Free Mixed Bag 11 Practice Test - CAT

10000!=(100!)KP, where K and P are integers. Then what is the maximum value of K?

A. 103
B. 104
C. 105
D. 102
E. 100

SOLUTION

Solution : A

option (A)

The key here are the prime numbers. K should be selected in such a way that it will depict the value of the highest power of the largest prime number before 100 which is a factor of 10000!. The largest prime number below 100 is 97 and the highest power of 97 in 10000! is 104.
But the highest power of 297 in 10000! is 103
This step needs to kept in mind
hence 103 is the highest power of 100! in 10000!

A, B and C start running simultaneously from point P, Q and R respectively on a circular track. The distance between any two of the three points P, Q and R is L and the ratio of the speeds of A, B and C are 1:2:3. If A and B run in opposite directions while B and C run in the same direction, what is the distance run by A before A, B and C meet for the 3rd time?

A. 10 L
B. A, B and C never meet
C. 403L
D. 12L

SOLUTION

Solution : B

Option (b)

A, B, C are separated by distance L each from each other. For all of them to meet, A should meet B and C at the meeting point of B and C. If we assume that C takes 3 units of time to cover the whole circle once, then B and C meets at the point R for the 1st time after 3 units of time. Then they keep meeting at the same point 9 units of time. That is at times 3, 12, 21 etc. But A will reach R for the 1st  time 6 units of time after the start and keeps doing that after every 9 units of time after that. That is at times 6, 15, 24 etc.

Find the number of solutions of 13A-18B=2 where A and B are natural numbers between 1 and 100.

A. 6
B. 5
C. 3
D. 1
E. 7

SOLUTION

Solution : B

option (b)

Based on Chinese Remainder theorem

The number can be re-written as 13A=18B+2-------(1)

Reduce the number to 13A=5B+2 (removing 13 from RHS)

Now find the first value of B for which A is an integer. In this case, B=10 is the first value for which A is also an integer.

B=10 and A=14 are the first integer solutions.

A will follow the AP 14 + 18d; d can take values 0 to 4 B will follow the AP 10 + 13d; d can take values 0 to 6A will be the limiting value and hence the solution is 5 and the answer option b)

In a quadratic function f(x) = ax2 + bx + c. It is known that f(1) = f(4) and f(2) = 0, then what is the second root of the expression?

A. 3
B. 5
C. 6
D. –1
E. –2

SOLUTION

Solution : A

option (a)

Solution: Let 'a' be the second root with '2' being the first root.

Then the given equation can be written as: (x-2)(x-a)=0--------------(1)

Now, we know that f(1)=f(4)-------------------(2)

Using (2) in (1):

(1-2)(1-a)=(4-2)(4-a)

Therefore, a=3 is the correct answer.

In a triangle ABC, 2 sides AB and AC are of length 5 units each. If a perpendicular of length 3 units is dropped from A to BC, then what is the length of the radius of the circle that can be inscribed in the triangle?

A. 3
B. 1.33
C. 2
D. 2.5

SOLUTION

Solution : B

Option (b)

3 sides of the triangle along with the perpendicular forms 2 right angled triangles. The 3rd side(base of triangle) is hence (5232) + (5232) = 8.
Area of the triangle = 12.3.8 =
rs = (a + b + c) r2, where s is the semiperimeter and r is the inradius of the triangle. Solving r = 2418 = 1.33

f(x) is a real valued function where f(xy) = [f(x)] + [f(y)] + f(x)f(y) for all real x,y. If f(2) = 1, then what is the value of f(1/2) ?

A. 1
B. 2
C. 0
D. –1
E. -1/2

SOLUTION

Solution : E

Option (e)

f(2) = f(2 x 1) = f(2) + f(1) + f(2).f(1) = 1

f(1) = 0

The integers 1,2,.........64 are written on a blackboard. The following operation is repeated 63 times: Any 2 numbers are chosen out of these 64 numbers and replaced with a number equal to 1 minus the sum of 2 numbers. What will be the number left over on the board after 63 operations?

A. 2017
B. 1009
C. 505
D. 253
E. 2016

SOLUTION

Solution : A

Option (a)

They key here is any number, so basically choose the easiest pair of numbers and solve the question. My choice will be the numbers distanced equally from the middle value such as  1 & 64, 2 & 63  etc. All of them will be replaced by 64. So after the 1st32 operations, you have 32 64s left. The pattern will look like this:-

After 32 32 64s

After 16 more 16 127s

After 8 more 8 253s

After 4 more 4 505s

After 2 more 2 1009s

After 1 more 1 2017

Total number of operations = 32 + 16 + 8 + 4 + 2 + 1 = 63

A plane is flying 2400 km from source A to destination B. Speed of the plane is 600 km/hr and speed of the wind tailing it is 40 km/hr. How many hours after take off would it be faster to go to B than to return to A in case of an emergency

A. 1.65 hours
B. 1.5 hours
C. 1.75 hours
D. 2 hours

SOLUTION

Solution : D

From A to B speed of flight = 640 km/hr , while for the reverse it is = 560 km/hr.

Since, ‘x’ is the distance from A, hence, as per the question:

x560(2400x)640

x1120  km.

Minimum such distance  = 1120 km and time taken to cover 1120 km =1120560=2 hr.

The sum of the all the products of the first n positive integers taken two at a time is?

A.
B.
C.
D.
E.

SOLUTION

Solution : C

Option (c)

Shortcut: Assumption

xy when n=2 is 1.2

Sum= 2

Substitute n=2 in answer options and see where 2 is obtained. Option c

Given that p and q are the roots of the equation x2 – ax +b =0 and Dn= pn+qn. Find the value of Dn+1

C. bDn
E. pDn

SOLUTION

Solution : A

Option (a)

Shortcut: Assumption

Assume a quadratic equation. Let’s take x2+5x-6=0.

Thus, a = sum of roots = -5 and b= product of roots = -6

Roots are p = 1 and q=-6

D1= p1+q1 = -5

D0 = 2

D2 = 37

Assume n=1

We need to find Dn+1 = D2 = 37

Look in the answer options for 37

Option a is the only one which gives (-5)(-5)-(-6)(2) = 37