S.P. of each buffalo = ₹ 20,000 S.P. of two buffaloes = ₹ 20,000 x 2 = ₹ 40,000 One buffalo is sold at 5% gain. Let C.P. be ₹ 100, then S.P. = 100 + 5 = ₹105 ∵ When S.P. is ₹ 105, then C.P. = ₹ 100 ∴ When S.P. is ₹ 1, then C.P. = 100/105 ∴ When S.P. is ₹ 20,000, then C.P. = 100/105 x 20000 = ₹ 19,04Read more
S.P. of each buffalo = ₹ 20,000
S.P. of two buffaloes = ₹ 20,000 x 2 = ₹ 40,000
One buffalo is sold at 5% gain.
Let C.P. be ₹ 100, then S.P. = 100 + 5 = ₹105
∵ When S.P. is ₹ 105, then C.P. = ₹ 100
∴ When S.P. is ₹ 1, then C.P. = 100/105
∴ When S.P. is ₹ 20,000, then C.P. = 100/105 x 20000 = ₹ 19,047.62
Another buffalo is sold at 10% loss.
Let C.P. be ₹ 100, then S.P. = 100 – 10 = ₹ 90
∵ When S.P. is ₹ 90, then C.P. = ₹ 100
∴ When S.P. is ₹ 1, then C.P. = 100/90
∴ When S.P. is ₹ 20,000, then C.P. = 100/90×20000 = ₹ 22,222.22
Total C.P. = ₹ 19,047.62 + ₹ 22,222.22 = ₹ 41,269.84
Since C.P. > S.P. Therefore here it is loss.
Loss = C.P. – S.P. = ₹ 41,269.84 – ₹ 40,000.00 = ₹ 1,269.84
Class 8 Maths Chapter 8 Exercise 8.2 Solution in Video
Class 8 Maths Chapter 8 Exercise 8.2 Solution in Video
Rate of discount on all items = 10% Marked Price of a pair of jeans = ₹ 1450 and Marked Price of a shirt = ₹ 850 Discount on a pair of jeans = Rate X M.P./100 = 10 X 1450/100 = ₹ 145 ∴ S.P. of a pair of jeans = ₹ 1450 – ₹ 145 = ₹ 1305 Marked Price of two shirts = 2 x 850 = ₹ 1700 Discount on two shiRead more
Rate of discount on all items = 10%
Marked Price of a pair of jeans = ₹ 1450 and Marked Price of a shirt = ₹ 850
Discount on a pair of jeans = Rate X M.P./100 = 10 X 1450/100 = ₹ 145
∴ S.P. of a pair of jeans = ₹ 1450 – ₹ 145 = ₹ 1305
Marked Price of two shirts = 2 x 850 = ₹ 1700
Discount on two shirts = Rate X M.P./100 = 10X1700/100 = ₹ 170
∴ S.P. of two shirts = ₹ 1700 – ₹ 170 = ₹ 1530
Therefore, the customer had to pay = 1305 + 1530 = ₹ 2,835
Class 8 Maths Chapter 8 Exercise 8.2 Solution in Video
Cost price of VCR = ₹ 8000 and Cost price of TV = ₹ 8000 Total Cost Price of both articles = ₹ 8000 + ₹ 8000 = ₹ 16,000 Now VCR is sold at 4% loss. Let C.P. of each article be ₹ 100, then S.P. of VCR = 100 – 4 = ₹ 96 ∵ When C.P. is ₹ 100, then S.P. = ₹ 96 ∴ When C.P. is ₹ 1, then S.P. = 96/100 ∴ WheRead more
Cost price of VCR = ₹ 8000 and Cost price of TV = ₹ 8000
Total Cost Price of both articles = ₹ 8000 + ₹ 8000 = ₹ 16,000
Now VCR is sold at 4% loss.
Let C.P. of each article be ₹ 100, then S.P. of VCR = 100 – 4 = ₹ 96
∵ When C.P. is ₹ 100, then S.P. = ₹ 96
∴ When C.P. is ₹ 1, then S.P. = 96/100
∴ When C.P. is ₹ 8000, then S.P. = 96/100×8000 =₹ 7,680
And TV is sold at 8% profit, then S.P. of TV = 100 + 8 = ₹ 108
∵ When C.P. is ₹ 100, then S.P. = ₹ 108
∴ When C.P. is ₹ 1, then S.P. = 108/100
∴ When C.P. is ₹ 8000, then S.P. = 108/100 = ₹ 8,640
Then, Total S.P. = ₹ 7,680 + ₹ 8,640 = ₹ 16,320
Since S.P. > C.P.,
Therefore Profit = S.P. – C.P. = 16320 – 16000 = ₹ 320
And Profit% = Profit/Cost Price x 100 = 320/16000 x 100 = 2%
Class 8 Maths Chapter 8 Exercise 8.2 Solution in Video
Here, C.P. = ₹ 15,500 and Repair cost = ₹ 450 Therefore Total Cost Price = 15500 + 450 = ₹ 15,950 Let C.P be ₹ 100, then S.P. = 100 + 15 = ₹ 115 ∵ When C.P. is ₹ 100, then S.P. = ₹ 115 ∴ When C.P. is ₹ 1, then S.P. = 115/100 ∴ When C.P. is ₹ 15950, then S.P. = 115/100x15950 = ₹ 18,342.50 Class 8 MatRead more
Here, C.P. = ₹ 15,500 and Repair cost = ₹ 450
Therefore Total Cost Price = 15500 + 450 = ₹ 15,950
Let C.P be ₹ 100, then S.P. = 100 + 15 = ₹ 115
∵ When C.P. is ₹ 100, then S.P. = ₹ 115
∴ When C.P. is ₹ 1, then S.P. = 115/100
∴ When C.P. is ₹ 15950, then S.P. = 115/100×15950 = ₹ 18,342.50
Class 8 Maths Chapter 8 Exercise 8.2 Solution in Video
On Sunday, people went to the Zoo = 845 On Monday, people went to the Zoo = 169 Number of decrease in the people = 845 – 169 = 676 Decrease percent = 676/845×100 = 80% Hence decrease in the people visiting the Zoo is 80%. Class 8 Maths Chapter 8 Exercise 8.2 Solution in Video for more answers vist tRead more
On Sunday, people went to the Zoo = 845
On Monday, people went to the Zoo = 169
Number of decrease in the people = 845 – 169 = 676
Decrease percent = 676/845×100 = 80%
Hence decrease in the people visiting the Zoo is 80%.
Class 8 Maths Chapter 8 Exercise 8.2 Solution in Video
Let original salary be ₹ 100. Therefore New salary i.e., 10% increase = 100 + 10 = ₹ 110 ∵ New salary is ₹ 110, when original salary = ₹ 100 ∴ New salary is ₹ 1, when original salary = 100/110 ∴ New salary is ₹ 1,54,000, when original salary = 100/110 x 154000 = ₹1,40,000 Hence original salary is ₹Read more
Let original salary be ₹ 100.
Therefore New salary i.e., 10% increase = 100 + 10 = ₹ 110
∵ New salary is ₹ 110, when original salary = ₹ 100
∴ New salary is ₹ 1, when original salary = 100/110
∴ New salary is ₹ 1,54,000, when original salary = 100/110 x 154000 = ₹1,40,000
Hence original salary is ₹ 1,40,000.
Class 8 Maths Chapter 8 Exercise 8.2 Solution in Video
Number of people who like cricket = 60% Number of people who like football = 30% Number of people who like other games = 100% – (60% + 30%) = 10% Now Number of people who like cricket = 60% of 50,00,000 = 60/100 x 50,00,000 = 30,00,000 And Number of people who like football = 30% of 50,00,000 = 30/1Read more
Number of people who like cricket = 60%
Number of people who like football = 30%
Number of people who like other games = 100% – (60% + 30%) = 10%
Now Number of people who like cricket = 60% of 50,00,000
= 60/100 x 50,00,000 = 30,00,000
And Number of people who like football = 30% of 50,00,000
= 30/100 x 50,00,000 = 15,00,000
∴ Number of people who like other games = 10% of 50,00,000
= 10/100 x 50,00,000 = 5,00,000
Hence, number of people who like other games are 5 lakh.
Class 8 Maths Chapter 8 Exercise 8.1 Solution in Video
Let her money in the beginning be ₹ x. According to question, x-75% of x=600 ⇒ x-75/100 X x=600 ⇒ x-(3/4)x = 600 ⇒ x (1- 3/4) = 600 ⇒ x(4-3/4) = 600 ⇒ x = 600 x 4 = ₹ 2400 Hence the money in the beginning was ₹ 2,400. Class 8 Maths Chapter 8 Exercise 8.1 Solution in Video for more answers vist to: hRead more
Let her money in the beginning be ₹ x.
According to question,
x-75% of x=600
⇒ x-75/100 X x=600
⇒ x-(3/4)x = 600
⇒ x (1- 3/4) = 600
⇒ x(4-3/4) = 600
⇒ x = 600 x 4 = ₹ 2400
Hence the money in the beginning was ₹ 2,400.
Class 8 Maths Chapter 8 Exercise 8.1 Solution in Video
Let total number of matches be x According to question, 40% of total matches = 10 ⇒ 40% of x = 10 ⇒ 40/100 X x ⇒ x=10x100/40 = 25 Hence, the total number of matches are 25. Class 8 Maths Chapter 8 Exercise 8.1 Solution in Video for more answers vist to: https://www.tiwariacademy.com/ncert-solutions/Read more
Let total number of matches be x
According to question, 40% of total matches = 10
⇒ 40% of x = 10 ⇒ 40/100 X x ⇒ x=10×100/40 = 25
Hence, the total number of matches are 25.
Class 8 Maths Chapter 8 Exercise 8.1 Solution in Video
A milkman sold two of his buffaloes for ₹ 20,000 each. On one he made a gain of 5% and on the other a loss of 10%. Find his overall gain or loss. (Hint: Find CP of each)
S.P. of each buffalo = ₹ 20,000 S.P. of two buffaloes = ₹ 20,000 x 2 = ₹ 40,000 One buffalo is sold at 5% gain. Let C.P. be ₹ 100, then S.P. = 100 + 5 = ₹105 ∵ When S.P. is ₹ 105, then C.P. = ₹ 100 ∴ When S.P. is ₹ 1, then C.P. = 100/105 ∴ When S.P. is ₹ 20,000, then C.P. = 100/105 x 20000 = ₹ 19,04Read more
S.P. of each buffalo = ₹ 20,000
S.P. of two buffaloes = ₹ 20,000 x 2 = ₹ 40,000
One buffalo is sold at 5% gain.
Let C.P. be ₹ 100, then S.P. = 100 + 5 = ₹105
∵ When S.P. is ₹ 105, then C.P. = ₹ 100
∴ When S.P. is ₹ 1, then C.P. = 100/105
∴ When S.P. is ₹ 20,000, then C.P. = 100/105 x 20000 = ₹ 19,047.62
Another buffalo is sold at 10% loss.
Let C.P. be ₹ 100, then S.P. = 100 – 10 = ₹ 90
∵ When S.P. is ₹ 90, then C.P. = ₹ 100
∴ When S.P. is ₹ 1, then C.P. = 100/90
∴ When S.P. is ₹ 20,000, then C.P. = 100/90×20000 = ₹ 22,222.22
Total C.P. = ₹ 19,047.62 + ₹ 22,222.22 = ₹ 41,269.84
Since C.P. > S.P. Therefore here it is loss.
Loss = C.P. – S.P. = ₹ 41,269.84 – ₹ 40,000.00 = ₹ 1,269.84
Class 8 Maths Chapter 8 Exercise 8.2 Solution in Video
Class 8 Maths Chapter 8 Exercise 8.2 Solution in Video
See lessDuring a sale, a shop offered a discount of 10% on the marked prices of all the items. What would a customer have to pay for a pair of jeans marked at ₹ 1450 and two shirts marked at ₹ 850 each?
Rate of discount on all items = 10% Marked Price of a pair of jeans = ₹ 1450 and Marked Price of a shirt = ₹ 850 Discount on a pair of jeans = Rate X M.P./100 = 10 X 1450/100 = ₹ 145 ∴ S.P. of a pair of jeans = ₹ 1450 – ₹ 145 = ₹ 1305 Marked Price of two shirts = 2 x 850 = ₹ 1700 Discount on two shiRead more
Rate of discount on all items = 10%
Marked Price of a pair of jeans = ₹ 1450 and Marked Price of a shirt = ₹ 850
Discount on a pair of jeans = Rate X M.P./100 = 10 X 1450/100 = ₹ 145
∴ S.P. of a pair of jeans = ₹ 1450 – ₹ 145 = ₹ 1305
Marked Price of two shirts = 2 x 850 = ₹ 1700
Discount on two shirts = Rate X M.P./100 = 10X1700/100 = ₹ 170
∴ S.P. of two shirts = ₹ 1700 – ₹ 170 = ₹ 1530
Therefore, the customer had to pay = 1305 + 1530 = ₹ 2,835
Class 8 Maths Chapter 8 Exercise 8.2 Solution in Video
for more answers vist to:
See lesshttps://www.tiwariacademy.com/ncert-solutions/class-8/maths/chapter-8/
A VCR and TV were bought for ₹ 8,000 each. The shopkeeper made a loss of 4% on the VCR and a profit of 8% on the TV. Find the gain or loss percent on the whole transaction.
Cost price of VCR = ₹ 8000 and Cost price of TV = ₹ 8000 Total Cost Price of both articles = ₹ 8000 + ₹ 8000 = ₹ 16,000 Now VCR is sold at 4% loss. Let C.P. of each article be ₹ 100, then S.P. of VCR = 100 – 4 = ₹ 96 ∵ When C.P. is ₹ 100, then S.P. = ₹ 96 ∴ When C.P. is ₹ 1, then S.P. = 96/100 ∴ WheRead more
Cost price of VCR = ₹ 8000 and Cost price of TV = ₹ 8000
Total Cost Price of both articles = ₹ 8000 + ₹ 8000 = ₹ 16,000
Now VCR is sold at 4% loss.
Let C.P. of each article be ₹ 100, then S.P. of VCR = 100 – 4 = ₹ 96
∵ When C.P. is ₹ 100, then S.P. = ₹ 96
∴ When C.P. is ₹ 1, then S.P. = 96/100
∴ When C.P. is ₹ 8000, then S.P. = 96/100×8000 =₹ 7,680
And TV is sold at 8% profit, then S.P. of TV = 100 + 8 = ₹ 108
∵ When C.P. is ₹ 100, then S.P. = ₹ 108
∴ When C.P. is ₹ 1, then S.P. = 108/100
∴ When C.P. is ₹ 8000, then S.P. = 108/100 = ₹ 8,640
Then, Total S.P. = ₹ 7,680 + ₹ 8,640 = ₹ 16,320
Since S.P. > C.P.,
Therefore Profit = S.P. – C.P. = 16320 – 16000 = ₹ 320
And Profit% = Profit/Cost Price x 100 = 320/16000 x 100 = 2%
Class 8 Maths Chapter 8 Exercise 8.2 Solution in Video
for more answers vist to:
See lesshttps://www.tiwariacademy.com/ncert-solutions/class-8/maths/chapter-8/
The cost of an article was ₹ 15,500, ₹ 450 were spent on its repairs. If it sold for a profit of 15%, find the selling price of the article.
Here, C.P. = ₹ 15,500 and Repair cost = ₹ 450 Therefore Total Cost Price = 15500 + 450 = ₹ 15,950 Let C.P be ₹ 100, then S.P. = 100 + 15 = ₹ 115 ∵ When C.P. is ₹ 100, then S.P. = ₹ 115 ∴ When C.P. is ₹ 1, then S.P. = 115/100 ∴ When C.P. is ₹ 15950, then S.P. = 115/100x15950 = ₹ 18,342.50 Class 8 MatRead more
Here, C.P. = ₹ 15,500 and Repair cost = ₹ 450
Therefore Total Cost Price = 15500 + 450 = ₹ 15,950
Let C.P be ₹ 100, then S.P. = 100 + 15 = ₹ 115
∵ When C.P. is ₹ 100, then S.P. = ₹ 115
∴ When C.P. is ₹ 1, then S.P. = 115/100
∴ When C.P. is ₹ 15950, then S.P. = 115/100×15950 = ₹ 18,342.50
Class 8 Maths Chapter 8 Exercise 8.2 Solution in Video
for more answers vist to:
See lesshttps://www.tiwariacademy.com/ncert-solutions/class-8/maths/chapter-8/
A shopkeeper buys 80 articles for ₹ 2,400 and sells them for a profit of 16%. Find the selling price of one article.
No. of articles = 80 Cost Price of articles = ₹ 2,400 And Profit = 16% ∵ Cost price of articles is ₹ 100, then selling price = 100 + 16 = ₹ 116 ∴ Cost price of articles is ₹ 1, then selling price = 116/100 ∴ Cost price of articles is ₹ 2400, then selling price = 116/100x2400 = ₹ 2784 Hence, SellingRead more
No. of articles = 80
Cost Price of articles = ₹ 2,400
And Profit = 16%
∵ Cost price of articles is ₹ 100, then selling price = 100 + 16 = ₹ 116
∴ Cost price of articles is ₹ 1, then selling price = 116/100
∴ Cost price of articles is ₹ 2400, then selling price = 116/100×2400 = ₹ 2784
Hence, Selling Price of 80 articles = ₹ 2784
Therefore Selling Price of 1 article = 2784/80 = ₹ 34.80
Class 8 Maths Chapter 8 Exercise 8.2 Solution in Video
for more answers vist to:
See lesshttps://www.tiwariacademy.com/ncert-solutions/class-8/maths/chapter-8/
On Sunday 845 people went to the Zoo. On Monday only 169 people went. What is the percent decrease in the people visiting the Zoo on Monday?
On Sunday, people went to the Zoo = 845 On Monday, people went to the Zoo = 169 Number of decrease in the people = 845 – 169 = 676 Decrease percent = 676/845×100 = 80% Hence decrease in the people visiting the Zoo is 80%. Class 8 Maths Chapter 8 Exercise 8.2 Solution in Video for more answers vist tRead more
On Sunday, people went to the Zoo = 845
On Monday, people went to the Zoo = 169
Number of decrease in the people = 845 – 169 = 676
Decrease percent = 676/845×100 = 80%
Hence decrease in the people visiting the Zoo is 80%.
Class 8 Maths Chapter 8 Exercise 8.2 Solution in Video
for more answers vist to:
See lesshttps://www.tiwariacademy.com/ncert-solutions/class-8/maths/chapter-8/
A man got 10% increase in his salary. If his new salary is ₹ 1,54,000, find his original salary.
Let original salary be ₹ 100. Therefore New salary i.e., 10% increase = 100 + 10 = ₹ 110 ∵ New salary is ₹ 110, when original salary = ₹ 100 ∴ New salary is ₹ 1, when original salary = 100/110 ∴ New salary is ₹ 1,54,000, when original salary = 100/110 x 154000 = ₹1,40,000 Hence original salary is ₹Read more
Let original salary be ₹ 100.
Therefore New salary i.e., 10% increase = 100 + 10 = ₹ 110
∵ New salary is ₹ 110, when original salary = ₹ 100
∴ New salary is ₹ 1, when original salary = 100/110
∴ New salary is ₹ 1,54,000, when original salary = 100/110 x 154000 = ₹1,40,000
Hence original salary is ₹ 1,40,000.
Class 8 Maths Chapter 8 Exercise 8.2 Solution in Video
for more answers vist to:
See lesshttps://www.tiwariacademy.com/ncert-solutions/class-8/maths/chapter-8/
If 60% people in a city like cricket, 30% like football and the remaining like other games, then what percent of the people like other games? If the total number of people are 50 lakh, find the exact number who like each type of game.
Number of people who like cricket = 60% Number of people who like football = 30% Number of people who like other games = 100% – (60% + 30%) = 10% Now Number of people who like cricket = 60% of 50,00,000 = 60/100 x 50,00,000 = 30,00,000 And Number of people who like football = 30% of 50,00,000 = 30/1Read more
Number of people who like cricket = 60%
Number of people who like football = 30%
Number of people who like other games = 100% – (60% + 30%) = 10%
Now Number of people who like cricket = 60% of 50,00,000
= 60/100 x 50,00,000 = 30,00,000
And Number of people who like football = 30% of 50,00,000
= 30/100 x 50,00,000 = 15,00,000
∴ Number of people who like other games = 10% of 50,00,000
= 10/100 x 50,00,000 = 5,00,000
Hence, number of people who like other games are 5 lakh.
Class 8 Maths Chapter 8 Exercise 8.1 Solution in Video
for more answers vist to:
See lesshttps://www.tiwariacademy.com/ncert-solutions/class-8/maths/chapter-8/
If Chameli had ₹ 600 left after spending 75% of her money, how much did she have in the beginning?
Let her money in the beginning be ₹ x. According to question, x-75% of x=600 ⇒ x-75/100 X x=600 ⇒ x-(3/4)x = 600 ⇒ x (1- 3/4) = 600 ⇒ x(4-3/4) = 600 ⇒ x = 600 x 4 = ₹ 2400 Hence the money in the beginning was ₹ 2,400. Class 8 Maths Chapter 8 Exercise 8.1 Solution in Video for more answers vist to: hRead more
Let her money in the beginning be ₹ x.
According to question,
x-75% of x=600
⇒ x-75/100 X x=600
⇒ x-(3/4)x = 600
⇒ x (1- 3/4) = 600
⇒ x(4-3/4) = 600
⇒ x = 600 x 4 = ₹ 2400
Hence the money in the beginning was ₹ 2,400.
Class 8 Maths Chapter 8 Exercise 8.1 Solution in Video
for more answers vist to:
See lesshttps://www.tiwariacademy.com/ncert-solutions/class-8/maths/chapter-8/
A football team won 10 matches out of the total number of matches they played. If their win percentage was 40, then how many matches did they play in all?
Let total number of matches be x According to question, 40% of total matches = 10 ⇒ 40% of x = 10 ⇒ 40/100 X x ⇒ x=10x100/40 = 25 Hence, the total number of matches are 25. Class 8 Maths Chapter 8 Exercise 8.1 Solution in Video for more answers vist to: https://www.tiwariacademy.com/ncert-solutions/Read more
Let total number of matches be x
According to question, 40% of total matches = 10
⇒ 40% of x = 10 ⇒ 40/100 X x ⇒ x=10×100/40 = 25
Hence, the total number of matches are 25.
Class 8 Maths Chapter 8 Exercise 8.1 Solution in Video
for more answers vist to:
See lesshttps://www.tiwariacademy.com/ncert-solutions/class-8/maths/chapter-8/