Algebra Solutions

8/12/2019 Algebra Solutions

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Page 1Algebra

CAT 1999

1. a The difference between two integers will be 1, only ifone is even and the other one is odd. 4x will alwaysbe 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 17closest 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.Alternate Solution:

4x17y = 1 and x 1000so, 17y 1 4000 i.e. y 235+ and moreover every 4thvalue of y with give value of x.

So, number of values =235

584

=

So, total number of terms will be 58 + 1 = 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. d 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

erofeb

gnireffo

forebmuN

srewolfdereffo

forebmuN

srewolftfel

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 movingbackwards, 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 that

can 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 for6 is (b).


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Page 2 Algebra

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 V

1 are the

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

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

2(1 V1). Now consider option (b).

Actual empty volume > 2(1 V2). Therefore, for this

situation m(1 V1) 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

1and option (c)

gives the answer as 1. But the actual answer can begreater than 1 as the volume of the vessel is 2 l.

Hence, (b) is the answer.

CAT 2000

9. b 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 + x 2.

10. d a1= 1, a

2= 7, a

3= 19, a

4= 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 5

r1 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

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


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Page 3Algebra

20. c Let the 6th and the 7th terms be x and y.Then 8th term = x + yAlso y2x2= 517(y + x)(yx) = 517=47 11So y + x = 47yx = 11Taking y = 29 and x = 18, we have 8th term = 47,9th term = 47 + 29 = 76 and 10th term = 76 + 47 = 123.

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

Taking x = y =2

1, value of

221 1

x yx y

+ + +

2 21 1

2 22 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 MBA

2because

121 nnn + .

From BA and MBA1, we get 1BA MBA because

++ 11

2

2

2

2

1

2

1

1

1

2

1

1

n

r

n

r

,0maxr

n

n

r

n

r

n

r

n

r

.

Now from MBA1and MBA

2, 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 50nr

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 45BA 50 50

n 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 is

ax2

+ 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 x27b + 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, 1

Hence, 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).

CAT 2002

25. 2 f(x) + f(y) = log

1 x 1 y

log1x 1y

+ +

+

(1 x) (1 y)

log(1x)(1y)

+ +=

1 x y xy

log1 xyxy

+ + += +


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Page 4 Algebra

1 xy x y

log1 xy(x y)

+ + += + +

x y1

1 xylog

x y1

1 xy

++ + = + +

x y

f1 xy

+= +

26. 4 x0= xx1=xx2=xx3= xx4= xx5=xx6=x..Choices (1), (2), (3) are incorrect.

27. 3 xy + yz + zx = 3

+ + =xy (y x)z 3

+ + =xy (y x)(5 x y) 3

+ + + =2 2x y xy 5x 5y 3 0

+ + + =2 2y (x 5) y x 5x 3 0As it is given that y is a real number, the discriminantfor above equation must be greater than or equal tozero.

Hence, + 2 2(x 5) 4(x 5x 3) 0

23x 10x 13 0

+ 23x 13x 3x 13 0

13

x 1,3

Largest value that x can have is13

3.

28. 1 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. 3 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. 2 Arithmetic mean is more by 1.8 means sum is more by18. So baab = 18b > a because sum has gone up, e.g. 31 13 = 18Hence, ba = 2

31. 4 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 (2) is wrong, otherwise choice (1) andchoice (3) are also not correct. Choose the numbersto check.

32. 42 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 whenboth roots are equal).

33. 3 Let the largest piece = 3x

Middle = xShortest = 3x23or 3x + x + (3x 23) = 40or 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. 3 If p = q = r = 1, then expression = 1Check the choice only, one choice gives the value ofexpression = 1.

CAT 2003 (Leak)

35. 3 2xx1 = 02x1 = 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.


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Page 5Algebra

36. 2 For the curves to intersect, log10x = x1

Thus,x

101

log x or x 10x

= =

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

37. 1 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. 1 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, (3) and (4) do not satisfy thesecond equation.Choice (2) 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. 3 Let the 1st term be aand common difference be dthen we have 3rdterm = a + 2d15thterm = a + 14d6thterm = a + 5d11thterm = a + 10d13thterm = a + 12dSince sum of 3rdand 15thterm = sum of 6th, 11thand13th term, therefore we have2a + 16d = 3a + 27d

a + 11d = 0Which is the 12thterm.

40. 4 We can see that x + 2 is an increasing function and

5x 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. 2 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 i.e. just less

than 2.5. In this case y will be just 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:If x 3.6 , theny = x2 + x2.5 + x3.6 = 3x8.1The minimum value of this will be at x = 3.6 and y = 2.7

Hence the minimum value of y is attained at x = 2.5

Alternate method:At x = 2, f(x) = 2.1At x = 2.5, f(x) = 1.6At x = 3.6, f(x) = 2.7Hence, at x = 2.5, f(x) will be minimum.

42. 2 Solution can be found using Statement A as we know

both the roots for the equation (viz. 12

and 12

).

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 1b

2+ b = 0 or

b =2. Thus c =2.

43. 1 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. 4 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 288thterm will be the 24thletter which is x.


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Page 6 Algebra

45. 4 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 + 5p2+ q2= (1)2+ 5Thus, minimum value of p2+ q2is 5.

46. 2 (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+ d2will 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 + 0.25Since it is an integer, the actual minimum value

= 4m2+ 2m + 1

47. 3 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) 843612+ + =

n(n 1)(n 2)8436

6

+ + =

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

48. 3 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 2

as log of negative number is not defined (see thesecond expression).

49. 2 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 smallest

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

Hence the maximum value of vzu

= 20.5 = 4.

50. 2 GRRRRR, RGRRRR, RRGRRR, RRRGRR, RRRRGR,RRRRRGGGRRRR, RGGRRR, RRGGRR, RRRGGR, RRRRGGGGGRRR, RGGGRR, RRGGGR, RRRGGGGGGGRR, RGGGGR, RRGGGGGGGGGR, RGGGGGGGGGGGHence 21 ways.

51. 4 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= 5

When 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. 1 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 more

questions 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. 3 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


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Page 7Algebra

54. 1 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.

CAT 2003 (Re-TEST)

55. 2 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 two

zones.

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 = 4C

29 = 6 9 = 54

Therefore, total number of lines in all = 54 + 36 = 90

56. 2 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, 4a = 3, 4a = 12,

b = 4

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

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

10 xx

log log 100x

=

1010

10

log 100log x

log 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. 2 3 3 0.0081

log M 3log N 1 log 53

+ = +

( )1/ 3 33

log10log2log (M N ) 1

log8log1000

= +

( )

( )1/ 3 3

3

1log2log (M N ) 1

3 1log2=

1/ 3 3

3

1 2log (M N ) 1

3 3= =

M1/3N3 = 32/3

MN9= 32 N9= 9/M.

59. 3 5x + 19y = 64We see that if y =1, we get an integer solution forx = 9, now if y changes (increases or decreases) by

5x will change (decrease or increase) by 19.Looking at the options, if x = 256, we get y = 64.Using these values we see options (1), (2) and (4)are eliminated and also that these exists a solution for250 < x 300.

60. 4 Sum of2 3

2

m mlog m log log

n n

+ + +

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 be2 2 3m m m m

log m log log logn n n

+ =

Now look at the option and put number of terms as 2,only option (4) validates the above mentioned answer.

As

n1

(n 1) 3 32

(n 1)

m m mlog log log

n nn

+


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Page 8 Algebra

61. 2 xyz = 4yx = zy2y = 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. 3 Let2 3 4

4 9 16 25S 1

7 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 2S 1 S 1 1

7 7 7 7 7 7 7

= + + + +

2

1 1 2 1 1S 1 1 1 1

7 7 7 7 7

= + + + +

2

1 2 1S 1 117 7

17

= +

26 2 7

S 17 7 6

= +

36 1S 1

49 3 = +

49 4S

36 3 =

49

S 27=

63. 1 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. 31 1

1 x 3n n

< +

Put n = 1

0 x 4 <

65. 4 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. 4 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. 2 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 1b2canbe zero also), i.e. axb 0Therefore, in both cases a xb 0.


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Page 9Algebra

68. 1 2g g g h= =

3 2g g g h g f= = =

4 3g g g f g e= = =

n 4 =

69. 4 f [f {f (f f )}]

= f [f {f h}]]

= f [f e}]

= f [f ]]= h

70. 1 8 2 2 2e e e e= = e e e = e

If we observe a anything = aa10= a

10 10 9 8{a (f g )} e

= a e= e

71. 4 It will go by elimination.97 = 2 is even, therefore option (1) not possible.2 9 = 18 is even, therefore option (2) not possible.

3 9 124

3 3

+= = is even, therefore option (3) is not

possible.The correct option is (4).

CAT 2004

72. 1 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 0

2

+ =

30termsS 0 =

73. 2 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.

74. 1 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. 3 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 12(a b c) 2

+ += =+ +(ii)If a+ b + c = 0, then

b + 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. 41

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.


10/15

Page 10 Algebra

77. 4 When a > 0, b < 0,ax2andb |x| are non negative for all x,i.e. ax2b|x| 0ax2b |x| is minimum at x = 0 when a > 0, b < 0.

78. 4

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 condition

to 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 conditions

for 3 families, then all three conditions will be satisfied.

79. 3 Given equation is x + y = xyxyxy + 1 = 1(x 1)(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. 3 Given a1= 81.33; a

2=19

Also:a

j= a

j1a

j2, for j 3

a3= a

2a

1=100.33

a4

= a3

a2

=81.33a

5= a

4a

3= 19

a6= a

5a

4= +100.33

a7= a

6a

5= +81.33

a8= a

7a

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+ a

2only i.e. 81.3319 = 62. 33.

81. 2 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 = =

82. 1 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 totalnumber of questions are 100, Group (A) will have 76questions, the corresponding weightage being 76marks. This satisfies all conditions and hence is the

correct option. It can be easily observed that no otheroption will fit the bill.

83. 3 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.

CAT 2005

84. 4 R =( )

( )

6565 65

6565

64 646464 64

130 30 1

30 30 1 30

30 30 1 130 30 1

30

=+ +

or R =

65

65

64 64

11 1

36 30

30 11 1

30

+

R = 30( )

( )

65

64

1 0.96

1 0.96

+

in( )

( )

65

64

1 0.96

1 0.96

+

Nris only slightly less then 1.& Dris only slightly more than 1. R is slightly less than36 but certainly greater than 1.

85. 4 If p = 1! = 1

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


11/15

Page 11Algebra

86. 3 a1= 1, an+13an+ 2 = 4nan+1= 3an+ 4n2when n = 2 then a2= 3 + 42 = 5when n = 3 then a3= 3 5 + 4 22 = 21from the options, we get an idea that an can beexpressed in a combination of some power of3 & some multiple of 100.(1) 399200; tells us that ancould be: 3

n12 n; but it does not fit a1or a2or a3

(2) 3

99

+ 200; tells us that ancould be: 3

n1

+ 2 n; again, not valid for a1, a2etc.(3) 3100200; tells 3n2n: valid for all a1, a2, a3.(4) 3100+ 200; tells 3n+ 2n: again not valid. so, (3) is the correct answer.

87. 4 P = x yx y

log logy x

+

= x x y ylog xlog y log ylog x+

= x y2log ylog x

Let,x

t log y=

21 1

p 2 t tt t

= =

Which can never be positive, out of given option itcant assume a value of +1. So (4) is ans.

88. 3 2x 4 4x x 4 4x= + = +

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

89. 4 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 = 12 and

m = 10 then A = 18100Therefore the number of males will be 10.

CAT 2006

90. 1 t3t4t5....t53

= =

3 4 5 51 52 53 3 4 2.......

5 6 7 53 54 55 54 55 495

Hence option (1).

91. 1 + 2 / 3 1/ 3x x 2 0

+ 2 / 3 1/ 3 1/ 3x 2x x 2 0

( )( ) + 1/ 3 1/ 3x 1 x 2 0

1/ 32 x 1 8 x 1

92. 4 Let number of elements in progression be n, then

( )= + 1000 1 n 1 d ( ) = = 3n 1 d 999 3 37Possible values of d = 3, 37, 9, 111, 27, 333, 999Hence 7 progressions.

93. 4 From the graph of (y - x) Vs. (y + x), it is obvious thatinclination is more than 45.

Slope of line = = + +

y xtan(45 );

y x

+ =+

y x 1 tan

y x 1 tan

By componendo-dividendo, + =+

y x 1 tan

y x 1 tan which

is nothing but the slope of the line that shows thegraph of y Vs. x.And as 0 < < 45, absolute value of tan is lessthan 1.

1

tan is negative and also, greater than 1.

The slope of the graph y Vs. x must be negativeand greater than 1. Accordingly, only option (4)

satisfies.You can also try by putting the values of (y + x) =2(say) and (y - x) = 4(anything more than 2 for thatmatter). You can solve for values of y and x andcross check with the given options.Alternate method:In the normal X-Y coordinate plane the X-axiscorresponds to y = 0And Y-axis corresponds to x = 0y + x = 0 and y - x = 0 are perpendicular lines on thisplane.

y

x

y+x=0(Y

X) axis

(yx)=0

(Y+X) axis


12/15

Page 12 Algebra

And y-x = 0 is the axis Y+X and y+x = 0 is the axisY-XSo, the dotted line is the graph drawn in the question.When you observe w.r.t to X-axis it looks like

x

y

94. 2 2x + y = 40x yy = 402xValues of x and y that satisfy the equation

x y

1 382 36

3 34

. .

. .

. .

13 14

13 values of (x, y) satisfy the equation such that xy

95. 5 = = = yx zy z xlog a.log b.log a b

= =x xy y

y yz z

log loga and b

log log

=

x xy y

y yz z

log loga b

log log

{

=

x xk k

y yk k

y yk kz zk k

log log

log logFor some base k

log log

log log

( ) ( ) = = =

3x 3 3xk

yyk

loglog ab

log

So, =3 3ab a b 0

Or, ( ) =2 2a b 1 a b 0

= ab 1Only option (5) does not satisfy.Hence (5).

96. 5 Equation (ii) can be written as

( ) = 1/ 50.3x 0.2y4 9 8 81

= 2 0.3x 2 0.2y 1/ 5(2 ) (3 ) 8 (81)

= = 0.6x 0.4y 3 4 1/ 5 3 4 / 52 3 2 (3 ) 2 30.6x = 3 x = 5

and 0.4y =4

5

y =2

If we put the values of x and y in first equation these

values satisfy the first equation also.So the answer is x = 5, y = 2Hence, option (5)

97. 5 f(x) = max (2x + 1, 3 - 4x)So, the two equations are y = 2x + 1 and y = 3 - 4xy2x = 1

+ =

y x1

1 1/ 2

Similarly y + 4x = 3

+ =y x 13 3 / 4

Their point of intersection would be2x + 1 = 3 - 4x6x = 2

= 1x3

(1/3,5/3)

(0,3)

(-1 /2,0) (0 ,3/4)

y=34x

y=2x+1

x

y


13/15

Page 13Algebra

So, when = max1

x then f(x) 3 4x3

And when = +max1

x then f(x) 2x 13

Hence the min. of this would be at x =1

3

i.e. = 5y3

Alternative method:As f(x) = max (2x + 1, 3 - 4x)We know that f(x) would be min at the point ofintersection of these curvesi.e. 2x + 1 = 3 - 4x6x = 2

i.e. =1

x3

Hence min f(x) is5

3

CAT 2007

98. 1 f(1) + f(2) + f(3) + . + f(n) = 2n f (n) , f(1) = 3600.

For n = 2:

f(1) + f(2) = 22 f(2) f(2) = 2f(1)

(2 1)

For n = 3:

( )2

2

13600 1 f(3) 3 f(3)

2 1

+ + =

2

2 2

2 1f(3) 3600

2 1 3 1

= Similarly

( )( )( ) ( )

2 2 2 2

2 2 2 2

2 3 4 ... 8f(9) 3600

2 1 3 1 4 1 ... 9 1

=

Therefore, f(9) = 80

99. 3 Let the number of currency 1 Miso, 10 Misos and 50Misos be x, y and z respectively.

x + 10y + 50z = 107

Now the possible values of z could be 0, 1 and 2.For z = 0: x + 10y = 107Number of integral pairs of values of x and y thatsatisfy the equation x + 10y = 107 will be 11. Thesevalues of x and y in that order are(7, 10); (17, 9); (27, 8)(107, 0).For z = 1: x + 10y = 57Number of integral pairs of values of x and y thatsatisfy the equation x + 10y = 57 will be 6. These

values of x and y in that order are (7, 5); (17, 4); (27,3);(37, 2); (47, 1) and (57, 0).For z = 2: x + 10y = 7There is only one integer value of x and y that

satisfies the equation x + 10y = 7 in that order is(7, 0).

Therefore total number of ways in which you canpay a bill of 107 Misos = 11 + 6 + 1 = 18

100. 51 4 1

, n 60m n 12

+ =

Substituting n = 9, n na b 0.01+

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