Quant Boosters by Hemant Malhotra  Set 4

hemant_malhotra
Director at ElitesGrid  CAT 2016  QA : 99.94, LRDI  99.70% / XAT 2017  QA : 99.975
A shipping clerk has five boxes of different but unknown weights each weighing less than 100 kg.The clerk weights the boxes in pairs. The weights obtained are 110, 112, 113, 114, 115, 116, 117,118, 120 and 121 kg. What is the weight of the heaviest box?
a. 60 kg
b. 62 kg
c. 64 kg
d. Cannot be determinedlet the weight of the 5 boxes be: a < b < c < d < e:
GIven, a+b = 110 and d+e = 121
Adding all the 10 eqns will give: a+b+c+d+e = 289
==> c= 58
Now, next min combination is 112 (a+c)==> b = 56, a = 54
Now, max = 121 > 59 + 62 or 60 + 61 (does not satisfy other combinations)
==> 62 KgWhat is the sum of all prime factors of 3^20 + 3^19 – 12
3^20+3^1912
so 4(3^19)4(3)
so 12(3^181)
so 12(3^91)(3^9+1)
so 12(3^31)(3^6+3^3+1)(3^3+1)(3^63^3+1)
so 12(26)(757)(28)(703)
so (2^5)(3)(7)(13)(19)(37)(757).
so sum=838A point is selected in a square of perimeter of 48 cm. it is equidistant from two adjacent vertices and to an opposite side. Find the length of the common distance
(12x)^2+6^2=x^2
so x=7.5At the beginning of a year ,the owner of a jewel shop raised the price of all the jewels in his shop by x% and lowered them by x%. The price of one jewel after this up and down cycle reduced by Rs.100.The owner carried out the same procedure after a month.After this second updown cycle,the price of that jewel was Rs.2304.Find the original price of that jewel(in Rs.)
A. 2500
B. 2550
C. 2600
D. 2650
E. None of thesea * (1+x/100)*(1x/100))=a * (1x^2/100^2)) = a  100
a  ax^2/10^4 = a  100
so ax^2/10^4 = 10^2
so ax^2 = 10^6
now after month he carried out same thing
a * (1+x/100))^2((1x/100))^2=2304
so a((1x^2/10^4)^2=2304
a * (110^6/10^4a)) = 2304
a * (1100/a))^2=2304
a * (1100/a))^2=2304
so a * (1+100/a^2200/a))=2304
so a=2500Suppose that a and b are two digit prime numbers such that a^2 – b^2 = 2a + 6b + 8
Find the largest possible value of a + ba^2b^2=2a+6b+8
a^22a+1b^26b9=0
(a1)^2(b+3)^2=0
(a1)^2=(b+3)^2
so a1=+(b+3)
so ab=4
or a+b=2 which is not possible
so ab=4
here a and b both prime
a=83 and b=79
so sum=162Babu rao goes for a 7day trip in India . There are four cities A, B, C, and D in the India . Each day Babu visits one city, and he will not visit the same city on two consecutive days. Given that he visits City A on both Day 1 and Day 7, how many different possible itineraries are there for his 7day trip?
Method1 Total number of itineraries starting at A and ending at A or B or C or D is 3^6=729
The number of itineraries standing at A and ending at B or C or D is the same,
(no advantage of one of these cities on the others . reason is simple B/C/D have equal chances) and the number of itineraries starting at A and ending at A say a is greater by 1 than b
so a+3b=729
then 4b=728 so b=182 and a=183Method2 Let f(n) be the possible itineraries that he visits City A on both Day 1 and Day n, where n >= 3
so we have f(3) = 3.
[Possibilties are ABA, ACA, ADA]
f(4) = 6
[Possibilties : ABCA, ABDA, ACBA, ACDA, ADBA, ADCA]
For f(5), there are two cases.
Case 1= Add BA, CA and DA after all the cases in f(3)
so there are 3 f(3) cases.
Case 2. Add one city between the last two city in f(4)
so there are 2 f(4) cases.
so we have f(5) = 3f(3) + 2f(4) = 21.
so we have f(6) = 3f(4) + 2f(5) = 60
f(7) = 3 f(5) + 2f(6) = 183.A square in the coordinate plane has vertices whose y coordinates are 0,1, 4, and 5. What is the area of the square?
A(a,0)
B(b,1)
C(C,4)
D(d,5)
here yCyB=4 then xaxb will be 4 means difference between y coordinate of BC will be same as difference between x coordinate of AB so dis will be sqrt(4^2 + 1^2)=sqrt(17)A threedigit positive number ‘X’ is in the decimal system. ‘X’ is equal to cube of its unit digit as well as square of a two digit number formed by the remaining two digits(in any order). Find the sum of the digits of ‘X’.
(100a+10b+c)=c^3
so 100a+10b=c * (c^21))
10*(10a+b)=c * (c^21))
10*(10a+b)=c * (c1)(c+1))
c can't be 10 so c+1=10
so c=9
now
10a+b=72
so a=7 and b=2 easy to findLet a, b, c, d belongs to [1/2, 2]. Suppose abcd=1.
Find the maximum value of (a + 1/b)(b + 1/c)(c + 1/d)(d + 1/a).(ab+1)(bc+1)(cd+1)(ad+1)
(ab+1)(cd+1) = (ab+1)(1/ab + 1)
Now: (x+1)(1/x +1) = 1+x+1/x+1
We know, that: x+1/x >=2 (2 @ x=1)
==> (ab+1)(1/ab + 1) > 2+2 =4Similarly: (bc+1)(ad+1)>=4
Since, in the given domain: a=b=c=d=1 satisfies:
==> Min. values= 4 * 4 = 16Find all the integer solutions of the equation ab + 2b – 3a = 25
a(b3) + 2(b3) = 19
(a+2)(b3) = 19
a+2 =19 , a=17
b3 =1 , a=4
a+2 = 1 , a =1
b3 = 19 , a =22
a+2 = 1 , a=3
b3 = 19 , b = 16
a+2 = 19 , a=21
b3 = 1 , b=2
4 solutions..