Here, Principal (P) = ₹56,000, Simple Interest (S.I.) = ₹280, Time (T) = 2 years Simple Interest = PXRXT/100 ⇒ 280 = 56000 X R X 2/100 ⇒ R= 280X100/5600x2 ⇒ R = 0.25% Hence, the rate of interest on sum is 0.25%. lass 7 Maths Chapter 8 Exercise 8.3 for more answers vist to: https://www.tiwariacademy.Read more
Here, Principal (P) = ₹56,000, Simple Interest (S.I.) = ₹280, Time (T) = 2 years
Simple Interest = PXRXT/100
⇒ 280 = 56000 X R X 2/100
⇒ R= 280X100/5600×2
⇒ R = 0.25%
The cost of a book = ₹275 Loss percent = 15% Loss = Loss% of C.P. = 15% of ₹275 = 15/100 x 275 = ₹41.25 Therefore, S.P. = C.P. – Loss = ₹275 – ₹41.25 = ₹233.75 Hence, Amina sells a book for ₹233.75. lass 7 Maths Chapter 8 Exercise 8.3 for more answers vist to: https://www.tiwariacademy.com/ncert-solRead more
The cost of a book = ₹275
Loss percent = 15%
Loss = Loss% of C.P. = 15% of ₹275
= 15/100 x 275 = ₹41.25
Therefore, S.P. = C.P. – Loss = ₹275 – ₹41.25 = ₹233.75
Hence, Amina sells a book for ₹233.75.
(i) Given ratio = 10 : 3 : 12 Total part = 10 + 3 + 12 = 25 Part of Carbon = 3/25 Percentage of Carbon part in chalk = 3/25x100 = 12% (ii) Quantity of Carbon in chalk stick = 3 g Let the weight of chalk be x g. Then, 12% of x = 3 ⇒ 12/100 X x=3 ⇒ x = 3x100/12 = 25 g Hence, the weight of chalk stickRead more
(i) Given ratio = 10 : 3 : 12
Total part = 10 + 3 + 12 = 25
Part of Carbon = 3/25
Percentage of Carbon part in chalk = 3/25×100 = 12%
(ii) Quantity of Carbon in chalk stick = 3 g
Let the weight of chalk be x g.
Then, 12% of x = 3
⇒ 12/100 X x=3
⇒ x = 3×100/12 = 25 g
Hence, the weight of chalk stick is 25 g.
Selling price of washing machine = ₹13,500 Loss percent = 20% Let the cost price of washing machine be ₹ x. Since, Loss = Loss% of C.P. ⇒ Loss = 20% of ₹ x = 20/100X x 20/100 X x =x/5 Therefore, S.P. = C.P. – Loss ⇒ 13500 = x - x/5 ⇒ 13500 = 4x/5 ⇒ x= 13500X5/4 = ₹16,875 Hence, the cost price of wasRead more
Selling price of washing machine = ₹13,500
Loss percent = 20%
Let the cost price of washing machine be ₹ x.
Since, Loss = Loss% of C.P.
⇒ Loss = 20% of ₹ x = 20/100X x 20/100 X x =x/5
Therefore, S.P. = C.P. – Loss
⇒ 13500 = x – x/5
⇒ 13500 = 4x/5
⇒ x= 13500X5/4 = ₹16,875
Hence, the cost price of washing machine is ₹16,875.
Increased in price of a car from ₹3,50,000 to ₹3,70,000. Amount change = ₹3,70,000 – ₹3,50,000 = ₹20,000. Therefore, Increased percentage = Amount of change/Original amount x 100 = 20000/350000 x 100 = (5)5/7% Hence, the percentage of price increased is (5)5/7%. lass 7 Maths Chapter 8 Exercise 8.3 fRead more
Increased in price of a car from ₹3,50,000 to ₹3,70,000.
Amount change = ₹3,70,000 – ₹3,50,000 = ₹20,000.
Therefore, Increased percentage = Amount of change/Original amount x 100
= 20000/350000 x 100 = (5)5/7%
Hence, the percentage of price increased is (5)5/7%.
(a) 3 : 1 Total part = 3 + 1 = 4 Therefore, Fractional part = 3/1 : 1/4 ⇒ Percentage of parts = 3/4 x 100 : 1/4 x 100 ⇒ Percentage of parts = 75% : 25% (b) 2 : 3 : 5 Total part = 2 + 3 + 5 = 10 Therefore, Fractional part = 2/10 : 3/10 : 5/10 ⇒ Percentage of parts = 2/10 x 100 : 3/10 x 100 : 5/10 x 1Read more
(a) 3 : 1
Total part = 3 + 1 = 4
Therefore, Fractional part = 3/1 : 1/4
⇒ Percentage of parts = 3/4 x 100 : 1/4 x 100
⇒ Percentage of parts = 75% : 25%
(b) 2 : 3 : 5
Total part = 2 + 3 + 5 = 10
Therefore, Fractional part = 2/10 : 3/10 : 5/10
⇒ Percentage of parts = 2/10 x 100 : 3/10 x 100 : 5/10 x 100
⇒ Percentage of parts = 20% : 30% : 50%
(c) 1 : 4
Total part = 1 + 4 = 5
Therefore, Fractional part = 1/5 : 4/5
⇒ Percentage of parts = 1/5 x 100 : 4/5 x 100
⇒ Percentage of parts = 20% : 80%
(d) 1 : 2 : 5
Total part = 1 + 2 + 5 = 8
Therefore, Fractional part = 1/8 : 2/8 : 5/8
⇒ Percentage of parts = 1/8 x 100: 2/8 x 100 : 5/8 x 100
⇒ Percentage of parts = 12.5% : 25% : 62.5%
If Meena gives an interest of ₹45 for one year at 9% rate p.a. What is the sum she has borrowed?
Simple Interest = ₹45, Rate (R) = 9% p.a., Time (T) = 1 years Simple Interest = PXRXT/100 ⇒ 45 = PX9X1/100 ⇒ P = 45x100/9X1 ⇒ P = ₹ 500 Hence, she borrowed ₹500. lass 7 Maths Chapter 8 Exercise 8.3 for more answers vist to: https://www.tiwariacademy.com/ncert-solutions/class-7/maths/chapter-8/
Simple Interest = ₹45, Rate (R) = 9% p.a., Time (T) = 1 years
Simple Interest = PXRXT/100
⇒ 45 = PX9X1/100
⇒ P = 45×100/9X1
⇒ P = ₹ 500
Hence, she borrowed ₹500.
lass 7 Maths Chapter 8 Exercise 8.3
for more answers vist to:
See lesshttps://www.tiwariacademy.com/ncert-solutions/class-7/maths/chapter-8/
What rate gives ₹ 280 as interest on a sum of ₹ 56,000 in 2 years?
Here, Principal (P) = ₹56,000, Simple Interest (S.I.) = ₹280, Time (T) = 2 years Simple Interest = PXRXT/100 ⇒ 280 = 56000 X R X 2/100 ⇒ R= 280X100/5600x2 ⇒ R = 0.25% Hence, the rate of interest on sum is 0.25%. lass 7 Maths Chapter 8 Exercise 8.3 for more answers vist to: https://www.tiwariacademy.Read more
Here, Principal (P) = ₹56,000, Simple Interest (S.I.) = ₹280, Time (T) = 2 years
Simple Interest = PXRXT/100
⇒ 280 = 56000 X R X 2/100
⇒ R= 280X100/5600×2
⇒ R = 0.25%
Hence, the rate of interest on sum is 0.25%.
lass 7 Maths Chapter 8 Exercise 8.3
for more answers vist to:
See lesshttps://www.tiwariacademy.com/ncert-solutions/class-7/maths/chapter-8/
Find the amount to be paid at the end of 3 years in each case: (a) Principal = ₹1,200 at 12% p.a. (b) Principal = ₹ 7,500 at 5% p.a.
(a) Here, Principal (P) = ₹1,200, Rate (R) = 12% p.a., Time (T) = 3 years Simple Interest = PXRXT/100 = 1200 X 12 X 3/100 = ₹ 432 Now, Amount = Principal + Simple Interest = ₹1200 + ₹432 = ₹1,632 (b) Here, Principal (P) = ₹7,500, Rate (R) = 5% p.a., Time (T) = 3 years Simple Interest = PXRXT/10 = 75Read more
(a) Here, Principal (P) = ₹1,200, Rate (R) = 12% p.a., Time (T) = 3 years
Simple Interest = PXRXT/100 = 1200 X 12 X 3/100
= ₹ 432
Now, Amount = Principal + Simple Interest
= ₹1200 + ₹432
= ₹1,632
(b) Here, Principal (P) = ₹7,500, Rate (R) = 5% p.a., Time (T) = 3 years
Simple Interest = PXRXT/10 = 7500x5x3/100
= .1,125
Now, Amount = Principal + Simple Interest
= ₹7,500 + ₹1,125
= ₹ 8,625
lass 7 Maths Chapter 8 Exercise 8.3
for more answers vist to:
See lesshttps://www.tiwariacademy.com/ncert-solutions/class-7/maths/chapter-8/
Amina buys a book for ₹275 and sells it at a loss of 15%. How much does she sell it for?
The cost of a book = ₹275 Loss percent = 15% Loss = Loss% of C.P. = 15% of ₹275 = 15/100 x 275 = ₹41.25 Therefore, S.P. = C.P. – Loss = ₹275 – ₹41.25 = ₹233.75 Hence, Amina sells a book for ₹233.75. lass 7 Maths Chapter 8 Exercise 8.3 for more answers vist to: https://www.tiwariacademy.com/ncert-solRead more
The cost of a book = ₹275
Loss percent = 15%
Loss = Loss% of C.P. = 15% of ₹275
= 15/100 x 275 = ₹41.25
Therefore, S.P. = C.P. – Loss = ₹275 – ₹41.25 = ₹233.75
Hence, Amina sells a book for ₹233.75.
lass 7 Maths Chapter 8 Exercise 8.3
for more answers vist to:
See lesshttps://www.tiwariacademy.com/ncert-solutions/class-7/maths/chapter-8/
(i) Chalk contains Calcium, Carbon and Oxygen in the ratio 10:3:12. Find the percentage of Carbon in chalk. (ii) If in a stick of chalk, Carbon is 3 g, what is the weight of the chalk stick?
(i) Given ratio = 10 : 3 : 12 Total part = 10 + 3 + 12 = 25 Part of Carbon = 3/25 Percentage of Carbon part in chalk = 3/25x100 = 12% (ii) Quantity of Carbon in chalk stick = 3 g Let the weight of chalk be x g. Then, 12% of x = 3 ⇒ 12/100 X x=3 ⇒ x = 3x100/12 = 25 g Hence, the weight of chalk stickRead more
(i) Given ratio = 10 : 3 : 12
Total part = 10 + 3 + 12 = 25
Part of Carbon = 3/25
Percentage of Carbon part in chalk = 3/25×100 = 12%
(ii) Quantity of Carbon in chalk stick = 3 g
Let the weight of chalk be x g.
Then, 12% of x = 3
⇒ 12/100 X x=3
⇒ x = 3×100/12 = 25 g
Hence, the weight of chalk stick is 25 g.
lass 7 Maths Chapter 8 Exercise 8.3
for more answers vist to:
See lesshttps://www.tiwariacademy.com/ncert-solutions/class-7/maths/chapter-8/
Juhi sells a washing machine for ₹13,500. She loses 20% in the bargain. What was the price at which she bought it?
Selling price of washing machine = ₹13,500 Loss percent = 20% Let the cost price of washing machine be ₹ x. Since, Loss = Loss% of C.P. ⇒ Loss = 20% of ₹ x = 20/100X x 20/100 X x =x/5 Therefore, S.P. = C.P. – Loss ⇒ 13500 = x - x/5 ⇒ 13500 = 4x/5 ⇒ x= 13500X5/4 = ₹16,875 Hence, the cost price of wasRead more
Selling price of washing machine = ₹13,500
Loss percent = 20%
Let the cost price of washing machine be ₹ x.
Since, Loss = Loss% of C.P.
⇒ Loss = 20% of ₹ x = 20/100X x 20/100 X x =x/5
Therefore, S.P. = C.P. – Loss
⇒ 13500 = x – x/5
⇒ 13500 = 4x/5
⇒ x= 13500X5/4 = ₹16,875
Hence, the cost price of washing machine is ₹16,875.
lass 7 Maths Chapter 8 Exercise 8.3
for more answers vist to:
See lesshttps://www.tiwariacademy.com/ncert-solutions/class-7/maths/chapter-8/
I buy a T.V. for ₹10,000 and sell it at a profit of 20%. How much money do I get for it?
The cost price of T.V. = ₹ 10,000 Profit percent = 20% Now, Profit = Profit% of C.P. = 20/100 x 10000 = ₹ 2,000 = Selling price = C.P. + Profit = .10,000 + .2,000 = . 12,000 Hence, he gets .12,000 on selling his T.V. lass 7 Maths Chapter 8 Exercise 8.3 for more answers vist to: https://www.tiwariacaRead more
The cost price of T.V. = ₹ 10,000
Profit percent = 20%
Now, Profit = Profit% of C.P.
= 20/100 x 10000
= ₹ 2,000
= Selling price = C.P. + Profit = .10,000 + .2,000 = . 12,000
Hence, he gets .12,000 on selling his T.V.
lass 7 Maths Chapter 8 Exercise 8.3
for more answers vist to:
See lesshttps://www.tiwariacademy.com/ncert-solutions/class-7/maths/chapter-8/
Arun bought a car for ₹3,50,000. The next year, the price went up to ₹3,70,000. What was the percentage of price increase?
Increased in price of a car from ₹3,50,000 to ₹3,70,000. Amount change = ₹3,70,000 – ₹3,50,000 = ₹20,000. Therefore, Increased percentage = Amount of change/Original amount x 100 = 20000/350000 x 100 = (5)5/7% Hence, the percentage of price increased is (5)5/7%. lass 7 Maths Chapter 8 Exercise 8.3 fRead more
Increased in price of a car from ₹3,50,000 to ₹3,70,000.
Amount change = ₹3,70,000 – ₹3,50,000 = ₹20,000.
Therefore, Increased percentage = Amount of change/Original amount x 100
= 20000/350000 x 100 = (5)5/7%
Hence, the percentage of price increased is (5)5/7%.
lass 7 Maths Chapter 8 Exercise 8.3
for more answers vist to:
See lesshttps://www.tiwariacademy.com/ncert-solutions/class-7/maths/chapter-8/
Convert each part of the ratio to percentage: (a) 3 : 1 (b) 2 : 3 : 5 (c) 1 : 4 (d) 1 : 2 : 5
(a) 3 : 1 Total part = 3 + 1 = 4 Therefore, Fractional part = 3/1 : 1/4 ⇒ Percentage of parts = 3/4 x 100 : 1/4 x 100 ⇒ Percentage of parts = 75% : 25% (b) 2 : 3 : 5 Total part = 2 + 3 + 5 = 10 Therefore, Fractional part = 2/10 : 3/10 : 5/10 ⇒ Percentage of parts = 2/10 x 100 : 3/10 x 100 : 5/10 x 1Read more
(a) 3 : 1
Total part = 3 + 1 = 4
Therefore, Fractional part = 3/1 : 1/4
⇒ Percentage of parts = 3/4 x 100 : 1/4 x 100
⇒ Percentage of parts = 75% : 25%
(b) 2 : 3 : 5
Total part = 2 + 3 + 5 = 10
Therefore, Fractional part = 2/10 : 3/10 : 5/10
⇒ Percentage of parts = 2/10 x 100 : 3/10 x 100 : 5/10 x 100
⇒ Percentage of parts = 20% : 30% : 50%
(c) 1 : 4
Total part = 1 + 4 = 5
Therefore, Fractional part = 1/5 : 4/5
⇒ Percentage of parts = 1/5 x 100 : 4/5 x 100
⇒ Percentage of parts = 20% : 80%
(d) 1 : 2 : 5
Total part = 1 + 2 + 5 = 8
Therefore, Fractional part = 1/8 : 2/8 : 5/8
⇒ Percentage of parts = 1/8 x 100: 2/8 x 100 : 5/8 x 100
⇒ Percentage of parts = 12.5% : 25% : 62.5%
lass 7 Maths Chapter 8 Exercise 8.3
for more answers vist to:
See lesshttps://www.tiwariacademy.com/ncert-solutions/class-7/maths/chapter-8/
Tell what is the profit or loss in the following transactions. Also find profit percent or loss percent in each case. (a) Gardening shears bought for ₹ 250 and sold for ₹ 325. (b) A refrigerator bought ₹12,000 and sold at ₹ 13,500. (c) A cupboard bought for ₹ 2,500 and sold at ₹ 3,000. (d) A skirt bought for ₹ 250 and sold at ₹ 150.
(a) Cost price of gardening shears = ₹ 250 Selling price of gardening shears = ₹ 325 Since, S.P. > C.P., therefore here is profit. ∴ Profit = S.P. – C.P. = ₹325 – ₹250 = ₹75 Now Profit% = Profit/C.P. x 100 = 75/250 x 100 = 30% Therefore, Profit = ₹75 and Profit% = 30% (b) Cost price of refrigeratRead more
(a) Cost price of gardening shears = ₹ 250
Selling price of gardening shears = ₹ 325
Since, S.P. > C.P., therefore here is profit.
∴ Profit = S.P. – C.P. = ₹325 – ₹250 = ₹75
Now Profit% = Profit/C.P. x 100
= 75/250 x 100 = 30%
Therefore, Profit = ₹75 and Profit% = 30%
(b) Cost price of refrigerator = ₹12,000
Selling price of refrigerator = ₹13,500
Since, S.P. > C.P., therefore here is profit.
∴ Profit = S.P. – C.P. = ₹13500 – ₹12000 = ₹1,500
Now Profit% = Profit/C.P x 100
= 1500/12000 x 100 = 12.5%
Therefore, Profit = ₹1,500 and Profit% = 12.5%
(c) Cost price of cupboard = ₹ 2,500
Selling price of cupboard = ₹ 3,000
Since, S.P. > C.P., therefore here is profit.
∴ Profit = S.P. – C.P. = ₹3,000 – ₹2,500 = ₹ 500
Now Profit% = Profit/CP x 100
= 500/2500 x 100 = 20%
Therefore, Profit = ₹ 500 and Profit% = 20%
(d) Cost price of skirt = ₹ 250
Selling price of skirt = ₹ 150
Since, C.P. > S.P., therefore here is loss.
∴ Loss = C.P. – S.P. =₹250 – ₹150 = ₹100
Now Loss% = Loss/C.P. x 100
= 100/250 x 100 = 40%
Therefore, Profit = ₹ 100 and Profit% = 40%
lass 7 Maths Chapter 8 Exercise 8.3
for more answers vist to:
See lesshttps://www.tiwariacademy.com/ncert-solutions/class-7/maths/chapter-8/