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 ) 7 , b ) 1 , c ) 2 , d ) 9 , e ) 8 | c | divide(add(sqrt(subtract(power(8, const_2), multiply(multiply(5, subtract(55, 59)), const_4))), 8), multiply(5, 2)) | calculate the value of y : 5 y ^ 2 - 8 y + 55 = 59 | using the elimination method substitute options for x and find the correct option . answer : c | a = 8 ** 2
b = 55 - 59
c = 5 * b
d = c * 4
e = a - d
f = math.sqrt(e)
g = f + 8
h = 5 * 2
i = g / h
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a ) 8 , b ) 9 , c ) 10 , d ) 11 , e ) 20 | e | divide(subtract(multiply(90, 8), multiply(85, 8)), subtract(92, 90)) | the average ( arithmetic mean ) of all scores on a certain algebra test was 90 . if the average of the 8 male students β grades was 85 , and the average of the female students β grades was 92 , how many female students took the test ? | "total marks of male = m total marks of female = f number of males = 8 number of females = f given : ( m + f ) / ( 8 + f ) = 90 - - - - - - - - - - - - - 1 also given , m / 8 = 85 thus m = 680 - - - - - - - - - 2 also , f / f = 92 thus f = 92 f - - - - - - - - - 3 put 2 and 3 in 1 : we get ( 680 + 92 f ) / ( 8 + f ) = 90 solving this we get f = 20 ans : e" | a = 90 * 8
b = 85 * 8
c = a - b
d = 92 - 90
e = c / d
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a ) 5 min , b ) 2 min , c ) 3 min , d ) 1 min , e ) 10 min | a | multiply(55, divide(1, 11)) | a pipe can fill a cistern in 55 minutes . find the time in which 1 / 11 part of the cistern will be filled ? | "full cistern filled in = 55 minutes 1 / 11 part filled in = 55 * 1 / 11 = 5 minutes answer is a" | a = 1 / 11
b = 55 * a
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['a ) 286 m', 'b ) 899 m', 'c ) 200 m', 'd ) 205 m', 'e ) 187 m'] | d | subtract(divide(600, const_2), 95) | if the perimeter of a rectangular garden is 600 m , its length when its breadth is 95 m is ? | 2 ( l + 95 ) = 600 = > l = 205 m answer : d | a = 600 / 2
b = a - 95
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a ) 50 , b ) 26 , c ) 27 , d ) 28 , e ) 100 | e | divide(multiply(72, divide(multiply(subtract(46, 36), const_1000), const_3600)), const_2) | two trains of equal are running on parallel lines in the same direction at 46 km / hr and 36 km / hr . the faster train passes the slower train in 72 sec . the length of each train is ? | "let the length of each train be x m . then , distance covered = 2 x m . relative speed = 46 - 36 = 10 km / hr . = 10 * 5 / 18 = 25 / 9 m / sec . 2 x / 72 = 25 / 9 = > x = 100 . answer : e" | a = 46 - 36
b = a * 1000
c = b / 3600
d = 72 * c
e = d / 2
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a ) 3 , b ) 4 , c ) 5 , d ) 6 , e ) 7 | a | divide(subtract(multiply(6, 8), multiply(3, 6)), add(6, 4)) | 6 workers should finish a job in 8 days . after 3 days came 4 workers join them . how many days d do they need to finish the same job ? | "let rate of one worker be r = > ( 6 * r ) * 8 = 1 ( rate * time = work ) = > r = 1 / 48 = > work remaining after 3 days 1 - ( 3 * 6 ) / 48 = 30 / 48 after 4 ppl joined in ( ( 6 + 4 ) * time ) / 48 = 30 / 48 time d = 3 days to finish the task imo a" | a = 6 * 8
b = 3 * 6
c = a - b
d = 6 + 4
e = c / d
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a ) 3 : 8 , b ) 3 : 6 , c ) 3 : 7 , d ) 11 : 1 , e ) 3 : 3 | d | divide(11, divide(11, 11)) | what is the ratio between perimeters of two squares one having 11 times the diagonal then the other ? | "d = 11 d d = d a β 2 = 11 d a β 2 = d a = 11 d / β 2 a = d / β 2 = > 11 : 1 answer : d" | a = 11 / 11
b = 11 / a
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a ) 500 , b ) 650 , c ) 400 , d ) 575 , e ) 610 | a | divide(add(150, 50), divide(40, const_100)) | a student has to obtain 40 % of the total marks to pass . he got 150 marks and failed by 50 marks . the maximum marks are ? | "let the maximum marks be x then , 40 % of x = 150 + 50 40 x / 100 = 200 x = 500 answer is a" | a = 150 + 50
b = 40 / 100
c = a / b
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a ) $ 200 , b ) $ 500 , c ) $ 600 , d ) $ 800 , e ) $ 1,200 | b | subtract(divide(add(add(multiply(const_100, const_10), 800), 800), const_2), 800) | david has $ 1,800 at the beginning of his trip , after spending money , he still has exactly $ 800 less than he spent on the trip . how much money does john still have ? | suppose total money spent = x not spend ( money he still has ) = x - 800 x + x - 800 = 1800 x = 1300 money not spend = 1300 - 800 = 500 answer : b | a = 100 * 10
b = a + 800
c = b + 800
d = c / 2
e = d - 800
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a ) 138 , b ) 225 , c ) 300 , d ) 324 , e ) 336 | c | divide(75, subtract(const_1, divide(const_1, const_4))) | what number is 75 more than three - fourth of itself ? | "3 / 4 x + 75 = x that means 75 = 1 / 4 x x = 75 * 4 = 300 c is the answer" | a = 1 / 4
b = 1 - a
c = 75 / b
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a ) 30 , b ) 35 , c ) 40 , d ) 45 , e ) 50 | c | add(multiply(divide(multiply(4, 4), const_2), const_4), 8) | kamal was 4 times as old as his son 8 years ago . after 8 years , kamal will be twice as old as his son . find out the present age of kamal . | let age of the son before 8 years = x then , age of kamal before 8 years ago = 4 x after 8 years , kamal will be twice as old as his son β 4 x + 16 = 2 ( x + 16 ) β x = 8 present age of kamal = 4 x + 8 = 4 Γ 8 + 8 = 40 answer is c . | a = 4 * 4
b = a / 2
c = b * 4
d = c + 8
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a ) 600 , b ) 288 , c ) 1687 , d ) 997 , e ) 121 | a | divide(120, divide(120, const_100)) | 80 % of a number is added to 120 , the result is the same number . find the number ? | "( 80 / 100 ) * x + 120 = x 2 x = 1200 x = 600 answer : a" | a = 120 / 100
b = 120 / a
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a ) 5 , b ) 6 , c ) 7 , d ) 8 , e ) 9 | b | floor(add(divide(log(const_2), log(add(const_1, divide(13, const_100)))), const_1)) | find the least number of complete years in which a sum of money put out at 13 % compound interest will be more than double of itself ? | "6 years answer : b" | a = math.log(2)
b = 13 / 100
c = 1 + b
d = math.log(c)
e = a / d
f = e + 1
g = math.floor(f)
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a ) 1900 , b ) 3500 , c ) 4000 , d ) 2000 , e ) 2500 | b | multiply(subtract(85, divide(subtract(5000, multiply(50, 85)), 50)), 50) | kiran has 85 currency notes in all , some of which were of rs . 100 denomination and the remaining of rs . 50 denomination . the total amount of all these currency notes was rs . 5000 . how much amount did she have in the denomination of rs . 50 ? a . b . c . none of these d . answer : option b explanation : | let the number of 50 β rupee notes be x . then , the number of 100 - rupee notes = ( 85 β x ) 50 x + 100 ( 85 β x ) = 5000 x + 2 ( 85 β x ) = 100 = x = 70 so , required amount = rs . ( 50 x 70 ) = rs . 3500 answer : b | a = 50 * 85
b = 5000 - a
c = b / 50
d = 85 - c
e = d * 50
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a ) rs . 21136.36 , b ) rs . 31136.36 , c ) rs . 51136.36 , d ) rs . 61136.36 , e ) none of these | c | divide(multiply(divide(multiply(54000, const_100), add(const_100, 20)), const_100), subtract(const_100, 12)) | a man sells a car to his friend at 12 % loss . if the friend sells it for rs . 54000 and gains 20 % , the original c . p . of the car was : | "explanation : s . p = rs . 54,000 . gain earned = 20 % c . p = rs . [ 100 / 120 Γ£ β 54000 ] = rs . 45000 this is the price the first person sold to the second at at loss of 12 % . now s . p = rs . 45000 and loss = 12 % c . p . rs . [ 100 / 88 Γ£ β 45000 ] = rs . 51136.36 correct option : c" | a = 54000 * 100
b = 100 + 20
c = a / b
d = c * 100
e = 100 - 12
f = d / e
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a ) 50 , b ) 55 , c ) 60 , d ) 65 , e ) 70 | c | multiply(sqrt(divide(subtract(1202, add(power(4, const_2), power(4, const_2))), 3)), 3) | there are 3 workers in the factory a , b , c . the age of a is 4 more than the age of b and the age of b is 4 more than the age of c . if the sum of the square of the ages is 1202 . what is the sum of their ages ? | let age of b = x ; age of a = x + 4 ; age of c = x - 4 ; sum of squares of ages = > 3 x ^ 2 + 32 = > x = 19.7 = > 20 sum of ages = 60 answer : c | a = 4 ** 2
b = 4 ** 2
c = a + b
d = 1202 - c
e = d / 3
f = math.sqrt(e)
g = f * 3
|
a ) 186 , b ) 188 , c ) 810 , d ) 167 , e ) 861 | c | multiply(120, multiply(divide(54, 12), divide(54, 36))) | if 12 men can reap 120 acres of land in 36 days , how many acres of land can 54 men reap in 54 days ? | "12 men 120 acres 36 days 54 men ? 54 days 120 * 54 / 12 * 54 / 36 10 * 54 * 3 / 2 54 * 15 = 810 answer : c" | a = 54 / 12
b = 54 / 36
c = a * b
d = 120 * c
|
a ) 343 , b ) 232 , c ) 216 , d ) 484 , e ) none of these | a | divide(volume_cube(28), volume_cube(divide(4, const_100))) | how many cubes of 4 cm edge can be cut out of a cube of 28 cm edge | "explanation : number of cubes = ( 28 x 28 x 28 ) / ( 4 x 4 x 4 ) = 343 answer : a" | a = volume_cube / (
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a ) rs . 500 , b ) rs . 1500 , c ) rs . 2200 , d ) rs . 2500 , e ) none of the above | c | multiply(multiply(subtract(4, 3), 1100), 3) | a sum of money is to be distributed among a , b , c , d in the proportion of 5 : 2 : 4 : 3 . if c gets rs . 1100 more than d , what is b ' s share ? | "let the shares of a , b , c and d be rs . 5 x , rs . 2 x , rs . 4 x and rs . 3 x respectively . then , 4 x - 3 x = 1100 x = 1100 . b ' s share = rs . 2 x = rs . ( 2 x 1100 ) = rs . 2200 . answer = c" | a = 4 - 3
b = a * 1100
c = b * 3
|
a ) 221 , b ) 287 , c ) 453 , d ) 288 , e ) 171 | c | divide(500, power(add(const_1, divide(5, const_100)), const_2)) | find the sum lend at c . i . at 5 p . c per annum will amount to rs . 500 in 2 years ? | explanation : 500 = p ( 21 / 20 ) 2 p = 453.51 answer : c | a = 5 / 100
b = 1 + a
c = b ** 2
d = 500 / c
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a ) 270 , b ) 277 , c ) 200 , d ) 257 , e ) 271 | c | multiply(divide(subtract(72, 36), const_3_6), 20) | two trains are moving in the same direction at 72 kmph and 36 kmph . the faster train crosses a man in the slower train in 20 seconds . find the length of the faster train ? | "relative speed = ( 72 - 36 ) * 5 / 18 = 2 * 5 = 10 mps . distance covered in 27 sec = 20 * 10 = 200 m . the length of the faster train = 200 m . answer : c" | a = 72 - 36
b = a / const_3_6
c = b * 20
|
a ) 3 , b ) 6 , c ) 1 / 9 , d ) 1 / 3 , e ) 9 / 16 | e | divide(multiply(1, power(3, const_2)), power(4, const_2)) | x varies inversely as square of y . given that y = 3 for x = 1 . the value of x for y = 4 will be equal to : | "explanation : solution : given x = k / y ^ 2 , where k is constant . now , y = 3 and x = 1 gives k = 9 . . ' . x = 9 / y ^ 2 = > x = 9 / 4 ^ 2 = 9 / 16 answer : e" | a = 3 ** 2
b = 1 * a
c = 4 ** 2
d = b / c
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a ) β 7 , b ) 8 , c ) 10 , d ) 12 , e ) 14 | b | multiply(8, 4) | the sum of all solutions for x in the equation x ^ 2 β 4 x + 14 = | x β 4 | + 8 is equal to : | "x ^ 2 - 4 x + 6 = | x - 4 | rhs can be - ve or + ve x ^ 2 - 5 x + 10 = 0 x ^ 2 - 3 x + 2 = 0 x = 5 , 2,1 we test all 3 values in original equation , all ok . thus , sum = 5 + 2 + 1 = 8 ans ( b )" | a = 8 * 4
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a ) none , b ) two , c ) four , d ) five , e ) seven | c | subtract(subtract(50, const_4), const_4) | r is the set of positive odd integers less than 50 , and s is the set of the squares of the integers in r . how many elements does the intersection of r and s contain ? | "squares < 50 { 1 , 4,9 , 16,25 , 36,49 } s = { 1,9 , 25,49 } r = { 1,3 , . . . . . 49 } hence c ." | a = 50 - 4
b = a - 4
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a ) 7 , b ) 8 , c ) 9 , d ) 10 , e ) 11 | e | subtract(add(add(17, 19), 2), 27) | in a sports club with 27 members , 17 play badminton and 19 play tennis and 2 do not play either . how many members play both badminton and tennis ? | "let x play both badminton and tennis so 17 - x play only badminton and 19 - x play only tennis . 2 play none and there are total 30 students . hence , ( 17 - x ) + ( 19 - x ) + x + 2 = 27 38 - 2 x + x = 27 38 - x = 27 x = 11 so 8 members play both badminton and tennis . e" | a = 17 + 19
b = a + 2
c = b - 27
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a ) 10 days , b ) 20 days , c ) 30 days , d ) 40 days , e ) 50 days | e | divide(inverse(subtract(divide(subtract(const_1, multiply(add(divide(const_1, 20), divide(const_1, 15)), 6)), 5), divide(const_1, 20))), const_2) | a & b can separately do a piece of work in 20 & 15 days . they worked together for 6 days , after which b was replaced by c . the work was finished in next 5 days . the no . of days in which c alone could do the work is | b worked for 6 days = > in 6 days b completed = 6 x 1 15 = 2 5 th work a worked for first 6 days and later 4 days = > in 10 days a completed = 10 x 1 20 = 1 2 of the work total work done by a and b = 2 5 + 1 2 = 4 + 5 10 = 9 10 balance work = 1 - 9 10 = 1 10 th of total work balance work was completed by c in 5 days c can do 1 / 10 of the work in 5 days . therefore c alone can do the work in 50 days . e | a = 1 / 20
b = 1 / 15
c = a + b
d = c * 6
e = 1 - d
f = e / 5
g = 1 / 20
h = f - g
i = 1/(h)
j = i / 2
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a ) 0.02 , b ) ( 0.02 ) ^ 13 , c ) 1 - ( 0.02 ) ^ 13 , d ) ( 0.98 ) ^ 13 , e ) 1 - ( 0.98 ) ^ 13 | e | subtract(const_1, power(subtract(const_1, 0.02), 13)) | a string of 13 light bulbs is wired in such a way that if any individual light bulb fails , the entire string fails . if for each individual light bulb the probability of failing during time period t is 0.02 , what is the probability that the string of light bulbs will fail during the time period t ? | "the string of light bulbs will fail if at least one light bulb fails . so , let ' s find the probability of the opposite event and subtract that value from 1 . the opposite event is when none of the 13 light bulbs fails , since the probability of each light bulb not to fail is 1 - 0.02 = 0.98 the the probability that none of the 13 light bulbs fails is 0.98 ^ 13 . hence , the probability that at least one light bulb fails is 1 - 0.98 ^ 13 . answer : e . now , you should have spotted that your reasoning was not right because of one simple thing , consider the case when we have 100 light bulbs instead of 13 , then according to your logic the probability that the string of light bulbs will fail would be 100 * 0.02 = 2 , which is not possible since the probability of an event can not be more than 1 ( 100 % ) ." | a = 1 - 0
b = a ** 13
c = 1 - b
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a ) 12 liters , b ) 32 liters , c ) 41 liters , d ) 50 liters , e ) 58 liters | e | multiply(divide(145, add(3, 2)), 2) | 145 liters of a mixture of milk and water contains in the ratio 3 : 2 . how much water should now be added so that the ratio of milk and water becomes 3 : 4 ? | "milk = 3 / 5 * 145 = 87 liters water = 58 liters 87 : ( 58 + p ) = 3 : 4 174 + 3 p = 348 = > p = 58 58 liters of water are to be added for the ratio become 3 : 4 . answer : e" | a = 3 + 2
b = 145 / a
c = b * 2
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a ) 91 seconds , b ) 102 seconds , c ) 177 seconds , d ) 200 seconds , e ) 270 seconds | a | divide(add(add(divide(270, 3), divide(274, 3)), divide(276, 3)), 3) | edward ' s home have 3 taps , one tap , working alone at its maximum constant rate , can fill a certain sink in 270 seconds . a second tap working alone at its maximum constant rate , can fill a certain sink in 274 seconds . a third tap , working alone at its maximum constant rate , can fill the same sink in 276 seconds . if all taps work together at their respective maximum constant rates , the time it will take to fill the sink is closest to | tap 1 : 270 secs tap 2 : 274 secs tap 3 : 276 secs considering the average of these 3 taps : 273.33 secs . so 1 tap can fill the tank in 273.33 secs , so 3 taps can fill the tank in 273.33 / 3 = 91.11 secs . closest answer is a . | a = 270 / 3
b = 274 / 3
c = a + b
d = 276 / 3
e = c + d
f = e / 3
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a ) 11 , b ) 15 , c ) 12 , d ) 17 , e ) 19 | c | divide(12, const_1) | the overall age of x and y is 12 year greater than the overall age of y and z . z is how many decades younger that x ? | "c 12 ( x + y ) β ( y + z ) = 12 x β z = 12" | a = 12 / 1
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a ) 1 : 30 , b ) 1 : 5 , c ) 1 : 7 , d ) 1 : 1 , e ) 1 : 6 | a | divide(divide(multiply(1, 1), multiply(3, 2)), divide(multiply(1, 1), multiply(2, 5))) | the compound ratio of 1 : 3 , 1 : 2 and 1 : 5 ? | 1 / 3 * 1 / 2 * 1 / 5 = 1 / 30 1 : 30 answer : a | a = 1 * 1
b = 3 * 2
c = a / b
d = 1 * 1
e = 2 * 5
f = d / e
g = c / f
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a ) 10 , b ) 17.5 , c ) 25 , d ) 12 , e ) 14.5 | b | multiply(divide(140, multiply(400, 2)), const_100) | what is rate of interest if principal . amount be 400 , simple interest 140 and time 2 year . | "s . i = ( p * r * t ) / 100 140 = 800 r / 100 r = 140 / 8 = 17.5 % answer b" | a = 400 * 2
b = 140 / a
c = b * 100
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a ) 14567 , b ) 18451 , c ) 18789 , d ) 18786 , e ) 14400 | b | multiply(12000, power(add(const_1, divide(24, const_100)), 2)) | the population of a village is 12000 . it increases annually at the rate of 24 % p . a . what will be its population after 2 years ? | "formula : ( after = 100 denominator ago = 100 numerator ) 12000 Γ 124 / 100 Γ 124 / 100 = 18451 b" | a = 24 / 100
b = 1 + a
c = b ** 2
d = 12000 * c
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a ) $ 200 , b ) $ 177.78 , c ) $ 100 , d ) $ 277.78 , e ) $ 77.78 | e | divide(multiply(divide(multiply(divide(40, const_100), 400), divide(40, const_100)), divide(40, const_100)), divide(40, const_100)) | a school has received 40 % of the amount it needs for a new building by receiving a donation of $ 400 each from people already solicited . people already solicited represent 40 % of the people from whom the school will solicit donations . how much average contribution is requited from the remaining targeted people to complete the fund raising exercise ? | "let the amount school needs = x let total people school plans to solicit = t school has received 60 % of x = > ( 3 / 5 ) x people already solicited = 40 % of t = > ( 2 / 5 ) t now , as per the information given in the question : ( 3 / 5 ) x = $ 400 . ( 2 / 5 ) . t - - - - - - - - - - - - - - - - - - - - - - - - - - - 1 remaning amount is 40 % i . e ( 2 / 5 ) x - - - - - - because school has already received 60 % and the remaining people are 60 % i . e ( 3 / 5 ) . t - - - - - because 40 % of the people are already solicited so , average contribution required from the remaining targeted people is ( 2 / 5 ) x = ( amount required ) . ( 3 / 5 ) . t - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - 2 divide eqn 1 by eqn 2 amount required = $ 77.78 e" | a = 40 / 100
b = a * 400
c = 40 / 100
d = b / c
e = 40 / 100
f = d * e
g = 40 / 100
h = f / g
|
a ) 2 , b ) 6 , c ) 7 , d ) 8 , e ) 9 | a | divide(divide(16, const_2), const_2) | an engineer designed a ball so that when it was dropped , it rose with each bounce exactly one - half as high as it had fallen . the engineer dropped the ball from a 16 - meter platform and caught it after it had traveled 40 meters . how many times did the ball bounce ? | "division of total distance travelled will be 16 + 16 + 8 ans : 2" | a = 16 / 2
b = a / 2
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a ) $ 245 , b ) $ 150 , c ) $ 260 , d ) $ 220 , e ) $ 250 | e | divide(multiply(80, const_100), multiply(subtract(add(const_1, divide(5, const_100)), subtract(const_1, divide(35, const_100))), const_100)) | a trader sold an item at a loss of 35 % . if he had increased the price by $ 80 he would have made a gain of 5 % . what is the cost price of the item ? | "let c . p . be $ x then 105 % of x - 65 % of x = 100 40 % of x = 100 40 / 100 * x = 100 40 x = 10 000 x = 250 answer is e" | a = 80 * 100
b = 5 / 100
c = 1 + b
d = 35 / 100
e = 1 - d
f = c - e
g = f * 100
h = a / g
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a ) 70 days , b ) 55 days , c ) 24 days , d ) 40 days , e ) 11 days | d | divide(multiply(10, const_3), subtract(divide(add(divide(multiply(10, const_3), 12), add(divide(multiply(10, const_3), 10), divide(multiply(10, const_3), 15))), const_2), divide(multiply(10, const_3), 10))) | a and b can do a piece of work in 10 days , b and c in 15 days , c and a in 12 days . how long will c take to do it ? | "2 c = 1 / 15 + 1 / 12 β 1 / 10 = 1 / 20 c = 1 / 40 = > 40 days answer : d" | a = 10 * 3
b = 10 * 3
c = b / 12
d = 10 * 3
e = d / 10
f = 10 * 3
g = f / 15
h = e + g
i = c + h
j = i / 2
k = 10 * 3
l = k / 10
m = j - l
n = a / m
|
a ) 15.60 , b ) 16.0 , c ) 17.8 , d ) 18.7 , e ) 19.1 | d | add(add(multiply(0.4, subtract(5, 1)), multiply(multiply(0.4, subtract(8, 1)), 5)), 3.1) | a certain telephone company company charges 3.10 cents for the first 1 / 5 of a minute plus 0.40 cents for each additional 1 / 5 of a minute . what would this company charge for a talk time of 8 minutes long ? | first 1 / 5 minute charge = 3.10 cents rest of the minute = 8 - ( 1 / 5 ) = 39 / 5 charge for the rest of the minute = 39 * 0.4 = 15.6 total charge = 3.10 + 15.6 = 18.7 cents answer is d . | a = 5 - 1
b = 0 * 4
c = 8 - 1
d = 0 * 4
e = d * 5
f = b + e
g = f + 3
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a ) 30 , b ) 32 , c ) 36 , d ) 40 , e ) 45 | d | divide(multiply(8, subtract(60, 10)), 10) | a certain number of men complete a piece of work in 60 days . if there were 8 men more , the work could be finished in 10 days less . how many men were originally there ? | originally 1 et there be x men . more men , less days ( x + 8 ) : x β· 60 : 50 so , x + 8 / x = 60 / 50 or x = 40 . answer : d | a = 60 - 10
b = 8 * a
c = b / 10
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a ) 6000 , b ) 6410 , c ) 6440 , d ) 6500 , e ) 6800 | a | divide(multiply(multiply(multiply(7.5, const_100), multiply(6, const_100)), 22.5), multiply(multiply(25, 11.25), 6)) | how many bricks , each measuring 25 cm x 11.25 cm x 6 cm , will be needed to build a wall of 7.5 m x 6 m x 22.5 cm ? | "number of bricks = volume of wall / volume of bricks = 750 x 600 x 22.5 / 25 x 11.25 x 6 = = 6000 answer : a" | a = 7 * 5
b = 6 * 100
c = a * b
d = c * 22
e = 25 * 11
f = e * 6
g = d / f
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a ) 1 / 9 , b ) 2 / 9 , c ) 4 / 9 , d ) 3 / 9 , e ) 2 / 27 | b | divide(4, multiply(2, power(const_3, 2))) | if the function q is defined by the formula q = 5 w / ( 4 vf ( z ^ 2 ) ) , by what factor will q be multiplied if w is quadrupled , f is doubled , and z is tripled ? | "we just need to find the factor thats all , w - > quadrupled - > 4 w f - > doubled - > 2 f z - > tripled - > 3 z hence , z ^ 2 = 9 z ^ 2 w is in numerator , and f * z in denominator . hence , additional factor being introduced = 4 / 2 * 9 = 4 / 18 = 2 / 9 = b" | a = 3 ** 2
b = 2 * a
c = 4 / b
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a ) 0 , b ) 1 , c ) 2 , d ) 3 , e ) 4 | c | subtract(divide(5, const_2), multiply(43, 43)) | what is the remainder when 43 ^ 91 is divided by 5 ? | "the units digit of the exponents of 3 cycle in a group of 4 : { 3 , 9 , 7 , 1 } 91 has the form 4 k + 3 so the units digit of 43 ^ 91 is 7 . the remainder when dividing by 5 is 2 . the answer is c ." | a = 5 / 2
b = 43 * 43
c = a - b
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a ) 16 , b ) 32 , c ) 64 , d ) 128 , e ) 256 | a | multiply(divide(64, power(const_2, 5)), const_2) | if a and b are integers and ( a * b ) ^ 5 = 64 y , y could be : | "distribute the exponent . a ^ 5 * b ^ 5 = 64 y find the prime factorization of 64 . this is 2 ^ 5 * 2 ^ 1 we need 2 ^ 4 ( or some other power of 2 that will give us a multiple of 2 ^ 5 as our second term ) . 2 ^ 4 = 16 the answer is a ." | a = 2 ** 5
b = 64 / a
c = b * 2
|
a ) 6 rs , b ) 2 rs , c ) 4 rs , d ) 9 rs , e ) 3 rs | 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 β s wages are ? | "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 . answer : a" | a = 3 * 5
b = 90 / a
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a ) 1 / 6 , b ) 5 / 42 , c ) 1 / 8 , d ) 3 / 16 , e ) 1 / 4 | a | divide(divide(choose(40, const_1), 40), power(const_3, const_2)) | each factor of 200 is inscribed on its own plastic ball , and all of the balls are placed in a jar . if a ball is randomly selected from the jar , what is the probability that the ball is inscribed with a multiple of 40 ? | "200 = the # of factors 200 has is = 12 ( see below ) ; so out of 12 factors only two are multiples of 40 : 40 and 200 , itself ; so , the probability is 2 / 12 = 1 / 6 . answer : a ." | a = math.comb(40, 1)
b = a / 40
c = 3 ** 2
d = b / c
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a ) 0.9 , b ) 0.75 , c ) 0.6 , d ) 0.8 , e ) 0.5 | a | divide(add(add(50, 30), 10), 100) | a certain bag contains 100 balls β 50 white , 30 green , 10 yellow , 7 red , and 3 purple . if a ball is to be chosen at random , what is the probability that the ball will be neither red nor purple ? | "according to the stem the ball can be white , green or yellow , so the probability is ( white + green + yellow ) / ( total ) = ( 50 + 30 + 10 ) / 100 = 90 / 100 = 0.9 . answer is a" | a = 50 + 30
b = a + 10
c = b / 100
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a ) 38 , b ) 35 , c ) 49 , d ) 43 , e ) 34 | c | subtract(add(const_4, const_4), const_1) | if ( n + 2 ) ! = 2550 ( n ! ) ; find β n β | "( n + 2 ) ! = 2550 ( n ! ) ( n + 2 ) ( n + 1 ) ( n ! ) = 2550 ( n ! ) ( n + 2 ) ( n + 1 ) = 2550 2 n + 3 n + 2 - 2550 = 0 2 n + 3 n - 2548 = 0 2 n + 52 n - 49 n - 2548 = 0 n ( n - 49 ) ( n + 52 ) = 0 n = 49 or n = - 52 answer : c" | a = 4 + 4
b = a - 1
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a ) a ) 31 , b ) b ) 41 , c ) c ) 51 , d ) d ) 67 , e ) e ) 71 | d | subtract(lcm(lcm(2, 3), 11), const_1) | a heap of coconuts is divided into groups of 2 , 3 and 11 and each time one coconut is left over . the least number of coconuts in the heap is ? a . 31 b . 41 c . 51 d . 61 | lcm = 66 = > 66 + 1 = 67 answer : d | a = math.lcm(2, 3)
b = math.lcm(a, 11)
c = b - 1
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a ) 18 , b ) 19 , c ) 20 , d ) 21 , e ) 12 | d | divide(441, subtract(22, const_1)) | along a yard 441 metres long , 22 trees are palnted at equal distances , one tree being at each end of the yard . what is the distance between two consecutive trees | "explanation : 22 trees have 21 gaps between them , required distance ( 441 / 21 ) = 21 option d" | a = 22 - 1
b = 441 / a
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a ) 145 cm 2 , b ) 105 cm 2 , c ) 125 cm 2 , d ) 115 cm 2 , e ) 135 cm 2 | b | multiply(divide(21, const_2), 10) | calculate the area of a triangle , if the sides are 30 cm , 21 cm and 10 cm , what is its area ? | "the triangle with sides 30 cm , 21 cm and 10 cm is right angled , where the hypotenuse is 30 cm . area of the triangle = 1 / 2 * 21 * 10 = 105 cm 2 answer : b" | a = 21 / 2
b = a * 10
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a ) 20 ft , b ) 25 ft , c ) 750 ft , d ) 900 ft , e ) 2400 ft | e | multiply(30, add(divide(multiply(30, 5), subtract(32, 30)), 5)) | the circumference of the front wheel of a cart is 30 ft long and that of the back wheel is 32 ft long . what is the distance traveled by the cart , when the front wheel has done 5 more revolutions than the rear wheel ? | point to note : both the wheels would have traveled the same distance . now consider , no . of revolutions made by back wheel as x , which implies that the number of revolutions made by the front wheel is ( x + 5 ) . equating the distance traveled by front wheel to back wheel : ( x + 5 ) * 30 = x * 32 . ( formula for calculating the distance traveled by each wheel is : # of revolutions * circumference . ) solving this eqn . gives x = 75 . sub x = 75 either in ( x + 5 ) * 30 or in x * 32 to get the distance , which is 2400 . so the correct choice is e . | a = 30 * 5
b = 32 - 30
c = a / b
d = c + 5
e = 30 * d
|
a ) 6 , b ) 8 1 / 1 , c ) 8 1 / 4 , d ) 8 1 / 2 , e ) 8 3 / 7 | a | add(multiply(12, subtract(const_1, multiply(add(inverse(12), inverse(15)), 4))), 4) | two pipes a and b can fill a cistern in 12 and 15 minutes respectively . both are opened together but after 4 minutes a is turned off . after how much more time will the cistern be filled ? | "4 / 12 + ( 4 + x ) / 15 = 1 x = 6 answer : a" | a = 1/(12)
b = 1/(15)
c = a + b
d = c * 4
e = 1 - d
f = 12 * e
g = f + 4
|
a ) 550 , b ) 2500 , c ) 250 , d ) 3000 , e ) 400 | b | divide(multiply(6800, 50), 136) | find the annual income derived by investing $ 6800 in 50 % stock at 136 . | "by investing $ 136 , income obtained = $ 50 . by investing $ 6800 , income obtained = $ [ ( 50 / 136 ) * 6800 ] = $ 2500 . answer b ." | a = 6800 * 50
b = a / 136
|
a ) 1 kmph , b ) 0.5 kmph , c ) 3 kmph , d ) 2.5 kmph , e ) 3.5 kmph | b | divide(subtract(multiply(divide(1, 15), const_60), multiply(divide(1, 20), const_60)), const_2) | a boat moves upstream at the rate of 1 km in 20 minutes and down stream 1 km in 15 minutes . then the speed of the current is : | "rate upstream = ( 1 / 20 * 60 ) = 3 kmph rate down stream = 1 / 15 * 60 = 4 kmph rate of the current = Β½ ( 4 - 3 ) = 0.5 kmph answer : b" | a = 1 / 15
b = a * const_60
c = 1 / 20
d = c * const_60
e = b - d
f = e / 2
|
a ) 10 min , b ) 20 min , c ) 30 min , d ) 40 min , e ) none of these | c | divide(multiply(const_2, const_60), 4) | a motorcyclist x drives along a circular fence at a rate of 2 rounds per hour and another motor cyclist y at a rate of 4 rounds per hour . after how long they will cross each other for the first time ? | explanation : since x and y move in the same direction they cross each other when there is a difference of two round between the two . relative speed of x and y = 4 - 2 = 2 . time taken to complete one round at this speed = 1 / 2 hour = 30 minutes answer : c | a = 2 * const_60
b = a / 4
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a ) 1.5 , b ) 1.75 , c ) 2.14 , d ) 2.34 , e ) 2.64 | c | multiply(divide(subtract(divide(7, add(8, 7)), divide(const_2, add(const_2, const_3))), divide(7, add(8, 7))), add(8, 7)) | a solution contains 8 parts of water for every 7 parts of lemonade syrup . how many parts of the solution should be removed and replaced with water so that the solution will now contain 30 % lemonade syrup ? | "let the total solution is 150 l with 80 l water 70 l syrup . to make 30 % syrup solution , the result solution must have 105 l syrup and 45 l syrup . therefore we are taking 25 l of syrup from initial solution and replacing with water . using urinary method : 70 l syrup in 150 l solution 25 l syrup in 53.6 l solution we started by multiplying 10 now to get to the result we need to divide by 25 = > amount of solution to be replaced with water = ( 53.6 / 25 ) = 2.14 . correct option : c" | a = 8 + 7
b = 7 / a
c = 2 + 3
d = 2 / c
e = b - d
f = 8 + 7
g = 7 / f
h = e / g
i = 8 + 7
j = h * i
|
a ) 140,90 , b ) 180,60 , c ) 160,90 , d ) 150,80 , e ) none of them | d | add(divide(subtract(230, 10), add(const_1, divide(const_1, const_2))), 10) | a third of arun Γ’ β¬ β’ s marks in mathematics exceed a half of his marks in english by 10 . if he got 230 marks in two subjects together how many marks did he got in english ? | "let arun Γ’ β¬ β’ s marks in mathematics and english be x and y then ( 1 / 3 ) x - ( 1 / 2 ) y = 10 2 x - 3 y = 60 Γ’ β¬ Β¦ Γ’ β¬ Β¦ > ( 1 ) x + y = 230 Γ’ β¬ Β¦ Γ’ β¬ Β¦ . > ( 2 ) solving ( 1 ) and ( 2 ) x = 150 and y = 80 answer is d ." | a = 230 - 10
b = 1 / 2
c = 1 + b
d = a / c
e = d + 10
|
a ) 202 , b ) 220 , c ) 210 , d ) 212 , e ) 222 | b | divide(multiply(multiply(multiply(const_12, const_2), 5), subtract(multiply(const_12, const_4), const_4)), multiply(const_12, const_2)) | how many times are the hands of a clock at right angle in 5 days ? | in 1 day , they are at right angles 44 times . in 5 days , they are at right angles 220 times . answer : option b | a = 12 * 2
b = a * 5
c = 12 * 4
d = c - 4
e = b * d
f = 12 * 2
g = e / f
|
a ) $ 31.80 , b ) $ 29.91 , c ) $ 30.15 , d ) $ 30.60 , e ) $ 30.85 | a | subtract(36.57, multiply(36.57, divide(15, const_100))) | sebastian bought a meal at a restaurant and left a 15 % tip . with the tip , he paid exactly $ 36.57 . how much did the meal cost without the tip ? | the tip is a percent increase of 15 % , which is 115 % . let x equal the price before the tip . thus , 115 % of this price equals $ 36.57 : 1.15 x = 36.57 divide both sides by 1.15 : = > x = 36.57 / 1.15 = 31.80 correct answer a ) $ 31.80 | a = 15 / 100
b = 36 * 57
c = 36 - 57
|
a ) 1 , b ) 17 , c ) 18 , d ) 77 , e ) 26 | a | subtract(9, reminder(4499, 9)) | what should be the least number to be added to the 4499 number to make it divisible by 9 ? | "answer : 1 option : a" | a = 9 - reminder
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a ) 155 , b ) 100 / 3 , c ) 61 , d ) 47 , e ) 43 | b | add(divide(3, const_10), power(const_2, add(const_2, const_4))) | let c be defined as the sum of all prime numbers between 0 and 28 . what is c / 3 | "prime numbers between 0 and 30 - 2 , 3 , 5 , 7 , 11 , 13 , 17 , 19 , 23 sum , c = 100 c / 3 = 11 / 3 answer b" | a = 3 / 10
b = 2 + 4
c = 2 ** b
d = a + c
|
a ) 1 / 2 , b ) 1 / 3 , c ) 2 / 3 , d ) 1 / 4 , e ) 1 / 5 | a | divide(divide(2020, 2020), add(divide(2020, 2020), divide(2020, 2020))) | we define that k @ j is the product of j number from k in increasing order for positive integers k , j . for example , 6 @ 4 = 6 * 7 * 8 * 9 . if a = 2020 and b = 2120 , what is the value e of the ratio a / b ? | "e - > a / b = 20 * 21 * β¦ β¦ * 39 / 21 * 22 * β¦ . * 39 * 40 = 20 / 40 = 1 / 2 . therefore , the answer is a ." | a = 2020 / 2020
b = 2020 / 2020
c = 2020 / 2020
d = b + c
e = a / d
|
['a ) 100', 'b ) 120', 'c ) can not be determined', 'd ) 180', 'e ) 200'] | c | multiply(4000, const_1) | how many books each of volume 100 meter cube can be packed into a crate of volume 4000 meter cube ? | gud question with a simple concept . in geo if we want to insert one shape into another we need to know the dimensions of the two shapes . in above with volume given , we can come up with different shapes , so we cant know the answer for ex : 4000 m 3 can be 400 * 10 or 40 * 100 or just 4000 * 1 we do n ' t know , so we cant calculate answer : c | a = 4000 * 1
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a ) 20 , b ) 21 , c ) 22 , d ) 23 , e ) 24 | c | add(divide(20, const_2), subtract(32, 20)) | jane started baby - sitting when she was 20 years old . whenever she baby - sat for a child , that child was no more than half her age at the time . jane is currently 32 years old , and she stopped baby - sitting 10 years ago . what is the current age of the oldest person for whom jane could have baby - sat ? | check two extreme cases : jane = 20 , child = 10 , years ago = 32 - 20 = 12 - - > child ' s age now = 10 + 12 = 22 ; jane = 22 , child = 11 , years ago = 32 - 22 = 10 - - > child ' s age now = 11 + 10 = 21 . answer : c . | a = 20 / 2
b = 32 - 20
c = a + b
|
a ) 5 , b ) 7 , c ) 11 , d ) 13 , e ) 17 | b | divide(add(divide(subtract(multiply(floor(divide(615, 25)), 25), multiply(floor(divide(295, 25)), 25)), 25), const_1), const_2) | if integer k is equal to the sum of all even multiples of 25 between 295 and 615 , what is the greatest prime factor of k ? | if we break down what the stem is asking what is the sum of all mult of 50 between 300 and 600 . using arithmetic progression to find n : 600 = 300 + ( n - 1 ) 50 300 + 50 = 50 n 350 = 50 n = > n = 7 the sum would be : 11 * mean mean = [ 600 + 300 ] / 2 = 450 7 * 450 = 4950 b | a = 615 / 25
b = math.floor(a)
c = b * 25
d = 295 / 25
e = math.floor(d)
f = e * 25
g = c - f
h = g / 25
i = h + 1
j = i / 2
|
a ) 9680 , b ) 4280 , c ) 2789 , d ) 7892 , e ) 2792 | a | add(multiply(multiply(add(divide(2, const_100), divide(divide(subtract(9200, 8000), 3), 8000)), 8000), 3), 8000) | sonika deposited rs . 8000 which amounted to rs . 9200 after 3 years at simple interest . had the interest been 2 % more . she would get how much ? | "( 8000 * 3 * 2 ) / 100 = 480 9200 - - - - - - - - 9680 answer : a" | a = 2 / 100
b = 9200 - 8000
c = b / 3
d = c / 8000
e = a + d
f = e * 8000
g = f * 3
h = g + 8000
|
a ) a ) 10,700 , b ) b ) 10,800 , c ) c ) 10,900 , d ) d ) 11,000 , e ) e ) 5,000 | e | multiply(multiply(const_4, const_2), const_100) | a certain city with a population of 60,000 is to be divided into 11 voting districts , and no district is to have a population that is more than 10 percent greater than the population of any other district what is the minimum possible population that the least populated district could have ? | "let x = number of people in smallest district x * 1.1 = number of people in largest district x will be minimised when the number of people in largest district is maximised 10 * x * 1.1 = 11 x = total number of people in other districts so we have 11 x + x = 60 k x = 5,000 answer : e" | a = 4 * 2
b = a * 100
|
a ) 8 , b ) 1 , c ) 5 , d ) 7 , e ) 3 | e | subtract(const_4, const_3) | the perimeter of a rectangular yard is completely surrounded by a fence that measures 10 meters . what is the length of the yard if the area of the yard is 6 meters squared ? | "perimeter of rectangular yard = 2 ( l + b ) = 10 - - > l + b = 5 area = l * b = 6 b = 5 - l l ( 5 - l ) = 6 5 l - l ^ 2 = 6 l ^ 2 - 5 l + 6 = 0 upon simplifying we get l = 2 or 3 . only 3 is there in the answer choice . answer : e" | a = 4 - 3
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a ) 28.8 sec , b ) 16.8 sec , c ) 15.2 sec , d ) 25.4 sec , e ) 16.2 sec | a | divide(add(150, 250), multiply(50, const_0_2778)) | how long does a train 150 m long traveling at 50 kmph takes to cross a bridge of 250 m in length ? | "d = 150 + 250 = 400 m s = 50 * 5 / 18 = 125 / 9 t = 400 * 125 / 9 = 28.8 sec answer : a" | a = 150 + 250
b = 50 * const_0_2778
c = a / b
|
a ) 2 , b ) 3 , c ) 7 , d ) 14 , e ) 56 | c | lcm(7, 6) | if 7 and 6 are factors of 60 n , what is the minimum value of n ? | "60 n / 7 * 6 should be integer = > 2 * 2 * 3 * 5 * n / 7 * 2 * 3 = 2 * 5 * n / 7 must be an integer for this to be true n must multiple of 7 , thus min of n = 7 hence c" | a = math.lcm(7, 6)
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a ) 150 , b ) 50 , c ) 10 , d ) 15 , e ) 12 | d | divide(add(50, 100), 10) | a train 50 m long passes a platform 100 m long in 10 seconds . the speed of the train in m / sec is ? | speed of train = distance covered / time . = ( 50 + 100 ) / 10 = 15 m / sec . answer : d | a = 50 + 100
b = a / 10
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a ) 30 , b ) 32 , c ) 34 , d ) 36 , e ) 38 | a | subtract(50, multiply(divide(50, const_100), 10)) | how many liters of water must be evaporated from 50 liters of a 4 - percent sugar solution to get a 10 - percent solution ? | "4 % of a 50 liter solution is 2 l which is 10 % of the solution at the end . the solution at the end must be 20 l . we need to evaporate 30 liters . the answer is a ." | a = 50 / 100
b = a * 10
c = 50 - b
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a ) 23 , b ) 25 , c ) 27 , d ) 37 , e ) 38 | e | subtract(subtract(70, 10), subtract(subtract(70, 5), 3)) | in a class of 70 students , 10 did not opt for math . 5 did not opt for science and 3 did not opt for either . how many students of the class opted for both math and science ? | "70 students total 10 did not opt for math 5 did not opt for science 3 did not opt for either total of 40 students in math and 2 did not opt for sci but did for math 40 - 2 = 38 38 students of the class opted for both math and science answer : e" | a = 70 - 10
b = 70 - 5
c = b - 3
d = a - c
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a ) 55684 , b ) 55432 , c ) 55638 , d ) 44530 , e ) 44538 | c | multiply(divide(562, 99), const_100) | 562 Γ£ β 99 = ? | "562 Γ£ β ( 100 - 1 ) 56200 - 562 = 55638 answer c" | a = 562 / 99
b = a * 100
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a ) 298 , b ) 237 , c ) 342 , d ) 396 , e ) 291 | d | subtract(subtract(550, divide(multiply(550, 18), const_100)), divide(multiply(subtract(550, divide(multiply(550, 18), const_100)), 12), const_100)) | the sale price sarees listed for rs . 550 after successive discount is 18 % and 12 % is ? | "550 * ( 88 / 100 ) * ( 82 / 100 ) = 396 answer : d" | a = 550 * 18
b = a / 100
c = 550 - b
d = 550 * 18
e = d / 100
f = 550 - e
g = f * 12
h = g / 100
i = c - h
|
a ) 3 / 5 , b ) 5 / 11 , c ) 1 / 24 , d ) 4 / 9 , e ) 2 / 5 | b | divide(subtract(divide(12, const_2), const_1), subtract(12, const_1)) | an empty wooden vessel weighs 12 % of its total weight when filled with paint . if the weight of a partially filled vessel is one half that of a completely filled vessel , what fraction of the vessel is filled . | "an empty wooden vessel weighs 12 % of its total weight when filled with paint : vessel = 0.12 ( vessel + paint ) ; 12 v = v + p ( so the weight of completely filled vessel is 12 v ) p = 11 v ( so the weight of the paint when the vessels is completely filled is 11 v ) . the weight of a partially filled vessel is one half that of a completely filled vessel : v + p ' = 1 / 2 * 12 v ; p ' = 5 v ( so the weight of the paint when the vessels is partially filled is 5 v ) . what fraction of the vessel is filled ? so , we need to find the ratio of the weight of the paint when the vessel iscompletely filledto the weight of the paint when the vessel ispartially filled : p ' / p = 5 v / 11 v = 5 / 11 . answer : b ." | a = 12 / 2
b = a - 1
c = 12 - 1
d = b / c
|
a ) 191 , b ) 355 , c ) 737 , d ) 840 , e ) 1,500 | e | divide(multiply(750, const_100), subtract(const_100, 50)) | a side of beef lost 50 percent of its weight in processing . if the side of beef weighed 750 pounds after processing , how many pounds did it weigh before processing ? | "let weight of side of beef before processing = x ( 50 / 100 ) * x = 750 = > x = ( 750 * 100 ) / 50 = 1500 answer e" | a = 750 * 100
b = 100 - 50
c = a / b
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a ) 20 % , b ) 30 % , c ) 40 % , d ) 50 % , e ) 70 % | d | multiply(50, const_1) | 50 % of major airline companies equip their planes with wireless internet access . 70 % of major airlines offer passengers free on - board snacks . what is the greatest possible percentage of major airline companies that offer both wireless internet and free on - board snacks ? | "to maximize the percentage of companies offering both , let ' s assume that all 50 % of companies which offer wireless internet also offer snacks . the answer is d ." | a = 50 * 1
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a ) 22 , b ) 20 , c ) 88 , d ) 71 , e ) 10 | e | divide(multiply(multiply(3, const_2), 10), subtract(12, multiply(3, const_2))) | a work which could be finished in 12 days was finished 3 days earlier after 10 more men joined . the number of men employed was ? | x - - - - - - - 12 ( x + 10 ) - - - - 6 x * 12 = ( x + 10 ) 6 x = 10 \ answer : e | a = 3 * 2
b = a * 10
c = 3 * 2
d = 12 - c
e = b / d
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a ) 12 % , b ) 23 % , c ) 14 % , d ) 15 % , e ) 16 % | b | divide(multiply(subtract(multiply(23, add(5, 3)), add(multiply(5, 18), multiply(3, 20))), const_100), add(multiply(5, 18), multiply(3, 20))) | a producer of tea blends two varieties of tea from two tea gardens one costing rs 18 per kg and another rs 20 per kg in the ratio 5 : 3 . if he sells the blended variety at rs 23 per kg , then his gain percent is | explanation : suppose he bought 5 kg and 3 kg of tea . cost price = rs . ( 5 x 18 + 3 x 20 ) = rs . 150 . selling price = rs . ( 8 x 23 ) = rs . 184 . profit = 184 - 150 = 34 so , profit % = ( 34 / 150 ) * 100 = 23 % option b | a = 5 + 3
b = 23 * a
c = 5 * 18
d = 3 * 20
e = c + d
f = b - e
g = f * 100
h = 5 * 18
i = 3 * 20
j = h + i
k = g / j
|
a ) 70 m , b ) 60 m , c ) 80 m , d ) 65 m , e ) 92 m | e | add(multiply(4, divide(divide(69, 4), subtract(4, const_1))), 69) | a can run 4 times as fast as b and gives b a start of 69 m . how long should the race course be so that a and b might reach in the same time ? | "speed of a : speed of b = 4 : 1 means in a race of 4 m a gains 3 m . then in a race of 69 m he gains 69 * ( 4 / 3 ) i . e 92 m answer : e" | a = 69 / 4
b = 4 - 1
c = a / b
d = 4 * c
e = d + 69
|
a ) 32 kmph , b ) 33 kmph , c ) 34 kmph , d ) 35 kmph , e ) 36 kmph | b | divide(add(100, 50), add(divide(100, 30), divide(50, 40))) | a car travels uphill at 30 km / hr and downhill at 40 km / hr . it goes 100 km uphill and 50 km downhill . find the average speed of the car ? | "avg speed = total distance / total time . total distance traveled = 100 + 50 = 150 km ; time taken for uphill journey = 100 / 30 = 10 / 3 ; time taken for down hill journey = 50 / 40 = 5 / 4 ; avg speed = 150 / ( 10 / 3 + 5 / 4 ) = 33 kmph answer : b" | a = 100 + 50
b = 100 / 30
c = 50 / 40
d = b + c
e = a / d
|
a ) $ 4.50 , b ) $ 5.10 , c ) $ 5.30 , d ) $ 7.50 , e ) $ 5.60 | d | add(3.20, multiply(divide(subtract(7.80, 3.20), add(5, subtract(sqrt(2), 1))), 5)) | the price of a bushel of corn is currently $ 3.20 , and the price of a peck of wheat is $ 7.80 . the price of corn is increasing at a constant rate of 5 x cents per day while the price of wheat is decreasing at a constant rate of x ( 2 ^ 1 / 2 ) - x cents per day . what is the approximate price when a bushel of corn costs the same amount as a peck of wheat ? | "i tried using time / rate approach : - initial price difference = 7.80 - 3.20 = 4.60 price of corn increasing by 5 x price of wheat decreasing by x ( 1.4 ) - x = . 4 x since both the quantities are moving towards reducing the price gap hence : - relative increase = 5 x + . 4 x let t be the time by which gap is filled so , 4.6 = t ( 5.4 x ) - > t = ( 4.6 ) / 5.4 x final price = 3.20 + 5 x * t - > 3.20 + 5 * 4.6 / 5.4 = 7.5 answer d ." | a = 7 - 80
b = math.sqrt(2)
c = b - 1
d = 5 + c
e = a / d
f = e * 5
g = 3 + 20
|
a ) 8 % , b ) 10 % , c ) 2 % , d ) 2.4 % , e ) 6 % | d | add(floor(multiply(divide(10, add(10, 400)), const_100)), const_1) | a person saved $ 10 in buying an item on sale . if he spent $ 400 for the item , approximately how much percent he saved in the transaction ? | "actual price = 400 + 10 = $ 410 saving = 10 / 410 * 100 = 100 / 41 = 2.4 % approximately answer is d" | a = 10 + 400
b = 10 / a
c = b * 100
d = math.floor(c)
e = d + 1
|
['a ) 78 : 89', 'b ) 5 : 48', 'c ) 49 : 81', 'd ) 74 : 36', 'e ) 25 : 49'] | c | divide(divide(power(7, const_2), const_2), divide(power(9, const_2), const_2)) | the diagonals of the two squares are in the ratio of 7 : 9 find the ratio of their area | let the diagonals of the square be 7 x and 9 x respectively . ratio of their areas = 1 / 2 ( 7 x ) ^ 2 : 1 / 2 ( 9 x ) = 49 x ^ 2 : 81 x ^ 2 = 49 : 81 answer ( c ) | a = 7 ** 2
b = a / 2
c = 9 ** 2
d = c / 2
e = b / d
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a ) 30 % , b ) 40 % , c ) 50 % , d ) 60 % , e ) 70 % | c | divide(subtract(const_1, divide(80, const_100)), divide(10, const_100)) | in a certain parking lot , 10 % of the cars are towed for parking illegally . however 80 % of the cars which are parked illegally are not towed . what percentage of cars in the parking lot are parked illegally ? | "let x be the number of cars and let y be the number of cars parked illegally . 10 % * x = 20 % * y y / x = 1 / 2 = 50 % the answer is c ." | a = 80 / 100
b = 1 - a
c = 10 / 100
d = b / c
|
a ) 2 / 7 , b ) 16 / 49 , c ) 16 / 25 , d ) 7 / 1 , e ) 7 / 2 | b | divide(power(2, add(2, 2)), power(7, 2)) | for what value of Γ’ β¬ Ε k Γ’ β¬ Β will the equation ( 2 kx 2 + 7 kx + 2 ) = 0 have equal roots ? | "for a 2 nd degree equation ax 2 + bx _ c = 0 has equal roots the condition is b 2 - 4 ac = 0 in the given equation ( 7 k ) ^ 2 - 4 * 2 k * 2 = 0 by solving this equation we get k = 0 , k = 16 / 49 answer : b" | a = 2 + 2
b = 2 ** a
c = 7 ** 2
d = b / c
|
a ) 6720 , b ) 7720 , c ) 7820 , d ) 7950 , e ) 8120 | a | multiply(multiply(multiply(multiply(4, 5), add(const_4, const_2)), add(add(const_4, const_2), const_1)), add(add(add(const_4, const_2), const_1), const_1)) | in how many ways can 5 different rings be worn in 4 particular fingers ? ( some fingers may get more than one ring and some may get no rings . ) can somebody explain ? | first ring can be worn in 4 ways ( on any of the four fingers ) ; second ring can be worn in 5 ways ( as it can go on any of four fingers - 4 ways ; plus it can go below the first one - 1 ) ; third ring can be worn in 6 ways ( the same logic as for the second ring ) ; fourth ring can be worn in 7 ways ; fifth ring can be worn in 8 ways ; total : 4 * 5 * 6 * 7 * 8 = 6720 . ans : a | a = 4 * 5
b = 4 + 2
c = a * b
d = 4 + 2
e = d + 1
f = c * e
g = 4 + 2
h = g + 1
i = h + 1
j = f * i
|
a ) 1 , b ) 5 , c ) 4 , d ) 6 , e ) 8 | b | divide(50, add(multiply(3, 2), 4)) | bag a contains red , white and blue marbles such that the red to white marble ratio is 1 : 3 and the white to blue marble ratio is 2 : 3 . bag b contains red and white marbles in the ratio of 1 : 4 . together , the two bags contain 50 white marbles . how many red marbles could be in bag a ? | 6 is the answer . bag a - r : w : b = 2 : 6 : 9 let w in bag a be 6 k bab b - r : w = 1 : 4 let w in bag b be 4 k w = 50 = 6 k + 4 k = > k = 5 total red ' s in bag a will be 2 k = 10 b | a = 3 * 2
b = a + 4
c = 50 / b
|
a ) 1 / 11 , b ) 1 / 20 , c ) 1 / 19 , d ) 1 / 10 , e ) 1 / 9 | a | divide(const_1, subtract(12, const_1)) | a box contains 6 pairs of shoes ( 12 shoes in total ) . if two shoes are selected at random , what it is the probability that they are matching shoes ? | the problem with your solution is that we do n ' t choose 1 shoe from 12 , but rather choose the needed one after we just took one and need the second to be the pair of it . so , the probability would simply be : 1 / 1 * 1 / 11 as after taking one at random there are 11 shoes left and only one is the pair of the first one ) = 1 / 11 answer : a . | a = 12 - 1
b = 1 / a
|
a ) 20 , b ) 30 , c ) 40 , d ) 32 , e ) 90 | d | multiply(8, divide(lcm(add(5, 3), add(8, 6)), add(8, 6))) | the number of stamps that kaye and alberto had were in the ration of 5 : 3 respectively . after kaye gave alberto 12 of her stamps , the ration of the number of kaye had to the number of alberto had was 8 : 6 . as a result of the gift , kaye had how many more stamps than alberto ? | c k 1 = 5 x a 1 = 3 x k 2 = 5 x - 12 a 2 = 3 x + 12 k 2 / a 2 = 8 / 6 ( 5 x - 12 ) / ( 3 x + 12 ) = 8 / 6 6 * ( 5 x - 12 ) = 8 * ( 3 x + 12 ) 30 x - 72 = 24 x + 96 6 x = 168 x = 28 k 2 = 5 * 28 - 12 = 128 a 2 = 3 * 28 + 12 = 96 k 2 - a 2 = 32 answer : d | a = 5 + 3
b = 8 + 6
c = math.lcm(a, b)
d = 8 + 6
e = c / d
f = 8 * e
|
a ) 27.5 % , b ) 27.4 % , c ) 27.6 % , d ) 27.7 % , e ) 28.4 % | b | subtract(30, divide(30, 2)) | you hold some gold in a vault as an investment . over the past year the price of gold increases by 30 % . in order to keep your gold in the vault , you must pay 2 % of the total value of the gold per year . what percentage has the value of your holdings changed by over the past year . | "( 100 % + 30 % ) * ( 100 % - 2 % ) = 130 * 0.98 = 127.4 % an increase of 27.4 % your gold holdings have increased in value by 27.4 % . the answer is b" | a = 30 / 2
b = 30 - a
|
a ) 12.5 % , b ) 90.9 , c ) 91.7 , d ) 11 % , e ) 1 % | b | multiply(divide(subtract(11, const_1), 11), const_100) | a number x is 11 times another number y . the percentage that y is less than x is | "say y = 1 and x = 11 . then y = 1 is less than x = 11 by ( 11 - 1 ) / 11 * 100 = 10 / 11 * 100 = 90.9 % . answer : b ." | a = 11 - 1
b = a / 11
c = b * 100
|
a ) 2 , b ) 4.3 , c ) 1.68 , d ) 1.5 , e ) 3.2 | c | subtract(1025.68, subtract(floor(1025.68), const_1)) | ravi ' s 4 days average income is 1025.68 . but in a form he should write his average income as the greatest positive even integer less than or equal to his 4 days average income . what is the difference between his real average and form filling average ? | since ravi ' s 4 days average income for form filling is defined as the greatest positive even integer less than or equal to his avg income , then avg income for form filling = 1024 ( the greatest positive even integer less than or equal to 1025.68 is 1024 ) . hence the difference = 1025.68 - 1024 = 1.68 answer : c . | a = math.floor(1025, 68)
b = a - 1
c = 1025 - 68
|
a ) 60 min , b ) 50 min , c ) 70 min , d ) 65 min , e ) 80 min | a | subtract(multiply(5, 15), 15) | roy was suffering from severe headaches . he went to see his doctor and the doctor gave him 5 tablets asking him to take one tablet every 15 minutes . how much time will it take roy to consume all the 5 tablets ? | a 60 min roy will be able to consume all the five tablets in an hour . tablet 1 will be taken in 0 min . tablet 2 will be taken in 15 min . tablet 3 will be taken in 30 min . tablet 4 will be taken in 45 min . tablet 5 will be taken in 60 min . | a = 5 * 15
b = a - 15
|
a ) 100 , b ) 110 , c ) 119 , d ) 120 , e ) 112 | c | add(multiply(2, power(7, const_2)), multiply(3, 7)) | if one root of the equation 2 x ^ 2 + 3 x β k = 0 is 7 , what is the value of k ? | we just enter this root into the equation in order to recieve an equation to find the answer ! 2 * 7 ^ 2 + 3 * 7 - k = 0 k = 98 + 21 = 119 the answer is c | a = 7 ** 2
b = 2 * a
c = 3 * 7
d = b + c
|
a ) s . 9920.736 , b ) s . 10823.20 , c ) s . 10123.20 , d ) s . 10123.29 , e ) s . 10123.28 | a | subtract(multiply(multiply(multiply(const_4, const_100), const_100), power(add(const_1, divide(12, const_100)), 3)), multiply(multiply(const_4, const_100), const_100)) | what will be the compound interest on a sum of rs . 24,500 after 3 years at the rate of 12 % p . a . ? | "explanation : amount = [ 24500 * ( 1 + 12 / 100 ) 3 ] = 24500 * 28 / 25 * 28 / 25 * 28 / 25 = rs . 35123.20 c . i . = ( 34420.74 - 24500 ) = rs . 9920.736 answer : a" | a = 4 * 100
b = a * 100
c = 12 / 100
d = 1 + c
e = d ** 3
f = b * e
g = 4 * 100
h = g * 100
i = f - h
|
a ) 33 , b ) 34 , c ) 35 , d ) 36 , e ) 37 | b | subtract(add(21, 14), const_1) | at a garage sale , all of the items were sold at different prices . if the price of a radio sold at the garage sale was both the 14 th highest price and the 21 st lowest price among the prices of the items sold , how many items were sold at the garage sale ? | there were 13 items sold at a higher price than the radio and 20 items sold at a lower price than the radio . including the radio , there were 13 + 20 + 1 = 34 items sold . the answer is b . | a = 21 + 14
b = a - 1
|
a ) 200 m , b ) 300 m , c ) 600 m , d ) 500 m , e ) 400 m | c | multiply(multiply(72, const_0_2778), 30) | what distance will be covered by a city bus moving at 72 kmph in 30 seconds ? | 72 kmph = 72 * 5 / 18 = 20 mps dist = speed * time = 20 * 30 = 600 m . answer c | a = 72 * const_0_2778
b = a * 30
|
a ) rs . 200 , b ) rs . 250 , c ) rs . 300 , d ) rs . 780 , e ) none of these | d | multiply(multiply(const_3, const_4), divide(260, add(3, 1))) | the cost of 3 pens and 5 pencils is rs . 260 . also the cost of one pen and one pencil is in the ratio of 5 : 1 respectively . what is the cost of one dozen pens ? | explanation : let the cost of one pen is β 5 x β and pencil is β x β 3 x 5 x + 5 x = rs . 260 15 x + 5 x = rs . 260 x = 260 / 20 = 13 : . cost of 1 pen = 5 x = 5 x 13 = 65 : . cost of 12 pens , i . e . ( one dozen ) = 65 x 12 = rs . 780 answer : option d | a = 3 * 4
b = 3 + 1
c = 260 / b
d = a * c
|
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