class 6 maths

NCERT Exemplar Class 6 Maths Solutions Pdf free download was designed by expert teachers from the latest edition of NCERT Exemplar Books to get good marks in board exams.

NCERT Exemplar Class 6 Maths Book PDF Download Chapter 7 Algebra Solutions

Multiple Choice Questions (MGQs)

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:

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:

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:

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:

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NCERT Exemplar Class 6 Maths Solutions Pdf free download was designed by expert teachers from the latest edition of NCERT Exemplar Books to get good marks in board exams.

NCERT Exemplar Class 6 Maths Book PDF Download Chapter 5 Data Handling Solutions

Question 1:
Using tally marks, which one of the following represents the number eight?

Solution:

Question 2:
The marks (out of 10) obtained by 28 students in a Mathematics test are listed as below:
8,1, 2, 6, 5, 5, 5,0,1, 9, 7, 8, 0, 5, 8, 3, 0, 8, 10, 10, 3, 4, 8, 7, 8, 9, 2, 0 The number of students who obtained marks more than or equal to 5, is (a) 13 (b) 15 (c) 16 (d) 17
Solution:
(d) In order to find out how many students obtained marks more than or equal to 5, we will arrange the given data in a table, using tally marks.
Marks in Mathematics test Tally marks Number of students

The number of students who obtained marks more than or equal to 5
=4 + 1 + 2 + 6 + 2 + 2
=17

Question 3:
In question 2, the number of students who scored marks less than 4 is (a) 15 (b) 13 (c) 12 (d) 10
Solution:
(d) From the table of question 2, we can calculate the number of students who scored marks less than 4.
Number of students who scored less than 4 marks = 4 + 2 + 2 + 2 = 10

Question 4:
The choices of the fruits of 42 students in a class are as follows:

where, A, B, G, M and 0 stands for the fruits Apple, Banana, Grapes, Mango and Orange, respectively.
Which two fruits are liked by an equal number of students?
(a) A and M (b) M and B
(c) B and O (d) B and G
Solution:
(d) Table with tally marks from the given information is shown below

From the table, we can observe that banana and grapes are liked by equal number of students, i.e. 8.

Question 5:
According to the data of question 4, which fruit is liked by most of the students?
(a) O (b) G (c) M (d) A
Solution:
(c) From the table of question 4, we can observe that mango is liked by most of the students.

True / False

In questions 6 to 13, state whether the given statements are True or False.

Question 6:
In a bar graph, the width of bars may be unequal.
Solution:
False
Since, a bar graph is a pictorial representation of the numerical data by a number of bars of uniform width erected horizontally or vertically with equal spacing between them.

Question 7:
In a bar graph, bars of uniform width are drawn vertically only.
Solution:
False
Bars of uniform width can be drawn vertically or horizontally.

Question 8:
In a bar graph, the gap between two consecutive bars may not be the same.
Solution:
False
In a bar graph, the gap between two consecutive bars must be equal.

Question 9:
In a bar graph, each bar (rectangle) represents only one value of the numerical data.
Solution:
True
In a bar graph, each bar (rectangle) represents only one value of the numerical data.

Question 10:
To represent the population of different towns using bar graph, it is convenient to take one unit length to represent one person.
Solution:
False
We take scale as per the given data.

Question 11:
Pictographs and bar graphs are pictorial representations of the numerical data.
Solution:
True
Pictographs and bar graphs are pictorial representations of the numerical data.

Question 12:
An observation, occurring five times in a data is recorded as | | | | | using tally marks.
Solution:
False
The numerical data five represented as using
tally marks.

Question 13:

Solution:
if picture of 1 book represents 50 books in a library, then picture of 1/2 book will represent 25 books in a library.

Fill in the Blanks

In questions 14 to 21, fill in the blanks to make the statements true.

Question 14:
A …………… is a collection of numbers gathered to give some meaningful information.
Solution:
A data is a collection of numbers gathered to give some meaningful information.

Question 15:
The data can be arranged in a tabular form using ……… marks.
Solution:
The data can be arranged in a tabular form using tally marks.

Question 16:
A ………. represents data through pictures of objects.
Solution:
A pictograph represents data through pictures of objects.

Question 17:
In a bar graph,………. can be drawn horizontally or vertically.
Solution:
In a bar graph, bars can be drawn horizontally or vertically.

Question 18:
In a bar graph, bars of width can be drawn horizontally or vertically
with spacing between them.
Solution:
In a bar graph, bars of uniform width can be drawn horizontally or vertically with equal spacing between them.

Question 19:
An observation occurring seven times in a data is represented as
using tally marks.
Solution:
An observation occurring seven times in a data is represented as

using tally marks.

Question 20:

Solution:
In a pictograph, if a symbol represents 20 flowers in a basket, then three symbols will represents (20 + 20 + 20)flowers i.e. 60 flowers.

Question 21:
On the scale of 1 unit length =10 crore, the bar of length 6 units will
represent crore and*of units will represent 75 crore.
Solution:
Given, 1 unit length =10 crore

Question 22:
In an examination, the grades achieved by 30 students of a class are
given below. Arrange these grades in a table using tally marks.
B, C, C, E, A, C, B, B, D, D, D, D, B, C, C, C, A, C, B, E, A, D, C, B, E, C, B, E, C, D.
Solution:
On arranging the given data in a table using tally marks, as shown below table

Question 23:
The number of two wheelers owned individually by each of 50 families is listed below. Make a table using tally marks.
1, 1, 2, 1, 1, 1, 2, 1, 2, 1, 0, 1, 1, 2, 3, 1, 2, 1, 1, 2, 1, 2, 3, 1, 0, 2, 1, 0, 2, 1, 2, 1, 2, 1, 1, 4, 1, 3, 1, 1, 2, 1, 1, 1, 1, 2, 3, 2, 1, 1
Find the number of families having two or, more two wheelers.
Solution:
On arranging the given data in a table using tally marks, as shown below table

From the table, the number of families having two or more two wheelers = 14+4+1 = 19

Question 24:
The lengths in centimetres (to the nearest centimetre) of 30 carrots are given as follows:
15, 22, 21, 20, 22, 15, 15, 20, 20, 15, 20, 18, 20, 22, 21,
20, 21, 18, 21, 18, 20, 18, 21, 18, 22, 20, 15, 21, 18, 20
Arrange the data given above in a table using tally marks and answer the following questions:
(a) What is the number of carrots, which have length more than 20 cm?
(b) Which length of the carrots occur maximum and miniumum number of times?
Solution:
On arranging the given data in a table using tally marks, as shown below table

(a) Number of carrots having length more than 20 cm = 6 + 4 = 10
(b) From the table, maximum number of carrots = 9 and minimum number of carrots = 4
Length of carrots occurring maximum number of times = 20 cm and length of carrot occurring minimum number of times = 22 cm

Question 25:
Thirty students were interviewed to find out what they want to be in future. Their responses are listed as below: .
doctor, engineer, doctor, pilot, officer, doctor, engineer, doctor, pilot, officer, pilot, engineer, officer, pilot, doctor, engineer, pilot, officer, doctor, officer, doctor, pilot, engineer, doctor, pilot, officer, doctor, pilot, doctor, engineer.
Arrange the data in a table using tally marks.
Solution:
On arranging the given data in a table using tally marks, we get the following table

Question 26:
Following are the choices of games of 40 students of Class VI:
football, cricket, football, kho-kho, hockey, cricket, hockey, kho-kho, tennis, tennis, cricket, football, football, hockey, kho-kho, football, cricket, tennis, football, hockey, kho-kho, football, cricket, cricket, football, hockey, kho-kho, tennis, football, hockey, cricket, football, hockey, cricket, football, kho-kho, football, cricket, hockey, football.
(a) Arrange the choices of games in a table using tally marks.
(b) Which game is liked by most of the students?
(c) Which game is liked by minimum number of students?
Solution:
(a) On arranging the given data in a table using tally marks, as shown below

(b) From the table, we can observe that football is liked by most of the students.
(c) From the table, we can observe that tennis is liked by minimum number of students.

Question 27:
Fill in the blanks in the following table which represents shirt size of 40 students of a school. ‘

Solution:

Question 28:
Following pictograph represents some surnames of people listed in the telephone directory of a city.

Observe the pictograph and answer the following questions:
(a) How many people have surname ‘Roy?
(b) Which surname appears the maximum number of times in the telephone directory?
(c) Which surname appears the least number of times in the directory?
(d) Which two surnames appear an equal number
Solution:

(b) From the pictograph, we can observe that most of people have surname ‘Patel’.
(c) From the pictograph, we can observe that surname ‘Saikia’ appears the least number of times in the directory.
(d) ‘Rao’ and ‘Roy’ surnames appear an equal number of times.

Question 29:
Students of Class VI in a school were given a task to count the number of articles made of different materials in the school. The information collected by them is represented as follows:

Observe the pictograph and answer the following questions:
(a) Which material is used in maximum number of articles?
(b) Which material is used in minimum number of articles?
(c) Which material is used in exactly half the number of articles as those made up of metal?
(d) What is the total number of articles counted by the students?
Solution:
The number of articles of different materials is depicted (represented) by the following table

(a) Metal is used in the maximum number of articles.
(b) Glass is used in the minimum number of articles.
(c) From the pictograph, we observe that 50 articles were made of metal whereas 25 articles were made of rubber, which is exactly half the number of articles as those made up of metal.
(d) Total number of articles = Articles made of (wood + glass + metal + rubber + plastic)
= 30+20+ 50+25+ 35 = 160

Question 30:
The number of scouts in a school is depicted by the following pictograph:

Observe the pictograph and answer the following questions:
(a) Which class has the minimum number of scouts?
(b) Which class has the maximum number of scouts?
(c) How many scouts are there in Class VI?
(d) Which class has exactly four times the scouts as that of Class X?
(e) What is the total number of scouts in the Classes VI to X?
Solution:
(a) Class X has minimum number of scouts, i.e. 10.
(b) Class VIII has maximum number of scouts i.e. 60.
(c) Number of scouts in Class VI =40
(d) Number of scouts in Class X =10 Number of scouts in Class VI = 4 x 10 = 40
Hence, class VI has exactly four times the scouts as that of Class X.
(e) Total number of scouts in Classes VI to X = 40 + 20 + 60 + 30 + 10 = 160.

Question 31:
A survey was carried out in a certain school to find out the popular school subjects among students of Classes VI to VIII. The data in this regard is displayed as pictograph given below:

(a) Which subject is most popular among the students?
(b) How many students like Mathematics?
(c) Find the number of students who like subjects other than Mathematics and Science.
Solution:
(a) Hindi is most popular subject among the students.
(b) 175 students like Mathematics.
(c) Subjects other than Mathematics and Science are Hindi, English and Social Studies. Therefore, the number of students who liked Hindi, English and Social Studies
= 200+ 150+75=425

Question 32:
The following pictograph depicts the information about the areas in a square kilometre (to nearest hundred) of some districts of Chhattisgarh state:

(a) What is the area of Koria district?
(b) Which two districts have the same area?
(c) How many districts have area more than 5000 sq km?
Solution:
(a) Area of Koria district =6000 sq km
(b) Raigarh and Jashpur have the same area.
(c) Four districts have area more than 5000 sq km.

Question 33:
The number of bottles of cold drink sold by a shopkeeper on six consecutive days is as follows:

Prepare a pictograph of the data using one symbol to represent 50 bottles.
Solution:

For Monday,
200 cold drink bottles can be represented by 4 complete symbols. For Tuesday,
300 cold drink bottles can be represented by 6 complete symbols. For Wednesday,
250 cold drink bottles can be represented by 5 complete symbols. For Thursday,
100 cold drink bottles can be represented by 2 complete symbols. For Friday,
150 cold drink bottles can be represented by 3 complete symbols. Hence, the required pictograph of given data is shown below

Question 34:
The following table gives information about the circulation of newspaper (dailies) in a town in five languages:

Prepare a pictograph of the above data using a symbol of your choice, each representing 1000 newspapers.
Solution:

Question 35:
Annual expenditure of a company in the year 2007-08 is given below:

Prepare a pictograph of the above data using an appropriate symbol to represent Rs. 10 lakh.
Solution:
Rs. 10 lakh can be represented by[T] and Rs. 5 lakh can be represented by
For salaries of employees,
Rs. 65 lakh can be represented by 6 complete and 1 incomplete symbol.
For advertisement,
110 lakh can be represented by 1 complete symbol.
For purchase of machinery,
Rs. 85 lakh can be represented by 8 complete and 1 incomplete symbol.
For electricity and water,
Rs. 15 lakh can be represented by 1 complete and 1 incomplete symbol.
For transportation,
Rs. 25 lakh can be represented by 2 complete and 1 incomplete symbol.
For other expenses,
Rs. 30 lakh can be represented by 3 complete symbol.
Hence, the required pictograph of given data is shown below, Take [T] =10 lakh and [f =5 lakh]

Question 36:
The following bar graph shows the number of houses (out of 100) in a town using different types of fuels for cooking.

Read the bar graph and answer of the following questions:
(a) Which fuel is used in maximum number of houses?
(b) How many houses are using coal as fuel?
(c) Suppose that the total number of houses in the town is 1 lakh. From the above graph, estimate the number of houses using electricity.
Solution:
(a) On observing the bar graph, we can conclude that LPG is used in maximum number of houses.
(b) 10 houses out of 100 are using coal as fuel.
(c) Number of houses using electricity out of 100 =5 Number of houses using electricity out of 1 lakh

Question 37:
The following bar graph represents the data for different sizes of shoes worn by the students in a school.

Read the graph and answer the following questions:
(a) Find the number of students whose shoes sizes have been collected.
(b) What is the number of students wearing shoe size 6?
(c) What are the different sizes of the shoes worn by the students?
(d) Which shoe size is worn by the maximum number of students?
(e) Which shoe size is worn by minimum number of students?
(f) State whether true or false:
The total number of students wearing shoe sizes 5 and 8 is the same as the number of students wearing shoe size 6.
Solution:
(a) Total number of students whose shoes sizes have been collected = 250 + 200 + 300 + 400+150= 1300
(b) Shoe size 6 is worn by 300 students.
(c) Different number of shoes worn by the students are 4, 5, 6, 7 and 8.
(d) Shoe number 7 is worn by maximum number of students.
(e) Shoe number 8 is worn by minimum number of students.
(f) False, since total 350 students wore shoe numbers 5 and 8, whereas only 300 students worn shoe number 6.

Question 38:
The following graph gives the information about the number of railway tickets sold for different cities on a railway ticket counter between 6.00 am to 10.00 am

Read the bar graph and answer the following questions:
(a) How many tickets were sold in all?
(b) For which city were the maximum number of tickets sold?
(c) For which city were the minimum number of tickets sold?
(d) Name the cities for which the number of tickets sold is more than 20.
(e) Fill in the blanks
Number of tickets sold for Delhi and Jaipur together exceeds the total number of tickets sold for Patna and Chennai by;
Solution:
(a) Total tickets sold = 80 + 50+ 100 + 20 + 40 = 290
(b) Maximum number of tickets were sold for Delhi, i.e. 100 tickets.
(c) Minimum number of tickets were sold for Chennai, i.e. 20 tickets.
(d) The cities for which the number of tickets sold is more than 20 are Patna, Jaipur, Delhi and Guwahati.
(e) Number of tickets sold for Delhi and Jaipur together = 100 + 50 = 150
and number of tickets sold for Patna and Chennai together = 80 + 20 = 100 Required difference = 150 -100 = 50

Question 39:
The bar graph given below represents approximate length (in km) of some national highways in India.

Study the bar graph and answer the following questions:
(a) Which National Highway (NH) is the longest among the above?
(b) Which National Highway is the shortest among the above?
(c) What is the length of National Highway 9?
(d) Length of which National Highway is about three times the National Highway 10?
Solution:
(a) National Highway 2 is the longest among the above shown highways. It is 1500 km long.
(b) National Highway 10 is the shortest among the above shown highways. It is 500 km long.
(c) The length of National Highway 9 is 900 km.
(d) Length of NH 10 = 500 km
Length of NH 9 = 900 km ‘
Length of NH 8 = 1400 km
Length of NH 3 = 1200 km and length of NH 2 = 1500 km
Clearly, the length of NH 2 is about three times the length of NH 10.

Question 40:
The bar graph below represents the circulation of newspapers in different languages in a town.

Study the bar graph and answer the following questions:
(a) What is the circulation of English newspaper?
(b) Name the two languages in which circulation of newspaper is the same.
(c) By how much is the circulation of newspaper in Hindi more than the newspaper in Bengali?
Solution:
(a) From the graph, we can observe that about 1000 English newspapers are in circulation.
(b) Marathi and Bengali are the two languages in which circulation of newspapers is the same.
(c) Given, circulation of newspapers in Hindi = 1400 and circulation of newspapers in Bengali = 600 So, difference = 1400 – 600 = 800
The circulation of newspapers in Hindi is 800 more than the newspapers in Bengali.

Question 41:
Read the bar graph given below and answer the following questions.

(a) What information is given by the bar graph?
(b) In which year is the number of students maximum?
(c) In which year is the number of students twice as that of 2001-02?
(d) In which year did the number of students decreases as compared to previous year?
(e) In which year is the increases in number of students maximum as compared to the previous year?
Solution:
(a) The bar graph shows the information about the number of students in each academic year from 2001 to 2006.
(b) Given, number of students in 2001-02 = 150 Number of students in 2002-03 = 200 Number of students in 2003-04 = 150 Number of students in 2004-05 = 300 Number of students in 2005-06 = 350
The number of students was maximum in the academic year 2005-06.
(c) Clearly, the number of students in academic year 2004-05 is twice as that of 2001 -02.
(d) From the graph, we can observe that, the number of students decreased to 150 in 2003-04 from 200 of previous year 2002-03.
(e) In the year 2004-05, the increase in number of students is maximum as compared to the previous year.

Question 42:
The lengths in km (rounded to nearest hundred) of some major rivers of India is given below:

Draw a bar graph to represent the above information.
Solution:
In order to construct a bar graph representing the above data, we follow the following steps:
Step I Take a graph paper and draw two mutually perpendicular lines OX and OY. Let OX as the horizontal axis and OY as the vertical axis.
Step II Along OX, mark river’s name and along OY, mark length in kilometres.
Step III Along OX, choose the uniform (equal) width of the bars and the uniform gap between them, according to the space available for the graph.
Step IV Choose a suitable scale to determine the heights of the bars, according to the availability of space. Here, we choose 1 unit length represents 200 km.
Hence, the required bar graph of given data is shown below

Question 43:
The number of ATMs of different banks in a city is shown below:

Draw a bar graph to represent the above information by choosing the scale of your choice.
Solution:
In order to construct a bar graph representing the above data, we follow the following steps:
Step I Take a graph paper and draw two mutually perpendicular lines OX and OY. Let OX as the horizontal axis and OY as the vertical axis.
Step II Along OX, mark bank names and along OY, mark number of ATMs.
Step III Along OX, choose the uniform (equal) width of the bars and the uniform gap between them, according to the space available for the graph.
Step IV Choose a suitable scale to determine the heights of the bars, according to the availability of space. Here, we choose 1 unit length represents 5 ATMs.
Step V Calculate the height of various bars as follows:

Hence, the required bar graph of given data is shown below

Question 44:
Number of mobile phone users in various age groups in a city is listed below:

Draw a bar graph to represent the above information.
Solution:
In order to construct a bar graph representing the above data, we follow the following steps:
Step I Take a graph paper and draw two mutually perpendicular lines OX and OY. Let OX as the horizontal axis and OY as the vertical axis.
Step II Along OX, mark age groups and along OY, mark number of mobile users.
Step IV Choose a suitable scale to determine the heights of the bars, according to the availability of space. Here, we choose 1 unit length represents 5000 mobile users.
Step V Calculate the height of various bars as follows

Hence, the required bar graph of given data is shown below

Question 45:
The following table gives the number of vehicles passing through a toll gate, every hour from 8.00 am to 1.00 pm:

Draw a bar graph representing the above data.
Solution:
In order to construct a bar graph representing the above data, we follow the following steps:
Step I Take a graph paper and draw two mutually perpendicular lines OX and Oy. Let OX as the horizontal axis and OY as the vertical axis.
Step II Along OX, mark time interval and along OY, mark number of vehicles.
Step III Along OX, choose the uniform (equal) width of the bars and the uniform gap between them, according to the space available for the graph.
Step IV Choose a suitable scale to determine the heights of the bars, according to the availability of space. Here, we choose 1 unit length represents 50 vehicles.

Hence, the required bar graph for the given data is shown below

Question 46:
The following table represent income of a Gram Panchayat from different

Solution:
In order to construct a bar graph representing the above data, we follow the following steps:
Step I Take a graph paper and draw two mutually perpendicular lines OX and OY. Let OX as the horizontal axis and OY as the vertical axis.
Step II Along OX, mark sources and along OY, mark income.
Step III Along OX, choose the uniform (equal) width of the bars and the uniform gap between them, according to the space available for the graph.
Step IV Choose a suitable scale to determine the heights of the bars, according to the availability of spate. Here, we choose 1 unit length to represents Rs. 10000.
Step V Calculate the various bars as follows:

Hence,the required bar graph for the given data is shown below

Question 47:
The following table gives the data of number of schools (stage-wise) of a country in the year 2002.

Draw a bar graph to represent the above data.
Solution:
In order to construct a bar graph representing the above data, we follow the following steps:
Step I Take a graph paper and draw two mutually perpendicular lines OX and OXX. Let OX as the horizontal axis and OY as the vertical axis.
Step II Along OX, mark stages and along OY, mark number of schools.
Step IV Choose a suitable scale to determine the heights of the bars, according to the availability of space. Here, we choose 1 unit length represents 5 schools.
Step V Calculate the height of various bars as follows

Hence, the required bar graph for the given data is shown below

Question 48:
Home appliances sold by a shop in one month are given as below:

Draw a bar graph to represent the above information.
Solution:
In order to draw a bar graph representing the above data, we follow the following steps:
Step I Take a graph paper and draw two mutually perpendicular lines OX and OY. Let OX as the horizontal axis and OY as the vertical axis.
Step II Along OX, mark of home appliances and along OY, mark number of appliances.
Step IV Choose a suitable scale to determine the heights of the bars, according to the availability of space. Here, we choose 1 unit length represents 10 home appliances.
Step V Calculate the height of various bars as follows

Hence, the required bar graph for the given data is shown below

Question 49:
In a botanical garden, the number of different types of plants are founds as follows:

Draw a bar graph to represent the above information and answer the following questions:
(a) Which type of plant is maximum in number in the garden?
(b) Which type of plant is minimum in number in the garden?
Solution:
In order to construct a bar graph representing the above data, we follow the following steps:
Step I Take a graph paper and draw two mutually perpendicular lines OX and Oy. Let OX as the horizontal axis and OY as the vertical axis.
Step II Along OX, mark type of the plant and along OY, mark number of plants.
Step IV Choose a suitable scale to determine the heights of the bars, according to the availability of space. Here, we choose 1 unit length represents 10 plants.
Step V Calculate the height of various bars as follows

Hence, the required bar graph for the given data is shown below

(a) On studying bar graph, tree is maximum in number in the garden.
(b) On studying bar graph, creeper plants is minimum in number in the garden.

Question 50:
Prepare a bar graph of the data given in question 28.
Solution:
In order to construct a bar graph, first we have to make a table representing the pictograph in tabular form of question 28.

In order to construct a bar graph representing the above data, we follow the following steps:
Step I Take a graph paper and draw two mutually perpendicular lines OX and OY. Let OX as the horizontal axis and OY as the vertical axis.
Step II Along OX, mark surname and along OY, mark number of people.
Step III Along OX, choose the uniform (equal) width of the bars and the uniform gap between them, according to the space available for the graph.
Step IV Choose a suitable scale to determine the heights of the bars, according to the availability of space. Here, we choose 1 unit length represents 100 people.
Step V Calculate the height of various bars as follows

Hence, the required bar graph for the given data is shown below

Question 51:
Refer to question 39. Prepare a pictograph of the data by taking a suitable symbol to represent 200 km.
Solution:
Let 200 km can be represented by NH and 100 km represented by N.
For NH 10,
500 km can be represented by 2 complete symbols and 1 incomplete symbol.
For NH 9,
900 km can be represented by 4 complete symbols and 1 incomplete symbol.
For NH 8,
1400 km can be represented by 7 complete symbols.
For NH 3,
1200 km can be represented by 6 complete symbols.
For NH 2,
1500 km can be represented by 7 complete symbols and 1 incomplete symbol.
Hence, the required pictograph of given data is shown below

Question 52:
Prepare a pictograph of the information given in question 38.
Solution:
10 tickets can be represented by □.
For Patna,
80 tickets can be represented by 8 symbols.
For Jaipur,
50 tickets can be represented by 5 symbols.
For Delhi,
100 tickets can be represented by 10 symbols.
For Chennai,
20 tickets can be represented by 2 symbols.
For Guwahati,
40 tickets can be represented by 4 symbols.
Hence, the required pictograph of given data is shown below

Question 53:
Refer to question 23. Prepare a bar graph of the data.
Solution:
In order to construct a bar graph representing the above data, we follow the following steps: ,
Step I Take a graph paper and draw two mutually perpendicular lines OX and OY. Let OX as the horizontal axis and OY as the vertical axis.
Step II Along OX, mark number of two wheelers and along OY, mark number of families. Step III Along OX, choose the uniform (equal) width of the bars and the uniform gap between them, according to the space available for the graph.
Step V Calculate the height of various bars as follows

Hence, the required bar graph for the given data is shown below

Question 54:
The following table shows the area of the land on which different crops were grown:

Prepare a pictograph by choosing a suitable symbol to represent 10 million hectares.
Solution:
5 million hectares can be represented by ill symbol.
For rice,
50 million hectares can be represented by 10 symbols.
For wheat,
30 million hectares can be represented by 6 symbols.
For pulses,
20 million hectares can be represented by 4 symbols.
For sugarcane,
25 million hectares can be represented by 5 symbols.
For cotton,
15 million hectares can be represented by 3 symbols.
Hence, the required pictograph of the given data is shown below

Question 55:
Refer to question 54. Prepare a bar graph of the data.
Solution:
In order to construct a bar graph representing above data, we follow the following steps:
Step I Take a graph paper and draw two mutually perpendicular lines OX and OY. Let OX as the horizontal axis and OY as the vertical axis.
Step II Along OX, mark crop and along OY, mark area of the land.
Step III Along OX, choose the uniform (equal) width of the bars and the uniform gap between them, according to the space available for the graph.
Step IV Choose a suitable scale to determine the heights of the bars, according to the availability of space. Flere, we choose 1 unit length to represents 5 million hectares.
Step V Calculate the heights of various bars as follows

Hence, the required bar graph of the given data is shown below

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NCERT Exemplar Class 6 Maths Solutions Pdf free download was designed by expert teachers from the latest edition of NCERT Exemplar Books to get good marks in board exams.

NCERT Exemplar Class 6 Maths Book PDF Download Chapter 4 Fractions and Decimals Solutions

Multiple Choice Questions (MGQs)

Question 1:
The fraction which is not equal to 4/5 is

Solution:
(d) For finding the fraction which is not equal to 4/5, we will find the fraction from the option

Question 2:
The two consecutive integers between which the fraction 5/7 lies are
(a) 5 and 6 (b) 0 and 1
(c) 5 and 7 (d) 6 and 7
Solution:
(b) We know that, if the numerator is less than the denominator, then the value of fraction is less than 1.
Hence, the fraction 5/7 lies between 0 and 1.

Question 3:
When 1/4 is written with denominator as 12, its numerator is
(a) 3 (b) 8 (c) 24 (d) 12
Solution:
(a) Given, fraction = 1/4
In order to make the denominator as 12, we will multiply the denominator by 3 and we will also multiply the numerator by 3, to make it an equivalent fraction.

Hence, when denominator of 1/4 is 12, then its numerator will be 3.

Question 4:
Which of the following is not in the lowest form?

Solution:
(b) We know that, a fraction is in its lowest form, if the HCF of their numerator and denominator is 1. Now,

Question 5:
If 5/8 = 20/P then the value of p is
Solution:

Question 6: Which of the following is not equal to the others?

Solution:
(c) In order to find which of the given fraction is not equal to others, we will convert each of the given fraction in its lowest form. Now,

Question 7:
Which of the following fraction is the greatest?

Solution:
(b) In order to find the greatest fraction among the above given fractions, we will convert all the fractions to an equivalent fraction with denominator equal to the LCM of their denominator.

Question 8:
Which of the following fraction is the smallest?

Solution:
(c) Since, for comparing fractions with same denominators, fraction with smaller numerator is

Question 9:

Solution:
(a) Since, fractions with same denominators can be added by simply adding the numerators and writing the common denominator as it is

Question 10:

Solution:
(b) Since, fractions with same denominators can be subtracted by simply subtracting the numerators and writing the common denominator as it is

Question 11:
0.7499 lies between
(a) 0.7 and 0.74      (c) 0.749 and 0.75
(b) 0.75 and 0.79   (d) 0.74992 and 0.75
Solution: (c) Since, 0.7499 is greater than 0.749 and less than 0.75. Therefore, 0.7499 lies between 0.749 and 0.75.
0.749 < 0.7499 < 0.75

Question 12:
0.023 lies between
(a) 0.2 and 0.3        (c) 0.03 and 0.029
(b) 0.02 and 0.03  (d) 0.026 and 0.024
Solution:
(b) Since, 0.023 is greater than 0.02 and less than 0.03. Therefore, 0.023 lies between 0.02 and 0.03.
0.02 < 0.023 <0.03

Question 13:

Solution:

Question 14:

Solution:

Question 15:
0.07 + 0.008 is equal to
(a) 0.15 (b) 0.015 (c) 0.078 (d) 0.78
Solution:
(c) Converting the given decimals to like decimals, we have 0.070 and 0.008.

Note: Decimals having the same number of digits on the right of the decimal point are known as like decimals.

1/8 as a decimal? 1 0.125. 2 0.225. 3 0.325. 4 0.425.

Question 16:
Which of the following decimals is the greatest?
(a) 0.182 (b) 0.0925 (c) 0.29 (d) 0.038
Solution:
(c) Here, whole part of all numbers are same and tenths part of 0.0925 and 0.038 are same
i.e. 0 and tenths part of 0.182 =1/10
and tenths part of 0.29 = 2/10
Hence, 0.29 is the greatest.

Question 17:
Which of the following decimals is the smallest?
(a) 0.27 (b) 1.5 (c) 0.082 (d) 0.103
Solution:
(c) Here, whole part of numbers 0.27, 0.082 and 0.103 are same and is less than 1.5.
Now, we will compare the tenths part of 0.27, 0.082 and 0.103.
Tenths part of 0.27 = 2/10
Tenths part of 0.082 = 0/10
and tenths part of 0.103 =1/10
Clearly, tenths part of 0.082 is smallest.
Hence, 0.082 is the smallest decimal.

Question 18:
13.572 correct to the tenths place is ,
(a) 10 (b) 13.57 (c) 14.5 (d) 13.6
Solution:
(d) For rounding off to tenths place, we look at the hundredths place.
Here, the digit at hundredths place is 7 which is greater than 5. So, the digit at the tenths place (5) will be increased by 1 and digits at the hundredths and thousandths place will be written as equal to zero.
Hence, rounding off 13.572 to nearest tenths, we get 13.6.

What is formal notation for rounding to the nearest hundredth (notation, math)?.

Question 19:
15.8 – 6.73 is equal to
(a) 8.07 (b) 9.07 (c) 9.13 (d) 9.25
Solution:
(b) Converting the given decimals to like decimals, we have 15.80 and 6.73.

Note: Decimals having the same number of digits on the right of the decimal point are known as like decimals.

In algebra, 1/32 to decimal number can be defined as a number whose whole number part and the fractional part is separated by a decimal point.

Question 20:
The decimal 0.238 is equal to the fraction

Solution:
(a) We know that a decimal can be converted into a fraction by taking the numerator as the number obtained by removing the decimal point from the given decimal and taking the denominator as the number obtained by inserting as many zeroes with 1 as there are number of places in the decimal part.
Finally, converting the obtained fraction in its lowest form by dividing numerator and denominator by their HCF.

Fill in the Blanks

In questions from 21 to 44, fill in the blanks to make the statements true.

Question 21:
A number representing a part of a …………. is called a fraction.
Solution:
Whole
By definition, a fraction is a number representing a part of a whole.

Question 22:
A fraction with denominator greater than the numerator is called ………. fraction.
Solution:
Proper
It is a standard definition.

Question 23:
Fractions with the same denominator are called …………. fractions.
Solution: Like
It is a standard definition.

Question 24:
13 5/18 is a ………… fraction.
Solution:
Mixed
Since, a combination of a whole number and a proper fraction is called a mixed fraction.

Question 25:
18/5 is an fraction.
Solution:
Improper
Since, a fraction whose numerator is more than or equal to the denominator is called an improper fraction.

Question 26:
7/18 is a ………….. fraction.
Solution:
Proper
Since, a fraction whose numerator is less than the denominator is called a proper fraction.

Question 27:
5/8 and 3/8 are ……….. proper fractions.
Solution:
Like
Since, fractions having the same denominators are called like fractions and fractions whose numerator is less than the denominator is called a proper fraction.

Question 28:
6/11 and 6/13 are …………. proper fractions.
Solution:
Unlike
Since, fractions having different denominators are called unlike fractions and fractions whose numerator is less than the denominator is called a proper fraction.

Question 29:
The fraction 6/15 in simplest from is ……….
Solution:
In order to reduce a fraction to its lowest terms, we divide its numerator and denominator by their HCF.

Question 30:
The fraction 17/34 in simplest form is ……….
Solution:
1/2
In order to reduce a fraction to its lowest terms, we divide its numerator and denominator by their HCF.

Question 31:
18/135 and 90/675 are proper, unlike and ……….. fraction.
Solution:

Question 32: 8 2/7 is equal to the improper …………… fraction .
Solution:

Question 33:
87/7 is equal to the mixed fraction ………..
Solution:
12 3/7
In order to express an improper fraction as a mixed fraction, we first divide the numerator by denominator and obtain the quotient and remainder and then we write the mixed fraction as

Question 34:

Solution:
9.26
Here,

Question 35:
Decimal 16.25 is equal to the fraction ………..
Solution:
We know that, a decimal can be converted into a fraction by taking the numerator as the number obtained by removing the decimal point from the given decimal and taking the denominator as the number obtained by inserting as many zeroes with 1 as there are numbers of places in the decimal part.
Finally, converting the obtained fraction in its lowest form by dividing numerator and denominator by their HCF.

Question 36:
Fraction is equal to the decimal number ………….
Solution:
0.28
In order to convert fraction into decimal, we first convert it into an equivalent fraction with denominator 10 or 100 or 1000 and then write its numerator and mark decimal point after one place or two place or three places from right towards left if the denominator is 10 or 100 or 1000, respectively.
If the numerator is short of digits, insert zeroes at the left of the numerator.

Question 37:

Solution:

Question 38:

Solution: 

Question 39:

Solution:

Question 40:

Solution:
8

Question 41:
4.55 + 9.73 = …………
Solution:
14.28
We have,

Question 42:
8.76 – 2.68 = …………
Solution:
6.08
We have,

Question 43:
The value of 50 coins of 50 paisa = Rs ………..
Solution:
Value of 1 coin of 50 paise = 50 paise
Value of 50 coins of 50 paise = 50 paise x 50 = 2500 paise Now, we know that 1 Rs.= 100 paise
In order to convert paise to rupee, we divide the given value by 100.
2500 paise = Rs. = Rs.25

Question 44:
3 hundredths + tenths = ……….
Solution:
0.33
We have,

True/False

In questions 45 to 65, state whether the given statements are True or False.

Question 45:
Fractions with same numerator are called like fractions.
Solution: False
Fractions with same denominator are called like fractions.

Question 46:
Fraction 18/19 is in its lowest form.
Solution:
False
We have,

Note: A fraction is in its lowest terms, if its numerator and denominator have no common factor other than 1.

Question 47:
Fractions 15/39 and 45/117 are equivalent fractions.
Solution: True
We have,

Question 48:
The sum of two fractions is always a fraction.
Solution:
False

Question 49:
The result obtained by subtracting a fraction from another fraction is necessarily a fraction.
Solution:
False

Question 50:
If a whole or an object is divided into a number of equal parts, then each part represents a fraction.
Solution:
True
Fraction is a part of a whole.

Question 51:
The place value of a digit at the tenths place is 10 times the same digit at the ones place.
Solution:
False

Question 52:
The place value of a digit at the hundredths place is — times the same
digit at the tenths place.
Solution:
True

Question 53:
The decimal 3.725 is equal to 3.72 correct to two decimal places.
Solution:
False
For correcting 3.725 to two decimal places we look at the thousandths place.
Here, the digit at thousandths places is 5. So, the digit at hundredths place 2 will be increased by 1 and 5 will be written as equal to zero.
Hence, 3.725 = 3.73 (correct to two decimal places)

Question 54:
In the decimal form, fraction 55/8 = 3.125.
Solution:
True
In order to convert fraction into decimal, we first convert it into an equivalent fraction with denominator 10 or 100 or 1000 and then write its numerator and mark decimal point after one place or two place or three places from right towards left, if the denominator is 10 or 100 or 1000 respectively.
If the numerator is short of digits, insert zeroes at the left of the numerator.

Question 55:
The decimal 23.2 = 23 2/5
Solution:
False
We know that, a decimal can be converted into a fraction by taking the numerator as the number obtained by removing the decimal point from the given decimal and taking the denominator as the number obtained by inserting as many zeroes with 1 as there are number of place in the decimal part.
Finally, converting the obtained fraction in its lowest form by dividing numerator and denominator by their HCF and converting it to mixed fraction if required.

Question 56:
The fraction represented by the shaded portion in the following figure is 3/8

Solution:
True
Total equal parts in figure = 8 and shaded parts in figure = 3
Now, the fraction of shaded part to the total parts = 3/8

Question 57:
The fraction represented by the unshaded portion in the following figure

Solution:
False
Total equal parts in figure = 9 and unshaded parts in figure = 4
Now, the fraction of unshaded part to the total parts = 4/9

Question 58:

Solution:
False
Since, fractions with same denominators can be added by simply adding the numerators and writing the common denominator as it is.

Question 59:

Solution:
False

Question 60:

Solution:
True
Since,

Question 61:
3.03 + 0.016 = 3.019
Solution:
False
Converting the given decimals to like decimals and then adding. 3.030 + 0.016 = 3046

Question 62:
42.28 – 3.19 = 39.09
Solution:
True
Given, 42.28-3.19 = 39.09

Question 63:

Solution:
True
We know that, if two fractions have same denominator, then fraction with greater numerator is greater.

Question 64:
19.25 < 19.053
Solution:
False

Question 65:
13.730 = 13.73
Solution:
True
Since, the value of zero at the last place in decimal part is negligible.
In each of the questions 66 to 71, fill in the blanks using “<” or ‘=’,

Question 66:

Solution:
‘<‘
In order to compare fractions with different denominators, we will convert them to like fractions.

Question 67:

Solution:
‘<‘
In order to compare fraction with different denominators we will convert them to like fractions.

Question 68:

Solution:
‘=’
In order to compare fractions with different denominators, we will convert them to like fractions.

Question 69:
3.25 — 3.4
Solution:
‘<‘

Question 70:
18/15 — 1.3
Solution:
‘<’
For comparing a fraction and a decimal, we will convert both of them to either into like fractions or into like decimals.

Question 71:
6.25 — 25/4
Solution: 
‘=’
For comparing a fraction and a decimal, we will convert both of them to either into like fractions or into like decimals.

Question 72:
Write the fraction represented by the shaded portion of the following figure.

Solution:
Circle is divided into 8 equal parts and number of shaded parts are 7.

Question 73:
Write the fraction represented by the unshaded portion of the following figure.

Solution:
Rectangle is divided into 15 equal parts and number of unshaded parts are 4.
Fraction of unshaded portion to the total portion = 4/15.

Question 74:
Ali divided one fruit cake equally among six person. What part of the cake he gave to each person?
Solution:
Given, total number of fruit cake = 1
Ali divided one fruit cake equally among six persons.
The part of cake given to one person = 1/6
Hence, the required part is 1/6.

Question 75:
Arrange 12.142, 12.124, 12.104, 12.401 and 12.214 in ascending order.
Solution:

Question 76:
Write the largest four digit decimal number less than 1 using the digits 1, 5, 3 and 8 once.
Solution:
Given digits are 1,5, 3, 8 and the number should be less than 1. So, the whole part will 0.
Now, for making largest four digit decimal number we will arrange the given digits in descending order after decimal point.
Required decimal number = 0.8531

Question 77:
Using the digits 2, 4, 5 and 3 once, write the smallest four digit decimal number.
Solution:
Given digits are 2, 4, 5 and 3.
For making smallest decimal number we will arrange the given digits in ascending order after decimal point.
Required decimal number = 0.2345

Question 78:
Express 11/20 as a decimal.
Solution:
In order to convert fraction into decimal, we first convert it into an equivalent fraction with denominator 10 or 100 or 1000 and then write its numerator and mark decimal point after one place or two place or three places from right towards left, if the denominator is 10 or 100 or 1000 respectively.
If the numerator is short of digits, insert zeroes at the left of the numerator.
11/20 x 5/5 = = 0.55

Question 79:
Express 6 2/3 as an improper fraction.
Solution:

Question 80:
Express 3- as a decimal.
Solution:
In order to convert fraction into decimal, we first convert it into an equivalent fraction with denominator 10 or 100 or 1000 and then write its numerator and mark decimal point after one place on two place or three places from right towards left, if the denominator is 10 or 100 or 1000 respectively.
If the numerator is short of digits, insert zeroes at the left of the numerator.

Question 81:
Express 0.041 as a fraction.
Solution:
We know that, a decimal can be converted into a fraction by taking the numerator as the number obtained by removing the decimal point from the given decimal and taking the denominator as the number obtained by inserting as many zeroes with 1 as there are numbers of place in the decimal part.
Finally, converting the obtained fraction in its lowest form by dividing numerator and denominator by this HCF.

Question 82:
Express 6.03 as a mixed fraction.
Solution:
We know that, a decimal can be converted into a fraction by taking the numerator as the number obtained by removing the decimal point from the given decimal and taking the denominator as the number obtained by inserting as many zeroes with 1 as there are number of places in the decimal part. Finally, converting the obtained fraction in its lowest form by dividing numerator and denominator by their HCF.

Question 83:
Convert 5201 g to kg.
Solution:
We know that, 1 kg = 1000 g
Now, for converting g into kg, we have to divide the given value by 1000. 5201,

Question 84:
Convert 2009 paise to rupees and express the result as a mixed fraction.
Solution:
We know that, Rs. 1 = 100 paise
Now, for converting paise into rupees, we have to divide the given value by 100.

Question 85:
Convert 1537 cm to m and express the result as an improper fraction.
Solution:
We know that, 1 m = 100 cm
Now, for converting cm into m, we have to divide the given value by 100.

Question 86:
Convert 2435 m to km and express the result as mixed fraction.
Solution:
We know that, 1 km = 1000 m
Now, for converting m into km, we have divide the given value by 1000.

Question 87:
Arrange the fractions and – in ascending order.
Solution:
In order to arrange the given fractions in ascending order, we have to convert them into like fractions. So, LCM of the denominators, i.e. 3, 4, 2 and 6 = 2 x 2 x 3 = 12.

Question 88:
Arrange the fractions and 6/7, 7/8, 4/5 and 3/4 in descending order.
Solution:
In order to arrange the given fractions in descending order, we have to convert them into like fractions. So, LCM of the denominators, i.e. 7, 8, 5 and 4
= 2 x 2 x 2 x 5 x 7 = 280

Question 89:
Write 3/4 as a fraction with denominator 44.
Solution:
Given fraction = 3/4
In order to express it, as a fraction with denominator 44, we will multiply the denominator
numerator by 11, to make it an equivalent fraction of 3/4.

Question 90:
Write 5/6 as a fraction with numerator 60.
Solution:
Given fraction = 5/6
In order to express it, as the fraction with numerator 60, we will multiply the numerator denominator by 12, to make it an equivalent fraction of

Question 91:
Write 128/9 as a mixed fraction.
Solution:

Question 92:
Round off 20.83 to nearest tenths.
Solution:
For rounding off the tenths place, we look at the hundredths place.
Here, the digit at hundredths place is 3 which is less then 5. So, the digit at the tenths place 8 will not be increased by 1 and 3 will be written as equal to zero.
Hence, rounding off 20.83 to nearest tenths, we get 20.80.

Question 93:
Round off 75.195 to nearest hundredths.
Solution:
For rounding off to hundredths place, we look at the thousandths place.
Here, the digit at thousandths place is 5 which is equal to 5. So, the digit at the hundredths place 9 will be increased by 1 and 5 will be written as equal to zero.
Hence, rounding off 75.195 to nearest hundredths, we get 75.200.

Question 94:
Round off 27.981 to nearest tenths.
Solution:
For rounding off the tenths place, we look at the hundredths place.
Here, the digit at hundredths place is 8 which is greater than 5. So, the digit at tenths place 9 will be increased by 1 and digits at the hundredths and thousandths place will be written as equal to zero.
Hence, rounding off 27.981 to nearest tenths, we get 28.

Question 95:
Add the fractions 3/8 and 2/3.
Solution:

Question 96:
Add the fractions 3/8 and 6 3/4.
Solution:

Question 97:
Subtract 1/6 from 1/2.
Solution:

Question 98:
Subtract 8 3/8 from 100/9.
Solution:

Question 99:
Subtract 1 1/4 from 6 1/2.
Solution:

Question 100:
Add 1 1/4 and 6 1/2.
Solution:

Question 101:
Katrina rode her bicycle 6- km in the morning and 8 3/4 km in the evening. Find the distance traveled by her altogether on that day. ,
Solution:

Question 102:
A rectangle is divided into certain number of equal parts. If 16 of the parts, so formed represent the fraction 1/4 , find the number of parts in which the rectangle has been divided.
Solution:

Question 103:
Grip size of a tennis racquet is 11 9/80 cm. Express the size as an improper fraction.
Solution:

Question 104:
On an average 1/10 of the food eaten is turned into organism’s own body is available for the next level of consumers in a food chain.What fraction of food eaten is not available for the next user?
Solution:

Question 105:
Mr. Rajan got a job at the age of 24 yr and he got retired from the job at the age of 60 yr. What fraction of his age till retirement was he in the job?
Solution:
Given, Rajan’s age on the joining =24 yr and retirement age = 60 yr

Question 106:
The food we eat remains in the stomach for a maximum of 4 h. For what fraction of a day, does it remain there?
Solution:
Given, maximum hours for which food remains in the stomach = 4 We know that, 1 day = 24 h
The fraction of a day for which food remains in the stomach 4 h/24 h = 1/6
So, the food remains in the stomach for 1/6 past of the day.

Question 107:
What should be added to 25.5 to get 50?
Solution:
To get the required result, we have to subtract 25.5 from 50.

So, 24.5 should be added to 25.5 to get 50.

Question 108:
Alok purchased 1 kg 200 g potatoes, 250 g dhania, 5 kg 300 g onion, 500 g palak and 2 kg 600 g tomatoes. Find the total weight of his purchases in kilograms.
Solution:

Question 109:
Arrange in ascending order.
0.011, 1.001, 0.101, 0.110
Solution:

Question 110:
Add the following.
20.02 and 2.002
Solution:
Converting the given decimals to like decimals, we have 20.020 and 2.002. Now,

Question 111:
It was estimated that because of people switching to Metro trains, about 33000 tonne of CNG, 3300 tonne of diesel and 21000 tonne of petrol was saved by the end of year 2007. Find the fraction of
(i) the quantity of diesel saved to the quantity of petrol saved
(ii) the quantity of diesel saved to the quantity of CNG saved.
Solution:

Question 112:
Energy content of different foods are as follows:

Solution:

Question 113:
A cup is 1/3 full of milk, what part of the cup is still to be filled by milk to make it full?
Solution:

Question 114:
Mary bought 3 1/2 m of lace. She used 1 3/4 m of lace for her new dress.
How much lace is left with her?
Solution:

Question 115:
When Sunita weighed herself on Monday, she found that she had 1 3
gained 1 1/4 kg. Earlier her weight was 46 3/8 kg. What was her weight on 4 8 Monday?
Solution:

Quantity 116:
Sunil purchased 12 1/2 L of juice on Monday and 14 3/4 L of juice on Tuesday. How many litres of juice did he purchase together in two days?
Solution:

Question 117:
Nazima gave 2 3/4 L out of the 5 1/2 L of juice she purchased to her friends.
How many litres of juice is left with her?
Solution:

Question 118:
Roma gave a wooden board of length 150 1/4 cm to a carpenter for making a shelf. The Carpenter sawed off a piece of 40 1/5 cm from it.
What is the length of the remaining piece?
Solution:

Question 119:
Nasir travelled 3 1/2 km in a bus and then walked 1 1/8 km to reach a town.
How much did he travel to reach the town?
Solution:

Question 120:
The fish caught by Neetu was of weight 3 3/4 kg and the fish caught by Narendra was of weight 2 1/2 kg.
How much did Neetu’s fish weight than that of Narendra?
Solution:

Question 121:
Neelam’s father needs 1 3/4 m of cloth for the skirt of Neelam’s new dress and 1/2 m for the scarf. How much cloth must he buy in all?
Solution:

Question 122:
What is wrong in the following additions?

Solution:
(a) On observing the sum, we find that the denominators of like fractions are also added which is wrong. So, the correct answer will be 12 3/4.
(b) On observing the sum, we find that the fractions have different denominators which could not be added directly. For adding fractions with different denominators we have to convert them into like fractions.

Question 123:
Which one is greater?
1 m 40 cm + 60 cm or 2.6 m
Solution:

Question 124:
Match the fractions of Column I with the shaded or marked portion of figures of Column II

Solution:
On observing the figures given in Column II, we get
(a) Line is divided into 10 equal parts out of which 6 parts are shaded.
The fraction of shaded portion to the total parts = 6/10
(b) Square is divided into 16 equal parts out of which 6 parts are shaded.
The fraction of shaded portion to the total parts = 6/16
(c) Rectangle is divided into 7 equal parts out of which 6 parts are shaded.
The fraction of shaded portion to the total parts = 6/7
(d) Each of the two circle is divided in 4 equal parts out of which 4 parts of one circle and 2 parts of second circle are shaded.
The fraction of shaded portion to the total parts = 4+2 / 4+4 = 6/8
(e) Rectangle is divided into 6 equal parts out of which 6 are shaded.
The fraction of shaded portion to the total parts =
Hence, the (i) ->d, (ii) -> a, (iii) ->e, (iv) -> b, (v) -> c

Question 125:
Find the fraction that represents the number of natural numbers to total numbers in the collection 0,1, 2, 3, 4, 5. What fraction will it be for whole number?
Solution:
Given collection is 0, 1, 2, 3, 4, 5.
Natural numbers = 1, 2, 3, 4, 5
The fraction of natural numbers to the collection =5/6
Now, whole numbers = 0,1,2, 3, 4, 5, 6
The fraction of whole numbers to the collection = 6/6 = 1/1

Question 126:
Write the fraction representing the total number of natural numbers in the collection of numbers -3, -2,-1, 0,1, 2, 3. What fraction will it be for whole numbers? What fraction will it be for integers?
Solution:
Given collection is -3 -2,-1, 0,1,2, 3.
Natural numbers = 1,2, 3
The fraction of natural numbers to the collection = 3/7
Now, whole numbers = 0,1,2,3
The fraction of whole numbers to the collection = 4/7
and integers = – 3, -2, -1, 0,1,2, 3
The fraction of integers to the collection 7/7 = 1/1

Question 127:
Write a pair of fractions whose sum is 7/11 and difference is 2/11.
Solution:

Question 128:
What fraction of straight angle is a right angle.
Solution:
We know that, measures of right angle and straight angle are 90° and 180°

Question 129:
Put the right card in the right bag.

Solution:

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