algebra exp

8/14/2019 Algebra Exp

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Algebra - Actual CAT Problems 99-05 Page 1

Answers and Explanations

1. a The difference between two integers will be 1, only ifone is even and the other one is odd. 4x will always

be even, so 17y has to be odd and hence y has to beodd.Moreover, the number 17y should be such a numberthat is 1 less than a multiple of 4. In other words, wehave to find all such multiples of 17, which are 1 lessthan a multiple of 4. The first such multiple is 51. Nowyou will find that as the multiples of 17 goes onincreasing, the difference between it and its closesthigher multiple of 4 is in the following pattern, 0, 3, 2, 1,e.g. 52 51 = 1,68 68 = 0, 88 85 = 3, 104 102 = 2, 120 119 = 1,136 136 = 0So the multiples of 17 that we are interested in are 3,7, 11, 15 .

Now since, x 1000, 4x 4000 . The multiple of 17

closest and less than 4000 is 3995 (17 235). Andincidentally, 3996 is a multiple of 4, i.e. the differenceis 4.This means that in order to find the answer, we needto find the number of terms in the AP formed by3, 7, 11, 15 235, where a = 3, d = 4.Since, we know that Tn = a + (n 1)d,so 235 = 3 + (n1) 4. Hence, n = 59.

2. a Let x be the fixed cost and y the variable cost17500 = x + 25y (i)30000 = x + 50y (ii)Solving the equation (i) and (ii), we getx = 5000, y = 500Now if the average expense of 100 boarders be A.Then100 A = 5000 + 500 100

A = 550.

3. b r 6 11 r 6 11, r 17 = = =or(r6) = 11, r =52q12=8 2q12 = 8 , q = 10or 2q12 =8 , q =2

Hence, minimum value ofr

q=

5

10

=2.

For questions 4 to 6:

foecalP

pihsrow

forebmuN

srewolf

erofebgnireffo

forebmuN

srewolf

dereffo

forebmuN

srewolf

tfel

1 y)8/51( y y)8/7(

2 y)4/7( y y)4/3(

3 y)2/3( y 2/y

4 y y 0

Starting from the fourth place of worship and moving

backwards, we find that number of flowers before

entering the first place of worship is15

y8

.

Hence, number of flowers before doubling =15

y16

(but this is equal to 30)Hence, y = 32Answer for 4 is (c)

The minimum value of y so that15

y16

is a whole number

is 16.Therefore, 16 is the minimum number of flowers thatcan be offered.Answer for 5 is (c).

For y = 16, the value of15

y 1516

= .

Hence, the minimum number of flowers with whichRoopa leaves home is 15.Answer for 6 is (b).

1 a 2 a 3 d 4 c 5 c 6 b 7 d 8 b 9 c 10 b11 c 12 c 13 b 14 b 15 d 16 d 17 c 18 b 19 a 20 c

21 c 22 d 23 b 24 a 25 b 26 d 27 d 28 a 29 c 30 b

31 d 32 d 33 c 34 c 35 c 36 b 37 a 38 b 39 c 40 d

41 b 42 b 43 a 44 d 45 d 46 b 47 c 48 c 49 b 50 b

51 d 52 a 53 c 54 a 55 b 56 b 57 b 58 b 59 c 60 d

61 b 62 c 63 a 64 c 65 d 66 d 67 b 68 a 69 d 70 a

71 d 72 a 73 b 74 a 75 c 76 d 77 d 78 d 79 c 80 c

81 b 82 a 83 c 84 d 85 d 86 c 87 d 88 c 89 b


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Algebra - Actual CAT Problems 99-05Page 2

7. d Let m = 1. So, option (a) will give the answer as Vm

and option (c) will give the answer as V1. Both of

these cannot be the answers as Vm

and V1

are theamount of volume filled.Let m = 2. So, option (b) will give the answer as 2(1V

2) and option (d) will give the answer as

2(1V1). Now consider option (b).Actual empty volume > 2(1 V

2). Therefore, for this

situation m(1V1) is the only possible answer.

8. b Let m = 1 and n = 1. Option (a) gives the answer as4

1

and option (d) gives the answer as greatest integer

less than or equal to2

1. So, both of these cannot be

the answer. Option (b) gives the answer as smallest

integer greater than or equal to2

1 and option (c)

gives the answer as 1. But the actual answer can be

greater than 1 as the volume of the vessel is 2 l.Hence, (b) is the answer.

9. c The data is not linear. So check (b).Let the equation be y = a + bx + cx2.Putting the values of x and y, we get the followingresult.

4 = a + b + c, 8 = a + 2b + 4c and 14 = a + 3b + 9c.Solving these, we get a = 2, b = 1 and c = 1.So the equation is y = 2 + x + x2.

10. b a1

= 1, a2

= 7, a3

= 19, a4

= 43.The difference between successive terms is in series6, 12, 24, 48, ..., i.e. they are in GP. Hence,

( )( )

99n

99100 1

2 1r 1a a a 1 6 6 2 5r1 21 = + = + =

11. c1 1 1 1

...1.3 3.5 5.7 19.21

+ + + +

1 1 1 1 1 1 1 1 1 1 11 ...

2 3 2 3 5 2 5 7 2 19 21

= + + + ( )2111 1 20 10

2 42 42 42 21

= = = =

12. c The vehicle travels 19.5 km/L at the rate of 50 km/hr.

So it should travel19.5

1.3km/L at the rate of 70 km/hr

= 15 km/L.The distance covered at 70 km/hr with 10 L = 10 15

= 150 km

13. b Use choices. The answer is (b), becausex 5, y


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21. c x + y = 1 and x > 0 y > 0

Taking x = y =2

1, value of

221 1

x y

x y

+ + +

2 21 1

2 2

2 2

= + + +

25 25 25

4 4 2= + =

It can be easily verified as it is the least value amongoptions.

For questions 22 and 23:

1 2 1 22

1 1 2

r r r rBA , MBA

n n n

+ += =+ and

1 2 2 11

1 1 2 1

r n r rMBA max 0,

n n n n

= +

From BA and MBA2, we get BA MBA2 because

121 nnn + .

From BA and MBA1, we get 1BA MBA because

++1

1

2

2

2

2

1

2

1

1

1

2

1

1

n

r

n

r,0max

r

n

n

r

n

r

n

r

n

r.

Now from MBA1

and MBA2, we get

21

2

21

1

1

1

2

2

2

2

1

2

1

1

nn

r

nn

r

n

r

n

r,0max

r

n

n

r

n

r

++

+

+ .

22. d From the above information, 1 2BA MBA MBA

None of these is the right answer.

23. b BA = 50 where there is no incomplete innings means

r2

= n2

= 0 50n

r

1

1 =

+=1

1

2

2

1

2

1

11

n

r

n

r,0max

n

n

n

rMBA

=

+ 50

1

45,0max

n

150

1

= 50 + 0 = 50

1 2 1

1 1 1

r r 50n 45 45

BA 50 50n n n

+ +

= = = + >

1 2 12

1 2 1

r r 50n 45 5MBA 50

n n n 1 n 1

+ += = =+ + +

Hence, BA will increase, MBA2

will decrease.

24. a Equation of quadratic equation isax2 + bx + c = 0x2 + bx + c = 0First roots = (4, 3)

Sum of the rootsb

7 b 7a

= = = .

Product of the rootsc

12 c 12a

= = = .

Equation formed x2

7b + 12 = 0 ... (i)Another boy gets the wrong roots (2, 3).

Sum of the roots b 5 b 5a

= = = .

Product of the rootsc

6 c 6a

= = = .

Equation formed x25b + 6 = 0 ... (ii)

x2 + xb + c1 = 0

b = 2 + 3 c = 6Hence, x27x + 6 = 0

x26xx + 6 = 0 x(x6)1(x6) = 0 (x6)(x1) = 0 x = 6, 1Hence, the actual roots = (6, 1).

Alternate method:Since constant = 6[3 2] and coefficient ofx = [4x3x] =7Since quadratic equation isx2(Sum of roots)x + Product of roots = 0 orx27x + 6 = 0Solving the equation (x6)(x1) = 0 or x = (6, 1).

25. b f(x) + f(y) = log1 x 1 y

log1x 1y

+ + +

(1 x) (1 y)

log(1x)(1y)

+ +=

1 x y xy

log1 xyxy

+ + += +

1 xy x y

log1 xy(x y)

+ + += + +

x y1

1 xylog

x y1

1 xy

++ + = + +

x y

f1 xy

+= +


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26. d x0 = xx1 =xx2 =xx3 = xx4 = xx5 =x

x6 =x.. Choices (a), (b), (c) are incorrect.

27. d (x + y + z)2 = x2 + y2 + z2 + 2(xy + zy + zx) x2 + y2 + z2 = 19 y + z cannot be simultaneously = 0else xy + zy + zx = 0 x2 < 19 x < 19 ~ 4.4

28. a Coefficient of n1

x (n 1)(n 4)2

= + +

2 3

2

2 3

S 2 5x 9x 14x ....

xS 2x 5x .....

S(1x) 2 3x 4x 5x ....

= + + + +

= + +

= + + + +

Let 1S S(1x)= 21S 2 3x 4x ...= + + +

21

21

xS 2x 3x ...

S (1x) 2 x x ....

= + +

= + + +

1

xS (1 x) 2

1 x = +

2 xS(1 x ) 21 x

= + 3

2xS

(1x) =

29. c 2 2 2x 5y z 4yx 2yz+ + = +

2 2 2 2(x 4y 4yx) z y 2yz 0+ + + =

2 2(x2y) (zy) 0+ =It can be true only if x = 2y and z = y

30. b Arithmetic mean is more by 1.8 means sum is more by18. So baab = 18b > a because sum has gone up, e.g. 3113 = 18Hence, ba = 2

31. d x1 [x] x

2x 2y3 L(x,y) 2x 2y+ + a 3 L a

2x 2y2 R(x,y) 2x 2y+ + a 2 R a

Therefore, L RNote: Choice (b) is wrong, otherwise choice (a) andchoice (c) are also not correct. Choose the numbersto check.

32. d2 2A B

1x x1

+ = 2 2 2A (x1) B x x x + =

This is a quadratic equation.Hence, number of roots = 2 or 1 (1 in the case when

both roots are equal).

33. c Let the largest piece = 3xMiddle = xShortest = 3x23or 3x + x + (3x23) = 40

or x = 9or the shortest piece = 3(9)23 = 4Check choices:The shortest piece has to be < 20 cm.27 is wrong choice.The largest piece is a multiple of 3.Or (23 + Shortest) should be a multiple of 3.Answer = 4 cm (Among other choices)

34. c If p = q = r = 1, then expression = 1Check the choice only, one choice gives the value ofexpression = 1.

35. c 2xx1 = 0 2x1 = xIf we put x = 0, then this is satisfied and if we putx = 1, then also this is satisfied.Now we put x = 2, then this is not valid.

36. b For the curves to intersect, log10 x = x1

Thus,x

101

log x or x 10x

= =

This is possible for only one value of x (2 < x < 3).

37. a It is given that p q r 0+ + , if we consider the firstoption, and multiply the first equation by 5, second by2 and third by 1, we see that the coefficients of x,y and z all add up-to zero.Thus, 5p2qr = 0No other option satisfies this.

38. a Let 'x' be the number of standard bags and 'y' be thenumber of deluxe bags.Thus, 4x + 5y 700 and 6x + 10y 1250Among the choices, (c) and (d) do not satisfy thesecond equation.Choice (b) is eliminated as, in order to maximize profitsthe number of deluxe bags should be higher than thenumber of standard bags because the profit margin ishigher in a deluxe bag.

39. c Let the 1st term be a and common difference be dthen we have 3rd term = a + 2d15th term = a + 14d

6th term = a + 5d11th term = a + 10d13th term = a + 12dSince sum of 3rd and 15th term = sum of 6th, 11th and13th term, therefore we have2a + 16d = 3a + 27d

a + 11d = 0Which is the 12th term.


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40. d We can see that x + 2 is an increasing function and5x is a decreasing function. This system of equationwill have smallest value at the point of intersection ofthe two. i.e. 5x = x + 2 or x = 1.5.Thus smallest value of g(x) = 3.5

41. b Case 1: If x < 2, then y = 2 x + 2.5 x + 3.6 x= 8.13x.This will be least if x is highest i.e. just less than 2.In this case y will be just more than 2.1

Case 2: If 2 x 2.5 < , then y = x 2 + 2.5 x +3.6x = 4.1xAgain, this will be least if x is the highest case y will bejust more than 1.6.

Case 3: If 2.5 x 3.6 < , then y = x2 + x2.5 + 3.6x = x0.9This will be least if x is least i.e. X = 2.5.

Case 4: Ifx 3.6

, then

y = x2 + x2.5 + x3.6 = 3x8.1The minimum value of this will be at x = 3.6 and y = 2.7Hence the minimum value of y is attained at x = 2.5

42. b Solution can be found using Statement A as we know

both the roots for the equation (viz.1

2and

1

2 ).

Also statement B is sufficient.Since ratio of c and b = 1, c = b.

Thus the equation = 4x2 + bx + b = 0. Since x =1

2 is

one of the roots, substituting we get 1 b

2

+ b = 0 or

b =2. Thus c =2.

43. a Both the series are infinitely diminishing series.

For the first series: First term =2

1

aand r =

2

1

a

For the second series: First term =1

aand r =

2

1

a

The sum of the first series =2

2

2

1

1a1 a 11

a

=

The sum of the second series = 2

2

1

aa1 a 11

a

=

Now, from the first statement, the relation can beanything (depending on whether a is positive ornegative).But the second statement tells us, 4a24a + 1 = 0 or

a =1

2. For this value of a, the sum of second series

will always be greater than that of the first.

44. d The number of terms of the series forms the sum offirst n natural numbers i.e.

n(n 1)

2

+ .

Thus the first 23 letters will account for the first

23 24

2

= 276 terms of the series.

The 288th term will be the 24th letter which is x.

45. d p + q = 2 and pq =1(p + q)2 = p2 + q2 + 2pq,Thus (2)2 = p2 + q2 + 2(1)p2 + q2 = 24 + 4 + 2 + 2p2 + q2 = 22 + 6p2 + q2 = 22 + 1 + 5

p2

+ q2

= (1)2

+ 5Thus, minimum value of p2 + q2 is 5.

46. b (a + b + c + d)2 = (4m + 1)2

Thus, a2 + b2 + c2 + d2 + 2(ab + ac + ad + bc + bd + cd)= 16m2 + 8m + 1a2 + b2 + c2 + d2 will have the minimum value if (ab + ac+ ad + bc + bd + cd) is the maximum.This is possible if a = b = c = d = (m + 0.25) since a+ b + c + d = 4m + 1In that case 2((ab + ac + ad + bc + bd + cd)= 12(m + 0.25)2 = 12m2 + 6m + 0.75Thus, the minimum value of a2 + b2 + c2 + d2

= (16m2 + 8m + 1)2(ab + ac + ad + bc + bd + cd)= (16m2 + 8m + 1)(12m2 + 6m + 0.75)= 4m2 + 2m + 0.25

Since it is an integer, the actual minimum value= 4m2 + 2m + 1

47. c Assume the number of horizontal layers in the pile ben.

Son(n 1)

84362

+ =

21 [ n n] 84362

+ =

n(n 1) (2n 1) n(n 1)8436

12 4

+ + + + =

2n 4n(n 1) 8436

12

+

+ =

n(n 1)(n 2)8436

6

+ + =

n(n 1) (n 2) 36 37 38 + + = So n = 36


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48. c Using log alog b =a

logb

,2 y 5

y 5 y 3.5

=

, where

y = 2x

Solving we get y = 4 or 8 i.e. x = 2 or 3. It cannot be 2as log of negative number is not defined (see the

second expression).

49. b u is always negative. Hence, for us to have a minimum

value ofvz

u, vz should be positive. Also for the least

value, the numerator has to be the maximum positivevalue and the denominator has to be the smallestnegative value. In other words, vz has to be 2 and uhas to be0.5.

Hence the minimum value ofvz

u=

2

0.5 =4.

For us to get the maximum value, vz has to be thesmallest negative value and u has to be the highest

negative value. Thus, vz has to be2 and u has to be0.5.

Hence the maximum value ofvz

u=

2

0.5

= 4.

50. b GRRRRR, RGRRRR, RRGRRR, RRRGRR, RRRRGR,RRRRRGGGRRRR, RGGRRR, RRGGRR, RRRGGR, RRRRGGGGGRRR, RGGGRR, RRGGGR, RRRGGGGGGGRR, RGGGGR, RRGGGGGGGGGR, RGGGGGGGGGGGHence 21 ways.

51. d When we substitute two values of x in the above

curves, at x = 2 we gety =8 + 4 + 5 = 1y = 42 + 5 = 7Hence at x = 2 the curves do not intersect.At x = 2, y1 = 17 and y2 = 11At x =1, y1 = 5 and 2 and y2 = 5When x = 0, y1 = 5 and y2 = 5And at x = 1, y1 = 7 and y2 = 7Therefore, the two curves meet thrice when x =1, 0and 1.

52. a Let us say there are only 3 questions. Thus there are231 = 4 students who have done 1 or more questionswrongly, 232 = 2 students who have done 2 or morequestions wrongly and 233 = 1 student who musthave done all 3 wrongly. Thus total number of wronganswers = 4 + 2 + 1 = 7 = 231 = 2n1.In our question, the total number of wrong answers= 4095 = 2121. Thus n = 12.

53. c Here x, y, z are distinct positive real number

So2 2 2x (y z) y (x 2) z (x y)

xyz

+ + + + +

x x y y z z

y z x z x y= + + + + +

x y y z z x

y x z y x z

= + + + + + [We know that

a b2

b a+ > if a and b are distinct numbers

> 2 + 2 + 2> 6

54. a The least number of edges will be when one point isconnected to each of the other 11 points, giving a totalof 11 lines. One can move from any point to any otherpoint via the common point. The maximum edges willbe when a line exists between any two points. Twopoints can be selected from 12 points in 12C2 i.e. 66lines.

55. b Consider first zone. The number of telephone linescan be shown a follows.

= 9 lines

Therefore, total number of lines required for internalconnections in each zone = 9 4 = 36 lines.Now consider the connection between any twozones.

Each town in first zone can be connected to threetowns in the second zone.Therefore, the lines required = 3 3 = 9Therefore, total number of lines required for connectingtowns of different zones = 4C2 9 = 6 9 = 54Therefore, total number of lines in all = 54 + 36 = 90

56. b ax2 + bx + 1 = 0For real roots

2b 4ac 0

2b 4a(1) 0

2b 4a For a = 1, 4a = 4, b = 2, 3, 4a = 2, 4a = 8, b = 3, 4

a = 3, 4a = 12, b = 4a = 4, 4a = 16, b = 4 Number of equations possible = 7.


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57. b 10 10 xlog x log x 2log 10 =

10 xx

log log 100n

=

101010

log 100log xlog x

=

1010

1 2log x

2 log x =

( )2

10log x 4 =

10log x 2 =

10 10log x 2 or log x 2 = =

2 210 x or 10 x = =

1x 100 or x

100

= =

58. b 3 3 0.0081

log M 3log N 1 log 53

+ = +

( )1/ 33

log10log21log (M N) 1

3 log8log1000= +

( )

( )1/ 3

3

1log21log (M N) 1

3 3 1log2=

1/ 3

3

1 1 2log (M N) 1

3 3 3 = =

or M1/3N= 32/3

or MN9

= 32

or N9 = 9/M

59. c 5x + 19y = 64We see that if y =1, we get an integer solution forx = 9, now if y changes (increases or decreases) by5x will change (decrease or increase) by 19.Looking at the options, if x = 256, we get y = 64.Using these values we see options (a), (b) and (d)are eliminated and also that these exists a solution for250 < x 300.

60. d Sum of2 3

2

m mlog m log log

n n

+ + +

L n terms such

problem must be solved by taking the value of numberof terms. Lets say 2 and check the given option. If welook at the sum of 2 terms of the given series it comes

out to be

2 2 3m m m mlog m log log log

n n n

+ =

Now look at the option and put number of terms as 2,only option (d) validates the above mentioned answer.

As

n1

(n 1) 3 32

(n 1)

m m mlog log log

n nn

+

61. b xyz = 4yx = z y2y = x + zy is the AM of x, y, z.

Also

2

3 3xyz 4= 1

3 3xyz 2 =

AM GM

2

3y 2

Therefore, the minimum value of y is

2

32 .

62. c Let2 3 4

4 9 16 25S 17 7 7 7

= + + + + (i)

2 3 41 1 4 9 16

S7 7 7 7 7

= + + + (ii)

(i)(ii) gives,

2 3 41 3 5 7 9

S 1 17 7 7 7 7

= + + + +

(iii)

2 3 41 1 1 3 5 7

S 17 7 7 7 7 7

= + + +

(iv)

(iii)(iv) gives,

2 3 4

1 1 1 2 2 2 2

S 1 S 1 17 7 7 7 7 7 7

= + + + + L

2

1 1 2 1 1S 1 1 1 1

7 7 7 7 7

= + + + +

L

21 2 1

S 1 117 7

17

= +

26 2 7

S 17 7 6

= +

36 1S 1

49 3 = +

49 4S

36 3 =

49S

27=


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63. a Let is the common root.3 23 4 5 0 + + + =

3 32 7 3 0 + + + =

2 3 2 0

+ =2, 1 = =But the above values of do not satisfy any of theequations. Thus, no root is common.

64. c1 1

1 x 3n n

< +

Put n = 1

0 x 4 <

65. d 36 72n

2n 2 n(n 4) 16x

n 4 n 4

+ + +=

+ +Put x = 36.

2(36) 2 6 40 16x

36 24 4

+ + =

+ +Which is least value of n = 28

66. d 13x + 1 < 2z and z + 3 = 25y

13x + 1 < 2 2(5y 3)

213x 1 10y 6+ <

213x 7 10y+ < put x = 1

220 10y< 2y 2>2y 2> 2(y 2) 0 >

X = 1

2 0y

2

67. b x =|a| bNow axb = a(|a| b) b

= a + |a| b2

axb = a + ab2 a 0 OR axb= aab2a < 0= a(1 + b2) = a(1b2)

Consider first case:As a 0 and |b| 1, therefore (1 + b2) is positive. a (1 + b2) 0 axb 0Consider second case.As a < 0 and |b| 1, therefore (1b2) 0 a (1b2) 0 (Sinceve -ve = +ve and 1b2 canbe zero also), i.e. axb 0Therefore, in both cases a xb 0.

68. a 2g g g h= =

3 2g g g h g f= = =

4 3

g g g f g e= = =n 4 =

69. d f [f {f (f f )}]

= f [f {f h}]]

= f [f e}]

= f [f ]]= h

70. a 8 2 2 2e e e e= = e e e = e

If we observe a anything = a a10 = a10 10 9 8{a (f g )} e

= a e= e

71. d It will go by elimination.97 = 2 is even, therefore option (a) not possible.2 9 = 18 is even, therefore option (b) not possible.

3 9 124

3 3

+= = is even, therefore option (c) is not

possible. The correct option is (d).

72. a Given

1 2 11 1 2 19t t ... t t t .... t+ + + = + + + ( for an A.P.)

[ ] [ ]11 19

2a (11 1)d 2a (191)d2 2

+ = +

22 a + 110d = 28 a + 342 d16 a + 232 d = 02a + 29 d = 0

[ ]30

2a (301)d 02

+ =

30termsS 0 =73. b We have

3f(0) 0 4(0) p p= + =

3f(1) 1 4(1) p p3= + =If P and P3 are of opp. signs then p(p3) < 0Hence 0 < p < 3.


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74. a We have(1) 1010 < n < 1011

(2) Sum of the digits for 'n' = 2Clearly-(n)min = 10000000001 (1 followed by 9 zeros andfinally 1)

Obviously, we can form 10 such numbers by shifting'1' by one place from right to left again and again.Again, there is another possibility for 'n'n = 20000000000So finally : No. of different values for n = 10 + 1 = 11ans.

75. c Ifa b c

rb c c a a b

= = =+ + +

then there are only two possibilities.(i)

If a b c 0, then+ +

a b c a b c

b c c a a b (b c) (c a) (a b)

+ += = =

+ + + + + + + +a b c 1

2(a b c) 2

+ += =+ +

(ii)If a+ b + c = 0, thenb + c =ac + a =ba + b =c

Hencea a

1b c (a)

= =+

Similarly,b c

1c a a b

= =+ +

Therefore option (3) is the correct one 1/2 or 1

76. d1

y1

23 y

=+

+

3 yy

7 2y

+ =+

22y 6y3 0 + =

6 36 24

y4

+ =

6 60 3 15

4 2

= =

Since 'y' is a +ve number, therefore:

15 3y

2= ans.

77. d When a > 0, b < 0,ax2 andb |x| are non negative for all x,i.e. ax2b|x| 0 ax2b |x| is minimum at x = 0 when a > 0, b < 0.

78. d

Family Adults Children1 0, 1, 2 3, 4, 5, .II 0, 1, 2 3, 4, 5, .III 0, 1, 2 3, 4, 5, .

As per the question, we need to satisfy threeconditions namely:1. Adults (A) > Boys (B)2. Boys (B) > Girls (G)3. Girls (G) > Families (F)Clearly, if the number of families is 2, maximum numberof adults can only be 4. Now, for the second conditionto be satisfied, every family should have atleast twoboys and one girl each. This will result in non-compliance with the first condition because adults willbe equal to boys. If we consider the same conditionsfor 3 families, then all three conditions will be satisfied.

79. c Given equation is x + y = xy

xyxy + 1 = 1 (x1)(y1) = 1x1 1& y 1 1or x 1 1& y1 1= = = =Clearly (0, 0) and (2, 2) are the only pairs that willsatisfy the equation.

80. c Given a1

= 81.33; a2

=19Also:a

j= a

j1a

j2, for j 3

a3

= a2a

1=100.33

a4

= a3a

2=81.33

a5

= a4a

3= 19

a6

= a5a

4= +100.33

a7

= a6a

5= +81.33

a8

= a7a

6=19

Clearly onwards there is a cycle of 6 and the sum ofterms in every such cycle = 0. Therefore, when weadd a

1, a

2, a

3... upto a

6002, we will eventually be left with

a1

+ a2

only i.e. 81.3319 = 62. 33.

81. b u = (log2x)26log

2x + 12

xu = 256Let log

2x = y x = 2y

u 8 8x 2 uy 8 uy

= = =

2 3 28 y 6y 12 y 6y 12y 8 0

y= + + =

3(y 2) 0 y 2 = =

x 4, u 4 = =


10/10


82. a Since Group (B) contains 23 questions, the marksassociated with this group are 46.Now check for option (1). If Group (C ) has onequestion, then marks associated with this group willbe 3. This means that the cumulative marks for thesetwo groups taken together will be 49. Since total

number of questions are 100, Group (A) will have 76questions, the corresponding weightage being 76marks. This satisfies all conditions and hence is thecorrect option. It can be easily observed that no otheroption will fit the bill.

83. c Since Group (C) contains 8 questions, thecorresponding weightage will be 24 marks. This figureshould be less than or equal to 20% of the total marks.Check from the options . Option (3) provides 13 or 14questions in Group (B), with a correspondingweightage of 26 or 28 marks. This means that numberof questions in Group (A) will either be 79 or 78 andwill satisfy the desired requirement.

84. d65 65

64 64

30 291

30 29>

+

65 65 64 64as30 29 30 29> +

( ) ( )64 6430 301 29 29 1> +

64 6430 29 29 30 >

63 6330 29>Hence option (d)

85. d If p = 1! = 1Then p + 2 = 3 when divided by 2! remainder will be 1.If p = 1! + 2 2! = 5

Then p + 2 = 7 when divided by 3! remainder is still 1.Hence p = 1! + (2 2!) + (3 3!) + + (10 10!) whendivided by 11! leaves remainder 1

Alternative method:P = 1 + 2.2! + 3.3!+ .10.10!= (21)1! + ( 31)2! + (41)3! + .(111)10!=2!1! + 3!2! + .. 11!10!= 1 + 11!Hence the remainder is 1.

86. c a1 = 1, an+13an + 2 = 4nan+1 = 3an + 4n2when n = 2 then a2 = 3 + 42 = 5when n = 3 then a3 = 3 5 + 4 22 = 21So, it is satisfying 3n2 nHence a100 = 3

1002 100

87. d P = x yx y

log logy x

+

= x x y ylog xlog y log ylog x+

= x y2log ylog x

xt log y=

21 1

p 2 t tt t

= = +

Which can never be 1

88. c 2x 4 4x x 4 4x= + = +

( )2x 4 4x=Now put the values from options.Only 3rd option satisfies the condition.

89. d There are two equations to be formed 40 m + 50 f =1000250 m + 300 f + 40 15 m + 50 10 f = A850 m + 8000 f = Am and f are the number of males and females A isamount paid by the employer.Then the possible values of f = 8, 9, 10, 11, 12If f = 8

M = 15If f = 9, 10, 11 then m will not be an integer while f = 12then m will be 10.By putting f = 8 and m = 15, A = 18800. When f = 12and m = 10 then A = 18100Therefore the number of males will be 10.

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