Quant Boosters by Hemant Malhotra  Set 3

hemant_malhotra
Director at ElitesGrid  CAT 2016  QA : 99.94, LRDI  99.70% / XAT 2017  QA : 99.975
Ramu chooses five distinct numbers. In how many different ways can he assign these numbers to the variables p, q, r, s, and t so that p < s, q < s, r < t, and s < t?
(A) 4
(B) 5
(C) 6
(D) 8
(E) 15t is definitely the largest no..
now there are 2 cases when r is > s then there are 2 possible cases
t r s p q and trsqp
when r < s then we have t s _ _ _
we have 6 possible cases
hence total 8How many positive integral solution does the following equation has:
14x + 21y + 42z = 840
2x+3y+6z=120now x should be multiple of 3 and y should be multiple of 2
so x=3k and y=2m
so 6k+6m+6z=120
so k+m+z=20
so 19c2=19*9=171If x and y are integers with (y − 1)^(x + y) = 4^3, then the number of possible values for x is
(A) 8
(B) 3
(C) 4
(D) 5
(E) 6(1) (y1)^(x+y) = 64
y = 65 and x = 64
(2) (y1)^(x+y) = 8^2
y = 9 and x = 7
y = 7 and x = 9
(3) (y1)^(x+y) = 4^3
y = 5 and x = 2
(4) (y1)^(x+y) = 2^6
y = 3 and x = 3
y = 1 and x = 7
so 6 valuesWhat is the largest power of 2 that divides the number K = 75! – 71!
71! ((75 * 74 * 73 * 72  1))
75 * 74 * 73 * 72  1 is odd number so no power of 2
so we basically need power of 2 in 71!
[71/2]+ [71/4] + [71/8] + [71/16] + [71/32] + [71/64]
35+17+8+4+2+1=67The hypotenuse of a right triangle is 10 cm, and the radius of the inscribed circle is 1.5 cm. Find the perimeter of the triangle.
x  1.5 + y  1.5 = 10
x + y = 13,
x + y + 10 = 23Now mug this formula too
min(sqrt(a^2+(b+c)^2),sqrt(b^2+(a+c)^2),sqrt(c^2+(a+b)^2))
Here a=4, b=3, c=2 so min will be sqrt((4^2+(3+2))^2)=sqrt41Suppose a function f is defined by f(1)=1,f(2)=2
f(n)=2f(n1) + f(n2) for n≥3.
find f(100)  2f(99)  2f(97)  f(96)f(100) = 2f(99) + f(98)
f(98) = 2f(97) +f(96). Adding them =0In which of the base is the operation 61 × 61 = 4541 valid?
let base n
(61)n*(61)n=(4541)n
1+6n)(1+6n)=1+4n+5n^2+4n^3
1+12n+36n^2=4n^3+5n^2+4n+1
so n=8What is the remainder when 27^11+9^11+3^11+1 is divided by 3^101
3^10=x
so 27 * 3^30 + 3^2 * 3^20 + 3 * 3^11 + 1
so 27 * x^3 + 9x^2 + 3x + 1 put x=1
so 27 + 9 + 3 + 1 = 40What is the last digit of the LCM of 7^5678  1 and 7^5678 + 1 ?
HCF = 2
LCM * HCF = first number * 2nd number
LCM * 2 = ( 7^5678  1) * (7^5678 + 1)
unit digit of 7^(5678)) = 9
so unit digit of 7^(5678)  1 will be 8
so unit digit of LCM will be 7^(5678) + 1 will be 0
so * * * 0 * * * 8/2=0 (((but u have to check that 2nd digit is even or odd )) its even that's y OA+02nd method to remove this thing
LCM * 2=((7^5678 1) * (7^(5678+1)
= ((49^(5678)1))
(491)(1 + 49 + 49^2 + ... 49^5677)/2 = 24(1 + 49 + 49^2 + ... 49^5677
49^k is mod 10=1 when k is even and is 1 when k is odd
now check inside portion is zero