All questions of Radar Graphs for Bank Exams Exam

We need to find the decrease in sales for both the items given and that too during the same year. This only happens one time in the year 2001 where sales of TV decreases from 20 million to 10 million and sales of laptop decreases from 25 million to 22.5 million. So, the correct answer is A.

In this question, we need to find the increase in the percentage of the given items in comparison to previous years. Instead of checking all the years, we will only check percentage increase for the one given in the option. For laptop 2007, in the previous year, the sales were 30 million while in 2006 it was 25.5 so there is an increase of 4.5 million units. Therefore percentage increase will be 4.5/25.5 x 100 = 18%.
Similarly for TV 2007 it will be, 2.5/22.5 x 100 = 11.11 %. For Laptop 2006 it will be, 7.5/15 x 100 = 50 % and for TV 2009, it will be 5/25 x 100 = 20 %. So for laptop 2006, the percentage increase in sales is the highest. Thus, the correct answer is C.

Number of students in university 1 in 2010 = 20000
Number of students in university 2 in 2012 = 20000
Required difference = 20000 – 20000 = 0.

Difference between the students of university 1 and university 2 in 2007 = 2000 – 1000 = 1000
Difference in number of students in 2008 = 25000 – 15000 = 10000
Difference in number of students in 2009 = 35000 – 25000 = 10000
Difference in number of students in 2010 = 20000 – 15000 = 5000
Difference in number of students in 2011 = 30000 – 25000 = 5000
Difference in number of students in 2012 = 35000 – 20000 = 15000

It is clear from above that the difference between the number of students in university 1 and the number of students in university 2 is highest in the year 2012.

Let selling price of A and that C be Rs 40x & Rs 27x respectively.
So, marked price of C = 27x *100/90 = Rs 30x
And, cost price of A = 40x * 100/80 * 100/125 =  Rs 40x
Required% = 30x/40x *100 = 75%

Number of students in university 1 in 2007 = 10000 
Number of students in university 2 in 2011 = 30000
∴  Required sum = 10000 + 30000 = 40000.

We need to first calculate the total sales of both the items in each of the years given above. The total sales will be:

Now, from the table, you can see the decrease in sales occurs only thrice during 2005, 2008, and 2009. So we will calculate percentage decrease for only these three years. For 2005 decrease in sales are 7.5 million. So, the percentage decrease from the previous year will be 7.5/45 x 100 = 16.6 %. Sales decreased by 10 million from 2007 to 2008. Thus, the percentage decrease will be 10/55 x 100 = 18.18 %. For 2009 the decrease is 5 million and so the percentage decrease will be, 5/45 x 100 = 11.11 %. So, the highest percentage decrease was noted in 2008 from 2007. So, the correct answer is C.

Here, you only have to add the total number of televisions laptop sold in all these years and compare them to find the ratio. The total number of televisions sold from 2004 to 2009 are 25 + 22.5 + 22.5 + 25 + 25 + 30 = 150. And the total number of laptops sold are: 20 + 15 + 25 .5 + 30 + 20 + 10 = 120.5. Thus the ratio of the is 150: 120.5 = 5: 4. So, the correct answer is C.

Total students in university 2 in 2010 = 15000
∴  Number of girls = 15000 × 25/100 = 3750
and number of boys = 15000 – 3750 = 11250.

Here the sales of televisions and laptops of six different years are given. You need to see that in which year the sale of TV was greater than the sale of the laptop. The values of different sales of units are given to you in the radar diagram above.
From the figure you can see that sales of the laptops in 2004 were lower than the sales TVs in 2004, similarly, sales of a laptop in 2005, 2008, and 2009 was lower than the sale of TVs in that year. So, there are four years where the sale of the laptop was lower than the sale of TV. Thus, the correct answer is D.