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In a cricket match, the captain of the team played for 4 hours 28 minutes 13 second in the first innings and 3 hours 49 minutes 35 seconds in the second innings. Find out how much more time he played in the first innings.

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answer is B.

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Detailed Solution

1^{st}

inning time =

4 h rs 28 \min 13 \sec

2^{nd}

Innings time =

3 h rs 49 \min 35 \sec

So,

1^{st}

Innings time in seconds =

4

hrs

= 4 \times 60

= 240

mins

= 240 \times 60

= 14400

secs
Now,

28

minutes =

28 \times 60

= 1680

secs
So total time for first Innings

= 14400 + 1680 + 13

= 16093

secs
For,

2^{nd}

Innings time in seconds =

3

hrs

= 3 \times 60 \times 60

= 10800

secs
Now,

49

minutes =

49 \times 60

= 2940

secs
So total time for second Innings

= 10800 + 2940 + 35

= 13775

secs
Now we have to find out how much longer he played

1^{st}

Innings = Total

1^{st}

- Total

2^{nd}

Inning

= 16093 - 13775

= 2318

secs
Now, for example

1

min

= 60

sec

1^{st}

Innings

= 2318

secs

= \frac{2318}{60}

= 38.63 ≅ 39

mins

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