Solutions Class 6 Maths Chapter 8 Introduction to Algebra


 

Question 3:  Think of a number. Multiply by 5. Add 6 to the result. Subtract y from this result. What is the result? 

Solution 3:

Let the number be x

The number is multiplying by 5 = 5x_______(1)

Add 6 in Equation (1) we get,

= 5x + 6_______(2)

By subtracting y from (2) we get,

= 5x + 6 – y.

Therefore, the result is 5x + 6 – y.

 

Question 4:  The number of rooms on the ground floor of a building is 12 less than the twice of the number of rooms on first floor. If the first floor has x rooms, how many rooms does the ground floor has? 

Solution 4:

It is given that the number of the rooms on the first floor = x

Let the number of rooms on the ground floor be = y

Number of rooms on the ground floor of a building is 12 less than the twice of the number of rooms on first floor

Algebraically we can say that, 

y = 2x – 12 

Thus, the rooms on the ground floor is y = 2x – 12.

 

Question 5:  Binny spend Rs. a daily and saves Rs. b per week. What is her income for two weeks? 

Solution 5:

Given,

Amount spent by Binny daily = Rs. a

Total Amount spent by Binny in a week = 7a

Amount saved by Binny weekly= Rs. b

Binny’s income for one week = 7a + b

Binny’s income for 2 weeks = 2(7a + b) 

Binny’s income for 2 weeks = Rs. 14a + 2b

Thus, Two weeks income of Binny is Rs. 14a + 2b.

 

Question 6:  Rahul scores 80 marks in English and x marks in Hindi. What is his total score in the two subjects? 

Solution 6:

It is given that,

Rahul’s Marks in English = 80

Rahul’s Marks in Hindi = x

Total scores in the two subjects = x + 80

Thus, total score in two subjects getting by Rahul is x + 80.

 

Question 7:   Rohit covers x centimetres in one step. How much distance does he cover in y steps? 

Solution 7:

Rohit covered distance in one step = x cm

Distance covered by Rohit in y steps = xy cm

Thus, In y steps Rohit covers xy cm.

 

Question 8:   One apple weighs 75 grams and one orange weighs 40 grams. Determine the weight of x apples and y oranges. 

Solution 8:

It is given that,

Weight of one apple is = 75g

Weight of one orange is = 40g

Weight of x apples is = 75g ×x  

So the weight of y oranges = 40g ×y

Total weight of x apples and y oranges is = (75x + 40y)g

 

Question 9:  One pencil costs Rs. 2 and one fountain pen costs Rs. 15. What is the cost of x pencils and y fountain pens? 

Solution 9:

Cost of a pencil = Rs. 2

Cost of a fountain pen = Rs. 15

Cost of x pencils = 2×x=2x

Cost of y fountain pens = 15×y=15y

The cost of x pencils and y fountain pens = Rs.(2x + 15y)

 

Exercise 8.2

 

Question 1:  Write each of the following products in exponential form:

(i) a × a × a × a × …….. 15 times

(ii) 8 × b × b × b × a × a × a × a

(iii) 5 × a × a × a × b × b × c × c × c

(iv) 7 × a × a × a …….. 8 times × b × b × b × …… 5 times

(v) 4 × a × a × …… 5 times × b × b × ……. 12 times × c × c …… 15 times

Solution 1:

RD Sharma Solutions Class 6 Maths Chapter 8 Introduction to Algebra-

 

Question 2:   Write each of the following in the product form:
(i) a2  b5
(ii) 8x3
(iii) 7a3 b4
(iv) 15 a9 b8 c6
(v) 30x4 y4 z5
(vi) 43p10 q5 r15
(vii) 17p12 q20
 
Solution 2:
RD Sharma Solutions Class 6 Maths Chapter 8 Introduction to Algebra-1

 
Question 3:   Write down each of the following in exponential form:
(i) 4a3×6ab2×c2
(ii) 5xy×3x2 y×7y2
(iii) a3×3ab2×2a2 b2
 
Solution 3:
(i) 4a3×6ab2×c2 
4×6× a3×a×b2×c2 
Thus, Exponential form 24a4 b2 c2 
 
(ii) 5xy×3x2y×7y2
5×3×7×x×x2×y×y×y2
Thus, Exponential form 105x3 y4
 
(iii) a3×3ab2×2a2 b2
 3×2×a3×a×a2×b2×b2
Thus, Exponential form 6a6 b4
 
 
Question 4:  The number of bacteria in a culture is x now. It becomes square of itself after one week. What will be its number after two weeks?
 
Solution 4:
Number of bacteria in a culture = x
After one week bacteria becomes square itself = x2
Number of bacteria after two weeks = (x2)2 = x4
Thus, the number of bacteria after two weeks is x4.
 
 
Question 5:  The area of a rectangle is given by the product of its length and breadth. The length of a rectangle is two-third of its breadth. Find its area if its breadth is x cm.
 
Solution 5:
Breadth = x cm 
Length = 2/3 x cm
Area of rectangle = l × b
Area of the rectangle = 2/3 x× x = 2/3 x2 cm2 
Thus, the area of rectangle is 2/3 x2 cm2.
 
 
Question 6:  If there are x rows of chairs and each row contains x^2 chairs. Determine the total number of chairs. 
Solution 6:
Number of rows = x
Number of chairs in each row = x2 chairs
Total number of chairs = x × x2= x3
Thus, the total number of chairs is x3.
 
 
Objective Type Questions : :->

Mark the correct alternative in each of the following:
 
Question 1:   5 more than twice a number x is written as
(a) 5 + x + 2
(b) 2x + 5
(c) 2x − 5
(d) 5x + 2
 
Solution 1: (b) 
It is given that the number be x.
Twice of a number = 2x
5 more than twice a number is 2x + 5.
 
Question 2:  The quotient of x by 2 is added to 5 is written as
(a) x/2+5
(b) 2/(x+5)
(c) ((x+2))/5
(d) x/((2+5))
 
Solution 2: (a) 
Quotient of x by 2 = x/2
Quotient is added by 5 = x/2+5.
 
Question 3:  The quotient of x by 3 is multiplied by y is written as
(a) x/3y
(b) 3x/y
(c) 3y/x
(d) xy/3
 
Solution 3: (d)
Quotient of x by 3 = x/3
Quotient is multiplied by y = x/3×y = xy/3
 
Question 4:  9 taken away from the sum of x and y is
(a) x + y - 9
(b) 9 - (x+y)
(c) (x+y)/9
(d) 9/( x+y)
 
Solution 4: (a) 
Sum of x and y = x + y
Taken away 9 from the sum of x and y is = x+y-9.
 
Question 5:  The quotient of x by y added to the product of x and y is written as
(a) x/y+xy
(b) y/x+ xy
(c) xy+x/y
(d) xy+y/x
 
Solution 5: (a) 
Quotient of x and y =  x/y
Product of x and y = xy
Added both the equations and written as x/y+xy.
 
Question 6:  a2b3 × 2ab2 is equal to
(a) 2a3b4
(b) 2a3b5
(c) 2ab
(d) a3b5
 
Solution 6: (b) 
= a2b3 × 2ab
= 2 × a2 × a × b3 × b2
= 2a3b5.
 
Question 7:  4a2b3 × 3ab2 × 5a3b is equal to
(a) 60a3b5
(b) 60a6b5
(c) 60a6b6
(d) a6b6
 
Solution 7: (c) 
= 4a2b3 × 3ab2 × 5a3
= 4 × 3 × 5 × a2 × a × a3 × b3 × b2 × b 
= 60a6b6
 
Question 8:  If 2x2y and 3xy2 denote the length and breadth of a rectangle, then its area is
(a) 6xy
(b) 6x2y2
(c) 6x3y3
(d) x3y3
 
Solution 8: (c) 
Length = 2x2y
Breadth = 3xy2
Area of a rectangle = l × b
Area of a rectangle = 2x2 y × 3xy
Area of a rectangle = 2×3×x3y3
Area of a rectangle = 6x3y3
 
Question 9:   In a room there are x^2 rows of chairs and each two contains 2x^2 chairs. The total number of chairs in the room is
(a) 2x3
(b) 2x4
(c) x4
(d) x4/2
 
Solution 9: (b) 
Number of chairs in the room = Number of rows × Number of chairs
Total number of chairs in the room 
=   x2×2x
= 2x4
 
Question 10:  a3 × 2a2b × 3ab5 is equal to
(a) a6b6
(b) 23a6b6
(c) 6a6b6
(d) None of these
 
Solution 10: (c) 
= a3 × 2a2b × 3ab
= 2 × 3 × a3 × a2 × a × b × b
= 6a6b6
Thus, a3 × 2a2b × 3ab5 is equal to 6a6b6



RD Sharma Solutions Class 6 Maths
RD Sharma Solutions Class 6 Maths Chapter 1 Knowing our Numbers
RD Sharma Solutions Class 6 Maths Chapter 2 Playing with Numbers
RD Sharma Solutions Class 6 Maths Chapter 3 Whole Numbers
RD Sharma Solutions Class 6 Maths Chapter 4 Operations on Whole Numbers
RD Sharma Solutions Class 6 Maths Chapter 5 Negative Numbers and Integers
RD Sharma Solutions Class 6 Maths Chapter 6 Fractions
RD Sharma Solutions Class 6 Maths Chapter 7 Decimals
RD Sharma Solutions Class 6 Maths Chapter 8 Introduction to Algebra
RD Sharma Solutions Class 6 Maths Chapter 9 Ratio Proportion and Unitary Method
RD Sharma Solutions Class 6 Maths Chapter 10 Basic Geomatrical Concepts
RD Sharma Solutions Class 6 Maths Chapter 11 Angles
RD Sharma Solutions Class 6 Maths Chapter 12 Triangle
RD Sharma Solutions Class 6 Maths Chapter 13 Quadrilaterals
RD Sharma Solutions Class 6 Maths Chapter 14 Circles
RD Sharma Solutions Class 6 Maths Chapter 15 Pair of Lines and Transversal
RD Sharma Solutions Class 6 Maths Chapter 16 Understanding Three Dimensional Shapes
RD Sharma Solutions Class 6 Maths Chapter 17 Symmetry
RD Sharma Solutions Class 6 Maths Chapter 18 Basic Geometrical Tools
RD Sharma Solutions Class 6 Maths Chapter 19 Geometrical Constructions
RD Sharma Solutions Class 6 Maths Chapter 20 Mensuration
RD Sharma Solutions Class 6 Maths Chapter 21 Data Handling Presentation of Data
RD Sharma Solutions Class 6 Maths Chapter 22 Data Handling Pictographs
RD Sharma Solutions Class 6 Maths Chapter 23 Data Handling Bar Graphs