SBI PO Exam 2015- Data Interpretation Questions (With Solutions) Set-10

SBI PO Exam 2015- Data Interpretation Questions (With Solutions)
SBI PO Exam 2015- Data Interpretation Questions (With Solutions) Set-10:
List of Practice Data Interpretation Questions for Upcoming SBI PO Exam was given here with solutions, candidates those who are preparing for those exams can use this material.

1).In what time will Rs. 8500 amount to Rs. 1576.50 at 9% per annum simple interest?
a)   9 years
b)   9×1/2 years
c)   7 years
d)   8 years
e)   7×1/2 years

2).Arun’s expenditure and savings are in the ration 3:2. His income increases by 10%. His expenditure also increases by 12%. By what per cent do his savings increase?
a)   18
b)   12
c)   7
d)   8
e)   10

3).A train covers a certain distance in 40 minutes if the train runs at a speed of 60 kmph. Find the speed of the train at which it must run to reduce the time of journey to 30 minutes.
a)   50 kmph
b)   80 kmph
c)   60 kmph
d)   75 kmph
e)   None of these

4).The perimeter of the sector of a circle of radius 6.2 cm is 17.4 cm. Find the area of the sector.
a)   16 cm2
b)   15.5 cm2
c)   15.6 cm2
d)   15.6 cm2
e)   17.5 cm2

 
5).How many word can be formed with the letters of the word ORDINATE so that vowels occupy odd places?
a)   120
b)   586
c)   256
d)   576
e)   676

Directions (Q.6-10): Read the following table carefully to answer these questions:
Average marks obtained by 20 boys and 20 girls in five subjects of five different schools.
 
Subject
Maxi-
marks
P
Q
R
S
T
Boys
Girls
Boys
Girls
Boys
Girls
Boys
Girls
Boys
Girls
English
200
85
90
80
75
100
110
65
60
105
110
History
100
40
50
45
50
50
55
40
45
65
60
Geography
100
50
40
40
45
60
55
50
55
60
65
Maths
200
120
110
95
85
135
130
75
80
130
135
Science
200
105
125
110
120
125
115
85
90
140
135
 
6).What was the total marks obtained by boys in History from School Q?
a)   900
b)   980
c)   1300
d)   800
e)   1000

7).In which of the following subjects did the girls has the highest average percentage of marks from all the schools?
a)   Science
b)   Maths
c)   History
d)   English
e)   Geography

8).The pooled average marks of both boys and girls in all the subjects were minimum from which of the following schools?
a)   R
b)   T
c)   P
d)   S
e)   Q

9).In which of the following schools the total marks obtained by the girls in Mathematics was exactly 100% more than the total marks obtained by the boys in History?
a)   P
b)   S
c)   R
d)   T
e)   Q

10).What was the difference between the total marks obtained in Mathematics by the boys from School R and that obtained in the same subject by the girls from School S?
a)   1000
b)   1100
c)   1111
d)   1400
e)   1200

Answers:
1). b)   2). c)   3). b)   4). b)   5). d)   6). a)   7). a)   8). d)   9). b)   10). b)
Check here below for the detailed solutions of the above problems:
 
1).SI = 15767.50 – 8500 = 7267.5
Time = (7267.5×100)/(8500×9) = 9.5 years
= 9×1/2 years
Answer: b
 
2). Alligation Method:

 

 We get two values of x:7 and 13. But to get a viable answer we must keep in mind that the central value (10) must lie between x and 12. Thus the value of x should be 7 and not 13.
Alternate Method:
Let Arun’s income = 3x+2x=5x
Increase income = 5x×(110/100)=(11/2)x
Increased expenditure = (3x×112)/100= Rs. 84x/25
New savings=(11x/2)-(84x/25) = (107x/50)
Increase in savings = (107x/50)-2x=7x/50
Percentage increase in savings
=(7x/50×2x)×100 = 7%
Answer: c
 
3). Let the distance be D km.
D/60=40 minutes = 40/60 = 2/3
Or D=(2/3)×60 = 40 Km
Let the required speed be x kmph.
Now x×1/2=40(30 minutes=1/2 hours)
X=80 kmph
Answer: b
 
4). Let AOB be the sector. Then the perimeter of the sector AOB = 17.4 cm

 

Or, OA+OB+ arc AB = 17.4
Or, 6.2+6.2+ arc AB=17.4
Or, are AB = 17.4 – 12.4 = 5 cm
Area of the sector = 1/2×lentgh of arc × radius=1/2×5×6.2=15.5 cm2
Answer: b
 
5). There are 4 vowels are 4 consonants in the word ORDINATE.
We have to arrange the letters in a row such that vowels occupy odd places. There are 4 odd places 1, 3, 5 and 7. Four vowels can be arranged in these 4 odd places in ways = 4! Ways
Remaining 4 even places viz 2,4,6,8 are to be occupied by 4 consonants. This can be done in 4! Ways.
Total no. of words in which vowels can occupy odd places = 4! × 4! = 576
Answer: d
 
6). Required marks obtained by 20 boys in History = 45×20=900
Answer: a
 
7).
Average % of marks obtained by girls is as follows in all schools and subjects

Subject
School
P
Q
R
S
T
English
45
37.5
55
30
55
History
50
50
55
45
60
Geography
40
45
55
55
65
Maths
55
42.5
65
40
67.5
Science
62.5
60
57.5
45
67.5
Hence is Science the average % marks obtained from all the schools is maximum.
Answer: a
 
8). Pooled average marks of boys and girls.

Subject
School
P
Q
R
S
T
English
87.5
77.5
105
62.5
107.5
History
45
47.5
52.5
42.5
62.5
Geography
45
42.5
57.5
52.5
62.5
Maths
115
90
132.5
77.5
132.5
Science
115
115
120
87.5
137.5
Hence in school S pooled average marks are minimum.
Answer: d
 
9).Total marks obtained by girls in Maths in School P = 110×20=2200
Similarly, in School Q = 85 × 20 = 1700
In School R = 130 × 20 = 2600
In School S = 80 × 20 = 1600
In School T = 135 × 20 = 2700
Total marks obtained by boys in History in School P = 40 × 20 = 800
Similarly, in School Q = 45 × 20 = 900
In School R = 50 × 20 = 1000
In School S = 40 × 20 = 800
In School T = 65 × 20 = 800
In School T = 65 × 20 = 1300
Hence in School S the marks obtained by girls are twice the marks obtained by boys in History.
Answer: b
 
10).Required difference = 135 × 20 – 80 × 20 = 2700 – 1600 = 1100
Answer: b
 
 
 
 

 

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