Line chart Questions and Answers

In 2001 = 20 + 40 + 70 = 130

In 2002 = 50 + 30 + 80 = 160

In 2003 = 30 + 130 + 50 = 210

In 2004 = 60 + 80 + 50 = 190

In 2005 = 110 + 40 + 70 = 220

In 2006 = 40 + 70 + 110 = 220

In 2007 = 130 + 110 + 150 = 390

Therefore, in the years 2005 and 2006 the export from all the three companies was same.

In 2001: 40 – 20 = 20
In 2002: 80 – 50 = 30
In 2003: 130 – 30 = 100
In 2004: 80 – 60 = 20
In 2005: 110 – 40 = 70
In 2006: 70 – 40 = 30
In 2007: 130 – 110 = 20
On the basis of the option, the difference between the exports was minimum in 2007.

Average export of company A, B, and C is 2003 = \frac{30 + 130 + 50}{3} = 70

Average export of company A, B, and C is 2006 = \frac{40 + 70 + 110}{3} = \frac{220}{3}

Therefore, the difference = \frac{220}{3} - 70 = \frac{220-210}{3} = \frac{10}{3}

In 2008 company Q manufactured 13500 smartphones
In 2009 company Q manufactured 16900 smartphones
Therefore, the difference = 169000 – 135000 = 34000 smartphones.

Given, the production of smartphones by company P from 2005 to 2009 was: 121000 + 101000 + 145000 + 88000 + 180000 = 635000

The production of smartphones by company Q from 2005 to 2009 was: 141000 + 120000 + 102000 + 135000 + 169000 = 667000

Therefore, the difference = 667000 – 635000 = 32000

Average number of smartphones produced by company P

= \frac{121000 + 101000 + 145000 + 88000 + 180000}{6}

= \frac{635000}{6} = 105833.3

In 2008 the smartphones produced by company P = 88000

In 2008 the smartphones produced by company Q = 135000

Therefore, percentage = \frac{135000}{88000} \times 100 = 153%

The difference between 1989 -1990 = 60 – 35 = 25

1990-1991 = 70 – 60 = 10

1991-1992 = 70 – 60 = 10

1992-1993 = 90 – 60 = 30

1993-1994 = 90 – 90 = 0

1994 – 1995 = 90 – 70 = 20

Therefore, the maximum difference was form 1992-1993

Let us assume the number of candidate applied for the exam in 1989 as x

Given, the number of students qualified = 5600

35% of x = 5600

\frac{35}{100} \times x = 5600

35x = 560000

x = \frac{560000}{35} = 16000

Therefore, students applied in 1989 = 16000.

The total students qualified in 1992 and 1993 cannot be concluded till we don’t know that how many students appeared in any one of the year or the percentage of qualified candidate together in 1992 and 1993.

Therefore, the data provided is inadequate.

The students qualifies in 1990 = 60% of 20000

\frac{60}{100} \times 20000 = 12000

Therefore, students qualified in 1991 = 29900 – 12000 = 17900

Then, let us assume the students appeared in 1991 as x

Then, 70% of x = 17900

\frac{70}{100}\times x = 17900

70x = 1790000

x = \frac{1790000}{70} = 25571(approx.)