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Rathore

· started a discussion

· 1 Months ago

All are correct. (Question: 112217.0 )

1st condition: 10x+y=xy+16
2nd condition: (x-y)^2+xy=10x+y
from above (x-y)^2=16
thus (x-y)=+4,-4.

Question:
Consider a two digit no. If we divide the number by the product of its digit, the quotient is 1 and the remainder is 16. If we add the product of its digits to the square of the difference of its digits,we get a number in the form 10x+y, then the no. of correct statement among the given statements is:


I. (x-y)2 = 16

II. (x-y) = 4

III. (x-y) = - 4

Options:
A) 0
B) 1
C) 2
D) 3
Solution:
Ans: (c)

It is clear that the two digit number which satisfies the given condition is 37.

Here 37 = 10 \(\times\) 3 +7, so that x = 3, y = 7

Clearly, x and y satisfy the statement I and III.

Knowledge Expert

· commented

· 1 Months ago

Dear Student,
The given answer is correct.
As (x-y)^2 = 16
Then either x-y = 4
or x-y = - 4
Both cannot be true at a time.
So only 2 statements will be true.

Thanks and Regards
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