Problems.

[goyalsau]

Question - 1

What is the product of all factors of the number N = 6^4 x 10^2, which are divisible by 5?

(A) 2^210 × 3^102 × 5^140
(B) 2^140 × 3^102 × 5^210
(C) 2^210 × 3^140 × 5^105
(D) 2^140 × 3^210 × 5^102
(E) 2^102 × 3^210 × 5^140


Question - 2

A speaks the truth in 70% cases and B in 80% cases. What is the probability that they will contradict each other in describing a single event?

(A) 0.42
(B) 0.56
(C) 0.62
(D) 0.36
(E) 0.38

Question - 3

Find the maximum value of the function f(x) = 24x - 9x2 -11.

(A) 18
(B) 5
(C) -11
(D) 4
(E) None of these

Question - 4

[towerSpider]

2,

.7*.2+.8*.3
.14+.24
.38

[goyalsau]

2,

.7*.2+.8*.3
.14+.24
.38

Can you please explain why is .7 * .2 + .8 * .3

[beat_gmat_09]

A & B will contradict when one says truth and other says false,
Two cases
1) A - Truth; B-False
A= Truth = 0.7; B=false = 1-0.8 = 0.2
P(A & B) = P(A)*P(B) = 0.7*0.2 = 0.14

2) A-False; B-Truth
A=False = 1-0.7 = 0.3; B=Truth = 0.8
P(A & B) = P(A)*P(B) = 0.3*0.8 = 0.24

Both cases can occur, P(both occur either 1st case or 2nd case) = 0.14+0.24 = 0.38

[naremnaresh]

f(x) = 24x - (9x) * x -11
if x= 0 then f(0) = -11
x = 1 then f(1) = 24 - 9 -11 = 24 - 20 = 4
x = 2 then f(2) = 48 - 36 - 11 = 48 - 47 = 1
for x >= 3
9x > 24 so f(x) will be negative always
if x < 0 values will be negative always

Maximum value of f(x) is 4
Option D is correct answer.

[naremnaresh]

For question -1
for N=64x102 there will not be any factor 5
so I think answer is 0.

[goyalsau]

For question -1
for N=64x102 there will not be any factor 5
so I think answer is 0.

I am really sorry Naresh, I forgot to edit the question. its not 64 & 102 it was 6^4 and 10^2 , I have edited the question, I know its very annoying when somebody post a wrong question, and ask other to come up with the solution, .

Hope you don't mind and will share your solutions.

[naremnaresh]

question1 answer

6^4 * 10 ^2 = 5^2 * 3^4 * 2^6


so product of factors will be

(5)* (5*2)*(5 *2^1)*....*(5*2^6)* = 7 times 5 , 6*7/2 = 21 times 2 and 0 times 3
(5*3)*(5*2*3)*(5 *2^1*3)*....*(5*2^6*3)* = 7 times 5 , 6*7/2 = 21 times 2 and 7 times 3
(5*3^2)*(5*2*3^2)*(5 *2^1*3^2)*....*(5*2^6*3^2)* = 7 times 5 , 6*7/2 = 21 times 2 and 14 times 3
.......s0 on

(5^1)* (5^1*2)*(5^1 *2^1)*....*(5^1*2^6)* = 14 times 5 , 6*7/2 = 21 times 2 and 0 times 3
(5^1*3)*(5^1*2*3)*(5^1 *2^1*3)*....*(5^1*2^6*3)* = 14 times 5 , 6*7/2 = 21 times 2 and 7 times 3
(5^1*3^2)*(5^1*2*3^2)*(5^1 *2^1*3^2)*....*(5^1*2^6*3^2)* = 14 times 5 , 6*7/2 = 21 times 2 and 14 times 3

......s0 on


so total time 5 will be = 7 * 5 + 14 *5 = 105
so total time 2 will be = 21 * 5 + 21 *5 = 210
so total time 5 will be = 2 * (0+7+14+21+28) = 140

answer is 2^210 X 3^140 X 5^105

option C is correct

[goyalsau]

question1 answer

6^4 * 10 ^2 = 5^2 * 3^4 * 2^6


so product of factors will be

(5)* (5*2)*(5 *2^1)*....*(5*2^6)* = 7 times 5 , 6*7/2 = 21 times 2 and 0 times 3
(5*3)*(5*2*3)*(5 *2^1*3)*....*(5*2^6*3)* = 7 times 5 , 6*7/2 = 21 times 2 and 7 times 3
(5*3^2)*(5*2*3^2)*(5 *2^1*3^2)*....*(5*2^6*3^2)* = 7 times 5 , 6*7/2 = 21 times 2 and 14 times 3
.......s0 on

(5^1)* (5^1*2)*(5^1 *2^1)*....*(5^1*2^6)* = 14 times 5 , 6*7/2 = 21 times 2 and 0 times 3
(5^1*3)*(5^1*2*3)*(5^1 *2^1*3)*....*(5^1*2^6*3)* = 14 times 5 , 6*7/2 = 21 times 2 and 7 times 3
(5^1*3^2)*(5^1*2*3^2)*(5^1 *2^1*3^2)*....*(5^1*2^6*3^2)* = 14 times 5 , 6*7/2 = 21 times 2 and 14 times 3

......s0 on


so total time 5 will be = 7 * 5 + 14 *5 = 105
so total time 2 will be = 21 * 5 + 21 *5 = 210
so total time 5 will be = 2 * (0+7+14+21+28) = 140

answer is 2^210 X 3^140 X 5^105

option C is correct

Thanks Buddy ,Even I did the same way But it very lengthy At least for me it is,

I was searching on the net some tricks for these kind of problems, this is what i got,

2^6 * 3^4 * 5 *2

If we are ask to find the total no. of factors then it's 7 * 5 * 3 = 105

Why we add 1 to the power of prime factorization, because factors are from 0 to 7

2^6 = 2^0 , 2 ^1 , 2^2 , 2^ 3 ...... 2 ^ 6 ------ 7 different powers

same with 3 and 5


So the factorization should be 2^0 , 2^1,..... 2^6 , 3^0 , 3^1 ... 3^4 ,,, 5^1 , 5^2

Because 5^0 is one and we need at least one 5

sum of powers of 2 i.e. 0+1+2+3 .... +6 = 21

total no. of 3 is 5 and total no. of 5 is 2 .

21 * 5 * 2 = 210

same ways with 3

sum of powers of 3 is 0+1+2+3+4 = 10

total no. 2 is 7 and total no. of 5 is 2

10* 7 *2 = 140

same with 5

sum of power of 5 is 1+2 = 3

total no. of 2 is 7 & total no. of 3 is 5

7 * 5 * 3 = 105


2 ^ 210 * 3 ^140 * 5 ^ 105