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What is the remainder when X is divided by 40?

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

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18 Jul 2019, 02:27
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.

statement 1 - 3X+30 leaves remainder 93 when divided by 120
3X+3 can be written as 3(X+10)/120 = X+10/40 so this leaves remainder 93
So X/40 will leave remainder 93+10 = 103

Hence Statement one is sufficient

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

Suppose X = 13 then if we multiply both denominator and numerator by 2 then 10(X-2)/40 = 10*(13-2)/40 = 110/40 = 30 leaves remainder 30
Suppose X = 17 then if we multiply both denminator and numerator by 2 then 10(X-2)/40 = 10(17-2)/40 = 10*15/40 = 150/40 leaves remainder 30.

Hence 5X-10 leaves remainder 15 when divided by 20 , but it cannot be determined for sure what is the remainder when it X is divided by 40
Hence insufficient

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

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18 Jul 2019, 02:31
1
What is remainder when x/40? Formula for remainders is a/b=integer+r. We will apply this formula for both statements.
1. When 120 divides 3x+30, remainder is 93. Substitute number to formula above.
(3x+30)/120=integer+93
3x+30=120*i+93 , we can multiply by 1/3 all sides to get easier numbers to calculate
x=40*i+21,
x/40=i+21, remainder is 21 A is good
2. When 20 divides 5x-10, remainder is 15. Substitute number to formula
(5x-10)/20=integer+15
5x-10=20*integer+15 , multiply by 1/5
x=4*integer+5
x/4=integer+5, different values possible.
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, 02:31
1
(1) 3X + 30 leaves remainder 93 when divided by 120.
--> 3X + 30 = 120M + 93, for any integer M
--> 3X = 120M + 63
--> X = 40M + 21

So, Remainder = 21

Sufficient

(2) 5X - 10 leaves remainder 15 when divided by 20.
--> 5X - 10 = 20N + 15, for any integer N
--> 5X = 20N + 25
--> X = 4N + 5
--> Possible values of X = 5, 9, 13, 17, . . . . .

So, Possible Remainders = 5, 9, 13, 17 . . . . .

Insufficient

IMO Option A

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

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18 Jul 2019, 03:15
1
What is the remainder when positive integer X is divided by 40?

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

Let's imagine if we 3x+30 divide by 120 we have reminder of 83 are 21, 61, 101....
(arithmetic progression: add 40 to previous term)
21/40=21 is reminder,
61/40=21 is reminder and so on...
Reminder is always equal to 21.
Sufficient.

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

Let's imagine if we 5x-10 divide by 20 we have reminder of 15 are 5, 9, 13...
(arithmetic progression: add 4 to previous term)
We've always got different reminders, so insufficient.

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

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18 Jul 2019, 03:35
1
(1) 3X + 30 leaves remainder 93 when divided by 120.
(2) 5X - 10 leaves remainder 15 when divided by 20.

Statement1: (3x + 30 - 93)/120 is an integer
Implies (x - 21)/40 is an integer.
Therefore reminder = 21.
Sufficient.

Statement2: (5x - 10 -15)/20 is an integer
(5x-25)/20 is an integer
(x-5)/4 is an integer
Not sufficient since we cannot determine reminder when divided by 40.

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

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18 Jul 2019, 04:19
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.

Solution :

Question Stem analysis:

We are required to find out remainder when 40 is divided by X,
IMP property : Dividend = Quotient X divisor + Remainder.

Statement one analysis:

We can form the equation using the above stated property,
3X + 30 = 120Q + 93.
3X = 120Q + 63
X= 40Q + 21
If we divide 40Q + 21 by 40, we know that 40Q is divisible by 40, and 21 when divided by 40 leaves a remainder of 21.
Hence statement one alone is sufficient. we can eliminate C & E.

Statement two alone:

We can form the equation using the above stated property,
5X- 10 =20Q + 15,
5X= 20Q+ 25
X = 4Q + 5
In this, we don't know if 40Q is divisible by 40, for eg, if Q=10, then yes we have a remainder of 5, if Q= 1, then the remainder is different. Hence without knowing the value of Q, this statement is insufficient.

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

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18 Jul 2019, 04:25
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.

Solution:
We have to find
$$\frac{x}{40}$$ = Q +$$\frac{R}{40}$$

From the first statement:
$$\frac{(3x+30)}{120}$$ = Q +$$\frac{93}{120}$$

$$\frac{x}{40}$$ = Q +$$\frac{21}{40}$$ Hence sufficient.

From Statement 2;

$$\frac{(5x-10)}{20}$$ = Q + $$\frac{15}{20}$$

$$\frac{x}{4}$$ = Q +$$\frac{5}{4}$$

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

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18 Jul 2019, 04:48
1
Given : x=40a+r ..........................(1)

We need to find r.

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

3x+30 = 120b+93
3(x+10) = 120b+93
x+10=40b+31
x=40b+21..........................(2)

If we compare equation (1) & (2)
r=21
Sufficient.

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

5x-10 = 20c+15
x-2=4c+3
x=4c+5...................(3)

Equation (1) & (3) are not comparable.
Insufficient.

So, Ans should be (A)
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What is the remainder when X is divided by 40?  [#permalink]

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18 Jul 2019, 04:49
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.

Statement1:

3X+30=120*a+93 (a is quotient)
3X=120a+93-30
3X=120a+63
3X/3=120a/3+63/3
X=40a+21
Well, if we divided X by 40, remainder will be 21.
Sufficient

Statement2:

5X - 10=20a+15 (a is quotient)
5X=20a+15+10
5X=20a+25
5X/5=20a/5+25/5
X=4a+5
Well, it depends on 'a' if we divide X by 40.
if a≥8, the remainder will be 5.
if a<8, the remainder will be different from 5.
Insufficient

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

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18 Jul 2019, 06:57
1
What is the remainder when positive integer X is divided by 40?

X = 40*k + a
a = ?

(1) 3X + 30 leaves remainder 93 when divided by 120.
3X + 30 = 120*k1 + 93
3X=120*k1+63
We know that X = 40*k + a
120 / 40 = 3, so 40 can be included in 120 3 times
That's mean a = 23
SUFFICIENT
(2) 5X - 10 leaves remainder 15 when divided by 20.
5X - 10=20*k2 + 15
5X = 20*(k2+1) + 5
As: X = 40*k + a
20 can be included in 40 2 times, so exists 2 possible remainders of a:
a = 5 and a = 25
INSUFFICIENT

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

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18 Jul 2019, 07:23
1
Option A

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

here 3x when divided by 120 will leave a remainder of 63 (ie 3 x 21).. In that case x when divided by 120 will leave a remainder of 21

(2) 5X - 10 leaves remainder 15 when divided by 20.
here 5x will leave a remainder of 5 when divided by 20... the are many possible values for such cases.. X can be 5, 9 etc... hence it is insufficient !!
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Re: What is the remainder when X is divided by 40?  [#permalink]

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18 Jul 2019, 07:27
1
Statement i= sufficient
3x+30=120k+93
x=40k+21 remainder 21
statement ii not sufficient
5x-10=20I+15
x=4I+5, 5,13 are remaninders not sufficient
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Re: What is the remainder when X is divided by 40?  [#permalink]

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18 Jul 2019, 07:52
1
X=40k+r where r is the remainder

Statement 1:
3X+30= 120k+3r+30
3r+30 = 93
3r=63
r=21
Sufficient.

Statement 2:
5X-10=200k+5r-10
Wen 5X-10 is divided by 20, 200k is reduced to 10k. But we don't know about 5r-10
If 5r-10 <20, 5r-10=15; r=5
If 5r-10>20, lets say, 5r-10 = 35, r=9
Not sufficient.

Hence, option (A).
Re: What is the remainder when X is divided by 40?   [#permalink] 18 Jul 2019, 07:52

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