XAT 2012 Qn 63 - Percents, Ratio

This is a very easy 1 mark question that appeared in XAT 2012. The question is from the topic percents.

Question 63 of XAT 2012

Tina, Mina, Gina, Lina and Bina are 5 sisters, aged in that order, with Tina being the eldest. Each of them had to carry a bucket of water from a well to their house. Their buckets’ capacities were proportional to their ages. While returning, equal amount of water got splashed out of their buckets. Who lost maximum amount of water as a percentage of the bucket capacity?
A.    Tina
B.    Mina
C.    Gina
D.    Lina
E.    Bina

Correct Answer - Bina. Choice E.

Explanatory Answer

Tina is the oldest and Bina is the youngest.
So, Bina’s bucket would have been the smallest.
   
Each sister lost equal amount of water.

As a proportion of the capacity of their buckets Bina would have lost the most.

Alternatively, you could solve this question by assuming some numbers for the capacity of the buckets that are in descending order from Tina to Bina and a value for the quantity that was lost and compute the percentages and verify.

You could access more XAT Practice questions on Percentage by clicking here..

XAT 2012 Q59 : Mean, Median, Mode

This is a 1 - Mark question that appeared in XAT 2012 from Statistics. This is an interesting question that tests your basic understanding of how Mean, Median and Mode can be interpreted.

Question

Ramesh analysed the monthly salary figures of five vice presidents of his company. All the salary figures are in integer lakhs. The mean and the median salary figures are Rs. 5 lakhs, and the only mode is Rs. 8 lakhs. Which of the options below is the sum (in Rs. lakhs) of the highest and the lowest salaries?
A.    9
B.    10
C.    11
D.    12
E.    None of the above

Correct Answer – Choice A. Rs. 9 lakhs

Explanatory Answer

The mean salary of the five vice presidents is Rs.5 lakhs.
So, the sum of their salaries = 5 * 5 = 25 lakhs.

Let their salaries in ascending order be a, b, c, d and e.
So, a + b + c + d + e = 25.
The median salary is Rs.5 lakhs. So, C’s salary is Rs.5 lakhs.

The only  mode is Rs.8 lakhs.
So,  Rs.8 lakhs salary is drawn by the maximum number of VPs.
C’s salary is Rs.5 lakhs. So, d and e have to draw Rs. 8 lakhs each.

Therefore, a + b + 5 + 8 + 8 = 21
or a + b = 4

Their salaries are in integer lakhs.
Therefore, a can draw Rs.1 lakh and b can draw Rs.3 lakhs or a and b can both draw Rs. 2 lakhs each.

However, there is only one mode. So, a and b cannot draw Rs.2 lakhs each.
So, a draws Rs.1 lakh (the least salary) and e draws Rs.8 lakhs (the highest salary).

So, a + e = Rs.9 lakhs

Click for additional Mean, Median and Mode questions

XAT 2012 Quant - Number Properties: LCM

Please watch this blog for periodic updates on new posts of previous year XAT questions with answers and explanation.

You could register you mobile number at http://labs.google.co.in/smschannels/channel/XATQuestions to receive questions by SMS. This service is available to any Indian mobile number that is not registered with National Do Not Call registry and is offered free of cost.

XAT 2012 Question 58  : LCM Word Problem

 Three Vice Presidents (VP) regularly visit the plant on different days. Due to labor unrest, VP (HR) regularly visits the plant after a gap of 2 days. VP (Operations) regularly visits the plant after a gap of 3 days. VP (sales) regularly visits the plant after a gap of 5 days. The VPs do not deviate from their individual schedules. CEO of the company meets the VPs when all the three VPs come to the plant together. CEO is on leave from January 5th to January  28th, 2012. Last time CEO met the VPs on January 3, 2012. When is the next time CEO will meet all the VPs?
A.    February 6, 2012
B.    February 7, 2012
C.    February 8, 2012
D.    February 9, 2012
E.    None of the above

Correct Answer February 8, 2012.

Explanatory Answer

VP (HR) visits the plant every 3rd day (the question states the VP visits after a gap of 2 days). 

The most important point in this question is this point. As the VP visits the plant after a gap of 2 days - he visits the plant every third day - not every second day.


If VP (HR) visited the plant on day 0, he will visit again on day 3, day 6 and so on.
i.e., on days that are multiples of 3.

VP (Operations) visits the plant every 4th day (as he visits after a gap of 3 days).
If VP (Operations) visited the plant on day 0, he will visit again on day 4, day 8 and so on.
i.e., on days that are multiples of 4.

VP (Sales) visits the plant every 6th day (as the VP Sales visits after a gap of 5 days).
If VP (Sales) visited the plant on day 0, he will visit again on day 6, day 12 and so on.
i.e., on days that are multiples of 6.

So, all 3 VPs will visit the plant together again on a day that is a multiple of 3, 4 and 6 or on a day that is a common multiple of 3, 4 and 6.

The first such common multiple is the LCM of 3, 4 and 6 = 12.

The last date on which all three VPs met the CEO was on Jan 3, 2012.
The next dates on which their visits will coincide will be after 12 days and multiples therefore. 
i.e., on Jan 15, 2012, Jan 27, 2012, Feb 8, 2012.

The CEO is on leave from Jan 5, 2012 to Jan 28, 2012. 
So, the CEO will meet the VPs on February 8, 2012.