APPLICATION OF
PERCENTAGE
Q.1. By selling an article for Rs 225, the shopkeeper, loses 25%. Find the gain
or loss%, if the article be sold for Rs. 360.
Ans. SP = Rs 225 (Given)
Loss = 25% (Given)
We know that,
100
100


 100 
CP = 
× 225 = Rs 300
 × 225 = Rs
 × SP = Rs 
75
 100 − Loss% 
 100 − 25 
Thus, CP = Rs 360 , Now SP = Rs 360 ∴
Gain = SP − CP
= Rs (360 − 300) = Rs 60
60
Hence, gain% on the article =
× 100 = 20%
300
Q.2. A dealer sold two T.V sets for Rs 2,400 each, gaining 20% on one set and
losing 20% on the other set. Find his net gain or loss and also express it as
a percentage.
Ans. SP of each T.V. set = Rs 2,400
(Given)
Case I:
Gain on one T.V. set = 20%
(Given)
100


 100

CP = 
× 2400  = Rs 2,000
 × SP = Rs 
 100 + Gain% 
 120

Case II:
Loss on other T.V. set = 20%
100
 100

CP = 
× 2400 = Rs 3000
 × SP = Rs
80
 100 − Loss 
Thus, CP of the two sets = Rs (2000 + 3000) = Rs 5000
SP of the two sets = Rs 2400 × 2 = Rs 4800
Loss = CP − SP = Rs (5000 − 4800) = Rs 200
200
Hence, Loss% =
× 100 = 4%
5000
Math Class IX
1
Question Bank
Q.3. A dealer sold two almirahs for Rs 6090 each gaining 16% on one and
losing 16% on the other. Find his net gain or loss per cent in the whole
transaction.
Ans. SP of first almirah = Rs 6090
(Given)
Gain% on first almirah = 16%
SP × 100
6090 × 100
 6090 × 100 
Then, CP =
=
= Rs 
 = Rs 5250
(100 + Gain%)
100 + 16
 116

∴ SP of the second almirah = Rs 6090
Loss on the other = 16%
SP × 100
6090 × 100
 6090 × 100 
Then, CP =
=
= Rs 
 = Rs 7250
(100 − Loss%)
100 − 16
84


Thus, total CP of both almirahs = Rs (5250 + 7250) = Rs 12500
Total SP of both almirahs = Rs (6090 + 6090) = Rs 12180
∴ Loss = CP − SP = Rs (12500 − 12180) = Rs 320
Loss% × 100  320 × 100 
Loss% =
=
 % = 2.56%
CP
 12500 
Hence, loss percentage on whole transaction is 2.56%.
Q.4. By selling a book for Rs 115.20, a man loses 10%. At what price should he
sell it to gain 5% ?
(Given)
Ans. SP of the book = Rs 115.20
Loss on the book = 10% (Given)
SP × 100
 115.20 × 100 
Then, CP =
= Rs 

(100 − Loss%)
 100 − 10 
 11520 × 100 
= Rs 
 = Rs 128
 100 × 90 
If gain = 5%
CP (100 + gain%) 128 (100 + 5)
 128 × 105 
Then, SP =
=
= Rs 
 = Rs 134.40
100
100
 100 
Math Class IX
2
Question Bank
Q.5. The difference between selling an article at 7% profit and at 16% profit is
Rs 63. Find the cost price of the article and also the two selling prices.
Ans. Difference in per cent profit = (16% − 7%) = 9%
Thus, 9% of CP = Rs 63
63 × 100
= Rs 700
⇒ CP =
9
When gain is 7% then SP
CP(100 + Gain%)
 700(100 + 7) 
 700 × 107 
=
= Rs 
 = Rs 
 = Rs 749
100
100


 100 
700(100 + 16) 700 × 116
=
= Rs 812
When gain is 16% then SP =
100
100
Hence, two selling prices are Rs 749 and Rs 812.
Q.6. A shopkeeper makes a profit of 20% on selling a transistor radio for
Rs 840. For how much should he sell another transistor radio, whose cost
price is Rs 25 more than the first one, to make the same percentage profit?
Ans. SP of transistor radio = Rs 840 , Profit of transistor radio = 20%
100


 100 
Then, CP = 
 × SP = Rs
 × 840
 100 + Profit 
 100 + 20 
 100

= Rs 
× 840  = Rs 700
 120

Thus, CP of the second radio set = Rs (700 + 25) = Rs 725
Thus, required profit % = 20%
120
 100 + Profit 
 100 + 20 
Then, SP = 
× 725
 × CP = Rs 
 × 725 = Rs
100
100


 100 
6

= Rs  × 725  = Rs 6 × 145 = Rs 870
5

Hence, transistor radio should be sell at Rs 870.
Math Class IX
3
Question Bank
Q.7. The cost of manufacturing of an article is made up of material, labour and
overheads in the ratio 4 : 3 : 2. If the cost of labour is Rs 45. Find the total
cost of the article. Also, find the profit per cent, if the article is sold for
Rs 180.
Ans. Cost of material : Cost of labour : Cost of overheads = 4 : 3 : 2
Thus, sum of ratios = 4 + 3 + 2 = 9
3
Cost of labour = × cost of article
9
But, given that cost of labour = Rs 45
9

Cost of article = Rs  45 ×  = Rs (15 × 9 ) = Rs 135
3

Hence, the cost of article = Rs 135
SP of article = Rs 180
Profit = SP − CP = Rs (180 − 135) = 45
Profit
45
Hence, profit per cent =
× 100% =
× 100%
CP
135
1
100
1
= × 100% =
% = 33 %
3
3
3
Q.8. A owns an old car worth Rs 50,000. He sells it to B at a profit of 15%.
After some time, B sells it back to A at 15% loss. Find A’s loss or gain
percentage on the whole.
Ans. CP of A for the old car = Rs 50,000 , Profit of A = 15%
CP × (100 + Profit%)
 50000 × (100 + 15) 
Thus, SP of A for the old car =
= Rs 

100
100


 50000 × 115 
= Rs 
 = Rs 500 × 115 = Rs 57500
100


CP of B for an old car = Rs 57500 , Loss of B for = 15%
CP × (100 − Loss%)
 57500 × (100 − 15) 
Thus, SP of B for old car =
= Rs 

100
100


 57500 × 85 
= Rs 
 = Rs 575 × 85 = Rs 48875
100


Again, CP of A = Rs 48875
Profit of A = SP − CP = Rs (57500 − 48875) = Rs 8625
Math Class IX
4
Question Bank
Profit
 8625

 8625 
× 100% = 
× 100  % = 
%
CP
 50000

 500 
1725
% = 17.25%
=
100
Hence, gain per cent for A on the whole is 17.25%.
Q.9. A sells a watch to B at a gain of 20% and B sells it to C at a loss of 10%
and C sells it for Rs 1404 gaining 4%. How much did A pay for it?
Ans. SP of watch for C = Rs 1404
(Given), Gain for watch = 4%
SP × 100
 1404 × 100 
Thus, CP of watch for C =
= Rs 

100 + Gain%
 100 + 4 
 1404 × 100 
= Rs 
 = Rs 1350
 104

Loss on watch for B = 10%
SP × 100
 1350 × 100 
Thus, CP of watch for B =
= Rs 

100 − Loss%
 100 − 10 
 1350 × 100 
= Rs 
 = Rs 1500
90


So, SP of watch for A = Rs 1500 , Gain of watch for A = 20%
SP × 100
 1500 × 100 
Thus, CP of watch for A =
= Rs 
 = Rs 1250
100 + Gain%
 100 + 20 
Q.10. A man bought eggs at Rs 19.20 per dozen. At what price per hundred
must he sell them as to earn a profit of 15%?
Ans. CP of 1 dozen eggs = Rs 19.20 (Given)
 19.20 × 100 
∴ CP of 100 eggs = Rs 
 = Rs 160 , Profit% = 15%
12


CP (100 + profit%)
Hence, SP of 100 eggs =
100
160(100 + 15) 
 160 × 115 
= Rs 
= Rs 
 = Rs 184

100


 100 
Profit per cent of A for an old car =
Math Class IX
5
Question Bank
Q.11. A man purchased two radios for Rs 1500 each and sold one of them at
12% gain. He sold the other radio on such a price that there was a total
loss of Rs 120 in the whole transaction. Find the percentage loss made by
him on second radio.
Ans. CP of one radio = Rs 1500 , Gain % on one radio = 12%
CP(100 + Gain%)
Then, SP =
100
1500 (100 + 12)
1500 × 112
=
= Rs
= Rs 1680
100
100
Thus, CP of both radios = Rs 1500 × 2 = Rs 3000 ∴ Total loss = Rs 120
Thus, total SP = CP − Loss = Rs (3000 − 120) = Rs 2880
SP of first radio = Rs 1680
Thus, SP of second radio = Rs (2880 − 1680) = Rs 1200
Loss on second radio = CP − SP = Rs (1500 − 1200) = Rs 300
Loss× 100 300 × 100
Loss% on second radio =
=
% = 20%
CP
1500
Hence, loss percentage on second radio set is 20%.
Q.12. A man sells a T.V. set for Rs 6,900 and makes a profit of 15%. He sells a
second T.V. set at a loss of 10%. If on the whole he neither gains nor loses.
Find the cost price of the second T.V. set.
Ans. SP of first TV set = Rs 6,900
Profit = 15%
100
Thus, CP of first TV set = Rs
× 6900 = Rs 100 × 60 = Rs 6,000
115
∴ Profit on first T.V. set = SP − CP = Rs (6900 − 6000) = Rs 900
Loss% on second T.V. set = 10%
100
Thus, CP of second T.V. set = Rs
× 900 = 10 × 900 = Rs 9000
10
Hence, the cost price of the second T.V. set is Rs 9000.
Math Class IX
6
Question Bank
Q.13. A shopkeeper sells sugar in such a way that selling price of 950 gm is the
same as the cost price of one kg. Find his profit.
Ans. Let CP of 1 kg sugar be Re 1.
950
kg sugar is Re 1.
Thus, SP of
1000
100
20
1
= Rs
= Rs
∴ SP of 1 kg sugar = Rs
950
95
19
20
1
 20 1 
 20 − 19 
Profit = SP − CP = Rs
− Re 1 = Rs  −  = Rs 
 = Rs
19
19
 19 1 
 19 
1
1
Profit
100
5
× 100% =
Profit per cent =
× 100% = 19 × 100% =
%=5 %
CP
1
1 × 19
19
19
5
Hence, profit percent of shopkeeper is 5 %.
19
Q.14. A man buys two cycles for a total cost of Rs 900. By selling one cycle at a
loss of 20% and the other at a profit of 25%, he makes a profit of Rs 90 on
the whole transaction. Find the cost price of each cycle.
Ans. Let the cost of one cycle be Rs x.
Then cost of the other cycle be Rs (900 − x)
25
20
According to the question, we have (900 − x) ×
− x×
= 90
100
100
900 − x x
− = 90
⇒
4
5
5(900 − x) − 4 x
= 90 ⇒ 4500 − 5 x − 4 x = 90 × 20
⇒
20
⇒ 4500 − 9 x = 1800 ⇒ − 9 x = 1800 − 4500
⇒ − 9 x = −2700 ⇒ 9 x = 2700
2700
= Rs 300
∴ x = Rs
9
∴ CP of one cycle = Rs 300 and CP of the other cycle = Rs (900 − 300) = Rs 600
Math Class IX
7
Question Bank
Q.15. A dealer sold three-forth of his articles at a gain of 20% and the remaining
at cost price. Find the gain earned by him in the whole transaction.
Ans. Let cost of each article be Rs 100 and number of articles = 100
∴ Total cost price = Rs 100 × 100 = Rs 10,000
3
3
CP of th of total articles = Rs 100 × × 100 = Rs 7500
4
4
Gain = 20%
7500 × (100 + 20)
120
∴ SP = Rs
= Rs 7500 ×
= Rs 9000
100
100
1
Cost price of remaining th articles = Rs 100 × 25 = Rs 2500
4
SP = CP = Rs 2500
Total CP = Rs 10000 and total SP = Rs (9000 + 2500) = Rs 11500
Thus, total gain = SP − CP = Rs (11500 − 10000) = Rs 1500
Total gain × 100  1500 × 100 
Hence, gain % =
=
 = 15%
CP
 10000 
Q.16. A fruit seller bought 80 kg of apples at Rs 12.50 per kg. He sold 50 kg of it
at a loss of 10 per cent. At what price per kg should he sell the remaining
apples so as to gain 20 percent on the whole?
Ans. Total quantity of apples bought = 80 kg
CP of 1 kg apples = Rs 12.50
CP of 50 kg apples = Rs 12.50 × 50 = Rs 625.00
CP × (100 − Loss%)
When loss is 10% then SP of 50 kg apples =
100
625 × (100 − 10)
625 × 90
56250
= Rs
Rs = Rs
= Rs
= Rs 562.50
100
100
100
CP of 80 kg apples = Rs 12.50 × 80
= Rs 1000.00 = Rs 1000
CP × (100 + Profit%)
When profit is 20% then, SP of 80 kg apples =
100
1000 × (100 + 20)
= Rs
100
= Rs (10 × 120) = Rs 1200
Hence, SP of 30 kg apples = SP of 80 kg − SP of 50 kg
Math Class IX
8
Question Bank
Q.17.
Ans.
Q.18.
Ans.
Q.19.
Ans.
= Rs (1200 − 562.50) = Rs 637.50
637.50
SP of 1 kg apples = Rs
= Rs 21.25
30
Find the rate of discount being given on a sweater whose price has been
slashed down from Rs 975 to Rs 760.50.
Marked price on sweater = Rs 975
∴ Amount of discount = MP − SP = Rs (975.00 − 760.50) = Rs 214.50
Discount× 100 214.50 × 100
21450 × 100
%=
=
% = 22%
∴ Discount % =
MP
975
100 × 975
When a discount of 15% is allowed on the marked price of an article, it is
sold for Rs 2975.
(i) Calculate the marked price.
(ii) Given that the marked price is 40% above cost price, find the cost
price.
(iii) Find the gain per cent obtained on selling the article.
SP of an article = Rs 2975
Discount on market price = 15%
SP × 100
2975 ×100 2975 × 100
(i) ∴ MP =
= Rs
=
= Rs 3500
100 − Discount%
100 − 15
85
(ii) ∵ Marked price is above 40% on the cost price
MP × 100 3500 ×100
∴ Cost price =
=
= Rs 2500
100 + 40%
140
(iii) Gain per cent obtained on selling the article
Gain × 100 (2975 − 2500) × 100
Gain % =
=
= 19%
CP
2500
A dealer marks his goods 45% above cost price and gives a discount of
20% on the marked price. Find the gain per cent made by him.
Let CP be Rs 100. Then marked price = Rs (100 + 45) = Rs 145
Discount on marked price = 20%
145 × 80
4
MP × (100 − Discount %)
= Rs
= Rs 145 × = Rs 116
∴ SP =
100
100
5
Gain × 100  16 × 100 
Gain% of dealer =
=
 % = 16%
CP
 100 
Math Class IX
9
Question Bank
Q.20. An article was marked 40% above cost price and a discount of 35% was
allowed on the marked price. Find the gain or loss percent incurred on it.
Ans. Let CP of an article is Rs 100
∴ Marked price of an article = Rs (100 + 40) = Rs 140
Discount on market price = 35%
MP × (100 − Discount %)
140(100 − 35)
140 × 65
∴ SP =
= Rs
= Rs
= Rs 91
100
100
100
∴ Loss = CP − SP = Rs (100 − 91) = Rs 9
Loss × 100 9 × 100
=
% = 9%
Hence, loss% on article =
CP
100
Q.21. A sells an article priced at Rs 36,000. He gives a discount of 8% on the
first Rs 20,000 and 5% on the remaining Rs 16,000. B also sells another
article of the same kind priced at Rs 36,000. He gives a discount of 7% on
the total price. Calculate the actual price charged by A and B for the
articles.
Ans. Total SP of A = Rs 36000
Discount on first Rs 20000 = 8%
Discount on remaining amount Rs (36000 − 20000) i.e., Rs 16000 = 5%
8
5
× 20000 + Rs
× 16000
∴ Discount on Rs 36000 = Rs
100
100
= Rs (1600 + 800) = Rs 2400
Thus, actual price charged by A = Rs 36,000 − Rs 2,400 = Rs 33,600
Total SP of B = Rs 36,000
 7

Discount given = 7% , Discount = Rs 
× 36000  = Rs 2520
 100

Thus, actual price charged by B = Rs (36,000 − 2520) = Rs 33480
Hence, actual price charged by A and B are Rs 33,600 and Rs 33480
respectively.
Math Class IX
10
Question Bank
Q.22. A shopkeeper bought a sewing machine for Rs 2450 and marks some price
on it. After allowing a discount of 25% on the marked price, he makes a
loss of 10% on it. Find the marked price of the sewing machine.
Ans. CP of sewing machine is Rs 2450.
Loss % on sewing machine = 10%
CP (100 − Loss%)
2450 (100 − 10)
∴ SP of machine =
= Rs
100
100
2450 × 90
= Rs 2205
= Rs
100
Discount % on marked price = 25%
SP × 100
2205 × 100 2205 × 100
∴ MP of the machine =
=
=
= Rs 2940
100 − Discount%
100 − 25
75
Q.23. After allowing a discount of 10% on the marked price, a trader makes a
profit of 26%. By what percent is the marked price above cost price?
Ans. Let cost price (CP) be Rs 100.
Gain % = 26%
(Given)
⇒ SP = Rs (100 + 26) = Rs 126
Discount % = 10%
(Given)
126 × 100
126 × 100
SP × 100
Thus, MP =
= Rs
= Rs
100 − Discount%
100 − 10
90
= Rs 140
Discount between cost price and marked price = Rs (140 − 100) = Rs 40
Hence, percentage of price which is above the cost price is 40%.
Q.24. A shopkeeper fixes the marked price of an item 35% above its cost price.
Find the percentage of discount allowed so as to gain 8%.
Ans. Let cost price of the item be Rs 100.
∴ Marked price of the item = Rs (100 + 35) = Rs 135
Gain = 8%
CP (100 + Gain%)
100 (100 + 8)
100 × 108
Thus, SP =
= Rs
= Rs
= Rs 108
100
100
100
Thus, total discount = MP − SP = Rs (135 − 108) = Rs 27
Total discount × 100 27 × 100
Hence, per cent discount =
=
= 20%
MP
135
Math Class IX
11
Question Bank
Q.25. A firm, dealing in furniture, allows 4% discount on the marked price of
each item. What price must be marked on a dining table which costs Rs
400 to assemble, so as to make a profit of 20% ?
Ans. Manufacturer’s cost is Rs 400
120
Profit (Given) = 20% Thus, SP = Rs
× 400 = Rs 480
100
Discount = 4%
(Given)
100
100


 100 
Thus, MP = Rs 
× 480 = Rs 500
 × SP = 
 × 480 =
96
 100 − Discount% 
 100 − 4 
Hence, price to be marked is Rs 500.
Q.26. A shopkeeper marks the price of an article at Rs 800. Find in each case,
the selling price, if he allows:
(i) two successive discounts of 10% and 8%
(ii) three successive discounts of 10%, 5% and 3%.
Ans. (i) Marked price of an article = Rs 800
First discount = 10%
(Given)
90
SP after first discount = Rs
× 800 = Rs 720
100
Second discount = 8% (Given)
92 × 72
 92

SP after second discount = Rs 
= Rs 662.40
× 720  = Rs
10
 100

Hence, the selling price is Rs 662.40.
(ii) From (i) we find that SP after first discount of 10% = Rs 720
Second discount = 5% (Given)
95
SP after second discount = Rs
× 720 = Rs 19 × 36 = Rs 684
100
Third discount = 3% (Given)
97
97
∴ SP after third discount = Rs
× 684 = Rs
× 171 = Rs 663.48
100
25
Hence, the selling price is Rs 663.48.
Math Class IX
12
Question Bank
Q.27. A shopkeeper gives 12% additional discount on the discounted price, after
given an initial discount of 20% on the labelled price of a radio. If the final
sale price of the radio is Rs 704, then what is its labelled price ?
Ans. Let labelled price (MP) be Rs 100.
First discount = 20%
100 (100 − 20)
100 × 80
∴ First price after given discount 20% = Rs
= Rs
= Rs 80
100
100
Second discount = 12%
80 (100 − 12)
80 × 88
352
∴ Final selling price = Rs
= Rs
= Rs
100
100
5
352
If final selling price is Rs
, then labelled price = Rs 100
5
100 × 5 × 704
And if final price is Rs 704, then labelled price = Rs
= Rs 1000
352
Q.28. The difference between a discount of 30% and two successive discounts of
20% and 10% on the list price of an article is Rs 72. Find the list price of
the article.
Ans. Let list price of the article is Rs 100.
Case I:
100 (100 − 30)
70
SP of the article =
= Rs 100 ×
= Rs 70
100
100
Case II:
Two successive discounts = 20% and 10%
Thus, SP of the article
(100 − 20) (100 − 10)
80 90
= Rs 100 ×
×
= Rs 100 ×
×
= Rs 72
100
100
100 100
Thus, difference between the two SP = Rs (72 − 70) = Rs 2
If difference is Rs 2, then list price is Rs 100.
100 × 72
And if difference is Rs 72, then list price is Rs
= Rs 3600.
2
Math Class IX
13
Question Bank
Q.29. After getting two successive discounts, a shirt with a list price of Rs 150 is
1
available at Rs 105. If the second discount is 12 % , find the first
2
discount.
Ans. List price of the shirt = Rs 150
SP of the shirt = Rs 105
∴ Total discount = Rs (150 − 105) = Rs 45
Let first rate of discount is x%
1
25
Second discount = 12 % = %
2
2
25 

100 − 
(100 − x) 
2
×
∴ SP of the shirt = Rs 150 ×
100
100
(100 − x) 175
525 (100 − x)
Rs 105 = Rs 150 ×
×
⇒ 105 =
100
2 × 100
400
⇒ 52500 − 525 x = 42000 ⇒ 52500 − 42000 = 525 x
10500
= 20
⇒ 10500 = 525 x ⇒ 525 x = 10500 ⇒ x =
525
Hence, first discount is 20%.
Q.30. A man borrows Rs 1,200 at 10 per cent per annum compound interest. If
Hari pays Rs 250 at the end of each year, find the amount of loan
outstanding at the beginning of the fourth year.
Ans. Principal (P) = Rs 1,200
Rate of interest (R) = 10% p.a.
1200 × 10 × 1
∴ Interest for the first year =
= Rs 120
100
Amount after first year = Rs (1,200 + 120) = Rs 1,320
Amount he pays after 1 year = Rs 250
∴ Principal for the second year (P) = Rs (1,320 − 250) = Rs 1070
1,070 × 10 × 1
Interest for the second year =
= Rs 107
100
Amount he pays after 2nd year = Rs 250
∴ Principal for the third year (P) = Rs (1070 + 107 − 250)
= Rs 927
Rate of interest = 10% p.a.
Math Class IX
14
Question Bank
927 × 10 × 1
9270
= Rs
= Rs 92.70
100
100
Amount for the third year = Rs (927 + 92.70) = Rs 1019.70
Amount he pays after third year = Rs 250
Principal for the fourth year = Rs (1019.70 − 250) = Rs 769.70
Hence, amount outstanding at the beginning of fourth year is Rs 769.70
A borrowed Rs 2,500 from the B at 12% per annum compound interest.
After 2 years, A gave Rs 2,936 and a watch to B to clear the account. Find
the cost of the watch.
Principal (P) = Rs 2,500
Rate of interest (R) = 12% p.a.
Time period = 1
2,500 × 12 × 1
Interest for the first year =
= Rs 300
100
Amount after the first year = Rs (2,500 + 300) = Rs 2,800
Principal for the second year = Rs 2,800
2,800 × 12 × 1
Interest for the second year =
= Rs 336
100
Amount for the second year = Rs (2,800 + 336) = Rs 3,136
Amount paid in cash = Rs 2,936
Balance amount to be paid = Rs (3,136 − 2,936) = Rs 200
Hence, cost of the watch is Rs 200.
What sum will amount to Rs 6,593.40 in 2 years at C.I., if the rates are 10
per cent and 11 per cent for the successive years ?
Let the principal be Rs x.
Rate of interest for the first year (R) = 10% p.a.
Time period = 1 year
x × 10 × 1
x
Thus, interest =
= Rs
100
10
x
11

Amount after the first year = Rs  x +  = Rs
x
10 
10

11
Principal for the second year = x
10
∴ Interest for the third year =
Q.31.
Ans.
Q.32.
Ans.
Math Class IX
15
Question Bank
Q.33.
Ans.
Q.34.
Ans.
Rate of interest for the second year (R) = 11% p.a.
11
x × 11 × 1
121x
10
=
=
100
1000
11
121
1100 x + 121x 1221
Amount after second year = x +
=
x=
x
10
1000
1000
1000
But amount = Rs 6,593.40 (Given)
1221
6,59,340
6,59340 × 1000
x = 6,593.40 =
= 540 × 10 = 5,400
⇒
⇒ x=
1000
100
100 × 1221
Hence, principal is Rs 5,400.
A man saves every year Rs 3000 and invests it at the end of the year at
10% compound interest. Calculate the total amount of his savings at the
end of the third year.
Principal for the first year (P) = Rs 3000
Rate of interest (R) = 10%
Time period = 1 year
3000 × 10 × 1
Interest for the first year =
= Rs 300
100
Amount after the first year = Rs (3000 + 300) = Rs 3300
Amount of savings for the second year = Rs (3300 + 3000) = Rs 6300
6300 × 10 × 1
Interest for the second year =
= Rs 630
100
Amount after the second year = Rs (6300 + 630) = Rs 6930
Amount of saving for the third year = Rs 3000
Total principal for the third year = Rs (6930 + 3000) = Rs 9930
The value of a machine depreciated by 10% per year during the first two
years and 15% per year during the third year. Express the total
depreciation of the machine, as per cent, during the three years.
Let the value (Principal) of the machine (P) = Rs 100
Rate of depreciation for the first year (R) = 10%
∴ Value of machine at the end of first year
100 × 10 

= Rs 100 −
 = Rs (100 − 10) = Rs 90
100 

Rate of depreciation for the second year = 10
∴ Value of machine at the end of second year
Math Class IX
16
Question Bank
90 × 10 

= Rs  90 −
 = Rs (90 − 9) = Rs 81
100 

Rate of depreciation for the third year = 15%
∴ Value of machine at the end of third year
81 × 15 
1215 


= Rs  81 −
 = Rs  81 −
 = Rs (81.00 − 12.15) = Rs 68.85
100 
100 


Total depreciation = Rs (100.00 − 68.85) = Rs 31.15
31.15 × 100
% = 31.15%
Percentage of depreciation =
100
Q.35. A man borrows Rs 18000 at 5% per annum compound interest. If he
repays Rs 6000 at the end of the first year and Rs 8000 at the end of the
second year; how much he should pay at the end of the third year in order
to clear the account? Answer correctly to the nearest rupee.
Ans. Principal (P) = Rs 18000
Rate of Interest (R) = 5% p.a.
Time Period (T) = 1 year
P×R ×T
18000 × 5 × 1
= Rs
Interest =
= Rs 900
100
100
Thus, amount after one year = P + I = Rs (18000 + 900) = Rs 18900
Amount paid after one year = Rs 6000
∴ Balance to be paid = Rs (18900 − 6000) = Rs 12900
12900 × 5 × 1
Interest for the second year = Rs
= Rs 645
100
Amount after second year = Rs (12900 + 645) = Rs 13545
Amount paid after second year = Rs 8000
Balance to be paid = Rs (13545 − 8000) = Rs 5545
5545 × 5 × 1
27725
Interest for the third year = Rs
= Rs
= Rs 277.25
100
100
Hence, amount after third year = Rs (5545 + 277.25) = Rs 5822.25
Math Class IX
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Question Bank
Q.36. Rohit lends Rs 25000 at CI for 3 years. If the rate of interest for the first
two years is 15% per year and for the third year it is 16%, calculate the
sum Rohit will get at the end of the third year.
Ans. Principal (P) = Rs 25000
Rate for first two years (R1 ) = 15% , Rate for the third year (R 2 ) = 16%
n
n
R  1
R 2

Amount = P 1 + 1   1 + 2 
 100   100 
2
1
2
15  
16 

 23   29 
= 25000 1 +
 1 +
 = 25000 ×    
 100   100 
 20   25 
23 23 29
= 25000 × × ×
= Rs 38,352.50
20 20 25
Q.37. What principal will amount to Rs 9744 in two years, if the rates of interest
for successive two years are 16% and 20% respectively?
Ans. Let the principal (P) be Rs x.
Rate of interests for two successive years are (R1) 16% and (R2) 20%
R 
R 
29 6
174
16  
20 


x
⇒ A = P 1 + 1   1 + 2  = x 1 +
 1 +
 = x × × = Rs
25 5
125
 100   100 
 100   100 
But given amount = Rs 9744
174
9744 × 125
x = 9744 ⇒ x =
= 7000 , Hence, principal = Rs 7000
⇒
125
174
Q.38. In what time will Rs 1500 yield Rs 496.50 as compound interest at 10%
per year compounded annually ?
Ans. Principal (P) = Rs 1500
Compound Interest = Rs 496.50
∴ Amount (A) = Rs (1500 + 496.50) = Rs 1996.50
Rate of interest (R) = 10% p.a.
n
R 
10 


⇒ A = P 1 +
 ⇒ 1996.50 = 1500 1 +

 100 
 100 
n
n
n
199650
199650
1331  11 
 11 
 11 
⇒
= 1500   ⇒   =
=
= 
100
100 × 1500 1000  10 
 10 
 10 
n
3
 11   11 
⇒   = 
 10   10 
Hence, time period (n) = 3 years.
Math Class IX
18
3
Question Bank
Q.39. A sum of money, invested at compound interest, amounts to Rs 19360 in 2
years and to Rs 23425.60 in 4 years. Find the rate per cent and the original
sum of money.
Ans. Amount for 4 years = Rs 23425.60
Amount for 2 years = Rs 19360.00
Let rate of interest = R%
n
R 

∵ A = P 1 +

 100 
2
R 

∴ 23425.60 = 19360 1 +

 100 
2
R 
23425.60
2342560

⇒ 1 +
=
 =
19360
19360 × 100
 100 
2
2
R 
121  11 
R 11

=   ⇒ 1+
=
⇒ 1 +
 =
100 10
 100  100  10 
R 11
1
1
= −1 =
⇒
⇒ R = × 100 = 10%
100 10
10
10
Now Amount (A) = Rs 19360, Rate (R) = 10%, Time (T) = 2 years,
Let principal (P) = ?
n
R 

⇒ A = P 1 +

 100 
2
19360 × 10 × 10
11 11
 11 
= Rs 16000
⇒ 19360 = P   = P × × ⇒ P =
10 10
11× 11
 10 
Math Class IX
19
Question Bank
Q.40. A sum of money lent out at CI at a certain rate per annum becomes three
times of itself in 8 years. Find in how many years will the money become
twenty-seven times of itself at the same rate of interest p.a.
Amount (A) = 3 × Rs 100 = Rs 300
Ans. Let the sum (P) = Rs 100 ∴
Time period (n) = 8 years Let rate = R% p.a.
∴ A = P(1+ R) n
∴ 300 = 100(1 + R)8
300 3
(i)
=
⇒ (1 + R)8 =
100 1
Case II:
Amount (A) = 27 × Rs 100 = Rs 2700
Rate of Interest (R) = R% p.a.
∴ A = P (1 + R) n ⇒ 2700 = 100 (1 + R) n
n
2700 27
27
=
⇒ (1 + R) =
⇒ [(1 + R)8 ] 8 =
100
1
1
n
n
(3) 8
⇒
= 27 = (3)3
Comparing both sides, we get :
n
= 3 ⇒ n = 3 × 8 = 24
8
Math Class IX
Time period = 24 years
20
Question Bank