Download quintessence

INFOMATHS
MARATHON-5 SET-A SOLUTIONS
1.
Ans. (b)
Min. value = 10! 11!
 n  1!
r ! n  r  1!
2
Ans. (b)
(7C0 + 7C1) + (7C1 + 7C2) + ….. + (7C6 + 7C7)
= 8C1 + 8C2 + ….. + 8C7
= 8C0 + 8C1 + ….. + 8C7 + 8C8 – (8C0 + 8C8)
= 28 – (1 + 1) = 28 – 2
  k 2  3
3.
4.
 1   k 2  3 .
n
r 1
r

1
 k2 3 
n
Since n  0 and r  1  n 
r 0
n
r 1
n
0 
1 0  k2  3  1
r 1
 3  k2  4
 3  k  2  3  k  2 or k
1
1
1
1
1
 n  n  n  ...  n
cr
c0
c1
c2
cn
1
1
1
 an  1  n  n  ...  n
c1
c2
cn
n
Let Tn   n
r 0

n
…(1)
7.
Ans. (a)
Total number of letters = 6
Number of vowels A, E = 2
Alphabetical order of vowels is A, E
Total number of arrangements for letters of
GARDEN = 6!
Vowels can be arranged in 2! Ways among
themselves.
But actually there is only way A, E.
6!
 Reqd. no. of ways   360.
2!
8.
Ans. (d)
Reqd. number of triangles
= 8C3 – 5C3 – 3C3
8 7  6 5 4


1
1 2  6 1 2
= 56 – 10 – 1 = 45.
9.
Ans. (c)
Bold lines have shown one such square whose
diagonal is of length 2. From one set of lines, two
sides of a square can be chosen in 5 ways.
(i) r = 0, 2 (ii) r = 1, 3 (iii) r = 2, 4 (iv) r = 2, 5
(v) r = 4, 6
…(2)
…(3)

1 
 Tn   an  1    an  1  n 
c1 

 n  2
1
2
3
 n  n  n  ....  n
c3
c4
c5
cn
Proceeding in the same way, we have

n 1 n  2
1 
Tn  nan   n  n  n  ...  n
0
c
c
c
1
2
n 1 

 n
n 1 n  2
1 
or Tn  nan   n  n
n
 ...  n   0
cn 1
cn  2
c1 
 cn
or Tn – nan + Tn = 0
1
 Tn  nan
 2Tn = nan
2
5.
Ans. (b)
We know that in any triangle the sum of two sides is
always greater than the third side.
 the triangles will not be formed if we select
segments of lengths (2, 3, 5), (2, 3, 6) and (2, 4, 6)
Hence no. of triangles formed = 5c3 – 3
6.
Ans. (d)
By the given condition,

2  k   3
 k  3, 2 or k   2,  3 
 (d) is the correct answer.
r
cr
1
2
3
n


 ...  n
c1 n c2 n c3
cn
(2) – (1) gives
1
2
n 1
Tn   an  1  n  n  ...  n
c2
c3
cn

n
0
r 1
and 1 
Ans. (c)
n
where
0  r  n 1
Ans. (a)
n
Cr + 4. nCr-1 + 6. nCr-2+4. nCr-3 + nCr-4
= (nCr + nCr-1) + 3(nCr-1 + nCr-2)
+ 3(nCr-2 + nCr-3) + (nCr-3 + nCr-4)
n+1
n+1
= Cr + 3 Cr-1 + 3. n+1Cr-2 + n+1Cr-3
= (n+1Cr + n+1Cr-1) + 2(n+1Cr-1 + n+1Cr-2)
+(n+1Cr-2 + n+1Cr-3)
= n+2Cr + 2. (n+2Cr-1 + n+2Cr-2)
= (n+2Cr + n+2Cr-1) + (n+2Cr-1 + n+2Cr-2)
= n+3Cr + n+3Cr-1 = n+4Cr
an   n
n!
 r  1! n  r  1!
1
INFOMATHS/MCA/MATHS/
INFOMATHS
Again from the second set of lines, two sides can be
chosen in 5 ways.
 The reqd. no. of squares = 5  5 = 25
Joining 3 points on the same line gives no triangles,
such that ’s are mC3 + nC3 + kC3
Required number = m+n+kC3 – mC3 – nC3 – kC3
10. Ans. (c)
16. Ans. (a)
We have in all 12 points. Since 3 points form a
The number of words before the word CRICKET
triangle.
= 4  5! + 2  4! + 2! = 350
 The total no. of triangles, including the triangles
formed by collinear points on AB, BC and CA, is
17. Ans. (d)
Along horizontal side, one unit can be taken in (2m –
12

11

10
12
C3 
 220
1) ways and 3 unit side can be taken in 2m – 3 ways.
1 2  3
 The number of ways of selecting a side
But this includes the no. of triangles formed by 3
horizontally is
3
points on AB i.e. C3 = 1 and for 4 points on BC =
(2m – 1 + 2m – 3 + 2m – 5 + … + 3 + 1)
4
C3 = 4
Similarly, the number of ways along vertical side is
5 4
5
(2n – 1 + 2n – 3 + … + 5 + 3 + 1)
and 5 points on BC i.e. C3 
 10
1 2
2m-1
 reqd. number of triangles
= 220 – 1 – 4 – 10 = 205
11. Ans. (c) Number of lines from 6 points = 6C2 = 15
2n - 1
Points of intersection obtained from these lines =
15
C2 = 105
Now, we find the number of times, the original 6
points come.
Consider one point say A1 joining to remaining 5
 The total number of rectangles
points, we get 5 lines and any two lines from these 5
= [1+3+5 + … + (2m – 1)][1 + 3 + 5+…+(2n – 1)]
lines given A1 as pt. of intersection.
5
m 1  2m  1 n 1  2n  1
 A1 come to C2 = 10 times in 105 pts. Of


 m2 n2
intersection. Similar is the case with other five
2
2
points.
 6 original points come 6  10 = 60 times in points 18. Ans. (b)
of intersection.
Clearly, nC3 = Tn
Hence, the number of distinct points of intersection =
So, n+1C3 = 21
105 – 60 + 6 = 51
 (nC3 + nC2) – nC3 = 21
 nC2 = 21
12. Ans. (c)
or n(n – 1) = 42 = 7.6
Since 3, 3, 5, 5, i.e. odd digits occupy even positions
n=7
2, 4, 6, 8 and 2 2 8 8 8 occupy remaining 5 odd
places.
19. Ans. (a)
4! 5!
Case I When A is excluded.
.
 reqd. no. of ways 
Number of triangles = Selection of 2 points from AB
2!2! 2!3!
and one point from AC + Selection of one point from
24  120

 60
AB and two points from AC
8 6
1
 m C2 n C1  m C1 n C2   m  n  2  mn
…(i)
2
13. Ans. (a) 240 is divisible by 4n + 2
Case II When A is included.
or 120 is divisible by 2n + 1.
The triangles with one vertex at A = Selection of one
Numbers of the form (2n + 1), n  I are all odd
point from AB and one point from AC = mn.
natural numbers.
 Number of triangles
Thus we have to find all odd number dividing 120.
1
These numbers are 1, 3, 5, 15
 mn  mn  m  n  2 
Hence number of divisors = 4.
2
1
 mn  m  n 
…(ii)
14. Ans. (b)
2
For n = 6, 3  3  3  3  3  3 = 729 < 900
 m  n  2
For n = 7, 3  3  3  3  3  3  3 = 2187 > 900
 Required ratio 
 m  n
For n = 8,
Number of n-digits formed > 900
20. Ans. (d)
Since the least n required.  n = 7.
Prob. that one test is held
15. Ans. (b)
1 4 8
 2    . Prob. That test is held on both days
Total number of points are m + n + k, the ’s formed
5 5 25
by these points = m+n+kC3
2
INFOMATHS/MCA/MATHS/
INFOMATHS
From (i), (ii) and (iii), we have
1
1
1
2
  x  , 3  x  1 and   x  .
3
3
2
2
Hence
 1 13   1 2 
 1 1
x   ,     ,    3,1    , 
3
3
3
3

 

 2 2
1 1 
 x , 
3 2 
1 1 1
  
5 5 25
 prob. that the student misses atleast one test
8
1
9



25 25 25
21. Ans. (d)
The numbers should be divisible by 6.
 no. of favourable ways = 16C3
( there are 16 numbers in the first 100 natural 25. Ans. (a) Total numbers of ways of selecting two
numbers which are divisible by 6
horses is 5C2 = 10.
e.g. 6, 12, 18, 24, …., 96)
If Mr. ‘A’ selects the winning horse then he is to
16
choose one more horse from the remaining four
C3
 required probability  100
horses which he can do in 4C1 = 4 ways.
C3
4 2
16  15  14

 required probability 

10 5
100  99  98
4  15  1
4
4
26. Ans. (d)



25  99  7 5  33  7 1155
Let P(C) = x  P(B) = 2x.
P(A) = 2 P(B) = 4x
22. Ans. (a) P(B  C)
But P(A) + P(B) + P(C) = 1
 x + 2x + 4x = 1  7x = 1
1
4
x 
 P  A  4 x 
7
7
4 3
Hence P A  1  P  A   1  
7 7
 

 
27. Ans. (b)

 P  B  P A  B  C  P A  B  C 


3 2 98 1
  

4 3
12
12
3
1
and P  B  
8
2
3 1 3
P  A P  B   . 
8 2 16
2 1
and P  A  B     P  A  .P  B 
8 4
 A and B are dependent.
P  A 
23. Ans. (a)
No. of selections having both numbers greater than 4
is 3C2.
3
C
 required probability  1  6 2
28. Ans. (c)
C2
Let A denote prob. Of getting 53 Sundays and B
3
1 4
denote prob. Of getting 53 Tuesdays and C denote
 1  1 
15
5 5
prob. Of getting 53 Thursday s
Reqd. prob. = P(A + B + C)
= P(A) = P(B) + P(C)
24. Ans. (b)
Since probability of an event is a non-negative
[ events are independent]
number not more than 1,
In 365 days, no. of weeks = 52 and 1 day
 we must have
That one day can be any one of the seven days
3x  1
1
0
1
…(i)
 P  A   P  B   P C 
3
7
1 x
1 1 1 3
0
1
…(ii)
 reqd. prob.    
4
7 7 7 7
1  2x
1
and 0 
…(iii)
2
 A

B  1 P  A  B
29. Ans. (a) P    P  A   
Also 0  P(A) + P(B) + P(C)  1
B
B
P B
 

( Events are M.E.)
1 P  A  B
3x  1 1  x 1  2 x

0


1
P B
3
4
2
 0  13  3x  12  13  3x  1
1
13
30. Ans. (b)
 x
…(iv)
P(M) = m, P(P) = p, P© = c
3
3
 
 
3
INFOMATHS/MCA/MATHS/
INFOMATHS
The prob. Of at least one success
P(M  P  C)
 m  p  c  mp  mc  pc  mcp 
 reqd. prob. 
3
4
…(1)
34. Ans. (b)
1st Test 2nd test 3rd test 4th test.
Number of ways in which the two faulty machines
may be detected (depending upon the test done to
identify the faulty machines)
= 4c2 = 6
No. of favourable cases = 1
[When faulty machines are identified in the first and
the second test]
1
 Reqd. prob. 
6
OR
This is a problem of without replacement.
2 1 1 1 1
Reqd. Prob.     
4 3 2 3 6
The prob. Of at least two successes
 mc p  mcp  mcp  mcp
 mc 1  p   mp 1  c   1  m cp  mcp
1
2
The prob. of exactly two successes
 mc p  mcp  mcp
 mc  mp  cp  2mcp 
…(2)
 mc 1  p   mp 1  c   1  m
2
5
1 2 1
(2) – (3) gives mcp   
2 5 10
2 1 1 1 7
 mc  mp  cp     
10 2 5 2 10
7 1 3
From (1), m  p  c   
10 10 4
3 7 1
 m pc   
4 10 10
30  28  4 54 27



40
40 20
 mc  mp  cp  3mcp 
2  25  25 1

100  100 8
…(3)
35. Ans. (b)
1
3
P  A  , P  A  B  
3
4
Now P  A  B   P  A  P  B   P  A  B 
 P  A  P  B 
3 1
3 1
  P  B  or   P  B 
4 3
4 3
94
5
or
 P  B   P  B
12
12
Again B  A  B
31. Ans. (b)
Total number of equality likely cases = 26C3
Favourable cases = 1
[ there is only one way to appear in alphabetical
order]
1
 reqd. prob.  26
C3
 P  B  P  A  B 
Hence
3
4
5
3
 P  B 
12
4
36. Ans. (c)
32. Ans. (d)
P A B
 A P A B
Events favourable to E1 are
P   

(1, 5), (2, 4), (3, 3), (4, 2), (5, 1)
P B
P B
B
Events favourable to E2 are
1 P  A  B
(2, 1), (2, 2), (2, 3), (2, 4), (2, 5), (2, 6), (6, 2), (5, 2),

(4, 2), (3, 2), (1, 2)
P B
P (E2/E1) = Prob. of E2 when E1 has already occurred
2
 [Events favourable to E2 out of E1 = (2, 4), (4, 37. Ans. (a)
5
We must have 0.1 + 2k + k + 0.2 + 3k + 0.1
2)]
= or 6k = 1 – 0.4 = 0.6
Or
0.6 1
k 

 0.1
2
6 10
 E2  P  E1  E2  36 2
P  


5
P  E2 
5
 E1 
38. Ans. (a)
36
1 1
S.D.   npq  16    4  2
2 2
33. Ans. (c)
m
n
Number of possible cases = 100  100. For 7 + 7
to be divisible by 5, one of the term has to end with 9 39. Ans. (a)
p x 0.1  0.4  0.9  1.6
and other with 1 (7x cannot divide with 9).
3
Mean x  i i 
 m can be 2, 6, 10, 14, ……, 98 (25 values) and n
pi
1
can be 4, 8, 12, ….., 100 (25 values) Since m and n
2
Variance  xi2 pi  x
can interchange.

 


 

 

= 1 (0.1) + 2 (0.2) + 32 (0.3) + 42 (0.4) – 9
2
4
2
INFOMATHS/MCA/MATHS/
INFOMATHS
= 0.1 + 0.8 + 2.7 + 6.4 – 9 = 10 – 9 = 1
 S.D. = 1
 (b) holds
43. Ans. (d)
In the 22nd century there are 25 leap years viz. 2100,
2104, ….., 2196 and 75 non-leap years.
Consider the following events:
E1 = Selecting a leap year from 22nd century
E2 = Selecting a non-leap year from 22nd century
A = There are 53 Sundays in a year of 22 nd century
We have,
25
75
P  E1  
, P  E2  
100
100
40. Ans. (d)
We have,
X 1 Y
P   , P
Y  2 X
1
 1
  and, P  X  Y   6
3

1
P  X Y  6 1
P X  
  ,
Y
1 2
X
3
1
P  X Y  6 1
P Y  
 
1 3
X
P 
2
Y
 
Clearly, P(X  Y) = P(X) P (Y).
So, X and Y are independent events.
1 2 2
 P  X Y   1 P X P Y  1  
2 3 3
1 1 1
and,  P  X c  Y   P Y   P  X  Y    
3 6 6
Hence, option (d) in correct.
 A 2
 A 1
P    and P   
 E2  7
 E1  7
Required probability = P(A)
 P   A  E1    A  E2  
 P   A  E1    A  E2  
 P  E1  P  A / E1   P  E2  P  A / E2 

   
41. Ans. (c)

Bc 
P  Ac 

C 


 P  A  B / C
c
44. Ans. (c) The numbers may be divided upto 3 sets.
A = {1, 4, 7, 10, …….. 3N – 2}
B = {2, 5, 8, 11, ………3N – 1}
C = {3, 6, 9, 12, ………3N}
x2 – y2 is divisible by 3, if either x – y ax + y is
divisible by 3.
Method 1. In total, 2 numbers x and y may be chosen in
3N
C2 ways. For required no. of ways. Either both elements
are chosen from. Set A or B or. Both are chosen from set
C.
 2NC2 + NC2.
2N
C2  N C2
Required Prob. =
3N
C2

 1  P  A  B  / C 
 1
 1
 1
25 2 75 1 5
 
 
100 7 100 7 28
P  A  B   C 
P C 
2 N .  2 N  1 N .  N  1

2!
2!

3 N .  3 N  1
2!
2 N  2 N  1  N .  N  1

3N .  3N  1
P AC B C
P C 
P  A  C   P  B  C   P  A  B  C 
P C 
 P  A P  C   P  B  P  C   0 


 1 

P C 





4N 2  2N  N 2  N
3N  3N  1

5 N 2  3N
3N  3N  1

5N  3
3  3 N  1

5N  3
9N  3
 1   P  A   P  B   1  P  A   P  B 
 P  Ac   P  B 
42. Ans. (a)
P (X = 3) = Probability of getting 6 in third trial
5 5 1 25
   
6 6 6 216
5
INFOMATHS/MCA/MATHS/