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.
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/