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 ) a ) 4 , b ) b ) 5 , c ) c ) 6 , d ) d ) 3 , e ) e ) 7 | d | sqrt(51) | from below option 51 is divisible by which one ? | "51 / 3 = 17 d" | a = math.sqrt(51)
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a ) 16 , b ) 12 , c ) 18 , d ) 22 , e ) 08 | d | multiply(2, 11) | each child has 2 pencils and 13 skittles . if there are 11 children , how many pencils are there in total ? | 2 * 11 = 22 . answer is d . | a = 2 * 11
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a ) 104 , b ) 55 , c ) 66 , d ) 98 , e ) 100 | d | divide(390, 4) | . a car covers a distance of 390 km in 4 hours . find its speed ? | 390 / 4 = 98 kmph answer : d | a = 390 / 4
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a ) 61376 , b ) 54411 , c ) 612314 , d ) 64170 , e ) 64171 | a | multiply(add(add(const_100, const_4), subtract(multiply(const_100, const_10), 8)), divide(add(divide(subtract(subtract(multiply(const_100, const_10), 8), add(const_100, const_4)), 8), const_1), const_2)) | find the sum of all 3 digit natural numbers , which are divisible by 8 | "the three digit natural numbers divisible by 8 are 104 , 112 , 120 , β¦ . 992 . let sndenote their sum . that is , sn = 104 112 120 128 , 992 g + + + + + . now , the sequence 104 , 112 , 120 , g , 992 forms an a . p . a = 104 , d = 8 , l = 992 n = l - a / d n = 112 s 112 = n / 2 ( a + l ) = 61376 answer a 61376" | a = 100 + 4
b = 100 * 10
c = b - 8
d = a + c
e = 100 * 10
f = e - 8
g = 100 + 4
h = f - g
i = h / 8
j = i + 1
k = j / 2
l = d * k
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a ) 55 , b ) 82 , c ) 73 , d ) 75 , e ) 85 | d | divide(add(multiply(25, subtract(25, 20)), multiply(25, subtract(25, 30))), add(25, 25)) | a man buys 25 lts of liquid which contains 20 % of the liquid and the rest is water . he then mixes it with 25 lts of another mixture with 30 % of liquid . what is the % of water in the new mixture ? | "20 % in 25 lts is 5 . so water = 25 - 5 = 20 lts . 30 % of 25 lts = 7.5 . so water in 2 nd mixture = 25 - 7.5 = 17.5 lts . now total quantity = 25 + 25 = 50 lts . total water in it will be 20 + 17.5 = 37.5 lts . % of water = ( 100 * 37.5 ) / 50 = 75 . answer : d" | a = 25 - 20
b = 25 * a
c = 25 - 30
d = 25 * c
e = b + d
f = 25 + 25
g = e / f
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a ) 235 miles . , b ) 245 miles . , c ) 255 miles . , d ) 265 miles . , e ) 275 miles . | c | add(multiply(45, 2), multiply(3, 55)) | john left home and drove at the rate of 45 mph for 2 hours . he stopped for lunch then drove for another 3 hours at the rate of 55 mph to reach his destination . how many miles did john drive ? | "the total distance d traveled by john is given by d = 45 * 2 + 3 * 55 = 255 miles . answer c" | a = 45 * 2
b = 3 * 55
c = a + b
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a ) 8 , b ) 10 , c ) 15 , d ) 6 , e ) 19 | d | subtract(30, divide(add(multiply(7.50, 30), 555), add(7.50, 25))) | a contractor is engaged for 30 days on the condition thathe receives rs . 25 for each day he works & is fined rs . 7.50 for each day is absent . he gets rs . 555 in all . for how many days was he absent ? | "30 * 25 = 750 455 - - - - - - - - - - - 195 25 + 7.50 = 32.5 195 / 32.5 = 6 d" | a = 7 * 50
b = a + 555
c = 7 + 50
d = b / c
e = 30 - d
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a ) 23 days , b ) 37 days , c ) 37 Β½ days , d ) 40 days , e ) 41 days | c | inverse(subtract(inverse(multiply(3, 5)), inverse(multiply(divide(5, 4), 20)))) | a does 4 / 5 th of a work in 20 days . he then calls in b and they together finish the remaining work in 3 days . how long b alone would take to do the whole work ? | explanation : a can finish the whole work in 20 Γ 5 / 4 days = 25 days a and b together finish the whole work in 5 Γ 3 days = 15 days therefore , b can finish the whole work in 25 b / 25 + b = 15 25 b = 15 ( 25 + b ) = 375 + 15 b 10 b = 375 and b = 375 / 10 = 37 Β½ days . answer : option c | a = 3 * 5
b = 1/(a)
c = 5 / 4
d = c * 20
e = 1/(d)
f = b - e
g = 1/(f)
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a ) 11 , b ) 17 , c ) 10 , d ) 17 , e ) 18 | e | subtract(add(inverse(add(inverse(36), inverse(12))), 12), const_3) | a and b can do a work in 12 days and 36 days respectively . if they work on alternate days beginning with b , in how many days will the work be completed ? | the work done in the first two days = 1 / 12 + 1 / 36 = 1 / 9 so , 9 such two days are required to finish the work . i . e . , 18 days are required to finish the work . answer : e | a = 1/(36)
b = 1/(12)
c = a + b
d = 1/(c)
e = d + 12
f = e - 3
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a ) 270 m , b ) 245 m , c ) 235 m , d ) 220 m , e ) 240 m | c | subtract(multiply(multiply(45, const_0_2778), 30), 140) | a train , 140 meters long travels at a speed of 45 km / hr crosses a bridge in 30 seconds . the length of the bridge is | "explanation : assume the length of the bridge = x meter total distance covered = 140 + x meter total time taken = 30 s speed = total distance covered / total time taken = ( 140 + x ) / 30 m / s = > 45 Γ£ β ( 10 / 36 ) = ( 140 + x ) / 30 = > 45 Γ£ β 10 Γ£ β 30 / 36 = 140 + x = > 45 Γ£ β 10 Γ£ β 10 / 12 = 140 + x = > 15 Γ£ β ... | a = 45 * const_0_2778
b = a * 30
c = b - 140
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a ) 2 , b ) 4 , c ) 6 , d ) 8 , e ) 10 | c | divide(add(multiply(18, 12), 18), 9) | if 12 : 18 : : x : 9 , then find the value of x | explanation : treat 12 : 18 as 12 / 18 and x : 9 as x / 9 , treat : : as = so we get 12 / 18 = x / 9 = > 18 x = 108 = > x = 6 option c | a = 18 * 12
b = a + 18
c = b / 9
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a ) 30 days , b ) 65 days , c ) 86 days , d ) 45 days , e ) 17 days | a | divide(10, subtract(const_1, divide(add(10, 10), 30))) | a can do a piece of work in 30 days ; b can do the same in 30 days . a started alone but left the work after 10 days , then b worked at it for 10 days . c finished the remaining work in 10 days . c alone can do the whole work in ? | 10 / 30 + 10 / 30 + 10 / x = 1 x = 30 days answer : a | a = 10 + 10
b = a / 30
c = 1 - b
d = 10 / c
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a ) 22 , b ) 27 , c ) 236 , d ) 90 , e ) 81 | d | add(subtract(subtract(divide(subtract(300, 40), add(2, 2)), 10), 10), subtract(subtract(divide(subtract(300, 40), add(2, 2)), 10), 10)) | the distance between towns a and b is 300 km . one train departs from town a and another train departs from town b , both leaving at the same moment of time and heading towards each other . we know that one of them is 10 km / hr faster than the other . find the speeds of both trains if 2 hours after their departure the... | let the speed of the slower train be xx km / hr . then the speed of the faster train is ( x + 10 ) ( x + 10 ) km / hr . in 2 hours they cover 2 x 2 x km and 2 ( x + 10 ) 2 ( x + 10 ) km , respectively . therefore if they did n ' t meet yet , the whole distance from a to b is 2 x + 2 ( x + 10 ) + 40 = 4 x + 602 x + 2 ( ... | a = 300 - 40
b = 2 + 2
c = a / b
d = c - 10
e = d - 10
f = 300 - 40
g = 2 + 2
h = f / g
i = h - 10
j = i - 10
k = e + j
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a ) 12 , b ) 15 , c ) 10 , d ) 20 , e ) 14 | a | divide(subtract(sqrt(add(multiply(multiply(16, 2), const_4), power(16, const_2))), 16), const_2) | tom read a book containing 480 pages by reading the same number of pages each day . if he would have finished the book 2 days earlier by reading 16 pages a day more , how many days did tom spend reading the book ? | "actually u can set up 2 equation p - - stands for the pages d - - stands for the days 1 ) p * d = 480 ( we want to find the days , sop = 480 / d ) 2 ) ( p + 16 ) ( d - 2 ) = 480 = > pd - 2 p + 16 d - 32 = 480 as the 1 ) stated u can put 1 ) into 2 ) = > 480 - 2 p + 16 d - 32 = 480 = > 16 d - 2 p = 32 put the bold one ... | a = 16 * 2
b = a * 4
c = 16 ** 2
d = b + c
e = math.sqrt(d)
f = e - 16
g = f / 2
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a ) 13000 , b ) 7000 , c ) 10000 , d ) 5000 , e ) none of these | d | divide(subtract(75000, 50000), add(const_2, const_3)) | a textile manufacturing firm employees 70 looms . it makes fabrics for a branded company . the aggregate sales value of the output of the 70 looms is rs 00000 and the monthly manufacturing expenses is rs 1 , 50000 . assume that each loom contributes equally to the sales and manufacturing expenses are evenly spread over... | explanation : profit = 5 , 00,000 Γ’ Λ β ( 1 , 50,000 + 75,000 ) = rs . 2 , 75,000 . since , such loom contributes equally to sales and manufacturing expenses . but the monthly charges are fixed at rs 75,000 . if one loan breaks down sales and expenses will decrease . new profit : - = > 500000 Γ£ β ( 69 / 70 ) Γ’ Λ β 1500... | a = 75000 - 50000
b = 2 + 3
c = a / b
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a ) rs . 11.81 , b ) rs . 12 , c ) rs . 18.94 , d ) rs . 12.31 , e ) none | c | divide(multiply(14, add(const_100, 15)), subtract(const_100, 15)) | a fruit seller sells mangoes at the rate of rs . 14 per kg and thereby loses 15 % . at what price per kg , he should have sold them to make a profit of 15 % ? | "solution 85 : 14 = 115 : x x = ( 14 Γ£ β 115 / 85 ) = rs . 18.94 hence , s . p per kg = rs . 18.94 answer c" | a = 100 + 15
b = 14 * a
c = 100 - 15
d = b / c
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a ) 1 , b ) 2 , c ) 3 , d ) 9 , e ) 27 | a | subtract(3, 2) | for a 3 - digit number xyz , where x , y , and z are the digits of the number , f ( xyz ) = 5 ^ x 2 ^ y 3 ^ z . if f ( abc ) = 3 * f ( def ) , what is the value of abc - def ? | since f ( abc ) = 3 * f ( def ) , i would assume that f = c - 1 from the function above . the answer should be ( a ) | a = 3 - 2
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a ) 33 , b ) 44 , c ) 55 , d ) 77 , e ) 22 | c | divide(add(165, 660), multiply(54, const_0_2778)) | how long does a train 165 meters long running at the rate of 54 kmph take to cross a bridge 660 meters in length | "t = ( 660 + 165 ) / 54 * 18 / 5 t = 55 answer : c" | a = 165 + 660
b = 54 * const_0_2778
c = a / b
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a ) 6 hours , b ) 5 hours , c ) 7 hours , d ) 8 hours , e ) none | c | divide(add(392, 70), add(divide(392, 8), 18)) | a truck covers a distance of 392 km at a certain speed in 8 hours . how much time would a car take at an average speed which is 18 kmph more than that of the speed of the truck to cover a distance which is 70 km more than that travelled by the truck ? | "explanation : speed of the truck = distance / time = 392 / 8 = 49 kmph now , speed of car = ( speed of truck + 18 ) kmph = ( 48 + 18 ) = 66 kmph distance travelled by car = 392 + 70 = 462 km time taken by car = distance / speed = 462 / 66 = 7 hours . answer β c" | a = 392 + 70
b = 392 / 8
c = b + 18
d = a / c
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a ) 145646 , b ) 236578 , c ) 645353 , d ) 456546 , e ) 220070 | e | add(multiply(multiply(add(555, 445), 2), subtract(555, 445)), 70) | a no . when divided by the sum of 555 and 445 gives 2 times their difference as quotient & 70 as remainder . find the no . is ? | "( 555 + 445 ) * 2 * 110 + 70 = 220000 + 70 = 220070 e" | a = 555 + 445
b = a * 2
c = 555 - 445
d = b * c
e = d + 70
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a ) 600 , b ) 480 , c ) 750 , d ) 650 , e ) 560 | e | subtract(subtract(multiply(45, 10), multiply(22, 20)), multiply(22, 15)) | the average of 45 results is 10 . the average of first 22 of them is 15 and that of last 22 is 20 . find the 23 result ? | "23 th result = sum of 45 results - sum of 44 results 10 * 45 - 15 * 22 + 20 * 22 = 450 - 330 + 440 = 560 answer is e" | a = 45 * 10
b = 22 * 20
c = a - b
d = 22 * 15
e = c - d
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a ) $ 880 , b ) $ 990 , c ) $ 1,000 , d ) $ 1,100 , e ) $ 1,260 | e | subtract(multiply(140, divide(const_100, 10)), 140) | if a 10 percent deposit that has been paid toward the purchase of a certain product is $ 140 , how much more remains to be paid ? | "10 / 100 p = 140 > > p = 140 * 100 / 10 = 1400 1400 - 140 = 1260 answer : e" | a = 100 / 10
b = 140 * a
c = b - 140
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a ) - $ 56 , b ) - $ 6 , c ) $ 0 , d ) $ 6 , e ) $ 8 | e | subtract(multiply(add(4, 4), 21), multiply(add(7, 21), divide(const_1, add(divide(const_1, add(4, 4)), divide(const_1, 20))))) | a professional janitor can clean a certain high school in ( 4 + 4 ) hours , working at a constant rate . a student sentenced to detention can clean that same high school in 20 hours , also working at a constant rate . if the student is paid $ 7 total per hour and the janitor is paid $ 21 per hour , how much more would ... | a professional janitor can clean a certain high school in ( 4 + 4 ) or 8 hours so ( applying rule # 1 ) , the janitor can clean 1 / 8 of the school in one hour a student sentenced to detention can clean that same high school in 20 hours so ( applying rule # 1 ) , the student can clean 1 / 20 of the school in one hour s... | a = 4 + 4
b = a * 21
c = 7 + 21
d = 4 + 4
e = 1 / d
f = 1 / 20
g = e + f
h = 1 / g
i = c * h
j = b - i
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a ) 1 / 2 , b ) 1 / 10 , c ) 1 / 18 , d ) 1 / 16 , e ) 1 / 11 | a | divide(add(multiply(divide(subtract(8, 2), subtract(6, 3)), 2), 3), add(multiply(6, divide(subtract(8, 2), subtract(6, 3))), 2)) | 3 men and 8 women complete a task in same time as 6 men and 2 women do . how much fraction of work will be finished in same time if 2 men and 3 women will do that task . | "3 m + 8 w = 6 m + 2 w 3 m = 6 w 1 m = 2 w therefore 3 m + 8 w = 14 w 2 m + 3 w = 7 w answer is 7 / 14 = 1 / 2 answer : a" | a = 8 - 2
b = 6 - 3
c = a / b
d = c * 2
e = d + 3
f = 8 - 2
g = 6 - 3
h = f / g
i = 6 * h
j = i + 2
k = e / j
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a ) 105 , b ) 67 , c ) 80 , d ) 60 , e ) 100 | b | multiply(60, add(const_1, divide(12, const_100))) | a light has a rating of 60 watts , it is replaced with a new light that has 12 % higher wattage . how many watts does the new light have ? | "final number = initial number + 12 % ( original number ) = 60 + 12 % ( 60 ) = 60 + 7 = 67 answer b" | a = 12 / 100
b = 1 + a
c = 60 * b
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a ) 50 / 9 , b ) 50 / 13 , c ) 50 / 17 , d ) 25 / 9 , e ) 5 5 / 1 | d | multiply(divide(subtract(divide(7, 6), const_1), 6), const_100) | a sum of money becomes 7 / 6 of itself in 6 years at a certain rate of simple interest . the rate per annum is ? | "let sum = x . then , amount = 7 x / 6 s . i . = 7 x / 6 - x = x / 6 ; time = 6 years . rate = ( 100 * x ) / ( x * 6 * 6 ) = 25 / 9 % . answer : d" | a = 7 / 6
b = a - 1
c = b / 6
d = c * 100
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a ) 50 , b ) 45 , c ) 150 , d ) 450 , e ) 500 | a | divide(0.15, divide(0.3, const_100)) | find the missing figures : 0.3 % of ? = 0.15 | let 0.3 % of x = 0.15 . then , 0.30 * x / 100 = 0.15 x = [ ( 0.15 * 100 ) / 0.3 ] = 50 . answer is a . | a = 0 / 3
b = 0 / 15
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a ) rs . 435 , b ) rs . 350 , c ) rs . 275 , d ) rs . 425 , e ) none of these | a | divide(subtract(multiply(30, 350), multiply(15, 265)), 15) | the mean daily profit made by a shopkeeper in a month of 30 days was rs . 350 . if the mean profit for the first fifteen days was rs . 265 , then the mean profit for the last 15 days would be | average would be : 350 = ( 265 + x ) / 2 on solving , x = 435 . answer : a | a = 30 * 350
b = 15 * 265
c = a - b
d = c / 15
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a ) 26 , b ) 28 , c ) 30 , d ) 32 , e ) 34 | c | add(divide(subtract(multiply(multiply(5, 2), 3), multiply(5, 2)), 2), subtract(multiply(multiply(5, 2), 3), multiply(5, 2))) | the ratio of the number of females to males at a party was 1 : 2 but when 5 females and 5 males left , the ratio became 1 : 3 . how many people were at the party originally ? | "the total number of people are x females + 2 x males . 3 * ( x - 5 ) = 2 x - 5 x = 10 there were 3 x = 30 people at the party originally . the answer is c ." | a = 5 * 2
b = a * 3
c = 5 * 2
d = b - c
e = d / 2
f = 5 * 2
g = f * 3
h = 5 * 2
i = g - h
j = e + i
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a ) 427 , b ) 859 , c ) 869 , d ) 856 , e ) none of these | d | subtract(lcm(lcm(lcm(24, 32), 36), 54), 8) | the least number which when increased by 8 each divisible by each one of 24 , 32 , 36 and 54 is : | "solution required number = ( l . c . m . of 24 , 32 , 36 , 54 ) - 8 = 864 - 8 = 856 . answer d" | a = math.lcm(24, 32)
b = math.lcm(a, 36)
c = math.lcm(b, 54)
d = c - 8
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['a ) 15840', 'b ) 9280', 'c ) 2667', 'd ) 8766', 'e ) 66711'] | b | multiply(square_perimeter(square_edge_by_area(53824)), 10) | find the length of the wire required to go 10 times round a square field containing 53824 m 2 . | a 2 = 53824 = > a = 232 4 a = 928 928 * 10 = 9280 answer : b | a = square_perimeter * (
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a ) rs . 500 , b ) rs . 600 , c ) rs . 650 , d ) rs . 720 , e ) none | e | multiply(multiply(100, divide(add(add(multiply(const_10, const_1000), multiply(const_4, const_1000)), multiply(const_4, 100)), add(100, multiply(100, divide(20, 100))))), divide(4, 100)) | a man invested rs . 14,400 in rs . 100 shares of a company at 20 % premium . if the company declares 4 % 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 . ( 4 / 100 x 12000 ) = rs . 480 . answer e" | a = 10 * 1000
b = 4 * 1000
c = a + b
d = 4 * 100
e = c + d
f = 20 / 100
g = 100 * f
h = 100 + g
i = e / h
j = 100 * i
k = 4 / 100
l = j * k
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a ) 24 , b ) 25 , c ) 26 , d ) 27 , e ) 28 | d | subtract(multiply(add(add(add(add(const_10, const_10), add(const_10, const_10)), add(const_10, const_10)), add(const_4, const_3)), add(add(add(add(const_10, const_10), add(const_10, const_10)), add(const_10, const_10)), add(const_4, const_3))), 4329) | what should be added to 4329 so that it may become a perfect square ? | "66 x 66 = 4356 4356 - 4329 = 27 if added to 27 get perfect square answer = d" | a = 10 + 10
b = 10 + 10
c = a + b
d = 10 + 10
e = c + d
f = 4 + 3
g = e + f
h = 10 + 10
i = 10 + 10
j = h + i
k = 10 + 10
l = j + k
m = 4 + 3
n = l + m
o = g * n
p = o - 4329
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a ) 2220000 , b ) 2850000 , c ) 2121000 , d ) 1855000 , e ) none of these | e | divide(2664000, multiply(divide(add(20, const_100), const_100), divide(add(20, const_100), const_100))) | the income of a company increases 20 % per annum . if its income is 2664000 in the year 1999 what was its income in the year 1997 ? | let income in 1997 = x according to the question , income in 1998 = x + x β 5 = 6 x β 5 income in 1999 = 6 x β 5 + 6 x β 25 = 36 x β 25 but given , income in 1999 = 2664000 β΄ 36 x β 25 = 2664000 β x = 1850000 answer e | a = 20 + 100
b = a / 100
c = 20 + 100
d = c / 100
e = b * d
f = 2664000 / e
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a ) 11 , b ) 17 , c ) 13 , d ) 20 , e ) none of these | d | subtract(add(11, 18), 9) | when 242 is divided by a certain divisor the remainder obtained is 11 . when 698 is divided by the same divisor the remainder obtained is 18 . however , when the sum of the two numbers 242 and 698 is divided by the divisor , the remainder obtained is 9 . what is the value of the divisor ? | "let that divisor be x since remainder is 11 or 18 it means divisor is greater than 18 . now 242 - 11 = 231 = kx ( k is an integer and 234 is divisble by x ) similarly 698 - 18 = 680 = lx ( l is an integer and 689 is divisible by x ) adding both 698 and 242 = ( 231 + 680 ) + 11 + 18 = x ( k + l ) + 29 when we divide th... | a = 11 + 18
b = a - 9
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a ) 220 , b ) 420 , c ) 250 , d ) 700 , e ) 500 | e | divide(multiply(divide(multiply(1100, divide(10, divide(20, 10))), add(const_1, divide(const_1, 10))), 10), multiply(10, 10)) | anil brought a scooter for a certain sum of money . he spent 10 % of the cost on repairs and sold the scooter for a profit of rs . 1100 . how much did he spend on repairs if he made a profit of 20 % ? | e c . p . be rs . x . then , 20 % of x = 1100 20 / 100 * x = 1100 = > x = 5500 c . p . = rs . 5500 , expenditure on repairs = 10 % actual price = rs . ( 100 * 5500 ) / 110 = rs . 5000 expenditures on repairs = ( 5500 - 5000 ) = rs . 500 . | a = 20 / 10
b = 10 / a
c = 1100 * b
d = 1 / 10
e = 1 + d
f = c / e
g = f * 10
h = 10 * 10
i = g / h
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a ) 3 days , b ) 8 days , c ) 9 days , d ) 11 days , e ) 7 days | e | divide(divide(multiply(7, 7), divide(7, 9)), 9) | if 7 persons can do 7 times of a particular work in 7 days , then , 9 persons can do 9 times of that work in ? | that is , 1 person can do one time of the work in 7 days . therefore , 9 persons can do 9 times work in the same 7 days itself . option ' e ' | a = 7 * 7
b = 7 / 9
c = a / b
d = c / 9
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a ) $ 4 , b ) $ 0.1 , c ) $ 1 , d ) $ 3 , e ) $ 1.65 | b | multiply(divide(add(multiply(divide(add(100, 100), 100), divide(3, const_2)), multiply(divide(3, const_2), divide(subtract(100, 90), 100))), 3.15), divide(subtract(100, 90), 100)) | the cost per pound of green tea and coffee were the same in june . in july , the price of coffee shot up by 100 % and that of green tea dropped by 90 % . if in july , a mixture containing equal quantities of green tea and coffee costs $ 3.15 for 3 lbs , how much did a pound of green tea cost in july ? | "lets assume price of coffee in june = 100 x price of green tea in june = 100 x price of coffee in july = 200 x ( because of 100 % increase in price ) price of green tea in july = 10 x ( because of 90 % decrease in price ) price of 1.5 pound of coffee 1.5 pound of green tea in july will be = 300 x + 15 x = 315 x as per... | a = 100 + 100
b = a / 100
c = 3 / 2
d = b * c
e = 3 / 2
f = 100 - 90
g = f / 100
h = e * g
i = d + h
j = i / 3
k = 100 - 90
l = k / 100
m = j * l
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a ) 1 / 25 , b ) 1 / 6 , c ) 1 / 5 , d ) 1 / 3 , e ) 6 | d | divide(divide(4, 6), 6) | if xy > 0 , 1 / x + 1 / y = 4 , and 1 / xy = 6 , then ( x + y ) / 2 = ? | "( 1 / x + 1 / y ) = 4 canbe solved as { ( x + y ) / xy } = 6 . substituting for 1 / xy = 6 , we get x + y = 4 / 6 = = > ( x + y ) / 2 = 4 / ( 6 * 2 ) = 1 / 3 . d" | a = 4 / 6
b = a / 6
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a ) 0.12356 , b ) 1.2356 , c ) 12.356 , d ) 0.012356 , e ) 0.0012356 | b | divide(multiply(0.01, add(add(multiply(multiply(add(const_3, const_2), const_2), multiply(multiply(const_3, const_4), const_100)), multiply(multiply(add(const_3, const_4), add(const_3, const_2)), multiply(add(const_3, const_2), const_2))), add(const_3, const_3))), const_100) | what is 0.01 percent of 12,356 ? | "soln : - 0.01 % of 12,356 = 0 . 011000.01100 x 12,356 = 1100 β 1001100 β 100 x 12,356 = 12,356100 β 10012,356100 β 100 = 1.2356 answer : b" | a = 3 + 2
b = a * 2
c = 3 * 4
d = c * 100
e = b * d
f = 3 + 4
g = 3 + 2
h = f * g
i = 3 + 2
j = i * 2
k = h * j
l = e + k
m = 3 + 3
n = l + m
o = 0 * 1
p = o / 100
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a ) 1 , b ) 2 , c ) 3 , d ) 4 , e ) 5 | a | subtract(subtract(multiply(2500, power(add(const_1, divide(4, const_100)), 2)), 2500), multiply(multiply(2500, divide(4, const_100)), 2)) | indu gave bindu rs . 2500 on compound interest for 2 years at 4 % per annum . how much loss would indu has suffered had she given it to bindu for 2 years at 2 % per annum simple interest ? | "2500 = d ( 100 / 2 ) 2 d = 1 answer : a" | a = 4 / 100
b = 1 + a
c = b ** 2
d = 2500 * c
e = d - 2500
f = 4 / 100
g = 2500 * f
h = g * 2
i = e - h
|
a ) 9 % , b ) 11 % , c ) 15 % , d ) 25 % , e ) 90 % | d | subtract(const_100, multiply(divide(9, 12), const_100)) | a case of 12 rolls of paper towels sells for $ 9 . the cost of one roll sold individually is $ 1 . what is the percent e of savings per roll for the 12 - roll package over the cost of 12 rolls purchased individually ? | "cost of 12 paper towels individually = 1 * 12 = 12 cost of a set of 12 paper towels = 9 cost of one roll = 9 / 12 = 3 / 4 = 0.75 savings per roll = 1 - . 75 = 0.25 % of savings is e = . 25 / 1 * 100 = 25 % d is the answer ." | a = 9 / 12
b = a * 100
c = 100 - b
|
a ) 12 , b ) 9 , c ) 3 , d ) 7.5 , e ) 2.5 | b | divide(multiply(12, 3), const_4.0) | for what values of k will the pair of equations 3 ( 3 x + 4 y ) = 36 and kx + 12 y = 30 does not have a unique solution ? | "we have 2 equations 1 . 3 ( 3 x + 4 y ) = 36 - - > 3 x + 4 y = 12 - - > 9 x + 12 y = 36 2 . kx + 12 y = 30 substract 1 - 2 , we get ( 9 - k ) x = 6 i . e . x = 6 / ( 9 - k ) then , by looking at options , we get some value of x except for b . when we put k = 9 , x becomes 6 / 0 and hence answer is b" | a = 12 * 3
b = a / 4
|
a ) a . 10 , b ) b . 12 , c ) c . 14 , d ) d . 26 , e ) e . 24 | d | add(add(add(add(add(add(add(add(add(add(add(add(const_1, add(3, 5)), const_1), const_1), const_1), const_1), 5), const_1), const_1), const_1), const_1), const_1), const_1) | working at constant rate , pump x pumped out half of the water in a flooded basement in 5 hours . the pump y was started and the two pumps , working independently at their respective constant rates , pumped out rest of the water in 3 hours . how many hours would it have taken pump y , operating alone at its own constan... | "rate of x = 1 / 8 rate of x + y = 1 / 6 rate of y = 1 / 6 - 1 / 8 = 1 / 24 26 hours d" | a = 3 + 5
b = 1 + a
c = b + 1
d = c + 1
e = d + 1
f = e + 1
g = f + 5
h = g + 1
i = h + 1
j = i + 1
k = j + 1
l = k + 1
m = l + 1
|
a ) 3003 , b ) 3027 , c ) 3024 , d ) 3021 , e ) 3018 | a | add(3000, 3) | there 3 kinds of books in the library physics , chemistry and biology . ratio of physics to chemistry is 3 to 2 ; ratio of chemistry to biology is 4 to 3 , and the total of the books is more than 3000 . which one of following can be the total r of the book ? | "first , you have to find the common ratio for all 3 books . you have : p : c : b 3 : 2 - - > multiply by 2 ( gives you row 3 ) 4 : 6 6 : 4 : 3 hence : p : c : b : t ( total ) r 6 : 4 : 3 : 13 - - - - > this means , the total number must be a multiple of 13 . answer a is correct since 299 is divisible by 13 , hence is ... | a = 3000 + 3
|
a ) 5 , b ) 3 , c ) 4 , d ) 2 , e ) 6 | d | subtract(add(const_4, const_3), divide(divide(add(36, 14), const_2), 5)) | on rainy mornings , mo drinks exactly n cups of hot chocolate ( assume that n is an integer ) . on mornings that are not rainy , mo drinks exactly 5 cups of tea . last week mo drank a total of 36 cups of tea and hot chocolate together . if during that week mo drank 14 more tea cups than hot chocolate cups , then how ma... | "t = the number of cups of tea c = the number of cups of hot chocolate t + c = 36 t - c = 14 - > t = 25 . c = 11 . mo drinks 5 cups of tea a day then number of days that are not rainy = 25 / 5 = 5 so number of rainy days = 7 - 5 = 2 d is the answer ." | a = 4 + 3
b = 36 + 14
c = b / 2
d = c / 5
e = a - d
|
a ) 33.5 kg , b ) 37.25 kg , c ) 42.45 kg , d ) 55.12 kg , e ) 29.78 kg | a | divide(add(multiply(16, 20), multiply(14, 25)), add(16, 14)) | there are 2 sections a and b in a class , consisting of 16 and 14 students respectively . if the average weight of section a is 20 kg and that of section b is 25 kg , find the average of the whole class ? | "total weight of 36 + 44 students = 16 * 20 + 14 * 25 = 670 average weight of the class is = 670 / 20 = 33.5 kg answer is a" | a = 16 * 20
b = 14 * 25
c = a + b
d = 16 + 14
e = c / d
|
a ) 10 , b ) 20 , c ) 25 , d ) 30 , e ) 37.5 | b | divide(subtract(add(40, 60), 75), subtract(divide(add(40, 60), 40), divide(75, 60))) | a car traveled 75 % of the way from town a to town b at an average speed of 60 miles per hour . the car travels at an average speed of s miles per hour for the remaining part of the trip . the average speed for the entire trip was 40 miles per hour . what is s ? | total distance = 100 miles ( easier to work with % ) 75 % of the distance = 75 miles 25 % of the distance = 25 miles 1 st part of the trip β 75 / 60 = 1.25 2 nd part of the trip β 25 / s = t total trip β ( 75 + 25 ) / 40 = 1.25 + t Β» 100 / 40 = 1.25 + t Β» 2.5 = 1.25 + t Β» t = 1.25 back to 2 nd part of the trip formula ... | a = 40 + 60
b = a - 75
c = 40 + 60
d = c / 40
e = 75 / 60
f = d - e
g = b / f
|
a ) 70.9 , b ) 75 , c ) 48 , d ) 65 , e ) 67.5 | a | multiply(const_100, divide(subtract(const_100, subtract(subtract(const_100, 35), multiply(subtract(const_100, 35), divide(10, const_100)))), subtract(subtract(const_100, 35), multiply(subtract(const_100, 35), divide(10, const_100))))) | the price of a jacket is reduced by 35 % . during a special sale the price of the jacket is reduced another 10 % . by approximately what percent must the price of the jacket now be increased in order to restore it to its original amount ? | "1 ) let the price of jacket initially be $ 100 . 2 ) then it is decreased by 35 % , therefore bringing down the price to $ 65 . 3 ) again it is further discounted by 10 % , therefore bringing down the price to $ 58.5 4 ) now 58.5 has to be added byx % in order to equal the original price . 58.5 + ( x % ) 58.5 = 100 . ... | a = 100 - 35
b = 100 - 35
c = 10 / 100
d = b * c
e = a - d
f = 100 - e
g = 100 - 35
h = 100 - 35
i = 10 / 100
j = h * i
k = g - j
l = f / k
m = 100 * l
|
a ) 10 , b ) 25 , c ) 40 , d ) 65 , e ) 90 | a | divide(add(multiply(13, 5), multiply(13, 4)), add(5, 4)) | the number of stamps that p and q had were in the ratio of 7 : 3 respectively . after p gave q 13 stamps , the ratio of the number of p ' s stamps to the number of q ' s stamps was 5 : 4 . as a result of the gift , p had how many more stamps than q ? | "p started with 7 k stamps and q started with 3 k stamps . ( 7 k - 13 ) / ( 3 k + 13 ) = 5 / 4 28 k - 15 k = 117 k = 9 p has 7 ( 9 ) - 13 = 50 stamps and q has 3 ( 9 ) + 13 = 40 stamps . the answer is a ." | a = 13 * 5
b = 13 * 4
c = a + b
d = 5 + 4
e = c / d
|
a ) 14 , b ) 25 , c ) 63 , d ) 84 , e ) 252 | b | add(10, const_1) | if x and y are positive integers and 25 x = 10 y what is the least possible value of xy ? | "25 x = 10 y = > x / y = 2 / 5 = > 5 x = 2 y 5 ( 3 ) = 2 ( 3 ) = > x * y = 9 but it is not given 5 ( 5 ) = 2 ( 5 ) = > x * y = 25 b" | a = 10 + 1
|
a ) 1 , b ) 2 , c ) 3 , d ) 4 , e ) 5 | c | subtract(multiply(divide(27, add(const_1, divide(80, 100))), const_2), 27) | a retailer bought a shirt at wholesale and marked it up 80 % to its initial price of $ 27 . by how many more dollars does he need to increase the price to achieve a 100 % markup ? | "let x be the wholesale price . then 1.8 x = 27 and x = 27 / 1.8 = 15 . to achieve a 100 % markup , the price needs to be $ 30 . the retailer needs to increase the price by $ 3 more . the answer is c ." | a = 80 / 100
b = 1 + a
c = 27 / b
d = c * 2
e = d - 27
|
a ) 196 : 121 , b ) 81 : 127 , c ) 181 : 196 , d ) 81 : 161 , e ) 81 : 182 | a | power(divide(2744, 1331), divide(const_1, const_3)) | the ratio of the volumes of two cubes is 2744 : 1331 . what is the ratio of their total surface areas ? | "explanation : ratio of the sides = Β³ β 2744 : Β³ β 1331 = 14 : 11 ratio of surface areas = 14 ^ 2 : 11 ^ 2 = 196 : 121 answer : option a" | a = 2744 / 1331
b = 1 / 3
c = a ** b
|
a ) 224 , b ) 242 , c ) 252 , d ) 262 , e ) 282 | a | divide(28, divide(350, 28)) | evaluate 28 % of 350 + 45 % of 280 | "explanation : = ( 28 / 100 ) * 350 + ( 45 / 100 ) * 280 = 98 + 126 = 224 answer : option a" | a = 350 / 28
b = 28 / a
|
a ) 12000 , 20000 , b ) 9600 , 22400 , c ) 12000 , 20007 , d ) 12000 , 20006 , e ) 12000 , 20001 | b | multiply(subtract(7, const_2), divide(32000, add(3, subtract(7, const_2)))) | divide rs . 32000 in the ratio 3 : 7 ? | 3 / 10 * 32000 = 9600 7 / 8 * 32000 = 22400 answer : b | a = 7 - 2
b = 7 - 2
c = 3 + b
d = 32000 / c
e = a * d
|
a ) 2 second , b ) 4 second , c ) 6 second , d ) 8 second , e ) 10 second | c | divide(110, multiply(add(60, 6), const_0_2778)) | a train is running with a speed of 60 kmph and its length is 110 metres . calculate the time by which it will pass a man running opposite with speed of 6 kmph | explanation : from the given question , we will first calculate the speed of train relative to man , = > ( 60 + 6 ) = 66 km / hr ( we added 6 because man is running opposite ) convert it in metre / second = 66 Γ£ β 5 / 18 = 55 / 3 m / sec time it will take to pass man = 110 Γ£ β 3 / 55 = 6 seconds answer is c | a = 60 + 6
b = a * const_0_2778
c = 110 / b
|
a ) 71 , b ) 44 , c ) 54 , d ) 16 , e ) 18 | c | subtract(66, multiply(4, const_3)) | the total of the ages of amar , akbar and anthony is 66 years . what was the total of their ages 4 years ago ? | explanation : required sum = ( 66 - 3 x 4 ) years = ( 66 - 12 ) years = 54 years . answer : c | a = 4 * 3
b = 66 - a
|
a ) 2 / 5 , b ) 3 / 5 , c ) 8 / 15 , d ) 1 / 2 , e ) 7 / 5 | b | subtract(const_1, add(add(divide(const_3, multiply(add(const_2, const_3), 3)), divide(const_2, multiply(add(const_2, const_3), 3))), divide(const_1, multiply(add(const_2, const_3), 3)))) | kim finds a 3 - meter tree branch and marks it off in thirds and fifths . she then breaks the branch along all the markings and removes one piece of every distinct length . what fraction of the original branch remains ? | 3 pieces of 1 / 5 length and two piece each of 1 / 15 and 2 / 15 lengths . removing one piece each from pieces of each kind of lengths the all that will remain will be 2 pieces of 1 / 5 i . e 2 / 5 , 1 piece of 1 / 15 , and 1 piece of 2 / 15 which gives us 2 / 5 + 1 / 15 + 2 / 15 - - - - - > 3 / 5 answer is b | a = 2 + 3
b = a * 3
c = 3 / b
d = 2 + 3
e = d * 3
f = 2 / e
g = c + f
h = 2 + 3
i = h * 3
j = 1 / i
k = g + j
l = 1 - k
|
a ) 1 / 5 , b ) 1 / 3 , c ) 1 / 2 , d ) 2 / 3 , e ) 3 / 5 | e | subtract(1, divide(divide(8, 12), add(divide(8, 12), 1))) | a certain country is divided into 8 provinces . each province consists entirely of progressives and traditionalists . if each province contains the same number of traditionalists and the number of traditionalists in any given province is 1 / 12 the total number of progressives in the entire country , what fraction of t... | "let p be the number of progressives in the country as a whole . in each province , the number of traditionalists is p / 12 the total number of traditionalists is 8 p / 12 = 2 p / 3 . the total population is p + 2 p / 3 = 5 p / 3 p / ( 5 p / 3 ) = 3 / 5 the answer is e ." | a = 8 / 12
b = 8 / 12
c = b + 1
d = a / c
e = 1 - d
|
a ) 59 , b ) 49 , c ) 58 , d ) 113 , e ) 131 | b | subtract(multiply(const_10, 6), const_10) | n and m are each 3 - digit integers . each of the numbers 1 , 2 , 3 , 45 and 6 is a digit of either n or m . what is the smallest possible positive difference between n and m ? | you have 6 digits : 12 , 3 , 4 , 5 , 6 each digit needs to be used to make two 3 digit numbers . this means that we will use each of the digits only once and in only one of the numbers . the numbers need to be as close to each other as possible . the numbers can not be equal so the greater number needs to be as small a... | a = 10 * 6
b = a - 10
|
a ) 9,000 , b ) 4,500 , c ) 1,750 , d ) 1,000 , e ) 2,000 | b | divide(divide(50, subtract(divide(subtract(divide(const_60, 36), inverse(2)), const_60), inverse(multiply(const_60, 2)))), const_1000) | with both inlets open , a water tank will be filled with water in 36 minutes . the first inlet alone would fill the tank in 2 hours . if in every minutes the second inlet admits 50 cubic meters of water than the first , what is the capacity of the tank ? | "the work done by inlet a and b together in 1 min = 1 / 36 the work done by inlet a ( first inlet ) in 1 min = 1 / 120 the work done by inlet b ( second inlet ) in 1 min = ( 1 / 36 ) - ( 1 / 120 ) = 1 / 51 difference of work done by b and a = b - a = 50 cubic meter i . e . ( 1 / 51 ) - ( 1 / 120 ) = 50 cubic meter i . ... | a = const_60 / 36
b = 1/(2)
c = a - b
d = c / const_60
e = const_60 * 2
f = 1/(e)
g = d - f
h = 50 / g
i = h / 1000
|
a ) 625 deg , b ) 420 deg , c ) 145 deg , d ) 150 deg , e ) 300 deg | b | subtract(multiply(40, multiply(const_3, const_2)), 16) | what is the angle between the hands of a clock when time is 16 : 40 ? | "angle between two hands = 40 h - 11 / 2 m = 40 * 16 - 40 * 11 / 2 = 640 - 220 = 420 deg answer : b" | a = 3 * 2
b = 40 * a
c = b - 16
|
a ) - 3 , b ) - 2 , c ) - 1 , d ) 1 , e ) 2 | a | divide(add(2, 8), add(8, 8)) | solve the equation for x : 8 ( x + y + 2 ) = 8 y - 8 | "a - 3 8 ( x + y + 2 ) = 8 y - 8 8 x + 8 y + 16 = 8 y - 8 8 x + 16 = - 8 8 x = - 24 = > x = - 3" | a = 2 + 8
b = 8 + 8
c = a / b
|
a ) 80 % , b ) 105 % , c ) 115 % , d ) 120 % , e ) 140 % | b | multiply(divide(multiply(subtract(const_1, divide(add(divide(multiply(add(8, subtract(8, const_1)), subtract(const_100, 8)), const_100), 10), 25)), const_100), multiply(subtract(const_1, divide(add(10, divide(multiply(add(8, const_4), subtract(const_100, 8)), const_100)), 20)), const_100)), const_100) | during a special promotion , a certain filling station is offering a 8 percent discount on gas purchased after the first 10 gallons . if kim purchased 20 gallons of gas , and isabella purchased 25 gallons of gas , then isabella β s total per - gallon discount is what percent of kim β s total per - gallon discount ? | "kim purchased 20 gallons of gas . she paid for 4 + 0.9 * 16 = 18.4 gallons , so the overall discount she got was 1.6 / 20 = 8 % . isabella purchased 25 gallons of gas . she paid for 4 + 0.9 * 21 = 22.9 gallons , so the overall discount she got was 2.1 / 25 = 8.4 % . 8.4 / 8 * 100 = 105 % . answer : b ." | a = 8 - 1
b = 8 + a
c = 100 - 8
d = b * c
e = d / 100
f = e + 10
g = f / 25
h = 1 - g
i = h * 100
j = 8 + 4
k = 100 - 8
l = j * k
m = l / 100
n = 10 + m
o = n / 20
p = 1 - o
q = p * 100
r = i / q
s = r * 100
|
a ) rs . 6 , b ) rs . 6.5 , c ) rs . 8 , d ) rs . 5 , e ) rs . 2 | a | divide(90, multiply(const_3, 5)) | 5 men are equal to as many women as are equal to 8 boys . all of them earn rs . 90 only . men Γ’ β¬ β’ s wages are ? | "answer : option a 5 m = xw = 8 b 5 m + xw + 8 b - - - - - 90 rs . 5 m + 5 m + 5 m - - - - - 90 rs . 15 m - - - - - - 90 rs . = > 1 m = 6 rs ." | a = 3 * 5
b = 90 / a
|
a ) 9 , b ) 8 , c ) 11 , d ) 8.5 , e ) 6 | c | divide(33, 3) | stacy has a 33 page history paper due in 3 days . how many pages per day would she have to write to finish on time ? | "33 / 3 = 11 answer : c" | a = 33 / 3
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a ) 3 , b ) 4 , c ) 5 , d ) none , e ) 6 | b | add(2, 2) | the smallest value of n , for which 2 n + 1 is not a prime number , is | "solution ( 2 Γ 1 + 1 ) = 3 . ( 2 Γ 2 + 1 ) = 5 . ( 2 Γ 3 + 1 ) = 7 . ( 2 Γ 4 + 1 ) = 9 . which is not prime , n = 4 . answer b" | a = 2 + 2
|
a ) 150 , b ) 188 , c ) 250 , d ) 288 , e ) 300 | c | multiply(divide(multiply(50, const_1000), const_3600), 18) | a train running at the speed of 50 km / hr crosses a pole in 18 seconds . find the length of the train . | "speed = 50 * ( 5 / 18 ) m / sec = 125 / 9 m / sec length of train ( distance ) = speed * time ( 125 / 9 ) * 18 = 250 meter answer : c" | a = 50 * 1000
b = a / 3600
c = b * 18
|
a ) 4000 , b ) 8877 , c ) 2877 , d ) 2678 , e ) 1011 | a | divide(7000, power(add(subtract(divide(9261, 7000), const_1), const_1), 2)) | what sum of money put at c . i amounts in 2 years to rs . 7000 and in 3 years to rs . 9261 ? | "7000 - - - - 2261 100 - - - - ? = > 32.3 % x * 1323 / 100 * 1323 / 100 = 7000 x * 1.75 = 7000 x = 7000 / 1.75 = > 3999.25 answer : a" | a = 9261 / 7000
b = a - 1
c = b + 1
d = c ** 2
e = 7000 / d
|
a ) $ 245 , b ) $ 255 , c ) $ 265 , d ) $ 275 , e ) $ 285 | c | divide(583, add(divide(120, const_100), const_1)) | two employees m and n are paid a total of $ 583 per week by their employer . if m is paid 120 percent of the salary paid to n , how much is n paid per week ? | "1.2 n + n = 583 2.2 n = 583 n = 265 the answer is c ." | a = 120 / 100
b = a + 1
c = 583 / b
|
a ) 2 , b ) 4 , c ) 6 , d ) 14 , e ) 16 | b | divide(subtract(217, 1), 54) | if remainder is 1 , quotient is 54 and dividend is 217 then what is divisor ? | "we know dividend = divisor * quotient + remainder = = = > 217 = divisor * 54 + 1 = = = = = > 216 / 54 = divisor = = = > divisor = 4 ans - b" | a = 217 - 1
b = a / 54
|
a ) 22 , b ) 35 , c ) 97 , d ) 32 , e ) 25 | c | subtract(negate(25), multiply(subtract(7, 13), divide(subtract(7, 13), subtract(4, 7)))) | 4 , 7 , 13 , 25 , 49 , ( . . . ) | "explanation : 4 4 Γ 2 - 1 = 7 7 Γ 2 - 1 = 13 13 Γ 2 - 1 = 25 25 Γ 2 - 1 = 49 49 Γ 2 - 1 = 97 answer : option c" | a = negate - (
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a ) 1 , b ) 5 , c ) 7 , d ) 9 , e ) 13 | c | subtract(1387, multiply(subtract(const_100, multiply(const_2, const_4)), 15)) | find the least number that must be subtracted from 1387 so that the remaining number is divisible by 15 . | on dividing 1387 by 15 we get the remainder 7 , so 7 should be subtracted . the answer is c . | a = 2 * 4
b = 100 - a
c = b * 15
d = 1387 - c
|
a ) 20 % , b ) 24 % , c ) 30 % , d ) 32 % , e ) 79 % | a | divide(multiply(subtract(add(add(const_100, 5), multiply(add(const_100, 5), divide(20, const_100))), const_100), const_100), add(add(const_100, 5), multiply(add(const_100, 5), divide(20, const_100)))) | the output of a factory was increased by 5 % to keep up with rising demand . to handle the holiday rush , this new output was increased by 20 % . by approximately what percent would the output now have to be decreased in order to restore the original output ? | "the original output increases by 5 % and then 20 % . total % change = a + b + ab / 100 total % change = 5 + 20 + 5 * 20 / 100 = 26 % now , you want to change it to 0 , so , 0 = 26 + x + 26 x / 100 x = - 26 ( 100 ) / 126 = 20 % approximately answer is a" | a = 100 + 5
b = 100 + 5
c = 20 / 100
d = b * c
e = a + d
f = e - 100
g = f * 100
h = 100 + 5
i = 100 + 5
j = 20 / 100
k = i * j
l = h + k
m = g / l
|
a ) 6 , b ) 12 , c ) 24 , d ) 36 , e ) 48 | b | multiply(sqrt(divide(62, 2)), 2) | if n is a positive integer and n ^ 2 is divisible by 62 , then the largest positive integer that must divide n is | "the question asks aboutthe largest positive integer that must divide n , not could divide n . since the least value of n for which n ^ 2 is a multiple of 72 is 12 then the largest positive integer that must divide n is 12 . complete solution of this question is given above . please ask if anything remains unclear . i ... | a = 62 / 2
b = math.sqrt(a)
c = b * 2
|
a ) 4 : 1 , b ) 10 : 1 , c ) 3 : 2 , d ) 2 : 3 , e ) 2 : 5 | b | divide(multiply(multiply(multiply(const_3, const_2), const_100), const_100), divide(multiply(multiply(multiply(const_3, const_2), const_100), const_100), multiply(add(const_2, const_3), const_2))) | if a and b get profits of rs . 60,000 and rs . 6,000 respectively at the end of year then ratio of their investments are | ratio = 60000 / 6000 = 10 : 1 answer : b | a = 3 * 2
b = a * 100
c = b * 100
d = 3 * 2
e = d * 100
f = e * 100
g = 2 + 3
h = g * 2
i = f / h
j = c / i
|
a ) 1 , b ) 2 , c ) 4 , d ) 7 , e ) 9 | c | floor(multiply(divide(add(multiply(divide(99, 9), 2), multiply(divide(99, 11), 3)), 99), const_10)) | what is the 99 th digit after the decimal point in the decimal expansion of 2 / 9 + 3 / 11 ? | 2 / 9 = 0.22222 . . . . = = > 99 th digit is 2 3 / 11 = 0.27272727 . . . . = = > every odd digit is 2 . so , 99 th digit will be 2 . 2 + 2 = 4 answer : c | a = 99 / 9
b = a * 2
c = 99 / 11
d = c * 3
e = b + d
f = e / 99
g = f * 10
h = math.floor(g)
|
a ) 17 , b ) 25 , c ) 27 , d ) 35 , e ) 50 | e | add(multiply(divide(add(multiply(8, const_3), 76), add(add(5, 8), 7)), 8), 10) | the ratio of ages of aman , bren , and charlie are in the ratio 5 : 8 : 7 respectively . if 8 years ago , the sum of their ages was 76 , what will be the age of bren 10 years from now ? | let the present ages of aman , bren , and charlie be 5 x , 8 x and 7 x respectively . 5 x - 8 + 8 x - 8 + 7 x - 8 = 76 x = 5 present age of bren = 8 * 5 = 40 bren ' s age 10 years hence = 40 + 10 = 50 answer = e | a = 8 * 3
b = a + 76
c = 5 + 8
d = c + 7
e = b / d
f = e * 8
g = f + 10
|
['a ) 12300', 'b ) 14500', 'c ) 15400', 'd ) 16700', 'e ) 18200'] | c | divide(circle_area(140), const_4) | a trainer is standing in one corner of a square ground of side 25 m . his voice can be heard upto 140 m . find the area of the ground in which his voice can be heard ? | area covered by goat = pi * r ^ 2 / 4 ( here we divide by 4 because the trainer is standing in a corner of the ground and only in 1 / 4 part , the voice can be heard ) where r = 14 m = length reaching the voice so area = ( 22 / 7 ) * 140 * 140 / 4 = 15400 sq m answer : c | a = circle_area / (
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a ) 1 / 2 , b ) 7 / 13 , c ) 4 / 13 , d ) 8 / 29 , e ) 6 / 33 | b | divide(7, subtract(multiply(7, 2), const_1)) | machine m , n , o working simultaneously machine m can produce x units in 3 / 4 of the time it takes machine n to produce the same amount of units . machine n can produce x units in 2 / 7 the time it takes machine o to produce that amount of units . if all 3 machines are working simultaneously , what fraction of the to... | now ultimately the speed of every machine is given with respect to mach o . so lets assume the speed of o , say 12 hrs to make x units ( assuming 6 because we can see we will need to divide by 3 and 4 mach o makes x units in 12 hrs so , mach n = 2 / 7 of o = 2 / 7 * 12 = 24 / 7 hrs to make x units and mach m = 3 / 4 of... | a = 7 * 2
b = a - 1
c = 7 / b
|
a ) 12 / 28 , b ) 48 / 19 , c ) 16 / 48 , d ) 18 / 48 , e ) 12 / 64 | b | divide(const_1, divide(add(add(inverse(2), inverse(6)), inverse(8)), 2)) | a and b can do a work in 2 days , b and c in 6 days and c and a in 8 days . in how many days will the work be completed , if all three of them work together ? | "one day work of a and b = 1 / 2 one day work of b and c = 1 / 6 one day work of c and a = 1 / 8 2 ( a + b + c ) = 1 / 2 + 1 / 6 + 1 / 8 2 ( a + b + c ) = 19 / 24 ( a + b + c ) = 19 / 48 number of days required = 48 / 19 days . answer : b" | a = 1/(2)
b = 1/(6)
c = a + b
d = 1/(8)
e = c + d
f = e / 2
g = 1 / f
|
a ) 21 years , b ) 22 years , c ) 23 years , d ) 12 years , e ) 16 years | e | divide(subtract(18, subtract(multiply(const_2, const_2), const_2)), subtract(const_2, const_1)) | a man is 18 years older than his son . in two years , his age will be twice the age of his son . the present age of this son is | "explanation : let ' s son age is x , then father age is x + 18 . = > 2 ( x + 2 ) = ( x + 18 + 2 ) = > 2 x + 4 = x + 20 = > x = 16 years option e" | a = 2 * 2
b = a - 2
c = 18 - b
d = 2 - 1
e = c / d
|
a ) 950 , b ) 940 , c ) 980 , d ) 960 , e ) 990 | e | divide(multiply(0.0088, 4.5), multiply(multiply(0.05, 0.1), 0.008)) | ( 0.0088 ) ( 4.5 ) / ( 0.05 ) ( 0.1 ) ( 0.008 ) = | "( 0.0088 ) ( 4.5 ) / ( 0.05 ) ( 0.1 ) ( 0.008 ) = 0.0088 * 450 / 5 * ( 0.1 ) ( 0.008 ) = 0.088 * 90 / 1 * 0.008 = 88 * 90 / 8 = 11 * 90 = 990 answer : e" | a = 0 * 88
b = 0 * 5
c = b * 0
d = a / c
|
a ) s . 1000 , b ) s . 1009 , c ) s . 1007 , d ) s . 1006 , e ) s . 1500 | e | divide(multiply(210, const_100), subtract(add(const_100, const_4), subtract(const_100, 10))) | a watch was sold at a loss of 10 % . if it was sold for rs . 210 more , there would have been a gain of 4 % . what is the cost price ? | explanation : 90 % 104 % - - - - - - - - 14 % - - - - 210 100 % - - - - ? = > rs . 1500 answer : e | a = 210 * 100
b = 100 + 4
c = 100 - 10
d = b - c
e = a / d
|
a ) 1865113 , b ) 1775123 , c ) 1765013 , d ) 1675123 , e ) none of them | c | multiply(4300631, power(add(const_4, const_1), const_4)) | ( 4300631 ) - ? = 2535618 | "let 4300631 - x = 2535618 then x = 4300631 - 2535618 = 1765013 answer is c" | a = 4 + 1
b = a ** 4
c = 4300631 * b
|
a ) 10.6 , b ) 10.9 , c ) 10.4 , d ) 27 , e ) 10.1 | d | divide(add(250, 500), multiply(add(60, 40), const_0_2778)) | two trains 250 m and 500 m long run at the speed of 60 km / hr and 40 km / hr respectively in opposite directions on parallel tracks . the time which they take to cross each other is ? | "relative speed = 60 + 40 = 100 km / hr . = 100 * 5 / 18 = 250 / 9 m / sec . distance covered in crossing each other = 250 + 500 = 750 m . required time = 750 * 9 / 250 = 27 sec answer : d" | a = 250 + 500
b = 60 + 40
c = b * const_0_2778
d = a / c
|
a ) 12 , b ) 24 , c ) 20 , d ) 30 , e ) 36 | a | multiply(add(3, 1), add(1, 2)) | in a rectangular coordinate system , what is the area of a rectangle whose vertices have the coordinates ( - 3 , 1 ) , ( 1 , 1 ) , ( 1 , - 2 ) and ( - 3 , - 2 ) ? | "length of side 1 = 3 + 1 = 4 length of side 2 = 2 + 1 = 3 area of rectangle = 4 * 3 = 12 a is the answer" | a = 3 + 1
b = 1 + 2
c = a * b
|
a ) 45 % , b ) 125 % , c ) 145 % , d ) 158 % , e ) 225 % | d | divide(subtract(22,947, 8,902), 8,902) | in 1970 there were 8,902 women stockbrokers in the united states . by 1978 the number had increased to 22,947 . approximately what was the percent increase ? | "the percent increase is ( 22947 - 8902 ) / 8902 = 14045 / 8902 = 1.58 so the approximate answer is d" | a = 22 - 947
b = a / 8
|
a ) 4 , b ) 5 , c ) 6 , d ) 2 , e ) 8 | d | divide(subtract(const_1, add(multiply(divide(const_1, 4), const_2), multiply(divide(const_1, 8), const_2))), divide(const_1, 8)) | a can finish a piece of work in 4 days . b can do it in 8 days . they work together for two days and then a goes away . in how many days will b finish the work ? | "2 / 4 + ( 2 + x ) / 8 = 1 = > x = 2 days answer : d" | a = 1 / 4
b = a * 2
c = 1 / 8
d = c * 2
e = b + d
f = 1 - e
g = 1 / 8
h = f / g
|
a ) 80 / 6 , b ) 80 / 7 , c ) 80 / 9 , d ) 80 / 2 , e ) 80 / 1 | c | divide(100, multiply(add(90, 72), const_0_2778)) | two trains of length 100 m and 200 m are 100 m apart . they start moving towards each other on parallel tracks , at speeds 90 kmph and 72 kmph . in how much time will the trains cross each other ? | "relative speed = ( 90 + 72 ) * 5 / 18 = 45 mps . the time required = d / s = ( 100 + 100 + 200 ) / 45 = 400 / 45 = 80 / 9 sec . answer : c" | a = 90 + 72
b = a * const_0_2778
c = 100 / b
|
a ) 32 , b ) 43 , c ) 44 , d ) 45 , e ) 46 | a | add(subtract(65, multiply(12, 3)), 3) | a batsman in his 12 th innings makes a score of 65 and thereby increases his average by 3 runs . what is his average after the 12 th innings if he had never been β not out β ? | "let β x β be the average score after 12 th innings β 12 x = 11 Γ ( x β 3 ) + 65 β΄ x = 32 answer a" | a = 12 * 3
b = 65 - a
c = b + 3
|
a ) 14 , b ) 13 , c ) 9 , d ) 6 , e ) 5 | d | add(subtract(add(13, 18), subtract(30, 3)), subtract(18, 13)) | of 30 applicants for a job , 13 had at least 4 years ' experience , 18 had degrees , and 3 had less than 4 years ' experience and did not have a degree . how many of the applicants had at least 4 years ' experience and a degree ? | "d . 6 30 - 3 = 27 27 - 13 - 18 = - 6 then 6 are in the intersection between 4 years experience and degree . answer d" | a = 13 + 18
b = 30 - 3
c = a - b
d = 18 - 13
e = c + d
|
a ) 65 seconds , b ) 46 seconds , c ) 25 seconds , d ) 97 seconds , e ) 26 seconds | c | divide(add(360, 140), divide(multiply(72, const_1000), const_3600)) | a train is 360 meter long is running at a speed of 72 km / hour . in what time will it pass a bridge of 140 meter length ? | "speed = 72 km / hr = 72 * ( 5 / 18 ) m / sec = 20 m / sec total distance = 360 + 140 = 500 meter time = distance / speed = 500 * ( 1 / 20 ) = 25 seconds answer : c" | a = 360 + 140
b = 72 * 1000
c = b / 3600
d = a / c
|
a ) 42 , b ) 10 , c ) 40 , d ) 65 , e ) 15 | e | multiply(multiply(5, 3), 3) | a certain university will select 2 of 3 candidates eligible to fill a position in the mathematics department and 4 of 5 candidates eligible to fill 2 identical positions in the computer science department . if none of the candidates is eligible for a position in both departments , how many different sets of 3 candidate... | "3 c 2 * 5 c 4 = 3 * 5 = 15 the answer is ( e )" | a = 5 * 3
b = a * 3
|
a ) 5 days , b ) 7 days , c ) 8 days , d ) 9 days , e ) 6 days | b | multiply(1, 9) | if 9 spiders make 9 webs in 9 days , then how many days are needed for 1 spider to make 1 web ? | "let , 1 spider make 1 web in a days . more spiders , less days ( indirect proportion ) more webs , more days ( direct proportion ) hence we can write as ( spiders ) 9 : 1 } : : a : 9 ( webs ) 1 : 9 β 7 Γ 1 Γ 7 = 1 Γ 7 Γ a β a = 7 answer : b" | a = 1 * 9
|
a ) 23 sec , b ) 15 sec , c ) 12 sec , d ) 11 sec , e ) 16 sec | c | divide(220, divide(multiply(add(59, 7), const_1000), const_3600)) | a bullet train 220 m long is running with a speed of 59 kmph . in what time will it pass a man who is running at 7 kmph in the direction opposite to that in which the bullet train is going ? | "c 12 sec speed of the bullet train relative to man = ( 59 + 7 ) kmph = 66 * 5 / 18 m / sec = 55 / 3 m / sec . time taken by the bullet train to cross the man = time taken by it to cover 220 m at ( 55 / 3 ) m / sec = ( 220 * 3 / 55 ) sec = 12 sec" | a = 59 + 7
b = a * 1000
c = b / 3600
d = 220 / c
|
a ) 1,108 , b ) 2,100 , c ) 2,108 , d ) 2,160 , e ) 2,256 | d | multiply(divide(208, 22.95), 250) | at the wholesale store you can buy an 8 - pack of hot dogs for $ 1.55 , a 20 - pack for $ 3.05 , and a 250 - pack for $ 22.95 . what is the greatest number of hot dogs you can buy at this store with $ 208 ? | "i think it should be d . i can buy 8 250 - pack for rs 22.95 * 8 = $ 183.60 now , i can buy 8 20 - pack for 3.05 * 8 = $ 24.40 now , i am left with only $ 1.15 . i can not but anything with this . hence total hotdogs = 250 * 8 + 20 * 5 = 2160" | a = 208 / 22
b = a * 250
|
a ) 3 / 1 , b ) 9 / 1 , c ) 3 / 3 , d ) 3 / 5 , e ) 5 / 2 | b | divide(subtract(27, 26), subtract(26, 17)) | two trains running in opposite directions cross a man standing on the platform in 27 seconds and 17 seconds respectively and they cross each other in 26 seconds . the ratio of their speeds is ? | "let the speeds of the two trains be x m / sec and y m / sec respectively . then , length of the first train = 27 x meters , and length of the second train = 17 y meters . ( 27 x + 17 y ) / ( x + y ) = 26 = = > 27 x + 17 y = 26 x + 26 y = = > 1 x = 9 y = = > x / y = 9 / 1 . answer : b" | a = 27 - 26
b = 26 - 17
c = a / b
|
a ) 2.66 cm , b ) 2.5 cm , c ) 3 cm , d ) 3.5 cm , e ) 4 cm | b | multiply(power(subtract(subtract(power(divide(3, 2), const_3), power(divide(add(1, divide(1, 2)), 2), const_3)), 1), inverse(3)), const_2) | the spherical ball of lead 3 cm in diameter is melted and recast into 3 spherical balls . the diameters of two of these are 1 Β½ cm and 2 cm respectively . the diameter of third ball is ? | 4 / 3 Ο * 3 * 3 * 3 = 4 / 3 Ο [ ( 3 / 2 ) 3 + 23 + r 3 ] r = 1.25 d = 2.5 answer : b | a = 3 / 2
b = a ** 3
c = 1 / 2
d = 1 + c
e = d / 2
f = e ** 3
g = b - f
h = g - 1
i = 1/(3)
j = h ** i
k = j * 2
|
a ) 10 , b ) 12 , c ) 14 , d ) 16 , e ) 17 | c | divide(196, subtract(15, const_1)) | find the number which when multiplied by 15 is increased by 196 | "explanation : let the number be x . then , 15 x = x + 196 = βΊ 14 x = 196 = βΊ x = 14 . answer : option c" | a = 15 - 1
b = 196 / a
|
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