## Crack IBPS PO : Missing Data Interpretation : Day 48

Directions : 1-5) Refer to the following table and answer the questions.

Marks obtained by three students A, B and C in an entrance examination is provided in the table Given below:

 Sections Paper I marks MM (25) Paper II marks MM (40) Paper III marks MM (100) Maximum Marks MM (130) A B C A B C A B C A B C Reasoning 22 17 10 36 36 30 – 95 80 116.8 – – Verbal 19 18 21 33 – 36 81 84 75 – 104.55 – Quant 24 – 19 31 27 35 97 93 85 – 113.75 – Gk 20 16 13 23 34 38 73 89 92 – – – Computer – 14 18 38 37 40 65 – 79 95.4 105.35 –

Normalised marks is given by: Normalised marks (130) = paper I (10) + paper II (30) + paper III(90)

Suppose a student gets 10 out of 25 in paper I, 40 out of 40 in paper II and 60 out of 100 in paper III , then normalised marks = 4 + 30 + 54 = 88 (out of 130).

Total Marks obtained in the examination= Total of the Normalised Marks of all the sections

Percentage marks = ((Total Marks obtained in the examination)/650)×100

 Sections Normalized marks A                                                            B                                                  C Reasoning 8.8+27+81 = 116.8 6.8+27+85.5 = 119.3 4+22.5+72 = 98.5 Verbal 7.6+24.75+72.9 = 105.25 7.2+21.75+75.6 = 104.55 8.4+ 27 +67.5 = 102.9 Quant 9.6+23.25+87.3= 120.15 9.2+20.25+83.7 = 113.15 7.6+26.25+76.5 = 110.35 Gk 8+17.25+65.7 = 90.95 6.4+25.5+80.1 = 112 5.2+ 28.5 +82.8 = 116.5 Computer 8.4+28.5+58.5 = 95.4 5.6+27.75+72 = 105.35 7.2+30+71.1 = 108.3 Total 528.55 554.35 536.55

Q.1) What is the total percentage scored by student C in the entrance examination?

a) 82.5%

b) 75.5%

c) 85.5%

d) 78%

e) None of these.

Required percentage =  536.55/650×100=82.5%

Q.2) Due to illness a student A was unable to give the Paper I of sections Verbal and Gk then what is the total percentage scored by him in the examination?

a) 72.37%

b) 76.57%

c) 78.92%

d) 78.57%

e) None of these

Percentage =  (New marks)/(Total Marks) ×100=  512.95/650 ×100= 78.92%

Q.3) What is the difference between the average marks per section obtained by student A and B in paper I?

a) 3.2

b) 4

c) 3.8

d) 3.6

e) 4.6

Average of  total marks obtained by student A in paper I

= (22+19+24+20+21) /5 = 106/5  = 21.2

Average of total marks obtained by student B in paper I

= (17+18+23+16+14) /5 = 88/5  = 17.6

Required difference = 21.2 – 17.6 = 3.6

Q.4) What is the ratio of total marks in paper II scored by student A, B and C respectively?

a) 161 : 164 : 179

b) 161 : 163 : 178

c) 131 : 151 : 171

d) 131 : 161 : 179

e) 161 : 163 : 179

Total prelims marks scored by student A

= 36 + 33+31+23+38 = 161

Total prelims marks scored by Student B

= 36+29+27+34+37 = 163

Total prelims marks scored by student C

= 30+36+35+38+40 = 179

Ratio = 161 : 163 : 179

Q.5) What is the total normalized marks obtained by student A?

a) 511.25

b) 522.55

c) 516.55

d) 517.55

e) None of these

Required total = 116.8+105.25 +120.15+90.95+95.4 = 528.55

Directions : 6-10) Find the wrong number in the series.

Q.6)  90          132           132           181               182                  240

a) 181

b) 182

c) 90

d) 240

e) 130

102-10=90

112+11=132

122-12=132

132+13=182

142-14=182

152+15=240

Hence 181 is the wrong number

Q.7)  9           18            91                61                341

a) 341

b) 61

c) 91

d) 18

e) None of these

23+13=9

33-23=19

43+33=91

53-43=61

63+53=341

Hence 18 is the wrong number

Q.8) 2            6            18            40              145        756

a) 756

b) 145

c) 2

d) 6

e) None of these

1!+12=2

2!+22=6

3!+32=15

4!+42=40

5!+52=145

6!+62=75

Hence 18 is the wrong number.

Q.9)  4           6              13.5                46.56              230.66

a) 6

b) 46.56

c) 230.66

d) 4

e) None of these

4×1.5 = 6

6×2.25= 13.5

13.5×3.375 = 45.56

45.56×5.0625 = 230.65

Hence 46.56 is the wrong number

Q.10) 642            81           28           19          18

a) 642

b) 18

c) 81

d) 28

e) 19