It would seem that your guide is mistaken - I think you did this correctly. What is the source?

A sample of question from the Business Aptitude Test, page 21 of the link below,
https://www.isb.edu/sites/default/files/ ... ooklet.pdf

I see. The question is referring to the bar graph presented. We see there that he finished race 1 in 10.55 seconds. If he ran the first 50 meters in 5 seconds, he ran the second 50 meters in 5.55 seconds. So his rate for the second half was 50/5.55, or about 100/11. (My estimation here means the correct answer will be slightly greater than what I calculate.)

So that 20% increase doesn't mean that we want the scenario in which he ran the second half of the race 20% faster than he ran the first half of the race. It means that we want the scenario in which he ran the second half of the race 20% faster than his recorded speed for the last 50 meters, 100/11. A 20% increase would be 120/11.

Time for the last 50 meters:
120/11 * t = 50
t = 550/120 = 55/12 = 4 + 7/12 = 4.583
5 seconds for first 50 meters + 4.583 for second 50 meters = 9.583 total
(Again, when I estimated the rate, I came up with a number that was slightly greater than the actual rate, so my time is a little smaller than the actual time.)

Closest to D.

(For anyone else following along, the question refers to the bar graph below)

It's closer to 4.63 than that, although the answer is still approximate.

From the graph, Ross completed Race 1 in 10.55 seconds. Since he completed the first 50 meters in 5 seconds, he completed the second 50 meters in 5.55 seconds.

D = RT

50 = R * 5.55

5000 = 555R

5000/555 = 1000/111 ≈ 9 = R

So Ross's rate was very close to 9 meters per second. Increasing this by 20% gives us 10.8 meters per second. To find the time it would take to do the second half of the race at this rate, we have

D = RT
50 = 10.8T
500 = 108T
500/108 = 125/27 = 4 + 17/27 ≈ 4.63 seconds

In a 100m race, Ross completed the first 50m in 5 seconds. His total time to complete the race was 10.55 seconds. If he had increased his actual speed over the remaining 50m by 20 percent, what would have been his total time for the race?

A. 10.66 seconds
B. 10.55 seconds
C. 10.44 seconds
D. 9.63 seconds
E. 8.8 seconds

Actual time for remaining 50m = (total time) - (time for the first 50m) = 10.55 - 5 = 5.55 = 555/100 = 111/20 seconds.

Faster time:
Rate and time are RECIPROCALS.
If Ross increases his speed for the second 50m by 20% -- implying that he travels at 120% = 6/5 of his actual speed -- he will take 5/6 of his actual time for the second 50m:
(5/6)(111/20) = 111/24 = 37/8 = 4.625 seconds.
Thus:
Total faster time = (actual time for first 50m) + (faster time for second 50m) = 5 + 4.625 = 9.625 seconds.

The correct answer is D.

Note that phrasing the decimals as FRACTIONS makes the calculations easier.

Hi Vincelauret,
Do u have more sample practice papers?Really would b thankful.

Sandip.