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

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
by

1 Answer

Answer :

(A) 550 lbf.ft/second
by

Related questions

Steam at 1000 lbf/ft^2 pressure and 300˚R has specific volume of 6.5 ft^3/lbm and a specific enthalpy of 9800 lbf-ft/lbm. Find the internal energy per pound mass of steam.  A.2500 lbf-ft/lbm  B.3300 lbf-ft/lbm  C.5400 lbf-ft/lbm  D.6900 lbf-ft/lbm Formula: h= u+ pV u= h– pV

Description : Steam at 1000 lbf/ft^2 pressure and 300˚R has specific volume of 6.5 ft^3/lbm and a specific enthalpy of 9800 lbf-ft/lbm. Find the internal energy per pound mass of steam.  A.2500 lbf-ft/lbm  B.3300 lbf-ft/lbm  C.5400 lbf-ft/lbm  D.6900 lbf-ft/lbm Formula: h= u+ pV u= h– pV

Last Answer : 3300 lbf-ft/lbm

The marked price of a watch was Rs. 720.A man brought the same for Rs.550.80 after getting two successive discounts the first being 10% .What was the second discount rate? A. 12% B. 14% C. 15% D. 18%

Description : The marked price of a watch was Rs. 720.A man brought the same for Rs.550.80 after getting two successive discounts the first being 10% .What was the second discount rate? A. 12% B. 14% C. 15% D. 18%

Last Answer : Answer – C. 15% Explanation – Let the second discount rate be x% Then, (100 – x)% of 90% of 720 = 550.80 => (100 – x)/100 x 90/100 x 720 = 550.80 = (100-x) = [55080/(9 x 72) = 85 => x= 15 second discount rate =15%

The unit of velocity head is (A) ft-lb/sec (B) ft-lb/ft 3 (C) ft-lbf/lb (D) ft-lbf/sec

Description : The unit of velocity head is (A) ft-lb/sec (B) ft-lb/ft 3 (C) ft-lbf/lb (D) ft-lbf/sec

Last Answer : (C) ft-lbf/lb

The unit of gcis (A) lbm/lbf. ft/sec2 (B) lbf/lbm. ft/sec2 (C) ft/sec2 (D) lbm/lbf. sec2/ft

Description : The unit of gcis (A) lbm/lbf. ft/sec2 (B) lbf/lbm. ft/sec2 (C) ft/sec2 (D) lbm/lbf. sec2/ft

Last Answer : (A) lbm/lbf. ft/sec2

3.0 lbm of air are contained at 25 psia and 100 ˚F. Given that Rair = 53.35 ft-lbf/lbm- ˚F, what is the volume of the container?  A.10.7 ft^3  B.14.7 ft^3  C.15 ft^3  D.24.9 ft^3 Formula: use the ideal gas law pV = mRT T = (100 +460) ˚R V = mRT/p

Description : 3.0 lbm of air are contained at 25 psia and 100 ˚F. Given that Rair = 53.35 ft-lbf/lbm- ˚F, what is the volume of the container?  A.10.7 ft^3  B.14.7 ft^3  C.15 ft^3  D.24.9 ft^3 Formula: use the ideal gas law pV = mRT T = (100 +460) ˚R V = mRT/p

Last Answer : 24.9 ft^3

If air is at pressure, p, of 3200 lbf/ft2 , and at a temperature, T, of 800 ˚R, what is the specific volume, v? (R=5303 ft-lbf/lbm-˚R, and air can be modeled as an ideal gas.)  A.9.8 ft^3/lbm  B.11.2 ft^3/lbm  C.13.33 ft^3/lbm  D.14.2 ft^3/lbm Formula: pv = RT v = RT / p

Description : If air is at pressure, p, of 3200 lbf/ft2 , and at a temperature, T, of 800 ˚R, what is the specific volume, v? (R=5303 ft-lbf/lbm-˚R, and air can be modeled as an ideal gas.)  A.9.8 ft^3/lbm  B.11.2 ft^3/lbm  C.13.33 ft^3/lbm  D.14.2 ft^3/lbm Formula: pv = RT v = RT / p

Last Answer : 13.33 ft^3/lbm

Ravi borrowed some money at the rate of 4 p.c.p.a. for the first three years, at the rate of 8 p.c.p.a. for the next two years and at the rate of 9 p.c.p.a. for the period beyond 5 years. If he pays a total simple interest of 19,550 at the end of 7 years, how much money did he borrow? a) 39,500 b) 42,500 c) 41,900 d) 43,000 e) None of these

Description : Ravi borrowed some money at the rate of 4 p.c.p.a. for the first three years, at the rate of 8 p.c.p.a. for the next two years and at the rate of 9 p.c.p.a. for the period beyond 5 years. If he pays a ... 7 years, how much money did he borrow? a) 39,500 b) 42,500 c) 41,900 d) 43,000 e) None of these

Last Answer : Let the amount borrowed be Rs P Total amount of simple interest = (P×3×4)/100 + (p×2×8)/100 + (p×2×9)/100 = 19550 0.12P + 0.16P + 0.18P = 19550 0.46p= 19550 P = Rs. 42500 Answer: b)

The sale price of an article including the sale tax is Rs. 616. The rate of sale tax is 10% . If the shopkeeper has made a profit of 12%, the cost price of the article is [The Pearson Guide book] A. Rs 500 B. Rs 515 C. Rs 550 D. Rs 600

Description : The sale price of an article including the sale tax is Rs. 616. The rate of sale tax is 10% . If the shopkeeper has made a profit of 12%, the cost price of the article is [The Pearson Guide book] A. Rs 500 B. Rs 515 C. Rs 550 D. Rs 600

Last Answer : Answer – A. Rs 500 Explanation – 110% of S.P. = 616 = S.P. = Rs.(616 x 100/110) = Rs.560. C.P. = Rs (100/112 x 560) = Rs 500

The labelled price of a product is Rs 750. If it is sold at a 20% discount, the dealer earns a 25% profit. What is the cost price? 1. Rs 550 2. Rs 450 3. Rs 435 4. Rs 480

Description : The labelled price of a product is Rs 750. If it is sold at a 20% discount, the dealer earns a 25% profit. What is the cost price? 1. Rs 550 2. Rs 450 3. Rs 435 4. Rs 480

Last Answer : 1. Rs 550

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

Description : 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

Last Answer : Answer – C (308) Explanation – Other number =[11 x 7700]/275 = 308

Calculate the power output in horsepower of an 80-kg man that climbs a flight of stairs 3.8 m high in 4.0 s.  a) 744.8 hp  b) 0.998 hp  c) 746 hp  d) 1.998 hp Formula: Power = Fd/t = mgh/t F = W = mg d = h

Description : Calculate the power output in horsepower of an 80-kg man that climbs a flight of stairs 3.8 m high in 4.0 s.  a) 744.8 hp  b) 0.998 hp  c) 746 hp  d) 1.998 hp Formula: Power = Fd/t = mgh/t F = W = mg d = h

Last Answer : 0.998 hp

75%of a number is equal to four-fifth of another number. What is the ratio between first number and the second number? a) 4:3 b) 30:32 c) 7:9 d) 16:15 e) None of these

Description : 75%of a number is equal to four-fifth of another number. What is the ratio between first number and the second number? a) 4:3 b) 30:32 c) 7:9 d) 16:15 e) None of these

Last Answer : Let the two number be x & y. According to question, 75/100 x : 4/5 y 3/4 x : 4/5 y 15x = 16y x /y = 16/15 Answer: d)

One-fifth of a number is equal to 5/8th of another number. If 35 is added to the first number, it becomes four times of the second number. Find the second number. 1. 39 2. 70 3. 40 4. 25 5. None of these

Description : One-fifth of a number is equal to 5/8th of another number. If 35 is added to the first number, it becomes four times of the second number. Find the second number. 1. 39 2. 70 3. 40 4. 25 5. None of these

Last Answer : Answer- 3(40) Explanation:- (1/5)x =(5/8)y………..1 if 35 is added to first no. , x + 35= 4y so,x= 4y -35 put in eq 1… (1/5)( 4y-35) = (5/8)y y=40 =second number

Two trains of equal length are running on parallel lines in the same direction at 46 km/hr and 36 km/hr. The faster train passes the slower train in 36 seconds. The length of each train is: A.40 m B.55 m C.65 m D.50 m E.None of these

Description : Two trains of equal length are running on parallel lines in the same direction at 46 km/hr and 36 km/hr. The faster train passes the slower train in 36 seconds. The length of each train is: A.40 m B.55 m C.65 m D.50 m E.None of these

Last Answer : Answer- D (50) Explanation – Relative speed= total lengths/time (46-36) x 5/18 = [L + L ] / 36 10 x (5/18) x36 x (1/2)= L L=50 m

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 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.

Ashok can row upstream at 8kmph and downstream at 12kmph.What is the speed of the stream ? A.6km/hr B.3km/h C.2 km/hr D.4km/hr

Description : Ashok can row upstream at 8kmph and downstream at 12kmph.What is the speed of the stream ? A.6km/hr B.3km/h C.2 km/hr D.4km/hr

Last Answer : Answer -C Basic Formula: If the speed downstream is a kmph and the speed upstream is b kmph then Speed of the stream = ½ (a-b) kmph Explanation: Speed downstream a = 12kmph Speed upstream b = 8 kmph Speed of the stream = ½ (a-b) = ½ (12-8) = 4/2 = 2 kmph speed of the stream = 2kmph

Is it more profitable for Company M to produce Q? I. Product R sells at a price four times that of Q II. One unit of Q requires 2 units of labour, while one unit of R requires 5 units of labour. There is a no other constraint on production. If the question can be answered with the help of statement I alone, If the question can be answered with the help of statement II alone, If both,statement I and statement II are needed to answer the question If the question cannot be answered even with the help of both the statements. If the data in either statement I or statement II alone is sufficient to answer the question

Description : Is it more profitable for Company M to produce Q? I. Product R sells at a price four times that of Q II. One unit of Q requires 2 units of labour, while one unit of R requires 5 ... both the statements. If the data in either statement I or statement II alone is sufficient to answer the question

Last Answer : from statement I : Let the price of product Q be P' Then the price of product R is 4P From Statement II: Cost of product Q =2 Cost of product R=5 Therefore, Profit percentage of product ... )=80P-100 Hence, product R is more profitable. Both statements are required to answer the question Answer :c

Is it more profitable for Company M to produce Q? I. Product R sells at a price four times that of Q II. One unit of Q requires 2 units of labour, while one unit of R requires 5 units of labour. There is a no other constraint on production. If the question can be answered with the help of statement I alone, If the question can be answered with the help of statement II alone, If both,statement I and statement II are needed to answer the question If the question cannot be answered even with the help of both the statements. If the data in either statement I or statement II alone is sufficient to answer the question

Description : Is it more profitable for Company M to produce Q? I. Product R sells at a price four times that of Q II. One unit of Q requires 2 units of labour, while one unit of R requires 5 ... both the statements. If the data in either statement I or statement II alone is sufficient to answer the question

Last Answer : from statement I : Let the price of product Q be P' Then the price of product R is 4P From Statement II: Cost of product Q =2 Cost of product R=5 Therefore, Profit percentage of product ... )=80P-100 Hence, product R is more profitable. Both statements are required to answer the question Answer :c

A 2 kg object weighs 1.8 kgf on a spring balance. The value of 'g' at that location in m/sec2is (A) 8.82 (B) 9.81 (C) 10.88 (D) 0.95

Description : A 2 kg object weighs 1.8 kgf on a spring balance. The value of 'g' at that location in m/sec2is (A) 8.82 (B) 9.81 (C) 10.88 (D) 0.95

Last Answer : (A) 8.82

What is the specific weight of water at standard condition? (Formula: γ = ρg / k)  a. 1000 kgm/m3  b. 9.8066 m/s2  c. 1000 kgf/m3  d. None of the above

Description : What is the specific weight of water at standard condition? (Formula: γ = ρg / k)  a. 1000 kgm/m3  b. 9.8066 m/s2  c. 1000 kgf/m3  d. None of the above

Last Answer : 1000 kgf/m3

The rate of Simple Interest in UBI & BOI are in the ratio of 5:4. Mr.Naveen wants to deposit his total savings in two banks in such a way that he receive equal half-yearly interest from both banks.He should deposit in both banks UBI & BOI in the ratio of a) 2:5 b) 1:5 c) 4:5 d) 3:5 e) None of these

Description : The rate of Simple Interest in UBI & BOI are in the ratio of 5:4. Mr.Naveen wants to deposit his total savings in two banks in such a way that he receive equal half-yearly interest from both banks.He should deposit in both banks UBI & BOI in the ratio of a) 2:5 b) 1:5 c) 4:5 d) 3:5 e) None of these

Last Answer : R1 =5x R2=4x T1=T2=1/2 yr [P1*5x*(1/2)]/100 = [P2*4x*(1/2)]/100 P1:P2 = 4:5 Answer: c)

One barometric pressure or 1 atmospheric pressure is equal to  (a) 1 kgf/cnr2  (b) 1.033 kgf/cm2  (c) 0 kgf/cm2  (d) 1.0197 kgf/cm2  (e) 100 kgf/cm2.

Description : One barometric pressure or 1 atmospheric pressure is equal to  (a) 1 kgf/cnr2  (b) 1.033 kgf/cm2  (c) 0 kgf/cm2  (d) 1.0197 kgf/cm2  (e) 100 kgf/cm2.

Last Answer : Answer : b

I kgf/cm2 is equal to  (a) 760 mm Hg  (b) zero mm Hg  (c) 735.6 mm Hg  (d) 1 mm Hg  (e) l00 mm Hg.

Description : I kgf/cm2 is equal to  (a) 760 mm Hg  (b) zero mm Hg  (c) 735.6 mm Hg  (d) 1 mm Hg  (e) l00 mm Hg.

Last Answer : Answer : c

If w is the weight of water per cubic metre, Q is the discharge in cubic metres per sec and H is the total head, the required water horse power of the pump, is (A) wQH/15 (B) wQH/360 (C) wQH/220 (D) wQH/550

Description : If w is the weight of water per cubic metre, Q is the discharge in cubic metres per sec and H is the total head, the required water horse power of the pump, is (A) wQH/15 (B) wQH/360 (C) wQH/220 (D) wQH/550

Last Answer : (A) wQH/15

How many second are there in 2 days ?

Description : How many second are there in 2 days ?

Last Answer : There are 24 hours in a day, and there are 3600 seconds in an hour. Therefore, there are 24 * 3600 = 86400 seconds in a day. So there are 86400 * 2 = 172800 seconds in 2 days.

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

There are two sets of five numbers each – set P and set Q. P is a set of consecutive even numbers and Q is a set of consecutive numbers. The sum of the numbers of P is 230. The second least number in Q is 33 less than twice the least number in P. Find the sum of the numbers in Q. 1 : 260 2 : 250 3 : 240 4 : 270 5 : None of these

Description : There are two sets of five numbers each - set P and set Q. P is a set of consecutive even numbers and Q is a set of consecutive numbers. The sum of the numbers of P is 230. The second least number in Q is 33 less ... the sum of the numbers in Q. 1 : 260 2 : 250 3 : 240 4 : 270 5 : None of these

Last Answer : 1 : 260

Ashwin purchased a second hand machine for Rs.1 lakh. Over the first year after its purchase. He incurred an expense of Rs.20000 in maintaining it. At the end of the first year he sold it for Rs.80000. Find his profit/loss percentage. If 25% of the total cost price (i.e., purchased price + expenses) is to be deducted on account of depreciation. a)12.5% profit b)12.5% loss c)11 1/9% profit d)11 1/9% loss

Description : Ashwin purchased a second hand machine for Rs.1 lakh. Over the first year after its purchase. He incurred an expense of Rs.20000 in maintaining it. At the end of the first year he sold it for Rs.80000. Find his ... account of depreciation. a)12.5% profit b)12.5% loss c)11 1/9% profit d)11 1/9% loss

Last Answer : d)11 1/9% loss 

Of the four numbers, the first is twice the second, the second is one-third of the third and the third is 5 times the fourth. The average of the numbers is 24.75. The largest of these numbers is: A.45 B.25 C.30 D.45 E.None of these

Description : Of the four numbers, the first is twice the second, the second is one-third of the third and the third is 5 times the fourth. The average of the numbers is 24.75. The largest of these numbers is: A.45 B.25 C.30 D.45 E.None of these

Last Answer : Answer – D (45) Explanation – Let the fourth number be a then hen, third number = 5a, second number= 5a/3 and first=10a/3 a+ 5a+ 5a/3 + 10/3 = 24.75 x 4 so,a=9 So, the numbers are 9, 45, 15 and 30

P, Q and R can do a piece of work in 18,24 and 36 days on the second day and P on the third day, again R on the fourth day and so on. Then in how, many days will the work be completed? 1. 25 2. 24 3. 22 4. 20 5. 26

Description : P, Q and R can do a piece of work in 18,24 and 36 days on the second day and P on the third day, again R on the fourth day and so on. Then in how, many days will the work be completed? 1. 25 2. 24 3. 22 4. 20 5. 26

Last Answer : 2;

A train starts from Delhi and reaches Lucknow in 24.5 hours. If it travels the first half of journey at 30 kmph and second half at 40 kmph then what is the total distance it travelled? 1) 750 km 2) 800 km 3) 840 km 4) 900 km 5) 920 km

Description : A train starts from Delhi and reaches Lucknow in 24.5 hours. If it travels the first half of journey at 30 kmph and second half at 40 kmph then what is the total distance it travelled? 1) 750 km 2) 800 km 3) 840 km 4) 900 km 5) 920 km

Last Answer : 3) 840 km

The population of a city is 480000. It increases by 8% in the first year and decreased by 9% in the second year. What will be the population of city after two years? 1) 565056 2) 471744 3) 518400 4) 481344 5) 523200

Description : The population of a city is 480000. It increases by 8% in the first year and decreased by 9% in the second year. What will be the population of city after two years? 1) 565056 2) 471744 3) 518400 4) 481344 5) 523200

Last Answer : 2) 471744

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)

In racing over a given distance, A can beat B by 20m, B can beat C by 10m, A can beat C by 28m. The racing distance is? a) 58 m b) 116 m c) 100 m d) 120 m e) cannot be determined

Description : In racing over a given distance, A can beat B by 20m, B can beat C by 10m, A can beat C by 28m. The racing distance is? a) 58 m b) 116 m c) 100 m d) 120 m e) cannot be determined

Last Answer : Let the uniform speeds of A,B,C be vA,vB,vC, respectively. Let t1 be the time taken for the race between A and B and t2 be the time taken for race between B and C Since A covers the distance in time t1, race between A and C ... ) = VB/ VC = d/(d-10) Thus (d-20)(d-10)=d(d-28).Hence, d=100. Answer: c)

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)

When a ball bounces, it rises to 2/3 of the height from which it fell. If the ball is dropped from a height of 36m, how high will it rise at the third bounce? a) 10 2/3 m b) 13 2/3 m c) 12 1/3 m d) 12 2/3m e) None of these

Description : When a ball bounces, it rises to 2/3 of the height from which it fell. If the ball is dropped from a height of 36m, how high will it rise at the third bounce? a) 10 2/3 m b) 13 2/3 m c) 12 1/3 m d) 12 2/3m e) None of these

Last Answer : Height at third bounce =36 × (2/3)3 = 10 2/3 metre Answer: a)

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.

P walks around a circular field at the rate of two rounds per hour while Q runs around it at the rate of eight rounds per hour. They start in the opposite direction from the same point at 6.45 a.m. They shall first cross each other at? a) 6.45 a.m b) 6.50 a.m c) 7.51 a.m d) 6.51 a.m e) 7.45 am

Description : P walks around a circular field at the rate of two rounds per hour while Q runs around it at the rate of eight rounds per hour. They start in the opposite direction from the same point at 6.45 a.m. They shall first cross each other at? a) 6.45 a.m b) 6.50 a.m c) 7.51 a.m d) 6.51 a.m e) 7.45 am

Last Answer : d Since P and Q move in the opposite direction along the circular field, so they will first meet each other when there is a difference of one round between the two. Relative speed of P and Q = 8+2 = 10 rounds per hour. Time ... 1/10 hr = 6 min. 6 Hrs 45 min + 6 min = 6 hrs 51 min i.e, 6.51 a.m

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.

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 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

A train 125 m long passes a man, running at 5 kmph in the same direction in which the train is going, in 10 seconds. The speed of the train is: A.35 km/hr B.50 km/hr C.48 km/hr D.55 km/hr E.None of these

Description : A train 125 m long passes a man, running at 5 kmph in the same direction in which the train is going, in 10 seconds. The speed of the train is: A.35 km/hr B.50 km/hr C.48 km/hr D.55 km/hr E.None of these

Last Answer : Answer – B (50 km/hr) Explanation – Let speed be S Relative speed = total lengths/time [ in this case man length is neglible compare to train so neglected] (S – 5) x 5/18 = 125 /10 S- 5 = (125/10)x(18/5) S-5 = 45 S=50 km/hr

A 270 m long train running at the speed of 120 kmph crosses another train running in opposite direction at the speed of 80 kmph in 9 seconds. What is the length of the other train? A.180 B.230 C.245 D.235 E.None of these

Description : A 270 m long train running at the speed of 120 kmph crosses another train running in opposite direction at the speed of 80 kmph in 9 seconds. What is the length of the other train? A.180 B.230 C.245 D.235 E.None of these

Last Answer : Answer – B (230) Explanation – Relative speed = total lenghths /time [120 +80] x5/18 ={ 270 + other train length (say L) } / 9 sec (x5/18 to convert km/hr to m/sec ) 200 x (5/18) x9 = 270 + L 500 = 270 +L L=230 m Note: always do cutting ,avoid solving exact

A this usual rowing rate, Mohit can travel 12 miles downstream in a certain river in 6 hours less than it takes him to travel the same distance upstream. But if he could double his usual rowing rate for his 24 miles round trip, the downstream 12 miles would then take only one hour less than the upstream 12 miles. What is the speed of the current in miles per hour? A.2.5m/hr B.4 m/hr C.8/3 m/hr D. 5/3m/hr

Description : A this usual rowing rate, Mohit can travel 12 miles downstream in a certain river in 6 hours less than it takes him to travel the same distance upstream. But if he could double his usual rowing rate for his 24 miles round trip, ... current in miles per hour? A.2.5m/hr B.4 m/hr C.8/3 m/hr D. 5/3m/hr

Last Answer : Answer-C Basic Formula: Speed of the stream = ½ (a-b) km/hr Explanation: Let the speed in still water be x m/hr Speed of stream be y m/hr Then, speed upstream = x-y m/hr and Speed downstream = x+y m/hr 12/x-y - ... 1] 3y^2 = 8 y so, y = 8/3 speed of the current = 8/3 m/hr = 2 (2/3) m/hr

The loss made by selling 16 m of a cloth equals the selling price of 4 m of that cloth. Find the loss percentage a)15% b)20% c)33 1/3% d)25%

Description : The loss made by selling 16 m of a cloth equals the selling price of 4 m of that cloth. Find the loss percentage a)15% b)20% c)33 1/3% d)25%

Last Answer : b)20%

The profit made by selling 25 m of a cloth equals the selling price of 5 m of that cloth. find the profit percentage? a)25% b)20% c)16 2/3% d)30%

Description : The profit made by selling 25 m of a cloth equals the selling price of 5 m of that cloth. find the profit percentage? a)25% b)20% c)16 2/3% d)30%

Last Answer : a)25%

The profit made by selling 30 m of a cloth equals the cost price of 6 m of that cloth. find the profit percentage a)25% b)20% c)16 2/3% d)30%

Description : The profit made by selling 30 m of a cloth equals the cost price of 6 m of that cloth. find the profit percentage a)25% b)20% c)16 2/3% d)30%

Last Answer : b)20% 

The loss made by selling 20 m of a cloth equals the cost price of 5 m of that cloth. Find the loss percentage a)33 1/3% b)25% c)20% d)40%

Description : The loss made by selling 20 m of a cloth equals the cost price of 5 m of that cloth. Find the loss percentage a)33 1/3% b)25% c)20% d)40%

Last Answer : b)25%

A rectangular grassy plot 110 m by 65 m has a gravel path 2.5 m wide all round it on the inside. Find the cost of gravelling the path at 80 paise per sq. Meter 1).Rs.680 2).Rs.860 3).Rs.608 4).None.

Description : A rectangular grassy plot 110 m by 65 m has a gravel path 2.5 m wide all round it on the inside. Find the cost of gravelling the path at 80 paise per sq. Meter 1).Rs.680 2).Rs.860 3).Rs.608 4).None.

Last Answer : 1).Rs.680 Exp: overall area of the grassy plot is 110 x 65 =7150 sq. meters.The area of the grassy plot after applying the gravel path is 105 x 60=6300 sq. meters.Area of the gravel path is 7150 – 6300 = 850 sq.mtrs.Hence x 0.80=Rs.680 .