Problem stringlengths 5 967 | Rationale stringlengths 1 2.74k | options stringlengths 37 300 | correct stringclasses 5
values | annotated_formula stringlengths 7 6.48k | linear_formula stringlengths 8 925 | category stringclasses 6
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what is the greatest value of positive integer x such that 2 ^ x is a factor of 100 ^ 90 ? | "put in simple words , we need to find the highest power of 2 in 100 ^ 90 100 = 2 ^ 2 * 5 ^ 2 therefore 100 ^ 90 = ( 2 ^ 2 * 5 ^ 2 ) ^ 90 = 2 ^ 180 * 5 ^ 180 answer : a" | a ) 180 , b ) 160 , c ) 140 , d ) 150 , e ) 120 | a | subtract(100, subtract(power(2, divide(100, 90)), const_1)) | divide(n1,n2)|power(n0,#0)|subtract(#1,const_1)|subtract(n1,#2)| | general |
alex takes a loan of $ 9,000 to buy a used truck at the rate of 9 % simple interest . calculate the annual interest to be paid for the loan amount . | from the details given in the problem principle = p = $ 9,000 and r = 9 % or 0.09 expressed as a decimal . as the annual interest is to be calculated , the time period t = 1 . plugging these values in the simple interest formula , i = p x t x r = 9,000 x 1 x 0.09 = 810.00 annual interest to be paid = $ 810 answer : b | a ) 680 , b ) 810 , c ) 800 , d ) 730 , e ) 750 | b | divide(multiply(multiply(multiply(9, const_100), sqrt(const_100)), 9), const_100) | multiply(n1,const_100)|sqrt(const_100)|multiply(#0,#1)|multiply(n1,#2)|divide(#3,const_100) | gain |
last month , john rejected 0.7 % of the products that he inspected and jane rejected 0.8 percent of the products that she inspected . if total of 0.75 percent of the products produced last month were rejected , what fraction of the products did jane inspect ? | "x - fraction of products jane inspected ( 1 - x ) - fraction of products john inspected 0.8 ( x ) + 0.7 ( 1 - x ) = 0.75 0.1 x = 0.75 - 0.7 x = 0.05 / 0.1 x = 1 / 2 therefore the answer is b : 1 / 2 ." | a ) 1 / 6 , b ) 1 / 2 , c ) 5 / 8 , d ) 5 / 6 , e ) 15 / 16 | b | divide(subtract(0.75, 0.7), subtract(0.8, 0.7)) | subtract(n2,n0)|subtract(n1,n0)|divide(#0,#1)| | gain |
if the average ( arithmetic mean ) of x and y is 60 , and z – x = 80 , what is the average of y and z ? | "x + y / 2 = 60 = > x + y = 120 x = z - 80 . . . sub this value z - 80 + y = 120 = > z + y = 200 = > z + y / 2 = 100 answer : a" | a ) 100 , b ) 120 , c ) 125 , d ) 115 , e ) 90 | a | subtract(multiply(80, const_2), multiply(60, const_2)) | multiply(n1,const_2)|multiply(n0,const_2)|subtract(#0,#1)| | general |
what is the smallest integer z for which 27 ^ z > 3 ^ 24 ? | "27 ^ z > 3 ^ 24 converting into the same bases : 27 ^ z > 27 ^ 8 therefore for the equation to hold true , z > 8 or z = 9 option c" | a ) 7 , b ) 8 , c ) 9 , d ) 10 , e ) 12 | c | add(floor(divide(24, const_3)), const_1) | divide(n2,const_3)|floor(#0)|add(#1,const_1)| | general |
a is twice as good a work man as b and together they finish the work in 8 days . in how many days a alone can finish the work ? | "wc = 2 : 1 2 x + x = 1 / 8 = > x = 1 / 24 2 x = 1 / 12 a can do the work in 12 days . answer : b" | a ) 23 , b ) 12 , c ) 77 , d ) 92 , e ) 61 | b | inverse(divide(inverse(8), add(const_2, const_1))) | add(const_1,const_2)|inverse(n0)|divide(#1,#0)|inverse(#2)| | physics |
the sum of two numbers is 21 and the sum of their squares is 527 . find the product of the numbers . | let a and b be the two numbers ( a + b ) ^ 2 = a ^ 2 + 2 ab + b ^ 2 given ( a + b ) = 21 a ^ 2 + b ^ 2 = 527 so , 21 ^ 2 = 527 + 2 ab 2 ab = 527 - 441 2 ab = 86 ab = 43 ans d | a ) 48 , b ) 41 , c ) 46 , d ) 43 , e ) 96 | d | divide(subtract(527, power(21, const_2)), const_2) | power(n0,const_2)|subtract(n1,#0)|divide(#1,const_2) | general |
thabo owns exactly 280 books , and each book is either paperback fiction , paperback nonfiction , or hardcover nonfiction . if he owns 20 more paperback nonfiction books than hardcover nonfiction books , and twice as many paperback fiction books as paperback nonfiction books , how many hardcover books nonfiction books ... | "i think we can use double - matrix method and solve using only one variable . our goal is to find the number of hardcover nonfiction books . let that number be x . we are given that all 140 books are either paperback fiction , paperback nonfiction , or hardcover nonfiction . this implies that number of hardcover ficti... | a ) 10 , b ) 20 , c ) 30 , d ) 40 , e ) 55 | e | subtract(divide(add(280, 20), const_4), 20) | add(n0,n1)|divide(#0,const_4)|subtract(#1,n1)| | general |
in the number 100 ab , a and b represent the tens and units digits , respectively . if 110 ab is divisible by 55 , what is the greatest possible value of b × a ? | you should notice that 55 * 2 = 110 so 11,000 is divisible by 55 : 55 * 200 = 11,000 ( or you can notice that 11,000 is obviously divisible by both 5 and 11 so by 55 ) - - > b * a = 0 * 0 = 0 . next number divisible by 55 is 11,000 + 55 = 11,055 : b * a = 5 * 5 = 15 ( next number wo n ' t have 110 as the first 3 digits... | a ) 0 , b ) 5 , c ) 10 , d ) 15 , e ) 25 | d | multiply(divide(55, divide(110, const_10)), subtract(divide(55, divide(110, const_10)), const_2)) | divide(n1,const_10)|divide(n2,#0)|subtract(#1,const_2)|multiply(#1,#2) | general |
a train 350 m long , running with a speed of 63 km / hr will pass a tree in ? | "speed = 63 * 5 / 18 = 35 / 2 m / sec time taken = 350 * 2 / 35 = 20 sec answer : e" | a ) 22 sec , b ) 16 sec , c ) 17 sec , d ) 88 sec , e ) 20 sec | e | multiply(divide(350, multiply(63, const_1000)), const_3600) | multiply(n1,const_1000)|divide(n0,#0)|multiply(#1,const_3600)| | physics |
a car travelling with 5 / 7 km of its actual speed covers 42 km in 1 hr 40 min 48 sec find the actual speed of the car ? | time taken = 1 hr 40 min 48 sec = 126 / 75 hrs let the actual speed be x kmph then 5 / 7 x * 126 / 75 = 42 x = ( 42 * 7 * 75 ) / 5 * 126 = 35 kmph answer ( c ) | a ) 78 kmph , b ) 85 kmph , c ) 35 kmph , d ) 24 kmph , e ) 74 kmph | c | divide(multiply(divide(42, add(1, divide(add(multiply(40, const_60), 48), const_3600))), 7), 5) | multiply(n4,const_60)|add(n5,#0)|divide(#1,const_3600)|add(n3,#2)|divide(n2,#3)|multiply(n1,#4)|divide(#5,n0) | physics |
compare the rates of two trains , one travelling at 90 km / hr and other is at 50 m / s ? | speed of the 1 st train = 90 km / hr speed of the 2 nd train = 50 m / s = 50 * 18 / 5 = 180 km / hr ratio of the speeds of the train = 90 : 180 = 1 : 2 answer is b | a ) 2 : 1 , b ) 1 : 2 , c ) 1 : 3 , d ) 3 : 1 , e ) 2 : 3 | b | divide(90, multiply(divide(50, const_1000), const_3600)) | divide(n1,const_1000)|multiply(#0,const_3600)|divide(n0,#1) | gain |
a computer manufacturer produces a certain electronic component at a cost of $ 100 per component . shipping costs for delivering the components are $ 2 per unit . further , the manufacturer has costs of $ 10,000 a month related to the electronic component regardless of how many it produces . if the manufacturer produce... | "$ 10000 is a fixed cost each component is $ 102 ( $ 100 to produce , $ 2 to ship ) manufacturer will be producing and selling 1000 components so therefore the equation to find price would be 1000 * p = 10000 + ( 1000 * 100 ) + ( 1000 * 2 ) p = ( 10000 + 100000 + 2000 ) / 1000 p = 112 answer : b" | a ) 122 , b ) 112 , c ) 134 , d ) 108 , e ) 101 | b | divide(add(add(multiply(multiply(const_4, const_4), const_1000), multiply(2, const_100)), multiply(add(100, 2), 1000)), 1000) | add(n0,n1)|multiply(const_4,const_4)|multiply(n1,const_100)|multiply(#1,const_1000)|multiply(n3,#0)|add(#3,#2)|add(#5,#4)|divide(#6,n3)| | general |
sari and ken climb up a mountain . at night , they camp together . on the day they are supposed to reach the summit , sari wakes up at 08 : 00 and starts climbing at a constant pace . ken starts climbing only at 10 : 00 , when sari is already 700 meters ahead of him . nevertheless , ken climbs at a constant pace of 500... | "both sari and ken climb in the same direction . speed of sari = 700 / 2 = 350 meters / hr ( since she covers 700 meters in 2 hrs ) speed of ken = 500 meters / hr at 8 : 00 , distance between ken and sari is 700 meters . ken needs to cover this and another 50 meters . time he will take = total distance to be covered / ... | a ) 13 : 00 , b ) 13 : 30 , c ) 14 : 00 , d ) 15 : 00 , e ) 15 : 30 | d | add(divide(add(700, 50), subtract(500, divide(700, const_2))), 10) | add(n4,n6)|divide(n4,const_2)|subtract(n5,#1)|divide(#0,#2)|add(n2,#3)| | physics |
the average weight of a , b and c is 30 kg . if the average weight of a and b be 25 kg and that of b and c be 28 kg , then the weight of b is : | explanation let a , b , c represent their respective weights . then , we have : a + b + c = ( 30 x 3 ) = 90 â € ¦ . ( i ) a + b = ( 25 x 2 ) = 50 â € ¦ . ( ii ) b + c = ( 28 x 2 ) = 56 â € ¦ . ( iii ) adding ( ii ) and ( iii ) , we get : a + 2 b + c = 106 â € ¦ . ( iv ) subtracting ( i ) from ( iv ) , we get : b = 16 .... | a ) 14 kg , b ) 15 kg , c ) 18 kg , d ) 19 kg , e ) 16 kg | e | subtract(add(multiply(25, const_2), multiply(28, const_2)), multiply(30, const_3)) | multiply(n1,const_2)|multiply(n2,const_2)|multiply(n0,const_3)|add(#0,#1)|subtract(#3,#2) | general |
the cost of a one - family home was $ 120000 in 1980 . in 1988 , the price had increased to $ 192000 . what was the percent increase in the cost of the home ? | increase = 192000 - 120000 = 72000 % increase = 72000 * 100 / 120000 = 60 % answer : option a | a ) 60 % , b ) 50 % , c ) 55 % , d ) 40 % , e ) 33.3 % | a | multiply(subtract(divide(192000, 120000), const_1), const_100) | divide(n3,n0)|subtract(#0,const_1)|multiply(#1,const_100) | general |
one fourth of one third of two fifth of a number is 20 . what will be 40 % of that number | explanation : ( 1 / 4 ) * ( 1 / 3 ) * ( 2 / 5 ) * x = 20 then x = 20 * 30 = 600 40 % of 600 = 240 answer : option d | a ) a ) 140 , b ) b ) 150 , c ) c ) 180 , d ) d ) 240 , e ) e ) 220 | d | divide(multiply(divide(20, multiply(multiply(divide(const_1, const_4), divide(const_1, const_3)), divide(const_2, add(const_2, const_3)))), 40), const_100) | add(const_2,const_3)|divide(const_1,const_4)|divide(const_1,const_3)|divide(const_2,#0)|multiply(#1,#2)|multiply(#3,#4)|divide(n0,#5)|multiply(n1,#6)|divide(#7,const_100) | gain |
a card board of 34 * 14 has to be attached to a wooden box and a total of 35 pins are to be used on the each side of the cardbox . find the total number of pins used . | ( 35 * 4 ) - 4 = 136 answer : b | a ) 135 , b ) 136 , c ) 137 , d ) 138 , e ) 139 | b | multiply(35, const_4) | multiply(n2,const_4) | general |
a 200 meter long train crosses a platform in 50 seconds while it crosses a signal pole in 42 seconds . what is the length of the platform ? | "speed = [ 200 / 42 ] m / sec = 100 / 21 m / sec . let the length of the platform be x meters . then , x + 200 / 50 = 100 / 21 21 ( x + 200 ) = 5000 è x = 38.09 m . answer : e" | a ) 39.9 , b ) 37.88 , c ) 35 , d ) 38.88 , e ) 38.09 | e | subtract(multiply(divide(200, 42), 50), 200) | divide(n0,n2)|multiply(n1,#0)|subtract(#1,n0)| | physics |
how much water should be added to 11 liters of a 20 % - solution of alcohol to reduce the concentration of alcohol in the solution by 75 % ? | "let x ltr water to be added 2 ltr alcohol to be represented as ( 20 ( 1 - 3 / 4 ( new soln . = 11 + x ) ) ) 2 = 5 % * ( 11 + x ) - - - - - - - - > x = 29 ans b" | a ) 25 liters , b ) 29 liters , c ) 30 liters , d ) 32 liters , e ) 35 liters | b | subtract(divide(multiply(const_2, const_100), subtract(20, multiply(divide(75, const_100), 20))), 11) | divide(n2,const_100)|multiply(const_100,const_2)|multiply(n1,#0)|subtract(n1,#2)|divide(#1,#3)|subtract(#4,n0)| | general |
a van takes 5 hours to cover a distance of 600 km . what speed in kph should the van maintain to cover the same distance in 3 / 2 of the previous time ? | "( 3 / 2 ) * 5 = 7.5 hours 600 / 7.5 = 80 kph the answer is a ." | a ) 80 , b ) 84 , c ) 88 , d ) 92 , e ) 96 | a | divide(600, multiply(divide(3, 2), 5)) | divide(n2,n3)|multiply(n0,#0)|divide(n1,#1)| | physics |
the ratio of ages of aman , bren , and charlie are in the ratio 5 : 8 : 7 respectively . if 8 years ago , the sum of their ages was 76 , what will be the age of bren 12 years from now ? | "let the present ages of aman , bren , and charlie be 5 x , 8 x and 7 x respectively . 5 x - 8 + 8 x - 8 + 7 x - 8 = 76 x = 5 present age of bren = 8 * 5 = 40 bren ' s age 12 years hence = 40 + 12 = 52 answer = e" | a ) 17 , b ) 25 , c ) 27 , d ) 52 , e ) 60 | e | add(multiply(divide(add(multiply(8, const_3), 76), add(add(5, 8), 7)), 8), 12) | add(n0,n1)|multiply(n1,const_3)|add(n4,#1)|add(n2,#0)|divide(#2,#3)|multiply(n1,#4)|add(n5,#5)| | general |
a and b undertake to do a piece of work for $ 500 . a alone can do it in 5 days while b alone can do it in 10 days . with the help of c , they finish it in 2 days . find the share of c ? | "c ' s 1 day work = ( 1 / 2 ) - ( 1 / 5 + 1 / 10 ) = 1 / 5 a : b : c = 1 / 5 : 1 / 10 : 1 / 5 = 2 : 1 : 2 b ' s share = 500 * 2 / 5 = $ 200 answer is a" | a ) $ 200 , b ) $ 150 , c ) $ 300 , d ) $ 25 , e ) $ 100 | a | multiply(divide(multiply(multiply(2, 10), inverse(5)), add(add(multiply(multiply(2, 10), subtract(inverse(2), add(inverse(5), inverse(10)))), multiply(multiply(2, 10), inverse(5))), multiply(multiply(2, 10), inverse(10)))), 500) | inverse(n1)|inverse(n3)|inverse(n2)|multiply(n2,n3)|add(#0,#2)|multiply(#0,#3)|multiply(#2,#3)|subtract(#1,#4)|multiply(#3,#7)|add(#8,#5)|add(#9,#6)|divide(#5,#10)|multiply(n0,#11)| | physics |
a clock store sold a certain clock to a collector for 15 percent more than the store had originally paid for the clock . when the collector tried to resell the clock to the store , the store bought it back at 45 percent of what the collector had paid . the shop then sold the clock again at a profit of 85 percent on its... | "now , in the question above , lets say the original cost of the clock to store was c $ and then it sold the same to the collector at 15 % profit . this means the clocks ' selling price was c ( 1.15 ) and this becomes cost price for the collector . now , when the collector tries to sell the same clock to the store , th... | a ) $ 168.58 , b ) $ 187.54 , c ) $ 158.74 , d ) $ 120.56 , e ) $ 200.84 | c | divide(multiply(80, divide(multiply(add(80, 85), divide(add(80, 15), const_2)), 80)), subtract(divide(add(80, 15), const_2), 15)) | add(n2,n3)|add(n0,n3)|divide(#1,const_2)|multiply(#0,#2)|subtract(#2,n0)|divide(#3,n3)|multiply(n3,#5)|divide(#6,#4)| | general |
angelina walked 100 meters from her home to the grocery at a constant speed . she then walked 180 meters to the gym at double the speed . she spent 40 seconds less on her way from the grocery to the gym than on her way from home to the grocery . what was angelina ' s speed , in meters per second , from the grocery to t... | "let the speed be x . . . so time taken from home to grocery = 100 / x . . the speed to gym = 2 x . . so time taken = 180 / 2 x = 90 / x . . its given 100 / x - 90 / x = 40 . . 10 / x = 40 . . x = 0.25 m / secs . . so grocery to gym = 2 * 0.25 = 0.5 m / s . . . answer : c" | a ) 2 , b ) 3 , c ) 0.5 , d ) 6 , e ) 12 | c | multiply(divide(subtract(100, divide(180, const_2)), 40), const_2) | divide(n1,const_2)|subtract(n0,#0)|divide(#1,n2)|multiply(#2,const_2)| | physics |
if 45 % of z is 39 % of y and y is 75 % of x , what percent of x is z ? | "( 45 / 100 ) z = ( 39 / 100 ) y and y = ( 75 / 100 ) x i . e . y = ( 3 / 4 ) x i . e . ( 45 / 100 ) z = ( 39 / 100 ) * ( 3 / 4 ) x i . e . z = ( 39 * 3 ) x / ( 45 * 4 ) i . e . z = ( 0.65 ) x = ( 65 / 100 ) x i . e . z is 65 % of x answer : option d" | a ) 200 , b ) 160 , c ) 100 , d ) 65 , e ) 50 | d | multiply(divide(divide(75, const_100), divide(divide(45, const_100), divide(39, const_100))), const_100) | divide(n2,const_100)|divide(n0,const_100)|divide(n1,const_100)|divide(#1,#2)|divide(#0,#3)|multiply(#4,const_100)| | gain |
10 camels cost as much as 24 horses , 16 horses cost as much as 4 oxen and 6 oxen as much as 4 elephants . if the cost of 10 elephants is rs . 130000 , find the cost of a camel ? | "cost of the camel = p 10 camels = 24 horses 16 horses = 4 oxen 6 oxen = 4 elephants 10 elephants = rs . 130000 p = rs . [ ( 24 * 4 * 4 * 130000 ) / ( 10 * 16 * 6 * 10 ) ] p = rs . ( 49920000 / 9600 ) = > p = rs . 5200 answer : d" | a ) s . 9800 , b ) s . 3800 , c ) s . 9800 , d ) s . 5200 , e ) s . 6880 | d | divide(multiply(multiply(multiply(24, 4), 4), 130000), multiply(multiply(multiply(10, 16), 6), 10)) | multiply(n1,n3)|multiply(n0,n2)|multiply(n3,#0)|multiply(n4,#1)|multiply(n7,#2)|multiply(n0,#3)|divide(#4,#5)| | general |
a worker makes a toy in every 3 h . if he works for 150 h , then how many toys will he make ? | no . of toys = 150 / 3 = 50 answer : d | a ) 40 , b ) 54 , c ) 45 , d ) 50 , e ) none | d | divide(150, 3) | divide(n1,n0) | physics |
in one hour , a boat goes 11 km along the stream and 5 km against the stream . the speed of the boat in still water in ( km / hr ) is | "solution speed in still water = 1 / 2 ( 11 + 5 ) km / hr = 8 km / hr . answer c" | a ) 3 , b ) 5 , c ) 8 , d ) 9 , e ) 7 | c | divide(add(11, 5), const_2) | add(n0,n1)|divide(#0,const_2)| | physics |
45 workers work 8 hours to dig a hole 30 meters deep . how many extra workers should be hired to dig another hole 40 meters deep by working for 6 hours ? | 45 workers * 8 hours / 30 meters = x * 6 / 40 x = 80 total workers 80 - 45 = 35 the answer is d . | a ) 75 , b ) 80 , c ) 50 , d ) 35 , e ) 40 | d | subtract(multiply(multiply(45, divide(8, 6)), divide(40, 30)), 45) | divide(n3,n2)|divide(n1,n4)|multiply(n0,#1)|multiply(#0,#2)|subtract(#3,n0) | physics |
if an amount of rs 42,900 is distributed equally amongst 22 persons , how much amount would each person get ? | required amount = 42900 / 22 = rs 1950 answer a | a ) rs 1950 , b ) rs 2000 , c ) rs 745 , d ) rs 765 , e ) none | a | divide(subtract(multiply(add(multiply(const_4, const_10), const_3), const_1000), const_100), 22) | multiply(const_10,const_4)|add(#0,const_3)|multiply(#1,const_1000)|subtract(#2,const_100)|divide(#3,n1) | general |
what is the perimeter , in meters , of a rectangular garden 8 meters wide that has the same area as a rectangular playground 16 meters long and 12 meters wide ? | "say length of garden is l meters . then ( l * 8 ) = 16 * 12 [ given area is same . area = l * b ] therefore l works out to 24 meters . perimeter of garden = 2 * ( 24 + 8 ) = 64 meter . answer is c" | a ) 48 , b ) 56 , c ) 64 , d ) 76 , e ) 192 | c | rectangle_perimeter(divide(rectangle_area(16, 12), 8), 8) | rectangle_area(n1,n2)|divide(#0,n0)|rectangle_perimeter(n0,#1)| | geometry |
what is the value of : - 9 ^ 7 + 9 ^ 9 ? | - 9 ^ 7 + 9 ^ 9 = 9 ^ 7 ( - 1 + 9 ^ 2 ) = 9 ^ 7 ( - 1 + 81 ) = 80 ( 9 ^ 7 ) , answer e . | a ) 9 ^ 2 , b ) - 18 ^ 16 , c ) 63 ( 9 ^ 7 ) , d ) - 80 ( 9 ^ 7 ) , e ) 80 ( 9 ^ 7 ) | e | divide(multiply(add(add(const_100, const_60), const_1), 9), const_100) | add(const_100,const_60)|add(#0,const_1)|multiply(n0,#1)|divide(#2,const_100)| | general |
let the sides of a rectangular prism are consecutive multiples of 5 . which among the following could be the base area | length = 5 * a breadth = 5 * ( a + 1 ) or 5 * ( a + 2 ) base area - > multiple of 25 450 = 25 * 18 answer : b | ['a ) 360', 'b ) 450', 'c ) 560', 'd ) 670', 'e ) 780'] | b | multiply(multiply(multiply(5, const_3), const_2), multiply(5, const_3)) | multiply(n0,const_3)|multiply(#0,const_2)|multiply(#1,#0) | geometry |
if teena is driving at 55 miles per hour and is currently 7.5 miles behind joe , who is driving at 40 miles per hour in the same direction then in how many minutes will teena be 45 miles ahead of joe ? | "this type of questions should be solved without any complex calculations as these questions become imperative in gaining that extra 30 - 40 seconds for a difficult one . teena covers 55 miles in 60 mins . joe covers 40 miles in 60 mins so teena gains 15 miles every 60 mins teena need to cover 7.5 + 45 miles . teena ca... | a ) 15 , b ) 60 , c ) 75 , d ) 90 , e ) 210 | e | multiply(divide(add(subtract(55, 40), 7.5), subtract(55, 40)), const_60) | subtract(n0,n2)|add(n1,#0)|divide(#1,#0)|multiply(#2,const_60)| | physics |
the sum of ages of 4 children born at the intervals of 10 years each is 80 years . what is the age of the youngest child ? | "let the ages of children be x , ( x + 3 ) , ( x + 6 ) , ( x + 9 ) and ( x + 12 ) years . then , x + ( x + 10 ) + ( x + 20 ) + ( x + 30 ) = 80 4 x = 20 x = 5 . age of the youngest child = x = 5 years . answer : b" | a ) 4 , b ) 5 , c ) 6 , d ) 7 , e ) 8 | b | subtract(subtract(divide(80, 4), 10), 10) | divide(n2,n0)|subtract(#0,n1)|subtract(#1,n1)| | general |
a bag contains 3 blue and 5 white marbles . one by one , marbles are drawn out randomly until only two are left in the bag . what is the probability x that out of the two , one is white and one is blue ? | "the required probability x = probability of choosing 6 balls out of the total 8 in such a way that we remove 4 out of 5 white and 2 out of 3 blue balls . ways to select 6 out of total 8 = 8 c 6 ways to select 4 out of 5 white balls = 5 c 4 ways to select 2 out of 3 blue balls = 3 c 2 thus the required probability = ( ... | a ) 15 / 56 , b ) 41 / 56 , c ) 13 / 28 , d ) 15 / 28 , e ) 5 / 14 | d | divide(multiply(3, 5), divide(multiply(add(3, 5), add(5, const_2)), const_2)) | add(n0,n1)|add(n1,const_2)|multiply(n0,n1)|multiply(#0,#1)|divide(#3,const_2)|divide(#2,#4)| | probability |
one hour after yolanda started walking from x to y , a distance of 17 miles , bob started walking along the same road from y to x . if yolanda ' s walking rate was 3 miles per hour and bob т ' s was 4 miles per hour , how many miles had bob walked when they met ? | when b started walking y already has covered 3 miles out of 17 , hence the distance at that time between them was 17 - 3 = 14 miles . combined rate of b and y was 3 + 4 = 7 miles per hour , hence they would meet each other in 14 / 7 = 2 hours . in 6 hours b walked 2 * 4 = 8 miles . answer : e . | a ) 24 , b ) 23 , c ) 22 , d ) 21 , e ) 8 | e | multiply(divide(subtract(17, 3), add(3, 4)), 4) | add(n1,n2)|subtract(n0,n1)|divide(#1,#0)|multiply(n2,#2) | physics |
a dozen eggs and 10 pounds of oranges are currently at the same price . if the price of a dozen eggs rises by 9 percent and the price of oranges rises by 6 % . how much more will it cost to buy a dozen eggs and 10 pounds of oranges . | "say currently both a dozen eggs and 10 pounds of oranges cost $ 100 ( they are at the same price ) . so , to buy a dozen eggs and 10 pounds of oranges we need $ 100 . after the increase , the price of a dozen eggs will be $ 109 and the price of 10 pounds of oranges will be $ 106 . so after the increase , to buy a doze... | a ) 5 % , b ) 10 % , c ) 15 % , d ) 20 % , e ) 25 % | c | multiply(divide(multiply(subtract(add(add(10, multiply(divide(9, const_100), 10)), add(10, multiply(divide(6, const_100), 10))), add(10, 10)), const_100), add(10, 10)), const_2) | add(n3,n3)|divide(n1,const_100)|divide(n2,const_100)|multiply(#1,n3)|multiply(n3,#2)|add(n3,#3)|add(n3,#4)|add(#5,#6)|subtract(#7,#0)|multiply(#8,const_100)|divide(#9,#0)|multiply(#10,const_2)| | general |
at what price must an article costing rs . 47.50 be marked in order that after deducting 5 % from the list price . it may be sold at a profit of 40 % on the cost price ? | "cp = 47.50 sp = 47.50 * ( 140 / 100 ) = 66.5 mp * ( 95 / 100 ) = 66.5 mp = 70 answer : e" | a ) 62.5 , b ) 69.5 , c ) 67.5 , d ) 62.5 , e ) 70 | e | divide(multiply(add(47.50, divide(multiply(47.50, 40), const_100)), const_100), subtract(const_100, 5)) | multiply(n0,n2)|subtract(const_100,n1)|divide(#0,const_100)|add(n0,#2)|multiply(#3,const_100)|divide(#4,#1)| | gain |
two friends decide to get together ; so they start riding bikes towards each other . they plan to meet halfway . each is riding at 6 mph . they live 36 miles apart . one of them has a pet carrier pigeon and it starts flying the instant the friends start traveling . the pigeon flies back and forth at 19 mph between the ... | "b 57 it takes 3 hours for the friends to meet ; so the pigeon flies for 3 hours at 19 mph = 57 miles" | a ) 54 , b ) 57 , c ) 60 , d ) 36 , e ) 96 | b | multiply(19, divide(divide(36, 2), 6)) | divide(n1,n3)|divide(#0,n0)|multiply(n2,#1)| | physics |
in a river flowing at 2 km / hr , a boat travels 48 km upstream and then returns downstream to the starting point . if its speed in still water be 6 km / hr , find the total journey time . | "explanation : speed of the boat = 6 km / hr speed downstream = ( 6 + 2 ) = 8 km / hr speed upstream = ( 6 - 2 ) = 4 km / hr distance traveled downstream = distance traveled upstream = 48 km total time taken = time taken downstream + time taken upstream = ( 48 / 8 ) + ( 48 / 4 ) = 18 hr . answer : option d" | a ) 10 hours , b ) 12 hours , c ) 14 hours , d ) 18 hours , e ) none of these | d | add(divide(48, subtract(6, 2)), divide(48, add(6, 2))) | add(n0,n2)|subtract(n2,n0)|divide(n1,#1)|divide(n1,#0)|add(#2,#3)| | general |
in smithtown , the ratio of right - handed people to left - handed people is 3 to 1 and the ratio of men to women is 3 to 2 . if the number of right - handed men is maximized , then what x percent of all the people in smithtown are left - handed women ? | "looking at the ratio we can take total number of people = 20 . . ans 5 / 20 or 25 % c" | a ) 50 % , b ) 40 % , c ) 25 % , d ) 20 % , e ) 10 % | c | multiply(divide(subtract(multiply(2, divide(add(3, 1), add(3, 2))), subtract(3, multiply(3, divide(add(3, 1), add(3, 2))))), add(3, 1)), const_100) | add(n0,n1)|add(n0,n3)|divide(#0,#1)|multiply(n3,#2)|multiply(n0,#2)|subtract(n0,#4)|subtract(#3,#5)|divide(#6,#0)|multiply(#7,const_100)| | general |
the number of degrees that the hour hand of a clock moves through between noon and 2.30 in the afternoon of the same day is ? | explanation : the hour hand moves from pointing to 12 to pointing to half way between 2 and 3 . the angle covered between each hour marking on the clock is 360 / 12 = 30 . since the hand has covered 2.5 of these divisions the angle moved through is 75 . answer : c | a ) 38 , b ) 27 , c ) 75 , d ) 28 , e ) 21 | c | multiply(multiply(add(floor(2.3), multiply(divide(subtract(2.3, floor(2.3)), const_60), const_100)), const_3), const_10) | floor(n0)|subtract(n0,#0)|divide(#1,const_60)|multiply(#2,const_100)|add(#0,#3)|multiply(#4,const_3)|multiply(#5,const_10) | physics |
a rectangular tiled patio is composed of 30 square tiles . the rectangular patio will be rearranged so that there will be 2 fewer columns of tiles and 4 more rows of tiles . after the change in layout , the patio will still have 30 tiles , and it will still be rectangular . how many rows are in the tile patio before th... | "suppose there are c columns and there are r rows original situation so , number of tiles = c * r = 30 also . reach column has r tiles and each row has c tiles new situation number of tiles in each column is r - 2 and number of tiles in each row is c + 4 so , number of rows = r - 2 and number of columns is c + 4 so , n... | a ) 5 , b ) 6 , c ) 10 , d ) 13 , e ) 28 | a | divide(30, divide(add(negate(4), sqrt(add(power(4, 2), multiply(4, multiply(30, 2))))), 2)) | multiply(n0,n1)|negate(n2)|power(n2,n1)|multiply(n2,#0)|add(#3,#2)|sqrt(#4)|add(#1,#5)|divide(#6,n1)|divide(n0,#7)| | geometry |
a row of ants comprised of 102 ants . a man walked upon it and all except 42 were crushed beneath his foot . how many ants are left alive ? | solution : all except 42 were crushed means that all the ants except 42 were crushed beneath the man ' s foot . hence , 42 ants are alive now answer b | a ) 41 , b ) 42 , c ) 43 , d ) 44 , e ) 45 | b | multiply(42, const_1) | multiply(n1,const_1) | general |
at a certain food stand , the price of each apple is 40 ¢ and the price of each orange is 60 ¢ . mary selects a total of 10 apples and oranges from the food stand , and the average ( arithmetic mean ) price of the 10 pieces of fruit is 54 ¢ . how many oranges must mary put back so that the average price of the pieces o... | "if the average price is 54 , then mary has 7 oranges and 3 apples ( a ratio of 7 : 3 ) . for the average price to be 45 , the ratio should be 1 : 3 . mary must put back 6 oranges . the answer is e ." | a ) 2 , b ) 3 , c ) 4 , d ) 5 , e ) 6 | e | divide(subtract(add(multiply(divide(subtract(multiply(60, 10), multiply(54, 10)), subtract(60, 40)), 40), multiply(subtract(10, divide(subtract(multiply(60, 10), multiply(54, 10)), subtract(60, 40))), 60)), multiply(45, 10)), subtract(60, 45)) | multiply(n1,n2)|multiply(n2,n4)|multiply(n2,n5)|subtract(n1,n0)|subtract(n1,n5)|subtract(#0,#1)|divide(#5,#3)|multiply(n0,#6)|subtract(n2,#6)|multiply(n1,#8)|add(#7,#9)|subtract(#10,#2)|divide(#11,#4)| | general |
what percent is 50 gm of 1 kg ? | "1 kg = 1000 gm 50 / 1000 × 100 = 5000 / 1000 = 5 % e )" | a ) 0.5 % , b ) 1 % , c ) 1.5 % , d ) 2 % , e ) 5 % | e | multiply(divide(50, 1), const_100) | divide(n0,n1)|multiply(#0,const_100)| | gain |
the dimensions of a rectangular solid are 6 inches , 5 inches , and 8 inches . if a cube , a side of which is equal to one of the dimensions of the rectangular solid , is placed entirely within thespherejust large enough to hold the cube , what the ratio of the volume of the cube to the volume within thespherethat is n... | answer : d . | ['a ) 10 : 17', 'b ) 2 : 5', 'c ) 5 : 16', 'd ) 10 : 19', 'e ) 32 : 25'] | d | divide(const_10, add(add(multiply(8, const_2), const_2), const_1)) | multiply(n2,const_2)|add(#0,const_2)|add(#1,const_1)|divide(const_10,#2) | geometry |
john purchased some shirts and trousers for $ 1300 . he paid $ 500 less for the shirts than he did for the trousers . if he bought 5 shirts and the cost of a shirt is $ 20 less than that of a trouser , how many trousers did he buy ? | "given that the total purchase of two items cost 1300 so the average purchase of one item will cost 1300 / 2 = 650 . its given as total shirt cost 500 $ less . hence total shirt cost = 650 - 250 and total trouser cost = 650 + 250 5 shirts = 400 $ = = > one shirt = 80 $ one trouser = 80 + 20 = 100 $ total trousers = 900... | a ) 4 , b ) 5 , c ) 6 , d ) 7 , e ) 9 | e | divide(subtract(1300, multiply(5, add(20, 20))), add(add(20, 20), 20)) | add(n3,n3)|add(n3,#0)|multiply(n2,#0)|subtract(n0,#2)|divide(#3,#1)| | general |
a contractor undertakes to complete the construction of a tunnel 720 meters long in 240 days and employs 50 men for the purpose . after 120 days , he finds that only 240 meters of the tunnel is complete . how many more men should be employ in order to complete the work in time ? | "in 120 days , only 240 m of the tunnel is constructed by 50 men . the remaining 120 days , 480 m of the tunnel can be constructed by 120 men . additional number of men required = 120 - 50 = 70 men . answer : c" | a ) 75 men , b ) 80 men , c ) 70 men , d ) 60 men , e ) 50 men | c | subtract(120, 50) | subtract(n3,n2)| | physics |
for all integers x > 1 . < x > = 2 x + ( 2 x - 1 ) + ( 2 x - 2 ) + . . . . . . 2 + 1 . what is the value of < 2 > * < 4 > ? | since < x > = 2 x + ( 2 x - 1 ) + ( 2 x - 2 ) + . . . . . . 2 + 1 = 1 + 2 + . . + ( 2 x - 2 ) + ( 2 x - 1 ) + 2 x then < x > is basically the sum of all integers from 1 to 2 x , inclusive . hence < 2 > is the sum of all integers from 1 to 2 * 2 = 4 and < 4 > is the sum of all integers from 1 to 2 * 4 = 8 - > < 4 > = 36... | a ) 60 , b ) 116 , c ) 210 , d ) 263 , e ) 360 | e | multiply(add(add(add(add(add(add(add(multiply(const_4, 2), subtract(multiply(const_4, 2), const_1)), subtract(multiply(const_4, 2), const_2)), subtract(multiply(const_4, 2), const_3)), subtract(multiply(const_4, 2), const_4)), subtract(multiply(const_4, 2), subtract(multiply(const_4, 2), const_3))), subtract(multiply(c... | multiply(n1,const_4)|multiply(n1,const_2)|subtract(#0,const_1)|subtract(#0,const_2)|subtract(#0,const_3)|subtract(#0,const_4)|subtract(#1,const_1)|subtract(#1,const_2)|subtract(#1,const_3)|add(#0,#2)|add(#1,#6)|subtract(#0,#4)|subtract(#0,#3)|subtract(#0,#2)|add(#9,#3)|add(#10,#7)|add(#14,#4)|add(#15,#8)|add(#16,#5)|ad... | general |
two identical circles intersect such that their centers and the points at which they intersect form a square of side 4 cm . what is the area of the region where the two circles intersect ? | drawing a diagram , we see that the radius of the circle equals the side of the square . the area of intersection is : pi * r ^ 2 / 4 - ( r ^ 2 - pi * r ^ 2 / 4 ) = pi * r ^ 2 / 2 - r ^ 2 = 8 pi - 16 the answer is d . | ['a ) 16 - 4 pi', 'b ) 4 - pi', 'c ) 2 pi - 4', 'd ) 8 pi - 16', 'e ) pi ^ 2 - 4'] | d | add(divide(multiply(const_3, const_pi), const_2), 4) | multiply(const_3,const_pi)|divide(#0,const_2)|add(n0,#1) | geometry |
pradeep has to obtain 35 % of the total marks to pass . he got 150 marks and failed by 25 marks . the maximum marks are | "explanation : let their maximum marks be x . then , 35 % of x = 150 + 25 = > 35 / 100 x = 175 x = ( 17500 / 35 ) x = 500 . answer : d" | a ) 300 , b ) 600 , c ) 800 , d ) 500 , e ) 900 | d | divide(add(150, 25), divide(35, const_100)) | add(n1,n2)|divide(n0,const_100)|divide(#0,#1)| | general |
if 4 people are selected from a group of 6 married couples , what is the probability that none of them would be married to each other ? | "each couple can send only one ` ` representative ' ' to the committee . we can choose 4 couples ( as there should be 4 members ) to send only one ` ` representatives ' ' to the committee in c 46 c 64 # of ways . but these 4 chosen couples can send two persons ( either husband or wife ) : 2 ∗ 2 ∗ 2 ∗ 2 = 242 ∗ 2 ∗ 2 ∗ ... | a ) 1 / 33 , b ) 2 / 33 , c ) 1 / 3 , d ) 16 / 33 , e ) 11 / 12 | d | multiply(multiply(multiply(divide(multiply(6, const_2), multiply(6, const_2)), divide(multiply(4, 4), subtract(multiply(6, const_2), const_1))), divide(subtract(multiply(4, 4), const_2), multiply(4, 4))), divide(subtract(subtract(multiply(4, 4), const_2), const_2), subtract(multiply(4, 4), const_1))) | multiply(n0,n0)|multiply(n1,const_2)|divide(#1,#1)|subtract(#0,const_2)|subtract(#0,const_1)|subtract(#1,const_1)|divide(#3,#0)|divide(#0,#5)|subtract(#3,const_2)|divide(#8,#4)|multiply(#2,#7)|multiply(#6,#10)|multiply(#9,#11)| | probability |
if the length of a rectangle is increased by 20 % and the breadth is reduced by 20 % , what will be the effect on its area ? | percentage change = x - y - xy ⁄ 100 = 20 – 20 – 20 × 20 / 100 = − 4 % = 4 % decrease answer d | ['a ) 4 % increase', 'b ) 6 % increase', 'c ) 5 % decrease', 'd ) 4 % decrease', 'e ) none of these'] | d | multiply(subtract(const_1, multiply(add(const_1, divide(20, const_100)), subtract(const_1, divide(20, const_100)))), const_100) | divide(n0,const_100)|add(#0,const_1)|subtract(const_1,#0)|multiply(#1,#2)|subtract(const_1,#3)|multiply(#4,const_100) | geometry |
each light bulb at a hotel is either incandescent or fluorescent . at a certain moment , forty percent of the incandescent bulbs are switched on , and eighty percent of the fluorescent bulbs are switched on . if 75 percent of all the bulbs are switched on at this moment , what percent of the bulbs that are switched on ... | "let i be the number of incandescent bulbs . let f be the number of fluorescent bulbs . 0.4 i + 0.8 f = 0.75 ( i + f ) 0.05 f = 0.35 i f = 7 i this means that for every 1 incandescent bulb , there are 7 fluorescent bulbs . the percent of bulbs that are switched on which are incandescent is : 0.4 i / ( 0.4 i + 0.8 f ) =... | a ) 6.7 % , b ) 10.3 % , c ) 14.5 % , d ) 18.2 % , e ) 23.6 % | a | divide(multiply(subtract(const_100, 75), multiply(add(divide(const_2, const_0_25), const_2), const_4)), const_100) | divide(const_2,const_0_25)|subtract(const_100,n0)|add(#0,const_2)|multiply(#2,const_4)|multiply(#3,#1)|divide(#4,const_100)| | gain |
a and b can do a piece of work in 2 days and 6 days respectively . both work for 1 days and then a goes away . find how long will b take to complete the remaining work ? | 1 / 2 + ( 1 + x ) / 6 = 1 x = 2 days answer : b | a ) 1 days , b ) 2 days , c ) 3 days , d ) 4 days , e ) 5 days | b | divide(subtract(const_1, add(divide(1, 6), divide(1, 2))), divide(1, 6)) | divide(n2,n1)|divide(n2,n0)|add(#0,#1)|subtract(const_1,#2)|divide(#3,#0) | physics |
karen places a bet with tom that she will beat tom in a car race by 4 miles even if karen starts 4 minutes late . assuming that karen drives at an average speed of 60 mph and tom drives at an average speed of 45 mph , how many h miles will tom drive before karen wins the bet ? | "let k and t be the speeds of karen and tom respectively . t be the time that karen will travel - - - - > t + 4 / 60 will be the total time tom will travel by the time the distance between karen and tom is 4 miles . thus , per the question , k ( t ) - t ( t + 4 / 60 ) = 4 - - - > t = 7 / 15 hours thus the distance trav... | a ) 15 , b ) 18 , c ) 21 , d ) 24 , e ) 27 | d | subtract(divide(add(multiply(divide(45, 60), 4), 4), subtract(divide(60, 60), divide(45, 60))), 4) | divide(n3,n2)|divide(n2,n2)|multiply(n0,#0)|subtract(#1,#0)|add(n0,#2)|divide(#4,#3)|subtract(#5,n0)| | physics |
rs . 1210 were divided among 3 person p , q , r so that p : q = 5 : 4 and q : r = 9 : 10 . then r gets the amount . | "p : q = 5 : 4 , q : r = 9 : 10 = ( 9 x 4 / 9 ) : ( 10 x 4 / 9 ) = 4 : 40 / 9 . so , p : q : r = 5 : 4 : 40 / 9 = 45 : 36 : 40 sum of ratio terms is = ( 45 + 36 + 40 ) = 121 . r share of amount is rs ( 1210 x 40 / 121 ) = rs . 400 . b" | a ) rs . 300 , b ) rs . 400 , c ) rs . 480 , d ) rs . 490 , e ) rs . 500 | b | multiply(divide(1210, add(add(divide(5, 4), divide(10, 9)), const_1)), divide(10, 9)) | divide(n2,n3)|divide(n5,n4)|add(#0,#1)|add(#2,const_1)|divide(n0,#3)|multiply(#4,#1)| | general |
in what ratio should a variety of rice costing rs . 7 per kg be mixed with another variety of rice costing rs . 14.50 per kg to obtain a mixture costing rs . 9.75 per kg ? | "let us say the ratio of the quantities of cheaper and dearer varieties = x : y by the rule of allegation , x / y = ( 14.50 - 9.75 ) / ( 9.75 - 7 ) = 31 / 15 answer : b" | a ) 19 : 10 , b ) 19 : 11 , c ) 11 : 19 , d ) 12 : 18 , e ) 11 : 20 | b | divide(divide(subtract(14.50, 9.75), subtract(14.50, 7)), subtract(const_1, divide(subtract(14.50, 9.75), subtract(14.50, 7)))) | subtract(n1,n2)|subtract(n1,n0)|divide(#0,#1)|subtract(const_1,#2)|divide(#2,#3)| | other |
if 10 is added to thrice a number and this sum is multiplied by 5 , the result is the same as if the number is multiplied by 5 and 13 is added to the product . what is the number ? | "let the number be x ; 5 ( 10 + 3 x ) = 5 x + 13 ; x = - 3.7 answer : b" | a ) - 3.5 , b ) - 3.7 , c ) - 5 , d ) - 9 , e ) 11 | b | divide(subtract(multiply(10, 5), 13), subtract(5, multiply(const_3, 5))) | multiply(n0,n1)|multiply(n1,const_3)|subtract(#0,n3)|subtract(n2,#1)|divide(#2,#3)| | general |
if a 2 + b 2 + c 2 = 390 and ab + bc + ca = 5 , then a + b + c is | by formula , ( a + b + c ) ^ 2 = a ^ 2 + b ^ 2 + c ^ 2 + 2 ( ab + bc + ca ) , since , a ^ 2 + b ^ 2 + c ^ 2 = 390 and ab + bc + ca = 5 , ( a + b + c ) ^ 2 = 390 + 2 ( 5 ) = 400 = 20 ^ 2 therefore : a + b + c = 20 answer : b | a ) 19 , b ) 20 , c ) 25 , d ) 30 , e ) 64 | b | sqrt(add(390, multiply(5, const_2))) | multiply(n4,const_2)|add(n3,#0)|sqrt(#1) | general |
in an office in singapore there are 60 % female employees . 50 % of all the male employees are computer literate . if there are total 62 % employees computer literate out of total 1500 employees , then the no . of female employees who are computer literate ? | "solution : total employees , = 1500 female employees , 60 % of 1500 . = ( 60 * 1500 ) / 100 = 900 . then male employees , = 600 50 % of male are computer literate , = 300 male computer literate . 62 % of total employees are computer literate , = ( 62 * 1500 ) / 100 = 930 computer literate . thus , female computer lite... | a ) 630 , b ) 674 , c ) 672 , d ) 960 , e ) none | a | multiply(subtract(divide(62, const_100), multiply(subtract(const_1, divide(60, const_100)), divide(50, const_100))), 1500) | divide(n2,const_100)|divide(n1,const_100)|divide(n0,const_100)|subtract(const_1,#2)|multiply(#1,#3)|subtract(#0,#4)|multiply(n3,#5)| | gain |
what will be the remainder when ( 27 ^ 27 ) + 27 is divided by 28 ? | "x ^ n + 1 will be divisible by x + 1 only when n is odd 27 ^ 27 + 1 will be divisible by 27 + 1 ( 27 ^ 27 + 1 ) + 26 , when divided by 28 will give 26 as remainder answer is d" | a ) 22 , b ) 20 , c ) 24 , d ) 26 , e ) 28 | d | power(27, 27) | power(n0,n0)| | general |
a man has some hens and cows . if the number of heads be 48 and the number of feet equals 140 , then the number of hens will be | let number of hens = h number of cows = c number of heads = 48 ⇒ h + c = 48 ⋯ ( 1 ) number of feet = 140 ⇒ 2 h + 4 c = 140 ⇒ h + 2 c = 70 ⋯ ( 2 ) ( 2 ) − ( 1 ) gives 2 c − c = 70 − 48 ⇒ c = 22 substituting the value of c in ( 1 ) , we get h + 22 = 48 ⇒ h = 48 − 22 = 26 h + 22 = 48 ⇒ h = 48 − 22 = 26 i . e . , number of... | a ) 22 , b ) 24 , c ) 26 , d ) 20 , e ) 18 | c | divide(subtract(multiply(48, const_4), 140), const_2) | multiply(n0,const_4)|subtract(#0,n1)|divide(#1,const_2) | general |
the cash realised on selling a 14 % stock is rs . 120.50 , brokerage being 1 / 4 % is | "explanation : cash realised = rs . ( 120.50 - 0.25 ) = rs . 120.25 . answer : a" | a ) 120.25 , b ) 120.35 , c ) 106.5 , d ) 156 , e ) 240 | a | subtract(120.50, divide(1, 4)) | divide(n2,n3)|subtract(n1,#0)| | general |
a and b are two partially filled buckets of water . if 7 liters are transferred from a to b , then a would contain one - third of the amount of water in b . alternatively , if 7 liters are transferred from b to a , b would contain one - half of the amount of water in a . bucket a contains how many liters of water ? | "let bucket a be a and bucket b be b scenario 1 a - 7 = 1 / 3 ( b + 7 ) - - - - > 3 a - 21 = b + 7 scenario 2 b - 7 = 1 / 2 ( a + 7 ) - - - - - > 2 b - 14 = a + 7 from scenario 1 , b = 3 a - 28 substitute b with this information in stmt 2 2 ( 3 a - 28 ) - 14 = a + 7 - - - - - - > 6 a - 56 - 14 = a + 7 - - - - - - > 6 a... | a ) 77 / 5 , b ) 13 , c ) 17 , d ) 21 , e ) 23 | a | divide(multiply(add(add(7, const_3), const_2), divide(7, const_2)), add(const_2, divide(const_1, const_2))) | add(n0,const_3)|divide(n0,const_2)|divide(const_1,const_2)|add(#0,const_2)|add(#2,const_2)|multiply(#3,#1)|divide(#5,#4)| | general |
how many liters of water must be evaporated from 50 liters of a 2 percent sugar solution to get a 5 percent sugar solution ? | "let x be the amount that needs to be evaporated . 0.02 ( 50 ) = 0.05 ( 50 - x ) 0.05 x = 2.5 - 1 x = 1.5 / 0.05 = 30 liters the answer is c ." | a ) 20 , b ) 25 , c ) 30 , d ) 32 , e ) 35 | c | subtract(50, multiply(divide(50, const_100), 5)) | divide(n0,const_100)|multiply(n2,#0)|subtract(n0,#1)| | gain |
an equilateral triangle t 2 is formed by joining the mid points of the sides of another equilateral triangle t 1 . a third equilateral triangle t 3 is formed by joining the mid - points of t 2 and this process is continued indefinitely . if each side of t 1 is 50 cm , find the sum of the perimeters of all the triangles... | "we have 50 for first triangle , when we join mid - points of first triangle we get the second equilateral triangle then the length of second one is 25 and continues . so we have 50,25 , 12.5 , . . . we have ratio = 1 / 2 , and it is gp type . sum of infinite triangle is a / 1 - r = 50 / 1 - ( 1 / 2 ) = 100 equilateral... | a ) 180 cm , b ) 220 cm , c ) 240 cm , d ) 270 cm , e ) 300 cm | e | add(triangle_perimeter(50, 50, 50), triangle_perimeter(50, 50, 50)) | triangle_perimeter(n5,n5,n5)|add(#0,#0)| | geometry |
three walls have wallpaper covering a combined area of 300 square meters . by overlapping the wallpaper to cover a wall with an area of 180 square meters , the area that is covered by exactly two layers of wallpaper is 40 square meters . what is the area that is covered with three layers of wallpaper ? | "300 - 180 = 120 sq m of the wallpaper overlaps ( in either two layers or three layers ) if 36 sq m has two layers , 120 - 40 = 80 sq m of the wallpaper overlaps in three layers . 80 sq m makes two extra layers hence the area over which it makes two extra layers is 40 sq m . answer ( c ) ." | a ) 5 square meters , b ) 36 square meters , c ) 40 square meters , d ) 83.3 square meters , e ) 120 square meters | c | divide(subtract(subtract(300, 180), 40), const_2) | subtract(n0,n1)|subtract(#0,n2)|divide(#1,const_2)| | geometry |
a trained covered x km at 40 kmph and another 2 x km at 20 kmph . find the average speed of the train in covering the entire 4 x km . | "total time taken = x / 40 + 2 x / 20 hours = 5 x / 40 = x / 8 hours average speed = 4 x / ( x / 8 ) = 32 kmph answer : c" | a ) 16 , b ) 18 , c ) 32 , d ) 19 , e ) 12 | c | divide(multiply(40, 4), add(divide(40, 40), divide(multiply(2, 40), 20))) | divide(n0,n0)|multiply(n0,n3)|multiply(n0,n1)|divide(#2,n2)|add(#0,#3)|divide(#1,#4)| | general |
if n = 2.0254 and n * is the decimal obtained by rounding n to the nearest hundredth , what is the value of n * – n ? | "n * = 2.03 n * - n = 2.03 - 2.0254 0.0046 answer : d" | a ) - 0.0053 , b ) - 0.0003 , c ) 0.0007 , d ) 0.0046 , e ) 0.0153 | d | subtract(subtract(2.0254, divide(divide(add(multiply(const_2, const_10), const_3), const_100), const_100)), 2.0254) | multiply(const_10,const_2)|add(#0,const_3)|divide(#1,const_100)|divide(#2,const_100)|subtract(n0,#3)|subtract(#4,n0)| | general |
find the sum lend at c . i . at 5 p . c per annum will amount to rs . 240 in 2 years ? | "explanation : 240 = p ( 21 / 20 ) 2 p = 217 answer : a" | a ) 217 , b ) 287 , c ) 400 , d ) 288 , e ) 171 | a | divide(240, power(add(const_1, divide(5, const_100)), 2)) | divide(n0,const_100)|add(#0,const_1)|power(#1,n2)|divide(n1,#2)| | general |
a bag contains 7 green balls and 7 white balls . if two balls are drawn simultaneously , what is the probability that both balls are the same colour ? | "the total number of ways to draw two balls is 14 c 2 = 91 the number of ways to draw two green balls is 7 c 2 = 21 the number of ways to draw two white balls is 7 c 2 = 21 p ( two balls of the same colour ) = 42 / 91 = 6 / 13 the answer is b ." | a ) 4 / 13 , b ) 6 / 13 , c ) 8 / 21 , d ) 11 / 21 , e ) 17 / 42 | b | add(multiply(divide(7, add(7, 7)), divide(subtract(7, const_1), subtract(add(7, 7), const_1))), multiply(divide(subtract(7, const_1), subtract(add(7, 7), const_1)), divide(7, add(7, 7)))) | add(n0,n1)|subtract(n1,const_1)|subtract(n0,const_1)|divide(n1,#0)|divide(n0,#0)|subtract(#0,const_1)|divide(#1,#5)|divide(#2,#5)|multiply(#3,#6)|multiply(#7,#4)|add(#8,#9)| | other |
a vendor sells 60 percent of apples he had and throws away 15 percent of the remainder . next day he sells 50 percent of the remainder and throws away the rest . what percent of his apples does the vendor throw ? | explanation : let , the number of apples be 100 . on the first day he sells 60 % apples i . e . , 60 apples . remaining apples = 40 . he throws 15 % of the remaining i . e . , 15 % of 40 = 6 . now he has 40 − 6 = 34 apples the next day he throws 50 % of the remaining 34 apples i . e . , 17 . therefore , in total he thr... | a ) 17 , b ) 23 , c ) 77 , d ) 26 , e ) 30 | b | add(multiply(subtract(const_100, 60), divide(15, const_100)), multiply(divide(50, const_100), subtract(subtract(const_100, 60), multiply(subtract(const_100, 60), divide(15, const_100))))) | divide(n1,const_100)|divide(n2,const_100)|subtract(const_100,n0)|multiply(#0,#2)|subtract(#2,#3)|multiply(#1,#4)|add(#3,#5) | gain |
the radius of a cylinder is 8 m , height 14 m . the volume of the cylinder is : | "cylinder volume = ï € r ( power 2 ) h = 22 / 7 ã — 8 ã — 8 ã — 14 = 2816 m ( power 3 ) answer is a ." | a ) 2816 , b ) 5500 , c ) 3300 , d ) 1100 , e ) 4400 | a | multiply(circumface(8), 14) | circumface(n0)|multiply(n1,#0)| | geometry |
average of money that group of 4 friends pay for rent each month is $ 800 . after one persons rent is increased by 20 % the new mean is $ 880 . what was original rent of friend whose rent is increased ? | 0.2 x = 4 ( 880 - 800 ) 0.2 x = 320 x = 1600 answer c | a ) 800 , b ) 900 , c ) 1600 , d ) 1100 , e ) 1200 | c | divide(multiply(subtract(880, 800), 4), divide(20, const_100)) | divide(n2,const_100)|subtract(n3,n1)|multiply(n0,#1)|divide(#2,#0) | general |
when a student joe , weighing 43 kg , joins a group of students whose average weight is 30 kg , the average weight goes up by 1 kg . subsequently , if two students , excluding joe , leave the group the average weight comes back to 30 kg . what is the difference between the average weight of the two students who left an... | "after two persons leave the group the average remains the same . that means the weight of the two persons = 43 + 30 = 73 so , the average the two persons = 36.5 that gives the answer 43 - 36.5 = 6.5 answer d" | a ) 5.5 kg , b ) 11 kg , c ) 30 kg , d ) 6.5 kg , e ) 71 kg | d | subtract(43, divide(subtract(add(multiply(30, subtract(43, add(30, 1))), 43), multiply(subtract(subtract(43, add(30, 1)), 1), 30)), const_2)) | add(n1,n2)|subtract(n0,#0)|multiply(n1,#1)|subtract(#1,n2)|add(n0,#2)|multiply(n1,#3)|subtract(#4,#5)|divide(#6,const_2)|subtract(n0,#7)| | general |
while working alone at their respective constant rates , server p uploads 480 files in 4 hours and server y uploads 480 files in 8 hours . if all files uploaded by these servers are the same size , how long would it take the two servers , working at the same time and at their respective constant rates , to process a to... | server p processes 480 / 4 files per hour = 120 per hour server y processes 180 / 8 files per hour = 60 per hour total files processed per hour when p and y work together = 120 + 60 per hour = 180 files per hour 480 / 180 = 2 2 / 3 hours = c | a ) 2 hr , b ) 2 hr 20 min , c ) 2 hr 40 min , d ) 5 hr 40 min , e ) 6 hr | c | divide(480, const_2) | divide(n0,const_2) | physics |
0.0203 x 2.92 / 0.0073 x 14.5 x 0.7 = ? | "= 0.0203 x 2.92 / 0.0073 x 14.5 x 0.7 = 203 x 292 / 73 x 145 x 7 = 4 / 5 = 0.8 answer is c ." | a ) 0.7 , b ) 0 , c ) 0.8 , d ) 8 , e ) 7 | c | divide(subtract(multiply(0.0203, 2.92), multiply(0.0073, 14.5)), 0.7) | multiply(n0,n1)|multiply(n2,n3)|subtract(#0,#1)|divide(#2,n4)| | general |
the speed of a train is 90 kmph . what is the distance covered by it in 30 minutes ? | "90 * 30 / 60 = 45 kmph answer : e" | a ) 15 , b ) 66 , c ) 77 , d ) 52 , e ) 45 | e | multiply(divide(30, const_60), 90) | divide(n1,const_60)|multiply(n0,#0)| | physics |
what is the smallest positive integer x such that 280 - x is the cube of a positive integer | "given 280 - x is a perfect cube so we will take 216 = 6 * 6 * 6 280 - x = 216 x = 280 - 216 = 64 correct option is b" | a ) 5 , b ) 64 , c ) 62 , d ) 60 , e ) 12 | b | add(const_3, const_4) | add(const_3,const_4)| | general |
in a kilometer race , a beats b by 60 meters or 15 seconds . what time does a take to complete the race ? | "time taken by b run 1000 meters = ( 1000 * 15 ) / 60 = 250 sec . time taken by a = 250 - 15 = 235 sec . answer : b" | a ) 221 , b ) 235 , c ) 245 , d ) 289 , e ) 260 | b | subtract(divide(multiply(const_1, const_1000), divide(60, 15)), 15) | divide(n0,n1)|multiply(const_1,const_1000)|divide(#1,#0)|subtract(#2,n1)| | physics |
in a certain diving competition , 5 judges score each dive on a scale from 1 to 10 . the point value of the dive is obtained by dropping the highest score and the lowest score and multiplying the sum of the remaining scores by the degree of difficulty . if a dive with a degree of difficulty of 3.2 received scores of 7.... | "degree of difficulty of dive = 3.2 scores are 6.0 , 7.5 , 8.0 , 8.5 and 9.0 we can drop 6.0 and 9.0 sum of the remaining scores = ( 7.5 + 8.3 + 8.6 ) = 24 point of value of the dive = 24.4 * 3.2 = 78.08 answer a" | a ) 78.08 , b ) 73.6 , c ) 75.2 , d ) 76.8 , e ) 81.6 | a | multiply(add(add(7.5, 8.3), 8.6), 3.2) | add(n4,n5)|add(n8,#0)|multiply(n3,#1)| | general |
train starts from amritsar to bombay at 9 am . it reaches destination after 3 days at 9 : 30 am . every day a train starts . how many trains does it come across on the way ? | because one train taking 3 days so 1 train first day another at second day and 3 rd one at third day but as given train reaches at 9.30 am insted of 9 . oo am so d same day train already left that train will be 4 th train . answer : b | a ) 5 th , b ) 4 th , c ) 6 th , d ) 7 th , e ) 8 th | b | add(3, const_1) | add(n1,const_1) | physics |
if a ( a - 9 ) = 10 and b ( b - 9 ) = 10 , where a ≠ b , then a + b = | "i . e . if a = - 1 then b = 10 or if a = 10 then b = - 1 but in each case a + b = - 1 + 10 = 9 answer : option d" | a ) − 48 , b ) − 2 , c ) 2 , d ) 9 , e ) 48 | d | subtract(subtract(subtract(subtract(add(add(9, 10), subtract(9, 10)), const_1), const_1), const_1), const_1) | add(n0,n1)|subtract(n0,n1)|add(#0,#1)|subtract(#2,const_1)|subtract(#3,const_1)|subtract(#4,const_1)|subtract(#5,const_1)| | general |
there are 12 teams in a certain league and each team plays each of the other teams exactly twice . if each game is played by 2 teams , what is the total number of games played ? | "every team plays with 11 teams . . . so total no of matches = 12 x 11 = 132 . now , each match is played twice = > 132 x 2 but 2 teams play a match = > 132 x 2 / 2 = 132 . answer : e" | a ) 15 , b ) 16 , c ) 28 , d ) 56 , e ) 132 | e | divide(multiply(12, subtract(12, const_1)), const_2) | subtract(n0,const_1)|multiply(n0,#0)|divide(#1,const_2)| | general |
at what rate percent per annum will the simple interest on a sum of money be 1 / 5 of the amount in 10 years ? | "let sum = x . then , s . i . = x / 5 , time = 10 years . rate = ( 100 * x ) / ( x * 5 * 10 ) = 2 % answer : a" | a ) 2 % , b ) 7 % , c ) 9 % , d ) 3 % , e ) 1 % | a | divide(multiply(divide(1, 5), const_100), 10) | divide(n0,n1)|multiply(#0,const_100)|divide(#1,n2)| | gain |
a man can do a job in 15 days . his father takes 10 days and his son finishes it in 25 days . how long will they take to complete the job if they all work together ? | "1 day work of the three persons = ( 1 / 15 + 1 / 10 + 1 / 25 ) = 21 / 150 so , all three together will complete the work in 150 / 21 = 7.1 days . answer : c" | a ) 6.3 , b ) 6.9 , c ) 7.1 , d ) 6.1 , e ) 6.2 | c | divide(const_1, add(divide(const_1, 25), add(divide(const_1, 15), divide(const_1, 10)))) | divide(const_1,n0)|divide(const_1,n1)|divide(const_1,n2)|add(#0,#1)|add(#3,#2)|divide(const_1,#4)| | physics |
in a neighborhood having 90 households , 11 did not have either a car or a bike . if 22 households had a both a car and a bike and 44 had a car , how many had bike only ? | { total } = { car } + { bike } - { both } + { neither } - - > 90 = 44 + { bike } - 22 + 11 - - > { bike } = 57 - - > # those who have bike only is { bike } - { both } = 57 - 22 = 35 . answer : b . | a ) 30 , b ) 35 , c ) 20 , d ) 18 , e ) 10 | b | subtract(subtract(add(subtract(90, 11), 22), 44), 22) | subtract(n0,n1)|add(n2,#0)|subtract(#1,n3)|subtract(#2,n2) | other |
the 3 digits of a number add up to 11 . the number is divisible by 5 . the leftmost digit is double the middle digit . what is the product of the 3 digits ? | xyz is a 3 digit integer divisible by 5 . so z can take only 0 or 5 . we are also given that x = 2 y this means x + y ( 2 y + y ) must be a multiple of 3 xy 5 or xy 0 case 1 x + y + 0 = 11 x + y = 11 this ca n ' t be the case because 11 is not a multiple of 3 case 2 x + y + 5 = 11 x + y = 6 solving we get , x = 4 y = 2... | a ) 38 , b ) 40 , c ) 78 , d ) 88 , e ) 125 | b | multiply(multiply(multiply(divide(subtract(11, 5), 3), const_2), divide(subtract(11, 5), 3)), 5) | subtract(n1,n2)|divide(#0,n0)|multiply(#1,const_2)|multiply(#1,#2)|multiply(n2,#3) | general |
if a randomly selected positive single digit multiple of 3 is multiplied by a randomly selected prime number less than 20 , what is the probability t that this product will be a multiple of 45 ? | "there are 3 single digit multiple of 3 , that is , 3 , 6,9 . there are 8 prime nos less than 20 - 2,3 , 5,7 , 11,13 , 17,19 total outcome - 8 * 3 = 24 favourable outcome = 1 ( 9 * 5 ) hence required probability t = 1 / 24 . answer c ." | a ) 1 / 32 , b ) 1 / 28 , c ) 1 / 24 , d ) 1 / 16 , e ) 1 / 14 | c | divide(const_1, add(20, const_4)) | add(n1,const_4)|divide(const_1,#0)| | general |
uba capital recently bought brand new vehicles for office use . uba capital only went for toyota and honda and bought more of toyota than honda at the ratio of 8 : 2 . if 80 % of the toyota bought and 20 % of the honda bought were suv â € ™ s . how many suv â € ™ s did uba capital buy in the aforementioned purchase ? | "let total no of vehicles bought be 100 , toyota 80 and honda 20 , so total number of suv ' s bought for toyota and honda respectively 80 * 80 / 100 = 64 and 20 * 20 / 100 = 4 so total 68 suv ' s were bought out of 100 vehicles bought . . so required % is 68 % answer : c" | a ) 66 % , b ) 64 % , c ) 68 % , d ) 69 % , e ) 72 % | c | subtract(const_100, add(multiply(8, 2), multiply(2, 8))) | multiply(n0,n1)|add(#0,#0)|subtract(const_100,#1)| | other |
10 ^ ( 655 ) ã · 10 ^ ( 650 ) = ? | "10 ^ ( 655 ) ã · 10 ^ ( 651 ) = 10 ^ ( 655 - 651 ) = 10 ^ 5 = 100000 answer : d" | a ) 100 , b ) 10 , c ) 10000 , d ) 100000 , e ) 1000000 | d | power(10, subtract(655, 650)) | subtract(n1,n3)|power(n0,#0)| | general |
a corporation that had $ 115.19 billion in profits for the year paid out $ 250.10 million in employee benefits . approximately what percent of the profits were the employee benefits ? ( note : 1 billion = 10 ^ 9 ) | "required answer = [ employee benefit / profit ] * 100 = [ ( 250.10 million ) / ( 115.19 billion ) ] * 100 = [ ( 250.10 * 10 ^ 6 ) / ( 115.19 * 10 ^ 9 ) ] * 100 = ( 2.2 / 1000 ) * 100 = 0.22 % so answer is ( e )" | a ) 50 % , b ) 20 % , c ) 5 % , d ) 2 % , e ) 0.22 % | e | multiply(divide(multiply(250.10, power(10, add(const_3, const_3))), multiply(115.19, power(10, 9))), const_100) | add(const_3,const_3)|power(n3,n4)|multiply(n0,#1)|power(n3,#0)|multiply(n1,#3)|divide(#4,#2)|multiply(#5,const_100)| | general |
a reduction of 10 % in the price of salt enables a lady to obtain 10 kgs more for rs . 100 , find the original price per kg ? | "100 * ( 10 / 100 ) = 10 - - - 10 ? - - - 1 = > rs . 1 100 - - - 90 ? - - - 1 = > rs . 1.1 answer : a" | a ) 1.1 , b ) 2.2 , c ) 2.5 , d ) 2.1 , e ) 2.9 | a | multiply(divide(divide(multiply(divide(10, const_100), 100), 10), multiply(divide(10, const_100), 100)), const_100) | divide(n0,const_100)|multiply(n2,#0)|divide(#1,n1)|divide(#2,#1)|multiply(#3,const_100)| | gain |
the average ( arithmetic mean ) of 16 students first quiz scores in a difficult english class is 62.5 . when one student dropped the class , the average of the remaining scores increased to 62.0 . what is the quiz score of the student who dropped the class ? | "total score of 16 students is 16 * 62.50 = 1000 total score of 15 students is 15 * 62 = 930 so , the score of the person who left is 70 ( 1000 - 930 ) answer will be ( e )" | a ) a ) 10 , b ) b ) 25 , c ) c ) 40 , d ) d ) 55 , e ) e ) 70 | e | subtract(multiply(16, 62.5), multiply(subtract(16, const_1), 62.0)) | multiply(n0,n1)|subtract(n0,const_1)|multiply(n2,#1)|subtract(#0,#2)| | general |
the triplicate ratio of 2 : 3 is ? | 2 ^ 3 : 3 ^ 3 = 8 : 27 answer : d | a ) 1 : 64 , b ) 1 : 62 , c ) 1 : 34 , d ) 8 : 27 , e ) 5 : 64 | d | divide(power(2, const_3), power(3, const_3)) | power(n0,const_3)|power(n1,const_3)|divide(#0,#1) | other |
if 15 lions can kill 15 deers in 15 minutes how long will it take 100 lions to kill 100 deers ? | we can try the logic of time and work , our work is to kill the deers so 15 ( lions ) * 15 ( min ) / 15 ( deers ) = 100 ( lions ) * x ( min ) / 100 ( deers ) hence answer is x = 15 answer : b | a ) 1 minutes , b ) 15 minute , c ) 100 minutes , d ) 10000 minutes , e ) 1000 minutes | b | multiply(divide(15, 15), 15) | divide(n0,n0)|multiply(n0,#0) | physics |
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