Quant Boosters  Kamal Lohia  Set 4

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.

Q11) 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)!  1

Q12) 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.

Q13) 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 m

The 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'.

Q14) 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).

Q15) 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

Q16) 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 = 1/1004

Q17) Suppose real numbers x and y satisfy x^2 + 9y^2  4x + 6y + 4 = 0.what is the maximum value of 4x  9y?