Quant with Kamal Lohia  Part 10

Q1) A stack of 2000 cards is labelled with the integers from 1 to 2000, with different integers on
different cards. The cards in the stack are not in numerical order. The top card is removed from the
stack and placed on the table, and the next card in the stack is moved to the bottom of the stack.
The new top card is removed from the stack and placed on the table, to the right of the card already
there, and the next card in the stack is moved to the bottom of the stack. This process  placing the
top card to the right of the cards already on the table and moving the next card in the stack to the
bottom of the stack  is repeated until all cards are on the table. It is found that, reading from left to
right, the labels on the cars are now in ascending order: 1, 2, 3 … 2000. In the original stack
of cards how many cards were above the cards labelled 1999?Just take it like this that all 2000 cards are standing in a circle and every alternate card is being eliminated to be placed on the table in a row starting from the first card. So we need to locate a card which is just before the eliminated according to above rule. Also here we need to find the secondlast eliminated one.
Now let the calculations begun. Shortest method seems to be the last one i.e. using odd numbers. So position of last card to be eliminated = S(2000) 1 = S(2^10 + 976)  1 = 2 x 976 = 1,952 and position of the second last card to be eliminated = 1952  1024 = 928. Hence card labelled as 1999 must be at 928th position. So there are 927 cards above it.
Q2) Find the last non zero digit of 100!
Formula : LNZ(10n!) = LNZ(4^n) x LNZ(2n!)
LNZ(100!) = LNZ(4^10) x LNZ(20!)
= LNZ (6 x 4) = 4Q3) Find last nonzero digit of 125!
Now see that 125 is not multiple of 10, hence we can apply the above pedagogy up to 120! And then multiply the result with last nonzero digit of product of remaining numbers i.e. 121 to 125.
So, Last nonzero digit of 125! = LNZ(125!) =LNZ(4^12) x LNZ(24!) x LNZ(121 x 122 x 123 x 124 x 125)
= LNZ(6 x 4^2) x LNZ(4!) x LNZ(21 x 22 x 23 x 24) x LNZ(121 x 61 x 123 x 31) 
= LNZ(6 x 4 x 4 x 3) = 8Huff! Looks lengthy, but believe me it's not. :)
Q4) If 16 oranges are distributed among 4 children such that each gets at least 3 oranges, the number of ways of distributing them is
a. 30
b. 210
c. 15
d. 35I have four oranges OOOO and three partitions  i.e. seven objects are to be arranged which will take exactly seven places for their placement. Now number of ways of arrangement of 4 identical oranges and 3 identical partitions (i.e. OOOO) is = 7!/(4!)(3!) = C(7, 3)
This is similar to finding the number of ways to select three places out of these seven where three identical partitions will be arranged in one way and at the remaining four places four identical oranges will be arranged in one way. So these ways of selection are = C(7, 3).
Q5) If all the first 96 multiples of 4 are arranged one after the other the penultimate(second to last in a series or sequence) digit in this sequence is
a. 2
b. 4
c. 8
d. 9You just need to find the ten's digit of 96th multiple of 4 which is = 8.
Q6) How many 6digit numbers contain exactly 4 different digits ?
a) 4536
b) 2,94,840
c) 1,91,520
d) noneLet the four different digits be a, b, c, d, then the number can be of two forms i.e. aabbcd or aaabcd.
So to calculate the total numbers; first we need to find the number of ways to select 4 digits (which is C(10, 4)); then number of ways to select the digits to be repeated in each case (which is C(4,2) and C(4,1) respectively); and then the number of ways of arrangement in both the cases (which is 6!/(2!)2 and 6!/3! respectively).
Hence the number of ways to form such 6digit numbers become = C(10, 4)[C(4, 2)6!/(2!)2 + C(4, 1)6!/3!] = 10 * 9 * 8 * 7 * 5 * 13 = 327600 .
But is this the final answer? .....certainly not! :D
See in the initial selection for the digits, we have used all 10 digits (including 0) for selecting the four distinct digits. That means in the above arrangements there will be some numbers which would be starting with zero. Now thinking alternately, first place of the number can be filled with any 10 digits equally. So removing the cases starting with zero, we are left with (9/10) of previously calculated result.
Thus the answer will be 9 * 9 * 8 * 7 * 5 * 13 = 294840.
Q7) In an examination the maximum marks for each of the 3 papers are 50 each.Maximum marks for the fourth paper are 100. Find the number of ways in which the candidate can score 60% marks in the aggregate?
It is simply: 103C3  352C3 = 110551.
Now let's see the explanation part.
Out of total 250 marks, you want to score 150 marks. So number of ways of scoring 150 is equivalent to not scoring the remaining 100 marks. So let's say we have 4 boxes a, b, c, d such that capacity of a, b, c is 50 each and that of d is 100. And all 4 boxes are filled to its capacity and we are to remove 100 balls out of it. Does it sound better? :)
i.e. we are trying to find the whole number solution of: a + b + c + d = 100 with the restriction that none of a, b, c can be more than 50. OK. Now without restriction, number of whole number solution of above equation is given by 103C3. We need to subtract the cases when any one of a, b, c is more than 50 (i.e. 51 + x say, where x is a whole number). Remember no more than one of a, b, c can be greater than 50. Also I have taken it as 51 + x because that's how I ensure that one of a, b, c is certainly more than 50. And which one of a, b, c can be determined in 3 ways.
Q8) A starts from a point P on a circular track ,at 6:00 am and runs around the track in clockwise direction. At 6:15 am ,B starts from the same point P and runs around the track in counter clockwise direction and meets A for the first time at 7:09 am.If B started at 6 am then he would have met A for the first time at 7 am.If both A & B start from P at 7:00 am and run in the opposite direction,then find the time at which they would meet at the starting point for the first time?
Let's say 'a' unit/min and 'b' unit/min be the speeds of A and B respectively.
Equating total distance around the circle from given two conditions we get:
69a + 54b = 60a + 60b
=> 9a = 6b or a:b ≡ 2:3
So they will be meeting at same starting point for the first time in their 5th meeting. And they are meeting for the first time after 1hr of starting. Thus the required time is 12 noon.Q9) When processing flower nectar into honeybees extract , a considerable amount of water gets reduced.How much flower nectar must be processed to yield 1 kg of honey , if nectar contains 50% water and the honey obtained from this nectar contains 15% water ?
Look for the honey quantity which is not changing as water is evaporating. Now because of evaporation, honey quantity is not changing only its percentage composition is changing.
So we want to get 1kg of honey that contains 85% honey extract i.e. 850g.
So this 850g of honey extract which amounts to 85% of 1 kg honey should be 50% of total flower nectar, i.e. flower nectar must be 1700g i.e. 1.7kg.Q10) There are 25 horses each of which runs at constant speed different from other horses.Since the track has only 5 lanes , each race can have atmost 5 horses. If you need to find out 3 fastest horses , what is the minimum number of races needed to identify them?
7 races are sufficient.
In first five races, we get five top scorers of whom a horse is going to be topper which can be found out in 6th race.
Now as we want to find top three scorers, it depends on top three scorers of all the first five races. So we can eliminate bottom two of each i.e. 10 horses from the race.
Also who scored fourth and fifth place in sixth race, they along with their predecessors in their earlier races can be removed safely i.e. 6 more removed. Also who scored second and third in sixth race, we can remove one and two persons behind them in their earlier race respectively i.e. 3 more removed. The horse who topped sixth race is topper overall, so leave him also.
So finally we are left with 25  10  6  3  1 = 5 horses which can participate in seventh race and decide the second and third scorer of all the 25 horses.