CBSE Sample Papers For Class 7 Maths Mock Paper 1

CBSE Sample Papers for Class 7 Maths Mock Paper 1

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CBSE Class 7 Maths Sample Papers Mock Paper 1 (2023-24)

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CBSE Sample Papers for Class 7 Maths – Chapter Breakdown

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Benefits of Sample Paper for Class 7 Maths:

There are various benefits of using the CBSE Sample Papers For Class 7 Maths Mock Paper 1 offered by the Extramarks’ website. They are:

  1. The sample papers give students a wide range of numerical problems that are most likely to appear on their examination paper.
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  3. Those who want to increase their level of preparation for the exams can learn effectively by using these sample papers.

Q1. Express 65,950 in the standard forms.


65,950 = 6.595 × 10,000
= 6.595 × 104

Q2. Draw a net for a cuboid.

Ans: The net for a cuboid is shown below:

Q3. How many lines of symmetry are there in a circle?


A diameter divides a circle in two equal halves.

Since, a circle has infinite number of diameters,

Therefore, a circle has infinite lines of symmetry.

Q4. Evaluate 100 + 2a – 4b, when a = 3 and b = 2.


Given expression is: 100 + 2a – 4b
When a = 3, b = 2; we have:
100 + 2a – 4b = 100 + 2(3) – 4(2)
= 100 + 6 – 8
= 98

Q5.Solve the equation




Multiplying both sides by 3, we get

–m = 2 × 3 = 6

or, m = –6

Q6. What is the perimeter of a square having its side as 9a units?


Side of a square = 9a units
⸫ Perimeter of a square = 4(side)
= 4(9a)
= 36a units


What is the reciprocal of 23+52?


23+52=46+156        =196So, the reciprocal of  196 is 619.


What is the value of 15 of 255?


15 of 255 = 15 × 255 = 51

Q9. Find: (–7) × (2) × (–90)


(–7) × (2) × (–90) = –14 × (–90)
= 1260

Q10. A basket is full of fruits mangoes, oranges and apples. If 60% are mangoes, 10% are oranges than what is the percentage of apples.


Percentage of mangoes in basket = 60%

Percentage of oranges in basket = 10%

Percentage of apples = 100 – (60 + 10) = 30%

Q11. Find angles x and y in the following figure:


Exterior angle = Sum of two interior opposite angles
50º + x = 120º

or x = 70º

Now, sum of all the angles of a triangle = 180º

∴ 50º + y + 70º = 180º
y = 60º

Q12. A wire is broken into two pieces. If the wire is


m long and one piece is


m long, find the length of the other piece.


Let the length of other piece be x mLength of wire = 334mLength of one piece = 112mLength of other piece= 334112 m                         x=15432 m                               =   1564 m=9 4m=214 m

Q13. Two-third of a number is 3 more than one-third of the number. Find the number.


Let the number =xAccording to the given condition,23x=13x+323x13x=3213x=3x3=3x=9Thus, the number is 9.



Write the following rational numbers in ascending order.15, 37, 26, 53


15, 37, 26, 53The L.C.M of 5, 7, 6 and 3

Thus, L.C.M of 5, 7, 6 and 3 is 210.1×425×42=422103×307×30=902102×356×35=702105×703×70=350210Thus,90210<70210<42210<35021037<26<15<53


Simplify 2×34×259×42.


2×34×259×42=2×34×2532×222                =2×34×2532×22×2                =34×21+532×24                =34×2632×24                =264×342                =22×32                =4×9                =36

Q16. The highest marks in mathematics (out of 20) in different sections of Class VII are given below:

Class & Section












Draw a bar graph to represent this data.


Steps to draw the bar graph:
Step 1: Draw two perpendicular lines, one vertical and one horizontal.
Step 2: Along the horizontal line, mark Class and section
Step 3: Along the vertical line, mark the marks obtained, by taking a suitable scale.
Step 4: Take bars of the same width, keeping a uniform gap between them.
Step 5: Draw bars as per the given marks by calculating their height.

Q17. Identify the constant terms in the following expressions. Also, determine the type of algebraic expression.
1. −3+5x
2. −5a+20b−3c+4


1. Given expression is: −3+5x
Constant term in −3+5x is: −3
Since, there are two terms in −3+5x,
So, the given expression is a binomial.

2. Given expression is: −5a+20b−3c+4
Constant term in −5a+20b−3c+4 is: 4
As there are four terms in −5a+20b−3c+4,
So, the given expression is a quadrinomial.

Q18. In the following fig. , DA⊥AB, CB⊥AB and AC = BD. State the three pairs of equal parts in ΔABC an ΔDAB. Which of the following statements is meaningful?



The three pairs of equal parts are:
∠ABC = ∠BAD (= 90°)
AC = BD (Given)
AB = BA (Common side)
From the above, ΔABC≅ ΔBAD (By RHS congruence rule).
So, statement (i) is true
Statement (ii) is not meaningful, in the sense that the correspondence among the vertices is not satisfied.

Q19. Subtract the sum of (–545) and 125 from 1005.


According to the question,

1005 – {(–545) + 125}
= 1005 – (–420)
= 1005 + 420
= 1425

Q20. Determine the solid shapes for the following nets.




1. The given net has a square base and 4 triangular faces.
So, the given net is of a square based pyramid.
2. The given net has 3 rectangular faces and two triangular faces.
So, it represents the net of a triangular prism.
3. The given net has a circular base and a triangular curve.
Therefore, it is a net of a cone.

Q21. Multiply:
(i) –25 by 11
(ii) –15 by –12
(iii) 0 by –33


(i) (–25) × 11 = – (25 × 11) = –275
(ii) (–15) × (–12) = 15 × 12 = 180
(iii) 0 × (–33) = 0 (as multiplication of any number with 0 is always 0)


Plot25, 15 and 35 on a number line.


Q23. In the gven fig., ray AZ bisects ∠DAB as well as ∠DCB.
(i) State the three pairs of equal parts in triangles BAC and DAC.
(ii) Is ΔBAC≅ ΔDAC? Give reasons.
(iii) Is AB = AD? Justify your answer.
(iv) Is CD = CB? Give reasons.


(i)  In ΔBAC and ΔDAC, the three pairs of equal parts are as given below:
DAC = ∠CAB        (AZ bisects ∠BAD)∠DCA = ∠
ACB        (AZ bisects ∠BCD)
and AC = AC (Common in both)
(ii)  From (i) above, ΔBAC≅ ΔDAC      (By ASA congruence rule)
(iii)  AB = AD        (Corresponding parts of congruent triangles)
(iv)   CD = CB        (Corresponding parts of congruent triangles)

Q24. Find angles x and y in the following figure:


In triangle PQR,

∠P = x ( vertically opposite angles)

∠Q = x

∠R = y = x (vertically opposite angles)

Therefore, ∠P + ∠Q + ∠R = 1800

⇒ 3x = 1800

⇒ x = 600.

Since, x = y (Verically opposide triangle)

Hence, y = 600.

Q25. The length of a rectangular field is twice its breadth. If the perimeter of the field is 150 m, find its length and breadth.


Let the breadth be x m
Then the length = 2x m
Perimeter = 150 m
2(2x + x) = 150
⇒ 6x = 150
⇒ x = 25
Therefore, the breadth of the rectangular field is 25 m and the length is 50 m.

Q26. Simplify:

i32×45×x434×43×x9                     ii45×95×x723×36×x5



i32×45×x434×43×x9 =324×453×x49                        =32×42×x5                        =4232×x5ii45×95×x723×36×x5=225×325×x723×36×x5 =                   210×310×x723×36×x5                       =2103×3106×x75                        =27×34×x2

Q27. The figure given below, shows two circles with the same centre. The radius of the larger circle is 10 cm and the radius of the smaller circle is 4 cm.

(a) the area of the larger circle,

(b) the area of the smaller circle,

(c) the shaded area between the two circles. (Take = 3.14)


(i) Radius of the larger circle, R = 10cm
So,        area of larger circle, A = R2
= 3.14  10 10
= 314cm2
(ii) Radius of the smaller circle, r = 4cm
So,        area of smaller circle, a = r2
= 3.14  4 4
= 50.24cm2
(iii)          Area of shaded region = (314 – 50.24)cm2
= 263.76 cm2

Q28. Anand took a wire of length 44 cm and bent it into the shape of a circle. Find the radius of that circle. Also, find its area. If the same wire is bent into the shape of a square, what will be the length of each of its sides? Which figure encloses more area – the circle or the square?


Q29. Draw a line, say AB, take a point C outside it. Through C, draw a line parallel to AB using a ruler and compass only.


Steps of Construction:
(i) Draw a line AB. Take a point P on it and a point C outside this line. Join C to P.
(ii) Taking P as centre and with a convenient radius, draw an arc intersecting line AB at point D and PC at point E.
(iii) Taking C as centre and with the same radius as before, draw an arc FG intersecting PC at H.
(iv) Adjust the compass up to the length of DE. Without changing the opening of the compass and taking H as the centre, draw an arc to intersect the previously drawn arc FG at point I.
(v) Join the points C and I to draw a line ‘l’.

This is the required line that is parallel to line AB.

Q30. If A = 3x3 + x2 + 5 and B = 5y3 – 6y2 + 17 for x = 3 and y = –4, what is the value of 2A + 7B –AB and 12A – 5B?


For x = 3,
A = 3x3 + x2 + 5
= 3 (3)3 + (3)2 + 5
= 81 + 9 + 5
= 95
For y = – 4,
B =5y3 – 6y2 + 17
= 5(–4)3 – 6(–4)2 + 17
= –320 – 96 + 17
= –399
2A + 7B – AB = 2(95) + 7(–399) – (95) (–399)
= 190 – 2793 + 37,905
= 35,302
12A – 5B = 12 (95) – 5(–399)
= 1140 + 1995
= 3135

Q31. Five years ago, A father was seven times as old as his son. Five years hence, the father would be three times as old as his son. Find their present ages.


5 years ago, let the son’s age be x.
Then, father’s age = 7x

Present age of son = x + 5
Present age of father = 7x + 5

After 5 years, son’s age = (x + 5) + 5 = x + 10
After 5 years, father’s age = (7x + 5) + 5 = 7x + 10

According to the given condition
7x + 10 = 3(x + 10)
⇒ 4x = 20
⇒ x = 5

Therefore, the present age of son is 10 years and the father’s present age is 40 years.

Q32. The following table shows the points of each player scored in four games:

Player Game 1 Game 2 Game 3 Game 4
A 14 16 10 10
B 0 8 6 4
C 8 11 Did not play 13

Now, answer the following questions:
(i) Find the mean to determine A’s average number of points scored per game.
(ii) To find the mean number of points per game for C, would you divide the total points by 3 or by 4? Why?
(iii) B played in all four games. How would you find the mean?
(iv) Who was the best performer?


(i) As average number of points = 14+16+10+104 = 504=12.5(ii) To find the mean number of points per game for C,we will divide the total points by 3 because C played 3 games.(iii) Mean of Bs score = 0+8+6+44 = 184=4.5(iv) The best performer will have the greatest average among all.Now, we can observe that the average of A is 12.5, which is more than that of B and C.Therefore, A is the best performer among these three.

Q33. In the figure given below, ACD is a triangle such that AB is a median, and AD = DB = AB = 25 cm. What is the value of ∠ADB, ∠ABC and BC?


In ∆ACD,
AB is a median, so BD = BC = 25 cm.
Given: AD = DB = AB
So, ∆ADB is an equilateral triangle.
We know that the measure of each angle of an equilateral triangle is 60o.
So, ∠ADB = ∠DBA = ∠DAB = 60o
∠DBA and ∠ABC form a linear pair.
So, ∠DBA + ∠ABC = 180o
60o + ∠ABC = 180o
∠ABC = 120o

Q34. ABCD is a quadrilateral, diagonal AC bisects ∠A and
AB = AD,prove that DC = BC.


AD = AB              [Given]
∠DAC = ∠BAC          [Given]
AC = CA              [Common]
ΔACD ≅ ΔACB        [By SAS]
therefore, DC = BC       [By CPCT]

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2. Are the CBSE Sample Papers For Class 7 Maths Mock Paper 1 enough to get ready for the examination?

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