# Quant Boosters - Nitin Gupta - Set 2

• We require
(1) g + h + i = -g
(2) gh + hi + ig = h
(3) ghi = -i.

From (3), either i = 0, or gh = -1.

If i = 0, then (1) becomes h = -2g, and (2) becomes h(g - 1) = 0.
Hence either g = h = 0, or g = 1, h = -2.

Now assume i ≠ 0, and gh = -1.
(1) becomes i = - h – 2g.
Substituting in (2), we get: -1 - (2g+ h)(g + h) = h,
so -g^2 –2g^4 + 3g^2 - 1 = -g, or 2g^4 – 2g^2 - g + 1 = 0.
So g= 1, or 2g^3 + 2g^2 - 1 = 0 ….

The first possibility gives g = 1, h = -1, i = -1. Suppose g = m/n is a root of 2g^3 + 2g^2 - 1 = 0 with m, n relatively prime integers. Then 2m^3 + 2m^2n - n^3 = 0. So any prime factor of n must divide 2 and any prime factor of m must divide 1. Hence the only possibilities are g = 1, -1, 1/2, -1/2, and we easily check that these are not solutions. So 2g^3 + 2g^2 - 1 = 0 has no rational roots.

Hence option (c) is the right answer

• Q19) Let x, y, z be prime numbers in the arithmetic progression such that x > y > z. Which among the following is always true?
(a) x – z is divisible by 12
(b) x + y is divisible by 8
(c) xz - 1 is divisible by 7
(d) atleast two of the foregoing
(e) none of the foregoing

• Take x = 7, y = 5 and z = 3 rules out first 4 options.
However, if z were > 3 then x - z is always divisible by 12. Check yourself for that.
Hence, option (e) is the right answer

• Q20) What is the minimum value of the expression 6x^2 + 3y^2 - 4xy - 8x + 6y + 2?
(a) -13/7
(b) -2
(c) -8/5
(d) -7/3
(e) none of the foregoing

• The expression 6x^2 + 3y^2 - 4xy - 8x + 6y + 2 can be written as (14x^2 -12x - 3)/3 + 3(y - (4x-6)/6)^2
To minimize (y - (4x-6)/6) = 0 and 14x^2 -12x - 3 should be minimum.
14x^2 -12x - 3 is minimum for x = 3/7 => y = -5/7.
Hence, option (a) is the right answer

• Q21) Let P(x) = nx + a where n and a are integers with n > 0. If the solution to 2^P(x) = 5 is x = log8 (10 ) , then 4n - 3a is
(a) 11
(b) 12
(c) 13
(d) 14
(e) None of the foregoing

• Taking log on both the sides of 2^P(x) = 5 we get P(x) = log5/log2.
x = log10/log8 = 1/3(1 + log5/log2) => n = -3 and a = 1.
Hence, option (e) is the right answer

• Q22) he number of permutations of the set {1, 2, 3, 4} in which no two adjacent positions are filled by consecutive integers (increasing order)
(a) is a prime
(b) is a composite number divisible by 3
(c) does not exceed 17
(d) is at most 19
(e) None of the foregoing

• The permutations where there are consecutive numbers are 1234, 1243, 1342, 1423, 2134, 2341, 2314, 3124, 3412, 3421, 4123, 4312, 4231
Hence, choice (a) is the right answer

• Q23) Let f(x) be an algebric expression of odd degree ( > 1). If the degree of f(x) + f(1-x) is atleast two less than the degree of f(x) and the coefficients of x and x^2 are equal in magnitude but opposite in sign in the given expression, then the highest possible degree of f(x) is
(a) 9
(b) 13
(c) 5
(d) 7
(e) none of the foregoing

• Taking f(x) = ax^n + bx^n-1 + cx^n-2 + .... , 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(1-x) 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 one-way tunnel and after having ridden one-fourth 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 top-coats and hats in the lounge and pick them while returning back.The number of ways in which none of them picks up his own top-coat as well as his own hat is p. The number of ways in which exactly one of them picks up his own top-coat as well as his own hat is q. The number of ways in which a person picks up someone else’s top-coat 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 top-coat is 3. The 2nd person (whose top-coat the 1st one picked)can pick someone else's top-coat 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 top-coat is 9. Thus p = 9 * 9 = 81.
A person among the 4 who picks his own top-coat as well as hat can be selected in 4 ways. The other 3 falls in the category [not own top-coat or not own hat] = not own top-coat + not-own hat - (not won top-coat 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 top-coat 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 non-zero 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

61

62

61

61

58

48

63

64