Quant Boosters  Hemant Malhotra  Set 2

hemant_malhotra last edited by
Director at ElitesGrid  CAT 2017  QA 100 Percentile / CAT 2016  QA : 99.94, LRDI  99.70% / XAT 2017  QA : 99.975
Tn = T(n1)  T(n2) has cycle of 6 & sum of any six consecutive terms of this function is zero
so subscript should yield the same remainder when divided by 6. use this in options and only S100 = S160 satisfies. (Both 100 and 160 yield remainder as 4 when divided by 6)

hemant_malhotra last edited by
Director at ElitesGrid  CAT 2017  QA 100 Percentile / CAT 2016  QA : 99.94, LRDI  99.70% / XAT 2017  QA : 99.975
Q4) If f(x) = 1/(1 + x) then f(3x) when expressed in terms of f(x) will be
a) 3f(x)
b) f(x)/(3 + 2f(x))
c) f(x)/(3  2f(x))
d) f(x)/(2  3f(x))

hemant_malhotra last edited by
Director at ElitesGrid  CAT 2017  QA 100 Percentile / CAT 2016  QA : 99.94, LRDI  99.70% / XAT 2017  QA : 99.975
f(x)=1/(1+x)
f(3x)=1/(1+3x)
now Question is simple and date is not that much . if data is lengthy then we can apply Options approach
take x=1
so f(1)=1/2
and f(3)=1/4
now f(3)=f(1)/2
so f(1)=2f(3)
a) out of picture
b) f(1)/(3 + 2f(1) = 1/2/(3 + 1)) = 1/6 not equal to f(3)
c) f(1)/3  2f(1) = 1/2/((3  1)) = 1/4 so OA = C

hemant_malhotra last edited by
Director at ElitesGrid  CAT 2017  QA 100 Percentile / CAT 2016  QA : 99.94, LRDI  99.70% / XAT 2017  QA : 99.975
Q5) Five friends attended a party. At the end of the night, they each randomly grabbed a left shoe and a right shoe. Find the number of ways such that each person left with exactly one of their own shoes?
a) 104
b) 112
c) 120
d) 128

hemant_malhotra last edited by
Director at ElitesGrid  CAT 2017  QA 100 Percentile / CAT 2016  QA : 99.94, LRDI  99.70% / XAT 2017  QA : 99.975
Conventional method :
Each person left with exactly one of their own shoes means either they will get rite one or left one so derangement of 5
so D(5) = 44
so 2 * 44 = 88
now 4 of them will get right then obviously 5th one will get right
now 3 of them correct then
so 5c3 * 2 = 20
and 2 of them correct = 2 * 5c3 = 20
so 88 + 40 = 128Quicker method:
Every friends has two options (left or right) so ans must be multiple of 2^5 and only 128 satisfies.

hemant_malhotra last edited by
Director at ElitesGrid  CAT 2017  QA 100 Percentile / CAT 2016  QA : 99.94, LRDI  99.70% / XAT 2017  QA : 99.975
Q6) A contractor hired a group of men and women to do a piece of work in 12 days. But everyday for the first 12 days, one worker, either a man or a woman, was absent, and hence the work was delayed by 3 days. The contractor made an interesting observation that if the number of days for which women and men remained absent were interchanged, then the work would take 16 days to complete. Approximately how many days would it take for one man and one woman to complete the work?
a) 16
b) 20
c) 24
d) Cannot be determined

hemant_malhotra last edited by
Director at ElitesGrid  CAT 2017  QA 100 Percentile / CAT 2016  QA : 99.94, LRDI  99.70% / XAT 2017  QA : 99.975
Since its a 12 day work, in first case they lost 1/4th job
and in second case they lost 1/3rd job
12 men and 12 woman can do (1/3) + (1/4) = 7/12 work in 1 day
so They can finish the work in 12/7 days
so 1 woman and 1 man can finish the work in 12 * 12/7 = 144/7 or 20.57 days

hemant_malhotra last edited by
Director at ElitesGrid  CAT 2017  QA 100 Percentile / CAT 2016  QA : 99.94, LRDI  99.70% / XAT 2017  QA : 99.975
Q7) Let S(n) denote the sum of first n terms of an Ap. If S(2n) = 3 * Sn then find ratio of S(3n)/S(n)
a) 4
b) 6
c) 8
d) 10

hemant_malhotra last edited by
Director at ElitesGrid  CAT 2017  QA 100 Percentile / CAT 2016  QA : 99.94, LRDI  99.70% / XAT 2017  QA : 99.975
let first term = 1 and n = 1
so S(2) = 3 * S(1)
S(1) is sum of first term means first term itself.
S(2) = 3 and S(1) = 1
So sum of 2 terms = 3
Sum of first term = 1
So 2nd term = 3  1 = 2
First term = 1
Second term = 2
So d = 1
Means third term = 3
So S(3) = 1 + 2 + 3 = 6
S(3)/S(1) = 6

hemant_malhotra last edited by
Director at ElitesGrid  CAT 2017  QA 100 Percentile / CAT 2016  QA : 99.94, LRDI  99.70% / XAT 2017  QA : 99.975
Q8) There are 30 cards. Each card has a single number written on it, such that all the integers from 1 to 30 are represented in the cards. Two cards (which have numbers a and b written on them) are arbitrarily chosen, and are removed from the pack. The above two cards are replaced with a single card, on which the number (ab + a + b) is written. This constitutes one operation. After 29 such operations, what will be the number written on the sole surviving card?
a) 31! − 1
b) 15 × 31
c) 30! − 1
d) 31! + (15 × 31)
e) 30! + (15 × 31) – 1

hemant_malhotra last edited by
Director at ElitesGrid  CAT 2017  QA 100 Percentile / CAT 2016  QA : 99.94, LRDI  99.70% / XAT 2017  QA : 99.975
if we take out 1 and 2, then inserted card number 2 + 1 + 2 = 5 = 3!  1
so here in this case 31!  1

hemant_malhotra last edited by
Director at ElitesGrid  CAT 2017  QA 100 Percentile / CAT 2016  QA : 99.94, LRDI  99.70% / XAT 2017  QA : 99.975
Q9) Six jars of cookies contain 18 , 19 , 21 , 23 , 25 and 34 cookies respectively. One jar contains almond cookies only. The other 5 jars contain no almond cookies. A takes three of the jars and B takes two of the others. Only the jars of almond cookies remains. If A gets twice as many cookies as B , how many almond cookies are there ?

hemant_malhotra last edited by
Director at ElitesGrid  CAT 2017  QA 100 Percentile / CAT 2016  QA : 99.94, LRDI  99.70% / XAT 2017  QA : 99.975
Sum of all the jars is 140
it is of the form 3k + 2
to make it multiple of 3 (A got 2x and B got x = > Total 3x) we should subtract a number of the form 3k + 2.
23 is present here in form of 3k + 2Another approach:
A has even number of cookies
even = even + even + even
even = odd + odd + even
three even is not here so he will choose two odd and one even
so 34 + 19 + 21 or 34 + 19 + 23 or 34 + 19 + 25 or 34 + 21 + 23 or 34 + 23 + 25
so 74 or 76 or 78 or 78 or 82
now B has 37 or 38 or 39 or 41
37 = 18 + 19 (19 is already used so not possible)
38 is not pssible
39 = 18 + 21 (this is possible combination)
41= 23 + 18
so 23 almond cookies are there

hemant_malhotra last edited by
Director at ElitesGrid  CAT 2017  QA 100 Percentile / CAT 2016  QA : 99.94, LRDI  99.70% / XAT 2017  QA : 99.975
Q10) Ann and Anne are in bumper cars starting 50 m apart. Each one approaches the other at a constant ground speed of 10 km/hr. A ﬂy starts at Ann, ﬂies to Anne, then back to Ann, and so on, back and forth until it gets crushed when the two bumper cars collide. When going from Ann to Anne, the ﬂy ﬂies at 20 km/hr; when going in the opposite direction the ﬂy ﬂies at 30 km/hr (thanks to a breeze). What is the total distance covered by the fly?

hemant_malhotra last edited by
Director at ElitesGrid  CAT 2017  QA 100 Percentile / CAT 2016  QA : 99.94, LRDI  99.70% / XAT 2017  QA : 99.975
Method 1:
Understand Basic  average speed = total distance/total timecar 1__________________12d _______________ car2
let distance between them = 12d
speed of fly is double so when they both will cover 4d distance fly will cover 8d distance
now speed is triple so when both cars will travel d then fly will travel 3d and meets ann again so
so distance travelled by fly in forward direction is 8d and in backward direction = 3d
so avg speed = (8d + 3d)/(8d/20 + 3d/30)
11/(8/20 + 3/30)
11 * 60/30 = 22 km/hr = 110/18 m/s
time taken by ann and anne to collide = 50/(100/18) = 9 seconds
fly will cover= 9 * 110/18 = 55 mMethod 2 :
Suppose that at a given instant the fly is at Ann and the two cars are 12d apart. Then, while each of the cars travels 4d, the fly travels 8d and meets Anne. Then the fly turns around, and while each of the cars travels d, the fly travels 3d and meets Ann again. So, in this process described, each car travels a total of 5d while the
fly travels 11d.
So the fly will travel 11/5 times the distance traveled by each bumper car = 11/5 * 50/2 = 55 meters

hemant_malhotra last edited by
Director at ElitesGrid  CAT 2017  QA 100 Percentile / CAT 2016  QA : 99.94, LRDI  99.70% / XAT 2017  QA : 99.975
Q11) A father starts every day from home to pick up his son from school at 3:30. One day, son leaves the school at 2:30 and hence he starts walking towards home at 6km/hr. He meets his father (who starts at his normal time) on the way and they reach 24 min earlier than usual. Find father's speed.

hemant_malhotra last edited by
Director at ElitesGrid  CAT 2017  QA 100 Percentile / CAT 2016  QA : 99.94, LRDI  99.70% / XAT 2017  QA : 99.975
They saved 24 minutes from both sides it means save 12 minutes on one side
therefore father reached to his son 12 minutes before than he use to be at 3:18
His son walked for 48 minutes. (from 2:30 to 3:18)
so distance travelled by son in 48 minutes is equal to distance travelled by father in 12 minutes
hence ratio of their speed is 1 : 4 and thus 24 km/h.

hemant_malhotra last edited by
Director at ElitesGrid  CAT 2017  QA 100 Percentile / CAT 2016  QA : 99.94, LRDI  99.70% / XAT 2017  QA : 99.975
Q12) How many distinct real roots the following equation has
x^4 + 8x^2 + 16 = 4x^2  12x + 9
a) 0
b) 1
c) 2
d) 3
e) 4

hemant_malhotra last edited by
Director at ElitesGrid  CAT 2017  QA 100 Percentile / CAT 2016  QA : 99.94, LRDI  99.70% / XAT 2017  QA : 99.975
x^4 + 8x^2 + 16 = 4x^2  12x + 9
(x^2 + 4)^2 = (2x  3)^2
x^2 + 4 = ± (2x  3)
x^2  2x + 7 = 0 or x^2 + 2x + 1 = 0
First equation has no real roots and second has equal roots so only 1 different root

hemant_malhotra last edited by
Director at ElitesGrid  CAT 2017  QA 100 Percentile / CAT 2016  QA : 99.94, LRDI  99.70% / XAT 2017  QA : 99.975
Q13) if sum of the 6th powers of the roots of the equation x^2  x  a = 0 is 19721 and the sum of the 4th powers of the roots is 881 find the sum of the 7th power of the roots?
a) 60740
b) 64720
c) 62826
d) 61741