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Effective rate for half-yearly compounding = 20% ÷ 2 = 10% per 6 months.

For 1 year, there are 2 compounding periods.

Net effective interest rate = 10 + 10 + (10 × 10) ÷ 100 = 21%.

Compound Interest = 21% of ₹8,000 = 0.21 × 8,000 = ₹1,680.

∴ The correct answer is ₹1,680.

Alternate Method

Given:

Principal (P) = ₹8,000

Rate of Interest (R) = 20% per annum

Time (T) = 1 year

Formula Used:

For half-yearly compounding: Rate (r) = R ÷ 2, Time periods (n) = 2 × T

Amount (A) = P(1 + r/100)ⁿ

Compound Interest (CI) = A − P

Calculations:

⇒ Effective rate (r) = 20 ÷ 2 = 10% per half-year

⇒ Number of periods (n) = 1 × 2 = 2 periods

⇒ A = 8000 × (1 + 10/100)²

⇒ A = 8000 × (1.1)²

⇒ A = 8000 × 1.21

⇒ A = ₹9,680

⇒ CI = 9680 − 8000

⇒ CI = ₹1,680

∴ The correct answer is ₹1,680.

Additional Information

Effective Annual Rate (EAR)

When interest is compounded more than once a year, the effective annual rate is given by [(1 + r/n)ⁿ − 1] × 100, where n is the number of compounding periods.

Quarterly Compounding

If interest is compounded quarterly, the rate is divided by 4 and the time (in years) is multiplied by 4.

CI vs SI Difference

For 2 years (or 2 periods), the difference between CI and SI is P(R/100)².

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Use the direct ratio formula: MP / CP = (100 + Profit%) / (100 − Discount%)

MP / 600 = (100 + 25) / (100 − 20) = 125 / 80

MP / 600 = 25 / 16

MP = (25 × 600) ÷ 16 = 25 × 37.5 = 937.50

∴ The correct answer is Rs. 937.50.

Alternate Method

Given:

Cost Price (CP) = Rs. 600

Profit Percentage = 25%

Discount Percentage = 20%

Formula Used:

1. Selling Price (SP) = CP × (100 + Profit%) ÷ 100
2. Marked Price (MP) = SP × 100 ÷ (100 − Discount%)

Calculations:

⇒ Selling Price (SP) = 600 × (100 + 25) ÷ 100

⇒ SP = 600 × 1.25 = Rs. 750

⇒ Marked Price (MP) = 750 × 100 ÷ (100 − 20)

⇒ MP = 750 × 100 ÷ 80

⇒ MP = 750 × 1.25 = Rs. 937.50

∴ The correct answer is Rs. 937.50.

Additional Information

Markup Percentage

It is the percentage by which the cost price is raised to fix the marked price. Formula: Markup% = (MP − CP) ÷ CP × 100.

Effective Discount

When two successive discounts a% and b% are given, the single equivalent discount is (a + b − ab/100)%.

Relation between Markup, Profit and Discount

The relationship is given by: 1 + Profit% = (1 + Markup%) × (1 − Discount%) where percentages are in decimal form.

⇒ Average weight of 11 students = 37

⇒ Sum of weights of 11 students = 37 × 11 = 407

Now, average weights of 10 students = 35

⇒ Sum of weights of 10 students = 35 × 10 = 350

∴ Weight of the student who left = 407 – 350 = 57 kg

ALTERNATE SOLUTION:

Weight of student who left = (37 × 11) – 350 = 57

Given:

Varun = 3 × Sister

Concept used:

x² – (sum of roots)x + product of roots = 0

Calculation:

According to the question,

(V + 6) × (S + 6) = 231

⇒ (3S + 6)(S + 6) = 231

⇒ 3S² + 18S + 6S + 36 = 231

⇒ 3S2 + 24S - 195 = 0

⇒ S2 + 8S - 65 = 0

⇒ S2 + 13S - 5S - 65 = 0

⇒ S(S + 13) - 5(S + 13) = 0

⇒ (S + 13)(S - 5) = 0

⇒ S = - 13, 5

Age cannot be in negative

So age of sister = 5 years

So, age of Varun is 15 years

∴ Varun's present age is 15 years

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Convert percentage to fraction: 36(2/3)% = 11/30.

Let Total Income = 30 units, then Expenditure = 11 units.

Savings = 30 − 11 = 19 units.

Given: 19 units = ₹5,700 ⇒ 1 unit = ₹300.

Monthly Expenses = 11 units = 11 × 300 = ₹3,300.

∴ The correct answer is ₹3,300.

Alternate Method

Given:

Expenditure Percentage = 36(2/3)% = 110/3%

Monthly Savings = ₹5,700

Calculations:

⇒ Savings Percentage = 100% − 36(2/3)% = 63(1/3)%

⇒ 63(1/3)% = 190/3%

⇒ According to the question, (190/3)% of Total Income = 5,700

⇒ Total Income = (5,700 × 3 × 100) / 190

⇒ Total Income = 30 × 3 × 100 = ₹9,000

⇒ Monthly Expenses = Total Income − Savings

⇒ Monthly Expenses = 9,000 − 5,700 = ₹3,300

∴ The correct answer is ₹3,300.

Additional Information

Percentage to Fraction Conversion

Common fractions like 33(1/3)% = 1/3 and 66(2/3)% = 2/3 help in faster calculations during exams.

Income, Expenditure and Savings Relation

Income = Expenditure + Savings. Always ensure the sum of percentages equals 100%.

Base Change in Percentages

If A is x% more than B, then B is [x / (100+x)] × 100 % less than A.

Given

Speed of the boat in still water = 8 km/h

Speed of the stream = 2 km/h

Distance = 90 km

Concept:

The speed of the boat upstream (against the current) is (boat speed - stream speed), and downstream (with the current) is (boat speed + stream speed). Total time is the sum of the times for the upstream and downstream journeys.

Solution:

⇒ Upstream speed = 8 km/h - 2 km/h = 6 km/h

⇒ Downstream speed = 8 km/h + 2 km/h = 10 km/h

⇒ Time upstream = Distance / Speed = 90 km / 6 km/h = 15 hours

⇒ Time downstream = Distance / Speed = 90 km / 10 km/h = 9 hours

⇒ Total time = Time upstream + Time downstream = 15 hours + 9 hours = 24 hours

Therefore, the total time taken by him in the given two-way movement is 24 hours.

According to the question

Animals are 1/12^th part of the town's population [30°/360° = 1/12]

So Animals = 720000 × 1/12

⇒ 60000

∴ The number of animals in the town is 60000

Given Data:

The age of Sitamma 10 years ago was thrice the age of Radha

10 years hence, Sitamma age will be twice that of Radha

Calculation:

Let the present ages of Sitamma and Radha be x and y years respectively.

According to the question,

10 years ago, the age of Sitamma was 3 times the age of Radha.

(x – 10) = 3(y – 10)

⇒ 3y – x = 20      ----(1)

And, 10 years later, the age of Sitamma will be twice the age of Radha.

(x + 10) = 2(y + 10)

⇒ x – 2y = 10      ----(2)

Adding equation (1) and (2), we get

⇒ y = 30 years

Put y = 30 years in equation (1), we get

⇒ 3 × 30 – x = 20

⇒ 90 – 20 = x

⇒ x = 70 years

Now, we have to find the ratio of present age  i.e. x : y

⇒ x : y

⇒ 70 : 30

⇒ 7 : 3

∴ The ratio of the present age of Sitamma and Radha is 7 : 3

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Net deviation of 13 numbers from the overall average 30 = 13 × (32 − 30) = +26.

To balance the total average, the deviation of the remaining 2 numbers must be −26.

Individual deviation for each of the 2 numbers = −26 ÷ 2 = −13.

Value of each number = Overall Average + Individual Deviation = 30 − 13 = 17.

∴ The correct answer is 17.

Alternate Method

Given:

Average of 15 numbers = 30

Average of 13 numbers = 32

Remaining 2 numbers are equal (Let each be x).

Formula Used:

Sum of observations = Average × Number of observations

13 Numbers Avg = 32 2 Numbers Avg = x

Total 15 Numbers (Average = 30)

Calculations:

⇒ Total sum of 15 numbers = 15 × 30 = 450

⇒ Sum of 13 numbers = 13 × 32 = 416

⇒ Sum of remaining 2 numbers = Total sum − Sum of 13 numbers

⇒ x + x = 450 − 416

⇒ 2x = 34

⇒ x = 34 ÷ 2 = 17

∴ The correct answer is 17.

Additional Information

Weighted Average Formula

The overall average of two groups is given by: Avg_total = (n₁ × A₁ + n₂ × A₂) ÷ (n₁ + n₂).

Net Deviation Rule

The sum of deviations of all the observations from their arithmetic mean is always zero (∑(x_i − mean) = 0).

Inclusion/Exclusion in Average

Value of new observation = New Average + (Number of old observations × Change in Average).

From the given information:

1: The person who is sitting third to the left of Diksha is at one of the extreme ends.

We place Diksha third to the left of the extreme left.

2: Ekta is sitting fourth of the right to Gopi. Falak is sitting to the immediate right of Ekta.

Case 1:

Case 2:

3: Komal is an immediate neighbour of Bhawna and Diksha. Therefore Case 2 is eliminated.

Now we place Bhawna and Komal such that Komal becomes the immediate neighbor of Bhawna & Diksha.

Now we place Anmol at the only space left,

From the arrangement, we can see that Bhawna is the immediate neighbour of Gopi and Komal.

Hence the correct answer is option 2)