Quant Boosters  Swetabh Kumar  Set 3

Number of Questions  30
Topic  Quant Mixed Bag
Solved ?  Yes
Source  Compilation of my solutions from various CAT prep forums.

Q1) If N = 2^3 * 3^4 * 5^9, then find the number of trailing zeroes at the end of product of all factors of N which are not divisible by 12.

number formed from 2 * 3^3 * 5^9 = 80 factors.
so product of these = 12^80 * (2 * 3^3 * 5^9)^40 = 2^200 * 3^83 * 5^360.
total product = (2^3 * 3^4 * 5^9)^(100) = 2^300 * 3^400 * 5^900
so not divisible ke liye divide the above two.
2^100 * 3^x * 5^540 so 10^100
100 trailing zeros.

Q2) Let a secret three digit number be cba. If the sum of cab + bac + bca + abc + acb = 2536, what is c+b+a ?

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 = 7

Q3) Find 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 values

Q4) Find the maximum power of 12 in C(100,60) ?

100! / 60! * 40!
12 = 2^2 * 3
max power of 3 in 100! = 100/3+100/9+100/27+100/81 = 33+11+3+1 = 48
max power of 3 in 60! = 60/3 + 60/9 + 60/27 = 20 + 6 + 2 = 28
max power of 3 in 40! = 40/3 + 40/9 + 40/27 = 13 + 4 + 1 = 18
3^48 / 3^28 * 3^18 = ( 3^2 )
max power of 4 in 100! = 100/4 + 100/16 + 10/64 = 32
max power of 4 in 60! = 60/4 + 60/16 = 18
max power of 4! in 40! = 40/4 + 40/16 =12
2^32 / 2^18 * 2^12 = 2^2
so highest power is 1

Q5) Two motorists set out at the same time to go from A to B, a distance of 100 miles. They both followed the same route and travelled at different, though uniform speeds of an integral number of miles per hour. The difference in their speeds was a prime number of miles per hour and after they had been driving for 2 hours, the distance of the slower car from A was 5 times that of the faster car from B. At what speed did the two motorists drive?

Sp and Sq and  Sp  Sq  = prime number
Sp * 2 = b
Sq * 2 = 500  5b
10Sp + 2 Sq = 500
(42, 40) satisfies

Q6) If 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 ]

Q7) Sara 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 = 105

Q8) There are two vessels A and B, both containing vinegar solution of 40% concentration. I add some pure vinegar to A to bring the concentration to 50%. In the vessel B, I take out some quantity of the solution and replace it with an equal quantity of pure vinegar, to bring the concentration to 50%. What is the ratio of the amount of vinegar added in A and B, if the quantity of initial solutions in A and B are in the ratio of 1 : 2?

For A
vinegar : 40
water : 60
20 ml vinegar added
conc of vinegar = 50%For B
vinegar =80 + 3x/5
water = 1203x/5
80+3x/5 = 120  3x/5
x = 100/3 ml was added
20 : 100/3 = 3:5

Q9) If Ratio of Cost Price and Mark Price is 5:8 and that of % profit on sale to % discount is 2:3 then find % of Discount

profit on sale = profit on SP = Profit/SP *100%
CP=50, MP=80.
3 * (SP50) * 100/SP = 2 * D (D=discount%)
Also, SP = (1D/100) * 80 = 80 * (100D)/100
so 3 * {80(100D)/100 50 } * 100 = 2D * {80 * (100D)}/100
150 { 752D} = D {2002D}
D^2  250D + 5625 = 0
D= (250200)/2 = 25%

Q10) A faulty clock gains 10 minutes every hour. If the time is set correctly at 12 Noon on 1st Jan 2010,then how many times will its minutehand and hourhand meet in the next 24 hours ?
(a) 22
(b) 26
(c) 24
(d) 25