A man drives from his house to the station. If he drives at the rate of 10 kms per hour, he reaches the station at 6 pm. if he drives at 15 kms per hour, he would reach the station at 4 pm. At what speed, in kms per hour, should he drives so as to reach the station at 5 pm? a) 12 b) 12.33 c) 5√5 d) 12.5 e) 13

A man drives from his house to the station. If he drives at the rate of 10 kms per hour, he reaches the station at 6 pm. if he drives at 15 kms per hour, he would reach the station at 4 pm. At what speed, in kms per hour, should he drives so as to reach the station at 5 pm?
a) 12
b) 12.33
c) 5√5
d) 12.5
e) 13
by

1 Answer

Answer :

Let ‘x’ be the distance between house and station.
Time taken to travel a distance of x at the speed of 10km/hr, T1 = x/10 hours
Time taken to travel a distance of x at the speed of 15km/hr, T2 = x/15 hours
we know that T1 – T2 = 2 hours
therefore, we get x = 60 km
also, we can conclude that starting time is 12 pm [60/15 = 4, since he reach station at 4
pm travelling at 15 km/hr]
hence, to reach the station at 5 pm, speed = 60/5 = 12 km/hr [12 pm – 5pm, time = 5
hours]
Answer: a)
by

Related questions

Mr.Aneesh left for city 1 from city 2 at 3.00 pm. He travelled at the speed of 60km/hr for 1 and half hours. After that the speed was reduced to 15 km/hr. If the distance between two cities is 120 kms, at what time did Mr.Aneesh reach city 1 ? a) 6.30 pm. b) 5.30 pm. c) 7.30 pm. d) 8.30 pm.

Description : Mr.Aneesh left for city 1 from city 2 at 3.00 pm. He travelled at the speed of 60km/hr for 1 and half hours. After that the speed was reduced to 15 km/hr. If the distance between two cities is 120 kms, at what time did Mr.Aneesh reach city 1 ? a) 6.30 pm. b) 5.30 pm. c) 7.30 pm. d) 8.30 pm.

Last Answer : A Mr.Aneesh travelled 60 km/hr for 1 1/2 hours (3/2 hours). Distance covered in 3/2 hours = 60 x 3/2 = 90 km. Therefore, remaining distance = 120 -90 = 30 km After 3/2 hours, the speed was ... village 2 at 3.00 pm + 3 hours 30 minutes = 6.30pm Hence the required answer is 6.30 pm.

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Last Answer : CP of the article  

A reduction of 20% in the price of sugar enables a person to get 7.8 kg more sugar for Rs.3.90. The original price of sugar per kg is a) Rs. 10 b) Rs.11.50 c) Rs. 12.5 d) Rs. 15 , e) Rs. 10.5

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Last Answer : (4)1200

Diameter of Sun is-------? A. 12 Lakh kms B. 13 Lakh kms C. 14 Lakh kms (Answer) D. 15 Lakh kms

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Excluding stoppages, the speed of the bus is 54km/hr and including stoppages it is 45km/hr. For how many minutes does the bus stop per hour? a)11 minutes b)12minutes c)10minutes d) 13minutes

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Shan bought 30 liters of milk at the rate of Rs.8 per liter. He got it churned after spending Rs.10 and 5kg of cream and 30 liter of toned milk were obtained. If he sold the cream at Rs.30 per kg and toned milk at Rs.4 per liter, his profit in the transaction is: A. 20% B. 8% C. 30% D. 40%

Description : Shan bought 30 liters of milk at the rate of Rs.8 per liter. He got it churned after spending Rs.10 and 5kg of cream and 30 liter of toned milk were obtained. If he sold the cream at Rs.30 per kg and toned milk at Rs.4 per liter, his profit in the transaction is: A. 20% B. 8% C. 30% D. 40%

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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)

Speed of a boat in standing water is 9kmph and the speed of the stream is 1.5kmph. A man rows to a place at a distance of 10.5 km and comes back to the starting point. Find the total time taken by him. A.24 hours B.16 hours C.20 hours D.15 hours

Description : Speed of a boat in standing water is 9kmph and the speed of the stream is 1.5kmph. A man rows to a place at a distance of 10.5 km and comes back to the starting point. Find the total time taken by him. A.24 hours B.16 hours C.20 hours D.15 hours

Last Answer : Answer- A Basic Formula: i. speed = distance traveled / time taken ii. speed of the stream = ½ (a-b) km/hr iii. speed in still water = ½ (a+b) km/hr Explanation: Speed in still water= ½ (a+b) = 9km ... gives a = 10.5km/hr ; b=7.5 kmphr Total time taken by him = 105/10.5 + 105/7.5 = 24 hours

Three pipes P, Q and R can fill a tank from empty to full in 40 minutes, 15 minutes, and 30 minutes respectively. When the tank is empty, all the three pipes are opened. P, Q and R discharge chemical solutions X, Y and Z respectively. What is the proportion of the solution Y in the liquid in the tank after 5 minutes? a) 8 / 15 b) 7 / 15 c) 8 / 17 d) 6 / 13 e) 8 / 13

Description : Three pipes P, Q and R can fill a tank from empty to full in 40 minutes, 15 minutes, and 30 minutes respectively. When the tank is empty, all the three pipes are opened. P, Q and R discharge chemical solutions X, Y and Z respectively. ... 5 minutes? a) 8 / 15 b) 7 / 15 c) 8 / 17 d) 6 / 13 e) 8 / 13

Last Answer : a Part of the tank filled by pipe P in 1 minute = 1 / 40 Part of the tank filled by pipe Q in 1 minute = 1 / 15 Part of the tank filled by pipe R in 1 minute = 1/ 30 Here we have to find the proportion of ... together in 5 minute = 5 1/8 = 5/ 8 Required proportion = (1/3) / ( 5/8) = 8 / 15

How many Btus of heat are removed per hour when the water at a flow rate of 10 pounds per minute is cooled 15 degrees Fahrenheit (Water weighs 8.33 pounds per gallon)?

Description : How many Btus of heat are removed per hour when the water at a flow rate of 10 pounds per minute is cooled 15 degrees Fahrenheit (Water weighs 8.33 pounds per gallon)?

Last Answer : Feel Free to Answer

Busses start from a bus terminal with a speed of 20 km/hr at intervals of 10 minutes. What is the speed of a man coming from the opposite direction towards the bus terminal if he meets the buses at intervals of 8 minutes? a) 3 km/hr b) 4 km/hr c) 5 km/hr d) 7 km/hr e) None of these

Description : Busses start from a bus terminal with a speed of 20 km/hr at intervals of 10 minutes. What is the speed of a man coming from the opposite direction towards the bus terminal if he meets the buses at intervals of 8 minutes? a) 3 km/hr b) 4 km/hr c) 5 km/hr d) 7 km/hr e) None of these

Last Answer : Distance covered in 10 minutes at 20 kmph = distance covered in 8 minutes at (20+x) kmph 20× 10/60=8/60(20+x) 200 = 160 + 8x 8x = 40 x=40/8=5 kmph Answer: c)

At the time of marriage, a man was 6 years older to his wife. 12 years after their marriage, his age is 6/5 times the age of his wife. What was wife’s age at the time of marriage? a) 18 years b) 24 years c) 30 years d) 36 years e) None of these

Description : At the time of marriage, a man was 6 years older to his wife. 12 years after their marriage, his age is 6/5 times the age of his wife. What was wife’s age at the time of marriage? a) 18 years b) 24 years c) 30 years d) 36 years e) None of these

Last Answer : ratio of their ages, 12 years hence = 6 : 5  Difference in ages = 1 ratio = 6 years wife’s age (12 years hence) = 5 ratio = 5 × 6 years = 30 years wife’s age at the time of marriage = 30 – 12 = 18 years Answer: a)

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

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

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

A client-server system uses a satellite network, with the satellite at a height of 40,000 kms. What is the best-case delay in response to a request? (Note that the speed of light in air is 3,00,000 km/second). (A) 133.33 m sec (B) 266.67 m sec (C) 400.00 m sec (D) 533.33 m sec

Description : A client-server system uses a satellite network, with the satellite at a height of 40,000 kms. What is the best-case delay in response to a request? (Note that the speed of light in air is 3,00,000 km/second). (A) 133.33 m sec (B) 266.67 m sec (C) 400.00 m sec (D) 533.33 m sec

Last Answer : (D) 533.33 m sec Explanation: The request has to go up and down, and the response has to go up and down. The total path length traversed is thus 160.000 km. The speed of light in air and vacuum is 300,000 km/sec, so the propagation delay alone is 160,000/300,000 sec or about 533 msec. 

A shopkeeper declares that he sells rice at the cost price. However he uses a weight of 425 grams. Instead of 500 grams, what is his percentage of profit? a) a) 16 5/6 b) 15 c) c) 20 7/23 d) 25 e) None of these

Description : A shopkeeper declares that he sells rice at the cost price. However he uses a weight of 425 grams. Instead of 500 grams, what is his percentage of profit? a) a) 16 5/6 b) 15 c) c) 20 7/23 d) 25 e) None of these

Last Answer : % profit = Loss/Actual Weight x 100 = 500 – 425/500 × 100 = 75/500 x 100 = 15% Answer: b)

A bus can travel 25% faster than a jeep. Both start from point P at the same time and reach point Q, 225 kms away from P, at the same time. On the way, however, the bus lost about 37.5 minutes while stopping at the certain place. What is the speed of the jeep? a) 64 km/hr b) 72 km/hr c) 68 km/hr d) 56 km/hr

Description : A bus can travel 25% faster than a jeep. Both start from point P at the same time and reach point Q, 225 kms away from P, at the same time. On the way, however, the bus lost about 37.5 minutes while stopping at the ... is the speed of the jeep? a) 64 km/hr b) 72 km/hr c) 68 km/hr d) 56 km/hr

Last Answer : B Let speed of the jeep =x kmph Then, speed of the bus =(100+25)x/100 =125x/100 =5x/4 kmph Time taken by the jeep to travel from P to Q =225/x hours Time taken by the bus to travel from p to Q =225/(5x/4) + ... 225/x=900/5x+37.5/60 225/x=180/x+37.5/60 45/x =37.5/60 37.5x=45*60 X=72 km/hr

Sanki purchased 20 dozen notebook at Rs. 48 per dozen. He sold 8 dozen at 10% profit and remaining 12 dozen with 20% profit. What is his profit percentage in the transaction? A. 8% B. 15% C. 16% D. 19%

Description : Sanki purchased 20 dozen notebook at Rs. 48 per dozen. He sold 8 dozen at 10% profit and remaining 12 dozen with 20% profit. What is his profit percentage in the transaction? A. 8% B. 15% C. 16% D. 19% 

Last Answer : Answer – C (16%) Explanation – C.P of 20 dozen = Rs.(48 x 20) = Rs.960 C.P of 8 dozen = Rs. (48 x8) = Rs. 384. C.P of 12 dozen =Rs. (960 – 384) = Rs. 576 Total S.P = Rs. (110/100 x 384 +120/100 x 576) = Rs.1113.60 ∴ Profit % = (153.60/960 x 100)% =16%

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)

A man sells two flats at the rate of Rs. 2 lakhs each. On one he gains 3% and on the other, he loses 3%. His gain or loss % in the whole transaction is A. 9% loss B. 2% loss C. 3% loss D. 25% loss

Description : A man sells two flats at the rate of Rs. 2 lakhs each. On one he gains 3% and on the other, he loses 3%. His gain or loss % in the whole transaction is A. 9% loss B. 2% loss C. 3% loss D. 25% loss

Last Answer : Answer – A (0.09% loss) Explanation – Loss% = X^2/100=(3)^2/100 % =0.09%

If the compound interest accrued on an amount of Rs.15,000/- in two years is Rs.2,496/-.What is the rate of interest p.c.p.a.? a) 8 b) 12 c) 7.5 d) Cannot be determined e) None of these

Description : If the compound interest accrued on an amount of Rs.15,000/- in two years is Rs.2,496/-.What is the rate of interest p.c.p.a.? a) 8 b) 12 c) 7.5 d) Cannot be determined e) None of these

Last Answer : Let the rate of interest p.c.p.a. be x. 2496 =15000 * x /100 x = 16.64 16.64 is the rate of interest compounded for two year & for one year = 8% 8% C.I. table for two year. I year = 8% II year = 8.64 Total =16.64% Answer: a)