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# [Smart Math] Time Speed Distance Problem 13

Here’s and example of a SMART MATH problem for TIME SPEED DISTANCE.

### Problem

A monkey climbs 30 meters of a pole during the day and slips down by 10 meters at night. Assuming that days and nights are equal, in how many days will the monkey scale a 120 meters high pole?

1. $5\frac{1}{3}$
2. $5\frac{2}{3}$
3. $6\frac{1}{3}$
4. $6\frac{2}{3}$
5. $7\frac{1}{2}$

### The Usual Method

During the course of the whole day and whole night the monkey will climb 30 – 10 = 20 meters.

To scale 120 meters pole, the monkey would take $\frac{120}{20}=6$days.

However, on the end of the 6th day, the monkey would have actually climbed 120 + 10 = 130 meters.

We can form a table where the day number and night number are given with the position of the monkey at the end of that period (day / night).

Day # aaa Position aaaaaaaa Night # aaaaaaaaaa Position

1                      30                                1                                  20

2                      50                                2                                  40

3                      70                                3                                  60

4                      90                                4                                  80

5                      110                              5                                  100

6                      130                              6                                  120

As can be seen from the table, that at the end of 5th day the monkey climbs 110 meters and thereafter slips in the night by 10 meters, to reach 100 meters. Now to reach 120 meters during the 6th day, the monkey needs $\frac{120-100}{30}=\frac{20}{30}=\frac{2}{3}$ of the 6th day. Since the complete day has equal duration of both day time and night time, the 2/3 of day time is equivalent to 1/3 of the whole day (Day time and night time together). Hence total duration needed to reach 120 meters = $5+\frac{1}{3}=5\frac{1}{3}$days.

(Ans: 1)

Estimated Time to arrive at the answer = 100 seconds.

### Using Technique

During the course of a complete day, the monkey climbs 30 – 10 = 20 meters, so to climb 120 meters, it will take $\frac{120}{20}=6$days. We know that since the monkey has slipped during the 6th night, the total complete days needed to reach 120 meters has to be less than 6 days. This eliminates all options from ‘3’ to ‘5’. This means that the answer can be $5\frac{1}{3}$ or $5\frac{2}{3}$days. You can simply check that in 5 complete days, the monkey climbs 5 x 20 = 100 meters. Since during day time, the monkey climbs 30 meters, so to climb the balance (120 – 100) 20 meters, $\frac{2}{3}$ of the 6th day time will be used. Since the complete day has equal duration of both day time and night time, the 2/3 of day time is equivalent to 1/3 of the whole day (Day time and night time together). Hence total duration needed to reach 120 meters = $5+\frac{1}{3}=5\frac{1}{3}$days.
(Note: Students commit mistakes in considering day time as whole day and get the answer as $5\frac{2}{3}$days in stead of $5\frac{1}{3}$days).