Problem stringlengths 5 967 | Rationale stringlengths 1 2.74k | options stringlengths 37 164 | correct stringclasses 5
values | annotated_formula stringlengths 7 1.65k | linear_formula stringlengths 8 925 | category stringclasses 6
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selling an kite for rs . 30 , a shop keeper gains 30 % . during a clearance sale , the shopkeeper allows a discount of 10 % on the marked price . his gain percent during the sale is ? | "explanation : marked price = rs . 30 c . p . = 100 / 130 * 30 = rs . 23.07 sale price = 90 % of rs . 30 = rs . 27 required gain % = 3.92 / 23.07 * 100 = 17 % . answer : d" | a ) 8 % , b ) 10 % , c ) 11 % , d ) 17 % , e ) 20 % | d | multiply(divide(subtract(multiply(divide(30, const_100), subtract(const_100, 10)), divide(multiply(30, const_100), add(30, const_100))), divide(multiply(30, const_100), add(30, const_100))), const_100) | add(n1,const_100)|divide(n0,const_100)|multiply(n0,const_100)|subtract(const_100,n2)|divide(#2,#0)|multiply(#1,#3)|subtract(#5,#4)|divide(#6,#4)|multiply(#7,const_100)| | gain | D |
1000 men have provisions for 20 days . if 650 more men join them , for how many days will the provisions last now ? | "1000 * 20 = 1650 * x x = 12.1 answer : e" | a ) 12.9 , b ) 12.5 , c ) 12.6 , d ) 12.2 , e ) 12.1 | e | divide(multiply(20, 1000), add(1000, 650)) | add(n0,n2)|multiply(n0,n1)|divide(#1,#0)| | physics | E |
if n = 2 ^ 0.25 and n ^ b = 16 , b must equal | "25 / 100 = 1 / 4 n = 2 ^ 1 / 4 n ^ b = 2 ^ 4 ( 2 ^ 1 / 4 ) ^ b = 2 ^ 4 b = 16 answer : d" | a ) 3 / 80 , b ) 3 / 5 , c ) 4 , d ) 16 , e ) 80 / 3 | d | divide(log(16), log(power(2, 0.25))) | log(n2)|power(n0,n1)|log(#1)|divide(#0,#2)| | general | D |
if the sum and difference of two numbers are 10 and 19 respectively , then the difference of their square is : | let the numbers be x and y . then , x + y = 10 and x - y = 19 x 2 - y 2 = ( x + y ) ( x - y ) = 10 * 19 = 190 . answer : e | a ) 12 , b ) 28 , c ) 160 , d ) 180 , e ) 190 | e | subtract(power(divide(add(10, 19), const_2), const_2), power(subtract(10, divide(add(10, 19), const_2)), const_2)) | add(n0,n1)|divide(#0,const_2)|power(#1,const_2)|subtract(n0,#1)|power(#3,const_2)|subtract(#2,#4) | general | E |
two trains running in opposite directions cross a man standing on the platform in 27 seconds and 17 seconds respectively . if they cross each other in 24 seconds , what is the ratio of their speeds ? | "let the speed of the trains be x and y respectively length of train 1 = 27 x length of train 2 = 17 y relative speed = x + y time taken to cross each other = 24 s = ( 27 x + 17 y ) / ( x + y ) = 24 = ( 27 x + 17 y ) / = 24 ( x + y ) = 3 x = 7 y = x / y = 7 / 3 answer : c" | a ) 1 : 3 , b ) 3 : 1 , c ) 7 : 3 , d ) 3 : 2 , e ) 3 : 4 | c | divide(subtract(27, 24), subtract(24, 17)) | subtract(n0,n2)|subtract(n2,n1)|divide(#0,#1)| | physics | C |
the sum of digits of a two digit number is 10 , the difference between the digits is 6 . find the number | description : = > x + y = 10 , x - y = 6 adding these 2 x = 16 = > x = 8 , y = 2 . thus the number is 82 answer c | a ) 85 , b ) 94 , c ) 82 , d ) 72 , e ) none | c | add(multiply(divide(add(10, 6), const_2), 10), subtract(10, divide(add(10, 6), const_2))) | add(n0,n1)|divide(#0,const_2)|multiply(n0,#1)|subtract(n0,#1)|add(#2,#3) | general | C |
in a group of 250 readers who read science fiction or literacy works or both , 180 read science fiction and 88 read literacy works . how many read both science fiction and literacy works ? | "consider total number of reader n ( s u l ) = 250 people who read science fiction n ( s ) = 180 people who read literacy works n ( l ) = 88 both science fiction and literacy n ( s â ˆ © l ) = ? n ( s u l ) = n ( s ) + n ( l ) - n ( s â ˆ © l ) 250 = 180 + 88 - n ( s â ˆ © l ) n ( s â ˆ © l ) = 268 - 250 n ( s â ˆ © l ... | a ) 268 , b ) 180 , c ) 150 , d ) 18 , e ) 88 | d | subtract(add(180, 88), 250) | add(n1,n2)|subtract(#0,n0)| | other | D |
a baker makes chocolate cookies and peanut cookies . his recipes allow him to make chocolate cookie in batches of 7 and peanut cookies in batches of 6 . if he makes exactly 97 cookies , what is the minimum number of chocolate chip cookies he makes ? | "7 c + 6 p = 97 we need to maximize p to minimize c so that the eq is also satisfied try substitution for cp to solve so that eqn is satisfied the least value of c for which equation gets satisfied is 5 i . e . 7 * 1 + 6 * 15 = 7 + 90 = 97 hence a is the answer" | a ) 7 , b ) 14 , c ) 21 , d ) 35 , e ) 49 | a | multiply(divide(subtract(97, reminder(97, add(7, 6))), add(7, 6)), 7) | add(n0,n1)|reminder(n2,#0)|subtract(n2,#1)|divide(#2,#0)|multiply(n0,#3)| | general | A |
on the independence day , bananas were be equally distributed among the children in a school so that each child would get two bananas . on the particular day 350 children were absent and as a result each child got two extra bananas . find the actual number of children in the school ? | "explanation : let the number of children in the school be x . since each child gets 2 bananas , total number of bananas = 2 x . 2 x / ( x - 350 ) = 2 + 2 ( extra ) = > 2 x - 700 = x = > x = 700 . answer : e" | a ) 237 , b ) 287 , c ) 197 , d ) 287 , e ) 700 | e | multiply(350, const_2) | multiply(n0,const_2)| | general | E |
evaluate 49 / . 07 | "explanation : 49 / . 07 = 4900 / 7 = 700 option c" | a ) 400 , b ) 500 , c ) 700 , d ) none of these , e ) 506 | c | divide(const_100.0, divide(07, 49)) | divide(n1,const_100)|divide(n0,#0)| | general | C |
in a competitive examination in a state a , 6 % candidates got selected from the total appeared candidates . state b had an equal number of candidates appeared and 7 % candidates got selected with 80 more candidates got selected than a . what was the number of candidates appeared from each state ? | let the number of candidates appeared be x then , 7 % of x - 6 % of x = 80 1 % of x = 80 x = 80 * 100 = 8000 answer is a | a ) 8000 , b ) 7540 , c ) 6500 , d ) 9100 , e ) 6000 | a | multiply(multiply(subtract(7, 6), const_100), 80) | subtract(n1,n0)|multiply(#0,const_100)|multiply(n2,#1) | general | A |
if a , b , c , d , e and f are integers and ( ab + cdef ) < 0 , then what is the maximum number s of integers that can be negative ? | "minimuum should be 1 maximum should be 4 : 1 out of a or b to make the multiplication negative 3 out of c , d , e or f to make the multiplication negative . negative + negative < 0 answer : c maximum will be 5 . . you dont require both the multiplicatin to be negative for entire equation to be negative . . . any one a... | a ) 2 , b ) 3 , c ) 4 , d ) 5 , e ) 6 | d | add(divide(const_10, const_2), 0) | divide(const_10,const_2)|add(n0,#0)| | general | D |
a , b and c rents a pasture for rs . 435 . a put in 12 horses for 8 months , b 16 horses for 9 months and 18 horses for 6 months . how much should b pay ? | 12 * 8 : 16 * 9 = 18 * 6 8 : 12 : 9 12 / 29 * 435 = 180 answer : a | a ) 180 , b ) 227 , c ) 268 , d ) 198 , e ) 176 | a | multiply(divide(435, add(add(multiply(12, 8), multiply(16, 9)), multiply(18, 6))), multiply(16, 9)) | multiply(n1,n2)|multiply(n3,n4)|multiply(n5,n6)|add(#0,#1)|add(#3,#2)|divide(n0,#4)|multiply(#5,#1) | general | A |
a circle graph shows how the megatech corporation allocates its research and development budget : 14 % microphotonics ; 24 % home electronics ; 20 % food additives ; 29 % genetically modified microorganisms ; 8 % industrial lubricants ; and the remainder for basic astrophysics . if the arc of each sector of the graph i... | "14 % microphotonics ; 24 % home electronics ; 20 % food additives ; 29 % genetically modified microorganisms ; 8 % industrial lubricants ; 100 - ( 14 + 24 + 20 + 29 + 8 ) = 5 % basic astrophysics . 5 % of 360 ° is 18 ° . answer : c ." | a ) 8 ° , b ) 10 ° , c ) 18 ° , d ) 36 ° , e ) 52 ° | c | divide(multiply(subtract(const_100, add(add(add(add(14, 24), 20), 29), 8)), divide(const_3600, const_10)), const_100) | add(n0,n1)|divide(const_3600,const_10)|add(n2,#0)|add(n3,#2)|add(n4,#3)|subtract(const_100,#4)|multiply(#1,#5)|divide(#6,const_100)| | gain | C |
suzie ’ s discount footwear sells all pairs of shoes for one price and all pairs of boots for another price . on monday the store sold 22 pairs of shoes and 16 pairs of boots for $ 580 . on tuesday the store sold 8 pairs of shoes and 32 pairs of boots for $ 800 . how much more do pairs of boots cost than pairs of shoes... | "let x be pair of shoes and y be pair of boots . 22 x + 16 y = 580 . . . eq 1 8 x + 32 y = 800 . . . . eq 2 . now multiply eq 1 by 2 and sub eq 2 . 44 x = 1160 8 x = 800 . 36 x = 360 = > x = 10 . sub x in eq 2 . . . . we get 80 + 32 y = 800 . . . then we get 32 y = 720 then y = 22.50 differenece between x and y is 12.5... | a ) $ 10.50 , b ) $ 12.50 , c ) $ 11.50 , d ) $ 16.50 , e ) $ 9.50 | b | divide(subtract(580, multiply(22, divide(subtract(multiply(580, const_2), 800), subtract(multiply(22, const_2), 8)))), 16) | multiply(n2,const_2)|multiply(n0,const_2)|subtract(#0,n5)|subtract(#1,n3)|divide(#2,#3)|multiply(n0,#4)|subtract(n2,#5)|divide(#6,n1)| | general | B |
a certain junior class has 1000 students and a certain senior class has 800 students . among these students , there are 60 siblings pairs each consisting of 1 junior and 1 senior . if 1 student is to be selected at random from each class , what is the probability that the 2 students selected will be a sibling pair ? | "there are 60 siblings in junior class and 60 their pair siblings in the senior class . we want to determine probability of choosing one sibling from junior class and its pair from senior . what is the probability of choosing any sibling from junior class ? 60 / 1000 ( as there are 60 of them ) . what is the probabilit... | a ) 3 / 40000 , b ) 1 / 3600 , c ) 9 / 2000 , d ) 1 / 60 , e ) 1 / 15 | a | divide(1, const_3) | divide(n3,const_3)| | probability | A |
how many positive integers less than 250 are there such that they are multiples of 15 or multiples of 14 ? | "250 / 15 = 16 ( plus remainder ) so there are 16 multiples of 15 250 / 14 = 17 ( plus remainder ) so there are 17 multiples of 14 we need to subtract 1 because 15 * 14 is a multiple of both so it was counted twice . the total is 16 + 17 - 1 = 32 the answer is c ." | a ) 30 , b ) 31 , c ) 32 , d ) 33 , e ) 34 | c | divide(factorial(subtract(add(const_4, 15), const_1)), multiply(factorial(15), 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 | C |
a train 100 meters long completely crosses a 300 meters long bridge in 24 seconds . what is the speed of the train is ? | "s = ( 100 + 300 ) / 24 = 400 / 24 * 18 / 5 = 60 answer : a" | a ) 60 kmph , b ) 76 kmph , c ) 34 kmph , d ) 43 kmph , e ) 40 kmph | a | divide(divide(add(100, 300), const_1000), divide(24, const_3600)) | add(n0,n1)|divide(n2,const_3600)|divide(#0,const_1000)|divide(#2,#1)| | physics | A |
a train running at a speed of 36 kmph crosses an electric pole in 12 seconds . in how much time will it cross a 340 m long platform ? | "let the length of the train be x m . when a train crosses an electric pole , the distance covered is its own length . so , x = 12 * 36 * 5 / 18 m = 120 m . time taken to cross the platform = ( 120 + 340 ) / 36 * 5 / 18 = 46 min . answer : d" | a ) 19 , b ) 27 , c ) 29 , d ) 46 , e ) 28 | d | divide(add(340, multiply(multiply(const_0_2778, 36), 12)), multiply(const_0_2778, 36)) | multiply(n0,const_0_2778)|multiply(n1,#0)|add(n2,#1)|divide(#2,#0)| | physics | D |
one fourth of one third of two fifth of a number is 35 . what will be 40 % of that number | "explanation : ( 1 / 4 ) * ( 1 / 3 ) * ( 2 / 5 ) * x = 35 then x = 35 * 30 = 1050 40 % of 1050 = 420 answer : option b" | a ) a ) 140 , b ) b ) 420 , c ) c ) 180 , d ) d ) 200 , e ) e ) 220 | b | divide(multiply(divide(35, multiply(multiply(divide(const_1, const_4), divide(const_1, const_3)), divide(const_2, add(const_2, const_3)))), 40), const_100) | add(const_2,const_3)|divide(const_1,const_4)|divide(const_1,const_3)|divide(const_2,#0)|multiply(#1,#2)|multiply(#3,#4)|divide(n0,#5)|multiply(n1,#6)|divide(#7,const_100)| | gain | B |
two trains are moving in opposite directions at 60 km / hr and 90 km / hr . their lengths are 1.10 km and 1.4 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 + 1.4 = 2.5 km = 2500 m . required time = 2500 * 3 / 125 = 60 sec . answer : a" | a ) 60 , b ) 277 , c ) 48 , d ) 99 , e ) 22 | a | subtract(divide(multiply(1.10, const_1000), divide(multiply(60, const_1000), const_3600)), divide(multiply(1.4, 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 |
it takes joey the postman 1 hours to run a 7 mile long route every day . he delivers packages and then returns to the post office along the same path . if the average speed of the round trip is 8 mile / hour , what is the speed with which joey returns ? | "let his speed for one half of the journey be 7 miles an hour let the other half be x miles an hour now , avg speed = 8 mile an hour 2 * 7 * x / 7 + x = 8 14 x = 8 x + 56 = > x = 8 d" | a ) 11 , b ) 12 , c ) 13 , d ) 8 , e ) 15 | d | divide(7, subtract(divide(multiply(const_2, 7), 8), 1)) | multiply(n1,const_2)|divide(#0,n2)|subtract(#1,n0)|divide(n1,#2)| | physics | D |
car a runs at the speed of 65 km / hr & reaches its destination in 5 hr . car b runs at the speed of 70 km / h & reaches its destination in 4 h . what is the respective ratio of distances covered by car a & car b ? | sol . distance travelled by car a = 65 ã — 5 = 325 km distance travelled by car b = 70 ã — 4 = 280 km ratio = 325 / 280 = 65 : 56 e | a ) 11 : 6 , b ) 12 : 7 , c ) 13 : 7 , d ) 15 : 6 , e ) 65 : 56 | e | divide(multiply(65, 5), multiply(70, 4)) | multiply(n0,n1)|multiply(n2,n3)|divide(#0,#1) | physics | E |
steve traveled the first 2 hours of his journey at 40 mph and the last 3 hours of his journey at 60 mph . what is his average speed of travel for the entire journey ? | "answer average speed of travel = total distance travelled / total time taken total distance traveled by steve = distance covered in the first 2 hours + distance covered in the next 3 hours . distance covered in the first 2 hours = speed * time = 40 * 2 = 80 miles . distance covered in the next 3 hours = speed * time =... | a ) 52 mph , b ) 56.67 mph , c ) 53.33 mph , d ) 64 mph , e ) 66.67 mph | a | add(divide(add(multiply(60, 3), multiply(40, 2)), add(3, 2)), subtract(divide(const_100, 3), const_0_33)) | add(n0,n2)|divide(const_100,n2)|multiply(n2,n3)|multiply(n0,n1)|add(#2,#3)|subtract(#1,const_0_33)|divide(#4,#0)|add(#6,#5)| | physics | A |
how many multiples of 4 are there between 20 and 112 , inclusive ? | "the multiples of 4 are from 4 * 5 up to 4 * 28 . 28 - 5 + 1 = 24 . the answer is d ." | a ) 21 , b ) 22 , c ) 23 , d ) 24 , e ) 25 | d | add(divide(subtract(112, 20), 4), const_1) | subtract(n2,n1)|divide(#0,n0)|add(#1,const_1)| | general | D |
a and b can do a work in 4 days , b and c in 5 days , c and a in 7 days . if a , b and c work together , they will complete the work in ? | "a + b 1 day work = 1 / 4 b + c 1 day work = 1 / 5 c + a 1 day work = 1 / 7 adding we get 2 ( a + b + c ) = 1 / 4 + 1 / 5 + 1 / 7 = 83 / 140 a + b + c 1 day work = 83 / 280 a , b , c can finish the work in 280 / 83 days = 3 days approximately answer is a" | a ) 3 days , b ) 5 days , c ) 7 days , d ) 1 days , e ) 2 dasy | a | divide(const_1, divide(add(add(inverse(4), inverse(5)), inverse(7)), const_2)) | inverse(n0)|inverse(n1)|inverse(n2)|add(#0,#1)|add(#3,#2)|divide(#4,const_2)|divide(const_1,#5)| | physics | A |
let f ( x , y ) be defined as the remainder when ( x – y ) ! is divided by x . if x = 36 , what is the maximum value of y for which f ( x , y ) = 0 ? | "the question is finding y such that ( 36 - y ) ! is a multiple of 36 . that means we need to have 2 ^ 2 * 3 ^ 2 in ( 36 - y ) ! 6 ! is the smallest factorial number with 2 ^ 2 * 3 ^ 2 as a factor . 36 - y = 6 y = 30 the answer is e ." | a ) 9 , b ) 12 , c ) 18 , d ) 20 , e ) 30 | e | subtract(36, multiply(const_2, divide(divide(36, const_2), add(const_1, const_4)))) | add(const_1,const_4)|divide(n0,const_2)|divide(#1,#0)|multiply(#2,const_2)|subtract(n0,#3)| | general | E |
a fill pipe can fill 1 / 2 of cistern in 30 minutes . in how many minutes , it can fill 1 / 2 of the cistern ? | "required time = 30 * 2 * 1 / 2 = 30 minutes answer is d" | a ) 15 min , b ) 20 min , c ) 25 min , d ) 30 min , e ) 35 min | d | divide(30, 1) | divide(n2,n0)| | physics | D |
two stations a and b are 110 km apart on a straight line . one train starts from a at 5 a . m . and travels towards b at 20 kmph . another train starts from b at 8 a . m . and travels towards a at a speed of 25 kmph . at what time will they meet ? | "suppose they meet x hours after 5 a . m . distance covered by a in x hours = 20 x km . distance covered by b in ( x - 1 ) hours = 25 ( x - 1 ) km . therefore 20 x + 25 ( x - 1 ) = 110 45 x = 135 x = 3 . so , they meet at 8 a . m . answer : c" | a ) 11 , b ) 10 , c ) 8 , d ) 12 , e ) 15 | c | add(8, divide(subtract(110, 20), add(25, 20))) | add(n2,n4)|subtract(n0,n2)|divide(#1,#0)|add(n3,#2)| | physics | C |
a store has 10 bottles of juice , including 5 bottles of apple juice . in the evening , 6 bottles of juice are sold one by one . what is the probability of selling 4 bottles of apple juice among the 6 bottles ? assume that every bottle has an equal chance of being bought . | "the total number of ways to sell 6 bottles from 10 is 10 c 6 = 210 . the number of ways to sell 4 bottles of apple juice is 5 c 4 * 5 c 2 = 5 * 10 = 50 p ( selling 4 bottles of apple juice ) = 50 / 210 = 5 / 21 the answer is d ." | a ) 2 / 7 , b ) 3 / 11 , c ) 4 / 15 , d ) 5 / 21 , e ) 6 / 25 | d | divide(choose(5, 4), choose(10, 5)) | choose(n1,n3)|choose(n0,n1)|divide(#0,#1)| | probability | D |
doughnuts , cookies and muffins in a pastry shop are in the ratio of 5 : 3 : 1 . if there are 50 doughnuts , the number of muffins in the shop is : . | explanation : let doughnuts = 5 x , cookies = 3 x & muffins = 1 x . now , 5 x = 50 hence x = 10 . number of muffins = 1 x which is 10 . answer : c | a ) 25 , b ) 30 , c ) 10 , d ) 15 , e ) 20 | c | divide(50, 5) | divide(n3,n0) | other | C |
john was 19 years old when he married betty . they just celebrated their fifth wedding anniversary , and betty ' s age is now 7 / 8 of john ' s . how old is betty ? | "assume betty ' s age on marriage = x years . john ' s age on marriage = 19 john ' s age after 5 years = 24 years . betty ' s age after 5 years = x + 5 given : x + 5 = 7 / 8 ( 24 ) = 21 therefore betty ' s current age = 21 option a" | a ) 21 , b ) 26 , c ) 28 , d ) 30 , e ) 32 | a | multiply(divide(7, 8), add(19, const_4)) | add(n0,const_4)|divide(n1,n2)|multiply(#0,#1)| | general | A |
what is the sum of the integers from - 160 to 162 , inclusive ? | "in an arithmetic progression , the nth term is given by tn = a + ( n - 1 ) d here tn = 162 , a = - 160 , d = 1 hence , 162 = - 160 + ( n - 1 ) or n = 323 sum of n terms can be calculated by sn = n / 2 ( a + l ) a = first term , l = last term , n = no . of terms sn = 323 * ( - 160 + 162 ) / 2 sn = 323 * 2 / 2 = 323 ans... | a ) 325 , b ) 327 , c ) 323 , d ) 330 , e ) 350 | c | divide(multiply(160, 162), const_4) | multiply(n0,n1)|divide(#0,const_4)| | general | C |
five 6 faced dice are thrown together . the probability that all the three show the same number on them is ? | the three dice can fall in 6 * 6 * 6 * 6 * 6 = 7776 ways . hence the probability is 6 / 7776 = 1 / 1296 answer : b | a ) 1 / 32 , b ) 1 / 1296 , c ) 1 / 33 , d ) 1 / 38 , e ) 1 / 34 | b | multiply(multiply(multiply(divide(const_1, 6), divide(const_1, 6)), divide(const_1, 6)), divide(const_1, 6)) | divide(const_1,n0)|multiply(#0,#0)|multiply(#0,#1)|multiply(#0,#2) | probability | B |
a grocer has a sale of rs . 6435 , rs . 6927 , rs . 6855 , rs . 7230 and rs . 6562 for 5 consecutive months . how much sale must he have in the sixth month so that he gets an average sale of rs . 6700 ? | "total fr 5 mnths = ( 6435 + 6927 + 6855 + 7230 + 6562 ) = rs 34009 . reqd . sale = rs . [ ( 6700 * 6 ) - 34009 ] = rs . ( 40200 - 34009 ) = rs . 6191 . answer : d" | a ) s . 4991 , b ) s . 5991 , c ) s . 6001 , d ) s . 6191 , e ) s . 7991 | d | subtract(multiply(add(5, const_1), 6700), add(add(add(add(6435, 6927), 6855), 7230), 6562)) | add(n5,const_1)|add(n0,n1)|add(n2,#1)|multiply(n6,#0)|add(n3,#2)|add(n4,#4)|subtract(#3,#5)| | general | D |
at a certain food stand , the price of each apple is 40 ¢ and the price of each orange is 60 ¢ . mary selects a total of 10 apples and oranges from the food stand , and the average ( arithmetic mean ) price of the 10 pieces of fruit is 56 ¢ . how many oranges must mary put back so that the average price of the pieces o... | "if the average price is 56 , then mary has 8 oranges and 2 apples ( a ratio of 4 : 1 ) . for the average price to be 50 , the ratio should be 1 : 1 . mary must put back 6 oranges . the answer is e ." | a ) 2 , b ) 3 , c ) 4 , d ) 5 , e ) 6 | e | divide(subtract(add(multiply(divide(subtract(multiply(60, 10), multiply(56, 10)), subtract(60, 40)), 40), multiply(subtract(10, divide(subtract(multiply(60, 10), multiply(56, 10)), subtract(60, 40))), 60)), multiply(50, 10)), subtract(60, 50)) | multiply(n1,n2)|multiply(n2,n4)|multiply(n2,n5)|subtract(n1,n0)|subtract(n1,n5)|subtract(#0,#1)|divide(#5,#3)|multiply(n0,#6)|subtract(n2,#6)|multiply(n1,#8)|add(#7,#9)|subtract(#10,#2)|divide(#11,#4)| | general | E |
if an article is sold at 20 % profit instead of 14 % profit , then the profit would be rs . 180 more . what is the cost price ? | "explanation : let the cost price of an article be rs . x . ( 20 % of x ) - ( 14 % of x ) = 180 20 x / 100 - 14 x / 100 = 105 = > 6 x = 180 * 100 = > x = 3000 cost price = rs . 3000 answer : c" | a ) rs . 1000 , b ) rs . 2000 , c ) rs . 3000 , d ) rs . 4000 , e ) rs . 5200 | c | multiply(divide(180, 14), const_100) | divide(n2,n1)|multiply(#0,const_100)| | gain | C |
the speed at which a man can row a boat in still water is 9.5 kmph . if he rows downstream , where the speed of current is 8.5 kmph , what time will he take to cover 45.5 metres ? | "speed of the boat downstream = 9.5 + 8.5 = 18 kmph = 18 * 5 / 18 = 5 m / s hence time taken to cover 45.5 m = 45.5 / 5 = 9.1 seconds . answer : d" | a ) 18 seconds , b ) 8.4 seconds , c ) 5 seconds , d ) 9.1 seconds , e ) 9.5 econds | d | divide(45.5, multiply(add(9.5, 8.5), const_0_2778)) | add(n0,n1)|multiply(#0,const_0_2778)|divide(n2,#1)| | physics | D |
a and b started a business investing rs . 10,000 and rs 20,000 respectively . in what ratio the profit earned after 2 years be divided between a and b respectively ? | "a : b = 10000 : 20000 = 1 : 2 answer : d" | a ) 3 : 2 , b ) 9 : 2 , c ) 18 : 20 , d ) 1 : 2 , e ) 18 : 4 | d | divide(add(multiply(multiply(const_3, const_3), add(multiply(const_3, const_3), const_1)), 2), multiply(2, add(multiply(const_3, const_3), const_1))) | multiply(const_3,const_3)|add(#0,const_1)|multiply(#1,#0)|multiply(n2,#1)|add(n2,#2)|divide(#4,#3)| | gain | D |
a man invests some money partly in 9 % stock at 96 and partly in 12 % stock at 120 . to obtain equal dividends from both , he must invest the money in the ratio : | "solution for an income of rs . 1 in 9 % stock at 96 , investment = rs . ( 96 / 9 ) = rs . 32 / 3 . for an income of rs . 1 in 12 % stock at 120 , investment = rs . ( 120 / 12 ) = rs . 10 . ∴ ratio of investments = 32 / 3 : 10 = 32 : 30 = 16 : 15 answer d" | a ) 3 : 4 , b ) 3 : 5 , c ) 4 : 5 , d ) 16 : 15 , e ) none | d | divide(multiply(96, const_2), multiply(120, const_3)) | multiply(n1,const_2)|multiply(n3,const_3)|divide(#0,#1)| | other | D |
a train passes a station platform in 60 sec and a man standing on the platform in 20 sec . if the speed of the train is 54 km / hr . what is the length of the platform ? | "speed = 54 * 5 / 18 = 15 m / sec . length of the train = 15 * 20 = 300 m . let the length of the platform be x m . then , ( x + 300 ) / 36 = 15 = > x = 420 m answer : b" | a ) 615 m , b ) 420 m , c ) 168 m , d ) 197 m , e ) 691 m | b | multiply(20, multiply(54, const_0_2778)) | multiply(n2,const_0_2778)|multiply(n1,#0)| | physics | B |
on selling 15 balls at rs . 720 , there is a loss equal to the cost price of 5 balls . the cost price of a ball is : | "( c . p . of 15 balls ) - ( s . p . of 15 balls ) = ( c . p . of 5 balls ) c . p . of 10 balls = s . p . of 15 balls = rs . 720 . c . p . of 1 ball = rs . 720 / 10 = rs . 72 . answer : option e" | a ) s . 45 , b ) s . 50 , c ) s . 55 , d ) s . 60 , e ) s . 72 | e | divide(720, subtract(15, 5)) | subtract(n0,n2)|divide(n1,#0)| | gain | E |
tom , working alone , can paint a room in 6 hours . peter and john , working independently , can paint the same room in 3 hours and 5 hours , respectively . tom starts painting the room and works on his own for one hour . he is then joined by peter and they work together for an hour . finally , john joins them and the ... | "tom paints 1 / 6 of the room in the first hour . tom and peter paint 1 / 6 + 1 / 3 = 1 / 2 of the room in the next hour for a total of 4 / 6 . the three people then paint the remaining 2 / 6 in a time of ( 2 / 6 ) / ( 21 / 30 ) = 10 / 21 hours peter worked for 31 / 21 hours so he painted 31 / 21 * 1 / 3 = 31 / 63 of t... | a ) 19 / 35 , b ) 23 / 42 , c ) 27 / 55 , d ) 31 / 63 , e ) 35 / 74 | d | divide(const_4, add(multiply(const_4, 5), const_1)) | multiply(n2,const_4)|add(#0,const_1)|divide(const_4,#1)| | physics | D |
last year sandy saved 6 % of her annual salary . this year , she made 10 % more money than last year , and she saved 8 % of her salary . the amount saved this year was what percent of the amount she saved last year ? | "let last year ' s salary be x . last year , sandy save 0.06 x this year , sandy saved 0.08 * 1.1 x = 0.088 x 0.088 x / 0.06 x = 88 / 60 = 1.47 = 147 % the answer is b ." | a ) 132 % , b ) 147 % , c ) 158 % , d ) 176 % , e ) 188 % | b | multiply(divide(multiply(divide(8, const_100), add(const_100, 10)), divide(multiply(6, const_100), const_100)), const_100) | add(n1,const_100)|divide(n2,const_100)|multiply(n0,const_100)|divide(#2,const_100)|multiply(#0,#1)|divide(#4,#3)|multiply(#5,const_100)| | gain | B |
a can finish a piece of work in 15 days . b can do it in 30 days . they work together for 6 days and then a goes away . in how many days will b finish the work ? | "6 / 15 + ( 6 + x ) / 30 = 1 = > x = 12 days answer : b" | a ) 11 days , b ) 12 days , c ) 14 days , d ) 10 days , e ) 15 days | b | divide(subtract(const_1, add(multiply(divide(const_1, const_4.0), const_2), multiply(divide(const_1, 30), const_2))), divide(const_1, 30)) | divide(const_1,const_4.0)|divide(const_1,n1)|multiply(#0,const_2)|multiply(#1,const_2)|add(#2,#3)|subtract(const_1,#4)|divide(#5,#1)| | physics | B |
find the 25 % of rs . 800 . | "explanation : 25 % of 800 = > 25 / 100 * 800 = rs . 200 answer : b" | a ) s . 50 , b ) s . 200 , c ) s . 100 , d ) s . 80 , e ) s . 60 | b | multiply(800, divide(25, const_100)) | divide(n0,const_100)|multiply(n1,#0)| | gain | B |
what would be the perimeter of the square whose side is equal to the diameter of the circle whose perimeter is 52.5 cm ? | diameter of the circle = perimeter / ï € = 52.5 x 7 / 22 = 16.70 perimeter of the square = 16.70 x 4 = 66.8 cm answer : c | ['a ) 52.5', 'b ) 60.2', 'c ) 66.8', 'd ) 74.25', 'e ) 80.12'] | c | square_perimeter(divide(52.5, const_pi)) | divide(n0,const_pi)|square_perimeter(#0) | geometry | C |
at what price must an article costing rs . 51.50 be marked in order that after deducting 5 % from the list price . it may be sold at a profit of 25 % on the cost price ? | "cp = 51.50 sp = 51.50 * ( 125 / 100 ) = 64.375 mp * ( 95 / 100 ) = 64.375 mp = 67.7 answer : d" | a ) 60.5 , b ) 69.5 , c ) 67.5 , d ) 67.7 , e ) 22.5 | d | divide(multiply(add(51.50, divide(multiply(51.50, 25), const_100)), const_100), subtract(const_100, 5)) | multiply(n0,n2)|subtract(const_100,n1)|divide(#0,const_100)|add(n0,#2)|multiply(#3,const_100)|divide(#4,#1)| | gain | D |
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 / 4 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 / 4 of that new total price : p = 1 / 4 * ... | a ) 1500 , b ) 1600 , c ) 1750 , d ) 1900 , e ) 2000 | c | subtract(multiply(multiply(const_0_25, const_1000), const_10), divide(add(500, multiply(multiply(const_0_25, const_1000), const_10)), 4)) | multiply(const_0_25,const_1000)|multiply(#0,const_10)|add(n1,#1)|divide(#2,n3)|subtract(#1,#3)| | general | C |
increasing the original price of a certain item by 25 percent and then increasing the new price by 25 percent is equivalent to increasing the original price by what percent ? | "soln : - x * 1.25 * 1.25 = 1.5625 x so there is a net increase of 56.25 % . answer e ." | a ) 31.25 , b ) 37.5 , c ) 50.0 , d ) 52.5 , e ) 56.25 | e | multiply(subtract(multiply(add(divide(25, const_100), const_1), add(divide(25, const_100), const_1)), const_1), const_100) | divide(n1,const_100)|divide(n0,const_100)|add(#0,const_1)|add(#1,const_1)|multiply(#2,#3)|subtract(#4,const_1)|multiply(#5,const_100)| | gain | E |
the average of 20 results is 65 and the average of other 65 results is 20 . what is the average of all the results ? | "answer sum of 85 result = sum of 20 result + sum of 65 result . = 20 x 65 + 65 x 20 = 2600 correct option : a" | a ) 30 , b ) 25 , c ) 48 , d ) 50 , e ) none | a | divide(add(multiply(20, 65), multiply(65, 20)), add(20, 65)) | add(n0,n1)|multiply(n0,n1)|add(#1,#1)|divide(#2,#0)| | general | A |
a man can row upstream at 5 kmph and downstream at 25 kmph , and then find the speed of the man in still water ? | "us = 5 ds = 25 m = ( 5 + 25 ) / 2 = 15 answer : a" | a ) 15 , b ) 77 , c ) 30 , d ) 88 , e ) 34 | a | divide(add(5, 25), const_2) | add(n0,n1)|divide(#0,const_2)| | physics | A |
find compound interest on rs . 8000 at 15 % per annum for 2 years 4 months , compounded annually . | "time = 2 years 4 months = 2 ( 4 / 12 ) years = 2 ( 1 / 3 ) years . amount = rs ' . [ 8000 x ( 1 + ¬ ( 15 / 100 ) ) 2 x ( 1 + ( ( 1 / 3 ) * 15 ) / 100 ) ] = rs . [ 8000 * ( 23 / 20 ) * ( 23 / 20 ) * ( 21 / 20 ) ] = rs . 11109 . . : . c . i . = rs . ( 11109 - 8000 ) = rs . 3109 . answer c rs 3109" | a ) rs 400 , b ) rs 300 , c ) rs 3109 , d ) rs 250 , e ) rs 400 | c | subtract(add(add(8000, divide(multiply(8000, 15), const_100)), divide(multiply(add(8000, divide(multiply(8000, 15), const_100)), 15), const_100)), 8000) | multiply(n0,n1)|divide(#0,const_100)|add(n0,#1)|multiply(n1,#2)|divide(#3,const_100)|add(#2,#4)|subtract(#5,n0)| | gain | C |
a , b , c rent a pasture . if a puts 10 oxen for 7 months , b puts 12 oxen for 5 months and c puts 15 oxen for 3 months for grazing and the rent of the pasture is rs . 175 , then how much amount should c pay as his share of rent ? | a : b : c = 10 * 7 : 12 * 5 : 15 * 3 = 2 * 7 : 12 * 1 : 3 * 3 = 14 : 12 : 9 amount that c should pay = 175 * ( 9 / 35 ) = 5 * 9 = 45 answer is b | a ) 23 , b ) 45 , c ) 15 , d ) 28 , e ) 18 | b | multiply(3, 15) | multiply(n4,n5) | general | B |
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 ? | "explanation : 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 * 26 = 676 wayshaving chosen the first two letters , the third letter can be chosen in 26 ways . = > al... | a ) 56678 , b ) 16789 , c ) 17576 , d ) 35142 , e ) none of these | c | 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 | C |
find the number , difference between number and its 3 / 5 is 60 . | "explanation : let the number = x , then , x - ( 3 / 5 ) x = 60 , = > ( 2 / 5 ) x = 60 = > 2 x = 60 * 5 , = > x = 150 answer : option a" | a ) 150 , b ) 153 , c ) 154 , d ) 155 , e ) 156 | a | multiply(divide(60, const_2), 5) | divide(n2,const_2)|multiply(n1,#0)| | general | A |
how many positive integers less than 1960 are such that the product of their digits is 10 . | "1960 is divisible by 10 ( or 2 * 5 ) = 196 196 is divisible by 4 = > 49 49 = 7 * 7 hence 1960 is divisible by 2 , 4,5 , 7 , 7 if there were only one 7 , i . e if the problem were 280 c" | a ) 220 , b ) 230 , c ) 280 , d ) 220 , e ) 240 | c | divide(factorial(subtract(add(const_4, 10), const_1)), multiply(factorial(10), 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 | C |
jaime earned enough money by selling seashells at 35 cents each to buy several used paperback books at 55 cents each . if he spent all of the money he earned selling seashells to buy the books , what is the least number of seashells he could have sold ? | let ' s test answer d : 22 seashells . . . . with 22 seashells , jamie would have 22 ( 35 ) = 770 cents . this would allow him to buy 14 books for 770 cents total , with no money left over . this is an exact match for what we were told , so this must be the answer . final answer : [ reveal ] spoiler : d | a ) 5 , b ) 11 , c ) 17 , d ) 22 , e ) 30 | d | add(divide(lcm(35, 55), 35), const_10) | lcm(n0,n1)|divide(#0,n0)|add(#1,const_10) | general | D |
a paint store mixes 3 / 4 pint of red paint and 2 / 3 pint of white paint to make a new paint color called perfect pink . how many pints of red paint would be needed to make 34 pints of perfect pink paint ? | "3 / 4 pint is required to make 3 / 4 + 2 / 3 = 17 / 12 pint of perfect pink so 17 / 12 pint requires 3 / 4 pint of red . . 1 pint will require 3 / 4 * 12 / 17 = 9 / 17 . . 34 pints will require 9 / 17 * 34 = 18 points . . second way . . get both red and white paints in same denominator . . . 3 / 4 + 2 / 3 = ( 9 / 12 )... | a ) 9 , b ) 16 , c ) 18 , d ) 25 1 / 3 , e ) 28 1 / 2 | c | multiply(34, divide(multiply(3, 2), multiply(4, 3))) | multiply(n0,n2)|multiply(n0,n1)|divide(#0,#1)|multiply(n4,#2)| | general | C |
a man saves 20 % of his monthly salary . if an account of dearness of things he is to increase his monthly expenses by 20 % , he is only able to save rs . 260 per month . what is his monthly salary ? | "income = rs . 100 expenditure = rs . 80 savings = rs . 20 present expenditure 80 * ( 20 / 100 ) = rs . 96 present savings = 100 â € “ 96 = rs . 4 100 - - - - - - 4 ? - - - - - - - - - 260 = > 6500 answer : e" | a ) 5000 , b ) 2999 , c ) 2878 , d ) 2990 , e ) 6500 | e | divide(multiply(260, const_100), subtract(const_100, add(subtract(const_100, 20), multiply(subtract(const_100, 20), divide(20, const_100))))) | divide(n1,const_100)|multiply(n2,const_100)|subtract(const_100,n0)|multiply(#0,#2)|add(#3,#2)|subtract(const_100,#4)|divide(#1,#5)| | general | E |
thomas ' s age and matt ' s age are in the ratio 3 : 5 . two years ago the ratio of their ages was 5 : 9 . find the ratio of their ages 4 years hence ? | let ages of thomas and matt be x and y respectively . x / y = 3 / 5 = > x = 3 / 5 y ( x - 2 ) / ( y - 2 ) = 5 / 9 = > 9 x - 18 = 5 y - 10 = > 9 x = 5 y + 8 but x = 3 / 5 y 9 * 3 / 5 y = 5 y + 8 27 y = 25 y + 40 = > y = 20 years = > x = 3 * 20 / 5 = > 12 years ratio of their ages four years hence , ( 12 + 4 ) / ( 20 + 4... | a ) 1 : 2 , b ) 2 : 3 , c ) 3 : 4 , d ) 4 : 5 , e ) 5 : 6 | b | divide(add(divide(multiply(divide(multiply(add(5, 3), 5), const_2), 3), 5), 4), add(divide(multiply(add(5, 3), 5), const_2), 4)) | add(n0,n1)|multiply(n1,#0)|divide(#1,const_2)|add(n4,#2)|multiply(n0,#2)|divide(#4,n1)|add(n4,#5)|divide(#6,#3) | other | B |
in the johnsons ' monthly budget , the dollar amounts allocated to household expenses , food , and miscellaneous items are in the ratio 5 : 4 : 1 , respectively . if the total amount allocated to these 3 categories is $ 1800 , what is the amount allocated to food ? | to solve this question , it will be best to first express the given ratio using variable multipliers . thus , we have : household expenses : food : miscellaneous = 5 x : 4 x : x since we are told that the total amount allocated to these categories is $ 1,800 we can set up the equation : 5 x + 4 x + x = 1,800 10 x = 1,8... | a ) $ 900 , b ) $ 720 , c ) $ 675 , d ) $ 450 , e ) $ 225 | b | multiply(divide(1800, add(add(5, 4), 1)), 4) | add(n0,n1)|add(n2,#0)|divide(n4,#1)|multiply(n1,#2) | general | B |
the measurement of a rectangular box with lid is 25 cmx 4 cmx 18 cm . find the volume of the largest sphere that can be inscribed in the box ( in terms of π cm 3 ) . ( hint : the lowest measure of rectangular box represents the diameter of the largest sphere ) | "d = 4 , r = 2 ; volume of the largest sphere = 4 / 3 π r 3 = 4 / 3 * π * 2 * 2 * 2 = 10.66 π cm 3 answer : d" | a ) 11.12 , b ) 10.11 , c ) 72 , d ) 10.66 , e ) 9.2 | d | multiply(divide(const_4, 3), power(3, 3)) | divide(const_4,n3)|power(n3,n3)|multiply(#0,#1)| | geometry | D |
in a kilometer race , a beats b by 60 meters or 15 seconds . what time does a take to complete the race ? | "time taken by b run 1000 meters = ( 1000 * 15 ) / 60 = 250 sec . time taken by a = 250 - 15 = 235 sec . answer : b" | a ) 221 , b ) 235 , c ) 245 , d ) 289 , e ) 260 | b | subtract(divide(multiply(const_1, const_1000), divide(60, 15)), 15) | divide(n0,n1)|multiply(const_1,const_1000)|divide(#1,#0)|subtract(#2,n1)| | physics | B |
dan ’ s car gets 32 miles per gallon . if gas costs $ 4 / gallon , then how many miles can dan ’ s car go on $ 58 of gas ? | "58 / 4 = 14.5 gallons 14.5 * 32 = 464 miles the answer is c ." | a ) 242 , b ) 353 , c ) 464 , d ) 575 , e ) 686 | c | divide(multiply(58, 32), 4) | multiply(n0,n2)|divide(#0,n1)| | physics | C |
two whole numbers whose sum is 42 can not be in the ratio | d ) 3 : 8 | a ) 2 : 5 , b ) 1 : 6 , c ) 2 : 4 , d ) 3 : 8 , e ) 2 : 40 | d | divide(divide(subtract(divide(const_100.0, const_2), const_10), const_2), add(divide(42, const_2), const_10)) | divide(const_100.0,const_2)|add(#0,const_10)|subtract(#0,const_10)|divide(#2,const_2)|divide(#3,#1)| | other | D |
nhai employs 100 men to build a highway of 2 km in 50 days working 8 hours a day . if in 25 days , they completed 1 / 3 part of work . than how many more employees should nhai hire to finish it in time working 10 hours a day ? | explanation : the given problem can be written in a tabular form like below : - men days hour work planned 100 50 8 1 actual 100 25 8 1 / 3 remaining ? 25 10 2 / 3 we can apply chain rule now . total men required to complete the remaining work = 100 × ( 25 / 25 ) × ( 8 / 10 ) × ( 2 / 3 ) / ( 1 / 3 ) = 160 . so addition... | a ) 55 , b ) 60 , c ) 70 , d ) 50 , e ) 65 | b | subtract(divide(multiply(2, multiply(multiply(100, 25), 8)), multiply(25, 10)), 100) | multiply(n0,n4)|multiply(n4,n7)|multiply(n3,#0)|multiply(n1,#2)|divide(#3,#1)|subtract(#4,n0) | physics | B |
a tank holds x gallons of a saltwater solution that is 20 % salt by volume . one fourth of the water is evaporated , leaving all of the salt . when 7 gallons of water and 14 gallons of salt are added , the resulting mixture is 33 1 / 3 % salt by volume . what is the value of x ? | "nope , 150 . i can only get it by following pr ' s backsolving explanation . i hate that . original mixture has 20 % salt and 80 % water . total = x out of which salt = 0.2 x and water = 0.8 x now , 1 / 4 water evaporates and all salt remains . so what remains is 0.2 x salt and 0.6 x water . now 14 gallons salt is add... | a ) 37.5 , b ) 75 , c ) 105 , d ) 150 , e ) 175 | c | divide(subtract(multiply(14, const_2), 7), subtract(subtract(subtract(1, divide(20, const_100)), multiply(subtract(1, divide(20, const_100)), divide(1, const_4))), multiply(const_2, divide(20, const_100)))) | divide(n0,const_100)|divide(n4,const_4)|multiply(n2,const_2)|multiply(#0,const_2)|subtract(#2,n1)|subtract(n4,#0)|multiply(#1,#5)|subtract(#5,#6)|subtract(#7,#3)|divide(#4,#8)| | general | C |
if xy > 0 , 1 / x + 1 / y = 6 , and 1 / xy = 12 , then ( x + y ) / 6 = ? | "( 1 / x + 1 / y ) = 6 canbe solved as { ( x + y ) / xy } = 12 . substituting for 1 / xy = 12 , we get x + y = 6 / 12 = = > ( x + y ) / 6 = 6 / ( 12 * 6 ) = 1 / 12 . a" | a ) 1 / 12 , b ) 1 / 6 , c ) 1 / 5 , d ) 5 , e ) 6 | a | divide(divide(6, 12), 12) | divide(n3,n5)|divide(#0,n5)| | general | A |
a train covers a distance of 90 km in 20 min . if it takes 9 sec to pass a telegraph post , then the length of the train is ? | "speed = ( 90 / 20 * 60 ) km / hr = ( 270 * 5 / 18 ) m / sec = 75 m / sec . length of the train = 75 * 9 = 675 m . answer : option a" | a ) 675 , b ) 680 , c ) 685 , d ) 690 , e ) 695 | a | divide(90, subtract(divide(90, 20), 9)) | divide(n0,n1)|subtract(#0,n2)|divide(n0,#1)| | physics | A |
p and q invested in a shop . the profits were divided in the ratio of 2 : 3 respectively . if p invested rs . 40,000 , the amount invested by q is : | "suppose q invested rs . y . then , 40000 / y = 2 / 3 or y = [ 40000 x 3 / 2 ] = 60000 . answer d" | a ) 35000 , b ) 40000 , c ) 50000 , d ) 60000 , e ) none | d | multiply(divide(40,000, 2), 3) | divide(n2,n0)|multiply(n1,#0)| | gain | D |
if 4 : 6 : : x : 36 , then find the value of x | explanation : treat 4 : 6 as 4 / 6 and x : 36 as x / 36 , treat : : as = so we get 4 / 6 = x / 36 = > 6 x = 144 = > x = 24 option a | a ) 24 , b ) 22 , c ) 28 , d ) 30 , e ) 18 | a | divide(add(multiply(6, 4), 6), 36) | multiply(n0,n1)|add(n1,#0)|divide(#1,n2)| | general | A |
if a is thrice as fast as b and together can do a work in 24 days . in how many days a alone can do the work ? | "a ’ s one day ’ s work = 1 / x b ’ s one day ’ s work = 1 / 3 x a + b ’ s one day ’ s work = 1 / x + 1 / 3 x = 1 / 24 = 3 + 1 / 3 x = 4 / 3 x = 1 / 24 x = 24 * 4 / 3 = 32 answer : d" | a ) 36 , b ) 42 , c ) 28 , d ) 32 , e ) 45 | d | inverse(divide(const_3, multiply(24, const_4))) | multiply(n0,const_4)|divide(const_3,#0)|inverse(#1)| | physics | D |
the average of 9 observations was 9 , that of the 1 st of 5 being 10 and that of the last 5 being 8 . what was the 5 th observation ? | "1 to 9 = 9 * 9 = 81 1 to 5 = 5 * 10 = 50 5 to 9 = 5 * 8 = 40 5 th = 50 + 40 = 90 – 81 = 9 answer : b" | a ) 6 , b ) 9 , c ) 8 , d ) 2 , e ) 3 | b | subtract(add(multiply(10, 5), multiply(9, 5)), multiply(9, 9)) | multiply(n3,n4)|multiply(n1,n3)|multiply(n0,n1)|add(#0,#1)|subtract(#3,#2)| | general | B |
in a coconut grove , ( x + 3 ) trees yield 60 nuts per year , x trees yield 120 nuts per year and ( x – 3 ) trees yield 180 nuts per year . if the average yield per year per tree be 100 , find x . | "( x + 3 ) × 60 + x × 120 + ( x − 3 ) × 180 / ( x + 3 ) + x + ( x − 3 ) = 100 ⇒ 360 x − 360 / 3 x = 100 ⇒ 60 x = 360 ⇒ x = 6 answer d" | a ) 3 , b ) 4 , c ) 5 , d ) 6 , e ) none of the above | d | divide(subtract(multiply(180, 3), multiply(60, 3)), subtract(add(add(60, 120), 180), multiply(100, const_3))) | add(n1,n2)|multiply(n0,n4)|multiply(n0,n1)|multiply(n5,const_3)|add(n4,#0)|subtract(#1,#2)|subtract(#4,#3)|divide(#5,#6)| | general | D |
a box contains 20 electric bulbs , out of which 4 are defective . two bulbs are chosen at random form this box . the probability that at least one of these is defective is : | "p ( none is defective ) = 16 c 2 / 20 c 2 = 12 / 19 p ( at least one is defective ) = 1 - 12 / 19 = 7 / 19 answer b" | a ) 8 / 19 , b ) 7 / 19 , c ) 10 / 19 , d ) 20 / 19 , e ) 22 / 19 | b | subtract(const_1, divide(choose(subtract(20, 4), const_2), choose(20, const_2))) | choose(n0,const_2)|subtract(n0,n1)|choose(#1,const_2)|divide(#2,#0)|subtract(const_1,#3)| | probability | B |
youseff lives x blocks from his office . it takes him 1 minute per block to walk to work and 20 seconds per block to ride his bike to work . it is takes him exactly 4 minutes more to walk to work than to ride his bike to work , then x equals ? | "please follow posting guidelines , link is in my signatures . as for your question , x / 60 = blocks / time / block = block ^ 2 / time . this is not what you want . you are given x blocks and 60 seconds per block . thus you need to put it as 60 * x to give you units of seconds as you are equating this to 240 ( which i... | a ) 4 , b ) 6 , c ) 10 , d ) 15 , e ) 20 | b | divide(multiply(4, const_3), const_2) | multiply(n2,const_3)|divide(#0,const_2)| | physics | B |
a merchant marks his goods up by 50 % and then offers a discount of 20 % on the marked price . what % profit does the merchant make after the discount ? | "let the price be 100 . the price becomes 150 after a 50 % markup . now a discount of 20 % on 150 . profit = 120 - 100 20 % answer a" | a ) 20 % , b ) 25 % , c ) 69 % , d ) 31 % , e ) 19 % | a | subtract(subtract(add(50, const_100), divide(multiply(add(50, const_100), 20), const_100)), const_100) | add(n0,const_100)|multiply(n1,#0)|divide(#1,const_100)|subtract(#0,#2)|subtract(#3,const_100)| | gain | A |
a gardener wants to plant trees in his garden in such a way that the number of trees in each row should be the same . if there are 2 rows or 7 rows or 9 rows , then no tree will be left . find the least number of trees required | "explanation : the least number of trees that are required = lcm ( 2 , 7 , 9 ) = 126 answer : c" | a ) 22 , b ) 60 , c ) 126 , d ) 76 , e ) 21 | c | divide(multiply(multiply(9, 7), 2), const_2) | multiply(n1,n2)|multiply(n0,#0)|divide(#1,const_2)| | general | C |
when greenville state university decided to move its fine arts collection to a new library , it had to package the collection in 20 - inch by 20 - inch by 15 - inch boxes . if the university pays $ 1.20 for every box , and if the university needs 3.06 million cubic inches to package the collection , what is the minimum... | total no . of boxes = 3060000 / ( 20 × 20 × 15 ) = 510 total cost = 510 × $ 1.20 = $ 612 answer d | a ) $ 255 , b ) $ 275 , c ) $ 510 , d ) $ 612 , e ) $ 2,550 | d | multiply(divide(multiply(3.06, multiply(const_1000, const_1000)), multiply(multiply(20, 20), 15)), 1.2) | multiply(const_1000,const_1000)|multiply(n0,n0)|multiply(n4,#0)|multiply(n2,#1)|divide(#2,#3)|multiply(n3,#4) | general | D |
a 220 metres long train running at the speed of 120 kmph crosses another train running in opposite direction at the speed of 80 kmph in 9 seconds . what is the length of the other train ? | speed = ( 120 + 80 ) km / h ( because direction is opposite hence relative velocity is added ) = 500 / 9 m / s time = 9 sec let the lenght of second train is x total distance covered = 220 + x therefore , d = speed * time thus 220 + x = 500 / 9 * 9 x = 500 - 220 = 280 m answer : e | a ) 230 m , b ) 240 m , c ) 250 m , d ) 260 m , e ) 280 m | e | subtract(multiply(multiply(add(120, 80), const_0_2778), 9), 220) | add(n1,n2)|multiply(#0,const_0_2778)|multiply(n3,#1)|subtract(#2,n0) | physics | E |
john makes $ 60 a week from his job . he earns a raise and nowmakes $ 70 a week . what is the % increase ? | "increase = ( 10 / 60 ) * 100 = ( 1 / 6 ) * 100 = 16.66 % . b" | a ) 16 % , b ) 16.66 % , c ) 17 % , d ) 17.2 % , e ) 18 % | b | multiply(divide(subtract(70, 60), 60), const_100) | subtract(n1,n0)|divide(#0,n0)|multiply(#1,const_100)| | gain | B |
a bag contains 4 red , 5 blue and 3 green balls . if 3 ballsare picked at random , what is the probability that both are red ? | "p ( both are red ) , = 4 c 3 / 12 c 3 = 1 / 55 e" | a ) 1 / 33 , b ) 1 / 22 , c ) 1 / 12 , d ) 1 / 44 , e ) 1 / 55 | e | divide(choose(4, 3), choose(add(add(4, 5), 3), 3)) | add(n0,n1)|choose(n0,n3)|add(n2,#0)|choose(#2,n3)|divide(#1,#3)| | other | E |
what least number should be added to 1054 , so that the sum is completely divisible by 23 | "explanation : ( 1054 / 23 ) gives remainder 19 19 + 4 = 23 , so we need to add 4 answer : option a" | a ) a ) 4 , b ) b ) 1 , c ) c ) 2 , d ) d ) 3 , e ) e ) 5 | a | subtract(23, reminder(1054, 23)) | reminder(n0,n1)|subtract(n1,#0)| | general | A |
two trains are moving in the same direction at 144 kmph and 72 kmph . the faster train crosses a man in the slower train in 19 seconds . find the length of the faster train ? | "relative speed = ( 144 - 72 ) * 5 / 18 = 4 * 5 = 20 mps . distance covered in 19 sec = 19 * 20 = 380 m . the length of the faster train = 380 m . answer : d" | a ) 320 , b ) 340 , c ) 360 , d ) 380 , e ) 400 | d | multiply(divide(subtract(144, 72), const_3_6), 19) | subtract(n0,n1)|divide(#0,const_3_6)|multiply(n2,#1)| | physics | D |
at the wholesale store you can buy an 8 - pack of hot dogs for $ 1.55 , a 20 - pack for $ 3.05 , and a 250 - pack for $ 22.95 . what is the greatest number of hot dogs you can buy at this store with $ 300 ? | "to maximize number of hot dogs with 300 $ total number of hot dogs bought in 250 - pack = 22.95 * 13 = 298.35 $ amount remaining = 300 - 298.35 = 1.65 $ this amount is too less to buy any 8 - pack . greatest number of hot dogs one can buy with 300 $ = 250 * 13 = 3250 answer e" | a ) 3,108 , b ) 3,100 , c ) 3,108 , d ) 3,124 , e ) 3,250 | e | multiply(divide(300, 22.95), 250) | divide(n6,n5)|multiply(n4,#0)| | general | E |
the contents of a certain box consist of 80 apples and 30 oranges . how many oranges must be added to the box so that exactly 50 % of the pieces of fruit in the box will be apples ? | "apple = ( apple + orange + x ) * 0.5 80 = ( 30 + 80 + x ) * 0.5 x = 50 answer : b" | a ) 45 , b ) 50 , c ) 55 , d ) 60 , e ) 65 | b | subtract(add(80, 30), divide(80, divide(50, const_100))) | add(n0,n1)|divide(n2,const_100)|divide(n0,#1)|subtract(#0,#2)| | general | B |
can n and can в are both right circular cylinders . the radius of can n is twice the radius of can b , while the height of can n is half the height of can b . if it costs $ 4.00 to fill half of can b with a certain brand of gasoline , how much would it cost to completely fill can n with the same brand of gasoline ? | let x be the radius of b and 2 h be the height of b . therefore , radius of n = 2 x and height = h vol of b = 3.14 * x ^ 2 * 2 h vol of a = 3.14 * 4 x ^ 2 * h cost to fill half of b = $ 4 - - > cost to fill full b = $ 8 - - > 3.14 * x ^ 2 * 2 h = 8 - - > 3.14 * x ^ 2 * h = 4 - - > 4 * ( 3.14 * x ^ 2 * h ) = $ 16 ans e | ['a ) $ 1', 'b ) $ 2', 'c ) $ 4', 'd ) $ 8', 'e ) $ 16'] | e | multiply(multiply(4, const_2), divide(power(const_2, const_2), const_2)) | multiply(n0,const_2)|power(const_2,const_2)|divide(#1,const_2)|multiply(#2,#0) | geometry | E |
two cars are traveling in the same direction along the same route . the red car travels at a constant speed of 40 miles per hour , and the black car is traveling at a constant speed of 50 miles per hour . if the red car is 10 miles ahead of the black car , how many hours will it take the black car to overtake the red c... | "option c 10 + 40 t = 50 t t = 1" | a ) 0.1 , b ) 0.6 , c ) 1 , d ) 1.2 , e ) 2 | c | divide(10, subtract(50, 40)) | subtract(n1,n0)|divide(n2,#0)| | physics | C |
if p is the product of the integers from 1 to 38 , inclusive , what is the greatest integer k for which 3 ^ k is a factor of p ? | "38 ! has 3 , 6 , 9 , . . . . 30 , 33 , 36 as factors , which are 12 multiples of 3 . we need to add 5 more to these 12 because of 9 , 18 , 27 , and 36 . the greatest integer of k is 17 . the answer is c ." | a ) 13 , b ) 15 , c ) 17 , d ) 19 , e ) 21 | c | add(multiply(3, const_4), 3) | multiply(n2,const_4)|add(n2,#0)| | general | C |
during a sale of 20 % on everything in a store , a kid is successful in convincing the store manager to give him 40 candies for the discounted price of 14 candies . the store still makes a profit of 12 % on this sale . what is the mark up percentage on each candy ? | hi akhil , i can share the way i solved it . . let ' s say marked price = m so , there ' s a discount of 20 % on m so , new s . p . = 80 % of m now , the child convinces the owner to sell 20 candies for the price of 14 candies . let ' s say each candy after discount is 1 $ . so , s . p . of 20 candies = 20 $ . the chil... | a ) 100 % , b ) 80 % , c ) 75 % , d ) 66 + 2 / 3 % , e ) 55 % | e | multiply(subtract(divide(add(divide(12, const_100), const_1), multiply(subtract(const_1, divide(20, const_100)), divide(subtract(40, 14), 40))), const_1), const_100) | divide(n3,const_100)|divide(n0,const_100)|subtract(n1,n2)|add(#0,const_1)|divide(#2,n1)|subtract(const_1,#1)|multiply(#4,#5)|divide(#3,#6)|subtract(#7,const_1)|multiply(#8,const_100) | gain | E |
the ratio between the number of sheep and the number of horses at the stewart farm is 3 to 7 , if each horse is fed 230 ounces of horse food per day and the farm needs a total 12,880 ounces of horse food per day , what is the number of sheep in the farm ? | let the number of sheeps and horses be 3 x and 7 x . now total number of horses = total consumption of horse food / consumption per horse = 12880 / 230 = 56 , which is equal to 7 x . = > x = 8 sheeps = 3 x = 3 * 8 = 24 . hence a | a ) 24 , b ) 28 , c ) 32 , d ) 56 , e ) 60 | a | multiply(divide(divide(add(add(multiply(multiply(const_4, const_2), const_10), multiply(multiply(const_4, const_2), const_100)), multiply(const_12, const_1000)), 230), 7), 3) | multiply(const_2,const_4)|multiply(const_1000,const_12)|multiply(#0,const_10)|multiply(#0,const_100)|add(#2,#3)|add(#4,#1)|divide(#5,n2)|divide(#6,n1)|multiply(n0,#7) | other | A |
how many multiples of 5 are there between 50 and 500 ? | "it should be mentioned whether 50 and 500 are inclusive . if 50 and 500 are inclusive , then the answer is ( 500 - 50 ) / 5 + 1 = 91 . if 50 and 500 are not inclusive , then the answer is ( 495 - 55 ) / 5 + 1 = 89 . since oa is c , then we have not inclusive case ." | a ) 128 , b ) 91 , c ) 89 , d ) 61 , e ) 60 | c | add(divide(subtract(500, 50), 5), const_1) | subtract(n2,n1)|divide(#0,n0)|add(#1,const_1)| | general | C |
what is the least number to be subtracted from 800 to make it a perfect square ? | "the numbers less than 800 and are squares of certain number is 784 . the least number that should be subtracted from 800 to make it perfect square = 800 - 784 = 16 . answer : b" | a ) 1 , b ) 16 , c ) 20 , d ) 71 , e ) 60 | b | subtract(multiply(const_100, add(const_10, multiply(const_3, const_2))), 800) | multiply(const_2,const_3)|add(#0,const_10)|multiply(#1,const_100)|subtract(#2,n0)| | general | B |
evaluate : 30 - 12 * 3 * 2 = ? | "according to order of operations , 12 ? 3 ? 2 ( division and multiplication ) is done first from left to right 12 * * 2 = 4 * 2 = 8 hence 30 - 12 * 3 * 2 = 30 - 8 = 22 correct answer e" | a ) 62 , b ) 52 , c ) 32 , d ) 12 , e ) 22 | e | subtract(30, multiply(multiply(12, const_2.0), 2)) | multiply(n1,const_2.0)|multiply(n3,#0)|subtract(n0,#1)| | general | E |
a batsman scored 120 runs which included 3 boundaries and 8 sixes . what % of his total score did he make by running between the wickets | "number of runs made by running = 110 - ( 3 x 4 + 8 x 6 ) = 120 - ( 60 ) = 60 now , we need to calculate 60 is what percent of 120 . = > 60 / 120 * 100 = 50 % b" | a ) 30 % , b ) 50 % , c ) 60 % , d ) 80 % , e ) 90 % | b | multiply(divide(subtract(120, add(multiply(3, 8), multiply(8, 3))), 120), const_100) | multiply(n1,n2)|multiply(n1,n2)|add(#0,#1)|subtract(n0,#2)|divide(#3,n0)|multiply(#4,const_100)| | general | B |
excluding stoppages , the speed of a train is 48 kmph and including stoppages it is 36 kmph . of how many minutes does the train stop per hour ? | "t = 12 / 48 * 60 = 15 answer : c" | a ) 16 , b ) 17 , c ) 15 , d ) 18 , e ) 12 | c | subtract(const_60, multiply(const_60, divide(36, 48))) | divide(n1,n0)|multiply(#0,const_60)|subtract(const_60,#1)| | physics | C |
this topic is locked . if you want to discuss this question please re - post it in the respective forum . matt and peter can do together a piece of work in 20 days . after they have worked together for 12 days matt stops and peter completes the remaining work in 14 days . in how many days peter complete the work separa... | together they complete the job in 20 days means they complete 12 / 20 of the job after 12 days . peter completes the remaining ( 8 / 20 ) of the job in 14 days which means that the whole job ( 1 ) can be completed in x days . < = > 8 / 20 - > 14 < = > x = 14 / ( 8 / 20 ) = 35 thus the answer is d . | a ) 26 days , b ) 27 days , c ) 23 days , d ) 35 days , e ) 24 days | d | add(multiply(divide(const_1, const_2), 14), multiply(14, const_2)) | divide(const_1,const_2)|multiply(n2,const_2)|multiply(n2,#0)|add(#2,#1) | physics | D |
if the given two numbers are respectively 6 % and 30 % of a third number , then what percentage is the first of the second ? | here , l = 6 and m = 30 therefore , first number = l / m x 100 % of second number = 6 / 30 x 100 % of second number = 20 % of second number answer : a | a ) 20 % , b ) 25 % , c ) 18 % , d ) 30 % , e ) none of these | a | multiply(divide(divide(6, const_100), divide(30, const_100)), const_100) | divide(n0,const_100)|divide(n1,const_100)|divide(#0,#1)|multiply(#2,const_100) | gain | A |
how many 4 x 5 x 8 cubes could fit in a 10 x 16 x 84 box ? | the answer is a ) 84 . the 10 x 16 x 84 box has an area of 13,440 . the 4 x 5 x 8 cubes have an area of 160 . if you divide 13,440 by 160 , you get 84 . | a ) 84 , b ) 90 , c ) 85 , d ) 77 , e ) 64 | a | volume_rectangular_prism(10, 16, 84) | volume_rectangular_prism(n3,n4,n5)| | geometry | A |
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