An ant moved for several seconds and covered 3 mm in the first second and 4 mm more in each successive second than in its predecessor. If the ant had covered 1 mm in the first second and 8 mm more in each successive second, then the difference between the path it would cover during the same time and the actual path would be more than 6 mm but less than 30 mm. Find the time for which the ant moved (in seconds).
Answers
Thank you for asking this question. The options for this question are missing. Here are the missing options:
A. 5 s
B. 4 s
C. 6 s
D. 2 s
E. None of these
Answer:
3+7+11+15+ (this will be an equation 1)
1+9+17+25+ (this will be an equation 2)
So the final answer for this question would be 4 seconds which is option B.
If there is any confusion please leave a comment below.
Answer:
Let ant move for t second
Case I: S
t
when ant move 3 m min first second and 4 m min in each second.
S
n = 2 t [2(3)+(t−1)4]
Case II: S
t
when ant move 1 mm in first second and 8 mm in each second
S
t = 2 t [2(v)+(t−1)8]
ATQ
6< 2 t [2+(t−1)8]− 2 t [6+(t−1)4]<30
6< 2 t [−8+4t]<30
6<[−4t+2t 2 ]<30
From option putting t=4 Only condition is satisfied
6<−16+32<30
6<16<30
THEREFORE TIME WOULD BE 4s
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