The H.C.F. of two numbers is 11 and their L.C.M. is 7700. If one of the numbers is 275, then the other is: A.269 B.275 C.308 D.310 E.None of these

The H.C.F. of two numbers is 11 and their L.C.M. is 7700. If one of the numbers is 275, then the other is: A.269 B.275 C.308 D.310 E.None of these
by

1 Answer

Answer :

Answer – C (308)

Explanation –

Other number =[11 x 7700]/275 = 308
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Related questions

The product of two numbers is 4107. If the H.C.F. of these numbers is 37, then the greater number is: A.124 B.100 C.111 D.175 E.None of these

Description : The product of two numbers is 4107. If the H.C.F. of these numbers is 37, then the greater number is: A.124 B.100 C.111 D.175 E.None of these

Last Answer : Answer – C (111) Explanation – Let the numbers be 37a and 37b. Then, 37a x 37b = 4107 ab = 3. Now, co-primes with product 3 are (1, 3). So, the required numbers are (37 x 1, 37 x 3) i.e., (37, 111). Greater number = 111.

The H.C.F of two numbers is 11 and their L.C.M is 7700. If one of these numbers is 275, then find the other number.

Description : The H.C.F of two numbers is 11 and their L.C.M is 7700. If one of these numbers is 275, then find the other number.

Last Answer : Ans 2. Product of two number s = product of their H.C.F. and L.C.M. required number = 11 X 7700/275 = 308

The product of two numbers is 2028 and their H.C.F. is 13. The number of such pairs is: A.1 B.2 C.3 D.5 E.None of these

Description : The product of two numbers is 2028 and their H.C.F. is 13. The number of such pairs is: A.1 B.2 C.3 D.5 E.None of these

Last Answer : Answer – B (2) Explanation – Let the numbers 13a and 13b. Then, 13a x 13b = 2028 ab = 12. Now, the co-primes with product 12 are (1, 12) and (3, 4).

The L.C.M. of two numbers is 48. The numbers are in the ratio 2 : 3. Then sum of the number is: A.30 B.22 C.40 D.60 E.None of these

Description : The L.C.M. of two numbers is 48. The numbers are in the ratio 2 : 3. Then sum of the number is: A.30 B.22 C.40 D.60 E.None of these

Last Answer : Answer – C (40) Explanation – Let the numbers be 2x and 3x. Then, their L.C.M. = 6x. So, 6x = 48 or x = 8. The numbers are 16 and 24. Hence, required sum = (16 + 24) = 40.

A, B and C start at the same time in the same direction to run around a circular stadium. A completes a round in 252 seconds, B in 308 seconds and C in 198 seconds, all starting at the same point. After what time will they meet again at the starting point? A.15 minutes 15 seconds B.42 minutes 30 seconds C.42 minutes D.46 minutes 12 seconds E.None of these

Description : A, B and C start at the same time in the same direction to run around a circular stadium. A completes a round in 252 seconds, B in 308 seconds and C in 198 seconds, all starting at the same point ... A.15 minutes 15 seconds B.42 minutes 30 seconds C.42 minutes D.46 minutes 12 seconds E.None of these

Last Answer : Answer – D (46 minutes 12 seconds) Explanation – L.C.M. of 252, 308 and 198 = 2772.So, A, B and C will again meet at the starting point in 2772 see i.e., 46 min. 12 sec

Six bells commence tolling together and toll at intervals of 2, 4, 6, 8, 10 and 12 seconds respectively. In 30 minutes, how many times do they toll together? A.8 B.11 C.13 D.16 E.None of theseSix bells commence tolling together and toll at intervals of 2, 4, 6, 8, 10 and 12 seconds respectively. In 30 minutes, how many times do they toll together? A.8 B.11 C.13 D.16 E.None of these

Description : Six bells commence tolling together and toll at intervals of 2, 4, 6, 8, 10 and 12 seconds respectively. In 30 minutes, how many times do they toll together? A.8 B.11 C.13 D.16 E.None of these

Last Answer : Answer – D (16) Explanation – L.C.M. of 2, 4, 6, 8, 10, 12 is 120.So, the bells will toll together after every 120 seconds, i.e, 2 minutes.In 30 minutes, they will toll together 30/2 + 1 = 16

The least number which should be added to 2497 so that the sum is exactly divisible by 5, 6, 4 and 3 is: A.10 B.14 C.23 D.30 E.None of these

Description : The least number which should be added to 2497 so that the sum is exactly divisible by 5, 6, 4 and 3 is: A.10 B.14 C.23 D.30 E.None of these

Last Answer : Answer – C (23) Explanation – L.C.M. of 5, 6, 4 and 3 = 60. On dividing 2497 by 60, the remainder is 37. Number to be added = (60 – 37) = 23

The least number, which when divided by 12, 15, 20 and 54 leaves in each case a remainder of 8 is: A.534 B.486 C.544 D.548 E.None of these

Description : The least number, which when divided by 12, 15, 20 and 54 leaves in each case a remainder of 8 is: A.534 B.486 C.544 D.548 E.None of these

Last Answer : Answer – D (548) Explanation – Required number = (L.C.M. of 12, 15, 20, 54) + 8 = 540 + 8 = 548.

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

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.

1/8 × 3/5 × 4/7 x 7897 = ? a) 310 b) 325 c) 375 d) 275 e) 338

Description : 1/8 × 3/5 × 4/7 x 7897 = ? a) 310 b) 325 c) 375 d) 275 e) 338

Last Answer : ? = 7897 × 4 × 3 / 8 × 5 × 7 = 338.31 = 338 (approx.) Answer: e)

One number is 3 more than another and the sum of their square is 269.Find the number. 1.10 2.12 3.15 4.20

Description : One number is 3 more than another and the sum of their square is 269.Find the number. 1.10 2.12 3.15 4.20

Last Answer : Answer-1 (10) Explanation:- x is one number x+3 the other x^2 +x^2 +6x+9 ; (that is (x+3)(x+3) =269 2x^2+6x -260=0 x^2+3x-130=0 , dividing by 2. (x+13)(x-10)=0 one number is 10, the other is 13 squared 10=100 square 13=169 sum is 269

A 300 metre long train crosses a platform in 39 seconds while it crosses a signal pole in 18 seconds. What is the length of the platform? A.310 m B.350 m C.600 m D.490 m E.None of these

Description : A 300 metre long train crosses a platform in 39 seconds while it crosses a signal pole in 18 seconds. What is the length of the platform? A.310 m B.350 m C.600 m D.490 m E.None of these

Last Answer : Answer - B (350 m) Explanation - when it cross a pole actually it is crossing itself so, S = 300/18 let length of platform be p relative speed = total lengths/time speed of ... pole,man and other is of considerable length like platform ,bridge. when train crosses pole,man it crosses itself

264 262 271 243 ? a) 302 b) 304 c) 306 d) 308 e) None of these

Description : 264 262 271 243 ? a) 302 b) 304 c) 306 d) 308 e) None of these

Last Answer : 264 – 1³ + 1 = 262 262 + 2³ + 1 = 271 271 – 3³ + 1 = 243 Answer: d)

The L.C.M of two numbers is 14 times their H.C.F . The sum of the L.C.M. and the H.C.F . is 600. If one number is 280 , then the other number is (a) 40 (b) 60 (c) 80 (d) 100

Description : The L.C.M of two numbers is 14 times their H.C.F . The sum of the L.C.M. and the H.C.F . is 600. If one number is 280 , then the other number is (a) 40 (b) 60 (c) 80 (d) 100

Last Answer : 80

How many students studying Dentistry know only either English or Hindi? (1) 166 (2) 162 (3) 308 (4) 198 (5) 248

Description : How many students studying Dentistry know only either English or Hindi? (1) 166 (2) 162 (3) 308 (4) 198 (5) 248

Last Answer : 4) 198

175 × 28 + 275 × 28 =? a) 11800 b) 12600 c) 12800 d) 11600 e) 12200

Description : 175 × 28 + 275 × 28 =? a) 11800 b) 12600 c) 12800 d) 11600 e) 12200

Last Answer : Answer: b)

(84111)^1/2= ? 1) 240 2) 270 3) 330 4) 290 5) 310

Description : (84111)^1/2= ? 1) 240 2) 270 3) 330 4) 290 5) 310

Last Answer : 4) 290

Two pipes fills a tank in 12 hours and 15 hours respectively while a third pipe empties the tank in 18 hours. !f all three pipes operate simultaneously, in what time will the tank be full? a) 10 hours b) 10 10/17 hours c) 11 hours d) 9 hours e) 12 1/17 hours

Description : Two pipes fills a tank in 12 hours and 15 hours respectively while a third pipe empties the tank in 18 hours. !f all three pipes operate simultaneously, in what time will the tank be full? a) 10 hours b) 10 10/17 hours c) 11 hours d) 9 hours e) 12 1/17 hours

Last Answer : b) 10 10/17 hours

The LCM of two numbers is 45 times their HCF. If one of the numbers is 125 and the sum of HCF and LCM is 1150, then the other number is? a) 100 b) 125 c) 250 d) 225 e) 180

Description : The LCM of two numbers is 45 times their HCF. If one of the numbers is 125 and the sum of HCF and LCM is 1150, then the other number is? a) 100 b) 125 c) 250 d) 225 e) 180

Last Answer : LCM = 45× HCF LCM + HCF = 46× HCF = 1150 HCF = 25, LCM =1125. product of 2 numbers = LCM×HCF required number = (25× 1125) / 125 = 225. Answer: d)

The distance between two cities A and B is 330km. A train starts from A at 8 (a)m. and travels towards B at 60 km/hr. Another train starts from B at 9 (a)m. and travels towards A at 75 km/hr. At what time do they meet? a) 10 am. b) 10 : 30 am. c) 11 am. d) 11 : 30 am. e) None of these

Description : The distance between two cities A and B is 330km. A train starts from A at 8 (a)m. and travels towards B at 60 km/hr. Another train starts from B at 9 (a)m. and travels towards A at 75 km/hr. At what time do they meet? a) 10 am. b) 10 : 30 am. c) 11 am. d) 11 : 30 am. e) None of these

Last Answer : Distance travelled by first train in one hour = 60 x 1 = 60 km Therefore, distance between two train at 9 a.m. = 330 – 60 = 270 km Now, Relative speed of two trains = 60 + 75 = 135 km/hr Time of meeting of two trains =270/135=2 hrs. Therefore, both the trains will meet at 9 + 2 = 11 A.M. Answer: c)

Two stations X and Y are 170 km apart on a straight line. One train starts from X at 6 a.m. and travels towards Y at 25 kmph. Another train starts from Y at 8 a.m. and travels towards X at a speed of 35 kmph. At what time will they meet? a) 11. 30 a.m b) 10.30 a.m c) 11 a.m d) 9 a.m e) 10 am

Description : Two stations X and Y are 170 km apart on a straight line. One train starts from X at 6 a.m. and travels towards Y at 25 kmph. Another train starts from Y at 8 a.m. and travels towards X at a speed of 35 kmph. At what time will they meet? a) 11. 30 a.m b) 10.30 a.m c) 11 a.m d) 9 a.m e) 10 am

Last Answer : e Suppose they meet z hours after 6 a.m. Distance covered by X in z hours = 25× z km. Distance covered by Y in (z - 2) hours = 35(z - 2) km. Therefore 25z + 35(z- 2) = 170 60z = 240 x = 4. So, they meet at 10 a.m.

The distance between two cities A and B is 330 km. A train starts from A at 8 a.m. and travels towards B at 60 km/hr. Another train starts from B at 9 a.m. and travels towards A at 75 km/hr. At what time do they meet? A.10:30 am B.10:45 am C.11 am D.11:25 am E.None of these

Description : The distance between two cities A and B is 330 km. A train starts from A at 8 a.m. and travels towards B at 60 km/hr. Another train starts from B at 9 a.m. and travels towards A at 75 km/hr. At what time do they meet? A.10:30 am B.10:45 am C.11 am D.11:25 am E.None of these

Last Answer : Answer – C (11 am) Explanation – Suppose they meet x hrs after 8 a.m. Then, (Distance moved by first in x hrs) + [Distance moved by second in (x-1) hrs] = 330 60x + 75(x – 1) = 330 x = 3 So, they meet at (8 + 3), i.e. 11 a.m

The distance between two cities A and B is 330 km. A train starts from A at 8 a.m. and travels towards B at 60 km/hr. Another train starts from B at 9 a.m. and travels towards A at 75 km/hr. At what time do they meet? A.10:30 am B.10:45 am C.11 am D.11:25 am E.None of these

Description : The distance between two cities A and B is 330 km. A train starts from A at 8 a.m. and travels towards B at 60 km/hr. Another train starts from B at 9 a.m. and travels towards A at 75 km/hr. At what time do they meet? A.10:30 am B.10:45 am C.11 am D.11:25 am E.None of these

Last Answer : Answer – C (11 am) Explanation – Suppose they meet x hrs after 8 a.m. Then, (Distance moved by first in x hrs) + [Distance moved by second in (x-1) hrs] = 330 60x + 75(x – 1) = 330 x = 3 So, they meet at (8 + 3), i.e. 11 a.m

A thief is noticed by a policeman from a distance of 200 m. The thief starts running and the policeman chases him. The thief and the policeman run at the rate of 10 km and 11 km per hour respectively. What is the distance between them after 6 minutes? A.100 m B.150 m C.190 m D.200 m E.None of these

Description : A thief is noticed by a policeman from a distance of 200 m. The thief starts running and the policeman chases him. The thief and the policeman run at the rate of 10 km and 11 km per hour respectively. What is the distance between them after 6 minutes? A.100 m B.150 m C.190 m D.200 m E.None of these

Last Answer : Answer- A(100m) Explanation: Relative speed of the thief and policeman = (11 – 10) km/hr = 1 km/hr Distance covered in 6 minutes =[(1/60)*6] km = (1/10)km = 100 m. Distance between the thief and policeman = (200 – 100) m = 100 m.

One horsepower is equal to (A) 550 lbf.ft/second (B) 550 kgf.m/second (C) Both (A) and (B) (D) 550 lbf.ft./h

Description : One horsepower is equal to (A) 550 lbf.ft/second (B) 550 kgf.m/second (C) Both (A) and (B) (D) 550 lbf.ft./h

Last Answer : (A) 550 lbf.ft/second

vIf the price of Mango is Rs. 130 per kg, Apple is Rs. 140 per kg and Orange is Rs. 80 per kg, then what is the ratio of their cost at vendor A? 1) 11 : 14 : 16 2) 39 : 28: 20 3) 39 : 28 : 21 4) 15 : 14 : 11 5) 39 : 26 : 20

Description : If the price of Mango is Rs. 130 per kg, Apple is Rs. 140 per kg and Orange is Rs. 80 per kg, then what is the ratio of their cost at vendor A? 1) 11 : 14 : 16 2) 39 : 28: 20 3) 39 : 28 : 21 4) 15 : 14 : 11 5) 39 : 26 : 20

Last Answer : 2) 39 : 28: 20

A manufacture undertakes to supply 2000 pieces of a particular component at Rs.25 per piece. According to his estimates, even if 5% fail to pass the quality tests, then he will make a profit of 25%. However as it turned out, 50% of the components were rejected. What is the loss to the manufacture? [TCS recruitment exam question] A. Rs 12,000 B. Rs 13,000 C. Rs 14,000 D. Rs 15,000 E. None f these

Description : A manufacture undertakes to supply 2000 pieces of a particular component at Rs.25 per piece. According to his estimates, even if 5% fail to pass the quality tests, then he will make a profit of 25%. However as it turned out, ... A. Rs 12,000 B. Rs 13,000 C. Rs 14,000 D. Rs 15,000 E. None f these

Last Answer : Answer - B. Rs 13,000 Explanation: 5% of 2000=100 so 2000-100=1900 so,if he sells 1900 he will get 25% profit cost per piece rs 25 so 25x 1900 don't solve here if 125% . ... /125=38000=CP if 50% rejected,only 1000 pieces sold so 1000 25=25000=SP loss=cp-sp= 38000-25000=13000

The average age of the whole class is 12.05 years. The average age of all girls is 12.5 years and the average age of boys is 11.75 years. If the total number of boys is 45 then what is the total number of girls in the class? a) 20 b) 25 c) 30 d) 35 e) 40

Description : The average age of the whole class is 12.05 years. The average age of all girls is 12.5 years and the average age of boys is 11.75 years. If the total number of boys is 45 then what is the total number of girls in the class? a) 20 b) 25 c) 30 d) 35 e) 40

Last Answer : Let the number of girls be x. The number of boys is 45. :. Total age of boys and girls in the class = (x+45) × 12.05 or, (x+45) × 12.05 = 45 × 11.75 + x × 12.5 or, 12.05x + 542.25 = 528.75 + 12.5x or, 13.5 = 0.45x or, x = 13.5/0.45 = 30 Answer is: c)

Two trains starting at the same time from two stations 200 km apart and going in opposite directions cross each other at a distance of 110 km from one of the stations. What is the ratio of their speeds? A.9 : 28 B.11 : 9 C.11 : 8 D.9 : 22 E.None of these

Description : Two trains starting at the same time from two stations 200 km apart and going in opposite directions cross each other at a distance of 110 km from one of the stations. What is the ratio of their speeds? A.9 : 28 B.11 : 9 C.11 : 8 D.9 : 22 E.None of these

Last Answer : Answer – B (11 : 9) Explanation – In the same time, they cover 110 km and 90 km respectively.Ratio of their speeds = 110 : 90 = 11 : 9 

Two pipes A and B can fill a tank in 5 and 6hrs.Pipe C can empty it in 12hrs.If all the 3 pipes are opened together,then the tank will be filled in 1)1(13/17)hrs 2)2(8/11)hours. 3)3(9/17)hrs 4)4(1/2)hrs

Description : Two pipes A and B can fill a tank in 5 and 6hrs.Pipe C can empty it in 12hrs.If all the 3 pipes are opened together,then the tank will be filled in 1)1(13/17)hrs 2)2(8/11)hours. 3)3(9/17)hrs 4)4(1/2)hrs

Last Answer : 3)3(9/17)hrs Exp:Net part filled in 1hr=>(1/5+1/6-1/12)=17/60 Tank filled in=60/17=3(9/17).

The average price of 10 books is Rs. 12 while the average price of 8 of these books is Rs. 11.75. Of the remaining two books, if the price of one book is 60% more than the price of the other, what is the price of each of these two books? A.Rs 10 and Rs 16 B.Rs 12 and Rs 24 C.Rs 24 and Rs 18 D.Rs 28 and Rs 12 E.None of these

Description : The average price of 10 books is Rs. 12 while the average price of 8 of these books is Rs. 11.75. Of the remaining two books, if the price of one book is 60% more than the price of the other, what is the price of ... and Rs 16 B.Rs 12 and Rs 24 C.Rs 24 and Rs 18 D.Rs 28 and Rs 12 E.None of these

Last Answer : Answer - A (Rs 10 and Rs 16) Explanation - Total price of the two books = Rs. [(12 x 10) - (11.75 x 8)] = Rs. (120 - 94) = Rs. 26 Let the price of one book be Rs.x Then, the price of other book = Rs. (x ... /5)x= (8/5)x so, x +(8/5)x=26 , x=10 The prices of the two books are Rs. 10 and Rs. 16

Calculate: a. Mass flow rate in lb/hr. b. The velocity at section 2 in fps  a. 800,000lb/hr;625ft/s  b. 900,000lb/hr;625 ft/s  c. 888,000lb/hr;269 ft/s  d. 700,000lb/hr;269 ft/s m = A1V!/V1

Description : Calculate: a. Mass flow rate in lb/hr. b. The velocity at section 2 in fps  a. 800,000lb/hr;625ft/s  b. 900,000lb/hr;625 ft/s  c. 888,000lb/hr;269 ft/s  d. 700,000lb/hr;269 ft/s m = A1V!/V1

Last Answer : 900,000 lb/hr;625 ft/s

The ages of A and B are in the ratio 8 : 5. If the sum of their ages is 39 years, what will be the ratio of their ages after 9 years? a) 3 : 2 b) 8 : 7 c) 10: 7 d) 11 : 8 e) None of these

Description : The ages of A and B are in the ratio 8 : 5. If the sum of their ages is 39 years, what will be the ratio of their ages after 9 years? a) 3 : 2 b) 8 : 7 c) 10: 7 d) 11 : 8 e) None of these

Last Answer : Sum of ratios = 8 +5 = 13. Sum of ages = 39 years 13 ratio = 39 years  1 ratio = 3 years 9 years = 3 ratio new ratio = (8+3) : (5+3) = 11 : 8 Answer: d)

Present ages of Simmi and Anu are in the ratio of 5 : 4 respectively. Three years hence, the ratio of their ages will become 11 : 9 respectively, What is Anu’s present age in years? A.20 B.24 C.28 D.35 E.None of these

Description : Present ages of Simmi and Anu are in the ratio of 5 : 4 respectively. Three years hence, the ratio of their ages will become 11 : 9 respectively, What is Anu’s present age in years? A.20 B.24 C.28 D.35 E.None of these

Last Answer : Answer – B (24) Explanation – Let the present ages of Simmi and Anu be 5x years and 4x years respectively Then, [5X + 3] / [4X + 3] =11/9 x=6 Anu’s present age = 4X = 24 years

The sum of 2 numbers is 216, and their HCF is 27. How many such pairs of numbers exist? (A pair a unique set of 2 numbers, so (a, b) and (b, a) will be considered as 1 pair). a) 0 b) 1 c) 2 d) 3 e) 4

Description : The sum of 2 numbers is 216, and their HCF is 27. How many such pairs of numbers exist? (A pair a unique set of 2 numbers, so (a, b) and (b, a) will be considered as 1 pair). a) 0 b) 1 c) 2 d) 3 e) 4

Last Answer : Let the numbers be a,b A = 27m, b = 27n where m, n are co-prime 27(m + n) = 216, or m+n = 8 M = 1, n = 7 and m = 3, n = 5 are the only 2 co – prime pairs possible. So we have 2 pairs: (27 , 189) and ( 81, 135) Answer: c)

Two train travel in opposite directions at 36 kmph and 45 kmph and a man sitting in slower train passes the faster train in 8 seconds. Then length of the faster train is: A.120 m B.140 m C.160 m D.180 m E.None of these

Description : Two train travel in opposite directions at 36 kmph and 45 kmph and a man sitting in slower train passes the faster train in 8 seconds. Then length of the faster train is: A.120 m B.140 m C.160 m D.180 m E.None of these

Last Answer : Answer – D (180 m) Explanation – When OPP. direction-PLUS Relative speed = (36 + 45) km/hr = [81 x 5/18] m/sec = [45/2] m/sec. Length of train = [45/2 x 8] m = 180 m.

The width of a rectangular park is 10/21 of its length. If the area of the park is 3360 sq m, then what is the difference between the length and the width of the park? a) 40m b) 44m c) 48m d) 52m e) 56m

Description : The width of a rectangular park is 10/21 of its length. If the area of the park is 3360 sq m, then what is the difference between the length and the width of the park? a) 40m b) 44m c) 48m d) 52m e) 56m

Last Answer : Let the length of the park be x metres. Width = x × 10 / 21 = 10x / 21 Area = x × 10x / 21 = 3360 or, x^2 = 3360 × 21 / 10 = 7056 or, x = 84 Hence, length = x = 84m and width = 84 × 10 / 21 = 40 metres Difference = 84 – 40 = 44 metres Answer: b)

What is the pressure of 4 moles of helium in a 50 L tank at 308 K Use PV = nRT.?

Description : What is the pressure of 4 moles of helium in a 50 L tank at 308 K Use PV = nRT.?

Last Answer : 2.02 atm (apex)

A shopkeeper sells two watches for Rs 308 each. On one watch he earns 12% profit and on the others he suffers 12% loss. His profit or loss in the entire transaction was: 1. 1(11/25)% loss 2. 1(11/25)% gain 3. 3(2/25)% loss 4. 3(2/25)% gain

Description : A shopkeeper sells two watches for Rs 308 each. On one watch he earns 12% profit and on the others he suffers 12% loss. His profit or loss in the entire transaction was: 1. 1(11/25)% loss 2. 1(11/25)% gain 3. 3(2/25)% loss 4. 3(2/25)% gain

Last Answer : 3. 3(2/25)% loss

.A shopkeeper sells two watches for Rs 308 each. On one watch he earns 12% profit and on the others he suffers 12% loss. His profit or loss in the entire transaction was: 1. 1(11/25)% loss 2. 1(11/25)% gain 3. 3(2/25)% loss 4. 3(2/25)% gain

Description : .A shopkeeper sells two watches for Rs 308 each. On one watch he earns 12% profit and on the others he suffers 12% loss. His profit or loss in the entire transaction was: 1. 1(11/25)% loss 2. 1(11/25)% gain 3. 3(2/25)% loss 4. 3(2/25)% gain

Last Answer : . 3(2/25)% loss

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

A man who makes a profit of 25 percent by selling sugar at Rs. 4.25/kg., lowers his price so as to gain only 5 paise per Kg. In what ratio must his sales be increased so that his total profit may be the same as before? a) 1 : 5 b) 1 : 11 c) 1 : 13 d) 1 : 17 e) 1 : 21

Description : A man who makes a profit of 25 percent by selling sugar at Rs. 4.25/kg., lowers his price so as to gain only 5 paise per Kg. In what ratio must his sales be increased so that his total profit may be the same as before? a) 1 : 5 b) 1 : 11 c) 1 : 13 d) 1 : 17 e) 1 : 21

Last Answer : 125% C.P. of sugar per kg = Rs. 4.25 C.P of the sugar per kg = [4.25×100] / 125 = Rs. 3.40. thus, the profit on 1 kg 85 paise by reducing price to 5 paise, to make same profit he must sell, 85/5 = 17kg so required ratio = 1 : 17. Answer: d)

A and B working alone can finish a job in 5 days and 7 days respectively. They work at it alternately for a day. If A starts the work, find in how many days the job will be finished? a) 29/5 b) 11/5 c) 24/5 d) 21/5 e) None of these

Description : A and B working alone can finish a job in 5 days and 7 days respectively. They work at it alternately for a day. If A starts the work, find in how many days the job will be finished? a) 29/5 b) 11/5 c) 24/5 d) 21/5 e) None of these

Last Answer : Work done by A in one day = 1/5 and work done by B in one day = 1/7 They are working alternately. Therefore, Work done in first two day = (1/5 + 1/7) = 12/35 Work done in first four day = 24/35 Work done in ... = 4/35/1/7 = 4/5 of the day. Therefore Total days taken = 5 + 4/5 = 29/5 days Answer: b)

A,B and C can do piece of work in 11 days,20 days and 55 days respectively, working alone. How soon can the work be done if a is assisted by B and C on alternate days/ a) 6 b) 7 c) 8 d) 9 e) 10

Description : A,B and C can do piece of work in 11 days,20 days and 55 days respectively, working alone. How soon can the work be done if a is assisted by B and C on alternate days/ a) 6 b) 7 c) 8 d) 9 e) 10

Last Answer : Work done by A and B in one day = 1/11 + 1/20 =31/220 Work done by A and C in one day = 1/11 + 1/55 =6/55 Work done in 2 days = 31 /220+6/55=55/220 = ¼ Hence, 1/4 of the work is done in 2 days Hence, it takes 8 days to finish the work Answer : c

A,B and C can do piece of work in 11 days,20 days and 55 days respectively, working alone. How soon can the work be done if a is assisted by B and C on alternate days/ a) 6 b) 7 c) 8 d) 9 e) 10

Description : A,B and C can do piece of work in 11 days,20 days and 55 days respectively, working alone. How soon can the work be done if a is assisted by B and C on alternate days/ a) 6 b) 7 c) 8 d) 9 e) 10

Last Answer : Work done by A and B in one day = 1/11 + 1/20 =31/220 Work done by A and C in one day = 1/11 + 1/55 =6/55 Work done in 2 days = 31 /220+6/55=55/220 = ¼ Hence, 1/4 of the work is done in 2 days Hence, it takes 8 days to finish the work Answer : c

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)

24 14 26 ? 28 16 30 a) 13 b) 17 c) 15 d) 11 e) 20

Description : 24 14 26 ? 28 16 30 a) 13 b) 17 c) 15 d) 11 e) 20

Last Answer : The series consists of two series 1 and 2: Series 1 : 24 26 28 30 +2 +2 +2 Series 2 : 14 15 16 +1 +1 Answer: c)

If x:y =3:4,in the ratio of (7x-4y) : ( 3x + y) a) 2:11 b) 4:9 c) 5:13 d) 6:5 e) 13:5

Description : If x:y =3:4,in the ratio of (7x-4y) : ( 3x + y) a) 2:11 b) 4:9 c) 5:13 d) 6:5 e) 13:5

Last Answer : An easy way to solve this question is use x = 3, y=4 7x – 4y = 21 – 16 = 5 3x + y = 9+4 = 13 Required ratio = 5:13 Answer: c)

11,66,462,3696,? a) 32364 b) 33364 c) 32264 d) 33264 e) 32664

Description : 11,66,462,3696,? a) 32364 b) 33364 c) 32264 d) 33264 e) 32664

Last Answer : 11×6=66,66×7=462,462×8=3696,3696×9=33264 Answer: d)