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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in the coordinate plane a slope of the line k is 0.5 times the y - intercept of the line k . what is the x - intercept of the line k ? | as y = 0.5 mx + m , from 0 = 1 / 2 mx + m we get x = - 2 . hence , the correct answer choice is e . | a ) - 4 , b ) 4 , c ) - 2 , d ) 1 / 2 , e ) - 2 | e | divide(const_1, 0.5) | divide(const_1,n0)| | general |
a metallic sheet is of rectangular shape with dimensions 48 m x 36 m . from each of its corner , a square is cut off so as to make an open box . if the length of the square is 8 m , the volume of the box ( in m 3 ) is | "sol . clearly , l = ( 48 - 16 ) m = 32 m , b = ( 36 - 16 ) m = 20 m , h = 8 m . ∴ volume of the box = ( 32 x 20 x 8 ) m 3 = 5120 m 3 answer c" | a ) 6420 , b ) 8960 , c ) 5120 , d ) 4830 , e ) none | c | volume_rectangular_prism(subtract(48, multiply(8, const_2)), subtract(36, multiply(8, const_2)), 8) | multiply(n2,const_2)|subtract(n0,#0)|subtract(n1,#0)|volume_rectangular_prism(n2,#1,#2)| | geometry |
a motorboat can go at 21 kmph and still water . it takes 7 hours to row from a to b and back . what is the distance between a and b if the speed of the stream is 3 kmph ? | let the distance between a and b be x km . total time = x / ( 21 + 3 ) + x / ( 21 - 3 ) = 7 = > x / 24 + x / 18 = 7 = > ( 4 x + 3 x ) / 72 = 7 = > x = 72 km . answer : d | a ) 65 km , b ) 85 km , c ) 55 km , d ) 72 km , e ) 40 km | d | divide(7, add(divide(const_1, add(21, 3)), divide(const_1, subtract(21, 3)))) | add(n0,n2)|subtract(n0,n2)|divide(const_1,#0)|divide(const_1,#1)|add(#2,#3)|divide(n1,#4) | physics |
the sum of all the digits of the integers from 18 to 21 inclusive is 24 ( 1 + 8 + 1 + 9 + 2 + 0 + 2 + 1 = 24 ) . what is the sum r of all the digits of the integers from 0 to 99 inclusive ? | "we want the sum of the digits from 0 to 99 , so i approximated : 0 - 9 - > 45 - > ( 9 + 0 ) * 10 / 2 40 - 49 - > 85 ( 13 + 4 ) * 10 / 2 90 - 99 - > 135 ( 18 + 9 ) * 10 / 2 we can see at a glance that theweightgoes up as the numbers go up ( meaning the difference between 85 and 45 is 40 , while 135 - 85 is 50 , this me... | a ) 450 , b ) 810 , c ) 900 , d ) 1000 , e ) 1100 | c | multiply(divide(multiply(9, add(9, 1)), 2), multiply(2, const_10)) | add(n6,n3)|multiply(const_10,n7)|multiply(n6,#0)|divide(#2,n7)|multiply(#3,#1)| | general |
a total of $ 20,000 was invested in two certificates of deposit at simple annual interest rates of 4 percent and 8 percent , respectively . if the total interest on the two certificates was $ 1,440 at the end of one year , what fractional part of the $ 20.000 was invested at the higher rate ? | "x * 4 / 100 * 1 + ( 20000 - x ) * 8 / 100 * 1 = 1440 4 x - 8 x = 144000 - 160000 = > - 4 x = - 16000 = > x = 4000 / 4 so 16000 / 20000 = 4 / 5 answer - b" | a ) 3 / 8 , b ) 4 / 5 , c ) 1 / 2 , d ) 3 / 5 , e ) 3 / 4 | b | divide(subtract(add(divide(divide(add(add(multiply(const_100, multiply(const_2, add(const_1, const_4))), multiply(const_4, const_100)), multiply(const_4, multiply(const_2, add(const_1, const_4)))), 20.000), 8), 8), 4), 20.000) | add(const_1,const_4)|multiply(const_100,const_4)|multiply(#0,const_2)|multiply(#2,const_100)|multiply(#2,const_4)|add(#3,#1)|add(#5,#4)|divide(#6,n4)|divide(#7,n2)|add(n2,#8)|subtract(#9,n1)|divide(#10,n4)| | gain |
in traveling from a dormitory to a certain city , a student went 1 / 5 of the way by foot , 2 / 3 of the way by bus , and the remaining 6 kilometers by car . what is the distance , in kilometers , from the dormitory to the city ? | "i believe there is a better way to do it . basically one of the options should satisfy the given criteria . 60 did 1 / 5 * 60 = 12 2 / 3 * 60 = 40 so total distance 52 + remaining 6 = 58 answer b" | a ) 30 , b ) 58 , c ) 60 , d ) 90 , e ) 120 | b | multiply(6, inverse(subtract(1, add(divide(1, 5), divide(2, 3))))) | divide(n0,n1)|divide(n2,n3)|add(#0,#1)|subtract(n0,#2)|inverse(#3)|multiply(n4,#4)| | physics |
when positive integer x is divided by positive integer y , the remainder is 9 . if x / y = 96.12 , what is the value of y ? | "x = qy + 9 , where q is the quotient x / y = ( q y + 9 ) / y = q + ( 9 / y ) = 96.12 = 96 + 0.12 = > q = 96 9 / y = 0.12 = > y = 75 answer is b ." | a ) 96 , b ) 75 , c ) 48 , d ) 25 , e ) 12 | b | divide(9, subtract(96.12, floor(96.12))) | floor(n1)|subtract(n1,#0)|divide(n0,#1)| | general |
tom read a book containing 1000 pages by reading the same number of pages each day . if he would have finished the book 5 days earlier by reading 15 pages a day more , how many days did tom spend reading the book ? | "actually u can set up 2 equation p - - stands for the pages d - - stands for the days 1 ) p * d = 480 ( we want to find the days , sop = 480 / d ) 2 ) ( p + 15 ) ( d - 5 ) = 1000 = > pd - 5 p + 15 d - 75 = 1000 as the 1 ) stated u can put 1 ) into 2 ) = > 1000 - 5 p + 15 d - 75 = 1000 = > 15 d - 5 p = 75 put the bold ... | a ) 20 , b ) 18 , c ) 16 , d ) 15 , e ) 15.39 | e | divide(subtract(sqrt(add(multiply(multiply(15, 5), const_4), power(15, const_2))), 15), const_2) | multiply(n1,n2)|power(n2,const_2)|multiply(#0,const_4)|add(#2,#1)|sqrt(#3)|subtract(#4,n2)|divide(#5,const_2)| | general |
find the compound interest on $ 16,000 at 20 % per annum for 9 months , compounded quarterly | "principal = $ 16000 ; time = 9 months = 3 quarters ; rate = 20 % per annum = 5 % per quarter . amount = $ [ 16000 x ( 1 + ( 5 / 100 ) ) ^ 3 ] = $ 18522 . ci . = $ ( 18522 - 16000 ) = $ 2522 . answer d ." | a ) 2532 , b ) 2552 , c ) 2524 , d ) 2522 , e ) 2512 | d | subtract(multiply(16,000, power(add(const_1, divide(9, const_100)), 20)), 16,000) | divide(n2,const_100)|add(#0,const_1)|power(#1,n1)|multiply(n0,#2)|subtract(#3,n0)| | gain |
the number of students in each section of a school is 24 . after admitting new students , three new sections were started . now , the total number of sections is 16 and there are 21 students in each section . the number of new students admitted is : | original number of sections = 16 - 3 = 13 original number of students = 24 x 13 = 312 present number of students = 21 x 16 = 336 number of new students admitted = 336 - 312 = 24 so the answer is option c ) 24 . | ['a ) 12', 'b ) 42', 'c ) 24', 'd ) 28', 'e ) 26'] | c | subtract(multiply(16, 21), multiply(24, subtract(16, const_3))) | multiply(n1,n2)|subtract(n1,const_3)|multiply(n0,#1)|subtract(#0,#2) | physics |
a man crosses a 600 m long street in 5 minutes . what is his speed in km per hour ? | speed = 600 / 5 * 60 = 2 m / sec converting m / sec to km / hr . = 2 * 18 / 5 = 7.2 km / hr answer is b | a ) 7 km / hr , b ) 7.2 km / hr , c ) 5.2 km / hr , d ) 10 km / hr , e ) 5 km / hr | b | divide(divide(600, const_1000), divide(5, const_60)) | divide(n0,const_1000)|divide(n1,const_60)|divide(#0,#1) | physics |
if the sum of the 4 th term and the 12 th term of an arithmetic progression is 24 , what is the sum of the first 15 terms of the progression ? | "4 th term + 12 th term = 24 i . e . , ( a + 3 d ) + ( a + 11 d ) = 24 now , sum of first 15 terms = ( 15 / 2 ) * [ 2 a + ( 15 - 1 ) d ] = ( 15 / 2 ) * [ 2 a + 14 d ] = ( 15 / 2 ) * 24 - - - - - - - - - - - - - - - from ( 1 ) = 180 answer : e" | a ) 300 , b ) 240 , c ) 220 , d ) 200 , e ) 180 | e | multiply(divide(15, const_2), 24) | divide(n3,const_2)|multiply(n2,#0)| | general |
how many 4 - digit positive integers are there , where each digit is positive , and no 3 adjacent digits are same ? | first digit . . 9 posibilities second digit , 8 possibilities third digit , 7 possibilities fourth digit , 7 possibilities . 9 * 8 * 7 * 7 = 3528 . b | a ) 1236 , b ) 3528 , c ) 4096 , d ) 4608 , e ) 6561 | b | multiply(multiply(add(const_4, const_3), add(const_4, const_3)), multiply(add(4, 4), multiply(3, 3))) | add(const_3,const_4)|add(n0,n0)|multiply(n1,n1)|multiply(#0,#0)|multiply(#1,#2)|multiply(#3,#4) | general |
15 lts are taken of from a container full of liquid a and replaced with liquid b . again 15 more lts of the mixture is taken and replaced with liquid b . after this process , if the container contains liquid a and b in the ratio 9 : 16 , what is the capacity of the container w ? | "if you have a 37.5 liter capacity , you start with 37.5 l of a and 0 l of b . 1 st replacement after the first replacement you have 37.5 - 15 = 22.5 l of a and 15 l of b . the key is figuring out how many liters of a and b , respectively , are contained in the next 15 liters of mixture to be removed . the current rati... | a ) a : 45 , b ) b : 25 , c ) c : 37.5 , d ) d : 36 , e ) e : 42 | c | divide(15, subtract(const_1, sqrt(divide(9, add(9, 16))))) | add(n2,n3)|divide(n2,#0)|sqrt(#1)|subtract(const_1,#2)|divide(n0,#3)| | general |
a number when divided by 45 , gives 30 as quotient and 0 as remainder . what will be the remainder when dividing the same number by 15 | explanation : p ÷ 45 = 30 = > p = 30 * 45 = 1350 p / 15 = 1350 / 15 = 90 , remainder = 0 answer : option a | a ) a ) 0 , b ) b ) 3 , c ) c ) 4 , d ) d ) 6 , e ) e ) 7 | a | divide(multiply(45, 30), 15) | multiply(n0,n1)|divide(#0,n3) | general |
a vendor buys 10 t - shirts at an average price of $ 14 per t - shirt . he then buys 15 more t - shirts at an average price of $ 11 per t - shirt . what is the average price s per t - shirt that the vendor paid for these purchases ? | "correct answer : a explanation : the relevant formula for this problem is average s = ( sum ) / ( number of terms ) . another way to look at the formula is sum = average x number of terms . for the first purchase , the vendor ' s sum ( total cost ) was $ 140 , since 14 x 10 = 140 . for the second purchase , the vendor... | a ) $ 12.20 , b ) $ 12.50 , c ) $ 12.55 , d ) $ 12.70 , e ) $ 13.00 | a | add(divide(add(10, 14), const_2), add(const_0_25, const_0_25)) | add(const_0_25,const_0_25)|add(n0,n1)|divide(#1,const_2)|add(#0,#2)| | general |
of the families in city x in 1904 , 50 percent owned a personal computer . the number of families in city x owning a computer in 1908 was 30 percent greater than it was in 1904 , and the total number of families in city x was 5 percent greater in 1908 than it was in 1904 . what percent of the families in city x owned a... | "say a 100 families existed in 1904 then the number of families owning a computer in 1904 - 50 number of families owning computer in 1908 = 50 * 130 / 100 = 65 number of families in 1908 = 105 the percentage = 65 / 105 * 100 = 61.90 % . option : a" | a ) 61.90 % , b ) 52.20 % , c ) 56.20 % , d ) 70.45 % , e ) 74.41 % | a | multiply(const_100, divide(divide(multiply(add(30, const_100), 50), const_100), add(const_100, 5))) | add(n3,const_100)|add(n5,const_100)|multiply(n1,#0)|divide(#2,const_100)|divide(#3,#1)|multiply(#4,const_100)| | general |
a merchant marks his goods up by 40 % and then offers a discount of 20 % on the marked price . what % profit does the merchant make after the discount ? | "let the price be 100 . the price becomes 140 after a 40 % markup . now a discount of 20 % on 140 . profit = 112 - 100 12 % answer e" | a ) 21 % , b ) 25 % , c ) 69 % , d ) 31 % , e ) 12 % | e | subtract(subtract(add(40, const_100), divide(multiply(add(40, const_100), 20), const_100)), const_100) | add(n0,const_100)|multiply(n1,#0)|divide(#1,const_100)|subtract(#0,#2)|subtract(#3,const_100)| | gain |
the perimeter of a triangle is 36 cm and the in radius of the triangle is 2.5 cm . what is the area of the triangle ? | "area of a triangle = r * s where r is the in radius and s is the semi perimeter of the triangle . area of triangle = 2.5 * 36 / 2 = 45 cm 2 answer : e" | a ) 76 , b ) 88 , c ) 66 , d ) 55 , e ) 45 | e | triangle_area(2.5, 36) | triangle_area(n0,n1)| | geometry |
during a certain two - week period , 70 percent of the movies rented from a video store were comedies , and of the remaining movies rented , there were 4 times as many dramas as action movies . if no other movies were rented during that two - week period and there were a action movies rented , then how many comedies , ... | "total movies = 100 . comedies = 70 . action + drama = 30 . since there were 5 times as many dramas as action movies , then action + 4 * action = 30 - - > action = a = 6 . comedies = 70 = 14 a . answer : d ." | a ) 16 a , b ) 13 a , c ) 14 a , d ) 11 2 / 3 a , e ) 12 a | d | floor(divide(70, add(4, const_1))) | add(n1,const_1)|divide(n0,#0)|floor(#1)| | general |
there were two candidates in an election . winner candidate received 58 % of votes and won the election by 288 votes . find the number of votes casted to the winning candidate ? | "w = 58 % l = 42 % 58 % - 42 % = 16 % 16 % - - - - - - - - 288 58 % - - - - - - - - ? = > 1044 answer : b" | a ) 776 , b ) 1044 , c ) 299 , d ) 257 , e ) 125 | b | divide(multiply(divide(288, divide(subtract(58, subtract(const_100, 58)), const_100)), 58), const_100) | subtract(const_100,n0)|subtract(n0,#0)|divide(#1,const_100)|divide(n1,#2)|multiply(n0,#3)|divide(#4,const_100)| | gain |
4 / 15 of 5 / 7 of a number is greater than 4 / 9 of 2 / 5 of the same number by 8 . what is half of that number ? | let the number be x . then 4 / 15 of 5 / 7 of x - 4 / 9 of 2 / 5 of x = 8 4 / 21 x - 8 / 45 x = 8 ( 4 / 21 - 8 / 45 ) x = 8 ( 60 - 56 ) / 315 x = 8 4 / 315 x = 8 x = ( 8 * 315 ) / 4 = 630 1 / 2 x = 315 hence required number = 315 answer is b . | a ) 305 , b ) 315 , c ) 345 , d ) 325 , e ) 335 | b | divide(divide(8, subtract(multiply(divide(4, 15), divide(5, 7)), multiply(divide(2, 5), divide(4, 9)))), 2) | divide(n0,n1)|divide(n2,n3)|divide(n6,n2)|divide(n0,n5)|multiply(#0,#1)|multiply(#2,#3)|subtract(#4,#5)|divide(n8,#6)|divide(#7,n6) | general |
a producer of tea blends two varieties of tea from two tea gardens one costing rs 18 per kg and another rs 20 per kg in the ratio 5 : 3 . if he sells the blended variety at rs 20 per kg , then his gain percent is | "explanation : suppose he bought 5 kg and 3 kg of tea . cost price = rs . ( 5 x 18 + 3 x 20 ) = rs . 150 . selling price = rs . ( 8 x 20 ) = rs . 160 . profit = 160 - 150 = 10 so , profit % = ( 10 / 150 ) * 100 = 7 % option a" | a ) 7 % , b ) 13 % , c ) 14 % , d ) 15 % , e ) 16 % | a | divide(multiply(subtract(multiply(20, add(5, 3)), add(multiply(5, 18), multiply(3, 20))), const_100), add(multiply(5, 18), multiply(3, 20))) | add(n2,n3)|multiply(n0,n2)|multiply(n1,n3)|add(#1,#2)|multiply(n4,#0)|subtract(#4,#3)|multiply(#5,const_100)|divide(#6,#3)| | gain |
what is the total number of y integers between 100 and 200 that are divisible by 3 ? | yes there is a different way of arriving at that answer . . . . u can also use airthmetic progression to get the answer since the first term to be divisble by 3 is 102 . . take that as a . . the starting no and since 198 is the last digit to be divisible by 3 take that as n . . . since the difference is 3 take that as ... | a ) 33 , b ) 32 , c ) 31 , d ) 30 , e ) 29 | a | add(divide(subtract(subtract(200, const_2), add(100, const_2)), 3), const_1) | add(n0,const_2)|subtract(n1,const_2)|subtract(#1,#0)|divide(#2,n2)|add(#3,const_1) | general |
if each participant of a chess tournament plays exactly one game with each of the remaining participants , then 136 games will be played during the tournament . find the number of participants . | let p be the number of participants . pc 2 = 136 ( p ) ( p - 1 ) = 272 = 17 * 16 p = 17 the answer is c . | a ) 15 , b ) 16 , c ) 17 , d ) 18 , e ) 19 | c | divide(add(sqrt(add(multiply(multiply(136, const_2), const_4), const_1)), const_1), const_2) | multiply(n0,const_2)|multiply(#0,const_4)|add(#1,const_1)|sqrt(#2)|add(#3,const_1)|divide(#4,const_2) | general |
a , b and c can do a piece of work in 4 days , 8 days and 20 days respectively . how long will they taken , if all the three work together ? | "1 / 4 + 1 / 8 + 1 / 20 = 17 / 40 = > 40 / 17 = > 2 6 / 17 days answer : e" | a ) 2 , b ) 4 , c ) 5 , d ) 67 , e ) 2 6 / 17 | e | inverse(add(inverse(20), add(inverse(4), inverse(8)))) | inverse(n0)|inverse(n1)|inverse(n2)|add(#0,#1)|add(#3,#2)|inverse(#4)| | physics |
for the past n days , the average ( arithmetic mean ) daily production at a company was 40 units . if today ' s production of 90 units raises the average to 45 units per day , what is the value of n ? | "( average production for n days ) * n = ( total production for n days ) - - > 40 n = ( total production for n days ) ; ( total production for n days ) + 90 = ( average production for n + 1 days ) * ( n + 1 ) - - > 40 n + 90 = 45 * ( n + 1 ) - - > n = 9 answer : d ." | a ) 30 , b ) 18 , c ) 10 , d ) 9 , e ) 7 | d | subtract(divide(subtract(90, 40), subtract(45, 40)), const_1) | subtract(n1,n0)|subtract(n2,n0)|divide(#0,#1)|subtract(#2,const_1)| | general |
in the johnsons ' monthly budget , the dollar amounts allocated to household expenses , food , and miscellaneous items are in the ratio 5 : 4 : 1 , respectively . if the total amount allocated to these 3 categories is $ 1800 , what is the amount allocated to food ? | to solve this question , it will be best to first express the given ratio using variable multipliers . thus , we have : household expenses : food : miscellaneous = 5 x : 4 x : x since we are told that the total amount allocated to these categories is $ 1,800 we can set up the equation : 5 x + 4 x + x = 1,800 10 x = 1,8... | a ) $ 900 , b ) $ 720 , c ) $ 675 , d ) $ 450 , e ) $ 225 | b | multiply(divide(1800, add(add(5, 4), 1)), 4) | add(n0,n1)|add(n2,#0)|divide(n4,#1)|multiply(n1,#2) | general |
one ton has 2600 pounds , and one pound has 16 ounces . how many packets containing wheat weighing 16 pounds and 4 ounces each would totally fill a gunny bag of capacity 13 tons ? | "16 pounds and 4 ounces = 16 * 16 + 4 = 260 ounces . 13 tons = 13 * 2600 pound = 13 * 2600 * 16 ounces . hence the answer is ( 13 * 2600 * 16 ) / 260 = 2080 . answer : a ." | a ) 2080 , b ) 1700 , c ) 2350 , d ) 2500 , e ) 8000 | a | divide(multiply(multiply(13, 2600), 16), add(multiply(16, 16), 4)) | multiply(n0,n4)|multiply(n1,n1)|add(n3,#1)|multiply(n1,#0)|divide(#3,#2)| | general |
what is the remainder when 8 ^ 381 is divided by 5 ? | i also agree that the remainder is ' 3 ' ( using the last digit of the powers of 7 ) . could we have the official answer please ? d | a ) 0 , b ) 1 , c ) 2 , d ) 3 , e ) 4 | d | subtract(divide(5, const_2), multiply(8, 8)) | divide(n2,const_2)|multiply(n0,n0)|subtract(#0,#1)| | general |
the average age of students of a class is 15.8 years . the average age of boys in the class is 16.4 years and that of the girls is 15.4 years , the ratio of the number of boys to the number of girls in the class is | "explanation : let the ratio be k : 1 . then , k * 16.4 + 1 * 15.4 = ( k + 1 ) * 15.8 < = > ( 16.4 - 15.8 ) k = ( 15.8 - 15.4 ) < = > k = 0.4 / 0.6 = 2 / 3 . required ratio = 2 / 3 : 1 = 2 : 3 . answer : b" | a ) 7 : 3 , b ) 2 : 3 , c ) 9 : 3 , d ) 6 : 3 , e ) 2 : 5 | b | divide(subtract(15.8, 15.4), subtract(16.4, 15.8)) | subtract(n0,n2)|subtract(n1,n0)|divide(#0,#1)| | general |
in a certain boys camp , 20 % of the total boys are from school a and 30 % of those study science . if there are 35 boys in the camp that are from school a but do not study science then what is the total number of boys in the camp ? | "since 30 % of the boys from school a study science , then 70 % of the boys from school a do not study science and since 20 % of the total number of boys are from school a , then 0.2 * 0.7 = 0.14 , or 14 % of the boys in the camp are from school a and do not study science . we are told that this number equals to 35 , s... | a ) 70 , b ) 245 , c ) 150 , d ) 35 , e ) 250 | e | divide(35, multiply(divide(subtract(const_100, 30), const_100), divide(20, const_100))) | divide(n0,const_100)|subtract(const_100,n1)|divide(#1,const_100)|multiply(#2,#0)|divide(n2,#3)| | gain |
of the female students at barkely university , 5 / 6 are on the honor roll . of the male students , 2 / 3 are on the honor roll . if 2 / 5 of the students are female , what fraction of all the students are on the honor roll ? | let the total students be 30 given 2 / 5 of the students are females = 12 then males = 3 / 5 = 18 5 / 6 of the females are on honor roll = 10 males on the honor roll = 2 / 3 = 12 total students on honor roll = 12 + 10 = 22 fraction = 22 / 30 d | a ) 5 / 9 , b ) 3 / 15 , c ) 23 / 30 , d ) 22 / 30 , e ) 53 / 90 | d | divide(add(multiply(multiply(5, 2), 2), 2), multiply(2, multiply(3, 5))) | multiply(n0,n2)|multiply(n0,n3)|multiply(n2,#0)|multiply(n2,#1)|add(n2,#2)|divide(#4,#3) | general |
a train running at the speed of 60 km / hr crosses a pole in 18 seconds . find the length of the train . | ": speed = 60 * ( 5 / 18 ) m / sec = 50 / 3 m / sec length of train ( distance ) = speed * time ( 50 / 3 ) * 18 = 300 meter answer : d" | a ) 150 , b ) 278 , c ) 179 , d ) 300 , e ) 191 | d | multiply(divide(multiply(60, const_1000), const_3600), 18) | multiply(n0,const_1000)|divide(#0,const_3600)|multiply(n1,#1)| | physics |
the length of rectangle is thrice its breadth and its perimeter is 88 m , find the area of the rectangle ? | 2 ( 3 x + x ) = 88 l = 33 b = 11 lb = 33 * 11 = 363 answer : b | ['a ) 432 sq m', 'b ) 363 sq m', 'c ) 452 sq m', 'd ) 428 sq m', 'e ) 525 sq m'] | b | multiply(multiply(divide(88, add(multiply(const_3, const_2), multiply(const_1, const_2))), const_3), divide(88, add(multiply(const_3, const_2), multiply(const_1, const_2)))) | multiply(const_2,const_3)|multiply(const_1,const_2)|add(#0,#1)|divide(n0,#2)|multiply(#3,const_3)|multiply(#3,#4) | geometry |
the ratio of the volumes of two cubes is 729 : 2197 . what is the ratio of their total surface areas ? | "ratio of the sides = 3 √ 729 : 3 √ 2197 = 9 : 13 ratio of surface areas = 9 ^ 2 : 13 ^ 2 = 81 : 169 answer : option a" | a ) 81 : 169 , b ) 81 : 122 , c ) 81 : 124 , d ) 81 : 126 , e ) 81 : 129 | a | power(divide(729, 2197), divide(const_1, const_3)) | divide(n0,n1)|divide(const_1,const_3)|power(#0,#1)| | geometry |
a monkey ascends a greased pole 22 meters high . he ascends 2 meters in the first minute and then slips down 1 meter in the alternate minute . if this pattern continues until he climbs the pole , in how many minutes would he reach at the top of the pole ? | "the money is climbing 1 meter in 2 min . this pattern will go on till he reaches 20 meters . i mean this will continue for first 20 * 2 = 40 mins . he would have reached 20 meters . after that he will climb 2 meters and he will reach the pole . so total time taken = 40 + 1 = 41 mins . so , asnwer will be b" | a ) 40 th minute , b ) 41 st minute , c ) 43 rd minute , d ) 42 nd minute , e ) 45 th minute | b | add(multiply(multiply(const_4, 2), 2), 1) | multiply(n1,const_4)|multiply(n1,#0)|add(n2,#1)| | physics |
a can run 256 metre in 28 seconds and b in 32 seconds . by what distance a beat b ? | "clearly , a beats b by 4 seconds now find out how much b will run in these 4 seconds speed of b = distance / time taken by b = 256 / 32 = 8 m / s distance covered by b in 4 seconds = speed ã — time = 8 ã — 4 = 32 metre i . e . , a beat b by 32 metre answer is a" | a ) 32 metre , b ) 28 metre , c ) 23 metre , d ) 15 metre , e ) 28 metre | a | subtract(256, multiply(divide(256, 32), 28)) | divide(n0,n2)|multiply(n1,#0)|subtract(n0,#1)| | physics |
if | x | = 4 x - 2 , then x = ? | "answer : approach : substituted option a i . e x = 2 . inequality satisfied . b" | a ) 1 , b ) 2 , c ) 1 and 1 / 2 , d ) - 1 / 2 , e ) - 1 | b | divide(2, subtract(4, const_1)) | subtract(n0,const_1)|divide(n1,#0)| | general |
two circular signs are to be painted . if the diameter of the larger sign is 7 times that of the smaller sign , how many times more paint is needed to paint the larger sign ? ( we can assume that a given amount of paint covers the same area on both signs . ) | let r be the radius of the smaller sign . then the diameter of the smaller sign is 2 r , the diameter of the larger sign is 14 r , and the radius of the larger sign is 7 r . the area a of the smaller sign is a = pir ^ 2 . the area of the larger sign is pi ( 7 r ) ^ 2 = 49 pir ^ 2 = 49 a . since the area is 49 times lar... | a ) 7 , b ) 14 , c ) 42 , d ) 49 , e ) 98 | d | power(7, const_2) | power(n0,const_2) | geometry |
when a merchant imported a certain item , he paid a 9 percent import tax on the portion of the total value of the item in excess of $ 1,000 . if the amount of the import tax that the merchant paid was $ 76.50 , what was the total value of the item ? | "let x be the value in excess of $ 1,000 . 0.09 x = 76.5 x = $ 850 the total value was $ 850 + $ 1,000 = $ 1,850 . the answer is d ." | a ) $ 1,250 , b ) $ 1,450 , c ) $ 1,650 , d ) $ 1,850 , e ) $ 2,050 | d | floor(divide(add(divide(76.50, divide(9, const_100)), 1,000), 1,000)) | divide(n0,const_100)|divide(n2,#0)|add(#1,n1)|divide(#2,n1)|floor(#3)| | general |
a man swims downstream 40 km and upstream 30 km taking 5 hours each time , what is the speed of the man in still water ? | "40 - - - 5 ds = 8 ? - - - - 1 30 - - - - 5 us = 6 ? - - - - 1 m = ? m = ( 8 + 6 ) / 2 = 7 answer : e" | a ) 12 , b ) 8 , c ) 36 , d ) 5 , e ) 7 | e | divide(add(divide(30, 5), divide(40, 5)), const_2) | divide(n1,n2)|divide(n0,n2)|add(#0,#1)|divide(#2,const_2)| | physics |
a train 125 m long passes a man , running at 8 km / hr in the same direction in which the train is going , in 10 seconds . the speed of the train is ? | "speed of the train relative to man = ( 125 / 10 ) m / sec = ( 25 / 2 ) m / sec . [ ( 25 / 2 ) * ( 18 / 5 ) ] km / hr = 45 km / hr . let the speed of the train be x km / hr . then , relative speed = ( x - 8 ) km / hr . x - 8 = 45 = = > x = 53 km / hr . answer : b" | a ) 36 , b ) 53 , c ) 88 , d ) 66 , e ) 22 | b | divide(divide(subtract(125, multiply(multiply(8, const_0_2778), 8)), 8), const_0_2778) | multiply(n1,const_0_2778)|multiply(n1,#0)|subtract(n0,#1)|divide(#2,n1)|divide(#3,const_0_2778)| | physics |
85 % of a number is added to 24 , the result is the same number . find the number ? | "4 a = 48 4 a = 20 a = 12 a = 5 a 2 = 144 a 2 = 25 combined area = a 2 = 169 = > a = 13 d = 13 √ 2 answer : b" | a ) 13 √ 5 , b ) 13 √ 2 , c ) 13 √ 9 , d ) 12 √ 9 , e ) 13 √ 1 | b | divide(24, divide(24, const_100)) | divide(n1,const_100)|divide(n1,#0)| | gain |
the ratio of 2 numbers is 1 : 2 and their h . c . f . and l . c . m are 10 and 20 . find the numbers . | "let the numbers be x and 2 x their h . c . f . = 10 l . c . m = 20 ie ) 2 x ^ 2 = 200 x ^ 2 = 100 , x = 10 t the numbers are x and 2 x , ie ) 10 , 20 respectively . answer is b" | a ) 20 , 10 , b ) 10 , 20 , c ) 20 , 30 , d ) 30 , 20 , e ) 10 , 10 | b | sqrt(divide(10, add(power(2, 1), add(power(2, 1), power(1, 1))))) | power(n0,n1)|power(n1,n1)|power(n2,n1)|add(#0,#1)|add(#3,#2)|divide(n3,#4)|sqrt(#5)| | other |
if x is 20 percent greater than 88 , then x = | "x is 20 % greater than 88 means x is 1.2 times 88 ( in other words 88 + 20 / 100 * 88 = 1.2 * 88 ) therefore , x = 1.2 * 88 = 105.6 answer : d" | a ) 68 , b ) 70.4 , c ) 86 , d ) 105.6 , e ) 108 | d | add(88, multiply(divide(20, const_100), 88)) | divide(n0,const_100)|multiply(n1,#0)|add(n1,#1)| | general |
set a consists of all the prime numbers between 15 and 36 . what is the range of set a ? | "the range of a set of data is the difference between the highest and lowest values in the set in this set , highest number = 31 lowest number = 17 range = highest - lowest = 31 - 17 = 14 option c" | a ) 12 , b ) 13 , c ) 14 , d ) 17 , e ) 23 | c | subtract(subtract(36, const_2), add(add(15, const_4), const_1)) | add(n0,const_4)|subtract(n1,const_2)|add(#0,const_1)|subtract(#1,#2)| | general |
in one hour , a boat goes 21 km along the stream and 9 km against the stream . the speed of the boat in still water ( in km / hr ) is : | "solution speed in still water = 1 / 2 ( 21 + 9 ) kmph . = 15 kmph . answer e" | a ) 3 , b ) 5 , c ) 10 , d ) 9 , e ) 15 | e | divide(add(21, 9), const_2) | add(n0,n1)|divide(#0,const_2)| | physics |
in the faculty of reverse - engineering , 226 second year students study numeric methods , 423 second year students study automatic control of airborne vehicles and 134 second year students study them both . how many students are there in the faculty if the second year students are approximately 70 % of the total ? | "answer is e : 736 solution : total number of students studying both are 423 + 226 - 134 = 515 ( subtracting the 134 since they were included in the both the other numbers already ) . so 70 % of total is 515 , so 100 % is approx . 736 ." | a ) 515 . , b ) 545 . , c ) 618 . , d ) 644 . , e ) 736 . | e | add(226, 423) | add(n0,n1)| | general |
machines x and y produce bottles at their respective constant rates . machine x produces k bottles in 6 hours and machine y produces k bottles in 12 hours . how many hours does it take machines x and y , working simultaneously , to produce 24 k bottles ? | "x rate = k / 6 y rate = k / 12 k / 6 + k / 12 = 24 k / t solving t = 96 answer e" | a ) 48 , b ) 42 , c ) 45 , d ) 98 , e ) 96 | e | divide(multiply(24, divide(add(6, 12), 12)), 12) | add(n0,n1)|divide(#0,n1)|multiply(n2,#1)|divide(#2,n1)| | physics |
on the number line , if x is halfway between - 4 and 4 , and if y is halfway between - 2 and 6 , what number is halfway between x and y ? | "x = 0 and y = 2 . the answer is e ." | a ) - 1 , b ) - 1 / 2 , c ) 0 , d ) 1 / 2 , e ) 1 | e | add(negate(4), 4) | negate(n0)|add(n1,#0)| | general |
5 % people of a village in sri lanka died by bombardment , 15 % of the remainder left the village on account of fear . if now the population is reduced to 3294 , how much was it in the beginning ? | x * ( 95 / 100 ) * ( 85 / 100 ) = 3294 x = 4079 answer : c | a ) 2988 , b ) 2776 , c ) 4079 , d ) 2871 , e ) 881 | c | floor(divide(3294, multiply(divide(subtract(const_100, 5), const_100), divide(subtract(const_100, 15), const_100)))) | subtract(const_100,n0)|subtract(const_100,n1)|divide(#0,const_100)|divide(#1,const_100)|multiply(#2,#3)|divide(n2,#4)|floor(#5) | general |
tanks a and b are each in the shape of a right circular cylinder . the interior of tank a has a height of 6 meters and a circumference of 8 meters , and the interior of tank b has a height of 8 meters and a circumference of 10 meters . the capacity of tank a is what percent of the capacity of tank b ? | "for a , r = 8 / 2 pi . its capacity = ( 4 pi ) ^ 2 * 6 = 96 pi for b , r = 10 / pi . its capacity = ( 5 pi ) ^ 2 * 8 = 200 pi a / b = 96 pi / 200 pi = 0.48 a" | a ) 48 % , b ) 80 % , c ) 100 % , d ) 120 % , e ) 125 % | a | multiply(multiply(power(divide(8, 10), const_2), divide(6, 8)), const_100) | divide(n0,n2)|divide(n1,n3)|power(#1,const_2)|multiply(#0,#2)|multiply(#3,const_100)| | physics |
the number of singles that a baseball player hit increased 12 percent from her first season to her second season , and the number of doubles that she hit in the same time period decreased by 2 percent . if the total number of singles and doubles that she hit increased 2 percent from her first season to her second seaso... | "soln : - 1.12 s + 0.98 d = 1.02 [ s + d ] 0.10 s = 0.04 d s / d = 2 / 5 answer : b" | a ) 1 : 2 , b ) 2 : 5 , c ) 1 : 3 , d ) 2 : 3 , e ) 3 : 5 | b | divide(subtract(const_100, 2), add(const_100, 12)) | add(n0,const_100)|subtract(const_100,n1)|divide(#1,#0)| | general |
on the independence day , bananas were be equally distributed among the children in a school so that each child would get two bananas . on the particular day 350 children were absent and as a result each child got two extra bananas . find the actual number of children in the school ? | "explanation : let the number of children in the school be x . since each child gets 2 bananas , total number of bananas = 2 x . 2 x / ( x - 350 ) = 2 + 2 ( extra ) = > 2 x - 700 = x = > x = 700 . answer : e" | a ) 237 , b ) 287 , c ) 197 , d ) 287 , e ) 700 | e | multiply(350, const_2) | multiply(n0,const_2)| | general |
every digit of a number written in binary is either 0 or 1 . to translate a number from binary , multiply the nth digit ( reading from right to left ) by 2 ^ ( n - 1 ) what is the largest prime number ( written in binary ) that is a factor of both 10010000 and 100100000 ? | binary divison can provide a quick answer if you are comfortable with it . as option e is the biggest binary number we try with it first : 100010000 / 1001 = 10000 1000100000 / 1001 = 100000 so answer is option is c | a ) 10 , b ) 11 , c ) 1001 , d ) 1011 , e ) 10001 | c | add(divide(100100000, 10010000), const_1000) | divide(n5,n4)|add(#0,const_1000) | general |
a cistern is normally filled in 8 hours but takes two hours longer to fill because of a leak in its bottom . if the cistern is full , the leak will empty it in ? | "1 / 8 - 1 / x = 1 / 10 x = 40 answer : c" | a ) 27 , b ) 20 , c ) 40 , d ) 73 , e ) 72 | c | inverse(subtract(divide(const_1, 8), divide(const_1, add(8, const_2)))) | add(n0,const_2)|divide(const_1,n0)|divide(const_1,#0)|subtract(#1,#2)|inverse(#3)| | physics |
find the distance covered by a man walking for 24 min at a speed of 10 km / hr ? | "distance = 10 * 24 / 60 = 4 km answer is c" | a ) 1 km , b ) 3 km , c ) 4 km , d ) 5 km , e ) 6 km | c | divide(multiply(24, divide(multiply(10, const_1000), const_60)), const_1000) | multiply(n1,const_1000)|divide(#0,const_60)|multiply(n0,#1)|divide(#2,const_1000)| | physics |
the diagonals of a rhombus are 30 cm and 20 cm . find its area ? | "1 / 2 * 30 * 20 = 300 answer : e" | a ) 176 , b ) 190 , c ) 150 , d ) 278 , e ) 300 | e | rhombus_area(30, 20) | rhombus_area(n0,n1)| | geometry |
at what price must an article costing rs . 50 be marked in order that after deducting 10 % from the list price . it may be sold at a profit of 25 % on the cost price ? | "cp = 50 sp = 50 * ( 125 / 100 ) = 62.50 mp * ( 90 / 100 ) = 62.50 mp = 69.44 answer : a" | a ) 69.44 , b ) 62.0 , c ) 62.6 , d ) 62.1 , e ) 62.2 | a | divide(multiply(add(50, divide(multiply(50, 25), const_100)), const_100), subtract(const_100, 10)) | multiply(n0,n2)|subtract(const_100,n1)|divide(#0,const_100)|add(n0,#2)|multiply(#3,const_100)|divide(#4,#1)| | gain |
every letter in the alphabet has a number value that is equal to its place in the alphabet . thus , the letter a has a value of 1 , the letter b has a value of 2 , the letter c has a value of 3 , etc . . . the number value of a word is obtained by adding up the value of the letters in the word and then multiplying that... | "` ` cat ' ' = ( 3 + 1 + 20 ) * 3 = 72 . the answer is d ." | a ) 45 , b ) 54 , c ) 63 , d ) 72 , e ) 81 | d | multiply(add(add(subtract(multiply(2, const_10), 2), 1), multiply(2, const_10)), 3) | multiply(n1,const_10)|subtract(#0,n1)|add(n0,#1)|add(#2,#0)|multiply(n2,#3)| | general |
a man two flats for $ 512456 each . on one he gains 20 % while on the other he loses 20 % . how much does he gain or lose in the whole transaction ? | in such a case there is always a loss loss % = ( 20 / 10 ) ^ 2 = 4 % answer is e | a ) 10 % , b ) 5 % , c ) 12 % , d ) 6 % , e ) 4 % | e | multiply(divide(subtract(add(multiply(divide(const_100, add(const_100, 20)), 512456), multiply(divide(const_100, subtract(const_100, 20)), 512456)), add(512456, 512456)), add(multiply(divide(const_100, add(const_100, 20)), 512456), multiply(divide(const_100, subtract(const_100, 20)), 512456))), const_100) | add(n1,const_100)|add(n0,n0)|subtract(const_100,n1)|divide(const_100,#0)|divide(const_100,#2)|multiply(n0,#3)|multiply(n0,#4)|add(#5,#6)|subtract(#7,#1)|divide(#8,#7)|multiply(#9,const_100) | gain |
a train travels 225 km in 3.5 hours and 370 km in 5 hours . find the average speed of train . | "as we know that speed = distance / time for average speed = total distance / total time taken thus , total distance = 225 + 370 = 595 km thus , total speed = 8.5 hrs or , average speed = 595 / 8.5 or , 70 kmph . answer : c" | a ) 80 kmph , b ) 60 kmph , c ) 70 kmph , d ) 90 kmph , e ) none of these | c | divide(add(225, 370), add(3.5, 5)) | add(n0,n2)|add(n1,n3)|divide(#0,#1)| | physics |
the cash difference between the selling prices of an article at a profit of 4 % and 8 % is rs 3 . the ratio of two selling prices is | "explanation : let the cost price of article is rs . x required ratio = ( 104 % of x ) / ( 108 % of x ) = 104 / 108 = 52 / 54 = 52 : 54 . answer : b" | a ) 51 : 52 , b ) 52 : 54 , c ) 53 : 54 , d ) 54 : 55 , e ) none of these | b | divide(add(const_100, 4), add(const_100, 8)) | add(n0,const_100)|add(n1,const_100)|divide(#0,#1)| | gain |
if | 4 x + 6 | = 36 , then find the sum of the values of x ? | "| 4 x + 6 | = 36 4 x + 6 = 36 or 4 x + 6 = - 36 4 x = 30 or 4 x = - 42 x = 7.5 or x = - 10.5 sum = 7.5 - 10.5 = - 3 answer is a" | a ) - 3 , b ) 4 , c ) 3 , d ) - 5 , e ) 6 | a | divide(subtract(36, 6), 4) | subtract(n2,n1)|divide(#0,n0)| | general |
the speed of a boat in still water in 42 km / hr and the rate of current is 8 km / hr . the distance travelled downstream in 44 minutes is : | "speed downstream = ( 42 + 8 ) = 50 kmph time = 44 minutes = 44 / 60 hour = 11 / 15 hour distance travelled = time × speed = 11 / 15 × 50 = 36.6 km answer : c" | a ) 86.6 km , b ) 46.6 km , c ) 36.6 km , d ) 35.6 km , e ) 26.6 km | c | multiply(add(42, 8), divide(44, const_60)) | add(n0,n1)|divide(n2,const_60)|multiply(#0,#1)| | physics |
an article is sold at a certain price . by selling it at 2 / 3 of that price one loses 10 % . find the gain percent at original price . | "let the original s . p be rs . x . then new s . p = rs . 2 / 3 x , loss = 10 % so c . p = rs . [ 100 / 90 * 2 / 3 x ] = 20 x / 27 . now c . p = rs . 20 x / 27 , s . p = rs . x , gain = rs . [ x - 20 x / 27 ] = rs . 7 x / 27 . gain % = [ 7 x / 27 * 27 / 20 x * 100 ] % = 35 % answer : b" | a ) 15 % , b ) 35 % , c ) 25 % , d ) 50 % , e ) 30 % | b | subtract(divide(subtract(const_100, 10), divide(2, 3)), const_100) | divide(n0,n1)|subtract(const_100,n2)|divide(#1,#0)|subtract(#2,const_100)| | gain |
120 + 520 ÷ 5.00 = ? | answer given expression = 120 + 520 ÷ 5.00 = 120 + 104 = 224 correct option : e | a ) 100 , b ) 150 , c ) 120 , d ) 520 , e ) none | e | multiply(120, divide(520, 5)) | divide(n1,n2)|multiply(n0,#0) | general |
if 50 apprentices can finish a job in 4 hours , and 30 journeymen can finish the same job in 6 hours , how much of the job should be completed by 10 apprentices and 15 journeymen in one hour ? | 50 apprentices can finish the job in 4 hours , thus : 10 apprentices can finish the job in 4 * 5 = 20 hours ; in 1 hour 10 apprentices can finish 1 / 20 of the job . 30 journeymen can finish the same job in 4,5 hours , thus : 15 journeymen can finish the job in 6 * 2 = 12 hours ; in 1 hour 15 journeymen can finish 1 / ... | a ) 1 / 9 , b ) 29 / 180 , c ) 26 / 143 , d ) 2 / 15 , e ) 39 / 121 | d | add(divide(const_1, divide(multiply(50, 4), const_10)), divide(const_1, divide(multiply(30, 6), 15))) | multiply(n0,n1)|multiply(n2,n3)|divide(#0,const_10)|divide(#1,n5)|divide(const_1,#2)|divide(const_1,#3)|add(#4,#5) | physics |
a shopkeeper sells 600 metres of cloth for rs . 36000 at a loss of rs . 10 per metre . find his cost price for one metre of cloth ? | "sp per metre = 36000 / 600 = rs . 60 loss per metre = rs . 10 cp per metre = 60 + 10 = rs . 70 answer : d" | a ) s . 59 , b ) s . 58 , c ) s . 09 , d ) s . 70 , e ) s . 13 | d | add(divide(36000, 600), 10) | divide(n1,n0)|add(n2,#0)| | gain |
how long will take a leak at the bottom of a tank to empty it if it will take 8 hours to fill it without the leak , but due to the leak it will take 9 hours to be filled ? | part filled without leak in 1 hour = 1 / 8 part filled with leak in 1 hour = 1 / 9 work done by leak in 1 hour = 1 / 8 â ˆ ’ 1 / 9 = 72 hours answer : c | a ) 50 hours , b ) 33 hours , c ) 72 hours , d ) 82 hours , e ) 92 hours | c | multiply(8, 9) | multiply(n0,n1) | physics |
on a sum of money , the simple interest for 2 years is rs . 325 , while the compound interest is rs . 340 , the rate of interest being the same in both the cases . the rate of interest is | "explanation : the difference between compound interest and simple interest on rs . p for 2 years at r % per annum = ( r ã — si ) / ( 2 ã — 100 ) difference between the compound interest and simple interest = 340 - 325 = 15 ( r ã — si ) / ( 2 ã — 100 ) = 15 ( r ã — 325 ) / ( 2 ã — 100 ) = 15 r = 9.23 % answer : option ... | a ) 15 % , b ) 14.25 % , c ) 9.23 % , d ) 10.5 % , e ) 11.5 % | c | divide(multiply(const_100, subtract(subtract(340, divide(325, 2)), divide(325, 2))), divide(325, 2)) | divide(n1,n0)|subtract(n2,#0)|subtract(#1,#0)|multiply(#2,const_100)|divide(#3,#0)| | gain |
a reduction of 20 % in the price of salt enables a lady to obtain 10 kgs more for rs . 400 , find the original price per kg ? | 100 * ( 20 / 100 ) = 20 - - - 10 ? - - - 1 = > rs . 2 400 - - - 80 ? - - - 2 = > rs . 10 answer : c | a ) s . 7 , b ) s . 9 , c ) s . 10 , d ) s . 11 , e ) s . 19 | c | multiply(divide(divide(multiply(divide(20, const_100), 400), 10), multiply(divide(20, const_100), 400)), const_100) | divide(n0,const_100)|multiply(n2,#0)|divide(#1,n1)|divide(#2,#1)|multiply(#3,const_100) | gain |
the product of a and b is equal to 1 more than twice the sum of a and b . if b = 7 , what is the value of b - a ? | "ab = 1 + 2 ( a + b ) 7 a = 1 + 2 a + 14 5 a = 15 a = 3 b - a = 7 - 3 = 4 c is the answer" | a ) 2 , b ) 5 , c ) 4 , d ) 24 , e ) 35 | c | subtract(7, divide(add(multiply(7, const_2), 1), subtract(7, const_2))) | multiply(n1,const_2)|subtract(n1,const_2)|add(n0,#0)|divide(#2,#1)|subtract(n1,#3)| | general |
he greatest possible length which can be used to measure exactly the lengths 7 m , 3 m 85 cm , 12 m 95 cm is : | required length = h . c . f . of 700 cm , 385 cm and 1295 cm = 35 cm . answer : option c | a ) 15 cm , b ) 25 cm , c ) 35 cm , d ) 42 cm , e ) 45 cm | c | multiply(add(const_4, const_1), 7) | add(const_1,const_4)|multiply(n0,#0) | physics |
12 is 6 % of a , and 6 is 12 % of b . c equals b / a . what is the value of c ? | 6 a / 100 = 12 a = 200 12 b / 100 = 6 b = 50 c = b / a = 50 / 200 = 1 / 4 the answer is c . | a ) 1 / 2 , b ) 1 / 3 , c ) 1 / 4 , d ) 1 / 6 , e ) 1 / 12 | c | divide(multiply(divide(6, 12), const_100), multiply(divide(12, 6), const_100)) | divide(n1,n0)|divide(n0,n1)|multiply(#0,const_100)|multiply(#1,const_100)|divide(#2,#3) | general |
( 17 ) 4.5 x ( 17 ) ? = 178 | "solution let ( 17 ) 4.5 * ( 17 ) x = 178 . then , ( 17 ) 4.5 + x = ( 17 ) 8 . ∴ 4.5 + x = 8 ⇔ x = ( 8 - 4.5 ) ⇔ x = 3.5 answer a" | a ) 3.5 , b ) 2.75 , c ) 4.25 , d ) 4.5 , e ) none of these | a | multiply(17, 17) | multiply(n0,n2)| | general |
working alone , printers x , y , and z can do a certain printing job , consisting of a large number of pages , in 15 , 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... | "the time it takes printer x is 15 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 15 / ( 20 / 3 ) = 3 * 15 / 20 = 9 / 4 the answer is c ." | a ) 5 / 2 , b ) 8 / 3 , c ) 9 / 4 , d ) 10 / 3 , e ) 12 / 5 | c | divide(15, 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 |
excluding stoppages , the speed of a train is 35 kmph and including stoppages it is 22 kmph . of how many minutes does the train stop per hour ? | "explanation : t = 13 / 35 * 60 = 22 answer : option c" | a ) 82 , b ) 17 , c ) 22 , d ) 82 , e ) 18 | c | subtract(const_60, multiply(const_60, divide(22, 35))) | divide(n1,n0)|multiply(#0,const_60)|subtract(const_60,#1)| | physics |
jane and thomas are among the 7 people from which a committee of 3 people is to be selected . how many different possible committees of 3 people can be selected from these 7 people if at least one of either jane or thomas is to be selected ? | "the total number of ways to choose 3 people from 7 is 7 c 3 = 35 . the number of committees without jane or thomas is 5 c 3 = 10 . there are 35 - 10 = 25 possible committees which include jane and / or thomas . the answer is b ." | a ) 20 , b ) 25 , c ) 30 , d ) 35 , e ) 40 | b | add(add(choose(7, const_1), choose(7, const_1)), choose(const_4, const_1)) | choose(n0,const_1)|choose(const_4,const_1)|add(#0,#0)|add(#2,#1)| | probability |
the rate of interest on a sum of money is 8 % p . a . for the first 3 years , 4 % p . a . for the next 4 years , and 5 % for the period beyond 7 years . if the s . i , occured on the sum for the total period of 8 years is rs . 540 / - , the sum is | explanation : i 1 = ( p x 3 x 8 ) / 100 = 6 p / 25 i 2 = ( p x 4 x 4 ) / 100 = 4 p / 25 i 3 = ( p x 1 x 5 ) / 100 = p / 20 6 p / 25 + 4 p / 25 + p / 20 = 540 9 p / 20 = 540 p = 1200 answer : option a | a ) 1,200 , b ) 2,000 , c ) 2,100 , d ) 2,250 , e ) 2,540 | a | divide(540, 540) | divide(n7,n7) | general |
a team of 8 persons joins in a shooting competition . the best marksman scored 85 points . if he had scored 92 points , the average score for the team would have been 84 . the number of points , the team scored was | "let the total score be x . ( x + 92 - 85 ) / 8 = 84 . so , x + 7 = 672 = > x = 665 . answer : a" | a ) 665 , b ) 127 , c ) 272 , d ) 287 , e ) 227 | a | subtract(add(multiply(84, 8), 85), 92) | multiply(n0,n3)|add(n1,#0)|subtract(#1,n2)| | general |
if tim had lunch at $ 50.50 and he gave 15 % tip , how much did he spend ? | "the tip is 20 % of what he paid for lunch . hence tip = 15 % of 50.50 = ( 15 / 100 ) * 50.50 = $ 7.575 total spent 50.50 + 7.575 = $ 58 correct answer a" | a ) $ 58 , b ) $ 60.60 , c ) $ 70.60 , d ) $ 40.60 , e ) $ 50.60 | a | add(50.50, divide(multiply(50.50, 15), const_100)) | multiply(n0,n1)|divide(#0,const_100)|add(n0,#1)| | gain |
if the numbers 1 to 98 are written on 98 pieces of paper , ( one on each ) and one piece is picked at random , then what is the probability that the number drawn is neither prime nor composite ? | "there are 25 primes , 72 composite numbers from 1 to 98 . the number which is neither prime nor composite is 1 . therefore , required probability = 1 / 98 . answer : c" | a ) 1 / 50 , b ) 1 / 25 , c ) 1 / 98 , d ) 1 , e ) 2 | c | divide(1, 98) | divide(n0,n1)| | other |
if ( x + 3 ) ^ 2 / ( 3 x + 65 ) = 2 , then the difference between the two possible values of x is : | "( x + 3 ) ^ 2 / ( 3 x + 65 ) = 2 ( x + 3 ) ^ 2 = 2 ( 3 x + 65 ) x ^ 2 + 6 x + 9 = 6 x + 130 x ^ 2 - 121 = 0 ( x - 11 ) ( x + 11 ) = 0 x = 11 or x = - 11 the answer is d ." | a ) 16 , b ) 18 , c ) 20 , d ) 22 , e ) 24 | d | multiply(sqrt(subtract(multiply(65, const_3.0), multiply(3, 3))), 2) | multiply(n4,n3)|multiply(n0,n0)|subtract(#0,#1)|sqrt(#2)|multiply(#3,n1)| | general |
if there are only 2 wheelers and 4 wheelers parked in a school located at the heart of the city , find the number of 4 wheelers parked there if the total number of wheels is 70 ? | "four wheeler = 17 * 4 = 68 ( max ) 2 wheel = 1 so no of 4 wheeler = 17 answer : e" | a ) 11 , b ) 12 , c ) 13 , d ) 14 , e ) 17 | e | divide(subtract(70, 2), 4) | subtract(n3,n0)|divide(#0,n1)| | general |
on a purchase of $ 112 , a store offered a payment plan consisting of a $ 12 down payment and 12 monthly payments of $ 10 each . what percent of the purchase price , to the nearest tenth of a percent , did the customer pay in interest by using this plan ? | "12 * 10 + 12 = 132 ( 12 / 112 ) * 100 = 10.7 answer : a" | a ) 10.7 % , b ) 11.7 % , c ) 12.7 % , d ) 13.7 % , e ) 14.7 % | a | multiply(divide(subtract(add(multiply(10, 12), 12), multiply(10, 12)), 112), const_100) | multiply(n2,n3)|add(n1,#0)|subtract(#1,#0)|divide(#2,n0)|multiply(#3,const_100)| | gain |
john bought 9.25 m of cloth for $ 425.50 . find the cost price per metre . | cloth bought by john = 9.25 m cost of 9.25 m = $ 425.50 cost per metre = 425.50 ÷ 9.25 the cost of the cloth per metre = $ 46 answers : a | a ) 46 , b ) 47 , c ) 58 , d ) 56 , e ) 54 | a | divide(425.5, 9.25) | divide(n1,n0) | physics |
a dog is tied to a tree by a long nylon cord . if the dog runs from the due north side of the tree to the due south side of the tree with the cord extended to its full length at all items , and the dog ran approximately 30 feet , what was the approximate length of the nylon cord e , in feet ? | "because the cord was extended to its full length at all items , the dog ran along a semi - circular path , from north to south . the circumference of a full circle is 2 * pi * r , but since we only care about the length of half the circle , the semi - circle path is pi * r . e = pi * r = 30 . round pi = 3 , then r = 1... | a ) 30 , b ) 25 , c ) 15 , d ) 10 , e ) 5 | d | divide(30, const_3) | divide(n0,const_3)| | general |
a can go round a circular path 8 times in 40 minutes . if the diameter of the circle is increased to 10 times the original diameter , then the time required by a to go round the new path once , travelling at the same speed as before is | solution let original diameter be d metres . then , its circumference = ( π d ) metres . time taken to cover ( 8 π d ) m = 40 min . new diameter = ( 10 d ) m . then , its circumference = ( π x 10 d ) m . time taken to go round it once ‹ = › [ 40 / 8 π d x 10 π d ] m ‹ = › 50 min . answer c | a ) 20 min , b ) 35 min , c ) 50 min , d ) 70 min , e ) none | c | divide(multiply(40, 10), 8) | multiply(n1,n2)|divide(#0,n0) | physics |
find compound interest on rs . 8000 at 15 % per annum for 2 years 4 months , compounded annually . | "time = 2 years 4 months = 2 ( 4 / 12 ) years = 2 ( 1 / 3 ) years . amount = rs ' . [ 8000 x ( 1 + ¬ ( 15 / 100 ) ) 2 x ( 1 + ( ( 1 / 3 ) * 15 ) / 100 ) ] = rs . [ 8000 * ( 23 / 20 ) * ( 23 / 20 ) * ( 21 / 20 ) ] = rs . 11109 . . : . c . i . = rs . ( 11109 - 8000 ) = rs . 3109 . answer c rs 3109" | a ) rs 400 , b ) rs 300 , c ) rs 3109 , d ) rs 250 , e ) rs 400 | c | subtract(add(add(8000, divide(multiply(8000, 15), const_100)), divide(multiply(add(8000, divide(multiply(8000, 15), const_100)), 15), const_100)), 8000) | multiply(n0,n1)|divide(#0,const_100)|add(n0,#1)|multiply(n1,#2)|divide(#3,const_100)|add(#2,#4)|subtract(#5,n0)| | gain |
x is the product of each integer from 1 to 75 , inclusive and y = 100 ^ k , where k is an integer . what is the greatest value of k for which y is a factor of x ? | "the number of trailing zeros in the decimal representation of n ! , the factorial of a non - negative integer n , can be determined with this formula : n 5 + n 52 + n 53 + . . . + n 5 k , where k must be chosen such that 5 k ≤ n x = 1 * 2 * 3 . . . . * 75 = 50 ! no . of trailing zeros in 75 ! = 75 / 5 + 75 / 5 ^ 2 = 1... | a ) 8 , b ) 7 , c ) 6 , d ) 9 , e ) 10 | d | add(divide(75, const_10), 1) | divide(n1,const_10)|add(#0,n0)| | general |
a will was written such that an estate would be divided among 3 guys with the remainder donated to charity . one stipulation in the will is that no one from the guys can receive an amount within 30 % of another ' s amount . if one of the guys received $ 20000 , what is the smallest possible range between the highest an... | 1 st - 20000 . 2 nd - 0.7 * 20000 = 14000 3 rd - 0.7 * 14000 = 9800 range = 20000 - 9800 = 10200 . answer : d . | a ) $ 4096 , b ) $ 5904 , c ) $ 7892 , d ) $ 10052 , e ) $ 20736 | d | divide(20000, const_2) | divide(n2,const_2) | general |
find 40 % of 440 | "we know that r % of m is equal to r / 100 × m . so , we have 40 % of 440 40 / 100 × 440 = 176 answer : d" | a ) 96 , b ) 94 , c ) 86 , d ) 176 , e ) 110 | d | divide(40, 440) | divide(n0,n1)| | gain |
in a class of 20 students , 2 students did not borrow any books from the library , 8 students each borrowed 1 book , 5 students each borrowed 2 books , and the rest of the students each borrowed at least 3 books . if the average ( arithmetic mean ) number of books borrowed per student was 2 , what is the maximum number... | "the average number of books per student was 2 means that total of 2 * 20 = 40 books were borrowed ; 2 + 8 + 5 = 15 students borrowed total of 2 * 0 + 8 * 1 + 5 * 2 = 18 books ; so 40 - 18 = 22 books are left to distribute among 20 - 15 = 5 students , these 5 arethe rest who borrowed at least 3 books ; tomaximizethe nu... | a ) 3 , b ) 5 , c ) 10 , d ) 13 , e ) 15 | c | add(subtract(multiply(20, 2), add(add(8, multiply(5, 2)), multiply(2, 3))), 3) | multiply(n0,n5)|multiply(n4,n5)|multiply(n1,n6)|add(n2,#1)|add(#3,#2)|subtract(#0,#4)|add(n6,#5)| | general |
mukesh borrowed a sum of rs . 10000 at simple interest from a bank . after two years he takes rs . 12000 more as loan and discharge all the debt at the end of next 3 years depositing rs . 27160 to the bank . what is the rate of interest per annum ? | si = 27160 - ( principal amount ) = 27160 - 22000 = 5160 . now , ( 10000 * 5 * r / 100 ) + ( 12000 * 3 * r / 100 ) = 5160 . 500 r + 360 r = 5160 . . . . . rate = 6 % . answer : b | a ) 5 % , b ) 6 % , c ) 8 % , d ) 10 % , e ) none of these | b | multiply(divide(subtract(27160, add(10000, 12000)), add(multiply(10000, add(3, const_2)), multiply(12000, 3))), const_100) | add(n0,n1)|add(n2,const_2)|multiply(n1,n2)|multiply(n0,#1)|subtract(n3,#0)|add(#3,#2)|divide(#4,#5)|multiply(#6,const_100) | general |
a player holds 13 cards of 4 suits , of which 7 are black and 6 are red . there are twice as many diamonds as spades and twice as many hearts as diamonds . how many clubs does he hold | explanation : clearly , the black cards are either clubs or spades while the red cards are either diamonds or hearts . let the number of spades be x . then , number of clubs = ( 7 - x ) . number of diamonds = 2 x number of spades = 2 x ; number of hearts = 2 x number of diamonds = 4 x . total number of cards = x + 2 x ... | a ) 4 , b ) 5 , c ) 6 , d ) 7 , e ) 8 | c | subtract(7, divide(subtract(13, 7), add(4, const_2))) | add(n1,const_2)|subtract(n0,n2)|divide(#1,#0)|subtract(n2,#2) | probability |
a worker can load one truck in 6 hours . a second worker can load the same truck in 8 hours . if both workers load one truck simultaneously while maintaining their constant rates , approximately how long , in hours , will it take them to fill one truck ? | "the workers fill the truck at a rate of 1 / 6 + 1 / 8 = 14 / 48 = 7 / 24 of the truck per hour . then the time to fill one truck is 24 / 7 which is about 3.4 hours . the answer is e ." | a ) 2.6 , b ) 2.8 , c ) 3.0 , d ) 3.2 , e ) 3.4 | e | inverse(add(divide(const_1, 6), divide(const_1, 8))) | divide(const_1,n0)|divide(const_1,n1)|add(#0,#1)|inverse(#2)| | physics |
here is a simple arithmetical puzzle in his own flowery , poetical language . if one - fifth of a hive of bees flew to the badamba flower , one - third flew to the slandbara , 3 times the difference of these two numbers flew to an arbour , and one bee continued to fly about , attracted on each side by the fragrant keta... | b 15 the number of bees was 15 | a ) 11 , b ) 15 , c ) 18 , d ) 17 , e ) 19 | b | divide(const_1, subtract(const_1, add(add(divide(const_1, add(const_1, const_4)), divide(const_1, const_3)), multiply(subtract(divide(const_1, const_3), divide(const_1, add(const_1, const_4))), 3)))) | add(const_1,const_4)|divide(const_1,const_3)|divide(const_1,#0)|add(#2,#1)|subtract(#1,#2)|multiply(n0,#4)|add(#3,#5)|subtract(const_1,#6)|divide(const_1,#7) | general |
the largest 5 digit number exactly divisible by 91 is ? | "largest 5 - digit number = 99999 91 ) 99999 ( 1098 91 - - - 899 819 - - - - 809 728 - - - 81 - - - required number = ( 99999 - 81 ) = 99918 . option a" | a ) 99918 , b ) 99919 , c ) 99921 , d ) 99923 , e ) 99929 | a | square_area(const_pi) | square_area(const_pi)| | general |
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