Question 1
Rs.1000
Rs.1100
Rs.1050
Rs.1200
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1120 = P [1 + (5*12/5)/100] P= 1000
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Question 2
Rs.525
Rs.500
Rs.550
Rs.555
70 = (P*4*7/2)/100 P = 500
Question 3
20 %
10 %
15 %
22 1/2%
9 = (1*40*R)/100 R = 22 1/2 %
Question 4
Rs.750
Rs.600
(x*5*1)/100 + [(1500  x)*6*1]/100 = 85 5x/100 + 90 – 6x/100 = 85 x/100 = 5 => x = 500
Question 5
6 2/3 years; 6 2/3 %
5 1/3 years; 5 1/3 %
4 2/3 years; 4 2/3 %
None
4/9 P = (P*R*R)/100 R = 20/3 = 6 2/3%
Question 6
2109
3109
4109
6109
Time = 2 years 4 months = 2(4/12) years = 2(1/3) years. Amount = Rs\\\'. [8000 X (1+(15/100))^2 X (1+((1/3)*15)/100)] =Rs. [8000 * (23/20) * (23/20) * (21/20)] = Rs. 11109. . :. C.I. = Rs. (11109  8000) = Rs. 3109.
Question 7
4360
4460
4560
4660
Let the sum be Rs.P.then P(1+R/100)^3=6690…(i) and P(1+R/100)^6=10035…(ii) On dividing,we get (1+R/100)^3=10025/6690=3/2. Substituting this value in (i),we get: P*(3/2)=6690 or P=(6690*2/3)=4460 Hence,the sum is rs.4460.
Question 8
2 years
2.5 years
3 years
4 years
Amount = Rs.(30000+4347) = Rs.34347 let the time be n years Then,30000(1+7/100)^n = 34347 (107/100)^n = 34347/30000 = 11449/10000 = (107/100)^2 n = 2years
Question 9
1261
1271
1281
1291
Clearly, Rate = 5% p.a., Time = 3 years, S.I.= Rs. 1200. . . So principal=RS [100*1200]/3*5=RS 8000 Amount = Rs. 8000 x [1 +5/100]^3  = Rs. 9261. .. C.I. = Rs. (9261  8000) = Rs. 1261.
Question 10
Rs.48
Rs.98.56
Rs.66.56
None of these
difference in C.I and S.I in 2years =Rs.32 S.I for 1year =Rs.400 S.I for Rs.400 for one year =Rs.32 rate=[100*32)/(400*1)%=8% difference between in C.I and S.I for 3rd year =S.I on Rs.832= Rs.(832*8*1)/100=Rs.66.56
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November 20, 2020
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