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CBSE Sample Papers for Class 6 Math Mock Paper 1 (202122)
The foundation for higher classes is laid in Mathematics Class 6. Getting good grades depends on having solid concepts. Instead of memorising the answer key, students should concentrate on comprehending the subjects. Students should practice with the CBSE Sample Papers For Class 6 Maths Mock Paper 1 for preparing for the CBSE Class 6 Maths exam. Students learn about the many question formats that can be used in exams. Also, it provides them with a general overview of the exam format, scoring system, and question kinds. Extramarks experts have published the CBSE Sample Papers For Class 6 Maths Mock Paper 1 to assist students in better preparing for the Mathematics examination.
A wonderful resource for getting high marks in the Mathematics exams is the CBSE Sample Papers For Class 6 Maths Mock Paper 1. Students can evaluate their current academic progress by completing these sets of the CBSE Sample Papers For Class 6 Maths Mock Paper 1. They become familiar with the exam’s syllabus and the types of questions that will be asked. To perform well in the exams, students must practice as many sample papers as they can.
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CBSE Class 6 Math Sample Papers Mock Paper 1
Solving the CBSE Sample Papers For Class 6 Maths Mock Paper 1 will boost students’ confidence by allowing them to practice different types of questions and help them remember all the formulas. Students gain insight into the various ways that a specific question might be solved. After completing the sample papers, students must review their test score in order to discover their mistakes. The questions provided at Extramarks are organised systematically and adhere to the CBSE Class 6 Syllabus.
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Tips for Solving Math Sample Papers
It would be beneficial for students if they began solving the CBSE Sample Papers For Class 6 Maths Mock Paper 1 once they have finished the entire course. When students feel ready to take the exam, they should attempt the sample papers. This will give them more selfassurance. The CBSE Sample Papers For Class 6 Maths Mock Paper 1 were produced by subjectmatter experts to help students comprehend the concepts. Students should thoroughly study all the important concepts and topics covered in the sample papers. Additionally, by doing this, students will be able to effectively get rid of all of their worries and do well on their final examinations. The CBSE Sample Papers For Class 6 Maths Mock Paper 1 are offered in a certain way so that students can more easily access the chapters as needed. The sample papers are essential for assisting students in creating a strong foundation for both their future academic work and exam preparation. Students can access study materials including the NCERT Books, CBSE Revision Notes, CBSE Important Question, CBSE Syllabus, CBSE Sample Papers by using the Extramarks website. They can also get the CBSE Extra Questions, Formulas, and Previous Year Question Papers by using Extramarks.
Q1In Δ ABC, AB = 13 cm , BC = 13 cm , CA = 5 cm. Classify the triangle on the basis of its sides.
”
Ans“In Δ ABC two sides (AB and BC) are equal so it satisfies the property of isosceles triangle. Therefore, on the basis of sides triangle is isosceles.
”
Q2Draw the line(s) of symmetry in the following Equilateral triangle.
AnsThe dotted lines show the line of symmetry. Therefore, the figure has three lines of symmetry.
Q3Which is greatest: 0.0903, 0.00903, 0.000903?
”
AnsIn this case, the three numbers have same parts upto tenth. But the hundredth part of 0.0903 is the greatest of three.
∴ 0.0903 is the greatest.
”
Q4Which number is greater −8 or −11?
AnsSince −8 lies to the right of −11.
Therefore, −8 > −11
Q5“Find the equivalent fraction of
$\frac{5}{9}$with denominator 63.
”
Ans“
”
Q6“Ahmed walks at a speed of 5 km per hour while Ali walks at the rate of 10 km per hour. Find the ratio of speed of Ahmed to the speed of Ali.
”
Ans“Ahmed’s speed = 5 km per hour
Ali’s speed = 10 km per hour.
Required ratio = 5 : 10 or 1 : 2.
”
Q7“Find 14 × 36 using distributivity.
”
Ans“14 × 36
= (10 + 4) × 36
= 10 × 36 + 4 × 36
[Using distributive property of multiplication over addition]
= 360 + 144
= 504
”
Q8“Which one is greater: (–35) – (–15) or (–15) – (–35)?
Ans“(–35) – (–15) = (–35) + (additive inverse of –15)
= –35 + 15
= –20
(–15) – (–35) = (–15) + (additive inverse of –35)
= –15 + 35
= 20
Thus, (–35) – (–15) < (–15) – (–35)
”
Q9a) If a clock hand starts from 12 and stops at 9. How many right angles has it moved?
b) Where will the hand of a clock stop if starts at 3 and makes one fourth of a revolution clockwise?
”
ANs“a) 3 right angles. i.e. from 12 to 3, 3 to 6, 6 to 9.
b) At 6.
”
Q10“Subtract the sum of 23.876 and 19.243 from the sum of 32.12 and 14.24.
”
Ans
Sum of 32.12 and 14.24 = 32.12 + 14.24 = 46.36
$\begin{array}{}\mathrm{23}.876\\ \underset{\xaf}{+19.243}\\ \mathrm{\hspace{0.33em}\hspace{0.33em}}43.119\end{array}$Sum of 23.876 and 19.243 = 23.876 + 19.243 = 43.119
$\begin{array}{}\text{46.360}\\ \underset{\xaf}{43.119}\\ \text{\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}}3.241\end{array}$⸫ Difference of both the sums = 46.36 – 43.119 = 3.241
”
Q11Check whether 7656429 is divisible by 11 or not?
AnsDivisibility rule of 11: If the difference between the sum of the digits at odd places (from the right) and the sum of the digits at even places (from the right) of the number is either 0 or divisible by 11, then the number is divisible by 11.
The sum of the digits at odd places (from the right)
= 9 + 4 + 5 + 7
=25
The sum of the digits at even places (from the right)
= 2 + 6 + 6
=14
So, difference of sum of digits at odd and even places is
25 − 14= 11 which is divisible by 11.
So, 7656429 is divisible by 11.
”
Q12Find the number which gives 78 when added to 24.
Ans“Let the number be y.
Then we add 24 with y to get 78 ⇒ y + 24 = 78
y = 78 – 24
= 54.
”
Q13“Fill in the blanks:
(i) 6 +— = 0
(ii) 19 +— = 0
(iii) — 215 = 64
”
Ans“(i) – 6 + 6 = 0
(ii) 19 + (19) = 0
(iii) 151 – 215 = – 64
”
Q14“Find the area of figure.
Ans“Area of the figure = Area of I + Area of II + Area of III
Now, Area of figure I = 3.4 × 2.5
= 8.5 cm^{2}
Area of figure II = (6 – 3) × (3.4 – 2)
= 3 × 1.4
= 4.2 cm^{2}
Area of figure III = 3 × 1
= 3 cm^{2}
Total area = 8.5 + 4.2 + 3
= 15.7 cm^{2}
”
Q15“A truck is travelling at 50km/h.
(a) How long does it take to go 8 km?
(b) How far does it go in 8 minutes?
”
Ans“
”
Q16“State the following statements as true or false. Correct the statement if false.
(a) An obtuse angle < 90º .
(b) A measure of complete angle is 360º.
(c) Straight angle lies between one fourth and half of a revolution.
”
Ans“a) False. It is greater 90°.
b) True.
c) False. Straight angle is equal to half of a revolution.
”
Q17“Sam collected a data for weights (in kg) of 30 students in class 6^{th}. He recorded his findings as shown below. Arrange the weights in a table using tally marks and answer the questions given below.
40, 34, 45, 32, 36, 39, 42, 45, 40, 34, 32, 32, 45, 45, 45, 39, 39, 40, 36, 32, 42, 42, 39, 40, 34, 45, 32, 42, 42, 39
a. How many students have the maximum weight?
b. How many students have weight less than 35 kg?
”
Ans“
Weight (in kg)  Number of Students  Tally mark 
32  5 
$\overline{)\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}$

34  3 
$\mathrm{\hspace{0.17em}\hspace{0.17em}\hspace{0.33em}}$

36  2 
$\hspace{0.17em}\hspace{0.17em}$

39  5 
$\overline{)\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}$

40  4 
$\hspace{0.17em}\hspace{0.17em}\hspace{0.33em}\hspace{0.33em}$

42  5 
$\overline{)\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}$

45  6 
$\overline{)\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}$

Total  30 
a. The maximum weight is 45 kg, and 6 students weigh 45 kgs. Thus 6 students have the maximum weight.
b. Weights less than 35 kg are 34 kg and 32 kg. 5 students have weight 32 kg, and 3 students have weight 34 kg. So, 5 + 3 = 8 students have weight less than 35 kg.
”
Q18“
A carpet 6.20 m long and 3.40 m broad is bounded by a tape around the edges. How much tape is needed? If the cost of the tape is ₹4 per meter, find the cost of the required tape.
”
ANs“ Required tape = Perimeter of the carpet
Perimeter of carpet = 2 × (length + breadth)
= 2 × ( 6.20 m + 3.40 m)
= 2 × 9.60 m
= 19.20 m
Perimeter cost of the tape
= ₹ 4
Total cost of Tape = 4 × 19.20
= ₹ 76.80
”
Q19“A shopkeeper sold 40 notebooks one day. The next day he sold 50 notebooks. If the notebook costs Rs 12, how much did he earn in all?
”
Ans“Money earned one day = Rs 40 × 12 = Rs 480
Money earned the next day = Rs 50 × 12 = Rs 600
Total money earned = Rs (480 + 600)= Rs 1080
Alternative Method:
Total notebooks sold = (40 + 50)
Money earned = Rs 12 × (40 + 50) = Rs (12 × 90) = Rs 1080
Q20“Draw a line segment of length 6.2 cm and construct its perpendicular bisector.
”
Ans“Step 1: Draw line segment
of length = 6.2 cm.
Step 2: With A as the centre, using compasses, draw a circle with radius more than half the length of
.
Step 3: With the same radius and B as centre, cut the previous circle at points C and D.
Step 4: Join
. It cuts
$\overline{)\mathrm{AB}}$at O. Note that O is the midpoint of
$\overline{)\mathrm{AB}}$ and
∠AOD = ∠BOD = 90°. Thus
is the perpendicular bisector of
$\overline{)\mathrm{AB}}$.
”
Q21“Fill in the blanks:
a. When the sum of the measures of two angles is that of a right angle, then each one of them is an_____ angle.
b. When the sum of the measure of two angles is that of a straight angle and one of them is acute then the other will be_____.
c. Line joining the opposite vertices of a polygon is called a ____.
”
Ans“a. acute
b. obtuse
c. diagonal
”
Q22“Find the L.C.M of 112, 160 and 188.
”
Ans“L.C.M of 112, 160 and 188 is given by
$\begin{array}{cc}2& 112,\text{\hspace{0.17em}}160,\text{\hspace{0.17em}}188\\ 2& 56,\text{\hspace{0.17em}}80,\text{\hspace{0.17em}}94\\ 2& 28,\text{}40,\text{}47\\ 2& 14,\text{\hspace{0.17em}}20,\text{\hspace{0.17em}}47\\ 2& 7,\text{\hspace{0.17em}}10,\text{}47\\ 5& 7,\text{\hspace{0.17em}}5,\text{\hspace{0.17em}}47\\ 7& 7,\text{\hspace{0.17em}}1,\text{\hspace{0.17em}}47\\ 47& 1,\text{\hspace{0.17em}}1,\text{\hspace{0.17em}}47\\ & 1,\text{\hspace{0.17em}}1,\text{\hspace{0.17em}}1\end{array}$
L.C.M. = 2 × 2 × 2 × 2 × 2 × 5 × 7 × 47
”
Q23“Arrange the following in descending order.
”
ANs“
$\begin{array}{}\left(\mathrm{ii}\right)\text{\hspace{0.17em}}\frac{8}{17},\frac{8}{9},\frac{8}{5},\frac{8}{13}\\ \mathrm{Since},\text{twofractionswithsame}\\ \text{numeratorsarecompared,thefraction}\\ \text{withthegreaterdenominatorwillbe}\\ \text{thesmallerfraction.}\\ \text{So,theascendingorderofdenominatorsis}\\ 5\text{91317}\\ \text{Thus,descendingorderoffractionsis}\\ \frac{8}{5}\frac{8}{9}\frac{8}{13}\frac{8}{17}\end{array}$
”
Q24“Solve the following.
(i) The sum of two integers is – 20. If one integer is 20, find the other integer.
(ii) The sum of two integers is 147. If one integer is – 59, find the other integer.
”
Ans”
(i) Sum of two integers = – 20
One integer = 20
Second integer = – 20 – 20
= – 40
(ii) Sum of two integers = 147
One integer = – 59
Second integer = 147 – (– 59)
= 147 + 59
= 206
”
Q25“The length, breadth and height of a hall are 3675 cm, 2100 cm and 1050 cm respectively. Find the maximum length of a tape which can measure the the three dimensions of the hall exactly.
”
Ans“We have to find the H.C.F of 3675, 2100 and 1050.
H.C.F. of 3675 and 2100
$\begin{array}{}\left.2100\right)\overline{3675}\left(1\right.\\ \underset{\xaf}{2100}\\ \left.1575\right)2100\left(1\right.\\ \underset{\xaf}{1575}\\ \left.525\right)1575\left(3\right.\\ \underset{\xaf}{1575}\\ 0\end{array}$$\begin{array}{}{H}.\mathrm{CF}\text{\hspace{0.17em}}\mathrm{of}\text{\hspace{0.17em}}3675\text{\hspace{0.33em}}\mathrm{and}\text{\hspace{0.33em}}2100\text{\hspace{0.33em}}=\text{\hspace{0.33em}}525\\ {H}.{C}.{F}\text{\hspace{0.33em}}\mathrm{of}\text{\hspace{0.33em}}525\text{\hspace{0.33em}}\mathrm{and}\text{\hspace{0.33em}}1050\\ \left.525\right)1050\left(2\right.\\ \underset{\xaf}{1050}\\ 0\end{array}$
H.C.F of 525 and 1050 = 525
Therefore, HCF of 3675, 2100 and 1050 = 525
The length of the tape which can measure the three dimensions of the hall exactly is 525 cm.
”
Q26“Draw a bar graph representing the number of literate people from 2015 to 2021.
Year  Number of literate people (in thousands) 
2015  16 
2016  24 
2017  40 
2018  36 
2019  12 
2020  74 
2021  6 
a) In which year the number of literate people is the highest?
b) In which year the number of literate people is the least?
”
ANs“Draw a pair of perpendicular lines with number of literate people on vertical axis and year on horizontal axis.
Choose a suitable scale: 1 unit = 10000 literate people
Draw bars with height of each bar representing the number of literate people corresponding to each year.
The graph will be obtained as
a) The highest bar is of the year 2020. So, it has the highest number of literate people.
b) The smallest bar is of the year 2021. So, it has the least number of people.
”
Q27“Mohan has a rectangular field of length 240 m
and breadth 170 m. He wants to fence it with
4 rounds of rope. What is the total length of
rope he must use?
”
Ans“Length of rectangular field = 240 m
Breadth of rectangular field = 170 m
Perimeter of rectangular field = 2(240 + 170) m
= 2(410) m
= 820 m
Length of rope in one round = 820 m
Length of rope in four round = 4×820 m
= 3280 m
Thus, the total length of required rope is 3240 m.
”
Q28“Subtract the sum of x + y^{2} and x^{2} + 2xy from x^{2} + y^{2} + xy + y.
”
Ans“First we find the sum of x + y^{2} and x^{2} + 2xy
= (x + y^{2}) + (x^{2} + 2xy)
= x^{2} + y^{2} + 2xy + x
Now, (x^{2} + y^{2} + xy + y) – (x^{2} + y^{2} + 2xy + x)
= x^{2} + y^{2} + xy + y – x^{2} – y^{2} – 2xy – x
= y – xy – x.
”
Q29“In some parts of a country, the temperature recorded in the month of November 2017 is –25°C, it decreases by 2°C every year till 2019 and then increases by 5°C every year till 2021. Based on the information given answer the following questions.
 What was the temperature in November 2018?
 –29°C
 –27°C
 –24°C
 –19°C
 What was the temperature in November 2020?
 –29°C
 –27°C
 –24°C
 –19°C
 What is the difference between the highest and the lowest temperature?
 –10°C
 –12°C
 10°C
 12°C
 What is the sum of the temperature recorded in the year 2017 and 2020?
 –1°C
 –49°C
 1°C
 49°C
 Which of the following is the coldest year?
 2017
 2018
 2019
 2020
”
Ans“Based on the given information, we know that temperature in the year 2017 = –25°C
 Temperature in the year 2018 = –25°C –2°C
= –27°C
The correct answer is option b).
 Temperature in the year 2019 = –27°C –2°C
= –29°C
Temperature in the year 2020 = –29°C + 5°C
= –24°C
The correct answer is option c).
 Temperature in the year 2021 = –24°C + 5°C
= –19°C
The highest temperature is –19°C and the lowest temperature is –29°C
Difference = Highest temperature – Lowest temperature
= –19°C – (–29°C)
= –19°C + 29°C
= 10°C
The correct answer is option c).
 Sum = Temperature in 2017 + Temperature in 2020
= –25°C + (–24°C)
= –25°C – 24°C
= –49°C
The correct answer is option b).
 The coldest year is 2019 because the temperature in the year 2019 is –29°C which is lowest among all.
The correct answer is option c).
”
Q30Ramesh bought vegetables from the market weighing 15 kg in all. Out of this, 10.250 kg is potatoes and 3.500 kg is tomatoes and the rest are onions. Find the weight of onions.
ANs”
Weight of onions = total weight – (weight of potatoes + weight of tomatoes)
= 15.000 – (10.250 + 3.500) kg
= 1.250 kg
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