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# x = ab.cd, where a, b, c and d are the tens, units, tenths and hundred

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x = ab.cd, where a, b, c and d are the tens, units, tenths and hundred  [#permalink]

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Updated on: 20 May 2018, 04:32
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Question Stats:

40% (02:36) correct 60% (02:23) wrong based on 101 sessions

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x = ab.cd, where a, b, c and d are the tens, units, tenths and hundredths digits respectively in the decimal representation of x. If the number obtained when x is rounded to the nearest integer is subtracted from the number 10a + b + 1, what is the result?

(1) 10a + b and 10c + d are prime numbers less than 50

(2) When the number 10b+a+d/10+c/100 is rounded to the nearest integer, the result is 10b +a +1

Originally posted by kratigupta on 19 May 2018, 11:36.
Last edited by gmatbusters on 20 May 2018, 04:32, edited 2 times in total.
Edited a typo.
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Re: x = ab.cd, where a, b, c and d are the tens, units, tenths and hundred  [#permalink]

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19 May 2018, 12:13
1
1
kratigupta wrote:
x = ab.cd, where a, b, c and d are the tens, units, tenths and hundredths digits respectively in the decimal representation of x. If the number obtained when x is rounded to the nearest integer is subtracted from the number 10a + b + 1, what is the result?

(1) 10a + b and 10c + d are prime numbers less than 50

(2) When the number 10b+a+d10+c100 is rounded to the nearest integer, the result is 10b +a +1

x can be represented as $$10a+b+\frac{c}{10}+\frac{d}{100}$$.

Statement 1 - 10a + b and 10c + d are prime numbers less than 50.

It means that cd < 50. In terms of decimal .cd is less than 0.5.

Hence the number will not be adjusted to next integer. It will remain same 10a+b. Difference can be found. Sufficient.

Statement 2 - When the number 10b+a+d10+c100 is rounded to the nearest integer, the result is 10b +a +1.

100c+10(b+d)+a is rounded. the result is 10b + a +1.

I think there is typo in this. Could you please check it again.
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Re: x = ab.cd, where a, b, c and d are the tens, units, tenths and hundred  [#permalink]

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20 May 2018, 00:31
1
Statement 1 correct.

Two digit Prime number (since 10a +b and 10c_d will be 2-digit) less than 50 are 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47.

from the given number c can have value either 1, 2, 3, 4.
so ab.cd is of the form ab.(1/2/3/4)d which when rounded off becomes ab only which is equal to 10a+b.
so when subtracted from 10a+b+1 answer will always be 1.
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Re: x = ab.cd, where a, b, c and d are the tens, units, tenths and hundred  [#permalink]

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20 May 2018, 04:38
2
1
x=100a+10b+c/10+d/100
Now after rounding off x can be 100a+10b or 100a+10b+1 depending on the value of c . So the question reduces to is c > or =5

(1) 10a + b and 10c + d are prime numbers less than 50
Since 10c+d<50. hence c <5.
SUFFICIENT.

(2) When the number 10b+a+d/10+c/100 is rounded to the nearest integer, the result is 10b +a +1
Since after rounding off 10b+a+d/10+c/10, we get 10b+a+1, it means d> or =5.
but value of c can not be determined. It can either be < or > or = 5.
So NOT SUFFICIENT.

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Re: x = ab.cd, where a, b, c and d are the tens, units, tenths and hundred  [#permalink]

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20 May 2018, 05:21
Please note that there is no need to find all the prime numbers less than 50, since 10c+d< 50, we get c<0
Finding prime numbers will eat a lot of precious time in exam.

nightvision wrote:
Statement 1 correct.

Two digit Prime number (since 10a +b and 10c_d will be 2-digit) less than 50 are 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47.

from the given number c can have value either 1, 2, 3, 4.
so ab.cd is of the form ab.(1/2/3/4)d which when rounded off becomes ab only which is equal to 10a+b.
so when subtracted from 10a+b+1 answer will always be 1.

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Re: x = ab.cd, where a, b, c and d are the tens, units, tenths and hundred  [#permalink]

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28 May 2018, 06:44
gmatbusters wrote:
x=100a+10b+c/10+d/100
Now after rounding off x can be 100a+10b or 100a+10b+1 depending on the value of c . So the question reduces to is c > or =5

(1) 10a + b and 10c + d are prime numbers less than 50
Since 10c+d<50. hence c <5.
SUFFICIENT.

(2) When the number 10b+a+d/10+c/100 is rounded to the nearest integer, the result is 10b +a +1
Since after rounding off 10b+a+d/10+c/10, we get 10b+a+1, it means d> or =5.
but value of c can not be determined. It can either be < or > or = 5.
So NOT SUFFICIENT.

Hi

Regarding statement 2, if we get 10b + a + 1 , doesn't that mean the number is rounded up and the number can be rounded up only when c is greater than/ equal to 5 ?
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Re: x = ab.cd, where a, b, c and d are the tens, units, tenths and hundred  [#permalink]

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28 May 2018, 06:52
Hi @s
in the question stem x = ab.cd

But as per the statement 2:
(2) When the number 10b+a+d/10+c/100 (Number is ba.dc) is rounded to the nearest integer, the result is 10b +a +1
Since after rounding off 10b+a+d/10+c/10, we get 10b+a+1, it means d> or =5.
but value of c can not be determined. It can either be < or > or = 5.
So NOT SUFFICIENT.

It seems you have considered the number as ba.cd.

I hope it is clear now, Please tag me if you have any further query.

@s wrote:
gmatbusters wrote:
x=100a+10b+c/10+d/100
Now after rounding off x can be 100a+10b or 100a+10b+1 depending on the value of c . So the question reduces to is c > or =5

(1) 10a + b and 10c + d are prime numbers less than 50
Since 10c+d<50. hence c <5.
SUFFICIENT.

(2) When the number 10b+a+d/10+c/100 is rounded to the nearest integer, the result is 10b +a +1
Since after rounding off 10b+a+d/10+c/10, we get 10b+a+1, it means d> or =5.
but value of c can not be determined. It can either be < or > or = 5.
So NOT SUFFICIENT.

Hi

Regarding statement 2, if we get 10b + a + 1 , doesn't that mean the number is rounded up and the number can be rounded up only when c is greater than/ equal to 5 ?

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Re: x = ab.cd, where a, b, c and d are the tens, units, tenths and hundred  [#permalink]

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28 May 2018, 10:36
gmatbusters wrote:
Hi @s
in the question stem x = ab.cd

But as per the statement 2:
(2) When the number 10b+a+d/10+c/100 (Number is ba.dc) is rounded to the nearest integer, the result is 10b +a +1
Since after rounding off 10b+a+d/10+c/10, we get 10b+a+1, it means d> or =5.
but value of c can not be determined. It can either be < or > or = 5.
So NOT SUFFICIENT.

It seems you have considered the number as ba.cd.

I hope it is clear now, Please tag me if you have any further query.

@s wrote:
gmatbusters wrote:
x=100a+10b+c/10+d/100
Now after rounding off x can be 100a+10b or 100a+10b+1 depending on the value of c . So the question reduces to is c > or =5

(1) 10a + b and 10c + d are prime numbers less than 50
Since 10c+d<50. hence c <5.
SUFFICIENT.

(2) When the number 10b+a+d/10+c/100 is rounded to the nearest integer, the result is 10b +a +1
Since after rounding off 10b+a+d/10+c/10, we get 10b+a+1, it means d> or =5.
but value of c can not be determined. It can either be < or > or = 5.
So NOT SUFFICIENT.

Hi

Regarding statement 2, if we get 10b + a + 1 , doesn't that mean the number is rounded up and the number can be rounded up only when c is greater than/ equal to 5 ?

Got it now thank you.
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Posts: 616
Re: x = ab.cd, where a, b, c and d are the tens, units, tenths and hundred  [#permalink]

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11 Jun 2018, 08:54
(1) 10a + b and 10c + d are prime numbers less than 50

Means CD<50
So not incremented so
10a+b+1-(10a+b)= 1
Sufficient

(2) When the number 10b+a+d/10+c/100 is rounded to the nearest integer, the result is 10b +a +1

C+d >0.50
But we don't know whether c is greater or d is greater

C is greater than value get incremented
D is greater than value does not get incremented
So multiple cases exist
Insufficient

Give kudos if it helps

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Intern
Joined: 03 Mar 2014
Posts: 2
x = ab.cd, where a, b, c and d are the tens, units, tenths and hundred  [#permalink]

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26 Jun 2018, 18:47
nightvision wrote:
Statement 1 correct.

Two digit Prime number (since 10a +b and 10c_d will be 2-digit) less than 50 are 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47.

from the given number c can have value either 1, 2, 3, 4.
so ab.cd is of the form ab.(1/2/3/4)d which when rounded off becomes ab only which is equal to 10a+b.
so when subtracted from 10a+b+1 answer will always be 1.

If cd = .47, then d will make c as 5 i.e. ab.47 = ab.5 --> 10a + b + 1
But if If cd = .41, then we have ab.41 = ab --> 10a + b
Hence statement 1 is insufficient.

Please correct me if I am wrong.
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Re: x = ab.cd, where a, b, c and d are the tens, units, tenths and hundred  [#permalink]

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26 Jun 2018, 22:26
1
Hitman47 wrote:
nightvision wrote:
Statement 1 correct.

Two digit Prime number (since 10a +b and 10c_d will be 2-digit) less than 50 are 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47.

from the given number c can have value either 1, 2, 3, 4.
so ab.cd is of the form ab.(1/2/3/4)d which when rounded off becomes ab only which is equal to 10a+b.
so when subtracted from 10a+b+1 answer will always be 1.

If cd = .47, then d will make c as 5 i.e. ab.47 = ab.5 --> 10a + b + 1
But if If cd = .41, then we have ab.41 = ab --> 10a + b
Hence statement 1 is insufficient.

Please correct me if I am wrong.

Hello

If ab.cd= ab.47, then to round off to the nearest integer, we will just look at the value of c (and not consider the value of d).
So here we see that c=4, so we will remove 47 after decimal and just write ab.

(we dont have to change the value of c by looking at the value of d first, thats the rule of rounding off numbers)
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Joined: 03 Mar 2014
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Re: x = ab.cd, where a, b, c and d are the tens, units, tenths and hundred  [#permalink]

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27 Jun 2018, 06:31
amanvermagmat wrote:
Hitman47 wrote:
nightvision wrote:
Statement 1 correct.

Two digit Prime number (since 10a +b and 10c_d will be 2-digit) less than 50 are 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47.

from the given number c can have value either 1, 2, 3, 4.
so ab.cd is of the form ab.(1/2/3/4)d which when rounded off becomes ab only which is equal to 10a+b.
so when subtracted from 10a+b+1 answer will always be 1.

If cd = .47, then d will make c as 5 i.e. ab.47 = ab.5 --> 10a + b + 1
But if If cd = .41, then we have ab.41 = ab --> 10a + b
Hence statement 1 is insufficient.

Please correct me if I am wrong.

Hello

If ab.cd= ab.47, then to round off to the nearest integer, we will just look at the value of c (and not consider the value of d).
So here we see that c=4, so we will remove 47 after decimal and just write ab.

(we dont have to change the value of c by looking at the value of d first, thats the rule of rounding off numbers)

I got it. Thanks a lot!
Re: x = ab.cd, where a, b, c and d are the tens, units, tenths and hundred &nbs [#permalink] 27 Jun 2018, 06:31
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