5 CAT Profit and Loss Questions you should solve

Let the total property of the Alphonso be Rs.x.

After Alphonso’s death, money possessed by the family members would be

Wife = x/2,  Ben = x/6, Carl =x/6, Dave =x/6

After Ben’s death,  money possessed by each of them would be

Alphonso’s wife = x/2, Ben = 0,Ben’s wife = x/6, Carl = x/6 + x/24 = 5x/24, Dave = x/6 + x/24 = 5x/24

After Carl’s death, money possessed by them

Alphonso’s wife has  x/2, Ben has 0, Ben’s wife has x/12, Carl has 0, Carl’s wife has 5x/48, Dave has

5x/24 + 5x/48 = 15x/48

After Dave’s death, money possessed by them is:

Alphonso’s wife has x/2 + 15x/96 = 63x/96, Ben has 0, Ben’s wife has x/12, Carl’s has 0, Carl’s wife has

5x/48, Dave has 0 and Dave’s wife has  15x/96

Now, given that 63x/96 = 1575000

x= 2400000

Alternative Method:

You can also solve this question by using options.

If we take total amount to be Rs 2400000, then after Alphonso’s death, the money with the family members will be:

Alphonso’s wife = 1200000, Ben = Carl = Dave = 400000, Ben will leave 100000 each for Carl and Dave.

So, Carl and Dave have 50000 each, Carl will leave 250000 for Dave, so Dave has 750000.

Dave left 750000/2 = 375000 for his mother, so his mother has 1200000 + 375000 = 1575000, which is given in the question. hence option 4 is the answer.

It is given that

2O + 3B + 4A = 15 …..(1)

3O + 2B + A = 10…….(2)

The answer to this question seems to be cannot be determined as we are given three variables but we can form two equations only. But the question is not asking about the individual price of 3 oranges, 3 bananas and 3 apples but it asks the cost of 3O + 3B + 3A. For that, if we add the two equations, we get

5O+5B+5A=25

O+B+A=5

Therefore 3O +3B+3A = 3×5 = 15

Total cost to produce 1500 watches = (1500 × 150 + 30000) = Rs. 2,55,000

Let he sells x watches during the season, therefore

number of watches sold after the season = (1500 – x)

∴Revenue earned on the sale of 1500 watches

= 250 × x + (1500 – x) × 100 = 150x + 150000

Now, break-even is achieved if production cost is equal to the selling price.

∴ 150x + 150000 = 2,55,000 ⇒x = 700

We will solve this question by taking all the cases one by one.

In the first case it is given that the profit is 10%.

For second case, let the CP of 1 kg of sugar be Rs. 100

Then CP of 900 g of sugar= (100/1000 )x 900 = Rs. 90

Hence, profit % in Case II= [{(100-90)/90}x100] = 11.11%

For case III, If he adds 10% impurity then his CP for 1 kg

= {(100/1100) x 100} = Rs. 90.90

Hence, profit % in Case III = [{(100-90.90)/90.90} x 100] = 10.01%

and in the last case, If he reduces weight by 5%

Then cost price of 950 g = {(100/1000) x 950} = Rs. 95 and SP = Rs. 105

Hence, profit % in Case IV = {(105 – 95 )/95} X 100 = 10.52%

Hence, the profit is maximum in second case.

Let the CP of the article be Rs. x, since he earns a profit of 20%, hence SP = 1.2x.

It is given that he is selling 16 articles to a dozen, so he a incurs loss by

selling 16 articles at the cost of 12 articles [loss = {(16-12)/16} x 100 = 25%]

∴ His selling price = SP × 0.75

Now SP × 0.75 = 1.2 x⇒ SP = (1.2/0.75)x = 1.6x.

This SP is arrived after giving a discount of 20% on MP.

Hence, MP = (1.6/0.8)x = 2x

It means that article has been marked 100% above the cost price.

Alternative Method:

Let the cost price = Rs 100. Since the profit is 20%, so the SP = Rs 120.

This SP = Rs 120 is arrived after giving a discount of 20%, i.e. MP = 120/0.8 = Rs 150.

Now he is selling 16 goods to a dozen, so his loss in this case = {(16-12)/16} x100 = 25%.

It means that Rs 150 were arrived after losing 25%. Hence the actual MP = 150/0.75 = Rs 200.

Hence, he has marked the MP 100% above the CP.