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

-: Select Topics :-

Section-A

Section-B

Section-C

Section-D

### Paper-6 Class-7 ML Aggarwal

(Based on Chapters 10 to 17)

Time allowed: 2 $\frac { 1 }{ 2 }$ 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 ∠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°

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

In a right angled triangle
Two legs are 8 cm and 15 cm
Hypotenuse = $\sqrt { { 8 }^{ 2 }+{ 15 }^{ 2 } }$
= √(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

∆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

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

Area of circle = Circumference (numerical)
πr2 = 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

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

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.

Let required angle = x
Its complement angle = 90° – x
According to the condition,
x = $\frac { 90-x }{ 2 }$ + 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. 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

A die is thrown
Total number of outcomes = 6
(i) Probability of a prime number = (2, 3, 5) = $\frac { 3 }{ 6 }$ = $\frac { 1 }{ 2 }$
(ii) Probability of a composite number= (4, 6) = $\frac { 2 }{ 6 }$ = $\frac { 1 }{ 3 }$

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 = ……… ∆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. 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= 120 cm2
If base BC = 10 cm
Then height DF = $\frac { Area }{ Base }$ = $\frac { 120 }{ 10 }$ = 12 cm

#### Question 14.

Find the mean age of six students whose ages (in years) are:
15, 13, 16, 13, 14, 16.

Mean age of 15, 13, 16, 13, 14, 16 $\frac { 15+13+16+13+14+16 }{ 6 }$ $\frac { 87 }{ 6 }$
= 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. 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. 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?

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 $\bar { BC }$.
(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. In ∆ABC, AD is perpendicular bisector of BC Now in ∆ABD and ∆ACD
BD = DC (D is the mid-point of BC)
∆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.

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 = $\frac { 22 }{ 7 }$. In the given figure,
ABCD is a rectangle whose length (l) = 20 cm
Two semicircles have been cut out from
the breadth sides whose radius = $\frac { 14 }{ 2 }$ = 7 cm
= Area of rectangle – Area of two semicircle
= l × b – 2 × $\frac { 1 }{ 2 }$ πr2
= 20 × 14 – 2 × $\frac { 1 }{ 2 }$ × $\frac { 22 }{ 7 }$ × 7 × 7
= 280 – 154= 126 cm2
(ii) Perimeter of shaded portion = 2 × l + 2πr
= 2 × 20 + 2 × $\frac { 22 }{ 7 }$ × 7
= 40 + 44
= 84 cm

#### Question 21.

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