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35 divided by 7 = 5 answer is c.

40 - 5 = 35 then 35/7 =5

on the second one

L = 2*W therfore L=x and W=2x if the perimeter(the sum of all sides) is 60 then 6x=60

so x=10 and given that A=L*W the area would be 200m^2

For the first problem:

let the son's age be x.

5 years earlier the father must have been 5 years younger, or 40-5=35 years old. 35 y.o. is 7 times his son's age 5 YEARS AGO, so

35=7(x-5)

5= x-5

x=10

So the answer to the first problem is a)10

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For the second problem, let L be the length and W be the width. L is twice of W, so L=2W. The perimeter (P) is equal to 2W+2L and is equal to 60, so

60 = 2W + 2L

60 = L + 2L

60 = 3L

L = 20.

Plug this into L=2W, so 20=2W and W=10

Area is simply A=LW, so A=(20)(10)=200

So the answer is d)200m^2

if the father is 40 now, five years ago, he was 35, making his son 5 at the time (35 divided by 7). now, all you have to do is add the five years to the son's age and you will get his present age... (a. 10)

perimeter is length plus width plus length plus width....

so if we use X to represent the width, the length would be 2X

so:

X+X+2X+2X=60

6X=60

X=60/6

X=10

so the width is 10 and the length is 20

area is 10x20, which will bring you to (d. 200 m2)

first question is A

40 - 5 = 35 (five years earlier)

35 / 7 = 5 (he is seven times his sons age. His son was 5, 5 years ago)

5 + 5 = 10 (his son is 10 in the present)

second question is D

y = 2x

2x + 2y = 60

solve and x = 10 and y = 20

so the area is x*y which is 200

40 - 5 = 5*(x - 5)

35/ 5 = x - 5

7 = x - 5

12 = x

The son is 12 years old.

L = 2w

60 = 2w + 2L

60 = 2w + 4w

60 = 6w

10 = w

A = Lw

A = 10*20

A = 200

easy

40-5=35

35 divided by 7 =5

then u have to add back the 5 years which makes 12(d

f = 40

35 = 7s

5 = s 5 years earlier

Present age equals s + 5

l = 2w

2l + 2 w = 60m

2 (2w) + 2w = 60

Area = l * w

b) 10

d) 200 m^2

d/

d/