options stringlengths 37 300 | correct stringclasses 5 values | annotated_formula stringlengths 7 727 | problem stringlengths 5 967 | rationale stringlengths 1 2.74k | program stringlengths 10 646 |
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a ) 12 , b ) 39 , c ) 43 , d ) 52 , e ) 34 | b | add(subtract(87, multiply(17, 3)), 3) | a batsman makes a score of 87 runs in the 17 th inning and thus increases his avg by 3 . find his average after 17 th inning . | "let the average after 17 th inning = x . then , average after 16 th inning = ( x - 3 ) . : . 16 ( x - 3 ) + 87 = 17 x or x = ( 87 - 48 ) = 39 . ans : b" | a = 17 * 3
b = 87 - a
c = b + 3
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a ) 2000 , b ) 4500 , c ) 5000 , d ) 8000 , e ) 10000 | e | divide(120000, 12) | a company produces 120000 bottles of water everyday . if a case can hold 12 bottles of water . how many cases are required by the company to hold its one day production | number of bottles that can be held in a case = 12 number of cases required to hold 120000 bottles = 120000 / 12 = 10000 cases . so the answer is e = 10000 | a = 120000 / 12
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a ) 34 , b ) 35 , c ) 36 , d ) 37 , e ) 38 | c | subtract(add(22, 15), const_1) | at a garage sale , all of the items were sold at different prices . if the price of a radio sold at the garage sale was both the 15 th highest price and the 22 nd lowest price among the prices of the items sold , how many items were sold at the garage sale ? | "there were 14 items sold at a higher price than the radio and 21 items sold at a lower price than the radio . including the radio , there were 14 + 21 + 1 = 36 items sold . the answer is c ." | a = 22 + 15
b = a - 1
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a ) 50 : 53 , b ) 58 : 53 , c ) 52 : 53 , d ) 42 : 36 , e ) 38 : 53 | c | divide(add(const_100, 4), add(const_100, 6)) | the cash difference between the selling prices of an book at a profit of 4 % and 6 % is $ 3 . the ratio of the two selling prices is : | "c 52 : 53 let c . p . of the book be $ x . then , required ratio = 104 % of x / 106 % of x = 104 / 106 = 52 / 53 = 52 : 53" | a = 100 + 4
b = 100 + 6
c = a / b
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a ) 16 % , b ) 25 % , c ) 33 % , d ) 40 % , e ) 52 % | c | multiply(divide(subtract(120, 80), 120), const_100) | in town x , 120 percent of the population are employed , and 80 percent of the population are employed males . what percent of the employed people in town x are females ? | "total employed people 12 % , out of which 80 are employed males , hence 40 % are employed females . ( employed females ) / ( total employed people ) = 40 / 120 = 1 / 3 = 33 % answer : c ." | a = 120 - 80
b = a / 120
c = b * 100
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a ) 256 , b ) 512 , c ) 768 , d ) 1024 , e ) 1536 | d | multiply(add(multiply(multiply(multiply(multiply(multiply(multiply(multiply(multiply(multiply(multiply(0.1, const_2), const_2), const_2), const_2), const_2), const_2), const_2), const_2), const_2), const_2), 0.1), const_10) | if an average hard drive had a capacity of 0.1 tb in 2000 , and average hard drive capacities double every 5 years , what will be the average hard drive capacity in 2050 ? | "0.1 * 2 ^ 10 = 0.1 * 1024 = 102.4 the answer is d ." | a = 0 * 1
b = a * 2
c = b * 2
d = c * 2
e = d * 2
f = e * 2
g = f * 2
h = g * 2
i = h * 2
j = i * 2
k = j + 0
l = k * 10
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a ) 1 , b ) 2 , c ) 3 , d ) 4 , e ) 5 | b | subtract(divide(15, 3), 3) | ( x + 3 ) is a factor in x ^ 2 - mx - 15 . what is the value of m ? | "i solved the second degree equation and found it like this : x ^ 2 - mx - 15 = 0 ( x - 5 ) ( x + 3 ) = 0 x = 5 or x = - 3 substituting both values for x in the equation we find : x ^ 2 - mx - 15 = > ( - 3 ) ^ 2 - m ( - 3 ) = 15 = > 9 + 3 m = 15 = > 3 m = 15 - 9 = 6 = > m = 2 and with 5 , using a similar process we end up with : ( 5 ) ^ 2 - m ( 5 ) = 15 - 5 m = 15 - 25 = - 10 m = 2 ao , ans b" | a = 15 / 3
b = a - 3
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a ) 3 / 13 , b ) 1 / 5 , c ) 1 / 10 , d ) 1 / 3 , e ) 1 / 7 | a | inverse(add(inverse(3), 4)) | a small water pump would take 3 hours to fill an empty tank . a larger pump would take 1 / 4 hour to fill the same tank . how many hours would it take both pumps , working at their respective constant rates , to fill the empty tank if they began pumping at the same time ? | rate of the small pump is 1 / 3 tank / hour rate of the larger pump is 1 / ( 1 / 4 ) or 4 tank / hour ; combined rate of the two pumps is 1 / 3 + 4 = 13 / 3 tank / hour , together they will fill the empty tank in 1 / ( 13 / 3 ) = 3 / 13 hours ( time = job / rate ) . answer : a | a = 1/(3)
b = a + 4
c = 1/(b)
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a ) 1125 % , b ) 956 % , c ) 417 % , d ) 3788 % , e ) 219 % | a | multiply(divide(subtract(7, divide(4, 7)), divide(4, 7)), const_100) | by approximately what percent is x greater than 4 / 7 if ( 1 / 7 ) ( x ) = 1 ? | what percent is x greater than 4 / 7 if ( 1 / 7 ) ( x ) = 1 ? = > x = 7 % change = [ ( 7 - 4 / 7 ) / ( 4 / 7 ) ] * 100 = ( 49 / 4 - 1 ) * 100 = 45 * 25 = 1125 % ans , a | a = 4 / 7
b = 7 - a
c = 4 / 7
d = b / c
e = d * 100
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a ) 0 , b ) 50 , c ) 450 , d ) 397 , e ) 500 | d | subtract(multiply(multiply(add(const_3, const_4), const_1000), divide(1, 10)), multiply(divide(divide(1, 10), const_100), multiply(add(const_3, const_4), const_1000))) | when 1 / 10 percent of 3,000 is subtracted from 1 / 10 of 3,000 , the difference is | "we can break this problem into two parts : 1 ) what is 1 / 10 percent of 3,000 ? 2 ) what is 1 / 10 of 3,000 ? to calculate 1 / 10 percent of 3,000 we must first remember to divide 1 / 10 by 100 . so we have : ( 1 / 10 ) / ( 100 ) to divide a number by 100 means to multiply it by 1 / 100 , so we have : 1 / 10 x 1 / 100 = 1 / 1,000 thus , 1 / 10 percent of 3,000 = 1 / 1,000 x 3,000 = 3 . now let ' s concentrate on part 2 . we need to calculate 1 / 10 of 3,000 . to do this we simply multiply 1 / 10 by 3,000 . 1 / 10 x 3,000 = 300 the answer to part 1 is 3 , and the answer to part 2 is 300 . their difference is 300 – 3 = 397 . answer d ." | a = 3 + 4
b = a * 1000
c = 1 / 10
d = b * c
e = 1 / 10
f = e / 100
g = 3 + 4
h = g * 1000
i = f * h
j = d - i
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a ) 17 , b ) 16 , c ) 18 , d ) 19 , e ) 20 | b | add(divide(subtract(89, 10), 5), const_1) | how many multiples of 5 are there between 10 and 89 , 10 and 89 inclusive ? | "5 multiples are . . . 10,15 , 20,25 , 30,35 , 40,45 , 50,55 , 60,65 , 70,75 , 80,85 , . . . , . . . , the answer is = 16 answer is b" | a = 89 - 10
b = a / 5
c = b + 1
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a ) 2.5 , b ) 2.7 , c ) 2.9 , d ) 3.1 , e ) 3.3 | d | inverse(add(divide(const_1, 5), divide(const_1, 8))) | a worker can load one truck in 5 hours . a second worker can load the same truck in 8 hours . if both workers load one truck simultaneously while maintaining their constant rates , approximately how long , in hours , will it take them to fill one truck ? | "the workers fill the truck at a rate of 1 / 5 + 1 / 8 = 13 / 40 of the truck per hour . then the time to fill one truck is 40 / 13 which is about 3.1 hours . the answer is d ." | a = 1 / 5
b = 1 / 8
c = a + b
d = 1/(c)
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a ) 125 / 64 , b ) 25 / 16 , c ) 1 , d ) 4 / 5 , e ) 64 / 125 | c | divide(multiply(4, 5), multiply(5, 4)) | if 4 a = 5 b and ab ≠0 , what is the ratio of a / 5 to b / 4 ? | a nice fast approach is the first find a pair of numbers that satisfy the given equation : 4 a = 5 b here ' s one pair : a = 5 and b = 4 what is the ratio of a / 5 to b / 4 ? in other words , what is the value of ( a / 5 ) / ( b / 4 ) ? plug in values to get : ( a / 5 ) / ( b / 4 ) = ( 5 / 5 ) / ( 4 / 4 ) = 1 / 1 = 1 c | a = 4 * 5
b = 5 * 4
c = a / b
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a ) 1 , b ) 2 , c ) 3 , d ) 4 , e ) 5 | c | divide(subtract(add(add(add(13, 5), 1), 1), subtract(13, 5)), subtract(5, 1)) | what number is that , to which if 1 , 5 , and 13 , be severally added , the first sum shall be to the second , as the second to the third ? | solution let x = the number required . by the conditions , x + 1 : x + 5 = a + 5 : x + 13 subtracting terms , ( art . 385,2 . ) x + 1 : 4 = x + 5 : 8 therefore 8 x + 8 = 4 x + 20 . and x = 3 . answer c | a = 13 + 5
b = a + 1
c = b + 1
d = 13 - 5
e = c - d
f = 5 - 1
g = e / f
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a ) 45 % , b ) 55 % , c ) 35 % , d ) 32.5 % , e ) 30 % | b | subtract(const_100, multiply(divide(subtract(const_100, 40), const_100), subtract(const_100, 25))) | the price of a cycle is reduced by 25 per cent . the new price is reduced by a further 40 per cent . the two reductions together are equal to a single reduction of | "price = p initially price reduced by 25 % which means new price is 3 / 4 p now on this new price further 40 percent is reduced which means the new price is merely 60 percent of 3 / 4 p = = > ( 3 / 4 ) x ( 3 / 5 ) p = 9 / 20 p is the new price after both deduction which is 45 percent of the original value p . this implies this entire series of deduction is worth having discounted 55 % of p . so answer is b = 55 %" | a = 100 - 40
b = a / 100
c = 100 - 25
d = b * c
e = 100 - d
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a ) 0.35 , b ) 0.3505 , c ) 0.25 , d ) 0.351 , e ) 0.3527 | c | multiply(divide(2.505, 10.02), const_100) | 2.505 / 10.02 = | "2.505 / 10.02 2505 / 1002 = 2.5 move the comma two places to the left as we have 2 decimal places from the divisor : 0.25 . answer : c" | a = 2 / 505
b = a * 100
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a ) 12 , b ) 12.5 , c ) 13 , d ) 13.5 , e ) 14 | b | divide(multiply(1000, 15), add(1000, 200)) | 1000 boys have provisions for 15 days . if 200 more men join them , for how many days will the provisions last now ? | explanation : 1000 * 15 = 1200 * x x = 12.5 b | a = 1000 * 15
b = 1000 + 200
c = a / b
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a ) 1 : 3 , b ) 1 : 9 , c ) 2 : 5 , d ) 5 : 3 , e ) 6 : 1 | a | divide(const_2, 6) | in a certain neighborhood , there are half as many beige houses as white houses , and 6 times as many white houses as brown houses . what is the ratio of the number of brown houses to the number of beige houses ? | the ratio of beige : white : brown = 3 : 6 : 1 the answer is a . | a = 2 / 6
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['a ) 2', 'b ) 2 √ 2', 'c ) 4', 'd ) 2 π', 'e ) π · √ 8'] | b | sqrt(multiply(4, 2)) | the surface area of a sphere is 4 π r ^ 2 , where r is the radius of the sphere . if the surface area of sphere a is eight times as large as the surface area of sphere b , then the radius of sphere a is how many times the radius of sphere b ? | r 1 : radius of sphere a r 2 : radius of sphere b as per question , 4 π r 1 ^ 2 = 8 ∗ 4 π r 2 ^ 2 or r 12 = 8 ∗ r 2 ^ 2 or r 1 = 2 √ 2 r 2 . hence , b | a = 4 * 2
b = math.sqrt(a)
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a ) 10 % , b ) 0 % , c ) 30 % , d ) 25 % , e ) 28 % | b | multiply(divide(subtract(50, 50), 50), const_100) | a man buy a book in rs 50 & sale it rs 50 . what is the rate of profit ? ? ? | "cp = 50 sp = 50 profit = 50 - 50 = 0 % = 0 / 50 * 100 = 0 % answer : b" | a = 50 - 50
b = a / 50
c = b * 100
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a ) 550 , b ) 400 , c ) 700 , d ) 600 , e ) 850 | d | multiply(20, multiply(72, const_0_2778)) | a train passes a station platform in 50 sec and a man standing on the platform in 20 sec . if the speed of the train is 72 km / hr . what is the length of the platform ? | "speed = 72 * 5 / 18 = 20 m / sec . length of the train = 20 * 20 = 400 m . let the length of the platform be x m . then , ( x + 400 ) / 50 = 20 = > x = 600 m . answer : d" | a = 72 * const_0_2778
b = 20 * a
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a ) 50.81 % , b ) 15.11 % , c ) 20.7 % , d ) 48.8 % , e ) 40 % | d | multiply(divide(subtract(divide(const_100, divide(subtract(const_100, 40), const_100)), divide(const_100, divide(add(const_100, 15), const_100))), divide(const_100, divide(subtract(const_100, 40), const_100))), const_100) | certain stocks in january were 40 % less than they were in february and 15 % greater than they were in march . what was the percentage decrease in the stocks from february to march ? | "let stocks value in feb = 100 = > in jan = 60 and march = 60 * 100 / 115 = 51.2 % thus , percent decrease in stocks from feb to march = 100 - 51.2 = 48.8 % hence answer is d" | a = 100 - 40
b = a / 100
c = 100 / b
d = 100 + 15
e = d / 100
f = 100 / e
g = c - f
h = 100 - 40
i = h / 100
j = 100 / i
k = g / j
l = k * 100
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a ) 75 kgs , b ) 64 kgs , c ) 100 kgs , d ) 65 kgs , e ) 70 kgs | c | add(divide(multiply(20, subtract(const_100, 25)), const_100), 85) | fresh grapes contain 85 % water by weight and raisins obtained by drying fresh grapes contain 25 % water by weight . how many kgs of fresh grapes are needed to get 20 kgs of raisins ? | the weight of non - water in 20 kg of dried grapes ( which is 100 - 25 = 75 % of whole weight ) will be the same as the weight of non - water in x kg of fresh grapes ( which is 100 - 85 = 15 % of whole weight ) , so 20 * 0.75 = x * 0.15 - - > x = 100 . answer : c | a = 100 - 25
b = 20 * a
c = b / 100
d = c + 85
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a ) 15 , b ) 16 , c ) 21 , d ) 12 , e ) 19 | b | subtract(divide(subtract(multiply(6, 24), 8), const_4), 8) | the average age of a family of 6 members is 24 years . if the age of the youngest member is 8 years , what was the average age of the family at the birth of the youngest member ? | "present age of total members = 6 x 24 = 144 8 yrs back their ages were = 6 x 8 = 48 ages at the birth of youngest member = 144 - 48 = 96 therefore , avg age at the birth of youngest member = 96 / 6 = 16 . answer : b" | a = 6 * 24
b = a - 8
c = b / 4
d = c - 8
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a ) $ 600 , b ) $ 760 , c ) $ 850 , d ) $ 980 , e ) $ 1,140 | b | multiply(add(add(subtract(subtract(45, 20), 10), multiply(subtract(const_1, divide(40, const_100)), 10)), multiply(subtract(const_1, divide(15, const_100)), 20)), 20) | tickets to a certain concert sell for $ 20 each . the first 10 people to show up at the ticket booth received a 40 % discount , and the next 20 received a 15 % discount . if 45 people bought tickets to the concert , what was the total revenue from ticket sales ? | price of 1 ticket = 20 $ revenue generated from sales of first 10 tickets = 10 * ( 60 / 100 * 20 ) = 10 * 12 = 120 revenue generated from sales of next 20 tickets = 20 * ( 85 / 100 * 20 ) = 20 * 17 = 340 revenue generated from sales of last 15 tickets = 20 * 15 = 300 revenue generated from sales of 45 tickets = 120 + 340 + 300 = 760 $ answer b | a = 45 - 20
b = a - 10
c = 40 / 100
d = 1 - c
e = d * 10
f = b + e
g = 15 / 100
h = 1 - g
i = h * 20
j = f + i
k = j * 20
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a ) 6 , b ) 8 , c ) 10 , d ) 12 , e ) 14 | a | add(divide(subtract(multiply(38, 4), multiply(35, 4)), subtract(44, 38)), 4) | a car averages 35 miles per hour for the first 4 hours of a trip and averages 44 miles per hour for each additional hour of travel time . if the average speed for the entire trip is 38 miles per hour , how many hours long is the trip ? | "let t be the total time of the trip . 35 * 4 + 44 ( t - 4 ) = 38 t 6 t = 176 - 140 t = 6 the answer is a ." | a = 38 * 4
b = 35 * 4
c = a - b
d = 44 - 38
e = c / d
f = e + 4
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a ) 288 , b ) 350 , c ) 889 , d ) 276 , e ) 234 | b | subtract(multiply(speed(300, 18), 39), 300) | a 300 m long train crosses a platform in 39 sec while it crosses a signal pole in 18 sec . what is the length of the platform ? | "speed = 300 / 18 = 50 / 3 m / sec . let the length of the platform be x meters . then , ( x + 300 ) / 39 = 50 / 3 3 x + 900 = 1950 = > x = 350 m . answer : b" | a = speed * (
b = a - 39
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a ) 0.0016 , b ) 0.0625 , c ) 0.0001 , d ) 0.25 , e ) 0.5 | c | power(divide(1, 4), 5) | what is the decimal equivalent of ( 1 / 4 ) ^ 5 ? | "( 1 / 4 ) ^ 5 = 1 / 1024 = 0.0001 answer : c" | a = 1 / 4
b = a ** 5
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a ) 3 , b ) 4 , c ) 5 , d ) 6 , e ) 7 | e | add(const_3, const_3) | a palindrome is a number that reads the same front - to - back as it does back - to - front ( e . g . 202 , 575 , 1991 , etc . ) p is the smallest integer greater than 800 that is both a prime and a palindrome . what is the sum of the digits of p ? | "given that p is smallest integer greater than 800 - assume there is a 3 - digit that satisfies the above conditions . let the number be xyx ; question asks us the values of 2 x + y we can straight away cross out options a ) and d ) - sum of digits 3 or 6 implies it is divisible by 3 - - - > we know that p is a prime number coming to option b ) 2 x + y = 4 - - > only x = 2 and y = 0 satisfy this equation ( x > 2 will never give sum of digits = 4 ) ; but 202 is divisible by 2 ; we know that p is a prime number similarly option c ) 2 x + y = 5 - - > only x = 2 and y = 1 satisfy this equation ( x > 2 will never give sum of digits = 5 ) ; but 212 is divisible by 2 ; we know that p is a prime number therefore answer option should be e - - - > can be verified by taking 2 x + y = 7 - - - > x = 3 and y = 1 ; gives 313 e" | a = 3 + 3
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a ) 1.7 , b ) 3.5 , c ) 6.3 , d ) 7.1 , e ) 10.4 | a | multiply(divide(divide(multiply(35, 20), const_100), 42), const_12) | a reduction of 35 % in the price of apples would enable a man to obtain 42 more for rs . 20 , what is reduced price per dozen ? | "a 20 * ( 35 / 100 ) = 6 - - - 42 ? - - - 12 = > rs . 1.7" | a = 35 * 20
b = a / 100
c = b / 42
d = c * 12
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a ) 3 , b ) 4 , c ) 6 , d ) 7 , e ) 8 | c | divide(add(multiply(factorial(52), factorial(5)), multiply(factorial(52), factorial(3))), 52) | what is the units digit of ( 52 ^ 5 ) ( 37 ^ 3 ) ( 71 ^ 9 ) ? | "the units digit of 52 ^ 5 is the units digit of 2 ^ 5 which is 2 . the units digit of 37 ^ 3 is the units digit of 7 ^ 3 which is 3 . the units digit of 71 ^ 9 is the units digit of 1 ^ 9 which is 1 . the units digit of 2 * 3 * 1 is 6 . the answer is c ." | a = math.factorial(52)
b = math.factorial(5)
c = a * b
d = math.factorial(52)
e = math.factorial(3)
f = d * e
g = c + f
h = g / 52
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a ) 25 , b ) 30 , c ) 40 , d ) 11 , e ) 60 | d | divide(220, multiply(const_0_2778, subtract(80, 8))) | how many seconds will a 220 metre long train take to cross a man running with a speed of 8 km / hr in the direction of the moving train if the speed of the train is 80 km / hr ? | "explanation : speed of train relatively to man = ( 80 - 8 ) km / hr = 72 km / hr = ( 72 x 5 / 18 ) m / sec = 20 m / sec time taken to pass the man = ( 220 / 20 ) sec = 11 sec . answer : d" | a = 80 - 8
b = const_0_2778 * a
c = 220 / b
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a ) 18 , b ) 27 , c ) 98 , d ) 27 , e ) 21 | e | multiply(subtract(divide(30, 16), const_1), 24) | pipe a can fill a tank in 16 minutes and pipe b cam empty it in 24 minutes . if both the pipes are opened together after how many minutes should pipe b be closed , so that the tank is filled in 30 minutes ? | "let the pipe b be closed after x minutes . 30 / 16 - x / 24 = 1 = > x / 24 = 30 / 16 - 1 = 14 / 16 = > x = 14 / 16 * 24 = 21 . answer : e" | a = 30 / 16
b = a - 1
c = b * 24
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a ) 1 , b ) 3 , c ) 5 , d ) 7 , e ) 9 | e | add(add(const_4, const_3), const_2) | what is the units digit of ( 147 ^ 25 ) ^ 50 ? | the units digit of the exponents of 7 repeat in a cycle of four , which is { 7,9 , 3,1 } . the number 25 has the form 4 n + 1 so the units digit is 7 inside the bracket . the exponent 50 has the form 4 n + 2 , so the units digit is 9 . the answer is e . | a = 4 + 3
b = a + 2
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a ) 9 % , b ) 10 % , c ) 11 % , d ) 12 % , e ) 65 % | e | multiply(subtract(divide(33, const_100), divide(subtract(9, multiply(divide(33, const_100), multiply(18, 2))), subtract(multiply(15, 3), multiply(18, 2)))), const_100) | fox jeans regularly sell for $ 15 a pair and pony jeans regularly sell for $ 18 a pair . during a sale these regular unit prices are discounted at different rates so that a total of $ 9 is saved by purchasing 5 pairs of jeans : 3 pairs of fox jeans and 2 pairs of pony jeans . if the sum of the two discounts rates is 33 percent , what is the discount rate on pony jeans ? | "x discount on pony jeans , ( 0.33 - x ) discount on fox jeans . set the equation : 3 * 15 ( 0.33 - x ) + 2 * 18 x = 9 - - > x = 0.65 = 65 % answer : e ." | a = 33 / 100
b = 33 / 100
c = 18 * 2
d = b * c
e = 9 - d
f = 15 * 3
g = 18 * 2
h = f - g
i = e / h
j = a - i
k = j * 100
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a ) 2.5 min , b ) 3 min , c ) 3.2 min , d ) 3.5 min , e ) 4 min | b | multiply(divide(add(divide(1, 4), divide(add(const_3, const_4), 2)), 75), const_60) | a train travelling at a speed of 75 mph enters a tunnel 31 / 2 miles long . the train is 1 / 4 mile long . how long does it take for the train to pass through the tunnel from the moment the front enters to the moment the rear emerges ? | total distance = ( 7 / 2 + 1 / 4 ) miles = 15 / 4 miles time taken = 15 / 4 * 75 hrs = 1 / 20 hrs = 3 min answer b | a = 1 / 4
b = 3 + 4
c = b / 2
d = a + c
e = d / 75
f = e * const_60
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a ) 1 / 3 , b ) - 3 , c ) - 1 / 3 , d ) 3 , e ) 0.3 | a | log(divide(log(subtract(7, multiply(add(const_4, const_1), const_1000))), log(add(const_4, const_1)))) | the value of log 343 7 is | "solution let log 3437 = n . then , 343 n = 7 = ( 73 ) n = 7 . ‹ = › 3 n = 1 ‹ = › n = 1 / 3 . log 343 7 = 1 / 3 . answer a" | a = 4 + 1
b = a * 1000
c = 7 - b
d = math.log(c)
e = 4 + 1
f = math.log(e)
g = d / f
h = math.log(g)
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a ) 700 , b ) 882 , c ) 799 , d ) 778 , e ) 901 | a | divide(multiply(140, const_100), subtract(add(const_100, 4), subtract(const_100, 16))) | a watch was sold at a loss of 16 % . if it was sold for rs . 140 more , there would have been a gain of 4 % . what is the cost price ? | "84 % 104 % - - - - - - - - 20 % - - - - 140 100 % - - - - ? = > rs . 700 answer : a" | a = 140 * 100
b = 100 + 4
c = 100 - 16
d = b - c
e = a / d
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a ) 92 , b ) 82 , c ) 72 , d ) 62 , e ) 52 | a | subtract(multiply(10, 83), multiply(9, 82)) | robin ' s average ( arithmetic mean ) test score on 9 tests is 82 . what must be robin ' s score on a 10 th test for his average score on the 10 tests to be 83 ? | ( 9 * 82 + x ) / 10 = 83 x = ( 10 * 83 ) - ( 9 * 82 ) x = 830 - 738 total score required 830 - 738 = 92 correct answer is a | a = 10 * 83
b = 9 * 82
c = a - b
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a ) 5 : 8 , b ) 5 : 5 , c ) 5 : 9 , d ) 3 : 2 , e ) 5 : 1 | d | divide(3, 4) | the simple form of the ratio 9 / 8 : 3 / 4 is ? | "9 / 8 : 3 / 4 = 3 : 2 answer : d" | a = 3 / 4
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a ) 382 , b ) 782 , c ) 278 , d ) 310 , e ) 881 | d | subtract(multiply(multiply(divide(72, const_3600), const_1000), 26), 210) | a goods train runs at the speed of 72 km / hr and crosses a 210 m long platform in 26 sec . what is the length of the goods train ? | "speed = 72 * 5 / 18 = 20 m / sec . time = 26 sec . let the length of the train be x meters . then , ( x + 210 ) / 26 = 20 x = 310 m . answer : d" | a = 72 / 3600
b = a * 1000
c = b * 26
d = c - 210
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a ) $ 900 , b ) $ 880 , c ) $ 1000 , d ) $ 700 , e ) $ 1500 | b | divide(220, subtract(const_1, divide(3, 4))) | linda spent 3 / 4 of her savings on furniture and the rest on a tv . if the tv cost her $ 220 , what were her original savings ? | "if linda spent 3 / 4 of her savings on furnitute , the rest 4 / 4 - 3 / 4 = 1 / 4 on a tv but the tv cost her $ 220 . so 1 / 4 of her savings is $ 220 . so her original savings are 4 times $ 220 = $ 880 correct answer b" | a = 3 / 4
b = 1 - a
c = 220 / b
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a ) 39 , b ) 95 , c ) 47 , d ) 55 , e ) 76 | c | divide(95, 2) | i have only 1 $ and 2 $ note . i want to give 95 $ to john . how many maximum number of 2 $ note i need ? | check whether 95 is divisible by 2 then that is maximum no of 2 $ note else check next maximum term before 95 so 94 / 2 = 47 ans is c | a = 95 / 2
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a ) 0.01 , b ) 1 , 000,000 , c ) 0.001 , d ) 1,000 , e ) 1.0 e - 05 | b | multiply(divide(1, multiply(const_100, const_100)), multiply(const_100, const_100)) | if 10,000 microns = 1 decimeter , and 10,000 , 000,000 angstroms = 1 decimeter , how many angstroms equal 1 micron ? | given that 10,000 microns = 1 decimeter = 10,000 , 000,000 angstroms so , 1 micron = 10,000 , 000,000 / 10,000 = 1 , 000,000 answer : b | a = 100 * 100
b = 1 / a
c = 100 * 100
d = b * c
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a ) 87 kmph , b ) 64 kmph , c ) 54 kmph , d ) 16 kmph , e ) 18 kmph | b | subtract(multiply(14, multiply(125, const_0_2778)), 125) | a train 125 m long crosses a platform 125 m long in 14 sec ; find the speed of the train ? | "d = 125 + 125 = 250 t = 14 s = 250 / 14 * 18 / 5 = 64 kmph answer : b" | a = 125 * const_0_2778
b = 14 * a
c = b - 125
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a ) 50 % , b ) 40 % , c ) 30 % , d ) 20 % , e ) 10 % | e | divide(subtract(30, 10), subtract(12, 10)) | salad dressing p is made up of 30 % vinegar and 70 % oil , and salad dressing q contains 10 % vinegar and 90 % oil . if the two dressings are combined to produce a salad dressing that is 12 % vinegar , dressing p comprises what percentage of the new dressing ? | let x be the percentage of dressing p in the new dressing . 0.3 x + 0.1 ( 1 - x ) = 0.12 0.2 x = 0.02 x = 0.1 = 10 % the answer is e . | a = 30 - 10
b = 12 - 10
c = a / b
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a ) 630 , b ) 735 , c ) 825 , d ) 600 , e ) none of these | b | subtract(multiply(3500, power(add(const_1, divide(10, const_100)), 2)), 3500) | what is the compound interest paid on a sum of rs . 3500 for the period of 2 years at 10 % per annum . | "solution = interest % for 1 st year = 10 interest % for 2 nd year = 10 + 10 % of 10 = 10 + 10 * 10 / 100 = 11 total % of interest = 10 + 11 = 21 total interest = 21 % 3500 = 3500 * ( 21 / 100 ) = 735 answer b" | a = 10 / 100
b = 1 + a
c = b ** 2
d = 3500 * c
e = d - 3500
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a ) 22 , b ) 15 , c ) 28 , d ) 26 , e ) 9 | b | add(add(multiply(3, 3), 3), const_3) | a grand father has 3 grand children . age difference of two children among them is 3 . eldest child age is 3 times the youngest child ’ s age and the eldest child age is two year more than the sum of age of other two children . what is the age of the eldest child ? | as the eldest son ' s age is 3 times that of the youngest , eldest son ' s age should be a multiple of 3 . from the given options take 15 as the eldest son ' s age . then youngest son ' s age becomes 5 . but eldest sons age is 2 more than the sum of the remaining two sons . so sum of the remaining two sons is 13 . so the age of the middle son is 13 - 5 = 8 . which satisfies another condition in the question that the difference between the two sons age is 3 . so answer is 15 . answer : b | a = 3 * 3
b = a + 3
c = b + 3
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a ) 228 , b ) 299 , c ) 300 , d ) 500 , e ) 188 | c | multiply(add(5, 6), const_100) | rs . 1500 is divided into two parts such that if one part is invested at 6 % and the other at 5 % the whole annual interest from both the sum is rs . 87 . how much was lent at 5 % ? | "( x * 5 * 1 ) / 100 + [ ( 1500 - x ) * 6 * 1 ] / 100 = 87 5 x / 100 + 90 â € “ 6 x / 100 = 87 x / 100 = 3 = > x = 300 answer : c" | a = 5 + 6
b = a * 100
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a ) 77 , b ) 270 , c ) 1600 , d ) 840 , e ) 980 | d | lcm(4, 7) | what is the lowest positive integer that is divisible by 4 through 7 , inclusive ? | "the integer should be divisible by : 4 , 5 , 6 and 7 . the least common multiple of these integers is lcm = 4 * 5 * 6 * 7 = 840 answer : d" | a = math.lcm(4, 7)
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a ) 80 , b ) 150 , c ) 240 , d ) 360 , e ) 480 | a | add(divide(60, 3), 60) | of the people who responded to a market survey , 60 preferred brand x and the rest preferred brand y . if the respondents indicated a preference for brand x over brand y by ratio of 3 to 1 , how many people responded to the survey ? | "ratio = 3 : 1 = > 3 x respondents preferred brand x and x preferred brand y since , no . of respondents who preferred brand x = 60 = > 3 x = 60 = > x = 20 hence total no . of respondents = 60 + 20 = 80 hence a is the answer ." | a = 60 / 3
b = a + 60
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a ) 22.3 , b ) 33.2 , c ) 22.2 , d ) 11.3 , e ) 37.5 | e | divide(subtract(multiply(50, 38), add(45, 55)), subtract(50, const_2)) | the average of 50 numbers id 38 . if two numbers , namely 45 and 55 are discarded , the average of the remaining numbers is : | "explanation : total of 50 numbers = ( 50 × 38 ) = 1900 ( 50 × 38 ) = 1900 total of 48 numbers = ( 1900 - ( 45 + 55 ) ] = 1800 required average = 180048 = 2256 = 37.5 answer : e" | a = 50 * 38
b = 45 + 55
c = a - b
d = 50 - 2
e = c / d
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a ) 1 / 512 , b ) 511 / 512 , c ) 9 / 512 , d ) 503 / 51 , e ) 503 / 510 | b | subtract(const_1, power(divide(const_1, const_2), 9)) | when fair coin is tossed 9 times find the probability of getting head at least once . | 1 - 1 / 2 ^ 9 = 1 - 1 / 512 = 511 / 512 answer : b | a = 1 / 2
b = a ** 9
c = 1 - b
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a ) 8600 litres , b ) 200 litres , c ) 21600 litres , d ) 11200 litres , e ) 13200 litres | c | multiply(divide(multiply(multiply(16, const_60), add(10, 8)), 8), 10) | an outlet pipe empties a tank which is full in 10 hours . if the inlet pipe is kept open , which lets water in at the rate of 16 litres / min then outlet pipe would take 8 hours longer . find the capacity of the tank . | "let the rate of outlet pipe be x liters / hour ; rate of inlet pipe is 16 litres / min , or 16 * 60 = 960 liters / hour ; net outflow rate when both pipes operate would be x - 960 liters / hour . capacity of the tank = x * 10 hours = ( x - 960 ) * ( 10 + 8 ) hours 10 x = ( x - 960 ) * 18 - - > x = 2160 - - > capacity = 10 x = 21600 liters . answer : c ." | a = 16 * const_60
b = 10 + 8
c = a * b
d = c / 8
e = d * 10
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a ) 80 , b ) 95 , c ) 105 , d ) 156 , e ) 210 | d | multiply(divide(120, add(subtract(divide(const_1, const_3), multiply(divide(const_1, const_3), divide(50, const_100))), multiply(subtract(const_1, divide(const_1, const_3)), divide(40, const_100)))), add(multiply(divide(const_1, const_3), divide(50, const_100)), subtract(subtract(const_1, divide(const_1, const_3)), multiply(subtract(const_1, divide(const_1, const_3)), divide(40, const_100))))) | in a certain company , a third of the workers do not have a retirement plan . 50 % of the workers who do not have a retirement plan are women , and 40 % of the workers who do have a retirement plan are men . if 120 of the workers of that company are men , how many of the workers are women ? | "set up equation : x = total number of workers 120 = 0,4 * 2 / 3 * x + 0,5 * 1 / 3 * x 120 = 13 / 30 x x = 276 276 - 120 = 156 answer d" | a = 1 / 3
b = 1 / 3
c = 50 / 100
d = b * c
e = a - d
f = 1 / 3
g = 1 - f
h = 40 / 100
i = g * h
j = e + i
k = 120 / j
l = 1 / 3
m = 50 / 100
n = l * m
o = 1 / 3
p = 1 - o
q = 1 / 3
r = 1 - q
s = 40 / 100
t = r * s
u = p - t
v = n + u
w = k * v
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a ) 44 , b ) 43 , c ) 42 , d ) 41 , e ) 48 | e | multiply(92, subtract(add(floor(divide(1782452, 92)), const_1), divide(1782452, 92))) | which number need to add to 1782452 to get a number exactly divisible by 92 ? | "1782452 / 92 = 19374 and reminder = 44 . 92 - 44 = 48 so , the next number divisible by 92 is 48 places in front of 1782452 which means 48 + 1782452 = 1782500 48 should be added to 1782452 e" | a = 1782452 / 92
b = math.floor(a)
c = b + 1
d = 1782452 / 92
e = c - d
f = 92 * e
|
a ) 180 km , b ) 190 km , c ) 220 km , d ) 230 km , e ) none | a | divide(19, add(inverse(multiply(subtract(14, 4), const_2)), inverse(add(4, 14)))) | a boat takes 19 hours for travelling downstream fro poin a to point b and coming back to a point c midway between a and b . if the velocity of the stream is 4 kmph and the speed of the boat in still water is 14 kmph , what is the distance between a and b ? | sol . speed downstream = ( 14 + 4 ) km / hr = 18 km / hr ; speed upstream = ( 14 - 4 ) km / hr = 10 km / hr . let the distance between a and b be x km . then , x / 18 + ( x / 2 ) / 10 = 19 ⇔ x / 18 + x / 20 = 19 ⇒ x = 180 km . answer a | a = 14 - 4
b = a * 2
c = 1/(b)
d = 4 + 14
e = 1/(d)
f = c + e
g = 19 / f
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a ) 1 : 8 , b ) 1 : 4 , c ) 2 : 1 , d ) 14 : 1 , e ) 18 : 1 | d | multiply(7, const_2) | city x has a population 7 times as great as the population of city y , which has a population twice as great as the population of city z . what is the ratio of the population of city x to the population of city z ? | x = 7 y , y = 2 * z x : y , y : z 7 : 1 , 2 : 1 14 : 2 , 2 : 1 so , x : z = 14 : 1 ( d ) | a = 7 * 2
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a ) 14 , b ) 16 , c ) 18 , d ) 20 , e ) 22 | b | add(multiply(add(3, 3), 3), floor(divide(subtract(150, multiply(divide(multiply(3, subtract(10, 1)), const_2), 10)), 10))) | in a certain supermarket , a triangular display of cans is arranged in 10 rows , numbered 1 through 10 from top to bottom . each successively numbered row contains 3 more cans than the row immediately above it . if there are fewer than 150 cans in the entire display , how many cans are in the sixth row ? | "let x be the number of cans in row 1 . the total number of cans is x + ( x + 3 ) + . . . + ( x + 27 ) = 10 x + 3 ( 1 + 2 + . . . + 9 ) = 10 x + 3 ( 9 ) ( 10 ) / 2 = 10 x + 135 since the total is less than 150 , x must equal 1 . the number of cans in the 6 th row is 1 + 3 ( 5 ) = 16 the answer is b ." | a = 3 + 3
b = a * 3
c = 10 - 1
d = 3 * c
e = d / 2
f = e * 10
g = 150 - f
h = g / 10
i = math.floor(h)
j = b + i
|
['a ) 8', 'b ) 9', 'c ) 10', 'd ) 11', 'e ) 16'] | a | add(divide(subtract(148, const_1), add(20, const_1)), const_1) | in a new housing development , trees are to be planted along the sidewalk of a certain street . each tree takes up one square foot of sidewalk space , and there are to be 20 feet between each tree . how many trees can be planted if the road is 148 feet long ? | let t be the number of trees . then the length required for trees on the sidewalk will be 1 * t = t to maximize the number of trees , the number of 20 feet spaces between trees should be 1 less than total number of trees . for example , if there are 3 trees , then there should be 2 spaces between them . so the number of 20 feet spaces will be t - 1 . then , the length of sidewalk required for 20 feet spaces will be 20 * ( t - 1 ) it is given that total length of sidewalk is 148 feet . or 20 ( t - 1 ) + t = 148 or 20 t - 20 + t = 148 or 21 t = 168 or t = 8 answer : - a | a = 148 - 1
b = 20 + 1
c = a / b
d = c + 1
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a ) 7 / 30 , b ) 3 / 14 , c ) 9 / 38 , d ) 4 / 21 , e ) 5 / 26 | c | multiply(divide(add(const_4, const_4), 20), divide(subtract(add(const_4, const_4), 1), subtract(20, 1))) | 20 balls are numbered 1 to 20 . a ball is drawn and then another ball is drawn without replacement . what is the probability that both balls have even numbers ? | "p ( 1 st ball is even ) = 10 / 20 p ( 2 nd ball is also even ) = 9 / 19 p ( both balls are even ) = 10 / 20 * 9 / 19 = 9 / 38 the answer is c ." | a = 4 + 4
b = a / 20
c = 4 + 4
d = c - 1
e = 20 - 1
f = d / e
g = b * f
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a ) 15.8 sec , b ) 14.9 sec , c ) 12.4 sec , d ) 16.8 sec , e ) 17.4 sec | e | divide(add(120, 170), multiply(60, const_0_2778)) | how long does a train 120 m long travelling at 60 kmph takes to cross a bridge of 170 m in length ? | "e 17.4 sec d = 120 + 170 = 290 m s = 60 * 5 / 18 = 50 / 3 t = 290 * 3 / 50 = 17.4 sec answer is e" | a = 120 + 170
b = 60 * const_0_2778
c = a / b
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a ) 21 / 25 , b ) 17 / 25 , c ) 4 / 25 , d ) 8 / 25 , e ) 10 / 25 | b | divide(multiply(subtract(multiply(const_6, const_3), const_1), const_2), 50) | the probability that a number selected at random from first 50 natural numbers is a composite number is | "the number of exhaustive events = 50 c 1 = 50 . we have 15 primes from 1 to 50 . number of favourable cases are 34 . therefore , required probability = 34 / 50 = 17 / 25 . answer : b" | a = 6 * 3
b = a - 1
c = b * 2
d = c / 50
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a ) 15 , b ) 66 , c ) 22.5 , d ) 52 , e ) 42 | c | multiply(divide(15, const_60), 90) | the speed of a train is 90 kmph . what is the distance covered by it in 15 minutes ? | "90 * 15 / 60 = 22.5 kmph answer : c" | a = 15 / const_60
b = a * 90
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a ) 9 , b ) 12 , c ) 13 , d ) 14 , e ) 17 | c | add(add(2, 2), subtract(power(subtract(multiply(const_12, const_4), const_1), 2), add(multiply(const_2, multiply(subtract(const_12, const_2), multiply(subtract(const_12, const_2), subtract(const_12, const_2)))), multiply(const_2, multiply(subtract(const_12, const_2), subtract(const_12, const_2)))))) | 225 , or 15 ^ 2 , is the first perfect square that begins with two 2 s . what is the sum of the digits of the next perfect square to begin with two 2 s ? | because 16 ^ 2 = 256 , y the next perfect square to begin with two 2 s can not be 3 - digit . let ' s try 4 - digit . 40 ^ 2 = 1600 and 50 ^ 2 = 2500 . so , if there is a 4 - digit perfect square to begin with two 2 s it must be between 41 ^ 2 and 49 ^ 2 . test the middle number 45 ^ 2 = 2025 . test larger number : 46 ^ 2 = 2116 . test next number : 47 ^ 2 = 2209 . bingo ! the sum of the digits = 2 + 2 + 0 + 9 = 13 . answer : c . | a = 2 + 2
b = 12 * 4
c = b - 1
d = c ** 2
e = 12 - 2
f = 12 - 2
g = 12 - 2
h = f * g
i = e * h
j = 2 * i
k = 12 - 2
l = 12 - 2
m = k * l
n = 2 * m
o = j + n
p = d - o
q = a + p
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a ) 15 degree , b ) 45 degree , c ) 55 degree , d ) 75 degree , e ) 65 degree | d | subtract(multiply(30, multiply(const_3, const_2)), 8) | what is the angle between the two hands of a clock when time is 8 : 30 | "use the formula i . e 1 / 2 ( 60 h - 11 m ) / / h - hour and m - minute 1 / 2 ( 60 * 8 - 11 * 30 ) 1 / 2 ( 150 ) 75 degree answer : d" | a = 3 * 2
b = 30 * a
c = b - 8
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a ) 6 sec , b ) 7 sec , c ) 7 2 / 2 sec , d ) 8 sec , e ) 9 sec | a | divide(add(150, 100), multiply(add(80, 70), const_0_2778)) | two trains are moving at 80 kmph and 70 kmph in opposite directions . their lengths are 150 m and 100 m respectively . the time they will take to pass each other completely is ? | 70 + 80 = 150 * 5 / 18 = 125 / 3 mps d = 150 + 100 = 250 m t = 250 * 3 / 125 = 6 sec answer : a | a = 150 + 100
b = 80 + 70
c = b * const_0_2778
d = a / c
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a ) 6 % , b ) 25 % , c ) 37 1 / 2 % , d ) 60 % , e ) 75 % | a | multiply(add(divide(const_1, 50), divide(const_1, 25)), const_100) | if x > 0 , x / 50 + x / 25 is what percent of x ? | just plug and chug . since the question asks for percents , pick 100 . ( but any number will do . ) 100 / 50 + 100 / 25 = 2 + 4 = 6 6 is 6 % of 100 . answer : a | a = 1 / 50
b = 1 / 25
c = a + b
d = c * 100
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a ) 2008 , b ) 2009 , c ) 2010 , d ) 2011 , e ) 2012 | d | add(2001, const_10) | the price of commodity p increases by 40 paise every year , while the price of commodity q increases by 15 paise every year . if in 2001 , the price of commodity p was rs . 4.20 and that of q was rs . 6.30 , in which year commodity p will cost 40 paise more than the commodity q ? | let the commodity p costs 40 paise more than the commodity q after n years . price of the commodity p in 2001 = rs . 4.20 since the price of the commodity p increases by rs 0.40 every year , price of the commodity p after n years from 2001 = rs . 4.20 + ( n × . 40 ) price of the commodity q in 2001 = rs . 6.30 since the price of the commodity q increases by rs 0.15 every year , price of the commodity q after n years from 2001 = rs . 6.30 + ( n × . 15 ) since the commodity p costs rs . 0.40 more that the commodity q after n years from 2001 , 4.20 + ( n × . 40 ) = 6.30 + ( n × . 15 ) + 0.40 = > ( 40 n - . 15 n ) = 6.30 - 4.20 + 0.40 = 2.5 = > . 25 n = 2.5 = > n = 2.5 / . 25 = 10 = > commodity p costs rs . 0.40 more that the commodity q after 10 years from 2001 . i . e . , in 2011 answer : d | a = 2001 + 10
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a ) 287 , b ) 398 , c ) 323 , d ) 245 , e ) 267 | c | subtract(multiply(30, multiply(52, const_0_2778)), 110) | a train 110 m long running at 52 kmph crosses a platform in 30 sec . what is the length of the platform ? | "d = 52 * 5 / 18 = 30 = 433 â € “ 110 = 323 answer : c" | a = 52 * const_0_2778
b = 30 * a
c = b - 110
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a ) 0.005 , b ) 0.002 , c ) 0.001 , d ) 0.0005 , e ) 0.0002 | c | divide(1, 1,000) | when magnified 1,000 times by an electron microscope , the image of a certain circular piece of tissue has a diameter of 1 centimeter . the actual diameter of the tissue , in centimeters , is | "it is very easy if x is the diameter , then the magnified length is 1000 x . ince 1000 x = 1 then x = 1 / 1000 = 0.001 . the answer is c" | a = 1 / 1
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a ) 42 , b ) 44 , c ) 46 , d ) 48 , e ) 50 | d | divide(120, subtract(divide(120, 30), divide(90, const_60))) | the distance from city a to city b is 120 miles . while driving from city a to city b , cara drives at a constant speed of 30 miles per hour . dan leaves city a 90 minutes after cara . what is the minimum constant speed in miles per hour that dan must exceed in order to arrive in city b before cara ? | "the time it takes cara to drive to city b is 120 / 30 = 4 hours . dan needs to take less than 2.5 hours for the trip . dan needs to exceed a constant speed of 120 / 2.5 = 48 miles per hour . the answer is d ." | a = 120 / 30
b = 90 / const_60
c = a - b
d = 120 / c
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a ) 175 , b ) 288 , c ) 277 , d ) 877 , e ) 361 | a | multiply(divide(630, add(add(multiply(2000, 8), multiply(subtract(2000, 1000), subtract(const_12, 8))), add(multiply(4000, 8), multiply(add(4000, 1000), subtract(const_12, 8))))), add(multiply(2000, 8), multiply(subtract(2000, 1000), subtract(const_12, 8)))) | a and b began business with rs . 2000 and rs . 4000 after 8 months , a withdraws rs . 1000 and b advances rs . 1000 more . at the end of the year , their profits amounted to rs . 630 find the share of a . | "( 2 * 8 + 1 * 4 ) : ( 4 * 8 + 5 * 4 ) 5 : 13 5 / 18 * 630 = 175 answer : a" | a = 2000 * 8
b = 2000 - 1000
c = 12 - 8
d = b * c
e = a + d
f = 4000 * 8
g = 4000 + 1000
h = 12 - 8
i = g * h
j = f + i
k = e + j
l = 630 / k
m = 2000 * 8
n = 2000 - 1000
o = 12 - 8
p = n * o
q = m + p
r = l * q
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a ) 1 / 2 , b ) 63 / 128 , c ) 4 / 7 , d ) 61 / 256 , e ) 63 / 64 | a | divide(add(add(add(choose(9, const_2), choose(9, const_3)), choose(9, const_4)), choose(9, 9)), power(const_2, 9)) | a fair coin is tossed 9 times . what is the probability of getting more heads than tails in 9 tosses ? | "on each toss , the probability of getting a head is 1 / 2 and the probability of getting a tail is 1 / 2 . there is no way to get the same number of heads and tails on an odd number of tosses . there will either be more heads or more tails . then there must be more heads on half of the possible outcomes and more tails on half of the possible outcomes . p ( more heads ) = 1 / 2 the answer is a ." | a = math.comb(9, 2)
b = math.comb(9, 3)
c = a + b
d = math.comb(9, 4)
e = c + d
f = math.comb(9, 9)
g = e + f
h = 2 ** 9
i = g / h
|
a ) 71 , b ) 12 , c ) 18 , d ) 11 , e ) 29 | d | divide(multiply(multiply(660, 10), 18), multiply(900, 12)) | if 18 bindres bind 900 books in 10 days , how many binders will be required to bind 660 books in 12 days ? | explanation : let the required no of binders be x . less books , less binders ( direct proportion ) more days , less binders ( indirect proportion ) \ inline \ fn _ jvn \ left . \ begin { matrix } books \ : 900 : 600 \ \ days \ : \ : 12 : 10 \ end { matrix } \ right \ } : : 18 : x \ inline \ fn _ jvn \ therefore \ inline \ fn _ jvn ( 900 \ times 12 \ times x ) = ( 600 \ times 10 \ times 18 ) \ inline \ fn _ jvn \ leftrightarrow x = \ inline \ fn _ jvn \ frac { 600 \ times 10 \ times 18 } { 900 \ times 12 } = 11 answer : d | a = 660 * 10
b = a * 18
c = 900 * 12
d = b / c
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a ) 1 , b ) 2 , c ) 3 , d ) 4 , e ) 5 | a | add(add(subtract(multiply(2, 1), multiply(3, 2)), multiply(1, 2)), add(subtract(multiply(2, 2), multiply(3, 1)), multiply(2, 1))) | if a ã — b = 2 a - 3 b + ab , then 1 ã — 2 + 2 ã — 1 is equal to : | explanation : 1 ã — 2 + 2 ã — 1 = ( 2 ã — 1 - 3 ã — 2 + 1 ã — 2 ) + ( 2 ã — 2 - 3 ã — 1 + 2 ã — 1 ) = ( 2 - 6 + 2 + 4 - 3 + 2 ) = 1 . answer : a | a = 2 * 1
b = 3 * 2
c = a - b
d = 1 * 2
e = c + d
f = 2 * 2
g = 3 * 1
h = f - g
i = 2 * 1
j = h + i
k = e + j
|
a ) 25 km , b ) 50 km , c ) 60 km , d ) 85 km , e ) 95 km | a | multiply(const_60, divide(multiply(divide(15, const_60), 50), 50)) | when a train travels at a speed of 80 kmph , it reaches the destination on time . when the same train travels at a speed of 50 kmph , it reaches its destination 15 min late . what is the length of journey ? | "let x be the time reached with the speed 80 km / h 50 km / h - - - - > x + 15 distance is equal so 80 ( km / h ) × xhr = 50 ( km / h ) × ( x + 15 ) hr so 80 x = 50 x + 750 so the would be in km and x = 25 answer : a" | a = 15 / const_60
b = a * 50
c = b / 50
d = const_60 * c
|
a ) 66.67 , b ) 882 , c ) 772 , d ) 252 , e ) 121 | a | multiply(divide(multiply(60, const_1000), const_3600), 4) | a train running at the speed of 60 km / hr crosses a pole in 4 seconds . find the length of the train . | "speed = 60 * ( 5 / 18 ) m / sec = 50 / 3 m / sec length of train ( distance ) = speed * time ( 50 / 3 ) * 4 = 66.67 meter . answer : a" | a = 60 * 1000
b = a / 3600
c = b * 4
|
a ) 82.5 , b ) 52.9 , c ) 52.2 , d ) 52.5 , e ) 52.7 | d | divide(add(multiply(30, 40), multiply(50, 60)), add(30, 50)) | the average marks of a class of 30 students is 40 and that of another class of 50 students is 60 . find the average marks of all the students ? | "sum of the marks for the class of 30 students = 30 * 40 = 1200 sum of the marks for the class of 50 students = 50 * 60 = 3000 sum of the marks for the class of 80 students = 1200 + 3000 = 4200 average marks of all the students = 4200 / 80 = 52.5 answer : d" | a = 30 * 40
b = 50 * 60
c = a + b
d = 30 + 50
e = c / d
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a ) 75 , b ) 45 , c ) 55 , d ) 65 , e ) 35 | d | subtract(divide(multiply(divide(multiply(75, 8), 50), 70), 6), 75) | 75 men working 8 hours per day dig 50 m deep . how many extra men should be put to dig to a depth of 70 m working 6 hours per day ? | ( 75 * 8 ) / 50 = ( x * 6 ) / 70 = > x = 140 140 â € “ 65 = 65 answer : d | a = 75 * 8
b = a / 50
c = b * 70
d = c / 6
e = d - 75
|
a ) 15 , b ) 16 , c ) 18 , d ) 20 , e ) 24 | a | divide(log(divide(multiply(const_3, const_10), add(const_4, const_1))), log(power(divide(multiply(const_2, const_10), add(const_4, const_1)), divide(const_1, 12)))) | on a certain date , pat invested $ 5,000 at x percent annual interest , compounded annually . if the total value of the investment plus interest at the end of 12 years will be $ 20,000 , in how many years , the total value of the investment plus interest will increase to $ 30,000 ? | if i were to choose during the test , would go for 18 or 20 . probably 18 cuz it wont take too long to get the value doubled . . . . i found a method : rule of 72 . given an x % return , it takes 5,000 to quadralope 12 years . so according to the rule : 72 / x is the no of years 5 , 000.00 took to double 10 , 000.00 . again , 10 , 000.00 took to double 20 , 000.00 same ( 72 / x ) no of years . 72 / x + 72 / x = 12 x = 12 % ( though rate here is not very much required ) . again , 20 , 000.00 takes the same ( 72 / x ) no of years to double 30 , 000.00 . 72 / x = 6 years . so altogather : 5,000 - 10,000 = 6 years 10,000 - 20,000 = 6 years 20,000 - 30,000 = 3 years answer a 15 years . | a = 3 * 10
b = 4 + 1
c = a / b
d = math.log(c)
e = 2 * 10
f = 4 + 1
g = e / f
h = 1 / 12
i = g ** h
j = math.log(i)
k = d / j
|
a ) $ 270 , b ) $ 250 , c ) $ 240 , d ) $ 220 , e ) $ 200 | a | divide(multiply(100, divide(multiply(add(100, 80), divide(add(100, 20), const_2)), 100)), subtract(divide(add(100, 20), const_2), 20)) | a clock store sold a certain clock to a collector for 20 percent more than the store had originally paid for the clock . when the collector tried to resell the clock to the store , the store bought it back at 50 percent of what the collector had paid . the shop then sold the clock again at a profit of 80 percent on its buy - back price . if the difference between the clock ' s original cost to the shop and the clock ' s buy - back price was $ 100 , for how much did the shop sell the clock the second time ? | original price = x sold to collector at 1.2 x re - buy price = 1.2 x * 0.5 = 0.6 x therefore , x - 0.6 x = 100 ; x = 250 second selling price = 1.8 * 0.6 * 250 = 270 answer : a | a = 100 + 80
b = 100 + 20
c = b / 2
d = a * c
e = d / 100
f = 100 * e
g = 100 + 20
h = g / 2
i = h - 20
j = f / i
|
a ) 16 , b ) 10 , c ) 12 , d ) 19 , e ) 17.8 | c | add(3, const_1) | the average of first six prime numbers greater than 3 is ? | "5 + 7 + 11 + 13 + 17 + 19 = 72 / 6 = 12 answer : c" | a = 3 + 1
|
a ) 50 , b ) 100 , c ) 140 , d ) 1,000 , e ) 2,000 | c | multiply(multiply(divide(50, 2), divide(50, 2)), const_1000) | positive integer y is 50 percent of 50 percent of positive integer x , and 2 y percent of x equals 100 . what is the value of x ? | "y = 50 % of 50 % 0 f x = x / 4 and 2 y / 100 of x = 100 2 y / 100 * 4 y = 5000 y = 35 and x = 140 answer - c" | a = 50 / 2
b = 50 / 2
c = a * b
d = c * 1000
|
a ) 22 , b ) 60 , c ) 28 , d ) 11 , e ) 218 | b | subtract(divide(multiply(2, multiply(multiply(100, 25), 8)), multiply(25, 10)), 100) | nhai employs 100 men to build a highway of 2 km in 50 days working 8 hours a day . if in 25 days they completed 1 / 3 part of work . than how many more employees should nhai hire to finish it in time working 10 hours a day ? | "here 2 km is immaterial . the given problem can be written in a tabular form like below . we can apply chain rule now . total men required to complete the remaining work = 100 × 2525 × 810 × 2313100 × 2525 × 810 × 2313 = 160 so additional men required = 160 - 100 = 60 answer : b" | a = 100 * 25
b = a * 8
c = 2 * b
d = 25 * 10
e = c / d
f = e - 100
|
a ) 1500 , b ) 2777 , c ) 2889 , d ) 27670 , e ) 2771 | a | divide(1680, add(divide(multiply(divide(add(multiply(2, 5), 2), 5), 5), const_100), const_1)) | find the principle on a certain sum of money at 5 % per annum for 2 2 / 5 years if the amount being rs . 1680 ? | "1680 = p [ 1 + ( 5 * 12 / 5 ) / 100 ] p = 1500 . answer : a" | a = 2 * 5
b = a + 2
c = b / 5
d = c * 5
e = d / 100
f = e + 1
g = 1680 / f
|
a ) 5 : 6 , b ) 7 : 8 , c ) 8 : 9 , d ) 9 : 10 , e ) 11 : 12 | c | divide(add(power(2, 3), 3), add(add(1, 2), 3)) | if x : y = 1 : 2 , find the value of ( 2 x + 3 y ) : ( x + 4 y ) | "x : y = 1 : 2 means x / y = 1 / 2 now , ( 2 x + 3 y ) : ( x + 4 y ) = ( 2 x + 3 y ) / ( x + 4 y ) [ divide numerator and denominator by y . ] = [ ( 2 x + 3 y ) / y ] / [ ( x + 4 y ) / 2 ] = [ 2 ( x / y ) + 3 ] / [ ( x / y ) + 4 ] , put x / y = 1 / 2 we get = [ 2 ( 1 / 2 ) + 3 ) / ( 1 / 2 + 4 ) = ( 1 + 3 ) / [ ( 1 + 8 ) / 2 ] = 4 / ( 9 / 2 ) = 4 / 1 × 2 / 9 = 8 / 9 therefore the value of ( 2 x + 3 y ) : ( x + 4 y ) = 8 : 9 answer is c" | a = 2 ** 3
b = a + 3
c = 1 + 2
d = c + 3
e = b / d
|
a ) 22 , b ) 50 , c ) 88 , d ) 65 , e ) 12 | d | multiply(13, 5) | what number has a 5 : 1 ratio to the number 13 ? | "5 : 1 = x : 13 x = 65 answer : d" | a = 13 * 5
|
a ) 15 sec , b ) 16 sec , c ) 28 sec , d ) 20 sec , e ) 25 sec | c | multiply(divide(490, multiply(63, const_1000)), const_3600) | a train 490 m long , running with a speed of 63 km / hr will pass a tree in ? | "speed = 63 * 5 / 18 = 35 / 2 m / sec time taken = 490 * 2 / 35 = 28 sec answer : c" | a = 63 * 1000
b = 490 / a
c = b * 3600
|
a ) 1 m , b ) 5 m , c ) 4 m , d ) 2 m , e ) 3 m | d | divide(multiply(5, 10), subtract(rectangle_area(20, 5), rectangle_area(4, 10))) | the dimensions of a field are 20 m by 5 m . a pit 10 m long , 4 m wide and 3 m deep is dug in one corner of the field and the earth removed has been evenly spread over the remaining area of the field . what will be the rise in the height of field as a result of this operation ? | "the volume of the earth removed is 10 * 4 * 3 = 120 m ^ 3 . the remaining area of the field is 20 * 5 - 10 * 4 = 60 m ^ 2 . 120 m ^ 3 of the earth evenly spread over the area of 60 m ^ 2 will rise the height by ( height ) = ( volume ) / ( area ) = 120 / 60 = 2 m . answer : d" | a = 5 * 10
b = rectangle_area - (
c = a / b
|
a ) 30 , b ) 40 , c ) 50 , d ) 60 , e ) 70 | a | inverse(divide(subtract(subtract(const_1, divide(3, 5)), multiply(4, divide(const_1, multiply(5, const_3)))), const_4)) | george does 3 / 5 th of a piece of work in 9 days . he then calls in paul , and they finish the work in 4 days . how long would paul take to do the work by himself ? | 100 % work will be done by george in 15 days ( 1 / 15 ) 9 + ( 1 / 15 + 1 / x ) 4 = 1 solving this eqn . we get x = 30 hence paul take to do the work in 30 days answer : a | a = 3 / 5
b = 1 - a
c = 5 * 3
d = 1 / c
e = 4 * d
f = b - e
g = f / 4
h = 1/(g)
|
a ) 2 , b ) 3 , c ) 4 , d ) 5 , e ) 6 | b | divide(15, 5) | a boat sails 15 km of a river towards upstream in 5 hours . how long will it take to cover the same distance downstream , if the speed of current is one - fourth the speed of the boat in still water : | "explanation : upstream speed = b - s downstream speed = b + s b - s = 15 / 5 = 3 km / h again b = 4 s therefore b - s = 3 = 3 s = > s = 1 and b = 4 km / h therefore b + s = 5 km / h therefore , time during downstream = 15 / 5 = 3 h answer : b ) 3 h" | a = 15 / 5
|
a ) 3 : 4 , b ) 4 : 3 , c ) 17 : 16 , d ) 16 : 17 , e ) none of these | b | divide(divide(20, const_100), divide(15, const_100)) | if 15 % of a is the same as 20 % of b , then a : b is : | "expl : 15 % of a i = 20 % of b = 15 a / 100 = 20 b / 100 = 4 / 3 = 4 : 3 answer : b" | a = 20 / 100
b = 15 / 100
c = a / b
|
a ) 10 , b ) 8 , c ) 9 , d ) 7.2 , e ) 6 | d | divide(add(7.5, subtract(7.5, 1.5)), const_2) | a man whose speed is 7.5 kmph in still water rows to a certain upstream point and back to the starting point in a river which flows at 1.5 kmph , find his average speed for the total journey ? | "m = 7.5 s = 1.5 ds = 9 us = 6 as = ( 2 * 9 * 6 ) / 15 = 7.2 answer : d" | a = 7 - 5
b = 7 + 5
c = b / 2
|
a ) 2372 , b ) 2572 , c ) 4542 , d ) 2343 , e ) 5296 | e | subtract(multiply(add(multiply(const_100, const_100), multiply(multiply(const_100, divide(60, 6)), 3)), multiply(multiply(add(const_1, divide(divide(60, 6), const_100)), add(const_1, divide(divide(60, 6), const_100))), add(const_1, divide(divide(60, 6), const_100)))), add(multiply(const_100, const_100), multiply(multiply(const_100, divide(60, 6)), 3))) | there is 60 % increase in an amount in 6 years at si . what will be the ci of rs . 16,000 after 3 years at the same rate ? | "let p = rs . 100 . then , s . i . rs . 60 and t = 6 years . r = 100 x 60 = 10 % p . a . 100 x 6 now , p = rs . 16000 . t = 3 years and r = 10 % p . a . c . i . = rs . 16000 x 1 + 10 3 - 1 100 = rs . 16000 x 331 1000 = 5296 . e" | a = 100 * 100
b = 60 / 6
c = 100 * b
d = c * 3
e = a + d
f = 60 / 6
g = f / 100
h = 1 + g
i = 60 / 6
j = i / 100
k = 1 + j
l = h * k
m = 60 / 6
n = m / 100
o = 1 + n
p = l * o
q = e * p
r = 100 * 100
s = 60 / 6
t = 100 * s
u = t * 3
v = r + u
w = q - v
|
a ) 18 , b ) 15 , c ) 10 , d ) 11 , e ) 12 | b | divide(255, subtract(18, const_1)) | along a yard 255 metres long , 18 trees are palnted at equal distances , one tree being at each end of the yard . what is the distance between two consecutive trees | "explanation : 18 trees have 17 gaps between them , required distance ( 281 / 17 ) = 15 option b" | a = 18 - 1
b = 255 / a
|
a ) 9 % , b ) 11 % , c ) 13 % , d ) 21 % , e ) 19 % | a | multiply(divide(subtract(const_100, 92), 92), const_100) | if the cost price is 92 % of the selling price , then what is the profit percent ? | "let s . p . = $ 100 c . p . = $ 92 profit = $ 8 profit % = 8 / 92 * 100 = 25 / 6 = 9 % approximately answer is a" | a = 100 - 92
b = a / 92
c = b * 100
|
a ) 50 , b ) 99 , c ) 64 , d ) 76 , e ) 12 | c | divide(add(280, 360), multiply(subtract(72, 36), const_0_2778)) | how much time will a train of length 280 m moving at a speed of 72 kmph take to cross another train of length 360 m , moving at 36 kmph in the same direction ? | "the distance to be covered = sum of their lengths = 280 + 360 = 640 m . relative speed = 72 - 36 = 36 kmph = 36 * 5 / 18 = 10 mps . time required = d / s = 640 / 10 = 64 sec . answer : c" | a = 280 + 360
b = 72 - 36
c = b * const_0_2778
d = a / c
|
a ) 4 : 9 , b ) 9 : 4 , c ) 4 : 5 , d ) 4 : 1 , e ) 4 : 2 | b | divide(sqrt(81), sqrt(16)) | two trains , one from howrah to patna and the other from patna to howrah , start simultaneously . after they meet , the trains reach their destinations after 16 hours and 81 hours respectively . the ratio of their speeds is ? | "let us name the trains a and b . then , ( a ' s speed ) : ( b ' s speed ) = √ b : √ a = √ 81 : √ 16 = 9 : 4 answer : b" | a = math.sqrt(81)
b = math.sqrt(16)
c = a / b
|
a ) 27 , b ) 36 , c ) 45 , d ) 54 , e ) 63 | c | multiply(divide(7, subtract(9, 7)), 10) | sandy is younger than molly by 10 years . if their ages are in the respective ratio of 7 : 9 , how old is molly ? | "s = m - 10 s / m = 7 / 9 9 s = 7 m 9 ( m - 10 ) = 7 m m = 45 the answer is c ." | a = 9 - 7
b = 7 / a
c = b * 10
|
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