CAT 1991 Question Paper

Let's say the principal amount is P, Rate is R (i.e. R = 10%)

1st Interest = $$PR$$

2nd Interest = $$\frac{4PR}{5}$$

3rd Interest = $$\frac{2PR}{5}$$

4th Interest = $$\frac{PR}{5}$$

For having the greatest percentage change, the slope of the line should be the highest. From the graph, we can see that slope is highest for A between Feb and March. So A is the correct answer.

To answer this question, let us calculate the highest increase in market value of each share.

For share A, the highest increase in market value is from Feb to March and equals 110 - 90 = 20
For share B, the market value increases constantly over all the months and is equal to 75-70 = 5
For share C, the highest increase in market value is from Feb to March and equals 60 - 55 = 5
For share D, the highest increase in market value is from Jan to Feb and equals 50 - 40 = 10

Hence, the correct answer is March and the required share is share A

To answer this question, let us calculate the highest percentage increase in market value of each share.

For share A, the highest percentage increase in market value is from Feb to March and equals 110/90 -1 = 22.22%
For share B, the highest percentage increase in market value is from Jan to Feb and equals 75/70 -1 = 7.14%
For share C, the highest percentage increase in market value is from Feb to March and equals 60/55 -1 = 9.09%
For share D, the highest percentage increase in market value is from Jan to Feb and equals 50/40 -1 = 25%

Hence, the required answer is February and the required share is share D

Loss from the transaction from Ram's standpoint is Sum of share prices of C and D - Share price of A.

So, loss from the transaction at the end of January is (40+60) - 100 - = 0
So, loss from the transaction at the end of February is (50+55) - 90= 15
So, loss from the transaction at the end of March is (60+50)-110 = 0
So, loss from the transaction at the end of April is (65+40) - 105 = 0
So, loss from the transaction at the end of May is (60+45) -100 = 5
So, loss from the transaction at the end of June is (55+45) - 110 = 0

So, the correct answer is February.

Loss from the transaction from Ram's standpoint is Sum of share prices of C and D - Share price of A.

So, loss from the transaction at the end of January is (40+60) - 100 - = 0
So, loss from the transaction at the end of February is (50+55) - 90= 15
So, loss from the transaction at the end of March is (60+50)-110 = 0
So, loss from the transaction at the end of April is (65+40) - 105 = 0
So, loss from the transaction at the end of May is (60+45) -100 = 5
So, loss from the transaction at the end of June is (55+45) - 110 = 0

So, the correct answer is 15

We should consider three possible cases as 1. when he is using test 1
2. When he is using test 2
3. When he is using no test
Now we will choose a method where the total expenditure will be least.
So for option A, let's consider that number of defective pieces are 50.
Hence, while using test 1:
Cost of testing = $$1000 \times 2 = 2000$$
Correcting 80% of pieces = $$40 \times 25 = 1000$$
Penalty for 20% of pieces = $$10 \times 50 = 500$$
Total = 3500

For test 2:
Cost of testing = $$1000 \times 3 = 3000$$
Correcting all 50 pieces = $$ 50 \times 25 = 1250 $$
Total= 4250

For no test:
Penalty for all 50 pieces = $$ 50 \times 50 = 2500$$

Hence cost is least when he is using no test while number of defective pieces is less than 100.

 

When there are 120 widgets defective:

Using 1st test:
Cost of testing = $$1000 \times 2 = 2000$$
Cost of correcting 80% pieces = $$96 \times 25 = 2400$$
Cost of penalty on 20% pieces = $$24 \times 50 = 1200$$
Total cost = 5600

Using 2nd test:

Cost of testing = $$1000 \times 3 = 3000$$
Cost of correcting all 120 pieces = $$120 \times 25 = 3000$$
Total cost = 6000

Using no test:

Total cost = 120 x 50 = 6000

Hence, he should use 1st test only.

Let's say defective pieces are 300
Using test 1
Cost of testing= $$1000 \times 2 = 2000$$
Cost of correcting 80% pieces = $$240 \times 25 = 6000$$
Cost of penalty on 20% pieces = $$60 \times 50 = 3000$$
Total cost = 11000

Using test 2
Cost of testing = $$1000 \times 3 = 3000$$
Cost of correcting all 300 pieces = $$300 \times 25 = 7500$$
Total cost = 10500

Using no test:
Cost of penalty on all pieces = $$300 \times 50 = 15000$$

Hence, he should use 2nd test.

For 160 defective pieces:
Using 1st test:
Cost of testing = $$1000 \times 2 = 2000$$
Cost correcting 80% pieces = $$128 \times 25 = 3200$$
Cost of penalty on 20% pieces = $$32 \times 50 = 1600$$
Total = 6800

Using 2nd test:
Cost of testing = $$1000 \times 3 = 3000$$
Cost of correcting all pieces = $$160 \times 25 = 4000$$
Total = 7000

Using no test:
Cost of penalty on all pieces = $$160 \times 50 = 8000$$

Hence, he should use first test only.