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 ) 20 , b ) 21 , c ) 22 , d ) 23 , e ) 24 | c | add(floor(divide(99, add(const_1, const_4))), floor(divide(99, power(add(const_1, const_4), const_2)))) | 99 ! how many zero ' s ? | 99 ! = 99 / 5 = 19 = 19 / 5 = 3 so 99 ! has ( 19 + 3 = 22 zeroes ) answer : c | a = 1 + 4
b = 99 / a
c = math.floor(b)
d = 1 + 4
e = d ** 2
f = 99 / e
g = math.floor(f)
h = c + g
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a ) 221 , b ) 190 , c ) 192 , d ) 191 , e ) 121 | c | subtract(divide(multiply(const_1, const_1000), divide(40, 8)), 8) | in a kilometer race , a beats b by 40 meters or 8 seconds . what time does a take to complete the race ? | "time taken by b run 1000 meters = ( 1000 * 8 ) / 40 = 200 sec . time taken by a = 200 - 8 = 192 sec . answer : c" | a = 1 * 1000
b = 40 / 8
c = a / b
d = c - 8
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a ) a . 40 , b ) b . 100 , c ) c . 400 , d ) d . 1,000 , e ) e . 2,000 | e | divide(const_180, const_1000) | the volume of a sphere with radius r is ( 4 / 3 ) * pi * r ^ 3 and the surface area is 4 * pi * r ^ 3 . if a sperical balloon has a volume of 4500 pi cubic centimeters , what is hte surface area of the balloon in square centimeters ? | "the surface area is 4 . pi . r ^ 2 ( its area remember not volume ) as 4 / 3 . pi . r ^ 3 = 4500 pi r = 15 so area = 4 . pi . r ^ 2 = 900 . pi = 900 x 3.14 = 2000 ( approx ) e" | a = const_180 / 1000
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a ) 2.5 sec , b ) 9.7 sec , c ) 3.5 sec , d ) 2.9 sec , e ) 7.5 sec | e | divide(300, multiply(144, const_0_2778)) | in what time will a train 300 m long cross an electric pole , it its speed be 144 km / hr ? | "speed = 144 * 5 / 18 = 40 m / sec time taken = 300 / 40 = 7.5 sec . answer : e" | a = 144 * const_0_2778
b = 300 / a
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a ) 3776 , b ) 810.27 , c ) 815.21 , d ) 688.12 , e ) 267.1 | b | multiply(1380, divide(3200, add(2250, 3200))) | anand and deepak started a business investing rs . 2250 and rs . 3200 respectively . out of a total profit of rs . 1380 , deepak ' s share is : | ratio of their shares = 2250 : 3200 = 45 : 64 deepak ' s share = 13800 * 64 / 109 = rs . 810.27 . answer : b | a = 2250 + 3200
b = 3200 / a
c = 1380 * b
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a ) 77.14 cm , b ) 25.14 cm , c ) 84.92 cm , d ) 94.94 cm , e ) 23.57 cm | b | divide(circumface(divide(square_edge_by_perimeter(rectangle_perimeter(18, 14)), const_2)), const_2) | the parameter of a square is equal to the perimeter of a rectangle of length 18 cm and breadth 14 cm . find the circumference of a semicircle whose diameter is equal to the side of the square . ( round off your answer to two decimal places ) | "let the side of the square be a cm . parameter of the rectangle = 2 ( 18 + 14 ) = 64 cm parameter of the square = 64 cm i . e . 4 a = 64 a = 16 diameter of the semicircle = 16 cm circimference of the semicircle = 1 / 2 ( ∏ ) ( 16 ) = 1 / 2 ( 22 / 7 ) ( 16 ) = 352 / 14 = 25.14 cm to two decimal places answer : b" | a = square_edge_by_perimeter / (
b = circumface / (
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a ) 6 / 7 , b ) 5 / 7 , c ) 4 / 7 , d ) 3 / 7 , e ) 2 / 7 | b | divide(const_2, add(const_3, const_4)) | find the probability that a leap year selected at random will not have 53 sundays | "there are 366 days in a leap year : 52 weeks and 2 more days . so , 52 sundays and 2 days . these 2 days can be : { mon , tue } , { tue , wed } , { wed , thu } , { thu , fri } , { fri , sat } , { sat , sun } and { sun , mon } ( 7 cases ) . in order to have 53 sundays we should have either { sat , sun } or { sun , mon } case . the probability of that is 2 / 7 . so , probability of not having 53 sundays is 1 - 2 / 7 = 5 / 7 answer : b ." | a = 3 + 4
b = 2 / a
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a ) 12 , b ) 9 , c ) 8 , d ) 6 , e ) 5 | d | divide(const_1, subtract(divide(const_1, 3), subtract(divide(const_1, 2), divide(const_1, 3)))) | a can do a piece of work in 3 hours ; b and c together can do it in 3 hours , while a and c together can do it in 2 hours . how long will b alone take to do it ? | "a ' s 1 hour ' s work = 1 / 3 ; ( b + c ) ' s 1 hour ' s work = 1 / 3 ; ( a + c ) ' s 1 hour ' s work = 1 / 2 . ( a + b + c ) ' s 1 hour ' s work = ( 1 / 3 + 1 / 3 ) = 2 / 3 . b ' s 1 hour ' s work = ( 2 / 3 - 1 / 2 ) = 1 / 6 . therefore a alone will take 6 hours to do the work . d" | a = 1 / 3
b = 1 / 2
c = 1 / 3
d = b - c
e = a - d
f = 1 / e
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a ) 100 seconds , b ) 112.5 seconds , c ) 77.5 seconds , d ) 87.5 seconds , e ) none | c | divide(subtract(200, 45), const_2) | in a 200 m race , if a gives b a start of 25 metres , then a wins the race by 10 seconds . alternatively , if a gives b a start of 45 metres the race ends in a dead heat . how long does a take to run 200 m ? | explanatory answer a gives b a start of 25 metres and still wins the race by 10 seconds . alternatively , if a gives b a start of 45 metres , then the race ends in a dead heat . therefore , the additional 20 metres start given to b compensates for the 10 seconds . i . e . , b runs 20 metres in 10 seconds . hence , b will take 100 seconds to run 200 metres . we know that a gives b a start of 45 metres . b will take 22.5 seconds to run this 45 metres as b runs 20 metres in 10 seconds or at the speed of 2 m / s . hence , a will take 22.5 seconds lesser than b or 100 - 22.5 = 77.5 seconds to complete the race . answer c | a = 200 - 45
b = a / 2
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a ) 280 , b ) 490 , c ) 720 , d ) 435 , e ) 679 | a | multiply(divide(add(40, multiply(2, 85)), subtract(340, 85)), 340) | if a farmer wants to plough a farm field on time , he must plough 340 hectares a day . for technical reasons he ploughed only 85 hectares a day , hence he had to plough 2 more days than he planned and he still has 40 hectares left . what is the area of the farm field and how many days the farmer planned to work initially ? | "let x be the number of days in the initial plan . therefore , the whole field is 340 â ‹ … x hectares . the farmer had to work for x + 2 days , and he ploughed 85 ( x + 2 ) hectares , leaving 40 hectares unploughed . then we have the equation : 340 x = 85 ( x + 2 ) + 40 255 x = 210 x = 0.823 so the farmer planned to have the work done in 6 days , and the area of the farm field is 340 ( 0.823 ) = 280 hectares correct answer a" | a = 2 * 85
b = 40 + a
c = 340 - 85
d = b / c
e = d * 340
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a ) 12 , b ) 14 , c ) 16 , d ) 18 , e ) 20 | c | divide(multiply(add(divide(60, const_100), const_1), 26), add(add(divide(60, const_100), const_1), const_1)) | p is 60 % more efficient than q . p can complete a work in 26 days . if p and q work together , how many days will it take to complete the same work ? | the work done by p in 1 day = 1 / 26 let work done by q in 1 day = q q × ( 160 / 100 ) = 1 / 26 q = 100 / ( 26 × 160 ) = 10 / ( 26 × 16 ) the work done by p and q in 1 day = 1 / 26 + 10 / ( 26 × 16 ) = 26 / ( 26 × 16 ) = 1 / 16 p and q together can do the work in 16 days . the answer is c . | a = 60 / 100
b = a + 1
c = b * 26
d = 60 / 100
e = d + 1
f = e + 1
g = c / f
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a ) 8.70 % , b ) 8.60 % , c ) 8.10 % , d ) 8.30 % , e ) 8.27 % | a | multiply(divide(subtract(const_100, 92), 92), const_100) | if the cost price is 92 % of sp then what is the profit % | "sol . sp = rs 100 : then cp = rs 92 : profit = rs 8 . profit = { ( 8 / 92 ) * 100 } % = 8.70 % answer is a ." | a = 100 - 92
b = a / 92
c = b * 100
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a ) s . 500 , b ) s . 1000 , c ) s . 1500 , d ) s . 2000 , e ) s . 2500 | a | multiply(divide(divide(350, 7), 2), const_100) | a sum was put a simple interest at a certain rate for 7 years . had it been put at 2 % higher rate , it would have fetched rs . 350 more . the sum is : | "explanation : let the sub be rs . x and the initial rate be r % . then x ã — ( r + 2 ) ã — 7 / 100 â ˆ ’ x ã — r ã — 7 / 100 = 350 â ‡ ’ x ã — 2 ã — 7 / 100 = 350 â ‡ ’ x ã — 7 / 100 = 175 â ‡ ’ 7 x = 17500 â ‡ ’ x = 2500 answer : option a" | a = 350 / 7
b = a / 2
c = b * 100
|
a ) 106.75 , b ) 213.5 , c ) 427 , d ) 729 , e ) 1,156 | c | subtract(divide(power(34, const_2), const_2), divide(power(27, const_2), const_2)) | the size of a flat - screen television is given as the length of the screen ’ s diagonal . how many square inches greater is the screen of a square 34 - inch flat - screen television than a square 27 - inch flat - screen television ? | "diagonal of a square = √ 2 * side of square = > side = diagonal / ( √ 2 ) area of square = s * s = ( diagonal ^ 2 ) / 2 difference of both tv ' s = ( 34 ^ 2 - 27 ^ 2 ) / 2 = 427 answer : c" | a = 34 ** 2
b = a / 2
c = 27 ** 2
d = c / 2
e = b - d
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a ) 1.05 , b ) 1.25 , c ) 1.45 , d ) 1.55 , e ) 1.65 | b | add(divide(1.5625, 1.5625), const_0_25) | â ˆ š 1.5625 = ? | explanation : 1 | 1.5625 ( 1.25 | 1 | - - - - - - - 22 | 56 | 44 | - - - - - - - 245 | 1225 | 1225 | - - - - - - - | x | - - - - - - - 1.5625 = 1.25 . answer is b | a = 1 / 5625
b = a + const_0_25
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a ) 20 % , b ) 18 % , c ) 4 % , d ) 12 % , e ) 14 % | c | subtract(const_100, divide(multiply(add(const_100, 20), subtract(const_100, 20)), const_100)) | the tax on a commodity is diminished by 20 % but its consumption is increased by 20 % . find the decrease percent in the revenue derived from it ? | "100 * 100 = 10000 80 * 120 = 9600 10000 - - - - - - - 400 100 - - - - - - - ? = 4 % answer : c" | a = 100 + 20
b = 100 - 20
c = a * b
d = c / 100
e = 100 - d
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a ) 24 , b ) 25 , c ) 26 , d ) 27 , e ) 29 | b | divide(add(multiply(multiply(divide(3, 2), 2), 7), 4), subtract(multiply(divide(3, 2), 2), 2)) | find ( 7 x + 4 y ) / ( x - 2 y ) if x / 2 y = 3 / 2 ? | "x / 2 y = 3 / 2 = > x = 6 y / 2 = 3 y = > ( 7 x + 4 y ) / ( x - 2 y ) = ( ( 7 * ( 3 y ) ) + 4 y ) / ( 3 y - 2 y ) = > 25 y / y = 25 answer : b" | a = 3 / 2
b = a * 2
c = b * 7
d = c + 4
e = 3 / 2
f = e * 2
g = f - 2
h = d / g
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a ) 11.73 , b ) 12 , c ) 13.8 , d ) 14 , e ) 15 | e | divide(divide(multiply(207, subtract(const_100, 15)), const_100), 12) | the price of lunch for 12 people was $ 207 including a 15 % gratuity for service . what was the average price per person , excluding the gratuity ? | "clearly e is the answer i used poe here lets consider option ( e ) 12 * 15 = 180 now 180 ( 115 / 100 ) = 207 = > possible answer imo e" | a = 100 - 15
b = 207 * a
c = b / 100
d = c / 12
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a ) rs . 960 , b ) rs . 760 , c ) rs . 840 , d ) rs . 1020 , e ) rs . 1615 | d | add(270, divide(multiply(270, const_100), multiply(12, 3))) | the banker ' s gain on a sum due 3 years hence at 12 % per annum is rs . 270 . the banker ' s discount is : | t . d = ( b . gx 100 ) / ( r / t ) = ( 270 x 100 ) / ( 12 x 3 ) = rs . 750 b . d . = rs . ( 750 + 270 ) = rs . 1020 . answer : d | a = 270 * 100
b = 12 * 3
c = a / b
d = 270 + c
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a ) s . 14028 , b ) s . 14000 , c ) s . 14003 , d ) s . 14029 , e ) s . 24029 | b | subtract(26000, multiply(const_60, const_100)) | a started a business with an investment of rs . 70000 and after 6 months b joined him investing rs . 120000 . if the profit at the end of a year is rs . 26000 , then the share of a is ? | "ratio of investments of a and b is ( 70000 * 12 ) : ( 120000 * 6 ) = 7 : 6 total profit = rs . 26000 share of b = 7 / 13 ( 26000 ) = rs . 14000 answer : b" | a = const_60 * 100
b = 26000 - a
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a ) 30 % , b ) 58 % , c ) 25 % , d ) 60 % , e ) 12 % | c | multiply(divide(30, subtract(150, 30)), const_100) | by selling 150 apples , a fruit - seller gains the selling price of 30 apples . find the gain percent ? | "option c explanation : sp = cp + g 150 sp = 150 cp + 30 sp 120 sp = 150 cp 120 - - - 30 cp 100 - - - ? = > 25 %" | a = 150 - 30
b = 30 / a
c = b * 100
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a ) 10 , b ) 12 , c ) 14 , d ) 16 , e ) 18 | c | divide(70, add(const_4, divide(const_2, const_2))) | a certain number of horses and an equal number of men are going somewhere . half of the owners are on their horses ' back while the remaining ones are walking along leading their horses . if the number of legs walking on the ground is 70 , how many horses are there ? | "explanation : let number of horses = number of men = x . then , number of legs = 4 x + 2 x ( x / 2 ) = 5 x . so , 5 x = 70 or x = 14 . answer : c" | a = 2 / 2
b = 4 + a
c = 70 / b
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a ) 720578 , b ) 80578 , c ) 80698 , d ) 81268 , e ) none of them | a | multiply(617, power(617, 583)) | 617 x 617 + 583 x 583 = ? | "= ( 617 ) ^ 2 + ( 583 ) ^ 2 = ( 600 + 17 ) ^ 2 + ( 600 - 17 ) ^ 2 = 2 [ ( 600 ) ^ 2 + ( 17 ) ^ 2 ] = 2 [ 360000 + 289 ] = 2 x 360289 = 720578 answer is a" | a = 617 ** 583
b = 617 * a
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a ) 64 , b ) 32 , c ) 8 , d ) 6 , e ) 2 | d | sqrt(power(power(15, divide(const_1, const_2)), const_3)) | what positive number , when squared , is equal to the cube of the positive square root of 15 ? | "let the positive number be x x ^ 2 = ( ( 15 ) ^ ( 1 / 2 ) ) ^ 3 = > x ^ 2 = 4 ^ 3 = 36 = > x = 6 answer d" | a = 1 / 2
b = 15 ** a
c = b ** 3
d = math.sqrt(c)
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a ) 0 , b ) 1 / 12 , c ) 5 / 12 , d ) 7 / 18 , e ) 4 / 9 | c | multiply(add(const_12, const_3), power(divide(1, 6), const_2)) | a cube with its sides numbered 1 through 6 is rolled twice , first landing on a and then landing on b . if any roll of the cube yields an equal chance of landing on any of the numbers 1 through 6 , what is the probability t that a + b is prime ? | "total # of outcomes is 6 * 6 = 36 ; favorable outcomes : a - b - - > prime 1 - 1 - - > 2 ; 1 - 2 - - > 3 ; 2 - 1 - - > 3 ; 1 - 4 - - > 5 ; 4 - 1 - - > 5 ; 2 - 3 - - > 5 ; 3 - 2 - - > 5 ; 1 - 6 - - > 7 ; 6 - 1 - - > 7 ; 2 - 5 - - > 7 ; 5 - 2 - - > 7 ; 3 - 4 - - > 7 ; 4 - 3 - - > 7 ; 6 - 5 - - > 11 ; 5 - 6 - - > 11 . total of 15 favorable outcomes t = 15 / 36 . answer : c ." | a = 12 + 3
b = 1 / 6
c = b ** 2
d = a * c
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a ) rs . 500 , b ) rs . 840 , c ) rs . 650 , d ) rs . 720 , e ) none | b | divide(multiply(multiply(divide(14400, add(100, 20)), 100), 7), 100) | a man invested rs . 14400 in rs . 100 shares of a company at 20 % premium . if the company declares 7 % dividend at the end of the year , then how much does he get ? | solution number of shares = ( 14400 / 120 ) = 120 . face value = rs . ( 100 x 120 ) = rs . 12000 . annual income = rs . ( 7 / 100 x 12000 ) = rs . 840 . answer b | a = 100 + 20
b = 14400 / a
c = b * 100
d = c * 7
e = d / 100
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a ) 3.6 secs , b ) 4.00 secs , c ) 40 secs , d ) 45 secs , e ) none of these | b | add(divide(1200, 330), divide(multiply(multiply(const_3, const_10), divide(1200, 330)), 330)) | a boy standing idle sounds a whistle to his friend at a distance of 1200 m moving away from him in a speeding car at 108 kms / hr . find the duration after which his friend is going to hear him . ( speed of sound = 330 m / sec ) . | if friend hears sound after t secs , then 330 * t = 1200 + ( 108 * 5 / 18 ) * t 300 t = 1200 t = 4 secs answer : b | a = 1200 / 330
b = 3 * 10
c = 1200 / 330
d = b * c
e = d / 330
f = a + e
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a ) 5 , b ) 6 , c ) 7 , d ) 8 , e ) 12 | b | divide(6, 1) | if ' a ' and ' b ' are non - zero numbers such that their sum is 6 times the product , what is the value of 1 / a + 1 / b ? | a + b = 6 ab 1 / a + 1 / b = ( b + a ) / ( ab ) = 6 . the answer is b . | a = 6 / 1
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a ) 8 , b ) 16 , c ) 32 , d ) 48 , e ) 54 | d | multiply(divide(multiply(2, 12), subtract(4, 2)), 4) | two integers are in the ratio of 1 to 4 . if 12 is added to the smaller number , the ratio becomes 1 to 2 . find the larger integer . | one option is to set up the equations and solve : if the ratio of two integers x and y is 1 to 4 , then 4 x = y , where x is the smaller integer . if adding 12 to the smaller integer makes the ratio 1 to 2 , then 2 ( x + 12 ) = y . substituting y = 4 x into the second equation yields 2 x + 24 = 4 x . so , x = 12 ( smaller integer ) and , y = 4 x = 48 ( larger integer ) so d is the correct answer . another option is to test the answer choices . a ) the larger number is 8 given : the two integers are in the ratio of 1 to 4 so , the smaller number must be 2 if 12 is added to the smaller number , the ratio becomes 1 to 2 if we add 12 to the smaller number ( 2 ) , we get 14 . so , the new ratio is 14 to 8 which is the same as 7 to 4 . no good . we want a resulting ratio of 1 to 2 eliminate a b ) the larger number is 16 given : the two integers are in the ratio of 1 to 4 so , the smaller number must be 4 if 12 is added to the smaller number , the ratio becomes 1 to 2 if we add 12 to the smaller number ( 4 ) , we get 16 so , the new ratio is 16 to 16 which is 1 to 1 . no good . we want a resulting ratio of 1 to 2 eliminate b c ) the larger number is 32 given : the two integers are in the ratio of 1 to 4 so , the smaller number must be 8 if 12 is added to the smaller number , the ratio becomes 1 to 2 if we add 12 to the smaller number ( 8 ) , we get 20 so , the new ratio is 20 to 32 which is 5 to 8 . no good . we want a resulting ratio of 1 to 2 eliminate c d ) the larger number is 48 given : the two integers are in the ratio of 1 to 4 so , the smaller number must be 12 if 12 is added to the smaller number , the ratio becomes 1 to 2 if we add 12 to the smaller number ( 12 ) , we get 24 so , the new ratio is 24 to 48 . this is the same as the ratio of 1 to 2 therefore ( d ) is correct e ) the larger number is 54 given : the two integers are in the ratio of 1 to 4 so , the smaller number must be 13.5 ( which is not an integer ) since both numbers are integers , the larger number can not be 54 . eliminate e therefore d is the only correct choice . | a = 2 * 12
b = 4 - 2
c = a / b
d = c * 4
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a ) 12100 , b ) 15240 , c ) 12456 , d ) 11452 , e ) 13310 | e | multiply(multiply(divide(add(10, const_100), const_100), 11000), divide(add(10, const_100), const_100)) | if the annual increase in the population of a town is 10 % and the present number of people is 11000 , what will the population be in 2 years ? | "the required population is = 11000 ( 1 + 10 / 100 ) ^ 2 = 11000 * 11 / 10 * 11 / 10 = 13310 answer is e" | a = 10 + 100
b = a / 100
c = b * 11000
d = 10 + 100
e = d / 100
f = c * e
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a ) 20 , b ) 34 , c ) 30 , d ) 40 , e ) 46 | c | divide(subtract(multiply(const_3, 60), 60), const_3) | shannon and maxine work in the same building and leave work at the same time . shannon lives due north of work and maxine lives due south . the distance between maxine ' s house and shannon ' s house is 60 miles . if they both drive home at the rate 2 r miles per hour , maxine arrives home 30 minutes after shannon . if maxine rider her bike home at the rate of r per hour and shannon still drives at a rate of 2 r miles per hour , shannon arrives home 2 hours before maxine . how far does maxine live from work ? | "nice question + 1 we have that x / 24 - ( 60 - x ) / 2 r = 30 also x / r - ( 60 - x ) / 2 r = 120 so we get that 2 x - 60 = 80 r 3 x - 60 = 240 r get rid of r 120 = 3 x x = 30 hence answer is c" | a = 3 * 60
b = a - 60
c = b / 3
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a ) rs . 5565 , b ) rs . 6298 , c ) rs . 6290 , d ) rs . 6725 , e ) rs . 6708 | a | divide(6400, add(const_1, divide(15, const_100))) | the owner of a furniture shop charges his customer 15 % more than the cost price . if a customer paid rs . 6400 for a computer table , then what was the cost price of the computer table ? | "cp = sp * ( 100 / ( 100 + profit % ) ) = 6400 ( 100 / 115 ) = rs . 5565 . answer : a" | a = 15 / 100
b = 1 + a
c = 6400 / b
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a ) 190 , b ) 200 , c ) 210 , d ) 220 , e ) 230 | c | multiply(140, subtract(const_2, const_1)) | a train speeds past a pole in 15 seconds and a platform 140 meters long in 25 seconds . what is the length of the train ( in meters ) ? | "let the length of the train be x meters . the speed of the train is x / 15 . then , x + 140 = 25 * ( x / 15 ) 10 x = 2100 x = 210 meters the answer is c ." | a = 2 - 1
b = 140 * a
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a ) 4 : 30 , b ) 4 : 45 , c ) 5 : 00 , d ) 5 : 15 , e ) 5 : 30 | e | subtract(divide(multiply(3, 15), 15), const_0_33) | it takes 3 workers a total of 15 hours to build a giant machine with each worker working at the same rate . if 5 workers start to build the machine at 11 : 00 am , and one extra worker per hour is added beginning at 1 : 00 pm , at what time will the machine be complete ? | "3 workers build 1 / 15 of a machine in one hour . 1 worker builds 1 / 45 of a machine in one hour . in the first 2 hours , 5 workers build 5 * ( 1 / 45 ) * 2 = 10 / 45 of a machine . from 1 : 00 to 2 : 00 , 6 workers build another 6 / 45 . the total is 16 / 45 . from 2 : 00 to 3 : 00 , 7 workers build another 7 / 45 . the total is 23 / 45 . from 3 : 00 to 4 : 00 , 8 workers build another 8 / 45 . the total is 31 / 45 . from 4 : 00 to 5 : 00 , 9 workers build another 9 / 45 . the total is 40 / 45 . to build another 5 / 45 , 10 workers need ( 5 / 10 ) * 60 minutes which is 30 minutes . the machine is complete at 5 : 30 . the answer is e ." | a = 3 * 15
b = a / 15
c = b - const_0_33
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a ) - 1.5 , b ) - 0.5 , c ) 0.5 , d ) 1.5 , e ) 2.5 | b | add(negate(11), 5) | on the number line , if x is halfway between - 11 and 5 , and if y is halfway between - 3 and 7 , what number is halfway between x and y ? | "x = - 3 and y = 2 . the answer is b ." | a = negate + (
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a ) 1 / 14 , b ) 1 / 20 , c ) 1 / 13 , d ) 1 / 7 , e ) 1 / 6 | e | multiply(2, multiply(divide(const_1, subtract(4, const_1)), divide(const_1, 4))) | john and david work at a hospital with 4 other workers . for an internal review , 2 of the 4 workers will be randomly chosen to be interviewed . what is the probability that john and david will both be chosen ? | "probability that john and david will both be chosen out of 8 workers = ( 2 / 4 ) * ( 1 / 3 ) = 1 / 6 answer e" | a = 4 - 1
b = 1 / a
c = 1 / 4
d = b * c
e = 2 * d
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a ) 1 / 2 , b ) 7 / 3 , c ) 6 / 7 , d ) 1 / 5 , e ) 11 / 13 | b | add(divide(divide(4, 5), divide(7, 7)), const_1) | if w / x = 5 / 7 and w / y = 4 / 7 , then ( x + y ) / y = | "ratio 1 : 7 w = 5 x ratio 2 : 7 w = 4 y 5 x = 4 y x = 4 y / 3 ( x + y ) / y = ( ( 4 y / 3 ) + y ) / y = y ( 4 / 3 + 1 ) / y = 7 / 3 answer is b" | a = 4 / 5
b = 7 / 7
c = a / b
d = c + 1
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a ) 1 / 6 , b ) 1 / 5 , c ) 1 / 4 , d ) 1 / 3 , e ) 1 / 2 | d | inverse(add(divide(const_12, 6), const_1)) | a worker ' s take - home pay last year was the same each month , and she saved the same fraction of her take - home pay each month . the total amount of money that she had saved at the end of the year was 6 times the amount of that portion of her monthly take - home pay that she did not save . if all the money that she saved last year was from her take - home pay , what fraction of her take - home pay did she save each month ? | "let x be the fraction of her take - home pay that the worker saved . let p be the monthly pay . 12 xp = 6 ( 1 - x ) p 12 xp = 6 p - 6 xp 18 xp = 6 p x = 1 / 3 the answer is d ." | a = 12 / 6
b = a + 1
c = 1/(b)
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a ) 101 , b ) 107 , c ) 111 , d ) 112 , e ) 113 | c | sqrt(add(power(sqrt(subtract(37, multiply(const_2, 4107))), const_2), multiply(const_4, 4107))) | the product of two numbers is 4107 . if the h . c . f of these numbers is 37 , then the greater number is : | "let the numbers be 37 a and 37 b . then , 37 a * 37 b = 4107 = > ab = 3 now , co - primes with product 3 are ( 1 , 3 ) . so , the required numbers are ( 37 * 1 , 37 * 3 ) i . e . , ( 1 , 111 ) . greater number = 111 . answer : c" | a = 2 * 4107
b = 37 - a
c = math.sqrt(b)
d = c ** 2
e = 4 * 4107
f = d + e
g = math.sqrt(f)
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a ) 124 % , b ) 120 % , c ) 96 % , d ) 80 % , e ) 64 % | c | multiply(divide(add(const_100, 60), multiply(divide(const_100, subtract(const_100, 40)), const_100)), const_100) | marts income is 60 percent more than tims income and tims income is 40 percent less than juans income . what percentage of juans income is marts income | "tim = 100 mary ' s income is 60 percent more than tims income ; mary ' s income = t + 0.6 t = 1.6 t = 1.6 * 100 = 160 tims income is 40 percent less than juans income . 100 is 40 % less than juan ' s income 100 = j - 0.4 j 100 = 0.6 j j = 100 / 0.6 = 1000 / 6 what percentage of juans income is marts income ( 160 / ( 1000 / 6 ) ) * 100 = ( 160 * 6 * 100 ) / 1000 = 96 % answer : c" | a = 100 + 60
b = 100 - 40
c = 100 / b
d = c * 100
e = a / d
f = e * 100
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a ) 9 , b ) 10 , c ) 11 , d ) 12 , e ) 13 | a | divide(add(12, 24), subtract(7.5, add(divide(12, 8), divide(24, 12)))) | on a trip , a cyclist averaged 8 miles per hour for the first 12 miles and 12 miles per hour for the remaining 24 miles . if the cyclist returned immediately via the same route and took a total of 7.5 hours for the round trip , what was the average speed ( in miles per hour ) for the return trip ? | "the time to go 36 miles was 12 / 8 + 24 / 12 = 1.5 + 2 = 3.5 hours . the average speed for the return trip was 36 miles / 4 hours = 9 mph . the answer is a ." | a = 12 + 24
b = 12 / 8
c = 24 / 12
d = b + c
e = 7 - 5
f = a / e
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a ) 19002 , b ) 19028 , c ) 27788 , d ) 19000 , e ) 27711 | d | subtract(20000, 1000) | the price of a t . v . set worth rs . 20000 is to be paid in 20 installments of rs . 1000 each . if the rate of interest be 6 % per annum , and the first installment be paid at the time of purchase , then the value of the last installment covering the interest as well will be ? | "money paid in cash = rs . 1000 balance payment = ( 20000 - 1000 ) = rs . 19000 answer : d" | a = 20000 - 1000
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a ) 132 , b ) 138 , c ) 144 , d ) 150 , e ) 156 | c | multiply(inverse(add(divide(const_1, 4), divide(const_1, 6))), const_60) | working alone , pump a can empty a pool in 4 hours . working alone , pump b can empty the same pool in 6 hours . working together , how many minutes will it take pump a and pump b to empty the pool ? | pump a can empty ( 1 / 4 ) of the pool per hour . pump b can empty ( 1 / 6 ) of the pool per hour . together the pumps can empty 1 / 4 + 1 / 6 = 5 / 12 of the pool per hour . 1 pool / ( 5 / 12 ) pool per hour = 12 / 5 hours = 144 minutes . the answer is c . | a = 1 / 4
b = 1 / 6
c = a + b
d = 1/(c)
e = d * const_60
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a ) a ) 3623216 , b ) b ) 2903616 , c ) c ) 3624316 , d ) d ) 3625116 , e ) e ) 3625216 | b | multiply(divide(1704, 1704), const_100) | 1704 x 1704 = ? | "1704 x 1704 = ( 1704 ) 2 = ( 1700 + 4 ) 2 = ( 1700 ) 2 + ( 4 ) 2 + ( 2 x 1700 x 4 ) = 2890000 + 16 + 13600 . = 2903616 . b )" | a = 1704 / 1704
b = a * 100
|
a ) 32.52 , b ) 32.47 , c ) 34.97 , d ) 32.92 , e ) 32.33 | c | add(divide(circumface(6.83), const_2), multiply(6.83, const_2)) | the radius of a semi circle is 6.83 cm then its perimeter is ? | "36 / 7 r = 6.3 = 34.97 answer : c" | a = circumface / (
b = a + 2
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a ) 7 , b ) 8 , c ) 10 , d ) 11 , e ) 13 | a | divide(log(add(divide(multiply(109300, const_2), 100), const_1)), log(const_3)) | joan took out a mortgage from hel local bank . each monthly mortgage payment she makes must be triple the amount of the previous month ' s payment . if her first payment is $ 100 , and the total amount she must pay back is $ 109300 , how many months will it take joan to pay back her mortgage ? | "joan starts off with 100 $ . . which is to be tripled every month her monthly payments look like this : 100 , 300 , 900 , 2700 . . . . . . . . . upto 109300 this can be re written as : 100 x 1 , 100 x 3 , 100 x 9 , 100 x 27 . . . . . . 100 x 1093 so we have 1 , 3 , 9 , 27 . . . . . 109300 in gp we know that a = 1 , and r = 3 ( its easy to figure it out by looking at the question , but regardless of it being mentioned in the question we can still compute the value of r using the formula tn = a 3 ^ n - 1 . . . ) therefore to find the sum of n terms of a gp we use this formula : sn = a ( 1 - r ^ n ) / 1 - r using this and plugging in the information we get . . . 1093 = 1 - 3 ^ n / 1 - 3 ; 1 - 3 ^ n / - 2 cross multiplying we get 1093 x - 2 = 1 - 3 ^ n - 2186 = 1 - 3 ^ n - 2187 = - 3 ^ n 2187 = 3 ^ n ( negatives cancel out ) 2187 can also be re written as 3 ^ 7 therefore ; 3 ^ 7 = 3 ^ n thus n = 7 ( a )" | a = 109300 * 2
b = a / 100
c = b + 1
d = math.log(c)
e = math.log(3)
f = d / e
|
a ) 35 , b ) 55 , c ) 39 , d ) 40 , e ) 60 | a | divide(multiply(30, add(3, divide(30, const_60))), 3) | jerry began driving from home on a trip averaging 30 miles per hour . how many miles per hour must carla drive on average to catch up to him in exactly 3 hours if she leaves 30 minutes after jerry ? | carla starts 30 minutes later and it takes 3 hr for carla to meet jerry so jerry total time travelled = 3 hr + 30 minutes jerry distance = 30 * ( 3 1 / 2 ) = 105 so carla need to travle 105 to meet jerry in 3 hrs speed of carla = 105 / 3 = 35 miles per hour answer is a | a = 30 / const_60
b = 3 + a
c = 30 * b
d = c / 3
|
a ) 28 , b ) 27 , c ) 26 , d ) 22 , e ) 25 | e | add(subtract(97, multiply(19, 4)), 4) | having scored 97 runs in the 19 th inning , a cricketer increases his average score by 4 . what will be his average score after 19 innings ? | "explanation : let the average score of the first 18 innings be n 18 n + 97 = 19 ( n + 4 ) = > n = 21 so , average score after 19 th innings = x + 4 = 25 . answer : e" | a = 19 * 4
b = 97 - a
c = b + 4
|
a ) 462 cm 2 , b ) 478 cm 2 , c ) 490 cm 2 , d ) 268 cm 2 , e ) 668 cm 2 | a | multiply(22, 21) | find the area of a parallelogram with base 22 cm and height 21 cm ? | "area of a parallelogram = base * height = 22 * 21 = 462 cm 2 answer : a" | a = 22 * 21
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a ) 4.37 % , b ) 4 % , c ) 2.3 % , d ) 3.75 % , e ) none | c | add(multiply(divide(subtract(divide(subtract(subtract(subtract(multiply(multiply(const_10, const_1000), const_10), const_1000), const_1000), multiply(add(1, const_3), const_100)), multiply(add(multiply(add(const_3, const_4), const_10), add(1, const_3)), const_1000)), 1), const_10), const_100), const_4) | the population of a town increased from 1 , 24,300 to 1 , 62,400 in a decade . the average percent increase of population per year is : | "explanation : increase in 10 years = ( 162400 - 124300 ) = 87500 . increase % = ( 38100 / 162400 x 100 ) % = 23 % . required average = ( 23 / 10 ) % = 2.3 % . answer : option c" | a = 10 * 1000
b = a * 10
c = b - 1000
d = c - 1000
e = 1 + 3
f = e * 100
g = d - f
h = 3 + 4
i = h * 10
j = 1 + 3
k = i + j
l = k * 1000
m = g / l
n = m - 1
o = n / 10
p = o * 100
q = p + 4
|
a ) 5 hours , b ) 1 hours , c ) 3 hours , d ) 2 hours , e ) 4 hours | c | divide(6, divide(add(multiply(divide(1, 20), const_60), divide(4, 4)), const_2)) | a boatman goes 4 km against the current of the stream in 4 hour and goes 1 km along the current in 20 minutes . how long will it take to go 6 km in stationary water ? | speed upstream = 4 / 4 = 1 km / hr speed downstream = 1 / ( 20 / 60 ) = 3 km / hr speed in still water = 1 / 2 ( 3 + 1 ) = 2 km / hr time taken to travel 6 km in still water = 6 / 2 = 3 hours answer : c | a = 1 / 20
b = a * const_60
c = 4 / 4
d = b + c
e = d / 2
f = 6 / e
|
a ) 0 , b ) 2 , c ) 3 , d ) 6 , e ) 7 | a | multiply(subtract(divide(power(divide(7899, const_3), 2), 9), add(add(add(multiply(multiply(const_1000, subtract(9, 2)), const_100), multiply(multiply(const_1000, subtract(9, 2)), const_10)), multiply(2, const_100)), subtract(const_100, 2))), 9) | if z is a multiple of 7899 , what is the remainder when z ^ 2 is divided by 9 ? | the sum of the digits is 7 + 8 + 9 + 9 = 33 . thus 3 is a factor of 7899 , so 3 is a factor of z . then 3 ^ 3 = 9 is a factor of z ^ 2 . then the remainder when z ^ 2 is divided by 9 is 0 . the answer is a . | a = 7899 / 3
b = a ** 2
c = b / 9
d = 9 - 2
e = 1000 * d
f = e * 100
g = 9 - 2
h = 1000 * g
i = h * 10
j = f + i
k = 2 * 100
l = j + k
m = 100 - 2
n = l + m
o = c - n
p = o * 9
|
a ) 5 / 3 , b ) 7 / 4 , c ) 9 / 5 , d ) 11 / 6 , e ) 13 / 7 | c | divide(12, divide(const_1, add(divide(const_1, 10), divide(const_1, 20)))) | working alone , printers x , y , and z can do a certain printing job , consisting of a large number of pages , in 12 , 10 , and 20 hours , respectively . what is the ratio of the time it takes printer x to do the job , working alone at its rate , to the time it takes printers y and z to do the job , working together at their individual rates ? | "the time it takes printer x is 12 hours . the combined rate of y and z is 1 / 10 + 1 / 20 = 3 / 20 the time it takes y and z is 20 / 3 the ratio of times is 12 / ( 20 / 3 ) = 3 * 12 / 20 = 9 / 5 the answer is c ." | a = 1 / 10
b = 1 / 20
c = a + b
d = 1 / c
e = 12 / d
|
a ) 8 m , b ) 10 m , c ) 12 m , d ) 15 m , e ) 17 m | c | divide(sqrt(divide(288, divide(const_1, const_2))), const_2) | the width of a rectangular hall is ½ of its length . if the area of the hall is 288 sq . m , what is the difference between its length and breadth ? | "let the length of the hall be x m breadth of the hall = 1 x / 2 m area of the hall = length * breadth 288 = x * 1 x / 2 x ² = 576 x = 24 difference between the length and breadth of the hall = x - 1 x / 2 = x / 2 = 24 / 2 = 12 m answer : c" | a = 1 / 2
b = 288 / a
c = math.sqrt(b)
d = c / 2
|
a ) 10 , b ) 20 , c ) 21 , d ) 25 , e ) 27 | c | multiply(7, divide(multiply(add(7, 9), subtract(9, multiply(divide(5, add(7, 5)), 9))), subtract(multiply(9, 7), multiply(7, 5)))) | a can contains a mixture of liquids a and b is the ratio 7 : 5 . when 9 litres of mixture are drawn off and the can is filled with b , the ratio of a and b becomes 7 : 9 . how many liter of liquid a was contained by the can initially ? | using the values here might be the most straightforward way for this question . as a : b : : 7 : 5 - - - > only option c is a multiple of 7 and hence it is a good place to start . also a : b : : 7 : 5 means that , a = ( 712 ) * total and b = ( 5 / 12 ) * total if a = 21 , b = 15 - - - > remove 9 litres - - - > you remove ( 7 / 12 ) * 9 of a - - - > a remaining = 21 - ( 7 / 12 ) * 9 = 63 / 4 similarly , for b , you remove ( 5 / 12 ) * 9 - - - > b remaining = 15 - ( 5 / 12 ) * 9 = 45 / 4 and then add 9 more litres of b - - - > 9 + 45 / 4 = 81 / 4 thus a / b ( final ratio ) = ( 45 / 4 ) / ( 81 / 4 ) = 7 : 9 , the same as the final ratio mentioned in the question . hence c is the correct answer . fyi , the algebraic equations will become : a / b = 7 / 9 = ( 7 x - ( 7 / 12 ) * 9 ) / ( 5 x - ( 5 / 12 ) * 9 + 9 ) , where 7 x and 5 x are initial quantities of a and b respectively . thus , 7 / 9 = ( 7 x - ( 7 / 12 ) * 9 ) / ( 5 x - ( 5 / 12 ) * 9 + 9 ) - - - > giving you x = 3 . thus a ( original ) = 7 * 3 = 21 . answer : c | a = 7 + 9
b = 7 + 5
c = 5 / b
d = c * 9
e = 9 - d
f = a * e
g = 9 * 7
h = 7 * 5
i = g - h
j = f / i
k = 7 * j
|
a ) 1 : 5 , b ) 2 : 5 , c ) 3 : 5 , d ) 4 : 5 , e ) 1 : 1 | b | divide(divide(const_100, add(const_100, 25)), divide(const_100, add(const_100, 50))) | bert and rebecca were looking at the price of a condominium . the price of the condominium was 25 % more than bert had in savings , and separately , the same price was also 50 % more than rebecca had in savings . what is the ratio of what bert has in savings to what rebecca has in savings . | "suppose bert had 100 so price becomes 125 , this 125 = 1.5 times r ' s saving . . so r ' s saving becomes 250 so required ratio is 100 : 250 = 2 : 5 answer : b" | a = 100 + 25
b = 100 / a
c = 100 + 50
d = 100 / c
e = b / d
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a ) 5 days , b ) 6 days , c ) 7 days , d ) 9 days , e ) 8 days | d | divide(subtract(const_1, divide(23, 40)), add(divide(const_1, 45), divide(const_1, 40))) | a and b can do a piece of work in 45 days and 40 days respectively . they began to do the work together but a leaves after some days and then b completed the remaining work in 23 days . the number of days after which a left the work was | "does the job in 40 days = 1 / 40 a & b together completes the job = ? after a leaves , b completes the remaining job in 23 days . i . e b ' s part of job in 23 days = 23 / 40 remaining part 17 / 40 is done by a & b together in x no . of days . i . e ( x / 45 ) + ( x / 40 ) = 17 / 40 = > x = 9 days answer : d" | a = 23 / 40
b = 1 - a
c = 1 / 45
d = 1 / 40
e = c + d
f = b / e
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a ) 23 , b ) 34 , c ) 35 , d ) 41 , e ) 44 | b | add(divide(subtract(75, add(add(9, 9), 9)), const_3), add(9, 9)) | virginia , adrienne , and dennis have taught history for a combined total of 75 years . if virginia has taught for 9 more years than adrienne and for 9 fewer years than dennis , for how many years has dennis taught ? | "let number of years taught by virginia = v number of years taught by adrienne = a number of years taught by dennis = d v + a + d = 75 v = a + 9 = > a = v - 9 v = d - 9 = > a = ( d - 9 ) - 9 = d - 18 d - 9 + d - 18 + d = 96 = > 3 d = 75 + 27 = 102 = > d = 34 answer b" | a = 9 + 9
b = a + 9
c = 75 - b
d = c / 3
e = 9 + 9
f = d + e
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a ) 68 , b ) 72 , c ) 76 , d ) 80 , e ) 84 | d | divide(28, subtract(divide(3, 4), divide(40, const_100))) | a big container is 40 % full with water . if 28 liters of water is added , the container becomes 3 / 4 full . what is the capacity of the big container in liters ? | "28 liters is 35 % of the capacity c . 28 = 0.35 c c = 28 / 0.35 = 80 liters . the answer is d ." | a = 3 / 4
b = 40 / 100
c = a - b
d = 28 / c
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a ) 1 : 2 , b ) 2 : 1 , c ) 8 : 3 , d ) 3 : 5 , e ) 8 : 1 | e | divide(multiply(54000, const_12), multiply(27000, add(const_4, const_3))) | x starts a business with rs . 54000 . y joins in the business after 9 months with rs . 27000 . what will be the ratio in which they should share the profit at the end of the year ? | "explanation : ratio in which they should share the profit = ratio of the investments multiplied by the time period = 54000 * 12 : 27000 * 3 = 54 * 12 : 27 * 3 = 2 * 4 : 1 = 8 : 1 . answer : option e" | a = 54000 * 12
b = 4 + 3
c = 27000 * b
d = a / c
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a ) 2 / 3 , b ) 4 / 5 , c ) 6 / 2 , d ) 5 / 2 , e ) 4 / 9 | e | divide(add(multiply(2, const_0_33), multiply(1, divide(const_2, const_3))), add(2, 1)) | at an international conference , “ red ” world countries and “ blue ” world countries are the only participants . the ratio of “ red ” world participants to “ blue ” world participants is 2 : 1 . if one - third of “ red ” world participants are left - handed and two - thirds of “ blue ” world participants are left - handed , then what is the fraction of the participants who are left - handed ? | "red : blue = 2 : 1 let red = 2 x and blue = 1 x 1 / 3 of red are left handed = > 1 / 3 * 2 x = 2 x / 3 red left handed 2 / 3 of blue are left handed = > 2 / 3 * 1 x = 2 x / 3 blue left handed fraction of participants who are left handed = total left handed / total participants = ( red left handed + blue left handed ) / total participants = ( 2 x / 3 + 2 x / 3 ) / ( 2 x + 1 x ) = 4 / 9 answer : e" | a = 2 * const_0_33
b = 2 / 3
c = 1 * b
d = a + c
e = 2 + 1
f = d / e
|
a ) 1 day , b ) 2 days , c ) 3 days , d ) 4 days , e ) 5 days | b | divide(72, multiply(divide(48, multiply(4, 2)), 6)) | if 4 men can colour 48 m long cloth in 2 days , then 6 men can colour 72 m long cloth in | "the length of cloth painted by one man in one day = 48 / 4 × 2 = 6 m no . of days required to paint 36 m cloth by 6 men = 72 / 6 × 6 = 2 day . b" | a = 4 * 2
b = 48 / a
c = b * 6
d = 72 / c
|
a ) 1720 , b ) 2160 , c ) 2240 , d ) 2460 , e ) 2520 | c | add(subtract(subtract(const_1000, const_10), multiply(multiply(const_10, multiply(4, 4)), multiply(const_4, const_2))), const_10) | how many 4 digit even numbers do not use any digit more than once | "a b c d ( thousand , hundreds , tens , units ) d can be 0 2 4 6 8 ( any of the 5 digits ) a can be anything except ( d or 0 ) so 8 possibilities c can be anything except a and b so 8 possibilities b can be anything except ( a d c ) so 7 possibilities total ways are 8 * 7 * 8 * 5 = 2240 ans : c" | a = 1000 - 10
b = 4 * 4
c = 10 * b
d = 4 * 2
e = c * d
f = a - e
g = f + 10
|
a ) 7 kmph , b ) 8 kmph , c ) 9 kmph , d ) 7 kmph , e ) 6 kmph | a | divide(add(10, 4), const_2) | calculate the speed of a boat in still water ( in km / hr ) if in one hour , the boat goes 10 km / hr downstream and 4 km / hr upstream . | speed in still water = ( 10 + 4 ) 1 / 2 kmph = 7 kmph . answer : a | a = 10 + 4
b = a / 2
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a ) 100 , b ) 90 , c ) 85 , d ) 110 , e ) 120 | b | add(add(divide(subtract(1000, 100), const_10), multiply(add(const_10, const_1), add(const_10, const_1))), multiply(10, const_2)) | how many times digit 10 is used while writing numbers from 100 to 1000 ? | "in 100 to 300 there are 20 ten ' s in 300 to 500 there are 20 ten ' s in 500 to 700 there are 20 ten ' s in 700 t 0 900 there are 20 ten ' s in 900 to 1000 there are 10 ten ' s so total is 90 ten ' s correct option : b" | a = 1000 - 100
b = a / 10
c = 10 + 1
d = 10 + 1
e = c * d
f = b + e
g = 10 * 2
h = f + g
|
a ) 13 , b ) 14 , c ) 15 , d ) 16 , e ) 17 | b | multiply(10, 5) | jacob is 10 years old . he is 5 times as old as his brother . how old will jacob be when he is twice as old ? | "j = 10 ; j = 5 b ; b = 10 / 5 = 2 ; twice as old so b = 2 ( now ) + ( 2 ) = 4 ; jacob is 10 + 4 = 14 answer : b" | a = 10 * 5
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a ) 51 : 52 , b ) 52 : 53 , c ) 53 : 54 , d ) 54 : 55 , e ) none of these | b | divide(add(const_100, 4), add(const_100, 6)) | the cash difference between the selling prices of an article at a profit of 4 % and 6 % is rs 3 . the ratio of two selling prices is | "explanation : let the cost price of article is rs . x required ratio = 104 % of x / 106 % of x = 104 / 106 = 52 / 53 = 52 : 53 option b" | a = 100 + 4
b = 100 + 6
c = a / b
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['a ) 7.5 cm rise', 'b ) 10 cm rise', 'c ) 15 cm rise', 'd ) 30 cm rise', 'e ) none of these'] | c | divide(power(30, const_3), multiply(60, 30)) | a steel vessel has a base of length 60 cm and breadth 30 cm . water is poured in the vessel . a cubical steel box having edge of 30 cm is immersed completely in the vessel . how much will the water rise ? | explanation : volume of cube = a 3 = a * a * a 60 cm * 30 cm * h = 30 cm * 30 cm * 30 cm h = 15 cm answer is c | a = 30 ** 3
b = 60 * 30
c = a / b
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a ) 10 , b ) 11 , c ) 12 , d ) 15 , e ) 16 | a | add(multiply(3, 3), const_1) | a 3 - digit number 4 a 3 is added to another 3 - digit number 984 to give a 4 - digit number 13 b 7 , which is divisible by 11 . then , ( a + b ) = ? | 4 a 3 | 9 8 4 } = = > a + 8 = b = = > b - a = 8 13 b 7 | also , 13 b 7 is divisible by 11 = > ( 7 + 3 ) – ( b + 1 ) = ( 9 – b ) ( 9 – b ) = 0 b = 9 ( b = 9 and a = 1 ) = > ( a + b ) = 10 . answer a | a = 3 * 3
b = a + 1
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a ) 8 , b ) 9 , c ) 10 , d ) 11 , e ) 12 | b | multiply(9, const_1) | if 3 ^ x * 4 ^ y = 19,683 and x – y = 9 , then x = ? | since 19,683 is an odd number , it is not a multiple of 4 . this means that y must equal zero . since x - y = 9 and y = 0 , then x = 9 . the answer is b . | a = 9 * 1
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a ) 3 , b ) 5 , c ) 4 , d ) 6 , e ) 7 | b | divide(64, multiply(46, const_0_2778)) | in what time will a railway train 64 m long moving at the rate of 46 kmph pass a telegraph post on its way ? | "t = 64 / 46 * 18 / 5 = 5 sec answer : b" | a = 46 * const_0_2778
b = 64 / a
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a ) 9 , b ) 4 , c ) 5 , d ) 6 , e ) 7 | a | divide(subtract(1,600, const_100), subtract(400, subtract(400, 150))) | matt gets a $ 1,600 commission on a big sale . this commission alone raises his average commission by $ 150 . if matt ' s new average commission is $ 400 , how many sales has matt made ? | "let , average commission = x no . of items sold = y total commission = xy new commission = xy + 1600 new average = ( xy + 1600 ) / ( y + 1 ) = 150 + x i . e . ( xy + 1600 ) = ( y + 1 ) * ( 150 + x ) i . e . ( xy + 1600 ) = ( xy + x + 150 y + 150 ) i . e . ( 1450 ) = ( x + 150 y ) new commission = 400 = 150 + x i . e . x = 250 i . e . y = 8 new sales = y + 1 = 9 answer : option a" | a = 1 - 600
b = 400 - 150
c = 400 - b
d = a / c
|
a ) 196 cm 2 , b ) 187 cm 2 , c ) 178 cm 2 , d ) 189 cm 2 , e ) 176 cm 2 | a | divide(multiply(28, 14), const_2) | if the sides of a triangle are 30 cm , 28 cm and 14 cm , what is its area ? | "the triangle with sides 30 cm , 28 cm and 14 cm is right angled , where the hypotenuse is 30 cm . area of the triangle = 1 / 2 * 28 * 14 = 196 cm 2 answer : a" | a = 28 * 14
b = a / 2
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a ) 1000 , b ) 2217 , c ) 1100 , d ) 2777 , e ) 2991 | c | divide(1232, 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 . 1232 ? | "1232 = p [ 1 + ( 5 * 12 / 5 ) / 100 ] p = 1100 answer : c" | a = 2 * 5
b = a + 2
c = b / 5
d = c * 5
e = d / 100
f = e + 1
g = 1232 / f
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a ) 500 , b ) 450 , c ) 250 , d ) 350 , e ) 525 | a | divide(add(30, 40), divide(subtract(const_100, 86), const_100)) | a group of students was interviewed for that if it was asked whether or not they speak french and / or english . among those who speak french , 30 speak english well , while 40 of them do not speak english . if 86 % of students do not speak french , how many students were surveyed ? | "number of students who speak french are 30 + 40 = 70 of total students , the percentage of students who do not speak french was 86 % - - > percentage of who do is 14 % 70 - - - - - - - 14 % x - - - - - - - 100 % x = 70 * 100 / 14 = 500 = number of all students answer is a" | a = 30 + 40
b = 100 - 86
c = b / 100
d = a / c
|
a ) 4.51 , b ) 4.71 , c ) 4.32 , d ) 4.21 , e ) 4.68 | b | divide(circumface(divide(square_edge_by_perimeter(rectangle_perimeter(4, 2)), const_2)), const_2) | the parameter of a square is equal to the perimeter of a rectangle of length 4 cm and breadth 2 cm . find the circumference of a semicircle whose diameter is equal to the side of the square . ( round off your answer to two decimal places ) ? | "let the side of the square be a cm . parameter of the rectangle = 2 ( 4 + 2 ) = 12 cm parameter of the square = 12 cm i . e . 4 a = 12 a = 3 diameter of the semicircle = 3 cm circimference of the semicircle = 1 / 2 ( â ˆ ) ( 3 ) = 1 / 2 ( 22 / 7 ) ( 3 ) = 66 / 14 = 4.71 cm to two decimal places answer : b" | a = square_edge_by_perimeter / (
b = circumface / (
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a ) 80 , b ) 160 , c ) 720 , d ) 2100 , e ) 2940 | e | multiply(divide(divide(factorial(15), factorial(subtract(15, 2))), 2), divide(divide(factorial(8), factorial(subtract(8, 2))), 2)) | 15 different biology books and 8 different chemistry books lie on a shelf . in how many ways can a student pick 2 books of each type ? | "no . of ways of picking 2 biology books ( from 15 books ) = 15 c 2 = ( 15 * 14 ) / 2 = 105 no . of ways of picking 2 chemistry books ( from 8 books ) = 8 c 2 = ( 8 * 7 ) / 2 = 28 total ways of picking 2 books of each type = 105 * 28 = 2940 ( option e )" | a = math.factorial(15)
b = 15 - 2
c = math.factorial(b)
d = a / c
e = d / 2
f = math.factorial(8)
g = 8 - 2
h = math.factorial(g)
i = f / h
j = i / 2
k = e * j
|
a ) 33 , b ) 878 , c ) 30 , d ) 10 , e ) 11 | d | inverse(subtract(5, divide(5, 10))) | a and b can do a piece of work in 10 days . with the help of c they finish the work in 5 days . c alone can do that piece of work in ? | "c = 1 / 5 â € “ 1 / 10 = 1 / 10 = > 10 days answer : d" | a = 5 / 10
b = 5 - a
c = 1/(b)
|
a ) 2.25 . , b ) 3.125 . , c ) 4.5 . , d ) 5.225 . , e ) 6.25 . | c | divide(divide(6048, 28), multiply(subtract(28, const_4), const_2)) | a computer factory produces 6048 computers per month at a constant rate , how many computers are built every 30 minutes assuming that there are 28 days in one month ? | "number of hours in 28 days = 28 * 24 number of 30 mins in 28 days = 28 * 24 * 2 number of computers built every 30 mins = 6048 / ( 28 * 24 * 2 ) = 4.5 answer c" | a = 6048 / 28
b = 28 - 4
c = b * 2
d = a / c
|
a ) 125 , b ) 127 , c ) 129 , d ) 131 , e ) 141 | c | gcd(subtract(1428, 9), subtract(2206, 13)) | find the greatest number which on dividing 1428 and 2206 , leaves a reminder of 9 and 13 respectively | "explanation : in this type of question , its obvious we need to calculate the hcf , trick is hcf of ( 1428 - 9 ) and ( 2206 - 13 ) = hcf ( 1419 , 2193 ) = 129 option c" | a = 1428 - 9
b = 2206 - 13
c = math.gcd(a, b)
|
a ) 520,000 , b ) 540,000 , c ) 525,000 , d ) 530,000 , e ) 550,000 | b | multiply(300, divide(subtract(2007, 2003), const_2)) | during 2003 , a company produced an average of 3,000 products per month . how many products will the company need to produce from 2004 through 2007 in order to increase its monthly average for the period from 2003 through 2007 by 300 % over its 2003 average ? | "company produced 12 * 3000 = 36,000 products in 2003 . if company produces x products from 2004 to 2007 , then total amount of product produced in 4 years ( 2003 through 2007 ) is x + 36,000 . the gives the average of ( x + 36,000 ) / 4 . this average needs to be 300 % higher than that in 2003 . in math terms , 36,000 + 300 % ( 36,000 ) = 144,000 . so : ( x + 36,000 ) / 4 = 144,000 x + 36,000 = 576,000 x = 540,000 the answer is b ." | a = 2007 - 2003
b = a / 2
c = 300 * b
|
a ) 22 , b ) 67 , c ) 26 , d ) 18 , e ) 16 | a | add(add(const_10, const_10), const_2) | what is the ratio of arun ' s monthly income in the year 2006 , suman ' s monthly income in the year 2007 and jyothi ' s monthly income in the year 2005 ? | answer : a | a = 10 + 10
b = a + 2
|
a ) 8 , b ) 9 , c ) 10 , d ) 11 , e ) 12 | c | divide(40, divide(divide(40, const_2), 5)) | a runner runs the 40 miles from marathon to athens at a constant speed . halfway through the run she injures her foot , and continues to run at half her previous speed . if the second half takes her 5 hours longer than the first half , how many hours did it take the runner to run the second half ? | the runner runs the first 20 miles at speed v and the second 20 miles at speed v / 2 . the time t 2 to run the second half must be twice the time t 1 to run the first half . t 2 = 2 * t 1 = t 1 + 5 t 1 = 5 and so t 2 = 10 . the answer is c . | a = 40 / 2
b = a / 5
c = 40 / b
|
a ) 50.78 kg , b ) 49.32 kg , c ) 46.66 kg , d ) 69.15 kg , e ) 70.89 kg | c | divide(add(multiply(40, 50), multiply(20, 40)), add(40, 20)) | there are 2 sections a and b in a class , consisting of 40 and 20 students respectively . if the average weight of section a is 50 kg and that of section b is 40 kg , find the average of the whole class ? | "total weight of 40 + 20 students = 40 * 50 + 20 * 40 = 2000 + 800 average weight of the class is = 2800 / 60 = 46.66 kg answer is c" | a = 40 * 50
b = 20 * 40
c = a + b
d = 40 + 20
e = c / d
|
a ) 3 , b ) 6 , c ) 9 , d ) 12 , e ) 15 | a | floor(divide(15, 4)) | on dividing 15 by a number , the quotient is 4 and the remainder is 3 . find the divisor ? | "d = ( d - r ) / q = ( 15 - 3 ) / 4 = 12 / 4 = 3 a )" | a = 15 / 4
b = math.floor(a)
|
a ) 2 , b ) - 2 , c ) 4 , d ) - 5 , e ) 6 | d | divide(subtract(46, 10), 4) | if | 4 x + 10 | = 46 , what is the sum of all the possible values of x ? | there will be two cases 4 x + 10 = 46 or 4 x + 10 = - 46 = > x = 9 or x = - 14 sum of both the values will be - 14 + 9 = - 5 answer is d | a = 46 - 10
b = a / 4
|
a ) 1 ⁄ 3 , b ) 4 ⁄ 3 , c ) 80 , d ) 120 , e ) 300 | e | divide(multiply(2, const_60), divide(2, 5)) | if it takes a machine 2 ⁄ 5 minute to produce one item , how many items will it produce in 2 hours ? | "1 item takes 2 / 5 min so it takes 120 min to produce x 2 x / 5 = 120 the x = 300 answer : e" | a = 2 * const_60
b = 2 / 5
c = a / b
|
a ) 2 , 400,000 , b ) 2 , 800,000 , c ) 55,000 , d ) 28,000 , e ) 280 | a | subtract(add(multiply(multiply(divide(volume_cube(100), const_10), 1.5), 1.5), multiply(multiply(divide(volume_cube(100), const_10), multiply(const_2, 3)), 3)), volume_cube(100)) | a specialized type of sand consists of 40 % mineral x by volume and 60 % mineral y by volume . if mineral x weighs 1.5 grams per cubic centimeter and mineral y weighs 3 grams per cubic centimeter , how many grams does a cubic meter of specialized sand combination weigh ? ( 1 meter = 100 centimeters ) | "let the volume be 1 m ^ 3 = 1 m * 1 m * 1 m = 100 cm * 100 cm * 100 cm = 1 , 000,000 cm ^ 3 by volume 40 % is x = 400,000 cm ^ 3 60 % is y = 600,000 cm ^ 3 by weight , in 1 cm ^ 3 , x is 1.5 gms in 400,000 cm ^ 3 , x = 1.5 * 400,000 = 600,000 grams in 1 cm ^ 3 , y is 3 gms in 600,000 cm ^ 3 , y = 3 * 600,000 = 1 , 800,000 gms total gms in 1 m ^ 3 = 600,000 + 1 , 800,000 = 2 , 400,000 answer : a" | a = volume_cube / (
b = a * 10
c = b * 1
d = c + 1
e = volume_cube / (
f = e * 10
g = 2 * 3
h = f * g
i = d - h
|
a ) 5.12 , b ) 4.13 , c ) 6.15 , d ) 7.5 , e ) none of these | a | multiply(const_60, divide(subtract(82, 75), 82)) | excluding stoppages , the speed of a bus is 82 kmph and including stoppages , it is 75 kmph . for how many minutes does the bus stop per hour ? | "due to stoppages , it covers 7 km less . time taken to cover 7 km = ( ( 7 / 82 ) ã — 60 ) = 5.12 min . option ( a ) is correct" | a = 82 - 75
b = a / 82
c = const_60 * b
|
a ) 60 , b ) 60.75 , c ) 58 , d ) 31.528 , e ) 30.375 | e | subtract(negate(13.50), multiply(subtract(6, 9), divide(subtract(6, 9), subtract(4, 6)))) | 4 , 6 , 9 , 13.50 , 20.25 , ( . . . ) | "4 ( 4 ã — 3 ) ã · 2 = 6 ( 6 ã — 3 ) ã · 2 = 9 ( 9 ã — 3 ) ã · 2 = 13.50 ( 13.50 ã — 3 ) ã · 2 = 20.25 ( 40.5 ã — 3 ) ã · 2 = 30.375 answer is e" | a = negate - (
|
a ) 180 , b ) 175 , c ) 156 , d ) 150 , e ) 105 | b | add(100, divide(subtract(85, 70), subtract(divide(70, 100), divide(50, 100)))) | during a certain season , a team won 85 percent of its first 100 games and 50 percent of its remaining games . if the team won 70 percent of its games for the entire season , what was the total number of games that the team played ? | "we are first given that a team won 85 percent of its first 100 games . this means the team won 0.85 x 100 = 85 games out of its first 100 games . we are next given that the team won 50 percent of its remaining games . if we use variable t to represent the total number of games in the season , then we can say t – 100 equals the number of remaining games in the season . thus we can say : 0.5 ( t – 100 ) = number of wins for remaining games 0.5 t – 50 = number of wins for remaining games lastly , we are given that team won 70 percent of all games played in the season . that is , they won 0.7 t games in the entire season . with this we can set up the equation : number of first 100 games won + number of games won for remaining games = total number of games won in the entire season 85 + 0.5 t – 50 = 0.7 t 35 = 0.2 t 350 = 2 t 175 = t answer is b ." | a = 85 - 70
b = 70 / 100
c = 50 / 100
d = b - c
e = a / d
f = 100 + e
|
a ) 26 , b ) 30 , c ) 36 , d ) 38 , e ) 40 | b | multiply(add(subtract(8, const_1), 8), divide(22, add(subtract(6, const_1), 6))) | a certain clock marks every hour by striking a number of times equal to the hour , and the time require for a stroke is exactly equal to the time interval between strokes . at 6 : 00 the time lapse between the beginning of the first stoke and the end of the last stroke is 22 seconds . at 8 : 00 , how many seconds elapse between the beginning of the first stroke and the end of the last stroke ? | "at 6 ' o clock , there would be 6 strikes . first strike , then a short interval , the second strike , then a short interval and so on till the 6 th strike . so there would be in all 5 intervals between 6 strikes . similarly , between 8 strikes , there would be 7 intervals . according to the question , the time spent in the strike and the interval is same . at 6 ' o clock , the 6 strikes and the 5 intervals together take 22 sec so each strike and each interval takes 2 secs . at 8 ' o clock , the 8 strikes and 7 intervals will take 2 * ( 8 + 7 ) = 30 secs b" | a = 8 - 1
b = a + 8
c = 6 - 1
d = c + 6
e = 22 / d
f = b * e
|
a ) 190 metres , b ) 160 metres , c ) 200 metres , d ) 120 metres , e ) 150 metres | c | multiply(1818, multiply(4040, const_0_2778)) | a train is running at a speed of 4040 km / hr and it crosses a post in 1818 seconds . what is the length of the train ? | "explanation : speed = 40 km / hr = 40 km / hr = 40 ã — 518 = 1009 m / s = 40 ã — 518 = 1009 m / s time = 18 = 18 seconds distance covered = 1009 ã — 18 = 200 m = 1009 ã — 18 = 200 m therefore , length of the train = 200 m answer is c" | a = 4040 * const_0_2778
b = 1818 * a
|
a ) 1 , b ) 10 , c ) 121 , d ) 100 , e ) 1000 | e | power(subtract(sqrt(121), 1), const_3) | if m and n are whole numbers such that mn = 121 , the value of ( m 1 ) n + 1 is : | if m and n are whole numbers such that mn = 121 , the value of ( m - 1 ) n + 1 is : a . 1 b . 10 c . 121 d . 1000 sol . - we know that 112 = 121 . putting m = 11 and n = 2 , we get : ( m - 1 ) n + 1 = ( 11 - 1 ) ( 2 + 1 ) = 103 = 1000 . answer : e | a = math.sqrt(121)
b = a - 1
c = b ** 3
|
a ) 120 cm 2 , b ) 112 cm 2 , c ) 776 cm 2 , d ) 130 cm 2 , e ) 886 cm 2 | d | divide(multiply(26, 10), const_2) | if the sides of a triangle are 28 cm , 26 cm and 10 cm , what is its area ? | "the triangle with sides 28 cm , 26 cm and 10 cm is right angled , where the hypotenuse is 28 cm . area of the triangle = 1 / 2 * 26 * 10 = 130 cm 2 answer : d" | a = 26 * 10
b = a / 2
|
a ) 76 seconds , b ) 18 seconds , c ) 52 seconds , d ) 19 seconds , e ) 45 seconds | c | divide(add(510, 140), divide(multiply(45, const_1000), const_3600)) | a train is 510 meter long is running at a speed of 45 km / hour . in what time will it pass a bridge of 140 meter length ? | "speed = 45 km / hr = 45 * ( 5 / 18 ) m / sec = 25 / 2 m / sec total distance = 510 + 140 = 650 meter time = distance / speed = 650 * ( 2 / 25 ) = 52 seconds answer : c" | a = 510 + 140
b = 45 * 1000
c = b / 3600
d = a / c
|
a ) 16.67 , b ) 30 , c ) 50 , d ) 60.33 , e ) 70 | c | divide(subtract(multiply(divide(40, const_100), 50), multiply(divide(20, const_100), 50)), subtract(divide(40, const_100), divide(20, const_100))) | how many ounces of a 60 % salt solution must be added to 50 ounces of a 20 percent salt solution so that the resulting mixture is 40 % salt ? | "for 60 % salt solution ( x ) and 20 % salt solution ( y ) to give 40 % = ( 60 + 20 ) / 2 salt solution , we should mix equal amounts of x and y . answer : c ." | a = 40 / 100
b = a * 50
c = 20 / 100
d = c * 50
e = b - d
f = 40 / 100
g = 20 / 100
h = f - g
i = e / h
|
a ) 15.6 , b ) 16.0 , c ) 17.5 , d ) 18.7 , e ) 19.1 | e | add(3.50, multiply(subtract(divide(8, divide(1, 5)), 1), 0.40)) | a certain taxi company charges $ 3.50 for the first 1 / 5 of a mile plus $ 0.40 for each additional 1 / 5 of a mile . what would this company charge for a taxi ride that was 8 miles long ? | "a certain taxi company charges $ 3.50 for the first 1 / 5 of a mile plus $ 0.40 for each additional 1 / 5 of a mile . what would this company charge for a taxi ride that was 8 miles long ? a . 15.60 b . 16.00 c . 17.50 d . 18.70 e . 19.10 1 / 5 miles = 0.2 miles . the cost of 8 miles long ride would be $ 3.50 for the first 0.2 miles plus ( 8 - 0.2 ) / 0.2 * 0.4 = $ 3.5 + $ 15.6 = $ 18.7 . answer : e ." | a = 1 / 5
b = 8 / a
c = b - 1
d = c * 0
e = 3 + 50
|
a ) 44 % , b ) 40 % , c ) 45 % , d ) 42 % , e ) 48 % | d | divide(add(multiply(3, 70), multiply(7, 30)), add(3, 7)) | the ratio of red balls to blue balls is 3 : 7 . 70 % of the red balls and 30 % of the blue balls are tennis balls . how many of the balls are tennis balls ? | let total no of balls 100 , red ball 30 and blue ball 70 , so tennis ball for red and blue ball respectively 30 * 70 / 100 = 21 and 70 * 30 / 100 = 21 so total 42 tennis balls out of 100 balls . . so required % is 42 % answer : d | a = 3 * 70
b = 7 * 30
c = a + b
d = 3 + 7
e = c / d
|
a ) 0.5 , b ) 1 , c ) 1.5 , d ) 2 , e ) 2.5 | d | multiply(subtract(5, 3), 3) | if ( c - a ) / ( c - b ) = 1 , then ( 5 b - 3 a ) / ( c - a ) = | "let ' s say c = 2 , b = 1 , a = 1 so that our 1 st expression holds true . now , ibsert those numbers in the second expression and we ' ll get 2 answer d ( hopefully ) ) )" | a = 5 - 3
b = a * 3
|
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