| Assertion (A): A two digit number, where tens digit is greater than ones digit is obtained by either multiplying sum of the digits by 8 and adding 1 or by multiplying the difference of digits by 13 and adding 2. The number is 41. |
| Reason (R): The linear equations used are \[\text{7x}-\text{2y}+\text{1}=0\]and \[\text{l2x}+\text{23y}+\text{2}=0.\] |
A) Both assertion (A) and reason (R) are true and reason (R) is the correct explanation of assertion (A).
B) Both assertion (A) and reason (R) are true but reason (R) is not the correct explanation of assertion (A).
C) Assertion (A) is true but reason (R) is false.
D) Assertion (A) is false but reason (R) is true.
Correct Answer: C
Solution :
| [c] Let the digit at units place be x and the digit at tens place be y. Then number \[=10y+x\] |
| According to the question, we have |
| \[10y+x=8(x+y)+1\] |
| \[\Rightarrow \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,7x-2y+1=0\] ...(1) |
| and \[10y+x=13(y-x)+2\] |
| \[\Rightarrow \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,14x-3y-2=0\] ...(2) |
| By using elimination method, we have eq. (2) \[-2x\] eq. (1), we get |
| \[(14x-3y-2)-2x(7x-2y+1)=0\] |
| \[\Rightarrow \,\,\,\,\,\,\,\,\,14x-3y-2-14x+4y-2=0\] |
| \[\Rightarrow \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,y-4=0\] |
| \[\Rightarrow \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,y=4\] |
| Put the value of y in eq. (1), we get |
| \[7x-2\times 4+1=0\] |
| \[\Rightarrow \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,7x=7\] |
| \[\Rightarrow \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,x=1\] |
| Hence, the number \[=10y+x=10\times 4+1=41\] |
| So, Assertion: True; Reason: False |
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