What is the remainder when X is divided by 40?

The question states that what is the remainder when X is divided by 40

Statement 1: 3X + 30 leaves remainder of 93 when divided by 120

Let X = 21, therefore, 3(21) + 30 = 93
Thus, 93/120 gives remainder of 93
therefore, 93/40 gives remainder of 13

Let X = 61, therefore, 3(61) + 30 = 213
thus, 213/120 gives remainder of 93
therefore, 213/40 gives remainder of 13

similarly, let X = 101, therefore, 3(101) + 30 = 333
thus, 333/120 gives remainder of 93
therefore, 333/40 gives remainder of 13

Therefore this statement is sufficient (AD)

Statement 2: (5X - 10)/20 gives remainder of 15
let X = 5, therefore (5x5 - 10)/20 = 15/20 gives remainder of 15
therefore, 5/40 gives remainder of 5

let X = 9, therefore (5x9 - 10)/20 = 35/20 gives remainder of 15
therefore, 9/40 gives remainder of 9

Not sufficient

Hence answer choice A.

IMO A

Have to find: Remainder of \(X/40\)

Statement 1 -> \(\frac{(3X + 30)}{120}\) gives a remainder of 93. Substitute for X based on this statement. For X = 21, \(\frac{(3X + 30)}{120}\) gives the remainder of 93 and \(X/40\) gives a remainder of 21. For X = 101, \(\frac{(3X + 30)}{120}\) gives the remainder of 93 and \(X/40\) gives a remainder of 21 again. ---> Sufficient

Statement 2 -> \(\frac{(5X - 10)}{20}\) gives a remainder of 15. Substitute for X based on this statement. For X = 5, \(\frac{(5X - 10)}{20}\) gives the remainder 15 and \(X/40\) gives the remainder 5. For X = 9, \(\frac{(5X - 10)}{20}\) gives the remainder 15, but \(X/40\) gives the remainder 9. ---> Insufficient

To find,

Remainder of X when X is divided by 40.

We know,

Dividend = Divisor * Quotient + Remainder

Let us check the two options -

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

=> 3*X + 30 = 120 * Quotient + 93
=> X + 10 = 40 * Quotient + 31
=> X = 40 * Quotient + 21

Hence, X whenever divided by 40 will give a remainder of 21.

Hence option 1 is sufficient.

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

=> 5*X - 10 = 20 * Quotient + 15
=> X - 2 = 4 * Quotient + 3
=> X = 4 * Quotient + 5
=> X = 4 * Quotient + 4 + 1
=> X = 4 * (Quotient + 1) + 1

Hence, X whenever divided by 4 will give a remainder of 1.

Now, if X is 5 (satisfying above condition in option 2) then it will give a remainder of 5 when divided by 40.
Now, if X is 9 (satisfying above condition in option 2) then it will give a remainder of 9 when divided by 40.
Now, if X is 45 (satisfying above condition in option 2) then it will give a remainder of 5 when divided by 40.
Now, if X is 53 (satisfying above condition in option 2) then it will give a remainder of 13 when divided by 40.

Hence, option 2 is insufficient.

Answer: A

IMO-A

Remainder when positive integer X is divided by 40

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

3X+30= 120K + 93
=> X+10= 40K + 31 [K = integer]
=> X-21 = 40K
=> X-21= { 0, 40, 80, ........40K}
=> X= 21, 61, 81,....40K+21
Remainder (X/40)= 21

Sufficient

(2) 5X - 10 leaves remainder 15 when divided by 20.
=>.5X-10=20K+15
=> 5X= 20K+25
=> X= 4K+5
=> X= {5,9,13,17...........4K+5}
Remainder (X/40)= {5,9,13,17.....}

Not Sufficient

What is the remainder when positive integer X is divided by 40?

40q+r = X ..... q is quotient and r is remainder. We have to find r.

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

120q+93 = 3X + 30
120q+63 = 3X
40q+21 = X ... This is in the same form as 40q+r=X

Therefore r = 21

(1) IS SUFFICIENT

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

20q+15 = 5X-10

20q + 25 = 5X

4q + 5 = X

It is not possible to represent X in the form of 40q+r....

X could be 9 with remainder 9, or x could be 45 with remainder 5.

(2) IS NOT SUFFICIENT

ANSWER: A - 1 Alone is SUFFICIENT

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.

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.

The answer is therefore A.

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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


Answer is (A)

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
Answer is A

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

Answer A

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"

First can answer the question.

(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.



ANSWER : A

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

(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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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.

Answer must be A

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