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 ) 2 , b ) 4 , c ) 6 , d ) 8 , e ) 10 | d | multiply(subtract(const_1, divide(3, 15)), 10) | dan can do a job alone in 15 hours . annie , working alone , can do the same job in just 10 hours . if dan works alone for 3 hours and then stops , how many hours will it take annie , working alone , to complete the job ? | "dan can complete 1 / 15 of the job per hour . in 3 hours , dan completes 3 ( 1 / 15 ) = 1 / 5 of the job . annie can complete 1 / 10 of the job per hour . to complete the job , annie will take 4 / 5 / 1 / 10 = 8 hours . the answer is d ." | a = 3 / 15
b = 1 - a
c = b * 10
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a ) 850 , b ) 1,100 , c ) 1,900 , d ) 2,550 , e ) 3,400 | c | divide(add(add(divide(subtract(add(200, const_3600), multiply(add(add(const_1, const_2), const_3), 200)), const_4), 200), add(add(divide(subtract(add(200, const_3600), multiply(add(add(const_1, const_2), const_3), 200)), const_4), 200), 200)), subtract(multiply(const_2, const_10), const_1)) | alan buys 4 tvs , a 26 inch , a 28 inch , a 30 inch , and a 32 inch , for his new house . each tv costs $ 200 more than the size below it . alan spends a total of $ 3,800 . how much would he have spent if he had bought only the 28 inch and 30 inch tvs ? | assume the cost of the least sized ( 26 inch ) tv = x cost of 28 inches tv = x + 200 cost of 30 inches tv = x + 400 cost of 32 inches tv = x + 600 total cost = 4 x + 1200 = 3800 therefore x = 2600 / 4 = 650 price of 28 inch + 30 inch = 850 + 1050 = 1900 option c | a = 200 + 3600
b = 1 + 2
c = b + 3
d = c * 200
e = a - d
f = e / 4
g = f + 200
h = 200 + 3600
i = 1 + 2
j = i + 3
k = j * 200
l = h - k
m = l / 4
n = m + 200
o = n + 200
p = g + o
q = 2 * 10
r = q - 1
s = p / r
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a ) 1234 , b ) 1367 , c ) 1789 , d ) 1568 , e ) 2500 | e | subtract(negate(50), multiply(subtract(3,5, 7,9), divide(subtract(3,5, 7,9), subtract(1, 3,5)))) | 1 , 3,5 , 7,9 , . 50 find term of sequnce | "this is an arithmetic progression , and we can write down a = 1 a = 1 , d = 2 d = 2 , n = 50 n = 50 . we now use the formula , so that sn = 12 n ( 2 a + ( n β 1 ) l ) sn = 12 n ( 2 a + ( n β 1 ) l ) s 50 = 12 Γ 50 Γ ( 2 Γ 1 + ( 50 β 1 ) Γ 2 ) s 50 = 12 Γ 50 Γ ( 2 Γ 1 + ( 50 β 1 ) Γ 2 ) = 25 Γ ( 2 + 49 Γ 2 ) = 25 Γ ( 2 + 49 Γ 2 ) = 25 Γ ( 2 + 98 ) = 25 Γ ( 2 + 98 ) = 2500 = 2500 . e" | a = negate - (
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a ) 94 , b ) 95 , c ) 93 , d ) 97 , e ) 98 | c | add(divide(subtract(358, 81), 3), const_1) | how many multiples of 3 are there between 81 and 358 ? | "3 * 27 = 81 3 * 119 = 357 total multiples of 3 = ( 119 - 27 ) + 1 = 93 answer is c ." | a = 358 - 81
b = a / 3
c = b + 1
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a ) 118 , b ) 110 , c ) 112 , d ) 126 , e ) 115 | d | multiply(subtract(divide(divide(multiply(subtract(const_100, 10), add(const_100, 40)), const_100), const_100), const_1), const_100) | a trader bought a car at 10 % discount on its original price . he sold it at a 40 % increase on the price he bought it . what percent of profit did he make on the original price ? | "original price = 100 cp = 80 s = 80 * ( 140 / 100 ) = 126 100 - 126 = 26 % answer : d" | a = 100 - 10
b = 100 + 40
c = a * b
d = c / 100
e = d / 100
f = e - 1
g = f * 100
|
a ) 9 , b ) 16.66 , c ) 47 , d ) 48 3 / 5 , e ) 59 | b | divide(multiply(30, 5), 9) | according to the formula f = 9 / 5 ( c ) + 32 , if the temperature in degrees farenheit ( f ) increases by 30 , by how much does the temperature in degrees celsius ( c ) increase ? | "you can plug in values . c = 5 / 9 * ( f - 32 ) f = 32 - - > c = 0 ; f = 32 + 30 = 62 - - > c = 5 / 9 * 30 = 16.66 . increase = 16.66 degrees . answer : b ." | a = 30 * 5
b = a / 9
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a ) 120 m , b ) 180 m , c ) 270 m , d ) 220 m , e ) 280 m | c | divide(18, subtract(divide(18, 10), 9)) | a train covers a distance of 18 km in 10 min . if it takes 9 sec to pass a telegraph post , then the length of the train is ? | "speed = ( 18 / 10 * 60 ) km / hr = ( 108 * 5 / 18 ) m / sec = 30 m / sec . length of the train = 30 * 9 = 270 m . answer : c" | a = 18 / 10
b = a - 9
c = 18 / b
|
a ) 5 , b ) 6 , c ) 8 , d ) 3 , e ) 10 | d | add(1, 1) | the smallest value of n , for which n + 1 is not a prime number , is | "( 1 + 1 ) = 2 . ( 2 + 1 ) = 3 . ( 3 + 1 ) = 4 . ( 4 + 1 ) = 5 . which is not prime , n = 3 . answer : d" | a = 1 + 1
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a ) 9 , b ) 7 , c ) 5 , d ) 3 , e ) 1 | d | divide(subtract(56, 7), subtract(56, 7)) | a number divided by 24 leaves remainder 56 what is the remainder when same number divided by 7 | "add 24 + 56 = 80 now 80 divided by 7 so we get 3 as reaminder answer : d" | a = 56 - 7
b = 56 - 7
c = a / b
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a ) 10.22 % , b ) 20.22 % , c ) 21.22 % , d ) 42.5 % , e ) ca n ' t be calculated | d | divide(multiply(subtract(add(const_100, 14), subtract(const_100, 20)), const_100), subtract(const_100, 20)) | a shop owner professes to sell his articles at certain cost price but he uses false weights with which he cheats by 14 % while buying and by 20 % while selling . what is his percentage profit ? | the owner buys 100 kg but actually gets 114 kg ; the owner sells 100 kg but actually gives 80 kg ; profit : ( 114 - 80 ) / 80 * 100 = ~ 42.5 % answer : d . | a = 100 + 14
b = 100 - 20
c = a - b
d = c * 100
e = 100 - 20
f = d / e
|
a ) 8300 , b ) 8400 , c ) 13500 , d ) 8700 , e ) 9000 | c | subtract(add(9548, 7314), 3362) | 9548 + 7314 = 3362 + ( ? ) | 9548 16862 = 3362 + x + 7314 x = 16862 - 3362 - - - - - = 13500 16862 - - - - - c ) | a = 9548 + 7314
b = a - 3362
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a ) 54.84 % , b ) 36.5 % , c ) 46.5 % , d ) 56.5 % , e ) 66.5 % | a | multiply(divide(subtract(620, add(add(add(50, 82), add(60, 48)), 40)), 620), const_100) | john had a stock of 620 books in his bookshop . he sold 50 on monday , 82 on tuesday , 60 on wednesday , 48 on thursday and 40 on friday . what percentage of the books were not sold ? | "let n be the total number of books sold . hence n = 50 + 82 + 64 + 78 + 135 = 280 let m be the books not sold m = 620 - n = 620 - 280 = 340 percentage books not sold / total number of books = 340 / 620 = 0.55 = 54.84 % correct answer a" | a = 50 + 82
b = 60 + 48
c = a + b
d = c + 40
e = 620 - d
f = e / 620
g = f * 100
|
a ) $ 1500 , b ) $ 150 , c ) $ 375 , d ) $ 200 , e ) $ 450 | c | divide(divide(1500, subtract(divide(6, 2), const_1)), 2) | a sum of money is distributed among w , x , y , z in the proportion of 1 : 6 : 2 : 4 . if x gets $ 1500 more than y , what is the w ' s share ? | let the shares of w , x , y , z are 1 a , 6 a , 2 a , 4 a 6 a - 2 a = 1500 4 a = $ 1500 , a = 375 w ' s share = 1 a = $ 375 answer is c | a = 6 / 2
b = a - 1
c = 1500 / b
d = c / 2
|
a ) 500 , b ) 387 , c ) 298 , d ) 267 , e ) 191 | a | divide(add(125, 40), divide(33, const_100)) | a student has to obtain 33 % of the total marks to pass . he got 125 marks and failed by 40 marks . the maximum marks are : | "explanation : given that the student got 125 marks and still he failed by 40 marks = > the minimum pass mark = 125 + 40 = 165 given that minimum pass mark = 33 % of the total mark answer : a ) 500" | a = 125 + 40
b = 33 / 100
c = a / b
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a ) 6.9 , b ) 7.4 , c ) 8.7 , d ) 9.3 , e ) 10.4 | c | divide(multiply(15, 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 15 percent of 58 ? | "( 15 / 100 ) * 58 = 8.7 the answer is c ." | 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 = 15 * n
p = o / 100
|
a ) 20 , b ) 50 , c ) 100 , d ) 400 , e ) 200 | e | add(power(multiply(const_2.0, 4), 4), power(multiply(4, 4), 4)) | if equation | x / 4 | + | y / 4 | = 2.5 encloses a certain region on the coordinate plane , what is the area of this region ? | the equation can be reduced to intercept form as | x / 10 | + | y / 10 | = 1 , so these are lines in four quadrants with x and y intercept as 10 , so it is a rhombus with diagonals of 20 each and hence area = 1 / 2 * d 1 * d 2 = 1 / 2 * 20 * 20 = 200 . answer e . | a = 2 * 0
b = a ** 4
c = 4 * 4
d = c ** 4
e = b + d
|
a ) 1 , b ) 3.5 , c ) 20 , d ) 49 , e ) 60 | e | divide(1, divide(1, 60)) | if 60 honey bees make 60 grams of honey in 60 days , then 1 honey bee will make 1 gram of honey in how many days ? | "explanation : let the required number days be x . less honey bees , more days ( indirect proportion ) less honey , less days ( direct proportion ) honey bees 1 : 60 : : 60 : x honey 60 : 1 = > 1 x 60 x x = 60 x 1 x 60 = > x = 60 . answer : e" | a = 1 / 60
b = 1 / a
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a ) 1 / 2 , b ) 1 / 3 , c ) 1 / 4 , d ) 1 / 6 , e ) 1 / 7 | a | divide(7, divide(multiply(7, subtract(7, const_3)), const_2)) | what is the probability of randomly selecting one of the shortest diagonals from all the diagonals of a regular 7 - sided polygon ) ? | "from any vertex , there are two vertices on sides , which do not make a diagonal but a side . so the remaining n - 3 vertices make diagonals . there are 2 of these diagonals which are the shortest . the probability of choosing one of the shortest diagonals is 2 / 4 = 1 / 2 . the answer is a ." | a = 7 - 3
b = 7 * a
c = b / 2
d = 7 / c
|
a ) 36 days , b ) 24 days , c ) 46 days , d ) 16 days , e ) 14 days | b | inverse(divide(inverse(16), add(const_2, const_1))) | a is twice as good a work man as b and together they finish a piece of work in 16 days . the number of days taken by a alone to finish the work is : | "solution ( a β s 1 day β s work ) : ( b β s 1 day β s work ) = 2 : 1 . ( a + b ) ' s 1 day β s work = 1 / 16 divide 1 / 14 in the ratio 2 : 1 . β΄ a β s 1 day β s work = ( 1 / 14 x 2 / 3 ) = 1 / 24 hence , a alone can finish the work in 24 days . answer b" | a = 1/(16)
b = 2 + 1
c = a / b
d = 1/(c)
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['a ) 11.25 cm .', 'b ) 12.25 cm', 'c ) 13.25 cm', 'd ) 14.25 cm', 'e ) none of these'] | a | divide(volume_cube(15), multiply(15, 20)) | a cube of edge 15 cm is imersed completely in a rectangular vessel containing water . if the dimensions of the base of vessel are 20 cm * 15 cm , find the rise in waer level . | sol . increase in volume = volume of the cube = ( 15 * 15 * 15 ) cm Β³ . β΄ rise in water level = [ volume / area ] = [ 15 * 15 * 15 / 20 * 15 ] cm = 11.25 cm . answer a | a = volume_cube / (
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a ) 16.16 % , b ) 15.55 % , c ) 14.44 % , d ) 13.33 % , e ) 12.22 % | d | subtract(const_100, multiply(multiply(add(const_1, divide(7, const_100)), subtract(const_1, divide(19, const_100))), const_100)) | a volunteer organization is recruiting new members . in the fall they manage to increase their number by 7 % . by the spring however membership falls by 19 % . what is the total change in percentage from fall to spring ? | "( 100 % + 7 % ) * ( 100 % - 19 % ) = 1.07 * . 81 = 0.8667 1 - 0.8667 = 13.33 % lost = - 13.33 % the answer is d the organization has lost 13.33 % of its total volunteers from fall to spring ." | a = 7 / 100
b = 1 + a
c = 19 / 100
d = 1 - c
e = b * d
f = e * 100
g = 100 - f
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a ) 0.33 , b ) 0.44 , c ) 0.55 , d ) 44 , e ) 55 | d | multiply(divide(divide(8, divide(const_1, const_2)), const_12), multiply(0.33, 100)) | in a market , a dozen eggs cost as much as a pound of rice , and a half - liter of kerosene costs as much as 8 eggs . if the cost of each pound of rice is $ 0.33 , then how many e cents does a liter of kerosene cost ? [ one dollar has 100 cents . ] | "main thing to remember is answer is asked in cents , however when we calculate , it comes up as 0.44 $ just multiply by 100 , answer e = 44 . d" | a = 1 / 2
b = 8 / a
c = b / 12
d = 0 * 33
e = c * d
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a ) 13 , b ) 12 , c ) 9 , d ) 8 , e ) 7 | b | divide(subtract(subtract(divide(150, const_2), 1), subtract(150, add(divide(150, const_2), divide(150, 6)))), const_2) | half of the workers in palabras bookstore have read the latest book by j . saramago , and 1 / 6 of the workers have read the latest book by h . kureishi . the number of workers that have read neither book is one less than the number of the workers that have read the latest saramago book and have not read the latest kureishi book . if there are 150 workers in the palabras bookstore , how many of them have read both books ? | "there are total 150 workers . half of the workers in palabras bookstore have read the latest book by j . saramago , so 75 have read saramago . 1 / 6 of the workers have read the latest book by h . kureishi . so ( 1 / 6 ) * 150 = 25 have read kureishi the number of workers that have read neither book is one less than the number of the workers that have read the latest saramago book and have not read the latest kureishi book if b workers have read both books , 20 - b have read saramago but not kureishi . so , ( 75 - b - 1 ) have read neither . total = n ( a ) + n ( b ) - both + neither 150 = 75 + 25 - b + ( 75 - b - 1 ) b = 12 answer ( b )" | a = 150 / 2
b = a - 1
c = 150 / 2
d = 150 / 6
e = c + d
f = 150 - e
g = b - f
h = g / 2
|
a ) 0.0414 , b ) 0.144 , c ) 0.0144 , d ) 0.414 , e ) none of them | c | divide(multiply(14.4, 0.144), 144) | if 144 / 0.144 = 14.4 / x , then the value of x is : | = 144 / 0.144 = 14.4 / x = 144 x 1000 / 144 = 14.4 / x = x = 14.4 / 1000 = 0.0144 answer is c . | a = 14 * 4
b = a / 144
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a ) 2 : 12 , b ) 5 : 12 , c ) 7 : 12 , d ) 1 : 12 , e ) 3 : 12 | d | divide(6000, 72000) | ravi and kavi start a business by investing Γ’ β ΒΉ 6000 and Γ’ β ΒΉ 72000 , respectively . find the ratio of their profits at the end of year . | "ratio of profit = ratio of investments = 6000 : 72000 = 1 : 12 answer : d" | a = 6000 / 72000
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a ) 20 % , b ) 30 % , c ) 40 % , d ) 70 % , e ) 60 % | e | subtract(100, 40) | john want to buy a $ 100 trouser at the store , but he think it β s too expensive . finally , it goes on sale for $ 40 . what is the percent decrease ? | "the is always the difference between our starting and ending points . in this case , it β s 100 β 40 = 60 . the β original β is our starting point ; in this case , it β s 100 . ( 60 / 100 ) * 100 = ( 0.6 ) * 100 = 60 % . e" | a = 100 - 40
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a ) 11 , b ) 12 , c ) 13 , d ) 8 , e ) 15 | d | divide(7, subtract(divide(multiply(const_2, 7), 8), 1)) | it takes joey the postman 1 hours to run a 7 mile long route every day . he delivers packages and then returns to the post office along the same path . if the average speed of the round trip is 8 mile / hour , what is the speed with which joey returns ? | "let his speed for one half of the journey be 7 miles an hour let the other half be x miles an hour now , avg speed = 8 mile an hour 2 * 7 * x / 7 + x = 8 14 x = 8 x + 56 = > x = 8 d" | a = 2 * 7
b = a / 8
c = b - 1
d = 7 / c
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a ) 14 , b ) 15 , c ) 61 , d ) 17 , e ) 18 | b | multiply(divide(const_100, 12), multiply(divide(9, const_100), 20)) | a man buys rs . 20 shares paying 9 % dividend . the man wants to have an interest of 12 % on his money . the market value of each share is ? | "dividend on rs . 20 = rs . ( 9 / 100 ) x 20 = rs . 9 / 5 . rs . 12 is an income on rs . 100 . rs . 9 / 5 is an income on rs . [ ( 100 / 12 ) x ( 9 / 5 ) ] = rs . 15 ? answer : b" | a = 100 / 12
b = 9 / 100
c = b * 20
d = a * c
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a ) 40 , b ) 44 , c ) 48 , d ) 52 , e ) 56 | b | divide(110, multiply(add(6, 3), const_0_2778)) | the speed at which a man can row a boat in still water is 6 km / hr . if he rows downstream , where the speed of current is 3 km / hr , how many seconds will he take to cover 110 meters ? | "the speed of the boat downstream = 6 + 3 = 9 km / hr 9 km / hr * 5 / 18 = 2.5 m / s the time taken to cover 110 meters = 110 / 2.5 = 44 seconds . the answer is b ." | a = 6 + 3
b = a * const_0_2778
c = 110 / b
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a ) 30 , b ) 36 , c ) 60 , d ) data inadequate , e ) none of these | b | divide(multiply(13.5, 24), 9) | 24 buckets of water fill a tank when the capacity of each bucket is 13.5 litres . how many buckets will be required to fill the same tank if the capacity of each bucket is 9 litres ? | "capacity of the tank = 24 Γ£ β 13.5 = 324 litres when the capacity of each bucket = 9 litres , then the required no . of buckets = 324 Γ’ Β β 9 = 36 answer b" | a = 13 * 5
b = a / 9
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a ) 5 / 21 , b ) 3 / 7 , c ) 4 / 7 , d ) 5 / 7 , e ) 2 / 7 | e | add(multiply(divide(4, 7), divide(subtract(7, 1), subtract(7, 1))), multiply(divide(4, 7), divide(subtract(7, 4), subtract(7, 1)))) | in a room filled with 7 people , 4 people have exactly 1 sibling in the room and 3 people have exactly 1 siblings in the room . if two individuals are selected from the room at random , what is the probability that those two individuals are not siblings ? | "there are suppose a b c d e f g members in the room 4 people who have exactly one sibling . . . . a b c d . . . . ( a is bs β sssibl β g β ssand β ssviceversa ) β ss ( c β ssis β ssds β sssibl β g β ssand β ssviceversa ) β ss ( c β ssis β ssdssibl β gandviceversa ) ( cisds sibling and viceversa ) ( c is ds sibling and viceversa ) . . . now remaning efg are 3 people who have exactly 1 siblings . . . . ( e has f and g as his / her sibling and so on . . ) there are now 3 different set of siblings ( a and b ) ( c and d ) ; ( efg ) now first selecting 1 people out of 7 is 7 c 1 = 7 first sibling pair - - - - ( a and b ) - - selecting 1 people - - 1 c 1 = 1 second sibling pair ( c and d ) - - selecting 1 people - - 1 c 1 = 1 third sibling pair ( e f g ) - - selecting 1 out of 3 - - 3 c 1 = 3 total = 1 + 1 + 3 = 5 but , a / c to formula p ( success ) - 1 - p ( fail ) here , p ( failure ) is selecting 1 people who are siblings = 5 / 7 ( 7 is 7 c 1 ) = 1 - 5 / 7 = 2 / 7 ans e" | a = 4 / 7
b = 7 - 1
c = 7 - 1
d = b / c
e = a * d
f = 4 / 7
g = 7 - 4
h = 7 - 1
i = g / h
j = f * i
k = e + j
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a ) 7 liters , b ) 15 liters , c ) 10 liters , d ) 9 liters , e ) none of these | c | divide(subtract(multiply(divide(25, const_100), 150), multiply(divide(20, const_100), 150)), subtract(const_1, divide(25, const_100))) | a mixture of 150 liters of wine and water contains 20 % water . how much more water should be added so that water becomes 25 % of the new mixture ? | "explanation : number of liters of water in 150 liters of the mixture = 20 % of 150 = 20 / 100 * 150 = 30 liters . p liters of water added to the mixture to make water 25 % of the new mixture . total amount of water becomes ( 30 + p ) and total volume of mixture is ( 150 + p ) . ( 30 + p ) = 25 / 100 * ( 150 + p ) 120 + 4 p = 150 + p = > p = 10 liters . answer is c" | a = 25 / 100
b = a * 150
c = 20 / 100
d = c * 150
e = b - d
f = 25 / 100
g = 1 - f
h = e / g
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a ) 2 , b ) 4 , c ) 1 , d ) 5 , e ) 3 | c | subtract(25, 20) | a factory producing tennis balls stores them in either big boxes , 25 balls per box , or small boxes , 20 balls per box . if 126 freshly manufactured balls are to be stored , what is the least number of balls that can be left unboxed ? | "we have to work with multiples of 20 and 25 . first , we must know the limits of this multiples , so : 126 / 25 = 5 . . . . so the max is 5 126 / 20 = 6 . . . so the max is 6 126 - 125 = 1 ( 5 big box or 5 small box + 1 big box ) answer : c" | a = 25 - 20
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a ) 25 , b ) 30 , c ) 40 , d ) 45 , e ) 54 | e | divide(900, multiply(const_0_2778, subtract(63, 3))) | how many seconds will a 900 metre long train take to cross a man walking with a speed of 3 km / hr in the direction of the moving train if the speed of the train is 63 km / hr ? | "relative speed of the train = 63 - 3 = 60 kmph = 60 * 5 / 18 = 50 / 3 m / sec t = 900 * 3 / 50 = 54 sec answer : e" | a = 63 - 3
b = const_0_2778 * a
c = 900 / b
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a ) 3 / 5 , b ) 4 / 5 , c ) 3 / 2 , d ) 3 / 7 , e ) 3 / 8 | b | divide(circle_area(divide(0.8, const_2)), const_2) | what will be the vulgar fraction of 0.8 | "explanation : 0.8 = 80 / 100 = 4 / 5 option b" | a = 0 / 8
b = circle_area / (
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a ) 120 , b ) 134 , c ) 127 , d ) 30 , e ) 650 | d | lcm(5, 6) | jenny can divide her sweets equally to 5 people and also to 6 people equally but not to 12 people . what could be the number ? | option ( b ) and ( c ) are ruled out as they are not divisilble by 5 and 6 . 120 is both divisible by 5 and 6 but also by 12 , so it wrong . 650 is divisible by 5 but not by 6 . hence , ( d ) is the answer , as it is both divisible by 5 or 6 but not by 12 | a = math.lcm(5, 6)
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a ) 11.25 , b ) 11.52 , c ) 12.5 , d ) 9 , e ) 7.2 | c | divide(inverse(divide(inverse(add(const_3, const_2)), add(const_1, 1.5))), 1.5) | one pipe can fill a pool 1.5 times faster than a second pipe . when both pipes are opened , they fill the pool in five hours . how long would it take to fill the pool if only the slower pipe is used ? | "say the rate of the slower pipe is r pool / hour , then the rate of the faster pipe would be 1.5 r = 3 r / 2 . since when both pipes are opened , they fill the pool in five hours , then their combined rate is 1 / 5 pool / hour . thus we have that r + 3 r / 2 = 1 / 5 - - > r = 2 / 25 pool / hour - - > time is reciprocal of rate thus it ' s 25 / 2 = 12.5 hours . answer : c ." | a = 3 + 2
b = 1/(a)
c = 1 + 1
d = b / c
e = 1/(d)
f = e / 1
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a ) 18 , b ) 21.6 , c ) 22 , d ) 22.5 , e ) 27.7 | e | multiply(multiply(15, divide(add(const_100, 35), const_100)), divide(add(const_100, 20), const_100)) | a distributor sells a product through an on - line store , which take a commission of 20 % of the price set by the distributor . the distributor obtains the product from a producer at the price of $ 15 per item . what is the price that the buyer observers on - line if the distributor wants to maintain a 35 % profit on the cost of the item ? | "producer price = $ 15 ; the distributor wants to maintain a 20 % profit on the cost of the item , thus he must get $ 15 * 1.2 = $ 18 after the store takes a commission of 35 % of the final price - - > ( final price ) * 0.65 = $ 18 - - > ( final price ) = $ 27.7 . answer : e ." | a = 100 + 35
b = a / 100
c = 15 * b
d = 100 + 20
e = d / 100
f = c * e
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a ) 6 , b ) 7 , c ) 8 , d ) 9 , e ) 10 | b | divide(add(11, 10), add(2, 1)) | at a certain committee meeting only associate professors and assistant professors are present . each associate professor has brought 2 pencils and 1 chart to the meeting , while each assistant professor has brought 1 pencil and 2 charts . if a total of 10 pencils and 11 charts have been brought to the meeting , how many people are present ? | "say there are ' a ' associate professors . so we have 2 a pencils and a charts . say there are ' b ' assistant professors . so we have b pencils and 2 b charts . total pencils are 10 so 2 a + b = 10 total charts are 11 so a + 2 b = 11 add both : 3 a + 3 b = 21 so a + b = 7 total number of people = 7 answer : b" | a = 11 + 10
b = 2 + 1
c = a / b
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a ) 72 , b ) 27 , c ) 34 , d ) 28 , e ) 20 | c | divide(divide(multiply(684, 20), const_100), 4) | a reduction of 20 % in the price of oil enables a house wife to obtain 4 kgs more for rs . 684 , what is the reduced price for kg ? | "684 * ( 20 / 100 ) = 136 - - - - 4 ? - - - - 1 = > rs . 34 answer : c" | a = 684 * 20
b = a / 100
c = b / 4
|
a ) 1 , b ) 6 , c ) 4 , d ) 7 , e ) 9 | b | divide(add(multiply(522, const_10), 6), divide(add(multiply(522, const_10), 6), 6)) | if 522 x is a 3 digit no . with as a digit x . if the no . is divisible by 6 , what is the value of the digit x is ? | if a number is divisible by 6 , it must be divisible by both 2 and 3 in 522 x , to this number be divisible by 2 , the value of x must be even . so it can be 2,4 or 6 from given options 552 x is divisible by 3 , if sum of its digits is a multiple of 3 . 5 + 5 + 2 + x = 12 + x , if put x = 2 , 12 + 2 = 14 not a multiple of 3 if put x = 4 , 12 + 6 = 18 is a multiple of 3 if put x = 6 , 12 + 2 = 14 not a multiple of 3 the value of x is 6 . b | a = 522 * 10
b = a + 6
c = 522 * 10
d = c + 6
e = d / 6
f = b / e
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a ) 4 / 3 , b ) 2 / 3 , c ) 2 / 6 , d ) 7 / 8 , e ) 9 / 7 | e | divide(multiply(add(add(const_100, const_60), const_1), 8), const_100) | what is the value of ( p + q ) / ( p - q ) if p / q is 8 ? | "( p + q ) / ( p - q ) = [ ( p / q ) + 1 ] / [ ( p / q ) - 1 ] = ( 8 + 1 ) / ( 8 - 1 ) = 9 / 7 answer : e" | a = 100 + const_60
b = a + 1
c = b * 8
d = c / 100
|
a ) 2 : 5 , b ) 3 : 5 , c ) 4 : 7 , d ) 2 : 6 , e ) 1 : 4 | b | divide(subtract(5, 2), subtract(10, 5)) | cereal a is 10 % sugar by weight , whereas healthier but less delicious cereal b is 2 % sugar by weight . to make a delicious and healthy mixture that is 5 % sugar , what should be the ratio of cereal a to cereal b , by weight ? | "2 % is 3 % - points below 5 % and 10 % is 5 % - points above 5 % . the ratio of a : b should be 3 : 5 . the answer is b ." | a = 5 - 2
b = 10 - 5
c = a / b
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a ) 326.2 , b ) 324.2 , c ) 328.2 , d ) 338.2 , e ) 328.9 | c | divide(multiply(12, sqrt(subtract(power(multiply(const_2, 28), const_2), power(12, const_2)))), const_2) | the side of a rhombus is 28 m and length of one of its diagonals is 12 m . the area of the rhombus is ? | "area of the rhombus = 1 / 2 * p * β 4 ( a ) 2 - ( p ) 2 a = 28 ; p = 12 a = 1 / 2 * 12 * β 4 ( 28 ) 2 - ( 12 ) 2 = 1 / 2 * 12 * β 3136 - 144 = 1 / 2 * 12 * β 2992 a = 328.2 answer : c" | a = 2 * 28
b = a ** 2
c = 12 ** 2
d = b - c
e = math.sqrt(d)
f = 12 * e
g = f / 2
|
a ) 1 , b ) 2 , c ) 5 , d ) 23 , e ) 25 | b | sqrt(add(power(sqrt(subtract(13, multiply(const_2, 2028))), const_2), multiply(const_4, 2028))) | the product of two numbers is 2028 and their hcf is 13 . what are the number of such pairs ? | "let the numbers be 13 x and 13 y ( hcf of the numbers = 13 ) 13 x Γ 13 y = 2028 xy = 12 co - primes with product 12 are ( 1 , 12 ) and ( 3 , 4 ) ( we need to take only co - primes with product 12 . if we take two numbers with product 12 , but not co - prime , the hcf will not remain as 13 ) hence the numbers with hcf 13 and product 2028 = ( 13 Γ 1 , 13 Γ 12 ) and ( 13 Γ 3 , 13 Γ 4 ) = ( 13 , 156 ) and ( 39 , 52 ) so , there are 2 pairs of numbers with hcf 13 and product 2028 answer is b" | a = 2 * 2028
b = 13 - a
c = math.sqrt(b)
d = c ** 2
e = 4 * 2028
f = d + e
g = math.sqrt(f)
|
a ) 9 , b ) 12 , c ) 15 , d ) 16 , e ) 18 | b | subtract(multiply(divide(multiply(24, divide(6, add(6, 3))), divide(4, add(6, 3))), divide(5, add(6, 3))), subtract(24, multiply(24, divide(6, add(6, 3))))) | at deluxe paint store , fuchsia paint is made by mixing 6 parts of red paint with 3 parts of blue paint . mauve paint is made by mixing 4 parts of red paint with 5 parts blue paint . how many liters of blue paint must be added to 24 liters of fuchsia to change it to mauve paint ? | in 24 liters , red = 6 / 9 * 24 = 16 and blue = 8 so , 16 / ( 8 + x ) = 4 / 5 or , x = 12 ( answer b ) | a = 6 + 3
b = 6 / a
c = 24 * b
d = 6 + 3
e = 4 / d
f = c / e
g = 6 + 3
h = 5 / g
i = f * h
j = 6 + 3
k = 6 / j
l = 24 * k
m = 24 - l
n = i - m
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a ) 4 / 99 , b ) 2 / 25 , c ) 8 / 99 , d ) 49 / 100 , e ) 86 / 99 | e | divide(subtract(add(multiply(divide(const_100, 4), const_2), multiply(divide(const_100, 5), const_2)), 4), subtract(const_100, 1)) | if x is a positive integer with fewer than 3 digits , what is the probability w that x * ( x + 1 ) is a multiple of either 4 or 5 ? | "interesting question ! also one that we should be able to answer very quickly be keeping an eye on our best friends , the answer choices . we know that x belongs to the set { 1 , 2 , 3 , . . . , 99 } . we want to know the probability w that x ( x + 1 ) is a multiple of either 4 or 5 . when will this happen ? if either x or ( x + 1 ) is a multiple of 4 or 5 . since 4 * 5 is 20 , let ' s look at the first 20 numbers to get a rough idea of how often this happens . out of the numbers from 1 to 20 : 4 , 5 , 6 , 8 , 9 , 10 , 11 , 12 , 13 , 15 , 16 , 17 , 20 so , 14 out of the first 20 numbers match our criteria . since : probability = ( # of desired outcomes ) / ( total # of possibilities ) , we guesstimate the answer to be 14 / 20 . since ( e ) is the only answer greater than 1 / 2 , we go with ( e ) ." | a = 100 / 4
b = a * 2
c = 100 / 5
d = c * 2
e = b + d
f = e - 4
g = 100 - 1
h = f / g
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a ) 32.5 % , b ) 34 % , c ) 22.5 % , d ) 33.3 % , e ) 37.5 % | d | multiply(const_100, divide(add(divide(40, const_100), multiply(2, divide(30, const_100))), add(const_1, 2))) | because he β s taxed by his home planet , mork pays a tax rate of 40 % on his income , while mindy pays a rate of only 30 % on hers . if mindy earned 2 times as much as mork did , what was their combined tax rate ? | "say morks income is - 100 so tax paid will be 40 say mindys income is 2 * 100 = 200 so tax paid is 30 % * 200 = 60 total tax paid = 40 + 60 = 100 . combined tax % will be 100 / 100 + 200 = 33.3 %" | a = 40 / 100
b = 30 / 100
c = 2 * b
d = a + c
e = 1 + 2
f = d / e
g = 100 * f
|
a ) rs . 220 , b ) rs . 250 , c ) rs . 280 , d ) rs . 300 , e ) rs . 320 | b | add(divide(divide(subtract(multiply(550, const_2), 120), const_2), const_2), divide(const_10, const_2)) | two employees m and n are paid a total of rs . 550 per week by their employer . if m is paid 120 percent of the sum paid to n , how much is n paid per week ? | let the amount paid to m per week = x and the amount paid to n per week = y then x + y = 550 but x = 120 % of y = 120 y / 100 = 12 y / 10 β΄ 12 y / 10 + y = 550 β y [ 12 / 10 + 1 ] = 550 β 22 y / 10 = 550 β 22 y = 5500 β y = 5500 / 22 = 500 / 2 = rs . 250 b ) | a = 550 * 2
b = a - 120
c = b / 2
d = c / 2
e = 10 / 2
f = d + e
|
a ) 5 / 36 , b ) 5 / 24 , c ) 1 / 12 , d ) 1 / 6 , e ) 1 / 4 | c | subtract(divide(const_1, const_2), divide(5, 12)) | a certain bag of gemstones is composed of two - thirds diamonds and one - third rubies . if the probability of randomly selecting two diamonds from the bag , without replacement , is 5 / 12 , what is the probability of selecting two rubies from the bag , without replacement ? | 2 / 3 * ( 2 x - 1 ) / ( 3 x - 1 ) = 5 / 12 = > x = 3 so total gems = 9 and probability of ruby = 1 / 3 * 2 / 8 = 1 / 12 answer 1 / 12 c | a = 1 / 2
b = 5 / 12
c = a - b
|
a ) 4 : 5 , b ) 9 : 11 , c ) 2 : 3 , d ) 4 : 7 , e ) 3 : 7 | b | divide(divide(10, add(5, 10)), divide(20, add(4, 20))) | one man adds 5 liters of water to 10 liters of milk and another 4 liters of water to 20 liters of milk . what is the ratio of the strengths of milk in 2 mixtures ? | strength of milk in the mixture = quantity of milk / total quantity of mixture strength of milk in the 1 st mixture = 10 / 15 strength of the milk 2 nd mixture = 20 / 24 the ratio of their strengths = 10 / 15 : 20 / 24 = 4 : 5 answer is b | a = 5 + 10
b = 10 / a
c = 4 + 20
d = 20 / c
e = b / d
|
a ) 36 , b ) 12 , c ) 15 , d ) 17 , e ) 18 | a | subtract(add(multiply(10, 5), multiply(6, 5)), multiply(6, 9)) | the average of 9 observations was 6 , that of the 1 st of 5 being 10 and that of the last 5 being 8 . what was the 5 th observation ? | "explanation : 1 to 9 = 9 * 6 = 54 1 to 5 = 5 * 10 = 50 5 to 9 = 5 * 8 = 40 5 th = 50 + 40 = 90 β 54 = 36 option a" | a = 10 * 5
b = 6 * 5
c = a + b
d = 6 * 9
e = c - d
|
a ) 72 , b ) 80 , c ) 125 , d ) 150 , e ) 100 | c | divide(add(60, multiply(60, divide(2, 3))), subtract(const_1, divide(20, const_100))) | in a certain school , 20 % of students are below 8 years of age . the number of students above 8 years of age is 2 / 3 of the number of students of 8 years of age which is 60 . what is the total number of students in the school ? | explanation : let the number of students be x . then , number of students above 8 years of age = ( 100 - 20 ) % of x = 80 % of x . 80 % of x = 60 + 2 / 3 of 60 80 / 100 x = 100 x = 125 . answer : option c | a = 2 / 3
b = 60 * a
c = 60 + b
d = 20 / 100
e = 1 - d
f = c / e
|
a ) 12 , b ) 16 , c ) 19 , d ) 20 , e ) 22 | b | add(multiply(24, const_100), multiply(multiply(subtract(const_1, multiply(add(divide(const_1, 24), divide(const_1, 48)), const_2)), 24), const_60)) | two pipes a and b can fill a tank in 24 and 48 minutes respectively . if both the pipes are used together , then how long will it take to fill the tank ? | "part filled by a in 1 min . = 1 / 24 part filled by b in 1 min . = 1 / 48 part filled by ( a + b ) in 1 min . = 1 / 24 + 1 / 48 = 1 / 16 . both the pipes can fill the tank in 16 minutes . answer : b" | a = 24 * 100
b = 1 / 24
c = 1 / 48
d = b + c
e = d * 2
f = 1 - e
g = f * 24
h = g * const_60
i = a + h
|
a ) rs . 5000 , b ) rs . 6000 , c ) rs . 4000 , d ) rs . 3000 , e ) rs . 2000 | a | multiply(1000, add(const_4, const_1)) | rohan spends 40 % of his salary on food , 20 % on house rent , 10 % on entertainment and 10 % on conveyance . if his savings at the end of a month are rs . 1000 . then his monthly salary is | "sol . saving = [ 100 - ( 40 + 20 + 10 + 10 ] % = 20 % . let the monthly salary be rs . x . then , 20 % of x = 1000 Γ’ β‘ β 20 / 100 x = 1000 Γ’ β‘ β x = 1000 Γ£ β 5 = 5000 . answer a" | a = 4 + 1
b = 1000 * a
|
a ) 12 kmph , b ) 13 kmph , c ) 14 kmph , d ) 15 kmph , e ) 16 kmph | a | subtract(36, divide(multiply(36, const_2), const_3)) | the time taken by a man to row his boat upstream is twice the time taken by him to row the same distance downstream . if the speed of the boat in still water is 36 kmph , find the speed of the stream ? | "the ratio of the times taken is 2 : 1 . the ratio of the speed of the boat in still water to the speed of the stream = ( 2 + 1 ) / ( 2 - 1 ) = 3 / 1 = 3 : 1 speed of the stream = 36 / 3 = 12 kmph . answer : a" | a = 36 * 2
b = a / 3
c = 36 - b
|
a ) 10 , b ) 15 , c ) 12.5 , d ) 13 , e ) 20 | e | divide(add(add(add(7, const_1), add(add(7, const_1), const_2)), add(subtract(5, 7), subtract(5, const_2))), 7) | find the average of first 7 multiples of 5 ? | "average = ( 5 + 10 + 15 + 20 + 25 + 30 + 35 ) / 7 = 20 answer is e" | a = 7 + 1
b = 7 + 1
c = b + 2
d = a + c
e = 5 - 7
f = 5 - 2
g = e + f
h = d + g
i = h / 7
|
a ) 16 % , b ) 20 % , c ) 25 % , d ) 30 % , e ) 50 % | a | multiply(divide(subtract(multiply(const_100, const_100), multiply(subtract(const_100, 40), add(const_100, 40))), multiply(const_100, const_100)), const_100) | robert ' s salary was decreased by 40 % and subsequently increased by 40 % . how much percentage does he lose ? | "let original salary be $ 100 salary after decreasing 40 % = 100 - 100 x 40 / 100 = $ 60 salary after increasing 40 % on $ 60 = 60 + 60 x 40 / 100 = $ 84 percentage of loss = 100 - 84 = 16 % answer : a" | a = 100 * 100
b = 100 - 40
c = 100 + 40
d = b * c
e = a - d
f = 100 * 100
g = e / f
h = g * 100
|
a ) s . 528 , b ) s . 542 , c ) s . 528 , d ) s . 720 , e ) s . 549 | d | multiply(4, divide(1560, add(add(4, 2), const_3))) | rs . 1560 is divided so that 4 times the first share , thrice the 2 nd share and twice the third share amount to the same . what is the value of the third share ? | "a + b + c = 1560 4 a = 3 b = 2 c = x a : b : c = 1 / 4 : 1 / 3 : 1 / 2 = 3 : 4 : 6 6 / 13 * 1560 = rs . 720 answer : d" | a = 4 + 2
b = a + 3
c = 1560 / b
d = 4 * c
|
a ) 12 , b ) 142 , c ) 164 , d ) 178 , e ) 84 | d | subtract(multiply(add(12, const_1), add(48, 10)), multiply(48, 12)) | the average runs of a cricket player of 12 innings was 48 . how many runs must he make in his next innings so as to increase his average of runs by 10 ? | "explanation : average after 13 innings = 58 required number of runs = ( 58 * 13 ) β ( 48 * 12 ) = 754 β 576 = 178 answer d" | a = 12 + 1
b = 48 + 10
c = a * b
d = 48 * 12
e = c - d
|
a ) 1 / 5 , b ) 2 / 5 , c ) 3 / 10 , d ) 3 / 7 , e ) 7 / 22 | e | divide(divide(22, 3), 22) | tickets numbered from 1 to 22 are mixed and then a ticket is selected randomly . what is the probability that the selected ticket bears a number which is a multiple of 3 ? | "here , s = [ 1 , 2 , 3 , 4 , β¦ . , 19 , 20 , 21 , 22 ] let e = event of getting a multiple of 3 = [ 3 , 6 , 9 , 12 , 15 , 18 , 21 ] p ( e ) = n ( e ) / n ( s ) = 7 / 22 the answer is e ." | a = 22 / 3
b = a / 22
|
a ) $ 1500 , b ) $ 1200 , c ) $ 1600 , d ) $ 1000 , e ) $ 1350 | c | multiply(900, power(subtract(const_1, divide(25, const_100)), 2)) | a present value of a machine is $ 900 . its value depletiation rate is 25 % per annum then find the machine value before 2 years ? | "p = $ 900 r = 25 % t = 2 years machine value before 2 years = p / [ ( 1 - r / 100 ) ^ t ] = 900 * 4 / 3 * 4 / 3 = $ 1600 answer is c" | a = 25 / 100
b = 1 - a
c = b ** 2
d = 900 * c
|
a ) a ) 30 , b ) b ) 31 , c ) c ) 32 , d ) d ) 53 , e ) e ) 34 | d | divide(factorial(subtract(add(const_4, 3), const_1)), multiply(factorial(3), factorial(subtract(const_4, const_1)))) | how many positive integers less than 100 are neither multiples of 3 or 5 . | "to answer this q we require to know 1 ) multiples of 5 till 100 = 100 / 5 = 20 2 ) multiples of 3 till 100 = 100 / 3 = 33.33 = 33 add the two 20 + 33 = 53 ; subtract common terms that are multiple of both 5 and 3 . . lcm of 5 and 3 = 15 multiples of 15 till 100 = 100 / 15 = 6 so total multiples of 5 and 3 = 53 - 6 = 47 ans = 100 - 47 = 53 d" | a = 4 + 3
b = a - 1
c = math.factorial(b)
d = math.factorial(3)
e = 4 - 1
f = math.factorial(e)
g = d * f
h = c / g
|
a ) 15310 , b ) 18870 , c ) 24510 , d ) 18510 , e ) 42510 | d | multiply(122, 252) | 122 * 252 - 12234 = ? | "d ? = 122 * 252 - 12234 = 30744 - 12234 = 18510" | a = 122 * 252
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a ) 22 , b ) 23 , c ) 24 , d ) 25 , e ) 26 | c | divide(multiply(72, 36), add(72, 36)) | a company has two models of computers , model x and model y . operating at a constant rate , a model x computer can complete a certain task in 72 minutes and a model y computer can complete the same task in 36 minutes . if the company used the same number of each model of computer to complete the task in 1 minute , how many model x computers were used ? | let ' s say 1 work is processing 72 gb of data . model x : 1 gb per min model y : 2 gb per min working together , 1 x and 1 y = 3 gb per min so , 24 times as many computers would work at 72 gb per min . so no . of x = 24 answer is c | a = 72 * 36
b = 72 + 36
c = a / b
|
a ) 700 , b ) 710 , c ) 600 , d ) 730 , e ) 740 | c | add(360, multiply(360, divide(40, const_100))) | a fruit seller had some oranges . he sells 40 % oranges and still has 360 oranges . how many oranges he had originally ? | "60 % of oranges = 360 100 % of oranges = ( 360 Γ 100 ) / 6 = 600 total oranges = 700 answer : c" | a = 40 / 100
b = 360 * a
c = 360 + b
|
a ) 800 , b ) 840 , c ) 880 , d ) 920 , e ) 960 | a | divide(divide(divide(120, subtract(const_1, divide(2, 5))), divide(2, 4)), divide(1, 2)) | of the goose eggs laid at a certain pond , 1 / 2 hatched and 3 / 4 of the geese that hatched from those eggs survived the first month . of the geese that survived the first month , 3 / 5 did not survive the first year . if 120 geese survived the first year and if no more than one goose hatched from each egg , how many goose eggs were laid at the pond ? | "let x be the number of eggs that were laid . ( 2 / 5 ) ( 3 / 4 ) ( 1 / 2 ) x = 120 ( 6 / 40 ) x = 120 x = 800 the answer is a ." | a = 2 / 5
b = 1 - a
c = 120 / b
d = 2 / 4
e = c / d
f = 1 / 2
g = e / f
|
a ) 34 % , b ) 24 % , c ) 22 % , d ) 18 % , e ) 8.5 % | a | multiply(divide(subtract(multiply(const_100, divide(16, const_100)), multiply(subtract(const_100, multiply(divide(const_1, const_4), const_100)), divide(10, const_100))), multiply(divide(const_1, const_4), const_100)), const_100) | one fourth of a solution that was 10 % sugar by weight was replaced by a second solution resulting in a solution that was 16 percent sugar by weight . the second solution was what percent sugar by weight ? | "et 100 be the total volume . total salt = 10 salt taken out = 10 / 4 = 2.5 to make solution 16 % , total salt = 16 salt added = 16 - 7.5 = 8.5 solution had = 8.5 / 25 * 100 = 34 % sugar answer : a" | a = 16 / 100
b = 100 * a
c = 1 / 4
d = c * 100
e = 100 - d
f = 10 / 100
g = e * f
h = b - g
i = 1 / 4
j = i * 100
k = h / j
l = k * 100
|
a ) 5 , b ) 7 , c ) 9 , d ) 11 , e ) 12 | b | divide(divide(divide(divide(420, const_2), const_3), const_4), divide(const_10, const_2)) | if n is a positive integer and the product of all integers from 1 to n , inclusive , is a multiple of 420 , what is the least possible value of n ? | "420 = 2 * 2 * 3 * 5 * 7 , so n must be at least 7 . the answer is b ." | a = 420 / 2
b = a / 3
c = b / 4
d = 10 / 2
e = c / d
|
a ) 0 , b ) 1 , c ) - 1 , d ) - 8 , e ) 8 | b | add(1, subtract(5, 5)) | if x = 5 - q and y = 3 q - 1 , then for what value of q , x is three times to y ? | "explanation : x = 3 y < = > 5 - q = 3 ( 3 q - 1 ) < = > 5 - q = 9 q - 3 < = > 8 q = 8 < = > q = 1 . answer : b" | a = 5 - 5
b = 1 + a
|
a ) 10 mps , b ) 8 mps , c ) 4 mps , d ) 7 mps , e ) 2 mps | c | multiply(const_0_2778, 16) | express a speed of 16 kmph in meters per second ? | "16 * 5 / 18 = 4 mps answer : c" | a = const_0_2778 * 16
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a ) 30 , b ) 32 , c ) 30 , d ) 36 , e ) 40 | d | add(add(add(12, 12), 12), 12) | xavier starts from p towards q at a speed of 90 kmph and after every 12 mins increases his speed by 10 kmph . if the distance between p and q is 51 km , then how much time does he take to cover the distance ? | "first 12 min = 90 * 12 / 60 = 18 km 2 nd 12 min = 100 * 12 / 60 = 20 km 3 rd 12 min = 110 * 12 / 60 = 22 km total time 12 * 3 = 36 min d" | a = 12 + 12
b = a + 12
c = b + 12
|
a ) 166 , b ) 105 , c ) 178 , d ) 177 , e ) 245 | e | add(add(multiply(divide(const_100, 45), 63), multiply(divide(30, 45), 63)), 63) | a sun is divided among x , y and z in such a way that for each rupee x gets , y gets 45 paisa and z gets 30 paisa . if the share of y is rs . 63 , what is the total amount ? | "x : y : z = 100 : 45 : 30 20 : 9 : 6 9 - - - 63 35 - - - ? = > 245 answer : e" | a = 100 / 45
b = a * 63
c = 30 / 45
d = c * 63
e = b + d
f = e + 63
|
a ) 48 , b ) 47 , c ) 46 , d ) 45 , e ) 49 | e | add(divide(subtract(100, 0), 2), const_1) | how many multiples of 2 are there between 0 and 100 , exclusive ? | "2 * 50 = 100 total multiples = ( 50 - 1 ) + 1 = 50 exclude 0 and 100 = 50 - 1 = 49 answer is e ." | a = 100 - 0
b = a / 2
c = b + 1
|
a ) 58 kgs , b ) 58.85 kgs , c ) 58.95 kgs , d ) 59 kgs , e ) 59.85 kgs | d | divide(add(multiply(58.4, 20), subtract(68, 56)), 20) | the average weight of a class of 20 boys was calculated to be 58.4 kgs and it was later found that one weight was misread as 56 kg instead of 68 kg . what is the correct weight ? | "actual total weight is ( 20 x 58.4 - 56 + 68 ) = 1180 kgs actual average weight is 1180 / 20 = 59 kgs d" | a = 58 * 4
b = 68 - 56
c = a + b
d = c / 20
|
a ) 2 / 15 , b ) 4 / 15 , c ) 1 / 15 , d ) 4 / 90 , e ) 5 / 60 | b | divide(subtract(multiply(multiply(5, const_2), 3), multiply(3, const_2)), multiply(multiply(5, const_2), multiply(3, 3))) | what is the probability that a two digit number selected at random will be a multiple of ' 3 ' and not a multiple of ' 5 ' ? | solution : there are a total of 90 two digit numbers . every third number will be divisible by ' 3 ' . therefore , there are 30 of those numbers that are divisible by ' 3 ' . of these 30 numbers , the numbers that are divisible by ' 5 ' are those that are multiples of ' 15 ' . i . e . numbers that are divisible by both ' 3 ' and ' 5 ' . there are 6 such numbers - - 15 , 30,45 , 60,75 and 90 . we need to find out numbers that are divisible by ' 3 ' and not by ' 5 ' , which will be : 30 Γ’ Λ β 6 = 24 24 out of the 90 numbers are divisible by ' 3 ' and not by ' 5 ' . the required probability is therefore , = 24 / 90 = 4 / 15 answer is b | a = 5 * 2
b = a * 3
c = 3 * 2
d = b - c
e = 5 * 2
f = 3 * 3
g = e * f
h = d / g
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a ) 5 / 3 , b ) 1 / 3 , c ) 7 / 3 , d ) 4 / 9 , e ) 3 / 5 | d | divide(const_4, add(multiply(const_4, const_2), const_1)) | tom , working alone , can paint a room in 12 hours . peter and john , working independently , can paint the same room in 6 hours and 2 hours , respectively . tom starts painting the room and works on his own for two hour . he is then joined by peter and they work together for two hour . finally , john joins them and the three of them work together to finish the room , each one working at his respective rate . what fraction of the whole job was done by peter ? | let the time when all three were working together be t hours . then : tom worked for t + 4 hour and has done 1 / 12 * ( t + 4 ) part of the job ; peter worked for t + 2 hour and has done 1 / 6 * ( t + 2 ) part of the job ; john worked for t hours and has done 1 / 2 * t part of the job : 1 / 12 * ( t + 4 ) + 1 / 6 * ( t + 2 ) + 1 / 2 * t = 1 - - > multiply by 12 - - > ( t + 4 ) + ( 2 t + 2 ) + 6 t = 12 - - > t = 2 / 3 ; hence peter has done 1 / 6 * ( 2 / 3 + 2 ) = 1 / 6 * 8 / 3 = 8 / 18 = 4 / 9 answer : d | a = 4 * 2
b = a + 1
c = 4 / b
|
a ) 12 : 5 , b ) 17 : 3 , c ) 10 : 6 , d ) 17 : 7 , e ) 10 : 3 | e | divide(50000, 15000) | p and q started a business investing rs 50000 and rs 15000 resp . in what ratio the profit earned after 2 years be divided between p and q respectively . | "explanation : in this type of question as time frame for both investors is equal then just get the ratio of their investments . p : q = 50000 : 15000 = 50 : 15 = 10 : 3 option e" | a = 50000 / 15000
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a ) 87 , b ) 54 , c ) 29 , d ) 30 , e ) 70 | b | add(add(add(multiply(multiply(3, 3), const_4), const_12), const_4), const_2) | there are many numbers that can be expressed as the sum of 3 squares in 3 different ways . can you find out the smallest of such number ? | b 54 7 ^ 2 + 2 ^ 2 + 1 ^ 1 6 ^ 2 + 3 ^ 2 + 3 ^ 2 2 ^ 2 + 5 ^ 2 + 5 ^ 2 | a = 3 * 3
b = a * 4
c = b + 12
d = c + 4
e = d + 2
|
a ) 16 seconds , b ) 20 seconds , c ) 14 seconds , d ) 12 seconds , e ) 15 seconds | b | divide(100, multiply(add(15, 3), const_0_2778)) | the speed at which a man can row a boat in still water is 15 kmph . if he rows downstream , where the speed of current is 3 kmph , what time will he take to cover 100 metres ? | "speed of the boat downstream = 15 + 3 = 18 kmph = 18 * 5 / 18 = 5 m / s hence time taken to cover 100 m = 100 / 5 = 20 seconds . answer : b" | a = 15 + 3
b = a * const_0_2778
c = 100 / b
|
a ) 12000 , b ) 22000 , c ) 24000 , d ) 26000 , e ) 32000 | a | multiply(multiply(subtract(1, divide(5, 2)), multiply(multiply(multiply(const_0_25, const_100), const_100), const_10)), add(divide(1, 5), 1)) | in a mayoral election , candidate x received 1 / 5 more votes than candidate y , and candidate y received 1 / 2 fewer votes than z . if z received 20,000 votes how many votes did candidate x received ? | "z = 20 - - > y received 1 / 2 fewer votes than z - - > y = z - 1 / 2 * z = 10 ; x received 1 / 5 more votes than y - - > x = y + 1 / 5 * y = 12 . answer : a ." | a = 5 / 2
b = 1 - a
c = const_0_25 * 100
d = c * 100
e = d * 10
f = b * e
g = 1 / 5
h = g + 1
i = f * h
|
a ) 12,28 , b ) 14,30 , c ) 17,39 , d ) 18,34 , e ) 19,34 | c | subtract(divide(subtract(add(multiply(6, 6), 22), 6), subtract(6, const_1)), const_1) | ages of two persons differ by 22 years . if 6 year ago , the elder one be 3 times as old the younger one , find their present age | "explanation : let the age of younger person is x , then elder person age is ( x + 22 ) = > 3 ( x - 6 ) = ( x + 22 - 6 ) [ 6 years before ] = > 3 x - 18 = x + 16 = > x = 17 . so other person age is x + 22 = 39 answer : option c" | a = 6 * 6
b = a + 22
c = b - 6
d = 6 - 1
e = c / d
f = e - 1
|
a ) 20 . , b ) 21 . , c ) 22 . , d ) 23 . , e ) 24 . | c | add(divide(1430, add(divide(1430, add(add(const_10, const_10), const_2)), 45)), 9) | a basket of 1430 apples is divided equally among a group of apple lovers . if 45 people join the group , each apple lover would receive 9 apples less . how many w apples did each person get before 45 people joined the feast ? | "before solving it algebraically , let us prime factorize 1430 = 2 * 5 * 11 * 13 . since number of apples per person * total persons w = 1430 , the answer should be a factor of 1430 . only c is . and that ' s your answer . c" | a = 10 + 10
b = a + 2
c = 1430 / b
d = c + 45
e = 1430 / d
f = e + 9
|
a ) s : 2000 , b ) s : 1067 , c ) s : 1278 , d ) s : 1028 , e ) s : 1027 | a | divide(multiply(280, const_100), subtract(add(const_100, 4), subtract(const_100, 10))) | a watch was sold at a loss of 10 % . if it was sold for rs . 280 more , there would have been a gain of 4 % . what is the cost price ? | "90 % 104 % - - - - - - - - 14 % - - - - 280 100 % - - - - ? = > rs : 2000 answer : a" | a = 280 * 100
b = 100 + 4
c = 100 - 10
d = b - c
e = a / d
|
a ) 5 kmph , b ) 2 kmph , c ) 7 kmph , d ) 8 kmph , e ) 3 kmph | a | divide(subtract(18, 8), const_2) | a man can row his boat with the stream at 18 km / h and against the stream in 8 km / h . the man ' s rate is ? | "ds = 18 us = 8 s = ? s = ( 18 - 8 ) / 2 = 5 kmph answer : a" | a = 18 - 8
b = a / 2
|
a ) 432 , b ) 640 , c ) 256 , d ) 512 , e ) none of these | d | multiply(multiply(divide(16, const_2), divide(16, const_2)), divide(16, const_2)) | free notebooks were distributed equally among children of a class . the number of notebooks each child got was one - eighth of the number of children . had the number of children been half , each child would have got 16 notebooks . how many notebooks were distributed in all ? | "in case i : let the no . of children = x . hence , total no . of notebooks distributed 1 β 8 x . x or x 2 β 8 . . . . . . . ( i ) in case ii : no . of children = x β 2 now , the total no . of notebooks = 16 Γ x β 2 . . . . . . . ( ii ) comparing ( i ) & ( ii ) , we get x 2 β 8 = 8 x or , x = 64 hence , total no . of notebooks = 64 Γ 64 / 8 = 512 answer d" | a = 16 / 2
b = 16 / 2
c = a * b
d = 16 / 2
e = c * d
|
a ) 160 , b ) 787 , c ) 144 , d ) 128 , e ) 121 | a | divide(40, divide(const_1, add(3, const_1))) | one pipe can fill a tank 3 times as fast as another pipe . if together the two pipes can fill tank in 40 min , then the slower pipe alone will be able to fill the tank in ? | let the slower pipe alone fill the tank in x min . then , faster pipe will fill it in x / 3 min . 1 / x + 3 / x = 1 / 40 4 / x = 1 / 40 = > x = 160 min . answer : a | a = 3 + 1
b = 1 / a
c = 40 / b
|
a ) 1500 , b ) 1800 . , c ) 1900 , d ) 1700 , e ) 1660 | b | add(1500, multiply(1500, divide(20, const_100))) | 1500 is increased by 20 % . find the final number . | final number = initial number + 20 % ( original number ) = 1500 + 20 % ( 1500 ) = 1500 + 300 = 1800 . answer b | a = 20 / 100
b = 1500 * a
c = 1500 + b
|
a ) 5 , b ) 10 , c ) 15 , d ) 20 , e ) 25 | b | subtract(subtract(50, divide(60, const_3)), divide(60, const_3)) | the perimeter of an equilateral triangle is 60 . if one of the sides of the equilateral triangle is the side of an isosceles triangle of perimeter 50 , then how long is the base of isosceles triangle ? | "the base of the isosceles triangle is 50 - 20 - 20 = 10 units the answer is b ." | a = 60 / 3
b = 50 - a
c = 60 / 3
d = b - c
|
a ) 480 , b ) 120 , c ) 289 , d ) 270 , e ) 927 | b | multiply(24, 20) | a cistern has a leak which would empty the cistern in 20 minutes . a tap is turned on which admits 1 liters a minute into the cistern , and it is emptied in 24 minutes . how many liters does the cistern hold ? | "1 / x - 1 / 20 = - 1 / 24 x = 120 120 * 1 = 120 answer : b" | a = 24 * 20
|
a ) 0.8 , b ) 0.4 , c ) 0.96 , d ) 0.69 , e ) 0.76 | b | divide(subtract(power(0.5, 3), power(0.1, 3)), add(add(power(0.5, 2), 0.05), power(0.1, 2))) | ( 0.5 ) ( power 3 ) - ( 0.1 ) ( power 3 ) / ( 0.5 ) ( power 2 ) + 0.05 + ( 0.1 ) ( power 2 ) is : | "given expression = ( 0.5 ) ( power 3 ) - ( 0.1 ) ( power 3 ) / ( 0.5 ) ( power 2 ) + ( 0.5 x 0.1 ) + ( 0.1 ) ( power 2 ) = a ( power 3 ) - b ( power 3 ) / a ( power 2 ) + ab + b ( power 2 ) = ( a - b ) = ( 0.5 - 0.1 ) = 0.4 answer is b" | a = 0 ** 5
b = 0 ** 1
c = a - b
d = 0 ** 5
e = d + 0
f = 0 ** 1
g = e + f
h = c / g
|
a ) 6 , b ) 1 , c ) 0 , d ) 8 , e ) 7 | a | subtract(add(17, power(add(const_1, const_4), 2)), multiply(12, 3)) | if p is a prime number greater than 3 , find the remainder when p ^ 2 + 17 is divided by 12 . | "take any prime number greater than 3 and check - say p = 5 so , p ^ 2 + 17 = 25 + 17 = > 42 42 / 12 will have remainder as 6 say p = 7 so , p ^ 2 + 17 = 49 + 17 = > 66 66 / 12 will have remainder as 6 so answer will be ( a ) 6" | a = 1 + 4
b = a ** 2
c = 17 + b
d = 12 * 3
e = c - d
|
a ) 90.5 % , b ) 87.5 % , c ) 80.5 % , d ) 77.5 % , e ) 70.5 % | b | multiply(divide(subtract(8, const_1), 8), const_100) | 9.8 , 9.8 , 9.9 , 9.9 , 10.0 , 10.0 , 10.1 , 10.5 the mean and the standard deviation of the 8 numbers shown above is 10 and 0.212 respectively . what percent of the 8 numbers are within 1 standard deviation of the mean ? | within 1 standard deviation of the mean - means in the range { mean - 1 * sd ; mean + 1 * sd } = { 10 - 1 * 0.212 ; 10 + 0.3 } = { 9.788 ; 10.212 } . from the 8 listed numbers , 7 are within this range so 6 / 8 = 87.5 % . answer : b . | a = 8 - 1
b = a / 8
c = b * 100
|
a ) 1.4 % , b ) 5.9 % , c ) 12 % , d ) 12.5 % , e ) 23.6 % | c | multiply(subtract(divide(28, 25), const_1), const_100) | at the opening of a trading day at a certain stock exchange , the price per share of stock m was $ 25 . if the price per share of stock m was $ 28 at the closing of the day , what was the percent increase in the price per share of stock m for that day ? | "opening = 25 closing = 28 rise in price = 3 so , percent increase = 3 / 25 * 100 = 12 answer : c" | a = 28 / 25
b = a - 1
c = b * 100
|
a ) 43 , b ) 30 , c ) 35 , d ) 45 , e ) 50 | a | subtract(multiply(8, 61), subtract(multiply(13, 60), multiply(57, 7))) | the average of 13 numbers is 60 . average of the first 7 of them is 57 and that of the last 7 is 61 . find the 8 th number ? | "sum of all the 13 numbers = 13 * 60 = 780 sum of the first 7 of them = 7 * 57 = 399 sum of the last 7 of them = 7 * 61 = 427 so , the 8 th number = 427 + 399 - 780 = 46 . answer a" | a = 8 * 61
b = 13 * 60
c = 57 * 7
d = b - c
e = a - d
|
a ) 3 , b ) 4 , c ) 5 , d ) 6 , e ) 2 | e | subtract(multiply(10, 2.00), multiply(1.60, 10)) | martin bought 10 concert tickets , some at the full price of $ 2.00 per ticket , and some at a discounted price of $ 1.60 per ticket . if he spent a total of $ 19.20 , how many discounted tickets did he buy ? | "let x be the number of tickets he bought at $ 2 per ticket . then 2 x + ( 10 - x ) 1.6 = 19.2 0.4 x = 3.2 = > x = 8 discounted tickets = 10 - x = 2 ans : e" | a = 10 * 2
b = 1 * 60
c = a - b
|
a ) 50 , b ) 60 , c ) 70 , d ) 80 , e ) 75 | e | divide(360, add(add(divide(7, 5), divide(7, 5)), const_2)) | the length of a rectangular field is 7 / 5 its width . if the perimeter of the field is 360 meters , what is the width of the field ? | "let l be the length and w be the width . l = ( 7 / 5 ) w perimeter : 2 l + 2 w = 360 , 2 ( 7 / 5 ) w + 2 w = 360 solve the above equation to find : w = 75 m and l = 105 m . correct answer e ) 75" | a = 7 / 5
b = 7 / 5
c = a + b
d = c + 2
e = 360 / d
|
a ) 2277 , b ) 2999 , c ) 1000 , d ) 2651 , e ) 2000 | e | divide(2090, multiply(add(const_1, divide(10, const_100)), subtract(const_1, divide(5, const_100)))) | the salary of a typist was at first raised by 10 % and then the same was reduced by 5 % . if he presently draws rs . 2090 . what was his original salary ? | "x * ( 110 / 100 ) * ( 95 / 100 ) = 2090 x * ( 11 / 10 ) * ( 1 / 100 ) = 22 x = 2000 answer : e" | a = 10 / 100
b = 1 + a
c = 5 / 100
d = 1 - c
e = b * d
f = 2090 / e
|
a ) 0 , b ) 2 , c ) 4 , d ) 6 , e ) 8 | e | divide(subtract(add(multiply(2, 3), 21), add(multiply(2, 4), 3)), subtract(multiply(2, 2), multiply(2, const_1))) | given f ( x ) = 2 x β 3 , for what value of x does 2 * [ f ( x ) ] β 21 = f ( x β 4 ) ? | "2 ( 2 x - 3 ) - 21 = 2 ( x - 4 ) - 3 2 x = 16 x = 8 the answer is e ." | a = 2 * 3
b = a + 21
c = 2 * 4
d = c + 3
e = b - d
f = 2 * 2
g = 2 * 1
h = f - g
i = e / h
|
a ) 280 meter , b ) 240 meter , c ) 200 meter , d ) 300 meter , e ) none of these | d | multiply(multiply(const_0_2778, 54), subtract(40, 20)) | a train passes a platform in 40 seconds . the same train passes a man standing on the platform in 20 seconds . if the speed of the train is 54 km / hr , the length of the platform is | "explanation : speed of the train = 54 km / hr = ( 54 Γ 10 ) / 40 m / s = 15 m / s length of the train = speed Γ time taken to cross the man = 15 Γ 20 = 300 m let the length of the platform = l time taken to cross the platform = ( 300 + l ) / 15 = > ( 300 + l ) / 15 = 40 = > 300 + l = 15 Γ 40 = 600 = > l = 600 - 300 = 300 meter answer : option d" | a = const_0_2778 * 54
b = 40 - 20
c = a * b
|
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