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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x ^ 2 + sx + 72 = 0 has two distinct integer roots ; how many values are possible for s ? | for a quadratic equation ax ^ 2 + sx + c = 0 , we know that - s / a is sum of roots and c / a is product of roots . the quadratic equation here is x ^ 2 + sx + 72 = 0 , where product of roots is 72 . if we find all the factors of 72 , we have the answer . by prime factorization , we get 72 = 2 ^ 3 * 3 ^ 2 . we know tha... | a ) 3 , b ) 6 , c ) 8 , d ) 12 , e ) 24 | d | multiply(add(const_3, const_1), add(const_2, const_1)) | add(const_1,const_3)|add(const_1,const_2)|multiply(#0,#1) | general |
the sides of a square region , measured to the nearest centimeter , are 5 centimeters long . the least possible value of the actual area of the square region is | "though there might be some technicalities concerning the termnearest ( as 4.5 is equidistant from both 4 and 5 ) the answer still should be : 4.5 ^ 2 = 20.25 . answer : b" | a ) 25.25 sq cm , b ) 20.25 sq cm , c ) 30.25 sq cm , d ) 36.25 sq cm , e ) 40.25 sq cm | b | power(subtract(subtract(5, const_0_25), const_0_25), const_2) | subtract(n0,const_0_25)|subtract(#0,const_0_25)|power(#1,const_2)| | geometry |
an outlet pipe empties a tank which is full in 10 hours . if the inlet pipe is kept open , which lets water in at the rate of 4 litres / min then outlet pipe would take 6 hours longer . find the capacity of the tank . | "let the rate of outlet pipe be x liters / hour ; rate of inlet pipe is 4 litres / min , or 4 * 60 = 240 liters / hour ; net outflow rate when both pipes operate would be x - 240 liters / hour . capacity of the tank = x * 10 hours = ( x - 240 ) * ( 10 + 6 ) hours 10 x = ( x - 240 ) * 16 - - > x = 640 - - > capacity = 1... | a ) 8600 litres , b ) 6400 litres , c ) 12800 litres , d ) 11200 litres , e ) 13200 litres | b | multiply(divide(multiply(multiply(4, const_60), add(10, 6)), 6), 10) | add(n0,n2)|multiply(n1,const_60)|multiply(#0,#1)|divide(#2,n2)|multiply(n0,#3)| | physics |
a , b and c invest in the ratio of 3 : 4 : 5 . the percentage of return on their investments are in the ratio of 6 : 5 : 4 . find the total earnings , if b earns rs . 350 more than a : | "explanation : a b c investment 3 x 4 x 5 x rate of return 6 y % 5 y % 4 y % return \ inline \ frac { 18 xy } { 100 } \ inline \ frac { 20 xy } { 100 } \ inline \ frac { 20 xy } { 100 } total = ( 18 + 20 + 20 ) = \ inline \ frac { 58 xy } { 100 } b ' s earnings - a ' s earnings = \ inline \ frac { 2 xy } { 100 } = 350 ... | a ) 10150 , b ) 7250 , c ) 2767 , d ) 1998 , e ) 2771 | a | multiply(add(add(multiply(3, 6), multiply(4, 5)), multiply(5, 4)), divide(350, subtract(multiply(4, 5), multiply(3, 6)))) | multiply(n0,n3)|multiply(n1,n2)|add(#0,#1)|subtract(#1,#0)|add(#2,#1)|divide(n6,#3)|multiply(#4,#5)| | general |
a train 50 m long is running with a speed of 68 kmph . in what time will it pass a man who is running at 8 kmph in the same direction in which the train is going ? | "solution speed of the train relative to man = ( 68 - 8 ) = 60 kmph = 60 x 5 / 18 = 50 / 3 m / sec . time taken by it to cover 50 m at ( 50 / 3 ) m / sec = ( 112 - 50 ) kmph = ( 50 x 3 / 50 ) sec = 3 sec . answer a" | a ) 3 sec . , b ) 7 sec . , c ) 9 sec . , d ) 11 sec . , e ) none | a | divide(50, multiply(add(68, 8), const_0_2778)) | add(n1,n2)|multiply(#0,const_0_2778)|divide(n0,#1)| | physics |
indu gave bindu rs . 4375 on compound interest for 2 years at 4 % per annum . how much loss would indu has suffered had she given it to bindu for 2 years at 4 % per annum simple interest ? | "4375 = d ( 100 / 4 ) 2 d = 7 answer : b" | a ) s . 10 , b ) s . 7 , c ) s . 5 , d ) s . 3 , e ) s . 4 | b | subtract(subtract(multiply(4375, power(add(const_1, divide(4, const_100)), 2)), 4375), multiply(multiply(4375, divide(4, const_100)), 2)) | divide(n2,const_100)|add(#0,const_1)|multiply(n0,#0)|multiply(n1,#2)|power(#1,n1)|multiply(n0,#4)|subtract(#5,n0)|subtract(#6,#3)| | gain |
how many books each of volume 100 meter cube can be packed into a crate of volume 3000 meter cube ? | "gud question with a simple concept . in geo if we want to insert one shape into another we need to know the dimensions of the two shapes . in above with volume given , we can come up with different shapes , so we cant know the answer for ex : 3000 m 3 can be 300 * 10 or 30 * 100 or just 3000 * 1 we do n ' t know , so ... | a ) 50 , b ) can not be determined , c ) 150 , d ) 300 , e ) 350 | b | multiply(3000, const_1) | multiply(n1,const_1)| | geometry |
an article is bought for rs . 675 and sold for rs . 700 , find the gain percent ? | "675 - - - - 25 100 - - - - ? = > = 4 % answer : c" | a ) 2 % , b ) 3 % , c ) 4 % , d ) 5 % , e ) 6 % | c | subtract(const_100, divide(multiply(700, const_100), 675)) | multiply(n1,const_100)|divide(#0,n0)|subtract(const_100,#1)| | gain |
having scored 98 runs in the 19 th inning , a cricketer increases his average score by 4 . what will be his average score after 19 innings ? | "explanation : let the average score of the first 18 innings be n 18 n + 98 = 19 ( n + 4 ) = > n = 22 so , average score after 19 th innings = x + 4 = 26 . answer : c" | a ) 28 , b ) 27 , c ) 26 , d ) 22 , e ) 24 | c | add(subtract(98, multiply(19, 4)), 4) | multiply(n1,n2)|subtract(n0,#0)|add(n2,#1)| | 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 50 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 / 50 work done by p , q and r in 1 day = 1 / 10 + 1 / 50 = 6 / 50 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 × 2 = 6 / 50 = > work done by p in 1 day = 3 / 50 = > wor... | a ) 30 days , b ) 25 days , c ) 20 days , d ) 15 days , e ) 24 days | b | divide(const_1, subtract(divide(add(divide(const_1, 10), divide(const_1, 50)), const_2), divide(const_1, 50))) | divide(const_1,n0)|divide(const_1,n1)|add(#0,#1)|divide(#2,const_2)|subtract(#3,#1)|divide(const_1,#4)| | physics |
dhoni spent 40 percent of his earning last month on rent and 20 percent less than what he spent on rent to purchase a new dishwasher . what percent of last month ' s earning did dhoni have left over ? | "say dhoni ' s earning last month was $ 100 . dhoni spent 40 percent of his earning last month on rent - - > $ 40 on rent ; 20 percent less than what he spent on rent to purchase a new dishwasher - - > $ 40 * 0.8 = $ 32 on the dishwasher . left over amount 100 - ( 40 + 32 ) = $ 28 answer : b" | a ) 30 % , b ) 28 % , c ) 32 % , d ) 33 % , e ) 34 % | b | subtract(const_100, add(40, multiply(40, subtract(const_1, divide(20, const_100))))) | divide(n1,const_100)|subtract(const_1,#0)|multiply(n0,#1)|add(n0,#2)|subtract(const_100,#3)| | gain |
a man whose bowling average is 12.4 , takes 7 wickets for 26 runs and there by decreases his average by 0.4 . the number of wickets taken by him before his last match is ? | 12.4 * x + 26 = ( 7 + x ) 12 solve equation x = 145 answer : c | a ) 143 , b ) 144 , c ) 145 , d ) 146 , e ) 147 | c | divide(subtract(multiply(floor(12.4), 7), 26), subtract(12.4, floor(12.4))) | floor(n0)|multiply(n1,#0)|subtract(n0,#0)|subtract(#1,n2)|divide(#3,#2) | general |
137 + 276 = 435 how much is 731 + 672 ? | 137 - 731 and 276 - 672 similarly 435 - 534 answer : a | a ) 534 , b ) 1403 , c ) 1623 , d ) 1513 , e ) 1613 | a | add(add(multiply(reminder(435, const_10), const_100), multiply(reminder(divide(subtract(435, reminder(435, const_10)), const_10), const_10), const_10)), divide(subtract(divide(subtract(435, reminder(435, const_10)), const_10), reminder(divide(subtract(435, reminder(435, const_10)), const_10), const_10)), const_10)) | reminder(n2,const_10)|multiply(#0,const_100)|subtract(n2,#0)|divide(#2,const_10)|reminder(#3,const_10)|multiply(#4,const_10)|subtract(#3,#4)|add(#1,#5)|divide(#6,const_10)|add(#7,#8) | general |
a batsman scored 120 runs which included 6 boundaries and 4 sixes . what % of his total score did he make by running between the wickets | number of runs made by running = 110 - ( 6 x 4 + 4 x 6 ) = 120 - ( 48 ) = 72 now , we need to calculate 60 is what percent of 120 . = > 72 / 120 * 100 = 60 % b | a ) 40 % , b ) 60 % , c ) 65 % , d ) 70 % , e ) 75 % | b | multiply(divide(subtract(120, add(multiply(6, const_4), multiply(4, 6))), 120), const_100) | multiply(n1,const_4)|multiply(n1,n2)|add(#0,#1)|subtract(n0,#2)|divide(#3,n0)|multiply(#4,const_100) | general |
a bag contains 3 red , 2 blue and 4 green balls . if 2 balls are picked at random , what is the probability that both are red ? | "p ( both are red ) , = 3 c 2 / 9 c 2 = 1 / 12 a" | a ) 1 / 12 , b ) 1 / 13 , c ) 1 / 14 , d ) 1 / 15 , e ) 1 / 18 | a | divide(choose(3, 2), choose(add(add(3, 2), 4), 2)) | add(n0,n1)|choose(n0,n3)|add(n2,#0)|choose(#2,n3)|divide(#1,#3)| | other |
two pipes can separately fill a tank in 20 and 30 hours respectively . both the pipes are opened to fill the tank but when the tank is full , a leak develops in the tank through which one - third of water supplied by both the pipes goes out . what is the total time taken to fill the tank ? | 1 / 20 + 1 / 30 = 1 / 12 1 + 1 / 3 = 4 / 3 1 - - - 12 4 / 3 - - - ? 4 / 3 * 12 = 16 hrs answer : b | a ) 17 hrs , b ) 16 hrs , c ) 15 hrs , d ) 13 hrs , e ) 12 hrs | b | add(inverse(add(inverse(20), inverse(30))), divide(inverse(add(inverse(20), inverse(30))), const_3)) | inverse(n0)|inverse(n1)|add(#0,#1)|inverse(#2)|divide(#3,const_3)|add(#4,#3) | physics |
the area of playground is 5900 sq . meters . what will be the cost of covering it with grass sheet 1 cm deep , if cost of grass sheet is $ 2.80 per cubic meter . | total volume * unit cost = total cost or , 5900 * 0.01 * 2.8 = total cost = 165 = c | ['a ) a ) $ 144', 'b ) b ) $ 150.50', 'c ) c ) $ 165', 'd ) d ) $ 158.60', 'e ) e ) $ 160.70'] | c | multiply(2.8, multiply(5900, divide(1, const_100))) | divide(n1,const_100)|multiply(n0,#0)|multiply(n2,#1) | physics |
hcf and lcm two numbers are 12 and 396 respectively . if one of the numbers is 48 , then the other number is ? | "12 * 396 = 48 * x x = 99 answer : a" | a ) 99 , b ) 66 , c ) 132 , d ) 264 , e ) 364 | a | divide(multiply(12, 396), 48) | multiply(n0,n1)|divide(#0,n2)| | physics |
the total number of plums that grow during each year on a certain plum tree is equal to the number of plums that grew during the previous year , less the age of the tree in years ( rounded down to the nearest integer ) . during its 4 th year , the plum tree grew 50 plums . if this trend continues , how many plums will ... | "1 st year : 0 - 1 ( age ) , we take age = 0 ( as the question says that we have to ( rounded down to the nearest integer ) ) 2 ndyear : 1 - 2 ( age ) , we take age = 1 3 rd year : 2 - 3 ( age ) , we take age = 2 4 th year : 3 - 4 ( age ) , we take age = 3 5 th year : 4 - 5 ( age ) , we take age = 4 6 th year : 5 - 6 (... | a ) 36 , b ) 41 , c ) 38 , d ) 40 , e ) 42 | b | add(divide(add(subtract(subtract(4, const_1), const_1), add(50, subtract(4, const_1))), const_3), const_3) | subtract(n0,const_1)|add(n1,#0)|subtract(#0,const_1)|add(#1,#2)|divide(#3,const_3)|add(#4,const_3)| | general |
a person buys an article at rs . 500 . at what price should he sell the article so as to make a profit of 20 % ? | explanation : cost price = rs . 500 profit = 20 % of 500 = rs . 100 selling price = cost price + profit = 500 + 100 = 600 answer : a | a ) 600 , b ) 2877 , c ) 208 , d ) 1882 , e ) 191 | a | add(500, multiply(500, divide(20, const_100))) | divide(n1,const_100)|multiply(n0,#0)|add(n0,#1) | gain |
if x ^ 2 is divisible by 240 what is the least possible value of integer x ? | 240 can be written as ( 2 ^ 4 ) * 3 * 5 . for x ^ 2 to be divisible by 240 it should contain at least 2 ^ 4 and 3 and 5 in its factors . we can leave out option e because 12 doesnt have 5 as one of its factor . now if we check for option b , 30 can be written as 2 * 3 * 5 , hence 30 ^ 2 will have 2 as the maximum power... | a ) 120 , b ) 30 , c ) 90 , d ) 60 , e ) 12 | d | divide(divide(240, 2), 2) | divide(n1,n0)|divide(#0,n0) | general |
machine t can produce x units in 3 / 4 of the time it takes machine n to produce x units . machine n can produce x units in 2 / 3 the time it takes machine o to produce x units . if all 3 machines are working simultaneously , what fraction of the total output is produced by machine n ? | let the following be true : t makes x in time t then the following follows : n makes x in 4 t / 3 o makes x in 3 / 2 ( 4 t / 3 ) = 2 t m : n : o = 1 : 4 / 3 : 2 = 3 : 4 : 6 so n = 4 / ( 3 + 4 + 6 ) = 4 / 13 = c | a ) 1 / 2 , b ) 1 / 3 , c ) 4 / 13 , d ) 8 / 29 , e ) 6 / 33 | c | multiply(inverse(add(divide(2, 3), add(const_1, multiply(divide(2, 3), divide(3, 4))))), divide(2, 3)) | divide(n2,n0)|divide(n0,n1)|multiply(#0,#1)|add(#2,const_1)|add(#3,#0)|inverse(#4)|multiply(#0,#5) | general |
( ( 1 ^ m ) / ( 5 ^ m ) ) ( ( 1 ^ 16 ) / ( 4 ^ 16 ) ) = 1 / ( 2 ( 10 ) ^ 31 ) what is m ? | ( ( 1 ^ m ) / ( 5 ^ m ) ) ( ( 1 ^ 16 ) / ( 4 ^ 16 ) ) = 1 / ( 2 ( 10 ) ^ 31 ) ( ( 1 / 5 ) ^ m ) * ( ( 1 / 2 ) ^ 32 ) = 1 / ( 2 * ( 2 * 5 ) ^ 31 ) ) 2 ^ 36 will cancel out , since 1 can be written as 1 ^ 35 , so ( 1 / 5 ) ^ m = ( 1 / 5 ) ^ 31 ( ( 1 / 5 ) ^ m ) * ( ( 1 / 2 ) ^ 32 ) = 1 / [ ( 2 ^ 32 ) * ( 5 ^ 31 ) ] so , ... | a ) 17 , b ) 18 , c ) 34 , d ) 31 , e ) 36 | d | inverse(multiply(inverse(31), divide(divide(divide(divide(1, 16), divide(4, 16)), divide(1, multiply(2, 10))), 5))) | divide(n0,n3)|divide(n4,n3)|inverse(n9)|multiply(n7,n8)|divide(#0,#1)|divide(n0,#3)|divide(#4,#5)|divide(#6,n1)|multiply(#7,#2)|inverse(#8) | general |
in a recent head - to - head run - off election , 12000 absentee ballets were cast . 1 / 6 of the absentee ballets were thrown out and 3 / 5 of the remaining absentee ballets were cast for candidate a . how many absentee votes did candidate b receive ? | 5 / 6 * 2 / 5 ( total absentee votes ) = 1 / 3 ( total votes ) = 1 / 3 * 12000 = 4000 answer is c | a ) 2,000 , b ) 3,000 , c ) 4,000 , d ) 8,000 , e ) 9,000 | c | subtract(divide(subtract(multiply(divide(multiply(divide(12000, 6), 5), 5), 3), divide(multiply(divide(12000, 6), 5), 5)), const_1000), 1) | divide(n0,n2)|multiply(n4,#0)|divide(#1,n4)|multiply(n3,#2)|subtract(#3,#2)|divide(#4,const_1000)|subtract(#5,n1) | general |
one - third of rahul ' s savings in national savings certificate is equal to one - half of his savings in public provident fund . if he has rs . 1 , 00,000 as total savings , how much has he saved in public provident fund ? | "let savings in n . s . c and p . p . f . be rs . x and rs . ( 100000 - x ) respectively . then , = 1 / 3 x = 1 / 2 ( 100000 - x ) = x / 3 + x / 2 = 50000 = 5 x / 6 = 50000 = x = 50000 x 6 / 5 = 60000 savings in public provident fund = rs . ( 100000 - 60000 ) = rs . 40000 answer is b ." | a ) 80000 , b ) 40000 , c ) 50000 , d ) 75000 , e ) 90000 | b | multiply(add(multiply(multiply(const_100, const_10), const_100), subtract(multiply(multiply(const_100, const_10), const_100), multiply(multiply(const_2, const_100), const_100))), divide(1, add(divide(const_3, const_2), 1))) | divide(const_3,const_2)|multiply(const_10,const_100)|multiply(const_100,const_2)|add(n0,#0)|multiply(#1,const_100)|multiply(#2,const_100)|divide(n0,#3)|subtract(#4,#5)|add(#4,#7)|multiply(#8,#6)| | general |
what is the probability that company g ' s advertisement will be one of the first two be shown during the first commercial break ? | g 1 / 15 + ( 14 / 15 ) * 1 / 14 = 2 / 15 d | a ) 1 / 225 , b ) 1 / 30 , c ) 1 / 15 , d ) 2 / 15 , e ) 4 / 15 | d | divide(const_2, multiply(const_3, const_5)) | multiply(const_3,const_5)|divide(const_2,#0) | probability |
at what rate percent on simple interest will rs . 750 amount to rs . 900 in 10 years ? | "150 = ( 750 * 10 * r ) / 100 r = 2 % answer : b" | a ) 6 % , b ) 2 % , c ) 4 % , d ) 5 % , e ) 3 % | b | multiply(divide(divide(subtract(900, 750), 750), 10), const_100) | subtract(n1,n0)|divide(#0,n0)|divide(#1,n2)|multiply(#2,const_100)| | gain |
one drier dries certain quantity of material in 24 minutes . another drier does the same work in 2 minutes how much time will it take to do the same job when both driers are put to work ? | by guess it is clear that the time taken will be less than 2 minutes and more than 1.5 mintes therefore , answer 1.85 minutes will be correct . answer - d | a ) 1.00 minutes , b ) 1.20 minutes , c ) 1.50 min , d ) 1.85 min , e ) 2.00 minutes | d | divide(const_1, add(divide(const_1, 24), divide(const_1, 2))) | divide(const_1,n0)|divide(const_1,n1)|add(#0,#1)|divide(const_1,#2) | physics |
if the lcm and hcf of 8 and another number is 24 and 4 respectively . find the other number ? | "hcf x lcm = product of numbers 4 x 24 = 8 x the other number other number = ( 4 x 24 ) / 8 other number = 12 answer : b" | a ) 11 , b ) 12 , c ) 14 , d ) 15 , e ) 16 | b | divide(multiply(24, 4), 8) | multiply(n1,n2)|divide(#0,n0)| | physics |
the average age of an adult class is 40 years . 18 new students with an avg age of 32 years join the class . therefore decreasing the average by 4 year . find what was theoriginal strength of class ? | "let original strength = y then , 40 y + 18 x 32 = ( y + 18 ) x 36 â ‡ ’ 40 y + 576 = 36 y + 648 â ‡ ’ 4 y = 72 â ˆ ´ y = 18 e" | a ) 8 , b ) 12 , c ) 15 , d ) 17 , e ) 18 | e | divide(subtract(multiply(18, subtract(40, 4)), multiply(18, 32)), 4) | multiply(n1,n2)|subtract(n0,n3)|multiply(n1,#1)|subtract(#2,#0)|divide(#3,n3)| | general |
a number whose fifth part increased by 7 is equal to its fourth part diminished by 7 is ? | "answer let the number be n . then , ( n / 5 ) + 7 = ( n / 4 ) - 7 â ‡ ’ ( n / 4 ) - ( n / 5 ) = 14 â ‡ ’ ( 5 n - 4 n ) / 20 = 14 â ˆ ´ n = 280 option : d" | a ) 160 , b ) 180 , c ) 200 , d ) 280 , e ) none | d | divide(add(7, 7), subtract(divide(const_1, const_4), divide(const_1, add(const_1, const_4)))) | add(n0,n1)|add(const_1,const_4)|divide(const_1,const_4)|divide(const_1,#1)|subtract(#2,#3)|divide(#0,#4)| | general |
a tradesman sold an article at a loss of 25 % . if the selling price had been increased by $ 500 , there would have been a gain of 15 % . what was the cost price of the article ? | let c . p . be $ x then 125 % of x - 85 % of x = 500 40 % of x = 500 2 x / 5 = 500 x = $ 1250 answer is e | a ) $ 1000 , b ) $ 1200 , c ) $ 1120 , d ) $ 1350 , e ) $ 1250 | e | divide(500, divide(add(25, 15), const_100)) | add(n0,n2)|divide(#0,const_100)|divide(n1,#1) | gain |
find the area , diameter = 9 m . | "diameter = 9 meter . radius = diameter / 2 . = 9 / 2 . = 4.5 meter . area of a circle = ï € r 2 . here , pi ( ï € ) = 3.14 meter , radius ( r ) = 4.5 . area of a circle = 3.14 ã — 4.5 ã — 4.5 . = 3.14 ã — 20.25 . = 63.64 m 2 . answer : b" | a ) 63.00 square meter , b ) 63.64 square meter , c ) 63.08 square meter , d ) 63.24 square meter , e ) 63.43 square meter | b | circle_area(divide(9, const_2)) | divide(n0,const_2)|circle_area(#0)| | physics |
if 1.5 x = 0.04 y then the value of ( y - x ) / ( y + x ) is | x / y = 0.04 / 1.5 y / x = 1.5 / 0.04 by componendo dividendo rule ( y + x ) / ( y - x ) = 1.54 / 1.46 ( y - x ) / ( y + x ) = 1.46 / 1.54 = 73 / 77 answer : b | a ) 730 / 77 , b ) 73 / 77 , c ) 7.3 / 77 , d ) 7.3 / 770 , e ) 7.3 / 77 | b | divide(subtract(divide(multiply(1.5, const_100), const_2), const_2), add(divide(multiply(1.5, const_100), const_2), const_2)) | multiply(n0,const_100)|divide(#0,const_2)|add(#1,const_2)|subtract(#1,const_2)|divide(#3,#2) | general |
the distance from the x - axis to point p is half the distance from the y - axis to point p . if the coordinates of p are ( x , - 8 ) , how many units is p from the y - axis ? | the x - axis is 8 units from the point p . thus the y - axis is 16 units from the point p . the answer is b . | a ) 20 , b ) 16 , c ) 8 , d ) 4 , e ) 2 | b | multiply(8, const_2) | multiply(n0,const_2) | general |
150 liters of a mixture of milk and water contains in the ratio 3 : 2 . how much water should now be added so that the ratio of milk and water becomes 3 : 4 ? | "milk = 3 / 5 * 150 = 90 liters water = 50 liters 90 : ( 50 + p ) = 3 : 4 150 + 3 p = 360 = > p = 70 50 liters of water are to be added for the ratio become 3 : 4 . answer : c" | a ) 90 liters , b ) 20 liters , c ) 50 liters , d ) 20 liters , e ) 70 liters | c | multiply(divide(150, add(3, 2)), 2) | add(n1,n2)|divide(n0,#0)|multiply(n2,#1)| | general |
two years ago , john put $ 5000 into a savings account . at the end of the first year , his account had accrued $ 500 in interest bringing his total balance to $ 5500 . the next year , his account balance increased by 10 % . at the end of the two years , by what percent has john ' s account balance increased from his i... | investment 5000 dollars 1 st year total gained = 500 total amount end of first year = 5500 second year account increased by 10 % = 5500 * 0.1 = 550 therefore total amount by second year end = 6050 so total percentage increase in money = ( 6050 - 5000 ) * 100 / 5000 = 21 % correct answer b = 21 % | a ) 19 % , b ) 21 % , c ) 20 % , d ) 22 % , e ) 25 % | b | add(add(divide(5000, 500), 10), const_1) | divide(n0,n1)|add(n3,#0)|add(#1,const_1) | gain |
a bucket full of nuts was discovered by the crow living in the basement . the crow eats a fifth of the total number of nuts in 6 hours . how many hours in total will it take the crow to finish a quarter of the nuts ? | "in one hour , the crow eats 1 / 30 of the nuts . ( 1 / 4 ) / ( 1 / 30 ) = 7.5 hours the answer is c ." | a ) 6.5 , b ) 7.0 , c ) 7.5 , d ) 8.0 , e ) 8.5 | c | divide(divide(const_1, const_4), divide(divide(const_1, add(const_2, const_3)), 6)) | add(const_2,const_3)|divide(const_1,const_4)|divide(const_1,#0)|divide(#2,n0)|divide(#1,#3)| | general |
how much is 65 % of 40 is greater than 4 / 5 of 25 ? | "( 65 / 100 ) * 40 – ( 4 / 5 ) * 25 26 - 20 = 6 answer : e" | a ) 29 , b ) 776 , c ) 66 , d ) 12 , e ) 6 | e | subtract(multiply(40, divide(65, const_100)), multiply(divide(4, 5), 25)) | divide(n0,const_100)|divide(n2,n3)|multiply(n1,#0)|multiply(n4,#1)|subtract(#2,#3)| | general |
in a graduating class of 232 students , 144 took geometry and 119 took biology . what is the difference between the greatest possible number r and the smallest possible number of students that could have taken both geometry and biology ? | "official solution : first of all , notice that since 144 took geometry and 119 took biology , then the number of students who took both geometry and biology can not be greater than 119 . { total } = { geometry } + { biology } - { both } + { neither } ; 232 = 144 + 119 - { both } + { neither } ; { both } = 31 + { neith... | a ) 144 , b ) 119 , c ) 113 , d ) 88 , e ) 31 | d | subtract(119, subtract(add(144, 119), 232)) | add(n1,n2)|subtract(#0,n0)|subtract(n2,#1)| | other |
if the numerator of a fraction be increased by 12 % and its denominator decreased by 2 % the value of the fraction becomes 6 / 7 . thus , the original fraction is : | if original fraction is x / y , then 1.12 x / 0.98 y = 6 / 7 ( 8 / 7 ) * ( x / y ) = 6 / 7 x / y = 6 / 8 = 3 / 4 answer : c | a ) 1 / 4 , b ) 2 / 4 , c ) 3 / 4 , d ) 2 / 3 , e ) 1 / 3 | c | divide(multiply(subtract(const_100, 2), 6), multiply(add(const_100, 12), 7)) | add(n0,const_100)|subtract(const_100,n1)|multiply(n2,#1)|multiply(n3,#0)|divide(#2,#3) | general |
3 - fourth of two - third of 3 - seventh of a number is 27 . what is 10 % of that number ? | explanation : solution : assume the number be x . then , 3 / 4 of 2 / 3 of 3 / 7 of x = 27 . x = 27 * 7 / 3 * 3 / 2 * 4 / 3 . x = 126 . ' . 10 % of 126 = 10 / 100 * 126 = 12.6 answer : a | a ) 12.6 % , b ) 6.3 % , c ) 27 % , d ) 25.2 % , e ) none of these | a | divide(multiply(27, 10), multiply(multiply(divide(3, add(const_3, const_4)), multiply(divide(3, const_4), divide(const_2, const_3))), const_100)) | add(const_3,const_4)|divide(n0,const_4)|divide(const_2,const_3)|multiply(n2,n3)|divide(n0,#0)|multiply(#1,#2)|multiply(#4,#5)|multiply(#6,const_100)|divide(#3,#7) | gain |
a shipment of 250 smartphones contains 67 that are defective . if a customer buys two smartphones at random from the shipment , what is the approximate probability that both phones are defective ? | probability of chosing one defective phone from a lot of 250 which ontains 67 defective phones is = ( 67 / 250 ) probability of chosing one defective phone from a lot of 249 ( we already picked one ) which ontains 66 ( we already picked one ) defective phones is = ( 66 / 249 ) combined probability of series of events =... | a ) 1 / 250 , b ) 1 / 14 , c ) 1 / 11 , d ) 1 / 9 , e ) 1 / 3 | b | divide(multiply(67, subtract(67, const_1)), multiply(250, subtract(250, const_1))) | subtract(n1,const_1)|subtract(n0,const_1)|multiply(n1,#0)|multiply(n0,#1)|divide(#2,#3) | other |
tabby is training for a triathlon . she swims at a speed of 1 mile per hour . she runs at a speed of 11 miles per hour . she wants to figure out her average speed for these two events . what is the correct answer for her ? | "( 1 mph + 11 mph ) / 2 = 6 mph correct option is : a" | a ) 6 mph , b ) 5.25 mph , c ) 3.5 mph , d ) 4 mph , e ) 0.5 mph | a | divide(add(1, 11), const_2) | add(n0,n1)|divide(#0,const_2)| | physics |
( 17 ) 4.25 x ( 17 ) ? = 178 | "solution let ( 17 ) 4.25 * ( 17 ) x = 178 . then , ( 17 ) 4.25 + x = ( 17 ) 8 . ∴ 4.25 + x = 8 ⇔ x = ( 8 - 4.25 ) ⇔ x = 3.75 answer d" | a ) 2.29 , b ) 2.75 , c ) 4.25 , d ) 3.75 , e ) none of these | d | multiply(17, 17) | multiply(n0,n2)| | general |
cole drove from home to work at an average speed of 75 kmh . he then returned home at an average speed of 105 kmh . if the round trip took a total of 2 hours , how many minutes did it take cole to drive to work ? | "let the distance one way be x time from home to work = x / 75 time from work to home = x / 105 total time = 2 hrs ( x / 75 ) + ( x / 105 ) = 2 solving for x , we get x = 175 / 2 time from home to work in minutes = ( 175 / 2 ) * 60 / 75 = 70 minutes ans = b" | a ) 66 , b ) 70 , c ) 72 , d ) 75 , e ) 78 | b | multiply(divide(multiply(105, 2), add(75, 105)), const_60) | add(n0,n1)|multiply(n1,n2)|divide(#1,#0)|multiply(#2,const_60)| | physics |
65 % of x = 20 % of 487.50 . find the value of x ? | 65 % of x = 20 % of 487.50 then , 65 / 100 * x = 20 / 100 * 4875 / 10 x = 150 answer is b | a ) 100 , b ) 150 , c ) 160 , d ) 180 , e ) 199 | b | divide(multiply(multiply(divide(487.5, const_100), 20), const_100), 65) | divide(n2,const_100)|multiply(n1,#0)|multiply(#1,const_100)|divide(#2,n0) | general |
m = { - 6 , - 5 , - 4 , - 3 , - 2 } t = { - 3 , - 2 , - 1 , 0 , 1 , 2 , 3 , 4 , 5 } if an integer is to be randomly selected from set m above and an integer is to be randomly selected from set t above , what is the probability that the product of the two integers will be negative ? | we will have a negative product only if 1 , 2 , 3 , 4 , or 5 are selected from set t . p ( negative product ) = 5 / 9 the answer is e . | a ) 0 , b ) 3 / 7 , c ) 2 / 5 , d ) 1 / 2 , e ) 5 / 9 | e | divide(5, add(4, 5)) | add(n1,n2)|divide(n1,#0) | general |
how many pieces of 85 cm length can be cut from a rod of 42.5 meters long ? | "number of pieces = 4250 / 85 = 850 / 17 = 50 answer is a ." | a ) 50 , b ) 40 , c ) 30 , d ) 20 , e ) 10 | a | divide(multiply(42.5, const_100), 85) | multiply(n1,const_100)|divide(#0,n0)| | physics |
if xerox paper costs 5 cents a sheet and a buyer gets 10 % discount on all xerox paper one buys after the first 2000 papers and 20 % discount after first 10000 papers , how much will it cost to buy 45000 sheets of xerox paper ? | "30 sec approach - solve it using approximation 45000 sheet at full price , 5 cent = 2250 45000 sheet at max discount price , 4 cent = 2000 your ans got to be between these two . ans e it is ." | a ) $ 1250 , b ) $ 1060 , c ) $ 1350 , d ) $ 900 , e ) $ 2100 | e | multiply(subtract(10000, 2000), multiply(subtract(const_1, divide(const_1, 10)), divide(const_1, 10))) | divide(const_1,n1)|subtract(n4,n2)|subtract(const_1,#0)|multiply(#0,#2)|multiply(#3,#1)| | gain |
a man buys an article for $ 10 . and sells it for $ 15 . find the gain percent ? | c . p . = $ 10 s . p . = $ 15 gain = $ 5 gain % = 5 / 10 * 100 = 50 % answer is b | a ) 25 % , b ) 50 % , c ) 20 % , d ) 15 % , e ) 30 % | b | subtract(divide(15, divide(10, const_100)), const_100) | divide(n0,const_100)|divide(n1,#0)|subtract(#1,const_100) | gain |
there are 10 fictions and 6 non - fictions . how many cases are there such that 2 fictions and 2 non - fictions are selected from them ? | "number of ways of selecting 2 fiction books = 10 c 2 number of ways of selecting 2 non fiction books = 6 c 2 10 c 2 * 6 c 2 = 45 * 15 = 675 answer : c" | a ) 90 , b ) 120 , c ) 675 , d ) 180 , e ) 200 | c | divide(multiply(multiply(10, const_4), multiply(6, 10)), power(factorial(2), 2)) | factorial(n2)|multiply(n0,const_4)|multiply(n0,n1)|multiply(#1,#2)|power(#0,n2)|divide(#3,#4)| | general |
if a * b * c = ( √ ( a + 2 ) ( b + 3 ) ) / ( c + 1 ) , find the value of 6 * 15 * 5 . | 6 * 15 * 5 = ( √ ( 6 + 2 ) ( 15 + 3 ) ) / ( 5 + 1 ) = ( √ 8 * 18 ) / 6 = ( √ 144 ) / 6 = 12 / 6 = 2 answer is a | a ) 2 , b ) 5 , c ) 11 , d ) 3 , e ) 4 | a | divide(sqrt(multiply(add(6, 2), add(15, 3))), add(5, 1)) | add(n0,n3)|add(n1,n4)|add(n2,n5)|multiply(#0,#1)|sqrt(#3)|divide(#4,#2) | general |
the speed of a car is 120 km in the first hour and 60 km in the second hour . what is the average speed of the car ? | "s = ( 120 + 60 ) / 2 = 90 kmph c" | a ) 89 kmph , b ) 92 kmph , c ) 90 kmph , d ) 65 kmph , e ) 77 kmph | c | divide(add(120, 60), const_2) | add(n0,n1)|divide(#0,const_2)| | physics |
if a ( a + 6 ) = 7 and b ( b + 6 ) = 7 , where a ≠ b , then a + b = | "a ( a + 6 ) = 7 = > we have a = 1 or - 7 also b ( b + 6 ) = 7 = > b = 1 or - 7 given a ≠ b 1 ) when a = 1 , b = - 7 and a + b = - 6 1 ) when a = - 7 , b = 1 and a + b = - 6 answer choice b" | a ) − 48 , b ) − 6 , c ) 2 , d ) 46 , e ) 48 | b | add(divide(7, const_10), divide(7, divide(7, const_10))) | divide(n1,const_10)|divide(n1,#0)|add(#0,#1)| | general |
fred and sam are standing 50 miles apart and they start walking in a straight line toward each other at the same time . if fred walks at a constant speed of 5 miles per hour and sam walks at a constant speed of 5 miles per hour , how many miles has sam walked when they meet ? | "relative distance = 50 miles relative speed = 5 + 5 = 10 miles per hour time taken = 50 / 10 = 5 hours distance travelled by sam = 5 * 5 = 25 miles = c" | a ) 5 , b ) 9 , c ) 25 , d ) 30 , e ) 45 | c | multiply(5, divide(50, add(5, 5))) | add(n1,n2)|divide(n0,#0)|multiply(n2,#1)| | physics |
when a number is divided by 6 & then multiply by 12 the answer is 9 what is the no . ? | "if $ x $ is the number , x / 6 * 12 = 9 = > 2 x = 9 = > x = 4.5 a" | a ) 4.5 , b ) 5 , c ) 5.5 , d ) 5.8 , e ) 6 | a | multiply(divide(9, 12), 6) | divide(n2,n1)|multiply(n0,#0)| | general |
31 of the scientists that attended a certain workshop were wolf prize laureates , and 12 of these 31 were also nobel prize laureates . of the scientists that attended that workshop and had not received the wolf prize , the number of scientists that had received the nobel prize was 3 greater than the number of scientist... | "lets solve by creating equation . . w = 31 . . total = 50 . . not w = 50 - 31 = 19 . . now let people who were neither be x , so out of 19 who won nobel = x + 3 . . so x + x + 3 = 19 or x = 8 . . so who won nobel but not wolf = x + 3 = 11 . . but people who won both w and n = 12 . . so total who won n = 11 + 12 = 23 .... | a ) a ) 11 , b ) b ) 23 , c ) c ) 24 , d ) d ) 29 , e ) d ) 36 | b | add(add(3, divide(subtract(subtract(50, 31), 3), const_2)), 12) | subtract(n4,n0)|subtract(#0,n3)|divide(#1,const_2)|add(n3,#2)|add(n1,#3)| | physics |
of the female students at barkely university , 5 / 6 are on the honor roll . of the male students , 2 / 3 are on the honor roll . if 3 / 5 of the students are female , what fraction of all the students are on the honor roll ? | "for total students take lcm of factions = 6 * 5 = 30 let female students = x , so male students = 30 - x now , female stu . on honor roll = 5 / 6 ( x ) and male stu . on honor roll = 2 / 3 ( 30 - x ) as given total female stu . = 3 / 5 ( 30 ) = 18 female stu . = ( 5 / 6 ) * 18 = 15 and male stu . = ( 2 / 3 ) * 12 = 8 ... | a ) 5 / 9 , b ) 3 / 15 , c ) 23 / 30 , d ) 43 / 60 , e ) 53 / 90 | c | divide(add(multiply(multiply(5, 2), 2), 2), multiply(2, multiply(3, 5))) | multiply(n0,n2)|multiply(n0,n3)|multiply(n2,#0)|multiply(n2,#1)|add(n2,#2)|divide(#4,#3)| | general |
the average of 10 numbers is 23 . if each number is increased by 3 , what will the new average be ? | "sum of the 10 numbers = 230 if each number is increased by 3 , the total increase = 3 * 10 = 40 the new sum = 230 + 30 = 260 the new average = 260 / 10 = 26 . answer : b" | a ) 36 , b ) 26 , c ) 72 , d ) 29 , e ) 22 | b | multiply(23, 3) | multiply(n1,n2)| | general |
in an election , candidate a got 60 % of the total valid votes . if 15 % of the total votes were declared invalid and the total numbers of votes is 560000 , find the number of valid vote polled in favor of candidate . | "total number of invalid votes = 15 % of 560000 = 15 / 100 × 560000 = 8400000 / 100 = 84000 total number of valid votes 560000 – 84000 = 476000 percentage of votes polled in favour of candidate a = 60 % therefore , the number of valid votes polled in favour of candidate a = 60 % of 476000 = 60 / 100 × 476000 = 28560000... | a ) 285600 , b ) 340000 , c ) 347000 , d ) 356000 , e ) 357000 | a | multiply(multiply(560000, subtract(const_1, divide(15, const_100))), divide(60, const_100)) | divide(n0,const_100)|divide(n1,const_100)|subtract(const_1,#1)|multiply(n2,#2)|multiply(#0,#3)| | gain |
the duplicate ratio of 2 : 6 is ? | "2 ^ 2 : 6 ^ 2 = 4 : 36 = 1 : 9 answer : c" | a ) 1 : 2 , b ) 1 : 4 , c ) 1 : 9 , d ) 1 : 18 , e ) 1 : 13 | c | divide(power(2, const_2), power(6, const_2)) | power(n0,const_2)|power(n1,const_2)|divide(#0,#1)| | other |
the least number , which when divided by 12 , 15 , 20 and 54 leaves in each case a remainder of 5 is : | required number = ( l . c . m . of 12 , 15 , 20 , 54 ) + 5 = 540 + 5 = 545 . answer : a | a ) 545 , b ) 488 , c ) 542 , d ) 548 , e ) 560 | a | multiply(54, const_10) | multiply(n3,const_10) | general |
excluding stoppages , the speed of a bus is 50 kmph and including stoppages , it is 43 kmph . for how many minutes does the bus stop per hour ? | "d 8.4 min due to stoppages , it covers 7 km less . time taken to cover 9 km = ( 7 / 50 x 60 ) min = 8.4 min" | a ) 70 min , b ) 16 min , c ) 20 min , d ) 8.4 min , e ) 40 min | d | multiply(const_60, divide(subtract(50, 43), 50)) | subtract(n0,n1)|divide(#0,n0)|multiply(#1,const_60)| | physics |
a & b started a partnership business . a ' s investment was thrice the investment of b and the period of his investment was two times the period of investments of b . if b received rs 7000 as profit , what is their total profit ? | "explanation : suppose b ' s investment = x . then a ' s investment = 3 x suppose bs period of investment = y , then a ' s period of investment = 2 y a : b = 3 x * 2 y : xy = 6 : 1 total profit * 1 / 7 = 7000 = > total profit = 7000 * 7 = 49000 . answer : option e" | a ) 28000 , b ) 30000 , c ) 32000 , d ) 34000 , e ) 49000 | e | divide(7000, divide(multiply(const_1, const_1), add(multiply(const_3, const_2), multiply(const_1, const_1)))) | multiply(const_1,const_1)|multiply(const_2,const_3)|add(#1,#0)|divide(#0,#2)|divide(n0,#3)| | general |
a man has rs . 10350 in the form of rs . 50 notes and rs . 500 notes . the total number of notes are 36 . find the number of notes of rs . 50 denomination . | "total money = rs . 10350 . let 50 rupees note was x . then 500 rupees note = 36 - x now , 50 * x + 500 * ( 36 - x ) = 10350 50 x + 18000 - 500 x = 10350 - 450 x = - 7650 x = 17 . no . of 50 rupees note = 17 . answer : option c" | a ) 15 , b ) 21 , c ) 17 , d ) 19 , e ) 21 | c | divide(subtract(multiply(500, 36), 10350), subtract(500, 50)) | multiply(n2,n3)|subtract(n2,n1)|subtract(#0,n0)|divide(#2,#1)| | general |
a part of certain sum of money is invested at 8 % per annum and the rest at 12 % per annum , if the interest earned in each case for the same period is equal , then ratio of the sums invested is ? | "12 : 8 = 3 : 2 answer : a" | a ) 3 : 2 , b ) 4 : 8 , c ) 4 : 3 , d ) 4 : 0 , e ) 4 : 9 | a | multiply(divide(12, const_100), 8) | divide(n1,const_100)|multiply(n0,#0)| | gain |
the ratio of the arithmetic mean of two numbers to one of the numbers is 3 : 5 . what is the ratio of the smaller number to the larger ? | "let the numbers be a , b acc . to problem , ( a + b ) / 2 ∗ a ( a + b ) / 2 ∗ a = 3 / 53 / 5 1 / 21 / 2 + b / 2 ∗ ab / 2 ∗ a = 3 / 53 / 5 b / ab / a = 1 / 51 / 5 ans . a" | a ) 1 : 5 , b ) 1 : 4 , c ) 1 : 3 , d ) 1 : 2 , e ) 2 : 3 | a | multiply(subtract(divide(3, 5), divide(const_1, const_2)), const_2) | divide(n0,n1)|divide(const_1,const_2)|subtract(#0,#1)|multiply(#2,const_2)| | other |
what ratio must a shopkeepermix peas and soybean of rs . 16 and rs . 25 / kg , as to obtain a mixture of rs . 20 ? | correct option : ( c ) use rule of alligation , to determine the ratio the required ratio of soybean and peas = 5 : 4 | a ) 10 : 7 , b ) 9 : 8 , c ) 5 : 4 , d ) 13 : 11 , e ) 14 : 8 | c | divide(subtract(25, 20), subtract(20, 16)) | subtract(n1,n2)|subtract(n2,n0)|divide(#0,#1) | general |
a sum of money deposited at c . i . amounts to rs . 3000 in 3 years and to rs . 3600 in 4 years . find the rate percent ? | "3000 - - - 600 100 - - - ? = > 20 % answer : e" | a ) 25 % , b ) 5 % , c ) 10 % , d ) 15 % , e ) 20 % | e | multiply(divide(subtract(3600, 3000), 3000), const_100) | subtract(n2,n0)|divide(#0,n0)|multiply(#1,const_100)| | gain |
the product z of two prime numbers is between 15 and 36 . if one of the prime numbers is greater than 2 but less than 6 and the other prime number is greater than 8 but less than 24 , then what is z ? | the smallest possible product is 33 which is 3 * 11 . all other products are too big . the answer is b . | a ) 35 , b ) 33 , c ) 28 , d ) 21 , e ) 15 | b | multiply(subtract(15, const_4), const_3) | subtract(n0,const_4)|multiply(#0,const_3) | general |
a sum amounts to rs . 3087 in 2 years at the rate of 5 % p . a . if interest was compounded yearly then what was the principal ? | "ci = 3087 , r = 5 , n = 2 ci = p [ 1 + r / 100 ] ^ 2 = p [ 1 + 5 / 100 ] ^ 2 3087 = p [ 21 / 20 ] ^ 2 3087 [ 20 / 21 ] ^ 2 2800 answer : e" | a ) s . 4000 , b ) s . 5000 , c ) s . 4500 , d ) s . 4800 , e ) s . 2800 | e | divide(3087, power(add(divide(5, const_100), const_1), 2)) | divide(n2,const_100)|add(#0,const_1)|power(#1,n1)|divide(n0,#2)| | gain |
in a certain company , the ratio of the number of managers to the number of non - managers in any department must always be greater than 7 : 37 . in the company , what is the maximum number of non - managers in a department that has 11 managers ? | "11 / 7 * 37 = 58.1 the answer is c ." | a ) 56 , b ) 57 , c ) 58 , d ) 59 , e ) 60 | c | floor(divide(multiply(37, 11), 7)) | multiply(n1,n2)|divide(#0,n0)|floor(#1)| | general |
rebecca ' s yearly income is $ 15000 and jimmy ' s annual income is $ 18000 . by how much must rebecca ' s yearly income increase so that it constitutes 55 % of rebecca and jimmy ' s combined income ? | total rebecca = x + 15000 ; total = x + 15000 + 18000 ; x + 15000 / x + 33000 = 55 / 100 therefore x = 7000 a | a ) 7000 , b ) 8000 , c ) 9000 , d ) 10000 , e ) 11000 | a | add(multiply(subtract(18000, 15000), const_2), const_1000) | subtract(n1,n0)|multiply(#0,const_2)|add(#1,const_1000) | general |
a car traveling at a certain constant speed takes 5 seconds longer to travel 1 km than it would take to travel 1 km at 75 km / hour . at what speed , in km / hr , is the car traveling ? | "time to cover 1 kilometer at 80 kilometers per hour is 1 / 75 hours = 3,600 / 75 seconds = 48 seconds ; time to cover 1 kilometer at regular speed is 48 + 5 = 53 seconds = 53 / 3,600 hours = 1 / 70 hours ; so , we get that to cover 1 kilometer 1 / 70 hours is needed - - > regular speed 70 kilometers per hour ( rate is... | a ) 70 , b ) 72 , c ) 74 , d ) 75 , e ) 78 | a | divide(1, divide(add(multiply(const_3600, divide(1, 75)), 5), const_3600)) | divide(n1,n3)|multiply(#0,const_3600)|add(n0,#1)|divide(#2,const_3600)|divide(n1,#3)| | physics |
by investing in 1623 % stock at 64 , one earns rs . 1900 . the investment made is | explanation : market value = rs . 64 face value is not given and hence take it as rs . 100 16 2 / 3 % of the face value = 50 / 3 ie , to earn 50 / 3 , investment = rs . 64 hence , to earn rs . 1500 , investment needed = 64 × 3 × 1500 / 50 = 5760 answer : option d | a ) s . 9600 , b ) s . 7500 , c ) s . 5640 , d ) s . 5760 , e ) - 7296 | d | multiply(divide(multiply(multiply(multiply(const_4, const_2), multiply(const_4, const_2)), const_3), divide(const_100, const_2)), subtract(1900, multiply(const_4, const_100))) | divide(const_100,const_2)|multiply(const_2,const_4)|multiply(const_100,const_4)|multiply(#1,#1)|subtract(n2,#2)|multiply(#3,const_3)|divide(#5,#0)|multiply(#6,#4) | gain |
the average of 10 numbers is 23 . if each number is increased by 5 , what will the new average be ? | "sum of the 10 numbers = 230 if each number is increased by 5 , the total increase = 5 * 10 = 50 the new sum = 230 + 50 = 280 the new average = 280 / 10 = 28 . answer : b" | a ) 36 , b ) 28 , c ) 72 , d ) 29 , e ) 22 | b | multiply(23, 5) | multiply(n1,n2)| | general |
how many days per calendar year is the month a factor of the day | solution : 124 all 31 days in january , because 1 is a factor of all integers 14 of the 28 days in february ( because every other day is divisible by 2 ) 10 of the 31 days in march ( the multiples of 3 between 3 and 30 ) 7 days in april ( the multiples of 4 between 4 and 28 ) 6 days in may ( the multiples of 5 between ... | a ) 0 , b ) 1 , c ) 59 , d ) 90 , e ) 124 | e | add(floor(divide(add(multiply(const_3, const_10), const_1), multiply(const_3, const_4))), add(floor(divide(multiply(const_3, const_10), add(const_10, const_1))), add(floor(divide(add(multiply(const_3, const_10), const_1), const_10)), add(floor(divide(multiply(const_3, const_10), multiply(const_3, const_3))), add(floor(... | add(const_1,const_4)|add(const_1,const_10)|multiply(const_10,const_3)|multiply(const_2,const_4)|multiply(const_3,const_3)|multiply(const_3,const_4)|add(#2,const_1)|add(#0,const_1)|divide(#2,const_4)|divide(#2,#4)|divide(#2,#1)|subtract(#2,const_2)|add(#6,#11)|add(#7,const_1)|divide(#6,const_3)|divide(#2,#7)|divide(#6,#... | other |
a contractor undertakes to do a job within 100 days and hires 10 people to do it . after 20 days , he realizes that one fourth of the work is done so he fires 2 people . in how many more days m will the work get over ? | "we can also use the concept of man - days here 100 days - - > 10 men so the job includes 100 * 10 = 1000 man - days after 20 days 1 / 4 of job is completed so 1 / 4 x 1000 man - days = 250 man - days job is done now the balance job = 1000 - 250 = 750 man - days worth of job since 2 men are fired so b / l men = 8 there... | a ) 60 , b ) 70 , c ) m = 75 , d ) 80 , e ) 100 | c | divide(multiply(divide(multiply(10, 20), const_0_25), subtract(const_1, const_0_25)), subtract(10, 2)) | multiply(n1,n2)|subtract(const_1,const_0_25)|subtract(n1,n3)|divide(#0,const_0_25)|multiply(#3,#1)|divide(#4,#2)| | physics |
a train 100 m long crosses a platform 100 m long in 12 sec ; find the speed of the train ? | "d = 100 + 100 = 200 t = 12 s = 200 / 12 * 18 / 5 = 60 kmph answer : a" | a ) 60 kmph , b ) 65 kmph , c ) 54 kmph , d ) 16 kmph , e ) 18 kmph | a | subtract(multiply(12, multiply(100, const_0_2778)), 100) | multiply(n1,const_0_2778)|multiply(n2,#0)|subtract(#1,n0)| | physics |
if circles x and y have the same area and circle x has a circumference of 14 π , half of the radius of circle y is : | x be radius of circle x y be radius of circle y given : pi * x ^ 2 = pi * y ^ 2 also , 2 * pi * x = 14 * pi x = 7 thus y = 7 y / 2 = 3.5 ans : d | ['a ) 16', 'b ) 8', 'c ) 4', 'd ) 3.5', 'e ) 1'] | d | divide(divide(14, const_2), const_2) | divide(n0,const_2)|divide(#0,const_2) | geometry |
the sum of three consecutive integers is 93 . what are the integers ? | "first x make the first number x second x + 1 to get the next numberwe go up one or + 1 third x + 2 add another 1 ( 2 total ) to get the third f + s + t = 93 first ( f ) plus second ( s ) plusthird ( t ) equals 93 ( x ) + ( x + 1 ) + ( x + 2 ) = 93 replace f with x , s with x + 1 , and t with x + 2 x + x + 1 + x + 2 = ... | a ) 20 , 21,22 , b ) 10 , 11,12 , c ) 30 , 31,32 , d ) 40 , 41,42 , e ) 30 , 31,32 | e | add(add(power(add(add(divide(subtract(subtract(93, const_10), const_2), const_4), const_2), const_2), const_2), power(add(add(add(divide(subtract(subtract(93, const_10), const_2), const_4), const_2), const_2), const_2), const_2)), add(power(divide(subtract(subtract(93, const_10), const_2), const_4), const_2), power(add... | subtract(n0,const_10)|subtract(#0,const_2)|divide(#1,const_4)|add(#2,const_2)|power(#2,const_2)|add(#3,const_2)|power(#3,const_2)|add(#5,const_2)|add(#4,#6)|power(#5,const_2)|power(#7,const_2)|add(#9,#10)|add(#11,#8)| | physics |
a store owner estimates that the average price of type a products will increase by 25 % next year and that the price of type b products will increase by 20 % next year . this year , the total amount paid for type a products was $ 4500 and the total price paid for type b products was $ 8300 . according to the store owne... | "cost of type a products next year = 1.25 * 4500 = 5625 cost of type b products next year = 1.2 * 8300 = 9960 total 5625 + 9960 = 15585 option b" | a ) $ 14,755 , b ) $ 15,585 , c ) $ 16,000 , d ) $ 16,225 , e ) $ 17,155 | b | multiply(divide(const_3, const_4), const_1000) | divide(const_3,const_4)|multiply(#0,const_1000)| | general |
when n is divided by 20 , the remainder is 6 . what is the remainder when n + 16 is divided by 5 ? | "assume n = 14 remainder ( n / 20 ) = 6 n + 16 = 36 remainder ( 36 / 5 ) = 1 option a" | a ) 1 , b ) 3 , c ) 4 , d ) 5 , e ) 6 | a | subtract(6, reminder(16, 5)) | reminder(n2,n3)|subtract(n1,#0)| | general |
on a sum of money , the s . i . for 2 years is $ 660 , while the c . i . is $ 693 , the rate of interest being the same in both the cases . the rate of interest is ? | "difference in c . i . and s . i for 2 years = $ 693 - $ 660 = $ 33 s . i for one year = $ 330 s . i . on $ 330 for 1 year = $ 33 rate = ( 100 * 33 ) / ( 330 ) = 10 % the answer is a ." | a ) 10 % , b ) 32 % , c ) 72 % , d ) 14 % , e ) 82 % | a | divide(multiply(const_100, subtract(693, 660)), divide(660, 2)) | divide(n1,n0)|subtract(n2,n1)|multiply(#1,const_100)|divide(#2,#0)| | gain |
in a group of ducks and cows , the total number of legs are 32 more than twice the number of heads . find the total number of cows . | "let the number of ducks be d and number of cows be c then , total number of legs = 2 d + 4 c = 2 ( d + 2 c ) total number of heads = c + d given that total number of legs are 32 more than twice the number of heads = > 2 ( d + 2 c ) = 32 + 2 ( c + d ) = > d + 2 c = 16 + c + d = > 2 c = 16 + c = > c = 16 i . e . , total... | a ) 12 , b ) 14 , c ) 16 , d ) 18 , e ) 20 | c | divide(32, const_2) | divide(n0,const_2)| | general |
a is half good a work man as b and together they finish a job in 12 days . in how many days working alone b finish the job ? | "c 18 wc = 1 : 2 2 x + x = 1 / 12 = > x = 1 / 36 2 x = 1 / 18 = > 18 days" | a ) 23 , b ) 22 , c ) 18 , d ) 36 , e ) 48 | c | multiply(12, divide(const_3, const_2)) | divide(const_3,const_2)|multiply(n0,#0)| | physics |
a cyclist traveled for two days . on the second day the cyclist traveled 4 hours longer and at an average speed 10 mile per hour slower than she traveled on the first day . if during the two days she traveled a total of 280 miles and spent a total of 10 hours traveling , what was her average speed on the second day ? | "solution : d = 280 mi t = 12 hrs day 1 time = t 1 day 2 time = t 2 t 2 - t 1 = 4 hrs - - - - - ( i ) t 1 + t 2 = 12 hrs - - - - - ( ii ) adding i and ii , t 2 = 8 hrs and t 1 = 4 hrs day 1 rate = r 1 day 2 rate = r 2 r 1 - r 2 = 10 mph i . e . r 1 = 10 + r 2 280 = 8 r 2 + 4 r 1 i . e . 280 = 8 r 2 + 4 ( 10 + r 2 ) i .... | a ) 5 mph , b ) 10 mph , c ) 20 mph , d ) 30 mph , e ) 40 mph | e | divide(subtract(280, multiply(divide(subtract(10, 4), const_2), 10)), add(divide(subtract(10, 4), const_2), add(divide(subtract(10, 4), const_2), 4))) | subtract(n3,n0)|divide(#0,const_2)|add(n0,#1)|multiply(n1,#1)|add(#2,#1)|subtract(n2,#3)|divide(#5,#4)| | physics |
a pair of articles was bought for $ 50 at a discount of 40 % . what must be the marked price of each of the article ? | s . p . of each of the article = 50 / 2 = $ 25 let m . p = $ x 60 % of x = 25 x = 25 * . 6 = $ 15 answer is b | a ) $ 25 , b ) $ 15 , c ) $ 29.65 , d ) $ 35.95 , e ) $ 45.62 | b | divide(multiply(subtract(const_100, 40), divide(50, const_2)), const_100) | divide(n0,const_2)|subtract(const_100,n1)|multiply(#0,#1)|divide(#2,const_100) | gain |
find the area of trapezium whose parallel sides are 28 cm and 20 cm long , and the distance between them is 21 cm ? | "area of a trapezium = 1 / 2 ( sum of parallel sides ) * ( perpendicular distance between them ) = 1 / 2 ( 28 + 20 ) * ( 21 ) = 504 cm 2 answer : c" | a ) 288 cm 2 , b ) 277 cm 2 , c ) 504 cm 2 , d ) 510 cm 2 , e ) 287 cm 2 | c | quadrilateral_area(21, 20, 28) | quadrilateral_area(n2,n1,n0)| | physics |
by selling 16 pencils for a rupee a man loses 5 % . how many for a rupee should he sell in order to gain 15 % ? | "85 % - - - 16 105 % - - - ? 85 / 105 * 16 = 8 answer : a" | a ) 12.95 , b ) 19 , c ) 17.56 , d ) 16.56 , e ) 15 | a | multiply(divide(const_1, multiply(add(const_100, 15), divide(const_1, subtract(const_100, 5)))), 16) | add(n2,const_100)|subtract(const_100,n1)|divide(const_1,#1)|multiply(#0,#2)|divide(const_1,#3)|multiply(n0,#4)| | gain |
if 7 : 13 : : 301 : x then the value of ‘ x ’ is : | "given the question ; 7 : 13 : : 301 : x 7 / 13 = 301 / x 7 x = 301 * 13 x = 301 * 13 / 7 x = 559 answer : c" | a ) 493 , b ) 537 , c ) 559 , d ) 587 , e ) 567 | c | divide(add(multiply(13, 7), 13), 301) | multiply(n0,n1)|add(n1,#0)|divide(#1,n2)| | general |
two goods trains each 500 m long are running in opposite directions on parallel tracks . their speeds are 45 km / hr and 45 km / hr respectively . find the time taken by the slower train to pass the driver of the faster one ? | "relative speed = 45 + 45 = 90 km / hr . 90 * 5 / 18 = 25 m / sec . distance covered = 500 + 500 = 1000 m . required time = 1000 / 25 = 40 sec . answer : a" | a ) 40 , b ) 93 , c ) 26 , d ) 23 , e ) 12 | a | add(45, 45) | add(n1,n2)| | physics |
the difference between c . i . and s . i . on an amount of $ 10,000 for 2 years is $ 49 . what is the rate of interest per annum ? | $ 49 is the interest on the first year of interest . let x be the interest rate . the interest after the first year is 10000 * x . the interest on the first year ' s interest is 10000 * x * x 10000 * x ^ 2 = 49 x = 0.07 the answer is c . | a ) 5 % , b ) 6 % , c ) 7 % , d ) 8 % , e ) 9 % | c | sqrt(49) | sqrt(n2) | gain |
find the value of x . ( 47 % of 1442 - 36 % of 1412 ) + 66 = x ? | d ) 6 | a ) 1 , b ) 2 , c ) 3 , d ) 6 , e ) 5 | d | floor(add(reminder(add(subtract(multiply(divide(47, const_100), 1442), multiply(divide(36, const_100), 1412)), 66), const_10), const_1)) | divide(n0,const_100)|divide(n2,const_100)|multiply(n1,#0)|multiply(n3,#1)|subtract(#2,#3)|add(n4,#4)|reminder(#5,const_10)|add(#6,const_1)|floor(#7)| | general |
9823 + x = 13200 , then x is ? | "answer x = 13200 - 9823 = 3377 option : d" | a ) 3327 , b ) 3237 , c ) 3337 , d ) 3377 , e ) none of these | d | subtract(13200, 9823) | subtract(n1,n0)| | general |
a store ’ s selling price of $ 2500 for a certain printer would yield a profit of 40 percent of the store ’ s cost for the printer . what selling price would yield a profit of 50 percent of the printer ’ s cost ? | 1.4 x = 2500 x = 2500 / 1.4 so , 1.5 x = 2500 * 1.5 / 1.4 = 2478 answer : - d | a ) $ 2400 , b ) $ 2464 , c ) $ 2650 , d ) $ 2478 , e ) $ 2800 | d | multiply(add(const_1, divide(40, const_100)), original_price_before_gain(40, 2500)) | divide(n1,const_100)|original_price_before_gain(n1,n0)|add(#0,const_1)|multiply(#2,#1) | gain |
in a soccer bet , we play 4 teams . first team odd 1.28 , second 5.23 , third 3.25 , fourth 2.05 . we place the bet with 5.00 euros . how much money we expect to win ? | in the case we won the bet , we have : 1.28 * 5.23 * 3.25 * 2.0 * 5.00 = 219.50 we will win 223 so correct answer is b | a ) 219.5 , b ) 223.0 , c ) 235.01 , d ) 266.74 , e ) 669.0 | b | multiply(multiply(multiply(multiply(5, 2.05), 3.25), 5.23), 1.28) | multiply(n4,n5)|multiply(n3,#0)|multiply(n2,#1)|multiply(n1,#2) | general |
sum of two numbers prime to each other is 16 and their l . c . m . is 63 . what are the numbers ? | as two numbers are prime , only options satisfy all but option c will not make the product of numbers i . e 63 answer : e | a ) 9 and 10 , b ) 8 and 9 , c ) 11 and 5 , d ) 7 and 8 , e ) 7 and 9 | e | add(63, 16) | add(n0,n1) | physics |
find the area of trapezium whose parallel sides are 22 cm and 18 cm long , and the distance between them is 15 cm . | "area of a trapezium = 1 / 2 ( sum of parallel sides ) * ( perpendicular distance between them ) = 1 / 2 ( 22 + 18 ) * ( 15 ) = 300 cm 2 answer : c" | a ) 227 , b ) 299 , c ) 300 , d ) 161 , e ) 212 | c | quadrilateral_area(15, 18, 22) | quadrilateral_area(n2,n1,n0)| | physics |
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