CAT 2000 QA | Previous Year CAT Paper

Answer: Option C

Explanation :

D = 0.a₁a₂a₁a₂a₁a₂.....

∴100D = a₁a₂∙a₁a₂.....

100D – D = 99D = a₁a₂, which is an integer.

∴ If D is multiplied by 99 or any multiple of 99 it will result in an integer.

From the options given, it is clear that 198 (= 99 × 2) will necessarily produce an integer, when multiplied by D.

Hence, option (c).

Answer: Option B

Explanation :

In the above table, it can be seen that the values are taken at equal intervals. Also, the second differences i.e. values of ∆²y are constant and hence, the function is actually a polynomial of degree 2.

Thus, y = a + bx + cx² best describes the relation between y and x.

Hence, option (b).

Alternatively,

The relation between x and is not a line since $\frac{y_2-y_1}{x_2-x_1}$ is not a constant.

Also, e^a+bx will not be a whole number. Hence, by substituting values from the table, it can be checked that the relation is given by y = a + bx + cx².

Hence, option (b).

Answer: Option D

Explanation :

a₁ = 1 = 6 – 5

a₂ = 7 = 12 – 5 = 6 × 2 – 5

a₃ = 19 = 24 – 5 = 6 × 2² – 5

a₄ = 43 = 48 – 5 = 6 × 2³ – 5

and so on.

Thus, a₁₀₀ = 6 × 2⁹⁹ – 5

Hence, option (d).

Answer: Option C

Explanation :

$\frac{1}{2^2-1}+\frac{1}{4^2-1}+\frac{1}{6^2-1}+...+\frac{1}{{20}^2-1}$

$=\frac{1}{1\times 3}+\frac{1}{3\times 5}+\frac{1}{5\times 7}+...+\frac{1}{19\times 21}$

$=\frac{1}{2}\left[\left(1-\frac{1}{3}\right)+\left(\frac{1}{3}-\frac{1}{5}\right)+\left(\frac{1}{5}-\frac{1}{7}\right)...+\left(\frac{1}{19}-\frac{1}{21}\right)\right]$

$=\frac{1}{2}\left[\left(1-\frac{1}{21}\right)\right]=\frac{1}{2}\left[\frac{20}{21}\right]=\frac{10}{21}$

Hence, option (c).

Answer: Option C

Explanation :

Given, that the truck can travel 19.5 kilometres on a litre of diesel at 50 kilometres per hour.
∴ At 50 km/hr, truck covers 19.5 km in 1 litre diesel.

∴ The truck will use 30% more diesel i.e., 1.3 litre diesel to cover 19.5 km at 70 km/hr.

∴ At 70 km/hr, in 10 litres of diesel, the truck will cover 19.5/1.3 × 10 = 150 km

Hence, option (c).

Answer: Option B

Explanation :

Average of first five integers is n.

∴ The integers are (n – 2), (n – 1), n, (n +1), (n + 2).

∴ The last two numbers are (n + 3), (n + 4).

∴ The average of the seven numbers = (7n + 7)/7 = n + 1

Hence, option (b).

Answer: Option B

Explanation :

 y > –1

∴ –2y < 2 < x

∴ –x < 2y

Hence, option (b).

Answer: Option A

Explanation :

First arrange one red and two blue flags in 3 ways. (i.e. BBR, BRB, RBB)

Now there are four positions (say 1, 2, 3, 4) to arrange 3 white flags. Since the flags at the ends are of different colours, two white flags can’t be at positions 1 and 4 simultaneously. Thus, the three flags can be arranged at 1, 2, 3 or 2, 3, 4 in 2 ways.

Thus, the six flags can be arranged in 3 × 2 = 6 ways.

Hence, option (a).

Answer: Option D

Explanation :

100 ≤ x ≤ 200 and x is odd multiple of 3 which is not divisible by 21.

Number of x those are odd multiples of 3 = 16

Number of x those are odd multiples of 21 = 3

∴ Number of elements in S = 16 − 3 = 13

Hence, option (d).

Answer: Option D

Explanation :

Option 1:

Product of odd numbers is always odd.

∴ x × y × z × z  is always odd

Option 2:

(x − y) is even.

∴  (x − y)² z  is even.

Option 3:

(x + y) is always even.

∴  (x + y − z)² (x + y) is even.

Option 4:

(x − y) is even, (y + z) is even, and (x + y − z) is odd.

∴  (x − y) (y + z) (x + y − z) is even.

Hence, option (d).

Answer: Option A

Explanation :

There is only one even prime number and one prime number ending with 5 (i.e 5 itself) in set S.

When all elements of set S are multiplied, there will be only 1 zero at the end of the product.

Hence, option (a).

Answer: Option C

Explanation :

Let the sides of the triangle be a, b and c then, a + b > c.

a + b + c = 14

Case 1: Largest side is 6
Possible sets of sides are  (6, 4, 4), (6, 5, 3) and (6, 6, 2).

Case 2: Largest side is 5
Possible set of sides is (5, 5, 4).

∴ 4 triangles are possible with integral sides and perimeter 14.

Hence, option (c).

Answer: Option C

Explanation :

1421 ≡ 5(mod 12); 1423 ≡ 7(mod 12); 1425 ≡ 9(mod 12)

∴ N ≡ 5(mod 12) × 7(mod 12) × 9(mod 12)

    = 315(mod 12) = 3(mod 12)

∴ The remainder is 3.
 
Hence, option (c).

Alternatively,

N = 1421 × 1423 × 1425

= (1422 – 1)(1422 + 1)(1428 – 3) = (1422² – 12)(1428 – 3)  

Both 1422² and 1428 are divisible by 3 as well as 4.

N = (1422² – 1)(1428 – 3)
 
(1422² – 1) divided by 12 leaves a remainder of −1 and (1428 – 3) divided by 12 leaves a remainder of −3
 
Effective remainder = −1 × −3 = 3
 
∴ N when divided by 12 leaves a remainder of 3.
 
∴ N = X + 3; where X is divisible by 12

Thus, when N = 1421 × 1423 × 1425 is divided by 12, the remainder will be 3.

Hence, option (c).

Answer: Option D

Explanation :

Let 34041 and 32506 when divided by n leave remainder r.

34041 = nk₁ + r       …(i)     

32506 = nk₂ + r       …(ii)      

Subtract equation (ii) from equation (i)

n (k₁ –  k₂) = 1535

∵ n has to be a factor of 1535.

From the given options, only 307 is a factor of 1535.

Hence, option (d).

Answer: Option A

Explanation :

The terms in the given expression are x₁,  x₂, ... x_n-1, x_n.

There are n terms in the expression.

The only possible value of each of the terms is either 1 or –1.

For the given expression to be zero there should be even number of terms of which half the terms are –1 and the remaining are 1.

∴ n is even.

Hence, option (a).

Answer: Option B

Explanation :

Population below 35 years of age = 30 + 17.75 + 17 = 64.75% of the total population = 400 million

∴ 30% of the total population $=\frac{30\times 400}{64.75}$ million ≈ 185 million

The ratio of females to males in the ‘below 15 years’ age group is 0.96.

i.e. if the total population is 196, then there are 96 females.

Approximately, the number of females (in millions) in the ‘below 15 years’ age group

$=\frac{185\times 96}{196}=90.6$

Hence, option (b).

Answer: Option C

Explanation :

Case i: The last digit is 9.

∴ The remaining 3 digits can be filled by any of the 9 digits (i.e., 0 to 8) in 9³ ways.

∴ Total number of ways = 729

Case ii: The last digit is not 9.

∴ The last digit can be filled in 4 ways.

Let any of the other 3 digits be 9. This can happen in 3 ways.

∴ The remaining two digits can be filled in 9², i.e. 81 ways.

∴ The total number of ways = 9² × 3 × 4 = 972

∴ Number of trials for last four digits = 729 + 972 = 1701

Also, for the first three digits he has two options.

The minimum number of trials he has to make before he can be certain to succeed = 2 × 1701 = 3402

Hence, option (c).

Answer: Option A

Explanation :

Sum of A and B = (Average of A and B) × 2

=  /(@(A, B), 2)

∴ Hence, option (a).

Answer: Option D

Explanation :

We need to find average of A, B and C i.e. (A + B + C)/3

The only operator that can give 3 in the denominator is “X”

Option 1 cannot be the answer, as there is no “X”.

Option 2 also cannot be the answer as X(@(/(@(/(B, A), 2), C), 3), 2) gives 2 in the denominator and not 3.

Similarly option 3 is not the answer.

Hence, option (d) must be the answer.

Option 4:

/ ( X (@ (/ (@ (B, A), 2), C), 3), 2)

$=/\left(x\left(@\left(/\left(\frac{A+B}{2},2\right),c\right),3\right),2\right)$

=/(X(@((A + B), C), 3), 2)

$=/\left(x\left(\left(\frac{A+B+C}{2}\right),3\right),2\right)=/\left(\frac{A+B+C}{6},2\right)$

$=\frac{2(A+B+C)}{6}=\frac{A+B+C}{3}$

Hence, option (d).

Answer: Option D

Explanation :

If (x + y) > 0, (x + y)^0.5 is real.

∴ For (x + y) > 0;

f(x, y) = Positive square-root of (x + y) > 0

g(x, y) = (x + y)² > 0

Otherwise,(i.e. for (x + y) < 0);

f(x, y) = (x + y)² > 0

g(x, y) = –(x + y) > 0

Now it is clear that f(x, y) + g(x, y) will always be positive.

Hence, option (d).

Answer: Option A

Explanation :

If 1 > (x + y) > 0; f(x, y) > g(x, y) and if (x + y) < –1; f(x, y) > g(x, y).

If both x and y are less than –1, (x + y) will also be less than –1.

Hence, option (a).

Answer: Option D

Explanation :

x, y and z  are distinct real numbers.

∴ Without loss of generality, let x < y < z

Then,

f(x, y, z) = y; g(x, y, z) = y; h(x, y, z) = z

j(x, y, z) = x; m(x, y, z) = z; n(x, y, z) = x

Substituting these values in the given options,

Option 1 = (z – y)/x may or may not be greater than 1.

Option 2 = x/z < 1

Option 3 = y/y = 1

Option 4 = (y + z − y)/x > 1

Hence, option (d).

Answer: Option A

Explanation :

Using the values from previous answer,

Option 1 = (y − z)/(y − z)

Option 2 = (z − y)/(y − x)

Option 3 = (x − y)/z

Option 4 = (y − z)/y

It can be easily concluded that the expression in option 1 is necessarily equal to 1.

Hence, option (a).

Answer: Option B

Explanation :

Using the values found previously,

Option 1 = (y − z)/(y − x)

Option 2 = (y + z + y + x)/(x + z – z − x)

Option 3 = (y − x)/(y – z)

Option 4 = (z − y)/(x – y)

The denominator in the expression in option 2 is zero.

Hence, the expression in option 2 is indeterminate.

Hence, option (b).

Answer: Option A

Explanation :

Machines A and D sent message at the beginning of the first second.

∴ Round 1: Origin to Machine A to Machine D to Origin

Distance covered by the robot = 80 m (40 m + 40 m)

Time taken to return to origin = 8 seconds

At the end of 8th second, messages from C and B are already received,

∴ Round 2: Origin to Machine B to Machine C to Origin

Distance covered by the robot = 60 m (30 m + 30 m)

Time taken to return to origin = 6 seconds

Total Time elapsed = 8 + 6 = 14 seconds

At the end of 14th second, message from E is already there.

The robot has travelled 140 m (80 m + 60 m), when it notices message from E.

Hence, option (a).

Answer: Option A

Explanation :

Machines A and D sent message at the beginning of the first second.

∴ Round 1: Origin 1 to Machine A to Machine D to Origin 2

Distance covered by the robot = 10 m + 30 m + 20 m = 60 m

Time taken for round 1 = 6 seconds

At the end of 6th second, messages from C and B are already received,

∴ Round 2: Origin 2 to Machine C to Machine B to Origin 1

Distance covered by the robot = 30 m + 10 m + 20 m = 60 m

Time taken to return to origin = 6 seconds

Total Time elapsed = 6 + 6 = 12 seconds

At the end of 12th second, message from E is already there.

The robot has travelled 120 m (60 m + 60 m), when it notices message from E.

Hence, option (a).

Answer: Option C

Explanation :

This is a graph of straight line that is parallel to x-axis passing through (0, 1)

∴ for every x, f(x) = 1

∴ f(x) = f(−x)

Hence, option (c).

Answer: Option D

Explanation :

As per the given figure,

f(1) = 2,  f(−1) = 1

∴ (1), (2) and (3) cannot be correct choices.

 f(1) = 2f(−1)

i.e., 3 f(1) = 6 f(−1)

∴ 3f(x) = 6 f(−x)

Hence, option (d).

Answer: Option B

Explanation :

In the given figure, f(3) = −1,  f(−3) = 1 

i.e. f(3) = −f(−3)

∴ (1), (3) and (4) cannot be correct choices.

If we check with remaining values we conclude that f(x) = −f(−x)

Hence, option (b).

Answer: Option B

Explanation :

Instruction 1: FILL (C, A)

2 litres of water is transferred from A to C

∴ A has 3 litres of water left and C has 2 litres of water.

Instruction 3: FILL (C, A)

2 litres of water is transferred from A to C

∴ A has 1 litres of water left and C has 2 litres of water.

Since the third instruction again transfers water from A to C, C should be empty before the third operation. Thus, instruction 2 should be either “EMPTY (C, B)” or “DRAIN (C)”.

Hence, option (b).

Answer: Option C

Explanation :

After third instruction A has 1 litre, B has 2 litres and C has 2 litres of water.

Instruction 4: DRAIN A

Instruction 5: EMPTY (B, A)

Instruction 6: EMPTY (C, A)

After 6^th instruction, C has no water as all 2 litres have been transferred to A.

Hence, option (c).

Answer: Option C

Explanation :

When x = 2, f(x) = 1/(1 + x)

$f^3\left(2\right)=\frac{1}{1+2}=\frac{1}{3},f^2\left(2\right)=f\left(f\right(2\left)\right)=f\left(\frac{1}{3}\right)=f\left(\frac{1}{3}\right)=\frac{1}{1+{\displaystyle \frac{1}{3}}}=\frac{3}{4}$

$f^3\left(2\right)=f\left(\frac{3}{4}\right)=\frac{1}{1+{\displaystyle \frac{3}{4}}}=\frac{4}{7}$

$f^4\left(2\right)=f\left(\frac{4}{7}\right)=\frac{1}{1+{\displaystyle \frac{4}{7}}}=\frac{7}{11}$

$f^5\left(2\right)=f\left(\frac{7}{11}\right)=\frac{1}{1+{\displaystyle \frac{7}{11}}}=\frac{11}{18}$

∴ f(2)f²(2)f³(2)f⁴(2)f⁵(2)=$\frac{1}{3}\times \frac{3}{4}\times \frac{4}{7}\times \frac{7}{11}\times \frac{11}{18}=\frac{1}{18}$

Hence, option (c).

Hence, option (c).

Answer: Option B

Explanation :

f(–r) = 1 + (–r) = 1 – r < 0 (∵ r ≥ 2)

f²(–r) = f(f(–r) = f(1 – r) = 1 + (1 – r) = 2 – r

f²(–r) = 2 – r = 0,     if r = 2

                   < 0,     if r > 2

Similarly,

f³(–r) = 3 – r = 0,     if r = 3

                   < 0,     if r > 3

and so on.

i.e. f^r–1(–r) < 0 and fr(–r) = 0

∴ f^r+1(–r) = 1/(1+0) = 1

∴ f^{r – 1} (–r) +  f^r (–r) + f ^{r + 1} (–r)  = –1 + 0 + 1 = 0.

Hence, option (b).

Answer: Option C

Explanation :

There are 8 teams in each group.

Stage 1:

Within a group each team played with every other team.

∴ Total number of matches = 7 + 6 + 5 + 4 + 3 + 2 + 1 = (7 × 8) / 2 = 28 matches in one group

∴ Total number of matches in both the groups = 56 matches

Stage 2:

Number of matches = 4 matches (Round 1) + 2 matches (Round 2) + 1 Match (Round 3) = 7

∴ Total number of matches played in the tournament = 63

Hence, option (c).

Answer: Option B

Explanation :

28 matches are played between 8 teams.

Consider option 1:

There is a possibility that 5 teams win exactly 5 matches each (5 × 5 = 25 matches played), one team wins 2 matches, one team wins exactly one match and one team lose all.

{For example: Let A, B, C, D, E, F, G, and H are the eight teams. A wins matches against B, C, D, E, F; B wins matches against C, D, E, F, G; C wins matches against D, E, F, G, H; G wins matches against A, D, E, F, H and H wins matches against A, B, D, E, F. Also E wins match against D. F wins matches against D and E. D loses all the matches}

Thus, for a team, there is no guarantee of its advancement to the next stage if it wins 5 matches.

 But if a team wins six matches, there is no possibility that 5 teams will win 6 matches each. (As there are only 28 matches) Hence, the team will definitely advance to the next stage.

 Hence, option (b).

Answer: Option A

Explanation :

We need to find, maximum value of x, such that despite winning x matches, a team gets definitely eliminated at the end of the first stage.

If x = 4; the team can reach in second stage.

{For example: Let A, B, C, D, E, F, G, and H are the eight teams. A wins matches against B, C, D, E, F, G(i.e. 6 matches); B wins matches against C, D(i.e. wins 2 matches);; C wins matches against D, E(i.e. wins 2 matches); D wins match against E; E wins match against B; F wins matches against B, C, D, E(i.e. wins 4 matches); G loose only one match against A (i.e. wins 6 matches);  and H loose only one match against G (i.e. wins 6 matches).}

If x = 3; the team can reach in second stage.

{For example: Let A, B, C, D, E, F, G, and H are the eight teams. A wins matches against B, C, D, E, F, G(i.e. 6 matches); B wins matches against C, D; C wins matches against D, E; D wins match against E and F(i.e. wins 2 matches); E wins match against B; F wins matches against B, C, E(i.e. wins 3 matches); G loose only one match against A (i.e. wins 6 matches);  and H loose only one match against G (i.e. wins 6 matches).}

If x = 2; the team can reach in second stage.

{For example: Let A, B, C, D, E, F, G, and H are the eight teams. A wins matches against B, C, D, E, F, G(i.e. 6 matches); B wins matches against C, D; C wins matches against D, E; D wins match against E and F(i.e. wins 2 matches); E wins match against B and F(i.e. wins 2 matches); F wins matches against B, C,(i.e. wins 2 matches); G loose only one match against A (i.e. wins 6 matches);  and H loose only one match against G (i.e. wins 6 matches).}

Thus, for x = 2, 3 and 4; we definitely cannot say that the team will not reach the second stage.

We can say that the value of x must be 1, because only four teams reach the second stage and it is not possible that more than two teams win exactly one match each.

Hence, option (a).

Answer: Option C

Explanation :

When the second stage starts, the first round will have 4 matches between 8 teams. From this round, only 4 teams will advance to the second round. So, second round will have 2 matches. In the third round, two teams will play 1 match and after that the winner will be declared.

Thus, the number of rounds in the second stage of the tournament = 3

Hence, option (c).

Answer: Option C

Explanation :

We have seen that even if a team wins 2 matches, it can be in the second stage of the tournament. Consider Team A wins 2 matches and enters the second round and wins the tournament. Then its total score will be 5 points. While there can be a team having more than 6 points but not the winner of the tournament. Thus, option 1 may not be true.

We have seen that even if a team wins 5 matches is can be eliminated and a team winning only two matches can enter second round. Thus, option 2 may not be true.

Consider a team in group I eliminated and the score is 5. A team in group II advances to the second stage with score 2. If this team (i.e. the team with score 2) is a winner then the score will be 2 + 3 = 5. Thus, option 3 is true.

Eight teams enter second stage. Of these, 4 teams advances to the second round scoring 1 point each. In the second round, only two teams win and their score in the second round so far will be 2. Thus, only two teams will have score 1. Thus, option 4 is not true.

Hence, option (c).

Answer: Option D

Explanation :

N = 55³ + 17³ − 72³ = 55³ + 17³ +(−72³)

It is known that if a + b + c = 0 then a³ + b³ + c³ = 3abc.

In the given expression, 55 + 17 + (−72) = 0

∴ N = 3 × 55 × 17 × (−72)

∴ N is divisible by both 3 and 17.

Hence, option (d)

Answer: Option B

Explanation :

x² + y² = 0.1                          …(i)

|x – y| = 0.2

Squaring both the sides, we get,

x² + y² – 2xy = 0.04             …(ii)

From (i) and (ii),

2xy = 0.1 – 0.04 = 0.06

Thus, x and y both are positive or both are negative.

∴ 2xy = 2|x||y|= 0.06

(|x| + |y|)² = x² + y² + 2|x||y| = 0.1 + 0.06 = 0.16

|x| + |y| > 0

∴ |x| + |y| = 0.4

Hence, option (b).

Answer: Option A

Explanation :

Equation of line AD is x + y = 1.

∴ AD intersects x-axis at (1, 0) and y-axis at (0, 1).

ABCD is a rhombus, with diagonals AC and BD intersecting at origin.

The four vertices are (1, 0), (0, 1), (−1, 0) and (0, −1).

Since AD pass through (1, 0) and (0, 1), BC passed through (−1, 0) and (0, −1).

Equation of BC $\frac{x}{-1}+\frac{y}{-1}=1$

i.e. x + y = −1

Hence, option (a).

Answer: Option A

Explanation :

Let the area of sector S₁ be x units.

The area of the sectors S₂, S₃, S₄, S₅, S₆, S₇ will be 2x, 4x, 8x, 16x, 32x and 64x

∴ The total area of 7 sectors = 127x units = (1/8) × total area of circle = (1/8) π

∴ 127x = π/8 units

A circle subtends an angle of 2π at the centre.

Hence, (1/8)th of the circle will subtend an angle of π/4 at the centre.

i.e. area of the seven sectors i.e. 127x will subtend an angle of π/4 at the centre.

Sector S₁, whose area is x will subtend an angle of π/(127 × 4) at the centre.

∴The required angle = π/508 radians

Hence, option (a).

Answer: Option D

Explanation :

In four moves the task can be accomplished.
One of the ways is given below.

Initial state: Table A: 3(on top) 2(in middle) 1(bottom)

Step 1: Move book 3 from table A to table B

Step 2: Move book 2 from table A to table B

Step 3: Move books 2(on top) and 3(below book 2) from table B to table A on book 1(bottommost)

Step 4: Move books 2, 3 and 1 from table A to table B.

Hence, option (d)

Answer: Option B

Explanation :

We need to find area bounded by |x + y| = 1, |x| = 1, and |y| = 1

i.e. all points P(x, y) in the plane such that −1≤ x + y ≤ 1, −1≤ x ≤ 1, and −1≤ y ≤ 1

−1≤ x + y will be the origin side of line x + y = −1

x + y ≤ 1 will be the origin side of line x + y = 1

−1≤ x ≤ 1 will be the area between the lines x = −1 and x = 1.

−1≤ y ≤ 1 will be the area between the lines y = −1 and y = 1.

Tracing these curves, we get the area shown in the graph below

∴ Shaded Area = Area of 3 squares of side 1 unit = 3 × (1)2 = 3

Hence, option (b).

Answer: Option B

Explanation :

x³ – ax² + bx – a = x³ + bx – ax² – a = x(x² + b) – a(x² + 1)

∴ x(x² + b) – a(x² + 1) = 0

If b = 1 then x(x² + 1) – a(x² + 1) = (x – a)(x² + 1) = 0

∴ (x – a) = 0 or (x² + 1) = 0

But then (x² + 1) = 0 does not have real roots.

∴ b ≠ 1

Hence, option (b).

Answer: Option A

Explanation :

a, b, c are the sides of a triangle.

It is given that a² + b² + c² = bc + ca + ab

⇒ a² + b² + c² - bc - ca - ab = 0

⇒ 2a² + 2b² + 2c² - 2bc - 2ca - 2ab = 0

⇒ (a² -2ab + b²) + (b² - 2bc + c²) + (c² - 2ca + a²) = 0

⇒ (a - b)² + (b - c)² + (c - a)² = 0

This is possible only if a = b = c

i.e., the triangle is equilateral.

Hence, option (a).

Answer: Option D

Explanation :

Consider the given figure,

Let ∠DAE = x
In ΔABC, AB = BC
∴ m∠ACB =  m∠BAC = x
∴ Being an exterior angle of ΔABC, m∠CBF = 2x

In ΔAGF, FG = GA
∴ m∠AFG =  m∠GAF = x
∴ Being an exterior angle of ΔAGF, m∠CGF = 2x

In ΔBCD, BC = CD
∴ m∠CBD = m∠CDB = 2x
∴ Being an exterior angle of ΔACD, m∠DCE = 3x
∵ DE = CD, in ΔDCE, m∠CED = m∠ECD = 3x

In ΔGFE, EF = FG
∴ m∠FGE = m∠FEG = 2x
∴ Being an exterior angle of ΔAFE, m∠EFD = 3x
∵ DE = EF, in ΔDFE, m∠EFD = m∠EDF = 3x

In ΔADE,
x + 3x + 3x = 7x = 180°
∴ x ≡ 25°
∴ m∠DAE ≡ 25°

Hence, option (d).

Answer: Option B

Explanation :

Let a, b, c, d and e be the weights, in kg ,of the five boxes with the shipping clerk, where,

a < b < c < d < e.

110 = a + b < a + c < ….. < c + e < d + e = 121

i.e. a + c = 112 and c + e = 120

Each box is weighed 4 times.

∴ 4a + 4b + 4c + 4d + 4e = 110 + 112 + 113 + 114 + 115 + 116 + 117 + 118 + 120 + 121 = 1156

∴ a + b + c + d + e = 289

Now it is clear that a + b = 110 and d + e = 121

∴ 110 + c + 121 = 289

∴ c = 58

Substituting this value in c + e = 120

∴ e = 62

Hence, option (b).

Answer: Option A

Explanation :

Let there be,

x roads connecting A and B directly,

y roads connecting B and C directly and

z roads connecting C and A directly

∴ Total number of routes connecting A and B: x + yz = 33           …(i)

∴ Total number of routes connecting B and C: y + xz = 23           …(ii)

Subtracting (ii) from (i)

(x − y) + z(y − x) = 10

−1(y − x) + z(y − x) = 10

 (y − x)(z − 1) = 10

 (y − x)(z − 1) = 5 × 2                                                                           …(iii)

From the options, the possible values for z are 3 and 6.

Consider z = 3

∴ y – x = 5                                                             …(iv)

From equations (i), (ii) and (iv) we get the values as x = 4.5 and y = 9.5 which is not possible

Consider z = 6

∴ y – x = 2                                                                …(v)  

Solving (i), (ii) and (v), we get y = 5, x = 3

Thus, there are 6 direct roads between A and C.

Hence, option (a).

Answer: Option B

Explanation :

The functions f(n) and g(n) are disjoint sets and union of these two sets is the set of all positive integers.

∵ g(n) = f(f(n)) + 1 for all n ≥ 1

and f (1) < f(2) <...< f(n) ....., and g(1) < g(2) <...< g(n) .......,

∴ f(1) = 1 or 2

If f(1) = 1

g(1) = f(f(1)) + 1

∴ g(1) = f (1) + 1 = 1 + 1 = 2

If f(1) = 2

g(1) = f(f(1)) + 1

∴ g(1) = f (2) + 1

∴ g(1) is greater than f(1), i.e. it is greater than 2.

But the set of all positive integers is the union of these two disjoint sets.

∴ This set has to include 1 which is not possible in this case as f(1) is 2 and g(1) will be greater than f(1).

∴ f(1) = 1 and g(1) = 2

Hence, option (b).

Answer: Option A

Explanation :

The frog has to jump at least four times to reach E.

i.e. a₄ = 2

(i.e. (i) A-B, B-C,C-D and D-E; (ii) A-H, H-G, G-F and F-E).

If the frog keeps jumping on left (right) hand side vertices it will not take more than four jumps.

If it jumps on right vertex and then keeps on jumping left vertices then it will take 6 jumps to reach E.

∴ We will not find any path such that the frog takes 5 jumps and reaches E.

∴ a₅ = 0

Similarly, we find that only for even values of n the frog can reach E.

2n − 1 is an odd number

∴ No route is possible with odd number of jumps.

Hence, option (a).

Answer: Option B

Explanation :

f (1, 2) = f(0 + 1, 1 + 1) = f(0, f(0 + 1, 1)) = f(0, f (1, 1))

f (1, 1) = f(0, f(1, 0))

f(1, 0) = f(0, 1) = 2

∴ f(1, 1) = f(0, 2) = 2 + 1 = 3

f (1, 2) = f(0 + 1, 1 + 1) = f(0, f(0 + 1, 1)) = f(0, f (1, 1)) = f(0, 3)

f(0, 3) = 3 + 1 = 4

Hence, option (b).

Answer: Option C

Explanation :

To convert 1982 from base 10 to 12:

Divide 1982 by 12, write the quotient in the column below 1982 and the remainder in the next column. Repeat the process till the quotient is 0.

After that, write the remainders in the reverse order i.e. upward direction.

1982₁₀ = 1192₁₂

Hence, option (c).

Answer: Option D

Explanation :

Let the volume of the cylindrical tank be x.

∴ Volume of the conical tank = x − 500                    

When 200 litres of fuel is pumped out of each tank, the amount in the cylindrical tank is twice that in the cone.

∴ (x − 200) = 2(x − 500 − 200)

Solving this equation, we get x = 1200

∴ The cylindrical tank had 1200 litres of jet fuel when it was full.

Hence, option (d).

Answer: Option D

Explanation :

Let the number of posts to be bought be n.

The length of the side of the property, when posts are at 6 meters distance = 6(n − 1)                …(i)

The length of the side of the property, when posts are at 8 meters distance = 8(n − 5− 1)          …(ii)

From (i) and (ii),

∴ 6(n − 1) = 8(n − 5− 1)

Solving this, we get n = 21

∴ Length of property = 6(21 − 1) = 120

∴ Number of posts the farmer bought = 21 − 5 = 16

Hence, option (d).

Answer: Option C

Explanation :

From statement A, there are four possibilities.

X > Y > Z 

X > Z > Y 

Z > X > Y 

Y > X > Z

∴ This statement alone is not sufficient.

From statement B, there are four possibilities.

Y > X > Z 

Y > Z > X 

X > Y > Z

Z > Y > X

∴ This statement alone is not sufficient.

Using both statements together we get,

Either X > Y > Z or Y > X > Z. 

In both the cases Z is smallest.

Hence, option (c).

Answer: Option A

Explanation :

From statement A

X (X + 3) < 0

∴ –3 < X  < 0

∴ |X| < 3

∴ Statement A alone is sufficient.

From statement B

X(X – 3) > 0

X < 0 or X > 3

This statement does not give a unique answer.

Hence, option (a).

Answer: Option C

Explanation :

From Statement A

Q = 1000

P ∩ Q = 100

This doesn’t gives any information regarding how many people are watching TV programme P

From Statement B

P ∪ Q = 1500

This statement alone is not sufficient.

Let’s combine the information in both the statements.

∴ P ∪ Q = P + Q − P ∩ Q

∴ 1500 = P + 1000 – 100

∴ P = 600

Hence, option (c).

Alternatively

By combining both the statements together we can arrive at the above Venn diagram which gives the number of people watching programme P.

Hence, option (c).

Answer: Option C

Explanation :

From statement A

PF = PR – FR = PR – OD = PR – 5

QD = QR – DR = QR – OF = QR – 5

∵ PE = PF and QE = QD

∴ PE = PR – 5 and QE = QR – 5

∴ PQ = PE + QE = PR – 5 + QR – 5

∴ PR + QR = PQ + 10

This statement alone is not sufficient to find PR + RQ.

From Statement B

Diameter of the circumscribing circle = hypotenuse of the triangle

∴ PQ = 26

∵ PR² + RQ² = PQ²

But this does not give the value of PR + RQ.

∴ This statement alone is not sufficient.

Let’s combine both the statements.

∴ PR + QR = PQ + 10 = 26 + 10 = 36

Hence, option (c).

Answer: Option A

Explanation :

From Statement A

The cost of buying the shares for Harshad is

CP + 0.01 CP = 1.01 CP

The cost of selling the shares for Harshad is

SP – 0.01 SP = 0.99 SP

∴ Profit = Cost of selling – Cost of buying

             = 0.99 SP – 1.01 CP

∵ SP = 1.05 CP

Profit = (0.99 × 1.05 CP) – 1.01 CP

         = 0.0295 CP

∴ Profit earned per rupee spent on buying the shares is Rs 0.0295.

From the number of shares given in the statement B, we cannot conclude anything.

Hence, option (a).

Answer: Option D

Explanation :

Statement A implies that a, b, c, d, e, f are distinct real numbers.

But if a/d = b/e = c/f then the lines may be parallel and might not intersect at all.

∴ Statement A alone is not sufficient.

Statement B implies that the equations are not homogenous equations.

∴ Statement B is also not sufficient.

The condition for intersecting lines is a/d ≠ b/e

Even after combining both the statements the above condition is not clear.

∴ We cannot be sure whether the lines are intersecting or parallel.

Hence, option (d).

Answer: Option B

Explanation :

From statement A

bc = cd

If c is the midpoint of bd it would also be midpoint of ae because circles are concentric.

∴ ac = ce

∴ The question can be answered using statement A alone.

From Statement B

If c is the point on the line joining the two centres, it has to bisect the chord bd.

∴ c will also bisect the chord ae as the circles are concentric.

∴ ac = ce

∴ The question can be answered using statement B alone also.

Hence, option (b).

Answer: Option D

Explanation :

By combining both the statements we can get the duration of the flight. But for the arrival time we should have the information regarding the time zone difference of Mumbai and No man’s land.

Hence, option (d).

Answer: Option D

Explanation :

From statement A

X – Y = 6

From statement B

XY is divisible by 6.

∵ There are many possible values for example (12, 6), (18, 12), (24, 18) ...

∴ The question cannot be answered even by combining both the statements.

Hence, option (d).