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

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
by

1 Answer

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)
by

Related questions

(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)

The least number, which when divided by 48, 60, 72, 108 and 140 leaves 38, 50, 62, 98 and 130 as remainders respectively, is: A.11115 B.15110 C.15130 D.15310 E.None of these

Description : The least number, which when divided by 48, 60, 72, 108 and 140 leaves 38, 50, 62, 98 and 130 as remainders respectively, is: A.11115 B.15110 C.15130 D.15310 E.None of these

Last Answer : Answer – B (15110) Explanation – Here (48 – 38) = 10, (60 – 50) = 10, (72 – 62) = 10, (108 – 98) = 10 & (140 – 130) = 10. Required number = (L.C.M. of 48, 60, 72, 108, 140) – 10 = 15120 – 10 = 15110

Reshma purchased 120 chairs at price of Rs. 110 per chair. He sold 30 chairsat a profit of Rs. 12 per chair and 75 chairs at profit of Rs. 14 per chairs. The remaining chairs were sold at a loss of Rs. 7 per chairs. What is the average profit per table? A. Rs 10.56 B. Rs 10.87 C. Rs 12.123 D. Rs 12.67

Description : Reshma purchased 120 chairs at price of Rs. 110 per chair. He sold 30 chairsat a profit of Rs. 12 per chair and 75 chairs at profit of Rs. 14 per chairs. The remaining chairs were sold at a loss of Rs. 7 per chairs ... the average profit per table? A. Rs 10.56 B. Rs 10.87 C. Rs 12.123 D. Rs 12.67 

Last Answer : Answer – B (Rs 10.875) Explanation – Total C.P = Rs. (120 x 110) =Rs. 13200. Total S.P = Rs.[(30 x 110 + 30 x 12) + (75 x 110 + 75 x 14) + (15 x 110 – 15 x 7) =Rs..14505 Average profit = Rs. (14505 – 13200) /120 = Rs. 1305/120= 10.875

19 + (12 ÷ 4 × (240 × 1 ÷ 180)) × 6 ÷ (101 + 98 ÷ 7 - 113) – (7 ÷ 2 + 14 – 3.5 – 9 × 8.5) × 4 a) 201 b) 241 c) -241 d) -281 e) 281

Description : 19 + (12 ÷ 4 × (240 × 1 ÷ 180)) × 6 ÷ (101 + 98 ÷ 7 - 113) – (7 ÷ 2 + 14 – 3.5 – 9 × 8.5) × 4 a) 201 b) 241 c) -241 d) -281 e) 281

Last Answer : 240 1 180 = 4/3 12 4 (240 1 180) = 3 (4/3) = 4 101 + 98 7 - 113 = 101 + 14 - 113 = 2 19 + (12 4 (240 1 180)) 6 (101 + 98 7 - 113) = 19 + 4 6 2 = 19 + 4 3 = 31 7 2 ... 180)) 6 (101 + 98 7 - 113) - (7 2 + 14 - 3.5 - 9 8.5) 4 = 31 - (-250) = 281 Answer: e)

23.98 × 62.49 × 12.03 = ? × 29.97 a) 500 b) 600 c) 900 d) 450 e) 750

Description : 23.98 × 62.49 × 12.03 = ? × 29.97 a) 500 b) 600 c) 900 d) 450 e) 750

Last Answer : 24× 62.5× 12 = ? × 30 24× 750 = ? × 30 ? = (24× 750) / 30 = 600 Answer: b)

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)

49/55 of 365/517 of 5111 = ? (Take 4 digits after the decimal point) a) 3015.6001 b) 3148.8123 c) 3214.2379 d) 3348.6179 e) 3358.6279

Description : 49/55 of 365/517 of 5111 = ? (Take 4 digits after the decimal point) a) 3015.6001 b) 3148.8123 c) 3214.2379 d) 3348.6179 e) 3358.6279

Last Answer : 49/55 of 365/517 of 5111 = ? Sol: 0.8909 x 0.7059 x 5111 = ? 0.62888631 x 5111 =? ? = 3214.2379 Answer: c)

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 digits of a two-digit number are in the ratio of 2 : 3 and the number obtained by interchanging the digits is bigger than the original number by 27.What is the original number? 1. 63 2. 48 3. 96 4. 69 5. 66

Description : The digits of a two-digit number are in the ratio of 2 : 3 and the number obtained by interchanging the digits is bigger than the original number by 27.What is the original number? 1. 63 2. 48 3. 96 4. 69 5. 66

Last Answer : Answer- 4 (69) Explanation:- lets digit at tens and ones place be 2x and 3x So,number is 2x*10 + 3x After interchanging number will be 3x*10 +2x so, [3x*10 +2x] – [2x*10 + 3x]27 we get,x=3 so number =69

The age of vaishu is 18 times that of her daughter raiza. If the age of raiza is 6years, what is the age of vaishu? a) 98 b) 110 c) 90 d) 108

Description : The age of vaishu is 18 times that of her daughter raiza. If the age of raiza is 6years, what is the age of vaishu? a) 98 b) 110 c) 90 d) 108

Last Answer : D vaishu’s present age = x vaishu’a age is 18 times her daughter’s age. Daughter’s age = 6. Therefore, 18 times of 6 = x 18 x 6 = x x= 108 years = vaishu’s age.

Convert binary 1001 to octal. 131) A) 1018 B) 98 C) 108 D) 118

Description : Convert binary 1001 to octal. 131) A) 1018 B) 98 C) 108 D) 118

Last Answer : D) 118

What is the frequency range of an aircraft’s Very High Frequency (VHF) communications? A. 118.000 MHz to 136.975 MHz (worldwide up to 151.975 MHz) B. 108.000 MHz to 117.95 MHz C. 329.15 MHz to 335.00 MHz D. 2.0000 MHz to 29.999 MHz

Description : What is the frequency range of an aircraft’s Very High Frequency (VHF) communications? A. 118.000 MHz to 136.975 MHz (worldwide up to 151.975 MHz) B. 108.000 MHz to 117.95 MHz C. 329.15 MHz to 335.00 MHz D. 2.0000 MHz to 29.999 MHz

Last Answer : A. 118.000 MHz to 136.975 MHz (worldwide up to 151.975 MHz)

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)

3, 5, 9, 15, 25, 41 , 67, ? (a) 108 (b) 52 (c) 110 (d) 111

Description : 3, 5, 9, 15, 25, 41 , 67, ? (a) 108 (b) 52 (c) 110 (d) 111

Last Answer : The difference between the number in this series is the sum of two previous differences .

5, 8, 12, 18, 24, 30, 36, 42, ____ a) 52 b) 46 c) 48 d) 54 e) 56

Description : 5, 8, 12, 18, 24, 30, 36, 42, ____ a) 52 b) 46 c) 48 d) 54 e) 56

Last Answer : Each term in the series is the addition of successive prime numbers. Like Prime numbers are2, 3, 5, 7, 11, 13, 17, 19, 23, 29 and so on. So first term is addition of (2 + 3). Ie 5, then second term is (3 + 5) ie ... +17)ie. 30, then (17+19)ie.36, then (19+23)ie., 42, then (23+29)i.e. 52. 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)

?% 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)

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

By selling 45 lemons for Rs 40, a man loses 20%.How many should he sell for Rs 24 to gain 20% in the transaction? a) 19 b) 22 c) 20 d) 18 e) 14

Description : By selling 45 lemons for Rs 40, a man loses 20%.How many should he sell for Rs 24 to gain 20% in the transaction? a) 19 b) 22 c) 20 d) 18 e) 14

Last Answer : Selling price = Rs. 40 Loss = 20% Therefore, cost price = 40×100 /(100 - 20) = Rs. 50 Therefore, cost price of each lemon = Rs. 50/45 Now, selling price = Rs. 24 Profit = 20% Cost price = 24×100/120 = Rs. 20 Number of lemons for Rs. 20 = 20/(50/45) = 900/50 = 18 Answer: d)

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)

The average weight of a group of 75 women was calculated as 47 kg. It was later discovered that the weight of one of the women was read as 45 kg, whereas her actual weight was 25kg. What is the actual average weight of the group of 75 women? (rounded off to two digits after decimal) (a) 46.73 kg (b) 46.64 kg (c) 45.96 kg (d) Cannot be determined (e) None of these

Description : The average weight of a group of 75 women was calculated as 47 kg. It was later discovered that the weight of one of the women was read as 45 kg, whereas her actual weight was 25kg. What is the actual average weight ... kg (b) 46.64 kg (c) 45.96 kg (d) Cannot be determined (e) None of these

Last Answer : (a) 46.73 kg

Suresh and Ramesh are twins. In a form by mistake, Suresh reverses the last 2 digits of his year of birth, and this makes him 9 years older than Ramesh. With this mistake, the sum of their ages in 2015 becomes 43. How old will Ramesh be 2021? a) 17 b) 19 c) 21 d) 23 e) 25

Description : Suresh and Ramesh are twins. In a form by mistake, Suresh reverses the last 2 digits of his year of birth, and this makes him 9 years older than Ramesh. With this mistake, the sum of their ages in 2015 becomes 43. How old will Ramesh be 2021? a) 17 b) 19 c) 21 d) 23 e) 25

Last Answer : Let the correct ages of Ramesh and Suresh be X in 2015 (since they are twins, they will have the same age) Given: 2x + 9 = 43 Or x = 17 years in 2015. Age in 2021 = 17 + 6 = 23 Note: year of birth = 2015 – 17 = 1998. Reversing, Suresh = 1989 or 9 year older. Answer: d)

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)

2 3 5 8 14 23 41 69 a) 5 b) 8 c) 14 d) 41 e) 69

Description : 2 3 5 8 14 23 41 69 a) 5 b) 8 c) 14 d) 41 e) 69

Last Answer : The series is an alternate series, having S 1 = 2 5 14 41; ×3 – 1 in each term S 2 = 3 8 23 69; ×3 – 1 in each term Answer: e)

What is the per cent increase in the number of products manufactured by the company in the year 2006 from the previous year ? (Rounded off to two digits after decimal) (1) 19·25 (2) 33·33 (3) 10·45 (4) 42·66 (5) None of these

Description : What is the per cent increase in the number of products manufactured by the company in the year 2006 from the previous year ? (Rounded off to two digits after decimal) (1) 19·25 (2) 33·33 (3) 10·45 (4) 42·66 (5) None of these

Last Answer : (5) None of these

The number of products sold by the company in the year 2004 is what per cent of the number of products manufactured by it in that year ? (Rounded off to two digits after decimal) (1) 71·43 (2) 67·51 (3) 81·67 (4) 56·29 (5) None of these

Description : The number of products sold by the company in the year 2004 is what per cent of the number of products manufactured by it in that year ? (Rounded off to two digits after decimal) (1) 71·43 (2) 67·51 (3) 81·67 (4) 56·29 (5) None of these

Last Answer : 1) 71·43 v

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

108 54 36 18 9 6 4 a) 54 b) 36 c) 18 d) 9 e) 6

Description : 108 54 36 18 9 6 4 a) 54 b) 36 c) 18 d) 9 e) 6

Last Answer : The series is ÷2, ÷1.5 alternately. Answer: d)

A, B and C play cricket. A's runs are to B's runs and B's runs are to C's as 3:2 , 3:2. They get altogether 342 runs. How many runs did A make? A) 162 B) 108 C) 72 D) 78 E) None

Description : A, B and C play cricket. A's runs are to B's runs and B's runs are to C's as 3:2 , 3:2. They get altogether 342 runs. How many runs did A make? A) 162 B) 108 C) 72 D) 78 E) None

Last Answer : Answer: A A:B = 3:2 = 9:6; B:C = 3:2 = 6:4 (making B equal) Therefore, A:B:C = 9:6:4 Therefore, the runs made by A = (9/19) X 342 = 162.

Two trains of equal lengths take 10 seconds and 15 seconds respectively to cross a telegraph post. If the length of each train be 120 metres, in what time (in seconds) will they cross each other travelling in opposite direction? A.12 sec B.18 sec C.14 sec D.25 sec E.None of these

Description : Two trains of equal lengths take 10 seconds and 15 seconds respectively to cross a telegraph post. If the length of each train be 120 metres, in what time (in seconds) will they cross each other travelling in opposite direction? A.12 sec B.18 sec C.14 sec D.25 sec E.None of these

Last Answer : Answer – A (12 sec) Explanation – Speed of the first train = [120 / 10] m/sec = 12 m/sec. Speed of the second train = [120 / 15] m/sec = 8 m/sec. Relative speed = (12 + 8) = m/sec = 20 m/sec. ∴ Required time = (120 + 120) / 20 secc = 12 sec

1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20, 21, 23, 24, 25, 26, 27, 28, 29, 30, 31, 32, 33, 34, 35, 36, 37, 38, 39, 40, 41, 42, 43, 44, 45, 46, 47, 48, 49, 50'. What number is missing? -Riddles

Description : 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20, 21, 23, 24, 25, 26, 27, 28, 29, 30, 31, 32, 33, 34, 35, 36, 37, 38, 39, 40, 41, 42, 43, 44, 45, 46, 47, 48, 49, 50'. What number is missing? -Riddles

Last Answer : 22

A national random sample of 20 ACT scores from 2010 is listed below. Calculate the sample mean and standard deviation. 29, 26, 13, 23, 23, 25, 17, 22, 17, 19, 12, 26, 30, 30, 18, 14, 12, 26, 17, 18 a. 20.50, 5.79 b. 20.50, 5.94 c. 20.85, 5.79 d. 20.85, 5.94

Description : A national random sample of 20 ACT scores from 2010 is listed below. Calculate the sample mean and standard deviation. 29, 26, 13, 23, 23, 25, 17, 22, 17, 19, 12, 26, 30, 30, 18, 14, 12, 26, 17, 18 a. 20.50, 5.79 b. 20.50, 5.94 c. 20.85, 5.79 d. 20.85, 5.94

Last Answer : Katie earned 84,92,84,75 and 70 on her first 5 tests. What is the minimum grade Katie needs to earn on the next test to have a mean of 84?

(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)

12 22 69 272 1365 ? a) 8196 b) 8184 c) 8195 d) 6830 e) None of these

Description : 12 22 69 272 1365 ? a) 8196 b) 8184 c) 8195 d) 6830 e) None of these

Last Answer : 12 × 2 - 2 = 22 22 × 3 + 3 = 69 69 × 4 - 4 = 272 272 × 5 + 5 = 1365 1365 × 6 - 6 = 8184 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.

The least multiple of 7, which leaves a remainder of 4, when divided by 6, 9, 15 and 18 is: A.68 B.98 C.180 D.364 E.None of these

Description : The least multiple of 7, which leaves a remainder of 4, when divided by 6, 9, 15 and 18 is: A.68 B.98 C.180 D.364 E.None of these

Last Answer : Answer – D (364) Explanation – L.C.M. of 6, 9, 15 and 18 is 90. Let required number be 90k + 4, which is multiple of 7. Least value of k for which (90k + 4) is divisible by 7 is k = 4. Required number = (90 x 4) + 4 = 364.

A two digit number is such that the sum of the digits is 11.When the number with the same digits is reversed is subtracted from this number, the difference is 9.What is the number? 1. 23 2. 24 3. 65 4. 14 5. 32

Description : A two digit number is such that the sum of the digits is 11.When the number with the same digits is reversed is  subtracted from this number, the difference is 9.What is the number? 1. 23 2. 24 3. 65 4. 14 5. 32

Last Answer : Answer- 3(65) Explanation:- let the number be written as xy x-ten’s place y units place x+y=11…………………1 (10x+y)-(10y+x)=9 10x+y-10y-x=9 9x-9y=9 x-y=1………………….2 Add (1) & (2) 2x=12 x=6 y=5 the number is 65

The ages of Gaurav and Goutham will be in the ratio 5 : 7 after ten years from now and will be in the ratio 13 : 18 after twelve years from now. Find the ratio of the sum of their ages 10 years hence to the sum of their ages 12 years hence. 1 : 25 : 27 2 : 13 : 15 3 : 14 : 17 4 : 18 : 23 5 : 30 : 31

Description : The ages of Gaurav and Goutham will be in the ratio 5 : 7 after ten years from now and will be in the ratio 13 : 18 after twelve years from now. Find the ratio of the sum of their ages 10 years hence to the sum of their ages 12 ... hence. 1 : 25 : 27 2 : 13 : 15 3 : 14 : 17 4 : 18 : 23 5 : 30 : 31

Last Answer : 5 : 30 : 31

The definition of Money bill is provided in which article of the Indian Constitution- a) Article 10 b) Article -19 c) Article 110 d)Article-111

Description : The definition of Money bill is provided in which article of the Indian Constitution- a) Article 10 b) Article -19 c) Article 110 d)Article-111

Last Answer : c) Article 110

Area of a rectangle is equal to the area of circle whose radius is 14 cms. If the breadth of the rectangle is 22 cms. What is its length? a) 27 cms b) 28 cms c) 25 cms. d) 29 cms e) None of these

Description : Area of a rectangle is equal to the area of circle whose radius is 14 cms. If the breadth of the rectangle is 22 cms. What is its length? a) 27 cms b) 28 cms c) 25 cms. d) 29 cms e) None of these

Last Answer : According to question, l×b = ¶r2 l ×22 = 22/7 ×14 × 14 l = 22/7 ×14 × 14/22 cms = 28 cms. Answer: b)

One year ago the ratio between Sam and Ash’s age was 4:3. One year hence the ratio of their ages will be 5:4. What is the sum of their present ages in years? A.14 B.16 C.18 D.12 E.None of these

Description : One year ago the ratio between Sam and Ash’s age was 4:3. One year hence the ratio of their ages will be 5:4. What is the sum of their present ages in years? A.14 B.16 C.18 D.12 E.None of these

Last Answer : Answer - B (16) Explanation - Let one year ago Sam's age be 4X years And, Ash's age be 3X years Present age of Sam = (4X + 1) years Present age of Ash= (3X + 1) years One year hence Sam's age = (4X + 2) ... = 2 Sum of their present ages = 4X + 1 + 3X + 1 = 7X + 2 = 7 x 2 + 2 = 16 years.

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)

A two digit number exceeds forty percent of another two digit number by 16. If the sum of these two digit numbers is 72, which of the following is the difference between these two digit numbers ? a) 10 b) 6 c) 8 d) 14 e) None of these

Description : A two digit number exceeds forty percent of another two digit number by 16. If the sum of these two digit numbers is 72, which of the following is the difference between these two digit numbers ? a) 10 b) 6 c) 8 d) 14 e) None of these

Last Answer : c) 8 Let first two digit no. and second two digit no. be x and y respectively. x = y × 40/100 +16 x - 2y/5 = 16....(i) x + y = 72...(ii) From equation (i) and (ii) x = 32, y = 40 Difference = 8 Answer: c)

Sum of Rs.118 was shared among 50 boys and girls,each girl receive Rs.2.60 and boy receive Rs.1.80.Find the number of girls. A) 15 B)25 C)30 D)35

Description : Sum of Rs.118 was shared among 50 boys and girls,each girl receive Rs.2.60 and boy receive Rs.1.80.Find the number of girls. A) 15 B)25 C)30 D)35

Last Answer : D)35

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)

39.98 × 85.01 – 79.9 × 34.9 + 175.03 × 12.08 = ? a) 4800 b) 4300 c) 2700 d) 2100 e) 3400

Description : 39.98 × 85.01 – 79.9 × 34.9 + 175.03 × 12.08 = ? a) 4800 b) 4300 c) 2700 d) 2100 e) 3400

Last Answer : ? = 40 × 85 – 80 × 35 + 175 × 12 = 3400 – 2800 + 2100 = 2700 Answer: c)

I. 56x^2 – 127x + 72 = 0 II. 6y^2 – 17y + 12 = 0 1 : if x ≥ y 2 : if x ≤ y 3 : if x > y 4 : if x < y 5 : if x = y or relationship cannot be established

Description : I. 56x^2 – 127x + 72 = 0 II. 6y^2 – 17y + 12 = 0 1 : if x ≥ y 2 : if x ≤ y 3 : if x > y 4 : if x < y 5 : if x = y or relationship cannot be established

Last Answer : 4 : if x < y