## Saturday, 27 October 2012

### Jogging & Math: "Jogging – Circular path"

One morning while I was jogging around my apartment complex, I got a doubt. Let us suppose that I am jogging in a circular path and another person is also jogging in a parallel circular path (with a different radius). What is the difference between the distances travelled by me and the other person? Even if the difference between the radii is small, the difference between total distances travelled is more. why so?  what is the underlying concept here?

Let us consider that person ‘A’ is jogging in outer circle of radius ‘R’ and ‘B’ is jogging in an inner circle of radius ‘r’. Obviously R > r. The distance travelled by A in one round is nothing but the length of circumference of the circular path he travelled.

The distance travelled by A = 2πR
Similarly, the distance travelled by B = 2πr
Difference of the distances = 2π(R-r).

The difference of distances travelled in one round is multiplied by a factor ‘2π’ to the difference of radii of the two circular paths. If we consider the approximate value of ‘π’ as 3.14, the multiplying factor becomes ‘6.28’. This means, if B is jogging at 2 ft lesser radial length to that of A, the total distance travelled by B in one round is around 12 ft less than that of A. This is only for one round. If we consider for 10 rounds, the difference becomes 120 ft.

## Monday, 22 October 2012

### CAT-2012 My Experience

For the benefit of this blog, to be equipped with the experience of real time testing environment, I have decided to give CAT every year from this time onwards. My last attempt at CAT was in 2008 and that was the last PBT conducted by IIMs. From 2009 onwards, it has been conducted computer based. The very difference between PBT and CBT is that, in PBT we can have our own section-wise timelines. If you are strong in  Quant and if you feel that you have already cleared the cut-off in Verbal section, then you can give some extra time to review Quant section and get some points there. But here in CBT, there are clear time lines defined for each section. We have 70 min time for each of QA-DI and VA-LR sections. We can't have 50 min here and 90 min there.

Date 21/10/2012
Time slot 10.00 AM
Candidates shall reach the venue 1.30 hrs before the allotted time slot. I reached by 8.15 AM. Initially finger prints of point-fingers of both hands and live-photo snap (half-size) are taken. Seat number is allotted. 42-systems are arranged in each of the three test-rooms of the Centre. I occupied the allotted seat at around 9.00 AM. Screen appeared with login/password prompt.
Test admin has entered login and password at 10 AM. Two options are provided on screen: Choosing a tutorial of 15 min duration or going straight-away to test. As most do, I opted for tutorial. Even if we know the procedures thoroughly, it is better to go through the tutorial for knowing about any surprise elements if exist.

Tutorial is a comprehensive document on test structure, screens, time duration etc., It explained how to go back and forth of the pages, where to mark for review, how to check the total Q's attempted/mark for review/ left unanswered etc.. It explained the use of scroll bar in browsing through questions of a multiple question-set. It explained about the prompts such as test/section completion prompt, 1-minute pre-completion indication prompt. One point to be noted is: What happens if you answer a question and mark it for review and in hurry it was left like that (forgot to remove "review" tag)?  Need not worry, it will be considered while calculating score.
In simple words the summary of the tutorial is: we can go back to any question and answer it or we can change the answer of any attempted question in a section before the closure of the time allotted for that section.

Section-I: Quantitative Ability and Data Interpretation 30 Q's 70 min
Section-II: Verbal Ability and Logical Reasoning 30 Q's 70 min

Section-I:

Difficulty level:
DI medium to tough
QA easy to medium

Section-II:

Difficulty level:
LR medium to tough
VA easy to medium

As per the rule, we should not reveal the questions given in the examination and the same commitment is taken online before proceeding to the first screen of the test.

Given the rules, I could not go beyond this point. This may not be comprehensive but I thought it may be a little bit helpful for the guys taking the exam... All The Best...

## Saturday, 6 October 2012

### Puzzle-34

Four sisters – Suvarna,Tara,Uma and Vibha are playing a game such that the looser doubles the money of each of the other players from her share.They played four games and each sister lost one game in alphabetical order. At the end of fourth game each sister had Rs.32. How much money did Suvarna start with?
(i)Rs. 60(ii) Rs. 34(iii) Rs. 66   (iv) Rs. 28
Solution:

## Friday, 5 October 2012

### Detailed solutions for competitive exam for ACIO Grade-II in IB

Assistant Central Intelligent Officer Grade-II (Exec) in Intelligence Beuro was conducted on 23.09.2012. Detailed solutions for the math and logic related problems are provided here:
1.                  Find the last two digits of : 15*37*63*51*97*17
(i)35     (ii)45    (iii)55   (iv)85
Solution:
Taking two numbers at a time and finding last two digits of their product by criss-cross method.
Last two digits of 15*37=> 7*5=35, take 5 as units digit. 7*1=7 add this to 3 to get 10, consider 0 here and add it to unit digit of 3*5=15 ie., 5 to get 5 as tens digit ===> 55 last two digits
Last two digits of 63*51 ===> 13
Last two digits of 97*17 ===> 49
Now we have come down to three numbers: 55,13,49
Now take 55 and 13 at a time to get the last two digits of their product. The result is : 15
Now we have come down to two numbers: 15,49 and the final result is :35

2.                  A man decides to travel 80 kilometers in 8 hours partly by foot and partly on a bicycle. If his speed on foot is 8 km/hr and on bicycle 16 km/hr, what distance would he travel on foot?
(i)20     (ii)30    (iii)48   (iv)60
Solution:
Let the distance travelled by foot be f and that by bicycle be b => f+b = 80 –--(1)
Time taken for foot ride = f/8, as the speed is given as 8 km/hr
Time taken for bicycle ride = b/16, as the speed is given as 16 km/hr
Total time taken = f/8 + b/16 = 8  –--(2)
Solving (1) and (2), we get f = 48

3.                  Due to 25% increase in the price of rice of rice per kilogram, a person is able to purchase 20 kilograms less for Rs400. What is the increased price of rice per kilogram?
(i)Rs 5  (ii) Rs 6            (iii) Rs 10         (iv) Rs 4
Solution:
Arithmetic Method:
“25% increase in price” => Let x be the increased price per kilogram. As the price increase is 25%, for an original price of 4x there is an increase of x per kilogram.
“a person is able to purchase 20 kilograms less for Rs400” => for a Rs 400 worth rice there is a price hike of 20x.
Now compare the two statements:
For 400 he is loosing 20x due to price hike
For 4x there is an increase of x => for 5x he is loosing x
By proportion, 5x/400 = x/20x => x = 4
Algebraic Method:
Let x be original price per kilogram. With an increase of 25%, the new price per kilogram is 1.25x.
For Rs400, he could have purchased 400/x for the original price but he can purchase 400/1.25x now. The difference is 20 kilogram. => 400/x – 400/1.25x = 20
=> 1/x(1-1/1.25) = 20/400=1/20 =>1/x(1-4/5)=1/20 => x = 4

4.                  There is a 7-digit telephone number with all different digits. If the digit at extreme right and extreme left are 5 and 6 respectively, find how many such telephone numbers are possible?
(i)120   (ii) 100000       (iii) 6720          (iv) 30240
Solution:
There are 5 more digits are to be selected and arranged from 8 digits (0,1,2,3,4,7,8,9).
As this involves selection and arrangement, it is a nPr problem but not nCr.
8P5 = 8!/3! = 8*7*6*5*4 = 6720

5.                  A speaks the truth 3 out of 4 times, and B 5 out of 6 times. What is the probability that they will contradict each other in stating the same fact?
(i)2/3   (ii) 1/3 (iii) 5/6            (iv) 1/21
Solution:
Probability that A speaks truth = A = 3/4
Probability that A speaks false = A’ = 1-3/4 = 1/4
Probability that B speaks truth = B = 5/6
Probability that B speaks false = B’ = 1-5/6 = 1/6
Probability that they will contradict each other in stating the same fact = AB’+A’B
=3/4 * 1/6 + 1/4*5/6 = 8/24 = 1/3

6.                  A circle is inscribed inside an equilateral triangle touching all the three sides. If the radius of the circle is 2cm, then find the area of the triangle?
(i)15√3 (ii) 18√3           (iii) 12√2          (iv) 12√3
Solution:
It is an incircle. For an equilateral triangle incentre and centroid coincide. Centroid divides the median in the ration 2:1. AG:GD = 2:1. Given inradius is 2, which is nothing but GD.
GD = 2 => AG = 4 => AD = AG+GD = 6.
For an equilateral triangle height h = √3/2 a , where a is side
=> a = 2h/√3 = 12/√3
Area of triangle = ½*a*h = ½*12/√3*6 = 12√3

7.                  Three bells chime at intervals of 48,60 and 90 minutes respectively. If all the three bells chime together at 10 AM, at what time will all the three chime again that day?
(i)1 PM            (ii) 2 PM          (iii) 8 PM         (iv) 10 PM
Solution:
This is an LCM problem. The three bells chime together in a regularly interval of LCM of 48,60,90 ie., 720 minutes=12 Hrs. So after 10 AM, the next term will be at 10PM.

8.                  After striking the floor, a ball rebounds to 4/5th of height from which it has fallen. Find the total distance that it travels before coming to rest if it has been gently dropped from a height of 120 meters?
(i)540(ii) 960   (iii) 1080          (iv) 1120
Solution:
After the drop, it travels down for h meters and rebounds back to 4h/5 and travels down 4h/5 and rebounds back 4/5(4h/5) etc...
The total distance it travels = h + 4h/5+ 4h/5+ 4/5(4h/5)+ 4/5(4h/5) + 4/5*4/5(4h/5)+...
S= h + 2*4h/5 + 2(4/5)(4h/5) + ....
= h + 2{4h/5 + 4/5(4h/5) + ....}
= h + 2(Sum of infinite GP with initial term 4h/5 and common ratio 4/5)
Formula: Sum of Infinite GP with initial term ‘a’ and common ratio ‘r’ is a/(1-r)
S = h + 2{(4h/5)/(1-4/5)}
= h + 8h = 9h = 9*120 = 1080
Note: two things required to solve this problem are,
first- deciding how much distances it travels before and after each rebound
second- formula for infinite GP.

9.                  A toy weighing 24 grams of an alloy of two metals is worth Rs174/-, but if the weights of the two metals be interchanged, the toy would be worth Rs 162/-. If the price of one metal be Rs 8 per gram, find the price of other metal used to make the toy?
(i)Rs 10/gram(ii) Rs 6/gram   (iii) Rs 4/gram (iv) Rs 5/gram
Solution:
Logic:
If the weight of one metal be A and per-gram-price be x and the weight of other metal be B and per-gram-price by y, then the total cost is Ax+By. If the weights are interchanged, the total cost becomes Ay+Bx. So here the point is if we add up both these expressions, it becomes Ax+By+Ay+Bx and be factored as (A+B)(x+y). As the total weight is given, it can be substituted in A+B and the problem can be easily solved.
Now let us do the problem:
The per-gram price of one metal is given as 8. So let x=8.
Total weight of alloy = A+B = 24
8A+By = 174;
yA+8B = 162;
add up the two equations=> 8A+By+yA+8B = 336 => 8(A+B)+y(A+B) = 336
=> (8+y)(A+B) = 336 => 8+y = 336/24 = 14 => y = 6

10.              Indira is three times older than Yogesh while Zaheer is half the age of Wahida. If Yogesh is older than Zaheer, then which of the following statements can be inferred?
(i)Yogesh is older than Wahida(ii) Indira is older than Wahida
(iii) Indira may be younger than Wahida       (iv) None of the above
Solution:
I = 3Y => I > Y;
Z = W/2 => W > Z;
Y > Z
One thing surely we can infer here is: I > Z. (as I > Y and Y > Z)
I > Z and W > Z: from these two statements, we can’t infer about relation b/w I and W.
W > Z and Y > Z: from these two statements, we can’t infer about relation b/w Y and
None of the given statements can be inferred. Answer is (iv)

11.              Four sisters – Suvarna,Tara,Uma and Vibha are playing a game such that the looser doubles the money of each of the of the other players from her share.They played four games and each sister lost one game in alphabetical order. At the end of fourth game each sister had Rs.32. How much money did Suvarna start with?
(i)Rs. 60(ii) Rs. 34(iii) Rs. 66   (iv) Rs. 28
Solution:
I like this problem. It is based on logic.

At the end of four games, each had Rs.32 => total amount with all the four is 4*32 = 128

“The point is even if some money changes hands with the outcome of each game, the total money (Rs128) with all the four does not vary at any point”.

As the games were lost by the four sisters in alphabetical order, Suvarna lost the first game and after that in the remaining three games her money got doubled three times ie., “The point here is she lost only in the first game and from this point onwards, her money got multiplied 8 times after completion of the remaining 3 games”.

Let the money that Suvarna initially had with be ‘X’. So the total amount with all the other three sisters is 128-X before starting the first game. After first game, in which Suvarna lost, Suvarna must be left with X-(128-X) as she needed to give (128-X) to make double the amounts with the other three.
 Suvarna's money Total Money with other three At the start X 128-X After 1st game X-(128-X) = 2X-128 2(28-X) After 2nd game 2(2X-128) Not required to find After 3rd game 4(2X-128) Not required to find After 4th game 8(2X-128) Not required to find

Finally the amount left with Suvarna is 8(2X-128). But it is given that at the end of four games, each had Rs.32. It means Suvarna ended up with Rs 32.
=> 8(2X-128) = 32 => X = 66
12.              It was Saturday on 17th December, 1982. What will be the day on 22nd December, 1984?
(i)Monday(ii) Tuesday(iii) Wednesday           (iv) Sunday
Solution:
Here we have to check for any leap years. 1984 is a leap year.
From 17th December, 1982 to 16th December,1983 => 365 days
From 17th December, 1983 to 16th December,1984 => 366 days as Feb’1984 has 29 days instead of 28
From 17th December, 1984 to 22nd December,1984 => 6 days
22nd Dec’1984 is (365+366+6)h day counting from 17th Dec’1982 => 737th day
As  on week consists of 7 days, we want remainder when 737 is divided by 7 => it is 2
=> that means 22nd Dec’1984 is 2nd  day ahead of some weeks of gap counting from 17th Dec’1982
=> if 17th Dec’1982 is Saturday, 21st Dec’1984 is Sunday and 22nd Dec’1984 is Monday.
13.              One term in the following number series is not correct. 15,16,22,29,45,70
Find out the wrong term?
(i)16(ii) 22(iii) 45         (iv) 70
Solution:
Find out the consecutive differences:
=> 1,6,7,16,25
It seems like series of squares of consecutive numbers excepting 6 and 7. The difference 6 is from 22-16 and the difference 7 is from 29-22. Had it been 20 instead of 22 in the given series of numbers, the differences will be in the following way:
1,4,9,16,25. This is perfect. So the wrong number is 22.
14.              A man is facing west. He turns 450 in the clockwise direction and another 1800 in the same direction and then 2700 in the anti-clockwise direction. Which direction is he facing now?
(i)South-West(ii) South(iii) North-West          (iv) West
Solution:
Let us represent clock-wise as negative and anti-clockwise as positive.
The total angle turns out to be = -45-180+270 = +450
Originally he is facing West and the final turn-out angle is 450 positive ie., 450 anti-clockwise => 450 towards South => the final position is : he is looking towards South-West.