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

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

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17 Jul 2019, 13:14
1
What is the remainder when positive integer X is divided by 40?

(1) 3X + 30 leaves remainder 93 when divided by 120.
3x + 30
= 3(x + 10) / 120 --- remainder 93
= x + 10 / 40 ---- remainder 13
= x / 40 ---remainder 3
Sufficient.

(2) 5X - 10 leaves remainder 15 when divided by 20.
If divided by 20 we get remainder 15
but we do not know the number.
the number can be 15 or 35
if divided by 40 to the above numbers we get different remainders but if divided by 20 we get the same remainder.
Insufficient.

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

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17 Jul 2019, 13:27
1
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
3x- 63 =120q —> (3x -63)/120=q
Quotient =Integer ,Only when x= 21 , (3(21)-63)/120=Integer
So 21/40 have a remainder of 21
(Sufficient)

(2) 5X - 10 leaves remainder 15 when divided by 20
5x-10 =20q —> (5x-10)/20= q
(5(2)-10)/20=Integer
(5(6)-10)/20=Integer .: x=2,6...
(Not sufficient) as 2/40 and 6/40 gives different remainders

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

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17 Jul 2019, 13:30
1
Statement 1
3x+30= 120k+93, where k is no-negative integer.
x+10=40k+31
x= 40k+21

We will always get 21 as a remainder, when x is divided by 40.
Sufficient

Statement 2
5x-10=20a+15, where a is non-negative integer
x-2=4a+3
x=4a+5

Hence, x can be 5, 9, 13, 17...and so on
If x=5, we will get 5 as a remainder, when x is divided by 40.
If x=9, we will get 9 as a remainder, when x is divided by 40.

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

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17 Jul 2019, 14:57
solving first and 2nd equation we derive a equation which is enough to tell the solution that both are sufficient to tell remainder
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Re: What is the remainder when X is divided by 40?  [#permalink]

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17 Jul 2019, 15:06
1
What is the remainder when positive integer X is divided by 40?

Remainder of X/40 = ?

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

Using remainder theorem, 3X + 30 = Q120 + 93, --> 3X = Q120 + 63 or X = 40Q + 21,
X = 61, 101, 141, etc for Q = 1, 2, 3, etc. In each of these cases, Remainder of X/40 will be equal to 21. Hence, sufficient.

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

5X - 10 = Q20 + 15, --> 5X = Q20 + 25 or X = (Q20 + 25)/5

X = 9, 13, 17, 21, etc...for Q = 1,2,3,4, etc, --> Different remainder values possible for X/40. 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, 15:17
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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Re: What is the remainder when X is divided by 40?  [#permalink]

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17 Jul 2019, 18:31
#1
3X + 30 leaves remainder 93 when divided by 120.
x=21,61 remainder would be when divided by 40 ; 19,21 insufficient

#2 5X - 10 leaves remainder 15 when divided by 20.
X=5,13 remainder when divided by 40 ; 5,13
insufficient
from 1&2 we dont have anything in common
IMO E

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.
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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, 19:48
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

1) 3x+30 divided by 120
remainder of 30 is 30 when divided by 120.
remainder of 3x is 63 when divided by 120.
remainder of x/40 =21
sufficient

2)remainder of 10/20 is 10
remainder of 5x is 5 when divided by 20
remainder of x/4 is 5
remainder of x/40 is 50
sufficient
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Re: What is the remainder when X is divided by 40?  [#permalink]

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17 Jul 2019, 20:01
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] 17 Jul 2019, 21:06

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