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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'.
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
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.
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
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
(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,
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),
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.