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Given

A man buy two watches at Rs.4000

He sold one watch at 20% profit and second at 10% loss.

The overall profit or loss will be 5%

Formula used

CP(1 + P/100) = SP

CP(1 - L/100) = SP

SP = Selling Price

CP = Cost Price

P = Profit

L = Loss

Calculation

Let CP of first watch be Rs.x, then CP of second watch be Rs.(4000 - x)

⇒ Total Selling Price = 4000 × 105 /100

⇒ 4200

⇒ According to question

⇒ x × 120/100 + (4000 - x) × 90/100 = 4200

⇒ 12x/10 + 3600 - 9x/10 = 4200

⇒ 3x/10 = 600

⇒ x = (600 × 10)/3

⇒ x = Rs.2000

∴ Cost price of first watch be Rs.2000

Alternate Method

(1 + 1) unit = 4000

2 unit = 4000

1 unit = 2000

Cost price of each watch is Rs.2000

Given:

The initial quantity of mixture containing milk and water is 156 liters.

Calculation:

Let the water to be added be x litres.

Then, initial quantity of milk = {7/(7 + 6)} × 156 liters

⇒ (7/13) × 156 litres

⇒ 84 litres

And, initial quantity of water = {6/(7 + 6)} × 156 liters

⇒ (6/13) × 156 litres

⇒ 72 litres

Now, 84/(72 + x) = 14/13

⇒ 84 × 13 = 14 × (72 + x)

⇒ 6 × 13 = 72 + x

⇒ x = 78 – 72

⇒ x = 6

∴ The quantity of water added is 6 litres.

Given:

Speed of the first train = 72 km/hr = 20 m/sec

Speed of the second train = 54 km/hr = 15 m/sec

First train crosses a post in 20 seconds and crosses the second train going in same direction in 1 min 45 sec = 105 seconds

Concept:

This question uses concepts of both crossing of trains and relative velocity.

Formula used:

Relative velocity when going in same direction = difference in velocities of the two trains

And, s = v × t  where, v = velocity of the train, t = time taken to cross, s = distance travelled

Calculation:

When the first train is crossing the post, it is actually crossing its own length and while it crosses the second train, it crosses the sum of the lengths of both the trains.

So, length of the train = 20 × 20 = 400 m

Let’s take length of the second train = x m

Relative speed between the two trains = 20 – 15 = 5 m/sec

ATQ,

(400 + x) = 5 × 105

⇒ 400 + x = 525

⇒ x = 125

Difference between their lengths = 400 – 125 = 275 m

Calculation:

The average of 13 papers = 80

∴ the sum of the marks = 13 × 80

⇒ 1040

Average for first 7 papers = 84

∴ The sum = 7 × 84

⇒ 588

Average of last 7 papers = 7 × 70

⇒ 490

∴ the marks obtained in the 7^th paper = (588 + 490) – 1040

⇒ 38

∴ the marks obtained in the 7^th paper are 38.

Given:

Total valid votes = 60%

Invalid votes = 30%

Total numbers of vote = 8000.

Concept used:

Total valid vote = total votes - invalid votes

Calculation:

Total number of votes = 8000 

Given that 30% of Percentage votes were invalid

⇒ Valid votes = 70%

Total valid votes = 8000 × (70/100) 

⇒ Valid vote = 5600

1^st candidate got 60% of the total valid votes. 

Hence the 2^nd candidate should have got 40% of the total valid votes 

⇒ Valid votes that 2^nd candidate got = total valid votes x (40/100) 

⇒ 5600 × (40/100)

⇒ 2240

∴ The valid votes of 2^nd candidate is 2240

Pattern of the given series can be explained as,

Calculation:

The series follows the following pattern

31 + 1² = 32

32 + 2² = 36

36 + 3² = 45

45 + 4² = 61

61 + 5² = 86

∴ The missing number is 45.

The logic behind the following series is as follows:

3 × 2 = 6

6 × 3 = 18

18 × 5 = ? = 90

90 × 7 = 630

630 × 11 = 6930

∴ The value of ‘?’ is 90.

Given:

60, 40, 50, ?, 180, 460

Calculation:

The series follows the following pattern

60 × 0.5 + 10 = 40

40 × 1 + 10 = 50

50 × 1.5 + 10 = 85

85 × 2 + 10 = 180

180 × 2.5 + 10 = 460

∴ The Required term in the series will be 85

Calculation:

The series follows the following pattern

13 × 1 + 1 = 14

14 × 2 + 2 = 30

30 × 3 + 3 = 93

93 × 4 + 4 = 376

376 × 5 + 5 = 1885

∴ The missing number is 93.