Quant Boosters  Sagar Gupta  Set 7

Let a secret three digit number be cba. If the sum of cab + bac + bca + abc + acb = 2536, what is c+b+a ?
222a +222b+222c a10b100c = 2536
221a + 212b + 122c = 2536
122(a+b+c) + (99a + 90b) = 2536
2536 mod 9 = 7
99a + 90b mod 9 = 0
122(a+b+c) mod 9 = 7
5(a+b+c) mod 9 = 7
so : a+b+c = 14 as 70 mod 9 = 7The average height of a group of n persons is 90. Two persons with heights 60 and 65 respectively leave the group. A third person with height between 92 and 97 joins the group. The new average height of the group is a prime number. Find the possible height of the person who joined the group if it is known that the new average height is less than 120?
A. 92
B.93
C.97
D.94
E. CBDa) 90n 125 + 92 = 90n  33 / n1 = 33 + 57n / n1 => n=4 / n=20 / n=58 > new avg age => 109 or 93 or 90 : 109 possible
b) 90n  125 + 93 = 90n  32/n1 = 32 + 58n/n1 => n=59 > new avg age => 91 : not possible
c)90n  125 + 97 = 90n  28/n1 = 28 + 62n/n1 => n=32 > new avg age = 92 : not possible
d) 90n 125 + 94 = 90n 31 / n1 = 31 + 59n/n1 => n=60 > new avg age = 91 : not possible
only 109 is possible : which comes from option AThere are 20 people in your applicant pool, including 5 pairs of identical twins. If you hire 5 people randomly, what are the chances you will hire at least 1 pair of identical twins?
10C5 + 10C4 * 10C1 + 10C3 * (10C2  5 ) + 10C2 * ( 10C3  5 * 8 ) + 10C1 * ( 10C4  5C2  5C1 * (8C24 ) ) + 10C0 * ( 10C5  5C2 * 6C1  5C1 * (8C3 4 * 6 )
252 + 10 * 210 + 120 * ( 40) + 45 * (80) + 10(210 10120 ) + ( 192  160 ) = 11584Find the number of integer solutions for x + 2 * y + z = 4
x +2y +z = 4
y=0 =>x +z = 4 : 16
y=+/1 =>x + z =2 : 8
y=+/2 => x + z = 0 : 1
16 + 2 [ 8 + 1 ] = 34Find the minimum value of x^2 * y^2  y^2 * z^2  z^2 * x^2 where x^2 + y^2 + z^2 = 5 and x,y,z are real numbers
since u need minimum value...put x=0
y^2 * z^2 y^2+z^2 = 5
min when y=z = root ( 5/2 )
y^2 z^2 = 25/4 = OASara has just joined Facebook. She has 5 friends. Each of her five friends has twenty five friends. It is found that at least two of Sara’s friends are connected with each other. On her birthday, Sara decides to invite her friends and the friends of her friends. How many people did she invite for her birthday party?
a0 , b0 , c0 , d0 , e0
a0 : a1 to a24 + sara
b0 : b1 to b24+ sara
c0 : c1 to c24 + sara
d0 : d1 to d24 + sara
e0 : e1 to e24+ sara
2 friends of sara are connected to each other : say ab
24 * 5 + 3 = 123
all five friends are mutual friends
each has 20 distinct friends and give people a0,b0,c0,d0,e0 : 20 * 5 + 5 =105A man sells two houses at the rate of Rs. 1.995 lakh each. On one he gains 5% and on the other, he loses 5%. His gain or loss percent in the whole transaction is ?
General Formula : if equal rate of interest ( one + and other  ) , the total loss % is given by r^2 / 100.
cp = a
sp = y
y = a ( 1 + r/100 ) => a = y / ( 1 + r/100 )
cp = b
sp = y
y = b ( 1  r/100 ) => b = y / ( 1  r/100 )
cp = ( 2y / 1  r^2 / 10000 )
sp = 2y
2y ( 1  r^2 / 10000 ) = 2y ( 1 + x / 100 )
( 1  r^2 / 10000 ) = 1  x/100
r^2 / 10000 = x/100
x = r^2 / 100 = loss %Given N = 35 x 36 x 37 .......... x 67, what is the remainder left when N is divided by 289?
35 x 36 x 37 .......... x 67 mod 289
take out 17 common
( 35 * 36 *...50 * 3 * 52 * 53 *....67 ) mod 17
1 * 2 *....16 * 3 * 1 * 2 * ... 16 mod 17
16! * 16 ! * 3 mod 17
16 * 16 * 3 = 1 * 3 mod 17 = 3
17 * 3 = 51
35 x 36 x 37 .......... x 67 mod 289
take out 17 common
( 35 * 36 *...50 * 3 * 52 * 53 *....67 ) mod 17
1 * 2 *....16 * 3 * 1 * 2 * ... 16 mod 17
16! * 16 ! * 3 mod 17
16 * 16 * 3 = 1 * 3 mod 17 = 3
17 * 3 = 51Find the number of ordered pairs of integers for : x^2 + y^2  xy = 727
x^2 + y^2  xy = 727
x^2  yx + (y^2  727 ) = 0
D = y^2  4(y^2  727)
D = 2908  3y^2
29073y^2+1 should be a perfect square
3(969y^2)+1 should be a perfect square
clearly y=31 , which gives 2 possible x = 13 and x= 18
values can be interchanged here :
y =13 will give x=31 , x=18
y=18 will also give x=31 , x=13
same will be possible when all these values are of negative sign
so total 3 * 2 * 2 = 12 valuesIf Ap is the sum to the first p terms of the series A = 12^144 + 12^143 + 12^142 + ………, then find Bp, which is the sum to the first p terms of the series A1 + A2 + A3 ...
sum of P terms :
12^144 + 12^143 +....12^145P
12^144 [ 1  (1/12)^P ] / [ 11/12 ]
12^145 [ 1  1/12^P ] / 11 = Ap
Bp = A1+A2+....Ap
12^145 [ 1  1/12^P ] / 11
12^145 / 11 [ 1  1/12 + 11/12^2+....1/12^P ]
12^145 / 11 [ 11P/11  1/11 + (1/12)^P/11 ]
12^145/121 [ 11P 1 + (1/12)^P ]or else...put p=2 or 3...and check !