Quant Boosters by Hemant Malhotra  Set 1

hemant_malhotra
Director at ElitesGrid  CAT 2016  QA : 99.94, LRDI  99.70% / XAT 2017  QA : 99.975
Find the sum of 3/2 + 5/4 + 9/8 + 17/16 + ... to 99 terms
 100  1/2^{99}
 101  1/2^{99}
 100 + 1/2^{99}
 100 + 1/2^{100}
We are asked for 99 terms (n = 99) and is Option A is of form (n+1)  1/2^{n}
now if this is true then we can represent other numbers also in this form
3/2 = (2  1/2^1) = 3/2 ( n = 1)
3/2+5/4= (3  1/2^2) = 11/4 ( n = 2) and so on.
so OA=1001/2^99
A and B working together can build a wall, 221 m long in 11 1/9 days. If they work on alternate days with A starting the work it will take 22 1/4 days to build it. If A and B work together and build a wall of twice the length and earn a total of Rs 1800 for it, then B's share of earning will be
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 workso in 1 day A will do 1/25 th work
so work done by B =9/1001/25=5/100=1/20
so ratio=4:5
so B's sharing will be 5/9 *1800=1000 rsThe nth term and the sum of the first n terms of a sequence are Tn and Sn respectively. If Tn = Tn1  Tn2 and Tn≠0, then which of the following is definitely true?
1) S88 = S188
2) S66 = S160
3) S100 = S160
4) S120 = S142
[note:digits after S is in subscript]Tn=Tn1Tn2 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)
If f(x) = 1 / (1+x) then f(3x) when expressed in terms of f(x) will be
 3f(x)
 f(x)/(3 + 2f(x))
 f(x)/(3  2f(x))
 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)/32f(1)=1/2/((31))=1/4 so OA =CFive 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) 128Conventional 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.
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?
1) 16
2) 20
3) 24
4) Cannot be determinedSince 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 daysLet Sn denote the sum of first n terms of an Ap. If S 2n=3*Sn then find ratio of S3n /Sn
a) 4
b) 6
c) 8
d) 10
let first term=1 and n=1
so S2=3*S1
(S1 IS SUM OF FIRST TERM MEANS 1ST TERM)
S2=3
AND S1=1
SO SUM OF 2 TERMS =3
SUM OF FIRST TERM=1
SO 2ND TERM=2
T1=1
T2=2
SO d=1
MEANS THIRD TERM=3
SO S3=1+2+3=6
SO S3/S1=6There 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?
1) 31! − 1
2) 15 × 31
3) 30! − 1
4) 31! + (15 × 31)
5) 30! + (15 × 31) – 1if we take out 1 and 2, then inserted card number 2+1+2=5=3!1
so here in this case 31!1
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+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 (( thi is possible combination))
41=23+18
so 23 almond cookies are there