12 Days of Christmas GMAT Competition - Day 11: If p is a positive mul

Answer B
p=10x
10x/15=some remainder R
from statment 1
p=10,20,40,50...
r=10,5,10,5,...
from statement 2
p1=10,20,30,40,50,60,70,80,....
p2+7=17,27,37,47,57,67,77,87...
R1=10,5,0,10..
R2=2,12,7,2..
so the remainder is constant 10 sufficient

If p is a positive integer divisible by 10, and r is the remainder when p is divided by 15, what is the value of r?
p = 10k1 means 10,20, 30, 40, 50......
p = 15 k2 + r , Means r = 10, 5,0, ...........

(1) p is not a multiple of 6

means 30, 60, 90 ........ not possible. Means r= 10 or 5 not sufficient

(2) When p + 7 is divided by 15, the remainder is less than r.
17, 27,37,47.........
the possible remainder - 2, 12, 7 is id less than r = 10, 5, 0

Which possible in the case where P= 10, 40, 70.......... therefore r= 10

Sufficient

Correct Answer B

If p is a positive integer divisible by 10, and r is the remainder when p is divided by 15, what is the value of r?
p=15n+r
p=10m
if n is odd then r =5
if n is even then r =10 or 0

(1) p is not a multiple of 6
p is not divisible by 3
Insufficient

(2) When p + 7 is divided by 15, the remainder is less than r.
if p+7 remainder is less than r then r >7
ie r =10 and n is even
Sufficient

IMO B

We can generalize that P is of the form 15n+5; 15n-5; 15n (when n is an even integer).

Case-1: If P = 15n+5 the remainder (r) will be 5 (n is odd)
Case-2:If P = 15n-5 the remainder (r) will be 10 (-5+15) (n is odd)
Case-3:If P = 15n (when n is an even integer) the remainder (r) will be 0

Statement-1
P is not a multiple of 6, so Case 3 will be eliminated. But we still have 2 values of r.
Insufficient.

Statement-2
In Case 1 - r will be 12
In Case 2 - r will be 2
In Case 3 - r will be 7

Only in Case 2 the remainder is less than r. So P will have a remainder 10 when divided by 15.

Sufficient.

IMO Option B

If p is a positive integer divisible by 10, and r is the remainder when p is divided by 15, what is the value of r?

Thus r = 0 or 5 or 10

(1) p is not a multiple of 6
I: p = 10; r = 10
II: p = 20; r = 5

INSUFFICIENT.

(2) When p + 7 is divided by 15, the remainder is less than r.
Let remainder is r'.
p = 10
p + 7 = 17; r'= 2 and r = 10

p = 20
p + 7 = 27; r'= 12 and r = 5

p = 30
p + 7 = 37; r'= 7 and r = 0

p = 40
p + 7 = 47; r'= 2 and r = 10

p = 50
p + 7 = 57; r'= 12 and r = 5

p = 60
p + 7 = 67; r'= 7 and r = 0

So, a pattern is formed where out of 3 remainders possible only 1 is applicable.
Hence r = 10

SUFFICIENT.

Answer B.

IMO Option B.

1. Insufficient
If p = 10, r = 10 and if p = 20, r = 5. We don't have a single value for r.

2. Sufficient
The pattern can be seen in the below table:

p r p+7 r2 r<r2?
10 10 17 2 No
20 5 27 12 No
30 0 37 7 Yes
40 10 47 2 No
50 5 57 12 No
60 0 67 7 Yes

As can be seen, statement 2 is true for values of p = 30, 60, 90, etc. Hence Sufficient.

Answer = B

The value of r is 10
P is divisible by 10 = 10, 20, 30, 40, 50, 60, 70, 80, 90......
P when divided by 15 leaves a remainder = SO for remainder P will be excluded with (N*10 - LCM of 10, 15)
So, P consist of 10, 20, 40, 50, 70, 80

Statement 1: No not a multiple of 6, which we have already excluded during removal of LCM of 10,15
Statement 2: P+7/15, remainder is smaller than r

Cases1: 10/15 = r 10, 17/15 = 2
True
Case 2: 20/15 = r 5, 27/15 = 12
False
Case 3: 40/15 = r 10, 47/15 = 2
True
Case 4: 50/15 = r 5, 57/15 = 12

Pattern, only P as 10, 40, 70.... will satisfy
So, r = 10

Answer 10

Asked: If p is a positive integer divisible by 10, and r is the remainder when p is divided by 15, what is the value of r?

p = 10k; where k is an integer
r = 0 when p=10k is divided by 15 when k = 3m
r = 10 when p=10k is divided by 15 when k = 3m + 1
r = 5 when p=10k is divided by 15 when k = 3m + 2

(1) p is not a multiple of 6
k is not = 3m
r = 10 when p=10k is divided by 15 when k = 3m + 1
r = 5 when p=10k is divided by 15 when k = 3m + 2
NOT SUFFICIENT

(2) When p + 7 is divided by 15, the remainder is less than r.
r = 0 when p=10k is divided by 15 when k = 3m: Not feasible since remainder can not be less than r=0: NOT FEASIBLE
r = 10 when p=10k is divided by 15 when k = 3m + 1; Remainder = 2 < r=10 when p + 7 = 30m + 17 when divided by 15 : FEASIBLE
r = 5 when p=10k is divided by 15 when k = 3m + 2; Remainder = 12 > r=5 when p + 7 = 30m + 27 when divided by 15: NOT FEASIBLE
r = 10
SUFFICIENT

IMO B

If p is a positive integer divisible by 10, and r is the remainder when p is divided by 15, what is the value of r?

p % 10 = 0
p % 15 = r
r = ?

(1) p is not a multiple of 6
p !% = 6
p = 10 r = 10
p = 20 r = 5
p = 40 r = 10
p = 50 r = 5 etc..
Hence, r = 5 or 10.
Not sufficient.

(2) When p + 7 is divided by 15, the remainder is less than r.
p + 7 % 15 = R < r
p = 10 R = 2 < r, r = 10 Yes.
P = 20 R = 12 < r, r = 5 No.
Therefore, r = 10
sufficient.

IMO B

If p is a positive integer divisible by 10, and r is the remainder when p is divided by 15, what is the value of r?

(1) p is not a multiple of 6
Example :
a. p=20 and r=5
b. p=70 and r=10

Value of r is not consistent hence (A) is not sufficient

(2) When p + 7 is divided by 15, the remainder is less than r.
Example:
a. p=77 , r=2 (here p was 70)
b. p=107, r=2( here, p was 100)
c. p=47, r =2 (here, p was 40)
d. p=130 , r=2 (here, p was 130)

Value of r is consistent hence (B) is sufficient

If p is a positive integer divisible by 10, and r is the remainder when p is divided by 15, what is the value of r?

(1) p is not a multiple of 6
(2) When p + 7 is divided by 15, the remainder is less than r.


[/quote]


p = 10, 20, 30, .......
p = 15 x + r,
p =20, r = 5;
p=30, r=0;
p=40, r= 10
p=50, r=5,
...........


Statement 1:

p cant be 30, 60, 90.....
r can be 10 or 5
Not Sufficient

Statement 2:

p=27, r =12
p=37, r= 7
p=47, r=2
p=57, r=12

Sufficient
r=10


Answer B

1) p can be 40 or 50. In the first case, r is 10 and in the second, r is 5. Not sufficient.

2) When p = 10, p + 7 = 17, so the remainder when it is divided by 15 is 2, which is less than r = 10. Again when, p = 20, p + 7 = 27, which will violate the condition as 12 is greater than r = 5. Again when p =30, p + 7 = 37, which will produce remainder of 7, which will not fulfill the condition. So, p + 7 will produce a remainder less than r only when p is 10, 100, 1000 etc. r will be 10 then. Sufficient.

B is the answer.

p = 10, 20, 30, ...
p : 15 -- remainder r = ?
Cycle of r is 10, 5, 0

1) p is not a multiple of 6
p = 10 then r = 10
p = 20 then r = 5
Insuff.

2) r((p + 7)/15) (r') < r(p/15) (r)
. r = 10 then r' = 2 < r
. r = 5 then r' = 12 > r. Then r cannot be 5
. r = 0 then r' = 7 > r. Then r cannot be 0
Finally, r can only be 10.
Suff.

Correct answer: B

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