I would say no.
Let me know if you got it right
Answer:
<h3>
9, 11, 13, 15</h3>
Step-by-step explanation:
{k - some integer}
2k+1 - the first odd integer (the least)
5(2k+1) - five times the least
5(2k+1)+3 -<u> three more than five times the least</u>
2k+1+2 = 2k+3 - the odd integer consecutive to 2k+1
2k+3+2 = 2k+5 - the next odd consecutive integer (third)
2k+5+2 = 2k+7 - the last odd consecutive integer (fourth)
2k+1+2k+3+2k+5+2k+7 - <u>the sum of four odd consecutive integers</u>
2k+1 + 2k+3 + 2k+5 + 2k+7 = 5(2k+1) + 3
8k + 16 = 10k + 5 + 3
- 10k -10k
-2k + 16 = 8
-16 - 16
-2k = -8
÷(-2) ÷(-2)
k = 4
2k+1 = 2•4+1 = 9
2k+3 = 2•4+3 = 11
2k+5 = 2•4+5 = 13
2k+7 = 2•4+7 = 15
Check: 9+11+13+15 = 48; 48-3 = 45; 45:5 = 9 = 2k+1
At this ratio, he would have counted 45 sedans if he had passed 18 pickups. The ratio between the pickups and the sedan is two by five, therefore he already has faced 9 times of this situation if he had already seen 18 pickups passed over. The number of sedans counted by him can be obtained by multiplying the sedans number of the ratio which is 5 with the number of the situation that had passed which is 9.
