Quant Boosters  Nitin Gupta  Set 2

Taking f(x) = ax^n + bx^n1 + cx^n2 + .... , where the third last coefficient and the second last coefficient are equal in magnitude but opposite in sign and n is odd.
f(x) + f(1x) has atleast first 2 terms as 0 => 2b + na = 0. Thus, we can have such f(x) for any n, thus n> Infinity.
Hence, choice (e) is the right answer.

Q24) A car gives the mileage of 60km/L, 50km/L and 40km/L when driven at the speeds of 40km/hr, 50km/hr and 60km/hr respectively. Assume that each car is driven only at the three speeds mentioned above. The car is driven for 3 hours using 2 Litres of petrol. The distance covered by the car
(a) > 120 km
(b) < 120 km
(c) = 120 km
(d) can not be determined
(e) none of the foregoing

Q25) One day Vikram was out bicycling. After entering a oneway tunnel and after having ridden onefourth of the distance through it, he looked back over his shoulder and saw a bus approaching the tunnel entrance at a speed of 80 miles/hr. Doing a quick mental exercise, Vikram realized that if he accelerated immediately to his top speed, he could just escape with his life, whichever direction he rode. What is Vikram's top biking speed in miles/hr?
a) 32
(b) 36
(c) 40
(d) 48
(e) none of the foregoing

Let d = distance truck is in front of tunnel entrance, L= length of tunnel, x = Vikram's speed.
Case 1: Vikram turns around and heads for entrance, a distance of L/4.
Vikram and truck get to entrance at same time T1 = d/80 = (L/4)/x.Case 2: Vikram streaks for exit, a distance of 3L/4. Vikram and truck get to exit at same time
T2 = (L+d)/80 = 3L/4)/x.Solving both equations for x and setting them equal, x = (L/4) * 80/d = (3L/4) * 80/(L+d)
After simplifying, d = L/2, hence x = 40.
Hence, choice (c) is the right answer.

Q26) Four persons go to a birthday party. They leave their topcoats and hats in the lounge and pick them while returning back.The number of ways in which none of them picks up his own topcoat as well as his own hat is p. The number of ways in which exactly one of them picks up his own topcoat as well as his own hat is q. The number of ways in which a person picks up someone else’s topcoat and yet someone else’s hat is r. Then p+q+r is
(a) 117
(b) 104
(c) 113
(d) 108
(e) none of the foregoing

The no. of ways in which the 1st person doesn't pick up his own topcoat is 3. The 2nd person (whose topcoat the 1st one picked)can pick someone else's topcoat in 3 more ways. For the 3rd and 4rth person we have just 1 choice. Thus, in all when none among the 4 pick his/her own topcoat is 9. Thus p = 9 * 9 = 81.
A person among the 4 who picks his own topcoat as well as hat can be selected in 4 ways. The other 3 falls in the category [not own topcoat or not own hat] = not own topcoat + notown hat  (not won topcoat and not own hat) = 2 * 6 + 2 * 6  2 * 2 = 20. Thus, q = 80.As explained in the 1st case, each person can pick someone else's topcoat in 9 ways. To each of these 9 ways we have 2 ways where he picks yet someone else's hat. Thus, r = 9 * 2 = 18
Hence, choice (e) is the right answer.

Let the distance travelled with 40km/hr, 50km/hr, 60km/hr is A, B and C respectively.
=> A/40 + B/50 + C/60 = 3 and A/60 + B/50 + C/40 = 2.
Eliminating A, we get 2B/5 + 5C/6 = 0 => B = C = 0 => A = 120 km
Hence, choice (c) is the right answer.

Q27) Five containers each have one litre of 10%, 20%, 30%, 40%, 50% milk in them. By mixing the contents of just two of the containers, one litres each of w%, x%, y%, z% and 42% (22 < w < = x < = y < = z < 42) milk is prepared, contents of no two containers are mixed twice in different proportions to get different concentrations. Also each original container is used exactly twice in making new mixtures. If the containers are mixed only in the multiples of 100 ml then z + y  x  w = ?
a) can not be determined
(b) 4
(c) 16
(d) 20
(e) none of the foregoing

Lets start with 42% milk preparation.
10% milk can be mixed with 50% milk to produce 42% milk but on doing so we are left with 800 ml of 10% milk which cant give any possible combination with others to produce milk more than 22% milk. Only 30% milk in quantity 400 ml with 50% milk in 600 ml gives us a 42% combination.
Now we are left with 600 ml of 30% and 400ml of 50%.Now since we are combining 600 and 400 ml of 2 hence all the solution have to be broken in two parts each of 400 ml and 600 ml.
Proceeding further from here, we find only 1 valid combination so that all the values are satisfied in the given range.
400ml of C + 600 ml of E > [120 + 300] > 420 out of 1000 > 42%
400ml of E + 600 ml of A > [200 + 60 ] > 260 out of 1000 > 26%
400ml of A + 600 ml of D > [40 + 240] > 280 out of 1000 > 28%
600ml of B + 400 ml of D > [120+ 160] > 280 out of 1000 > 28%
400ml of B + 600 ml of C > [80 + 180] > 260 out of 1000 > 26%Hence, z + y  x  w = 28 + 28  26  26 = 4
Hence, choice (b) is the right answer.

Q28) Let a, b, c be distinct nonzero integers such that 5 < = a, b, c < = 5. How many solutions (a, b, c) does the equation 1/a + 1/b + 1/c = 1/(a+b+c) have?
a) 72
(b) 120
(c) 192
(d) 240
(e) none of the foregoing

1/a + 1/b + 1/c = 1/(a+b+c) reduces to (a+b)(b+c)(c+a) = 0;
If a + b = 0 then we can chose b in 10 ways (integers from 5 to 5 excluding 0) and c in 10  2 = 8 ways (since a, b and c are different). So total 80 ways
Similarly, b + c = 0, and c + a = 0 gives 80 solutions each.
So total 80 + 80 + 80 = 240
Hence, choice (d) is the right answer.

Q29) A biologist catches a random sample of 60 fish from a lake, tags them and releases them. Six months later she catches a random sample of 70 fish and finds 3 are tagged. She assumes 25% of the fish in the lake on the earlier date have died or moved away and that 40% of the fish on the later date have arrived (or been born) since. What does she estimate as the number of fish in the lake on the earlier date?
(a) 420
(b) 560
(c) 630
(d) 720
(e) 840

Let no. of Fishes on earlier day = x; 60 fishes are tagged, out of which 25% have died/moved remaining fishes that are tagged = 45.
No. of fishes on day of second observation = 5/4x
No. of fishes caught = 70
No. of Fishes tagged = 3 = 4.2857% (3/70)
Total No. fishes on Second day of Observation = 45/.042857 (45 * 70/3) = 1050
No of Fishes on earlier day = x = 4/5 * 1050 = 840
Hence, choice (e) is the correct option

Q30) Consider a regular polygon of p sides .The number of values of p for which the polygon will have angles whose values in degrees can be expressed in integers?
(a) 24
(b) 23
(c) 22
(d) 20
(e) 21

Each angle is 180(p2)/p.
180{360}/{p} = k
So 360/p has to be an integer.
360 = 2^3 * 3^2 * 5^1
So there are 4 * 3 * 2 = 24 possibilities, but we exclude 1 and 2, because p > = 3
So , 24 2 = 22
Hence, choice (c) is the right answer