EXERCISE 8 (A)
Q1 | Ex-8A | Class 8 | S.Chand | Composite maths | Profit,Loss and Discount |Ch-8 | myhelper
QUESTION 1
A peon purchased a chair for ₹ 700, spent ₹ 170 on its repair and ₹ 30 on the cartage.If he sold the chair for ₹ 1080, what is his gain percent?
Sol :
Cost price of chair = 700 + 170 + 30= 900
Selling price of chair = 1080
Gain = Selling price - Cost price
= 1080 - 900
= 180
Gain Percent $=\dfrac{\text{gain}}{\text{cost price}}\times 100$
$=\dfrac{180}{900}\times 100$
$=\dfrac{180}{9}$
= 20 %
Q2 | Ex-8A | Class 8 | S.Chand | Composite maths | Profit,Loss and Discount |Ch-8 | myhelper
QUESTION 2
Ramesh bought 10 cycles for ₹ 500 each.He spent ₹ 2000 on the repair of all the cycles.He sold five of them for ₹ 750 each and the remaining for ₹ 550each.What is total gain or loss %?
Sol :
Cost price of 1 cycle = 500 ₹
Cost price of 10 cycle = 500 × 10
= 5000 ₹
Total spending on repair = 2000 ₹
Total cost price = 5000 ₹ + 2000 ₹
= 7000 ₹
Selling price of 5 cycles for ₹ 750 = 5 × 750
= 3750 ₹
Selling price of 5 cycles for ₹ 550 = 5 × 550
= 2750 ₹
Selling price of 10 cycles = S.P of 5 cycles for 750 + S.P of 5 cycles for 550
= 3750 + 2750
= 6500 ₹
Loss = Cost price - Selling price
= 7000 - 6500
= 500 ₹
Loss percentage $=\dfrac{\text{loss}}{\text{Total cost price}}$
$=\dfrac{500}{7000}\times 100$
$=\dfrac{50}{7}$
$=7\dfrac{1}{7}$
Q3 | Ex-8A | Class 8 | S.Chand | Composite maths | Profit,Loss and Discount |Ch-8 | myhelper
QUESTION 3
By selling an article for ₹ 960 a man incurrs a loss of 4%.What was the cost price of the article?
Sol :
Cost price of article $=\dfrac{100}{100-\text{loss percent}}\times\text{Selling Price}$
$=\dfrac{100}{100-4}\times 960$
= 1000
Q4 | Ex-8A | Class 8 | S.Chand | Composite maths | Profit,Loss and Discount |Ch-8 | myhelper
QUESTION 4
If the selling price of 20 articles is equal to the cost price of 15articles, then what is the loss percentage ?
Sol :
Quantity=15
Let the CP of 15 articles=100
SP of 20 articles=CP of 15 articles=100
SP of 20 articles=100
20×(SP of 1 article)=100
(SP of 1 article)$=\frac{100}{20}$
∴SP of 15 articles=15×(SP of 1 article)
$=15 \times \frac{100}{20}=15\times \frac{10}{2}$
=75
As SP<CP . So, its
Loss=CP-SP
=100-75=25
Loss %$=\frac{\text{Loss}}{\text{CP}}\times 100$
$=\frac{25}{100}\times 100$
=25%
Q5 | Ex-8A | Class 8 | S.Chand | Composite maths | Profit,Loss and Discount |Ch-8 | myhelper
QUESTION 5
A hawker gains the selling price of 4 ball-point pens on selling 1 dozen pens.What is his gain percentage?
Sol :
As we know 1 dozen = 12 pens
Let the selling price of of one ball point pen be x then the profit will be equal to 4x
And the selling price of 12 pens = 12 x
Then cost price of 12 pens is equals to (C.P = S.P - profit)
Cost price of 12 pens = Selling price - Profit
C.P = 12x - 4x
C.P = 8x
Then , $\text{profit }\% = \dfrac{\text{profit}}{C.P}\times 100$
$=\dfrac{4x}{8x}\times 100$
$=\dfrac{1}{2}\times 100$
= 50 %
Q6 | Ex-8A | Class 8 | S.Chand | Composite maths | Profit,Loss and Discount |Ch-8 | myhelper
QUESTION 6
A man bought pencils at the rate of 6 for ₹ 4 and sold them at the rate of 4 for ₹ 6. What is his gain percentage in transaction?
Sol :
6 pencils are bought for ₹ 4
So, 1 is bought for $=\dfrac{4}{6}$
4 are sold for ₹ 6
So, 1 is sold for $=\dfrac{6}{4}$
Here, S.P > C.P , so , it is a gain
gain = S.P - C.P
$=\frac{6}{4}-\frac{4}{6} $
LCM of 4 and 6 is 12
$=\frac{6\times 3-4\times 2}{12}$
$=\frac{18-8}{12}=\frac{10}{12}=\frac{5}{6}$
∴ gain percent $=\dfrac{\text{gain}}{\text{cost price}}\times 100$$=\dfrac{\frac{5}{6}}{\frac{4}{6}}\times 100$
$=\frac{5}{4}\times 100$
= 125%
Q7 | Ex-8A | Class 8 | S.Chand | Composite maths | Profit,Loss and Discount |Ch-8 | myhelper
QUESTION 7
By selling a basket of mangoes for ₹ 105, a fruit seller loses 30%. For how much should he sell it to gain 40%?
Sol :
As we know
$\text{S.P}=\left(\dfrac{100-\text{loss} \%}{100}\right)\times\text{C.P}$ ....(i)
And here we know S.P = 105 ,
Loss % = 30 % and putting these in (i)
$\text{105}=\left(\dfrac{100-30}{100}\right)\times \text{C.P}$
$\text{105}=\left(\dfrac{70}{100}\right)\times \text{C.P}$
$\text{105}=\dfrac{7}{10}\times \text{C.P}$
$\text{C.P}=105 \times \dfrac{10}{7}$
C.P = 150
Gain % = 40 and putting these two in (ii) , we get
$\text{S.P}=\left(\dfrac{100+\text{Gain}\%}{100}\right)\times \text{C.P}$.....(ii)
$\text{S.P}=\left(\dfrac{100+40}{100}\right)\times 150$
$\text{S.P}=\dfrac{140}{100}\times 150$
S.P = 14 × 15
S.P = 210
Q8 | Ex-8A | Class 8 | S.Chand | Composite maths | Profit,Loss and Discount |Ch-8 | myhelper
QUESTION 8
A man sold a book at a loss of 20%.If he had sold the books for ₹ 12 more,he would have gained 10% .Find the cost price of the book.
Sol :
We know that
Loss % = 20 % and in this case let the Selling Price bex
Putting these in (i)
$\text{S.P}=\left(\dfrac{100-\text{loss} \%}{100}\right)\times\text{C.P}$ .....(i)
$x=\dfrac{100-20}{100}\times \text{C.P}$
$x=\dfrac{80}{100}\times \text{C.P}$
$x=\dfrac{8}{10}\times \text{C.P}$
$\text{C.P}=\dfrac{10x}{8}$ ...(ii)
$\text{S.P}=\left(\dfrac{100+\text{Gain}\%}{100}\right)\times \text{C.P}$...(iii)
And here we know gain % = 10 % , in this case we have Selling Price = x + 12
Putting these two in (iii) we get
$x + 12 = \dfrac{100+10}{100}\times \text{C.P}$
$x + 12 = \dfrac{110}{100}\times \text{C.P}$
$x + 12 = \dfrac{11}{10}\times \text{C.P}$
$\text{C.P}=\dfrac{10}{11}\times (x+12)$ ...(iv)
From (ii) and (iv) , we get
$\text{C.P}=\dfrac{10x}{8}$
$\text{C.P}=\dfrac{10}{11}\times (x+12)$
Which means
$\dfrac{10x}{8}=\dfrac{10}{11}\times (x+12)$
$\dfrac{x}{8}=\dfrac{1}{11}\times (x+12)$
$\dfrac{x}{8}=\dfrac{x}{11}+\dfrac{12}{11}$
$\dfrac{x}{11}-\dfrac{x}{8}=-\dfrac{12}{11}$
$\dfrac{8x-11x}{88}=-\dfrac{12}{11}$
$-\dfrac{3x}{88}=-\dfrac{12}{11}$
$\dfrac{3x}{8}=12$
3x = 12 × 8
$x=\dfrac{96}{3}$
x = 32 ...(v)
putting (v) in (ii)
$\text{C.P}=\dfrac{10\times 32}{8}$
$\text{C.P}=\dfrac{320}{8}$
C.P = 40
Q9 | Ex-8A | Class 8 | S.Chand | Composite maths | Profit,Loss and Discount |Ch-8 | myhelper
QUESTION 9
A trader sells two cycles at ₹ 1188 each, gaining 10% on the first cycle and losing 10% on the second cycle.What is the profit or loss percentage in the whole transaction ?
Sol :
Case 1 (First cycle)
$\text{C.P}=\dfrac{100}{100+\text{gain} \%}\times \text{S.P}$
$\text{C.P}=\dfrac{100}{100+10}\times 1188$
$\text{C.P}=\dfrac{100}{110}\times 1188$
$\text{C.P}=\dfrac{11880}{11}$
C.P = 1080
Case 2 (Second cycle)
$\text{C.P}=\dfrac{100}{100-\text{loss} \%}\times \text{S.P}$
$\text{C.P}=\dfrac{100}{100-10}\times 1188$
$\text{C.P}=\dfrac{100}{90}\times 1188$
$\text{C.P}=\dfrac{11880}{9}$
C.P = 1320
S.P = 1188 + 1188
S.P = 2376
C.P = 1080 + 1320
C.P = 2400
Here , we come to know that S.P < C.P . So , it is a loss
Loss = C.P - S.P
Loss = 2400 - 2376
Loss = 24
$\text{Loss} \% = \dfrac{\text{loss}}{C.P}\times 100$
$\text{Loss }\% = \dfrac{24}{2400}\times 100$
Loss % = 1 %
Q10 | Ex-8A | Class 8 | S.Chand | Composite maths | Profit,Loss and Discount |Ch-8 | myhelper
QUESTION 10
A merchant bought two calculator which together cost him ₹ 4800.He sold one of them at a loss 15% and the other at a gain of 19%.If the selling price of both the calculator is equal, find the cost price of the lower priced calculator.
Sol :
Let the S.P. be x
We know that $\text{C.P}=\left(\dfrac{100}{100-loss\%}\right)\times\text{S.P}$
also $\text{C.P}=\left(\dfrac{100}{100+gain\%}\right)\times \text{S.P}$
C.P. of 1st calculator $=\left(\dfrac{100}{100-15}\right)\times x$
$=\dfrac{20x}{17}$
C.P. of 2nd calculator $=\left(\dfrac{100}{100+19}\right)\times x$
$=\dfrac{100x}{119}$
Now , C.P. of 1st calculator + C.P. of 2nd calculator = Total cost
$=\dfrac{20x}{17} + \dfrac{100x}{119}=4800$
$=\dfrac{240x}{119}=4800$
x = 2380
C.P. Of 1st calculator $=\dfrac{20}{17}\times 2380$= 2800
C.P. of 2nd calculator $=\dfrac{100}{119}\times 2380$ = 2000
C.P. of lower priced calculator = Rs. 2000
Q11 | Ex-8A | Class 8 | S.Chand | Composite maths | Profit,Loss and Discount |Ch-8 | myhelper
QUESTION 11
A manufacturer sells an article to a wholesale dealer at a profit of10%.The wholesale dealer sells it to a shopkeeper at 20% profit.The shopkeeper sells it to the customer for ₹ 56,100 at a loss of 15%.What is the cost price of the article to the manufacturer?
Sol :
Given : Selling price of article = 56,100 and at loss of 15 %
We know that $\text{C.P}=\left(\dfrac{100}{100-loss\%}\right)\times\text{S.P}$
$\text{C.P}=\left(\dfrac{100}{100-15}\right)\times 56100$
$\text{C.P}=\left(\dfrac{100}{85}\right)\times 56100$
Cost price of article for shopkeeper = 66,000 (Shopkeeper's cost price)
Now , lets find C.P of article for the wholesale dealer . In this case S.P =66000 and at a gain of 20 %
$\text{C.P}=\left(\dfrac{100}{100+gain\%}\right)\times \text{S.P}$
$\text{C.P}=\left(\dfrac{100}{100+20\%}\right)\times \text{66000}$
= 55000
Now , lets find C.P of article for manufacturer . In this case S.P = 55000 andat a gain of 10 %
$\text{C.P}=\left(\dfrac{100}{100+ gain\%}\right)\times \text{S.P}$
$\text{C.P}=\left(\dfrac{100}{100+10\%}\right)\times \text{55000}$
= 50000
Cost price of the article to the manufacturer = 50,000
Q12 | Ex-8A | Class 8 | S.Chand | Composite maths | Profit,Loss and Discount |Ch-8 | myhelper
QUESTION 12
A man buys a plot of agricultural land for ₹ 3,60,000. He sells one third of it at a loss of 25% and two fifth at a gain of 20%.At what price must he sell the remaining field so as to make an overall profit of 10% ?
Sol :
C.P = 360000
To gain 10% on whole land ,
S.P = 360000 + 10% of 360000
= 396000
$\dfrac{1}{3}$ of the land sold on 20% loss
S.P of $\dfrac{1}{3}$ land
$=\left(\dfrac{360000}{3}\right)-20\% \text{ of }\left(\dfrac{360000}{3}\right)$
= 96000
S.P of $\dfrac{2}{5}$ of the land
$=\dfrac{360000\times 2}{5}+25\% \text{ of } \dfrac{360000\times 2}{5}$
= 180000
Thus , S.P of the remaining land
= 39600 - 96000 - 180000
= 120000
ALTERNATE METHOD
S.P of total agricultural field at a profit of 10% $=\dfrac{360000\times110}{100}$
= 396000
So , S.P of $\dfrac{1}{3}$ of field
$=\left(\dfrac{360000}{3}\right) \times \left(\dfrac{80}{100}\right)$
= 96000
S.P of $\dfrac{2}{5}$ of the field
$=\dfrac{2\times360000\times 125}{5\times 100}$
= 180000
Hence ,
S.P of the remaining field
= 396000-96000-180000
= 120000
Supreeee
ReplyDeleteThanks for such a nice help
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