1. A lends Rs. 1500 to B and a certain sum to C at the same time at 8% per annum simple interest. If after 4 years, A altogether receives Rs. 1400 as interest from B and C, then the sum lent to C is

1. Rs.2875

2. Rs.1885

3. Rs.2245

4. Rs.2615

Answer & Explanation

Answer:- 1

Explanation :-

Lets suppose Money lent to C be Rs. x

SI (for B) = (1500*8*4)/100 = 480

SI (for C) = (x*8*4)/100 = (32x)/100

Total Interest = 1400

Therefore, 480+(32x/100) = 1400

32x/100 = 920

x = Rs. 2875

2. A sum of Rs. 10 is given as a loan to be returned in 6 monthly installments at Rs.3. What is the rate of interest?

1. 820%

2. 620%

3. 780%

4. 640%

Answer & Explanation

Answer:- 4

Explanation :-

3. If the simple interest on a certain sum of money after 3 1⁄8 years is 1⁄4 of the principal, what is the rate of interest per annum?

1. 6%

2. 4%

3. 8%

4. 12%

Answer & Explanation

Answer:- 3

Explanation :-

Let the sum of money be Rs. x

Time = 3 ^{1}⁄_{8} Years = ^{25}⁄_{8} Years

SI = ^{x}⁄_{4}

Therefore, SI = ((P*R*T)/100))

x/4 = (x*R*25/8)/100

R = (x*100*8)/(x*R*25)

R = 8%

4. If a sum of Rs. 9 is lent to be paid back in 10 equal monthly installments of re. 1 each, then the rate of interest is

1. 11.33%

2. 11%

3. 266.67%

4. 26.67%

Answer & Explanation

Answer:- 4

Explanation :-

5. Divide Rs. 2379 into 3 parts so that their amount after 2,3 and 4 years respectively may be equal, the rate of interest being 5% per annum at simple interest. The first part is

1. Rs. 828

2. Rs. 746

3. Rs. 248

4. Rs. 1024

Answer & Explanation

Answer:- 1

Explanation :-

Let the parts be x, y and z

R = 5%

x + SI on x (for 2 years) = y + SI on y (for 3 years) = z + SI on z (for 4 years)

(x+(x*5*2)/100) = (y+(y*5*3)/100) = (z+(z*5*4)/100))

(x+x/10) = (y+3y/20) = (z+z/5)

(11x/10) = (23y/20) = (6z)/5

Let (11x/10) = (23y/20)=(6z/5) = k (where k is a constant)

Therefore, x=(10k/11) , y=(20k/23) , z=(5k/6)

we know that x + y + z = 2379

(10k/11) + (20k/23) + (5k/6)=2379

(10k×23×6) + (20k×11×6) + (5k×11×23) = (2379×11×23×6)

1380k + 1320k + 1265k = 2379×11×23×6

3965k = 2379×11×23×6

k=(2379×11×23×6)/3965

first part, x = 10k/11 = (10/11)×(2379×11×23×6)/3965

x=828

OR

Let the parts be x (for2 years), y (for 3 years) and z(for 4 years)

R = 5%

Therefore

R on x = 10%. R on y = 15%. R on z = 20%.

110% of x =115% of y = 120% of z

x/y = 115/110 = 23/22 , x/z = 120/110 = 24/22

=>x : y : z = (23*24):(22*24):(22*23)

x : y : z = 276:264:253 & here x + y + z = (276+264+253) = 793

but x + y + z = 2379 (given)

Therefore

x = (2379/793)*276 = Rs.828