Quant Quiz for IBPS PO and RBI Grade B

Study the table and answer the given questions:
Data related to the number of employees in six companies during 6 years
Note: M= Male, F=Female





1).What is the difference between the average number of male employees in all the given companies in 2006 and the average number of female employees in all the given companies in 2009?
a)    68
b)    74
c)    52
d)    58
e)    63

2).The total number of female employees in Company L in 2004, 2005, 2006 and 2007 together is what per cent more than the total number of female employees in Company T in the same years together? (Rounded off to the nearest integer)
a)    16
b)    14
c)    12
d)    18
e)    20
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3).In which of the given years is the percentage increases in the number of male employees of Company T from previous year the highest?
a)    2009
b)    2006
c)    2005
d)    2008
e)    2007
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4).The total number of female employees in Company J during all the given years together is what per cent of the total number of male employees in Company I during all the given years together?
a)    95
b)    93
c)    87
d)    91
e)    89


5).What is the ratio of the total number of employees (male and female) in companies J, K and N together in 2005 to the total number of employees (male and female) in the same companies together in 2007?
a)    9:15
b)    11:18
c)    9:14
d)    13:16
e)    13:18

Answers
1).a  The average no. of male employees in 2006
= (13+164+152+178+212+241)/6
= 1086/6 = 181
The average no. of female employees in 2009
= (242+238+267+214+230+303)/6
= 1494/6 = 249
Difference = 249-181 = 68
Answer: a

2).  b Total no. of female employees in Company L in 2004, 2005, 2006 and 2007
= 145+183+191+234 = 753
Total no. of female employees in Company T in 2004, 2005, 2006 and 2007
= 121+178+162+198 = 659
Required % = [(753-659)/659]×100 = (94/659)×100
= 14.26 = 14%

3).  a In Company T the percentage increase in 2005
=[(180-148)/148]×100 = (32/148)×100
= 21.62%
Similarly, the percentage increase in 2006
=[(212-180)/180]×100 = (32/180)×100 = 17.77%
In 2007 = [(263-212)/212]×100 = (51/212)×100
= 24.05%
In 2008= [(310-263)/263] ×100 = (4700/263)
= 17.83%
In 2009 = [(396-310)/310] × 100 = (8600/310)
= 27.74%
Hence in 2009 the percentage increase is the highest.

4). b Total no. of female employees in Company J = 111+136+157+211+356+238 = 1209
Total no. of male employees in Company I
= 142+165+139+217+265+372 = 1300
Required % = (1209/1300)×100 = 93%

5). e Total no. of employees in 2005 in Company J, K and N together
= 161+136+194+134+212+177 = 1014
Total no. of employees in 2007 in Company J, K and N together
= 179+211+254+168+333+259 = 1404
Required ratio = 1014:1404 = 13:18
Study the table and answer the given questions:
Data related to the number of employees in six companies during 6 years
Note: M= Male, F=Female





1).What is the difference between the average number of male employees in all the given companies in 2006 and the average number of female employees in all the given companies in 2009?
a)    68
b)    74
c)    52
d)    58
e)    63

2).The total number of female employees in Company L in 2004, 2005, 2006 and 2007 together is what per cent more than the total number of female employees in Company T in the same years together? (Rounded off to the nearest integer)
a)    16
b)    14
c)    12
d)    18
e)    20
Show/Hide Answer

3).In which of the given years is the percentage increases in the number of male employees of Company T from previous year the highest?
a)    2009
b)    2006
c)    2005
d)    2008
e)    2007
Show/Hide Answer

4).The total number of female employees in Company J during all the given years together is what per cent of the total number of male employees in Company I during all the given years together?
a)    95
b)    93
c)    87
d)    91
e)    89


5).What is the ratio of the total number of employees (male and female) in companies J, K and N together in 2005 to the total number of employees (male and female) in the same companies together in 2007?
a)    9:15
b)    11:18
c)    9:14
d)    13:16
e)    13:18

Answers
1).a  The average no. of male employees in 2006
= (13+164+152+178+212+241)/6
= 1086/6 = 181
The average no. of female employees in 2009
= (242+238+267+214+230+303)/6
= 1494/6 = 249
Difference = 249-181 = 68
Answer: a

2).  b Total no. of female employees in Company L in 2004, 2005, 2006 and 2007
= 145+183+191+234 = 753
Total no. of female employees in Company T in 2004, 2005, 2006 and 2007
= 121+178+162+198 = 659
Required % = [(753-659)/659]×100 = (94/659)×100
= 14.26 = 14%

3).  a In Company T the percentage increase in 2005
=[(180-148)/148]×100 = (32/148)×100
= 21.62%
Similarly, the percentage increase in 2006
=[(212-180)/180]×100 = (32/180)×100 = 17.77%
In 2007 = [(263-212)/212]×100 = (51/212)×100
= 24.05%
In 2008= [(310-263)/263] ×100 = (4700/263)
= 17.83%
In 2009 = [(396-310)/310] × 100 = (8600/310)
= 27.74%
Hence in 2009 the percentage increase is the highest.

4). b Total no. of female employees in Company J = 111+136+157+211+356+238 = 1209
Total no. of male employees in Company I
= 142+165+139+217+265+372 = 1300
Required % = (1209/1300)×100 = 93%

5). e Total no. of employees in 2005 in Company J, K and N together
= 161+136+194+134+212+177 = 1014
Total no. of employees in 2007 in Company J, K and N together
= 179+211+254+168+333+259 = 1404
Required ratio = 1014:1404 = 13:18

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