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# What is the remainder when X is divided by 40?

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Senior Manager
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Re: What is the remainder when X is divided by 40?  [#permalink]

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17 Jul 2019, 15:17
1
We are to find the remainder when a positive integer x is divided by 40.

1. 3x +30 leaves a remainder of 93 when divided by 120.
(3x+30)/120 = 120m + 93
(X+10)/40=40m+31
If m=0 x+10=31 hence x=21
21/40 leaves R=21
If m=1, x+10=71 hence x=61
61/40 leaves R=21
If m=10, x+10=431 hence x=421.
421/40 leaves R=21. Therefore statement 1 on it’s own is sufficient.

2: 5x-10 leaves Remainder of 15 when divided by 20.
(5x-10)/20= 20m+15
(x-2)/4 = 4m + 3
When m=0 x-2=3 hence x=5
5/40 leaves R=5
When m=1, x-2=7 hence x=9
9/40 leaves a remainder of 9. Hence not sufficient.

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Re: What is the remainder when X is divided by 40?  [#permalink]

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17 Jul 2019, 15:17
1
We are to find the remainder when a positive integer x is divided by 40.

1. 3x +30 leaves a remainder of 93 when divided by 120.
(3x+30)/120 = 120m + 93
(X+10)/40=40m+31
If m=0 x+10=31 hence x=21
21/40 leaves R=21
If m=1, x+10=71 hence x=61
61/40 leaves R=21
If m=10, x+10=431 hence x=421.
421/40 leaves R=21. Therefore statement 1 on it’s own is sufficient.

2: 5x-10 leaves Remainder of 15 when divided by 20.
(5x-10)/20= 20m+15
(x-2)/4 = 4m + 3
When m=0 x-2=3 hence x=5
5/40 leaves R=5
When m=1, x-2=7 hence x=9
9/40 leaves a remainder of 9. Hence not sufficient.

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Re: What is the remainder when X is divided by 40?  [#permalink]

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17 Jul 2019, 15:23
1
What is the remainder when positive integer X is divided by 40?

(1) 3X + 30 leaves remainder 93 when divided by 120.
(2) 5X - 10 leaves remainder 15 when divided by 20.

x=40y+Z
we need to find Z

From the first statement we have:
3X+30 = 120K+93
X+10=40K+31
X= 40K+21

So the remainder will be 21. Sufficient

From the second statement we have:
5X-10 = 20F+15
X-2=4F+3
X=4F+5

We can't define the reminder of X when divided by 40. Insufficient

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What is the remainder when X is divided by 40?  [#permalink]

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Updated on: 18 Jul 2019, 15:53
1
1) 3*x+30 = 120*q + 93. Rearranging for x: x + 10 = 40*q + 31, x = 40*q + 21.
Therefore, the remainder when x/40 = (40*q+21)/40 is always 21.

2) 5*x - 10 = 20*q + 15. Rearranging for x: x - 2 = 4*q + 3, x = 4*q + 5
Therefore, the remainder when x/40 = (4*q + 5)/40 is could be 5, 9, etc..

Originally posted by leemoon on 17 Jul 2019, 18:25.
Last edited by leemoon on 18 Jul 2019, 15:53, edited 1 time in total.
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What is the remainder when X is divided by 40?  [#permalink]

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17 Jul 2019, 19:07
1
Question: What is the remainder when positive integer X is divided by 40?

(1) (3X + 30) leaves remainder 93 when divided by 120.
$$(3X + 30) = 120Q + 93$$, where $$93$$ is the remainder and positive integer quotient $$Q \geq{0}$$
<=> $$X = 40Q + 21$$
Therefore, when positive integer X is divided by $$40$$, the remainder is $$21$$
SUFFICIENT

(2) (5X - 10) leaves remainder 15 when divided by 20.
$$(5X - 10) = 20Q + 15$$, where $$15$$ is the remainder and positive integer quotient $$Q \geq{0}$$
<=> $$X = 4Q + 5$$
We only know that when positive integer X is divided by $$4$$, the remainder is $$5$$ . Unfortunately, we have no sufficient information on what the remainder is when positive integer X is divided by $$40$$.
NOT SUFFICIENT

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Re: What is the remainder when X is divided by 40?  [#permalink]

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17 Jul 2019, 19:37
1
What is the remainder when positive integer X is divided by 40?

(1) 3X + 30 leaves remainder 93 when divided by 120.---Sufficient

3X+30 /120 with remainder 93 , 3X+30 = 120A+93

Case 1: A =0
for remainder to be 93 minimum 3X+30 = 93
=> 3X = 63
=> X = 21
Remainder of X /40 = 21

Case 2: A=1
3x+30 = 21x ==> X = 61
Remainder of X/40 = 21
So remainder of X /40 = 21 ------> Sufficient

(2) 5X - 10 leaves remainder 15 when divided by 20.
5X-10 = 20B+15
Case 1: B = 0
5X-10 = 15 ===> X =25/5 = 5

Remainder for X /40 =5

Case 2 :B =1
5X-10 = 20+15
=> X =35/5 =7

Remainder for X/40 = 7

Case 3: B=2
5X-10 =20*2+15
=> X =55/5 = 11

Remainder for X/40 =11
So here remainder keeps changing So--- not Sufficient

Option A is the Correct Answer
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Re: What is the remainder when X is divided by 40?  [#permalink]

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17 Jul 2019, 20:01
1

St:1--3X+30=120k+93

3X=120k+63
X=40k+21

Therefore when divided by 40 remainder will be 21.

St:2-- 5X-10=20K+15
5X=20K+25
X=4K+5

Now when divided by 40 the remainder can be as follows :-

When K=1 ,X=9 and remainder =9
When K=10,X=45 and remainder=5

Therefore not sufficient
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Re: What is the remainder when X is divided by 40?  [#permalink]

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17 Jul 2019, 20:48
1
What is the remainder when positive integer X is divided by 40?

(1) 3X + 30 leaves remainder 93 when divided by 120.
(2) 5X - 10 leaves remainder 15 when divided by 20.

We need to find remainder when $$\frac{x}{40}.$$

St 1) $$\frac{3x+30}{120}$$= q+93 (get rid of 120)
3x+30=120q+93
3x=120q+63 (divide by 3)
x=40q+21
Now let's plug in some number for q.
If q=0, x is 21 (remainder when $$\frac{x}{40}$$)
If q=1, x is 61 ($$\frac{61}{40}$$ remainder is 21)
If q=2, x is 101 ($$\frac{101}{40}$$ remainder is again 21)
For any number substituted for q, we will always get remainder of 21 because q is multiple of 40 and thus will divide 40 evenly leaving remainder of 21 always. Thus, st 1 is sufficient

St 2) $$\frac{5x-10}{20}$$=q+15 (get rid of 20)
5x-10=20q+15
5x=20q+25 (divide by 5)
x=4q+5
If q=0, x is 5 ($$\frac{5}{40}$$ remainder is 5)
If q=1, x is 9 ($$\frac{9}{40}$$ remainder is 9)
We already have two different values, thus st 2 is NOT sufficient
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Re: What is the remainder when X is divided by 40?  [#permalink]

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17 Jul 2019, 20:53
1
X = 40k + r What is r?

1) 3X+30 = 120a +63

3X = 120a + 63
X=40a+21 . So remainder is 21, sufficient

2) 5X - 10 = 20b + 15
X = 4b+5 X can be 5,9,13,17 or 45 etc. Remainder of each of these numbers when divided by 40 is different. Insufficient.

So, A
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Re: What is the remainder when X is divided by 40?  [#permalink]

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17 Jul 2019, 20:55
1
What is the remainder when positive integer X is divided by 40?

(1) 3X + 30 leaves remainder 93 when divided by 120.
(2) 5X - 10 leaves remainder 15 when divided by 20.

to find
X = 40 K +R
where R is the remainder

now
using first
given 3X + 30 = 120K+93
or 3X = 120K+63
or X = (120k + 63)/ 3
now we need remainder when divided by 40
thus X/40 = (120k + 63)/ 3*40

this will 120k will be completely divided by 120 and 6 will be the remainder when divided by 120
thus A is suffiencient

for given eq
5X - 10 = 20K +15
or 5X = 20K+25
or X = (20K+25)/5
divided each side by 40
x/40 = (20K + 25)/5*50
now 20 is not completely divisible by (250)
hence we cant determine the exact remainder without knowing K
Thus B is not sufficient

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What is the remainder when X is divided by 40?  [#permalink]

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Updated on: 17 Jul 2019, 20:59
1
We are asked to find the remainder when positive number X is divided by 40

St 1: $$\frac{3X + 30}{120}$$= Q + $$\frac{93}{120}$$ (where Q is the quoitent) ==> X = 40Q + 21 ==> $$\frac{X}{40}$$ = $$\frac{40Q}{40}$$ + $$\frac{21}{40}$$ ==> sufficient since our remainder will always be 21 no matter the value of Q

St 2: $$\frac{5X - 10}{20}$$ = Q + $$\frac{15}{40}$$ ==> X = 4Q + 5 ==> $$\frac{X}{40}$$ = $$\frac{4Q}{40}$$ + $$\frac{5}{40}$$==> insufficient since we will getdifferent numbers for the remainder depending on the value of Q

Alternate solution

St 1: since our remainder is 93, it means that 3X + 30 could equal 93, 213, 333, etc ==> solve for X to find that X could be 21, 61, 101, etc, which all give a remainder of 21 when divided by 40 ==> sufficient

St 2: since our remainder is 15, it means that 5X - 10 could equal 15, 35, 55, 75, etc ===> solve for X to find that X could be 5, 9, 13, 17, etc, which all give different remainders when divided by 40 ==> insufficient

Originally posted by bebs on 17 Jul 2019, 20:55.
Last edited by bebs on 17 Jul 2019, 20:59, edited 1 time in total.
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Re: What is the remainder when X is divided by 40?  [#permalink]

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17 Jul 2019, 20:57
1
Remainder when $$\frac{X}{40}$$ ?

(1) 3X + 30 leaves remainder 93 when divided by 120.
3X+30 = 120 *Q +93
3X = 120 *Q +63
X = 40*Q + 21
This is exactly what is asked. when divided by 40 , remainder is 21 . so, sufficient

(2) 5X - 10 leaves remainder 15 when divided by 20.
5X-10 = 15Q + 20
5X = 15Q + 30
X = 3Q + 15
X = 15, 18, 21, 24, ... If these are divided by 40 , then remainder could be 15, 18, 21, 24... hence, not sufficient

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Re: What is the remainder when X is divided by 40?  [#permalink]

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17 Jul 2019, 21:06
1
Statement 1 is sufficient and statement 2 is not.
3X+30=120y+93
which gives
X=40Y+21
so 21 would be remainder.

by solving statement 2, we get
X=4Z+7.
So, we are not sure about the remainder here.

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Re: What is the remainder when X is divided by 40?  [#permalink]

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17 Jul 2019, 21:44
1
To find: Remainder when integer X is divided by 40 i.e. X=40q+R, R=?
Statement1: 3x+30 leaves a remainder 93 when divided by 120
i.e 3x+30=120q+93
=>3x=120q+63
=> x=40q+21
therefore, remainder = 21

Statement 2: 5x-10 leaves a remainder 15 when divided by 20
i.e 5x-10=20q+15
=> 5x=20q+25
=> x =4q+5
=>x=4(q+1)+1 (i.e remainder can't be greater than divisor)
so possible x=1,5,9,13,................................,41,45,49,53.............
for all above x values, remainders are different for each when divided by 40 i.e 1,5,9,13,...................,1,5,9,13,........
So, statement 2 is not sufficient.

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Re: What is the remainder when X is divided by 40?  [#permalink]

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17 Jul 2019, 21:58
1
Given, X is a positive integer.
We need to find the remainder of X/40.

(1) 3X + 30 leaves remainder 93 when divided by 120.
3X + 30 = 120 * A + 93, where A is the quotient
3X = 120*A + 63 -> (a)

Substituting values of A in (a):
A = 1
3X = 120 + 63 => X = 61
Remainder of $$\frac{X}{40}$$ = $$\frac{61}{40}$$ = 21 -> [1]

A = 2
3X = 240 + 63 => X = 101
Remainder of $$\frac{X}{40}$$ = $$\frac{101}{40}$$ = 21 -> [2]

From [1] and [2] we find the pattern will keep continuing. Thus, the Remainder of $$\frac{X}{40}$$is 21.

Sufficient

(2) 5X - 10 leaves remainder 15 when divided by 20.
5X – 10 = 20 * B + 15, where B is the quotient.
5X = 20*B + 25 -> (b)

Substituting values of B in (b):
B = 1
5X = 20 + 25 => X = 9
Remainder of $$\frac{X}{40}$$ = $$\frac{9}{40}$$ = 9 -> [3]

B = 2
5X = 40+ 25 => X = 13
Remainder of $$\frac{X}{40}$$ = $$\frac{13}{40}$$ = 13 -> [4]

From [3] and [4] we find that it does not give a unique solution.

Not Sufficient

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Re: What is the remainder when X is divided by 40?  [#permalink]

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17 Jul 2019, 22:43
1
(1) 3X + 30 leaves remainder 93 when divided by 120.
First number that satisfies this is 93, 3x+30=93 => x=21. Next number would be 120+93=213; 213=3x+30=> x=61.
possible values of x=21,61,101 so on. Each leave a remainder of 21 when divided by 40. Sufficient.

(2) 5X - 10 leaves remainder 15 when divided by 20.
First number that satisfies this is 15, 5x-10=15 => 5. Next number would be 15+20=35; 35=5x-10=> x=9.
x:5,9,13,17...... Rach when divided by 40 can leave different remainder. Insufficient

Ans. A
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Re: What is the remainder when X is divided by 40?  [#permalink]

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17 Jul 2019, 23:07
1
What is the remainder when positive integer X is divided by 40?

(1) 3X + 30 leaves remainder 93 when divided by 120.

Assume 'a' is the quotient.
We can write 3X+30 as:

3X + 30 = 120*a + 93

simplify this for X,
3X = 120*a + 63
X = 40*a + 21.....................when X is divided by 40, remainder will be "21"

(2) 5X - 10 leaves remainder 15 when divided by 20.

Just like we did in first part, assume 'b' is the quotient.

5X-10 = 20*b +15
5X = 20*b + 25 = 20*(b+1) + 5

X = 4*(b+1) + 1......................from this we can not answer the question.

So, second can not answer the question.

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Re: What is the remainder when X is divided by 40?  [#permalink]

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18 Jul 2019, 00:12
1
(1) After all cancellations, it is given
x= a*40+21,
x =21, 61, 101, 141, 181,....... divide 40, some quotient and remainder 21, sufficient
(2) After all cancellations, it is given
x=b*4+5
x=5, 9, 13, 17, 21, 25............ divide by 40, remainder 5, 9, 17, 21, 25, etc , insufficient
IMO A
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Re: What is the remainder when X is divided by 40?  [#permalink]

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18 Jul 2019, 00:12
1
What is the reminder of $$\frac{x}{40}$$ ?

ST1. $$3x + 30$$ leaves remainder $$93$$ when divided by $$120$$.

If $$3x + 30 = 120k + 93$$ is simplified, we get $$x = 40k + 21$$. Now the question is what is the reminder of $$\frac{(40k + 21)}{40}$$ ?

If simplified we get $$k + \frac{21}{40}$$, thus regardless of $$k$$ the reminder is $$21$$.

Sufficient

ST2. $$5x - 10$$ leaves remainder $$15$$ when divided by $$20$$.

If $$5x-10=20p + 15$$ is simplified, we get $$x=4p+5$$. Now the question is what is the reminder of $$\frac{(4p + 5)}{40}$$ ?

If $$p=1$$, then the remainder is $$9$$. If $$p=2$$, then the remainder is $$13$$.

Insufficient

Hence A
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Re: What is the remainder when X is divided by 40?  [#permalink]

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18 Jul 2019, 01:46
1
What is the remainder when x is divided by 40?

1) 3X + 30 leaves remainder 93 when divided by 120.

Let’s create the formula of this statement:

30x+30 = 120a + 93
30x = 120a + 63
x = 40a + 21

We can see that if 40a + 21 is divided by 40, then the remainder is 21.
Therefore 1) is sufficient.

2) 5X - 10 leaves remainder 15 when divided by 20.

Let’s create the formula of this statement:

5x – 10 = 20b + 15
5x = 20b +25
X = 4b + 5

We can see that if 4b + 5 is divided by 40, then the remainder is different depending on b.
Therefore 2) is insufficient.

IMO A
Re: What is the remainder when X is divided by 40?   [#permalink] 18 Jul 2019, 01:46

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