Answer:
The solution set can be calculatedby the following steps;
Step-by-step explanation:
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2 answers:
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Step-by-step explanation:
Ques 1)
abc: 15 = 16 rest 1
that means that if abc is divided by 15 than we get the quotient as 16 and remainder as 1.
Hence using : dividend=divisor×quotient+remainder we have:
abc=15×16+1
Hence, the value of abc is 241.
Ques 2)
a)
486: x = 17
i.e.
⇒ x=486×17=8262
b)
y: 18 = 35 rest 1
that means when y is divided by 18 we get the quotient as 35 and remainder as 1.
Hence using : dividend=divisor×quotient+remainder we have:
y=35×18+1=631
y=631
c)
730: z = 27 rest 1
that means when 730 is divided by z we get the quotient as 27 and remainder as 1.
Hence using : dividend=divisor×quotient+remainder we have:
730=27z+1
27z=730-1
27z=729
z=27 (dividing both side by 27)
Ques 3)
Let'a' and 'b' denote the two natural numbers such that a is greater than 'b'.
a-b=70----(1)
also when a is divided by b then we get the quotient as 8 and some remainder.
Let the remainder be 'r'.
this means a=8b+r
let us consider r=0.
a=8b
hence from (1) we have
7b=70
b=10 (on dividing both side by 7)
hence a=80.
when r=1,2,3,4,5,6
we do not get a natural number as value for a and b.
also when r=7 we get the value of b=9 and a=79
After r>7 we get the value of b<r such a condition is not possible.