## Ques 1.

After receiving two successive raises,Ravi’s salary become equal to 15/8 times of his initial salary. By how much percent was the salary raised the first time if the second raise was twice as high (in percent) as the first?

**Solution :**

Lets Initial salary = X

First raised salary % = Y

Second raised salary % = 2Y

As per problem

==> X * (100+Y)/100 * (100+2Y)/100
= 15/8 * X

==> (100+Y)*(100+2Y) = 15/8 * 100
*100

==> 2Y^2 + 300Y - 8750 = 0

==> Y = 25

## Ques 2.

A Merchant gives 3 consecutive discounts of 10%, 15% and 15% after which he sells his goods at a percentage profit of 30.05% on the C.P. Find the value of the percentage profit that the shopkeeper would have earned if he had given discount of 10% and 15% only.

**Solution :**

Assume Marked Priced = 100

After 3 discounts selling price =
100(90/100)*(85/100)*(85/100) = 65.025

Still he getting 30.05 % profit

Cost Price = (65.025*100)/130.05 = 50

If he Allows only 10% and 15% discount
Selling Price = 100 *(90/100) *(85/100) = 76.5

Then he gets % of profit = (26.5/50)
*100 = 53%

## Ques 3.

A sum of Rs. 1000 after 3 years at compound interest becomes a certain amount that is equal to the amount that is the result of a 3 year depreciation from Rs. 1728. Find the difference between the rates of CI and depreciation. (GivenCI is 10% p.a.).

**Solution :**

Amount after 3 years =
1000(1+(10/100)^3 = 1331

Depreciation Rate = X%

1728*(1-(X/100))^3 = 1331

==> X = 100/12

Difference = 10 - (100/12) = 20/12
=5/3%

## Ques 4.

A mixture can be prepared by sugar and salt. Price of sugar is thrice the price of salt. Siva sells the mixture at Rs. 4320 per 20g, thereby making a profit of 20%. If the ratio of sugar and salt in the mixture be 2 : 3, find the cost price of sugar.

**Solution :**

Sugar = 3X

Salt = X

Selling Price of 1 kg = 4320/20 = 216

Cost Price of 1 kg = 216 *100/120 =
180

3X
X

180

180-X 3X-180

==> (180-X)/(3X-180) = 2/3

==> X = 100

Sugar = 3*100 = 300

## Ques 5.

The probability that a contractor will get a plumbing contract is 2/3 and the probability that he will get an electric contract is 5/9. If the probability of getting at least one contract is 4/5, what is the probability that he will get both the contracts?

**Solution :**

Probability of getting At least one Contract = 4/5

Probability of getting both contracts
= X

==>2/3 + 5/9 -x = 4/5

==> X = 19/45

## Ques 6.

In an examination it is required to get 900 marks of the aggregate marks to pass. One of the students got 43% marks and he was declared failed by 40 marks. Find the maximum aggregate marks of the examination.

**Solution :**

As per problem

==> (43X/100) + 40 = 900

==> X = 2000

## Ques 7.

P, Q and R invested Rs 10000, Rs 12000 and Rs 1300 in a business. After the end of 4 months all of them withdrew Rs 1000. After a total of 8 months from the start of business all of them added Rs 1000 in their investments. Find the ratio of their shares in total profit at the end of a year.

**Solution :**

Ratio =5000*8 + 4500*4 : 6000*8 +
5500*4 : 6500*8 + 6000*4

==>50*2 + 45 : 60*2 + 55 : 65*2 + 60

==>10*2
+ 9 : 12*2 + 11 : 13*2 + 12

==>29 : 35 : 38

==>29 : 35 : 38

## Ques 8.

Two circles have their circumferences equal to 440 m and 528 m respectively. What is the difference between the area of the larger circle and the smaller circle?

**Solution :**

Circle 1 circumferences
= 440

==> 2*pi
*R1 = 440

==>
R1 = 70

Circle
2 circumferences = 528

==>
2*pi*R2 = 528

==>
R2 = 84

Difference
between Areas = Pi*R2^2 - Pi*R1^2 = 22/7 (84+70)(84-70)

==>
Difference areas = 6776

## Ques 9.

9000 1795 355 68 ? 1.32

**Solution :**

9000 /
5 – 5 = 1795

1795 / 5 – 4 = 355

355 / 5 – 3 = 68

68 / 5 – 2 = 11.6

1795 / 5 – 4 = 355

355 / 5 – 3 = 68

68 / 5 – 2 = 11.6

## Ques 10.

x^3=10648 ; y^2=484

**Solution :**

x^3
=10648

==>x = 22

==>x = 22

y^2= 484

==>y = + 22, – 22

**X ≥ Y**