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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 ) = 18 so people read both science fiction and literacy works are 18 answer : d" | 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 or b can be negative to make ab negative and it can still be more ( away from 0 ) than the multiplication of 4 other - ve numbers . . . actually by writing minimum required as 1 out of 6 , you are actually meaning s = 5 out of 6 also possible as you will see 5 or 1 will give you same equation . . ans d" | 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 is proportional to the percentage of the budget it represents , how many degrees of the circle are used to represent basic astrophysics research ? | "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 at suzie ’ s discount footwear ? | "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 12.50 answer : b" | 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 probability of choosing pair of chosen sibling in senior class ? as in senior class there is only one pair of chosen sibling it would be 1 / 800 ( as there is only one sibling pair of chosen one ) . so the probability of that the 2 students selected will be a sibling pair is : 60 / 1000 ∗ 1 / 800 = 3 / 40000 answer : a ." | 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 = 60 * 3 = 180 miles . therefore , total distance covered = 80 + 180 = 260 miles . total time taken = 2 + 3 = 5 hours . hence , average speed = total distance travelled / total time taken = 260 / 5 = 52 miles per hour . choice a" | 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 answer : c" | 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 of fruit that she keeps is 50 ¢ ? | "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 three of them work together to finish the room , each one working at his respective rate . what fraction of the whole job was done by peter ? | "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 the room . the answer is d ." | 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 * $ 3,000 = $ 750 . plug this value in the first equation : c + 750 = $ 2,500 - - > c = $ 1,750 answer : c ." | 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 . = > all the three letters can be chosen in 676 * 26 = 17576 ways . it implies that the maximum possible number of five letter palindromes is 17576 because the fourth letter is the same as the second letter and the fifth letter is the same as the first letter . answer : c" | 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 ) + ( 8 / 12 ) pints . . from here we get the ratio of red : white = 9 : 8 . . so in 34 pints red will be 9 / ( 9 + 8 ) * 34 = 18 . . answer : c" | 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 ) = 16 / 24 . = > 2 : 3 answer : b | 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 = 1,800 x = 180 thus , the total amount allocated to food is 4 x 180 = $ 720 . answer b . | 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 additional men required = 160 - 100 = 60 . answer is b | 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 added and 7 gallons of water is added . so salt now becomes - > ( 0.2 x + 14 ) and water - - > ( 0.6 x + 7 ) amount of salt is 33.33 % of total . so amount of water is 66.66 % . so salt is half of the volume of water . so ( 0.2 x + 14 ) = ( 0.6 x + 7 ) / 2 = > 0.2 x = 21 solving , x = 105 answer : c" | 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 is time in seconds . ) . thus the correct equation is : 60 * x - 20 * x = 240 - - - - > 40 x = 240 - - > x = 6 option b" | 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 amount the university must spend on boxes ? | 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 car ? | "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 child bought it for 14 $ so , he got a discount of 6 / 20 * 100 = 30 % so , the latest s . p . = 70 % of 80 % of m = 0.7 * 0.8 m now , we are given that the shopkeeper still makes a profit of 12 % . so we have , 0.7 * 0.8 * m = 1.12 c . p so , we get , m = 2 c . p . i . e . marked price was kept 100 % above c . p . e | 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 separately . | 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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