Find a smallest number which is exactly divisible by 8,10 leaves the remainder 3,9 respectively.
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The answer of the question is 29 as 29=3×8 and remainder is 3. 29=2×10 and remainder is 9
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here let the smallest number be =X
given , X/8=3 and. X/10=9
X/8-3=0. X/10-9=0
as here, in both RHS are equall so automatically LHS is equal to each other
X/8-3=X/10-9. (now rationalised the denominator)
8x-24=10x-90
now take X term on LHS
8x-10x= -90-24
-2x= -66
X=66/2
X=33
therefore the smallest number =X=33
given , X/8=3 and. X/10=9
X/8-3=0. X/10-9=0
as here, in both RHS are equall so automatically LHS is equal to each other
X/8-3=X/10-9. (now rationalised the denominator)
8x-24=10x-90
now take X term on LHS
8x-10x= -90-24
-2x= -66
X=66/2
X=33
therefore the smallest number =X=33
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