Word Problems - Proportions, Speed & Time: Problems with Solutions

Solution:
Let the first number be [tex]n[/tex].
Then the second one is [tex]n+2[/tex] and the third one is [tex]n+4[/tex].

[tex]n+(n+2)+(n+4)=78[/tex]
[tex]3n+6=78[/tex]
[tex]3n=72[/tex]
[tex]n=24[/tex]

So the three numbers are 24, 26 and 28.

Answer: 24, 26, 28

Solution:
First find how many steps she climbs in one second:
[tex]60:40=1.5[/tex] steps per second.

In 150 seconds she climbs
[tex]1.5\cdot 150=225[/tex] steps.

Answer: 225 steps

Solution:
The number of applicants went down by
[tex]120000-46200=73800[/tex].

A percent of decrease is always compared with the original amount, which is 120000:
[tex]\frac{73800}{120000}\cdot 100\%=61.5\%[/tex]

Answer: 61.5%

Solution:
Let [tex]x[/tex] be the number of pounds sold in the morning.
Then [tex]2x[/tex] pounds were sold in the afternoon.

Morning + afternoon = 360 pounds, so
[tex]x+2x=360[/tex]
[tex]3x=360[/tex]
[tex]x=120[/tex]

In the afternoon the store sold [tex]2x=240[/tex] pounds.

Answer: 240 pounds

Solution:
They can both be at the starting point only after each of them has ridden a whole number of laps.
So the time we are looking for must be a multiple of 3 and also a multiple of 4 — the least common multiple of 3 and 4:
[tex]\operatorname{LCM}(3,4)=12[/tex]

In 12 minutes the first cyclist rides [tex]12:3=4[/tex] laps and the second one [tex]12:4=3[/tex] laps, so both are back at the start.

Answer: 12 minutes

Solution:
Be careful: with 10 hurdles on the track there are only 9 gaps between them, not 10.
The gaps between the hurdles cover
[tex]9\cdot 10=90[/tex] yards.

Now add the run-up before the first hurdle and the run-in after the last one:
[tex]15+90+15=120[/tex] yards.

The height of the hurdles (42 inches) is extra information — a hurdle stands on the track, it does not make the track longer, so we do not use it.

There are 3 feet in a yard, so
[tex]120\text{ yd}=120\cdot 3=360\text{ ft}[/tex]

Answer: 360 feet

Solution:
At the end of the month the account holds the original deposit plus 0.8% of it, that is
[tex]100\%+0.8\%=100.8\%[/tex] of the deposit.

[tex]100.8\%\cdot 150000=1.008\cdot 150000=151200[/tex]

Answer: 151200 dollars

Solution:
Let the starting number be [tex]x[/tex].
Splitting it into 3 equal groups gives [tex]\frac{x}{3}[/tex] in each group.
Splitting each group in half gives
[tex]\frac{x}{3}:2=\frac{x}{6}[/tex].

[tex]\frac{x}{6}=13[/tex]
[tex]x=13\cdot 6=78[/tex]

Answer: 78

Solution:
Speed is distance divided by time:
[tex]375:3=125[/tex]

Answer: 125 mph

Solution:
From 9:15 to 10:35 the flight lasts 1 hour and 20 minutes.

In one hour the plane covers 180 miles.
20 minutes is [tex]\frac{20}{60}=\frac{1}{3}[/tex] of an hour, so in 20 minutes it covers
[tex]\frac{1}{3}\cdot 180=60[/tex] miles.

[tex]180+60=240[/tex] miles.

Answer: 240 miles

Solution:
The speed is given in miles per hour, so first write the time in hours:
[tex]15\text{ min}=\frac{15}{60}=\frac{1}{4}[/tex] hour.

Distance = speed [tex]\times[/tex] time:
[tex]12\cdot\frac{1}{4}=3[/tex] miles.

Answer: 3 miles

Solution:
Going up:
[tex]t=\frac{1}{6}[/tex] hour [tex]=\frac{60}{6}=10[/tex] minutes.

Going down the speed is [tex]2\cdot 6=12[/tex] mph over the same 1 mile, so the time is half as long:
[tex]\frac{1}{12}[/tex] hour [tex]=5[/tex] minutes.

Total:
[tex]10+5=15[/tex] minutes.

Answer: 15 minutes

Solution:
At 60 mph the train covers 60 miles in 60 minutes, that is 1 mile every minute.

So 3 miles take 3 minutes:
[tex]9{:}10+3\text{ min}=9{:}13[/tex]

Answer: 9:13

Solution:
First find the distance. 3 hours and 30 minutes is [tex]3.5[/tex] hours, so
[tex]60\cdot 3.5=210[/tex] miles.

The motorcyclist covers the same 210 miles in 5 hours:
[tex]210:5=42[/tex]

Answer: 42 mph

Solution:
A quarter of the way is
[tex]600:4=150[/tex] miles.

The car covers these 150 miles in
[tex]150:40=3.75[/tex] hours = 3 hours and 45 minutes.

The remaining [tex]600-150=450[/tex] miles are covered in
[tex]450:60=7.5[/tex] hours = 7 hours and 30 minutes.

Now add the two times:
3 hours 45 minutes + 7 hours 30 minutes = 10 hours 75 minutes.
75 minutes is 1 hour and 15 minutes, so this is 11 hours and 15 minutes.

Answer: 11 hours and 15 minutes

Solution:
John, if working alone, would paint the house in
$3\times24=72$ hours, since according to the problem, it would take him 3 times longer than if he worked with Joe (and together they complete the job in 24 hours).

Let $x$ represent the number of hours Joe needs to paint the entire house by himself.

In one hour, he would paint $\frac1x$ of the house, and John would paint $\frac{1}{72}$

Together, they paint $\frac1x+\frac{1}{72}$ of the house per hour.
They painted the entire house in 24 hours, so:
$\left(\frac1x+\frac{1}{72}\right)\cdot 24=1$

$\frac{24}{x}+\frac{24}{72}=1$

$\frac{24}{x}+\frac{1}{3}=1$

$\frac{24}{x}=1-\frac{1}{3}$

$\frac{24}{x}=\frac{2}{3}$

$\frac{12}{x}=\frac{1}{3}$

$x=36$

So Joe would finish the job alone in 36 hours.

Solution:
If all [tex]35[/tex] animals were pheasants, they would have [tex]35\times 2=70[/tex] legs. Each rabbit adds [tex]2[/tex] more legs:
[tex](94-70)\div 2=12[/tex]
(With an equation: [tex]4x+2(35-x)=94[/tex], [tex]x=12[/tex].) There are [tex]12[/tex] rabbits and [tex]23[/tex] pheasants.

Solution:
Let the daughter be [tex]x[/tex]; the mother is [tex]8x[/tex]. In [tex]4[/tex] years:
[tex]8x+4=4(x+4) \Rightarrow 8x+4=4x+16 \Rightarrow 4x=12 \Rightarrow x=3[/tex]
Check: [tex]24=8\times 3[/tex]; [tex]28=4\times 7[/tex].

Solution:
The third book costs exactly the [tex]10[/tex] left over plus the missing [tex]2[/tex]:
[tex]10+2=12[/tex]
(Equation: [tex]2x+10=3x-2 \Rightarrow x=12[/tex].)

Solution:
Let there be [tex]p[/tex] poles. Then the crows are [tex]p+1[/tex]; on the other hand, they sit two on each of [tex]p-1[/tex] poles:
[tex]p+1=2(p-1) \Rightarrow p=3[/tex]
There are [tex]3[/tex] poles and [tex]4[/tex] crows.

Solution:
Let the watermelon weigh [tex]x[/tex]. The other three quarters weigh [tex]3[/tex]:
[tex]x=3+\dfrac{x}{4} \Rightarrow \dfrac{3x}{4}=3 \Rightarrow x=4[/tex]
[tex]1.5\times 4=6[/tex]

Solution:
[tex]40-5=35[/tex] students do at least one of the two. The sum [tex]30+27=57[/tex] counts those who do both twice:
[tex]57-35=22[/tex]

Solution:
Half of the contents weighs [tex]120-64=56[/tex], so all the contents weigh [tex]112[/tex]:
[tex]120-112=8[/tex]

Solution:
Work backwards from Kircho's result to find the number:
[tex]54\times 4-5=211[/tex]
[tex](211+4)\div 5=43[/tex]

Solution:
The sum of all [tex]10[/tex] numbers minus the sum of the first [tex]9[/tex]:
[tex]10\times 79-9\times 80=790-720=70[/tex]

Solution:
Adding the three sums counts every girl twice:
[tex](38+45+49)\div 2=132\div 2=66[/tex]

Solution:
Let the three equal results be [tex]t[/tex]. Then the numbers are [tex]t+5[/tex], [tex]t-5[/tex] and [tex]\frac{t}{5}[/tex]:
[tex]t+5+t-5+\dfrac{t}{5}=88 \Rightarrow \dfrac{11t}{5}=88 \Rightarrow t=40[/tex]
The numbers are [tex]45[/tex], [tex]35[/tex] and [tex]8[/tex], and [tex]45-8=37[/tex].