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16 Nov 2021, 06:00
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The remainder obtained by the division of two positive integers by a common divisor are 33 and 47 respectively. However when the sum of these two numbers is divided by the same divisor, the remainder obtained is only 30. What is the divisor?
A. 20
B. 30
C. 40
D. 50
E. 65
A. 20
B. 30
C. 40
D. 50
E. 65
PyjamaScientist
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16 Nov 2021, 06:22
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Option (D) in my opinion.
Let the two positive no.s be x and y.
Let the common divisor be a.
Given:
x = a*k1 + 33. Implies x > 33
y = a*k2 + 47. Implies x > 47
Adding both gives, x + y = a (k1*k2) + 80.
But, given that the sum of two no.s leaves a remainder of 30 only when divided by a.
It can be possible only when a = 50.
Because, then, x + y = 50(k1*k2) + 50 + 30, will leave the remainder 30 when divided by 30 (for a > 47).
Let the two positive no.s be x and y.
Let the common divisor be a.
Given:
x = a*k1 + 33. Implies x > 33
y = a*k2 + 47. Implies x > 47
Adding both gives, x + y = a (k1*k2) + 80.
But, given that the sum of two no.s leaves a remainder of 30 only when divided by a.
It can be possible only when a = 50.
Because, then, x + y = 50(k1*k2) + 50 + 30, will leave the remainder 30 when divided by 30 (for a > 47).
30 Dec 2021, 05:36
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Given x = Dq1 + 33
and y = Dq2 + 47
and x + y = Dm + 80
Dq1 + Dq2 - Dm = 50
D(q1 + q2 -M) = 50
this means D is a multiple of 50, and by option only D is 50 and other are not a mutliple of 50. hence D.
and y = Dq2 + 47
and x + y = Dm + 80
Dq1 + Dq2 - Dm = 50
D(q1 + q2 -M) = 50
this means D is a multiple of 50, and by option only D is 50 and other are not a mutliple of 50. hence D.
