# SI

• Aug 12th 2009, 10:06 AM
bluffmaster.roy.007
SI
A person spends a part of amount of $20,000 for buying. 40% of the remaining amount is given at 6% interest and the rest money is given at 10% interest. If the total interest the person gets is$1250, then what is the amount he used for buying?
• Aug 12th 2009, 11:29 AM
Wilmer
I do not understand what you posted.
• Aug 12th 2009, 12:14 PM
e^(i*pi)
Quote:

Originally Posted by bluffmaster.roy.007
A person spends a part of amount of $20,000 for buying. 40% of the remaining amount is given at 6% interest and the rest money is given at 10% interest. If the total interest the person gets is$1250, then what is the amount he used for buying?

Quote:

Originally Posted by Wilmer
I do not understand what you posted.

My Interpretation

1. A person gets $20,000 2. This person then spends an amount of money 3. 40% of the remainder attracts 6% interest 4. 10% of the remainder (from step 2) attracts 10% interest. ---------- Let x be the amount he spends and y be the amount which gets interest $y = 20,000-x$ 40% of y gets 6% interest: $0.4y(1.06)^n$ 60% of y gets 10% interest: $0.6y(1.10)^n$ 0.4y(1.06)^n + 0.6y(1.10)^n = 1250 If we assume n is over one period of time (usually annually) so that will cancel $0.424y+0.66y = 1250$ $1.084y = 1250$ $ y = 1153.1365$ $x = 20000-y = 20000-1153.1365 = \18846.86$ • Aug 12th 2009, 06:49 PM Soroban Hello, bluffmaster.roy.007! I interpreted it differently . . . Quote: A person spends a part of amount of$20,000 for buying.
40% of the remaining amount is given at 6% interest
and the rest of the money is given at 10% interest.

If the total interest the person gets is $1250, then what is the amount he used for buying? The person has$20,000.
He spend $x$ dollars for buying.
He has $20,\!000-x$ dollars remaining.

40% of this remainder is: . $0.40(20,\!000-x)$ dollars.
This earns 6% interest: . $I_1 \:=\:(0.06)(0.40)(20,\!000-x)\:=\:480 - 0.024x$

60% of the remainder is: . $0.60(20,\!000-x)$ dollars.
This earns 10% interest: . $I_2 \:=\:(0.10)(0.60)(20,\!000-x) \:=\:1200 - 0.06x$

The total interest is $1250: . $(480-0.024x) + (1200 - 0.06x) \:=\:1250$ We have: . $0.084x \:=\:430 \quad\Rightarrow\quad x \:=\:\frac{430}{0.084} \:=\:5119.047619$ Therefore, he used$5,119.05 for buying.

• Aug 18th 2009, 01:46 PM
mobyn3d
alternate solution......
Mr sorobon is rite..........
lets see how eazy it is..........consider x be the amount remainin...
that is x=20,000-amount used for buying.

40% of remaining amount x gets 6%interest
that is......(0.4)(0.06)(x)

and for second interest (0.6)(0.1)(x)

together they make 1250$=0.024x+0.06x x=1250/0.084=14880.1 so amount used for buying is 20,000-x and that is 5119.9$....