Hello and welcome to exampundit. Here are the solutions of WBSEDCL 2017 Office Executive Practice Set.
Solutions
610:
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 C_{12} = 2.50 Lakhs
So, C_{12} = 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}
1115:
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 m^{2}
Floor Area of drawing room = 20 × 18 = 360 m^{2}
Floor Area of Dining room = 64 m^{2}
Floor Area of Kitchen room = 120 m^{2}
Floor Area of Bathroom = 48 m^{2}
So, Total = 952 m^{2}
Remaining area = 1000 – 952 = 48 m^{2}
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

Solutions:
16. Reqd Difference,
= (^{1911}/_{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
2125:
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.
3135:
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.
3135:
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
5155:
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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