Free Mixed Bag 11 Practice Test - CAT 

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.
$\to$ But the highest power of $2^{97}$ in 10000! is 103
This step needs to kept in mind
hence 103 is the highest power of 100! in 10000!

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 $1^{st}$ 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 $1^{st}$  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.

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)

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.

Option (b)

3 sides of the triangle along with the perpendicular forms 2 right angled triangles. The 3${}^{rd}$ side(base of triangle) is hence $\sqrt{(5^2-3^2)}$ + $\sqrt{(5^2-3^2)}$ = 8.
Area of the triangle = $\frac{1}{2}$.3.8 = rs = (a + b + c) $\frac{r}{2}$, where s is the semiperimeter and r is the inradius of the triangle. Solving r = $\frac{24}{18}$ = 1.33

Option (e)

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

f(1) = 0

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 $1^{st}32$ operations, you have 32 64s left. The pattern will look like this:-

After 32 $\to$ 32 64s

After 16 more $\to$ 16 127s

After 8 more $\to$ 8 253s

After 4 more $\to$ 4 505s

After 2 more $\to$2 1009s

After 1 more $\to$1 2017

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

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:

$\frac{x}{560}\ge \frac{(2400-x)}{640}$

$\Rightarrow x\ge 1120km.$

$\Rightarrow$  Minimum such distance  = 1120 km and time taken to cover 1120 km $=\frac{1120}{560}=2hr.$

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

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 D_n+1 = D₂ = 37

Look in the answer options for 37

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