DILR in CAT 2016 will now be a stand-alone section with 32 questions in MCA and Non MCQ format. Your high scores in other 2 sections and low scores in Data Interpretation & Logical Reasoning (DILR) section will drag you back and you cannot expect to be shortlisted by any of the IIMs if your DILR score is low.
Offering the opportunity to attempt the real type of CAT questions and assess yourself where you stand on CAT 2016 preparation in DILR section, Prof Nishit Sinha CAT expert, IIM Lucknow alumnus and author of titles on QA, LRDI and VARC, widely published by Pearson shares small Mock test to practice DILR questions well.
MBAUniverse.com suggests the CAT 2016 aspirants to complete the test in 12 minutes without taking any break. Only after completion of the test, you should cross check the answers and your scores. The answers are also followed by detailed solutions to understand the approach to solve the questions.
Each correct answer carries 3 marks and each wrong answer carries the penalty of 1 mark negative marking. So carefully attempt this small CAT 2016 DILR Mock. Before you arrive at your final score you should calculate and deduct the negative marks out of your credit marks.
CAT 2016: DILR MOCK TEST
Direction for questions 1 to 4: Read the information given below and solve the questions based on it.
Three children live at houses with different colored doors. Following are the details of the children:
Children Name – Govind, Jai, Sama Door colours – Blue, Green, Red
House Numbers – 1, 2, 3 Roads – Bridge Street, Cherry Road, Sandy Lane
Sama’s house no. is 1 lower than Jai’s. Govind does not live at House No.1, but does not have a green door. One child lives at No.2 at Cherry Road. House No.1 has a red door, but is not on Sandy lane.
Q1. Who lives at Bridge street?
(a) Govind (b) Jai
(c) Sama (d) More than one answer is possible
Q2. Which of the following is correctly matched door color and house number?
(a) Blue – House no. 1 (b) Blue – House no. 2
(c ) Red - House no. 1 (d) Red - House no. 2
Q3. Which of the following is the correct combination for Govind?
(a) Govind – Green – 2 – Cherry Road (b) Govind –Green– 3– Sandy Lane
(c) Govind – Blue – 2 – Cherry Road (d) Govind –Blue– 3 – Sandy Lane
Q4. For how many children, is it possible to exactly determine the house number, road and
color of road?
(a) 0 (b) 1 (c) 2 (d)3
Direction for questions 5-8: Go through the information given below and solve the questions based on it.
You are the Quality Control manager at a restaurant. Go through the case facts given below and solve the questions based on it:
Restaurant produces chicken nugget box (1,000 boxes at a time in a batch). Restaurant has to ensure that it does not supply undercooked nuggets (known as defective) to the customers. If a customer receives a defective nugget box, company has to pay a penalty of Rs. 150 per nugget box. To check the level of preparedness of nuggets, any of the two tests can be used:
Test 1 |
Test 2 |
||
Cost |
Rs. 6 per nugget box |
Cost |
Rs. 9 per nugget box |
Detection Rate |
80% |
Detection Rate |
100% |
Restaurant can re-cook the undercooked nuggets at a cost of Rs. 75 per nugget box. This box can be now sold to customer as it is now certainly properly cooked. All the boxes which are detected as defective must be re-cooked and after that, these boxes are properly cooked without any defect.
Let ‘x’ be the number of defective nugget boxes per batch.
Q5. Manager should not use any quality control test, if it is expected that
(a) x < 100 (b) 100 < x < 200 (c) x > 200 (d) None of these
Q6. If 200 < x < 400, then
(a) You should use Test 1 (b) You should use Test 2
(c) You can use Test 1 or Test 2 (d) None of these
Q7. If x = 1000, you should use
(a) Test 2 (b) Test 1 and Test 2 (c) Test 1 or No test (d) None
Q8. Company should be indifferent between Test 1 and No Test if it finds that the number of defective boxes is equal to
(a) 50 (b) 100 (c) 200 (d) Not possible
Answers
Q1 |
Q2 |
Q3 |
Q4 |
Q5 |
Q6 |
Q7 |
Q8 |
c |
B |
b |
d |
A |
b |
a |
b |
Explanation
Solutions and Explanations:
Solution to Questions 1 – 4:
Following is the arrangement:
Child |
Color of Door |
House Number |
Road |
Govind |
Green |
3 |
Sandy lane |
Jai |
Blue |
2 |
Cherry Road |
Sama |
Red |
1 |
Bridge Street |
Q1. Sama lives as Bridge Street. Hence option (c) is the answer.
Q2. House number 2 has blue colored door. Hence option (b) is the answer.
Q3. It can be seen from the table that option (b) is the answer.
Q4. As obvious from the above table, it can be determined for all three of them. Hence option (d) is the answer.
Solution to Questions 5 – 8:
Total cost = Cost of Testing + Re-cook charge + Penalty
It can be seen that if the number of defective nugget boxes per batch is very low, for example, 10, then it is better to pay the penalty on this, than to use any test.
Q5. Total cost = Cost of Testing + Re-cook charge + Penalty
Using Test 1, total cost of testing = Rs. 6 × 1000 = Rs. 6000
Using options, at x = 100, Cost of Test 1 = Penalty paid for x = 100.
Using Test 1, Detection = 80 % of 100 = 80. These 80 will be reworked upon and for remaining 20, company will pay penalty.
Total cost = Cost of Testing + Re-cook charge + Penalty = Rs.6000 + 80 × Rs. 75 + 20 × Rs. 150 = Rs. 15,000.
For x = 100, penalty = Rs. 150 × 100 = Rs. 15000. Hence option (a) is the answer.
Q6. Check for x = 300.
Total cost using Test 1:
Cost of testing = Rs. 6 × 1000 = Rs. 6000
Out of total defective 300 boxes, 80% will be detected. So, total detected boxes = 80% of 300 = 240 boxes. These 240 boxes will be reworked upon and remaining 60 boxes will go undetected. As per the rule, restaurant will have to pay penalty on these 60 boxes.
Re- cook charges = Rs. 75 × 240 = Rs. 18000
Penalty = Rs. 150 × 60 = Rs. 9000
So, total cost using Test 1 = Cost of Testing + Re-cook charge + Penalty = Rs.6000 + Rs. 18000 + Rs. 9000 = Rs. 33,000.
Total cost using Test 2:
Cost of testing = Rs. 9 × 1000 = Rs. 9000
Out of total defective 300 boxes, 100% will be detected. So, total detected boxes = 100% of 300 = 300 boxes. These 300 boxes will be reworked upon.
Re- cook charges = Rs. 75 × 300 = Rs. 22500
So, total cost using Test 2 = Cost of Testing + Re-cook charge + Penalty = Rs.9000 + Rs. 22500 + Rs. 0 = Rs. 31,500.
It can be seen that Total cost of Test 1 < Total cost of Test 2. Hence test 2 should be used. Hence option (b) is the answer.
Q7. Option (a) is the answer.
Q8. Using solution to Q(5), Penalty for x = 100 is equal to Total cost of Test 1 for x = 100.
Hence answer is – No test or Test 1 if x = 100.
Hence option (b) is the answer.
Mock is sourced from goo.gl/XRGgOW
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