Problem stringlengths 5 967 | Rationale stringlengths 1 2.74k | options stringlengths 37 300 | correct stringclasses 5
values | annotated_formula stringlengths 7 6.48k | linear_formula stringlengths 8 925 | category stringclasses 6
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the population of a town is 10000 . it increases annually at the rate of 20 % p . a . what will be its population after 3 years ? | "formula : ( after = 100 denominator ago = 100 numerator ) 10000 × 120 / 100 ^ 3 = 17280 b )" | a ) 14300 , b ) 17280 , c ) 14500 , d ) 14600 , e ) 15400 | b | add(10000, multiply(divide(multiply(10000, 20), const_100), 3)) | multiply(n0,n1)|divide(#0,const_100)|multiply(#1,n2)|add(n0,#2)| | gain |
a is twice as good a workman as b and they took 6 days together to do the work b alone can do it in ? | "wc = 2 : 1 2 x + x = 1 / 6 x = 1 / 18 = > 18 days answer : b" | a ) 25 days , b ) 18 days , c ) 21 days , d ) 11 days , e ) 13 days | b | multiply(divide(multiply(6, add(const_2, const_1)), const_2), const_2) | add(const_1,const_2)|multiply(n0,#0)|divide(#1,const_2)|multiply(#2,const_2)| | physics |
how many even multiples of 15 are there between 149 and 301 ? | "150 = 10 * 15 300 = 20 * 15 the even multiples are 15 multiplied by 10 , 12 , 14 , 16 , 18 , and 20 for a total of 6 . the answer is b ." | a ) 5 , b ) 6 , c ) 9 , d ) 10 , e ) 11 | b | add(divide(subtract(301, 149), multiply(15, const_2)), const_1) | multiply(n0,const_2)|subtract(n2,n1)|divide(#1,#0)|add(#2,const_1)| | general |
what least number should be added to 1101 , so that the sum is completely divisible by 24 | "explanation : ( 1056 / 24 ) gives remainder 21 21 + 3 = 24 , so we need to add 3 . answer : option d" | a ) a ) 4 , b ) b ) 1 , c ) c ) 2 , d ) d ) 3 , e ) e ) 5 | d | subtract(24, reminder(1101, 24)) | reminder(n0,n1)|subtract(n1,#0)| | general |
a and b can do a work in 6 days , b and c in 2 days and c and a in 8 days . in how many days will the work be completed , if all three of them work together ? | "one day work of a and b = 1 / 6 one day work of b and c = 1 / 2 one day work of c and a = 1 / 8 2 ( a + b + c ) = 1 / 6 + 1 / 2 + 1 / 8 2 ( a + b + c ) = 19 / 24 ( a + b + c ) = 19 / 48 number of days required = 48 / 19 days . answer : d" | a ) 12 / 24 , b ) 48 / 63 , c ) 45 / 56 , d ) 48 / 19 , e ) 28 / 12 | d | divide(const_1, divide(add(add(inverse(6), inverse(2)), inverse(8)), 6)) | inverse(n0)|inverse(n1)|inverse(n2)|add(#0,#1)|add(#3,#2)|divide(#4,n0)|divide(const_1,#5)| | physics |
working alone , mary can pave a driveway in 5 hours and hillary can pave the same driveway in 3 hours . when they work together , mary thrives on teamwork so her rate increases by 10 % , but hillary becomes distracted and her rate decreases by 10 % . if they both work together , how many hours will it take to pave the ... | "initial working rates : mary = 1 / 5 per hour hillary = 1 / 3 per hour rate when working together : mary = 1 / 5 + ( 1 / 10 * 1 / 5 ) = 2 / 9 per hour hillary = 1 / 3 - ( 1 / 10 * 1 / 3 ) = 3 / 10 per hour together they work 2 / 9 + 3 / 10 = 1 / 2 per hour so they will need 2 hours to complete the driveway . the corre... | a ) 2 hours , b ) 4 hours , c ) 5 hours , d ) 6 hours , e ) 7 hours | a | inverse(add(multiply(divide(const_1, 5), add(divide(10, const_100), const_1)), multiply(divide(const_1, 3), divide(10, const_100)))) | divide(n2,const_100)|divide(const_1,n0)|divide(const_1,n1)|divide(n3,const_100)|add(#0,const_1)|multiply(#2,#3)|multiply(#4,#1)|add(#6,#5)|inverse(#7)| | gain |
how many 3 digit no ' s are between 100 - 500 , where sum of two digit is 3 rd ? | total permutations possible = 18 + 16 + 14 + 12 = 60 answer : c | a ) 24 , b ) 54 , c ) 60 , d ) 84 , e ) 94 | c | multiply(divide(multiply(add(add(subtract(add(const_4, const_4), divide(500, const_100)), const_1), add(const_4, const_4)), divide(500, const_100)), subtract(3, const_1)), const_2) | add(const_4,const_4)|divide(n2,const_100)|subtract(n0,const_1)|subtract(#0,#1)|add(#3,const_1)|add(#4,#0)|multiply(#5,#1)|divide(#6,#2)|multiply(#7,const_2) | general |
find the perimeter and area of a square of side 19 cm . | "we know that the perimeter of square = 4 ã — side side = 19 cm therefore , perimeter = 4 ã — 19 cm = 76 cm now , area of the square = ( side ã — side ) sq . units = 19 ã — 19 cm â ² = 361 cm â ² answer : b" | a ) 324 , b ) 361 , c ) 400 , d ) 441 , e ) 484 | b | square_area(19) | square_area(n0)| | geometry |
in a mixture of 45 litres the ratio of milk to water is 4 : 1 . additional 21 litres of water is added to the mixture . find the ratio of milk to water in the resulting mixture . | "given that milk / water = 4 x / x and 4 x + x = 45 - - > x = 9 . thus milk = 4 x = 36 liters and water = x = 9 liters . new ratio = 36 / ( 9 + 21 ) = 36 / 30 = 6 / 5 . answer : c ." | a ) 2 / 1 , b ) 4 / 1 , c ) 6 / 5 , d ) 3 / 4 , e ) 3 / 2 | c | divide(subtract(45, divide(45, add(4, 1))), add(divide(45, add(4, 1)), 21)) | add(n1,n2)|divide(n0,#0)|add(n3,#1)|subtract(n0,#1)|divide(#3,#2)| | general |
the cross - section of a stream is a trapezium in shape . if the stream is 10 m wide at the top and the area of cross - section is 640 sq m , the depth of stream is 80 m and width at the bottom is ? | 1 / 2 * 80 ( 10 + b ) = 640 b = 6 m answer : b | a ) 2 m , b ) 6 m , c ) 4 m , d ) 8 m , e ) 9 m | b | divide(subtract(640, multiply(multiply(80, divide(const_1, const_2)), 10)), multiply(80, divide(const_1, const_2))) | divide(const_1,const_2)|multiply(n2,#0)|multiply(n0,#1)|subtract(n1,#2)|divide(#3,#1) | physics |
the cost of 2 books and 2 magazines is $ 26 . the cost of 1 book and 3 magazines is $ 27 . how much does 1 magazine cost ? | let the cost of 1 book = x , let the cost of 1 magazine = y 2 x + 2 y = 26 x = ( 26 - 2 y ) / 2 x = 13 - y again , ( 13 - y ) + 3 y = 27 13 + 2 y = 27 2 y = 14 y = 7 answer : c | a ) 8 , b ) 5 , c ) 7 , d ) 4 , e ) 6 | c | divide(subtract(multiply(2, 27), 26), subtract(multiply(2, 3), 2)) | multiply(n0,n5)|multiply(n0,n4)|subtract(#0,n2)|subtract(#1,n0)|divide(#2,#3) | general |
6,11 , 21 , 41 , 81 , ( . . . ) | "explanation : 6 6 × 2 - 1 = 11 11 × 2 - 1 = 21 21 × 2 - 1 = 41 41 × 2 - 1 = 81 81 × 2 - 1 = 161 answer : option e" | a ) 22 , b ) 35 , c ) 27 , d ) 32 , e ) 161 | e | subtract(negate(81), multiply(subtract(21, 41), divide(subtract(21, 41), subtract(6,11, 21)))) | negate(n3)|subtract(n1,n2)|subtract(n0,n1)|divide(#1,#2)|multiply(#3,#1)|subtract(#0,#4)| | general |
the organizers of a fair projected a 25 percent increase in attendance this year over that of last year , but attendance this year actually decreased by 20 percent . what percent of the projected attendance was the actual attendance ? | last year attendance = 100 ( assume ) ; projected attendance = 125 ; actual attendance = 80 . the actual attendance therefore was ( actual ) / ( project ) = 80 / 125 * 100 = 64 $ of the projected attendance . answer : c . | a ) 45 % , b ) 56 % , c ) 64 % , d ) 75 % , e ) 80 % | c | multiply(divide(multiply(20, const_4), add(const_100, 25)), const_100) | add(n0,const_100)|multiply(n1,const_4)|divide(#1,#0)|multiply(#2,const_100) | general |
the length of the longest tape in cm which can be used to measure exactly , the length 10 m ; 3 m 85 cm ; and 11 m 50 cm is : | "the three lengths in cm are 1000 , 385 & 1150 . hcf of 700 , 385 & 1295 is 5 . hence , the answer is 5 cm . answer : a" | a ) 5 , b ) 35 , c ) 34 , d ) 36 , e ) 38 | a | sqrt(85) | sqrt(n2)| | physics |
if x + y = 270 , x - y = 200 , for integers of x and y , y = ? | "x + y = 270 x - y = 200 2 x = 70 x = 35 y = 235 answer is b" | a ) 200 , b ) 235 , c ) 50 , d ) 115 , e ) 150 | b | divide(add(270, 200), const_2) | add(n0,n1)|divide(#0,const_2)| | general |
a and b together can do a piece of work in 6 days and a alone can do it in 11 days . in how many days can b alone can do it ? | "explanation : a and b can do work 1 / 6 in 1 day a alone can do 1 / 11 work in 1 day b alone can do ( 1 / 6 - 1 / 11 ) = 5 / 66 work in 1 day = > complete work can be done in 66 / 5 days by b answer : option c" | a ) 12 days , b ) 15 days , c ) 66 / 5 days , d ) 21 days , e ) 22 / 5 days | c | inverse(subtract(inverse(6), inverse(11))) | inverse(n0)|inverse(n1)|subtract(#0,#1)|inverse(#2)| | physics |
the average weight of 22 boys in a class is 50.25 kg and that of the remaining 8 boys is 45.15 kg . find the average weights of all the boys in the class . | explanation : average weight of 22 boys = 50.25 total weight of 22 boys = 50.25 × 22 average weight of remaining 8 boys = 45.15 total weight of remaining 8 boys = 45.15 × 8 total weight of all boys in the class = ( 50.25 × 16 ) + ( 45.15 × 8 ) total boys = 22 + 8 = 30 average weight of all the boys = ( ( 50.25 × 22 ) +... | a ) 48.89 , b ) 42.25 , c ) 50 , d ) 51.25 , e ) 52.25 | a | divide(add(multiply(22, 50.25), multiply(8, 45.15)), add(22, 8)) | add(n0,n2)|multiply(n0,n1)|multiply(n2,n3)|add(#1,#2)|divide(#3,#0) | general |
profits in the partnership of bess , bill and bob are shared in the ratio 1 : 2 : 3 . if bill ' s share of the profits is $ 300 , what is bob ' s share ? | ans is d given profit ratio , bess : bill : bob = 1 : 2 : 3 to make bill ' s portion $ 300 , multiply by $ 150 = > bess : bill : bob = $ 150 : $ 300 : $ 450 = > bob ' s share = $ 450 | a ) $ 150 , b ) $ 900 , c ) $ 600 , d ) $ 450 , e ) none | b | add(add(add(divide(300, 1), divide(300, 2)), divide(300, 2)), divide(300, 1)) | divide(n3,n0)|divide(n3,n1)|add(#0,#1)|add(#2,#1)|add(#3,#0) | other |
a sporting goods store sold 60 frisbees in one week , some for $ 3 and the rest for $ 4 each . if receipts from frisbee sales for the week totaled $ 204 , what is the fewest number of $ 4 frisbees that could have been sold ? | "in this question however , because we are told that exactly 64 frisbees have been sold and revenue was exactly $ 204 , there is only one possible solution for the number of $ 3 and $ 4 frisbees sold . to solve , we have 2 equations and 2 unknowns let x = number of $ 3 frisbees sold let y = number of $ 4 frisbees sold ... | a ) 24 , b ) 12 , c ) 8 , d ) 4 , e ) 2 | a | subtract(204, multiply(60, 3)) | multiply(n0,n1)|subtract(n3,#0)| | general |
patrick purchased 90 pencils and sold them at a loss equal to the selling price of 40 pencils . the cost of 90 pencils is how many times the selling price of 90 pencils ? | "say the cost price of 90 pencils was $ 90 ( $ 1 per pencil ) and the selling price of 1 pencil was p . selling at a loss : 90 - 90 p = 40 p - - > p = 9 / 13 . ( cost price ) / ( selling price ) = 1 / ( 9 / 13 ) = 13 / 9 = 1.44 . answer : e ." | a ) 0.75 , b ) 0.8 , c ) 1 , d ) 1.2 , e ) 1.44 | e | inverse(divide(90, add(90, 40))) | add(n0,n1)|divide(n0,#0)|inverse(#1)| | general |
tom purchased 8 kg of apples at the rate of 70 per kg and 9 kg of mangoes at the rate of 75 per kg . how much amount did he pay to the shopkeeper ? | cost of 8 kg apples = 70 × 8 = 560 . cost of 9 kg of mangoes = 75 × 9 = 675 . total cost he has to pay = 560 + 675 = 1235 . e ) | a ) a ) 1040 , b ) b ) 1045 , c ) c ) 1055 , d ) d ) 1060 , e ) e ) 1235 | e | add(multiply(8, 70), multiply(9, 75)) | multiply(n0,n1)|multiply(n2,n3)|add(#0,#1) | gain |
it takes joey the postman 1 hours to run a 6 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 6 miles an hour let the other half be x miles an hour now , avg speed = 8 mile an hour 2 * 6 * x / 6 + x = 8 12 x = 8 x + 48 = > x = 12 b" | a ) 11 , b ) 12 , c ) 13 , d ) 14 , e ) 15 | b | divide(6, subtract(divide(multiply(const_2, 6), 8), 1)) | multiply(n1,const_2)|divide(#0,n2)|subtract(#1,n0)|divide(n1,#2)| | physics |
if ( a - b - c + d = 18 ) and ( a + b - c - d = 6 ) , what is the value of ( b - d ) ^ 2 ? | "eq 1 : a - b - c + d = 18 eq 2 : a + b - c - d = 6 ( 1 ) subtract eq 1 from eq 2 a - b - c + d = 18 - a + b - c - d = 6 - - - - - - - - - - - - - - - - - - - - - - - - - 2 b + 2 d = 12 ( 2 ) simplify - b + d = 6 b - d = - 6 ( b - d ) ^ 2 = ( - 6 ) ^ 2 = 36 my answer : e" | a ) 4 . , b ) 8 . , c ) 12 . , d ) 16 . , e ) 36 . | e | power(subtract(6, divide(add(18, 6), 2)), 2) | add(n0,n1)|divide(#0,n2)|subtract(n1,#1)|power(#2,n2)| | general |
a car traveling at a certain constant speed takes 5 seconds longer to travel 1 km than it would take to travel 1 km at 60 km / hour . at what speed , in km / hr , is the car traveling ? | "time to cover 1 kilometer at 80 kilometers per hour is 1 / 60 hours = 3,600 / 60 seconds = 60 seconds ; time to cover 1 kilometer at regular speed is 60 + 5 = 65 seconds = 65 / 3,600 hours = 1 / 55 hours ; so , we get that to cover 1 kilometer 1 / 55 hours is needed - - > regular speed 55 kilometers per hour ( rate is... | a ) 70 , b ) 72 , c ) 74 , d ) 75 , e ) 55 | e | divide(1, divide(add(multiply(const_3600, divide(1, 60)), 5), const_3600)) | divide(n1,n3)|multiply(#0,const_3600)|add(n0,#1)|divide(#2,const_3600)|divide(n1,#3)| | physics |
what is the probability of getting 53 mondays in a leap year ? | "in a leap year contains 52 monday ' s since 52 weeks are present and 2 days extra it may be sunday and monday 2 . monday and tuesday 3 . tuesday and thursday 4 . thursday and friday similary 7 alternatives are possible . among this only 2 are possible cases so 2 / 7 answer : b" | a ) 1 / 7 , b ) 2 / 7 , c ) 3 / 7 , d ) 4 / 7 , e ) 5 / 7 | b | divide(const_2, choose(add(const_3, const_3), const_3)) | add(const_3,const_3)|choose(#0,const_3)|divide(const_2,#1)| | probability |
if a and b are positive integers and ( 5 ^ a ) ^ b = 5 ^ 2 , what is the value of 2 ^ a * 2 ^ b ? | "5 ^ ab = 5 ^ 2 therefore ab = 2 either a = 1 or 2 or b = 2 or 1 therefore 5 ^ a * 5 ^ b = 5 ^ ( a + b ) = 5 ^ 3 = 125 e" | a ) 3125 , b ) 625 , c ) 5 , d ) 25 , e ) 125 | e | multiply(power(5, 5), 5) | power(n2,n0)|multiply(n0,#0)| | general |
if w is the set of all the integers between 10 and 332 , inclusive , that are either multiples of 3 or multiples of 2 or multiples of both , then w contains how many numbers ? | "official solution : number of multiples of 3 step 1 . subtract the extreme multiples of 3 within the range ( the greatest is 330 , the smallest is 12 ) : 330 - 12 = 318 step 2 . divide by 3 : 318 / 3 = 106 step 3 . add 1 : 106 + 1 = 107 . so there are 107 multiples of 3 within the range : examples are 51 , 54 , 57 , 6... | a ) 111 , b ) 28 , c ) 160 , d ) 213 , e ) 107 | d | subtract(add(floor(divide(subtract(332, 10), 3)), divide(subtract(332, 10), 2)), floor(divide(subtract(332, 10), multiply(2, 3)))) | multiply(n3,n2)|subtract(n1,n0)|divide(#1,n3)|divide(#1,n2)|divide(#1,#0)|floor(#3)|floor(#4)|add(#2,#5)|subtract(#7,#6)| | other |
a tank with a volume of 30 cubic feet has one inlet pipe and 2 outlet pipes . the inlet pipe fills water into the tank at the rate of 3 cubic inches / min and the 2 outlet pipes empty it out at the rates of 12 cubic inches / min and 6 cubic inches / min respectively . if all 3 pipes are opened when the tank is full , h... | the tank is emptied at this rate : 12 + 6 - 3 = 15 cubic inches / min the tank has a volume of 30 * 12 * 12 * 12 = 51840 cubic inches . the time it takes to empty the tank is 51840 / 15 = 3456 minutes . the answer is b . | ['a ) 2345', 'b ) 3456', 'c ) 4567', 'd ) 5678', 'e ) 6789'] | b | divide(multiply(30, power(12, const_3)), subtract(add(12, 6), 3)) | add(n4,n5)|power(n4,const_3)|multiply(n0,#1)|subtract(#0,n2)|divide(#2,#3) | physics |
an investment yields an interest payment of $ 231 each month . if the simple annual interest rate is 9 % , what is the amount of the investment ? | "let the principal amount = p simple annual interest = 9 % simple monthly interest = ( 9 / 12 ) = ( 3 / 4 ) % ( 3 / 4 ) * ( p / 100 ) = 230 = > p = ( 231 * 4 * 10 ^ 2 ) / 3 = 77 * 4 * 10 ^ 2 = 308 * 10 ^ 2 = 30800 answer b" | a ) $ 28,300 , b ) $ 30,800 , c ) $ 31,300 , d ) $ 32,500 , e ) $ 35,100 | b | multiply(divide(231, divide(9, multiply(const_3, const_4))), const_100) | multiply(const_3,const_4)|divide(n1,#0)|divide(n0,#1)|multiply(#2,const_100)| | gain |
if the price of sugar rises from rs . 6 per kg to rs . 7.50 per kg , a person , to have no increase in the expenditure on sugar , will have to reduce his consumption of sugar by | "sol . let the original consumption = 100 kg and new consumption = x kg . so , 100 x 6 = x × 7.50 = x = 80 kg . ∴ reduction in consumption = 20 % . answer b" | a ) 15 % , b ) 20 % , c ) 25 % , d ) 30 % , e ) none | b | multiply(subtract(const_1, divide(multiply(const_1, 6), 7.50)), const_100) | multiply(n0,const_1)|divide(#0,n1)|subtract(const_1,#1)|multiply(#2,const_100)| | general |
there are cats got together and decided to kill the mice of 999975 . each cat kills equal number of mice and each cat kills more number of mice than cats there were . then what are the number of cats ? | 999975 can be written as 1000000 â € “ 25 = 10002 â € “ 52 ie of the form a 2 - b 2 = ( a + b ) ( a - b ) = ( 1000 + 5 ) * ( 1000 - 5 ) = ( 1005 ) * ( 995 ) given that number of cats is less than number if mice . so number of cats is 995 and number of mice were 1005 answer c | a ) 941,1009 , b ) 991,1001 , c ) 995,1005 , d ) 791,1009 , e ) 931,1009 | c | divide(999975, add(multiply(const_100, const_10), add(const_3, const_2))) | add(const_2,const_3)|multiply(const_10,const_100)|add(#0,#1)|divide(n0,#2) | general |
how many positive integers less than 50 have a reminder 01 when divided by 5 ? | "1 also gives the remainder of 1 when divided by 5 . so , there are total of 10 numbers . answer : a ." | a ) 10 , b ) 14 , c ) 15 , d ) 16 , e ) 17 | a | divide(factorial(subtract(add(const_4, 01), const_1)), multiply(factorial(01), 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 |
natasha climbs up a hill , and descends along the same way she went up . it takes her 4 hours to reach the top and 2 hours to come back down . if her average speed along the whole journey is 3.5 kilometers per hour , what was her average speed ( in kilometers per hour ) while climbing to the top ? | "let the distance to the top be x , so the total distance traveled by natasha is 2 x . the total time is 4 + 2 = 6 hours the average speed = total distance / total time taken = 2 x / 6 = x / 3 the average speed of the complete journey is 3.5 km / hour x / 3 = 3.5 x = 10.5 km the average speed while climbing = distance ... | a ) 2.125 , b ) 2.375 , c ) 2.625 , d ) 2.675 , e ) 2.825 | c | divide(divide(multiply(add(4, 2), 3.5), 2), 4) | add(n0,n1)|multiply(n2,#0)|divide(#1,n1)|divide(#2,n0)| | physics |
how many positive integers d between 200 and 300 ( both inclusive ) are not divisible by 2 , 3 or 5 ? | 1 ) i figured there are 101 integers ( 300 - 200 + 1 = 101 ) . since the set begins with an even and ends with an even , there are 51 evens . 2 ) question says integers are not divisible by 2 , leaving all of the odds ( 101 - 51 = 50 integers ) . 3 ) question says integers are not divisible by 5 , removing all the inte... | a ) 3 , b ) 16 , c ) 75 , d ) 24 , e ) 26 | e | subtract(add(const_100, const_1), subtract(add(add(add(add(divide(const_100, const_2), const_1), add(divide(subtract(const_100, const_1), const_3), const_1)), add(divide(const_100, 5), const_1)), add(divide(subtract(const_100, const_10), multiply(multiply(3, 5), 2)), const_1)), add(add(add(divide(subtract(const_100, mu... | add(const_1,const_100)|divide(const_100,const_2)|divide(const_100,n4)|divide(const_100,const_10)|multiply(n3,n4)|multiply(const_2,const_3)|subtract(const_100,const_1)|subtract(const_100,const_10)|add(#1,const_1)|add(#2,const_1)|add(#3,const_1)|divide(#6,const_3)|divide(#7,#4)|multiply(n2,#4)|subtract(const_100,#5)|add(... | other |
the number of the members of a club is more than 50 and less than 80 . when 5 people sit at a table , other people exactly divided to 8 - people groups ( 8 people sit at a table ) or 7 - people groups ( 7 people sit at a table ) . if the members are divided to 9 - people groups , how many people will be left ? | "the number of members is 7 k + 5 = 8 j + 5 the only number in this range which satisfies this is 61 . 61 / 9 = 6 ( 9 ) + 7 the answer is d ." | a ) 4 , b ) 5 , c ) 6 , d ) 7 , e ) 8 | d | reminder(add(lcm(8, 7), 5), 9) | lcm(n3,n5)|add(n2,#0)|reminder(#1,n7)| | general |
in a certain flower shop , which stocks 4 types of flowers , there are 1 / 3 as many violets as carnations , and 1 / 3 as many tulips as violets . if there are equal numbers of roses and tulips , what percent of the flowers in the shop are carnations ? | given : - violets - c / 3 carnations - c tulip - c / 9 rose - c / 9 total flowers in terms of c = c / 3 + c + c / 9 + c / 9 = 14 c / 9 percentage of carnations = c / 14 c / 9 * 100 = 64.28 % answer b | a ) 63 , b ) 64.28 , c ) 65 , d ) 66 , e ) 68 | b | divide(const_100, add(add(add(divide(const_1, multiply(3, 3)), divide(const_1, multiply(3, 3))), const_1), const_0_33)) | multiply(n2,n2)|divide(const_1,#0)|add(#1,#1)|add(#2,const_1)|add(#3,const_0_33)|divide(const_100,#4) | general |
a batsman scored 90 runs which included 4 boundaries and 6 sixes . what percent of his total score did he make by running between the wickets ? | "explanation : number of runs made by running , = > 90 − ( 4 × 4 + 6 × 6 ) . = > 90 − 52 = > 38 hence , the required percentage is : - = > 38 / 90 * 100 = > 42.22 % answer : c" | a ) 45 % , b ) 500 / 11 , c ) 42.22 % , d ) 55 % , e ) 35 % | c | multiply(divide(subtract(90, add(multiply(4, 6), multiply(6, 4))), 90), const_100) | multiply(n1,n2)|multiply(n1,n2)|add(#0,#1)|subtract(n0,#2)|divide(#3,n0)|multiply(#4,const_100)| | general |
a train sets off at 9 : 00 am at the speed of 70 km / h . another train starts at 9 : 30 am in the same direction at the rate of 84 km / h . at what time will the second train catch the first train ? | "in thirty minutes the first train travels 35 km . the second train catches the first train at a rate of 84 km / h - 70 km / h = 14 km / h . the second train will catch the first train in 35 / 14 = 2.5 hours , so at 12 : 00 noon . the answer is c ." | a ) 11 : 00 , b ) 11 : 30 , c ) 12 : 00 , d ) 12 : 30 , e ) 1 : 00 | c | divide(add(70, multiply(70, divide(const_1, const_2))), subtract(84, 70)) | divide(const_1,const_2)|subtract(n5,n2)|multiply(n2,#0)|add(n2,#2)|divide(#3,#1)| | physics |
two brother x and y appeared for an exam . the probability of selection of x is 1 / 5 and that of b is 2 / 7 . find the probability that both of them are selected . | "explanation : let a be the event that x is selected and b is the event that y is selected . p ( a ) = 1 / 5 , p ( b ) = 2 / 7 . let c be the event that both are selected . p ( c ) = p ( a ) ã — p ( b ) as a and b are independent events : = ( 1 / 5 ) ã — ( 2 / 7 ) = 2 / 35 answer : a ) 2 / 35" | a ) 2 / 35 , b ) 2 / 30 , c ) 2 / 63 , d ) 2 / 29 , e ) 2 / 10 | a | multiply(divide(1, 5), divide(2, 7)) | divide(n0,n1)|divide(n2,n3)|multiply(#0,#1)| | general |
the volumes of two cones are in the ratio 1 : 30 and the radii of the cones are in the ratio of 1 : 2 . what is the length of the wire ? | "the volume of the cone = ( 1 / 3 ) π r 2 h only radius ( r ) and height ( h ) are varying . hence , ( 1 / 3 ) π may be ignored . v 1 / v 2 = r 1 ^ 2 . h 1 / r 2 ^ 2 . h 2 = > 1 / 30 = ( 1 ) ^ 2 h 1 / ( 2 ) ^ 2 h 2 = > h 1 / h 2 = 6 / 5 i . e . h 1 : h 2 = 6 : 5 answer : a" | a ) 6 : 5 , b ) 6 : 9 , c ) 6 : 2 , d ) 2 : 2 , e ) 2 : 8 | a | divide(divide(1, 30), power(divide(1, 2), 2)) | divide(n0,n1)|divide(n0,n3)|power(#1,n3)|divide(#0,#2)| | geometry |
the average of runs of a cricket player of 10 innings was 42 . how many runs must he make in his next innings so as to increase his average of runs by 4 ? | "average after 11 innings = 46 required number of runs = ( 46 * 11 ) - ( 42 * 10 ) = 506 - 420 = 86 . answer : b" | a ) 87 , b ) 86 , c ) 28 , d ) 76 , e ) 80 | b | subtract(multiply(add(10, const_1), add(4, 42)), multiply(10, 42)) | add(n0,const_1)|add(n1,n2)|multiply(n0,n1)|multiply(#0,#1)|subtract(#3,#2)| | general |
in a survey of parents , exactly 9 / 10 of the mothers and 3 / 4 of the fathers held full - time jobs . if 40 percent of the parents surveyed were women , what percent of the parents did not hold full - time jobs ? | "fathers without full - time jobs are 1 / 4 * 3 / 5 = 3 / 20 of all the parents surveyed . mothers without full - time jobs are 1 / 10 * 2 / 5 = 2 / 50 of all the parents surveyed . the percent of parents without full - time jobs is 3 / 20 + 2 / 50 = 19 / 100 = 19 % the answer is c ." | a ) 27 % , b ) 21 % , c ) 19 % , d ) 18 % , e ) 16 % | c | add(subtract(subtract(const_100, 40), multiply(divide(3, 4), subtract(const_100, 40))), subtract(40, multiply(divide(9, 10), 40))) | divide(n2,n3)|divide(n0,n1)|subtract(const_100,n4)|multiply(#0,#2)|multiply(n4,#1)|subtract(#2,#3)|subtract(n4,#4)|add(#5,#6)| | general |
cereal a is 8 % sugar by weight , whereas healthier but less delicious cereal b is 2 % sugar by weight . to make a delicious and healthy mixture that is 4 % sugar , what should be the ratio of cereal a to cereal b , by weight ? | "( 8 / 100 ) a + ( 2 / 100 ) b = ( 4 / 100 ) ( a + b ) 4 a = 2 b = > a / b = 1 / 2 answer is c ." | a ) 2 : 9 , b ) 2 : 7 , c ) 1 : 2 , d ) 1 : 4 , e ) 1 : 3 | c | divide(subtract(4, 2), subtract(8, 4)) | subtract(n2,n1)|subtract(n0,n2)|divide(#0,#1)| | general |
the ratio of boarders to day students at a school was originally 2 to 5 . however , after a number of new boarders join the initial 120 boarders , the ratio changed to 1 to 2 . if no boarders became day students and vice versa , and no students left the school , how many new boarders joined the school ? | "let x be the number of new boarders . the ratio changed from 2 : 5 = 4 : 10 up to 1 : 2 = 5 : 10 . 120 / ( 120 + x ) = 4 / 5 x = 30 the answer is a ." | a ) 30 , b ) 50 , c ) 70 , d ) 80 , e ) 90 | a | subtract(divide(multiply(divide(120, 2), 5), 2), 120) | divide(n2,n0)|multiply(n1,#0)|divide(#1,n4)|subtract(#2,n2)| | other |
a 20 kg metal bar made of alloy of tin and silver lost 2 kg of its weight in the water . 10 kg of tin loses 1.375 kg in the water ; 5 kg of silver loses 0.375 kg . what is the ratio of tin to silver in the bar ? | "the bar lost certain percentage of its weight . we do n ' t know how much tin was lost and how much silver was lost but in all 2 kg was lost with is 10 % of its overall weight . tin loses 1.375 kg in 10 kg so 13.75 % of its weight when it is put in water . silver loses . 375 kg in 5 kg so . 375 / 5 * 100 = 7.5 % of it... | a ) 1 / 4 , b ) 2 / 5 , c ) 1 / 2 , d ) 3 / 5 , e ) 2 / 3 | e | divide(divide(subtract(2, multiply(divide(0.375, 5), 20)), subtract(divide(1.375, 10), divide(0.375, 5))), subtract(20, divide(subtract(2, multiply(divide(0.375, 5), 20)), subtract(divide(1.375, 10), divide(0.375, 5))))) | divide(n5,n4)|divide(n3,n2)|multiply(n0,#0)|subtract(#1,#0)|subtract(n1,#2)|divide(#4,#3)|subtract(n0,#5)|divide(#5,#6)| | other |
the radius of a cylinder is 10 m , height 12 m . the volume of the cylinder is : | "cylinder volume = ï € r ( power 2 ) h = 22 / 7 ã — 10 ã — 10 ã — 12 = 3771.4 m ( power 3 ) answer is d ." | a ) 2200 , b ) 5500 , c ) 3300 , d ) 3771.4 , e ) 4400 | d | multiply(circumface(10), 12) | circumface(n0)|multiply(n1,#0)| | geometry |
what amount does an investor receive if the investor invests $ 6000 at 10 % p . a . compound interest for two years , compounding done annually ? | a = ( 1 + r / 100 ) ^ n * p ( 1.1 ) ^ 2 * 6000 = 1.21 * 6000 = 7260 the answer is d . | a ) $ 6980 , b ) $ 7070 , c ) $ 7120 , d ) $ 7260 , e ) $ 7340 | d | multiply(6000, power(add(const_1, divide(10, const_100)), const_2)) | divide(n1,const_100)|add(#0,const_1)|power(#1,const_2)|multiply(n0,#2) | gain |
the racing magic takes 50 seconds to circle the racing track once . the charging bull makes 40 rounds of the track in an hour . if they left the starting point together , how many minutes will it take for them to meet at the starting point for the second time ? | "time taken by racing magic to make one circle = 50 seconds time taken bycharging bullto make one circle = 60 mins / 40 = 1.5 mins = 90 seconds lcm of 90 and 50 seconds = 450 seconds time taken for them to meet at the starting point for the second time = 450 * 2 = 900 seconds = 15 mins answer e" | a ) 3 , b ) 6 , c ) 9 , d ) 12 , e ) 15 | e | multiply(divide(multiply(multiply(50, divide(50, 40)), const_2), 50), const_2) | divide(n0,n1)|multiply(#0,n0)|multiply(#1,const_2)|divide(#2,n0)|multiply(#3,const_2)| | physics |
three pipes , a , b , & c are attached to a tank . a & b can fill it in 20 & 30 minutes respectively while c can empty it in 15 minutes . if a , b & c are kept open successively for 2 minute each , how soon will the tank be filled ? | "in three minute 1 / 20 + 1 / 30 - 1 / 15 = 1 / 60 part is filled 6 min - - - - - - - - 1 / 60 parts x min - - - - - - - - - 1 part ( full ) x = 360 min = 6 hours answer : e" | a ) 2 hours , b ) 4 hours , c ) 3 hours , d ) 5 hours , e ) 6 hours | e | add(add(2, 2), 2) | add(n3,n3)|add(n3,#0)| | physics |
a train 360 m long runs with a speed of 45 km / hr . what time will it take to pass a platform of 340 m long ? | "speed = 45 km / hr = 45 ã — ( 5 / 18 ) m / s = 150 / 12 = 50 / 4 = 25 / 2 m / s total distance = length of the train + length of the platform = 360 + 340 = 700 meter time taken to cross the platform = 700 / ( 25 / 2 ) = 700 ã — 2 / 25 = 56 seconds answer : e" | a ) 38 sec , b ) 35 sec , c ) 44 sec , d ) 40 sec , e ) 56 | e | multiply(divide(add(divide(340, const_1000), divide(360, const_1000)), 45), const_3600) | divide(n2,const_1000)|divide(n0,const_1000)|add(#0,#1)|divide(#2,n1)|multiply(#3,const_3600)| | physics |
in a certificate by mistake a candidate gave his height as 25 % more than actual height . in the interview panel , he clarified that his height was 5 feet 6 nches . find the % correction made by the candidate from his stated height to his actual height ? | his height was = 5 feet 6 inch = 6 + 60 = 66 inch . required % correction = 66 * ( 1.25 - 1 ) = 16.5 d | a ) 15 , b ) 20 , c ) 18 , d ) 16.5 , e ) 25 | d | multiply(divide(subtract(add(25, add(const_0_25, const_0_25)), 5), add(const_100, subtract(add(25, add(const_0_25, const_0_25)), 5))), const_100) | add(const_0_25,const_0_25)|add(n0,#0)|subtract(#1,n1)|add(#2,const_100)|divide(#2,#3)|multiply(#4,const_100) | gain |
there are 2 available positions and 50 candidates , one half of whom are democrats and another half are republicans . if it was decided that the positions would be filled at random , then what is the probability q that the both positions will be taken by members of just one party ? | "q probability of one party having both spots : ( 1 / 2 ) * ( 24 / 49 ) = 12 / 49 ( 1 / 2 ) or ( 25 / 50 ) because it does not matter which party or which person gets the first spot . ( 24 / 49 ) because after one person from a particular party is chosen , there are 24 members of the same party left out of 49 total can... | a ) 1 / 25 , b ) 12 / 49 , c ) 1 / 4 , d ) 24 / 49 , e ) 1 / 2 | d | multiply(multiply(divide(subtract(divide(50, 2), const_1), subtract(50, const_1)), divide(divide(50, 2), 50)), 2) | divide(n1,n0)|subtract(n1,const_1)|divide(#0,n1)|subtract(#0,const_1)|divide(#3,#1)|multiply(#4,#2)|multiply(n0,#5)| | other |
what is the smallest positive integer k such that the product of 3675 x k is a perfect square ? | "a perfect square , is just an integer that can be written as the square of some other integer . for example 16 = 4 ^ 2 , is a perfect square . now , 3675 = 5 ^ 2 * 7 ^ 2 * 3 , so if k = 3 then 3675 k = ( 5 * 7 * 3 ) ^ 2 , which is a perfect square ( basically the least positive value of k must complete only the power ... | a ) 3 , b ) 9 , c ) 15 , d ) 25 , e ) 63 | a | add(const_3, const_4) | add(const_3,const_4)| | general |
one pipe can fill a tank three times as fast as another pipe . if together the two pipes can fill the tank in 39 minutes , then the slower pipe alone will be able to fill the tank in | "explanation : let the slower pipe alone fill the tank in x minutes then faster will fill in x / 3 minutes . part filled by slower pipe in 1 minute = 1 / x part filled by faster pipe in 1 minute = 3 / x part filled by both in 1 minute = 1 / x + 3 / x = 1 / 39 = > 4 / x = 1 / 39 x = 39 ∗ 4 = 156 mins option d" | a ) 144 mins , b ) 140 mins , c ) 136 mins , d ) 156 minw , e ) none of these | d | multiply(add(const_1, const_4), 39) | add(const_1,const_4)|multiply(n0,#0)| | physics |
there are 7 non - collinear points . how many triangles can be drawn by joining these points ? | explanation : a triangle is formed by joining any three non - collinear points in pairs . there are 7 non - collinear points the number of triangles formed = \ inline { \ color { black } 7 c _ { 3 } } = 35 answer : c ) 35 | a ) 22 , b ) 38 , c ) 35 , d ) 29 , e ) 18 | c | multiply(7, add(const_4, const_1)) | add(const_1,const_4)|multiply(n0,#0) | geometry |
a man buys 58 pens at marked price of 46 pens from a whole seller . if he sells these pens giving a discount of 1 % , what is the profit percent ? | "explanation : let marked price be re . 1 each c . p . of 58 pens = rs . 46 s . p . of 58 pens = 99 % of rs . 58 = rs . 57.42 profit % = ( profit / c . p . ) x 100 profit % = ( 11.42 / 46 ) x 100 = 24.82 % answer : c" | a ) 7.6 % , b ) 7.7 % , c ) 24.82 % , d ) 13.6 % , e ) 27.82 % | c | multiply(const_100, divide(subtract(multiply(58, subtract(const_100, 1)), multiply(46, const_100)), multiply(46, const_100))) | multiply(n1,const_100)|subtract(const_100,n2)|multiply(n0,#1)|subtract(#2,#0)|divide(#3,#0)|multiply(#4,const_100)| | gain |
the sale price sarees listed for rs . 200 after successive discount is 10 % and 5 % is ? | "200 * ( 90 / 100 ) * ( 95 / 100 ) = 171 answer : e" | a ) 187 , b ) 169 , c ) 172 , d ) 178 , e ) 171 | e | subtract(subtract(200, divide(multiply(200, 10), const_100)), divide(multiply(subtract(200, divide(multiply(200, 10), const_100)), 5), const_100)) | multiply(n0,n1)|divide(#0,const_100)|subtract(n0,#1)|multiply(n2,#2)|divide(#3,const_100)|subtract(#2,#4)| | gain |
if josh , doug , and brad have a total of $ 68 between them , and josh has two times as much money as brad but only 3 - fourths as much as doug , how much money does doug have ? | josh + doug + brad = 68 ; josh = 2 brad , josh = 3 / 4 doug josh + 1 / 2 josh + 4 / 3 josh = 68 ( substituted the given values ) josh = 24 . 24 = 3 / 4 doug = > doug = 32 answer is d . | a ) $ 8 , b ) $ 9 , c ) $ 27 , d ) $ 32 , e ) $ 36 | d | multiply(divide(68, add(add(const_1, divide(const_4, const_3)), divide(const_1, const_2))), divide(const_4, const_3)) | divide(const_4,const_3)|divide(const_1,const_2)|add(#0,const_1)|add(#2,#1)|divide(n0,#3)|multiply(#4,#0) | general |
the bankers discount of a certain sum of money is rs . 144 and the true discount on the same sum for the same time is rs . 120 . the sum due is : | sum = ( b . d * t . d ) / ( b . d - t . d ) ( 144 * 120 ) / 144 - 120 ; 720 answer : d | a ) 715 , b ) 716 , c ) 718 , d ) 720 , e ) 722 | d | divide(multiply(120, 144), subtract(144, 120)) | multiply(n0,n1)|subtract(n0,n1)|divide(#0,#1) | gain |
tammy climbed a mountain in two days . she spent a total of 14 hours climbing the mountain . on the second day , she walked at an average speed that was half a kilometer per hour faster , but 2 hours less than what she walked on the first day . if the total distance she climbed during the two days is 52 kilometers , ho... | "ans : c total time = 14 hrs let time traveled during 1 st day = x let time traveled during 2 nd day = x - 2 total time = 14 x + x - 2 = 14 x = 8 speed * time = distance s * 8 + ( s + 0.5 ) ( 8 - 2 ) = 52 solving s = 4.5 now speed for 2 nd day is 0.5 less than the 1 st day which is 4.5 thus speed for 2 nd day = 4 its s... | a ) 3 , b ) 3.5 , c ) 4 , d ) 4.5 , e ) 6 | c | add(divide(subtract(52, divide(subtract(divide(add(14, 2), 2), 2), 2)), add(subtract(divide(add(14, 2), 2), 2), divide(add(14, 2), 2))), divide(const_1, 2)) | add(n0,n1)|divide(const_1,n1)|divide(#0,n1)|subtract(#2,n1)|add(#2,#3)|divide(#3,n1)|subtract(n2,#5)|divide(#6,#4)|add(#7,#1)| | physics |
two trains of length 120 m and 280 m are running towards each other on parallel lines at 42 kmph and 30 kmph respectively . in what time will they be clear of each other from the moment they meet ? | "relative speed = ( 42 + 30 ) * 5 / 18 = 4 * 5 = 20 mps . distance covered in passing each other = 120 + 280 = 400 m . the time required = d / s = 400 / 20 = 20 sec . answer : d" | a ) 22 , b ) 12 , c ) 67 , d ) 20 , e ) 81 | d | divide(add(120, 280), multiply(add(42, 30), const_0_2778)) | add(n0,n1)|add(n2,n3)|multiply(#1,const_0_2778)|divide(#0,#2)| | physics |
pipe a fills a tank of capacity 950 liters at the rate of 40 liters a minute . another pipe b fills the same tank at the rate of 30 liters a minute . a pipe at the bottom of the tank drains the tank at the rate of 20 liters a minute . if pipe a is kept open for a minute and then closed and pipe b is open for a minute a... | in one cycle they fill 40 + 30 - 20 = 50 liters 950 = 50 * n = > n = 19 here n = number of cycles . total time = 19 * 3 = 57 as in one cycle there are 3 minutes . thus 57 minutes answer : a | a ) 57 minutes , b ) 14 minutes , c ) 39 minutes , d ) 40 minutes 20 seconds , e ) none of these | a | multiply(divide(950, subtract(add(40, 30), 20)), const_3) | add(n1,n2)|subtract(#0,n3)|divide(n0,#1)|multiply(#2,const_3) | physics |
during a sale , the price of a pair of shoes is marked down 12 % from the regular price . after the sale ends , the price goes back to the original price . what is the percent of increase to the nearest percent from the sale price back to the regular price for the shoes ? | "assume the price = 100 price during sale = 88 price after sale = 100 percent increase = 12 / 88 * 100 = 14 % approx . correct option : d" | a ) 9 % , b ) 10 % , c ) 11 % , d ) 14 % , e ) 90 % | d | divide(multiply(12, const_100), subtract(const_100, 12)) | multiply(n0,const_100)|subtract(const_100,n0)|divide(#0,#1)| | gain |
if p / q = 5 / 4 , then 2 p + q = ? | "let p = 5 , q = 4 then 2 * 5 + 4 = 14 so 2 p + q = 14 . answer : b" | a ) 12 , b ) 14 , c ) 13 , d ) 15 , e ) 16 | b | add(multiply(5, 2), 4) | multiply(n0,n2)|add(n1,#0)| | general |
find the missing figures : 0.75 % of ? = 0.06 | "let 0.75 % of x = 0.06 . then , 0.75 * x / 100 = 0.06 x = [ ( 0.06 * 100 ) / 0.75 ] = 8 . answer is c ." | a ) 4 , b ) 6 , c ) 8 , d ) 10 , e ) 12 | c | divide(0.06, divide(0.75, const_100)) | divide(n0,const_100)|divide(n1,#0)| | gain |
m is 30 % of q . q is 20 % of p . n is 50 % of p . find m : n ratio | let p be 100 n = 50 % of 100 ( p = 100 ) = 50 q = 20 % of 100 ( p = 100 ) = 20 m = 30 % of 20 ( q = 20 ) = 6 m : n = 6 : 50 m : n = 3 : 25 answer : c | a ) 1 : 25 , b ) 2 : 25 , c ) 3 : 25 , d ) 4 : 25 , e ) 5 : 25 | c | divide(multiply(divide(30, const_100), divide(20, const_100)), divide(50, const_100)) | divide(n0,const_100)|divide(n1,const_100)|divide(n2,const_100)|multiply(#0,#1)|divide(#3,#2) | other |
a alone can do a piece of work in 6 days and b alone in 8 days . a and b undertook to do it for rs . 5000 . with the help of c , they completed the work in 3 days . how much is to be paid to c ? | "c ' s 1 day ' s work = 1 / 3 - ( 1 / 6 + 1 / 8 ) = 1 / 3 - 7 / 24 = 1 / 24 a ' s wages : b ' s wages : c ' s wages = 1 / 6 : 1 / 8 : 1 / 24 = 4 : 3 : 1 c ' s share ( for 3 days ) = rs . ( 3 * 1 / 24 * 5000 ) = rs . 625 answer = c" | a ) s . 375 , b ) s . 425 , c ) s . 625 , d ) s . 800 , e ) s . 850 | c | multiply(multiply(subtract(inverse(3), add(inverse(8), inverse(6))), 5000), 3) | inverse(n3)|inverse(n1)|inverse(n0)|add(#1,#2)|subtract(#0,#3)|multiply(n2,#4)|multiply(n3,#5)| | physics |
a box contains 9 pairs of shoes ( 18 shoes in total ) . if two shoes are selected at random , what it is the probability that they are matching shoes ? | "the problem with your solution is that we do n ' t choose 1 shoe from 18 , but rather choose the needed one after we just took one and need the second to be the pair of it . so , the probability would simply be : 1 / 1 * 1 / 17 ( as after taking one at random there are 17 shoes left and only one is the pair of the fir... | a ) 1 / 190 , b ) 1 / 17 , c ) 1 / 19 , d ) 1 / 10 , e ) 1 / 9 | b | divide(const_1, subtract(18, const_1)) | subtract(n1,const_1)|divide(const_1,#0)| | general |
a marching band of 240 musicians are to march in a rectangular formation with s rows of exactly t musicians each . there can be no less than 8 musicians per row and no more than 30 musicians per row . how many different rectangular formations x are possible ? | the combinations could be { ( 1,240 ) , ( 2,120 ) , ( 3,80 ) , ( 4,60 ) , ( 5,48 ) , ( 6,40 ) , ( 8,30 ) , ( 10,24 ) , ( 12,20 ) , ) 15,16 ) , ( 16,15 ) , ( 20,12 ) , ( 24,10 ) , ( 30,8 ) , ( 40,6 ) , ( 48,5 ) , ( 60,4 ) , ( 80,3 ) , ( 120,2 ) , ( 240,1 ) } of these we are told 8 < = t < = 30 so we can remove these pai... | a ) 3 , b ) 4 , c ) 5 , d ) 6 , e ) x = 8 | e | divide(240, 30) | divide(n0,n2) | general |
what is the lowest positive integer that is divisible by each of the integers 1 through 7 , inclusive ? | "the integer should be divisible by : 2 , 3 , 4 ( = 2 ^ 2 ) , 5 , 6 ( = 2 * 3 ) , and 7 . the least common multiple of these integers is lcm = 2 ^ 2 * 3 * 5 * 7 = 420 . answer : a ." | a ) 420 , b ) 840 , c ) 1,260 , d ) 2,520 , e ) 5,040 | a | lcm(1, 7) | lcm(n0,n1)| | general |
the salary of a typist was at first raised by 10 % and then the same was reduced by 5 % . if he presently draws rs . 5225 . what was his original salary ? | "x * ( 110 / 100 ) * ( 95 / 100 ) = 5225 x * ( 11 / 10 ) * ( 1 / 100 ) = 55 x = 5000 answer : b" | a ) 2277 , b ) 5000 , c ) 1000 , d ) 2651 , e ) 1971 | b | divide(5225, multiply(add(const_1, divide(10, const_100)), subtract(const_1, divide(5, const_100)))) | divide(n0,const_100)|divide(n1,const_100)|add(#0,const_1)|subtract(const_1,#1)|multiply(#2,#3)|divide(n2,#4)| | gain |
the regular price per can of a certain brand of soda is $ 0.15 . if the regular price per can is discounted 10 percent when the soda is purchased in 24 - can cases , what is the price of 75 cans of this brand of soda purchased in 24 - can cases ? | the discounted price of one can of soda is ( 0.9 ) ( $ 0.15 ) , or $ 0.135 therefore , the price of 75 cans of soda at the discounted price would be ( 75 ) ( $ 0.135 ) = 10.125 answer : c . | a ) $ 16.32 , b ) $ 18.00 , c ) $ 10.125 , d ) $ 24.48 , e ) $ 28.80 | c | multiply(divide(subtract(const_100, 10), const_100), multiply(0.15, 75)) | multiply(n0,n3)|subtract(const_100,n1)|divide(#1,const_100)|multiply(#2,#0) | gain |
solve : - 666 x 666 x 666 + 555 x 555 x 555 = ? ( 666 x 666 - 666 x 555 + 555 x 555 ) | given exp . = ( a 3 + b 3 ) = ( a + b ) = ( 666 + 555 ) = 1221 ( a 2 - ab + b 2 ) answer d | a ) 888 , b ) 333 , c ) 555 , d ) 1221 , e ) 889 | d | divide(add(multiply(multiply(666, 666), 666), multiply(multiply(555, 555), 555)), subtract(add(multiply(666, 666), multiply(555, 555)), multiply(666, 555))) | multiply(n0,n0)|multiply(n3,n3)|multiply(n0,n3)|add(#0,#1)|multiply(n0,#0)|multiply(n3,#1)|add(#4,#5)|subtract(#3,#2)|divide(#6,#7) | general |
when positive integer x is divided by positive integer y , the remainder is 11.52 . if x / y = 96.12 , what is the value of y ? | "when positive integer x is divided by positive integer y , the remainder is 11.52 - - > x = qy + 11.52 ; x / y = 96.12 - - > x = 96 y + 0.12 y ( so q above equals to 96 ) ; 0.12 y = 11.52 - - > y = 96 . answer : a ." | a ) 96 , b ) 75 , c ) 48 , d ) 25 , e ) 12 | a | divide(11.52, subtract(96.12, floor(96.12))) | floor(n1)|subtract(n1,#0)|divide(n0,#1)| | general |
if 213 × 16 = 3408 , then 0.016 × 2.13 is equal to : | "solution 0.016 × 2.13 = ( 16 / 1000 x 213 / 100 ) = ( 16 / 1000 x 213 / 100 ) = 3408 / 100000 = 0.03408 . answer e" | a ) 0.3408 , b ) 3.408 , c ) 34.08 , d ) 340.8 , e ) 0.03408 | e | multiply(0.016, 2.13) | multiply(n4,n3)| | general |
in a certain accounting class of 90 students , 70 % of the students took the final exam on the assigned day while the rest of the students took the exam on a make - up date . if the students on the assigned day had an average score of 65 % , and the students on the make - up date had an average score of 85 % , what was... | "70 % of the students took the final exam on the assigned day - - > 0.7 * 90 = 63 students took the final exam on the assigned day , with an average score of 65 . the rest of the students took the exam on a make - up date - - > 90 - 63 = 27 students took the exam on a make - up date , with an average score of 85 . ( av... | a ) 68 % , b ) 71 % , c ) 73 % , d ) 75 % , e ) 79 % | b | divide(add(multiply(70, 65), multiply(85, subtract(90, 70))), 90) | multiply(n1,n2)|subtract(n0,n1)|multiply(n3,#1)|add(#0,#2)|divide(#3,n0)| | general |
3 - 5 * 6 + 2 = | 3 - 5 * 6 + 2 = 5 - 30 = - 25 . . . so the answer is - 25 . . . e | a ) a ) 2 , b ) b ) - 13 , c ) c ) - 18 , d ) d ) - 17 , e ) e ) - 25 | e | multiply(subtract(3, 5), add(6, 2)) | add(n2,n3)|subtract(n0,n1)|multiply(#0,#1) | general |
75 boys can complete a work in 36 days . how many men need to complete twice the work in 20 days | one man can complete the work in 36 * 75 = 2700 days = one time work to complete the work twice it will be completed in let m be the no . of worker assign for this therefore the eqn becomes m * 20 = 2 * 2700 m = 270 workers answer : c | a ) 160 , b ) 170 , c ) 270 , d ) 190 , e ) 200 | c | divide(multiply(75, multiply(36, const_2)), 20) | multiply(n1,const_2)|multiply(n0,#0)|divide(#1,n2) | physics |
ratio of the earnings of a and b is 4 : 7 . if the earnings of a increases by 50 % and those of b decreased by 25 % , the new ratio of their earnings becomes 8 : 7 . what are a ' s earnings ? | let the original earnings of a and b be rs . 4 x and rs . 7 x . new earnings of a = 150 % 0 f rs . 4 x = ( 150 / 100 * 4 x ) = rs . 6 x new earnings of b = 75 % of rs . 7 x = ( 75 / 100 * 7 x ) = rs . 21 x / 4 6 x : 21 x / 4 = 8 : 7 this does not give x . so , the given data is inadequate . answer : d | a ) 8 : 6 , b ) 8 : 9 , c ) 8 : 3 , d ) 8 : 7 , e ) 8 : 1 | d | divide(8, 7) | divide(n4,n1) | general |
the expression x # y denotes the product of the consecutive multiples of 3 between x and y , inclusive . what is the sum of the exponents in the prime factorization of 21 # 30 ? | "first , let ' s translate the expression 21 # 30 , using the definition given : 21 # 30 = 21 × 24 × 27 × 30 we need the prime factorization of this product . let ' s factor out 3 from each multiple . 21 # 30 = 3 ^ 4 ( 7 × 8 × 9 × 10 ) now let ' s replace each consecutive integer with its prime factorization : 21 # 30 ... | a ) 10 , b ) 11 , c ) 12 , d ) 13 , e ) 14 | c | add(add(add(3, add(const_2, const_2)), add(const_2, const_2)), add(3, add(const_2, const_2))) | add(const_2,const_2)|add(n0,#0)|add(#1,#0)|add(#2,#1)| | general |
what is the square root of 144 , divided by 6 ? | "square root is a number times itself square root of 144 = 12 , 12 / 6 = 2 ( c ) 2" | a ) 9 , b ) 36 , c ) 2 , d ) 6 , e ) 1 | c | divide(sqrt(144), 6) | sqrt(n0)|divide(#0,n1)| | other |
the diameter of the driving wheel of a bus in 140 cm . how many revolutions per minute must the wheel make in order to keep a speed of 20 kmph ? | distance covered in 1 min = ( 20 * 1000 ) / 60 = 333.33 m circumference of the wheel = ( 2 * ( 22 / 7 ) * . 70 ) = 4.4 m no of revolution per min = 333.33 / 4.4 = 75.75 answer : d | a ) 21.21 , b ) 20.25 , c ) 32.25 , d ) 75.75 , e ) 62.61 | d | divide(divide(multiply(20, const_1000), const_60), multiply(multiply(divide(add(20, const_2), add(const_4, const_3)), const_2), divide(divide(140, const_100), const_2))) | add(n1,const_2)|add(const_3,const_4)|divide(n0,const_100)|multiply(n1,const_1000)|divide(#3,const_60)|divide(#2,const_2)|divide(#0,#1)|multiply(#6,const_2)|multiply(#5,#7)|divide(#4,#8) | physics |
a grocer has 400 pounds of coffee in stock , 30 percent of which is decaffeinated . if the grocer buys another 100 pounds of coffee of which 60 percent is decaffeinated , what percent , by weight , of the grocer ’ s stock of coffee is decaffeinated ? | "1 . 30 % of 400 = 120 pounds of decaffeinated coffee 2 . 60 % of 100 = 60 pounds of decaffeinated coffee 3 . wt have 180 pounds of decaffeinated out of 500 pounds , that means 180 / 500 * 100 % = 36 % . the correct answer is d ." | a ) 28 % , b ) 30 % , c ) 36 % , d ) 36 % , e ) 40 % | d | multiply(divide(add(multiply(divide(30, 100), 400), multiply(100, divide(60, 100))), add(400, 100)), 100) | add(n0,n2)|divide(n1,n2)|divide(n3,n2)|multiply(n0,#1)|multiply(n2,#2)|add(#3,#4)|divide(#5,#0)|multiply(#6,n2)| | gain |
a boat having a length 3 m and breadth 2 m is floating on a lake . the boat sinks by 1 cm when a man gets into it . the mass of the man is ? | "in this type of question , first we will calculate the volume of water displaces then will multiply with the density of water . volume of water displaced = 3 * 2 * 0.01 = 0.06 m cube mass of man = volume of water displaced * density of water = 0.06 * 1000 = 60 kg answer : b" | a ) 77 kg , b ) 60 kg , c ) 77 kg , d ) 55 kg , e ) 47 kg | b | multiply(multiply(multiply(3, 2), divide(1, const_100)), const_1000) | divide(n2,const_100)|multiply(n0,n1)|multiply(#0,#1)|multiply(#2,const_1000)| | physics |
a cyclist covers a distance of 500 m in 4 min 30 sec . what is the speed in km / hr of the cyclist ? | distance = 500 m time = 4 min 10 sec = 250 sec speed = ( 500 / 250 ) m / s = 2 m / s speed in km / hr = 2 * 18 / 5 = 3.2 km / hr correct option is c | a ) 2 , b ) 2.5 , c ) 3.2 , d ) 4.6 , e ) 2.8 | c | multiply(divide(500, add(multiply(4, const_60), 30)), const_3_6) | multiply(n1,const_60)|add(n2,#0)|divide(n0,#1)|multiply(#2,const_3_6) | physics |
the number n is 76 h 4 , where h represents the 10 ' s digit . if n is divisible by 9 , what is the value of h ? | integer is divisible by 9 - sum of digits is divisible by 9 answer : a | a ) 1 , b ) 3 , c ) 5 , d ) 7 , e ) 9 | a | subtract(9, reminder(add(multiply(76, const_100), 4), 9)) | multiply(n0,const_100)|add(n1,#0)|reminder(#1,n3)|subtract(n3,#2) | general |
if a boat is rowed downstream for 96 km in 3 hours and upstream for 88 km in 11 hours , what is the speed of the boat and the river ? | explanation : if x : speed of boats man in still water y : speed of the river downstream speed ( ds ) = x + y upstream speed ( us ) = x â € “ y x = ( ds + us ) / 2 y = ( ds â € “ us ) / 2 in the above problem ds = 32 ; us = 8 x = ( 32 + 8 ) / 2 = 40 / 2 = 20 km / hr y = ( 32 - 8 ) / 2 = 24 / 2 = 12 km / hr answer : d | a ) 12 , 20 , b ) 20 , 20 , c ) 20 , 11 , d ) 20 , 12 , e ) 12 , 12 | d | divide(add(divide(88, 11), divide(96, 3)), const_2) | divide(n2,n3)|divide(n0,n1)|add(#0,#1)|divide(#2,const_2) | physics |
117 x 217 + 83 x 83 = ? | "= ( 117 ) ^ 2 + ( 83 ) ^ 2 = ( 100 + 17 ) ^ 2 + ( 100 - 17 ) ^ 2 = 2 [ ( 100 ) ^ 2 + ( 17 ) ^ 2 ] = 2 [ 10000 + 289 ] = 2 x 10289 = 20578 answer is b" | a ) 79698 , b ) 20578 , c ) 80698 , d ) 81268 , e ) none of them | b | multiply(117, power(217, 83)) | power(n1,n2)|multiply(n0,#0)| | general |
a flagstaff 17.5 m high casts a shadow of length 40.25 m . the height of the building , which casts a shadow of length 28.75 m under similar conditions will be | "let the height of the building x metres . less lengthy shadow , less in the height ( direct proportion ) 40.25 : 28.75 : : 17.5 : x 40.25 * x = 28.75 * 17.5 x = ( 28.75 x 17.5 ) / 40.25 x = 12.5 answer is a ." | a ) 12.5 m , b ) 10 m , c ) 17.5 m , d ) 21.25 m , e ) 9.8 m | a | multiply(28.75, divide(17.5, 40.25)) | divide(n0,n1)|multiply(n2,#0)| | physics |
three rugs have a combined area of 196 square meters . by overlapping the rugs to cover floor area of 140 square meters , the area that is covered by exactly two layers of rug is 24 square meters . what is the area that is covered with three layers of rug ? | "total = rug 1 + rug 2 + rug 3 - { overlap of exactly 2 rugs } - 2 * { overlap of exactly 3 rugs } 140 = 194 - 24 - 2 * { overlap of exactly 2 rugs } - - > { overlap of exactly 3 rugs } = 15 . answer : a ." | a ) 15 square meters , b ) 20 square meters , c ) 24 square meters , d ) 28 square meters , e ) 30 square meters | a | divide(subtract(subtract(196, 140), 24), const_2) | subtract(n0,n1)|subtract(#0,n2)|divide(#1,const_2)| | geometry |
a line has a slope of 3 / 4 and intersects the point r ( - 12 , - 39 ) . at which point does this line intersect the x - axis ? | "assume that the equation of the line is y = mx + c , where m and c are the slope and y - intercept . you are also given that the line crosses the point ( - 12 , - 39 ) , this means that this point will also lie on the line above . thus you get - 39 = m * ( - 12 ) + c , with m = 3 / 4 as the slope is given to be 3 / 4 ... | a ) ( 40,0 ) , b ) ( 30,0 ) , c ) ( 0,40 ) , d ) ( 40,30 ) , e ) ( 0,30 ) | a | multiply(negate(divide(subtract(negate(39), multiply(negate(12), divide(3, 4))), divide(3, 4))), const_10) | divide(n0,n1)|negate(n3)|negate(n2)|multiply(#0,#2)|subtract(#1,#3)|divide(#4,#0)|negate(#5)|multiply(#6,const_10)| | general |
180 college students were asked in a survey if they preferred windows or mac brand computers . 50 students claimed that they preferred mac to windows brand computers . one third as many of the students who preferred mac to windows , equally preferred both brands . 60 of the students had no preference . how many of the ... | "180 = 50 ( mac ) + x ( window ) + 60 ( both ) = > x = 70 answer : b" | a ) 25 , b ) 70 , c ) 50 , d ) 60 , e ) 75 | b | subtract(subtract(subtract(180, 60), 50), divide(subtract(subtract(180, 60), 50), const_3)) | subtract(n0,n2)|subtract(#0,n1)|divide(#1,const_3)|subtract(#1,#2)| | other |
a train is 360 meter long is running at a speed of 45 km / hour . in what time will it pass a bridge of 140 meter length ? | "speed = 45 km / hr = 45 * ( 5 / 18 ) m / sec = 25 / 2 m / sec total distance = 360 + 140 = 500 meter time = distance / speed = 500 * ( 2 / 25 ) = 40 seconds answer : c" | a ) 15 , b ) 44 , c ) 40 , d ) 27 , e ) 12 | c | divide(add(360, 140), divide(multiply(45, const_1000), const_3600)) | add(n0,n2)|multiply(n1,const_1000)|divide(#1,const_3600)|divide(#0,#2)| | physics |
what is the dividend . divisor 14 , the quotient is 9 and the remainder is 5 ? | "d = d * q + r d = 14 * 9 + 5 d = 126 + 5 d = 131 answer b" | a ) a ) 130 , b ) b ) 131 , c ) c ) 148 , d ) d ) 158 , e ) e ) 160 | b | add(multiply(14, 9), 5) | multiply(n0,n1)|add(n2,#0)| | general |
shop offered 30 % offer for every shirt , smith bought a shirt at rs . 560 . and what was the shop ' s original selling price ? | "sp * ( 70 / 100 ) = 560 sp = 8 * 100 = > cp = 800 answer : b" | a ) 500 , b ) 800 , c ) 600 , d ) 700 , e ) 750 | b | divide(560, subtract(const_1, divide(30, const_100))) | divide(n0,const_100)|subtract(const_1,#0)|divide(n1,#1)| | gain |
if q = { 1 , 7 , 18 , 20 , 29 , 33 } , how much less is the mean of the numbers in q than the median of the numbers in q ? | this is a good question to understand the difference between mean and median . mean : average of all the numbers . ( sum of all the elements divided by the number of elements ) median : arrange the elements of the set in increasing order . if the number of terms is odd , the middle term is the median . if the number of... | a ) 1.0 , b ) 1.5 , c ) 2.0 , d ) 2.5 , e ) 3.0 | a | subtract(divide(add(18, 20), const_2), divide(add(add(add(add(add(1, 7), 18), 20), 29), 33), multiply(const_2, const_3))) | add(n2,n3)|add(n0,n1)|multiply(const_2,const_3)|add(n2,#1)|divide(#0,const_2)|add(n3,#3)|add(n4,#5)|add(n5,#6)|divide(#7,#2)|subtract(#4,#8) | general |
company kw is being sold , and both company a and company b were considering the purchase . the price of company kw is 20 % more than company a has in assets , and this same price is also 100 % more than company b has in assets . if companies a and b were to merge and combine their assets , the price of company kw woul... | "let the price of company a ' s assets be 100 price of assets of kw is 20 % more than company a ' s assets which is 120 price of assets of kw is 100 % more than company b ' s assets which means price of company b ' s assets is half the price of kw = 60 a + b = 160 kw = 120 kw / ( a + b ) * 100 = 120 / 160 * 100 = 75 % ... | a ) 66 % , b ) 75 % , c ) 86 % , d ) 116 % , e ) 150 % | b | multiply(divide(add(100, 20), add(100, divide(add(100, 20), const_2))), 100) | add(n0,n1)|divide(#0,const_2)|add(n1,#1)|divide(#0,#2)|multiply(#3,n1)| | gain |
a bowl of fruit contains 12 apples and 23 oranges . how many oranges must be removed so that 60 % of the pieces of fruit in the bowl will be apples ? | "number of apples = 14 number of oranges = 23 let number of oranges that must be removed so that 60 % of pieces of fruit in bowl will be apples = x total number of fruits after x oranges are removed = 12 + ( 23 - x ) = 35 - x 12 / ( 35 - x ) = 6 / 10 = > 20 = 35 - x = > x = 15 answer c" | a ) 3 , b ) 6 , c ) 15 , d ) 17 , e ) 20 | c | divide(subtract(multiply(add(12, 23), divide(60, const_100)), 12), divide(60, const_100)) | add(n0,n1)|divide(n2,const_100)|multiply(#0,#1)|subtract(#2,n0)|divide(#3,#1)| | gain |
if a town of 25,000 people is growing at a rate of approx . 1 % per year , the population of the town in 5 years will be closest to ? | "let population in the beginning = 100 at the end of 1 st year population = 100 + 1 % of 100 = 100 + 1 = 101 at the end of 2 nd year population = 101 + 1 % of 101 = 101 + 1 approx = 102 approx ( on the lower side because 1 % of 101 is greater than 1 ) at the end of 3 rd year population = 102 + 1 % of 102 = 102 + 1 appr... | a ) 26000 , b ) 27000 , c ) 28000 , d ) 29000 , e ) 30000 | a | multiply(multiply(multiply(5, add(const_3, const_4)), const_100), add(add(const_3, const_4), const_3)) | add(const_3,const_4)|add(#0,const_3)|multiply(n2,#0)|multiply(#2,const_100)|multiply(#1,#3)| | gain |
one ton has 2100 pounds , and one pound has 16 ounces . how many packets containing wheat weighing 16 pounds and 4 ounces each would totally fill a gunny bag of capacity 13 tons ? | "16 pounds and 4 ounces = 16 * 16 + 4 = 260 ounces . 13 tons = 13 * 2100 pound = 13 * 2100 * 16 ounces . hence the answer is ( 13 * 2100 * 16 ) / 260 = 1680 . answer : a ." | a ) 1680 , b ) 1700 , c ) 2350 , d ) 2500 , e ) 8000 | a | divide(multiply(multiply(13, 2100), 16), add(multiply(16, 16), 4)) | multiply(n0,n4)|multiply(n1,n1)|add(n3,#1)|multiply(n1,#0)|divide(#3,#2)| | general |
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