Problem stringlengths 5 967 | Rationale stringlengths 1 2.74k | options stringlengths 37 300 | correct stringclasses 5
values | annotated_formula stringlengths 7 6.48k | linear_formula stringlengths 8 925 | category stringclasses 6
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the length of the rectangular field is double its width . inside the field there is square shaped pond 8 m long . if the area of the pond is 1 / 18 of the area of the field . what is the length of the field ? | "explanation : a / 18 = 8 * 8 = > a = 8 * 8 * 18 x * 2 x = 8 * 8 * 18 x = 24 = > 2 x = 48 answer : option e" | a ) 23 , b ) 32 , c ) 34 , d ) 43 , e ) 48 | e | sqrt(divide(multiply(square_area(8), 18), inverse(const_2))) | inverse(const_2)|square_area(n0)|multiply(n2,#1)|divide(#2,#0)|sqrt(#3)| | geometry |
if jake loses 8 pounds , he will weigh twice as much as his sister . together they now weigh 290 pounds . what is jake ’ s present weight , in pounds ? | "lets say j is the weight of jack and s is the wt of his sister . if he loses 8 pounds , he s twice as heavy as his sister . j - 8 = 2 * s also , together they weight 290 pounds j + s = 290 solvong the 2 equation , we get j = 196 pounds ! d" | a ) 131 , b ) 135 , c ) 169 , d ) 196 , e ) 212 | d | add(multiply(divide(subtract(290, 8), const_3), const_2), 8) | subtract(n1,n0)|divide(#0,const_3)|multiply(#1,const_2)|add(n0,#2)| | general |
how many integers from 0 to 59 , inclusive , have a remainder of 1 when divided by 3 ? | "my ans is also c . 17 . explanation : 1 also gives 1 remainder when divided by 3 , another number is 4 , then 7 and so on . hence we have an arithmetic progression : 1 , 4 , 7 , 10 , . . . . . 58 , which are in the form 3 n + 1 . now we have to find out number of terms . tn = a + ( n - 1 ) d , where tn is the nth term... | a ) 15 , b ) 16 , c ) 17 , d ) 18 , e ) 20 | e | divide(59, const_10) | divide(n1,const_10)| | general |
david works at a science lab that conducts experiments on bacteria . the population of the bacteria multiplies at a constant rate , and his job is to notate the population of a certain group of bacteria each hour . at 1 p . m . on a certain day , he noted that the population was 2,500 and then he left the lab . he retu... | "let the rate be x , then population of the bacteria after each hour can be given as 2500 , 2500 x , 2500 ( x ^ 2 ) , 2500 ( x ^ 3 ) now population at 4 pm = 160,000 thus we have 2500 ( x ^ 3 ) = 160,000 = 64 thus x = 4 therefore population at 3 pm = 2500 ( 16 ) = 40,000 answer : b" | a ) 45,000 , b ) 40,000 , c ) 50,000 , d ) 55,000 , e ) 60,000 | b | multiply(multiply(power(divide(multiply(multiply(2, 4), 2,500), 2,500), const_0_33), 2,500), power(divide(multiply(multiply(2, 4), 2,500), 2,500), const_0_33)) | multiply(n2,n4)|multiply(n1,#0)|divide(#1,n1)|power(#2,const_0_33)|multiply(n1,#3)|multiply(#4,#3)| | physics |
on a trip covering 850 km , a bicyclist travels the first 400 km at 20 km per hour and the remainder of the distance at 15 km per hour . what is the average speed , in km per hour , for the entire trip ? | "time 1 = 400 / 20 = 20 hours time 2 = 450 / 15 = 30 hours total time = 50 hours average speed = 850 / 50 = 17 km / hr the answer is c ." | a ) 16.0 , b ) 16.5 , c ) 17.0 , d ) 17.5 , e ) 18.0 | c | speed(850, add(divide(400, 20), divide(subtract(850, 400), 15))) | divide(n1,n2)|subtract(n0,n1)|divide(#1,n3)|add(#0,#2)|speed(n0,#3)| | physics |
the manager of a produce market purchased a quantity of tomatoes for $ 0.80 per pound . due to improper handling , 15 percent of the tomatoes , by weight , were ruined and discarded . at what price per pound should the manager sell the remaining tomatoes if she wishes to make a profit on the sale of the tomatoes equal ... | "assume the manager bought 100 tomatoes . cost price = 80 given : 15 % are damaged - - > available tomatoes to sell = 85 85 * x - 80 = 0.08 * 80 85 x - 80 = 6.4 85 x = 86.64 x = 86.64 / 85 = 87 / 85 ( approx ) = 1.023 x is slightly under 1.023 = 1.02 answer : d" | a ) $ 0.94 , b ) $ 0.96 , c ) $ 0.98 , d ) $ 1.02 , e ) $ 1.20 | d | multiply(add(const_1, divide(8, const_100)), divide(0.80, divide(subtract(const_100, 15), const_100))) | divide(n2,const_100)|subtract(const_100,n1)|add(#0,const_1)|divide(#1,const_100)|divide(n0,#3)|multiply(#2,#4)| | gain |
the greatest number which on dividing 1642 and 1856 leaves remainders 6 and 4 respectively , is : | "explanation : required number = h . c . f . of ( 1642 - 6 ) and ( 1856 - 4 ) = h . c . f . of 1636 and 1852 = 4 . answer : d" | a ) 123 , b ) 127 , c ) 235 , d ) 4 , e ) 505 | d | gcd(subtract(1856, 4), subtract(1642, 6)) | subtract(n1,n3)|subtract(n0,n2)|gcd(#0,#1)| | general |
marcella has 27 pairs of shoes . if she loses 9 individual shoes , what is the greatest number of matching pairs she could have left ? | "marcella has 27 pairs of shoes and loses 9 shoes . to minimize the loss of identical pairs of shoes we want marcella to lose as many identical pairs as possible . this would yield 4 identical pairs and 1 additional shoe ( destroying 5 pairs of shoes ) . the 27 pairs of shoes minus the 5 ' destroyed ' pairs yields 22 p... | a ) 22 , b ) 20 , c ) 19 , d ) 16 , e ) 15 | a | subtract(27, add(floor(divide(9, const_2)), const_1)) | divide(n1,const_2)|floor(#0)|add(#1,const_1)|subtract(n0,#2)| | general |
how many 3 - digit numbers can be formed from the digits 2 , 3 , 5 , 6 , 7 and 9 , which are divisible by 5 and none of the digits is repeated ? | since each desired number is divisible by 5 , so we must have 5 at the unit place . so , there is 1 way of doing it . the tens place can now be filled by any of the remaining 5 digits ( 2 , 3 , 6 , 7 , 9 ) . so , there are 5 ways of filling the tens place . the hundreds place can now be filled by any of the remaining 4... | a ) 5 , b ) 4 , c ) 20 , d ) 25 , e ) 30 | c | multiply(5, const_4) | multiply(n3,const_4) | general |
a man invests some money partly in 12 % stock at 105 and partly in 8 % stock at 88 . to obtain equal dividends from both , he must invest the money in the ratio . | in case of stock 1 , if he invest rs . 105 , he will get a dividend of rs . 12 ( assume face value = 100 ) in case of stock 2 , if he invest rs . 88 , he will get a dividend of rs . 8 ( assume face value = 100 ) ie , if he invest rs . ( 88 * 12 ) / 8 , he will get a dividend of rs . 12 required ratio = 105 : ( 88 × 12 ... | a ) 35 : 44 , b ) 34 : 44 , c ) 22 : 44 , d ) 20 : 40 , e ) 50 : 45 | a | divide(multiply(105, const_2), multiply(88, const_3)) | multiply(n1,const_2)|multiply(n3,const_3)|divide(#0,#1) | other |
a woman invested $ 1,000 , part at 5 % and the rest at 6 % . her total investment with interest at the end of the year was $ 1,055 . how much did she invest at 5 % ? | "et x be the portion invested at 5 % and let ( 1 - x ) be the rest which is invested at 6 % the question states that the return after 1 year is ( 1055 / 1000 ) - 1 = 0.055 = 5.5 % we want to find the dollar amount invested in x using our defined variables , put together the equation and solve for x ( the percentage of ... | a ) $ 500 , b ) $ 600 , c ) $ 700 , d ) $ 900 , e ) $ 950 | a | subtract(multiply(multiply(5, const_2), const_100), subtract(multiply(subtract(add(multiply(multiply(5, const_2), const_100), add(multiply(multiply(5, const_2), 5), const_4)), multiply(multiply(5, const_2), const_100)), const_100), multiply(5, multiply(multiply(5, const_2), const_100)))) | multiply(n1,const_2)|multiply(#0,const_100)|multiply(n1,#0)|add(#2,const_4)|multiply(n1,#1)|add(#3,#1)|subtract(#5,#1)|multiply(#6,const_100)|subtract(#7,#4)|subtract(#1,#8)| | gain |
three numbers are in the ratio of 2 : 3 : 4 and their l . c . m . is 288 . what is their h . c . f . ? | "let the numbers be 2 x , 3 x , and 4 x . lcm of 2 x , 3 x and 4 x is 12 x . 12 x = 288 x = 24 hcf of 2 x , 3 x and 4 x = x = 24 the answer is b ." | a ) 18 , b ) 24 , c ) 36 , d ) 42 , e ) 48 | b | multiply(2, 4) | multiply(n0,n2)| | other |
a merchant sells an item at a 10 % discount , but still makes a gross profit of 20 percent of the cost . what percent of the cost would the gross profit on the item have been if it had been sold without the discount ? | original sp = x cost = c current selling price = . 9 x ( 10 % discount ) . 9 x = 1.2 c ( 20 % profit ) x = 1.2 / . 9 * c x = 4 / 3 c original selling price is 1.3 c which is 30 % profit answer a | a ) 30 % , b ) 40 % , c ) 50 % , d ) 60 % , e ) 75 % | a | subtract(const_100, subtract(subtract(const_100, 10), 20)) | subtract(const_100,n0)|subtract(#0,n1)|subtract(const_100,#1) | gain |
what decimal fraction is 70 ml of a litre ? | "answer required fraction = 70 / 1000 = 7 / 100 = . 07 correct option : b" | a ) . 7 , b ) . 07 , c ) . 05 , d ) 0.07 , e ) none of these | b | divide(70, const_1000) | divide(n0,const_1000)| | physics |
the average age of applicants for a new job is 20 , with a standard deviation of 8 . the hiring manager is only willing to accept applicants whose age is within one standard deviation of the average age . assuming that all applicants ' ages are integers and that the endpoints of the range are included , what is the max... | "minimum age = average - 1 standard deviation = 20 - 8 = 12 maximum age = average + 1 standard deviation = 20 + 8 = 28 maximum number of different ages of the applicants = 28 - 12 + 1 = 17 answer c" | a ) 8 , b ) 16 , c ) 17 , d ) 18 , e ) 34 | c | add(subtract(add(20, 8), subtract(20, 8)), const_1) | add(n0,n1)|subtract(n0,n1)|subtract(#0,#1)|add(#2,const_1)| | general |
p , q and r have $ 7000 among themselves . r has two - thirds of the total amount with p and q . find the amount with r ? | a 2800 let the amount with r be $ r r = 2 / 3 ( total amount with p and q ) r = 2 / 3 ( 7000 - r ) = > 3 r = 14000 - 2 r = > 5 r = 14000 = > r = 2800 . | a ) 2800 , b ) 2403 , c ) 3998 , d ) 2539 , e ) 1930 | a | divide(multiply(7000, multiply(const_2, const_2)), add(add(multiply(divide(multiply(const_2, const_2), const_3), const_3), multiply(const_1, const_3)), multiply(const_1, const_3))) | multiply(const_2,const_2)|multiply(const_1,const_3)|divide(#0,const_3)|multiply(n0,#0)|multiply(#2,const_3)|add(#4,#1)|add(#5,#1)|divide(#3,#6) | general |
in the xy - plane the point ( - 2 - 3 ) is the centre of a circle , the point ( - 2 , 2 ) lies inside the circle and the point ( 4 , - 3 ) lies outside the circle . if the radius r of the circle r is an integer then r = | "can be solved without much calculations . you are given that ( - 2 , - 3 ) is the center of the circle . point ( 4 , - 3 ) lies inside the circle - - - > the radius is lesser than distance of ( - 2 , - 3 ) from ( 4 , - 3 ) - - - > lesser than 6 units but the radius will also be greater than the distance of ( - 2 , - 3... | a ) 6 , b ) 5.5 , c ) 4 , d ) 3 , e ) 2 | b | divide(add(subtract(4, negate(2)), 4), 2) | negate(n0)|subtract(n4,#0)|add(n4,#1)|divide(#2,n0)| | general |
unit size cubes are stacked inside a big rectangular box having dimensions corresponding to three consecutive multiples of 5 . choose exact number of cubes that can completely fill the box . | let the dimensions of the box be , length = 5 * a , breadth = 5 * ( a + 1 ) , height = 5 * ( a + 2 ) hence , volume = 5 * 5 * 5 * a * ( a + 1 ) * ( a + 2 ) among any 3 consecutive positive integers , we will either have ( a number that is divisible by both 23 ) or ( a number divisible by 2 and another number divisible ... | ['a ) 750', 'b ) 850', 'c ) 950', 'd ) 1050', 'e ) none of the above'] | a | divide(volume_rectangular_prism(5, multiply(5, const_2), multiply(5, const_3)), volume_cube(const_1)) | multiply(n0,const_2)|multiply(n0,const_3)|volume_cube(const_1)|volume_rectangular_prism(n0,#0,#1)|divide(#3,#2) | geometry |
p and q started a business investing rs 85000 and rs 15000 resp . in what ratio the profit earned after 2 years be divided between p and q respectively . | "explanation : in this type of question as time frame for both investors is equal then just get the ratio of their investments . p : q = 85000 : 15000 = 85 : 15 = 17 : 3 option b" | a ) 17 : 5 , b ) 17 : 3 , c ) 17 : 6 , d ) 17 : 7 , e ) 17 : 8 | b | divide(85000, 15000) | divide(n0,n1)| | gain |
in one alloy there is 15 % chromium while in another alloy it is 8 % . 15 kg of the first alloy was melted together with 35 kg of the second one to form a third alloy . find the percentage of chromium in the new alloy . | "the amount of chromium in the new 15 + 35 = 50 kg alloy is 0.15 * 15 + 0.08 * 35 = 5.05 kg , so the percentage is 5.05 / 50 * 100 = 10.1 % . answer : c ." | a ) 8.8 % , b ) 9 % , c ) 10.1 % , d ) 8.6 % , e ) 8.4 % | c | multiply(divide(add(divide(multiply(15, 15), const_100), divide(multiply(8, 35), const_100)), add(15, 35)), const_100) | add(n2,n3)|multiply(n0,n2)|multiply(n1,n3)|divide(#1,const_100)|divide(#2,const_100)|add(#3,#4)|divide(#5,#0)|multiply(#6,const_100)| | gain |
working alone , sawyer finishes cleaning half the house in a third of the time it takes nick to clean the entire house alone . sawyer alone cleans the entire house in 6 hours . how many hours will it take nick and sawyer to clean the entire house if they work together ? | answer is 3.6 hours . sawyer does the complete house in 6 hours while nick does it in 9 hours . 1 / ( 1 / 6 + 1 / 9 ) = 3.6 answer is e | a ) 1.5 , b ) 2 , c ) 2.4 , d ) 3 , e ) 3.6 | e | inverse(add(divide(const_1, 6), divide(const_1, multiply(divide(6, const_2), const_3)))) | divide(const_1,n0)|divide(n0,const_2)|multiply(#1,const_3)|divide(const_1,#2)|add(#0,#3)|inverse(#4) | physics |
if x is 30 percent greater than 88 , then x = | "x = 88 * 1.3 = 114.4 so the answer is d ." | a ) 68 , b ) 70.4 , c ) 86 , d ) 114.4 , e ) 108 | d | add(88, multiply(divide(30, const_100), 88)) | divide(n0,const_100)|multiply(n1,#0)|add(n1,#1)| | general |
find value for x from below equation ? x + 1 = 5 | 1 . subtract 1 from both sides : x + 1 - 1 = 5 - 1 2 . simplify both sides : x = 4 c | a ) - 5 , b ) - 4 , c ) 4 , d ) 3 , e ) 2 | c | subtract(5, 1) | subtract(n1,n0) | general |
what least number should be added to 1015 , so that the sum is completely divisible by 25 ? | "1015 ã · 25 = 40 with remainder = 15 15 + 10 = 25 . hence 10 should be added to 1015 so that the sum will be divisible by 25 answer : option d" | a ) 4 , b ) 3 , c ) 2 , d ) 10 , e ) 5 | d | subtract(25, reminder(1015, 25)) | reminder(n0,n1)|subtract(n1,#0)| | general |
the population of a city increases @ 4 % p . a . there is an additional annual increase of 4 % of the population due to the influx of job seekers , find the % increase in population after 2 years ? | total annual increament in population = 4 + 4 = 8 % let the population be x population after 2 years = 1.08 x + . 0864 x population increase = 1.08 x + . 0864 x - x % increase = ( ( 1.08 x + . 0864 x - x ) / x ) * 100 = ( 1.08 + . 0864 - 1 ) * 100 = . 1664 * 100 = 16.64 % answer : c | a ) 14.64 % , b ) 15.64 % , c ) 16.64 % , d ) 17.64 % , e ) 18.64 % | c | multiply(add(4, 4), 2) | add(n0,n0)|multiply(n2,#0) | general |
the sum of even numbers between 1 and 31 is : | explanation : let sn = ( 2 + 4 + 6 + . . . + 30 ) . this is an a . p . in which a = 2 , d = 2 and l = 30 let the number of terms be n . then a + ( n - 1 ) d = 30 = > 2 + ( n - 1 ) x 2 = 30 n = 15 . sn = n / 2 ( a + l ) = 15 / 2 x ( 2 + 30 ) = ( 15 x 16 ) = 240 . answer : a | a ) 240 , b ) 789 , c ) 520 , d ) 879 , e ) 456 | a | multiply(add(31, 1), divide(floor(divide(31, const_2)), const_2)) | add(n0,n1)|divide(n1,const_2)|floor(#1)|divide(#2,const_2)|multiply(#0,#3) | general |
a whale goes on a feeding frenzy that lasts for 9 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... | "suppose food eaten in first hour : x the ap is : x , x + 3 , x + 6 , . . . . [ number of terms ' n ' = 9 ] therefore , 9 / 2 [ 2 x + ( 9 - 1 ) * 3 ] = 450 . solving for x , x = 38 . now , 6 th term will be : x + ( 6 - 1 ) * d = 38 + 5 * 3 = 53 . hence d ! !" | a ) 38 , b ) 47 , c ) 50 , d ) 53 , e ) 62 | d | add(divide(450, 9), 3) | divide(n2,n0)|add(n1,#0)| | general |
the average weight of 8 person ' s increases by 2.5 kg when a new person comes in place of one of them weighing 55 kg . what might be the weight of the new person ? | "c 75 kg total weight increased = ( 8 x 2.5 ) kg = 20 kg . weight of new person = ( 55 + 20 ) kg = 75 kg ." | a ) 56 kg , b ) 90 kg , c ) 75 kg , d ) data inadequate , e ) none of these | c | add(multiply(8, 2.5), 55) | multiply(n0,n1)|add(n2,#0)| | general |
income and expenditure of a person are in the ratio 5 : 4 . if the income of the person is rs . 17000 , then find his savings ? | "let the income and the expenditure of the person be rs . 5 x and rs . 4 x respectively . income , 5 x = 17000 = > x = 3400 savings = income - expenditure = 5 x - 4 x = x so , savings = rs . 3400 . answer : b" | a ) 3600 , b ) 3400 , c ) 3608 , d ) 3602 , e ) 3603 | b | subtract(17000, multiply(divide(4, 5), 17000)) | divide(n1,n0)|multiply(n2,#0)|subtract(n2,#1)| | other |
a store reduced the price of all items in the store by 8 % on the first day and by another 10 % on the second day . the price of items on the second day was what percent of the price before the first reduction took place ? | consider price of the all items as $ 100 after a initial reduction of 8 % price becomes = 0.92 * 100 = $ 92 after the final reduction of 10 % price becomes = 0.9 * 92 = $ 82.8 price of all items on second day is 82.8 % of price on first day correct answer option e | a ) 80.0 , b ) 80.9 , c ) 81.0 , d ) 81.1 , e ) 82.8 | e | multiply(multiply(divide(subtract(const_100, 8), const_100), divide(subtract(const_100, 10), const_100)), const_100) | subtract(const_100,n0)|subtract(const_100,n1)|divide(#0,const_100)|divide(#1,const_100)|multiply(#2,#3)|multiply(#4,const_100) | gain |
out of 470 students of a school , 325 play football , 175 play cricket and 50 neither play football nor cricket . how many students play both football and cricket ? | "n ( a ) = 325 , n ( b ) = 175 , n ( aub ) = 470 - 50 = 420 . required number = n ( anb ) = n ( a ) + n ( b ) - n ( aub ) = 325 + 175 - 420 = 80 . answer is a" | a ) 80 , b ) 150 , c ) 100 , d ) 180 , e ) 220 | a | subtract(add(175, 325), subtract(470, 50)) | add(n1,n2)|subtract(n0,n3)|subtract(#0,#1)| | other |
in a class of 60 students 41 are taking french , 22 are taking german . of the students taking french or german , 9 are taking both courses . how many students are not enrolled in either course ? | "formula for calculating two overlapping sets : a + b - both + not ( a or b ) = total so in our task we have equation : 41 ( french ) + 22 ( german ) - 9 ( both ) + not = 60 54 + not = 60 not = 60 - 54 = 6 so answer is a" | a ) 6 , b ) 15 , c ) 24 , d ) 33 , e ) 54 | a | subtract(60, subtract(add(41, 22), 9)) | add(n1,n2)|subtract(#0,n3)|subtract(n0,#1)| | other |
tom and linda stand at point a . linda begins to walk in a straight line away from tom at a constant rate of 4 miles per hour . one hour later , tom begins to jog in a straight line in the exact opposite direction at a constant rate of 9 miles per hour . if both tom and linda travel indefinitely , what is the positive ... | e is the answer . . . . d = ts where d = distance , t = time and s = speed to travel half distance , ( 2 + 4 t ) = 9 t = = > t = 2 / 5 = = > 24 minutes to travel double distance , 2 ( 2 + 4 t ) = 9 t = = > 4 = = > 240 minutes difference , 216 minutes e | a ) 60 , b ) 72 , c ) 84 , d ) 90 , e ) 216 | e | multiply(subtract(divide(multiply(const_2, const_2), subtract(9, multiply(const_2, 4))), divide(const_2, subtract(9, 4))), const_60) | multiply(const_2,const_2)|multiply(n0,const_2)|subtract(n1,n0)|divide(const_2,#2)|subtract(n1,#1)|divide(#0,#4)|subtract(#5,#3)|multiply(#6,const_60) | physics |
in 1998 the profits of company n were 10 percent of revenues . in 1999 , the revenues of company n fell by 30 percent , but profits were 10 percent of revenues . the profits in 1999 were what percent of the profits in 1998 ? | "0,07 r = x / 100 * 0.1 r answer a" | a ) 70 % , b ) 105 % , c ) 120 % , d ) 124.2 % , e ) 138 % | a | multiply(divide(multiply(subtract(const_1, divide(30, const_100)), divide(10, const_100)), divide(10, const_100)), const_100) | divide(n4,const_100)|divide(n3,const_100)|divide(n1,const_100)|subtract(const_1,#1)|multiply(#0,#3)|divide(#4,#2)|multiply(#5,const_100)| | gain |
p is able to do a piece of work in 15 days and q can do the same work in 20 days . if they can work together for 4 days , what is the fraction of work left ? | explanation : p ' s 1 - day work = 1 / 15 q ' s 1 - day work = 1 / 20 work done by ( p + q ) in 1 day = 1 / 15 + 1 / 20 = 7 / 60 . work done by them in 4 days = ( 7 / 60 ) * 4 = 7 / 15 . work left = 1 - ( 7 / 15 ) = 8 / 15 . answer is a | a ) 8 / 15 , b ) 7 / 15 , c ) 1 / 15 , d ) 3 / 15 , e ) none of these | a | subtract(const_1, multiply(add(divide(const_1, 20), divide(const_1, 15)), 4)) | divide(const_1,n1)|divide(const_1,n0)|add(#0,#1)|multiply(n2,#2)|subtract(const_1,#3) | physics |
of the five - digit positive integers that have no digits equal to zero , how many have two digits that are equal to each other and the remaining digit different from the other two ? | "of the five - digit positive integers that have no digits equal to zero , how many have two digits that are equal to each other and the remaining digit different from the other two ? a . 24 b . 36 c . 72 d . 144 e . 216 choosing the digit for p - 9 ways ; choosing the digit for q - 8 ways ; choosing the digit for r - ... | a ) 24440 , b ) 36440 , c ) 72440 , d ) 181440 , e ) 216440 | d | multiply(multiply(subtract(const_10, const_1), subtract(const_10, const_2)), const_1) | subtract(const_10,const_1)|subtract(const_10,const_2)|multiply(#0,#1)|multiply(#2,const_1)| | general |
when processing flower - nectar into honey bees ' extract , a considerable amount of water gets reduced . how much flower - nectar must be processed to yield 1 kg of honey , if nectar contains 50 % water , and the honey obtained from this nectar contains 30 % water ? | "explanation : flower - nectar contains 50 % of non - water part . in honey this non - water part constitutes 70 % ( 100 - 30 ) . therefore 0.5 x amount of flower - nectar = 0.70 x amount of honey = 0.70 x 1 kg therefore amount of flower - nectar needed = ( 0.70 / 0.51 ) kg = 1.4 kgs answer : c" | a ) 1.2 kg , b ) 1.5 kg , c ) 1.4 kg , d ) 1.9 kg , e ) none of these | c | divide(divide(subtract(const_100, 30), const_100), divide(50, const_100)) | divide(n1,const_100)|subtract(const_100,n2)|divide(#1,const_100)|divide(#2,#0)| | general |
for any integer m greater than 1 , $ m denotes the product of all the integers from 1 to m , inclusive . how many prime numbers are there between $ 7 + 2 and $ 7 + 10 , inclusive ? | $ is basically a factorial of a number . so , we are asked to find the number of primes between 7 ! + 2 and 7 ! + 10 , inclusive . from each number 7 ! + k were 2 ≤ k ≤ 102 ≤ k ≤ 10 we can factor out k , thus there are no pries in the given range . for example : 7 ! + 2 = 2 ( 3 * 4 * 5 * 6 * 7 + 1 ) - - > a multiple of... | a ) none , b ) one , c ) two , d ) three , e ) four | a | add(subtract(add(add(add(1, 1), 2), 10), add(7, 7)), const_2) | add(n0,n0)|add(n2,n2)|add(n3,#0)|add(n5,#2)|subtract(#3,#1)|add(#4,const_2) | general |
excluding stoppages , the speed of a train is 50 kmph and including stoppages it is 30 kmph . of how many minutes does the train stop per hour ? | "explanation : t = 20 / 50 * 60 = 24 answer : option d" | a ) 82 , b ) 17 , c ) 12 , d ) 24 , e ) 18 | d | subtract(const_60, multiply(const_60, divide(30, 50))) | divide(n1,n0)|multiply(#0,const_60)|subtract(const_60,#1)| | physics |
find the constant k so that : - x 2 - ( k + 9 ) x - 8 = - ( x - 2 ) ( x - 4 ) | - x 2 - ( k + 9 ) x - 8 = - ( x - 2 ) ( x - 4 ) : given - x 2 - ( k + 9 ) x - 8 = - x 2 + 6 x - 8 - ( k + 9 ) = 6 : two polynomials are equal if their corresponding coefficients are equal . k = - 15 : solve the above for k correct answer c | a ) 11 , b ) 12 , c ) 15 , d ) 14 , e ) 19 | c | add(9, add(4, 2)) | add(n0,n4)|add(n1,#0) | general |
a 230 m long train running at the speed of 120 km / hr crosses another train running in opposite direction at the speed of 80 km / hr in 9 sec . what is the length of the other train ? | "relative speed = 120 + 80 = 200 km / hr . = 200 * 5 / 18 = 500 / 9 m / sec . let the length of the other train be x m . then , ( x + 2340 ) / 9 = 500 / 9 = > x = 270 . answer : b" | a ) 230 , b ) 270 , c ) 260 , d ) 256 , e ) 298 | b | subtract(multiply(multiply(add(120, 80), const_0_2778), 9), 230) | add(n1,n2)|multiply(#0,const_0_2778)|multiply(n3,#1)|subtract(#2,n0)| | physics |
a candidate who gets 40 % of the marks fails by 40 marks . but another candidate who gets 60 % marks gets 20 marks more than necessary for passing . find the number of marks for passing ? | "40 % - - - - - - - - - - - - 40 60 % - - - - - - - - - - - - 20 - - - - - - - - - - - - - - - - - - - - - - 20 % - - - - - - - - - - - - - 60 40 % - - - - - - - - - - - - - - ? 120 + 40 = 160 marks answer : c" | a ) 100 marks , b ) 200 marks , c ) 160 marks , d ) 371 marks , e ) 827 marks | c | add(multiply(divide(add(40, 20), subtract(divide(60, const_100), divide(40, const_100))), divide(40, const_100)), 40) | add(n1,n3)|divide(n2,const_100)|divide(n0,const_100)|subtract(#1,#2)|divide(#0,#3)|multiply(#4,#2)|add(n1,#5)| | gain |
two numbers are in the ratio of 1 : 2 . if 10 be added to both , their ratio changes to 3 : 5 . the greater number is | "let the ratio be x : y , given x / y = 1 / 2 , ( x + 10 ) / ( y + 10 ) = 3 / 5 = > x = 20 and y = 40 answer : e" | a ) 20 , b ) 24 , c ) 28 , d ) 32 , e ) 40 | e | multiply(1, 10) | multiply(n0,n2)| | other |
the length of the bridge , which a train 100 metres long and travelling at 45 km / hr can cross in 30 seconds , is : | "speed = [ 45 x 5 / 18 ] m / sec = [ 25 / 2 ] m / sec time = 30 sec let the length of bridge be x metres . then , ( 100 + x ) / 30 = 25 / 2 = > 2 ( 100 + x ) = 750 = > x = 275 m . answer : option e" | a ) 230 , b ) 240 , c ) 245 , d ) 250 , e ) 275 | e | subtract(multiply(divide(multiply(45, speed(const_1000, const_1)), speed(const_3600, const_1)), 30), 100) | speed(const_1000,const_1)|speed(const_3600,const_1)|multiply(n1,#0)|divide(#2,#1)|multiply(n2,#3)|subtract(#4,n0)| | physics |
a train running at the speed of 90 km / hr crosses a pole in 9 sec . what is the length of the train ? | "speed = 90 * 5 / 18 = 25 m / sec length of the train = speed * time = 25 * 9 = 225 m answer : b" | a ) 288 , b ) 225 , c ) 277 , d ) 272 , e ) 150 | b | multiply(divide(multiply(90, const_1000), const_3600), 9) | multiply(n0,const_1000)|divide(#0,const_3600)|multiply(n1,#1)| | physics |
rs . 850 is divided so that 4 times the first share , twice the 2 nd share and twice the third share amount to the same . what is the value of the first share ? | "a + b + c = 850 4 a = 2 b = 2 c = x a : b : c = 1 / 4 : 1 / 2 : 1 / 2 = 1 : 2 : 2 1 / 5 * 850 = rs 170 answer : e" | a ) s 400 , b ) s 500 , c ) s 850 , d ) s 540 , e ) s 170 | e | multiply(4, divide(850, add(add(4, 2), const_3))) | add(n1,n2)|add(#0,const_3)|divide(n0,#1)|multiply(n1,#2)| | general |
we bought 85 hats at the store . blue hats cost $ 6 and green hats cost $ 7 . the total price was $ 550 . how many green hats did we buy ? | "let b be the number of blue hats and let g be the number of green hats . b + g = 85 . b = 85 - g . 6 b + 7 g = 550 . 6 ( 85 - g ) + 7 g = 550 . 510 - 6 g + 7 g = 550 . g = 550 - 510 = 40 . the answer is b ." | a ) a ) 36 , b ) b ) 40 , c ) c ) 41 , d ) d ) 42 , e ) e ) 44 | b | subtract(550, multiply(85, 6)) | multiply(n0,n1)|subtract(n3,#0)| | general |
solve for x and check : 6 x = 54 | "solution : dividing each side by 6 , we obtain ( 6 x / 6 ) = ( 54 / 6 ) therefore : x = 9 check : 6 x = 54 ( 6 * 9 ) = 54 54 = 54 answer : d" | a ) 12 , b ) 15 , c ) 5 , d ) 9 , e ) none of these | d | multiply(6, divide(54, 6)) | divide(n1,n0)|multiply(n0,#0)| | general |
jackie has two solutions that are 2 percent sulfuric acid and 12 percent sulfuric acid by volume , respectively . if these solutions are mixed in appropriate quantities to produce 60 liters of a solution that is 5 percent sulfuric acid , approximately how many liters of the 12 percent solution will be required ? | let a = amount of 2 % acid and b = amount of 12 % acid . now , the equation translates to , 0.02 a + . 12 b = . 05 ( a + b ) but a + b = 60 therefore . 02 a + . 12 b = . 05 ( 60 ) = > 2 a + 12 b = 300 but b = 60 - a therefore 2 a + 12 ( 60 - a ) = 300 = > 10 a = 420 hence a = 42 . , b = 60 - 42 = 18 answer : a | a ) 18 , b ) 20 , c ) 24 , d ) 36 , e ) 42 | a | multiply(const_3, divide(60, const_10)) | divide(n2,const_10)|multiply(#0,const_3) | gain |
excluding stoppages , the speed of a train is 45 kmph and including stoppages it is 36 kmph . of how many minutes does the train stop per hour ? | "t = 9 / 45 * 60 12 answer : c" | a ) 16 , b ) 17 , c ) 12 , d ) 16 , e ) 16 | c | subtract(const_60, multiply(const_60, divide(36, 45))) | divide(n1,n0)|multiply(#0,const_60)|subtract(const_60,#1)| | physics |
one - fourth of a number is greater than one - fifth of the number succeeding it by 1 . find the number . | number is 24 as 1 / 4 th of 24 = 6 1 / 5 th of 25 ( 24 + 1 ) = 5 6 = 5 + 1 answer : a | a ) 24 , b ) 42 , c ) 36 , d ) 48 , e ) 50 | a | divide(add(divide(const_2, const_10), 1), subtract(divide(1, const_4), divide(const_2, const_10))) | divide(const_2,const_10)|divide(n0,const_4)|add(n0,#0)|subtract(#1,#0)|divide(#2,#3) | general |
working at their respective constant rates , machine a makes 100 copies in 20 minutes and machine b makes 150 copies in 10 minutes . if these machines work simultaneously at their respective rates for 30 minutes , what is the total number of copies that they will produce ? | "machine a can produce 100 * 30 / 20 = 150 copies and , machine b can produce 150 * 30 / 10 = 450 copies total producing 600 copies . e is the answer" | a ) 250 , b ) 425 , c ) 675 , d ) 700 , e ) 600 | e | multiply(add(divide(100, 20), divide(150, 10)), 30) | divide(n0,n1)|divide(n2,n3)|add(#0,#1)|multiply(n4,#2)| | physics |
a car gets 40 kilometers per gallon of gasoline . how many gallons of gasoline would the car need to travel 160 kilometers ? | "each 40 kilometers , 1 gallon is needed . we need to know how many 40 kilometers are there in 160 kilometers ? 160 / 40 = 4 * 1 gallon = 4 gallons correct answer b" | a ) 8 gallons , b ) 4 gallons , c ) 6 gallons , d ) 5.5 gallons , e ) 10 gallons | b | divide(160, 40) | divide(n1,n0)| | physics |
a vessel of capacity 2 litre has 25 % of alcohol and another vessel of capacity 6 litre had 40 % alcohol . the total liquid of 8 litre was poured out in a vessel of capacity 10 litre and thus the rest part of the vessel was filled with the water . what is the new concentration of mixture ? | 25 % of 2 litres = 0.5 litres 40 % of 6 litres = 2.4 litres therefore , total quantity of alcohol is 2.9 litres . this mixture is in a 10 litre vessel . hence , the concentration of alcohol in this 10 litre vessel is 29 % answer : d | a ) 31 % . , b ) 71 % . , c ) 49 % . , d ) 29 % . , e ) 51 % . | d | multiply(divide(add(multiply(divide(25, const_100), 2), multiply(divide(40, const_100), 6)), 10), const_100) | divide(n1,const_100)|divide(n3,const_100)|multiply(n0,#0)|multiply(n2,#1)|add(#2,#3)|divide(#4,n5)|multiply(#5,const_100) | 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 6 hours , how many minutes did it take cole to drive to work ? | "first round distance travelled ( say ) = d speed = 75 k / h time taken , t 2 = d / 75 hr second round distance traveled = d ( same distance ) speed = 105 k / h time taken , t 2 = d / 105 hr total time taken = 6 hrs therefore , 6 = d / 75 + d / 105 lcm of 75 and 105 = 525 6 = d / 75 + d / 105 = > 6 = 7 d / 525 + 5 d / ... | a ) 84 , b ) 136 , c ) 172 , d ) 210 , e ) 478 | d | multiply(divide(multiply(105, 6), add(75, 105)), const_60) | add(n0,n1)|multiply(n1,n2)|divide(#1,#0)|multiply(#2,const_60)| | physics |
the average of temperatures at noontime from monday to friday is 30 ; the lowest one is 20 , what is the possible maximum range of the temperatures ? | "average = 30 , sum of temperatures = 30 * 5 = 150 as the min temperature is 20 , max would be 150 - 4 * 20 = 30 - - > the range = 30 ( max ) - 20 ( min ) = 10 answer : e" | a ) 20 , b ) 25 , c ) 40 , d ) 45 , e ) 10 | e | subtract(subtract(multiply(30, add(const_2, const_3)), multiply(20, const_4)), 20) | add(const_2,const_3)|multiply(n1,const_4)|multiply(n0,#0)|subtract(#2,#1)|subtract(#3,n1)| | general |
find the c . i . on a sum of rs . 10000 for 6 months at 20 % per annum , interest being compounded halfyearly ? | "c . i . = 10000 ( 11 / 10 ) ^ 1 - 10000 = 1000 answer : c" | a ) 3010 , b ) 2010 , c ) 1000 , d ) 3100 , e ) 2100 | c | subtract(multiply(10000, multiply(add(const_1, divide(const_0_25, const_4)), add(const_1, divide(const_0_25, const_4)))), 10000) | divide(const_0_25,const_4)|add(#0,const_1)|multiply(#1,#1)|multiply(n0,#2)|subtract(#3,n0)| | gain |
if the population of a certain country increases at the rate of one person every 15 seconds , by how many persons does the population increase in 40 minutes ? | "since the population increases at the rate of 1 person every 15 seconds , it increases by 4 people every 60 seconds , that is , by 4 people every minute . thus , in 40 minutes the population increases by 40 x 4 = 160 people . answer . c ." | a ) 80 , b ) 100 , c ) 160 , d ) 240 , e ) 300 | c | multiply(divide(const_60, 15), 40) | divide(const_60,n0)|multiply(n1,#0)| | physics |
the average of 10 result is 42 . average of the first 7 of them is 30 and that of the last 7 is 80 . find the 8 th result ? | "sum of all the 10 results = 10 * 42 = 420 sum of the first 7 of them = 7 * 30 = 210 sum of the last 7 of them = 7 * 80 = 560 so , the 8 th number = 420 + 210 - 560 = 70 . c" | a ) 35 , b ) 37 , c ) 70 , d ) 75 , e ) 50 | c | subtract(add(multiply(7, 30), multiply(7, 80)), multiply(10, 42)) | multiply(n2,n3)|multiply(n2,n5)|multiply(n0,n1)|add(#0,#1)|subtract(#3,#2)| | general |
a certain boxer has agreed to pay his opponent a fee of 3 % of his total purse for every pound over the specified weight limit he weighs in . if the boxer pays his opponent a fee of $ 26250 after weighing in 7 pounds over the specified limit , what was the boxer ' s purse ? | . 03 * 7 = . 21 26250 / . 21 = $ 125,000 answer : c | a ) $ 65,625 , b ) $ 105,000 , c ) $ 125,000 , d ) $ 183,750 , e ) $ 1 , 050,000 | c | subtract(subtract(divide(divide(26250, divide(multiply(7, 3), const_100)), const_1000), const_10), const_10) | multiply(n0,n2)|divide(#0,const_100)|divide(n1,#1)|divide(#2,const_1000)|subtract(#3,const_10)|subtract(#4,const_10) | general |
what is the maximum number q of 27 cubic centimetre cubes that can fit in a rectangular box measuring 8 centimetre x 9 centimetre x 12 centimetre ? | 27 cubic centimetre cubes gives side = 3 cm so if : l * w * h is 9 * 12 * 8 , then max . cube we can have are 3 * 4 * 2 = 24 l * w * h is 9 * 8 * 12 , then max . cube we can have are 3 * 2 * 4 = 24 l * w * h is 12 * 8 * 9 , then max . cube we can have are 4 * 2 * 3 = 24 l * w * h is 12 * 9 * 8 , then max . cube we can ... | ['a ) 36', 'b ) 32', 'c ) 24', 'd ) 21', 'e ) 15'] | c | divide(volume_rectangular_prism(multiply(const_2, const_3), 9, 12), 27) | multiply(const_2,const_3)|volume_rectangular_prism(n2,n3,#0)|divide(#1,n0) | geometry |
- 64 x 29 + 165 = ? | "given exp . = - 64 x ( 30 - 1 ) + 165 = - ( 64 x 30 ) + 64 + 165 = - 1830 + 229 = - 1601 answer is d" | a ) 2436 , b ) 2801 , c ) - 2801 , d ) - 1601 , e ) none of them | d | multiply(subtract(const_1, const_2), subtract(multiply(64, 29), 165)) | multiply(n0,n1)|subtract(const_1,const_2)|subtract(#0,n2)|multiply(#1,#2)| | general |
if ( c - a ) / ( c - b ) = 1 , then ( 5 b - 4 a ) / ( c - a ) = | "let ' s say c = 3 , b = 1 , a = 1 so that our 1 st expression holds true . now , ibsert those numbers in the second expression and we ' ll get 0.5 answer a ( hopefully ) ) )" | a ) 0.5 , b ) 1 , c ) 1.5 , d ) 2 , e ) 2.5 | a | multiply(subtract(5, 4), 4) | subtract(n1,n2)|multiply(n2,#0)| | general |
two trains of length 100 m and 120 m are running towards each other on parallel lines at 42 kmph and 30 kmph respectively . in what time will they be clear of each other from the moment they meet ? | "relative speed = ( 42 + 30 ) * 5 / 18 = 4 * 5 = 20 mps . distance covered in passing each other = 100 + 120 = 220 m . the time required = d / s = 220 / 20 = 11 sec . answer : a" | a ) 11 sec , b ) 70 sec , c ) 21 sec , d ) 20 sec , e ) 19 sec | a | divide(add(100, 120), multiply(add(42, 30), const_0_2778)) | add(n0,n1)|add(n2,n3)|multiply(#1,const_0_2778)|divide(#0,#2)| | physics |
in an examination , a student scores 3 marks for every correct answer and loses 2 marks for every wrong answer . if he attempts all 100 questions and secures 300 marks , the number of questions he attempts correctly , is : | "let the number of correct answers be x . number of incorrect answers = ( 100 â € “ x ) . 3 x â € “ 2 ( 100 â € “ x ) = 300 or 5 x = 500 or x = 100 . answer : e" | a ) 150 , b ) 250 , c ) 400 , d ) 300 , e ) 100 | e | divide(add(300, 100), add(3, 2)) | add(n2,n3)|add(n0,n1)|divide(#0,#1)| | physics |
what is the % change in the area of a rectangle when its length increases by 30 % and its width decreases by 30 % ? | "( 13 / 10 ) * ( 7 / 10 ) = 91 / 100 of original area 91 / 100 is a 9 % decrease from 100 / 100 - > d" | a ) 0 % , b ) 20 % increase , c ) 20 % decrease , d ) 9 % decrease , e ) insufficient data | d | subtract(const_100, divide(multiply(add(const_100, 30), subtract(const_100, 30)), const_100)) | add(n0,const_100)|subtract(const_100,n0)|multiply(#0,#1)|divide(#2,const_100)|subtract(const_100,#3)| | geometry |
what will be the cost of building a fence around a square plot with area equal to 289 sq ft , if the price per foot of building the fence is rs . 58 ? | let the side of the square plot be a ft . a 2 = 289 = > a = 17 length of the fence = perimeter of the plot = 4 a = 68 ft . cost of building the fence = 68 * 58 = rs . 3944 . answer : a | ['a ) rs . 3944', 'b ) rs . 3942', 'c ) rs . 3987', 'd ) rs . 3929', 'e ) rs . 3938'] | a | multiply(square_perimeter(sqrt(289)), 58) | sqrt(n0)|square_perimeter(#0)|multiply(n1,#1) | geometry |
points a , b , c , and d , in that order , lie on a line . if ab = 3 cm , ac = 4 cm , and bd = 9 cm , what is cd , in centimeters ? | "putting a value to each point , lets use the following : a - 0 b - 3 ( ab = 3 ) c - 4 ( ac = 4 ) d - 12 ( bd = 9 ) cd is 12 - 4 = 8 . ans b" | a ) 1 , b ) 8 , c ) 3 , d ) 4 , e ) 5 | b | subtract(4, 3) | subtract(n1,n0)| | physics |
robert is travelling on his cycle andhas calculated to reach point a at 2 pm . if he travels at 10 kmph , he will reach there at 12 pm if he travels at 15 kmph . at what speed musthe travel to reach a at 1 pm ? | "let the distance travelled by x km . then , x - x = 2 10 15 3 x - 2 x = 60 x = 60 km . time taken to travel 60 km at 10 km / hr = 60 hrs = 6 hrs . 10 so , robert started 6 hours before 2 p . m . i . e . , at 8 a . m . required speed = 60 kmph . = 12 kmph . 5 c" | a ) 8 kmph , b ) 10 kmph , c ) 12 kmph , d ) 14 kmph , e ) 16 kmph | c | divide(divide(2, subtract(divide(1, 10), divide(1, 15))), subtract(divide(divide(2, subtract(divide(1, 10), divide(1, 15))), 10), 1)) | divide(n4,n1)|divide(n4,n3)|subtract(#0,#1)|divide(n0,#2)|divide(#3,n1)|subtract(#4,n4)|divide(#3,#5)| | physics |
find the value of m 32519 x 9999 = m ? | "32519 x 9999 = 32519 x ( 10000 - 1 ) = 32519 x 10000 - 32519 x 1 = 325190000 - 32519 = 325157481 e" | a ) 724533811 , b ) 353654655 , c ) 545463251 , d ) 725117481 , e ) 325157481 | e | multiply(subtract(9999, const_4), 32519) | subtract(n1,const_4)|multiply(#0,n0)| | general |
sheila works 8 hours per day on monday , wednesday and friday , and 6 hours per day on tuesday and thursday . she does not work on saturday and sunday . she earns $ 252 per week . how much does she earn in dollars per hour ? | "let sheila earn x dollars per hour so , on monday , wednesday and friday , she earns 8 x each and , on tuesday and thursday , she earns 6 x each in total , over the week she should earn , 3 ( 8 x ) + 2 ( 6 x ) = 36 x she earns $ 252 per week 36 x = 252 x = 7 correct option : e" | a ) 11 , b ) 10 , c ) 9 , d ) 8 , e ) 7 | e | divide(252, add(multiply(8, const_3), multiply(6, const_2))) | multiply(n0,const_3)|multiply(n1,const_2)|add(#0,#1)|divide(n2,#2)| | physics |
a can run 4 times as fast as b and gives b a start of 66 m . how long should the race course be so that a and b might reach in the same time ? | speed of a : speed of b = 4 : 1 means in a race of 4 m a gains 3 m . then in a race of 66 m he gains 66 * ( 4 / 3 ) i . e 88 m answer : a | a ) 88 m , b ) 60 m , c ) 80 m , d ) 65 m , e ) 75 m | a | add(multiply(4, divide(divide(66, 4), subtract(4, const_1))), 66) | divide(n1,n0)|subtract(n0,const_1)|divide(#0,#1)|multiply(n0,#2)|add(n1,#3) | physics |
how many multiples of 5 are there between 5 to 105 ? | "explanation : as you know , multiples of 5 are integers having 0 or 5 in the digit to the extreme right ( i . e . the units place ) . so the numbers are 10 , 15 , 20 , 25 , 30 , 35 , 40 , 45 , 50 , 55 , 60 , 65 , 70 , 75 , 80 , 85 , 90 , 95 , 100 . answer : e" | a ) 9 , b ) 18 , c ) 17 , d ) none of these , e ) 19 | e | add(divide(subtract(105, 5), 5), const_1) | subtract(n2,n1)|divide(#0,n0)|add(#1,const_1)| | general |
a and b go around a circular track of length 600 m on a cycle at speeds of 30 kmph and 54 kmph . after how much time will they meet for the first time at the starting point ? | "time taken to meet for the first time at the starting point = lcm { length of the track / speed of a , length of the track / speed of b } = lcm { 600 / ( 30 * 5 / 18 ) , 600 / ( 54 * 5 / 18 ) } = 360 sec . answer : b" | a ) 120 sec , b ) 360 sec , c ) 178 sec , d ) 187 sec , e ) 189 sec | b | divide(600, subtract(multiply(54, const_0_2778), multiply(30, const_0_2778))) | multiply(n2,const_0_2778)|multiply(n1,const_0_2778)|subtract(#0,#1)|divide(n0,#2)| | physics |
the ratio of male to female in a class is 2 : 3 . the career preference of the students in the class are to be represented in a circle graph . if the area of the graph allocated to each career preference is to be proportional to the number of students who have that career preference , how many degrees of the circle sho... | let the common ratio be x . . so m = 2 x and f = 3 x and total = 5 x 1 of m = 2 x and 1 of f = 3 x . . total preferring that carrer = 5 x now 5 x = 360 , so x = 360 / 5 = 72 . . so x * 8 / 3 = 72 * 8 / 3 = 192 d | ['a ) a ) 160 degree', 'b ) b ) 168 degree', 'c ) c ) 191 degree', 'd ) d ) 72 degree', 'e ) e ) 204 degree'] | d | multiply(divide(const_1, 3), multiply(divide(3, add(2, 3)), const_360)) | add(n0,n1)|divide(const_1,n1)|divide(n1,#0)|multiply(#2,const_360)|multiply(#1,#3) | geometry |
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 20 % ? | sp = 800 profit = 25 % cp = ( sp ) * [ 100 / ( 100 + p ) ] = 800 * [ 100 / 125 ] = 640 loss = 25 % = 25 % of 640 = rs . 128 sp = cp - loss = 640 - 128 = rs . 512 answer : c | a ) 228 , b ) 480 , c ) 512 , d ) 288 , e ) 276 | c | subtract(divide(multiply(800, const_100), add(25, const_100)), divide(multiply(divide(multiply(800, const_100), add(25, const_100)), 20), const_100)) | add(n1,const_100)|multiply(n0,const_100)|divide(#1,#0)|multiply(n2,#2)|divide(#3,const_100)|subtract(#2,#4) | gain |
the length of a rectangular plot is 20 metres more than its breadth . if the cost of fencing the plot @ rs . 26.50 per metre is rs . 9540 , what is the length of the plot in metres ? | "let length of plot = l meters , then breadth = l - 20 meters and perimeter = 2 [ l + l - 20 ] = [ 4 l - 40 ] meters [ 4 l - 40 ] * 26.50 = 9540 [ 4 l - 40 ] = 9540 / 26.50 = 360 4 l = 400 l = 400 / 4 = 100 meters . answer : e" | a ) 20 , b ) 200 , c ) 300 , d ) 400 , e ) 100 | e | subtract(divide(divide(9540, 26.50), const_2), multiply(const_2, 20)) | divide(n2,n1)|multiply(n0,const_2)|divide(#0,const_2)|subtract(#2,#1)| | physics |
what is the greatest of 3 consecutive integers whose sum is 36 ? | "36 / 3 = 12 the three numbers are 11 , 12 , and 13 . the answer is b ." | a ) 12 , b ) 13 , c ) 14 , d ) 15 , e ) 16 | b | add(divide(subtract(36, 3), 3), const_2) | subtract(n1,n0)|divide(#0,n0)|add(#1,const_2)| | physics |
each of the 43 points is placed either inside or on the surface of a perfect sphere . if 16 % or fewer of the points touch the surface , what is the maximum number of segments which , if connected from those points to form chords , could be the diameter of the sphere ? | "maximum number of points on the surface is 16 % * 43 = 6.88 . . . or 6 since it has to be an integer now note that if two points form a diameter , they can not be part of any other diameter . so in the best case we can pair up the points we have 6 points , so at best we can form 3 pairs ( 6 ) . so , answer is ( a )" | a ) 3 , b ) 11 , c ) 13 , d ) 23 , e ) 38 | a | divide(multiply(43, divide(16, const_100)), const_2) | divide(n1,const_100)|multiply(n0,#0)|divide(#1,const_2)| | geometry |
xavier starts from p towards q at a speed of 60 kmph and after every 12 mins increases his speed by 10 kmph . if the distance between p and q is 60 km , then how much time does he take to cover the distance ? | "first 12 min = 60 * 12 / 60 = 12 km 2 nd 12 min = 70 * 12 / 60 = 14 km 3 rd 12 min = 80 * 12 / 60 = 16 km 4 th 12 min = 90 * 12 / 60 = 18 km total time 12.4 = 48 min a" | a ) 48 , b ) 59 , c ) 60 , d ) 56 , e ) 50 | a | add(add(add(12, 12), 12), 12) | add(n1,n1)|add(n1,#0)|add(n1,#1)| | 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 6000 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 = 6000 = > total profit = 6000 * 7 = 42000 . answer : option c | a ) 28000 , b ) 30000 , c ) 42000 , d ) 34000 , e ) none of these | c | divide(6000, 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 |
the average of 5 quantities is 11 . the average of 3 of them is 4 . what is the average of remaining 2 numbers ? | "answer : a ( 5 x 11 - 3 x 4 ) / 2 = 21.5" | a ) 21.5 , b ) 10.6 , c ) 8 , d ) 9.5 , e ) none of these | a | divide(subtract(multiply(5, 11), multiply(3, 4)), 2) | multiply(n0,n1)|multiply(n2,n3)|subtract(#0,#1)|divide(#2,n4)| | general |
10 % of a number is added to 45 , the result is the same number . find the number ? | "( 10 / 100 ) * x + 45 = x 9 x = 450 x = 50 answer : a" | a ) 50 , b ) 288 , c ) 270 , d ) 129 , e ) 281 | a | divide(45, divide(45, const_100)) | divide(n1,const_100)|divide(n1,#0)| | gain |
if 4 x + y + z = 80 , 2 x - y - z = 4003 x + y - z = 20 for integers of x , y and z , find x = ? | 4 x + y + z = 80 - - - - - - - - - - 1 ) 2 x - y - z = 40 - - - - - - - - - - - - - 2 ) 3 x + y - z = 20 - - - - - - - - - - - - 3 ) from 1 ) and 2 ) 6 x = 120 x = 20 answer is b | a ) 10 , b ) 20 , c ) 15 , d ) 26 , e ) 18 | b | divide(80, 4) | divide(n1,n0) | general |
calculate how much time could be saved if the train would run at its usual speed , given that it ran at 5 / 7 of its own speed and got to its destination in 15 hours ? | "new speed = 5 / 7 of usual speed new time = 5 / 7 of usual time 5 / 7 of usual time = 15 hrs usual time = 15 * 5 / 7 = 10.71 hrs time saved = 15 - 10.71 = 4.29 hrs answer is c" | a ) 1.29 hrs , b ) 2.29 hrs , c ) 4.29 hrs , d ) 5.29 hrs , e ) 3.29 hrs | c | add(divide(reminder(subtract(multiply(15, const_60), multiply(multiply(15, divide(5, 7)), const_60)), const_60), const_100), floor(divide(subtract(multiply(15, const_60), multiply(multiply(15, divide(5, 7)), const_60)), const_60))) | divide(n0,n1)|multiply(n2,const_60)|multiply(n2,#0)|multiply(#2,const_60)|subtract(#1,#3)|divide(#4,const_60)|reminder(#4,const_60)|divide(#6,const_100)|floor(#5)|add(#7,#8)| | physics |
find the average of first 10 multiples of 7 | "explanation : = 7 ( 1 + 2 + 3 + . . . + 10 ) 10 = 7 ( 10 ( 10 + 1 ) ) 10 × 2 = 7 ( 110 ) 10 × 2 = 38.5 answer : option c" | a ) 35.5 , b ) 37.5 , c ) 38.5 , d ) 40.5 , e ) none of these | c | divide(add(add(add(10, const_1), add(add(10, const_1), const_2)), add(subtract(7, 10), subtract(7, const_2))), 10) | add(n0,const_1)|subtract(n1,n0)|subtract(n1,const_2)|add(#0,const_2)|add(#1,#2)|add(#0,#3)|add(#5,#4)|divide(#6,n0)| | general |
two numbers are in the ratio 3 : 5 . if 9 be subtracted from each , they are in the ratio of 9 : 17 . the first number ? | "( 3 x - 9 ) : ( 5 x - 9 ) = 9 : 17 x = 12 = > 3 x = 36 answer : a" | a ) 36 , b ) 86 , c ) 27 , d ) 86 , e ) 27 | a | add(multiply(3, divide(9, multiply(3, 5))), multiply(5, divide(9, multiply(3, 5)))) | multiply(n0,n1)|divide(n2,#0)|multiply(n0,#1)|multiply(n1,#1)|add(#2,#3)| | other |
john makes $ 30 a week from his job . he earns a raise and now makes $ 40 a week . what is the % increase ? | "increase = ( 10 / 30 ) * 100 = 33.33 % . e" | a ) 16 % , b ) 16.66 % , c ) 18 % , d ) 21 % , e ) 33.33 % | e | multiply(divide(subtract(40, 30), 30), const_100) | subtract(n1,n0)|divide(#0,n0)|multiply(#1,const_100)| | gain |
a truck covers a distance of 550 metres in 1 minute whereas a bus covers a distance of 33 kms in 45 minutes . the ratio of their speeds is ? | ratio of speeds = ( 550 / 60 ) 18 / 5 : 60 * 33 / 45 = 33 : 44 = 3 : 4 answer is b | a ) 1 : 2 , b ) 3 : 4 , c ) 2 : 5 , d ) 3 : 7 , e ) 1 : 3 | b | divide(550, multiply(divide(33, 45), const_1000)) | divide(n2,n3)|multiply(#0,const_1000)|divide(n0,#1) | physics |
a particular library has 150 books in a special collection , all of which were in the library at the beginning of the month . these book are occasionally loaned out through an inter - library program . if , by the end of the month , 65 percent of books that were loaned out are returned and there are 122 books in the sp... | "the total number of books is 150 . let x be the number of books which were loaned out . 65 % of books that were loaned out are returned . 35 % of books that were loaned out are not returned . now , there are 122 books , thus the number of un - returned books is 150 - 122 = 28 books . 0.35 x = 28 x = 80 the answer is c... | a ) 40 , b ) 60 , c ) 80 , d ) 100 , e ) 120 | c | divide(subtract(150, 122), subtract(const_1, divide(65, const_100))) | divide(n1,const_100)|subtract(n0,n2)|subtract(const_1,#0)|divide(#1,#2)| | gain |
a batsman in his 12 th innings makes a score of 48 and thereby increases his average by 2 runs . what is his average after the 12 th innings if he had never been ‘ not out ’ ? | "let ‘ x ’ be the average score after 12 th innings ⇒ 12 x = 11 × ( x – 2 ) + 48 ∴ x = 26 answer a" | a ) 26 , b ) 43 , c ) 44 , d ) 45 , e ) 46 | a | add(subtract(48, multiply(12, 2)), 2) | multiply(n0,n2)|subtract(n1,#0)|add(n2,#1)| | general |
the average of 1 st 3 of 4 numbers is 7 and of the last 3 are 5 . if the sum of the first and the last number is 13 . what is the last numbers ? | "a + b + c = 21 b + c + d = 15 a + d = 13 a – d = 6 a + d = 13 2 d = 7 d = 3.5 answer : c" | a ) 4 , b ) 5 , c ) 3.5 , d ) 7 , e ) 8 | c | subtract(subtract(multiply(3, 7), add(subtract(13, 7), const_3.0)), 7) | multiply(n1,n3)|subtract(n6,n3)|add(n5,#1)|subtract(#0,#2)|subtract(#3,n3)| | general |
the size of a flat - screen television is given as the length of the screen ’ s diagonal . how many square inches greater is the screen of a square 24 - inch flat - screen television than a square 17 - inch flat - screen television ? | "if we take a square with side length x and draw a diagonal , we get two isosceles right triangles . if we focus on one such right triangle , we see that the legs have length x . square 24 - inch flat - screen television the diagonal ( hypotenuse ) = 24 so , we can apply the pythagorean theorem to get x ² + x ² = 24 ² ... | a ) 143.5 , b ) 154 , c ) 160 , d ) 148.75 , e ) 142.25 | a | subtract(divide(power(24, const_2), const_2), divide(power(17, const_2), const_2)) | power(n0,const_2)|power(n1,const_2)|divide(#0,const_2)|divide(#1,const_2)|subtract(#2,#3)| | geometry |
the average weight of 8 persons increases by 2.5 kg when a new person comes in place of one of them weighing 65 kg . what might be the weight of the new person ? | "sol . total weight increased = ( 8 × 2.5 ) kg = 20 kg . weight of new person = ( 65 + 20 ) kg = 85 kg . answer c" | a ) 76 kg , b ) 76.5 kg , c ) 85 kg , d ) 90 kg , e ) none | c | add(65, multiply(8, 2.5)) | multiply(n0,n1)|add(n2,#0)| | general |
a , b , and c are integers and a < b < c . s is the set of all integers from a to b , inclusive . q is the set of all integers from b to c , inclusive . the median of set s is ( 3 / 4 ) * b . the median of set q is ( 6 / 8 ) * c . if r is the set of all integers from a to c , inclusive , what fraction of c is the media... | "the answer isc : 11 / 16 . the key to this problem is remembering that the median for a consecutive set of numbers is equivalent to its mean . for example , the mean and median of a set consisting of x , x + 1 , x + 2 , . . . , y will always be ( x + y ) / 2 . for set s , consisting of numbers ( a , a + 1 , . . . , b ... | a ) 3 / 8 , b ) 1 / 2 , c ) 11 / 16 , d ) 5 / 8 , e ) 3 / 4 | d | divide(add(const_1, divide(subtract(multiply(divide(6, 8), const_2), const_1), divide(const_1, subtract(multiply(divide(3, 4), const_2), const_1)))), const_2) | divide(n2,n3)|divide(n0,n1)|multiply(#0,const_2)|multiply(#1,const_2)|subtract(#2,const_1)|subtract(#3,const_1)|divide(const_1,#5)|divide(#4,#6)|add(#7,const_1)|divide(#8,const_2)| | general |
line q has the equation 5 y – 3 x = 75 . if line s is perpendicular to q , has an integer for its y - intercept , and intersects q in the second quadrant , then how many possible line s ’ s exist ? ( note : intersections on one of the axes do not count . ) | "5 y - 3 x = 75 and so y = 3 x / 5 + 15 when x = 0 , then y = 15 . when y = 0 , then x = - 25 the slope is 3 / 5 , so the slope of line s is - 5 / 3 . through the point ( - 25 , 0 ) , 0 = - 5 ( - 25 ) / 3 + c the y - intercept is c = - 125 / 3 < - 41 . thus the perpendicular line s can have y - intercepts from - 41 up ... | a ) 48 , b ) 52 , c ) 56 , d ) 60 , e ) 64 | c | add(subtract(divide(75, 5), const_2), add(divide(75, 3), divide(75, 5))) | divide(n2,n1)|divide(n2,n0)|add(#0,#1)|subtract(#1,const_2)|add(#2,#3)| | general |
a rectangular tank needs to be coated with insulation . the tank has dimensions of 3 feet , 5 feet , and 2 feet . each square foot of insulation costs $ 20 . how much will it cost to cover the surface of the tank with insulation ? | "the total surface area is 2 ( 2 * 3 + 3 * 5 + 2 * 5 ) = 62 square feet the total cost is 62 * $ 20 = $ 1240 the answer is b ." | a ) $ 1100 , b ) $ 1240 , c ) $ 1360 , d ) $ 1480 , e ) $ 1550 | b | multiply(surface_rectangular_prism(3, 5, 2), 20) | surface_rectangular_prism(n0,n1,n2)|multiply(n3,#0)| | geometry |
two concentric circles form a ring . the inner and outer circumference of the ring are 352 / 7 m and 528 / 7 m respectively . find the width of the ring . | explanation : let the inner and outer radii be r and r meters then , \ inline { \ color { black } 2 \ prod r } = 352 / 7 = > r = ( 352 / 7 ) * ( 7 / 22 ) * ( 1 / 2 ) = 8 m \ inline { \ color { black } 2 \ prod r } = 528 / 7 = > r = ( 528 / 7 ) * ( 7 / 22 ) * ( 1 / 2 ) = 12 m width of the ring = r - r = 12 - 8 = 4 m ans... | ['a ) 2', 'b ) 4', 'c ) 5', 'd ) 6', 'e ) 78'] | b | subtract(divide(divide(divide(528, 7), const_2), const_pi), divide(divide(divide(352, 7), const_2), const_pi)) | divide(n2,n1)|divide(n0,n1)|divide(#0,const_2)|divide(#1,const_2)|divide(#2,const_pi)|divide(#3,const_pi)|subtract(#4,#5) | physics |
a shopkeeper sold an article for rs 2564.36 . approximately what was his profit percent if the cost price of the article was rs 2300 | "explanation : gain % = ( 264.36 * 100 / 2300 ) = 11.49 % = 12 % approx option e" | a ) 4 % , b ) 5 % , c ) 6 % , d ) 7 % , e ) 12 % | e | floor(multiply(const_100, divide(subtract(2564.36, 2300), 2300))) | subtract(n0,n1)|divide(#0,n1)|multiply(#1,const_100)|floor(#2)| | gain |
in a recent election , geoff received 0.5 percent of the 6000 votes cast . to win the election , a candidate needed to receive more than x % of the vote . if geoff needed exactly 3000 more votes to win the election , what is the value of x ? | word problems are tricky in somehow more than other problem because you have the additional step to translate . breaking the problem : geoff how many votes he receives ? ? 30 votes he needs 3571 more votes so : 30 + 3000 = 3030 now what ' s the problem wants ? ? a x % . . . . . . . . 3030 is what % of total votes 6000 ... | a ) 50 , b ) 50.5 , c ) 56 , d ) 60 , e ) 63 | b | multiply(divide(add(3000, divide(multiply(6000, 0.5), const_100)), 6000), const_100) | multiply(n0,n1)|divide(#0,const_100)|add(n2,#1)|divide(#2,n1)|multiply(#3,const_100) | gain |
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