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
values |
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a can do a piece of work in 15 days and b alone can do it in 10 days . b works at it for 5 days and then leaves . a alone can finish the remaining work in | "explanation : b ' s 5 days work = 1 / 10 * 5 = 1 / 2 remaining work = 1 â ˆ ’ 1 / 2 = 1 / 2 a can finish work = 15 â ˆ — 1 / 2 = 7.5 days answer is c" | a ) 5 days , b ) 6 days , c ) 7.5 days , d ) 8.5 days , e ) 8 days | c | divide(const_1, add(divide(const_1, 15), divide(const_1, 10))) | divide(const_1,n0)|divide(const_1,n1)|add(#0,#1)|divide(const_1,#2)| | physics |
a couple decides to have 3 children . if they succeed in having 3 children and each child is equally likely to be a boy or a girl , what is the probability that they will have exactly 1 girl and 2 boys ? | sample space = 2 ^ 3 = 8 favourable events = { bgg } , { bgb } , { bbb } , { ggg } , { gbg } , probability = 5 / 8 = 5 / 8 . ans ( c ) . | a ) 3 / 4 , b ) 1 / 2 , c ) 5 / 8 , d ) 1 , e ) 1 / 4 | c | subtract(1, multiply(divide(factorial(3), factorial(2)), power(divide(1, 2), 3))) | divide(n2,n3)|factorial(n0)|factorial(n3)|divide(#1,#2)|power(#0,n0)|multiply(#3,#4)|subtract(n2,#5) | general |
if money is invested at r percent interest , compounded annually , the amount of the investment will double in approximately 60 / r years . if pat ' s parents invested $ 7,000 in a long - term bond that pays 6 percent interest , compounded annually , what will be the approximate total amount of the investment 20 years ... | "since investment doubles in 60 / r years , then for r = 6 it ' ll double in 60 / 6 = ~ 10 years ( we are not asked about the exact amount so such an approximation will do ) . thus after 20 years investment will become $ 7,000 * 2 = $ 14,000 . answer : b" | a ) $ 20000 , b ) $ 14000 , c ) $ 12000 , d ) $ 10000 , e ) $ 9000 | b | divide(multiply(multiply(add(const_2, const_3), const_1000), 6), const_2) | add(const_2,const_3)|multiply(#0,const_1000)|multiply(n2,#1)|divide(#2,const_2)| | general |
find the numbers which are in the ratio 3 : 2 : 4 such that the sum of the first and the second added to the difference of the third and the second is 28 ? | "let the numbers be a , b and c . a : b : c = 3 : 2 : 4 given , ( a + b ) + ( c - b ) = 21 = > a + c = 28 = > 3 x + 4 x = 28 = > x = 4 a , b , c are 3 x , 2 x , 4 x a , b , c are 12 , 8 , 16 . answer : c" | a ) 4 , 3,22 , b ) 4 , 4,22 , c ) 12 , 6,16 , d ) 9 , 6,12 , e ) 9 , 2,23 | c | divide(multiply(4, 3), 3) | multiply(n2,n0)|divide(#0,n0)| | general |
eddy and freddy start simultaneously from city a and they travel to city b and city c respectively . eddy takes 3 hours and freddy takes 4 hours to complete the journey . if the distance between city a and city b is 570 kms and city a and city c is 300 kms . what is the ratio of their average speed of travel ? ( eddy :... | "distance traveled by eddy = 570 km time taken by eddy = 3 hours average speed of eddy = 570 / 3 = 190 km / hour distance traveled by freddy = 300 km time taken by freddy = 4 hours average speed of freddy = 300 / 4 = 75 km / hour ratio of average speed of eddy to freddy = 190 / 75 = 38 / 15 answer c" | a ) 8 / 3 , b ) 3 / 8 , c ) 38 / 15 , d ) 5 / 8 , e ) 5 / 3 | c | divide(divide(570, 3), divide(300, 4)) | divide(n2,n0)|divide(n3,n1)|divide(#0,#1)| | physics |
two trains 180 m and 360 m long run at the speed of 60 kmph and 30 kmph in opposite directions in parallel tracks . the time which they take to cross each other is ? | "relative speed = 60 + 30 = 90 kmph * 5 / 18 = 25 m / s distance covered in crossing each other = 180 + 360 = 540 m required time = 540 * 1 / 25 = 21.6 sec answer is a" | a ) 21.6 sec , b ) 8.9 sec , c ) 10.8 sec , d ) 12.6 sec , e ) 15 sec | a | divide(add(180, 360), multiply(add(60, 30), const_0_2778)) | add(n0,n1)|add(n2,n3)|multiply(#1,const_0_2778)|divide(#0,#2)| | physics |
if 5 ^ 10 x 2 ^ 10 = 10 ^ n what is the value of n ? | "5 ^ 10 * 2 ^ 10 = 10 ^ n or 10 ^ 10 = 10 ^ n n = 10 d" | a ) 16 , b ) 14 , c ) 12 , d ) 10 , e ) 18 | d | divide(log(multiply(power(2, 10), power(5, 10))), log(10)) | log(n4)|power(n2,n3)|power(n0,n1)|multiply(#1,#2)|log(#3)|divide(#4,#0)| | general |
the average of 55 results is 28 and the average of other 28 results is 55 . what is the average of all the results ? | "answer sum of 83 result = sum of 55 result + sum of 28 result . = 30 x 20 + 20 x 30 = 3080 correct option : a" | a ) 37 , b ) 25 , c ) 48 , d ) 50 , e ) none | a | divide(add(multiply(55, 28), multiply(28, 55)), add(55, 28)) | add(n0,n1)|multiply(n0,n1)|add(#1,#1)|divide(#2,#0)| | general |
jerry ’ s average ( arithmetic mean ) score on the first 3 of 4 tests is 90 . if jerry wants to raise his average by 2 points , what score must he earn on the fourth test ? | "total score on 3 tests = 90 * 3 = 270 jerry wants the average to be = 92 hence total score on 4 tests should be = 92 * 4 = 368 score required on the fourth test = 368 - 270 = 98 option d" | a ) 87 , b ) 89 , c ) 90 , d ) 98 , e ) 95 | d | subtract(multiply(4, add(90, 2)), multiply(90, 3)) | add(n2,n3)|multiply(n0,n2)|multiply(n1,#0)|subtract(#2,#1)| | general |
36 men can complete a piece of work in 18 days . in how many days will 9 men complete the same work ? | "explanation : less men , means more days { indirect proportion } let the number of days be x then , 9 : 36 : : 18 : x x = 72 answer : d ) 72 days" | a ) 24 , b ) 77 , c ) 88 , d ) 72 , e ) 21 | d | divide(multiply(18, 36), 9) | multiply(n0,n1)|divide(#0,n2)| | physics |
a tank contains 10,000 gallons of a solution that is 5 percent sodium chloride by volume . if 4,000 gallons of water evaporate from the tank , the remaining solution will be approximately what percent sodium chloride ? | "the amount of sodium chloride is 0.05 * 10,000 = 500 gallons 500 / 6000 = 1 / 12 which is about 8.33 % the answer is d ." | a ) 6.85 % , b ) 7.25 % , c ) 7.76 % , d ) 8.33 % , e ) 8.75 % | d | multiply(divide(multiply(multiply(const_100, const_100), divide(5, const_100)), subtract(multiply(const_100, const_100), add(multiply(add(const_2, const_3), multiply(multiply(add(const_2, const_3), const_2), const_100)), multiply(add(const_2, const_3), const_100)))), const_100) | add(const_2,const_3)|divide(n1,const_100)|multiply(const_100,const_100)|multiply(#1,#2)|multiply(#0,const_2)|multiply(#0,const_100)|multiply(#4,const_100)|multiply(#0,#6)|add(#7,#5)|subtract(#2,#8)|divide(#3,#9)|multiply(#10,const_100)| | gain |
the diameter of a garden roller is 1.4 m and it is 4 m long . how much area will it cover in 5 revolutions ? ( use ï € = 22 â „ 7 ) | "required area covered in 5 revolutions = 5 ã — 2 ï € rh = 5 ã — 2 ã — 22 â „ 7 ã — 0.7 ã — 4 = 88 m 2 answer c" | a ) 80 m 2 , b ) 84 m 2 , c ) 88 m 2 , d ) 86 m 2 , e ) none of these | c | multiply(5, multiply(multiply(multiply(4, divide(22, 7)), divide(1.4, 4)), 4)) | divide(n0,n1)|divide(n3,n4)|multiply(#1,n1)|multiply(#0,#2)|multiply(n1,#3)|multiply(n2,#4)| | physics |
at a certain high school , the senior class is twice the size of the junior class . if 7 / 8 of the seniors and 1 / 2 of the juniors study japanese , what fraction of the students in both classes study japanese ? | "start by deciding on a number of students to represent the number of students in the senior class . for this example i will choose 200 students . that would make the number of students in the junior class 100 . then we can find out how many students are taking japanese in each grade and add them together . ( 7 / 8 ) *... | a ) 3 / 4 , b ) 1 / 2 , c ) 1 / 4 , d ) 3 / 8 , e ) 5 / 16 | a | divide(2, add(8, 2)) | add(n1,n3)|divide(n3,#0)| | general |
a crow leaves its nest , and flies back and forth from its nest to a nearby ditch to gather worms . the distance between the nest and the ditch is 400 meters . in one and a half hours , the crow manages to bring worms to its nest 15 times . what is the speed of the crow in kilometers per hour ? | "the distance between the nest and the ditch is 400 meters . 15 times mean = a crow leaves its nest , and flies back ( going and coming back ) i . e . 2 times we get total 30 rounds . so the distance is 30 * 400 = 12000 . d = st 12000 / 1.5 = t , i think we can take 12000 meters as 12 km , then only we get t = 8 . ( 10... | a ) 2 , b ) 4 , c ) 6 , d ) 8 , e ) 10 | d | divide(divide(multiply(400, multiply(15, const_2)), const_1000), divide(15, const_10)) | divide(n1,const_10)|multiply(n1,const_2)|multiply(n0,#1)|divide(#2,const_1000)|divide(#3,#0)| | physics |
what is the total cost of 2 sandwiches at $ 2.44 each and 4 sodas at $ 0.87 each ? | "answer = c 2 * 2.44 + 4 * 0.87 = 2 ( 2.50 - 0.06 ) + 4 ( 1.00 - 0.13 ) = 5 + 4 - 0.12 - 0.52 = 9 - 0.64 = 8.36" | a ) $ 3.36 , b ) $ 6.85 , c ) $ 8.36 , d ) $ 10.08 , e ) $ 11.85 | c | add(multiply(2, 2.44), multiply(4, 0.87)) | multiply(n0,n1)|multiply(n2,n3)|add(#0,#1)| | general |
average of 5 numbers is 12 . if one number is excluded the average becomes 10 . the excluded number is | explanation : number is ( 5 * 12 ) - ( 4 * 10 ) = 60 - 40 = 20 answer : option b | a ) 35 , b ) 20 , c ) 40 , d ) 30 , e ) 67 | b | subtract(multiply(5, 12), multiply(10, const_4)) | multiply(n0,n1)|multiply(n2,const_4)|subtract(#0,#1) | general |
a watch was sold at a loss of 10 % . if it was sold for rs . 200 more , there would have been a gain of 4 % . what is the cost price ? | "90 % 104 % - - - - - - - - 14 % - - - - 200 100 % - - - - ? = > rs . 1429 answer : b" | a ) 1000 , b ) 1429 , c ) 1977 , d ) 2778 , e ) 2711 | b | divide(multiply(200, const_100), subtract(add(const_100, 4), subtract(const_100, 10))) | add(const_100,n2)|multiply(n1,const_100)|subtract(const_100,n0)|subtract(#0,#2)|divide(#1,#3)| | gain |
set a contains all the even numbers between 12 and 60 inclusive . set b contains all the even numbers between 62 and 110 inclusive . what is the difference between the sum of elements of set b and the sum of the elements of set a ? | "each term in set b is 50 more than the corresponding term in set a . the difference of the sums = 25 * 50 = 1250 . the answer is b ." | a ) 850 , b ) 1250 , c ) 1650 , d ) 2050 , e ) 2450 | b | multiply(subtract(62, 12), add(divide(subtract(60, 12), const_2), const_1)) | subtract(n1,n0)|subtract(n2,n0)|divide(#0,const_2)|add(#2,const_1)|multiply(#3,#1)| | general |
a worker earns $ 20 on the first day and spends $ 15 on the second day . the worker earns $ 20 on the third day and spends $ 15 on the fourth day . if this pattern continues , on which day will the worker first reach a net total of $ 90 ? | "every two days , the net total is $ 5 . after 28 days , the worker will have $ 70 . on day 29 , the worker will receive $ 20 for a net total of $ 90 . the answer is b ." | a ) 28 , b ) 29 , c ) 31 , d ) 34 , e ) 36 | b | add(multiply(divide(20, subtract(20, 15)), const_2), const_1) | subtract(n0,n1)|divide(n0,#0)|multiply(#1,const_2)|add(#2,const_1)| | physics |
log 3 n + log 9 n what is 3 digit number n that will be whole number | "no of values n can take is 1 9 ^ 3 = 729 answer : b" | a ) 629 , b ) 729 , c ) 829 , d ) 929 , e ) 727 | b | power(9, 3) | power(n1,n0)| | other |
an employee ’ s annual salary was increased $ 15,000 . if her new annual salary now equals $ 90,000 , what was the percent increase ? | "new annual salary = $ 90,000 salary increase = $ 15,000 . original salary = $ 90,000 - $ 15,000 . = $ 75,000 % increase = ( $ 15,000 / $ 75,000 ) * 100 = 20 % hence c ." | a ) 15 % , b ) 16 2 ⁄ 3 % , c ) 20 % , d ) 22 % , e ) 24 % | c | multiply(subtract(divide(multiply(subtract(const_100, const_10), const_1000), subtract(multiply(subtract(const_100, const_10), const_1000), multiply(multiply(const_0_25, const_100), const_1000))), const_1), const_100) | multiply(const_0_25,const_100)|subtract(const_100,const_10)|multiply(#1,const_1000)|multiply(#0,const_1000)|subtract(#2,#3)|divide(#2,#4)|subtract(#5,const_1)|multiply(#6,const_100)| | general |
how many zeros does 50 ! end with ? | "according to above 50 ! has 50 / 5 + 50 / 25 = 10 + 2 = 12 trailing zeros . answer : d ." | a ) 20 , b ) 24 , c ) 25 , d ) 12 , e ) 32 | d | add(add(divide(50, add(const_4, const_1)), divide(subtract(50, add(const_4, const_1)), power(add(const_4, const_1), const_2))), divide(subtract(50, add(const_4, const_1)), power(add(const_4, const_1), const_3))) | add(const_1,const_4)|divide(n0,#0)|power(#0,const_2)|power(#0,const_3)|subtract(n0,#0)|divide(#4,#2)|divide(#4,#3)|add(#1,#5)|add(#7,#6)| | other |
when 1 + 2 = 23 , 2 + 3 = 65 , 3 + 4 = 127 , then 4 + 5 = ? | "1 + 2 = > 1 x 2 = 2 & 1 + 2 = 3 = > 2 & 3 = > 23 2 + 3 = > 2 ã — 3 = 6 & 2 + 3 = 5 = > 6 & 6 = > 65 3 + 4 = > 3 ã — 4 = 12 & 3 + 4 = 7 = > 12 & 7 = > 127 then 4 + 5 = > 4 ã — 5 = 20 & 4 + 5 = 9 = > 20 & 9 = > 209 answer : a" | a ) 209 , b ) 250 , c ) 265 , d ) 280 , e ) 225 | a | add(multiply(multiply(4, 5), const_10), 1) | multiply(n7,n10)|multiply(#0,const_10)|add(n0,#1)| | general |
paul sells encyclopedias door - to - door . he earns $ 150 on every paycheck , regardless of how many sets he sells . in addition , he earns commission as follows : commission sales 10 % $ 0.00 - $ 1000.00 5 % $ 1000.01 - - - > he does not earn double commission . that is , if his sales are $ 12000 , he earns 10 % on t... | his pay check was $ 1320 . out of this , $ 150 was his fixed salary so the total commission he earned was $ 1320 - $ 150 = $ 1170 he earns 10 % on the sales of first $ 10,000 which gives a commission of $ 1000 . he earns 5 % on every subsequent dollar . since he earns another $ 170 , he must have had sales of another 1... | a ) 13,400 , b ) 14,800 , c ) 16,400 , d ) 15,800 , e ) 19,600 | a | add(subtract(add(10, const_4), const_1), divide(const_4, const_10)) | add(n1,const_4)|divide(const_4,const_10)|subtract(#0,const_1)|add(#1,#2) | general |
if an article is sold at 19 % profit instead of 12 % profit , then the profit would be $ 84 more . what is the cost price ? | "7 % * cost price = $ 84 1 % * cost price = $ 84 / 7 = $ 12 the cost price is $ 1200 . the answer is b ." | a ) $ 900 , b ) $ 1200 , c ) $ 1500 , d ) $ 1800 , e ) $ 2100 | b | multiply(divide(84, 12), const_100) | divide(n2,n1)|multiply(#0,const_100)| | gain |
if 213 × 16 = 3408 , then 16 × 21.3 is equal to : | "solution 16 × 21.3 = ( 16 x 213 / 10 ) = ( 16 x 213 / 10 ) = 3408 / 10 = 340.8 . answer d" | a ) 0.3408 , b ) 3.408 , c ) 34.08 , d ) 340.8 , e ) none of these | d | multiply(16, 21.3) | multiply(n4,n3)| | general |
a , b and c invest in a partnership in the ratio : 7 / 24 / 36 / 5 . after 4 months , a increases his share 50 % . if the total profit at the end of one year is rs . 21600 , then what is b ' s share in the profit ? | ratio of the initial investment = 7 / 2 : 4 / 3 : 6 / 5 = 105 : 40 : 36 therefore , let the initial investments of a , b and c be 105 x , 40 x and 36 x respectively a increases his share 50 % after 4 months . hence the ratio of their investments = ( 105 x × 4 ) + ( 105 x × 150 / 100 × 8 ) : 40 x × 12 : 36 x × 12 = 105 ... | a ) 1000 , b ) 2000 , c ) 3000 , d ) 4000 , e ) 5000 | d | divide(multiply(21600, const_10), subtract(subtract(add(36, 24), 5), const_1)) | add(n1,n2)|multiply(n6,const_10)|subtract(#0,n3)|subtract(#2,const_1)|divide(#1,#3) | gain |
if x and y are numbers such that ( x + 3 ) ( y - 3 ) = 0 , what is the smallest possible value of x ^ 2 + y ^ 2 | "from ( x + 3 ) ( y - 3 ) = 0 it follows that either x = - 3 or y = 3 . thus either x ^ 2 = 9 or y ^ 2 = 9 . now , if x ^ 2 = 9 , then the least value of y ^ 2 is 0 , so the least value of x ^ 2 + y ^ 2 = 9 + 0 = 9 . similarly if y ^ 2 = 9 , then the least value of x ^ 2 is 0 , so the least value of x ^ 2 + y ^ 2 = 0 +... | a ) 0 , b ) 3 , c ) 6 , d ) 9 , e ) 12 | d | power(3, 2) | power(n0,n3)| | general |
a bag contains 6 black and 3 white balls . one ball is drawn at random . what is the probability that the ball drawn is white ? | "let number of balls = ( 6 + 3 ) = 9 . number of white balls = 3 . p ( drawing a white ball ) = 3 / 9 = 1 / 3 . option b ." | a ) 3 / 4 , b ) 1 / 3 , c ) 1 / 7 , d ) 1 / 8 , e ) 4 / 3 | b | divide(add(divide(divide(factorial(3), factorial(subtract(3, const_2))), factorial(const_2)), divide(divide(factorial(6), factorial(subtract(6, const_2))), factorial(const_2))), divide(divide(factorial(add(6, 3)), factorial(subtract(add(6, 3), const_2))), factorial(const_2))) | add(n0,n1)|factorial(n1)|factorial(const_2)|factorial(n0)|subtract(n1,const_2)|subtract(n0,const_2)|factorial(#4)|factorial(#5)|factorial(#0)|subtract(#0,const_2)|divide(#1,#6)|divide(#3,#7)|factorial(#9)|divide(#10,#2)|divide(#11,#2)|divide(#8,#12)|add(#13,#14)|divide(#15,#2)|divide(#16,#17)| | probability |
find large number from below question the difference of two numbers is 1200 . on dividing the larger number by the smaller , we get 6 as quotient and the 15 as remainder | "let the smaller number be x . then larger number = ( x + 1365 ) . x + 1200 = 6 x + 15 5 x = 1185 x = 237 large number = 237 + 1365 = 1437 a" | a ) 1437 , b ) 1250 , c ) 1540 , d ) 1600 , e ) 1635 | a | multiply(divide(subtract(1200, 15), subtract(6, const_1)), 6) | subtract(n0,n2)|subtract(n1,const_1)|divide(#0,#1)|multiply(n1,#2)| | general |
due to construction , the speed limit along an 8 - mile section of highway is reduced from 55 miles per hour to 35 miles per hour . approximately how many minutes more will it take to travel along this section of highway at the new speed limit than it would have taken at the old speed limit ? | "old time in minutes to cross 8 miles stretch = 8 * 60 / 55 = 8 * 12 / 11 = 8.72 new time in minutes to cross 8 miles stretch = 8 * 60 / 35 = 8 * 12 / 7 = 13.71 time difference = 4.99 ans : a" | a ) a ) 4.99 , b ) b ) 8 , c ) c ) 10 , d ) d ) 15 , e ) e ) 24 | a | max(multiply(subtract(add(55, 8), const_1), subtract(divide(8, 35), divide(8, 55))), const_4) | add(n0,n1)|divide(n0,n2)|divide(n0,n1)|subtract(#0,const_1)|subtract(#1,#2)|multiply(#3,#4)|max(#5,const_4)| | physics |
set a consists of all the prime numbers between 2 and 13 . 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 = 11 lowest number = 3 range = highest - lowest = 11 - 3 = 8 option e" | a ) 12 , b ) 13 , c ) 16.6 , d ) 17 , e ) 8 | e | subtract(subtract(13, const_2), add(add(2, const_4), const_1)) | add(n0,const_4)|subtract(n1,const_2)|add(#0,const_1)|subtract(#1,#2)| | general |
if the sum of a number and its square is 210 , what is the number ? | "explanation : let the integer be x . then , x + x 2 = 210 x 2 + x - 210 = 0 ( x + 15 ) ( x – 14 ) = 0 x = 14 answer : b" | a ) 16 , b ) 14 , c ) 25 , d ) 87 , e ) 171 | b | floor(sqrt(210)) | sqrt(n0)|floor(#0)| | geometry |
a train of length 240 m crosses a bridge of length 150 m in 20 seconds . what is the speed of train ? | "sol : ( length of train + length of bridge ) = speed of train x time ( 240 + 150 ) = 20 x speed speed = 390 / 20 = 19.5 m / s = 70.2 km / h answer = d" | a ) 33 , b ) 27 , c ) 25 , d ) 70.2 , e ) 72 | d | multiply(const_3_6, divide(add(240, 150), 20)) | add(n0,n1)|divide(#0,n2)|multiply(#1,const_3_6)| | physics |
three interviewers , a , b , and c are interviewing 50 applicants . only with three interviewers ' admission can an applicant be admitted . if interviewer a admitted 20 applicants , b admitted 22 applicants , and c admitted 20 applicants , at least how many applicants get the admission ? | "if a admitted 20 are overlapping with b admission of 22 but c does not overlap with anybody . then no student will get nod from all the 3 . hence 0 student will get admission . answer : b" | a ) 4 , b ) 0 , c ) 6 , d ) 8 , e ) 12 | b | subtract(add(add(20, 22), 20), 50) | add(n1,n2)|add(n3,#0)|subtract(#1,n0)| | geometry |
find the sum of first 100 odd numbers | "explanation : n 2 = 1002 = 10000 answer : option c" | a ) 5500 , b ) 2005 , c ) 10000 , d ) 5800 , e ) 4960 | c | add(divide(divide(100, divide(divide(divide(divide(divide(100, const_2), const_2), const_2), const_2), const_2)), const_2), add(const_1, sqrt(divide(divide(100, divide(divide(divide(divide(divide(100, const_2), const_2), const_2), const_2), const_2)), const_2)))) | divide(n0,const_2)|divide(#0,const_2)|divide(#1,const_2)|divide(#2,const_2)|divide(#3,const_2)|divide(n0,#4)|divide(#5,const_2)|sqrt(#6)|add(#7,const_1)|add(#8,#6)| | general |
a man â € ™ s current age is ( 2 / 5 ) of the age of his father . after 6 years , he will be ( 1 / 2 ) of the age of his father . what is the age of father at now ? | "let , father â € ™ s current age is a years . then , man â € ™ s current age = [ ( 2 / 5 ) a ] years . therefore , [ ( 2 / 5 ) a + 6 ] = ( 1 / 2 ) ( a + 6 ) 2 ( 2 a + 30 ) = 5 ( a + 6 ) a = 30 e" | a ) 40 , b ) 45 , c ) 38 , d ) 50 , e ) 30 | e | divide(subtract(6, multiply(6, divide(1, 2))), subtract(divide(1, 2), divide(2, 5))) | divide(n3,n0)|divide(n0,n1)|multiply(n2,#0)|subtract(#0,#1)|subtract(n2,#2)|divide(#4,#3)| | general |
the toll t , in dollars , for a truck using a certain bridge is given by the formula t = 1.50 + 1.50 ( x − 2 ) , where x is the number of axles on the truck . what is the toll for an 18 - wheel truck that has 2 wheels on its front axle and 4 wheels on each of its other axles ? | "number of wheels in truck = 18 number of wheels on its front axle = 2 number of wheels remaining = 16 number of axles remaining axles = 16 / 4 = 4 total number of axles = 5 t = 1.50 + 1.50 ( x − 2 ) = 1.50 + 1.5 * 3 = 1.5 + 4.5 = 6 $ answer d" | a ) $ 2.50 , b ) $ 3.00 , c ) $ 3.50 , d ) $ 6.00 , e ) $ 5.00 | d | add(1.50, multiply(1.50, subtract(add(divide(subtract(18, 2), 4), const_1), 2))) | subtract(n3,n2)|divide(#0,n5)|add(#1,const_1)|subtract(#2,n2)|multiply(n1,#3)|add(n0,#4)| | general |
a fraction in reduced form is such that when it is squared and then its numerator is reduced by 33 ( 1 / 3 ) % and denominator is reduced to 20 % , its result is twice the original fraction . the sum of numerator and denominator is : | solution : let the fraction be x / y . when fraction is squared its numerator is reduced by 33 ( 1 / 3 ) and denominator is reduced by 20 % . according to question , ( x / y ) 2 * 33 ( 1 / 3 ) % / 20 % = 2 ( x / y ) . or , ( x / y ) 2 * ( 2 / 3 ) / ( 1 / 5 ) = 2 ( x / y ) . or , x / y = 3 / 5 . sum of numerator and den... | a ) 8 , b ) 13 , c ) 17 , d ) 15 , e ) none of these | a | add(3, divide(const_100, 20)) | divide(const_100,n3)|add(n2,#0) | general |
car a is 24 miles behind car b , which is traveling in the same direction along the same route as car a . car a is traveling at a constant speed of 58 miles per hour and car bis traveling at a constant speed of 50 miles per hour . how many hours will it take for car a to overtake and drive 8 miles ahead of car b ? | "relative speed of car a is 58 - 50 = 8 miles per hour , to catch up 24 miles and drive 8 miles ahead so to drive 32 miles it ' ll need 32 / 8 = 4 hours . answer : e ." | a ) 1.5 , b ) 2.0 , c ) 2.5 , d ) 3.0 , e ) 4.0 | e | divide(add(24, 8), subtract(58, 50)) | add(n0,n3)|subtract(n1,n2)|divide(#0,#1)| | physics |
a rainstorm increased the amount of water stored in state j reservoirs from 180 billion gallons to 200 billion gallons . if the storm increased the amount of water in the reservoirs to 60 percent of total capacity , approximately how many billion gallons of water were the reservoirs short of total capacity prior to the... | after the reservoir is filled to 200 gallons the amount of water is at 60 % - which means that 40 % of the reservoir is empty . to figure out what that 40 % is approximate : 200 gallons / 60 percent = x gallons / 40 percent , therefore , x = 133.33 gallons , answer choices a , b , c , d are below 133.33 . we know that ... | a ) 90 , b ) 114 , c ) 125 , d ) 130 , e ) 144 | e | divide(divide(multiply(200, const_100), 60), const_2) | multiply(n1,const_100)|divide(#0,n2)|divide(#1,const_2)| | general |
find the simple interest for rs . 2000 at 8 % p / a for 2 years . | "s . i = ( 2000 * 8 * 2 ) / 100 p = 320 answer : b" | a ) rs . 300 , b ) 3 rs . 20 , c ) rs . 420 , d ) rs . 520 , e ) rs . 460 | b | multiply(2000, divide(8, const_100)) | divide(n1,const_100)|multiply(n0,#0)| | gain |
16 business executives and 7 chairmen meet at a conference . if each business executive shakes the hand of every other business executive and every chairman once , and each chairman shakes the hand of each of the business executives but not the other chairmen , how many handshakes would take place ? | "there are 16 business exec and in each handshake 2 business execs are involved . hence 16 c 2 = 120 also , each of 16 exec will shake hand with every 7 other chairmen for total of 112 handshake . total = 120 + 112 = 232 ans : c" | a ) 144 , b ) 131 , c ) 232 , d ) 90 , e ) 45 | c | add(divide(multiply(16, subtract(16, const_1)), const_2), multiply(16, 7)) | multiply(n0,n1)|subtract(n0,const_1)|multiply(n0,#1)|divide(#2,const_2)|add(#3,#0)| | geometry |
if two trains are 120 miles apart and are traveling toward each other at constant rate of 30 miles per hour and 40 miles per hour , respectively , how far apart will they be 1 hour before they meet ? | they would meet in 120 / 70 = 12 / 7 hours 1 hour earlier 5 / 7 hours distance between trains = total distance - ( distance covered by train a + distance covered by train b ) = 120 - ( 5 / 7 ( 40 + 30 ) ) [ distance of a = 5 / 7 * 40 ; like wise distance by b ] = 70 km answer : e | a ) 10 , b ) 30 , c ) 40 , d ) 50 , e ) 70 | e | subtract(120, add(multiply(30, subtract(divide(120, add(40, 30)), 1)), multiply(40, subtract(divide(120, add(40, 30)), 1)))) | add(n1,n2)|divide(n0,#0)|subtract(#1,n3)|multiply(n1,#2)|multiply(n2,#2)|add(#3,#4)|subtract(n0,#5) | physics |
a computer is programmed to multiply consecutive even integers 2 * 4 * 6 * 8 * … * n until the product is divisible by 2211 , what is the value of n ? | "factorise 2211 . . 3 * 11 * 67 . . so n has to be a multiple of largest prime number , 61 . . so n = 2 * 61 = 134 . . ans : b" | a ) 22 , b ) 134 , c ) 62 , d ) 122 , e ) 672 | b | multiply(2, divide(divide(2211, add(const_10, const_1)), const_3)) | add(const_1,const_10)|divide(n4,#0)|divide(#1,const_3)|multiply(n0,#2)| | general |
if 30 honey bees make 30 grams of honey in 30 days , then 1 honey bee will make 1 gram of honey in how many days ? | "explanation : let the required number days be x . less honey bees , more days ( indirect proportion ) less honey , less days ( direct proportion ) honey bees 1 : 30 : : 30 : x honey 30 : 1 = > 1 x 30 x x = 30 x 1 x 30 = > x = 30 . answer : c" | a ) 1 , b ) 3.5 , c ) 30 , d ) 49 , e ) 30 | c | divide(1, divide(1, 30)) | divide(n3,n0)|divide(n3,#0)| | physics |
a basket of 1430 apples is divided equally among a group of apple lovers . if 45 people join the group , each apple lover would receive 9 apples less . how many q apples did each person get before 45 people joined the feast ? | "before solving it algebraically , let us prime factorize 1430 = 2 * 5 * 11 * 13 . since number of apples per person * total persons q = 1430 , the answer should be a factor of 1430 . only c is . and that ' s your answer . c" | a ) 20 . , b ) 21 . , c ) 22 . , d ) 23 . , e ) 24 . | c | add(divide(1430, add(divide(1430, add(add(const_10, const_10), const_2)), 45)), 9) | add(const_10,const_10)|add(#0,const_2)|divide(n0,#1)|add(n1,#2)|divide(n0,#3)|add(n2,#4)| | general |
if the sides of a triangle are 9 cm , 16 cm and 19 cm , what is its area ? | "the triangle with sides 9 cm , 16 cm and 19 cm is right angled , where the hypotenuse is 19 cm . area of the triangle = 1 / 2 * 9 * 16 = 72 cm 2 answer : option b" | a ) 69 , b ) 72 , c ) 75 , d ) 89 , e ) 93 | b | divide(multiply(16, 19), const_2) | multiply(n1,n2)|divide(#0,const_2)| | geometry |
the number of positive integers valued pairs ( x , y ) satisfying 4 x - 17 y = 1 and x < = 1000 ( x is less than or equal to 1000 ) is | "we are asked to find integer solutions for x , y . there is no need to apply euclidian algorithm here , we can get particular solutions simply by plugging in some values . as y reaches 3 we get : y ’ = 3 and x ’ = 13 . we need only to solve for x . general solution for x will be : x = x ’ + bn = 13 + 17 n hence lower ... | a ) 59 , b ) 57 , c ) 55 , d ) 58 , e ) 60 | a | divide(divide(add(floor(divide(subtract(multiply(4, 1000), 1), 17)), 1), const_2), const_2) | multiply(n0,n3)|subtract(#0,n2)|divide(#1,n1)|floor(#2)|add(#3,n2)|divide(#4,const_2)|divide(#5,const_2)| | general |
a monkey ascends a greased pole 20 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 10 meters . i mean this will continue for first 18 * 2 = 36 mins . he would have reached 18 meters . after that he will climb 2 meters and he will reach the pole . so total time taken = 36 + 1 = 37 mins . so , asnwer will be c" | a ) 30 th minute , b ) 31 st minute , c ) 37 th minute , d ) 32 nd minute , e ) 33 th minute | c | add(multiply(multiply(const_4, 2), 2), 1) | multiply(n1,const_4)|multiply(n1,#0)|add(n2,#1)| | physics |
a retailer sells 10 shirts . the first 3 he sells for $ 20 , $ 22 and $ 25 . if the retailer wishes to sell the 10 shirts for an overall average price of over $ 20 , what must be the minimum average price of the remaining 7 shirts ? | "first 3 shirts are sold for $ 20 , $ 22 and $ 25 = $ 67 . to get average price of $ 20 , total sale should be 10 * $ 20 = $ 200 so remaining 7 shirts to be sold for $ 200 - $ 67 = $ 133 answer should be 133 / 7 = $ 19.00 that is d" | a ) $ 14.00 , b ) $ 16.00 , c ) $ 17.00 , d ) $ 19.00 , e ) $ 23.00 | d | divide(subtract(multiply(10, const_100.0), add(add(20, 22), 25)), 7) | add(n2,n3)|multiply(n0,n6)|add(n4,#0)|subtract(#1,#2)|divide(#3,n7)| | general |
for any positive integer n , the sum of the first n positive integers equals n ( n + 1 ) / 2 . what is the sum z of all the even integers between 99 and 301 ? | "i would solve it in a different way . first of all , total number of even integers between 99 and 301 are , ( 301 - 99 ) / 2 = 202 / 2 = 101 average = ( 301 + 99 ) / 2 = 400 / 2 = 200 sum z = average * total count = 200 * 101 = 20,200 answer is b" | a ) 10,100 , b ) 20,200 , c ) 22,650 , d ) 40,200 , e ) 45,150 | b | add(divide(subtract(subtract(301, 1), add(99, 1)), 2), 1) | add(n2,n0)|subtract(n3,n0)|subtract(#1,#0)|divide(#2,n1)|add(n0,#3)| | general |
there are 300 employees in a room . 99 % are managers . how many managers must leave the room to bring down the percentage of managers to 98 % ? | there are 297 managers and 3 others . the 3 others would compose 2 % of the total number of people if there were 150 people in the room . thus 150 managers must leave . the answer is c . | a ) 200 , b ) 175 , c ) 150 , d ) 125 , e ) 100 | c | divide(subtract(multiply(300, divide(99, const_100)), multiply(300, divide(98, const_100))), subtract(const_1, divide(98, const_100))) | divide(n1,const_100)|divide(n2,const_100)|multiply(n0,#0)|multiply(n0,#1)|subtract(const_1,#1)|subtract(#2,#3)|divide(#5,#4) | gain |
the ratio of the number of red cars in a certain parking lot to the number of black cars is 3 to 8 . if there are 75 black cars in the lot , how many red cars are there in the lot ? | "b is correct r / b = 3 / 8 and b = 75 r = 75 * 3 / 8 = 28" | a ) 11 , b ) 28 , c ) 24 , d ) 29 , e ) 32 | b | floor(multiply(divide(3, 8), 75)) | divide(n0,n1)|multiply(n2,#0)|floor(#1)| | other |
a total of 50 percent of the geese included in a certain migration study were male . if some of the geese migrated during the study and 20 percent of the migrating geese were male , what was the ratio of the migration rate for the male geese to the migration rate for the female geese ? [ migration rate for geese of a c... | let ' take the number of geese to be 100 . male = 50 . female = 50 . now the second part of the q , let ' s take the number migrated to be 20 . so we have 20 geese that migrated and out of that 20 % are male i . e 20 / 100 * 20 = 4 geese ( males ) and now we know out of the total 20 geese , 4 are male , then 16 have to... | a ) 1 / 4 , b ) 7 / 12 , c ) 2 / 3 , d ) 7 / 8 , e ) 8 / 7 | a | divide(divide(divide(20, const_100), divide(50, const_100)), divide(divide(multiply(multiply(const_2, const_4), const_10), const_100), divide(50, const_100))) | divide(n1,const_100)|divide(n0,const_100)|multiply(const_2,const_4)|divide(#0,#1)|multiply(#2,const_10)|divide(#4,const_100)|divide(#5,#1)|divide(#3,#6) | general |
mukesh weight 56.7 kg . if he reduces his weight in the ratio 7 : 6 . new weight of mukesh is ? | given ratio is 7 : 6 ; let weight before and after reduction be 7 x and 6 x respectively . a / q 7 x = 56.7 = > x = 56.7 / 7 = > x = 8.1 reduced weight = 6 x = 6 * 8.1 = 48.6 ; hence reduced weight = 48.6 answer : b | a ) 47.6 , b ) 48.6 , c ) 49.6 , d ) 50.6 , e ) 51.6 | b | subtract(56.7, divide(56.7, 6)) | divide(n0,n2)|subtract(n0,#0) | other |
a shopkeeper fixes the marked price of an item 30 % above its cost price . the percentage of discount allowed to gain 8 % is | "explanation : let the cost price = rs 100 then , marked price = rs 130 required gain = 8 % , so selling price = rs 108 discount = 130 - 108 = 22 discount % = ( 22 / 130 ) * 100 = 16.92 % option b" | a ) 15.92 % , b ) 16.92 % , c ) 17.92 % , d ) 18.92 % , e ) none of these | b | subtract(const_100, multiply(divide(add(8, const_100), add(30, const_100)), const_100)) | add(n1,const_100)|add(n0,const_100)|divide(#0,#1)|multiply(#2,const_100)|subtract(const_100,#3)| | gain |
maxwell leaves his home and walks toward brad ' s house at the same time that brad leaves his home and runs toward maxwell ' s house . if the distance between their homes is 36 kilometers , maxwell ' s walking speed is 3 km / h , and brad ' s running speed is 6 km / h . what is the distance traveled by maxwell when the... | consider max starts from point a and brad starts from point b and move towards each other . assume they shall meet at point o after time ' t ' . the question asks us to find oa . from the question stem we can make out : - distance oa = 50 km - distance ob = > 3 xt = 36 - 6 xt ( i . e distance = speed x time ) = > 9 t =... | a ) 12 , b ) 14 , c ) 15 , d ) 13 , e ) 11 | a | multiply(3, divide(36, add(3, 6))) | add(n1,n2)|divide(n0,#0)|multiply(n1,#1) | physics |
a meal cost $ 32.50 and there was no tax . if the tip was more than 10 pc but less than 15 pc of the price , then the total amount paid should be : | "10 % ( 32.5 ) = 3.25 15 % ( 32.5 ) = 4.875 total amount could have been 32.5 + 3.25 and 32.5 + 4.875 = > could have been between 35.75 and 37.375 = > approximately between 36 and 37 answer is e ." | a ) 40 - 42 , b ) 39 - 41 , c ) 38 - 40 , d ) 37 - 39 , e ) 36 - 37 | e | add(multiply(32.50, divide(15, const_100)), 32.50) | divide(n2,const_100)|multiply(n0,#0)|add(n0,#1)| | general |
stacy and heather are 15 miles apart and walk towards each other along the same route . stacy walks at constant rate that is 1 mile per hour fast than heather ' s constant rate of 5 miles / hour . if heather starts her journey 24 minutes after stacy , how far from the original destination has heather walked when the tw... | "ss - stacy ' s speed = 6 m / hr sh - heather ' s speed = 5 m / hr in 24 minutes stacy will cover = ( 24 / 60 ) * 6 = 2.4 miles now since both are walking in opposite directions , add their speeds - 6 + 5 = 11 m / hr and distance to cover is 15 - 2.4 = 12.6 time taken = distance / speed = 12.6 / 11 = 1.145 hrs heather ... | a ) 6.5 mile , b ) 5.7 mile , c ) 7.0 mile , d ) 8.0 mile , e ) 12 mile | b | multiply(divide(subtract(15, multiply(24, divide(add(1, 5), const_60))), add(5, add(1, 5))), 5) | add(n1,n2)|add(n2,#0)|divide(#0,const_60)|multiply(n3,#2)|subtract(n0,#3)|divide(#4,#1)|multiply(n2,#5)| | physics |
a group of students decided to collect as many paise from each member of group as is the number of members . if the total collection amounts to rs . 13.69 , the number of the member is the group is : | "money collected = ( 13.69 x 100 ) paise = 1369 paise numbers of members = 1369 squareroot = 37 answer c" | a ) 57 , b ) 67 , c ) 37 , d ) 87 , e ) 97 | c | sqrt(multiply(13.69, const_100)) | multiply(n0,const_100)|sqrt(#0)| | general |
find the area of trapezium whose parallel sides are 10 cm and 18 cm long , and the distance between them is 15 cm . | "explanation : area of a trapezium = 1 / 2 ( sum of parallel sides ) * ( perpendicular distance between them ) = 1 / 2 ( 10 + 18 ) * ( 15 ) = 210 cm 2 answer : option e" | a ) 287 cm 2 , b ) 785 cm 2 , c ) 285 cm 2 , d ) 785 cm 2 , e ) 210 cm 2 | e | quadrilateral_area(15, 18, 10) | quadrilateral_area(n2,n1,n0)| | physics |
how many paying stones , each measuring 2 1 / 2 m * 2 m are required to pave a rectangular court yard 40 m long and 16 1 / 2 m board ? | "40 * 33 / 2 = 5 / 2 * 2 * x = > x = 132 answer : b" | a ) 99 , b ) 132 , c ) 16 , d ) 10 , e ) 15 | b | divide(multiply(40, add(16, divide(1, 2))), multiply(add(2, divide(1, 2)), 2)) | divide(n1,n0)|add(n5,#0)|add(n0,#0)|multiply(n4,#1)|multiply(n0,#2)|divide(#3,#4)| | general |
when a number is divided by 13 , the remainder is 11 . when the same number is divided by 17 , the remainder is 9 . what is the number ? | "x = 13 p + 11 and x = 17 q + 9 13 q + 11 = 17 q + 9 17 q - 13 q = 2 q = 2 + 13 q / 17 the least value of p for which q = 2 + 13 q / 17 is a whole number is p = 26 x = ( 13 * 26 * 11 ) = ( 338 + 11 ) = 349 . answer is e" | a ) 389 , b ) 356 , c ) 380 , d ) 350 , e ) 349 | e | add(multiply(13, add(add(const_10, const_12), const_4)), 11) | add(const_10,const_12)|add(#0,const_4)|multiply(n0,#1)|add(n1,#2)| | general |
if 1 / ( x + 5 ) + 1 / ( x - 5 ) = 1 / ( x - 5 ) , what is the value of x ? | "if we solve the question , we get x = 5 . option : e" | a ) - 1 , b ) 3 , c ) 1 , d ) 2 , e ) 5 | e | subtract(add(5, 1), 1) | add(n1,n0)|subtract(#0,n0)| | general |
how many liters of water must be added to 20 liters of milk and water containing 10 % water to make it 60 % water ? | "by rule of alligation : 60 % - 10 % = 50 % 100 % - 60 % = 40 % quantity of pure water : quantity of the mixture = 5 : 4 there are 20 liters of mixture , so we need to add 25 liters of pure water . the answer is e ." | a ) 12 , b ) 15 , c ) 18 , d ) 22 , e ) 25 | e | multiply(divide(20, 10), divide(60, 10)) | divide(n0,n1)|divide(n2,n1)|multiply(#0,#1)| | gain |
the value of a machine depreciates at 20 % per annum . if its present value is $ 1 , 50,000 , at what price should it be sold after two years such that a profit of $ 24,000 is made ? | "c $ 1 , 20,000 the value of the machine after two years = 0.8 * 0.8 * 1 , 50,000 = $ 96,000 sp such that a profit of $ 24,000 is made = 96,000 + 24,000 = $ 1 , 20,000" | a ) $ 250000 , b ) $ 430000 , c ) $ 120000 , d ) $ 170000 , e ) $ 150000 | c | add(multiply(multiply(subtract(1, divide(20, const_100)), subtract(1, divide(20, const_100))), add(multiply(multiply(const_100, const_100), sqrt(const_100)), multiply(multiply(divide(sqrt(const_100), const_2), const_100), const_100))), multiply(multiply(add(20, const_2), const_100), sqrt(const_100))) | add(n0,const_2)|divide(n0,const_100)|multiply(const_100,const_100)|sqrt(const_100)|divide(#3,const_2)|multiply(#2,#3)|multiply(#0,const_100)|subtract(n1,#1)|multiply(#4,const_100)|multiply(#7,#7)|multiply(#6,#3)|multiply(#8,const_100)|add(#5,#11)|multiply(#12,#9)|add(#13,#10)| | gain |
a certain university will select 1 of 7 candidates eligible to fill a position in the mathematics department and 2 of 12 candidates eligible to fill 2 identical positions in the computer science department . if none of the candidates is eligible for a position in both departments , how many different sets of 3 candidat... | "ans : 462 7 c 1 * 12 c 2 answer e )" | a ) 42 , b ) 70 , c ) 140 , d ) 165 , e ) 462 | e | multiply(multiply(12, 3), 7) | multiply(n3,n5)|multiply(n1,#0)| | other |
an inspector rejects 10 % of the meters as defective . how many will he examine to reject 20 ? | "then , 10 % of x = 20 ( 10 / 100 ) x = 20 x = ( 20 * 100 * ) / 10 = 200 answer is d" | a ) a ) 100 , b ) b ) 120 , c ) c ) 250 , d ) d ) 200 , e ) e ) 160 | d | divide(multiply(20, const_100), 10) | multiply(n1,const_100)|divide(#0,n0)| | gain |
pipe a can fill a tank in 3 hours . due to a leak at the bottom , it takes 6 hours for the pipe a to fill the tank . in what time can the leak alone empty the full tank ? | "let the leak can empty the full tank in x hours 1 / 3 - 1 / x = 1 / 6 = > 1 / x = 1 / 2 - 1 / 6 = 1 / 4.5 = > x = 4.5 answer : c" | a ) 67 , b ) 88 , c ) 4.5 , d ) 26 , e ) 12 | c | divide(multiply(6, 3), subtract(6, 3)) | multiply(n0,n1)|subtract(n1,n0)|divide(#0,#1)| | physics |
the area of a rectangular field is equal to 200 square meters . its perimeter is equal to 60 meters . find the width of this rectangle . | "l * w = 200 : area , l is the length and w is the width . 2 l + 2 w = 60 : perimeter l = 30 - w : solve for l ( 30 - w ) * w = 200 : substitute in the area equation w = 10 and l = 20 : correct answer b" | a ) 5 , b ) 10 , c ) 15 , d ) 20 , e ) 25 | b | divide(subtract(divide(60, const_2), sqrt(subtract(multiply(divide(60, const_2), divide(60, const_2)), multiply(const_4, 200)))), const_2) | divide(n1,const_2)|multiply(n0,const_4)|multiply(#0,#0)|subtract(#2,#1)|sqrt(#3)|subtract(#0,#4)|divide(#5,const_2)| | geometry |
carmen made a sculpture from small pieces of wood . the sculpture is 2 feet 10 inches tall . carmen places her sculpture on a base that is 2 inches tall . how tall are the sculpture andbase together ? | "we know 1 feet = 12 inch then 2 feet = 24 inch 24 + 10 = 34 then 34 + 2 = 36 36 / 12 = 3.0 feet answer : d" | a ) 3.1 feet , b ) 3.2 feet , c ) 3.3 feet , d ) 3.0 feet , e ) 3.5 feet | d | divide(add(add(multiply(add(2, 10), 2), 10), 2), add(2, 10)) | add(n0,n1)|multiply(n0,#0)|add(#1,n1)|add(n2,#2)|divide(#3,#0)| | geometry |
if the simple interest on a sum of money for 2 years at 5 % per annum is rs . 56 , what is the compound interest on the same sum at the rate and for the same time ? | "explanation : sum = ( 56 * 100 ) / ( 2 * 5 ) = rs . 560 amount = [ 560 * ( 1 + 5 / 100 ) 2 ] = rs . 617.4 c . i . = ( 617.4 - 560 ) = rs . 57.4 . answer : a" | a ) s . 57.40 , b ) s . 57.22 , c ) s . 51.219 , d ) s . 56.18 , e ) s . 53.11 | a | subtract(add(add(divide(multiply(divide(56, multiply(divide(5, const_100), 2)), 5), const_100), divide(56, multiply(divide(5, const_100), 2))), divide(multiply(add(divide(multiply(divide(56, multiply(divide(5, const_100), 2)), 5), const_100), divide(56, multiply(divide(5, const_100), 2))), 5), const_100)), divide(56, 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 |
18 men working 8 hours per day dig 30 m deep . how many extra men should be put to dig to a depth of 50 m working 6 hours per day ? | "( 18 * 8 ) / 30 = ( x * 6 ) / 50 = > x = 40 40 – 18 = 22 answer : a" | a ) 22 , b ) 66 , c ) 88 , d ) 100 , e ) 281 | a | subtract(divide(multiply(divide(multiply(18, 8), 30), 50), 6), 18) | multiply(n0,n1)|divide(#0,n2)|multiply(n3,#1)|divide(#2,n4)|subtract(#3,n0)| | physics |
the perimeter of a rectangle is equal to 280 meters . the ratio of its length to its width is 5 : 2 . find the area of the rectangle . | solution if the ratio of the length to the width is 5 : 2 , then the measure l of the length and and the measure w of the with can be written as l = 5 x and w = 2 x we now use the perimeter to write 280 = 2 ( 2 l + 2 w ) = 2 ( 5 x + 2 x ) = 14 x solve for x 280 = 14 x x = 280 / 14 = 20 the area a of the rectangle is gi... | ['a ) 400 square meters', 'b ) 4000 square meters', 'c ) 20 square meters', 'd ) 2000 square meters', 'e ) 1000 square meters'] | b | multiply(multiply(power(divide(280, multiply(add(5, 2), const_2)), const_2), 5), 2) | add(n1,n2)|multiply(#0,const_2)|divide(n0,#1)|power(#2,const_2)|multiply(n1,#3)|multiply(n2,#4) | geometry |
at the end of year x , automobile installment credit accounted for 36 % of all outstanding consumer installment credit . at that time automobile finance companies extended $ 35 billion of credit , or 1 / 3 of the automobile installment credit . how many billion dollars of consumer installment credit was outstanding at ... | "system of equations a = ( 36 / 100 ) c ( 1 / 3 ) a = 35 - - > a = 105 substitution 105 = ( 36 / 100 ) c c = ( 100 / 36 ) 105 - you can do the ugly calculation 105 / 36 * 100 the correct answer is a . the correct answer is 291.67" | a ) 291.67 , b ) 105 , c ) 301.05 , d ) 288 , e ) 249.55 | a | divide(multiply(35, 3), divide(36, const_100)) | divide(n0,const_100)|multiply(n1,n3)|divide(#1,#0)| | general |
the cross - section of a tunnel is a trapezium in shape . if the tunnel is 15 m wide at the top and 5 m wide at the bottom and the area of cross - section is 400 sq m , the depth of tunnel is ? | 1 / 2 * d ( 15 + 5 ) = 400 d = 400 answer : c | a ) 200 m , b ) 600 m , c ) 400 m , d ) 800 m , e ) 900 m | c | multiply(divide(400, divide(add(15, 5), const_2)), const_10) | add(n0,n1)|divide(#0,const_2)|divide(n2,#1)|multiply(#2,const_10) | physics |
if j is the least positive integer that is divisible by every integer from 10 to 15 inclusive , then j / 26 is = | the integer should be divisible by : 10 , 11 , 12 , 13 , 14 and 15 , that is 5 * 211 , 3 * 2 ^ 2 , 13 , 2 * 7 , and 3 * 5 . the least common multiple of these integers is the product of 13 , 11 , 7 , 5 , 3 and 2 ^ 2 . then , r / 26 is ( 13 * 11 * 7 * 5 * 3 * 2 ^ 2 ) / 2 * 13 = 11 * 7 * 5 * 3 * 2 = 2310 . answer : b . | a ) 1155 , b ) 2310 , c ) 4620 , d ) 30030 , e ) 60060 | b | divide(divide(multiply(multiply(multiply(10, add(10, const_1)), add(add(10, const_1), const_1)), add(add(add(add(10, const_1), const_1), const_1), const_1)), const_2), const_4) | add(n0,const_1)|add(#0,const_1)|multiply(n0,#0)|add(#1,const_1)|multiply(#1,#2)|add(#3,const_1)|multiply(#5,#4)|divide(#6,const_2)|divide(#7,const_4) | general |
john purchased a grinder and a mobile for rs . 15000 & rs . 8000 respectively . he sold the grinder at a loss of 4 % and the mobile phone at a profit of 20 % . overall how much he make aprofit . | "let the sp of the refrigerator and the mobile phone be rs . r and rs . m respectively . r = 15000 ( 1 - 4 / 100 ) = 15000 - 600 m = 8000 ( 1 + 20 / 100 ) = 8000 + 1600 total sp - total cp = r + m - ( 15000 + 8000 ) = - 600 + 1600 = rs . 1200 as this is positive , an overall profit of rs . 1200 was made . c" | a ) s . 190 , b ) s . 1120 , c ) s . 1200 , d ) s . 1250 , e ) s . 1290 | c | subtract(multiply(20, divide(8000, const_100)), multiply(4, divide(15000, const_100))) | divide(n1,const_100)|divide(n0,const_100)|multiply(n3,#0)|multiply(n2,#1)|subtract(#2,#3)| | gain |
the area of sector of a circle whose radius is 18 metro and whose angle at the center is 42 â ° is ? | "42 / 360 * 22 / 7 * 18 * 18 = 118.8 m 2 answer : d" | a ) 52.6 , b ) 52.9 , c ) 52.8 , d ) 118.8 , e ) 52.2 | d | multiply(multiply(power(18, 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 |
the amounts of time that three secretaries worked on a special project are in the ratio of 1 to 2 to 5 . if they worked a combined total of 120 hours , how many hours did the secretary who worked the longest spend on the project ? | "8 x = 120 = > x = 15 therefore the secretary who worked the longest spent 15 x 5 = 75 hours on the project option ( c )" | a ) 80 , b ) 70 , c ) 75 , d ) 16 , e ) 14 | c | multiply(divide(120, add(add(1, 2), 5)), 5) | add(n0,n1)|add(n2,#0)|divide(n3,#1)|multiply(n2,#2)| | physics |
a number of 44 marbles is to be divided and contain with boxes . if each box is to contain 3 , 4 , or 5 marbles , what is the largest possible number of boxes ? | "to maximize # of boxes we should minimize marbles per box : 13 * 3 + 1 * 5 = 44 - - > 13 + 1 = 14 . answer b" | a ) 10 , b ) 14 , c ) 15 , d ) 16 , e ) 17 | b | divide(44, 3) | divide(n0,n1)| | general |
danny is sitting on a rectangular box . the area of the front face of the box is half the area of the top face , and the area of the top face is 1.5 times the area of the side face . if the volume of the box is 192 , what is the area of the side face of the box ? | "lets suppose length = l , breadth = b , depth = d front face area = l * w = 1 / 2 w * d ( l = 1 / 2 d or d = 2 l ) top face area = w * d side face area = w * d = 1.5 d * l ( w = 1.5 l ) volume = l * w * d = 192 l * 1.5 l * 2 l = 192 l = 4 side face area = l * d = l * 2 l = 4 * 2 * 4 = 32 e is the answer" | a ) 13 , b ) 16 , c ) 18 , d ) 29 , e ) 32 | e | divide(power(multiply(const_3, power(192, const_2)), divide(const_1, const_3)), 1.5) | divide(const_1,const_3)|power(n1,const_2)|multiply(#1,const_3)|power(#2,#0)|divide(#3,n0)| | geometry |
diana is painting statues . she has 1 / 2 of a gallon of paint remaining . each statue requires 1 / 4 gallon of paint . how many statues can she paint ? | "number of statues = all the paint ÷ amount used per statue = 1 / 2 ÷ 1 / 4 = 1 / 2 * 4 / 1 = 2 answer is a" | a ) 2 , b ) 20 , c ) 28 , d ) 14 , e ) 19 | a | divide(divide(1, 2), divide(1, 4)) | divide(n0,n1)|divide(n2,n3)|divide(#0,#1)| | general |
find the average of first 15 natural numbers . | explanation : sum of first n natural numbers = n ( n + 1 ) / 2 hence , sum of first 15 natural numbers = ( 15 x 16 ) / 2 = 120 therefore , required average of = 120 / 15 = 8 answer : d | a ) 10 , b ) 120 , c ) 15 , d ) 8 , e ) 16 | d | divide(add(add(add(add(add(add(add(add(add(add(add(add(add(add(const_1, const_2), const_3), const_4), add(const_4, const_1)), add(add(const_1, const_2), const_3)), add(const_4, const_3)), add(const_4, const_4)), add(add(const_4, const_4), const_1)), const_10), add(const_10, const_1)), const_12), add(const_12, const_1))... | add(const_1,const_2)|add(const_1,const_4)|add(const_3,const_4)|add(const_4,const_4)|add(const_1,const_10)|add(const_1,const_12)|add(#0,const_3)|add(#3,const_1)|add(#5,const_1)|add(#6,const_4)|add(#9,#1)|add(#10,#6)|add(#11,#2)|add(#12,#3)|add(#13,#7)|add(#14,const_10)|add(#15,#4)|add(#16,const_12)|add(#17,#5)|add(#18,#... | general |
the “ a - number ” of a number x is defined as the ones digit of 2 ^ x . antony rolls a die with 6 sides labeled with the integers from 1 to 6 , each of which has an equal probability of landing face - up . he then takes 3 ^ a , where a is the a - number of the result of his die roll , and plots 3 ^ a on a number line ... | if you calculate 3 ^ a for 1 st roll , all 6 results will be 9 , 81 , 6561 , 729 , 9 , 81 . this result is the same for 2 nd roll . 9 , 81 , 6561 , 729 , 9 , 81 . about distance : if the first result is 9 and the second is also 9 , the distance is 9 - 9 = 0 which is smaller than 9 . if the first result is 9 and the sec... | a ) 3 / 8 , b ) 13 / 36 , c ) 17 / 36 , d ) 19 / 36 , e ) 23 / 36 | b | divide(add(add(add(add(const_4, 2), 1), 2), const_4), multiply(6, 6)) | add(n0,const_4)|multiply(n1,n1)|add(n2,#0)|add(n0,#2)|add(#3,const_4)|divide(#4,#1) | general |
you and your friend spent a total of $ 17 for lunch . your friend spent $ 3 more than you . how much did your friend spend on their lunch ? | "my lunch = l , my friends lunch = l + 3 ( l ) + ( l + 3 ) = 17 l + l + 3 - 3 = 17 - 3 2 l = 14 l = 7 my friends lunch l + 3 = 7 + 3 = $ 10 , the answer is d" | a ) $ 9 , b ) $ 3 , c ) $ 4 , d ) $ 10 , e ) $ 5 | d | add(divide(subtract(17, 3), const_2), 3) | subtract(n0,n1)|divide(#0,const_2)|add(n1,#1)| | general |
a man buy a book in rs 50 & sale it rs 60 . what is the rate of profit ? ? ? | "cp = 50 sp = 60 profit = 60 - 50 = 10 % = 10 / 50 * 100 = 20 % answer : b" | a ) 10 % , b ) 20 % , c ) 30 % , d ) 25 % , e ) 28 % | b | multiply(divide(subtract(60, 50), 50), const_100) | subtract(n1,n0)|divide(#0,n0)|multiply(#1,const_100)| | gain |
a man complete a journey in 1010 hours . he travels first half of the journey at the rate of 2121 km / hr and second half at the rate of 2424 km / hr . find the total journey in km . | "solution 1 average speed = 2 ã — 21 ã — 2421 + 24 = 22.4 km / hr = 2 ã — 21 ã — 2421 + 24 = 22.4 km / hr total distance = 22.4 ã — 10 = 224 km answer is c" | a ) 121 km , b ) 242 km , c ) 224 km , d ) 112 km , e ) 110 km | c | multiply(const_2, divide(multiply(multiply(2121, 2424), 1010), add(2121, 2424))) | add(n1,n2)|multiply(n1,n2)|multiply(n0,#1)|divide(#2,#0)|multiply(#3,const_2)| | physics |
the ratio of investments of two partners p and q is 7 : 5 and the ratio of their profits is 7 : 10 . if p invested the money for 20 months , find for how much time did q invest the money ? | "7 * 5 : 20 * x = 7 : 10 x = 40 answer : b" | a ) 11 , b ) 40 , c ) 18 , d ) 16 , e ) 12 | b | multiply(multiply(divide(7, 5), divide(10, 7)), 20) | divide(n0,n1)|divide(n3,n2)|multiply(#0,#1)|multiply(n4,#2)| | gain |
pipe a fills a tank in 56 minutes . pipe b can fill the same tank 7 times as fast as pipe a . if both the pipes are kept open when the tank is empty , how many minutes will it take to fill the tank ? | a ' s rate is 1 / 56 and b ' s rate is 1 / 8 . the combined rate is 1 / 56 + 1 / 8 = 1 / 7 the pipes will fill the tank in 7 minutes . the answer is d . | a ) 4 , b ) 5 , c ) 6 , d ) 7 , e ) 8 | d | inverse(add(divide(const_1, 56), divide(7, 56))) | divide(const_1,n0)|divide(n1,n0)|add(#0,#1)|inverse(#2) | physics |
a contractor undertakes to built a walls in 50 days . he employs 20 peoples for the same . however after 25 days he finds that only 40 % of the work is complete . how many more man need to be employed to complete the work in time ? | 20 men complete 0.4 work in 25 days . applying the work rule , m 1 × d 1 × w 2 = m 2 × d 2 × w 1 we have , 20 × 25 × 0.6 = m 2 × 25 × 0.4 or m 2 = 20 × 25 × 0.6 / 25 × 0.4 = 30 men answerd | a ) 25 , b ) 30 , c ) 35 , d ) 30 , e ) none of these | d | divide(multiply(20, divide(subtract(const_100, 40), const_100)), divide(const_4, const_10)) | divide(const_4,const_10)|subtract(const_100,n3)|divide(#1,const_100)|multiply(n1,#2)|divide(#3,#0) | physics |
find the value for m ? 19 ( m + n ) + 17 = 19 ( - m + n ) - 135 | 19 m + 19 n + 17 = - 19 m + 19 n - 135 38 m = - 152 = > m = - 4 e | a ) 0 , b ) - 1 , c ) 1 , d ) 2 , e ) - 4 | e | add(135, 17) | add(n1,n3) | general |
there are 720 students in a school . the ratio of boys and girls in this school is 3 : 5 . find the total of girls & boys are there in this school ? | "in order to obtain a ratio of boys to girls equal to 3 : 5 , the number of boys has to be written as 3 x and the number of girls as 5 x where x is a common factor to the number of girls and the number of boys . the total number of boys and girls is 720 . hence 3 x + 5 x = 720 solve for x 8 x = 720 x = 90 number of boy... | a ) 320 , b ) 345 , c ) 450 , d ) 380 , e ) 400 | c | multiply(divide(720, 5), 3) | divide(n0,n2)|multiply(n1,#0)| | other |
the length of a rectangle is two - fifths of the radius of a circle . the radius of the circle is equal to the side of the square , whose area is 2500 sq . units . what is the area ( in sq . units ) of the rectangle if the rectangle if the breadth is 10 units ? | "given that the area of the square = 2500 sq . units = > side of square = √ 2500 = 50 units the radius of the circle = side of the square = 50 units length of the rectangle = 2 / 5 * 50 = 20 units given that breadth = 10 units area of the rectangle = lb = 20 * 10 = 200 sq . units answer : option e" | a ) 140 , b ) 150 , c ) 160 , d ) 170 , e ) 200 | e | multiply(10, multiply(const_2, divide(sqrt(2500), divide(10, const_2)))) | divide(n1,const_2)|sqrt(n0)|divide(#1,#0)|multiply(#2,const_2)|multiply(n1,#3)| | geometry |
convert 1.6 hectares in ares | "1.6 hectares in ares 1 hectare = 100 ares therefore , 1.6 hectares = 1.6 × 100 ares = 160 ares . answer - b" | a ) 130 ares . , b ) 160 ares . , c ) 180 ares . , d ) 230 ares . , e ) 250 ares . | b | divide(multiply(multiply(multiply(add(const_3, const_2), const_2), multiply(add(const_3, const_2), const_2)), 1.6), 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 average of 7 numbers is 15 . if each number be multiplied by 5 . find the average of new set of numbers ? | explanation : average of new numbers = 15 * 5 = 75 answer : option e | a ) a ) 110 , b ) b ) 122 , c ) c ) 90 , d ) d ) 85 , e ) e ) 75 | e | multiply(15, 5) | multiply(n1,n2) | general |
calculate the speed of a boat in still water ( in km / hr ) if in one hour , the boat goes 15 km / hr downstream and 7 km / hr upstream . | "speed in still water = ( 15 + 7 ) 1 / 2 kmph = 11 kmph . answer : e" | a ) 12 kmph , b ) 13 kmph , c ) 14 kmph , d ) 15 kmph , e ) 11 kmph | e | divide(add(15, 7), const_2) | add(n0,n1)|divide(#0,const_2)| | physics |
a and b go around a circular track of length 400 m on a cycle at speeds of 36 kmph and 36 kmph . after how much time will they meet for the first time at the starting point ? | "time taken to meet for the first time at the starting point = lcm { length of the track / speed of a , length of the track / speed of b } = lcm { 400 / ( 36 * 5 / 18 ) , 400 / ( 36 * 5 / 18 ) } = lcm ( 40 , 40 ) = 40 sec . answer : a" | a ) 40 sec , b ) 198 sec , c ) 178 sec , d ) 665 sec , e ) 276 sec | a | divide(400, subtract(multiply(36, const_0_2778), multiply(36, const_0_2778))) | multiply(n2,const_0_2778)|multiply(n1,const_0_2778)|subtract(#0,#1)|divide(n0,#2)| | physics |
a circle graph shows how the megatech corporation allocates its research and development budget : 9 % microphotonics ; 14 % home electronics ; 10 % food additives ; 29 % genetically modified microorganisms ; 8 % industrial lubricants ; and the remainder for basic astrophysics . if the arc of each sector of the graph is... | "9 % microphotonics ; 14 % home electronics ; 10 % food additives ; 29 % genetically modified microorganisms ; 8 % industrial lubricants ; 100 - ( 9 + 14 + 10 + 29 + 8 ) = 30 % basic astrophysics . 30 % of 360 ° is 108 ° . answer : a ." | a ) 108 ° , b ) 10 ° , c ) 18 ° , d ) 36 ° , e ) 52 ° | a | divide(multiply(subtract(const_100, add(add(add(add(9, 14), 10), 29), 8)), divide(const_3600, const_10)), const_100) | add(n0,n1)|divide(const_3600,const_10)|add(n2,#0)|add(n3,#2)|add(n4,#3)|subtract(const_100,#4)|multiply(#1,#5)|divide(#6,const_100)| | gain |
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