# 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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**Q1. **Express 65,950 in the standard forms.

**Ans:**

65,950 = 6.595 × 10,000

= 6.595 × 10^{4}

**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?

**Ans:**

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.

**Ans:**

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

**Ans:**

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?

**Ans:**

Side of a square = 9a units

⸫ Perimeter of a square = 4(side)

= 4(9a)

= 36a units

**Q7. **

**Ans:**

**Q8. **

**Ans:**

**Q9. **Find: (–7) × (2) × (–90)

**Ans:**

(–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.

**Ans:**

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:

**Ans:**

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

$1\frac{1}{2}$m long, find the length of the other piece.

**Ans:**

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

**Ans:**

**Q14.**

**Ans:**

$\begin{array}{}\mathrm{Thus},{L}.{C}.{M}\mathrm{of}5,7,6\mathrm{and}3\mathrm{is}210.\\ \u2013\frac{1\times 42}{5\times 42}=\u2013\frac{42}{210}\\ \u2013\frac{3\times 30}{7\times 30}=\u2013\frac{90}{210}\\ \u2013\frac{2\times 35}{6\times 35}=\u2013\frac{70}{210}\\ \frac{5\times 70}{3\times 70}=\frac{350}{210}\\ \mathrm{Thus},\\ \u2013\frac{90}{210}\u2013\frac{70}{210}\u2013\frac{42}{210}\frac{350}{210}\\ \u2013\frac{3}{7}\u2013\frac{2}{6}\u2013\frac{1}{5}\frac{5}{3}\end{array}$

**Q15. **

**Ans:**

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

Class & Section |
VII A |
VII B |
VII C |
VII D |
VII E |

Marks |
18 |
18 |
17 |
19 |
20 |

Draw a bar graph to represent this data.

**Ans:**

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

**Ans:**

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?

(i) ΔABC≅ ΔBAD (ii) ΔABC≅ ΔABD

**Ans:**

∠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.

**Ans:**

According to the question,

1005 – {(–545) + 125}

= 1005 – (–420)

= 1005 + 420

= 1425

**Q20.** Determine the solid shapes for the following nets.

1.

2.

3.

**Ans:**

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

**Ans:**

(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)

**Q22. **

**Ans:**

**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.

** **

**Ans:**

(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:

**Ans:**

In triangle PQR,

∠P = x ( vertically opposite angles)

∠Q = x

∠R = y = x (vertically opposite angles)

Therefore, ∠P + ∠Q + ∠R = 180^{0}

⇒ 3x = 180^{0}

⇒ x = 60^{0}.

Since, x = y (Verically opposide triangle)

Hence, y = 60^{0}.

**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.

**Ans:**

Let the breadth be *x* m

Then the length = 2*x* m

Perimeter = 150 m

2(2*x* + *x*) = 150

⇒ 6*x* = 150

⇒ *x* = 25

Therefore, the breadth of the rectangular field is 25 m and the length is 50 m.

**Q26. **Simplify:

**Ans:**

**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.

Find:

(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)

**Ans:**

(i) Radius of the larger circle, R = 10cm

So, area of larger circle, A = R^{2}

= 3.14 10 10

= 314cm^{2}

(ii) Radius of the smaller circle, r = 4cm

So, area of smaller circle, a = r^{2}

= 3.14 4 4

= 50.24cm^{2}

(iii) Area of shaded region = (314 – 50.24)cm^{2}

= 263.76 cm^{2}

**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?

**Ans:**

**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.

**Ans:**

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 = 3x^{3} + x^{2 }+ 5 and B = 5y^{3} – 6y^{2} + 17 for x = 3 and y = –4, what is the value of 2A + 7B –AB and 12A – 5B?

**Ans:**

For x = 3,

A = 3x^{3} + x^{2} + 5

= 3 (3)^{3 }+ (3)^{2} + 5

= 81 + 9 + 5

= 95

For y = – 4,

B =5y^{3} – 6y^{2} + 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.

**Ans:**

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?

**Ans:**

**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?

**Ans:**

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 60^{o}.

So, ∠ADB = ∠DBA = ∠DAB = 60^{o}

∠DBA and ∠ABC form a linear pair.

So, ∠DBA + ∠ABC = 180^{o}

60^{o} + ∠ABC = 180^{o}

∠ABC = 120^{o}

**Q34. **ABCD is a quadrilateral, diagonal AC bisects ∠A and

AB = AD,prove that DC = BC.

**Ans:**

In ΔACD and ΔACB

AD = AB [Given]

∠DAC = ∠BAC [Given]

AC = CA [Common]

ΔACD ≅ ΔACB [By SAS]

therefore, DC = BC [By CPCT]

## FAQs (Frequently Asked Questions)

### 1. Are the CBSE Sample Papers For Class 7 Maths Mock Paper 1 challenging for students?

With consistent practice on their part and the right guidance from Extramarks, students can easily study the CBSE Sample Papers For Class 7 Maths Mock Paper 1 and excel at the important topics to do well in their examinations. For students to easily understand the topic, Extramarks offers in-depth solutions to the sample papers and NCERT questions. Students who are hesitant to ask questions in front of their peers at school can also participate in doubt sessions through Extramarks.

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

In order for students to improve their essential concepts and be prepared to handle any challenges connected to them, the CBSE Sample Papers For Class 7 Maths Mock Paper 1 are sufficient. For better preparation, students may also practice past years’ papers. According to the subject-matter experts at the Extramarks’ website, students who fully understand the concepts perform better in examination.