# Free Mixed Bag 12 Practice Test - CAT

What is the second last digit of 2564345 ?

A. 2
B. 4
C. 8
D. 0
E. None of these

#### 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)

If a1 = 1 and an+1 – 3an + 2 = 4n for every positive integer n, then a100 equals

A. 399 – 200
B. 399 + 200
C. 3100 – 200
D. 3100 + 200
E. 399-100

#### SOLUTION

Solution : C

option (c)

Write options as

a) 3n-1- 2n,

b) 3n-1+ 2n,

c) 3n- 2n,

d) 3n+ 2n,

e) 3n-1-n

Check for n = 1, an=1

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

A. 3
B. 9
C. 7
D. 1
E. 2

#### 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.

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

A. 6
B. 20
C. 15
D. 32
E. 34

#### 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

For what values of ‘p’ would the equation x2+2(p1)x+p+5=0 possess at least one positive root?

A. [,5]
B. [,1]
C. [1,]
D. [2,]
E. none of these.

#### SOLUTION

Solution : A

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

Now D from the equation =4(p1)24(1)(p+5)=4p212p16

=4(p23p4)=4(p4)(p+1)......(i)
x=[b+D]0
Db2
b=4(p1)2
p23p4p2+12p
p5..........(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).

The roots of the equation x4+mx3+71x2+px+q = 0 are in AP with common difference = 1. Find p+q, if m is negative.

A. –22
B. –34
C. –24
D. 34
E. None of these.

#### 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

Find 1 + (xy) + (xy)2 + (xy)3 + …… if a= 1+ 2x + 3x2 + 4x3+….. ( -1<a<1) and b= 1+3y + 6y2 + 10y3+…. (-1<b<1)

A. (a1/2b1/3) / (a1/2 + b1/3 -1)
B. (a1/2b1/3) / (a1/2 + b1/3 +1)
C. (a1/2 + b1/3 +1) / (a1/2b1/3)
D. (a1/2 + b1/3 -2) / (a1/2b1/3)
E. (a1/2 + b1/3 +2) / (a1/2b1/3)

#### 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

If a,b,c are the roots of the equation x3-4x+24=0, then the equation who’s roots are b+c,c+a and a+b is given by?

A. x3 -4x +24=0
B. x3 -6x +24=0
C. x3 +4x -24=0
D. x3 -4x -24=0
E. none of these

#### SOLUTION

Solution : D

Option (d)

a+b+c=0 ( coefficient of x2 is 0) therefore

b+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 (-x3)-4(-x)+24 = x3 -4x -24=0.

Given that 2m-1 is an odd number and 3n-1 is an even number. Which of the following are necessarily odd? 1) n24m+5 2) m22n+2 3) 6m2n1

A. Only 2
B. Both 1 and 2
C. Only 3
D. Only 1
E. None of the above options

#### 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

n24m+5 (n2) will be odd. Hence 4m is even and 5 is odd`x therefore its even

m22n+2 (m2) will be even/odd, therefore it cannot be determined

6m2n1 (6m2) is even, n is odd and 1 is odd. Therefore it is even

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.

A. 21
B. 36
C. 66
D. 26

#### 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 = 12C23×5C2=36