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

Model Question **Paper-6 Class-7 ML Aggarwal** ICSE Mathematics Solutions. APC Understanding Mathematics for ICSE Class-7 Model Question Paper-6 Solutions of Section-A, B, C and D based on Chapter-10 to 17. Visit official Website **CISCE **for detail information about ICSE Board Class-7 Mathematics.

## Model Question **Paper-6 Class-7 ML Aggarwal** ICSE Mathematics Solutions

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### Paper-6 Class-7 ML Aggarwal

(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 comprises of 10 questions of 4 marks each.
- Section 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**

(Model Question **Paper-6 Class-7 ML Aggarwal** ICSE Mathematics Solutions)

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

#### Answer

∠AOC and ∠COB = 180° (Linear pair)

⇒ 2x + x + 15 = 180°

⇒ 3x = 180°- 15°= 165°

⇒ x = 55° (b)

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°

#### Answer

One exterior angle of a triangle = 118°

One of interior opposite two angles = 54°

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

Question 3.

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

(a) 23 cm

(b) 20 cm

(c) 17 cm

(d) 17 m

#### Answer

In a right angled triangle

Two legs are 8 cm and 15 cm

Hypotenuse =

= √(64 + 225) = √289 = 17 cm (c)

Question 4.

If ΔABC = ΔPRQ, the the correct statement is

(a) AB = PQ

(b) ∠B = ∠Q

(c) ∠C = ∠R

(d) AC = PQ

#### Answer

∆ABC = ∆PRQ

AC = PQ is correct (d)

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

#### Answer

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

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

#### Answer

Area of circle = Circumference (numerical)

πr^{2} = πd = 2πr ⇒ r = 2

d = 2r = 4 units (b)

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

#### Answer

A quadrilateral has two lines of symmetry

and rotational symmetry of order 2 is a rhombus. (c)

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

#### Answer

Mode is the observation of the data

occurring maximum number of times. (c)

**Section – B**

( Model Question **Paper-6 Class-7 ML Aggarwal** ICSE Mathematics Solutions )

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.

#### Answer

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°

Question 10.

In the given figure, find the value of x.

#### Answer

In the given figure,

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

But Ext. ∠ACD = ∠BAC + ∠B

125° = 70° + x

⇒ x = 125° – 70° = 55°

x = 55°

Question 11.

A die is thrown once. Find the probability of getting

(i) a prime number

(ii) a composite number

#### Answer

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) = =

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 = ………

#### Answer

∆DEF = ∆PQR (SAS criterion)

∠E = ∠Q

EF = QR

PQ = DE

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.

#### Answer

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

#### Question 14.

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

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

#### Answer

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

=

=

= 14.5 years

**Section – C**

( Model Question **Paper-6 Class-7 ML Aggarwal** ICSE Mathematics Solutions )

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.

#### Answer

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°

#### Question 16.

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

#### Answer

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°

#### 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?

#### Answer

Length of two sides of a triangle are 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

#### 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.

#### Answer

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.

#### 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.

#### Answer

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.

#### 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 = .

#### Answer

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

#### Question 21.

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

#### Answer

Reflection of the letter E has been 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.

#### Answer

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

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