# CAT Model Paper 3 Questions and Answers with Explanation Part 1

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Q: 1. Ganesh & Gopal start simultaneously at 7 am from Girish Park & Dumdum respectively towards each other. Every time they reach one end, then return towards the other. Without any stoppage, Ganesh could reach Dumdum at 9 am, while Gopal would take an hour longer to reach Girish Park. However, every time they meet anywhere on the road between Dumdum & Girish Park, they stop by chat for ‘n’ hours before resuming their journey, ‘n’ being the number of meetings (when they meet 1st time, they chat for 1 hour, when they meet time, they chat for 2 hours and so on); in case they meet either right at Girish Park or Dumdum, they chat for hours. By what time does Ganesh start travelling after their meeting?

(A) 3 PM

(B) 5 PM

(C) 10 PM

(D) None of these

Ans: C

Solution

Since Ganesh & Gopal takes 2 hrs and 3 hrs to travel the same distance, their speed . Hence on their 1st meeting, Ganesh would travel of the entire distance, hence time taken 2 minutes.

By the time they meet 1st time, they complete 1 round of the entire path. By the time they meet 2^{nd} time, they complete 3 rounds of the entire path, hence would take another minutes after they start after 1^{st} meeting. When they meet 3^{rd} time, they’ll make two more rounds of the entire path, hence; will take another 144 minutes after they start after 2nd meeting. Hence total time taken (assuming no stoppage) minutes. Hence it is evident that while 1st and 2nd meetings were in the middle of the path, but 3^{rd} meeting happens at the extreme end of the path (at Dumdum), hence total time for chatting during these 3 meetings hours. Hence total time hours (from 7 am).

Q: 2. In how many ways can 5 boys and 6 girls sit such that the number of persons of both genders on either side of the middle person is same?

(A)

(B)

(C)

(D) None of these

Ans: C

Solution:

The middle position must be taken by a boy, can be done in 5 ways.

There must be boys and girls to the liftoff the boy in the middle. These can be selected in ways.

The remaining boys and girls must be seated to the right of the boy in the middle.

The boys and girls on either side can be arranged among themselves in 5! ways.

Total.

Q: 3. A teacher gives a strange punishment to a student for his lack of attention during the class. He is supposed to write 100 Xs and 100 Ys in his notebook. He then performs the following operation on these letters. He erases a pair of letters and replaces them with one letter such that if the erased letters were identical, he replaces them with X while if they were different, he replaces them with Y. What is the maximum number of X’s that can appear in his copy at any point in time during the entire operation?

(A)

(B)

(C)

(D) None of these

Ans: C

Solution

Since we need to maximize the number of X’s, we will start by erasing the Ys. For every pair of Y erased, one X is added. So, for Ys erased, Xs will be added. Hence, Maximum number of Xs is

Q: 4. If p is a prime number greater than 3, how many numbers are there of the form such that they are also prime numbers?

(A) Only one

(B) Only two

(C) Infinitely many

(D) None

Ans: D

Solution

We note that the remainder that is left on dividing a number by 3 is either 1.Hence when the square of a number is divided by 3 the remainder is either. Since p is a prime greater than 3, on dividing by 3 can leave a remainder of 1 only since a prime number is a natural number which has exactly two distinct natural numbers as factors. Further on dividing by 3 must leave a remainder -1 since p is a prime greater than 3 and hence it must be odd and also 2 when divided by leaves a remainder .Hence is divisible by 3 and is of the form 3k.

Hence, all such numbers are composite and never prime.

**Start Passage**

__Directions for questions 5-7: Answer the questions based on the given data__*.*

In Bomanalli Branch of SitiBank, there are & 1 counters for handling Transactions,

Customer Complaints and Loan Applications respectively. The following table provides

Information on Turnaround time (time taken to resolve customer request) taken by the person allocated to the respective counter along with unproductive time.

C1 | C2 | C3 | C4 | C5 | Loans | Complaints | |

TAT (in mins) | 10 | 12 | 15 | 10 | 12 | 45 | 30 |

Unproductive Time | 20% | 10% | 30% | 10% | 20% | 20% | 20% |

Unproductive Time (UPT) is defined as time required for organizing cash (if applicable) and/or logging the details of a customer after customer leaves. The same is calculated as a percentage of TAT.

Idle time (IT) is when the counter is empty without any customer.

Bank opens at 9 am and each employee works for 8 hours a day, including 1 hour of recess time from 1 pm to 2 pm.

The following table provides the data of footfall during last 3 days of the April 2013:

Day 1 | Day 2 | Day3 | |

Transaction | 145 | 142 | 140 |

Loans | 5 | 7 | 6 |

Complaints | 8 | 9 | 10 |

Q: 5. Based on the information on unproductive time, if it is known that were utilized to the extent possible, then on which day did the bank observe maximum unproductive + idle time?

(A) Day 1

(B) Day 2

(C) Day 3

(D) Data insufficient

Ans: A

Solution

TAT (in mins) | Total time per customer [TAT+UPT] | Maximum no of customers (420 minutes/TAT+UPT) | |

C1 | 10 | 12 | 35 |

C2 | 12 | 13.2 | 31 |

C3 | 15 | 19.5 | 21 |

C4 | 10 | 11 | 38 |

C5 | 12 | 14.4 | 29 |

Loans | 45 | 54 | 7 |

Complaints | 30 | 36 | 11 |

Thus, as per the above table total no of customers handled by. Thus it is sufficient if we look at, Complaints and Loans:

D1 | D2 | D3 | ||

C5 | 20 | 17 | 15 | |

Loans | 5 | 7 | 6 | |

Complaints | 8 | 9 | 10 | |

C5 | UPT during customer handling | |||

IT=UPT during idle time (no customer) | ||||

Total UPT+IT | 180 | 216 | 240 | |

Loans | UPT during customer handling | |||

IT=UPT during idle time (no customer) | ||||

Total UPT +IT | ||||

Complaints | UPT during customer handling | |||

UPT during idle time (no customer) | ||||

Total UPT+IT | ||||

Total UPT |

UPT during customer handling = Number of customers

IT = UPT during idle time minutes – [Number of customers

Option (A)

Q: 6. Which counter had a capacity of handling maximum number of customers?

(A)

(B)

(C)

(D)

Ans: C

Solution:

TAT (in mins) | Total time per customer [TAT+UPT] | Maximum no of customers (420 minutes/TAT+UPT) | |

C1 | 10 | 12 | 35 |

C2 | 12 | 13.2 | 31 |

C3 | 15 | 19.5 | 21 |

C4 | 10 | 11 | 38 |

C5 | 12 | 14.4 | 29 |

Loans | 45 | 54 | 7 |

Complaints | 30 | 36 | 11 |

Clearly, from the table above, C4 had maximum capacity.

Option (C).

Q: 7. If no customer visits the bank after 3:30 pm on DAY 1, then what is the least possible time by which all customers are handled?

(A)

(B)

(C)

(D)

Ans: C

Solution

Since, C3 is least productive; we can start by utilizing that counter as less as possible. But in that case, after 234 minutes, it is idle. Hence, we can shift 2 customers from each of the other counters to that one to best utilize the time. The calculation will be as follows:

C1 | C2 | C3 | C4 | C5 | |

Number of customer | 35 | 31 | 12 | 38 | 29 |

Total time (TAT+UPT) | 420 | 409.2 | 234 | 418 | 417.6 |

Shifting | -2 | -2 | 8 | -2 | -2 |

Number of customers | 33 | 29 | 20 | 36 | 27 |

Total time(TAT+UPT) | 396 | 382.8 | 390 | 396 | 388.8 |

Thus maximum time minutes. Considering break of 1 hour, earliest time by which all customer can be handled pm.

**End Passage**