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Olympiad Test: Comparing Quantities - Class 8 MCQ


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20 Questions MCQ Test - Olympiad Test: Comparing Quantities

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Olympiad Test: Comparing Quantities - Question 1

A sum of money, at compound interest, yields Rs. 200 and Rs. 220 at the end of first and second years respectively. What is the rate percent?

Detailed Solution for Olympiad Test: Comparing Quantities - Question 1

Given:

  • The amount at the end of the first year = Rs. 200
  • The amount at the end of the second year = Rs. 220

We need to calculate the rate percent.

Step 1: Find the Compound Interest for the second year

The difference between the amount at the end of the second year (Rs. 220) and the amount at the end of the first year (Rs. 200) is the compound interest earned in the second year:

Compound Interest for 2nd year = 220 - 200 = Rs. 20

This Rs. 20 is the interest earned on Rs. 200 (the amount at the end of the first year).

Step 2: Use the formula for Compound Interest

Let the principal be P and the rate be r% per annum.

At the end of the first year, the amount is given by:

A1 = P * (1 + r/100)

We are given that A1 = 200, so:

P * (1 + r/100) = 200

At the end of the second year, the amount is given by:

A2 = P * (1 + r/100)2

We are given that A2 = 220, so:

P * (1 + r/100)2 = 220

Step 3: Solve the equations

We already know that the compound interest for the second year is Rs. 20. The interest for the second year is the difference between the amount at the end of the second year and the amount at the end of the first year:

A2 - A1 = 220 - 200 = 20

Now, substitute A1 = P * (1 + r/100) into the equation:

P * (1 + r/100) = 200

This gives:

P = 200 / (1 + r/100)

Now substitute P into A2 = P * (1 + r/100)2 = 220:

(200 / (1 + r/100)) * (1 + r/100)2 = 220

Simplifying:

200 * (1 + r/100) = 220

Solving for r:

1 + r/100 = 220 / 200

1 + r/100 = 1.1

r/100 = 0.1

r = 10

Correct Option: The correct option is: C: 10%

Olympiad Test: Comparing Quantities - Question 2

The sale price of a shirt is Rs.176. If a discount of 20% is allowed on its marked price, what is the marked price of the shirt?

Detailed Solution for Olympiad Test: Comparing Quantities - Question 2

Given:

  • Sale price of the shirt = Rs. 176
  • Discount = 20%

Let the marked price of the shirt be M.

Since a 20% discount is allowed, the sale price is 80% of the marked price. So, we can write:

Now, solve for M:

Thus, the marked price of the shirt is Rs. 220.

Correct option: D: Rs. 220

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Olympiad Test: Comparing Quantities - Question 3

The difference in S.I. and C.I. on a certain sum of money in 2 years at 15% p.a. is Rs.144. Find the sum.

Detailed Solution for Olympiad Test: Comparing Quantities - Question 3

The problem involves finding the principal (sum of money) when the difference between Simple Interest (S.I.) and Compound Interest (C.I.) for 2 years at a rate of 15% p.a. is given as Rs. 144.

Formula for the difference between S.I. and C.I. for 2 years:

Where:

  • P = Principal (sum of money to find)
  • R = Rate of interest (15%=15/100=0.15)
  • Difference = Rs. 144 (given)

Substituting values into the formula:

Final Answer:

The principal (sum of money) is Rs. 6400, which matches Option d).

Olympiad Test: Comparing Quantities - Question 4

In what time will a sum of Rs. 800 at 5% p.a. C.I. amount to Rs. 882?

Detailed Solution for Olympiad Test: Comparing Quantities - Question 4

Given:

  • Principal (P) = Rs. 800
  • Rate (R) = 5% per annum
  • Amount (A) = Rs. 882

We can use the formula for compound interest:

A = P * (1 + R/100)t

Where:

  • A is the amount after t years,
  • P is the principal,
  • R is the rate of interest,
  • t is the time in years.

Substitute the given values:

882 = 800 * (1 + 5/100)t

Now simplify:

882 = 800 * (1.05)t

Now divide both sides by 800:

882 / 800 = (1.05)t

1.1025 = (1.05)t

Take the logarithm of both sides:

log(1.1025) = t * log(1.05)

Using a calculator:

log(1.1025) ≈ 0.0436 and log(1.05) ≈ 0.0212

Now solve for t:

t = 0.0436 / 0.0212 ≈ 2.06

So, the time is approximately 2 years.

Correct Answer: D: 2 years

Olympiad Test: Comparing Quantities - Question 5

What is the C.I. on Rs. 8000 for 1 year at 5% p.a. payable half-yearly?

Detailed Solution for Olympiad Test: Comparing Quantities - Question 5

Given:

  • Principal (P) = Rs. 8000
  • Rate (R) = 5% per annum
  • Time (T) = 1 year
  • Interest is compounded half-yearly, so the rate for each half-year = 5%/2 = 2.5% per half year.
  • Number of half-years (n) = 2

Formula for Compound Interest:

Where:

  • A is the amount after time t,
  • P is the principal,
  • R is the annual interest rate,
  • t is the time in years,
  • n is the number of compounding periods per year.

Step-by-step Calculation:

  1. Substitute the values into the formula:
  2. Now calculate the Compound Interest:
    C.I. = A − P
    C.I. = 8405 − 8000 = 405

Therefore, the Compound Interest is Rs. 405.

Correct Answer: D: Rs. 405

Olympiad Test: Comparing Quantities - Question 6

Raghu borrowed Rs. 25000 at 20% p.a. compounded half-yearly. What amount of money will clear his debt after ?

Detailed Solution for Olympiad Test: Comparing Quantities - Question 6

Given:

  • Principal (P) = Rs. 25,000
  • Rate (R) = 20% per annum
  • Time (T) = 1.5 years
  • Compounded half-yearly, so the rate for each half-year = 20/2 = 10% per half year.
  • Number of compounding periods (n) = 2 compounding periods per year, so for 1.5 years, the total number of periods (nt) = 3 periods.

Formula for Compound Interest:

Where:

  • A is the amount after time t,
  • P is the principal,
  • R is the annual interest rate,
  • t is the time in years,
  • n is the number of compounding periods per year.

Step-by-step Calculation:

Substitute the values into the formula:

Therefore, the amount Raghu will need to pay after 1.5 years is Rs. 33,275.

Correct Answer: C: Rs. 33275​​​​​​​

Olympiad Test: Comparing Quantities - Question 7

In what time will Rs.1000 amount to Rs.1331 at 10% p.a. compounded annually?

Detailed Solution for Olympiad Test: Comparing Quantities - Question 7

Using formula for compound interest,

According to the question,

Amount = 1331, P = 1000

R = 10%

Substituting values in the above formula,

∴ It will take 3 years for principal amount of Rs. 1000 to become Rs. 1331 when compounded annually at 10% per annum.

Olympiad Test: Comparing Quantities - Question 8

The present population of a town is 25000. It grows at 4%, 5% and 8% during first year, second year and third year respectively. Find the population after 3 years.

Detailed Solution for Olympiad Test: Comparing Quantities - Question 8

Population after n years is calculated as:

Pn = P0 × (1 + r1) × (1 + r2) × (1 + r3)

Where:

  • P0 is the initial population,
  • r1, r2, r3 are the growth rates for each year.

Given:

  • Initial population P0 = 25000
  • Growth rates:
    • Year 1: 4% = 0.04
    • Year 2: 5% = 0.05
    • Year 3: 8% = 0.08

Now, we can calculate the population after each year:

Year 1:

Population after Year 1 = 25000 × (1 + 0.04) = 25000 × 1.04 = 26000

Year 2:

Population after Year 2 = 26000 × (1 + 0.05) = 26000 × 1.05 = 27300

Year 3:

Population after Year 3 = 27300 × (1 + 0.08) = 27300 × 1.08 = 29484

Olympiad Test: Comparing Quantities - Question 9

The value of an article which was purchased 2 years ago, depreciates at 12% per annum. If its present value is Rs.9680, what is the price at which it was purchased?

Detailed Solution for Olympiad Test: Comparing Quantities - Question 9

Present Value = Original Value × (1 - R/100)t

  • Present Value = Rs. 9680
  • Rate of depreciation (R) = 12% per annum
  • Time (t) = 2 years

Step-by-Step Calculation:

9680 = Original Value × (1 - 12/100)2 9680 = Original Value × (0.88)2 9680 = Original Value × 0.7744

Original Value = 9680 / 0.7744 Original Value = 12500

  1. Substitute the known values into the formula:
  2. Solve for the Original Value:

The price at which it was purchased is Rs. 12,500.

Correct Answer: B: Rs. 12500

Olympiad Test: Comparing Quantities - Question 10

The cost of a vehicle is Rs.175000. If its value depreciates at the rate of 20% per annum, find the total depreciation after 3 years.

Detailed Solution for Olympiad Test: Comparing Quantities - Question 10

Value of the vehicle after 3 years:

Value = Rs. 175000 × (1 - 20/100)3
Value = Rs. 175000 × (0.80)3
Value = Rs. 175000 × 0.512
Value = Rs. 89600

Therefore, the total depreciation is:

Total Depreciation = Rs. (175000 - 89600) Total Depreciation = Rs. 85400

The total depreciation after 3 years is Rs. 85400.

Olympiad Test: Comparing Quantities - Question 11

The C.I. on a certain sum for 2 years is Rs. 410 and S.I. is Rs. 400. What is the rate of interest per annum?

Detailed Solution for Olympiad Test: Comparing Quantities - Question 11

Given that Simple interest in 2 years = Rs. 400

⇒ Simple interest for 1 year = Rs. 400/2 = Rs. 200

Hence Compound Interest for 1styear = Rs 200

Givencompound interest for 2 years = Rs. 410

Therefore compound interest for 2 years = CI for 1st year + CI for 2nd year

⇒ Compound interest for 2nd year = Rs. 410 – Rs. 200 = Rs. 210

that is interest on Rs. 200 for 1 year = Rs. 210 – Rs 200 = Rs. 10

⇒ 2R = 10

∴ R = 5%

Olympiad Test: Comparing Quantities - Question 12

The C.I. on a certain sum for 2 years at 10% per annum is Rs. 525. Calculate the S.I. on the same sum for double the time at half the rate percent per annum.

Detailed Solution for Olympiad Test: Comparing Quantities - Question 12

Given:

Time = 2 years

Rate = 10%

C.I. = Rs. 525

Calculation:

C.I. = P[1 + (r/100)]t - P

⇒ 525 = P[(11/10)2 - 1]

⇒ 525 = P[21/100]

⇒ 2500 = P

New rate = 10% = 5%

Time = 4 years

S.I. = PRT/100

⇒ 2500 × 5 × 4/100

⇒ 25 × 20

⇒ 500

∴ S.I. will be Rs.500

Olympiad Test: Comparing Quantities - Question 13

The S.I. on a sum of money for two years is Rs. 660, while C.I. is Rs.696.30, the rate of interest being the same in both the cases. Find the rate of interest.

Detailed Solution for Olympiad Test: Comparing Quantities - Question 13

Difference between CI and SI for 2 years = 696.30 – 660 = 36.30

SI for 2 years = 660

SI for 1 year = 660/2 = 330

∴ Interest rate = (36.3/330) × 100 = 11%

Olympiad Test: Comparing Quantities - Question 14

A sum of money doubles itself in 3 years at C.I., when the interest is compounded annually. In how many years will it amount to 16 times of itself?

Detailed Solution for Olympiad Test: Comparing Quantities - Question 14

⇒ 16 = 2 × 2 × 2 × 2 = 24

⇒ n = 4

⇒ Amount doubles in  3 years

⇒ x = 3

Time = n × x

⇒ Time = 4 × 3 =12 years

∴ The amount will become 16 times in 12 years.

Olympiad Test: Comparing Quantities - Question 15

The C.I. on a certain sum at 5% for 2 years is Rs. 328. Calculate the S.I. for the sum at the same rate and for the same period.

Detailed Solution for Olympiad Test: Comparing Quantities - Question 15

Given:

Compound interest = Rs. 328, Rate = 5% and Time = 2 years

Formula used:

If P = Principal, R = Rate of Interest, T = Time period, then:

CI = P(1 + R/100)T - P

SI = (P × R × T) /100

Calculation:

328 = P(1 + 5/100)2 - P

⇒ 328 = (41/400)P

⇒ P = Rs. 3200

Now S.I = (3200 × 5 × 2)/100 = Rs. 320

∴ The simple interest of the sum is Rs. 320

Olympiad Test: Comparing Quantities - Question 16

Find the amount on Rs.12500 for 2 years compounded annually, the rate of interest being 15% for the first year and 16% for the second year.

Detailed Solution for Olympiad Test: Comparing Quantities - Question 16

The formula for compound interest is:

A = P * (1 + R/100)t

Where:

  • A = Amount after t years
  • P = Principal amount
  • R = Rate of interest
  • t = Time in years

Step-by-step Solution:

Step 1: Calculate the amount after the first year.

For the first year, the interest rate is 15%. Using the compound interest formula:

A1 = P * (1 + 15/100)
A1 = 12500 * (1 + 0.15)
A1 = 12500 * 1.15
A1 = Rs. 14375

So, after the first year, the amount is Rs. 14375.

Step 2: Calculate the amount after the second year.

For the second year, the interest rate is 16%. The amount at the end of the first year becomes the principal for the second year.

A2 = A1 * (1 + 16/100)
A2 = 14375 * (1 + 0.16)
A2 = 14375 * 1.16
A2 = Rs. 16675

So, after the second year, the amount is Rs. 16675.

Answer: The amount after 2 years is Rs. 16675.

Answer: C: Rs. 16675

Olympiad Test: Comparing Quantities - Question 17

Sugar is bought at Rs.16.20 per kg and sold at Rs.17.28  per kg. What is the gain percent?

Detailed Solution for Olympiad Test: Comparing Quantities - Question 17

Given:

  • Cost price (C.P.) = Rs. 16.20 per kg
  • Selling price (S.P.) = Rs. 17.28 per kg

Formula for Gain Percent:

Where:

Gain = Selling Price − Cost Price

Step 1: Calculate the Gain

Gain = 17.28 − 16.20 = 1.08 Rs.

Step 2: Calculate the Gain Percent

Final Answer: The gain percent is 6.67% or .

Olympiad Test: Comparing Quantities - Question 18

A woman bought two packs of toffees, with the same of number of toffees in each pack. She bought the first pack at 25 paise per toffee and the second pack at 3 toffees for 65 paise. She mixed them together and sold at Rs.3.50 a dozen. What is her gain percent?

Detailed Solution for Olympiad Test: Comparing Quantities - Question 18

Given:

  • Cost of the first pack: 25 paise per toffee.
  • Cost of the second pack: 3 toffees for 65 paise.
  • Selling price: Rs. 3.50 a dozen (12 toffees).

Step 1: Calculate the cost price of each pack

Cost of the first pack:

Cost per toffee = 25 paise

Cost of the first pack = 25 × n paise (where n is the number of toffees in the pack)

Cost of the second pack:

Cost of 3 toffees = 65 paise

Cost of 1 toffee = 65 ÷ 3 = 21.67 paise

Cost of the second pack = (65 ÷ 3) × n = 21.67 × n paise

Step 2: Total cost price of the two packs

Total cost = 25n + (65n ÷ 3) = (75n ÷ 3) + (65n ÷ 3) = (140n ÷ 3) paise

Step 3: Selling price of the two packs

Selling price of 1 toffee = 350 ÷ 12 = 29.17 paise

Total selling price for 2n toffees = 29.17 × 2n = 58.34n paise

Step 4: Calculate the gain

Gain = Total selling price − Total cost price

Gain = 58.34n − (140n ÷ 3)

Gain = (175.02n ÷ 3) − (140n ÷ 3) = (35.02n ÷ 3) paise

Step 5: Calculate the gain percent

Gain percent = (Gain ÷ Total cost price) × 100

Gain percent = (35.02n ÷ 140n) × 100

Gain percent = (35.02 ÷ 140) × 100 = 25%

Final Answer: The gain percent is 25%.

Answer: B: 25

Olympiad Test: Comparing Quantities - Question 19

A man bought 542 kg of sugar for Rs. 7560.90and sold it so as to gain 20%. What is the selling price per kilogram of sugar?

Detailed Solution for Olympiad Test: Comparing Quantities - Question 19

Given:

  • Cost price of 542 kg of sugar = Rs. 7560.90
  • Gain percentage = 20%

To find: Selling price per kilogram of sugar

Step 1: Find the Cost Price per kilogram

Cost price per kg = (Total cost price) / (Total quantity)

Cost price per kg = 7560.90 / 542 = 13.96 Rs.

Step 2: Find the Selling Price per kilogram

The man gains 20% on the cost price. So,

Selling price per kg = Cost price per kg + (20% of Cost price per kg)

Selling price per kg = 13.96 + (0.20 × 13.96)

Selling price per kg = 13.96 + 2.792 = 16.752 Rs.

Rounding to two decimal places:

Selling price per kg = 16.74 Rs.

Answer: A: Rs. 16.74

Olympiad Test: Comparing Quantities - Question 20

A trader marks his goods 30% above the cost price but makes a reduction of 25/4 on the marked price for ready money. What is his gain percent?

Detailed Solution for Olympiad Test: Comparing Quantities - Question 20

Let cost price = x
initial marked price = cost price + 30% of cost price
= X + 30%X = 130 % X
Final marked price = initial marked price - 25/4% of initial marked price
= 130%X - 25/4 % of 130 % X = 121.875 %
Therefore gain percentage = (121.875 - 100)% = 21.875%

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