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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mixture contains alcohol and water in the ratio 4 : 3 . if 4 liters of water is added to the mixture , the ratio becomes 4 : 5 . find the quantity of alcohol in the given mixture . | "let the quantity of alcohol and water be 4 x litres and 3 x litres respectively 16 x = 4 ( 3 x + 5 ) 4 x = 20 x = 5 quantity of alcohol = ( 4 x 5 ) litres = 20 litres . answer : a" | a ) 20 , b ) 99 , c ) 27 , d ) 22 , e ) 29 | a | multiply(divide(multiply(4, 5), subtract(multiply(4, 4), multiply(3, 4))), 4) | multiply(n0,n4)|multiply(n0,n2)|multiply(n0,n1)|subtract(#1,#2)|divide(#0,#3)|multiply(n0,#4)| | general |
the sum of the fourth and twelfth term of an arithmetic progression is 20 . what is the sum of the first 12 terms of the arithmetic progression ? | "n th term of a . p . is given by a + ( n - 1 ) d 4 th term = a + 3 d 12 th term = a + 11 d given a + 3 d + a + 11 d = 20 - - > 2 a + 14 d = 20 - - > a + 7 d = 10 sum of n term of a . p = n / 2 [ 2 a + ( n - 1 ) d ] subsitiuing n = 12 . . . we get 12 / 2 [ 2 a + 14 d ] = 15 [ a + 7 d ] = 12 * 10 = 120 . . . answer is b... | a ) 300 , b ) 120 , c ) 150 , d ) 170 , e ) 270 | b | divide(multiply(20, 12), const_2) | multiply(n0,n1)|divide(#0,const_2)| | general |
sonika bought a v . c . r . at the list price of 18,700 . if the rate of sales tax was 8 % , find the amount she had to pay for purchasing the v . c . r . | "sol . list price of v . c . r . = 18,700 rate of sales tax = 8 % ∴ sales tax = 8 % of 18,700 = 8 ⁄ 100 × 18700 = 1496 so , total amount which sonika had to pay for purchasing the v . c . r . = 18,700 + 1496 = 20,196 . answer a" | a ) 20,196 , b ) 19,780 , c ) 19,680 , d ) 19,380 , e ) none of these | a | floor(divide(divide(multiply(add(multiply(multiply(add(multiply(add(const_1, const_4), const_2), 8), const_100), multiply(add(const_1, const_4), const_2)), multiply(add(const_1, const_4), const_100)), add(const_100, 8)), const_100), multiply(multiply(add(const_1, const_4), const_100), const_2))) | add(const_1,const_4)|add(n1,const_100)|multiply(#0,const_2)|multiply(#0,const_100)|add(n1,#2)|multiply(#3,const_2)|multiply(#4,const_100)|multiply(#6,#2)|add(#7,#3)|multiply(#8,#1)|divide(#9,const_100)|divide(#10,#5)|floor(#11)| | gain |
if the speed of a man is 54 km per hour , then what is the distance traveled by him in 25 seconds ? | the distance traveled in 25 sec = 54 * ( 5 / 18 ) * 25 = 375 m answer : c | a ) 275 m , b ) 360 m , c ) 375 m , d ) 420 m , e ) 440 m | c | multiply(multiply(54, const_0_2778), 25) | multiply(n0,const_0_2778)|multiply(n1,#0) | physics |
soja finished 2 / 3 of a book . she calculated that she finished 100 more pages than she has yet to read . how long is her book ? | let x be the total number of pages in the book , then she finished 2 / 3 * x pages . then she has x − 2 / 3 * x = 1 / 3 * x pages left . 2 / 3 * x − 1 / 3 * x = 100 1 / 3 * x = 100 x = 300 so the book is 300 pages long . answer is a . | a ) a ) 300 , b ) b ) 150 , c ) c ) 130 , d ) d ) 90 , e ) e ) 210 | a | multiply(100, 3) | multiply(n1,n2) | general |
a monkey ascends a greased pole 18 metres high . he ascends 2 metres in first minute and slips down 1 metre in the alternate minute . in which minute , he reaches the top ? | "in 2 minutes , he ascends = 1 metre â ˆ ´ 16 metres , he ascends in 32 minutes . â ˆ ´ he reaches the top in 33 rd minute . answer c" | a ) 21 st , b ) 22 nd , c ) 33 rd , d ) 24 th , e ) none of these | c | subtract(multiply(2, 18), 1) | multiply(n0,n1)|subtract(#0,n2)| | physics |
the average score of a cricketer in 10 matches is 60 and in other 15 matches is 70 . then find the average score in all the 25 matches ? | "average in 25 matches = ( 10 * 60 + 15 * 70 ) / 10 + 15 = 66 answer is d" | a ) 50 , b ) 48 , c ) 65 , d ) 66 , e ) 72 | d | divide(add(multiply(10, 60), multiply(15, 70)), 25) | multiply(n0,n1)|multiply(n2,n3)|add(#0,#1)|divide(#2,n4)| | general |
one - third less than 30 % of 180 is equal to : | "lots of ways to tackle this . 30 % of 180 = 54 1 / 3 of 45 = 18 so , 1 / 3 less than 54 is equal to 54 - 18 = 36 answer : c" | a ) 15 , b ) 30 , c ) 36 , d ) 40 , e ) 45 | c | subtract(multiply(divide(30, const_100), 180), multiply(divide(const_1, const_3), multiply(divide(30, const_100), 180))) | divide(n0,const_100)|divide(const_1,const_3)|multiply(n1,#0)|multiply(#1,#2)|subtract(#2,#3)| | general |
the average of marks obtained by 120 boys was 35 . if the average of marks of passed boys was 39 and that of failed boys was 15 , the number of boys who passed the examination is ? | "let the number of boys who passed = x . then , 39 x x + 15 x ( 120 - x ) = 120 x 35 24 x = 4200 - 1800 = > x = 2400 / 24 x = 100 . hence , the number of boys passed = 100 . answer : a" | a ) 100 , b ) 110 , c ) 120 , d ) 130 , e ) 140 | a | divide(subtract(multiply(35, 120), multiply(120, 15)), subtract(39, 15)) | multiply(n0,n1)|multiply(n0,n3)|subtract(n2,n3)|subtract(#0,#1)|divide(#3,#2)| | general |
a whale goes on a feeding frenzy that lasts for 5 hours . for the first hour he catches and eats x kilos of plankton . in every hour after the first , it consumes 3 kilos of plankton more than it consumed in the previous hour . if by the end of the frenzy the whale will have consumed a whopping accumulated total 450 ki... | "if you list the amount eaten each hour , you ' ll get an equally spaced list , increasing by 3 each hour . in any equally spaced list , the median equals the mean . here , the mean is 450 / 5 = 90 , so the median is also 90 , and that is the amount eaten in the 2 nd hour . we need to add 3 to find the total eaten in t... | a ) 38 , b ) 47 , c ) 50 , d ) 53 , e ) 93 | e | add(divide(450, 5), 3) | divide(n2,n0)|add(n1,#0)| | general |
a sum of money is to be distributed among a , b , c , d in the proportion of 5 : 2 : 4 : 3 . if c gets $ 500 more than d , what is b ' s share ? | "let the shares of a , b , c and d be 5 x , 2 x , 4 x and 3 x respectively . then , 4 x - 3 x = 500 x = $ 500 b ' s share = 2 x = 2 * $ 500 = $ 1000 the answer is b ." | a ) $ 500 , b ) $ 1000 , c ) $ 1500 , d ) $ 2000 , e ) $ 2500 | b | multiply(multiply(subtract(4, 3), 500), 3) | subtract(n2,n3)|multiply(n4,#0)|multiply(n3,#1)| | general |
if 5 spiders make 5 webs in 5 days , then 1 spider will make 1 web in how many days ? | "let the required number days be x . less spiders , more days ( indirect proportion ) less webs , less days ( direct proportion ) spiders 1 : 5 webs 5 : 1 1 x 5 x x = 5 x 1 x 5 = > x = 5 answer is c" | a ) 7 , b ) 6 , c ) 5 , d ) 4 , e ) 3 | c | multiply(1, 5) | multiply(n0,n3)| | physics |
10 camels cost as much as 24 horses , 16 horses cost as much as 4 oxen and 6 oxen as much as 4 elephants . if the cost of 10 elephants is rs . 150000 , find the cost of a camel ? | "cost of the camel = p 10 camels = 24 horses 16 horses = 4 oxen 6 oxen = 4 elephants 10 elephants = rs . 150000 p = rs . [ ( 24 * 4 * 4 * 150000 ) / ( 10 * 16 * 6 * 10 ) ] p = rs . ( 57600000 / 9600 ) = > p = rs . 6000 answer : d" | a ) s . 9800 , b ) s . 3800 , c ) s . 9800 , d ) s . 6000 , e ) s . 6880 | d | divide(multiply(multiply(multiply(24, 4), 4), 150000), multiply(multiply(multiply(10, 16), 6), 10)) | multiply(n1,n3)|multiply(n0,n2)|multiply(n3,#0)|multiply(n4,#1)|multiply(n7,#2)|multiply(n0,#3)|divide(#4,#5)| | general |
a candidate got 31 % of the votes polled and he lost to his rival by 2451 votes . how many votes were cast ? | "let x be the total number of votes . 0.31 x + 2451 = 0.69 x 0.38 x = 2451 x = 2451 / 0.38 = 6450 the answer is d ." | a ) 6150 , b ) 6250 , c ) 6350 , d ) 6450 , e ) 6550 | d | divide(2451, subtract(subtract(const_1, divide(31, const_100)), divide(31, const_100))) | divide(n0,const_100)|subtract(const_1,#0)|subtract(#1,#0)|divide(n1,#2)| | gain |
find the simple interest on $ 2000 for 3 years at 10 % per annum ? | "si = ptr / 100 = 2000 * 3 * 10 / 100 = $ 600 answer is d" | a ) $ 250 , b ) $ 300 , c ) $ 500 , d ) $ 600 , e ) $ 1000 | d | multiply(2000, divide(3, const_100)) | divide(n1,const_100)|multiply(n0,#0)| | gain |
an inspector rejects 10 % of the meters as defective . how many will he examine to reject 10 ? | "then , 10 % of x = 10 ( 10 / 100 ) x = 10 x = ( 10 * 100 * ) / 10 = 100 answer is a" | a ) 100 , b ) 120 , c ) 250 , d ) 200 , e ) 160 | a | divide(multiply(10, const_100), 10) | multiply(n1,const_100)|divide(#0,n0)| | gain |
the difference between the place value and the face value of 4 in the numeral 854973 is | "( place value of 4 ) - ( face value of 4 ) = ( 4000 - 4 ) = 3996 answer : option c" | a ) 973 , b ) 6973 , c ) 3996 , d ) 6084 , e ) none of these | c | subtract(multiply(const_10, 4), 4) | multiply(n0,const_10)|subtract(#0,n0)| | general |
the mean of 50 observations was 36 . it was found later that an observation 29 was wrongly taken as 23 . the corrected new mean is : | "explanation : correct sum = ( 36 * 50 + 29 - 23 ) = 1806 . correct mean = = 1806 / 50 = 36.12 answer : d ) 36.12" | a ) 36.1 , b ) 36.5 , c ) 36.22 , d ) 36.12 , e ) 36.18 | d | divide(add(multiply(36, 50), subtract(subtract(50, const_2), 23)), 50) | multiply(n0,n1)|subtract(n0,const_2)|subtract(#1,n3)|add(#0,#2)|divide(#3,n0)| | general |
shawn invested one half of his savings in a bond that paid simple interest for 2 years and received $ 550 as interest . he invested the remaining in a bond that paid compound interest , interest being compounded annually , for the same 2 years at the same rate of interest and received $ 605 as interest . what was the v... | "si for 1 st year = 550 / 2 = 275 275 + 275 + si on ( 275 ) = 605 = = > si on 275 = 55 = = > si on 275 = 275 * r * 1 / 100 = 55 r = 20 % now using si for 2 years on t / 2 of the money ( t / 2 ) * 20 * 2 / 100 = 550 t = 550 * 5 = 2750 answer : a" | a ) $ 2,750 , b ) $ 5,500 , c ) $ 11,000 , d ) $ 22,000 , e ) $ 44,000 | a | multiply(divide(multiply(divide(550, 2), divide(550, 2)), subtract(605, 550)), 2) | divide(n1,n0)|subtract(n3,n1)|multiply(#0,#0)|divide(#2,#1)|multiply(n0,#3)| | gain |
what percent is 350 gm of 1 kg ? | "1 kg = 1000 gm 350 / 1000 ã — 100 = 35000 / 1000 = 35 % answer is c" | a ) 25 % , b ) 40 % , c ) 35 % , d ) 8 % , e ) 12 % | c | multiply(divide(350, 1), const_100) | divide(n0,n1)|multiply(#0,const_100)| | gain |
7 people average age is 30 . youngest person age is 3 . find average of the people when youngest was born . | "average age of people = 30 so have total age = 210 before 7 years we have to deduct each person age by seven years 210 - 21 = 189 so average age would be 189 / 7 = 27 answer : e" | a ) 23 , b ) 24 , c ) 25 , d ) 26 , e ) 27 | e | divide(subtract(multiply(30, 7), multiply(3, 7)), 7) | multiply(n0,n1)|multiply(n0,n2)|subtract(#0,#1)|divide(#2,n0)| | general |
the average of first five multiples of 8 is : | "solution average = 8 ( 1 + 2 + 3 + 4 + 5 ) / 5 = 120 / 5 = 24 answer e" | a ) 3 , b ) 9 , c ) 12 , d ) 15 , e ) 24 | e | add(8, const_1) | add(n0,const_1)| | general |
what is value of ( ( 2 ^ 5 ) * ( 9 ^ 2 ) ) / ( ( 8 ^ 2 ) * ( 3 ^ 5 ) ) = ? | = > ( ( 2 ^ 5 ) * ( ( 3 ^ 2 ) ^ 2 ) / ( ( 2 ^ 3 ) ^ 2 ) * ( 3 ^ 5 ) ) = > ( ( 2 ^ 5 ) * ( 3 ^ 4 ) ) / ( ( 2 ^ 6 ) * ( 3 ^ 5 ) ) = > 1 / ( 2 * 3 ) = > 1 / 6 option a is answer | a ) 1 / 6 , b ) 1 / 3 , c ) 2 / 3 , d ) 1 / 4 , e ) 4 | a | divide(multiply(power(2, 5), power(9, 2)), multiply(power(8, 2), power(3, 5))) | power(n0,n1)|power(n2,n0)|power(n4,n0)|power(n6,n1)|multiply(#0,#1)|multiply(#2,#3)|divide(#4,#5) | general |
two circular frames are kept one above the other . frame x has a diameter of 16 cm and frame y has a diameter of 12 cm . what fraction of the surface of frame x is not covered by frame y ? | the questions asks us to find the surface which is not covered by the frame y i . e . , area of the surface not covered . where as circumference is the length along the edge of the circle , 2 * pi * r implies the length of the curve pi * r ^ 2 implies area enclosed by that curve . . hence area of the circle is consider... | ['a ) 3 / 5', 'b ) 7 / 16', 'c ) 12 / 16', 'd ) 24 / 12', 'e ) 2 / 3'] | b | divide(subtract(circle_area(divide(16, const_2)), circle_area(divide(12, const_2))), circle_area(divide(16, const_2))) | divide(n0,const_2)|divide(n1,const_2)|circle_area(#0)|circle_area(#1)|subtract(#2,#3)|divide(#4,#2) | geometry |
a car gets 40 kilometers per gallon of gasoline . how many gallons of gasoline would the car need to travel 140 kilometers ? | "each 40 kilometers , 1 gallon is needed . we need to know how many 40 kilometers are there in 140 kilometers ? 140 ÷ 40 = 3.5 × 1 gallon = 3.5 gallons correct answer e" | a ) 4.5 , b ) 5.5 , c ) 6.5 , d ) 7.5 , e ) 3.5 | e | divide(140, 40) | divide(n1,n0)| | physics |
in some quantity of ghee , 60 % of pure ghee and 40 % of is vanaspati . if 10 kg of pure ghee is added , then the strength of vanaspati ghee becomes 20 % . the original quantity was ? | let the original quantity be x kg vanaspati ghee in x kg = 40 / 100 * x = 2 x / 5 kg ( 2 x / 5 ) / ( x + 10 ) = 20 / 100 2 x / ( 5 x + 50 ) = 1 / 5 5 x = 50 x = 10 kg answer is b | a ) 5 kg , b ) 10 kg , c ) 15 kg , d ) 20 kg , e ) 25 kg | b | divide(multiply(divide(20, const_100), 10), subtract(divide(40, const_100), divide(20, const_100))) | divide(n3,const_100)|divide(n1,const_100)|multiply(n2,#0)|subtract(#1,#0)|divide(#2,#3) | gain |
jack and jill are marathon runners . jack can finish a marathon ( 42 km ) in 6 hours and jill can run a marathon in 4.2 hours . what is the ratio of their average running speed ? ( jack : jill ) | "average speed of jack = distance / time = 42 / 6 = 7 average speed of jill = 42 / ( 4.2 ) = 10 ratio of average speed of jack to jill = 7 / 10 answer a" | a ) 7 / 10 , b ) 15 / 14 , c ) 4 / 5 , d ) 5 / 4 , e ) can not be determined | a | divide(divide(42, 6), divide(42, 4.2)) | divide(n0,n1)|divide(n0,n2)|divide(#0,#1)| | physics |
( 128.5 x 4 ) + ( 13.8 x 5 ) = ? x 10 | "explanation : ? = ( 128.5 x 4 ) + ( 13.8 x 5 ) / 10 = 514 + 69 / 10 = 58.3 answer : option c" | a ) 524.48 , b ) 556.02 , c ) 58.3 , d ) 585.64 , e ) 595.46 | c | multiply(128.5, power(add(const_4, const_1), const_4)) | add(const_1,const_4)|power(#0,const_4)|multiply(n0,#1)| | general |
when a natural number n is successively divided by 12 , 34 the remainders are 56 , 78 . what will be the sum of the remainders if the order of the division is reversed ? | 1 2 3 4 5 6 7 8 leave the top right - most number 8 start with bottom right - most number 5 8 * 3 + 7 = 31 31 * 2 + 6 = 68 68 * 1 + 5 = 73 this is the number required now , do the successive division in the reverse order the sum of the remainders is 10 hence , the correct option is c | a ) 11 , b ) 12 , c ) 10 , d ) 2 , e ) 7 | c | subtract(56, add(12, 34)) | add(n0,n1)|subtract(n2,#0) | general |
speed of a boat in still water is 5 km / hr and speed of the stream is 2 kmph . a man rows to a place at a distance of 252 km and comes back to the starting point the total time taken by him is . | explanation : speed up stream = 5 - 2 = 3 km / hr speed down stream = 5 + 2 = 7 km / hr total time taken = 252 / 3 + 252 / 7 = 84 + 36 = 120 hr answer : option b | a ) 160 hr , b ) 200 hr , c ) 120 hr , d ) 140 hr , e ) 180 hr | b | multiply(divide(add(divide(252, add(5, 2)), divide(252, subtract(5, 2))), const_60), const_100) | add(n0,n1)|subtract(n0,n1)|divide(n2,#0)|divide(n2,#1)|add(#2,#3)|divide(#4,const_60)|multiply(#5,const_100) | physics |
each week , harry is paid x dollars per hour for the first 30 hours and 1.5 x dollars for each additional hour worked that week . each week , james is paid x dollars per per hour for the first 40 hours and 2 x dollars for each additional hour worked that week . last week james worked a total of 42 hours if harry and ja... | "james worked for 41 hours hence he earned 40 * x + 2 * 2 x = 44 x dollars ; we know that harry also earned the same 44 x dollars , out of which he earned 30 x dollars for thefirst 30 hoursplus 14 x additional dollars . since for each additional hour he gets 1.5 x dollars then he worked for 14 x / 1.5 x = 9 additional ... | a ) 35 , b ) 36 , c ) 37 , d ) 38 , e ) 39 | e | add(divide(subtract(add(40, 2), 30), 1.5), 30) | add(n2,n3)|subtract(#0,n0)|divide(#1,n1)|add(n0,#2)| | general |
some of 50 % - intensity red paint is replaced with 25 % solution of red paint such that the new paint intensity is 35 % . what fraction of the original paint was replaced ? | "35 % is 10 % - points above 25 % and 15 % - points below 50 % . thus the ratio of 25 % - solution to 50 % - solution is 3 : 2 . 3 / 5 of the original paint was replaced . the answer is d ." | a ) 1 / 35 , b ) 1 / 5 , c ) 2 / 3 , d ) 3 / 5 , e ) 4 / 5 | d | divide(subtract(divide(35, const_100), divide(50, const_100)), subtract(divide(25, const_100), divide(50, const_100))) | divide(n2,const_100)|divide(n0,const_100)|divide(n1,const_100)|subtract(#0,#1)|subtract(#2,#1)|divide(#3,#4)| | gain |
the shopkeeper increased the price of a product by 25 % so that customer finds it difficult to purchase the required amount . but somehow the customer managed to purchase only 64 % of the required amount . what is the net difference in the expenditure on that product ? | "quantity x rate = price 1 x 1 = 1 0.64 x 1.25 = 0.8 decrease in price = ( 0.2 / 1 ) × 100 = 20 % e )" | a ) 12.5 % , b ) 13 % , c ) 15 % , d ) 17 % , e ) 20 % | e | divide(multiply(subtract(multiply(const_100, const_100), multiply(add(const_100, 25), 64)), const_100), multiply(const_100, const_100)) | add(n0,const_100)|multiply(const_100,const_100)|multiply(n1,#0)|subtract(#1,#2)|multiply(#3,const_100)|divide(#4,#1)| | general |
an engineer undertakes a project to build a road 10 km long in 30 days and employs 30 men for the purpose . after 10 days , he finds only 2 km of the road has been completed . find the ( approximate ) number of extra men he must employ to finish the work in time . | "30 workers working already let x be the total men required to finish the task in next 20 days 2 km done hence remaining is 8 km also , work has to be completed in next 20 days ( 30 - 10 = 20 ) we know that , proportion of men to distance is direct proportion and , proportion of men to days is inverse proportion hence ... | a ) 30 , b ) 40 , c ) 50 , d ) 60 , e ) 70 | a | subtract(divide(multiply(multiply(30, subtract(10, 2)), 10), multiply(2, subtract(30, 10))), 30) | subtract(n0,n4)|subtract(n1,n3)|multiply(n2,#0)|multiply(n4,#1)|multiply(n3,#2)|divide(#4,#3)|subtract(#5,n2)| | physics |
the number 155 can be written as the sum of the squares of 3 different positive integers . what is the sum of these 3 integers ? | "7 ^ 2 + 5 ^ 2 + 9 ^ 2 = 49 + 25 + 81 = 155 7 + 5 + 9 = 21 hence answer is e" | a ) 17 , b ) 16 , c ) 15 , d ) 14 , e ) 21 | e | add(add(add(const_4, 3), add(3, const_2)), 3) | add(n1,const_4)|add(const_2,n1)|add(#0,#1)|add(n1,#2)| | geometry |
two digits in john ' s age are the same as the digit in the wilson ' s age , but in reverse order . in 21 years john will be twice as old as wilson will be then . what is the difference in their current age ? | john ' s age = 10 x + y so wilson ' s age = 10 y + x . . after 21 years , 10 x + y + 21 = 2 * ( 10 y + x + 21 ) . . . . so 21 + 19 y = 8 x . . . check for odd values of y , y = 1 satisfies the eqn with x = 5 . . . so ages are 51 and 15 and ans as found correctly by u is d . . 36 | a ) 25 , b ) 43 , c ) 14 , d ) 36 , e ) 12 | d | subtract(multiply(add(21, const_3), const_2), multiply(const_3, const_4)) | add(n0,const_3)|multiply(const_3,const_4)|multiply(#0,const_2)|subtract(#2,#1) | general |
a cab driver 5 days income was $ 45 , $ 50 , $ 60 , $ 65 , $ 70 . then his average income is ? | avg = sum of observations / number of observations avg income = ( 45 + 50 + 60 + 65 + 70 ) / 5 = 58 answer is a | a ) $ 58 , b ) $ 42 , c ) $ 65 , d ) $ 48 , e ) $ 62 | a | divide(add(add(add(add(45, 50), 60), 65), 70), 5) | add(n1,n2)|add(n3,#0)|add(n4,#1)|add(n5,#2)|divide(#3,n0) | general |
if a : b is 3 : 4 and b : c is 4 : 5 then a : c is equal to | the two ratios given are having the same number 4 for b in both the ratios . hence - a : b = 3 : 4 b : c = 4 : 5 = > a : c = 3 : 5 answer b | a ) 1 : 3 , b ) 3 : 5 , c ) 3 : 4 , d ) 3 : 7 , e ) 1 : 5 | b | multiply(divide(4, 5), divide(3, 4)) | divide(n1,n3)|divide(n0,n1)|multiply(#0,#1) | general |
in a box of 9 pencils , a total of 2 are defective . if a customer buys 3 pencils selected at random from the box , what is the probability that neither pencils will be defective ? | first , there are 7 c 3 ways you can select 3 good pencils from 4 good ones . second , there are 9 c 3 ways you select 3 pencils from 6 ones in the box . then , the probability that neither pen will be defective is : 7 c 3 / 9 c 3 = 35 / 84 = 5 / 12 answer is a | a ) 5 / 12 , b ) 1 / 15 , c ) 2 / 13 , d ) 2 / 15 , e ) 1 / 17 | a | divide(divide(factorial(subtract(9, 2)), multiply(factorial(subtract(subtract(9, 2), 3)), factorial(3))), divide(factorial(9), multiply(factorial(subtract(9, 3)), factorial(3)))) | factorial(n2)|factorial(n0)|subtract(n0,n1)|subtract(n0,n2)|factorial(#2)|factorial(#3)|subtract(#2,n2)|factorial(#6)|multiply(#5,#0)|divide(#1,#8)|multiply(#7,#0)|divide(#4,#10)|divide(#11,#9) | general |
what profit percent is made by selling an article at a certain price , if by selling at 2 / 3 rd of that price , there would be a loss of 18 % ? | sp 2 = 2 / 3 sp 1 cp = 100 sp 2 = 82 2 / 3 sp 1 = 82 sp 1 = 123 100 - - - 23 = > 23 % answer : e | a ) 20 % , b ) 26 % , c ) 42 % , d ) 27 % , e ) 23 % | e | subtract(divide(subtract(const_100, 18), divide(2, 3)), const_100) | divide(n0,n1)|subtract(const_100,n2)|divide(#1,#0)|subtract(#2,const_100) | gain |
8 identical machines , working alone and at their constant rates , take 6 hours to complete a job lot . how long would it take for 6 such machines to perform the same job ? | "let each machine do 1 unit of work for 1 hour 8 machines - - > 8 units of work in 1 hour for 6 hours = 8 * 6 = 48 units of total work is done . now this 48 units of total work must be done by 6 machines 6 units of work ( 6 machines ) - - - > 1 hour for 48 units of work 6 * 8 - - - > 1 * 8 hours b 8 hours" | a ) 2.25 hours , b ) 8 hours , c ) 12 hours , d ) 14.25 hours , e ) 16 hours | b | divide(multiply(6, 8), 6) | multiply(n0,n1)|divide(#0,n2)| | physics |
by selling an article at rs . 800 , a shopkeeper makes a profit of 25 % . at what price should he sell the article so as to make a loss of 40 % ? | "sp = 800 profit = 25 % cp = ( sp ) * [ 100 / ( 100 + p ) ] = 800 * [ 100 / 125 ] = 640 loss = 40 % = 40 % of 640 = rs . 256 sp = cp - loss = 640 - 256 = rs . 384 answer : e" | a ) s . 429 , b ) s . 480 , c ) s . 429 , d ) s . 128 , e ) s . 384 | e | subtract(divide(multiply(800, const_100), add(25, const_100)), divide(multiply(divide(multiply(800, const_100), add(25, const_100)), 40), const_100)) | add(n1,const_100)|multiply(n0,const_100)|divide(#1,#0)|multiply(n2,#2)|divide(#3,const_100)|subtract(#2,#4)| | gain |
milk contains 10 % water . what content of pure milk should be added to 20 liters of milk to reduce this to 5 % ? | quantity of water in 20 liters = 10 % of 20 liters = 2 liters let x liters of pure milk be added . then , 2 / ( 20 + x ) = 5 / 100 5 x = 100 x = 20 liters answer is a | a ) 20 liters , b ) 25 liters , c ) 30 liters , d ) 15 liters , e ) 18 liters | a | subtract(divide(multiply(divide(10, const_100), 20), divide(5, const_100)), 20) | divide(n0,const_100)|divide(n2,const_100)|multiply(n1,#0)|divide(#2,#1)|subtract(#3,n1) | gain |
it takes avery 3 hours to build a brick wall while tom can do it in 2 hours . if the two start working together and after an hour avery leaves , how much time will it take tom to complete the wall on his own ? | "avery takes 3 hours tom takes 2 hours efficiency of avery is 1 / 3 units / hr efficiency of tom is 1 / 2 units / hr combined efficiency of tom and avery is 1 / 3 + 1 / 2 = 5 / 6 units / hr since they worked for 1 hour they completed 5 / 6 units of work and 1 / 6 units of work is left which is to be completed by tom ( ... | a ) 20 , b ) 30 , c ) 40 , d ) 50 , e ) 60 | a | multiply(divide(subtract(const_1, add(divide(const_1, 3), divide(const_1, 2))), divide(const_1, 2)), const_60) | divide(const_1,n0)|divide(const_1,n1)|add(#0,#1)|subtract(const_1,#2)|divide(#3,#1)|multiply(#4,const_60)| | physics |
salesperson a ' s compensation for any week is $ 360 plus 6 percent of the portion of a ' s total sales above $ 2,000 for that week . salesperson b ' s compensation for any week is 8 percent of a ' s total sales for that week . for what amount of total weekly sales would both salepeople earn the same compensation ? | "sometime , setting up an equation is an easy way to go with : 360 + 0.06 ( x - 2000 ) = 0.08 x x = 12000 ans : a" | a ) $ 12,000 , b ) $ 18,000 , c ) $ 15,000 , d ) $ 4,500 , e ) $ 4,000 | a | divide(add(divide(subtract(360, multiply(divide(6, const_100), 2,000)), subtract(divide(8, const_100), divide(6, const_100))), divide(subtract(360, multiply(divide(6, const_100), 2,000)), subtract(divide(8, const_100), divide(6, const_100)))), 2,000) | divide(n1,const_100)|divide(n3,const_100)|multiply(#0,n2)|subtract(#1,#0)|subtract(n0,#2)|divide(#4,#3)|add(#5,#5)|divide(#6,n2)| | general |
a car gets 40 kilometers per gallon of gasoline . how many gallons of gasoline would the car need to travel 200 kilometers ? | "each 40 kilometers , 1 gallon is needed . we need to know how many 40 kilometers are there in 180 kilometers ? 200 ÷ 40 = 5 × 1 gallon = 5 gallons correct answer b" | a ) 4.5 , b ) 5 , c ) 6.5 , d ) 7.5 , e ) 8.5 | b | divide(200, 40) | divide(n1,n0)| | physics |
at a monthly meeting , 3 / 5 of the attendees were males and 7 / 8 of the male attendees arrived on time . if 9 / 10 of the female attendees arrived on time , what fraction of the attendees at the monthly meeting did not arrive on time ? | males who did not arrive on time are 1 / 8 * 3 / 5 = 3 / 40 of the attendees . females who did not arrive on time are 1 / 10 * 2 / 5 = 2 / 50 of the attendees . the fraction of all attendees who did not arrive on time is 3 / 40 + 2 / 50 = 23 / 200 the answer is a . | a ) 23 / 200 , b ) 11 / 100 , c ) 7 / 50 , d ) 3 / 20 , e ) 4 / 25 | a | add(multiply(subtract(const_1, divide(9, 10)), subtract(const_1, divide(3, 5))), multiply(subtract(const_1, divide(7, 8)), divide(3, 5))) | divide(n4,n5)|divide(n0,n1)|divide(n2,n3)|subtract(const_1,#0)|subtract(const_1,#1)|subtract(const_1,#2)|multiply(#3,#4)|multiply(#1,#5)|add(#6,#7) | general |
there are 15 boys and 10 girls in a class . if three students are selected at random , in how many ways that 1 girl and 2 boys are selected ? | "n ( s ) = sample space = 25 c 3 = 2300 e = event that 1 girl and 2 boys are selected n ( e ) = we have to select 2 boys from 15 and 1 girl from 10 = 15 c 2 * 10 c 1 = 1050 ans - b" | a ) 950 , b ) 1050 , c ) 2150 , d ) 2050 , e ) 1000 | b | multiply(choose(15, 2), choose(10, 1)) | choose(n0,n3)|choose(n1,n2)|multiply(#0,#1)| | probability |
12 men can complete a piece of work in 25 days . in how many days can 20 men complete that piece of work ? | "12 * 25 = 20 * x = > x = 15 days answer : e" | a ) 55 days , b ) 77 days , c ) 20 days , d ) 88 days , e ) 15 days | e | divide(multiply(25, 12), 20) | multiply(n0,n1)|divide(#0,n2)| | physics |
consider a lady took a loan from a bank at the rate of 12 % p . a . simple interest . after 3 years she had to pay rs . 7200 interest only for the period . the principal amount borrowed by her was | explanation : principal = rs . ( 100 × 7200 / 12 × 3 ) = > rs . 20,000 . answer : d | a ) rs . 2000 , b ) rs . 10000 , c ) rs . 15000 , d ) rs . 20000 , e ) none of these | d | divide(7200, divide(multiply(3, 12), const_100)) | multiply(n0,n1)|divide(#0,const_100)|divide(n2,#1) | gain |
how many pages do you need if you have to write 150 lines of information into pages that each hold 30 lines ? | sol . 30 lines per page total 150 lines = 150 / 30 = 5 answer : b | a ) a ) 7 , b ) b ) 5 , c ) c ) 9 , d ) d ) 8 , e ) e ) none of the above | b | divide(150, 30) | divide(n0,n1) | general |
a dog takes 10 leaps for every 2 leaps of a hare . if one leap of the dog is equal to 2 leaps of the hare , the ratio of the speed of the dog to that of the hare is : | "explanation : dog : hare = ( 10 * 2 ) leaps of hare : 2 leaps of hare = 20 : 2 . answer : b ) 10 : 1" | a ) 5 : 1 , b ) 10 : 1 , c ) 15 : 1 , d ) 20 : 1 , e ) 25 : 1 | b | divide(multiply(10, 2), 2) | multiply(n0,n2)|divide(#0,n1)| | other |
a train running at the speed of 60 km / hr crosses a pole in 9 seconds . what is the length of the train ? | "speed = ( 60 * 5 / 18 ) m / sec = ( 50 / 3 ) m / sec length of the train = ( speed x time ) = ( 50 / 3 * 9 ) m = 150 m . answer : option d" | a ) 100 , b ) 120 , c ) 130 , d ) 150 , e ) 160 | d | multiply(divide(multiply(60, const_1000), const_3600), 9) | multiply(n0,const_1000)|divide(#0,const_3600)|multiply(n1,#1)| | physics |
triangle xyz is an isosceles right triangle . if side xy is longer than side yz , and the area of the triangle is 64 , what is the measure of side xy ? | "ans b . . 16 . xy being larger means it is the hyp . . area = ( 1 / 2 ) * ( yz ) ^ 2 = 64 or yz = 8 * \ sqrt { 2 } . . therefore hyp = xy = 16" | a ) 4 , b ) 16 , c ) 8 , d ) 8 √ 2 , e ) can not be determined from the information provided | b | sqrt(add(power(multiply(sqrt(64), sqrt(const_2)), const_2), power(multiply(sqrt(64), sqrt(const_2)), const_2))) | sqrt(n0)|sqrt(const_2)|multiply(#0,#1)|power(#2,const_2)|add(#3,#3)|sqrt(#4)| | geometry |
a sum of money lent out at s . i . amounts to rs . 780 after 2 years and to rs . 1020 after a further period of 5 years . the sum is ? | "s . i for 5 years = ( 1020 - 780 ) = rs . 240 . s . i . for 2 years = 240 / 5 * 2 = rs . 96 . principal = ( 780 - 96 ) = rs . 684 answer : a" | a ) s . 684 , b ) s . 500 , c ) s . 540 , d ) s . 740 , e ) s . 840 | a | subtract(780, multiply(divide(subtract(1020, 780), 5), 2)) | subtract(n2,n0)|divide(#0,n3)|multiply(n1,#1)|subtract(n0,#2)| | general |
there are 40 balls which are red , blue or green . if 15 balls are green and the sum of red balls and green balls is less than 25 , at most how many red balls are there ? | "r + g + b = 40 g = 15 r + g < 25 = > r + 15 < 25 = > r < 10 = > at most 6 red balls answer : e" | a ) 4 , b ) 5 , c ) 6 , d ) 7 , e ) 9 | e | subtract(15, const_1) | subtract(n1,const_1)| | general |
the sale price of an article including the sales tax is rs . 616 . the rate of sales tax is 10 % . if the shopkeeper has made a profit of 13 % , then the cost price of the article is : | "110 % of s . p . = 616 s . p . = ( 616 * 100 ) / 110 = rs . 560 c . p = ( 110 * 560 ) / 113 = rs . 545 answer : option a" | a ) 545 , b ) 334 , c ) 555 , d ) 664 , e ) 5598 | a | divide(multiply(divide(multiply(616, const_100), add(const_100, 10)), add(const_100, 10)), add(const_100, 13)) | add(n1,const_100)|add(n2,const_100)|multiply(n0,const_100)|divide(#2,#0)|multiply(#0,#3)|divide(#4,#1)| | gain |
in order to obtain an income of rs . 580 from 15 % stock at rs . 100 , one must make an investment of | "to obtain rs . 10 , investment = rs . 100 . to obtain rs . 580 , investment = = rs . 8700 . answer : c" | a ) 5363 , b ) 6240 , c ) 8700 , d ) 5600 , e ) 2732 | c | multiply(divide(100, 15), 580) | divide(n2,n1)|multiply(n0,#0)| | gain |
a shopkeeper sold an article offering a discount of 4 % and earned a profit of 20 % . what would have been the percentage of profit earned if no discount was offered ? | "let c . p . be rs . 100 . then , s . p . = rs . 120 let marked price be rs . x . then , 96 / 100 x = 120 x = 12000 / 96 = rs . 125 now , s . p . = rs . 125 , c . p . = rs . 100 profit % = 25 % . answer : a" | a ) 25 , b ) 28 , c ) 30 , d ) 32 , e ) 34 | a | subtract(divide(multiply(add(const_100, 20), const_100), subtract(const_100, 4)), const_100) | add(n1,const_100)|subtract(const_100,n0)|multiply(#0,const_100)|divide(#2,#1)|subtract(#3,const_100)| | gain |
the surface area of a cube is 1734 sq . cm . find its volume | explanation : let the edge of the cube bea . then , 6 a 2 = 1734 = > a = 17 cm . volume = a 3 = 173 = 4193 cm 3 answer : c | ['a ) 2334 cubic . cm', 'b ) 3356 cubic . cm', 'c ) 4913 cubic . cm', 'd ) 3478 cubic . cm', 'e ) none of these'] | c | volume_cube(sqrt(divide(1734, multiply(const_2, const_3)))) | multiply(const_2,const_3)|divide(n0,#0)|sqrt(#1)|volume_cube(#2) | geometry |
apple street begins at peach street and runs directly east for 3.2 kilometers until it ends when it meets cherry street . apple street is intersected every 200 meters by a perpendicular street , and each of those streets other than peach street and cherry street is given a number beginning at 1 st street ( one block ea... | 3.2 km / 200 m = 16 . however , the street at the 3.2 - km mark is not 16 th street ; it is cherry street . therefore , the highest numbered street is 15 th street . the answer is b . | a ) 14 th , b ) 15 th , c ) 16 th , d ) 17 th , e ) 18 th | b | subtract(divide(multiply(3.2, const_1000), 200), const_1) | multiply(n0,const_1000)|divide(#0,n1)|subtract(#1,const_1) | physics |
a bag contains 8 red , 6 blue and 4 green balls . if 2 ballsare picked at random , what is the probability that both are red ? | "p ( both are red ) , = 8 c 216 c 2 = 8 c 216 c 2 = 28 / 120 = 7 / 30 d" | a ) 1 / 13 , b ) 2 / 23 , c ) 5 / 26 , d ) 7 / 30 , e ) 3 / 23 | d | divide(choose(8, 2), choose(add(add(8, 6), 4), 2)) | add(n0,n1)|choose(n0,n3)|add(n2,#0)|choose(#2,n3)|divide(#1,#3)| | other |
a , b and c completed a piece of work , a worked for 6 days , b for 9 days and c for 4 days . their daily wages were in the ratio of 3 : 4 : 5 . find the daily wages of c , if their total earning was $ 1480 ? | a $ 100 3 x 4 x 5 x 6 9 4 18 x + 36 x + 20 x = 1480 74 x = 1480 = > x = 20 5 x = 100 rs . | a ) $ 100 , b ) $ 150 , c ) $ 125 , d ) $ 134 , e ) $ 243 | a | multiply(5, divide(1480, add(add(multiply(6, 3), multiply(9, 4)), multiply(4, 5)))) | multiply(n0,n3)|multiply(n1,n2)|multiply(n2,n5)|add(#0,#1)|add(#3,#2)|divide(n6,#4)|multiply(n5,#5) | physics |
the average salary of the employees in a office is rs . 120 / month . the avg salary of officers is rs . 440 and of non officers is rs 110 . if the no . of officers is 15 , then find the no of nonofficers in the office . | "let no . of non - officers be x 15 * 440 + x * 110 = ( x + 15 ) 120 x = 480 c" | a ) 400 , b ) 420 , c ) 480 , d ) 450 , e ) 510 | c | divide(subtract(multiply(15, 440), multiply(15, 120)), subtract(120, 110)) | multiply(n1,n3)|multiply(n0,n3)|subtract(n0,n2)|subtract(#0,#1)|divide(#3,#2)| | general |
what is the sum of the multiples of 5 from 80 to 140 , inclusive ? | "the formula we want to use in this type of problem is this : average * total numbers = sum first , find the average by taking the sum of the f + l number and divide it by 2 : a = ( f + l ) / 2 second , find the total numbers in our range by dividing our f and l numbers by 5 and add 1 . ( 140 / 5 ) - ( 80 / 5 ) + 1 mul... | a ) 560 , b ) 780 , c ) 990 , d ) 1008 , e ) 1200 | c | multiply(divide(add(subtract(140, const_3), add(80, const_2)), const_2), add(divide(subtract(subtract(140, const_3), add(80, const_2)), 5), const_1)) | add(n1,const_2)|subtract(n2,const_3)|add(#0,#1)|subtract(#1,#0)|divide(#3,n0)|divide(#2,const_2)|add(#4,const_1)|multiply(#6,#5)| | general |
a rectangular field is to be fenced on three sides leaving a side of 30 feet uncovered . if the area of the field is 720 sq . feet , how many feet of fencing will be required ? | "explanation we have : l = 30 ft and lb = 720 sq . ft . so , b = 24 ft . length of fencing = ( l + 2 b ) = ( 30 + 48 ) ft = 78 ft . answer a" | a ) 78 , b ) 40 , c ) 68 , d ) 88 , e ) none | a | add(multiply(divide(720, 30), const_2), 30) | divide(n1,n0)|multiply(#0,const_2)|add(n0,#1)| | geometry |
jack and jill are marathon runners . jack can finish a marathon ( 42 km ) in 5 hours and jill can run a marathon in 4.2 hours . what is the ratio of their average running speed ? ( jack : jill ) | "average speed of jack = distance / time = 42 / 5 average speed of jill = 42 / ( 4.2 ) = 10 ratio of average speed of jack to jill = ( 42 / 5 ) / 10 = 42 / 50 = 21 / 25 answer d" | a ) 14 / 15 , b ) 15 / 14 , c ) 4 / 5 , d ) 21 / 25 , e ) can not be determined | d | divide(divide(42, 5), divide(42, 4.2)) | divide(n0,n1)|divide(n0,n2)|divide(#0,#1)| | physics |
a bottle contains a certain solution . in the bottled solution , the ratio of water to soap is 3 : 2 , and the ratio of soap to salt is three times this ratio . the solution is poured into an open container , and after some time , the ratio of water to soap in the open container is halved by water evaporation . at that... | "water : soap = 3 : 2 soap : salt = 9 : 2 = > for 9 soap , salt = 2 = > for 2 soap , salt = ( 2 / 9 ) * 2 = 4 / 9 so , water : soap : salt = 3 : 2 : 4 / 9 = 27 : 18 : 4 after open container , water : soap : salt = 13.5 : 18 : 4 so , water : salt = 13.5 : 4 = 27 : 8 answer : e" | a ) 1 : 1 , b ) 2 : 3 , c ) 3 : 2 , d ) 9 : 4 , e ) 27 : 8 | e | divide(multiply(multiply(2, 3), 3), multiply(multiply(2, 2), 2)) | multiply(n0,n1)|multiply(n1,n1)|multiply(n0,#0)|multiply(n1,#1)|divide(#2,#3)| | other |
at a party , 13 guests brought wine and 14 guests brought beer . if exactly 11 guests brought only one of the two types of drinks , how many guests brought both types of drinks ? | say x guests brought both drinks . ( 13 - x ) + ( 14 - x ) = 11 - - > x = 8 . answer : d . | a ) 5 , b ) 6 , c ) 7 , d ) 8 , e ) 14 | d | divide(subtract(add(13, 14), 11), const_2) | add(n0,n1)|subtract(#0,n2)|divide(#1,const_2) | other |
a sum of money is put out at compound interest for 2 years at 20 % . it would earn rs . 482 more if the interest were paid half - yearly . calculate the sum . | p ( 11 / 10 ) 4 - p ( 6 / 5 ) 2 = 482 p = 2000 answer : d | a ) 2878 , b ) 279 , c ) 279 , d ) 482 , e ) 661 | d | add(multiply(add(20, const_1), add(20, const_1)), add(20, const_1)) | add(n1,const_1)|multiply(#0,#0)|add(#0,#1) | general |
a broker invested her own money in the stock market . during the first year , she increased her stock market wealth by 75 percent . in the second year , largely as a result of a slump in the stock market , she suffered a 30 percent decrease in the value of her stock investments . what was the net increase or decrease o... | "the actual answer is obtained by multiplying 175 % by 70 % and subtracting 100 % from this total . that is : 175 % × 70 % = 122.5 % ; 122.5 % − 100 % = 122.5 % . answer : e" | a ) − 5 % , b ) 5 % , c ) 15 % , d ) 20 % , e ) 22.5 % | e | multiply(subtract(multiply(add(const_1, divide(75, const_100)), subtract(const_1, divide(30, const_100))), const_1), const_100) | divide(n0,const_100)|divide(n1,const_100)|add(#0,const_1)|subtract(const_1,#1)|multiply(#2,#3)|subtract(#4,const_1)|multiply(#5,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 . william . mr . william paid only $ 480 as farm tax . the percentage of total land of mr . willam over the total taxable land of the village is : | "this will be equal to the percentage of total cultivated land he holds over the total cultivated land in the village . that leads to ( 480 / 3840 ) x 100 = 12.5 % in percentage terms . but the question asks ratio between his total land to total cultivated land . hence the answer is 12.5 % x ( 100 / 50 ) = 25 % the cor... | 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 total number of digits used in numbering the pages of a book having 266 pages is | "total number of digits = ( no . of digits in 1 - digit page nos . + no . of digits in 2 - digit page nos . + no . of digits in 3 - digit page nos . ) = ( 1 x 9 + 2 x 90 + 3 x 167 ) = ( 9 + 180 + 501 ) = 690 . answer : e" | a ) 732 , b ) 990 , c ) 109 , d ) 130 , e ) 690 | e | subtract(subtract(multiply(266, const_3), subtract(const_100, const_1)), subtract(const_10, const_1)) | multiply(n0,const_3)|subtract(const_100,const_1)|subtract(const_10,const_1)|subtract(#0,#1)|subtract(#3,#2)| | general |
the ratio of incomes of two person p 1 and p 2 is 5 : 4 and the ratio of their expenditures is 3 : 2 . if at the end of the year , each saves rs . 1400 , then what is the income of p 1 ? | "let the income of p 1 and p 2 be rs . 5 x and rs . 4 x respectively and let their expenditures be rs . 3 y and 2 y respectively . then , 5 x – 3 y = 1400 … ( i ) and 4 x – 2 y = 1400 … … . . ( ii ) on multiplying ( i ) by 2 , ( ii ) by 3 and subtracting , we get : 2 x = 1400 - > x = 700 p 1 ’ s income = rs 5 * 700 = r... | a ) s . 800 , b ) s . 2400 , c ) s . 4000 , d ) s . 3500 , e ) s . 4200 | d | multiply(subtract(multiply(divide(1400, 2), 3), 1400), 5) | divide(n6,n1)|multiply(n4,#0)|subtract(#1,n6)|multiply(n2,#2)| | other |
the value of a machine depreciates at the rate of 10 % every year . it was purchased 3 years ago . if its present value is rs 8,748 , its purchase price was ? | "we have p = 8748 , r = - 10 and n = 3 therefore , the purchase price of the machine = p / ( 1 + ( r / 100 ) ) n = 8748 / ( 1 - ( 10 / 100 ) ) 3 = 8748 x ( 100 / 90 ) x ( 100 / 90 ) x ( 100 / 90 ) = 12,000 . answer : c" | a ) rs 16,000 , b ) rs 18,000 , c ) rs 12,000 , d ) rs 14,000 , e ) none of these | c | divide(8,748, subtract(const_1, multiply(divide(10, const_100), 3))) | divide(n0,const_100)|multiply(n1,#0)|subtract(const_1,#1)|divide(n2,#2)| | gain |
a new home buyer pays 4 % annual interest on her first mortgage and 9 % annual interest on her second mortgage . if she borrowed a total of $ 340,000 , 80 % of which was in the first mortgage , what is her approximate monthly interest payment ? | "0.04 x + 0.09 y = 340000 [ 1 ] 0.04 x = 0.80 * 340000 = 272000 [ 2 ] 272000 + 0.09 y = 340000 - - > 0.09 y = 64000 [ 3 ] 272000 / 12 = 22666.67 [ 4 ] 64000 / 12 = 5666.67 [ 5 ] adding [ 4,5 ] we get : 28333.33 [ 6 ] dividing [ 6 ] / 2 to get an average we get 1.416 , ans a" | a ) $ 1,416 , b ) $ 1,733 , c ) $ 3,466 , d ) $ 13,333 , e ) $ 20,796 | a | subtract(9, multiply(4, const_2)) | multiply(n0,const_2)|subtract(n1,#0)| | general |
a mixture of 150 liters of wine and water contains 10 % water . how much more water should be added so that water becomes 25 % of the new mixture ? | "number of liters of water in 150 liters of the mixture = 10 % of 150 = 10 / 100 * 150 = 15 liters . p liters of water added to the mixture to make water 25 % of the new mixture . total amount of water becomes ( 15 + p ) and total volume of mixture is ( 150 + p ) . ( 15 + p ) = 25 / 100 * ( 150 + p ) 60 + 4 p = 150 + p... | a ) 10 , b ) 30 , c ) 20 , d ) 15 , e ) 16 | b | divide(subtract(multiply(divide(25, const_100), 150), multiply(divide(10, const_100), 150)), subtract(const_1, divide(25, const_100))) | divide(n2,const_100)|divide(n1,const_100)|multiply(n0,#0)|multiply(n0,#1)|subtract(const_1,#0)|subtract(#2,#3)|divide(#5,#4)| | general |
what is the difference between the local values of 8 in the number 58408 ? | explanation : 8000 â € “ 8 = 7992 answer : b | a ) 2992 , b ) 7992 , c ) 2877 , d ) 2077 , e ) 6229 | b | subtract(multiply(8, const_1000), 8) | multiply(n0,const_1000)|subtract(#0,n0) | general |
three numbers are in the ratio of 2 : 3 : 4 . if the sum of the squares of the extremes is 180 , then the middle number is : | "number be = 2 x , 3 x , 4 x , ( 2 x ) 2 + ( 4 x ) 2 = 180 20 x = 180 ⇒ x = 9 ⇒ x = 3 , midline number = 3 × 3 = 9 answer : d" | a ) 6 , b ) 12 , c ) 15 , d ) 9 , e ) 7 | d | multiply(2, 4) | multiply(n0,n2)| | general |
the area of a parallelogram is 147 sq m and its base is thrice the corresponding height . then the length of the base is ? | b * h = 147 b * h = 147 = > b ^ 2 = 49 m . b = 7 answer : ( d ) | a ) 14 m , b ) 29 m , c ) 49 m , d ) 7 , e ) none | d | sqrt(divide(147, const_2)) | divide(n0,const_2)|sqrt(#0)| | geometry |
a contractor undertakes to complete the construction of a tunnel 720 meters long in 240 days and employs 60 men for the purpose . after 120 days , he finds that only 240 meters of the tunnel is complete . how many more men should be employ in order to complete the work in time ? | in 120 days , only 240 m of the tunnel is constructed by 60 men . the remaining 120 days , 480 m of the tunnel can be constructed by 120 men . additional number of men required = 120 - 60 = 60 men . answer : b | a ) 22 , b ) 27 , c ) 60 , d ) 88 , e ) 12 | b | subtract(120, 60) | subtract(n3,n2) | physics |
a certain automobile company ’ s best - selling model is the speedster . the speedster , like all of their other models , comes in coupe and convertible styles . 2 / 3 of the current inventory is speedsters , of which 4 / 5 are convertibles . if there are 50 vehicles that are not speedsters , how many speedster convert... | "total vehicle = 2 / 3 of speedster + 1 / 3 of others . speedster convertibles = 2 / 3 total vehicle * 4 / 5 given : 1 / 3 constitutes 50 vehicles . hence 2 / 3 constitutes 100 speedster convertibls = 100 * 4 / 5 = 80 d" | a ) 30 , b ) 48 , c ) 60 , d ) 80 , e ) 90 | d | multiply(multiply(divide(50, subtract(const_1, divide(2, 3))), divide(2, 3)), divide(4, 5)) | divide(n2,n3)|divide(n0,n1)|subtract(const_1,#1)|divide(n4,#2)|multiply(#3,#1)|multiply(#0,#4)| | physics |
if there are 4 peanuts in a box and mary puts 12 more peanuts inside , how many peanuts are in the box ? | "12 + 4 = 16 correct answer is b ) 16" | a ) 8 , b ) 16 , c ) 10 , d ) 11 , e ) 12 | b | add(4, 12) | add(n0,n1)| | general |
4 dice are thrown and the sum of the numbers noted is 10 . find the probability that all the numbers lie between 2 and 5 ( both inclusive ) ? | sum of numbers noted on 4 dice ( with six faces ) to be 10 will include combinations : ( 1,1 , 2,6 ) ( 1,1 , 3,5 ) ( 1,1 , 4,4 ) ( 1,1 , 5,3 ) ( 1,1 , 6,2 ) ( 1,2 , 1,6 ) ( 1,2 , 2,5 ) ( 1,2 , 3,4 ) ( 1,2 , 4,3 ) ( 1,2 , 5,2 ) ( 1,2 , 6,1 ) ( 1,3 , 1,5 ) ( 1,3 , 2,4 ) ( 1,3 , 3,3 ) ( 1,3 , 4,2 ) ( 1,3 , 5,1 ) ( 1,4 , 1... | a ) 11 / 79 , b ) 10 / 79 , c ) 12 / 79 , d ) 13 / 79 , e ) 14 / 79 | b | divide(10, subtract(multiply(multiply(2, 4), 10), const_1)) | multiply(n0,n2)|multiply(n1,#0)|subtract(#1,const_1)|divide(n1,#2) | other |
a truck covers a distance of 200 metres in 1 minute whereas a bus covers a distance of 100 kms in 50 minutes . the ratio of their speeds is ? | "ratio of speeds = ( 200 / 60 ) 18 / 5 : 60 * 100 / 50 = 12 : 120 = 1 : 10 answer is c" | a ) 2 : 3 , b ) ratio 1 : 5 , c ) ratio 1 : 10 , d ) ratio 9 : 13 , e ) ratio 7 : 15 | c | divide(200, multiply(divide(100, 50), const_1000)) | divide(n2,n3)|multiply(#0,const_1000)|divide(n0,#1)| | physics |
a reduction of 30 % in the price of oil enables a house wife to obtain 9 kgs more for rs . 900 , what is the reduced price for kg ? | "900 * ( 30 / 100 ) = 270 - - - - 9 ? - - - - 1 = > rs . 30 answer : d" | a ) 72 , b ) 27 , c ) 40 , d ) 30 , e ) 20 | d | divide(divide(multiply(900, 30), const_100), 9) | multiply(n0,n2)|divide(#0,const_100)|divide(#1,n1)| | gain |
in how many ways 6 boys and 4 girls can be seated in a row so that they are alternative ? | "6 boys can be seated in 6 ! 4 girls can be seated in 4 ! required number = 6 ! 4 ! = 17280 answer is c" | a ) 14420 , b ) 12620 , c ) 17280 , d ) 21200 , e ) 24560 | c | multiply(factorial(6), factorial(4)) | factorial(n0)|factorial(n1)|multiply(#0,#1)| | probability |
total dinning bill of 7 people was $ 139.00 and 10 % tip divided the bill evenly ? what is the bill amount each person shared . | "dinner bill of 7 person = 139 + 10 % tip so , 10 % of 139 = ( 139 * 10 ) / 100 = 13.9 so , the actual total amount = 139 + 13.9 = $ 152.9 so per head bill = 152.9 / 7 = $ 21.84 answer : a" | a ) 21.84 , b ) 22.84 , c ) 23.84 , d ) 24.84 , e ) 25.84 | a | divide(multiply(139.00, add(divide(const_1, 10), const_1)), 7) | divide(const_1,n2)|add(#0,const_1)|multiply(n1,#1)|divide(#2,n0)| | general |
city x has a population 3 times as great as the population of city y , which has a population twice as great as the population of city z . what is the ratio of the population of city x to the population of city z ? | "x = 3 y , y = 2 * z x : y , y : z 3 : 1 , 2 : 1 6 : 2 , 2 : 1 so , x : z = 6 : 1 ( d )" | a ) 1 : 8 , b ) 1 : 4 , c ) 2 : 1 , d ) 6 : 1 , e ) 8 : 1 | d | multiply(3, const_2) | multiply(n0,const_2)| | general |
in how many ways can 22 books on english and 20 books on hindi be placed in a row on a shelf so that two books on hindi may not be together ? | "in order that two books on hindi are never together , we must place all these books as under : h e h e h e h … . . h e h where h denotes the position of hindi book and e that of english book . since there are 22 books on english , the number of places marked e are 23 . now , 20 places out of 23 can be chosen in 23 c 2... | a ) 4586 , b ) 5896 , c ) 2415 , d ) 1771 , e ) 1881 | d | add(add(choose(subtract(22, const_1), const_3), permutation(subtract(22, const_1), const_2)), choose(subtract(22, const_1), const_1)) | subtract(n0,const_1)|choose(#0,const_3)|choose(#0,const_1)|permutation(#0,const_2)|add(#1,#3)|add(#4,#2)| | probability |
what is the dividend ? the divisor is 800 , the quotient is 594 , and the remainder is - 968 . | dividend = ? ? = d * q + r ? = 800 * 594 + - 968 ? = 475,200 + - 968 ? = 474,232 | a ) 6623 , b ) 43036 , c ) 51515 , d ) 474232 , e ) 2025 | d | subtract(multiply(800, 594), 968) | multiply(n0,n1)|subtract(#0,n2) | general |
working independently , tina can do a certain job in 12 hours . working independently , ann can do the same job in 9 hours . if tina works independently at the job for 8 hours and then ann works independently , how many hours will it take ann to complete the remainder of the jobs ? | work done by tina in 1 hr = 1 / 12 work done by tina in 8 hr = 1 / 12 * 8 which is 2 / 3 now , the remaining work is 1 / 3 which is completed by ann in 1 / 3 * 9 which is 3 hr answer : e | a ) 2 / 3 , b ) 3 / 4 , c ) 1 , d ) 2 , e ) 3 | e | divide(9, divide(12, subtract(12, 8))) | subtract(n0,n2)|divide(n0,#0)|divide(n1,#1) | physics |
sum of 24 odd numbers is ? | "sum of 1 st n odd no . s = 1 + 3 + 5 + 7 + . . . = n ^ 2 so , sum of 1 st 24 odd numbers = 24 ^ 2 = 576 answer : e" | a ) 572 , b ) 573 , c ) 574 , d ) 575 , e ) 576 | e | multiply(multiply(24, const_2), divide(24, const_2)) | divide(n0,const_2)|multiply(n0,const_2)|multiply(#0,#1)| | general |
set x consists of all two - digit primes and set y consists of all positive odd multiples of 9 less than 100 . if the two sets are combined into one , what will be the range of the new set ? | "set x = { 11 , 13 , 17 , . . . . . . . . . . . . . , 83 , 89 , 97 } set y = { 9 , 27 , 45 , . . . . . . . . . . . . . . . , 63 , 81 , 99 } combining two sets , say set z set z = { 9 , 11 , 13 , 17 , 27 , . . . . . . . . . . . . . . . . . . . , 63 , 81 , 83 , 89 , 97,99 } range = max value - min value range ( z ) = 99 ... | a ) 84 , b ) 89 , c ) 90 , d ) 92 , e ) 95 | c | subtract(subtract(100, const_3), 9) | subtract(n1,const_3)|subtract(#0,n0)| | general |
the duplicate ratio of 6 : 4 is ? | "36 ^ 2 : 4 ^ 2 = 36 : 16 = 9 : 4 answer : b" | a ) 1 : 2 , b ) 9 : 4 , c ) 1 : 8 , d ) 1 : 18 , e ) 1 : 13 | b | divide(power(6, const_2), power(4, const_2)) | power(n0,const_2)|power(n1,const_2)|divide(#0,#1)| | other |
find 4 + 7 + 10 + 13 + 16 + . . . up to 20 terms | a = 4 d = 7 – 4 = 3 sum of first 20 terms , s 20 answer is a | a ) 650 , b ) 640 , c ) 630 , d ) 620 , e ) 610 | a | add(add(multiply(20, 16), multiply(13, 10)), multiply(const_10, subtract(multiply(7, 4), subtract(10, const_2)))) | multiply(n4,n5)|multiply(n2,n3)|multiply(n0,n1)|subtract(n2,const_2)|add(#0,#1)|subtract(#2,#3)|multiply(#5,const_10)|add(#4,#6) | general |
there are 10 slate rocks , 11 pumice rocks , and 4 granite rocks randomly distributed in a certain field . if 2 rocks are chosen at random and without replacement , what is the probability that both rocks will be slate rocks ? | "10 / 25 * 9 / 24 = 3 / 20 the answer is a ." | a ) 3 / 20 , b ) 5 / 26 , c ) 7 / 34 , d ) 9 / 37 , e ) 11 / 49 | a | multiply(divide(10, add(add(10, 11), 4)), divide(subtract(10, const_1), subtract(add(add(10, 11), 4), const_1))) | add(n0,n1)|subtract(n0,const_1)|add(n2,#0)|divide(n0,#2)|subtract(#2,const_1)|divide(#1,#4)|multiply(#3,#5)| | other |
if 100 % of j is equal to 25 % of k , 150 % of k is equal to 50 % of l , and 125 % of l is equal to 75 % of m , then 20 % of m is equal to what percent of 200 % of j ? | "imo answer should be 350 . . . consider j = 10 , then k = 50 , l = 150 and m = 350 . . . . 20 % of 350 , comes out to be 70 . . . . 200 % of 10 is 20 . . . . ( 70 * 100 ) / 20 = 350 . . . . ans : c" | a ) 0.35 , b ) 3.5 , c ) 350 , d ) 35 , e ) 3500 | c | multiply(divide(multiply(divide(multiply(multiply(100, 150), 125), multiply(multiply(25, 50), 75)), 20), 200), const_100) | multiply(n0,n2)|multiply(n1,n3)|multiply(n4,#0)|multiply(n5,#1)|divide(#2,#3)|multiply(n6,#4)|divide(#5,n7)|multiply(#6,const_100)| | gain |
a room of 5 m 44 cm long and 3 m 74 cm broad is to be paved with squre tiles . find the least number of squre tiles required to cover the floor . | "explanation : area of the room = 544 * 374 sq cm size of largest square tile = h . c . f of 544 cm and 374 cm = 34 cm area of 1 tile = 34 * 34 sq cm no . of tiles required = ( 544 * 374 ) / ( 34 * 34 ) = 176 answer : d ) 176" | a ) 233 , b ) 877 , c ) 279 , d ) 176 , e ) 291 | d | divide(multiply(add(multiply(5, const_100), 44), add(multiply(3, const_100), 74)), power(divide(add(multiply(5, const_100), 44), power(const_2, const_4)), const_2)) | multiply(n0,const_100)|multiply(n2,const_100)|power(const_2,const_4)|add(n1,#0)|add(n3,#1)|divide(#3,#2)|multiply(#3,#4)|power(#5,const_2)|divide(#6,#7)| | physics |
5 % of men have unlisted phone numbers and 4 % of women have unlisted phone numbers . population ratio of men : women in the city of gpuzzles is 4 : 3 . if we select randomly 100 people from the phone directory , can you tell us how many people selected will definitely have unlisted phone numbers ? | solution : 0 % explanation : in the number is in the phone directory , it means the number is listed and therefore if we pick any number from phone directory than number is surely listed . answer a | a ) 0 % , b ) 1 % , c ) 2 % , d ) 3 % , e ) 4 % | a | add(multiply(multiply(const_0_25, const_100), divide(4, const_100)), multiply(multiply(divide(3, 4), const_100), divide(5, const_100))) | divide(n1,const_100)|divide(n0,const_100)|divide(n3,n1)|multiply(const_0_25,const_100)|multiply(#0,#3)|multiply(#2,const_100)|multiply(#1,#5)|add(#4,#6) | other |
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