Quant Boosters  Nitin Gupta  Set 2

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

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 semicircle 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 semicircle 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 = bd => 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

Let O be the point of intersection of diagonals PR and QS,and T be the point of intersection of PR and the circumcircle of triangle PQS.Extend SQ to meet the circumcircle of triangle PRS at U.
PO.OT=QO.OS and SO.OU=PO.OR
Let PR=2a and QS=2b.Because PT is a diameter of the circumcircle of triangle PQS,and SU is a diameter of the circumcircle of triangle PRS,the above equalities can be rewritten as
a(25a)=b^2 and b(50b)=a^2
Hence, a = 2b.
It follows that 50b=5b^2=>b=10,a=20.
Thus area of PQRS = 1/2 * PR.QS = 2ab = 400
Hence, choice (d) is the right answer.

Q18) Let there be Z number of rational triplets (g, h, i) for which g, h, i are the roots of x^3 + gx^2 + hx + i = 0. Then which of the following is the value of Z ?
(a) 0
(b) 1
(c) 2
(d) 3
(e) Cannot be determined