Quant with Kamal Lohia  Part 11

Q1) Find the sum of the series 1(1!) + 2 (2!) + ... + n(n!)
1(1!) + 2(2!) + 3(3!) + ... + n(n!)
= {2  1}(1!) + {3  1}(2!) + {4  1}(3!) + ... + {(n + 1)  1}(n!)
= {2(!)  1(1!)} + {3(2!)  1(2!)} + {4(3!)  1(3!)} + ... + {(n + 1)(n!)  1(n!)}
= {2!  1!} + {3!  2!} + {4!  3!} + ... + {(n + 1)!  n!}
= (n + 1)!  1Q2) S is a set of three numbers whose sum and HCF are 168 and 8 respectively. How many values are possible for S if at least one of the numbers is not more than 24?
Let the three numbers be a, b, c such that a = hx, b = hy and c = hz where HCF (a, , b, c) = h and HCF (x, y, z) = 1
Now it is given that h = 8 and a + b + c = 168.
So 8x + 8y + 8z = 168
i.e. x + y + z = 21Further it is given that at least one of a, b, c is not more than 24 i.e. one of x, y, z is not more than 3.
Let x be that number, whose value can not be more than 3.
Now just make the cases:
(i) If x = 1, then y + z = 20 i.e. 10 cases {(1, 1, 19), (1, 2, 18), ..., (1, 10, 10)}
(ii) If x = 2, then y + z = 19 i.e. 8 cases {(2, 2, 17), (2, 3, 16), ...., (2, 8, 11) (2, 9, 10)}
(iii) If x = 3, then y + z = 18 i.e. 4 cases {(3, 4, 14), (3, 5, 13), (3, 7, 11), (3, 8, 10)}Thus exactly 10 + 8 + 4 = 22 sets of three such numbers are possible.
Remember that we are not to considered the ordered pairs here (i.e. 1, 2, 18 and 2, 1, 18 are to be taken as same sets) and further HCF (x, y, z) also need to be 1.Two boats start towards each other, from the two points exactly opposite of each other on the opposite banks of a river, simultaneously. They meet at a distance of 410 m from one of the banks and continue sailing further till they reach the opposite banks. They take rest for 1 hr each and start off the return journey taking the same route. Now they meet at a distance of 230 m from the other bank. Find the distance between the two banks. (Assume that river water is still.)
a. 750 m
b. 840 m
c. 1100 m
d. 1000 mThe distance between the two banks is = 3(410)  230 = 1000m. (option d)
Just observe that the moment the two boats meet for the first time, sum of distance travelled by the two boats is 'd' {if I assume the distance between two banks to be 'd'}.
Now next time when they meet, sum of the distance travelled by the two boats together becomes '3d' as both boats have reached to the opposite ends (i.e. travelled a distance of 'd' individually) and then turned back to meet at a point.
Clearly the ratio of time taken to meet for the first time and that to meet for second time is in the ratio of distances travelled i.e. 1 : 3. Also the distances travelled by any individual boat is also in the same ratio 1 : 3 as they also have travelled the distances for same time.
So the boat which had travelled 410m till first meeting, has travelled (d + 230)m for the second meeting. And that's how we get the above value for 'd'.Q3) How many real numbers are solutions to the equation x^4 + 4x = 10?
x^4 + 4x = 10
i.e. x^4 = 10  4x
i.e. we just need to find the number of intersections of y = x^4 and y = 10  4x which is simply 2 as first curve draws a steeper parabola closer to Y=axis and the second curve draws an inverted V having a vertex at (0, 10).Q4) if 2^(2^x) + 4^(2^x) = 56 then what is the value of 2^2^2^x?
2^(2^x) + 4^(2^x) = 56
Let 2^(2^x) = tSo given equation becomes t + t^2 = 56
i.e. (t  7)(t + 8 ) = 0
i.e. t = 7 as t is positive.Now we need to calculate 2^t = 2^7 = 128
Q5) let f(x) be a polynomial of degree 2006 satisfying f(k)=1/k, 1 < = k < = 2007
what is the value of f(2008)?It is given that f(k) = 1/k for some particular values of x = k.
So for those values of x = k, we have k * f(k) = 1 i.e. k * f(k)  1 = 0
i.e. for some particular values of x, we get x * f(x)  1 as zero. And this x * f(x)  1 is also a polynomial of one higher degree than that of f(x). And the values of x at which this new polynomial becomes zero are its roots.Let xf(x)  1 be a new polynomial of degree 1 more than that of f(x) i.e. 2007. So it'll have 2007 roots and all of those are known to us as 1, 2, 3, ...., 2007.
So we get that xf(x)  1 = A(x  1)(x  2)(x  3).....(x  2007)
Putting x = 0, we get that 1 = A(1)(2)(3)...(2007)
i.e. 1 = A(2007!)
i.e. A = 1/2007!Thus xf(x)  1 = [(x  1)(x  2)(x  3).....(x  2007)]/2007!
and 2008f(2008)  1 = [(2008  1)(2008  2)(2008  3).....(2008  2007)]/2007! = 1
i.e. f(2008) = 2/2008 = 1/1004Q6) Suppose real numbers x and y satisfy x^2 + 9y^2  4x + 6y + 4 = 0.what is the maximum value of 4x  9y?
x^2 + 9y^2  4x + 6y + 4 = 0
i.e. {x^2  2(x)(2) + 2^2} + {(3y)^2 + 2(3y)(1) + 1} = 1
i.e. (x  2)^2 + (3y + 1)^2 = 1Now we want to maximise 4x  9y. Let 4x  9y = k, i.e. x = (k + 9y)/4
So eliminating x, above equation becomes: (k + 9y  8 )^2 + 16(3y + 1)^2 = 16
i.e. 225y^2 + {18(k  8 ) + 96}y + (k  8 )^2 = 0
i.e. 225y^2 + (18k  48)y + (k  8 )^2 = 0Now as y is a real number, so discriminant of the above equation should be greater than or equal to zero.
So, (18k  48)^2  4(225)(k  8 )^2 ≥ 0
i.e. (18k  48)^2  (30k  240)^2 ≥ 0
i.e. (3k  8 )^2  (5k  40)^2 ≥ 0
i.e. (2k + 32)(8k  48) ≥ 0
i.e. (k  6)(k  16) ≤ 0
i.e. 6 ≤ k ≤ 16.Thus required maximum value of 'k' is 16.
Alternate approach is geometric one
We know that (x  2)^2 + (3y + 1)^2 = 1 and we need to maximise 4x  9y.
Now, let's replace 3y by Y, so that know equation turns into a circle
(x  2)^2 + (Y + 1)^2 = 1 with center at (2, 1) and radius 1.
And now we need to find the maximum value of 4x  3Y which will be equal to 4x  3Y = k (say) and will be achieved when this line is tangential to the circle.Rest part is simple. I hope you know how to find the perpendicular distance of a point from a line.
Using the relation for the distance, we get that {4(2)  3(1)  k}/√(4^2 + 3^3) = 1
i.e. 11  k = 5
i.e. k = 6 or 16.Thus maximum value of k = 16 as found earlier too.
Q7) Find the remainder when 12437681243768.... upto 700 digits is divided by 101?
Say our number is N = abcd = ab*10^2 + cd
Now we know that 10^2 when divided by 101 (i.e. 10^2 + 1) gives remainder 1.
So N = {ab(1) + cd} mod 101 = {ab  cd} mod 101If you had another group of 2digits before ab, then that'd have been multiple of 10^4 which can be written as 10^2 * 10^2 = (1)(1) mod 101 = 1 mod 101. So that group would have been positive. Hence the alternate +/ pattern comes in picture here.
Now to enlarge the concept for this question, you can easily observe that in given number the pattern of digits repeat after every 7 digits. So if I consider group of 14 digits collectively, then each group would be multiple of powers of 10^14..i.e.
N = 12437681243768 * (10^14)^49 + 12437681243768 * (10^14)^48 + 12437681243768 * (10^14)^47 + ... + 12437681243768 * (10^14)^1 + 12437681243768
Now just observe that as 10^2 = 1 mod 101, so 10^14 = (10^2)^7 = (1)^7 mod 101 = 1 mod 101.
Thus again we are going to get alternate +/ groups of these 14 digits, while dividing with 101, with first group from right most side being positive.
As we have exactly 50 groups of 14 digits each here and half of which are positive and half negative. Thus required remainder is simply zero as all the groups will be cancelled.
Q8) A dishonest shopkeeper cheats both his supplier and the customer. He cheats his supplier by 25% and his customer by 40% and claims to sell his articles at the cost price. Find his total profit.
a. 186%
b. 147.3%
c. 132.4%
d. 124%
e. 108.3%There are various ways to solve this.
See when the shopkeeper is cheating the supplier by 25%, that means he is paying for 100 articles and he is taking 125 by means of cheating (which directly means that he is making a profit of 25%, if he doesn't indulge in any further cheating while selling). Now when he cheats the customer by 40%, that means he is charging for 100 articles but actually giving him 40% lesser i.e. 60 articles only (which means he is making a profit of 66.66% profit, if he is not indulged in any other fardulent activity).
Let his CP be 100, but because of cheating he purchases goods worth 125. Right?
Now while selling, he sells all goods worth 125, but he shows them as worth (5/3)*125 = 208.3
Clearly his total CP = 100, total SO = 208.3 and thus profit% = 108.3% smileAlternately, you can make use of Successive change formula also (i.e. a% + b% successive changes yield a net change of {a + b + ab/100}%)
So here we have two successive profit percentages involved, first is 25% while buying and 66.66% while selling.
Thus net profit made = (25 + 66.66 + 25*66.66/100)% = 108.3% smileHere is a third way too, involving equal quantity's buy and sell.
Let each article costs 1 Re and he purchases 100 articles for Rs 100. But by cheating he takes 25 more articles. So total articles in his inventory is 125.
Now while selling, he sells all the 100 articles at same as cost rice for Rs 100 but because of cheating, he gives 40% lesser articles i.e. delivers only 60 articles saving 40 for himself as more proft.
In all, what he paid he got back by making the sale. And he saved 25 + 40 = 65 articles for himself. (Mind that he sold only 60 articles).
So profit earned is = (65/60)*100 = 108.3%Q9) A train stops at exactly six intermediate stations  A, B, C, D, E and F  in that order, between its originating station and destination station. At each of the intermediate stations, twice as many people get in as those that get down. The number of people getting down at the intermediate stations are all prime numbers, one each between 0 and 10, 10 and 20 and so on upto between 50 and 60, in the order of the stations given above. The difference between the number of people getting in at any two consecutive intermediate stations is at least 20. The total number of passengers getting down in all the intermediate stations together is an even number. Also no person gets in and gets down (or vice versa) at the same station. Which of the following cannot be the number of people getting down at any intermediate station?
a) 47
b) 3
c) 17
d) 23We are to just pick 6 prime numbers, one each from the given ranges (i.e. 0 to 10, 11 to 20, 21 to 30, 31 to 40, 41 to 50 and 51 to 60) such that difference between two consecutively selected prime numbers is at least 10.
Also as sum of all 6 prime numbers is even, so each of six prime numbers must be ODD.
It can be easily checked that selected prime numbers can be: 31323374759 only.
Thus required answer is (c) 17.Q 10) The sum of 20 distinct numbers is 801. What is their minimum LCM possible?
a. 360
b. 840
c. 480
d. 42We are looking for the smallest number which have 20 or more divisors (as all of the divisors of a number are distinct and also their LCM is same as the number itself) such that sum of 20 of its divisors is 801. Check the options. Correct answer is 360