67,74,78,?,92,94 a) 86 b) 85 c) 83 d) 84 e) 88

67,74,78,?,92,94
a) 86
b) 85
c) 83
d) 84
e) 88
by

1 Answer

Answer :

The next term is obtained by adding the unit’s digit to the number.
Hence the missing number =78+8=86
Answer: a)
by

Related questions

In a class there are 15 boys and 10 girls. Three students are selected at random. The difference between the probability that 2 boys and 1 girl are selected compared to 1 boy and 2 girls are selected is: a) 23/78 b) 19/88 c) 15/92 d) 4/23 e) 7/46

Description : In a class there are 15 boys and 10 girls. Three students are selected at random. The difference between the probability that 2 boys and 1 girl are selected compared to 1 boy and 2 girls are selected is: a) 23/78 b) 19/88 c) 15/92 d) 4/23 e) 7/46

Last Answer : 2 boys, 1 gi(Prl = (15c2×10c1) / 25c3 = 1050/2300 1 boy, 2 girls = (15c1×10c2) / 25c3 = 675/2300 Difference = (1050 - 675)/2300 = 375/2300 = 15/92 Answer: c)

(331/30)+( 661/60) +(704/11) -35.013 + 36.026 =? a) 69 b) 67 c) 83 d) 89 e) 85

Description : (331/30)+( 661/60) +(704/11) -35.013 + 36.026 =? a) 69 b) 67 c) 83 d) 89 e) 85

Last Answer : ?=331/30 + 661/60 + 704/11 – 35.013 + 36.026 =11 + 11 + 64 -1.013 =86-1 ≈ 85 Answer: e)

The average age of Anu, banu and tonu is 74 yrs. 10years hence the average age of anu and tonu is 86 yrs.6yrs ago the average age of aarthi and banu was 72 yrs. Find the present age of aarthi. A) 80 B) 86 C) 84 D) 83

Description : The average age of Anu, banu and tonu is 74 yrs. 10years hence the average age of anu and tonu is 86 yrs.6yrs ago the average age of aarthi and banu was 72 yrs. Find the present age of aarthi. A) 80 B) 86 C) 84 D) 83

Last Answer : B) anu+ banu+ tonu = 74 * 3 = 222yrs 10yrs hence, anu + 10 + tonu +10 = 86 * 2 anu + tonu= 152 banu = 222-152 = 70 yrs aarthi -6 + banu - 6= 72 * 2 aarthi + banu = 144 + 12 = 156 aarthi age = 156-70 aarthi age = 86yrs

180 sweets are divided among friends A, B, C and D in which B and C are brothers also such that sweets divided between A and B are in the ratio 2 : 3, between B and C in the ratio 2 : 5 and between C and D in ratio 3 : 4. What is the number of sweets received by the brothers together? A) 78 B) 84 C) 92 D) 102 E) 88

Description : 180 sweets are divided among friends A, B, C and D in which B and C are brothers also such that sweets divided between A and B are in the ratio 2 : 3, between B and C in the ratio 2 : 5 and between C ... . What is the number of sweets received by the brothers together? A) 78 B) 84 C) 92 D) 102 E) 88

Last Answer : Answer: B A : B : C : D 2*2*3 : 3*2*3 : 3*5*3 : 3*5*4 4 : 6 : 15 : 20 B and C together = [(6+15)/(4+6+15+20)] * 180 = 84

For heavier fuels the cooling system of an internal combustion engine, the temperature is maintained at a) 78 to 80°C b) 80 to 84 °C c) 85 to 88°C d) 88 to 90°C

Description : For heavier fuels the cooling system of an internal combustion engine, the temperature is maintained at a) 78 to 80°C b) 80 to 84 °C c) 85 to 88°C d) 88 to 90°C

Last Answer : d) 88 to 90°C

3, 9, 17, 27, 39, 53, ? a) 64 b) 69 c) 74 d) 78 e) 81

Description : 3, 9, 17, 27, 39, 53, ? a) 64 b) 69 c) 74 d) 78 e) 81

Last Answer : Series is 1× 2 + 1, 2 × 3 + 3, 3 × 4 + 5, 4 × 5 + 7, 5 × 6 + 9 Answer: b)

Total literate population of city E is what percentage of total population of city D? 1) 67.5% 2) 71.5% 3) 74.5% 4) 77.5% 5) 82.5%

Description : Total literate population of city E is what percentage of total population of city D? 1) 67.5% 2) 71.5% 3) 74.5% 4) 77.5% 5) 82.5%

Last Answer : 2) 71.5%

Number of girls enrolled in only Dancing classes is what per cent of the boys enrolled in the same? (Rounded off to two digits after decimal) (1) 38•67 (2) 35•71 (3) 41•83 (4) 28•62 (5) None of these

Description : Number of girls enrolled in only Dancing classes is what per cent of the boys enrolled in the same? (Rounded off to two digits after decimal) (1) 38•67 (2) 35•71 (3) 41•83 (4) 28•62 (5) None of these

Last Answer : (2) 35•71

(37) 0.75 x [(7056) 0.25] ÷ (64) – 2 = (8) ? (Take 3 digits after the decimal point) a) 95.357 b) 69.567 c) 96.257 d) 78.584 e) 88.598

Description : (37) 0.75 x [(7056) 0.25] ÷ (64) – 2 = (8) ? (Take 3 digits after the decimal point) a) 95.357 b) 69.567 c) 96.257 d) 78.584 e) 88.598

Last Answer : (37) 0.75 x [(7056) 0.25] ÷ (64) – 2 = (8) ? Sol: 27.75 x 1764 ÷ 64 – 2 = 8 x ? 27.75 x 27.5625 – 2 = 8 x ? 762.859/8=? ? = 95.357 Answer: a)

55 60 67 78 91 108 ? a) 125 b) 121 c) 127 d) 89 e) None of these

Description : 55 60 67 78 91 108 ? a) 125 b) 121 c) 127 d) 89 e) None of these

Last Answer : 55 + 5 = 60 60 + 7 = 67 67 + 11 = 78 78 + 13 = 91 91 + 17 = 108+108 + 19 = 127 Answer: c)

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)

(37÷44) + (78÷34) x (480÷85) = ? a) 11.769 b) 12.756 c) 13.796 d) 12.796 e) none

Description : (37÷44) + (78÷34) x (480÷85) = ? a) 11.769 b) 12.756 c) 13.796 d) 12.796 e) none

Last Answer : (37÷44) + (78÷34) x (480÷85) = ? Sol: 0.841 + 2.29412 x 5.64706 =? 0.841 + 12.955 =? ? = 13.796 Answer: c)

(37÷44) + (78÷34) x (480÷85) = ? a) 11.769 b) 12.756 c) 13.796 d) 12.796 e) none

Description : (37÷44) + (78÷34) x (480÷85) = ? a) 11.769 b) 12.756 c) 13.796 d) 12.796 e) none

Last Answer : (37÷44) + (78÷34) x (480÷85) = ? Sol: 0.841 + 2.29412 x 5.64706 =? 0.841 + 12.955 =? ? = 13.796 Answer: c)

What is the approximate percentage of candidates successful to enrolled from institute A? a) 92 b) 89 c) 86 d) 96 e) 88

Description : What is the approximate percentage of candidates successful to enrolled from institute A? a) 92 b) 89 c) 86 d) 96 e) 88

Last Answer : c) 86

A moist soil sample of volume 60 cc. weighs 108 g and its dried weight is 86.4 g. If its absolute density is 2.52, the degree of saturation is (A) 54 % (B) 64 % (C) 74 % (D) 84 %

Description : A moist soil sample of volume 60 cc. weighs 108 g and its dried weight is 86.4 g. If its absolute density is 2.52, the degree of saturation is (A) 54 % (B) 64 % (C) 74 % (D) 84 %

Last Answer : (D) 84 %

If D is the diameter of a circular sewer and D’ is the top horizontal diameter of an equivalent egg shaped section, the relationship which holds good, is A. D’ = 0.64 D B. D’ = 0.74 D C. D’ = 0.84 D D. D’ = 0.94 D

Description : If D is the diameter of a circular sewer and D’ is the top horizontal diameter of an equivalent egg shaped section, the relationship which holds good, is A. D’ = 0.64 D B. D’ = 0.74 D C. D’ = 0.84 D D. D’ = 0.94 D

Last Answer : ANS: C

A sample of natural gas containing 80% methane (CH4 ) and rest nitrogen (N2 ) is burnt with 20% excess air. With 80% of the combustibles producing CO2 and the remainder going to CO, the Orsat analysis in volume percent is (A) CO2 : 6.26, CO : 1.56, O2 : 3.91, H2O :15.66, N2 : 72.60 (B) CO2 : 7.42, CO : 1.86, O2 : 4.64, N2 :86.02 (C) CO2 : 6.39, CO : 1.60, O2 : 3.99, H2O:25.96, N2 :72.06 (D) CO2 : 7.60, CO : 1.90, O2 : 4.75, N2 : 85.74

Description : A sample of natural gas containing 80% methane (CH4 ) and rest nitrogen (N2 ) is burnt with 20% excess air. With 80% of the combustibles producing CO2 and the remainder going to CO, the Orsat analysis in volume percent is (A) CO2 : 6 ... 96, N2 :72.06 (D) CO2 : 7.60, CO : 1.90, O2 : 4.75, N2 : 85.74

Last Answer : (B) CO2 : 7.42, CO : 1.86, O2 : 4.64, N2 :86.02

In the following partial sequence of mRNA, a mutation of the template DNA results in a change in codon 91 to UAA. The type of mutation is 88 89 90 91 92 93 94 GUC GAC CAG UAG GGC UAA CCG (A) Missene (B) Silent (C) Nonsense (D) Frame shit

Description : In the following partial sequence of mRNA, a mutation of the template DNA results in a change in codon 91 to UAA. The type of mutation is 88 89 90 91 92 93 94 GUC GAC CAG UAG GGC UAA CCG (A) Missene (B) Silent (C) Nonsense (D) Frame shit

Last Answer : Answer : B

Biological value of egg white protein is (A) 94 (B) 83 (C) 85 (D) 77

Description : Biological value of egg white protein is (A) 94 (B) 83 (C) 85 (D) 77

Last Answer : Answer : B

A grouped frequency distribution table with classes of equal sizes using 63-72 (72 included) as one of the class is constructed for the following data 30, 32, 45, 54, 74, 78, 108, 112, 66, 76, 88, -Maths 9th

Description : A grouped frequency distribution table with classes of equal sizes using 63-72 (72 included) as one of the class is constructed for the following data 30, 32, 45, 54, 74, 78, 108, 112, 66, 76, 88, -Maths 9th

Last Answer : NEED ANSWER

A grouped frequency distribution table with classes of equal sizes using 63-72 (72 included) as one of the class is constructed for the following data 30, 32, 45, 54, 74, 78, 108, 112, 66, 76, 88, -Maths 9th

Description : A grouped frequency distribution table with classes of equal sizes using 63-72 (72 included) as one of the class is constructed for the following data 30, 32, 45, 54, 74, 78, 108, 112, 66, 76, 88, -Maths 9th

Last Answer : (b) We arrange the given data into groups like 13-22,23-32 103-112. (since, our data is from 14 to 112). The class width in this case is 9. Now, the given data can be arranged in tabular form as follows. Hence, the number of classes in distribution will be 10.

There are two groups of a class consisting of 72 and 88 students respectively. If the average weight of 1st group is 80kg and the weight of 2nd group is 70kg. Find the average weight of whole class? A) 64kg B) 74.5kg C) 84.5kg D) 54kg

Description : There are two groups of a class consisting of 72 and 88 students respectively. If the average weight of 1st group is 80kg and the weight of 2nd group is 70kg. Find the average weight of whole class? A) 64kg B) 74.5kg C) 84.5kg D) 54kg

Last Answer :  B)  Total weight of (72 + 88) = (72*80) + (88 * 70)kg = (5760 + 6160)kg  = 11,920kg  Average weight of the whole class = 11920 / 160  = 74.5 kg

Which one of the following longitudes determines the Indian Standard Time? [CDS 2009] (a) 85.5ºE (b) 86.5ºE (c) 84.5ºE (d) 82.5ºE

Description : Which one of the following longitudes determines the Indian Standard Time? [CDS 2009] (a) 85.5ºE (b) 86.5ºE (c) 84.5ºE (d) 82.5ºE

Last Answer : Ans: (d)

Presence of nitrogen in high concentration in contaminated air reduces partial pressure of oxygen in lungs, thereby causing asphyxia (suffocation) leading to death from oxygen deficiency. Concentration of N2 in contaminated air at which it acts as a natural asphyxiate is ≥ __________ percent. (A) 84 (B) 88 (C) 80 (D) 92

Description : Presence of nitrogen in high concentration in contaminated air reduces partial pressure of oxygen in lungs, thereby causing asphyxia (suffocation) leading to death from oxygen deficiency. Concentration of N2 in contaminated air at ... asphyxiate is ≥ __________ percent. (A) 84 (B) 88 (C) 80 (D) 92

Last Answer : (A) 84

The percentage of Nitrogen in Air is - (1) 0.94 (2) 0.03 (3) 78.03 (4) 85.02

Description : The percentage of Nitrogen in Air is - (1) 0.94 (2) 0.03 (3) 78.03 (4) 85.02

Last Answer : (3) 78.03 Explanation: By volume, thy air contains 78.09% nitrogen, 20.95% oxygen, 0.93% argon, 0.039% carbon dioxide, and small amounts of other gases. Air also contains a variable amount of water vapor, on average around 1%.

The percentage of litrogen in Air is (1) 0.94 (2) 0.03 (3) 78.03 (4) 85.02

Description : The percentage of litrogen in Air is (1) 0.94 (2) 0.03 (3) 78.03 (4) 85.02

Last Answer : 78.03

In 10 years time, Ram will be 2 times older than his son. 5 years ago he was 9 times as old as his son. What will be the sum of the ages of Ram and his son 3 years from now? a) 46 b) 66 c) 86 d) 76 e) None of these

Description : In 10 years time, Ram will be 2 times older than his son. 5 years ago he was 9 times as old as his son. What will be the sum of the ages of Ram and his son 3 years from now? a) 46 b) 66 c) 86 d) 76 e) None of these

Last Answer : Let current age of Ram and his son be R,S (R + 10) = 3(S + 10) [2 times older = 3 times the age] R = 3S + 20 (R - 5) = 9(S - 5) 3S + 20 – 5= 9S – 45 3S + 15 = 9S – 45 or S = 10., R = 50 sum of ages 3 years from now = 10 + 3 + 50 + 3 = 66. Answer: b)

A contract is to be completed in 46 days and 117 men were set to work, each working 8 hours a day. After 33 days, 4/7 of the work completed. How many additional men may be employed so that the work may be completed in time, each man now working 9 hours a day? a) 79 b) 80 c) 81 d) 82 e) 83

Description : A contract is to be completed in 46 days and 117 men were set to work, each working 8 hours a day. After 33 days, 4/7 of the work completed. How many additional men may be employed so that the work may be completed in time, each man now working 9 hours a day? a) 79 b) 80 c) 81 d) 82 e) 83

Last Answer : work finished by 33 days = 33 117 8 = 30888 man hours 4/7th of the job is finished. remaining work = 1-(4/7) = 3/7 if 4/7th of the work is 30888 hours, then 3/7th of the work = [30888 ... 9 number of men = 198 since, 117 men are already working, additional men required = 198 - 117 = 81 Answer: c)

Evaluate: 48×52 + 61×59 + 77×83 a) 12486 b) 10996 c) 13406 d) 13206 e) None of these

Description : Evaluate: 48×52 + 61×59 + 77×83 a) 12486 b) 10996 c) 13406 d) 13206 e) None of these

Last Answer : 48×52 = (50 - 2) × (50 + 2) = 502 – 22 = 2496 61 × 59 = (60 + 1) × (60 - 1) = 602 – 12 = 3599 77 × 83 = (80 - 3) × (80 + 3) = 802 – 32 = 6391 Sum = 2496 + 3599 + 6391 = 12486 Answer: a)

3.25 × 16 + 7.25 × 8 = ? + 75 a) 83 b) 75 c) 61 d) 53 e) 35

Description : 3.25 × 16 + 7.25 × 8 = ? + 75 a) 83 b) 75 c) 61 d) 53 e) 35

Last Answer : ? + 75 = 52 + 58 = 110 ? = 110 – 75 = 35 Answer: e)

3, 0, -4, -17, -74, ? a) -309 b) -302 c) -363 d) -290 e) -377

Description : 3, 0, -4, -17, -74, ? a) -309 b) -302 c) -363 d) -290 e) -377

Last Answer : The pattern is 3×1-3 = 0 ,0×2-4 = -4 ,-4×3-5 = -17,-17×4-6 =-74,-74×5-7=-377 Answer: e)

Mr. Xavier invested a certain amount in Debit and Equity funds in the ratio of 4 : 5 respectively. At the end of one year, he earned a total dividend of 30% on his investment. After one year he reinvested the amount including dividend in the ratio of 6 : 7 in Debt and Equity Funds. If the amount reinvested in Equity Funds was Rs. 94,500/-, what was the original amount invested in Equity Funds? (a) Rs. 75,000/- (b) Rs. 81,000/- (c) Rs. 60,000/- (d) Rs. 65,000/- (e) None of these

Description : Mr. Xavier invested a certain amount in Debit and Equity funds in the ratio of 4 : 5 respectively. At the end of one year, he earned a total dividend of 30% on his investment. After one year he reinvested the amount including ... . 81,000/- (c) Rs. 60,000/- (d) Rs. 65,000/- (e) None of these

Last Answer : (a) Rs. 75,000/-

12 17 32 57 ? 137 a) 98 b) 95 c) 37 d) 92 e) none of these

Description : 12 17 32 57 ? 137 a) 98 b) 95 c) 37 d) 92 e) none of these

Last Answer : The pattern of the number series is 12+5×1=17, 17+5×3=32, 32+5×5=57 57+5×7=92, 92+5×9=137 Answer: d)

112 111 119 92 156 31 ? a) 375 b) 287 c) 387 d) 247 e) None of these

Description : 112 111 119 92 156 31 ? a) 375 b) 287 c) 387 d) 247 e) None of these

Last Answer : Answer: d)

15, 18, 21.6, 25.92, ? a) 31.104 b) 31.014 c) 31.114 d) 30.114 e) 30.004

Description : 15, 18, 21.6, 25.92, ? a) 31.104 b) 31.014 c) 31.114 d) 30.114 e) 30.004

Last Answer : The series is a geometric progression with a common ratio 1.2. Hence the next term is 25.92 × 1.2 = 31.104 Answer: a)

In what ratio must a person mix three kinds of Oats costing him Rs 1.20,Rs 1.44 and Rs 1.74 per Kg so that the mixture may be worth Rs 1.41 per Kg? A)11:77:7 B) 11:45:7 C)25:45:8 D) 27:23:6

Description : In what ratio must a person mix three kinds of Oats costing him Rs 1.20,Rs 1.44 and Rs 1.74 per Kg so that the mixture may be worth Rs 1.41 per Kg? A)11:77:7 B) 11:45:7 C)25:45:8 D) 27:23:6

Last Answer : A)11:77:7

What is the total number of literate people in city 'A'? 1) 21.62 lakh 2) 22.18 lakh 3) 23.94 lakh 4) 24.63 lakh 5) 25.02 lakh

Description : What is the total number of literate people in city 'A'? 1) 21.62 lakh 2) 22.18 lakh 3) 23.94 lakh 4) 24.63 lakh 5) 25.02 lakh

Last Answer : 3) 23.94 lakh

5287 – 176.22 – 78.584 = ? a) 5023.196 b) 5032.196 c) 5302.196 d) 5203.196 e) None of these

Description : 5287 – 176.22 – 78.584 = ? a) 5023.196 b) 5032.196 c) 5302.196 d) 5203.196 e) None of these

Last Answer : 5287 – 176.22 – 78.584 = 5032.196 Answer: b)

1 15 16 31 47 78 125 ? a) 172 b) 203 c) 139 d) 167 e) None of these

Description : 1 15 16 31 47 78 125 ? a) 172 b) 203 c) 139 d) 167 e) None of these

Last Answer : 15 + 1 = 16 15 + 16 = 31 16 + 31 = 47 31 + 47 = 78 47 + 78 = 125 78 + 125 = 203 Answer: b)

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)

A man buys milk at Rs. 2.40 per litre. He adds one-third water to it and sells the mixture at Rs. 2.88 per litre. Find his profit percentage. (Assume water is free). a) 21% b) 30% c) 40% d) 50% e) 60%

Description : A man buys milk at Rs. 2.40 per litre. He adds one-third water to it and sells the mixture at Rs. 2.88 per litre. Find his profit percentage. (Assume water is free). a) 21% b) 30% c) 40% d) 50% e) 60%

Last Answer : Let say he purchases 3 litres of milk whose cost is Rs. 7.20 He adds 1 litre of water and the mixture of 4 litres is sold at Rs 2.88 × 4 = Rs11.52 Profit percentage = (11.52 – 7.20) × 100.0/7.20 = 4.32 × 100.0/7.2 = 60% Answer: e)

11 6 7 12 26 67.5 ? a) 205.5 b) 195.5 c) 216.5 d) 200.5 e) 222.5

Description : 11 6 7 12 26 67.5 ? a) 205.5 b) 195.5 c) 216.5 d) 200.5 e) 222.5

Last Answer : The series is: 11 × 0.5 + 0.5 = 6; 6 × 1 + 1 = 7; 7 × 1.5 + 1.5 = 12; 12 × 2 + 2 = 26; 26 × 2.5 + 2.5 = 67.5; 67.5 × 3 + 3 = 205.5. Answer: a)

987.67 × 123.35 ÷ 9 = ? a) 13411 b) 13621 c) 13489 d) 13551 e) 13721

Description : 987.67 × 123.35 ÷ 9 = ? a) 13411 b) 13621 c) 13489 d) 13551 e) 13721

Last Answer : ? = (987 × 123) / 9 = 13489 Answer: c)

A box contains 4 green, 5 red and 6 white balls. Three balls are drawn randomly. What is the probability that the balls drawn are of different colours? a) 24/91 b) 67/91 c) 21/91 d) 70/91 e) 3/13

Description : A box contains 4 green, 5 red and 6 white balls. Three balls are drawn randomly. What is the probability that the balls drawn are of different colours? a) 24/91 b) 67/91 c) 21/91 d) 70/91 e) 3/13

Last Answer : Answer is: a)

If 6 years are subtracted from the present age of Anuj and the remainder is divided by 18, then the present age of his grandson Gopal is obtained.If Gopal is 2 years younger to Mohan whose age is 5 years, then what is the age of Anuj? A.44 B.60 C.80 D.92

Description : If 6 years are subtracted from the present age of Anuj and the remainder is divided by 18, then the present age of his grandson Gopal is obtained.If Gopal is 2 years younger to Mohan whose age is 5 years, then what is the age of Anuj? A.44 B.60 C.80 D.92

Last Answer : Answer : B (60) Explanation - Let Anuj's age be X Gopal is 2 years younger than Mohan, so Gopal is 3 years (i.e 5 - 2 = 3) If Arun had born 6 years before, his age would had been X - 6. As per the ... as that of Gokul's age. i.e. (X - 6) /18= 3 X-6 =3 x 18 x = 60

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)

22.222 + 2220.2 + 202.20 + 0.0202 – 2.0022 = ? a) 2244.24 b) 2442.64 c) 2424.44 d) 2444.64 e) 2224.84

Description : 22.222 + 2220.2 + 202.20 + 0.0202 – 2.0022 = ? a) 2244.24 b) 2442.64 c) 2424.44 d) 2444.64 e) 2224.84

Last Answer : b) 2442.64

28, 42, 84, 210, 630, ? a) 2675 b) 2515 c) 2445 d) 2375 e) 2205

Description : 28, 42, 84, 210, 630, ? a) 2675 b) 2515 c) 2445 d) 2375 e) 2205

Last Answer : Series is × 1.5, ×2, ×2.5, ×3, ×3.5 Answer: e)

√(4.9 − 0.49) = ? a) 10.42 b) 5.84 c) 12.84 d) 2.1 e) None of these

Description : √(4.9 − 0.49) = ? a) 10.42 b) 5.84 c) 12.84 d) 2.1 e) None of these

Last Answer : Answer: d)

(54.78)^2= ? 1) 3000 2) 3300 3) 3500 4) 3700 5) 3900

Description : (54.78)^2= ? 1) 3000 2) 3300 3) 3500 4) 3700 5) 3900

Last Answer : 1) 3000