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, D b CD Question Set No. : 106

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Instructions:

1. All the questions are to be answered and maximum time 90 minutes.

2. This question paper contains 55 questions.

3. Each question carries equal mark.

4. No negative marks

5. Maximum mark 55.

6. Please do not write any marks on the question paper.

7. Ask for rough sheets if required.

8. Directions for answering the questions are given in the test questionpaper before each group of questions to which they apply. Read thesedirections carefully and answer the questions by darkening theappropriate ovals.

9. Follow the instructions of the invigilator. Candidates found violatingthe instructions will be disqualified.

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1. The radius of a circle is increased by 100%. The area of the circle will increases by

a. 100% b. 200% c. 300% d. 400%

2. A 10 hectares field is reaped by 2 men, 3 women and 4 children in 10 days. If 1 man, 1 womanand 1 child work in the ratio 5 : 4 : 2, then 16 hectares field will be reaped by 6 men, 4 womenand 7 children in

a. 5 days b. 6.5 days c. 7 days d. 8 days

3. Find a and b such that (x + 1) and (x + 2) are factors of the polynomial x3+ ax2-bx + 10.

a. 8 and -17 b. 17 and -8 c. -9 and -1 d. None of these

4. If '+' is coded as 'x', 'x' is coded as '+', '+' is coded as '-' and '-' is coded as '+',then= 3 + 5/3+(7/3- 2)x (213- 1)= ?

a.2.5 b. -42 c. 12/5 d. -2.5

5. If R be a relation from A = { 1,2,3, 4} to B = {1, 3, 5}, i.e. (a, b) E R <=> a < b, then R 0 R-1is

a. {(1, 3), (1, 5), (2, 3), (2, 5), (3, 5), (4, 5)}c. {(3, 3), (3, 5), (5, 3), (5, 5)}

b. {(3, 1), (5, 1), (3, 2), (5, 2), (5, 3), (5, 4)]d. {(3, 3), (3, 4), (4, 5)}

6. If one root of quadratic equation px2+qx+ r=O is equal to the 4th power of the other, then

a. p4l5~5+p1/5r4l5+q=0 b. (pr) 1/5+(pr}4I5+q=0 c. p4l5+r4l5=q4l5 d. None of these

Direction for questions 7 and 8: Read the following letter sequence carefully and answer thequestions that follow. Y B Z A R 5 H I J K L M T U V G FEW XeD Q P 0 N.

7. If every alternate letter starting with Y is replaced with a natural number starting from I, thenwhich character will be 14th to the left of 11th to the right of 5th character from your left?

a.2 b. B C.~13 d. N

8. If letters of the series given above are written in reverse order, then which letter will be5th letter to the left of 17th letter to the right of 20th letter from your right?

a. W b. J c. H d.1

9. The points of intersection of three lines 2X + 3Y - 5 = 0, 5X - 7Y + 2 = 0 and 9X - 5Y - 4= 0

a. form a trianglec. are on lines parallel to each other

b. are on lines perpendicular to each otherd. are coincident

10. A horse is tied by a rope to one of the vertices of a square of a side 15 m. The length ofthe rope is 10m, what per cent of the field is grazed by the horse?

a. 32.4% b. 34.8% c.50% .. d. None of these

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Direction for questions 11 and 12: Answer the questions based on the followinginformation.A salesman enters the quantity sold and the price into the 'computer. Both the numbers are two-digitnumbers. But, by mistake, both the numbers were entered withtheir digits interchanged. Thetotal sales value remained the same, Le. Rs. 1,148, but the inventoryreduced by 54. (Inventory isthe quantity leftover)

11. What is the actual price per piece?

a. Rs. 82 b. Rs. 41 c. Rs. 6 d. Rs. 28

12. What is the actual quantity sold?

a.28 b.14 c.82 d.41

13. Sanjay Veer is a land surveyor. One day he records the followingdetails in his notebook.What is theareaofthe landsurveyed?(Allreadingsare in metres.)

ToZ150

ToK50 I 100

50 I50ToYTol36 ~

FromX14. Completethe followingseries. 1, 1,4, 12,27, 51, -'

a. 7,000 m2

c. 8,250 m2

b. 7,500 m2

d. 6,000 m2

a.59 b.86 c.107 d. 114

15. What is the difference between the smallest and the largest six-digitnumbers formedusing the digits 0,1,2,3,4 and 5?

a.440865 b.419760 c.502343 d. None of these

16. Value of diamond is directly proportional to square of its weight. Diamond piece breaksinto 3 pieces inweight ratio of 3 : 4 : 5. Find the percentage reduction in value ofdiamond.

a. 55.62% b.70% c. 65.27% d. None of these

17. In a watch, the minute hand crosses the hour handior the third time exactly after every 3 hr18 min and 15 s of watch time. What is the time gained or lost by this watch in one day?

a. 14 min 10 s lost b. 13 min 50 s lost c. 13 min 20 s gained d. 14 min 40 s gained

18. The average salary of female employees in a company is Rs. 5,000 and that of male isRs. 6,000. The mean salary of all the employees is Rs. 5,700. Find the percentage ofmale and female employees respectively.

a. 30% and 70% b. 70% and 30% c. 55% and 45% d. None of these

19. Which ofthe following divides 1442+ 1692+ 144 x 169?

a.157 b.144 c.313 d. None of these

20. If ax = by = cz, b/a = clb and a, band c are notequal, thenwhat is the valueof 2z I x+z?

a. y/x b. xIy c. xIz .. d.zlx

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Direction for questions 21 to 25: Answer the questions based on the following information.When they hold a meeting, seven company executives - T, U, V, W, X, Y and Z sit around arectangular table. Three executives sit along one side of the table, and three sit along the otherside, each directlyopposite one of the other three. The seventh sits at the head of the table; thereis no seat at the foot of the table. U always sits in one of the two seats farthest from the head ofthe table. Y and V always sit next to each other. V never sits next to Z. If Z does not sit at thehead of the table, W sits there.

21. Which of the followingis an acceptable seating arrangement of the executives, startingwith U, moving toward the head of the table, and coJltinuingaround the table?

a. U, X,T, Z, V, Y,W b. U,T, X, Z, Y, V,W c. U, X, Z, Y, V,W, Td.U,Z,S,X,V, Y,T

22. IfW sits directly opposite to T, X must sit next to which of the following executives?

a. T b. U c. V d. Y

23. IfT sits directly opposite to Z and next to V, which executive must sit directly opposite to U?a. Y b. W c. X d. V

24. IfW sits directlyopposite to U and next to T, the two executives immediatelyon eitherside of X must bea. Y and V b. Y and W c. T and Z d. T and V

25. If Z sits at the head of the table, Y directly opposite to U, and V immediately on X's left,what is the total number of possible seating arrangements of the executives?

a. 1 b. 2 c. 3 d. 4

26. The average speed of a train in the onward journey~is25% more than that of the returnjourney. The train halts for one hour on reaching the destination. The total time taken forthe complete to and frojourney is 17 hours covering a distance of 800 km. The speed ofthe train in the onward journey is

a. 45kmper hour b. 47.06kmper hour c. 50.00kmper hour d. 56.25kmper hour.

27. In a town 25% families own a phone and 15% own a car. 65% families own neither aphone nor a car. 2000 families own both a car and a phone. Consider the followingstatementsin this regard:

I. 10% families own both a car and a phone II. 35% families own either a car or a phoneIII. 40,000 families live in the town.Which of the above statements are correct?

a. I and II b. I and III c. II and III d. I, II and III

28. In a code language, 'SOLID' is written as 'WPSLPIMFHA'. What does the code word'ATEXXQIBVO' refer to?

a. EAGER b. WAFER c.WAGER d. WATER

29. Ravi bought some candies at the rate of Rs. 50 each, some chocolates at the rate of Rs.40 each, some fruits at the rate of Rs. 25 each and some doughnuts at the rate of Rs. 10each. The average price of the items was Rs. 30 per item. Thus, he bought at least

a. 3 candies, 1 chocolate, 3 fruits and 5 doughnutsb. 1 candy, 2 chocolates, 2 fruits and 1 doughnutsc. 1 candy, 1 chocolate, 1 fruit and 2 doughnutsd. 1 candy, 1 chocolate, 2 fruits and 1 doughnut f':

lob ~30. I bought 5 pens, 7 pencils and 4 erasers. Rajan bought 6 pens, 8 erasers and 14 pencils

for an amount which was half more than what I had paid. What per cent of the totalamount paid by me was paid for the pens?a. 37.5% b. 62.5% c. 50% d. None of these

Direction for questions 31 to 34: Answer the questions based on the following information.Mulayam Software Co., before selling a package to its clients, follows the given schedule.

Month1-23-45-89-1011-15

r man/month4020101510

31. Due to overrun in 'design', the design stage took 3 months, i.e. months 3, 4 and 5. Thenumber of people working on design in the fifth month was 5. Calculate the percentagechange in the cost incurred in the fifth month. (Due to improvement in 'coding' technique,this stage was completed in months 6-8 only.)a. 225% b. 150% c. 275% d. 240%

32. With reference to the above question, what is the cost incurred in the new 'coding' stage?(Under the new technique, 4 people work in the sixth month and 5 in the eighth.)

a. Rs. 1,40,000 b. Rs. 1,50,000 c. Rs. 1,60,000 d. Rs. 1,70,000

33. What is the difference in cost between the old and the new techniques?

a. Rs. 30,000 b. Rs. 60,000 c. Rs. 70,000 d. Rs. 40,000

34. Under the new technique, which stage of software development is most expensive forMulayam Software Co.?

a. Testing b. Specification c. Coding d. Design

35. In an office, the distribution of work hours is as shown in the following table:

Consider the following inferences drawn from the table:I. The average number of hours worked by a staff member is about 30.II. The percentage of those who worked 35 or more hours is less than 25.III. At least 5 staff members worked more than 44 hours.Which of these inferences is/are valid?

a. I alone b. II alone c. I and IIf,

d. I, II and III4

No.of StaffMembers No.of hoursWorked5 0-1912 20-2425 25-2940 30-3415 35-398 40-45

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36. Amar, Akbar and Anthony are friends, being looked after by a matron Farah. Amarweights 50% more than Akbar and Anthony weights 25% less than Amar. Farah weightsa third fo the combined weigh of the three boys. All four together weigh 232 kg. Thecorrect arrangement of the persons in the ascending order of their weights is

a. Anthony, Akbar, Farah, Amar b. Anthony, Akbar, Amar, Farahc. Akbar, Anthony, Amar, Farah d. Akbar, Anthony, Farah, Amar

37. Mr. Prasad, a mechanical engineer, has made three devices which he claims can reducefuel consumption in four-wheelers by 25%, 30% and 45%, respectively. Mr. Gorakh buysthese three devices and fits them into his car. The exact saving of fuel consumption willwork out to (given that the three devices are independent in effect)

a.100% b.71.125% c.65.25% d.75%

38. Let Un+1= 2 un+ 1 (n = 0,1, 2, ...) and Uo= O. Then u1Qisnearest to

a.1023 b. 2047 c. 4095 d.8192

39. If 1st, 3rd, 5th, 6th and 8th letters of the word 'REG!JLATE' are used to form a newmeaningful word, then what will be the first letter of the word so formed? Mark 'X' as theanswer if no meaningful word is formed out of these letters and 'M' if more than twomeaningful words can be formed.

a. R b. L c. M d. X

40. The wages received by A, Band C together amount to Rs. 144. If A works for 10 days, Bfor 12 days and C for 15 days, then the amount received by A, when the proportion oftheir daily wages is 1/2, 1/3, 1/5, is

a. Rs. 60 b. Rs. 48 c. Rs. 90 d. None of these

41. There are two women participating in a chess tournament. Each participant played twogames with every other participants. The number of games played by the men amongthemselves exceeded by 66 than the number of games played by men against women.How many participants were there in the tournament?

a.16 b. 15 c.14 d.13

42. Manju and Anju solved a quadratic equation. While~solving,Manju made a mistake in theconstant term and got the roots as 6 and 2, whereas Anju made a mistake in thecoefficient of x only and got the roots as -7 and -1. The correct roots of the equation are

a. 6, -1 b. 7, 1 c. -7, -2 d. -6, 2

43. How many meaningful English words can be made using the first, the fourth, the eighth and theeleventh letters of the word INTERMEDIATE using each letter only once in each word?

a. One b. Two c. Three d. More than three

Direction for questions 44 and 45: Each of the following number series contains a wrongnumber. Find out that number.

44. 17,29,53,101,199,389,773

a.29 b.53 c. 101 d.199

45. 18,46, 102,212,438,886, 1782

a.46 b.102 c.212 d.438f;

\ b b &Direction for questions 46 to 50: Find the odd word out from each of the following sets of four word.

Direction for questions 51 to 55: Choose the alternative that is closest in meaning to thecapitalised word.

51. PRECURSOR

a. Attacker b. Pursuer c. Forerunner d. Torturer

52. AVOWa. To disclaim b. To declare openly c. To curse d. To guarantee

53. APOCRYPHAL

a. Priestly b. Ancient c. Ambiguous d. Counterfeit

54. INTREPID

a. Fearless b. Dull c. Cowardly d. Complacent

55. EXIGUOUS

a. Scanty b. Distinct c. Going d. Extinct

46. a. Impetuosity b. Equanimity c. Zealousness d. Effervescence

47. a. Drip b. Intrusion c. Percolation d. Effluence

48. a. Duplicity b. Guilelessness c'.Artfulness d. Shrewdness

49. a. Taxi b. Cruiser c. Amble d. Cab

50. a. Hiatus b. Break c. Pause d.End

Name:

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Question Answer

No. a bed

1 0000

2 0000

3 0000

4 0000

5 0000

6 0000

7 0000

8 0000

9 0000

10 0000

11 000012 0000

13 0000

14 0000

15 0000

16 000017 0000

18 0000

19 0000

20 000021 0000

22 000023 0000

24 0000

25 0000

26 0000

27 0000

28 0000

29 00000("\ 0000

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Roll No:

ANSWER CODING SHEET I Question Set No.1061

Question

No.

31

38

39

40

41

42

43

44

45

46

47

48

49

50

51

52

53

54

55

Answer

32

33

a bed

00000000

0000

000000000000

00000000

0000

000000000000

0000

00000000000000000000

0000

000000000000

0000

00000000

34

35

36

37

E3RR~DRi-"iS

Name:

/55

Question Answer

No. a bed

1 0080

2 0008

3 8000

4 0080

5 0080

6 0008

7 0800

8 0008

9 0008

10 0800

11 080012 8000

13 0080

14 0800

15 8000

16 008017 0800

18 0800

19 8000

20 800021 0800

22 800023 8000

24 0080

25 0800

26 0008

27 0080

28 0008

29 0008

30 0800

f~\ {) b @) Roll No:

ANSWER CODING SHEET I Question Set No.1061

Question

No.

31

32

33

34

35

36

37

38

39

40

41

42

43

44

45

46

47

. Answer

48

49

a bed

08008000

0800

080000808000

08008000

0080

80000008

0800

0008

00080080080008000800

0080

000800800800

0008

80008000

*:

50

51

52

53

54

55

BRR~DRI-"i5

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ANSWER KEY FOR QUESTION SET NO 106

1. Let radius of circle be R1 = r.

:. Area of circle A1= 1t r2

Radius is increased by 100%

:. R2 = 2r

A2 = 1tx (2r)2Area of the circle increased = A2 - A1 I A1 X 100

= 41t2- 1tr2 I 1t r"' x 100= 300%

2. Since the ratio in which they work is 5 : 4 : 2,5 women = 4 men, therefore, 3 women = 1215 men and 4 women = 16/55 children = 2 men, therefore, 4 children = 8/5men and 7 children = 14/5men.Therefore, if 2+12/5+8/5 = 6 men reap 10 hectares in 10 days, then 6+16/5+14/5= 12 men will reap 16hectares in 10(6/12)(16/10) = 8 days.

3. Since (x + 1) and (x + 2) are factors of the polynomial, ifwe put x = -1 and x = -2 in the expression, the valuewould come to be zero.Hence, if P(x) = X3+ ax2- bx + 10, thenP(-1)=0=-1+a+b+10=>a + b = -9 (i)and P(-2) = 0 = -8 + 4a + 2b + 10=>2a+b=-1 (ii)Now solving (i) and (ii), a = 8 and b = -17.

4. Since '+' ~ 'x', 'x' ~ '-;-', '-;-' ~ '-' and '-' ~ '+', the given

expression will be coded as 3 x 5/3 - (7/3+2) -;-(2/3+1)=5-13/3-;-5/3= 5-13x3/5 = 25-13/5 = 12/5

5. We haveR = {(1, 3), (1, 5), (2, 3), (2, 5), (3, 5), (4, 5)}Therefore,R-1 = {(3, 1), (5, 1), (3, 2), (5,2), (5, 3), (5, 4)}Hence, R 0 R-1 = {(3, 3), (3, 5), (5, 3), (5, 5)}

6. Assume one root of the quadratic equation be a . Thenthe other root is a 4 .

.. a+a4=-q/p..and a. a4 = rip=>a=(r/p)1/s (ii)Substituting the value of a from (ii) in (i), we get(r/p)1/s = (rip) 4/5= - q/p=> p 1-1/5r1/5+p1-4/5r4l5_q=0=>p4/5r1/5+p1/5r4/5+q= 0

(i)

7. The changed series is1 B 2 A 3 S 4 I 5 K 6 M 7 LJ8 G 9 E 10 X 11 D 12 P 13 N~ ~ ~

Fourteenth Fifth Eleventhcharacter character characterto the from the to theleft left right

8. The reversed series isNO PO D CXW E F GV UT M LKJ I H S RAZ B Y

~ ~ ~Fifth Seventeenthletter letterto the to theleft right

t,TwentiethLetterfrom theright

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9. The three lines can be expressed as Y = 5/3 - 2x / 3, Y = 5X / 7 + 2/7 and Y = 9X /5 - 4/5. Hence, the slopes ofthe three lines are -2/3, 5/7 and 9/5 respectively and their Y intercepts are 5/3, 2/7 and -4/5 respectively. For any twolines to be perpendicular to each other, the product of their slopes = -1. We find that the product of none of theslopes is -1. For any two be parallel, their slopes should be the same. This is again not the case. And finally forthe three lines to be intersecting at the same point, there should be one set of values of (x, y) that should satisfythe equations of 3 lines. Solving the first two equations, we get x = 1 and y = 1. If we substitute this in the thirdequation, we find that it also satisfies that equation. Hence, the solution set (1, 1) satisfies all three equations,suggesting that the three lines intersect at the same point, viz. (1, 1), hence they are coincident.

10. Area of square of side 15 m = 225 m2Area of field grazed = Y. x Jr~= Y.x 22/7 x 100 = 78.57 m2Percentage of area grazed by the horse= 78.57 / 225 x 100 = 34.8%

Questions 11 and 12:Hint: Students, please note that this sum could be intelligently solved by looking at both the questions togetherand also the answer choices. We know that the inventory has reduced by 54 units. This means two things: (i)actual quantity sold was less than the figure that was entered the computer (i.e. after interchanging digits), so theunit's place digit of the actual quantity sold should be less than its ten's place digit; and (ii) the difference betweenthe actual quantity sold and the one that was entered in the computer is 54. From question 125, we can figure outthat the only answer choice that supports both these conditions is (a), as (82 - 28 = 54).So the actual quantity sold = 28. Now since the total sales is Rs. 1,148, actual price per piece = 1148 / 28= Rs. 41. Hence, the answer to question 124 is (b).

13. The readings give rise to the following diagram

Z150

K

y

x

Total area LlXQY + LlQZY + LlKZP + Trapezium KPRL + LlXLR = 8250 m2

14. The series increases by addition of (x2- 1) to theprevious number, where x = 1, 2,3, ....1+W-1)=11 + (22_1) = 44+(32-1)=1212 + W-1)=2727+W-1)=5151 + w- 1) = 86

15. Largest six-digit number using the given digits = 543210. Smallest six-digit number using the given digits =102345. The difference = 440865.

16. Value of weight,V=kW2Weight ratio =>3 : 4 : 5i.e. 3x : 4x : 5x

Total value initially = 144 kiValue of broken pieces = 9 ki + 16 ki + 25 ki=50kiPercentage reduction in value = 94 /144 x 100 = 65.27%

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17. In a watch that is running correct, the minute hand should cross the hour hand once in every65 + 5/11 min. So they should ideally cross three times once in 3x (720/11) = 2060 /11 min = 196.36 min. But in thewatch under consideration they meet after every 3 hr, 18 min and 15 s, i.e. (3 x 60 + 18 + 15/60) = 793/4 min =198.25 min. In other words, our watch is actually losing time (as it is slower than the normal watch). Hence, whenour watch elapsed 198.25 min, it actually should have elapsed 196.36 min. So in a day, when our watch will elapse(60x 24) = 1440, it should actually elapse (1440 x 196.36/198.25 = 1426.27. Hence, the amount of time lost byour watch in one day = (1440 - 1426.27) = 13.73, i.e. 13 min and 50 s (approximately).

18. Assume a and b be the number of male and female employees respectively and x and y be the averagesalary of male and female respectively. Assume n be the average salary of all the employees in the company.Then n = a x + b y 1a+b~ 5700 = ax6000+bx5000 1a+b~ 57 (a+b) = ax60+bx50~ 3a = 7b~ a 1b = 7/3So, percentage of male employees in the company = 7/10 x 100 = 70% and female = 3/10 x 100= 30 %

19.1442+ 1692+ (144 x 169)124+ 134+ (122x 132)= {122+ (12 x 13) + 132}{122_(12 x 13) + 132}= 469 x 157

This is as per the formulaa4+ b4+ a2b2= (a2+ ab + b2)(a2- ab + b2).

20. Let us assume

ax = bY= CZ= a1/x 1/y 1/z

:. a = a , b = a and c = a

So b/a = clb

~a1/Y-1/x=a 1/z-1/y

~ 1/y - 1/x = 1/z - 1/y

~ x+z 1xz = 2/y ~ 2z 1x+z = y/x

21. In alternative a, V and Z are not supposed to sit together.Alternative b, is correct.In alternative c, Y sits at the head of table.In alternative d, X sits at the head of table.

22. Since V and Y sit together Wand T sit near Z leaving Xnext to T.

23.

t=:jw

25. ~z~

26. (d) : Suppose speed of train in the return journey is x km/hr Average speed of onward journey = (x+25/100x) km/hr= 125/100x km/hr = 5/4 km/hr

Distance of one-way journey = 800/2 = 400 km/hrBy hypothesis, 400/x + 400/(5/4x) = 17-1 or 400/x + 320/x = 16.Or 720/x = 16 x = 45

Hence, required speed of onward journey = 5/4x = 5/4 x 45 = 56.25 km/hr

..

3

24. L V Y

I Izw 1 X

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27. (c) Suppose number of families in the town = 100Families having neither phone nor car = 100 x 65/100 = 65Families having either phone or car = 100-65 = 35 = 35%Now families having a phone = 25Families not having a phone = 100 - 25 = 75Again families having car = 15Families having no car = 100-15 = 85Again, families having neither car nor phone = 65Families having no car only = 85-65 = 20And families having no phone only = 75-65 = 10Hence, families having neither phone nor car = 20+10+65 = 95Families having both car and phones = 100 - 95 = 5 = 5%Now families having both car and phone = 2000Hence, total families = 2000/5 x 100 = 40,000.

28. (d) : In the given code, S is replaced by Wand P.But S occupies 19th place, W23rd and P 16th place, so that 23+16-1/2 = 19Similarly, other letters have been coded.In the coded word, position of A is 1 or 27 and that of Tis 20. So that 27+20-1/2 = 23 (W)Next E (position 5 or 31) and X(24) are to be replaced by 31+24-1/2 = 27 (A)X (24) and Q (17) are to be replaced by 24+17-1/2 = 20 (T)And I (9) and B (2 or 28) are to be replaced by 9 + 2-1/2 = 5(E)22+15-1/2 = 18(R)Hence, the original word for the coded word is WATER

29. Use options to solve.

30. Let us look at the two equations. Let (5 pens + 7 pencils + 4 erasers) cost Rs. x (1).Hence, (6 pens + 14 pencils + 8 erasers) will cost Rs. 1.5x ... (2).Had, in the second case, Rajan decided to buy 10 pens instead of 6,the quantity of each one of them would have doubled over the first case and hence it would have cost me Rs.2x. So (10 pens + 14 pencils + 8 erasers) = Rs. 2x (3)Now subtracting the second equation from the third, we get 4 pens cost Rs. 0.5x. Since 4 pens cost Re 0.5x, 5 ofthem will cost Re 0.625x. This is the amount that I spent on pens. Hence, fraction of the total amount paid = 0.625= 62.5%.

31. Originally for the fifth month, 4 people were scheduled to do coding. This would have cost them (10000 x 4)= Rs. 40,000. Now there are 5 people who are working on design in the fifth month.The total cost for this would be (20000 x 5) = Rs.1 ,00,000.Hence, percentage change in the cost incurred in the fifth month = (100000 - 40000) /40000 x 100 = 150%.

32. As given in the previous question, it can be seen that the coding stage is now completed in 6th, 7th and 8thmonths.Number of people employed in the 6th month is 4 and in the 8th month is 5. In the 7th month also there are 5people employed (from previous data). Hence, if we were to combine these months, we find that the totalcost incurred in the coding stage = (5 + 5 + 4) x 10000 = Rs.1 ,40,000.

33. The difference in the cost will arise only because of the following months: 5, 6 and 8. And we can compare thecosts as given below

It can be clearly seen that the difference in the cost between the old and the new technique is Rs. 60,000.

4

Original scheme Niew scheme

Cost Total cost Cost per Tot.al cost forMonth People per man! for the People man!

the monthmonth month month

5 4 100DO 40000 5 2DOOO 1.oo,OOtJ

6 5 10000 50000 4 10000 40,000

8 4 10000 40000 5 10000 50,000

Total cost Rs.1,30,OOO Total cost Rs. 1,90,000

t Db r~@

34. The cost incurred in various stages under the present scheme is as given below.

Hence, the mosle'JCPensi¥E steQ1!?is speciliiCJ1ltion.

35. (c):

Average = 3208/105 = 30.3 = 30 (approx.) or statement I is correct.Again staff which worked for 35 hours or more = 15+8=23Their percentage = 23/105 x 100 = 21.9 < 25Hence, statement II is also correct.

36. (a) Let x, y, z, t be the weights of Amar, Akbar, Anthony and Farah respectively.By hypothesis, x = 3/2y or y = 2/3xZ = 3/4x

t = x+y+z 13 or x+y+z = 3tx+y+z+t = 232Putting the values of y and z in (iii), x+2/3x+3/4x = 3t or 3t=29/12x or t=29/36xSubstituting these values in (iv), x+2/3x+3/4x+29/36x = 232 or 87/36x = 232x = 232 * 36/87 = 96kgSo that, y = 2/3x = 2/3*96 = 64kgz= 3/4x = 3/4 * 96 = 72kgt = 29/36 *96 = 77 .3kgHence, the persons in ascending order of their weights are Akbar < Anthony < Farah < Amar

5

Cost per Tobl cost Tobl costMonth Peopre man! for the for the

_th month stage

1 2 40000 00000Specification As. 2.00,000

2 3 40000 '120000

3 4 20000 00000Design Rs.1.4O,000

.( :3 20000 00000

5 4 10000 40000

Coding5 10000 50000

A!>.'1,00,0(;'0

7 5 10000 50000

.8 4 10000 50000

Q 4 15000 00000

Testing Rs. 75,00010 1 15000 15000

11 3: 10000 30000

12 3 10000 30000

Maintenance 13 10000 10000 Rs. 00,000

14 10000 100nO

15 loonn 10000

,-,

Mid-valueNo. of staff

No. of Hours(x)

members fx(f)

0-19 10 5 5020-24 22 12 26425-29 27 25 67530-34 33 40 132035-39 37 15 55540-45 43 8 344

f - 105 fx = 3208

-

1 D b f~@

37. Let us assume, before using the given three devices, four-wheelers take x unit fuel.By using all the three devices, fuel consumption will reduced to= xx (100-25/100) x (100-30/100) x (100-45/100)=0.28875 xHence, percentage saving in fuel consumption =x-0.28875x/xx 100= 71.125%

38. U1= 2uo + 1 = 1. U2= 2U1+ 1 = 2 + 1

U3= 2U2+ 1 = 2(2 + 1) + 1 = 22 + 2 + 1

U4 = 2U3+ 1 = 2 (22 + 2 + 1) + 1 = 23 + 22 + 2 + 1.

..........

SO U1Q=29+ 28+... + 2 + 1 (a GPwith a = 1, r = CR = 2, n = 10):::::> U1Q

=1(210_1)/2-1 =21°-1 =1024-1 =1023

39. Letters at first, third, fifth, sixth and eighth positions are R, G, L, A and E. The words that can be formed usingthese letters are REGAL, GLARE, and LARGE. So we mark M as the answer

40. The ratio of their daily wages is %:1/3:1/5. So the ratio of their wages for the full work is 10/2:12/3:15/5.i.e.5:4:3.Hence A's amount = 5/12 x 144 = Rs. 60

41. x men played 2 x 2x games with women and x(x - 1) games among themselves the difference being 66.So x(x-1) = 4x + 66:::::>2 x - 5x - 66 = 0:::::>x = 11, so x + 2 = 13

42. ; + ax + b = 0

For Manju,; + ax + b1= 0 Sum of roots = -a6+2 =-a:::::>a =-8For Anju, ; + a1x + b = 0Multiplication of roots = b(-7) x (-1)=b:::::>b=7:. Equation is ; - 8x + 7 = 0:::::>;- 7x - x + 7 = 0or (x - 7) (x - 1) = 0:::::>x =7 or 1

43. TIDE, DIET, TIED, EDIT

44. Answer is 199. As we can see that

17 t 29 t 53+12 +24 +48

101 t 199 t 389 t 773+96 +192 +384

Thus, 101 + 96 = 197 (and not 199)

45. Answer is 212. As we can see that18+ 5 = 23, 23 x 2 = 46,46 + 5 = 51, 51 x 2 = 102,102 + 5 = 107, 107 x 2 = 214,214 + 5 = 219,219 x 2 = 438,438 + 5 = 443,443 x 2 = 886, 886 + 5 = 891, 891 x 2 = 1782Thus, 107 x 2 = 214 (and not 212)

46. All others have a quality of excitement in them.47. All others refer to a flow of a liquid.48. All others refer to deception in some form.49. All others are modes of transport.50. All others refer to a break in a continuous action.51. A precursor comes beforehand.52. Avow means to admit, confess, declare openly.53. Apocryphal is something of questionable authenticity.54. Intrepid is to be brave.55. Exiguous is to be scanty in number.

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