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**amrutha735** · (This post was last modified: 10-12-2017, 10:39 AM by amrutha735.)

BANK PO - Model Question Paper

92.The subject of the sentence is The development' which is ' in the singular. Hence, the verb corresponding to this is has been’ and not ’have been'.
Choice (3)
93. The sentence is grammatically correct
   Choice (5)
94. The subject is 'the officer' which is in the singular and hence the verb has to be was' and not 'were'.
   Choice (3)
95. In this sentence, part (4) is erroneous. We say things are placed in some category or things come under some category. Hence ‘under' is the correct word and not 'inside'.
Choice (4)
Solutions for questions 96 to 100: 96.
96.Rights can be ‘asserted’, ‘demanded' or 'exercised' but not ‘utilized’ or ‘thrown'. In this context, options (4) and (5) can be ruled out. The words for the second blank in choices (1) and (3), 'perseverance' and 'justice', render them inapt. Women silently put up with ‘atrocities’.
Choice (2)
97.Only ‘battered’ collocates with 'physically' in the sentence. The word for the second blank too fits in, thus rendering choice (1) most appropriate. Choice (1)
98.The word that collocates with ‘apron’ is ‘strings' and not ‘relations’, alliance’, ‘association’ or ‘organization’. Strings are either ‘tightened’ or ‘severed', and contextually, option (3) is the right choice.
Choice (3)
99.Gluttony and excess' are negative words, and hence gifts’, ’boons’ or ‘attractions’ are inapt. The word 'dangers' seems a possible choice, but can be ruled out because beneficial’ doesn’t go with it contextually.
Choice (3)
100.'Muscular strength’ is an important ‘component’, ‘part’ or aspect*, but never ‘segment' or ‘complement’. It should be an ‘integral’ part of everyone’s life. 'Equal', ‘efficient', effective' and ‘equivalent’ are absurd here. Choice (1) tits.
Choice (1)
TEST -III
Solutions for questions 101 to 105:
101.343/98 + 15/2 6 * x/6 +5

7/2 + 15/2 = x + 5 (1)
= x + 5 -» x= 6

102. Let N =
5.4 x 5.4/0.18x0.18x0.18 *10^2
54 X 54 X 10^6 X 10^2/10^2 * 10*18*18
■x10*
= 5 x 10s
102 x 18 x 18 x 18 Let D = (82 + S2) / (52x 2) / 5 = 100 +5O*5 = 2/5 N 5x10*
Let D = (8^2 +6^2)+(5^2 + 2)
= 100 + 50 + 5 = 2/5
N/D = 5*10^5/2 *5 = 125 * 10^4
103. Let N = 12^3 + x * 121 - 1/11
and D = 729-512-216 = 1
...123^3/x *11^2 - 1/11 = 1
123^3/x + 11^2 -1/11 = 1
12^3 /x * 11^2 = 12/11 = x * 12^2 *11^3

104. Let N = —- (513/450 )+ (43/50)
= 1 (7/50 )+ (43/50 )=2
Let D =612/360 + 16/10 = 33/10
N/D = 20/33 ; Given that, 20/33 = x/11
x = 6 2/3

105.Let N = 261(99)-1089
= 261 (99) - (99 x 11) = 99 X 250
Let D = 10^4 + 8 x 5 = 250
N/D = 99 x250 /250 = 99 = 10^2 -1
... ?=10.
99 = 10^2-1
Choice (2)
Solutions for questions'106 to 110:
106.4x2+2,10x2, 20*2*3,43*2+4 90x2+5, 185*2*6, 376
In the series 10 is not maintaining the proper relation with
,x2+1, 9*2+2 20x3*3, 43.
Choice (3)
4 and 20, the correct series is 4*2+1 So the wrong number is 10.
107.16x2*2, 34.108. 212x3*3. 639*2*2, 1280*3*3, 3843 108 is not maintaining the proper relation with 34 and 212. The correct series is 16*2+2 , 34*3*0. 105*2*2, 212*3+3, 639*2+2, 1280. So the wrong number is 108.
Choice (2)
108. 18*1+3, 21*2+4. 46*2+9, 101*3+16, 218*2+36,461*2+3, 958.
18 is not maintaining the proper relation with 21. So the wrong number is 18. Choice (1)
109.15*1+1,16*2*1, 33x3+6,105*4+4, 424*5+5, 2125*6+8,12756 By observation the last four numbers, those have a relation x4 + 4, x5 + 5, x6 + 6. So the correct series is
15*1+1 , 16*2+2 34x3+3 105*4+4 424.... So the wrong
number is 33. Choice (5)
110. 121, 169, 225, and 289 are the squares of consecutive odd integers. So the correct series is 72, 92, 112,132,152, 173. So the wrong number is 80.
Choice (4)
Solutions for questions 111 to 120:
111. Let the three numbers be 2x, 3x, 5x respectively. Their HCF = x = 24 =* Sum = 10x = 240
Choice (1)
112. Percentage increase in price = 20%
Percentage decrease in Consumption such that the
expenditure remains the same = , 100*20/
(100 + 20) 3
Choice (1)
113. The number of different signals that can be made by using one or more of 4 flags is 4P1 + 4P2 + 4P3 + 4P4-= 4+12 + 24 + 24 = 64 Choice (3)
114. Let the boat’s speed in still water be u and the speed of current be v.
Time taken to cover a distance down stream =
u +v
= 60 minutes. Time taken to cover the same distance up stream =
= 100minutes. ... u+V/u -v = 100/60 = 5/3 u-v U-V . 60 3
u 5 + 3 v~5-3
Choice (2)
115. A alone can complete the work in 20 days and B alone in 30 days.
A and B work together for six days and B works for the remaining days, say x days.
6 6+x 1 15
6/20+6*x/30 =1 x =15
Choice (3)
116. Let the cost of a tube of toothpaste, a brush and tongue cleaner be p, q and t respectively.
3p = 9b => 2p = 6b (1) and 6b = 4t=»3b = 2t->(2)
2p + 3b + 2t= 120
=* 6b + 3b + 3b = 120 12b = 120 => b = *10
From (2). 3b =2t
=>t = 15 Choice (2)
117. If Charan won 5 matches, he would have 25 chocolates with him but at the end he has only 15 chocolates. So, he lost 10 matches. Hence, totally they have playe15 games together.
Choice (2)
118.Let the sum borrowed be P
Then P x 12/100 x 3 + P x 16/100 x 5 + P x 20/100 x 3 = 6840 + P
=>P = *9,000
Choice (4)
119.First arrange the 5 boys in a row.
They can be arranged in 5! ways, x B1 x B2 X Bs3 X B4 x B5X
Now there are 6 gaps, denoted above by X, the 6 girls can be arranged in these gaps in 6! ways.
... 5!. x 6! = 86.400
  Choice (2)
120. Let the property distributed be x.
Then x----------x = 60,000
30
10
X
6x
>——=60.000 => x = ?3,00,000 30
Total property = 3 x 3,00,000 = *9,00,000.
Choice (3)
Solutions for questions 121 to 125:
121. 7168 + 295 - 2674+ 180-1128 = ? Applying B.O.D.M.A.S. rule,
= 7463 - 2674 + 180-1128 = 7463+180 - 2674-1128 = 7643 - 2674-1128=4969- 1128 = 3841.
  Choice (1)
122.1859x19 - 2299x11
11 + 13
11x169x19-19x121x11
11 + 13 _ 11x19(189 -121)
11 + 13
- 11x 19(13- 11)(13 + 11) (11 + 13)
= 11 x10x2=418
Choice (1)
123.128 + 7 3/4x8 4 / 80 - 20 x 3/2
(128 + 62)/(80-30) =(190/50) =19/5
Choice (4)
124. 3180 /4 x16+ (3728-1680) = ?
= 3180 * 4 X 16 + 2048 = 795 x 16 + 2048 = 12720 + 2048= 14768
Choice (2)
125. 372 + ?=75^2+112
=> 372 + ? = 5625 + 121 =» 5746 = 372 + ?
? = 5746- 1369 = 4377
Choice (3)
Solutions for questions 126 to 130:
126.We know, average speed of train
Total distance covered /Total time taken
Data in statement I alone is not sufficient since particulars of joumey from B to
c are also required in order to find average speed of train
Data in statement II alone is not sufficient since in this case also,particulars of journey from A to B are also required to find average speed of train
Obviously,statement (I) alone or (II) alone is not sufficient since in this case also, particulars of journey from A to B as well as from B to C are required i.e.data in statements I and II together are not sufficient to answer the question
To find total distance,additional data regarding time taken from A to B as well as from B to C are required i.e data in statement I and II together are not sufficient to answer the question Choice (4)

127.Given,
upstream distance (A to B) = downstream distance (B to A)
=> tup.Vup = tdown-V down
where Vup is upstream rate and Vdown is downstream rate
Let Vb be the speed of boat in still water and Vs be the
speed of still water
We have Vup = Va - Vs
Vdown = VB +Vs
tup-(VB - Vs) = tdown.(VB + Vs)
= (( tdown +tup) /tup - tdown)= Vb/Vs -(A)
from data in statement I VB = 16 kmph which alone is not
from data in statement II, Vs = 4 Kmph which alone is not sufficient
Obviously, data in I alone or II alone are not sufficient Data in I or II are sufficient and necessary to answer the question which is clear from the relation (A)
i.e.(tdown + tup)/(tup - tdown) = 16/4
Choice(5)

128. (1)We have 7x2 + 5xy - 2^ = 0 where x, y are the ages of A and 8
Further, 7x2 + 5xy - 2y^2= 0
=> (7x - 2y) (x + y) = 0
= 7x = 2y or x = -y (not considered)
...7x = 2y => x: y = 2:7 I alone is sufficient.
B be x and y respectively
Let the present ages of A and B be x and y respectively.
As per the data, y + 5 = 3x + 10
From which it is not possible to compute x:y.
i.e. ’I alone is sufficient and 'IT alone is not suffident.
    Choice(1)

129.(1)Earnings of A and B would be in the ratio of their efficiencies (i.e. amount of work done in a day)Ratio of the number of days A and B take to complete the work individually
= Inverse ratio of the ratio of the efficiencies of A and
B = 1/2 : 1/3 = 3:2
(2)Let A and B take x and y days to complete the work respectively
We have, x= y + 5
from which it is not possible to compute x: y II alone is not sufficient
i.e. I alone is sufficient and II alone is not sufficient to answer the question Choice(1)
130.We know, A =P( 1 + TR/100)
But I =PTR/100

130. We know, A = P (1+TR/100)
But I = PTR/100
= A = 100I/TR [ 1+TR/100]
=A = (100/TR +1) =I (1+100/TR)

Given that interest earned in five years is *1700.
This data is not suffident to find rate percent per annum at SI. i.e, I alone is not suffident.
Data from II is not suffident to find r% p.a. SI.
Obviously, I alone or II alone are not sufficient
Both I and II are necessary and sufficient to answer the question.
Because A = I (1+100/TR)
=25500 = 1700(1 + 100/5R)
Using this equation we can find the value of R.
  Choice (5)

Solutions for questions 131 to 135:
131.Total number of gas stoves produced by all companies on Monday = 680 + 580 + 670 + 940 + 980 = 3850
The total number of gas stoves rejected by all companies on Monday = 180 + 160 + 180 + 190 + 200 = 910
Required percentage = 910/3850x100 = 24%.
Choice (4)

132.Among companies C, D and E, for company E the production it more but rejected units is less compared to C and D. So the percentage of rejection Is not less for company C and D when compared to that of E. The percentage of rejection for company
A = 1200/4361 *100 = 27%
B = 1265/4680 *100 = 27%
E = 1280/5065*100 = 25%
Choice (4)
133.Total number of gas stoves not rejected by company D from Monday to wednesday
= (940 - 190)+(840 - 310)+(670 - 173) = 1777
   Choice (1)
134.Total number of gas stoves produced by all the companies on saturday =690+620+900+880+780 = 3870
Tuesday =725+690+840+640 = 3735
3870 is more than 3735, the answer should be more than 1.
Choice (5)
135.By observation we can say that the percentage of rejection of gas stoves for company B on Saturday is the highest.
  Choice (2)
136. Let the percentage increase in the production of company A, B, C, D, E and Ffrom 1999 to 2000 be denoted by a, b. c, d, e and f.
a = (35-28)/28 * 100 = 25%
b = (60 - 50)/50 * 100 = 20%
c = (80 - 70 )/70 * 100 = 20%
d = (85-80)/80 * 100 = 1/16 * 100 = 6.25%
e = (56 - 45)/45 * 100 = 118/45 * 100 = 11 * 20/9 = 11*2.2
= 24.2%
f = (36-30)/30 * 100 = 20%
137.The average production for all three years is minimum for the companies with the least total production. By observation we can say that it must be the least for A or F
A = 28+35+50 = 113
F =30+36+45 = 111.
The average production for company F is the least.
Choice (5)

138.The total production of all the 6 companies in 1999 = 28 + 50 + 70 + 80 + 45 + 30 = 303 lakh tonnes. The total production of all the 6 companies in 2000 = 35 + 60 + 80 + 85 + 56 + 36 = 352 lakh tonnes.
Hence the required percentage =(352-303)/303 *100
= 49/303 * 100 = 16%.
Choice (5)

139.The total production of company A exceeds the total production of company F over the years by
(113-111)/11 x100=200/111 = 1.8%

Choice (1)
140. The average annual growth in the production of steel for company D is (5+5)/2 = 5 Choice (3)
Solutions for questions 141 to 145:
141.6000/7500 X 100 = 80%
142. For school
Choice (1)
A.12000/27000 x 100 = 44.44%
B. 6000/15000 x 100 = 40%
C.21000/45000 x 100 = 46.67%
D.9000/21000 x 100 = 42. 86%
E-12000/30000 X 100 = 40%
  Choice (3)
143. By observation there is only one such school (school D).
Choice (2)

144. The ratio of grants to tuition fees for School A, 6000/12000 = 0.5
School B. 5400/6000 = 0.9
School C. 12000/21000 = 0.571
School D, 4200/9000 = 0.466
School E, 7500 /12000 = 0.625
Choice (4)
Alternate method:
Only in school D, the tuition fee Vs more than twice the tunds collected grants.

145. The ratio of the term fees to total income for schoold A,
2400/27000 = 8.88%
B,1200/15000 = 8%
C,4500/45000 = 10%
D,2400/21000 = 11.43%
E, 3000/30000 = 10%
Choice (4)

Solutions for questions 146 to 150:
146.Total number of workers in department S = (18/100) * 5000 = 900
Skilled workers in department S
18/100 * 60/100*5000 = 540
Unskilled workers = 900 - 540 = 360
     Choice (1)
147. Total number of skilled workers in department T
(24/100) * (60/100)*500 = 720
—X5000 =720 100 100
Choice (2)
148. Total number of skilled workers in the departments P, Q and U
= (20% +15% + 8%) of 60% of 5000
= (43/100) * (60/100)* 5000 = 1290
Choice (4)
149. Ratio of skilled workers to total workers, department wise:
Skilled workers in department P = 20% of 60% = 12%
.*. The ratio for P = 12/13 = 0.92
Similarly,
in Q = 9/24 = 0375
in S = 10.8/18 = 0.6
in U = 4.8/15 = 0.32
in T = 14.4/19 = 0.75
Choice (1)
150. Total number of unskilled workers in all the departments
= 40/100 X 5000 = 2000
Choice (1)