Examples on Calculating Profit or Loss

Solved examples on calculating profit or loss using the basic facts and important formula on gain or loss.

Let’s observe the fully solved examples on calculating profit or loss with detailed description to solve the answers step-by-step.

1. Henry sold a bicycle at 8% gain. Had it been sold for $ 75 more, the gain would have been 14%. Find the cost price of the bicycle.

Solution:

Let the cost price of the bicycle be $ x.

SP of the bicycle at 8% gain = $ [{(100 + gain %) /100} × CP]

= $ [{(100 + 8)/100} × x]

= $ {(108/100) × x}

= $ (27x/25)

SP of the bicycle at 14% gain = $ [{(100 + 14)/100} × x]

= $ {(114/100) × x}

= $ (57 x/50)

Therefore, (57 x /50) - (27 x/25) = 75

⇔ (57 x – 54 x)/50 = 75

⇔ 3 x = (50 × 75)

⇔ x = (50 × 25)

⇔ x = 1250

Hence the CP of the bicycle is $ 1250.

2. Mike sold a watch at 5% loss. Had he sold it for $ 104 more, he would have gained 8%. Find the selling price of the watch.

Solution:

Let the selling price of the watch be $ x.

Loss% = 5%.

Therefore, CP of the watch = {100/(100 - loss %) × SP}

= $ {100/(100 - 5) × x}

= $ {(100/95) × x}

= $ (20x /19)

Now, CP = $ (20x /19) and gain % = 8%.

Then, SP = [{(100 + gain %)/100} × CP]

= $ [{(100 + 8)/100} ×(20 x /19)]

= $ {(108/100)×(20x /19)}

= $ (108x /95)

Therefore, (108x /95) - x = 104

⇔ (108x - 95x) = (104 × 95)

⇔ 13x = (104 × 95)

⇔ x = (104 × 95)/13

⇔ x = 760.

Hence, the selling price of the watch is $ 760.

More worked-out examples on calculating profit or loss to get the basic concepts to solve the questions and answers with explanation.

Examples on Calculating Profit or Loss

3. Greg sells two watches for $ 1955 each, gaining 15% on one and losing 15% on the other. Find her gain or loss per cent in the whole transaction.

Solution:

SP of the first watch = $ 1955.

Gain% = 15%.

Therefore, CP of the first watch = [{100/(100 + gain %)} × SP]

= $ [{100/(100 + 15)} × 1955]

= $ {(100/115) × 1955}

= $ 1700.

SP of the second watch = $ 1955.

Loss% = 15%.

CP of the second watch = [{100/(100 - loss %)} × SP]

= $ [{100/(100 - 15)} × 1955]

= $ {(100/85) × 1955}

= $ 2300

Total CP of the two watches = $ (1700 + 2300) = $ 4000.

Total SP of the two watches = $ (1955 × 2) = $ 3910.

Since (SP) < (CP), there is a loss in the whole transaction.

Loss = $ (4000 - 3910) = $ 90.

Therefore, Loss% = {(90/4000) × 100} % = 2¹/₄%

Hence, Greg loses 2¹/₄% in the whole transaction.

4. Nick purchased two hand bags for $ 750 each. He sold these bags, gaining 6% on one and losing 4% on the other. Find his gain or loss per cent in the whole transaction.

Solution:

CP of the first handbag = $ 750.

Gain% = 6%.

SP of the first handbag = [{(100 + gain %)/100} × CP]

= $ [{(100 + 6)/100} × 750]

= $ {(106/100) × 750}

= $ 795.

CP of the second handbag = $ 750.

Loss% = 4%.

SP of second handbag = [{(100 - loss %)/100} × CP]

= $ [{(100 - 4)/100} × 750]

= $ {(96/100) × 750}

= $ 720.

Total CP of the two handbags = $ (750 × 2) = $ 1500.

Total SP of the two handbags = $ (795 + 720) = $ 1515.

Since (SP) > (CP), there is a gain in the whole transaction

Gain = $ (1515 - 1500) = $ 15.

Gain % = {(gain/Total CP) × 100}%

= {(15/1500) × 100}%

= 1%.

Hence, Nick gains 1% in the whole transaction.

5. A reduction of20% in the price of sugar enables Mrs. Jones to buy an extra 5 kg of it for $ 320.

Find:

(i) the original rate, and

(ii) the reduced rate per kg.

Solution:

Let the original rate be $ x per kg.

Reduced rate = (80% of $ x) per kg

= $ (x × 80/100) per kg

Quantity of sugar for $ 320 at original rate = 320/x kg

Quantity of sugar for $ 320 at the new rate = 320/(4x/5) kg

= (320 × 5)/4x kg

= 400/x kg.

Therefore, (400/x) - (320/x) = 5

⇔ 5x = (400 - 300)

⇔ 5x = 80

⇔ x = 16

(i) Original rate = $ 16 per kg

(ii) Reduced rate = (4/5 × 16) per kg

= $ 64/5 per kg

= $ 12.80 per kg.