## RD Sharma Solutions for Class 6 Chapter 8 Introduction to Algebra Free Online

Exercise 8.1 page: 8.7

**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:**

(i) Consider d as the diameter and r as the radius of the circle

Hence, we get d = 2r.

(ii) Consider A as the area, l as the length and b as the breadth of a rectangle

Hence, we get A = l × b.

(iii) Consider S.P as the selling price, C.P as the cost price and P as the profit

Hence, we get S.P = C.P + P

(iv) Consider A as the amount, P as the principal and I as the interest

Hence, we get A = P + I

(v) Consider P as the perimeter, l as the length and b as the breadth of a rectangle

Hence, P = 2 (l + b)

(vi) Consider P as the perimeter and a as the side of a square

Hence, P = 4a

**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:**

(i) The sum of 6 and x can be written as 6 + x.

(ii) 3 more than a number y can be written as y + 3.

(iii) One-third of a number x can be written as x/3.

(iv) One-half of the sum of number x and y can be written as (x + y)/ 2.

(v) Number y less than a number 7 can be written as 7 – y.

(vi) 7 taken away from x can be written as x – 7.

(vii) 2 less than the quotient of x and y can be written as x/y – 2.

(viii) 4 times x taken away from one-third of y can be written as y/3 – 4x.

(ix) Quotient of x by 3 is multiplied by y can be written as xy/3.

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

**Solution:**

Consider x as the number.

Multiplying the number by 5 = 5x

Again add 6 to the number = 5x + 6

By subtracting y from the above equation = 5x + 6 – y.

Hence, the result is 5x + 6 – y.

**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:**

Consider y as the number of rooms on the ground floor

We know that

The number of rooms on the first floor = x

It is given that number of rooms on the ground floor of a building is 12 less than the twice of the number of rooms on first floor

So we get

y = 2x – 12

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

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

**Solution:**

Amount spent by Binny = Rs a

Amount saved by Binny = Rs b

Amount spent by Binny in one week = 7a

So the total income for one week = Amount spent by Binny in one week + Amount saved by Binny

Substituting the values

Total income for one week = 7a + b

We get Binny’s income for 2 weeks = 2 (7a + b) = Rs 14a + 2b

Hence, the income of Binny for two weeks is Rs 14a + 2b.

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

**Solution:**

Marks scored by Rahul in English = 80

Marks scored by Rahul in Hindi = x

So the total scores in the two subjects = x + 80

Hence, the total score of Rahul in two subjects is x + 80.

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

**Solution:**

Distance covered by Rohit in one step = x cm

So the distance covered by Rohit in y steps = xy cm

Hence, Rohit covers xy cm in y steps.

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

**Solution:**

Weight of one apple = 75 g

Weight of one orange = 40 g

So the weight of x apples = 75x g

So the weight of y oranges = 40y g

We get the weight of x apples and y oranges = (75x + 40y) g

Hence, the weight of x apples and y oranges is (75x + 40y) g.

**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:**

Cost of one pencil = Rs 2

Cost of one fountain pen = Rs 15

Cost of x pencils = 2x

Cost of y fountain pens = 15y

So the cost of x pencils and y fountain pens = Rs (2x + 15y)

Hence, the cost of x pencils and y fountain pens is Rs (12x + 15y).

Exercise 8.2 page: 8.11

**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:**

(i) a × a × a × a × …….. 15 times is written in exponential form as a15.

(ii) 8 × b × b × b × a × a × a × a is written in exponential form as 8a4b3.

(iii) 5 × a × a × a × b × b × c × c × c is written in exponential form as 5a3b2c3.

(iv) 7 × a × a × a …….. 8 times × b × b × b × …… 5 times is written in exponential form as 7a8b5.

(v) 4 × a × a × …… 5 times × b × b × ……. 12 times × c × c …… 15 times is written in exponential form as 4a5b12c15.

**2. Write each of the following in the product form:**

**(i) a2 b5**

**(ii) 8x3**

**(iii) 7a3b4**

**(iv) 15 a9b8c6**

**(v) 30x4y4z5**

**(vi) 43p10q5r15**

**(vii) 17p12q20**

**Solution:**

(i) a2 b5 is written in the product form as a × a × b × b × b × b × b.

(ii) 8x3 is written in the product form as 8 × x × x × x.

(iii) 7a3b4 is written in the product form as 7 × a × a × a × b × b × b × b.

(iv) 15 a9b8c6 is written in the product form as 15 × a × a …… 9 times × b × b × … 8 times × c × c × ….. 6 times.

(v) 30x4y4z5 is written in the product form as 30 × x × x × x × x × y × y × y × y × z × z × z × z × z.

(vi) 43p10q5r15 is written in the product form as 43 × p × p …. 10 times × q × q …. 5 times × r × r × …. 15 times.

(vii) 17p12q20 is written in the product form as 17 × p × p …. 12 times × q × q × ….. 20 times.

**3. Write down each of the following in exponential form:**

**(i) 4a3 × 6ab2 × c2**

**(ii) 5xy × 3x2y × 7y2**

**(iii) a3 × 3ab2 × 2a2b2**

**Solution:**

(i) 4a3 × 6ab2 × c2 is written in exponential form as 24a4b2c2.

(ii) 5xy × 3x2y × 7y2 is written in exponential form as 105x3y4.

(iii) a3 × 3ab2 × 2a2b2 is written in exponential form as 6a6b4.

**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:**

Number of bacteria in a culture = x

It is given that

Number of bacteria becomes square of itself in one week = x2

So the number of bacteria after two weeks = (x2)2 = x4

Hence, the number of bacteria after two weeks is x4.

**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:**

It is given that

Area of rectangle = l × b

Breadth = x cm

Length = 2/3 x cm

So the area of the rectangle = 2/3 x × x = 2/3 x2 cm2

Hence, the area of rectangle is 2/3 x2 cm2.

**6. If there are x rows of chairs and each row contains x2 chairs. Determine the total number of chairs.**

**Solution:**

Number of rows of chairs = x

Each row contains = x2 chairs

So the total numbers of chairs = number of rows of chairs × chairs in each row

We get

Total number of chairs = x × x2 = x3

Hence, the total number of chairs is x3.

Objective Type Questions PAGE: 8.13

**Mark the correct alternative in each of the following:**

**1. 5 more than twice a number x is written as**

(a) 5 + x + 2

(b) 2x + 5

(c) 2x − 5

(d) 5x + 2

(a) 5 + x + 2

(b) 2x + 5

(c) 2x − 5

(d) 5x + 2

**Solution:**

The option (b) is correct answer.

5 more than twice a number x is written as 2x + 5.

**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)

(a) x/2 + 5

(b) 2/x+5

(c) (x+2)/ 5

(d) x/ (2+5)

**Solution:**

The option (a) is correct answer.

The quotient of x by 2 is added to 5 is written as x/2 + 5.

**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

(a) x/3y

(b) 3x/y

(c) 3y/x

(d) xy/3

**Solution:**

The option (d) is correct answer.

It can be written as

x/3 × y = xy/3

**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

(a) x + y − 9

(b) 9 − (x+y)

(c) x+y/ 9

(d) 9/ x+y

**Solution:**

The option (a) is correct answer.

9 taken away from the sum of x and y is x + y – 9.

**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

(a) x/y + xy

(b) y/x + xy

(c) xy+x/ y

(d) xy+y/ x

**Solution:**

The option (a) is correct answer.

The quotient of x by y added to the product of x and y is written as x/y + xy.

**6. a**

(a)

(b)

(c)

(d) a

*2*b*3*× 2ab*2*is equal to(a)

*2*a*3*b*4*(b)

*2*a*3*b*5*(c)

*2*ab(d) a

*3*b*5***Solution:**

The option (b) is correct answer.

It can be written as

a

*2*b*3*× 2ab*2*= 2a2 × a × b3 × b2 = 2a3b5.**7. 4a2b3 × 3ab2 × 5a3b is equal to**

(a) 60a3b5

(b) 60a6b5

(c) 60a6b6

(d) a6b6

(a) 60a3b5

(b) 60a6b5

(c) 60a6b6

(d) a6b6

**Solution:**

The option (c) is correct answer.

It can be written as

4a2b3 × 3ab2 × 5a3b = 4 × 3 × 5 × a2 × a × a3 × b3 × b2 × b = 60a6b6

**8. If 2x2y and 3xy2 denote the length and breadth of a rectangle, the its area is**

(a) 6xy

(b) 6x2y2

(c) 6x3y3

(d) x3y3

(a) 6xy

(b) 6x2y2

(c) 6x3y3

(d) x3y3

**Solution:**

The option (c) is correct answer.

We know that area of a rectangle = length × breadth

By substituting the values

Area = 2x2y × 3xy2 = 6x3y3

**9. In a room there are x2 rows of chairs and each two contains 2x2 chairs. The total number of chairs in the room is**

(a) 2x3

(b) 2x4

(c) x4

(d) x4/2

(a) 2x3

(b) 2x4

(c) x4

(d) x4/2

**Solution:**

The option (b) is correct answer.

We know that

Total number of chairs in the room = Number of rows × Number of chairs

By substituting the values

Total number of chairs in the room = x2 × 2x2 = 2x4

**10. a3 × 2a2b × 3ab5 is equal to**

(a) a6b6

(b) 23a6b6

(c) 6a6b6

(d) None of these

(a) a6b6

(b) 23a6b6

(c) 6a6b6

(d) None of these

**Solution:**

The option (c) is correct answer.

It can be written as

a3 ×

*2*a2b × 3ab5 = 2 × 3a3 × a2 × a × b × b2 = 6a6b6