jake borrowed $5000 from two different banks. One bank charges 7% simple interest, and other charges 8.5% simple interest. if the total interest payment is $380, how much did he borrow from each bank?

Question

solve the problem

Answer 1

Let the amounts borrowed be A and B=5000-A.

Let X be the interest on A, so X=7AT/100 and 380-X=8.5(5000-A)T/100, and T=time over which the money was borrowed.

AT=100X/7, so 380-X=(8.5*5000T-8.5AT)/100.

38000-100X=42500T-850X/7; 266000-700X=297500T-850X; 150X=297500T-266000. 3X=5950T-5320=5949T+T-5319-1. X=(1983T-1773)+(T-1)/3.

If T=1, X=$210 and A=$3000 (21000/7) making B=$2000.

If T=2, X=$2193.33 which is clearly in excess of total interest of $380.

5950T>5320 so that X>0, therefore T>76/85. When T=76/85 yr X=0, which implies no interest, when we know the interest accumulated is $380. So it seems reasonable to suppose T=1 year.

 

Answer 2

K1+K2=5000

p1=7% ;  p2=8.5%

P1+P2=380

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P1=(K1 x p1)/100 ;     P2=(K2 x p2)/100

(K1 x p1)/100 + (K2 x p2)/100 = 380

(K1 x 7)/100 + (K2 x 8.5)/100 = 380

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7K1 + 8.5K2 = 38 000

K1 + K2 = 5000                       / x (-7)

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7K1 + 8.5K2 = 38 000                /

-7K1 - 7K2 = - 35 000                 / (+)

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1.5 K2 = 3000         ==>       K2 = 3000/ 1.5 = 2 000 $

 K1 + K2 = 5000     ==>        K1 = 3 000 $

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P1=(K1 x p1)/100 ;       P2=(K2 x p2)/100

P1=(3 000 x 7)/100 ;     P2=(2 000 x 8.5)/100

P1= 210 $                    P2=170 $

Easy 

Answer 3

If x is borrowed at 7%, and y is borrowed at 8.5%, then

0.07 x + 0.085 y = 380

and x + y = 5000

Substituting y= 5000 - x into the first equation

0.07 x + 0.085( 5000 - x) = 380,

-0.015 x = 380 - 425 = - 45,

so x = 45/ 0.015 = 45000/15 = 3000

Therefore y = 5000 - 3000 = 2000

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