Problem stringlengths 5 967 | Rationale stringlengths 1 2.74k | options stringlengths 37 164 | correct stringclasses 5 values | annotated_formula stringlengths 7 1.65k | linear_formula stringlengths 8 925 | category stringclasses 6 values | answer stringclasses 5 values |
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if 25 % of a number exceeds 8 % of it by 11.9 , then find the number ? | "use the elimination method to find the correct option . of all the options only 70 fits 25 % of 70 = 17.5 8 % of 70 = 5.6 17.5 - 5.6 = 11.9 required number is 70 . answer : b" | a ) 76 , b ) 70 , c ) 55 , d ) 65 , e ) 22 | b | add(divide(11.9, divide(subtract(25, 8), const_100)), 8) | subtract(n0,n1)|divide(#0,const_100)|divide(n2,#1)|add(n1,#2)| | gain | B |
the length of a rectangle is five times of the radius of a circle . the radius of the circle is equal to the side of the square , whose area is 16 sq . units . what is the area ( in sq . units ) of the rectangle if the rectangle if the breadth is 11 units ? | given that the area of the square = 16 sq . units = > side of square = Γ’ Λ Ε‘ 16 = 4 units the radius of the circle = side of the square = 4 units length of the rectangle = 5 * 4 = 20 units given that breadth = 11 units area of the rectangle = lb = 20 * 11 = 220 sq . units answer : e | ['a ) 180 sq . units', 'b ) 11 sq . units', 'c ) 110 sq . units', 'd ) 140 sq . units', 'e ) 220 sq . units'] | e | multiply(multiply(sqrt(16), add(const_2, const_3)), 11) | add(const_2,const_3)|sqrt(n0)|multiply(#0,#1)|multiply(n1,#2) | geometry | E |
in 2008 , the profits of company n were 8 percent of revenues . in 2009 , the revenues of company n fell by 20 percent , but profits were 15 percent of revenues . the profits in 2009 were what percent of the profits in 2008 ? | "the profit 0 f 2009 interms of 2008 = 0.8 * 15 / 8 * 100 = 150 % a" | a ) 150 % , b ) 105 % , c ) 120 % , d ) 124.2 % , e ) 138 % | a | multiply(divide(multiply(15, subtract(const_1, divide(20, const_100))), 8), const_100) | divide(n3,const_100)|subtract(const_1,#0)|multiply(n4,#1)|divide(#2,n1)|multiply(#3,const_100)| | gain | A |
a box contains 9 slips that are each labeled with one number : 1 , 3 , 5 , 8 , 13 , 21 , 34 and 55 . two of the slips are drawn at random from the box without replacement . what is the probability that the sum of the numbers on the two slips is equal to one of the numbers left in the box ? | probability = no : of desired outcomes / total no : of outcomes . you are picking two slips out of 8 slips . so total no : of outcomes = 8 c 2 = 28 desired outcome : sum of the numbers on the two slips is equal to one of the numbers left in the box . how many such outcomes are there ? if you look at the numbers closely , you will see that the following pair of numbers will give you the desired outcome . ( 3,5 ) ( 5,8 ) ( 8,13 ) ( 13,21 ) ( 21,34 ) . there are 7 such pairs . if the two numbers which i pick is from any of these 5 pairs , then i get my desired outcome . so no : of desired outcomes = 5 probability = 5 / 28 answer : c | a ) 7 / 72 , b ) 1 / 6 , c ) 5 / 28 , d ) 15 / 36 , e ) 21 / 36 | c | divide(subtract(8, 3), divide(multiply(8, subtract(8, 1)), const_2)) | subtract(n4,n2)|subtract(n4,n1)|multiply(n4,#1)|divide(#2,const_2)|divide(#0,#3) | general | C |
a certain quantity of 40 % solution is replaced with 25 % solution such that the new concentration is 35 % . what is the fraction of the solution that was replaced ? | "original quantity = a substituted quantity = b then : ( a * 0.4 + 0.25 * b Ρ Π° Ρ 0.4 * b ) / a = 0.35 0.4 + ( b / a ) * ( - 0.15 ) = 0.35 b / a = - 0.05 / - 0.15 = 1 / 3 answer : b" | a ) 1 / 4 , b ) 1 / 3 , c ) 1 / 2 , d ) 2 / 3 , e ) 3 / 4 | b | inverse(add(divide(subtract(35, 25), subtract(40, 35)), const_1)) | subtract(n2,n1)|subtract(n0,n2)|divide(#0,#1)|add(#2,const_1)|inverse(#3)| | other | B |
in order to obtain an income of rs . 500 from 30 % stock at rs . 90 , one must make an investment of | "explanation : market value = rs . 90 required income = rs . 500 . here face value is not given . take face value as rs . 100 if it is not given in the question to obtain rs . 30 ( ie , 30 % of the face value 100 ) , investment = rs . 90 to obtain rs . 15000 , investment = 90 / 30 Γ£ β 500 = rs . 1500 answer : option e" | a ) rs . 1550 , b ) rs . 1430 , c ) rs . 1450 , d ) rs . 1400 , e ) rs . 1500 | e | multiply(divide(90, 30), 500) | divide(n2,n1)|multiply(n0,#0)| | gain | E |
how many of the positive factors of 19 are not factors of 29 ? | "factors of 19 - 1 , 19 factors of 29 - 1 , 29 comparing both , we have three factors of 19 which are not factors of 29 - 19 , the answer is b" | a ) 0 , b ) 1 , c ) 2 , d ) 3 , e ) 4 | b | divide(29, 19) | divide(n1,n0)| | other | B |
x , y , and z are different prime numbers . the product x ^ 3 * y ^ 2 * z ^ 2 is divisible by how many different positive numbers ? | "the exponents of x ^ 3 * y ^ 2 * z ^ 2 are 3 , 2 , and 2 . the number of factors is ( 3 + 1 ) ( 2 + 1 ) ( 2 + 1 ) = 36 the answer is b ." | a ) 24 , b ) 36 , c ) 48 , d ) 60 , e ) 72 | b | subtract(power(3, const_4), const_4) | power(n0,const_4)|subtract(#0,const_4)| | general | B |
the ratio of the present age of sandy to that of molly is 7 : 2 . twelve years from now , the ratio of the ages of sandy to molly will be 5 : 2 . what was sandy ' s age 6 years ago ? | "let the present age of sandy be 7 x years and that of molly be 2 x years . ( 7 x + 12 ) / ( 2 x + 12 ) = 5 / 2 4 x = 36 x = 9 six years ago , sandy ' s age was 7 ( 9 ) - 6 = 57 the answer is d ." | a ) 39 , b ) 45 , c ) 51 , d ) 57 , e ) 63 | d | subtract(divide(multiply(subtract(multiply(5, add(const_10, 6)), multiply(2, add(const_10, 6))), 7), add(2, 2)), 6) | add(n4,const_10)|add(n1,n1)|multiply(n2,#0)|multiply(n1,#0)|subtract(#2,#3)|multiply(n0,#4)|divide(#5,#1)|subtract(#6,n4)| | other | D |
what is the average ( arithmetic mean ) of the numbers 100 , 150 , 200 , 200 , 250 , and 300 ? | "{ 100 , 150 , 200 , 200 , 250 , 300 } = { 200 - 100,200 - 50 , 200 , 200,200 + 50,200 + 100 } - - > the average = 200 . answer : d ." | a ) 100 , b ) 150 , c ) 140 , d ) 200 , e ) 250 | d | divide(divide(multiply(add(100, 200), add(divide(subtract(200, 100), 100), const_1)), const_2), add(divide(subtract(200, 100), 100), const_1)) | add(n0,n2)|subtract(n2,n0)|divide(#1,n0)|add(#2,const_1)|multiply(#0,#3)|divide(#4,const_2)|divide(#5,#3)| | general | D |
if a rectangular room measures 11 meters by 6 meters by 4 meters , what is the volume of the room in cubic centimeters ? ( 1 meter = 100 centimeters ) | "e . 264 , 000,000 11 * 100 * 6 * 100 * 4 * 100 = 264 , 000,000" | a ) 24,000 , b ) 240,000 , c ) 2 , 400,000 , d ) 24 , 000,000 , e ) 264 , 000,000 | e | multiply(multiply(multiply(multiply(4, 100), divide(1, const_10)), multiply(6, 100)), multiply(11, 100)) | divide(n3,const_10)|multiply(n2,n4)|multiply(n1,n4)|multiply(n0,n4)|multiply(#0,#1)|multiply(#4,#2)|multiply(#5,#3)| | geometry | E |
a , b and c rent a pasture . if a puts 10 oxen for 7 months , b puts 12 oxen for 5 months and c puts 15 oxen for 3 months for grazing and the rent of the pasture is rs . 210 , then how much amount should c pay as his share of rent ? | "a : b : c = 10 Γ 7 : 12 Γ 5 : 15 Γ 3 = 2 Γ 7 : 12 Γ 1 : 3 Γ 3 = 14 : 12 : 9 amount that c should pay = 210 Γ 9 / 35 = 6 Γ 9 = 54 answer is a" | a ) 54 , b ) 45 , c ) 25 , d ) 15 , e ) 55 | a | multiply(210, divide(multiply(15, 3), add(add(multiply(10, 7), multiply(12, 5)), multiply(15, 3)))) | multiply(n4,n5)|multiply(n0,n1)|multiply(n2,n3)|add(#1,#2)|add(#3,#0)|divide(#0,#4)|multiply(n6,#5)| | general | A |
in the manufacture of a certain product , 9 percent of the units produced are defective and 4 percent of the defective units are shipped for sale . what percent of the units produced are defective units that are shipped for sale ? | percent of defective produced = 9 % percent of the defective units that are shipped for sale = 4 % percent of units produced are defective units that are shipped for sale = ( 4 / 100 ) * ( 9 / 100 ) * 100 % = ( 36 / 10000 ) * 100 % = ( 36 / 100 ) % = . 36 % answer b | a ) 0.125 % , b ) 0.36 % , c ) 0.8 % , d ) 1.25 % , e ) 2.0 % | b | multiply(9, divide(4, const_100)) | divide(n1,const_100)|multiply(n0,#0) | gain | B |
a cubical block of metal weighs 7 pounds . how much will another cube of the same metal weigh if its sides are twice as long ? | "for example our cube have a side 1 meter , so we have 1 cubical meter in this cube and this cubical meter weigth 7 pounds if we take cube with side 2 meters we will have 8 cubical meters in this cube 8 meters * 7 pounds = 56 pounds so answer is d and similar but more theoretical approach : if we have sides a and b than they have equal ration with their areas : a / b = a ^ 2 / b ^ 2 and they have equal ration with their volumes : a / b = a ^ 3 / b ^ 3 we have two sides 1 / 2 so their volume will be in ratio 1 / 8 weight of one cube * volume of another cube 7 * 8 = 56 so answer is d" | a ) 48 , b ) 32 , c ) 24 , d ) 56 , e ) 12 | d | multiply(7, multiply(const_2, const_4)) | multiply(const_2,const_4)|multiply(n0,#0)| | geometry | D |
a number when multiplied by 3 is odd and divisible by 9 . this number when multiplied by 4 is 108 . what is the original number ? | working backward , 108 divided by 4 is 27 . this number fits the rules of three and nine , where the digits add up to a sum divisible by nine . answer is e . | a ) 10 , b ) 18 , c ) 13 , d ) 21 , e ) 27 | e | divide(108, 4) | divide(n3,n2) | general | E |
a fifth of arun Γ’ β¬ β’ s marks in mathematics exceed a third of his marks in english by 20 . if he got 260 marks in two subjects together how many marks did he got in english ? | let arun Γ’ β¬ β’ s marks in mathematics and english be x and y then ( 1 / 5 ) x - ( 1 / 3 ) y = 20 3 x - 5 y = 300 Γ’ β¬ Β¦ Γ’ β¬ Β¦ > ( 1 ) x + y = 260 Γ’ β¬ Β¦ Γ’ β¬ Β¦ . > ( 2 ) solving ( 1 ) and ( 2 ) x = 200 and y = 60 answer is c . | a ) 12080 , b ) 18060 , c ) 20060 , d ) 20040 , e ) none of them | c | multiply(multiply(multiply(260, 20), const_2), const_2) | multiply(n0,n1)|multiply(#0,const_2)|multiply(#1,const_2) | general | C |
the average age of a group of 10 persons was decreased by 3 years when one person , whose age was 45 years , was replaced by a new person . find the age of the new person ? | initial average age of the 10 persons be p . age of the new person q . sum of the ages of the initial 10 persons = 10 p new average = ( p - 3 ) 10 ( p - 3 ) = 10 p - 45 + q = > q = 15 answer : d | a ) 18 , b ) 56 , c ) 12 , d ) 15 , e ) 14 | d | subtract(45, multiply(10, 3)) | multiply(n0,n1)|subtract(n2,#0) | general | D |
an escalator moves towards the top level at the rate of 15 ft . sec and its length is 180 feet . if a person walks on the moving escalator at the rate of 3 feet per second towards the top level , how much time does he take to cover the entire length . | "explanation : time taken to cover the entire length = tot . dist / resultant speed = 180 / ( 15 + 3 ) = 10 sec answer : e" | a ) 12 sec , b ) 20 sec , c ) 15 sec , d ) 18 sec , e ) 10 sec | e | divide(180, add(15, 3)) | add(n0,n2)|divide(n1,#0)| | gain | E |
a box contains 10 tablets of medicine a and 14 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 . | the worst case scenario will be if we remove all 14 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 14 + 2 = 16 tablets . answer : b . | a ) 12 , b ) 16 , c ) 17 , d ) 19 , e ) 21 | b | add(14, const_2) | add(n1,const_2) | general | B |
if 0.5 % of a = 75 paise , then the value of a is ? | "answer β΅ 0.5 / 100 of a = 75 / 100 β΄ a = rs . ( 75 / 0.5 ) = rs . 150 correct option : a" | a ) rs . 150 , b ) rs . 17 , c ) rs . 1.70 , d ) rs . 4.25 , e ) none | a | divide(75, 0.5) | divide(n1,n0)| | gain | A |
the diagonals of a rhombus are 22 cm and 30 cm . find its area ? | "1 / 2 * 22 * 30 = 330 answer : d" | a ) 358 , b ) 329 , c ) 350 , d ) 330 , e ) 317 | d | rhombus_area(22, 30) | rhombus_area(n0,n1)| | geometry | D |
a textile manufacturing firm employees 125 looms . it makes fabrics for a branded company . the aggregate sales value of the output of the 125 looms is rs 5 , 00,000 and the monthly manufacturing expenses is rs 1 , 50,000 . assume that each loom contributes equally to the sales and manufacturing expenses are evenly spread over the number of looms . monthly establishment charges are rs 75000 . if one loom breaks down and remains idle for one month , the decrease in profit is : | "explanation : profit = 5 , 00,000 Γ’ Λ β ( 1 , 50,000 + 75,000 ) = rs . 2 , 75,000 . since , such loom contributes equally to sales and manufacturing expenses . but the monthly charges are fixed at rs 75,000 . if one loan breaks down sales and expenses will decrease . new profit : - = > 500000 Γ£ β ( 124 / 125 ) Γ’ Λ β 150000 Γ£ β ( 124 / 125 ) Γ’ Λ β 75000 . = > rs 2 , 72,200 . decrease in profit = > 2 , 75,000 Γ’ Λ β 2 , 72,200 = > rs . 2,800 . answer : d" | a ) 13000 , b ) 7000 , c ) 10000 , d ) 2800 , e ) none of these | d | subtract(subtract(multiply(multiply(5, const_1000), const_100), add(multiply(75000, const_2), 75000)), subtract(subtract(multiply(divide(subtract(125, 1), 125), multiply(multiply(5, const_1000), const_100)), multiply(multiply(75000, const_2), divide(subtract(125, 1), 125))), 75000)) | multiply(n2,const_1000)|multiply(n6,const_2)|subtract(n0,n4)|add(n6,#1)|divide(#2,n0)|multiply(#0,const_100)|multiply(#4,#5)|multiply(#4,#1)|subtract(#5,#3)|subtract(#6,#7)|subtract(#9,n6)|subtract(#8,#10)| | general | D |
amar takes as much time in running 18 meters as a car takes in covering 48 meters . what will be the distance covered by amar during the time the car covers 1.2 km ? | "c 300 m distance covered by amar = 18 / 4.8 ( 1.6 km ) = 3 / 8 ( 1200 ) = 300 m answer is c" | a ) 600 m , b ) 200 m , c ) 300 m , d ) 400 m , e ) 100 m | c | divide(multiply(18, multiply(1.2, const_1000)), 48) | multiply(n2,const_1000)|multiply(n0,#0)|divide(#1,n1)| | physics | C |
working alone , printers x , y , and z can do a certain printing job , consisting of a large number of pages , in 16 , 10 , and 20 hours , respectively . what is the ratio of the time it takes printer x to do the job , working alone at its rate , to the time it takes printers y and z to do the job , working together at their individual rates ? | "the time it takes printer x is 16 hours . the combined rate of y and z is 1 / 10 + 1 / 20 = 3 / 20 the time it takes y and z is 20 / 3 the ratio of times is 16 / ( 20 / 3 ) = 3 * 16 / 20 = 12 / 5 the answer is c ." | a ) 7 / 4 , b ) 10 / 3 , c ) 12 / 5 , d ) 15 / 7 , e ) 18 / 11 | c | divide(16, divide(const_1, add(divide(const_1, 10), divide(const_1, 20)))) | divide(const_1,n1)|divide(const_1,n2)|add(#0,#1)|divide(const_1,#2)|divide(n0,#3)| | general | C |
the probability of two events a and b are 0.25 and 0.35 respectively . the probability that both a and b occur is 0.15 . the probability that neither a nor b occur is _________ | "we are apply that formula . . . . . . . . . . . . . . p ( aorb ) = p ( a ) + p ( b ) - p ( a and b ) = . 25 + . 35 - . 15 = . 45 but the probability of neither a nor b = 1 - . 45 = 0.55 answer : a" | a ) 0.55 , b ) 0.4 , c ) 0.5 , d ) 0.05 , e ) 0.6 | a | subtract(const_1, subtract(add(0.25, 0.35), 0.15)) | add(n0,n1)|subtract(#0,n2)|subtract(const_1,#1)| | other | A |
what number has a 20 : 1 ratio to the number 10 ? | "20 : 1 = x : 10 x = 10 * 20 x = 150 answer : c" | a ) 130 , b ) 100 , c ) 200 , d ) 150 , e ) 120 | c | multiply(10, 20) | multiply(n0,n2)| | other | C |
if p # q denotes the least common multiple of p and q , then t = ( ( 12 # 16 ) # ( 18 # 24 ) ) = ? | "there are several ways to find the least common multiple of two numbers . in this case , the most efficient method is to use the greatest common factor : ( a * b ) / ( gcf ab ) = lcm ab the greatest common factor of 12 and 16 is 4 . so , 12 # 16 = 12 * 16 / 4 = 48 . the greatest common factor of 18 and 24 is 6 . so , 18 # 24 = 18 * 24 / 6 = 72 finally , the greatest common factor of 48 and 72 is 24 . so , t = ( ( 12 # 16 ) # ( 18 # 24 ) ) = 48 # 72 = ( 48 * 72 ) / 24 = 2 * 72 = 144 the correct answer is c ." | a ) 216 , b ) 180 , c ) 144 , d ) 108 , e ) 72 | c | add(divide(subtract(multiply(18, 24), multiply(12, 16)), const_2), 24) | multiply(n2,n3)|multiply(n0,n1)|subtract(#0,#1)|divide(#2,const_2)|add(n3,#3)| | general | C |
at 3.40 , the hour hand and the minute hand of a clock form an angle of | sol : use formula ΞΈ = β£ β£ β£ 30 h β 112 m β£ β£ β£ ΞΈ = | 30 h β 112 m | angle = 30 Γ 3 β 11 / 2 Γ 40 = 90 β 220 = 130 Β° answer : e | a ) 378 , b ) 277 , c ) 297 , d ) 267 , e ) 130 | e | subtract(multiply(const_10, multiply(const_3, power(const_2, const_3))), multiply(subtract(subtract(multiply(3.4, add(const_4, const_1)), add(const_4, const_1)), const_1), const_10)) | add(const_1,const_4)|power(const_2,const_3)|multiply(#1,const_3)|multiply(n0,#0)|multiply(#2,const_10)|subtract(#3,#0)|subtract(#5,const_1)|multiply(#6,const_10)|subtract(#4,#7) | physics | E |
a dealer offers a cash discount of 20 % and still makes a profit of 20 % when he further allows 10 articles to be sold at the cost price of 9 articles to a particular sticky bargainer . how much percent above the cost price were his articles listed ? | "given cash discount - 20 % profit - 20 % items sold - 10 price sold at = list price of 9 assume list price = $ 10 total invoice = $ 90 - 20 % cash discount = $ 72 let cost price of 10 items be x so total cost = 10 * x given the shopkeeper had a profit of 20 % 10 * x * 120 / 100 = 72 or x = $ 6 which means his products were listed at $ 10 which is a 66 + ( 2 / 3 ) % markup over $ 6 answer d" | a ) 50 % , b ) 55 % , c ) 60 % , d ) 66 + ( 2 / 3 ) % , e ) 80 % | d | multiply(subtract(divide(divide(divide(add(const_100, 20), const_100), subtract(const_1, divide(subtract(10, 9), 10))), divide(subtract(const_100, 20), const_100)), const_1), const_100) | add(n1,const_100)|subtract(n2,n3)|subtract(const_100,n0)|divide(#0,const_100)|divide(#1,n2)|divide(#2,const_100)|subtract(const_1,#4)|divide(#3,#6)|divide(#7,#5)|subtract(#8,const_1)|multiply(#9,const_100)| | gain | D |
how many multiples of 5 are there between 5 to 95 ? | "as you know , multiples of 5 are integers having 0 or 5 in the digit to the extreme right ( i . e . the unit β s place ) . so the numbers are 10 , 15 , 20 , 25 , 30 , 35 , 40 , 45 , 50 , 55 , 60 , 65 , 70 , 75 , 80 , 85 , 90 . answer c" | a ) 9 , b ) 18 , c ) 17 , d ) 16 , e ) 15 | c | add(divide(subtract(95, 5), 5), const_1) | subtract(n2,n1)|divide(#0,n0)|add(#1,const_1)| | general | C |
a 450 m long train crosses a platform in 39 sec while it crosses a signal pole in 18 sec . what is the length of the platform ? | "speed = 450 / 18 = 25 m / sec . let the length of the platform be x meters . then , ( x + 450 ) / 39 = 25 = > x = 975 m . l = 975 - 450 = 525 answer : option b" | a ) 600 , b ) 525 , c ) 360 , d ) 370 , e ) 380 | b | subtract(multiply(speed(450, 18), 39), 450) | speed(n0,n2)|multiply(n1,#0)|subtract(#1,n0)| | physics | B |
a lady grows cabbage in her garden that is in the shape of a square . each cabbage takes 1 square foot of area in her garden . this year , she has increased her output by 127 cabbages when compared to last year . the shape of the area used for growing the cabbage has remained a square in both these years . how many cabbages did she produce this year ? | "explanatory answer the shape of the area used for growing cabbage has remained a square in both the years . let the side of the square area used for growing cabbage this year be x ft . therefore , the area of the ground used for cultivation this year = x 2 sq . ft . let the side of the square area used for growing cabbage last year be y ft . therefore , the area of the ground used for cultivation last year = y 2 sq . ft . as the number of cabbage grown has increased by 127 , the area would have increased by 127 sq ft because each cabbage takes 1 sq ft space . hence , x 2 - y 2 = 127 ( x + y ) ( x - y ) = 127 . 127 is a prime number and hence it will have only two factors . i . e . , 127 and 1 . therefore , 127 can be expressed as product of 2 numbers in only way = 127 * 1 i . e . , ( x + y ) ( x - y ) = 127 * 1 so , ( x + y ) should be 127 and ( x - y ) should be 1 . solving the two equations we get x = 64 and y = 63 . therefore , number of cabbage produced this year = x 2 = 642 = 4096 . alternative approach : use answer choices the area in both the years are squares of two numbers . that rules out choice b , c and d . as 4098,5000 and 5096 are not the square of any number . check choice a : if this year ' s produce is 4096 , last year ' s produce would have been 4096 - 127 = 3969 3969 is the square of 63 . so , 4096 is the answer . choice a" | a ) 4096 , b ) 4098 , c ) 5000 , d ) 5096 , e ) can not be determined | a | power(add(divide(subtract(127, 1), const_2), 1), const_2) | subtract(n1,n0)|divide(#0,const_2)|add(n0,#1)|power(#2,const_2)| | geometry | A |
a particular store purchased a stock of turtleneck sweaters and marked up its cost by 20 % . during the new year season , it further marked up its prices by 25 % of the original retail price . in february , the store then offered a discount of 8 % . what was its profit on the items sold in february ? | "assume the total price = 100 x price after 20 % markup = 120 x price after 25 % further markup = 1.25 * 120 x = 150 x price after the discount = 0.92 * 150 x = 138 x hence total profit = 38 % option e" | a ) 27.5 % , b ) 30 % , c ) 35 % , d ) 37.5 % , e ) 38 % | e | subtract(multiply(divide(subtract(const_100, 8), const_100), multiply(add(const_100, 20), divide(add(const_100, 25), const_100))), const_100) | add(n0,const_100)|add(n1,const_100)|subtract(const_100,n2)|divide(#2,const_100)|divide(#1,const_100)|multiply(#0,#4)|multiply(#3,#5)|subtract(#6,const_100)| | gain | E |
what no . should be subtracted from x ^ 3 + 4 x 2 β 7 x + 12 x ^ 3 + 4 x ^ 2 β 7 x + 12 , if it is to be perfectly divisible by x + 3 x + 3 ? | according to remainder theorem when dfracf ( x ) x + adfracf ( x ) x + a , then the remainder is f ( β a ) f ( β a ) . in this case , as x + 3 x + 3 divides x 3 + 4 x 2 β 7 x + 12 β kx 3 + 4 x 2 β 7 x + 12 β k perfectly ( kk being the number to be subtracted ) , the remainder is 0 when the value of xx is substituted by - 3 . i . e . , ( β 3 ) 3 + 4 ( β 3 ) 2 β 7 ( β 3 ) + 12 β k = 0 ( β 3 ) 3 + 4 ( β 3 ) 2 β 7 ( β 3 ) + 12 β k = 0 or β 27 + 36 + 21 + 12 = k β 27 + 36 + 21 + 12 = k or k = k = 42 b | a ) 38 , b ) 42 , c ) 46 , d ) 49 , e ) 62 | b | add(subtract(add(power(negate(3), 3), multiply(4, power(negate(3), 2))), multiply(7, negate(3))), 12) | negate(n0)|multiply(n3,#0)|power(#0,n2)|power(#0,n0)|multiply(n1,#2)|add(#4,#3)|subtract(#5,#1)|add(n4,#6) | general | B |
if x is to be chosen at random from the set { 1 , 2 , 3 , 4 } and y is to be chosen at random from the set { 5 , 6 , 7 } , what is the probability that xy will be even ? | "in order to make even nos . by multiplication , we should have even * odd , odd * even or even * even total even nos . possible by multiplying nos . from the 2 sets : ( 1 * 6 ) ; 2 * any of the three from set b ; 3 * 6 ; & 4 * any of the three from set b 1 + 3 + 1 + 3 = 8 total possibilities = 4 * 3 = 12 p ( event ) = 8 / 12 or 2 / 3 ans d" | a ) 1 / 6 , b ) 1 / 3 , c ) 1 / 2 , d ) 2 / 3 , e ) 5 / 6 | d | divide(power(2, 2), multiply(choose(4, 2), choose(2, 1))) | choose(n3,n1)|choose(n1,n0)|power(n1,n1)|multiply(#0,#1)|divide(#2,#3)| | probability | D |
if 4 cats can kill 4 rats in 4 minutes , how long will it take 100 cats to kill 100 rats ? | "it will take 4 minutes for 100 cats to kill 100 rats . 1 cat can kill 1 rat in 4 minutes , so 100 cats can kill 100 rats in 4 minutes answer d" | a ) 1 minutes , b ) 2 minutes , c ) 3 minutes , d ) 4 minutes , e ) 5 minutes | d | multiply(4, const_1) | multiply(n0,const_1)| | physics | D |
rs . 705 is divided amongst a , b , c so that 3 times a ' s share , 5 times b ' s share and 4 times c ' s share are all equal . find b ' s share ? | a + b + c = 705 3 a = 5 b = 4 c = x a : b : c = 1 / 3 : 1 / 5 : 1 / 4 = 20 : 12 : 15 12 / 47 * 705 = rs . 180 answer : c | a ) 177 , b ) 150 , c ) 180 , d ) 716 , e ) 616 | c | divide(705, add(add(divide(4, 3), divide(4, 5)), const_1)) | divide(n3,n1)|divide(n3,n2)|add(#0,#1)|add(#2,const_1)|divide(n0,#3)| | general | C |
the average weight of 8 person ' s increases by 3.5 kg when a new person comes in place of one of them weighing 65 kg . what might be the weight of the new person ? | "total weight increased = ( 8 x 3.5 ) kg = 28 kg . weight of new person = ( 65 + 28 ) kg = 93 kg . e )" | a ) kg , b ) 70 kg , c ) 80 kg , d ) 90 kg , e ) 93 kg | e | add(multiply(8, 3.5), 65) | multiply(n0,n1)|add(n2,#0)| | general | E |
the average salary of 15 people in the shipping department at a certain firm is $ 20,000 . the salary of 5 of the employees is $ 25,000 each and the salary of 4 of the employees is $ 15,000 each . what is the average salary of the remaining employees ? | "total salary . . . 15 * 20 k = 300 k 5 emp @ 25 k = 125 k 4 emp @ 15 k = 60 k remaing 6 emp sal = 300 k - 125 k - 60 k = 115 k average = 115 k / 6 = 19100 ans : d" | a ) $ 19,250 , b ) $ 18,500 , c ) $ 18,000 , d ) $ 19,100 , e ) $ 12,300 | d | divide(subtract(subtract(multiply(multiply(5, 4), multiply(4, 4)), multiply(multiply(5, 5), 5)), multiply(4, 15)), add(const_2, 5)) | add(n2,const_2)|multiply(n2,n4)|multiply(n4,n4)|multiply(n2,n2)|multiply(n0,n4)|multiply(#1,#2)|multiply(n2,#3)|subtract(#5,#6)|subtract(#7,#4)|divide(#8,#0)| | general | D |
pipe a that can fill a tank in an hour and pipe b that can fill the tank in half an hour are opened simultaneously when the tank is empty . pipe b is shut 10 minutes before the tank overflows . when will the tank overflow ? | the last 10 minutes only pipe a was open . since it needs 1 hour to fill the tank , then in 10 minutes it fills 1 / 5 th of the tank , thus 5 / 6 of the tank is filled with both pipes open . the combined rate of two pipes is 1 + 2 = 3 tanks / hour , therefore to fill 5 / 6 th of the tank they need ( time ) = ( work ) / ( rate ) = ( 5 / 6 ) / 3 = 5 / 18 hours = 27 minutes . total time = 10 + 27 = 37 minutes . answer : a . | a ) 37 mins , b ) 35 mins , c ) 40 mins , d ) 32 mins , e ) 36 mins | a | subtract(add(divide(multiply(add(divide(multiply(multiply(const_1, const_60), divide(multiply(const_1, const_60), const_2)), divide(multiply(const_1, const_60), const_2)), divide(multiply(multiply(const_1, const_60), divide(multiply(const_1, const_60), const_2)), multiply(const_1, const_60))), 10), divide(multiply(multiply(const_1, const_60), divide(multiply(const_1, const_60), const_2)), multiply(const_1, const_60))), 10), const_3) | multiply(const_1,const_60)|divide(#0,const_2)|multiply(#1,#0)|divide(#2,#1)|divide(#2,#0)|add(#3,#4)|multiply(n0,#5)|divide(#6,#4)|add(n0,#7)|subtract(#8,const_3) | physics | A |
average of money that group of 4 friends pay for rent each month is $ 800 . after one persons rent is increased by 20 % the new mean is $ 880 . what was original rent of friend whose rent is increased ? | 0.2 x = 4 ( 880 - 800 ) 0.2 x = 320 x = 1600 answer c | a ) 800 , b ) 900 , c ) 1600 , d ) 1100 , e ) 1200 | c | divide(multiply(subtract(880, 800), 4), divide(20, const_100)) | divide(n2,const_100)|subtract(n3,n1)|multiply(n0,#1)|divide(#2,#0) | general | C |
list i : 3 , 4 , 8 , 19 list ii : x , 3 , 4 , 8 , 19 | we start by calculating the median of the numbers of list i : 3 , 4 , 8 , 19 . we see that the numbers in the list are in order already and , since we have an even number of numbers , the median is the average of the two middle numbers . median = ( 4 + 8 ) / 2 median = 12 / 2 median = 6 the median of list i is 6 . looking at list ii : x , 3 , 4 , 8 , 19 , we see that we have an odd number of terms . thus , when the list is ordered from least to greatest the median must be the middle term . since the medians of the two lists must be equal , we know that the median of list ii must be 6 and therefore x is 6 . the answer is a . | a ) 6 , b ) 7 , c ) 8 , d ) 9 , e ) 10 | a | multiply(subtract(const_3, const_1), 3) | subtract(const_3,const_1)|multiply(n0,#0) | general | A |
the youngest of 4 children has siblings who are 3 , 6 , and 7 years older than she is . if the average ( arithmetic mean ) age of the 4 siblings is 30 , what is the age of the youngest sibling ? | "x + ( x + 3 ) + ( x + 6 ) + ( x + 7 ) = 120 4 x + 16 = 120 4 x = 104 x = 26 the answer is e ." | a ) 22 , b ) 23 , c ) 24 , d ) 25 , e ) 26 | e | divide(subtract(multiply(4, 30), add(add(4, 6), 7)), 4) | add(n0,n2)|multiply(n0,n5)|add(n3,#0)|subtract(#1,#2)|divide(#3,n0)| | general | E |
renu can do a piece of work in 5 days , but with the help of her friend suma , she can do it in 4 days . in what time suma can do it alone ? | "renu Γ’ β¬ β’ s one day Γ’ β¬ β’ s work = 1 / 5 suma Γ’ β¬ β’ s one day Γ’ β¬ β’ s work = 1 / 4 - 1 / 5 = 1 / 20 suma can do it alone in 20 days . answer : e" | a ) 10 , b ) 12 , c ) 14 , d ) 15 , e ) 20 | e | inverse(subtract(divide(const_1, 4), divide(const_1, 5))) | divide(const_1,n1)|divide(const_1,n0)|subtract(#0,#1)|inverse(#2)| | physics | E |
in feb mrs bil ' s earnings were 70 percent of the bil family ' s total income . in june mrs bil earned 10 percent more than in may . if the rest of the family ' s income was the same both months , then , in june , mrs bil ' s earnings were approximately what percent of the bil family ' s total income ? | lets say the family income is 100 in may , bil earned 70 family income is 30 in june , bil earned 10 % more than may , so it is ( 70 + 10 * 70 / 100 = 77 ) family income is same 30 in june bil ' s income percent is 77 * 100 / 107 ~ 72 ans is a | a ) 72 , b ) 34 , c ) 78 , d ) 37 , e ) 27 | a | multiply(divide(add(70, divide(70, 10)), add(const_100, divide(70, 10))), const_100) | divide(n0,n1)|add(n0,#0)|add(#0,const_100)|divide(#1,#2)|multiply(#3,const_100) | general | A |
if 3 x = 8 y = z , what is x + y , in terms of z ? | "3 x = 8 y = z x = z / 3 and y = z / 8 x + y = z / 3 + z / 8 = 11 z / 24 answer is c" | a ) z / 2 , b ) 2 z , c ) 11 z / 24 , d ) 3 z / 5 , e ) z / 9 | c | divide(subtract(divide(multiply(3, const_100), const_2), const_2), add(divide(multiply(3, const_100), const_2), const_2)) | multiply(n0,const_100)|divide(#0,const_2)|add(#1,const_2)|subtract(#1,const_2)|divide(#3,#2)| | general | C |
the total marks obtained by a student in mathematics and physics is 60 and his score in chemistry is 20 marks more than that in physics . find the average marks scored in mathamatics and chemistry together ? | "let the marks obtained by the student in mathematics , physics and chemistry be m , p and c respectively . given , m + c = 60 and c - p = 20 m + c / 2 = [ ( m + p ) + ( c - p ) ] / 2 = ( 60 + 20 ) / 2 = 40 . answer : a" | a ) 40 , b ) 99 , c ) 88 , d ) 77 , e ) 66 | a | divide(add(60, 20), const_2) | add(n0,n1)|divide(#0,const_2)| | general | A |
think of a number , divide it by 5 and add 6 to it . the result is 65 . what is the number thought of ? | "explanation : 65 - 6 = 59 59 x 5 = 295 answer : c" | a ) 24 , b ) 77 , c ) 295 , d ) 267 , e ) 29 | c | multiply(subtract(65, 6), 5) | subtract(n2,n1)|multiply(n0,#0)| | general | C |
a person borrows rs . 4000 for 2 years at 4 % p . a . simple interest . he immediately lends it to another person at 6 p . a for 2 years . find his gain in the transaction per year . | "gain in 2 years = [ ( 4000 * 6 * 2 ) / 100 ] - [ ( 4000 * 4 * 2 ) / 100 ] 480 - 320 = 160 gain in 1 year = ( 160 / 2 ) = 80 rs answer : b" | a ) 100 rs , b ) 80 rs , c ) 160 rs , d ) 180 rs , e ) 200 rs | b | divide(subtract(divide(multiply(multiply(4000, 6), 2), const_100), divide(multiply(multiply(4000, 4), 2), const_100)), 2) | multiply(n0,n3)|multiply(n0,n2)|multiply(n1,#0)|multiply(n1,#1)|divide(#2,const_100)|divide(#3,const_100)|subtract(#4,#5)|divide(#6,n1)| | gain | B |
the speed of a car increases by 2 kms after every one hour . if the distance travelling in the first one hour was 50 kms . what was the total distance traveled in 12 hours ? | "explanation : total distance travelled in 12 hours = ( 50 + 52 + 54 + . . . . . upto 12 terms ) this is an a . p with first term , a = 50 , number of terms , n = 12 , d = 2 . required distance = 12 / 2 [ 2 x 50 + { 12 - 1 ) x 2 ] = 6 ( 122 ) = 732 kms . answer : c" | a ) 252 kms , b ) 152 kms , c ) 732 kms , d ) 752 kms , e ) 152 kms | c | multiply(add(multiply(2, 50), multiply(subtract(12, const_1), 2)), divide(12, 2)) | divide(n2,n0)|multiply(n0,n1)|subtract(n2,const_1)|multiply(n0,#2)|add(#1,#3)|multiply(#4,#0)| | physics | C |
if two times of the daughter β s age in years is included to the mother β s age , the total is 70 and if two times of the mother β s age is included to the daughter β s age , the total is 95 . so the mother β s age is , | b 40 let daughter β s age = a and mother β s age = b given : 2 a + b = 70 and , a + 2 b = 95 solving b , we will get b = 40 . | a ) 30 , b ) 40 , c ) 50 , d ) 60 , e ) 39 | b | divide(subtract(multiply(95, const_2), 70), const_3) | multiply(n1,const_2)|subtract(#0,n0)|divide(#1,const_3) | general | B |
the sum of all the integers k such that β 20 < k < 24 is | "- 19 - - - - - - - - - - - - - - - - - - 0 - - - - - - - - - - - - - - - - - 23 values upto + 23 cancels outwe are left with only - 19 - 18 sum of which is - 37 . hence option d . e" | a ) 0 , b ) - 2 , c ) - 25 , d ) - 49 , e ) - 37 | e | add(add(negate(20), const_1), add(add(negate(20), const_1), const_1)) | negate(n0)|add(#0,const_1)|add(#1,const_1)|add(#1,#2)| | general | E |
7 , 10.5 , 15.75 , 23.63 , 35.43 , ( . . . ) | "7 ( 7 Γ£ β 3 ) Γ£ Β· 2 = 10.5 ( 10.5 Γ£ β 3 ) Γ£ Β· 2 = 15.75 ( 15.75 Γ£ β 3 ) Γ£ Β· 2 = 23.63 ( 23.63 Γ£ β 3 ) Γ£ Β· 2 = 35.43 ( 35.43 Γ£ β 3 ) Γ£ Β· 2 = 53.156 answer is b" | a ) 52.112 , b ) 53.156 , c ) 54 , d ) 89 , e ) 88 | b | subtract(negate(23.63), multiply(subtract(10.5, 15.75), divide(subtract(10.5, 15.75), subtract(7, 10.5)))) | negate(n3)|subtract(n1,n2)|subtract(n0,n1)|divide(#1,#2)|multiply(#3,#1)|subtract(#0,#4)| | general | B |
a train leaves mumabai at 9 am at a speed of 45 kmph . after one hour , another train leaves mumbai in the same direction as that of the first train at a speed of 90 kmph . when and at what distance from mumbai do the two trains meet ? | "when the second train leaves mumbai the first train covers 45 * 1 = 45 km so , the distance between first train and second train is 45 km at 10.00 am time taken by the trains to meet = distance / relative speed = 45 / ( 90 - 45 ) = 1 hours so , the two trains meet at 2 p . m . the two trains meet 1 * 90 = 90 km away from mumbai . answer : c" | a ) 287 , b ) 279 , c ) 90 , d ) 278 , e ) 379 | c | multiply(divide(multiply(45, const_1), subtract(90, 45)), 90) | multiply(n1,const_1)|subtract(n2,n1)|divide(#0,#1)|multiply(n2,#2)| | physics | C |
if ( 1 β 1.25 ) n = 5 , then n = | "( 1 β 1.25 ) n = 5 simplify to get : - 0.25 n = 5 rewrite as ( - 1 / 4 ) n = 5 multiply both sides by - 4 to get : n = - 20 answer : a" | a ) β 20 , b ) β 140 , c ) β 4 , d ) 4 , e ) 400 | a | negate(multiply(5, const_4)) | multiply(n2,const_4)|negate(#0)| | general | A |
on my sister ' s birthday , she was 154 cm in height , having grown 10 % since the year before . how tall was she the previous year ? | "let the previous year ' s height be x . 1.1 x = 154 x = 140 the answer is b ." | a ) 150 cm , b ) 140 cm , c ) 142 cm , d ) 148 cm , e ) 146 cm | b | subtract(154, divide(multiply(154, 10), const_100)) | multiply(n0,n1)|divide(#0,const_100)|subtract(n0,#1)| | physics | B |
bob wants to run a mile in the same time as his sister . if bob β s time for a mile is currently 10 minutes 40 seconds and his sister β s time is currently 10 minutes 8 seconds , by what percent does bob need to improve his time in order run a mile in the same time as his sister ? | "bob ' s time = 640 secs . his sis ' time = 608 secs . percent increase needed = ( 640 - 608 / 640 ) * 100 = 32 / 640 * 100 = 5 % . ans ( b ) ." | a ) 3 % , b ) 5 % , c ) 8 % , d ) 10 % , e ) 12 % | b | multiply(multiply(10, 10), subtract(const_1, divide(add(multiply(10, const_60), 8), add(multiply(10, const_60), 40)))) | multiply(n0,n0)|multiply(n2,const_60)|multiply(n0,const_60)|add(n3,#1)|add(n1,#2)|divide(#3,#4)|subtract(const_1,#5)|multiply(#0,#6)| | physics | B |
the age of father 7 years ago was 5 times the age of his son . 5 years hence , father ' s age will be thrice that of his son . the ratio of their present ages is : | let the ages of father and son 7 years ago be 5 x and x years respectively . then , ( 5 x + 7 ) + 5 = 3 [ ( x + 7 ) + 5 ] 5 x + 12 = 3 x + 36 x = 12 . required ratio = ( 5 x + 7 ) : ( x + 7 ) = 67 : 19 . answer : option a | a ) 67 : 19 , b ) 65 : 17 , c ) 61 : 19 , d ) 63 : 15 , e ) 67 : 17 | a | divide(subtract(multiply(5, divide(subtract(add(multiply(5, const_3), subtract(multiply(7, 5), 7)), 5), subtract(5, const_3))), subtract(multiply(7, 5), 7)), divide(subtract(add(multiply(5, const_3), subtract(multiply(7, 5), 7)), 5), subtract(5, const_3))) | multiply(n1,const_3)|multiply(n0,n1)|subtract(n1,const_3)|subtract(#1,n0)|add(#0,#3)|subtract(#4,n1)|divide(#5,#2)|multiply(n1,#6)|subtract(#7,#3)|divide(#8,#6) | general | A |
susan can type 10 pages in 5 minutes . mary can type 5 pages in 10 minutes . working together , how many pages can they type in 30 minutes ? | susan can type 2 pages in 1 min mary can type 0.5 pages in 1 min so , both of them work together they type 2.5 pages in 1 min so , in 30 min they type ( 30 * 2.5 ) = 75 pages answer : e | a ) 15 , b ) 20 , c ) 25 , d ) 65 , e ) 75 | e | add(multiply(divide(30, 10), 5), multiply(10, divide(30, 5))) | divide(n4,n0)|divide(n4,n1)|multiply(n1,#0)|multiply(n0,#1)|add(#2,#3) | physics | E |
if x + y = 280 , x - y = 200 , for integers of x and y , y = ? | "x + y = 280 x - y = 200 2 x = 80 x = 40 y = 240 answer is b" | a ) 200 , b ) 240 , c ) 50 , d ) 115 , e ) 150 | b | divide(add(280, 200), const_2) | add(n0,n1)|divide(#0,const_2)| | general | B |
joe went on a diet 4 months ago when he weighed 222 pounds . if he now weighs 198 pounds and continues to lose at the same average monthly rate , in approximately how many months will he weigh 180 pounds ? | 222 - 198 = 24 pounds lost in 4 months 24 / 4 = 6 , so joe is losing weight at a rate of 6 pounds per month . . . . in approximately how many months will he weigh 180 pounds ? a simple approach is to just list the weights . now : 198 lbs in 1 month : 192 lbs in 2 months : 186 lbs in 3 months : 180 lbs answer : c | a ) 3 , b ) 3.5 , c ) 3 , d ) 4.5 , e ) 5 | c | divide(subtract(198, 180), divide(subtract(222, 198), 4)) | subtract(n2,n3)|subtract(n1,n2)|divide(#1,n0)|divide(#0,#2) | general | C |
how many integers between 400 and 1000 are there such that their unit digit is odd ? | there are 600 numbers from 401 to 1000 ( inclusive ) . half of the numbers are odd , so there are 300 odd numbers . the answer is c . | a ) 250 , b ) 150 , c ) 300 , d ) 400 , e ) 450 | c | divide(subtract(1000, 400), const_2) | subtract(n1,n0)|divide(#0,const_2) | general | C |
there were two candidates in an election . winner candidate received 70 % of votes and won the election by 280 votes . find the number of votes casted to the winning candidate ? | "w = 70 % l = 30 % 70 % - 30 % = 40 % 40 % - - - - - - - - 280 70 % - - - - - - - - ? = > 280 / 40 * 70 = 490 answer : c" | a ) 228 , b ) 744 , c ) 490 , d ) 199 , e ) 231 | c | divide(multiply(divide(280, divide(subtract(70, subtract(const_100, 70)), const_100)), 70), const_100) | subtract(const_100,n0)|subtract(n0,#0)|divide(#1,const_100)|divide(n1,#2)|multiply(n0,#3)|divide(#4,const_100)| | gain | C |
dan β s car gets 32 miles per gallon . if gas costs $ 4 / gallon , then how many miles can dan β s car go on $ 42 of gas ? | "42 / 4 = 10.5 gallons 10.5 * 32 = 336 miles the answer is c ." | a ) 196 , b ) 284 , c ) 336 , d ) 412 , e ) 572 | c | divide(multiply(42, 32), 4) | multiply(n0,n2)|divide(#0,n1)| | physics | C |
a bag contains 3 red , 5 yellow and 4 green balls . 3 balls are drawn randomly . what is the probability that balls drawn contain exactly two green balls ? | total number of balls = 3 + 5 + 4 = 12 n ( s ) = 12 c 3 = 12 * 11 * 10 / 3 * 2 = 220 n ( e ) = 4 c 2 * 8 c 1 = 6 * 8 = 48 probability = 48 / 220 = 12 / 55 answer is c | a ) 9 / 16 , b ) 7 / 18 , c ) 12 / 55 , d ) 14 / 67 , e ) 19 / 87 | c | divide(multiply(choose(4, const_2), choose(add(3, 5), const_1)), choose(add(add(3, 5), 4), 3)) | add(n0,n1)|choose(n2,const_2)|add(n2,#0)|choose(#0,const_1)|choose(#2,n0)|multiply(#1,#3)|divide(#5,#4) | probability | C |
36 men can complete a piece of work in 18 days . in how many days will 9 men complete the same work ? | "explanation : less men , means more days { indirect proportion } let the number of days be x then , 9 : 36 : : 18 : x x = 72 answer : d ) 72 days" | a ) 24 , b ) 77 , c ) 88 , d ) 72 , e ) 21 | d | divide(multiply(18, 36), 9) | multiply(n0,n1)|divide(#0,n2)| | physics | D |
a car travels at a speed of 65 miles per hour . how far will it travel in 6 hours ? | "during each hour , the car travels 65 miles . for 6 hours it will travel 65 + 65 + 65 + 65 + 65 + 65 = 6 Γ 65 = 390 miles correct answer is c ) 390 miles" | a ) 125 miles , b ) 225 miles , c ) 390 miles , d ) 425 miles , e ) 525 miles | c | multiply(65, 6) | multiply(n0,n1)| | physics | C |
he average weight of 8 persons increases by 2.5 kg when a new person comes in place of one of them weighing 65 kg . what might be the weight of the new person ? | explanation : total weight increased = ( 8 x 2.5 ) kg = 20 kg . weight of new person = ( 65 + 20 ) kg = 85 kg . answer : d | a ) 75 kg , b ) 55 kg , c ) 45 kg , d ) 85 kg , e ) 25 kg | d | add(multiply(2.5, 8), 65) | multiply(n0,n1)|add(n2,#0) | general | D |
if 6 x ^ 2 + x - 12 = ( hx + b ) ( cx + d ) , then | h | + | b | + | c | + | d | = for a complete solution and more practice problems , see this blog : http : / / magoosh . com / gmat / 2012 / algebra - on . . . to - factor / | 6 x ^ 2 + x - 12 = 6 x ^ 2 + 9 x - 8 x - 12 = > 3 x ( 2 x + 3 ) - 4 ( 2 x + 3 ) = > ( 2 x + 3 ) ( 3 x - 4 ) = ( hx + b ) ( cx + d ) hence h = 2 , b = c = 3 , d = - 4 so , 2 + 3 + 3 + | - 4 | = 2 + 3 + 3 + 4 = 12 answer b . | a ) 10 , b ) 12 , c ) 15 , d ) 18 , e ) 20 | b | add(add(add(2, const_3), const_3), const_4) | add(n1,const_3)|add(#0,const_3)|add(#1,const_4) | general | B |
diana took out a charge account at the general store and agreed to pay 7 % simple annual interest . if she charges $ 75 on her account in january , how much will she owe a year later , assuming she does not make any additional charges or payments ? | 1.07 * $ 75 = $ 80.25 the answer is e . | a ) $ 79.25 , b ) $ 79.50 , c ) $ 79.75 , d ) $ 80.00 , e ) $ 80.25 | e | add(75, multiply(divide(7, const_100), 75)) | divide(n0,const_100)|multiply(n1,#0)|add(n1,#1) | general | E |
the perimeter of an equilateral triangle is 60 . if one of the sides of the equilateral triangle is the side of an isosceles triangle of perimeter 50 , then how long is the base of isosceles triangle ? | "the base of the isosceles triangle is 50 - 20 - 20 = 10 units the answer is b ." | a ) 5 , b ) 10 , c ) 15 , d ) 20 , e ) 25 | b | subtract(subtract(50, divide(60, const_3)), divide(60, const_3)) | divide(n0,const_3)|subtract(n1,#0)|subtract(#1,#0)| | geometry | B |
when positive integer x is divided by positive integer y , the remainder is 5 . if x / y = 96.2 , what is the value of y ? | "guys , one more simple funda . 5 / 2 = 2.5 now . 5 x 2 = 1 is the remainder 25 / 4 = 6.25 now . 25 x 4 = 1 is the remainder 32 / 5 = 6.4 now . 4 x 5 = 2 is the remainder given x / y = 96.2 and remainder is 5 so . 2 x y = 5 hence y = 25 ans d" | a ) 96 , b ) 75 , c ) 48 , d ) 25 , e ) 12 | d | divide(5, subtract(96.2, floor(96.2))) | floor(n1)|subtract(n1,#0)|divide(n0,#1)| | general | D |
carmen made a sculpture from small pieces of wood . the sculpture is 2 feet 10 inches tall . carmen places her sculpture on a base that is 4 inches tall . how tall are the sculpture andbase together ? | we know 1 feet = 12 inch then 2 feet = 24 inch 24 + 10 = 34 then 34 + 4 = 38 38 / 12 = 3.17 feet answer : a | ['a ) 3.17 feet', 'b ) 3.2 feet', 'c ) 3.3 feet', 'd ) 3.4 feet', 'e ) 3.5 feet'] | a | divide(add(add(multiply(add(2, 10), 2), const_10), 4), add(2, 10)) | add(n0,n1)|multiply(n0,#0)|add(#1,const_10)|add(n2,#2)|divide(#3,#0) | geometry | A |
a train covers a distance of 20 km in 20 min . if it takes 9 sec to pass a telegraph post , then the length of the train is ? | "speed = ( 20 / 20 * 60 ) km / hr = ( 60 * 5 / 18 ) m / sec = 50 / 3 m / sec . length of the train = 50 / 3 * 9 = 150 m . answer : a" | a ) 150 m , b ) 200 m , c ) 120 m , d ) 225 m , e ) 160 m | a | divide(20, subtract(divide(20, 20), 9)) | divide(n0,n1)|subtract(#0,n2)|divide(n0,#1)| | physics | A |
a call center has two teams . each member of team a was able to process 7 / 5 calls as compared to each member of team b . if team a has 5 / 8 as many number of call center agents as team b , what fraction of the total calls was processed by team b ? | "let team b has 8 agents , so team a has 5 agents let each agent of team b picked up 5 calls , so total calls by team b = 40 so , each agent in team a picked up 7 calls , so total calls for team a = 35 fraction for team b = 40 / ( 40 + 35 ) = 8 / 15 = answer = c" | a ) 3 / 2 , b ) 3 / 4 , c ) 8 / 15 , d ) 1 / 2 , e ) 1 / 5 | c | divide(multiply(8, 5), add(multiply(8, 5), multiply(5, 7))) | multiply(n1,n3)|multiply(n0,n1)|add(#0,#1)|divide(#0,#2)| | general | C |
how many odd prime numbers are there less than 125 ? | odd prime number less than 125 : 3 , 5 , 7 , 11 , 13 , 17 , 19 , 23 , 29 , 31 , 37 , 41 , 43 , 47 , 53 , 59 , 61 , 67 , 71 , 73 , 79 , 83 , 89 , 97 , 101 , 103 , 107 , 109 , 113 there is 29 the odd prime number answer is a | a ) 29 , b ) 34 , c ) 44 , d ) 54 , e ) 64 | a | add(subtract(125, const_100), const_4) | subtract(n0,const_100)|add(#0,const_4) | general | A |
a train 550 m long is running with a speed of 60 km / hr . in what time will it pass a man who is running at 6 km / hr in the direction opposite to that in which the train is going ? | "speed of train relative to man = 60 + 6 = 66 km / hr . = 66 * 5 / 18 = 55 / 3 m / sec . time taken to pass the men = 550 * 3 / 55 = 30 sec . answer : option e" | a ) 5 , b ) 6 , c ) 7 , d ) 8 , e ) 30 | e | divide(550, multiply(add(60, 6), const_0_2778)) | add(n1,n2)|multiply(#0,const_0_2778)|divide(n0,#1)| | physics | E |
a can do a work in 30 days , b can do it in 20 days . they together under took to do a piece of work for rs . 1000 , what is the share of b ? | explanation : share of their work days = 30 : 20 share of their work = 20 : 30 share of b β s money = 3 / 5 * 1000 = 600 answer : option d | a ) rs . 500 , b ) rs . 400 , c ) rs . 800 , d ) rs . 600 , e ) rs . 200 | d | multiply(30, 20) | multiply(n0,n1) | physics | D |
a polling company surveyed a certain country , and it found that 35 % of that country β s registered voters had an unfavorable impression of both of that state β s major political parties and that 20 % had a favorable impression only of party a . if one registered voter has a favorable impression of both parties for every two registered voters who have a favorable impression only of party b , then what percentage of the country β s registered voters have a favorable impression of both parties ( assuming that respondents to the poll were given a choice between favorable and unfavorable impressions only ) ? | "assume the total pool of registered voters = 100 , so 35 of the country β s registered voters had an unfavorable impression of both of that state β s major political parties and 20 had a favorable impression only of party a let x = # of voters with a favorable impression of both parties let 2 x = # of voters with a favorable impression only of party b so unfavorable a and favorable b # of voters = 2 x - x = x 35 + x = number of unfavorable a 20 + 35 + x = 100 55 + x = 100 x = 45 answer : e" | a ) 15 , b ) 20 , c ) 30 , d ) 35 , e ) 45 | e | divide(subtract(const_100, add(35, 20)), const_3) | add(n0,n1)|subtract(const_100,#0)|divide(#1,const_3)| | gain | E |
there are 8 executives , including the ceo and cfo , that are asked to form a small team of 5 members . however , the ceo and cfo may not both be assigned to the team . given this constraint , how many ways are there to form the team ? | "the total number of ways to form a team of 5 is 8 c 5 = 56 . we need to subtract the number of teams that have both the ceo and the cfo . the number of teams with both the ceo and cfo is 6 c 3 = 20 . the number of ways to form an acceptable team is 56 - 20 = 36 . the answer is c ." | a ) 34 , b ) 35 , c ) 36 , d ) 37 , e ) 38 | c | subtract(factorial(subtract(8, 5)), add(multiply(5, const_3), const_3)) | multiply(n1,const_3)|subtract(n0,n1)|add(#0,const_3)|factorial(#1)|subtract(#3,#2)| | other | C |
for how many integer values of n will the value of the expression 4 n + 7 be an integer greater than 1 and less than 200 ? | from first constraint 4 n + 7 > 1 4 n > - 6 n > - ( 3 / 2 ) n > - 1.5 ( n = - 1 , 0 , 1 , 2 3 . . . . . . . . upto infinity ) from second constraint 4 n + 7 < 200 4 n < 193 n < 48 . 25 n = ( - infinity , . . . . . . . - 3 , - 2 , - 1 , 0 , 1 , 2 , . . . . . . . . . upto 48 ) combining the two - 1.5 < n < 48.25 n = 1 to 48 ( 48 integers ) and n = - 1 and 0 so 50 integers . c is the answer | a ) 48 , b ) 49 , c ) 50 , d ) 51 , e ) 52 | c | subtract(floor(divide(subtract(200, 7), 4)), floor(divide(subtract(1, 7), 4))) | subtract(n3,n1)|subtract(n2,n1)|divide(#0,n0)|divide(#1,n0)|floor(#2)|floor(#3)|subtract(#4,#5) | general | C |
the salaries of a and b together amount to $ 6000 . a spends 95 % of his salary and b , 85 % of his . if now , their savings are the same , what is a ' s salary ? | "let a ' s salary is x b ' s salary = 6000 - x ( 100 - 95 ) % of x = ( 100 - 85 ) % of ( 6000 - x ) x = $ 4500 answer is d" | a ) $ 1000 , b ) $ 1250 , c ) $ 2500 , d ) $ 4500 , e ) $ 5200 | d | subtract(6000, divide(6000, add(divide(subtract(const_100, 85), subtract(const_100, 95)), const_1))) | subtract(const_100,n2)|subtract(const_100,n1)|divide(#0,#1)|add(#2,const_1)|divide(n0,#3)|subtract(n0,#4)| | gain | D |
what will be the compound interest on a sum of rs . 25,000 after 3 years at the rate of 12 % p . a . ? | "amount = [ 25000 * ( 1 + 12 / 100 ) 3 ] = 25000 * 28 / 25 * 28 / 25 * 28 / 25 = rs . 35123.20 c . i . = ( 35123.20 - 25000 ) = rs . 10123.20 answer : c" | a ) rs . 10123.77 , b ) rs . 10123.21 , c ) rs . 10123.20 , d ) rs . 10123.28 , e ) rs . 10123.21 | c | 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 | C |
for the past n days , the average ( arithmetic mean ) daily production at a company was 60 units . if today ' s production of 90 units raises the average to 65 units per day , what is the value of n ? | "( average production for n days ) * n = ( total production for n days ) - - > 60 n = ( total production for n days ) ; ( total production for n days ) + 90 = ( average production for n + 1 days ) * ( n + 1 ) - - > 60 n + 90 = 65 * ( n + 1 ) - - > n = 5 . or as 30 extra units increased the average for n + 1 days by 5 units per day then 30 / ( n + 1 ) = 5 - - > n = 5 . answer : e ." | a ) 30 , b ) 18 , c ) 10 , d ) 9 , e ) 5 | e | subtract(divide(subtract(90, 60), subtract(65, 60)), const_1) | subtract(n1,n0)|subtract(n2,n0)|divide(#0,#1)|subtract(#2,const_1)| | general | E |
john took a bus from home to market , that travels at 50 kmph . while walking back at 5 kmph , halfway through , he suddenly realized he was getting late and he cycled back the remaining distance in 35 kmph . find the average speed . | let the distance be 2 x ( one way ) time taken by bus = 2 x / 50 , by walking = x / 5 , by cycling = x / 35 hours : . average speed = total distance / total time = 5 x / x / 25 + x / 5 + x / 35 = 5 * 60 / 2.4 + 12 + 1.7 = 18.6 answer : d | a ) 8.5 kmph , b ) 16.0 kmph , c ) 22.5 kmph , d ) 18.6 kmph , e ) none of these | d | add(const_0_25, divide(35, const_2)) | divide(n2,const_2)|add(#0,const_0_25) | physics | D |
a portion of the 85 % solution of chemicals was replaced with an equal amount of 30 % solution of chemicals . as a result , 40 % solution of chemicals resulted . what part of the original solution was replaced ? | "this is a weighted average question . say x % of the solution was replaced - - > equate the amount of chemicals : 0.85 ( 1 - x ) + 0.3 * x = 0.4 - - > x = 9 / 11 . answer : d ." | a ) 5 / 11 , b ) 6 / 11 , c ) 7 / 11 , d ) 9 / 11 , e ) 13 / 6 | d | divide(subtract(85, 40), subtract(85, 30)) | subtract(n0,n2)|subtract(n0,n1)|divide(#0,#1)| | gain | D |
the average mark of a class of thirty two students is 75 . if 3 students whose marks are 28 and 34 are removed , then find the approximate average mark of the remaining students of the class . | exp . total mark of 32 students = 75 * 32 = 2400 , total mark after the removal of 2 students = 2400 β ( 28 + 34 ) = 2400 β 62 = 2338 approximate average mark = 2338 / ( 32 - 2 ) = 2338 / 30 = 78 answer : c | a ) 71 , b ) 74 , c ) 78 , d ) 70 , e ) 80 | c | divide(subtract(subtract(multiply(75, subtract(34, const_2)), 28), 34), subtract(subtract(34, const_2), 3)) | subtract(n3,const_2)|multiply(n0,#0)|subtract(#0,n1)|subtract(#1,n2)|subtract(#3,n3)|divide(#4,#2) | general | C |
two trains are moving in opposite directions with speed of 210 km / hr and 90 km / hr respectively . their lengths are 1.10 km and 0.9 km respectively . the slower train cross the faster train in - - - seconds | explanation : relative speed = 210 + 90 = 300 km / hr ( since both trains are moving in opposite directions ) total distance = 1.1 + . 9 = 2 km time = 2 / 300 hr = 1 / 150 hr = 3600 / 150 seconds = 24 seconds answer : option e | a ) 56 , b ) 48 , c ) 47 , d ) 26 , e ) 24 | e | multiply(divide(add(1.1, 0.9), add(210, 90)), const_3600) | add(n2,n3)|add(n0,n1)|divide(#0,#1)|multiply(#2,const_3600) | physics | E |
the salary of a worker is first increased by 25 % and afterwards reduced by 25 % . what is the net change in the worker ' s salary ? | "let x be the original salary . the final salary is 0.75 ( 1.25 x ) = 0.9375 x the answer is d ." | a ) 3.25 % increase , b ) 3.25 % decrease , c ) 6.25 % increase , d ) 6.25 % decrease , e ) no change | d | subtract(const_100, subtract(add(25, const_100), divide(multiply(add(25, const_100), 25), const_100))) | add(n0,const_100)|multiply(n0,#0)|divide(#1,const_100)|subtract(#0,#2)|subtract(const_100,#3)| | gain | D |
if ( 1 β 1.5 ) n = 1 , then n = | "( 1 β 1.5 ) n = 1 simplify to get : - 0.5 n = 1 rewrite as ( - 1 / 2 ) n = 1 multiply both sides by - 2 to get : n = - 2 answer : c" | a ) β 400 , b ) β 140 , c ) β 2 , d ) 4 , e ) 400 | c | negate(multiply(1, const_4)) | multiply(n2,const_4)|negate(#0)| | general | C |
out of 470 students of a school , 325 play football , 175 play cricket and 50 neither play football nor cricket . how many students play both football and cricket ? | "n ( a ) = 325 , n ( b ) = 175 , n ( aub ) = 470 - 50 = 420 . required number = n ( anb ) = n ( a ) + n ( b ) - n ( aub ) = 325 + 175 - 420 = 80 . answer is a" | a ) 80 , b ) 150 , c ) 100 , d ) 180 , e ) 220 | a | subtract(add(175, 325), subtract(470, 50)) | add(n1,n2)|subtract(n0,n3)|subtract(#0,#1)| | other | A |
how many even multiples of 55 are there between 549 and 1101 ? | "550 = 10 * 55 1100 = 20 * 55 the even multiples are 55 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(1101, 549), multiply(55, const_2)), const_1) | multiply(n0,const_2)|subtract(n2,n1)|divide(#1,#0)|add(#2,const_1)| | general | B |
there are 60 supermarkets in the fgh chain . all of them are either in the us or canada . if there are 22 more fgh supermarkets in the us than in canada , how many fgh supermarkets are there in the us ? | "x + ( x - 22 ) = 60 - - > x = 41 . answer : c ." | a ) 20 , b ) 31 , c ) 41 , d ) 53 , e ) 64 | c | divide(add(60, 22), const_2) | add(n0,n1)|divide(#0,const_2)| | general | C |
a certain industrial loom weaves 0.128 meters of cloth every second . approximately how many seconds will it take for the loom to weave 25 meters of cloth ? | "explanation : let the required number of seconds be x more cloth , more time , ( direct proportion ) hence we can write as ( cloth ) 0.128 : 25 : : 1 : x β 0.128 x = 25 β x = 25 / 0.128 = 25000 / 128 = 3125 / 16 β 195 answer : option d" | a ) 205 , b ) 200 , c ) 180 , d ) 195 , e ) 175 | d | divide(25, 0.128) | divide(n1,n0)| | physics | D |
tim and Γ© lan are 150 miles away from one another . they are starting to move towards each other simultaneously , tim at a speed of 10 mph and Γ© lan at a speed of 5 mph . if every hour they double their speeds , what is the distance that tim will pass until he meets Γ© lan ? | "tim and elan will meet at the same time while their ratio of speed is 2 : 1 respectively . so their individual distance traveled ratio will be same . plugging in the answer choice only answer choice e meet the 2 : 1 ( tim : elan = 100 : 50 ) ratio of maintaining total distance traveled 150 miles socorrect answer e" | a ) 30 miles . , b ) 35 miles . , c ) 45 miles . , d ) 90 miles . , e ) 100 miles . | e | multiply(divide(10, add(5, 10)), 150) | add(n1,n2)|divide(n1,#0)|multiply(n0,#1)| | physics | E |
3 pumps , working 8 hours a day , can empty a tank in 2 days . how many hours a day must 8 pumps work to empty the tank in 1 day ? | "3 pumps take 16 hrs total ( 8 hrs a day ) if 1 pump will be working then , it will need 16 * 3 = 48 hrs 1 pump need 48 hrs if i contribute 8 pumps then 48 / 8 = 6 hrs . answer : a" | a ) 6 , b ) 10 , c ) 11 , d ) 12 , e ) 13 | a | divide(multiply(multiply(3, 8), 2), 8) | multiply(n0,n1)|multiply(n2,#0)|divide(#1,n3)| | physics | A |
an empty tank be filled with an inlet pipe β a β in 42 minutes . after 12 minutes an outlet pipe β b β is opened which can empty the tank in 30 minutes . after 6 minutes another inlet pipe β c β opened into the same tank , which can fill the tank in 35 minutes and the tank is filled . find the time taken to fill the tank ? | assume total tank capacity = 210 liters now capacity of pipe a = 210 / 42 = 5 liters capacity of b = 210 / 30 = - 7 liters capacity of c = 210 / 35 = 6 min assume tank gets filled in x min after the third pipe got opened . so x Γ 5 + 6 Γ ( β 2 ) + 4 x = 210 x Γ 5 + 6 Γ ( β 2 ) + 4 x = 210 β 48 + 4 x = 210 β 4 x = 162 β x = 40.5 β 48 + 4 x = 210 β 4 x = 162 β x = 40.5 total time taken to fill the tank = 40.5 + 12 + 6 = 51.5 answer : b | a ) 51.6 , b ) 51.5 , c ) 51.7 , d ) 51.1 , e ) 51.0 | b | divide(subtract(add(add(30, 35), 42), const_2), const_2) | add(n2,n4)|add(n0,#0)|subtract(#1,const_2)|divide(#2,const_2) | physics | B |
two pipes a and b can fill a tank in 24 hours and 36 hours respectively . if both the pipes are opened simultaneously , how much time will be taken to fill the tank ? | "part filled by a in 1 hour = 1 / 24 part filled by b in 1 hour = 1 / 36 part filled by ( a + b ) in 1 hour = 1 / 24 + 1 / 36 = 5 / 72 both the pipes together fill the tank in 72 / 5 = 14 2 / 5 hours answer is c" | a ) 20 hours , b ) 15 hours , c ) 14 2 / 5 hours , d ) 12 hours , e ) 8 hours | c | divide(const_1, add(divide(const_1, 24), divide(const_1, 36))) | divide(const_1,n0)|divide(const_1,n1)|add(#0,#1)|divide(const_1,#2)| | physics | C |
a train running at a speed of 36 km / h passes an electric pole in 14 seconds . in how many seconds will the whole train pass a 370 - meter long platform ? | let the length of the train be x meters . when a train crosses an electric pole , the distance covered is its own length x . speed = 36 km / h = 36000 m / 3600 s = 10 m / s x = 14 * 10 = 140 m . the time taken to pass the platform = ( 140 + 370 ) / 10 = 51 seconds the answer is d . | a ) 45 , b ) 47 , c ) 49 , d ) 51 , e ) 53 | d | divide(add(multiply(multiply(36, const_0_2778), 14), 370), multiply(36, const_0_2778)) | multiply(n0,const_0_2778)|multiply(n1,#0)|add(n2,#1)|divide(#2,#0) | physics | D |
of 300 surveyed students , 20 % of those who read book a also read book b and 50 % of those who read book b also read book a . if each student read at least one of the books , what is the difference between the number of students who read only book a and the number of students who read only book b ? | "say the number of students who read book a is a and the number of students who read book b is b . given that 20 % of those who read book a also read book b and 50 % of those who read book b also read book a , so the number of students who read both books is 0.2 a = 0.5 b - - > a = 2.5 b . since each student read at least one of the books then { total } = { a } + { b } - { both } - - > 300 = 2.5 b + b - 0.5 b - - > b = 100 , a = 2.5 b = 250 and { both } = 0.5 b = 50 . the number of students who read only book a is { a } - { both } = 250 - 50 = 200 ; the number of students who read only book b is { b } - { both } = 100 - 50 = 50 ; the difference is 200 - 50 = 150 . answer : e ." | a ) 120 , b ) 125 , c ) 130 , d ) 135 , e ) 150 | e | subtract(multiply(multiply(divide(300, subtract(add(divide(divide(50, const_100), divide(20, const_100)), const_1), divide(50, const_100))), divide(divide(50, const_100), divide(20, const_100))), subtract(const_1, divide(20, const_100))), multiply(divide(300, subtract(add(divide(divide(50, const_100), divide(20, const_100)), const_1), divide(50, const_100))), subtract(const_1, divide(50, const_100)))) | divide(n2,const_100)|divide(n1,const_100)|divide(#0,#1)|subtract(const_1,#1)|subtract(const_1,#0)|add(#2,const_1)|subtract(#5,#0)|divide(n0,#6)|multiply(#7,#2)|multiply(#7,#4)|multiply(#8,#3)|subtract(#10,#9)| | gain | E |
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