Given:

Speed of boat in still water = 4 kmph

Speed against stream = 2 kmph

Formula used:

Speed of boat against the stream = speed of boat in still water - speed of stream

Calculations:

Let Speed of boat in still water be = X and,

Speed of stream be = Y

So, Speed of boat against the stream = X - Y

=> 4 - Y = 2

=> Y = 4 - 2

=> Y = 2

∴ The answer is None of these.

Given-

Two taps A and B fill the tank individually in 10 min and 20 min respectively.

A leakage C at the bottom which can empty it in 40 min.

Solution-

Let the capacity of the tank = LCM of (10, 20, 40) = 40 unit

⇒ Efficiency of A = 40/10 = 4 unit/min

⇒ Efficiency of B = 40/20 = 2 unit/min

⇒ Efficiency of C = 40/40 = 1 unit/min

⇒ (A + B - C) = (4 + 2 - 1) unit/min = 5 unit/min

⇒ time = 40/5 = 8 min

∴ time taken by Both the taps fill the tank with leakage is 8 min.

Now, on applying componendo and dividendo, we get

Alternate Method

Calculation:

Given:

M + 7W + 5C = 18 hours

2M + 10C = 30 hours

2M + 3W = 45 hours

Concept used:

Total work = LCM

Work done in unit value of time is known as efficiency

Formula used:

Efficiency = total work/total days

Calculations:

Let total work be x

Total efficiency of M + 7W + 5C = x/18     ----(1)

Total efficiency of 2M + 10C = x/30     ----(2)

Total efficiency of 2M + 3W = x/45     ----(3)

Equation (2) is divided by 2

Total efficiency of M + 5C = x/(30 × 2)

⇒ x/60

Put the efficiency value of M + 5C in equation (1)

M + 7W + 5C = x/18

⇒ x/60 + 7W = x/18

⇒ 7W = (x/18) – (x/60)

⇒ 7W = (10x – 3x)/180

⇒ 7W = 7x/180

⇒ W = x/180

Efficiency of one woman = x/180

Put the value of efficiency of one woman in equation (3)

2M + 3W = x/45

⇒ 2M + 3(x/180) = x/45

⇒ 2M + x/60 = x/45

⇒ 2M = (x/45) – (x/60)

⇒ 2M = (4x – 3x)/180

⇒ 2M = x/180

⇒ M = x/360

Put the value of efficiency of one man in equation (2)

2M + 10C = x/30

⇒ 2(x/360) + 10C = x/30

⇒ x/180 + 10C = x/30

⇒ 10C = (x/30) – (x/180)

⇒ 10C = (6x – x)/180

⇒ 10C = 5x/180

⇒ 2C = x/180

⇒ C = x/360

Efficiency of 3M + 9C = 3(x/360) + 9(x/360)

⇒ x/120 + x/40

⇒ (x + 3x)/120

⇒ 4x/120

⇒ x/30

Time taken 3M + 9C to complete the work = Total work/efficiency

⇒ x/(x/30)

⇒ 30 hours

∴ Time taken 3M + 9C to complete the work is 30 hours

Alternate method:

Total work = LCM

⇒ 180

Total efficiency of M + 7W + 5C = 180/18 = 180     ----(1)

Total efficiency of 2M + 10C = 180/30 = 6     ----(2)

Total efficiency of 2M + 3W = 180/45 = 4     ----(3)

Equation (2) is divided by 2

Total efficiency of M + 5C = 180/(30 × 2)

⇒ 180/60 = 3

Put the efficiency value of M + 5C in equation (1)

M + 7W + 5C = 10

⇒ 3 + 7W = 10

⇒ 7W = 10 – 3

⇒ 7W = 7

⇒ W = 1

Efficiency of one woman = 1

Put the value of efficiency of one woman in equation (3)

2M + 3W = x/45

⇒ 2M + 3(1) = 4

⇒ 2M + 3 = 4

⇒ 2M = 4 – 3

⇒ 2M = 1

⇒ M = 1/2

Put the value of efficiency of one man in equation (2)

2M + 10C = 6

⇒ 2(1/2) + 10C = 6

⇒ 1 + 10C = 6

⇒ 10C = 6 – 1

⇒ 10C = 5

⇒ C = 5/10

⇒ C = 1/2

Efficiency of 3M + 9C = 3(1/2) + 9(1/2)

⇒ (3/2) + (9/2)

⇒ 12/2 = 6

Time taken 3M + 9C to complete the work = Total work/efficiency

⇒ 180/6

⇒ 30 hours

∴ Time taken 3M + 9C to complete the work is 30 hours

Calculation:

Key Points

Let the sides of the rectangle = a and b

Area of the rectangle = a x b =  30 cm² ... (1)

and perimeter = 2(a+b) = 26 cm ... (2)

By putting the options in the equations, we get a = 10 cm and b = 3 cm

Additional Information

a + b = 13 cm

a + (30/a) = 13

⇒ a² + 30 = 13a

⇒ a2 - 13a + 30 = 0

⇒ a2 - 10a - 3a + 30 = 0

⇒ a(a - 10) -3 (a - 10) = 0

⇒ (a - 3)(a - 10) = 0

Hence a = 3 or a =10

By putting the value of 'a' into equation 1, we get

b = 10 or b =3

Given:

The ratio of the their marks = 8 : 7

Combined average of their percentage = 85

Sum of their marks = 255

Formula  Used:

Concept of Ratios and Percentage.

Calculation:

Let Bholu's marks be 8x and Golu's marks be 7x.

So, the sum of the marks = 8x + 7x = 15x

But it is given that,

15x = 255

⇒ x = 17

Bholu marks = 8x = 8 × 17 = 136

Golu marks = 7x = 7 × 17 = 119

∴ The combined average of their marks = (136 + 119)/2 = 255/2 = 127.5.

If the total marks of the exam is 100 then their combined average of their percentage is 85.

If their combined average of their percentage is 127.5, then the total marks = (127.5/85) × 100 = 150

∴ The maximum mark are 150.

Given:

An item of marked price Rs. 80 was sold in Rs. 68.

Formula used:

Selling Price = (100% - Discount%) of Marked Price

Calculation:

According to the question,

⇒ 68 = (100% - D%) of 80

⇒ 17 = (1 - D/100) of 20

⇒ 17 = 20 - 20D/100

⇒ D/5 = 20 - 17

⇒ D = 15%

The discount = 15%

Therefore, "15%" is the required answer.

Rate of Interest = 18% per annum

Time period of Loan = 4 years

As per the formulae of Simple Interest,

S.I = (P × N × R)/100      --- (1)

where, S.I = Interest incurred on the principal amount, P = Principal amount, N = Time period of the loan, R = Rate of interest

From (1), we get

12,240 = (P × 4 × 18)/100

⇒ P = (12240 × 100)/(4 × 18)

⇒ P = 17000

Thus, Raghu has borrowed Rs.17,000/-

MP of the article = 60

SP after discount = 60 - 0.05 × 60 = 57

⇒ SP after gift = 57 - 3 = 54

∴ Final SP = Rs. 54

Profit % = 20

? × 65/0.31 = (961)^–0.5 – (441)^0.5

⇒ ? × 65/0.31 = 961^–1/2 – 441^1/2

⇒ ? × 65/0.31 = (1/961)^1/2 – 21

⇒ ? × 65/0.31 = (1/31) – 21

⇒ ? × 65/0.31 = (1 - 651)/31

⇒ ? × 65/0.31 = -650/31

⇒ ? = -650/31 × 0.31/65

∴ ? = -0.1