CAT 2018 Question Paper Slot 1

It is given that a satellite serving either B, or C, or S does not serve O. So we can say that it's basically 3 satellites broadcasting (B), communication (C), surveillance (S) which can have intersections. Those satellites which are not part of any category are placed in others. We can draw the Venn diagram as follows. 

1. The numbers of satellites serving B, C, and S (though may be not exclusively) are in the ratio 2:1:1. 2. The number of satellites serving all three of B, C, and S is 100. 3. The number of satellites exclusively serving C is the same as the number of satellites exclusively serving S. This number is 30% of the number of satellites exclusively serving B. 4. The number of satellites serving O is the same as the number of satellites serving both C and S but not B.

Let '10x' be the number of satellites exclusively serving B. Then, the number of satellites exclusively serving C and S = 0.30*10x = 3x

Let 'y' be the number of satellites serving others(O).

Let 'z' be the number of satellites serving B, C but not S. Since the numbers of satellites serving B, C, and S (though may be not exclusively) are in the ratio 2:1:1. Therefore, we can can say that number of satellites serving B, S but not C = z.

It is given that 

$$\Rightarrow$$ 10x+2z+2y+6x+100 = 1600

$$\Rightarrow$$ 8x+z+y = 750 ... (1)

The numbers of satellites serving B, C, and S (though may be not exclusively) are in the ratio 2:1:1.

$$\Rightarrow$$ $$\dfrac{10x+2z+100}{z+100+3x+y} = \dfrac{2}{1}$$

$$\Rightarrow$$ $$10x+2z+100=2(z+100+3x+y)$$

$$\Rightarrow$$ $$4x=100+2y$$

$$\Rightarrow$$ $$2x=50+y$$

$$\Rightarrow$$ $$y=2x-50$$   ... (2)

We can substitute this in equation (1)

$$\Rightarrow$$ 8x+z+2x - 50 = 750

$$\Rightarrow$$ z = 800 - 10x    ... (3)

Let us define boundary condition for x, 

$$\Rightarrow$$ 2x - 50 $$\geq$$ 0

$$\Rightarrow$$ x $$\geq$$ 25

Also,  800 - 10x $$\geq$$ 0

$$\Rightarrow$$ x $$\leq$$ 80

Therefore, we can say that x $$\epsilon$$ [25, 80].

It is given that at least 100 of the 1600 satellites were serving O.

$$\Rightarrow$$ 2x - 50 $$\geq$$ 100

$$\Rightarrow$$ x $$\geq$$ 75

The number of satellites serving S = 100 + 800 - 10x + 2x - 50 + 3x = 850 - 5x

At x$$_{min}$$ = 75, the number of satellites serving S = 850 - 5*75 = 475

At x$$_{max}$$ = 80, the number of satellites serving S = 850 - 5*80 = 450

Hence, we can say that the number of satellites serving S must be from 425 to 475. Therefore, we can say that option A is the correct answer.

It is given that a satellite serving either B, or C, or S does not serve O. So we can say that it's basically 3 satellites broadcasting (B), communication (C), surveillance (S) which can have intersections. Those satellites which are not part of any category are placed in others. We can draw the Venn diagram as follows. 

1. The numbers of satellites serving B, C, and S (though may be not exclusively) are in the ratio 2:1:1. 2. The number of satellites serving all three of B, C, and S is 100. 3. The number of satellites exclusively serving C is the same as the number of satellites exclusively serving S. This number is 30% of the number of satellites exclusively serving B. 4. The number of satellites serving O is the same as the number of satellites serving both C and S but not B.

Let '10x' be the number of satellites exclusively serving B. Then, the number of satellites exclusively serving C and S = 0.30*10x = 3x

Let 'y' be the number of satellites serving others(O).

Let 'z' be the number of satellites serving B, C but not S. Since the numbers of satellites serving B, C, and S (though may be not exclusively) are in the ratio 2:1:1. Therefore, we can can say that number of satellites serving B, S but not C = z.

It is given that 

$$\Rightarrow$$ 10x+2z+2y+6x+100 = 1600

$$\Rightarrow$$ 8x+z+y = 750 ... (1)

The numbers of satellites serving B, C, and S (though may be not exclusively) are in the ratio 2:1:1.

$$\Rightarrow$$ $$\dfrac{10x+2z+100}{z+100+3x+y} = \dfrac{2}{1}$$

$$\Rightarrow$$ $$10x+2z+100=2(z+100+3x+y)$$

$$\Rightarrow$$ $$4x=100+2y$$

$$\Rightarrow$$ $$2x=50+y$$

$$\Rightarrow$$ $$y=2x-50$$   ... (2)

We can substitute this in equation (1)

$$\Rightarrow$$ 8x+z+2x - 50 = 750

$$\Rightarrow$$ z = 800 - 10x    ... (3)

Let us define boundary condition for x, 

$$\Rightarrow$$ 2x - 50 $$\geq$$ 0

$$\Rightarrow$$ x $$\geq$$ 25

Also,  800 - 10x $$\geq$$ 0

$$\Rightarrow$$ x $$\leq$$ 80

Therefore, we can say that x $$\epsilon$$ [25, 80].

It is given that the number of satellites serving at least two among B, C, and S is 1200.

$$\Rightarrow$$ 800 - 10x + 800 - 10x + 2x -50 + 100 = 1200

$$\Rightarrow$$ 18x = 450

$$\Rightarrow$$ x = 25

We can determine number of satellites in each of the following category. Hence, option C is definitely false. Therefore, we can say that option C is incorrect.  

It is given that Indu and Jatin both scored 100% in exactly one section. We can say that Jatin scored 100% marks in DI. Therefore, Jatin's composite score = 2*20+16+14 = 70

It is given that Jatin’s composite score was 10 more than Indu’s. Therefore, we can say that Indu's composite score = 70 - 10 = 60.

Indu also scored 100% in exactly one section. 

Case 1: Indu scored 100% marks in DI.

If Indu scored 100% marks in DI, then Indu's score in GA = 60 - 2*20 - 8 = 12 which is less than 70% of maximum possible marks. Indu already has less than 70% in WE, therefore we Indu can't be recruited . Hence, we can reject this case. 

Consequently, we can say that Indu scored 100% marks in WE. Therefore, Indu's score in DI = $$\dfrac{60-8-20}{2}$$ = 16

It is also given that Danish, Harini, and Indu had scored the same marks the in GA. 

We are given that, among the four recruited, Geeta had the lowest composite score. 

Maximum composite score that Geeta can get = 2*14 + 6 + 20  = 54 {Assuming 100% marks in WE}. Since, Geeta was recruitedat a composite score of 54 or less we can say that Ester was definitely recruited. 

It is given that no two candidates had the same composite score. We can see that Chetna's composite score is 54. Hence, Geeta can't have a composite score of 54. Therefore, we can say that Geeta's composite score is 53 or less. 

We already know the four people(Jatin, Indu, Geeta, Ester) which were recruited. Hence, we cab say that Danish was rejected at a composite score of 51. Hence, we can say that Geeta's composite score in 52 or more. 

Consequently, we can say that Geeta's composite score if either 52 or 53. Therefore we can say that Geeta scored either 18 {52-(2*14+6)} or 19 {53-(2*14+6)} marks in WE. 

Ajay was the unique highest scorer in WE. 

Case 1: Geeta scored 19 marks in WE.

We can say that if Geeta scored 19 marks in WE, then Ajay scored 20 marks in DI. In that case Ajay's composite score = 2*8 + 20 + 16 = 52. Which is a possible case. 

Case 1: Geeta scored 18 marks in WE.

We can say that if Geeta scored 18 marks in WE, then Ajay can score either 19 or 20 marks in DI. 

If Ajay scored 20 marks in DI then in that case Ajay's composite score = 2*8 + 20 + 16 = 52 which will be same as Geeta's composite score. Hence, we can say that in this case Ajay can't score 20 marks.

If Ajay scored 19 marks in DI then in that case Ajay's composite score = 2*8 + 19 + 16 = 51 which will be same as Danish's composite score. Hence, we can say that in this case Ajay can't score 19 marks.

Therefore, we can say that case 2 is not possible at all.

Let us check all the statement one by one. 

Statement 1: Jatin's composite score was more than that of Danish. We can see that this statement is correct. 
Statement 2:  Indu scored less than Chetna in DI. We can see that Indu scored 16 marks in DI whereas Chetna scored 19 marks in DI. Hence, we can say that this statement is also correct. 
Statement 3: Jatin scored more than Indu in GA. We can see that Jatin scored 14 marks in GA whereas Indu scored 20 marks in GA. Hence, we can say that this statement is incorrect. 

Hence, we can say that option D is the correct answer. 

It is given that Indu and Jatin both scored 100% in exactly one section. We can say that Jatin scored 100% marks in DI. Therefore, Jatin's composite score = 2*20+16+14 = 70

It is given that Jatin’s composite score was 10 more than Indu’s. Therefore, we can say that Indu's composite score = 70 - 10 = 60.

Indu also scored 100% in exactly one section. 

Case 1: Indu scored 100% marks in DI.

If Indu scored 100% marks in DI, then Indu's score in GA = 60 - 2*20 - 8 = 12 which is less than 70% of maximum possible marks. Indu already has less than 70% in WE, therefore we Indu can't be recruited . Hence, we can reject this case. 

Consequently, we can say that Indu scored 100% marks in WE. Therefore, Indu's score in DI = $$\dfrac{60-8-20}{2}$$ = 16

It is also given that Danish, Harini, and Indu had scored the same marks the in GA. 

We are given that, among the four recruited, Geeta had the lowest composite score. 

Maximum composite score that Geeta can get = 2*14 + 6 + 20  = 54 {Assuming 100% marks in WE}. Since, Geeta was recruitedat a composite score of 54 or less we can say that Ester was definitely recruited. 

It is given that no two candidates had the same composite score. We can see that Chetna's composite score is 54. Hence, Geeta can't have a composite score of 54. Therefore, we can say that Geeta's composite score is 53 or less. 

We already know the four people(Jatin, Indu, Geeta, Ester) which were recruited. Hence, we cab say that Danish was rejected at a composite score of 51. Hence, we can say that Geeta's composite score in 52 or more. 

Consequently, we can say that Geeta's composite score if either 52 or 53. Therefore we can say that Geeta scored either 18 {52-(2*14+6)} or 19 {53-(2*14+6)} marks in WE. 

Ajay was the unique highest scorer in WE. 

Case 1: Geeta scored 19 marks in WE.

We can say that if Geeta scored 19 marks in WE, then Ajay scored 20 marks in DI. In that case Ajay's composite score = 2*8 + 20 + 16 = 52. Which is a possible case. 

Case 1: Geeta scored 18 marks in WE.

We can say that if Geeta scored 18 marks in WE, then Ajay can score either 19 or 20 marks in DI. 

If Ajay scored 20 marks in DI then in that case Ajay's composite score = 2*8 + 20 + 16 = 52 which will be same as Geeta's composite score. Hence, we can say that in this case Ajay can't score 20 marks.

If Ajay scored 19 marks in DI then in that case Ajay's composite score = 2*8 + 19 + 16 = 51 which will be same as Danish's composite score. Hence, we can say that in this case Ajay can't score 19 marks.

Therefore, we can say that case 2 is not possible at all.

Let us check all the statement one by one.

Option A: Bala scored same as Jatin in DI. We can say that Bala scored 20 marks in DI. In that Bala's composite score = 2*20 + 9 + 11 = 60 which is sam as Indu's composite score. Therefore, we can say that this is a false statement. Hence, option A is the correct answer. 

It is given that Indu and Jatin both scored 100% in exactly one section. We can say that Jatin scored 100% marks in DI. Therefore, Jatin's composite score = 2*20+16+14 = 70

It is given that Jatin’s composite score was 10 more than Indu’s. Therefore, we can say that Indu's composite score = 70 - 10 = 60.

Indu also scored 100% in exactly one section. 

Case 1: Indu scored 100% marks in DI.

If Indu scored 100% marks in DI, then Indu's score in GA = 60 - 2*20 - 8 = 12 which is less than 70% of maximum possible marks. Indu already has less than 70% in WE, therefore we Indu can't be recruited . Hence, we can reject this case. 

Consequently, we can say that Indu scored 100% marks in WE. Therefore, Indu's score in DI = $$\dfrac{60-8-20}{2}$$ = 16

It is also given that Danish, Harini, and Indu had scored the same marks the in GA. 

We are given that, among the four recruited, Geeta had the lowest composite score. 

Maximum composite score that Geeta can get = 2*14 + 6 + 20  = 54 {Assuming 100% marks in WE}. Since, Geeta was recruitedat a composite score of 54 or less we can say that Ester was definitely recruited. 

It is given that no two candidates had the same composite score. We can see that Chetna's composite score is 54. Hence, Geeta can't have a composite score of 54. Therefore, we can say that Geeta's composite score is 53 or less. 

We already know the four people(Jatin, Indu, Geeta, Ester) which were recruited. Hence, we cab say that Danish was rejected at a composite score of 51. Hence, we can say that Geeta's composite score in 52 or more. 

Consequently, we can say that Geeta's composite score if either 52 or 53. Therefore we can say that Geeta scored either 18 {52-(2*14+6)} or 19 {53-(2*14+6)} marks in WE. 

Ajay was the unique highest scorer in WE. 

Case 1: Geeta scored 19 marks in WE.

We can say that if Geeta scored 19 marks in WE, then Ajay scored 20 marks in DI. In that case Ajay's composite score = 2*8 + 20 + 16 = 52. Which is a possible case. 

Case 1: Geeta scored 18 marks in WE.

We can say that if Geeta scored 18 marks in WE, then Ajay can score either 19 or 20 marks in DI. 

If Ajay scored 20 marks in DI then in that case Ajay's composite score = 2*8 + 20 + 16 = 52 which will be same as Geeta's composite score. Hence, we can say that in this case Ajay can't score 20 marks.

If Ajay scored 19 marks in DI then in that case Ajay's composite score = 2*8 + 19 + 16 = 51 which will be same as Danish's composite score. Hence, we can say that in this case Ajay can't score 19 marks.

Therefore, we can say that case 2 is not possible at all.

It is given that all the candidates except Ajay and Danish had different marks in DI and Bala's composite score was less than Chetna's composite score. 

Let us assume that Bala scored 'x' marks in DI.

$$\Rightarrow$$ 2*x + 9 + 11 < 54

$$\Rightarrow$$ x < 17

We can see that Bala's score will be less than 17. Bala's maximum score in DI will be the largest possibel number less than 17 which is not same as any other candidate's score in DI. From, the table we can see that 16, 15 and 14 are already taken by Indu, Falak and Geeta respectively. 

Therefore, we can say that Bala can score a maximum of 13 marks in DI.

It is given that Indu and Jatin both scored 100% in exactly one section. We can say that Jatin scored 100% marks in DI. Therefore, Jatin's composite score = 2*20+16+14 = 70

It is given that Jatin’s composite score was 10 more than Indu’s. Therefore, we can say that Indu's composite score = 70 - 10 = 60.

Indu also scored 100% in exactly one section. 

Case 1: Indu scored 100% marks in DI.

If Indu scored 100% marks in DI, then Indu's score in GA = 60 - 2*20 - 8 = 12 which is less than 70% of maximum possible marks. Indu already has less than 70% in WE, therefore we Indu can't be recruited . Hence, we can reject this case. 

Consequently, we can say that Indu scored 100% marks in WE. Therefore, Indu's score in DI = $$\dfrac{60-8-20}{2}$$ = 16

It is also given that Danish, Harini, and Indu had scored the same marks the in GA. 

We are given that, among the four recruited, Geeta had the lowest composite score. 

Maximum composite score that Geeta can get = 2*14 + 6 + 20  = 54 {Assuming 100% marks in WE}. Since, Geeta was recruitedat a composite score of 54 or less we can say that Ester was definitely recruited. 

It is given that no two candidates had the same composite score. We can see that Chetna's composite score is 54. Hence, Geeta can't have a composite score of 54. Therefore, we can say that Geeta's composite score is 53 or less. 

We already know the four people(Jatin, Indu, Geeta, Ester) which were recruited. Hence, we cab say that Danish was rejected at a composite score of 51. Hence, we can say that Geeta's composite score in 52 or more. 

Consequently, we can say that Geeta's composite score if either 52 or 53. Therefore we can say that Geeta scored either 18 {52-(2*14+6)} or 19 {53-(2*14+6)} marks in WE. 

Ajay was the unique highest scorer in WE. 

Case 1: Geeta scored 19 marks in WE.

We can say that if Geeta scored 19 marks in WE, then Ajay scored 20 marks in DI. In that case Ajay's composite score = 2*8 + 20 + 16 = 52. Which is a possible case. 

Case 1: Geeta scored 18 marks in WE.

We can say that if Geeta scored 18 marks in WE, then Ajay can score either 19 or 20 marks in DI. 

If Ajay scored 20 marks in DI then in that case Ajay's composite score = 2*8 + 20 + 16 = 52 which will be same as Geeta's composite score. Hence, we can say that in this case Ajay can't score 20 marks.

If Ajay scored 19 marks in DI then in that case Ajay's composite score = 2*8 + 19 + 16 = 51 which will be same as Danish's composite score. Hence, we can say that in this case Ajay can't score 19 marks.

Therefore, we can say that case 2 is not possible at all.

It is given that all the candidates scored different marks in WE.

We can see that Ajay, Geeta and Ester has already scored 20, 19 and 18 marks in WE. Therefore, Harini can score a maximum of 17 marks in WE. If Harini's score in WE is 17, then Harini's composite score = 2*5+17+20 = 47 which is same as Falak's composite score. Hence, we can say that Harini can't score 17 marks in WE. Jatin and Danish have already scored 16 and 15 marks respectively. 

Therefore, we can say that the maximum marks that Harini could have scored in WE = 14. 

Givne that: $$2^{x}=3^{\log_{5}{2}}$$

$$\Rightarrow$$ $$2^{x}=2^{\log_{5}{3}}$$

$$\Rightarrow$$ $$x=\log_{5}{3}$$

$$\Rightarrow$$ $$x=\log_{5}{\dfrac{3*5}{5}}$$

$$\Rightarrow$$ $$x=\log_{5}{5}+\log_{5}{\dfrac{3}{5}}$$

$$\Rightarrow$$ $$x=1+\log_{5}{\dfrac{3}{5}}$$. Hence, option D is the correct answer. 

Given that two parallel chords on the same side of a diameter have lengths 4 cm and 6 cm.

In the diagram we can see that AB = 6 cm, CD = 4 cm and MN = 1 cm.

We can see that M and N are the mid points of AB and CD respectively. 

AM = 3 cm and CD = 2 cm. Let 'OM' be x cm. 

In right angle triangle AMO,

$$AO^2 = AM^2 + OM^2$$ 

$$\Rightarrow$$ $$AO^2 = 3^2 + x^2$$  ... (1)

In right angle triangle CNO,

$$CO^2 = CN^2 + ON^2$$ 

$$\Rightarrow$$ $$CO^2 = 2^2 + (OM+MN)^2$$ 

$$\Rightarrow$$ $$CO^2 = 2^2 + (x+1)^2$$  ... (2)

We know that both AO and CO are the radius of the circle. Hence $$AO = CO$$

Therefore, we can equate equation (1) and (2)

$$3^2+x^2$$=$$2^2+(x+1)^2$$

$$\Rightarrow$$ x = 2 cm

Therefore, the radius of the circle 

$$AO = \sqrt{AM^2 + OM^2}$$ 

$$\Rightarrow$$ $$AO=\sqrt{3^2+2^2}=\sqrt{13}$$. Hence, option A is the correct answer. 

Let the efficiency of humans be 'h' and the efficiency of robots be 'r'.
In the first case, 
Total work = (15h + 5r) * 30......(i)
In the second case,
Total work = (5h + 15r) * 60......(ii)
On equating (i) and (ii), we get
(15h + 5r) * 30 = (5h + 15r) * 60
Or, 15h + 5r = 10h + 30r
Or, 5h = 25r
Or, h = 5r
Total work = (15h + 5r) * 30 = (15h + h) * 30 = 480h
Time taken by 15 humans = $$\dfrac{\text{480h}}{\text{15h}}$$ days= 32 days.
Hence, option C is the correct answer.

Let ABCD be a rectangle inscribed in a circle of radius 13 cm. Which one of the following pairs can represent, in cm, the possible length and breadth of ABCD?

We know that AC is the diameter and $$\angle$$ ABC = 90°. AC = 2*13 = 26 cm

In right angle triangle ABC,

$$AC^2 = AB^2 + BC^2$$

$$\Rightarrow$$ $$AB^2+BC^2=26^2$$

$$\Rightarrow$$ $$AB^2+BC^2=676$$

Let us check with the options.

Option (A): $$24^2+10^2 = 676$$. Hence, this is a possible answer. 

Option (B): $$25^2+9^2 = 706 \neq 676$$.  Hence, this is an incorrect pair.

Option (C): $$25^2+10^2 = 725 \neq 676$$.  Hence, this is an incorrect pair.

Option (D): $$24^2+12^2 = 720 \neq 676$$. Hence, this is an incorrect pair.

Therefore, we can say that option A is the correct answer.