FMS 2004 Analysis

FMS 2004 - Analysis and Sample Questions Page 1

Analysis of FMS’04

Overview:

Total Time: 120 minTotal No. Of Questions: 175Marking Scheme: +4 for correct answer and –1 for wrong.

Sr. No. Area No. of Qs

Total Marks (MM=700)

1. Reading Comprehension 43 200 2. English Usage 32 100 3. Quantitative Aptitude 50 200 4. Reasoning + DI + GK 50 200 Total 175 700

Sectional Analysis:

English Usage + RC

Sr. No.

Topic No. Of Qs

Level of Difficulty/Remarks

1

Grammar (Sentence correction) 10 Average

2. Analogy 5 Difficult 3. Fill in the blanks 11 Simple 4. Sentence Arrangements 6 Simple 5. Short Passages 6 Average+ 6. Reading Comprehension

(4 passages) 37 Average+


Quantitative Aptitude

Sr. No.

Topic No. Of Qs


1

Arithmetic Number System

%, Ratio, PLD and Mix & Sol TSD and Time & Work

26 12 6 8

Average

2. Algebra Function

Progression Probability + P & C

16 6 4 6

Average+

3. Geometry & Mensuration Mensuration

Triangle

4 2 2

Average

4. Trigonometry + Sets

4

Simple

Reasoning

Sr. No.

Topic No. Of Qs


1.

Mathematical Reasoning 4 Average

2. Analytical Reasoning (3 Caselets + Individual Qs)

10 Difficult

3. Critical Reasoning 10 Average 4. Logical Reasoning 9 Simple 5.

Data Interpretation (1 sets and 1 individual question)

7 Average

6. General Knowledge 2 Average 7. Series (Alphanumeric) 5 Simple 8. Coding-decoding 3 Simple


Strategy to attempt paper:

Quantitative section was difficult when compared to last year’s paper. So, students preparing for CATmight have noticed that FMS these days is trying to move towards CAT. The approach in this sectionwill be very similar to what we do in CAT. Of course, selection was paramount and going for your strongareas would have helped a lot. But if you had followed the strategy of going for one question after theother, then, that would have definitely affected your performance. If you had revised your CAT materialwell, then shortcuts for the quant questions should not be a problem. Moreover, an options basedapproach was apt for this kind of section. There, a higher focus in arithmetic esp. number system, whichalone constituted majority of this section.

Reasoning section:

This section was a good blend of different kinds of reasoning which includes analytical reasoning, logicalreasoning, verbal, coding decoding, blood relation, inference based questions in addition to DI. So agood performance in this section will be crucial.

English section:

EU section was an easy one and time saving as well. Except for the some of the analogies, all thequestions could be attempted in a very short time span.

RC was as usual with comparatively long passages but most of them were inference based. So, at leasttwo passages should have been attempted. Overall this section was of average difficulty considering thelength of the passages and the time it took to do them.

In brief:

Sr. No. Area No. of Qs

Total Marks (MM=700)

Possible No. of Attempts

1. Reading Comprehension 43 172 22-24 2. English Usage 32 128 24-26 3. Quantitative Aptitude 50 200 25-28 4. Reasoning + DI + GK 50 200 25-28 Total 175 700 102-105

Overall Cut offs

If you expect a net score of 340+ in FMS, you hold a good chance of getting a call.


Sample Paper

1. ‘?’ in the expression will be best replaced by :

2 287 87 87 87?83 83 89 89

(a) Greater than (b) Less Than (c) equal to (d) Less than equal to

Ans. (a)

2. Average runs scored by Ganguly, Dravid, Agarkar, Sehwag, and Sachin is 39 .What will be total runsscored by Ganguly and Dravid if average runs scored by Agarkar, Sehwag, and Sachin is 36?

(a) 85 (b) 86 (c) 90 (d) 87

Ans. (d)5 x 39 – 36 x 3 = 195 – 108 = 87

3. What is the sum of the prime numbers lying between 10 and 40? (a) 178 (b) 180 (c) 181 (d) none of these

Ans. (b)Prime numbers between 10 to 40 are 11, 13, 17, 19, 23, 29, 31, 37 and their sum is 180.

4. Average of 7 consecutive odd numbers is x. What is the total of the 11 consecutive odd numbers inthe form of x where first odd number is same in both series?

(a) 11x (b) 11x + 11 (c) 11x + 44 (d) 11x +34

Ans. 7 odd numbers essentially are (x – 6), (x – 4), (x – 2), x, (x + 2), (x + 4), (x + 6) then 11 odd numberswill be = (x – 6), (x – 4), (x – 2), x, (x + 2), (x + 4), (x + 6), (x + 8), (x + 10), (x + 12), (x + 14)Which is equal to 11x + 44

5. If the given numbers a, b, c are even

and x, y, z are odd then × × ×a b y zc is:

(a) Always Even (b) Always odd (c) Even or Odd (d) none of these

Ans. (c)

6. 125 divided between A, B, C and D

If A – 4 = B + 4 = C

4D4

= then what is the share recieved by C?

(a) 10 (b)9 (c) 50 (d) 5

Ans. (d)

Let A – 4 = B + 4 = C

4D4

= = k

Representing it in the form of k we have A + B + C + D = k + 4 + k – 4 + 4k + k/4 = 25k/4 = 125 Therefore share recieved by D = 5


7. Which is the next term in the series:AJKTU, BILSV, CHMRW, DGNQX

(a) EFOPY (b) EJOPX (c) EJOQY (d) EJOPY

Ans. (a)

8.

Shaded area in 11 days. How many days for unshaded.

Ans.

If circles are drawn in the figure as shown with vertices of a square as the center of the triangle thenhow many days it will take for the unshaded area If work done in covering the shaded area is 11 days?

Ans– Asuming side of the square as 2 cm we have radius of the circle = 1cm. Area of the shadedregion in side the triangle = pie x 1

Unshaded area = 4 – pie Number of days needed in doing the work done for unshaded area = 1/ pie x ( 4 – pie)2

= 1.099

9. At 10% compound interest annual Rs 10,000 put in every year for 3 years. What will be the amountat the end of 4 years?

Amount after one year = 10000 x 1.1+ 10000 = 11000 + 10000 = 21000 Total amount after two years = 21000 x 1.1 + 10000 = 33100 Total amount after three years = 33100 x 1.1 + 10000 = 46410

10. Present time of the clock is 7:40:06 sec. If clock loses 4 sec/hr, what would be the time after6.5 hrs ?(a) 2:10:34 (b) 2:09:40 (c) 14:10:34 (d) 14:09:40

Ans. Total time lost in 6.5 hours =24 + 2 = 26 secondsTherefore Time after 6.5 hours = 7:40:06 + 6:30:00 – 00:00:26

= 14:09:40

11. 1.5 people construct 1.5 m long wall in 1.5 days. What will be the length of the wall constructed if6 people construct it for 6 days?(a) 4 m (b) 1.5 m (c) 24 m (d) 6m

Ans. Length of the wall = 1.5 x 6/1.5 x 6/1.5 =24 m


12. Given 3 3

3 3a b 14

,13a b

+= find

a ba b

+−

(a) (1+ 131/3 )/ (1 – 131/3 ) (b) (131/3 +1)/ ( 131/3 –1)(c) both option (a) and (b) (d) none of these

Ans. (c)

13. If a,b,c,are three digits and N2 = ccb, and M2 = bcc and a = 2b then value of (M +N+a) is: (a) 35 (b) 36 (c) 34 (d) 33

Ans. N2 = 441 and M2 = 144 thus b = 1 c = 4 and a = 2 thus value of M + N + a = 12 + 21 + 2 = 35

14. If the ratio of Protestors : Police = 9 : 1 and 135 are caught by police at the rate of 5 prostutators forevery police, find the total number of protestors.

(a) 243 (b) 343 (c) 512 (d) 216

Ans. Ratio of Protestors : Police : Prostutators = 9:1:5Thus option (a)

15. What is the last digit of the expression (483)82

(a) 3 (b) 9 (c) 7 (d) 1

Ans. (b)

16. Which of the following is incorrect?(a) 17 Jan 2004 sunday (b) 21 February 2004 sunday(c) 19 March 2004 sunday (d) 10 April 2004 sunday

Ans. (c)

17. How many numbers < 9999 which are perfect cube and multiple of 9? (a) 6 (b) 7 (c) 5 (d) 4

Ans.(b)All the numbers which are multiple of 3 and have their values less than the cube of 24 are the answersand they are 3,6,9,12,15,18 and 21

18. What is the sum of the numbers that can be formed by 1, 2, 3 and 4? (a) 720 (b) 14400 (c) 7999200 (d) none of these

Ans. (1 + 2 + 3 + 4) x 6! x 1111 = 7999200

19. Which of the following is a prime numbers?(a) 899 (b) 17223 (c) 14641 (d) 1111

Ans. (a)

20. Which of the following is largest?(a) 129 (b) 1011 (c) 1110 (d) All are equal

Ans. (b)


21. Which of the following is the smallest?S

(a) 2 2 21 1 1

2 2 7+ + (b) 2 2 2

1 1 1

3 2 6+ + (c) 2 2 2

1 1 1

3 4 5+ + (d) 2 2 2

1 1 1

7 3 5+ +

Ans. (d)

22. In series, find the missing term2745, 2198, 1729, 1332, ____

(a) 1111 (b)1000 (c) 999 (d)1001

Ans. (d)

23. 21 22 23 247 7 7 7+ + + divide by 25, find remainder..

(a) 0 (b) 6 (c)24 (d) none of these

Ans. (a)Taking 721 common and solving the expression we get 721 x 400 which is clearly divsible by 25 . Hencethe answer.

24. If N = a2 + b2 AND A = 10x + y, b = 10x + z, where z is the smallest prime number, and a + b = 31,Find N. (x, y, z are digits)

(a) 468 (b) 505 (c) 521 (d) none of these

Ans. x will be one which can be clearly seen from the given equation. z is 2 so y will 9.N = 192 + 122 = 505

25. 2 2 2A B Ctan tan tan ?2 2 3

+ +

(a) ≥ 1 (b) ≤ 1 (c) > 1 (d) < 1

Ans. (a)

26. The unit digit of a two digit number is one more than the digit of tens place. If the number is more thanthe five times of the sum of the two digits of the number. Find sum of all such possible numbers.(a) 246 (b) 275 (c) 290 (d) 301

Ans. (c)The only possible options of two digits are 56 , 67, 78 and 89. Sum of these is 290.

27. For what value of x given function defined?

2f(x) x 1 2 1 x x 1= − + − + +

(1) x ≥ 1 (b) x ≤ 1 (c) –∞ < x < ∞ (d) x = 1

Ans. (d)

28. Four usual dices are thrown on the ground. Total of these faces are 13 as the top faces shows 4, 3,1 and 5. What is the total of 4 faces touching the ground?(a) 12 (b) 13 (c) 15 (d) Cannot determined

Ans. (d)


29. If 20 × 21 × 22 × ... 30 = A. If A is divisible by 10x. find the maximum value of x. (a) 4 (b)5 (c) 6 (d) 3

Ans. (a)

30. A, B and C are three angles such that tanA + tanB + tanC = tanA tanB tanC, which of the folowing istrue?(a) A, B, C is a triangle i.e. A + B + C = π(b) A = B = C i.e. ABC is equilateral triangle(c) A + B = C i.e. ABC is a right angle triangle(d) None of these

Ans. (a)

31. 1− is not defined but donated by i find n

n

n 1(i)

=∑

(a) i (b) 0 (c) 1 (d) depends on n

Ans. (d)

32. In a series of which, the first three terms are (a + b)2, (a2 + b2) and (a – b)2, we add the first n termsof the series and call it S(a, b). If a = 7, b = 3 then find S(7, 3) for n = 20.

Ans. S(7,3) = 20a2 + 20b2 + (2 + 0 – 2 – 4 – 6 – 8....(20 terms) )ab = 20 * 49 + 20 * 9 – 340 * 7 * 3

= –5980

33. Value of 100

n

n 1(2)

=∑ is:

(a) 2101 – 1 (b) 2100 – 2 (c) 2101 – 2 (d) 2100 – 1

Ans. (c)

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FMS 2004 Analysis