Latest Bank Exam Aptitude Question SOLUTION: If 37/13 = 2+1/(x+1/(y+1/z)),where x,y,z are natural number..what are x,y,z=?

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(#M40034960) BANK EXAM Numerical Ability question ------ Keep an EYE

If 37/13 = 2+1/(x+1/(y+1/z)),where x,y,z are natural number..what are x,y,z=?

Asked In Bank Exam Nidhi Bajaj (10 years ago)

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Ann Theressa  (10 years ago)

x = 1, y=5, z=2

37/13 = 2+1/(x+1/(y+1/z))
2 + 11/13 = 2 + 1/(x+1/(y+1/z))
===>11/13 = 1/(x+1/(y+1/z))

Taking reciprocal on both sides
13/11 = x+1/(y+1/z)
1 + 2/11 = x + 1/(y+1/z)
===> x=1 and 1/(y+1/z) = 2/11

1/(y+1/z) = 2/11
Taking reciprocal

y+1/z = 11/2
y + 1/z = 5 + 1/2
===> y = 5

1/z = 1/2
z = 2

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