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16 Apr 2025, 00:30
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C
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Difficulty:
35%
(medium)
Question Stats:
80% (02:32) correct
20%
(02:57)
wrong
based on 127
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If the digits of a positive two-digit number are reversed, the new number becomes 3/8 of the original number. What is the sum of the squares of the digits of the original number?
A. 48
B. 51
C. 53
D. 56
E. 62
A. 48
B. 51
C. 53
D. 56
E. 62
16 Apr 2025, 02:55
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Original number => AB, Reversed = BA
Question: \(A^2 + B^2 = ? \)
\(10A+B = \frac{3}{8} (10B+ A)\) (Given)
77A = 22 B
=> B = 7/2 A
\(A^2 + B^2 = B^2 * (\frac{53}{4})\) (Replacing A from above)
=> Answer must be factor of 53 (Since 53 is prime)
C is correct
Question: \(A^2 + B^2 = ? \)
\(10A+B = \frac{3}{8} (10B+ A)\) (Given)
77A = 22 B
=> B = 7/2 A
\(A^2 + B^2 = B^2 * (\frac{53}{4})\) (Replacing A from above)
=> Answer must be factor of 53 (Since 53 is prime)
C is correct
Bunuel
16 Apr 2025, 04:07
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10y+x= 3/8(10x+y)
=> 77x =22y
=> 7x=2y
=>x=2 ,y=7
so 49+4 = 53
C
=> 77x =22y
=> 7x=2y
=>x=2 ,y=7
so 49+4 = 53
C
