Simple & Compound Interest Questions for CAT with Answers PDF

Let equal installment = Rs. x

21,000 × 1.3 = 26.3x
⇒ x ≈ 1,038

Time (T) = 1 year 7 months =
Simple Interest = 

A = P

Now, 8P = P
⇒ 8P = P (2)^t
Or, 2³ = 2^t
∴ t = 3
Required time = 3 years

After 1 year, the debt will be 28,000 - 10,000 = 18,000
Note : Now Interest for first year will not be counted and subtracted as it was the interest levied on Rs. 28,000.
Interest for next 6 months = = Rs. 1080
Total money to be paid = 18,000 + 1080 = Rs. 19,080

Let the rate % be x.

10% of interest paid
⇒ 90% of interest left = Rs. 900
So, complete interest (100%) = Rs. 1000

= 12.5%

Difference in C.I. of two successive periods = S.I. on C.I. of the earlier period
11.25 is S.I. on Rs. 225 at the same rate for six months.

⇒ r = 5%
Annual rate of interest = 10%

Let the two parts be P₁ and P₂.
According to the question,

⇒ P₁ : P₂ = 2 : 1
∴ P1 = 6000 × 2/3 = 4000 and P₂ = 2000

Principal = Rs 5000
Rates = 10% for yearly and 5% for half yearly
Times = 4 periods for compounded half yearly and 2 periods for compounded yearly
Difference = 5,000 = 5,000 (1.215 – 1.210) = Rs. 28

Let the amount invested at 4% be x and that at 5% be (8000 - x).

⇒ 4x + 40,000 - 5x = 35,000
Therefore, x = 5,000
Hence, amounts invested by Dhiraj at 4% and 5% are Rs. 5,000 and Rs. 3,000 respectively.

Let the sum be P, Rate of interest be r.
So, according to the question:
P × r × 4 + 200 = P × (r + 0.05) × 4
4Pr + 200 = 4Pr + 0.2P
200 = 0.2P
P = Rs. 1000

SI = 
840 - 700 = 700 × 2 × 
140 = 14R
R = 10% per year
Then, total amount after 3 years = 700 + 700 × (10 + 2) × 3/10
Total amount after 3 year at a rate of (10 + 2)% per year = 700 + 252 = Rs. 952

Amount he has to pay after 2 years

= 13,200 + 17,400 + 24,800 = Rs. 55,400
He lent Rs. 12,000 + Rs. 15,000 + Rs. 20,000 at 10% compound interest.
So, after 2 years, he will get back 47,000=Rs. 56,870
So, money gained = Rs. 1470

There should be no loss if
22% ≤ 5k%
k ≥ 22/5
k ≥ 4.4
For minimum amount, k = 4.4
So,
(k - 1) × 10⁴ ≤ P
(4.4 - 1) × 10⁴ ≤ P
34000 ≤ P
Therefore, minimum amount of loan is Rs. 34,000

As in the first case, she is paying Rs. 10,000 + Rs. 18,200 = Rs. 28,200
As in the second case, she is paying Rs. 15,000 + Rs. 12,800 = Rs. 27,800
So, the difference of Rs. 400 (Rs. 28,200 - Rs. 27,800) is nothing but the interest of Rs. 5,000 (Rs. 15,000 - Rs. 10,000) for 1 year.

⇒ r = 8%

SI (to be paid) = = 0.1P
CI (to be received) = A - P = P - P = 0.0816P
So, he has to pay 0.1P and he gets 0.0816P.
So, loss = 0.1P - 0.0816P = 0.0184P
Loss% =× 100 = 1.84%

A → B = Rs. 900
Let time = T years
Rate of interest = r%

⇒ r = 4%

B → A = Rs. P
Rate of interest = r = 4%
Time = T = 2 years

P = Rs. 600

Let the principal be P.
Suppose R = T = x

Let the original value of the property be 'P'.
Using the formula: 

Hence, required answer = Rs. 25,600

Let principal = P. rate = R
According to question,
For 2 years,
C.I. - S.I. = 20

For 3 years,
C.I. - S.I. = 61

Dividing (1) and (2), we get

R = 5%

Let the principal = x
Simple interest =
Compound interest = x 
So,