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07 Dec 2021, 03:45
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B
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
75%
(hard)
Question Stats:
59% (02:54) correct
41%
(02:49)
wrong
based on 119
sessions
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Not Attempted Yet
Two sea trawlers left a sea port simultaneously in two mutually perpendicular directions, half in an hour later the shortest distance between them was 17 kilometer and another 15 minutes later, one sea trawler was 10.5 kilometer further from the origin then the other. find the speed of each sea trawler?
A. 18 km/h, 24 km/h
B. 16 km/h, 30 km/h
C. 14 km/h, 20 km/h
D. 17 km/h, 32 km/h
E. None of these
Are You Up For the Challenge: 700 Level Questions
A. 18 km/h, 24 km/h
B. 16 km/h, 30 km/h
C. 14 km/h, 20 km/h
D. 17 km/h, 32 km/h
E. None of these
Are You Up For the Challenge: 700 Level Questions
10 Dec 2021, 01:30
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Let the speeds of the faster trawler (T1) and the slower trawler (T2) be x and y kph respectively. Then the distance covered by T1 and T2 in 45 minutes (30+15) will be 3x/4 kms and 3y/4. Then, 3x/4 = 3y/4 + 21/2....> 3x = 3y + 42....> x = y + 14. Luckily, there is only one option (B) where the difference in speeds is 14 but because of option (E) "None of these", we can't straightaway opt for (B). So we must test this by ascertaining whether, basing on this values, T1 does indeed cover 10.5 kms more than T2 in 45 minutes. Distance covered by T1 = 15 + 7.5 = 22.5. Distance covered by T2 = 8 + 4 = 12. So, T1 does indeed cover 10.5 kms more than T2 in the 45 minutes after starting.
ANS: B
P.S. If there were more than one option with 14 km speed difference then each answer would have to be tested. Alternatively, you could substitute the value of x in terms of y (or vice versa) from (x=y+14) in the equation (x/2)^2 + (y/2)^2 = (17)^2 (Pythagoras) and get the value of x or y. But this is a cumbersome and time-consuming process mainly because of the large numbers involved.
ANS: B
P.S. If there were more than one option with 14 km speed difference then each answer would have to be tested. Alternatively, you could substitute the value of x in terms of y (or vice versa) from (x=y+14) in the equation (x/2)^2 + (y/2)^2 = (17)^2 (Pythagoras) and get the value of x or y. But this is a cumbersome and time-consuming process mainly because of the large numbers involved.
02 May 2025, 12:44
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Bunuel
Please refer to the fig below :
\((\frac{1}{2}*a)^2 + (\frac{1}{2}*b)^2 = 289\)
\(a^2 +b^2= 1156\) ... (i)
\(\frac{3}{4}a - \frac{3}{4}b = 10.5\)
\(a-b= 14\) ... (ii)
\( a^2 +b^2-2ab = 196 \)
\(1156 -196 = 2ab\)... from (i)
\(ab= 480 \)
Here we can see only one option satisfies:
Ans B
Hope it helped.
