How many times does 30 go into 36?

1 Answer

Answer :

44717

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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)

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Description : how many times does12 go in to 36?

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Description : How many times does 18 to into 36?

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Description : Box A contains wheat worth Rs.30 per kg and box B contains wheat worth Rs.40 per kg.If both A and B are mixed in the the ratio 4:7 then the price of mixture per kg is A)36.36 B)35.80 C)42.50 D)31.30

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Description : Ten years ago, Khush was thrice as old as Sam was but 10 years hence, he will be only twice as old. Find Khush’s present age? A.30 B.35 C.40 D.36 E.None of these

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Description : 3 , 8 , 15 , 24 , ? 48 , 63 (a) 30 (b) 32 (c) 35 (d) 36

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Description : Manish purchased 25 kg of rice @ Rs32 per kg and 15 kg of rice @ Rs 36 per kg. He mixed the two varieties of rice and sold it @ Rs 40.20 per kg. What is the percent profit earned? 1. 25 2. 40 3. 30 4. 20

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Description : If an integer occupies 4 bytes and a character occupies 1 byte of memory, each element of the following structure would occupy how many bytes ? struct name { int age; char name[30]; }; A) 30 B) 32 C) 34 D) 36

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Description : What is the shear stress acting along the neutral axis of triangular beam section, with base 60 mm and height 150 mm, when shear force of 30 kN acts? a. 15.36 N/mm2 b. 10.6 N/mm2 c. 8.88 N/mm2 d. Insufficient data

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Description : What will be the output? numberGames = {} numberGames[(1,2,4)] = 8 numberGames[(4,2,1)] = 10 numberGames[(1,2)] = 12 sum = 0 for k in numberGames: sum += numberGames[k] print len(numberGames) + sum Page No 36 a) 30 b) 24 c) 33 d) 12

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Description : The missing term in the series 1, 4, 27 16, ? , 36, 343, ……is (A) 30 (B) 49 (C) 125 (D) 81

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Description : The mean of the ages of father and his son is 27 years. After 18 years, father will be twice as old as his son. Their present ages are (A) 42, 12 (B) 40, 14 (C) 30, 24 (D) 36, 18

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Description : Two numbers are in the ration 3:5. If 9 is subtracted from the numbers, the ratio becomes 12:23. The numbers are (A) 30, 50 (B) 36, 60 (C) 33, 55 (D) 42, 70

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Description : The average of four consecutive even numbers is 27. The largest of these numbers is: (A) 36 (B) 32 (C) 30 (D) 28 

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Description : The height of a geo-stationary satellite from the Earth’s surface is approximately : (1) 36,000 km (2) 42,000 km (3) 30,000 km (4) None of these

Last Answer :  36,000 km

Description : P and Q finished a work in 36 days; Q and R can do it in 60 days; P and R can do it in 45 days. In what time can P alone do it. a) 30 days b) 60 days c) 90 days d) 120 days

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Description : If Gowtham ride a bike at 30 km/hr, then he arrives at a certain place at 3 p.m. If he ride at 45 km/hr, he will arrive at the same place at 1 P.m. At what speed must he ride to get there at noon? a) 18 km/hr b) 24 km/hr c) 36 km/hr d) 28 km/hr

Last Answer : C When speed of gowtham = 30 km/hr =d/t and, When speed of the gowtham = 45 km/hr =d/(t−2) Equating the value of d: 30t =45(t-2) 30t=45t-90 15t=90 t=6 hours Finally desired speed =d/(t-1) =30t/(t-1) =30*6/(6-1) =180/5 =36 km/hr

Description : Ganesh is travelling on his motorbike and has calculated to reach golden temple at 8 p.m if he travels at 30 km/hr; he will reach there at 6 p.m if he travels at 45 km/hr. At what speed must he travel to reach the golden temple at 7 p.m? a) 36 b) 46 c) 32 d) 30

Last Answer : A Let the distance travelled be x km x/30 – x/45 = 2 (3x – 2x) / 90 = 2 x/90 = 2 x = 180km Time taken to travel 180 km at 30 km/hr = 180 / 30 = 6 hrs So ganesh started 6 hrs before 8 pm That is 2 pm Required speed = distance / time =(180/5) =36 km/hr

Description : Eight years ago, the ratio of the ages of micky and Donald was 6:5, 12years hence, the ratio of their ages will be 11:10. What is Donald age at present a) 30 years b) 28 years c) 34 years d) 36 years

Last Answer : B Let 8 years ago the age of Micky and Donald be 6x and 5x resp. then, ((6x+8)+12) / ((5x+8)+12) = 11/10 10(6x+20) = 11(5x+20) 5x = 20 => x = 4 So Donald age is (5x+8) = 28

Description : 12 years ago M was half of N in age. If the ratio of their present ages is 4 : 5, what will be the total of their present ages a) 30 b) 35 c) 36 d) 32

Last Answer : C Let M's age 12 years ago = x years. Then, N's age 12 years ago = 2x years. (x + 12) / (2x+ l2) = 4/5 => x = 4. So, the total of their present ages =(x + 12 + 2x + 12) = (3x + 24) = 36 years.

Description : Two pumps M and N can separately fill a dumper in 36 minutes and 45 minutes respectively. Both the pumps are opened together but 12 minutes after the start the pump M is turned off. How much time will it take to fill the dumper? A) 9 min B) 10 min C) 30 min D) 32 min

Last Answer : C 12/36 + x/45 = 1 (12*5)/180+4x/180=1 60+4x=180 4x=120 X=120/4=30min

Description : Two pipes, Xand Y can fill a tank in 36 and 30 minutes respectively. Both are opened together, but at the end of 3minutes, X is turned off. In how many more minutes will Y fill the cistern? A) 7(24/5) B) 7(1/2) C) 24(1/3) D) 8(1/4) 

Last Answer : C X can fill cistern in 36 minutes. X fills cistern in 1 minute = 1/36 Y can fill cistern in 30 minutes. Y fills cistern in 1 minute = 1/30 part. Xand Y together can fill cistern in 1 minute, = {(1/ ... part. 147/180 part cistern must be filled by Y in, [(147/180)/(1/30)] = 24(1/3) minutes.

Description : The average age of 160 boys in a class is 58 yrs. The average group of 30 boys in the class is 42 yrs and the average of another group of 50 boys in the class is 36 years. What is the average age of the remaining boys? A) 72.58 B) 74.25 C) 77.75 D) 75.68 

Last Answer : C) Total age of 160 boys = 160* 58= 9280 total age of 30 boys = 30 * 42= 1260 total age of next 50boys = 50 * 36= 1800 average of the remaining boys = [(9280-{1260+1800})/[160 - (30 + 50)] =>9280-3060/80 =>6220/80 =77.75yrs