GMAT Club Olympics 2024 (Day 10): A car is traveling on a straight

The answer is 10 for both x and y
IF we input the value of 10 then the time comes out to be 5 and if we put the value of 10 in place of y then the time comes out to be 5. Since we know the total time taken is 10 hours and the condition that x>=y. Hence x=y

­We are given that,
A car takes 10 hours to complete the journey of 100 miles.
One half of the travel time, speed of car was constant rate of x mph
Second half of the travel time, speed of car was constant rate of y mph

Half travel time = \(\frac{10}{2} = 5 hours\)

Then distance covered in first half time = 5x
Distance covered in second half time = 5y

\(5x + 5y = 100\)
\(x + y = 20­\)

Looking at the optoins,
Only value which satisfies the equation is 10 for both x and y.
Any other value of x and we cannot find value of y in options which satisfies the equation.

\(x = 10\) and \(y = 10\)
And condition was \(x >= y\).

So,
x, y = 10­

­

­Eqn is 5x + 5y = 100
x + y = 20

out of options
it seems x=y

x= 10
y= 10

5x+5y=100
x+y=20
x=10 and y=10
1-4 2-4

5x+5y=100
x+y=20

only x=y=10 satisfies

Ans 10, 10­

Total distance traveled = 100m
Total time taken = 10 hrs
Speed for 5 hrs = x m/hr
Speed for next 5 hrs = y m/hr
x>=y----(1)
Avg speed for the journey = 100/10=10 m/hr

10=(5x+5y)/10
x+y=20 -----(2)
From the options given and from statements (1) and (2), values of x and y are x=10 and y=10

ANs :

x= 10 ; y =10

Car takes 10 hours to complete 100 miles: avg speed is 10

As speed x takes one half time : hence x take 5 hours and

similary y will take 5 hours

As time taken is same : avg speed = AM of speeds

x+y/2 = 10

x+y = 20

So x = 10 and y = 10

10, 10

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Total time =10 hours
at speed of x car travels half of the total time
Distance travelled with speed x = 5x
Distance travelled with speed y = 5y
­5x+5y=100
x+y=20 Given x>=y
Only possible solution among options
x=10 and y=10
1-4 2-4­

­A car is traveling on a straight stretch of a road at a constant rate of x mph, for exactly one half of the travel time. The remaining time, the car travels at a constant rate of y mph. The car does not stop and takes exactly 10 hours to complete the journey of 100 miles. It is also known that x is greater than or equal to y. Select values of x and y that are jointly consistent with the information provided. Make only two selections, one in each column.

x = 10
y = 10

x >= y
average rate = 100 miles/10 hours = 10 mph
x mph for 5 hours => 50miles/x + 50 miles/y => x = y = 10 mph­

To solve this problem, we need to use the given information to find values of
x and y that satisfy the conditions:

The car travels for exactly 10 hours.
The total distance traveled is 100 miles.

The car travels at a constant rate of
x mph for exactly one half of the travel time and at a constant rate of
y mph for the remaining time.
x is greater than or equal to y.
Let's denote the total travel time as
T (in hours). Since the car travels for half of the travel time at x mph and
the other half at y mph, each segment of travel time is T/2 .

Given:
T=10 hours

The distance traveled in each segment is:
Distance=Speed×Time

For the first half of the travel time:
D1 = x*T/2 = 5x

For the second half of the travel time:
D2 =y * T/2 = 5y

The total distance traveled is: D1 + D2 =100

So,
5x+5y=100
x+y=20

We need to find pairs of values for x and
y from the given lists that satisfy this equation, and where
x is greater than or equal to y.

Let's test the given values:

If x=10, then:
x+y=20
10+y=20
y=10

If x=20, then:
x+y=20
20+y=20
y=0 (which is not in the given list)

If x=50, then:
x+y=20
50+y=20
y=−30 (which is not valid)

If x=7, then:
x+y=20
7+y=20
y=13 (which is not in the given list)

If x=5, then:
x+y=20
5+y=20
y=15 (which is not in the given list)

If x=3, then:
x+y=20
3+y=20
y=17 (which is not in the given list)

So, the only valid pair from the given list is:
x=10 and y=10

Thus, the selections that are jointly consistent with the information provided are: x=10 and y=10

Distance = Rate*time
5*rate x + 5*rate Y =100
X+Y =20

Hence, 10;10 satisfies the equation­