In this chapter, we provide NCERT Exemplar Problems Solutions for Class 6 Maths Chapter 7 Algebra for English medium students, Which will very helpful for every student in their exams. Students can download the latest NCERT Exemplar Problems Solutions for Class 6 Maths Chapter 7 Algebra pdf, free NCERT Exemplar Problems Solutions for Class 6 Maths Chapter 7 Algebra book pdf download. Now you will get step by step solution to each question.

Textbook | NCERT |

Class | Class 6 |

Subject | Maths |

Chapter | Chapter 7 |

Chapter Name | Algebra |

Category | NCERT Exemplar |

## NCERT Exemplar Class 6 Maths Chapter 7 Algebra

**Question 1:**

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

(a) 50 + n (b) 50 n (c) 50 + n (d) 50 – n**Solution:**

(b) Given, each matchbox contains 50 matchsticks.

Then, total number of matchsticks in n boxes = Matchsticks in one box x Total boxes

= 50 x n = 50 n

Hence, (b) is correct option.

**Question 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) year**Solution:**

(c) Given, Amulya’s present age = x

5 years ago, Amulya’s age = (x – 5) years Hence, (c) is correct option.

**Question 3:**

Which of the following represent 6 x b?

(a) 6 b (b) 6/b (c) 6 + b (d) 6 – b**Solution:**

(a) Given that, 6 x b = 6b

Hence, (a) is correct option.**Note:** In algebra multiplication, sign does not show in the product (result).

**Question 4:**

Which of the following is an equation?

(a) x +1 (b) x -1 (c) x -1 = 0 (d) x +1 > 0**Solution:**

(c) We know that, an expression with a variable, constants and the sign of equality (=) is called an equation.

So, x -1 = 0 is an equation.

Hence, (c) is correct option.

**Question 5:**

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

(a) 20 (b) 12 (c) 5 (d) 8**Solution:**

(b) Given, expression = x + 10 On substituting x = 2, we get x + 10 = 2 + 10 = 12 Hence, (b) is correct option.

**Question 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**Solution:**

(b) Given, perimeter of regular hexagon is x metres, Number of sides in regular hexagan = 6

Length of each sides

Hence, (b) is correct option.

**Question 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**Solution:**

(b) To get solution as x = 2, solve each equation. .

**Question 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**Solution:**

(a) Let a and b be two integers, then in commutative property

a + b = b + a

Here, x and y are integers.

Then, x + y= y+ x

Hence, (a) is correct option.

**Question 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**Solution:**

**Question 10:**

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

(a) 35 (b) 53 (c) 15 (d) 8**Solution:**

(c) Given, in algebra, a x b = ab, which means a is multiplied by b.

Also, in arithmetic, 3 x 5 means 3 is multiplied by 5.

3 x 5 = 15

Hence, (c) is correct option.

**Question 11:**

In algebra, letters may stand for

(a) known quantities (b) unknown quantities

(c) fixed numbers (d) None of these**Solution:**

(b) In algebra, letters may stand for unknown quantities.

Hence, (b) is correct option.

**Question 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**Solution:**

(a) Since, the value of a variable is not fixed.

So, variable means that it can take different values.

Hence, (a) is correct option.

**Question 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**Solution:**

(c) 10 – x means x is subtracted from 10.

Hence, (c) is correct option.

**Question 14:**

Savitri has a sum of Rs. x. She spent 11000 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**Solution:**

(a) Given,

Savitri has total money = Rs. x Spent on grocery = Rs. 1000 Spent on clothes = Rs. 500 Spent on education = Rs. 400 Received as a gift = Rs. 200

Then, money left with her = Rs. {x – [1000 + 500 + 400 – 200]}

= Rs.{x-[1900 – 200]}

= Rs.{x – 1700}

Hence, (a) is correct option.

**Question 15:**

The perimeter of the triangle as shown in the figure is

(a) 2x + y (b) x + 2y (c) x + y (d) 2x – y**Solution:**

(a) We know that, perimeter of the triangle = Sum of all sides of triangle

Here, sides are x , x and y.

Perimeter of the triangle = x + x + y = 2 x+y

Hence, (a) is correct option.

**Question 16:**

The area of a square having each side x is

(a) x x x (b) 4x (c) x + x (d)4 + x**Solution:**

(a) Here, side = x

We know that, area of square = Side x Side

Area of square = x x x

Hence, (a) correct option.

**Question 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**Solution:**

(c) First x is multiplied by 2.

2x x – 2x

Now, 2x is subtracted from 3 = 3 – 2x

Hence, (c) is correct option.

**Question 18:**

q/2= 3 has a solution

(a) 6 (b) 8 (c)3 (d)2**Solution:**

**Question 19:**

x – 4 = – 2 has a solution

(a) 6 (b) 2 (c)-6 (d)-2**Solution:**

(b) Given equation is

x- 4 = -2

=> x = -2 + 4

=> x = 2

Hence, (b) is correct option.

**Question 20:**

4/2 = 2 denotes a

(a) numerical equation (b) algebraic expression

(c) equation with a variable (d) false statement**Solution:**

(a) We know that, an equation which contains only numbers is called a numerical equation.

Since, equation 4/2 = 2 contains only numbers, so it is a numerical equation.

Hence, (a) is correct option.

**Question 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**Solution:**

(a) Given, pencils in Kanta’s box = p

When q more pencils are put in the box, then total number of pencils = p + q Hence, (a) is correct option.

**Question 22:**

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

(a) 4 (b) 2 (c) 12 (d) -12**Solution:**

**Question 23:**

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

(a) x-27 =13 (b) x -13= 27 (c) x+ 27 =13 (d) x +13 = 27**Solution:**

(d) Let the number be x.

According to the question, x + 13 = 27 Hence, (d) is correct option.

**Fill in the Blanks**

In questions 24 to 40, fill in the blanks to make the statements true.

**Question 24:**

The distance (in km) traveled in h hours at a constant speed of 40 km per hours is ………….**Solution:**

40 h km

**Question 25:**

p kg of potatoes are bought for Rs. 70. Cost of 1 kg of potatoes (in Rs.) is ……………**Solution:**

Rs: 70/p

Cost of p kg potatoes = Rs. 70

Cost of 1 kg potatoes = Rs. 70/p

**Question 26:**

An autorickshaw charges Rs. 10 for the first kilometer, then Rs. 8 for each such subsequent kilometre.

The total charge (in Rs.) for d kilometres is …………**Solution:**

(8d + 2)

Given, autorickshaw charges for first km = Rs. 10

After that, each such subsequent km charge = Rs. 8 Total distance = d km

∴ Total charge = 10 + (d -1) 8

= 10+ 8d – 8= Rs. 8d + 2

**Question 27:**

If 7x + 4 = 25, then the value of x is ………..**Solution:**

x = 3

**Question 28:**

The solution of the equation 3x + 7 = – 20 is …………**Solution:**

-9

**Question 29:**

‘x exceeds y by T can be expressed as …………..**Solution:**

‘x exceeds y by 7′ can be expressed as x = y + 7.

**Question 30:**

8 more than three times the number ‘x’ can be written as ………..**Solution:**

Given, number = x

Now, three times of x = 3x

Then, according to the question = 3x + 8

**Question 31:**

Number of pencils for Rs. x at the rate of r 2 per pencil is ………….**Solution:**

**Question 32:**

The number of days in W weeks is …………**Solution:**

We know that, number of days in 1 week = 7 So, number of days in W weeks = 7 x W =7W

**Question 33:**

Annual salary at r rupees per month along with a festival bonus of Rs. 2000 is ……….**Solution:**

Given, bonus = Rs. 2000

Total months in one year =12 Per months salary = Rs. r

∴Annual salary = Rs. (12 r + 2000)

**Question 34:**

The two digit number whose ten’s digit is t and unit digit is u is ………….**Solution:**

The number is 10 f + u.

Here, ten’s digit = t, unit digit = u

∴ Required number = 10 x Ten’s digit + 1 x Unit digit

=10 x t + u = 10t + u

**Question 35:**

The variable used in the equation 2p + 8 = 18 is ………….**Solution:**

Given equation is 2p + 8 = 18.

Here, variable p is used in the equation 2p+ 8 = 18 Note Value of p is not fixed.

**Question 36:**

p metres = ………. centimetres**Solution:**

100 p. We know that, 1 metre = 100 cm

So, p metres = 100 x p = 100 p cm

**Question 37:**

p litres = ……….. millilitres**Solution:**

1000 p

We know that, 1 litre = 1000 millilitres

So, p litres = 1000 x p millilitres = 1000 p millilitres

**Question 38:**

rupees = ………… paise**Solution:**

100 r

We know that, 1 rupee = 100 paise

So, r rupees = 100 x r paise = 100 r paise

**Question 39:**

If the present age of Ramandeep is n years, then her age after 7 years will be …………..**Solution:**

(n + 7) years

Given, present age of Ramandeep = n years Age after 7 yr = (Present age + 7) years = (n + 7) years

**Question 40:**

If I spend / rupees from 100 rupees, money left with me is rupees.**Solution:**

Rs. (100-f)

Given, total money = Rs. 100

Spend money = Rs. f

Left money with me = Total money – Money spent = Rs. (100 – f)

**True/False**

In questions 41 to 55, state whether the given statements are True or False.

**Question 41:**

0 is a solution of the equation x + 1 = 0.**Solution:**

False

Given, equation x + 1 = 0

=> x = -1 [transposing+1 to RHS]

So, -1 is the solution of the given equation.

Hence, the given statement is false.

**Question 42:**

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

True

**Question 43:**

If m is a whole number, then 2 m denotes a multiple of 2.**Solution:**

True

Given, m is a whole number.

Put m=0,1,2, 3,…

2 m = 2 x 0 = 0,

2 m = 2 x 1 = 2

2 m = 2 x 2 = 4,

2 m = 2 x 3 = 6

Clearly, 2 m is the multiple of 2.

So, multiples of 2 are 2, 4, 6,…

Hence, the statement is true.

**Question 44:**

The additive inverse of an integer x is Z x.**Solution:**

False

Given, integer = x Additive inverse of x = – x

Then, the sum of integer and its additive inverse = x + (- x)

= x – x = 0

But according to question, additive inverse is 2 x. So, the statement is false.

**Question 45:**

If x is a negative integer, – x is a positive integer.**Solution:**

True

If at is a negative integer, then positive integer = – (x)

= – x

So, the statement is true.

**Question 46:**

2x – 5 > 11 is an equation.**Solution:**

False

2x – 5 > 11 is not an equation because it has no equality (=) sign.

So, the statement is false.

**Question 47:**

In an equation, the LHS is equal to the RHS.**Solution:**

True

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

So, the statement is true.

**Question 48:**

In the equation 7k – 7 = 7, the variable is 7.**Solution:**

False

Equation is 7k — 7=7 Here, variable is k.

So, the statement is false.

**Question 49:**

a = 3 is a solution of the equation 2a – 1 = 5.**Solution:**

**Question 50:**

The distance between New Delhi and Bhopal is not a variable.**Solution:**

True

Distance between New Delhi and Bhopal is fixed.

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

**Question 51:**

1 minutes are equal to 601 seconds.**Solution:**

True

We know that, 1 minute = 60 seconds [transposing +2 to RHS]

minutes = 60 x f seconds = 601 seconds So, the statement is true.

**Question 52:**

x = 5 is the solution of the equation 3x + 2 = 20.**Solution:**

False

**Question 53:**

‘One-third of a number added to itself gives 8’, can be expressed as

(x/3) + 8 = x.**Solution:**

**Question 54:**

The difference between the ages of two sisters Leela and Yamini is a variable.**Solution:**

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.

**Question 55:**

The number of lines that can be drawn through a point is a variable.**Solution:**

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.

**Question 56:**

One more than twice the number.**Solution:**

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.

**Question 57:**

20°C less than the present temperature.**Solution:**

Let the present temperature be f°C.

∴ Required expression

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

**Question 58:**

The successor of an integer.**Solution:**

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.

**Question 59:**

The perimeter of an equilateral triangle, if side of the triangle is m.**Solution:**

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.

**Question 60:**

Area of the rectangle with length k units and breadth n units.**Solution:**

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.

**Question 61:**

Omar helps his mother 1 hour more than his sister does.**Solution:**

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

**Question 62:**

Two consecutive odd integers.**Solution:**

Any odd integer can be written as 2n + 1, where n is an integer.

So, next odd integer will be (2n + 1) + 2, i.e. 2n + 3.

Hence, two consecutive odd integers are 2n + 1 and 2n + 3.**Note:** A sequence of consecutive even or odd integer is a list of two or more integers which increase by 2 from one integer to the next consecutive integer. They have a difference of 2 between every two integers.

**Question 63:**

Two consecutive even integers.**Solution:**

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.

**Question 64:**

Multiple of 5.**Solution:**

The multiples of a whole number are found by taking the product of any counting number and that whole number.

Multiples of 5 are Multiply 5 by 1 -+ 5 x 1 = 5 Multiply 5 by 2 -+ 5 x 2 and so on.

Hence, multiple of 5 = 5n, where n is any whole number.

**Question 65:**

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

**Question 66:**

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

Let height of Empire State be h metre.

Then, height of Mount Everest = 20 x h = 20h metre Hence, the required expression is 20 h.

**Question 67:**

If a notebook costs t p and a pencil costs Rs. 3, then the total cost (in Rs.) of two notebooks and one pencil.**Solution:**

Given,

Cost of one notebook = Rs. p Cost of 2 notebooks = 2 x p=Rs.2p Similarly, cost of one pencil = Rs. 3 Now, total cost = Cost of 2 notebooks + Cost of one pencil = Rs.(2p+3)

Hence, the required expression is 2p + 3.

**Question 68:**

z is multiplied by -3 and the result is subtracted from 13.**Solution:**

According to the question,

z is multiplied by – 3 = (- 3) x z

Now, result is subtracted from 13 = 13 – (- 3) z = 13 + 3z

Hence, the required expression is 13 + 3z.

**Question 69:**

p is divided by 11 and the result is added to 10.**Solution:**

**Question 70:**

x times of 3 is added to the smallest natural number.**Solution:**

According to the question, x times of 3 = 3x and smallest natural number = 1 Now, according to question,

Resulting expression = 3x + 1 Hence, the required expression is 3x + 1.

**Question 71:**

6 times q is subtracted from the smallest two digit number.**Solution:**

6 times of q = 6g

and smallest two digit number = 10

Then, according to question, resulting expression = 10 – 6g

Hence, the required expression = 10 – 6g

**Question 72:**

Write two equations for which 2 is the solution.**Solution:**

**Question 73:**

Write an equation for which 0 is a solution.**Solution:**

**Question 74:**

Write an equation whose solution is not a whole number.**Solution:**

We know that, whole numbers are 0,1,2, 3,…

Now, let the one number be x whose solution is not a whole number.

For getting equation, the number x will be added to 1 which results into 0. Then,

x + 1 = 0 [transposing +1 to RHS]

On solving x = -1

which is not a whole number.

Hence, required equation is x + 1 = 0.

In questions 75 to 84, change the statements, converting expression into statements in ordinary language.

**Question 75:**

A pencil costs Rs. p and a pen costs 15p.**Solution:**

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

**Question 76:**

Leela contributed Rs. y towards the Prime Minister’s Relief Fund. Leela is now left with Rs. (y + 10000).**Solution:**

Amount left with Leela is Rs. 10000 more than the amount she contributed towards Prime Minister’s Relief Fund.

**Question 77:**

Kartik is n years old. His father is In years old.**Solution:**

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

**Question 78:**

The maximum temperature on a day in Delhi was p°C. The minimum temperature was (p – 10)°C.**Solution:**

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

**Question 79:**

John planted t plants last year. His friend Jay planted 2t + 10 plants that year.**Solution:**

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

**Question 80:**

Sharad used to take p cups of tea in a day. After having some health problem, he takes p – 5 cups of tea a day.**Solution:**

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

**Question 81:**

The number of students dropping out of school last year was m. Number of students dropping out of school this year is m – 30.**Solution:**

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

**Question 82:**

Price of petrol was Rs. p per litre last month. Price of petrol now is Rs.(p – 5) per litre.**Solution:**

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

**Question 83:**

Khader’s monthly salary was Rs. p in the year 2005. His salary in 2006 was

Rs.(p+1000).**Solution:**

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

**Question 84:**

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.**Solution:**

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

**Question 85:**

Translate each of the following statements into an equation, using x as the variable.

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

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

(c) Two-third of a number is 12.

(d) 9 added to twice a number gives 13.

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

**Question 86:**

Translate each of the following statements into an equation.

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

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

(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.

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

(a) Given,

Perimeter of an equilateral triangle = p

Side of an equilateral triangle = a Then, three times of side (a) = 3a Then, according to the question, p = 3a

(b) Given,

Diameter of a circle = d

Radius of a circle = r

Twice of radius, r = 2r

Then, according to the question, d = 2r

(c) Given, selling price of an item = Rs. s Cost price of an item = Rs. c

Profit = Rs. p

Then, according to question, s = c + p

(d) Given, amount = Rs. a Principal = Rs. p Interest = Rs.

Then, according to the question, a = p + i

**Question 87:**

Let Kanika’s present age be x years. Complete the following table, showing ages of her relatives.

(i) Her brother is 2 years younger.

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

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

(iv) Her grandfather’s age is 8 times of her age.**Solution:**

Given, Kanika’s present age =xyr

(i) Her brother’s age = x – 2 y r [v her brother is 2 years younger]

(ii) Kanika’s present age = x yr

Father’s age = Kanika’s present age + 35 = x + 35 yr

(iii) Kanika’s present age = x yr

Father’s age = (x + 35) yr [from part (ii)]

and Mother’s age = Father’s age – 3 = x + 35 – 3

= (x + 32)yr

(iv) Kanika’s present age = x yr

Now, grandfather’s age = 8 times of Kanika’s age = 8 x x = 8x yr

**Question 88:**

If m is a whole number less than 5, complete the table and by inspection of the table, find the solution of the equation 2m – 5 = – 1.

Solution:

Since, m is a whole number which is less than 5, then solution of the equation is given by putting the value of

**Question 89:**

A class with p students has planned a picnic. x 50 per student is collected, out of which x 1800 is paid in advance for transport. How much money is left with them to spend on other items?**Solution:**

Total number of students = p

Money collect from per student = X 50 Total money collected = X 50p Advance money paid for transport = X1800 Money left with them = X (50p -1800)

**Question 90:**

In a village, there are 8 water tanks to collect rain water. On a particular day, x litres of rain water is collected per tank. If 100 litres of water was already there in one of the tanks, then what is the total amount of water in the tanks on that day?**Solution:**

According to the question,

Number of tanks to collect rain water = 8 Rain water collected in per tank (in L)= x Then, total rain water in tanks (in L)

= Number of tanks x Rain water collected per tank =8 x x=8x

But in the one tank, already 100 L of water exist,.

Then, total amount of water in the tank

= 100 + [Total rain water in L ]

= (100+ 8x)

**Question 91:**

What is the area of a square whose side is m cm?**Solution:**

Given,

Side of a square = m cm

∴ Area of a square = Sidex Side = mxmsq cm

**Question 92:**

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. Write the rule that is expressed by this formula in words.**Solution:**

In this question, given formula for getting perimeter of triangle is p = a+ b + c.

Here, a, b and c are the length of sides of the triangle.

Hence, the perimeter of a triangle is equal to the sum of all sides of a triangle.

**Question 93:**

Perimeter of a rectangle is found by using the formula p = 2(l + w), where / and w are respectively the length and breadth of the rectangle. Write the rule that is expressed by this formula in words.**Solution:**

Perimeter of a rectangle = 2 (Length of the rectangle + Breadth of the rectangle)

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

**Question 94:**

On my last birthday, I weighed 40 kg. If I put on m kg of weight after a year, what is my present weight?**Solution:**

Given, my weight on my last birthday = 40 kg

Weight increase after a year -m kg

Present weight = 40 kg + m kg = (40 + m) kg

**Question 95:**

Length and breadth of a bulletin board are r cm and t cm, respectively.

(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.

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

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

(iv) What will be the expenditure for making 23 boards, if the carpenter charges ? x per board?**Solution:**

Given, length of bulletin board = r cm and breadth of bulletin board = t cm Then, perimeter of bulletin board = 2(r + f) cm and area of bulletin board = rt cm

(i) Required length of aluminium strip = [2 (r +1) + 10] cm

(ii) To repair one board, number of nails required = x For 15 boards, number of nails required = 15x

(iii) Area of one board = rt sq cm

Area of eight boards = 8 x Area of one board = 8 rt sq cm

Extra cloth for one board = 500 sq cm [given]

Extra cloth for 8 boards = 500 x 8 = 4000 sq cm Required area of the cloth to cover 8 boards = (8rt + 4000) sq cm

(iv) Charges for one board = Rs. x Charges of 23 boards = Rs. 23 x

Hence, expenditure for making 23 boards is T 23x.

**Question 96:**

Sunita is half the age of her mother Geeta. Find their ages (i) after 4 years? (ii) before 3 years?

Solution:

Let the age of Sunita’s mother = 2x yr

Then, according to the question,

Sunita’s age = (½) x Age of Sunita’s mother = 2x/2

After 4 yr,

Sunita’s age = (x + 4)yr

∴ Her mother’s age = (2x + 4)yr

Note After 4 years means, 4 years is added in present age.

(ii) Before 3 yr,

Sunita’s age = (x – 3) yr

and her mother’s age = (2x – 3)yr**Note:** Before 3 years means, 3 years is subtracted from present age.

**Question 97:**

Match the items of Column I with that of Column II**Solution:**

**All Chapter NCERT Exemplar Problems Solutions For Class 6 maths**

—————————————————————————–**All Subject NCERT Exemplar Problems Solutions For Class 6**

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