1.

When a number is subtracted from the number 8,12 and 20, the remainders are in continued proportion, Find the number ?

2.

Aditya and Bhushan invested 10000 each in scheme A and scheme B respectively for 3 years. Scheme A offers Simple interest @ 12% per annum and scheme B offers compound interest @ 10%. After 3 years, who will have larger amount and by how much?

**Answer: Option 'B'**

**Aditya, 290**

Lets first calculate the total rate % that Aditya will have after 3 years:

As per the question Aditya invested at rate of 12% pa simple interst

So, for 3 years tenure he will get = 12 * 3 = 36%

And the amount that Bhushan invested at rate of 10% pa compound interest

By net% effect formula, we can calculate the total perecntage for 3 years tenure = 33.1% (sub details)

So, the difference between SI and CI = 36% – 33.1% = 2.9% (SI is more)

Here Aditya will get, 2.9% of 10000 = 290

So Aditya will have Rs. 290 more than Bhushan.

---------------------------------------------------------------------------------

Sub-details:-

Net% effect = x + y = xy/100 %

For the first 2 years: Here, x = y = 10%

= 10 + 10 = 10* 10 = 21 /100 %

And for the next year: Here x = 21% and y = 10%

= 21 + 10 = 21 * 10 = 33.1/100 %

**Hence, option C is correct.**

3.

Shantanu borrowed Rs. 2.5 lakh from a bank to purchase one car. If the rate of interest be 6% per annum compounded annually, what payment he will have to make after 2 years 6 months?

**Answer: Option 'C'**

**2,89,325**

----> CI for 2 years 6 months at the rate of 6, applying the net% effect for first 2 years

----> = 6 + 6 + (6 * 6)/100 = 12.36%

----> Rate of interest for 6 months = (6/12) * 6 = 3%

----> For next 6 months = 12.36 + 3 + (12.36 * 3) /100 = 15.36 + 0.37% = 15.73%

----> Here, we can see that in 2 years 6 months the given compound rate of interest is approximate 15.73%.

----> Now, 115.73% of 250000 = (115.73 * 250000)/100 = 289,325.

----> **Hence, option D is correct.**

4.

In a mixture, the ratio of the alchohol and water is 6 : 5. When 22 litre mixture are replaced by water, the ratio becomes 9 : 13. Find the quantity of water after replacement.

5.

A person closes his account in an investment scheme by withdrawing Rs 10000. One year ago, he had withdrawn Rs 6000. Two years ago he had withdrawn Rs 5000. Three years ago he had not withdrawn any money. How much money had he deposited approximately at the time of opening the account 4 years ago, if the annual rate of compound interest is 10%.

**Answer: Option 'A'**

**Rs. 15470**

Suppose the person has deposited Rs. X at the time of opening account.

---> After one year, he had

----> Rs. (x + x * 10 * 1/100) = Rs. (11x/10)

----> After two years, he had

-----> Rs.((11x/10)+(11x/10) * (10 * 1/100)) = (121x/100)

----> After withdrawing Rs 5000 from Rs. (121x/100) ,the balance

----> = Rs. (121x - 500000/100)

---> After 3 years, he had

----> ((121x - 500000/100) + (121x - 500000/100) * (10 * 1/100))

----> = 11(121x - 500000/1000)

----> After withdrawing 6000 from above, the balance

----> = Rs.(1331x/1000) â€“ 11500

----> After 4 years, he had

----> Rs.(11/10) ((1331x/1000) â€“ 11500) - 10000 = 0

----> Rs 15470

6.

In a mixture of milk and water of volume 30 litres, the ratio of milk and water is 7 : 3. How much quantity of water is to be added to the mixture to make the ratio of milk and water 1 : 2?

**Answer: Option 'C'**

**33 liters**

In 30 liter of mixture quantity of milk = (7*30)/10 = 21 liter

Hence quantity of water = 30 - 21 = 9 liter

Now let x liter of water is added in mixture to make the ratio of milk and water 1 : 2.

Hence now quantity of water = (9 + x) L

Milk = 21 L

Therefore (21 : (9 + x)) = 1 : 2

21/(9 + x) = 1/2

42 = 9 + x

x = 42 - 9 = 33

Hence 33 liters of water is added in the mixture

7.

A container contains pure milk. From these 4 litres of milk taken out and replaced by water. This process is repeated for one more time and the remaining milk in the container is 12.8 litres. What is the initial quantity of milk in the container?

**Answer: Option 'B'**

**20 litres**

Let us take initial quantity of a container be x

Remaining milk = Initial (1 - Replaced/Initial)n

12.8 = x (1 - 4/x)^{2}

12.8 = x (1 + 16/x2 - 8x)

12.8 = x [(x^{2} + 16 - 8x) / x^{2}]

12.8x = x^{2} + 16 - 8x

5x2 - 104x + 80 = 0

Simplify the above equation, we get x= 20 and 0.8 (Eliminate)

8.

Pankaj borrowed a total amount of Rs.32500 from his three friends Raj, Akash and Suresh. All of his friends apply different rates of interest in such a way that Raj applies 12%, Akash applies 16% and Suresh applies 18% interest rate respectively and total he gives Rs.5090 as interest. If the amount that Pankaj had taken from Raj is (18/25) of the amount taken from Suresh, then find that what amount Pankaj has taken from Akash?

Rs. 10000

9.

Ratio of numerical value of rate of interest and time period is 4 : 1. Man invested Rs. 2400 and gets Rs. 864 as simple interest. Find the value of X, if man invested Rs. (2400 + X) at same rate of interest on C.I. for two years and get Rs. 814.08 as interest ?

**Answer: Option 'C'**

**Rs. 800**

Let man invested at the rate of 4x % per annum and for the period of time is x yr

ATQ

(2400 * 4x * x)/100 = 864

4x2 = 864/24

4x2 = 36

x2 = 9

x = 3

So, Rate of interest = 4 * 3 = 12% per annum

Time of period = 3 years.

Equivalent C.I. of two year at the rate of 12% per annum

CI = P[(1 + r/100)n - 1]

814.08 = (2400 + x) [(1 + 12/100)2 - 1]

814.08 = (2400 + x) [(112/100)(112/100) - 1]

814.08 = (2400 + x) [(28/25)(28/25) - 1]

814.08 = (2400 + x) [784/625 - 1]

814.08 = (2400 + x) [(784 - 625)/625]

814.08 = (2400 + x) (159/625)

814.08*(625/159) = 2400 + x

3200 = 2400 + x

x = 3200 - 2400 = 800

10.

Rs. 39,030 is divide between A and B in such a way that amount given to A on C.I. in 7 years is equal to amount given to B on C.I. in 9 years. Find the part of A. If the rate of interest is 4%.

**Answer: Option 'B'**

**Rs. 20,280**

Given:

Let the principal of A and B be 'a' and 'b' respectively.

Amount = Rs. 39,030

r = 4%

Amount given to A for 7 yrs = Amount given to B for 9 yrs

WKT, Amount = P[1+ (r/100)]^{n}

a[1 + (4/100)]7 = b[1 + (4/100)]^{9}

a/b = [1 + (4/100)]9/[1 + (4/100)]^{7}

a/b = [1 + (4/100)]^{2}/1

a/b = [26/25]^{2}/1

a : b = 676/625

a : b = 676 : 625

Part of A = 39030*(676/1301) = Rs. 20,280.

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