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if ( 6 ) ( x ^ 2 ) has 3 different prime factors , at most how many different prime factors does x have ? | "without any discount sam should pay 10 * 10 = $ 100 . now , the overall discount would be slightly less than 38 % , thus he must pay slightly more than $ 62 . answer : e ." | a ) $ 92.00 , b ) $ 88.00 , c ) $ 87.04 , d ) $ 80.96 , e ) $ 65.00 | e | multiply(subtract(10, divide(multiply(30, 8), const_100)), 10) | multiply(n2,n4)|divide(#0,const_100)|subtract(n0,#1)|multiply(#2,n0)| | gain |
an item is being sold for $ 10 each . however , if a customer will β buy at least 3 β they have a promo discount of 30 % . also , if a customer will β buy at least 10 β items they will deduct an additional 8 % to their β buy at least 3 β promo price . if sam buys 10 pcs of that item how much should he pay ? | each 40 kilometers , 1 gallon is needed . we need to know how many 40 kilometers are there in 180 kilometers ? 180 Γ£ Β· 40 = 9.5 Γ£ β 1 gallon = 9.5 gallons correct answer is b ) 9.5 gallons | a ) 3.5 gallons , b ) 9.5 gallons , c ) 8.7 gallons , d ) 4.5 gallons , e ) 9.2 gallons | b | divide(190, 20) | divide(n1,n0) | physics |
a car gets 20 kilometers per gallon of gasoline . how many gallons of gasoline would the car need to travel 190 kilometers ? | "amount = [ 35000 * ( 1 + 12 / 100 ) 3 ] = 35000 * 28 / 25 * 28 / 25 * 28 / 25 = rs . 49172.48 c . i . = ( 49172.48 - 35000 ) = rs : 14172.48 answer : b" | a ) s : 10123.19 , b ) s : 14172.48 , c ) s : 10123.20 , d ) s : 10123.28 , e ) s : 10123.12 | b | subtract(multiply(multiply(multiply(const_4, const_100), const_100), power(add(const_1, divide(12, const_100)), 3)), multiply(multiply(const_4, const_100), const_100)) | divide(n2,const_100)|multiply(const_100,const_4)|add(#0,const_1)|multiply(#1,const_100)|power(#2,n1)|multiply(#3,#4)|subtract(#5,#3)| | gain |
what will be the compound interest on a sum of rs . 35,000 after 3 years at the rate of 12 % p . a . ? | "48 / 3 = 16 a" | a ) a ) 3 , b ) b ) 5 , c ) c ) 9 , d ) d ) 7 , e ) e ) 11 | a | sqrt(48) | sqrt(n0)| | general |
from below option 48 is divisible by which one ? | "66 * ( 90 / 100 ) * ( ( 100 - x ) / 100 ) = 56.16 x = 5.45 % answer : c" | a ) 3.45 % , b ) 4.45 % , c ) 5.45 % , d ) 6.45 % , e ) 7.45 % | c | multiply(divide(subtract(subtract(66, multiply(66, divide(10, const_100))), 56.16), subtract(66, multiply(66, divide(10, const_100)))), const_100) | divide(n2,const_100)|multiply(n0,#0)|subtract(n0,#1)|subtract(#2,n1)|divide(#3,#2)|multiply(#4,const_100)| | gain |
the list price of an article is rs . 66 . a customer pays rs . 56.16 for it . he was given two successive discounts , one of them being 10 % . the other discount is ? | "sol . let c . p . = rs . 100 . then , marked price = rs . 160 , s . p . = rs . 99 . β΄ discount % = [ 11 / 160 * 100 ] % = 6.8 % answer c" | a ) 10 % , b ) 10.5 % , c ) 6.8 % , d ) 12.5 % , e ) none | c | multiply(const_100, divide(add(multiply(add(const_2, const_3), const_2), 1), add(const_100, 60))) | add(const_2,const_3)|add(n0,const_100)|multiply(#0,const_2)|add(#2,n1)|divide(#3,#1)|multiply(#4,const_100)| | gain |
a trader marked the selling price of an article at 60 % above the cost price . at the time of selling , he allows certain discount and suffers a loss of 1 % . he allowed a discount of : | "explanation : 243 + 26 = 269 / 14 = 3 ( remainder ) answer : e" | a ) 7 , b ) 6 , c ) 9 , d ) 2 , e ) 3 | e | subtract(26, multiply(14, const_2)) | multiply(n2,const_2)|subtract(n1,#0)| | general |
a number when divided by 243 gives a remainder 26 , what remainder will be obtained by dividing the same number 14 ? | "actual price = 200 + 10 = $ 210 saving = 10 / 210 * 100 = 100 / 21 = 5 % approximately answer is d" | a ) 8 % , b ) 10 % , c ) 2 % , d ) 5 % , e ) 6 % | d | add(floor(multiply(divide(10, add(10, 200)), const_100)), const_1) | add(n0,n1)|divide(n0,#0)|multiply(#1,const_100)|floor(#2)|add(#3,const_1)| | general |
a person saved $ 10 in buying an item on sale . if he spent $ 200 for the item , approximately how much percent he saved in the transaction ? | "average age of 36 students in a group is 14 sum of the ages of 36 students = 36 Γ 14 when teacher ' s age is included to it , the average increases by one = > average = 15 sum of the ages of 36 students and the teacher = 37 Γ 15 hence teachers age = 37 Γ 15 - 36 Γ 14 = 37 Γ 15 - 14 ( 37 - 1 ) = 37 Γ 15 - 37 Γ 14 + 14 = 37 ( 15 - 14 ) + 14 = 37 + 14 = 51 answer is e ." | a ) 50 , b ) 49 , c ) 53 , d ) 54 , e ) 51 | e | add(36, const_1) | add(n0,const_1)| | general |
the average age of 36 students in a group is 14 years . when teacher ' s age is included to it , the average increases by one . find out the teacher ' s age in years ? | "8 x = 128 = > x = 16 therefore the secretary who worked the longest spent 16 x 5 = 90 hours on the project option ( a )" | a ) 90 , b ) 70 , c ) 56 , d ) 16 , e ) 14 | a | multiply(divide(128, add(add(1, 2), 5)), 5) | add(n0,n1)|add(n2,#0)|divide(n3,#1)|multiply(n2,#2)| | physics |
the amounts of time that three secretaries worked on a special project are in the ratio of 1 to 2 to 5 . if they worked a combined total of 128 hours , how many hours did the secretary who worked the longest spend on the project ? | a can finish 1 work in 20 days b can finish 1 / 1.5 work in 20 days - since a is 1.5 faster than b this means b can finish 1 work in 20 * 1.5 days = 30 days now using the awesome gmat formula when two machines work together they can finish the job in = ab / ( a + b ) = 20 * 30 / ( 20 + 30 ) = 20 * 30 / 50 = 12 days so answer is c | a ) 23 , b ) 22 , c ) 12 , d ) 24 , e ) 25 | c | divide(const_1, add(divide(const_1, 20), divide(divide(const_1, 20), 1.5))) | divide(const_1,n1)|divide(#0,n0)|add(#0,#1)|divide(const_1,#2) | physics |
a is 1.5 times as fast as b . a alone can do the work in 20 days . if a and b working together , in how many days will the work be completed ? | "explanation : probability that a speaks truth is 65 / 100 = 0.65 probability that b speaks truth is 60 / 100 = 0.6 since both a and b are independent of each other so probability of a intersection b is p ( a ) Γ p ( b ) = 0.65 Γ 0.6 = 0.39 answer : a" | a ) 0.39 , b ) 0.48 , c ) 0.41 , d ) 0.482 , e ) 0.411 | a | multiply(divide(65, multiply(multiply(const_4, const_5), const_5)), divide(60, multiply(multiply(const_4, const_5), const_5))) | multiply(const_4,const_5)|multiply(#0,const_5)|divide(n0,#1)|divide(n1,#1)|multiply(#2,#3)| | gain |
if a speaks the truth 65 % of the times , b speaks the truth 60 % of the times . what is the probability that they tell the truth at the same time | "c 1 loses 15 minutes every hour . so after 60 minutes have passed , c 1 displays that 60 - 15 = 45 minutes have passed . c 2 gains 15 minutes for every 60 minutes displayed on c 1 . thus , the time displayed on c 2 is 75 / 60 = 5 / 4 the time displayed on c 1 . so after 60 minutes have passed , c 2 displays the passing of ( 5 / 4 * 45 ) minutes . c 3 loses 20 minutes for every 60 minutes displayed on c 2 . thus , the time displayed on c 3 is 40 / 60 = 2 / 3 the time displayed on c 2 . so after 60 minutes have passed , c 3 displays the passing of ( 2 / 3 * 5 / 4 * 45 ) minutes . c 4 gains 20 minutes for every 60 minutes displayed on c 3 . thus , the time displayed on c 4 is 80 / 60 = 4 / 3 the time displayed on clock 3 . so after 60 minutes have passed , c 4 displays the passing of 4 / 3 * 2 / 3 * 5 / 4 * 45 = 50 minutes . c 4 loses 10 minutes every hour . in 6 hours , c 4 will lose 6 * 10 = 60 minutes = 1 hour . since the correct time after 6 hours will be 6 pm , c 4 will show a time of 6 - 1 = 6 : 24 pm . the correct answer is e ." | a ) 5 : 00 , b ) 5 : 34 , c ) 5 : 42 , d ) 6 : 00 , e ) 6 : 24 | e | subtract(multiply(6, const_10), multiply(multiply(multiply(divide(add(const_60, 15), const_60), divide(subtract(const_60, 20), const_60)), divide(add(const_60, 20), const_60)), subtract(const_60, 35))) | add(n16,const_60)|add(n3,const_60)|multiply(n23,const_10)|subtract(const_60,n16)|subtract(const_60,n1)|divide(#0,const_60)|divide(#1,const_60)|divide(#3,const_60)|multiply(#6,#7)|multiply(#5,#8)|multiply(#9,#4)|subtract(#2,#10)| | physics |
a not - so - good clockmaker has four clocks on display in the window . clock # 1 loses 35 minutes every hour . clock # 2 gains 15 minutes every hour relative to clock # 1 ( i . e . , as clock # 1 moves from 12 : 00 to 1 : 00 , clock # 2 moves from 12 : 00 to 1 : 15 ) . clock # 3 loses 20 minutes every hour relative to clock # 2 . finally , clock # 4 gains 20 minutes every hour relative to clock # 3 . if the clockmaker resets all four clocks to the correct time at 12 noon , what time will clock # 4 display after 6 actual hours ( when it is actually 6 : 00 pm that same day ) ? | if a = 1 , then putting values in equation = - [ ( 1 ) ^ 2 + ( 1 ) ^ 3 + ( 1 ^ 4 ) + ( 1 ^ 5 ) ] = - [ 1 + 1 + 1 + 1 ] = - 4 answer = b = - 4 | a ) - 14 , b ) - 4 , c ) 0 , d ) 4 , e ) 14 | b | negate(add(add(add(power(1, 2), power(1, 3)), power(1, 4)), power(1, 5))) | power(n0,n1)|power(n0,n2)|power(n0,n3)|power(n0,n4)|add(#0,#1)|add(#4,#2)|add(#5,#3)|negate(#6) | general |
if a = 1 , what is the value of β ( a ^ 2 + a ^ 3 + a ^ 4 + a ^ 5 ) ? | "sp 2 = 2 / 3 sp 1 cp = 100 sp 2 = 70 2 / 3 sp 1 = 70 sp 1 = 105 100 - - - 105 = > 5 % answer : e" | a ) 20 % , b ) 29 % , c ) 70 % , d ) 27 % , e ) 5 % | e | subtract(divide(subtract(const_100, 30), divide(2, 3)), const_100) | divide(n0,n1)|subtract(const_100,n2)|divide(#1,#0)|subtract(#2,const_100)| | gain |
what profit percent is made by selling an article at a certain price , if by selling at 2 / 3 rd of that price , there would be a loss of 30 % ? | we can determine quickly that total number should range between 49 / 8 < = n < = 49 / 1.5 , so ans should be between 6 and 33 . now solving the expression 8 a + 1.5 b = 49 decreasing 49 in multiple of 8 and checking divisibility of that number by 1.5 . this way we get 2 red giants , 22 white dwarfs we get 49 , but 2 + 22 = 24 and 24 is not an option . next we get 5 red giants and 6 white dwarfs to get 49 , 5 * 8 + 6 * 1.5 = 49 hence total number is 5 + 6 = 11 ans b | a ) 10 , b ) 11 , c ) 12 , d ) 13 , e ) 14 | b | add(divide(subtract(49, multiply(1.5, add(const_2, const_4))), 8), add(const_2, const_4)) | add(const_2,const_4)|multiply(n1,#0)|subtract(n2,#1)|divide(#2,n0)|add(#0,#3) | general |
an astronomer noted that a grouping of red giant stars had an average solar mass of 8 m each , and a grouping of white dwarf stars had an average solar mass of 1.5 m each . if the astronomer calculated the total solar mass of both groupings to be 49 m , what total number of red giant stars and white dwarf stars did the astronomer note ? | a = b + 300 . a = 4 ( b - 600 ) . 4 ( b - 600 ) = b + 300 . 3 b = 2700 . b = 900 . the answer is a . | a ) 900 , b ) 1000 , c ) 1100 , d ) 1200 , e ) 1300 | a | divide(add(multiply(600, 4), 300), subtract(4, const_1)) | multiply(n1,n2)|subtract(n2,const_1)|add(n0,#0)|divide(#2,#1) | general |
village a β s population is 300 greater than village b ' s population . if village b β s population were reduced by 600 people , then village a β s population would be 4 times as large as village b ' s population . what is village b ' s current population ? | "explanation : 2 c + 3 t = 1500 - - - ( 1 ) 3 c + 2 t = 1200 - - - ( 2 ) subtracting 2 nd from 1 st , we get - c + t = 300 = > t - c = 300 answer : e" | a ) 228 , b ) 287 , c ) 277 , d ) 188 , e ) 300 | e | subtract(divide(subtract(multiply(3, 1500), multiply(2, 3)), subtract(multiply(3, 3), multiply(2, 2))), divide(subtract(3, multiply(2, divide(subtract(multiply(3, 1500), multiply(2, 3)), subtract(multiply(3, 3), multiply(2, 2))))), 3)) | multiply(n2,const_3)|multiply(n3,const_2)|multiply(n1,const_3)|multiply(n0,const_2)|subtract(#0,#1)|subtract(#2,#3)|divide(#4,#5)|multiply(n0,#6)|subtract(n3,#7)|divide(#8,n1)|subtract(#6,#9)| | general |
the cost of 2 chairs and 3 tables is rs . 1500 . the cost of 3 chairs and 2 tables is rs . 1200 . the cost of each table is more than that of each chair by ? | "ci = 4851 , r = 5 , n = 2 ci = p [ 1 + r / 100 ] ^ 2 = p [ 1 + 5 / 100 ] ^ 2 4851 = p [ 21 / 20 ] ^ 2 4851 [ 20 / 21 ] ^ 2 4400 answer : c" | a ) s . 4000 , b ) s . 5000 , c ) s . 4400 , d ) s . 4800 , e ) s . 5800 | c | divide(4851, power(add(divide(5, const_100), const_1), 2)) | divide(n2,const_100)|add(#0,const_1)|power(#1,n1)|divide(n0,#2)| | gain |
a sum amounts to rs . 4851 in 2 years at the rate of 5 % p . a . if interest was compounded yearly then what was the principal ? | "manager ' s monthly salary = rs . ( 1700 * 11 - 1600 * 10 ) = rs . 2700 answer : c" | a ) rs . 3601 , b ) rs . 3618 , c ) rs . 2700 , d ) rs . 3619 , e ) rs . 3610 | c | subtract(multiply(add(1600, 100), add(10, const_1)), multiply(1600, 10)) | add(n1,n2)|add(n0,const_1)|multiply(n0,n1)|multiply(#0,#1)|subtract(#3,#2)| | general |
the average monthly salary of 10 employees in an organisation is rs . 1600 . if the manager ' s salary is added , then the average salary increases by rs . 100 . what is the manager ' s monthly salary ? | "( 1200 * 8 * 2 ) / 100 = > 192 answer : c" | a ) 190 , b ) 188 , c ) 192 , d ) 145 , e ) 188 | c | subtract(divide(multiply(multiply(1200, 18), 2), const_100), divide(multiply(multiply(1200, 10), 2), const_100)) | multiply(n0,n2)|multiply(n0,n1)|multiply(#0,n3)|multiply(n3,#1)|divide(#2,const_100)|divide(#3,const_100)|subtract(#4,#5)| | gain |
if a lends rs . 1200 to b at 10 % per annum and b lends the same sum to c at 18 % per annum then the gain of b in a period of 2 years is ? | i get 5 / 8 as well 1 to 96 inclusive means we have 48 odd and 48 even integers e o e / 6 = integer , therefore we have 48 / 96 numbers divisible by 6 o e o / 6 = not integer we can not forget multiples of 6 from 1 to 96 we have 16 numbers that are multiple of 8 therefore , 48 / 96 + 16 / 96 = 64 / 96 = 2 / 3 answer : a | a ) 2 / 3 , b ) 3 / 8 , c ) 1 / 2 , d ) 5 / 8 , e ) 3 / 4 | a | divide(add(divide(96, 2), divide(96, 6)), 96) | divide(n1,n3)|divide(n1,n4)|add(#0,#1)|divide(#2,n1) | general |
if an integer n is to be chosen at random from the integers 1 to 96 , inclusive , what is the probability that n ( n + 1 ) ( n + 2 ) will be divisible by 6 ? | "the numbers should be of the form 5 c + 3 . the minimum is 3 when c = 0 . the maximum is 48 when c = 9 . there are 10 such numbers . the answer is e ." | a ) 6 , b ) 7 , c ) 8 , d ) 9 , e ) 10 | e | divide(const_100.0, const_10) | divide(const_100.0,const_10)| | general |
how many integers from 0 to 50 inclusive have a remainder of 3 when divided by 5 ? | explanation total marks of 5 students = ( 65 + 75 + 55 + 72 + 69 ) = 336 required marks = [ ( 70 x 6 ) β 336 ] = ( 420 β 336 ) = 84 answer a | a ) 84 , b ) 68 , c ) 85 , d ) 75 , e ) 42 | a | subtract(multiply(70, 6), add(add(add(add(65, 75), 55), 72), 69)) | add(n3,n4)|multiply(n0,n1)|add(n5,#0)|add(n6,#2)|add(n7,#3)|subtract(#1,#4) | general |
6 students wrote science exam . their average marks are 70 . 5 students got 65 , 75 , 55 , 72 and 69 marks respectively . therefore what is the marks of the sixth student ? | "a relative speed = ( 54 + 90 ) * 5 / 18 = 8 * 5 = 40 mps . the time required = d / s = ( 100 + 100 + 200 ) / 35 = 400 / 40 = 10 sec ." | a ) 10 sec , b ) 11 sec , c ) 12 sec , d ) 60 / 7 sec , e ) 90 / 7 sec | a | divide(100, multiply(add(54, 90), const_0_2778)) | add(n3,n4)|multiply(#0,const_0_2778)|divide(n2,#1)| | physics |
two trains of length 100 m and 200 m are 100 m apart . they start moving towards each other on parallel tracks , at speeds 54 kmph and 90 kmph . in how much time will the trains cross each other ? | "total amount of water evaporated each day during a 50 - day period = . 008 * 50 = . 010 * 100 / 2 = 1.0 / 2 = . 5 percent of the original amount of water evaporated during this period = ( . 5 / 10 ) * 100 % = 5 % answer d" | a ) 0.004 % , b ) 0.04 % , c ) 0.40 % , d ) 5 % , e ) 40 % | d | multiply(divide(multiply(50, 0.010), 10), const_100) | multiply(n1,n2)|divide(#0,n0)|multiply(#1,const_100)| | gain |
a bowl was filled with 10 ounces of water , and 0.010 ounce of the water evaporated each day during a 50 - day period . what percent of the original amount of water evaporated during this period ? | "prime factors β number , as i assume , for a number x = a ^ n * b ^ m * c ^ o * d ^ p . . . is = n + m + o + p . . . so , 28 = 2 ^ 2 * 7 ^ 1 prime factors β number will be 2 + 1 = 3 . hence , answer is b ." | a ) 2 , b ) 3 , c ) 4 , d ) 5 , e ) 6 | b | add(add(add(const_1, add(const_1, const_1)), const_1), const_1) | add(const_1,const_1)|add(#0,const_1)|add(#1,const_1)|add(#2,const_1)| | other |
what is the prime factors β number of 28 ? | t 7 + t 23 = t 8 + t 15 + t 13 = > a + 6 d + a + 22 d = a + 7 d + a + 14 d + a + 12 d = > a + 5 d = 0 = > t 6 = 0 i . e . 6 th term is zero . answer : a | a ) 6 , b ) 8 , c ) 10 , d ) 12 , e ) 14 | a | subtract(add(13, add(8, 15)), add(7, 23)) | add(n2,n3)|add(n0,n1)|add(n4,#0)|subtract(#2,#1) | general |
the sum of 7 th and 23 rd term of a . p . is equal to the sum of 8 th , 15 th and 13 th term . find the term which is 0 | "hope this might be useful to you . let the number of people who have opted only to register = x now since the registration cost is 50 $ per person , the total amount sums to = 50 x $ as per the information given in the question , the number of registrants who paid for lunch was 30 more than the number who did not . that means , total number of people who registered and paid for lunch = 30 + x . for the people who registered for lunch the cost is 50 $ ( for the event registration ) + 22 $ ( for lunch ) = 72 $ . total amount in this case sums to = 72 ( 30 + x ) = 2160 + 72 x now , total amount received was 75360 . thus , from the above data , 50 x + 2160 + 72 x = 75360 122 x = 75360 - 2160 122 x = 73200 x = 600 . hence the correct ans is c" | a ) 700 , b ) 800 , c ) 600 , d ) 1,500 , e ) 1,800 | c | multiply(const_1, const_1) | multiply(const_1,const_1)| | general |
the cost of registration at a professional association meeting was $ 50 per person ; a lunch for registrants only was available for an additional $ 22 per person . if the number of registrants who paid for lunch was 30 more than the number who did not , and if receipts for registration and lunch totaled $ 75,360 , how many people paid just for registration at the meeting ? | "sol : breadth of the rectangular plot is = 5 ^ 2 - 4 ^ 2 = 3 m therefore , perimeter of the rectangular plot = 2 ( 4 + 3 ) = 14 m c ) 14 m" | a ) 20 m , b ) 15 m , c ) 14 m , d ) 10 m , e ) 25 m | c | divide(add(add(sqrt(subtract(power(5, const_2), power(4, const_2))), 4), add(sqrt(subtract(power(5, const_2), power(4, const_2))), 4)), 4) | power(n0,const_2)|power(n1,const_2)|subtract(#0,#1)|sqrt(#2)|add(n1,#3)|add(#4,#4)|divide(#5,n1)| | geometry |
what is the perimeter of a rectangular field whose diagonal is 5 m and length is 4 m ? | "on dividing 709 by 9 , we get remainder = 7 therefore , required number to be subtracted = 7 answer : c" | a ) a ) 2 , b ) b ) 3 , c ) c ) 7 , d ) d ) 5 , e ) e ) 6 | c | subtract(709, multiply(add(multiply(add(const_4, const_1), const_10), add(const_4, const_2)), 9)) | add(const_2,const_4)|add(const_1,const_4)|multiply(#1,const_10)|add(#0,#2)|multiply(n1,#3)|subtract(n0,#4)| | general |
the least number which must be subtracted from 709 to make it exactly divisible by 9 is : | prime numbers between 0 and 30 - 2 , 3 , 5 , 7 , 11 , 13 , 17 , 19 , 23 , 29 , 31 , 33 sum , c = 193 c / 3 = 64.3 answer c | a ) 155 , b ) 129 , c ) 64.3 , d ) 47 , e ) 43 | c | add(divide(3, const_10), power(const_2, add(const_2, const_4))) | add(const_2,const_4)|divide(n2,const_10)|power(const_2,#0)|add(#1,#2) | general |
let c be defined as the sum of all prime numbers between 0 and 35 . what is c / 3 | speed downstream = ( 22 + 5 ) = 27 kmph time = 24 minutes = 24 / 60 hour = 2 / 5 hour distance travelled = time Γ speed = 2 / 5 Γ 27 = 10.8 km answer is c . | a ) 10.6 , b ) 10.2 , c ) 10.8 , d ) 10.4 , e ) 10.0 | c | multiply(add(22, 5), divide(24, const_60)) | add(n0,n1)|divide(n2,const_60)|multiply(#0,#1) | physics |
the speed of a boat in still water in 22 km / hr and the rate of current is 5 km / hr . the distance travelled downstream in 24 minutes is : | "volume of water displaced = ( 3 x 3 x 0.01 ) m 3 = 0.09 m 3 . mass of man = volume of water displaced x density of water = ( 0.09 x 1000 ) kg = 90 kg . answer : d" | a ) 100 kg , b ) 120 kg , c ) 89 kg , d ) 90 kg , e ) 110 kg | d | multiply(multiply(multiply(3, 3), divide(1, const_100)), const_1000) | divide(n2,const_100)|multiply(n0,n1)|multiply(#0,#1)|multiply(#2,const_1000)| | physics |
a boat having a length 3 m and breadth 3 m is floating on a lake . the boat sinks by 1 cm when a man gets on it . the mass of the man is : | "80 % - - - 12 120 % - - - ? 80 / 120 * 12 = 8 answer : a" | a ) 8 , b ) 76 , c ) 17 , d ) 7 , e ) 77 | a | multiply(divide(const_1, multiply(add(const_100, 20), divide(const_1, subtract(const_100, 20)))), 12) | add(n2,const_100)|subtract(const_100,n1)|divide(const_1,#1)|multiply(#0,#2)|divide(const_1,#3)|multiply(n0,#4)| | gain |
by selling 12 pencils for a rupee a man loses 20 % . how many for a rupee should he sell in order to gain 20 % ? | "old time in minutes to cross 10 miles stretch = 10 * 60 / 55 = 10 * 12 / 11 = 10.9 new time in minutes to cross 10 miles stretch = 10 * 60 / 35 = 10 * 12 / 7 = 17.14 time difference = 6.24 ans : a" | a ) 6.24 , b ) 8 , c ) 10 , d ) 15 , e ) 24 | a | max(multiply(subtract(add(55, 10), const_1), subtract(divide(10, 35), divide(10, 55))), const_4) | add(n0,n1)|divide(n0,n2)|divide(n0,n1)|subtract(#0,const_1)|subtract(#1,#2)|multiply(#3,#4)|max(#5,const_4)| | physics |
due to construction , the speed limit along an 10 - mile section of highway is reduced from 55 miles per hour to 35 miles per hour . approximately how many minutes more will it take to travel along this section of highway at the new speed limit than it would have taken at the old speed limit ? | "speed ratio = 1 : 4 / 5 = 5 : 4 time ratio = 4 : 51 - - - - - - - - 7 4 - - - - - - - - - ? Γ¨ 28 answer : c" | a ) 16 min , b ) 26 min , c ) 28 min , d ) 20 min , e ) 12 min | c | multiply(divide(7, divide(5, 4)), 5) | divide(n1,n0)|divide(n2,#0)|multiply(n1,#1)| | physics |
walking with 4 / 5 of my usual speed , i miss the bus by 7 minutes . what is my usual time ? | "p ( r / 100 ) ^ 2 = c . i - s . i p ( 10 / 100 ) ^ 2 = 150 15000 answer : a" | a ) s . 15000 , b ) s . 15100 , c ) s . 15800 , d ) s . 16000 , e ) s . 16200 | a | divide(150, multiply(divide(10, const_100), divide(10, const_100))) | divide(n0,const_100)|multiply(#0,#0)|divide(n2,#1)| | gain |
if difference between compound interest and simple interest on a sum at 10 % p . a . for 2 years is rs . 150 then sum is | "solution speed in still water = 1 / 2 ( 8 + 2 ) km / hr = 5 kmph . answer b" | a ) 3 , b ) 5 , c ) 8 , d ) 9 , e ) 10 | b | divide(add(8, 2), const_2) | add(n0,n1)|divide(#0,const_2)| | gain |
in one hour , a boat goes 8 km along the stream and 2 km against the stream . the sped of the boat in still water ( in km / hr ) is : | c . p . for one coconut = 150 β 100 = 3 β 2 s . p . for one coconut = 2 profit on one coconut = 2 - 3 β 2 = 1 β 2 β΄ profit on 2000 coconut = 1 β 2 Γ 2000 = 1000 answer b | a ) 500 , b ) 1000 , c ) 1500 , d ) 2000 , e ) none of these | b | multiply(2000, subtract(2, divide(150, 100))) | divide(n0,n1)|subtract(n2,#0)|multiply(n3,#1) | gain |
coconuts were purchased at 150 per 100 and sold at 2 per coconut . if 2000 coconuts were sold , what was the total profit made ? | "c 120 120 ( 5 x 4 x 3 x 2 x 1 ) ." | a ) 1 , b ) 60 , c ) 120 , d ) 130 , e ) 180 | c | circle_area(divide(5, multiply(const_2, const_pi))) | multiply(const_2,const_pi)|divide(n0,#0)|circle_area(#1)| | other |
what is the factorial of 5 ? | "rate * time = work let painter w ' s rate be w and painter x ' s rate be x r * t = work w * 2 = 1 ( if the work done is same throughout the question then the work done can be taken as 1 ) = > w = 1 / 2 x * e = 1 = > x = 1 / e when they both work together then their rates get added up combined rate = ( w + x ) r * t = work ( w + x ) * 3 / 4 = 1 = > w + x = 4 / 3 = > 1 / 2 + 1 / e = 4 / 3 = > 1 / e = ( 8 - 3 ) / 6 = 5 / 6 = > e = 6 / 5 = 1 [ 1 / 5 ] answer b" | a ) 3 / 4 , b ) 1 [ 1 / 5 ] , c ) 1 [ 2 / 5 ] , d ) 1 [ 3 / 4 ] , e ) 2 | b | add(subtract(4, 2), divide(const_1, add(2, 3))) | add(n0,n1)|subtract(n2,n0)|divide(const_1,#0)|add(#2,#1)| | physics |
when working alone , painter w can paint a room in 2 hours , and working alone , painter x can paint the same room in e hours . when the two painters work together and independently , they can paint the room in 3 / 4 of an hour . what is the value of e ? | "explanation : speed = 45 km / hr = 45 * ( 5 / 18 ) m / sec = 25 / 2 m / sec total distance = 435 + 140 = 575 meter time = distance / speed = 575 β 2 / 25 = 46 seconds option c" | a ) 20 seconds , b ) 30 seconds , c ) 46 seconds , d ) 50 seconds , e ) none of these | c | divide(add(435, 140), divide(multiply(45, const_1000), const_3600)) | add(n0,n2)|multiply(n1,const_1000)|divide(#1,const_3600)|divide(#0,#2)| | physics |
a train is 435 meter long is running at a speed of 45 km / hour . in what time will it pass a bridge of 140 meter length | "h . c . f = ( product of the numbers ) / ( their l . c . m ) = 22500 / 450 = 50 . answer : a" | a ) 50 , b ) 30 , c ) 125 , d ) 25 , e ) none of these | a | divide(22500, 450) | divide(n1,n0)| | physics |
if the l . c . m of two numbers is 450 and their product is 22500 , find the h . c . f of the numbers . | let base = x cm height = 2 x cm area = x Γ£ β 2 x = 2 x ^ 2 area = x Γ£ β 2 x = 2 x ^ 2 area is given as 72 cm ^ 2 2 x ^ 2 = 72 x ^ 2 = 36 x = 6 cm answer : c | ['a ) 1 cm', 'b ) 3 cm', 'c ) 6 cm', 'd ) 4 cm', 'e ) 2 cm'] | c | sqrt(divide(72, const_2)) | divide(n0,const_2)|sqrt(#0) | geometry |
the area of a parallelogram is 72 cm ^ 2 and its altitude is twice the corresponding base . what is the length of the base ? | pencil + notebook = 80 notebook + eraser = 115 pencil + eraser = 75 let ' s add all three equations . 2 pencils + 2 notebooks + 2 erasers = 270 cents the cost to buy 3 of each would be ( 3 / 2 ) ( 270 ) = 405 the answer is e . | a ) 325 , b ) 345 , c ) 365 , d ) 385 , e ) 405 | e | multiply(divide(add(add(multiply(1.15, const_100), 80), 75), const_2), 3) | multiply(n1,const_100)|add(n0,#0)|add(n2,#1)|divide(#2,const_2)|multiply(n3,#3) | gain |
martin buys a pencil and a notebook for 80 cents . at the same store , gloria buys a notebook and an eraser for $ 1.15 cents , and zachary buys a pencil and an eraser for 75 cents . how many cents would it cost to buy 3 pencils , 3 notebooks , and 3 erasers ? ( assume that there is no volume discount . ) | "f 200 / x leaves a reminder 3 then ( 200 - 3 ) i . e . 197 is divisible by x so ( 200 + 197 ) / x leaves a reminder rem ( 200 / x ) + rem ( 197 / x ) = > 3 + 0 = 3 answer : b" | a ) 2 , b ) 3 , c ) 4 , d ) 6 , e ) 8 | b | subtract(const_100.0, subtract(297, 200)) | subtract(n2,const_100.0)|subtract(n0,#0)| | general |
when 200 is divided by positive integer x , the remainder is 3 . what is the remainder when 297 is divided by x ? | we can determine quickly that total number should range between 1020 / 160 < = n < = 1020 / 90 , so ans should be between 6 and 12 . now solving the expression 160 a + 90 b = 1020 decreasing 1020 by multiples of 160 and checking divisibility of that number by 9 , we get fast song plays for 3 minutes and slow somg plays for 6 minutes , 3 * 160 + 6 * 90 = 1020 hence total number of minutes stream of music plays is 3 + 6 = 9 minutes ans d | a ) 6 , b ) 7 , c ) 8 , d ) 9 , e ) 10 | d | add(floor(multiply(divide(1020, add(160, 90)), const_2)), const_1) | add(n0,n1)|divide(n2,#0)|multiply(#1,const_2)|floor(#2)|add(#3,const_1) | physics |
if a fast song has 160 beats per minute , and a slow song has 90 beats per minute , how many minutes total would you play a fast and a slow song to have a stream of music that had a total of 1020 beats ? | "explanation : let the number of persons be n Γ’ Λ Β΄ total handshakes = nc 2 = 190 n ( n - 1 ) / 2 = 190 Γ’ Λ Β΄ n = 20 answer : option e" | a ) 15 , b ) 16 , c ) 17 , d ) 18 , e ) 20 | e | divide(add(sqrt(add(multiply(multiply(190, const_2), const_4), const_1)), const_1), const_2) | multiply(n0,const_2)|multiply(#0,const_4)|add(#1,const_1)|sqrt(#2)|add(#3,const_1)|divide(#4,const_2)| | general |
in a party every person shakes hands with every other person . if there were a total of 190 handshakes in the party then what is the number of persons present in the party ? | "sol . required ratio = 6 * 1 * 1 / 6 * 7 * 7 = 1 / 49 = 1 : 49 . answer b" | a ) 1 : 25 , b ) 1 : 49 , c ) 1 : 52 , d ) 1 : 522 , e ) none | b | divide(const_4, const_100) | divide(const_4,const_100)| | geometry |
a cube of edge 7 cm is cut into cubes each of edge 1 cm . the ratio of the total surface area of one of the small cubes to that of the large cube is equal to : | "present age is 4 x and 3 x , = > 4 x + 6 = 30 = > x = 6 so deepak age is = 3 ( 6 ) = 18 answer : a" | a ) 18 , b ) 15 , c ) 77 , d ) 266 , e ) 182 | a | divide(multiply(subtract(30, 6), 3), 4) | subtract(n3,n2)|multiply(n1,#0)|divide(#1,n0)| | other |
ratio between rahul and deepak is 4 : 3 , after 6 years rahul age will be 30 years . what is deepak present age ? | "12 * 8 : 16 * 9 = 18 * 6 8 : 12 : 9 9 / 29 * 480 = 149 answer : d" | a ) 270 , b ) 199 , c ) 676 , d ) 149 , e ) 122 | d | multiply(divide(480, 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 , b and c rents a pasture for rs . 480 . a put in 12 horses for 8 months , b 16 horses for 9 months and 18 horses for 6 months . how much should c pay ? | "remainder will be number / 100 here as the divisor is two digit number = 12 . hence checking for the last two digits = 5 * 7 * 9 = 15 thus remainder = 3 . answer : d" | a ) 0 , b ) 1 , c ) 2 , d ) 3 , e ) 4 | d | subtract(multiply(multiply(1525, 1527), 1529), subtract(multiply(multiply(1525, 1527), 1529), const_3)) | multiply(n0,n1)|multiply(n2,#0)|subtract(#1,const_3)|subtract(#1,#2)| | general |
what is remainder of the division ( 1525 * 1527 * 1529 ) / 12 ? | solution : this is a percent decrease problem . we will use the formula : percent change = ( new β old ) / old x 100 to calculate the final answer . we first set up the ratios of royalties to sales . the first ratio will be for the first 20 million in sales , and the second ratio will be for the next 108 million in sales . because all of the sales are in millions , we do not have to express all the trailing zeros in our ratios . first 20 million royalties / sales = 5 / 20 = 1 / 4 next 108 million royalties / sales = 9 / 108 = 1 / 12 because each ratio is not an easy number to use , we can simplify each one by multiplying each by the lcm of the two denominators , which is 60 . keep in mind that we are able to do this only because our answer choices are expressed in percents . first 20 million royalties / sales = ( 5 / 20 ) x 12 = 3 next 108 million royalties / sales = 9 / 108 = ( 1 / 12 ) x 12 = 1 we can plug 15 and 5 into our percent change formula : ( new β old ) / old x 100 [ ( 1 β 3 ) / 3 ] x 100 - 200 / 3 x 100 at this point we can stop and consider the answer choices . since we know that 200 / 3 is just a bit less than Β½ , we know that - 200 / 3 x 100 is about a 67 % decrease . answer e . | a ) 8 % , b ) 15 % , c ) 45 % , d ) 52 % , e ) 67 % | e | multiply(divide(subtract(multiply(divide(5, 20), const_100), multiply(divide(9, 108), const_100)), multiply(divide(5, 20), const_100)), const_100) | divide(n0,n1)|divide(n2,n3)|multiply(#0,const_100)|multiply(#1,const_100)|subtract(#2,#3)|divide(#4,#2)|multiply(#5,const_100) | general |
a pharmaceutical company received $ 5 million in royalties on the first $ 20 million in sales of the generic equivalent of one of its products and then $ 9 million in royalties on the next $ 108 million in sales . by approximately what percent did the ratio of royalties to sales decrease from the first $ 20 million in sales to the next $ 108 million in sales ? | "10 overs - run rate = 5.2 runs scored in first 10 overs = 52 remaining overs 40 total runs to be scored = 282 52 runs already scored 282 - 52 = 230 230 runs to be scored in 40 overs let required runrate be x 40 * x = 230 x = 230 / 40 x = 5.75 the required runrate is 5.75 answer : d" | a ) 6.25 , b ) 6.5 , c ) 6.75 , d ) 5.75 , e ) 8 | d | divide(subtract(282, multiply(10, 5.2)), 40) | multiply(n0,n1)|subtract(n3,#0)|divide(#1,n2)| | gain |
in the first 10 overs of a cricket game , the run rate was only 5.2 . what should be the run rate in the remaining 40 overs to reach the target of 282 runs ? | "{ total } = { french } + { russian } - { both } + { neither } { total } = 30 + ( { total } - 32 ) - ( 0.1 * { total } ) + 0.2 * { total } solving gives { total } = 20 . answer : a ." | a ) 20 , b ) 96 , c ) 108 , d ) 120 , e ) 150 | a | divide(subtract(32, 30), subtract(divide(20, const_100), divide(10, const_100))) | divide(n2,const_100)|divide(n3,const_100)|subtract(n1,n0)|subtract(#0,#1)|divide(#2,#3)| | other |
of the diplomats attending a summit conference , 30 speak french , 32 do not speak russian , and 20 % of the diplomats speak neither french nor russian . if 10 % of the diplomats speak both languages , then how many diplomats attended the conference ? | let number of cars sold in 1 st quarter = x number of cars sold in 2 nd quarter = 32 % greater than the number sold during the first quarter = ( 1 + 32 / 100 ) x = 1.32 x 1.32 x = 3 , 976,000 = > x = 3 , 012,121 so , answer will be d | a ) 714,240 , b ) 2 , 261,760 , c ) 2 , 400,000 , d ) 3 , 012,121 , e ) 3 , 915,790 | d | multiply(multiply(divide(divide(divide(add(multiply(3, multiply(const_1000, const_1000)), 976000), add(divide(32, const_100), const_1)), const_1000), const_100), 3), 3) | divide(n3,const_100)|multiply(const_1000,const_1000)|add(#0,const_1)|multiply(n1,#1)|add(n2,#3)|divide(#4,#2)|divide(#5,const_1000)|divide(#6,const_100)|multiply(n1,#7)|multiply(n1,#8) | gain |
during the second quarter of 1984 , a total of 3 , 976000 domestic cars were sold . if this was 32 % greater than the number sold during the first quarter of 1984 , how many were sold during the first quarter ? | "0.2 a = 0.08 b - > a / b = 0.08 / 0.20 = 8 / 20 = 2 / 5 : . a : b = 2 : 5 answer : c" | a ) 2 : 3 , b ) 3 : 4 , c ) 2 : 5 , d ) 20 : 3 , e ) 30 : 7 | c | divide(multiply(0.08, const_100), multiply(0.2, const_100)) | multiply(n1,const_100)|multiply(n0,const_100)|divide(#0,#1)| | other |
if 0.2 of a number is equal to 0.08 of another number , the ratio of the numbers is : | explanation : let distance = x km and usual rate = y kmph . then , x / y - x / ( y + 3 ) = 40 / 60 - - > 2 y ( y + 3 ) = 9 x - - - - - ( i ) also , x / ( y - 2 ) - x / y = 40 / 60 - - > y ( y - 2 ) = 3 x - - - - - - - - ( ii ) on dividing ( i ) by ( ii ) , we get : x = 40 km . answer : c | a ) 27 , b ) 87 , c ) 40 , d ) 18 , e ) 17 | c | multiply(multiply(divide(multiply(multiply(2, 3), 2), subtract(3, 2)), divide(40, const_60)), add(const_1, divide(divide(multiply(multiply(2, 3), 2), subtract(3, 2)), 3))) | divide(n1,const_60)|multiply(n0,n2)|subtract(n0,n2)|multiply(n2,#1)|divide(#3,#2)|divide(#4,n0)|multiply(#4,#0)|add(#5,const_1)|multiply(#7,#6) | physics |
a man covered a certain distance at some speed . had he moved 3 kmph faster , he would have taken 40 minutes less . if he had moved 2 kmph slower , he would have taken 40 minutes more . the distance ( in km ) is | w + b + c + 14 + 15 = 12 * 5 = 60 = > w + b + c = 60 - 29 = 31 w + b + c + 29 = 31 + 29 = 60 average = 60 / 4 = 15 answer d | a ) 12 , b ) 13 , c ) 14 , d ) 15 , e ) 16 | d | divide(add(subtract(subtract(multiply(add(const_1, const_4), 12), 15), 14), 29), const_4) | add(const_1,const_4)|multiply(n2,#0)|subtract(#1,n1)|subtract(#2,n0)|add(n3,#3)|divide(#4,const_4) | general |
if the average of w , b , c , 14 and 15 is 12 . what is the average value of w , b , c and 29 | this is maximum - minimum . hence , 40 - ( 1 + 1 + 1 + 1 + 1 + 1 + 1 ) = 32 and 40 - ( 1 + 2 + 3 + 4 + 5 + 6 + 7 ) = 11 . so , 32 - 11 = 21 . the correct answer is a . | a ) 21 , b ) 29 , c ) 23 , d ) 25 , e ) 26 | a | subtract(40, add(add(8, const_2), 8)) | add(n0,const_2)|add(n0,#0)|subtract(n1,#1)| | general |
set a of 8 positive integers may have the same element and have 40 . and set b of 8 positive integers must have different elements and have 40 . when m and n are the greatest possible differences between 40 and other elements β sums in set a and set b , respectively , m - n = ? | "( 6 * 65 ) + 47 + x > 500 390 + 47 + x > 500 437 + x > 500 = > x > 63 option d" | a ) 85 , b ) 74 , c ) 67 , d ) 63 , e ) 28 | d | subtract(500, add(multiply(6, 65), 47)) | multiply(n0,n1)|add(n2,#0)|subtract(n4,#1)| | general |
after 6 games , team b had an average of 65 points per game . if it got only 47 points in game 7 , how many more points does it need to score to get its total above 500 ? | "solution : mabel handled 90 transactions anthony handled 10 % more transactions than mabel anthony = 90 + 90 Γ 10 % = 90 + 90 Γ 0.10 = 90 + 9 = 99 cal handled 2 / 3 rds of the transactions than anthony handled cal = 2 / 3 Γ 99 = 66 jade handled 16 more transactions than cal . jade = 66 + 16 = 82 jade handled = 82 transactions . answer : c" | a ) 80 , b ) 81 , c ) 82 , d ) 83 , e ) 84 | c | add(divide(multiply(multiply(divide(90, const_100), add(10, const_100)), 2), 3), 16) | add(n1,const_100)|divide(n0,const_100)|multiply(#0,#1)|multiply(n2,#2)|divide(#3,n3)|add(n4,#4)| | general |
on thursday mabel handled 90 transactions . anthony handled 10 % more transactions than mabel , cal handled 2 / 3 rds of the transactions that anthony handled , and jade handled 16 more transactions than cal . how much transactions did jade handled ? | "c . p . = $ 100 s . p . = $ 125 gain = $ 25 gain % = 25 / 100 * 100 = 25 % answer is c" | a ) 10 % , b ) 15 % , c ) 25 % , d ) 20 % , e ) 30 % | c | subtract(divide(125, divide(100, const_100)), const_100) | divide(n0,const_100)|divide(n1,#0)|subtract(#1,const_100)| | gain |
a man buys an article for $ 100 . and sells it for $ 125 . find the gain percent ? | "this problem can be solved easily if we just use approximation : 35 % is a little over 1 / 3 , while 4 / 13 is a little less than 4 / 12 , which is 1 / 3 . thus , the answer is about 1 / 3 of 1 / 3 of 780 , or 1 / 9 of 780 . since the first 1 / 3 is a slight underestimate and the second 1 / 3 is a slight overestimate , the errors will partially cancel each other out . our estimate will be relatively accurate . the number 780 is between 720 and 810 , so ( 1 / 9 ) * 780 will be between 80 and 90 . keeping track not only of your current estimate , but also of the degree to which you have overestimated or underestimated , can help you pinpoint the correct answer more confidently . the closest answer is 84 , so this is the answer to choose . the answer is c ." | a ) 62 , b ) 73 , c ) 84 , d ) 95 , e ) 106 | c | divide(multiply(35, add(add(multiply(multiply(add(const_3, const_2), const_2), multiply(multiply(const_3, const_4), const_100)), multiply(multiply(add(const_3, const_4), add(const_3, const_2)), multiply(add(const_3, const_2), const_2))), add(const_3, const_3))), const_100) | add(const_2,const_3)|add(const_3,const_4)|add(const_3,const_3)|multiply(const_3,const_4)|multiply(#0,const_2)|multiply(#3,const_100)|multiply(#1,#0)|multiply(#4,#5)|multiply(#6,#4)|add(#7,#8)|add(#9,#2)|multiply(n0,#10)|divide(#11,const_100)| | gain |
what is 35 % of 4 / 13 of 780 ? | "let v be the volume of the tank . the rate per minute at which the tank is filled is : v / 20 + v / 12 - v / 10 = v / 30 per minute the tank will be filled in 30 minutes . the answer is d ." | a ) 24 , b ) 26 , c ) 28 , d ) 30 , e ) 32 | d | subtract(add(divide(const_1, 20), divide(const_1, 12)), divide(const_1, 10)) | divide(const_1,n0)|divide(const_1,n1)|divide(const_1,n2)|add(#0,#1)|subtract(#3,#2)| | physics |
two pipes can fill a tank in 20 minutes and 12 minutes . an outlet pipe can empty the tank in 10 minutes . if all the pipes are opened when the tank is empty , then how many minutes will it take to fill the tank ? | "in one resolution , the distance covered by the wheel is its own circumference . distance covered in 500 resolutions . = 500 * 2 * 22 / 7 * 20 = 31428.5 cm = 314.3 m answer : c" | a ) 708 m , b ) 704 m , c ) 314.3 m , d ) 714 m , e ) 744 m | c | divide(multiply(multiply(multiply(divide(add(multiply(add(const_3, const_4), const_3), const_1), add(const_3, const_4)), 20), const_2), 500), const_100) | add(const_3,const_4)|multiply(#0,const_3)|add(#1,const_1)|divide(#2,#0)|multiply(n0,#3)|multiply(#4,const_2)|multiply(n1,#5)|divide(#6,const_100)| | physics |
the radius of a wheel is 20 cm . what is the distance covered by the wheel in making 500 resolutions ? | since - 4 is the least integer in list a , then 7 is the largest integer in that list . thus the range of the positive integers in the list is 7 - 1 = 6 . answer : b . | a ) 5 , b ) 6 , c ) 7 , d ) 11 , e ) 12 | b | subtract(subtract(12, add(4, const_1)), const_1) | add(n1,const_1)|subtract(n0,#0)|subtract(#1,const_1) | general |
list a consists of 12 consecutive integers . if - 4 is the least integer in list a , what is the range of positive integers in list a ? | "total paint initially = 360 gallons paint used in the first week = ( 1 / 2 ) * 360 = 180 gallons . remaning paint = 180 gallons paint used in the second week = ( 1 / 5 ) * 180 = 36 gallons total paint used = 216 gallons . option d" | a ) 18 , b ) 144 , c ) 175 , d ) 216 , e ) 250 | d | add(multiply(divide(360, 2), 1), divide(subtract(360, multiply(divide(360, 2), 1)), 5)) | divide(n0,n2)|multiply(n1,#0)|subtract(n0,#1)|divide(#2,n4)|add(#3,#1)| | physics |
joe needs to paint all the airplane hangars at the airport , so he buys 360 gallons of paint to do the job . during the first week , he uses 1 / 2 of all the paint . during the second week , he uses 1 / 5 of the remaining paint . how many gallons of paint has joe used ? | "1 feet = 12 inches 1 mile = 5280 feet 100 mile = 5280 * 12 * 100 = 6336000 ans : a" | a ) 6336000 , b ) 6542000 , c ) 5462300 , d ) 6213000 , e ) 6120330 | a | divide(multiply(multiply(multiply(add(const_3, const_2), const_2), multiply(add(const_3, const_2), const_2)), 100), multiply(multiply(add(const_3, const_2), const_2), multiply(add(const_3, const_2), const_2))) | add(const_2,const_3)|multiply(#0,const_2)|multiply(#1,#1)|multiply(n0,#2)|divide(#3,#2)| | physics |
convert 100 miles into inches ? | "sum of decimal places = 7 . since the last digit to the extreme right will be zero ( since 5 x 4 = 20 ) so there will be 6 significant digits to the right of the decimal point . answer is e ." | a ) 2 , b ) 3 , c ) 4 , d ) 5 , e ) 6 | e | subtract(subtract(const_100, 95.75), const_1) | subtract(const_100,n0)|subtract(#0,const_1)| | general |
how many digits will be there to the right of the decimal point in the product of 95.75 and . 02554 ? | "i get 5 / 8 as well 1 to 96 inclusive means we have 48 odd and 48 even integers e o e / 4 = integer , therefore we have 48 / 96 numbers divisible by 8 o e o / 4 = not integer we can not forget multiples of 8 from 1 to 96 we have 24 numbers that are multiple of 4 therefore , 48 / 96 + 24 / 96 = 72 / 96 = 3 / 4 answer : e" | a ) 1 / 4 , b ) 3 / 8 , c ) 1 / 2 , d ) 5 / 8 , e ) 3 / 4 | e | divide(add(divide(96, 2), divide(96, 4)), 96) | divide(n1,n3)|divide(n1,n4)|add(#0,#1)|divide(#2,n1)| | general |
if an integer n is to be chosen at random from the integers 1 to 96 , inclusive , what is the probability that n ( n + 1 ) ( n + 2 ) will be divisible by 4 ? | "increase in house value = $ 24,000 - $ 20,000 = $ 4000 so , tax increase = 12 % of $ 4000 = $ 480 answer : e" | a ) $ 32 , b ) $ 50 , c ) $ 320 , d ) $ 400 , e ) $ 480 | e | divide(multiply(subtract(multiply(multiply(add(const_3, const_4), const_4), const_1000), multiply(multiply(add(const_4, const_1), const_4), const_1000)), 12), const_100) | add(const_3,const_4)|add(const_1,const_4)|multiply(#0,const_4)|multiply(#1,const_4)|multiply(#2,const_1000)|multiply(#3,const_1000)|subtract(#4,#5)|multiply(n0,#6)|divide(#7,const_100)| | general |
in township k each property is taxed at 12 percent of its assessed value . if the assessed value of a property in township k is increased from $ 20,000 to $ 24,000 , by how much will the property tax increase ? | y = 288 * a + 44 = ( 24 * 12 ) * a + 24 + 20 the answer is a . | a ) 20 , b ) 21 , c ) 23 , d ) 25 , e ) 26 | a | reminder(44, 24) | reminder(n1,n2) | general |
when y is divided by 288 , the remainder is 44 . what is the remainder when the same y is divided by 24 ? | "1 of 5 will be chosen for the math 2 of 10 will be chosen for the computer none of the 3 chosen people can be in more than one departments . we can choose any of the 5 candidates for the math dep . , which gives as 5 selections . we can choose 2 of the 10 candidates for the computer dep . , which gives us 2 selections and 8 rejections . so , the way to find how many different selections of 2 candidates we can have for the computer dep . , we do : 10 ! / 2 ! * 8 ! = ( 9 * 10 ) / 2 = 90 / 2 = 45 . we are multiplying our individual selections : 5 * 45 = 225 in the bolded part , we do n ' t have to multiply all of the numbers , as those in 8 ! are included in 10 ! , so we simplify instead . ans e" | a ) 42 , b ) 70 , c ) 140 , d ) 165 , e ) 225 | e | multiply(multiply(10, 3), 5) | multiply(n3,n5)|multiply(n1,#0)| | other |
a certain university will select 1 of 5 candidates eligible to fill a position in the mathematics department and 2 of 10 candidates eligible to fill 2 identical positions in the computer science department . if none of the candidates is eligible for a position in both departments , how many different sets of 3 candidates are there to fill the 3 positions ? | "1 / 2 * d ( 14 + 4 ) = 380 d = 42 answer : d" | a ) 39 , b ) 28 , c ) 27 , d ) 42 , e ) 71 | d | divide(divide(divide(380, divide(add(14, 4), const_2)), 4), const_2) | add(n0,n1)|divide(#0,const_2)|divide(n2,#1)|divide(#2,n1)|divide(#3,const_2)| | physics |
the cross - section of a cannel is a trapezium in shape . if the cannel is 14 m wide at the top and 4 m wide at the bottom and the area of cross - section is 380 sq m , the depth of cannel is ? | "cp * ( 76 / 100 ) = 1140 cp = 15 * 100 = > cp = 1500 answer : a" | a ) 1500 , b ) 6789 , c ) 1200 , d ) 6151 , e ) 1421 | a | divide(1140, subtract(const_1, divide(24, const_100))) | divide(n0,const_100)|subtract(const_1,#0)|divide(n1,#1)| | gain |
after decreasing 24 % in the price of an article costs rs . 1140 . find the actual cost of an article ? | "60 students total 20 did not opt for math 15 did not opt for science 5 did not opt for either total of 40 students in math and 10 did not opt for sci but did for math 40 - 10 = 30 30 students of the class opted for both math and science answer : d . 30" | a ) 23 , b ) 25 , c ) 27 , d ) 30 , e ) 48 | d | subtract(subtract(60, 20), subtract(subtract(60, 15), 5)) | subtract(n0,n1)|subtract(n0,n2)|subtract(#1,n3)|subtract(#0,#2)| | other |
in a class of 60 students , 20 did not opt for math . 15 did not opt for science and 5 did not opt for either . how many students of the class opted for both math and science ? | "time = 6 distance = 540 3 / 2 of 6 hours = 6 * 3 / 2 = 9 hours required speed = 540 / 9 = 60 kmph c )" | a ) 48 kmph , b ) 52 kmph , c ) 6 o kmph , d ) 63 kmph , e ) 65 kmph | c | divide(540, multiply(divide(3, 2), 6)) | divide(n2,n3)|multiply(n0,#0)|divide(n1,#1)| | physics |
a jeep takes 6 hours to cover a distance of 540 km . how much should the speed in kmph be maintained to cover the same direction in 3 / 2 th of the previous time ? | "length = 5 m 44 cm = 544 cm breadth = 3 m 74 cm = 374 cm area = 544 * 374 hcf = 34 area of square = 34 * 34 cm 2 no of tiles req = 544 * 374 / 34 * 34 = 16 * 11 = 176 answer a" | a ) 176 , b ) 124 , c ) 224 , d ) 186 , e ) 190 | a | divide(multiply(add(multiply(5, const_100), 44), add(multiply(3, const_100), 74)), multiply(subtract(44, add(multiply(const_2, const_4), const_2)), subtract(44, add(multiply(const_2, const_4), const_2)))) | multiply(n0,const_100)|multiply(n2,const_100)|multiply(const_2,const_4)|add(n1,#0)|add(n3,#1)|add(#2,const_2)|multiply(#3,#4)|subtract(n1,#5)|multiply(#7,#7)|divide(#6,#8)| | physics |
a room 5 m 44 cm long and 3 m 74 cm broad needs to be paved with square tiles . what will be the least number of square tiles required to cover the floor ? | "let the numbers be 2 x and 3 x their h . c . f . = 10 so the numbers are 2 * 10 , 3 * 10 = 20,30 l . c . m . = 60 answer is b" | a ) 30 , b ) 60 , c ) 20 , d ) 10 , e ) 40 | b | sqrt(divide(10, add(power(3, 2), add(power(2, 2), power(2, 2))))) | power(n0,n1)|power(n1,n1)|power(n2,n1)|add(#0,#1)|add(#3,#2)|divide(n3,#4)|sqrt(#5)| | other |
the ratio of 2 numbers is 2 : 3 and their h . c . f . is 10 . their l . c . m . is ? | "oa is ' c ' . oe : take the remainder from each of 1250 / 18 , 1090 / 18 and so on . . 1250 / 18 gives remainder = 8 1090 / 18 gives remainder = 10 1045 / 18 gives remainder = 1 1055 / 18 gives remainder = 11 the net remainder is the product of above individual remainders . i . e = 8 * 10 * 1 * 11 break them into pairs 8 * 10 / 18 gives remainder 8 and 1 * 11 / 18 gives remainder 11 so 8 * 11 / 18 gives remainder 16 . answer : c" | a ) 34 , b ) 19 , c ) 16 , d ) 14 , e ) 10 | c | reminder(multiply(1090, 1250), 1045) | multiply(n0,n1)|reminder(#0,n2)| | general |
what is the remainder when 1250 * 1090 * 1045 * 1055 is divided by 18 ? | "explanation : required number = ( l . c . m . of 12 , 15 , 20 , 54 ) + 8 = 540 + 8 = 548 . answer : option d" | a ) 504 , b ) 536 , c ) 544 , d ) 548 , e ) none of these | d | multiply(54, const_10) | multiply(n3,const_10)| | general |
the least number , which when divided by 12 , 15 , 20 and 54 leaves in each case a remainder of 8 , is : | "( i ) 7 a - 13 b = 50 ( ii ) 2 a + 22 b = - 50 adding ( i ) and ( ii ) : 9 a + 9 b = 0 the answer is c ." | a ) - 9 , b ) - 6 , c ) 0 , d ) 6 , e ) 9 | c | divide(const_0_33, const_1000) | divide(const_0_33,const_1000)| | general |
if 7 a - 3 b = 10 b + 50 = - 12 b - 2 a , what is the value of 9 a + 9 b ? | "one monkey takes 20 min to eat 1 banana , so in 80 mins 1 monkey will eat 4 bananas , so for 80 bananas in 80 min we need 80 / 4 = 20 monkeys answer : d" | a ) 9 , b ) 10 , c ) 11 , d ) 20 , e ) 13 | d | divide(const_3.0, divide(80, 10)) | divide(n3,n1)|divide(n3,#0)| | physics |
suppose 10 monkeys take 20 minutes to eat 10 bananas . how many monkeys would it take to eat 80 bananas in 80 minutes ? | "if x < 0 and y < 0 , then we ' ll have x - x + y = 7 and x - y - y = 6 . from the first equation y = 7 , so we can discard this case since y is not less than 0 . if x > = 0 and y < 0 , then we ' ll have x + x + y = 7 and x - y - y = 6 . solving gives x = 4 > 0 and y = - 1 < 0 - - > x + y = 3 . since in ps questions only one answer choice can be correct , then the answer is c ( so , we can stop here and not even consider other two cases ) . answer : c . adding both eqn we get 2 x + ixi + iyi = 13 now considering x < 0 and y > 0 2 x - x + y = 13 we get x + y = 11 hence answer should be a" | a ) 11 , b ) - 1 , c ) 3 , d ) 5 , e ) 13 | a | multiply(6, const_2) | multiply(n1,const_2)| | general |
if x + | x | + y = 4 and x + | y | - y = 6 what is x + y = ? | "3 x + y = 40 2 x - y = 20 5 x = 60 x = 12 y = 4 4 y ^ 2 = 4 * 16 = 64 answer is e" | a ) 2 , b ) 4 , c ) 0 , d ) 10 , e ) 64 | e | multiply(3, power(subtract(40, multiply(divide(add(40, 20), add(3, 2)), 3)), 2)) | add(n1,n3)|add(n0,n2)|divide(#0,#1)|multiply(n0,#2)|subtract(n1,#3)|power(#4,n2)|multiply(n0,#5)| | general |
if 3 x + y = 40 , 2 x - y = 20 , for integers of x and y , 4 y ^ 2 = ? | "you may set up common equation like this : job / a + job / b + job / c = job / x memorize this universal formula , you will need it definitely for gmat . and find x from this equation in this specific case , the equation will look like this : 30 / 40 + 30 / 30 + 30 / 24 = 30 / x if you solve this equation , you get the same answer b ( 10 )" | a ) 5 minutes , b ) 10 minutes , c ) 15 minutes , d ) 18 minutes , e ) 20 minutes | b | divide(30, add(divide(30, 24), add(divide(30, 40), divide(30, 30)))) | divide(n0,n1)|divide(n0,n2)|divide(n0,n3)|add(#0,#1)|add(#3,#2)|divide(n0,#4)| | physics |
jonathan can type a 30 page document in 40 minutes , susan can type it in 30 minutes , and jack can type it in 24 minutes . working together , how much time will it take them to type the same document ? | "original perimeter = x hence original side = x / 4 new side = 7 x / 4 new perimeter = 4 * 7 x / 4 = 7 x correct option : c" | a ) 3 x , b ) 4 x , c ) 7 x , d ) 12 x , e ) 27 x | c | square_perimeter(multiply(7, const_4)) | multiply(n0,const_4)|square_perimeter(#0)| | geometry |
the measure of the side of a square is multiplied by 7 . if x represents the perimeter of the original square , what is the value of the new perimeter ? | "the worst case scenario will be if we remove all 17 tablets of medicine b first . the next 2 tablets we remove have to be of medicine a , so to guarantee that at least two tablets of each kind will be taken we should remove minimum of 17 + 2 = 19 tablets . answer : d ." | a ) 12 , b ) 15 , c ) 17 , d ) 19 , e ) 21 | d | add(17, const_2) | add(n1,const_2)| | general |
a box contains 10 tablets of medicine a and 17 tablets of medicine b . what is the least number of tablets that should be taken from the box to ensure that at least two tablets of each kind are among the extracted . | "6 inches = 1 / 2 feet ( there are 12 inches in a foot . ) , so 60 * 25 * 1 / 2 = 750 feet ^ 3 of milk must be removed , which equals to 750 * 7.5 = 5625 gallons . answer : d ." | a ) 100 , b ) 250 , c ) 750 , d ) 5625 , e ) 5635 | d | multiply(multiply(multiply(60, 25), divide(1, const_2)), 7.5) | divide(n3,const_2)|multiply(n0,n1)|multiply(#0,#1)|multiply(n4,#2)| | general |
the milk level in a rectangular box measuring 60 feet by 25 feet is to be lowered by 6 inches . how many gallons of milk must be removed ? ( 1 cu ft = 7.5 gallons ) | "the total number of ways to choose 2 children from 8 is 8 c 2 = 28 the number of ways to choose 1 boy and 1 girl is 4 * 4 = 16 p ( 1 boy and 1 girl ) = 16 / 28 = 4 / 7 the answer is d ." | a ) 1 / 2 , b ) 2 / 3 , c ) 3 / 5 , d ) 4 / 7 , e ) 5 / 9 | d | divide(multiply(choose(4, const_2), choose(4, const_2)), choose(add(4, 4), 2)) | add(n0,n0)|choose(n0,const_2)|choose(n0,const_2)|choose(#0,n2)|multiply(#1,#2)|divide(#4,#3)| | probability |
from a group of 4 boys and 4 girls , 2 children are to be randomly selected . what is the probability that 1 boy and 1 girl will be selected ? | since jill owns 5 of the pen , the subset from which the 2 pens hould be chosen are the 2 pens not owned by jill fom the universe of 7 . the first pen can be one of the 2 from the 7 with probability 2 / 7 . the second pen can be one of the 1 from the 6 remaining with probability 1 / 6 , the total probability will be 2 / 7 Γ 1 / 6 . on cancellation , this comes to 2 / 42 . thus , the answer is b - 2 / 42 . | a ) 5 / 42 , b ) 2 / 42 , c ) 7 / 42 , d ) 2 / 7 , e ) 5 / 7 | b | multiply(divide(subtract(7, 5), 7), divide(subtract(subtract(7, 5), const_1), subtract(7, const_1))) | subtract(n1,n2)|subtract(n1,const_1)|divide(#0,n1)|subtract(#0,const_1)|divide(#3,#1)|multiply(#2,#4) | probability |
rhonda picked 2 pen from the table , if there were 7 pens on the table and 5 belongs to jill , what is the probability that the 2 pen she picked does not belong to jill ? . | "let the market price of the product is mp . let the original cost price of the product is cp . selling price ( discounted price ) = 100 % of mp - 20 % mp = 80 % of mp . - - - - - - - - - - - - - - - - ( 1 ) profit made by selling at discounted price = 20 % of cp - - - - - - - - - - - - - - ( 2 ) apply the formula : profit w = selling price - original cost price = > 20 % of cp = 80 % of mp - 100 % cp = > mp = 120 cp / 80 = 3 / 2 ( cp ) now if product is sold without any discount , then , profit = selling price ( without discount ) - original cost price = market price - original cost price = mp - cp = 3 / 2 cp - cp = 1 / 2 cp = 50 % of cp thus , answer should bec ." | a ) 20 % , b ) 40 % , c ) 50 % , d ) 60 % , e ) 75 % | c | subtract(const_100, subtract(subtract(const_100, 20), 20)) | subtract(const_100,n0)|subtract(#0,n1)|subtract(const_100,#1)| | gain |
a merchant sells an item at a 20 % discount , but still makes a gross profit of 20 percent of the cost . what percent w of the cost would the gross profit on the item have been if it had been sold without the discount ? | "equation is correct , so math must be a problem . 0.13 * 40,000 + 0.2 * ( x - 40,000 ) = 8,000 - - > 5,200 + 0.2 x - 8,000 = 8,000 - - > 0.2 x = 10,800 - - > x = 54,000 . answer : a ." | a ) $ 54,000 , b ) $ 56,000 , c ) $ 64,000 , d ) $ 66,667 , e ) $ 80,000 | a | add(multiply(multiply(const_4, const_10), const_1000), divide(subtract(multiply(multiply(const_4, const_2), const_1000), multiply(divide(13, const_100), multiply(multiply(const_4, const_10), const_1000))), divide(20, const_100))) | divide(n0,const_100)|divide(n2,const_100)|multiply(const_2,const_4)|multiply(const_10,const_4)|multiply(#2,const_1000)|multiply(#3,const_1000)|multiply(#0,#5)|subtract(#4,#6)|divide(#7,#1)|add(#8,#5)| | general |
country x taxes each of its citizens an amount equal to 13 percent of the first $ 40,000 of income , plus 20 percent of all income in excess of $ 40,000 . if a citizen of country x is taxed a total of $ 8,000 , what is her income ? | "explanation : p Γ· 4 = 40 = > p = 40 * 4 = 160 p / 3 = 160 / 3 = 53 , remainder = 1 answer : option a" | a ) a ) 1 , b ) b ) 3 , c ) c ) 4 , d ) d ) 6 , e ) e ) 7 | a | divide(multiply(4, 40), 3) | multiply(n0,n1)|divide(#0,n3)| | general |
a number when divided by 4 , gives 40 as quotient and 0 as remainder . what will be the remainder when dividing the same number by 3 | "6 a 2 = 294 = 6 * 49 a = 7 = > a 3 = 343 cc answer : d" | a ) 8 cc , b ) 9 cc , c ) 2 cc , d ) 343 cc , e ) 6 cc | d | volume_cube(sqrt(divide(294, add(const_2, const_4)))) | add(const_2,const_4)|divide(n0,#0)|sqrt(#1)|volume_cube(#2)| | geometry |
the surface of a cube is 294 sq cm . find its volume ? | "set a : people with more than 4 years exp set b : people with degree aub = total - ( less than 4 exp and no degree ) aub = 30 - 5 = 25 aub = a + b - aib aib = 21 + 16 - 25 = 12 answer a" | a ) 12 , b ) 10 , c ) 9 , d ) 7 , e ) 5 | a | add(subtract(add(16, 21), subtract(30, 5)), subtract(21, 16)) | add(n1,n3)|subtract(n0,n4)|subtract(n3,n1)|subtract(#0,#1)|add(#3,#2)| | general |
of 30 applicants for a job , 16 had at least 4 years ' experience , 21 had degrees , and 5 had less than 4 years ' experience and did not have a degree . how many of the applicants had at least 4 years ' experience and a degree ? | "correct avg marks = 100 + ( 10 - 50 ) / 20 avg = 100 - 2 = 98 answer is c" | a ) 78 , b ) 82 , c ) 98 , d ) 91 , e ) 85 | c | divide(add(subtract(multiply(100, 20), 50), 10), 20) | multiply(n0,n1)|subtract(#0,n2)|add(n3,#1)|divide(#2,n0)| | general |
the average marks of 20 students in a class is 100 . but a student mark is wrongly noted as 50 instead of 10 then find the correct average marks ? | "10 $ per week ! an year has 52 weeks . annual charges per year = 52 * 10 = 520 $ 30 $ per month ! an year has 12 months . annual charges per year = 12 * 20 = 240 $ 520 - 240 = 280 ans e" | a ) $ 140 , b ) $ 160 , c ) $ 220 , d ) $ 240 , e ) $ 280 | e | subtract(multiply(add(multiply(10, add(const_3, const_2)), const_2), 10), multiply(20, const_12)) | add(const_2,const_3)|multiply(n1,const_12)|multiply(#0,n0)|add(#2,const_2)|multiply(n0,#3)|subtract(#4,#1)| | general |