Q66. What should come in place of question mark in the following question?
(4 × 4)3 ÷ (4)5 × (2 × 8)2 = (16)?
1. 5
2. 6
3. 3
4. 8
5. None of these
Q67. What should come in place of question mark in the following question?
√529 ÷ 46 × 6.4 + (8)3 – 252= ?
1. 265.1
2. 246.1
3. 256.4
4. 285.4
5. None of these
Q68. What should come in place of question mark in the following question?
14% of 250 + ? % of 300 = 125
1. 22
2. 24
3. 36
4. 30
5. None of these
Q69. Find the approximate value of the expression:
150.08% of 1168.03 ÷ 7.96 + 18.956 = 29.231 ×?
1. 10
2. 9
3. 8
4. 6
5. 11
Q70. Find the approximate value of the expression:
179.86 × 18.01 - ?2 = 2984
1. 16
2. 12
3. 14
4. 18
5. 19
Q71-72) Directions: Study the following table to answer these questions.
Number of Officers in Various Departments of an Organization called MNMO in Different Scales
Q71. What percentage of Scale IV officers are deployed in ‘operations’ department?
1. 25.5
2. 26.04
3. 26.69
4. 25
5. 26.40
Q72. Out of the total number of employees in ‘Personnel’ department, approximately what per cent of employees are in Scale II?
1. 20.52
2. 25.17
3. 22.29
4. 20
5. 24.36
Q73. In public Relations department, the number of employees in Scale II is less than that in Scale I by what per cent? (Rounded off to two digits after decimal)
1. 14.67
2. 13.65
3. 13.89
4. 14.27
5. 14.03
Q74. Total number of employees in Scale VI is what per cent of the total number of employees in Scale I?
1. 2.5
2. 3.25
3. 3.6
4. 3.88
5. 4.12
Q75. What is the ratio between the total number of employees in Scale III and Scale IV respectively? 1. 2 : 1
2. 9 : 1
3. 1 : 2
4. 2 : 3
5. None of these
Q76. In a partnership, profit earned by Matt is Rs. 300 less than that earned by Allan. Allan invested three fourth of the money that Matt invested and for twice the time as Matt. How much profit Allan and Matt earned in total?
1. Rs. 600
2. Rs. 1200
3. Rs. 1500
4. Rs. 1800
5. Rs. 900
Q77. Nancy’s brother is 5 years elder to her. The sum of their ages is three fourth of their father’s age. After 4 years, father’s age will be twice the age of Nancy’s brother. What will be the age of Nancy 7 years from now?
1. 14 years
2. 16 years
3. 18 years
4. 21 years
5. 24 years
Q78. A 140 metre long train crosses another 210 metre long train running in the opposite direction in 7 seconds. If the speed of the first train is 70 km/hr, what is the speed of the second train in km/h?
1. 100
2. 110 3. 140
4. Cannot be determined
5. None of these
Q79. What will come in place of the question mark (?) in the following number series?
24 50 98 198 394 (?)
1. 800
2. 788
3. 790
4. 810
5. None of these
Q80. What will come in place of the question mark (?) in the following number series?
100 49 47 67.5 131 (?)
1. 300
2. 322.5
3. 350
4. 327.5
5. None of these
Q81. What will come in place of the question mark (?) in the following number series?
33 43 65 99 145 (?)
1. 201
2. 203
3. 205 4. 211
5. None of these
Q82. In the following number series, only one number is wrong. Find out the wrong number.
19 68 102 129 145 154
1. 154
2. 129
3. 145
4. 102
5. None of these
Q83. Which one of the following numbers in the series is wrong?
6, 26, 81, 166, 172
1. 81
2. 26
3. 6
4. 166
5. 172
Q84. In what ratio should water be mixed with soda costing Rs. 13 per litre so as to make a profit of 20% by selling the diluted liquid at Rs. 14.40 per litre?
1. 10 : 1
2. 11 : 1
3. 1 : 11
4. 12 : 1
5. 1 : 12
Q85. Shankar bought two boxes for Rs. 1035, sold first box at a profit of 25% and the second box at a loss of 18%. What will be the cost prices of the two boxes respectively, if their selling price are the same?
1. Rs. 625, Rs. 410
2. Rs. 710, Rs. 325
3. Rs. 410, Rs. 625
4. Rs. 755, Rs. 280
5. Rs. 325, Rs. 710
Q86. The average age of family of 4 members 5 years ago was 16 years. With the birth of a baby, the average of family is 1 more than what it was 5 years ago. Find the age of the baby.
1. 1 year
2. 2 years
3. 3 years
4. 4 years
5. 5 years
Q87. In the following question, two equations are given. You have to solve these equations and find out the values of x and y and determine the relation between them.
I. 16x2 + 20x + 6 = 0
II. 10y2 + 38y + 24 = 0
1. x < y 2. x > y
3. x ≥ y
4. x ≤ y
5. x = y or the relationship cannot be established
Q88. In the following question, two equations numbered I and II are given. You have to solve both the equations and determine the relation between x and y.
I. x2 + 11x + 30 = 0
II. y2 + 7y + 12 = 0
1. x > y
2. x ≥ y
3. x < y 4. x ≤ y 5. x = y or the relationship cannot be established. Q89. In the following question, two equations numbered I and II are given. You have to solve both the equations and determine the relation between x and y. I. x2 – 16 = 0
II. y2 – 9y + 20 = 0
1. x > y
2. x ≥ y
3. x < y 4. x ≤ y 5. x = y or the relationship cannot be established. Q90. In the following question, two equations numbered I and II are given. You have to solve both the equations and determine the relation between x and y. I. x – √121 = 0 II. y2 – 121 = 0
1. x > y
2. x ≥ y
3. x < y 4. x ≤ y 5. x = y or the relationship cannot be established. Q91. In the following question, two equations numbered I and II are given. You have to solve both the equations and determine the relation between x and y. I. x2- x -12 = 0
II. y2 + 5y + 6 = 0
1. x > y
2. x ≥ y
3. x < y 4. x ≤ y 5. x = y or the relationship cannot be established. Q91-96) Directions: Study the following Graph carefully and answer the question given below: Total Number of Arts, Science and Commerce Students in Various Colleges
Q92. What is the average numbers of students (in thousands) taking up Commerce from all the six colleges together? (Rounded off to two digits after decimal)
1. 44.90
2. 43.33
3. 51.33
4. 38.09
5. 30.83
Q93. What is the respective ratio of the number of the students taken Commerce to the number of students taking Science in college E?
1. 14 : 19
2. 12 : 17
3. 17 : 19
4. 15 : 11
5. 13 : 18
Q94. How many candidates (in thousands) have taken Arts from all the colleges?
1. 265
2. 237.5
3. 185
4. 260.5
5. 232
Q95. In college B, the students taking Science is what percent of the total number of students taking Arts, Science and Commerce in college B?
1. 45
2. 55
3. 25
4. 70
5. 37.5
Q96. The number of students taking Commerce in college B, is how many thousand fewer than the number of student taking Commerce in college A and college C together?
1. 12.5
2. 27.5
3. 50
4. 10
5. 32.5
Q97. Aditya can do 50% more work than Radhika can do in the same time. Radhika alone do a piece of work in 30 hours. Aditya with the help of Radhika, can finish twice the same work in how many hours?
1. 24
2. 20
3. 22.5
4. 20.5
5. None of these
Q98. A rectangular sheet of length 44 cm and breadth 10 cm is taken. It is folded along its length in a circle to form a right circular cylinder. What will be the volume of the cylinder hence formed (in cubic cm)? (Take π = 22/7)
1. 1540
2. 1400
3. 1280
4. 1640
5. 1720
Q99. In how many ways the letters of word TESTBOOK can be arranged so that there are vowels at first and last positions?
1. 360
2. 540
3. 720
4. 1080
5. 1440
Q100. There is 60% increase in an amount in 4 years at simple interest. What will be the compound interest of Rs. 48,000 after 2 years at the same rate?
1. 10080
2. 15480
3. 14850
4. 33120
5. None of these
Answers&Explanations
66. Laws of Indices:-
1 -: a m × a n = a {m+n}
2 -: a m ÷ a n = a {m-n}
3 -: [(am ) n ] = a mn
4 -: (a) (1/m) = m √a
5 -: (a) (-m) =1/a m
6 -: (a) (m/n) = n √a m
7 -: (a) 0 = 1
Now, the given expression:
(4 × 4) 3 ÷ (4) 5 × (2 × 8) 2 = (4) ?
⇒ (2 2 × 2 2 ) 3 ÷ (2 2 ) 5 × (2 × 2 4 ) 2 = (16) ?
⇒ (2) 12 ÷ (2) 10 × (2) 8 = (2) ? × 4
⇒ (2) ? = 2 10 ÷ 4
⇒ ? = 2.5
67. Follow BODMAS rule to solve this question, as per the order given below,
Step -1-Parts of an equation enclosed in 'Brackets' must be solved first,
Step -2-Any mathematical 'Of' or 'Exponent' must be solved next,
Step -3-Next, the parts of the equation that contain 'Division' and 'Multiplication' are calculated,
Step -4-Last but not least, the parts of the equation that contain 'Addition' and 'Subtraction' should be calculated.
Now, the given expression,
√529 ÷ 46 × 6.4 + (8) 3 – 252 = ?
⇒ 23 ÷ 46 × 6.4 + (8) 3 – 252 = ?
⇒ ? = 3.2 + 512 – 252
⇒ ? = 263.2
68 . Follow BODMAS rule to solve this question, as per the order given below,
Step -1-Parts of an equation enclosed in 'Brackets' must be solved first,
Step -2-Any mathematical 'Of' or 'Exponent' must be solved next,
Step -3-Next, the parts of the equation that contain 'Division' and 'Multiplication' are calculated,
Step -4-Last but not least, the parts of the equation that contain 'Addition' and 'Subtraction' should be calculated.
Now, the given expression,
14% of 250+ ?% of 300 = 125
⇒ 35 + ?% of 300 = 125
⇒ 3× ? = 90
⇒ ? = 30
69. 150.08% of 1168.03 ÷ 7.96 + 18.956 = 29.231 ×?
Here, 150.08 ≈ 150; 1168.03 ≈ 1168; 7.96 ≈ 8; 18.956 ≈ 19; 29.231 ≈ 29
Taking the approximate values we have,
150% of 1168 ÷ 8 + 19 = 29 × ?
⇒ 1168 × 1.5 ÷ 8 + 19 = 29 × ?
⇒ 219 + 19 = 29 × ?
⇒ 29 × ? = 238
⇒ ? = 238/29 = 8.20 ≈ 8
Hence, the value is 8.
70 . 179.86 × 18.01 - ? 2 = 2984
Here, 179.86 ≈ 180 and 18.01 ≈ 18
Taking the approximate values, we have
⇒ 180 × 18 - ? 2 = 2984
⇒ 3240 - ? 2 = 2984
⇒ ? 2 = 3240 – 2984
⇒ ? 2 = 256
⇒ ? = 16
71 . From the table,
The total number of employees in Scale IV = 45 + 125 + 155 + 65 + 35 + 55 = 480
The number of employees are deployed in ‘operations’ department in Scale IV = 125
The required percentage = 125 480 × 100 = 26.04
∴ 26.04% of Scale IV officers are deployed in ‘operations’ department.
7 2. From the table,
Total number of employees in ‘Personnel’ department = 220 + 125 + 85 + 45 + 30 + 8 = 513
In Personnel department, the number of employees in Scale II = 125
The required p ercentage = 125 513 × 100 = 24.36
∴ Out of the total number of employees in ‘Personnel’ department, approximately 24.36% employees are in Scale II.
73 . From the table,
In public Relations department, the number of employees in Scale II = 155
In public Relations department, the number of e mployees in Scale I = 180
So, in public Relations department, the number of employees in Scale II is (180 – 155) = 25 less than that in scale I.
The required percentage = 25 180 × 100 = 13.89
∴ In public Relations department, the number of employees in Scale II is less than that in Scale I by 13.89%.
74 . From the table,
Total number of employees in Scale VI = 8 + 15 + 30 + 20 + 10 + 8 = 91
Total number of employees in Scale I = 220 + 625 + 650 + 350 + 320 + 180 = 2345
The required percentage = 91 2345 × 100 = 3.88
∴ Total number of employees in Scale VI is 3.88% of the total number of employees in Scale I.
75 . From the table,
The total number of employees in Scale III = 85 + 280 + 225 + 150 + 100 + 120 = 960
The total number of employees in Scale IV = 45 + 125 + 15 5 + 65 + 35 + 55 = 480
∴ The ratio between the total number of employees in Scale III and Scale IV respectively = 960:480 = 2:1
76 . Allan invested three fourth of the money that Matt invested and for twice the time as Matt.
Suppose Matt invested Rs. T fo r N years. So, Allan invested Rs. 3T/4 for 2N years.
⇒ Ratio of amounts invested multiplied by time for Matt and Allan = (TN) : (6TN/4) = 2 : 3
Let profit earned by Matt be Rs. P and that by Allan be Rs. (P + 300).
We know, ratio of amounts invested multip lied by time is same as ratio of profits.
⇒ P/(P + 300) = 2/3
⇒ 3P = 2P + 600
⇒ P = 600
∴ Total profit = P + P + 300 = Rs. 1500
77 . Let the age of Nancy be N years.
⇒ Age of Nancy’s brother = (N + 5) years
The sum of their ages is three fourth of their fa ther’s age.
⇒ Father’s age is 4/3 of sum of ages of Nancy and Nancy’s brother.
⇒ Father’s age = (4/3)(N + N + 5) years = (4/3)(2N + 5) years
After 4 years, father’s age will be twice the age of Nancy’s brother.
⇒ [(4/3)(2N + 5) + 4] = 2(N + 5 + 4)
⇒ 8N + 20 + 12 = 6N + 54
⇒ N = 22/2 = 11
∴ Age of Nancy 7 years from now = (11 + 7) years = 18 years
78 . Let the speed of the 210 m long train = x km/h
As the trains are travelling in opposite directions,
Relative speed of the first train w.r.t the second train = Speed of train 1 + Speed of train 2
⇒ Relative speed = (70 + x)km/h = (70 + x) × 5/18 m/s ---(1)
Thus,
Distance covered by train 1 = Length of Train 1 + Length of train 2
⇒ Distance covered by train 1 = 140 + 210 = 350 m
Time taken = 7 seconds
∴ Relative speed × (Time taken) = Total distance covered
⇒ (70 + x) × (5/18) × 7 = 350
⇒ (70 + x) × (5/18) = 50
⇒ (70 + x)/18 = 10
⇒ x = 110 km/h
Thus,
The speed of the second train is 110 km/h
79 . The pattern of given series is:
→ 24,
→ 50 = (24 × 2) + 2,
→ 98 = (50 × 2) – 2,
→ 198 = (98 × 2) + 2,
→ 394 = (198 × 2) – 2,
→ ? = (394 × 2) + 2,
→ ? = 790
80 . The pattern of given series is:
→ 100,
→ 49 = (100 × 0.5) – 1,
→ 47 = (49 × 1) – 2,
→ 67.5 = (47 × 1.5) – 3,
→ 131= (67.5 × 2) – 4,
→ ? = (131 × 2.5) – 5,
→ ? = 322.5
81 . The pattern of given series is:
→ 33,
→ 43 = 33 + 10,
→ 65 = 43 + (10 + 12),
→ 99 = 65 + (10 + 24),
→ 145 = 99 + (10 + 36),
→ ? = 145 + (10 + 48),
→ ? = 203
82 . The pattern of given series is:
→ 68 – 19 = 49 (7 x 7)
→ 102 – 68 = 34 (*36 = 6 x 6 )
→ 129 – 102 = 27 (*25 = 5 x 5)
→ 145 - 129 = 16 (4 x 4)
→ 154 – 145 = 9 (3 x 3)
If we replace 102 by 104, then 104 – 68 = 36 & 129 – 104 = 25.
So, the wrong number is 102.
8 3. The pattern of given series is:
→ 6 = 1 × 5 + 1
→ 26 = 6 × 4 + 2
→ 81 = 26 × 3 + 3
→ 166 = 81 × 2 + 4
→ 171 = 166 × 1 + 5
Hence, 172 is wrong in the series.
84 . Let’s assume that x litres of water is mixed with y litres of soda.
Cost price of soda = Rs. 13/Litre
Cost of y litre soda = Rs. 13y
Total quantity of the mi xture = (x + y) litre
The cost price of the mixture = Rs. 13y [∵ Water is assumed available free of cost]
The selling price of (x + y) litres of mixture = Rs. 14.40 × (x + y)
The cost price of the mixture = SellingPrice×100 100+Profit% = 13푦 ⇒ (14.40) × (x + y) × 100 100 + 20 = 13y
⇒ 1440x + 1440y = 13y × 120
⇒ 1440x = 1560y – 1440y
⇒ 1440x = 120y
⇒ x/y = 1/12
∴ Water and soda should be mixed in 1 : 12 ratio.
85 . Let’s assume the cost price of first box to be Rs. x
And, the cost price of second box to be Rs. y
A ccording to the given information,
⇒ x + y = 1035 -------Equation (1)
∵ first box is sold at 25% profit,
Selling price of first box = x + (25% of x) = 1.25x
∵ second box is sold at 18% loss,
Selling price of second box = y – (18% of y) = 0.82y
As per the given information, selling prices of both the boxes are same.
⇒ 1.25x = 0.82y
⇒ x = 0.82y/1.25
Substituting the value of x in Equation (1), we get, ⇒ 82푦 125 + 푦 = 1035 ⇒ 207푦 125 = 1035 ⇒ 푦 = 1035 × 125 207 = 625
Substituting i n Equation (1), we get,
x = 1035 – 625 = 410
∴ the selling prices of two boxes are Rs. 625 and Rs. 410 respectively.
86 . We know that, average = Sum of all quantities/Number of quantities
Average age of family = Sum of ages of all family members/Number o f members
∴ 16 = Sum of ages of 4 family members five years ago/4
⇒ Sum of ages of 4 family members 5 years ago = 4 × 16 = 64
In five years, each of these 4 members’ ages have increased by 5.
∴ Sum of present ages of 4 family members = Sum of ages of 4 family members 5 years ago + (5 × 4)
∴ Sum of present ages of 4 family members = 64 + 20 = 84
Let the age of the baby be x years.
∴ Sum of present ages of all family members including baby = 84 + x
According to the information given in the question, average p resent age of family members including baby, is 1 more than what it was 5 years ago excluding baby, i.e. 16 + 1 = 17
∴ Average = 17
⇒ 17 = (84 + x)/5
⇒ 84 + x = 17 × 5 = 85
⇒ x = 1
∴ the age of the baby is 1 year.
87 . I. 16x 2 + 20x + 6 = 0
⇒ 8x 2 + 10x + 3 = 0 [Dividing both sides by 2 ]
⇒ 8x 2 + 6x + 4x + 3 = 0
⇒ 2x(4x + 3) + 1(4x + 3) = 0
⇒ (4x + 3)(2x + 1) = 0
Then, x = - ¾ or x = - ½
II. 10y 2 + 38y + 24 = 0
⇒ 5y 2 + 19y + 12 = 0 [Dividing both sides by 2 ]
⇒ 5y 2 + 15y + 4y + 12 = 0
⇒ 5y(y + 3) + 4(y + 3) = 0
⇒ (y + 3)(5y + 4) = 0
Then, y = - 3 or y = - 4/5
So, when x = - ¾ , x > y for y = - 3 and x > y for y = - 4/5
And when x = - ½ , x > y for y = - 3 and x > y for y = - 4/5
∴ We can clearly observe that x > y.
88 . I. x 2 + 11x + 30 = 0
⇒ x 2 + 5x + 6x + 30 = 0
⇒ x(x + 5) + 6(x + 5) = 0
⇒ (x + 5)(x + 6) = 0
Then, x = - 5 or x = - 6
II. y 2 + 7y + 12 = 0
⇒ y 2 + 4y + 3y + 12 = 0
⇒ y(y + 4) + 3(y + 12) = 0
⇒ (y + 4)(y + 3) = 0
Then, y = - 4 or y = - 3
So, when x = - 5, x < y for y = - 4 and x < y for y = - 3 And when x = - 6, x < y for y = - 4 and x < y for y = - 3 ∴ We can clearly see that x < y.
90 . I. x - √121= 0
⇒ x = √121 = 11
Then, x = + 11
II. y 2 – 121 = 0
⇒ y 2 = 121
⇒ y = ± 11
Then, y = + 11 or y = - 11
So, when x = + 11, x = y for y = + 11 and x > y for y = - 11
∴ So, we can observe that x ≥ y.
91 . I. x 2 – x – 12 = 0
⇒ x 2 – 4x + 3x – 12 = 0
⇒ x(x – 4) + 3(x – 4) = 0
⇒ (x – 4)(x + 3) = 0
Then, x = + 4 or x = - 3
II. y 2+ 5y + 6 = 0
⇒ y 2 + 3y + 2y + 6 = 0
⇒ y(y + 3) + 2(y + 3) = 0
⇒ (y + 3)(y + 2) = 0
Then, y = - 3 or y = - 2
So, when x = + 4, x > y for y = - 3 and x > y for y = - 2
And when x = - 3, x = y for y = - 3 and x < y for y = - 2 ∴ So, we can observe that no clear relationship cannot be determined between x and y.
92 . From the given data,
Number of students taking Commerce from all six colleges (in thousands) = 185
Total number of colleges = 6
Average number of students taking Commerce (in thousands) = 185/6 = 30.83
93 . From the given data,
Number of students taken Commerce in college E (in thousands) = 37.5
Number of students taken Science in college E (in thou sands) = 27.5
Required ratio = 37.5 : 27.5 = 15 : 11
94 . From the given data,
95 . From the given data,
Number of students taking Science in college B (in thousands) = 45
Total Number of students in college B (in thousands) = 120
Required % = 45 120 × 100 = 37.5%
96 . From the given data,
Number of students taking Commerce in college B (in thousand) = 25
Number of students taking Commerce in college A (in thousand) = 40
Number of students taking Commerce in college C (in thousand) = 17.5
Total number of students taking Commerce in college A and C together (in thous and) = 17.5 + 40 = 57.5
Required number of students (in thousand) = 57.5 – 25 = 32.5
97. Aditya can do 50% more work than Radhika in the same time.
⇒ Aditya’s efficiency is 1.5 times of Radhika’s efficiency
Radhika alone do a piece of work in 30 hrs.
∴ Part of work done by Radhika in one hour = 1/30
∴ Part of work done by Aditya in one hour = (2/3) × (1/30) = 1/20
Working together, part of work finish ed by both in one hour = 1 30 + 1 20 = 2+3 60 = 1 12
∴ they’ll take 12 hours to finish the entire work.
⇒ In order to finish twice the original work, they’ll require 12 × 2 = 24 hours.
98 . If rectangular sheet is folded along its length to from a right circular cylinder, the height of cylinder will be same as breadth of rectangle, and circumference of base will be same as length of rectangle.
If height of cylinder is H cm, and radius is R cm, then
H = breadth = 10 cm
And, 2πR = length = 44 cm
⇒ 2 × (22/7) × R = 44
⇒ R = 7 cm
∴ Volume of cylinder = πR 2 H = (22/7) × 7 × 7 × 10 = 1540 cubic cm
99 . There are 3 vowels in word TESTBOOK, E, O and O. If there are vowels at first and last position, this is possible in 3 ways.
E at first and O at last, O at first a nd E at last, and O at both first and last.
In each of the three cases, the remaining six letters that have to come in between will be different except that T will be occur twice.
⇒ Remaining 6 letters can be arranged in 6!/2! Ways, i.e., 360 ways.
∴ Total number of ways in which this can be done = 3 × 360 = 1080
100 . Simple Interest = (P × R × T)/100 Where, P = Principal, R = % rate of interest, T = time period in year
As per given information, Simple interest = 60% of principal
∴ 0.6P = (P × R × 4)/100
⇒ R = 15
Now let's find out the compound interest of Rs. 48,000 after 2 years at 10% P = Rs. 48,000 T = 2 years
(4 × 4)3 ÷ (4)5 × (2 × 8)2 = (16)?
1. 5
2. 6
3. 3
4. 8
5. None of these
Q67. What should come in place of question mark in the following question?
√529 ÷ 46 × 6.4 + (8)3 – 252= ?
1. 265.1
2. 246.1
3. 256.4
4. 285.4
5. None of these
Q68. What should come in place of question mark in the following question?
14% of 250 + ? % of 300 = 125
1. 22
2. 24
3. 36
4. 30
5. None of these
Q69. Find the approximate value of the expression:
150.08% of 1168.03 ÷ 7.96 + 18.956 = 29.231 ×?
1. 10
2. 9
3. 8
4. 6
5. 11
Q70. Find the approximate value of the expression:
179.86 × 18.01 - ?2 = 2984
1. 16
2. 12
3. 14
4. 18
5. 19
Q71-72) Directions: Study the following table to answer these questions.
Number of Officers in Various Departments of an Organization called MNMO in Different Scales
| Dept. | Operations | Systems | Accounts | Maintenance | Public Relations | Personnel |
|---|---|---|---|---|---|---|
| 1 | 220 | 625 | 650 | 350 | 320 | 180 |
| 2 | 125 | 425 | 575 | 290 | 260 | 155 |
| 3 | 85 | 280 | 225 | 150 | 100 | 120 |
| 4 | 45 | 125 | 155 | 65 | 35 | 55 |
| 5 | 30 | 50 | 75 | 40 | 25 | 45 |
| 6 | 8 | 15 | 30 | 20 | 10 | 18 |
Q71. What percentage of Scale IV officers are deployed in ‘operations’ department?
1. 25.5
2. 26.04
3. 26.69
4. 25
5. 26.40
Q72. Out of the total number of employees in ‘Personnel’ department, approximately what per cent of employees are in Scale II?
1. 20.52
2. 25.17
3. 22.29
4. 20
5. 24.36
Q73. In public Relations department, the number of employees in Scale II is less than that in Scale I by what per cent? (Rounded off to two digits after decimal)
1. 14.67
2. 13.65
3. 13.89
4. 14.27
5. 14.03
Q74. Total number of employees in Scale VI is what per cent of the total number of employees in Scale I?
1. 2.5
2. 3.25
3. 3.6
4. 3.88
5. 4.12
Q75. What is the ratio between the total number of employees in Scale III and Scale IV respectively? 1. 2 : 1
2. 9 : 1
3. 1 : 2
4. 2 : 3
5. None of these
Q76. In a partnership, profit earned by Matt is Rs. 300 less than that earned by Allan. Allan invested three fourth of the money that Matt invested and for twice the time as Matt. How much profit Allan and Matt earned in total?
1. Rs. 600
2. Rs. 1200
3. Rs. 1500
4. Rs. 1800
5. Rs. 900
Q77. Nancy’s brother is 5 years elder to her. The sum of their ages is three fourth of their father’s age. After 4 years, father’s age will be twice the age of Nancy’s brother. What will be the age of Nancy 7 years from now?
1. 14 years
2. 16 years
3. 18 years
4. 21 years
5. 24 years
Q78. A 140 metre long train crosses another 210 metre long train running in the opposite direction in 7 seconds. If the speed of the first train is 70 km/hr, what is the speed of the second train in km/h?
1. 100
2. 110 3. 140
4. Cannot be determined
5. None of these
Q79. What will come in place of the question mark (?) in the following number series?
24 50 98 198 394 (?)
1. 800
2. 788
3. 790
4. 810
5. None of these
Q80. What will come in place of the question mark (?) in the following number series?
100 49 47 67.5 131 (?)
1. 300
2. 322.5
3. 350
4. 327.5
5. None of these
Q81. What will come in place of the question mark (?) in the following number series?
33 43 65 99 145 (?)
1. 201
2. 203
3. 205 4. 211
5. None of these
Q82. In the following number series, only one number is wrong. Find out the wrong number.
19 68 102 129 145 154
1. 154
2. 129
3. 145
4. 102
5. None of these
Q83. Which one of the following numbers in the series is wrong?
6, 26, 81, 166, 172
1. 81
2. 26
3. 6
4. 166
5. 172
Q84. In what ratio should water be mixed with soda costing Rs. 13 per litre so as to make a profit of 20% by selling the diluted liquid at Rs. 14.40 per litre?
1. 10 : 1
2. 11 : 1
3. 1 : 11
4. 12 : 1
5. 1 : 12
Q85. Shankar bought two boxes for Rs. 1035, sold first box at a profit of 25% and the second box at a loss of 18%. What will be the cost prices of the two boxes respectively, if their selling price are the same?
1. Rs. 625, Rs. 410
2. Rs. 710, Rs. 325
3. Rs. 410, Rs. 625
4. Rs. 755, Rs. 280
5. Rs. 325, Rs. 710
Q86. The average age of family of 4 members 5 years ago was 16 years. With the birth of a baby, the average of family is 1 more than what it was 5 years ago. Find the age of the baby.
1. 1 year
2. 2 years
3. 3 years
4. 4 years
5. 5 years
Q87. In the following question, two equations are given. You have to solve these equations and find out the values of x and y and determine the relation between them.
I. 16x2 + 20x + 6 = 0
II. 10y2 + 38y + 24 = 0
1. x < y 2. x > y
3. x ≥ y
4. x ≤ y
5. x = y or the relationship cannot be established
Q88. In the following question, two equations numbered I and II are given. You have to solve both the equations and determine the relation between x and y.
I. x2 + 11x + 30 = 0
II. y2 + 7y + 12 = 0
1. x > y
2. x ≥ y
3. x < y 4. x ≤ y 5. x = y or the relationship cannot be established. Q89. In the following question, two equations numbered I and II are given. You have to solve both the equations and determine the relation between x and y. I. x2 – 16 = 0
II. y2 – 9y + 20 = 0
1. x > y
2. x ≥ y
3. x < y 4. x ≤ y 5. x = y or the relationship cannot be established. Q90. In the following question, two equations numbered I and II are given. You have to solve both the equations and determine the relation between x and y. I. x – √121 = 0 II. y2 – 121 = 0
1. x > y
2. x ≥ y
3. x < y 4. x ≤ y 5. x = y or the relationship cannot be established. Q91. In the following question, two equations numbered I and II are given. You have to solve both the equations and determine the relation between x and y. I. x2- x -12 = 0
II. y2 + 5y + 6 = 0
1. x > y
2. x ≥ y
3. x < y 4. x ≤ y 5. x = y or the relationship cannot be established. Q91-96) Directions: Study the following Graph carefully and answer the question given below: Total Number of Arts, Science and Commerce Students in Various Colleges
Q92. What is the average numbers of students (in thousands) taking up Commerce from all the six colleges together? (Rounded off to two digits after decimal)
1. 44.90
2. 43.33
3. 51.33
4. 38.09
5. 30.83
Q93. What is the respective ratio of the number of the students taken Commerce to the number of students taking Science in college E?
1. 14 : 19
2. 12 : 17
3. 17 : 19
4. 15 : 11
5. 13 : 18
Q94. How many candidates (in thousands) have taken Arts from all the colleges?
1. 265
2. 237.5
3. 185
4. 260.5
5. 232
Q95. In college B, the students taking Science is what percent of the total number of students taking Arts, Science and Commerce in college B?
1. 45
2. 55
3. 25
4. 70
5. 37.5
Q96. The number of students taking Commerce in college B, is how many thousand fewer than the number of student taking Commerce in college A and college C together?
1. 12.5
2. 27.5
3. 50
4. 10
5. 32.5
Q97. Aditya can do 50% more work than Radhika can do in the same time. Radhika alone do a piece of work in 30 hours. Aditya with the help of Radhika, can finish twice the same work in how many hours?
1. 24
2. 20
3. 22.5
4. 20.5
5. None of these
Q98. A rectangular sheet of length 44 cm and breadth 10 cm is taken. It is folded along its length in a circle to form a right circular cylinder. What will be the volume of the cylinder hence formed (in cubic cm)? (Take π = 22/7)
1. 1540
2. 1400
3. 1280
4. 1640
5. 1720
Q99. In how many ways the letters of word TESTBOOK can be arranged so that there are vowels at first and last positions?
1. 360
2. 540
3. 720
4. 1080
5. 1440
Q100. There is 60% increase in an amount in 4 years at simple interest. What will be the compound interest of Rs. 48,000 after 2 years at the same rate?
1. 10080
2. 15480
3. 14850
4. 33120
5. None of these
Answers&Explanations
66. Laws of Indices:-
1 -: a m × a n = a {m+n}
2 -: a m ÷ a n = a {m-n}
3 -: [(am ) n ] = a mn
4 -: (a) (1/m) = m √a
5 -: (a) (-m) =1/a m
6 -: (a) (m/n) = n √a m
7 -: (a) 0 = 1
Now, the given expression:
(4 × 4) 3 ÷ (4) 5 × (2 × 8) 2 = (4) ?
⇒ (2 2 × 2 2 ) 3 ÷ (2 2 ) 5 × (2 × 2 4 ) 2 = (16) ?
⇒ (2) 12 ÷ (2) 10 × (2) 8 = (2) ? × 4
⇒ (2) ? = 2 10 ÷ 4
⇒ ? = 2.5
67. Follow BODMAS rule to solve this question, as per the order given below,
Step -1-Parts of an equation enclosed in 'Brackets' must be solved first,
Step -2-Any mathematical 'Of' or 'Exponent' must be solved next,
Step -3-Next, the parts of the equation that contain 'Division' and 'Multiplication' are calculated,
Step -4-Last but not least, the parts of the equation that contain 'Addition' and 'Subtraction' should be calculated.
Now, the given expression,
√529 ÷ 46 × 6.4 + (8) 3 – 252 = ?
⇒ 23 ÷ 46 × 6.4 + (8) 3 – 252 = ?
⇒ ? = 3.2 + 512 – 252
⇒ ? = 263.2
68 . Follow BODMAS rule to solve this question, as per the order given below,
Step -1-Parts of an equation enclosed in 'Brackets' must be solved first,
Step -2-Any mathematical 'Of' or 'Exponent' must be solved next,
Step -3-Next, the parts of the equation that contain 'Division' and 'Multiplication' are calculated,
Step -4-Last but not least, the parts of the equation that contain 'Addition' and 'Subtraction' should be calculated.
Now, the given expression,
14% of 250+ ?% of 300 = 125
⇒ 35 + ?% of 300 = 125
⇒ 3× ? = 90
⇒ ? = 30
69. 150.08% of 1168.03 ÷ 7.96 + 18.956 = 29.231 ×?
Here, 150.08 ≈ 150; 1168.03 ≈ 1168; 7.96 ≈ 8; 18.956 ≈ 19; 29.231 ≈ 29
Taking the approximate values we have,
150% of 1168 ÷ 8 + 19 = 29 × ?
⇒ 1168 × 1.5 ÷ 8 + 19 = 29 × ?
⇒ 219 + 19 = 29 × ?
⇒ 29 × ? = 238
⇒ ? = 238/29 = 8.20 ≈ 8
Hence, the value is 8.
70 . 179.86 × 18.01 - ? 2 = 2984
Here, 179.86 ≈ 180 and 18.01 ≈ 18
Taking the approximate values, we have
⇒ 180 × 18 - ? 2 = 2984
⇒ 3240 - ? 2 = 2984
⇒ ? 2 = 3240 – 2984
⇒ ? 2 = 256
⇒ ? = 16
71 . From the table,
The total number of employees in Scale IV = 45 + 125 + 155 + 65 + 35 + 55 = 480
The number of employees are deployed in ‘operations’ department in Scale IV = 125
The required percentage = 125 480 × 100 = 26.04
∴ 26.04% of Scale IV officers are deployed in ‘operations’ department.
7 2. From the table,
Total number of employees in ‘Personnel’ department = 220 + 125 + 85 + 45 + 30 + 8 = 513
In Personnel department, the number of employees in Scale II = 125
The required p ercentage = 125 513 × 100 = 24.36
∴ Out of the total number of employees in ‘Personnel’ department, approximately 24.36% employees are in Scale II.
73 . From the table,
In public Relations department, the number of employees in Scale II = 155
In public Relations department, the number of e mployees in Scale I = 180
So, in public Relations department, the number of employees in Scale II is (180 – 155) = 25 less than that in scale I.
The required percentage = 25 180 × 100 = 13.89
∴ In public Relations department, the number of employees in Scale II is less than that in Scale I by 13.89%.
74 . From the table,
Total number of employees in Scale VI = 8 + 15 + 30 + 20 + 10 + 8 = 91
Total number of employees in Scale I = 220 + 625 + 650 + 350 + 320 + 180 = 2345
The required percentage = 91 2345 × 100 = 3.88
∴ Total number of employees in Scale VI is 3.88% of the total number of employees in Scale I.
75 . From the table,
The total number of employees in Scale III = 85 + 280 + 225 + 150 + 100 + 120 = 960
The total number of employees in Scale IV = 45 + 125 + 15 5 + 65 + 35 + 55 = 480
∴ The ratio between the total number of employees in Scale III and Scale IV respectively = 960:480 = 2:1
76 . Allan invested three fourth of the money that Matt invested and for twice the time as Matt.
Suppose Matt invested Rs. T fo r N years. So, Allan invested Rs. 3T/4 for 2N years.
⇒ Ratio of amounts invested multiplied by time for Matt and Allan = (TN) : (6TN/4) = 2 : 3
Let profit earned by Matt be Rs. P and that by Allan be Rs. (P + 300).
We know, ratio of amounts invested multip lied by time is same as ratio of profits.
⇒ P/(P + 300) = 2/3
⇒ 3P = 2P + 600
⇒ P = 600
∴ Total profit = P + P + 300 = Rs. 1500
77 . Let the age of Nancy be N years.
⇒ Age of Nancy’s brother = (N + 5) years
The sum of their ages is three fourth of their fa ther’s age.
⇒ Father’s age is 4/3 of sum of ages of Nancy and Nancy’s brother.
⇒ Father’s age = (4/3)(N + N + 5) years = (4/3)(2N + 5) years
After 4 years, father’s age will be twice the age of Nancy’s brother.
⇒ [(4/3)(2N + 5) + 4] = 2(N + 5 + 4)
⇒ 8N + 20 + 12 = 6N + 54
⇒ N = 22/2 = 11
∴ Age of Nancy 7 years from now = (11 + 7) years = 18 years
78 . Let the speed of the 210 m long train = x km/h
As the trains are travelling in opposite directions,
Relative speed of the first train w.r.t the second train = Speed of train 1 + Speed of train 2
⇒ Relative speed = (70 + x)km/h = (70 + x) × 5/18 m/s ---(1)
Thus,
Distance covered by train 1 = Length of Train 1 + Length of train 2
⇒ Distance covered by train 1 = 140 + 210 = 350 m
Time taken = 7 seconds
∴ Relative speed × (Time taken) = Total distance covered
⇒ (70 + x) × (5/18) × 7 = 350
⇒ (70 + x) × (5/18) = 50
⇒ (70 + x)/18 = 10
⇒ x = 110 km/h
Thus,
The speed of the second train is 110 km/h
79 . The pattern of given series is:
→ 24,
→ 50 = (24 × 2) + 2,
→ 98 = (50 × 2) – 2,
→ 198 = (98 × 2) + 2,
→ 394 = (198 × 2) – 2,
→ ? = (394 × 2) + 2,
→ ? = 790
80 . The pattern of given series is:
→ 100,
→ 49 = (100 × 0.5) – 1,
→ 47 = (49 × 1) – 2,
→ 67.5 = (47 × 1.5) – 3,
→ 131= (67.5 × 2) – 4,
→ ? = (131 × 2.5) – 5,
→ ? = 322.5
81 . The pattern of given series is:
→ 33,
→ 43 = 33 + 10,
→ 65 = 43 + (10 + 12),
→ 99 = 65 + (10 + 24),
→ 145 = 99 + (10 + 36),
→ ? = 145 + (10 + 48),
→ ? = 203
82 . The pattern of given series is:
→ 68 – 19 = 49 (7 x 7)
→ 102 – 68 = 34 (*36 = 6 x 6 )
→ 129 – 102 = 27 (*25 = 5 x 5)
→ 145 - 129 = 16 (4 x 4)
→ 154 – 145 = 9 (3 x 3)
If we replace 102 by 104, then 104 – 68 = 36 & 129 – 104 = 25.
So, the wrong number is 102.
8 3. The pattern of given series is:
→ 6 = 1 × 5 + 1
→ 26 = 6 × 4 + 2
→ 81 = 26 × 3 + 3
→ 166 = 81 × 2 + 4
→ 171 = 166 × 1 + 5
Hence, 172 is wrong in the series.
84 . Let’s assume that x litres of water is mixed with y litres of soda.
Cost price of soda = Rs. 13/Litre
Cost of y litre soda = Rs. 13y
Total quantity of the mi xture = (x + y) litre
The cost price of the mixture = Rs. 13y [∵ Water is assumed available free of cost]
The selling price of (x + y) litres of mixture = Rs. 14.40 × (x + y)
The cost price of the mixture = SellingPrice×100 100+Profit% = 13푦 ⇒ (14.40) × (x + y) × 100 100 + 20 = 13y
⇒ 1440x + 1440y = 13y × 120
⇒ 1440x = 1560y – 1440y
⇒ 1440x = 120y
⇒ x/y = 1/12
∴ Water and soda should be mixed in 1 : 12 ratio.
85 . Let’s assume the cost price of first box to be Rs. x
And, the cost price of second box to be Rs. y
A ccording to the given information,
⇒ x + y = 1035 -------Equation (1)
∵ first box is sold at 25% profit,
Selling price of first box = x + (25% of x) = 1.25x
∵ second box is sold at 18% loss,
Selling price of second box = y – (18% of y) = 0.82y
As per the given information, selling prices of both the boxes are same.
⇒ 1.25x = 0.82y
⇒ x = 0.82y/1.25
Substituting the value of x in Equation (1), we get, ⇒ 82푦 125 + 푦 = 1035 ⇒ 207푦 125 = 1035 ⇒ 푦 = 1035 × 125 207 = 625
Substituting i n Equation (1), we get,
x = 1035 – 625 = 410
∴ the selling prices of two boxes are Rs. 625 and Rs. 410 respectively.
86 . We know that, average = Sum of all quantities/Number of quantities
Average age of family = Sum of ages of all family members/Number o f members
∴ 16 = Sum of ages of 4 family members five years ago/4
⇒ Sum of ages of 4 family members 5 years ago = 4 × 16 = 64
In five years, each of these 4 members’ ages have increased by 5.
∴ Sum of present ages of 4 family members = Sum of ages of 4 family members 5 years ago + (5 × 4)
∴ Sum of present ages of 4 family members = 64 + 20 = 84
Let the age of the baby be x years.
∴ Sum of present ages of all family members including baby = 84 + x
According to the information given in the question, average p resent age of family members including baby, is 1 more than what it was 5 years ago excluding baby, i.e. 16 + 1 = 17
∴ Average = 17
⇒ 17 = (84 + x)/5
⇒ 84 + x = 17 × 5 = 85
⇒ x = 1
∴ the age of the baby is 1 year.
87 . I. 16x 2 + 20x + 6 = 0
⇒ 8x 2 + 10x + 3 = 0 [Dividing both sides by 2 ]
⇒ 8x 2 + 6x + 4x + 3 = 0
⇒ 2x(4x + 3) + 1(4x + 3) = 0
⇒ (4x + 3)(2x + 1) = 0
Then, x = - ¾ or x = - ½
II. 10y 2 + 38y + 24 = 0
⇒ 5y 2 + 19y + 12 = 0 [Dividing both sides by 2 ]
⇒ 5y 2 + 15y + 4y + 12 = 0
⇒ 5y(y + 3) + 4(y + 3) = 0
⇒ (y + 3)(5y + 4) = 0
Then, y = - 3 or y = - 4/5
So, when x = - ¾ , x > y for y = - 3 and x > y for y = - 4/5
And when x = - ½ , x > y for y = - 3 and x > y for y = - 4/5
∴ We can clearly observe that x > y.
88 . I. x 2 + 11x + 30 = 0
⇒ x 2 + 5x + 6x + 30 = 0
⇒ x(x + 5) + 6(x + 5) = 0
⇒ (x + 5)(x + 6) = 0
Then, x = - 5 or x = - 6
II. y 2 + 7y + 12 = 0
⇒ y 2 + 4y + 3y + 12 = 0
⇒ y(y + 4) + 3(y + 12) = 0
⇒ (y + 4)(y + 3) = 0
Then, y = - 4 or y = - 3
So, when x = - 5, x < y for y = - 4 and x < y for y = - 3 And when x = - 6, x < y for y = - 4 and x < y for y = - 3 ∴ We can clearly see that x < y.
89 . I. x 2 –16 = 0
⇒ x 2 = 16
⇒ x = ± 4
Then, x = + 4 or x = - 4
II. y 2 - 9y + 20 = 0
⇒ y 2 – 5y – 4y + 20 = 0
⇒ y(y – 5) – 4(y – 5) = 0
⇒ (y – 5)(y – 4) = 0
Then, y = + 5 or y = + 4
So, when x = + 4, x < y for y = + 5 and x = y for y = + 4
And when x = - 4, x < y for y = + 5 and x < y for y = + 4
∴ We can clearly observe that x ≤ y.
90 . I. x - √121= 0
⇒ x = √121 = 11
Then, x = + 11
II. y 2 – 121 = 0
⇒ y 2 = 121
⇒ y = ± 11
Then, y = + 11 or y = - 11
So, when x = + 11, x = y for y = + 11 and x > y for y = - 11
∴ So, we can observe that x ≥ y.
91 . I. x 2 – x – 12 = 0
⇒ x 2 – 4x + 3x – 12 = 0
⇒ x(x – 4) + 3(x – 4) = 0
⇒ (x – 4)(x + 3) = 0
Then, x = + 4 or x = - 3
II. y 2+ 5y + 6 = 0
⇒ y 2 + 3y + 2y + 6 = 0
⇒ y(y + 3) + 2(y + 3) = 0
⇒ (y + 3)(y + 2) = 0
Then, y = - 3 or y = - 2
So, when x = + 4, x > y for y = - 3 and x > y for y = - 2
And when x = - 3, x = y for y = - 3 and x < y for y = - 2 ∴ So, we can observe that no clear relationship cannot be determined between x and y.
92 . From the given data,
| College | A | B | C | D | E | F | Total |
|---|---|---|---|---|---|---|---|
| Number of students taking Commerce (in thousand) | 40 | 25 | 17.5 | 35 | 37.5 | 30 | 185 |
Total number of colleges = 6
Average number of students taking Commerce (in thousands) = 185/6 = 30.83
93 . From the given data,
Number of students taken Commerce in college E (in thousands) = 37.5
Number of students taken Science in college E (in thou sands) = 27.5
Required ratio = 37.5 : 27.5 = 15 : 11
94 . From the given data,
| College | A | B | C | D | E | F | Total |
|---|---|---|---|---|---|---|---|
| Number of students taking Commerce (in thousand) | 22.5 | 50 | 40 | 35 | 50 | 40 | 237.5 |
95 . From the given data,
| Stream | Number of student in college B (in thousands) |
|---|---|
| Arts | 50 |
| Science | 45 |
| Commerce | 25 |
| Total | 120 |
Total Number of students in college B (in thousands) = 120
Required % = 45 120 × 100 = 37.5%
96 . From the given data,
Number of students taking Commerce in college B (in thousand) = 25
Number of students taking Commerce in college A (in thousand) = 40
Number of students taking Commerce in college C (in thousand) = 17.5
Total number of students taking Commerce in college A and C together (in thous and) = 17.5 + 40 = 57.5
Required number of students (in thousand) = 57.5 – 25 = 32.5
97. Aditya can do 50% more work than Radhika in the same time.
⇒ Aditya’s efficiency is 1.5 times of Radhika’s efficiency
Radhika alone do a piece of work in 30 hrs.
∴ Part of work done by Radhika in one hour = 1/30
∴ Part of work done by Aditya in one hour = (2/3) × (1/30) = 1/20
Working together, part of work finish ed by both in one hour = 1 30 + 1 20 = 2+3 60 = 1 12
∴ they’ll take 12 hours to finish the entire work.
⇒ In order to finish twice the original work, they’ll require 12 × 2 = 24 hours.
98 . If rectangular sheet is folded along its length to from a right circular cylinder, the height of cylinder will be same as breadth of rectangle, and circumference of base will be same as length of rectangle.
If height of cylinder is H cm, and radius is R cm, then
H = breadth = 10 cm
And, 2πR = length = 44 cm
⇒ 2 × (22/7) × R = 44
⇒ R = 7 cm
∴ Volume of cylinder = πR 2 H = (22/7) × 7 × 7 × 10 = 1540 cubic cm
99 . There are 3 vowels in word TESTBOOK, E, O and O. If there are vowels at first and last position, this is possible in 3 ways.
E at first and O at last, O at first a nd E at last, and O at both first and last.
In each of the three cases, the remaining six letters that have to come in between will be different except that T will be occur twice.
⇒ Remaining 6 letters can be arranged in 6!/2! Ways, i.e., 360 ways.
∴ Total number of ways in which this can be done = 3 × 360 = 1080
100 . Simple Interest = (P × R × T)/100 Where, P = Principal, R = % rate of interest, T = time period in year
As per given information, Simple interest = 60% of principal
∴ 0.6P = (P × R × 4)/100
⇒ R = 15
Now let's find out the compound interest of Rs. 48,000 after 2 years at 10% P = Rs. 48,000 T = 2 years
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