# Quant Boosters - Nitin Gupta - Set 2

• 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

• 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^2-n+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

• Let M be the position of the ship at the moment the missile is fired. Let V be the point a quarter of the way around the circle from M (in the direction the ship is moving).Take the point at which Lighthouse is situated to be U. Then the missile moves along the semi-circle on diameter UV and hits the ship at V.

To see this take a point T on the quarter circle and let the line UT meet the semi-circle at R. Let Z be the center of the semicircle. The angle VZR is twice the angle VUR, so the arc VT is the same length as the arc VR. Hence also the arc UR is the same length as the arc MT.

So time needed= 10 * pi * 3600 /1500 = 75.43

Hence, choice (c) is the right answer

• Q16) Three positive integers a, b, and c are consecutive terms in an arithmetic progression. Given that n is also a positive integer, for how many values of n below 1000 does the equation a^2 - b^2 - c^2 = n have no solutions?
(a) 458
(b) 493
(c) 524
(d) 559
(e) 562

• Let a = b+d, and c = b-d => n = b(4d - b)
put b = 4d - 1, it gives n = (4d - 1)*1 = 4d - 1 = 3,7,11,...,999 for d = 1,2,...,250
put b = 4d - 2 it gives n = 8d - 4 = 4,12,...,996 for d = 1,2,...,125
put b = 4d -3, n = 12d - 9 which is of the form 4k - 1 and already counted
put b = 4d - 4 it gives n = 16d - 16 = 16,32,...,992 for d = 2,3,...,63
it can be seen that b = 4d - 3, 4d - 5,... etc. give these already counted values of n.
=> total n are 999 - 250 - 125 - 62 = 562.

• Q17) Consider two triangles PRS and PQS.Let two circles circumscribe these two triangles and their radii be 25 and 25/2.Also,PQRS is a rhombus whose area is equal to A. Then the value of A is
(a) 200
(b) 125
(c) 375
(d) 400
(e) 325

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