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So, the wrong number is 32.Right number= 9 × 2 + 2 × 6= 18 + 12= 30
45. (2)
So, the wrong number is 34650.Right number = 17325 × 3 = 51975
46. (1) According to the question,Average age of man and his twin sons = 30Total age = 30 × 3 = 90 yrAccording to the question,Ratio of father and one son = 5 : 2Ratio of father and both thesons = 5 : 2 : 2(As sons are born on the same day)5x + 2x + 2x = 90 or 9x = 90x = 10Age of father = 5 × 10 = 50 yr
47. (4) Let usual speed and usual timetaken by the jogger are x and t,respectively. Let his new speed x.Then
xt = x' 34 t x =
34 x' x =
43 x
Thus, he has to increase hisspeed by43 100%x x
x
i.e. 1
33 %3
48. (4) Let quantitites of two types of teacosting ` 18 per kg and ` 20 perkg be 5x kg and 3x kg respectively.Then,CP of tea = 18 × 5x + 20 × 3x
= 90x + 60x= 150x
SP of tea = 21 × 8x = 168xProfit per cent
= 168 150100%
150x x
x
= 18150× 100% = 12%
49. (2) Let sum = P, then for 20 yr,SI = (3P – P) = 2P
2P = P × R ×20
100
2 = R5 R = 10%
Again, for another time (T),SI = (2P – P) = P
P = P ×10 × T
100 T = 10 yrsFast track methed:-Here, n = 3, m = 2, T1 = 20 yr
T2 = m –1n –1 × T1
= 2 13 1
× 20 = 10 yr
50. (3) Given, P = 185220, R = 5%(increases)and n = 3 yrAcording to the formula,
Population n yr ago = 2
PR
1+100
(by Technique 5)
= 3
1855205
1100
= 185220
21 21 2120 20 20
= 85220 20 20 2021 21 21
= 20 × 20 × 20 × 20= 160000
51. (4) 4 leaps of cat = 3 leaps of dog
1 leap of cat = 34 leap of dog
Cat takes 5 leaps feor every 4 leaps of dogRequired ratio= (5 × Cat's leap) : (4 × Dog's leap)
(120 – x) = 64x = 120 – 64 = 56Now, 56 men can discharged tofinish the work in time.
55. (2) Let the certain distance be d andtime t.Now, by given condition,d3 = (t + 15) min = (t +15)
60 h
20d = t + 15 t = 20d – 15
and d4 = (t – 15) min
= (t –15)
60 h
15d = t – 15 t = 15d + 15From Eqs (i) and (ii), we get20d – 15 = 5d + 15 5d = 30d = 6 km
56. (3) Relative speed of train= Speed of train + Speed of car 90 = Speed of train + 15Speed of train= 90 – 15 = 75 km/hr
57. (1) Given,Summer's speed in still water= x = 9 km/hRate of stream = y = 6 km/hSpeed of downstream= x + y = 9 + 6 = 15 km/h
58. (3) There is a 7 digit telephonenumber but exreme right andextreme left positions are fixed.i.e. 6 × × × × × 5i.e. Required number of ways= 8 × 7 × 6 × 5 × 4 = 6720
59. (2) Required probabilityGiven, n = 5 and r = 3
60. (c) The bells will toll together aftertime in seconds equal to the LCMof 9, 6, 4, 10 and 8.LCM of 9, 6, 5, 10 and 8 is
LCM = 2 × 2 × 3 × 3 × 5 × 2 = 360In one hour, the rings will toll
together 3600360 times = 10 times
61. (1) Total length of roads in 2006 fromthe bar garp is 3000,000 kms. Fromthe pie chart the proportion oflength of highways in 2006 is 86%.Hence the length of highways in2006 =3000,00 × 86%=2580,000 kms
62. (2) From the pie-chart, it can be seenthat the highways constitute 86%and urban roads constitute 7% ofthe total road lengths. Hence, theratio of Highways to Urban Roads
is 786
12 : 163. (3) Assessment value of Highways in 2006
= 0.86 × 3,0000 × 1.5 lakh= ` 3870000 lakh
Assessment value of Urban Roads in 2006=0.07 × 3,000,000 × 1 lakh= ` 210000 lakh
Assessment value of Project Roads in 2006= 0.07 × 3,000,000 × 0.75 lakh= ` 157500 lakh
Total value= 3870000 + 210000 + 157500= ` 4237500 lakh
64. (1) In the year 2003, the total length was25,00,000 kms and the highwaysformed 80% of the same, which isequal to 20,00,000 kms. The highwayswere 25,80,000 kms in the year 2006.Hence, the increase in the length ofhighways was (25,80,000 – 20,00,000)= 5,80,000 kms.Thus, the increase in percentageterms
= 000,00,20000,80,5
× 100 = 29%
65. (2) One knows that for every % theangle substended is 3.6 degreesin a pie-chart. Thus, theproportion of the Urban Roads
from the year would be 6.372
= 20%.
In the year 2002, the total roadlength as can be seen from thebar chart = 24,00,000 kms. Thus20% of 24,00,000 = 480,000 kms.
REASONING(66-70):66. (1) Find the corresponding letters
from the right.67. (5) The first letter follows +1, +2, +3,
+4, ... The second follows +2, +3,+4, +5, ...
68. (2) Move two letters backward in thealphabetical series for eachcorresponding letter.
From (ii), (vii), (viii) and (x), fe = made... (xi)
From (iii), (v), (vii) and (x), da = beFrom (iv), (vi), (vii) and (vii), we = grace
... (xiii)71. (3) 72. (5) 73. (1) 74 (4)(75-79):75. (5) R < P < U < K
U < LI. K > R is true.II. L > R is true.
76. (3) H = I < R < MHence M > II. M = III. M > IEither I or II is true.
77. (2) D > H > NH > I < SI. N < S is not true.II. I < D is true.
78. (2) W < Y < P < O < II. Y < I is false.II. O > W is true.
79. (5) A > B > C > FE > D > C > ZI. A > Z is true.II. F < E is true.
(80-84): On the basis of the giveninformation we have the followingtable:
80. (1) From the table C's compulsorysubject is History.
81. (4) The female member is the personwho teaches English i.e. D.
82. (5) None of the teachers have the samecompulsory and optional subjects.
83. (3) E and F have Physics andMathematics as the compulsoryand optional subjects.
84. (4) A, B and C have History as compulsorysubject, from the above table.
(85-90):85. (5) All squares are rings + All rings are
triangles = A + A = A = All squaresare triangles conversion Sometriangles are squares. Henceconclusion II ('At least sometriangles are squares', i.e. ('Sometriangles are squares') follows.Again, No rectangle is a triangles+ Some triangles are rings(converse of 'All rings are triangles')= E + I = O* = Some rings are notrectangle. Now, draw the allpossible venn diagrams for theabove conclusion as given below:
(Some rings are rectangles.)We don't need to draw otherpossibilities. Therefore, the possibility,i.e. conclusion I follows.
86.(4) No flat is an appartment + Someapartments are bunglaows(conversion of 'Some bunglaowsare apartments) = E + I = O* =Some bungalows are not flats.Therefore neither conclusion Inor conclusion II follows.
conversion Some alphabetsare XYZ (I) + No XYZ is a day (E) =I + E = O = Some alphabets arenot days. Therefore conclusion Idoes not follow. Again, All ABCare XYZ + All XYZ are alphabets= A + A = A = All ABC arealphabets conversion Somealphabets are ABC (I). Therefore,conclusion II follows.
89. (1) From the statement 'Someteachers are professors' there arepossibility that 'All teachers areprofessors' and 'All professors areteachers'. Similarly from thestatement 'Some lecturers areteachers' there is a possibilitythat 'All teachers are lecturers'.Again, All professors are teachers+ All teachers are lecturers = A +A = A = All professors arelecturers is a possibility. Butconclusion II does not follow.
90. (2) From the above conclusion I doesnot follow. Again, from the above,we have the following possibility:'All lecturers are teachers' and 'Allteachers are professors'. Now, Alllecturers are teachers + Allteachers are professors = A + A =A = 'All lecturers are professors.Therefore conclusion II, i.e. 'Alllecturers being professor is apossibility' follows.