Prakash's Widget Testing Optimization

CAT 1991 Slot 1 · DILR · Hard · Data Interpretation

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Passage / data set

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: Costs Rs. 2 per widget. It sends 20% of the bad ones to the buyer as good. (b) Use 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 of 2

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

  1. A.

    less than 100

  2. B.

    more than 200

  3. C.

    between 120 & 190

  4. D.

    Cannot be found out

Answer

A

Explanation

Let xx be the number of defective widgets in 1000.

  • Cost without testing =50x= 50x.
  • Cost with Test I =2000+25(0.8x)+50(0.2x)=2000+30x= 2000 + 25(0.8x) + 50(0.2x) = 2000 + 30x. Comparing No Test vs Test I: 50x<2000+30x  ⟹  20x<2000  ⟹  x<10050x < 2000 + 30x \implies 20x < 2000 \implies x < 100. Thus, Prakash should not test if x<100x < 100.

Question 2 of 2

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

  1. A.

    should either use Test I or not test.

  2. B.

    should either use Test II or not test.

  3. C.

    should use Test I or Test II.

  4. D.

    should use Test I only.

Answer

D

Explanation

For x=120x = 120:

  • No test cost =50×120=6000= 50 \times 120 = 6000
  • Test I cost =2000+30(120)=5600= 2000 + 30(120) = 5600
  • Test II cost =3000+25(120)=6000= 3000 + 25(120) = 6000 Test I yields the lowest cost (5600), so he should use Test I only.

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