CBSE 8th Mathematics A square and A cube - New Model Questions Papers Study Material - QB365 Set B

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Questions:

Long Answer Type Questions

7 x 5 = 35

Mother wants to divide ₹ 36 between her daughters Shreya and Bhoomika in the ratio of their ages. If age of Shreya is 15 yr and age of Bhoomika is 12 yr find, how much Shreya and Bhoomika will get.

Find the shaded area in the given figure, if PT = TD.

Each student of Class VIll contributed some money for a picnic. The money contributed by each student was equal to square of the total number of students. If the total collected amount was 2209 then find the total number of students.

The area of a square plot is 101

﻿\(\frac{1}{400} \mathrm{~m}^2\)﻿

find the length of one side of the plot

The given table shows the crop production of a state in the year 2008 and 2009. Observe, the table given below and answer the given questions.

(i) For which crop(s) did the production decrease?
(ii) Write the production of all the crops in 2009 in their standard form.
(iii) Assuming the same decrease in rice production each year as in 2009, how many acres will be harvested in 2015? Write in standard form.

Find three consecutive numbers such that the first number is a multiple of 2, the second number is a
multiple of 3 and the third number is a multiple of 4. Are there more such numbers? How often do they Occur?

Draw a quadrilateral whose diagonals have equal lengths of 8 cm that bisect each other and intersect at an angle of
(i) 30°
(ii) 40°
(iii) 90°
(iv) 140°

Multiple Choice Question

16 x 1 = 16

A mixture of 30 kg contains sand and cement in the ratio of 2 : 1. How much Sand should be added to the mixture to make the ratio of sand to cement 3 : 1?

(a)

5 kg

(b)

2 kg

(c)

10 kg

(d)

6 kg

If (3a + 2b): (5a +3b) =18:29 then the value of a : b is

(a)

3 : 4

(b)

4 : 3

(c)

2 : 3

(d)

3 : 2

10.

The product of 632 and 999 is

(a)

631368

(b)

631386

(c)

624624

(d)

626424

11.

The value of (5 + a) (5 - b) is

(a)

5a + 5b + ab + 25

(b)

5a - 5b - ab - 25

(c)

5b - 5a + ab - 25

(d)

5a - 5b - ab + 25

12.

The value of 21+23+25+27+29 is equal to

(a)

3³

(b)

6³

(c)

4³

(d)

5³

13.

The cube of 24 is

(a)

12167

(b)

15625

(c)

13824

(d)

10648

14.

Which of the following is a square of an even number?

(a)

144

(b)

169

(c)

441

(d)

625

15.

If Ax3 =1A,then A is equal to

(a)

(b)

(c)

(d)

16.

If a number is divisible by both 5 and 6 then it may be divisible by

(a)

(b)

(c)

(d)

17.

The expression 119x332 x 6 is a/an

(a)

even number

(b)

odd number

(c)

prime number

(d)

None of these

18.

Which one has all the properties of a kite and a parallelogram?

(a)

Trapezium

(b)

Rhombus

(c)

Rectangle

(d)

Parallelogram

19.

In a parallelogram ABCD,

﻿\(\angle A\)﻿

and

\angle B

are in the ratio 2 : 3. Find the

\angle A

(a)

30°

(b)

72°

(c)

60°

(d)

90°

20.

Vanita folds a rectangular sheet. The folded sheet is shown below.

What is the length of the sheet?

(a)

8 cm

(b)

﻿\(\sqrt{10} \mathrm{~cm}\)﻿

(c)

10 cm

(d)

12 cm

21.

The scientific notation of 51.26000 is

(a)

5.126 × 10⁴

(b)

5.126 × 10¹

(c)

5.126 × 10^-4

(d)

5.126 × 10^-6

22.

The standard form of 0.000072 is

(a)

72 × 10⁴

(b)

72 × 10^-4

(c)

7.2 × 10⁵

(d)

7.2 × 10^-5

23.

The exponential form of 625 is

(a)

5³

(b)

5²

(c)

5⁴

(d)

5⁵

Very Short Answer Type Questions

5 x 1 = 5

24.

Find the product of 56 × 101.

25.

Express (6³) as the sum of odd numbers

26.

Find the value of x in the given parallelogram.

27.

Expand the number 4050 in power of 10.

28.

Find the value of [2^-1 × 3^-1]^-1.

Short Answer Type Questions

7 x 2 = 14

29.

Find the ratio of 80 cm to 1.2 m

30.

Using prime factorisation, find the square root of 11025.

31.

Compute these products using the suggested identity.
(i) 46² using Identity 1A for (a + b)²
(ii) 397 × 403 using Identity 1C for (a + b) (a - b)
(iii) 91² using Identity 1B for (a - b)²
(iv) 43 × 45 using Identity 1C for (a + b) (a - b)

32.

Construct a square whose length of diagonal is 6 cm.

33.

Find the values of A and Bin the following addition.

34.

In the given figure, ABCD and BDCE are parallelograms with common base DC. If BC ⊥ BD, then find

\angle B E C .

35.

Express

\frac{1.5 \times 10^6}{2.5 \times 10^4}

in the standard form.

Assertion and reason

4 x 1 = 4

36.

Assertion (A) 8, 32, 6, 36 are in proportion.
Reason (R) If two ratios are equal then it is called proportion.
(a) Both Assertion and Reason are true and Reason is the correct explanation of Assertion.
(b) Both Assertion and Reason are true but Reason is not the correct explanation of Assertion.
(c) Assertion is true but Reason is false.
(d) Assertion is false but Reason is true.

37.

Assertion (A) If x + y = 17 and x - y = 3 then the value of x² + y² is 149.
Reason (R) (a + b)² - (a - b)² = 4ab
Codes:
(a) Both A and R are correct and R is the correct explanation of A.
(b) Both A and R are correct but R is not the correct explanation of A.
(c) A is true but R is false.
(d) A is false but R is true.

38.

Assertion (A) If 575a is divisible by 11, then the value of a is 3.
Reason (R) If the difference between the sum of the digits at odd places (from the right to left) and the sum of the digits at even places (from the right to left) of the number is either O or divisible by 11, then the number is divisible by 11.
CODES
(a) Both A and R are true and R is the correct explanation of A.
(b) Both A and R are true, but R is not the correct explanation of A.
(c) A is false but R is true.
(d) A is true but R is false.

39.

Assertion (A) : In a trapezium, the non-parallel sides are equal in length always.
Reason (R) : Trapeziums have one pair of opposite sides that are parallel.
CODES
(a) Both A and R are true and R is the correct explanation of A
(b) Both A and R are true but R is not the correct explanation of A
(c) A is true but R is false
(d) A is false but R is true

Case Study Questions

3 x 4 = 12

40.

Mayank has a square garden with its side (2a + b) m. There is a path 'a' m wide inside the garden along its sides.

Base on the above information, answer the following questions.
(i) What is the area of the garden?
(ii) What is the area of the path?

41.

A teacher writes a 5-digit number A69BC on the board and says it is divisible by 3. Based on the above information, answer the following the questions.
(i) If C=2Aand C=8, then find the possible values of B.
(ii) If C=8 and B=3, then find the possible values of A
(iii) If A=4andB=3,then find the possible values of C.

42.

A playground is in the form of a rectangle ABCD. Two players Ram and Vishnu are standing at the points B and E respectively, where EC = BC.
(i) Find the value of x.
(ii) Find the value of 2x + 3y.
(iii) What kind of values do you depict from sports person?

2. A rectangle MORE is shown below.
Answer the following questions by giving appropriate reason.
(i) Is RE=0M?

﻿\(\text { (ii) Is } \angle M Y O=\angle R X E \text { ? }\)﻿

(iii)

﻿\(\text { Is } \angle M O Y=\angle R E X ?\)﻿

(iv)

﻿\(\text { Is } \triangle M Y O \cong \triangle R X E \text { ? }\)﻿

(v) Is MY = RX?

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Answers Key:

Long Answer Type Questions

7 x 5 = 35

Given, Shreya's age = 15 yr
and Bhoomika's age = 12 yr
∴ Ratio of their ages =

\frac{\text { Shreya's age }}{\text { Bhoomika's age }}=\frac{15 \mathrm{yr}}{12 \mathrm{yr}}=\frac{15}{12}

\frac{15+3}{12+3}=\frac{5}{4}=5: 4

[∵ HCF (15, 12) = 3]
Now, mother wants to divide ₹ 36 between her daughters in the ratio of their ages.
∴ Sum of the parts of ratios = 5 + 4 = 9
Amount which has to be divided = ₹ 36
Here, we can say that Shreya gets 5 parts and Bhoomika gets 4 parts out of every 9 parts
∴

\text { Shreya's share }=\frac{5}{9} \times 36=₹ 20

and

\text { ∴ Shreya's share }=\frac{5}{9} \times 36=₹ 20

Hence, Shreya gets ₹ 20 and Bhoomika gets ₹ 16.

We have, AB = a + b and QR = b
∴ AR = AB - BR
= a + b - b [∵ BR = QR]
= a
∴ DV = a [∵ DV = AR]
Also, AD = a + b
⇒ AP + PD = a + b
⇒ PD = a + b - AP = a + b - b [∵ AP = QR = b]
⇒ PD = a ⇒ DT + PT = a
⇒ DT + PT = a (∵ PT = TD, given)
⇒ 2DT = a
⇒

D T=\frac{a}{2}

Now, shaded area = Area of square ABCD - Area of square QSBR - Area of rectangle TDVU
= AB × AB - QR × QR - DV × DT
= (a + b) × (a + b) – b × b - a ×

\frac{a}{2}

= (a + b)² – b² -

\frac{a^2}{2}

= a² + 2ab + b² - b² -

\frac{a^2}{2}

\begin{aligned} & =\left(a^2-\frac{a^2}{2}\right)+2 a b \\ & =\left(\frac{a^2}{2}+2 a b\right) \text { sq units } \end{aligned}

Given, total amount = 2209
According to the question, Money contributed by each student
=Square of the total number of students
Total number of students
Square root of the money contributed by each student

﻿\(=\sqrt{2209}=\sqrt{47 \times 47}=47\)﻿

Let length of the square plot be a then the area of square = a²
According to the question

﻿\(\begin{aligned}\text { Area } & =101 \frac{1}{400} \mathrm{~m}^2 \Rightarrow a^2=101 \frac{1}{400} \Rightarrow a^2=\frac{40401}{400} \\a & =\frac{201}{20}=10 \frac{1}{20}\end{aligned}\)﻿

Hence, length of one side of the plot is

﻿\(10 \frac{1}{20} \mathrm{~m} .\)﻿

(i) Production decreased for bajra, jowar and rice.
(ii) Production of bajra in 2008 = 1.4 × 10³
and decrease in production of bajra in 2009
= 100 = 0.1 × 10³
So, production of bajra in 2009
= 1.4 × 10³ -0.1 × 10³
= (1.4 - 0.1) × 10³ = 1.3 × 10³
Production of jowar in 2008 = 1.7 × 10⁶
and decrease in production of jowar in 2009
= 440000 = 0.44 × 10⁶
So, production of jowar in 2009
= 1.7 × 10⁶ - 0.44 × 10⁶
= (1.7- 0.44) × 10⁶
= 1.26 × 10⁶
Production of rice in 2008 = 3.7 × 10³
and decrease in production of rice in 2009
= 100 = 0.1 × 10³
So, production of rice in 2009
= 3.7 × 10³- 0.1 × 10³
= (3.7-0.1) × 10³ = 3.6 × 10³
Production of wheat in 2008 = 5.1 × 10⁵
and increase in production of wheat in 2009
= 190000 = 1.9 × 10⁵
So, production of wheat in 2009
= 5.1 × 10⁵ + 1.9 × 10⁵
= (5.1+1.9) × 10⁵
= 7.0 × 10⁵
(iii) Decrease in production of rice in 2009
= 0.1 × 10³
The number of years from 2009 to 2015 is 6.
∴ The total decrease = 6 × 0.1 × 10³
= 0.6 × 10³
So, production of rice in 2015
= 3.7 × 10³- 0.6 × 10³
= (3.7 - 0.6) × 10³
= 3.1 × 10³ hec
Hence, 3.1 × 10³ hec will be harvested in 2015.

Let the three consecutive numbers be n, n + 1and n + 2.
Since, first number is a multiple of 2.
So, n=2m₁ for some whole number m₁
Also given, second number is a multiple of 3.
So, n+ 1=2m₁ +1=3m₂, for some whole number m₂
⇒2m₁ - 3m₂ -1
⇒Numbers of the form 2m are one less than a multiple
of 3
⇒m₁ is one more than the multiples of 3.
∴m₁ =3p+1 for some whole number p
Also, given, third number is a multiple of 4.
So, n +2 = 2m +2 = 2(m₁+ 1) is a multiple of 4,
which is possible only when m +1 is even i.e. m₁ is odd.
Now, on combining m₁ =3p + l and m₁is odd, we get 3p is even.
⇒ p is even Let us say p =2k
Then, m₁=3(2k) +1=6k +1
Hence, n=2m₁ = 2(6k + 1) = 12k +2
Therefore, the required consecutive numbers are
12k + 2,12k +3, 12k + 4, where k=0, 1, 2, ......
Such triples exist infinitely many times and appear after
every 12 numbers: (2, 3, 4), (14, 15, 16), (26, 27, 28), ......

The steps of construction are as follows.
I. Draw a line segment AC of length 8 cm.
II. Mark the mid-point O of AC.
III. Draw another line segment BD of length 8 cm, passing through O, such that O is also the mid-point of BD and ensure that angle

\angle C O D \text { (or } \angle A O B \text { or }

\angle B O C \text { or } \angle D O A \text { ) is the specified angle. }

IV. Connect the end points A, B, C and D to form the quadrilateral ABCD.
(i) Angle of 30° Construct the quadrilateral as described above, ensuring the diagonals intersect at 30° angle. This will result in a rectangle

(ii) Angle of 40° Construct the quadrilateral as described above, ensuring the diagonals intersect at 40° angle. This will result in a rectangle

(iii) Angle of 90° Construct the quadrilateral as described above, ensuring the diagonal intersect at 90° angle. This will result in a square

(iv) Angle of 140° Construct the quadrilateral as described above, ensuring the diagonal intersect at 140° angle. This will result into a rectangle.

Multiple Choice Question

16 x 1 = 16

(b)
4 : 3

10.

(a)
631368

11.

(d)
5a - 5b - ab + 25

12.

(d)
5³

13.

14.

(a)
144

15.

(b)
5

16.

(a)
30

17.

(a)
even number

18.

(b)
Rhombus

19.

(b)
72°

20.

21.

(b)
5.126 × 10¹

22.

(d)
7.2 × 10^-5

23.

Very Short Answer Type Questions

5 x 1 = 5

24.

We know that a × 101 = a × 100 + a
∴ 56 × 101 = 56 × 10² + 56 [∵a = 56]
= 5600 + 56 = 5656

25.

Given, number is 6³
So, n =6and(1 – 1)= (6-)5
We will start with (6 x 5) + l=30 + 1=31
So, odd numbers from 31 are 31, 33, 35, 37, 39, 41.
Now, the sum of 6 consecutive odd numbers starting
from 31=31+33+35 +37+39 + 41=216

26.

In a parallelogram

\angle Q=100^{\circ}

Since, opposite angles are equal.
∴

\angle P=\angle R=x

and

\angle Q=\angle S=100^{\circ}

Now, by angles sum property of a quadrilateral,
⟹ x + 100° + x + 100° = 360°
⟹ 2x = 360° - 200°
⟹

﻿\(x=\frac{160^{\circ}}{2}=80^{\circ}\)﻿

27.

We have, 4050 = 4 × 1000 + 0 × 100 + 5 × 10 + 0
= 4 × 10³ + 0 × 10² + 5 × 10¹ + 0 × 10⁰

28.

We have, [2^-1 × 3^-1]^-1

\because 2^{-1}=\frac{1}{2} \text { and } 3^{-1}=\frac{1}{3} \quad\left[\because a^{-n}=\frac{1}{a^n}\right]

﻿\(\therefore\left[2^{-1} \times 3^{-1}\right]^{-1}=\left(\frac{1}{2} \times \frac{1}{3}\right)^{-1}=\left(\frac{1}{6}\right)^{-1}=6\)﻿

Short Answer Type Questions

7 x 2 = 14

29.

Since, both the quantities are not in the same unit.
Firstly, we have to convert them into same unit.
We know that 1 m =100 cm
⇒ 1.2 m = 1.2 X 100 cm = 120 cm
∴ Ratio of 80 cm to 1.2 m, i.e. 80 cm to 120cm

﻿\(=\frac{80}{120}=\frac{2}{3}, \text { hence the required ratio is } 2: 3 .\)﻿

30.

Factors of 15 are 1, 3, 5, 15.
Here, 1x 15 = 15 and 3 X5 = 15
(1, 15) and (3, 5) are pairs of partner

31.

(i) We have, 46 = (40 + 6)²
= (40)² + 2 × 40 × 6 + (6)² [∵ (a + b)² = a² + 2ab + b²)
= 1600 + 480 + 36 = 2116
(ii) We have, 397 × 403 = (400 - 3) (400 + 3)
= (400)² - 3²[∵ (a - b) (a + b) = a² - b²]
= 160000 – 9 = 159991
(iii) We have, 91² = (100 - 9)²
= (100)² - 2 × 100 x 9 + (9)² [∵ (a - b)² = a² - 2ab + b²]
= 10000 - 1800 + 81
= 10081 - 1800 = 8281
(iv) 43 × 45 = (44 - 1) (44 + 1)
= (44)² - (1)² [∵ (a - b) (a + b) = a² - b²]
= (40 + 4)² - 1
= (40)² + 2 × 40 × 4 + (4)² - 1
= 1600 + 320 + 16 - 1 = 1936 - 1 = 1935

32.

The steps of construction are as follows.
(i) Draw a line segment AC of length 6 cm
(ii) Mark the mid-point O of A C.
(iii) Draw another line segment BD of length 6 cm, passing through 0, such that O is also the mid point of BD.
Ensure the angles

\angle A O B=\angle B O C=\angle C O D .

=\angle D O A=90^{\circ}

(iv) Connect the end-points A, B, C and D to form the quadrilateral ABCD as square.

33.

We have,

Here, we have two letters A and B whose values are to be found. Studying the addition in one's column, we have 1 + B and we get 0 from this i.e. a number whose one's digit is 0. So, B must be 9.
Then, for addition often' s column, we have 1+A+1=9Then,A must be 7 and then the puzzle is solved as shown below.

Hence, A =7andB=9.

34.

We have, parallelograms ABCD and BDCE and BC ⊥ BD.
∴

\angle A B C=180^{\circ}-30^{\circ}=150^{\circ}

\text { [since, } \angle D A B \text { and } \angle A B C \text { are adjacent angles of }

parallelogram ABCD]
∴

\angle C B E=180^{\circ}-\angle A B C=180^{\circ}-150^{\circ}=30^{\circ}

Also, it is given that

\angle D B C=90^{\circ}

∴

\angle D B E=90^{\circ}+30^{\circ}=120^{\circ}

But

\angle D B E=\angle D C E=120^{\circ}

∴

\angle B E C=180^{\circ}-120^{\circ}=60^{\circ}

[since,

\angle B E C \text { and } \angle D C E

are adjacent angles of parallelogram DCEB]

35.

We have,

\frac{1.5 \times 10^6}{2.5 \times 10^4}=\frac{1.5 \times 10^2 \times 10^4}{2.5 \times 10^4}=\frac{150}{2.5} \times 10^{4-4}

=\frac{150}{2.5} \times 10^0=60 \times 1=60

= 6.0 × 10¹ = 6.0 × 10

Assertion and reason

4 x 1 = 4

36.

(d) Assertion is false but Reason is true.

37.

(b) Both A and R are correct but R is not the correct explanation of A.

38.

(a) Both A and R are true and R is the correct explanation of A.

39.

(d) A is false but R is true.

Case Study Questions

3 x 4 = 12

40.

(i) Given, side of square garden ABCD = 2a + b

∴ Area of square garden ABCD = (2a + b)²
= 4a² + 2 × 2a × b + b²[∵(a + b)² = a² + 2ab + b²]
= (4a² + 4ab + b²) m²
(ii) Side of square garden ABCD = 2a + b
∴ Side of square PQRS = 2a + b - (a + a)
= 2a + b - 2a = b
∴ Area of square PQRS = b² m²
∴ Area of path = Area of square ABCD - Area of square PQRS
= 4a² + 4ab + b² - b²
= (4a² + 4ab) m²

41.

i Given, 5-digit number is A69BC

∴ A69BC= 469B8
Since, 469B8 is divisible by 3.
∴ Sum of its digits is divisible by 3.
⇒4+6+9 +B+8 is divisible by 3.
⇒27+B is divisible by 3.
∴ So, B can be equal to 0, 3, 6,9.
(ii)∵ C=8 and B=3
∴ A69BC=A6938
Since, A6938 is divisible by3
∴ Sum of its digits is divisible by 3.
⇒A+6+9+3+8 is divisible by 3.
⇒A +26 is divisible by 3.
So, A can be equal to 1, 4, 7.
iii. ∵ A=4 and B=3
∴ A69BC= 4693C
Since, 4693C is divisible by 3
∴ Sum of its digits is divisible by 3.
⇒4 +6 +9+3+Cis divisible by 3.
⇒22 +C is divisible by 3
So, C can be equal to 2, 5, 8.

42.

1. In

﻿\(\triangle B C E, \angle B C E=90^{\circ}\)﻿

and CE = BC
∴

﻿\(\angle C B E=\angle B E C=z \text { (say) }\)﻿

Using angle sum property of the triangle, z + z + 90° = 180°
⟹ 2z = 180°- 90°
⟹

﻿\(z=\frac{90^{\circ}}{2}=45^{\circ}\)﻿

(i) x = 180° – 45° = 135°

﻿\(\left[\because \angle D E C=180^{\circ} \text { and } \angle C E B=45^{\circ}\right]\)﻿

and y = 90°– 45° = 45°

﻿\(\left[\because \angle C B A=90^{\circ}\right]\)﻿

(ii) Here, 2x + 3y = 2 x 135 + 3 x 45 = 135 (2 +1) = 405°
(iii) A sports person has energy, stamina and a healthy body as well as he has a team spirit to motivate others in a team.
2. (i) Yes, RE = OM
This is because the opposite sides are equal in any rectangle.
(ii) Yes,

﻿\(\angle M Y O=\angle R X E\)﻿

Here, MY and RX are perpendicular to OE.
Since,

﻿\(\angle R X O=90^{\circ} \Rightarrow \angle R X E=90^{\circ}\)﻿

and

﻿\(\angle M Y E=90^{\circ} \Rightarrow \angle M Y O=90^{\circ}\)﻿

(iii) Yes,

﻿\(\angle M O Y=\angle R E X\)﻿

Since, these are alternate interior angles, as RE || OM and EO is a transversal
(iv)

﻿\(\text { Yes, } \triangle M Y O \cong \triangle R X E \text {. }\)﻿

﻿\(\text { In } \triangle M Y O \text { and } \triangle R X E \text {, }\)﻿

we see that
MO = RE [proved]

﻿\(\angle M O Y=\angle R E X\)﻿

[proved]

﻿\(\angle M Y O=\angle R X E\)﻿

[proved]

﻿\(\therefore \triangle M Y O \cong \triangle R X E\)﻿

[by AAS]
(v) Yes, MY = RX
Since, these are corresponding part of congruent triangles

CBSE 8th Mathematics A square and A cube - New Model Questions Papers Study Material - QB365 Set B

QB365 Student Learing Managemnet System

Std: 8th Standard CBSE - Mathematics

chapter: A square and A cube - New - Model Questions Papers

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