1068:11=97.090909090909
243:11=22.0909090909
1200:11=109.090909091
2025:11=184.090909091
x - 3h kolesnie velosipedi;
y - 2h kolesnie;
togda po usloviu mozhno sostavit' sistemu:
x+y=20;togda x=20-y;
3x+2y=55;ili 3(20-y)+2y=55;
60-3y+2y=55,togda y=5.
Otvet:5 2h kolesnih velosipedov.
Esli reshat' bez uravnenii:
1)dopustim vse velosipedi trehkolesnie,togda bi bilo 20*3=60 koles;
no ih 55,togda,2)esli uvelichivat' chislo dvuhkolesnih velosipedov na 1, budet umen'shatsia chislo trehkolesnih na 1,a chislo koles umenshitsia na 3-2=1.
3) no nam nado umenshit chislo koles s 60 do 55,znachit dvuhkolesnih velosipedov 60-55=5(no etot sposob reshenia bolshoi i ne ochen' udobnii dlia ponimania).
1068:11=97.090909090909
243:11=22.0909090909
1200:11=109.090909091
2025:11=184.090909091