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13 Oct 2018, 10:14
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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) \(2q^2 + 144 = 36q\)
(1) p is the sum of three consecutive odd multiples of 5.
(2) \(2q^2 + 144 = 36q\)
Weekly Quant Quiz #4 Ques 4
Updated on: 13 Oct 2018, 10:33
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.
Last edited by swapnil611 on 13 Oct 2018, 10:33, edited 1 time in total.
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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
(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
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13 Oct 2018, 10:22
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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.
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.
