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For positive integers p and q, p/q when expressed as a decimal, is a terminating decimal?
(1) p is the sum of three consecutive odd multiples of 5.
(2) $$2q^2 + 144 = 36q$$

Weekly Quant Quiz #4 Ques 4

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Re: For positive integers p and q, p/q when expressed as a decimal, is a t  [#permalink]

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For positive integers p and q, p/q when expressed as a decimal, is a terminating decimal?
(1) p is the sum of three consecutive odd multiples of 5.
(2) 2q2+144=36q2q2+144=36q

For a decimal to be terminating the denominator must be a multiple of 5 or 2 in its base form..
from A p is of the form 15(2n+3) ...45,75,105.... so its in sufficient
from B we know q can be 12 or 6... but cannot be sure if the reduced form of fraction will have only 2 or 5 hence insufficient

From A & B we know P is multiple of 3....so if q is 6 or 12 then the fraction will will have denominator of the form 2 hence termination hence answer is C

Originally posted by swapnil611 on 13 Oct 2018, 10:19.
Last edited by swapnil611 on 13 Oct 2018, 10:33, edited 1 time in total.
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From statement 1:

P is the sum of three consecutive odd multiples of 5. P will be 45n+30.
therefore P is odd.
But terminating decimal depends on the denominator. A decimal will be terminating only if the denominator consists only of 2s or 5s. Hence 1 is insufficient.

From statement 2:
Solving for q gives q = 12 or 6.
Both contain 3 as a factor. Hence We need the value of P.
2 is insufficient.

Combining both: If p = 45 and q =12 then p/q is terminating. even when q =6.
So combining works.

C is the answer.
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I feel Option (C) should be correct, please find my solution attached.

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1) Value of p may be 5, 15, 25, and likewise. This says nothing about q( sum 5+15+25=45)Now 45/11 may be non terminating while 45/6 is terminating, Hence NOT SUFFICIENT
2) Value of q may be 12 or 6 upon solving the quadratic equation. (15/6 is non terminating, while 15/12 is terminating) , Hence this too is NOT SUFFICIENT

But when we take both statements together, It becomes sufficient as 45/6 or 45/12 is ALWAYS terminating. Hence 1) and 2) together are SUFFICIENT

Hence the C should be correct
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Originally posted by saukrit on 13 Oct 2018, 10:25.
Last edited by saukrit on 13 Oct 2018, 21:11, edited 1 time in total.
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Re: For positive integers p and q, p/q when expressed as a decimal, is a t  [#permalink]

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For positive integers p and q, p/q when expressed as a decimal, is a terminating decimal?
(1) p is the sum of three consecutive odd multiples of 5.
(2) 2q2+144=36q2q2+144=36q

IMO C is right.

Stmnt 1; sum 5+15+25=45
stmnt 2: quadratic we get q= 12,6

45/12= 3.75 and 45/6=7.5

C should be correct
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(1) p is the sum of three consecutive odd multiples of 5. - NS ( no info for q)
(2) 2q^2+144=36q NS (no info for p)
2q^2+144=36q ==> q^2-18Q+72=0 => (q-6)(q-12)=0 ==> q=2*3 or 4*3

1+2) Suff..........p= 5(2a+1)+5(2a+3)+5(2a+5)=30a+45 = 3(10a+15) has a factor of 3 and others of q could only be 2 or 4.... thus p/q terminatic decimal

Ans C
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Re: For positive integers p and q, p/q when expressed as a decimal, is a t  [#permalink]

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Statement 1 : p is the sum of three consecutive odd multiples of 5.

p could be { 5+10+15 = 30 } or { 10+15+20 = 45}

This doesn't given us anything about q so===> it is insufficient

Statement 2 : this quadratic equation gives us q as values 12 and 6

doent give us anything about value of p so this alone is insufficient

Both considered together ==>
so when lets say p = 12 , q = 30

==> p/q = 12/30 is finite

but when q = 45 , p/q is not finite ... so both together is also insufficient ==> OA is E
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Re: For positive integers p and q, p/q when expressed as a decimal, is a t  [#permalink]

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swapnil611 wrote:
For positive integers p and q, p/q when expressed as a decimal, is a terminating decimal?
(1) p is the sum of three consecutive odd multiples of 5.
(2) 2q2+144=36q2q2+144=36q

For a decimal to be terminating the denominator must be a multiple of 5 or 2 in its base form..
from A p is of the form 15(2n+3) ...45,75,105.... so its in sufficient
from B we know q can be 12 or 6... but cannot be sure if the reduced form of fraction will have only 2 or 5 hence insufficient

From A & B we know P is multiple of 3....so if q is 6 or 12 then the fraction will will have denominator of the form 2 hence termination hence answer is C

Hi,

How have you reduced A) to 15(2n+3)?
shouldn't it be 5*n+5*(n+2)+5*(n+4) = 5(n+n+2+n+4) = 5(3n+6) = 15(n+2) ?
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For positive integers p and q, p/q when expressed as a decimal, is a t  [#permalink]

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I assumed numbers as 5×(2n +1),5×(2n+3),5×(2n+5) hence the sum 5×(6n+9)=15(2n+3)...If you assume number as n it can be both even or odd but in the question it's mentioned odd multiple of 5.
Hope am clear

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Re: For positive integers p and q, p/q when expressed as a decimal, is a t  [#permalink]

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Can someone please explain why only fractions with denominators of 2 or 5 are terminating? Thanks....
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Re: For positive integers p and q, p/q when expressed as a decimal, is a t  [#permalink]

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Hi
You can observe that if the numerator is not a multiple of denominator and denominator is multiple of 3,7 or any number other than 2 or 5, the result will be non terminating decimals.

For example:
1/3 = 0.333333333333...
1/7 = 0.1428571428...

RichardSaunders wrote:
Can someone please explain why only fractions with denominators of 2 or 5 are terminating? Thanks....

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_________________ Re: For positive integers p and q, p/q when expressed as a decimal, is a t   [#permalink] 19 Feb 2019, 20:22
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