Problem stringlengths 5 967 | Rationale stringlengths 1 2.74k | options stringlengths 37 164 | correct stringclasses 5 values | annotated_formula stringlengths 7 1.65k | linear_formula stringlengths 8 925 | category stringclasses 6 values | answer stringclasses 5 values |
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we run a business that rents out canoes and kayaks . a canoe rental costs $ 12 per day , and a kayak rental costs $ 18 dollars per day . one day , our business rents out 3 canoes for every 2 kayaks and receives a total of $ 504 in revenue . how many more canoes than kayaks were rented out ? | "let x be the number of canoes . then 2 x / 3 is the number of kayaks . 12 x + ( 2 x / 3 ) * 18 = 504 12 x + 12 x = 504 24 x = 504 x = 21 ( canoes ) 2 x / 3 = 14 ( kayaks ) there were 21 - 14 = 7 more canoes rented out . the answer is c ." | a ) 5 , b ) 6 , c ) 7 , d ) 8 , e ) 9 | c | multiply(subtract(3, 2), divide(504, add(multiply(3, 12), multiply(2, 18)))) | multiply(n0,n2)|multiply(n1,n3)|subtract(n2,n3)|add(#0,#1)|divide(n4,#3)|multiply(#4,#2)| | general | C |
in a certain pond , 60 fish were caught , tagged , and returned to the pond . a few days later , 60 fish were caught again , of which 2 were found to have been tagged . if the percent of tagged fish in the second catch approximates the percent of tagged fish in the pond , what ` s the approximate number of fish in the pond ? | "the percent of tagged fish in the second catch is 2 / 60 * 100 = 3.33 % . we are told that 3.33 % approximates the percent of tagged fish in the pond . since there are 60 tagged fish , then we have 0.033 x = 60 - - > x = 1,818 . answer : c ." | a ) 400 , b ) 625 , c ) 1818 , d ) 2500 , e ) 10 000 | c | divide(60, divide(2, 60)) | divide(n2,n1)|divide(n0,#0)| | gain | C |
a grocer has 400 pounds of coffee in stock , 20 percent of which is decaffeinated . if the grocer buys another 100 pounds of coffee of which 60 percent is decaffeinated , what percent , by weight , of the grocer ’ s stock of coffee is decaffeinated ? | "1 . 20 % of 400 = 80 pounds of decaffeinated coffee 2 . 60 % of 100 = 60 pounds of decaffeinated coffee 3 . wt have 140 pounds of decaffeinated out of 500 pounds , that means 140 / 500 * 100 % = 28 % . the correct answer is a ." | a ) 28 % , b ) 30 % , c ) 32 % , d ) 34 % , e ) 40 % | a | multiply(divide(add(multiply(divide(20, 100), 400), multiply(100, divide(60, 100))), add(400, 100)), 100) | add(n0,n2)|divide(n1,n2)|divide(n3,n2)|multiply(n0,#1)|multiply(n2,#2)|add(#3,#4)|divide(#5,#0)|multiply(#6,n2)| | gain | A |
a can finish a piece of work in 5 days . b can do it in 16 days . they work together for two days and then a goes away . in how many days will b finish the work ? | "2 / 4 + ( 2 + x ) / 16 = 1 = > x = 6 days answer : c" | a ) 4 , b ) 5 , c ) 6 , d ) 7 , e ) 8 | c | divide(subtract(const_1, add(multiply(divide(const_1, const_4.0), const_2), multiply(divide(const_1, 16), const_2))), divide(const_1, 16)) | divide(const_1,const_4.0)|divide(const_1,n1)|multiply(#0,const_2)|multiply(#1,const_2)|add(#2,#3)|subtract(const_1,#4)|divide(#5,#1)| | physics | C |
if 4 men working 10 hours a day earn rs . 1200 per week , then 9 men working 6 hours a day will earn how much per week ? | "explanation : ( men 4 : 9 ) : ( hrs / day 10 : 6 ) : : 1200 : x hence 4 * 10 * x = 9 * 6 * 1200 or x = 9 * 6 * 1200 / 4 * 10 = 1620 answer : c" | a ) rs 840 , b ) rs 1320 , c ) rs 1620 , d ) rs 1680 , e ) none of these | c | multiply(divide(multiply(9, 6), multiply(4, 10)), 1200) | multiply(n3,n4)|multiply(n0,n1)|divide(#0,#1)|multiply(n2,#2)| | physics | C |
he total marks obtained by a student in physics , chemistry and mathematics is 140 more than the marks obtained by him in physics . what is the average mark obtained by him in chemistry and mathematics ? | "let the marks obtained by the student in physics , chemistry and mathematics be p , c and m respectively . p + c + m = 140 + p c + m = 140 average mark obtained by the student in chemistry and mathematics = ( c + m ) / 2 = 140 / 2 = 70 . answer : b" | a ) 55 , b ) 70 , c ) 75 , d ) 85 , e ) 95 | b | divide(140, const_2) | divide(n0,const_2)| | general | B |
if n = 2.0823 and n * is the decimal obtained by rounding n to the nearest hundredth , what is the value of n * – n ? | "n * = 2.08 n * - n = 2.08 - 2.0823 - 0.0023 answer : b" | a ) - 0.0053 , b ) - 0.0023 , c ) 0.0007 , d ) 0.0047 , e ) 0.0153 | b | subtract(subtract(2.0823, divide(divide(add(multiply(const_2, const_10), const_3), const_100), const_100)), 2.0823) | multiply(const_10,const_2)|add(#0,const_3)|divide(#1,const_100)|divide(#2,const_100)|subtract(n0,#3)|subtract(#4,n0)| | general | B |
the ratio between the sale price and the cost price of an article is 6 : 4 . what is the ratio between the profit and the cost price of that article ? | "c . p . = rs . 4 x and s . p . = rs . 6 x . then , gain = rs . 2 x required ratio = 2 x : 4 x = 1 : 2 b" | a ) 23 , b ) 1 : 2 , c ) 2 : 5 , d ) 3 : 5 , e ) 25 | b | divide(subtract(6, 4), 4) | subtract(n0,n1)|divide(#0,n1)| | other | B |
an army ’ s recruitment process included n rounds of selection tasks . for the first a rounds , the rejection percentage was 60 percent per round . for the next b rounds , the rejection percentage was 50 percent per round and for the remaining rounds , the selection percentage was 70 percent per round . if there were 100000 people who applied for the army and 1400 were finally selected , what was the value of n ? | "since 70 % is accepted and figure will be 1400 therefore in last stage tested / interviewed no . candidates should be 2000 so from here i will work for those two steps of decrements of 100000 candidates . fastly i reduce 60 % till it gets closer to our required 2000 candidates step ( 1 ) 40000 accepted . step ( 2 ) another 40 % of 40000 = 16000 accepted . here it is quiet observable that if we further deduct candidate by 60 % it would change our probablity of easy going 2000 candidate . so i would get to second stage of recruitment where 50 % is accepted step ( 3 ) 50 % of 16000 = 8000 step ( 4 ) 50 % of 8000 = 4000 step ( 5 ) 50 % of 4000 = 2000 . . . here we are done with total 5 steps and last step of accepting 70 % of 2000 = 1400 ( our target ) total 6 steps required . answer : c" | a ) 4 , b ) 5 , c ) 6 , d ) 8 , e ) 10 | c | add(add(const_2, add(const_1, const_4)), const_2) | add(const_1,const_4)|add(#0,const_2)|add(#1,const_2)| | general | C |
if x gets 25 % more than y and y gets 20 % more than z , the share of z out of rs . 740 will be : | "z share = z , y = 1.2 z x = 1.25 × 1.2 z , x + y + z = 740 ( 1.25 × 1.2 + 1.2 + 1 ) z = 74 3.7 z = 740 , z = 200 answer : . b ." | a ) rs . 300 , b ) rs . 200 , c ) rs . 240 , d ) rs . 350 , e ) none of these | b | divide(740, add(add(multiply(add(const_1, divide(25, const_100)), add(const_1, divide(20, const_100))), add(const_1, divide(20, const_100))), const_1)) | divide(n1,const_100)|divide(n0,const_100)|add(#0,const_1)|add(#1,const_1)|multiply(#3,#2)|add(#2,#4)|add(#5,const_1)|divide(n2,#6)| | general | B |
the number which exceeds 16 % of it by 42 is : | "solution solution let the number be x . x - 16 % of x = 42 x - 16 / 100 x = 42 x - 4 / 25 x = 42 21 / 25 x = 42 x = ( 42 x 25 / 21 ) = 50 answer a" | a ) 50 , b ) 52 , c ) 58 , d ) 60 , e ) 62 | a | divide(multiply(42, const_100), subtract(const_100, 16)) | multiply(n1,const_100)|subtract(const_100,n0)|divide(#0,#1)| | gain | A |
for any positive integer n , the sum of the first n positive integers equals n ( n + 1 ) / 2 . what is the sum of odd all the integers between 1 and 50 ? | "numbers are 1 , 3,5 - - - - - - - , 47,49 average of the set : ( largest + smallest ) / 2 = ( 49 + 1 ) / 2 = 25 ; # of terms : ( largest - smallest ) / 2 + 1 = ( 49 - 1 ) / 2 + 1 = 25 the sum = 25 * 25 = 625 answer : b" | a ) 500 , b ) 625 , c ) 750 , d ) 550 , e ) 600 | b | add(divide(subtract(subtract(50, 1), add(1, 1)), 2), 1) | add(n2,n0)|subtract(n3,n0)|subtract(#1,#0)|divide(#2,n1)|add(n0,#3)| | general | B |
a box contains 100 balls , numbered from 1 to 100 . if 3 balls are selected at random and with replacement from the box . if the 3 numbers on the balls selected contain two odd and one even . what is the probability j that the first ball picked up is odd numbered ? | "answer - d selecting the balls either even or odd is having probability 50 / 100 = 1 / 2 we have already selected 3 balls with 2 odd numbers and 1 even number . so we have 3 combinations ooe , oeo , eoo . we have 3 outcomes and 2 are favourable as in 2 cases 1 st number is odd . so probability j is 2 / 3 . d" | a ) 0 , b ) 1 / 3 , c ) 1 / 2 , d ) 2 / 3 , e ) 1 | d | divide(const_2, 3) | divide(const_2,n3)| | probability | D |
find large number from below question the difference of two numbers is 1395 . on dividing the larger number by the smaller , we get 6 as quotient and the 15 as remainder | "let the smaller number be x . then larger number = ( x + 1395 ) . x + 1395 = 6 x + 15 5 x = 1380 x = 276 large number = 276 + 1395 = 1671 e" | a ) 1235 , b ) 1345 , c ) 1678 , d ) 1767 , e ) 1671 | e | multiply(divide(subtract(1395, 15), subtract(6, const_1)), 6) | subtract(n0,n2)|subtract(n1,const_1)|divide(#0,#1)|multiply(n1,#2)| | general | E |
chris mixed 3 pounds of raisins with 4 pounds of nuts . if a pound of nuts costs 4 times as much as a pound of raisins , then the total cost of the raisins was what fraction of the total cost of the mixture ? | "1 lbs of raisin = $ 1 3 lbs of raisin = $ 3 1 lbs of nuts = $ 4 4 lbs of nuts = $ 16 total value of mixture = 16 + 3 = 19 fraction of the value of raisin = 3 / 19 ans : e" | a ) 1 / 7 , b ) 1 / 5 , c ) 1 / 4 , d ) 1 / 3 , e ) 3 / 19 | e | divide(3, add(multiply(4, 4), 3)) | multiply(n1,n2)|add(n0,#0)|divide(n0,#1)| | general | E |
two cubes of tire sides ratio 2 : 3 . find its cube volumes ratio ? | a ( power 3 ) : b ( power 3 ) = 2 ( power 3 ) : 3 ( power 3 ) = 8 : 27 answer is d . | ['a ) 2 : 27', 'b ) 3 : 27', 'c ) 1 : 27', 'd ) 8 : 27', 'e ) 5 : 27'] | d | divide(volume_cube(2), volume_cube(3)) | volume_cube(n0)|volume_cube(n1)|divide(#0,#1) | geometry | D |
a sum of money at simple interest amounts to rs . 825 in 3 years and to rs . 850 in 4 years . the sum is ? | "s . i . for 1 year = ( 850 - 825 ) = rs . 25 s . i . for 3 years = 25 * 3 = rs . 75 principal = ( 825 - 75 ) = rs . 750 . answer : b" | a ) s . 738 , b ) s . 750 , c ) s . 650 , d ) s . 730 , e ) s . 735 | b | subtract(825, divide(multiply(subtract(850, 825), 3), 4)) | subtract(n2,n0)|multiply(n1,#0)|divide(#1,n3)|subtract(n0,#2)| | gain | B |
a mixture of 20 kg of spirit and water contains 10 % water . how much water must be added to this mixture to raise the percentage of water to 25 % | exp . water in the given mixture = 10 * 20 / 100 = 2 kg , and spirit = ( 20 - 2 ) = 18 kg let x kg of water added , then , x + 2 / 20 + x * 100 = = 25 4 x + 8 = 20 + x , or x = 4 kg answer : a | a ) 4 kg , b ) 5 kg , c ) 8 kg , d ) 30 kg , e ) 35 kg | a | divide(subtract(multiply(divide(25, const_100), 20), divide(20, 10)), subtract(const_1, divide(25, const_100))) | divide(n2,const_100)|divide(n0,n1)|multiply(n0,#0)|subtract(const_1,#0)|subtract(#2,#1)|divide(#4,#3) | general | A |
if in a race of 70 m , a covers the distance in 20 seconds and b in 25 seconds , then a beats b by : | "explanation : the difference in the timing of a and b is 5 seconds . hence , a beats b by 5 seconds . the distance covered by b in 5 seconds = ( 70 * 5 ) / 25 = 14 m hence , a beats b by 14 m . answer c" | a ) 20 m , b ) 16 m , c ) 14 m , d ) 10 m , e ) 15 m | c | multiply(divide(subtract(25, 20), 25), 70) | subtract(n2,n1)|divide(#0,n2)|multiply(n0,#1)| | physics | C |
the faces of storage box are to be painted by 6 different colors . in how many ways can be this be done ? | if i have to paint 6 sides with 6 different colour . . . first face can have 6 c 1 options , 2 nd would have 5 c 1 , and subsequent ones would have 4 c 1 , 3 c 1 , 2 c 1 and 1 options respectively . total options = 6 c 1 x 5 c 1 x 4 c 1 x 3 c 1 x 2 c 1 x 1 = 720 distinct ways correct answer - a | a ) a . 720 , b ) b . 256 , c ) c . 1 , d ) d . 12 , e ) e . 36 | a | multiply(multiply(multiply(multiply(multiply(6, subtract(6, const_1)), const_4), const_3), const_2), const_1) | subtract(n0,const_1)|multiply(n0,#0)|multiply(#1,const_4)|multiply(#2,const_3)|multiply(#3,const_2)|multiply(#4,const_1) | general | A |
light glows for every 13 seconds . how many times did it between 1 : 57 : 58 and 3 : 20 : 47 am | the diff in sec between 1 : 57 : 58 and 3 : 20 : 47 is 4969 sec , 4969 / 13 = 382 . so total 383 times light ll glow answer : c | a ) 381 , b ) 382 , c ) 383 , d ) 384 , e ) 385 | c | divide(add(add(const_2, 47), multiply(add(20, add(const_2, const_60)), const_60)), 13) | add(n6,const_2)|add(const_2,const_60)|add(n5,#1)|multiply(#2,const_60)|add(#0,#3)|divide(#4,n0) | physics | C |
a group of workers had agreed on a very strange payment schedule . each of them would get $ 100 for every day that they had worked , but for every day they did not work will have to return $ 25 . after 30 days they realised they did not earn any money . how many days the workers did not work ? | the workers earn 4 times more than they have to return per day for not working . so the number of days they did not work is 4 times the number of days they worked . that means 24 days of not working in 30 days . correct answer b | a ) 15 , b ) 24 , c ) 30 , d ) 20 , e ) 10 | b | subtract(30, divide(multiply(25, 30), add(100, 25))) | add(n0,n1)|multiply(n1,n2)|divide(#1,#0)|subtract(n2,#2) | physics | B |
what is the angle between the hands of a clock when time is 16 : 40 ? | "angle between two hands = 40 h - 11 / 2 m = 40 * 16 - 40 * 11 / 2 = 640 - 220 = 420 deg answer : b" | a ) 625 deg , b ) 420 deg , c ) 145 deg , d ) 150 deg , e ) 300 deg | b | subtract(multiply(40, multiply(const_3, const_2)), 16) | multiply(const_2,const_3)|multiply(n1,#0)|subtract(#1,n0)| | geometry | B |
according to a recent student poll , 4 / 5 out of 25 members of the finance club are interested in a career in investment banking . if two students are chosen at random , what is the probability that at least one of them is interested in investment banking ? | "20 students are interested , 5 are not interested prob = 1 - 5 c 2 / 25 c 2 = 1 - ( 5 * 4 / ( 25 * 24 ) ) = 1 - 1 / 30 = 29 / 30 answer : a" | a ) 29 / 30 , b ) 4 / 49 , c ) 2 / 7 , d ) 45 / 49 , e ) 13 / 14 | a | divide(subtract(choose(25, const_2), choose(subtract(25, multiply(25, divide(4, 5))), const_2)), choose(25, const_2)) | choose(n2,const_2)|divide(n0,n1)|multiply(n2,#1)|subtract(n2,#2)|choose(#3,const_2)|subtract(#0,#4)|divide(#5,#0)| | gain | A |
if 20 typists can type 42 letters in 20 minutes , then how many letters will 30 typists working at the same rate complete in 1 hour ? | "20 typists can type 42 letters , so 30 typists can type = 42 * 30 / 20 42 * 30 / 20 letters can be typed in 20 mins . in 60 mins typist can type = 42 * 30 * 60 / 20 * 20 = 189 a is the answer" | a ) 189 , b ) 72 , c ) 144 , d ) 216 , e ) 400 | a | multiply(divide(multiply(42, const_3), 20), 30) | multiply(n1,const_3)|divide(#0,n0)|multiply(n3,#1)| | physics | A |
the number which exceeds 16 % of it by 42 is ? | "let the number be x . then , x - 16 % of x = 42 . x - 16 / 100 x = 42 x = ( 42 * 25 ) / 21 = 50 answer : a" | a ) 50 , b ) 25 , c ) 52 , d ) 58 , e ) 60 | a | divide(multiply(42, const_100), subtract(const_100, 16)) | multiply(n1,const_100)|subtract(const_100,n0)|divide(#0,#1)| | gain | A |
the area of a rectangular field is equal to 300 square meters . its perimeter is equal to 70 meters . find the length and width of this rectangle . | l * w = 300 : area , l is the length and w is the width . 2 l + 2 w = 70 : perimeter l = 35 - w : solve for l ( 35 - w ) * w = 300 : substitute in the area equation w = 15 and l = 20 : solve for w and find l using l = 35 - w . correct answer a | ['a ) w = 15 and l = 20', 'b ) w = 25 and l = 30', 'c ) w = 35 and l = 40', 'd ) w = 45 and l = 50', 'e ) w = 55 and l = 60'] | a | divide(subtract(divide(70, const_2), sqrt(subtract(power(divide(70, const_2), const_2), multiply(const_4, 300)))), const_2) | divide(n1,const_2)|multiply(n0,const_4)|power(#0,const_2)|subtract(#2,#1)|sqrt(#3)|subtract(#0,#4)|divide(#5,const_2) | geometry | A |
the bankers discount of a certain sum of money is rs . 72 and the true discount on the same sum for the same time is rs . 60 . the sum due is : | "sum = ( b . d * t . d ) / ( b . d - t . d ) ( 72 * 60 ) / 72 - 60 ; 360 answer : a" | a ) rs . 360 , b ) rs . 432 , c ) rs . 540 , d ) rs . 1080 , e ) rs . 1280 | a | divide(multiply(60, 72), subtract(72, 60)) | multiply(n0,n1)|subtract(n0,n1)|divide(#0,#1)| | gain | A |
paul completes a piece of work in 80 days , rose completes the same work in 120 days . if both of them work together , then the number of days required to complete the work is ? | if a can complete a work in x days and b can complete the same work in y days , then , both of them together can complete the work in x y / x + y days . that is , the required no . of days = 80 × 120 / 200 = 48 days answer is d | a ) 42 , b ) 44 , c ) 46 , d ) 48 , e ) 50 | d | inverse(add(divide(const_1, 80), divide(const_1, 120))) | divide(const_1,n0)|divide(const_1,n1)|add(#0,#1)|inverse(#2) | physics | D |
the sum of two consecutive number is 97 . which is the larger number ? | "let consecutive number be x , x + 1 therefore sum of the consecutive number is x + x + 1 = 97 2 x + 1 = 97 2 x = 96 x = 48 therefore larger number is x + 1 = 49 answer : b" | a ) 42 , b ) 49 , c ) 44 , d ) 45 , e ) 46 | b | add(add(power(add(add(divide(subtract(subtract(97, const_10), const_2), const_4), const_2), const_2), const_2), power(add(add(add(divide(subtract(subtract(97, const_10), const_2), const_4), const_2), const_2), const_2), const_2)), add(power(divide(subtract(subtract(97, const_10), const_2), const_4), const_2), power(add(divide(subtract(subtract(97, const_10), const_2), const_4), const_2), const_2))) | subtract(n0,const_10)|subtract(#0,const_2)|divide(#1,const_4)|add(#2,const_2)|power(#2,const_2)|add(#3,const_2)|power(#3,const_2)|add(#5,const_2)|add(#4,#6)|power(#5,const_2)|power(#7,const_2)|add(#9,#10)|add(#11,#8)| | physics | B |
in a fort , there are 1200 soldiers . if each soldier consumes 3 kg per day , the provisions available in the fort will last for 30 days . if some more soldiers join , the provisions available will last for 25 days given each soldier consumes 2.5 kg per day . find the number of soldiers joining the fort in that case . | assume x soldiers join the fort . 1200 soldiers have provision for 1200 ( days for which provisions last them ) ( rate of consumption of each soldier ) = ( 1200 ) ( 30 ) ( 3 ) kg . also provisions available for ( 1200 + x ) soldiers is ( 1200 + x ) ( 25 ) ( 2.5 ) k as the same provisions are available = > ( 1200 ) ( 30 ) ( 3 ) = ( 1200 + x ) ( 25 ) ( 2.5 ) x = [ ( 1200 ) ( 30 ) ( 3 ) ] / ( 25 ) ( 2.5 ) - 1200 = > x = 528 . answer : b | a ) 627 , b ) 528 soldiers , c ) 626 , d ) 657 , e ) 673 | b | subtract(divide(divide(multiply(multiply(1200, 3), 30), 25), 2.5), 1200) | multiply(n0,n1)|multiply(n2,#0)|divide(#1,n3)|divide(#2,n4)|subtract(#3,n0) | general | B |
a group of men decided to do a work in 10 days , but 5 of them became absent . if the rest of the group did the work in 12 days , find the original number of men ? | original number of men = 5 * 12 / ( 12 - 10 ) = 30 answer is c | a ) 15 , b ) 20 , c ) 30 , d ) 25 , e ) 18 | c | divide(multiply(5, 12), subtract(12, 10)) | multiply(n1,n2)|subtract(n2,n0)|divide(#0,#1) | physics | C |
sophia finished 23 of a book . she calculated that she finished 90 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 23 ⋅ x pages . then she has x − 23 ⋅ x = 13 ⋅ x pages left . 23 ⋅ x − 13 ⋅ x = 90 13 ⋅ x = 90 x = 270 . so answer is d ." | a ) 180 , b ) 200 , c ) 250 , d ) 270 , e ) 300 | d | divide(90, subtract(divide(const_2, const_3), subtract(const_1, divide(const_2, const_3)))) | divide(const_2,const_3)|subtract(const_1,#0)|subtract(#0,#1)|divide(n1,#2)| | general | D |
if y is 40 % greater than x , than x is what % less than y ? | "y = 1.4 x x = y / 1.4 = 10 y / 14 = 5 y / 7 x is 2 / 7 less which is 28.6 % less than y . the answer is b ." | a ) 24.4 % , b ) 28.6 % , c ) 32.3 % , d ) 36.5 % , e ) 40.9 % | b | multiply(divide(40, add(40, const_100)), const_100) | add(n0,const_100)|divide(n0,#0)|multiply(#1,const_100)| | general | B |
the difference between a two - digit number and the number obtained by interchanging the digits is 9 . what is the difference between the two digits of the number ? | "suppose the two - digit number be 10 x + y . then we have been given l 0 x + y – ( 10 y + x ) = 9 ⇒ 9 x – 9 y = 9 ⇒ x – y = 1 hence , the required difference = 1 note that if the difference between a two - digit number and the number obtained by interchanging the digits is d , then the difference between the two digits of the number = d ⁄ 9 answer e" | a ) 8 , b ) 2 , c ) 7 , d ) can not be determined , e ) none of these | e | divide(9, subtract(const_10, const_1)) | subtract(const_10,const_1)|divide(n0,#0)| | general | E |
jackie has two solutions that are 4 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 4 percent solution will be required ? | "let a = amount of 4 % acid and b = amount of 12 % acid . now , the equation translates to , 0.04 a + . 12 b = . 05 ( a + b ) but a + b = 60 therefore . 04 a + . 12 b = . 05 ( 60 ) = > 4 a + 12 b = 300 but b = 60 - a therefore 4 a + 12 ( 60 - a ) = 300 = > 16 a = 420 hence a = 26.25 . answer : e" | a ) 18 , b ) 20 , c ) 24 , d ) 36 , e ) 26.25 | e | multiply(const_3, divide(60, const_10)) | divide(n2,const_10)|multiply(#0,const_3)| | gain | E |
evaluate : 37 - 18 ÷ 3 × 2 = | "according to order of operations , 18 ÷ 3 × 2 ( division and multiplication ) is done first from left to right 18 ÷ 3 × 2 = 6 × 2 = 12 hence 37 - 18 ÷ 3 × 2 = 37 - 12 = 25 correct answer b ) 25" | a ) 20 , b ) 25 , c ) 27 , d ) 22 , e ) 17 | b | subtract(37, multiply(multiply(18, 3), 2)) | multiply(n1,n2)|multiply(n3,#0)|subtract(n0,#1)| | general | B |
6 friends went to a hotel and decided to pay the bill amount equally . but 5 of them could pay rs . 100 each as a result 6 th has to pay rs . 100 extra than his share . find the amount paid by him . | "explanation : average amount paid by 5 persons = rs . 100 increase in average due to rs . 120 paid extra by the 6 th men = rs . 100 / 5 = rs . 20 therefore , average expenditure of 6 friends = rs . 100 + rs . 20 = rs . 120 therefore , amount paid by the 6 th men = rs . 120 + rs . 100 = rs . 220 correct option : b" | a ) 156 , b ) 220 , c ) 130 , d ) 240 , e ) none | b | add(100, divide(add(multiply(100, 5), 100), 5)) | multiply(n1,n2)|add(n4,#0)|divide(#1,n1)|add(n4,#2)| | general | B |
two pipes p and q can fill a cistern in 12 and 10 minutes respectively . both are opened together , but at the end of 3 minutes the first is turned off . how much longer will the cistern take to fill ? | "3 / 12 + x / 10 = 1 x = 8 1 / 2 answer : b" | a ) 1 / 8 , b ) 1 / 2 , c ) 2 / 4 , d ) 1 / 4 , e ) 1 / 4 | b | multiply(subtract(const_1, multiply(add(divide(const_1, 12), divide(const_1, 10)), 3)), 10) | divide(const_1,n0)|divide(const_1,n1)|add(#0,#1)|multiply(#2,n2)|subtract(const_1,#3)|multiply(n1,#4)| | physics | B |
what is the sum of all even numbers from 1 to 801 ? | "explanation : 800 / 2 = 400 400 * 401 = 160400 answer : a" | a ) 160400 , b ) 281228 , c ) 281199 , d ) 122850 , e ) 128111 | a | divide(multiply(1, 801), const_4) | multiply(n0,n1)|divide(#0,const_4)| | general | A |
an exam consists of 8 true / false questions . brian forgets to study , so he must guess blindly on each question . if any score above 60 % is a passing grade , what is the probability that brian passes ? | "if you have 8 t or f and brian is going to guess then each question he has a 50 % chance of getting correct . if a passing score is 70 % it means brian needs to get 6 / 8 = 75 % , 7 / 8 = 87.5 % , or 8 / 8 = 100 % to pass . each is a possibility . if brian gets a 5 / 8 ( = 62.5 % ) or below he fails . so first figure out the number of ways that brian can get 6 out of 8 , 7 out of 8 , and 8 out of 8 questions correct . which is 8 choose 6 , equals is 28 , 8 choose 7 , equals 8 , and 8 choose 8 , equals 1 . this sums to 37 . the number of possible questions outcomes - the sum of 8 choose 8 , 7 choose 8 , 6 choose 8 … . 2 choose 8 , 1 choose 8 , and 0 choose 8 is 256 , so the chance of him passing is 35 / 256 . d" | a ) 1 / 16 , b ) 37 / 256 , c ) 1 / 2 , d ) 35 / 256 , e ) 15 / 16 | d | add(divide(subtract(const_1, add(add(power(divide(const_1, const_2), 8), multiply(8, power(divide(const_1, const_2), 8))), multiply(multiply(8, const_3), power(divide(const_1, const_2), 8)))), const_10), subtract(const_1, add(add(power(divide(const_1, const_2), 8), multiply(8, power(divide(const_1, const_2), 8))), multiply(multiply(8, const_3), power(divide(const_1, const_2), 8))))) | divide(const_1,const_2)|multiply(n0,const_3)|power(#0,n0)|multiply(n0,#2)|multiply(#1,#2)|add(#3,#2)|add(#5,#4)|subtract(const_1,#6)|divide(#7,const_10)|add(#8,#7)| | general | D |
if ( n + 2 ) ! / n ! = 110 , n = ? | "( n + 2 ) ! / n ! = 110 rewrite as : [ ( n + 2 ) ( n + 1 ) ( n ) ( n - 1 ) ( n - 2 ) . . . . ( 3 ) ( 2 ) ( 1 ) ] / [ ( n ) ( n - 1 ) ( n - 2 ) . . . . ( 3 ) ( 2 ) ( 1 ) ] = 132 cancel out terms : ( n + 2 ) ( n + 1 ) = 110 from here , we might just test the answer choices . since ( 11 ) ( 10 ) = 110 , we can see that n = 9 b" | a ) 2 / 131 , b ) 9 , c ) 10 , d ) 11 , e ) 12 | b | subtract(add(const_4, const_4), const_1) | add(const_4,const_4)|subtract(#0,const_1)| | general | B |
if the weight of 12 meters long rod is 14 kg . what is the weight of 6 meters long rod ? | "answer ∵ weight of 12 m long rod = 14 kg ∴ weight of 1 m long rod = 14 / 12 kg ∴ weight of 6 m long rod = 14 x 6 / 12 = 7 kg option : c" | a ) 8 kg . , b ) 10.8 kg . , c ) 7 kg . , d ) 18.0 kg , e ) none | c | divide(multiply(6, 14), 12) | multiply(n1,n2)|divide(#0,n0)| | physics | C |
a train travelling at a speed of 75 mph enters a tunnel 31 / 2 miles long . the train is 1 / 4 mile long . how long does it take for the train to pass through the tunnel from the moment the front enters to the moment the rear emerges ? | total distance = ( 7 / 2 + 1 / 4 ) miles = 15 / 4 miles time taken = 15 / 4 * 75 hrs = 1 / 20 hrs = 3 min answer b | a ) 2.5 min , b ) 3 min , c ) 3.2 min , d ) 3.5 min , e ) 4 min | b | multiply(divide(add(divide(1, 4), divide(add(const_3, const_4), 2)), 75), const_60) | add(const_3,const_4)|divide(n3,n4)|divide(#0,n2)|add(#1,#2)|divide(#3,n0)|multiply(#4,const_60) | physics | B |
carl is facing very difficult financial times and can only pay the interest on a $ 20,000 loan he has taken . the bank charges him a quarterly compound rate of 5 % . what is the approximate interest he pays annually ? | "usually , you are given the annual rate of interest and it is mentioned that it is annual rate . the bank charges him a quarterly compounded annual rate of 20 % . here you find per quarter rate as ( 20 / 4 ) % = 5 % i have actually never seen a question with quarter rate given but since this question did not mentionannual rate of interestand since the options did not make sense with 5 % annual rate of interest , it is apparent that the intent was a 5 % quarterly rate . so the bank charges 5 % every quarter and compounds it in the next quarter . had it been a simple quarterly rate , we would have just found 4 * 5 % of 20,000 = $ 4000 as our answer . but since , the interest is compounded , it will be a bit more than $ 2000 . option ( d ) looks correct ." | a ) $ 1200 , b ) $ 2000 , c ) $ 2150 , d ) $ 4000 , e ) $ 12000 | d | subtract(multiply(multiply(const_100, const_100), power(add(const_1, divide(5, const_100)), const_4)), multiply(const_100, const_100)) | divide(n1,const_100)|multiply(const_100,const_100)|add(#0,const_1)|power(#2,const_4)|multiply(#1,#3)|subtract(#4,#1)| | gain | D |
if the sum of two positive integers is 18 and the difference of their squares is 36 , what is the product of the two integers ? | "let the 2 positive numbers x and y x + y = 18 - - 1 x ^ 2 - y ^ 2 = 36 = > ( x + y ) ( x - y ) = 36 - - 2 using equation 1 in 2 , we get = > x - y = 2 - - 3 solving equation 1 and 3 , we get x = 10 y = 8 product = 10 * 8 = 80 answer b" | a ) 108 , b ) 80 , c ) 128 , d ) 135 , e ) 143 | b | multiply(divide(subtract(18, divide(36, 18)), divide(36, 18)), add(divide(subtract(18, divide(36, 18)), divide(36, 18)), divide(36, 18))) | divide(n1,n0)|subtract(n0,#0)|divide(#1,#0)|add(#2,#0)|multiply(#3,#2)| | general | B |
nicky and cristina are running a 100 meter race . since cristina is faster than nicky , she gives him a 12 second head start . if cristina runs at a pace of 5 meters per second and nicky runs at a pace of only 3 meters per second , how many seconds will nicky have run before cristina catches up to him ? | "the distance traveled by both of them is the same at the time of overtaking . 3 ( t + 12 ) = 5 t t = 18 . cristina will catch up nicky in 18 seconds . so in 18 seconds cristina would cover = 18 * 5 = 90 meter . now time taken my nicky to cover 90 meter = 90 / 3 = 30 seconds . a" | a ) 30 , b ) 35 , c ) 40 , d ) 25 , e ) 60 | a | add(divide(multiply(3, 12), 3), divide(multiply(3, 12), subtract(5, 3))) | multiply(n1,n3)|subtract(n2,n3)|divide(#0,n3)|divide(#0,#1)|add(#2,#3)| | physics | A |
the grade point average of the entire class is 85 . if the average of one third of the class is 97 , what is the average of the rest of the class ? | "let x be the number of students in the class . let p be the average of the rest of the class . 85 x = ( 1 / 3 ) 97 x + ( 2 / 3 ) ( p ) x 255 = 97 + 2 p 2 p = 158 p = 79 . the answer is d ." | a ) 76 , b ) 77 , c ) 78 , d ) 79 , e ) 80 | d | divide(subtract(multiply(85, const_4), 97), subtract(const_4, const_1)) | multiply(n0,const_4)|subtract(const_4,const_1)|subtract(#0,n1)|divide(#2,#1)| | general | D |
a pipe takes a hours to fill the tank . but because of a leakage it took 5 times of its original time . find the time taken by the leakage to empty the tank | "pipe a can do a work 60 min . lets leakage time is x ; then 1 / 60 - 1 / x = 1 / 300 x = 75 min answer : e" | a ) 50 min , b ) 60 min , c ) 90 min , d ) 80 min , e ) 75 min | e | multiply(const_10, multiply(const_1, 5)) | multiply(n0,const_1)|multiply(#0,const_10)| | physics | E |
a , b and c have rs . 1000 between them , a and c together have rs . 700 and b and c rs . 600 . how much does c have ? | "a + b + c = 1000 a + c = 700 b + c = 600 - - - - - - - - - - - - - - a + b + 2 c = 1300 a + b + c = 1000 - - - - - - - - - - - - - - - - c = 300 answer : c" | a ) 100 , b ) 200 , c ) 300 , d ) 400 , e ) 500 | c | subtract(add(700, 600), 1000) | add(n1,n2)|subtract(#0,n0)| | general | C |
the number 523 rbc is divisible by 7 , 89 . then what is the value of r * b * c | lcm of 7 , 8 and 9 is 504 , thus 523 rbc must be divisible by 504 . 523 rbc = 523000 + rbc 523000 divided by 504 gives a remainder of 352 . hence , 352 + rbc = k * 504 . k = 1 rbc = 152 - - > r * b * c = 10 k = 2 rbc = 656 - - > r * b * c = 180 as rbc is three digit number k can not be more than 2 . two answers ? well only one is listed in answer choices , so d . answer : d . | a ) 504 , b ) 532 , c ) 210 , d ) 180 , e ) 280 | d | multiply(multiply(multiply(const_3, const_2), add(const_2, const_3)), multiply(const_3, const_2)) | add(const_2,const_3)|multiply(const_2,const_3)|multiply(#0,#1)|multiply(#2,#1) | general | D |
solution a is 10 % salt and solution b is 80 % salt . if you have 30 ounces of solution a and 60 ounces of solution b , in what ratio could you mix solution a with solution b to produce 50 ounces of a 50 % salt solution ? | "forget the volumes for the time being . you have to mix 20 % and 80 % solutions to get 50 % . this is very straight forward since 50 is int he middle of 20 and 80 so we need both solutions in equal quantities . if this does n ' t strike , use w 1 / w 2 = ( a 2 - aavg ) / ( aavg - a 1 ) w 1 / w 2 = ( 80 - 50 ) / ( 50 - 10 ) = 3 / 4 so the volume of the two solutions will be equal . answer has to be 3 : 4 . c" | a ) 6 : 4 , b ) 6 : 14 , c ) 3 : 4 , d ) 4 : 6 , e ) 3 : 7 | c | divide(divide(subtract(multiply(50, divide(80, const_100)), multiply(50, divide(50, const_100))), subtract(divide(80, const_100), divide(10, const_100))), subtract(50, divide(subtract(multiply(50, divide(80, const_100)), multiply(50, divide(50, const_100))), subtract(divide(80, const_100), divide(10, const_100))))) | divide(n1,const_100)|divide(n4,const_100)|divide(n0,const_100)|multiply(n4,#0)|multiply(n4,#1)|subtract(#0,#2)|subtract(#3,#4)|divide(#6,#5)|subtract(n4,#7)|divide(#7,#8)| | other | C |
a motorcyclist goes from bombay to pune , a distance of 192 kms at an average of 32 kmph speed . another man starts from bombay by car 2 ½ hours after the first , and reaches pune ½ hour earlier . what is the ratio of the speed of the motorcycle and the car ? | b 1 : 2 t = 192 / 32 = 6 h t = 6 - 3 = 3 time ratio = 6 : 3 = 2 : 1 speed ratio = 1 : 2 | a ) 2 : 8 , b ) 1 : 2 , c ) 2 : 5 , d ) 6 : 9 , e ) 6 : 2 | b | divide(divide(192, divide(192, 32)), divide(192, subtract(divide(192, 32), const_3))) | divide(n0,n1)|divide(n0,#0)|subtract(#0,const_3)|divide(n0,#2)|divide(#1,#3) | physics | B |
a cyclist rides a bicycle 8 km at an average speed of 10 km / hr and again travels 10 km at an average speed of 8 km / hr . what is the average speed for the entire trip ? | "distance = 18 km time = 8 / 10 + 10 / 8 = ( 64 + 100 ) / 80 = 164 / 80 = 41 / 20 hours average speed = ( 18 * 20 ) / 41 = 8.78 km / h the answer is c ." | a ) 9.28 , b ) 8.48 , c ) 8.78 , d ) 8.98 , e ) 9.18 | c | divide(add(8, 10), add(divide(8, 10), divide(10, 8))) | add(n0,n1)|divide(n0,n1)|divide(n1,n0)|add(#1,#2)|divide(#0,#3)| | general | C |
0.0004 ? = 0.02 | "explanation : required answer = 0.0004 / 0.02 = 0.04 / 2 = 0.02 . answer : option a" | a ) 0.02 , b ) 0.2 , c ) 2 , d ) 20 , e ) none of these | a | multiply(divide(0.0004, 0.02), const_100) | divide(n0,n1)|multiply(#0,const_100)| | general | A |
two pipes a and b can fill a tank in 25 hours and 35 hours respectively . if both the pipes are opened simultaneously , how much time will be taken to fill the tank ? | "part filled by a in 1 hour = 1 / 25 part filled by b in 1 hour = 1 / 30 part filled by ( a + b ) in 1 hour = 1 / 25 + 1 / 30 = 11 / 150 both the pipes together fill the tank in 150 / 11 = 14 6 / 11 hours answer is b" | a ) 20 hours , b ) 14 6 / 11 hours , c ) 10 hours , d ) 12 hours , e ) 8 hours | b | divide(const_1, add(divide(const_1, 25), divide(const_1, 35))) | divide(const_1,n0)|divide(const_1,n1)|add(#0,#1)|divide(const_1,#2)| | physics | B |
a room is 7 meters 20 centimeters in length and 4 meters 32 centimeters in width . find the least number of square tiles of equal size required to cover the entire floor of the room . | "let us calculate both the length and width of the room in centimeters . length = 7 meters and 20 centimeters = 720 cm width = 4 meters and 32 centimeters = 432 cm as we want the least number of square tiles required , it means the length of each square tile should be as large as possible . further , the length of each square tile should be a factor of both the length and width of the room . hence , the length of each square tile will be equal to the hcf of the length and width of the room = hcf of 720 and 432 = 144 thus , the number of square tiles required = ( 720 x 432 ) / ( 144 x 144 ) = 5 x 3 = 15 answer : b" | a ) 10 , b ) 15 , c ) 17 , d ) 19 , e ) 21 | b | divide(multiply(add(multiply(7, const_100), 20), add(multiply(4, const_100), 32)), multiply(gcd(add(multiply(7, const_100), 20), add(multiply(4, const_100), 32)), gcd(add(multiply(7, const_100), 20), add(multiply(4, const_100), 32)))) | multiply(n0,const_100)|multiply(n2,const_100)|add(n1,#0)|add(n3,#1)|gcd(#2,#3)|multiply(#2,#3)|multiply(#4,#4)|divide(#5,#6)| | general | B |
the diagonal of a square is twice the side of equilateral triangle then the ratio of area of the triangle to the area of square is ? | let the side of equilateral triangle = 1 unit . we know that area of an equilateral triangle = 3 – √ 4 a 234 a 2 as side = 1 unit area of the equilateral triangle = 3 – √ 434 now diagonal of the square = 2 ( side of the equilateral triangle ) = 2 we know that area of the square = 12 d 212 d 2 where d = diagonal so area of the square = 12 ( 22 ) = 212 ( 22 ) = 2 ratio of the areas of equilateral triangle and square = 3 – √ 434 : 2 ⇒ ⇒ 3 √ 8 answer : b | ['a ) 3 √ 9', 'b ) 3 √ 8', 'c ) 3 √ 7', 'd ) 3 √ 5', 'e ) 3 √ 1'] | b | divide(triangle_area_three_edges(const_1, const_1, const_1), square_area(sqrt(divide(power(const_2, const_2), add(const_1, const_1))))) | add(const_1,const_1)|power(const_2,const_2)|triangle_area_three_edges(const_1,const_1,const_1)|divide(#1,#0)|sqrt(#3)|square_area(#4)|divide(#2,#5) | geometry | B |
a business executive and his client are charging their dinner tab on the executive ' s expense account . the company will only allow them to spend a total of 75 $ for the meal . assuming that they will pay 7 % in sales tax for the meal and leave a 15 % tip , what is the most their food can cost ? | "let x is the cost of the food 1.07 x is the gross bill after including sales tax 1.15 * 1.07 x = 75 x = 60.95 hence , the correct option is c" | a ) 69.55 $ , b ) 50.63 $ , c ) 60.95 $ , d ) 52.15 $ , e ) 53.15 $ | c | divide(75, add(divide(add(7, 15), const_100), const_1)) | add(n1,n2)|divide(#0,const_100)|add(#1,const_1)|divide(n0,#2)| | general | C |
31 of the scientists that attended a certain workshop were wolf prize laureates , and 13 of these 31 were also nobel prize laureates . of the scientists that attended that workshop and had not received the wolf prize , the number of scientists that had received the nobel prize was 3 greater than the number of scientists that had not received the nobel prize . if 51 of the scientists attended that workshop , how many of them were nobel prize laureates ? | "lets solve by creating equation . . w = 31 . . total = 52 . . not w = 52 - 31 = 21 . . now let people who were neither be x , so out of 19 who won nobel = x + 3 . . so x + x + 3 = 21 or x = 9 . . so who won nobel but not wolf = x + 3 = 12 . . but people who won both w and n = 13 . . so total who won n = 12 + 13 = 25 . . d" | a ) a ) 11 , b ) b ) 18 , c ) c ) 24 , d ) d ) 25 , e ) d ) 36 | d | add(add(3, divide(subtract(subtract(51, 31), 3), const_2)), 13) | subtract(n4,n0)|subtract(#0,n3)|divide(#1,const_2)|add(n3,#2)|add(n1,#3)| | physics | D |
a person spent rs . 5,040 from his salary on food and 5,000 on house rent . after that he was left with 20 % of his monthly salary . what is his monthly salary ? | total money spent on food and house rent = 5,040 + 5,000 = 10,040 which is 100 - 20 = 80 % of his monthly salary ∴ his salary = 10040 x 100 / 80 = 12550 answer : e | a ) 22,550 , b ) 32,550 , c ) 52,550 , d ) 62,550 , e ) 12,550 | e | divide(add(multiply(add(const_4, const_1), const_100), 20), sqrt(const_100)) | add(const_1,const_4)|sqrt(const_100)|multiply(#0,const_100)|add(n2,#2)|divide(#3,#1) | gain | E |
if a person walks at 15 km / hr instead of 10 km / hr , he would have walked 10 km more . the actual distance traveled by him is ? | "let the actual distance traveled be x km . then , x / 10 = ( x + 10 ) / 15 5 x - 150 = > x = 30 km . answer : b" | a ) 50 km , b ) 30 km , c ) 18 km , d ) 16 km , e ) 97 km | b | multiply(10, divide(10, subtract(15, 10))) | subtract(n0,n1)|divide(n2,#0)|multiply(n1,#1)| | general | B |
if the sum of two numbers is 24 and the sum of their squares is 400 , then the product of the numbers is | "according to the given conditions x + y = 24 and x ^ 2 + y ^ 2 = 400 now ( x + y ) ^ 2 = x ^ 2 + y ^ 2 + 2 xy so 24 ^ 2 = 400 + 2 xy so xy = 176 / 2 = 88 answer : d" | a ) 40 , b ) 44 , c ) 80 , d ) 88 , e ) 48 | d | divide(subtract(power(24, const_2), 400), const_2) | power(n0,const_2)|subtract(#0,n1)|divide(#1,const_2)| | general | D |
a train with a length of 100 meters , is traveling at a speed of 72 km / hr . the train enters a tunnel 2.3 km long . how many minutes does it take the train to pass through the tunnel from the moment the front enters to the moment the rear emerges ? | 72 km / hr = 1.2 km / min the total distance is 2.4 km . 2.4 / 1.2 = 2 minutes the answer is a . | a ) 2.0 , b ) 2.5 , c ) 3.0 , d ) 3.5 , e ) 4.0 | a | multiply(divide(add(2.3, divide(100, const_1000)), 72), const_60) | divide(n0,const_1000)|add(n2,#0)|divide(#1,n1)|multiply(#2,const_60) | physics | A |
a train 225 m long passes a man , running at 10 km / hr in the same direction in which the train is going , in 10 seconds . the speed of the train is : | "speed of the train relative to man = 225 / 10 m / sec = 45 / 2 m / sec . = 45 / 2 x 18 / 5 km / hr = 81 km / hr . let the speed of the train be x km / hr . then , relative speed = ( x - 5 ) km / hr . x - 10 = 81 = 91 km / hr . answer : c" | a ) 81 , b ) 90 , c ) 91 , d ) 85 , e ) 96 | c | divide(divide(subtract(225, multiply(multiply(10, const_0_2778), 10)), 10), const_0_2778) | multiply(n1,const_0_2778)|multiply(n1,#0)|subtract(n0,#1)|divide(#2,n1)|divide(#3,const_0_2778)| | physics | C |
set a of 8 positive integers may have the same element and have 37 . and set b of 8 positive integers must have different elements and have 37 . when m and n are the greatest possible differences between 37 and other elements ’ sums in set a and set b , respectively , m - n = ? | this is maximum - minimum . hence , 37 - ( 1 + 1 + 1 + 1 + 1 + 1 + 1 ) = 30 and 37 - ( 1 + 2 + 3 + 4 + 5 + 6 + 7 ) = 9 . so , 30 - 9 = 21 . the correct answer is b . | a ) 21 , b ) 22 , c ) 23 , d ) 25 , e ) 26 | b | subtract(37, add(add(8, const_2), 8)) | add(n0,const_2)|add(n0,#0)|subtract(n1,#1)| | general | B |
a girl was asked to multiply a certain number by 43 . she multiplied it by 34 and got his answer less than the correct one by 1242 . find the number to be multiplied . | "let the required number be x . then , 43 x – 34 x = 1242 or 9 x = 1242 or x = 138 . required number = 138 . answer : e" | a ) 130 , b ) 132 , c ) 134 , d ) 136 , e ) 138 | e | divide(1242, subtract(43, 34)) | subtract(n0,n1)|divide(n2,#0)| | general | E |
machine t can produce x units in 3 / 4 of the time it takes machine n to produce x units . machine n can produce x units in 2 / 3 the time it takes machine o to produce x units . if all 3 machines are working simultaneously , what fraction of the total output is produced by machine n ? | let the following be true : t makes x in time t then the following follows : n makes x in 4 t / 3 o makes x in 3 / 2 ( 4 t / 3 ) = 2 t m : n : o = 1 : 4 / 3 : 2 = 3 : 4 : 6 so n = 4 / ( 3 + 4 + 6 ) = 4 / 13 = c | a ) 1 / 2 , b ) 1 / 3 , c ) 4 / 13 , d ) 8 / 29 , e ) 6 / 33 | c | multiply(inverse(add(divide(2, 3), add(const_1, multiply(divide(2, 3), divide(3, 4))))), divide(2, 3)) | divide(n2,n0)|divide(n0,n1)|multiply(#0,#1)|add(#2,const_1)|add(#3,#0)|inverse(#4)|multiply(#0,#5) | general | C |
there are 840 male and female participants in a meeting . half the female participants and one - quarter of the male participants are democrats . one - third of all the participants are democrats . how many of the democrats are female ? | "female = x male = 840 - x x / 2 + 840 - x / 4 = 1 / 3 * ( 840 ) = 280 x = 280 x / 2 = 140 is supposed to be the answer m is missing something correct option a" | a ) 140 , b ) 100 , c ) 125 , d ) 175 , e ) 225 | a | divide(subtract(multiply(divide(840, const_3), const_4), 840), const_2) | divide(n0,const_3)|multiply(#0,const_4)|subtract(#1,n0)|divide(#2,const_2)| | general | A |
( 753 x 753 + 247 x 247 - 753 x 247 ) / ( 753 x 753 x 753 + 247 x 247 x 247 ) = ? | given exp . = ( a ^ 2 + b ^ 2 - ab ) / ( a ^ 3 + b ^ 3 ) = 1 / ( a + b ) = 1 / ( 753 + 247 ) = 1 / 1000 answer : a | a ) 1 / 1000 , b ) 1 / 506 , c ) 253 / 500 , d ) 253 / 1000 , e ) none of these | a | divide(subtract(add(multiply(753, 753), multiply(247, 247)), multiply(753, 247)), add(multiply(multiply(753, 753), 753), multiply(multiply(247, 247), 247))) | multiply(n0,n0)|multiply(n2,n2)|multiply(n0,n2)|add(#0,#1)|multiply(n0,#0)|multiply(n2,#1)|add(#4,#5)|subtract(#3,#2)|divide(#7,#6) | general | A |
last month , john rejected 0.5 % of the products that he inspected and jane rejected 0.7 percent of the products that she inspected . if total of 0.75 percent of the products produced last month were rejected , what fraction of the products did jane inspect ? | x - fraction of products jane inspected ( 1 - x ) - fraction of products john inspected 0.7 ( x ) + 0.5 ( 1 - x ) = 0.75 0.2 x = 0.75 - 0.5 x = 0.25 / 0.2 x = 5 / 4 therefore the answer is d : 5 / 4 . | a ) 1 / 6 , b ) 1 / 2 , c ) 5 / 8 , d ) 5 / 4 , e ) 15 / 16 | d | divide(subtract(0.75, 0.5), subtract(0.7, 0.5)) | subtract(n2,n0)|subtract(n1,n0)|divide(#0,#1) | gain | D |
two persons a and b can complete a piece of work in 30 days and 45 days respectively . if they work together , what part of the work will be completed in 5 days ? | "a ' s one day ' s work = 1 / 30 b ' s one day ' s work = 1 / 45 ( a + b ) ' s one day ' s work = 1 / 30 + 1 / 45 = 1 / 18 the part of the work completed in 5 days = 5 ( 1 / 18 ) = 5 / 18 . answer a" | a ) 5 / 18 , b ) 1 / 6 , c ) 1 / 4 , d ) 1 / 9 , e ) 2 / 6 | a | multiply(5, add(divide(const_1, 30), divide(const_1, 45))) | divide(const_1,n0)|divide(const_1,n1)|add(#0,#1)|multiply(n2,#2)| | physics | A |
a train traveling with constant speed crosses a 90 m long platform in 12 seconds and a 120 m long platform in 15 seconds . find the length of the train and its speed . | let the length of the train be x m and its speed be y m / sec . then , x / y = 12 = > y = x / 12 ( x + 90 ) / 12 = x + 120 / 15 = > x = 30 m . answer : a | a ) 30 , b ) 40 , c ) 60 , d ) 50 , e ) none | a | divide(subtract(multiply(12, 120), multiply(90, 15)), subtract(15, 12)) | multiply(n1,n2)|multiply(n0,n3)|subtract(n3,n1)|subtract(#0,#1)|divide(#3,#2) | physics | A |
if ( 2 to the x ) - ( 2 to the ( x - 2 ) ) = 3 ( 2 to the 11 ) , what is the value of x ? | "( 2 to the power x ) - ( 2 to the power ( x - 2 ) ) = 3 ( 2 to the power 11 ) 2 ^ x - 2 ^ ( x - 2 ) = 3 . 2 ^ 11 hence x = 13 . answer is c" | a ) 9 , b ) 11 , c ) 13 , d ) 15 , e ) 17 | c | add(11, 2) | add(n0,n5)| | general | C |
sandy buys an old scooter for $ 800 and spends $ 200 on its repairs . if sandy sells the scooter for $ 1200 , what is the gain percent ? | selling price / total cost = 1200 / 1000 = 1.2 the gain percent is 20 % . the answer is c . | a ) 15 % , b ) 18 % , c ) 20 % , d ) 22 % , e ) 25 % | c | multiply(divide(subtract(1200, add(200, 800)), add(200, 800)), const_100) | add(n0,n1)|subtract(n2,#0)|divide(#1,#0)|multiply(#2,const_100) | gain | C |
the present age of a father is 3 years more than 3 times the age of his son . 5 years hence , father ' s age will be 10 years more than twice the age of the son . find the present age of the father . | if the present age be x years . father ' s will be ( 3 x + 3 ) years . . so , ( 3 x + 3 + 5 ) = 2 ( x + 3 ) + 10 or , x = 8 so the fathers present age = ( 3 x + 3 ) = ( 3 * 8 + 3 ) years = 27 years . . answer : option c | a ) 33 , b ) 38 , c ) 27 , d ) 40 , e ) 48 | c | subtract(subtract(add(3, multiply(3, subtract(subtract(add(multiply(const_2, 5), 10), 5), 3))), const_10), const_1) | multiply(n2,const_2)|add(n3,#0)|subtract(#1,n2)|subtract(#2,n0)|multiply(n0,#3)|add(n0,#4)|subtract(#5,const_10)|subtract(#6,const_1) | general | C |
a contest will consist of n questions , each of which is to be answered eithertrueorfalse . anyone who answers all n questions correctly will be a winner . what is the least value of n for which the probability is less than 1 / 100 that a person who randomly guesses the answer to each question will be a winner ? | "a contest will consist of n questions , each of which is to be answered eithertrueorfalse . anyone who answers all n questions correctly will be a winner . what is the least value of n for which the probability is less than 1 / 1000 that a person who randomly guesses the answer to each question will be a winner ? a . 5 b . 10 c . 50 d . 100 e . 1000 soln : ans is b probability that one question is answered right is 1 / 2 . now for minimum number of questions needed to take probability less than 1 / 1000 is = > ( 1 / 2 ) ^ n < 1 / 100 n = 50 satisfies this . c" | a ) 5 , b ) 10 , c ) 50 , d ) 100 , e ) 1000 | c | multiply(const_1000, divide(1, 100)) | divide(n0,n1)|multiply(#0,const_1000)| | general | C |
in a certain pond , 40 fish were caught , tagged , and returned to the pond . a few days later , 50 fish were caught again , of which 2 were found to have been tagged . if the percent of tagged fish in the second catch approximates the percent of tagged fish in the pond , what is the approximate number of fish in the pond ? | "this is a rather straight forward ratio problem . 1 . 40 fish tagged 2 . 2 out of the 50 fish caught were tagged thus 2 / 50 2 / 50 = 40 / x thus , x = 1000 think of the analogy : 2 fish is to 50 fish as 50 fish is to . . . ? you ' ve tagged 50 fish and you need to find what that comprises as a percentage of the total fish population - we have that information with the ratio of the second catch . e" | a ) 400 , b ) 625 , c ) 1,250 , d ) 2,500 , e ) 1,000 | e | divide(40, divide(2, 50)) | divide(n2,n1)|divide(n0,#0)| | gain | E |
in a sports club with 30 members , 17 play badminton and 21 play tennis and 2 do not play either . how many members play both badminton and tennis ? | "let x play both badminton and tennis so 17 - x play only badminton and 19 - x play only tennis . 2 play none and there are total 30 students . hence , ( 17 - x ) + ( 21 - x ) + x + 2 = 30 40 - 2 x + x = 30 40 - x = 30 x = 10 so 10 members play both badminton and tennis . d" | a ) 7 , b ) 8 , c ) 9 , d ) 10 , e ) 11 | d | subtract(add(add(17, 21), 2), 30) | add(n1,n2)|add(n3,#0)|subtract(#1,n0)| | other | D |
a shopkeeper sells 20 % of his stock at 10 % profit ans sells the remaining at a loss of 5 % . he incurred an overall loss of rs . 250 . find the total worth of the stock ? | "let the total worth of the stock be rs . x . the sp of 20 % of the stock = 1 / 5 * x * 1.1 = 11 x / 50 the sp of 80 % of the stock = 4 / 5 * x * 0.95 = 19 x / 25 = 38 x / 50 total sp = 11 x / 50 + 38 x / 50 = 49 x / 50 overall loss = x - 49 x / 50 = x / 50 x / 50 = 250 = > x = 12500 answer : a" | a ) 12500 , b ) 20000 , c ) 20289 , d ) 20027 , e ) 20026 | a | divide(250, subtract(multiply(divide(5, const_100), divide(subtract(const_100, 20), const_100)), multiply(divide(10, const_100), divide(20, const_100)))) | divide(n2,const_100)|divide(n1,const_100)|divide(n0,const_100)|subtract(const_100,n0)|divide(#3,const_100)|multiply(#1,#2)|multiply(#0,#4)|subtract(#6,#5)|divide(n3,#7)| | gain | A |
virginia , adrienne , and dennis have taught history for a combined total of 93 years . if virginia has taught for 9 more years than adrienne and for 9 fewer years than dennis , for how many years has dennis taught ? | "let number of years taught by virginia = v number of years taught by adrienne = a number of years taught by dennis = d v + a + d = 96 v = a + 9 = > a = v - 9 v = d - 9 = > a = ( d - 9 ) - 9 = d - 18 d - 9 + d - 18 + d = 93 = > 3 d = 93 + 27 = 120 = > d = 40 answer c" | a ) 23 , b ) 32 , c ) 40 , d ) 41 , e ) 44 | c | add(divide(subtract(93, add(add(9, 9), 9)), const_3), add(9, 9)) | add(n1,n2)|add(n1,#0)|subtract(n0,#1)|divide(#2,const_3)|add(#0,#3)| | general | C |
there is a 30 % chance sandy will visit china this year , while there is a 60 % chance that she will visit malaysia this year . what is the probability that sandy will visit either china or malaysia this year , but not both ? | p ( china and not malaysia ) = 0.3 * 0.4 = 0.12 p ( malaysia and not china ) = 0.6 * 0.7 = 0.42 total probability = 0.12 + 0.42 = 0.54 = 54 % the answer is c . | a ) 42 % , b ) 48 % , c ) 54 % , d ) 60 % , e ) 66 % | c | divide(add(30, 60), multiply(multiply(const_5, const_5), const_4)) | add(n0,n1)|multiply(const_5,const_5)|multiply(#1,const_4)|divide(#0,#2) | probability | C |
arnold and danny are two twin brothers that are celebrating their birthday . the product of their ages today is smaller by 9 from the product of their ages a year from today . what is their age today ? | "ad = ( a + 1 ) ( d + 1 ) - 9 0 = a + d - 8 a + d = 8 a = d ( as they are twin brothers ) a = d = 4 b is the answer" | a ) 2 . , b ) 4 . , c ) 5 . , d ) 7 , e ) 9 . | b | divide(subtract(9, const_1), const_2) | subtract(n0,const_1)|divide(#0,const_2)| | general | B |
a family x went for a vacation . unfortunately it rained for 13 days when they were there . but whenever it rained in the mornings , they had clear afternoons and vice versa . in all they enjoyed 11 mornings and 12 afternoons . how many days did they stay there totally ? | explanation : clearly 11 mornings and 12 afternoons = 23 half days = > since 13 days raining means 13 half days . = > so 23 - 13 = 10 halfdays ( not effecetd by rain ) = > so 10 halfdays = 5 fulldays hence , total no . of days = 13 + 5 = 18 days answer : b | a ) 16 , b ) 18 , c ) 20 , d ) 13 , e ) 14 | b | add(13, divide(subtract(add(11, 12), 13), const_2)) | add(n1,n2)|subtract(#0,n0)|divide(#1,const_2)|add(n0,#2) | general | B |
in an office , totally there are 5200 employees and 45 % of the total employees are males . 50 % of the males in the office are at - least 50 years old . find the number of males aged below 50 years ? | "number of male employees = 5200 * 45 / 100 = 2340 required number of male employees who are less than 50 years old = 4160 * ( 100 - 50 ) % = 2340 * 50 / 100 = 1170 . answer : b" | a ) 1040 , b ) 1170 , c ) 1150 , d ) 4160 , e ) none of these | b | multiply(divide(multiply(5200, 45), const_100), divide(subtract(const_100, 50), const_100)) | multiply(n0,n1)|subtract(const_100,n2)|divide(#0,const_100)|divide(#1,const_100)|multiply(#2,#3)| | general | B |
jar a has 16 % more marbles than jar b . what percent of marbles from jar a need to be moved into jar b so that both jars have equal marbles ? | an easy way to solve this question is by number plugging . assume there are 100 marbles in jar b then in jar a there will be 116 marbles . now , for both jars to have equal marbles we should move 8 marbles from a to b , which is 8 / 116 = ~ 7 % of a . answer : b . | a ) 5 % , b ) 7 % , c ) 9 % , d ) 11 % , e ) 13 % | b | multiply(divide(divide(16, const_2), add(16, const_100)), const_100) | add(n0,const_100)|divide(n0,const_2)|divide(#1,#0)|multiply(#2,const_100) | gain | B |
an association of mathematics teachers has 1,260 members . only 640 of these members cast votes in the election for president of the association . what percent of the total membership voted for the winning candidate if the winning candidate received 60 percent of the votes cast ? | "total number of members = 1260 number of members that cast votes = 640 since , winning candidate received 60 percent of the votes cast number of votes for winning candidate = ( 60 / 100 ) * 640 = 384 percent of total membership that voted for winning candidate = ( 384 / 1260 ) * 100 = 30.47 % answer a" | a ) 30.47 % , b ) 58 % , c ) 42 % , d ) 34 % , e ) 25 % | a | multiply(divide(multiply(divide(60, const_100), 640), multiply(const_100, power(const_4, const_2))), const_100) | divide(n2,const_100)|power(const_4,const_2)|multiply(n1,#0)|multiply(#1,const_100)|divide(#2,#3)|multiply(#4,const_100)| | gain | A |
john and jacob set out together on bicycle traveling at 12 and 9 miles per hour , respectively . after 40 minutes , john stops to fix a flat tire . if it takes john one hour to fix the flat tire and jacob continues to ride during this time , how many hours will it take john to catch up to jacob assuming he resumes his ride at 12 miles per hour ? ( consider john ' s deceleration / acceleration before / after the flat to be negligible ) | "john ' s speed - 12 miles / hr jacob ' s speed - 9 miles / hr after 40 min ( i . e 2 / 3 hr ) , distance covered by john = 12 x 2 / 3 = 8 miles . jacob continues to ride for a total of 1 hour and 40 min ( until john ' s bike is repaired ) . distance covered in 1 hour 40 min ( i . e 5 / 3 hr ) = 9 x 5 / 3 = 15 miles . now , when john starts riding back , the distance between them is 7 miles . jacob and john are moving in the same direction . for john to catch jacob , the effective relative speed will be 12 - 9 = 3 miles / hr . thus , to cover 7 miles at 3 miles / hr , john will take 7 / 3 = 2 1 / 3 hours answer b" | a ) 3 , b ) 2 1 / 3 , c ) 3 1 / 2 , d ) 4 , e ) 4 1 / 2 | b | divide(add(subtract(12, 9), subtract(9, subtract(multiply(divide(40, const_60), 12), multiply(divide(40, const_60), 9)))), subtract(multiply(divide(40, const_60), 12), multiply(divide(40, const_60), 9))) | divide(n2,const_60)|subtract(n0,n1)|multiply(n0,#0)|multiply(n1,#0)|subtract(#2,#3)|subtract(n1,#4)|add(#1,#5)|divide(#6,#4)| | physics | B |
a batsman in his 12 th innings makes a score of 115 and thereby increases his average by 3 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 – 4 ) + 115 ∴ x = 82 answer d" | a ) 42 , b ) 43 , c ) 44 , d ) 82 , e ) 46 | d | add(subtract(115, multiply(12, 3)), 3) | multiply(n0,n2)|subtract(n1,#0)|add(n2,#1)| | general | D |
how many zeros are there in 50 ! | "to find the number of trailing zeroes in 50 ! factorial we must find the number of integral values in : 50 / 5 + 50 / 5 ^ 2 + 50 / 5 ^ 3 + . . . . . . . . . = 10 + 2 + 0 + 0 . . . . = 12 answer : c" | a ) 10 , b ) 11 , c ) 12 , d ) 13 , e ) 14 | c | add(add(divide(50, add(const_4, const_1)), divide(subtract(50, add(const_4, const_1)), power(add(const_4, const_1), const_2))), divide(subtract(50, add(const_4, const_1)), power(add(const_4, const_1), const_3))) | add(const_1,const_4)|divide(n0,#0)|power(#0,const_2)|power(#0,const_3)|subtract(n0,#0)|divide(#4,#2)|divide(#4,#3)|add(#1,#5)|add(#7,#6)| | other | C |
after successive discounts of 20 % , 10 % and 5 % a certain good is sold for rs . 6840 . find the actual price of the good . | "let actual price was 100 . after three successive discount this will become , 100 = = 20 % discount = > 80 = = 10 % discount = > 72 = = 5 % discount = 68.4 now compare , 68.4 = 6840 1 = 6840 / 68.4 100 = ( 6840 * 100 ) / 68.4 = rs . 10,000 . answer : option d" | a ) rs . 6000 , b ) rs . 9000 , c ) rs . 10800 , d ) rs . 10000 , e ) rs . 9980 | d | divide(multiply(6840, const_100), subtract(subtract(subtract(const_100, 20), divide(multiply(subtract(const_100, 20), 10), const_100)), divide(multiply(subtract(subtract(const_100, 20), divide(multiply(subtract(const_100, 20), 10), const_100)), 5), const_100))) | multiply(n3,const_100)|subtract(const_100,n0)|multiply(n1,#1)|divide(#2,const_100)|subtract(#1,#3)|multiply(n2,#4)|divide(#5,const_100)|subtract(#4,#6)|divide(#0,#7)| | gain | D |
the ratio of the volumes of two cubes is 729 : 1000 . what is the ratio of their total surface areas ? | "ratio of the sides = ³ √ 729 : ³ √ 1000 = 9 : 10 ratio of surface areas = 92 : 102 = 46 : 51 answer : c" | a ) 1 : 12 , b ) 81 : 13 , c ) 46 : 51 , d ) 86 : 16 , e ) 56 : 13 | c | power(divide(729, 1000), divide(const_1, const_3)) | divide(n0,n1)|divide(const_1,const_3)|power(#0,#1)| | geometry | C |
5000 - 4500 ÷ 10.00 = ? | "answer given expression = 5000 - 4500 ÷ 10.00 = 5000 - 450 = 4550 correct option : a" | a ) 4550 , b ) 5000 , c ) 0.5 , d ) 4000 , e ) none | a | subtract(multiply(divide(5000, const_100), 4500), multiply(divide(const_1, const_3), multiply(divide(5000, const_100), 4500))) | divide(n0,const_100)|divide(const_1,const_3)|multiply(n1,#0)|multiply(#1,#2)|subtract(#2,#3)| | general | A |
fox jeans regularly sell for $ 15 a pair and pony jeans regularly sell for $ 18 a pair . during a sale these regular unit prices are discounted at different rates so that a total of $ 7 is saved by purchasing 5 pairs of jeans : 3 pairs of fox jeans and 2 pairs of pony jeans . if the sum of the two discounts rates is 18 percent , what is the discount rate on pony jeans ? | "you know that fox jeans costs $ 15 , and pony jeans costs $ 18 , you also know that 3 pairs of fox jeans and 2 pairs of pony jeans were purchased . so 3 ( 15 ) = 45 - fox 2 ( 18 ) = 36 - pony the total discount discount is $ 3 and you are asked to find the percent discount of pony jeans , so 45 ( 18 - x ) / 100 + 36 ( x ) / 100 = 7 or 45 * 18 - 45 * x + 36 * x = 7 * 100 or 9 x = - 7 * 100 + 45 * 18 x = 110 / 9 = 12.2 % e" | a ) 9 % , b ) 10 % , c ) 11 % , d ) 12 % , e ) 15 % | e | multiply(subtract(divide(18, const_100), divide(subtract(7, multiply(divide(18, const_100), multiply(18, 2))), subtract(multiply(15, 3), multiply(18, 2)))), const_100) | divide(n6,const_100)|multiply(n1,n5)|multiply(n0,n4)|multiply(#0,#1)|subtract(#2,#1)|subtract(n2,#3)|divide(#5,#4)|subtract(#0,#6)|multiply(#7,const_100)| | gain | E |
a city with a population of 126,160 is to be divided into 7 voting districts , and no district is to have a population that is more than 10 percent greater than the population of any other district . what is the minimum possible population that the least populated district could have ? | "the minimum possible population occurs when all the other districts have a population that is 10 % greater than the least populated district . let p be the population of the least populated district . then 126,160 = p + 6 ( 1.1 ) p 7.6 p = 126,160 p = 16,600 the answer is b ." | a ) 16,500 , b ) 16,600 , c ) 16,700 , d ) 16,800 , e ) 16,900 | b | divide(divide(subtract(multiply(const_1000, const_100), subtract(subtract(const_3600, const_100), const_1000)), const_1000), add(multiply(add(const_1, divide(10, const_100)), subtract(7, const_1)), const_1)) | divide(n2,const_100)|multiply(const_100,const_1000)|subtract(const_3600,const_100)|subtract(n1,const_1)|add(#0,const_1)|subtract(#2,const_1000)|multiply(#4,#3)|subtract(#1,#5)|add(#6,const_1)|divide(#7,const_1000)|divide(#9,#8)| | general | B |
the average ( arithmetic mean ) of 20 , 40 , and 60 is 4 more than the average of 10 , 70 , and what number ? | "a 1 = 120 / 3 = 40 a 2 = a 1 - 4 = 36 sum of second list = 36 * 3 = 108 therefore the number = 108 - 80 = 28 answer : a" | a ) 28 , b ) 25 , c ) 35 , d ) 45 , e ) 55 | a | subtract(add(add(20, 40), 60), add(add(multiply(4, const_3), 10), 70)) | add(n0,n1)|multiply(n3,const_3)|add(n2,#0)|add(n4,#1)|add(n5,#3)|subtract(#2,#4)| | general | A |
if p / q = 5 / 4 , then 2 p + q = ? | "let p = 5 , q = 4 then 2 * 5 + 4 = 14 so 2 p + q = 14 . answer : b" | a ) 12 , b ) 14 , c ) 13 , d ) 15 , e ) 16 | b | add(multiply(5, 2), 4) | multiply(n0,n2)|add(n1,#0)| | general | B |
the floor of a rectangular room is 17 m long and 12 m wide . the room is surrounded by a veranda of width 2 m on all its sides . the area of the veranda is : | "area of the outer rectangle = 21 ã — 16 = 336 m 2 area of the inner rectangle = 17 ã — 12 = 204 m 2 required area = ( 336 â € “ 204 ) = 132 m 2 answer b" | a ) 124 m 2 , b ) 132 m 2 , c ) 148 m 2 , d ) 152 m 2 , e ) none of these | b | subtract(rectangle_area(add(multiply(2, 2), 17), add(12, multiply(2, 2))), rectangle_area(17, 12)) | multiply(n2,n2)|rectangle_area(n0,n1)|add(n0,#0)|add(n1,#0)|rectangle_area(#2,#3)|subtract(#4,#1)| | geometry | B |
there are 870 male and female participants in a meeting . half the female participants and one - quarterof the male participants are democrats . one - third of all the participants are democrats . how many of the democrats are female ? | "let m be the number of male participants and f be the number of female articipants in the meeting . thetotal number of participants is given as 870 . hence , we have m + f = 870 now , we have that half the female participants and one - quarter of the male participants are democrats . let d equal the number of the democrats . then we have the equation f / 2 + m / 4 = d now , we have that one - third of the total participants are democrats . hence , we have the equation d = 870 / 3 = 290 solving the three equations yields the solution f = 290 , m = 580 , and d = 290 . the number of female democratic participants equals half the female participants equals 290 / 2 = 145 . answer : e" | a ) 75 , b ) 100 , c ) 125 , d ) 175 , e ) 145 | e | divide(subtract(multiply(divide(870, const_3), const_4), 870), const_2) | divide(n0,const_3)|multiply(#0,const_4)|subtract(#1,n0)|divide(#2,const_2)| | general | E |
if two integers x , y ( x > y ) are selected from - 5 to 6 ( inclusive ) , how many cases are there ? | "there are 12 integers from - 5 to 6 inclusive . 12 c 2 = 66 . the answer is b ." | a ) 62 , b ) 66 , c ) 70 , d ) 74 , e ) 78 | b | add(add(add(add(add(add(add(6, 5), add(6, const_2)), add(6, const_1)), 6), 5), const_2), const_1) | add(n0,n1)|add(n1,const_2)|add(n1,const_1)|add(#0,#1)|add(#3,#2)|add(#4,n1)|add(#5,n0)|add(#6,const_2)|add(#7,const_1)| | general | B |
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