1.A:

To make visualization easy, it is convenient to conceptually open out the bark of the tree trunk and flatten it. The cylindrical surface will then be a rectangle.
It may be noted that:

Width of the rectangle = Circumference of the cylinder = 32 inches.

Height of the rectangle = Vertical distance on the cylinder = 60 inches (in one twist).

Using Pythagoras' Theorem for a right-angled triangle, Length of the hypotenuse = (322 + 602) 1/2 = 68 inches.

Now, the number of twists the creeper makes around the tree trunk is 8 (= 480 / 60). If the length of the creeper (as given by the hypotenuse) is 68 inches in one twist, then the total length of the creeper in 8 twists is 544 inches.

2.A:

If 15 teams participated, then the first team plays matches against the other 14 teams. The second team has already played against the first team, and so has to play matches against only the other 13 teams. In this manner, the second-last team has to play against only one team, and the last team has already played against all the teams. Thus, the total number of matches is

14 + 13 + ........ + 2 + 1 = 105.

If 105 matches are totally played, then 15 teams participated.

3.A:

In one hour,

the large inlet pipe fills 1 / 1 of the tank;
the small inlet pipe fills 1 / 5 of the tank;
the outlet pipe empties 1 / 6 of the tank;
and therefore all three pipes together fill [ (1 / 1) + (1 / 5) - (1 / 6) ] of the tank.

Fraction of the tank that will be filled in 0.97 hours = 0.97 [ (1 / 1) + (1 / 5) - (1 / 6) ] = 1.00.

Alternative Solution through Fundamental Equations:

It it important to note that Flow Rate = Volume / Time ... equation (1) Rate of Accumulation = Input Rate - Output Rate ... equation (2) Let V be the total volume of the tank.

From equation (1), Flow Rate (large inlet pipe) = V / 1 Flow Rate (small inlet pipe) = V / 5 Flow Rate (outlet pipe) = V / 6. Substituting in equation (2), Rate of Accumulation in tank = (V / 1) + (V / 5) - (V / 6).

Using the above result in equation (1), Time required to fill the complete tank = V / [ (V / 1) + (V / 5) - (V / 6) ].

Note that V cancels out on simplifying the above expression.

Fraction of the tank that will be filled in

0.97 hours = 0.97 [ (1 / 1) + (1 / 5) - (1 / 6) ] = 1.00.

4.A:

"Bodo weighs 150 pounds" is the correct answer.

His sister weighs 10 pounds. Thus, Bodo weighs 140 pounds more than his sister and their combined weight is 160 as per the problem requirements.

The common mistake is to quickly jump to the conclusion that "Bodo weighs 140 pounds and his sister weighs 20 pounds." But, Bodo would then erroneously weigh 120 pounds more than his sister.

5.A:

If there are 16 tangent lines, then there are 32 unbounded regions.

It is not very difficult to visualize that each tangent line drawn creates two unbounded regions.

6.A:

Total effort required by 8 painters for 30 days = 8 x 30 = 240 man-days.

Initial effort by 8 painters for 9 days = 8 x 9 = 72 man-days.

Job still requires 240 - 72 = 168 man-days, and there are 24 painters available to do it.

Additional effort required by 24 painters for 168 man-days = 168 / 24 = 7 days.

Thus, 7 more days are required to complete the painting job.

7.A:

Total marks in 6 subjects = 6 x 81 = 486.

Total marks in 2 English subjects = 2 x 69 = 138.

Total marks in remaining 4 subjects = 486 - 138 = 348.

Average marks in remaining 4 subjects = 348 / 4 = 87.

8.A:

On the first day, the snail climbs up 6 ft and slips down 4 ft while sleeping. So, next morning, it is 2 ft from where it started. The snail thus travels 2 ft upwards every day. Therefore, in 6 days, it has traveled a distance of 12 ft from the bottom.

Here lies the catch to the problem! On the last day, the snail travels 6 ft upwards and hence reaches the top of the wall in a total of 7 days.

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