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03 Apr 2020, 03:17
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C
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
55%
(hard)
Question Stats:
70% (03:05) correct
30%
(03:22)
wrong
based on 240
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A person invested a total amount of $1,500,000. A part of it was invested in a fixed deposit earning 6% annual interest, and the remaining amount was invested in two other deposits in the ratio 2 : 1, earning annual interest at the rates of 4% and 3%, respectively. If the total annual interest income is $76,000 then what was the amount invested in the fixed deposit?
A. 700,000
B. 800,000
C. 900,000
D. 1,000,000
E. 1,100,000
Are You Up For the Challenge: 700 Level Questions
A. 700,000
B. 800,000
C. 900,000
D. 1,000,000
E. 1,100,000
Are You Up For the Challenge: 700 Level Questions
03 Apr 2020, 04:24
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Let the amount deposit in fd and other two parts are x, 2y and y. Also removing last three zeros from 1500000 and 76000 to simplify the calculation.
x +3y = 1500
6/100*x + 4/100*2y+3/100*y=76
6x+11y=76
Solving these two equations
X=900
So answer is 900000 that is option c
Posted from my mobile device
x +3y = 1500
6/100*x + 4/100*2y+3/100*y=76
6x+11y=76
Solving these two equations
X=900
So answer is 900000 that is option c
Posted from my mobile device
General Discussion
03 Apr 2020, 03:54
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Bunuel
Let amounts invested at 4% & 3% rate of interest = \(2p\) & \(p\) respectively
--> Amount invested in fixed deposit at 6% rate of interest = \(1,500,000 - 3p\)
Total interest = \(76,000\)
--> \(4\)% of \(2p\) + \(3\)% of \(p\) + \(6\)% of \((1,500,000 - 3p)\) = \(76,000\)
--> \(\frac{4}{100}*2p + \frac{3}{100}*p + \frac{6}{100}*(1,500,000 - 3p) = 76,000\)
--> \(\frac{8p + 3p + 9,000,000 - 18p}{100} = 76,000\)
--> \(9,000,000 - 7p = 7,600,000\)
--> \(7p = 1,400,000\)
--> \(p = 200,000\)
Amount invested in fixed deposit = \(1,500,000 - 3*200,000 = $900,000\)
Option C
