If John deposited 1000$ in a certain bank account, where the interest

Question

Weekly Quant Quiz Question -8

If John deposited 1000$ in a certain bank account, where the interest earned in a year is given by Interest = Principal deposited    , where x is annual interest paid by the bank. Is the annual interest greater than 7%?
a) The interest earned in a two-year period is 210$.
b)  (1+\frac{x}{100})^n >1.14

Only the TEXT Solution is Allowed

GMATBusters · Accepted Answer

Official Solution: 
 Statement1 : The interest earned in a two-year period is 210$.
Since we don't know whether the interest earned is in first 2-year period or 10-12 year period or so, nothing can be said about the rate of interest.
 NOT SUFFICIENT 
 Statement 2 :  (1+\frac{x}{100})^n >1.14 
Since value of n is not given, we can not say anything about rate of interest.
 NOT SUFFICIENT 
Even after combining statement 1 & 2, we can not say anything about rate of interest.
 NOT SUFFICIENT 
 Answer E

saukrit · Answer

Answer is A

As per first statement for a period of 2 year interest is 210....With 7% interest Interest will be arounf 140...hence 210 is much higher hence A is SUFFICIENT

As per second statement b) (1+/100)^n>1.14 we have no clue about number of years or interst rate...if it is 1 year then yes if it is 2 years or more than no...Hence INSUFFICIENT

Aashutoshkumar · Answer

Solving for equation 1 we have rate as 10, statement 1 is therefore sufficient, statement isn't sufficient since values of x and n can be varied to satisfy tje equation

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Gladiator59 · Answer

St a)

210 = 1000*((1+x/100)^2 -1)

On simplifying a bit... divide by 1000 and take 1 to LHS...
1.21 = (1+x/100)^2
1.21 = (1+x%)^2
1.21 = 1 + 2x + x^2 ( we can ignore x^2 as it will be a negligible amount)
1.21 = 1 + 2x

x = 10.5%

St a) is sufficient.

St b) From above simplification -
1.14 < 1+2x
x > 7%

Hence St b) is also sufficient.

hence Option (D) is correct.

Thanks,
Gladi

u1983 · Answer

E.

1) NS ....... no info about n.... the time frame of compounding
2) NS ....... no info about n.... the time frame of compounding

rahul16singh28 · Answer

Statement I:

P (1+x/100)^2−1] = 210

P = 1000

x = 10. Hence, Sufficeient.

Statement II

(1+x/100)^n]  > 1.14
Here, X can be 7 or greater than 7 for n>= 2.

honneeey · Answer

from 1 st equation 210/1000=

solving we get
x^2+200x-2100=0

Solving we get,
x=10,-210
answer is 10 therefore sufficient
 
solving for second prompt given
 (1+x/100)^n>1.14
solving we get nx/100=14/100
nx=14 therefore n=2,7 x=7,2 not sufficient

Answer=A

mongmat27 · Answer

Ans A

Principal = 1000

210 = 1000((1+(x/100)raised to power n) -1)

Here n=2

Thus, upon solving,

We get,

121 = 100(1 + (x/100))raises to power 2

Taking square root on both sides, we get

11 = 10+ (x/10)

Thus, x =10.

Solving for n =1, we get x >14, so yes

For n =2, 1+(x/100) > 1.5

X>50 so yes.

So ans is D , both are sufficient

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Su_bha · Answer

there is nothing mention about compound annually so we can't  use compound intrest formula also in question a equation for a year interest is given in that equation it is nothing said about n so  any of the statement and combination is not sufficient to answer so option E is best answer

If John deposited 1000$ in a certain bank account, where the interest

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