ren-6-1.pdf

7/27/2019 ren-6-1.pdf

1/11

Page1

Section I (QA + DI)

1. If A is the sum of the squares of the first n natural numbers (where n < 100), then for how many

values of n will A be divisible by 5?(a) 40 (b) 60 (c) 59 (d) 39

2. In the figure given below, AB is the diameter of the larger circle while three smaller circles are drawninside this circle such that their diameters are along AB. The radius of each of these three circles is1 cm and the length of AB is 6 cm. Another circle with center at R is drawn which touches the twosmaller circles and the larger circle. What is the length of the radius (in cm) of this circle?

O

R

A

B

(a)3

2(b)

1

2(c) 1 (d) None of these

3. From the first 20 natural numbers how many Arithmetic Progressions of five terms can be formedsuch that the common difference is a factor of the fifth term?

(a) 16 (b) 22 (c) 25 (d) 26

4. ( )4x 5

5f(x) 4f 9 2x 1 , where x R and x 4.x 4

+ + = + What is the value of f(7)?

(a) 17 (b) 8 (c) 7 (d) None of these

5. There were 4 parcels all of whose weights were integers (in kg). The weights of all the possible pairsof parcels were noted down and amongst these the distinct values observed were 94 kg, 97 kg,101 kg and 104 kg. Which of the following can be the weight of one of the parcels?(a) 40 kg (b) 45 kg (c) 48 kg (d) 53 kg

Mock CAT 6 2013

7/27/2019 ren-6-1.pdf

2/11

Page2

6. In the figure given below, P is a point inside the triangle ABC. Line segments DE, FG and HI aredrawn through P, parallel to the sides AB, BC and CA respectively. The areas of the three trianglesDPG, FPI and EPH are 1, 9, and 25 respectively. What is the area of the triangle ABC? (All the areasare in sq cm).

A

D

G

C

I

P

H

F

B E

(a) 81 (b) 144 (c) 16 (d) 64

7. Guppy has a watch that shows the date without the month and the year. By default, the watch

displays 31 days in each month. Therefore, at the end of all the months with less than 31 days thedate on the watch needs to be readjusted. On 10 thMarch 2001 it showed the right date as 10.What date would it show on 15thMay 2002, if it is known that Guppy never readjusted his watchduring this period?(a) 23 (b) 7 (c) 8 (d) 22

Directions for questions 8 to 10: Answer the following questions on the basis of information given below:The graph given below shows the statistics of 12 Cricket players. Each point on the graph indicates theaverage score per match of a player and the number of matches played by that player. Each of the playersplays for one of the four teams - Team 1, Team 2, Team 3 and Team 4.

AverageScorepermatch

50 400

25

30

35

40

45

50

55

100 150 200 250 300 350

Num ber of matches played

Team 1

Team 2

Team 3

Team 4

7/27/2019 ren-6-1.pdf

3/11

Page3

8. If only the runs scored by the above mentioned 12 players are considered then which team has gotthe maximum aggregate score?(a) Team 1 (b) Team 2 (c) Team 3 (d) Team 4

9. How many players are there who have played more than 200 matches and have scored less than9,000 runs?(a) 2 (b) 3 (c) 4 (d) 5

10. What is the overall average score of those players of Team 4, whose averages are better than theaverage of that player who has played the second highest number of matches for Team 3?(a) 41.16 (b) 42.50 (c) 42.07 (d) 43.40

11. Let P be a point on the side AB of a triangle ABC. Lines drawn parallel to PC, through the pointsA and B, meet BC and AC extended at X and Y respectively. The lengths of AX, BY and PC area units, b units and c units respectively. Then c will be equal to the half of(a) Arithmetic Mean of a and b (b) Geometric Mean of a and b(c) Harmonic Mean of a and b (d) None of these

12. A game consisting of 50 rounds is played among P, Q and R as follows:

Two players play in each round and the player who loses in that round is replaced by the third playerin the next round. If the only rounds in which P played against Q are the 3rd, 14th, 25thand 36th, thenwhat can be the maximum number of games won by R?(a) 40 (b) 42 (c) 41 (d) 36

13. A is the set of the first 100 natural numbers. What is the minimum number of elements that shouldbe picked from A to ensure that atleast one pair of numbers whose difference is 10 is picked?(a) 51 (b) 55 (c) 20 (d) 11

14.(X + 3) (X + 8) (X + 15) (X + 24) (X + 80)

, , , ...... is a sequence where X 1.

3 4 5 6 10

What is the least value of X for which HCF (Numerator, Denominator) = 1 for each term of the givensequence?(a) 17 (b) 13 (c) 11 (d) None of these

15. What is the number of non-negative integer solutions for the equation x2 xy + y2= x + y?(a) 3 (b) 4 (c) 1 (d) None of these

16. A sequence of non-negative integers is given such that t1 = 150 and n n - 2 n - 1t = t t for n > 2. For

what value of t2 would the sequence have the maximum possible number of terms? (a) 90 (b) 97 (c) 93 (d) 75

17. Anshul and Nitish run between point A and point B which are 6 km apart. Anshul starts at 10 a.m.from A, reaches B, and returns to A. Nitish starts at 10:30 a.m. from B, reaches A, and comes backto B. Their speeds are constant with Nitishs speed being twice that of Anshuls. While returning totheir starting points they meet at a point which is exactly midway between A and B. When do theymeet for the first time?

(a)1

10 : 33 a.m.3

(b)2

10 : 37 a.m.3

(c) 10 : 33 a.m. (d)2

10 : 33 a.m.3

7/27/2019 ren-6-1.pdf

4/11

Page4

18. The graph of y = ax2+ bx + c is shown below. If it is given that | | | | < , then which of the followingis true?

( 0), (, 0)Ox-axis

y-axis

(a) a < 0, b < 0, c < 0 (b) a < 0, b > 0, c > 0(c) a < 0, b < 0, c > 0 (d) a < b, b > 0, c < 0

19. A and B are moving along the circumference of a circle with speeds that are in the ratio 1 : K. Theystart simultaneously from a point P in the clockwise direction. They meet for the first time at a pointQ which is at a distance of one-third the circumference from P, in the clockwise direction. K cannotbe equal to

(a)1

4(b)

4

7(c) 4 (d) None of these

20. In triangle PQR, PQ = PR = 10 cm. Points S, T and U lie on PQ, QR and PR respectively such thatST is parallel to PR and UT is parallel to PQ. What is the perimeter (in cm) of the quadrilateralPSTU?(a) 18 (b) 20 (c) 24 (d) Data Insufficient

21. If x is a real number then what is the number of solutions for the equation ( )4 2x 16 x 4+ = ?(a) 0 (b) 1 (c) 2 (d) 3

22. N! is completely divisible by 1352. What is sum of the digits of the smallest such number N?(a) 11 (b) 15 (c) 16 (d) 19

23. The lengths of the three edges of a cuboid are increased by a%, b% and c%. The volume increasesby V%, where V is an integer. How many values can V take if a, b, c are real numbers and

10 a, b, c 20?(a) 11 (b) 39 (c) 41 (d) Cannot be determined

7/27/2019 ren-6-1.pdf

5/11

Page5

Directions for questions 24 to 26:Answer the following questions on the basis of information given below:The following table shows the marks obtained by four students in three Mock CAT tests. The four studentsAbhishek, Akshay, Amitabh and Aamir are disguised as A, B, C and D in no particular order.

Mock CAT A B C D

Mock 1 47 51 53

Mock 2 62 59 10

Mock 3 72 25 28

Total score a b c d

Additional information:

The maximum possible marks in each Mock CAT were 75. In each Mock CAT, a correct answer carried +3 marks and an incorrect answer carried -1 mark. Abhishek scored the lowest marks in Mock 1 and Aamir scored the highest marks in Mock 2 among

the four students. Abhishek's total score is more than Amitabh's total score. Akshay's total score is less than Amitabh's total score.

As total score is a multiple of 6.

24. How many different total scores are possible for Akshay?(a) 101 (b) 97 (c) 98 (d) 100

25. What is the name of the person disguised as B?(a) Aamir (b) Abhishek (c) Akshay (d) Either (a) or (b)

26. How many different values are possible for a?(a) 7 (b) 8 (c) 9 (d) 15

Directions for questions 27 to 29:Answer the following questions on the basis of information given below:Five actresses Careena, Catrina, Carishma, Coena and Cashmira are disguised as U, V, X, Y and Z, notnecessarily in this order. Director Bhakti Kapoor interviewed the actresses in the order Z, Y, X, U and V forhis film Bhoot Aunty.

Additional Information:

Catrina was interviewed after Cashmira. Careena was interviewed before Carishma. The fee (in Rs.) demanded by X, U, V were 2397 lakhs, 2379 lakhs and 1213 lakhs respectively. One of the remaining two actress demanded Rs. 1123 lakhs and the other demanded a sum

(in Rs.) between 1201 lakhs and 2288 lakhs. The sum of the fees demanded by Careena and Carishma is the same as the sum of the fees

demanded by Cashmira and Catrina.

27. Who is disguised as Y?(a) Coena or Cashmira (b) Coena or Careena(c) Coena or Cashmira or Careena (d) Coena or Cashmira or Catrina

7/27/2019 ren-6-1.pdf

6/11

Page6

28. What is the fee (in Rs.) demanded by Coena?(a) 2397 or 1213 lakhs (b) 1213 lakhs(c) 1213 or 2379 lakhs (d) 1123 lakhs

29. Who was interviewed last ?(a) Coena or Catrina (b) Coena or Carishma(c) Carishma or Catrina (d) None of these

30. The question given below is followed by two statements, A and B. Mark the answer using thefollowing instructions:Mark (a) if the question can be answered by using either statement alone.Mark (b) if the question can be answered by using one of the statements alone, but cannot beanswered by using the other statement alone.Mark (c) if the question cannot be answered even by using both the statements together.Mark (d) if the question can be answered by using both the statements together, but cannot beanswered by using either statement alone.

Q. In a class of 200 students, the highest and the lowest scores in a test are 98 and 18 respectively.

Is 50 the average score of the class in the test?A. 100 students score above 50 and the remaining 100 students score below 50 in the test.B. If the highest score and the lowest score in the test are excluded, the sum of the top 99 scores

is exactly double of the sum of the bottom 99 scores.

7/27/2019 ren-6-1.pdf

7/11

Page1

1. c For n = 2, n= 4 and n = 5 the values that A assumesare 12 + 22, 12 + 22 + 32 + 42, 12 + 22 + 32 + 42 + 52

respectively. Each of these is divisible by 5.

For n = 1 or 3, A takes values 12 and 12 + 22 + 32

respectively both of which are not divisible by 5.

So in the set of the 1st5 natural numbers, 3 numbersare divisible by 5.

For n = 6, 7, 8, 9, 10 A behaves in exactly the samemanner as for n = 1, 2, 3, 4, 5 respectively.This pattern repeats for the next set of 5 naturalnumbers and so on.

So for n = 1 to n = 100, A is divisible by 5, in three-fifthsof cases. So for 60 values of n A would be divisibleby 5.Since n < 100 and for n = 100, A is divisible by 5, thetotal number of values that satisfy the condition wouldbe 59.

Alternate solution:

Sum of the squares of first n natural numbers is

n(n 1)(2n 1)A.

6

+ +=

Now n can take 5 types of values i.e. 5k, 5k + 1,5k + 2, 5k 2 and 5k 1.Lets put all the values in A:If n = 5k, A will be divisible by 5.

If n = 5k + 1, A =(5k 1)(5k 2)(10k 3)

.6

+ + +

So A is not divisible by 5.

If n = 5k + 2, A =(5k 2)(5k 3)(10k 5)

.

6

+ + +

So A is divisible by 5.

If n = 5k2, A =(5k 2)(5k 1)(10k 3)

.6

So A is not divisible by 5.

If n = 5k1, A =(5k 1)(5k)(10k 1)

.6

So A is divisible by 5.

So all the numbers of the type 5k, 5k + 2 and 5k 1i.e. 3 numbers out of every 5 consecutive numberswill satisfy the given condition. So 57 out of the first95 natural numbers will satisfy the condition. 97 and99 also satisfy the given condition. So total numbersare 57 + 2 = 59.

2. c

O

R

T

S

Let the radius of the circle with center R be rcm.

Note: If two circles touch each other (internally orexternally) then the line joining their centers will alwayspass though the point of contact.The circle with center R and the smaller circle withcenter O touch each other externally.Hence, OR = OS + SR = 1 + r ...(i)Also, OT must pass through R as the circle with centerR and the larger circle with center O touch each otherinternally.Hence, OT = 3 = OR + RT = 1 + r + r =1 + 2r ...from (i)

r = 1 cm.

3. c Let d be the common difference and a be the first termof AP. The fifth term of the series will be a + 4d.If a + 4d is divisible by d then a should also be divisible

by d. Hence the cases are:d = 1, a = 1, 2, 3,.16d = 2, a = 2, 4, 6,.12d = 3, a = 3, 6d = 4, a = 4So the answer is 16 + 6 + 2 + 1 = 25.

4. a ( )

4x 5

5f(x) 4f 9 2x 1x 4

+

+ = + ...(i)Putting x = 7 in (i):5f(7) + 4f(11) = 9 (2 7 + 1)

5f(7) + 4f(11) = 135 ...(I)

Put x = 11 in (i):5f (11) + 4f(7) = 9 x (2 11 + 1)

5f(11) + 4f(7) = 207 ...(II)

Solving (I) & (II) we get:

f(7) =153

179

=

5. b There are 4 parcels, which would result in 4C2 = 6pairs but it is given that there are only 4 distinctweights. This can only happen when there are someweights which are identical.Out of the 4 numbers here, 2 are odd and 2 are even.So the weight of the identical pair must be either94 kg or 104 kg.If it is 94 kg, the equal weights must be 47 kg each.

Section I (QA + DI) Solutions

7/27/2019 ren-6-1.pdf

8/11

Page2

This means that the other two weights must be 50 kgand 54 kg. So the 4 parcels will weigh 47 kg, 47 kg,50 kg and 54 kg.If it is 104 kg, the equal weights must be 52 kg each.This means that the other two weights must be 45 kgand 49 kg. So the 4 parcels will weigh 45 kg, 49 kg,52 kg and 52 kg.

6. a FG BC, DE AB and IH AC.

As FP BE and BF EP , FBEP is a parallelogram.

Similarly, ADPI & PGCH are also parallelograms.

A

D

G

C

I

P

H

F

B E

3x

5x

x

DPG , IFP and PEH are similar to ABC.

If the area (in sq. cm) of DPG , IFP and PEHare 1, 9 and 25 respectively then we can say theircorresponding sides are in the ratio 1 : 3 : 5. Let thelengths (in units) be x, 3x and 5x for the sides PG, FPand EH respectively.

Also BC = BE + EH + HC = FP + EH + PGBC = 3x + 5x + x = 9x

DPG is similar to ABC and the ratio of the areasof similar triangles is equal to the ratio of the squaresof their corresponding sides,

So

2

2

Area( DPG) (x)

Area( ABC) (9x)

=

21 1

Area( ABC) 9

= Area( ABC) 81 = sq. cm

7. b On 1stApril 2001, Guppys watch will correctly showthe date as 1as March has 31 days only.From 1st April 2001 to 30th April 2002 a total of13 months or 365 + 30 = 395 days would havepassed.

So the date shown by Guppys watch on 30

th

April,2002 must be 395 - 12 31 = 23.Subsequently his watch will show the date as 24on 1stMay, 1on 9thMay and 7on 15thMay, 2002.

For questions 8 to 10:

Let the player who has played the maximum number ofmatches be Player 1 and the second highest number ofmatches be Player 2 and so on for each team.

Total runs scored by each of the players are:

Team 1:

Player 1 = 350 42 = 14700Player 2 = 170 43 = 7310Player 3 = 140 52 = 7280Total score by the three players = 14700 + 7310 + 7280= 29290

Team 2:


Team 3:


Team 4:

Player 1 = 340 40 = 13600Player 2 = 240 45 = 10800Player 3 = 230 34 = 7820Total score by the three players = 13600 + 10800 + 7820= 32220.

8. d Team 4 has scored the maximum runs through thementioned players.

9. b Only 3 players have scored less than 9,000 runs evenafter playing more than 200 matches.

10. c Player 1 and Player 2 of Team 4 have better averagesthan Player 2 of Team 3.Runs scored by Player 1 and Player 2 of Team 4

= 13600 + 10800 = 24400Matches played by them = 340 + 240 = 580

Average score =24400

42.07580

=

11. c

a

c

C

bY

Xx

yP

A

B

Let AP = x , PB = y

Now, PC is similar to Y.

Thereforex c

x y b=

+...(i)

Similarly, BX is similar to PBC.

7/27/2019 ren-6-1.pdf

9/11

Page3

Hencey c

x y a=

+...(ii)

(i) & (ii) gives

1 1 1

c a b= +

abc

a b=

+

1c (Harmonic mean of a andb)

2=

12. b If P played against Q in the 3rd, 14th, 25thand 36th

rounds, then R must have lost in the 2nd, 13th, 24th

and 35throunds. So games won by R are 508 = 42,if R has won in the 50th round and games won byR are 509 = 41, if R has lost in the 50thround.

13. a Lets divide the first 100 natural numbers in five sets

of 20 numbers each:{1, 2, 3.20}, {21, 22, 23.40}, ......{81, 82,

83100}. If we pick the first ten numbers fromeach set we will not get any pair of two numberswhose difference is 10.However, if we pick just one more number from any ofthe sets, it would have a difference of 10 with one ofthe numbers which has already been picked.So the answer is 10 5 + 1 = 51.

14. d The general term is of the form(X + n(n + 2))

.(n + 2)

n(n + 2) is always divisible by (n + 2). So we can say

that n(n + 2) 1would never be divisible by (n + 2). If

we put X = 1, the numerator and denominator of allthe terms would be co-prime.

15. d (x2xy + y2) = (x + y)Multiplying both sides by 2:2(x2xy + y2) = 2(x + y)(xy)2 + x2 + y2= 2x + 2y

2 2 2(x y) (x 1) (y 1) 2

0 1 1 (A)

1 0 1 (B)

1 1 0 (C)

+ + =

Integer solutions for (x, y):

Case 1: (0, 0) and (2, 2)Case 2: (1, 2) and (1, 0)Case 3: (2, 1) and (0, 1)So there are six non-negative integer solutions.

16. c All the terms of the sequence have to be non-negativeintegers. As soon as we get a negative term it wouldmean that the sequence terminates at the previousterm.Lets write the first few terms:t3 =150t2 t4 =2t2150t5 =3003t2 t6 =5t2450t7 =7508t2 t8 =13t21200t9 =195021t2 t10 =34t23150

Now lets try to make as many of them positive aspossible:

150t2 0 or 150 t22t2150 0 or t2 753003t2 0 or 100 t25t2450 0 or t2 907508t2 0 or 93.75 t213t21200 0 or t2 92.30

So t2must be greater than 92 and less than 94, for thefirst 8 terms to be positive.So when t2= 93, the sequence would have exactly

8 terms.For every other value of t2the number of terms wouldbe less than 8.So the answer is 93.

17. a Let the speed of Anshul be v km/hr.So the speed of Nitish would be 2v km/hr.Time taken by Anshul and Nitish to reach exactlymidway between A and B, while returning to their

starting points9 9

hrs and hrsv 2v

= respectively.

Anshul started

1

hr2 early. Hence,

9 9 1v 9 km/hr

v 2v 2= + =

Distance covered by Anshul till 10:30 a.m.

19 4.5 km

2= =

Time required by Anshul and Nitish to cover theremaining 64.5 = 1.5 km for their first meeting

1.5 1.5 1 60 1

hrs hrs minutes 3 minutesv 2v 9 18 18 18 3

= = = = =+ +

Time when they meet first1 1

10 : 30 3 10 : 33 a.m.3 3

= + =

7/27/2019 ren-6-1.pdf

10/11

Page4

18. c As the graph is downward open, so a < 0.

Also, sum of the rootsb

a

+ = is negative (or less

than zero).

[ ]Since | | | | and is less than zero <

b0

a

or x > 51 which would mean that Abhishekcould not score the lowest in Mock 1. Hence, the assumptionis incorrect and so Abhishek must be disguised as A andAkshay as B.

24. b Lets assume that Akshay, who is disguised as D,score p marks in Mock 3.So his total score d = 63 + pAs we know that d < c63 + p < 138or p < 75All the scores from 25 to +75 are possible in any ofthe tests, except 70, 73 and 74.Therefore, different scores possible for Akshay= 10131 = 97

25. a

26. a Abhishek is disguised as A. Lets assume that Abhishekscores q marks in Mock 2. As Aamir scored the highestin Mock 2, q < 62 and 47 + 72 + q > 51 + 59 + 28.

19 q 62 < 2288, it cannotbe taken.

27. c

28. d

29. c

30. c From Statement A: The average score of the classcannot be calculated as neither the total scores northe average scores of the two groups are known.

From Statement B:The data is insufficient to calculatethe exact average score of the class.

From Statements A and B together: Combining

the two statements also does not result in anythingconclusive about the average score of the class.

ren-6-1.pdf

Documents