Quiz 1

Time = 2 years 4 months = 2(4/12) years = 2(1/3) years.
Amount = Rs'. [8000 X (1+(15/100))^2 X (1+((1/3)*15)/100)]
=Rs. [8000 * (23/20) * (23/20) * (21/20)]
= Rs. 11109. .
:. C.I. = Rs. (11109 - 8000) = Rs. 3109.

Amount=[7500×(1+4/100)2]=(7500×26/25×26/25)=8112

Given, SI = $ 6750, R = 203% p.a. and T = 3 years.
sum = 100 × SI / R × T
= $ (100 × 6750 × ³/20 × 1/3 ) = $ 33750.
Now, P = $ 33750, R = 203% p.a. and T = 3 years.
Therefore, amount after 3 years
= $ {33750 × (1 + (20/3 × 100)}³ [using A = P (1 + R/100)]
= $ (33750 × 16/15 × 16/15 × 16/15) = $ 40960
Thus, amount = $ 40960
Hence, compound interest = $ (40960 - 33750) = $ 7210

Let P = Rs.100
Simple Interest = Rs. 80 ( ? 80% increase is due to the simple interest)
Rate of interest=(100×SI)/PT=(100×80)/(100×8) =10% per annum
Now let's find out the compound interest of Rs. 14,000 after 3 years at 10%
P = Rs.14000
T = 3 years
R = 10%
Amount after 3 years =P(1+R/100)^T
=14000(1+10/100)^3
=14000(110/100)^3
=14000(11/10)^3
=14×11^3=18634.
Compound Interest = Rs.18634 - Rs.14000 = Rs.4634

Sum = Rs. 100 x 90/20 = Rs. 450
C.I. = Rs.450 x (1+ 10/100)^2- 450 = Rs. 94.50

P(1 + R/100)10 = 3P
=> P(1 + R/100)10 = 3
Let P(1 + R/100)n = 9P
=> (1 + R/100)n = 9
=> 32 = [(1 + R/100)10]2
=> (1 + R/100)n => (1 + R/100)20
=> n = 20 Years.

Principal = Rs.1000;
Amount = Rs.1331;
Rate = Rs.10%p.a.
Let the time be n years then,
1000(1+10/100)^n=1331
(11/10)^n=1331/1000
(11/10)^3=1331/1000
So answer is 3 years

Let the required rate be R% per annum
Here, A = $ 2205, P = $ 2000 and n = 2 years
Using the formula A = P(1 + R/100)n,
2205 = 2000 × ( 1 + R/100)²
(1 + R/100)² = 2205/2000 = 441/400 = (21/20)²
( 1 + R/100) = 21/20
R/100 = (21/20 – 1) = 1/20
R = (100 × 1/20) = 5
Hence, the required rate of interest is 5% per annum

Let the sum be $ 100. Then,
SI = $ (100 × 6 × 2/100) = $ 12
and compound interest = $ {100 × (1 + 6/100)² - 100}
= $ {(100 × 53/50 × 53/50) - 100} = $ (2809/25 - 100) = $ 309/25
Therefore, (CI) - (SI) = $ (309/25 – 100) = $ 9/25
If the difference between the CI and SI is $ 9/25, then the sum = $ 100
If the difference between the CI and SI is $ 18, then the sum = $ (100 × 25/9 × 18 ) = $ 5000
Hence, the required sum is $ 5000.

Increase% = (1/8) x 100% = 12.5%
Height after 2 years =64 x {1 +25/(2 x 100)}2
= 64 x 9/8 x 9/8
= 81 cm

Time = 2 years 4 months = 2(4/12) years = 2(1/3) years.
Amount = Rs'. [8000 X (1+(15/100))^2 X (1+((1/3)*15)/100)]
=Rs. [8000 * (23/20) * (23/20) * (21/20)]
= Rs. 11109. .
:. C.I. = Rs. (11109 - 8000) = Rs. 3109.

Amount=[7500×(1+4/100)2]=(7500×26/25×26/25)=8112

Given, SI = $ 6750, R = 203% p.a. and T = 3 years.
sum = 100 × SI / R × T
= $ (100 × 6750 × ³/20 × 1/3 ) = $ 33750.
Now, P = $ 33750, R = 203% p.a. and T = 3 years.
Therefore, amount after 3 years
= $ {33750 × (1 + (20/3 × 100)}³ [using A = P (1 + R/100)]
= $ (33750 × 16/15 × 16/15 × 16/15) = $ 40960
Thus, amount = $ 40960
Hence, compound interest = $ (40960 - 33750) = $ 7210

Let P = Rs.100
Simple Interest = Rs. 80 ( ? 80% increase is due to the simple interest)
Rate of interest=(100×SI)/PT=(100×80)/(100×8) =10% per annum
Now let's find out the compound interest of Rs. 14,000 after 3 years at 10%
P = Rs.14000
T = 3 years
R = 10%
Amount after 3 years =P(1+R/100)^T
=14000(1+10/100)^3
=14000(110/100)^3
=14000(11/10)^3
=14×11^3=18634.
Compound Interest = Rs.18634 - Rs.14000 = Rs.4634

Sum = Rs. 100 x 90/20 = Rs. 450
C.I. = Rs.450 x (1+ 10/100)^2- 450 = Rs. 94.50

P(1 + R/100)10 = 3P
=> P(1 + R/100)10 = 3
Let P(1 + R/100)n = 9P
=> (1 + R/100)n = 9
=> 32 = [(1 + R/100)10]2
=> (1 + R/100)n => (1 + R/100)20
=> n = 20 Years.

Principal = Rs.1000;
Amount = Rs.1331;
Rate = Rs.10%p.a.
Let the time be n years then,
1000(1+10/100)^n=1331
(11/10)^n=1331/1000
(11/10)^3=1331/1000
So answer is 3 years

Let the required rate be R% per annum
Here, A = $ 2205, P = $ 2000 and n = 2 years
Using the formula A = P(1 + R/100)n,
2205 = 2000 × ( 1 + R/100)²
(1 + R/100)² = 2205/2000 = 441/400 = (21/20)²
( 1 + R/100) = 21/20
R/100 = (21/20 – 1) = 1/20
R = (100 × 1/20) = 5
Hence, the required rate of interest is 5% per annum

Let the sum be $ 100. Then,
SI = $ (100 × 6 × 2/100) = $ 12
and compound interest = $ {100 × (1 + 6/100)² - 100}
= $ {(100 × 53/50 × 53/50) - 100} = $ (2809/25 - 100) = $ 309/25
Therefore, (CI) - (SI) = $ (309/25 – 100) = $ 9/25
If the difference between the CI and SI is $ 9/25, then the sum = $ 100
If the difference between the CI and SI is $ 18, then the sum = $ (100 × 25/9 × 18 ) = $ 5000
Hence, the required sum is $ 5000.

Increase% = (1/8) x 100% = 12.5%
Height after 2 years =64 x {1 +25/(2 x 100)}2
= 64 x 9/8 x 9/8
= 81 cm