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 a1 = 1 and an+1 – 3an + 2 = 4n for every positive integer n, then a100 equals
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
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 x4+mx3+71x2+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 + 3x2 + 4x3+….. ( -1<a<1) and b= 1+3y + 6y2 + 10y3+…. (-1<b<1)
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 x3-4x+24=0, then the equation who’s roots are b+c,c+a and a+b is given by?
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.
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