A person walks from Rugby to London in four days, and back in five day

Suppose the distance travelled by him on the first day = x+4

D = (x+4)+(x+3)+(x+2)+(x+1) = 4x+10

Also, D = (x-4)+(x-3)+(x-2)+(x-1)+x= 5x-10

4x+10 = 5x-10

x=20

D = 4x+10 = 4*20+10 = 90

Ans: D

Let speed from Rugby to London
x (x-1) (x-2) (x-3)
Appplying Total Distance = Avg. Speed * Time
D1 (suppose) = (4x-6)/4*4 = 4x-6

Similarly D2 (suppose distance from London to Rugby) = (5y-10)/5*5 = 5y-10

Now: D1=D2
4x-6=5y-10

Only D fits in....

Hope it helps..

First, from Rugby to London
1st Day - x
2nd Day - x-1
3rd Day - x-2
4th Day - x-3

From London to Rugby
Day
5th - x-4
6th - x-5
7th - x-6
8th - x-7
9th - x-8

Distance from Rugby to London = Distance from London to Rugby
Therefore,
x+x-1+x-2+x-3 = x-4+x-5+x-6+x-7+x-8
==> 4x-6 = 5x-30
x=24

4(24)-6 = 90

Let the distance cover by the person on first day = n
Succeeding day = n-1
Next day = n-2
And so on.........
Therefore
He covered n+(n-1)+(n-2)+(n-3) miles from rugby to London in four days.
That is 4n-6 ------------(1)
Now it took him 5 days to return back,
The distance covered by him in 5 days = (n-4)+(n-5)+(n-6)+(n-7)+(n-8) => 5n-30 -----------(2)
Distance between two cities is same
4n-6 = 5n-30
24 = n
So the person cover 24 miles on first day.
He covered 4n-6 miles to cover the distance between rugby and London.
Substituting value;
4(24)-6 = 96-6 = 90 miles.(LHS)
Also,
5(24)-30 = 120-30 = 90 miles.(RHS)
Answer is D

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r+(r-1)+(r-2)+(r-3)=(r-4)+(r-5)+(r-6)+(r-7)+(r-8)
4r-6=5r-30
r=-6+30, r=24;
4(24)-6=90

Ans (D)

Let the distance he walks on the 1st day be x.
=> on 2nd day he walks = x -1
on 3rd day he walks = x -2,
on 4th day he walks = x- 3
=> total distance b/w Rugby and London = 4x -7.

Now, he walks back to London.
The distance covered on 5th day till 10th day is as: x-4, x-5, x-6, x-7,x-8.
=> the total distance = 5x - 30.

Now,
4x- 7 = 5x -30
=> x = 23.

Now 4x - 7 = 85.
=> 85 miles.

Answer C.

Each day he walks one mile less than on the preceding day.

Let's first day the boy works for x miles

so X + X-1 + ...X-3 = X-4 + ...+X-8
so 4X-6 = 5X-30 => X = 24 .
Total distance = 4X-6 = 24 * 4 -6 = 90 miles .

Ans is D ..
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A person walks from Rugby to London in four days, and back in five days. Each day he walks one mile less than on the preceding day. What is the distance between the two cities?

A. 75 miles
B. 80 miles
C. 85 miles
D. 90 miles
E. 100 miles

consider he started with x1 miles on day 1
distance travelled on 4 days equal to dist travelled on 5 days

so, x1+x1-1+x1-2+x1-3 = x1-4+x1-5+x1-6+x1-7+x1-8
on solving we get x1=24

therefore,
on first day he walked 24 miles

total = 24+23+22+21
90 miles

Ans D

Let distance covered on Day1 be x
Therefore, Distance covered each day in terms of miles is:
Day1: x
Day2: x-1
Day3: x-2
Day4: x-3
Day5: x-4
Day6: x-5
Day7: x-6
Day8: x-7
Day9:x-8

Day1 to Day4 is distance to cover journey from Rugby to London and Day5 to Day9 is for the return journey.
Speed = Distance / Time. Also, distance for onward and return journey must be the same.
Therefore,
(x)+(x-1)+(x-2)+(x-3)=(x-4)+(x-5)+(x-6)+(x-7)+(x-8)
Solving the above will give the value of x (24), which is the distance run on the first day.
Therefore, total distance from Rugby to London will be = 24+23+22+21=90 miles

ANS: D

9 +10 + 9 + 10 + 9 + 10 + 9 + 10+ 9 = 85

Posted from my mobile device

Let n be the number of miles per day and X be total distance therefore 5(n-1) + 4n = X
=> 9n - 5 = X. From options X = 85 satisfied for n = 10.

Posted from my mobile device

Solution:

The individual walks for 9 days. Let x = be the number of miles he walks on the fifth day. We can create the equation:

Going distance (4 days) = Returning distance (5 days)

x + 4 + x + 3 + x + 2 + x + 1 = x + x - 1 + x - 2 + x - 3 + x - 4

4x + 10 = 5x - 10

20 = x

Therefore, the distance between the two cities is 4(20) + 10 = 90 miles.

Answer: D
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