Quant Boosters  Nitin Gupta  Set 2

Number of Questions  30
Topic  Quant Mixed Bag
Solved ?  Yes
Source  Selected questions/solutions from Nitin Gupta Sir (AlphaNumeric)

Q1) Consider a set P={1,2,3...,11,12} of natural numbers.We define another set Q such that it contains no more than one out of any three consecutive natural numbers.How many subsets Q of P including the empty set are possible?
(a) 114
(b) 117
(c) 129
(d) 136
(e) 130

Method 1
Subsets with 0 element = 1.
Subsets with 1 element = 12.
Subsets with 2 element = 10C2 = 45.
Subsets with 3 element = 8C3 = 56.
Subsets with 4 element = 6C4 = 15
So total subsets = 1 + 12 + 45 + 56 + 15 = 129.Method 2
Let f(n) be number of subsets of {1,2, .....n} which contain no more than one out of any three consecutive natural numbers.
then f(n) = f(n1) + f(n3)
Logic is to include f(n1) subsets and add n to all the subsets for f(n3) to get all possible subsets for n
with f(1) =2, f(2) = 3 , f(3) =4.
solving we get f(12)=129.Hence, choice (c) is the right answer.

Q2) The cost of 20 oranges and 1 kg of apple is Rs 60 while their selling price is Rs 72. The cost of 10 apples and 1 kg of orange is Rs 50 while their selling price is Rs 60. Given that the profit percent on the sale of two fruits are different, then the sum of the selling price of 5 oranges and 3 apples and the costprice of 6 kg oranges and 5 kg apples (is)
(a) can not be determined
(b) Rs 318
(c) Rs 375
(d) Rs 384
(e) none of the foregoing

20 Oranges and 1KG of apple = 12 Apples and 1.2 KG of Oranges (50*1.2)
If the profit %ages are different, then it must be clear that the ratio of apples : Oranges in the 60Rs. sale and the 72Rs. sale must be the same.
Hence, we can safely say that 1.2 KG of oranges = 20 Oranges and 1KG of apples = 12 apples.
SP of 20 Oranges and 12 Apples = 72 => SP of 5 Oranges and 3 apples = 18
CP of 1.2KG of Oranges and 1KG of Apples = 60 => CP of 6KG Oranges and 6KG apples = 300Hence, choice (b) is the correct option

Q3) A circle passes through the vertex C of rectangle ABCD and touches its sides AB and AD at P and Q respectively. If the distance from C to the line segment PQ is equal to 4 units, then the area of the rectangle ABCD in sq. units (is)
(a) 20
(b) can not be determined
(c) 16
(d) greater than 20
(e) none of the foregoing

Method 1
Let M be the feet of perpendicular from C to PQ. Now, by alternate segment theorem we have
=> < CQM = < CPB and < CPM = < CQD. Thus, right triangles CQM and CPB are similar and also CPM and CQD. Thus, CQ/CP = CM/CB and CP/CQ = CM/CD. Thus, CB * CD = CM^2 = 16.Method 2
Assume the vertices of A, B, C, D as (0, b), (a, b), (a, 0) and (0,0).
Since, AP = AQ, let AQ = m. Thus P = (0, bm) and Q = (m, b).
Now x = 0 is tangent to the circle at P.
The eq. of circle is x^2 + (y  b+m)^2 + kx = 0, here k is variable. But, it's also tangent at Q = (m, b).
Thus we get k = 2m.
Now, the circle x^2 + (yb+m)^2 2mx passes through (a, 0) => a^2 + (bm)^2 = 2am.
Equation of PQ is xy = bm.
Also (am+b)/v2 = 4.
Squaring and using the relation we get 2ab = 32 => ab = 16.
Hence, choice (c) is the correct option

Q4) Given p < = 4 is a positive real . Let A be the area of the bounded region enclosed by the curves y = 1  1  x and y = 2x  p. Then which among the following best describes A?
(a) 0 < A < = 1/2
(b) 1/6 < = A < = 1/4
(c) 0 < A < = 1/3
(d) 0 < A < 2/3
(e) 1/6 < = A < = 1/3

Case 1: 0 < p < = 1
The area formed is of a triangle with vertices (p/3, p/3), (p/2, 0) and (p, p).
Thus, the area is p^2/6 sq. units.
Case 2: 1 < = p < = 3
The figures formed is a quadrilateral (p/3 ,p/3),(p/2,0), ((p+2)/3 , (4p)/3) and (1,1).
Thus, the area formed is 1/3  1/6*(p2)^2
Case 3: 3 < = p < = 4
The area formed is the image of case 1. Area = (4p)^2/6
Thus, min(A) = 0 at p = 4 and max(A) = 1/3 at p = 2
Hence, option (c) is the right answer

Q5) If x,y,z are positive real numbers representing the sides of a triangle,then (x^2+y^2+z^2)/(xy+yz+xz) is necessarily less than which of the following?
(a) 1.80
(b) 2.0
(c) 1.85
(d) 1.95
(e) None of the foregoing

As x, y and z are sides of the triangle hence,
x  y < z (1)
z  x < y (2)
y  z < x (3)
squaring and adding all this three equation we get:
(xy)^2 + (zx)^2 + (yz)^2 < x^2 + y^2 + z^2
=> 2 * (x^2 + y^2 + z^2)  2 * (xy + yz + zx) < x^2 + y^2 + z^2
=> x^2 + y^2 + z^2 < 2 * (xy + yz + zx)
=>(x^2 + y^2 + z^2)/(xy + yz + xz) < 2
Hence, choice (b) is the right option

Q6) If x, y, z are natural numbers such that y ≠ 1 and (xy√3)/(yz√3) is a rational number. Then which among the following is always true?
(a) x^2 + 4z^2 is composite
(b) x^2 + y^2 + z^2 is nonprime
(c) x^2 + 4z^2 is prime
(d) both (a) and (b)
(e) none of the foregoing

(xy√3)/(yz√3) = (xy√3) * (y+z√3)/(y^2  3z^2)
= (xy +√3(xz  y^2) 3yz)/(y^2  3z^2) is rational
=> y^2 = xz.Let y = xr, z = xr^2
=> x^2 + 4z^2 = x^2 * (1+4r^4)
= x^2 * (2r^2 + 2r+1)(2r^22r+1).If x = 1 then r ≠ 1;
Thus, x^2 + 4z^2 can be written as product of 2 natural numbers > 1 which are x(2r^2 + 2r+1) and x(2r^2  2r + 1).
Similarly, x^2 + y^2 + z^2 = x(r^2 + r + 1) * x(r^2  r + 1) and can be written as product of 2 natural numbers > 1.
Hence, option (d) is the right answer

Q7) In a sports tournament, the number of European teams is 9 more than the number of Asian teams. Each team plays the other exactly once, the winner gets 1 point and the loser gets nothing. There were no draws during the tournament. The total points earned by European teams is 9 times the total points earned by the Asian teams. What is the maximum possible points of the best Asian team?
(a) 9
(b) 11
(c) 21
(d) not be uniquely determined
(e) none of the foregoing

Let the no. of Asian teams be x. Therefore no. of European teams will be x+9. Matches can be divided into three groups: 1) Those played b/w two Asians 2) Those played b/w two Europeans 3) Those played b/w an Asian and a European
In the 1st group no. of points won (by Asians) = x(x1)/2
In the 2nd group no. of points won (by Europeans) = (x+9)( x+8 )/2
Let k be the points Asians gain in the 3rd group. Therefore those won by Europeans = x(x+9)  k
Using the given conditions we have
9*[x(x1)/2 + k] = (x+9)( x+8 )/2 + x(x+9)  k
This gives 3x^2  22x + 10k  36 = 0
=> Factorize it to get the following: (3x+5)(x9) = 10k  9
Since RHS is negative, LHS too should be negative. Hence, x lies between 0 and 9. On checking, we find that the only integer solutions of the above equation occur when either x=8 and k=2 or when x=6 and k=6.
The maximum no. of matches an Asian team can win if x = 6 and k = 6 is
5 (won from other 5 Asian teams in 1st Group) + 6 (If the same Asian team is the one which wins all the AsianEuropean matches in 3rd Group) = 11
and if x = 8 and k = 2 it is 7 + 2 = 9
So the maximum no. of points an Asian team can have is 11.

Q8) Consider a positive integer Z which when represented in decimal base,does not end in zero. Z and the number obtained by reversing the digits of Z are both multiples of seven. Let the number of such Zs in the set {10 ,11,12... 998,999,1000} be K. Then k is given by which of the following?
(a) 13
(b) 14
(c) 15
(d) 16
(e) None of the foregoing .

There is only one 2digit number, namely 77.
Indeed, if 10a + b is (where a, b are integers 1–9), then 7 divides 10a + b and 7 divides 10b + a.
Thus, 7 divides{ 3(10a + b) − (10b + a)} = 29a − 7b = a + 7(4a − b),
so that 7 divides a, and similarly 7 divides b, so we’d better have a = b = 7.There are 16 3digit numbers. First consider the 12 palindromic ones (ones where the hundreds and units digits are the same): 161, 252, 343, 434, 525, 595, 616, 686, 707, 777, 868, and 959.
Now consider the general case: suppose 100a + 10b + c is, where a, b, c are integers 1–9. Then 7 divides 100a + 10b + c and 7 divides 100c + 10b + a, so 7 divides (100a+10b+c)−(100c+10b+a) = 99(a−c), and so 7 divides a−c. In order for this not to result in a palindromic integer, we must have a − c = +/ 7 and, moreover, both 100a + 10b + a and 100c + 10b + c must be palindromic integers.
Consulting our list above, we find 4 more integers: 168, 259, 861, and 952.
So K = 1 + 12 + 4 = 17
Hence, choice (e) is the right answer

Q9) Let a, b, c be real such that a+b+c ≠ 0. If (a^5+b^5+c^5)/((a^2+b^2+c^2)(a^3+b^3+c^3)) = 5/6, then ab(a+b) + bc(c+a) + ca(c+a)  3abc equals
(a) 1/2*(a^3 + b^3 + c^3)
(b) (a^2 + b^2 + c^2)^2/(a+b+c)
(c) a^3 + b^3 + c^3
(d) 3*(a^3 + b^3 + c^3)
(e) (a^4 + b^4 + c^4)/(a + b +c)

6(a^5 + b^5 + c^5)  5(a^2 + b^2 + c^2)(a^3 + b^3 + c^3) = [a+b+c]^2 * [a^3 + b^3 + c^3  2ab(a+b)  2bc(b+c)  2ca(c+a) + 6abc]
Otherwise, put a = 1 and b = 0 the equation (a^5+b^5+c^5)/((a^2+b^2+c^2)(a^3+b^3+c^3)) = 5/6 reduces to (c+1)^3*(c^2  3c + 1) = 0 => c^2  3c + 1 = 0 as a+b+c ≠ 0.
We are asked to find ab(a+b) + bc(b+c) + ca(c+a)  3abc = c(1+c)
Looking at the choices (a) becomes 1/2 * (1+c^3).
Multiplying both sides of c^2  3c + 1 = 0 by c + 1 we get c(1+c) = 1/2*(1+c^3)
Hence, option (a) is the right answer

Q10) Two triangles of sides (2,2,a) and (3,3,b) are inscribed in a circle. If a^4 + 36b^2 = b^4 + 16a^2, then what is the ratio of a/b?