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 floor of a rectangular room is 20 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 = 24 ã — 16 = 384 m 2 area of the inner rectangle = 20 ã — 12 = 240 m 2 required area = ( 304 â € “ 180 ) = 144 m 2 answer a" | a ) 144 m 2 , b ) 120 m 2 , c ) 108 m 2 , d ) 158 m 2 , e ) none of these | a | subtract(rectangle_area(add(multiply(2, 2), 20), add(12, multiply(2, 2))), rectangle_area(20, 12)) | multiply(n2,n2)|rectangle_area(n0,n1)|add(n0,#0)|add(n1,#0)|rectangle_area(#2,#3)|subtract(#4,#1)| | geometry |
the average weight of a class is x pounds . when a new student weighing 90 pounds joins the class , the average decreases by 1 pound . in a few months the student ’ s weight increases to 110 pounds and the average weight of the class becomes x + 4 pounds . none of the other students ’ weights changed . what is the valu... | when the student weighs 90 pounds the average weight is x - 1 pounds ; when the student weighs 110 pounds the average weight is x + 4 pounds . so , the increase in total weight of 110 - 90 = 20 pounds corresponds to the increase in average weight of ( x + 4 ) - ( x - 1 ) = 5 pounds , which means that there are 20 / 5 =... | a ) 85 , b ) 86 , c ) 88 , d ) 90 , e ) 99 | e | add(add(90, 4), add(4, 1)) | add(n0,n3)|add(n1,n3)|add(#0,#1) | general |
a sum was put at simple interest at a certain rate for 10 years . had it been put at 5 % higher rate , it would have fetched rs . 200 more . what was the sum ? | "at 5 % more rate , the increase in s . i for 10 years = rs . 200 ( given ) so , at 5 % more rate , the increase in si for 1 year = 200 / 10 = rs . 20 / - i . e . rs . 20 is 5 % of the invested sum so , 1 % of the invested sum = 20 / 5 therefore , the invested sum = 20 × 100 / 5 = rs . 400 answer : e" | a ) s . 550 , b ) s . 1000 , c ) s . 600 , d ) s . 200 , e ) s . 400 | e | divide(200, multiply(divide(5, const_100), 10)) | divide(n1,const_100)|multiply(n0,#0)|divide(n2,#1)| | gain |
the sum of a certain number and its reciprocal is equal to 29 . what is the absolute value of the difference of this number and its reciprocal ? | x + 1 / x = 2.9 x - 1 / x = y add two equations 2 x = 2.9 + y so when 2.9 and correct answer choice are added , the final digit should be even so eliminate a , b and d substitute y = 0.3 x = 1.6 1.6 + 1 / 1.6 is not equal to 2.9 substitute y = 2.1 so x = 2.5 2.5 + 1 / 2.5 = 2.9 answer : e | a ) 22 , b ) 12 , c ) 03 , d ) 04 , e ) 21 | e | subtract(divide(29, const_10), divide(power(const_2, const_3), const_10)) | divide(n0,const_10)|power(const_2,const_3)|divide(#1,const_10)|subtract(#0,#2) | general |
what is the total number of positive integers that are less than 500 and that have no positive factor in common with 500 other than 1 ? | since 500 = 2 ^ 2 * 5 ^ 3 then a number can not have 2 and / or 5 as a factor . the odd numbers do not have 2 as a factor and there are 250 odd numbers from 1 to 500 . we then need to eliminate the 50 numbers that end with 5 , that is 5 , 15 , 25 , . . . , 495 . there are a total of 250 - 50 = 200 such numbers between ... | a ) 50 , b ) 100 , c ) 200 , d ) 300 , e ) 400 | c | subtract(subtract(500, const_1), subtract(add(add(divide(subtract(subtract(500, const_2), const_2), const_2), const_1), add(divide(subtract(subtract(500, divide(500, const_100)), divide(500, const_100)), divide(500, const_100)), const_1)), add(divide(subtract(subtract(500, const_10), const_10), const_10), 1))) | divide(n0,const_100)|subtract(n0,const_1)|subtract(n0,const_2)|subtract(n0,const_10)|subtract(#2,const_2)|subtract(n0,#0)|subtract(#3,const_10)|divide(#4,const_2)|divide(#6,const_10)|subtract(#5,#0)|add(#7,const_1)|add(n2,#8)|divide(#9,#0)|add(#12,const_1)|add(#10,#13)|subtract(#14,#11)|subtract(#1,#15) | general |
a cistern can be filled by a tap in 4 hours while it can be emptied by another tap in 8 hours . if both the taps are opened simultaneously , then after how much time will the cistern get filled ? | "net part filled in 1 hour = ( 1 / 4 - 1 / 9 ) = 1 / 8 the cistern will be filled in 8 / 1 hrs i . e . , 8 hrs . answer : e" | a ) 7.5 , b ) 7.4 , c ) 7.9 , d ) 7.2 , e ) 8.0 | e | divide(const_1, subtract(divide(const_1, 4), divide(const_1, 8))) | divide(const_1,n0)|divide(const_1,n1)|subtract(#0,#1)|divide(const_1,#2)| | physics |
in a simultaneous throw of pair of dice . find the probability of getting the total more than 7 . | "explanation : here n ( s ) = ( 6 x 6 ) = 36 let e = event of getting a total more than 7 = { ( 2,6 ) , ( 3,5 ) , ( 3,6 ) , ( 4,4 ) , ( 4,5 ) , ( 4,6 ) , ( 5,3 ) , ( 5,4 ) , ( 5,5 ) , ( 5,6 ) , ( 6,2 ) , ( 6,3 ) , ( 6,4 ) , ( 6,5 ) , ( 6,6 ) } therefore , p ( e ) = n ( e ) / n ( s ) = 15 / 36 = 5 / 12 . answer : b ) 5 ... | a ) 5 / 19 , b ) 5 / 12 , c ) 5 / 28 , d ) 5 / 10 , e ) 5 / 11 | b | divide(add(add(7, const_4), const_4), multiply(add(const_4, const_2), add(const_4, const_2))) | add(n0,const_4)|add(const_2,const_4)|add(#0,const_4)|multiply(#1,#1)|divide(#2,#3)| | general |
krishan and nandan jointly started a business . krishan invested 6 times as nandan did and invested his money for double time as compared to nandan . nandan earned rs . 6000 . if the gain is proportional to the money invested and the time for which the money is invested then the total gain was ? | 6 : 1 2 : 1 - - - - - - 12 : 1 1 - - - - - 6000 13 - - - - - ? = > rs . 78,000 answer : a | a ) rs . 78000 , b ) rs . 48000 , c ) rs . 6000 , d ) rs . 82000 , e ) rs . 32000 | a | multiply(6000, add(multiply(6, const_2), const_1)) | multiply(n0,const_2)|add(#0,const_1)|multiply(n1,#1) | general |
consider the word rotor . whichever way you read it , from left to right or from right to left , you get the same word . such a word is known as palindrome . find the maximum possible number of 5 - letter palindromes | the first letter from the right can be chosen in 26 ways because there are 26 alphabets . having chosen this , the second letter can be chosen in 26 ways . = > the first two letters can be chosen in 26 ã — 26 = 67626 ã — 26 = 676 ways having chosen the first two letters , the third letter can be chosen in 26 ways . = >... | a ) 1700 , b ) 7570 , c ) 1576 , d ) 17576 , e ) 500 | d | power(add(add(add(const_10, const_10), const_3), const_3), subtract(5, const_2)) | add(const_10,const_10)|subtract(n0,const_2)|add(#0,const_3)|add(#2,const_3)|power(#3,#1) | general |
a glass was filled with 25 ounces of water , and 0.04 ounce of the water evaporated each day during a 10 - day period . what percent of the original amount of water evaporated during this period ? | "in 10 days 10 * 0.04 = 0.4 ounces of water evaporated , which is 0.4 / 25 â ˆ — 100 = 1.6 of the original amount of water . answer : d ." | a ) 0.004 % , b ) 0.04 % , c ) 0.4 % , d ) 1.6 % , e ) 20 % | d | multiply(divide(multiply(0.04, 10), 25), const_100) | multiply(n1,n2)|divide(#0,n0)|multiply(#1,const_100)| | gain |
if teena is driving at 55 miles per hour and is currently 7.5 miles behind joe , who is driving at 40 miles per hour in the same direction then in how many minutes will teena be 37.5 miles ahead of joe ? | "this type of questions should be solved without any complex calculations as these questions become imperative in gaining that extra 30 - 40 seconds for a difficult one . teena covers 55 miles in 60 mins . joe covers 40 miles in 60 mins so teena gains 15 miles every 60 mins teena need to cover 7.5 + 37.5 miles . teena ... | a ) 15 , b ) 60 , c ) 180 , d ) 90 , e ) 105 | c | multiply(divide(add(subtract(55, 40), 7.5), subtract(55, 40)), const_60) | subtract(n0,n2)|add(n1,#0)|divide(#1,#0)|multiply(#2,const_60)| | physics |
vijay lent out an amount rs . 10000 into two parts , one at 8 % p . a . and the remaining at 10 % p . a . both on simple interest . at the end of the year he received rs . 850 as total interest . what was the amount he lent out at 8 % pa . a ? | let the amount lent out at 8 % p . a . be rs . a = > ( a * 8 ) / 100 + [ ( 10000 - a ) * 10 ] / 100 = 850 = > a = rs . 15000 . answer : a | a ) rs . 15000 , b ) rs . 6000 , c ) rs . 25000 , d ) rs . 10000 , e ) rs . 18000 | a | multiply(subtract(add(multiply(divide(10, const_100), 10000), divide(8, const_100)), 850), const_100) | divide(n1,const_100)|divide(n2,const_100)|multiply(n0,#1)|add(#0,#2)|subtract(#3,n3)|multiply(#4,const_100) | gain |
find k if 64 ÷ k = 4 . | "since 64 ÷ k = 4 and 64 ÷ 16 = 4 , then k = 16 correct answer b" | a ) 8 , b ) 16 , c ) 24 , d ) 36 , e ) 40 | b | divide(64, 4) | divide(n0,n1)| | general |
what is the area of square field whose side of length 12 m ? | "12 * 12 = 144 sq m answer : b" | a ) 225 sq m , b ) 144 sq m , c ) 586 sq m , d ) 287 sq m , e ) 296 sq m | b | square_area(12) | square_area(n0)| | geometry |
the mean of 50 observations is 200 . but later he found that there is decrements of 9 from each observations . what is the the updated mean is ? | "191 answer is d" | a ) 165 , b ) 185 , c ) 190 , d ) 191 , e ) 199 | d | subtract(200, 9) | subtract(n1,n2)| | general |
a couple spent $ 184.80 in total while dining out and paid this amount using a credit card . the $ 184.80 figure included a 20 percent tip which was paid on top of the price which already included a sales tax of 10 percent on top of the price of the food . what was the actual price of the food before tax and tip ? | "let the price of the meal be x . after a 10 % sales tax addition , the price is 1.1 * x after a 20 % tip on this amount , the total is 1.2 * 1.1 * x = 1.32 x 1.32 x = 184.80 x = 140 the correct answer is c ." | a ) $ 124 , b ) $ 132 , c ) $ 140 , d ) $ 148 , e ) $ 156 | c | divide(multiply(divide(multiply(184.80, const_100), add(const_100, 20)), const_100), add(const_100, 10)) | add(n2,const_100)|add(n3,const_100)|multiply(n0,const_100)|divide(#2,#0)|multiply(#3,const_100)|divide(#4,#1)| | general |
for any integer k greater than 1 , the symbol k * denotes the product of all the fractions of the form 1 / t , where t is an integer between 1 and k , inclusive . what is the value of 5 * / 6 * ? | "when dealing with ' symbolism ' questions , it often helps to ' play with ' the symbol for a few moments before you attempt to answer the question that ' s asked . by understanding how the symbol ' works ' , you should be able to do the latter calculations faster . here , we ' re told that k * is the product of all th... | a ) 5 , b ) 5 / 4 , c ) 4 / 5 , d ) 1 / 4 , e ) 1 / 5 | d | divide(divide(divide(1, const_3), const_3), add(1, const_4)) | add(n0,const_4)|divide(n2,const_3)|divide(#1,const_3)|divide(#2,#0)| | general |
running at the same rate , 8 identical machines can produce 560 paperclips a minute . at this rate , how many paperclips could 30 machines produce in 6 minutes ? | "8 machines produce 560 in 1 min 8 machines produce 560 * 6 in 6 min 30 machine produce 560 * 6 * ( 30 / 8 ) in 6 minutes 560 * 6 * 30 / 8 = 12600 answer is d ." | a ) 1344 , b ) 3360 , c ) 8400 , d ) 12600 , e ) 67200 | d | multiply(multiply(560, divide(30, 8)), 6) | divide(n2,n0)|multiply(n1,#0)|multiply(n3,#1)| | gain |
abel can complete a work in 10 days , ben in 12 days and carla in 15 days . all of them began the work together , but abel had to leave after 2 days and ben 5 days before the completion of the work . how long did the work last ? | "abel in the 2 days that he worked completed 1 / 5 of the job = 4 / 5 remains then if ben had to leave 5 days before the completion , this means that carla had to work alone for these 5 days in which she completed 1 / 3 of the job . now together , ben and carla completed the job in ( 1 / 12 + 1 / 15 ) ( t ) = 7 / 15 3 ... | a ) 6 , b ) 7 , c ) 8 , d ) 8 1 / 9 , e ) 10 | d | multiply(add(2, 5), 5) | add(n3,n4)|multiply(n4,#0)| | physics |
a man can do a piece of work in 5 days , but with the help of his son , he can do it in 3 days . in what time can the son do it alone ? | "son ' s 1 day ' s work = ( 1 / 3 ) - ( 1 / 5 ) = 2 / 5 the son alone can do the work in 15 / 2 days answer is c" | a ) 13 / 2 , b ) 7 , c ) 15 / 2 , d ) 8 , e ) 9 | c | divide(multiply(5, 3), subtract(5, 3)) | multiply(n0,n1)|subtract(n0,n1)|divide(#0,#1)| | physics |
annika hikes at a constant rate of 10 minutes per kilometer . she has hiked 2.75 kilometers east from the start of a hiking trail when she realizes that she has to be back at the start of the trail in 45 minutes . if annika continues east , then turns around and retraces her path to reach the start of the trail in exac... | "set up two r x t = d cases . 1 . 1 / 10 km / min x t = 2.75 from which t = 27.5 mins . we know total journey time now is 45 + 27.5 = 72.5 the rate is the same ie 1 / 10 km / min . set up second r x t = d case . 1 / 10 km / min x 72.5 = 7.25 km now the total journey would be halved as distance would be same in each dir... | a ) 3.625 , b ) 3.5 , c ) 4 , d ) 4.5 , e ) 5 | a | add(divide(subtract(divide(45, 10), 2.75), const_2), 2.75) | divide(n2,n0)|subtract(#0,n1)|divide(#1,const_2)|add(n1,#2)| | physics |
the average of 10 numbers is calculated as 20 . it is discovered later on that while calculating the average , one number namely 36 was wrongly read as 26 . the correct average is ? | "10 * 20 + 36 – 26 = 200 / 10 = 20 answer : e" | a ) 12 , b ) 26 , c ) 16 , d ) 97 , e ) 20 | e | add(20, divide(subtract(36, 26), 10)) | subtract(n2,n3)|divide(#0,n0)|add(n1,#1)| | general |
two trains 161 meters and 165 meters in length respectively are running in opposite directions , one at the rate of 80 km and the other at the rate of 65 kmph . in what time will they be completely clear of each other from the moment they meet ? | "t = ( 161 + 165 ) / ( 80 + 65 ) * 18 / 5 t = 8.09 answer : d" | a ) 6.18 , b ) 7.12 , c ) 7.1 , d ) 8.09 , e ) 8.11 | d | divide(add(161, 165), multiply(add(80, 65), const_0_2778)) | add(n0,n1)|add(n2,n3)|multiply(#1,const_0_2778)|divide(#0,#2)| | physics |
there is 60 % increase in an amount in 6 years at s . i . what will be the c . i . of rs . 12,000 after 3 years at the same rate ? | "explanation : let p = rs . 100 . then , s . i . rs . 60 and t = 6 years . r = ( 100 * 60 ) / ( 100 * 6 ) = 10 % p . a . now , p = rs . 12000 , t = 3 years and r = 10 % p . a . c . i . = [ 12000 * { ( 1 + 10 / 100 ) 3 - 1 } ] = 12000 * 331 / 1000 = rs . 3972 answer : option c" | a ) rs . 3932 , b ) rs . 3972 , c ) rs . 3372 , d ) rs . 3942 , e ) rs . 3772 | c | subtract(multiply(add(multiply(const_100, const_100), multiply(multiply(const_100, divide(60, 6)), 3)), multiply(multiply(add(const_1, divide(divide(60, 6), const_100)), add(const_1, divide(divide(60, 6), const_100))), add(const_1, divide(divide(60, 6), const_100)))), add(multiply(const_100, const_100), multiply(multip... | divide(n0,n1)|multiply(const_100,const_100)|divide(#0,const_100)|multiply(#0,const_100)|add(#2,const_1)|multiply(#3,n3)|add(#1,#5)|multiply(#4,#4)|multiply(#4,#7)|multiply(#6,#8)|subtract(#9,#6)| | gain |
in a mixed college 160 students are there in one class . out of this 160 students 1 / 2 students are girls . how many boys are there ? | "total number of students : 160 total girls : 160 * 1 / 2 = 80 total boys : 160 - 80 = 80 answer is c" | a ) a ) 40 , b ) b ) 60 , c ) c ) 80 , d ) d ) 120 , e ) e ) 140 | c | multiply(divide(160, 2), const_3) | divide(n0,n3)|multiply(#0,const_3)| | general |
the average weight of 8 person ' s increases by 4.2 kg when a new person comes in place of one of them weighing 65 kg . what is the weight of the new person ? | "explanation : total increase in weight = 8 ã — 4.2 = 33.6 if x is the weight of the new person , total increase in weight = x â ˆ ’ 65 = > 33.6 = x - 65 = > x = 33.6 + 65 = 98.6 answer : option e" | a ) 75 kg , b ) 50 kg , c ) 85 kg , d ) 80 kg , e ) 98.6 kg | e | add(multiply(8, 4.2), 65) | multiply(n0,n1)|add(n2,#0)| | general |
on sunday morning , pugsley and wednesday are trading pet spiders . if pugsley were to give wednesday two of his spiders , wednesday would then have 9 times as many spiders as pugsley does . but , if wednesday were to give pugsley 6 of her spiders , pugsley would now have 6 fewer spiders than wednesday had before they ... | if pugsley were to give wednesday two of his spiders , wednesday would then have nine times as many spiders as pugsley does : ( w + 2 ) = 9 ( p - 2 ) if wednesday were to give pugsley six of her spiders , pugsley would now have six fewer spiders than wednesday had before they traded : p + 6 = w - 6 solving gives p = 4 ... | a ) 4 , b ) 6 , c ) 8 , d ) 10 , e ) 20 | a | subtract(divide(add(add(add(add(6, 6), 6), multiply(add(6, 6), 9)), const_2), add(6, const_2)), add(6, 6)) | add(n1,n1)|add(n1,const_2)|add(n1,#0)|multiply(n0,#0)|add(#2,#3)|add(#4,const_2)|divide(#5,#1)|subtract(#6,#0) | physics |
rates for having a manuscript typed at a certain typing service are $ 10 per page for the first time a page is typed and $ 5 per page each time a page is revised . if a certain manuscript has 100 pages , of which 30 were revised only once , 20 were revised twice , and the rest required no revisions , what was the total... | "50 pages typed 1 x 30 pages typed 2 x ( original + one revision ) 20 pages typed 3 x ( original + two revisions ) 50 ( 10 ) + 30 ( 10 + 5 ) + 20 ( 10 + 5 + 5 ) = 500 + 450 + 400 = 1350 answer - b" | a ) $ 1250 , b ) $ 1350 , c ) $ 1150 , d ) $ 1360 , e ) $ 1370 | b | add(add(multiply(100, 10), multiply(30, 5)), multiply(multiply(20, 5), const_2)) | multiply(n0,n2)|multiply(n1,n3)|multiply(n1,n4)|add(#0,#1)|multiply(#2,const_2)|add(#3,#4)| | general |
shekhar started a business investing rs . 25,000 in 1999 . in 2000 , he invested an additional amount of rs . 10,000 and rajeev joined him with an amount of rs . 35,000 . in 2001 , shekhar invested another additional amount of rs . 10,000 and jatin joined them with an amount of rs . 35,000 . what will be rajeev ’ s sha... | shekhar : rajeev : jatin = ( 25000 x 12 + 35000 x 12 + 45000 x 12 ) : ( 35000 x 24 ) : ( 35000 x 12 ) = 1260000 : 840000 : 420000 = 3 : 2 : 1 . rajeev ’ s share = rs . ( 150000 × 2 / 6 ) = rs . 50000 answer e | a ) rs . 10000 , b ) rs . 20000 , c ) rs . 30000 , d ) rs . 40000 , e ) rs . 50000 | e | multiply(multiply(multiply(multiply(add(const_3, const_4), add(const_2, const_3)), multiply(multiply(add(const_2, const_3), const_2), const_100)), const_2), divide(multiply(multiply(const_3, add(const_2, const_3)), multiply(multiply(add(const_2, const_3), const_2), multiply(multiply(add(const_2, const_3), const_2), con... | add(const_2,const_3)|add(const_3,const_4)|multiply(#0,const_3)|multiply(#0,const_2)|multiply(#0,#0)|multiply(#1,#0)|multiply(#3,const_100)|multiply(#3,#6)|multiply(#4,#6)|multiply(#5,#6)|multiply(#2,#7)|multiply(#8,const_3)|multiply(#9,const_2)|multiply(#7,const_2)|add(#11,#12)|add(#14,#13)|add(#15,#7)|add(#16,#9)|divi... | gain |
if x dollars is invested at 10 percent for one year and y dollars is invested at 8 percent for one year , the annual income from the 10 percent investment will exceed the annual income from the 8 percent investment by $ 74 . if $ 2,000 is the total amount invested , how much is invested at 8 percent ? | "0.1 x = 0.08 ( 2000 - x ) + 74 0.18 x = 234 x = 1300 then the amount invested at 8 % is $ 2000 - $ 1300 = $ 700 the answer is b ." | a ) $ 600 , b ) $ 700 , c ) $ 800 , d ) $ 900 , e ) $ 1000 | b | subtract(multiply(multiply(const_100, 10), const_2), divide(add(multiply(multiply(10, 8), const_2), 74), add(divide(10, const_100), divide(8, const_100)))) | divide(n0,const_100)|divide(n1,const_100)|multiply(n0,const_100)|multiply(n0,n1)|add(#0,#1)|multiply(#2,const_2)|multiply(#3,const_2)|add(n4,#6)|divide(#7,#4)|subtract(#5,#8)| | general |
at the opening of a trading day at a certain stock exchange , the price per share of stock k was $ 28 . if the price per share of stock k was $ 29 at the closing of the day , what was the percent increase in the price per share of stock k for that day ? | "opening = 28 closing = 29 rise in price = 1 so , percent increase = 1 / 28 * 100 = 3.57 answer : a" | a ) 3.57 % , b ) 5.9 % , c ) 11.1 % , d ) 12.5 % , e ) 23.6 % | a | multiply(subtract(divide(29, 28), const_1), const_100) | divide(n1,n0)|subtract(#0,const_1)|multiply(#1,const_100)| | general |
the population of a town is 20000 . it increases annually at the rate of 20 % p . a . what will be its population after 2 years ? | "explanation : formula : ( after = 100 denominator ago = 100 numerator ) 20000 * 120 / 100 * 120 / 100 = 120 * 120 * 2 = 28800 answer : option d" | a ) 24440 , b ) 24800 , c ) 28400 , d ) 28800 , e ) 28600 | d | add(20000, multiply(divide(multiply(20000, 20), const_100), 2)) | multiply(n0,n1)|divide(#0,const_100)|multiply(#1,n2)|add(n0,#2)| | gain |
bookman purchased 55 copies of a new book released recently , 10 of which are hardback and sold for $ 20 each , and rest are paperback and sold for $ 10 each . if 14 copies were sold and the total value of the remaining books was 360 , how many paperback copies were sold ? | the bookman had 10 hardback ad 55 - 10 = 45 paperback copies ; 14 copies were sold , hence 55 - 14 = 41 copies were left . let # of paperback copies left be p then 10 p + 20 ( 41 - p ) = 560 - - > 10 p = 260 - - > p = 26 # of paperback copies sold is 45 - 26 = 19 answer : d | a ) 8 , b ) 10 , c ) 12 , d ) 19 , e ) 16 | d | divide(subtract(subtract(add(multiply(subtract(55, 10), 10), multiply(10, 20)), 360), multiply(gcd(55, 10), 20)), 10) | gcd(n0,n1)|multiply(n1,n2)|subtract(n0,n1)|multiply(n1,#2)|multiply(n2,#0)|add(#3,#1)|subtract(#5,n5)|subtract(#6,#4)|divide(#7,n1) | general |
in a company with 48 employees , some part - time and some full - time , exactly ( 1 / 3 ) of the part - time employees and ( 1 / 4 ) of the full - time employees take the subway to work . what is the greatest possible number r of employees who take the subway to work ? | "p / 3 + f / 4 = p / 3 + ( 48 - p ) / 4 = 12 + p / 2 p / 3 + f / 3 = ( p + f ) / 3 = 48 / 3 = 16 p / 4 + f / 4 = 12 p / 3 + f / 3 > p / 3 + f / 4 > p / 4 + f / 4 - - > 16 > 12 + p / 12 > 12 greatest possible r : 12 + p / 12 = 15 - - > p = 36 ( integer - - > good ) 15 or d is the answer" | a ) 12 , b ) 13 , c ) 14 , d ) 15 , e ) 16 | d | add(divide(multiply(48, 3), multiply(3, 4)), divide(subtract(48, multiply(3, 4)), divide(multiply(48, 3), multiply(3, 4)))) | multiply(n0,n2)|multiply(n2,n4)|divide(#0,#1)|subtract(n0,#1)|divide(#3,#2)|add(#2,#4)| | general |
in an electric circuit , three resistors with resistances 2 ohms , 5 ohms and 6 ohms are connected in parallel . in this case , if r is the combined resistance of these three resistors , then the reciprocal of r is equal to the sum of the reciprocals resistors . what is r value ? | the wording is a bit confusing , though basically we are told that 1 / r = 1 / 2 + 1 / 5 + 1 / 6 , from which it follows that r = 13 / 15 ohms option : a | ['a ) 13 / 15 ohms', 'b ) 15 / 13 ohms', 'c ) 5 / 8 ohms', 'd ) 3 / 7 ohms', 'e ) 45 / 18 ohms'] | a | add(divide(const_1, 6), add(divide(const_1, 2), divide(const_1, 5))) | divide(const_1,n0)|divide(const_1,n1)|divide(const_1,n2)|add(#0,#1)|add(#3,#2) | geometry |
if the weight of 13 meters long rod is 23.4 kg . what is the weight of 6 meters long rod ? | "answer ∵ weight of 13 m long rod = 23.4 kg ∴ weight of 1 m long rod = 23.4 / 13 kg ∴ weight of 6 m long rod = 23.4 x 6 / 13 = 10.8 kg option : b" | a ) 7.2 kg . , b ) 10.8 kg . , c ) 12.4 kg . , d ) 18.0 kg , e ) none | b | divide(multiply(6, 23.4), 13) | multiply(n1,n2)|divide(#0,n0)| | physics |
if the simple interest on a certain sum of money for 7 years is one – fifth of the sum , then the rate of interest per annum is | "explanation : let the principal ( p ) be x then , simple interest ( si ) = x / 5 time ( t ) = 7 years rate of interest per annum ( r ) = ( 100 × si ) / pt = ( 100 × ( x / 5 ) / ( x × 7 ) = 20 / 7 = 2.85 % answer : option e" | a ) 4 % , b ) 2.50 % , c ) 6.20 % , d ) 5 % , e ) 2.85 % | e | divide(divide(const_100, add(const_1, const_4)), 7) | add(const_1,const_4)|divide(const_100,#0)|divide(#1,n0)| | gain |
in may , the grounds keeper at spring lake golf club built a circular green with an area of 64 π square feet . in august , he doubled the distance from the center of the green to the edge of the green . what is the total area of the renovated green ? | "area of circle 64 pi sq feet = pi r ^ 2 therefore r = 8 now green radi doubled i . e r = 16 area = 256 pi d" | a ) 100 pi , b ) 144 pi , c ) 196 pi , d ) 256 pi , e ) 78 | d | circle_area(multiply(sqrt(64), const_2)) | sqrt(n0)|multiply(#0,const_2)|circle_area(#1)| | geometry |
how many positive integers less than 60 are equal to the product of a positive multiple of 5 and an even number ? | "number of positive multiples of 5 less than 60 = 11 5 * 1 = 5 5 * 2 = 10 5 * 3 = 15 5 * 4 = 20 5 * 5 = 25 5 * 6 = 30 5 * 7 = 35 5 * 8 = 40 5 * 9 = 45 5 * 10 = 50 5 * 11 = 55 only 5 of the above are a product of a positive multiple of 5 and an even number - 10 , 20,30 , 40,50 answer b" | a ) 4 , b ) 5 , c ) 9 , d ) 10 , e ) 11 | b | divide(factorial(subtract(add(const_4, 5), const_1)), multiply(factorial(5), factorial(subtract(const_4, const_1)))) | add(n1,const_4)|factorial(n1)|subtract(const_4,const_1)|factorial(#2)|subtract(#0,const_1)|factorial(#4)|multiply(#1,#3)|divide(#5,#6)| | general |
alex has to take a quiz , which has 10 true false - type question each carry 1 mark . what is the probability that alex ' s can score more than 8 mark in the quiz . given that he decides to guess randomly on the quiz . | each question has 2 options ( true or false ) . total number of ways of answering all 10 questions = ( 2 ) 10 ( 2 ) 10 more than 8 correct means either 9 correct or all correct . number of ways such that 9 correct = 10 ! / 9 ! = 10 number of ways of all correct = 1 therefore , probability of more than 8 correct = 11 / ... | a ) 12 / 1024 , b ) 11 / 1024 , c ) 11 / 256 , d ) 12 / 256 , e ) 14 / 256 | b | add(divide(const_1, power(const_2, 10)), divide(10, power(const_2, 10))) | power(const_2,n0)|divide(const_1,#0)|divide(n0,#0)|add(#1,#2) | general |
if x and y are the two digits of the number 653 xy such that this number is divisible by 80 , then x + y = ? | "80 = 2 x 5 x 8 since 653 xy is divisible by 2 and 5 both , so y = 0 . now , 653 x is divisible by 8 , so 13 x should be divisible by 8 . this happens when x = 6 . x + y = ( 6 + 0 ) = 6 . answer : a" | a ) 6 , b ) 4 , c ) 8 , d ) 8 , e ) none of these | a | divide(reminder(multiply(653, const_100), 80), const_10) | multiply(n0,const_100)|reminder(#0,n1)|divide(#1,const_10)| | general |
if x and y are integers , what is the least positive number of 24 x + 9 y ? | "24 x + 9 y = 3 ( 8 x + 3 y ) which will be a minimum positive number when 8 x + 3 y = 1 . 8 ( - 1 ) + 3 ( 3 ) = 1 then 3 ( 8 x + 3 y ) can have a minimum positive value of 3 . the answer is b ." | a ) 1 , b ) 3 , c ) 4 , d ) 6 , e ) 9 | b | subtract(24, 9) | subtract(n0,n1)| | general |
the average earning of a mechanic for the first - 4 days of a week is rs . 18 and for the last 4 days is rs . 22 . if he earns rs . 13 on the fourth day , his average earning for the whole week is ? | answer total earning for the week = sum of earning of first four days + sum of earning of last four days - earning of 4 th day = 4 x 18 + 4 x 22 - 13 = rs . 147 â ˆ ´ average earning = 147 / 7 = rs . 21 correct option : c | a ) rs . 18.95 , b ) rs 16 , c ) rs . 21 , d ) rs . 25.71 , e ) none of these | c | divide(subtract(add(multiply(4, 18), multiply(4, 22)), 13), add(const_3, const_4)) | add(const_3,const_4)|multiply(n0,n1)|multiply(n0,n3)|add(#1,#2)|subtract(#3,n4)|divide(#4,#0) | general |
six bells commence tolling together and toll at intervals of 4 , 6 , 8 , 10 , 12 and 14 seconds respectively . in 70 minutes , how many times do they toll together ? | "lcm of 4 , 6 , 8 10 , 12 and 14 is 840 . so , after each 840 seconds , they would toll together . hence , in 70 minutes , they would toll 70 * 60 seconds / 840 seconds = 5 times but then the question says they commence tolling together . so , they basically also toll at thebeginning ( 0 second ) . so , total tolls tog... | a ) 2 , b ) 3 , c ) 4 , d ) 5 , e ) 6 | e | divide(12, divide(multiply(multiply(4, multiply(multiply(add(4, const_3), 4), const_3)), 4), 12)) | add(n0,const_3)|multiply(n0,#0)|multiply(#1,const_3)|multiply(n0,#2)|multiply(n0,#3)|divide(#4,n4)|divide(n4,#5)| | physics |
can you please walk me through how to best approach this problem ? thanks if # p # = ap ^ 3 + bp – 1 where a and b are constants , and # - 5 # = 10 , what is the value of # 5 # ? | # p # = ap ^ 3 + bp - 1 # - 5 # = 10 putting p = - 5 in above equation - 125 a - ( 5 b + 1 ) = 10 or # - 5 # = ( 125 a + 5 b + 1 ) = - 10 therefore 125 a + 5 b = - 11 . . . . . ( 1 now putting p = 5 # 5 # = 125 a + 5 b - 1 using equation 1 ( 125 a + 5 b = - 11 ) # 5 # = - 11 - 1 = - 12 hence b | a ) 5 , b ) - 12 , c ) - 2 , d ) - 3 , e ) - 5 | b | subtract(negate(3), const_2) | negate(n0)|subtract(#0,const_2) | general |
what is the unit digit in the product ( 3 ^ 68 x 6 ^ 59 x 7 ^ 71 ) ? | "unit digit in 3 ^ 4 = 1 = > unit digit in ( 3 ^ 4 ) ^ 16 = 1 therefore , unit digit in 3 ^ 65 = unit digit in [ ( 3 ^ 4 ) ^ 16 x 3 ] = ( 1 x 3 ) = 3 unit digit in 6 ^ 59 = 6 unit digit in 7 ^ 4 = 1 = > unit digit in ( 7 ^ 4 ) ^ 17 is 1 unit digit in 7 ^ 71 = unit digit in [ ( 7 ^ 4 ) ^ 17 x 7 ^ 3 ] = ( ( 1 x 3 ) = 3 t... | a ) 1 , b ) 2 , c ) 4 , d ) 6 , e ) none of them | c | subtract(multiply(multiply(3, 6), 3), subtract(multiply(multiply(3, 6), 3), const_4)) | multiply(n0,n2)|multiply(n0,#0)|subtract(#1,const_4)|subtract(#1,#2)| | general |
two goods trains each 500 m long are running in opposite directions on parallel tracks . their speeds are 45 km / hr and 15 km / hr respectively . find the time taken by the slower train to pass the driver of the faster one ? | "relative speed = 45 + 15 = 60 km / hr . 60 * 5 / 18 = 50 / 3 m / sec . distance covered = 500 + 500 = 1000 m . required time = 1000 * 3 / 50 = 60 sec . answer : b" | a ) 78 , b ) 60 , c ) 26 , d ) 23 , e ) 12 | b | add(45, 15) | add(n1,n2)| | physics |
if two integers x , y ( x > y ) are selected from - 10 to 11 ( inclusive ) , how many possible cases are there ? | "if two integers x , y ( x > y ) are selected from - 10 to 9 ( inclusive ) , how many possible cases are there ? a . 150 b . 180 c . 190 d . 210 e . 240 - - > 22 c 2 = 22 * 21 / 2 = 231 . therefore , the answer is e ." | a ) 150 , b ) 180 , c ) 190 , d ) 210 , e ) 231 | e | add(add(add(add(add(add(add(11, 10), add(11, const_2)), add(11, const_1)), 11), 10), 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)| | probability |
an empty fuel tank with a capacity of 200 gallons was filled partially with fuel a and then to capacity with fuel b . fuel a contains 12 % ethanol by volume and fuel b contains 16 % ethanol by volume . if the full fuel tank contains 24 gallons of ethanol , how many gallons of fuel a were added ? | "say there are a gallons of fuel a in the tank , then there would be 200 - a gallons of fuel b . the amount of ethanol in a gallons of fuel a is 0.12 a ; the amount of ethanol in 200 - a gallons of fuel b is 0.16 ( 200 - a ) ; since the total amount of ethanol is 24 gallons then 0.12 a + 0.16 ( 200 - a ) = 24 - - > a =... | a ) 160 , b ) 150 , c ) 100 , d ) 80 , e ) 200 | e | divide(subtract(multiply(200, divide(16, const_100)), 24), subtract(divide(16, const_100), divide(12, const_100))) | divide(n2,const_100)|divide(n1,const_100)|multiply(n0,#0)|subtract(#0,#1)|subtract(#2,n3)|divide(#4,#3)| | gain |
a take twice as much time as b or thrice as much time to finish a piece of work . working together , they can finish the work in 2 days . b can do the work alone in ? | "b 6 hours suppose a , b and c take x , x / 2 and x / 3 respectively to finish the work . then , ( 1 / x + 2 / x + 3 / x ) = 1 / 2 6 / x = 1 / 2 = > x = 12 so , b takes 6 hours to finish the work ." | a ) 3 hours , b ) 6 hours , c ) 7 hours , d ) 4 hours , e ) 8 hours | b | multiply(add(add(const_1, const_2), const_3), 2) | add(const_1,const_2)|add(#0,const_3)|multiply(n0,#1)| | physics |
a circle in the coordinate plane passes through points ( - 3 , - 2 ) and ( 1 , 3 ) . what is the smallest possible area of that circle ? | "the distance between the two points is sqrt ( 41 ) . radius = sqrt ( 41 ) / 2 area = pi * ( sqrt ( 41 ) / 2 ) ^ 2 a . 10.5 π" | a ) 10.5 π , b ) 26 π , c ) 262 √ π , d ) 52 π , e ) 64 π | a | square_area(divide(sqrt(add(multiply(add(3, 1), add(3, 1)), multiply(add(2, 3), add(2, 3)))), 2)) | add(n0,n2)|add(n1,n3)|multiply(#0,#0)|multiply(#1,#1)|add(#2,#3)|sqrt(#4)|divide(#5,n1)|square_area(#6)| | geometry |
a certain candy manufacturer reduced the weight of candy bar m by 10 percent buy left the price unchanged . what was the resulting percent increase in the price per ounce of candy bar m ? | "assume 1 oz candy cost $ 1 before . now price remain same $ 1 but weight of candy reduces to 0.9 oz new price of candy = 1 / 0.9 = 1.11 price increase 11 % b" | a ) 10 , b ) 11 , c ) 12 , d ) 14 , e ) 9 | b | subtract(multiply(divide(const_1, subtract(const_1, divide(10, const_100))), const_100), const_100) | divide(n0,const_100)|subtract(const_1,#0)|divide(const_1,#1)|multiply(#2,const_100)|subtract(#3,const_100)| | gain |
a cyclist traveled for two days . on the second day the cyclist traveled 6 hours longer and at an average speed 10 mile per hour slower than she traveled on the first day . if during the two days she traveled a total of 280 miles and spent a total of 12 hours traveling , what was her average speed on the second day ? | "solution : d = 280 mi t = 12 hrs đ â y 1 time = t 1 d â y 2 time = t 2 t 2 - t 1 = 4 hrs - - - - - ( i ) t 1 + t 2 = 12 hrs - - - - - ( ii ) adding i and ii , t 2 = 8 hrs and t 1 = 4 hrs d à y 1 rate = r 1 d â y 2 rate = r 2 r 1 - r 2 = 10 mph í . ẹ . r 1 = 10 + r 2 280 = 8 r 2 + 4 r 1 í . ẹ . 280 = 8 r 2 + 4 ( 10 + r... | a ) 5 mph , b ) 10 mph , c ) 20 mph , d ) 30 mph , e ) 40 mph | a | divide(subtract(280, multiply(divide(subtract(12, 6), const_2), 10)), add(divide(subtract(12, 6), const_2), add(divide(subtract(12, 6), const_2), 6))) | subtract(n3,n0)|divide(#0,const_2)|add(n0,#1)|multiply(n1,#1)|add(#2,#1)|subtract(n2,#3)|divide(#5,#4)| | physics |
a person buys an article at rs . 500 . at what price should he sell the article so as to make a profit of 60 % ? | "cost price = rs . 500 profit = 60 % of 500 = rs . 300 selling price = cost price + profit = 500 + 300 = 800 answer : a" | a ) 800 , b ) 882 , c ) 772 , d ) 652 , e ) 271 | a | add(500, multiply(500, divide(60, const_100))) | divide(n1,const_100)|multiply(n0,#0)|add(n0,#1)| | gain |
a man purchased 2 blankets @ rs . 100 each , 5 blankets @ rs . 150 each and two blankets at a certain rate which is now slipped off from his memory . but he remembers that the average price of the blankets was rs . 150 . find the unknown rate of two blankets ? | "10 * 150 = 1500 2 * 100 + 5 * 150 = 950 1500 – 1050 = 550 answer : e" | a ) 278 , b ) 277 , c ) 278 , d ) 450 , e ) 550 | e | subtract(multiply(const_10, 150), add(multiply(2, 100), multiply(5, 150))) | multiply(n3,const_10)|multiply(n0,n1)|multiply(n2,n3)|add(#1,#2)|subtract(#0,#3)| | general |
a trader bought a car at 5 % discount on its original price . he sold it at a 40 % increase on the price he bought it . what percent of profit did he make on the original price ? | "original price = 100 cp = 95 s = 95 * ( 140 / 100 ) = 133 100 - 133 = 5 % answer : b" | a ) 118 , b ) 133 , c ) 112 , d ) 113 , e ) 115 | b | multiply(subtract(divide(divide(multiply(subtract(const_100, 5), add(const_100, 40)), const_100), const_100), const_1), const_100) | add(n1,const_100)|subtract(const_100,n0)|multiply(#0,#1)|divide(#2,const_100)|divide(#3,const_100)|subtract(#4,const_1)|multiply(#5,const_100)| | gain |
what will be the difference between simple and compound interest at 6 % per annum on a sum of rs . 1000 after 4 years ? | "s . i . = ( 1000 * 6 * 4 ) / 100 = rs . 240 c . i . = [ 1000 * ( 1 + 6 / 100 ) 4 - 1000 ] = rs . 262.5 difference = ( 262.5 - 240 ) = rs . 22.5 answer : e" | a ) 24.19 , b ) 24.12 , c ) 22.1 , d ) 24.1 , e ) 22.5 | e | subtract(subtract(multiply(1000, power(add(divide(6, const_100), const_1), 4)), 1000), multiply(multiply(1000, divide(6, const_100)), 4)) | divide(n0,const_100)|add(#0,const_1)|multiply(n1,#0)|multiply(n2,#2)|power(#1,n2)|multiply(n1,#4)|subtract(#5,n1)|subtract(#6,#3)| | gain |
zinc and copper are melted together in the ratio 9 : 11 . what is the weight of melted mixture , if 28.8 kg of zinc has been consumed in it ? | sol . for 9 kg zinc , mixture melted = ( 9 + 11 ) kg . for 28.8 kg zinc , mixture , melted = [ 20 / 9 x 28.8 ] kg = 64 kg . answer c | a ) 58 kg , b ) 60 kg , c ) 64 kg , d ) 70 kg , e ) none | c | add(multiply(divide(11, 9), 28.8), 28.8) | divide(n1,n0)|multiply(n2,#0)|add(n2,#1) | general |
what is the speed of the stream if a canoe rows upstream at 9 km / hr and downstream at 12 km / hr | "sol . speed of stream = 1 / 2 ( 12 - 9 ) kmph = 1.5 kmph . answer e" | a ) 1 kmph , b ) 4 kmph , c ) 3 kmph , d ) 2 kmph , e ) 1.5 kmph | e | divide(subtract(12, 9), const_2) | subtract(n1,n0)|divide(#0,const_2)| | physics |
. a car covers a distance of 495 km in 5 hours . find its speed ? | "495 / 5 = 99 kmph answer : a" | a ) 99 , b ) 100 , c ) 102 , d ) 288 , e ) 2761 | a | divide(495, 5) | divide(n0,n1)| | physics |
if a number p is prime , and 2 p + 1 = q , where q is also prime , then the decimal expansion of 1 / q will produce a decimal with q - 1 digits . if this method produces a decimal with 166 digits , what is the units digit of the product of p and q | "3 / 7 = 0.428571 . . . ( a repeating pattern one digit long ) b" | a ) 1 , b ) 3 , c ) 5 , d ) 7 , e ) 9 | b | subtract(divide(subtract(add(166, 1), 1), 2), subtract(divide(subtract(add(166, 1), 1), 2), 1)) | add(n2,n4)|subtract(#0,n1)|divide(#1,n0)|subtract(#2,n2)|subtract(#2,#3)| | general |
a man is walking at a speed of 10 km per hour . after every kilometre , he takes rest for 3 minutes . how much time will be take to cover a distance of 3 kilometres ? | "rest time = number of rest ã — time for each rest = 2 ã — 3 = 6 minutes total time to cover 3 km = ( 3 â „ 10 ã — 60 ) minutes + 6 minutes = 24 minutes answer e" | a ) 48 min . , b ) 50 min . , c ) 45 min . , d ) 55 min . , e ) 24 min . | e | add(multiply(divide(3, 10), speed(const_60, const_1)), multiply(const_4, 3)) | divide(n1,n0)|multiply(n1,const_4)|speed(const_60,const_1)|multiply(#0,#2)|add(#3,#1)| | physics |
the average weight of 5 person ' s increases by 5.5 kg when a new person comes in place of one of them weighing 68 kg . what might be the weight of the new person ? | "total weight increased = ( 5 x 5.5 ) kg = 27.5 kg . weight of new person = ( 68 + 27.5 ) kg = 95.5 kg option b" | a ) 60 kg , b ) 95.5 kg , c ) 80 kg , d ) 85 kg , e ) 90 kg | b | add(multiply(5, 5.5), 68) | multiply(n0,n1)|add(n2,#0)| | general |
in the first 10 overs of a cricket game , the run rate was only 3.2 . what should be the rate in the remaining 40 overs to reach the target of 292 runs ? | "required run rate = [ 292 - ( 3.2 * 10 ) ] / 40 = 260 / 40 = 6.5 answer : a" | a ) 6.5 , b ) 6.22 , c ) 6.29 , d ) 6.39 , e ) 6.13 | a | divide(subtract(292, multiply(10, 3.2)), 40) | multiply(n0,n1)|subtract(n3,#0)|divide(#1,n2)| | gain |
the radius of a semi circle is 5.2 cm then its perimeter is ? | "36 / 7 r = 5.2 = 26.74 answer : c" | a ) 32.52 , b ) 32.48 , c ) 26.74 , d ) 32.9 , e ) 32.31 | c | add(divide(circumface(5.2), const_2), multiply(5.2, const_2)) | circumface(n0)|multiply(n0,const_2)|divide(#0,const_2)|add(#2,#1)| | physics |
p and q invested in a business . they earned some profit which they divided in the ratio of 2 : 3 . if p invested rs . 50 , 000 , the amount invested by q is : | q invested = 50000 / 2 * 3 = 75000 answer : c | a ) rs . 65,000 , b ) rs . 70,000 , c ) rs . 75,000 , d ) rs . 90,000 , e ) rs . 60,000 | c | multiply(multiply(add(add(const_3, const_2), 2), add(const_3, const_2)), 2) | add(const_2,const_3)|add(n0,#0)|multiply(#1,#0)|multiply(n0,#2) | gain |
if log 102 = 0.3010 , log 10 3 = 0.4771 , then the number of zeros between the decimal point and the first significant figure in ( 0.0432 ) 10 is ? | x = ( 0.0432 ) 10 = ( 432 / 10000 ) 10 = ( 33.24 / 104 ) 10 therefore , log 10 x = 10 ( log 103 + 4 log 102 - 4 ) = 10 ( 1.4313 + 1.2040 - 4 ) = 10 ( - 1.3647 ) = - 13.647 = - 14.353 therefore , x = antilog ( - 14.053 ) thus , number of zeros between the decimal and the first significant figure = 13 answer : b | a ) 10 , b ) 13 , c ) 14 , d ) 15 , e ) 16 | b | negate(multiply(subtract(add(multiply(3, 0.4771), multiply(const_4, 0.301)), const_4), 10)) | multiply(n3,n4)|multiply(n1,const_4)|add(#0,#1)|subtract(#2,const_4)|multiply(n2,#3)|negate(#4) | other |
eden drove an average speed of 15 miles per hour for the first 20 miles of a tripthen at a average speed of 30 miles / hr for the remaining 20 miles of the trip if she made no stops during the trip what was eden ' s avg speed in miles / hr for the entire trip | avg . speed = total distance / total time total distance = 40 miles total time = 20 / 15 + 15 / 30 = 2 avg . speed = 20 . answer - b | a ) 35 , b ) 20 , c ) 45 , d ) 50 , e ) 55 | b | divide(add(20, 20), add(divide(20, 15), divide(20, 30))) | add(n1,n1)|divide(n1,n0)|divide(n1,n2)|add(#1,#2)|divide(#0,#3) | physics |
the average temperature for monday , tuesday , wednesday and thursday was 48 degrees and for tuesday , wednesday , thursday and friday was 46 degrees . if the temperature on monday was 44 degrees . find the temperature on friday ? | "m + tu + w + th = 4 * 48 = 192 tu + w + th + f = 4 * 46 = 184 m = 44 tu + w + th = 192 - 44 = 148 f = 184 – 148 = 36 answer : c" | a ) 65 degrees , b ) 73 degrees , c ) 36 degrees , d ) 34 degrees , e ) 74 degrees | c | subtract(44, subtract(multiply(48, const_4), multiply(46, const_4))) | multiply(n0,const_4)|multiply(n1,const_4)|subtract(#0,#1)|subtract(n2,#2)| | general |
if w is the set of all the integers between 49 and 88 , inclusive , that are either multiples of 3 or multiples of 2 or multiples of both , then w contains how many numbers ? | "official solution : number of multiples of 3 step 1 . subtract the extreme multiples of 3 within the range ( the greatest is 87 , the smallest is 51 ) : 87 - 51 = 36 step 2 . divide by 3 : 36 / 3 = 12 step 3 . add 1 : 12 + 1 = 13 . so there are 13 multiples of 3 within the range : examples are 51 , 54 , 57 , 60 , etc ... | a ) 11 , b ) 44 , c ) 48 , d ) 28 , e ) 26 | e | subtract(add(floor(divide(subtract(88, 49), 3)), divide(subtract(88, 49), 2)), floor(divide(subtract(88, 49), multiply(2, 3)))) | multiply(n3,n2)|subtract(n1,n0)|divide(#1,n3)|divide(#1,n2)|divide(#1,#0)|floor(#3)|floor(#4)|add(#2,#5)|subtract(#7,#6)| | other |
in the xy - coordinate plane , the graph of y = - x ^ 2 + 9 intersects line l at ( p , - 5 ) and ( t , - 7 ) . what is the least possible value of the slope of line l ? | we need to find out the value of p and l to get to the slope . line l and graph y intersect at point ( p , - 5 ) . hence , x = p and y = - 5 should sactisfy the graph . soliving 5 = - p 2 + 9 p 2 = 4 p = + or - 2 simillarly point ( t , - 7 ) should satisfy the equation . hence x = t and y = - 7 . - 7 = - t 2 + 9 t = + ... | a ) 6 , b ) - 1 , c ) - 2 , d ) - 6 , e ) - 10 | b | divide(subtract(7, 5), subtract(sqrt(add(9, 7)), sqrt(add(5, 9)))) | add(n1,n3)|add(n1,n2)|subtract(n3,n2)|sqrt(#0)|sqrt(#1)|subtract(#3,#4)|divide(#2,#5) | general |
two numbers n and 16 have lcm = 52 and gcf = 8 . find n . | "the product of two integers is equal to the product of their lcm and gcf . hence . 16 * n = 52 * 8 n = 52 * 8 / 16 = 26 correct answer b" | a ) 35 , b ) 26 , c ) 76 , d ) 87 , e ) 24 | b | divide(multiply(52, 8), 16) | multiply(n1,n2)|divide(#0,n0)| | physics |
an ant walks an average of 1000 meters in 30 minutes . a beetle walks 10 % less distance at the same time on the average . assuming the beetle walks at her regular rate , what is its speed in km / h ? | the ant walks an average of 1000 meters in 30 minutes 1000 meters in 1 / 2 hours the beetle walks 10 % less distance = 1000 - 10 = 900 meters in 30 minutes 0.900 km in 30 / 60 = 1 / 2 hours speed = 0.900 * 2 = 1.8 km / h i guess option a should be 1.8 | a ) 1.8 , b ) 1.6 , c ) 2.775 . , d ) 1.9 , e ) 3.5 . | a | multiply(divide(divide(1000, const_1000), divide(30, const_60)), subtract(const_1, divide(10, const_100))) | divide(n0,const_1000)|divide(n1,const_60)|divide(n2,const_100)|divide(#0,#1)|subtract(const_1,#2)|multiply(#3,#4) | general |
two trains 151 meters and 165 meters in length respectively are running in opposite directions , one at the rate of 80 km and the other at the rate of 65 kmph . in what time will they be completely clear of each other from the moment they meet ? | "t = ( 151 + 165 ) / ( 80 + 65 ) * 18 / 5 t = 7.84 answer : d" | a ) 7.19 , b ) 7.18 , c ) 7.16 , d ) 7.84 , e ) 7.12 | d | divide(add(151, 165), multiply(add(80, 65), const_0_2778)) | add(n0,n1)|add(n2,n3)|multiply(#1,const_0_2778)|divide(#0,#2)| | physics |
joe has candys on his pocket , 4 of those candies have lemon flavor , and 3 have caramel flavor . if joe take one , what is the probability that he picks one caramel flavor ? | the total number of candies is 7 and the number of candies with lemon flavor is 3 then , in one chance the probability to pick one candie with lemon flavor is 3 / 7 answer d | a ) 1 , b ) 2 , c ) none , d ) 3 / 7 , e ) 3 / 4 | d | divide(3, add(4, 3)) | add(n0,n1)|divide(n1,#0) | probability |
john bought a total of 24 mangoes and oranges . each mango costs 80 cents and each orange costs 60 cents . if the average price of the 24 mangoes and oranges that john originally purchased was 65 cents , then how many oranges needs to return to raise the average price of his purchase to 72 cents ? | "let number of mangoes be x , number of oranges be 24 - x 0.80 x + ( 24 - x ) 0.60 / 24 = 0.65 solving for x , we get x = 6 - - > mangoes 6 , oranges 18 now , number of oranges to be returned be y 0.80 * 6 + ( 18 - y ) * 0.60 / 24 - y = 0.72 solving for y , y = 14 ans : e" | a ) 4 , b ) 5 , c ) 6 , d ) 7 , e ) 14 | e | multiply(divide(65, const_100), 24) | divide(n4,const_100)|multiply(n0,#0)| | general |
the difference between the ages of two persons is 10 years . fifteen years ago , the elder one was twice as old as the younger one . the present age of the elder person is | "sol . let their ages of x years and ( x + 10 ) years respectively . then , ( x + 10 ) - 15 = 2 ( x - 15 ) ⇔ x - 5 = 2 x - 30 ⇔ x = 25 . ∴ present age of the elder person = ( x + 10 ) = 35 years . answer c" | a ) 25 years , b ) 30 years , c ) 35 years , d ) 40 years , e ) none | c | add(subtract(10, subtract(10, add(const_3, const_2))), multiply(subtract(10, add(const_3, const_2)), const_2)) | add(const_2,const_3)|subtract(n0,#0)|multiply(#1,const_2)|subtract(n0,#1)|add(#2,#3)| | general |
how many 3 - digit numerals begin with a digit that represents a prime and end with a digit that represents a odd prime number ? | "prime digits 2 , 3,5 and 7 . three digit numbers _ _ _ 1 st place can be filled in 4 ways 2 nd place can be filled in 10 ways 3 rd place can be filled in 3 ways total = 4 * 10 * 3 = 120 ans : b" | a ) 16 , b ) 120 , c ) 160 , d ) 180 , e ) 240 | b | add(multiply(const_100, 3), const_100) | multiply(n0,const_100)|add(#0,const_100)| | general |
an automobile parts supplier charges $ 20 per package of gaskets . when a customer orders more than 10 packages of gaskets , the supplier charges 4 / 5 the price for each package in excess of 10 . during a certain week , the supplier sold 60 packages of gaskets . if 20 percent of the gaskets went to company x , 15 perc... | "$ 25 per packet of gasket in case a customer orders less than 10 in case a customer orders > 10 price per gasket = 20 * 4 / 5 = 16 a certain week the supplier sold 60 gasket 1 . he sold 20 % of the gaskets to x = 12 gaskets = 20 * 10 + 16 * 2 = 200 + 32 = 232 2 . he sold 15 % of the gaskets to y = 9 gaskets = 20 * 9 =... | a ) 1005 , b ) 1045 , c ) 1075 , d ) 1015 , e ) 1096 | e | add(add(multiply(multiply(20, divide(4, 5)), subtract(subtract(subtract(60, multiply(60, divide(20, const_100))), multiply(60, divide(20, const_100))), 10)), multiply(20, 10)), add(multiply(20, multiply(60, divide(20, const_100))), multiply(20, multiply(60, divide(20, const_100))))) | divide(n2,n3)|divide(n6,const_100)|multiply(n0,n1)|multiply(n0,#0)|multiply(n5,#1)|multiply(n0,#4)|subtract(n5,#4)|add(#5,#5)|subtract(#6,#4)|subtract(#8,n1)|multiply(#3,#9)|add(#10,#2)|add(#11,#7)| | general |
the length of a rectangle is increased by 25 % and its breadth is decreased by 15 % . what is the effect on its area ? | "100 * 100 = 10000 125 * 85 = 10625 answer : c" | a ) 1288 , b ) 1299 , c ) 10625 , d ) 10000 , e ) 2887 | c | multiply(add(25, const_100), subtract(const_100, 15)) | add(n0,const_100)|subtract(const_100,n1)|multiply(#0,#1)| | geometry |
suraj has a certain average of runs for 8 innings . in the 9 th innings he scores 90 runs thereby increasing his average by 6 runs . what is his average after the 9 th innings ? | "to improve his average by 6 runs per innings he has to contribute 8 x 6 = 48 runs for the previous 8 innings . thus , the average after the 9 th innings = 90 - 48 = 42 . answer : a" | a ) 42 , b ) 64 , c ) 36 , d ) 72 , e ) 27 | a | divide(subtract(90, multiply(8, 6)), subtract(9, 8)) | multiply(n0,n3)|subtract(n1,n0)|subtract(n2,#0)|divide(#2,#1)| | general |
20 men or 25 women can finish a work in 25 days . how many days 38 men and 15 women will take to finish this job ? | 20 men = 25 women therefore , 1 women = 20 / 25 men = 4 / 5 men 15 women = 4 / 5 * 15 men = 12 men i . e . 38 men + 15 women = 38 men + 12 men = 50 men 1 work done by men = 20 * 25 20 * 25 = 50 * ? number of days = 20 * 25 / 50 = 10 days . b | a ) 12 days , b ) 10 days , c ) 5 days , d ) 6 days , e ) 15 days | b | divide(multiply(25, 25), add(multiply(divide(25, 20), 38), 15)) | divide(n1,n0)|multiply(n1,n1)|multiply(n3,#0)|add(n4,#2)|divide(#1,#3) | physics |
two heavily loaded sixteen - wheeler transport trucks are 940 kilometers apart , sitting at two rest stops on opposite sides of the same highway . driver a begins heading down the highway driving at an average speed of 90 kilometers per hour . exactly one hour later , driver b starts down the highway toward driver a , ... | "i ' ve been reading the website for a while and i ' m always keen to see different approaches so i would like to share one that works for me : short version : truck a travels for an hour . distance remaining = 940 - 90 = 850 k ratio of speeds 9 : 8 - > 850 / 17 = 50 truck a = 90 + 50 * 9 = 540 truck b = 50 * 8 = 400 d... | a ) 90 , b ) 130 , c ) 150 , d ) 140 , e ) 450 | d | add(90, multiply(divide(subtract(940, 90), add(90, 80)), subtract(90, 80))) | add(n1,n2)|subtract(n0,n1)|subtract(n1,n2)|divide(#1,#0)|multiply(#3,#2)|add(n1,#4)| | general |
in an examination , 32 % of total students failed in hindi , 56 % failed in english and 12 % in both . the percentage of these who passed in both the subjects is : | "explanation : formula n ( a â ˆ ª b ) = n ( a ) + n ( b ) â ˆ ’ n ( a â ˆ © b ) fail in hindi or english = 32 + 56 â € “ 12 = 76 therefore students who passed = 100 â € “ 76 = 24 . answer : a" | a ) 24 , b ) 37 , c ) 28 , d ) 40 , e ) 81 | a | subtract(const_100, subtract(add(32, 56), 12)) | add(n0,n1)|subtract(#0,n2)|subtract(const_100,#1)| | general |
of the 250 employees at company x , 80 are full - time , and 175 have worked at company x for at least a year . there are 20 employees at company x who aren ’ t full - time and haven ’ t worked at company x for at least a year . how many full - time employees of company x have worked at the company for at least a year ... | "250 employees 80 are full - time 175 have worked at company x for at least a year 20 employees at company x who aren ’ t full - time and haven ’ t worked at company x for at least a year . how many full - time employees of company x have worked at the company for at least a year ? 250 - 80 = 170 employees not full tim... | a ) 25 , b ) 30 , c ) 50 , d ) 80 , e ) 100 | a | subtract(subtract(250, 80), 20) | subtract(n0,n1)|subtract(#0,n3)| | general |
a farmer has an apple orchard consisting of fuji and gala apple trees . due to high winds this year 10 % of his trees cross pollinated . the number of his trees that are pure fuji plus the cross - pollinated ones totals 221 , while 3 / 4 of all his trees are pure fuji . how many of his trees are pure gala ? | "let f = pure fuji , g = pure gala and c - cross pollinated . c = 10 % of x where x is total trees . c = . 1 x also 3 x / 4 = f and c + f = 221 = > . 1 x + 3 / 4 x = 221 = > x = 260 260 - 221 = pure gala = 39 . e" | a ) 22 , b ) 33 , c ) 55 , d ) 77 , e ) 39 | e | subtract(divide(221, add(divide(10, const_100), divide(3, 4))), 221) | divide(n0,const_100)|divide(n2,n3)|add(#0,#1)|divide(n1,#2)|subtract(#3,n1)| | general |
the ratio between the perimeter and the breadth of a rectangle is 5 : 1 . if the area of the rectangle is 216 sq . cm , what is the length of the rectangle ? | 2 ( l + b ) = 5 b 1 2 l + 2 b = 5 b 3 b = 2 l b = 2 l 3 then , area = 216 cm 2 l x b = 216 l x 2 l = 216 3 l 2 = 324 l = 18 cm . a | ['a ) 18 cm', 'b ) 20 cm', 'c ) 22 cm', 'd ) 24 cm', 'e ) 25 cm'] | a | sqrt(multiply(216, divide(const_3, const_2))) | divide(const_3,const_2)|multiply(n2,#0)|sqrt(#1) | geometry |
60 percent of movie theatres in town x have 3 screens or less . 20 % of those theatres sell an average of more than $ 300 worth of popcorn per showing . 50 percent of all the movie theatres in town x sell $ 300 or less of popcorn per showing . what percent of all the stores on the street have 4 or more screens and sell... | "lets take numbers here . assume that the total number of movie theaters in the town = 100 then number of movie theaters with 3 screens or less = 60 = > number of movie theaters with 4 screens or more = 40 movie theaters with 3 screens or less selling popcorn at more than $ 300 = 20 % of 60 = 12 number of movie theater... | a ) 12 , b ) 18 , c ) 32 , d ) 40 , e ) 44 | e | multiply(divide(20, const_100), multiply(divide(20, 300), 300)) | divide(n2,const_100)|multiply(n3,#0)|multiply(#0,#1)| | general |
on a two - dimensional coordinate plane , the line q = x ^ 2 - x ^ 3 touches the x - axis in how many places ? | "apparently it ' s q = x ^ 2 - x ^ 3 instead of q = x ^ 2 - q ^ 3 . in this case : the x - intercept is the value ( s ) of x for q = 0 . 0 = x ^ 2 - x ^ 3 ; 0 = x ^ 2 ( 1 - x ) ; x = 0 or x = 1 . answer : c ." | a ) 0 , b ) 1 , c ) 2 , d ) 3 , e ) 4 | c | subtract(max(3, 2), const_1) | max(n0,n1)|subtract(#0,const_1)| | general |
if x and y are integers such that | y + 3 | ≤ 3 and 2 y – 3 x + 6 = 0 , what is the least possible value w of the product xy ? | "how to deal with inequalities involving absolute values ? first example shows us the so callednumber case in this case we have | y + 3 | ≤ 3 which is generalized | something | ≤ some number . first we solve as if there were no absolute value brackets : y + 3 ≤ 3 y ≤ 0 so y is 0 or negative second scenario - remove the... | a ) - 12 , b ) - 3 , c ) 0 , d ) 2 , e ) none of the above | c | multiply(divide(add(6, multiply(negate(add(3, 3)), 2)), 3), negate(add(3, 3))) | add(n0,n0)|negate(#0)|multiply(n2,#1)|add(n4,#2)|divide(#3,n0)|multiply(#4,#1)| | general |
there is 60 lit of milk and water in which milk forms 84 % . howmuch water must be added to this solution to make it solution in which milk forms 60 % | "60 * 84 / 100 = 50.40 lit milk that is 9.60 lit water let x lit water will be added then ( 60 + x ) * 60 / 100 = 50.40 so x = 18.75 answer : e" | a ) 18.75 , b ) 19.75 , c ) 20.75 , d ) 21.75 , e ) 24 | e | subtract(multiply(divide(const_100, 60), divide(multiply(60, 84), const_100)), 60) | divide(const_100,n2)|multiply(n0,n1)|divide(#1,const_100)|multiply(#0,#2)|subtract(#3,n0)| | gain |
the average ( arithmetic mean ) of the 5 positive integers k , m , r , s , and t is 10 , and k < m < r < s < t . if t is 20 , what is the greatest possible value of the median of the 5 integers ? | "we need to find the median which is the third value when the numbers are in increasing order . since k < m < r < s < t , the median would be r . the average of the positive integers is 10 which means that in effect , all numbers are equal to 10 . if the largest number is 20 , it is 10 more than 20 . we need r to be ma... | a ) 16 , b ) 18 , c ) 13 , d ) 20 , e ) 22 | c | subtract(divide(subtract(multiply(10, 5), 20), const_2), const_2) | multiply(n0,n1)|subtract(#0,n2)|divide(#1,const_2)|subtract(#2,const_2)| | general |
if a person walks at 15 km / hr instead of 10 km / hr , he would have walked 15 km more . the actual distance traveled by him is ? | "let the actual distance traveled be x km . then , x / 10 = ( x + 15 ) / 15 x - 30 = > x = 30 km . answer : a" | a ) 30 , b ) 50 , c ) 15 , d ) 12 , e ) 20 | a | multiply(10, divide(15, subtract(15, 10))) | subtract(n0,n1)|divide(n2,#0)|multiply(n1,#1)| | general |
the population of a bacteria culture doubles every 1 minutes . approximately how many minutes will it take for the population to grow from 1,000 to 500,000 bacteria | "this one ' s easy . 1000 * 2 ^ t = 500,000 2 ^ t = 500 now gauging , since 2 ^ 8 = 256 , then 2 ^ 9 = 512 so t = 9 but be careful , ' t ' is in time intervals of 1 minutes so answer is 9 * 1 = 9 minutes answer ( a )" | a ) 9 , b ) 12 , c ) 14 , d ) 16 , e ) 18 | a | multiply(log(divide(multiply(multiply(add(const_4, const_1), 1,000), const_100), 1,000)), 1) | add(const_1,const_4)|multiply(#0,n1)|multiply(#1,const_100)|divide(#2,n1)|log(#3)|multiply(n0,#4)| | general |
a rectangular tiled patio is composed of 48 square tiles . the rectangular patio will be rearranged so that there will be 2 fewer columns of tiles and 4 more rows of tiles . after the change in layout , the patio will still have 48 tiles , and it will still be rectangular . how many rows are in the tile patio before th... | "suppose there are c columns and there are r rows original situation so , number of tiles = c * r = 48 also . reach column has r tiles and each row has c tiles new situation number of tiles in each column is r - 2 and number of tiles in each row is c + 4 so , number of rows = r - 2 and number of columns is c + 4 so , n... | a ) 5 , b ) 6 , c ) 10 , d ) 13 , e ) 28 | b | divide(48, divide(add(negate(4), sqrt(add(power(4, 2), multiply(4, multiply(48, 2))))), 2)) | multiply(n0,n1)|negate(n2)|power(n2,n1)|multiply(n2,#0)|add(#3,#2)|sqrt(#4)|add(#1,#5)|divide(#6,n1)|divide(n0,#7)| | geometry |
what is the units digit of 28 ! + 50 ! + 3 ! + 4 ! ? | "for all n greater than 4 , the units digit of n ! is 0 . the sum of the four units digits is 0 + 0 + 6 + 4 = 10 the units digit is 0 . the answer is a ." | a ) 0 , b ) 2 , c ) 4 , d ) 6 , e ) 8 | a | add(add(const_4, const_3), const_2) | add(const_3,const_4)|add(#0,const_2)| | other |
a shopkeeper loses 15 % , if an article is sold for rs . 119 . what should be the selling price of the article to gain 20 % ? | "given that sp = rs . 119 and loss = 15 % cp = [ 100 ( sp ) ] / ( 100 - l % ) = ( 100 * 119 ) / 85 = rs . 140 . to get 20 % profit , new sp = [ ( 100 + p % ) cp ] / 100 = ( 140 * 120 ) / 100 = rs . 168 answer : b" | a ) s . 177 , b ) s . 168 , c ) s . 144 , d ) s . 129 , e ) s . 112 | b | add(divide(119, subtract(const_1, divide(15, const_100))), multiply(divide(119, subtract(const_1, divide(15, const_100))), divide(20, const_100))) | divide(n0,const_100)|divide(n2,const_100)|subtract(const_1,#0)|divide(n1,#2)|multiply(#3,#1)|add(#3,#4)| | gain |
eight women can do a work in 10 days . ten men can complete the same work in 6 days . what is the ratio between the capacity of a man and a woman ? | "explanation : ( 8 ã — 10 ) women can complete the work in 1 day . â ˆ ´ 1 woman ' s 1 day ' s work = 1 / 80 ( 10 ã — 6 ) men can complete the work in 1 day . â ˆ ´ 1 man ' s 1 day ' s work = 1 / 60 so , required ratio = 1 / 80 : 1 / 60 = 4 : 3 answer : d" | a ) 1 : 2 , b ) 2 : 1 , c ) 2 : 3 , d ) 4 : 3 , e ) none of these | d | divide(divide(const_1, multiply(10, 6)), divide(const_1, multiply(6, const_10))) | multiply(n0,n1)|multiply(n1,const_10)|divide(const_1,#0)|divide(const_1,#1)|divide(#2,#3)| | physics |
x , y and z , each working alone can complete a job in 6 , 8 and 12 days respectively . if all three of them work together to complete a job and earn $ 2350 , what will be z ' s share of the earnings ? | "the dollars earned will be in the same ratio as amount of work done 1 day work of z is 1 / 12 ( or 2 / 24 ) 1 day work of the combined workforce is ( 1 / 6 + 1 / 8 + 1 / 12 ) = 9 / 24 z ' s contribution is 2 / 9 of the combined effort translating effort to $ = 2 / 9 * 2350 = $ 522.3 hence : d" | a ) $ 525 , b ) $ 550.50 , c ) $ 1080.02 , d ) $ 522.3 , e ) $ 1020.2 | d | multiply(divide(2350, add(add(inverse(6), inverse(8)), inverse(12))), inverse(6)) | inverse(n0)|inverse(n1)|inverse(n2)|add(#0,#1)|add(#3,#2)|divide(n3,#4)|multiply(#5,#0)| | physics |
the sides of a cube measures 6.5 cm . if the surface area of the cube is the same as a ball , what is the radius of the ball ? round to the nearest whole number . | first calculate the surface area of the cube . the cube is made of 6 squares with the sides measuring 6.5 cm each . the surface area of 1 side of the cube is 42.25 cm . multiply the surface area of the 1 side by 6 will give you the cube ' s total surface area . the cubes surface area is 253.5 cm . now we know that the ... | ['a ) 9', 'b ) 7', 'c ) 4', 'd ) 8', 'e ) 6'] | c | floor(sqrt(divide(divide(surface_cube(6.5), const_4), const_pi))) | surface_cube(n0)|divide(#0,const_4)|divide(#1,const_pi)|sqrt(#2)|floor(#3) | geometry |
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