Quant Boosters by Hemant Malhotra - Set 5
Director at ElitesGrid | CAT 2016 - QA : 99.94, LR-DI - 99.70% / XAT 2017 - QA : 99.975
There is a group of nine children, which includes two pairs of identical twins A,B and C,D. In how many ways can the nine children be seated in a row, so that A and B do not sit next to each other and C and D do not sit next to each other.
Total-(when A and B together)-When (C and D together))
when total -(A and B together)=when A and B are not together
now this will include case when C and D are also together
so remove that case when C and D together from this
so 9!-((8! * 2!)-((8! * 2!))
8!((9-2-2)=5 * 8!
now this will include case when ALL are together so we have to add that case
ABCD or CDAB like this so arrange them 2!*2!*7!
so 5 * 8!+2! * 2! * 7!
7!((40+4))=44 * 7!
Find the GCF of all positive integers of the form a^4 -1 where a is prime and a > 5
6k * 2 * (3k+1)(36k^2+1+12k+1))
24 * k * (3k+1)((18k^2+6k+1))
k * (3k+1)(6k((3k+1)+1) mod 10=0
so one more common factor will be there always 10
so 24 * 10 = 240 will be hcf
The first term of a given sequence is 1, and each successive term is the sum of all the previous terms of the sequence. What is the value of the first term which exceeds 5000?
This is a simple geometric progression (leaving first term) with common ratio of 2. So first term greater than 5000 will be next term after 4096 so 8192
Number of 6 digits numbers that can be made with the digits 1, 2, 3, 4 and having exactly two pairs of digits.
Approach= exactly 2 pair of digits and 2 different digits
so form will be like aabbcd
so select those two pairs out of 4 (1,2,3,4))
so 4c2 * (( now we have left with 2 different digits ))
now 2 different digits out of 2 digits =2c2
now arrange aabbcd
6!/2! * 2!
4c2 * 2c2 * 6!/2! * 2! = 1080
how many non congruent triangles are possible with perimeter 19,whose two sides are even and one side is odd??
Method1- Total - all odd sides
total: 22^2/48 =10
max side can be 9
(9-2a)+(9-2b)+(9-2c) = 19
a+b+c=4 so 6C2 =15
aab: (0,0,4), (1,1,2),(2,2,0)
so unordered: (15-3*3)/3! +3 = 1+3=4
so 2 even and one odd= 10-4= 6
Two tanks, T1 and T2, of equal capacities, are provided with inlet taps, A and B, of different flow rates, respectively. Tap A begins to fill the first tank. Seven minutes later, tap B is opened. Nine minutes after that, the total quantity of water in the two tanks is just enough to fill one tank completely. Exactly t minutes after that both the tanks are full. What is the value of t?
let capacity =V
Let flow per minute of A=x
and flow per minute of B=y
so in 7 minutes A will provide 7x
and in next 9 minutes A will give 9x
and in 9 min B will give 9y
so total in T1=16x
and total in T2=9y
some of those =16x+9y=V
now after T minutes
A will give Tx
and B will give Ty
so 16x+Tx=16x+9y so T=9y/x
so min value of y=4
so x=3 and y=4
so T=16*3/4=12 minute
if r and s are the roots of the equation 2x^2 – 3x + 4 =0, what is 4r^2 + 7rs + 4s^2
If (fog)(x) = x^2 – 4x + 5 and f(x) = x^2 + 2x + 2; find g(x)
gx = ax+b
(ax+b)^2 + 2ax + 2b + 2
comparing with x^2 -4x + 5
a = 1
2b + 2a = -4
b = -3
There are three cities: A, B and C. Each of these cities is connected with the other two cities by at least one direct road. If a traveler wants to go from one city (origin) to another city (destination), she can do so either by traversing a road connecting the two cities directly, or by traversing two roads, the first connecting the origin to the third city and the second connecting the third city to the destination. In all there are 33 routes from A to B (including those via C). Similarly, there are 23 routes from B to C (including those via A). How many roads are there from A to C directly?
x + yz = 33
z + yx = 23
(z - x ) ( y - 1 ) = 10
so y-1 = 2 or 5
y = 3 or 6
now check at y=6 , a and c will be integer but at y =3 they will not be integer so 6 roads are possible
Find the sum of the first 125 terms of the sequence 1,2,1,3,2,1,4,3,2,1..
n(n+1)/2 = 125 , n = 15 , then 120
then 16 +15 + 14 + 13 + 12 = 70
1 + 3 + 6 + 10 ... 120
n^2/2 + n/2
n(n+1)(2n+1)/12 + n(n+1)/4
n = 15 , so 620 + 60 = 680
680 + 70 = 750