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SimCAT - 7Explanatory Answers

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SECTION - I

1. | logx+8(3x – 1)| = 1

\ logx+8(3x – 1) = 1 or –1

\ 3x – 1 = x + 8 or 3x – 1 =8x

1

+

\ 2x = 9 or 3x2 + 23x – 8 = 1

\ x =2

9 or 3x2 + 23x – 9 = 0

\ x =2

9 or x =

6

10852923 +±- =

6

63723 ±-

Now, 25 < 637 < 26.

\ 6

63723 ±- »

6

AB.2523 ±-

Or the two roots of quadratic equation are6

AB.2 and

6

AB.48- or slightly greater

than3

1 and slightly less than –8.

When x >3

1, both x + 8 and 3x – 1 are positive. Therefore, the logarithm is defined.

However, when x < –8, x + 8 < 0 and therefore, the logarithm is not defined.

Therefore, there are two real values of x that satisfy the given equation.Hence, [3].

2. Since k and 4 are the roots of the equation, the equation is of the form m(x – 4)

(x – k)

\ ax2 + bx + c = m(x – 4)(x – k)

\ ax2 + bx + c = m(x2 – 4x – kx + 4k)

\ ax2 + bx + c = mx2 + mx(–k – 4) + m(4k)

Since both a > 0 and c > 0, both m > 0 and k > 0.

\ (–k – 4) < 0

\ m(–k – 4) < 0 for all values of k.Hence, [1].

EXPLANATORY ANSWERS



3. Members of set A form an AP with a = 4 and d = 5.

\ Total number of members in set A = 15

4199+

- = 40

If p = 4, the number of members of A that qualify the requirement is 40.If p = 9, the number of members of A that qualify the requirement is 39 (all except

199) and so on.

\ Required number of pairs = 1 + 2 + 3 + ... + 40 =2

4140´ = 820

Therefore, the required answer is 820.

4. Ratio of the number of gold coins with Arnab and Rajdeep is 4 : 3.

Ratio of the number of gold coins with Rajdeep and Barkha is 5 : 8.Therefore, the ratio of the number of gold coins with Arnab, Rajdeep and Barkha

is 20 : 15 : 24.Since sum of the number of gold coins with the three is less than 100, the number

of coins with Arnab is 20, the number of coins with Rajdeep is 15 and the number of coins with Barkha is 24. Therefore, the total number of coins with them is

20 + 15 + 24 = 59.Statements (1), (2) and (4) are correct. However, statement (3) is incorrect.

Hence, [3].

5. For a cubic equation ax 3 + bx2 + cx + d = 0 with roots p, q and r,

p + q + r =a

b- ; pq + qr + pr =

a

c; pqr =

a

d-

\ r

1

q

1

p

1++ =

pqr

pqpr qr ++ =

d

c-

\ 3 =m

24 – Þ m = 8


6. Suppose the marks of the five students are 1, 2, 3, 4 and 5. If the marks of the

first four students from the left are in descending order and the marks of first twostudents from the right are in the descending order, the marks of 4th student has

to be the least among the five. The marks of the three students to his left are higher

than his. These three marks can be selected out of the remaining 4 in

4

C3 = 4 differentways. The 5th number is the remaining number and that can be selected in onlyone way.

\ Required number of arrangement of the five students = 4.The five students can be arranged in 5! = 120 different ways.



\ Required probability =30

1

120

4= .

Hence, [1].

7. We first draw the hexagon ABCDEF and then locate the midpoint of BE (O).

A B

F CO

E D

D AFE @ D BOD and OD is the median of D BDE.

\ Area ( D AFE) : Area ( D BDE) = Area ( D BOD) : Area ( D BDE) = 1 : 2

Hence, [1].

8. Suppose the set consists of ‘n’ numbers, N 1, N2, N3, ...., Nn where N1 = A.

\ Nn = N

1 + n – 1

\ Sum of all the elements of the set:

2

n × (N1 + Nn) =

2

n × (N1 + N1 + n – 1) =

2

n(2N1 + n – 1)

\ 2

n × (2N1 + n – 1) = 50

\ n × (2N1 + n – 1) = 100

Number ‘n’ has to be the factor of 100. Therefore possible values of ‘n’ are 2, 4,5, 10, 20, 25, 50 and 100.

When n = 2, 2 × (2N1 + 1) = 100 or 2N 1 + 1 = 50. N1 is not a natural number.When n = 4, 4 × (2N 1 + 3) = 100 or 2N1 + 3 = 25. \ N1 = 11

When n = 5, 5 × (2N 1 + 4) = 100 or 2N1 + 4 = 20. \ N1 = 8.When n = 10, 10 × (2N 1 + 9) = 100 or 2N 1 + 9 = 10. N1 is not a natural number.

For all the values of n greater than 10, the values of N 1 are not natural numbers.Therefore, there are two possible values of N 1 or A. The sum of the two values of

A is 11 + 8 = 19.Therefore, the required answer is 19.



9. Effectively, the merchant charged amount for 1,080 gm for 900 gm sugar. Therefore,

his profit percent =900

9001080 - × 100 =

900

180 × 100 = 20%.

Hence, [3].

10. Let the number that needs to be subtracted/added be ‘x’.

\ (30 – x), (40 – x) and (44 + x) are in GP.

\ (40 – x)2 = (30 – x) × (44 + x)

\ 1600 – 80x + x 2 = 1320 – 14x – x 2

\ 2x2 – 66x + 280 = 0

\ x2 – 33x + 140 = 0

\ (x – 28)(x – 5) = 0

\ x = 28 or x = 5

\ The numbers in GP are {2, 12, 72} or {25, 35, 49}.

\ Difference between the largest and the smallest numbers is 70 or 24.Hence, [4].

11. Suppose the amount of annual instalment = Rs. x crores. Amount at the end of first year = 200 × 1.1 = 220 Crores. Out of this, the company

repays Rs. x crores. Therefore, the outstanding amount at the end of the first year = 220 – x

Therefore, the amount at the end of the second year = 1.1(220 – x) = 242 – 1.1x.

Since the amount paid by the company at the end of the second year = x,242 – 1.1x = x

\ 2.1x = 242

\ x = 1.2

242 = 115.24 Crores




12. A –B

1 = 1

\ A = 1 +B

1 =

B

B1+

B –C

1 = 1

\ B – 1 =C

1

\ C =1B

1

-

\ ABC = B

B1+

× B × 1B

1

- = 1B

B1

-

+

= – B1

B1

-

+

= B1

1

-

-

– B1

B

-

\ A

1 AC + = C +

A

1 =

1B

1

- +

B1

B

+ =

B1

B

+ –

1

1 B-

\ ABC – A

1 AC + =

B1

1

-

- –

B1

B

- –

B1

B

+ +

B1

1

- = –B ÷

ø

öçè

æ

++

- B1

1

B1

1 = 2B1

B2

-

-

Hence, [4].

13. Suppose Manoj’s speed is x steps per second and the speed of the escalator isy steps per second.

Manoj has to take 40 steps on the escalator that goes up while he has to take 120

steps on the escalator that comes down.

\ downcomesthatescalator anonhimbytakenTime

upgoesthatesclator anonhimbytakenTime =

3

1

On the escalator that goes up, he takes 40 steps. Suppose ‘x’ steps are contributedby the escalator that goes up. On an escalator that comes down, he takes 120 steps.

Since now, the time taken is 3 times as much as on an escalator going up, theescalator will now contribute 3x steps.

\ 40 + x = 120 – 3x

\ x = 20

Therefore, the number of steps on the escalator is 40 + 20 = 60.Hence, [2].



14. We can consider the following cases:

Case 1: All the six letters in the same boxThis can be done in only one way.

Case 2: Five letters in one box and one letter in the second boxThis can be done in 6C1 = 6 different ways.

Case 3: Four letters in one box and two letters in the second boxThis can be done in 6C2 = 15 different ways.

Case 4: Three letters in each box

We can divide 6 letters into two groups of 3 letters each in6

3C

2! =

2

20 = 10 different

ways.Therefore, required number of ways = 1 + 6 + 15 + 10 = 32.

Hence [3].

15.

Area of segment ACB = Area of sector OACB – Area of D OAB

= ÷ ø

öçè

æ ´´ r

2

1 2 – ÷

ø

öçè

æ ´´ sinr

2

1 2

Area of shaded region = Area of circle – 12 × [twice the area of segment ACB]

= p r 2 – 12 × 2 × úûùê

ëé ÷

ø öç

è æ ´´-÷

ø öç

è æ ´´ sinr

21r

21 22 = r 2 [ p – 12 ( q – sin q )]

= 72 úû

ùêë

é÷ ø

öçè

æ -

p-p

2

1

612 = 72 [6 – p ] = 72 ú

û

ùêë

é-

7

226 = 140 cm2

Hence [2].



16. The two-digit number is of the form 10x + y. We are looking for the numbers 10x+ y such that y is not the factor of the numbers.

When y = 0, unit's place in the number = 0. No natural number has 0 as its factor.Therefore, 10, 20, 30 and 40 satisfy the requirement (total 4).

When y = 1, unit's place in the number = 1. All natural numbers have 1 as their factor.

When y = 2, unit's place in the number = 2 i.e. the number is even. 2 is the factor of all even numbers.

When y = 3, unit's place in the number = 3. Numbers 13, 23 and 43 satisfy therequirement. (total 3)

When y = 4, unit's place in the number = 4. Numbers 14 and 34 satisfy the requirement(total 2).

When y = 5, unit's place in the number = 5. All numbers ending in 5 have 5 astheir factor.

When y = 6, unit's place in the number = 6. Numbers 16, 26 and 46 satisfy therequirement (total 3).

When y = 7, unit's place in the number = 7. Numbers 17, 27, 37 and 47 satisfythe requirement (total 4).

When y = 8, unit's place in the number = 8. Numbers 18, 28 and 38 satisfy therequirement (total 3).

When y = 9, unit's place in the number = 9. Numbers 19, 29, 39 and 49 satisfythe requirement (total 4).

Therefore, in all, 4 + 3 + 2 + 3 + 4 + 3 + 4 = 23 numbers satisfy the requirement.


17. Total distance travelled by the jeep = 2000 km. There are 6 working tyres. Therefore,the total distance travelled by the working tyres = 2000 X 6 = 12000 km. This distance

was covered by 8 tyres.

Therefore, distance travelled by each tyre =8

12000 = 1500 km. Hence, [3].

18. 4x2 + 16x + 16 = (2x + 4) 2. Therefore, |4x2 + 16x + 16| touches the x-axis at

x = –2. The nature of the curve is a quadratic curve lying wholly above the x-axisfor other values of x.

Curve – |x + 2| touches the x-axis at (–2, 0). The curve lies wholly below thex-axis for other values of x. Therefore, the curve 2 – |x + 2| is the curve shifted

up by two units. Therefore, the two curves intersect each other in two points, asshown below:



–6 –4 –2 0 2 4

2

4

–2

4

Hence, [3].

19. We have, f(x) = 2f(x – 1) + 3\ f(2) = 2f(1) + 3 = 5

\ f(3) = 2f(2) + 3 = 13

\ f(4) = 2f(3) + 3 = 29

\ f(5) = 2f(4) + 3 = 61

... and so on.

\ f(3) = f(2) + 8 = f(2) + 2 3

\ f(4) = f(3) + 16 = f(3) + 2 4

\ f(5) = f(4) + 32 = f(4) + 2 5

... and so on.

\ f(50) = f(49) + 250

Also, f(49) = f(48) + 249

\ f(50) = f(48) + 249 + 250

\ f(50) – f(48) = 2 49(2 + 1) = 3 × 2 49

Hence, [3].



20. Line x – y3 = 34 intersects x-axis in point A º ( )0,34 and y-axis at point

B º (0, –4).Using Pythagoras Theorem, OA 2 + OB2 = AB2.

\ AB2 = ( )234 + (–4)2 = 64

\ AB = 8

O

B

P

84

A34

We have, OA × OB = AB × OP

\ OP = AB

OBOA ´ =

8

434 ´ = 32

Hence, [1].

21. Suppose q

2

= p

2

+ 231\ q2 – p2 = 231

\ (q + p)(q – p) = 231

Now, 231 = 1 × 231 = 3 × 77 = 7 × 33 = 11 × 21Since q and p both are natural numbers, q + p > q – p

Therefore we haveq + p = 231 and q – p = 1, which gives q = 116 and p = 115,

q + p = 77 and q – p = 3, which gives q = 40 and p = 37,q + p = 33 and q – p = 7, which gives q = 20 and p = 13, and

q + p = 21 and q – p = 11, which gives q = 16 and p = 5Thus there are four such values of p.

Hence, [3].

22. 3sinA + 4cosB =7

24.... (I)

4sinA + 3cosB =7

25.... (II)



(I) + (II) gives

7sin A + 7cosB = 7 .... (III)

\ sinA + cosB = 1

(II) – (I) gives

sinA – cosB =7

1.... (IV)

From (III) & (IV)

sinA =7

4, cosB =

7

3

\ cosA =

2

7

4

1 ÷ ø

öçè

æ -

= 49

33

=

33

7

\ sinB =

2

7

31 ÷

ø

öçè

æ - =

49

40 =

40

7

\ cosA

sinB =

33

40

Hence, [2].

23. Number of whole number solutions to the equation x + y + z = 48 =48!2!50! = 1225

Case I : When all three numbers are equal

x = y = z = 16.Number of ways = 1

Case II: (0,24,24), (2,23,23), (4,22,22),... (14,17,17) (18,15,15) (20,14,14).... (48,0,0)Number of groups = 24

Each of these order groups can be arranged in2!

3! = 3 ways.

Total number of ways = 24 × 3 = 72 ways.

Thus, the number of cases in which x ¹ y ¹ z is 1225 – 72 – 1 = 1152.

Three different variables x, y, z can be arranged in 6 different ways.Out of these, only one way has x < y < z.

\ Required number of solutions =1152

6 = 192

Hence, [2].



24. If a3 = 1, a3 – 1 = 0 or (a – 1)(1 + a + a 2) = 0.

Since a ¹ 1, a – 1 ¹ 0Therefore 1 + a + a 2 = 0.

1 + a + a2 + a3 + a4 + ... + a30 = (1 + a + a2) + (a3 + a4 + a5) + ... + (a27 +a28 + a29) + a30

= (1 + a + a 2) + a3(1 + a + a2) + ... + a27(1 + a + a2) + (a3)10

Since (1 + a + a 2) = 0, all the brackets in the expression above are zero and (a 3)10

= 1. Therefore, the required sum is equal to 1. Hence, [4].

25. The first solution has 85% honey and the second solution has 40% honey. Therequirement is that we need 20% honey in the resultant mixture. No matter what

ratio we choose to mix the two solutions, we cannot obtain a mixture with 20% honeyas the minimum percentage of honey in the mixture will be greater than 40%.

Hence, [4].

26. The line x + y = 6 intersects the x axis at point (6,0) and y axis at point (0, 6),as shown below.

1 2 3 4 5 60

1

2

3

4

5

6

When x = 0, y = 6 and when x = 1, y = 5. Therefore, total 5 squares can be accommodatedbetween the y-axis and the line y = 1. Similarly, 4 squares can be accommodated

between the lines y = 1 and y = 2, and so on. Therefore, the total number of squaresthat can be accommodated is: 5 + 4 + 3 + 2 + 1 = 15.




27. The diameter of the sphere is equal to length of the side of the cube. Therefore,the radius of the sphere = 5 cm.

\ Volume of sphere =34 × p × 53 = p

3500

\ Volume of sand = 1000 – p3

500 =

3

5003000 p-

\ Ratio of the volume of the sand to the volume of cube =3000

5003000 p-

= 1 –6

p =

6

6 p-

Hence, [2].

28. The number of times he will ring the bell= Number of arrangements of 4 digits – Number of derangements of 4 digits

= 4! – 4! úû

ùêë

é+-+-

!4

1

!3

1

!2

1

!1

11 = 24 – 24 ú

û

ùêë

é+-

24

1

6

1

2

1

= 24 – 24 ×24

9 = 15

Hence, [4].

29. Suppose the time taken to fill the first half of the tank in the first situation = a hours

and the time taken to fill the second half of the tank in the first situation = b hours.\ a + b = 18

If the time taken by only the inlet pipe to fill half part of the tank is 'a' hours, then the

time taken by the inlet pipe to fill rd3

2 part of the tank =

3

4 a hours. If the time taken

by the inlet and outlet pipes together to fill half part of the tank is 'b' hours, then the

time taken by the both of them to fill the remaining3

1rd part of the tank in the second

scenario =3

2b hours.

\ 34 a +

32 b = 16

Solving the two equations simultaneously, we get a = 6 and b = 12.

\ The time taken by the inlet pipe to fill half the tank = 6 hours.

\ The time taken by the inlet pipe to fill the tank completely = 12 hours.Hence, [3].



30. Suppose M is the center of the semicircle. Join MP. Since side AC is a tangent

to semicircle at P, angle MPC is 90 degrees.

6

A

BC

M

r

8 – r

P

10

r

8

Now, D ABC ~ D MPC using A-A-A test.

\ PC

BC

MC

AC

MP

AB==

AB = 6, BC = 8, MP = r and MC = 8 – r. Also, AC = 22 86 + = 10

\ r 8

10

r

6

-=

\ r = 3

Hence, [4].

31. The word VLADIVOSTOK has 11 letters, out of which 4 are vowels and 7 are consonants.The 4 vowels are 2 Os, 1 A and 1 I. The 7 consonants are 2 Vs, 1 L, 1 D, 1 S,

1 T and 1 K.Positions 2, 3, 5, 7 and 11 are prime-numbered positions (total 5), out of which 4

can be occupied by vowels. The prime-numbered positions occupied by vowels canbe selected from 5 in 5C4 = 5 different ways. The four vowels can be arranged in

2!

4! = 12 ways. The 7 consonants can be arranged in

2!

7! different ways.

Therefore, the number of ways the letters of VLADIVOSTOK can be arranged is

5 × 12 ×2!7! = 30 × 7! ways.

Hence, [2].



32. Since a and b are roots of the equation x 2 – kx + 72 = 0, a + b = k.Similarly, since c and d are roots of the equation x 2 – 8x + k = 0, c + d = 8.

Therefore, a + b + c + d = k + 8.Since the equation x 2 – kx + 72 = 0 has real roots, k 2 > 4 × 72 or k 2 > 288... (I)

Similarly, since the equation x 2 – 8x + k = 0 has real roots, 64 > 4k or k < 16...(II)There is no value of k that satisfies both inequalities (I) and (II) simultaneously.

Therefore, k + 8 cannot take any real value.Hence, [4].

33. We have,

17

x 21 – xP B

y

21 – y

M

RD

A

10

C

Therefore, using Pythagoras theorem, AP 2 + PM2 = AM2 and PB2 + PM2 = BM2

x2 + y2 = 289 ... (I)(21 – x)2 + y2 = 100 ... (II)

\ 441 – 42x + x 2 + y2 = 100Using the value of x2 + y2 from equation (I), we get

441 + 289 – 42x = 100

\ 730 – 42x = 100

\ x =42

630 = 15

From equation (I), we get, y = 8.

We have, MC2 = (21 – x)2 + (21 – y)2 = 62 + 132 = 205

\ MC = 205

Hence, [1].

34. Suppose A = 3333 33. Therefore, log A = (33)log 3333. Now, 3 < log 3333 < 4.Therefore, 99 < log A < 128

Suppose B = 333 333. Therefore, log B = (333)log 333. Now, 2 < log 333 < 3.Therefore, 666 < log B < 999

Suppose C = 33 3333. Therefore, log C = (3333)log 33.



Now, 1 < log 33 < 2. Therefore, 3333 < log C < 6666Suppose D = 333333. Therefore, log D = (33333)log 3. Now, 10 0.25 < 3 < 100.5

Therefore, 0.25 < log3 < 0.50. Therefore, 8333.25 < log D < 16666.50.Clearly, log D is the largest among all the logarithms taken.

Therefore D is the largest among A, B, C and D.Hence, [4].



SECTION - II

35. Total stars = (30 + 50) + [(110 + 80) × 2] + [(150 + 120) × 3] + [(120 + 140) × 4]

+ [(140 + 130) × 5] = 3660Total reviews = 550 + 520 = 1070

Average rating =1070

3660 = 3.42 stars

Hence, [3].

36. Proportion of people who gave a rating not less than 3 stars for:

Lenovo =80150

81738654525

+

+++++ =

230

198 = 0.86

Dell =

90120

15527403320

+

+++++ =

210

140 = 0.66

Sony =110110

512013152518

+

+++++ =

220

142 = 0.64

Acer =100100

114215181237

+

+++++ =

200

135 = 0.67

Toshiba =14070

4556272550

+

+++++ =

210

185 = 0.88

Hence, [3].



37. (1) Percentage of 5 stars ratings w.r.t its total ratings for:

Sony =÷ ø

öçè

æ

+

+

110110

5115

× 100 = 30%

Lenovo = ÷ ø

öçè

æ

+

+

80150

865 × 100 = 31.7%

Hence, brand Sony won’t be selected.Thus, Criterion 1 is not valid.

(2) Percentage of 1 or 2 stars ratings w.r.t its total ratings for:

Philipkart = ÷ ø

öçè

æ +

550

11030 × 100 > 25%

Xnapdeal = ÷ ø öç

è æ +

5208050 × 100 = 25%

Within Xnapdeal, percentage of 5 stars ratings for:

Toshiba = ÷ ø

öçè

æ

140

45 × 100 » 32%

Sony = ÷ ø

öçè

æ

110

51 × 100 » 46%

It can be seen from the table that Acer, Dell and Lenovo have much lower percentageof 5 stars ratings.

Hence, Sony from Xnapdeal is bought.

Thus, Criterion 2 is valid.

(3) Acer and Sony received equal number of ratings from both websites.

Among them, Sony got maximum number of 4 or 5 star ratings.Difference in the total number of 5 star and 1 star ratings for:

Philipkart = 140 – 30 = 110Xnapdeal = 130 – 50 = 80.

Hence, Sony from Xnapdeal is bought.Thus, Criterion 3 is valid.

Hence, [4].



38. For top 5 brands on Philipkart:Total change in ratings = 30 × (3 – 1) + 110 × (3 – 2) = 170

Total ratings = 550

Average change in ratings =550

170 = 0.309 star.

Statement 1 is true.

For top 5 brands on Xnapdeal:Total change in ratings = 50 × (3 – 1) + 80 × (3 – 2) = 180

Total ratings = 520

Average change in ratings =520

180 = 0.346 star.

Statement 2 is true.

Average ratings of the top 5 brands increased by more than 0.3 stars for each site,hence the same is true for both sites combined.

Statement 3 is true. Hence, [4].

39. Naval spending (in million $) of:Mexico = 15% of 984 = 148

Indonesia = 55% of 739 = 406

Canada = 35% of 672 = 235France = 40% of 1387 = 555

Russia = 30% of 899 = 270Maximum percentage difference:

÷ ø öç

è æ -

148148555

×100= 275%

Hence, [3].



40. Country Calculations Comments

Indonesia is clearly less than 100. So GDP of

Indonesia is definitely less than 10,000

Canada is clearly greater than 100. So GDP ofCanada is definitely greater than 10,000

Russia is clearly less than 100. So GDP of

Russia is definitely less than 10,000

France is clearly greater than 100. So GDP of

France is definitely greater than 10,000

1005.7

739´

1005.6

672´

10023

899 ´

1005.12

1387´

5.7

739

5.6

672

23

899

5.12

1387

So we need to check for GDP of Canada and France.

Canada:

5.6

672 =

5.6

)22650( + = 100 +

22

6.5, which is slightly greater than 103.

So GDP of Canada is slightly greater than 10,300.

France:

5.12

1387 =

5.12

)1371250( + = 100 +

137

12.5, which is definitely greater than 110.

So GDP of France is greater than 11,000. Hence, [4].

41. From Q5, Naval spending = 148 + 406 + 235 + 555 + 270 = 1614 million $ Army spending = 25% of 984 + 15% of 739 + 15% of 672 + 25% of 1387 + 35%

of 899 = 1119 million $ Air force spending = 55% of 984 + 30% of 739 + 30% of 672 + 25% of 1387 +

20% of 899 = 1491 million $We don't have to check for “Others”, since each of its data points lay below that

of Naval/Air force spending. Hence, [1].



42. Combined GDP of Mexico and Canada in 2040 = 20,696 Million $Combined GDP of Mexico and Canada in 2037 would be maximum, if the growth

rate is exactly 8% for each of the preceding three years.

Thus, combined GDP of Mexico and Canada in 2037 = 308.1

20696 = 16,429.15 Million $.

Hence, [2].

Answers to questions 43 to 46:

Number of students who passed = 45% of 120 = 54

\Number of ATKT students = 120 - (54 + 12) = 54

The number of students who did not clear Physics but cleared Chemistry or Biology= 37 – 12 = 25.

The number of students who did not clear only Chemistry = 10 = The number of students

who cleared both Physics and Biology, but not ChemistryThe number of students who did not clear exactly two subjects Physics and Chemistry= 6 = The number of students who cleared only Biology

Thus, we can create the following Venn diagram.

Cleared in Physics Cleared in Chemistry

Cleared in Biology

19 – y 19 – xy

54x10

6

12

Using this, all questions can be answered.

43. Required answer = 19 – y + y = 19. Hence, [2].



44. Total number of students who cleared Physics = 19 – y + y + 10 + 54 = 83Out of these, the number of students who cleared at most one other subject

= 19 – y + y + 10 = 29

\ Required percentage =83

29 × 100 » 35%

Hence, [3].

45. Required percentage =)x70(

)x54(

+

+ × 100

But 'x' is unknown. Hence, [4].

46. Number of students who cleared Chemistry = y + 19 – x + 54 + x = 88

Þ y = 15

\ The number of students who cleared exactly two subjects such that one of thesubjects is Physics = 15 + 10 = 25



The total distance travelled by each individual can be represented as follows:

Motorcycle Train Car

A (3) × 2 = 6 (5 + 2 + 2) × 2 = 18 (3 + 7 + 2) × 2 = 24

B (1 + 1 + 1 + 1 + 1) × 2 = 10 (2 + 4) × 2 = 12 (5 + 2) × 2 = 14

C (2 + 2 + 2 + 1) × 2 = 14 (2 + 5 + 2 + 1) × 2 = 20 (10 + 2 + 1) × 2 = 26D (1 + 3) × 2 = 8 (1 + 2 + 5) × 2 = 16 (13) × 2 = 26

Distance travelled (km)

Dividing individual distances by the individual speeds, we get the following table:

Motorcycle Train Car Total

A

B

C

D

Time taken (hours)

9

1

54

6=

7

1

70

10=

4

1

56

14=

8

1

64

8=

5

1

90

18=

15

2

90

12=

4

1

80

20=

5

1

80

16=

4

1

96

24=

6

1

84

14=

5

2

65

26=

3

1

78

26=

180

101

70

31

10

9

120

79



47. From the table, total time taken by D =120

79 hours =

120

79 × 60 minutes = 39.5

minutes = 39 minutes 30 seconds. Hence, [3].

48. Average speed of C =timeTotal

distance!otal =

10

9

262014 ++ =

9

600 = 66.66 km/hr.


49. Total distance travelled by B = 10 + 12 + 14 = 36 km.Therefore, the required answer is 36.

50. Option 1:

Time taken by C in motorcycle =4

1 hours

Time taken by A in car =4

1 hours

So, this option is true.

Option 2:

Time taken by B in train =15

2 hours

Time taken by D in motorcycle =8

1 hours

15

2 >

8

1, so this option is false.

Option 3:

Time taken by C in car =5

2 hours

Time taken by A in train =5

1 hours

52 >

51 , so this option is true.

Option 4:

Time taken by D in train =5

1 hours

Time taken by A in train =5

1 hours

So, this option is true. Hence, [2].




Actual feet sizes are 7.5, 8, 8.5, 9 and 9.5 (in no part icular order).Received shoe sizes are 7.5, 8.5, 9, 9.5 and 9.5 (in no particular order).

Thus, we can conclude that the shoe size 7.5 can only correspond to the feet size 7.5and the feet size 9.5 corresponds to the shoe size 9.5, such that both of them are narrow

feet. Now, Paulos' feet size cannot be 9.5, otherwise the shoes that he received couldfit him. Therefore, his feet size is 7.5 and Nebu's feet size is 9.5. Thus, we get the following

table:

Person Feet size Feet typeShoe size

ordered

Shoe size

received

Lazarus Broad 9.5

Mordecai Broad 7.5

Nebu 9.5 Narrow 9.5 9Onesimus Broad 8.5

Paulos 7.5 Narrow 7.5 9.5

The persons with broad feet have feet sizes 8, 8.5 and 9. Therefore, to meet the given

conditions, they must have ordered shoes of sizes 8.5, 9 and 9.5 respectively. Now, if Lazarus' feet size is 8.5 or 9, the shoes received by him would fit him, which is contradictory.

Therefore, his feet size is 8. Thus, we get the final table as follows:

Person Feet size Feet typeShoe size

ordered

Shoe size

received

Lazarus 8 Broad 8.5 9.5

Mordecai 8.5/9 Broad 9/9.5 7.5

Nebu 9.5 Narrow 9.5 9

Onesimus 9/8.5 Broad 9.5/9 8.5

Paulos 7.5 Narrow 7.5 9.5

Based on this, all questions can be answered:

51. Hence, [1].

52. The maximum difference was for Paulos (9.5 – 7.5 = 2). Hence, [3].

53. Hence, [2].

54. Hence, [4].




Using statements 1 and 4, we get the following arrangements for both the rounds.

From statement 2, we get two possibilities:

Case I: Case II:

But using the third statement, Case II can be eliminated. Therefore, we get the final

arrangement as follows:



55. H is sitting opposite E in round 2. Hence, [3].

56. A is sitting opposite E in round 1. Hence, [1].

57. Either B or F sits two places to the right of D in round 2. Hence, [4].

58. If E is sitting 3 places to the left D in round 1, then G is sitting 3 places to theright of B in round 2. Therefore, the required answer is G.


Using statements 1 and 4, we get the following cases:

From statement 5, case 1 can be eliminated. There are 3 possible cases for the positionsof R and P using statements 2 and 5:



Using statement 6, case 2c can be eliminated. Using statements 3, 6 and 7, the finaltwo cases can be determined as follows:

Using this, all questions can be answered.

59. Hence, [2].

60. Hence, [4].

61. Therefore, the required answer is 0.

62. Hence, [4].


We need 2 batswomen, 2 bowlers and 1 wicket keeper. There are 3 batswomen, namelySupriya, Sumaya and Sanjana. If Sumaya is selected, Sanjana has to be selected. If Sumayais not selected, the batters will be Supriya and Sanjana. Therefore, Sanjana must be selectedto the team. Moreover, if Sanjana is selected, Savita cannot be selected. Therefore, wehave to select from the following:

Batswomen: Sanjana and Supriya/Sumaya

Bowlers: Sujata and Suprita

Wicket keeper: Sunita or Sarita

Now, all the questions can be answered.

63. Hence, [3].

64. Hence, [4].

65. As seen above, Savita cannot be selected. Therefore, none of the three statementsis correct. Hence, [4].

66. Hence, [3].



SECTION - III

67. Sentences [1], [3] and [4] all talk about eating habits of the felids: [4] introducesthe felids and their carnivorous habits; [1] states how extreme this habit is; and [3]compares them to other mammals in terms of carnivorousness. Only [2] does notfit into this sequence, as it merely provides information about which animals makeup the felid family, and does not mention their eating habits at all. Hence, [2].

68. Sentences [1], [2] and [4] discuss what virtue is and when one can be said to havelived a virtuous life. [4] and [2] give Socrates' and Plato's views on virtue, respectively.[1] continues on the same theme. [3], on the other hand, does not involve a def initionof virtue, but simply a contrast between Socrates' beliefs about his own virtue andignorance. So it does not fit into the sequence. Hence, [3].

69. Sentences [1], [2] and [3] are about the history and ancestry of apples: [3] introducesthe topic; [1] continues this point, and adds the point about the addition of genesfrom other species; and [2] states which ancestral species modern apples are closestto. Only [4] does not fit into this sequence, as it talks about research being doneon breeding apples, which looks to the future rather than to the past, unlike other sentences. Hence, [4].

70. In A, the verb 'combine' is in the wrong form - it should be 'combining'. Alternatively,'Mandelbrot set' should be followed by a comma and the word 'which'. The unnecessaryrepetition of 'used' in B results in a parallelism error - the second 'used' should beremoved. In D, the correct word should be the adjective 'simple' (as a descriptionof the noun 'definition'), not its adverb form 'simply'. Thus, A, B & D are incorrect,and the remaining two statements are correct. Hence, ABD.

71. There is a subject-verb disagreement in A: the verb 'present' should be in the singular - i.e. 'presents' - as its subject is 'The Invisible Man', which is the name of a singlenovel, and not 'novels'. In C, the correct preposition after 'account' should be 'of',not 'on'. Thus, A and C are incorrect, and the rest of the statements are correct.Hence, AC.

72. Statement A suffers from a dangling modifier: there is no hint as to who is doingthe 'attempting'. Something like 'we find' should be added before 'almost' in order to provide a subject for the modifier to refer to. 'A little' in B is incorrect, as it means'a small amount', and has a positive connotation, whereas the sentences precedingand following this one suggest that the references do not add much, so the correct

word here would be simply 'little', which has a negative connotation. The rest of thestatements are correct, and only A and B are incorrect. Hence, AB.

73. According to Milan Kundera, there is no equivalent to the Czech word lítost in other languages, and so he claims that it is untranslatable. However, if people who do



not know Czech can still figure out what the word means based on the context inKundera's books, then the word cannot be truly called 'untranslatable'. So [1] weakensKundera's claim. Similarly, if [2] is true, it also means that lítost is not really untranslatable - it just doesn't have a single-word translation in other languages. So[2] also weakens the argument. Only [3] is not a suitable answer: since the translationof Kundera's statement in this paragraph does not explain what lítost really means,it does not undermine his claim that that particular word is untranslatable.

Hence, [4].

74. Option [4] is clearly incorrect, as it contradicts what is stated in the fourth sentenceof the paragraph - i.e. that there are hundreds of genes that differ from person toperson. On the other hand, neither can [3] be inferred, as there is no suggestionthat all genes differ from person to person. The author's objection to the HumanGenome project is not that the average of 200 people's genome is not sufficient.Rather, he claims that it is the point of the ABO gene to be different in everybody,

so there can be no 'average' form of it. Hence, [1].

75. There is no suggestion in the paragraph that the author views standardization of a language as 'futile', 'difficult' or 'undesirable'. Rather, he seems to consider it aworthy and achievable goal. So [1] and [2] are incorrect. His argument in this paragraphis that concentrating mainly or only on vocabulary is, however, not a good way topromote standardization, as contentious words are encountered rarely. He concludesthat it is better to focus on aspects of language that are more frequently encountered.Thus, his assumption is that there are aspects of language other than vocabularythat play a bigger role in daily language usage (and that these are the aspects thatshould be the focus of standardization efforts). So the correct answer is [4]. Notethat [3] is incorrect, as it still focuses on vocabulary, which the author argues against.

Hence, [4].

76. Statement B, which states the result of the studies mentioned in 1, links to it. Dlinks to B, as it continues the point about people's memories of familiar things. A,with 'for instance' gives an example of how people retain certain aspects of thingsin their mind, and thus elaborates upon C. Therefore, CA is a link. A also links to6, as they both provide examples of how people are better at remembering certainaspects of familiar things than others. Therefore, the correct sequence is BDCA.Hence, [1].

77. The term 'scientific revolution' links D to 1 as 1 talks about the more modern facetsof scientific development while D harks back to the past in saying that scientificinvestigations occurred in ancient times as well. 'These earlier periods' in C are the'ancient and medieval times' mentioned in D, so we get a DC link. B and 6 takentogether illustrate the point made in A, so there is an AB6 link. Thus, the correctsequence is DCAB. Hence, [4].

78. C connects better to 1 than D does, as C continues the point about translators beingeager to translate Agatha Christie's books, an example being Thailand. A links to



6, as both these sentences talk about the conclusion of a particular court case. Boththese links are found only in option [2]: CDBA. Hence, [2].

79. 'Pound' in [1] correctly refers to an 'enclosure for stray animals'. 'Pound out' in [2]means to 'produce by striking or thumping something'. In [3], 'pound' means to 'beator crush'. So the usage of 'pound' in all these sentences is correct. Only [4] is incorrect:'pound' is a unit of weight - it is not used for measuring liquids such as milk.Hence, [4].

80. 'Picked up' in [2] means 'increased in speed or tempo'. In [3], 'picked at' means'ate just a little bit of'. In [4], the same phrase means 'found fault with' or 'nagged'.Only in [1] is 'picked' used wrongly - the correct phrase should be 'pricked up', meaning'became attentive'. Hence, [1].

81. An 'open book' is 'something or someone that can be readily examined or understood'. An 'open secret' is 'something that is supposed to be secret but which is in factwidely known'. An 'open season' is a 'period of unrestrained criticism or attack onsomething or someone'. So options [1], [2] and [4] are correct. Only the usage of 'open' in [3] is incorrect: the correct phrase should be to 'keep an eye open (or aneye out)' - meaning to 'be especially alert or vigilant' - and not to 'keep an openeye'. Hence, [3].

82. 'Brio' means 'vigour', 'liveliness' or 'spirit'. While the rest of the options seem asif they could fit into the sentence, none of them means the same as 'brio'.

Hence, [4].

83. 'Blight' means 'something that causes ruin or destruction'. 'Bane', which means

'something that ruins or spoils', is therefore the closest in meaning. 'Defect', whichmeans 'flaw', does not have quite the requisite meaning, and nor does 'dearth', whichmeans 'lack or scarcity'. 'Salve', which has a positive connotation, does not fit atall. Hence, [1].

84. 'Surprisingly' and 'otherwise' suggest that 'baroque' should be roughly the oppositeof 'lucid' (clear or simple). 'Eloquent' (highly expressive) and 'ebullient' (enthusiasticor excited) can therefore be ruled out at once since they are not opposite in meaningto 'lucid'. 'Oblique', meaning 'indirect', could fit, but it is not really close in meaningas to the question word 'baroque', which means 'ornate' or 'extravagantly convoluted'.Hence, [3].

85. Option [1] involves a fact that has not been stated in the passage, so it can be ruledout at once. [2] makes no sense unless you know that 'Diana' is the name of agoddess - in which case, like [1], it involves external knowledge, and so can be ruledout as well. While [4] is the author's main point in this passage, it does not explainwhy he refers to the Moon as 'Diana'. The author begins the passage by mentioningthe 'dragons' that scare people from attempting a mission to Mars, and a 'Siren'



that tempts them away, i.e. the Moon. As a part of this analogy, he describes theMoon as a person: 'A Siren is there too, named Diana, the Moon Goddess, and her songs can be heard calling the Martian mariners to divert their ships once moreto a barren destination.' Hence, [3].

86. The author is not completely averse to going to the Moon - in the last passage, hedoes acknowledge that the Moon has some uses. So [1] is not quite correct; also,it fails to mention that the author wants people to go to Mars rather than the Moon.The author does not discuss the fears keeping people from planning a mission toMars in any great detail (it is only mentioned in paragraph 2), so [3] cannot be theanswer. While the author does want to do [2], that is not his main purpose in writingthis passage. Rather, his main purpose is to point out that people should not be'seduced' by the Moon's 'Siren song', i.e. they should not be distracted by the thoughtof going to Mars via or after going to the Moon instead of going to Mars directly.Hence, [4].

87. Refer to paragraph 5: B and E are clearly stated. C is stated in paragraph 4. It ispossible to infer from the penultimate sentence of paragraph 5 that Mars has a volcanichistory (like the Earth), whereas the Moon does not, so D is also correct. Only Ais incorrect: according to paragraph 5, 'Mars' gravity is 2.4 times that of the Moon',so the Moon's gravity would be 1/2.4 that of Mars. Hence, BCDE.

88. The sentence after the Napoleon Bonaparte quote - 'Well, if you want to go to Mars,go to Mars!' - taken in connection with the author's message in the rest of the passage,suggests that the quote means that if you want to do something, do it directly. Thatis, if you want to take Vienna, take Vienna directly, rather than wasting time tryingto conquer other places first; and if you want to go to Mars, go to Mars directly,

rather than allowing yourself to be distracted by the Moon as a connecting destinationto Mars. So the answer is [2]. While it is possible that the quote may imply whatis stated in options [1], [3] or [4] instead/also, these options are not inferable fromthe passage. Hence, [2].

89. In the first paragraph, the author explains what Occam's Razor is; in the rest of the passage, he talks about its applications, especially cases in which it is misappliedor of limited value. Therefore, the author can best be said to be trying to do [3].Note that [1] and [4] are incomplete answers, as the former does not mention theapplications and the latter does not mention the explanation. The author does notdebate anything - in fact, he shows a distinct aversion to arguing (see the thirdparagraph). So [2] is not a suitable answer either. Hence, [3].

90. Refer to paragraph 3. While the author does suggest that the concept of the supernaturalis not parsimonious and that postulating the existence of God as the creator of theuniverse is extravagant, he does not use either of these points to show that Occam'sRazor does not help in proving or disproving the existence of God, so [1] and [2]can be ruled out. Rather, the author simply points out that Occam's Razor is a rule



of thumb, not a 'Metaphysical Principle or Fundamental Requirement of Rationality',so it does not make sense to use it for something as major as proving or disprovingthe existence of God. Note that while he does claim that he doesn't want to argueabout this point, he does go on to state [3], so [4] is incorrect. Hence, [3].

91. The author reluctantly admits that the student's interpretation of Parmenides' claimseems to be valid, so [3] is incorrect. However, this is not the main point of theparagraph, but rather just an example, so [4] is incorrect as well. [2] is the pointof the last sentence, not the whole paragraph, and it seems to be mentioned mainlyas an aside, so it can be eliminated as well. Only [1] is the main point that the author is making in the last paragraph: he provides the example of Parmenides to showhow 'some thinkers have carried Occam's Razor to drastic extremes'. Hence, [1].

92. While the author does mention certain shortcomings of Occam's Razor in this passage,there is no suggestion that he dislikes it personally - his attitude is completely objective.

In the second paragraph, he mentions that Conwy Lloyd Morgan's ideas can beproblematic when overused and taken to extreme lengths, but that does not meanthat he disagrees with the basic ideas themselves. So [1] and [3] are incorrect. However,[2] is easily inferable: in the last paragraph, the author mentions his student, andcalls himself a philosopher. Hence, [2].

93. While the points made in Loos's essay do make up a substantial part of the passage,the author cites it primarily in order to discuss the general issue of ornamentationin architecture, so [1] is too specific. On the other hand, [2] and [3] are too general:the author does not discuss all aspects of architecture in recent centuries; nor doeshe compare and contrast 19th- and 20th-century architecture in general. Rather, hetalks specifically about the attitude towards ornamentation in architecture, starting

from 1828 till the present day, and differing views about the same. Hence, [4].

94. According to the first two paragraphs, Heinrich Hübsch considered ornamentationto be an important part of architecture, whereas Adolf Loos detested it, so [1] canbe eliminated at once. According to the fifth paragraph, Joseph Rykwert does notconsider ornamentation to be a problem in itself, so [4] too is incorrect. While Reyner Banham clearly admires Loos's contribution to architecture (see paragraph 4 as wellas the last sentence of the passage), he also criticizes Loos's essay, so it is notclear whether he fully agrees with Loos's ideas. In the second paragraph, Le Corbusier is stated to be one of Loos's fellow Modernists, who rebelled against ornamentationin architecture, so it is likely that he shared Loos's views on the same. Hence, [2].

95. Refer to the last paragraph. The author does not discuss 21st-century architecturein terms of any particulars. All he states is that four decades after Joseph Rykwert'sessay (i.e. in 2015), architecture is once again dealing with questions that HeinrichHübsch raised in the 19th century. We can thus infer that the attitude of 21st-centuryarchitecture towards ornamentation is closer to the more ornamental style of 19th-century architecture than that of the highly austere style of 20th-century Modernist



architecture. Thus [1] and [2] are incorrect, and [4] is correct. [3] may be true, butit cannot be inferred with any degree of certainty from this passage. Hence, [4].

96. Option [1] is a paraphrase of Banham's statement in paragraph 4. [2] is inferablefrom his statement in the last paragraph. [4] is stated in paragraph 3. Only [3] isincorrect: the preference for Rundbogenstil is that of Heinrich Hübsch, whose viewsare criticized in 'Ornament and Crime'. Hence, [3].

97. Option [1] is stated in the second paragraph; [3] in the first paragraph; and [4] inthe third paragraph. But [2] is not stated - rather, in the last paragraph, the author claims that he worries about science-based superstition as well. Hence, [2].

98. Since the author does provide several examples of the misuse of science as aninstitution in the last paragraph (with no indication as to whether these occurred insituations where science was centralized), [1] cannot be inferred. According to theauthor, science can be a kind of superstition when it's based on an argument fromauthority - so [3] is a misinterpretation of the author's point. While the author doesstate [4], it is more as evidence for his point, rather than his point itself. The correctanswer is [2]: in the last paragraph, the author expresses his concern about onemore kind of superstition, i.e. scientific superstition, and states that it occurs whenscience is based on faith, a phenomenon that has been increasing over the pastcentury. Hence, [2].

99. The author's argument in the third and fourth paragraphs is that fundamentalists willsoon outnumber moderate/secular people, because the former are breeding at a faster rate than the latter (whose birthrate is in fact declining). However, if A is true, thehigh birthrate among fundamentalists would still not result in them outnumbering the

moderate/secular people (at least, not for a long time) implying that superstition willnot keep growing so quickly. Thus A weakens the argument. The author takes intoconsideration the possibility of B, so it cannot be said to weaken his argument; also,without any precise statistics regarding the proportion of people who turn away fromfundamentalism later in l ife, we cannot say whether it will make a substantial differenceto the overall number of fundamentalists. If C is true, more countries may turn awayfrom fundamentalism - but their population will also decline, as secular people breedless, according to the author. So C does not weaken the author's argument. Additionally,there seems to be no link between 'superstition' and 'economic prosperity' in thethird and fourth paragraphs. Hence, A.

100. Statement A can be easily inferred based on the author's worry about the misuse

of science in the last paragraph. Considering that the author spends a significantportion of the passage worrying about religious people outbreeding secular ones,it is very likely that B is true. However, there is no basis for inferring C - the author is talking about a general demographic trend (about secular people not having children),which may not necessarily be a personal experience. According to the first paragraph,the author considers banning genetically modified foods to be a 'stupid thing', so



D is incorrect as well. There is no reason to infer E - the only reference to a declinein living standards is regarding past civilizations (see the second paragraph). Therefore,only A and B can be inferred about the author. Hence, AB.

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