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PROFIT & LOSS
STANDARD LEVEL
C.S.VEERARAGAVAN
COST PRICE A)₹ 300 B)₹ 350 C)₹ 375 D)₹ 400
Let the cost price of table = ₹ N
selling price with 15% gain = ₹
115N
100
CP = [(75%) x CP]/100 = ₹ 75N/100
New SP = ₹ (115N/100) – 60 AND
Profit 32%
115N
100
−60 −
75N
100
75N
100
X 100 = 32
115N
100
− 60 −
75N
100
=
32 N
100
∗
3
4
40N – 6000 = 24 N
16N = 6000
N = 375
A person sold a table at
a gain of 15%.
Had he bought it for 25% less and
sold it for ₹ 60 less,
he would have made a profit of 32%.
The cost price of table was
SHORTCUT
GIVEN GAIN 15% = 100 + 15 = 115
CP = (100 – 25)% = 75
GIVEN PROFIT = 75% OF 32 = 24
DIFFERENCE = 115 – (75 + 24)
= 16
SOLD LESS = 60 * 100 = 6000
COST PRICE = 6000/16 = 375
COST PRICE
A)₹ 25 B)₹ 75
C)₹ 50 D)₹ 40
Let the cost price of first bicycle be ₹ P.
The cost price of second bicycle = ₹ (1600 - P)
[20% of P + 10% of (1600 - P )] - [10% of P +
20% of (1600 - P)] = 5
2P
10 +
1600 −P
10 −
P
10 +
2 1600 − P
10 = 5
P
10
−
1600 −P
10
= 5
⇒ 2P = 1600 + 50
∴ P = 1650/2 = 825
Cost of second bicycle = (1600 - 825) = ₹ 775
∴ Required difference = 825 - 775 = ₹ 50
A person bought two bicycles
for ₹ 1600 and
sold the first at 10% profit and
the second at 20% profit.
If he sold the first at 20% profit
and the second at 10% profit,
he would get ₹ 5 more.
The difference in the cost price
of the two bicycles was
SELLING PRICE
Cost of a packet of coffee powder
and a litre of milk
are ₹ 20 and ₹ 30 respectively.
10 cups of coffee is made with
one packet coffee powder and for
each cup 200 ml of milk is used.
If coffee is sold at 25% profit, the
selling price of each cup of coffee
is
A)₹ 12.50 B)₹ 6.25
C) ₹ 8 D)₹ 10
Cost of coffee powder used
in one cup = 20/10 = ₹ 2
Cost of milk used in one cup
= (30/1000) x 200 = ₹ 6
∴ Cost of each cup coffee
= 2 + 6 = ₹ 8
To gain 25% profit,
sale price of each cup of coffee
= 125% of 8
= ₹ 10
COST PRICE
A shopkeeper sells transistors at
15% above its cost price.
If he had bought it at 5% more
than what he paid for it
and sold it for ₹ 6 more, he would
have gained 10%.
The cost price of the transistor is ?
A)₹ 800 B)₹ 1000
C)₹ 1200 D)₹ 1400
Let cost price of transistor = ₹ N
According to the question,
CP of transistor = (N x 105)/100
SP of transistor = (115 x N)/100 + 6
Profit percentage = (SP - CP)/CP x100
115N
100
+ 6 −105N
100
105N
100
= 10
⇒ 10 = [(10N + 600) x 100]/ 105N
⇒ 105N = 100N + 6000
⇒ 5N = 6000
⇒ N= ₹ 1200
MAXIMUM PROFIT
Teenagers shoe company sells the shoes whose prices i.e.,cost price and
selling price are the multiples of either 13,14,15,16,17,18 or 19, starting
from Rs. 399 to Rs.699
(i.e, 399 ≤ CP/SP ≤ 699).
What can be the maximum profit of the company?
a)Rs. 292 b)Rs. 398 c)Rs. 298 d)Rs. 300
The maximum possible profit = maximum possible difference in SP and CP.
It means SP be maximum and CP be minimum
CP (min) = Rs. 399 19 x m = 399, where m is an integer.
Again SP (max) = Rs. 697, which is very close to 699
Here 697 = 17 k, k is a positive integer.
So, the maximum profit = 697 - 399 = Rs. 298
NET PROFIT / LOSS
Jhun Jhunwala makes 1000 toys and incurs a cost of Rs. 1.2 for each toy.
He marks-up the price in such a way that if he sells only 70% of
manufactured toys he will realize 16.66% overall profit.
He sells only 750 articles at the marked price since rest of the toys are
found to be defective so can't be sold.
What is the net profit or loss of Jhun Jhunwala?
a) 14.44% loss b)25% profit c)33.33% profit d)none of these
Total cost price = 1000 x 1.2 = Rs. 1200
Expected selling price = 700 x selling price per toys
= 1200 X 1.1666 = 1400
→ Selling price per toys = Rs. 2 per toy
Now the real selling price = 750 x 2 = Rs.1500
∴ Profit = (1500 - 1200) = Rs. 300
∴ Profit % = ( 300 / 1200 ) x 100 = 25%
Percentage Increase in CP
• A retailer increase the selling price by 25% due to which his profit
percentage increases from 20% to 25%.
• What is the percentage increase in cost price ?
• A)20% B)30% C)25% D)50%
• In beginning
Cost Price (CP) = Rs. 100 Profit % = 20 Selling Price (SP) =
120
When profit increases from 20% to 25%
Cost Price (CP) = y Profit % = 25
• Selling Price (SP) after 25% increase = 120 + 25% = Rs. 150
y + 25 % profit = 150
1.25 y = 150
so y = 120
% change in cost price = [ (120 - 100) / 100 ] x 100 = 20%
Successive discounts
The price of an article reduces to
576 after two successive
discounts.
The markup is 80% above the cost
price of Rs. 500.
What is the new profit percentage
if instead of two successive
discount
the markup price was further
increased successively two times
by the same percentage?
a)259.2% b)59.2%
c)159.2%
d)can’t be determined
Cost price (CP) = 500
Selling Price (SP) = 576
Markup price (MP) = 900
SP = MP 1 −
𝑅
100
2
576 = 900 1 −
𝑅
100
2
24 = 30 1 −
𝑅
100
1
Rate of Discount = 20%
New SP = 900 1 +
20
100
2
= 900 * (36/25)
= 1296
New Profit% =
1296 −500
500
∗100
= 796/5
= 159.2%
PERCENTAGE DISCOUNT
I wanted to purchase 10 chairs for
the class room whose cost was Rs.
200 each.
the trader offered me a discount if
I were to purchase a set of 12
chairs.
So I calculated that if I assume the
normal price of 10 chairs then
we can purchase 2 extra chairs
which cost me only Rs. 80 each of
two chairs at the cost price of 12
chairs after discount.
What is the percentage discount?
a)6% b)8%
c)12% d)10%
Price of 10 chairs = 10 x 200 = 2000
Price of 12 chairs (without discount)
= 12 x 200 = 2400
Price of 12 chairs (with discount)
= 10 x 200 + 2 x 80 = 2160
Therefore discount
= 2400 - 2160 = 240
Hence discount %
= (240 / 2400) x 100 = 10%
Profit % & Discount %
Cost price of 12 oranges is
equal to the selling price of 9
oranges and
the discount on 10 oranges is
equal to the profit on 5
oranges.
What is the percentage point
difference between the profit
percentage and discount
percentage?
A)20 B)22.22
C)16.66 D)15
Ratio of selling price and Cost Price,
SP:CP = 12:9 =4:3
Profit of 3 oranges = Re 1 (Let CP = Re 1)
Profit = 1/3 = 33.33%
and, Discount = 11.11%
Since, CP : SP : MP = 3 : 4 : 4.5
Profit doubles that of discount.
So, % point discount = 33.33%-11.11%
= 22.22% point.

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Profit and loss standard level

  • 1. PROFIT & LOSS STANDARD LEVEL C.S.VEERARAGAVAN
  • 2. COST PRICE A)₹ 300 B)₹ 350 C)₹ 375 D)₹ 400 Let the cost price of table = ₹ N selling price with 15% gain = ₹ 115N 100 CP = [(75%) x CP]/100 = ₹ 75N/100 New SP = ₹ (115N/100) – 60 AND Profit 32% 115N 100 −60 − 75N 100 75N 100 X 100 = 32 115N 100 − 60 − 75N 100 = 32 N 100 ∗ 3 4 40N – 6000 = 24 N 16N = 6000 N = 375 A person sold a table at a gain of 15%. Had he bought it for 25% less and sold it for ₹ 60 less, he would have made a profit of 32%. The cost price of table was SHORTCUT GIVEN GAIN 15% = 100 + 15 = 115 CP = (100 – 25)% = 75 GIVEN PROFIT = 75% OF 32 = 24 DIFFERENCE = 115 – (75 + 24) = 16 SOLD LESS = 60 * 100 = 6000 COST PRICE = 6000/16 = 375
  • 3. COST PRICE A)₹ 25 B)₹ 75 C)₹ 50 D)₹ 40 Let the cost price of first bicycle be ₹ P. The cost price of second bicycle = ₹ (1600 - P) [20% of P + 10% of (1600 - P )] - [10% of P + 20% of (1600 - P)] = 5 2P 10 + 1600 −P 10 − P 10 + 2 1600 − P 10 = 5 P 10 − 1600 −P 10 = 5 ⇒ 2P = 1600 + 50 ∴ P = 1650/2 = 825 Cost of second bicycle = (1600 - 825) = ₹ 775 ∴ Required difference = 825 - 775 = ₹ 50 A person bought two bicycles for ₹ 1600 and sold the first at 10% profit and the second at 20% profit. If he sold the first at 20% profit and the second at 10% profit, he would get ₹ 5 more. The difference in the cost price of the two bicycles was
  • 4. SELLING PRICE Cost of a packet of coffee powder and a litre of milk are ₹ 20 and ₹ 30 respectively. 10 cups of coffee is made with one packet coffee powder and for each cup 200 ml of milk is used. If coffee is sold at 25% profit, the selling price of each cup of coffee is A)₹ 12.50 B)₹ 6.25 C) ₹ 8 D)₹ 10 Cost of coffee powder used in one cup = 20/10 = ₹ 2 Cost of milk used in one cup = (30/1000) x 200 = ₹ 6 ∴ Cost of each cup coffee = 2 + 6 = ₹ 8 To gain 25% profit, sale price of each cup of coffee = 125% of 8 = ₹ 10
  • 5. COST PRICE A shopkeeper sells transistors at 15% above its cost price. If he had bought it at 5% more than what he paid for it and sold it for ₹ 6 more, he would have gained 10%. The cost price of the transistor is ? A)₹ 800 B)₹ 1000 C)₹ 1200 D)₹ 1400 Let cost price of transistor = ₹ N According to the question, CP of transistor = (N x 105)/100 SP of transistor = (115 x N)/100 + 6 Profit percentage = (SP - CP)/CP x100 115N 100 + 6 −105N 100 105N 100 = 10 ⇒ 10 = [(10N + 600) x 100]/ 105N ⇒ 105N = 100N + 6000 ⇒ 5N = 6000 ⇒ N= ₹ 1200
  • 6. MAXIMUM PROFIT Teenagers shoe company sells the shoes whose prices i.e.,cost price and selling price are the multiples of either 13,14,15,16,17,18 or 19, starting from Rs. 399 to Rs.699 (i.e, 399 ≤ CP/SP ≤ 699). What can be the maximum profit of the company? a)Rs. 292 b)Rs. 398 c)Rs. 298 d)Rs. 300 The maximum possible profit = maximum possible difference in SP and CP. It means SP be maximum and CP be minimum CP (min) = Rs. 399 19 x m = 399, where m is an integer. Again SP (max) = Rs. 697, which is very close to 699 Here 697 = 17 k, k is a positive integer. So, the maximum profit = 697 - 399 = Rs. 298
  • 7. NET PROFIT / LOSS Jhun Jhunwala makes 1000 toys and incurs a cost of Rs. 1.2 for each toy. He marks-up the price in such a way that if he sells only 70% of manufactured toys he will realize 16.66% overall profit. He sells only 750 articles at the marked price since rest of the toys are found to be defective so can't be sold. What is the net profit or loss of Jhun Jhunwala? a) 14.44% loss b)25% profit c)33.33% profit d)none of these Total cost price = 1000 x 1.2 = Rs. 1200 Expected selling price = 700 x selling price per toys = 1200 X 1.1666 = 1400 → Selling price per toys = Rs. 2 per toy Now the real selling price = 750 x 2 = Rs.1500 ∴ Profit = (1500 - 1200) = Rs. 300 ∴ Profit % = ( 300 / 1200 ) x 100 = 25%
  • 8. Percentage Increase in CP • A retailer increase the selling price by 25% due to which his profit percentage increases from 20% to 25%. • What is the percentage increase in cost price ? • A)20% B)30% C)25% D)50% • In beginning Cost Price (CP) = Rs. 100 Profit % = 20 Selling Price (SP) = 120 When profit increases from 20% to 25% Cost Price (CP) = y Profit % = 25 • Selling Price (SP) after 25% increase = 120 + 25% = Rs. 150 y + 25 % profit = 150 1.25 y = 150 so y = 120 % change in cost price = [ (120 - 100) / 100 ] x 100 = 20%
  • 9. Successive discounts The price of an article reduces to 576 after two successive discounts. The markup is 80% above the cost price of Rs. 500. What is the new profit percentage if instead of two successive discount the markup price was further increased successively two times by the same percentage? a)259.2% b)59.2% c)159.2% d)can’t be determined Cost price (CP) = 500 Selling Price (SP) = 576 Markup price (MP) = 900 SP = MP 1 − 𝑅 100 2 576 = 900 1 − 𝑅 100 2 24 = 30 1 − 𝑅 100 1 Rate of Discount = 20% New SP = 900 1 + 20 100 2 = 900 * (36/25) = 1296 New Profit% = 1296 −500 500 ∗100 = 796/5 = 159.2%
  • 10. PERCENTAGE DISCOUNT I wanted to purchase 10 chairs for the class room whose cost was Rs. 200 each. the trader offered me a discount if I were to purchase a set of 12 chairs. So I calculated that if I assume the normal price of 10 chairs then we can purchase 2 extra chairs which cost me only Rs. 80 each of two chairs at the cost price of 12 chairs after discount. What is the percentage discount? a)6% b)8% c)12% d)10% Price of 10 chairs = 10 x 200 = 2000 Price of 12 chairs (without discount) = 12 x 200 = 2400 Price of 12 chairs (with discount) = 10 x 200 + 2 x 80 = 2160 Therefore discount = 2400 - 2160 = 240 Hence discount % = (240 / 2400) x 100 = 10%
  • 11. Profit % & Discount % Cost price of 12 oranges is equal to the selling price of 9 oranges and the discount on 10 oranges is equal to the profit on 5 oranges. What is the percentage point difference between the profit percentage and discount percentage? A)20 B)22.22 C)16.66 D)15 Ratio of selling price and Cost Price, SP:CP = 12:9 =4:3 Profit of 3 oranges = Re 1 (Let CP = Re 1) Profit = 1/3 = 33.33% and, Discount = 11.11% Since, CP : SP : MP = 3 : 4 : 4.5 Profit doubles that of discount. So, % point discount = 33.33%-11.11% = 22.22% point.