Capgemini Sample Dimensions Based Questions

Question 1
A manufacturer reduces the size of his machine of Material1 from 30" to 28" and that of Material2 from 24" to 23". He normally spends Rs.1440 for Material1 and Material2. 3/4th of the amount spent would be for Material1. How much will he save under the size of new size (Assuming that the costs incurred by the machines are directly proportional to their sizes)?
a)Rs.85 b)Rs.87 c)Rs.86 d)Rs.88
Answer : b)Rs.87
Solution:
The manufacturer totally spends Rs.1440 for both material 1 and 2.
Out of this 3/4th is entirely for Material1.
Amount spent on Material1 = 1440 x 3/4 = Rs.1080
i.e, he spent Rs.1080 for Material1 of 30" and he reduces 2".
 Size  Amount
 30"  Rs.1080
 2"    ?
Amount saved on Material1 = 1080 x 2/30 = Rs.72
Amount spent on Material2 = total amount - amount for Material1 = 1440 - 1080 = Rs.360.
i.e., he spent Rs.360 for Material2 of 24" and he reduces 1".
 Size  Amount
 24"  Rs.360
 1"    ?
Amount saved on Material2 = 360 x 1/24 = Rs.15
Therefore total amount saved after size reduction = 72 + 15 = Rs.87
Question 2
The dimensions of a certain machine are X" x Y" x Z". If the average of its dimensions equals 680" and none of the dimensions is less than 640", what is the greatest possible length of one of the dimensions?
a)760" b)800" c)720" d)745"
Answer : a) 760"
Solution:
Given (X+Y+Z)/3 = 680".
Therefore the total length of the dimensions is X + Y + Z = 3 x 680 = 2040.
Let us assume Z is the dimension with greatest possible length.
From above equation, Z = 2040 - X - Y ...(1)
Based on eq (1), for Z to be of maximum value, X and Y should be as low as possible. Since it is given that none of the dimensions is lesser than 640", the lowermost value that X and Y can get is 640".
Substituting X = Y = 640" in eq (1), we get,
Z = 2040 - 640 -640 = 760".
Question 3
What will be the ratio of the shortest side of new dimensions to the greatest side of the old dimensions if the size of a certain machine is increased proportionally until the sum of its dimensions equals 800" and the dimensions of old size are 130" x 60" x 360" ?.
a)1:2 b)3:5 c)5:18 d)18:3
Answer : c) 5:18
Solution:
From the given data, we have
The dimensions of the machine before size increment = 130" x 70" x 360".
Sum of the dimensions = 130+70+360 = 560"
Note that the greatest side = 360" & smallest side = 70". ...(A)
It is given that the sum of the dimensions of new size = 800"
Since dimensions are increased proportionately, the smallest side of the old dimensions before increment will be the smallest side of new dimension after size increment.
 Sum  Side
 560"   70"
 800"    ?
Smallest side will be increased to 800 x 70 / 560 = 100"
i.e., the smallest side of the new dimensions is 100". 
We have already found that the the greatest side of old dimensions is 360".
Required Ratio = Smallest side of the New Dimensions : Greatest side of the Old 
Dimensions
= 100 : 360 = 5 : 18
Hence the answer is 5 : 18.

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10th 3i InfoTech Academic Books Accenture ACIO ActiveX ADT Agricultural AIEEE Air Force Algebraic Amdocs Android Answers Application Development Aptitude Aptitude Questions Architectures ASP ATOS B.Sc B.Tech.B.E. Bank Exam BCA BE Board Exam Books break-continue Business Plan C C Programming C# C++ Campus campus interview Candidate profile Capgemini Career CDS Certification CET Challenge Circle Cisco class code Cognizant communication Company Company Profile Competitive Exams computer Computer Networks concentric circles constructor Course Credit Suisse CSS CTS Data Structure DBMS DC Deloitte difficult interview questions dimensions Distributed Computing do while dotNet Download ds Dynamic Web Development e-Admit card Educational engineering entertainment Even Odd Events exam schedule exception for loop fresher GATE general Discussion general knowledge Get Placed Government Job Hall Ticket HCL how to answer How to Prepare HR HR Interview HSC hypertext preprocessor IB IBM IBPS IIT Indian Army Information infosys Intelligence Bureau Internship interview Experience interview questions Interview Tips IntroC IntroC# IntroJava IntroPHP IT J2EE J2ME Java JavaScript jobs Language Books Language Tutorial Languages limit number of objectsJava Limit the number of objects being created in JAVA Linux Linux Administrator Linux Developer Logical Questions loops M Tech M.E. M.Tech M.Tech AND B.Tech Management Management Skills Matrices MBA mca ME microsoft mistakes Mixture Mobile Computing mock questions mock test MySql naukri NDA OOP opening Operating System Oracle paper Persistent PHP php programming php string variables PL/SQL Placement placement guide Placement Paper Placement Process preparing for placement presentation probability Problems Professional program Programming Project Engineer project idea Projects Puzzle qualities Question of the day Questions Quiz Question Recruitment Recruitment Pattern Requirement Result Resume Reviews Screen Sizes Scripting Session Skills Software Software Engineering solved papers Source Code Speed time and distance SQL SSC story Stress Interview Study Material study tips submit resume Synonym TCS Tech Mahindra tech news Technical Books Technical Interview Testing thank you letter Thought Time Table TutC++ Unix questions asked in aptitude and inteviews for MCA UPSC verbal Web Designing Web Developer Website Development What to read while loop Wipro Writing Resume