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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if 28 a + 30 b + 31 c = 545 . then a + b + c = ? . a , b , c are natural numbers | have a look on your calender since , we all knw a year consist of 365 days february is d only month which has 28 days 4 months in a year has 30 days and , rest 7 months has 31 days . . so , following d given eq . we can write 28 * 1 + 30 * 10 + 31 * 7 . . hence values of a , b and c are 1 , 10 and 7 respectively . . a ... | a ) 12 , b ) 14 , c ) 8 , d ) 25 , e ) 18 | e | divide(545, 31) | divide(n3,n2) | general |
how many different ways can 2 students be seated in a row of 3 desks , so that there is always at least one empty desk between the students ? | "1 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 1 x 2 = 2 a" | a ) 2 , b ) 3 , c ) 4 , d ) 6 , e ) 12 | a | permutation(subtract(3, const_1), 2) | subtract(n1,const_1)|permutation(#0,n0)| | probability |
p and q started a business investing rs 75000 and rs 15000 resp . in what ratio the profit earned after 2 years be divided between p and q respectively . | "explanation : in this type of question as time frame for both investors is equal then just get the ratio of their investments . p : q = 75000 : 15000 = 75 : 15 = 5 : 1 option a" | a ) 5 : 1 , b ) 17 : 3 , c ) 5 : 6 , d ) 17 : 7 , e ) 3 : 8 | a | divide(75000, 15000) | divide(n0,n1)| | gain |
evaluate : 55 - 12 * 3 * 2 = ? | "according to order of operations , 12 ? 3 ? 2 ( division and multiplication ) is done first from left to right 12 * * 2 = 4 * 2 = 8 hence 55 - 12 * 3 * 2 = 55 - 8 = 47 correct answer a" | a ) 47 , b ) 52 , c ) 32 , d ) 12 , e ) 22 | a | subtract(55, multiply(multiply(12, const_2.0), 2)) | multiply(n1,const_2.0)|multiply(n3,#0)|subtract(n0,#1)| | general |
the area of sector of a circle whose radius is 12 metro and whose angle at the center is 42 ° is ? | "42 / 360 * 22 / 7 * 12 * 12 = 52.8 m 2 answer : c" | a ) 56.8 m 2 , b ) 82.9 m 2 , c ) 52.8 m 2 , d ) 17.9 m 2 , e ) 72.9 m 2 | c | multiply(multiply(power(12, const_2), divide(add(multiply(const_2, const_10), const_2), add(const_4, const_3))), divide(42, divide(const_3600, const_10))) | add(const_3,const_4)|divide(const_3600,const_10)|multiply(const_10,const_2)|power(n0,const_2)|add(#2,const_2)|divide(n1,#1)|divide(#4,#0)|multiply(#6,#3)|multiply(#5,#7)| | geometry |
if 2 x = 4 y = z , what is x - y , in terms of z ? | "2 x = 4 y = z i . e . x = z / 2 and y = z / 4 x - y = z / 2 - z / 4 = z / 4 answer : option a" | a ) z / 4 , b ) z / 2 , c ) z , d ) 2 z , e ) 4 z | a | divide(subtract(divide(multiply(2, const_100), const_2), const_2), add(divide(multiply(2, const_100), const_2), const_2)) | multiply(n0,const_100)|divide(#0,const_2)|add(#1,const_2)|subtract(#1,const_2)|divide(#3,#2)| | general |
the average weight of a group of boys is 35 kg . after a boy of weight 40 kg joins the group , the average weight of the group goes up by 1 kg . find the number of boys in the group originally ? | let the number off boys in the group originally be x . total weight of the boys = 35 x after the boy weighing 40 kg joins the group , total weight of boys = 35 x + 40 so 35 x + 40 = 36 ( x + 1 ) = > x = 5 . answer : b | a ) 4 , b ) 5 , c ) 6 , d ) 7 , e ) 8 | b | add(subtract(40, add(35, 1)), const_1) | add(n0,n2)|subtract(n1,#0)|add(#1,const_1) | general |
how many 13 - digits number are palindromic numbers ? a palindromic number reads the same forward and backward , example 1234567654321 . | "take the task of building palindromes and break it intostages . stage 1 : select the 13 th digit we can choose 1 , 2 , 3 , 4 , 5 , 6 , 7 , 8 , or 9 so , we can complete stage 1 in 9 ways stage 2 : select the 12 th 11 th 10 th , 9 th , 8 th , 7 th , we can choose 0 , 1 , 2 , 3 , 4 , 5 , 6 , 7 , 8 , or 9 so , we can com... | a ) 100 , b ) 610 , c ) 729 , d ) 900 , e ) 9000000 | e | multiply(multiply(multiply(multiply(13, const_10), const_10), const_10), const_10) | multiply(n0,const_10)|multiply(#0,const_10)|multiply(#1,const_10)|multiply(#2,const_10)| | general |
5216 × 51 = ? | "normal way of multiplication may take time . here is one alternative . 5216 × 51 = ( 5216 × 50 ) + 5216 = ( 5216 × 1002 ) + 5216 = 5216002 + 5216 = 260800 + 5216 = 266016 answer is c ." | a ) 266015 , b ) 266014 , c ) 266016 , d ) 266116 , e ) 266226 | c | multiply(divide(5216, 51), const_100) | divide(n0,n1)|multiply(#0,const_100)| | general |
points x , y , and z lie , in that order , on a straight railroad track . the distance from point x to point y is twice the distance from point y to point z . a train traveled from point x to point z without stopping . the train ' s average speed when traveling from point x to point y was 100 miles per hour and the tra... | "average speed = distance / time because we are looking for average speed we can pick a distance for the variable d . speed x - y = 100 speed y - z = 75 average speed = total distance / total rate rate = distance / time x = = = = = = = = = = = = = = = = = = = = y = = = = = = = = = = z if x - y is twice the length of y ... | a ) 90 , b ) 85 , c ) 85.5 , d ) 70 , e ) 90 | a | multiply(divide(const_3, add(divide(const_2, divide(100, 75)), const_1)), 75) | divide(n0,n1)|divide(const_2,#0)|add(#1,const_1)|divide(const_3,#2)|multiply(n1,#3)| | physics |
if there are 200 questions in a 3 hr examination . among these questions are 20 type a problems , which requires twice as much as time be spent than the rest of the type b problems . how many minutes should be spent on type a problems ? | "x = time for type b prolems 2 x = time for type a problem total time = 3 hrs = 180 min 180 x + 20 * 2 x = 180 x = 180 / 220 x = 0.818182 time taken for type a problem = 20 * 2 * 0.818182 = 32.72727 min answer : d" | a ) 72 min , b ) 62 min , c ) 70 min , d ) 32.72727 min , e ) 76 min | d | multiply(multiply(const_2, divide(multiply(3, const_60), add(subtract(200, 20), multiply(const_2, 20)))), 20) | multiply(n1,const_60)|multiply(n2,const_2)|subtract(n0,n2)|add(#1,#2)|divide(#0,#3)|multiply(#4,const_2)|multiply(n2,#5)| | general |
an integer n between 1 and 100 , inclusive , is to be chosen at random . what is the probability that n ( n + 2 ) will be divisible by 2 ? | "n ( n + 2 ) to be divisible by 2 either n or n + 2 must be a multiples of 2 . in each following group of numbers : { 1 , 2 , 3 , 4 } , { 5 , 6 , 7 , 8 } , . . . , { 97 , 98 , 99 , 100 } there are exactly 2 numbers out of 4 satisfying the above condition . for example in { 1 , 2 , 3 , 4 } n can be : 2 , or 4 . thus , t... | a ) 1 / 9 , b ) 1 / 3 , c ) 1 / 2 , d ) 2 / 3 , e ) 5 / 6 | c | multiply(divide(divide(100, 2), 100), 2) | divide(n1,n3)|divide(#0,n1)|multiply(n2,#1)| | general |
a certain characteristic in a large population has a distribution that is symmetric about the mean m . if 36 percent of the distribution lies within one standard deviation d of the mean , what percent of the distribution is less than m + d ? | "this is easiest to solve with a bell - curve histogram . m here is equal to µ in the gaussian normal distribution and thus m = 50 % of the total population . so , if 36 % is one st . dev , then on either side of m we have 36 / 2 = 18 % . so , 18 % are to the right and left of m ( = 50 % ) . in other words , our value ... | a ) 16 % , b ) 32 % , c ) 68 % , d ) 84 % , e ) 92 % | c | subtract(const_100, divide(subtract(const_100, 36), const_2)) | subtract(const_100,n0)|divide(#0,const_2)|subtract(const_100,#1)| | general |
every student in a room is either a junior or a senior . there is at least one junior and at least one senior in the room . if 4 / 5 of the juniors is equal to 1 / 2 of the seniors , what fraction of the students in the room are juniors ? | "let total number of juniors = j total number of seniors = s ( 4 / 5 ) j = ( 1 / 2 ) s = > s = 8 / 5 j total number of students = j + s = ( 13 / 5 ) j fraction of the students in the room are juniors = j / ( j + s ) = j / [ ( 13 / 5 ) j ] = 5 / 13 answer c" | a ) 13 / 20 , b ) 11 / 3 , c ) 5 / 13 , d ) 12 / 17 , e ) 17 / 20 | c | divide(multiply(5, const_2.0), add(2, multiply(5, 5))) | multiply(n2,n2)|add(n3,#0)|divide(#0,#1)| | general |
a certain car dealership sells economy cars , luxury cars , and sport utility vehicles . the ratio of economy to luxury cars is 3 : 2 . the ratio of economy cars to sport utility vehicles is 4 : 1 . what is the ratio of luxury cars to sport utility vehicles ? | "the ratio of economy to luxury cars is 3 : 2 - - > e : l = 3 : 2 = 12 : 8 . the ratio of economy cars to sport utility vehicles is 4 : 1 - - > e : s = 4 : 1 = 12 : 3 . thus , l : s = 8 : 3 . answer : b ." | a ) 9 : 8 , b ) 8 : 3 , c ) 3 : 2 , d ) 2 : 3 , e ) 1 : 2 | b | divide(divide(multiply(const_4, 1), multiply(1, 1)), divide(multiply(1, const_4), multiply(2, const_4))) | multiply(n3,const_4)|multiply(n3,n3)|multiply(n1,const_4)|divide(#0,#1)|divide(#0,#2)|divide(#3,#4)| | other |
if the simple interest on a sum of money for 2 years at 5 % per annum is rs . 40 , what is the compound interest on the same sum at the rate and for the same time ? | "explanation : sum = ( 40 * 100 ) / ( 2 * 5 ) = rs . 400 amount = [ 400 * ( 1 + 5 / 100 ) 2 ] = rs . 441 c . i . = ( 441 - 400 ) = rs . 41 answer : a" | a ) s . 41.00 , b ) s . 51.22 , c ) s . 51.219 , d ) s . 41.18 , e ) s . 51.11 | a | subtract(add(add(divide(multiply(divide(40, multiply(divide(5, const_100), 2)), 5), const_100), divide(40, multiply(divide(5, const_100), 2))), divide(multiply(add(divide(multiply(divide(40, multiply(divide(5, const_100), 2)), 5), const_100), divide(40, multiply(divide(5, const_100), 2))), 5), const_100)), divide(40, m... | divide(n1,const_100)|multiply(n0,#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 |
income and expenditure of a person are in the ratio 5 : 4 . if the income of the person is rs . 18000 , 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 = 18000 = > x = 3600 savings = income - expenditure = 5 x - 4 x = x so , savings = rs . 3600 . answer : a" | a ) rs . 3600 , b ) rs . 3629 , c ) rs . 3279 , d ) rs . 3629 , e ) rs . 3283 | a | subtract(18000, multiply(divide(4, 5), 18000)) | divide(n1,n0)|multiply(n2,#0)|subtract(n2,#1)| | other |
in a certain diving competition , 5 judges score each dive on a scale from 1 to 10 . the point value of the dive is obtained by dropping the highest score and the lowest score and multiplying the sum of the remaining scores by the degree of difficulty . if a dive with a degree of difficulty of 3.2 received scores of 7.... | "degree of difficulty of dive = 3.2 scores are 6.0 , 7.5 , 8.8 , 8.5 and 9.0 we can drop 6.0 and 9.0 sum of the remaining scores = ( 7.5 + 8.8 + 8.5 ) = 24.8 point of value of the dive = 24 * 3.2 = 79.36 answer e" | a ) 68.8 , b ) 73.6 , c ) 75.2 , d ) 76.8 , e ) 79.36 | e | multiply(add(add(7.5, 8.8), 8.5), 3.2) | add(n4,n5)|add(n8,#0)|multiply(n3,#1)| | general |
at the faculty of aerospace engineering , 310 students study random - processing methods , 232 students study scramjet rocket engines and 112 students study them both . if every student in the faculty has to study one of the two subjects , how many students are there in the faculty of aerospace engineering ? | "310 + 232 - 112 ( since 112 is counted twice ) = 430 a is the answer" | a ) 430 , b ) 432 , c ) 440 , d ) 442 , e ) 446 | a | add(subtract(310, divide(112, const_2)), subtract(232, divide(112, const_2))) | divide(n2,const_2)|subtract(n0,#0)|subtract(n1,#0)|add(#1,#2)| | other |
gold is 19 times as heavy as water and copper is 9 times as heavy as water . in what ratio should these be mixed to get an alloy 12 times as heavy as water ? | "g = 19 w c = 9 w let 1 gm of gold mixed with x gm of copper to get 1 + x gm of the alloy 1 gm gold + x gm copper = x + 1 gm of alloy 19 w + 9 wx = x + 1 * 12 w 19 + 9 x = 12 ( x + 1 ) x = 7 / 3 ratio of gold with copper = 1 : 7 / 3 = 3 : 7 answer is b" | a ) 1 : 2 , b ) 3 : 7 , c ) 4 : 1 , d ) 5 : 2 , e ) 6 : 5 | b | divide(subtract(12, 9), subtract(19, 12)) | subtract(n2,n1)|subtract(n0,n2)|divide(#0,#1)| | general |
in 2008 , the profits of company n were 10 percent of revenues . in 2009 , the revenues of company n fell by 20 percent , but profits were 14 percent of revenues . the profits in 2009 were what percent of the profits in 2008 ? | "x = profits r = revenue x / r = 0,1 x = 10 r = 100 2009 : r = 80 x / 80 = 0,14 = 14 / 100 x = 80 * 14 / 100 x = 11.2 11.2 / 10 = 1,12 = 112 % , answer b" | a ) 80 % , b ) 112 % , c ) 120 % , d ) 124.2 % , e ) 138 % | b | multiply(divide(multiply(14, subtract(const_1, divide(20, const_100))), 10), const_100) | divide(n3,const_100)|subtract(const_1,#0)|multiply(n4,#1)|divide(#2,n1)|multiply(#3,const_100)| | gain |
what is the 100 th digit of ( 3 ! * 5 ! + 4 ! * 3 ! ) / 2 ? | ( 3 ! * 5 ! + 4 ! * 3 ! ) / 2 = 3 ! ( 5 ! + 4 ! ) / 2 = 6 ( 120 + 24 ) / 2 = 432 100 th digit of the above product will be equal to 4 answer a | a ) 4 , b ) 3 , c ) 2 , d ) 7 , e ) 6 | a | floor(divide(divide(add(multiply(factorial(3), factorial(5)), multiply(factorial(4), factorial(3))), 2), 100)) | factorial(n1)|factorial(n2)|factorial(n3)|multiply(#0,#1)|multiply(#2,#0)|add(#3,#4)|divide(#5,n5)|divide(#6,n0)|floor(#7) | general |
the sides of a rectangular field are in the ratio 3 : 4 . if the area of the field is 9408 sq . m , the cost of fencing the field @ 25 paise per metre is | solution let length = ( 3 x ) metres and breadth = ( 4 x ) metres . then , 3 x × 4 x = 9408 ⇔ 12 x 2 = 9408 ⇔ x 2 = 784 ⇔ x = 28 . so , length = 84 m and breadth = 112 m . perimeter = [ 2 ( 84 + 112 ) ] m = 392 m . ∴ cost of fencing = rs . ( 0.25 × 392 ) = rs . 98.00 . answer d | ['a ) rs . 55.50', 'b ) rs . 67.50', 'c ) rs . 86.50', 'd ) rs . 98.00', 'e ) none of these'] | d | divide(multiply(rectangle_perimeter(multiply(3, sqrt(divide(9408, multiply(3, 4)))), multiply(4, sqrt(divide(9408, multiply(3, 4))))), 25), const_100) | multiply(n0,n1)|divide(n2,#0)|sqrt(#1)|multiply(n0,#2)|multiply(n1,#2)|rectangle_perimeter(#3,#4)|multiply(n3,#5)|divide(#6,const_100) | physics |
an error 2 % in excess is made while measuring the side of a square . what is the percentage of error in the calculated area of the square ? | "formula for calculation of percentage error is : = ( x − y − ( xy ) / 100 ) % = ( 2 + 2 + ( 2 × 2 ) / 100 ) % = 4.04 % answer is a ." | a ) 4.04 , b ) 4.02 , c ) 4.01 , d ) 4.05 , e ) 4.06 | a | divide(multiply(subtract(square_area(add(const_100, 2)), square_area(const_100)), const_100), square_area(const_100)) | add(n0,const_100)|square_area(const_100)|square_area(#0)|subtract(#2,#1)|multiply(#3,const_100)|divide(#4,#1)| | gain |
at a certain university , 70 % of the professors are women , and 70 % of the professors are tenured . if 90 % of the professors are women , tenured , or both , then what percent of the men are tenured ? | "answer is 75 % total women = 70 % total men = 40 % total tenured = 70 % ( both men and women ) therefore , women tenured + women professors + men tenured = 90 % men tenured = 20 % but question wants to know the percent of men that are tenured 20 % / 40 % = 50 % c" | a ) 25 , b ) 37.5 , c ) 50 , d ) 62.5 , e ) 75 | c | add(subtract(const_100, 70), subtract(90, 70)) | subtract(const_100,n0)|subtract(n2,n0)|add(#0,#1)| | gain |
if p and t are positive integers such that p > t > 1 , what is the remainder when 92 p × 5 p + t + 11 t × 6 pt is divided by 10 ? | 92 p × 5 p + t + 11 t × 6 pt - - 1 = b | a ) a . 0 , b ) b . 1 , c ) c . 3 , d ) d . 4 , e ) e . 5 | b | subtract(6, 5) | subtract(n4,n2) | general |
mary passed a certain gas station on a highway while traveling west at a constant speed of 40 miles per hour . then , 15 minutes later , paul passed the same gas station while traveling west at a constant speed of 50 miles per hour . if both drivers maintained their speeds and both remained on the highway for at least ... | "d = rt m : r = 40 mph , t = t + 1 / 4 hr d = 40 ( t + 1 / 4 ) p : r = 50 , t = t d = 50 t since they went the same distance : 40 t + 40 / 4 = 50 t 10 t = 10 t = 1 hr , b" | a ) 1 / 2 , b ) 1 , c ) 3 / 4 , d ) 2 , e ) 3 | b | multiply(const_60, divide(multiply(40, divide(15, const_60)), subtract(50, 40))) | divide(n1,const_60)|subtract(n2,n0)|multiply(n0,#0)|divide(#2,#1)|multiply(#3,const_60)| | physics |
a person buys an article at rs . 350 . at what price should he sell the article so as to make a profit of 30 % ? | "cost price = rs . 350 profit = 30 % of 350 = rs . 105 selling price = cost price + profit = 350 + 105 = 455 answer : a" | a ) 455 , b ) 277 , c ) 269 , d ) 261 , e ) 281 | a | add(350, multiply(350, divide(30, const_100))) | divide(n1,const_100)|multiply(n0,#0)|add(n0,#1)| | gain |
two trains 140 m and 200 m long run at the speed of 60 km / hr and 40 km / hr respectively in opposite directions on parallel tracks . the time which they take to cross each other is ? | "relative speed = 60 + 40 = 100 km / hr . = 100 * 5 / 18 = 250 / 9 m / sec . distance covered in crossing each other = 140 + 200 = 340 m . required time = 340 * 9 / 250 = 12.24 sec . answer : e" | a ) 10.9 sec , b ) 10.1 sec , c ) 10.6 sec , d ) 10.8 sec , e ) 12.24 sec | e | divide(add(140, 200), multiply(add(60, 40), const_0_2778)) | add(n0,n1)|add(n2,n3)|multiply(#1,const_0_2778)|divide(#0,#2)| | physics |
how many cubes of 5 cm edge can be put in a cubical box of 1 m edge . | "number of cubes = 100 â ˆ — 100 â ˆ — 100 / 5 * 5 * 5 = 8000 note : 1 m = 100 cm answer : b" | a ) 1770 cm , b ) 8000 cm , c ) 7860 cm , d ) 6170 cm , e ) 1870 cm | b | divide(volume_cube(1), volume_cube(divide(5, const_100))) | divide(n0,const_100)|volume_cube(n1)|volume_cube(#0)|divide(#1,#2)| | physics |
the average of first seven multiples of 4 is : | "explanation : ( 4 ( 1 + 2 + 3 + 4 + 5 + 6 + 7 ) / 7 = 4 x 28 / 7 = 16 answer : b" | a ) 9 , b ) 16 , c ) 15 , d ) 8 , e ) 10 | b | add(4, const_1) | add(n0,const_1)| | general |
there are 5 chess amateurs playing in villa ' s chess club tournament . if each chess amateur plays with exactly 4 other amateurs , what is the total number of chess games possible to be played in the tournament ? | "method 1 : take the first amateur . he plays a game with each of the other four i . e . 4 games . now take the second one . he has already played a game with the first one . he plays 3 games with the rest of the 3 amateurs i . e . 3 more games are played . now take the third amateur . he has already played a game each... | a ) 10 , b ) 20 , c ) 40 , d ) 60 , e ) 120 | a | divide(multiply(5, 4), const_2) | multiply(n0,n1)|divide(#0,const_2)| | general |
in a urban village of india named ` ` owlna ' ' , 75 % people have refrigerator , 90 % people have television , 85 % people got computers and 75 % got air - conditionor . how many people ( minimum ) got all these luxury . | "e 10 % 100 - [ ( 100 - 75 ) + ( 100 - 90 ) + ( 100 - 85 ) + ( 100 - 75 ) ] = 100 - ( 25 + 10 + 15 + 25 ) = 100 - 75" | a ) 23 % , b ) 17 % , c ) 11 % , d ) 10 % , e ) 25 % | e | subtract(const_100, add(add(add(subtract(const_100, 75), subtract(const_100, 90)), subtract(const_100, 85)), subtract(const_100, 75))) | subtract(const_100,n0)|subtract(const_100,n1)|subtract(const_100,n2)|subtract(const_100,n3)|add(#0,#1)|add(#4,#2)|add(#5,#3)|subtract(const_100,#6)| | general |
a bowl of nuts is prepared for a party . brand p mixed nuts are 20 % almonds and brand q ' s deluxe nuts are 25 % almonds . if a bowl contains a total of 67 ounces of nuts , representing a mixture of both brands , and 15 ounces of the mixture are almonds , how many ounces of brand q ' s deluxe mixed nuts are used ? | "lets say x ounces of p is mixed with q . = > 67 - x ounces of q is present in the mixture ( as the total = 67 ounces ) given total almond weight = 15 ounces ( 20 x / 100 ) + ( 25 / 100 ) ( 67 - x ) = 15 = > x = 35 = > 67 - 35 = 32 ounces of q is present in the mixture . answer is c ." | a ) 16 , b ) 20 , c ) 32 , d ) 44 , e ) 48 | c | divide(subtract(15, multiply(divide(20, const_100), 67)), subtract(divide(25, const_100), divide(20, const_100))) | divide(n0,const_100)|divide(n1,const_100)|multiply(n2,#0)|subtract(#1,#0)|subtract(n3,#2)|divide(#4,#3)| | general |
if $ 5,000 is invested in an account that earns 8 % interest compounded semi - annually , then the interest earned after one year would be how much greater than if the $ 5,000 had been invested at 8 % simple yearly interest ? | "solution amount ( ci ) = p + ( 1 + r / n ) ^ nt = 5000 + ( 1 + 0.08 / 2 ) ^ 2 = 5408 amount ( si ) = p + ptr / 100 = 5000 + ( 5000 * 1 * 8 / 100 ) = 5400 difference = 5408 - 5400 = 8 $ answer : b" | a ) $ 4 , b ) $ 8 , c ) $ 12 , d ) $ 16 , e ) $ 432 | b | multiply(8, const_1) | multiply(n1,const_1)| | gain |
the length of a rectangular garden is three times its width . if the area of the rectangular garden is 675 square meters , then what is the width of the rectangular garden ? | "let x be the width of the garden . 3 x ^ 2 = 675 x ^ 2 = 225 x = 15 the answer is b ." | a ) 14 , b ) 15 , c ) 16 , d ) 17 , e ) 18 | b | sqrt(divide(675, const_3)) | divide(n0,const_3)|sqrt(#0)| | geometry |
how many odd 3 - digit integers smaller than 500 are there such that all their digits are different ? | case 1 : numbers starting with 1 1 _ _ the unit digit can take 4 numbers ( 3,5 , 7,9 ) to be odd so 4 ways . the tens digit can take remaining 8 numbers left after using 2 numbers so 8 ways . total = 8 * 4 = 32 case 2 : numbers starting with 2 2 _ _ the unit digit can take 5 numbers ( 1 , 3,5 , 7,9 ) to be odd so 5 way... | a ) 72 , b ) 88 , c ) 112 , d ) 122 , e ) 144 | e | divide(multiply(divide(factorial(subtract(const_10, const_1)), factorial(subtract(subtract(const_10, const_1), subtract(3, const_1)))), divide(subtract(500, const_100), const_100)), const_2) | subtract(const_10,const_1)|subtract(n0,const_1)|subtract(n1,const_100)|divide(#2,const_100)|factorial(#0)|subtract(#0,#1)|factorial(#5)|divide(#4,#6)|multiply(#7,#3)|divide(#8,const_2) | general |
a van takes 5 hours to cover a distance of 435 km . what speed in kph should the van maintain to cover the same distance in 3 / 2 of the previous time ? | "( 3 / 2 ) * 5 = 7.5 hours 435 / 7.5 = 58 kph the answer is e ." | a ) 50 , b ) 52 , c ) 54 , d ) 56 , e ) 58 | e | divide(435, multiply(divide(3, 2), 5)) | divide(n2,n3)|multiply(n0,#0)|divide(n1,#1)| | physics |
if the price of a tv is first decreased by 20 % and then increased by 50 % , then the net change in the price will be : | "explanation : solution : let the original price be rs . 100 . new final price = 150 % of ( 80 % of 100 ) = rs . 150 / 100 * 80 / 100 * 100 = rs . 120 . . ' . increase = 20 % answer : a" | a ) 20 % increase , b ) 10 % increase , c ) 10 % decrease , d ) 6 % increase , e ) none of these | a | subtract(divide(multiply(subtract(const_100, 20), add(const_100, 50)), const_100), const_100) | add(n1,const_100)|subtract(const_100,n0)|multiply(#0,#1)|divide(#2,const_100)|subtract(#3,const_100)| | general |
18 % of 680 - ? = 68.7 | "e 53.7 ( 18 * 680 ) / 100 - ? = 68.7 ? = 122.4 - 68.7 ? = 53.7" | a ) 44.1 , b ) 32.4 , c ) 66.5 , d ) 75.2 , e ) 53.7 | e | divide(multiply(divide(multiply(68.7, 680), const_100), 18), const_100) | multiply(n1,n2)|divide(#0,const_100)|multiply(n0,#1)|divide(#2,const_100)| | general |
a sum of money is put out at compound interest for 2 years at 20 % . it would fetch rs . 241 more if the interest were payable half - yearly , then it were pay able yearly . find the sum . | "p ( 11 / 10 ) ^ 4 - p ( 6 / 5 ) ^ 2 = 241 p = 10000 answer : d" | a ) 12200 , b ) 11897 , c ) 18799 , d ) 10000 , e ) 12782 | d | add(multiply(add(20, const_1), add(20, const_1)), add(20, const_1)) | add(n1,const_1)|multiply(#0,#0)|add(#0,#1)| | general |
a , b and c can do a piece of work in 24 , 30 and 40 days respectively . they start the work together but c leaves 4 days before the completion of the work . in how many days is the work done ? | x / 24 + x / 30 + x / 40 = 1 x = 11 days answer : d | a ) 11 , b ) 5 , c ) 70 , d ) 11 , e ) 13 | d | add(subtract(subtract(24, subtract(40, 30)), 4), const_1) | subtract(n2,n1)|subtract(n0,#0)|subtract(#1,n3)|add(#2,const_1) | physics |
of the 120 passengers on flight 750 , 45 % are female . 10 % of the passengers sit in first class , and the rest of the passengers sit in coach class . if 1 / 3 of the passengers in first class are male , how many females are there in coach class ? | "number of passengers on flight = 120 number of female passengers = . 45 * 120 = 54 number of passengers in first class = ( 10 / 100 ) * 120 = 12 number of passengers in coach class = ( 90 / 100 ) * 120 = 108 number of male passengers in first class = 1 / 3 * 12 = 4 number of female passengers in first class = 12 - 4 =... | a ) 44 , b ) 48 , c ) 46 , d ) 52 , e ) 56 | c | subtract(multiply(120, divide(45, const_100)), subtract(multiply(120, divide(10, const_100)), divide(multiply(120, divide(10, const_100)), 3))) | divide(n2,const_100)|divide(n3,const_100)|multiply(n0,#0)|multiply(n0,#1)|divide(#3,n5)|subtract(#3,#4)|subtract(#2,#5)| | gain |
what is the smallest positive integer x , such that 6000 x is a perfect cube ? | "take out the factors of 6000 that will come 2 * 3 * 10 ^ 3 . for perfect cube you need every no . raise to the power 3 . for 6000 x to be a perfect cube , need two 2 and 2 3 that means 36 or 6 ^ 2 c is the answer ." | a ) 4 , b ) 6 , c ) 36 , d ) 12 , e ) 18 | c | add(const_3, const_4) | add(const_3,const_4)| | geometry |
if x is equal to the sum of the integers from 10 to 20 , inclusive , and y is the number of even integers from 10 to 20 , inclusive , what is the value of x + y ? | sum s = n / 2 { 2 a + ( n - 1 ) d } = 11 / 2 { 2 * 10 + ( 11 - 1 ) * 1 } = 11 * 15 = 165 = x number of even number = ( 20 - 10 ) / 2 + 1 = 6 = y x + y = 165 + 6 = 171 a | a ) 171 , b ) 175 , c ) 180 , d ) 190 , e ) 200 | a | add(multiply(divide(add(10, 20), const_2), add(subtract(20, 10), const_1)), add(divide(subtract(20, 10), const_2), const_1)) | add(n0,n1)|subtract(n1,n0)|add(#1,const_1)|divide(#1,const_2)|divide(#0,const_2)|add(#3,const_1)|multiply(#2,#4)|add(#5,#6) | general |
49 * 49 * 49 * 49 = 7 ? | "sol . 49 * 49 * 49 * 49 = ( 72 * 72 * 72 * 72 ) = 7 ( 2 + 2 + 2 + 2 ) = 78 . so , the correct answer is 8 . answer d" | a ) 4 , b ) 5 , c ) 7 , d ) 9 , e ) 8 | d | multiply(49, 49) | multiply(n0,n1)| | general |
find the least number which when divided by 33 and 8 leaves a remainder of 2 in each case . | "the least number which when divided by different divisors leaving the same remainder in each case = lcm ( different divisors ) + remainder left in each case . hence the required least number = lcm ( 33 , 8 ) + 2 = 266 . answer : a" | a ) 266 , b ) 268 , c ) 270 , d ) 272 , e ) 274 | a | add(2, lcm(33, 8)) | lcm(n0,n1)|add(n2,#0)| | general |
50 % of a number is added to 120 , the result is the same number . find the number ? | "( 50 / 100 ) * x + 120 = x x = 240 answer : a" | a ) 240 , b ) 288 , c ) 1687 , d ) 997 , e ) 121 | a | divide(120, divide(120, const_100)) | divide(n1,const_100)|divide(n1,#0)| | gain |
there are 4 runners on a track team who run an average of 4.5 seconds per 40 yards . if another runner joins their team who runs 4.3 seconds per 40 yards , what will the new average 40 yard time be ? | ( sum of the 4 times ) / 4 = 4.5 sum of the 4 times = 18 new sum = 18 + 4.3 = 22.3 new average = 22.3 / 5 = 4.46 ans : a | a ) 4.46 , b ) 4.39 , c ) 4.42 , d ) 4.49 , e ) 4.33 | a | divide(add(4.5, 4.3), const_2) | add(n1,n3)|divide(#0,const_2) | general |
a train is moving at 4 / 5 of its usual speed . the train is 30 minutes too late . what is the usual time ( in hours ) for the train to complete the journey ? | "new time = d / ( 4 v / 5 ) = 5 / 4 * usual time 30 minutes represents 1 / 4 of the usual time . the usual time is 2 hours . the answer is c ." | a ) 1 , b ) 1.5 , c ) 2 , d ) 2.5 , e ) 3 | c | divide(multiply(multiply(30, divide(4, 5)), inverse(subtract(const_1, divide(4, 5)))), const_60) | divide(n0,n1)|multiply(n2,#0)|subtract(const_1,#0)|inverse(#2)|multiply(#3,#1)|divide(#4,const_60)| | physics |
the speed of a car increases by 2 kms after every one hour . if the distance travelling in the first one hour was 50 kms . what was the total distance traveled in 12 hours ? | "explanation : total distance travelled in 12 hours = ( 50 + 52 + 54 + . . . . . upto 12 terms ) this is an a . p with first term , a = 50 , number of terms , n = 12 , d = 2 . required distance = 12 / 2 [ 2 x 50 + { 12 - 1 ) x 2 ] = 6 ( 122 ) = 732 kms . answer : c" | a ) 252 kms , b ) 152 kms , c ) 732 kms , d ) 752 kms , e ) 152 kms | c | multiply(add(multiply(2, 50), multiply(subtract(12, const_1), 2)), divide(12, 2)) | divide(n2,n0)|multiply(n0,n1)|subtract(n2,const_1)|multiply(n0,#2)|add(#1,#3)|multiply(#4,#0)| | physics |
what is the least integer greater than – 3 + 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 : ( - 3 ) + ( 0.5 ) = - 2.5 on a number line , since we ' re adding + . 5 to a number , the total moves to the right ( so we ' re moving from -... | a ) – 2 , b ) – 1 , c ) 0 , d ) 1 , e ) 2 | a | add(0.5, negate(3)) | negate(n0)|add(n1,#0)| | general |
if greg buys 3 shirts , 4 trousers and 2 ties , the total cost is $ 90 . if greg buys 7 shirts , 2 trousers and 2 ties , the total cost is $ 50 . how much will it cost him to buy 3 trousers , 5 shirts and 2 ties ? | "solution : 3 x + 4 y + 2 z = 90 7 x + 2 y + 2 z = 50 adding both the equations = 10 x + 6 y + 4 z = 140 5 x + 3 y + 2 z = 70 ans b" | a ) $ 60 , b ) $ 70 , c ) $ 75 , d ) $ 96 , e ) can not be determined | b | divide(add(50, 90), 2) | add(n3,n7)|divide(#0,n2)| | general |
in a certain candy store , 22 % of the customers are caught sampling the candy and are charged a small fine , but 8 % of the customers who sample the candy are not caught . what is the total percent of all customers who sample candy ? | "since 8 % of the customers who sample the candyare notcaught , then 88 % of the customers who sample the candyarecaught : { % of customers who sample candy } * 0.92 = 0.22 ; { % of customers who sample candy } = 0.239 . answer : b ." | a ) 22 % , b ) 23.9 % , c ) 24 % , d ) 25 % , e ) 34 % | b | divide(22, divide(subtract(const_100, 8), const_100)) | subtract(const_100,n1)|divide(#0,const_100)|divide(n0,#1)| | gain |
| x + 3 | – | 4 - x | = | 8 + x | how many u solutions will this equation have ? | "| x | = x when x > = 0 ( x is either positive or 0 ) | x | = - x when x < 0 ( note here that you can put the equal to sign here as well x < = 0 because if x = 0 , | 0 | = 0 = - 0 ( all are the same ) so the ' = ' sign can be put with x > 0 or with x < 0 . we usually put it with ' x > 0 ' for consistency . a" | a ) 0 , b ) 1 , c ) 2 , d ) 3 , e ) 4 | a | divide(multiply(add(4, 3), const_2), 8) | add(n0,n1)|multiply(#0,const_2)|divide(#1,n2)| | general |
the area of an isosceles trapezoid with sides of length 5 and bases of length 6 and 12 is ? | "isosceles trapezoid indicates that if we draw perpendicular from two vertex of the smaller side , the longer side with side 12 will be divided into 3 parts = 6 , 3 and 3 ( making one square and two right triangles ) for the right triangle , height will be = ( 5 ^ 2 - 3 ^ 2 ) ^ 1 / 2 = 4 and the area of trapezoid = 1 /... | a ) 44 , b ) 42 , c ) 40 , d ) 38 , e ) 36 | e | quadrilateral_area(const_4, 12, 6) | quadrilateral_area(const_4,n2,n1)| | geometry |
a sum of money is sufficient to pay a ' s wages for 21 days and b ' s wages for 28 days . the same money is sufficient to pay the wages of both for ? | let total money be rs . x a ' s 1 day ' s wages = rs . x / 21 , b ' s 1 day ' s wages = rs . x / 28 ( a + b ) ' s 1 day ' s wages = rs . ( x / 21 + x / 28 ) = rs . x / 12 ; money is sufficient to pay the wages of both for 12 days . correct option : a | a ) 12 days , b ) 12 1 / 4 days , c ) 14 days , d ) 24 1 / 2 days , e ) none of these | a | divide(const_1, add(divide(const_1, 21), divide(const_1, 28))) | divide(const_1,n0)|divide(const_1,n1)|add(#0,#1)|divide(const_1,#2) | general |
in a certain school , 20 % of students are below 8 years of age . the number of students above 8 years of age is 2 / 3 of the number of students of 8 years of age which is 24 . what is the total number of students in the school ? | "explanation : let the number of students be x . then , number of students above 8 years of age = ( 100 - 20 ) % of x = 80 % of x . 80 % of x = 24 + 2 / 3 of 24 80 / 100 x = 40 x = 50 . answer : option a" | a ) 50 , b ) 80 , c ) 120 , d ) 150 , e ) 100 | a | divide(add(24, multiply(24, divide(2, 3))), subtract(const_1, divide(20, const_100))) | divide(n3,n4)|divide(n0,const_100)|multiply(n6,#0)|subtract(const_1,#1)|add(n6,#2)|divide(#4,#3)| | general |
the l . c . m . of two numbers is 48 . the numbers are in the ratio 2 : 3 . then sum of the number is : | "let the numbers be 2 x and 3 x . then , their l . c . m . = 6 x . so , 6 x = 48 or x = 8 . the numbers are 16 and 24 . hence , required sum = ( 16 + 24 ) = 40 . answer : option c" | a ) 28 , b ) 32 , c ) 40 , d ) 53 , e ) 54 | c | add(multiply(divide(48, multiply(2, 3)), 3), multiply(divide(48, multiply(2, 3)), 2)) | multiply(n1,n2)|divide(n0,#0)|multiply(n2,#1)|multiply(n1,#1)|add(#2,#3)| | other |
if x + ( 1 / x ) = 5 , what is the value of r = x ^ 2 + ( 1 / x ) ^ 2 ? | "squaring on both sides , x ^ 2 + ( 1 / x ) ^ 2 + 2 ( x ) ( 1 / x ) = 5 ^ 2 x ^ 2 + ( 1 / x ) ^ 2 = 23 answer : c" | a ) r = 21 , b ) r = 22 , c ) r = 23 , d ) 24 , e ) 27 | c | subtract(power(5, 2), 2) | power(n1,n2)|subtract(#0,n2)| | general |
how many multiples of 2 are there between 1 and 56 , exclusive ? | "27 multiples of 2 between 1 and 56 exclusive . from 2 * 1 upto 2 * 27 , ( 1,2 , 3,4 , . . . , 27 ) . hence , 27 multiples ! correct option is c" | a ) 21 , b ) 22 , c ) 27 , d ) 26 , e ) 28 | c | add(divide(subtract(56, 1), 2), const_1) | subtract(n2,n1)|divide(#0,n0)|add(#1,const_1)| | general |
the average of marks obtained by 120 boys was 39 . if the average of marks of passed boys was 39 and that of failed boys was 15 , the number of boys who passed the examination is ? | "let the number of boys who passed = x . then , 39 x x + 15 x ( 120 - x ) = 120 x 39 24 x = 4680 - 1800 = > x = 2880 / 24 x = 120 . hence , the number of boys passed = 120 . answer : c" | a ) 100 , b ) 110 , c ) 120 , d ) 130 , e ) 140 | c | divide(subtract(multiply(39, 120), multiply(120, 15)), subtract(39, 15)) | multiply(n0,n1)|multiply(n0,n3)|subtract(n2,n3)|subtract(#0,#1)|divide(#3,#2)| | general |
the area of a circular place is 17.56 hectares . find the cost of fencing it at the rate of rs . 3 / meter approximately . | area = ( 17.56 x 10000 ) m 2 = 175600 m 2 . π r 2 = 175600 ⇔ ( r ) 2 = ( 175600 x ( 7 / 22 ) ) ⇔ r = 236.37 m . circumference = 2 π r = ( 2 x ( 22 / 7 ) x 236.37 ) m = 1485.78 m . cost of fencing = rs . ( 1485.78 x 3 ) = rs . 4457 . c | ['a ) rs . 4233', 'b ) rs . 4350', 'c ) rs . 4457', 'd ) rs . 4670', 'e ) rs . 4756'] | c | multiply(circumface(sqrt(divide(multiply(multiply(const_1000, const_10), 17.56), const_pi))), const_3) | multiply(const_10,const_1000)|multiply(n0,#0)|divide(#1,const_pi)|sqrt(#2)|circumface(#3)|multiply(#4,const_3) | geometry |
from a pack of 52 cards , two cards are drawn at random together at random what is the probability of both the cards being kings ? | "let s be the sample space then n ( s ) = 52 c 2 = ( 52 * 51 ) / ( 2 * 1 ) = 1326 let e be the event of getting 2 kings out of 4 n ( e ) = 4 c 2 = ( 4 * 3 ) / ( 2 * 1 ) = 6 p ( e ) = n ( e ) / n ( s ) = 6 / 1326 = 1 / 221 answer ( a )" | a ) 1 / 221 , b ) 8 / 221 , c ) 4 / 589 , d ) 4 / 587 , e ) 7 / 654 | a | divide(multiply(divide(52, const_4), divide(52, const_4)), choose(52, const_2)) | choose(n0,const_2)|divide(n0,const_4)|multiply(#1,#1)|divide(#2,#0)| | probability |
the length of a rectangle is halved , while its breadth is tripled . wat isthe % change in area ? | "let original length = x and original breadth = y . original area = xy . new length = x . 2 new breadth = 3 y . new area = x x 3 y = 3 xy . 2 2 increase % = 1 xy x 1 x 100 % = 50 % . 2 xy c" | a ) 40 % , b ) 45 % , c ) 50 % , d ) 60 % , e ) 67 % | c | multiply(divide(subtract(multiply(const_3, divide(const_1, const_2)), const_1), const_1), const_100) | divide(const_1,const_2)|multiply(#0,const_3)|subtract(#1,const_1)|divide(#2,const_1)|multiply(#3,const_100)| | geometry |
3 , 7,12 , 18,25 . . . . . . . . . . . . . . 10 th terms | "3 + 4 = 7 7 + 5 = 12 12 + 6 = 18 18 + 7 = 25 25 + 8 = 33 33 + 9 = 42 42 + 10 = 52 52 + 11 = 63 63 + 11 = 74 answer : e" | a ) 43 , b ) 53 , c ) 63 , d ) 65 , e ) 74 | e | subtract(negate(10), multiply(subtract(7,12, 18,25), divide(subtract(7,12, 18,25), subtract(3, 7,12)))) | negate(n3)|subtract(n1,n2)|subtract(n0,n1)|divide(#1,#2)|multiply(#3,#1)|subtract(#0,#4)| | general |
10 men and 15 women together can complete a work in 5 days . it takes 100 days for one man alone to complete the same work . how many days will be required for one woman alone to complete the same work ? | "1 man ' s 1 day work = 1 / 100 ( 10 men + 15 women ) ' s 1 day work = 1 / 5 15 women ' s 1 day work = ( 1 / 5 - 10 / 100 ) = 1 / 10 1 woman ' s 1 day work = 1 / 150 1 woman alone can complete the work in 150 days . answer : b" | a ) 127 days , b ) 150 days , c ) 177 days , d ) 187 days , e ) 225 days | b | multiply(divide(multiply(const_1, 100), subtract(multiply(const_1, 100), multiply(10, 5))), multiply(15, 5)) | multiply(n3,const_1)|multiply(n0,n2)|multiply(n1,n2)|subtract(#0,#1)|divide(#0,#3)|multiply(#4,#2)| | physics |
two trains running in opposite directions cross a man standing on the platform in 27 seconds and 17 seconds respectively . if they cross each other in 25 seconds , what is the ratio of their speeds ? | "let the speed of the trains be x and y respectively length of train 1 = 27 x length of train 2 = 17 y relative speed = x + y time taken to cross each other = 25 s = ( 27 x + 17 y ) / ( x + y ) = 25 = ( 27 x + 17 y ) / = 25 ( x + y ) = 2 x = 8 y = x / y = 8 / 2 = 4 / 1 i . e 4 : 1 answer : b" | a ) 1 : 3 , b ) 4 : 1 , c ) 2 : 3 , d ) 3 : 2 , e ) 3 : 4 | b | divide(subtract(27, 25), subtract(25, 17)) | subtract(n0,n2)|subtract(n2,n1)|divide(#0,#1)| | physics |
on selling 20 balls at rs . 720 , there is a loss equal to the cost price of 5 balls . the cost price of a ball is : | "( c . p . of 20 balls ) - ( s . p . of 20 balls ) = ( c . p . of 5 balls ) c . p . of 15 balls = s . p . of 20 balls = rs . 720 . c . p . of 1 ball = rs . 720 / 15 = rs . 48 . answer : option a" | a ) s . 48 , b ) s . 50 , c ) s . 55 , d ) s . 60 , e ) s . 70 | a | divide(720, subtract(20, 5)) | subtract(n0,n2)|divide(n1,#0)| | gain |
the average of first 13 natural numbers is ? | "sum of 13 natural no . = 182 / 2 = 91 average = 91 / 13 = 7 answer : c" | a ) 5.2 , b ) 5.5 , c ) 7 , d ) 9 , e ) 10 | c | add(13, const_1) | add(n0,const_1)| | general |
tom opened a shop investing rs . 30000 . jose joined him 2 months later , investing rs . 45000 . they earned a profit of rs . 54000 after completion of one year . what will be jose ' s share of profit ? | sol = ~ s - so anju ’ s share = [ 5 / 9 ] x 54000 = 30000 e | a ) 20000 , b ) 24000 , c ) 34000 , d ) 35000 , e ) 30000 | e | multiply(54000, subtract(const_1, divide(multiply(30000, multiply(const_2, multiply(const_2, const_3))), add(multiply(45000, subtract(multiply(const_2, multiply(const_2, const_3)), 2)), multiply(30000, multiply(const_2, multiply(const_2, const_3))))))) | multiply(const_2,const_3)|multiply(#0,const_2)|multiply(n0,#1)|subtract(#1,n1)|multiply(n2,#3)|add(#4,#2)|divide(#2,#5)|subtract(const_1,#6)|multiply(n3,#7) | gain |
fresh grapes contain 75 % water by weight and raisins obtained by drying fresh grapes contain 25 % water by weight . how many kgs of fresh grapes are needed to get 25 kgs of raisins ? | "the weight of non - water in 20 kg of dried grapes ( which is 100 - 25 = 75 % of whole weight ) will be the same as the weight of non - water in x kg of fresh grapes ( which is 100 - 75 = 25 % of whole weight ) , so 25 * 0.75 = x * 0.25 - - > x = 75 . answer : a ." | a ) 75 kgs , b ) 64 kgs , c ) 72 kgs , d ) 65 kgs , e ) 70 kgs | a | add(divide(multiply(25, subtract(const_100, 25)), const_100), 75) | subtract(const_100,n1)|multiply(n2,#0)|divide(#1,const_100)|add(n0,#2)| | gain |
every year an amount increases by 1 / 8 th of itself . how much will it be after two years if its present value is rs . 64000 ? | 64000 * 9 / 8 * 9 / 8 = 81000 answer : a | a ) 81000 , b ) 22887 , c ) 26877 , d ) 26771 , e ) 17121 | a | add(add(64000, multiply(divide(1, 8), 64000)), multiply(divide(1, 8), add(64000, multiply(divide(1, 8), 64000)))) | divide(n0,n1)|multiply(n2,#0)|add(n2,#1)|multiply(#2,#0)|add(#2,#3) | general |
a train covers a distance of 12 km in 10 min . if it takes 5 sec to pass a telegraph post , then the length of the train is ? | "speed = ( 12 / 10 * 60 ) km / hr = ( 72 * 5 / 18 ) m / sec = 20 m / sec . length of the train = 20 * 8 = 160 m . answer : d" | a ) 298 m , b ) 188 m , c ) 120 m , d ) 160 m , e ) 189 m | d | divide(12, subtract(divide(12, 10), 5)) | divide(n0,n1)|subtract(#0,n2)|divide(n0,#1)| | physics |
victor gets 95 % marks in examinations . if these are 285 marks , find the maximum marks . | let the maximum marks be m then 95 % of m = 285 ⇒ 95 / 100 × m = 285 ⇒ m = ( 285 × 100 ) / 95 ⇒ m = 28500 / 95 ⇒ m = 300 therefore , maximum marks in the examinations are 300 . answer : e | a ) 334 , b ) 500 , c ) 376 , d ) 288 , e ) 300 | e | divide(285, divide(95, const_100)) | divide(n0,const_100)|divide(n1,#0) | gain |
last month , john rejected 0.5 % of the products that he inspected and jane rejected 0.9 percent of the products that she inspected . if total of 0.75 percent of the products produced last month were rejected , what fraction of the products did jane inspect ? | "x - fraction of products jane inspected ( 1 - x ) - fraction of products john inspected 0.9 ( x ) + 0.5 ( 1 - x ) = 0.75 0.4 x = 0.75 - 0.5 x = 0.25 / 0.4 x = 5 / 8 therefore the answer is c : 5 / 8 ." | a ) 1 / 6 , b ) 1 / 2 , c ) 5 / 8 , d ) 5 / 6 , e ) 15 / 16 | c | divide(subtract(0.75, 0.5), subtract(0.9, 0.5)) | subtract(n2,n0)|subtract(n1,n0)|divide(#0,#1)| | gain |
find the circumference and area of radius 7 cm . | "circumference of circle = 2 π r = 2 × 22 / 7 × 7 = 44 cm area of circle = π r ² = 22 / 7 × 7 × 7 cm ² = 154 cm ² answer : c" | a ) 151 cm ² , b ) 152 cm ² , c ) 154 cm ² , d ) 155 cm ² , e ) 157 cm ² | c | circle_area(7) | circle_area(n0)| | geometry |
a company wants to spend equal amounts of money for the purchase of two types of computer printers costing $ 375 and $ 150 per unit , respectively . what is the fewest number of computer printers that the company can purchase ? | the smallest amount that the company can spend is the lcm of 375 and 150 , which is 750 for each , which is total 1500 . the number of 1 st type of computers which costing $ 375 = 750 / 375 = 2 . the number of 2 nd type of computers which costing $ 150 = 750 / 150 = 5 . total = 2 + 5 = 7 answer is e . | a ) 3 , b ) 4 , c ) 5 , d ) 6 , e ) 7 | e | add(divide(lcm(375, 150), 375), divide(lcm(375, 150), 150)) | lcm(n0,n1)|divide(#0,n0)|divide(#0,n1)|add(#1,#2) | general |
how many integerskgreater than 100 and less than 1200 are there such that if the hundreds and the units digits ofkare reversed , the resulting integer is k + 99 ? | "numbers will be like 102 = > 201 = 102 + 99 203 = > 302 = 103 + 99 so the hundereth digit and units digit are consecutive where unit digit is bigger than hundred digit . there will be ten pairs of such numbers for every pair there will 10 numbers like for 12 = > 102 , 112,132 , 142,152 , 162,172 , 182,192 . total = 10... | a ) 50 , b ) 60 , c ) 70 , d ) 80 , e ) 100 | e | add(multiply(divide(100, const_10), multiply(const_2, const_4)), divide(100, const_10)) | divide(n0,const_10)|multiply(const_2,const_4)|multiply(#0,#1)|add(#0,#2)| | general |
a batch of cookies was divided amomg 3 tins : 2 / 3 of all the cookies were placed in either the blue or the green tin , and the rest were placed in the red tin . if 1 / 5 of all the cookies were placed in the blue tin , what fraction of the cookies that were placed in the other tins were placed in the green tin | "this will help reduce the number of variables you have to deal with : g + b = 2 / 3 r = 1 / 3 b = 1 / 5 we can solve for g which is 7 / 10 what fraction ( let it equal x ) of the cookies that were placed in the other tins were placed in the green tin ? so . . x * ( g + r ) = g x * ( 7 / 10 + 1 / 3 ) = 7 / 10 x = 21 / ... | a ) 21 / 31 , b ) 9 / 4 , c ) 5 / 9 , d ) 7 / 5 , e ) 9 / 7 | a | add(subtract(1, divide(2, 3)), subtract(divide(2, 3), divide(1, 5))) | divide(n1,n2)|divide(n3,n4)|subtract(n3,#0)|subtract(#0,#1)|add(#2,#3)| | general |
a grocer is storing soap boxes in cartons that measure 30 inches by 42 inches by 60 inches . if the measurement of each soap box is 7 inches by 6 inches by 5 inches , then what is the maximum number of soap boxes that can be placed in each carton ? | "however the process of dividing the volume of box by the volume of a soap seems flawed but it does work in this case due to the numbers dimensions of the box = 30 * 42 * 60 dimensions of the soap = 5 * 6 * 7 we get = 6 x 6 x 10 = 360 so the question is why this particular arrangement , in order to maximize number of s... | a ) 360 , b ) 252 , c ) 280 , d ) 300 , e ) 420 | a | divide(multiply(multiply(30, 42), 60), multiply(multiply(7, 6), 5)) | multiply(n0,n1)|multiply(n3,n4)|multiply(n2,#0)|multiply(n5,#1)|divide(#2,#3)| | general |
if ( a - b - c + d = 13 ) and ( a + b - c - d = 5 ) , what is the value of ( b - d ) ^ 2 ? | a - b - c + d = 13 - - equation 1 a + b - c - d = 5 - - equation 2 adding 1 and 2 , we get 2 a - 2 c = 18 = > a - c = 9 - - equation 3 using equation 3 in 2 , we get b - d = 5 - 9 = - 4 = > ( b - d ) ^ 2 = 16 answer d | a ) 4 . , b ) 8 . , c ) 12 . , d ) 16 , e ) 64 | d | power(subtract(5, divide(add(13, 5), const_2)), const_2) | add(n0,n1)|divide(#0,const_2)|subtract(n1,#1)|power(#2,const_2) | general |
a hollow iron pipe is 21 cm long and its external diameter is 8 cm . if the thickness of the pipe is 1 cm and iron weighs 8 g / cm 3 , then the weight of the pipe is | "explanation : external radius = 4 cm , internal radius = 3 cm . volume of iron = [ ( 22 / 7 ) × ( 42 - 32 ) × 21 ] cm 3 = 462 cm 3 weight of iron = ( 462 x 8 ) gm = 3696 gm = 3.696 kg answer : b" | a ) 3.6 kg , b ) 3.696 kg , c ) 36 kg , d ) 36.9 kg , e ) none of these | b | divide(multiply(subtract(volume_cylinder(divide(8, const_2), 21), volume_cylinder(subtract(divide(8, const_2), 1), 21)), 8), const_1000) | divide(n1,const_2)|subtract(#0,n2)|volume_cylinder(#0,n0)|volume_cylinder(#1,n0)|subtract(#2,#3)|multiply(n1,#4)|divide(#5,const_1000)| | physics |
a train 225 m in length crosses a telegraph post in 15 seconds . the speed of the train is ? | "s = 225 / 25 * 18 / 5 = 32 kmph answer : c" | a ) 37 kmph , b ) 34 kmph , c ) 32 kmph , d ) 38 kmph , e ) 76 kmph | c | multiply(const_3_6, divide(225, 15)) | divide(n0,n1)|multiply(#0,const_3_6)| | physics |
find the value of 3 x [ ( 3.6 x 0.48 x 2.50 ) / ( 0.12 x 0.09 x 0.5 ) ] | "answer 3 x [ ( 3.6 x 0.48 x 2.50 ) / ( 0.12 x 0.09 x 0.5 ) ] = 3 x [ ( 36 x 48 x 250 ) / ( 12 x 9 x 5 ) ] = 3 x 4 x 4 x 50 = 2400 correct option : e" | a ) 800 , b ) 500 , c ) 900 , d ) 1600 , e ) 2400 | e | multiply(divide(multiply(multiply(3.6, 0.48), 2.50), multiply(multiply(0.12, 0.09), 0.5)), 3) | multiply(n1,n2)|multiply(n4,n5)|multiply(n3,#0)|multiply(n6,#1)|divide(#2,#3)|multiply(n0,#4)| | general |
sara bought both german chocolate and swiss chocolate for some cakes she was baking . the swiss chocolate cost $ 3.5 per pound , and german chocolate cost $ 2.8 per pound . if the total the she spent on chocolate was $ 28 and both types of chocolate were purchased in whole number of pounds , how many total pounds of ch... | "if there were all the expensive ones , 3.5 . . . . there would be 28 / 3.5 or 8 of them but since 2.8 $ ones are also there , answer has to be > 8 . . if all were 2.8 $ ones , there will be 28 / 2.8 or 10 . . . so only 9 is left answer a . ." | a ) 9 , b ) 8 , c ) 10 , d ) 12 , e ) 15 | a | divide(28, divide(add(3.5, 2.8), const_2)) | add(n0,n1)|divide(#0,const_2)|divide(n2,#1)| | other |
the length of the bridge , which a train 110 metres long and travelling at 45 km / hr can cross in 30 seconds , is ? | "speed = [ 45 x 5 / 18 ] m / sec = [ 25 / 2 ] m / sec time = 30 sec let the length of bridge be x metres . then , ( 130 + x ) / 30 = 25 / 2 = > 2 ( 110 + x ) = 750 = > x = 265 m . answer : d" | a ) 10 m , b ) 16 m , c ) 245 m , d ) 265 m , e ) 27 m | d | subtract(multiply(divide(multiply(45, speed(const_1000, const_1)), speed(const_3600, const_1)), 30), 110) | speed(const_1000,const_1)|speed(const_3600,const_1)|multiply(n1,#0)|divide(#2,#1)|multiply(n2,#3)|subtract(#4,n0)| | physics |
if 16 men can reap 80 hectares in 24 days , then how many hectares can 36 men reap in 30 days ? | "explanation : let the required no of hectares be x . then men - - - hectares - - - days 16 - - - - - - - - - 80 - - - - - - - - - 24 36 - - - - - - - - - x - - - - - - - - - 30 more men , more hectares ( direct proportion ) more days , more hectares ( direct proportion ) x = 36 / 16 * 30 / 24 * 80 x = 225 answer : b" | a ) 127 , b ) 225 , c ) 287 , d ) 450 , e ) 281 | b | divide(multiply(36, 30), divide(multiply(16, 24), 80)) | multiply(n3,n4)|multiply(n0,n2)|divide(#1,n1)|divide(#0,#2)| | physics |
a shop sells 1 chocolate at the rate rupee 1 each . you can exchange 3 warppers for one chocolate . if you have rs . 15 , what is the maximum number of chocolates you can get ? | rs . 15 = 15 chocolates 15 wrappers = 5 chocolates 5 wrappers = 1 chocolate ( u have 2 wrappers with you . . ) count this wrapper with already existing wrappers = 15 + 5 + 1 + 1 = 22 chocolates . . answer : b | a ) 21 , b ) 22 , c ) 23 , d ) 24 , e ) 25 | b | add(add(add(divide(15, 3), 1), 1), 15) | divide(n3,n2)|add(n0,#0)|add(n0,#1)|add(n3,#2) | gain |
jim is now twice as old as fred , who is 9 years older than sam . 6 years ago , jim was 5 times as old as sam . how old is jim now ? | j = 2 f = 2 ( s + 9 ) = 2 s + 18 j - 6 = 5 * ( s - 6 ) ( 2 s + 18 ) - 6 = 5 s - 30 s = 14 and so j = 46 the answer is b . | a ) 42 , b ) 46 , c ) 50 , d ) 54 , e ) 58 | b | multiply(add(divide(add(subtract(multiply(6, 5), 6), multiply(9, const_2)), subtract(9, 6)), 9), const_2) | multiply(n0,const_2)|multiply(n1,n2)|subtract(n0,n1)|subtract(#1,n1)|add(#0,#3)|divide(#4,#2)|add(n0,#5)|multiply(#6,const_2) | general |
a person lent a certain sum of money at 4 % per annum at simple interest and in 8 years the interest amounted to rs . 204 less than the sum lent . what was the sum lent ? | p - 204 = ( p * 4 * 8 ) / 100 p = 300 answer : c | a ) 228 , b ) 278 , c ) 300 , d ) 500 , e ) 821 | c | divide(204, subtract(const_1, divide(multiply(4, 8), const_100))) | multiply(n0,n1)|divide(#0,const_100)|subtract(const_1,#1)|divide(n2,#2) | gain |
in the coordinate plane , a circle centered on point ( 3 , - 4 ) passes through point ( 1 , 1 ) . what is the area of the circle ? | "r ^ 2 = ( 3 - 1 ) ^ 2 + ( - 4 - 1 ) ^ 2 = 4 + 25 = 29 area of circle = π r ^ 2 = 29 π answer : a" | a ) 29 π , b ) 18 π , c ) 25 π , d ) 37 π , e ) 41 π | a | circle_area(sqrt(add(power(subtract(3, 1), const_2), power(add(1, 4), const_2)))) | add(n1,n2)|subtract(n0,n2)|power(#1,const_2)|power(#0,const_2)|add(#2,#3)|sqrt(#4)|circle_area(#5)| | geometry |
the l . c . m . of 2 numbers is 48 . the numbers are in the ratio 2 : 3 . find their sum ? | "let the numbers be 2 x and 3 x l . c . m . = 6 x 6 x = 48 x = 8 the numbers are = 16 and 24 required sum = 16 + 24 = 40 answer is b" | a ) 36 , b ) 40 , c ) 48 , d ) 32 , e ) 56 | b | add(multiply(divide(2, multiply(48, 2)), 2), multiply(divide(2, multiply(48, 2)), 48)) | multiply(n1,n2)|divide(n0,#0)|multiply(n2,#1)|multiply(n1,#1)|add(#2,#3)| | other |
the credit card and a global payment processing companies have been suffering losses for some time now . a well known company recently announced its quarterly results . according to the results , the revenue fell to $ 42.0 billion from $ 69.0 billion , a year ago . by what percent did the revenue fall ? | "$ 69 - $ 42 = 27 $ ( 27 / 69 ) * 100 = 39.13 % answer : d" | a ) 20.8 , b ) 30.4 , c ) 31.8 , d ) 39.1 , e ) 33.8 | d | multiply(divide(subtract(69.0, 42.0), 69.0), const_100) | subtract(n1,n0)|divide(#0,n1)|multiply(#1,const_100)| | gain |
how many positive integers less than 100 have a remainder of 2 when divided by 13 ? | "take the multiples of 13 and add 2 0 x 13 + 2 = 2 . . . . 13 x 7 + 2 = 93 there are 14 numbers answer c" | a ) 5 , b ) 7 , c ) 8 , d ) 9 , e ) 6 | c | divide(factorial(subtract(add(const_4, 2), const_1)), multiply(factorial(2), factorial(subtract(const_4, const_1)))) | add(n1,const_4)|factorial(n1)|subtract(const_4,const_1)|factorial(#2)|subtract(#0,const_1)|factorial(#4)|multiply(#1,#3)|divide(#5,#6)| | general |
convert 60 miles into yards ? | "1 mile = 1760 yards 60 miles = 60 * 1760 = 105600 yards answer is a" | a ) 105600 yards , b ) 35200 yards , c ) 39520 yards , d ) 42560 yards , e ) 41520 yards | a | divide(multiply(multiply(multiply(add(const_3, const_2), const_2), multiply(add(const_3, const_2), const_2)), 60), multiply(multiply(add(const_3, const_2), const_2), multiply(add(const_3, const_2), const_2))) | add(const_2,const_3)|multiply(#0,const_2)|multiply(#1,#1)|multiply(n0,#2)|divide(#3,#2)| | physics |
the price of a bushel of corn is currently $ 3.20 , and the price of a peck of wheat is $ 10.80 . the price of corn is increasing at a constant rate of 5 x cents per day while the price of wheat is decreasing at a constant rate of x ( 2 ^ 1 / 2 ) - x cents per day . what is the approximate price when a bushel of corn c... | i tried using time / rate approach : - initial price difference = 10.80 - 3.20 = 7.60 price of corn increasing by 5 x price of wheat decreasing by x ( 1.4 ) - x = . 4 x since both the quantities are moving towards reducing the price gap hence : - relative increase = 5 x + . 4 x let t be the time by which gap is filled ... | a ) $ 4.50 , b ) $ 10.2 , c ) $ 5.30 , d ) $ 5.50 , e ) $ 5.60 | b | add(3.2, multiply(divide(subtract(10.8, 3.2), add(5, subtract(sqrt(const_2), 1))), 5)) | sqrt(const_2)|subtract(n1,n0)|subtract(#0,n4)|add(n2,#2)|divide(#1,#3)|multiply(n2,#4)|add(n0,#5) | general |
how many positive integers e between 200 and 300 ( both inclusive ) are not divisible by 2 , 3 or 5 ? | 1 ) i figured there are 101 integers ( 300 - 200 + 1 = 101 ) . since the set begins with an even and ends with an even , there are 51 evens . 2 ) question says integers are not divisible by 2 , leaving all of the odds ( 101 - 51 = 50 integers ) . 3 ) question says integers are not divisible by 5 , removing all the inte... | a ) 3 , b ) 16 , c ) 75 , d ) 24 , e ) 26 | e | subtract(add(subtract(300, 200), const_1), add(add(add(divide(subtract(300, 200), 2), const_1), add(floor(subtract(add(divide(subtract(300, 200), 3), const_1), add(multiply(5, 3), 2))), const_1)), subtract(add(divide(subtract(300, 200), 5), const_1), multiply(5, 3)))) | multiply(n3,n4)|subtract(n1,n0)|add(#1,const_1)|add(n2,#0)|divide(#1,n2)|divide(#1,n3)|divide(#1,n4)|add(#4,const_1)|add(#5,const_1)|add(#6,const_1)|subtract(#8,#3)|subtract(#9,#0)|floor(#10)|add(#12,const_1)|add(#7,#13)|add(#14,#11)|subtract(#2,#15) | other |
the hcf of two numbers is 23 and the other two factors of their lcm are 13 and 14 . what is the largest number ? | hcf of the two numbers = 23 since hcf will be always a factor of lcm , 23 is a factor of the lcm . given that other two factors in the lcm are 13 and 14 . hence factors of the lcm are 23 , 13 , 14 so , numbers can be taken as ( 23 × 13 ) and ( 23 × 14 ) = 299 and 322 hence , largest number = 322 e ) | a ) 280 , b ) 284 , c ) 292 , d ) 320 , e ) 322 | e | multiply(23, 14) | multiply(n0,n2) | other |
in a throw of dice what is the probability of ge æ « ng number greater than 2 | explanation : number greater than 2 is 3 , 4 , 5 & 6 , so only 4 number total cases of dice = [ 1,2 , 3,4 , 5,6 ] so probability = 4 / 6 = 2 / 3 answer : b | a ) 1 / 2 , b ) 2 / 3 , c ) 1 / 5 , d ) 1 / 6 , e ) none of these | b | divide(subtract(const_6, 2), const_6) | subtract(const_6,n0)|divide(#0,const_6) | probability |
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