# Setting up equations to solve worded problems #2

• Sep 12th 2009, 05:45 PM
teddybear67
Setting up equations to solve worded problems #2
I have like a few really basic but idontknow how to solve questions. hope someone can help! thanks in advance! ^^

1) Sini was 12years oldr than Sana 4 years ago
He will be twice as old as sana in 4 years time
Find the sum of their ages 3 years ago

2) The length of a rectangle is increased by 16% and its width is decreased by 16%. Find the percentage increase or decrease in the area.

3) During a cross-country race, Newson ran 50% of the journey at 10km/h, the next 35% at 7km/h and remaning journey at 5km/h. Calculate the average speed in km/h of Newson for the whole journey

Thankssss!(Rofl)
• Sep 12th 2009, 06:34 PM
HallsofIvy
Quote:

Originally Posted by teddybear67
I have like a few really basic but idontknow how to solve questions. hope someone can help! thanks in advance! ^^

1) Sini was 12years oldr than Sana 4 years ago
He will be twice as old as sana in 4 years time
Find the sum of their ages 3 years ago

Let "A" be Sini's present age and let "B" be Sana's present age. Four years ago, Sini's age was A- 4 and Sana's age was B- 4.

"Sini was 12years older than Sana 4 years ago": A- 4= B- 4+ 12 which reduces to A= B+ 12. (If Sini was twelve years older than Sana at any time he alway will be 12 years older!)

In four years time, Sini's age will be A+ 4 and Sini's age will be B+ 4.

"He will be twice as old as Sana in 4 years time." A+ 4= 2(B+ 4) which is the same as A+ 4= 2B+ 8 or A= 2B+ 4. Subtract A= B+ 12 from that to get 0= B- 8 so B= 8. Sana is 8 years old and Sini is 8+ 12= 20 years old. Three years ago Sana was 8- 5= 3 years old and Sini was 20-3= 17 years old The sum of there ages three years ago is 17+ 3= 20.

Quote:

2) The length of a rectangle is increased by 16% and its width is decreased by 16%. Find the percentage increase or decrease in the area.
Let the length of the rectangle be "L" and the initial width be "W". Then its initial area is LW. If the length is increased by 16% it will become L+ 0.16L= 1.16L. If its width is decreased by 16% it will become W- 0.16W= 0.84W. The new area will be (1.16L)(0.84W)= 0.9744LW so that the change is 0.9744LW- Lw= -0.0256 LW or a decrease of 2.56%.

3) During a cross-country race, Newson ran 50% of the journey at 10km/h,
Quote:

the next 35% at 7km/h and remaning journey at 5km/h. Calculate the average speed in km/h of Newson for the whole journey.
Suppose the entire race was of length "L" km. Then he ran .5L at 10 km/h which took .5L/10= L/20 hours. He ran .25L at 7km/h which took .25L/7= L/28 hours. He ran the remaining .25L at 5 km/h which took .25L/5= L/20 hours. That is, he ran the total L km in L/20+ L/28+ L/20= 19L/140 hours or an average speed of L/(19L/140)= 140/19 km/h.

Quote:

Thankssss!(Rofl)
• Sep 12th 2009, 06:35 PM
Finley
Here's assistance with the first:

Sini = x
Sana = y

Equation 1:
$\displaystyle y+12-4=x-4$
$\displaystyle y+8=x-4$

Equation 2:
$\displaystyle 2(y+4)=x+4$
$\displaystyle 2y+8=x+4$

This brings us to a matter of solving equations simultaneously:
2y+8=x+4
y+8=x-4

Do you know how to do this?

OOPS!.. was beaten to it! :P (my bad)
• Sep 12th 2009, 07:04 PM
teddybear67

@Hallsofivy
with regard to the third question, "He ran .25L at 7km/h which took .25L/7= L/28" its supposed to be .35L right? since he ran the next 35%
so it will be .35L/7 = L/20 hrs
remaining 15%,
.15L/5 = 3L/100
In total, it would be L/20 + L/20 + 3L/100 = 13L/100 hours
average speed would be L / (13L/100) = $\displaystyle \frac{7}{9}{13} h$ right?
• Sep 12th 2009, 07:16 PM
Finley
Yes, there was a slight error in the arithmetic:

$\displaystyle L/13L/100$
$\displaystyle = 100/13$
$\displaystyle \approx 7.692 km/h$