8th Class Mathematics Playing with Numbers Question Bank Playing with Numbers

  • question_answer 18) Match the following.
    Column - I Column - II
    (P) If \[213x27\] is divisible by 9, then \[x=\] (i) 2
    (Q) If \[2415x\]is divisible by 6, then \[x=\] (ii) 8
    (R) If \[22135x\] is divisible by 4 and 3, then \[x=\] (iii) 3
    (S) If \[7251x93\] is divisible by 11, then \[x=\] (iv) 6

    A)   (P)\[\to \](iii); (Q)\[\to \](ii); (R)\[\to \](iv); (S)\[\to \](i)

    B)  (P)\[\to \](ii); (Q)\[\to \](iv); (R)\[\to \](i); (S)\[\to \](iii)

    C)  (P)\[\to \](iii); (Q)\[\to \](iv): (R)\[\to \](i); (S)\[\to \](ii)

    D)  (P)\[\to \](ii); (Q)\[\to \](iii); (R)\[\to \](i); (S)\[\to \](iv)

    Correct Answer: C

    Solution :

    (P) Since, \[213\times 27\] is divisible by 9. So, 2+1+3+x+2+7=15+xis divisible by 9. \[\therefore \] x=3 (Q) \[2415x\] is divisible by 6, if it is divisible by both 2 and 3. \[\therefore x=6\] (R) \[23245x\] is divisible by 4 and 3 \[\Rightarrow 5x\] is divisible by 4 \[\therefore \] Possible values of x are 2, 6             Also, \[2+3+2+4+5+x=16+x\] is divisible by 3. \[\therefore \] Possible values of \[x\] is 2. (S) We have, \[7251x\text{ }93\] is divisible by 11 \[\therefore [(7\text{ }+\text{ }5\text{ }+\text{ }x\text{ }+\text{ }3)\text{ }-\text{ }(2\text{ }+\text{ }1\text{ }+\text{ }9)]\] is divisible by 11 \[\Rightarrow 15+x-12=3+x\] is divisible by 11 \[\therefore x\] can be equal to 8.

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