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Updated on: 21 Nov 2018, 03:57
Originally posted by EgmatQuantExpert on 24 Oct 2018, 07:03.
Last edited by EgmatQuantExpert on 21 Nov 2018, 03:57, edited 7 times in total.
Last edited by EgmatQuantExpert on 21 Nov 2018, 03:57, edited 7 times in total.
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Difficulty:
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38%
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Common Mistakes One Must Avoid in Remainders – Practice question 2
P is a two-digit integer, which can be written as 30a + b, where a and b are positive integers. Find the remainder, when P is divided by 3.
A) Statement (1) ALONE is sufficient, but statement (2) alone is not sufficient.
B) Statement (2) ALONE is sufficient, but statement (1) alone is not sufficient.
C) BOTH statements TOGETHER are sufficient, but NEITHER statement ALONE is sufficient.
D) EACH statement ALONE is sufficient.
E) Statements (1) and (2) TOGETHER are NOT sufficient.
To solve question 3: Question 3
To read the article: Common Mistakes One Must Avoid in Remainders
P is a two-digit integer, which can be written as 30a + b, where a and b are positive integers. Find the remainder, when P is divided by 3.
- (1) a = 3
(2) \(b^3 – 5b^2 – 14b = 0\)
A) Statement (1) ALONE is sufficient, but statement (2) alone is not sufficient.
B) Statement (2) ALONE is sufficient, but statement (1) alone is not sufficient.
C) BOTH statements TOGETHER are sufficient, but NEITHER statement ALONE is sufficient.
D) EACH statement ALONE is sufficient.
E) Statements (1) and (2) TOGETHER are NOT sufficient.
To solve question 3: Question 3
To read the article: Common Mistakes One Must Avoid in Remainders
24 Oct 2018, 08:32
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P is a two-digit integer, which can be written as 30a + b, where a and b are positive integers. Find the remainder, when P is divided by 3.
30a+b.... Now 30a will always be divisible by 3, so it will depend on b.
(1) a =4
Not required. Nothing about b
Insufficient
(2) \(b^3 – 5b^2 – 14b = 0\)
\(b^3 – 5b^2 – 14b = 0..........b(b^2-5b-14)=0.........b(b^2-7b+2b-14)=0........b(b-7)(b+2)=0\), so b can be -2, 0 or 7...
But b is a positive integer so ONLY possible value 7..
30a+7 divided by 3 will leave 1 as remainder
Sufficient
B
30a+b.... Now 30a will always be divisible by 3, so it will depend on b.
(1) a =4
Not required. Nothing about b
Insufficient
(2) \(b^3 – 5b^2 – 14b = 0\)
\(b^3 – 5b^2 – 14b = 0..........b(b^2-5b-14)=0.........b(b^2-7b+2b-14)=0........b(b-7)(b+2)=0\), so b can be -2, 0 or 7...
But b is a positive integer so ONLY possible value 7..
30a+7 divided by 3 will leave 1 as remainder
Sufficient
B
General Discussion
P is a two-digit integer, which can be written as 30a + b, where a and b are positive integers. Find the remainder, when P is divided by 3.
(1) a =4
(2) b3–5b2–14b=0
Given : P=30a + b ; divide by 3
\(\frac{P}{3}\) = \(\frac{30a}{3}\) + \(\frac{b}{3}\)
Now, 30 is divisible by 3, hence 30a will leave a remainder 0 upon division by 3; so actual remainder will be given by value of \(\frac{b}{3}\)
Question now becomes b=?
Stat1) a=4 , gives nothing about b, hence Not suff
Stat2) \(b^3–5b^2–14b=0\), take common b out => \(b*(b^2-5b-14)=0\) so either b=0 or \(b^2-5b-14=0\) ; further solving the later we get b= 7,-2. Since it is given that b is positive so in all we get b is 7; So remainder will be 1 Hence Suff
B
Edited : Corrected my mistake of missing that b was positive so 0 wouldnt count anyway .. silly me
(1) a =4
(2) b3–5b2–14b=0
Given : P=30a + b ; divide by 3
\(\frac{P}{3}\) = \(\frac{30a}{3}\) + \(\frac{b}{3}\)
Now, 30 is divisible by 3, hence 30a will leave a remainder 0 upon division by 3; so actual remainder will be given by value of \(\frac{b}{3}\)
Question now becomes b=?
Stat1) a=4 , gives nothing about b, hence Not suff
Stat2) \(b^3–5b^2–14b=0\), take common b out => \(b*(b^2-5b-14)=0\) so either b=0 or \(b^2-5b-14=0\) ; further solving the later we get b= 7,-2. Since it is given that b is positive so in all we get b is 7; So remainder will be 1 Hence Suff
B
Edited : Corrected my mistake of missing that b was positive so 0 wouldnt count anyway .. silly me
