1 Mohan sold an item for 4510/- and incurred a loss … = Marked Price – Selling Price Discount % = 𝒊 𝑷 𝒊 × 100 Marked Price ... 1 Mohan sold an item for 4510/- and incurred

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Author : J. Maha Laxamaiah Mailid : [email protected]

Profit = Selling price (S. P) – Cost price (C. P)

Loss = Cost price (C. P) – Selling price (S. P)

Profit percentage (P %) = 𝑷𝒓𝒐𝒇𝒊𝒕

𝑪𝒐𝒔𝒕 𝑷𝒓𝒊𝒄𝒆 × 𝟏𝟎𝟎

Loss percentage (L %) = 𝑳𝒐𝒔𝒔


Selling price (S. P) = 𝟏𝟎𝟎+𝑷𝒓𝒐𝒇𝒊𝒕 %

𝟏𝟎𝟎 × 𝑪𝒐𝒔𝒕 𝒑𝒓𝒊𝒄𝒆

Selling price (S. P) = 𝟏𝟎𝟎−𝑳𝒐𝒔𝒔 %

𝟏𝟎𝟎 × 𝑪𝒐𝒔𝒕 𝒑𝒓𝒊𝒄𝒆

NOTE: Profit% and Loss% are always calculated on Cost price

Discount = Marked Price – Selling Price

Discount % = 𝑫𝒊𝒔𝒄𝒐𝒖𝒏𝒕

𝑴𝒂𝒓𝒌𝒆𝒅 𝑷𝒓𝒊𝒄𝒆 × 100

Marked Price = (𝟏𝟎𝟎+𝑷𝒓𝒐𝒇𝒊𝒕 %

𝟏𝟎𝟎−𝑫𝒊𝒔𝒄𝒐𝒖𝒏𝒕 %) × Cost Price

Above formula is useful for solving several problems in Profit and Discounts.

If X1 and X2 both are the rates of gain or both are the rates of loss, then

C. P = (100

X1−X2) × Amount of difference between S. P

If in X1 and X2 one is the rate of gain and another one is rate of loss, then

C. P = (100

X1 + X2) × Amount of difference between S. P

1 Mohan sold an item for 4510/- and incurred a loss of 45%. At what price

should he have sold the item to have gained profit of 45%. ?

SOLUTION:

Short method for selling price

S. P × (𝑷𝒓𝒐𝒇𝒊𝒕 %

𝑳𝒐𝒔𝒔 %)

4510 × (𝟏𝟒𝟓

𝟓𝟓) = 11890

Discount is always calculated on Marked Price / Labelled Price.

Shopkeeper will label his price with certain % above CP.

SP = Marked Price – Discount.

Profit / Loss is always calculated on Cost Price.

SP = CP + Profit

Now SP = MP [ 100 − 𝐷%

100]

SP = CP [100 + 𝑃%

100]

These two are the basic formulae and the other formulae can

be derived from these.

Usual sign convention is + for profit and – for loss and always

– fordiscount.

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Here, loss% => 100 – 45 = 55%

Profit% = 100 + 45 = 145%

= 11890/- (answer)

2 Ramya purchased an item for 46000/- and sold it at loss of 12% with that

amount she purchased another item and sold it a gain of 12%. What was her

overall gain or loss ?

SOLUTION:

Short method for cost price

C. P × LOSS% × PROFIT%

46000 = 𝟖𝟖

𝟏𝟎𝟎 ×

𝟏𝟏𝟐

𝟏𝟎𝟎 => 45337. 6

Here, loss% => 100 – 12 = 88%

Profit% => 100 + 12 = 112%

= 45337.6

Therefore, 46000 – 45337.6

= 662. 40 loss (answer)

3 In a sale perfumes are available at a discount of 25% on the selling price.

If a perfumes cost 5895/- in the sale, what is the selling price of the perfume ?

SOLUTION:

Let selling price of perfume = X

According to question

X × 𝟕𝟓

𝟏𝟎𝟎 = 5895 (discount = 25%)

X = 7860 (answer)

4 A man purchased 25 apples for 20/- and sold 20 apples for 25/-. Find

profit % ?

SOLUTION:

SHORT METHOD:

Profit % = 252 − 202

202 ∗ 100

= 56. 25% (answer)

5 On selling 17 balls at 720/- there is a loss equal to the cost price of 5 balls.

Then find the cost price …….

SOLUTION:

C.P – S.P = LOSS

C.P of 17 balls – S.P of 17 balls = C.P of 5 balls

C.P of 12 balls = S.P of 17 balls = 720

Therefore, C.P of 12 balls = 720

One ball = 𝟕𝟐𝟎

𝟏𝟐

= 60 (answer)

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6 A man purchased some eggs at for 5/- and sold them at 5 for 12/-. Thus he

gained 143 in all. The no. of eggs he bought is…

SOLUTION:

C.P of one egg = 𝟓

𝟑 rupee

S.P of one egg = 𝟏𝟐

𝟓 rupee

Gain in one egg => 𝟏𝟐

𝟓−

𝟓

𝟑

Let no. of eggs = X


(𝟏𝟐

𝟓−

𝟓

𝟑) × 𝑿 = 𝟏𝟒𝟑

X = 195/- (answer)

7 By selling an article a man makes a profit of 25% of its selling price. His

profit % is ……….

SOLUTION:

Let S.P = X, then profit % = X × 𝟐𝟓

𝟏𝟎𝟎

= X/4

C.P = S.P – PROFIT

C.P = 𝒙 − 𝒙

𝟒

= 𝟑𝒙

𝟒

Therefore, profit % = 𝑃𝑟𝑜𝑓𝑖𝑡

𝐶.𝑃 × 100

= 𝑋

4⁄

3𝑋4⁄

× 100

= 33 1/3 % (answer)

8 An article is listed at 920/-. A customer pays 742. 90/- for it after getting

two successive discounts. If the rate of first discount is 15%, the rate of second

discount is …….

SOLUTION:

Marked price = 920/-

After getting two successive discounts it becomes 742. 90/-


920 × 85

100 × (

100 − x

100) = 742.90

X = 5% (answer)

9 By selling an article at 80% of the marked price, there is a loss of 10%. If

the article is sold at the marked price, the profit % will be

SOLUTION:

Let C.P = X & Marked price = Y

Then S.P = Y × 𝟖𝟎

𝟏𝟎𝟎

For loss %, y × 80

100=

90

100 × x

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y = 9x

8

Required Profit % = (y−x

x) × 100

= (

9x8

− x

x) × 100

X = 12.5% (answer)

10 The marked price of a Radio set is 480/-. The shopkeeper allows a

discount of 10% and gains 8%. If no discount is allowed his gain % would be

……..

SOLUTION:

After discount, S.P = 480 × 𝟗𝟎

𝟏𝟎𝟎

= 432/-

As profit is 8%, then

C.P => 432 = 𝟏𝟎𝟖

𝟏𝟎𝟎 × 𝑪. 𝑷

C.P = 400/-

So without discount, Profit => 480 – 400 = 80/-

Profit % = 𝟖𝟎

𝟒𝟎𝟎 × 𝟏𝟎𝟎

= 20% (answer)

11 How much percent more than C.P should a shopkeeper mark his goods

so that after allowing a discount of 25% on the marked price he gains 20%. ?

SOLUTION:

Formula = X + Y + 𝑿𝒀

𝟏𝟎𝟎

X – 25 – 𝑿∗𝟐𝟓

𝟏𝟎𝟎= 𝟐𝟎

X = 60% (answer)

12 A shopkeeper earns a profit of 12% on selling a book at 10% discount on

printed. The ratio of the C.P to printed price of the book…….

SOLUTION:

Formula

Marked Price = (𝟏𝟎𝟎 + 𝑷𝒓𝒐𝒇𝒊𝒕 %)

𝟏𝟎𝟎 − 𝑫𝒊𝒔𝒄𝒐𝒖𝒏𝒕 %) × 𝑪𝒐𝒔𝒕 𝑷𝒓𝒊𝒄𝒆

Here, profit % (P %) = 112% (100 + 12)

Discount % (D %) = 90% (100 - 10)

M. P = 𝟏𝟏𝟐

𝟗𝟎 × 𝑪. 𝑷

Required ratio 𝐂.𝐏

𝐌.𝐏=

𝟒𝟓

𝟓𝟔 (answer)

13 A trade man sold an article at a loss of 20%. If the S.P had been

increased by 100/-, there would have been a gain of 5%. The C.P of the article

is ………

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SOLUTION:

Let C.P = X, then S.P = X × 𝟖𝟎

𝟏𝟎𝟎

And new S.P = X × 𝟏𝟎𝟓

𝟏𝟎𝟎


105

100 × X −

80

100 × Y = 100

X = 400/- (answer)

14 20% loss on selling price is what percent loss on the cost price ?

SOLUTION:

Let S.P = 100, then C.P = 120

Then, loss => 120 – 100 = 20/-

Therefore, loss % = 𝟐𝟎

𝟏𝟐𝟎∗ 𝟏𝟎𝟎

= 16 2/3 % (answer)

15 By selling an article for 144/- a person gained such that the percentage

gain equals the cost price of the article. The cost price of the article ……..

SOLUTION:

Let C.P = X & profit % = C.P


Formula

S. P = C. P × (𝟏𝟎𝟎 + 𝑷𝒓𝒐𝒇𝒊𝒕 %

𝟏𝟎𝟎)

144 = X × (𝟏𝟎𝟎 + 𝑿

𝟏𝟎𝟎)

X = 80/- (answer)

16 If the selling price of the article is doubled, then its loss % is converted

into equal profit %. The loss % of the article is ………

SOLUTION:

Let loss = X, & C.P = 100/-, then S.P = 100-X

New S.P = 2(100-X) and profit % = X %


2 × (100 – X) = 100 + X

X = 33𝟏

𝟑 % (answer)

17 A trader sells an article at a profit of 30%. Had he purchased it at 10%

less and sells for 330/- less he would have gained 20%. Find the C.P …….

SOLUTION:

Let C.P = X, then S.P = X × 𝟏𝟑𝟎

𝟏𝟎𝟎

New C.P = X × 𝟗𝟎

𝟏𝟎𝟎 & New S.P = X ×

𝟏𝟑𝟎

𝟏𝟎𝟎 - 330

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X × 130

100− 330 = X ∗

90

100∗

120

100

X = 1500/- (answer)

18 By selling toffees at rate of 20 for 10/- , a man loses 4%. To gain 20%,

how many toffees must be sold for 10/- ?

SOLUTION:

Short method

If 𝟏𝟎

𝟐𝟎 …………. = 96% (loss% => 100 – 4 = 96%)

𝟏𝟎

𝐱 ………. = 120% (Profit% => 100 + 20 = 120%)

10

X × 96 ×

20

10= 120

X = 16 toffees (answer)

19 If on a marked price the difference of selling price with a discount of

30% and two successive discounts of 20% and 10% is 72, then the Marked

Price is ….

SOLUTION:

Successive discount

20 + 10 – 𝟐𝟎 × 𝟏𝟎

𝟏𝟎𝟎

= 28%

Let marked price = X


X × 30

100 – X ×

28

100= 72

X = 3600/- (answer)

20 A reduction of 20% in the price of an apple enable a man to buy 10

apples more for 250/-. Te reduced price of apples per dozen…….

SOLUTION:

Formula

Reduced price of an apple

= 𝑿 × 𝒀

𝟏𝟎𝟎 × 𝒁

Here, X = 20% (reduction), Y = 250/- and Z = 10 (more apples)

= 𝟐𝟎 × 𝟐𝟓𝟎

𝟏𝟎𝟎 × 𝟏𝟎

= 5/-

Therefore, reduced price of an apple for dozen

= 12 × 5

= 60/- (answer)

21 While selling a watch, a shopkeeper gives a discount of 5%. If he gives a

discount of 6%, he earns 15/- less as profit. What is the marked price of

watch?

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SOLUTION:

Let marked price = X


95

100∗ X −

94

100∗ X = 15

X = 1500/- (answer)

22 Krishna purchased a no. of articles at 10/- for each and the same number

for 14/- each. He mixed them together and sold them for 13/- each. Then his

gain or loss% is ?

SOLUTION:

Total C. P of articles = 𝟏𝟎+𝟏𝟒

𝟐 = 12/-

S. P = 13/-

Then profit => S. P – C. P = 13 – 12

= 1/-

Profit% = 𝟏

𝟏𝟐 × 100

= 8𝟏

𝟑 % (answer)

23 A man sells an article at 5% above its C. P. If he had bought it at 5% less

than what he paid for it and sold it for 2/- less, he would have gained 10%.

Find the C. P of the article

SOLUTION:

Let C. P = X, then new C. P = 𝒙 × 𝟗𝟓

𝟏𝟎𝟎

And S. P when it sells 2/- less =

105

100 × X − 2

According to formula, S. P = C. P × (100+ Profit %) /100 𝟏𝟎𝟓

𝟏𝟎𝟎 × 𝐗 − 𝟐 =

𝟗𝟓

𝟏𝟎𝟎 × 𝐗 ×

𝟏𝟏𝟎

𝟏𝟎𝟎X

= 400/- (answer)

24 Pure ghee costs 100/- per kg. After adulterating it with vegetable oil

costing 50/- per kg a shopkeeper sells the mixture at the rate of 96/- per kg

thereby making a profit of 20%. In what ratio does he mix the two ?v

SOLUTION:

Mean price (C. P) = 96 × 𝟏𝟎𝟎

𝟏𝟐𝟎

= 80/- per kg

By the rule of Allegation & Mixtures

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Required ratio => 30 : 20

= 3 : 2 (answer)

25 A fruit seller has 24 kg of apples. He sells a part of these at 20% gain and

the balance at a loss of 5%. If on the whole he earns a profit of 10%, the

amount of apples sold at loss is

SOLUTION:

By the rule of Allegation & Mixtures

Required ratio => 15: 10 = 3: 2

Therefore, the amount of apples sold at loss is => 𝟐𝟒 × 𝟐

𝟓

= 9.6/- (answer)

26 If the selling price of 24/- results in a 20% discount o the list price, what

selling price would result in a 30% discount on list price?

SOLUTION:

SHORT METHOD

If 80% (100 – 20) …………….. = 24

70% (100 – 30) ……………. =?

= 𝟕𝟎 × 𝟐𝟒

𝟖𝟎

= 21/- (answer)

27 What percentage on profit on cost price equals 30% of profit on selling

price?

SOLUTION:

If C. P = 100/- And Gain = X

Then S. P = (100 + X)/-

Here, formula is

Gain% = 𝑮𝒂𝒊𝒏

𝑺𝒆𝒍𝒍𝒊𝒏𝒈 𝑷𝒓𝒊𝒄𝒆 × 𝟏𝟎𝟎

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30 = 𝑿

𝟏𝟎𝟎+𝑿 × 𝟏𝟎𝟎

X = 42 6/7 % (answer)

28 A dealer marks his goods 30% above his C. P and then allows 15%

discount on it. What is the C. P of an article on which he gains 84/-?

SOLUTION:

Let C. P of articles = X/-

M. P => X × 𝟏𝟑𝟎

𝟏𝟎𝟎 =

𝟏𝟑𝒙

𝟏𝟎

Selling Price = Marked Price × (𝟏𝟎𝟎−𝑫𝒊𝒔𝒄𝒐𝒖𝒏𝒕%

𝟏𝟎𝟎)

S. P = 𝟏𝟑𝑿

𝟏𝟎 ×

𝟖𝟓

𝟏𝟎𝟎

S. P = 𝟐𝟐𝟏𝒙

𝟐𝟎𝟎

Gain = S. P – C. P

84 = 𝟐𝟐𝟏𝑿

𝟐𝟎𝟎− 𝑿

X = 800/- (answer)

29 A man purchased 150 pens at the rate of 12/- per pen. He sold 50 pens at

a gain of 10%. The percentage gain at which he must sell the remaining pens

so as to gain 15% on the whole outlay is

SOLUTION:

Required S. P of 150 pens

= 150 × 12 × 𝟏𝟏𝟓

𝟏𝟎𝟎

= 2070/-

S. P of first 50 pens =

= 50 × 12 × 𝟏𝟏𝟎

𝟏𝟎𝟎

= 660/-

Required S. P of 100 pens => 2070 – 660 = 1410/-

C. P of 100 pens => 100 × 12 = 1200/-


= 1410 – 1200

Gain = 210/-

Gain% = 𝑮𝒂𝒊𝒏


Gain% = 𝟐𝟏𝟎

𝟏𝟐𝟎𝟎 × 𝟏𝟎𝟎

= 17 ½ % (answer)

30 A sells an article to B making a profit of 1/5 of his outlay. B sells it to C

gaining 20%. If C sells it for 600/- and incurs a loss of 1/6 of his outlay, then C.

P of A is

SOLUTION:

Let C. P of article = X

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SHORT METHOD:

600 = 𝟔

𝟓 ×

𝟏𝟐𝟎

𝟏𝟎𝟎 ×

𝟓

𝟔 × 𝒙

Here, profit = 1 + 𝟏

𝟓 =

𝟔

𝟓, loss = 1 –

𝟏

𝟔 =

𝟓

𝟔

X = 500/- (answer)

31 Nitin bought some oranges at 40/- a dozen and an equal numbers at 30/-

a dozen. He sold them at 45/- a dozen and made a profit of 480/-. The no. of

oranges, he bought was

SOLUTION:

Let the no. of oranges bought be 2X (X + X)

dozens

Total C. P => 40X + 30X = 70X

Now total S. P => 45 × 2 = 90X

Profit = S. P – C. P

480 = 90X – 70X

X = 24/-

Therefore, no. of oranges bought = 2×24

= 48 dozens (answer)

32 A shopkeeper marks his goods at 40% above their C. P. He is able to sell

3/4th of his goods at this price, and the remaining at 40% discount. Assuming

that the shopkeeper is able to sell the goods he buys, find his loss or gain as %

on the whole transaction

SOLUTION:

Let total C. P = 100/- (100 articles)

Then, total S. P

= 75 × 140

100+ 25 ×

60

100 × 1.4

= 126/-

Gain 26 (126- 100)

Therefore, Gain% = 26 (answer)

33 An article is sold at a gain of 15%. Had it been sold for 27/- more, the

profit would have been 20%. The C. P of the article is

SOLUTION:

SHORT METHOD:

PERCENTAGE METHOD:

Difference => 20- 15 = 5%

5% (if) ……………. 27

100% ………………?

= 𝟏𝟎𝟎 × 𝟐𝟕

𝟓

= 540/- (answer)

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34 Half of 100 articles were sold at a profit of 20% and the rest of a profit of

40%. If all the articles had been sold at a profit of 25%, the total profit would

have been 100 less than earlier profit. The C. P of each article was

SOLUTION:

Let C. P of an article = X (Total C. P = 100X)

S. P of 50 articles

= 5X × 𝟏𝟐𝟎

𝟏𝟎𝟎

= 60X

S. P of remain 50 articles

= 50X × 𝟏𝟒𝟎

𝟏𝟎𝟎

= 70X

S. P of 100 articles => 60X + 70X = 130X

And, new S. P => 100X × 𝟏𝟐𝟓

𝟏𝟎𝟎

= 125X


130X – 125X = 100

X = 20/- (answer)

35 A certain no. of articles were purchased for 90/-. Three more articles

could have been purchased for the same amount if each article was cheaper by

1/-. Find the no. of articles purchased

SOLUTION:

Let the no. of articles = X

90

X−

90

X + 3= 1

X = 15/- (answer)

36 A shopkeeper allows a discount of 10% on the marked price of an item

but charges a sales tax of 8% o the discounted price. If a customer pays 680.

40/- as the price of the item including sales tax, find the marked price of the

item

SOLUTION:

Let the M. P = X, discount = 10%

Discounted price => X × 𝟗𝟎

𝟏𝟎𝟎

= 9X/10

Sales tax = 8%

Therefore, S. P = 𝟗𝑿

𝟏𝟎 ×

𝟏𝟎𝟖

𝟏𝟎𝟎


9X

10 ×

108

100= 680. 40

X = 700/- (answer)

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37 A bicycle agent allows 20% discount on his M. P and then makes a profit

of 20% on his outlay (C. P). What is the M. P of the bicycle on which he gains

40/-?

SOLUTION:

Let M. P = X, C. P = Y

Discount = 20% (given), profit = 20% (given) and Gain = 40/- (given)

S. P = X × 𝟖𝟎

𝟏𝟎𝟎 [∴ 𝑺. 𝑷 = 𝑴. 𝑷 ∗

𝟏𝟎𝟎 −𝑫𝒊𝒔𝒄𝒐𝒖𝒏𝒕%

𝟏𝟎𝟎]

= 𝟒𝒙

𝟓

And

4X

5= Y ×

120

100 [S. P = C. P ∗

100 + Profit%

100]

Y = 𝟐𝒙

𝟓


40 = 𝟒𝑿

𝟓−

𝟐𝑿

𝟓

X = 300/- (answer)

38 The marked price of an article is 50% above cost price. When marked

price is increased by 20%, the profit doubles. If original marked price is 300/-,

then original selling price is

SOLUTION:

Let original S. P = X

C. P of article => 300 × 𝟏𝟎𝟎

𝟏𝟓𝟎 = 200/- (from question)


X × 120

100− 200 = 2(X − 200)

X = 250/- (answer)

39 If books bought at prices ranging from 100/- to 350/- are sold at prices

ranging from 300/- to 425/-. What is the greatest possible profit that might be

made in selling 8 books?

SOLUTION:

Least C. P => 200 × 8 = 1600/-

Greatest S. P => 425 × 8 = 3400/-

Required profit => 3400 – 1600

= 1800/- (answer)

40 The price of a sugar is increased by 20%. If the expenditure on sugar has

to be kept the same as earlier, the ratio between the reduction in consumption

and the original consumption is

SOLUTION:

The raised price = 𝟏𝟐𝟎

𝟏𝟎𝟎 of the former pric

So, the house holder must now consume 100/ 120 of the original amount

So, the reduction in consumption

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= (𝟏 − 𝟏𝟎𝟎

𝟐𝟎𝟎) of the original consumption

= 1/6 of the original consumption (answer)

41 A man bought some oranges at 10/- per dozen and bought the same no.

of oranges at 8 per dozen. He sold these oranges 11/- per dozen and gained

120/-. The total no. of oranges bought by him was

SOLUTION:

C. P of 2- dozen oranges => (10 + 8) = 18/-

S. P of 2- dozen oranges => (11 + 11) = 22/-

If profit is 4/- (22- 18), then oranges bought = 2 dozen

If profit is 120/-, then oranges bought = ?

= 𝟏𝟐𝟎 × 𝟐

𝟒

= 60 dozens (answer)

42 A girl buys 2- pigeons for 182/-. She sells one at a loss of 5% and another

at a profit of 8%. But she neither gains nor loss on the whole. Find the price of

pigeon which has sold at a profit

SOLUTION:

The ratio of price of pigeons sold at loss and profit respectively are in the ratio = 8

: 5

Cost price of profitable pigeon is = 𝟓

𝟓+𝟖 × 𝟏𝟖𝟐

= 70/- (answer)

43 How many kilograms of rice costing 18. 30/- per kg must be mixed with

126 kg of rice costing 8. 55/- per kg so that 20% may be gained by selling the

mixture at 14. 40/- per kg?

SOLUTION:

S. P of one kg mixture = 14. 40/-

Gain = 20%

Then, C. P = 14. 40 × 𝟏𝟎𝟎

𝟏𝟐𝟎

= 12/-

Dearer price: Cheaper price => 345: 630

=23. 42

Let X- kg of dearer may be mixed with 126 kg of cheaper

Then, 𝑿

𝟏𝟐𝟔=

𝟐𝟑

𝟒𝟐

X = 69 kg (answer)

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44 In a big bazaar, a shopkeeper mixed two varieties of Daal at 20/- per kg

an 30/- per kg in the ratio 2: 3 and sell the mixture at 10% profit. Find the

price per kg at which he sold the mixture

SOLUTION:

C. P of mixture

= 𝟐𝟎 × 𝟐+𝟑𝟎 × 𝟑

𝟐+𝟑

= 26/-

S. P of the mixture => 26 × 𝟏𝟏𝟎

𝟏𝟎𝟎

= 28. 60/- (answer)

45 Gauri went to stationary and bought things worth 25/- out of which 30

paise went on sales tax on taxable purchases. If the tax rate was 6%, then

what was the cost of the tax free items?

SOLUTION:

Let the amount taxable purchase = X/-

Then 6% of X = 𝟑𝟎

𝟏𝟎𝟎

X = 5/-

Cost of tax free items = [25- (5+ 0.30)]

= 19. 70/- (answer)

46 A person buys certain no. of marbles at 20 per rupee are equal no. of

marbles at 30 per rupee. He mixes them and sells them at 25 per rupee. His

gain or loss in the transaction is

SOLUTION:

C. P of each marble @ 20 per rupee = 𝟏

𝟐𝟎

C. P of each marble @ 30 per rupee = 𝟏

𝟑𝟎

Average cost price of each marble

=

𝟏

𝟐𝟎+

𝟏

𝟑𝟎

𝟐

According to the question

S. P of each marble @ per rupee = 𝟏

𝟐𝟓

Here, 𝟏

𝟐𝟒 >

𝟏

𝟐𝟓 => means, loss

Therefore, loss %

= (𝟏

𝟐𝟓−

𝟏

𝟐𝟒𝟏

𝟐𝟒

) × 𝟏𝟎𝟎

Loss % = 4% (answer)

47 A man purchased a table and a chair for 1300/-. He sold the table at a

profit of 20% and the chair at a profit of 25%. In this way, his total profit was

23 1/3 %. The cost price of the table

SOLUTION:

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C. P of the table = X

C. P of the chair = 1300- X


Profit on table + profit on chair = total profit

X × 20

100+ (1300 − X) ∗

25

100= 233

1% of 1300

X = 433. 3/- (answer)

48 Titan sells a wrist watch to a wholesaler making a profit of 10%. The

wholesaler in turn, sells it to the retailer making a profit of 10% a customer

purchases it by paying 990/-. Thus, the profit f retailer is 25/11 %. What is the

cost incurred by the titan to produce it?

SOLUTION:

Let C. P of Titan = X

X × 110

100 ×

110

100 ×

100+ 25

11

100= 900

X = 800/- (answer)

49 A shopkeeper sold 12 cameras at a profit of 20% and 8 cameras at a

profit of 10%. If he had sold all the 20 cameras at a profit of 15%, then his

profit would have been reduced by 36/-. What is the cost price of each

camera?

SOLUTION:

Let C. P of each camera = X

S. P of 12 cameras sold at 20% =

X × 12 × 𝟏𝟐𝟎

𝟏𝟎𝟎

= 𝟏𝟒𝟒𝑿

𝟏𝟎

S. P of 8 cameras sold at 10% =

X × 8 × 𝟏𝟏𝟎

𝟏𝟎𝟎

= 𝟖𝟖𝑿

𝟏𝟎


36 = 144x

10+

88x

10− 23x

X = 180/- (answer)

50 A cycle is sold at a profit 15%. If both cost price and selling price are

decreased by 200/- the profit would be 5% more. What is the original cost

price of cycle?

SOLUTION:

Let the original C. P = X

So, S. P = X × 𝟏𝟏𝟓

𝟏𝟎𝟎

= 𝟐𝟑𝒙

𝟐𝟎

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Now, new C. P = X – 200

Then, new S. P = 𝟐𝟑𝑿

𝟐𝟎− 𝟐𝟎𝟎

Profit = S. P – C. P

= 23X

20− 200 − (X − 200)

Profit = 𝟑𝑿

𝟐𝟎

Profit % = 𝑷𝒓𝒐𝒇𝒊𝒕

𝑪𝒐𝒔𝒕 𝒑𝒓𝒊𝒄𝒆 × 𝟏𝟎𝟎

20 = (𝟑𝑿

𝟐𝟎𝟑𝑿

𝟐𝟎− 𝟐𝟎𝟎

) × 𝟏𝟎𝟎

X = 800/- (answer)

51 A sells an article to B making a profit of 1/5th his outlay. B sells it to C,

gaining 20%. If C sells it for 600/- and incurs a loss of 1/6th his outlay, the C.

P of A is

SOLUTION:

Let the C. P of A = X


X × (1 + 𝟏

𝟓) × 𝟏𝟐𝟎 × (𝟏 −

𝟏

𝟔) = 𝟔𝟎𝟎

X = 500/- (answer)

52 A man sold an article for 75/- and lost something. Had he sold it for 96/-

his gain would have been double the former loss. The cost price of the article

is

SOLUTION:

Let C. P = X


96 – X = 2 × ( X – 75 )

X = 82/- (answer)

53 A reduction of 25% in the price of eggs will enable one to buy 4 dozens

more eggs for 96/-. What is the price per dozen?

SOLUTION:

Let original price = X

New price per dozen = X × 𝟕𝟓

𝟏𝟎𝟎 =>

𝟑𝑿

𝟒


96

3X4

− 96

X= 4

X = 8/- (answer)

54 An article sold at a gain of 15%. Had it been sold for 27 more, the profit

would have been 20%. The C. P of the article is

SOLUTION:

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PERCENTAGE METHOD:

If 5% (20% - 15%) ………………… = 27

100% …………………………………… = ?

= (𝟏𝟎𝟎 × 𝟐𝟕

𝟓)

= 540/- (answer)

55 A retailer bought 90 pens for 40/- each. He sells 30 of them at a gain of

12%. What must be the gain% of the remaining pens so as to get 15% gain on

the whole?

SOLUTION:

Let the required gain% be X%. Then,


30 × 12 + (90 – 30) × X = 90 × 15

X = 16.5% (answer)

56 The cost price of 5 articles is equal to the selling price of 4 articles. What

is the profit or loss %?

SOLUTION:

If we assume CP of 5 articles = SP of 4 articles = Rs. 100

CP of one article = Rs. 20

SP of one article = Rs. 25

Profit = Rs. 5

Profit % = 𝟓

𝟐𝟎 × 100

= 25% (answer)

57 A shopkeeper sold two Televisions of different make for Rs. 12000 each.

On one, he got profit of 20% and on other, he attained loss of 10%. What is

the net profit percentage?

SOLUTION:

Here P = 20% = x% ,

L = 10% = y%

Overall Profit / Loss % = [𝟐 (𝟏𝟎𝟎+𝒙)(𝟏𝟎𝟎+𝒚)

(𝟏𝟎𝟎+𝒙)+ (𝟏𝟎𝟎+𝒚)− 𝟏𝟎𝟎]%

(Use + when there is profit and – when there is loss)

= [𝟐 ×𝟏𝟐𝟎 ×𝟗𝟎

𝟐𝟏𝟎− 𝟏𝟎𝟎]%

= 2 𝟔

𝟕 % (profit since the result is positive)

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58 A shopkeeper sold some articles at Rs.35 per article and gained 40% on

it. What would be the selling price of each article to get 60% profit?

SOLUTION:

Use 𝐒𝐏𝟏

𝟏𝟎𝟎+𝒙 =

𝐒𝐏𝟐

𝟏𝟎𝟎+𝒚 analogy for such type of problems

𝟑𝟓

𝟏𝟒𝟎 =

𝐒𝐏𝟐

𝟏𝟔𝟎 => 𝐒𝐏𝟐 = Rs. 40

𝐎𝐑

PERCENTAGE METHOD:

If 35/- per article ………….. 140 (gain 40%)

Then ‘X/-‘ per article ……... 160 (gain 60%)

X

35 × 140 = 160

X = 40/- (answer)

59 A trader bought two horses for 19, 500/-. He sold one at a loss of 20%

and the other at a profit of 15%. If the selling price of each horse is the same,

then their cost prices are respectively ….

SOLUTION:

Let Cost Price of one horse = x/-, then of other 19, 500 – x

One horse sold at loss of 20%, other at 15% gain.

Selling Price is same

80% of x = 115% of (19, 500 – x)

80

100 × x =

115

100 × (19, 500 − x)

x = 11, 500/− (answer)

Cost price of other = 19, 500 – 11, 500

= 8, 000/- (answer)

60 A shopkeeper gave an additional 20% concession on the reduced price

after giving 30% standard concession on an article. If Arun bought that

article for 1, 120/-, what was the original price?

SOLUTION:

Let original price = x/-


x × 100 − 30

100 ×

100 − 20

100= 1120

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X = 2, 000/- (answer)

61 A reduction of 20% in the price of sugar enables a person to obtain 40 kg

more for 210/-. What is the reduced price per kg?

SOLUTION:

Owing to the fall in price, there is a saving of 20% on 210/- i.e 42/-

So, for 42/-, a person purchases 40 kg of sugar.

Hence, reduced price per kg = 𝟒𝟐

𝟒𝟎

= 1.05/- (answer)

62 A article sold at 30% profit. Had it been sold at 155/- more than previous

selling price and the cost price were also increased by 100/- then profit would

have been 5% more. Then find the cost price of the article…

SOLUTION:

Let cost price of article = 100x

Selling price of article = 130x

Increased cost price = (100x + 100)

Increased selling price = (130x + 155)

Profit = 𝑺𝒆𝒍𝒍𝒊𝒏𝒈 𝒑𝒓𝒊𝒄𝒆−𝑪𝒐𝒔𝒕 𝒑𝒓𝒊𝒄𝒆

𝑪𝒐𝒔𝒕 𝒑𝒓𝒊𝒄𝒆∗ 𝟏𝟎𝟎

Profit = (𝟏𝟑𝟎𝒙 + 𝟏𝟓𝟓)− (𝟏𝟎𝟎𝒙 + 𝟏𝟎𝟎)

(𝟏𝟎𝟎𝒙+𝟏𝟎𝟎) × 𝟏𝟎𝟎

By solving, we get x = 4

Therefore, Cost price of article = 100 × 4

= 400 (answer)

63 Ram sells onions in the streets of Varanasi. Due to recent shortfall in the

supply of onions, he doubles his selling price despite the cost price remains

same for him due to a fixed contract. He realizes that his profit has tripled.

Find the profit percent ….

SOLUTION:

Let original Selling price be SP and Cost price be CP

Then, 2SP – CP = 3(SP – CP)

SP = 2CP

Hence, original profit % = 100% (answer)

64 A loss of 19% gets converted into a profit on 17% when the selling price

is increased by 162/-. The cost price of the article is ………..

SOLUTION:

SHORT METHOD

Loss (-19%) Profit ( +17%)

Difference => 17 – (- 19)

= 36%

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If 36% ………………. 162

Then 100% ………. ?

=100 × 162

36

= 450/- (answer)

65 On selling a Pen at 5% loss and a book at 15% gain, Rajini gains 7/-. If

he sells the pen at 5% gain and the book at 10% gain, then he gains 13/-. The

actual price of the book is ………….

SOLUTION:

Let the C. P of a pen = x/- and that of a book = y/-


15y – 5x = 700 ………………. (1)

And, 10y + 5x = 130 ………. (2)

From equations (1) and (2), we get

y = 80/-

Therefore, C. P of a book = 80/- (answer)

66 A article is sold at profit of 25%. Its CP and SP are more by 20 and 4

respectively. The percentage of profit decreased by 15%. Find the Cost price

…………

SOLUTION:

Let Cost price = x

Selling price = 5𝑥

4


(5𝑥

4+ 4) − (𝑥 + 20) = (𝑥 + 20) ×

10

100

x = 120/- (answer)

Documents

1 Mohan sold an item for 4510/- and incurred a loss … = Marked Price – Selling Price Discount % = 𝒊 𝑷 𝒊 × 100 Marked Price ... 1 Mohan sold an item for 4510/- and incurred