# Ncert exemplar Solutions Class 6 Mathematics Algebra

## Ncert exemplar Solutions Class 6 Mathematics Algebra

Welcome to NCTB Solutions. Here with this post we are going to help 6th class students for the Solutions of NCERT Class 6 Mathematics Book, Unit 7, Algebra. Here students can easily find step by step solutions of all the problems for Algebra, Also here our mathematics teacher’s are solved all the problems with easily understandable methods with proper guidance so that all the students can understand easily. Here in this post students will get Unit 7 solutions.

Algebra Unit 7 Solution :

Multiple Choice Questions :

Question no – (1)

If each match box contains 50 matchsticks, the number of matchsticks required to fill n such boxes is

Solution :

Match box contains 50 matchsticks.

The number of matchsticks required to fill n such boxes is

= 50 x n = 50n

Hence, correct answer is option – (B) 50n

Question no – (2)

Amulya is x years of age now. 5 years ago her age was

Solution :

Amulya is x years of age now.

5 years ago her age was (x – 5) years

So, the correct answer is option – (C) (x – 5) years.

Question no – (3)

Which of the following represents 6 × x

Solution :

The correct option is – (A) 6x

Because, we represent 6 × x as 6x

Question no – (4)

Which of the following is an equation?

Solution :

Equation containing variables and numbers.

x – 1 = 0 is the equation because after putting value of x we get the solution of the equation.

Thus, the correct answer is option – (C) x – 1 = 0

Question no – (5)

If x takes the value 2, then the value of x + 10 is

Solution :

Given, x + 10

We take x = 2 then value of equation is

x + 10

= 2 + 10

= 12

Therefore, the correct answer is option – (B) 12

Question no – (6)

If the perimeter of a regular hexagon is x metres, then the length of each of its sides is

Solution :

Regular hexagon has 6 sides.

Perimeter of a regular hexagon = 6 x Length of side.

Perimeter of a regular hexagon is x metres.

Length of side = x / 6 metre.

Hence, correct answer is option – (B) (x ÷ 6) metres

Question no – (7)

Which of the following equations has x = 2 as a solution?

Solution :

We put x = 2 in all these equations.

x + 2 = 5

x – 2 = 0

2 + 2 = 4 which is not equal to 5.

2 – 2 = 0 which is equal to 0.

x = 2 is a solution of x – 2 = 0.

Therefore, the correct answer is option – (B) x – 2 = 0

Question no – (8)

For any two integers x and y, which of the following suggests that operation of addition is commutative?

Solution :

If X + Y is the equation.

We interchange this equation as Y + X there is no effect on result is called Operation of addition is commutative.

Thus, the correct answer is option – (A) x + y = y + x

Question no – (9)

Which of the following equations does not have a solution in integers?

Solution :

We taking all equations and solve it.

(i) x + 1 = 1

x = 1-1 = 0 equation have a solution in integers.

(ii) x – 1 = 3

x = 3 + 1 = 4 equation have a solution in integers.

(iii) 2x + 1 = 6

2x = 6 – 1 = 5

X = 5/2 equation have a solution is not in integers.

Therefore, the correct answer is option (C) 2x + 1 = 6.

Question no – (10)

In algebra, a × b means ab, but in arithmetic 3 × 5 is

Solution :

In algebra, a × b means ab.

In arithmetic 3 × 5 is

= 3 × 5 = 15

Hence, correct answer is option (C) 15

Question no – (11)

In algebra, letters may stand for

Solution :

In algebra, letters may stand for unknown quantities.

So, the correct answer is option – (B) Unknown quantities.

Question no – (12)

“Variable” means that it

Solution :

“Variable” is a letter which we take different values of that variables to satisfy solution of equation.

Hence, correct answer is option – (A) Can take different values.

Question no – (13)

10 – x means

Solution :

10 – x means x is subtracted from 10

Thus, the correct answer is option – (C) x is subtracted from 10

Question no – (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?

Solution :

Savitri has a sum of Rs. x.

She spent Rs 1000 on grocery, Rs 500 on clothes and Rs 400 on education.

Total amount she spent = Rs.1900

Rs.1900 – Rs.200 = Rs.1700

Money (in Rs) is left with her = Total amount – Rs.1700

Money (in Rs) is left with her = X – Rs.1700

Therefore, the correct answer is option – (A) x – 1700

Question no – (15)

The perimeter of the triangle shown in Fig. 7.1 is

Solution :

Perimeter is the sum of all exterior sides.

Sides of triangle are x, x and y.

Perimeter of the triangle = x + x + y

Perimeter of the triangle = 2x + y

Hence, correct answer is option – (A) 2x + y

Question no – (16)

The area of a square having each side x is

Solution :

The area of a square = side x side

The area of a square having each side x.

The area of a square = X x X

Thus, the correct answer is option (A) x × x

Question no – (17)

The expression obtained when x is multiplied by 2 and then subtracted from 3 is

Solution :

The expression obtained when x is multiplied by 2 and then subtracted from 3

x is multiplied by 2 = 2 x X = 2x

Then subtracted from 3 = 3 – 2x

The expression obtained is 3 – 2x

Thus, the correct answer is option – (C) 3 – 2x

Question no – (18)

q/2 = 3 has a solution

Solution :

q/2 = 3

q = 3 x 2

q = 6

Solution of q is 6.

Therefore, the correct answer is option – (A) 6

Question no – (19)

x – 4 = -2 has a solution

Solution :

Given, x – 4 = – 2

x = – 2 + 4 (when there is equal sign between two and there is shifting of numbers its sign changes.)

x = 2 (when signs are different and addition is there, we subtracted smaller number from larger number and give larger number sign)

Hence, correct answer is option – (B) 2

Alternative Solution,

Given, x – 4 = – 2

= x = – 2 + 4

x = 2

The correct option is – (B) 2

Question no – (20)

4/2 = 2 denotes a

Solution :

Given, 4/2 = 2

The Equation containing all numerical terms.

Hence, It is called as numerical equation.

Therefore, the correct answer is option – (A) Numerical equation.

Question no – (21)

Kanta has p pencils in her box. She puts q more pencils in the box. The total number of pencils with her are

Solution :

Kanta has p pencils in her box.

She puts q more pencils in the box.

The total number of pencils with her = we have to add p pencils and q pencils.

The total number of pencils with her = p + q

Thus, the correct answer is option – (A) p + q

Question no – (22)

The equation 4x = 16 is satisfied by the following value of x

Solution :

The equation 4x = 16

We have to find value of x.

4x = 16

x = 16/4 (when there is equal sign between two and there is shifting of numbers its sign changes.)

x = 4

Hence, correct answer is option – (A) 4

Alternative Solution,

Given, 4x = 16

= x = 16/4

x = 4

The appropriate option is – (A) 4

Question no – (23)

I think of a number and on adding 13 to it, I get 27. The equation for this is

Solution :

Consider the number is X.

Adding 13 to it, I get 27.

= x + 13 = 27

Thus, the correct answer is option (D) x + 13 = 27

Fill the blanks :

Question no – (24)

Solution :

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

Explanation :

The distance (in km) travelled in h hours at a constant speed of 40 km per hour is Constant speed of 40 km

The distance (in km) travelled in h hours = we have to multiply h hours with 40

The distance (in km) travelled in h hours = 40 x h

Question no – (25)

Solution :

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

Explanation :

p kg of potatoes are bought for Rs 70.

Price of p kg of potatoes is Rs.70

Cost of 1 kg of potatoes (in Rs) = we have to divide Rs.70 by p kg of potatoes

Cost of 1 kg of potatoes (in Rs) = 70 / p

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

Question no – (26)

Solution :

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 8d + 2.

An auto rickshaw charges Rs 10 for the first kilometre.

8 for each such subsequent kilometre.

The total charge (in Rs) for d kilometres = 10 + 8 x (d – 1)

The total charge (in Rs) for d kilometres = 10 + 8d – 8

The total charge (in Rs) for d kilometres = 8d + 2

Question no – (27)

Solution :

If 7x + 4 = 25, then the value of x is 3.

Given, 7x + 4 = 25

= 7x = 25 – 4

= 7x = 21

= x = 21/7

x = 3

Alternative Solution,

7x + 4 = 25

7x = 25 – 4 (when there is equal sign between two and there is shifting of numbers its sign changes.)

7x = 21

X = 21/7

X = 3

If 7x + 4 = 25, then the value of x is 3

Question no – (28)

Solution :

The solution of the equation 3x + 7 = -20 is -3

Given, 3x + 7 = – 20

= 3x = – 20 – 7

= x = -27/9

x = – 3

Alternative Solution,

Equation 3x + 7 = -20

3x = – 20 – 7 (when there is equal sign between two and there is shifting of numbers its sign changes.)

3x = -27

X = – 27/3

X = – 9

Hence, the solution of the equation 3x + 7 = -20 is -9.

Question no – (29)

Solution :

‘x exceeds y by 7’ it is expressed as

x – y = 7 or x = y + 7

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

Question no – (30)

Solution :

8 more than three times the number x’ can be written as 3x + 8.

Question no – (31)

Solution :

Number of pencils bought for Rs x at the rate of Rs 2 per pencil is x/2.

Explanation :

Number of pencils bought for Rs x at the rate of Rs 2 per pencil

We have to divide total Rs. X by Rs. 2

Number of pencils bought for Rs x at the rate of Rs 2 per pencil = X / 2

Number of pencils bought for Rs x at the rate of Rs 2 per pencil is X / 2.

Question no – (32)

Solution :

The number of days in w weeks is 7w.

Explanation :

The number of days in a week is 7.

The number of days in w weeks is we have to multiply 7 with w.

The number of days in w weeks = 7 x w = 7w.

Question no – (33)

Solution :

Annual salary at r rupees per month along with a festival bonus of Rs 2000 is, = 12r + 2000

Explanation :

Annual salary at r rupees per month.

There are total 12 months in a year.

Annual salary = 12 x r

Festival bonus of Rs 2000. =

We have to add Rs.2000 in annual salary.

Annual salary at r rupees per month along with a festival bonus of Rs 2000 = 12r + 2000

Annual salary at r rupees per month along with a festival bonus of Rs 2000 is 12r + 2000

Question no – (34)

Solution :

The two digit number whose ten’s digit is ‘t’ and units’s digit is ‘u’ is, tu

Question no – (35)

Solution :

The variable used in the equation 2p + 8 = 18 is P.

Explanation :

The variable is a letter.

Equation 2p + 8 = 18

The variable used is p.

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

Question no – (36)

Solution :

x metres = 100x centimetres.

Explanation :

We know, 1 m = 100 cm

1 x X m = 100 x X cm

X m = 100x cm.

X metres = 100x centimetres

Question no – (37)

Solution :

p litres = 1000P millilitres.

Explanation :

We know, 1 litre = 1000 ml

P litre = 1000 x p ml

p litres = 1000 x p millilitre

Question no – (38)

Solution :

r rupees = 100r paise

Explanation :

We know, 1 rupee = 100 paise.

r rupees = 100 x r paise.

r rupees = 100 x r paise

Question no – (39)

Solution :

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

Explanation :

If the present age of Ramandeep is n years.

Then her age after 7 years = we have to add 7 in n years.

Her age after 7 years = n + 7

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

Question no – (40)

Solution :

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

Explanation :

Total 100 rupees.

I spend f rupees from 100 rupees.

The money left with me = we have to subtract f rupees from 100 rupees.

The money left with me = 100 – f

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

True or False :

Question no – (41)

Solution :

0 is a solution of the equation x + 1 = 0

This statement is False.

Because,

We put value of x = 0 in equation x + 1 = 0

x + 1 = 0 0 + 1 = 1 which is not solution of equation.

Question no – (42)

Solution :

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

This statement is – True.

Because,

x + 1 = 0

x = -1

We put x = -1 in 2x + 2 = 0

2x + 2 = 0

2 x (-1) + 2

– 2 + 2 = 0

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

Question no – (43)

Solution :

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

This statement is True.

Because, m is a whole number

Whole number is starting from 0,1,2,3…………….

2m denotes a multiple of 2.

Alternative Solution,

If m is a whole number

Value of m may be = 1, 2, 3,

2m = 2 × 1, = 2

= 2 × 2 = 4

= 2 × 3 = 6

2, 4, 6 are multiple of 2.

Question no – (44)

Solution :

The additive inverse of an integer x is 2x

This statement is False.

Because,

The additive inverse of an integer x is -x.

Question no – (45)

Solution :

If x is a negative integer, – x is a positive integer

This statement is True.

Because, x is a negative integer, – x is a positive integer.

Alternative Solution,

If is negative integer such as x = -5

= (-x) – (-5)

= + 5

Then (- x) is a positive integer.

Question no – (46)

Solution :

2x – 5 > 11 is an equation

This statement is False.

Because,

2x – 5 = 11 is an equation.

2x – 5 > 11 comparison symbol not used in equation.

Alternative Solution,

2x – 50 > 11 is not an equation because from here we can’t the exact solution and value of x.

Question no – (47)

Solution :

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

This statement is True.

In a equation L.H.S = R.H.S

Such as 2x + 2 = 0

The value of x = – 1

= 2 (- 1) + 2

= – 2 + 2

= 0

L.H.S = R.H.S

Question no – (48)

Solution :

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

This statement is False.

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

Question no – (49)

Solution :

a = 3 is a solution of the equation 2a – 1 = 5

This statement is True.

Because,

2a – 1 = 5

We put a = 3

2 x 3 – 1

= 6 – 1 = 5

a = 3 is a solution of the equation 2a – 1 = 5

Alternative Solution,

Given statement is True.

Given, 2a – 1 = 5

= 2a = 5 + 1

= a = 6/2

= a = 3

a = 3 is the solution of the equation.

Question no – (50)

Solution :

The distance between New Delhi and Bhopal is not a variable

This statement is True.

Because, The distance between New Delhi and Bhopal is not a variable. It is a numerical value.

Question no – (51)

Solution :

t minutes are equal to 60t seconds

This statement is True.

Because we know,

1 minute = 60 seconds

t minutes = 60t seconds

Alternative Solution,

Given statement is True.

t minutes = 1 × t minutes

= 60 × t seconds

60 t seconds

Question no – (52)

Solution :

x = 5 is the solution of the equation 3x + 2 = 20

This statement is False.

Because,

Equation 3x + 2 = 20

We put x = 5

3x + 2

= 3 x 5 + 2

= 17

LHS is not equal to RHS.

x = 5 is not the solution of the equation 3x + 2 = 20

Alternative Solution,

Given statement is False.

Given, 3x + 2 = 20

= 3x = 20 – 2

= 3x = 18

= x = 6

x = 6 is the solution of the equation not x = 5

Question no – (53)

Solution :

One third of a number added to itself gives 8’, can be expressed as x/3+8 = x

This statement is False.

Because,

Let the number is x.

One third of a number = x/3

‘One third of a number added to itself gives 8’ = x / 3 + x = 8

‘One third of a number added to itself gives 8’, cannot be expressed as x/3+8 = x

Alternative Solution,

Given statement is False.

Let, the number = x

One third of it = 1/3 × x = x/3

It expressed as, x/3 + x = 8

Question no – (54)

Solution :

The difference between the ages of two sisters Leela and Yamini is a variable –

This statement is False.

Because, The difference between the ages of two sisters Leela and Yamini is not a variable. It is numerical value.

Question no – (55)

Solution :

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

This statement is False.

Because, The number of lines that can be drawn through a point is not a variable. It is numerical value.

Write the corresponding expressions :

Question no – (56)

Solution :

Let the number is x.

Twice the number = 2x

One more than twice the number = we are adding 1 in 2x

One more than twice the number = 2x + 1.

Alternative Solution,

The corresponding expression,

= 2 × x + 1

= 2x + 1

Question no – (57)

Solution :

Present temperature be x °C.

20°C less than the present temperature = x°C – 20°C

20°C less than the present temperature. = x – 20

Alternative Solution,

20°C less than the present temperature,

The corresponding expressions,

= t – 20

Question no – (58)

Solution :

Successor of any number is the next number of that number.

Let integer is p.

The successor of an integer = we are adding 1 to integer p.

The successor of an integer = p + 1

Alternative Solution,

The successor of an integer,

Let, an integer = n

The successor of an integer is,

= n + 1

Question no – (59)

Solution :

Side of the triangle is m.

Perimeter is the sum of all exterior sides.

Equilateral triangle having all sides are equal length.

Perimeter of an equilateral triangle = m + m + m

Perimeter of an equilateral triangle = 3m

Alternative Solution,

The side of the triangle is m,

The perimeter of an equilateral triangle = 3m

Question no – (60)

Solution :

Area of the rectangle = Length x Breadth

Length of the rectangle = k units

Breadth of the rectangle = n units.

Area of the rectangle = k units x n units.

Area of the rectangle = k x n

Alternative Solution,

The area of the rectangle = Length × Breadth

= k × n

= kn sq. unit

Question no – (61)

Solution :

Omar sister does ‘x’ hour work.

Omar helps his mother 1 hour more than his sister does = X + 1

Alternative Solution,

Let, Oman’s sister helps his mother = x hour.

Omar helps his mother

= x + 1 hour.

Question no – (62)

Solution :

Let, x be the integer.

1st odd integer = 2 x x + 1

2nd odd integer = 2 x x + 3

Two consecutive odd integers = 2x + 1 and 2x + 3

Alternative Solution,

Two consecutive odd integers are

2n + 1 and 2n + 3

Question no – (63)

Solution :

Let, p be the integer.

1st even integer = 2 x p

2nd even integer = 2 x p + 2

Two consecutive odd integers = 2p and 2p + 2

Alternative Solution,

Two consecutive even integer are 2m and 2m + 2

Question no – (64)

Solution :

Let, p is the numbers starting from 1, 2,………

Multiple of 5 = we multiply p with 5

Multiple of 5 = 5p

Alternative Solution,

Multiple of 5 is ,

= 5n

Question no – (65)

Solution :

Let, numerator of fraction is m.

The denominator of a fraction is 1 more than its numerator.

Denominator of fraction = m + 1

Fraction is m / m + 1

Alternative Solution,

Let, the numerator = x

The denominator = x + 1

The fraction = x/x + 1

Question no – (66)

Solution :

Let, height of Empire State building is q.

The height of Mount Everest is 20 times the height of Empire State building.

The height of Mount Everest = we multiply 20 with q.

The height of Mount Everest = 20q

Alternative Solution,

Let, the height of Empire state bolding = x

The Height of Mount Everest is

= 20x

Question no – (67)

Solution :

If a note book costs Rs p.

Cost of two notebooks = Rs. 2p

Cost of 1 pencil = Rs.3

The total cost (in Rs) of two note books and one pencil = Rs 2p + Rs.3

The total cost (in Rs) of two note books and one pencil = 2p + 3

Alternative Solution,

Total cost of two notebooks and one pencil,

= 2p + 3

Question no – (68)

Solution :

z is multiplied by –3 = z x -3 = -3z.

And the result is subtracted from 13 = 13 – ( -3z) = 13 + 3z

z is multiplied by –3 and the result is subtracted from 13 = 13 + 3z

Question no – (69)

Solution :

p is divided by 11 = p / 11

And the result is added to 10 = p / 11 + 10

p is divided by 11 and the result is added to 10 = p / 11 + 10

Alternative Solution,

p is divided by 11 and the result is added to 10

= P/11 + 10

Question no – (70)

Solution :

Smallest natural number is 1.

x times of 3 = 3x

x times of 3 is added to the smallest natural number = 3x + 1

Alternative Solution,

x times of 3 is added to the smallest natural number,

= 3x + 1

Question no – (71)

Solution :

Smallest two digit number is 10.

6 times q = 6 x q = 6q

6 times q is subtracted from the smallest two digit number = 1 – 6q.

Alternative Solution,

6 times q is subtracted from the smallest two digit number,

= 10 – 6q

Question no – (72)

Solution :

We have to form 2 equations having solution is 2.

(i) 6x + 8 = 20

We solve equation 1 the solution of equation is 2.

(ii) 4p – 7 = 1

We solve equation 2 the solution of equation is 2.

Alternative Solution,

Two equations for which 2 is the solution,

= x – 2 = 0,

= 2x – 3 = 1

Question no – (73)

Solution :

We have to form an equation having solution is 0.

4z + 4 = 4

We solve equation, the solution of equation is 0.

Alternative Solution,

An equation for which 0 is a solution,

2x + 5 = 5

= 2x = 5 – 5

x = 0

Question no – (74)

Solution :

Whole number is the numbers starting from 0.

Solution is not a whole number means solution is negative integer.

2y + 2 = 0

We solve equation, the solution of equation is -1, which is not a whole number.

Alternative Solution,

An equation whose solution is not a whole number,

= x + 1 = 0

Converting expressions into statements in ordinary language :

Question no – (75)

Solution :

A pencil costs Rs. P

A pen costs Rs. 5p.

We making statement from above expression,

The cost of pen is 5 times the cost of a pencil.

Question no – (76)

Solution :

Leela contributed Rs y towards the Prime Minister’s Relief Fund.

Leela is now left with Rs (y + 10000)

We making statement from above expression,

Amount left with Leela is Rs 10,000 more than the amount she contributed towards Prime Minister’s Relief fund.

Question no – (77)

Solution :

Kartik is n years old.

His father is 7n years old.

We making statement from above expression,

Age of Kartik’s Father is seven times the age of Kartik.

Question no – (78)

Solution :

The maximum temperature on a day in Delhi was p°C.

The minimum temperature was (p – 10)°C.

We making statement from above expression,

The difference between maximum and minimum temperature on a day in Delhi was 10°C.

Question no – (79)

Solution :

John planted t plants last year.

His friend Jay planted 2t + 10 plants that year.

We making statement from above expression,

Last year Jay planted 10 more plants than twice the number of plants planted by John.

Question no – (80)

Solution :

Sharad used to take p cups tea a day.

After having some health problem, he takes p – 5 cups of tea a day.

We making statement from above expression,

Sharad reduced the consumption of tea per day by 5 cups after having some health problem.

Question no – (81)

Solution :

The number of students dropping out of school last year was m.

Number of students dropping out of school this year is m – 30.

We making statement from above expression,

The number of students dropping out this year is 30 less than the number of students dropped last year.

Question no – (82)

Solution :

Price of petrol was Rs p per litre last month

Price of petrol now is Rs (p – 5) per litre.

We making statement from above expression,

The price of petrol per litre reduced this month by Rs.5 than its price last month.

Question no – (83)

Solution :

Khader’s monthly salary was Rs P in the year 2005.

Salary in 2006 was Rs (P + 1000).

We making statement from above expression,

Khader’s monthly salary increased by Rs 1000 in the year 2006 than in 2005.

Question no – (84)

Solution :

The number of girls enrolled in a school last year was g.

The number of girls enrolled this year in the school is 3g – 10.

We making statement from above expression,

The number of girls enrolled this year was 10 less than 3 times the girls enrolled last year.

Question no – (85)

Solution :

(a) 13 subtracted from twice a number gives 3.

= 3x – 13 = 3

(b) One fifth of a number is 5 less than that number.

= x/5 = x – 5

(c) Two-third of number is 12.

= 2x/3 = 12

(d) 9 added to twice a number gives 13

= 2x + 9 = 13

(e) 1 subtracted from one-third of a number gives 1.

= x/3 – 1 = 1

Question no – (86)

Solution :

(a) The perimeter (p) of an equilateral triangle is three times of its side (a).

Equation = P = 3a

(b) The diameter (d) of a circle is twice its radius (r).

Equation = d = 2r

(c) The selling price (s) of an item is equal to the sum of the cost price (c) of an item and the profit (p) earned.

Equation = s = c + p

(d) Amount (a) is equal to the sum of principal (p) and interest (i).

Equation = a = p + I

Question no – (87)

Solution :

Kanika’s present age be x years.

(i) her brother is 2 years younger.

Her brother is 2 year smaller than her age.

We transform this statement into equation is,

x – 2

(ii) Her father’s age exceeds her age by 35 years.

Her father’s age 35 years more than her age.

x + 35

(iii) Mother’s age is 3 years less than that of her father.

Her father’s age = x + 35

Mother’s age is 3 years less than that of her father.

Her Mother’s age = x + 35 – 3

Her Mother’s age = x – 32

(iv) Her grandfather’s age is 8 times of her age

Her grandfather’s age = we multiply his age by 8

Her grandfather’s age = 8x

= 8x

Alternative Solution,

 Situation (described in ordinary language) : Expressions : (i) Her brother is 2 years younger. (i) x – 2 (ii) Her father’s age exceeds her age by 35 years. (ii) x + 35 (iii) Mother’s age is 3 years less than that of her father. (iii) x + 32 (iv) Her grand father’s age is 8 times of her age. (iv) 8x

Question no – (88)

Solution :

m is a whole number less than 5

Whole number is starting from 0.

(i) We put first m = 0,

2m – 5

= 2 x 0 – 5

= – 5

Not satisfy solution of equation.

(ii) We put first m = 1,

2m – 5

= 2 x 1 – 5

= 2 – 5

= -3

Not satisfy solution of equation.

(iii) We put first m = 2,

2m – 5

= 2 x 2 – 5

= 4 – 5

= -1

m = 2 Satisfy solution of equation.

Alternative Solution,

 M 0 1 2 3 4 2m – 5 – 5 – 3 – 1 1 3

Question no – (89)

Solution :

A class with p students has planned a picnic.

Rs 50 per student is collected.

Total amount collected by p students = 50p

Rs 1800 is paid in advance for transport.

Money is left with them to spend on other items

We have to subtract Rs.1800 from 50p.

Money is left with them to spend on other items

= 50p – 1800

Alternative Solution,

As per the question,

Class with p students has planned a picnic

Rs 50 per student is collected

Out of which Rs 1800 is paid

= 50p – 1800

Question no – (90)

Solution :

In a village, there are 8 water tanks to collect rain water.

x litres of rain water is collected per tank.

Total litres of water collected in 8 water tanks = 8x Litres.

100 litres of water was already there in one of the tanks.

Total amount of water in the tanks on that day =

We have to add 100 litres of water from total 8x Litres.

Total amount of water in the tanks on that day = (8x + 100) Litres.

Alternative Solution,

The total amount of water in the tanks on that day,

= (8x + 100)^2

Question no – (91)

Solution :

We know, Area of a square = side x side

Side of a square is m cm.

Area of a square = m x m sq.cm

Alternative Solution,

The area of square = m × m sq. cm
= m^2 sq. cm

The area of square is m^2 sq. cm.

Question no – (92)

Solution :

Perimeter of a triangle is found by using the formula,

P = a + b + c.

Where a, b and c are the sides of the triangle.

We are writing statements from the expression is,

The perimeter of a triangle is the sum of all its sides.

Question no – (93)

Solution :

Perimeter of a rectangle is found by using the formula,

P = 2 ( l + w)

Where l and w are respectively the length and breadth of the rectangle.

We are writing statements from the expression is,

The perimeter of a rectangle is twice the sum of its length and breadth.

Question no – (94)

Solution :

On my last birthday, I weighed 40 kg.

I put on m kg of weight after a year.

My present weight,

We are adding m kg weight to the previous year 40 kg weight.

My present weight = (m + 40) kg.

Question no – (95)

Solution :

(i) What will be the length (in cm) of the aluminium strip required to frame the board, if 10 cm extra strip is required to fix it properly.

We have to find the perimeter of bulletin board.

Bulletin board is of rectangular shaped.

Perimeter of rectangular bulletin board = 2 (Length + Breadth)

Perimeter of rectangular bulletin board =2 (r + t)

10 cm extra strip is required to fix it properly.

Length (in cm) of the aluminium strip required to frame the board = 2 (r + t) + 10 cm

(ii) If x nails are used to repair one board, how many nails will be required to repair 15 such boards?

X nails are used to repair one board.

Nails will be required to repair 15 such boards

We have to multiply 15 with x nails.

Nails required to repair 15 such boards = 15x

(iii) If 500 sq.cm extra cloth per board is required to cover the edges, what will be the total area of the cloth required to cover 8 such boards?

500 sq.cm extra cloth per board is required to cover the edges.

Area of rectangular bulletin board = Length x Breadth

Area of rectangular bulletin board = r x t

There are 8 boards.

Area of rectangular 8 bulletin board = 8(r x t)

500 sq.cm extra cloth per board is required to cover the edges

Total 8 board

= 500 x 8

= 4000 sq.cm

Total area of the cloth required to cover 8 such boards

= 8(r x t) + 4000 sq.cm

(iv) What will be the expenditure for making 23 boards, if the carpenter charges Rs x per board.

The carpenter charges Rs x per board.

Expenditure for making 23 boards =

We are multiplying 23 with Rs.x

Expenditure for making 23 boards = 23x

Question no – (96)

Solution :

(i) required length aluminium strip,

= 2 (r + t) + 10

(ii) = 15 × x = 15x nails will be required.

(iii) Total area of cloth required,

= (8rt + 4000) sq. m

(iv) The expenditure will be,

= 23 × x

= 23x Rupees.

Question no – (96)

Solution :

(i) After 4 years?

Sunita is half the age of her mother Geeta.

Let, mother geeta age is ‘x’ year.

Sunita’s age = x / 2 years.

(i) Mother Geeta age after 4 years = x + 4 years

(ii) Sunita’s age after 4 years = x / 2 + 4 years.

(ii) Before 3 years?

(i) After 4 years?

Sunita is half the age of her mother Geeta.

Let, mother Geeta’s age is ‘x’ year.

Sunita’s age = x / 2 years.

(i) Mother Geeta age after 4 years = x + 4 years

(ii) Sunita’s age after 4 years = x / 2 + 4 years.

(ii) Before 3 years?

(i) Mother Geeta age before 3 years = x – 3 years

(ii) Sunita’s age before 3 years = x/2 – 3 years.

(i) Mother Geeta age before 3 years = x – 3 years

(ii) Sunita’s age before 3 years = x/2 – 3 years

Alternative Solution,

Let, Sunita’s age x

Her Mathers Geeta’s age = 2x

(i) After 4 years?

After 4 year,

Sunita’s age = x + 4

Geeta’s age = 2x + 4

(ii) Before 3 years?

Sunita’s age = x – 3

Geeta’s age = 2x – 3

Question no – (97)

Solution :

 Column I : Column II : (i) The number of corners of a quadrilateral (B) constant (ii) The variable in the equation 2p + 3 = 5 (E) p (iii) The solution of the equation x + 2 = 3 (C) +1 (iv) solution of the equation 2p + 3 = 5 (C) +1 (v) A sign used in an equation (A) =

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Updated: June 27, 2023 — 7:34 am