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Sunday, July 26, 2015

UPSC Maths Challenge : A Problem A Day! (Day 23)

I am feeling quite spirited today (not sure why) and I have posted six problems in this post instead of one! Hope that they help. All the problems are from the syllabus suggested by UPSC.





Today's Problem(25/07/2015)


Questions Credit : http://careerbless.com


1) A water tank is hemispherical below and cylindrical at the top. If the radius is 12m and capacity is 3312*pi cubic metre the height of the cylindrical portion in metres is?

2) If three metallic spheres of radii 6 cms, 8 cms and 10 cms are melted to from a single sphere, the diameter of the new sphere will be?

3) Sixty men can build a wall in 40 days, but though they begin the work together, 5 men quit every ten days. The time needed to build the wall is?

4) Shyam is travelling on his cycle and has calculated to reach point 'A' at 2 PM If he travels at 10 kmph. he will reach there at 12 noon if he travels at 15 kmph. At what speed must he travel to reach point 'A' at 1 PM?

5) An aeroplane travelling at 700 km/hr in level flight drops a bomb from a height of 1000 metres on a target. The time taken from releasing the bomb to hitting the target is nearest to the figure?

6) In a History examination, the average for the entire class was 80 marks. If 10% of the students scored 95 marks and 20% scored 90 marks. What was the average marks of the remaining students of the class?


1. Volume of a cylinder = πr^2h cubic metre = (144π x h) cubic metre
Volume of a hemisphere = 2/3πr^2h cubic metre = (144π x 8) cubic metre
Volume of the water tank = (πr^2h + 2/3πr^2h) cubic units = 144π (8 + h) cubic metre
According to the question,we have 144π (8 + h) = 3312π
(8 + h) = 23 or h = (23 - 8) metre = 15 metre 
So,the height of the cylinder is 15 metre.

2. Volume of sphere with radius 6 cm = 4/3πr^3 cm^3 = 288π cm^3
Volume of sphere with radius 8 cm = 4/3πr^3 cm^3 = 2048/3π cm^3
Volume of sphere with radius 10 cm = 4/3πr^3 cm^3 = 4000/3π cm^3

Volume of the new sphere formed by melting the other spheres = 4/3πr^3 cm^3  
Therefore,4/3πr^3  = π(288  + 2048/3 + 4000/3) = π(864/3 + 2048/3 + 4000/3) 
or 4r^3 = 6912 or r = 1728 or r = 12 cm.So, d = 2r = 2(12) cm = 24 cm.
Hence,the diameter of the sphere measures 24 cm.

3. Work done in one day by one person is equal to x/2400. [Considering x to be an arbitary variable].
Now,the number of workers that reduces in every 10 day interval is 5.
Therefore,the number of extra days required would be 
(10)[60 + 55 + 50 + 45]/2400 + (x/2400)(40) = 1
or 21/24 + x/60 = 1 or 52.5 + x = 60 or x = 7.5.
So,the total number of days required is (40 + 7.5) days = 47.5 days.
(Solution Credit to this particular problem: Dinesh,careerbless.com)

4. Since we do not know the at what time Shyam actually sets on his journey,we will consider any one of the time limits provided (either 12 p.m or 2 p.m) to be taking x hours from the time he started,that is we have consider either duration to be x (or y or z or whatever you wish to take) hours.Now, if we consider this for 2 p.m,then by time = distance/speed,we have
 x = d/10 ......Eqn 1.
Now,(x - 2) = d/15 or 15x - 30 = d or 15x = 30 + d or x = (d + 30)/15......Eqn 2.
From eqns 1 and 2,we have (d + 30)/15 = d/10 or 10d + 300 = 15d or 5d = 300 or d = 60 km.
Now, x -1 = d/10 - 1 or x - 1 = 60/10 - 1 = 5 [We are talking about 1 p.m and we have chosen the time duration of upto to p.m to be x,so upto 1 p.m,it is (x -1) hours]
So,the time taken upto 1 p.m is 5 hours.Therefore,the speed = distance/time = 60/5 km/hr
 = 12 km/hr

5.I hate the storyline of the problem.Anyhow,I am solving it just for the quantitative part.
The problem is absolutely the simplest one in the list.We only have to convert 700km/hr into either m/hr or m/s,which looks conventional.
Now,700km/hr = (700 x 1000)m/(60 x 60)s = 1750/9 m/s 
Now, speed x time = distance.Therefore,time = (1000)/(1750/9) = (1000 x 9)/1750 = 36/7 m/s
= 5.1428 m/s,which is approximately equal to 5.2 m/s.

6.Let the total number of students be 10.
Therefore,10/100 x 10 = 1,i.e 1 out of them scored 95.
Also,20/100 x 1 = 2,i.e 2 out of them scored 90.
Now,let x1 = 95,x2=x3=90.Again,(x1 + x2 + x3 + x4 +...... + x10) /10 = 80  [As given in the question]
So, x1 + x2 + x3 + x4 +...... + x10 = 800
or 95 + 90 + 90 + x4 +...... + x10 = 800
or x4 +...... + x10 = 800 - 275 = 525
or (x4 +...... + x10)/7 =  525/7 = 75
So,the average marks of the remaining students of the class is 75.
(This particular problem is partially inspired by the solution given by Munirasu,www.m4maths.com)



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