187.5, 182, 171, ?, 132.5, 105, 72 a) 145.5 b) 155.5 c) 154.5 d) 144.5 e) None of these

187.5, 182, 171, ?, 132.5, 105, 72
a) 145.5
b) 155.5
c) 154.5
d) 144.5
e) None of these
by

1 Answer

Answer :

The pattern is -5.5, -11, -16.5, -22 and so on.
Answer: c)
by

Related questions

210, 154, 182, 168, 175,171.5, ? a) 172.75 b) 173.25 c) 172.25 d) 173.75 e) 174.75

Description : 210, 154, 182, 168, 175,171.5, ? a) 172.75 b) 173.25 c) 172.25 d) 173.75 e) 174.75

Last Answer : Starting from the third term, every term is the average of the previous two terms. Hence the next term is (175 + 171.5)/ 2 = 173.25 Answer: b)

The ratio between two number is 3:4 ,if each number be increased by 9, the ratio becomes 18:23 find the sum of the number a) 135 b) 105 c) 155 d) 165 e) None of these

Description : The ratio between two number is 3:4 ,if each number be increased by 9, the ratio becomes 18:23 find the sum of the number a) 135 b) 105 c) 155 d) 165 e) None of these

Last Answer : Let the two numbers be 3x and 4x. When they are increased by 9 they become 3x + 9 and 4x + 9. It is given that the ratio is 18:23 Thus, 3x + 9/4x + 9 = 18:23 23(3x + 9) = 18(4x + 9) 69x + 207 = 72x ... = -45 X = 15 Thus two numbers are 3x15 = 45 and 4 x 15 = 60 And the sum is 60+45 = 105 Answer: b)

97,122,?,307,503,792 a) 171 b) 158 c) 186 d) 203 e) 178

Description : 97,122,?,307,503,792 a) 171 b) 158 c) 186 d) 203 e) 178

Last Answer : Answer: c)

501/58 * 291/101 ÷ 31/155 = ? a) 140 b) 125 c) 95 d) 85 e) 110

Description : 501/58 * 291/101 ÷ 31/155 = ? a) 140 b) 125 c) 95 d) 85 e) 110

Last Answer : Answer: b)

3599 ÷ 224.95 × 23.04 = ? + 16.98 × 11.03 a) 130 b) 210 c) 180 d) 115 e) 155

Description : 3599 ÷ 224.95 × 23.04 = ? + 16.98 × 11.03 a) 130 b) 210 c) 180 d) 115 e) 155

Last Answer : 3600 ÷ 225 × 23 = ? +17 × 11 =>16 × 23 = ? + 187 => ? = 368 – 187 =181 ≈ 180 Answer: c)

A three-phase alternator is developing 300 amps, with a 0.8 power factor, at 450 volts. The true indicated power on the kilowatt meter, located on the main switchboard, will be ____________. A. 133 kw B. 155 kw C. 187 kw D. 212 kw

Description : A three-phase alternator is developing 300 amps, with a 0.8 power factor, at 450 volts. The true indicated power on the kilowatt meter, located on the main switchboard, will be ____________. A. 133 kw B. 155 kw C. 187 kw D. 212 kw

Last Answer : Answer: C

3/8th of 4/9th of 1092 = ? a) 182 b) 728 c) 364 d) 218 e) None of these

Description : 3/8th of 4/9th of 1092 = ? a) 182 b) 728 c) 364 d) 218 e) None of these

Last Answer : 1092 × 4/9 × 3/8 = ? 1092 × 1/3 × 1/2 = ? ? = 182 Answer: a)

Three consecutive numbers such that twice the first, 3 times the second and 4 times the third together make 182. The numbers in question are 1. 18, 22 and 23 2. 18, 19 and 20 3. 19, 20 and 21 4. 20, 21 and 22 5. 21, 22 and 23

Description : Three consecutive numbers such that twice the first, 3 times the second and 4 times the third together make 182. The numbers in question are 1. 18, 22 and 23 2. 18, 19 and 20 3. 19, 20 and 21 4. 20, 21 and 22 5. 21, 22 and 23

Last Answer : Answer- 3( 19,20,21) Explanation:- let number be x,x+1,x+2 2x + 3(x+1)+4(x+2)=180 x=19 so numbers are 19,20,21

The appointment of a Governor in a state is made as per the provision in the Constitution under article. (1) 153 (2) 154 (3) 155 (4) 156

Description : The appointment of a Governor in a state is made as per the provision in the Constitution under article. (1) 153 (2) 154 (3) 155 (4) 156

Last Answer : (3) 155 Explanation: Appointment of the Governor of Indian state is described under Article 155 of the Indian constitution.

13 6 7 9 20 47.5 ? a) 149.5 b) 140.5 c) 150.5 d) 147.5 e) 145.5

Description : 13 6 7 9 20 47.5 ? a) 149.5 b) 140.5 c) 150.5 d) 147.5 e) 145.5

Last Answer : The series is × 0.5 – 0.5, × 1 + 1, ×1.5 – 1.5, ×2 + 2, ... Answer: e)

What is thr hcf of 84 154 and 182?

Description : What is thr hcf of 84 154 and 182?

Last Answer : It is: 14

The area of a circle is 154 sq cm. Its circumference is a) 44cm b) 88cm c) 154cm d) 176cm e) 164 cm

Description : The area of a circle is 154 sq cm. Its circumference is a) 44cm b) 88cm c) 154cm d) 176cm e) 164 cm

Last Answer : a) 44cm

A goat is tied to one corner of a square plot of side 12 meter by a rope 7 meter long. Find the area it can graze? (a) 38.5 sq.m (b) 155 sq.m (c) 144 sq.m (d) 19.25 sq.m

Description : A goat is tied to one corner of a square plot of side 12 meter by a rope 7 meter long. Find the area it can graze? (a) 38.5 sq.m (b) 155 sq.m (c) 144 sq.m (d) 19.25 sq.m

Last Answer : (a) 38.5 sq.m

16, 32, 35, 140, 145, 870, 877, _____ 1 : 7016 2 : 885 3 : 5020 4 : 6018 5 : None of these

Description : 16, 32, 35, 140, 145, 870, 877, _____ 1 : 7016 2 : 885 3 : 5020 4 : 6018 5 : None of these

Last Answer : Answer- 1(7016) Explanation16×2 =32 32+3=35 35×4=140 140+5=145 145×6=870 870+7=877 877×8=7016

Vijay purchased two kg of sugar and three kg of rice for Rs.145. Ravi purchased three kg of sugar and two kg of rice for Rs.130. What is the price of one kg of sugar and one kg of rice? 1 : Rs.45 2 : Rs.50 3 : Rs.55 4 : Rs.60 5 : None of these

Description : Vijay purchased two kg of sugar and three kg of rice for Rs.145. Ravi purchased three kg of sugar and two kg of rice for Rs.130. What is the price of one kg of sugar and one kg of rice? 1 : Rs.45 2 : Rs.50 3 : Rs.55 4 : Rs.60 5 : None of these

Last Answer : 3 : Rs.55

Equal quantities of 3 mixtures of milk and water are mixed in the ratio 1:3, 2:3 and 3:4.The ratio of water and milk in the new mixture is A) 45:76 B) 151:269 C) 123:154 D) 145:245 E) None of these

Description : Equal quantities of 3 mixtures of milk and water are mixed in the ratio 1:3, 2:3 and 3:4.The ratio of water and milk in the new mixture is A) 45:76 B) 151:269 C) 123:154 D) 145:245 E) None of these

Last Answer : Answer: B Milk = 1/4 : 2/5 :3/7 = 35/140 :56/140 : 60/140 Quantity of milk in new mix = 35+56+60 = 151 Quantity of water in new mix = 140*3 = 420-151 = 269 M:W = 151:269

Find the magnetic field intensity of a toroid of turns 40 and radius 20cm. The current carried by the toroid be 3.25A. a) 103.45 b) 102 c) 105.7 d) 171

Description : Find the magnetic field intensity of a toroid of turns 40 and radius 20cm. The current carried by the toroid be 3.25A. a) 103.45 b) 102 c) 105.7 d) 171

Last Answer : a) 103.45

A trader sells 145 m of cloth for Rs 12325 at the profit of Rs 10 per metre of cloth. What is the cost price of 1 m of cloth? 1. Rs 65 2. Rs 75 3. Rs 95 4. Rs 85

Description : A trader sells 145 m of cloth for Rs 12325 at the profit of Rs 10 per metre of cloth. What is the cost price of 1 m of cloth? 1. Rs 65 2. Rs 75 3. Rs 95 4. Rs 85

Last Answer : 2. Rs 75

Calculate the free torsional vibrations of a single motor system from the following data: C = 8 GN/m 2 , L=9m, I = 600 Kg-m 2 , J = 8×10 4 m 4 a) 162,132 b) 172,132 c) 182,132 d) 192,132

Description : Calculate the free torsional vibrations of a single motor system from the following data: C = 8 GN/m 2 , L=9m, I = 600 Kg-m 2 , J = 8×10 4 m 4 a) 162,132 b) 172,132 c) 182,132 d) 192,132

Last Answer : b) 172,132

3 4 12 45 ? 1005 a) 152 b) 198 c) 144 d) 192 e) None of these

Description : 3 4 12 45 ? 1005 a) 152 b) 198 c) 144 d) 192 e) None of these

Last Answer : Answer: e)

What is the ratio between money spent by company Q on tax and money spent by P on salary? a) 23:121 b) 33:144 c) 47:144 d) 144:157 e) None of these

Description : What is the ratio between money spent by company Q on tax and money spent by P on salary? a) 23:121 b) 33:144 c) 47:144 d) 144:157 e) None of these

Last Answer : e) None of these

(138)2 + (511)2 – (132)2 = (85)2 + ? a) 410416 b) 255516 c) 265918 d) 345622 e) 546056

Description : (138)2 + (511)2 – (132)2 = (85)2 + ? a) 410416 b) 255516 c) 265918 d) 345622 e) 546056

Last Answer : (138)2 + (511)2 – (132)2 = (85)2 + ? Sol: 19044 + 261121 – 17424 = 7225 + ? 19044 + 261121 – 17424 – 7225 = ? ? = 255516 Answer: b)

McCullum scored ______ on 58 balls in T20 cricket and set a world record against Bangladesh at Pallekele on 21 September 2012. A. 187 B. 123 C. 152 D. 132

Description : McCullum scored ______ on 58 balls in T20 cricket and set a world record against Bangladesh at Pallekele on 21 September 2012. A. 187 B. 123 C. 152 D. 132

Last Answer : ANSWER: B

McCullum scored ______ on 58 balls in T20 cricket and set a world record against Bangladesh at Pallekele on 21 September 2012. A. 187 B. 123 C. 152 D. 132

Description : McCullum scored ______ on 58 balls in T20 cricket and set a world record against Bangladesh at Pallekele on 21 September 2012. A. 187 B. 123 C. 152 D. 132

Last Answer : ANSWER: B

The octal equivalent of 1100101.001010 is ______ a) 624.12 b) 145.12 c) 154.12 d) 145.21

Description : The octal equivalent of 1100101.001010 is ______ a) 624.12 b) 145.12 c) 154.12 d) 145.21

Last Answer : Answer: b Explanation: The octal equivalent is obtained by grouping the numbers into three, from right to left before decimal and from right to left after the decimal place. Here, i.e. 145.12 is the octal equivalent of the number

24,48,144,576,2880,? 1 : 17280 2 : 16640 3 : 14400 4 : 20160 5 : 14240

Description : 24,48,144,576,2880,? 1 : 17280 2 : 16640 3 : 14400 4 : 20160 5 : 14240

Last Answer : Answer- 1 (17280) Explanation- 24 x 2=48 48×3=144 144×4=576 576×5=2880

.Total how many students in the college know all three languages? (1)108 (2)132 (3)169 (4)137 (5)142

Description : .Total how many students in the college know all three languages? (1)108 (2)132 (3)169 (4)137 (5)142

Last Answer : (3)169

136% of 105 – 84% of 165 = ? a) 4.2 b) 5.7 c) 6.4 d) 7.8 e) 8.6

Description : 136% of 105 – 84% of 165 = ? a) 4.2 b) 5.7 c) 6.4 d) 7.8 e) 8.6

Last Answer : a) 4.2

Hemant sold 10 sarees for a total profit of Rs. 460 and 12 sarees for a total profit of Rs. 144. At what profit per saree should he sell the remaining 20 sarees so that he gets an average profit of Rs. 18 per sarees? A. Rs 7.10 B. Rs 7.60 C. Rs 7.99 D. Rs 8

Description : Hemant sold 10 sarees for a total profit of Rs. 460 and 12 sarees for a total profit of Rs. 144. At what profit per saree should he sell the remaining 20 sarees so that he gets an average profit of Rs. 18 per sarees? A. Rs 7.10 B. Rs 7.60 C. Rs 7.99 D. Rs 8

Last Answer : Answer – B (Rs 7.60) Explanation – Total profit required = Rs.(42 x 18) = Rs.756 Profit on 22 sarees = Rs. (460 + 144) = Rs. 604 Profit on 20 sarees = Rs. (756 – 604) = Rs. 152 Average profit on these sarees = Rs.(152/20) = Rs. 7.60

Vinoth could get equal number of Rs. 55, Rs. 85 and Rs. 105 ticket for a Zoo. He spend Rs. 3920 for all the tickets. How many of each did he buy? a) 15 b) 18 c) 16 d) 26 e) None of these

Description : Vinoth could get equal number of Rs. 55, Rs. 85 and Rs. 105 ticket for a Zoo. He spend Rs. 3920 for all the tickets. How many of each did he buy? a) 15 b) 18 c) 16 d) 26 e) None of these

Last Answer : Total money = 55x + 85x + 105x = 3920 245x = 3920 x = 16 Answer: c)

2 9 30 105 ? 2195 a) 432 b) 426 c) 440 d) 436 e) None of these

Description : 2 9 30 105 ? 2195 a) 432 b) 426 c) 440 d) 436 e) None of these

Last Answer : The series is ×1 + 1 × 7, ×2 + 2 × 6, ×3 + 3 × 5... Answer: d)

2 9 ? 105 436 2195 13182 a) 32 b) 45 c) 40 d) 30 e) 25

Description : 2 9 ? 105 436 2195 13182 a) 32 b) 45 c) 40 d) 30 e) 25

Last Answer : The series is (2 + 7) × 1 = 9; (9 + 6) × 2 = 30; (30 + 5) × 3 = 105; (105 + 4) × 4 = 436; (436 + 3) × 5 = 2195; (2195 + 2) × 6 = 13182 Answer: d)

The class marks of a frequency distribution are 104, 114, 124, 134, 144, 154, 164. -Maths 9th

Description : The class marks of a frequency distribution are 104, 114, 124, 134, 144, 154, 164. -Maths 9th

Last Answer : Since the class marks are equally spaced. ∴ Class size = 114 - 104 = 10 If a is a class mark and h is size of class interval, then lower limit and upper limit of the class interval area a - h / 2 and a + h / ... are 99 - 109, 109 - 119, 119 - 129, 129 - 139, 139 - 149, 149 - 159, 159 - 169.

The class marks of a frequency distribution are 104, 114, 124, 134, 144, 154, 164. -Maths 9th

Description : The class marks of a frequency distribution are 104, 114, 124, 134, 144, 154, 164. -Maths 9th

Last Answer : Since the class marks are equally spaced. ∴ Class size = 114 - 104 = 10 If a is a class mark and h is size of class interval, then lower limit and upper limit of the class interval area a - h / 2 and a + h / ... are 99 - 109, 109 - 119, 119 - 129, 129 - 139, 139 - 149, 149 - 159, 159 - 169.

Calculate the natural frequency of transverse vibrations if the static deflection is 0.01mm. a) 157.6 b) 144.8 c) 173.2 d) 154.1

Description : Calculate the natural frequency of transverse vibrations if the static deflection is 0.01mm. a) 157.6 b) 144.8 c) 173.2 d) 154.1

Last Answer : a) 157.6

Correct viscosity of furnace oil at the burner tip for proper atomisation is about 25 centistokes. To reduce the viscosity of high viscosity furnace oil (250 centistokes) to the correct atomisation viscosity (i.e. 25 cst), it should be preheated to about ___________°C. (A) 70 (B) 85 (C) 105 (D) 145

Description : Correct viscosity of furnace oil at the burner tip for proper atomisation is about 25 centistokes. To reduce the viscosity of high viscosity furnace oil (250 centistokes) to the correct atomisation viscosity (i.e. 25 cst), it should be preheated to about ___________°C. (A) 70 (B) 85 (C) 105 (D) 145

Last Answer : (C) 105

A foot ball team lost 75 % of the matches it played. If it won 45 matches, find the number of matches it played. a) 120 b) 145 c) 105 d) 140 e) 180

Description : A foot ball team lost 75 % of the matches it played. If it won 45 matches, find the number of matches it played. a) 120 b) 145 c) 105 d) 140 e) 180

Last Answer : Answer: E Percentage of matches lost = 75 % Therefore Percentage of matches won (100 - 75) % = 25 % Let the number of matches played be x. Then 25 % of x = 45 25/100 × x =45  x = (45 × 100)/25  x = (4500)/25  x = 180 Therefore, the total number of matches played is 180.

Noise emitted by a ventilation fan at a distance of 3 metres is about __________ decibels. (A) 85 (B) 105 (C) 125 (D) 145

Description : Noise emitted by a ventilation fan at a distance of 3 metres is about __________ decibels. (A) 85 (B) 105 (C) 125 (D) 145

Last Answer : (B) 105

If a voltage of 132 VDC is applied to the illustrated circuit where the resistance is 12 ohms, then current will be ___________. EL-0018 A. 0.090 amps B. 1.090 amps C. 11 amps D. 144 milliamps

Description : If a voltage of 132 VDC is applied to the illustrated circuit where the resistance is 12 ohms, then current will be ___________. EL-0018 A. 0.090 amps B. 1.090 amps C. 11 amps D. 144 milliamps

Last Answer : Answer: C

The average weight of four boys A, B, C, and D is 75 kg. The fifth boy E is included and the average weight decreases by 4 kg. A is replaced by F. The weight of F is 6 kg more than E. Average weight decreases because of the replacement of A and now the average weight is 72 kg. Find the weight of A. 1) 57 kg 2) 54 kg 3) 56 kg 4) 60 kg 5) 58 kg

Description : The average weight of four boys A, B, C, and D is 75 kg. The fifth boy E is included and the average weight decreases by 4 kg. A is replaced by F. The weight of F is 6 kg more than E. Average weight decreases because of ... . Find the weight of A. 1) 57 kg 2) 54 kg 3) 56 kg 4) 60 kg 5) 58 kg

Last Answer : 3) 56 kg

24+(72 ÷ 48×8+14) × 36 ÷ (3×4) + 36 ÷ 3 ×4 a) 120 b) 140 c) 160 d) 180 e) None of these

Description : 24+(72 ÷ 48×8+14) × 36 ÷ (3×4) + 36 ÷ 3 ×4 a) 120 b) 140 c) 160 d) 180 e) None of these

Last Answer : 72÷48×8+14 = (72/48) × 8 + 14 = (3/2) × 8 + 14 = 12 +14 = 26 24 + (72 ÷ 48 × 8 + 14) ×36 ÷ (3 × 4) + 36 ÷ 3 × 4 = 24 + 26 × 36÷12 + 36 ÷ 3 × 4 = 24 + 26 × 3 + 12 × 4 = 24 + 78 + 48 = 150 Answer: e)

72× 25 + 45× 96 = 163 + ? a) 1912 b) 2024 c) 1844 d) 1908 e) 1960

Description : 72× 25 + 45× 96 = 163 + ? a) 1912 b) 2024 c) 1844 d) 1908 e) 1960

Last Answer : 1800 + 4320 = 4096 + ? ? = 1800 + 4320 – 4096 = 2024 Answer: b)

1824 ÷ 32 + 5040 ÷ ? – 5600 ÷ 35 = 37 a) 36 b) 48 c) 60 d) 72 e) 84

Description : 1824 ÷ 32 + 5040 ÷ ? – 5600 ÷ 35 = 37 a) 36 b) 48 c) 60 d) 72 e) 84

Last Answer : 57 + 5040 ÷ ? – 160 = 37 5040 ÷ ? = 37 – 57 + 160 = 140 = 5040 ÷ 140 = 36 Answer: a)

53 × 47 + 64 × 56 -72 × 68 = ? a) 1221 b) 1403 c) 1249 d) 1337 e) 1179

Description : 53 × 47 + 64 × 56 -72 × 68 = ? a) 1221 b) 1403 c) 1249 d) 1337 e) 1179

Last Answer : Answer: e)

72, 115, 244, ?, 760, 1147 a) 501 b) 431 c) 473 d) 459 e) 445

Description : 72, 115, 244, ?, 760, 1147 a) 501 b) 431 c) 473 d) 459 e) 445

Last Answer : The pattern is: 72+43×1=115, 115+43×3=244, 244+43×5=459, 459+43×7=760, 760+43×9=1147 Answer: d)

?% of 725 + 14% of 2100 = 700 a) 64 b) 80 c) 72 d) 56 e) 48

Description : ?% of 725 + 14% of 2100 = 700 a) 64 b) 80 c) 72 d) 56 e) 48

Last Answer : ?% × 725 +14 × 21=700 =>?% × 725 +294 =700 => ?% × 725 =406 =>?=406/725 × 100 =56 Answer: d)

19/14 x 12/72 x 18/29 x 767 = ? (Take 2 digits after the decimal point) a) 110.69 b) 118.78 c) 111.56 d) None e) 108.98

Description : 19/14 x 12/72 x 18/29 x 767 = ? (Take 2 digits after the decimal point) a) 110.69 b) 118.78 c) 111.56 d) None e) 108.98

Last Answer : 19/14 x 12/72 x 18/29 x 767 =? Sol: 1.35 x 0.16 x 0.62 x 767 = ? 102.71 = ? Answer: d)

The average of weight of three men A,B and C is 84 kg. Another man D joins the group and the average now becomes 80 kg. If another man E, whose weight is 3 kg.more than that of D, replaces A, then the average weight of B,C,D and E becomes 79 kg. The weight of A is : a) 70 kg b) 72 kg c) 75 kg d) 80 kg e) 78 kg

Description : The average of weight of three men A,B and C is 84 kg. Another man D joins the group and the average now becomes 80 kg. If another man E, whose weight is 3 kg.more than that of D, replaces A, then the average weight of B,C ... 79 kg. The weight of A is : a) 70 kg b) 72 kg c) 75 kg d) 80 kg e) 78 kg

Last Answer : A+B+C = 3×84=252 A+B+C+D= (4×80)=320 D = (320-252)=68 and E = (68+3)=71 Now, B+C+D+E = (4×79)=316 B+C+D=(316-71)=245 kg So, A = (320-245)=75 kg Answer: c)

Sum of present age of A, B and C is 72 years. If 4 years ago, their ages were in the ratio 1 : 2 : 3, find A’s present age. a) 7 years b) 10 years c) 12 years d) 14 years e) None of these

Description : Sum of present age of A, B and C is 72 years. If 4 years ago, their ages were in the ratio 1 : 2 : 3, find A’s present age. a) 7 years b) 10 years c) 12 years d) 14 years e) None of these

Last Answer : Sum of present ages of A, B and C is 72 years. sum of their ages (4 years ago) = 72 – (3 × 4) = 60 years 4 years ago, ratio of ages of A, B and C was 1 : 2 :3. A’s age, 4 years ago = (1/6) ×60 = 10 years A’s present age = 10 + 4 = 14 years Answer: d)

In how many different ways can the letters of the word DESIGN be arranged so that the vowels are at the two ends? a) 48 b) 66 c) 12 d) 72 e) 96

Description : In how many different ways can the letters of the word DESIGN be arranged so that the vowels are at the two ends? a) 48 b) 66 c) 12 d) 72 e) 96

Last Answer : Total ways = 4 ! x 2 ! = 4 × 3 × 2 x 2= 48 Answer: a)