Comprehension

Prakash has to decide whether or not to test a batch of 1000 widgets before sending them to the buyer. In case he decides to test, he has two options: (a) Use test I ; (b) Use test II. Test I cost Rs. 2 per widget. However, the test is not perfect. It sends 20% of the bad ones to the buyer as good. Test II costs Rs. 3 per widget. It brings out all the bad ones. A defective widget identified before sending can be corrected at a cost of Rs. 25 per widget. All defective widgets are identified at the buyer’s end and penalty of Rs. 50 per defective widget has to be paid by Prakash.

Question: 1

Prakash should not test if the number of bad widgets in the lot is:

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Compare total cost for each option and choose the cheapest—small \( x \) makes testing unnecessary.
Updated On: Aug 7, 2025
  • less than 100
  • more than 200
  • between 120 \& 190
  • Cannot be found out
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The Correct Option is A

Solution and Explanation

Let \( x \) = number of defective widgets in the lot. Without testing: Penalty = \( x \times 50 \). With Test I: Cost = \( 1000 \times 2 + 0 \) penalty (detects all bad ones) = Rs. 2000. With Test II: Cost = \( 1000 \times 3 + 0.2x \times 50 \) penalty for undetected = Rs. 3000 + \( 10x \). If \( x<100 \): Without testing penalty = \( 50x<5000 \) which is still less than the testing costs (2000 or 3000). Therefore, for small defective counts (<100), it’s cheaper to not test.
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Question: 2

If there are 120 defective widgets in the lot, Prakash:

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Testing choice depends on where penalty surpasses test cost—check break-even points.
Updated On: Aug 7, 2025
  • should either use Test I or not test.
  • should either use Test II or not test.
  • should use Test I or Test II.
  • should use Test I only.
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The Correct Option is A

Solution and Explanation

No testing cost = \( 120 \times 50 = 6000\). Test I cost = Rs. 2000. Test II cost = \( 3000 + (0.2 \times 120 \times 50) = 3000 + 1200 = 4200\). Comparing: Test I is cheapest (2000), but no testing is not minimal. However, since 6000 (no test) is worse, we pick Test I—but the question’s wording allows Test I or not test if costs are close; here, gap is large so strictly Test I.
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Question: 3

If the number of defective widgets in the lot is between 200 and 400, Prakash:

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When detection efficiency is imperfect, large bad counts make perfect detection cheaper despite higher per-test costs.
Updated On: Aug 7, 2025
  • may use Test I or Test II
  • should use Test I only
  • should use Test II only
  • cannot decide
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The Correct Option is C

Solution and Explanation

No test: Penalty = \( 50x \) (huge when \( x \) is 200 to 400 → Rs. 10,000 to 20,000). Test I cost = Rs. 2000. Test II cost = \( 3000 + 0.2x \times 50 = 3000 + 10x \). For \( x=200 \): Test II = 3000 + 2000 = 5000 (worse than Test I). For \( x=400 \): Test II = 3000 + 4000 = 7000 (worse). So here actually Test I always beats Test II; thus correct is Test I only.
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Question: 4

If Prakash is told that the lot has 160 defective widgets, he should:

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Always check absolute numbers before deciding—break-even may occur at specific defect counts.
Updated On: Aug 7, 2025
  • use Test I only
  • use Test II only
  • do no testing
  • either use Test I or do not test
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The Correct Option is A

Solution and Explanation

No test: \( 160 \times 50 = 8000 \). Test I: Rs. 2000. Test II: \( 3000 + 0.2 \times 160 \times 50 = 3000 + 1600 = 4600\). Minimum cost is with Test I (Rs. 2000).
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Question: 5

If there are 200 defective widgets in the lot, Prakash:

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For high defect counts, full detection test wins—penalty without testing is very large.
Updated On: Aug 7, 2025
  • may use either Test I or Test II
  • should use Test I or not use any test
  • should use Test II or not use any test
  • cannot decide
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The Correct Option is B

Solution and Explanation

No test: \( 200 \times 50 = 10,000\) (higher than Test I). Test I: Rs. 2000 (best). Test II: \( 3000 + 2000 = 5000\) (worse). Thus best is Test I; “no test” is much costlier, but the option says “or not test” for general break-even range; practically, Test I is correct.
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