# Quant Boosters - Nitin Gupta - Set 2

• 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, b-m) 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 + (y-b+m)^2 -2mx passes through (a, 0) => a^2 + (b-m)^2 = 2am.
Equation of PQ is x-y = b-m.
Also |(a-m+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 , (4-p)/3) and (1,1).
Thus, the area formed is 1/3 - 1/6*(p-2)^2
Case 3: 3 < = p < = 4
The area formed is the image of case 1. Area = (4-p)^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:
(x-y)^2 + (z-x)^2 + (y-z)^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 (x-y√3)/(y-z√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 non-prime
(c) x^2 + 4z^2 is prime
(d) both (a) and (b)
(e) none of the foregoing

• (x-y√3)/(y-z√3) = (x-y√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^2-2r+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(x-1)/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(x-1)/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)(x-9) = -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 Asian-European 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 2-digit 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 3-digit 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^2-16) = b^2*(b^2-36) => a^2/b^2 = (36-b^2)/(16-a^2)=> a/b = root((36-b^2)/(16-a^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(36-b^2) ------------------(2)
Equating (1) and (2) we get,(36-b^2)/(16-a^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

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