# Free Mixed Bag 12 Practice Test - CAT

### Question 1

What is the second last digit of 2564345 ?

#### SOLUTION

Solution :A

Option (a)

64345=22070=(210)207

this is (210)odd power. It will end with 24. The second last digit is 2. option (a)

### Question 2

If a_{1} = 1 and a_{n+1} – 3a_{n} + 2 = 4n for every positive integer n, then a_{100} equals

^{99}– 200

^{99}+ 200

^{100}– 200

^{100}+ 200

^{99}-100

#### SOLUTION

Solution :C

option (c)

Write options as

a) 3

^{n-1}- 2n,b) 3

^{n-1}+ 2n,c) 3

^{n}- 2n,d) 3

^{n}+ 2n,e) 3

^{n-1}-nCheck for n = 1, a

_{n}=1

### Question 3

Find the rightmost non-zero integer of the expression 1430343+1470367?

#### SOLUTION

Solution :C

option (c)

It can be easily observed that 1470367 has more number of zeroes as compared to

1430343. hence, the rightmost non-zero integer will depend on the unit digit of only 1430343, which in turn will depend on the unit digit of 3343 343 can written as 4k+3

In the power cycle of 3 (3,9,7,1), the third digit is 7. Hence, the last non-zero integer in the expression is 7.

### Question 4

The remainder of k69=1 how many values are possible for k if k is a natural number less than 50

#### SOLUTION

Solution :E

Option (e)

6 is the Euler’s number of 9 ,if you divide the 6th power of any number which is relatively prime with 9 , the remainder will be 1. answer = all numbers which are non-multiples of 3. there will be 503 multiples of 3. 50-16= 34 non-multiples of 3, which is the answer

### Question 5

For what values of ‘p’ would the equation x2+2(p−1)x+p+5=0 possess at least one positive root?

#### SOLUTION

Solution :A

For an equation to have positive roots D must be greater than '0'.

Now D from the equation =4(p−1)2−4(1)(p+5)=4p2−12p−16

=4(p2−3p−4)=4(p−4)(p+1)......(i)

x=[−b+√D]≥0

D≥b2

b=4(p−1)2

p2−3p−4≥p2+1−2p

p≤−5..........(2)

Taking intersection of equation 1 and 2

pE[−∞,−5]So roots for eq. (i) are : -1 and 4 and comparing with ax2+bx+c , we have 'a' as>0 so function will be open ended in the upward direction.

So, , D is positive in region [−∞,−1] and [4,∞]. So for equation to posses at least one positive 'p' should either lie in [−∞,−1] or in [4,∞]. So answer is option (b).

### Question 6

The roots of the equation x^{4}+mx^{3}+71x^{2}+px+q = 0 are in AP with common difference = 1. Find p+q, if m is negative.

#### SOLUTION

Solution :B

Option (b)

Sum of product of roots taken 2 at a time = 71

Roots =2, 3, 4, 5

p = - sum of product of roots taken 3 at a time = -154

q = product of roots = 120

p+q = -34

### Question 7

Find 1 + (xy) + (xy)^{2} + (xy)^{3} + …… if a= 1+ 2x + 3x^{2} + 4x^{3}+….. ( -1<a<1) and b= 1+3y + 6y^{2} + 10y^{3}+…. (-1<b<1)

^{1/2}b

^{1/3}) / (a

^{1/2}+ b

^{1/3}-1)

^{1/2}b

^{1/3}) / (a

^{1/2}+ b

^{1/3}+1)

**(a**

^{1/2}+ b^{1/3}+1) / (a^{1/2}b^{1/3})**(a**

^{1/2}+ b^{1/3}-2) / (a^{1/2}b^{1/3})**(a**

^{1/2}+ b^{1/3}+2) / (a^{1/2}b^{1/3})#### SOLUTION

Solution :A

Option (a)

Assume x=0 and y=0. Then the equation gives a value of 1. Put the values of x

and y to get a and b. a=1 and b=1

Go from answer options. Only answer option a gives 1

### Question 8

If a,b,c are the roots of the equation x^{3}-4x+24=0, then the equation who’s roots are b+c,c+a and a+b is given by?

^{3}-4x +24=0

^{3}-6x +24=0

^{3}+4x -24=0

^{3}-4x -24=0

#### SOLUTION

Solution :D

Option (d)

a+b+c=0 ( coefficient of x

^{2}is 0) thereforeb+c=-a a+b=-c and a+c=-b

the equation who’s roots are –a,-b and –c is obtained by replacing x by –x in the equation

the required equation is (-x

^{3})-4(-x)+24 = x^{3}-4x -24=0.

### Question 9

Given that 2m-1 is an odd number and 3n-1 is an even number. Which of the following are necessarily odd? 1) n2−4m+5 2) m2−2n+2 3) 6m2−n−1

#### SOLUTION

Solution :E

if 2m-1 is odd, m can be odd or even

If 3y-1 is even , then n has to be odd

n2−4m+5 (n2) will be odd. Hence 4m is even and 5 is odd`x therefore its even

m2−2n+2 (m2) will be even/odd, therefore it cannot be determined

6m2−n−1 (6m2) is even, n is odd and 1 is odd. Therefore it is even

### Question 10

How many ways a person can buy 13 fruits of three types, mangoes ,apples and oranges. He can buy a maximum of 7 of any type and minimum of 1.

#### SOLUTION

Solution :B

Option (b)

A+B+C=13

Applying the lower limit of 1,

A+B+C= 10 with A,B,C<6

Using the one zero method,

Total number of cases = 12C2−3×5C2=36