Author name: Bhagya

Haryana State Board HBSE 8th Class Maths Solutions Chapter 14 Factorization Ex 14.1 Textbook Exercise Questions and Answers.

Haryana Board 8th Class Maths Solutions Chapter 14 Factorization Exercise 14.1

Question 1.
Find the common factors of the given terms:
(i) 12x, 36
(ii) 2y, 22xy
(iii) 14pq, 28 p²q²
(iv) 2x, 3x², 4
(v) 6abc, 24 ab², 12a²b
(vi) 16x³, -4x², 32x
(vii) 10pq, 20qr, 30rp
(viii) 3x²y³, 10x³y², 6x²y²z
Solution:
(i) 12x = 2 × 2 × 3 × x
36 = 2 × 2 × 3 × 3
∴ Highest common factor, HCF = 2 × 2 × 3 = 12

(ii) 2y = 2 × y
22xy = 2 × 11 × x × y
∴ HCF = 2 × y = 2y

(iii) 14pq = 2 × 7 × p x q
28 p²q² = 2 × 2 × 7 × p × p × q × q
∴ HCF = 2 × 7 × p × q = 14 pq

(iv) 2x = 2 × x
3x² = 3 × x × x
4 = 2 × 2
∴ HCF = 1

(v) 6abc = 2 × 3 × a × b × c
24 ab² = 2 × 2 × 2 × 3 × a × b × b
12a²b = 2 × 2 × 3 × a × a × b
∴ HCF = 2 × 3 × a × b
= 6ab

(vi) 16x³ = 2 × 2 × 2 × 2 × x × x × x
-4x² = (-2) × 2 × x
32x = 2 × 2 × 2 × 2 × 2 × x
∴ HCF = 2 × 2 × x
= 4x

(vii) 10pq = 2 × 5 × p × q
20qr = 2 × 2 × 5 × q × r
30rp = 2 × 3 × 5 × r × p
∴ HCF = 2 × 5
= 10

(viii) 3x²y³ = 3 × x × x × y × y × y
10 x³ y2² = 2 × 5 × x × x × x y × y
6x²y²z = 2 × 3 × x × x × y × y × z
∴ HCF = x × x × y × y
= x²y²

Question 2.
Factorise the following ex-pressions
(i) 7x – 42
(ii) 6p – 12q
(iii) 7a² + 14a
(iv) -16z + 20z³
(v) 20l²m + 30alm
(vi) 5x²y – 15xy²
(vii) 10a² – 15b² + 20c²
(viii) -4a² + 4ab – 4ca
(ix) x²yz + xy²z + xyz²
(x) ax²y + bxy² + cxyz
Solution:
(i) 7a – 42 = 7(x – 6)
(ii) 6p – 12q = 6 (p – 2q)
(iii) 7a² + 14a = 7a (a + 2)
(iv) -16z + 20z³ = 42 (-4z + 5z²)
= 4z (5z² – 4z)
(v) 20l²m + 30alm = 10lm (2l + 3a)
(vi) 5x²y – 15xy² = 5xy (x – 3y)
(vii) 10a² – 15b² + 20c² = 5 (2a² – 3b² + 4c²)
(viii) -4a² + 4ab – 4ca = 4a(-a + b – c)
(ix) x²yz + xy²z + xyz² = xyz (x + y + z)
(x) ax²y + bxy² + cxyz = xy (ax + by + cz)

Question 3.
Factorise:
(i) x² + xy + 8x + 8y
(ii) 15xy – 6x + 5y – 2
(iii) ax + bx – ay – by
(iv) 15pq + 15 + 9q + 25p
(v) z – 7 + 7xy – xyz
Solution:
(i) x² + xy + 8x + By
= x (x + y) + 8 (x + y)
= (x + y)(x + 8)

(ii) 15xy – 6x + 5y – 2
= 3x (5y – 2) + 1 (5y – 2)
= (5y – 2) (3x + 1)

(iii) ax + bx – ay – by
= x(a + b) – y (a + b)
= (a + b) (x – y)

(iv) 15pq + 15 + 9q + 25p
= 15pq + 9q + 25p + 15
= 3q + (5p + 3) + 5 (5p + 3)
= (5p + 3) (3q + 5)

(v) z – 7 + 7xy – xyz
= z – 7 – xyz + 7xy
= 1 (z – 7) – xy (z – 7)
= (z – 7)(1 – xy)

Haryana State Board HBSE 8th Class Maths Solutions Chapter 13 Direct and Inverse Proportions InText Questions and Answers.

Haryana Board 8th Class Maths Solutions Chapter 13 Direct and Inverse Proportions InText Questions

Try These (Page 204)

Question 1.
Observe the following tables and find if x and y are directly proportional.

Solution:
\(\frac{x}{y}\) = \(\frac{x_{1}}{y_{1}}\) = \(\frac{x_{2}}{y_{2}}\) ………..

Question 2.
Principal = Rs. 1000, Rate = 8% per annum. Fill in the following table and find which type of interest (simple or compound) changes in direct proportion with time period.

Solution:

Try These (Page 211)

Question 3.
Observe the following tables and find which pair of variables (here x and y) are in inverse proportion.

Solution:
(i) Two quantities are said to vary inversely if the increase in one quantity causes the decrease in the other quantity.
In this case if y₁, y₂ are the values of y corresponding to the values x₁, x₂ of x respectively then
x₁y₁ = x₂y₂
\(\frac{x_{1}}{x_{2}}\) = \(\frac{y_{1}}{y_{2}}\)
x × y = x₁y₁ = x₂y₂ = ………
x × y ≠ 50 × 5 ≠ 40 × 6 ≠ 30 × 7 ≠ 20 × 8
Hence, table (i) is not inversely proportional.

(ii) x × y = x₁ × y₁ = x₂ × y₂
= x₃ × y₃…………
= 100 × 60 = 200 × 30
= 300 × 20 = 400 × 15
= 600 = 6000 = 6000 = 6000
Hence table (ii) is inversely proportional.

(iii) x × y = x₁y₁= x₂ × y₂ = x₃ × y₃ = ……..
= 90 × 10 = 60 × 15
= 45 × 20 = 30 × 25
= 20 × 30 = 5 × 35
= 900 = 900 = 900 ≠ 750
≠ 600 ≠ 175
Hence, in table (iii) 90, 10; 60, 15; 45, 20 are inversely proportion but other are not inversely proportional.

Haryana State Board HBSE 8th Class Maths Solutions Chapter 13 Direct and Inverse Proportions Ex 13.2 Textbook Exercise Questions and Answers.

Haryana Board 8th Class Maths Solutions Chapter 13 Direct and Inverse Proportions Exercise 13.2

Question 1.
Which of the following are in inverse proportion ?
(i) The number of workers on a job and the time to complete the job.
(ii) The time taken for a journey and the distance travelled in a uniform speed.
(iii) Area of cultivated land and the crop harvested.
(iv) The time taken,for a fixed journey and the speed of the vehicle.
(v) The population of a country and the area of land per person.
Answer:
Two quantities are said to vary inversely proportional if the increase (or decrease) in one quantity causes the decrease (or increase) in the other quantity.
Hence,
(i) is inverse proportional.
(ii) is direct proportional.
(iii) is inverse proportional.
(iv) is inverse pooportional.
(v) is inverse proportional.

Question 2.
In a Television game show, the prize money of Rs. 1,00,000 is to be divided equally amongst the winners. Complete the following table and find whether the prize money given to an individual winner is directly or inversely proportional to the number of winners ?

Solution:
If xy = K, then x and y are said to vary inversely.
xy = x₁y₁ = x₂y₂ = x₃y₃ = ….
= 1 × 1,00,000
= 2 × 50,000 = 1,00,000
or \(\frac{x_{1}}{x_{2}}\) = \(\frac{y_{1}}{y_{2}}\)
⇒ \(\frac{1}{4}\) = \(\frac{x}{1,00,000}\)
⇒ 4x = 1,00,000
∴ x = \(\frac{1,00,000}{4}\) = 25000
Similarly, y = 20000, z = 12500
a = 10000, b = 6000.

Question 3.
Rehman is making a wheel using spokes. He wants to fix equal spokes in such a way that the angles between any pair of consecutive spokes are equal. Help him by completing the following table :


(i) Are the number of spokes and the angles formed between the pairs of consecutive spokes in inverse proportion ?
(ii) Calculate the angle between a pair of consecutive spokes on a wheel with 15 spokes.
(iii) How many spokes would be needed, if the angle between a pair of consecutive spokes is 40° ?
Solution:
In inversely proportional,

(i) Since, number of spokes are increases and angle between a pair of consecutive spokes are decreases. Hence it is inversely proportional.
(ii) \(\frac{4}{15}\) = \(\frac{a}{90}\) ⇒ a = \(\frac{4 \times 90}{15}\) = 24°
⇒ Angle = 24°
(iii) \(\frac{4}{x}\) = \(\frac{40}{90}\) ⇒ 40 × x = 4 × 90
x = \(\frac{4 \times 90}{40}\) = 9
No. of spokes = 9.

Question 4.
If a box of sweets is divided among 24 children, they will get 5 sweets each. How many would each get, if the number of the children is reduced by 4 ?
Solution:
Let each get sweets = x

Now, \(\frac{5}{x}\) = \(\frac{20}{24}\)
⇒ 20x = 5 × 24
⇒ x = \(\frac{5 \times 24}{20}\) = 6
Hence required each get sweets = 6.

Question 5.
A farmer has enough food to feed 20 animals in his cattle for 6 days. How long would the food last if there were 10 more animals in his cattle ?
Solution:
Let the No. of days = x

Now, \(\frac{20}{30}\) = \(\frac{x}{6}\)
⇒ 30 × x = 20 × 6
⇒ x = \(\frac{20 \times 6}{30}\) = 4
Hence required days = 4.

Question 6.
A contractor estimates that 3 persons could rewire Jasminder’s house in 4 days. If, he uses 4 persons instead of three, how long should they take to complete the job ?
Solution:
Let the days = x

Now, \(\frac{3}{4}\) = \(\frac{x}{4}\)
⇒ 4 × x = 3 × 4
⇒ x = \(\frac{3 \times 4}{4}\) = 3 days
Hence, required days = 3.

Question 7.
A batch of bottles were packed in 25 boxes with 12 bottles in each box. If the same batch is packed using 20 bottles in each box, how many boxes would be filled ?

Solution:
Let the no. of boxes = x

Now, \(\frac{25}{x}\) = \(\frac{20}{12}\)
⇒ 20 × x = 25 × 12
⇒ x = \(\frac{25 \times 12}{20}\) = 15 boxes
Hence, required boxes = 15.

Question 8.
A factory requires 42 machines to produce a given number of articles in 63 days. How many machines would be required to produce the same number of articles in 54 days ?
Solution:
Let the number of machines = x

Now, \(\frac{42}{x}\) = \(\frac{54}{63}\)
⇒ 54 × x = 42 × 63
⇒ x = \(\frac{42 \times 63}{54}\) = 49
Hence, the required machines = 49.

Question 9.
A car takes 2 hours to reach a destination by travelling at the speed of 60 km/h. How long will it take when the car travels at the speed of 80 km/h ?
Solution:
Let the time = x

Now, \(\frac{2}{x}\) = \(\frac{80}{60}\)
⇒ 80 × x = 2 × 60
⇒ x = \(\frac{2 \times 60}{80}\) = 1.5 hours
Hence required time = 1.5 hours.

Question 10.
Two persons could fit new windows in a house in 3 days.
(i) One of the persons fell ill before the work started. How long would the job take now ?
(ii) How many persons would be needed to fit the window in one day ?
Solution:
(i) Let the no. of days = x

Now, \(\frac{2}{1}\) = \(\frac{x}{3}\)
⇒ 1 × x = 2 × 3
⇒ x = 6 days
Hence required days = 6.

(ii) Let the no. of person = y

Now, \(\frac{2}{x}\) = \(\frac{1}{3}\)
⇒ x × 1 = 2 × 3 = 6 persons
Hence required persons = 6.

Question 11.
A school has 8 periods a day each of 45 minutes duration. How long would each period be, if the school has 9 periods a day, assuming the number of school hours to be the same ?
Solution:
Let the no. of duration = x

Now, \(\frac{45}{x}\) = \(\frac{9}{8}\)
⇒ 9 × x = 8 × 45
⇒ x = \(\frac{8 \times 45}{9}\) = 40
Hence, requied time duration = 40 minutes.

Haryana State Board HBSE 8th Class Maths Solutions Chapter 13 Direct and Inverse Proportions Ex 13.1 Textbook Exercise Questions and Answers.

Haryana Board 8th Class Maths Solutions Chapter 13 Direct and Inverse Proportions Exercise 13.1

Question 1.
Following are the car parking charges near a railway station upto
4 hours — Rs. 60
8 hours — Rs. 100
12 hours — Rs. 140
24 hours — Rs. 180

Check if the parking charges are direct proportion to the parking time.
Solution:
Suppose the parking time of car is x hours and its charges in Rs. is y.

If x and y are direct proportion, then

Question 2.
A mixture of paint is prepared by mixing 1 part of red pigments with 8 parts of base. In the following table, find the parts of base that need to be added.

Solution:
Let the parts of red pigment = x and
part of base = y.
∴ \(\frac{x}{y}\) = \(\frac{1}{8}\)
\(\frac{x}{y}\) = \(\frac{1}{32}\) = \(\frac{4}{32}\) = \(\frac{7}{56}\) = \(\frac{12}{96}\) = \(\frac{20}{160}\)

Question 3.
In Question 2 above, if 1 part of a red pigment requires 75 mL of base, how much red pigment should we mix with 1800 mL of base ?
Solution:
Let red pigment = x

∴ \(\frac{1}{75}\) = \(\frac{x}{1800}\)
⇒ x × 75 = 1 × 1800
⇒ x = \(\frac{1800}{75}\)
∴ x = 24
Hence required red pigment = 24.

Question 4.
A machine in a soft drink factory fills 840 bottles in six hours. How many bottles will in fill in five hours ?
Solution:
Let the no. of bottles = x
Number of time and number of bottles vary directly.

Now, \(\frac{6}{840}\) = \(\frac{5}{x}\)
⇒ 6 × x = 5 × 840
⇒ x = \(\frac{5×840}{6}\)
⇒ x = 700
Hence, required bottles = 700.

Question 5.
A photograph of a bacteria enlarged 50,000 times attains a length of 5 cm as shown in the diagram. What is the actual length of the bacteria ? If the photograph is enlarged 20,000 times only, what would be enlarged length ?

Solution:
Let the length of bacteria = x. Length of bacteria and enlarged times of bacteria vary directly.

Now, \(\frac{50,000}{5}\) = \(\frac{20,000}{x}\)
⇒ 50,000 × x = 5 × 20,000
⇒ x = \(\frac{5×20,000}{50,000}\)
⇒ x = 2
∴ Enlarged length = 2 cm
and Actual length, x₁ = \(\frac{2}{20,000}\)
= \(\frac{1}{10,000}\) = 0.0001 cm.

Question 6.
In a model of a ship, the mast is 9 cm high, while the mast of the actual ship is 12 m high. If the length of the ship is 28 m, how long is the model ship ?

Solution:
Try Yourself.

Question 7.
Suppose 2 kg of sugar contains 9 × 10⁶ crystals. How many sugar crystals are there in
(i) 5 kg of sugar ?
(ii) 1.2 kg of sugar ?
Solution:
Let 5 kg of sugar contains = x crystals
and 1.2 kg of sugar contains = y crystals.
Sugar and crystals vary directly.

Now, (i) \(\frac{2}{9 \times 10^{6}}\) = \(\frac{5}{x}\)
⇒ 2 × x = 5 × 9 × 10
⇒ x = \(\frac{5 \times 9 \times 10 \times 10^{5}}{2}\)
⇒ x = 225 × 10⁵
= 2.25 × 10⁷
∴ 5 kg of sugar contains 2.25 × 10⁷ crystals.

(ii) \(\frac{2}{9 \times 10^{6}}\) = \(\frac{1.2}{y}\)
⇒ 2 × y = 1.2 × 9 × 10⁶
⇒ y = \(\frac{1.2 \times 9 \times 10 \times 10^{5}}{2}\)
= 54 × 10⁶
= 5.4 × 10⁶
∴ 1.2 kg of sugar contains 5.4 × 10⁶ crystals.

Question 8.
Rashmi has a road map with a scale of 1 cm representing 18 km. She drives on a road for 72 km. What would be her distance covered in the map ?
Solution:
Let the distance covered in the map = x.
Map with a scale and distance covered in the map vary directly.

Now, \(\frac{1}{18}\) = \(\frac{x}{72}\)
⇒ 18 × x = 72 × 1
⇒ x = \(\frac{72}{18}\) = 4
Hence required distance covered in the map = 4 cm.

Question 9.
A 5 m 60 cm high vertical pole casts a shadow 3 m 20 cm long. Find at the same time
(i) the length of the shadow cast by another pole 10 m 50 cm high,
(ii) the height of a pole which casts a shadow 5 m long.
Solution:
Let the length of shadow = x
and the height of a pole = y
Length of the shadow and height of pole vary directly.

Now, (i) \(\frac{5.60}{3.20}\) = \(\frac{10.50}{x}\)
⇒ 5.60 × x = 3.20 × 10.50
⇒ x = \(\frac{3.20 \times 10.50}{5.60}\) = 6 m
Hence required length of shadow 6m.

(ii) \(\frac{5.60}{3.20}\) = \(\frac{y}{5}\)
⇒ 3.20 × y = 5.60 × 5
⇒ y = \(\frac{5.60 \times 5}{3.20}\) = 8.75
Hence required height of pole = 8.75 m or 8 m 75 cm.

Question 10.
A loaded truck travels 14 km in 25 minutes. If the speed remains the same, how far can it travel in 5 hours ?
Solution:
60 minutes = 1 hour
∴ 25 minutes = \(\frac{1}{60}\) × 25 = \(\frac{5}{12}\) hour
Let distance of travelling truck = x
Distance of travelling truck and time vary directly.

Now, \(\frac{14}{5 / 12}\) = \(\frac{x}{5}\)
⇒ \(\frac{5}{12}\) × x = 14 × 5
⇒ x = \(\frac{14 \times 5}{5 / 12}\) = \(\frac{14 \times 5 \times 12}{5}\)
⇒ x = 168 km
Hence distance of travelling truck = 168 km.

Haryana State Board HBSE 8th Class Maths Solutions Chapter 12 Exponents and Powers InText Questions and Answers.

Haryana Board 8th Class Maths Solutions Chapter 12 Exponents and Powers InText Questions

Try These (Page 194)

Question 1.
Find the multiplicative inverse of the following :
(i) 2^-4
(ii) 10^-5
(iii) 7^-2
(iv) 5^-3
(v) 10^-100
Solution:
(i) ∵ a^-m = \(\frac{1}{a^{m}}\)
where m is a positive integer and a^-m is the multiplicative inverse of a^m.
∴ Multiplicative inverse of 2^-4 = \(\frac{1}{2^{4}}\)

(ii) Similarly, multiplicative inverse of 10^-5 = \(\frac{1}{10^{5}}\)
(iii) Multiplicative inverse of 7^-2 = \(\frac{1}{7^{2}}\)
(iv) Multiplicative inverse of 5^-3 = \(\frac{1}{5^{3}}\)
(v) Multiplicative inverse of 10^-100 = \(\frac{1}{10^{100}}\)

Question 2.
Expand the following numbers using exponents:
(i) 1025.63
(ii) 1256.249.
Solution:
(i) We learnt how to write numbers like 1025 in expanded form using exponents as 1 × 10³ + 0 × 10² + 2 × 10¹ + 5 × 10⁰.
Similarly we have
1025.63 = 1 × 1000 + 0 × 100 + 2 × 10 + 5 × 1 + \(\frac{6}{10}\) + \(\frac{3}{100}\)
= 1 × 10³ + 0 × 10² + 2 × 10¹ + 5 + 6 × 10^-1 + 3 × 10^-2.

(ii) 1256.249 = 1 × 1000 + 2 × 100 + 5 × 10 + 6 × 1 + \(\frac{2}{10}\) + \(\frac{4}{100}\) + \(\frac{9}{1000}\)
= 1 × 10³ + 2 × 10² + 5 × 10¹ + 6 × 10⁰ + 2 × 10^-1 + 4 × 10^-2 + 9 × 10^-3.

Try These (Page 195)

Question 1.
Simplify and write in exponential form :
(i) (-2)^-3 × (-2)^-4
(ii) p³ × p^-10
(iii) 3² × 3^-5 × 3⁶.
Solution:
Since, a^m × aⁿ = a^m+n
(i) (-2)^-3 × (-2)^-4 = (-2)^(-3)+(4)
= (-2)^-7

(ii) p³ × p^-10 = (p)^3+(-10)
= p^-7

(iii) 3² × 3^-5 × 3⁶ = (3)^2+(-5)+6
= (3)^8-5 = 3³.

Try These (Page 199)

Question 1.
Write the following numbers in standard form :
(ii) 0.000000564
(ii) 0.0000021
(iii) 21600000
(iv) 15240000.
Solution:
(i) 0.000000564
= \(\frac{564}{1000000000}\) = \(\frac{5.64 \times 100}{10^{9}}\)
= 5.64 × 10² × 10^-9
= 5.64 × 10^-7.

(ii) 0.0000021 = \(\frac{21}{10000000}\) = \(\frac{2.1 \times 10}{10^{7}}\)
= 2.1 × 10¹ × 10^-7
= 2.1 × 10^-6

(iii) 21600000 = 2.16 × 10⁷
(iv) 15240000 = 1.524 × 10⁷.

Question 2.
Write all the facts given in the standard form.
Solution:
(i) Decimal is moved 7 places to the right.
(ii) Decimal is moved 6 places to the right.
(iii) Decimal is moved 7 places to the left.
(iv) Decimal is moved 7 places to the left.
We conclude the above number can be written in the form K × 10ⁿ where n is some integer and K is a terminating decimal lying between 1 and 10, i.e.,
1 ≤ K< 10.
That is the fact of standard form.

Haryana State Board HBSE 8th Class Maths Solutions Chapter 12 Exponents and Powers Ex 12.2 Textbook Exercise Questions and Answers.

Haryana Board 8th Class Maths Solutions Chapter 12 Exponents and Powers Exercise 12.2

Question 1.
Express the following numbers in standard form :
(i) 0.0000000000085
(ii) 0.00000000000942
(iii) 6020000000000000
(iv) 0.00000000837
(v) 31860000000.
Solution:
(i) 0.0000000000085
= \(\frac{85}{10000000000000}\)
= \(\frac{8.5 \times 10}{10^{13}}\)
= 8.5 × 10¹ × 10^-13
= 8.5 × 10^-12

(ii) 0.00000000000942
= \(\frac{942}{100000000000000}\)
= \(\frac{9.42 \times 100}{10^{14}}\)
= 9.42 × 10² × 10^-14
= 9.42 × 10^-12

(iii) 6020000000000000 = 6.02 × 10¹⁵
(iv) 0.00000000837
= \(\frac{837}{100000000000}\)
= \(\frac{8.37 \times 100}{10^{11}}\)
= 8.37 × 10^-12 × 10^-11
= 8.37 × 10^-9

(v) 31860000000 = 3.186 × 10¹⁰.

Question 2.
Express the fallowing number in usual form :
(i) 3.02 × 10^-6
(ii) 4.5 × 10⁴
(iii) 3 × 10^-8
(iv) 1.0001 × 10⁹
(v) 5.8 × 10¹²
(vi) 3.61492 × 10⁶.
Solution:

= 361492 × 10^6-5
= 361492 × 10 = 3614920.

Question 3.
Express the number appearing in the following statements in standard form :
(i) 1 micron is equal to \(\frac{1}{1000000}\) m.
(ii) Charge of an electron is 0.000,000,000,000,000,000,16coulom b.
(iii) Size of a bacteria is 0.0000005 m.
(iv) Size of a plant cell is 0.00001275 m.
(v) Thickness of a thick paper is 0.07 mm.
Solution:

= 1.275 × 10^3-8
= 1.275 × 10^-5 m.
(v) Thickness of a thick paper
= \(\frac{7}{100}\) = \(\frac{7.0 \times 10}{10^{2}}\)
= 7.0 × 10^1-2
= 7.0 × 10^-1 mm.

Question 4.
In a stack there are 5 books each of thickness 20 mm and 5 paper sheets each of thickness 0.016 mm. What is the total thickness of the stack.
Solution:
Total thickness of stack
= 20 m + 0.016
= 20.016
= \(\frac{20016}{1000}\) = \(\frac{2.0016×1000}{1000}\)
= \(\frac{2.0016 \times 10^{4}}{10^{3}}\)
= 2.0016 × 10.

Haryana State Board HBSE 8th Class Maths Solutions Chapter 12 Exponents and Powers Ex 12.1 Textbook Exercise Questions and Answers.

Haryana Board 8th Class Maths Solutions Chapter 12 Exponents and Powers Exercise 12.1

Question 1.
Evaluate :
(i) 3^-2
(ii) (-4)^-2
(iii) \(\left(\frac{1}{2}\right)^{-5}\)
Solution:

Question 2.
Simplify and express the result in power notation with positive exponent.
(i) (-4)⁵, (-4)⁶
(ii) \(\left(\frac{1}{2^{3}}\right)^{2}\)
(iii) (-3)⁴ × \(\left(\frac{5}{3}\right)^{4}\)
(iv) (3^-7 ÷ 3^-10) × 3^-5
(v) 2^-5 × (-7)^-3.
Solution:
(i) ∵ a^m ÷ aⁿ = a^m-n
∴ (-4)⁵ ÷ (-4)⁸ = (-4)^5-8

= +1 × 5⁴ = 5⁴
= 5 × 5 × 5 × 5
= 625.

(iv) ∵ a^m ÷ aⁿ = a^m-n
and a^m × aⁿ = a^m+n
∴ (3^-7 ÷ 3^-10) × 3^-5 = 3^-7-(-10) × 3^-5
= 3^-7+10 × 3^-5
= 3³ × 3^-5

Question 3.
Find the value of :
(i) (3⁰ + 4^-1) × 2²
(ii) (2^-1 × 4^-1), 2^-2
(iii) \(\left(\frac{1}{2}\right)^{-2}\) + \(\left(\frac{1}{3}\right)^{-2}\) + \(\left(\frac{1}{4}\right)^{-2}\)
(iv) (3^-1 + 4^-1 + 5^-1)⁰
(v) \(\left\{\left(\frac{-2}{3}\right)^{-2}\right\}^{2}\)
Solution:

Question 4.
Evaluate:
(i) \(\frac{8^{-1} \times 5^{3}}{2^{-4}}\)
(ii) (5^-1 × 2^-1) × 6^-1
Solution:

Question 5.
Find the value of m for which 5^m ÷ 5^-3 = 5⁵
Solution:
∵ a^m ÷ a^m = a^m-n
∴ 5^m ÷ 5^-3 = 5⁵
or 5^m-(-3) = 5⁵
or 5^m+3 = 5⁵ [∵ a^m = aⁿ]
or m + 3 = 5 [∵ m = n]
or m = 5 – 3
∴ m = 2

Question 6.
Evaluate:
(i) \(\left\{\left(\frac{1}{3}\right)^{-1}-\left(\frac{1}{4}\right)^{-1}\right\}^{-1}\)
(ii) \(\left(\frac{5}{8}\right)^{-7} \times\left(\frac{8}{5}\right)^{-4}\)
Solution:

Question 7.
Simplify:
(i) \(\frac{25 \times t^{-4}}{5^{-3} \times 10 \times t^{-8}}\) (t ≠ 0)
(ii) \(\frac{3^{-5} \times 10^{-5} \times 125}{5^{-7} \times 6^{-5}}\)
Solution:


= 5^10-5 = 5⁵
= 5 × 5 × 5 × 5 × 5
= 3125.

Haryana State Board HBSE 8th Class Maths Solutions Chapter 11 Mensuration InText Questions and Answers.

Haryana Board 8th Class Maths Solutions Chapter 11 Mensuration InText Questions

Try These (Page 170)

Question 1.
(a) Match the following figures with their respective areas in the box :

(b) Write the perimeter of each shape.
Solution:

(b) (i) Perimter of parallelogram = 2(a + b) = 2(14 + b)
(ii) Perimeter of semicircle = \(\frac{2 \pi r}{2}\) = πr = \(\frac{22}{7}\) × 7 = 2 cm.
(iii) Perimeter of triangle = 14 + 11 + 9 = 34 cm.
(iv) Perimeter of square = 4 × side = 4 × 7 = 28 cm.
(v) Perimeter of rectangle = 2(l + b) = 2(14 + 7) = 2 × 21 = 42 cm.

Try These (Page 172)

Question 1.
Nazma’s sister also has a trapezium shaped plot. Divide it into three parts as shown (Fig. 11.6). Show that the area of trapezium WXYZ = h \(\frac{(a+b)}{2}\)
Solution:

Area of trapezium = Area of △(l) + Area of rectangle + Area of △(2)

Hence, area of trapezium = \(\frac{1}{2}\) h (a + b).
Hence verified

Question 2.
If h = 10 cm, c = 6 cm, 6 = 12 cm, d = 4 cm, find the values of each of its parts separately and add to find the area WXYZ. Verify it by putting the values of h, a and b in the expression \(\frac{h(a+b)}{2}\)
Solution:
∵ Area of trapezium
= \(\frac{1}{2}\)h(a + b) = \(\frac{1}{2}\) × h(c + d + 2b)
= \(\frac{1}{2}\) × 10(6 + 4 + 2 × 12)
= \(\frac{1}{2}\) × 10(10 + 24)
= \(\frac{1}{2}\) × 10 × 34 = 170 cm²

Try These (Page 173)

Question 1.
Find the area of the following trapeziums (Fig. 11.7).

Solution:
(i) Area of trapezium
= \(\frac{1}{2}\) h (a + b)= \(\frac{1}{2}\) × 3 (9 + 7)
= \(\frac{1}{2}\) × 3 × 16 = 24 cm³.

(ii) Area of trapezium= \(\frac{1}{2}\)h(a + b)
= \(\frac{1}{2}\) × 6 × (10 + 5)
= \(\frac{1}{2}\) × 6 × 15
= 45 cm³

Try These (Page 174)

Question 1.
We know that parallelogram is also a quadrilateral. Let us also split such a quadrilateral into two triangles, find their areas and hence that of the parallelogram. Does this agree with the formula that you know already ? (Fig. 11.8)

Solution: Area of parallelogram= a × h
L.H.S. = Area of parallelogram
= Area 6f triangle (l) + Area of triangle (2)
= \(\frac{1}{2}\) × b ×h + \(\frac{1}{2}\) × b × h
= \(\frac{b h+b h}{2}\) = \(\frac{2(b h)}{2}\)
= b × h
= R.H.S.
Hence verified.

Try These (Page 175)

Question 1.
Find the area of these quadrilaterals (Fig. 11.9).

Solution:
(i) Area of quadrilateral ABCD = Area of △ABC + Area of △ACD
= \(\frac{1}{2}\) × AC × 3 + \(\frac{1}{2}\) × AC × 5
= \(\frac{1}{2}\) × 6 × 3 + \(\frac{1}{2}\) × 6 × 5
= 9 + 15 = 24 cm²

(ii) Area of rhombus
= \(\frac{1}{2}\) × d₁ × d₂ = \(\frac{1}{2}\) × 7 × 6
= 21 cm²

(iii) Area of parallelogram
= 2 × Area of triangle
= 2 × \(\frac{1}{2}\) × 8 × 2 = 16 cm².

Try These (Page 176)

Question 1.
(i) Divide the following polygons (Fig. 11.19) into parts (triangles and trapezium) to find out its area.

(ii) Polygon ABODE is divided into parts as shown in (Fig. 11.11). Find its area if AD = 8 cm, AH = 6 cm, AG = 4 cm, AF = 3 cm and perpendiculars BF = 2 cm, OH = 3 cm, EG = 2.5 cm.
Area of polygon ABCDE = Area of DAFB + …..
Area of △AFB = \(\frac{1}{2}\) × AF × BF
= \(\frac{1}{2}\) × 3 × 2 = …

Area of trapezium FBCH
= FH × \(\frac{(\mathrm{BF}+\mathrm{CH})}{2}\)
= 3 × \(\frac{(\mathrm{2}+\mathrm{3})}{2}\)
[FH = AH – AF]
Area of △CHD = \(\frac{1}{2}\) × HD × CH = ………..
Area of △ADE = \(\frac{1}{2}\) × AD × GE = ……………..
So, the area of polygon ABODE = ………..

(iii) Find the area of polygon MNOPQR (Fig. 11.12) if MP = 9 cm, MD = 7 cm, MO = 6 cm, MB = 4 cm, MA = 2 cm.

NA, OC, QD and RB are perpendiculars to diagonal MP.
Solution:
(i) (a)

Area of polygon EFGHI
= Area of DFGI + Area of DGHI + Area of DEFI
= \(\frac{1}{2}\) FI × h₁ + \(\frac{1}{2}\) GI × h₂ + \(\frac{1}{2}\) FI × h₃.

(b)
Area of polygon MNOPQR
= Area of △NOQ + Area of △OQP + Area of △MRQ + Area of △MNQ
= \(\frac{1}{2}\) NQ × h₁ + \(\frac{1}{2}\) OQ × h₂ + \(\frac{1}{2}\) MQ × h₃ + \(\frac{1}{2}\) NQ × h₄

(ii) Area of polygon ABODE = Area of △AFB + Area of trapezium FBCH + Area of △CHD + Area of △ADE
= \(\frac{1}{2}\) × AF × BF + \(\frac{1}{2}\) (BF + CH) × FH + \(\frac{1}{2}\) × HD × CH + \(\frac{1}{2}\) × AD × EG
= \(\frac{1}{2}\) × 3 × 2 + \(\frac{1}{2}\)(2 + 3) × 3 + \(\frac{1}{2}\) × 2 × 3 + \(\frac{1}{2}\) × 8 × 2.5
[where, FH = AH – FH = 6-3 = 3 cm, HD = AD – AH = 8 – 6 = 2 cm]
= 3 + \(\frac{1}{2}\) + 3 + 10.0
= \(\frac{16}{1}\) + \(\frac{15}{2}\) = \(\frac{32+15}{2}\)
= \(\frac{47}{2}\) = 23.5 cm².

(iii) Area of polygon MNOPQR
= ar △AMN + ar trapezium ANOC + ar △COP + ar △MBR + ar, trapezium BDQR + ar △DPQ
= \(\frac{1}{2}\) × MA × AN + \(\frac{1}{2}\) (AN + CO) × AC + \(\frac{1}{2}\) × CP × CO + \(\frac{1}{2}\) × MB × BR + \(\frac{1}{2}\) × (BR + DQ) × BD + \(\frac{1}{2}\) × DP × QD
= \(\frac{1}{2}\) × 2 × 2.5 + \(\frac{1}{2}\) (2.5 + 3) × 4 + \(\frac{1}{2}\) × 3 × 3 + \(\frac{1}{2}\) × 4 × 2.5 + \(\frac{1}{2}\) × (2.5 + 2) × 3 + \(\frac{1}{2}\) × 2 × 2
where, AC = MC – MA
= 6 – 2 = 4 cm
CP = MP – MC
= 9 – 6 = 3 cm
BD = MD – MB
= 7 – 4 = 3 cm
DP = MP – MD
= 9 – 7 = 2 cm
∴ Area of polygon MNOPQR
= 2.5 + 11 + \(\frac{9}{2}\) + 5 + \(\frac{13.5}{2}\) + 2
= 2.5 + 11 + 4.5 + 5 + 6.25 + 2
= 31.25 cm².

Try These (Page 181)

Question 1.
Find the total surface area of the following cuboids (Fig. 11.30).

Solution:
(i) Total surface area of Cuboid = 2 (lb + bh + hl)
= 2(6 × 4 + 4 × 2 + 2 × 6)
= 2(24 + 8 + 12)
= 2 × 44 = 88 cm²

(ii)Total surface area of cuboid = 2 (lb + bh + hl)
= 2(10 × 4 + 4 × 4 + 4 × 10)
= 2(40 + 16 + 40)
= 2 × 96 = 192 cm².

Try These (Page 182)

Question 1.
Find the surface area of cube A and lateral surface area of cube B (Fig. 11.32).

Solution:
(A) Total surface area of cube, = 6l² = 6 × (10)²
= 6 × 100 = 600 cm²
(B) Lateral surface area of cube
= 4 × (side)² = 4l²
= 4 × (8)² = 4 × 64
= 256 cm².

Try These (Page 184)

Question 1.
Find total surface area of the following cylinders (Fig. 11.33).

Solution:
(i) Total surface area of cylinder
= 2πr (r + h) [where r = 14, h = 8]
= 2 × \(\frac{22}{7}\) × 14(14 + 8)
= 2 × 22 × 2 × 22 = 1936 cm².

(ii) Total surface area of cylinder
= 2πr (r + h) = 2 × \(\frac{22}{7}\) × 1(1 + 2)
[∵ d = 2, r = \(\frac{d}{2}\)
where r = 1 m, h = 2 m]
= 2 × \(\frac{22}{7}\) × 1 × 3
= \(\frac{132}{7}\)
= 18.86 m²

Try These (Page 188)

Question 1.
Find the volume of the following cuboids (Fig. 11.40).

Solution:
(i) l = 8 cm, b = 3 cm, h = 2 cm
Volume of cuboid = l × b × h
= 8 × 3 × 2 = 48 cm³

(ii) l × b = 24 m2 and h = 3 cm
Volume of cuboid = l × b × h
= 24 × 3 = 72 cm³.

Try These (Page 189)

Question 1.
Find the volume of the following cubes:
(a) with a side 4 cm
(b) with a side 1.5 m.
Solution:
Volume of cube
= l³ = (4 cm)³
= 4 × 4 × 4 = 64 cm³

(ii) Volume of cube
= l³ = (1.5 cm)³
= 1.5 × 1.5 × 1.5
= 3.375 cm³.

Try These (Page 189)

Question 1.
Find the volume of the following cylinders :

Solution:
(i) Volume of cylinder = πr²h
= \(\frac{22}{7}\) × (7)² × 10
= \(\frac{22}{7}\) × 7 × 7 × 10
= 1540 cm³.

(ii)Volume of cylinder
= Area of base ×height
= πr² × h
= 250 m² × 2 m
= 500 m³.

Haryana State Board HBSE 8th Class Maths Solutions Chapter 11 Mensuration Ex 11.4 Textbook Exercise Questions and Answers.

Haryana Board 8th Class Maths Solutions Chapter 11 Mensuration Exercise 11.4

Question 1.
Given a cylindrical tank, in which situation will you find surface area and in which situation volume.
(a) To find how much it can hold.
(b) Number of cement bags required to plaster it.
(c) To find the number of smaller tanks that can be filled with water from it.

Solution:
(a) Volume of cylinder
= Area of base × height.
(b) Total surface area of cylinder = 2πr (r + h).
(c) Volume of cylinder = πr²h.

Question 2.
Diameter of cylinder A is 7 cm, and the height is 14 cm. Diameter of cylinder B is 14 cm and height is 7 cm. Without doing any calculations can you suggest whose volume is greater ? Verify it by finding the volume of both the cylinders. Check whether the cylinder with greater volume also has greater surface area ?

Solution:
(A) Radius = \(\frac{d}{2}\) = \(\frac{7}{2}\) cm
Height = 14 cm
Volume of cylinder = πr²h

= \(\frac{22}{7}\) × 7 × 7 × 7
= 1078 cm³.

(A) Curved surface area of cylinder = 2πrh
= 2 × \(\frac{22}{7}\) × \(\frac{7}{2}\) × 14
= 308 cm²

(B) Curved surface area of cylinder
= 2πrh
= 2 × \(\frac{22}{7}\) × 7 × 7
= 308 cm²
Hence the volume of cylinder (B) is greater than cylinder (A)
and surface area of cylinder (A) and (B) are equal.
If radius is greater, then volume of cylinder is greater.

Question 3.
Find the height of a cuboid whose base area is 180 cm2 and volume is 900 cm³?
Solution:
Volume of cuboid = l × b × h
Base area = l × b = 180 cm²
Volume of cuboid = 180 × h
900 cm³ = 180 × h
\(\frac{900}{180}\) = h
5 cm = h
∴ Height = 5 cm.

Question 4.
A cuboid is of dimensions 60 cm × 54 cm × 30 cm. How many small cubes with side 6 cm can be placed in the given cuboid ?
Solution:
Volume of cuboid
= 60 × 54 × 30
= 97200 cm²
Volume of cube (6 cm)³ = 216 cm³
Number of small cubes
= \(\frac{97200}{216}\) = 450 cubes.

Question 5.
Find the height of the cylinder whose volume is 1.54 m³ and diameter of the base is 140 cm ?
Solution:
Volume of cylinder = 1.54 m³
Diameter of cylinder = 140 m

1 = h
Hence, height of cylinder = 1 m or 100 cm.

Question 6.
A milk tank is in the form of cylinder whose radius is 1.5 m and length is 7 m. Find the quantity of milk in litres that can be stored in the tank ?
Solution:
Radius = 1.5 m
Height = 7 m
Volume of cylinder = πr²h
= \(\frac{22}{7}\) × 1.5 × 1.5 × 7
= 49.5 cm³
We know that
1 m³ = 1000 litres
Hence 49.5 m³ = 1000 × 49.5 m³
= 49500 litres
Hence, the required quantity of milk in litres in tank = 49500 litres.

Question 7.
If each edge of a cube is doubled,
(a) how many times will its surface area increase ?
(b) how many times will its volume increase ?
Solution:

(a) Let the edge of a cube = x
If each edge of a cube is doubled then edge
= 2x
If l = x
Now, surface area of cube
= 6 × (x)²
= 6x²
If l = 2x, then surface area of cube
= 6 × (2x)²
= 6 × 4x²
= 24x².
6x² × 3 = 24x²
Hence, three times will its surface area increase.

(b) If l = x,
Volume of cube = (l)³ = x³
If l = 2x,
Volume of cube = (2x)³ = 8x³
x³ × 8 = 8x³
Hence, eight times will its volume increase.

Question 8.
Water is pouring into a cuboidal reservoir at the rate of 60 litres per minute. If the volume of reservoir is 108 m³, find the number of hours it will take to fill the reservoir.

Solution:
Volume of cuboidal reservoir = 108 m³
Volume of cuboidal reservoir in litres
= 108 m³ = 108 × 1000 litres
(∵ 1 cm³ = 1000 l)
= 108000 litres
Since water is pouring into a cuboidal reservoir at the rate of 60 litres per minute.
Hence no. of the times take to fill the water
= \(\frac{10800}{60}\) = 1800 minutes
= \(\frac{10800}{60}\) = 30 hours.

Haryana State Board HBSE 8th Class Maths Solutions Chapter 11 Mensuration Ex 11.3 Textbook Exercise Questions and Answers.

Haryana Board 8th Class Maths Solutions Chapter 11 Mensuration Exercise 11.3

Question 1.
There are two cuboidal boxes as shown in the adjoining figure. Which box requires the lesser amount of material to make ?

Solution:
(a) Length = 60 cm
Breadth = 40 cm
Height = 50 cm
Total surface area of cuboidal boxes
= 2 (lb + bh + hl)
= 2 × (60 × 40 + 40 × 50 + 50 × 60)
= 2 × (240 + 200 + 300
= 2 × (740) = 1480 cm².

(b) l = b = h = 50 cm
Total surface area of cuboidal boxes = 6l²
= 6 × (50)² = 6 × 2500
= 15000 cm²
Hence (a) box requires the lesser amount of material to make.

Question 2.
A suitcase with measures 80 cm × 48 cm × 24 cm is to be covered with a tarpaulin cloth. How many metres of tarpaulin of width 96 cm is required to cover 100 such suitcases ?
Solution:
Total surface area of cuboidal suitcase
= 2 (lb + bh + hl)
= 2(80 × 48 + 48 × 24 + 24 × 80)
= 2 × (3840 + 1152 + 1920)
= 2 × 6912
= 13824 cm².

Question 3.
Find the side of a cube whose surface area is 600 cm².
Solution:
The total surface area of cube = 6l²
600 cm² = 6l²
\(\frac{600}{6}\) = l²
100 = l²
\(\sqrt {100}\) = l
10 = l
Hence the required side of a cube = 10 cm.

Question 4.
Rukhsar painted the outside of the cabinet of measure 1 m × 2 m × 1.5 m. How much surface area did she cover if she painted all except the bottom of the cabinet.

Solution:
l = 2 m, b = 1 m, h = 1.5 m
Lateral surface area of cuboidal cabinet
= 2 (l + b) × h
= 2(2 + 1) × 1.5
= 2 × 3 × 1.5 = 9 m²
Hence, required surface area covered = 9 m².

Question 5.
Daniel is painting the walls and ceiling of a cuboidal hall with length, breadth and height of 15 m, 10 m and 7 m respectively. From each can of paint 100 m2 of area is painted. How many cans of paint will ‘she need to paint the room ?
Solution:
l = 15 m, b = 10 m, h = 7 m
Lateral surface area of cuboidal classrom
= 2 (l + b) × h
= 2(15 + 10) × 7
= 2 × 25 × 7 = 350 m²
Area of ceiling = l × b
= 15 × 10 = 150 m²
Hence, the total area of painted wall and ceiling
= 350+ 150 = 500 m²
Since each can of paint covers 100 m² of area
Hence 500 m² of area needed can = \(\frac{500}{100}\) = 5
Hence, the required 5 cans of paint cover 500 m².

Question 6.
Describe how the two figures at the right are alike and how they are different. Which box surface area ?

Solution:
Lateral surface area of cylinder
= 2 × \(\frac{22}{7}\) × \(\frac{7}{22}\) × 7
[where = 7 cm
∴ r = \(\frac{7}{2}\) cm
h = 7 cm]
= 154 cm²
Lateral surface area of cube
= 4l² [where l = side = 7 cm]
= 4 × (7)²
= 4 × 49 = 196 cm²
Hence cubical box has larger lateral surface area.

Question 7.
A closed cylindrical tank of radius 7 m and height 3 m is made from a sheet of metal. How much sheet of metal is required ?

Solution:
Total surface area of cylindrical tank
= 2πr (r + h)
= 2 × \(\frac{22}{7}\) × 7 (7 + 3)
= 2 × 22 × 10 = 440 m²
Hence, the required metal sheet = 440 m².

Question 8.
The lateral surface area of a hollow cylinder is 4224 cm². It is cut along its height and formed a rectangular sheet of width 33 cm. Find the perimeter of rectangular sheet ?
Solution:
Width of rectangular sheet = 33 cm
Hence, height of cylinder = 33 cm
The lateral surface area = 2πrh
4224 = 2 × \(\frac{22}{7}\) × r × h
4224 = 2 × \(\frac{22}{7}\) × r × 33
\(\frac{4224 \times 7}{2 \times 22 \times 33}\) = r
20.36 = r
∴ Radius = 20.36
d = 2 × 20.36 = 40.72 cm
Now l = 40.72 and b = 33 cm
Hence, perimeter of rectangular sheet
= 2(l + b)
= 2 × (40.72 + 33)
= 2 × 73.22 = 147.44 cm.

Question 9.
A road roller takes 750 complete revolutions to move once over to level a road. Find the area of the road if the diameter of a road roller is 84 cm and length is 1 m.

Solution:
Diameter of road roller = 84 cm
∴ r = \(\frac{84}{2}\) = 2 cm
l = h = 1 m = 100 cm
Total surface area of road roller = 2πr(r + h)
= 2 × \(\frac{22}{7}\) × 42 × (42 + 100)
= 2 × 22 × 6 × 142
= 37488 cm²
Hence, the area of road
= 37488 × 750
= 28116000 cm²
\(\frac{28116000}{100×100}\) = 2811.6 m²
Hence, the required area of road = 2811.6 m².

Question 10.
A company packages its milk powder in cylindrical container whose base has a diameter of 14 cm and height 20 cm. Company places a label around the surface of the container (as shown in figure). If the label is placed 2 cm from top and bottom, what is the area of the label ?

Solution:
r = \(\frac{d}{2}\) = \(\frac{14}{2}\) = 7 cm
h = 20 cm
Top and bottom height = 2 + 2 = 4m
Hence required height of cylindrical powdered milk
= 20 – 4 = 16 cm
Lateral surface area of cylindrical powdered milk
= 2πrh = 2 × \(\frac{22}{7}\) × 7 × 16
= 2 × 22 × 16 = 704 cm²
Hence the required area of label = 704 cm².