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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the ratio of three numbers is 2 : 3 : 5 and their sum is 150 . the second number of the three numbers is ? | "2 : 3 : 5 total parts = 10 10 parts - - > 150 1 part - - - - > 15 the second number of the three numbers is = 3 * 15 = 45 answer : e" | a ) 24 , b ) 26 , c ) 27 , d ) 29 , e ) 45 | e | sqrt(divide(150, add(power(5, 3), add(power(2, 3), power(3, 3))))) | power(n0,n1)|power(n1,n1)|power(n2,n1)|add(#0,#1)|add(#3,#2)|divide(n3,#4)|sqrt(#5)| | other |
10 % people of a village in sri lanka died by bombardment , 15 % of the remainder left the village on account of fear . if now the population is reduced to 3213 , how much was it in the beginning ? | "x * ( 90 / 100 ) * ( 85 / 100 ) = 3213 x = 4200 answer : b" | a ) a ) 3800 , b ) b ) 4200 , c ) c ) 4400 , d ) d ) 4500 , e ) e ) 4600 | b | floor(divide(3213, multiply(divide(subtract(const_100, 10), const_100), divide(subtract(const_100, 15), const_100)))) | subtract(const_100,n0)|subtract(const_100,n1)|divide(#0,const_100)|divide(#1,const_100)|multiply(#2,#3)|divide(n2,#4)|floor(#5)| | general |
a jogger running at 9 km / hr along side a railway track is 180 m ahead of the engine of a 120 m long train running at 45 km / hr in the same direction . in how much time will the train pass the jogger ? | "speed of train relative to jogger = 45 - 9 = 36 km / hr . = 36 * 5 / 18 = 10 m / sec . distance to be covered = 180 + 120 = 300 m . time taken = 300 / 10 = 30 sec . answer : option e" | a ) 89 , b ) 20 , c ) 36 , d ) 88 , e ) 30 | e | divide(add(180, 120), multiply(subtract(45, 9), divide(divide(const_10, const_2), divide(subtract(45, 9), const_2)))) | add(n1,n2)|divide(const_10,const_2)|subtract(n3,n0)|divide(#2,const_2)|divide(#1,#3)|multiply(#4,#2)|divide(#0,#5)| | general |
4 mat - weavers can weave 4 mats in 4 days . at the same rate , how many mats would be woven by 6 mat - weavers in 6 days ? | "let the required number of bottles be x . more weavers , more mats ( direct proportion ) more days , more mats ( direct proportion ) wavers 4 : 6 : : 4 : x days 4 : 6 4 * 4 * x = 6 * 6 * 4 x = ( 6 * 6 * 4 ) / ( 4 x 4 ) x = 9 . answer is b ." | a ) 25 , b ) 9 , c ) 39 , d ) 61 , e ) 16 | b | multiply(multiply(4, divide(6, 4)), divide(6, 4)) | divide(n3,n0)|multiply(n0,#0)|multiply(#0,#1)| | gain |
nitin ranks 12 th in a class of 56 students . what is rank from the last ? | "explanation : number students behind the nitin in rank = ( 56 - 12 ) = 44 nitin is 45 th from the last answer : c ) 45" | a ) 33 , b ) 38 , c ) 45 , d ) 28 , e ) 19 | c | subtract(56, 12) | subtract(n1,n0)| | other |
find the cost of fencing around a circular field of diameter 40 m at the rate of rs . 3 a meter ? | "2 * 22 / 7 * 20 = 125.66 125.66 * 3 = rs . 376.98 answer : c" | a ) 400 , b ) 370.4 , c ) 376.98 , d ) 340.9 , e ) 350.03 | c | multiply(circumface(divide(40, const_2)), 3) | divide(n0,const_2)|circumface(#0)|multiply(n1,#1)| | physics |
tickets to a certain concert sell for $ 20 each . the first 10 people to show up at the ticket booth received a 40 % discount , and the next 20 received a 15 % discount . if 50 people bought tickets to the concert , what was the total revenue from ticket sales ? | "price of 1 ticket = 20 $ revenue generated from sales of first 10 tickets = 10 * ( 60 / 100 * 20 ) = 10 * 12 = 120 revenue generated from sales of next 20 tickets = 20 * ( 85 / 100 * 20 ) = 20 * 17 = 340 revenue generated from sales of last 20 tickets = 20 * 20 = 400 revenue generated from sales of 50 tickets = 120 + ... | a ) $ 600 , b ) $ 740 , c ) $ 860 , d ) $ 980 , e ) $ 1,140 | c | multiply(add(add(subtract(subtract(50, 20), 10), multiply(subtract(const_1, divide(40, const_100)), 10)), multiply(subtract(const_1, divide(15, const_100)), 20)), 20) | divide(n2,const_100)|divide(n4,const_100)|subtract(n5,n0)|subtract(const_1,#0)|subtract(#2,n1)|subtract(const_1,#1)|multiply(n1,#3)|multiply(n0,#5)|add(#6,#4)|add(#8,#7)|multiply(n0,#9)| | gain |
x + ( 1 / x ) = 1.5 find x ^ 2 + ( 1 / x ^ 2 ) | "squaring on both sides ( x + 1 / x ) ^ 2 = 1.5 ^ 2 x ^ 2 + 1 / x ^ 2 = 2.25 - 2 x ^ 2 + 1 / x ^ 2 = 0.25 answer : e" | a ) 2.25 , b ) 3.25 , c ) 4.25 , d ) 5.25 , e ) 0.25 | e | subtract(power(1.5, 2), 2) | power(n1,n2)|subtract(#0,n2)| | general |
the volume of a sphere with radius r is ( 4 / 3 ) * pi * r ^ 3 and the surface area is 4 * pi * r ^ 3 . if a sperical balloon has a volume of 36 pi cubic centimeters , what is hte surface area of the balloon in square centimeters ? | "the surface area is 4 . pi . r ^ 2 ( its area remember not volume ) as 4 / 3 . pi . r ^ 3 = 36 pi r = 3 so area = 4 . pi . r ^ 2 = 36 . pi = 36 x 3.14 = 100 ( approx ) b" | a ) a . 40 , b ) b . 100 , c ) c . 400 , d ) d . 1,000 , e ) e . 10,000 | b | divide(const_180, const_1000) | divide(const_180,const_1000)| | geometry |
a soccer store typically sells replica jerseys at a discount of 30 percent to 40 percent off list price . during the annual summer sale , everything in the store is an additional 20 percent off the original list price . if a replica jersey ' s list price is $ 80 , approximately what percent of the list price is the low... | "let the list price be 2 x for min sale price , the first discount given should be 50 % , 2 x becomes x here now , during summer sale additional 20 % off is given ie sale price becomes 0.8 x it is given lise price is $ 80 = > 2 x = 80 = > x = 30 and 0.8 x = 32 so lowest sale price is 32 , which is 40 % of 80 hence , c ... | a ) 20 , b ) 25 , c ) 30 , d ) 40 , e ) 50 | c | divide(80, const_2) | divide(n3,const_2)| | general |
at a certain paint store forest green is made by mixing 4 parts blue paint with 3 parts yellow paint . verdant green is made by mixing 4 parts yellow paint with 3 parts blue paint . how many liters of yellow paint must be added to 35 liters of forest green to change it to verdant green ? | "35 liter of forset green have 20 liter of blue and 15 liter of yellow suppose we add x liter of yellow to make it a verdant green so the ratio of blue to yellow in verdant green is ¾ so the equation is blue / yellow = 20 / ( 15 + x ) = ¾ 45 + 3 x = 80 = > x = 35 / 3 answer : d" | a ) 1 / 3 , b ) 2 / 3 , c ) 4 / 3 , d ) 35 / 3 , e ) 10 / 3 | d | subtract(multiply(divide(4, 3), 4), 3) | divide(n0,n1)|multiply(n0,#0)|subtract(#1,n1)| | general |
find the value of a / b + b / a , if a and b are the roots of the quadratic equation x 2 + 12 x + 3 = 0 ? | "a / b + b / a = ( a 2 + b 2 ) / ab = ( a 2 + b 2 + a + b ) / ab = [ ( a + b ) 2 - 2 ab ] / ab a + b = - 12 / 1 = - 12 ab = 3 / 1 = 3 hence a / b + b / a = [ ( - 12 ) 2 - 2 ( 3 ) ] / 3 = 126 / 3 = 42 . e )" | a ) 6 , b ) 7 , c ) 23 , d ) 17 , e ) 42 | e | subtract(divide(power(negate(12), 2), 3), 2) | negate(n1)|power(#0,n0)|divide(#1,n2)|subtract(#2,n0)| | general |
p can do a work in the same time in which q and r together can do it . if p and q work together , the work can be completed in 10 days . r alone needs 60 days to complete the same work . then q alone can do it in | "work done by p and q in 1 day = 1 / 10 work done by r in 1 day = 1 / 60 work done by p , q and r in 1 day = 1 / 10 + 1 / 60 = 7 / 60 but work done by p in 1 day = work done by q and r in 1 day . hence the above equation can be written as work done by p in 1 day ã — 2 = 7 / 60 = > work done by p in 1 day = 7 / 120 = > ... | a ) 20 , b ) 24 , c ) 25 , d ) 27 , e ) 30 | b | divide(const_1, subtract(divide(add(divide(const_1, 10), divide(const_1, 60)), const_2), divide(const_1, 60))) | divide(const_1,n0)|divide(const_1,n1)|add(#0,#1)|divide(#2,const_2)|subtract(#3,#1)|divide(const_1,#4)| | physics |
the total age of a and b is 12 years more than the total age of b and c . c is how many year younger than a ? | "given that , a + b = 12 + b + c ⇒ a - c = 12 therefore , c is younger than a by 12 years answer is c ." | a ) 10 , b ) 11 , c ) 12 , d ) 13 , e ) 15 | c | multiply(12, const_1) | multiply(n0,const_1)| | general |
there are 1000 students in a school and among them 25 % of them attends chess class . 50 % of the students who are in the chess class are also enrolled for swimming . no other students are interested in swimming so how many will attend the swimming class if all enrolled attends ? | "25 % of 1000 gives 250 . so 250 attends chess and 50 % of 250 gives 125 so 125 enrolled for swimming answer : c" | a ) 1 , b ) 10 , c ) 125 , d ) 50 , e ) 20 | c | divide(multiply(divide(multiply(25, 1000), const_100), 50), const_100) | multiply(n0,n1)|divide(#0,const_100)|multiply(n2,#1)|divide(#2,const_100)| | gain |
the temperature of a certain cup of coffee 10 minutes after it was poured was 120 degrees fahrenheit . if the temperature f of the coffee t minutes after it was poured can be determined by the formula f = 120 ( 2 ^ - at ) + 60 , where f is in degrees fahrenheit and a is a constant . then the temperature of the coffee 6... | answer : b the temperature of coffee 10 minutes after it was poured ( 120 f ) will help in solving the constant “ a ” . 120 = 120 ( 2 ^ 10 a ) + 60 2 ^ - 1 = 2 ^ 10 a a = - 1 / 10 the temperature of coffee 60 minutes after it was poured is : f = 120 ( 2 ^ - 60 / 10 ) + 60 f = 120 * 1 / 64 + 60 f = 15 / 8 + 60 f = 495 /... | a ) 125 , b ) 95 , c ) 80 , d ) 61.8 , e ) 50 | d | add(multiply(power(2, multiply(divide(60, 10), subtract(const_1, const_2))), 120), 60) | divide(n4,n0)|subtract(const_1,const_2)|multiply(#0,#1)|power(n3,#2)|multiply(n1,#3)|add(n4,#4) | general |
the edges of three iron cubes are 6 cm ; 8 cm ; 10 cm respectively . a new cube was made by melting them . find the edge of the new cube ? | 63 + 83 + 103 = a 3 = > a = 12 answer : d | ['a ) 17', 'b ) 19', 'c ) 16', 'd ) 12', 'e ) 14'] | d | divide(add(add(add(const_100, const_3), add(multiply(8, const_10), const_3)), add(multiply(10, 6), const_3)), subtract(add(const_1, multiply(const_2, 10)), const_0_25)) | add(const_100,const_3)|multiply(n1,const_10)|multiply(n0,n2)|multiply(n2,const_2)|add(#1,const_3)|add(#2,const_3)|add(#3,const_1)|add(#0,#4)|subtract(#6,const_0_25)|add(#7,#5)|divide(#9,#8) | physics |
the average weight of 8 person ' s increases by 2.5 kg when a new person comes in place of one of them weighing 65 kg . what might be the weight of the new person ? | "explanation : total weight increased = ( 8 x 2.5 ) kg = 20 kg . weight of new person = ( 65 + 20 ) kg = 85 kg . answer : c" | a ) 76 , b ) 76.5 , c ) 85 , d ) 80 , e ) none of these | c | add(multiply(8, 2.5), 65) | multiply(n0,n1)|add(n2,#0)| | general |
if a and b get profits of rs . 5,000 and rs . 2,000 respectively at the end of year then ratio of their investments are | "ratio = 5000 / 2000 = 5 : 2 answer : c" | a ) 4 : 1 , b ) 1 : 4 , c ) 5 : 2 , d ) 2 : 3 , e ) 2 : 5 | c | divide(multiply(multiply(multiply(const_3, const_2), const_100), const_100), divide(multiply(multiply(multiply(const_3, const_2), const_100), const_100), multiply(add(const_2, const_3), const_2))) | add(const_2,const_3)|multiply(const_2,const_3)|multiply(#1,const_100)|multiply(#0,const_2)|multiply(#2,const_100)|divide(#4,#3)|divide(#4,#5)| | gain |
someone on a skateboard is traveling 5 miles per hour . how many feet does she travel in 45 seconds ? ( 1 mile = 5280 feet ) | "per second = > 5 * 5280 ft / 60 * 60 = 7.33 ft 45 seconds = > 7.33 * 45 = 329.85 ft answer : d" | a ) 380 ft , b ) 400 ft , c ) 410.05 ft , d ) 329.85 ft , e ) 420.85 ft | d | multiply(45, divide(multiply(5, 5280), const_3600)) | multiply(n0,n3)|divide(#0,const_3600)|multiply(n1,#1)| | physics |
a computer manufacturer produces a certain electronic component at a cost of $ 70 per component . shipping costs for delivering the components are $ 4 per unit . further , the manufacturer has costs of $ 14,500 a month related to the electronic component regardless of how many it produces . if the manufacturer produces... | "$ 14500 is a fixed cost each component is $ 74 ( $ 70 to produce , $ 4 to ship ) manufacturer will be producing and selling 250 components so therefore the equation to find price would be 250 * p = 14500 + ( 250 * 70 ) + ( 250 * 4 ) p = ( 14500 + 17500 + 1000 ) / 250 p = 132 answer : a" | a ) 132 , b ) 138 , c ) 144 , d ) 128 , e ) 116 | a | divide(add(add(multiply(multiply(const_4, const_4), const_1000), multiply(4, const_100)), multiply(add(70, 4), 250)), 250) | 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 |
two trains running in opposite directions cross a man standing on the platform in 27 seconds and 17 seconds respectively and they cross each other in 20 seconds . the ratio of their speeds is ? | "let the speeds of the two trains be x m / sec and y m / sec respectively . then , length of the first train = 27 x meters , and length of the second train = 17 y meters . ( 27 x + 17 y ) / ( x + y ) = 20 = = > 27 x + 17 y = 20 x + 20 y = = > 7 x = 3 y = = > x / y = 3 / 7 . answer : a" | a ) 3 / 7 , b ) 3 / 2 , c ) 3 / 5 , d ) 3 / 1 , e ) 3 / 3 | a | divide(subtract(27, 20), subtract(20, 17)) | subtract(n0,n2)|subtract(n2,n1)|divide(#0,#1)| | physics |
a bucket contains 2 mixtures a and b in the ratio 7 : 5 . if 9 liters of the mixture is replaced by 9 liters of liquid b , then the ratio of the two liquids becomes 7 : 9 . how much of the liquid a was there in the bucket ? | 1 st ratio = 7 : 5 2 nd ratio = 7 : 9 difference of cross products ratio = 7 * 9 - 7 * 5 = 28 common factor of the 1 st ratio = ( quantity replaced / sum of terms in 1 st ratio ) + ( quantity replaced * term a in 2 nd ratio / difference ) = ( 9 / 7 + 5 ) + ( 9 * 7 / 28 ) = 3 quantity of a = 7 * 3 = 21 liters answer is ... | a ) 15 liters , b ) 21 liters , c ) 12 liters , d ) 18 liters , e ) 25 liters | b | multiply(divide(add(multiply(9, multiply(divide(7, add(7, 5)), 9)), multiply(7, subtract(9, multiply(divide(5, add(7, 5)), 9)))), subtract(multiply(9, 7), multiply(5, 7))), 7) | add(n1,n2)|multiply(n1,n3)|multiply(n1,n2)|divide(n1,#0)|divide(n2,#0)|subtract(#1,#2)|multiply(n3,#3)|multiply(n3,#4)|multiply(n3,#6)|subtract(n3,#7)|multiply(n1,#9)|add(#8,#10)|divide(#11,#5)|multiply(n1,#12) | general |
jane makes toy bears . when she works with an assistant , she makes 53 percent more bears per week and works 10 percent fewer hours each week . having an assistant increases jane ’ s output of toy bears per hour by what percent ? | "let ' s assume just jane 40 bears per 40 / hrs a week , so that is 1 bear / hr . with an assistant she makes 61.2 bears per 36 hours a week or 1.7 bears / hr ( [ 40 bears * 1.53 ] / [ 40 hrs * . 90 ] ) . [ ( 1.7 - 1 ) / 1 ] * 100 % = 70 % answer : b" | a ) 20 % , b ) 70 % , c ) 100 % , d ) 180 % , e ) 200 % | b | multiply(divide(10, subtract(subtract(const_100, 53), 10)), const_100) | subtract(const_100,n0)|subtract(#0,n1)|divide(n1,#1)|multiply(#2,const_100)| | physics |
two numbers are in the ratio of 1 : 2 . if 5 be added to both , their ratio changes to 3 : 5 . the greater number is | "let the ratio be x : y , given x / y = 1 / 2 , ( x + 5 ) / ( y + 5 ) = 3 / 5 = > x = 10 and y = 20 answer : a" | a ) 20 , b ) 24 , c ) 28 , d ) 32 , e ) 36 | a | multiply(1, 5) | multiply(n0,n2)| | other |
if 60 % of 3 / 5 of a number is 72 , then the number is ? | "let the number be x . then 60 % of 3 / 5 of x = 72 60 / 100 * 3 / 5 * x = 72 x = ( 72 * 25 / 9 ) = 200 required number = 200 . correct option : e" | a ) 80 , b ) 100 , c ) 75 , d ) 90 , e ) none of these | e | divide(72, multiply(divide(60, const_100), divide(3, 5))) | divide(n0,const_100)|divide(n1,n2)|multiply(#0,#1)|divide(n3,#2)| | gain |
the cost to park a car in a certain parking garage is $ 12.00 for up to 2 hours of parking and $ 1.75 for each hour in excess of 2 hours . what is the average ( arithmetic mean ) cost per hour to park a car in the parking garage for 9 hours ? | "total cost of parking for 9 hours = 12 $ for the first 2 hours and then 1.75 for ( 9 - 2 ) hours = 12 + 7 * 1.75 = 24.25 thus the average parking price = 24.25 / 9 = 2.69 $ b is the correct answer ." | a ) $ 1.09 , b ) $ 2.69 , c ) $ 2.25 , d ) $ 2.37 , e ) $ 2.50 | b | divide(add(12.00, multiply(1.75, subtract(9, 2))), 9) | subtract(n4,n1)|multiply(n2,#0)|add(n0,#1)|divide(#2,n4)| | general |
in a certain large company , the ratio of college graduates with a graduate degree to non - college graduates is 1 : 8 , and ratio of college graduates without a graduate degree to non - college graduates is 2 : 3 . if one picks a random college graduate at this large company , what is the probability t this college gr... | "in believe the answer is d . please see below for explanation . 0 ) we are told the following ratios cgd - college graduate with degree ncg - non college graduate cgn - college graduate no degree cgd ncg cgn 1 8 3 2 in order to make cgd and cgn comparable we need to find the least common multiple of 8 and 3 and that i... | a ) 1 / 11 , b ) 1 / 12 , c ) 1 / 13 , d ) 3 / 19 , e ) 3 / 43 | d | divide(divide(divide(1, 8), divide(2, 3)), add(divide(divide(1, 8), divide(2, 3)), 1)) | divide(n0,n1)|divide(n2,n3)|divide(#0,#1)|add(#2,n0)|divide(#2,#3)| | other |
walking with 4 / 5 of my usual speed , i arrive at the bus stop 6 minutes later than normal . how many minutes does it take to walk to the bus stop at my usual speed ? | "let t = usual time = distance / usual speed t + 6 = distance / ( 4 * usual speed / 5 ) = ( 5 * distance ) / ( 4 * usual speed ) = 5 t / 4 t = 24 the answer is e ." | a ) 20 min , b ) 22 min , c ) 15 min , d ) 18 min , e ) 24 min | e | divide(6, subtract(divide(5, 4), const_1)) | divide(n1,n0)|subtract(#0,const_1)|divide(n2,#1)| | physics |
with both valves open , the pool will be filled with water in 48 minutes . the first valve alone would fill the pool in 2 hours . if the second valve emits 50 cubic meters of water more than the first every minute , then what is the capacity t of the pool ? | "d . 12000 cubic meters . if both hte valves fill the pool in 48 minutes and valve 1 only fills in 120 minutes then valve 2 alone will fill the pool in ( 48 * 120 ) / ( 120 - 48 ) = 80 minutes . now , if valve 1 admits x cubic meter of water per minute then the capacity of pool will be 120 x and also 80 ( x + 50 ) . or... | a ) 9000 cubic meters , b ) 10500 cubic meters , c ) 11750 cubic meters , d ) 12000 cubic meters , e ) 12500 cubic meters | d | multiply(50, multiply(const_60, multiply(divide(48, const_60), add(const_4, const_1)))) | add(const_1,const_4)|divide(n0,const_60)|multiply(#0,#1)|multiply(#2,const_60)|multiply(n2,#3)| | physics |
convert 500 miles to km ? | "800 km 500 * 1.6 = 800 answer : c" | a ) 600 km , b ) 700 km , c ) 800 km , d ) 900 km , e ) 1000 km | c | divide(multiply(multiply(multiply(add(const_3, const_2), const_2), multiply(add(const_3, const_2), const_2)), 500), multiply(multiply(add(const_3, const_2), const_2), multiply(add(const_3, const_2), const_2))) | add(const_2,const_3)|multiply(#0,const_2)|multiply(#1,#1)|multiply(n0,#2)|divide(#3,#2)| | physics |
audrey 4 hours to complete a certain job . ferris can do the same job in 3 hours . audrey and ferris decided to collaborate on the job , working at their respective rates . while audrey worked continuously , ferris took 4 breaks of equal length . if the two completed the job together in 2 hours , how many minutes long ... | "audery and ferris collective work rate : 1 / 4 + 1 / 3 = 7 / 12 collective work time = 12 / 7 = 1.7 hrs job was actually done in = 2 ( includes breaks ) breaks = actual time taken - collective work time = 2 - 1.7 = . 3 hrs = 1 / 2 so ferrais took 4 breaks = . 3 / 4 = . 07 hrs = 7 m so answer is b ) 10 mins" | a ) 5 , b ) 7 , c ) 15 , d ) 20 , e ) 25 | b | divide(subtract(multiply(2, multiply(4, const_10)), multiply(divide(3, 2), multiply(4, const_10))), 4) | divide(n1,n3)|multiply(n2,const_10)|multiply(n3,#1)|multiply(#0,#1)|subtract(#2,#3)|divide(#4,n2)| | physics |
each week a restaurant serving mexican food uses the same volume of chili paste , which comes in either 25 - ounce cans or 15 - ounce cans of chili paste . if the restaurant must order 30 more of the smaller cans than the larger cans to fulfill its weekly needs , then how manysmallercans are required to fulfill its wee... | "let x be the number of 25 ounce cans . therefore ( x + 30 ) is the number of 15 ounce cans . total volume is same , therefore 25 x = 15 ( x + 30 ) 10 x = 450 x = 45 therefore , number of 15 ounce cans = 45 + 30 = 75 ans - c" | a ) 60 , b ) 70 , c ) 75 , d ) 100 , e ) 120 | c | add(15, 30) | add(n1,n2)| | general |
a man purchased 3 blankets @ rs . 100 each , 5 blankets @ rs . 150 each and two blankets at a certain rate which is now slipped off from his memory . but he remembers that the average price of the blankets was rs . 160 . find the unknown rate of two blankets ? | "10 * 160 = 1600 3 * 100 + 5 * 150 = 1050 1600 – 1050 = 550 answer : b" | a ) 420 , b ) 550 , c ) 490 , d ) 450 , e ) 457 | b | subtract(multiply(const_10, 160), add(multiply(3, 100), multiply(5, 150))) | multiply(n4,const_10)|multiply(n0,n1)|multiply(n2,n3)|add(#1,#2)|subtract(#0,#3)| | general |
a 300 meter long train crosses a platform in 30 seconds while it crosses a signal pole in 18 seconds . what is the length of the platform ? | "speed = [ 300 / 18 ] m / sec = 50 / 3 m / sec . let the length of the platform be x meters . then , x + 300 / 30 = 50 / 3 3 ( x + 300 ) = 1500 è x = 200 m . answer : e" | a ) 227 , b ) 266 , c ) 230 , d ) 288 , e ) 200 | e | subtract(multiply(divide(300, 18), 30), 300) | divide(n0,n2)|multiply(n1,#0)|subtract(#1,n0)| | physics |
how many different values of positive integer x , for which | x + 6 | < x , are there ? | "answer d i opted to put the random value option . i used 0 , 6 , - 6 and the the extreme of 20 and - 20 . . i was able to solve it in 1 : 09 d" | a ) 0 , b ) 2 , c ) 3 , d ) 8 , e ) 16 | d | add(6, 6) | add(n0,n0)| | general |
two trains of equal length are running on parallel lines in the same direction at 46 km / hr and 36 km / hr . if the faster train passes the slower train in 36 seconds , what is the length of each train ? | "assume the length of each train = x total distance covered for overtaking the slower train = x + x = 2 x relative speed = 46 - 36 = 10 km / hr = ( 10 × 10 ) / 36 = 100 / 36 m / s time = 36 seconds 2 x / ( 100 / 36 ) = 36 = > ( 2 x × 36 ) / 100 = 36 = > x = 50 meter answer is c ." | a ) 25 , b ) 30 , c ) 50 , d ) 75 , e ) 40 | c | divide(multiply(divide(multiply(subtract(46, 36), const_1000), const_3600), 36), const_2) | subtract(n0,n1)|multiply(#0,const_1000)|divide(#1,const_3600)|multiply(n2,#2)|divide(#3,const_2)| | general |
convert the 9 / 36 m / s into kilometers per hour ? | "9 / 36 m / s = 9 / 36 * 18 / 5 = 9 / 10 = 0.9 kmph . answer : a" | a ) 0.9 kmph , b ) 9.9 kmph , c ) 1.3 kmph , d ) 1.2 kmph , e ) 5.7 kmph | a | multiply(const_3_6, divide(9, 36)) | divide(n0,n1)|multiply(#0,const_3_6)| | physics |
in what time will a train 120 m long cross an electric pole , it its speed be 160 km / hr ? | "speed = 160 * 5 / 18 = 44 m / sec time taken = 120 / 44 = 2.7 sec . answer : c" | a ) 2.5 sec , b ) 2.9 sec , c ) 2.7 sec , d ) 8.7 sec , e ) 8.5 sec | c | divide(120, multiply(160, const_0_2778)) | multiply(n1,const_0_2778)|divide(n0,#0)| | physics |
all the faces of cubes are painted with red colour . it is then cut into 64 equal small cubes . find how many small cubes are there whose two adjacent faces are coloured red ? | "there are 64 small cubes , hence one side side of the big cube is 3 √ 64 = 4 cm number of small cubes having two adjacent faces coloured red = ( x - 2 ) x no . of edges = ( 4 - 2 ) x 12 = 24 answer : d" | a ) 0 , b ) 8 , c ) 16 , d ) 24 , e ) 4 | d | divide(divide(64, const_4), const_2) | divide(n0,const_4)|divide(#0,const_2)| | geometry |
a 300 metre long train crosses a platform in 39 seconds while it crosses a signal pole in 18 seconds . what is the length of the platform ? | "solution speed = ( 300 / 18 ) m / sec = ( 50 / 3 ) m / sec . length of the platform be x metres . then = x + 300 / 39 = 50 / 3 m / sec = 3 ( x + 300 ) = 1950 . = 350 m answer b" | a ) 320 m , b ) 350 m , c ) 650 m , d ) none of these , e ) can not be determined | b | divide(subtract(multiply(divide(300, divide(18, const_3)), 39), multiply(const_3, 300)), const_3) | divide(n2,const_3)|multiply(n0,const_3)|divide(n0,#0)|multiply(n1,#2)|subtract(#3,#1)|divide(#4,const_3)| | physics |
marketing executives for a certain chewing gum company projected a 30 percent increase in revenue this year over that of last year , but revenue this year actually decreased by 25 % . what percent of the projected revenue was the actual revenue ? | "last year revenue = 100 ( assume ) ; this year revenue = 75 ; projected revenue = 130 . actual / projected * 100 = 75 / 130 * 100 = 57.7 % . answer : b ." | a ) 53 % , b ) 57.7 % , c ) 62.5 % , d ) 64 % , e ) 75 % | b | multiply(divide(subtract(const_100, 25), add(30, const_100)), const_100) | add(n0,const_100)|subtract(const_100,n1)|divide(#1,#0)|multiply(#2,const_100)| | general |
a train 360 m long runs with a speed of 45 km / hr . what time will it take to pass a platform of 190 m long ? | "explanation : speed = 45 km / hr = 45 × ( 10 / 36 ) m / s = 150 / 12 = 50 / 4 = 25 / 2 m / s total distance = length of the train + length of the platform = 360 + 190 = 550 meter time taken to cross the platform = 550 / ( 25 / 2 ) = 550 × 2 / 25 = 44 seconds answer : option c" | a ) 38 sec , b ) 35 sec , c ) 44 sec , d ) 40 sec , e ) 50 sec | c | multiply(divide(add(divide(190, const_1000), divide(360, const_1000)), 45), const_3600) | divide(n2,const_1000)|divide(n0,const_1000)|add(#0,#1)|divide(#2,n1)|multiply(#3,const_3600)| | physics |
find large number from below question the difference of two numbers is 2500 . on dividing the larger number by the smaller , we get 6 as quotient and the 15 as remainder | "let the smaller number be x . then larger number = ( x + 1365 ) . x + 2500 = 6 x + 15 5 x = 2485 x = 497 large number = 497 + 1365 = 2997 d" | a ) 3345 , b ) 1250 , c ) 1540 , d ) 2997 , e ) 1635 | d | multiply(divide(subtract(2500, 15), subtract(6, const_1)), 6) | subtract(n0,n2)|subtract(n1,const_1)|divide(#0,#1)|multiply(n1,#2)| | general |
248 : 3 : : 328 : ? | "248 - - - - 24 / 8 . . . 3 328 - - - 32 / 8 . . . 4 answer c" | a ) 7 , b ) 5 , c ) 4 , d ) 6 , e ) none of the above | c | multiply(328, divide(248, 3)) | divide(n0,n1)|multiply(n2,#0)| | general |
which number should replace both the asterisks in ( * / 18 ) x ( * / 72 ) = 1 ? | "let ( y / 18 ) x ( y / 72 ) = 1 y ^ 2 = 18 x 72 = 18 x 18 x 4 y = ( 18 x 2 ) = 36 the answer is c ." | a ) 24 , b ) 30 , c ) 36 , d ) 42 , e ) 48 | c | sqrt(multiply(72, 18)) | multiply(n0,n1)|sqrt(#0)| | general |
the triplicate ratio of 1 : 5 is ? | "1 ^ 3 : 5 ^ 3 = 1 : 125 answer : e" | a ) 1 : 7 , b ) 1 : 8 , c ) 1 : 3 , d ) 1 : 1 , e ) 1 : 125 | e | divide(power(1, 5), power(5, 5)) | power(n0,n1)|power(n1,n1)|divide(#0,#1)| | other |
in a 500 m race , the ratio of the speeds of two contestants a and b is 3 : 4 . a has a start of 200 m . then , a wins by : | "to reach the winning post a will have to cover a distance of ( 500 - 200 ) m , i . e . , 300 m . while a covers 3 m , b covers 4 m . while a covers 300 m , b covers 4 x 300 / 3 m = 400 m . thus , when a reaches the winning post , b covers 400 m and therefore remains 100 m behind . a wins by 100 m . answer : b" | a ) 60 m , b ) 100 m , c ) 43 m , d ) 20 m , e ) 23 m | b | subtract(500, divide(multiply(subtract(500, 200), 4), 3)) | subtract(n0,n3)|multiply(n2,#0)|divide(#1,n1)|subtract(n0,#2)| | physics |
a certain shade of gray paint is obtained by mixing 3 parts of white paint with 5 parts of black paint . if 2 gallons of the mixture is needed and the individual colors can be purchased only in one gallon or half gallon cans , what is the least amount q of paint , in gallons , that must be purchased in order to measure... | "given w : b = 3 : 5 that means say 3 gallons of white paint + 5 gallons of black paint = 8 gallons of paint mixture . but we want least amount of whiteblack paints for minimum of 2 gallons of mixture , so lets reduce keeping same ratio , 1.5 : 2.5 gives 1.5 + 2.5 = 4 gallons of mixture , but we want only 2 gallons , l... | a ) 2 , b ) 2 1 / 2 , c ) 3 , d ) 3 1 / 2 , e ) 4 | b | add(multiply(const_0_25, 2), multiply(2, 5)) | multiply(const_0_25,n2)|multiply(n1,n2)|add(#0,#1)| | general |
a and b are two multiples of 14 , and q is the set of consecutive integers between a and b , inclusive . if q contains 9 multiples of 14 , how many multiples of 7 are there in q ? | "lets q ' s first number be 14 as per question , q contains 9 multiples of 14 , so let q ' s last number be 126 . ( 14 ∗ 9 ) we have , a n = a + ( n − 1 ) d then 126 = 14 + ( n - 1 ) 7 solving for n , we get n = 17 answer : d" | a ) 20 , b ) 19 , c ) 18 , d ) 17 , e ) 16 | d | subtract(multiply(9, const_2), const_1) | multiply(n1,const_2)|subtract(#0,const_1)| | physics |
express a speed of 66 kmph in meters per second ? | "66 * 5 / 18 = 18.33 mps answer : e" | a ) 10.33 mps , b ) 17 mps , c ) 97 mps , d ) 17.33 mps , e ) 18.33 mps | e | multiply(const_0_2778, 66) | multiply(n0,const_0_2778)| | physics |
a man buys a cycle for rs . 1600 and sells it at a loss of 15 % . what is the selling price of the cycle ? | "s . p . = 85 % of rs . 1600 = rs . 85 / 100 x 1600 = rs . 1360 answer : b" | a ) s . 1090 , b ) s . 1360 , c ) s . 1190 , d ) s . 1202 , e ) s . 1256 | b | divide(multiply(subtract(const_100, 15), 1600), const_100) | subtract(const_100,n1)|multiply(n0,#0)|divide(#1,const_100)| | gain |
the parameter of a square is equal to the perimeter of a rectangle of length 8 cm and breadth 6 cm . find the circumference of a semicircle whose diameter is equal to the side of the square . | "let the side of the square be a cm . parameter of the rectangle = 2 ( 8 + 6 ) = 28 cm parameter of the square = 28 cm i . e . 4 a = 28 a = 7 diameter of the semicircle = 7 cm circimference of the semicircle = 1 / 2 ( â ˆ ) 7 = 1 / 2 ( 22 / 7 ) ( 7 ) = 11 cm answer : e" | a ) 23.56 , b ) 23.59 , c ) 23.55 , d ) 23.53 , e ) 11.0 | e | divide(circumface(divide(square_edge_by_perimeter(rectangle_perimeter(8, 6)), const_2)), const_2) | rectangle_perimeter(n0,n1)|square_edge_by_perimeter(#0)|divide(#1,const_2)|circumface(#2)|divide(#3,const_2)| | geometry |
a tempo is insured to an extent of 5 / 7 of its original value . if the premium on it at the rate of 3 % amounts to $ 300 , the original value of the tempo is ? | "let the original value of the tempo is $ x 3 % of 5 / 7 of x = 300 ( 3 / 100 ) * ( 5 / 7 ) x = 300 x = $ 14000 answer is d" | a ) 12500 , b ) 13280 , c ) 17520 , d ) 14000 , e ) 12560 | d | divide(300, divide(multiply(divide(5, 7), 3), const_100)) | divide(n0,n1)|multiply(n2,#0)|divide(#1,const_100)|divide(n3,#2)| | gain |
last year department store x had a sales total for december that was 3 times the average ( arithmetic mean ) of the monthly sales totals for january through november . the sales total for december was what fraction of the sales total for the year ? | let avg for 11 mos . = 10 therefore , dec = 30 year total = 11 * 10 + 30 = 140 answer = 30 / 140 = 3 / 14 = c | a ) 1 / 4 , b ) 4 / 15 , c ) 3 / 14 , d ) 4 / 11 , e ) 4 / 5 | c | divide(3, add(subtract(const_12, const_1), 3)) | subtract(const_12,const_1)|add(n0,#0)|divide(n0,#1) | general |
if a - b = 3 and a ^ 2 + b ^ 2 = 29 , find the value of ab | "2 ab = ( a ^ 2 + b ^ 2 ) - ( a - b ) ^ 2 = 29 - 9 = 20 = > ab = 10 . answer : a" | a ) 10 , b ) 12 , c ) 15 , d ) 18 , e ) 19 | a | multiply(multiply(add(3, divide(subtract(sqrt(29), 3), 2)), divide(subtract(sqrt(29), 3), 2)), 2) | sqrt(n3)|subtract(#0,n0)|divide(#1,n1)|add(n0,#2)|multiply(#3,#2)|multiply(n1,#4)| | general |
the ratio of the cost price and selling price is 4 : 5 . the profit percent is ? | "let the c . p . = $ 4 x then s . p . = $ 5 x then , gain = 5 x - 4 x = $ x gain % = x / 4 x * 100 = 25 % answer is b" | a ) 20 % , b ) 25 % , c ) 30 % , d ) 50 % , e ) 15 % | b | multiply(subtract(divide(5, 4), const_1), const_100) | divide(n1,n0)|subtract(#0,const_1)|multiply(#1,const_100)| | gain |
farm tax is levied on the 50 % of the cultivated land . the tax department collected total $ 3840 through the farm tax from the village of mr . willam . mr . willam paid only $ 480 as farm tax . the percentage of total land of mr . willam over the total taxable land of the village is : | only trick n this question is to ignore 50 % information as farm tax is levied uniformly in the village and that includes mr william ' s land . what percentage of tax mr william paid ? this will be equal to the percentage of total cultivated land he holds over the total cultivated land in the village . that leads to ( ... | a ) 15 % , b ) 25 % , c ) 0.125 % , d ) 0.2083 % , e ) none | b | divide(multiply(multiply(divide(480, 3840), const_100), const_100), 50) | divide(n2,n1)|multiply(#0,const_100)|multiply(#1,const_100)|divide(#2,n0) | general |
the difference between the local value and the face value of 7 in the numeral 32675149 is | "explanation : ( local value of 7 ) - ( face value of 7 ) = ( 70000 - 7 ) = 69993 b )" | a ) 69990 , b ) 69993 , c ) 70000 , d ) 77000 , e ) 78000 | b | subtract(multiply(7, const_1000), 7) | multiply(n0,const_1000)|subtract(#0,n0)| | general |
how many even 4 - digit numbers can be formed , so that the numbers are divisible by 4 and no two digits are repeated ? | number is divisible by 4 if the last two digits form a number divisible by 4 . therefore last two digits can be : 00 ; 04 ; 08 ; 12 16 ; . . . 96 . basically multiples of 4 in the range 0 - 96 , inclusive . multiples of 4 in a range 0 - 96 , inclusive are last multiple in the range − first multiple in the range 4 + 1 =... | a ) 336 , b ) 784 , c ) 1120 , d ) 1804 , e ) 1936 | c | add(multiply(multiply(subtract(subtract(const_10, const_2), const_1), subtract(subtract(const_10, const_2), const_1)), subtract(subtract(divide(const_100, 4), const_3), multiply(const_2, const_3))), multiply(multiply(subtract(const_10, const_2), subtract(subtract(const_10, const_2), const_1)), multiply(const_2, const_3... | divide(const_100,n0)|multiply(const_2,const_3)|subtract(const_10,const_2)|subtract(#2,const_1)|subtract(#0,const_3)|multiply(#3,#3)|multiply(#2,#3)|subtract(#4,#1)|multiply(#5,#7)|multiply(#6,#1)|add(#8,#9) | general |
317 x 317 + 283 x 283 = ? | "= ( 317 ) ^ 2 + ( 283 ) ^ 2 = ( 300 + 17 ) ^ 2 + ( 300 - 17 ) ^ 2 = 2 [ ( 300 ) ^ 2 + ( 17 ) ^ 2 ] = 2 [ 90000 + 289 ] = 2 x 90289 = 180578 answer is d" | a ) 79698 , b ) 80578 , c ) 80698 , d ) 180578 , e ) none of them | d | multiply(317, power(317, 283)) | power(n1,n2)|multiply(n0,#0)| | general |
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 36 seconds , the speed of each train ( in km / hr ) is | explanation : distance covered = 120 + 120 = 240 m time = 36 s let the speed of each train = v . then relative speed = v + v = 2 v 2 v = distance / time = 240 / 36 = 6.667 m / s speed of each train = v = 6.667 / 2 = 3.33 m / s = 3.33 × 36 / 10 km / hr = 12 km / hr answer : option a | a ) 12 , b ) 36 , c ) 28 , d ) 20 , e ) 24 | a | multiply(const_3_6, divide(divide(add(120, 120), 36), const_2)) | add(n0,n0)|divide(#0,n1)|divide(#1,const_2)|multiply(#2,const_3_6) | physics |
two different primes may be said torhymearound an integer if they are the same distance from the integer on the number line . for instance , 3 and 7 rhyme around 5 . what integer r between 1 and 20 , inclusive , has the greatest number of distinct rhyming primes around it ? | since we are concerned with integers between 1 and 20 , write down the primes till 40 . 2 , 3 , 5 , 7 , 11 , 13 , 17 , 19 , 23 , 29 , 31 , 37 ( you should be very comfortable with the first few primes . . . ) 2 , 3 , 5 , 7 , 1112 , 13 , 17 , 19 , 23 , 29 , 31 , 37 - three pairs ( 1113 ) , ( 717 ) , ( 5 , 19 ) 2 , 3 , 5... | a ) 12 , b ) 15 , c ) 17 , d ) r = 18 , e ) 20 | d | multiply(multiply(const_3, const_2), 3) | multiply(const_2,const_3)|multiply(n0,#0) | physics |
pumps a , b , and c operate at their respective constant rates . pumps a and b , operating simultaneously , can fill a certain tank in 2 hours ; pumps a and c , operating simultaneously , can fill the tank in 3 / 2 hours ; and pumps b and c , operating simultaneously , can fill the tank in 2 hours . how many hours does... | a + b = 2 ; a + c = 3 / 2 , b + c = 2 ; add then 2 * ( a + b + c ) = 4 + 3 / 2 = 11 / 2 a + b + c = 11 / 4 hrs c | a ) 1 / 3 , b ) 1 / 2 , c ) 1 / 4 , d ) 1 , e ) 5 / 6 | c | divide(subtract(add(3, 3), add(divide(3, 2), add(2, 2))), 2) | add(n1,n1)|add(n0,n0)|divide(n1,n0)|add(#1,#2)|subtract(#0,#3)|divide(#4,n0) | physics |
how many even number in the range between 8 to 100 inclusive are not divisible by 3 | "we have to find the number of terms that are divisible by 2 but not by 6 ( as the question asks for the even numbers only which are not divisible by 3 ) for 2 , 8,10 , 12,14 . . . 100 using ap formula , we can say 100 = 10 + ( n - 1 ) * 2 or n = 47 . for 6 , 12,18 , . . . 96 using ap formula , we can say 96 = 12 + ( n... | a ) 15 , b ) 30 , c ) 32 , d ) 33 , e ) 46 | c | subtract(divide(subtract(subtract(100, 8), const_2), const_2), divide(divide(subtract(subtract(subtract(subtract(100, const_2), multiply(3, const_4)), 3), 3), 3), const_2)) | multiply(n2,const_4)|subtract(n1,n0)|subtract(n1,const_2)|subtract(#1,const_2)|subtract(#2,#0)|divide(#3,const_2)|subtract(#4,n2)|subtract(#6,n2)|divide(#7,n2)|divide(#8,const_2)|subtract(#5,#9)| | general |
in a group of 100 people , 50 like reading books , 20 like both reading books and listening songs . how many like listening songs ? | make a venn diagram , and enter your data . let the number of people who like listening songs be x x + 20 + 30 = 100 x = 50 so number who like listening songs = 20 + 50 = 70 answer c | a ) 40 , b ) 50 , c ) 70 , d ) 60 , e ) none | c | add(add(20, subtract(50, 20)), 20) | subtract(n1,n2)|add(n2,#0)|add(n2,#1) | other |
a man rows his boat 90 km downstream and 72 km upstream , taking 3 hours each time . find the speed of the stream ? | "speed downstream = d / t = 90 / 3 = 30 kmph speed upstream = d / t = 72 / ( 3 ) = 24 kmph the speed of the stream = ( 30 - 24 ) / 2 = 3 kmph answer : d" | a ) 6 , b ) 5 , c ) 7 , d ) 3 , e ) 2 | d | divide(subtract(divide(90, 3), divide(72, 3)), const_2) | divide(n0,n2)|divide(n1,n2)|subtract(#0,#1)|divide(#2,const_2)| | physics |
there are 97 lights which are functional and each is controlled by a separate on / off switch . two children a and b start playing with the switches . a starts by pressing every third switch till he reaches the end . b , thereafter , presses every fifth switch till he too reaches the end . if all switches were in off p... | editing my solution : number of switches = 97 number of switches turned on by a : 3 , 6 , . . . 96 = 32 number of switches turned on by b : 5 , 10 , . . . . 95 = 19 few switches are turned on by a and later turned off by b : lcm ( 3,5 ) = 15 x = 15 , 30 , . . . . 90 = 6 . subtract the above 6 switches from both a and b... | a ) 29 , b ) 49 , c ) 39 , d ) 59 , e ) 69 | c | subtract(add(floor(divide(97, const_3)), floor(divide(97, add(const_1, const_4)))), multiply(floor(divide(97, multiply(const_3, add(const_1, const_4)))), const_2)) | add(const_1,const_4)|divide(n0,const_3)|divide(n0,#0)|floor(#1)|multiply(#0,const_3)|divide(n0,#4)|floor(#2)|add(#3,#6)|floor(#5)|multiply(#8,const_2)|subtract(#7,#9) | other |
the product of two consecutive odd no is - 1 , then what is the value of the sum of them . | "two odd numbers are - 1 and + 1 . - 1 + 1 = 0 answer : c" | a ) - 2 , b ) - 1 , c ) 0 , d ) 1 , e ) 2 | c | multiply(power(const_2, 1), factorial(1)) | factorial(n0)|power(const_2,n0)|multiply(#0,#1)| | general |
to fill a tank , 10 buckets of water is required . how many buckets of water will be required to fill the same tank if the capacity of the bucket is reduced to two - fifth of its present ? | "let the capacity of 1 bucket = x . then , the capacity of tank = 10 x . new capacity of bucket = 2 / 5 x therefore , required number of buckets = ( 10 x ) / ( 2 x / 5 ) = ( 10 x ) x 5 / 2 x = 50 / 2 = 25 answer is a ." | a ) 25 , b ) 50 , c ) 75 , d ) 100 , e ) 125 | a | divide(multiply(10, add(const_4, const_1)), const_2) | add(const_1,const_4)|multiply(n0,#0)|divide(#1,const_2)| | physics |
if 30 oxen can plough 1 / 7 th of a field in 2 days , how many days 18 oxen will take to do the remaining work ? | solution : we will use work equivalence method , 30 / 18 = ( 1 / 7 ) / ( 6 / 7 ) * x / 2 ; 5 / 3 = ( 1 / 6 ) * x / 2 ; or , x = 60 / 3 = 20 days . answer : option b | a ) 30 days , b ) 20 days , c ) 15 days , d ) 18 days , e ) 21 days | b | divide(divide(30, 18), divide(divide(divide(1, 7), divide(subtract(7, const_1), 7)), 2)) | divide(n0,n4)|divide(n1,n2)|subtract(n2,const_1)|divide(#2,n2)|divide(#1,#3)|divide(#4,n3)|divide(#0,#5) | physics |
how many two - element subsets of { 6,7 , 8,9 } are there that do not contain the pair of elements 8 and 9 ? | required subsets are = { 6,7 } , { 6,8 } , { 6,9 } , { 7,8 } , { 7,9 } = 5 answer : a | a ) 5 , b ) 4 , c ) 3 , d ) 6 , e ) 7 | a | subtract(9, const_4) | subtract(n3,const_4) | general |
one fourth of a solution that was 10 % sugar by weight was replaced by a second solution resulting in a solution that was 14 percent sugar by weight . the second solution was what percent sugar by weight ? | say the second solution ( which was 1 / 4 th of total ) was x % sugar , then 3 / 4 * 0.1 + 1 / 4 * x = 1 * 0.14 - - > x = 0.26 . alternately you can consider total solution to be 100 liters and in this case you ' ll have : 75 * 0.1 + 25 * x = 100 * 0.14 - - > x = 0.26 . answer : e . | a ) 74 % , b ) 64 % , c ) 42 % , d ) 28 % , e ) 26 % | e | multiply(divide(subtract(multiply(const_100, divide(14, const_100)), multiply(subtract(const_100, multiply(divide(const_1, const_4), const_100)), divide(10, const_100))), multiply(divide(const_1, const_4), const_100)), const_100) | divide(n1,const_100)|divide(n0,const_100)|divide(const_1,const_4)|multiply(#0,const_100)|multiply(#2,const_100)|subtract(const_100,#4)|multiply(#1,#5)|subtract(#3,#6)|divide(#7,#4)|multiply(#8,const_100) | gain |
two - fifth of one - third of 3 - seventh of a number is 15 . what is 40 % of that number ? | explanation : let the number be x . then , 2 / 5 of 1 / 3 of 3 / 7 of x = 15 = > x = ( 15 * 5 / 3 * 3 * 5 / 2 ) = 525 / 2 40 % of 525 / 2 = ( 40 / 100 * 525 / 2 ) = 105 answer is e | a ) 72 , b ) 84 , c ) 136 , d ) 140 , e ) none of these | e | multiply(divide(15, multiply(multiply(divide(const_2, divide(15, 3)), divide(const_1, const_4)), divide(const_3, add(const_3, const_4)))), divide(40, const_100)) | add(const_3,const_4)|divide(n1,n0)|divide(const_1,const_4)|divide(n2,const_100)|divide(const_3,#0)|divide(const_2,#1)|multiply(#5,#2)|multiply(#4,#6)|divide(n1,#7)|multiply(#8,#3) | gain |
when 52416 is divided by 312 , the quotient is 168 . what will be the quotient when 522.416 is divided by 0.0168 ? | "for the 1 st no . there are 3 digits after decimal for the 2 nd no . there are 5 digits after decimal total no . of decimals = 8 req . no . of digits = ( n - 1 ) = ( 8 - 1 ) = 7 answer : e" | a ) 5 , b ) 6 , c ) 4 , d ) 3 , e ) 7 | e | divide(522.416, 0.0168) | divide(n3,n4)| | general |
there are 69 people that own pets . 15 people own only dogs , 10 people own only cats , 5 people own only cats and dogs , 3 people own cats , dogs and snakes . how many total snakes are there ? | "lets assign variables to all the areas in venn diagram of three . three different units are dog , cat , snake = total = 69 only dog = d = 15 only cat = c = 10 only snake = s exactly dog and cat = 5 exactly dog and snake = x exactly cat and snake = y all three = 3 so 69 = 15 + 10 + 5 + 3 + x + y + s we need to know tot... | a ) 2 , b ) 4 , c ) 8 , d ) 39 , e ) 32 | d | add(subtract(69, add(add(add(15, 10), 5), 3)), 3) | add(n1,n2)|add(n3,#0)|add(n4,#1)|subtract(n0,#2)|add(n4,#3)| | general |
an employee ’ s annual salary was increased $ 25,000 . if her new annual salary now equals $ 90,000 , what was the percent increase ? | new annual salary = $ 90,000 salary increase = $ 25,000 . original salary = $ 90,000 - $ 25,000 . = $ 75,000 % increase = ( $ 25,000 / $ 65,000 ) * 100 = 38.46 % hence d . | a ) 15 % , b ) 162 ⁄ 3 % , c ) 20 % , d ) 38.46 % , e ) 24 % | d | multiply(subtract(divide(multiply(subtract(const_100, const_10), const_1000), subtract(multiply(subtract(const_100, const_10), const_1000), multiply(multiply(const_0_25, const_100), const_1000))), const_1), const_100) | multiply(const_0_25,const_100)|subtract(const_100,const_10)|multiply(#1,const_1000)|multiply(#0,const_1000)|subtract(#2,#3)|divide(#2,#4)|subtract(#5,const_1)|multiply(#6,const_100) | general |
30 men can do a work in 40 days . when should 15 men leave the work so that the entire work is completed in 40 days after they leave the work ? | total work to be done = 30 * 40 = 1200 let 15 men leave the work after ' p ' days , so that the remaining work is completed in 40 days after they leave the work . 40 p + ( 15 * 40 ) = 1200 40 p = 600 = > p = 15 days answer : c | a ) 87 days , b ) 10 days , c ) 15 days , d ) 44 days , e ) 22 days | c | divide(subtract(multiply(30, 40), multiply(40, 15)), 40) | multiply(n0,n1)|multiply(n1,n2)|subtract(#0,#1)|divide(#2,n1) | physics |
a car averages 70 mph for the first 4 hours of a trip and averages 60 mph for each additional hour . the average speed for the entire trip was 65 mph . how many hours long is the trip ? | "let the time for which car averages 60 mph = t 65 * ( t + 4 ) = 70 * 4 + 60 t = > 5 t = 20 = > t = 4 total duration of the trip = 4 + 4 = 8 answer a" | a ) 8 , b ) 9 , c ) 10 , d ) 11 , e ) 12 | a | add(divide(subtract(multiply(65, 4), multiply(70, 4)), subtract(60, 65)), 4) | multiply(n3,n1)|multiply(n0,n1)|subtract(n2,n3)|subtract(#0,#1)|divide(#3,#2)|add(n1,#4)| | general |
the speed of the boat in still water in 12 kmph . it can travel downstream through 39 kms in 3 hrs . in what time would it cover the same distance upstream ? | "still water = 12 km / hr downstream = 39 / 3 = 13 km / hr upstream = > > still water = ( u + v / 2 ) = > > 12 = u + 13 / 2 = 11 km / hr so time taken in upstream = 39 / 11 = 3.5 hrs answer : a" | a ) 3.5 hours , b ) 6 hours , c ) 4 hours , d ) 5 hours , e ) 6 hours | a | divide(39, subtract(12, subtract(divide(39, 3), 12))) | divide(n1,n2)|subtract(#0,n0)|subtract(n0,#1)|divide(n1,#2)| | physics |
if n = 7 ^ 11 – 7 , what is the units digit of n ? | always divide the power ( incase 11 ) by 4 and use the remainder as the new power . the question now becomes 7 ^ 3 - 7 . now 7 ^ 3 has last digit 3 . since 7 ^ 11 ( or for that matter 7 ^ 3 ) is greater than 7 , we subtract 7 from 13 ( the 10 + 3 - - > 10 has come from carry over from the tenth place ) . thus 13 - 7 = ... | a ) 0 , b ) 1 , c ) 4 , d ) 6 , e ) 8 | d | divide(log(7), log(power(7, 11))) | log(n2)|power(n0,n1)|log(#1)|divide(#0,#2)| | general |
in a fuel station the service costs $ 1.75 per car , every liter of fuel costs 0.65 $ . assuming that a company owns 14 cars and that every fuel tank contains 55 liters and they are all empty , how much money total will it cost to fuel all cars ? | "total cost = ( 1.75 * 14 ) + ( 0.65 * 14 * 55 ) = 24.5 + 500.5 = > 525 hence answer will be ( c ) 525" | a ) 320 $ , b ) 380 $ , c ) 525 $ , d ) 456 $ , e ) 480 $ | c | multiply(multiply(0.65, 55), 14) | multiply(n1,n3)|multiply(n2,#0)| | general |
in a new housing development , trees are to be planted along the sidewalk of a certain street . each tree takes up one square foot of sidewalk space , and there are to be 14 feet between each tree . how many trees can be planted if the road is 151 feet long ? | "let t be the number of trees . then the length required for trees on the sidewalk will be 1 * t = t to maximize the number of trees , the number of 14 feet spaces between trees should be 1 less than total number of trees . for example , if there are 3 trees , then there should be 2 spaces between them . so the number ... | a ) 8 , b ) 9 , c ) 10 , d ) 11 , e ) 12 | d | add(divide(subtract(151, const_1), add(14, const_1)), const_1) | add(n0,const_1)|subtract(n1,const_1)|divide(#1,#0)|add(#2,const_1)| | geometry |
the maitre ' d at an expensive manhattan restaurant has noticed that 60 % of the couples order dessert and coffee . however , 20 % of the couples who order dessert do n ' t order coffee . what is the probability q that the next couple the maitre ' d seats will not order dessert ? | "could you use a venn diagram and just go with the number 100 . 60 people order dessert and coffee . . . which is the union of d and c . q = 2 / 10 of d are n ' t in d u c = so 8 / 10 of d are in duc which means = 60 = 8 / 10 d . so d in total = 75 , and 15 d ' s are n ' t in d union c . which means 25 people are in c ... | a ) 20 % , b ) 25 % , c ) 40 % , d ) 60 % , e ) 75 % | b | multiply(subtract(const_1, divide(divide(60, multiply(multiply(const_2, const_5), multiply(const_2, const_5))), subtract(const_1, divide(20, multiply(multiply(const_2, const_5), multiply(const_2, const_5)))))), multiply(multiply(const_2, const_5), multiply(const_2, const_5))) | multiply(const_2,const_5)|multiply(#0,#0)|divide(n0,#1)|divide(n1,#1)|subtract(const_1,#3)|divide(#2,#4)|subtract(const_1,#5)|multiply(#1,#6)| | probability |
the product of two numbers is 120 and the sum of their squares is 289 . the sum of the number is : | "let the numbers be x and y . then , xy = 120 and x 2 + y 2 = 289 . ( x + y ) 2 = x 2 + y 2 + 2 xy = 289 + ( 2 x 120 ) = 529 x + y = 529 = 23 . answer a" | a ) 23 , b ) 20 , c ) 12 , d ) 128 , e ) 171 | a | sqrt(add(power(sqrt(subtract(289, multiply(const_2, 120))), const_2), multiply(const_4, 120))) | multiply(n0,const_4)|multiply(n0,const_2)|subtract(n1,#1)|sqrt(#2)|power(#3,const_2)|add(#0,#4)|sqrt(#5)| | general |
when tom works alone he chops 2 lb . salad in 3 minutes , and when tammy works alone she chops 3 lb . salad in 2 minutes . they start working together , and after some time finish chopping 65 lb . of salad . of those 80 lb . , the salad quantity chopped by tammy is what percent greater than the quantifying chopped by t... | tom chops 4 lbs in 6 minutes tammy chops 9 lbs in 6 minutes so in the same amount of time , tammy chops 125 % more than tom , since 9 is 125 % greater than 4 . so 125 % is the answer . note that the actual time does n ' t matter . if you multiply the time each work by x , you ' ll multiply the work each does by x , and... | a ) 44 % , b ) 125 % , c ) 105 % , d ) 225 % , e ) 400 % | b | multiply(divide(subtract(divide(3, 2), divide(2, 3)), divide(2, 3)), const_100) | divide(n1,n0)|divide(n0,n1)|subtract(#0,#1)|divide(#2,#1)|multiply(#3,const_100) | physics |
the population of a town is 8000 . it decreases annually at the rate of 10 % p . a . what will be its population after 3 years ? | "formula : ( after = 100 denominator ago = 100 numerator ) 8000 ã — 90 / 100 ã — 90 / 100 x 90 / 100 = 5832 answer : e" | a ) 5100 , b ) 5120 , c ) 5200 , d ) 5400 , e ) 5832 | e | subtract(subtract(8000, multiply(8000, divide(10, const_100))), multiply(subtract(8000, multiply(8000, divide(10, const_100))), divide(10, const_100))) | divide(n1,const_100)|multiply(n0,#0)|subtract(n0,#1)|multiply(#0,#2)|subtract(#2,#3)| | gain |
a reduction of 30 % in the price of salt enables a lady to obtain 10 kgs more for rs . 100 , find the original price per kg ? | "100 * ( 30 / 100 ) = 30 - - - 10 ? - - - 1 = > rs . 3 100 - - - 70 ? - - - 3 = > rs . 4.3 answer : e" | a ) 2.8 , b ) 2.3 , c ) 2.5 , d ) 2.1 , e ) 4.3 | e | multiply(divide(divide(multiply(divide(30, const_100), 100), 10), multiply(divide(30, const_100), 100)), const_100) | divide(n0,const_100)|multiply(n2,#0)|divide(#1,n1)|divide(#2,#1)|multiply(#3,const_100)| | gain |
what is the sum of all the even numbers between 31 and 99 , inclusive ? | "to figure this problem out , you just have to add all the even numbers , starting with 32 . 32 + 34 + 36 + 38 and so on , all the way up to 98 . you ' ll get 2,180 , so final answer : d" | a ) 3,000 , b ) 2,562 , c ) 2,500 , d ) 2,180 , e ) 2,766 | d | add(add(add(add(add(add(const_12, const_2), const_1), add(add(const_12, const_2), add(add(add(add(add(const_2, const_4), const_4), subtract(const_10, const_1)), add(add(const_2, const_4), const_4)), add(const_10, const_2)))), add(add(add(const_12, const_2), const_1), const_1)), 31), add(const_2, const_4)) | add(const_12,const_2)|add(const_2,const_4)|add(const_10,const_2)|subtract(const_10,const_1)|add(#0,const_1)|add(#1,const_4)|add(#5,#3)|add(#4,const_1)|add(#6,#5)|add(#8,#2)|add(#0,#9)|add(#4,#10)|add(#11,#7)|add(n0,#12)|add(#13,#1)| | general |
a certain farmer pays $ 30 per acre per month to rent farmland . how much does the farmer pay per month to rent a rectangular plot of farmland that is 1080 feet by 605 feet ? ( 43,560 square feet = 1 acre ) | basically the question an error . 1 acre = 43,560 square feet and if it is then the answer is 450 ( c ) | a ) $ 5,330 , b ) $ 3,360 , c ) $ 450 , d ) $ 360 , e ) $ 150 | c | multiply(30, divide(multiply(1080, 605), divide(multiply(1080, 605), const_10))) | multiply(n1,n2)|divide(#0,const_10)|divide(#0,#1)|multiply(n0,#2)| | geometry |
according to the formula f = 9 / 5 ( c ) + 32 , if the temperature in degrees farenheit ( f ) increases by 25 , by how much does the temperature in degrees celsius ( c ) increase ? | "you can plug in values . c = 5 / 9 * ( f - 32 ) f = 32 - - > c = 0 ; f = 32 + 25 = 57 - - > c = 5 / 9 * 25 = 13.88 . increase = 13.88 degrees . answer : b ." | a ) 9 , b ) 13.88 , c ) 47 , d ) 48 3 / 5 , e ) 59 | b | divide(multiply(25, 5), 9) | multiply(n1,n3)|divide(#0,n0)| | general |
kamal obtained 76 , 60 , 72 , 65 and 82 marks ( out of 100 ) in english , mathematics , physics , chemistry and biology . what are his average marks ? | "sol . average = 76 + 60 + 72 + 65 + 82 / 5 ) = ( 355 / 5 ) = 71 . answer a" | a ) 71 , b ) 69 , c ) 72 , d ) 75 , e ) none | a | divide(add(add(add(add(76, 60), 72), 65), 82), add(const_1, const_4)) | add(n0,n1)|add(const_1,const_4)|add(n2,#0)|add(n3,#2)|add(n4,#3)|divide(#4,#1)| | general |
a person buys an article at rs . 550 . at what price should he sell the article so as to make a profit of 30 % ? | "cost price = rs . 550 profit = 30 % of 550 = rs . 165 selling price = cost price + profit = 550 + 165 = 715 answer : b" | a ) 600 , b ) 715 , c ) 772 , d ) 662 , e ) 521 | b | add(550, multiply(550, divide(30, const_100))) | divide(n1,const_100)|multiply(n0,#0)|add(n0,#1)| | gain |
an article with cost price of 280 is sold at 30 % profit . what is the selling price ? | "sp = 1.30 * 280 = 364 answer : b" | a ) 198 , b ) 364 , c ) 369 , d ) 207 , e ) 210 | b | add(280, multiply(280, divide(30, const_100))) | divide(n1,const_100)|multiply(n0,#0)|add(n0,#1)| | gain |
john can type a set of pages in 5 hours at a constant rate . if john types for 3 hours and jack does the rest of the job , who works at 2 / 5 constant rate of john . how long will it take for jack alone to do the rest of the job ? | john will do 3 / 5 in 3 hrs , so john does 3 / 5 of the work . . therefore , jack will do the remaining 2 / 5 th work alone . . as the speed of jack is 2 / 5 rate of john , jack will do the 2 / 5 th work in same time that john takes to complete full job . . . ans 5 d | a ) 1 hrs , b ) 3 hrs , c ) 4 hrs , d ) 5 hrs , e ) 7 hrs | d | divide(multiply(const_1, divide(2, 5)), multiply(inverse(5), subtract(const_1, divide(3, 5)))) | divide(n2,n0)|divide(n1,n0)|inverse(n0)|multiply(#0,const_1)|subtract(const_1,#1)|multiply(#2,#4)|divide(#3,#5) | physics |
a trader mixes 80 kg of tea at 15 per kg with 20 kg of tea at cost price of 20 per kg . in order to earn a profit of 30 % , what should be the sale price of the mixed tea ? | "c . p . of mixture = 80 × 15 + 20 × 20 / 80 + 20 = 16 ∴ s . p . = ( 100 + 30 ) / 100 × 16 = 20.8 answer c" | a ) 23.75 , b ) 22 , c ) 20.8 , d ) 19.2 , e ) none of these | c | add(divide(multiply(divide(add(multiply(80, 15), multiply(20, 20)), add(80, 20)), 30), const_100), divide(add(multiply(80, 15), multiply(20, 20)), add(80, 20))) | add(n0,n2)|multiply(n0,n1)|multiply(n2,n2)|add(#1,#2)|divide(#3,#0)|multiply(n4,#4)|divide(#5,const_100)|add(#6,#4)| | gain |
two persons a and b can complete a piece of work in 15 days and 30 days respectively . if they work together , what part of the work will be completed in 5 days ? | "a ' s one day ' s work = 1 / 15 b ' s one day ' s work = 1 / 30 ( a + b ) ' s one day ' s work = 1 / 15 + 1 / 30 = 1 / 10 the part of the work completed in 5 days = 5 ( 1 / 10 ) = 1 / 2 . answer : d" | a ) 1 / 8 , b ) 1 / 3 , c ) 1 / 6 , d ) 1 / 2 , e ) 5 / 3 | d | multiply(5, add(divide(const_1, 15), divide(const_1, 30))) | divide(const_1,n0)|divide(const_1,n1)|add(#0,#1)|multiply(n2,#2)| | physics |
the temperature of a certain cup of coffee 10 minutes after it was poured was 120 degrees fahrenheit . if the temperature f of the coffee t minutes after it was poured can be determined by the formula f = 120 ( 2 ^ - at ) + 60 , where f is in degrees fahrenheit and a is a constant . then the temperature of the coffee 5... | "answer : b the temperature of coffee 10 minutes after it was poured ( 120 f ) will help in solving the constant “ a ” . 120 = 120 ( 2 ^ 10 a ) + 60 2 ^ - 1 = 2 ^ 10 a a = - 1 / 10 the temperature of coffee 50 minutes after it was poured is : f = 120 ( 2 ^ - 50 / 10 ) + 60 f = 120 * 1 / 32 + 60 f = 15 / 4 + 60 f = 255 ... | a ) 63.75 , b ) 75 , c ) 80 , d ) 85 , e ) 90 | a | add(multiply(power(2, multiply(divide(60, 10), subtract(const_1, 2))), 120), 60) | divide(n4,n0)|subtract(const_1,n3)|multiply(#0,#1)|power(n3,#2)|multiply(n1,#3)|add(n4,#4)| | general |
water boils at 212 ° f or 100 ° c and ice melts at 32 ° f or 0 ° c . if the temperature of a pot of water is 40 ° c , what is the temperature of the pot of water in ° f ? | "let f and c denote the temperature in fahrenheit and celsius respectively . ( f - 32 ) / ( 212 - 32 ) = ( c - 0 ) / ( 100 - 0 ) f = 9 c / 5 + 32 f = 9 ( 40 ) / 5 + 32 = 104 ° f the answer is e ." | a ) 85 ° f , b ) 92 ° f , c ) 96 ° f , d ) 99 ° f , e ) 104 ° f | e | add(multiply(divide(subtract(212, 32), 100), 40), 32) | subtract(n0,n2)|divide(#0,n1)|multiply(n4,#1)|add(n2,#2)| | physics |
the number of natural numbers n such that ( n + 1 ) ^ 2 / ( n + 7 ) is an integer , is ? | we can use ( n + 1 ) ^ 2 greater than ( n + 7 ) which gives n > 2 , - 3 . . . . . . . . now there are four integers - 2 . - 1 , 01 in between - 3 and 2 . so answer is 4 . answer : a | a ) 4 , b ) 5 , c ) 6 , d ) 7 , e ) 8 | a | subtract(7, const_3) | subtract(n2,const_3) | general |
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