## ML Aggarwal Class 7 ICSE Maths Model Question Paper 6

(Based on Chapters 10 to 17)

**Time allowed: 2 Hours**

**Maximum Marks: 90**

**General Instructions**

- All questions are compulsory.
- The question paper consists of 29 questions divided into four sections A, B, C, and D.
- Section A comprises of 8 questions of 1 mark each.
- Section B comprises of 6 questions of 2 marks each.
- Section C comparises of 10 questions of 4 marks each.
- Sectioni D comprises of 5 questions of 6 marks each.
- Question numbers 1 to 8 in Section A is multiple choice questions where you are to select one correct option out of the given four.

**Section – A**

Question numbers 1 to 8 is of 1 mark each.

Question 1.

In the given figure, if ∠AOC and ∠COB form a linear pair, then the value of x

(a) 60

(b) 55

(c) 50

(d) 45

Question 2.

An exterior angle of a triangle is 118°. If one of the two interior opposite angle is 54°, then the other interior opposite angle is

(a) 62°

(b) 54°

(c) 64°

(d) 59°

Question 3.

In a right angled triangle, the lengths of two legs are 8 cm and 15 cm. The lenghts of the hypotenuse is

(a) 23 cm

(b) 20 cm

(c) 17 cm

(d) 17 m

Question 4.

If ΔABC = ΔPRQ, the the correct statement is

(a) AB = PQ

(b) ∠B = ∠Q

(c) ∠C = ∠R

(d) AC = PQ

Question 5.

The number of lines that can be drawn parallel to a given line l through a point outside the line l is

(a) 0

(b) 1

(c) 2

(d) infinitely many

Question 6.

If the area of a circle is numerically equal to its circumference, then the diameter of the circle is

(a) 2 units

(b) 4 units

(c) 6 units

(d) 8 units

Question 7.

A quadrilateral having exactly two lines of symmetry and rotational symmetry of order 2 is a

(a) square

(b) parallelogram

(c) rhombus

(d) kite

Question 8.

A mode is the observation of the data

(a) whose position is in the middle

(b) having maximum value

(c) occurring a maximum number of times

(d) occurring a minimum number of times

**Section – B**

Question numbers 9 to 14 are of 2 marks each.

Question 9.

An angle is 30° more than one-half of its complement. Find the angle.

Question 10.

In the given figure, find the value of x.

Question 11.

A die is thrown once. Find the probability of getting

(i) a prime number

(ii) a composite number

Question 12.

You want to show that ΔDEF = ΔPQR by SAS congruence rule.

It is given that ∠E = ∠Q, you need to have

(i) EF = …….

(ii) PQ = ………

Question 13.

In the given figure, two adjacent sides of a parallelogram are 15 cm and 10 cm. If the distance between the longer sides is 8 cm, find the

(i) area of parallelogram

(ii) the distance between the shorter sides.

Question 14.

Find the mean age of six students whose ages (in years) are:

15, 13, 16, 13, 14, 16.

**Section – C**

Question numbers 15 to 24 are of 4 marks each.

Question 15.

In the given figure, lines l and m are parallel. Find the values of x, y, and z.

Question 16.

In the given figure, BC = AC. Find the value of x.

Question 17.

If the lengths of the two sides of a triangle are 6 cm and 8.5 cm, then what can be the length of the third side?

Question 18.

In the given figure, AD is perpendicular bisector of .

(i) State three pairs of equal parts in ∆ABD and ∆ACD.

(ii) Is ∆ABD = ∆ACD? Give reasons.

(iii) Is ABC an isosceles triangle? Give reasons.

Question 19.

Draw a line, say AB, take a point P outside line AB. Through P, draw a line parallel to line AB using ruler and compasses only.

Question 20.

In the given figure, ABCD is a rectangle with AB = 20 cm and BC = 14 cm. Two semicircles are cut from each of two breadths as diameters. Find

(i) the area of the shaded region

(ii) the perimeter of the shaded region.

Take π = .

Question 21.

Draw the reflection of the letter E in the given mirror line shown dotted.

Question 22.

Three cubes each with edge 2 units are placed side by side to form a cuboid. Find the dimensions of the cuboid so formed and draw an isometric sketch of this cuboid.

Question 23.

Draw two nets of a regular tetrahedron.

Question 24.

A boy scored the following marks in various class tests during a year, each test is marked out of 20:

15, 17, 16, 7, 9, 12, 14, 16, 3, 19, 12, 16

(i) Arrange the marks in ascending order.

(ii) What are his modal marks?

(iii) What are his median marks?

(iv) What are his mean marks?

**Section – D**

Question numbers 25 to 29 are of 6 marks each.

Question 25.

In the given figure, all measurements are in centimeters. If AD is perpendicular to BC, find the length of B.

Question 26.

In the adjoining figure, show that ΔABC = ΔDBC. Hence, find the values of x and y.

Question 27.

By using ruler and compasses only, construct a triangle ABC with BC = 7.5 cm, ∠B = 60° and ∠A = 90°.

Question 28.

Anjali took a wire of length 88 cm and bent it into the shape of a circle. Find the area enclosed by that circle. If the same wire is bent in the shape of a square, then find the area enclosed by that square. Which shape encloses more area and by how much?

Take π =

Question 29.

Given below is the data of school going students (boys and girls):

Mode of transport | School bus | Walking | Bicycle | Other vehicles |

Boys | 75 | 120 | 240 | 150 |

Girls | 135 | 60 | 180 | 90 |

Draw a double bar graph to represent the above data and hence answer the following questions:

(i) Which mode of transport is used by more students?

(ii) Which mode of transport is used by more number of girls that the number of boys?

What values are being promoted by using a bicycle as a mode of transport?

**Solutions**

Solution 1:

∠AOC and ∠COB = 180° (Linear pair)

⇒ 2x + x + 15 = 180°

⇒ 3x = 180°- 15°= 165°

⇒ x = 55° (b)

Solution 2:

One exterior angle of a triangle = 118°

One of interior opposite two angles = 54°

Second inner angle = 118° – 54° = 64° (c)

Solution 3:

In a right angled triangle

Two legs are 8 cm and 15 cm

Hypotenuse =

= √(64 + 225)

= √289

= 17 cm (c)

Solution 4:

∆ABC = ∆PRQ

AC = PQ is correct (d)

Solution 5:

Number of lines drawn from a point parallel to a given line is only one. (b)

Solution 6:

Area of circle = Circumference (numerical)

πr^{2} = πd = 2πr ⇒ r = 2

d = 2r = 4 units (b)

Solution 7:

A quadrilateral has two lines of symmetry and a rotational symmetry of order 2 is a rhombus. (c)

Solution 8:

Mode is the observation of the data occurring maximum number of times. (c)

Solution 9:

Let required angle = x

Its complement angle = 90° – x

According to the condition,

x = + 30

⇒ 2x = 90 – x + 60

⇒ 2x + x = 150

⇒ 3x = 150 150°

⇒ x = 50°

Hence required angle = 50°

Solution 10:

In the given figure,

∠BAC = ∠EAF (Vertically opposite angles) = 70°

But Ext. ∠ACD = ∠BAC + ∠B

125° = 70° + x

⇒ x = 125° – 70° = 55°

x = 55°

Solution 11:

A die is thrown

Total number of outcomes = 6

(i) Probability of a prime number = (2, 3, 5) = =

(ii) Probability of a composite number= (4, 6) = =

Solution 12:

∆DEF = ∆PQR (SAS criterion)

∠E = ∠Q

EF = QR

PQ = DE

Solution 13:

In ||gm ABCD

AB = 15 cm, BC = 10 cm

Perpendicular DE = 8 cm and DF = ?

Now area of ||gm ABCD = Base × Height

= AB × DE

= 15 × 8 cm^{2}

= 120 cm^{2}

If base BC = 10 cm

Then height DF = = = 12 cm

Solution 14:

Mean age of 15, 13, 16, 13, 14, 16

=

=

= 14.5 years

Solution 15:

In the given l || m

x = 65° (Alternate Angles)

Similarly z = y

But x + 45° + z = 180° (Angle on one side of a line)

⇒ 65° + 45° + z = 180°

⇒ 110° + z = 180°

⇒ z = 180°- 110° = 70°

y = z = 70°

x = 65°, y = 70°, z = 70°

Solution 16:

In the given figure,

BC = AC, ∠C = x, ∠ABD = 155°

∠CAB = ∠CBA

But ∠ABC + ∠ABD = 180° (Linear pair)

⇒∠ABC + 155°= 180°

⇒ ∠ABC = 180° – 155° = 25°

⇒ ∠CAB = ∠ABC = 25°

But ∠CAB + ∠ABC + ∠ACB = 180° (Angles of a triangle)

⇒ 25° + 25° + x° = 180°

⇒ x + 50° = 180°

⇒ x = 180° – 50° = 130°

Solution 17:

Length of two sides of a triangle an 6 cm and 8.5 cm.

Third side will be less than (6 + 8.5) = 14.5 cm

and greater than (8.5 – 6) = 2.5 cm

Solution 18:

In ∆ABC, AD is perpendicular bisector of BC

BD = DC and ∠ADB = ∠ADC = 90°

Now in ∆ABD and ∆ACD

BD = DC (D is the mid-point of BC)

∠ADB = ∠ADC (Each 90°)

AB = AD (Common)

∆ABD = ∆ACD (SAS criterion)

AB = AC (c.p.c.t.)

∆ABC is an isosceles triangle.

Solution 19:

Steps of construction :

(i) Draw a line AB.

(ii) Take a point P outside AB.

(iii) Take a point Q on AB and join PQ.

(iv) From P, construct ∠OPX = ∠PQB and product XP to Y.

Then XY is parallel to AB.

Solution 20:

In the given figure,

ABCD is a rectangle whose length (l) = 20 cm

Breadth (b) = 14 cm

Two semicircles have been cut out from the breadth sides whose radius = = 7 cm

(i) Area of shaded portion = Area of rectangle – Area of two semicircle

= l × b – 2 × πr^{2}

= 20 × 14 – 2 × × × 7 × 7

= 280 – 154= 126 cm^{2}

(ii) Perimeter of shaded portion = 2 × l + 2πr

= 2 × 20 + 2 × × 7

= 40 + 44

= 84 cm

Solution 21:

Reflection of the letter E has been shown dotted.

Solution 22:

Three cubes, every 2 units are placed together side by side forming a cuboid.

Then length = 2 × 3 = 6 units

Breadth = 2 units

Height = 2 units

Solution 23:

Two nets of regular tetrahedron are given below:

Solution 24:

Marks scored by a boy during a year:

15, 17, 16, 7, 9, 12, 14, 16, 3, 19, 12, 16

(i) Arranging in ascending order:

3, 7, 9, 12, 12, 14, 15, 16, 16, 16, 17, 19

(ii) Model marks are 16

(iii) Median : Here n = 12 which is even

Solution 25:

In the figure, AC = 25, BC = 28,

AD ⊥ BC and AD = 15

Now in right ∆ADC

AC^{2} = AD^{2} + DC^{2} (Pythagoras Theorem)

⇒ (25)^{2} = 15^{2} + DC^{2}

⇒ 625 = 225 + DC^{2}

⇒ DC^{2} = 625 – 225 = 400

⇒ DC^{2} = (20)^{2}

⇒ DC = 20 units

BC = 28 units

BD = BC – CD = 28 – 20 = 8 units

Now in right ∆ABD

AB^{2} = AD^{2} + BD^{2} = 15^{2} + 8^{2} = 225 + 64 = 289 = (17)^{2}

⇒ AB = 17 units

Solution 26:

In the given figure

∆ABC = ∆DBC

∠ABC = ∠DBC

⇒ 2y +6 = 56°

⇒ 2y = 56° – 6° = 50°

y = = 25°

and ∠ABC = 56°

∠ACB = ∠BCD = 40°

But in ∆DBC

∠CBD + ∠BCD + ∠BDC = 180° (Angle of a triangle)

56° + x + 40° = 180°

⇒ x + 96° = 180°

⇒ x = 180° – 96° = 84°

x = 84°, y = 25°

Solution 27:

Steps of construction:

In ∆ABC

∠B = 60°, ∠A = 90°;

then ∠C = 180° – (60° + 90°)

⇒∠C = 180° – 150° = 30°

(i) Take a line segment BC = 7.5 cm

(ii) At B, draw a ray BX making an angle of 60° and at C, draw a ray CY making an angle of 30° which intersect each other at A.

∆ABC is the required triangle whose ∠A will be 90°.

Solution 28:

Length of wire = 88 cm

By bending it into a shape of a circle, the circumference of circle = 88 cm

By bending the wire into the shape of a square.

Perimeter of square = 88 cm

Then side = = = 22 cm

and area = (Side)^{2} = (22)^{2} cm^{2} = 484 cm^{2}

We see that area of circle is greater than that of square.

Difference = 616 – 484 = 132 cm^{2}

Solution 29:

Mode of transport | School bus | Walking | Bicycle | Other vehicles |

Boys | 75 | 120 | 240 | 150 |

Girls | 135 | 60 | 180 | 90 |

(i) Bicycle is used by more students.

(ii) Most of the girls used the school bus.

(iii) Use of the bicycle is good for health.

So, it should be promoted by the students.

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