A two-digit locker code has digits p and q, where pq represents the number with pas tens digit and q as units digit. This number is 60% of a certain number y. If the digits are reversed to form the number qp, the new number becomes 60% more than y. What is the value of y?

(a) 45

(b) 54

(c) 62

(d) 26

Question

A two-digit locker code has digits p and q, where pq represents the number with pas tens digit and q as units digit. This number is 60% of a certain number y. If the digits are reversed to form the number qp, the new number becomes 60% more than y. What is the value of y?

(a) 45

(b) 54

(c) 62

(d) 26

TutorArc · Accepted Answer

Let the original two-digit number be 10p + q, which equals 60% of y, so 10p + q= 0.6y.

When the digits are reversed, the number becomes 10q + p, which is 60% more than y, 

so 10q + p = 1.6y.

Now solve the system:
From equation 1: 3y = 50p + 5q

From equation 2: 8y = 50q + 5p

p = 2 and q = 7.


Then the original number = 10p + q = 10 &times; 2 + 7 = 27, and 

since 27 = 0.6y, we get y = 27/0.6 = 45

Hence, option a is correct.

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