I have trouble figuring out some word problems. Can someone please help?

Question

Question 1:
A train covered a certain distance at a uniform speed. If the train would have been 10 km/h faster, it would have taken 2 hours less than the scheduled time. And, if the train were slower by 10 km/h, it would have taken 3 hours more than the scheduled time. Find the distance covered by the train.

Question 2:
The students of a class are made to stand in rows. If 3 students are extra in a row, there would be 1 row less. If 3 students are less in a row, there would be 2 rows more. Find the number of students in the class.

Answer 1

Question 1

Distance = Speed * Time

d=st
d=(s+10)(t-2)=st-2s+10t-20
d=(s-10)(t+3)=st+3s-10t-30

Substitute d=st into the second and third equations

st=st-2s+10t-20
st=st+3s-10t-30

Subtract st from both sides of these equations and rearrange:

-2s+10t=20
3s-10t=30

Solve these simultaneous equations.
s=50 km/h
t=12 h
d = st = 50 km/h *12h = 600km



Question 2

Total Students = Students per row * Rows
t = sr
t = (s+3)(r-1) = sr - s + 3r - 3
t = (s-3)(r+2) = sr +2s - 3r - 6

Substituting t=sr into the second and third equations.
sr = sr - s + 3r - 3
sr = sr +2s - 3r - 6

Subtracting sr from both sides and rearranging
-s + 3r = 3
2s - 3r = 6

Solve the simultaneous equations:
s = 9 students per row
r = 4 rows

t= sr = 36 students

There are 36 students in the class

Answer 2

Question 1
Let the speed be s distance be d and time be t then
Distance = Speed × Time
or
d = st
In first condition speed increased by 10 km/h then speed = s + 10 and time required is 2 hr less i.e. (t – 2) so distance,
d= (s+10)( t – 2)=st – 2s+10t – 20
In second condition speed decreased by 10 km/h then speed = (s – 10) and time required is 3 hr more i.e. (t + 3) so distance,
d=(s –10)(t+3)=st+3s –10t – 30

equating both the equations

st-2s+10t-20 =st+3s–10t–30

solving the equation
– 2s+10t = 20
3s–10t=30

Solving equations.
s=50 km/h
t=12 h
d = st = 50 km/h *12h = 600km

Question 2

Total Students = Students per row × Rows
Let total students be t students per row be s and number of rows be r, then
t = sr
t = (s+3)(r-1) = sr - s + 3r - 3
t = (s-3)(r+2) = sr +2s - 3r - 6

sr = sr - s + 3r - 3
sr = sr +2s - 3r - 6

-s + 3r = 3
2s - 3r = 6
s = 9 students per row
r = 4 rows
t= sr = 36 students

Answer 3

1) rate * time = distance
regular rate = x
regular time = t
distance = xt

faster train:
rate = x + 10
time = t - 2
distance = (x + 10)(t - 2)

slower train:
rate = x - 10
time = t + 3
distance = (x - 10)(t + 3)

so...
(x + 10)(t - 2) = xt
xt - 2x + 10t - 20 = xt
-2x + 10t - 20 = 0
-2x + 10t = 20
x - 5t = -10

AND
(x - 10)(t + 3) = xt
xt + 3x - 10t - 30 = xt
3x - 10t - 30 = 0
3x - 10t = 30

Now you have two equations and two variables.
x - 5t = -10 ===> x = 5t - 10
3x - 10t = 30

3(5t - 10) - 10t = 30
5t - 30 = 30
5t = 60
t = 12
x = 5(12) - 10 = 50

The distance covered is xt = (12)(50) = 600 miles

Answer 4

overlook approximately concerning to the words and check out the physics. the engine burns for ten seconds offering sufficient thrust to hold the rocket to a distinctive top. After inertia burns off, the rocket falls at 32fps so on so forth... Plotting approx how long the rocket will take to effect earth lies therein.

Answer 5

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