!~CAT 2007 Solution ~!

devan

Lets have this thread for solving 2007 CAT paper,

The questions are available here:

http://www.mbaguys.net/t271/

Key will be posted here, discussions & detailed answering will follow in this thread

Section I:

1. Option 2
2. Option 5
3. Option 1
4. Option 3
5. Option 4 is a valid statement
6. Option 3
7. Option 3 is possible
8. Option 2
9. Option 1
10. Option 1
11. Option 4
12. Option 2
13. Option 3
14. Option 2
15. Option 4
16. Option 4
17. Option 1
18. Option 5
19. Option 2
20. Option 4
21. Option 3
22. Option 3
23. Option 2
24. Option 5
25. Option 4

Section II

26. Option 1
27. Option 2
28. Option 5
29. Option 5
30. Option 1
31. Option 4
32. Option 3
33. Option 5
34. Option 2
35. Option 3
36. Option 5
37. Option 1
38. Option 2
39. Option 5
40. Option 1
41. –
42. Option 3
43. Option 1
44. Option 4
45. Option 2
46. Option 4
47. Option 2
48. Option 3
49. Option 2
50. Option 4

Section III

51. Option 5
52. Option 2
53. –
54. Option 2
55. Option 5
56. Option 3
57. Option 2
58. Option 1
59. –
60. Option 5
61. Option 1
62. Option 2
63. Option 1
64. Option 3
65. Option 5
66. Option 3
67. Option 5
68. Option 2
69. Option 4
70. Option 2
71. Option 2
72. Option 4
73. Option 1
74. Option 5
75. Option 3


Section- 1

Solution 1.


Given S= {2,3, 4, ----2n + 1}
X= Average of the odd integers of S
So, X= Average of 3, 5, 7, ---- 2n +1
Since 3, 5, 7, ---- 2n + 1 are in Arithmetic series,
Average of the series= average of 3 and 2n + 1

So, X= {3+ (2n+1)}/ 2
= n + 2
Given, Y= Average of 2,4, 6--- 2n
Since 2, 4, 6---2n are in Arithmetic series
Y= average of 2 and 2n
So, Y = n+1
Hence X- Y = (n+2) – (n+1) = 1

Note: This question is independent of the value of n

devan

2.

Sum of the ages of the eight members ten years ago = 231 years
After three years, their sum would be 231 + 8 (3) i.e, 255 years. But at this point one of the members aged 60 years died and a new born entered the family. So, the sum of the ages of the members of the family will now = 255- 60 i.e., 195 years.
After 3 more years, another 60 years old is replaced by a new born. So the sum of the ages of all the eight members now = 195 + 8 (3) – 60=159 years.
After 4 more years, i.e., at present the sum of their ages= 159 + 8 (4) i.e., 191 years.

Therefore, the average age at present is = 194/ 8 = 24 years.

priyad

nice thread

let me also contribute,

Solution for Question No. 3,

3. Given that f(1) + f(2) + f(3) + ---------- f(n) = (n^2) [f(n)]

→ f(1) + f(2) + f(3) + ----------+ (n-1) f(n-1) = [(n^2) – 1] [f(n)]
→, [(n-1)^2] f(n-1) = [(n^2)- 1] [f(n)]
→, f(n) = {(n-1)/ (n+1)} / [f (n-1)]


Therefore
f(9) = (8/10) * (7/9) * (6/ 8) * (5/ 7) * (4/ 6) * (3/ 5) * ( 2/ 4) * (1/ 3) * f(1) = {(2*1) / (10 * 9)} * f(1)
= 3600/ 45
= 80

devan

4.

Let the number of 1 Miso, 10 Misos and 50 Misos used to pay the bill of 107 Misos be x, y and z respectively.
So, x + 10y + 50z = 107 ----------(i)
Cleary z can take values 0, 1 and 2 only.
Case 1: z = 0
(i) becomes x + 10y = 107
→, y = (107 – x) / 10
Therefore Minimum value of y is 0 and maximum value of y is 10
So there are 11 possibilities when z = 0

Case 2: z = 1
(i) becomes x + 10y = 57
→, y = (57 – x )/ 10
So the minimum value of y is 0 and maximum value of y is 5
Therefore there are 6 possibilities when z = 1

Case 3: z= 2
(i) becomes x + 10y = 7
Clearly x = 7 and y = 0 is the only solution when z = 2

Hence total number of ways in which 107 Misos can be paid,
= 11 + 6 + 1
= 18 ways.

devan

5.

Let the amount on the cheque be Rs. x and y paise.
Value of the amount with him initially = (100y + x) paise
Value of the amount with him after buying a toffee
= (100y + x – 50) paise
Its given that 100y + x – 50 = 3 (100x + y)
→ 97y – 299x = 50
→ 97 (y – 3x) = 8x + 50
To satisfy the above equation, 8x + 50 must be divisible by 97 and also it is even. Hence the least possible value of 8x + 50 = 97 (2) = 194
In this case x = 18
Therefore choice (4) is a valid statement.

devan

6.

(1/ m) + (4/ n) = (1/ 2)
→ 12n + 48m = mn
→ m(n- 48) – 12(n – 48) = (12) (48)
→ (m- 12_ (n – 48) = (2 ^ 6)* (3 ^ 2)
Given that n is odd and less than 60
Hence, n – 48 is odd and less than 12
So n- 48 = 1 or 3 or (3 ^ 2)
Therefore, (m, n) has 3 possibilities

devan

7.

The average weights of the 2 sections and the number of students in the two sections are :

Section I II
Av. Wt W 1 = 45 – x W 2 = 45 + x
No 50 50
( The total deviation from 45 has to be 0)
After the two changes, Deepak moving to section I and Poonam to section II, the averages get switched, as follows,
Section I II
Av. Wt 45 + x 45 - x
No 50 50
In section I, Poonam is replaced by Deepak and the av. Wt increases by 2x, i.e., the total weight changes from 50 (45 – x) to 50 (45 + x)
Therefore D’s weight (d) excepts P’s weight (p) by 100x
i.e., d – p= 100x….(1)

From (A), x = 0.5, i.e., d – p= 50
But we can not get p
From (B) alone, the averages are as follows,

Section I II
Av. Wt 45 45
No 50 Deepak 49
Initially total weight of section I is 50( 45 – x)
After D joins, it is 51(45) = 50(45) + 45
So, d = 45 + 50 x
We can say from (1) that p = 50x – 5, but we don’t know x

By combining A, B we can conclude that p = 50(0.5) – 5 = 20 i.e.,
Choice (3)

devan

8. The outer radius R= 5m
From (A), the inner radius r > 4m
The capacity (V) = (4/3) (22/7) (64) (m^3)
= (268) (m^3)
If r=5m, V is approx = (4/3) (22/7) (125) (m^3)
= (22/21) (500) (m^3)
So we cant not say whether V is ≥ 400 (m^3) or not
From (B), the volume of the material of the tank,
= {[3(10^3)]/ 3} cm^3 = 10 (cm^3)
The total volume occupied by the tank is known to be 524 (m^3).
Therefore the tank capacity is 514 (m^3) which is adequate. (option 2)

devan

9.

From (A), the minimum value of (x^2) + (y^2) + (z^2) is attained when x, y, z are as close to each other as possible, i.e, two of them are 30 and third is 29.
Hence, the minimum value of (x^2) +(y^2) + (z^2)= (30^2) + (30^2) + (29^2)= 2641
B alone is not a sufficient criterion to arrive at a unique answer. (Option 1)

10.
Given segment JK and JM is perpendicular to JK. From (A), there is a point O on JK such that OM = 2 OL. If we suppose, that it is possible for Rahim to draw a square JKLM, the perpendicular distance of M from JK should be equal to JK.
Let P be the midpoint of JK and O be a variable point on JK. If O = P, OM= OL and (OM/ OL)= 1.
As O moves down towards K, (OM/ OL) would be square root of 2. If O moves up towards J, (OM/ OL) decreases.
When O = J, (OM/ OL) = 1 / square root of 2.
Hence (1/ sq.root 2) ≤ (OM/ OL) ≤ sq. root 2.
Therefore OM can be 2 times of OL. Hence from (A), we can find why Rahim is unable to draw the intended square. (Choice 1)

devan

11.
Let T hr be the time gap and V kmph be the plane cruising speed.
Therefore, [3000/ (V- 50)] + t = 7 → (1) [when traveling from B to A]
[3000/ (v- 50)] – t = 7 → (1) [when travelling from A to B]
(1) + (2) gives, {[3000/ (V-50)] + [3000/(V+50)]} = 11
3000 (2V) = 11[(V^2) – 2500]
11(V^2) – 6000V- 27500 = 0
11(V^2) – 6050V + 50V – 27500 = 0
11(V-550) + 50 (V-550) = 0
V = 550 or –(50/11)
As V is positive, V=550
Therefore t = 7 – (3000/500) = 1
Hence, t = 1 hour.