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Bunuel
If the price of a computer were reduced by 16 percent, which of the following could not be the final price of the computer? (Assume initial price to be integer in cents)

A. $844.10
B. $896.70
C. $1,056.30
D. $1,136.10
E. $1,264.20
Let X be the initial price of computer without discount
then price (final) after discount should be=X(1-16/100)------->X(21/25)=A(say)
means X=A*(25/21).....
so initial price to be integer(As per stem) final price must be multiple of 21(3 or 7)
if we check options all were divisible by 3 except option (A)..which is Ans.
Ans A
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Bunuel
If the price of a computer were reduced by 16 percent, which of the following could not be the final price of the computer? (Assume initial price to be integer in cents)

A. $844.10
B. $896.70
C. $1,056.30
D. $1,136.10
E. $1,264.20

Price is reduced by 16% i.e the new price is 84% of the old price. Hence the new price is multiple of 2X2X3X7
Check the divisibility by 3 as this is the easiest to do. It will give 'A' as the right choice.
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Bunuel
If the price of a computer were reduced by 16 percent, which of the following could not be the final price of the computer? (Assume initial price to be integer in cents)

A. $844.10
B. $896.70
C. $1,056.30
D. $1,136.10
E. $1,264.20

Let price of the computer P$ = 100*P cents.
If the price were reduced by 16 % . The final price will be 84P cents.

So, the options *100 must be a multiple of 84

Now 84 = 2*2*3*7

All the options are divisible by 2
Lets check divisibility by 3
A. 8+4+4+1+0 = 17 not divisible by 3
B. 8+9+6+7+0 = 30 which is divisible by 3
C. 1+0+5+6+3+0 = 15 which is divisible by 3
D. 1+1+3+6+1+0 = 12 which is divisible by 3
E. 1+2+6+4+2+0 = 15 which is divisible by 3.

So Correct Answer A
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chetan2u
Balajikarthick1990
Bunuel
If the price of a computer were reduced by 16 percent, which of the following could not be the final price of the computer? (Assume initial price to be integer in cents)

A. $844.10
B. $896.70
C. $1,056.30
D. $1,136.10
E. $1,264.20

I go with E. Basically i checked for the divisibility since 84 = 4*3*7

Hi Balaji,

Your method is correct but application is NOT..
since the final price has to be a multiple of 84 or 4*3*7, easiest is to check for div by 3..

E is correct as the total is 1+2+6+4+2 = 15, div by 3..
ans will be A 844.10..SUM = 8+4+4+1 = 17, NOT div by 3..
ans A

Hi there,

Do divisibility rules apply to decimals? I was under the impression that they did not apply, but I think I am wrong now on this.
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chetan2u
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Bunuel
If the price of a computer were reduced by 16 percent, which of the following could not be the final price of the computer? (Assume initial price to be integer in cents)

A. $844.10
B. $896.70
C. $1,056.30
D. $1,136.10
E. $1,264.20

I go with E. Basically i checked for the divisibility since 84 = 4*3*7

Hi Balaji,

Your method is correct but application is NOT..
since the final price has to be a multiple of 84 or 4*3*7, easiest is to check for div by 3..

E is correct as the total is 1+2+6+4+2 = 15, div by 3..
ans will be A 844.10..SUM = 8+4+4+1 = 17, NOT div by 3..
ans A

Hi there,

Do divisibility rules apply to decimals? I was under the impression that they did not apply, but I think I am wrong now on this.

When you talk about factors and multiples, it is always integers.
But if I say 1.34567*3, then whatever the answer be in decimals, it should be divisible by 3. Again don’t get into the literal meaning of factors and multiples as this is to ease calculations.
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Hi,
Let the price of the computer be some C.
Now:
the new price is 0.84C.
So we write: 0.84 = 21/25
That means the final price needs to be a multiple of 21.
Only A is not a multiple of 21.
Hope this helps
:D
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Since the last digit is 4 and the price would be 0.84*x so shouldn't the last digit be among 0,4,6,8?
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