Free Mixed Bag 2 Practice Test - CAT 

option d

$7^{3001}-1$ and $7^{3001}+1$ are consecutive even natural numbers

Hence, their $HCF=2$

$LCM=\frac{\text{product of numbers}}{\text{HCF}}=\frac{(7^{8001}{)}^2-1}{2}=\frac{7^{6002}-1}{2}$

$7^{6002}$ ends with 9 (use the concept of power cycle)

So $7^{6002}-1$ ends in 8. We also know that as $7^{6002}$ is a perfect square of an odd number, its tens digit will be even. So the required unit digit is 4.

option (c)

Finding the sum of even multiples of 15 is equivalent to finding the sum of multiples of 30.

By observation, the first multiple of 30 greater than 295 will be equal to 300 and the last multiple of

30 smaller than 615 will be equal to 600.

The sum of a set = (the mean of the set) $\times$ (the number of terms in the set)

There are 11 terms in the set

The mean of the set = (the first term + the last term) divided by 2: $\frac{(300+600)}{2}=450$

If k= the sum of this set

$k=450\times 11$

Therefore, the largest prime factor of k is 11. The correct answer is C.

Using the technique for inverse function at x=1, f(x) = -1

Replace x by f(x) in the answer options and

Look for an answer options where x= -1, & f(x)=1

Answer is option (d)

2nd method: - Let y = x² - 2x 

 x² - 2x - y = 0

option c

Solution:

Let first two numbers be x and y.. so the sequence will be

$x,y,\frac{y}{x},\frac{1}{x},\frac{1}{y},\frac{x}{y},x,y,..$ the sequence repeats after 6 numbers. Also the product of first six numbers is 1.

So now, product of first 50 numbers will be xy=27 while product of first 100 numbers will be $\frac{y^2}{x}=27$. On solving we get y=9 and x=3.

So their sum = x+y=12.

Alternatively

Look at the product, it is 27 which is nothing but $3^3$and the series is all about products. So the numbers in the series will be full of threes. Look at the answer options for numbers which can be expressed as sum of 2 powers of 3. Only $12=3+3^2$is the possibility.

Series is $3,3^2,3^3,\frac{1}{3},\frac{1}{3^2},\frac{1}{3},3,3^2.............$

Two varieties @ Rs. 126 and Rs. 130 are mixed in equal ratio of 1: 1. So, their average price can be taken as $\frac{(126+130)}{2}=Rs.128$ per kg. Now we have two varieties: one at the rate Rs. 128 per kg and other at the rate Rs. x per kg which are mixed in the ratio 2: 2 or 1: 1 to give a resulting mixture of Rs. 172 per kg.

Applying Alligation:

$\frac{X-172}{44}=\frac{1}{1}$

 x = 216.

So, the price of the third variant is Rs. 216 per kg.

As it takes 4 mins to rescue the passengers, in 16 mins the rescue vessel should reach the ship. Now the ship too is traveling towards the shore with 8kmph.

Let the speed of the rescue vessel be X.. Then  $X\times \frac{16}{60}+8\times \frac{16}{60}=12$

$\Rightarrow$ X = 37.

Ans: d. 2:5

Using variables

In the given figure, we draw QT || VW.

$\Delta$ PTQ and $\Delta$ PVS are congruent (A,A,A and side)

Hence PT=Y, QT=X=UT (also since $\Delta$UTQ and $\Delta$UVW are similar).

Thus UV=UW=2X+Y. Area of square: $X^2+Y^2$; Area of  $\Delta$UVW= $\frac{1}{2}$ *(2X+Y)${}^2$.

Given X=2Y, $X^2+Y^2$: ( $\frac{1}{2}$ *(2X+Y)${}^2$) = 5Y${}^2$: ( 25Y${}^2$* $\frac{1}{2}$ ); = $\frac{2}{5}$.

 Using numbers,You can solve the problem faster as follows:

PV=2, SV=1 => PS= $\sqrt{5}$ $\Rightarrow$ Area of square =5

PTQ congruent to PVS

PT=1 and QT=2

UT=TQ (45-45-90) $\Rightarrow$ UT= 2

Now UV= 5 $\Rightarrow$ WV=5 (45-45-90)

UW=5$\sqrt{2}$

option (d)

f(x) = [(x + 1)/x] [(x + 2)/(x +1)] … [(x +n + 1)/(x + n)] = [(x + n + 1)/x]

g(x) = [(x – 1)/x] [(x -2)/(x – 1)] … [(x-n-1)/(x – n)] = [(x – n -1)/x]

 f(x) + g(x) = (x + n + 1 + x - n – 1)/x = 2

Hence option (d)

Shortcut:- Single Substitution

Put x=1, n=0

f(x)= 2

g(x)=0 => f(x) +g(x) =2

put x=1 and n=0 and look for 2 in the answer options. Answer is option (d)

In 1 hour Ram can do $10\%$ of the work, Ajit can do  $\frac{100}{12}\%$ of the work and Suresh can do  $\frac{100}{15}\%$  of the work.

So, total number of hours = 2 + 2 + 3 = 7 hours.

Option (e)

Assume a=1, b=2, c=3

X=-3 Y=-5 Z=2

XY+YZ+XZ=15-10-6=-1

Assume a=2, b=3 and c=4

X=-5 Y=-7 Z=3

XY+YZ+XZ= 35-21-15= -1

Hence answer =e

option (c )

Divisibility test for 3- sum of digits should be divisible by 3.

Case 1:- When 0 is not included $\to$ sum = 15 divisible by 3

Number of combinations = 5! = 120

Case 2:- When 3 is not used. $\to$ Sum = 12 divisible by 3

Number of combinations - 4.4..3.2 =96

Total = 120+96= 216

option d

For the product (2A5)(13B) to be divisible by 36, we need it to be divisible by both 4 and 9. Since 2A5 is odd, it does not contain a factor of 2.Therefore, 13B must be divisible by 4.

For a positive integer to be divisible by 4, the number formed by its last two digits must

be divisible by 4, i.e. 3B is divisible by 4, i.e. B= 2 or B = 6.

Case 1: B = 2

In this case, 132 is divisible by 3, but not by 9. Therefore, for the original product to be

divisible by 9, we need 2A5 to be divisible by 3.

For a positive integer to be divisible by 3, the sum of its digits is divisible by 3, i.e.

2 + A + 5 = A + 7 is divisible by 3.

Therefore, A = 2 or 5 or 8.

Case 2: B = 6

In this case, 136 contains no factors of 3, so for the original product to be divisible by 9,

we need 2A5 to be divisible by 9.

For a positive integer to be divisible by 9, the sum of its digits is divisible by 9, i.e.

2 +A+5 is divisible by 9. Therefore, A is 2.

Therefore, the four possible ordered pairs are $\Rightarrow$A, B = 2,2, 8,2, 5,2, 2,6.

Soln:

Option (b)

f₄ + f₅ = 2f₃ + f₄ = 2f₂ + 2f₃ = 4f₁ + 2f₂ = 4f₀ + 4f₁ = 4f₀ + 4f₀ = 8f₀ = 8

Hence option (e)

OR starting with

f(0) = 1

F(1) = 1

F(2) = 2

F( 3) = 2

F(4) = 4

F (5) = 4

F(6) = 8

F(7) = 8

……………..

…………….

F(N) =2^N/2

F(N+1) = 2^N/2

Therefore f( 20) = 2¹⁰and f(21) = 2¹⁰

f(20) + f(21) =1024+1024 = 2048

As we know, S = I + 2 II + 3 III

and X = I + II + III

S - X = II + 2 III

If we want to maximize exactly III. We need to put exactly II = 0

$\frac{(S-X)}{2}\Rightarrow \frac{(224-120)}{2}=52$

Short Cut: - Maximum of all can be directly calculated as

$\frac{S-X}{n-1}$, Where n is number of things.

Answer is $\frac{(224-120)}{2}=52.$ Option (a)

1st method: -

80 people like Shahrukh, it means at least 20 people like Amir or Akshay or Amir and Akshay both.

Similarly atleast 36 people like Shahrukh or Akshay or Shahrukh and Akshay both.

Again atleast 20 people like people like Shahrukh or Amir or Shahrukh and Amir both.

40 + 56 + 40 = 136

It means that if there is no interection in these three courses, 136 would be maximum number of people who like Shahrukh, Amir, Akshay alone or 2 at a time.

Thus it gives atleast 0% people like all three.

2nd method: - Use minimum of all technique taught in class

120-80= 40

120-64=56

120-80=40

Sum of all differences = 136

$120-136<0$. Hence, the minimum is 0. Option (e)