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Given :

There are four consecutive even numbers, such that the average of the last and the first number is 11.

Calculations :

Let the numbers be 2x, (2x + 2), (2x + 4) and (2x + 6)             (numbers are even)

According to the question 

Average of the first and the last is 11 

So, 

(2x + 2x + 6)/2 = 11 

⇒ 4x + 6 = 22 

⇒ 4x = 16 

⇒ x = 4 

So the number are 8, 10, 12 and 14 

Difference between the second and the third number = 12 - 10 

⇒ 2 

∴ The correct answer will be option number 2 

Alternate Method

Calculations:

Here, all four numbers are consecutive even number and the difference between two consecutive even number is 2.

So, the difference between the second and the third number is 2

∴ The correct answer will be option number 2

Shortcut Trick

Savings of A = 100% − 85% = 15% of salary A.

Savings of B = 100% − 95% = 5% of salary B.

Since Savings are equal: 15% of A = 5% of B ⇒ 3A = 1B ⇒ A : B = 1 : 3.

Total Ratio parts = 1 + 3 = 4 units → ₹40000. So, 1 unit = ₹10000.

∴ The correct answer is ₹10,000.

Alternate Method

Given: Total salary = ₹40,000. Expenditure of A = 85%, Expenditure of B = 95%. Savings are equal.

⇒ Savings of A = (100 − 85)% of A = 15% of A

⇒ Savings of B = (100 − 95)% of B = 5% of B

⇒ 15% × A = 5% × B

⇒ A / B = 5 / 15 = 1 / 3

⇒ Sum of ratios = 1 + 3 = 4

⇒ Salary of A = (1 / 4) × 40000

⇒ Salary of A = ₹10,000

∴ The correct answer is ₹10,000.

Additional Information

Income Expenditure Formula

Income = Expenditure + Savings. This basic identity is used to calculate missing components when percentages are given.

Ratio of Salaries

If savings are equal, the ratio of salaries is inversely proportional to the savings percentage: Salary₁ × Savings%₁ = Salary₂ × Savings%₂.

Percentage to Fraction Conversion

To simplify calculations, convert percentages to fractions: 5% = 1/20, 15% = 3/20.

( A + B + C ) in 25 days 

Alone C in 50 days 

Total work in LCM of 25 and 50 is 50 unit 

So, efficiency of A + B + C is 2 unit and C is 1 unit 

All A, B and C working for 20 days 

Then work is during 20 days = 20 × 2 unit = 40 unit 

Then A and B left the work,

Remaining work to be completed by C = ( 50 - 40 ) = 10 unit 

Days taken by C to complete this work = 10 /1 = 10 days 

Hence 10 days is the correct answer.

Given:

Distance covered in 4 hours in upstream = 28 km

Distance covered in 2 hours in downstream = 50 km

Formula used:

Speed of an object in downstream = x + y

Speed of an object in upstream = x – y

Speed of object in still water, x = (D + U)/2

Speed of stream, y = (D - U)/2

where,

D = Speed of object in downstream

U = Speed of object in upstream

Calculation:

Speed of object in upstream = 28/4

⇒ 7 km/h

Speed of object in downstream = 50/2

⇒ 25 km/h

Speed of object in still water, x = (D + U)/2

⇒ x = (7 + 25)/2

⇒ 16 km/h

Time taken to cover 32 km in still water

⇒ 32/16

⇒ 2 hours

∴ Ujjwal will take 2 hours to cover 32 km in still water.

Given :

Ratio of investment of A : B = 5 : 6

Ratio of investment of B : C = 3 : 7

A, B, and C invested for 12 months, 10 months, and 10 months respectively.

Profit share of A = 1740 Rs.

Calculation :

Ratio of investment for A : B : C = 5 : 6 : 7 × 2 = 5 : 6 : 14

Ratio of time period for A : B : C = 12 : 10 : 10 = 6 : 5 : 5

Ratio of profit earned by A : B : C = 5 × 6 : 6 × 5 : 14 × 5 = 30 : 30 : 70 = 3 : 3 : 7

Let the profit share of C = x

ATQ, 1740/x = 3/7

⇒ x = 4060 Rs.

Hence, the correct answer is 4060 Rs.

Given:

Eight boxes - P, Q, R, S, T, U, V and W

The bottommost box is numbered 1 while the box above it is 2, and so on.

Now,

1) Box U is a prime-numbered box but not box 5.

2. The number of boxes above W is equal to the number of boxes below box U.

3. Box number of box S is twice of box number of box V.

4. Box P is just above box U.

5. Box Q is two boxes above box T. This is not possible in Cases II and III.

Thus, the final arrangement is -

Box U is just above box S.

The correct answer is U.

Given:

Eight boxes - P, Q, R, S, T, U, V and W

The bottommost box is numbered 1 while the box above it is 2, and so on.

Now,

1) Box U is a prime-numbered box but not box 5.

2. The number of boxes above W is equal to the number of boxes below box U.

3. Box number of box S is twice of box number of box V.

4. Box P is just above box U.

5. Box Q is two boxes above box T. This is not possible in Cases II and III.

Thus, the final arrangement is -

Seven boxes are below box R.

The correct answer is Seven.

Given:

Eight boxes - P, Q, R, S, T, U, V and W

The bottommost box is numbered 1 while the box above it is 2, and so on.

Now,

1) Box U is a prime-numbered box but not box 5.

2. The number of boxes above W is equal to the number of boxes below box U.

3. Box number of box S is twice of box number of box V.

4. Box P is just above box U.

5. Box Q is two boxes above box T. This is not possible in Cases II and III.

Thus, the final arrangement is -

Box W is just above box T.

The correct answer is W.

Given:

Eight boxes - P, Q, R, S, T, U, V and W

The bottommost box is numbered 1 while the box above it is 2, and so on.

Now,

1) Box U is a prime-numbered box but not box 5.

2. The number of boxes above W is equal to the number of boxes below box U.

3. Box number of box S is twice of box number of box V.

4. Box P is just above box U.

5. Box Q is two boxes above box T. This is not possible in Cases II and III.

Thus, the final arrangement is -

V is the bottommost box.

The correct answer is V.

Given:

Eight boxes - P, Q, R, S, T, U, V and W

The bottommost box is numbered 1 while the box above it is 2, and so on.

Now,

1) Box U is a prime-numbered box but not box 5.

2. The number of boxes above W is equal to the number of boxes below box U.

3. Box number of box S is twice of box number of box V.

4. Box P is just above box U.

5. Box Q is two boxes above box T. This is not possible in Cases II and III.

Thus, the final arrangement is -

There are four boxes above box P.

The correct answer is Four.