CBSE Sample Papers For Class 6 Maths Mock Paper 1

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CBSE Sample Papers for Class 6 Math Mock Paper 1 (2021-22)

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.

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CBSE Class 6 Math Sample Papers Mock Paper 1

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Tips for Solving Math Sample Papers

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Q1-In 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.

”
Q2-Draw the line(s) of symmetry in the following Equilateral triangle.
Ans-The dotted lines show the line of symmetry. Therefore, the figure has three lines of symmetry.

Q3-Which is greatest: 0.0903, 0.00903, 0.000903?

”
Ans-In 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.

”
Q4-Which number is greater −8 or −11?

Ans-Since −8 lies to the right of −11.
Therefore, −8 > −11

Q5-“Find the equivalent fraction of

\frac{5}{9}

with denominator 63.

”
Ans-“

\frac{5}{9} = \frac{5 \times 7}{9 \times 7} = \frac{35}{63}

”
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)

”
Q9-a) 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-

\begin{matrix} 32 .12 \\ \underline{+ 14 .24} \\ \underline{46 .36} \end{matrix}

Sum of 32.12 and 14.24 = 32.12 + 14.24 = 46.36

\begin{matrix} 23.876 \\ \underline{+ 19 .243} \\ 43 .119 \end{matrix}

Sum of 23.876 and 19.243 = 23.876 + 19.243 = 43.119

\begin{matrix} 46.360 \\ \underline{- 43 .119} \\ 3 .241 \end{matrix}

⸫ Difference of both the sums = 46.36 – 43.119 = 3.241

”
Q11-Check whether 7656429 is divisible by 11 or not?

Ans-Divisibility 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.

”
Q12-Find 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 cm²
Area of figure II = (6 – 3) × (3.4 – 2)
= 3 × 1.4
= 4.2 cm²
Area of figure III = 3 × 1
= 3 cm²
Total area = 8.5 + 4.2 + 3
= 15.7 cm²

”
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	$\| \| \| \|$
34	3	$\| \| \|$
36	2	$\| \|$
39	5	$\| \| \| \|$
40	4	$\| \| \| \|$
42	5	$\| \| \| \|$
45	6	$\| \| \| \| \|$
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

AB

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

AB

.
Step 3: With the same radius and B as centre, cut the previous circle at points C and D.
Step 4: Join

CD

. It cuts

AB

at O. Note that O is the midpoint of

AB

and
∠AOD = ∠BOD = 90°. Thus

CD

is the perpendicular bisector of

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{matrix} 2 & 112, 160, 188 \\ 2 & 56, 80, 94 \\ 2 & 28, ​ 40, ​ 47 \\ 2 & 14, ​​ 20, 47 \\ 2 & 7, 10, ​ 47 \\ 5 & 7, ​​ 5, 47 \\ 7 & 7, ​ 1, 47 \\ 47 & 1, 1, 47 \\ 1, 1, 1 \end{matrix}

L.C.M. = 2 × 2 × 2 × 2 × 2 × 5 × 7 × 47

= 52640

”
Q23-“Arrange the following in descending order.

(i) \frac{4}{6}, \frac{5}{8}, \frac{7}{12}, \frac{5}{16} (ii) \frac{8}{17}, \frac{8}{9}, \frac{8}{5}, \frac{8}{13}

”
ANs-“

\begin{matrix} (i) \frac{4}{6}, \frac{5}{8}, \frac{7}{12}, \frac{5}{16} \\ LCM of 6, 8, 12 and 16 \\ 2 6, 8, 12, 16 \\ 2 3, 4, 6, 8 \\ 2 3, 2, 3, 4 \\ 2 3, 1, 3, 2 \\ 3 3, 1, 3, 1 \\ 1, 1, 1, 1 \end{matrix}

\begin{matrix} LCM of 6, 8, 12 and 16 = 2 \times 2 \times 2 \times 2 \times 3 \\ = 48 \\ So, \frac{4}{6}, \frac{7}{12}, \frac{5}{8}, \frac{5}{16} = \frac{4 \times 8}{6 \times 8}, \frac{7 \times 4}{12 \times 4}, \frac{5 \times 6}{8 \times 6}, \frac{5 \times 3}{16 \times 3} \\ = \frac{32}{48}, \frac{28}{48}, \frac{30}{48}, \frac{15}{48} \\ Descending order is: \\ \frac{32}{48}, \frac{30}{48}, \frac{28}{48}, \frac{15}{48} = \frac{4}{6}, \frac{5}{8}, \frac{7}{12}, \frac{5}{16} \end{matrix}

\begin{matrix} (ii) \frac{8}{17}, \frac{8}{9}, \frac{8}{5}, \frac{8}{13} \\ Since, two fractions with same \\ numerators are compared, the fraction \\ with the greater denominator will be \\ the smaller fraction. \\ So, the ascending order of denominators is \\ 5 < 9< 13< 17 \\ Thus, descending order of fractions is \\ \frac{8}{5} > \frac{8}{9} > \frac{8}{13} > \frac{8}{17} \end{matrix}

”
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{matrix} 2100) \bar{3675} (1 \\ \underline{- 2100} \\ 1575) 2100 (1 \\ \underline{- 1575} \\ 525) 1575 (3 \\ \underline{- 1575} \\ 0 \end{matrix}

\begin{matrix} H . CF of 3675 and 2100 = 525 \\ H . C . F of 525 and 1050 \\ 525) 1050 (2 \\ \underline{- 1050} \\ 0 \end{matrix}

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 + y² and x² + 2xy from x² + y² + xy + y.

”
Ans-“First we find the sum of x + y² and x² + 2xy
= (x + y²) + (x² + 2xy)
= x² + y² + 2xy + x
Now, (x² + y² + xy + y) – (x² + y² + 2xy + x)
= x² + y² + xy + y – x² – y² – 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).

”
Q30-Ramesh 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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FAQs (Frequently Asked Questions)

1. What is the advantage of the Extramarks’ CBSE Sample Papers For Class 6 Maths Mock Paper 1?

Extramarks’ subject-matter experts have years of classroom experience. They have published the CBSE sample paper solutions for Class 6. Everything contained in the study guide provided by Extramarks as per the NCERT syllabus will allow students to answer any question in unit tests, half-yearly exams and final exams. The concept has been presented in a very easy style by experts so that students can readily understand it. Students can get enough practice of the concepts by using the CBSE Sample paper created by Extramarks.

2. How can students use the Extramarks website to raise their score?

The Extramarks website assists students in solidifying their knowledge of the subject and acquiring the skills necessary to answer various questions available in textbooks. The website offers PDF files organised by subject-matter experts. The CBSE Sample Papers For Class 6 Maths Mock Paper 1 will unquestionably aid students in greatly enhancing their academic performance.

CBSE Sample Papers For Class 6 Maths Mock Paper 1

CBSE Sample Papers for Class 6 Math Mock Paper 1 (2021-22)

CBSE Class 6 Math Sample Papers Mock Paper 1

Benefits of CBSE Class 6 Math Sample Paper

Tips for Solving Math Sample Papers

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FAQs (Frequently Asked Questions)

1. What is the advantage of the Extramarks’ CBSE Sample Papers For Class 6 Maths Mock Paper 1?

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CBSE Sample Papers For Class 6 Maths Mock Paper 1

CBSE Sample Papers for Class 6 Math Mock Paper 1 (2021-22)

CBSE Class 6 Math Sample Papers Mock Paper 1

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