  Question:

In a two-digit number, the digit in the unit's place is more than twice the digit in ten's place by 1. If the digits in the unit's place and the ten's place are interchanged, difference between the newly formed number and the original number is less than the original number by 1. How do I find the original number ?

• A

35

• B

36

• C

37

• D

39

Posted by: Parool B. on 03.10.2017 Suhrita S. from Kolkata
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S
Sumit B. from Patna
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The correct option is given by option (C)

Explanation:

Let the digit at unit's place be u and the digit at ten's place be t. Thus, the original number is

o = 10t + u

The number obtained on interchanging the digits is

n = 10u + t

Now, according to the first condition given in the problem,

u = 2t +1 .......... (1)

And according to the second condition given in the problem,

10u + t - (10t + u) = 10t + u -1 ............... (2)

Solving eq.(1) and eq.(2) simultaneously, we find,

t = 3 and u = 7

Hence the number is 37.

Therefore, the correct answer is given by option (C) Pritha C. from Kolkata
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C.37 Let the original no. be 'xy'. Hence its value is (10x+y) Given, y is 1 greater than twice of x. => y=2x+1....(i) New no. 'yx' has value (10y+x) Given, new no. - original no. = original no. -1 => (10y+x) - (10x+y) = (10x+y) -1......(ii) Putting value of y from (i) in (ii): We get, x=3 Therefore, y=2*3+1 =7 Hence original no. 'xy'=37. S.Chandra from Kolkata
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Taking original number as (10x + y), we get,  (10y + x) - (10x + y) = (10x + y) - 1.   => 19x - 8y = 1 .... (i)

Again, y = 2x + 1 ... (ii)

Solving the eqns, x = 3 and y = 7 .

Hence, the original no. is 37 . Ans. (option C) Atul G. from Navi Mumbai
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Let the ten's digit be x. Then, unit's digit = 2x + 1.

[10x + (2x + 1)] - [{10 (2x + 1) + x} - {10x + (2x + 1)}] = 1

<=> (12x + 1) - (9x + 9) = 1 <=> 3x = 9, x =  3.

So, ten's digit = 3 and unit's digit = 7. Hence, original number = 37.

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