Quant Boosters  Nitin Gupta  Set 2

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

Each term in the summation is n/(n^4 + n^2 + n).
Had each term be n/(n^4 + n^2 + 1), then the summation would be 1/2(1  1/3 + 1/3  1/7 + .... upto infinity) = 1/2.
Note that n^4 + n^2 + 1 can be factored as (n^2+1)^2  n^2 = (n^2+n+1)*(n^2n+1).
n/(n^4 + n^2 + n) < n/(n^4 + n^2 + 1) for all n > 1.
Thus, our answer is < 1/2. All options among the first 4 are > = 1/2.

Q15) A cargo ship circles a lighthouse at a distance 20 km with speed 1500 km/h. A torpedo launcher fires a missile towards the ship from the lighthouse at the same speed and which moves so that it is always on the line between the lighthouse and the ship. How long does it take to hit?
(a) 37.7 secs
(b) 56.57 secs
(c) 75.43 secs
(d) 94.29 secs
(e) 113.14 secs