XAT FLT 1 Explanations

8/7/2019 XAT FLT 1 Explanations

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Page1 XAT FLT - 1MBATest PrepMBATest Prep

XAT .LT 1

Answers and Explanations

FLT-0008/10

Total

Questions

Time

Taken

(Min)

Total

Attempts

Correct

Attempts

Incorrect

AttemptsNet Score

TOTAL 104 120

31

MY PERFORMANCE

Analytical Reasoning and

Decision Making

Verbal and Logical Ability

Quantitative Ability and

Data Interpretation38

35

1 D 2 B 3 C 4 C 5 C 6 D 7 A 8 A 9 B 10 C

11 E 12 B 13 A 14 C 15 E 16 A 17 B 18 E 19 E 20 C

21 E 22 B 23 D 24 A 25 D 26 B 27 C 28 D 29 A 30 B

31 E 32 A 33 D 34 D 35 B 36 C 37 A 38 D 39 E 40 D

41 A 42 C 43 B 44 B 45 A 46 E 47 C 48 B 49 E 50 A

51 B 52 B 53 E 54 D 55 D 56 C 57 C 58 E 59 B 60 B

61 E 62 C 63 A 64 E 65 D 66 B 67 D 68 E 69 C 70 D

71 E 72 A 73 B 74 D 75 A 76 C 77 E 78 B 79 D 80 C

81 E 82 B 83 B 84 E 85 C 86 C 87 A 88 D 89 E 90 E

91 A 92 B 93 D 94 E 95 D 96 A 97 C 98 C 99 B 100 A

101 E 102 D 103 B 104 E


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Page2 XAT FLT - 1MBATest Prep

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For questions 8 to 10:From Statement 1:(A + B + C + D + E) + (F + G + C + H + I)= (A + B + C + D + E + F + G + H + I) + C = 45 + C = 2 25 = 50.Hence, C = 5.So, (A + B) + (D + E) = (F + G) + (H + I) = 20.

From Statement 2:We need to express 20 as the sum of two prime numbers in 2 differentways.

13 + 7 = 17 + 3 = 20.3 and 17 can only be written as (2 + 1) and (8 + 9) respectively.Hence, 7 and 13 can only be written as (3 + 4) and (7 + 6) respectively.Now, A and F are even and respectively less than B and G.So, A = 8 and F = 6(as A > F).Substituting the correct values from A to I, we get that

6

7

8 9 5 2 1

4

3

8. A The value of H is 4.

9. B We can conclude that D = 2, H = 4, G = 7 and B = 9.Therefore, option (B) is the correct choice.

10. C The average of A, F, E and I is8 6 1 3

4.5.4

+ + +=

11. E Let, the cost price of a cow and a buffalo be Rs. X and Rs. 3Xrespectively.Total Cost = Rs. (2X + 3X) = Rs. 5X.Marked price of a cow = Rs. 1.2X..Marked price of a buffalo = Rs. 4.2X..Selling price of a cow = 0.6 Rs. 1.2X = Rs. 0.72X.Selling price of a buffalo = 0.75 Rs. 4.2X = Rs. 3.15X.Total amount got by the sale of two cows and a buffalo = 2x

Rs. 0.72X + Rs. 3.15X = Rs. 4.59X.Loss = Rs. 5XRs. 4.59X = Rs. 0.41X.Loss % = 0.41 20 = 8.2%.

For questions 12 to 14:

12. B Let the break even point be attained on the sale of x units ofGuitar.Therefore, 10000000 + 500x = 3000x.Or, x = 4000.So, the total sales = 4000 Rs. 3000 = Rs. 1.2 Crore.

13. A Total cost when 10000 Guitars were sold when the variablecost has increased by 5% = Rs.1 Crore + 1.05 Rs. 50 lac= Rs.152.5 lac.Initial total cost = Rs. 1 Crore + Rs. 50 lac = Rs. 150 lac.Initial Profit at normal variable cost = Rs. 300 lacRs. 150 lac= Rs. 150 lac.New Profit = Rs. 300 lacRs. 152.5 lac = Rs. 147.5 lac.Therefore percentage decrease in the profit

2.5= 100 = 1.66%.

150

14. C Total number of Guitars sold by Jagdish by the end of thesecond month = 7500.Total cost of manufacturing 7500 Guitars= Rs. 1 Crore + Rs. 37.5 lac = Rs. 137.5 lac.Total sales = 3000 7500 = Rs. 225 Lac.Profit earned = Rs. (225137.5) Lac = 87.5 Lac.Total number of Guitars sold by Jagdish by the end of the thirdmonth = 10000.Total cost of manufacturing 10000 Guitars= Rs. 1 Crore + Rs. 50 lac = Rs. 150 lac.

Total sales = 3000 10000 = Rs. 300 Lac.Profit earned = Rs.300Rs.150 = Rs. 150 Lac

Therefore, the required ratio87.5

= 7 : 12.150

=

15. E The sum of all the digits of the four-digit number should be amultiple of 9 and the maximum possible sum has to be lessthan 36.It means the sum of the digits can be 9, 18 or 27.Now, let N = abcd, a + b = x + 1 and c + d = x.As per the condition given: x + x + 1 = 9, 18 or 27.

2x + 1 = 9, 18 or 27.

Since, 18 is an even number so x 18.So we have x = 4 or 13.

When x = 4, the smallest value of N is 2304.When x = 13, the largest value of N is 9576.Therefore, the required difference = 95762304 = 7272.

16. A Co-ordinates of the point2 5 3 7 3

M , , 5 .2 2 2

+ + = =

Co-ordinates of the point P =4 + 5 6 + 7 1 13

, , .1 + 2 1 + 2 3 3

=

Distance between the points M and P

2 23 1 13

= + 52 3 3

65= units.

6

Hence, the area of the triangle MNP is

5 65

square units.12

For questions 17 to 19:

17. B Let the total number of mobile hand sets sold in India in theyear 2003 = 100N. (N is a real number)The following table provides details about the number of mobilehand sets sold by each of the companies X, Y and Z in theWest and the Central region.

X Y Z

West 0.4 22N = 8.8N 0.3 22N = 6.6N 0.3 22N = 6.6N

Central 0.25 20N = 5N 0.3 20N = 6N 0.45 20N = 9N

Total 13.8N 12.6N 15.6N

So, we can conclude from the table given above that Z soldthe maximum number of mobile hand sets in the West and theCentral region combined together.


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18. E The following table provides information about the leaders ineach of the five regions.

Region Leaders

North X and Z

East Z

West X

South Y and Z

Central Z

Therefore, the required ratio is 2 : 4 : 1.

19. E Let the total number of mobile hand sets sold in India in theyear 2003 = 100N. (N is a real number)Number of mobile hand sets sold by X in South = 0.3 23N= 6.9N.Number of mobile hand sets sold by Y in North = 0.3 17N= 5.1N.Number of mobile hand sets sold by Z in East = 0.4 18N= 7.2N.Therefore, the required percentage

6.9N100 56%.

5.1N 7.2N= =

+

20. C Given that2 3 4

5 6 7

log 3 log 4 log 5

log 6 log 7 log 8

2x = .

5

Since, eae

log blog b = ,

log atherefore we have

e e e2 3 4 2

e e e

log 3 log 4 log 5log 3log 4log 5 = = log 5.

log 2 log 3 log 4

Similarly, the expression log56log

67log

78

= log

58.

So,log 52

log 852 5x = = = 0.625.

85

21. E Novice Expert

Initially 99 1

After 3 hrs 98 2

After 6 hrs 96 4

After 9 hrs 92 8. . .. . .. . .

After 3n 1002n

2n

After 18 hrs 36 64

After 21 hrs 0 100

Note: 36 out of 64 expert will teach 36 novice karate betweenthe 18h and 21st hour.

22. B Let the ages (in increasing order) be a, b and c.Hence, (a + 6) + (b + 6) + (c + 6) = 2(a + b + c) a + b + c = 18 ...(i)Also, 2(10 + a) = 10 + c 2a + 10 = c ...(ii)From (i) and (ii): 3a + b = 8Either, a = 2 and b = 2 (this is not possible)Or, a = 1, b = 5 (this is possible)Hence, out of the options given, the age of one of my sonscould be 5 years.

23. D Let the number of members who play both cricket and tennisbe x.Therefore, 50x + x + 50x = 80.Or, x = 20.It is given that all the members who know French also playcricket.Let, the number of members who know French and play onlycricket be y.Therefore, the number of members who know French and

play both these games3y

.7

=

To minimize the number of members who play both thesegames and do not know French, we need to maximize the

value of

3y

.7

Maximum possible value of y for which3y

7is an integer is

28.Therefore, the minimum possible number of members whoplay both these games and do not know French

3x 28 20 12 8.

7= = =

24. A Probability of picking:

2 red balls from Box 1 =2

24

2

C 1.

6C= ...(i)

2 black balls from Box 1 =2

24

2

C 1.

6C= ...(ii)

1 red and 1 black ball from Box 1 = =2

2 11 4

2

C 4C .

6C...(iii)

If 2 red balls are picked, then 3 new red balls are added to Box2. So we have 5 red balls only.

Probability of picking one red ball in this case =5

15

1

C1.

C= ...(iv)

If we have 2 black balls in Box 2, then 1 red ball is added to it.

Probability of picking one red ball =

11

3 1

C 1

.3C = ...(v)

If we have 1 red and 1 black ball in Box 2, then 2 red balls areadded to it.

Probability of picking one red ball3

14

1

C 3.

4C= = ...(vi)

The required probability is = (i) (iv) + (ii) (v) + (iii) (vi)

1 1 1 4 3 131 .

6 6 3 6 4 18

= + + =


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25. D Distance between A and B = 10u kms.Distance travelled by the train in 2 hours = 2u kms.New speed of the train = (u + 30) km/hr.Time wasted in repair = 2 hours.Time left to cover the distance (10u2u = 8u kms) is1022 = 6 hours.

Therefore,8u

6.u 30

=+

Or, 8u = 6u + 180.

Or, u = 90 km/hr.Therefore, the distance between A and B is 10 90= 900 kms.

26. B Let the requirement be that A speaks before B and C butspeaks after D.Also, if there are only four persons A, B, C and D that have tospeak in the rally, they can do so in 4! ways, if no restriction isapplied.But as per the condition given in the question, there are onlytwo possible ways in which A, B, C and D can speak in theelection rally. The two ways are DABC or DACB.Total ways in which 10 leaders can speak in the election rallywithout any constraint = 10!.

Probability that A speaks after D and before B and C2

.

4!

=

Required number of ways2 10!

10! 2 .4! 4!

= =

For questions 27 and 28:

Given that f(a, b) = 4a2 + 9b24a6b +21

.2

f(a, b) = 4a24a + 1 + 9b26b + 1 +21

2.2

f(a, b) = (2a1)2 + (3b1)2 +17

.2

Minimum possible value of the function f(a, b) will be at a =

1

2 and

b =1

.3

27. C So, the minimum possible integral value of the function f(a, b)

will be17

2

+ 1 = 9.

28. D For maximizing the value of a, the value of (3b1)2 should be0.

Therefore, ( )2 17 1

2a 1 9 .2 2

= =

Or, 2 1a .2 2

+=

29. A A

B

C

D

By the symmetry of the quadrilateral ABCD, we can concludethat AC is the diameter of the circle circumscribing thequadrilateral ABCD.(Triangles ABC and ADC are congruent)Let, AC = 8x and BC = 2x.

2 2AB AC BC 2 15x. = =

Required Ratio2 15 x

15 :1.2x

= =

30. B The equation is a2 + 10ab2 = 104. (a + 5)2b2 = 129. (a + 5b) (a + 5 + b) = 43 3 = 1 129.

As both a and b are positive integers we can say that(a + 5 + b) > (a + 5 b). Thus, the following two cases arepossible:

Case I:a + 5 + b = 43 and a + 5b = 3 a = 18 and b = 20.

Case II:a + 5 + b = 129 and a + 5b = 1 a = 60 and b = 64The values from case II will be rejected as it s given that b isless than 25.So, there is only one such value of a.

31. E Given that 3 < M < 6 and 9 < N < 15, therefore 27 < MN < 90and 81 < N2 < 225.So, 108 < MN + N2 < 315.

2108 MN N 315.

19 O 11

+

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