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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if 15 % of 40 is greater than 25 % of a number by 2 , the number is | "explanation : 15 / 100 * 40 - 25 / 100 * x = 2 or x / 4 = 4 so x = 16 option b" | a ) 14 , b ) 16 , c ) 18 , d ) 20 , e ) 22 | b | divide(subtract(multiply(divide(15, const_100), 40), 2), divide(25, const_100)) | divide(n0,const_100)|divide(n2,const_100)|multiply(n1,#0)|subtract(#2,n3)|divide(#3,#1)| | gain |
john purchased 1325 large bottles at $ 1.89 per bottle and 750 small bottles at $ 1.38 per bottle . what was the approximate average price paid per bottle ? | "( 1325 * 1.89 + 750 * 1.38 ) / ( 1325 + 750 ) = ~ 1.71 option ( b )" | a ) $ 1.63 , b ) $ 1.71 , c ) $ 1.68 , d ) $ 1.72 , e ) $ 1.76 | b | divide(add(multiply(1325, 1.89), multiply(750, 1.38)), add(1325, 750)) | add(n0,n2)|multiply(n0,n1)|multiply(n2,n3)|add(#1,#2)|divide(#3,#0)| | general |
find the smallest number of six digits exactly divisible by 25 , 35,45 and 15 . | "smallest number of six digits is 100000 . required number must be divisible by l . c . m . of 25,35 , 45,15 i . e 1575 , on dividing 100000 by 1575 , we get 800 as remainder . therefore , required number = 100000 + ( 1575 Γ’ β¬ β 800 ) = 100775 . answer is b ." | a ) 100555 , b ) 100775 , c ) 100885 , d ) 100995 , e ) 100665 | b | add(subtract(multiply(const_10, multiply(const_100, const_100)), const_100), 35,45) | multiply(const_100,const_100)|multiply(#0,const_10)|subtract(#1,const_100)|add(n1,#2)| | general |
the moon revolves around the earth at a speed of approximately 1.05 kilometers per second . this approximate speed is how many kilometers per hour ? | moon revolves around the earth at a speed of 1.02 kilometers per second . one hour equal to 60 minutes . one minute equals to 60 seconds . so one hour equals to 3600 seconds . so one hour , speed = 1.05 * 3600 = 3780 kilometers per hour . option d is correct | a ) 60 , b ) 61.2 , c ) 62.5 , d ) 3,780 , e ) 3,672 | d | multiply(1.05, const_3600) | multiply(n0,const_3600) | physics |
the speed of a car is 100 km in the first hour and 60 km in the second hour . what is the average speed of the car ? | "s = ( 100 + 60 ) / 2 = 80 kmph answer : c" | a ) 12 , b ) 75 , c ) 80 , d ) 54 , e ) 15 | c | divide(add(100, 60), const_2) | add(n0,n1)|divide(#0,const_2)| | physics |
a sum of money is put out at compound interest for 2 years at 20 % . it would fetch rs . 1446 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 = 1446 p = 60000 answer : d" | a ) 32200 , b ) 41897 , c ) 58799 , d ) 60000 , e ) 62782 | 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 man buys a cycle for rs . 2000 and sells it at a loss of 10 % . what is the selling price of the cycle ? | "s . p . = 90 % of rs . 2000 = 90 / 100 x 2000 = rs . 1800 answer : a" | a ) 1800 , b ) 1420 , c ) 1430 , d ) 1440 , e ) 1540 | a | divide(multiply(subtract(const_100, 10), 2000), const_100) | subtract(const_100,n1)|multiply(n0,#0)|divide(#1,const_100)| | gain |
a runs twice as fast as b and gives b a start of 71 m . how long should the racecourse be so that a and b might reach in the same time ? | "ratio of speeds of a and b is 2 : 1 b is 71 m away from a but we know that a covers 1 meter ( 2 - 1 ) more in every second than b the time taken for a to cover 71 m is 71 / 1 = 71 m so the total time taken by a and b to reach = 2 * 71 = 142 m answer : d" | a ) 75 m . , b ) 80 m . , c ) 150 m . , d ) 142 m . , e ) none of the above | d | multiply(71, const_2) | multiply(n0,const_2)| | physics |
a and b began business with rs . 3000 and rs . 4000 after 8 months , a withdraws rs . 1000 and b advances rs . 1000 more . at the end of the year , their profits amounted to rs . 630 find the share of b . | "( 3 * 8 + 2 * 4 ) : ( 4 * 8 + 5 * 4 ) 8 : 13 13 / 21 * 630 = 390 answer : d" | a ) 240 , b ) 288 , c ) 277 , d ) 390 , e ) 361 | d | multiply(divide(630, add(add(multiply(3000, 8), multiply(subtract(3000, 1000), subtract(const_12, 8))), add(multiply(4000, 8), multiply(add(4000, 1000), subtract(const_12, 8))))), add(multiply(3000, 8), multiply(subtract(3000, 1000), subtract(const_12, 8)))) | add(n1,n3)|multiply(n0,n2)|multiply(n1,n2)|subtract(n0,n3)|subtract(const_12,n2)|multiply(#3,#4)|multiply(#0,#4)|add(#1,#5)|add(#2,#6)|add(#7,#8)|divide(n5,#9)|multiply(#7,#10)| | gain |
a reduction of 20 % in the price of salt enables a lady to obtain 10 kgs more for rs . 500 , find the original price per kg ? | "100 * ( 20 / 100 ) = 20 - - - 10 ? - - - 1 = > rs . 2 500 - - - 80 ? - - - 2 = > rs . 12.5 answer : c" | a ) s . 12.7 , b ) s . 12.4 , c ) s . 12.5 , d ) s . 12.1 , e ) s . 12.9 | c | multiply(divide(divide(multiply(divide(20, const_100), 500), 10), multiply(divide(20, const_100), 500)), const_100) | divide(n0,const_100)|multiply(n2,#0)|divide(#1,n1)|divide(#2,#1)|multiply(#3,const_100)| | gain |
evaluate : 59 - 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 59 - 12 * 3 * 2 = 59 - 8 = 51 correct answer b" | a ) 42 , b ) 51 , c ) 62 , d ) 72 , e ) 82 | b | subtract(59, multiply(multiply(12, 3), 2)) | multiply(n1,n2)|multiply(n3,#0)|subtract(n0,#1)| | general |
the average age of a class of 20 students is 22 years . the average increased by 2 when the teacher ' s age also included . what is the age of the teacher ? | "if age of the teacher was 22 , average would not have changed . since average increased by 2 , age of the teacher = 19 + 22 Γ 1 = 44 answer : e" | a ) 40 years , b ) 41 years , c ) 42 years , d ) 43 years , e ) 44 years | e | subtract(multiply(add(20, 2), add(22, 2)), multiply(20, 22)) | add(n0,n2)|add(n1,n2)|multiply(n0,n1)|multiply(#0,#1)|subtract(#3,#2)| | general |
the length of a rectangle is 18 cm and its breadth is 10 cm . when the length is increased to 25 cm , what will be the breadth of the rectangle if the area remains the same ? | solution let the breadth be b . then , β 25 b = 18 Γ 10 β b = ( 18 x 10 / 25 ) cm = 7.2 cm answer c | ['a ) 7 cm', 'b ) 7.1 cm', 'c ) 7.2 cm', 'd ) 7.3 cm', 'e ) none of these'] | c | divide(multiply(18, 10), 25) | multiply(n0,n1)|divide(#0,n2) | geometry |
a whale goes on a feeding frenzy that lasts for 9 hours . for the first hour he catches and eats x kilos of plankton . in every hour after the first , it consumes 3 kilos of plankton more than it consumed in the previous hour . if by the end of the frenzy the whale will have consumed a whopping accumulated total 540 ki... | "if you list the amount eaten each hour , you ' ll get an equally spaced list , increasing by 3 each hour . in any equally spaced list , the median equals the mean . here , the mean is 540 / 9 = 60 , so the median is also 60 , and that is the amount eaten in the 5 th hour . we need to add 3 to find the total eaten in t... | a ) 38 , b ) 47 , c ) 50 , d ) 53 , e ) 63 | e | add(divide(540, 9), 3) | divide(n2,n0)|add(n1,#0)| | general |
walking at 70 % of his usual speed a man takes 24 minutes more to cover a distance . what is his usual time to cover this distance ? | "speed is inversly proprtional to time walking at 70 % of speed meand 7 / 10 s takes 10 / 7 t . it takes 24 minutes extra to cover the distance . then 10 / 7 t = t + 24 10 t = 7 t + 168 3 t = 168 t = 56 . option e is correct" | a ) 30 , b ) 36 , c ) 42 , d ) 48 , e ) 56 | e | divide(24, subtract(divide(const_1, divide(70, const_100)), const_1)) | divide(n0,const_100)|divide(const_1,#0)|subtract(#1,const_1)|divide(n1,#2)| | physics |
the profit earned by selling an article for rs . 832 is equal to the loss incurred when the same article is sold for rs . 448 . what should be the sale price for making 40 % profit ? | "c . p . = rs . x . then , 832 - x = x - 448 2 x = 1280 = > x = 640 required s . p . = 140 % of rs . 640 = 140 / 100 * 640 = rs . 896 . e" | a ) 220 , b ) 230 , c ) 960 , d ) 780 , e ) 896 | e | multiply(subtract(832, divide(subtract(832, 448), const_2)), add(const_1, divide(40, const_100))) | divide(n2,const_100)|subtract(n0,n1)|add(#0,const_1)|divide(#1,const_2)|subtract(n0,#3)|multiply(#2,#4)| | gain |
two employees x and y are paid a total of rs . 880 per week by their employer . if x is paid 120 percent of the sum paid to y , how much is y paid per week ? | "let the amount paid to x per week = x and the amount paid to y per week = y then x + y = 880 but x = 120 % of y = 120 y / 100 = 12 y / 10 Γ’ Λ Β΄ 12 y / 10 + y = 880 Γ’ β‘ β y [ 12 / 10 + 1 ] = 880 Γ’ β‘ β 22 y / 10 = 880 Γ’ β‘ β 22 y = 8800 Γ’ β‘ β y = 8800 / 22 = 800 / 2 = rs . 400 e )" | a ) s . 150 , b ) s . 200 , c ) s . 250 , d ) s . 350 , e ) s . 400 | e | divide(multiply(880, multiply(add(const_1, const_4), const_2)), multiply(add(multiply(add(const_1, const_4), const_2), const_1), const_2)) | add(const_1,const_4)|multiply(#0,const_2)|add(#1,const_1)|multiply(n0,#1)|multiply(#2,const_2)|divide(#3,#4)| | general |
two trains start from p and q respectively and travel towards each other at a speed of 50 km / hr and 40 km / hr respectively . by the time they meet , the first train has travelled 1 ookm more than the second . the distance between p and q is | "explanation : at the time of meeting , let the distance travelled by the second train be x km then , distance covered by the first train is ( x + 100 ) km = > x / 40 = ( x + 100 ) / 50 = > 50 x = 40 x + 4000 = > x = 400 km so , distance between p and q = ( x + x + 100 ) km = 900 km answer : option d" | a ) 500 km , b ) 600 km , c ) 700 km , d ) 900 km , e ) 400 km | d | add(add(divide(multiply(const_100.0, 40), subtract(50, 40)), 1), divide(multiply(1, 40), subtract(50, 40))) | multiply(n1,const_100.0)|subtract(n0,n1)|divide(#0,#1)|add(n2,#2)|add(#3,#2)| | physics |
a certain store sold pens for $ 0.35 each and pencils for $ 0.25 each . if a customer purchased both pens and pencils from the store for a total of $ 3.50 , what total number of pens and pencils did the customer purchase ? | "answer : algebraically , the question looks like this : 2.5 = 0.35 x + 0.25 y there are six possible numbers of pens that fit that requirement : 1 pen : $ 0.35 2 pens : $ 0.70 3 pens : $ 1.05 4 pens : $ 1.40 5 pens : $ 1.75 balance can fetch 7 pencils for $ 0.25 each . that ' s 5 pens and 7 pencils for a total of 12 p... | a ) 8 , b ) 9 , c ) 10 , d ) 11 , e ) 12 | e | multiply(divide(3.50, add(0.35, 0.25)), const_2) | add(n0,n1)|divide(n2,#0)|multiply(#1,const_2)| | other |
dan β s car gets 32 miles per gallon . if gas costs $ 4 / gallon , then how many miles can dan β s car go on $ 42 of gas ? | "42 / 4 = 10.5 gallons 10.5 * 32 = 336 miles the answer is c ." | a ) 196 , b ) 284 , c ) 336 , d ) 412 , e ) 572 | c | divide(multiply(42, 32), 4) | multiply(n0,n2)|divide(#0,n1)| | physics |
aarti can do a piece of work in 8 days . in how many days will she complete three time of work of same type ? | "we have the important relation , more work , more time ( days ) a piece of work can be done in 8 days . three times of work of same type can be done in 8 x 3 = 24 days answer c" | a ) 6 days , b ) 19 days , c ) 24 days , d ) 3 days , e ) 13 days | c | multiply(const_3, 8) | multiply(n0,const_3)| | physics |
in a box of 11 pencils , a total of 2 are defective . if a customer buys 3 pencils selected at random from the box , what is the probability that neither pencils will be defective ? | "first , there are 9 c 3 ways you can select 3 good pencils from 4 good ones . second , there are 11 c 3 ways you select 3 pencils from 6 ones in the box . then , the probability that neither pen will be defective is : 9 c 3 / 11 c 3 = 84 / 165 = 28 / 55 answer is b" | a ) 1 / 25 , b ) 28 / 55 , c ) 2 / 35 , d ) 2 / 55 , e ) 1 / 75 | b | divide(divide(factorial(subtract(11, 2)), multiply(factorial(subtract(subtract(11, 2), 3)), factorial(3))), divide(factorial(11), multiply(factorial(subtract(11, 3)), factorial(3)))) | factorial(n2)|factorial(n0)|subtract(n0,n1)|subtract(n0,n2)|factorial(#2)|factorial(#3)|subtract(#2,n2)|factorial(#6)|multiply(#5,#0)|divide(#1,#8)|multiply(#7,#0)|divide(#4,#10)|divide(#11,#9)| | general |
the average salary of a person for the months of january , february , march and april is rs . 8000 and that for the months february , march , april and may is rs . 8300 . if his salary for the month of may is rs . 6500 , find his salary for the month of january ? | sum of the salaries of the person for the months of january , february , march and april = 4 * 8000 = 32000 - - - - ( 1 ) sum of the salaries of the person for the months of february , march , april and may = 4 * 8300 = 33200 - - - - ( 2 ) ( 2 ) - ( 1 ) i . e . may - jan = 1200 salary of may is rs . 6500 salary of janu... | a ) 2177 , b ) 2876 , c ) 4500 , d ) 5300 , e ) 6711 | d | subtract(multiply(8000, const_4), subtract(multiply(8300, const_4), 6500)) | multiply(n0,const_4)|multiply(n1,const_4)|subtract(#1,n2)|subtract(#0,#2) | general |
by investing rs . 1800 in 9 % stock , syam earns rs . 120 . find out the stock is then quoted at which rate ? | "assume that face value = rs . 100 . dividend per share = rs . 9 ( as it is a 9 % stock ) by investing rs . 1800 , he earns rs . 120 investment needed to earn rs . 9 = 1800 Γ 9 / 120 = rs . 135 ie , stock is then quoted ( then market value ) = rs . 135 answer is b ." | a ) 130 , b ) 135 , c ) 140 , d ) 150 , e ) 145 | b | multiply(multiply(divide(1800, 120), divide(9, const_100)), const_100) | divide(n0,n2)|divide(n1,const_100)|multiply(#0,#1)|multiply(#2,const_100)| | gain |
65 % of x = 20 % of 552.50 . find the value of x ? | "65 % of x = 20 % of 552.50 then , 65 / 100 * x = 20 / 100 * 5525 / 10 x = 170 answer is b" | a ) 100 , b ) 170 , c ) 150 , d ) 180 , e ) 199 | b | divide(multiply(multiply(divide(552.50, const_100), 20), const_100), 65) | divide(n2,const_100)|multiply(n1,#0)|multiply(#1,const_100)|divide(#2,n0)| | general |
find the value of y from ( 12 ) ^ 3 x 6 ^ 3 Γ· 432 = y ? | "5184 e" | a ) 2134 , b ) 2234 , c ) 2540 , d ) 2560 , e ) 864 | e | divide(multiply(power(12, 3), power(6, 3)), 432) | power(n0,n1)|power(n2,n3)|multiply(#0,#1)|divide(#2,n4)| | general |
a train 105 m long passes a man , running at 5 km / hr in the same direction in which the train is going , in 10 seconds . the speed of the train is : | speed of the train relative to man = ( 105 / 10 ) m / sec = ( 21 / 2 ) m / sec . [ ( 21 / 2 ) * ( 18 / 5 ) ] km / hr = 37.8 km / hr . let the speed of the train be x km / hr . then , relative speed = ( x - 5 ) km / hr . x - 5 = 37.8 = = > x = 42.8 km / hr . answer : c | a ) 22 , b ) 50 , c ) 42.8 , d ) 288 , e ) 12 | c | divide(divide(subtract(105, multiply(multiply(5, const_0_2778), 5)), 10), const_0_2778) | multiply(n1,const_0_2778)|multiply(n1,#0)|subtract(n0,#1)|divide(#2,n2)|divide(#3,const_0_2778)| | physics |
find the greatest number that will divide 256 , 193 and 210 leaving 6 , 18 and 100 as remainders respectively | "explanation : answer will be hcf of ( 256 - 6 , 193 - 18 , 210 - 10 ) hcf of ( 250 , 175 , 200 ) = 25 option b" | a ) 19 , b ) 25 , c ) 23 , d ) 20 , e ) 26 | b | add(const_12, multiply(18, const_4)) | multiply(n4,const_4)|add(#0,const_12)| | general |
the area of a circle is increased by 1500 % . by what percent has the radius of the circle increased ? | "the area of the circle is increased by 1500 % , thus the area is increased 16 times . the area of a circle it proportional to the square of the radius ( area = Ο r ^ 2 ) , therefore the radius must increase 4 times ( diameter increase 4 times = area increase 16 times ) , which is increase by 500 % . answer : c" | a ) 100 % , b ) 200 % , c ) 300 % , d ) 600 % , e ) 800 % | c | multiply(const_100, divide(const_2, const_2)) | divide(const_2,const_2)|multiply(#0,const_100)| | gain |
the total price of a basic computer and printer are $ 2,500 . if the same printer had been purchased with an enhanced computer whose price was $ 500 more than the price of the basic computer , then the price of the printer would have been 1 / 6 of that total . what was the price of the basic computer ? | let the price of basic computer be c and the price of the printer be p : c + p = $ 2,500 . the price of the enhanced computer will be c + 500 and total price for that computer and the printer will be 2,500 + 500 = $ 3,000 . now , we are told that the price of the printer is 1 / 6 of that new total price : p = 1 / 6 * $... | a ) 1500 , b ) 1600 , c ) 1750 , d ) 1900 , e ) 2000 | e | subtract(multiply(multiply(const_0_25, const_1000), const_10), divide(add(500, multiply(multiply(const_0_25, const_1000), const_10)), 6)) | multiply(const_0_25,const_1000)|multiply(#0,const_10)|add(n1,#1)|divide(#2,n3)|subtract(#1,#3) | general |
a certain car ' s price decreased by 2.5 % ( from the original price ) each year from 1996 to 2002 , during that time the owner of the car invested in a new carburetor and a new audio system for the car , which increased car ' s price by $ 1,000 . if the price of the car in 1996 was $ 22,000 , what is the car ' s price... | "price in 96 = 22000 price decrease each year = 2.5 / 100 * 22000 = 550 price in 97 = 22000 - 550 price in 98 = 22000 - 2 * 550 price in 99 = 22000 - 3 * 550 price in 00 = 22000 - 4 * 550 price in 01 = 22000 - 5 * 550 price in 02 = 22000 - 6 * 550 = 18700 investment in the car = 1000 net price of the car in 02 = 18700 ... | a ) $ 18,400 , b ) $ 19,700 , c ) $ 20,200 , d ) $ 20,400 , e ) $ 21,100 | b | multiply(const_2, const_10) | multiply(const_10,const_2)| | gain |
last week john spent 30 percent of his wages on recreation . this week , his wages are 25 percent less than last week ΚΌ s wages and he spent 20 percent of his wages on recreation . the amount he spends on recreation this week is what percent of the amount he spent on recreation last week ? | say john ' s wages last week were $ 100 , so he spent 0.30 * 100 = $ 30 on recreation ; this week ' s wages is 0.75 * 100 = $ 75 , so he spends 0.2 * 75 = $ 15 on recreation ; 15 / 30 = 0.50 , hence the amount he spends on recreation this week is 50 % of the amount he spent on recreation last week : 30 * 0.5 = 15 answe... | a ) 60 % , b ) 70 % , c ) 80 % , d ) 90 % , e ) 50 % | e | multiply(divide(multiply(divide(20, const_100), multiply(divide(subtract(const_100, 25), const_100), const_100)), multiply(divide(30, const_100), const_100)), const_100) | divide(n2,const_100)|divide(n0,const_100)|subtract(const_100,n1)|divide(#2,const_100)|multiply(#1,const_100)|multiply(#3,const_100)|multiply(#0,#5)|divide(#6,#4)|multiply(#7,const_100) | general |
in 2000 there were 10000 women stockbrokers in the united states . by 2005 the number had increased to 20000 . approximately what was the percent increase ? | the percent increase is ( 20000 - 10000 ) / 10000 = 10000 / 10000 = 1 % so the approximate answer is a | a ) 1 % , b ) 2 % , c ) 1.5 % , d ) 3 % , e ) 4 % | a | divide(subtract(20000, 10000), 10000) | subtract(n3,n1)|divide(#0,n1) | general |
a sum of money at simple interest amounts to rs . 2240 in 2 years and to rs . 2600 in 5 years . what is the principal amount | "explanation : si for 3 year = 2600 - 2240 = 360 si for 2 year 360 / 3 * 2 = 240 principal = 2240 - 240 = 2000 answer : option c" | a ) 1000 , b ) 1500 , c ) 2000 , d ) 2500 , e ) 3500 | c | subtract(2240, multiply(subtract(2600, 2240), 5)) | subtract(n2,n0)|multiply(n3,#0)|subtract(n0,#1)| | gain |
if β + β means β Γ β , β - β means β + β , β Γ β means β Γ· β and β Γ· β means β - β then find the value of ; 6 β 9 + 8 Γ 3 Γ· 25 = | d 5 | a ) 6 , b ) 10 , c ) 12 , d ) 5 , e ) 8 | d | subtract(add(divide(multiply(9, 8), 3), 6), 25) | multiply(n1,n2)|divide(#0,n3)|add(n0,#1)|subtract(#2,n4) | general |
a dog takes 3 leaps for every 5 leaps of a hare . if one leap of the dog is equal to 3 leaps of the hare , the ratio of the speed of the dog to that of the hare is : | "explanation : dog : hare = ( 3 * 3 ) leaps of hare : 5 leaps of hare = 9 : 5 . answer : a" | a ) 9 : 5 , b ) 2 : 3 , c ) 4 : 7 , d ) 5 : 6 , e ) none of these | a | divide(multiply(3, 3), 5) | multiply(n0,n2)|divide(#0,n1)| | other |
what positive number , when squared , is equal to the cube of the positive square root of 12 ? | "let the positive number be x x ^ 2 = ( ( 12 ) ^ ( 1 / 2 ) ) ^ 3 = > x ^ 2 = 4 ^ 3 = 16 = > x = 4 answer d" | a ) 64 , b ) 32 , c ) 8 , d ) 4 , e ) 2 | d | sqrt(power(power(12, divide(const_1, const_2)), const_3)) | divide(const_1,const_2)|power(n0,#0)|power(#1,const_3)|sqrt(#2)| | geometry |
if 20 men can build a wall 112 metres long in 6 days , what length of a similar wall can be built by 10 men in 3 days ? | "20 men is 6 days can build 112 metres 10 men in 3 days can build = 112 * ( 10 / 20 ) x ( 3 / 6 ) = 28 meters answer : b ." | a ) 65 mtr . , b ) 28 mtr , c ) 70 mtr . , d ) 78 mtr . , e ) 17 mtr . | b | multiply(112, divide(multiply(10, 3), multiply(20, 6))) | multiply(n3,n4)|multiply(n0,n2)|divide(#0,#1)|multiply(n1,#2)| | physics |
a alone can finish a work in 10 days which b alone can finish in 15 days . if they work together and finish it , then out of a total wages of rs . 3200 , a will get : | "explanation : ratio of working days of a : b = 10 : 15 therefore , their wages ratio = reverse ratio = 15 : 10 therefore , a will get 15 units of ratio total ratio = 25 1 unit of ratio = 3000 / 25 = 128 so , a β s amount = 120 Γ 15 = rs . 1920 . answer : option c" | a ) rs . 1200 , b ) rs . 1500 , c ) rs . 1920 , d ) rs . 2000 , e ) none of these | c | multiply(divide(divide(multiply(15, const_2), 10), add(divide(multiply(15, const_2), 10), divide(multiply(15, const_2), 15))), 3200) | multiply(n1,const_2)|divide(#0,n0)|divide(#0,n1)|add(#1,#2)|divide(#1,#3)|multiply(n2,#4)| | physics |
triathlete dan runs along a 3 - mile stretch of river and then swims back along the same route . if dan runs at a rate of 10 miles per hour and swims at a rate of 6 miles per hour , what is his average rate for the entire trip in miles per minute ? | "dan travels 6 miles round trip . running part : ( 3 / 10 * 60 = 18 minutes ) swimming part : ( 3 / 6 * 60 = 30 minutes ) 6 miles in ( 18 + 30 ) minutes 6 / 48 = 1 / 8 mile per minute answer : 1 / 8 mile per minute a" | a ) 1 / 8 , b ) 2 / 15 , c ) 3 / 15 , d ) 1 / 4 , e ) 3 / 8 | a | divide(add(3, 3), add(multiply(divide(3, 6), const_60), multiply(divide(3, 10), const_60))) | add(n0,n0)|divide(n0,n2)|divide(n0,n1)|multiply(#1,const_60)|multiply(#2,const_60)|add(#3,#4)|divide(#0,#5)| | physics |
i flew my tiny seaplane to visit my mother . on the flight up , i flew at 140 mph . on the way home , i flew 88 mph . what was my average speed for the trip ? | "( 140 mph + 88 mph ) / 2 = 114 mph correct answer is : a" | a ) 114 mph , b ) 110 mph , c ) 88 mph , d ) 100 mph , e ) 99 mph | a | divide(add(140, 88), const_2) | add(n0,n1)|divide(#0,const_2)| | physics |
how many even integers from 1 to 50 ( both inclusive ) have even number of factors ? | "integers having even number of factors will be perfect squares . even numbers will have even perfect squares . thus , the possible values for the perfect squares are : 4 , 16,36 and the corresponding integers are 2 , 4,6 ( more than 2 ) . thus a is the correct answer ." | a ) 3 , b ) 14 , c ) 5 , d ) 6 , e ) 7 | a | divide(sqrt(50), const_2) | sqrt(n1)|divide(#0,const_2)| | other |
the average age of 25 students in a group is 12 years . when teacher ' s age is included to it , the average increases by one . what is the teacher ' s age in years ? | "age of the teacher = ( 26 * 13 - 25 * 12 ) = 38 years . answer : b" | a ) 31 , b ) 38 , c ) 39 , d ) 40 , e ) 57 | b | add(25, const_1) | add(n0,const_1)| | general |
a $ 84.95 lawn chair was sold for $ 59.95 at a special sale . by approximately what percent was the price decreased ? | "listed selling price of chair = 84.95 $ discounted selling price of chair = 59.95 $ discount = 84.95 - 59.95 = 25 $ % decrease in price of chair = ( 25 / 84.95 ) * 100 % = 29 % approx answer d" | a ) 15 % , b ) 20 % , c ) 25 % , d ) 29 % , e ) 80 % | d | multiply(divide(subtract(84.95, 59.95), 84.95), const_100) | subtract(n0,n1)|divide(#0,n0)|multiply(#1,const_100)| | general |
what is 35 % of 4 / 13 of 845 ? | "this problem can be solved easily if we just use approximation : 35 % is a little over 1 / 3 , while 4 / 13 is a little less than 4 / 12 , which is 1 / 3 . thus , the answer is about 1 / 3 of 1 / 3 of 845 , or 1 / 9 of 845 . since the first 1 / 3 is a slight underestimate and the second 1 / 3 is a slight overestimate ... | a ) 58 , b ) 69 , c ) 80 , d ) 91 , e ) 102 | c | divide(multiply(35, add(add(multiply(multiply(add(const_3, const_2), const_2), multiply(multiply(const_3, const_4), const_100)), multiply(multiply(add(const_3, const_4), add(const_3, const_2)), multiply(add(const_3, const_2), const_2))), add(const_3, const_3))), const_100) | add(const_2,const_3)|add(const_3,const_4)|add(const_3,const_3)|multiply(const_3,const_4)|multiply(#0,const_2)|multiply(#3,const_100)|multiply(#1,#0)|multiply(#4,#5)|multiply(#6,#4)|add(#7,#8)|add(#9,#2)|multiply(n0,#10)|divide(#11,const_100)| | gain |
a grocer has a sale of rs . 5700 , rs . 8550 , rs . 6855 , rs . 3850 for 4 consecutive months . how much sale must he have in the fifth month so that he gets an average sale of rs . 7800 ? | total sale for 4 months = rs . ( 5700 + 8550 + 6855 + 3850 ) = rs . 24955 required sale = rs . [ ( 7800 x 5 ) - 24955 ] = rs . ( 39000 - 24955 ) = rs . 14045 option c | a ) s . 14991 , b ) s . 49930 , c ) s . 14045 , d ) s . 14999 , e ) s . 14578 | c | subtract(multiply(7800, add(4, const_1)), add(3850, add(add(5700, 8550), 6855))) | add(n4,const_1)|add(n0,n1)|add(n2,#1)|multiply(n5,#0)|add(n3,#2)|subtract(#3,#4) | general |
a shop produces sarongs . the daily average production is given by 5 n + 20 , where n is the number of workers aside from the owner . in the first k days , 500 units are produced , and then 5 workers are added to the team . after another k days , the cumulative total is 1200 . how many workers were part of the latter p... | "the daily average production is given by 5 n + 20 - given in the first k days , 500 units are produced = ( 5 n + 20 ) k = 500 k = 500 / 5 n + 20 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1 5 workers were added = 5 ( n + 5 ) + 20 = 5 n + 45 cumulative is 1200 . . thus for the current period ... | a ) a ) 6 , b ) b ) 14 , c ) c ) 11 , d ) d ) 15 , e ) e ) 23.5 | b | add(20, 5) | add(n0,n1)| | general |
the cash difference between the selling prices of an book at a profit of 4 % and 8 % is $ 3 . the ratio of the two selling prices is : | "let c . p . of the book be $ x . then , required ratio = 104 % of x / 108 % of x = 104 / 108 = 26 / 27 = 26 : 27 b" | a ) 55 : 23 , b ) 26 : 27 , c ) 52 : 53 , d ) 45 : 53 , e ) 51 : 34 | b | divide(add(const_100, 4), add(const_100, 8)) | add(n0,const_100)|add(n1,const_100)|divide(#0,#1)| | gain |
the ratio of ducks and frogs in a pond is 35 : 39 respectively . the average number of ducks and frogs in the pond is 152 . what is the number of frogs in the pond ? | solution : ratio of ducks and frogs in pond , = 35 : 39 . average of ducks and frogs in pond , = 152 . so , total number of ducks and frogs in the pond , = 2 * 152 = 304 . therefore , number of frogs , = ( 304 * 39 ) / 74 = 160 . answer : option a | a ) 160 , b ) 152 , c ) 156 , d ) 144 , e ) none | a | multiply(39, divide(152, divide(add(add(35, 39), const_2), const_2))) | add(n0,n1)|add(#0,const_2)|divide(#1,const_2)|divide(n2,#2)|multiply(n1,#3) | general |
suraj has a certain average of runs for 14 innings . in the 15 th innings he scores 140 runs thereby increasing his average by 8 runs . what is his average after the 15 th innings ? | "to improve his average by 8 runs per innings he has to contribute 14 x 8 = 112 runs for the previous 14 innings . thus , the average after the 15 th innings = 140 - 112 = 28 . answer : b" | a ) 48 , b ) 28 , c ) 36 , d ) 72 , e ) 27 | b | divide(subtract(140, multiply(14, 8)), subtract(15, 14)) | multiply(n0,n3)|subtract(n1,n0)|subtract(n2,#0)|divide(#2,#1)| | general |
a cistern is fille in 8 hours and it takes 12 hours when there is a leak in its bottom . if the cistern is full , in what time shall the leak empty it ? | explanation : work done by 1 hour by filling pipe = 1 / 8 work done by 1 hour by the leak and the filling pipe = 1 / 12 work done by 1 hour by the leak = 1 / 8 β 1 / 12 = 1 / 24 hence the leak can empty it in 24 hours . answer : option c | a ) 22 hours , b ) 21 hours , c ) 24 hours , d ) 23 hours , e ) 25 days | c | inverse(subtract(inverse(8), inverse(12))) | inverse(n0)|inverse(n1)|subtract(#0,#1)|inverse(#2) | physics |
if 12 men do a work in 80 days , in how many days will 16 men do it ? | 12 * 80 = 16 * x x = 60 days answer : d | a ) 22 , b ) 87 , c ) 77 , d ) 60 , e ) 85 | d | divide(multiply(12, 80), 16) | multiply(n0,n1)|divide(#0,n2) | physics |
how many points ( x , y ) lie on the line segment between ( 22 , 12 2 / 3 ) and ( 16 , 17 2 / 3 ) such that x and y are both integers ? | "slope = ( 17 2 / 3 - 12 2 / 3 ) / ( 16 - 22 ) = - 5 / 6 y = mx + b = > 12 2 / 3 = - 22 * 5 / 6 + b = > b = 31 y = - 5 x / 6 + 31 only integer values work , and the only multiples of 6 between 16 and 22 for x values is 18 , thus 1 point . a" | a ) 1 , b ) 5 , c ) 7 , d ) 8 , e ) 9 | a | divide(16, 16) | divide(n4,n4)| | general |
a certain class of students is being divided into teams . the class can either be divided into 15 teams with an equal number of players on each team or 24 teams with an equal number of players on each team . what is the lowest possible number of students in the class ? | "let total no of students in the class be n so , we are told that n is divisible by both 15 24 so , lets find the least common multiple of 15 24 , ie 120 so our answer is ( e ) 120" | a ) 6 , b ) 36 , c ) 48 , d ) 60 , e ) 120 | e | lcm(15, 24) | lcm(n0,n1)| | general |
a department store perceived a run on a lower priced garment and raised the price by 20 % to keep stock available until another shipment arrived . customers fumed and the department store manager allowed them to purchase the garments at 80 % of the new price to save face . what was the difference in the new retail pric... | quantity x rate = price 1 x 1 = 1 0.8 x 1.20 = 0.960 decrease in price = ( 0.040 / 1 ) Γ£ β 100 = 4.0 % answer = option a | a ) 4.0 % , b ) 5.0 % , c ) 6.0 % , d ) 7.0 % , e ) 8.0 % | a | multiply(subtract(const_1, multiply(divide(80, const_100), divide(add(const_100, 20), const_100))), const_100) | add(n0,const_100)|divide(n1,const_100)|divide(#0,const_100)|multiply(#1,#2)|subtract(const_1,#3)|multiply(#4,const_100) | general |
the price of a mobile was increased by 40 percent . the new price was then decreased by 15 percent . a single increase of what percent is equivalent to these two successive changes ? | consider base price - $ 100 25 % increase = 1.40 * 100 = $ 140 then a 15 % decrease on new price = 0.85 * 140 = $ 119 so final price of radio - $ 119 therefore a 19 % increase correct option - b | a ) 25 % , b ) 19 % , c ) 30 % , d ) 22.5 % , e ) 15 % | b | multiply(subtract(multiply(divide(add(const_100, 40), const_100), divide(subtract(const_100, 15), const_100)), const_1), const_100) | add(n0,const_100)|subtract(const_100,n1)|divide(#0,const_100)|divide(#1,const_100)|multiply(#2,#3)|subtract(#4,const_1)|multiply(#5,const_100) | general |
a , b and c completed a piece of work , a worked for 16 days , b for 9 days and c for 4 days . their daily wages were in the ratio of 3 : 4 : 5 . find the daily wages of c , if their total earning was rs . 1480 ? | "3 x 4 x 5 x 16 9 4 48 x + 36 x + 20 x = 1480 104 x = 1480 = > x = 14.23 5 x = 71.15 rs . answer : c" | a ) rs . 80 , b ) rs . 120 , c ) rs . 71.15 , d ) rs . 51.15 , e ) rs . 180 | c | multiply(5, divide(1480, add(add(multiply(16, 3), multiply(9, 4)), multiply(4, 5)))) | multiply(n0,n3)|multiply(n1,n2)|multiply(n2,n5)|add(#0,#1)|add(#3,#2)|divide(n6,#4)|multiply(n5,#5)| | physics |
there are 20 students in a class . in how many different ways can a committee of 3 students be formed ? | "20 c 3 = 20 * 19 * 18 / 6 = 1140 the answer is d ." | a ) 650 , b ) 730 , c ) 980 , d ) 1140 , e ) 1480 | d | multiply(subtract(const_1, divide(3, 20)), 20) | divide(n1,n0)|subtract(const_1,#0)|multiply(#1,n0)| | probability |
the closest approximation of w ( 69.28 Γ 0.004 ) / 0.03 is | "w ( 69.28 Γ 0.004 ) / 0.03 1 . 0.004 = 4 Γ 10 ^ ( - 3 ) 2 . 0.03 = 3 Γ 10 ^ ( - 2 ) 3 . ( a Γ b ) / c = a Γ ( b / c ) 4 . 0.004 / 0.03 = 4 Γ 10 ^ ( - 3 ) / ( 3 Γ 10 ^ ( - 2 ) ) = 4 Γ 10 ^ ( - 3 - ( - 2 ) ) / 3 = 4 Γ 10 ^ ( - 1 ) / 3 = ( 4 / 3 ) Γ 10 ^ ( - 1 ) = 1.333 Γ 10 ^ ( - 1 ) therefore , ( 69.28 Γ 0.004 ) / 0.03... | a ) 0.092 , b ) 0.92 , c ) 9.2 , d ) 92 , e ) 920 | c | multiply(divide(0.004, 0.03), 69.28) | divide(n1,n2)|multiply(n0,#0)| | general |
the least number by which 72 must be multiplied in order to produce a multiple of 112 , is : | required number is divisible by 72 as well as by 112 , if it is divisible by their lcm , which is 1008 . now , 1008 when divided by 72 , gives quotient = 14 . therefore , required number = 14 . answer : c | a ) 6 , b ) 12 , c ) 14 , d ) 18 , e ) 20 | c | divide(112, gcd(72, 112)) | gcd(n0,n1)|divide(n1,#0) | general |
if there are 200 questions in a 3 hr examination . among these questions are 50 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 150 x + 50 * 2 x = 180 x = 180 / 250 x = 0.72 time taken for type a problem = 50 * 2 * 0.72 = 72 min answer : a" | a ) 72 min , b ) 62 min , c ) 70 min , d ) 74 min , e ) 76 min | a | multiply(multiply(const_2, divide(multiply(3, const_60), add(subtract(200, 50), multiply(const_2, 50)))), 50) | 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 |
a person spends 1 / 3 rd of the money with him on clothes , 1 / 5 th of the remaining on food and 1 / 4 th of the remaining on travel . now , he is left with rs 100 . how much did he have with him in the beginning ? | "suppose the amount in the beginning was rs β x β money spent on clothes = rs 1 x / 3 balance = rs 2 x / 3 money spent on food = 1 / 5 of 2 x / 3 = rs 2 x / 15 balance = 2 x / 3 - 2 x / 15 = rs 8 x / 15 money spent on travel = 1 / 4 of 8 x / 15 = rs 2 x / 15 = 8 x / 15 - 2 x / 15 = 6 x / 15 = rs 2 x / 5 therefore 2 x /... | a ) rs 200 , b ) rs 250 , c ) rs 300 , d ) rs 450 , e ) rs 550 | b | divide(100, subtract(subtract(subtract(1, divide(1, 3)), divide(subtract(1, divide(1, 3)), 5)), divide(subtract(subtract(1, divide(1, 3)), divide(subtract(1, divide(1, 3)), 5)), 4))) | divide(n0,n1)|subtract(n0,#0)|divide(#1,n3)|subtract(#1,#2)|divide(#3,n5)|subtract(#3,#4)|divide(n6,#5)| | general |
marginal cost is the cost of increasing the quantity produced ( or purchased ) by one unit . if the fixed cost for n products is $ 10,000 and the marginal cost is $ 80 , and the total cost is $ 16,000 , what is the value of n ? | "total cost for n products = fixed cost for n products + n * marginal cost - - > $ 16,000 = $ 10,000 + n * $ 80 - - > n = 75 . answer : d ." | a ) 30 , b ) 50 , c ) 60 , d ) 75 , e ) 100 | d | divide(80, const_10) | divide(n1,const_10)| | general |
find the area , diameter = 12 m . | "diameter = 12 meter . radius = diameter / 2 . = 12 / 2 . = 6 meter . area of a circle = Ο r 2 . here , pi ( Ο ) = 3.14 meter , radius ( r ) = 6 . area of a circle = 3.14 Γ 6 Γ 6 . . = 3.14 Γ 36 . = 113.04 square meter or 113.04 m 2 . answer : b" | a ) 113.00 square meter , b ) 113.04 square meter , c ) 113.08 square meter , d ) 113.24 square meter , e ) 113.43 square meter | b | circle_area(divide(12, const_2)) | divide(n0,const_2)|circle_area(#0)| | physics |
765.765 - ? + 46.82 = 437.785 | "explanation : 374.80 answer : option a" | a ) a ) 374.80 , b ) b ) 752.804 , c ) c ) 714.642 , d ) d ) 629.906 , e ) of these | a | subtract(multiply(divide(765.765, const_100), 46.82), multiply(divide(const_1, const_3), multiply(divide(765.765, const_100), 46.82))) | divide(n0,const_100)|divide(const_1,const_3)|multiply(n1,#0)|multiply(#1,#2)|subtract(#2,#3)| | general |
sam taken a loan rs . 15000 / - from co - operative society with an interest @ 11.5 % per month . at the same time he deposited rs . 10000 / - as fixed deposit with an interest @ 9.5 % per month . after one week sam asked the manager to calculate the interest to be paid . what is the interest amount for 7 days ? | loan amount : rs . 15000 / - @ 11.5 % interest per month = 15000 / - * 11.5 % = rs . 1725 interest for one day = 1725 / 30 = 57.50 interest for 7 days is = 57.50 * 7 = 403 fd amount is = rs . 10000 / - @ 9.5 % interest per month = 10000 * 9.5 % = 950 / - interest for 7 days = 950 / 30 * 7 = 222 interest amount to be pa... | a ) a ) 165 , b ) b ) 220 , c ) c ) 310 , d ) d ) 185 , e ) e ) 181 | e | multiply(7, divide(subtract(multiply(15000, divide(11.5, const_100)), multiply(10000, divide(9.5, const_100))), multiply(const_10, const_3))) | divide(n1,const_100)|divide(n3,const_100)|multiply(const_10,const_3)|multiply(n0,#0)|multiply(n2,#1)|subtract(#3,#4)|divide(#5,#2)|multiply(n4,#6) | general |
how many integers from 20 to 150 , inclusive , are divisible by 3 but not divisible by 7 ? | "we should find # of integers divisible by 3 but not by 3 * 7 = 21 . # of multiples of 21 in the range from 20 to 150 , inclusive is ( 147 - 21 ) / 21 + 1 = 7 ; 44 - 7 = 37 . answer : a ." | a ) 37 , b ) 40 , c ) 42 , d ) 45 , e ) 55 | a | divide(150, const_10) | divide(n1,const_10)| | general |
on a two - dimensional coordinate plane , the line d = x ^ 2 - x ^ 3 touches the x - axis in how many places ? | "apparently it ' s d = x ^ 2 - x ^ 3 instead of d = x ^ 2 - d ^ 3 . in this case : the x - intercept is the value ( s ) of x for d = 0 . 0 = x ^ 2 - x ^ 3 ; 0 = x ^ 2 ( 1 - x ) ; x = 0 or x = 1 . answer : c ." | a ) 0 , b ) 1 , c ) 2 , d ) 3 , e ) 4 | c | subtract(max(3, 2), const_1) | max(n0,n1)|subtract(#0,const_1)| | general |
how many even integers from 1 to 100 ( both inclusive ) have even number of factors ? | integers having even number of factors will be perfect squares . even numbers will have even perfect squares . thus , the possible values for the perfect squares are : 4 , 16,36 , 64,100 and the corresponding integers are 2 , 4,6 , 8,10 ( more than 4 ) . thus c is the correct answer . | a ) 13 , b ) 14 , c ) 5 , d ) 6 , e ) 7 | c | divide(sqrt(100), const_2) | sqrt(n1)|divide(#0,const_2) | other |
the ratio of the number of females to males at a party was 1 : 2 but when 4 females and 4 males left , the ratio became 1 : 3 . how many people were at the party originally ? | "the total number of people are x females + 2 x males . 3 * ( x - 4 ) = 2 x - 4 x = 8 there were 3 x = 24 people at the party originally . the answer is c ." | a ) 20 , b ) 22 , c ) 24 , d ) 26 , e ) 28 | c | add(divide(subtract(multiply(multiply(4, 2), 3), multiply(4, 2)), 2), subtract(multiply(multiply(4, 2), 3), multiply(4, 2))) | multiply(n1,n2)|multiply(n5,#0)|subtract(#1,#0)|divide(#2,n1)|add(#3,#2)| | other |
if 7875 / 5.25 = 1500 , then 787.5 / 52.5 is equal to ? | "answer given expression 787.5 / 52.5 = 7875 / 525 = 7875 / ( 525 x 100 ) = 1500 / 100 15 correct option : a" | a ) 15 , b ) 19 , c ) 12 , d ) 14 , e ) 16 | a | divide(multiply(1500, 5.25), 7875) | multiply(n1,n2)|divide(#0,n0)| | general |
two trains , one from howrah to patna and the other from patna to howrah , start simultaneously . after they meet , the trains reach their destinations after 18 hours and 8 hours respectively . the ratio of their speeds is | let us name the trains as a and b . then , ( a ' s speed ) : ( b ' s speed ) = b : a = 8 : 18 = 4 : 9 . answer : a | a ) 4 : 9 , b ) 4 : 3 , c ) 4 : 12 , d ) 4 : 8 , e ) 4 : 5 | a | divide(sqrt(8), sqrt(18)) | sqrt(n1)|sqrt(n0)|divide(#0,#1)| | physics |
when the price of an article is reduced by 30 % , the sales increases by 50 % . the percentage change in the total amount of receipts is ? | we have , receipts = price x sales therefore , net % change in receipts = ( x + y + xy / 100 ) % = [ - 30 + 50 + ( - 30 x 50 ) / 100 ] % = 5 % increase . answer : b | a ) 5 % decrease , b ) 5 % increase , c ) 10 % decrease , d ) 15 % decrease , e ) none of these | b | subtract(divide(multiply(subtract(const_100, 30), 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 |
find the ratio in which rice at rs . 7.20 a kg be mixed with rice at rs . 5.60 a kg to produce a mixture worth rs . 6.30 a kg | "by the rule of alligation : cost of 1 kg rice of 1 st kind cost of 1 kg rice of 2 nd kind required ratio = 70 : 90 = 7 : 9 answer : e" | a ) 2 : 0 , b ) 2 : 3 , c ) 2 : 1 , d ) 2 : 2 , e ) 7 : 9 | e | divide(subtract(6.30, 5.60), subtract(7.20, 6.30)) | subtract(n2,n1)|subtract(n0,n2)|divide(#0,#1)| | other |
two trains of equal lengths take 10 sec and 15 sec respectively to cross a telegraph post . if the length of each train be 140 m , in what time will they cross other travelling in opposite direction ? | "speed of the first train = 120 / 10 = 12 m / sec . speed of the second train = 120 / 5 = 8 m / sec . relative speed = 12 + 8 = 20 m / sec . required time = ( 140 + 140 ) / 20 = 14 sec . answer : d" | a ) 17 sec , b ) 12 sec , c ) 16 sec , d ) 14 sec , e ) 18 sec | d | divide(multiply(140, const_2), add(speed(140, 15), speed(140, 10))) | multiply(n2,const_2)|speed(n2,n1)|speed(n2,n0)|add(#1,#2)|divide(#0,#3)| | physics |
at a special sale , 5 tickets can be purchased for the price of 3 tickets . if 5 tickets are purchased at the sale , the amount saved will be what percent of the original price of the 5 tickets ? | "let the price of a ticket be rs . 100 , so 3 tickets cost 300 & 5 tickets cost 500 5 tickets purchased at price of 3 tickets ie . , for 300 , so amount saved s rs . 200 , % of 5 tickets = ( 200 / 500 ) * 100 = 40 % answer : c" | a ) 20 % , b ) 33.3 % , c ) 40 % , d ) 60 % , e ) 66.6 % | c | divide(multiply(subtract(multiply(5, 5), multiply(3, 5)), const_100), multiply(5, 5)) | multiply(n0,n0)|multiply(n0,n1)|subtract(#0,#1)|multiply(#2,const_100)|divide(#3,#0)| | gain |
the average salary of all the workers in a workshop is rs . 8000 . the average salary of 7 technicians is rs . 20000 and the average salary of the rest is rs . 6000 . the total number of workers in the workshop is | "sol . let the total number of workers be x . then , 8000 x = ( 20000 Γ 7 ) + 6000 ( x β 7 ) βΉ = βΊ 2000 x = 98000 βΉ = βΊ x = 49 . answer b" | a ) 20 , b ) 49 , c ) 22 , d ) 23 , e ) none | b | add(7, divide(multiply(7, subtract(20000, 8000)), subtract(8000, 6000))) | subtract(n2,n0)|subtract(n0,n3)|multiply(n1,#0)|divide(#2,#1)|add(n1,#3)| | general |
find the average of all the numbers between 6 and 34 which are divisible by 7 . | "solution average = ( 7 + 14 + 21 + 28 ) / 4 ) = 70 / 4 = 17.5 . answer c" | a ) 18 , b ) 20 , c ) 17.5 , d ) 30 , e ) 32 | c | divide(add(add(6, const_4), subtract(34, const_4)), const_2) | add(n0,const_4)|subtract(n1,const_4)|add(#0,#1)|divide(#2,const_2)| | general |
at what price must an article costing rs . 47.50 be marked in order that after deducting 6 % from the list price . it may be sold at a profit of 25 % on the cost price ? | "cp = 47.50 sp = 47.50 * ( 125 / 100 ) = 59.375 mp * ( 94 / 100 ) = 59.375 mp = 63.16 answer : a" | a ) 63.16 , b ) 62.6 , c ) 62.1 , d ) 62.7 , e ) 62.2 | a | divide(multiply(add(47.50, divide(multiply(47.50, 25), const_100)), const_100), subtract(const_100, 6)) | multiply(n0,n2)|subtract(const_100,n1)|divide(#0,const_100)|add(n0,#2)|multiply(#3,const_100)|divide(#4,#1)| | gain |
if a mixture is 3 β 5 alcohol by volume and 4 β 5 water by volume , what is the ratio of the volume of alcohol to the volume of water in this mixture ? | "should be a sub - 600 level q . . volume = { 3 / 5 } / { 4 / 5 } = 3 / 4 c" | a ) 1 / 3 , b ) 3 / 2 , c ) 3 / 4 , d ) 3 , e ) 3 / 5 | c | divide(divide(3, 5), divide(4, 5)) | divide(n0,n1)|divide(n2,n1)|divide(#0,#1)| | general |
what is difference between biggest and smallest fraction among 2 / 3 , 3 / 4 , 4 / 5 and 5 / 6 | "explanation : 2 / 3 = . 66 , 3 / 4 = . 75 , 4 / 5 = . 8 and 5 / 6 = . 833 so biggest is 5 / 6 and smallest is 2 / 3 their difference is 5 / 6 - 2 / 3 = 1 / 6 option c" | a ) 2 / 5 , b ) 3 / 5 , c ) 1 / 6 , d ) 1 / 7 , e ) none of these | c | subtract(divide(4, 5), divide(2, 3)) | divide(n3,n5)|divide(n0,n1)|subtract(#0,#1)| | general |
a reduction of 25 % in the price of salt enables a lady to obtain 10 kgs more for rs . 800 , find the original price per kg ? | "answer : 800 * ( 25 / 100 ) = 200 - - - 10 ? - - - 1 = > rs . 20 800 - - - 600 ? - - - 20 = > rs . 26.67 . answer : b" | a ) 25.67 , b ) 26.67 , c ) 27.2 , d ) 26.98 , e ) 27.76 | b | multiply(divide(divide(multiply(divide(25, const_100), 800), 10), multiply(divide(25, const_100), 800)), const_100) | divide(n0,const_100)|multiply(n2,#0)|divide(#1,n1)|divide(#2,#1)|multiply(#3,const_100)| | gain |
the compound and the simple interests on a certain sum at the same rate of interest for two years are rs . 11730 and rs . 10200 respectively . find the sum . | "the simple interest for the first year is 10200 / 2 is rs . 5100 and compound interest for first year also is rs . 5100 . the compound interest for second year on rs . 5100 for one year so rate of the interest = ( 100 * 1530 ) / ( 5100 * 1 ) = 30 % p . a . so p = ( 100 * 10200 ) / ( 30 * 2 ) = rs . 17000 answer : b" | a ) 33498 , b ) 17000 , c ) 2665 , d ) 28876 , e ) 313 | b | divide(multiply(const_100, 10200), multiply(divide(multiply(subtract(11730, 10200), const_100), divide(10200, const_2)), const_2)) | divide(n1,const_2)|multiply(n1,const_100)|subtract(n0,n1)|multiply(#2,const_100)|divide(#3,#0)|multiply(#4,const_2)|divide(#1,#5)| | gain |
the length of a room is 5.5 m and width is 3.75 m . what is the cost of paying the floor by slabs at the rate of $ 300 per sq . metre . | "area = 5.5 Γ 3.75 sq . metre . cost for 1 sq . metre . = $ 300 hence , total cost = 5.5 Γ 3.75 Γ 300 = $ 6187.50 d" | a ) $ 2587.50 , b ) $ 3587.50 , c ) $ 4187.50 , d ) $ 6187.50 , e ) $ 8587.50 | d | multiply(300, multiply(5.5, 3.75)) | multiply(n0,n1)|multiply(n2,#0)| | physics |
of the 150 people at a party , 70 were women , and 30 women tried the appetizer . if 60 people did not try the appetizer , what is the total number of men who tried the appetizer ? | "total people at party = 150 women = 70 so men 150 - 70 = 80 no . of pple who tried appetizer = 150 - 60 ( given info ) = 90 no of women who tried appetizer = 30 so remaining ppl ( men ) who tried the appetizer = 90 - 30 = 60 correct option c" | a ) 40 , b ) 50 , c ) 60 , d ) 70 , e ) 80 | c | subtract(subtract(150, 60), 30) | subtract(n0,n3)|subtract(#0,n2)| | general |
a man swims downstream 28 km and upstream 12 km taking 2 hours each time , what is the speed of the man in still water ? | "28 - - - 2 ds = 14 ? - - - - 1 12 - - - - 2 us = 6 ? - - - - 1 m = ? m = ( 14 + 6 ) / 2 = 10 answer : a" | a ) 10 , b ) 8 , c ) 5 , d ) 2 , e ) 4 | a | divide(add(divide(12, 2), divide(28, 2)), const_2) | divide(n1,n2)|divide(n0,n2)|add(#0,#1)|divide(#2,const_2)| | physics |
in a restaurant , the profit is 130 % of the cost . if the cost increases by 12 % but the selling price remains constant , approximately what percentage of the selling price is the profit ? | "explanation : let c . p . = rs . 100 . then , profit = rs . 130 , s . p . = rs . 230 . new c . p . = 112 % of rs . 100 = rs . 112 new s . p . = rs . 230 . profit = rs . ( 230 - 112 ) = rs . 118 . required percentage = ( 118 / 230 * 100 ) % = 51 % appox answer : b" | a ) 30 % , b ) 51 % , c ) 90 % , d ) 100 % , e ) none of these | b | multiply(const_100, divide(subtract(add(multiply(const_100, divide(130, const_100)), const_100), multiply(const_100, add(const_1, divide(12, const_100)))), add(multiply(const_100, divide(130, const_100)), const_100))) | divide(n0,const_100)|divide(n1,const_100)|add(#1,const_1)|multiply(#0,const_100)|add(#3,const_100)|multiply(#2,const_100)|subtract(#4,#5)|divide(#6,#4)|multiply(#7,const_100)| | gain |
a dog takes 5 leaps for every 6 leaps of a hare . if one leap of the dog is equal to 3 leaps of the hare , the ratio of the speed of the dog to that of the hare is : | "explanation : dog : hare = ( 5 * 3 ) leaps of hare : 6 leaps of hare = 15 : 3 - - > 5 : 1 answer : d" | a ) 1 : 6 , b ) 5 : 6 , c ) 6 : 1 , d ) 5 : 1 , e ) 9 : 1 | d | divide(multiply(5, 3), 6) | multiply(n0,n2)|divide(#0,n1)| | other |
a mathematics teacher tabulated the marks secured by 35 students of 8 th class . the average of their marks was 72 . if the marks secured by reema was written as 36 instead of 66 then find the correct average marks up to two decimal places . | total marks = 35 x 72 = 2520 corrected total marks = 2520 - 36 + 66 = 2550 correct average = 2550 / 35 = 72.85 answer : a | a ) 72.85 , b ) 74.31 , c ) 72.43 , d ) 73.43 , e ) can not be determined | a | divide(subtract(multiply(35, 72), subtract(66, 36)), 35) | multiply(n0,n2)|subtract(n4,n3)|subtract(#0,#1)|divide(#2,n0) | general |
a bag contains 6 green balls and 7 white balls . if two balls are drawn simultaneously , what is the probability that both balls are the same colour ? | "the total number of ways to draw two balls is 13 c 2 = 78 the number of ways to draw two green balls is 6 c 2 = 15 the number of ways to draw two white balls is 7 c 2 = 21 p ( two balls of the same colour ) = 36 / 78 = 6 / 13 the answer is d ." | a ) 2 / 7 , b ) 3 / 10 , c ) 5 / 11 , d ) 6 / 13 , e ) 11 / 21 | d | add(multiply(divide(7, add(6, 7)), divide(subtract(7, const_1), subtract(add(6, 7), const_1))), multiply(divide(subtract(6, const_1), subtract(add(6, 7), const_1)), divide(6, add(6, 7)))) | add(n0,n1)|subtract(n1,const_1)|subtract(n0,const_1)|divide(n1,#0)|divide(n0,#0)|subtract(#0,const_1)|divide(#1,#5)|divide(#2,#5)|multiply(#3,#6)|multiply(#7,#4)|add(#8,#9)| | other |
in the biology lab of ` ` jefferson ' ' high school there are 0.036 * 10 ^ 5 germs , equally divided among 75000 * 10 ^ ( - 3 ) petri dishes . how many germs live happily in a single dish ? | "0.036 * 10 ^ 5 can be written as 3600 75000 * 10 ^ ( - 3 ) can be written as 75 required = 3600 / 75 = 48 answer : e" | a ) 10 , b ) 20 , c ) 30 , d ) 40 , e ) 48 | e | divide(multiply(multiply(const_1000, const_100), 0.036), divide(75000, const_1000)) | divide(n3,const_1000)|multiply(const_100,const_1000)|multiply(n0,#1)|divide(#2,#0)| | general |
set x consists of the integers from 1 to 12 , inclusive , while set y consists of the integers from 0 to 20 , inclusive . how many distinct integers do belong to the both sets at the same time ? | x = { 1,2 , 3,4 , 5,6 , 7 , 8 , 9 , 10 , 11 , 12 } y = { 0 , 1,2 , 3,4 , 5,6 , 7 , 8 , 9 , 10 , 11,12 , 13,14 , 15,16 , 17,18 , 19,20 } common elements = { 1,2 , 3,4 , 5,6 , 7,8 , 9,10 , 11,12 } = 12 elements answer : option a . | a ) 12 , b ) 15 , c ) 10 , d ) 20 , e ) 8 | a | add(subtract(12, 1), const_1) | subtract(n1,n0)|add(#0,const_1) | other |
a train 160 m long running at 72 kmph crosses a platform in 25 sec . what is the length of the platform ? | "c 340 c = ( 72 * 5 / 18 ) * 25 - 160 = 340" | a ) 443 m , b ) 354 m , c ) 340 m , d ) 350 m , e ) 250 m | c | subtract(multiply(25, multiply(72, const_0_2778)), 160) | multiply(n1,const_0_2778)|multiply(n2,#0)|subtract(#1,n0)| | physics |
a cube of edge 5 cm is cut into cubes each of edge 1 cm . the ratio of the total surface area of one of the small cubes to that of the large cube is equal to : | sol . required ratio = 6 * 1 * 1 / 6 * 5 * 5 = 1 / 25 = 1 : 25 . answer a | ['a ) 1 : 25', 'b ) 1.225', 'c ) 1 : 52', 'd ) 1 : 522', 'e ) none'] | a | divide(const_4, const_100) | divide(const_4,const_100) | geometry |
the difference b / w the compound interest payble half yearly and the simple interest on a certain sum lent out at 10 % p . a for 1 year is rs 25 . what is the sum ? | in simple words s . i = ptr / 100 = ( p * 1 * 10 ) / 100 p / 10 here amount a = p + s . i = p + p / 10 = 11 p / 10 . . . . now for compound interest when calculated half yearly amount = p [ 1 + r / 200 ] ^ 2 t = p [ 1 + 10 / 200 ] ^ 2 * 1 = 441 p / 400 here its given in the question dat diff btwn simple interest and co... | a ) rs . 15000 , b ) rs . 12000 , c ) rs . 10000 , d ) rs . 20000 , e ) none of these | c | divide(25, subtract(subtract(power(add(divide(divide(const_10, const_2), const_100), 1), const_2), const_1), divide(10, const_100))) | divide(const_10,const_2)|divide(n0,const_100)|divide(#0,const_100)|add(n1,#2)|power(#3,const_2)|subtract(#4,const_1)|subtract(#5,#1)|divide(n2,#6) | general |
how many bricks , each measuring 25 cm x 11 cm x 6 cm , will be needed to build a wall of 2 m x 3 m x 2 cm ? | "number of bricks = volume of the wall / volume of 1 brick = ( 200 x 300 x 2 ) / ( 25 x 11 x 6 ) = 72.7 = 73 . answer : option c" | a ) 5600 , b ) 6000 , c ) 73 , d ) 7200 , e ) 8600 | c | divide(multiply(multiply(multiply(2, const_100), multiply(3, const_100)), 2), multiply(multiply(25, 11), 6)) | multiply(n3,const_100)|multiply(n4,const_100)|multiply(n0,n1)|multiply(#0,#1)|multiply(n2,#2)|multiply(n5,#3)|divide(#5,#4)| | physics |
60 percent of movie theatres in town x have 2 screens or less . 20 % of those theatres sell an average of more than $ 300 worth of popcorn per showing . 56 percent of all the movie theatres in town x sell $ 300 or less of popcorn per showing . what percent of all the stores on the street have 4 or more screens and sell... | "lets take numbers here . assume that the total number of movie theaters in the town = 100 then number of movie theaters with 3 screens or less = 60 = > number of movie theaters with 4 screens or more = 40 movie theaters with 3 screens or less selling popcorn at more than $ 300 = 20 % of 60 = 12 number of movie theater... | a ) 12 , b ) 18 , c ) 32 , d ) 40 , e ) 44 | d | multiply(divide(20, 300), multiply(divide(20, 300), 300)) | divide(n2,n7)|multiply(n3,#0)|multiply(#0,#1)| | general |
if w is the set of all the integers between 69 and 99 , inclusive , that are either multiples of 3 or multiples of 2 or multiples of both , then w contains how many numbers ? | "multiples of 2 from 69 to 99 = multiples of 2 from 1 to 99 - multiples of 2 from 1 to 68 = [ 99 / 2 ] - [ 68 / 2 ] = 49 - 34 = 15 multiples of 3 from 69 to 99 = multiples of 3 from 1 to 99 - multiples of 3 from 1 to 68 = [ 99 / 3 ] - [ 68 / 3 ] = 33 - 23 = 10 multiples of 2 and 3 bothi . e . 6 from 69 to 99 = multiple... | a ) 20 , b ) 32 , c ) 33 , d ) 34 , e ) 35 | a | subtract(add(floor(divide(subtract(99, 69), 3)), divide(subtract(99, 69), 2)), floor(divide(subtract(99, 69), multiply(2, 3)))) | multiply(n3,n2)|subtract(n1,n0)|divide(#1,n3)|divide(#1,n2)|divide(#1,#0)|floor(#3)|floor(#4)|add(#2,#5)|subtract(#7,#6)| | other |
75 boys can complete a work in 26 days . how many men need to complete twice the work in 20 days | "one man can complete the work in 26 * 75 = 1950 days = one time work to complete the work twice it will be completed in let m be the no . of worker assign for this therefore the eqn becomes m * 20 = 2 * 1950 m = 195 workers answer : b" | a ) 160 , b ) 195 , c ) 180 , d ) 190 , e ) 200 | b | divide(multiply(75, multiply(26, const_2)), 20) | multiply(n1,const_2)|multiply(n0,#0)|divide(#1,n2)| | physics |
a trader bought a car at 20 % discount on its original price . he sold it at a 55 % increase on the price he bought it . what percent of profit did he make on the original price ? | "original price = 100 cp = 80 s = 80 * ( 155 / 100 ) = 124 100 - 124 = 24 % answer : b" | a ) 17 % , b ) 24 % , c ) 12 % , d ) 82 % , e ) 16 % | b | multiply(subtract(divide(divide(multiply(subtract(const_100, 20), add(const_100, 55)), const_100), const_100), const_1), const_100) | add(n1,const_100)|subtract(const_100,n0)|multiply(#0,#1)|divide(#2,const_100)|divide(#3,const_100)|subtract(#4,const_1)|multiply(#5,const_100)| | gain |
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