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MCA NIMCET Profit And Loss PYQ


MCA NIMCET PYQ 2019
A dealer offers a cash discount of 20% and still makes a profit of 20%, when he further allows 16 articles to a dozen to a particularly sticky bargainer. How much percent above the actual price price were his was the listed price of the article?





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MCA NIMCET Previous Year PYQMCA NIMCET NIMCET 2019 PYQ

Solution

Problem: Calculate Percent Above Cost for Listed Price

Given:

  • Cash discount = 20%
  • Profit = 20%
  • Offers 16 articles for the price of 12 (i.e. sells 16 for price of 12)

Step 1: Let the Cost Price per article = C

Step 2: Dealer makes 20% profit, so S.P. per article (normal) = 1.2C

Step 3: But dealer gives 16 articles for price of 12, so effectively:

Effective S.P. per article=12×Listed Price per article×(10.20)16 Here, listed price per article = L, and after 20% discount, price per article = 0.8L So, Effective S.P. per article=12×0.8L16=0.6L

Step 4: Since effective S.P. per article = 1.2C (profit price), we have:

0.6L=1.2CL=1.2C0.6=2C

Step 5: The listed price L is 2 times the cost price C, so:

Percent above cost=(LCC)×100=(21)×100=100%

Answer: The listed price was 100% above the actual cost price.


MCA NIMCET PYQ 2019
A dealer offers sells half of the eggs that he has and another half an egg to Anurag. Then he sells half of the balance eggs and another half an egg to Deepak. Then he sells half of the balance eggs and another half an egg to Sivani. In the end he is left with just 7 eggs and he claims that he never broke an egg. How many eggs did he start with?





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MCA NIMCET Previous Year PYQMCA NIMCET NIMCET 2019 PYQ

Solution

Solution:

Let the initial number of eggs = x.

First sale (to Anurag):
Sells half of the eggs plus half an egg:
Eggs sold=x2+12 Remaining eggs: x(x2+12)=x212
Second sale (to Deepak):
Sells half of remaining eggs plus half an egg:
Remaining eggs: (x212)(12(x212)+12)=x454
Third sale (to Shivani):
Sells half of remaining eggs plus half an egg:
Remaining eggs: (x454)(12(x454)+12)=x98
Given that after all sales, 7 eggs are left: x98=7 Multiplying both sides by 8: x9=56 x=65
Verification:
Since dealer never broke an egg, and he must sell whole number of eggs each time:

Checking for x=65:
- After first sale: Remaining eggs = 32
- After second sale: Remaining eggs = 16
- After third sale: Remaining eggs = 7 (Not matching without breaking eggs). ❌

Checking for x=63:
- After first sale: Sells 632+12=32 eggs, Remaining = 31 eggs ✅
- After second sale: Sells 312+12=16 eggs, Remaining = 15 eggs ✅
- After third sale: Sells 152+12=8 eggs, Remaining = 7 eggs ✅

Thus, the dealer originally had: 63 eggs


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