Directions: A question is followed by five statements given in five o...
Let invested sum = P
Interest amount after 3 years = P [(1 + R/100)3-1] = 15250 …………. (1)
Statement (a):
According to the statement:
(P x R x 4)/100=16000
PR = 400000 …………. (2)
From (1) and (2):
( P[(1+R/100)3-1])/PR = 15250/400000 = 61/1600
((1 + R/100)3 - 1)/R = 61/1600
1600(1 + R/100)3-1600 = 61R
(1 + R/100)3 = (61R + 1600)/1600 = 61R/1600+1
1+ R3/1003 + 3R/100 + (3R2)/1002 = 61R/1600+1
(R3 + 3R x 1002 + 300R2)/1003 = 61R/1600
(R3 + 3R x 1002 + 300R2)/1002 = 61R/16
16R3 + 480000R + 4800R2 = 610000R
16R2 + 480000 + 4800R = 610000
16R2 – 130000 + 4800R = 0
R2 – 8125 + 300R = 0
After solving:
R = 25
Statement (a) alone is sufficient.
Statement (b):
SI must be less than CI for same sum at same rate of interest.
SI amount after 3 years = 15250 – 3250 = Rs12000
According to the question:
According to the statement:
(P x R x 3)/100 = 12000
PR = 400000 …………. (3)
Since statement (2) and (3) are similar which means by using equations (1) and (3) we can calculate the value of ‘R’.
Statement (b) alone is sufficient.
Statement (c):
According to the question:
P[(1 + R/100)2 - 1]=9000…………….. (4)
From (1) and (4):
( P[(1 + R/100)3 - 1])/(P [(1 + R/100)2 - 1] ) = 15250/9000 = 61/36
[(1 + R/100)3 - 1]/[(1 + R/100)2 - 1] = 61/36
[{(1 + R/100) - 1}{(1 + R/100)2 + 1 + (1 + R/100) } ]/[{(1 + R/100) - 1}{(1 + R/100) + 1} ] = 61/36
((1 + R/100)2 + 1 + (1 + R/100))/((1 + R/100) + 1) = 61/36
36(1 + R/100)2 + 36 +3 6(1 + R/100) = 61(1 + R/100) + 61
36(1 + R/100)2 - 25(1 + R/100) - 25 = 0
36(1 + R/100)2 - 25(1 + R/100) - 25 = 0
36(1 + R/100)2 - 45(1 + R/100) + 20(1 + R/100) - 25=0
9(1 + R/100)[4(1 + R/100) - 5] + 5[4(1 + R/100) - 5] = 0
[4(1 + R/100) - 5][9(1 + R/100) + 5] = 0
[4(1 + R/100) - 5] = 0
1+ R/100 = 5/4 ⇒ R/100 = 1/4 ⇒ R = 25
[9(1 + R/100) + 5] = 0
1 + R/100 = -5/9 ⇒ R/100 = -14/9 ⇒ R = -1400/9(Invalid)
Statement (c) alone is sufficient.
Statement (d):
According to the question:
P(1 + R/100)3 = 31250……………(5)
From (1) and (5):
31250 – P = 15250
P = 16000
From equation (5):
16000(1 + R/100)3 = 31250
(1 + R/100)3 = 31250/16000 = 125/64 = (5/4)3
1+ R/100 = 5/4 ⇒ R/100=1/4 ⇒ R = 25
Statement (d) alone is sufficient.
Statement (e):
According to the question:
2P [(1 + R/100)3 - 1] = 30500
P [(1 + R/100)3 - 1] = 15250………….. (6)
Here equations (1) and (6) are similar which means we cannot determine the value of ‘R’.