SBI-PO 2005 (Quantitative Aptitude)
QUANTITATIVE APTITUDE
Directions (76-78) What approximate value will come in place of the questions mark (?) in the following equations?
76. 125% of 4875 + 88.005 × 14.995 = ?
(a) 7395
(b) 7490
(c) 7510
(d) 7375
(e) 7415
Ans.(e) ? = 125% of 4875 + 88.005 × 14.995
≈ 125% of 4900 + 88 × 15
| ≈ | 125 × 4900 100 | + 1320 |
≈ 6125 + 1320
≈ 7445 ≈ 7415
(a) 1440
(b) 1400
(c) 1000
(d) 1040
(e) 1140
(d)ANS. ? = 127.001 × 7.998 + 6.05 × 4.001
= 127 × 8 + 6 × 4
≈ 1016 + 24 ≈ 1040
78. 1010 ÷ 36 + 187 × 20.05 = ?
(a) 3650
(b) 3770
(c) 3825
(d) 3800
(e) 3700
(b) Ans. ? = 1010 ÷ 36 + 187 × 20.05
≈ 1008 ÷ 36 + 187 × 20
[1010 ≈ 1008 which is a multiple of 36]
≈ 28 + 3740
≈ 3768 ≈ 3770
79 What will come in place of question mark (?) min the following equation ?
167.5 ÷ 83.5 ÷ 27.5 = ?
(a) 84
(b) 164
(c) 215
(d) 227
(e) None of these
(a)Ans. ? = 167.5 ÷ 8 3.5 ÷ 27.5
| = | 167.5 83.5 × 27.5 | = | (24)7.5 (23)3.5 × 27.5 |
| = | (2)4 × 7.5 (2)3×3.5 × 27.5 | = | 230 (2)10.5 × 27.5 |
| = | 230 (2)10.5+7.5 | = | 230 218 | = 230-18 = 212 = 84 |
80. Present ages of Seema and Naresh are in the respective ratio of 5 : 7 Five years hence the ratio of their ages becomes 3 : 4 respectively. What is Naresh's present age in years ?
(a) 25
(b) 40
(c) 30
(d) can not be determined
(e) None of these
(e) Ans. Let the present ages of Seema and Naresh be 5 x and 7x years respectively.
According to the question,
| 5x + 5 7x + 5 | = | 3 4 |
⇒ 21x + 15 = 20x + 20
⇒ x = 20 - 15 = 5
∴ Naresh's presents age
= 7x years
= 7 × 5 = 35 years
81. In a two digit number the digit in the unit's place is twice the digit in the ten's place and the number obtained by interchanging the digits is more than the original number by 27. What is 50% of the original number ?
(a) 36
(b) 63
(c) 48
(d) 18
(e) None of these
(a) Ans. Let the ten's digit be x.
∴ Unit's digit = 2x
∴ Original number
= 10x + 2x = 12x
On interchanging the digits, the new number
= 10 × 2x + x = 21x
According to question,
21x - 12x = 27 ⇒ 9x = 27
⇒ x = (27/9) = 3
∴ Original number = 12x
= 12 × 3 = 36
82. A committee of 3 members is to be formed out of 3 men and 4 women. In how many different ways can it be done so that at least one members is a woman ?
(a) 34
(b) 12
(c) 30
(d) 36
(e) None of these
(a) Ans. The committee can be formed in the following ways :
(1) By selecting 2 men and 1 women
(2) By selecting 1 man and 2 women
(3) By selecting 3 women
∴ Total number of ways of forming the committee
= 3C2 × 4C1 + 3C1 × 4C2 + 4C3
= 3 × 4 + 3 × 6 + 4
= 12 + 18 + 4 = 34
83. 4 boys and three girls are to be seated in a row in such a way that no two boys sit adjacent to each other. In how many different ways can it be done ?
(a) 5040
(b) 30
(c) 144
(d) 72
(e) None of these
(c) Ans. 3 girls can be seated in a row in 3! ways. Now, in the 4 gaps 4 boys can be seated in 4! ways Hence, the number of ways in which no two boys sit adjacent to each other = 31 × 41 = 6 × 24 = 144
84. Mr. 'X' invested certain amounts in two different schemes 'A'&'B', Scheme 'A' offers simple interest at 12 p.c.p.a. and Scheme 'B' offers compound interest at 10 p.c.p.a. Interest accrued on the amount invested in Scheme A in 2 years was Rs. 3,600 and the total amount invested was Rs. 35,000. What was interest accrued on the amount invested in Scheme 'B' ?
(a) Rs.4,800
(b) Rs. 4,200
(c) 4,000
(d) Can not b determined
(e) None of these
(b) Ans. Let the amount invested in scheme A be Rs. x.
| ∴ | x × 12 × 2 100 | = 3600 |
| x = | 36 × 100 24 | = Rs. 15000 |
Total investment = Rs. 35000
∴ Amount invested in scheme B
= Rs. (35000 -15000)
= Rs. 20000
| ∴ C.I. = P | [ | ( | 1 + | R 100 | ) | T | - 1 | ] |
| = 20000 | [ | ( | 1 + | R 100 | ) | 2 | - 1 | ] |
= 20000 (2.21 - 1)
= 20000 × 0.21 = Rs. 4200
85. The salary of an employee increases consistently by 50% every year. If his salary total today is Rs. 10,000, what will be the salary after another 4 years.?
(a) Rs. 62,500
(b) Rs. 26,500
(c) Rs. 50,625
(d) Rs. 33, 750
(e) None of these
(c) Ans. Required salary
| = 10000 | ( | 1 + | 500 100 | ) | 4 |
= 10000 × (1.5)4 = Rs. 50625
Directions (86-90) : In each of the following questions a number series is given. After the series a number is given followed by (a), (b), (c), (d) and (e). You have to complete the series starting with the given number, following the sequence of original series and answer and answer the questions that follow the series.
86. 3 19 103 439 l38l 2887 5
(a) (b) (c) (d) (e)
What will come in place of (b) ?
(a) 139
(b) 163
(c) 161
(d) 157
(e) None of these
(b)Ans.
Similarly,
87. 4 13 40 135 552 2765
2 (a) (b) (c) (d) (e)
What will come in place of (c) ?
(a) 123
(b) 133
(c) 127
(d) 131
(e) None of these
(a) Ans. The given series is based on the following pattern :
13 = 4 × 1 + 1 × 9
40 = 13 × 2 + 2 × 7
135 = 40 × 3 + 3 × 5
552 = 135 × 4 + 4 × 3
2765 = 552 × 5 + 5 × 1
Similarly,
(a) = 2 × 1 + 1 × 9 = 11
(b) = 11 × 2 + 2 × 7 = 36
(c) = 36 × 3 + 3 × 5 = 123
Hence, 123 will come in place of (c).
88. 5 12 4 10 3 8 6
(a) (b) (c) (d) (e)
What will come in place of (d) ?
(a) 3
(b) 5
(c) 4
(d) 7
(e) None of these
Ans. (c) The given series is based on the following pattern :
Similarly
hence , 4 will come in place of (d) .
89. 3 13 37 87 181 401 6
(a) (b) (b) (d) (e)
What will come in place of (d) ?
(a) 169
(b) 161
(c) 171
(d) 159
(e) None of these
Ans. (d) The given series is based on the following pattern:
[7,11,13,17,17 are consecutive prime numbers]
Similarly ,
Hence , 159 will come in place of (d)
90. 8 4 6 15 52.5 236.25 12 (a) (b) (c) (d) (e)
What will come in place of (c) ?
(a) 18.25
(b) 19
(c) 22.5
(d) 20.75
(e) None of these
Ans. (c) The given series is based on the following pattern.
Similar,
Hence , 22.5 will come in place of (c)
Directions (91-95) In each of the following questions, a question is followed by information given in three statements. you have to study the question along with the statements and decide the information given in which of the statement(s) is necessary and sufficient to answer the question.
91. What is the average weight of girls in the class ?
I. Average weight of all the 60 students is 42kg.
II. Average weight of boys is 43 kg.
III. Total weight of all girls together is 1144 kg.
(a) Any two of three
(b) All I, II & III
(c) I & II only
(d) II & III only
(e) Question cannot be answered even with information in all three statements.
(b) Ans. From statement I, Total weight of 60 students
= 60 × 42 = 2520 kg
From statement III,
Total weight of all the girls = 1144 kg
∴ Total weight of all the boys
= 2520 - 1144 = 1376 kg.
From statement II,
Number of boys = (1376/32) = 32
∴ Number of girls = 60 - 32 = 28
From statement III,
Average weight of girls
= (1144/28) = 40.86 kg
92. What is the selling price of the T.V. set if no discount is offered ?
I Profit earned was 20%.
II Had 10% discount been offered on selling price the profit would have been Rs. 1,200.
III. Cost price is Rs. 15,000.
(a) Any two of the three
(b) Only I & II
(c) Only I & III
(d) Only II & III
(e) None of these
(a) Ans. From statement I and III, S.P.can be obtained.
From statement II and III, Let the SP be Rs. x.
After 10% discount, SP = 90% of x = (1/10)x
∴ (9/10) x - 15000 = 1200
⇒ 9x = 150000 + 1200
⇒ x (162000/9) = Rs. 18000
From statement I and II, Let the CP = Rs. x
| ∴ S.P = | 120 100 | x = Rs. | 6x 5 |
Rs. (54x/50)
| ∴ | 54x 50 | - x = 1200 |
⇒ 54x - 50x = 50 × 1200
| ⇒ x | 50 × 1200 4 | = 15000 |
93. What is the speed of the train?
I. Length of the platform is 150% of the length of the train.
II. The train crosses the platform in 25 seconds.
III. The train crosses the signal pole in 19 seconds.
(a) All I, & III
(b) I and either II or III
(c) Only II& III
(d) Any two of the three
(e) Question cannot be answered even with the information in all three statements.
(e) Ans. From statement I,
Let the length of the train be x meter.
∴ Length of plat from
= (3x/2) meter
From statement II,
| Speed of train = | (3x/2) +x 25 |
| = | 5x 50 | x 10 | ............(i) |
But x is not know. Hence, we proceed.
From statement III,
Speed of train = (x/10) ........... (ii)
Clearly, we reach at no unique conclusion.
94. How many children are there in the class ?
I. 20% children are there in the class ?
II. 44 children can speak languages other than Hindi.
III. There are 30 boys in the class.
(a) All I,II & III
(b) Any two of the three
(c) II and either I or III
(d) I and II only
(e) None of these
(d) Ans. From statement I, 80% children speak language other than Hindi.
From statement II,
∴ Total Number of children
| = | 44 × 100 80 | = 55 |
95. What is the volume of the cylindrical tank ?
I. Area of the base is X square metes.
II. Height of the tank is Y metes.
III. Diameter of the base is equal to height of the tank.
(a) Only I & II
(b) Only II & III
(c) Only I & III
(b) All I,II & III
(e) Any two of the three
(e) Ans. From statement I and II, Volume of tank = XY cubic meters
From statement II and III,
Radius of base = (Y/2)
∴ Volume = π × (radius)2 × height
From statements I and II,
Area of base = sq. meters.
Hence, radius can be determined.
Height = 2 × radius.
Clearly, after knowing height and area of base, volume can be determined.
Directions (96-100): In each of these questions two equations numbered I & II are given. you have to solve both the equations and give answer
(a) if a < b
(b) if a > b
(c) if relationship between a & b to cannot be established
(d) if a > b
(e) if a < b
96. I. a2 + 5a + 6 = 0
II. b2 + 3b + 2 = 0
(e) Ans. I. a2 + 5a + 6 = 0
⇒ a2 + 2a + 3a + 6 = 0
⇒ a(a + 2) + 3 (a + 2) = 0
⇒ (a + 2) (a + 3) = 0
⇒ a = - 2 or -3
II. b2 + 3b + 2 = 0
⇒ b2 + 2b + b + 2 = 0
⇒ b(b + 2) + 1 (b + 2) = 0
⇒ (b + 1) (b + 2) = 0
⇒ b = -1 or - 2
Clearly, a ≤ b
97. I. 2a2 + 3a + 1 = 0
II. 12b2 + 7b + 1 = 0
(a) Ans. I. 2a2 + 3a + 1 = 0
⇒ 2a2 + 2a + a + 1 = 0
⇒ 2a(a + 1) + 1 (a + 1) = 0
⇒ (a + 1) (2a + 1) = 0
⇒ a = - 1 or - (1/2)
II. 12b2 + 7b + 1 = 0
⇒ 3b(4b + 1) + 1 (4b + 1) = 0
⇒ (3b + 1) (4b + 1) = 0
⇒ b = - (1/3) or - (1/4)
Clearly, a < b
98. I. a2 = 4
II. a2 = 9
(c) Ans. I. a2 = 4 ⇒ a = + 2
II. b2 = 9 ⇒ b = + 3
99. I. 6a2 - 25a + 25 = 0
II. 15b2 - 16b + 4 = 0
(b) Ans. I. 6a2 - 25a + 25 = 0
⇒ 6a2 - 15a - 10a + 25 = 0
⇒ 3a(2a - 5) - 5 (2a - 5) = 0
⇒ (3a - 5) (2a - 5) = 0
⇒ a = (5/3) or - (5/2)
II. 15b2 - 16b + 4 = 0
⇒ 15b2 - 10b - 6b + 4 = 0
⇒ 15b(3b - 2) -2 (3b - 2) = 0
⇒ (5b - 2) (3b - 2) = 0
⇒ b = (2/5) or (2/3)
Clearly, a > b
100. I. 4a2 - 20a + 21 = 0
II. 2b2 - 5b + 3 = 0
(d) Ans. I. 4a2 - 20a + 21 = 0
⇒ 4a2 - 14a - 6a + 21 = 0
⇒ 2a(2a - 7) -3 (2a - 7) = 0
⇒ (2a - 7) (2a - 3) = 0
⇒ a = (7/2) or (3/2)
II. 2b2 - 5b + 3 = 0
⇒ 2b2 - 3b - 2b + 3 = 0
⇒ b(2b - 3) - 1 (2b - 3) = 0
⇒ (b - 1) (2b - 3) = 0
⇒ b = 1 or 2/3
Clearly, a ≥ b
Directions (101-105) Study the following table carefully to answer these questions :
PERCENTAGE OF MARKS OBTANED BY SIX STUDENTS IN SIX DIFFERENT SUBJECTS
| Subject | Student | |||||
| P (70) | Q (80) | R (120) | S (125) | T (75) | U (150) | |
| A | 68 | 84 | 77 | 72 | 64 | 82 |
| B | 49 | 79 | 62 | 85 | 56 | 76 |
| C | 56 | 81 | 68 | 60 | 58 | 68 |
| D | 75 | 85 | 82 | 88 | 72 | 78 |
| E | 70 | 66 | 65 | 76 | 77 | 83 |
| F | 72 | 70 | 79 | 68 | 68 | 71 |
Note. Figure in bracket below each subject indicates the maximum marks allotted
101. Which is the average percentage of marks obtained by all students S, T & U together ?
(a) 65.6
(b) 66.5
(c) 66.8
(d) 65.4
(e) None of these
(e) Ans. Required average percentage
| = | 64 + 56 + 58 + 72 + 77 + 68 6 | = | 395 6 | = 65.8 |
102. Marks obtained by 'A' in subjects P, Q & R together are approximately what per cent of the marks obtained by F in subjects S, T & U together ?
(a) 80
(b) 75
(c) 85
(d) 105
(e) 115
(c) Ans. Marks obtained by A in ;
Subject P → 68% of 70
≈ (70/100) × 70 ≈ 49
Subject Q → (84/100) × 80 ≈ 67
Subject R → (77/100) × 120 ≈ 92
Total marks
= 49 + 67 + 92 = 208
Marks obtained by F in :
Subject S → 68% of 125
= (68/100) × 125 = 85
| Subject T → | 68 × 75 100 | = 51 |
| Subject U → | 71 × 150 100 | ≈ 105 |
∴ Total marks
= 85 + 51 + 105 = 241
∴ Required percentage
= (208/241) × 100 ≈ 85
103. What is the overall percentage of marks obtained by 'B' in all the subjects together (rounded off to two digits after decimal)?
(a) 70.02
(b) 72.51
(c) 67.83
(d) 71.50
(e) None of these
(c) Ans. Required overall percentage of marks
| = | 49 + 79+ 62+ 85+ 56+ 76 6 |
= (407/6) = 67.83
104. What are the marks obtained by D in all the subjects together ?
(a) 449.8
(b) 499.9
(c) 480
(d) 490
(e) None of these
(b) Ans. Marks obtained by D in :
| Subject P → | 75 × 70 100 | = 52.50 |
| Subject Q → | 85 × 80 100 | = 68 |
| Subject R → | 82 × 120 100 | = 98.40 |
| Subject S → | 88 × 125 100 | = 110 |
| Subject T → | 72 × 75 100 | = 54 |
| Subject U → | 68 × 150 100 | = 117 |
∴ Total marks = 5250 + 68 + 98.40 +110 + 54 + 117 = 499.9
105. What are average marks obtained out of 80 by all the six students together in subject 'Q' ?
(a) 68
(b) 77.5
(c) 88.57
(d) 62
(e) None of these
(d) Ans. Average percentage
| = | 84 + 79 + 81 + 85 + 66 + 70 6 |
= 465/6
∴ Required average marks
| = | 465 6 | % of 80 = | 465 × 80 6 × 100 | = 62 |
Directions (106-110) Study the following graph carefully to answer these questions :
NUMBER OF ITEMS PRODUCED (IN THOUSADS) AND COST ( IN RUPEES) PER HUNDRED ITEMS IN SIX COMPANIES
106. What will be the total cost of items produced by Company C?
(a) Rs. 32 lakhs
(b) Rs. 24 lakhs
(c) Rs. 27 lakhs
(d) Rs. 36 lakhs
(e) None of these
(c) Ans. Item product = 45000
Cost per hundred = Rs. 6000
∴ Total cost
| = Rs. | 45000 × 6000 100 |
107. What is the average cost per hundred items for all the given companies ?
(a) Rs. 4183 (2/3)
(b) Rs. 4283 (1/3)
(c) Rs. 43331 (2/3)
(d) Rs. 4333 (1/3)
(e) None of these
(e) Ans. Required average cost
| = Rs. | 4500 + 3500 + 6000 + 5500 + 5000 + 4000 6 |
= Rs. (28500/6) = Rs. 4750
108. What is total cost of the items produced by companies A & B together ?
(a) Rs. 17.50 lakhs
(b) Rs. 33.25 lakhs
(c) Rs. 15.75 lakhs
(d) Rs. 32.75 lakhs
(e) None of these
(b) Ans. Cost of items produced by company :
| A → Rs. | ( | 35000 × 4500 100 | ) | = Rs. 15.75 lakhs |
| B → Rs. | ( | 50000 × 3500 100 | ) | = Rs. 17.5 lakhs |
∴ Total cost
= Rs. (15.75 + 17.5) lakhs
= Rs. 33.25 lakhs.
109. What was the total number of items produced by all the companies together ?
(a) 28500
(b) 258000
(c) 25800
(d) 2850000
(e) None of these
(e) Ans. Total number of items produced by all companies = (35 + 50 + 45 + 30 + 40 + 60) thousands = 260000
110. If the number of items produced by Company 'D' increase by 30%, what will be the total cost of items produced ?
(a) Rs. 21.45 lakhs
(b) Rs. 22. 45 lakhs
(c) Rs. 24.25 lakhs (d) 22.25 lakhs
(e) None of these
(a) Ans. = New number of items produced by company
| D = | ( | 130 100 | × 30 | ) | thousands |
= 39 thousands
Cost per hundreds = Rs. 5500
∴ Total cost
| = Rs. | ( | 39000 × 5500 100 | ) |
= Rs. 21.45 lakhs.
Directions (111-115) : These questions are based on the following information :
Children in a class play only one or two or all the three games badminton football and cricket. 5 children play only cricket, 8 children play only football and 7 children play only badminton 3 children play only two games badminton and football, 4 children play only two games cricket and football and another 4 children play only two games badminton and cricket. 2 children play all the three grams.
Venn - Diagram for questions 111 - 115:
111. In all how many children play football ?
(a) 8
(17)
(c) 15
(d) 14
(e) None of these
(b) Ans. Number of football players = 3 + 2 + 4 + 8 = 17
112. How many children play badminton as.well as cricket ?
(a) 9
(b) 10
(c) 4
(d) 6
(e) None of these
(d) Ans. Number of badminton as well as cricket players = 4 + 2 = 6
113. Total how many children are there in the class ?
(a) 33
(b) 31
(c) 36
(d) 35
(e) None of these
(a) Ans. Number of children = 7 + 3 + 2 + 4 + 8 + 4 + 5 = 33
114. In all how many children play badminton ?
(a) 14
(b) 17
(c) 12
(d) 13
(e) None of these
(e) Ans. Number of badminton players = 7 + 3 + 2 + 4 = 16
115. How many children play football as well cricket ?
(a) 7
(b) 4
(c) 6
(d) 15
(e) None of these
(c) Ans. Number of football as well as cricket players = 2 + 4 = 6
DETAILS ABOUT THE DISTRIBUTION OF EMPLOYEES EXPENDITURE OF AN ORGANIZATION (DISTRYBUTED PROPORTIONATELY ACROSS THE DEPARTMENTS)
Annual expenditure on different items
Total Expenditure = Rs.12 crore's
DEPARTMENTWISE DISTRIBUTION OF EMPLOYEES
Total number of employees = 1200
116. What was the total expenditure on Accounts Department ?
(a) Rs. 16.8 lakhs
(b) Rs. 1,680 lakhs
(c) Rs. 18.6 millions
(d) Rs. 16.8 millions
(e) None of these
(d) Ans. Total expenditure on Accounts department = 14 % 0f Rs. 12 crore's
| = Rs | 14×12 100 | crore's |
= Rs. 1.68 crore's
= Rs. 16.8 millions
117. What was per employee expenditure on Medical ?
(a) Rs. 12,000
(b) Rs. 13,000
(c) Rs. 12,500
(d) Rs. 13.500
(e) None of these
(b) Ans. Total expenditure on medical = 13% of Rs. 12 crore's
| = Rs. | 13×12 100 | crore's |
= Rs. 1.56 crore's
= Rs. 15600000
∴ Expenditure per employee
= 15600000/1200
= Rs. 13000
118. What was the total expenditure on salary of employees in Marketing Department ?
(a) Rs. 6.12 lakhs
(b) Rs. 61.2 lakhs
(c) Rs. 6.12 millions
(d) Rs. 176 lakhs
(e) None of these
(c) Ans. Total expenditure on salary of employees = 30% of Rs. 12 crore's
| = Rs. | 30 × 120000000 100 |
= Rs. 36000000
∴ Expenditure on Marketing employees
= 17% of 36000000
= Rs. (17 × 360000)
= Rs. 6120000
= Rs. 6120000
= Rs. 6.12 millions
119. What was the amount spent on Electricity ?
(a) Rs. 132 millions
(b) Rs. 13.2 lakhs
(c) Rs. 126 millions
(d) Rs. 12.6 lakhs
(e) None of these
(e) Ans. Amount spent on electricity = 11% of Rs. 12 crore's
| = Rs. | 11 × 120 100 | millions |
= Rs. 13.2 millions
120. What was the expenditure on telephone for employees in computer Department ?
(a) Rs. 11.52 lakhs
(b) Rs. 11.52 millions
(c) Rs. 10.72 lakhs
(d) Rs. 10.72 millions
(e) None of these
(a) Ans. Expenditure on telephone in computer department = 12% of 8% Rs. 12 crore's
| = Rs. | 12 100 | × | 8 100 | 120 millions |
= Rs. 1.152 millions
= Rs. 11.52 lakhs
Directions (121-125) : Study the following table carefully to answer these questions.
SALES OF PRODUCT (IN MILLION TONS) FOR SIX STATES OVER THE YEARS
| Year | STATE | |||||
| A | B | C | D | E | F | |
| 1998 | 25 | 45 | 38 | 52 | 47 | 55 |
| 1999 | 32 | 39 | 40 | 55 | 46 | 67 |
| 2000 | 41 | 50 | 43 | 57 | 39 | 64 |
| 2001 | 37 | 48 | 43 | 58 | 32 | 72 |
| 2002 | 28 | 53 | 46 | 62 | 37 | 58 |
| 2003 | 43 | 55 | 49 | 63 | 42 | 62 |
121.. If the cost of product per thousand tons in 1998 was Rs. 1.8 lakhs, what was the cost of average sales for the given States in that year
(a) Rs. 7860000 lakhs
(b) Rs. 786 laksh
(c) Rs. 7860 lakhs
(d) Rs. 78600 lakhs
(d) None of these
(d) Ans. Average sales
| = | ( | 25 + 45 + 38 + 52 + 47 + 55 6 | ) | millions tons |
= (262/6) = 43.7 million tons
∴ Required cost
= Rs. (1.8 × 43700) lakhs
= Rs. 78660 lakhs
= Rs. 78600 lakhs
122. Total sales in year 2000 were what per cent of the total sales in year 2003? (rounded off to two digits after decimal)
(a) 93.63
(b) 92.65
(c) 106.80
(d) 93.23
(e) None of these
(a) Ans. Total sales in year 2000 = (41 + 50 + 43 + 57 + 39 + 64) million tons
= 264 million tons
Total sales in 2003
= (43 + 55 + 49 + 63 + 42 + 62) million tons
= 314 million tons
∴ Required percentage
| = | 294 314 | × 100 = 93.63 % |
123. What was the percentage increase in total sales in 2003 from 1998 ? (rounded off to nearest integer)
(a) 19
(b) 20
(c) 16
(d) 17
(e) None of these
(a) Ans. Total sales in year 1998 = (25 + 45 + 38 + 52 + 47 + 55) million tons
= 262 million tons
Total sales in 2003
= 314 million tons
∴ % Increase
| = | 314 - 262 262 | × 100 |
= 19.85 ≈ 20
124. Approximately what was the ratio between average sales of States B & C respectively ?
(a) 31:25
(b) 26:31
(c) 29:24
(d) 26:29
(e) 29:26
(e) Ans. Average sales of state :
| B → | 45 + 39+ 50 + 48 + 53 + 55 6 |
| = | 290 6 | millions tons |
∴ Required ratio
| = | 290 6 | : | 259 6 | ≈ 290 : 260 |
≈ 90 : 60
125. For which of the following years the percentage increase/decrease in sales from the previous years was highest for State 'E'?
(a) 1999
(b) 2000
(c) 2001
(d) 2002
(e) 2003
(c) Ans. Percentage decrease in 1999
| = | 47 - 46 47 | × 100 |
= 15.22
Percentage decrease in 2001
| = | 37 - 32 32 | × 100 |
= 17.95
Percentage increase in 2002
| = | 37 - 32 32 | × 100 |
= 15.625
Percentage increase in 2003
| = | 42 - 37 37 | × 100 |
= 13.51
Hence, the percentage decrease was maximum in the year 2001.
SBI-PO 2005 (Quantitative Aptitude)
QUANTITATIVE APTITUDE
Directions (76-78) What approximate value will come in place of the questions mark (?) in the following equations?
76. 125% of 4875 + 88.005 × 14.995 = ?
(a) 7395
(b) 7490
(c) 7510
(d) 7375
(e) 7415
Ans.(e) ? = 125% of 4875 + 88.005 × 14.995
≈ 125% of 4900 + 88 × 15
| ≈ | 125 × 4900 100 | + 1320 |
≈ 6125 + 1320
≈ 7445 ≈ 7415
(a) 1440
(b) 1400
(c) 1000
(d) 1040
(e) 1140
(d)ANS. ? = 127.001 × 7.998 + 6.05 × 4.001
= 127 × 8 + 6 × 4
≈ 1016 + 24 ≈ 1040
78. 1010 ÷ 36 + 187 × 20.05 = ?
(a) 3650
(b) 3770
(c) 3825
(d) 3800
(e) 3700
(b) Ans. ? = 1010 ÷ 36 + 187 × 20.05
≈ 1008 ÷ 36 + 187 × 20
[1010 ≈ 1008 which is a multiple of 36]
≈ 28 + 3740
≈ 3768 ≈ 3770
79 What will come in place of question mark (?) min the following equation ?
167.5 ÷ 83.5 ÷ 27.5 = ?
(a) 84
(b) 164
(c) 215
(d) 227
(e) None of these
(a)Ans. ? = 167.5 ÷ 8 3.5 ÷ 27.5
| = | 167.5 83.5 × 27.5 | = | (24)7.5 (23)3.5 × 27.5 |
| = | (2)4 × 7.5 (2)3×3.5 × 27.5 | = | 230 (2)10.5 × 27.5 |
| = | 230 (2)10.5+7.5 | = | 230 218 | = 230-18 = 212 = 84 |
80. Present ages of Seema and Naresh are in the respective ratio of 5 : 7 Five years hence the ratio of their ages becomes 3 : 4 respectively. What is Naresh's present age in years ?
(a) 25
(b) 40
(c) 30
(d) can not be determined
(e) None of these
(e) Ans. Let the present ages of Seema and Naresh be 5 x and 7x years respectively.
According to the question,
| 5x + 5 7x + 5 | = | 3 4 |
⇒ 21x + 15 = 20x + 20
⇒ x = 20 - 15 = 5
∴ Naresh's presents age
= 7x years
= 7 × 5 = 35 years
81. In a two digit number the digit in the unit's place is twice the digit in the ten's place and the number obtained by interchanging the digits is more than the original number by 27. What is 50% of the original number ?
(a) 36
(b) 63
(c) 48
(d) 18
(e) None of these
(a) Ans. Let the ten's digit be x.
∴ Unit's digit = 2x
∴ Original number
= 10x + 2x = 12x
On interchanging the digits, the new number
= 10 × 2x + x = 21x
According to question,
21x - 12x = 27 ⇒ 9x = 27
⇒ x = (27/9) = 3
∴ Original number = 12x
= 12 × 3 = 36
82. A committee of 3 members is to be formed out of 3 men and 4 women. In how many different ways can it be done so that at least one members is a woman ?
(a) 34
(b) 12
(c) 30
(d) 36
(e) None of these
(a) Ans. The committee can be formed in the following ways :
(1) By selecting 2 men and 1 women
(2) By selecting 1 man and 2 women
(3) By selecting 3 women
∴ Total number of ways of forming the committee
= 3C2 × 4C1 + 3C1 × 4C2 + 4C3
= 3 × 4 + 3 × 6 + 4
= 12 + 18 + 4 = 34
83. 4 boys and three girls are to be seated in a row in such a way that no two boys sit adjacent to each other. In how many different ways can it be done ?
(a) 5040
(b) 30
(c) 144
(d) 72
(e) None of these
(c) Ans. 3 girls can be seated in a row in 3! ways. Now, in the 4 gaps 4 boys can be seated in 4! ways Hence, the number of ways in which no two boys sit adjacent to each other = 31 × 41 = 6 × 24 = 144
84. Mr. 'X' invested certain amounts in two different schemes 'A'&'B', Scheme 'A' offers simple interest at 12 p.c.p.a. and Scheme 'B' offers compound interest at 10 p.c.p.a. Interest accrued on the amount invested in Scheme A in 2 years was Rs. 3,600 and the total amount invested was Rs. 35,000. What was interest accrued on the amount invested in Scheme 'B' ?
(a) Rs.4,800
(b) Rs. 4,200
(c) 4,000
(d) Can not b determined
(e) None of these
(b) Ans. Let the amount invested in scheme A be Rs. x.
| ∴ | x × 12 × 2 100 | = 3600 |
| x = | 36 × 100 24 | = Rs. 15000 |
Total investment = Rs. 35000
∴ Amount invested in scheme B
= Rs. (35000 -15000)
= Rs. 20000
| ∴ C.I. = P | [ | ( | 1 + | R 100 | ) | T | - 1 | ] |
| = 20000 | [ | ( | 1 + | R 100 | ) | 2 | - 1 | ] |
= 20000 (2.21 - 1)
= 20000 × 0.21 = Rs. 4200
85. The salary of an employee increases consistently by 50% every year. If his salary total today is Rs. 10,000, what will be the salary after another 4 years.?
(a) Rs. 62,500
(b) Rs. 26,500
(c) Rs. 50,625
(d) Rs. 33, 750
(e) None of these
(c) Ans. Required salary
| = 10000 | ( | 1 + | 500 100 | ) | 4 |
= 10000 × (1.5)4 = Rs. 50625
Directions (86-90) : In each of the following questions a number series is given. After the series a number is given followed by (a), (b), (c), (d) and (e). You have to complete the series starting with the given number, following the sequence of original series and answer and answer the questions that follow the series.
86. 3 19 103 439 l38l 2887 5
(a) (b) (c) (d) (e)
What will come in place of (b) ?
(a) 139
(b) 163
(c) 161
(d) 157
(e) None of these
(b)Ans.
Similarly,
87. 4 13 40 135 552 2765
2 (a) (b) (c) (d) (e)
What will come in place of (c) ?
(a) 123
(b) 133
(c) 127
(d) 131
(e) None of these
(a) Ans. The given series is based on the following pattern :
13 = 4 × 1 + 1 × 9
40 = 13 × 2 + 2 × 7
135 = 40 × 3 + 3 × 5
552 = 135 × 4 + 4 × 3
2765 = 552 × 5 + 5 × 1
Similarly,
(a) = 2 × 1 + 1 × 9 = 11
(b) = 11 × 2 + 2 × 7 = 36
(c) = 36 × 3 + 3 × 5 = 123
Hence, 123 will come in place of (c).
88. 5 12 4 10 3 8 6
(a) (b) (c) (d) (e)
What will come in place of (d) ?
(a) 3
(b) 5
(c) 4
(d) 7
(e) None of these
Ans. (c) The given series is based on the following pattern :
Similarly
hence , 4 will come in place of (d) .
89. 3 13 37 87 181 401 6
(a) (b) (b) (d) (e)
What will come in place of (d) ?
(a) 169
(b) 161
(c) 171
(d) 159
(e) None of these
Ans. (d) The given series is based on the following pattern:
[7,11,13,17,17 are consecutive prime numbers]
Similarly ,
Hence , 159 will come in place of (d)
90. 8 4 6 15 52.5 236.25 12 (a) (b) (c) (d) (e)
What will come in place of (c) ?
(a) 18.25
(b) 19
(c) 22.5
(d) 20.75
(e) None of these
Ans. (c) The given series is based on the following pattern.
Similar,
Hence , 22.5 will come in place of (c)
Directions (91-95) In each of the following questions, a question is followed by information given in three statements. you have to study the question along with the statements and decide the information given in which of the statement(s) is necessary and sufficient to answer the question.
91. What is the average weight of girls in the class ?
I. Average weight of all the 60 students is 42kg.
II. Average weight of boys is 43 kg.
III. Total weight of all girls together is 1144 kg.
(a) Any two of three
(b) All I, II & III
(c) I & II only
(d) II & III only
(e) Question cannot be answered even with information in all three statements.
(b) Ans. From statement I, Total weight of 60 students
= 60 × 42 = 2520 kg
From statement III,
Total weight of all the girls = 1144 kg
∴ Total weight of all the boys
= 2520 - 1144 = 1376 kg.
From statement II,
Number of boys = (1376/32) = 32
∴ Number of girls = 60 - 32 = 28
From statement III,
Average weight of girls
= (1144/28) = 40.86 kg
92. What is the selling price of the T.V. set if no discount is offered ?
I Profit earned was 20%.
II Had 10% discount been offered on selling price the profit would have been Rs. 1,200.
III. Cost price is Rs. 15,000.
(a) Any two of the three
(b) Only I & II
(c) Only I & III
(d) Only II & III
(e) None of these
(a) Ans. From statement I and III, S.P.can be obtained.
From statement II and III, Let the SP be Rs. x.
After 10% discount, SP = 90% of x = (1/10)x
∴ (9/10) x - 15000 = 1200
⇒ 9x = 150000 + 1200
⇒ x (162000/9) = Rs. 18000
From statement I and II, Let the CP = Rs. x
| ∴ S.P = | 120 100 | x = Rs. | 6x 5 |
Rs. (54x/50)
| ∴ | 54x 50 | - x = 1200 |
⇒ 54x - 50x = 50 × 1200
| ⇒ x | 50 × 1200 4 | = 15000 |
93. What is the speed of the train?
I. Length of the platform is 150% of the length of the train.
II. The train crosses the platform in 25 seconds.
III. The train crosses the signal pole in 19 seconds.
(a) All I, & III
(b) I and either II or III
(c) Only II& III
(d) Any two of the three
(e) Question cannot be answered even with the information in all three statements.
(e) Ans. From statement I,
Let the length of the train be x meter.
∴ Length of plat from
= (3x/2) meter
From statement II,
| Speed of train = | (3x/2) +x 25 |
| = | 5x 50 | x 10 | ............(i) |
But x is not know. Hence, we proceed.
From statement III,
Speed of train = (x/10) ........... (ii)
Clearly, we reach at no unique conclusion.
94. How many children are there in the class ?
I. 20% children are there in the class ?
II. 44 children can speak languages other than Hindi.
III. There are 30 boys in the class.
(a) All I,II & III
(b) Any two of the three
(c) II and either I or III
(d) I and II only
(e) None of these
(d) Ans. From statement I, 80% children speak language other than Hindi.
From statement II,
∴ Total Number of children
| = | 44 × 100 80 | = 55 |
95. What is the volume of the cylindrical tank ?
I. Area of the base is X square metes.
II. Height of the tank is Y metes.
III. Diameter of the base is equal to height of the tank.
(a) Only I & II
(b) Only II & III
(c) Only I & III
(b) All I,II & III
(e) Any two of the three
(e) Ans. From statement I and II, Volume of tank = XY cubic meters
From statement II and III,
Radius of base = (Y/2)
∴ Volume = π × (radius)2 × height
From statements I and II,
Area of base = sq. meters.
Hence, radius can be determined.
Height = 2 × radius.
Clearly, after knowing height and area of base, volume can be determined.
Directions (96-100): In each of these questions two equations numbered I & II are given. you have to solve both the equations and give answer
(a) if a < b
(b) if a > b
(c) if relationship between a & b to cannot be established
(d) if a > b
(e) if a < b
96. I. a2 + 5a + 6 = 0
II. b2 + 3b + 2 = 0
(e) Ans. I. a2 + 5a + 6 = 0
⇒ a2 + 2a + 3a + 6 = 0
⇒ a(a + 2) + 3 (a + 2) = 0
⇒ (a + 2) (a + 3) = 0
⇒ a = - 2 or -3
II. b2 + 3b + 2 = 0
⇒ b2 + 2b + b + 2 = 0
⇒ b(b + 2) + 1 (b + 2) = 0
⇒ (b + 1) (b + 2) = 0
⇒ b = -1 or - 2
Clearly, a ≤ b
97. I. 2a2 + 3a + 1 = 0
II. 12b2 + 7b + 1 = 0
(a) Ans. I. 2a2 + 3a + 1 = 0
⇒ 2a2 + 2a + a + 1 = 0
⇒ 2a(a + 1) + 1 (a + 1) = 0
⇒ (a + 1) (2a + 1) = 0
⇒ a = - 1 or - (1/2)
II. 12b2 + 7b + 1 = 0
⇒ 3b(4b + 1) + 1 (4b + 1) = 0
⇒ (3b + 1) (4b + 1) = 0
⇒ b = - (1/3) or - (1/4)
Clearly, a < b
98. I. a2 = 4
II. a2 = 9
(c) Ans. I. a2 = 4 ⇒ a = + 2
II. b2 = 9 ⇒ b = + 3
99. I. 6a2 - 25a + 25 = 0
II. 15b2 - 16b + 4 = 0
(b) Ans. I. 6a2 - 25a + 25 = 0
⇒ 6a2 - 15a - 10a + 25 = 0
⇒ 3a(2a - 5) - 5 (2a - 5) = 0
⇒ (3a - 5) (2a - 5) = 0
⇒ a = (5/3) or - (5/2)
II. 15b2 - 16b + 4 = 0
⇒ 15b2 - 10b - 6b + 4 = 0
⇒ 15b(3b - 2) -2 (3b - 2) = 0
⇒ (5b - 2) (3b - 2) = 0
⇒ b = (2/5) or (2/3)
Clearly, a > b
100. I. 4a2 - 20a + 21 = 0
II. 2b2 - 5b + 3 = 0
(d) Ans. I. 4a2 - 20a + 21 = 0
⇒ 4a2 - 14a - 6a + 21 = 0
⇒ 2a(2a - 7) -3 (2a - 7) = 0
⇒ (2a - 7) (2a - 3) = 0
⇒ a = (7/2) or (3/2)
II. 2b2 - 5b + 3 = 0
⇒ 2b2 - 3b - 2b + 3 = 0
⇒ b(2b - 3) - 1 (2b - 3) = 0
⇒ (b - 1) (2b - 3) = 0
⇒ b = 1 or 2/3
Clearly, a ≥ b
Directions (101-105) Study the following table carefully to answer these questions :
PERCENTAGE OF MARKS OBTANED BY SIX STUDENTS IN SIX DIFFERENT SUBJECTS
| Subject | Student | |||||
| P (70) | Q (80) | R (120) | S (125) | T (75) | U (150) | |
| A | 68 | 84 | 77 | 72 | 64 | 82 |
| B | 49 | 79 | 62 | 85 | 56 | 76 |
| C | 56 | 81 | 68 | 60 | 58 | 68 |
| D | 75 | 85 | 82 | 88 | 72 | 78 |
| E | 70 | 66 | 65 | 76 | 77 | 83 |
| F | 72 | 70 | 79 | 68 | 68 | 71 |
Note. Figure in bracket below each subject indicates the maximum marks allotted
101. Which is the average percentage of marks obtained by all students S, T & U together ?
(a) 65.6
(b) 66.5
(c) 66.8
(d) 65.4
(e) None of these
(e) Ans. Required average percentage
| = | 64 + 56 + 58 + 72 + 77 + 68 6 | = | 395 6 | = 65.8 |
102. Marks obtained by 'A' in subjects P, Q & R together are approximately what per cent of the marks obtained by F in subjects S, T & U together ?
(a) 80
(b) 75
(c) 85
(d) 105
(e) 115
(c) Ans. Marks obtained by A in ;
Subject P → 68% of 70
≈ (70/100) × 70 ≈ 49
Subject Q → (84/100) × 80 ≈ 67
Subject R → (77/100) × 120 ≈ 92
Total marks
= 49 + 67 + 92 = 208
Marks obtained by F in :
Subject S → 68% of 125
= (68/100) × 125 = 85
| Subject T → | 68 × 75 100 | = 51 |
| Subject U → | 71 × 150 100 | ≈ 105 |
∴ Total marks
= 85 + 51 + 105 = 241
∴ Required percentage
= (208/241) × 100 ≈ 85
103. What is the overall percentage of marks obtained by 'B' in all the subjects together (rounded off to two digits after decimal)?
(a) 70.02
(b) 72.51
(c) 67.83
(d) 71.50
(e) None of these
(c) Ans. Required overall percentage of marks
| = | 49 + 79+ 62+ 85+ 56+ 76 6 |
= (407/6) = 67.83
104. What are the marks obtained by D in all the subjects together ?
(a) 449.8
(b) 499.9
(c) 480
(d) 490
(e) None of these
(b) Ans. Marks obtained by D in :
| Subject P → | 75 × 70 100 | = 52.50 |
| Subject Q → | 85 × 80 100 | = 68 |
| Subject R → | 82 × 120 100 | = 98.40 |
| Subject S → | 88 × 125 100 | = 110 |
| Subject T → | 72 × 75 100 | = 54 |
| Subject U → | 68 × 150 100 | = 117 |
∴ Total marks = 5250 + 68 + 98.40 +110 + 54 + 117 = 499.9
105. What are average marks obtained out of 80 by all the six students together in subject 'Q' ?
(a) 68
(b) 77.5
(c) 88.57
(d) 62
(e) None of these
(d) Ans. Average percentage
| = | 84 + 79 + 81 + 85 + 66 + 70 6 |
= 465/6
∴ Required average marks
| = | 465 6 | % of 80 = | 465 × 80 6 × 100 | = 62 |
Directions (106-110) Study the following graph carefully to answer these questions :
NUMBER OF ITEMS PRODUCED (IN THOUSADS) AND COST ( IN RUPEES) PER HUNDRED ITEMS IN SIX COMPANIES
106. What will be the total cost of items produced by Company C?
(a) Rs. 32 lakhs
(b) Rs. 24 lakhs
(c) Rs. 27 lakhs
(d) Rs. 36 lakhs
(e) None of these
(c) Ans. Item product = 45000
Cost per hundred = Rs. 6000
∴ Total cost
| = Rs. | 45000 × 6000 100 |
107. What is the average cost per hundred items for all the given companies ?
(a) Rs. 4183 (2/3)
(b) Rs. 4283 (1/3)
(c) Rs. 43331 (2/3)
(d) Rs. 4333 (1/3)
(e) None of these
(e) Ans. Required average cost
| = Rs. | 4500 + 3500 + 6000 + 5500 + 5000 + 4000 6 |
= Rs. (28500/6) = Rs. 4750
108. What is total cost of the items produced by companies A & B together ?
(a) Rs. 17.50 lakhs
(b) Rs. 33.25 lakhs
(c) Rs. 15.75 lakhs
(d) Rs. 32.75 lakhs
(e) None of these
(b) Ans. Cost of items produced by company :
| A → Rs. | ( | 35000 × 4500 100 | ) | = Rs. 15.75 lakhs |
| B → Rs. | ( | 50000 × 3500 100 | ) | = Rs. 17.5 lakhs |
∴ Total cost
= Rs. (15.75 + 17.5) lakhs
= Rs. 33.25 lakhs.
109. What was the total number of items produced by all the companies together ?
(a) 28500
(b) 258000
(c) 25800
(d) 2850000
(e) None of these
(e) Ans. Total number of items produced by all companies = (35 + 50 + 45 + 30 + 40 + 60) thousands = 260000
110. If the number of items produced by Company 'D' increase by 30%, what will be the total cost of items produced ?
(a) Rs. 21.45 lakhs
(b) Rs. 22. 45 lakhs
(c) Rs. 24.25 lakhs (d) 22.25 lakhs
(e) None of these
(a) Ans. = New number of items produced by company
| D = | ( | 130 100 | × 30 | ) | thousands |
= 39 thousands
Cost per hundreds = Rs. 5500
∴ Total cost
| = Rs. | ( | 39000 × 5500 100 | ) |
= Rs. 21.45 lakhs.
Directions (111-115) : These questions are based on the following information :
Children in a class play only one or two or all the three games badminton football and cricket. 5 children play only cricket, 8 children play only football and 7 children play only badminton 3 children play only two games badminton and football, 4 children play only two games cricket and football and another 4 children play only two games badminton and cricket. 2 children play all the three grams.
Venn - Diagram for questions 111 - 115:
111. In all how many children play football ?
(a) 8
(17)
(c) 15
(d) 14
(e) None of these
(b) Ans. Number of football players = 3 + 2 + 4 + 8 = 17
112. How many children play badminton as.well as cricket ?
(a) 9
(b) 10
(c) 4
(d) 6
(e) None of these
(d) Ans. Number of badminton as well as cricket players = 4 + 2 = 6
113. Total how many children are there in the class ?
(a) 33
(b) 31
(c) 36
(d) 35
(e) None of these
(a) Ans. Number of children = 7 + 3 + 2 + 4 + 8 + 4 + 5 = 33
114. In all how many children play badminton ?
(a) 14
(b) 17
(c) 12
(d) 13
(e) None of these
(e) Ans. Number of badminton players = 7 + 3 + 2 + 4 = 16
115. How many children play football as well cricket ?
(a) 7
(b) 4
(c) 6
(d) 15
(e) None of these
(c) Ans. Number of football as well as cricket players = 2 + 4 = 6
DETAILS ABOUT THE DISTRIBUTION OF EMPLOYEES EXPENDITURE OF AN ORGANIZATION (DISTRYBUTED PROPORTIONATELY ACROSS THE DEPARTMENTS)
Annual expenditure on different items
Total Expenditure = Rs.12 crore's
DEPARTMENTWISE DISTRIBUTION OF EMPLOYEES
Total number of employees = 1200
116. What was the total expenditure on Accounts Department ?
(a) Rs. 16.8 lakhs
(b) Rs. 1,680 lakhs
(c) Rs. 18.6 millions
(d) Rs. 16.8 millions
(e) None of these
(d) Ans. Total expenditure on Accounts department = 14 % 0f Rs. 12 crore's
| = Rs | 14×12 100 | crore's |
= Rs. 1.68 crore's
= Rs. 16.8 millions
117. What was per employee expenditure on Medical ?
(a) Rs. 12,000
(b) Rs. 13,000
(c) Rs. 12,500
(d) Rs. 13.500
(e) None of these
(b) Ans. Total expenditure on medical = 13% of Rs. 12 crore's
| = Rs. | 13×12 100 | crore's |
= Rs. 1.56 crore's
= Rs. 15600000
∴ Expenditure per employee
= 15600000/1200
= Rs. 13000
118. What was the total expenditure on salary of employees in Marketing Department ?
(a) Rs. 6.12 lakhs
(b) Rs. 61.2 lakhs
(c) Rs. 6.12 millions
(d) Rs. 176 lakhs
(e) None of these
(c) Ans. Total expenditure on salary of employees = 30% of Rs. 12 crore's
| = Rs. | 30 × 120000000 100 |
= Rs. 36000000
∴ Expenditure on Marketing employees
= 17% of 36000000
= Rs. (17 × 360000)
= Rs. 6120000
= Rs. 6120000
= Rs. 6.12 millions
119. What was the amount spent on Electricity ?
(a) Rs. 132 millions
(b) Rs. 13.2 lakhs
(c) Rs. 126 millions
(d) Rs. 12.6 lakhs
(e) None of these
(e) Ans. Amount spent on electricity = 11% of Rs. 12 crore's
| = Rs. | 11 × 120 100 | millions |
= Rs. 13.2 millions
120. What was the expenditure on telephone for employees in computer Department ?
(a) Rs. 11.52 lakhs
(b) Rs. 11.52 millions
(c) Rs. 10.72 lakhs
(d) Rs. 10.72 millions
(e) None of these
(a) Ans. Expenditure on telephone in computer department = 12% of 8% Rs. 12 crore's
| = Rs. | 12 100 | × | 8 100 | 120 millions |
= Rs. 1.152 millions
= Rs. 11.52 lakhs
Directions (121-125) : Study the following table carefully to answer these questions.
SALES OF PRODUCT (IN MILLION TONS) FOR SIX STATES OVER THE YEARS
| Year | STATE | |||||
| A | B | C | D | E | F | |
| 1998 | 25 | 45 | 38 | 52 | 47 | 55 |
| 1999 | 32 | 39 | 40 | 55 | 46 | 67 |
| 2000 | 41 | 50 | 43 | 57 | 39 | 64 |
| 2001 | 37 | 48 | 43 | 58 | 32 | 72 |
| 2002 | 28 | 53 | 46 | 62 | 37 | 58 |
| 2003 | 43 | 55 | 49 | 63 | 42 | 62 |
121.. If the cost of product per thousand tons in 1998 was Rs. 1.8 lakhs, what was the cost of average sales for the given States in that year
(a) Rs. 7860000 lakhs
(b) Rs. 786 laksh
(c) Rs. 7860 lakhs
(d) Rs. 78600 lakhs
(d) None of these
(d) Ans. Average sales
| = | ( | 25 + 45 + 38 + 52 + 47 + 55 6 | ) | millions tons |
= (262/6) = 43.7 million tons
∴ Required cost
= Rs. (1.8 × 43700) lakhs
= Rs. 78660 lakhs
= Rs. 78600 lakhs
122. Total sales in year 2000 were what per cent of the total sales in year 2003? (rounded off to two digits after decimal)
(a) 93.63
(b) 92.65
(c) 106.80
(d) 93.23
(e) None of these
(a) Ans. Total sales in year 2000 = (41 + 50 + 43 + 57 + 39 + 64) million tons
= 264 million tons
Total sales in 2003
= (43 + 55 + 49 + 63 + 42 + 62) million tons
= 314 million tons
∴ Required percentage
| = | 294 314 | × 100 = 93.63 % |
123. What was the percentage increase in total sales in 2003 from 1998 ? (rounded off to nearest integer)
(a) 19
(b) 20
(c) 16
(d) 17
(e) None of these
(a) Ans. Total sales in year 1998 = (25 + 45 + 38 + 52 + 47 + 55) million tons
= 262 million tons
Total sales in 2003
= 314 million tons
∴ % Increase
| = | 314 - 262 262 | × 100 |
= 19.85 ≈ 20
124. Approximately what was the ratio between average sales of States B & C respectively ?
(a) 31:25
(b) 26:31
(c) 29:24
(d) 26:29
(e) 29:26
(e) Ans. Average sales of state :
| B → | 45 + 39+ 50 + 48 + 53 + 55 6 |
| = | 290 6 | millions tons |
∴ Required ratio
| = | 290 6 | : | 259 6 | ≈ 290 : 260 |
≈ 90 : 60
125. For which of the following years the percentage increase/decrease in sales from the previous years was highest for State 'E'?
(a) 1999
(b) 2000
(c) 2001
(d) 2002
(e) 2003
(c) Ans. Percentage decrease in 1999
| = | 47 - 46 47 | × 100 |
= 15.22
Percentage decrease in 2001
| = | 37 - 32 32 | × 100 |
= 17.95
Percentage increase in 2002
| = | 37 - 32 32 | × 100 |
= 15.625
Percentage increase in 2003
| = | 42 - 37 37 | × 100 |
= 13.51
Hence, the percentage decrease was maximum in the year 2001.
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