sbi associate junior asst - 8 (ph-i) (solution) - amazon s3 · pdf file4 ===== centres at:...

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Centres at: MUKHERJEE NAGAR MUNIRKA UTTAM NAGAR DILSHAD GARDEN ROHINI BADARPUR JAIPUR GURGAON LAXMI NAGARGHAZIABAD NOIDA MEERUT VARANASI ROHTAK PANIPAT SONEPAT PATNA AGRA CHANDIGARH LUCKNOW ALLAHABAD

SBI ASSOCIATE JUNIOR ASSISTANT (PH-I) 2016 MOCK TEST PAPER - 8 (SOLUTION)

ENGLISH16. (1) Use 'sold' in place of 'sell'.17. (1) Use 'unlike' in place of 'unlikely'.18. (2) Remove 'very'.19. (1) Use 'was' in place of 'is'.20. (4) It should be 'left it immediately'.

MATHS31. (2) Using the BODMAS rule,

(585.65 – 36.25) ÷ 50 ×20 = ?(547.40) ÷ 50 × 20 = ?10.948 × 20 = ?? = 218.96

32. (4) 54.2 + 13.52 + 0.52 – 0.56 – 0.07 = ?The above expression can also bewritten as:(54.2 + 13.52 + 0.52) – (0.56 +0.07) = ?(68.24) – (0.63) = ?? = 67.61

33. (3) 89.61 + ? = 1811.2 – 127.84589.61 + ? = 1683.355? = 1683.355 – 89.61

= 1593.745

34. (2) 155 × 249696 = (? – 115) × 31

155 26

31

= ? – 115

130 + 115 = ? ? = 245

35. (4)2? 125 5776

100

= 225 + 3242?

100 × 125

= 225 + 324 + 2(76)2?

100 × 125 = 549 + 76

?2 = 625 100

125

?2 = 500? = 500

= 5 10 10 10 5

36. (1)9 2 27 9 ?

18 7.5 5 4

4.5 3 ?

2.4 1.25

? = 4.537. (3) (4 × 4)3 ÷ (512 ÷ 8)4 × (32 × 8)4 = (2 × 2)? + 4

46 ÷ 412 × 416 = (4)? + 4

(4)6 – 12 + 6 = (4)? + 4

? + 4 = 10 ? = 10 – 4 = 6

38. (1) 515.15 – 15.51– 1.51 – 5.11 – 1.11 = ? ? = 491.91

39. (5) ? = 3 2

896 56 8 125 (6 19)

? = 16 8 12

125 (36 19)

? = 128 12125 55

= 14070 = 2

40. (5) (18)2 – (5)3 – 121 – 19 = (?)2

324 – 125 – 11 – 19 = (x)2 199 – 30 = (x)2

169 = (x)2 (13)2 = (x)2 x = 13

41. (4) This is a decreasing series. Eachterm is half of its previous term.So, the required term is half of 420,i.e. 210.

42. (4) The terms of the series follow thepattern as:400 + 400 = 800

800 + 4005 = 880

880 + 40025 = 896

896 + 400125 = 899.2

899.2 + 400625 = 899.84

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43. (2) The pattern followed by the termsof the series :1664 × 2.5 = 41604160 ÷ 4 = 10401040 × 2.5 = 26002600 ÷ 4 = 650650 × 2.5 = 1625

44. (2) The pattern of the series is asfollows:12 × 1 = 1212 × 1.5 = 1818 × 2.5 = 4545 × 4 = 180180 × 6.5 = 11701170 × 10.5 = 12285

45. (2) The given series follows thepattern as:3 + 63 = 219219 + 53 = 344344 + 43 = 408408 + 33 = 435435 + 23 = 443443 + 13 = 444

46. (3) Quantity of water in 120 litres ofmixture

= 120 × 14

= 30 litres

Now, quntity of water in 20 litresmixture

= 20 × 14

= 5 litres

Remaining water in the remainingmixture= 30 – 5 = 25 litresNow, 3.8 litres of water is addedto the mixture. Then the quantityof water= 25 + 3.8 = 28.8 litresRequired percentage of water in the

new mixture = 28.8120 × 100 = 24%

47. (4) Ratio of profit of A to B to C= 18000 × 12 : 24000 × 8 + 24000× 2 : 4 × 15000 + 4 × 18000= 216000 : 192000 + 48000 : 60000+ 72000= 216000 : 240000 : 132000= 216 : 240 : 132= 18 : 20 : 11Share of profit of B

= 2049 × 12005 = ` 4900

48. (3) 20 men complete the work in16 days.Total work done by 20 men= 20 × 16 = 320 unitsNow, in 5 days work done by 20men= 20 × 5 = 100 unitsRemaining work = 320 – 100 =220 unitsNow, suppose x men left the workafter 5 days of beginning the work(20 – x) men finish the 220 units in

1 55183 3

days

So, (20 – x) × 553 = 220

or, 1100 – 55x = 660or, 55x = 440x = 8 men

49. (5) Let the present age of Bob be xyrs and that of Abby be y yrs.Then, x = y – 8 ... (i)Now, four years hence

4 54 4

xx

or, 4y + 16 = 5x + 20or, 4y + 16 = 5(y – 8) + 20or, 4y + 16 = 5y – 40 + 20

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or, y = 36 yrsHence Bob's present age= 36 – 8 = 28 years

50. (2) Total no. of balls = 3 + 5 + 7 = 15Now, n(S) = 15C1 = 15n(E) = 7C1 + 3C1 = 7 + 3 = 10Reqd probability

( ) 10 2( ) 15 3

n En S

51. (1) Initial amount of water = 18 litreAmount of milk = 80 litreAmount of mixutre = 18 + 80 = 98 litre49 litre or half of the mixture wassold. In the remaining 49 litre of

mixture, there is 182

9 litre of

water and 802

40 litre of milk.

Now, suppose 2x litres of milk andx litres of water are added.

40 2 49 1

xx

or, 40 + 2x = 36 + 4x2x = 4 litresor, x = 2 litres

52. (3) Let the total sum be ` x.

Then, A gets 2x

+ 40

B gets 38x

– 120

C gets ` 200

Now, 2x

+ 40 + 38x

– 120 + 200 = x

or4 320 3 960 1600

8x x x

or, 8x – 7x = 1920 – 960or, x = ` 960

53. (5) 28 men complete the work in 15days.1 man completes the work in 15× 28 days.30 men compelte the work in15 28

30

= 14 days

Again, 15 women can completethe work in 24 days.1 woman can complete the workin 24 × 15 days.18 women can complete the

work in 24 15

18

= 20 days

Reqd ratio = 1 1:

14 20 = 20 : 14 = 10

: 754. (4) Total expenditure on painting

= ` 10560Rate = ` 110 per square ft.

Area of wall = 10560110 = 96 square ft

Let the width of wall be x ft.Length = (x + 4) ftNow, according to the question,(x + 4)x = 96 = 12 × 8 (x + 4)x = (8 + 4) × 8 x = 8 footLength of wall= x + 4 = 8 + 4 = 12 ft

55. (2) Let the CP of article be ` x.Now, according to the question,85% of x = 6120

x × 85100 = 6120

x = 6120 10085 = ` 7200

New SP = ` (6120 + 1728) = `7848

Gain percent= 7848 7200

7200

× 100

= 64872 = 9%

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56. (5) Let the monthly salary of Pia andSom be 5x and 4x respectively.Then, money given by Pia to hermother

= 5x × 35 = 3x

Remaining amount = 5x – 3x = 2xMoney given by Pia as sister'stuition fees = 15% of 5x = 0.75xMoney given by Pia towards loan= 18% of 5x = 0.9xTotal money given= 3x + 0.75x + 0.90x = 4.65xRemaining amount= 5x – 4.65x = 0.35xNow, 0.35x = 2100

x = 21000.35 x =

21000035 = ` 6000

Monthly salary of Som= 4 × 6000= ` 24000

57. (3) Let the distance between B and Cbe x km and that between A and Bbe x + 4 km.Then, according to the question,Speed of the boat in still water = 24kmphSpeed of the stream = 4 kmphDownstream speed= 24 + 4 = 28 kmphUpstream speed= 24 – 4 = 20 kmph

Then, 4 36 3

20 28 60 5x x

or, 7 5 20 3

140 5x x

or, 2 20 3

140 5x

or, 10x – 100 = 420or, 10x = 420 + 100 = 520x = 52 kmDistance between A and B= (x + 4)= 52 + 4= 56 km

58. (4) Time taken in covering 7.5 km

upstream = 32 hours

Time taken in covering 7.5 km

downstream = 34

hours

Speed = Distance

TimeRate downstream of boat7.5 7.5 23 32

kmph = 5 kmph

Rate downstream of boat

= 7.5 4

3

= 10 kmph

59. (2) Let the amount got by B be ` x.

Expense on dinner = `310

x

Remaining amount

= x – 310

x = 10 310x x = `

710

x

Expense on book = ` 460

710

x× 2

7 = 460

5x

= 460

x = 5 × 460 = ` 2300Initial amount with A= ` (2300 × 4) = ` 9200

60. (3) Let the amount invested inscheme A be ` x.Amount invested in scheme B = ` 3xCI obtained from scheme A

= PR

1 1100

T

= ` x220

1 1100

= ` x21

1 15

= ` x26 1

5

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= ` x36 125

= `1125

x

SI from scheme B

= Principal × Time× Rate100

= 3 2 9

100x

= `54100

x

54 11 1200100 25

x x

54 44 1200

100x x

10 1200100

x

x = 1200 × 10= ` 12000

(61-65):61. (5) The given ratio of males to female

who qualify from State P in 2008is 11 : 7.Let the no. of male and femalequalified candidates be 11x and7x respectively.Then, according to the question,7x = 126x = 18Males = 18 × 11 = 198Total number of qualifiedcandidates= 198 + 126 = 324Let the total number of appearedcandidates be y.Then y × 60% = 324

y = 324 100

60

= 540

62. (3) Let the number of appeared candidatesfrom State Q in 2006 be x.Then, the number of appearedcandidates in 2007

= x + 100

100x

= 2x

According to the question,30

100x

+ 2 45

100x

= 408

or, 310

x+

910

x = 408

or, 12x = 408 × 10

x = 408 1012 = 340

Hence the number of candidatesfrom State Q in 2006 is 340.

63. (1) Difference = 450 60

100

– 600 43

100

= 270 – 258= 12

64. (4) Let the percentage of candidates whoqualify from state Q in 2010 be x%.Then,

Average = 280 60 550 50 400 %

100 1003

x

= 210or, 168 + 275 + 4x = 630or, 4x = 630 – 443 = 187

x = 1874

= 46.75%

The number of qualifiedcandidates from State Q in 2010 =

400 × 46.75100 = 187

65. (3) The ratio of qualified candidatesfrom State P in 2009 to that in2010 is 14 : 9.Now, in 2009 the number of qualified

candidates = 480 70

100

= 336

Let the number of qualifiedcandidates from State P in 2009be 14x and 2010 be 9x.

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Then, 14x = 336

x = 33614

= 24

In 2010 the no. of qualifiedcandidates= 9 × 24 = 216

REASONING(66-67):66. (2)67. (5)(68-69):

G and T are daughters of Y.N is son-in-law of Y.Y is father-in-law of N.

68. (5) N is father of A.Therefore, A is grandson or granddaughter of Y.

69. (1) If H is mother of T, then H is alsomother of G.G is the wife of N.Therefore, N is son-in-law of H.

(70-74):70. (4) Given statements:

N < A = T > Z ... (i)R > T ... (ii)Z < S ... (iii)Combining (i) and (ii), we getN < A = T < RThus, N < R or R > N is true.Hence conclusion I is true.Again, from (iii), Z < S or S > Z istrue. Hence conclusion II is true.

71. (5) Given statements:L < E = A > P ... (i)Y > E > R ... (ii)Combining both statements, we getL < E < YThus, L < Y or Y > L is true.Hence I (Y > L) is not true.Again, from (i) and (ii), we getA = E > RThus, A > R is true. Henceconclusion II is true.

72. (2) Given statements:D < S > L > U ... (i)Q < S ... (ii)Combining (i) and (ii), we getD < S > QThus, we can't compare D and Q.Hence I(Q < D) is not true.Again, from (i) and (ii), we getQ < S > L > UWe can't compare Q and U.Hence II (U > Q) is not true.

73. (5) Given statements:L < E = A > P ... (i)Y > E > R ... (ii)Combining (i) and (ii), we getR < E = A > PThus, we can't compare R and P.Hence, I (P > R) is not true.Again, from (i) and (ii), we get Y >E = AThus, Y > A or A < Y is true.Hence, conclusion II is true.

74. (3) Given statements:N < A = T > Z ... (i)R > T ... (ii)Z < S ... (iii)Combining (i) and (ii), we getR > T > ZThus, R > Z is true.It means either R > Z is true.or R = Z is true.Hence conclusion I and II make acomplementary pair.So, either conclusion I or II istrue.

(75-79):rural and urban divide na kuzu la ... (i)gap in rural infrastructre kt lavm pl ... (ii)urban planning more important ti na cu bu ... (iii)more divide than gap pl cu dmzu ... (iv)Now, from (i) and (ii),rural la ... (v)From (i) and (iii),

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urban na ... (vi)From (i) and (iv),divide zu ... (vii)From (i), (v), (vi) and (vii),and ku ... (viii)From (ii) and (iv),gap pl ... (ix)From (iii) and (iv),more cu ... (x)Now, from (ii), (v) and (ix)in/infrastructure kt/vm ... (xi)From (iii), (vi) and (x)planning/important ti/buFrom (iv), (vii), (ix) and (x)than dm

75. (1)76. (4)77. (3)78. (2)79. (3)80. (4) ONCE, CONE. Thus only two

words can be formed(81-85):

81. (2) Three persons - U, S, V - sitbetween W and T.Only X sits to the right of W. Wfaces O.W sits second to the right of S.

82. (3) K is facing V.83. (5) U is facing J.84. (4) Except J, all others are sitting the

extreme ends of the line.85. (1) T is third to the left of U.

(86-90):

86. (2) Cars M and X parked immediatelynext to Car Y.

87. (4) Car M is parked third to the leftor fifth to the right of Car W.Either two or four cars are parkedbetween Cars Y and Z. Car K isparked to the immediate left ofCar M. Three cars are parkedbetween Cars Y and L.

88. (3) Except KM, in all other pairsthere is one car between the twogiven cars. Car K is to theimmediate Left of Car M.

89. (1)

90. (5) Car W is parked fourth to the leftof Car K.

(91-95):91. (1)92. (1) 8th to the left of 17th from left

= (17 – 8 =) 9th from the left = M93. (5) In all others, the first and third

elements are consecutive ones.94. (3) 6th to the right 19th from the

right = (19 – 6 =)13th from theright = V

95. (4) M, F and T(96-100):96. (4) The govt provides security to

tourists across the country. Thatis why the govt has sent a policeteam to nab the culprits.

97. (3)98. (4) Extreme action (complete ban) is

not a solution to the problem.99. (4)100. (4)

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1. (4)2. (2)3. (1)4. (5)5. (2)6. (1)7. (2)8. (1)9. (1)10. (4)11. (1)12. (1)13. (2)14. (2)15. (4)16. (3)17. (5)18. (5)19. (2)20. (3)

21. (1)22. (3)23. (2)24. (5)25. (5)26. (4)27. (5)28. (1)29. (2)30. (4)31. (2)32. (4)33. (3)34. (2)35. (4)36. (1)37. (3)38. (1)39. (5)40. (5)

SBI ASSOCIATE JUNIOR ASSISTANT (PH-I) 2016 MOCK TEST PAPER - 8 (ANSWERKEY)

41. (4)42. (4)43. (2)44. (2)45. (2)46. (3)47. (4)48. (3)49. (5)50. (2)51. (1)52. (3)53. (5)54. (4)55. (2)56. (5)57. (3)58. (4)59. (2)60. (3)

61. (5)62. (3)63. (1)64. (4)65. (3)66. (2)67. (5)68. (5)69. (1)70. (4)71. (5)72. (2)73. (5)74. (3)75. (1)76. (4)77. (3)78. (2)79. (3)80. (4)

81. (2)82. (3)83. (5)84. (4)85. (1)86. (2)87. (4)88. (3)89. (1)90. (5)91. (1)92. (1)93. (5)94. (3)95. (4)96. (4)97. (3)98. (4)99. (4)100. (4)

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