CRE 4 - Profit & Loss

SP = 1800 = 90% of CP = 90/100 × CP.

⇒ CP  = 1800 × 100/90 = Rs. 2000.

Hence, option (d).

CP  = 80 × 100/80 = 100 Rs.

Hence, option (c).

Selling Price of a watch = Rs. 480

Profit = 20%

Hence, SP = CP(1 + P/100)

⇒ 480 = CP(1 + 20/100)

⇒ Cost Price of the watch = Rs. 400

Now, If the selling price of the watch is Rs. 320

Then, Loss % = (400 - 320) × 100/400

⇒ Loss % = 20%

Hence, option (c).

Selling price of 1 orange = Rs. 1/10

Loss % = 10%

Now, SP = CP(1 - L/100)

⇒ 1/10 = CP(1 - 10/100)

Therefore, Cost Price of 1 orange = Rs. 1/9

 Now, if he had to gain 80% then,

SP = 1/9(1 + 80/100)

⇒ SP = 1/9 × 18/10

⇒ SP = Rs. 1/5

Hence, he should sell 5 oranges for a Rs. to gain 80%.

Hence, option (d).

Let number of mangoes = 100

CP of 20 = 3  Rs.

∴ CP of 100 = 15 Rs

SP of 25 = 4 Rs

∴ SP of 100 = 16 Rs

∴ Profit =1/15 × 100 = 6.66%

Hence, option (a).

Net percentage change = 30 - 30 - (30 × 30)/100

⇒ Net change = -30²/100

⇒ There is a real loss of 9%.

Option B is the correct answer.

Hence, option (b).

Let no. of orange = 6

CP of 2 = 5

∴ CP of 6 = 15

SP of 3 = 7

SP of 6 = 14

∴ Loss% = 1/15 × 100 = $6\frac{2}{3}\%$

 Hence, option (a).

SP of 8 = CP of 10 = CP of 8 + CP of 2

∴ Profit% =2/8 × 100 = 25%

Hence, option (b).

SP of 10 = CP of 8 = CP of 10 – CP of 2

∴ loss % = 2/10 × 100 = 20%

Hence, option (b).

SP of 45m - CP of 45m = CP of 15m

⇒ SP of 45m = CP of 45m + CP of 15m

⇒ Gain % = 15 × 100/45

Therefore, profit % = 33.33%

Hence, option C is the correct answer.

Hence, option (c).

Selling Price = Cost Price + Profit

SP of 33 = CP of 33 + SP of 11

SP of 22 = CP of 33

∴ Profit% = 11/22 × 100 = 50%

Hence, option B is the correct answer.

Hence, option (b).

Let the cost price be Rs. x

Hence, P% = x%

SP = CP(1 + P/100)

⇒ 75 = x(1 + x/100)

⇒ x = 50

Hence, cost price of the watch is Rs. 50

Hence, option (a).

SP of 1st half = 400 × 105/100 = Rs. 420

SP of whole = 800 × 120/100 = Rs. 960

Therefore, Second half he should sell at = (540 – 400) × 100/400 = 140 × 100/400 = 35% profit.

Hence, option (d).

Recognize, for every 3 litre of milk purchased, 4 litres of mixture is sold.

CP for 3 litres milk = 3 × Rs. 2.40 = Rs. 7.20

SP for 4 litres mixture = 4 × Rs. 2.88 = Rs. 11.52

∴ Profit% = (11.52 - 7.20)/7.20 = 0.6 = 60%

Hence, option (d).

Let cost price of A be (CP)_A, of B be (CP)_B and that of C be (CP)_C.

Similarly, selling prices be designed as (SP)_A ; (SP)_B and (SP)_C

If (CP)_A = x 

Then, (CP)_B = (SP)_A = 1.1 x 

(CP)_C = (SP)_B = (1.2) (1.1) x = Rs. 924 ⇒ x = Rs. 700

Hence, option (b).

Let CP = Rs. x and SP = Rs. 1.25 x

But it is given that, $\frac{1.25x-(x-1.75)}{x-1.75}$ = 0.6

⇒ 0.25 x + 1.75 = 0.6 x – 1.05

⇒ x = Rs. 8

Hence, option (d).

CP : 18 eggs for Rs. 16

⇒ 1 egg for Rs. 16/18

SP : 22 egg for Rs. 20

⇒ 1 egg for Rs. 20/22

= $\left[\left(\frac{20}{22}-\frac{16}{18}\right)/\left(\raisebox{1ex}{$16$}\!\left/ \!\raisebox{-1ex}{$18$}\right.\right)\right]\times 100$

= $\left[\left(\frac{180-176}{198}\right)/\left(\raisebox{1ex}{$16$}\!\left/ \!\raisebox{-1ex}{$18$}\right.\right)\right]\times 100$

= $\frac{4}{198}\times \frac{18}{16}\times 100$ = $\frac{1}{44}\times 100$ = $2\frac{3}{11}$

Hence, option (b).

Let the CP’s of buffaloes be Rs. x and (1200 – x).

Then, it is given that 0.95 x + 1.07 (1200 – x) = 1200

⇒ 0.12x = 84

⇒ x = 700

∴ Cost of other buffalo = Rs. 500

Hence, option (c).

Easiest way to solve such problems is to assume some simple values

Let 120 apples be purchased at Rs. 12/dozen

Thus, total cost of purchase = Rs. 12 × 10 dozen = Rs. 120

Now, SP = Rs. 96 apples/100 

⇒ 96/100 × 120 = SP for 120 apples = Rs. 115.20

∴ Loss% = 4.8/120 × 100 = 4%

Hence, option (a).

Let the (CP)_Arun = x

Then, (SP)_Arun = (CP)_Babu = 1.1 x 

(SP)_Babu = (CP)_Chander = Rs. 60.5 = (1.1) (1.1)x 

⇒ x = 60.5/1.1² = Rs. 50

Hence, option (c).

Let the manufacturers cost price be m. Then,

(SP)_manufacturer = (CP)_wholesaler = 1.2 m

(SP)_wholesaler = (CP)_retailer = (1.1) (1.2) m

(CP)_retailer = (CP)_customer = (1.1) (1.1) (1.2) m = Rs. 1452

⇒ 1.452m = 1452

m = 1452/1.452 = 1000

Hence, option (b).

Let Manufacturer’s price be M. Then,

(SP)_manufacturer = (CP)_dealer = 1.1 m 

(SP)_dealer = (CP)_customer = (1.125) (1.1) m

i.e. (1.125) (1.1) m = 990 ⇒ m = 800

∴ customer paid Rs. (990 – 800) = Rs. 190 above the manufacturer’s prize.

Hence, option (c).

Compute selling price of 1 m cloth = Rs. 225/480 = Rs. 15/32

Let CP of 1 m cloth be x. Hence, CP of 480 m = 480 x

Then gain = 72x 

gain % = (72x)/(480x) × 100 = 15%

Hence, option (c).

Cost price of the car sold at 20% loss = 100 × SP/80 =100 × 20000/80 = 25000.

Now he has incurred a loss of Rs. 5000 on this.

Now as on the whole there is "no" net loss hence he must have gained Rs. 5000 in the second.

Selling price of the second product = 20000. Profit = Rs. 5000. 

So Cost Price of the second product = 20000 - 5000 = 15000.

Percentage Profit = 100 × 5000 / 15000 = 33.33%.

Hence, option (d).

Since the two houses are sold for 130000 each, Now use P + Q + PQ/100

We get 30 – 30 + (-30 × 30)/100 = -9%. So there will be a loss of 9%.

Hence, option (e).

SP of 25 magazines = CP of 15 magazines

SP of 25 = CP of 25 – CP of 10

Loss = 10/25 × 100 = 40%

Hence, option (d).

Loss = SP of 5 oranges on selling 40 oranges

CP of 40 oranges – SP of 40 oranges = SP of 5 oranges

SP of 45 oranges = CP of 40 oranges = CP of 45 oranges – CP of 5 Oranges

Loss = 100 × 5/45 = 11.11%

Hence, option (a).

Let CP = Rs. x and SP = y

Then,

If SP_new = 0.5 y, Loss% = (x - 0.5y)/x = 0.125 or x – 0.5y = 0.125x

or 0.875 x = 0.5 y ⇒ y = 0.875/0.5 x = 1.75 x

But when, CP = x, SP = y, profit % = (y - x)/x × 100

= $\frac{1.75x-x}{x}\times 100$ = $\frac{0.75x}{x}\times 100$ = 75%

Hence, option (d).

If x be the number of oranges at each of the prices, then,

Total purchase = x/11 + x/9 = (9x + 11x)/99 = 20x/99

Sales revenue = 2x/10 = x/5

% gain = $\frac{{\displaystyle \frac{x}{5}}-{\displaystyle \frac{20x}{99}}}{{\displaystyle \frac{20x}{99}}}\times 100$ = $\frac{{\displaystyle \frac{99x-100x}{495}}}{{\displaystyle \frac{20x}{99}}}\times 100$ = $\frac{-x}{495}\times \frac{99}{20x}\times 100$ = -1 or 1% loss.

Hence, option (c).

Let CP be Rs. x. 

Then, (119 - x)/x × 100 = x

⇒ x² = 11900 - 100x

⇒ x² + 100x - 11900 = 0

⇒ x = $\frac{-100\pm \sqrt{{100}^2+4\times 11900}}{2}$ = $\frac{-100+240}{2}$ = Rs. 70

(-ve value has no meaning, hence discarded)

Hence, option (c).

Let CP of sugar be Rs. x/kg. 

Then, $\frac{4.25-x}{x}$ = 0.25

⇒ 4.25 - x = 0.25x

⇒ 1.25x = 4.25

∴ x = 3.4

Old profit = 4.25 – 3.4 = 0.85

New Profit = 0.17 

Hence, 0.85 : 0.17 :: old sales : new sales.

⇒ Hence,  $\frac{old sales}{new sales}=\frac{1}{5}$

Hence, option (b).

4 Bags = 6 books, 4 books = 8 pencils, 4 pencils = 3 magazines, Magazine = Rs. 96

Pencil = 3/4 × 96 = Rs. 72

Book = 8/4 × 72 = Rs. 144

Bag = 6/4 × 144 = Rs. 216

Hence, option (a).

The maximum profit will be when C.P. is minimum and S.P. is maximum.

So maximum profit per chair = 950 – 400 = Rs. 550.

Hence, option (b).

CP of 30 articles = SP of 20 articles

SP of 20 = CP of 20 + CP of 10. Now SP = CP + Profit

So profit = 10/20 × 100 = 50%

Hence, option (e).

If S.P. is 7/5 times of C.P.,

SP = 7/5 CP = CP + 2/5 CP = CP + Profit.

So the profit will be 2/5 of C.P.

Therefore, Profit percentage = 2/5 × 100 = 40%

Hence, option (c).