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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there is a 30 % increase in the price of an article in the first year , a 20 % decrease in the second year and a 10 % increase in the next year . if the final price of the article is rs . 2288 , then what was the price of the article initially ? | let the price of the article , four years age be rs . 100 in the 1 st year , price of the article = 100 + 30 = rs . 130 . in the 2 nd year , price = 130 - 20 % of 130 = 130 - 26 = rs . 104 . in the 3 rd year , price = 104 + 10 % of 104 = 104 + 10.4 = rs . 114.40 . but present price of the article is rs . 2288 for 114.4... | a ) rs . 2008 , b ) rs . 2022 , c ) rs . 2000 , d ) rs . 2029 , e ) rs . 2021 | c | divide(2288, add(const_1, divide(add(subtract(30, 20), 10), const_100))) | subtract(n0,n1)|add(n2,#0)|divide(#1,const_100)|add(#2,const_1)|divide(n3,#3) | general |
if p ( a ) = 2 / 5 and p ( b ) = 2 / 5 , find p ( a n b ) if a and b are independent events . | "p ( a n b ) = p ( a ) . p ( b ) p ( a n b ) = 2 / 5 . 2 / 5 p ( a n b ) = 4 / 25 . b" | a ) 7 / 25 , b ) 4 / 25 , c ) 8 / 25 , d ) 2 / 13 , e ) 3 / 17 | b | multiply(divide(2, 5), divide(2, 5)) | divide(n0,n1)|divide(n2,n3)|multiply(#0,#1)| | general |
if x and y are integers such that ( x + 1 ) ^ 2 is less than or equal to 64 and ( y - 1 ) ^ 2 is less than 64 , what is the sum of the maximum possible value of xy and the minimum possible value of xy ? | "( x + 1 ) ^ 2 < = 64 x < = 7 x > = - 9 ( y - 1 ) ^ 2 < 64 y < 9 y > - 7 max possible value of xy is - 9 Γ - 6 = 54 minimum possible value of xy is - 9 Γ 8 = - 72 - 72 + 54 = - 18 answer : b" | a ) - 16 , b ) - 18 , c ) 0 , d ) 14 , e ) 16 | b | add(sqrt(64), sqrt(64)) | sqrt(n2)|sqrt(n5)|add(#0,#1)| | general |
if the difference between compound interest ( interest compounded yearly ) and simple interest on a sum for 2 years at 10 % p . a . is rs . 150 then sum is | compund interest = p [ 1 + r / 100 ] ^ t - p ci = p [ 21 / 100 ] simple interest = ptr / 100 si = p [ 20 / 100 ] difference p [ 21 / 100 ] - p [ 20 / 100 ] = 150 p = 15000 answer : b | a ) rs . 12000 , b ) rs . 15000 , c ) rs . 13000 , d ) rs . 10000 , e ) rs . 14000 | b | multiply(multiply(150, 10), 10) | multiply(n1,n2)|multiply(n1,#0) | gain |
a leak in the bottom of a tank can empty the full tank in 7 hours . an inlet pipe fills water at the rate of 6 litres a minute . when the tank is full , the inlet is opened and due to the leak , the tank is empty in 12 hours . how many litres does the cistern hold ? | "solution work done by the inlet in 1 hour = ( 1 / 7 - 1 / 12 ) = 5 / 84 work done by the inlet in 1 min . = ( 5 / 84 Γ 1 / 60 ) = 0.000992 volume of 0.000992 part = 6 litres . therefore , volume of whole = ( ( 1 / 0.000992 ) Γ 6 ) βΉ = βΊ 6048 litres . answer d" | a ) 7580 , b ) 7960 , c ) 8290 , d ) 6048 , e ) none | d | divide(multiply(6, multiply(12, const_60)), subtract(divide(multiply(12, const_60), multiply(7, const_60)), const_1)) | multiply(n2,const_60)|multiply(n0,const_60)|divide(#0,#1)|multiply(n1,#0)|subtract(#2,const_1)|divide(#3,#4)| | physics |
in a certain animal population , for each of the first 3 months of life , the probability that an animal will die during that month is 1 / 10 . for a group of 500 newborn members of the population , approximately how many would be expected to survive the first 3 months of life ? | number of newborns that can die in first month = 1 / 10 * 500 = 50 survived = 450 number of newborns that can die in second month = 1 / 10 * 450 = 45 survived = 405 number of newborns that can die in third month = 1 / 10 * 405 = 40 survived = 365 answer : d | a ) 340 , b ) 346 , c ) 352 , d ) 365 , e ) 370 | d | multiply(multiply(multiply(500, subtract(1, divide(const_1, 10))), subtract(1, divide(const_1, 10))), subtract(1, divide(const_1, 10))) | divide(const_1,n2)|subtract(n1,#0)|multiply(n3,#1)|multiply(#2,#1)|multiply(#3,#1) | probability |
kim has 5 pairs of shoes ; each pair is a different color . if kim randomly selects 2 shoes without replacement from the 10 shoes , what is the probability that she will select 2 shoes of the same color ? | "total pairs = 10 c 2 = 45 ; same color pairs = 5 c 1 * 1 c 1 = 5 ; prob = 1 / 9 or 2 / 10 * 1 / 9 * 5 = 1 / 9 ans c" | a ) 2 / 5 , b ) 1 / 5 , c ) 1 / 9 , d ) 1 / 10 , e ) 1 / 25 | c | divide(5, choose(10, 2)) | choose(n2,n1)|divide(n0,#0)| | probability |
the mean of 50 observations was 36 . it was found later that an observation 45 was wrongly taken as 23 . the corrected new mean is | "solution correct sum = ( 36 x 50 + 45 - 23 ) = 1822 . Γ’ Λ Β΄ correct mean = 1822 / 50 = 36.44 . answer d" | a ) 35.24 , b ) 36.14 , c ) 36.24 , d ) 36.44 , e ) none | d | divide(add(multiply(36, 50), subtract(subtract(50, const_2), 23)), 50) | multiply(n0,n1)|subtract(n0,const_2)|subtract(#1,n3)|add(#0,#2)|divide(#3,n0)| | general |
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 ? | "explanation : in this type of question , where we have one person work and together work done . then we can easily get the other person work just by subtracting them . as , son ' s one day work = ( 1 / 3 β 1 / 5 ) = ( 5 β 3 / 15 ) = 2 / 15 so son will do whole work in 15 / 2 days which is = 7 1 / 2 days option a" | a ) 7 1 / 2 days , b ) 6 1 / 2 days , c ) 5 1 / 2 days , d ) 4 1 / 2 days , e ) 3 1 / 2 days | a | divide(multiply(5, 3), subtract(5, 3)) | multiply(n0,n1)|subtract(n0,n1)|divide(#0,#1)| | physics |
a luxury liner , queen marry ii , is transporting several cats as well as the crew ( sailors , a cook , and one - legged captain ) to a nearby port . altogether , these passengers have 15 heads and 43 legs . how many cats does the ship host ? | "sa ' s + co + ca + cats = 15 . sa ' s + 1 + 1 + cats = 15 or sa ' s + cats = 13 . sa ' s ( 2 ) + 2 + 1 + cats * 4 = 43 sa ' s * 2 + cats * 4 = 40 or sa ' s + cats * 2 = 20 or 13 - cats + cat * 2 = 20 then cats = 7 d" | a ) 4 , b ) 5 , c ) 6 , d ) 7 , e ) 8 | d | multiply(divide(subtract(subtract(43, const_1), multiply(subtract(15, const_1), const_2)), subtract(multiply(subtract(15, const_1), const_4), multiply(subtract(15, const_1), const_2))), subtract(15, const_1)) | subtract(n1,const_1)|subtract(n0,const_1)|multiply(#1,const_2)|multiply(#1,const_4)|subtract(#0,#2)|subtract(#3,#2)|divide(#4,#5)|multiply(#6,#1)| | general |
find the amount on rs . 5000 in 2 years , the rate of interest being 20 % per first year and 25 % for the second year ? | "5000 * 120 / 100 * 125 / 100 = > 7500 answer : d" | a ) 3377 , b ) 2678 , c ) 5460 , d ) 7500 , e ) 1671 | d | divide(multiply(divide(multiply(5000, add(const_100, 20)), const_100), add(const_100, 25)), const_100) | add(n3,const_100)|add(n2,const_100)|multiply(n0,#1)|divide(#2,const_100)|multiply(#0,#3)|divide(#4,const_100)| | gain |
two trains are moving in opposite directions at 60 km / hr and 90 km / hr . their lengths are 1.10 km and 0.9 km respectively . the time taken by the slower train to cross the faster train in seconds is ? | "relative speed = 60 + 90 = 150 km / hr . = 150 * 5 / 18 = 125 / 3 m / sec . distance covered = 1.10 + 0.9 = 2 km = 2000 m . required time = 2000 * 3 / 125 = 48 sec . answer : c" | a ) 26 sec , b ) 76 sec , c ) 48 sec , d ) 27 sec , e ) 22 sec | c | subtract(divide(multiply(1.10, const_1000), divide(multiply(60, const_1000), const_3600)), divide(multiply(0.9, const_1000), divide(multiply(90, const_1000), const_3600))) | multiply(n2,const_1000)|multiply(n0,const_1000)|multiply(n3,const_1000)|multiply(n1,const_1000)|divide(#1,const_3600)|divide(#3,const_3600)|divide(#0,#4)|divide(#2,#5)|subtract(#6,#7)| | physics |
a , b and c are partners . a receives 2 / 3 of profits , b and c dividing the remainder equally . a ' s income is increased by rs . 300 when the rate to profit rises from 5 to 7 percent . find the capital of c ? | "a : b : c = 2 / 3 : 1 / 6 : 1 / 6 = 4 : 1 : 1 x * 2 / 100 * 2 / 3 = 300 c ' s capital = 22500 * 1 / 6 = 3750 answer : c" | a ) 3377 , b ) 2899 , c ) 3750 , d ) 2778 , e ) 1991 | c | divide(multiply(300, const_100), 2) | multiply(n2,const_100)|divide(#0,n0)| | general |
define a * by the equation a * = Ο - x . then ( ( β Ο ) * ) * = | for a * f ( f ( β Ο ) ) = f ( Ο β ( β Ο ) ) = f ( Ο + Ο ) = f ( 2 Ο ) = Ο β 2 Ο = β Ο = c | a ) β 2 Ο , b ) - 1 , c ) β Ο , d ) 2 Ο , e ) 4 Ο | c | subtract(add(divide(divide(add(multiply(add(const_10, const_4), const_100), add(add(const_10, const_4), const_2)), const_1000), const_10), const_3), subtract(add(divide(divide(add(multiply(add(const_10, const_4), const_100), add(add(const_10, const_4), const_2)), const_1000), const_10), const_3), negate(add(divide(divi... | add(const_10,const_4)|add(#0,const_2)|multiply(#0,const_100)|add(#1,#2)|divide(#3,const_1000)|divide(#4,const_10)|add(#5,const_3)|negate(#6)|subtract(#6,#7)|subtract(#6,#8) | general |
the average age of a group of n people is 15 years old . one more person aged 35 joins the group and the new average is 17 years old . what is the value of n ? | 15 n + 35 = 17 ( n + 1 ) 2 n = 18 n = 9 the answer is b . | a ) 8 , b ) 9 , c ) 10 , d ) 11 , e ) 12 | b | divide(subtract(35, 17), subtract(17, 15)) | subtract(n1,n2)|subtract(n2,n0)|divide(#0,#1) | general |
what is the measure of the radius of the circle inscribed in a triangle whose sides measure 4 , 11 and 12 units ? | sides are 4 , 11 and 12 . . . thus it is right angle triangle since 12 ^ 2 = 4 ^ 2 + 11 ^ 2 therefore , area = 1 / 2 * 11 * 4 = 22 we have to find in - radius therefore , area of triangle = s * r . . . . where s = semi - perimeter and r = in - radius now s = semi - perimeter = 12 + 11 + 4 / 2 = 13,5 thus , 22 = 13,5 * ... | ['a ) 1.6 units', 'b ) 6 units', 'c ) 3 units', 'd ) 5 units', 'e ) 12 units'] | a | divide(triangle_area_three_edges(4, 11, 12), divide(triangle_perimeter(4, 11, 12), const_2)) | triangle_area_three_edges(n0,n1,n2)|triangle_perimeter(n0,n1,n2)|divide(#1,const_2)|divide(#0,#2) | geometry |
what is x if x + 2 y = 10 and y = 4 ? | "x = 10 - 2 y x = 10 - 8 . x = 2 answer : e" | a ) a ) 10 , b ) b ) 8 , c ) c ) 6 , d ) d ) 4 , e ) e ) 2 | e | subtract(10, multiply(2, 4)) | multiply(n0,n2)|subtract(n1,#0)| | general |
we bought a total of 90 books at the store . math books cost $ 4 and history books cost $ 5 . the total price was $ 396 . how many math books did we buy ? | "m + h = 90 h = 90 - m 4 m + 5 h = 396 4 m + 5 * ( 90 - m ) = 396 m = 54 the answer is b ." | a ) 47 , b ) 54 , c ) 56 , d ) 61 , e ) 64 | b | subtract(90, subtract(396, multiply(90, 4))) | multiply(n0,n1)|subtract(n3,#0)|subtract(n0,#1)| | general |
what is the smallest positive perfect square that is divisible by 12 , 15 , and 18 ? | "take the lcm of 12 , 15,18 that will come 180 . the smallest positive perfect square given in the option that can be divided with 180 is 900 . so the ans is ( a )" | a ) 900 , b ) 1,600 , c ) 2,500 , d ) 3,600 , e ) 4,900 | a | add(multiply(multiply(multiply(12, power(const_3, const_2)), 15), const_2), multiply(15, 18)) | multiply(n1,n2)|power(const_3,const_2)|multiply(n0,#1)|multiply(n1,#2)|multiply(#3,const_2)|add(#4,#0)| | geometry |
find the probability that a leap year selected at random will have 53 mondays | there are 366 days in a leap year : 52 weeks and 2 more days . so , 52 mondays and 2 days . these 2 days can be : { mon , tue } , { tue , wed } , { wed , thu } , { thu , fri } , { fri , sat } , { sat , sun } and { sun , mon } ( 7 cases ) . in order to have 53 mondays we should have either { mon , tuesday } or { sun , m... | a ) 6 / 7 , b ) 5 / 7 , c ) 4 / 7 , d ) 3 / 7 , e ) 2 / 7 | e | divide(const_2, add(const_3, const_4)) | add(const_3,const_4)|divide(const_2,#0) | probability |
a train passes a station platform in 36 sec and a man standing on the platform in 20 sec . if the speed of the train is 81 km / hr . what is the length of the platform ? | "speed = 81 * 5 / 18 = 22.5 m / sec . length of the train = 22.5 * 20 = 450 m . let the length of the platform be x m . then , ( x + 450 ) / 36 = 22.5 = > x = 360 m . answer : d" | a ) 240 , b ) 288 , c ) 277 , d ) 360 , e ) 422 | d | multiply(20, multiply(81, const_0_2778)) | multiply(n2,const_0_2778)|multiply(n1,#0)| | physics |
solution x is 30 % chemical a and 70 % chemical b by volume . solution y is 40 % chemical a and 60 % chemical b by volume . if a mixture of x and y is 32 % chemical a , what percent of the mixture is solution x ? | the volume of the mixture be x + y . 0.3 x + 0.4 y = 0.32 ( x + y ) x = 4 y x / ( x + y ) = 4 / 5 = 80 % . the answer is b . | a ) 85 % , b ) 80 % , c ) 75 % , d ) 70 % , e ) 65 % | b | multiply(divide(divide(subtract(40, 32), subtract(32, 30)), add(divide(subtract(40, 32), subtract(32, 30)), const_1)), const_100) | subtract(n2,n4)|subtract(n4,n0)|divide(#0,#1)|add(#2,const_1)|divide(#2,#3)|multiply(#4,const_100) | gain |
the wages earned by robin is 30 % more than that earned by erica . the wages earned by charles is 50 % more than that earned by erica . how much percent is the wages earned by charles more than that earned by robin ? | "let wage of erica = 10 wage of robin = 1.3 * 10 = 13 wage of charles = 1.5 * 10 = 15 percentage by which wage earned by charles is more than that earned by robin = ( 15 - 13 ) / 13 * 100 % = 2 / 13 * 100 % = 15 % answer c" | a ) 18.75 % , b ) 23 % , c ) 15 % , d ) 50 % , e ) 100 % | c | multiply(divide(subtract(add(const_100, 50), add(const_100, 30)), add(const_100, 30)), const_100) | add(n1,const_100)|add(n0,const_100)|subtract(#0,#1)|divide(#2,#1)|multiply(#3,const_100)| | general |
8 x 5.4 - 0.6 x 10 / 1.2 = ? | "given expression = ( 43.2 - 6 ) / 1.2 = 37.2 / 1.2 = 31 answer is d ." | a ) 30 , b ) 45 , c ) 50 , d ) 31 , e ) 21 | d | divide(subtract(multiply(8, 5.4), multiply(0.6, 10)), 1.2) | multiply(n0,n1)|multiply(n2,n3)|subtract(#0,#1)|divide(#2,n4)| | general |
veena ranks 65 rd from the top in a class of 182 . what is her rank from the bottom if 22 students have failed the examination ? | "total student = 182 failed = 22 paasd student = 182 - 22 = 160 from bottom her rank is = 160 - 65 + 1 = 96 answer : a" | a ) 96 , b ) 108 , c ) 110 , d ) 90 , e ) 93 | a | subtract(subtract(subtract(182, 22), subtract(65, const_1)), const_4) | subtract(n1,n2)|subtract(n0,const_1)|subtract(#0,#1)|subtract(#2,const_4)| | other |
there were 36000 hardback copies of a certain novel sold before the paperback version was issued . from the time the first paperback copy was sold until the last copy of the novel was sold 9 times as many paperback copies as hardback copies were sold . if a total of 440000 copies of the novel were sold in all , how man... | say x was the # of hardback copies sold from the time the first paperback copy was sold . then the total # of paperback copies sold was 9 x ; hence the total # of copies sold was ( hardback ) + ( paperback ) = ( 36 + x ) + ( 9 x ) = 440 - - > x = 40.4 . so , the total # of paperback copies sold was 9 x = 9 * 40.4 = 363... | a ) 45,000 , b ) 360,000 , c ) 364,500 , d ) 363,600 , e ) 396,900 | d | multiply(9, divide(subtract(440000, 36000), add(9, const_1))) | add(n1,const_1)|subtract(n2,n0)|divide(#1,#0)|multiply(n1,#2) | general |
a certain car ' s price decreased by 2.5 % ( from the original price ) each year from 1996 to 2002 , during that time the owner of the car invested in a new carburetor and a new audio system for the car , which increased car ' s price by $ 1,500 . if the price of the car in 1996 was $ 22,000 , what is the car ' s price... | "important point to notice - 2.5 % decrease from the original price 2.5 % of 22,000 = 550 total reduction in 6 years = 550 * 6 = 3300 final price = 22,000 + 1500 - 3300 = 20,200 c is the answer" | a ) $ 18,400 , b ) $ 19,500 , c ) $ 20,200 , d ) $ 20,400 , e ) $ 21,100 | c | multiply(const_2, const_10) | multiply(const_10,const_2)| | gain |
in a covering a certain distance , the speeds of a and b are in the ratio of 3 : 4 . a takes 30 minutes more than b to reach the destination . the time taken by a to reach the destination is ? | ratio of speeds = 3 : 4 ratio of times taken = 4 : 3 suppose a takes 4 x hrs and b takes 3 x hrs to reach the destination . then , 4 x - 3 x = 30 / 60 = > x = 1 / 2 time taken by a = 4 x hrs = 4 * 1 / 2 = 2 hrs . answer : c | a ) 8 hrs , b ) 9 hrs , c ) 2 hrs , d ) 2 hrs , e ) 9 hrs | c | multiply(4, divide(30, const_60)) | divide(n2,const_60)|multiply(n1,#0) | physics |
in a group of ducks and cows , the total number of legs are 24 more than twice the no . of heads . find the total no . of buffaloes . | "let the number of buffaloes be x and the number of ducks be y = > 4 x + 2 y = 2 ( x + y ) + 24 = > 2 x = 24 = > x = 12 b" | a ) 10 , b ) 12 , c ) 14 , d ) 16 , e ) 19 | b | divide(24, const_2) | divide(n0,const_2)| | general |
rohan spends 40 % of his salary on food , 20 % on house rent , 10 % on entertainment and 10 % on conveyance . if his savings at the end of a month are rs . 2500 . then his monthly salary is | "sol . saving = [ 100 - ( 40 + 20 + 10 + 10 ] % = 20 % . let the monthly salary be rs . x . then , 20 % of x = 2500 Γ’ β‘ β 20 / 100 x = 2500 Γ’ β‘ β x = 2500 Γ£ β 5 = 12500 . answer a" | a ) rs . 12500 , b ) rs . 1000 , c ) rs . 8000 , d ) rs . 6000 , e ) rs . 15000 | a | multiply(2500, add(const_4, const_1)) | add(const_1,const_4)|multiply(n4,#0)| | gain |
a man complete a journey in 10 hours . he travels first half of the journey at the rate of 20 km / hr and second half at the rate of 20 km / hr . find the total journey in km . | "0.5 x / 20 + 0.5 x / 20 = 10 - - > x / 20 + x / 20 = 20 - - > 2 x = 20 x 20 - - > x = ( 20 x 20 ) / 2 = 200 km . answer : e ." | a ) 220 km , b ) 224 km , c ) 230 km , d ) 232 km , e ) 200 km | e | multiply(const_2, divide(multiply(multiply(20, 20), 10), add(20, 20))) | add(n1,n2)|multiply(n1,n2)|multiply(n0,#1)|divide(#2,#0)|multiply(#3,const_2)| | physics |
the probability of a student possessing a ball point pen in exam is 3 / 5 & possessing an ink pen is 2 / 3 . find his probability of possessing at least one of them | the probability of a student possessing a ball point pen in exam is 3 / 5 . the probability of a student not possessing a ball point pen in exam is 2 / 5 & possessing an ink pen is 2 / 3 . & not possessing an ink pen is 1 / 3 . his probability of possessing none of them = 2 / 5 * 1 / 3 = 2 / 15 his probability of posse... | a ) 10 / 15 , b ) 11 / 15 , c ) 12 / 15 , d ) 13 / 15 , e ) 14 / 15 | d | divide(add(const_12, const_1), multiply(3, 5)) | add(const_1,const_12)|multiply(n0,n1)|divide(#0,#1) | general |
if 200 ! / 10 ^ n is an integer , what is the largest possible value of n ? | "we have to basically count the number of 5 s withing 200 200 / 5 = 40 200 / 5 ^ 2 = 8 200 / 5 ^ 3 = 1 we know for sure that there will be at least 49 even numbers ( multiples of 2 ) within 200 so there are 49 10 s in 200 ! answer : e" | a ) 40 , b ) 42 , c ) 44 , d ) 48 , e ) 49 | e | add(divide(divide(200, add(const_4, const_1)), add(const_4, const_1)), divide(200, add(const_4, const_1))) | add(const_1,const_4)|divide(n0,#0)|divide(#1,#0)|add(#2,#1)| | general |
if you write down all the numbers from 1 to 50 , then how many times do you write 3 ? | "explanation : explanation : clearly , from 1 to 50 , there are ten numbers with 3 as the unit ' s digit - 3 , 13 , 23 , 33 , 43 , and ten numbers with 3 as the ten ' s digit - 30 , 31 , 32 , 33 , 34 , 35 , 36 , 37 , 38 , 39 . so , required number = 5 + 10 = 15 . answer : b" | a ) a ) 11 , b ) b ) 15 , c ) c ) 20 , d ) d ) 21 , e ) e ) 22 | b | divide(subtract(50, 1), 3) | subtract(n1,n0)|divide(#0,n2)| | general |
evaluate 248 + 64 β β β β β β β β β β β β β | explanation : = 248 + 64 β β β β β β β β β β β β β = 248 + 8 β β β β β β β = 256 β β β β = 16 answer : c | a ) 14 , b ) 26 , c ) 16 , d ) 36 , e ) 46 | c | multiply(const_2, subtract(multiply(64, const_4), 248)) | multiply(n1,const_4)|subtract(#0,n0)|multiply(#1,const_2) | general |
two employees x and y are paid a total of rs . 638 per week by their employer . if x is paid 120 percent of the sum paid to y , how much is y paid per week ? | "let the amount paid to x per week = x and the amount paid to y per week = y then x + y = 638 but x = 120 % of y = 120 y / 100 = 12 y / 10 β΄ 12 y / 10 + y = 638 β y [ 12 / 10 + 1 ] = 638 β 22 y / 10 = 638 β 22 y = 6380 β y = 6380 / 22 = 580 / 2 = rs . 290 c" | a ) s . 250 , b ) s . 280 , c ) s . 290 , d ) s . 299 , e ) s . 300 | c | divide(multiply(638, multiply(add(const_1, const_4), const_2)), multiply(add(multiply(add(const_1, const_4), const_2), const_1), const_2)) | add(const_1,const_4)|multiply(#0,const_2)|add(#1,const_1)|multiply(n0,#1)|multiply(#2,const_2)|divide(#3,#4)| | general |
for any integer k > 1 , the term β length of an integer β refers to the number of positive prime factors , not necessarily distinct , whose product is equal to k . for example , if k = 24 , the length of k is equal to 4 , since 24 = 2 Γ 2 Γ 2 Γ 3 . if x and y are positive integers such that x > 1 , y > 1 , and x + 3 y ... | "we know that : x > 1 , y > 1 , and x + 3 y < 1000 , and it is given that length means no of factors . for any value of x and y , the max no of factors can be obtained only if factor is smallest noall factors are equal . hence , lets start with smallest no 2 . 2 ^ 1 = 2 2 ^ 2 = 4 2 ^ 3 = 8 2 ^ 4 = 16 2 ^ 5 = 32 2 ^ 6 =... | a ) 14 , b ) 12 , c ) 10 , d ) 16 , e ) 18 | d | add(add(4, 3), add(add(4, 4), 1)) | add(n2,n7)|add(n2,n2)|add(n0,#1)|add(#0,#2)| | general |
a circular grass lawn of 35 metres in radius has a path 7 metres wide running around it on the outside . find the area of path . | radius of a circular grass lawn ( without path ) = 35 m β΄ area = Ο r 2 = Ο ( 35 ) 2 radius of a circular grass lawn ( with path ) = 35 + 7 = 42 m β΄ area = Ο r 2 = Ο ( 42 ) 2 β΄ area of path = Ο ( 42 ) 2 β Ο ( 35 ) 2 = Ο ( 422 β 352 ) = Ο ( 42 + 35 ) ( 42 β 35 ) answer a | ['a ) 1694 m 2', 'b ) 1700 m 2', 'c ) 1598 m 2', 'd ) 1500 m 2', 'e ) none of thes'] | a | subtract(circle_area(add(35, 7)), circle_area(35)) | add(n0,n1)|circle_area(n0)|circle_area(#0)|subtract(#2,#1) | geometry |
sonika deposited rs . 6000 which amounted to rs . 9200 after 3 years at simple interest . had the interest been 2 % more . she would get how much ? | "( 6000 * 3 * 2 ) / 100 = 360 9200 - - - - - - - - 9560 answer : d" | a ) 9660 , b ) 6560 , c ) 7560 , d ) 9560 , e ) 8560 | d | add(multiply(multiply(add(divide(2, const_100), divide(divide(subtract(9200, 6000), 3), 6000)), 6000), 3), 6000) | divide(n3,const_100)|subtract(n1,n0)|divide(#1,n2)|divide(#2,n0)|add(#0,#3)|multiply(n0,#4)|multiply(n2,#5)|add(n0,#6)| | gain |
excluding stoppages , the speed of a train is 42 kmph and including stoppages it is 27 kmph . of how many minutes does the train stop per hour ? | "explanation : t = 15 / 42 * 60 = 21.42 answer : option c" | a ) a ) 19.42 , b ) b ) 20.42 , c ) c ) 21.42 , d ) d ) 22.42 , e ) e ) 23.42 | c | subtract(const_60, multiply(const_60, divide(27, 42))) | divide(n1,n0)|multiply(#0,const_60)|subtract(const_60,#1)| | physics |
the prices of tea and coffee per kg were the same in june . in july the price of coffee shot up by 20 % and that of tea dropped by 20 % . if in july , a mixture containing equal quantities of tea and coffee costs 80 / kg . how much did a kg of coffee cost in june ? | "let the price of tea and coffee be x per kg in june . price of tea in july = 1.2 x price of coffee in july = 0.8 x . in july the price of 1 / 2 kg ( 800 gm ) of tea and 1 / 2 kg ( 800 gm ) of coffee ( equal quantities ) = 80 1.2 x ( 1 / 2 ) + 0.8 x ( 1 / 2 ) = 80 = > x = 80 thus proved . . . option c ." | a ) 50 , b ) 60 , c ) 80 , d ) 100 , e ) 120 | c | divide(80, multiply(subtract(const_1, divide(20, const_100)), add(divide(20, const_100), const_1))) | divide(n0,const_100)|divide(n1,const_100)|add(#0,const_1)|subtract(const_1,#1)|multiply(#2,#3)|divide(n2,#4)| | general |
working alone , mary can pave a driveway in 6 hours and hillary can pave the same driveway in 6 hours . when they work together , mary thrives on teamwork so her rate increases by 25 % , but hillary becomes distracted and her rate decreases by 20 % . if they both work together , how many hours will it take to pave the ... | "initial working rates : mary = 1 / 6 per hour hillary = 1 / 6 per hour rate when working together : mary = 1 / 6 + ( 1 / 4 * 1 / 6 ) = 1 / 5 per hour hillary = 1 / 6 - ( 1 / 5 * 1 / 6 ) = 2 / 15 per hour together they work 1 / 5 + 2 / 15 = 1 / 3 per hour so they will need 3 hours to complete the driveway . the correct... | a ) 3 hours , b ) 4 hours , c ) 5 hours , d ) 6 hours , e ) 7 hours | a | inverse(add(multiply(divide(const_1, 6), add(divide(25, const_100), const_1)), multiply(divide(const_1, 6), divide(20, const_100)))) | divide(n2,const_100)|divide(const_1,n0)|divide(const_1,n1)|divide(n3,const_100)|add(#0,const_1)|multiply(#2,#3)|multiply(#4,#1)|add(#6,#5)|inverse(#7)| | gain |
two boys starts from the same place walking at the rate of 4.5 kmph and 5.5 kmph respectively in the same direction . what time will they take to be 9.5 km apart ? | explanation : relative speed = 5.5 - 4.5 = 1 kmph ( because they walk in the same direction ) distance = 9.5 km time = distance / speed = 9.5 / 1 = 9.5 hr answer : a | a ) 9.5 , b ) 5.5 , c ) 8.5 , d ) 9.6 , e ) 9.7 | a | divide(9.5, subtract(5.5, 4.5)) | subtract(n1,n0)|divide(n2,#0) | gain |
a train is moving at 4 / 5 of its usual speed . the train is 45 minutes too late . what is the usual time ( in hours ) for the train to complete the journey ? | new time = d / ( 4 v / 5 ) = 5 / 4 * usual time 45 minutes represents 1 / 4 of the usual time . the usual time is 4 * 45 minutes = 3 hours . the answer is e . | a ) 1 , b ) 1.5 , c ) 2 , d ) 2.5 , e ) 3 | e | divide(multiply(multiply(45, divide(4, 5)), inverse(subtract(const_1, divide(4, 5)))), const_60) | divide(n0,n1)|multiply(n2,#0)|subtract(const_1,#0)|inverse(#2)|multiply(#3,#1)|divide(#4,const_60) | physics |
in a class of 52 students , 12 enrolled for both english and german . 22 enrolled for german . if the students of the class enrolled for at least one of the two subjects , then how many students enrolled for only english and not german ? | total = english + german - both + neither - - > 52 = english + 22 - 12 + 0 - - > english = 42 - - > only english = english - both = 42 - 12 = 30 answer : a . | a ) 30 , b ) 10 , c ) 18 , d ) 28 , e ) 32 | a | subtract(subtract(add(52, 12), 22), 12) | add(n0,n1)|subtract(#0,n2)|subtract(#1,n1) | other |
if the ratio of apples to bananas is 5 to 2 and the ratio of bananas to cucumbers is 1 to 4 , what is the ratio of apples to cucumbers ? | the ratio of bananas to cucumbers is 1 to 4 which equals 2 to 8 . the ratio of apples to bananas to cucumbers is 5 to 2 to 8 . the ratio of apples to cucumbers is 5 to 8 . the answer is d . | a ) 1 : 3 , b ) 2 : 5 , c ) 3 : 5 , d ) 5 : 8 , e ) 4 : 7 | d | divide(divide(5, 2), 4) | divide(n0,n1)|divide(#0,n3) | other |
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 l 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 l 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 |
the average of marks obtained by 120 candidates was 35 . if the avg of marks of passed candidates was 39 and that of failed candidates was 39 and that of failed candidates was 15 , the no . of candidates who passed the examination is ? | let the number of candidate who passed = y then , 39 y + 15 ( 120 - y ) = 120 x 35 β 24 y = 4200 - 1800 β΄ y = 2400 / 24 = 100 c | a ) 80 , b ) 90 , c ) 100 , d ) 120 , e ) 130 | c | divide(subtract(multiply(120, 35), multiply(120, 15)), subtract(39, 15)) | multiply(n0,n1)|multiply(n0,n4)|subtract(n2,n4)|subtract(#0,#1)|divide(#3,#2) | general |
in may , the groundskeeper at spring lake golf club built a circular green with an area of 90 Ο square feet . in august , the groundskeeper 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 = Ο r ^ 2 , so doubling the radius results in an area that is 4 times the original area . 4 ( 90 Ο ) = 360 Ο the answer is a ." | a ) 360 Ο , b ) 400 Ο , c ) 450 Ο , d ) 500 Ο , e ) 600 Ο | a | multiply(power(multiply(sqrt(90), const_2), const_2), const_pi) | sqrt(n0)|multiply(#0,const_2)|power(#1,const_2)|multiply(#2,const_pi)| | geometry |
the average of first six prime numbers which are between 40 and 80 is | "explanation : first six prime numbers which are between 40 and 80 = 41 , 43 , 47 , 53 , 59 , 61 average = ( 41 + 43 + 47 + 53 + 59 + 61 ) / 6 = 50.7 answer : b" | a ) 35.4 , b ) 50.7 , c ) 45.7 , d ) 57 , e ) 67 | b | add(40, const_1) | add(n0,const_1)| | general |
boy sells a book for rs . 720 he gets a loss of 10 % , to gain 10 % , what should be the sp ? | "cost price = 720 / 90 x 100 = 800 to gain 10 % = 800 x 10 / 100 = 80 sp = cp + gain = 800 + 80 = 880 answer : b" | a ) 430 , b ) 880 , c ) 550 , d ) 590 , e ) 600 | b | add(divide(720, subtract(const_1, divide(10, const_100))), multiply(divide(720, subtract(const_1, divide(10, const_100))), divide(10, const_100))) | divide(n1,const_100)|divide(n2,const_100)|subtract(const_1,#0)|divide(n0,#2)|multiply(#3,#1)|add(#3,#4)| | gain |
renu can do a piece of work in 5 days , but with the help of her friend suma , she can do it in 4 days . in what time suma can do it alone ? | "renu Γ’ β¬ β’ s one day Γ’ β¬ β’ s work = 1 / 5 suma Γ’ β¬ β’ s one day Γ’ β¬ β’ s work = 1 / 4 - 1 / 5 = 1 / 20 suma can do it alone in 20 days . answer : e" | a ) 10 , b ) 12 , c ) 14 , d ) 15 , e ) 20 | e | inverse(subtract(divide(const_1, 4), divide(const_1, 5))) | divide(const_1,n1)|divide(const_1,n0)|subtract(#0,#1)|inverse(#2)| | physics |
when positive integer n is divided by 5 , the remainder is 1 . when n is divided by 7 , the remainder is 5 . what is the smallest positive integer p , such that ( n + p ) is a multiple of 30 ? | "when positive integer n is divided by 5 , the remainder is 1 i . e . , n = 5 x + 1 values of n can be one of { 1 , 6 , 11 , 16 , 21 , 26 , 31 . . . . . . . . . . . . . 46 , 51 , 56,61 . . . . . . . . . . . . . . . . . . } similarly , when n is divided by 7 , the remainder is 5 . . i . e . , n = 7 y + 5 values of n can... | a ) 1 , b ) 2 , c ) 4 , d ) 19 , e ) 20 | c | subtract(30, reminder(5, 7)) | reminder(n3,n2)|subtract(n4,#0)| | general |
the compounded ratio of ( 2 : 3 ) , ( 6 : 11 ) and ( 11 : 2 ) is : | answer : option c 2 / 3 : 6 / 11 : 11 / 2 = 2 : 1 | a ) 1 : 2 , b ) 5 : 9 , c ) 2 : 1 , d ) 11 : 24 , e ) none | c | multiply(multiply(divide(2, 3), divide(6, 11)), divide(11, 2)) | divide(n3,n0)|divide(n0,n1)|divide(n2,n3)|multiply(#1,#2)|multiply(#0,#3) | other |
in a graduating class of 232 students , 144 took geometry and 119 took biology . what is the difference between the greatest possible number p and the smallest possible number of students that could have taken both geometry and biology ? | "official solution : first of all , notice that since 144 took geometry and 119 took biology , then the number of students who took both geometry and biology can not be greater than 119 . { total } = { geometry } + { biology } - { both } + { neither } ; 232 = 144 + 119 - { both } + { neither } ; { both } = 31 + { neith... | a ) 144 , b ) 119 , c ) 113 , d ) 88 , e ) 31 | d | subtract(119, subtract(add(144, 119), 232)) | add(n1,n2)|subtract(#0,n0)|subtract(n2,#1)| | other |
what is the remainder when 30 ^ 72 ^ 87 is divided by 11 . | "let us take n = 72 ^ 87 eq is . . 30 ^ n / 11 = 8 ^ n / 11 here cyclicity of 8 is 10 then n / 10 = 72 ^ 87 / 10 = 2 ^ 87 / 10 here cyclicity of 2 is then 87 % 4 = 3 ( 3 rd cycle of 2 ) . . . . i . e 2 ^ 3 % 10 = 8 ( 8 th cycle of 8 ) therefore final ans is 8 ^ 8 % 11 = 5 answer : a" | a ) 5 , b ) 7 , c ) 1 , d ) 3 , e ) 4 | a | reminder(multiply(72, 30), 87) | multiply(n0,n1)|reminder(#0,n2)| | general |
the number of people who purchased book a is twice the number of people who purchased book b . the number of people who purchased both books a and b is 500 , which is twice the number of people who purchased only book b . what is the number of people v who purchased only book a ? | "this is best solved using overlapping sets or a venn diagram . we know that a = 2 b , and that 500 people purchased both a and b . further , those purchasing both was double those purchasing b only . this gives us 250 people purchasing b only . with the 500 that pruchased both , we have a total of 750 that purchased b... | a ) 250 , b ) 500 , c ) 750 , d ) 1000 , e ) 1500 | d | subtract(multiply(add(500, divide(500, const_2)), const_2), 500) | divide(n0,const_2)|add(n0,#0)|multiply(#1,const_2)|subtract(#2,n0)| | other |
in a recent election , geoff received 1 percent of the 6,000 votes cast . to win the election , a candidate needed to receive more than x % of the vote . if geoff needed exactly 3000 more votes to win the election , what is the value of x ? | "word problems are tricky in somehow more than other problem because you have the additional step to translate . breaking the problem : geoff how many votes he receives ? ? 60 votes he needs 3571 more votes so : 60 + 3000 = 3060 now what ' s the problem wants ? ? a x % . . . . . . . . 3060 is what % of total votes 6000... | a ) 50 , b ) 54 , c ) 56 , d ) 51 , e ) 63 | d | add(divide(const_100, const_2), 1) | divide(const_100,const_2)|add(n0,#0)| | gain |
two mba admissions committees are to be formed randomly from 6 second year mbas with 3 members each . what is the probability e that jane will be on the same committee as albert ? | "total number of ways to choose 3 member committee - 6 c 3 = ( 6 ! / 3 ! 3 ! ) = 20 no . of ways albert n jane are in same committee : - ( 4 c 1 * 2 ) = 8 probability e = ( 8 / 20 ) * 100 = 40 % . + 1 for me . . : d" | a ) 12 % , b ) 20 % , c ) 33 % , d ) 40 % , e ) 50 % | d | multiply(divide(multiply(choose(const_4, const_1), const_2), choose(6, 3)), multiply(multiply(const_5, const_5), const_4)) | choose(const_4,const_1)|choose(n0,n1)|multiply(const_5,const_5)|multiply(#0,const_2)|multiply(#2,const_4)|divide(#3,#1)|multiply(#5,#4)| | probability |
if 40 % of a certain number is 160 , then what is 20 % of that number ? | "explanation : 40 % = 40 * 4 = 160 20 % = 20 * 4 = 80 answer : option e" | a ) 75 , b ) 100 , c ) 120 , d ) 30 , e ) 80 | e | multiply(divide(160, divide(40, const_100)), divide(20, const_100)) | divide(n0,const_100)|divide(n2,const_100)|divide(n1,#0)|multiply(#2,#1)| | gain |
workers decided to raise rs . 3 lacs by equal contribution from each . had they contributed rs . 50 eachextra , the contribution would have been rs . 3.15 lacs . how many workers were they ? | "n * 50 = ( 315000 - 300000 ) = 15000 n = 15000 / 50 = 300 a" | a ) 300 , b ) 230 , c ) 500 , d ) 560 , e ) 590 | a | divide(multiply(multiply(subtract(3.15, 3), const_1000), const_100), 50) | subtract(n2,n0)|multiply(#0,const_1000)|multiply(#1,const_100)|divide(#2,n1)| | general |
1 ^ 2 β 2 ^ 2 + 3 ^ 2 β 4 ^ 2 + 5 ^ 2 β 6 ^ 2 + 7 ^ 2 β 8 ^ 2 + 9 ^ 2 β 10 ^ 2 = ? | 1 - 4 + 9 - 16 + 25 - 36 + 49 - 64 + 81 - 100 = - 55 answer : d | a ) - 21 , b ) - 53 , c ) - 35 , d ) - 55 , e ) - 58 | d | power(5, 2) | power(n8,n1) | general |
there are 7 players in a bowling team with an average weight of 112 kg . if two new players join the team , one weighs 110 kg and the second weighs 60 kg , what will be the new average weight ? | "the new average will be = ( 112 * 7 + 110 + 60 ) / 9 = 106 kgs c is the answer" | a ) 115 kg . , b ) 110 kg . , c ) 106 kg . , d ) 105 kg . , e ) 100 kg . | c | divide(add(multiply(7, 112), add(110, 60)), add(7, const_2)) | add(n2,n3)|add(n0,const_2)|multiply(n0,n1)|add(#0,#2)|divide(#3,#1)| | general |
fresh grapes contain 90 % by weight while dried grapes contain 20 % water by weight . what is the weight of dry grapes available from 25 kg of fresh grapes ? | the weight of non - water in 25 kg of fresh grapes ( which is 100 - 90 = 10 % of whole weight ) will be the same as the weight of non - water in x kg of dried grapes ( which is 100 - 20 = 80 % of whole weight ) , so 25 Γ’ Λ β 0.1 = x Γ’ Λ β 0.8 - - > x = 3.12 answer : c . | a ) 2 kg , b ) 2.4 kg , c ) 3.12 kg , d ) 10 kg , e ) none of these | c | multiply(divide(divide(multiply(subtract(const_100, 90), 25), const_100), subtract(const_100, 20)), const_100) | subtract(const_100,n0)|subtract(const_100,n1)|multiply(n2,#0)|divide(#2,const_100)|divide(#3,#1)|multiply(#4,const_100) | gain |
anna left for city a from city b at 5.20 a . m . she traveled at the speed of 80 km / hr for 2 hrs 15 min . after that the speed was reduced to 60 km / hr . if the distance between two cities is 350 kms , at what time did anna reach city a ? | distance covered in 2 hrs 15 min i . e . , 2 1 / 4 hrs = 80 * 9 / 4 = 180 hrs . time taken to cover remaining distance = ( 350 - 180 ) / 60 = 17 / 6 hrs = 2 5 / 6 = 2 hrs 50 min total time taken = ( 2 hrs 15 min + 2 hrs 50 min ) = 5 hrs 5 min . so , anna reached city a at 10.25 a . m . answer : e | a ) 10.21 , b ) 10.27 , c ) 10.25 , d ) 10.23 , e ) 60.25 | e | add(60, divide(multiply(subtract(divide(divide(subtract(350, divide(multiply(80, multiply(const_3, const_3)), const_4)), 60), 2), const_1), const_60), const_100)) | multiply(const_3,const_3)|multiply(n1,#0)|divide(#1,const_4)|subtract(n5,#2)|divide(#3,n4)|divide(#4,n2)|subtract(#5,const_1)|multiply(#6,const_60)|divide(#7,const_100)|add(n4,#8) | physics |
at veridux corporation , there are 250 employees . of these , 90 are female , and the rest are males . there are a total of 40 managers , and the rest of the employees are associates . if there are a total of 155 male associates , how many female managers are there ? | "250 employees : 90 male , 160 female 40 managers , 210 associates 155 male associates implies 55 female associates which means the remaining 35 females must be managers e . 35" | a ) 15 , b ) 20 , c ) 25 , d ) 30 , e ) 35 | e | multiply(40, const_1) | multiply(n2,const_1)| | general |
pencils , pens and exercise books in a shop are in the ratio of 10 : 2 : 3 . if there are 120 pencils , the number of exercise books in the shop is : | "explanation : let pencils = 10 x , pens = 2 x & exercise books = 3 x . now , 10 x = 120 hence x = 12 . number of exercise books = 3 x = 36 . answer : b" | a ) 26 , b ) 36 , c ) 46 , d ) 56 , e ) 66 | b | multiply(divide(120, 10), 3) | divide(n3,n0)|multiply(n2,#0)| | other |
in a group of ducks and cows , the total number of legs are 26 more than thrice the number of heads . find the total number of cows . | "explanation : let the number of ducks be d and number of cows be c then , total number of legs = 2 d + 4 c = 2 ( d + 2 c ) total number of heads = c + d given that total number of legs are 26 more than twice the number of heads = > 2 ( d + 2 c ) = 26 + 2 ( c + d ) = > d + 2 c = 13 + c + d = > 2 c = 13 + c = > c = 13 i... | a ) a ) 14 , b ) b ) 13 , c ) c ) 16 , d ) d ) 8 , e ) e ) 6 | b | divide(26, const_2) | divide(n0,const_2)| | general |
working alone at its constant rate , machine a produces x boxes in 5 minutes and working alone at its constant rate , machine b produces 2 x boxes in 10 minutes . how many minutes does it take machines a and b , working simultaneously at their respective constant rates , to produce 3 x boxes ? | "rate = work / time given rate of machine a = 2 x / 10 min machine b produces 2 x boxes in 10 min hence , machine b produces 2 x boxes in 10 min . rate of machine b = 2 x / 10 we need tofind the combined time that machines a and b , working simultaneouslytakeat their respective constant rates let ' s first find the com... | a ) 3 minutes , b ) 7.5 minutes , c ) 5 minutes , d ) 6 minutes , e ) 12 minutes | b | divide(multiply(3, 5), add(speed(5, 5), speed(multiply(2, 5), 10))) | multiply(n0,n3)|multiply(n0,n1)|speed(n0,n0)|speed(#1,n2)|add(#2,#3)|divide(#0,#4)| | physics |
a factory that employs 1000 assembly line workers pays each of these workers $ 5 per hour for the first 40 hours worked during a week and 1 Β½ times that rate for hours worked in excess of 40 . what was the total payroll for the assembly - line workers for a week in which 30 percent of them worked 35 hours , 50 percent ... | "30 % of 1000 = 300 worked for 20 hours payment @ 5 / hr total payment = 300 * 35 * 5 = 52500 50 % of 1000 = 500 worked for 40 hours payment @ 5 / hr total payment = 500 * 40 * 5 = 100000 remaining 200 worked for 50 hours payment for first 40 hours @ 5 / hr payment = 200 * 40 * 5 = 40000 payment for next 10 hr @ 7.5 / ... | a ) $ 180,000 , b ) $ 185,000 , c ) $ 190,000 , d ) $ 207,500 , e ) $ 205,000 | d | multiply(add(divide(1, 35), 1), divide(multiply(1000, divide(add(add(multiply(add(multiply(divide(const_3, const_2), multiply(5, 35)), multiply(40, 5)), subtract(35, add(const_3, 5))), multiply(multiply(40, 5), 5)), multiply(multiply(5, 35), const_3)), 35)), 1000)) | add(n1,const_3)|divide(n3,n6)|divide(const_3,const_2)|multiply(n1,n6)|multiply(n1,n2)|add(n3,#1)|multiply(#2,#3)|multiply(n1,#4)|multiply(#3,const_3)|subtract(n6,#0)|add(#6,#4)|multiply(#10,#9)|add(#11,#7)|add(#12,#8)|divide(#13,n6)|multiply(n0,#14)|divide(#15,n0)|multiply(#5,#16)| | general |
solution a is 20 % sugar and solution b is 80 % sugar . 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 % sugar 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 - ... | a ) 6 : 4 , b ) 6 : 14 , c ) 4 : 4 , d ) 4 : 6 , e ) 3 : 7 | c | multiply(divide(subtract(divide(80, const_100), divide(50, const_100)), divide(60, const_100)), const_2) | divide(n1,const_100)|divide(n4,const_100)|divide(n3,const_100)|subtract(#0,#1)|divide(#3,#2)|multiply(#4,const_2) | other |
12 different biology books and 8 different chemistry books lie on a shelf . in how many ways can a student pick 2 books of each type ? | no . of ways of picking 2 biology books ( from 12 books ) = 12 c 2 = ( 12 * 11 ) / 2 = 66 no . of ways of picking 2 chemistry books ( from 8 books ) = 8 c 2 = ( 8 * 7 ) / 2 = 28 total ways of picking 2 books of each type = 66 * 28 = 1848 ( option e ) | a ) 80 , b ) 160 , c ) 720 , d ) 1600 , e ) 1848 | e | multiply(divide(divide(factorial(12), factorial(subtract(12, 2))), 2), divide(divide(factorial(8), factorial(subtract(8, 2))), 2)) | factorial(n0)|factorial(n1)|subtract(n0,n2)|subtract(n1,n2)|factorial(#2)|factorial(#3)|divide(#0,#4)|divide(#1,#5)|divide(#6,n2)|divide(#7,n2)|multiply(#8,#9) | other |
the total price of a basic computer and printer are $ 2,500 . if the same printer had been purchased with an enhanced computer whose price was $ 500 more than the price of the basic computer , then the price of the printer would have been 1 / 3 of that total . what was the price of the basic computer ? | "let the price of basic computer be c and the price of the printer be p : c + p = $ 2,500 . the price of the enhanced computer will be c + 500 and total price for that computer and the printer will be 2,500 + 500 = $ 3,000 . now , we are told that the price of the printer is 1 / 3 of that new total price : p = 1 / 3 * ... | a ) 1500 , b ) 1600 , c ) 1750 , d ) 1900 , e ) 2000 | a | subtract(multiply(multiply(const_0_25, const_1000), const_10), divide(add(500, multiply(multiply(const_0_25, const_1000), const_10)), 3)) | multiply(const_0_25,const_1000)|multiply(#0,const_10)|add(n1,#1)|divide(#2,n3)|subtract(#1,#3)| | general |
a baseball card decreased in value 20 % in its first year and 20 % in its second year . what was the total percent decrease of the card ' s value over the two years ? | "consider the initial value of the baseball card as $ 100 after first year price = 100 * 0.8 = 80 after second year price = 80 * 0.8 = 64 final decrease = [ ( 100 - 64 ) / 100 ] * 100 = 36 % correct answer - d" | a ) 28 % , b ) 30 % , c ) 32 % , d ) 36 % , e ) 72 % | d | subtract(const_100, multiply(multiply(subtract(const_1, divide(20, const_100)), subtract(const_1, divide(20, const_100))), const_100)) | divide(n1,const_100)|divide(n0,const_100)|subtract(const_1,#0)|subtract(const_1,#1)|multiply(#2,#3)|multiply(#4,const_100)|subtract(const_100,#5)| | gain |
2 is what percent of 50 | "explanation : 2 / 50 * 100 = 1 / 25 * 100 = 4 % option b" | a ) 2 % , b ) 4 % , c ) 6 % , d ) 8 % , e ) none of these | b | multiply(divide(2, 50), const_100) | divide(n0,n1)|multiply(#0,const_100)| | gain |
the contents of a certain box consist of 14 apples and 26 oranges . how many oranges must be removed from the box so that 70 percent of the pieces of fruit in the box will be apples ? | "the objective here is that 70 % of the fruit in the box should be apples . now , there are 14 apples at start and there is no talk of removing any apples , so number of apples should remain 14 and they should constitute 70 % of total fruit , so total fruit = 14 / 0.7 = 20 so we should have 20 - 14 = 6 oranges . right ... | a ) 3 , b ) 6 , c ) 14 , d ) 17 , e ) 20 | e | subtract(add(14, 26), divide(14, divide(70, const_100))) | add(n0,n1)|divide(n2,const_100)|divide(n0,#1)|subtract(#0,#2)| | general |
light glows for every 20 seconds . how many max . times did it glow between 1 : 57 : 58 and 3 : 20 : 47 am . | "time difference is 1 hr , 22 min , 49 sec = 4969 sec . so , light glows floor ( 4969 / 20 ) = 248 times . answer : d" | a ) 380 times , b ) 381 times , c ) 382 times , d ) 248 times , e ) 482 times | d | divide(add(add(const_2, 47), multiply(add(20, add(const_2, const_60)), const_60)), 20) | add(n6,const_2)|add(const_2,const_60)|add(n5,#1)|multiply(#2,const_60)|add(#0,#3)|divide(#4,n0)| | general |
each week , harry is paid x dollars per hour for the first 18 hours and 1.5 x dollars for each additional hour worked that week . each week , james is paid x dollars per per hour for the first 40 hours and 2 x dollars for each additional hour worked that week . last week james worked a total of 41 hours if harry and ja... | "42 x = 18 x + 1.5 x ( h - 18 ) = = > 42 = 18 + 1.5 ( h - 18 ) = = > h - 18 = 24 / 1.5 = 16 = = > h = 34 answer is a" | a ) 34 , b ) 36 , c ) 37 , d ) 38 , e ) 39 | a | add(divide(subtract(add(40, 2), 18), 1.5), 18) | add(n2,n3)|subtract(#0,n0)|divide(#1,n1)|add(n0,#2)| | general |
the average of first five multiples of 2 is ? | "average = 2 ( 1 + 2 + 3 + 4 + 5 ) / 5 = 30 / 5 = 6 . answer : a" | a ) 6 , b ) 8 , c ) 9 , d ) 5 , e ) 7 | a | add(2, const_1) | add(n0,const_1)| | general |
in an intercollegiate competition that lasted for 3 days , 157 students took part on day 1 , 111 on day 2 and 98 on day 3 . if 89 took part on day 1 and day 2 and 56 took part on day 2 and day 3 and 32 took part on all three days , how many students took part only on day 1 ? | "day 1 & 2 = 89 ; only day 1 & 2 ( 89 - 32 ) = 57 , day 2 & 3 = 56 ; only day 2 & 3 ( 56 - 32 ) = 24 , only day 1 = 157 - ( 57 + 24 + 32 ) = 44 answer : b" | a ) 40 , b ) 44 , c ) 35 , d ) 49 , e ) 38 | b | subtract(157, add(add(32, 89), divide(add(32, subtract(111, add(add(89, 56), 32))), 2))) | add(n7,n13)|add(n7,n10)|add(n13,#1)|subtract(n3,#2)|add(n13,#3)|divide(#4,n4)|add(#0,#5)|subtract(n1,#6)| | physics |
a student committee on academic integrity has 72 ways to select a president and vice president from a group of candidates . the same person can not be both president and vice president . how many candidates are there ? | "xc 1 * ( x - 1 ) c 1 = 72 x ^ 2 - x - 72 = 0 ( x - 9 ) ( x + 8 ) = 0 x = 9 , - 8 - 8 ca n ' t possible . c" | a ) 7 , b ) 8 , c ) 9 , d ) 10 , e ) 11 | c | divide(add(const_1, sqrt(add(multiply(const_4, 72), power(negate(const_1), const_2)))), const_2) | multiply(n0,const_4)|negate(const_1)|power(#1,const_2)|add(#0,#2)|sqrt(#3)|add(#4,const_1)|divide(#5,const_2)| | other |
a metallic sheet is of rectangular shape with dimensions 48 m x 36 m . from each of its corners , a square is cut off so as to make an open box . if the length of the square is 3 m , the volume of the box ( in m 3 ) is : | "clearly , l = ( 48 - 6 ) m = 42 m , b = ( 36 - 6 ) m = 30 m , h = 8 m . volume of the box = ( 42 x 30 x 3 ) m 3 = 3780 m 3 . answer : option a" | a ) 3780 , b ) 5120 , c ) 6420 , d ) 8960 , e ) 7960 | a | volume_rectangular_prism(subtract(48, multiply(3, const_2)), subtract(36, multiply(3, const_2)), 3) | multiply(n2,const_2)|subtract(n0,#0)|subtract(n1,#0)|volume_rectangular_prism(n2,#1,#2)| | geometry |
find the numbers which are in the ratio 3 : 2 : 4 such that the sum of the first and the second added to the difference of the third and the second is 21 ? | let the numbers be a , b and c . a : b : c = 3 : 2 : 4 given , ( a + b ) + ( c - b ) = 21 = > a + c = 21 = > 3 x + 4 x = 21 = > x = 3 a , b , c are 3 x , 2 x , 4 x a , b , c are 9 , 6 , 12 . answer : d | a ) 4 , 3,22 , b ) 4 , 4,22 , c ) 9 , 3,32 , d ) 9 , 6,12 , e ) 9 , 2,23 | d | divide(multiply(4, const_3), const_3) | multiply(n2,const_3)|divide(#0,const_3) | general |
in a certain village , 20 litres of water are required per household per month . at this rate , if there are 10 households in the village , how long ( in months ) will 2000 litres of water last ? | i find it much easier to understand with real numbers , so choose ( almost ) any numbers to replace m , n and p : in a certain village , m 20 litres of water are required per household per month . at this rate , if there aren 10 households in the village , how long ( in months ) willp 2000 litres of water last ? water ... | a ) 5 , b ) 2 , c ) 10 , d ) 1 , e ) 12 | c | divide(2000, multiply(20, 10)) | multiply(n0,n1)|divide(n2,#0) | gain |
if x + | x | + y = 5 and x + | y | - y = 6 what is x + y = ? | if x < 0 and y < 0 , then we ' ll have x - x + y = 7 and x - y - y = 6 . from the first equation y = 7 , so we can discard this case since y is not less than 0 . if x > = 0 and y < 0 , then we ' ll have x + x + y = 7 and x - y - y = 6 . solving gives x = 4 > 0 and y = - 1 < 0 - - > x + y = 3 . since in ps questions onl... | a ) 1 , b ) - 1 , c ) 3 , d ) 12 , e ) 13 | d | multiply(6, const_2) | multiply(n1,const_2) | general |
a certain clock rings two notes at quarter past the hour , 4 notes at half past , and 6 notes at 3 - quarters past . on the hour , it rings 8 notes plus an additional number of notes equal to whatever hour it is . how many notes will the clock ring from 1 : 00 p . m . through 5 : 00 p . m . , including the rings at 1 :... | form 1 pm to 5 pm . excluding the actual hour chime we have 20 ( 1 pm ) + 20 ( 2 pm ) + 20 ( 3 pm ) + 20 ( 4 pm ) + 8 ( 5 pm ) = 88 now the hour chimes are 1 + 2 + 3 + 4 + 5 = 15 total = 88 + 15 = 103 answer d . | a ) 87 , b ) 95 , c ) 102 , d ) 103 , e ) 115 | d | add(add(add(add(add(add(8, 1), add(add(8, 1), 1)), add(add(add(8, 1), 1), 1)), add(add(add(add(8, 1), 1), 1), 1)), add(add(add(add(add(8, 1), 1), 1), 1), 1)), add(add(add(add(add(add(add(8, 1), 1), 1), 1), add(add(add(add(8, 1), 1), 1), 1)), add(add(add(add(8, 1), 1), 1), 1)), add(add(add(add(8, 1), 1), 1), 1))) | add(n3,n4)|add(n4,#0)|add(#0,#1)|add(n4,#1)|add(#2,#3)|add(n4,#3)|add(#4,#5)|add(n4,#5)|add(#5,#5)|add(#6,#7)|add(#8,#5)|add(#10,#5)|add(#9,#11) | general |
on an order of 9 dozen boxes of a consumer product , a retailer receives an extra dozen free . this is equivalent to allowing him a discount of : | "clearly , the retailer gets 1 dozen out of 10 dozens free . equivalent discount = 1 / 10 * 100 = 10 % . answer a ) 10 %" | a ) 10 % , b ) 15 % , c ) 20 % , d ) 25 % , e ) 30 % | a | subtract(const_100, multiply(divide(9, const_4), const_100)) | divide(n0,const_4)|multiply(#0,const_100)|subtract(const_100,#1)| | general |
the total age of a and b is 13 years more than the total age of b and c . c is how many years younger than a ? | "solution [ ( a + b ) - ( b + c ) ] = 13 Γ’ β¬ ΒΉ = Γ’ β¬ ΒΊ a - c = 13 . answer a" | a ) 13 , b ) 24 , c ) c is elder than a , d ) data inadequate , e ) none | a | multiply(13, const_1) | multiply(n0,const_1)| | general |
rs . 1200 is divided so that 5 times the first share , 10 times the 2 nd share and fifteen times third share amount to the same . what is the value of the second share ? | "a + b + c = 1200 5 a = 10 b = 15 c = x a : b : c = 1 / 5 : 1 / 10 : 1 / 15 = 3 : 2 : 1 2 / 6 * 1200 = rs 400 answer : d" | a ) s 525 , b ) s 527 , c ) s 598 , d ) s 400 , e ) s 500 | d | multiply(5, divide(1200, add(add(5, 10), const_3))) | add(n1,n2)|add(#0,const_3)|divide(n0,#1)|multiply(n1,#2)| | general |
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 |
if the product of 6 integers is negative , at most how many of the integers can be negative ? | "product of even number of - ve numbers is positive so it cant be 6 . answer : d" | a ) 2 , b ) 3 , c ) 4 , d ) 5 , e ) 6 | d | subtract(6, const_1) | subtract(n0,const_1)| | general |
there is a 50 % chance jen will visit chile this year , while there is a 50 % chance that she will visit madagascar this year . what is the probability that jen will visit either chile or madagascar this year , but not both ? | "p ( chile and not madagascar ) = 0.5 * 0.5 = 0.25 p ( madagascar and not chile ) = 0.5 * 0.5 = 0.25 total probability = 0.25 + 0.25 = 0.5 = 50 % the answer is b ." | a ) 25.0 % , b ) 50.0 % , c ) 62.5 % , d ) 75.0 % , e ) 80.0 % | b | multiply(add(multiply(divide(subtract(power(multiply(const_2, const_5), const_2), 50), power(multiply(const_2, const_5), const_2)), divide(50, power(multiply(const_2, const_5), const_2))), multiply(divide(50, power(multiply(const_2, const_5), const_2)), divide(50, power(multiply(const_2, const_5), const_2)))), power(mu... | multiply(const_2,const_5)|power(#0,const_2)|divide(n1,#1)|divide(n0,#1)|subtract(#1,n0)|divide(#4,#1)|multiply(#3,#2)|multiply(#5,#2)|add(#7,#6)|multiply(#8,#1)| | probability |
a pipe takes a hours to fill the tank . but because of a leakage it took 7 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 / 420 x = 70 min answer : e | a ) 50 min , b ) 60 min , c ) 90 min , d ) 80 min , e ) 70 min | e | multiply(const_10, multiply(const_1, 7)) | multiply(n0,const_1)|multiply(#0,const_10) | physics |
at chennai it rained as much on tuesday as on all the others days of the week combined . if the average rainfall for the whole week was 3 cm . how much did it rain on tuesday ? | total rainfall = 3 x 7 = 21 cm hence , rainfall received on tuesday = 21 / 2 ( as it rained as much on tuesday as on all the others days of the week combined ) = 10.5 cm . answer : c | a ) 2.625 cm , b ) 3 cm , c ) 10.5 cm , d ) 15 cm , e ) none of these | c | divide(multiply(3, add(const_3, const_4)), const_2) | add(const_3,const_4)|multiply(n0,#0)|divide(#1,const_2) | general |
a shopkeeper buys mangoes at the rate of 4 a rupee and sells them at 2 a rupee . find his net profit or loss percent ? | "the total number of mangoes bought by the shopkeeper be 12 . if he buys 4 a rupee , his cp = 3 he selling at 2 a rupee , his sp = 4 profit = sp - cp = 4 - 2 = 2 profit percent = 2 / 2 * 100 = 100 % answer : a" | a ) 100 % , b ) 200 % , c ) 250 % , d ) 300 % , e ) 50 % | a | divide(multiply(2, const_100), 4) | multiply(n1,const_100)|divide(#0,n0)| | gain |
( 3 * 10 ^ 2 ) * ( 4 * 10 ^ - 2 ) = ? | 3 * 10 ^ 2 = 300 4 * 10 ^ - 2 = 0.04 ( 3 * 10 ^ 2 ) * ( 4 * 10 ^ - 2 ) = 300 * 0.04 = 12.00 the answer is option b | a ) 14 , b ) 12 , c ) 1200 , d ) 1.2 , e ) 14.11 | b | multiply(multiply(3, power(10, 2)), multiply(4, power(10, negate(2)))) | negate(n2)|power(n1,n2)|multiply(n0,#1)|power(n1,#0)|multiply(n3,#3)|multiply(#2,#4) | general |
the ratio of three numbers is 1 : 2 : 3 and their sum is 60 . the second number of the three numbers is ? | "1 : 2 : 3 total parts = 6 6 parts - - > 60 1 part - - - - > 10 the second number of the three numbers is = 2 * 10 = 20 answer : c" | a ) 24 , b ) 26 , c ) 20 , d ) 29 , e ) 30 | c | sqrt(divide(60, add(power(3, 2), add(power(1, 2), power(2, 2))))) | power(n0,n1)|power(n1,n1)|power(n2,n1)|add(#0,#1)|add(#3,#2)|divide(n3,#4)|sqrt(#5)| | other |
natasha climbs up a hill , and descends along the same way she went up . it takes her 3 hours to reach the top and 2 hours to come back down . if her average speed along the whole journey is 3 kilometers per hour , what was her average speed ( in kilometers per hour ) while climbing to the top ? | "lets assume distance to top as x , so the total distance travelled by natasha = 2 x total time taken = 3 + 2 = 5 hrs avg speed = total dist / total time taken = 2 x / 5 avg speed of complete journey is given as = 3 hrs 2 x / 5 = 3 x = 7.5 miles avg speed while climbing = distance / time = 7.5 / 3 = 2.5 option b" | a ) 1.5 , b ) 2.5 , c ) 3.75 , d ) 5 , e ) 7.5 | b | divide(divide(multiply(add(3, 2), 3), 2), 3) | add(n0,n1)|multiply(n2,#0)|divide(#1,n1)|divide(#2,n0)| | physics |
the annual birth and death rate in a country per 1000 are 39.4 and 19.4 respectively . the number of years k in which the population would be doubled assuming there is no emigration or immigration is | "suppose the population of the country in current year is 1000 . so annual increase is 1000 + 39.4 - 19.4 = 1020 hence every year there is an increase of 2 % . 2000 = 1000 ( 1 + ( 2 / 100 ) ) ^ n n = 35 answer is d ." | a ) 20 , b ) k = 25 , c ) 30 , d ) k = 35 , e ) 40 | d | divide(subtract(const_100, multiply(const_10, const_3)), multiply(divide(subtract(39.4, 19.4), 1000), const_100)) | multiply(const_10,const_3)|subtract(n1,n2)|divide(#1,n0)|subtract(const_100,#0)|multiply(#2,const_100)|divide(#3,#4)| | general |
susan made a block with small cubes of 8 cubic cm volume to make a block , 3 small cubes long , 9 small cubes wide and 5 small cubes deep . she realizes that she has used more small cubes than she really needed . she realized that she could have glued a fewer number of cubes together to lock like a block with same dime... | the total volume ( in terms of number of cubes ) of the solid = 3 * 9 * 5 = 135 the total volume ( in terms of number of cubes ) of the hollow = ( 3 - 2 ) * ( 9 - 2 ) * ( 5 - 2 ) = 21 so number of cubes required = 135 - 21 = 114 answer : b | ['a ) 113', 'b ) 114', 'c ) 115', 'd ) 116', 'e ) 117'] | b | add(multiply(multiply(const_1, 9), 5), add(subtract(multiply(multiply(const_1, 9), 5), multiply(add(const_4, const_3), const_3)), multiply(multiply(const_1, 9), 5))) | add(const_3,const_4)|multiply(n2,const_1)|multiply(n3,#1)|multiply(#0,const_3)|subtract(#2,#3)|add(#2,#4)|add(#5,#2) | geometry |
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