Question

Four children have small toys. The first child has 1/10 of the toys, the second child has 12 more toys than the first, the third child has one more toy of what the first child has and the fourth child has double the third child. How many toys are there?​

Answer 1

Answer:

a,b,c,d

x=number of toys total

a+b+c+d=x

a=(1/10)x

b=12+(1/10)x

c=1+(1/10)x

d=2(1+(1/10)x)=2+(2/10)x

add them

(1/10)x+12+(1/10)x+1+(1/10)x+2+(2/10)x=x

group like terms

(1/10)x)+(1/10)x+(1/10)x+(2/10)x+12+1+2=x

add like terms

(5/10)x+15=x

(1/2)x+15=x

subtract (1/2)x from both sides

15=(1/2)x

multily both sides by 2/1 to clear fraction

30=x

30 total toys

Step-by-step explanation:

Answer 2

Answer:

T = total toys

c1 = 1/10 * T

c2 = c1 + 12

c3 = c1 + 1

c4 = 2 * ( c1 + 1)

c1 +c2 +c3 +c4 = T

( 1/10 * T ) + ( 1/10 * T + 12 ) + ( 1/10 * T + 1 ) + 2 ( 1/10 * T + 1 ) = T

5/10 * T + 12 + 1 + 2 = T [ collecting factors of T ]

1/2 * T + 15 = T [ simplifying ]

15= 1/2 * T [ subtract 1/2 * T from both sides ]

30 = T [ multiply both sides by 2 ]

there are 30 toys