Profit and Loss Discount and Tax Class 8 RS Aggarwal Exe-7B Goyal Brothers ICSE Maths Solutions We provide step by step Solutions of council prescribe textbook / publication to develop skill and confidence. Visit official Website CISCE for detail information about ICSE Board Class-8 Mathematics.
Profit and Loss Class 8 RS Aggarwal Exe-7B Goyal Brothers ICSE Maths Solutions
| Board | ICSE |
| Publications | Goyal Brothers Prakshan |
| Subject | Maths |
| Class | 8th |
| writer | RS Aggarwal |
| Book Name | Foundation |
| Ch-7 | Profit and Loss Discount Tax |
| Exe-7B | Find SP/CP when Gain/Loss Percent given |
| Edition | 2024-2025 |
How to Find SP/CP When Gain/Loss Percent Given
In case to calculate SP when CP and Gain/Loss Percent given
- SP = {(100 + P%)/100} x CP.
- SP = {(100 – L%)/100} x CP.
In case to calculate CP when SP and Gain/Loss Percent given
- CP = {100/(100 + P%)} x SP.
- CP = {100/(100 – L%)} x SP
Exercise- 7B
(Profit and Loss Discount and Tax Class 8 RS Aggarwal Exe-7B Goyal Brothers ICSE Maths Solutions)
Que-1: Find the selling price when :
(i) C.P. = Rs7640, Gain = 15% (ii) C.P. = Rs4850, Loss = 12% (iii) C.P. = Rs720, Loss = 8*(3/4)% (iv) C.P. = Rs2652, Gain = 16*(2/3)%
Sol: (i) SP = {(100+Gain%)/100} × CP
= {(100+15)/100} x 7640
= (115/100) x 7640
= 1.15 x 7640
= Rs8786.
(ii) SP = {(100+Loss%)/100} × CP
= {(100-12)/100} x 4850
= (88/100) x 4850
= 0.88 x 4850
= Rs4268
(iii) C.P. = Rs720
Loss = 35/4 %
Loss = (35/4)/100 x 720
Loss = 35/4 x 720/100
Loss = Rs63
S.P. = C.P. – loss
S.P. = 720 – 63
S.P. = Rs657
(iv) C. P = Rs 2652
Gain = 50/3 %
Gain = {(50/3)/100} x 2652
Gain = (50/3) x (2652/100)
Gain = 442
S.P. = C.P. + Gain
S.P. = 2652 + 442
S.P. = Rs3094.
Que-2: Find the cost price when :
(i) S.P. = Rs207, Gain = 15% (ii) S.P. = Rs448.20, Loss = 17% (iii) S.P. = Rs1479, Gain = 6*(1/4)% (iv) S.P. = Rs611.80, Loss = 8%
Sol: (i) CP = (100*S.P)/(100+P%)
= (100*207)/(100+15%)
= (20700)/(115)
= 180.
(ii) CP = (100*S.P)/(100-L%)
= (100*448.20)/(100-17%)
= (44820)/(83)
= 540.
(iii) Given, S.P. = ₹ 1479
Profit = 6 + 1/4 = 25/4%
Hence, C.P. = 100×S.P. /(100 + P)
= 100×1479 /{100+(25/4)}
= 147900 / (425/4)
= 147900 × (4/425)
= 1392.
(iv) CP = (100*S.P)/(100-L%)
= (100*611.80)/(100-8%)
= (61180)/(92)
= 665.
Que-3: A sells a bicycle to B at a profit of 20% and B sells it to C at a profit of 5%. If C pays Rs3780, what did A pay for it?
Sol: Let the C.P of A be 100.
Given that A sells a bicycle to B at 20%. Then the cost price of B = 100 + 20% of 100 = 120%.
Given that B sells a bicycle to C at 5%.Then the cost price of C = 120 + 5% of 120 =120 + 6 = 126.
Given The cost price of C = 3780
If 126 = 3780
1 = 3780/126
100 = (3780 * 100)/126
= 3000.
Que-4: Raju sold a watch to Sonu at 12% gain and Sonu had to Sell it to Manu at a loss of 5%. If Manu paid Rs5320, how much did Raju pay for it?
Sol: Let CP of Raju = x
SP of Raju= CP of Sonu = x × (100+12)/100 = 112x/100
SP of Sonu= CP of Manu = 112x/100 * (100 – 5)/100 = 1064x/1000
Given, 1064x/1000 = 5320
On solving x = 5000
Hence, CP of Raju = Rs 5000
Que-5: A grocer purchase 80 kg of rice at Rs27 per kg and mixed it with 120 kg of rice purchased at Rs32 per kg. At what rate per kg should he sell the mixture to gain 16% ?
Sol: For the 80 kg of rice purchased at ₹27 per kg, the cost is 80 * 27 = ₹2160.
For the 120 kg of rice purchased at ₹32 per kg, the cost is 120 * 32 = ₹3840.
The total cost of both types of rice is ₹2160 + ₹3840 = ₹6000.
Profit Percentage = (Profit / Cost Price) * 100
16 = (Profit / ₹6000) * 100
Profit = (16/100) * ₹6000 = ₹960
Selling Price = Cost Price + Profit = ₹6000 + ₹960 = ₹6960
Selling Price per kg = ₹6960 / (80 kg + 120 kg) = ₹6960 / 200 kg = ₹34.80
Que-6: Mrs. Harjeet bought two bags for Rs1150 each. She sold one of them at a gain of 6% and other at a loss of 2%. How much did she gain ?
Sol: SP1 = (100+P%)/100*CP
= {(100+6)/100} * 1150
= (106/100) * 1150 = 1219
SP2 = {(100-L%)/100}*CP
= {(100-2)/100} * 1150
= (98/100) * 1150 = 1127
TOTAL CP = 2*1150 = 2300
TOTAL SP = 1127+1219 = 2346
GAIN = SP-CP = 2346-2300 = 46
Que-7: A trader purchased a wall clock and a watch for a sum of Rs5070. He sold them making a profit of 10% on the wall clock and 15% on the watch. He earns a profit of Rs669.50. Find the cost price of the wall clock and that of the watch.
Sol: Let the price of the watch is x
Then, The price of the clock is 5070−x
So,
(x×15%)+((5070−x)×10%) = 669.5
0.15x+507−0.1x = 669.5
0.05x = 669.5−507 = 162.5
x = 162.5/0.05
x = 3250
Thus the price of the watch is Rs3250
And the price of the clock is 5070−3250 = 1820
Que-8: Toffees are bought at 15 for Rs20. How many toffees would be sold for Rs20 so as to gain 25% ?
Sol: C.P of 15 toffees = Rs 20
C.P of 1 toffee = 20/ 15 => 4/3
So we get C.P of one toffee = 4/3
Now , we are given gain % = 25 %
We know that Formula of S.P = ( 100 + gain /100 ) * C.P
= {(100+25)/100} x 4/3
= {125/100} x (4/3)
= 5/3
So S.P of one toffee = Rs 5/3
Therefore , Number of toffee sold = Total price/ S.P of one toffee
= 20 / ( 5/3)
= 20 * 3 / 5
= 60 / 5
= 12 Toffees
Que-9: Two-thirds of a consignment was sold at a profit of 5% and the remainder at a loss of 2%. If the total profit was Rs4000, find the value at which the consignment was purchased.
Sol: Let the value of the consignment be Rs y
Profit on 2/3 of consignment = Rs (5 × 2 × y) / (100 × 3) = Rs 10y / 300
Loss on 1/3 of consignment = Rs (1 × y × 2) / (3 × 100) = Rs 2y / 300
According to problem
10y / 300 – 2y / 300 = 4000
On multiplying by 300 we get
10y – 2y = 1200000
8y = 1200000
y = 1200000/8
y = 150000
Que-10: A grocer bought sugar worth of Rs4500. He sold one-third of it at 10% gain. At what gain per cent, the remaining sugar be sold to have a 12% gain on the whole ?
Sol: C.P. of sugar = Rs. 4500
C.P. of 1/3 of sugar
= Rs.4500×1/3 = Rs.1500
Gain% = 10%
∴S.P. = {C.P.(100+gain%)}/100
= {Rs.1500×(100+10)}/100
= Rs.1500×(110/100)
= Rs.1650
Gain on total=12%
∴ Total S.P. = {C.P.(100+gain%)}/100
= {Rs.4500(100+12)}/100
= Rs.4500×(112/100)
= Rs,5040
Remaining C.P = Rs.4500−Rs.1500
= Rs.3000
and remaining S.P.= Rs.5040−Rs.1650
= Rs.3390
∴ Gain = S.P.−C.P
= Rs.3390−Rs.3000
= Rs.390
Gain% = (gain×100)/C.P
= (390×100)/3000
= 13% Ans.
Que-11: A man buys a piece of land for Rs384000. He sells two-fifth of it at a loss of 6%. At what gain per cent should he sell the remaining piece of land to gain 10% on the whole ?
Sol: The cost price of the land is Rs.384000.
Let the selling price of the land is Rs.x.
He sells two fifths of it at a loss of 6% .
S.P. = C.P.{(100-L%)/100}
x = (2/5) x 384000{(100-6)/100}
x = (2/5) x 384000 (94/100)
x = 144384
When there is 10% gain on the whole he should sell at Rs.384000.
So, The price is 384000 + (10/100) x 384000 = 422400
Remaining land is 1 – (2/5) = 3/5
Let g be the percentage gain to sell the remaining land
(3/5) x 384000 + (g/100) x (3/5) x 384000 = 278016
(g/100) x (3/5) x 384000 = 47616
g = (47616x5x100)/(384000×3)
g = 23808000/1152000
g = 62/3 = 20*(2/3)%
Que-12: By selling an almirah for Rs10416, a man gains 12%. What will be his gain or loss per cent if it is sold for Rs9914 ?
Sol: Let us assume Cost price be x.
Gain = 12%
Selling price = x+12% of x = 10416
i.e.
112x/100 = 10416
x = 10416*100/112 = 9300.
So, Cost price = Rs 9300.
Now,
Selling price = 9114
Cost price = 9300
So, Loss = 9300-9114 = 186.
Loss % = 186*100/9300
= 2%.
Que-13: A chair was sold for Rs2142 at a gain of 5%. At what price it should have been sold to gain 10% ?
Sol: Selling price = Rs.2142
Gain = 5%
Let the cost price be Rs.x
∴ CP = {100/(100+Profit%)} × SP
= {100/105} × 2142
= Rs.2040
So, the cost price of the table is Rs.2040
Now,
New gain = 10%
∴ SP = {(100+Profit%)/100} × CP
= {110/100} × 2040
= Rs.2244
Que-14: A television is sold for Rs9360 at a loss of 4%. For how much it should have been sold to gain 4% ?
Sol: SP = ₹9,360
Loss = 4 %
CP = SP × 100/100 – Loss %
= (9,360 × 100)/100 – 4
= (9,360 × 100)/96
= ₹9,750
Again , Gain = 4 %
CP = ₹9,750
Gain = 4 % of ₹9,750
= (4/100) × 9,750
= ₹394
SP = CP + Gain
= ₹(9,750 + 394)
= ₹10,144
Que-15: A shopkeeper sold two fans at Rs1980 each. On one he gained 10%, while on the other he lost 10%. Calculate the loss or gain per cent on the whole transaction.
Sol: Sp of 1st fan = ₹1980
g% = 10%
CP = {100/(100+10)} × 1980 = ₹1800
SP of 2nd fan = ₹1980
L% = 10%
CP = {100/(100-10)} × 1980 = ₹2200
total sp = 1980+1980 = ₹3960
total cp = 1800+2200 = ₹4000
loss = cp – sp = 4000 – 3960 = 40
L% = (loss/cp) × 100
= (40/4000) × 100
= 1%
Que-16: Shanti sold two cameras for Rs6555 each. On one she lost 5%, while on the other she gained 15%. Find the gain or loss per cent in the whole transaction.
Sol: Selling price of each camera is Rs 6555.
On first camera she lost 5%.
On second camera she gained 15%.
SP1 = CP1×(1−(5/100))
6555 = CP1 × 0.95
CP1 = 6555/0.95
CP1 = Rs6900
Let’s denote the Cost Price (CP) of the second camera as CP2.
SP2 = CP2×(1+(15/100))
6555 = CP2 × 1.15
CP2 = 6555/1.15
CP2 = Rs5700
Total Cost Price (CP) = CP1+CP2
= 6900 + 5700 = Rs12600
Total Selling Price (SP) = 6555+6555
= Rs13110
Gain = Total Selling Price – Total Cost Price
= 13110−12600 = Rs 510
Gain Percent = (Gain/Total Cost Price) × 100
= (510/12600) x 100
= 85/21 = 4*(1/21)%.
Que-17: By selling 45 lemons for Rs40, a man loses 20%. How many should he sell for Rs24 to gain 20% on the transaction.
Sol: Selling price of 45 lemons = 40
Loss % = 20%
Cost price of 45 lemons = (40/80) × 100 = 50
Cost price of 1 lemon = 50/45 = Rs. 10/9
If a man sells a lemons for Rs. 24 and gains 20%, then
Cost price = (24/120) × 100 = Rs. 20
If cost price of 1 lemon is Rs. 10/9 then number of lemons which can be brought in Rs. 20 = 20 × (9/10) = 18 lemons
Que-18: Rajini sold a pressure cooker at a loss of 8%. Had she bought it at 10% less and sold for Rs176 more, she would have gained 20%. Find the cost price of the pressure cooker.
Sol: Let as assume the cost price CP = 100
then selling price SP = 100*0.92 = 92
if bought at 10% loss then CP = 90
and SP = 90*1.20 = 108
the difference is 16(108-92) when CP 100
so difference will be 176 when CP will be 1100.
Que-19: A man sold a toaster at a profit of 10%. Had he purchased it for 5% less and sold it for Rs56 more, he could have gained 25%. For how much did he buy it ?
Sol: Let the cost price (CP) of the toaster be Rs. x
Profit % = 10%
S.P. = {(100+P%)/100} x C.P.
S.P. = {(100+10)/100} × x
S.P. = (110x/100)
S.P. = 11x/10
If cost price is 5% less
New C.P.
x – 5% of x = 95x/100
He still gained 25%
New S.P. = {(100+P%)/100} x C.P.
S.P. = {(100+25)/100} x (95x/100)
S.P. = (125/100) × (95x/100)
S.P. = 19x/16
Given, if he purchased it for 5% less and sold it for ₹56 more, he could have gained 25%.
So, the difference in SP is Rs.56
New S.P. – S.P. = 56
= (19x/16) – (11x/10) = 56
= {(95x-88x)/80} = 56
= 7x/80 = 56
x = (56×80)/7
x = 640.
Que-20: A shopkeeper sells each of his goods at a gain of 22*(1/2)%. If on any day, his total sale was of Rs9408, what was (i) the total cost of all goods sold on that day (ii) his profit of that day.
Sol: S.P. = 9408
Gain% = 45/2 %
(i) C.P. = (100*S.P)/(100+P%)
= (100*9408)/(100+(45/2))
= (940800)/(245/2)
= (940800) x (2/245)
= 1881600/245
= 7680
(ii) Profit = S.P. – C.P.
= 9408 – 7680
= 1728.
–: Profit and Loss Discount and Tax Class 8 RS Aggarwal Exe-7B Goyal Brothers ICSE Maths Solutions :–
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