WBSEDCL Office Executive 2017 – Full Practice Paper – Solutions

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7 years ago

Hello and welcome to exampundit. Here are the solutions of WBSEDCL 2017 Office Executive Practice Set.

Solutions

6-10:

Solutions:

6. Let the population of state ‘F’ in 2013 = 100

Then population of state ‘F’ in 2014 = 150 (Since growth is 50%)

So, Required % = 100/150 × 100 » 67%

7. Population of state ‘B’ in the year 2014 =

5 × 155/100 135/100

» 10.50 lakh

8. C12/D12 = 2/3 and C12 = 2.50 Lakhs

So, C12 = 2.5 x 140/100 = 3.5 Lakhs

So we get, the population of D in 2013 = 3 x 3.50/2 = 5.25 Lakhs

9. Population of state ‘B’ in 2012

= 4 × 100 × 100/155 × 135= Population of D in 2012

Population of state ‘D’ in 2014 = 1×100×100×160×170/100×100×155×135

= 5 lakhs

10. Suppose population of state E in 2012 = 100

Then population of state E in 2013 = 125

*since 25% growth

And population of state E in 2014 = 125 (120/100) = 150

*since 20% growth

So, required ratio = 100/150 = 2/3

11-15:

Solutions :

Area of 4 walls of a room = 2 (l + b) × h

Area of floor = l × b

Floor Area of bedroom = 15 × 12 × 2 = 360 m2

Floor Area of drawing room = 20 × 18 = 360 m2

Floor Area of Dining room = 64 m2

Floor Area of Kitchen room = 120 m2

Floor Area of Bathroom = 48 m2

So, Total = 952 m2

Remaining area = 1000 – 952 = 48 m2

Cost of carpeting = 48 × 16 = Rs. 768

Room	Area	Cost
2 Bedrooms	2x(2(15+12)×15)=810×2=1620	81000
Drawing	2(20+18)×15=1140	57000
Dining	2(8+8)×15=480	24000
Kitchen	2(10+12)×15=660	23100
Bathroom	2(8+6)×15=420	14700

16-20

Solutions:

16. Reqd Difference,

= (19-11/100) x 120000 = 9600

17. Reqd percentage,

= 10200/120000 x 100 = 8.5%

18. Estimated cost of furniture and miscellaneous expenditures

(13+8/100) x 120000 = 25200

Actual cost of Furniture,

88/100 x 13/100 x 120000 = 13728

Actual cost of furniture and miscellaneous expenditure

= 13728 + 10200 = 23928

Total expenditure of the family

= 120000 – 25200 + 23928 = 118728

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19. Reqd Estimated cost,

(15+14/100) x 120000 = 34800

20. Amount spent on furniture,

88/100 x 13/100 x 120000 = 13728

21-25:

S is the grandmother of T and is a housewife. So X who is a lawyer and grandfather of R must be married to S. Thus R and T must be brother or sister and be the two students. Q who is an engineer and father of T will be father of T and R and must be married to P who shall be the only teacher in the family. Thus the questions can be answered as follows.

31-35:

31. Solutions – The third letter of second term is the next letter according to alphabet to the third letter of first term.

32. Solutions – There is a gap of two letters between the two consecutive letters of each term.

33. Solutions – First, Second, and third each term is one more than the square of prime number. Hence the fourth term = (19)2 + 1

34. Solutions – Second term = First term + 1/8 First term.
Fourth term = Third term + 1/8 Third term

35:

51-55:
51. Solutions – No error
52. Solutions – Replace ‘founding’ with ‘find’.
53. Solutions – Replace ‘in’ with ‘of’.
54. Solutions – Replace ‘turn up ‘ with ‘turned up’.
55. Solutions – Replace ‘to’ with ‘on’.

56. 2; + 4 × 2, + 8 × 3, + 12 × 4

57. 1; The series is based on combination of two series. S1 = +13, +26, +39… and S2 = +7, +14, +21…

58. 4; +36, + 72, + 144, + 288…

59. 3; (22)², (27)², (32)², (37)²..

60. 3; The series is + 72 , + 142 , + 212 , + … 7. 5; The series is 40 × 1.

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