# Quant Boosters - Hemant Malhotra - Set 2

• Wall length = 221m ; Time taken in days 100/9
so in one day A and B can complete 221 * 9/100 of the work.
so when they work in alternate way they will do this work in 2 days
so in 22 days they will do 221 * 99/100 unit work so remaining work = 1/100
and days =89/4
so remaining days = 1/4
so in 1/4 days A wil do 1/100 work
so in 1 day A will do 1/25 th work
so work done by B = 9/100 - 1/25 = 5/100 = 1/20
so ratio = 4 : 5
so B's sharing will be 5/9 * 1800 = 1000 rs

• Q3) The nth term and the sum of the first n terms of a sequence are Tn and Sn respectively. If Tn = T(n-1) - T(n-2) and Tn ≠ 0, then which of the following is definitely true?
a) S88 = S188
b) S66 = S160
c) S100 = S160
d) S120 = S142
[note:digits after S is in subscript]

• Tn = T(n-1) - T(n-2) 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)

• 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))

• 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

• 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

• 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 = 128

Quicker method:
Every friends has two options (left or right) so ans must be multiple of 2^5 and only 128 satisfies.

• 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

• 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

• 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

• 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

• 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

• if we take out 1 and 2, then inserted card number 2 + 1 + 2 = 5 = 3! - 1
so here in this case 31! - 1

• 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 ?

• 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 + 2

Another 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

• 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?

• Method 1:
Understand Basic - average speed = total distance/total time

car 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 m

Method 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

• 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.

• 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.

• 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

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