Wiley Edge C/C++ Quiz 2

To find the LCM of these two numbers, we can calculate it as follows:
LCM(68, 62) = (68 * 62) / GCD(68, 62).
The GCF of 68 and 62 is 2.LCM(68, 62) = (68 * 62) / 2 = 2108.T

he LCM of 68 and 62 is 2108 seconds. 2108/60 = 35.1333 minutes;10:00 A.M. + 35.1333 minutes = 10:35 A.M.

In still water the time upstream will be the same as the time downstream provider boat is at the same speed.

At a speed of 6 km/h, John arrives 10 minutes late,At a speed of 8 km/h, John arrives 10 minutes early.
Setting up these equations:(d/6) = (work start time + 10 minutes) / 60 ;
(d/8) = (work start time - 10 minutes) / 60.work start time is 90 minutes or 1 hour and 30 minutes. we get :
d = (90 minutes + 10 minutes) / 10 ;
d = 10 kilometers.

In every two-hour cycle, the squirrel gains a net height of 16 - 7 = 9 feet.
To reach the top of the 80-foot tree, it would take a total of 8 cycles (80 / 9 = 8.888),
which corresponds to 16 hours and 40 minutes.

The monkey climbs 8 m in the first minute and slips down 2 m in the next minute. This means it covers a net distance of 8 m - 2 m = 6 m in every 2-minute interval. To reach the top of the 80 m tree, the monkey will need 80 m / 6 m = 13 intervals * 2 minutes per interval = 26 minutes.

: Since B runs twice as fast as C, it would cover the same distance in half the time. Therefore, B would cover the distance in 40 minutes. Similarly, since A runs four times as fast as B, it would cover the same distance in a quarter of the time that B took. Thus, A would cover the distance in 10 minutes.

The hiker climbs 1.5 meters per minute and slides down 80 centimeters, which is equivalent to 0.8 meters. So, he makes a net progress of 0.7 meters (1.5 - 0.8) every minute. To cover the 12-meter mountain, he will need 12 / 0.7 = 17.14 minutes, which is approximately 17 minutes and 9 seconds. Starting at 9 am, he will reach the mountain top at around 9:17 am.

The ratio of the speeds of the two cars can be determined by the ratio of the times it takes for them to reach their destinations after passing each other. In this case, the first car takes 2 hours to reach City Y, while the second car takes 6 hours to reach City X. Therefore, the ratio of their speeds is 2:6, which simplifies to 1:3. However, we take the reciprocal of the ratio, resulting in the answer 3:1.

Since the time it takes for the cars to reach their destinations is inversely proportional to their speeds, we can set up the following equation: (Time taken by car 1 to reach its destination) / (Time taken by car 2 to reach its destination) = (Speed of car 2) / (Speed of car 1)Substituting the given values, we have:2 / 5 = (Speed of car 2) / (Speed of car 1)
Cross-multiplying, we get:
5 * (Speed of car 2) = 2 * (Speed of car 1)
Speed of car 2 : Speed of car 1 = 2 : 5
Simplifying the ratio, we get:
Speed of car 1 : Speed of car 2 = 5 : 2

The cannons were fired 8 minutes apart so if the observer were to be stationary , he would have heard shots 8 minutes apart but as he was moving , he heard them 7 minutes 20 seconds apart. The observer traveled for 7 minutes and 20 seconds at Speed x before hearing the second shot. So whatever distance sound would have traveled in those other 40 seconds , the observer covered it in 7 min and 20 seconds. 7 min 20 sec = 440 sec; 440*x=340*40; x = 30.9 ;
approx x= 31m/s