p nations

T
R 676
a=P) 1+
100
-1
- 53000
×625
1|
8 53000 x 51
14144= =P||1+ 100 625
=74324.8

1414.4 =p x 0.1664 SIPrincipal xTimex Rate

4.4, (c)
1414.4 100
P=
0.1664
=*85000
30000 ×3x5
Amount = ? (8500 + 1414.4) 100
=4500
=79914.4 00601 Let x % of 22500 = 4500
Principal xTimex Rate01
(c) SI=.
100 22500 xx
= 4500
100
10000×2x3
=F600
Rsalqioig
100 4500 x 100
=20
. Khanna's monthly salary 22500
600x100
3
=20000

R )T.
100 3041.28 =

’ 3041.28 =

3041.28=
-
244 Simple Interest and
Compound Interst
216
3041.28 (a) Amount
625

304128x628
216
&SO)
o, I5800 +7716.72 =1580%0
Alternate Method:
Sucvessive inerease of l6% is happening twice

100 23516.72
34.56%-Interest obtain - 3041.28 15800
3041.28
1%
34.56 4884=1+ 100
3041.28 Taking square root of both the sides.
100%-Amount = 34.56 x 100% 8800
1.22 1+ r
6. () Interest 162%ofthe principal 100
Interest 162 ’ 122 100 +r
or, r= 22%
Principal 100
9. (e) The diference between CI and SI in three
Interest x100 162 100 years
Rate Principalx Time 100 9 Sum xr-(300+r)
-18% per annum 100
or. 381.888 x 100 sum + r (300 + r)
7. 381888x1000 381888 x1000
.:. Sum= 144(300+12)
2 144x312 =8500
R
’16250 + 5616 =16250 Pxrxt 18440 x 1Sx4
10. (e) SI=
100 -=11064
100
100
21866 11, (d) Quicker Method : Difference between CI and SI in
16250 -(" three years
841 Sum x r(300 + r) 16200 x25 (300 +25)
625 100
100lsoti 10
16200 x 625(325)
--(+ R 29
1000000
12. (e) Simple interest
=3290.625

lt _principle xtime xrate 16500x 4x16

100 2s =10560
100 100
R 29 4
-l=
100 25 25 13. (a) S.I, = principal xtimexrate
4
100
’R= 25
X 100 =16% Per annum 12000 x3 x12
Alternate Method: =4320
100
5616 time
=34.56% rate
16250 CI. =P|1+: 100/
In two years, at the rate of 16% p.a.
we get interest of 34.56%
Rate = 16%
 I n t e r e s t
and Compound Interest 245
3 CI=Total amount- sum
S l i m p l e

21952
-8112-7500=612
L1s625
18. (c) P+ PXrxt -=5428
= 7 4 8 5 9 . 1 3 6

100
difference = 4859. 136 - 4320
Required
=539.136 4600x3xt
=5428-4600=828
100
rein
in Equity Funds
vestedd
828× 100
Amount
’t= =6 years
l in debt + Equity Funds 4600x3
(a)
14
= 9 4 5 0 0

r e i n v e s t e d

Amount

= 94500x 13-z175500
7 19. (b) Required difference =
tinvested
earlier in Debt + Equity Funds
2
Amount.
6
175500 -135000
= 7300 x =26.28
1.3
100
funds
invested in equity Prt
amount
Original
-x135000=75000
20. (e) Using SI = 100
9
interest
Px7x4
simple 3584 =
Rate of 100
16 b) Interestx 100
’P=7 12800
Principal x time Now, amount got by CI
10800x 100 12%
22500x4
Compound interest
2
time 4
rate
- 12800( 1+7100/
-Principal|1* 100

= 12800 x1.04x 1.04

=713844.48
13844.48 12800 =
1044.48
Hence, C[=A-P= where
of interest,
21. (b) Let R,and R, be the two diferent rate
R > R,
2200 x R x4 2200x R x4 202.40

100 100
2200x 4
[R,- R] = 202.40
100

1595724
-22500%5 202.40x100 50.6
R- 7= 2200x 4 22
R-R=2,3%
l6. (d) We know compound interest =
SIx100
22. (a) Rate =
Principalx Time
5940x10012% per annun
6500 x3
= 7640 [529 7640 x 129
=72463.9
400 400

17. (b) Total

amount =
75001
= 7500
=78112
 246 Simple Interest aand
- 16500 [(112)-1] Alternate solution: Compound
= 16500 x0.404928 =6681.31
Total percentage incrcase is 9
rteren
23. (a) Letthe principal be x, then 1n a year.
Percentage increase in both 4
the
-X=
650052 = 9% (13l and IInd
scheme) + 2%%schemes
100 (second
r 2323.
650052 9%+%of
4 100000 scheme
20 20 20 100 = 9% (100000) + 2%
650052 x 8000 (second scheme)
’12167x-8000x= 3
100 O/% of 10000
650052 x 8000
4 =2% (second schemel
= 156x80 =12480
100x 4167 3
’ Second scheme %of 100000 =
SIx100 6500x 100 375O0
24. (a) P=
8x13
=6250 Amount invested = (62500, 37500)
RxT
27. (d) Let the parts be x, y and [2600 -(x +yl T
Ci-6250(1+ 100/ 6250 =R 1040
xx4x1 yx6x1_[2600-(x +y)]x8x|
25. (a) Let the population of males=x; then the population of 100 100 100
females =9000-r
Now, 5% ofx + 8% of (9000 x) 42
= (9600 -9000 ) =600 x 6 3 ory=x.
3
or 0.05 x + 720-0.08x=600
5

or,
720-600 =0.08x 0.05x
120=0.03x
So. Xx4x1 2600x
X=4000 100 100
Regd ratio of population of males and females
’ 4x= (7800- 5x) x8 ’ 52x = (7800x8)
4000 4000 3
=4:5
90004000 5000
7800×8
Alternate solution: ’X=| =1200.
52
Total percentage increase in population is:
Money invested at 4% =1200.
5% (male+ female) +3% female
5% (9000) +3% female = 600 28. (b) Let the sum of money be x.
’ 450+3% female =600
1% female = 50 Now,
100% female = 5000
100% male = 4000
Ratio =4:5 or, -(2) or 1+-2
100
26. (b) Let the sum invested at 9% bex and that invested at
Again, let the sum becomes 16 times inn years. Then,
11% be (100000 - x).
Then,
(100000-x) x11x1
100 16=2D or 2 = 2 or n=4
’
39 1) 29. (c) Let Principal = P
100224 100 3
-P=993 ( " , " " ] p - 9 8
9x +1100000-1lx 39000 10 10 10
=9750
100 4 993 x 1000-3000
(1331-1000 P=993 or, P=
2x= (1100000-975000) =125000 ’x=62500. 331
1000
Sum invested at 9%=62500.
(3000 x 3 >x 10) =900

Sum invested at 11% =<(100000-62500) =37500.

.:. Simple interest = 100
 Interest 247
Compound
tand
100x 160
I n t e r e s t

S l i m p l e

9261 :. Principal = 10%1

1600.
3
Population 3 yTs. ago Amount for 2 ycars compounded half yearly
(d) 4
30
-e 160-(1)-194431.
20x20x20 344.81.
.C,I, =(1944.81 - 1600) =
-=8000
9261x
21x21x21

amount (1600 x10x2)

31.
tthe principal
Let x be
(dd) w be the time to double
the money. x D i n
S.I, = 100
-* 320.
also be
interest will (344.81 - 320) =24.81.
Then .. (C.I.) -(S.I.) =
xx25 xy principal be ? P and rate
of interest be R% per
36. (d) Let the
4x100 annum.
years
400 = 25y Dífferecneof C.I. and S.I. for 3
y = 16 years rate PR(300 +R
discount final
successive
a (d) 1 104 100
xy
= - X 4 +

100 R
40×30
=-70 +12 =-58% Difference of C.I. and S.I.
for 2 years = P 100
= -40-30+

100
final rate
successive discount
2nd
95x 20 PR(300 + R
300+R 25
= -45- 20+
100 25
100 104 8
=-65+9=-56%
P R 2 R 100
be MP
Let marked price 104
58 --MP x 56 =12
then MP x
100 100 ’R=00
=12%.
MP × 2
=12 sum=Aand interest rate= r%
100 37. (d) Let
MP =600
A+
AxrX5852
33. (d) Final rate of interest for two
pens 1000
xy = 10+10+ 10x10 21% =5852 ... ()
=xtyt 100
100
Axrx7
Let principal be P. A+ = 7788
121 100
’Px- =12100
100
100000i
...(1)
P= 100 x 100 =
34. (d) Let the original rate be R%. Then, new rate = (2R)%. From equations (i) and (ii), r= 11%E
Interest x 100
(725xRx1 (362.50× 2R x1) =33.50 o 38. (b) Rate of simple interest Principal x time
100 100×3
0sEoo 10800x100
»(2175+ 725)R = 33.50x100x3= 10050 = 12%
22500x 4
time
10050 rate
’R=
2900
=3.46% Compound interest =Principal | 1+:100
3). (a) For first year, S.I. = C.I.ot
Now, ? 10 is S.I. on? 100.
:. 16 is S.I. on 100
x16=*160.
So, S.I. on principal for l year at 10% is 160
248 Simple Interest
and
2g)2
100
xx4
(45000-*)6 Compound
100
’ 2x= 45000 x 3-3x
= 22500x
159
=R5724 ’x=
45000 x3
625 =27000
Another part is 18000.
Let r= Average rate
of interest
Interest for Ist part in one
year =
Similarly, interest for : rest part
27000x4
100
Total interest =2160 in one
s94.5 = 5800 year =\080
45000 xr
100 =2160
594.5
5800 216
’r= = 4.8%
45
6394.5 2
5800 47. (b) Given R=10%, P=400 and
T= 3

1.05-(1+ Compounding is half-yearly, then,

3
I
T=x2=3 years
’ 1.05 -l= ’r=0.05 =5% 10
100 P= =5%
2
Principal x Timex Rate
40. (b) S.I.= 100 Amount, 4=P|1+:200R
4500x 2>x12 5600x 2 x9 3
Difference
100
= 1080 - 1008 =72
100 4-40[1+S
100
41. (d) 21,2121
= 400 x
42. (d) 50years 20 20 20 =7463.0s
We have, A,=*400, A,=7200, R, =10%, R, =4% 48. (a) Difference in rate of interest at 49%
Time (T) =[A, -A,] x 100 +A,R-A,R 16+16+16,+ 4x4x4
= [400 - 200] x 100 + [200 x 10 400 x 4] 100 d00x
= 20000/400 = 50 years.
48 64
43. (c) For three years =0.48+0.0064 =0,4864%
100 (100)2
Sum =Diference x (100)
r(300 + r)=48 x (100) Difference in amount =
5000x0.4864
=24.32,
20 (300 + 20) =375 100x 10000
49. (d) Let the person invest x and y at two different rates
44. (c) P=12000, T,=10 years, T,-3 years, R,=10%, R,=? 12% and 14% respectively.
[(12000 x10 x5)/100 -(12000 xR, x3)100] =3320 Px RxT)
xx12x1, yx10x1 =130
83 100 100 100
’ 50- 3R, = 3
’ 12x+ 10y = 13000 ...)

67 After inter changing invested amount.

yx12x1, xx10x1 =134
45. (a) Total amount = 500 x (1+10/100)3 + 500 %(1+10/ 100 100
100)2 + 500 x (1+10/100) = 1820.5 ’ 12y + 10x = 13400
. 1)
46. (b) Let a person invest 4% ofx. On solving equations (i) and (ii), we get
According to question x=500 and y = 700
 Interest 249
Compound
Interest
and
10
Simple

SLx100

Principal x Time 21
’401/100 -(520
50. (a) Rate

19
7200x100
=12% per annum
21
20000x3
-(5200-y 100
200% =S2021 5-5121
105%
200% = 546000-
305%= 546000
3 x=1790
ScheneA= 520-1790 =3410
R=13%, T=42 nonths
56. (a)
For halfyear
[(1.12)- 1]
- 20000 ( 1 . 4 0 4 9 2 8 - 1) 13 2=7 half-years
= 20000 >x
=78098.56
Interest
RI12
Required
Compound
20000x 7x 6.5 =9100
1. (e) SI=
100
= 35000 x A=P+SI= 20000+ 910029100

108x 108-10000
3years 9years =(3)´years
=35000 x 10000
35000x 1664 57. (d) 7200 ?
=5824 6000
10000

and interest= R% 7200

If theprincipal = P
$2. (c) Then, required ratio times
In 3 years, amount becomes G000
PxRx12
12 2
100
PxRx18

100
1 8 2 : 3

-9) tímes

. in9 years 7200 will become

PP
diference in CI and SI = 6 6
53. (e) For 2 years, 100 7200 X time = 10368

is
Pr(r+300) 58. (a) Principal = x, X, and x,
For 3 years difference 100
x x4x 10 =x, x 8 x 15 =x x 12 x 20
PP
16
X:=3:1;x:x =2:1
100
Prr(r+ 300) 49 X iX=6:2:1
100 59. (c) Let share ofP =I,
100 16
.. share ofQ = 3903 -x
So
r+300 49 . according to question
Solve, r = - (3103 - x)
r
4 (6) 210=Px u00 x3 04 )2
3903 - x

now, SI =4xpx3
S(= 4x210 = 840. So total SI for 6 years 3903 - x
55. (c) =840 + 210 =1050. ’ = ? 2028
Amountt invested in Scheme B=x
Amount invested in SchemeA=5200-x 7000 xrx8
4 60. (d) SI for ? 7000 for 8 years =
21 100
XX10x
100 -(5200 x)x 100 Again borrowed = 3000
 Simple Interest aand
250
Compound
SI=
3000 x r x5 ’ P
1760 x 100
32 x5
Interest
100
’ P= 1100
7000 xr x83000 xrx5
= 4615 Total investment
Total interest =
100 100 =4000+ 1100 =s100
560r + 150r = 4615 67. (b) Let Ramesh had sum of P
710r = 4615
r=6.5% Then, P2x6,1 10 )2
61. (c) Px2x10)
-248.10 48P
100
100 3 \100TP=S610
Pz 16000
62. (c) Amount after 2..5 yrs=3525, after 2 yrs =3420 48P 121P -100P
So SI for half yr =3525-3420 = 105,
So for 1 yr SI= 105 x 2 =210
100 300 -=5610o000
P+2x SI =3420 144P +21P
So P=3420 -2 x 210= 3000 300
-=5610
2 165P
So, 3000 ×rx = 420 -=5610
100 300

o2years ’ P= 5610 × 3000eL

l year
63. (a) 165 0001 000885
’ P= 34 x 300
P
35280 37044 ’ P= 10, 200
68. (c) Interest for 2
In 1year amount
increases by 37044
yrs = 10+ 10+
35280
times Interest for 3 yrs (10x10)/100=219%
=21+10+ (21x10)/100 =33.19%
Now, (33.1-21)% of
P=35280 x 35280 35280
Or, 12. 1% of P=24200P= 24200
37044 37044 32000
64. (d) According to question, 4% rate in 4 or,
adds 16% to the years 69. (e) LetP= (24200x100)/12.1
the sum invested at =21lakh
:. 116% of priniple Total sum =5500 6% bex
principle = 2900 The interest of one
’100% of principle = the interest of part at 6% for 4 years
now, (S) new = 2500 2500 another at 10% for 2 years.
is equal to

.:. 2500= 2500 x 4×T Simple interest PxRx T

100 [Where, P= Principal
’T=25 years.
100 amount,T= duration in years,
annually] R=Interest percentag?
ATQ,
65. (e) A=9400| 1+20 2
Xx 4 x 6%= (5500 - x)
×
100 6x/25 = (5500 x) x2x 10%
6x=(5500 x)
1/5o
= 27500 - × 25 x (1/5)
=9400 y 36 6x
25 5x
= 13536 11x =27500
66. (b) X=2500
According to question, 70. (b) Let theThe sum invested at 6% interest =2500.
Compound interest earned in 2 principal amount be ? x
years Calculating SI:
SI for 1 year at 12%
For 3 years it will berate is (12/100) x X.
1760 Calculating C1: (36/100) x x ---()
Let amount For 1 st year, Interest
invested in another =
For 2 nd year, interest =(12/100) XX
scheme is P. (12/100) ×+ (144/10000)
’ 1760 = 500% of Px8x4 x+(2 Xxx
100 For 3rd year, interest =(12/100) xx+(12/100) xx+
(12/100) ×x+ (144/10000) Xx+ (144/10000)
(1728/1000000) × X. --
Subtracting equation 2and 1 and solving further,
 254
Interest
Compound

Interestand 2R x 15000 2100

x / 1 0 0 0 0 0 0 = 112.32 77. (c) 100
S I m p l o

144 X R= 7%
x=2500, Now for x= 1
4800
× 100
-= 8000 R,= 8% for CI
64
Sum
3x 20
20% per
annum

Equívalent CI at rate of 8% for

2yrs = 8+8+ 109
years at
Cin2 20x20 = 44%
= 16.64%
20 + 16.64
+ 100
x 17100) =1664 171
= 20 for 2 yr =
CIat 8% 100
171 = 2845.44
ATQ
44 Approximately =
16.64 x
x = 4 4 0 0

(800+x) For 9%
10000

8000 + x= 18.81 17100 = 20.13

× 171
x = 200O CI = 100 x
bep
3216.
sum 171 =
Letthe Px2x8 = 2 2 5 Approx. = 20 x
2le) 21 3591
100
For 10% = -x 17100 =
3.
i.e, 1, 2 and
100
| 16P -= 225 values of x
are possible
So, 3 question,
According to
100
78. (a)
225x 100
P=.
5
Px15x3_4320
interest be
'r% per 100
’ P = 4 5 0 0

?'p'and rate
of 4320 x 100
be
2) Let that sum .. P=
45
annum

Amount = ? 1 . 6 p
P=79600

SI = 0.6p Required amount

ATQ,
-960+400)|
0 6 p = P *

100
r x s

-c«o0 40)(1
r = 12%
be x = 18515- 14000
the amount 10x10
4. (d) Let = 21%.
10+ =74515
in 2years = 10 + 100

CI at 10%
Xx15x3,
1.35Xx20x4
=1799892
79. (a) 100
ATQ, 100

Xx21-=1050 x=5000 45X 108X =1799892

100 100 100

And, 153X
5000 x 8xt-= 2000 ’t=5 years. 1799892
100
100 X = 1176400
|176400
3.16 x
15. (c) Let sum be P Requíred value=
.. =3717424

= 122 boavso
invested 251
80. (a) Let man years amount invested
And, after two
3A
Ot P(21-20) = 122
P=?12,200
= (A +
A
2
p.a.
100 two year at 10%
compounded
p.a Equivalent CI of
76. (b) 0verall rate for 2 years at 20%
20x 20
= 44% 10x10 = 21%
+
20+ 100 = 10 + 10
yearly is equivalent to 20 + 2

ATQ, ATQ, 10 = 1371.6

21A
44% of sum = 3432 21
100
100% of sum = 7800 A x 100
+(A t 100
7800 x15 x3 =3510
Simple interest earned = 100
 252
Simple Interest aand
0.210A + 0.171A = 1371.6
0.381A = 1371.6
84. (e) Let the sum be 100x Compound Interest
A = 3600
CI in first year = ? 20x
CI in two years =
15x3 44% of 100x = 44y
Required amount =3600 + 3600 x =7 5220 CI in 2nd year
100
ATQ
=44x-20x=24x
81. (d) x1420
100 = (r+528) 24x - 20x =836
X=209
36x Required sum = 20,900
25
= (x+ 528) 85. (d) Gita
Rupali
36x-25x = 13200 18000 x 10 24000 x9
llx = 13200 10
12
X=1200
6
Amount invested by Prem =3 x 1200=3600
Amount invested by Arun =2x 1200 =2400 .:. Rupali's share = 20000 × 6
10909
20 2 86. (a) By going with the options
2400x| 1+
100
Required ratio =
3600x 25 x3 Interest received at SI = 6000x3x15
100 100 =2700
:x+5= 20%
240036 -2400 Interest received after 2 years at CI
25 1056 88
36x25 x3 2700 225 6000 × 44
82. (a) SI= PNR/100 = 2640
100
30000 = px6x10 . Difference = 2700-2640 =60
100
x= 15%
(30000 x100) 87. (c) Compound interest = 2648
p=
60 -50000
Compound Interest:
8 8000 x = 8000+ 2648 = 10648
50000 x 4000
M00/
8 10648 1331
54000 x = 4320 8000 1000
100
C[= 4000 + 4320=78320
83. (d) Interest earned from scheme P = 3000×Rx4
=120R r 1
Sor o 100
100 10
Interest earned from scheme Q= (3000-x)x2Rx4 r=10%
100
2R(3000-x) Simple interest = Pxnxr 8000 x 10x 10
=8000
25 100 100
According to question, 88. (a) 12100
10000 100
120R =
3 2R(3000-x)
4 25
80 x 25 =3000 --x
2000 =3000 -x
X=1000

Thuson comparing, r= 10%

 253
CompoundInterest
Interest annd
for 2nd year is
2
Simple

a0 (e) Since
the compound interest
obtained
A-P{1+ 100 - 12300(1+
principle of
?1210
the simple interest on the =12800 x1.04×1.04 = 13844 4g
That means it is on 1st Hence,CI=A-P= 13844 48-12S=1A4 AZ
(Main principal + Interest
2ndyear which is Time 4 years,rate 7% p.a., I=3584
10/100 7x4- 3584
year)
= [Principal of 2nd year] x 1 ’ 126
’ 1210 12100
’ Principal of 2ndyear] = 100 ’ 12600
Principal=12600
We know that,
Principal t Interest
of
[Principal of 2ndyear]
= Main 12600
4% 4%
Ist year
Suppose the main
principal is P
12100
S12) DcI. CI=144 48
1]/100 =
’P+ [Px 10 x 4% 20.48
’1.1P= 12100
had sum of P
’ P = 11000 94. (b) Let Ramesh 2
amount =11000 10
5610
.. Principal
3 36P Then, 3 100
x=
SI =Px12 100 100
90. (c) Total
Effective rate of for CI at 15%
interest
annually for 2
years =15 +
p.a.
15 +
48P
100
-P=S610
compounding
48P 121P-100P =5610
(15x15) =32.25%
300
100 100
32.2532.25P
144P +21P = 5610
Total CI= Px = 100 100
300
36P 32.25P
=750 165P
ATQ, 100 100 =5610

P=20000
300
man=100x
91. (b) Let sum invested by 5610×300

140 = 140x P=
he got = 100x x 165
And amount 100 ’P= 34 x 300
per annum.
Let rate of interest be 'r% ’ P=10, 200
Sohit= x
According to the question, 95. (e) Let sum invested by
4 And, rate of interest =
100x x rx = 140r - 100x According tothe question,
100
r= 10 3 xXrxT
2
Required interest=
7200 x 10x 100-=72160
xXrx(T+5)
40000x 5x3
=76000 ..T=10.
92. (c) S.I. for 3 years =
100 10,277
C.I. for 3 years 96. (b) Given that principal P= Rs.
Rate R = 13%
Number of in stallments
’Value of eachinstallment
3 P
= 40000x0.157625
3
100 100
100+R.
=6305 =? 305.
6305-6000
Required difference = 10277
3
Prt 100 100
93. (e) Using SI = 100
100+13) 100+13)
Px7x4 ’ P=12800
3584 = 10277
100 = Rs4300.
Now, amount got by CI 2.39
 256

27750
=37500 Simple Interest
0.74 118. (e) od lot and
Suresh invested = 37500 +
115. (a) Rate = 10%
5000 = 42500
10
100
8000n12 Compound Intere
100 11 100
Let Principal = 100 100
Now, 800
100 880
1 88
1 960.00
10
Total C.I for 3 96.00
10 yrs= 105.60
119. (c) Amount after 3 yrS =2940.16 10.56
10 According to question, 80000+
Difference between C.I. & 10 3 2940.16=
For two years
1
(21-20)
SI
=6250 ’ 10,940.16
/>100 =6250
’ A= 6250x 64
According to question :. 2A =2 x 125
=3200
(110 100) > 10 ’ 100 3200 =
x 10 ’ 1000 6400
116. (a) SI on Hence, simple interest
scheme A= 2920x 5x5 = 6400x12x15
100 =Rs 730
S( on Scheme B= 2920x 5x8
120. (b) The
sum lent = 3720 x100 3720
100 -=7|1520
= Rs 1168 121. (d) S 100 +8 x3
x100 =3O00
Required differrence 100 scheme 124
both schemes = between
interest
1168 -730 =7438 received from 2
8800 = 20
100 -P’ 8800:
117. (b) Rate =
14%=
7 7 ’ P= 8800x 25
Let the principal = (7)= In I[nd scheme -11 -20000 (Principal amount)
ATQ. 343
Principal = 2 x 20000 =
40000
1 343 1 Compound Interest =4
7
= 40000 121
= 40000 >x 21
122.. (b) Interest 100 100 -8400
49 received in scheme A+
scheme B’ Total Interest Interest received in
49
7 7 ax15x2(a+1000)12x2
100
+ = 8448
7 100
1 30a 24a
100 +240 =8448nip
Interest for 3rd year = 64 100
i.e. 12 month = 64 54a
=8208
100
for 3 months 64
=x3=16 12 o bn
a =15,200|
123. (c) A=P+I
Total CI for 2.3 years = 49
+ 49 + Here, Sum = P and Interest =I
16+7=121
Total SI for 2.3 years = 49 + 49 + 343x1×3 SI=(p xrxt/100
A=p+(prt)/100
=110.25 712) =P+ (Px 10 x 4)/100 =P + 0.4P= 1.4P
Now, this amount invest in scheme in Bfor 5yearsat
Required difference = 10.75 8% simple rate of interest
In question A=p+(prt)/100 1.96P
0.56P =
10.75 ’ 32.25 =1.4P +(1.4P x 8 x5)/100 F1.4P+
=
1’3 According to question, 201.60
3433 x 343 Difference =1.96P=360
0002 P=201.60/0.56
- 1.40P= 0.56P=
1029 =|4x
360=
Amount invested in schemne B=1.4P

504
 Interest 257
Interest tannd Compound
Simple = 118013.95 84000 =34013.95
According to question, 130, (a) Let the Sum invested at simple interest =x
124. (d) 2
A. T. Q
(a+500) x 12 x
100
2
-|a(1-a=1s0 xx15x4
= 600
100
24(a + 500) 2la = 150 x=1000
100 100
-21a= 15000 (R+2)72
24a + 12000
’a=1000
Therefore, 10001+ = 1210
’ 3a =3000 K 100
invested in scheme
Hence, amount=R1500
1210
= 1000 + 500 30000 x25 x5-+30000
100 1000
from SI =
Amount received 100
125. (b) R+2
=67500 1+
10 )2 100 V100
67500×1+0
Amount at CI= 100/ R+2_11
100 10
121 =81657
= 67500x R+2 1
100 14175
= 81675 67500 =R
Hence, interest receivedscheme 100 10
P R+ 2=10
126. (d) Amount received from
x5x10
=(x + 500) +
(X +500) -x+ s00) R=8%
100
received from scheme
131. (c) According to question,
.:. Compound interest 2
Q= .3780 9000
o("i 9000 = 3960

R 3960
|1100-= 9000
3 21
’(x+500)x 100 =3780 2
1+=1.2
=0.44 + 1= 1.44’ 100
200 ’x+ 500 =12000
’x+ 500 =3780 x
63 1.2-1= 0.2 R=0.2 x 100 = 20%
>x=11,500| 100
invested in scheme P and O are
3x 9000 x 3x(20-5)
127. (c) Let the amount Hence, simple interest =
and 2x respectively. 100
According to question, = 90 x 3 x15 = 4050
3x x 4 x 4 2x x8x 2
-=1200 Pxtx10
100 132. (b) = 3000 (Given)
100 100
48x 32x 80x
=1200
’ Pxt= 30000 ...i)
= 1200
100 100 100 According to question,
1200 x 10
’x= -=1500 Pxtx12 6768
100
Hence, the amount Vaayun invested is = 3x+ 2x
=5 x 1500 -7500 30000 × 12 36
SIx100 6500 x 100 100
(r-1300) -(P+1200) -6768
6250
128.(a) Principal = RxT 13x8 11
’ (P+1200) x 25 =6768-3600
2
8
=77290
Amt = 25
100/ ’ P+1200 = 3168 x -=7200
-6250=71040
CI =Amt -Principal = 7290
11
’ P= 7200- 1200 = 76000
30240 x 100
129.(e) Rate = = 12%
84000x3 Hence, amount received from (P + 1200) = 7200>
84000
Compound interest = 100,