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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if there is an equal probability of a child being born a boy or a girl , what is the probability that a couple who have 3 children have two children of the same sex and one of the opposite sex ? | "let boy be represented by b and girl by g . the possible outcomes of two children of same sex and one of opposite sex can be : bbg or gbb or bgb or ggb or bgg or gbg ( 1 / 2 * 1 / 2 * 1 / 2 ) + ( 1 / 2 * 1 / 2 * 1 / 2 ) + ( 1 / 2 * 1 / 2 * 1 / 2 ) + ( 1 / 2 * 1 / 2 * 1 / 2 ) + ( 1 / 2 * 1 / 2 * 1 / 2 ) + ( 1 / 2 * 1 /... | a ) 1 / 3 , b ) 2 / 3 , c ) 1 / 4 , d ) 3 / 4 , e ) 3 / 5 | d | divide(const_2, 3) | divide(const_2,n0)| | general |
a man buys 12 lts of liquid which contains 20 % of the liquid and the rest is water . he then mixes it with 10 lts of another mixture with 30 % of liquid . what is the % of water in the new mixture ? | "20 % in 12 lts is 2.4 . so water = 12 - 2.4 = 9.6 lts . 30 % of 10 lts = 3 . so water in 2 nd mixture = 10 - 3 = 7 lts . now total quantity = 12 + 10 = 22 lts . total water in it will be 9.6 + 7 = 16.6 lts . % of water = ( 100 * 16.6 ) / 22 = 75.45 answer : a" | a ) 75.45 , b ) 76.45 , c ) 77.45 , d ) 74.45 , e ) 73.45 | a | divide(add(multiply(12, subtract(12, 20)), multiply(10, subtract(12, 30))), add(12, 10)) | add(n0,n2)|subtract(n0,n1)|subtract(n0,n3)|multiply(n0,#1)|multiply(n2,#2)|add(#3,#4)|divide(#5,#0)| | gain |
a and b can do a work in 20 days , b and c can do it in 30 days ; a , b and c together can finish it in 40 days . a and c together will do it in ? | a + b + c 1 day work = 1 / 40 a + b 1 day work = 1 / 20 b + c 1 day work = 1 / 30 a + c 1 day work = 2 * 1 / 40 - 1 / 20 + 1 / 30 = 1 / 30 a and c together will do the work in 30 days . answer is b | a ) 25 days , b ) 30 days , c ) 20 days , d ) 15 days , e ) 10 days | b | inverse(add(subtract(divide(const_1, 30), subtract(divide(const_1, 20), divide(const_1, 40))), subtract(divide(const_1, 20), divide(const_1, 40)))) | divide(const_1,n1)|divide(const_1,n0)|divide(const_1,n2)|subtract(#1,#2)|subtract(#0,#3)|add(#4,#3)|inverse(#5) | physics |
the least perfect square , which is divisible by each of 21 , 36 and 66 is : | "explanation : l . c . m . of 21 , 36 , 66 = 2772 . now , 2772 = 2 x 2 x 3 x 3 x 7 x 11 to make it a perfect square , it must be multiplied by 7 x 11 . so , required number = 22 x 32 x 72 x 112 = 213444 answer is a" | a ) 213444 , b ) 214344 , c ) 214434 , d ) 231444 , e ) 233444 | a | multiply(multiply(multiply(power(add(const_3, const_4), const_2), power(divide(divide(36, const_3), const_2), const_2)), power(const_2, const_2)), power(const_3, const_2)) | add(const_3,const_4)|divide(n1,const_3)|power(const_2,const_2)|power(const_3,const_2)|divide(#1,const_2)|power(#0,const_2)|power(#4,const_2)|multiply(#5,#6)|multiply(#7,#2)|multiply(#8,#3)| | geometry |
find the greatest number that will divide 54 , 87 and 172 so as to leave the same remainder in each case | explanation : required number = ( 87 - 54 ) , ( 172 - 87 ) , ( 172 - 54 ) = h . c . f of 33,85 and 118 is 1 answer : option d | a ) 13 , b ) 3 , c ) 15 , d ) 1 , e ) 2 | d | gcd(gcd(subtract(subtract(172, 87), 54), subtract(87, 54)), subtract(172, 87)) | subtract(n2,n1)|subtract(n1,n0)|subtract(#0,n0)|gcd(#2,#1)|gcd(#3,#0) | general |
sally has a gold credit card with a certain spending limit , and a platinum card with twice the spending limit of the gold card . currently , she has a balance on her gold card that is 1 / 3 of the spending limit on that card , and she has a balance on her platinum card that is 1 / 9 of the spending limit on that card ... | "let s assume the platinum card spending limit = x gold card spending limit will be = x / 2 balance on gold card is = x / 2 * 1 / 3 = x / 6 platinum card unspent limit is = x - 1 / 9 x = 8 / 9 x so if gold card balance is transferred then the rest unspent will be 8 / 9 x - x / 6 = 13 / 18 x so the ans is d" | a ) 11 / 30 , b ) 29 / 60 , c ) 17 / 30 , d ) 13 / 18 , e ) 11 / 15 | d | subtract(1, add(multiply(inverse(3), inverse(const_2)), inverse(9))) | inverse(n3)|inverse(n1)|inverse(const_2)|multiply(#1,#2)|add(#0,#3)|subtract(n0,#4)| | general |
the number which exceeds 12 % of it by 52.8 is : | "explanation : let the number be x . then , x – 12 % of x = 52.8 x – ( 12 / 100 ) x = 52.8 x ( 1 – 12 / 100 ) = 52.8 ( 88 / 100 ) x = 52.8 x = ( 100 x 52.8 ) / 88 = 60 answer : a" | a ) 60 , b ) 77 , c ) 269 , d ) 26 , e ) 91 | a | divide(multiply(52.8, const_100), subtract(const_100, 12)) | multiply(n1,const_100)|subtract(const_100,n0)|divide(#0,#1)| | gain |
according to the directions on the can of frozen orange juice concentrate , 1 can of concentrate is to be mixed with 3 cans of water to make orange juice . how many 12 ounces cans of the concentrate are required to prepare 360 6 ounces servings of orange juice ? | "its a . total juice rquired = 360 * 6 = 2160 ounce 12 ounce concentate makes = 12 * 4 = 48 ounce juice total cans required = 2160 / 48 = 45 . answer a" | a ) a ) 45 , b ) b ) 34 , c ) c ) 50 , d ) d ) 67 , e ) e ) 100 | a | divide(divide(multiply(360, 6), 12), const_4) | multiply(n3,n4)|divide(#0,n2)|divide(#1,const_4)| | general |
solution for 2.01 + . 3 + . 34 | "2.01 + . 3 + . 34 = 0 0 = 0 - 2.01 - 0.3 - 0.34 0 = - 2.65 answer : b" | a ) 2.91 , b ) 2.65 , c ) 2.938 , d ) 2.986 , e ) 2.999 | b | add(add(2.01, divide(3, const_1000)), divide(34, const_1000)) | divide(n1,const_1000)|divide(n2,const_1000)|add(n0,#0)|add(#2,#1)| | general |
two pipes p and q can fill a cistern in 12 and 15 minutes respectively . both are opened together , but at the end of 6 minutes the first is turned off . how many more minutes will it take for the cistern to fill after the first pipe is turned off ? | "let x be the total time it takes for the cistern to fill . 6 / 12 + x / 15 = 1 x / 15 = 1 / 2 x = 7.5 after the first pipe is turned off , it takes 1.5 more minutes to fill the cistern . the answer is b ." | a ) 1 , b ) 1.5 , c ) 2 , d ) 2.5 , e ) 3 | b | multiply(subtract(const_1, multiply(add(divide(const_1, 12), divide(const_1, 15)), 6)), 15) | divide(const_1,n0)|divide(const_1,n1)|add(#0,#1)|multiply(#2,n2)|subtract(const_1,#3)|multiply(n1,#4)| | physics |
in the xy - coordinate system , what is the slope of the line that goes through the point ( 1 , 1 ) and is equidistant from the two points p = ( 3 , 7 ) and q = ( 9 , 11 ) ? | "first , get the middle coordinate between ( 3,7 ) and ( 9,11 ) . . . x = 3 + ( 9 - 3 ) / 2 = 6 y = 7 + ( 11 - 7 ) / 2 = 9 second , get the slope of ( 9,6 ) and ( 1,1 ) . m = 6 - 1 / 9 - 1 = 5 / 8 = 0.625 answer : b" | a ) 0.5 , b ) 0.625 , c ) 0.75 , d ) 0.875 , e ) 1.0 | b | divide(divide(add(1, const_2.0), const_2), divide(add(1, 1), 7)) | add(n1,const_2.0)|add(const_4.0,n1)|divide(#0,n3)|divide(#1,n3)|divide(#2,#3)| | general |
a certain car can travel 40 minutes on a gallon of gasoline at 60 miles per hour . if the car had started with a full tank and had 12 gallons of gasoline left in its tank at the end , then what percent of the tank was used to travel 80 miles at 60 mph ? | "total time for travelling 80 miles @ 60 mph = 80 / 60 = 4 / 3 hour = 80 minutes . given , the car uses 1 gallon for every 40 minutes of driving @ 60 mph . thus in 80 minutes it will use = 2 gallons . thus , full tank = 2 + 12 = 14 gallons - - - > 2 / 14 = 14 % of the fuel used . e is the correct answer ." | a ) 15 % , b ) 20 % , c ) 25 % , d ) 30 % , e ) 14 % | e | divide(divide(multiply(40, 60), multiply(80, 60)), add(divide(multiply(40, 60), multiply(80, 60)), 12)) | multiply(n0,n1)|multiply(n3,n4)|divide(#0,#1)|add(n2,#2)|divide(#2,#3)| | physics |
the ratio of flour to water to sugar in a recipe is 11 : 5 : 2 . the ratio in a new recipe calls for a doubling of the ratio of flour to water from the original recipe and a halving of the ratio of flour to sugar . if the new recipe calls for 7.5 cups of water , how much sugar is required ? | the ratio of flour to water is 22 : 5 . the ratio of flour to sugar is 5.5 : 2 = 22 : 8 . the new ratio of flour to water to sugar is 22 : 5 : 8 if we need 7.5 cups of water , then we need 12 cups of sugar . the answer is e . | a ) 8 , b ) 9 , c ) 10 , d ) 11 , e ) 12 | e | multiply(divide(7.5, 5), multiply(const_4, 2)) | divide(n3,n1)|multiply(n2,const_4)|multiply(#0,#1) | other |
with a uniform speed a car covers the distance in 8 hours . had the speed been increased by 5 km / hr , the same distance could have been covered in 7 1 / 2 hours . what is the distance covered ? | "let the distance be x km . then , x / ( 7 1 / 2 ) - x / 8 = 5 2 x / 15 - x / 8 = 5 = > x = 600 km . answer : e" | a ) 187 km , b ) 480 km , c ) 278 km , d ) 297 km , e ) 600 km | e | divide(1, 2) | divide(n3,n4)| | physics |
a merchant marks goods up by 60 % and then offers a discount on the marked price . the profit that the merchant makes after offering the discount is 20 % . what % discount did the merchant offer ? | "let p be the original price of the goods and let x be the rate after the markup . ( 1.6 p ) * x = 1.2 p x = 1.2 / 1.6 = 0.75 which is a discount of 25 % . the answer is b ." | a ) 20 % , b ) 25 % , c ) 30 % , d ) 35 % , e ) 40 % | b | multiply(divide(subtract(subtract(add(const_1, divide(60, const_100)), const_1), divide(20, const_100)), add(const_1, divide(60, const_100))), const_100) | divide(n0,const_100)|divide(n1,const_100)|add(#0,const_1)|subtract(#2,const_1)|subtract(#3,#1)|divide(#4,#2)|multiply(#5,const_100)| | gain |
a student scored an average of 65 marks in 3 subjects : physics , chemistry and mathematics . if the average marks in physics and mathematics is 90 and that in physics and chemistry is 70 , what are the marks in physics ? | "given m + p + c = 65 * 3 = 195 - - - ( 1 ) m + p = 90 * 2 = 180 - - - ( 2 ) p + c = 70 * 2 = 140 - - - ( 3 ) where m , p and c are marks obtained by the student in mathematics , physics and chemistry . p = ( 2 ) + ( 3 ) - ( 1 ) = 180 + 140 - 195 = 125 answer : d" | a ) 86 , b ) 165 , c ) 76 , d ) 125 , e ) 26 | d | subtract(add(multiply(90, const_2), multiply(70, const_2)), multiply(65, 3)) | multiply(n2,const_2)|multiply(n3,const_2)|multiply(n0,n1)|add(#0,#1)|subtract(#3,#2)| | general |
find the number of trailing zeros in the product of ( 1 ^ 1 ) * ( 5 ^ 5 ) * ( 10 ^ 10 ) * ( 15 ^ 15 ) * ( 20 ^ 20 ) * ( 25 ^ 25 ) … … … . * ( 50 ^ 50 ) . | looking at the numbers it looks like ( 1 x 5 ) ^ 5 ( 2 x 5 ) ^ 10 . . . ( 10 x 5 ) ^ 50 1 . determine the limiting factor . is it 2 or is it 5 ? we know that all the numbers are multiple of 5 but not of 2 . thus , the limiting factor in this case is 2 . let ' s drop all the 5 . then , we count factors of 2 of even mult... | a ) 150 , b ) 200 , c ) 250 , d ) 245 , e ) 225 | c | add(multiply(15, 15), 25) | multiply(n6,n6)|add(n10,#0) | general |
how many numbers from 2 to 8 are exactly divisible by 2 ? | "2 / 2 = 1 and 8 / 2 = 4 4 - 1 = 3 3 + 1 = 4 numbers . answer : b" | a ) a ) 2 , b ) b ) 4 , c ) c ) 5 , d ) d ) 7 , e ) e ) 8 | b | add(divide(subtract(multiply(floor(divide(8, 2)), 2), multiply(add(floor(divide(2, 2)), const_1), 2)), 2), const_1) | divide(n1,n2)|divide(n0,n2)|floor(#0)|floor(#1)|add(#3,const_1)|multiply(n2,#2)|multiply(n2,#4)|subtract(#5,#6)|divide(#7,n2)|add(#8,const_1)| | general |
the length of a bridge in meters , which a train 80 - meters long and traveling at 45 km / hr can cross in 30 seconds is ? | "45 km / h = 45000 m / 3600 s = 12.5 m / s in 30 seconds , the train can go 30 ( 12.5 ) = 375 meters let x be the length of the bridge . x + 80 = 375 meters x = 295 meters the answer is e ." | a ) 215 , b ) 235 , c ) 255 , d ) 275 , e ) 295 | e | subtract(multiply(divide(multiply(45, const_1000), const_3600), 30), 80) | multiply(n1,const_1000)|divide(#0,const_3600)|multiply(n2,#1)|subtract(#2,n0)| | physics |
find the smallest number which should be divided with 3872 to make it a perfect square . | "3872 = 11 * 11 * 2 * 2 * 2 * 2 * 2 required smallest number = 2 2 is the smallest number which should be divided with 3872 to make it a perfect square . answer : c" | a ) 8 , b ) 6 , c ) 2 , d ) 4 , e ) 1 | c | divide(divide(divide(divide(divide(3872, const_3), const_3), const_4), const_4), const_4) | divide(n0,const_3)|divide(#0,const_3)|divide(#1,const_4)|divide(#2,const_4)|divide(#3,const_4)| | geometry |
x and y are both integers . if x / y = 59.80 , then what is the sum of all the possible two digit remainders of x / y ? | "remainder = 0.80 = - - > 80 / 100 - - > can be written as ( 80 / 4 ) / ( 100 / 4 ) = 20 / 25 so remainders can be 20 , 40 , 60 , 80 we need the sum of only 2 digit remainders - - > 20 + 40 + 60 + 80 = 200 answer : d" | a ) 560 , b ) 616 , c ) 672 , d ) 200 , e ) 1024 | d | add(multiply(divide(const_3, const_2), const_100), add(multiply(add(const_2, const_3), 59.80), const_3)) | add(const_2,const_3)|divide(const_3,const_2)|multiply(n0,#0)|multiply(#1,const_100)|add(#2,const_3)|add(#4,#3)| | general |
a , b , c , d and e are 5 consecutive points on a straight line . if bc = 2 cd , de = 4 , ab = 5 and ac = 11 , what is the length of ae ? | "a - - - - - b - - - - c - - - - - - - - d - - - - e given : de = 4 - ( i ) ab = 5 - ( ii ) ac = 11 - ( iii ) bc = 2 cd - ( iv ) from ( ii ) and ( iii ) , bc = 6 from ( iv ) , cd = 3 length of ae = ac + cd + de = 11 + 3 + 4 = 18 correct option : d" | a ) 21 , b ) 26 , c ) 30 , d ) 18 , e ) 16 | d | add(add(11, divide(subtract(11, 5), 2)), 4) | subtract(n4,n0)|divide(#0,n1)|add(n4,#1)|add(n2,#2)| | physics |
on a partly cloudy day , derek decides to walk back from work . when it is sunny , he walks at a speed of s miles / hr ( s is an integer ) and when it gets cloudy , he increases his speed to ( s + 1 ) miles / hr . if his average speed for the entire distance is 2.8 miles / hr , what fraction t of the total distance did... | if s is an integer and we know that the average speed is 2.8 , s must be = 2 . that meanss + 1 = 3 . this implies that the ratio of time for s = 2 is 1 / 4 of the total time . the formula for distance / rate is d = rt . . . so the distance travelled when s = 2 is 2 t . the distance travelled for s + 1 = 3 is 3 * 4 t or... | a ) 1 / 4 , b ) 4 / 5 , c ) 1 / 5 , d ) 1 / 6 , e ) 1 / 7 | e | subtract(divide(lcm(const_2, const_3), 2.8), const_2) | lcm(const_2,const_3)|divide(#0,n1)|subtract(#1,const_2)| | general |
a train leaves delhi at 9 a . m . at a speed of 35 kmph . another train leaves at 2 p . m . at a speed of 40 kmph on the same day and in the same direction . how far from delhi , will the two trains meet ? | "d = 35 * 5 = 175 rs = 40 – 35 = 5 t = 175 / 5 = 35 d = 40 * 35 = 1400 km answer : e" | a ) 229 , b ) 288 , c ) 600 , d ) 888 , e ) 1400 | e | multiply(divide(multiply(35, add(const_3, 2)), subtract(40, 35)), 40) | add(n2,const_3)|subtract(n3,n1)|multiply(n1,#0)|divide(#2,#1)|multiply(n3,#3)| | physics |
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 . 2000 . then his monthly salary is | sol . saving = [ 100 - ( 40 + 20 + 10 + 10 ] % = 20 % . let the monthly salary be rs . x . then , 20 % of x = 2000 â ‡ ” 20 / 100 x = 2000 â ‡ ” x = 2000 ã — 5 = 10000 . answer a | a ) rs . 10000 , b ) rs . 6000 , c ) rs . 8000 , d ) rs . 4000 , e ) rs . 2000 | a | multiply(2000, add(const_4, const_1)) | add(const_1,const_4)|multiply(n4,#0) | gain |
a survey of employers found that during 1993 employment costs rose 3.5 percent , where employment costs consist of salary costs and fringe - benefit costs . if salary costs rose 3 percent and fringe - benefit costs rose 7.5 percent during 1993 , then fringe - benefit costs represented what percent of employment costs a... | "the amount by which employment costs rose is equal to 0.035 ( salary costs + fringe benefit costs ) ; on the other hand the amount by which employment costs rose is equal to 0.03 * salary costs + 0.075 * fringe benefit costs ; so , 35 ( s + f ) = 30 s + 75 f - - > s = 9 f - - > f / s = 1 / 9 - - > f / ( s + f ) = 1 / ... | a ) 16.5 % , b ) 10 % , c ) 35 % , d ) 55 % , e ) 65 % | b | multiply(divide(subtract(3.5, 3), subtract(7.5, 3)), const_100) | subtract(n1,n2)|subtract(n3,n2)|divide(#0,#1)|multiply(#2,const_100)| | gain |
excluding stoppages , the speed of a train is 45 kmph and including stoppages it is 42 kmph . of how many minutes does the train stop per hour ? | "t = 3 / 45 * 60 = 4 answer : b" | a ) 16 , b ) 4 , c ) 15 , d ) 18 , e ) 12 | b | subtract(const_60, multiply(const_60, divide(42, 45))) | divide(n1,n0)|multiply(#0,const_60)|subtract(const_60,#1)| | physics |
. 008 / ? = . 01 | "let . 008 / x = . 01 ; then x = . 008 / . 01 = . 8 / 1 = . 8 answer is a" | a ) . 8 , b ) . 09 , c ) . 009 , d ) . 0009 , e ) none of them | a | divide(divide(008, const_1000), divide(01, const_100)) | divide(n0,const_1000)|divide(n1,const_100)|divide(#0,#1)| | general |
a man spend 1 / 5 of his salary on food , 1 / 10 of his salary on house rent and 3 / 5 salary on clothes . he still has $ 17000 left with him . find salary . . | "[ 1 / ( x 1 / y 1 + x 2 / y 2 + x 3 / y 3 ) ] * total amount = balance amount [ 1 - ( 1 / 5 + 1 / 10 + 3 / 5 ) } * total salary = $ 17000 , = [ 1 - 9 / 10 ] * total salary = $ 17000 , total salary = $ 17000 * 10 = $ 180000 , correct answer ( e )" | a ) $ 10700 , b ) $ 17000 , c ) $ 180000 , d ) $ 1700 , e ) $ 170000 | e | divide(17000, subtract(1, add(add(divide(1, 5), divide(1, 10)), divide(3, 5)))) | divide(n0,n1)|divide(n0,n3)|divide(n4,n1)|add(#0,#1)|add(#3,#2)|subtract(n0,#4)|divide(n6,#5)| | general |
what is the lowest positive integer that is divisible by 8 through 11 , inclusive ? | "the integer should be divisible by : 8 , 9 , 10 and 11 . the least common multiple of these integers is lcm = 8 * 9 * 10 * 11 = 7920 answer : e" | a ) 999 , b ) 7811 , c ) 6555 , d ) 9200 , e ) 7920 | e | lcm(8, 11) | lcm(n0,n1)| | general |
the sector of a circle has radius of 20 cm and central angle 135 o . find its perimeter ? | "perimeter of the sector = length of the arc + 2 ( radius ) = ( 135 / 360 * 2 * 22 / 7 * 20 ) + 2 ( 20 ) = 47.1 + 40 = 87.1 cm answer : b" | a ) 91.5 , b ) 87.1 , c ) 91.7 , d ) 91.3 , e ) 91.1 | b | multiply(multiply(const_2, divide(multiply(subtract(20, const_3), const_2), add(const_4, const_3))), 20) | add(const_3,const_4)|subtract(n0,const_3)|multiply(#1,const_2)|divide(#2,#0)|multiply(#3,const_2)|multiply(n0,#4)| | physics |
tara bought 19 cartons of ice cream and 4 cartons of yoghurt . each carton of ice cream cost $ 7 and each carton of yoghurt cost $ 1 . how much more did tara spend on ice cream than on yoghurt ? | step 1 : find the cost of the ice cream . 19 × $ 7 = $ 133 step 2 : find the cost of the yoghurt . 4 × $ 1 = $ 4 step 3 : find how much more the ice cream cost than the yoghurt . $ 133 – $ 4 = $ 129 tara spent $ 129 more on ice cream . answer is e . | a ) $ 144 , b ) $ 130 , c ) $ 80 , d ) $ 110 , e ) $ 129 | e | subtract(multiply(19, 7), multiply(4, 1)) | multiply(n0,n2)|multiply(n1,n3)|subtract(#0,#1) | general |
if 9 boys meet at a reunion and each boy shakes hands exactly once with each of the others , then what is the total number of handshakes | "n ( n - 1 ) / 2 = 9 * 8 / 2 = 36 answer : c" | a ) 41 , b ) 42 , c ) 36 , d ) 44 , e ) 45 | c | divide(factorial(9), multiply(factorial(subtract(9, const_2)), factorial(const_2))) | factorial(n0)|factorial(const_2)|subtract(n0,const_2)|factorial(#2)|multiply(#3,#1)|divide(#0,#4)| | general |
car a runs at the speed of 50 km / hr and reaches its destination in 8 hours . car b runs at the speed of 65 km / h and reaches its destination in 4 hours . what is the ratio of distances covered by car a and car b ? | "car a travels 50 × 8 = 400 km car b travels 65 × 4 = 260 km the ratio is 400 : 260 = 40 : 26 = 20 : 13 the answer is c ." | a ) 3 : 7 , b ) 4 : 9 , c ) 20 : 13 , d ) 5 : 7 , e ) 6 : 11 | c | divide(multiply(50, 8), multiply(65, 4)) | multiply(n0,n1)|multiply(n2,n3)|divide(#0,#1)| | physics |
a grocer is storing soap boxes in cartons that measure 25 inches by 48 inches by 60 inches . if the measurement of each soap box is 8 inches by 6 inches by 5 inches , then what is the maximum number of soap boxes that can be placed in each carton ? | "however the process of dividing the volume of box by the volume of a soap seems flawed but it does work in this case due to the numbers dimensions of the box = 25 * 48 * 60 dimensions of the soap = 5 * 6 * 8 we get = 5 x 6 x 10 = 300 so the question is why this particular arrangement , in order to maximize number of s... | a ) 210 , b ) 252 , c ) 280 , d ) 320 , e ) 300 | e | divide(multiply(multiply(25, 48), 60), multiply(multiply(8, 6), 5)) | multiply(n0,n1)|multiply(n3,n4)|multiply(n2,#0)|multiply(n5,#1)|divide(#2,#3)| | general |
what is the greatest 5 - digit number when divided by 6 , 7 , 8 , 9 , and 10 leaves a remainder of 4 , 5 , 6 , 7 , and 9 respectively ? | when you divide a positive integer by 10 , the remainder will just be the units digit . we know the remainder is 9 when we divide by 10 , so c is the only possible answer . | a ) 56780 , b ) 78910 , c ) 97479 , d ) 97918 , e ) 97920 | c | divide(multiply(const_1000, const_1000), const_10) | multiply(const_1000,const_1000)|divide(#0,const_10) | general |
a collection of books went on sale , and 2 / 3 of them were sold for $ 5 each . if none of the 50 remaining books were sold , what was the total amount received for the books that were sold ? | "if 50 books constitute 1 / 3 rd of the total , then 2 / 3 rd of the total = 100 books amount received for sold books = 100 * 5 = $ 500 answer : c" | a ) $ 550 , b ) $ 600 , c ) $ 500 , d ) $ 400 , e ) $ 450 | c | multiply(const_2.0, divide(multiply(50, divide(2, 3)), divide(const_1, 3))) | divide(n2,n1)|divide(const_1,n1)|multiply(n3,#0)|divide(#2,#1)|multiply(n0,#3)| | general |
a rower can row 6 km / h in still water . when the river is running at 2 km / h , it takes the rower 1 hour to row to big rock and back . how many kilometers is it to big rock ? | "let x be the distance to big rock . time = x / 4 + x / 8 = 1 x = 32 / 12 = 2.67 km the answer is c ." | a ) 2.21 , b ) 2.45 , c ) 2.67 , d ) 2.83 , e ) 2.95 | c | multiply(divide(subtract(6, 2), add(add(6, 2), subtract(6, 2))), add(6, 2)) | add(n0,n1)|subtract(n0,n1)|add(#0,#1)|divide(#1,#2)|multiply(#0,#3)| | physics |
5 men are equal to as many women as are equal to 8 boys . all of them earn rs . 75 only . men â € ™ s wages are ? | "5 m = xw = 8 b 5 m + xw + 8 b - - - - - 75 rs . 5 m + 5 m + 5 m - - - - - 75 rs . 15 m - - - - - - 75 rs . = > 1 m = 5 rs . answer : d" | a ) 6 rs , b ) 7 rs , c ) 8 rs , d ) 5 rs , e ) 1 rs | d | divide(75, multiply(const_3, 5)) | multiply(n0,const_3)|divide(n2,#0)| | general |
the radius of a cylindrical vessel is 7 cm and height is 6 cm . find the whole surface of the cylinder ? | "r = 7 h = 6 2 π r ( h + r ) = 2 * 22 / 7 * 7 ( 13 ) = 572 answer : e" | a ) 281 , b ) 771 , c ) 440 , d ) 767 , e ) 572 | e | surface_cylinder(7, 6) | surface_cylinder(n0,n1)| | geometry |
the value of ( 34.31 * 0.473 * 1.5 ) / ( 0.0673 * 23.25 * 7.57 ) is close to | "( 34.31 * 0.473 * 1.5 ) / ( 0.0673 * 23.25 * 7.57 ) = 24.343 / 11.845 = 2.06 answer : d" | a ) 2 , b ) 1.15 , c ) 2.01 , d ) 2.06 , e ) 2.35 | d | divide(divide(multiply(multiply(34.31, 0.473), 1.5), multiply(multiply(7.57, 23.25), 0.0673)), const_10) | multiply(n0,n1)|multiply(n4,n5)|multiply(n2,#0)|multiply(n3,#1)|divide(#2,#3)|divide(#4,const_10)| | general |
elvin ' s monthly telephone bill is the sum of the charge for the calls he made during the month and a fixed monthly charge for internet service . elvin ' s total telephone bill for january was $ 40 and elvin ' s total telephone bill for february was 76 $ . if elvin ' s charge for the calls he made in february was twic... | bill = fixed charge + charge of calls made in jan , bill = fixed charge ( let , y ) + charge of calls made in jan ( let , x ) = $ 40 in feb , bill = fixed charge ( let , y ) + charge of calls made in feb ( then , 2 x ) = $ 76 i . e . x + y = 40 and 2 x + y = 76 take the difference if two equations i . e . ( 2 x + y ) -... | a ) $ 5 , b ) $ 10 , c ) $ 14 , d ) $ 4 , e ) $ 28 | d | subtract(multiply(40, const_2), 76) | multiply(n0,const_2)|subtract(#0,n1) | general |
how many seconds does puja take to cover a distance of 500 m , if she runs at a speed of 30 km / hr ? | hint : time = distance / speed we see that the distance is given in metres while the speed is given in km / hr and the answer is asked in seconds . so , convert km / hr into m / s by multiplying 5 / 18 m / s to the given value of speed . 30 km / hr = 30 x 5 / 18 = 75 / 9 m / sec i . e . place these values in the formul... | a ) 60 sec , b ) 82 sec , c ) 95 sec , d ) 100 sec , e ) 120 sec | a | divide(500, multiply(30, const_0_2778)) | multiply(n1,const_0_2778)|divide(n0,#0) | physics |
the l . c . m . of 2 numbers is 40 . the numbers are in the ratio 4 : 5 . find their sum ? | "let the numbers be 4 x and 5 x l . c . m . = 20 x 20 x = 40 x = 2 the numbers are = 8 and 10 required sum = 8 + 10 = 18 answer is a" | a ) 18 , b ) 40 , c ) 48 , d ) 32 , e ) 56 | a | add(multiply(divide(2, multiply(40, 4)), 4), multiply(divide(2, multiply(40, 4)), 40)) | multiply(n1,n2)|divide(n0,#0)|multiply(n2,#1)|multiply(n1,#1)|add(#2,#3)| | other |
the sum of the ages 4 members of a family 5 years ago was 94 . today , when the daughter has been married off and replaced by a daughter - in - law , the sum of their ages is 100 . assuming that there has been no other change in the family structure and all members are alive , what is the difference in the ages of daug... | solution : sum of ages of 4 members 5 years ago = 94 = > sum of present ages of 4 members = 94 + 4 * 5 = 114 difference in the sum of the ages = difference in the ages of daughter and daughter - in - law difference in the sum of the ages = 114 - 100 = 14 = > difference in the ages of daughter and daughter - in - law = ... | a ) 22 , b ) 11 , c ) 25 , d ) 14 , e ) 16 | d | subtract(add(multiply(5, 4), 94), 100) | multiply(n0,n1)|add(n2,#0)|subtract(#1,n3) | general |
35 binders can bind 1400 books in 21 days . how many binders will be required to bind 1600 books in 20 days ? | "binders books days 35 1400 21 x 1600 20 x / 35 = ( 1600 / 1400 ) * ( 21 / 20 ) = > x = 42 answer : c" | a ) 87 , b ) 18 , c ) 42 , d ) 16 , e ) 10 | c | divide(divide(1600, 20), divide(1400, multiply(35, 21))) | divide(n3,n4)|multiply(n0,n2)|divide(n1,#1)|divide(#0,#2)| | physics |
for each month of a given year except december , a worker earned the same monthly salary and donated one - tenth of that salary to charity . in december , the worker earned n times his usual monthly salary and donated one - fifth of his earnings to charity . if the worker ' s charitable contributions totaled one - sixt... | "let monthly salary for each of the 11 months except december was x , then 11 x * 1 / 10 + nx * 1 / 5 = 1 / 6 ( 11 x + nx ) ; 11 / 10 + n / 5 = 1 / 6 ( 11 + n ) 11 + 2 n / 10 = 11 + n / 6 = > 66 + 12 n = 110 + 10 n = > 2 n = 44 n = 22 answer : b" | a ) 8 / 5 , b ) 22 , c ) 3 , d ) 11 / 3 , e ) 4 | b | divide(multiply(subtract(const_12, const_1), subtract(inverse(subtract(const_12, const_3)), inverse(const_10))), subtract(inverse(add(const_1, const_4)), inverse(subtract(const_12, const_3)))) | add(const_1,const_4)|inverse(const_10)|subtract(const_12,const_1)|subtract(const_12,const_3)|inverse(#3)|inverse(#0)|subtract(#4,#1)|subtract(#5,#4)|multiply(#2,#6)|divide(#8,#7)| | general |
what is the probability of getting exactly 4 heads in a single throw of five fair coins ? | "one possible case is hhhht . p ( hhhht ) = 1 / 2 * 1 / 2 * 1 / 2 * 1 / 2 * 1 / 2 = 1 / 32 there are 5 c 4 = 5 possible cases . p ( 4 heads ) = 5 * 1 / 32 = 5 / 32 the answer is d ." | a ) 1 / 4 , b ) 3 / 8 , c ) 3 / 16 , d ) 5 / 32 , e ) 7 / 32 | d | divide(const_2, choose(add(const_3, const_3), const_3)) | add(const_3,const_3)|choose(#0,const_3)|divide(const_2,#1)| | probability |
40 × ? + ( 12 + 8 ) × 3 / 5 = 1212 | "explanation : = > 40 × ? + ( 12 + 8 ) × 3 / 5 = 1212 = > 40 × ? = 1212 - ( 12 + 8 ) × 3 / 5 = > 40 × ? = 1212 - 20 × 3 / 5 = > 40 × ? = 1212 - 20 × 3 / 5 = 1200 = > ? = 1200 / 40 = 30 answer : option e" | a ) a ) 7 , b ) b ) 5 , c ) c ) 4 , d ) d ) 18 , e ) e ) 30 | e | divide(subtract(1212, multiply(add(12, 8), divide(3, 5))), 40) | add(n1,n2)|divide(n3,n4)|multiply(#0,#1)|subtract(n5,#2)|divide(#3,n0)| | general |
in how many ways 4 boys and 5 girls can be seated in a row so that they are alternate . | "solution : let the arrangement be , g b g b g b g b g 4 boys can be seated in 4 ! ways . girl can be seated in 3 ! ways . required number of ways , = 4 ! * 5 ! = 2880 . answer : option e" | a ) 144 , b ) 288 , c ) 12 , d ) 256 , e ) 2880 | e | multiply(factorial(4), factorial(5)) | factorial(n0)|factorial(n1)|multiply(#0,#1)| | probability |
in a manufacturing plant , it takes 48 machines 4 hours of continuous work to fill 8 standard orders . at this rate , how many hours of continuous work by 72 machines are required to fill 12 standard orders ? | "the choices give away the answer . . 48 machines take 4 hours to fill 8 standard orders . . in next eq we are doubling the machines from 48 to 72 , but the work is not doubling ( only 1 1 / 2 times ) , = 4 * 48 / 72 * 12 / 8 = 4 ans a" | a ) 3 , b ) 6 , c ) 8 , d ) 9 , e ) 12 | a | divide(divide(multiply(multiply(48, 12), 4), 72), 8) | multiply(n0,n4)|multiply(#0,n1)|divide(#1,n3)|divide(#2,n2)| | physics |
f ( x ) is a function such that f ( x ) + 3 f ( 8 - x ) = x for all real numbers x . find the value of f ( 2 ) | "f ( x ) + 3 f ( 8 - x ) = f ( 2 ) + 3 f ( 6 ) = 2 : x = 2 above f ( 6 ) + 3 f ( 2 ) = 6 : x = 6 above f ( 6 ) = 6 - 3 f ( 2 ) : solve equation c for f ( 6 ) f ( 2 ) + 3 ( 6 - 3 f ( 2 ) ) = 2 : substitute f ( 2 ) = 2 : solve above equation correct answer c" | a ) 0 , b ) 1 , c ) 2 , d ) 3 , e ) 4 | c | divide(subtract(2, multiply(multiply(3, 3), 2)), subtract(const_1, multiply(3, 3))) | multiply(n0,n0)|multiply(#0,n2)|subtract(const_1,#0)|subtract(n2,#1)|divide(#3,#2)| | general |
two cubes of their volumes in the ratio 64 : 1331 . the ratio of their surface area is : | "the ratio of their surface area is 64 : 1331 4 : 11 answer is d ." | a ) 1.5 : 5 , b ) 2 : 5 , c ) 3 : 5 , d ) 4 : 11 , e ) 4 : 5 | d | divide(power(64, const_0_33), power(1331, const_0_33)) | power(n0,const_0_33)|power(n1,const_0_33)|divide(#0,#1)| | geometry |
at joel ’ s bookstore , the current inventory is 30 % historical fiction . of the historical fiction books , 30 % are new releases , while 40 % of the other books are new releases . what fraction of all new releases are the historical fiction new releases ? | "let there be 100 books in all historic fiction books = 30 % of total = 30 other books = 70 new historic fiction = 30 % of 30 = 9 other new books = 40 % of 70 = 28 total new books = 37 fraction = 9 / 37 ans : b" | a ) 4 / 25 , b ) 9 / 37 , c ) 2 / 5 , d ) 8 / 15 , e ) 2 / 3 | b | divide(divide(multiply(30, 30), const_100), add(divide(multiply(30, 30), const_100), divide(multiply(40, subtract(const_100, 30)), const_100))) | multiply(n0,n1)|subtract(const_100,n0)|divide(#0,const_100)|multiply(n2,#1)|divide(#3,const_100)|add(#2,#4)|divide(#2,#5)| | gain |
a teacher grades students ’ tests by subtracting twice the number of incorrect responses from the number of correct responses . if student b answers each of the 100 questions on her test and receives a score of 73 , how many questions did student b answer correctly ? | "a score of 73 brings to mind that it is an odd number even though you are subtracting an even number ( twice the incorrect responses ) out of the correct responses score . so the correct responses score must be an odd number too ( odd - even = odd ) . since the overall score is 73 , the number of correct responses mus... | a ) 55 , b ) 60 , c ) 73 , d ) 82 , e ) 91 | e | subtract(100, divide(subtract(100, 73), const_3)) | subtract(n0,n1)|divide(#0,const_3)|subtract(n0,#1)| | general |
a box measuring 30 inches long by 48 inches wide by 12 inches deep is to be filled entirely with identical cubes . no space is to be left unfilled . what is the smallest number of cubes that can accomplish this objective ? | "least number of cubes will be required when the cubes that could fit in are biggest . 6 is the biggest number that could divide all three , 30 , 48 and 12 . thus side of cube must be 6 , and total number of cubes = 30 / 6 * 48 / 6 * 12 / 6 = 80 ans b ." | a ) 78 , b ) 80 , c ) 85 , d ) 88 , e ) 90 | b | divide(multiply(multiply(30, 48), 12), volume_cube(divide(12, const_2))) | divide(n2,const_2)|multiply(n0,n1)|multiply(n2,#1)|volume_cube(#0)|divide(#2,#3)| | geometry |
on a certain road 10 % of the motorists exceed the posted speed limit and receive speeding tickets , but 20 % of the motorists who exceed the posted speed limit do not receive speeding tickets . what percent of the motorists on the road exceed the posted speed limit ? | "say there are 100 motorists . { # of motorists who exceed speed & receive tickets } + { # of motorists who exceed speed & do n ' t receive tickets } = { total # of motorist who exceed speed } ; given : { # of motorists who exceed speed & receive tickets } = 10 ; also , if { total # of motorist who exceed speed } = x ,... | a ) 10.5 % , b ) 12.5 % , c ) 15 % , d ) 22 % , e ) 30 % | b | divide(const_100, multiply(multiply(divide(10, const_100), divide(20, const_100)), const_100)) | divide(n0,const_100)|divide(n1,const_100)|multiply(#0,#1)|multiply(#2,const_100)|divide(const_100,#3)| | gain |
a man can row downstream at the rate of 24 kmph and upstream at 7 kmph . find the man ’ s rate in still water and rate of current ? | "rate of still water = 1 / 2 ( down stream + upstream ) = 1 / 2 ( 24 + 7 ) = 15.5 kmph rate of current = 1 / 2 ( down stream - upstream ) = 1 / 2 ( 24 - 7 ) = 1 / 2 ( 17 ) = 8.5 kmph answer is a ." | a ) 8.5 , b ) 8.0 , c ) 9.5 , d ) 9.0 , e ) 8.25 | a | divide(subtract(24, 7), const_2) | subtract(n0,n1)|divide(#0,const_2)| | gain |
the average weight of 5 persons increases by 1.5 kg . if a person weighing 65 kg is replaced by a new person , what could be the weight of the new person ? | "total weight increases = 5 × 1.5 = 7.5 kg so the weight of new person = 65 + 7.5 = 72.5 kg answer c" | a ) 76 kg , b ) 77 kg , c ) 72.5 kg , d ) data inadequate , e ) none of these | c | add(65, multiply(5, 1.5)) | multiply(n0,n1)|add(n2,#0)| | general |
two trains , one from howrah to patna and the other from patna to howrah , start simultaneously . after they meet , the trains reach their destinations after 4 hours and 16 hours respectively . the ratio of their speeds is ? | "let us name the trains a and b . then , ( a ' s speed ) : ( b ' s speed ) = √ b : √ a = √ 16 : √ 4 = 4 : 2 answer : b" | a ) 4 : 5 , b ) 4 : 2 , c ) 4 : 4 , d ) 4 : 8 , e ) 4 : 1 | b | divide(sqrt(16), sqrt(4)) | sqrt(n1)|sqrt(n0)|divide(#0,#1)| | physics |
on july 1 of last year , total employees at company e was decreased by 10 percent . without any change in the salaries of the remaining employees , the average ( arithmetic mean ) employee salary was 10 percent more after the decrease in the number of employees than before the decrease . the total of the combined salar... | "the total number of employees = n the average salary = x total salary to all emplyoees = xn after the total number of employees = n - 0.1 n = 0.9 n the average salary = x + 10 % of x = 1.1 x total salary to all emplyoees = 0.9 n ( 1.1 x ) total salary after as a % of total salary before e = [ 0.9 n ( 1.1 x ) ] / xn = ... | a ) 90 % , b ) 99 % , c ) 100 % , d ) 101 % , e ) 110 % | b | multiply(10, 10) | multiply(n1,n1)| | general |
when positive integer n is divided by 5 , the remainder is 1 . when n is divided by 7 , the remainder is 3 . what is the smallest positive integer k such that k + n is a multiple of 70 . | "first , let us say i have a number n which is divisible by 5 and by 7 . we all agree that it will be divisible by 35 , the lcm of 5 and 7 . now , if i have a number n which when divided by 5 gives a remainder 1 and when divided by 7 gives a remainder 1 , we can say the number is of the form n = 5 a + 1 e . g . 5 + 1 ,... | a ) 3 , b ) 4 , c ) 12 , d ) 32 , e ) 35 | d | subtract(70, reminder(3, 7)) | reminder(n3,n2)|subtract(n4,#0)| | general |
the difference between the place values of 6 and 5 in the number 826533 is | sol . = ( place value of 6 ) – ( place value of 5 ) = ( 6000 - 500 ) = 5500 answer a | a ) 5500 , b ) 4500 , c ) 2500 , d ) 6970 , e ) none | a | subtract(multiply(const_10, 6), 6) | multiply(n0,const_10)|subtract(#0,n0)| | general |
a works twice as fast as b . if b can complete a work in 12 days independently , the number of days in which a and b can together finish the work in : | "explanation : ratio of rates of working of a and b = 2 : 1 . so , ratio of times taken = 1 : 2 . b ' s 1 day ' s work = 1 / 12 a ' s 1 day ' s work = 1 / 6 ; ( 2 times of b ' s work ) ( a + b ) ' s 1 day ' s work = ( 1 / 6 + 1 / 12 ) = 3 / 12 = 1 / 4 so , a and b together can finish the work in 4 days . answer is a" | a ) 4 days , b ) 6 days , c ) 8 days , d ) 18 days , e ) 28 days | a | inverse(add(divide(const_1, 12), multiply(const_2, divide(const_1, 12)))) | divide(const_1,n0)|multiply(#0,const_2)|add(#0,#1)|inverse(#2)| | physics |
if a student loses 5 kilograms , he will weigh twice as much as his sister . together they now weigh 104 kilograms . what is the student ' s present weight in kilograms ? | "let x be the weight of the sister . then the student ' s weight is 2 x + 5 . x + ( 2 x + 5 ) = 104 3 x = 99 x = 33 kg then the student ' s weight is 71 kg . the answer is b ." | a ) 70 , b ) 71 , c ) 72 , d ) 73 , e ) 74 | b | subtract(104, divide(subtract(104, 5), const_3)) | subtract(n1,n0)|divide(#0,const_3)|subtract(n1,#1)| | other |
a football team lost 5 yards and then gained 13 . what is the team ' s progress ? | "for lost , use negative . for gain , use positive . progress = - 5 + 13 = 8 yards e" | a ) 2 , b ) 4 , c ) 5 , d ) 6 , e ) 8 | e | subtract(13, 5) | subtract(n1,n0)| | gain |
find the last digit of 322 ^ 369 | for calculating the unit place of these type of question ( exponent ) as the power of 2 repeats after 4 , so we will divide the exponent by 4 e . g , 2 ^ 1 = 2 and 2 ^ ( 4 + 1 ) = 32 , both have same unit place 369 / 4 gives the remainder as 1 so we can write 322 ^ 369 = 2 ^ 1 = 2 so , the unit place will be 2 answer :... | a ) 1 , b ) 2 , c ) 3 , d ) 4 , e ) 5 | b | reminder(power(reminder(322, const_10), reminder(369, const_10)), const_10) | reminder(n0,const_10)|reminder(n1,const_10)|power(#0,#1)|reminder(#2,const_10) | general |
find the value of x from logx 100 = 2 | "solution : logb 1000 = 3 we can write it as , b 3 = 1000 b 3 = 103 so from the above equation b = 10 answer is a" | a ) 10 , b ) 12 , c ) 13 , d ) 14 , e ) 15 | a | multiply(subtract(2, const_4), 100) | subtract(n1,const_4)|multiply(#0,n0)| | general |
x starts a business with rs . 52000 . y joins in the business after 6 months with rs . 26000 . what will be the ratio in which they should share the profit at the end of the year ? | "explanation : ratio in which they should share the profit = ratio of the investments multiplied by the time period = 52000 * 12 : 26000 * 6 = 52 * 12 : 26 * 6 = 24 : 6 = 4 : 1 . answer : option d" | a ) 1 : 2 , b ) 2 : 3 , c ) 1 : 3 , d ) 4 : 1 , e ) none of these | d | divide(multiply(52000, const_12), multiply(26000, add(const_4, const_3))) | add(const_3,const_4)|multiply(n0,const_12)|multiply(n2,#0)|divide(#1,#2)| | other |
if x is equal to the sum of the even integers from 40 to 60 inclusive , and y is the number of even integers from 40 to 60 inclusive , what is the value of x + y ? | this is a perfect example of why you should not use formulas without understanding them properly . if you understand them , you will not make a mistake and will save time . the formula quoted by the original poster : n ( n + 1 ) is absolutely fine . but one needs to understand that n is the number of even terms startin... | a ) 550 , b ) 551 , c ) 560 , d ) 561 , e ) 572 | d | add(divide(multiply(add(divide(subtract(60, 40), const_2), const_1), add(multiply(const_2, 40), multiply(divide(subtract(60, 40), const_2), const_2))), const_2), add(divide(subtract(60, 40), const_2), const_1)) | multiply(n0,const_2)|subtract(n1,n0)|divide(#1,const_2)|add(#2,const_1)|multiply(#2,const_2)|add(#0,#4)|multiply(#3,#5)|divide(#6,const_2)|add(#3,#7) | general |
a monkey start climbing up a tree 17 ft tall . each hour it hops 3 ft and slips back 2 ft . how much time would it take the monkey to reach the top . | "if monkey hops 3 ft and slips back 2 ft in a hour , it means the monkey hops ( 3 ft - 2 ft ) = 1 ft / hr . similarly in 14 hrs it wil be 14 ft . bt since the height of the tree is 17 ft , so if the monkey hops up the tree in the next hr i . e 15 th hr then it reaches at the top of the tree . hence it takes 15 hrs for ... | a ) 15 hrs , b ) 18 hrs , c ) 19 hrs , d ) 17 hrs , e ) 16 hrs | a | subtract(divide(17, subtract(3, 2)), 2) | subtract(n1,n2)|divide(n0,#0)|subtract(#1,n2)| | physics |
in a competitive examination in state a , 6 % candidates got selected from the total appeared candidates . state b had an equal number of candidates appeared and 7 % candidates got selected with 79 more candidates got selected than a . what was the number of candidates appeared from each state ? | "state a and state b had an equal number of candidates appeared . in state a , 6 % candidates got selected from the total appeared candidates in state b , 7 % candidates got selected from the total appeared candidates but in state b , 79 more candidates got selected than state a from these , it is clear that 1 % of the... | a ) 7000 , b ) 7900 , c ) 6000 , d ) 5000 , e ) 4000 | b | divide(79, divide(subtract(7, 6), const_100)) | subtract(n1,n0)|divide(#0,const_100)|divide(n2,#1)| | general |
the price ratio of oranges to lemons is 2 : 3 . the price ratio of grapefruits to lemons is 3 : 5 . what is the ratio , by price , of grapefruit to oranges ? | let price of lemons be 30 ( since that is common ) so , price of oranges = 20 & price of lemons is = 30 further price of grapefruit will be 6 * 3 = 18 thus price , of grapefruit to oranges = 18 : 20 answer : a | a ) 18 : 20 , b ) 10 : 9 , c ) 2 : 5 , d ) 5 : 2 , e ) 3 : 2 | a | divide(multiply(divide(3, 5), 3), multiply(divide(const_2, const_3), const_3)) | divide(n1,n3)|divide(const_2,const_3)|multiply(n1,#0)|multiply(#1,const_3)|divide(#2,#3) | other |
20 carrots on a scale weigh 3.64 kg . when 4 carrots are removed from the scale , the average weight of the 16 carrots is 180 grams . what is the average weight ( in grams ) of the 4 carrots which were removed ? | "16 * 180 = 2880 . the other 4 carrots weigh a total of 760 grams . the average weight is 760 / 4 = 190 grams . the answer is d ." | a ) 160 , b ) 170 , c ) 180 , d ) 190 , e ) 200 | d | multiply(divide(subtract(3.64, divide(multiply(16, 180), const_1000)), 4), const_1000) | multiply(n3,n4)|divide(#0,const_1000)|subtract(n1,#1)|divide(#2,n2)|multiply(#3,const_1000)| | general |
a can do a particular work in 6 days . b can do the same work in 8 days . a and b signed to do it for rs . 3200 . they completed the work in 3 days with the help of c . how much is to be paid to c ? | "explanation : amount of work a can do in 1 day = 1 / 6 amount of work b can do in 1 day = 1 / 8 amount of work a + b can do in 1 day = 1 / 6 + 1 / 8 = 7 / 24 amount of work a + b + c can do = 1 / 3 amount of work c can do in 1 day = 1 / 3 - 7 / 24 = 1 / 24 work a can do in 1 day : work b can do in 1 day : work c can d... | a ) rs . 380 , b ) rs . 600 , c ) rs . 420 , d ) rs . 400 , e ) rs . 480 | d | multiply(multiply(3, subtract(divide(const_1, 3), add(divide(const_1, 6), divide(const_1, 8)))), 3200) | divide(const_1,n3)|divide(const_1,n0)|divide(const_1,n1)|add(#1,#2)|subtract(#0,#3)|multiply(n3,#4)|multiply(n2,#5)| | physics |
a train requires 10 seconds to pass a pole while it requires 40 seconds to cross a stationary train which is 300 mtrs long . find the speed of the train . | "in 10 s the train crosses the pole and in 40 sec the train crosses one more stationary train in 30 sec the train travels a distance of 300 mtrs speed = 300 / 30 = 10 m / s = 10 * 18 / 5 = 36 kmph answer : a" | a ) 36 kmph , b ) 42 kmph , c ) 48 kmph , d ) 52 kmph , e ) 64 kmph | a | multiply(divide(300, subtract(40, 10)), const_3_6) | subtract(n1,n0)|divide(n2,#0)|multiply(#1,const_3_6)| | physics |
two trains are running in opposite directions in the same speed . the length of each train is 120 meter . if they cross each other in 4 seconds , the speed of each train ( in km / hr ) is | "explanation : distance covered = 120 + 120 = 240 m time = 4 s let the speed of each train = v . then relative speed = v + v = 2 v 2 v = distance / time = 240 / 4 = 60 m / s speed of each train = v = 60 / 2 = 30 m / s = 30 × 36 / 10 km / hr = 108 km / hr answer : option c" | a ) 42 , b ) 36 , c ) 108 , d ) 120 , e ) 124 | c | multiply(const_3_6, divide(divide(add(120, 120), 4), const_2)) | add(n0,n0)|divide(#0,n1)|divide(#1,const_2)|multiply(#2,const_3_6)| | physics |
how many multiples of 7 are there between 700 and 1000 , inclusive ? | "there are exactly 301 numbers between 700 to 1000 . . 701 to 800 = 100 801 to 1000 = 200 . . . total 300 numbers , count 700 too . . 301 / 7 = 43 . . . ans option a ." | a ) 43 , b ) 42 , c ) 22 , d ) 21 , e ) 20 | a | add(divide(subtract(1000, 700), 7), const_1) | subtract(n2,n1)|divide(#0,n0)|add(#1,const_1)| | general |
an unbaised coin is tossed until it shows up the same face in 2 consicative throws wat is the probability that the no of tosses is not more than 4 . | using 0 and 1 for head and tail respectively and considering binary system , we find that out of 16 possibilities , 14 are desired options . then probability = 14 / 16 = 7 / 8 answer : c | a ) 3 / 4 , b ) 1 / 4 , c ) 7 / 8 , d ) 1 / 8 , e ) 6 / 8 | c | negate_prob(divide(const_1, multiply(4, 2))) | multiply(n0,n1)|divide(const_1,#0)|negate_prob(#1) | probability |
at a certain bowling alley , it costs $ 0.50 to rent bowling shoes for the day and $ 2 to bowl 1 game . if a person has $ 12.80 and must rent shoes , what is the greatest number of complete games that person can bowl in one day ? | after renting bowling shoes the person is left with $ 12.80 - $ 0.5 = $ 12.30 , which is enough for 12.3 / 2 < 7 = ~ 6 . answer : c . | a ) 7 , b ) 8 , c ) 6 , d ) 10 , e ) 11 | c | divide(subtract(12.8, 0.5), 2) | subtract(n3,n0)|divide(#0,n1) | physics |
if a * b * c = 130 , b * c * d = 65 , c * d * e = 750 and d * e * f = 250 the ( a * f ) / ( c * d ) = ? | "explanation : a â ˆ — b â ˆ — c / b â ˆ — c â ˆ — d = 130 / 65 = > a / d = 2 d â ˆ — e â ˆ — f / c â ˆ — d â ˆ — e = 250 / 750 = > f / c = 1 / 3 a / d * f / c = 2 * 1 / 3 = 2 / 3 answer : b" | a ) 1 / 2 , b ) 2 / 3 , c ) 3 / 4 , d ) 4 / 3 , e ) none of these | b | divide(multiply(130, 250), multiply(750, 65)) | multiply(n0,n3)|multiply(n1,n2)|divide(#0,#1)| | general |
a cistern can be filled by a tap in 4 hours while it can be emptied by another tap in 9 hours . if both the taps are opened simultaneously , then after how much time cistern will get filled ? | "explanation : when we have question like one is filling the tank and other is empting it , then we subtraction as , filled in 1 hour = 1 / 4 empties in 1 hour = 1 / 9 net filled in 1 hour = 1 / 4 - 1 / 9 = 5 / 36 so cistern will be filled in 36 / 5 hours i . e . 7.2 hours option c" | a ) 7 hours , b ) 7.1 hours , c ) 7.2 hours , d ) 7.3 hours , e ) none of these | c | divide(const_1, subtract(divide(const_1, 4), divide(const_1, 9))) | divide(const_1,n0)|divide(const_1,n1)|subtract(#0,#1)|divide(const_1,#2)| | physics |
evaluate : 986 x 237 - 986 x 37 | "986 x 237 - 986 x 37 = 986 x ( 237 - 37 ) = 986 x 200 = 197200 . answer is d ." | a ) 197600 , b ) 168000 , c ) 278000 , d ) 192700 , e ) none of them | d | subtract(986, multiply(multiply(237, 986), 37)) | multiply(n1,n2)|multiply(n3,#0)|subtract(n0,#1)| | general |
what is the sum of the odd integers from 25 to 55 , inclusive ? | "the mean is 40 . sum = mean ( # of elements ) there are 16 odd numbers between 25 - 55 inclusive . 16 * 40 = 640 e" | a ) 495 , b ) 550 , c ) 555 , d ) 600 , e ) 640 | e | divide(multiply(25, 55), const_4) | multiply(n0,n1)|divide(#0,const_4)| | general |
in a class of students , 2 / 3 of the number of girls is equal to 1 / 4 of the total number of students . what is the ratio of boys to girls in the class ? | "( 2 / 3 ) g = ( 1 / 4 ) ( b + g ) 8 g = 3 b + 3 g 5 g = 3 b b / g = 5 / 3 . the answer is d ." | a ) 1 / 3 , b ) 2 / 3 , c ) 4 / 3 , d ) 5 / 3 , e ) 7 / 3 | d | divide(subtract(divide(2, 3), divide(1, 4)), divide(1, 4)) | divide(n0,n1)|divide(n2,n3)|subtract(#0,#1)|divide(#2,#1)| | general |
a portion of the 85 % solution of chemicals was replaced with an equal amount of 20 % solution of chemicals . as a result , 40 % solution of chemicals resulted . what part of the original solution was replaced ? | "let the original amount of 85 percent solution = x and amount replaced be y 0.85 ( x - y ) + 0.2 y = 0.4 x 0.85 x - 0.85 y + 0.2 y = 0.4 x 0.45 x = 0.65 y 9 x = 13 y y = 9 x / 13 so , amount replaced = 9 / 13 of original amount answer : c" | a ) 5 / 13 , b ) 6 / 13 , c ) 9 / 13 , d ) 13 / 9 , e ) 13 / 6 | c | divide(subtract(85, 40), subtract(85, 20)) | subtract(n0,n2)|subtract(n0,n1)|divide(#0,#1)| | gain |
rain is falling at a rate of 10 centimeters per hour all over north carolina . somewhere downtown in north carolina a group of people are waiting for the rain to stop . if the rain filled a round puddle the with a base area of 300 square centimeters and a depth of 30 centimeters , how long did the people wait for the r... | answer is : d , 3 hours the volume of the puddle is irrelevant and only height matters since rain fell all over the city . thus , it takes only 30 / 10 = 3 hours of rain to fill the puddle | ['a ) 4 hours', 'b ) 11 hours', 'c ) 7 hours 15 minutes', 'd ) 3 hours', 'e ) 14 hours'] | d | divide(30, 10) | divide(n2,n0) | geometry |
the greatest number which on dividing 1442 and 1816 leaves remainders 12 and 6 respectively , is : | explanation : required number = h . c . f . of ( 1442 - 12 ) and ( 1816 - 6 ) = h . c . f . of 1430 and 1810 = 10 . answer : e | a ) 123 , b ) 127 , c ) 235 , d ) 305 , e ) 10 | e | gcd(subtract(1816, 6), subtract(1442, 12)) | subtract(n1,n3)|subtract(n0,n2)|gcd(#0,#1) | general |
if the average ( arithmetic mean ) of x + 1 , x + 3 , and x + 5 is 0 , then x = | "( x + 1 + x + 3 + x + 5 ) / 3 = 0 = > 3 x + 9 = 0 = > x = - 3 answer b" | a ) – 4 , b ) – 3 , c ) – 2 , d ) – 1 , e ) 0 | b | subtract(multiply(3, const_2), multiply(1, const_2)) | multiply(n1,const_2)|multiply(n0,const_2)|subtract(#0,#1)| | general |
a wooden box of dimensions 4 m x 2 m x 4 m is to carry rectangularboxes of dimensions 4 cm x 2 cm x 2 cm . the maximum number ofboxes that can be carried in the wooden box , is | explanation : number = ( 400 * 200 * 400 ) / 4 * 2 * 2 = 2000000 answer : b | a ) 9800000 , b ) 2000000 , c ) 7500000 , d ) 1200000 , e ) none of these | b | divide(multiply(multiply(multiply(4, const_100), multiply(2, const_100)), multiply(4, const_100)), multiply(multiply(4, 2), 2)) | multiply(n0,const_100)|multiply(n1,const_100)|multiply(n0,n1)|multiply(#0,#1)|multiply(n1,#2)|multiply(#3,#0)|divide(#5,#4) | physics |
find the simple interest on rs . 750 for 9 months at 6 paisa per month ? | "explanation : i = ( 750 * 9 * 6 ) / 100 = 405 answer : option b" | a ) rs . 415 , b ) rs . 405 , c ) rs . 450 , d ) rs . 345 , e ) rs . 564 | b | multiply(750, divide(9, const_100)) | divide(n1,const_100)|multiply(n0,#0)| | gain |
if two projectiles are launched at the same moment from 1182 km apart and travel directly towards each other at 460 km per hour and 525 km per hour respectively , how many minutes will it take for them to meet ? | the projectiles travel a total of 985 km per hour . the time to meet is 1182 / 985 = 1.2 hours = 72 minutes the answer is b . | a ) 68 , b ) 72 , c ) 76 , d ) 80 , e ) 84 | b | multiply(divide(1182, add(460, 525)), const_60) | add(n1,n2)|divide(n0,#0)|multiply(#1,const_60) | physics |
a and b complete a work in 20 days . a alone can do it in 40 days . if both together can do the work in how many days ? | "1 / 20 + 1 / 40 = 0.075 days answer : d" | a ) 1.0875 days , b ) 0.1875 days , c ) 0.0675 days , d ) 0.075 days , e ) 0.0175 days | d | inverse(add(inverse(20), inverse(40))) | inverse(n0)|inverse(n1)|add(#0,#1)|inverse(#2)| | physics |
a metallic sphere of radius 12 cm is melted and drawn into a wire , whose radius of cross section is 8 cm . what is the length of the wire ? | "volume of the wire ( in cylindrical shape ) is equal to the volume of the sphere . π ( 8 ) ^ 2 * h = ( 4 / 3 ) π ( 12 ) ^ 3 = > h = 36 cm answer : d" | a ) 16 cm , b ) 24 cm , c ) 28 cm , d ) 36 cm , e ) 39 cm | d | divide(multiply(const_4, divide(power(12, const_3), power(8, const_2))), const_3) | power(n0,const_3)|power(n1,const_2)|divide(#0,#1)|multiply(#2,const_4)|divide(#3,const_3)| | physics |
if 0.4 of a number is equal to 0.06 of another number , the ratio of the numbers is : | "expl : 0.4 a = 0.06 b - > a / b = 0.06 / 0.40 = 6 / 40 = 3 / 20 : . a : b = 3 : 20 answer : c" | a ) 2 : 3 , b ) 3 : 4 , c ) 3 : 20 , d ) 20 : 3 , e ) 22 : 1 | c | divide(multiply(0.06, const_100), multiply(0.4, const_100)) | multiply(n1,const_100)|multiply(n0,const_100)|divide(#0,#1)| | other |
a group of people participate in some curriculum , 25 of them practice yoga , 18 study cooking , 10 study weaving , 4 of them study cooking only , 5 of them study both the cooking and yoga , 4 of them participate all curriculums . how many people study both cooking and weaving ? | "both cooking and weaving = 18 - ( 4 + 4 + 5 ) = 5 so , the correct answer is e ." | a ) 1 , b ) 2 , c ) 3 , d ) 4 , e ) 5 | e | subtract(subtract(subtract(18, 5), 4), 4) | subtract(n1,n4)|subtract(#0,n5)|subtract(#1,n3)| | other |
a $ 500 investment and a $ 1,500 investment have a combined yearly return of 22 percent of the total of the two investments . if the $ 500 investment has a yearly return of 7 percent , what percent yearly return does the $ 1,500 investment have ? | "the equation we can form the question : return on total investment = sum of individual investments ( 500 + 1500 ) ( 22 ) = ( 500 â ˆ — 7 ) + ( 1500 x ) , where x is the return on investment of 1500 . solving the equation , we get x = 27 % ( option c ) answer : c" | a ) 29 % , b ) 20 % , c ) 27 % , d ) 25 % , e ) 12 % | c | multiply(divide(subtract(multiply(add(500, multiply(const_3, 500)), divide(22, const_100)), multiply(500, divide(7, const_100))), multiply(const_3, 500)), const_100) | divide(n2,const_100)|divide(n4,const_100)|multiply(n0,const_3)|add(n0,#2)|multiply(n0,#1)|multiply(#3,#0)|subtract(#5,#4)|divide(#6,#2)|multiply(#7,const_100)| | gain |
joe went on a diet 3 months ago when he weighed 222 pounds . if he now weighs 198 pounds and continues to lose at the same average monthly rate , in approximately how many months will he weigh 170 pounds ? | "222 - 198 = 24 pounds lost in 3 months 24 / 3 = 8 , so joe is losing weight at a rate of 8 pounds per month . . . . in approximately how many months will he weigh 170 pounds ? a simple approach is to just list the weights . now : 198 lbs in 1 month : 190 lbs in 2 months : 182 lbs in 3 months : 174 lbs in 4 months : 16... | a ) 3 , b ) 3.5 , c ) 4 , d ) 4.5 , e ) 5 | b | divide(subtract(198, 170), divide(subtract(222, 198), 3)) | subtract(n2,n3)|subtract(n1,n2)|divide(#1,n0)|divide(#0,#2)| | general |
two cars are traveling in the same direction along the same route . the red car travels at a constant speed of 40 miles per hour , and the black car is traveling at a constant speed of 50 miles per hour . if the red car is 30 miles ahead of the black car , how many hours will it take the black car to overtake the red c... | option e 20 + 40 t = 50 t t = 3 | a ) 0.1 , b ) 0.6 , c ) 1 , d ) 1.2 , e ) 3 | e | divide(30, subtract(50, 40)) | subtract(n1,n0)|divide(n2,#0) | physics |
in a certain city , 60 percent of the registered voters are democrats and the rest are republicans . in a mayoral race , if 70 percent of the registered voters who are democrats and 20 percent of the registered voters who are republicans are expected to vote for candidate a , what percent of the registered voters are e... | "say there are total of 100 registered voters in that city . thus 60 are democrats and 40 are republicans . 60 * 0.70 = 42 democrats are expected to vote for candidate a ; 40 * 0.20 = 8 republicans are expected to vote for candidate a . thus total of 42 + 8 = 50 registered voters are expected to vote for candidate a , ... | a ) 50 % , b ) 53 % , c ) 54 % , d ) 55 % , e ) 57 % | a | add(multiply(60, divide(70, const_100)), multiply(subtract(const_100, 60), divide(20, const_100))) | divide(n1,const_100)|divide(n2,const_100)|subtract(const_100,n0)|multiply(n0,#0)|multiply(#1,#2)|add(#3,#4)| | gain |
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