Quant Boosters  Shashank Prabhu, CAT 100 Percentiler  Set 7

shashank_prabhu last edited by
CAT 100%iler, 5 times AIR 1, Director  Learningroots, Ex ITC, Pagalguy, TAS
Number of Questions  30
Topic  Quant Mixed Bag
Solved ? : Yes
Source : Learningroots forum

shashank_prabhu last edited by
CAT 100%iler, 5 times AIR 1, Director  Learningroots, Ex ITC, Pagalguy, TAS
Q1) Sun Life Insurance Company issues standard, preferred, and ultrapreferred policies. Among the company's policy holders of a certain age, 50% are standard with a probability of 0.01 of dying in the next year, 30% are preferred with a probability 0.008 of dying in the next year, and 20% are ultrapreferred with a probability of 0.007 of dying in the next year. If a policy holder of that age dies in the next year, what is the probability of the deceased being a preferred policy holder?
a. 0.1591
b. 0.2727
c. 0.375
d. None of these

shashank_prabhu last edited by
CAT 100%iler, 5 times AIR 1, Director  Learningroots, Ex ITC, Pagalguy, TAS
We can simply consider the original sample set to be 10000 (as the values are in the form of 0.00x and so, we get nice values to work with). Standard are 5000 with a probability of 0.01 of dying so, it is probable that 50 among these will die. Similarly, among preferred, 24 will die and among ultra preferred, 14 will die. So, there are 88 deaths happening out of which, 24 are those of the preferred holders. So, 24/88 would be 0.2727

shashank_prabhu last edited by
CAT 100%iler, 5 times AIR 1, Director  Learningroots, Ex ITC, Pagalguy, TAS
Q2) A dessert recipe calls for 50% melted chocolate and 50% raspberry puree to make a particular sauce. A chef accidentally makes 15 cups of the sauce with 40% melted chocolate and 60% raspberry puree instead. How many cups of the sauce does he need to remove and replace with pure melted chocolate to make the sauce the proper 50% of each?
a. 1.5
b. 2.5
c. 3
d. 5
e. 5

shashank_prabhu last edited by
CAT 100%iler, 5 times AIR 1, Director  Learningroots, Ex ITC, Pagalguy, TAS
Use alligations. When mixing 40% chocolate and 100% chocolate to get 50% chocolate, the mix should be in the ratio 5:1. This means 5 parts 40% chocolate + 1 part 100% chocolate gives 6 parts 50% chocolate. So to get 15 parts 50% chocolate, we need 15/6 = 2.5 cups of pure chocolate.

shashank_prabhu last edited by
CAT 100%iler, 5 times AIR 1, Director  Learningroots, Ex ITC, Pagalguy, TAS
Q3) Twentyfour men can complete a work in sixteen days. Thirtytwo women can complete the same work in twentyfour days. Sixteen men and sixteen women started working for twelve days. How many more men are to be added to complete the work remaining work in 2 days?
a. 16
b. 24
c. 36
d. 48
e. 54

shashank_prabhu last edited by
CAT 100%iler, 5 times AIR 1, Director  Learningroots, Ex ITC, Pagalguy, TAS
Let work done by 1 man = m units
Let work done by 1 woman = w units
24 men can do 24m units of work in 1 day
In 16 days they can do 24m * 16 units which is the total work
Similarly, total work will be 32w * 24
Since work done is the same, we can equate the two
24m * 16 = 32w * 24
m=2w
16 m and 16 w working for 12 days should do (16m+16w) * 12 = (16m * 8m) * 12 = 24m * 12.
x men join and they work for 2 more days.
So 24m * 12 + 24m * 2 +xm * 2 = 24m * 16
Solving this, we get x = 24

shashank_prabhu last edited by
CAT 100%iler, 5 times AIR 1, Director  Learningroots, Ex ITC, Pagalguy, TAS
Q4) There are 140 students in a bschool. The number of students who specialize in Marketing, Finance and HR is 50, 80 and 70 respectively. The ratio of the number of students who have taken more than one of the specializations to the number of students who have taken up all three specializations is 3:2. If each student of the school specializes in at least one subject, then how many students specialize in exactly two subjects?

shashank_prabhu last edited by
CAT 100%iler, 5 times AIR 1, Director  Learningroots, Ex ITC, Pagalguy, TAS
It's better to do these using the a+2b+3c and a+b+c logic (people vs. combinations). In this case, a+b+c=140 and a+2b+3c=200. Also, (b+c)/c=3/2 and so, c=2b. On solving, you will get b=12. (a is the number of students who take exactly one specialization, b is the number of student who take exactly two specializations and c is the number of students who take all three)

shashank_prabhu last edited by
CAT 100%iler, 5 times AIR 1, Director  Learningroots, Ex ITC, Pagalguy, TAS
Q5) The sequence 1, 2, 4, 5, 7, 9, 10, 12, 14, 16, 17,…. has one odd number followed by the next two even numbers, then the next three odd numbers followed by the next four even numbers and so on. What is the 2017th term of the sequence?

shashank_prabhu last edited by
CAT 100%iler, 5 times AIR 1, Director  Learningroots, Ex ITC, Pagalguy, TAS
Just need to observe that the first series ends at 1, second series at 4, third series at 9 and so on. So, the 63rd series will end at 3969. The total number of terms till the end of the 63rd series will be 2016. So, the 2017th term will be the next even number i.e. 3970.

shashank_prabhu last edited by
CAT 100%iler, 5 times AIR 1, Director  Learningroots, Ex ITC, Pagalguy, TAS
Q6) How many points of intersection are there for the two functions defined as f(x) = 1/(x^2  1) and g(x) = f(x  1)?

shashank_prabhu last edited by
CAT 100%iler, 5 times AIR 1, Director  Learningroots, Ex ITC, Pagalguy, TAS
Equate the two functions. You will get
1/(x^21)=1/((x1)^21)
x^21=x^22x
X=1/2
So only one point

shashank_prabhu last edited by
CAT 100%iler, 5 times AIR 1, Director  Learningroots, Ex ITC, Pagalguy, TAS
Q7) To create an army comprising of a giant, 5 barbarians and 7 archers, it takes 84 gems. To create another army comprising of a giant, 7 barbarians and 10 archers, it takes 114 gems. If a player wants to create an army of 9 giants, 23 barbarians and 30 archers, how many gems would that player need to spend?

shashank_prabhu last edited by
CAT 100%iler, 5 times AIR 1, Director  Learningroots, Ex ITC, Pagalguy, TAS
Let the multiplier of the first one be x and that of the second one be y. Equate the coefficients. So we will get x+y=9, 5x+7y=23 and 7x+10y=30. If you get a common solution as is true in this case, you will get the answer. The other way is of course taking values for archers in terms of multiples of 10. 2 or 3 iterations should do it i suppose. (Answer is 426)

shashank_prabhu last edited by
CAT 100%iler, 5 times AIR 1, Director  Learningroots, Ex ITC, Pagalguy, TAS
Q8) A is a set having k elements. X and Y are subsets of A such that X is a subset of Y. In how many ways can we choose X and Y? Your answer will be in terms of n

shashank_prabhu last edited by
CAT 100%iler, 5 times AIR 1, Director  Learningroots, Ex ITC, Pagalguy, TAS
Take a set of some 3 elements and {1, 2, 3}.
If there are no elements in Y, there will be only one combination possible.
If there is 1 element in Y, there will be 3c1(1c0+1c1) combinations possible.
If there are 2 elements in Y, there will be 3c2(2c0+2c1+2c2) combinations possible.
If there are 3 elements in Y, there will be 3c3(3c0+3c1+3c2+3c3) combinations possible.
So, 1+6+12+8=27.
The basic point is, you will get nc0+nc1(2)+nc2(4)+nc3(8).... which will translate to 3^n

shashank_prabhu last edited by
CAT 100%iler, 5 times AIR 1, Director  Learningroots, Ex ITC, Pagalguy, TAS
Q9) x^4  12x^3 + ax^2  bx + c = 0. if it is known that its roots are +ve real numbers then what is maximum value of a + b + c ?

shashank_prabhu last edited by
CAT 100%iler, 5 times AIR 1, Director  Learningroots, Ex ITC, Pagalguy, TAS
p + q + r + s = 12... so, the sum of products of roots taken 2 at a time, 3 at a time and 4 at a time would be maximum when they are all equal to 3.
pq+pr+ps+qr+qs+rs+pqr+pqs+prs+qrs+pqrs will be 243.

shashank_prabhu last edited by
CAT 100%iler, 5 times AIR 1, Director  Learningroots, Ex ITC, Pagalguy, TAS
Q10) Chord PQ of a circle is the perpendicular bisector of chord XY of the same circle such that they intersect at point M inside the circle. XY = PM = 14. What is the area of the circle?