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Haryana State Board HBSE 8th Class Maths Solutions Chapter 16 Playing with Numbers Ex 16.2 Textbook Exercise Questions and Answers.

Haryana Board 8th Class Maths Solutions Chapter 16 Playing with Numbers Exercise 16.2

Question 1.
If 21y5 is a multiple of 9, where y is: a digit, what is the value of y ?
Solution:
Since 21y5 is a multiple of 9.
i.e. 21y5 is divisible by 9.
Therefore, 2 + 1 + y + 5 = (8 + y) must be a multiple of 9. i.e. (8 + y) should be divisible by 9.
This is possible when 8 + y = 9 or 18
⇒ 8 + y = 9 or 8 + y = 18
y = 1 or y = 10
But, since x is a digit, therefore y = 10 is not possible.
Thus y = 1.

Question 2.
If 31z5 is a multiple of 9, where z is a digit, what is the value of z ?
You will find that there are two answers for the last problem. Why is this so ?
Solution:
Since 31z5 is divisible by 9.
∴ 3 + 1 + z + 5 = a multiples of 9.
9 + z = 9, 18 or 27
⇒ 9 + z = 9 9 + z = 18
⇒ z = 0 z = 9
9 + z = 27
z = 21
which is not a one digit number.
∴ The values of z are (0 or 9).
There are two values of z as the sum of digits is a multiple of 9. So, it may be 9, 18, 27 … any one of them.

Question 3.
If 24x is a multiple of 3, where x is a digit, what is the value of x ?
[Hint: Since 24x is a multiple of 3, its sum of digits 6 + x is a multiple of 3; so 6 + x is one of these numbers : 0, 3, 6, 9, 12, 15, 18, … . But since a: is a digit, it can only be that 6 + x = 6 or 9 or 12 or 15. Therefore, x = 0 or 3 or 6 or 9. Thus, x can have any of four different values.]
Solution:
We have 24x is a multiple of 3.
∴ 2 + 4 + x = 6 + x
is a multiple of 3.
∴ 6 + x = 0, 3, 6, 9, 12, 15, …
But, x is a digit.
∴ It can only be that
6 + x = 6 or 9, 12 or 15
∴ x = 0, 3, 6 or 9.

Question 4.
If 31z5 is a multiple of 3, where z is a digit, what might be the values of z ?
Solution:
Since 31z5 is a multiple of 3.
∴ The sum of the digits (3 + 1 + z + 5) is a multiple of 3.
So, 9 + z is one of these numbers :
0, 3, 6, 9, 12, …
9 + z = 9 ⇒ z = 1
[Values less than 9 gives negative result]
9 + z = 12 ⇒ z = 3
9 + 2= 15 ⇒ z = 6
9 + z = 18 ⇒ z = 9
9 + z = 21 ⇒ z = 12
which is not a one digit number.
∴ The values of z are 1, 3, 6 or 9.

Haryana State Board HBSE 8th Class Maths Solutions Chapter 16 Playing with Numbers Ex 16.1 Textbook Exercise Questions and Answers.

Haryana Board 8th Class Maths Solutions Chapter 16 Playing with Numbers Exercise 16.1

Question 1.
Find the values of the letters in each of the following and give reasons for the steps involved.

Solution:
1. We have

Study the addition in the ones column: from A + 5, we get ‘2’ that is, a number whose ones digit is 2.
Thus, A should be 7. So the puzzle can be solved as shown below :

That is, B = 6.

2. We have

Here, A + 8 = 3 (a number whose ones digit is 3)
∴ ‘A’ should be 5.
Now,

Thus, A = 5, B = 4 and C = 1.

3. We have

Since the ones digit of, A × A = A, so, it must be 6, because since produce of a number and A is 9.
i.e. 1 × A = 9 ⇒ A = 9
but A = 9 does not satisfies.
So, A = 6
Now,

4. We have

Addition in ones column :
Here, B + 7 is A’ that is a number whose ones digit is ‘A’.
Let us put the values of ‘B’ starting from 0.
If B = 0, then A = 7 ∴ 7 + 6 ≠ 6
If B = 1, then A = 8 ∴ 8 + 3 ≠ 6
If B = 2, then A = 9 ∴ 9 + 3 ≠ 6
If B = 5, then A = 2 ∴ (2 + 3) + 1 = 6

So, B = 5 and A = 2 satisfies the given condition.

5. We have

Since the ones digit of B × 3 is B, it must be B = 0 or B = 5.
Now look at A.

So, B = 0 and A = 5 and C = 1.

6. We have

Since the ones digit of B × 5 is B
∴ B must be 0 or 5
Now, look at A.
Use hits and trial method :

∴ A = 7, B = 5 and C = 3.

7. We have

Since the ones digit of B × 6 is 6.
∴ Possible values of B are 0, 4, 6 and 8.
Now, look at A.
So, the value of BBB may be 444,666 or 888. [000 is not possible]
∴ 666 ÷ 6 = 111
which is a 3-digits number.
888 ÷ 6 = 148
which is a 3-digits number.
444 ÷ 6 = 74
which is a 2-digits number.

Thus, B = 4 and A = 7.

8. We have

If you study the addition in the ones column i.e. (1 + B), you get ‘0’.
∴ B should be 9.
Thus, A must be 7.

Thus, B = 9 and A = 7.

9. We have

If you study the addition in the ones column i.e. (B + 1) then you get ‘8’.
∴ ‘B’ should be 7 and A should be 4.

So, the given condition is satisfied.
Hence, B = 7 and A = 4.

10. We have

Since addition of ones column gives the value 9.
∴ A + B should be (1, 8) or (8, 1).
Now, let us apply hits and trial method.
If A = 1 and B = 8, then

If A = 8 and B = 1, then

Thus, A = 8 and B = 1 satisfies the given condition.

Haryana State Board HBSE 8th Class Maths Solutions Chapter 15 Introduction to Graphs InText Questions and Answers.

Haryana Board 8th Class Maths Solutions Chapter 15 Introduction to Graphs InText Questions

Try These (Page 244)

Question 1.
The following table gives the quantity of petrol and its cost:

Plot a graph to show the data.
Also, use the graph to find how much petrol can be purchased for Rs. 800.
Solution:
(i) Let us take a suitable scale on both the axes (Fig. 15.14).

(ii) Mark number of litres along the horizontal axis.
(iii) Mark cost of petrol along the vertical axis.
(iv) Plot the points : (10, 500), (15, 750), (20, 1000), (25, 1250).
(v) Join the points.
If we observe that graph and drop perpendiculars from 800. The point is Q. We draw perpendicular from this point on X-axis. The required value is 16 litre.

Try These (Page 245)

Question 1.
A bank gives 10% Simple Interest (S.I.) on deposits by senior citizens. Draw a graph to illustrate the relation between the sum deposited and simple interest earned. Find from your graph
(a) the annual interest obtainable for an investment of Rs. 250.
(b) the investment one has to make to get an annual simple interest of Rs. 70.
Is this, a case of direct variation ? Justify your answer.
Solution:

Steps to follow :
1. Find the quantities to be plotted as Deposit and S.I.
2. Decide the quantities to be taken on x-axis and on y-axis.
3. Choose a scale.
4. Plot points.
5. Join the points.
We get a table of values

(i) Scale : 1 unit = Rs. 100 on horizontal axis; 1 unit = Rs. 10 on vertical axis.
(ii) Mark Deposits along horizontal axis.
(iii) Mark Simple Interest along vertical axis.
(iv) Plot the points : (100, 10), (200, 20), (300, 30), (500, 50) etc.
(v) Join the points. We get a graph that is a line (Fig 15.15).

(a) Corresponding to Rs. 250 on horizontal axis, we get the interest to be Rs. 25 on the vertical axis.
(b) Corresponding to Rs. 70 on the vertical axis, we get the sum to be Rs. 700 on the horizontal axis.
Yes, this is a case of direct variation. More you deposit more you earn interest. Also, the graph is a straight line which indicates that the two quantities Simple Interest and amount deposited are in direct variation.

Haryana State Board HBSE 8th Class Maths Solutions Chapter 15 Introduction to Graphs Ex 15.3 Textbook Exercise Questions and Answers.

Haryana Board 8th Class Maths Solutions Chapter 15 Introduction to Graphs Exercise 15.3

Question 1.
Draw the graphs for the following tables of values, with suitable scales on the axes.
(a) Cost of apples

(b) Distance travelled by a car

(i) How much distance did the car cover during the period 7.30 a.m. to 8 a.m. ?
(ii) What was the time when the car had covered a distance of 100 km since it’s start ?
(c) Interest on deposits for a year

(i) Does the graph pass through the origin ?
(ii) Use the graph to find the interest on Rs. 2500 for a year.
(iii) To get an interest of Rs. 280 per year, how much money should be deposited ?
Solution:
(a)

Scale : Horizontal: 1 unit = 1 apple
Vertical: 1 unit = Rs. 5
Here, 10 small division = 1 unit
Steps to follow :
(i) Mark no. of apples along horizontal axis.
(ii) Mark cost (in Rs.) along vertical axis.
(iii) Plot the points as shown in graph.
(iv) Join the points (1, 5), (2, 10), (3, 15), (4, 20) and (5, 25).

(b)


Scale : 2 units = 1 hour (Horizontal line)
1 unit = 40 km (Vertical line)
(i) Corresponding to 7.30 a.m. on horizontal axis, we get the distance 100 km on vertical axis.
So, distance covered by car during 7.30 a.m. and 8 a.m. = 120 – 100 = 20 km.
(ii) 7.30 a.m.

(c)

Scale :
Horizontal : 1 unit = Rs. 80
Vertical: 1 unit = Rs. 1000
Hence, 10 small division = 1 unit.
Steps Of follow :
(i) Mark simple interest (in Rs.) along horizontal axis.
(ii) Mark deposite (in Rs.) along vertical axis
(iii) Plot the points as showin in graph.
(iv) Joint the points (80, 1000), (160, 2000), (240, 3000), (320, 4000), (400, 5000)

Question 2.
Draw a graph for the following :
(i)

Is it a linear graph ?
(ii)

Is it a linear graph ?
Solution:

Scale : 1 unit on horizontal line = 1 cm
1 unit on vertical line = 4 cm
Since graph is not a straight line.
∴ It is not a linear graph.
(ii)

Scale : 1 unit on X-axis = 1 cm
1 unit on Y-axis = 4 cm²

Haryana State Board HBSE 8th Class Maths Solutions Chapter 15 Introduction to Graphs Ex 15.2 Textbook Exercise Questions and Answers.

Haryana Board 8th Class Maths Solutions Chapter 15 Introduction to Graphs Exercise 15.2

Question 1.
Plot the following points on a graph sheet. Verify if they lie on a line
(a) A(4, 0), B(4, 2), C(4, 6), D(4, 2.5)
(b) P(1, 1), Q(2, 2), R(3, 3), S(4, 4)
(c) K(2, 3), L(5, 3), M(5, 5), N(2, 5).
Solution:
(i) (a)

These points A, B, C and D lie on a line. The line is parallel to Y-axis.
(b)

These points P, Q, R and S lie on a line. The line is a straight line passing through the origin.
(c)

These do not lie on a line. They form a rectangle.

Question 2.
Draw the line passing through (2, 3) and (3, 2). Find the coordinates of the points at which this line meets the x-axis and y-axis.
Solution:

∴ Coordinates of the points at which this line meets the x-axis and y-axis are (5, 0) and (0, 5).

Question 3.
Write the coordinates of the vertices of each of these adjoining figures.

Solution:
Co-ordinates of the vertices of rectangle OABC are :
A(2, 0), B(2, 3), C(0, 3) and O(0, 0)
Coordinates of the vertices of parallelogram PQRS are :
P(4, 3), Q(6, 1), R(6, 5) and S(4, 7)
Coordinates of the vertices of △LMK are :
L(7, 7), M(10, 8) and K(10, 5).

Question 4.
State whether True or False. Correct that are false.
(i) A point whose x-coordinate is zero and y-coordinate is non-zero will lie on the y-axis.
(ii) A point whose y-coordinate is zero and x-coordinate is 5 will lie on y-axis.
(iii) The coordinates of the origin are (0, 0).
Solution:
(i) True.
(ii) False. A point whose y-coordinate is zero and x-coordinate is 5 will lie on x-axis.
(iii) True.

Haryana State Board HBSE 8th Class Maths Solutions Chapter 15 Introduction to Graphs Ex 15.1 Textbook Exercise Questions and Answers.

Haryana Board 8th Class Maths Solutions Chapter 15 Introduction to Graphs Exercise 15.1

Question 1.
The following graph shows the temperature of a patient in a hospital, recorded every hour.

(a) What was the patient’s temperature at 1 p.m. ?
(b) When was the patient’s temperature 38.5°C ?
(c) The patient’s temperature was the same two times during the period given. What were these two times ?
(d) What was the temperature at 1.30 p.m. ? How did you arrive at your answer ?
(e) During which periods did the patients temperature showed an upward trend ?
Solution:
(a) The Patient’s temperature at 1 p.m. was 36.5°C.
(b) 12 noon.
(c) 1 p.m. and 2 p.m.
(d) The temperature at 1.30 p.m. was 36.5°. We observe the graph and finds that one fine indicates 30 minutes. So, taking a fine after 1 p.m. on X-axis move upward parallel to Y-axis where the line intersect the given line graph. It is obviously 36.5°C.
(e) 9 a.m. to 11 a.m.

Question 2.
The following line graph shows the yearly sales figures for a manufacturing company.
(a) What were the sales in (i) 2002 (ii) 2006 ?
(b) What were the sales in (i) 2003 (ii) 2005 ?
(c) Compute the difference between the sales in 2002 and 2006.
(d) In which year was there the greatest difference between the sales as compared to its previous year ?

Solution:
(a) (i) Sales in 2002 = 4 crores
(ii) Sales, in 2006 = 8 crores
(b) (i) Sales in 2003 = 7 crores
(ii) Sales in 2005 = 10 crores
(c) The difference between the sales in 2002 and 2006 = 8 – 4 = 4 crores.
(d) The greatest difference between the sales as compared to its previous year in 2004 – 2005.

Question 3.
For an experiment in Botany, two different plants, plant A and plant B were grown under similar laboratory conditions. Their heights were measured at the end of each week for 3 weeks. The results are shown by the following graph.

(a) How high was Plant A after (i) 2 weeks (ii) 3 weeks ?
(b) How high was Plant B after (i) 2 weeks (ii) 3 weeks ?
(c) How much did Plant A grow during the 3rd week ?
(d) How much did Plant B grow from the end of the 2nd week to the end of the 3rd week ?
(e) During which week did Plant A grow most ?
(f) During which week did Plant B grow least ?
(g) Were the two plants of the same height during any week shown here ? Specify.
Solution:
(a) (i) Height of Plant A after 2 weeks = 7 cm
(ii) Height of Plant A after 3 weeks = 9 cm
(b) (i) Height of Plant B after 2 weeks = 7 cm
(ii) Height of Plant B after 3 weeks = 10 cm
(c) The Plant A grows during the 3rd week = 9 – 7 = 2 cm.
(d) The Plant B grows from the end of the 2nd week to the end of the 3rd week
= 10 cm – 7 cm = 3 cm.
(e) 2nd week.
(f) First week.
(g) Yes, the two plants have same height in the end of 2nd week.

Question 4.
The following graph shows the temperature forecast and the actual temperature for each day of a week.
(a) On which days was the forecast temperature the same as the actual temperature ?
(b) What was the maximum forecast temperature during the week ?
(c) What was the minimum actual temperature during the week ?
(d) On which day did the actual temperature differ the most from the forecast temperature ?

Solution:
(a) Tuesday, Friday and Sunday.
(b) Sunday (35°C).
(c) 15°C.
(d) Thursday = 22.5°C – 15°C = 7.5°C.

Question 5.
Use the tables below to draw linear graphs :
(a) The number of days a hill side city received snow in different years.

(b) Population (in thousands) of men and women in a village in different years.

Solution:
(a) Show fall received in different days during the years in a hill side city.


(b) Population in village (in thousands)

Question 6.
A courier-person cycles from a town to a neighbouring suburban area to deliver a parcel to a merchant. His distance from the town at different times is shown by the following graph :
(a) What is the scale taken for the time axis ?
(b) How much time did the person take for the travel ?
(c) How far is the place of the merchant from the town ?
(d) Did the person stop on his way ? Explain.
(e) During which period did he ride fastest ?

Solution:
(a) 4 small division = 1 hour
Let the 4 small division = 1 cm
∴ 1 cm = 1 hour
(b) 3 hour 30 minutes or 3.30 hour
(c) 22 km
(d) Yes, he stops between 10 a.m. to 10.30 a.m. because the distance is not changing with the change in time. In other words, we can say that the person is at rest. Hence, if a body remains at rest the graph of its motion is a straight line parallel to time-axis i.e. x-axis.
(e) Between 8 a.m. to 9 a.m.

Question 7.
Can there be a time-temperature graph as follows ? Justify your answer.


Solution:
(i) Yes, it may be a time-temperature graph because with change in time temperature may increase for example, during day time in summer.
(ii) Yes, it may be possible that with change in time temperature may fall uniformaly. For example, during night in winter.
(iii) No, it can not be a time-temperature graph. Here, temperature is changing without any change of time, which is impossible.
(iv) No, temperature cannot be constant with change in time. However, it may be possible for a short span of time.

Haryana State Board HBSE 8th Class Maths Solutions Chapter 14 Factorization InText Questions and Answers.

Haryana Board 8th Class Maths Solutions Chapter 14 Factorization InText Questions

Try These (Page 219)

Question 1.
Factorise :
(i) 12x + 36
(ii) 22y – 33z
(iii) 14 pq + 35 pqr
Solution:
(i) 12x + 36 = 12(x + 3)
(ii) 22y – 33z = 11 (2y + 3z)
(iii) 14 pq + 35pqr = 7pq (2 + 5r)

Try These (Page 225)

Question 1.
Divide:
(i) 24xy²z³ by 6yz²
(ii) 63a²b⁴c⁶ by 7a²b²c²
Solution:
(i) 24xy²z³ + 6yz²
= \(\frac{4 \times 6 \times x \times y \times y \times z \times z^{2}}{6 \times y \times z^{2}}\)
= 4 × x × y × z
= 4xyz

(ii) 63a²b⁴c⁶ ÷ 7a²b²c³
= \(\frac{7 \times 9 \times a^{2} \times b^{2} \times b^{2} \times c^{3} \times c^{2}}{7 \times a^{2} \times b^{2} \times c^{3}}\)
= 9 x b² x c³ = 9b²c³

Haryana State Board HBSE 8th Class Maths Solutions Chapter 14 Factorization Ex 14.4 Textbook Exercise Questions and Answers.

Haryana Board 8th Class Maths Solutions Chapter 14 Factorization Exercise 14.4

Find the correct the errors in the following matematical statements :
1. 4(x – 5) = 4x – 5
2. x(3x + 2) = 3x² + 2
3. 2x + 3y = 5xy
4. x + 2x + 3x = 5x
5. 5y + 2y + y – 7y = 0
6. 3x + 2x = 5x²
7. (2x)² + 4(2x) + 7 = 2x² + 8x + 7
8. (2x)² + 5x = 4x + 5x = 9x
9. (3x + 2)² = 3x² + 6x + 4
10. Substituting x = -3 in
(a ) x² + 5x + 4 gives (-3)² + 5(-3) + 4 =
9 + 2 + 4 = 15
(b) x² – 5x + 4 gives (-3)² – 5(-3) + 4 =
9 – 15 + 4 = -2
(c) x² + 5x gives (-3)² + 5(-3) = – 9 – 15 = -24
11. (y – 3)² = y² – 9
12. (z + 5)² = z² + 25
13. (2a + 3b) (a – b) = 2a² – 3b²
14. (a + 4) (a + 2) = a² + 8
15. (a – 4) (a – 2) = a² – 8
16. \(\frac{3 x^{2}}{3 x^{2}}\) = 0
17. \(\frac{3 x^{2}+1}{3 x^{2}}\) = 1 + 1 = 2
18. \(\frac{3 x}{3 x+2}\) = \(\frac{1}{2}\)
19. \(\frac{3}{4 x+3}\) = \(\frac{1}{4x}\)
20. \(\frac{4 x+5}{4 x}\) = 5
21. \(\frac{7 x+5}{5}\) = 7x.
Solution:
4(x – 5) = 4x – 5
L.H.S. = 4(x – 5)
= 4x – 20
Hence, 4(x – 5) = 4x – 20

2. x(3x + 2) = 3x² + 2
L.H.S. = x(3x + 2)
= 3x² + 2x .
Hence, x(3x + 2) = 3x² + 2x

3. 2x + 3y = 5x
L.H.S. = 2x + 3y
Hence, 2x + 3y = 2x + 3y or 3y + 2x

4. x + 2x + 3x = 5x
L.H.S. = x + 2x + 3x
= 6x
Hence, x + 2x + 3x = 6x

5. 5y + 2y + y – 7y = 0
L.H.S. = 5y + 2y + y – 7y
= 7y – 7y + y
= y
Hence, 5y + 2y + y – 7y = y

6. 3x + 2x = 5x²
L.H.S. = 3x + 2x = 5x
Hence, 3x + 2x = 5x

7. (2x)² + 4 (2x) + 7 = 2x² + 8x + 7
L.H.S: = (2x)² + 4 (2x) + 7
= 4x² + 8x + 7
Hence,
(2x)² + 4(2x) + 7 = 4x² + 8x + 7

8. (2x)² + (5x) = 4x + 5x = 9x
L.H.S. = (2x)² + (5x)
= 4x² + 5x
Hence, (2x)² + (5x) = 4x² + 5x

9. (3x + 2)² = 3x² + 6x + 4
L.H.S. = (3x + 2)²
= (3x)² + 2 × 3x × 2 + (2)²
[∵ (a + b)² = a² + 2ab + b²]
= 9x² + 12x + 4
Hence, (3x + 2)² = 9x² + 12x + 4

10 (a). If x = -3
x² + 5x + 4 gives (-3)² + 5 (-3) + 4
= 9 + 2 + 4 = 15
L.H.S. = x² + 5x + 4
= (-3)² + 5 × (-3) + 4
= 9 + (-15) + 4
= 13 – 15 = -2
Hence, x² + 5x + 4 = -2

(b) If x = -3
x² – 5x + 4 = (-3)² + 5 (-3) + 4
= 9 + 15 + 4 = 28

(c) If x = -3
x² + 5x = =(-3)² + 5(-3)
= 9 + (-15)
= 9 – 15 = -6

11. (y – 3)² = y² – 9
= y² – 2 × y × 3 + (3)²
[∵ (a – b)² = a² – 2ab + b²]
= y² – 6y + 9
Hence, (y – 3)² = y² – 6y + 9

12. (z + 5)² = z² + 25
L.H.S. = (z + 5)²
= (z)² + 2 × z × 5 + (5)²
= z² + 10z + 25
Hence, (z + 5)² = z² + 10z + 25

13. (2a + 3b) (a – b) = 2a² – 3b²
L.H.S. = (2a + 3b) (a – b)
= 2a² – 2ab + 3ab – 3b²
= 2a² – ab – 3b²
Hence, (2a + 3b) (a – b) = 2a² + ab – 3b².

14. (a + 4)(a + 2) = a² + 8
(a + 4)(a + 2) = a² + 8
L.H.S. = (a + 4)(a + 2)
= a² + 2a + 4a + 8
= a² + 6a + 8
Hence, (a + 4) (a + 2) = a² + 6a + 8

15. (a – 4)(a – 2) = a² – 8
L.H.S. = (a – 4)(a – 2)
= a² – 2a – 4a + 8
= a² – 6a + 8
Hence, (a – 4) (a – 2) = a² – 6a + 8.

16.

17.

18.

19.

20.

21.

Haryana State Board HBSE 8th Class Maths Solutions Chapter 14 Factorization Ex 14.3 Textbook Exercise Questions and Answers.

Haryana Board 8th Class Maths Solutions Chapter 14 Factorization Exercise 14.3

Question 1.
Carry out the following-divisions :
(i) 28x⁴ ÷ 56x
(ii) -36y³ ÷ 9y²
(iii) 66pq²r³ ÷ 11qr²
(iv) 34x³y³z² ÷ 51xy²z³
(v) 12a⁸b⁸ ÷ (-6a⁶b⁴)
Solution:

Question 2.
Divide the given polynomial by the given monomial.
(i) (5x² – 6x) ÷ 3x
(ii) (3y⁸ – 4y⁶ + 5y⁴), y⁴
(iii) 8 (x⁴y²z² + x²y³z² + x²y²z²)
(iv) 9x²y²(3z – 24) ÷ 27xy(z – 8)
(v) 96abc(3a – 12) (5b – 30) ÷ 144 (a – 4) (b – 6)
Solution:

Question 3.
Work out the following divisions :
(i) (10x – 25) ÷ 5
(ii) (10x – 25) ÷ (2x – 5)
(iii) 10y(6y + 21) ÷ 5(2y + 7)
(iv) 9x²y²(3z – 24) ÷ 27xy (z – 8)
(v) 96abc (3a -12) (5b – 30) ÷ 144 (a – 4) (b – 6)
Solution:

Question 4.
Divide as directed :
(i) 5(2x + 1) (3x + 5) ÷ (2x + 1)
(ii) 26xy (x + 5) (y – 4) ÷ 13x (y – 4)
(iii) 52pqr (p + q) (q + r) (q + p) ÷ 104pq (q + r) (r + p)
(iv) 20 (y + 4) (y² + 5y + 3) ÷ 5(y + 4)
(v) x(x + 1) (x + 2) (x + 3) ÷ x(x + 1)
Solution:
(i) 5(2x + 1) (3x + 5) ÷ (2x + 1)

(ii) 26xy (x + 5) (y – 4) ÷ 13x (y – 4)

(iii) 52pqr (p + q) (q + r) (r + p) ÷ 104pq (q + r) (r + p)

= 4(y² + 5y + 3)
(iv) 20 (y + 4) (y² + 5y + 3) ÷ 5 (y + 4)

(v) x(x + 1) (x + 2) (x + 3) ÷ x(x + 1)

= (x + 2) (x + 3)

Question 5.
Factorise the expression and divide them as directed :
(i) (y² + 7y + 10) ÷ (y + 5)
(ii) (m² – 14m – 32) ÷ (m + 2)
(iii) (5p² – 25p + 20) ÷ (p – 1)
(iv) 4yz (z² + 6z – 16) ÷ 2y (z + 8)
(v) 5pq(p² – q²) ÷ 2p(p + q)
(vi) 12xy(9x² – 16y²) ÷ 4xy(3x + 4y)
(vii) 39y³(50y² – 98) ÷ 26y²(5y + 7)
Solution:
(i) (y² + 7y + 10) , (y + 5)
y² + 7x + 10 = y² + 5 + 2 + 10
= (y + 5) + 2(y + 5)
= (y + 5) (y + 2)

(ii) (m² – 14m – 32) ÷ (m + 2)
m² – 14m – 32 = m² – 16m + 2m – 32
= m (m – 16) + 2 (m – 16)
= (m – 16) (m + 2)

(iii) (5p² – 25p + 20) ÷ (p – 1)
5p² – 25p + 20 = 5p² – 20p – 5p + 20
= 5p (p – 4) – 5(p – 4)
= (p – 4) (5p – 5)
= (p – 4) × 5(p – 1)
= 5 (p – 4) (p – 1)

= 5 (p – 4)

(iv) 4yz(z² + 6z – 16) ÷ 2y (z + 8)
z² + 6z – 16 = z² + 8z – 2z – 16
= z(z + 8) – 2 (2 + 8)
= (z + 8) (z – 2)

= 2z(z – 2)

(v) 5pq(p² – q²) ÷ 2p(p + q)

(vi) 12xy (9x² – 16y²) + 4xy (3x + 4y)
9x² – 16y² = (3x)² – (4y)²
= (3x + 4y) (3x – 4y)

= 3(3x – 4y)

(vii) 39y³(50y² – 98) + 26y²(5y + 7)
50y² – 98 = 2(25y² – 49)
= 2(5y)² – (7)²]
= 2(5y + 7) (5y – 7)

= 3y(5y – 7)

Haryana State Board HBSE 8th Class Maths Solutions Chapter 14 Factorization Ex 14.2 Textbook Exercise Questions and Answers.

Haryana Board 8th Class Maths Solutions Chapter 14 Factorization Exercise 14.2

Question 1.
Factorise the following expressions :
(i) a² + 8a + 16
(ii) p² – 10p + 25
(iii) 25m² + 30m + 9
(iv) 49 y² + 84yz + 36z²
(v) 4x² – 8x + 4
(vi) 121b² – 88bc + 16c²
(vii) (l + m)² – 4lm
[HintExpand (l + m)² First]
(viii) a⁴ + 2a²b² + b⁴
Solution:
(i) a² + 8a + 16
= a² + 4a + 4a + 16
= a (a + 4) + 4 (a + 4)
= (a + 4) (a + 4)

(ii) p² – 10p + 25
= p² – 5p – 5p + 25
= p(p – b) – 5(p – 5)
= (p – 5)(p – b)

(iii) 25m² + 30m + 9
= 25m² + 15m + 15m + 9
= 5m (5m + 3) + 3 (5m + 3)
= (5m + 3) (5m + 3)

(iv) 49y² + 84yz + 36z²
= 49y² + 42yz + 42yz + 36z²
= 7y (7y + 6z) + 6z (7y + 6z)
= (7y + 6z) (7y + 6z)

(v) 4x² – 8x + 4
= 4x² – 4x – 4x + 4
= 4x (x – 1) – 4(x – 1)
= (x – 1) (4x – 4)

(vi) 121b² – 88bc + 16c²
= 121b² – 44bc – 44bc +16c²
= 11b (11b – 4c) – 4c(11b – 4c)
= (11b – 4c) (11b – 4c)

(vii) (l + m)² – 4lm
= l² + 2lm + m² – 4lm
= l² – 2lm + m²
= l² – lm – lm + m²
= l (l – m) – m(l – m)
= (l – m)(l – m)

(viii) a⁴ + 2a²b² + b⁴
= a⁴ + a²b² + a²b² + b⁴
= a²(a² + b²) + b²(a² + b²)
= (a² + b²) (a² + b²)

Question 2 .
Factorise;
(i) 4p² – 9q²
(ii) 63a² -112b²
(iii) 49x² – 36
(iv) 16x⁵ – 144x³
(v) (l + m)² (l – m)²
(vi) 9x²y² – 16
(vii) (x² – 2xy + y²)
(viii) 25a² – 4b² + 28bc – 49c²
Solution:
(i) ∵ a² – b² = (a + b)(a – b)
∴ 4p² – 9q² = (2p)² – (3q)²
= (2p + 3q) (2p – 3q)²

(ii) 63a² – 112b² = 7(9a² – 16b²)
= 7[(3a)² – (4b)²]
= 7[(3a + 4b) (3a – 4b)]
= 7 (3a + 4b) (3a – 4b)

(iii) 49x² – 36 = (7x)² – (6)²
= (7x + 6) (7x – 6)

(iv) 16x⁵ – 144x³ = x (16x⁴ – 144x²)
= x [(4x²)² – (12x)²]
= x [(4x² + 12x) (4x² – 12x)]
= x (4x² + 12x) (4x² – 12x)

(v) (l + m)² -(l – m)²
= [{(l + m) + (l – m)} – {(l + m) – (l – m)}]
= [{l + m + l – m}{l + m – l + m}]
= 2l × 2m = 4lm

(vi) 9x²y² – 16 = (3xy)² – (4)²
= (3xy + 4) (3xy – 4)

(vii) (x² – 2xy + y²) – z²
= (x – y)² – z²
= (x – y + z) (x – y – z)

(viii) 25a² – 4b² + 28bc – 49c²
= 25a² – (4b² – 28bc + 49c²)
= (5a)² – [(26)²– 2 x 2b x 7c + (7c)²]
= (5a)² – (2b – 7c)²
= [(5a) (2b – 7c)] [(5a) – (2b – 7c)]
= (5a + 2b – 7c) (5a – 2b + 7c)

Question 3.
Factorise the expressions:
(i) ax² + 6x
(ii) 7p² + 21q²
(iii) 2x³ + 2xy² + 2xz²
(iv) am² + bm² + bn² + an²
(v) (lm + l) + m + 1
(vi) y(y + z) + 9 (y + z)
(vii) 5y² – 20y – 8z + 2yz
(viii) 10ab + 4a + 5b + 2
(ix) 6xy – 4y + 6 – 9x
Solution:
(i) ax² + bx = x (ax + b)
(ii) 7p² + 21q² = 7(p² + 3q²)
(iii) 2x³ + 2xy² + 2xz² = 2x(x² + y² + z²)

(iv) am² + 6m² + bn² + an²
= m²(a + b) + n²(b + a)
= m²(a + b) + n²(a + b)
= (a + b) (m² + n²)

(v) (lm + l) + m + 1
= l(m + 1) + 1 (m + 1)
= (m + 1) (l + 1)

(vi) y(y + z) + 9(y + z)
= (y + z)(y + 9)

(vii) 5y² – 20y – 8z + 2yz
= 5y² – 20y + 2yz – 8z
= 5y (y – 4) + 2z (y – 4)
= (y – 4) (5y + 2z)

(viii) 10ab + 4a + 5b + 2
= 2a (5b + 2) + 1(5b + 2)
= (5b + 2) (2a + 1)

(ix) 6xy – 4y + 6 – 9x
= 6xy – 4y – 9x + 6
= 2y (3x – 2) – 3(3x – 2)
= (3x – 2) (2y – 3).

Question 4.
Factorise :
(i) a⁴ – b⁴
(ii) p⁴ – 81
(iii) x⁴ – (y + z)⁴
(iv) x⁴ – (x – z)⁴
(v) a⁴ – 2a²b² + b⁴
Solution:
(i) a⁴ – b⁴ = (a²)² – (b²)²
= (a² + b²) (a² – b²)

(ii) p⁴ – 81 = (p²)² – (9)²
= (p² + 9) (p² – 9)

(iii) x⁴ – (y + z)⁴
= (x²)² – [(y + z)²]²
= [x² + (y + z)²] [x² – (y + z)²]
= (x² + y² + 2yz + z²) [(x² – (y² + 2yz + z²)]
= (x² + y² + z² + 2yz) (x² – y² – z² – 2yz)

(iv) x⁴ – (x – z)⁴
= (x²)² – [(x – z)²]²
= [x² + (x – z)²] [x² – (x – z)²]
= (x² + x² – 2xz + z²) [x²(x² – 2xz + z)²]
= (2x² + z² – 2xz) (x² – x² + 2xz – z²)
= (2x² + z² – 2xz) (2xz – z²)

(v) a⁴ – 2a²b² + b⁴ = (a²)² – 2a²b² + (b²)²
= (a² – b²)²
= (a² – b²) (a² – b²)
= (a + b) (a – b) (a + b) (a – b)
= (a + b)² (a – b)²

Question 5.
Factorise the following expressions:
(i) p² + 6p + 8
(ii) q² – 10q + 21
(iii) p² + 6p – 16
Solution:
(i) p² + 6p + 8
= p² + 2 × p × 3 + 9 – 1
= p² + 2 × p × 3 + (3)² – 1
= (p + 3)² – 1
[∵ a² + 2ab + b² = (a + b)²]
= (p + 3)² – (1)²
[∵ a² – b² = (a + b) (a – b)]
= (p + 3 + 1)(p + 3 – 1)
= (p + 4) (p + 2)

(ii) q² – 10q + 21
= q² – 2 × q × 5+ (5)² – 4
= (q – 5)² – (2)²
[∵ a² – 2ab + b² = (a – b)² ]
= (q – 5 + 2) × (q – 5 – 2)
[∵ a² – b² = (a + b) (a – b)
= (q – 3)(q – 7)

(iii) p² + 6p – 16
= p² + 2 × p × 3 + (3)² – 25
= (p + 3)² – (5)²
= (p + 3 + 5)(p + 3 – 5)
= (p + 8)(p – 2)