If Sum of Two Numbers Is 1215 And Their HCF is 81, then the Possible Number of Pairs of Such Numbers Are

If sum of two numbers is 1215 and their HCF is 81, then the possible number of pairs of such numbers are  

(a) 2 

(b) 3 

(c) 4 

(d) 5

It is given that the sum of two numbers is $1215$ and their H.C.F is $81$.

Let the two numbers be $x$ and $y$.

Now,

$81x+81y=1215$    -------------[sum of two numbers are $1215$]

$=\; 81(x+y)=1215$

$=x+y=15$

So,

For, $x=1,y=14$,the numbers are 

$\mathrm{=\; 1}\times 81+14\times 81$

=81+1134
$1\times 81+14\times 81=81+1134=1215$

 = 1215

For, $x=7,y=8$,the numbers are 

= 7×81+8×81

=567+648

=1215$$7×81+8×81=567+648=1215

For, $x=2,y=13$,the numbers are 

$\mathrm{=\; 2}\times 81+13\times 81$

=162+1053

=1215$$2×81+13×81=162+1053=1215

$\mathrm{<span\; style="font-family:\; inherit;\; font-size:\; large;"><br></span>}$

For, $x=4,y=11$,the numbers are 

$\mathrm{=\; 4}\times 81+11\times 81$

=324+891

=1215$$4×81+11×81=324+891=1215

$\mathrm{<span\; style="font-family:\; inherit;\; font-size:\; large;"><br></span>}$

Therefore, the numbers of such pairs are $4$.

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