Time and Distance - 05

Explanation:-

$$\eqalign{ & {\text{Let}}\,{\text{the}}\,{\text{distance}}\,{\text{travelled}}\,{\text{on}}\,{\text{foot}}\,{\text{be}}\,x\,km \cr & {\text{Then,}}\,{\text{distance}}\,{\text{travelled}}\,{\text{on}}\,{\text{bicycle}} = \left( {61 - x} \right)km \cr & {\text{So}},\,\frac{x}{4} + \frac{{\left( {61 - x} \right)}}{9} = 9 \cr & \Rightarrow 9x + 4\left( {61 - x} \right) = 9 \times 36 \cr & \Rightarrow 5x = 80 \cr & \Rightarrow x = 16\,km \cr}$$

Explanation:-

Let length of each train be x meter.
Then, speed of 1^st train = ^x/₁₈ m/sec;
Speed of 2^nd train = ^x/₁₂ m/sec;

Now,
When both trains cross each other, time taken;
$$\eqalign{ & = \left[ {\frac{{2x}}{{\left\{ {\left( {\frac{x}{{18}}} \right) + \left( {\frac{x}{{12}}} \right)} \right\}}}} \right] \cr & = \frac{{2x}}{{\left[ {\frac{{\left( {2x + 3x} \right)}}{{36}}} \right]}} \cr & = \frac{{2x \times 36}}{{5x}} \cr & = \frac{{72}}{5} \cr & = 14.4\,{\text{seconds}} \cr}$$

Explanation:-

A →____________________← B
Ratio of the speed of trains is given by;
(Speed of train A/ Speed train of B) = ^√b/_√a
= ^√9/_√4
= ³/₂
= 3:2.

Explanation:-

Let the speed of the train be x km per hour.
Man is walking in same direction, so,
Relative speed becomes = (x-3) km/h.
In 33 secs train travels its own distance i.e. 300m.
Speed of train = ^{300 * 18}/_{33 * 5} km/hour;
Or, x-3 = 32.72 km/h;
Or, x = (32.72)+3
Or, x= 35.72 km/h.

Explanation:-

Since it climbs 7 meter in every 2 hr. At the end of 14 hours it would have climbed up to 49 mt. In 15th hour it will reach 61 mt. Then drops back to 56 mt in the 16th hour. It will take another 35 min to travel remaining 7 mt since it will be ascending in the 17th hour. Thus, 16 hr and 35 min.

Detail Explanation:

Given,
Ant ascends 12 meters in first hours and slips down 5 meters in second hour. This is repeated in every alternate hour. (12 - 5 = 7 meters)

Therefore, in ends of 14 hours,
Ant ascends = ^{(14 * 7)}/₂ = 49 meters.
So, In next hour i.e 15 hour, Ant will ascends another 12 meter.
Till end of 15^th hour, Ant climbed = 49 + 12 = 61 meters. But in 16^th hour, Ant will slips down 5 meters (61 - 5 = 56 meters).
Remaining pole to climb,
63 - 56 = 7

Explanation:-

1^st Method:
Man climbs 1 m and slips down 50 cm (0.5 m) in one minute
i.e. he climbs (1-0.5 = 0.5 m) in one minute.
But in the last minute he will be climbing 1 m as he gets on the top so no slip.
Time taken to climb 11 meter = [(¹⁰/_0.5) +1] = 21 minutes.

He climbs the wall at 5:21 pm.

2^nd Method (short-cut):

Reqd. time = [(ht. of pole - slipped distance) /(climbed distance - slipped distance)]*t;
$$\eqalign{ & = \left[ {\frac{{\left( {x - z} \right)}}{{y - z}}} \right] \times t \cr & {\text{Required}}\,{\text{time}}, \cr & = \left\{ {\frac{{\left( {11 - 0.5} \right)}}{{\left( {1 - 0.5} \right)}}} \right\} \times 1 \cr & = 21\,{\text{minutes}} \cr}$$

Explanation:-

Let the distance between school and home be x km.
The difference of time when A goes school to school with these two different speed is 10 min $$\eqalign{ & = \frac{{10}}{{60}}\,{\text{hour}} \cr & {\text{Now,}} \cr & \left( {\frac{x}{3}} \right) - \left( {\frac{x}{4}} \right) = \frac{{10}}{{60}} \cr & {\text{Or}},\,\frac{x}{{12}} = \frac{1}{6} \cr & {\text{Or}},\,x = \frac{{12}}{6} \cr & = 2\,{\text{km}} \cr}$$

Explanation:-

$$\eqalign{ & {\text{Speed of athlete}}, \cr & = \frac{{200\,{\text{m}}}}{{24\,{\text{secs}}}} \cr & = \frac{{200 \times 18}}{{5 \times 24}} \cr & = 30\,{\text{km/hour}} \cr}$$

Explanation:-

1 hour 45 minutes = 1+(⁴⁵/₆₀) = ⁷/₄ hours.
Speed of the motorboat up-stream,
= Distance /time taken
= ^{56 km}/_{(7/4) h} = ^{56 * 4}/₇ = 32 kmph
Let the speed of the current = x kmph
Hence, 36-x = 32
Or, x = 36-32 = 4 kmph
Speed of boat down the stream = 36 + 4 = 40 kmph.

Explanation:-

When A runs 1000 m, B runs 900 m.
Hence, when A runs 500 m, B runs 450 m.
Again, when B runs 400 m, C runs 360 m.
And, when B runs 450 m, C runs = 360 * ⁴⁵⁰/₄₀₀ = 405 m.
Required distance = 500 - 405 = 95 meter.
That means when A runs 500 meter then B can run 450m then C runs 405 m.