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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


-: Select Topics :-

Section-A

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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
ML Aggarwal Class 7 ICSE Maths Model Question Paper 6 Q1

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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 = \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

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

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Answer

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

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 = \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.
ML Aggarwal Class 7 ICSE Maths Model Question Paper 6 Q10

Answer

In the given figure,
∠BAC = ∠EAF (Vertically opposite angles) = 70°
ML Aggarwal Class 7 ICSE Maths Model Question Paper 6 S10
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) = \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 = ………
ML Aggarwal Class 7 ICSE Maths Model Question Paper 6 Q12

Answer

∆DEF = ∆PQR (SAS criterion)
∠E = ∠Q
EF = QR
PQ = DE
ML Aggarwal Class 7 ICSE Maths Model Question Paper 6 S12

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.
ML Aggarwal Class 7 ICSE Maths Model Question Paper 6 Q13

Answer

In ||gm ABCD
AB = 15 cm, BC = 10 cm
Perpendicular DE = 8 cm and DF = ?
ML Aggarwal Class 7 ICSE Maths Model Question Paper 6 S13
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.

Answer

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.
ML Aggarwal Class 7 ICSE Maths Model Question Paper 6 Q15

Answer

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In the given l || m
x = 65° (Alternate Angles)
Similarly z = y
ML Aggarwal Class 7 ICSE Maths Model Question Paper 6 S15
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.
ML Aggarwal Class 7 ICSE Maths Model Question Paper 6 Q16

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 \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.
ML Aggarwal Class 7 ICSE Maths Model Question Paper 6 Q18

Answer

In ∆ABC, AD is perpendicular bisector of BC
BD = DC and ∠ADB = ∠ADC = 90°
ML Aggarwal Class 7 ICSE Maths Model Question Paper 6 S18
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.
ML Aggarwal Class 7 ICSE Maths Model Question Paper 6 S19

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 }.
ML Aggarwal Class 7 ICSE Maths Model Question Paper 6 Q20

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 = \frac { 14 }{ 2 } = 7 cm
(i) Area of shaded portion
= 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.
ML Aggarwal Class 7 ICSE Maths Model Question Paper 6 Q21

Answer

Reflection of the letter E has been shown dotted.
ML Aggarwal Class 7 ICSE Maths Model Question Paper 6 S21

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
ML Aggarwal Class 7 ICSE Maths Model Question Paper 6 S22

Question 23.
Draw two nets of a regular tetrahedron.

Answer

Two nets of the regular tetrahedron are given below:
ML Aggarwal Class 7 ICSE Maths Model Question Paper 6 S23

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?

Answer

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
ML Aggarwal Class 7 ICSE Maths Model Question Paper 6 S24
ML Aggarwal Class 7 ICSE Maths Model Question Paper 6 S24.1


Section – D

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

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.
ML Aggarwal Class 7 ICSE Maths Model Question Paper 6 Q25

Answer

In the figure, AC = 25, BC = 28,
AD ⊥ BC and AD = 15
Now in right ∆ADC
AC2 = AD2 + DC2 (Pythagoras Theorem)
⇒ (25)2 = 152 + DC2
⇒ 625 = 225 + DC2
⇒ DC2 = 625 – 225 = 400
⇒ DC2 = (20)2
⇒ DC = 20 units
BC = 28 units
BD = BC – CD = 28 – 20 = 8 units
Now in right ∆ABD
AB2 = AD2 + BD2 = 152 + 82 = 225 + 64 = 289 = (17)2
⇒ AB = 17 units

Question 26.

In the adjoining figure, show that ΔABC = ΔDBC. Hence, find the values of x and y.
ML Aggarwal Class 7 ICSE Maths Model Question Paper 6 Q26

Answer

In the given figure
∆ABC = ∆DBC
∠ABC = ∠DBC
⇒ 2y +6 = 56°
⇒ 2y = 56° – 6° = 50°
y = \frac { 50 }{ 2 } = 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°

Question 27.

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

Answer

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°.
ML Aggarwal Class 7 ICSE Maths Model Question Paper 6 S27

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 = \frac { 22 }{ 7 }

Answer

Length of wire = 88 cm
By bending it into a shape of a circle, the circumference of circle = 88 cm
ML Aggarwal Class 7 ICSE Maths Model Question Paper 6 S28
By bending the wire into the shape of a square.
Perimeter of square = 88 cm
Then side = \frac { Perimeter }{ 4 } = \frac { 88 }{ 4 } = 22 cm
and area = (Side)2 = (22)2 cm2 = 484 cm2
We see that area of circle is greater than that of square.
Difference = 616 – 484 = 132 cm2

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?

Answer

Mode of transport School bus Walking Bicycle Other vehicles
Boys 75 120 240 150
Girls 135 60 180 90

ML Aggarwal Class 7 ICSE Maths Model Question Paper 6 S29
(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.

 

– : End of Model Question Paper-6 Solved Class-7 ML Aggarwal Solutions  :–

 

Return to –  ML Aggarwal Maths Solutions for ICSE Class -7


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