Quant Boosters  Nitin Gupta  Set 2

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?

a^4+36b^2 = b^4 +16a^2 => a^2*(a^216) = b^2*(b^236) => a^2/b^2 = (36b^2)/(16a^2)=> a/b = root((36b^2)/(16a^2))  (A)
Also , area of the triangle with side (2,2,a) = (a * 2 * 2)/(4 * R) = (a/4) * root(4 * 2^2  a^2) {Where R is the circumradius}
=> R = 4/root(16  a^2) (1)
Also from the second triangle with sides(3,3,b) we get
R = 9/root(36b^2) (2)
Equating (1) and (2) we get,(36b^2)/(16a^2) = 81/16
Using (A) we have a/b = 9/4

Q11) RST is a triangle and U, V, W are the feet of perpendiculars dropped from R, S, T on sides ST, TR and RS respectively. If SU = 2, SW = 3 and TV = 4, then UV is
(a) 9/4
(b) 7/3
(c) 7/2
(d) 3/2
(e) None of the foregoing

Triangles TVU & TSR are similar, 4/(TU+2)=UV/(3+RW)
Triangles SUW & SRT are similar, 2/(3+RW) = 3/(2+TU)
Hence UV = 8/3
Hence, choice (e) is the right option

Q12) Three articles are sold at the profit of 20%, 25% and 40% and the net profit comes out to be 30%. Had the article sold at 25% been sold at 15% profit and the article sold at 40% been sold at its cost price then the profit on the sale of three articles would have been

From 1st condition, C3 = C1 + 0.5C2 => new profit = (0.2C1+0.15C2)*100/(2C1+1.5C2) = 10%

Q13) Let the cost price of an article A is as much more than another article B as much that of B is more than that of another article C whose cost price is more than that of fourth article D by the same amount. If A, B, C, D are sold at 10%, 15%, 30% and 5% profit respectively, then the net profit percent on the sale of four articles together is
(a) 5%
(b) 10%
(c) 15%
(d) can not be determined
(e) none of the foregoing

If Cd is the Cp of D and x be the common difference in CPs, then net profit = (60Cd + 90x)/(4Cd + 6x) = 15%

Q14) The value of 1/(1+1^2+1^4)+ 2/(2+2^2+2^4) + 3/(3+3^2+3^4) + ... upto infinity is
(a) 1/2
(b) 2/3
(c) 1
(d) 3/2
(e) None of these