Problem stringlengths 5 967 | Rationale stringlengths 1 2.74k | options stringlengths 37 300 | correct stringclasses 5
values | annotated_formula stringlengths 7 6.48k | linear_formula stringlengths 8 925 | category stringclasses 6
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on a map , 3 centimeters represents 33 kilometers . two cities 209 kilometers apart would be separated on the map by how many centimeters ? | "1 centimeter represents 11 kilometers ( 33 / 3 ) x = 209 / 11 = 19 answer : a" | a ) 19 , b ) 7 , c ) 13 , d ) 110 , e ) 180 | a | multiply(divide(3, 33), 209) | divide(n0,n1)|multiply(n2,#0)| | gain |
a boat can travel with a speed of 5 km / hr in still water . if the speed of the stream is 5 km / hr , find the time taken by the boat to go 100 km downstream . | "speed downstream = ( 5 + 5 ) km / hr = 10 km / hr . time taken to travel 100 km downstream = 100 / 10 hrs = 10 hrs . answer : e" | a ) 1 hr , b ) 2 hrs , c ) 3 hrs , d ) 4 hrs , e ) 10 hrs | e | divide(100, add(5, 5)) | add(n0,n1)|divide(n2,#0)| | physics |
when f is divided by 5 , the remainder is 3 . when y is divided by 5 , the remainder is 4 . what is the remainder when f + y is divided by 5 ? | in my view the answer should be c f / 5 has remainder = 3 - > f = 5 x q + 3 y / 5 has a remainder = 4 - > y = 5 x q + 3 combining both ( ( 5 x q 1 + 3 ) + ( 5 x q 2 + 3 ) ) / 5 = 5 ( q 1 + q 2 ) / 5 + 7 / 5 = q 1 + q 2 + 7 / 5 7 / 5 = 1 + 2 / 5 = > remainder 2 answer c | a ) 0 , b ) 1 , c ) 2 , d ) 3 , e ) 4 | c | reminder(add(3, 4), 5) | add(n1,n3)|reminder(#0,n0) | general |
the side of a rhombus is 24 m and length of one of its diagonals is 18 m . the area of the rhombus is ? | "area of the rhombus = 1 / 2 * p * √ 4 ( a ) 2 - ( p ) 2 a = 24 ; p = 18 a = 1 / 2 * 18 * √ 4 ( 24 ) 2 - ( 18 ) 2 = 1 / 2 * 18 * √ 2304 - 324 = 1 / 2 * 18 * √ 1980 a = 400.47 answer : d" | a ) 410.47 , b ) 403.47 , c ) 420.47 , d ) 400.47 , e ) 300.47 | d | divide(multiply(18, sqrt(subtract(power(multiply(const_2, 24), const_2), power(18, const_2)))), const_2) | multiply(n0,const_2)|power(n1,const_2)|power(#0,const_2)|subtract(#2,#1)|sqrt(#3)|multiply(n1,#4)|divide(#5,const_2)| | geometry |
the average salary of all the workers in a workshop is rs . 8000 . the average salary of 7 technicians is rs . 12000 and the average salary of the rest is rs . 6000 . the total number of workers in the workshop is : | "let the total number of workers be x . then , 8000 x = ( 12000 * 7 ) + 6000 ( x - 7 ) = 2000 x = 42000 = x = 21 . answer a" | a ) 21 , b ) 20 , c ) 22 , d ) 25 , e ) 40 | a | add(7, divide(multiply(7, subtract(12000, 8000)), subtract(8000, 6000))) | subtract(n2,n0)|subtract(n0,n3)|multiply(n1,#0)|divide(#2,#1)|add(n1,#3)| | general |
one night a certain hotel rented 2 / 3 of its rooms , including 2 / 3 of their air conditioned rooms . if 3 / 5 of its rooms were air conditioned , what percent of the rooms that were not rented were air conditioned ? | "the rooms which were not rented is 1 / 3 the ac rooms which were not rented is ( 1 / 3 ) * ( 3 / 5 ) = 1 / 5 the percentage of unrented rooms which were ac rooms is ( 1 / 5 ) / ( 1 / 3 ) = 3 / 5 = 60 % the answer is b ." | a ) 50 % , b ) 60 % , c ) 65 % , d ) 70 % , e ) 75 % | b | multiply(divide(multiply(subtract(const_1, divide(2, 2)), multiply(divide(2, 3), const_100)), subtract(const_100, multiply(divide(2, 3), const_100))), const_100) | divide(n0,n1)|divide(n2,n0)|multiply(#0,const_100)|subtract(const_1,#1)|multiply(#2,#3)|subtract(const_100,#2)|divide(#4,#5)|multiply(#6,const_100)| | gain |
a car travels at a speed of 55 miles per hour . how far will it travel in 5 hours ? | "during each hour , the car travels 55 miles . for 5 hours it will travel 55 + 55 + 55 + 55 + 55 = 5 * 55 = 275 miles correct answer e" | a ) 425 miles , b ) 625 miles , c ) 325 miles , d ) 225 miles , e ) 275 miles | e | multiply(55, 5) | multiply(n0,n1)| | physics |
every second saturday and all sundays are holidays . how many working days will be there in a month of 30 days beginning on a saturday ? | explanation : mentioned month has begins on a saturday and has 30 days sundays = 2 nd , 9 th , 16 th , 23 rd , 30 th = > total sundays = 5 second saturdays = 8 th and 22 nd total second saturdays = 2 total holidays = total sundays + total second saturdays = 5 + 2 = 7 total days in the month = 30 total working days = 30... | a ) 24 , b ) 23 , c ) 18 , d ) 21 , e ) 22 | b | subtract(30, add(add(const_1, const_1), add(const_3, const_2))) | add(const_1,const_1)|add(const_2,const_3)|add(#0,#1)|subtract(n0,#2) | physics |
the present age of a father is 3 years more than 3 times the age of his son . 3 years hence , father ’ s age will be 10 years more than twice athe age of the son . find the present age of the father . | sol . let the son ’ s present age be x years . then , father ’ s present age = ( 3 x + 3 ) years . therefore ( 3 x + 3 + 3 ) = 2 ( x + 3 ) + 10 ‹ = › 3 x + 6 = 2 x + 16 ‹ = › x = 10 hence , father ’ s present age = ( 3 x + 3 ) = ( 3 x 10 + 3 ) years = 33 years . answer a | a ) 33 , b ) 34 , c ) 35 , d ) 36 , e ) 37 | a | add(3, multiply(10, 3)) | multiply(n0,n3)|add(n0,#0) | general |
find number which is 60 % less than 80 . | "explanation : 60 % less is 40 % of the given number therefore , 40 % of 80 is 32 . answer : b" | a ) 18 , b ) 32 , c ) 28 , d ) 26 , e ) 98 | b | divide(multiply(80, 60), const_100) | multiply(n0,n1)|divide(#0,const_100)| | gain |
a cricket bat is sold for $ 850 , making a profit of $ 230 . the profit percentage would be | "230 / ( 850 - 230 ) = 230 / 620 = 23 / 62 = 37 % answer : d ." | a ) 24 % , b ) 25 % , c ) 30 % , d ) 37 % , e ) 40 % | d | multiply(divide(230, subtract(850, 230)), const_100) | subtract(n0,n1)|divide(n1,#0)|multiply(#1,const_100)| | gain |
a factory produces 5500 toys per week . if the workers at this factory work 4 days a week and if these workers make the same number of toys everyday , how many toys are produced each day ? | "to find the number of toys produced every day , we divide the total number of toys produced in one week ( of 4 days ) by 4 . 5500 / 4 = 1375 toys correct answer e" | a ) 4436 toys , b ) 5487 toys , c ) 6113 toys , d ) 2354 toys , e ) 1375 toys | e | divide(5500, 4) | divide(n0,n1)| | physics |
if g is the smallest positive integer such that 3150 multiplied by g is the square of an integer , then g must be | solution : this problem is testing us on the rule that when we express a perfect square by its unique prime factors , every prime factor ' s exponent is an even number . let ’ s start by prime factorizing 3150 . 3150 = 315 x 10 = 5 x 63 x 10 = 5 x 7 x 3 x 3 x 5 x 2 3150 = 2 ^ 1 x 3 ^ 2 x 5 ^ 2 x 7 ^ 1 ( notice that the... | a ) 2 , b ) 5 , c ) 6 , d ) 7 , e ) 14 | e | multiply(const_2, divide(divide(divide(divide(divide(3150, const_2), add(const_2, const_3)), add(const_2, const_3)), const_3), const_3)) | add(const_2,const_3)|divide(n0,const_2)|divide(#1,#0)|divide(#2,#0)|divide(#3,const_3)|divide(#4,const_3)|multiply(#5,const_2) | geometry |
the total age of a and b is 13 years more than the total age of b and c . c is how many year younger than a | "explanation : given that a + b = 13 + b + c = > a – c = 13 + b – b = 13 = > c is younger than a by 13 years answer : option c" | a ) a ) 11 , b ) b ) 12 , c ) c ) 13 , d ) d ) 14 , e ) e ) 15 | c | multiply(13, const_1) | multiply(n0,const_1)| | general |
? % of 360 = 108.0 | "? % of 360 = 108.0 or , ? = 108.0 × 100 / 360 = 30 answer e" | a ) 277 , b ) 36 , c ) 64 , d ) 72 , e ) 30 | e | divide(multiply(108.0, const_100), 360) | multiply(n1,const_100)|divide(#0,n0)| | gain |
two numbers are in the ratio of 1 : 2 . if 7 be added to both , their ratio changes to 3 : 5 . the greater number is | "let the ratio be x : y , given x / y = 1 / 2 , ( x + 7 ) / ( y + 7 ) = 3 / 5 = > x = 14 and y = 28 answer : c" | a ) 20 , b ) 24 , c ) 28 , d ) 32 , e ) 36 | c | multiply(1, 7) | multiply(n0,n2)| | other |
find the numbers which are in the ratio 3 : 2 : 1 such that the sum of the first and the second added to the difference of the third and the second is 16 ? | "let the numbers be a , b and c . a : b : c = 3 : 2 : 1 given , ( a + b ) + ( c - b ) = 16 = > a + c = 16 = > 3 x + x = 16 = > x = 4 a , b , c are 3 x , 2 x , x a , b , c are 12 , 8 , 4 . answer : e" | a ) 4 , 3,22 , b ) 4 , 4,22 , c ) 9 , 3,32 , d ) 9 , 6,12 , e ) 12 , 8,4 | e | divide(multiply(1, 3), 3) | multiply(n2,n0)|divide(#0,n0)| | general |
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 . 175 , then how much amount should c pay as his share of rent ? | "a : b : c = 10 × 7 : 12 × 5 : 15 × 3 = 2 × 7 : 12 × 1 : 3 × 3 = 14 : 12 : 9 amount that c should pay = 175 × ( 9 / 35 ) = 5 × 9 = 45 answer is b ." | a ) 40 , b ) 45 , c ) 49 , d ) 51 , e ) 43 | b | multiply(175, 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 discount electronics store normally sells all merchandise at a discount of 10 percent to 30 percent off the suggested retail price . if , during a special sale , an additional 15 percent were to be deducted from the discount price , what would be the lowest possible price of an item costing $ 260 before any discount ... | "since the question is essentially just about multiplication , you can do the various mathstepsin a variety of ways ( depending on whichever method you find easiest ) . we ' re told that the first discount is 10 % to 30 % , inclusive . we ' re told that the next discount is 20 % off of the discounted price . . . . we '... | a ) $ 130.00 , b ) $ 145.60 , c ) $ 154.70 , d ) $ 182.00 , e ) $ 210.00 | c | multiply(260, divide(add(30, 15), const_100)) | add(n1,n2)|divide(#0,const_100)|multiply(n3,#1)| | gain |
a grocer purchased a quantity of bananas at 3 pounds for $ 0.50 and sold the entire quantity at 4 pounds for $ 1.00 . how many pounds did the grocer purchase if the profit from selling the bananas was $ 7.00 ? | cost price of 1 pound of bananas = 0.5 / 3 = 1 / 6 selling price of 1 pound of bananas = 1 / 4 profit per pound = ( 1 / 4 - 1 / 6 ) = ( 1 / 12 ) total profit is given as 7 ( 1 / 12 ) * x = 7 x = 84 answer : b | a ) 40 , b ) 84 , c ) 90 , d ) 120 , e ) 240 | b | divide(7, subtract(divide(1, 4), divide(0.5, const_3))) | divide(n3,n2)|divide(n1,const_3)|subtract(#0,#1)|divide(n4,#2) | gain |
in a triangle abc , point d is on side ab and point e is on side ac , such that bced is a trapezium . de : bc = 3 : 5 . calculate the ratio of the area of triangle ade and the trapezium bced . | area of tri ( ade ) / area of tri ( abc ) = de ^ 2 / bc ^ 2 since tri ( ade ) - tri ( abc ) area of tri ( ade ) / area of tri ( abc ) = 9 / 25 area of tri ( abc ) / area of tri ( ade ) = 25 / 9 area of tri ( abc ) = area of tri ( ade ) + area of trap ( bcde ) ( area of tri ( ade ) + area of trap ( bcde ) ) / area of tr... | ['a ) 9 / 16', 'b ) 8 / 16', 'c ) 7 / 16', 'd ) 6 / 16', 'e ) 5 / 16'] | a | divide(power(3, const_2), subtract(power(5, const_2), power(3, const_2))) | power(n0,const_2)|power(n1,const_2)|subtract(#1,#0)|divide(#0,#2) | geometry |
96 % of the population of a village is 23040 . the total population of the village is ? | "x * ( 96 / 100 ) = 23040 x = 240 * 100 x = 24000 answer : b" | a ) 26799 , b ) 24000 , c ) 26682 , d ) 29973 , e ) 12312 | b | multiply(divide(const_100, 96), 23040) | divide(const_100,n0)|multiply(n1,#0)| | general |
solve the equation for x : 6 x - 87 + 3 x = 4 + 9 - x | "c 10 9 x + x = 13 + 87 10 x = 100 = > x = 10" | a ) 12 , b ) 5 , c ) 10 , d ) 9 , e ) 3 | c | divide(add(87, 4), add(3, 6)) | add(n1,n3)|add(n2,n0)|divide(#0,#1)| | general |
water consists of hydrogen and oxygen , and the approximate ratio , by mass , of hydrogen to oxygen is 2 : 16 . approximately how many grams of oxygen are there in 117 grams of water ? | "( 16 / 18 ) * 117 = 104 grams the answer is c ." | a ) 100 , b ) 102 , c ) 104 , d ) 106 , e ) 108 | c | multiply(2, divide(117, add(2, 16))) | add(n0,n1)|divide(n2,#0)|multiply(n0,#1)| | other |
the average monthly salary of 20 employees in an organisation is rs . 1200 . if the manager ' s salary is added , then the average salary increases by rs . 100 . what is the manager ' s monthly salary ? | "explanation : manager ' s monthly salary rs . ( 1300 * 21 - 1200 * 20 ) = rs . 3300 . answer : c" | a ) 3600 , b ) 3890 , c ) 3300 , d ) 2789 , e ) 2891 | c | subtract(multiply(add(1200, 100), add(20, const_1)), multiply(1200, 20)) | add(n1,n2)|add(n0,const_1)|multiply(n0,n1)|multiply(#0,#1)|subtract(#3,#2)| | general |
the ages of patrick and michael are in the ratio of 3 : 5 and that of michael and monica are in the ratio of 3 : 5 . if the sum of their ages is 141 , what is the difference between the ages of patrick and monica ? | "ages of p and mi = 3 x + 5 x ages of mi and mo = 3 x : 5 x rationalizing their ages . ratio of their ages will be 9 x : 15 x : 25 x sum = 47 x = 141 x = 3 difference if ages of pa and mo = 25 x - 9 x = 16 x = 16 * 3 = 48 answer b" | a ) 27 , b ) 48 , c ) 45 , d ) 72 , e ) 18 | b | subtract(multiply(multiply(5, 5), divide(141, add(add(multiply(3, 3), multiply(3, 5)), multiply(5, 5)))), multiply(multiply(3, 3), divide(141, add(add(multiply(3, 3), multiply(3, 5)), multiply(5, 5))))) | multiply(n0,n0)|multiply(n1,n0)|multiply(n1,n3)|add(#0,#1)|add(#3,#2)|divide(n4,#4)|multiply(#5,#2)|multiply(#5,#0)|subtract(#6,#7)| | general |
there are 408 boys and 288 girls in a school which are to be divided into equal sections of either boys or girls alone . find the total number of sections thus formed . | explanation : hcf ( 408 , 288 ) = 24 the number of boys or girls that can be placed in a section = 24 . thus the total number of sections is given by 408 / 24 + 288 / 24 = 17 + 12 = 29 answer : e | a ) 31 , b ) 32 , c ) 35 , d ) 30 , e ) 29 | e | divide(add(408, 288), multiply(multiply(multiply(const_2, const_2), const_2), const_3)) | add(n0,n1)|multiply(const_2,const_2)|multiply(#1,const_2)|multiply(#2,const_3)|divide(#0,#3) | general |
what is the units digit of the expression 14 ^ 7 − 17 ^ 4 ? | i think answer on this one should be d too . since we know that 14 ^ 7 > 17 ^ 4 , as will said one should always check if the number is positive . | a ) 0 , b ) 3 , c ) 4 , d ) 8 , e ) 6 | d | add(add(const_4, const_3), const_2) | add(const_3,const_4)|add(#0,const_2)| | general |
a certain experimental mathematics program was tried out in 2 classes in each of 42 elementary schools and involved 55 teachers . each of the classes had 1 teacher and each of the teachers taught at least 1 , but not more than 3 , of the classes . if the number of teachers who taught 3 classes is n , then the least and... | "one may notice that greatest possible values differ in each answer choice in contrast to the least values , which repeat . to find out the greatest value you should count the total classes ( 42 * 2 = 84 ) , then subtract the total # of teachers since we know from the question that each teacher taught at least one clas... | a ) 0 and 13 , b ) 0 and 14 , c ) 1 and 10 , d ) 1 and 9 , e ) 2 and 8 | b | divide(subtract(multiply(42, 2), 55), 2) | multiply(n0,n1)|subtract(#0,n2)|divide(#1,n0)| | general |
if a no . when divided by 44 , gives 432 as quotient and 0 as remainder . what will be the remainder when dividing the same no . by 35 | "p ã · 44 = 432 = > p = 432 * 44 = 19008 p / 35 = 19008 / 35 = 543 , remainder = 3 b" | a ) 1 , b ) 3 , c ) 5 , d ) 7 , e ) 9 | b | reminder(multiply(432, 44), 35) | multiply(n0,n1)|reminder(#0,n3)| | general |
if a ^ 4 + b ^ 4 = 100 , then the greatest possible value of b is between | for the greatest possible value of b ^ 4 , we must minimize the value of b ^ 4 i . e . lets say a ^ 4 = 0 then we need to find a number b such that b ^ 4 < 100 . 3 ^ 4 = 81 and 4 ^ 4 = 256 so we can say that the maximum possible value of b can be a little more than 3 hence answer = between 3 and 4 hence d | a ) 0 and 1 , b ) 1 and 2 , c ) 2 and 3 , d ) 3 and 4 , e ) 4 and 5 | d | add(multiply(multiply(4, 4), const_2), const_2) | multiply(n0,n0)|multiply(#0,const_2)|add(#1,const_2) | general |
in how many years will a sum of money doubles itself at 5 % per annum on simple interest ? | "p = ( p * 5 * r ) / 100 r = 20 % answer : b" | a ) 33 , b ) 20 , c ) 88 , d ) 66 , e ) 21 | b | divide(const_100, 5) | divide(const_100,n0)| | gain |
how many different ways can 2 students be seated in a row of 7 desks , so that there is always at least one empty desk between the students ? | 15 ways to seat the students : with two empty seats between 1 empty w / one student on the left most 1 empty . . . . right most two students can be interchanged 15 x 2 = 30 c | a ) 2 , b ) 3 , c ) 30 , d ) 6 , e ) 12 | c | permutation(subtract(7, const_1), 2) | subtract(n1,const_1)|permutation(#0,n0) | probability |
the ratio of 2 numbers is 5 : 2 and their h . c . f . is 20 . their l . c . m . is ? | "let the numbers be 5 x and 2 x their h . c . f . = 20 so the numbers are 5 * 20 , 2 * 20 = 100 , 40 l . c . m . = 200 answer is c" | a ) 400 , b ) 300 , c ) 200 , d ) 100 , e ) 40 | c | sqrt(divide(20, add(power(2, 5), add(power(2, 5), power(5, 5))))) | power(n0,n1)|power(n1,n1)|power(n2,n1)|add(#0,#1)|add(#3,#2)|divide(n3,#4)|sqrt(#5)| | other |
what is the greatest possible length which can be used to measure exactly the lengths 180 m , 500 m and 15 m 20 cm ? | required length = hcf of 18000 cm , 50000 cm , 1520 cm = 80 cm answer is a | a ) 80 cm , b ) 50 cm , c ) 40 cm , d ) 10 cm , e ) 100 cm | a | multiply(20, const_4) | multiply(n3,const_4) | physics |
if radius of a circle is diminished by 10 % then its area is diminished by | explanation : let old radius = 10 units . new radius is diminshed by 10 % . so new radius = 90 % ( 10 ) = 9 units . old area = π × r 2 = 100 π new area = π × 92 = 81 π change = 19 π / 100 π × 100 = 19 % answer : b | ['a ) 22', 'b ) 19', 'c ) 877', 'd ) 26', 'e ) 271'] | b | multiply(divide(subtract(circle_area(const_1), circle_area(subtract(const_1, divide(10, const_100)))), circle_area(const_1)), const_100) | circle_area(const_1)|divide(n0,const_100)|subtract(const_1,#1)|circle_area(#2)|subtract(#0,#3)|divide(#4,#0)|multiply(#5,const_100) | geometry |
an order was placed for a carpet whose length and width were in the ratio of 3 : 2 . subsequently , the dimensions of the carpet were altered such that its length and width were in the ratio 4 : 3 but were was no change in its perimeter . what is the ratio of the areas of the carpets ? | "let the length and width of one carpet be 3 x and 2 x . let the length and width of the other carpet be 4 y and 3 y . 2 ( 3 x + 2 x ) = 2 ( 4 y + 3 y ) 5 x = 7 y ( 5 / 7 ) * x = y the ratio of the areas of the carpet in both cases : = 3 x * 2 x : 4 y * 3 y = 6 x ^ 2 : 12 y ^ 2 = x ^ 2 : 2 y ^ 2 = x ^ 2 : 2 * ( 25 / 49... | a ) 9 : 10 , b ) 19 : 20 , c ) 29 : 30 , d ) 39 : 40 , e ) 49 : 50 | e | divide(rectangle_area(3, 2), rectangle_area(divide(divide(rectangle_perimeter(3, 2), 2), add(4, 3)), multiply(divide(divide(rectangle_perimeter(3, 2), 2), add(4, 3)), 4))) | add(n2,n3)|rectangle_area(n0,n1)|rectangle_perimeter(n0,n1)|divide(#2,n1)|divide(#3,#0)|multiply(n2,#4)|rectangle_area(#4,#5)|divide(#1,#6)| | geometry |
the ratio of the areas of two squares , one having double its diagonal then the other is : | lenth of the diagonals be 2 x and x units . areas are 1 / 2 × ( 2 x ) power 2 and ( 1 / 2 × ( xpower 2 ) ) required ratio = 1 / 2 × 4 x 2 : 1 / 2 x 2 = 4 : 1 answer is e . | ['a ) 2 : 1', 'b ) 1 : 2', 'c ) 2 : 3', 'd ) 1 : 4', 'e ) 4 : 1'] | e | divide(multiply(inverse(const_2), const_4), inverse(const_2)) | inverse(const_2)|multiply(#0,const_4)|divide(#1,#0) | geometry |
a group of people participate in some curriculum , 25 of them practice yoga , 15 study cooking , 8 study weaving , 2 of them study cooking only , 7 of them study both the cooking and yoga , 3 of them participate all curriculums . how many people study both cooking and weaving ? | both cooking and weaving = 15 - ( 2 + 3 + 7 ) = 3 so , the correct answer is c . | a ) 1 , b ) 2 , c ) 3 , d ) 4 , e ) 5 | c | subtract(subtract(subtract(15, 7), 3), 2) | subtract(n1,n4)|subtract(#0,n5)|subtract(#1,n3) | other |
a car travels first 160 km at 70 km / hr and the next 160 km at 80 km / hr . what is the average speed for the first 320 km of the tour ? | "car travels first 160 km at 70 km / hr time taken to travel first 160 km = distancespeed = 160 / 70 car travels next 160 km at 80 km / hr time taken to travel next 160 km = distancespeed = 160 / 80 total distance traveled = 160 + 160 = 2 × 160 total time taken = 160 / 70 + 160 / 80 average speed = total distance trave... | a ) 71.11 , b ) 71.12 , c ) 71.1 , d ) 74.66 , e ) 71.13 | d | multiply(320, divide(const_1, add(divide(160, 70), divide(160, 80)))) | divide(n0,n1)|divide(n0,n3)|add(#0,#1)|divide(const_1,#2)|multiply(n4,#3)| | general |
in an election , candidate a got 75 % of the total valid votes . if 15 % of the total votes were declared invalid and the total numbers of votes is 560000 , find the number of valid vote polled in favor of candidate . | "total number of invalid votes = 15 % of 560000 = 15 / 100 × 560000 = 8400000 / 100 = 84000 total number of valid votes 560000 – 84000 = 476000 percentage of votes polled in favour of candidate a = 75 % therefore , the number of valid votes polled in favour of candidate a = 75 % of 476000 = 75 / 100 × 476000 = 35700000... | a ) 330000 , b ) 340000 , c ) 347000 , d ) 356000 , e ) 357000 | e | multiply(multiply(560000, subtract(const_1, divide(15, const_100))), divide(75, const_100)) | divide(n0,const_100)|divide(n1,const_100)|subtract(const_1,#1)|multiply(n2,#2)|multiply(#0,#3)| | gain |
joe ’ s average ( arithmetic mean ) test score across 4 equally weighted tests was 90 . he was allowed to drop his lowest score . after doing so , his average test score improved to 95 . what is the lowest test score that was dropped ? | "the arithmetic mean of 4 equally weighted tests was 90 . so what we can assume is that we have 4 test scores , each 90 . he dropped his lowest score and the avg went to 95 . this means that the lowest score was not 90 and other three scores had given the lowest score 5 each to make it up to 90 too . when the lowest sc... | a ) 20 , b ) 25 , c ) 55 , d ) 75 , e ) 80 | d | subtract(multiply(90, 4), multiply(95, const_3)) | multiply(n0,n1)|multiply(n2,const_3)|subtract(#0,#1)| | general |
john found that the average of 15 numbers is 40 . if 15 is added to each number then the mean of number is ? | "( x + x 1 + . . . x 14 ) / 15 = 40 55 option a" | a ) 55 , b ) 45 , c ) 65 , d ) 78 , e ) 64 | a | add(40, 15) | add(n1,n2)| | general |
357 : 73 : : ? | "357 : 73 to get 2 nd number , middle digit is removed , 1 st & last interchanged their position of 1 st no . only option a ) match this . answer : a" | a ) 429 : 94 , b ) 201 : 21 , c ) 138 : 38 , d ) 93 : 39 , e ) 95 : 39 | a | multiply(divide(357, 73), const_100) | divide(n0,n1)|multiply(#0,const_100)| | general |
scott ' s average ( arithmetic mean ) golf score on his first 4 rounds was 78 . what score does he need on his fifth round to drop his average score by 2 points ? | average score for first 4 rounds = 78 average score after 5 rounds = 76 score scott needs on his 5 th round to drop average score by 2 points = 78 - ( 2 * 5 ) = 68 answer a alternatively , sum of scores of first 4 rounds = 78 * 4 = 312 sum of scores after 5 rounds = 76 * 5 = 380 score scott needs on his 5 th round to d... | a ) 68 , b ) 72 , c ) 78 , d ) 88 , e ) 312 | a | subtract(multiply(subtract(78, 2), add(4, const_1)), multiply(4, 78)) | add(n0,const_1)|multiply(n0,n1)|subtract(n1,n2)|multiply(#0,#2)|subtract(#3,#1) | general |
in 15 year ’ s time the combined age of my 4 brothers will be 107 . what will it be in the 6 year ’ s time ? | c 71 combined age in 15 years = 107 . 4 × 15 = 60 , therefore combined age now is 107 – 60 = 47 . in 6 years time , therefore , combined age will be 47 + 24 ( 4 x 6 ) = 71 . | a ) 61 , b ) 78 , c ) 71 , d ) 68 , e ) 94 | c | add(subtract(107, multiply(15, 4)), multiply(6, 4)) | multiply(n1,n3)|multiply(n0,n1)|subtract(n2,#1)|add(#0,#2) | probability |
a , b and c have rs . 600 between them , a and c together have rs . 250 and b and c rs . 450 . how much does c have ? | "a + b + c = 600 a + c = 250 b + c = 450 - - - - - - - - - - - - - - a + b + 2 c = 700 a + b + c = 600 - - - - - - - - - - - - - - - - c = 100 answer : d" | a ) 70 , b ) 90 , c ) 120 , d ) 100 , e ) 140 | d | subtract(add(250, 450), 600) | add(n1,n2)|subtract(#0,n0)| | general |
a store reduced the price of all items in the store by 15 % on the first day and by another 10 % on the second day . the price of items on the second day was what percent of the price before the first reduction took place ? | "consider price of the all items as $ 100 after a initial reduction of 15 % price becomes = 0.85 * 100 = $ 85 after the final reduction of 10 % price becomes = 0.9 * 85 = $ 76.5 price of all items on second day is 76.5 % of price on first day correct answer option a" | a ) 76.5 , b ) 80.9 , c ) 81.0 , d ) 81.1 , e ) 81.9 | a | multiply(multiply(divide(subtract(const_100, 15), const_100), divide(subtract(const_100, 10), const_100)), const_100) | subtract(const_100,n0)|subtract(const_100,n1)|divide(#0,const_100)|divide(#1,const_100)|multiply(#2,#3)|multiply(#4,const_100)| | gain |
find the area of the quadrilateral of one of its diagonals is 40 cm and its off sets 9 cm and 6 cm ? | 1 / 2 * 40 ( 9 + 6 ) = 300 cm 2 answer : b | ['a ) 189 cm 2', 'b ) 300 cm 2', 'c ) 127 cm 2', 'd ) 177 cm 2', 'e ) 187 cm 2'] | b | multiply(multiply(divide(const_1, const_2), add(6, 9)), 40) | add(n1,n2)|divide(const_1,const_2)|multiply(#0,#1)|multiply(n0,#2) | geometry |
an equal number of desks and bookcases are to be placed along a library wall that is 15 meters long . each desk is 2 meters long , and each bookshelf is 1.5 meters long . if the maximum possible number of desks and bookcases are to be placed along the wall , then the space along the wall that is left over will be how m... | "let x be the number of desks and bookcases that are placed along the library wall . 2 x + 1.5 x < 15 3.5 x < 15 since x is a non negative integer , the largest number x can be is 4 . when x is 4 , the desks and bookcases take up 3.5 * 4 = 14 m = k , leaving 1 m of empty space . thus , i believe the answer is b ) 1" | a ) 0.5 , b ) 1 , c ) 1.5 , d ) 2 , e ) 3 | b | subtract(15, add(multiply(const_4, 2), multiply(const_4, 1.5))) | multiply(n1,const_4)|multiply(n2,const_4)|add(#0,#1)|subtract(n0,#2)| | general |
how many kg of pure salt must be added to 100 kg of 10 % solution of salt and water to increase it to a 25 % solution ? | "amount salt in 100 kg solution = 10 * 100 / 100 = 10 kg let x kg of pure salt be added then ( 10 + x ) / ( 100 + x ) = 25 / 100 40 + 4 x = 100 + x 3 x = 60 x = 20 answer is d" | a ) a ) 6.7 , b ) b ) 1.3 , c ) c ) 9.6 , d ) d ) 20 , e ) e ) 7.9 | d | divide(subtract(25, 10), subtract(const_1, divide(25, 100))) | divide(n2,n0)|subtract(n2,n1)|subtract(const_1,#0)|divide(#1,#2)| | general |
my grandson is about as many days as my son in weeks , and my grandson is as many months as i am in years . my grandson , my son and i together are 140 years . can you tell me my age in years ? | let the age of the man = x years so according to the given condition age of my grandson = x / 12 years and age of my son = x 30 / 52 = 15 x / 26 and x + ( x / 12 ) + ( 15 x / 26 ) = 140 or ( 156 x + 13 x + 90 x ) / 156 = 140 or 259 x = 156 * 140 or x = 156 * 140 / 259 = 84 answer : d | a ) 80 , b ) 90 , c ) 85 , d ) 84 , e ) 89 | d | multiply(divide(140, add(const_10, const_10)), const_12) | add(const_10,const_10)|divide(n0,#0)|multiply(#1,const_12) | physics |
if money is invested at r percent interest , compounded annually , the amount of the investment will double in approximately 50 / r years . if luke ' s parents invested $ 15,500 in a long term bond that pays 12 percent interest compounded annually , what will be the approximate total amount of the investment 12 years l... | answer equals e in 48 years . i thought by 50 th year it would reach 120,000 . options should have been separated more widely for clarity . | a ) 62,000 , b ) 85,500 , c ) 95,500 , d ) 100,500 , e ) 120,000 | e | divide(multiply(multiply(add(const_2, const_3), const_1000), 12), const_2) | add(const_2,const_3)|multiply(#0,const_1000)|multiply(n2,#1)|divide(#2,const_2)| | general |
there are 361 doctors and nurses in a hospital . if the ratio of the doctors to the nurses is 8 : 11 , then how many nurses are there in the hospital ? | "given , the ratio of the doctors to the nurses is 8 : 11 number of nurses = 11 / 19 x 361 = 209 answer : a" | a ) 152 , b ) 209 , c ) 57 , d ) 171 , e ) 181 | a | multiply(multiply(8, subtract(11, 8)), 11) | subtract(n2,n1)|multiply(n1,#0)|multiply(n2,#1)| | other |
20 is to be divided into two parts such that the sum of 10 times the first and 22 times the second is 780 . the bigger part is : | "explanation : let the two parts be ( 20 - x ) and x . then , 10 ( 20 - x ) + 22 x = 780 = > 12 x = 580 = > x = 48.33 . bigger part = ( 20 - x ) = 28.33 . answer : d ) 28.3" | a ) 33.33 , b ) 34.01 , c ) 26.32 , d ) 28.33 , e ) 19.21 | d | subtract(20, divide(subtract(780, multiply(20, 10)), subtract(22, 10))) | multiply(n0,n1)|subtract(n2,n1)|subtract(n3,#0)|divide(#2,#1)|subtract(n0,#3)| | general |
a chartered bank offers a 5 - year escalator guaranteed investment certificate . in successive years it pays annual interest rates of 4 % , 4.5 % , 5 % , 5.5 % , and 6 % , respectively , compounded at the end of each year . the bank also offers regular 5 - year gics paying a fixed rate of 5 % compounded annually . calc... | explanation : fv = $ 1000 ( 1.04 ) ( 1.045 ) ( 1.05 ) ( 1.055 ) ( 1.06 ) = $ 1276.14 the maturity value of the regular gic is fv = $ = $ 1276.28 answer : a ) 1276.28 | a ) 1276.28 , b ) 1276.22 , c ) 1276.29 , d ) 1276.21 , e ) 1276.23 | a | multiply(power(add(divide(5, const_100), const_1), 5), 1000) | divide(n0,const_100)|add(#0,const_1)|power(#1,n0)|multiply(n8,#2) | gain |
on a trip covering 350 km , a bicyclist travels the first 200 km at 20 km per hour and the remainder of the distance at 15 km per hour . what is the average speed , in km per hour , for the entire trip ? | "time 1 = 200 / 20 = 10 hours time 2 = 150 / 15 = 10 hours total time = 20 hours average speed = 350 / 20 = 17.5 km / hr the answer is c ." | a ) 15.5 , b ) 16.0 , c ) 17.5 , d ) 19.0 , e ) 18.5 | c | speed(350, add(divide(200, 20), divide(subtract(350, 200), 15))) | divide(n1,n2)|subtract(n0,n1)|divide(#1,n3)|add(#0,#2)|speed(n0,#3)| | physics |
how many seconds will a 1200 m long train take to cross a man walking with a speed of 3 km / hr in the direction of the moving train if the speed of the train is 63 km / hr ? | "speed of train relative to man = 63 - 3 = 60 km / hr . = 60 * 5 / 18 = 50 / 3 m / sec . time taken to pass the man = 1200 * 3 / 50 = 72 sec . answer : e" | a ) 26 sec , b ) 65 sec , c ) 55 sec , d ) 19 sec , e ) 72 sec | e | divide(1200, multiply(subtract(63, 3), const_0_2778)) | subtract(n2,n1)|multiply(#0,const_0_2778)|divide(n0,#1)| | physics |
the ratio of two quantities is 3 to 4 . if each of the quantities is increased 4 and 3 respectively , what is the ratio of these two new quantities ? | "3 / 4 = 3 x / 4 x we need to find out ( 3 x + 4 ) / ( 4 x + 3 ) off course we can not solve this to arrive at any rational number hence e ." | a ) 3 / 4 , b ) 8 / 9 , c ) 18 / 19 , d ) 23 / 24 , e ) it can not be determined from the information given | e | divide(add(3, 4), add(4, 4)) | add(n0,n2)|add(n1,n2)|divide(#0,#1)| | other |
the area of a square field 3136 sq m , if the length of cost of drawing barbed wire 3 m around the field at the rate of rs . 1.30 per meter . two gates of 1 m width each are to be left for entrance . what is the total cost ? | "answer : option c explanation : a 2 = 3136 = > a = 56 56 * 4 * 3 = 672 â € “ 6 = 666 * 1.3 = 865.8 answer : c" | a ) 399 , b ) 272 , c ) 865.8 , d ) 277 , e ) 311 | c | multiply(multiply(subtract(multiply(sqrt(3136), const_4), multiply(const_2, 1)), 1.30), 3) | multiply(n3,const_2)|sqrt(n0)|multiply(#1,const_4)|subtract(#2,#0)|multiply(n2,#3)|multiply(#4,n1)| | physics |
25 women can complete a work in 7 days and 10 children take 14 days to complete the work . how many days will 5 women and 10 children take to complete the work ? | "1 women ' s 1 day work = 1 / 175 1 child ' s 1 day work = 1 / 140 ( 5 women + 10 children ) ' s 1 day work = ( 5 / 175 + 10 / 140 ) = 1 / 10 5 women and 10 children will complete the work in 10 days . answer : e" | a ) 8 days , b ) 6 days , c ) 7 days , d ) 9 days , e ) 10 days | e | inverse(add(divide(5, multiply(25, 7)), divide(25, multiply(25, 14)))) | multiply(n0,n1)|multiply(n0,n3)|divide(n4,#0)|divide(n0,#1)|add(#2,#3)|inverse(#4)| | physics |
an aeroplane covers a certain distance at a speed of 240 kmph in 5 hours . to cover the same distance in 1 2 / 3 hours , it must travel at a speed of : | "explanation : distance = ( 240 x 5 ) = 1200 km . speed = distance / time speed = 1200 / ( 5 / 3 ) km / hr . [ we can write 1 2 / 3 hours as 5 / 3 hours ] required speed = 1200 x 3 / 5 km / hr = 720 km / hr . answer : option d" | a ) 300 kmph , b ) 360 kmph , c ) 600 kmph , d ) 720 kmph , e ) none | d | divide(divide(multiply(240, 5), add(const_1, divide(const_2, const_3))), const_2) | divide(const_2,const_3)|multiply(n0,n1)|add(#0,const_1)|divide(#1,#2)|divide(#3,const_2)| | physics |
in arun ' s company 60 % of the employees earn less than $ 50000 a year , 60 % of the employees earn more than $ 40000 a year , 11 % of the employees earn $ 43000 a year and 5 % of the employees earn $ 49000 a year . what is the median salary for the company ? | in arun ' s company 60 % of the employees earn less than $ 50,000 a year , 50 and 51 employee will be each 43 k . hence median = ( 43 k + 43 k ) / 2 | a ) 43.0 , b ) 45.5 , c ) 46.0 , d ) 49.0 , e ) 50.0 | a | divide(divide(divide(multiply(const_2, 43000), const_2), const_100), const_10) | multiply(n5,const_2)|divide(#0,const_2)|divide(#1,const_100)|divide(#2,const_10) | general |
50 square stone slabs of equal size were needed to cover a floor area of 72 sq . m . find the length of each stone slab ? | "area of each slab = 72 / 50 m 2 = 1.44 m 2 length of each slab √ 1.44 = 1.2 m = 120 cm" | a ) 120 cm , b ) 767 cm , c ) 88 cm , d ) 666 cm , e ) 776 cm | a | multiply(sqrt(divide(72, 50)), const_100) | divide(n1,n0)|sqrt(#0)|multiply(#1,const_100)| | geometry |
a certain company reported that the revenue on sales increased 40 % from 2000 to 2003 , and increased 90 % from 2000 to 2005 . what was the approximate percent increase in revenue for this store from 2003 to 2005 ? | "assume the revenue in 2000 to be 100 . then in 2003 it would be 140 and and in 2005 190 , so from 2003 to 2005 it increased by ( 190 - 140 ) / 140 = 50 / 140 = ~ 29 % . answer : b" | a ) 50 % , b ) 36 % , c ) 30 % , d ) 32 % , e ) 29 % | b | multiply(divide(subtract(add(const_1, divide(90, const_100)), add(const_1, divide(40, const_100))), add(const_1, divide(40, const_100))), const_100) | divide(n3,const_100)|divide(n0,const_100)|add(#0,const_1)|add(#1,const_1)|subtract(#2,#3)|divide(#4,#3)|multiply(#5,const_100)| | gain |
the number of the members of a club is more than 20 and less than 50 . when 5 people sit at a table , other people exactly divided to 6 - people groups ( 6 people sit at a table ) or 7 - people groups ( 7 people sit at a table ) . if the members are divided to 8 - people groups , how many people will be left ? | the number of members is 7 k + 5 = 6 j + 5 the only number in this range which satisfies this is 47 . 47 / 8 = 5 ( 8 ) + 7 the answer is e . | a ) 3 , b ) 4 , c ) 5 , d ) 6 , e ) 7 | e | reminder(add(lcm(6, 7), 5), 8) | lcm(n3,n5)|add(n2,#0)|reminder(#1,n7) | general |
a train 300 m long is running at a speed of 68 kmph . how long does it take to pass a man who is running at 8 kmph in the same direction as the train ? | "speed of the train relative to man = ( 68 - 8 ) kmph = ( 60 * 5 / 18 ) m / sec = ( 50 / 3 ) m / sec time taken by the train to cross the man = time taken by it to cover 300 m at 50 / 3 m / sec = 300 * 3 / 50 sec = 18 sec answer : d ." | a ) 5 sec , b ) 9 sec , c ) 12 sec , d ) 18 sec , e ) 19 sec | d | divide(300, multiply(subtract(68, 8), const_0_2778)) | subtract(n1,n2)|multiply(#0,const_0_2778)|divide(n0,#1)| | physics |
at the store opening , larry ' s grocery had 50 lemons and 60 oranges . by closing , the store at 20 lemons and 40 oranges left . by approximately what percent did the ratio of lemons to orange decrease from opening to closing . | opening : lemons / oranges = 50 / 60 = 100 / 120 closing : lemons / oranges = 20 / 40 = 60 / 120 aside : it ' s useful to write both ratios with the same denominator . this allows us to ignore the denominator and focus solely on the numerators . so , our ratio went from 100 / 120 to 60 / 120 ignoring the denominators ,... | a ) 50.0 % , b ) 40.0 % , c ) 30.0 % , d ) 20.0 % , e ) 10.0 % | b | multiply(divide(subtract(divide(50, 60), divide(20, 40)), divide(50, 60)), const_100) | divide(n0,n1)|divide(n2,n3)|subtract(#0,#1)|divide(#2,#0)|multiply(#3,const_100) | general |
what is the 27 th digit to the right of the decimal point in the decimal form of 5 / 11 ? | "5 / 11 = 0.45454545 . . . the odd numbered positions in the decimal expansion are all 4 . the answer is b ." | a ) 3 , b ) 4 , c ) 5 , d ) 6 , e ) 7 | b | divide(5, 11) | divide(n1,n2)| | general |
if x is a positive integer with fewer than 3 digits , what is the probability that z x * ( x + 1 ) is a multiple of either 4 or 5 ? | "interesting question ! also one that we should be able to answer very quickly be keeping an eye on our best friends , the answer choices . we know that x belongs to the set { 1 , 2 , 3 , . . . , 99 } . we want to know the probability z that x ( x + 1 ) is a multiple of either 4 or 5 . when will this happen ? if either... | a ) 4 / 99 , b ) 2 / 25 , c ) 8 / 99 , d ) 49 / 100 , e ) 86 / 99 | e | divide(subtract(add(multiply(divide(const_100, 4), const_2), multiply(divide(const_100, 5), const_2)), 4), subtract(const_100, 1)) | divide(const_100,n2)|divide(const_100,n3)|subtract(const_100,n1)|multiply(#0,const_2)|multiply(#1,const_2)|add(#3,#4)|subtract(#5,n2)|divide(#6,#2)| | general |
the cash realised on selling a 14 % stock is rs . 108.25 , brokerage being 1 / 4 % is | "explanation : cash realised = rs . ( 108.25 - 0.25 ) = rs . 108 . answer : c" | a ) 123 , b ) 106 , c ) 108 , d ) 156 , e ) 240 | c | subtract(108.25, divide(1, 4)) | divide(n2,n3)|subtract(n1,#0)| | general |
if y is 40 % greater than x , than x is what % less than y ? | "y = 1.4 x x = y / 1.4 = 10 y / 14 = 5 y / 7 x is 2 / 7 less which is 28.6 % less than y . the answer is b ." | a ) 24.4 % , b ) 28.6 % , c ) 32.3 % , d ) 36.5 % , e ) 40.9 % | b | multiply(divide(40, add(40, const_100)), const_100) | add(n0,const_100)|divide(n0,#0)|multiply(#1,const_100)| | general |
a luxury liner , queen marry ii , is transporting several cats as well as the crew ( sailors , a cook , and one - legged captain ) to a nearby port . altogether , these passengers have 16 heads and 41 legs . how many cats does the ship host ? | sa ' s + co + ca + cats = 16 . sa ' s + 1 + 1 + cats = 16 or sa ' s + cats = 14 . sa ' s ( 2 ) + 2 + 1 + cats * 4 = 41 sa ' s * 2 + cats * 4 = 38 or sa ' s + cats * 2 = 19 or 14 - cats + cat * 2 = 19 then cats = 5 a | a ) 5 , b ) 6 , c ) 4 , d ) 7 , e ) 8 | a | multiply(divide(subtract(subtract(41, const_1), multiply(subtract(16, const_1), const_2)), subtract(multiply(subtract(16, const_1), const_4), multiply(subtract(16, const_1), const_2))), subtract(16, const_1)) | subtract(n1,const_1)|subtract(n0,const_1)|multiply(#1,const_2)|multiply(#1,const_4)|subtract(#0,#2)|subtract(#3,#2)|divide(#4,#5)|multiply(#6,#1) | general |
in a fuel station the service costs $ 1.75 per car , every liter of fuel costs 0.85 $ . assuming that a company owns 12 cars and that every fuel tank contains 55 liters and they are all empty , how much money total will it cost to fuel all cars ? | 12 * 1.75 + 0.85 * 12 * 55 = 582 hence - c | a ) 320 $ , b ) 420 $ , c ) 582 $ , d ) 650 $ , e ) 780 $ | c | multiply(multiply(0.85, 55), 12) | multiply(n1,n3)|multiply(n2,#0)| | general |
what sum of money will produce rs . 75 as simple interest in 4 years at 3 1 / 2 percent ? | "75 = ( p * 4 * 7 / 2 ) / 100 p = 536 answer : e" | a ) 337 , b ) 500 , c ) 266 , d ) 288 , e ) 536 | e | divide(75, divide(multiply(4, add(3, divide(1, 2))), const_100)) | divide(n3,n4)|add(n2,#0)|multiply(n1,#1)|divide(#2,const_100)|divide(n0,#3)| | gain |
income and expenditure of a person are in the ratio 5 : 4 . if the income of the person is rs . 16000 , then find his savings ? | "let the income and the expenditure of the person be rs . 5 x and rs . 4 x respectively . income , 5 x = 16000 = > x = 3200 savings = income - expenditure = 5 x - 4 x = x so , savings = rs . 3200 . answer : c" | a ) 3600 , b ) 3607 , c ) 3200 , d ) 3602 , e ) 3603 | c | subtract(16000, multiply(divide(4, 5), 16000)) | divide(n1,n0)|multiply(n2,#0)|subtract(n2,#1)| | other |
a certain fraction has the same ratio to 1 / 33 , as 3 / 4 does to 7 / 11 . what is this certain fraction ? | "x / ( 1 / 33 ) = ( 3 / 4 ) / ( 7 / 11 ) x = 3 * 11 * 1 / 33 * 4 * 7 = 1 / 28 the answer is e ." | a ) 1 / 14 , b ) 1 / 18 , c ) 1 / 21 , d ) 1 / 25 , e ) 1 / 28 | e | divide(1, 33) | divide(n0,n1)| | other |
a certain car increased its average speed by 5 miles per hour in each successive 5 - minute interval after the first interval . if in the first 5 - minute interval its average speed was 38 miles per hour , how many miles did the car travel in the third 5 - minute interval ? | "in the third time interval the average speed of the car was 38 + 5 + 5 = 30 miles per hour ; in 5 minutes ( 1 / 12 hour ) at that speed car would travel 48 * 1 / 12 = 4 miles . answer : e ." | a ) 1.0 , b ) 1.5 , c ) 2.0 , d ) 2.5 , e ) 4.0 | e | multiply(add(add(38, 5), 5), divide(5, const_60)) | add(n0,n3)|divide(n0,const_60)|add(n0,#0)|multiply(#2,#1)| | physics |
two brothers ram and ravi appeared for an exam . the probability of selection of ram is 1 / 7 and that of ravi is 1 / 5 . find the probability that both of them are selected . | "let a be the event that ram is selected and b is the event that ravi is selected . p ( a ) = 1 / 7 p ( b ) = 1 / 5 let c be the event that both are selected . p ( c ) = p ( a ) x p ( b ) as a and b are independent events : = 1 / 7 x 1 / 5 = 1 / 35 answer : a" | a ) 1 / 35 , b ) 2 / 3 , c ) 7 / 35 , d ) 5 / 7 , e ) 7 / 5 | a | multiply(divide(1, 7), divide(1, 5)) | divide(n0,n1)|divide(n2,n3)|multiply(#0,#1)| | general |
if a and b are distinct positive integers . the units digit of a ^ 2 is equal to the units digit of a , and the units digit of b ^ 2 is equal to the units digit of b . if the units digit of a · b is equal to neither the units digit of a nor the units digit of b , then what is the units digit of a · b ? | there are four digits which when squared give the same units digits : 0 , 1 , 5 , 6 0 ^ 2 = 0 1 ^ 2 = 1 5 ^ 2 = 25 6 ^ 2 = 36 so a and b must take 2 of these values . they are positive integers so 0 is not possible . if one of the digits is 1 , when the other is multiplied by 1 , you will get the same digit . 1 * 5 = 5... | a ) 0 , b ) 1 , c ) 4 , d ) 5 , e ) 6 | a | multiply(multiply(2, const_3), add(const_2, const_3)) | add(const_2,const_3)|multiply(n0,const_3)|multiply(#0,#1) | general |
what is the next number : 3 , 27 , 627 , __ | "5 ^ 0 + 2 = 3 5 ^ 2 + 2 = 27 5 ^ 4 + 2 = 627 5 ^ 6 + 2 = 15,627 the answer is d ." | a ) 8847 , b ) 9837 , c ) 11,497 , d ) 15,627 , e ) 19,307 | d | subtract(subtract(subtract(multiply(627, const_10), const_100), 3), 3) | multiply(n2,const_10)|subtract(#0,const_100)|subtract(#1,n0)|subtract(#2,n0)| | general |
if the ratio of apples to bananas is 5 to 2 and the ratio of bananas to cucumbers is 1 to 2 , what is the ratio of apples to cucumbers ? | "the ratio of bananas to cucumbers is 1 to 2 which equals 2 to 4 . the ratio of apples to bananas to cucumbers is 5 to 2 to 4 . the ratio of apples to cucumbers is 5 to 4 . the answer is e ." | a ) 6 : 5 , b ) 1 : 4 , c ) 4 : 3 , d ) 2 : 3 , e ) 5 : 4 | e | divide(divide(5, 2), 2) | divide(n0,n1)|divide(#0,n3)| | other |
what is the least integer greater than 1 + 0.5 ? | "this question is just about doing careful arithmetic and remembering what makes a numberbiggerorsmallercompared to another number . first , let ' s take care of the arithmetic : ( 1 ) + ( 0.5 ) = 1.5 on a number line , since we ' re adding + . 5 to a number , the total moves to the right ( so we ' re moving from 1 to ... | a ) – 2 , b ) – 1 , c ) 0 , d ) 1 , e ) 2 | e | multiply(add(1, const_1), const_2) | add(n0,const_1)|multiply(#0,const_2)| | general |
when 242 is divided by a certain divisor the remainder obtained is 6 . when 698 is divided by the same divisor the remainder obtained is 13 . however , when the sum of the two numbers 242 and 698 is divided by the divisor , the remainder obtained is 5 . what is the value of the divisor ? | "let that divisor be x since remainder is 6 or 13 it means divisor is greater than 13 . now 242 - 6 = 236 = kx ( k is an integer and 234 is divisble by x ) similarly 698 - 13 = 685 = lx ( l is an integer and 689 is divisible by x ) adding both 698 and 242 = ( 236 + 685 ) + 6 + 13 = x ( k + l ) + 19 when we divide this ... | a ) 11 , b ) 14 , c ) 18 , d ) 23 , e ) none of these | b | subtract(add(6, 13), 5) | add(n1,n3)|subtract(#0,n6)| | general |
in a river flowing at 2 km / hr , a boat travels 56 km upstream and then returns downstream to the starting point . if its speed in still water be 6 km / hr , find the total journey time . | "explanation : speed of the boat = 6 km / hr speed downstream = ( 6 + 2 ) = 8 km / hr speed upstream = ( 6 - 2 ) = 4 km / hr distance travelled downstream = distance travelled upstream = 56 km total time taken = time taken downstream + time taken upstream = ( 56 / 8 ) + ( 56 / 4 ) = 21 hr . answer : option d" | a ) 10 hours , b ) 12 hours , c ) 14 hours , d ) 21 hours , e ) none of these | d | add(divide(56, subtract(6, 2)), divide(56, add(6, 2))) | add(n0,n2)|subtract(n2,n0)|divide(n1,#1)|divide(n1,#0)|add(#2,#3)| | general |
a certain rectangular crate measures 8 feet by 12 feet by 14 feet . a cylindrical gas tank is to be made for shipment in the crate and will stand upright when the crate is placed on one of its six faces . what should the radius of the tank be if it is to be of the largest possible volume ? | "ans is ` ` b ' ' . for max volume of cylinder ( pi * r ^ 2 * h ) we need to max out r ^ 2 * h . we do n ' t know what the dimensions of the crate refer to . so to maximize the above eqn , radius may be of 9 , 10,12 one of the base area ( 8 x 12 , 14 x 12 or 8 x 14 ) r is maximum for base 14 x 12 and 12 can be maximum ... | a ) 6 , b ) 5 , c ) 6 , d ) 8 , e ) 10 | a | sqrt(divide(volume_cylinder(divide(12, const_2), 8), multiply(const_pi, 8))) | divide(n1,const_2)|multiply(n0,const_pi)|volume_cylinder(#0,n0)|divide(#2,#1)|sqrt(#3)| | geometry |
there are 640 students in a school . the ratio of boys and girls in this school is 3 : 5 . find the total of girls & boys are there in this school ? | "in order to obtain a ratio of boys to girls equal to 3 : 5 , the number of boys has to be written as 3 x and the number of girls as 5 x where x is a common factor to the number of girls and the number of boys . the total number of boys and girls is 640 . hence 3 x + 5 x = 640 solve for x 8 x = 640 x = 80 number of boy... | a ) 320 , b ) 345 , c ) 375 , d ) 380 , e ) 400 | e | multiply(divide(640, 5), 3) | divide(n0,n2)|multiply(n1,#0)| | other |
a ratio between two numbers is 5 : 4 and their l . c . m . is 80 . the first number is | "sol . let the required numbers be 5 x and 4 x . then , their l . c . m . is 20 x . ∴ 20 x = 80 ⇔ x = 4 . hence , the first number is 20 . answer c" | a ) 60 , b ) 45 , c ) 20 , d ) 15 , e ) none | c | multiply(divide(80, multiply(5, 4)), 5) | multiply(n0,n1)|divide(n2,#0)|multiply(n0,#1)| | other |
if the simple interest on a sum of money for 3 years at 7 % per annum is rs . 570 , what is the compound interest on the same sum at the rate and for the same time ? | "sum = ( 570 * 100 ) / ( 7 * 3 ) = rs . 2 , 714.29 amount = [ 2 , 714.29 * ( 1 + 7 / 100 ) 3 ] = rs . 3 , 325.12 c . i . = ( 3 , 325.12 - 2 , 714.29 ) = rs . 610.83 . answer : e" | a ) rs . 610.83 , b ) rs . 810.83 , c ) rs . 617 , d ) rs . 618 , e ) rs . 610.83 | e | subtract(add(add(divide(multiply(divide(570, multiply(divide(7, const_100), const_2.0)), 7), const_100), divide(570, multiply(divide(7, const_100), 3))), divide(multiply(add(divide(multiply(divide(570, multiply(divide(7, const_100), 3)), 7), const_100), divide(570, multiply(divide(7, const_100), 3))), 7), const_100)), ... | divide(n1,const_100)|multiply(const_2.0,#0)|divide(n2,#1)|multiply(n1,#2)|divide(#3,const_100)|add(#4,#2)|multiply(n1,#5)|divide(#6,const_100)|add(#5,#7)|subtract(#8,#2)| | gain |
what is the least number that should be added to 1055 , so the sum of the number is divisible by 23 ? | "( 1055 / 23 ) gives a remainder 20 so we need to add 3 . the answer is c ." | a ) 1 , b ) 2 , c ) 3 , d ) 4 , e ) 5 | c | multiply(subtract(add(const_1, floor(divide(1055, 23))), divide(1055, 23)), 23) | divide(n0,n1)|floor(#0)|add(#1,const_1)|subtract(#2,#0)|multiply(n1,#3)| | general |
an entry in an unshaded square is obtained by adding the entries connected to it from the row above ( 11 is one such number ) . write the value of n ? | see the values row wise 5 6 n 7 11 6 + n 7 + n 11 + 6 + n 6 + n + 7 + n 60 hence 11 + 6 + n + 6 + n + 7 + n = 60 n = 10 answer : a | ['a ) 10', 'b ) 11', 'c ) 12', 'd ) 9', 'e ) 8'] | a | divide(subtract(multiply(11, const_3), const_3), const_3) | multiply(n0,const_3)|subtract(#0,const_3)|divide(#1,const_3) | geometry |
in the above number , a and b represent the tens and units digits , respectively . if the above number is divisible by 65 , what is the greatest possible value of b x a ? | "i also was confused when i was looking forabove number : d as far as i understood , 65 is a factor of ab . in other words , the values of b ( units digits can be 5 or 0 . better to have option for 5 in this case to havebigger result ) . now let ' s try 65 x 1 ( a = 6 , b = 5 respectively we have = 30 ) . this is the g... | a ) 0 , b ) 10 , c ) 15 , d ) 20 , e ) 30 | e | multiply(add(const_3, const_4), add(const_2, const_3)) | add(const_3,const_4)|add(const_2,const_3)|multiply(#0,#1)| | general |
a computer factory produces 3024 computers per month at a constant rate , how many computers are built every 30 minutes assuming that there are 28 days in one month ? | "number of hours in 28 days = 28 * 24 number of 30 mins in 28 days = 28 * 24 * 2 number of computers built every 30 mins = 3024 / ( 28 * 24 * 2 ) = 2.25 answer a" | a ) 2.25 . , b ) 3.125 . , c ) 4.5 . , d ) 5.225 . , e ) 6.25 . | a | divide(divide(3024, 28), multiply(subtract(28, const_4), const_2)) | divide(n0,n2)|subtract(n2,const_4)|multiply(#1,const_2)|divide(#0,#2)| | physics |
a man has rs . 400 in the denominations of one - rupee notes , 5 - rupee notes and 10 rupee notes . the number of notes of each denomination is equal . what is the total number of notes that he has ? | let number of notes of each denomination be x . then x + 5 x + 10 x = 400 16 x = 400 x = 25 . hence , total number of notes = 3 x = 75 answer : option c | a ) 45 , b ) 60 , c ) 75 , d ) 90 , e ) 95 | c | multiply(divide(400, add(add(const_1, 5), 10)), const_3) | add(n1,const_1)|add(n2,#0)|divide(n0,#1)|multiply(#2,const_3) | general |
in a sports club with 30 members , 17 play badminton and 17 play tennis and 2 do not play either . how many members play both badminton and tennis ? | "let x play both badminton and tennis so 17 - x play only badminton and 17 - x play only tennis . 2 play none and there are total 30 students . hence , ( 17 - x ) + ( 17 - x ) + x + 2 = 30 36 - 2 x + x = 30 36 - x = 30 x = 6 so 6 members play both badminton and tennis . b" | a ) 7 , b ) 6 , c ) 9 , d ) 10 , e ) 11 | b | subtract(add(add(17, 17), 2), 30) | add(n1,n2)|add(n3,#0)|subtract(#1,n0)| | other |
3 / 4 of 1 / 2 of 2 / 5 of 5000 = ? | "d 750 ? = 5000 * ( 2 / 5 ) * ( 1 / 2 ) * ( 3 / 4 ) = 750" | a ) 392 , b ) 229 , c ) 753 , d ) 750 , e ) 540 | d | multiply(multiply(multiply(divide(3, 4), divide(1, 2)), divide(2, 5)), 5000) | divide(n3,n5)|divide(n0,n1)|divide(n2,n3)|multiply(#1,#2)|multiply(#0,#3)|multiply(n6,#4)| | general |
if x / y = 2 / z , then 5 x ^ 2 = | "this question is most easily solved by isolating y in the equation and substituting into the expression 5 x ² : x / y = 2 / z x = 2 y / z if we substitute 2 y / z into the expression for x , we get : 5 ( 2 y / z ) ² = 5 ( 4 y ² / z ² ) = 20 y ² / z ² . the correct answer is choice ( b )" | a ) y / z , b ) 20 y ² / z ² , c ) y ² z ² , d ) 5 y ² / z ² , e ) yz | b | divide(add(2, 5), subtract(2, 5)) | add(n0,n1)|subtract(n0,n1)|divide(#0,#1)| | general |
sandy walked 20 meters towards south . then sandy turned to her left and walked 25 meters . she then turned to her left and walked 20 meters . she then turned to her right and walked 30 meters . what distance is she from the starting point and in which direction ? | "the net distance is 25 + 30 = 55 meters to the east . the answer is e ." | a ) 35 m east , b ) 35 m north , c ) 30 m west , d ) 45 m west , e ) 55 m east | e | add(20, 30) | add(n0,n3)| | physics |
in a sample of associates at a law firm , 25 percent are second - year associates , and 75 percent are not first - year associates . what percentage of the associates at the law firm have been there for more than two years ? | "75 percent are not first - year associates - - > this means 75 percent associates are either two or more than two years already , we know % of second - year associates = 25 % so , % of associates there for more than two years = 75 % - 25 % = 50 % . . = > ( c ) . . is it correct ?" | a ) 0 , b ) 25 , c ) 50 , d ) 75 , e ) 100 | c | subtract(75, 25) | subtract(n1,n0)| | general |
if ( 2 to the x ) - ( 2 to the ( x - 2 ) ) = 3 ( 2 to the 10 ) , what is the value of x ? | "( 2 to the power x ) - ( 2 to the power ( x - 2 ) ) = 3 ( 2 to the power 10 ) 2 ^ x - 2 ^ ( x - 2 ) = 3 . 2 ^ 10 hence x = 12 . answer is d" | a ) 9 , b ) 11 , c ) 13 , d ) 12 , e ) 17 | d | add(10, 2) | add(n0,n5)| | general |
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