Class 8

Balbharti Maharashtra State Board Class 8 Maths Solutions covers the Practice Set 9.1 8th Std Maths Answers Solutions Chapter 9 Discount and Commission.

Practice Set 9.1 8th Std Maths Answers Chapter 9 Discount and Commission

8th Standard Maths Practice Set 9.1 Question 1. If marked price = Rs 1700, selling price = Rs 1540, then find the discount.
Solution:
Here, Marked price = Rs 1700,
selling price = Rs 1540
Selling price = Marked price – Discount
∴ 1540 = 1700 – Discount
∴ Discount = 1700 – 1540
= Rs 160
∴ The amount of discount is Rs 160.

Discount and Commission Practice Set 9.1 Question 2. If marked price Rs 990 and percentage of discount is 10, then find the selling price.
Solution:
Here, marked price = Rs 990,
discount = 10%
Let the percentage of discount be x
∴ x = 10%
i. Discount

= Rs 99

ii. Selling price = Marked price – Discount
= 990 – 99
= Rs 891
∴ The selling price is Rs 891.

Practice Set 9.1 Question 3. If selling price Rs 900, discount is 20%, then find the marked price.
Solution:
Here, selling price = Rs 900, discount = 20%
Let the marked price be Rs 100
Since, the discount given = 20%
∴ Amount of discount = Rs 20
∴ Selling price = 100 – 20 – Rs 80
Let actual marked price be Rs x
∴ For marked price of Rs x, selling price is Rs 900
\(\frac{80}{100}=\frac{900}{x}\)
∴ 80 × x = 100 × 900
∴ \(x=\frac{100 \times 900}{80}\)
∴ x = Rs 1125
∴ The marked price is Rs 1125.

Discount and Commission Std 8 Question 4. The marked price of the fan is Rs 3000. Shopkeeper gave 12% discount on it. Find the total discount and selling price of the fan.
Solution:
Here, Marked price = Rs 3000, discount = 12%
Let the percentage of discount be x.
∴ x = 12%
i. Discount

= 30 × 12
= Rs 360

ii. Selling price = Marked price – Discount
= 3000 – 360
= Rs 2640
∴ The total discount is Rs 360 and the selling price of the fan is Rs 2640.

Discount and Commission 8th Standard Question 5. The marked price of a mixer is Rs 2300. A customer purchased it for Rs 1955. Find percentage of discount offered to the customer.
Solution:
Here, marked price = Rs 2300,
selling price = Rs 1955
i. Selling price = Marked price – Discount
∴ 1955 = 2300 – Discount
∴ Discount = 2300 – 1955
= Rs 345

ii. Let the percentage of discount be x


∴ x = 15%
∴ The percentage of discount offered to the customer is 15%.

Question 6.
A shopkeeper gives 11% discount on a television set, hence the cost price of it is Rs 22,250. Then find the marked price of the television set.
Solution:
Here, selling price = Rs 22,250, discount = 11%
Let marked price be Rs 100
Since, the discount given = 11%
∴ Amount of discount = Rs 11
∴ Selling price = 100 – 11 = Rs 89
∴ Let actual marked price be Rs x
∴ For marked price of Rs x, selling price is Rs 22,250

∴ x = Rs 25,000
∴ The marked price of the television set is Rs 25,000.

8th Std Maths Discount and Commission Question 7. After offering discount of 10% on marked price, a customer gets total discount of Rs 17. To find the cost price for the customer, fill in the following boxes with appropriate numbers and complete the activity.
Solution:
Suppose, marked price of the item = 100 rupees Therefore, for customer that item costs 100 – 10 = 90 rupees.
Hence, when the discount is [10] then the selling price is [90] rupees.
Suppose when the discount is [17] rupees, the selling price is x rupees.

∴ The customer will get the item for Rs 153.

Question 8.
A shopkeeper decides to sell a certain item at a certain price. He tags the price on the item by increasing the decided price by 25%. While selling the item, he offers 20% discount. Find how many more or less percent he gets on the decided price.
Solution:
Here, price increase = 25%,
discount offered = 20%
Let the decided price be Rs 100
∴ Increase in price = Rs 25
∴ Shopkeeper marks the price = 100 + 25
= Rs 125
∴ marked price = Rs 125
Let the percentage of discount be x
∴ x = 20%

∴ Selling price = Marked price – Discount
= 125 – 25
= Rs 100
∴ If the decided price is Rs 100, then shopkeeper gets Rs 100.
∴ The shopkeeper gets neither more nor less than the decided price i.e. he gets 0% more / less.

Maharashtra Board Class 8 Maths Chapter 9 Discount and Commission Practice Set 9.1 Intext Questions and Activities

Question 1.
Write the appropriate numbers in the following boxes. (Textbook pg. no. 51)

1. \(\frac { 12 }{ 100 }=\) __ percent = __%
2. 47% = __
3. 86% = __
4. 4% of 300 = 300 × __ = __
5. 15% of 1700 = 1700 × __= __

Solution:

1. \(\frac { 12 }{ 100 }=\) 12 percent = 12%
2. 47% = \(\frac { 47 }{ 100 }\)
3. 86% = \(\frac { 86 }{ 100 }\)
4. 4% of 300 = 300 × \(\frac { 4 }{ 100 }\) = 12
5. 15% of 1700 = 1700 × \(\frac { 15 }{ 100 }\) = 255

Question 2.
You may have seen advertisements like ‘Monsoon Sale’, ‘Stock Clearance Sale’ etc offering different discount. In such a sale, a discount is offered on various goods. Generally in the month of July, sales of clothes are declared. Find and discuss the purpose of such sales. (Textbook pg. no. 51)
Solution:
(Students should attempt the above activity on their own)

Maharashtra Board Class 8 Maths Solutions

Balbharti Maharashtra State Board Class 8 Maths Solutions covers the Practice Set 10.2 8th Std Maths Answers Solutions Chapter 10 Division of Polynomials.

Practice Set 10.2 8th Std Maths Answers Chapter 10 Division of Polynomials

Division of Polynomials Class 8 Practice Set 10.2 Question 1. Divide and write the quotient and the remainder.
i. (y² + 10y + 24) ÷ (y + 4)
ii. (p² + 7p – 5) ÷ (p + 3)
iii. (3x + 2x² + 4x³) ÷ (x – 4)
iv. (2m³ + m² + m + 9) ÷ (2m – 1)
v. (3x – 3x² – 12 + x⁴ + x³) ÷ (2 + x²)
vi. (a⁴ – a³ + a² – a + 1) ÷ (a³ – 2)
vii. (4x⁴ – 5x³ – 7x + 1) ÷ (4x – 1)
Solution:
i. (y² + 10y + 24) ÷ (y + 4)

∴ Quotient = y + 6
Remainder = 0

ii. (p² + 7p – 5) ÷ (p + 3)

∴ Quotient = p + 4
Remainder = -17

iii. (3x + 2x² + 4x³) ÷ (x – 4)
Write the dividend in descending order of their indices.
3x + 2x² + 4x³ = 4x³ + 2x² + 3x

∴ Quotient = 4x² + 18x + 75
Remainder = 300

iv. (2m³ + m² + m + 9) ÷ (2m – 1)

∴ Quotient = m² + m + 1
Remainder = 10

v. (3x – 3x² – 12 + x⁴ + x³) ÷ (2 + x²)
Write the dividend in descending order of their indices.
(x⁴ + x³ – 3x² + 3x – 12) ÷ (x² + 2)

∴ Quotient = x² + x – 5
Remainder = x – 2

vi. (a⁴ – a³ + a² – a + 1) ÷ (a³ – 2)

∴ Quotient = a – 1
Remainder = a² + a – 1

vii. (4x⁴ – 5x³ – 7x + 1) ÷ (4x – 1)
Write the dividend in descending order of their indices.
(4x⁴ – 5x³ – 7x + 1) = (4x⁴ – 5x³ + 0x² – 7x + 1)

∴ Quotient = \(x^{3}-x^{2}-\frac{x}{4}-\frac{29}{16}\)
Remainder = \(\frac { -13 }{ 16 }\)

Balbharti Maharashtra State Board Class 8 Maths Solutions covers the Practice Set 11.2 8th Std Maths Answers Solutions Chapter 11 Statistics.

Practice Set 11.2 8th Std Maths Answers Chapter 11 Statistics

practice set 11.2 8th class Question 1.
Observe the following graph and answer the questions.

i. State the type of the graph.
ii. How much is the savings of Vaishali in the month of April?
iii. How much is the total of savings of Saroj in the months March and April?
iv. How much more is the total savings of Savita than the total savings of Megha?
v. Whose savings in the month of April is the least?
Solution:
i. The given graph is a subdivided bar graph.
ii. Vaishali’s savings in the month of April is Rs 600.
iii. Total savings of Saroj in the months of March and April is Rs 800.
iv. Savita’s total saving = Rs 1000, Megha’s total saving = Rs 500
∴ difference in their savings = 1000 – 500 = Rs 500.
Savita’s saving is Rs 500 more than Megha.
v. Megha’s savings in the month of April is the least.

practice set 11.2 Question 2.
The number of boys and girls, in std 5 to std 8 in a Z.P. School is given in the table. Draw a subdivided bar graph to show the data. (Scale : On Y axis, 1cm = 10 students)

Standard	5th	6th	7th	8th
Boys	34	26	21	25
Girls	17	14	14	20

Solution:

Standard	5th	6th	7th	8th
Boys	34	26	21	25
Girls	17	14	14	20
Total	51	40	35	45

Statistics class 8 practice set 11.1 Question 3.
In the following table number of trees planted in the year 2016 and 2017 in four towns is given. Show the data with the help of subdivided bar graph.

Year\Town	karjat	Wadgaon	Shivapur	Khandala
2016	150	250	200	100
2017	200	300	250	150

Solution:

Year\Town	karjat	Wadgaon	Shivapur	Khandala
2016	150	250	200	100
2017	200	300	250	150
Total	350	550	450	250

Statistics class 8 Question 4.
In the following table, data of the transport means used by students in 8th standard for commutation between home and school is given. Draw a subdivided bar diagram to show the data.
(Scale: On Y axis: 1 cm = 500 students)

Means of commutation\Town	Paithan	Yeola	Shahapur
Cycle	3250	1500	1250
Bus and auto	750	500	500
On foot	1000	1000	500

Solution:

Means of commutation\Town	Paithan	Yeola	Shahapur
Cycle	3250	1500	1250
Bus and auto	750	500	500
On foot	1000	1000	500
Total	5000	3000	2250

Balbharti Maharashtra State Board Class 8 Maths Solutions covers the Practice Set 5.4 8th Std Maths Answers Solutions Chapter 5 Expansion Formulae.

Practice Set 5.4 8th Std Maths Answers Chapter 5 Expansion Formulae

Question 1.
Expand:
i. (2p + q + 5)²
ii. (m + 2n + 3r)²
iii. (3x + 4y – 5p)²
iv. (7m – 3n – 4k)²
Solution:
i. (2p + q + 5)² = (2p)² + (q)² + (5)² + 2(2p) (q) + 2(q) (5) + 2(2p) (5)
… [(a + b + c)² = a² + b² + c² + 2ab + 2bc + 2ac]
= 4p² + q² + 25 + 4pq + 10q + 20p

ii. (m + 2n + 3r)² = (m)² + (2n)² + (3r)² + 2(m) (2n) + 2(2n) (3r) + 2(m) (3r)
… [(a + b + c)² = a² + b² + c² + 2ab + 2bc + 2ac]
= m² + 4n² + 9r² + 4mn + 12nr + 6mr

iii. (3x + 4y – 5p)² = (3x)² + (4y)² + (- 5p)² + 2(3x) (4y) + 2(4y) (- 5p) + 2(3x) (- 5p)
… [(a + b + c)² = a² + b² + c² + 2ab + 2bc + 2ac]
= 9x + 16y² + 25p² + 24xy – 40py – 30px

iv. (7m – 3n – 4k)² = (7m)² + (- 3n)² + (- 4k)² + 2(7m) (- 3n) + 2 (- 3n) (- 4k) + 2 (7m) (- 4k)
… [(a + b + c)² = a² + b² + c² + 2ab + 2bc + 2ac]
= 49m² + 9n² + 16k² – 42mn + 24nk – 56km

Question 2.
Simplify:
i. (x – 2y + 3)² + (x + 2y – 3)²
ii. (3k – 4r – 2m)² – (3k + 4r – 2m)²
iii. (7a – 6b + 5c)² + (7a + 6b – 5c)²
Solution:
i. (x – 2y + 3)² + (x + 2y – 3)²
= [(x)² + (- 2y)² + (3)² + 2 (x) (- 2y) + 2 (- 2y) (3) + 2 (x) (3)] + [(x)² + (2y)² + (- 3)² + 2 (x) (2y) + 2 (2y) (- 3) + 2 (x) (- 3)]
… [(a + b + c)² = a² + b² + c² + 2ab + 2bc + 2ac]
= x² + 4y² + 9 – 4xy – 12y + 6x + x² + 4y² + 9 + 4xy – 12y – 6x
= x + x² + 4y² + 4y² + 9 + 9 – 4xy + 4xy – 12y – 12y + 6x – 6x
= 2x² + 8y² + 18 – 24y

ii. (3k – 4r – 2m)² – (3k + 4r – 2m)²
= [(3k)² + (- 4r)² + (- 2m)² + 2 (3k) (- 4r) + 2 (- 4r) (- 2m) + 2 (3k) (- 2m)] – [(3k)² + (4r)² + (- 2m)² + 2 (3k) (4r) + 2 (4r) (- 2m) + 2 (3k) (- 2m)]
… [(a + b + c)² = a² + b² + c² + 2ab + 2bc + 2ac]
= (9k² + 16r² + 4m² – 24kr + 16rm – 12km) – (9k² + 16r² + 4m² + 24kr – 16rm – 12km)
= 9k² + 16r² + 4m² – 24kr + 16rm – 12km – 9k² – 16r² – 4m² – 24kr + 16rm + 12km
= 9k² – 9k² + 16r² – 16r² + 4m² – 4m² – 24kr – 24kr + 16rm + 16rm – 12km + 12km
= 32rm – 48kr

iii. (7a – 6b + 5c)² + (7a + 6b – 5c)²
= [(7a)² + (- 6b)² + (5c)² + 2(7a) (-6b) + 2(-6b) (5c) + 2(7a) (5c)] + [(7a)² + (6b)² + (- 5c)² + 2 (7a) (6b) + 2 (6b) (- 5c) + 2 (7a) (- 5c)]
… [(a + b + c)² = a² + b² + c² + 2ab + 2bc + 2ac]
= 49a² + 36b² + 25c² – 84ab – 60bc + 70ac + 49a² + 36b² + 25c² + 84ab – 60bc – 70ac
= 49a² + 49a² + 36b² + 36b² + 25c² + 25c² – 84ab + 84ab – 60bc – 60bc + 70ac – 70ac
= 98a² + 72b² + 50c² – 120bc

Maharashtra Board Class 8 Maths Chapter 5 Expansion Formulae Practice Set 5.4 Intext Questions and Activities

Question 1.
Fill in the boxes with appropriate terms in the steps of expansion. (Textbook pg. no. 27)
(2p + 3m + 4n)²
= (2p)² + (3m)² + __ + 2 × 2p × 3m + 2 × __ × 4n + 2 × 2p × __
= __ + 9m² + __ + 12pm + __ + __
Solution:
(2p + 3m + 4n)²
= (2p)² + (3m)² + (4n)² + 2 x 2p x 3m + 2 x 3m x 4n + 2 x 2p x 4n
= 4p² + 9m² + 16n² + 12pm + 24mn + 16pn

Balbharti Maharashtra State Board Class 8 Maths Solutions covers the Practice Set 2.2 8th Std Maths Answers Solutions Chapter 2 Parallel Lines and Transversals.

Practice Set 2.2 8th Std Maths Answers Chapter 2 Parallel Lines and Transversals

Question 1.
Choose the correct alternative.
i. In the given figure, if line m || line n and line p is a transversal, then find x.

(A) 135°
(B) 90°
(C) 45°
(D) 40°
Solution:
(C) 45°

Hint:

line m || line n and line p is a transversal.
∴ m∠BFG + m∠FGD = 180°
…[Interior angles]
∴ 3x + x = 180°
∴ 4x = 180°
∴ x = \(\frac { 180 }{ 4 }\)
∴ x = 45°

ii. In the given figure, if line a || line b and line l is a transversal, then find x.

(A) 90°
(B) 60°
(C) 45°
(D) 30°
Solution:
(D) 30°

Hint:

line a || line b and line l is a transversal.
∴ m∠UVS = m∠PUV
…[Alternate angles]
= 4x
m∠UVS + m∠WVS = 180°
… [Angles in a linear pair]
∴ 4x + 2x = 180°
∴ 6x = 180°
∴ x = \(\frac { 180 }{ 6 }\)
∴ x = 30°

Question 2.
In the given figure, line p || line q. Line t and line s are transversals. Find measure of ∠x and ∠y using the measures of angles given in the figure.

Solution:

i. Consider ∠z as shown in figure.
line p || line q and line t is a transversal.
∴ m∠z = 40° …(i) [Corresponding angles]
m∠x + m∠z = 180° …[Angles in a linear pair]
∴ m∠x + 40o = 180° …[From(i)]
∴ m∠x= 180° – 40°
∴ m∠x = 140°

ii. Consider ∠w as shown in the figure.
m∠w + 70° = 180° …[Angles in a linear pair]
∴ m∠w = 180° – 70°
∴ m∠w = 110° …(ii)
line p || line q and line s is a transversal.
∴ m∠y = m∠w …[Alternate angles]
∴ m∠y =110° …[From (ii)]
∴ m∠x = 140°, m∠y = 110°

Question 3.
In the given figure, line p || line q, line l || line m. Find measures of ∠a, ∠b and ∠c, using the measures of given angles. Justify your answers.

Solution:
i. line p || line q and line l is a transversal.
∴ m∠a + 80° = 180° …[Interior angles]
∴ m∠a= 180° – 80°
∴ m∠a= 100°

ii. line l || line m and line p is a transversal.
∴ m∠c = 80° …(i) [Exterior alternate angles]

iii. line p || line q and line m is a transversal.
∴ m∠b = m∠c … [Corresponding angles]
m∠b = 80° …[From (i)]
∴ m∠a = 100°, m∠b = 80°, m∠c = 80°

Question 4.
In the given figure, line a || line b, line l is a transversal. Find the measures of ∠x, ∠y, ∠z using the given information.

Solution:
line a || line b and line l is a transversal.
∴ m∠x = 105° …(i) [Corresponding angles]

ii. m∠y = m∠x … [Vertically opposite angles]
∴ m∠y = 105° …[From (i)]

iii. m∠z + 105° = 180° …[Angles in a linear pair]
∴ m∠z = 180°- 105°
∴ m∠z = 75°
∴ m∠x = 105°, m∠y = 105°, m∠z = 75°

Question 5.
In the given figure, line p || line l || line q. Find ∠x with the help of the measures given in the figure.

Solution:

line p || line l and line IJ is a transversal.
m∠IJN = m∠JIH … [Alternate angles]
∴ m∠IJN = 40° …(i)
line l || line q and line MJ is a transversal.
m∠MJN = m∠JMK … [Alternate angles]
∴ m∠MJN = 30° …(ii)
Now, m∠x = m∠IJN + m∠MJN
…[Angle addition property]
= 40° + 30° …[From (i) and (ii)]
∴ m∠x = 70°

Maharashtra Board Class 8 Maths Chapter 2 Parallel Lines and Transversals Practice Set 2.2 Intext Questions and Activities

Question 1.
When two parallel lines are intersected by a transversal eight angles are formed. If the measure of one of these eight angles is given, can we find measures of remaining seven angles? (Textbook pg, no. 9)
Solution:
Yes, we can find the measures of the remaining angles.

In the given figure, line m || line n and line l is a transversal.
m∠a = 60°(say) …(i)
i. m∠a + m∠b = 180° …[Angles in a linear pair]
∴ 60° + m∠b =180° … [From (i)]
∴ m∠b = 180° – 60°
∴ m∠b = 120° …(ii)

ii. m∠c = m∠b …[Vertically opposite angles]
∴ m∠c = 120° .. .(iii) [From (ii)]

iii. m∠d = m∠a …[Vertically opposite angles]
∴ m∠d = 60° …(iv) [From (i)]

iv. m∠e = m∠d …[Alternate angles]
∴ m∠e = 60° … [From (iv)]

v. m∠f = m∠c …[Alternate angles]
∴ m∠f = 120° …[From (iii)]

vi. m∠g = m∠d …[Corresponding angles]
∴ m∠g = 60° … [From (iv)]

vii. m∠h = m∠c … [Corresponding angles]
∴ m∠h = 120° …[From (iii)]

Question 2.
As shown in the figure (A), draw two parallel lines and their transversal on a paper. Draw a copy of the figure on another blank sheet using a trace paper, as shown in the figure (B). Colour part Land part II with different colours. Cut out the two parts with a pair of scissors. Place, part I and part II on each angle in the figure A and answer the following questions. (Textbook pg. no. 9)

1. Which angles coincide with part I?
2. Which angles coincide with part II?

Solution:

1. ∠d, ∠f and ∠h coincide with part I.
2. ∠c, ∠e and ∠g coincide with part II.

Balbharti Maharashtra State Board Class 7 Maths Solutions covers the 7th Std Maths Practice Set 10 Answers Solutions Chapter 3 HCF and LCM.

HCF and LCM Class 7 Practice Set 10 Answers Solutions Chapter 3

Question 1.
Which number is neither a prime number nor a composite number?
Solution:
1

Question 2.
Which of the following are pairs of co-primes?
i. 8,14
ii. 4,5
iii. 17,19
iv. 27,15
Solution:
i. Factors of 8: 1, 2, 4, 8
Factors of 14: 1, 2, 7, 14
∴ Common factors of 8 and 14: 1,2
∴ 8 and 14 are not a pair of co-prime numbers.

ii. Factors of 4: 1, 4
Factors of 5: 1, 5
∴ Common factors of 4 and 5: 1
∴ 4 and 5 are a pair of co-prime numbers.

iii. Factors of 17: 1, 17
Factors of 19: 1, 19
∴ Common factors of 17 and 19: 1
∴ 17 and 19 are a pair of co-prime numbers.

iv. Factors of 27: 1, 3, 9, 27
Factors of 15: 1, 3, 5, 15 .
∴ Common factors of 27 and 15 : 1,3
∴ 27 and 15 are not a pair of co-prime numbers.

Question 3.
List the prime numbers from 25 to 100 and say how many they are.
Solution:
29, 31, 37, 41, 43, 47, 53, 59, 61, 67, 71, 73, 79, 83, 89, 97.
There are 16 prime numbers from 25 to 100.

Question 4.
Write all the twin prime numbers from 51 to 100.
Solution:

1. 59 and 61
2. 71 and 73

Question 5.
Write 5 pairs of twin prime numbers from 1 to 50.
Solution:

1. 3,5
2. 5,7
3. 11,13
4. 17,19
5. 29,31
6. 41,43

Question 6.
Which are the even prime numbers?
Solution:
2

Maharashtra Board Class 7 Maths Chapter 3 HCF and LCM Practice Set 10 Intext Questions and Activities

Question 1.
Answer the following questions. (Textbook pg. no. 15)
i. Which is the smallest prime number?
ii. List the prime numbers from 1 to 50. How many are they?
iii. Identify the prime numbers in the list below.
17, 15 ,4, 3, 1, 2, 12, 23, 27, 35, 41, 43, 58, 51, 72, 79, 91, 97.
Solution:
i. 2 is the smallest prime number.
ii. There are 15 prime numbers from 1 to 50.
They are: 2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47.
iii. [17], 15 ,4, [3], 1, [2], 12, [23], 27, 35, [41], [43], 58, 51, 72, [79], 91, [97].