Problem stringlengths 5 967 | Rationale stringlengths 1 2.74k | options stringlengths 37 300 | correct stringclasses 5
values | annotated_formula stringlengths 7 6.48k | linear_formula stringlengths 8 925 | category stringclasses 6
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the largest natural number by which the product of 3 consecutive even natural numbers is always divisible , is | solution required number = ( 2 Γ 4 Γ 6 ) = 48 . answer c | a ) 16 , b ) 24 , c ) 48 , d ) 96 , e ) 78 | c | multiply(multiply(subtract(3, const_1), multiply(subtract(3, const_1), subtract(3, const_1))), add(multiply(subtract(3, const_1), subtract(3, const_1)), subtract(3, const_1))) | subtract(n0,const_1)|multiply(#0,#0)|add(#1,#0)|multiply(#1,#0)|multiply(#2,#3) | general |
a sum of money is put out at compound interest for 2 years at 20 % . it would fetch rs . 482 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 = 482 p = 2000 answer : d" | a ) 2099 , b ) 6086 , c ) 2976 , d ) 2000 , e ) 3862 | 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 group of hikers is planning a trip that will take them up a mountain using one route and back down using another route . they plan to travel down the mountain at a rate of one and a half times the rate they will use on the way up , but the time each route will take is the same . if they will go up the mountain at a r... | "on the way down , the rate is 1.5 * 5 = 7.5 miles per day . the distance of the route down the mountain is 2 * 7.5 = 15 miles . the answer is e ." | a ) 4 , b ) 6 , c ) 8 , d ) 12 , e ) 15 | e | multiply(multiply(5, const_2), divide(const_3, const_2)) | divide(const_3,const_2)|multiply(n0,const_2)|multiply(#0,#1)| | physics |
in the xy - coordinate plane , the graph of y = - x ^ 2 + 4 intersects line l at ( p , 5 ) and ( t , - 7 ) . what is the least possible value of the slope of line l ? | "we need to find out the value of p and l to get to the slope . line l and graph y intersect at point ( p , 5 ) . hence , x = p and y = 5 should sactisfy the graph . soliving 5 = - p 2 + 4 p 2 = 1 p = + or - 1 simillarly point ( t , - 7 ) should satisfy the equation . hence x = t and y = - 7 . - 7 = - t 2 + 9 t = + or ... | a ) 6 , b ) 2 , c ) - 4 , d ) - 6 , e ) - 10 | c | divide(subtract(7, 5), subtract(sqrt(add(4, 7)), sqrt(add(5, 4)))) | add(n1,n3)|add(n1,n2)|subtract(n3,n2)|sqrt(#0)|sqrt(#1)|subtract(#3,#4)|divide(#2,#5)| | general |
the value of a machine depreciates at 25 % 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 ? | "the value of the machine after two years = 0.75 * 0.75 * 1 , 50,000 = $ 84,375 sp such that a profit of $ 24,000 is made = 84,375 + 24,000 = $ 1 , 08,375 d" | a ) $ 258375 , b ) $ 438375 , c ) $ 128375 , d ) $ 108375 , e ) $ 158375 | d | add(multiply(multiply(subtract(1, divide(25, const_100)), subtract(1, divide(25, 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(25, 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 |
triangle atriangle b are similar triangles with areas 2268 units square and 2527 units square respectively . the ratio of there corresponding height would be | "let x be the height of triangle a and y be the height of triangle of b . since triangles are similar , ratio of area of a and b is in the ratio of x ^ 2 / y ^ 2 therefore , ( x ^ 2 / y ^ 2 ) = 2268 / 2527 ( x ^ 2 / y ^ 2 ) = ( 18 * 18 * 7 ) / ( 19 * 19 * 7 ) ( x ^ 2 / y ^ 2 ) = 18 ^ 2 / 19 ^ 2 x / y = 18 / 19 ans = b" | a ) 9 : 10 , b ) 18 : 19 , c ) 23 : 27 , d ) 13 : 17 , e ) 15 : 23 | b | sqrt(divide(2268, 2527)) | divide(n0,n1)|sqrt(#0)| | geometry |
40 litres of diesel is required to travel 600 km using a 800 cc engine . if the volume of diesel required to cover a distance varies directly as the capacity of the engine , then how many litres of diesel is required to travel 800 km using 1200 cc engine ? | "explanatory answer to cover a distance of 800 kms using a 800 cc engine , the amount of diesel required = 800 / 600 * 40 = 53.33 litres . however , the vehicle uses a 1200 cc engine and the question states that the amount of diesel required varies directly as the engine capacity . i . e . , for instance , if the capac... | a ) 80 litres , b ) 90 litres , c ) 120 litres , d ) 170 litres , e ) none of these | a | multiply(1200, multiply(800, divide(40, multiply(800, 600)))) | multiply(n1,n2)|divide(n0,#0)|multiply(n2,#1)|multiply(n4,#2)| | physics |
a and b go around a circular track of length 600 m on a cycle at speeds of 24 kmph and 60 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 { 600 / ( 24 * 5 / 18 ) , 600 / ( 60 * 5 / 18 ) } = 180 sec . answer : d" | a ) 120 sec , b ) 176 sec , c ) 178 sec , d ) 180 sec , e ) 189 sec | d | divide(600, subtract(multiply(60, const_0_2778), multiply(24, const_0_2778))) | multiply(n2,const_0_2778)|multiply(n1,const_0_2778)|subtract(#0,#1)|divide(n0,#2)| | physics |
the captain of a cricket team of 11 members is 28 years old and the wicket keeper is 3 years older . if the ages of these two are excluded , the average age of the remaining players is one year less than the average age of the whole team . what is the average age of the team ? | "explanation : let the average age of the whole team by x years . 11 x Γ’ β¬ β ( 28 + 31 ) = 9 ( x - 1 ) 11 x Γ’ β¬ β 9 x = 50 2 x = 50 x = 25 so , average age of the team is 25 years . answer e" | a ) 20 years , b ) 21 years , c ) 22 years , d ) 23 years , e ) 25 years | e | divide(subtract(add(28, add(28, 3)), multiply(3, 3)), const_2) | add(n1,n2)|multiply(n2,n2)|add(n1,#0)|subtract(#2,#1)|divide(#3,const_2)| | general |
during 2003 , a company produced an average of 2,000 products per month . how many products will the company need to produce from 2004 through 2007 in order to increase its monthly average for the period from 2003 through 2007 by 150 % over its 2003 average ? | company produced 12 * 2000 = 24,000 products in 2003 . if company produces x products from 2004 to 2007 , then total amount of product produced in 4 years ( 2003 through 2007 ) is x + 24,000 . the gives the average of ( x + 24,000 ) / 4 . this average needs to be 300 % higher than that in 2003 . in math terms , 36,000 ... | a ) 287,000 , b ) 290,000 , c ) 284,000 , d ) 285,000 , e ) 286,000 | c | multiply(150, divide(subtract(2007, 2003), const_2)) | subtract(n3,n0)|divide(#0,const_2)|multiply(n6,#1) | general |
a library has an average of 510 visitors on sundays and 240 on other day . the average number of visitors in a month of 30 days starting with sunday is | "explanation : as the month begin with sunday , so there will be five sundays in the month . so result will be : = ( 510 Γ 5 + 240 Γ 25 / 30 ) = ( 8550 / 30 ) = 285 option b" | a ) 280 , b ) 285 , c ) 290 , d ) 295 , e ) 275 | b | divide(add(multiply(add(floor(divide(30, add(const_3, const_4))), const_1), 510), multiply(subtract(30, add(floor(divide(30, add(const_3, const_4))), const_1)), 240)), 30) | add(const_3,const_4)|divide(n2,#0)|floor(#1)|add(#2,const_1)|multiply(n0,#3)|subtract(n2,#3)|multiply(n1,#5)|add(#4,#6)|divide(#7,n2)| | general |
a train speeds past a pole in 15 sec and a platform 150 m long in 25 sec , its length is ? | "let the length of the train be x m and its speed be y m / sec . then , x / y = 15 = > y = x / 15 ( x + 150 ) / 25 = x / 15 = > x = 225 m . answer : option d" | a ) 100 , b ) 150 , c ) 160 , d ) 225 , e ) 180 | d | multiply(150, subtract(const_2, const_1)) | subtract(const_2,const_1)|multiply(n1,#0)| | physics |
there are 2 sections a and b in a class , consisting of 40 and 30 students respectively . if the average weight of section a is 50 kg and that of section b is 60 kg , find the average of the whole class ? | "total weight of 40 + 30 students = 40 * 50 + 30 * 60 = 2000 + 1800 average weight of the class is = 3800 / 70 = 54.28 kg answer is e" | a ) 50.78 kg , b ) 49.32 kg , c ) 61.67 kg , d ) 69.15 kg , e ) 54.28 kg | e | divide(add(multiply(40, 50), multiply(30, 60)), add(40, 30)) | add(n1,n2)|multiply(n1,n3)|multiply(n2,n4)|add(#1,#2)|divide(#3,#0)| | general |
a customer purchased a package of ground beef at a cost of $ 1.76 per pound . for the same amount of money , the customer could have purchased a piece of steak that weighed 36 percent less than the package of ground beef . what was the cost per pound of the steak ? | "for simplicity , let ' s assume the customer bought 1 pound of ground beef for $ 1.76 . let x be the price per pound for the steak . then 0.64 x = 176 x = 176 / 0.64 = $ 2.75 the answer is d ." | a ) $ 2.00 , b ) $ 2.25 , c ) $ 2.50 , d ) $ 2.75 , e ) $ 3.00 | d | divide(1.76, add(multiply(const_0_25, const_2), multiply(const_0_33, const_1))) | multiply(const_0_25,const_2)|multiply(const_0_33,const_1)|add(#0,#1)|divide(n0,#2)| | general |
a train 275 meters long is running with a speed of 60 kmph . in what time will it pass a man who is running at 6 kmph in the direction opposite to that in which the train is going ? | "speed of train relative to man = ( 60 + 6 ) km / hr = 66 km / hr [ 66 * 5 / 18 ] m / sec = [ 55 / 3 ] m / sec . time taken to pass the man = [ 275 * 3 / 55 ] sec = 15 sec answer : c" | a ) 7 sec , b ) 5 sec , c ) 15 sec , d ) 6 sec , e ) 12 sec | c | multiply(const_3600, divide(divide(275, const_1000), add(60, 6))) | add(n1,n2)|divide(n0,const_1000)|divide(#1,#0)|multiply(#2,const_3600)| | physics |
lagaan is levied on the 60 percent of the cultivated land . the revenue department collected total rs . 3 , 74,000 through the lagaan from the village of mutter . mutter , a very rich farmer , paid only rs . 480 as lagaan . the percentage of total land of mutter over the total taxable land of the village is : | "total land of sukhiya = \ inline \ frac { 480 x } { 0.6 } = 800 x \ therefore cultivated land of village = 374000 x \ therefore required percentage = \ inline \ frac { 800 x } { 374000 } \ times 100 = 0.21390 d" | a ) 0.20833 , b ) 0.14544 , c ) 0.25632 , d ) 0.2139 , e ) 0.63435 | d | multiply(divide(divide(480, divide(60, const_100)), add(multiply(multiply(3, const_100), const_1000), multiply(add(multiply(const_4, const_10), const_4), const_1000))), const_100) | divide(n0,const_100)|multiply(n1,const_100)|multiply(const_10,const_4)|add(#2,const_4)|divide(n3,#0)|multiply(#1,const_1000)|multiply(#3,const_1000)|add(#5,#6)|divide(#4,#7)|multiply(#8,const_100)| | general |
the average ( arithmetic mean ) of the even integers from 16 to 44 inclusive is how much greater than the average ( arithmetic mean ) of the even integers from 14 to 56 inclusive ? | "so , according to a mean of a set of even numbers from 16 to 44 = ( 16 + 44 ) / 2 = 30 and mean of a set of even numbers from 14 to 56 = ( 14 + 56 ) / 2 = 35 difference = 35 - 30 = 5 answer : b ." | a ) 10 , b ) 5 , c ) 15 , d ) 20 , e ) 7 | b | subtract(divide(add(56, 14), const_2), divide(add(44, 16), const_2)) | add(n2,n3)|add(n0,n1)|divide(#0,const_2)|divide(#1,const_2)|subtract(#2,#3)| | general |
if a card is drawn from a well shuffled pack of cards , the probability of drawing a spade or a king is ? | "p ( s α΄ k ) = p ( s ) + p ( k ) - p ( s β© k ) , where s denotes spade and k denotes king . p ( s α΄ k ) = 13 / 52 + 4 / 52 - 1 / 52 = 4 / 13 answer : d" | a ) 4 / 15 , b ) 4 / 10 , c ) 4 / 18 , d ) 4 / 13 , e ) 4 / 11 | d | add(divide(const_3, const_52), divide(divide(const_52, const_4), const_52)) | divide(const_3,const_52)|divide(const_52,const_4)|divide(#1,const_52)|add(#0,#2)| | probability |
two employees a and b are paid a total of rs . 570 per week by their employer . if a is paid 150 percent of the sum paid to b , how much is b paid per week ? | "let the amount paid to a per week = x and the amount paid to b per week = y then x + y = 570 but x = 150 % of y = 150 y / 100 = 15 y / 10 β΄ 15 y / 10 + y = 570 β y [ 15 / 10 + 1 ] = 570 β 25 y / 10 = 570 β 25 y = 5700 β y = 5700 / 25 = rs . 228 d )" | a ) s . 130 , b ) s . 140 , c ) s . 150 , d ) s . 228 , e ) s . 282 | d | divide(570, add(divide(150, const_100), const_1)) | divide(n1,const_100)|add(#0,const_1)|divide(n0,#1)| | general |
a company has two models of computers , model m and model n . operating at a constant rate , a model m computer can complete a certain task in 36 minutes and a model n computer can complete the same task in 18 minutes . if the company used the same number of each model of computer to complete the task in 1 minute , how... | "let ' s say 1 work is processing 36 gb of data . model m : 1 gb per min model n : 2 gb per min working together , 1 m and 1 n = 3 gb per min so , 12 times as many computers would work at 36 gb per min . so no . of m = 12 answer is a" | a ) 12 , b ) 11 , c ) 10 , d ) 9 , e ) 8 | a | inverse(add(inverse(18), inverse(36))) | inverse(n1)|inverse(n0)|add(#0,#1)|inverse(#2)| | physics |
a retailer sold an appliance for $ 70 . if the retailer ' s gross profit on the appliance was 25 percent of the retailer ' s cost for the appliance , how many dollars was the retailer ' s gross profit ? | "gross profit = selling price - cost = 70 - x 70 - x = 1 / 4 x x = 56 gp = 70 - 56 = 14" | a ) $ 14 , b ) $ 16 , c ) $ 20 , d ) $ 24 , e ) $ 25 | a | subtract(70, multiply(divide(const_1, add(divide(25, 70), const_1)), 70)) | divide(n1,n0)|add(#0,const_1)|divide(const_1,#1)|multiply(n0,#2)|subtract(n0,#3)| | gain |
a sporting goods store sold 60 frisbees in one week , some for $ 3 and the rest for $ 4 each . if receipts from frisbee sales for the week totaled $ 200 , what is the fewest number of $ 4 frisbees that could have been sold ? | the wording of the question is slightly misleading . when a question asks what is thefewestthatcouldhave been sold , it is suggesting that there could be many different quantities sold . in this question however , because we are told that exactly 64 frisbees have been sold and revenue was exactly $ 204 , there is only ... | a ) 24 , b ) 12 , c ) 20 , d ) 4 , e ) 2 | c | subtract(200, multiply(60, 3)) | multiply(n0,n1)|subtract(n3,#0) | general |
there are 16 bees in the hive , then 7 more fly . how many bees are there in all ? | 16 + 7 = 23 . answer is b . | a ) 7 , b ) 23 , c ) 12 , d ) 17 , e ) 25 | b | add(16, 7) | add(n0,n1)| | general |
laura took out a charge account at the general store and agreed to pay 6 % simple annual interest . if she charges $ 35 on her account in january , how much will she owe a year later , assuming she does not make any additional charges or payments ? | "principal that is amount taken by laura at year beginning = 35 $ rate of interest = 6 % interest = ( 6 / 100 ) * 35 = 2.10 $ total amount that laura owes a year later = 35 + 2.1 = 37.1 $ answer b" | a ) $ 2.10 , b ) $ 37.10 , c ) $ 37.16 , d ) $ 38.10 , e ) $ 38.80 | b | add(multiply(divide(6, const_100), 35), 35) | divide(n0,const_100)|multiply(n1,#0)|add(n1,#1)| | general |
with both inlets open , a water tank will be filled with water in 24 minutes . the first inlet alone would fill the tank in 2 hours . if in every minutes the second inlet admits 50 cubic meters of water than the first , what is the capacity of the tank ? | the work done by inlet a and b together in 1 min = 1 / 24 the work done by inlet a ( first inlet ) in 1 min = 1 / 120 the work done by inlet b ( second inlet ) in 1 min = ( 1 / 24 ) - ( 1 / 120 ) = 1 / 30 difference of work done by b and a = b - a = 50 cubic meter i . e . ( 1 / 30 ) - ( 1 / 120 ) = 50 cubic meter i . e... | a ) 2,000 , b ) 10,500 , c ) 11,750 , d ) 12,000 , e ) 13,000 | a | divide(divide(50, subtract(divide(subtract(divide(const_60, 24), inverse(const_2)), const_60), inverse(multiply(const_60, const_2)))), const_1000) | divide(const_60,n0)|inverse(const_2)|multiply(const_2,const_60)|inverse(#2)|subtract(#0,#1)|divide(#4,const_60)|subtract(#5,#3)|divide(n2,#6)|divide(#7,const_1000) | physics |
if 4 men can colour 48 m long cloth in 2 days , then 6 men can colour 36 m long cloth in ? | "the length of cloth painted by one man in one day = 48 / 4 Γ 2 = 6 m no . of days required to paint 36 m cloth by 6 men = 36 / 6 Γ 6 = 1 day . option ' a '" | a ) 1 day , b ) 2 days , c ) 3 days , d ) 4 days , e ) 5 days | a | divide(36, multiply(divide(48, multiply(4, 2)), 6)) | multiply(n0,n2)|divide(n1,#0)|multiply(n3,#1)|divide(n4,#2)| | physics |
find the area of a rhombus one side of which measures 20 cm and one diagonal is 23 cm . | "explanation : let other diagonal = 2 x cm . since diagonals of a rhombus bisect each other at right angles , we have : ( 20 ) 2 = ( 12 ) 2 + ( x ) 2 = > x = β ( 20 ) 2 β ( 12 ) 2 = β 256 = 16 cm . _ i so , other diagonal = 32 cm . area of rhombus = ( 1 / 2 ) x ( product of diagonals ) = ( 1 / 2 Γ 23 x 32 ) cm 2 = 368 ... | a ) 100 cm 2 , b ) 150 cm 2 , c ) 300 cm 2 , d ) 368 cm 2 , e ) 400 cm 2 | d | add(multiply(multiply(divide(const_1, const_2), 23), sqrt(subtract(multiply(multiply(20, 20), const_4), multiply(23, 23)))), 23) | divide(const_1,const_2)|multiply(n0,n0)|multiply(n1,n1)|multiply(n1,#0)|multiply(#1,const_4)|subtract(#4,#2)|sqrt(#5)|multiply(#3,#6)|add(n1,#7)| | geometry |
compound interest of rs . 6000 at 10 % per annum for 1 1 / 2 years will be ( interest compounded half yearly ) . | "10 % interest per annum will be 5 % interest half yearly for 3 terms ( 1 1 / 2 years ) so compound interest = 6000 [ 1 + ( 5 / 100 ) ] ^ 3 - 6000 = 6000 [ ( 21 / 20 ) ^ 3 - 1 ] = 6000 ( 9261 - 8000 ) / 8000 = 6 * 1261 / 8 = 945 answer : d" | a ) rs . 473 , b ) rs . 374 , c ) rs . 495 , d ) rs . 945 , e ) none of the above | d | subtract(multiply(6000, power(add(1, divide(divide(10, 2), const_100)), multiply(add(1, divide(1, 2)), 2))), 6000) | divide(n1,n4)|divide(n2,n4)|add(n2,#1)|divide(#0,const_100)|add(#3,n2)|multiply(#2,n4)|power(#4,#5)|multiply(n0,#6)|subtract(#7,n0)| | gain |
a and b started a business investing rs . 75,000 and rs 20,000 respectively . in what ratio the profit earned after 2 years be divided between a and b respectively ? | "a : b = 75000 : 20000 = 75 : 20 = 13 : 4 answer : e" | a ) 9 : 2 , b ) 3 : 2 , c ) 18 : 20 , d ) 18 : 4 , e ) 13 : 4 | e | divide(add(multiply(multiply(const_3, const_3), add(multiply(const_3, const_3), const_1)), 2), multiply(2, add(multiply(const_3, const_3), const_1))) | multiply(const_3,const_3)|add(#0,const_1)|multiply(#1,#0)|multiply(n2,#1)|add(n2,#2)|divide(#4,#3)| | gain |
a person borrows 5000 for 2 years at 4 % p . a . simple interest . he immediately lends it to another person at 61 β 4 % p . a . for 2 years . find his gain in the transaction per year . | "gain in 2 years = [ ( 5000 Γ 25 / 4 Γ 2 / 100 ) β ( 5000 Γ 4 Γ 2 / 100 ) ] = ( 625 β 400 ) = 225 . β΄ gain in 1 year = ( 225 β 2 ) = 112.50 answer a" | a ) 112.50 , b ) 125 , c ) 150 , d ) 167.5 , e ) none of these | a | multiply(5000, divide(const_1, const_100)) | divide(const_1,const_100)|multiply(n0,#0)| | gain |
a side of beef lost 30 percent of its weight in processing . if the side of beef weighed 500 pounds after processing , how many pounds did it weigh before processing ? | "let weight of side of beef before processing = x ( 70 / 100 ) * x = 500 = > x = ( 500 * 100 ) / 70 = 714.28 answer a" | a ) 714.28 , b ) 712 , c ) 714.88 , d ) 714.25 , e ) 714 | a | divide(multiply(500, const_100), subtract(const_100, 30)) | multiply(n1,const_100)|subtract(const_100,n0)|divide(#0,#1)| | gain |
if the price of a tv is first decreased by 20 % and then increased by 40 % , then the net change in the price will be : | "explanation : solution : let the original price be rs . 100 . new final price = 140 % of ( 80 % of 100 ) = rs . 140 / 100 * 80 / 100 * 100 = rs . 112 . . ' . increase = 12 % answer : b" | a ) 4 % increase , b ) 12 % increase , c ) 10 % decrease , d ) 6 % increase , e ) none of these | b | subtract(divide(multiply(subtract(const_100, 20), add(const_100, 40)), const_100), const_100) | add(n1,const_100)|subtract(const_100,n0)|multiply(#0,#1)|divide(#2,const_100)|subtract(#3,const_100)| | general |
two sets of 3 consecutive positive odd integers have exactly one integer in common . the sum of the integers in the set with greater numbers is how much greater than the sum of the integers in the other set ? | a = ( 1 , 3,5 ) , sum of this = 9 b = ( 5 , 7,9 ) , sum of this = 21 , the differenct between 21 - 9 = 12 hence , 12 is the answer i . e . d | a ) 4 , b ) 7 , c ) 8 , d ) 12 , e ) it can not be determined from the information given . | d | add(subtract(add(const_3, const_4), const_1), add(subtract(add(const_3, const_2), 3), subtract(add(const_2, const_4), subtract(add(const_3, const_2), 3)))) | add(const_2,const_3)|add(const_2,const_4)|add(const_3,const_4)|subtract(#0,n0)|subtract(#2,const_1)|subtract(#1,#3)|add(#3,#5)|add(#6,#4) | general |
a train running at 1 / 5 of its own speed reached a place in 20 hours . how much time could be saved if the train would have run at its own speed ? | "time taken if run its own speed = 1 / 5 * 20 = 4 hrs time saved = 20 - 4 = 16 hrs answer : d" | a ) 8 hrs , b ) 10 hrs , c ) 12 hrs , d ) 16 hrs , e ) 6 hrs | d | multiply(divide(1, 5), 20) | divide(n0,n1)|multiply(n2,#0)| | physics |
in a certain pond , 60 fish were caught , tagged , and returned to the pond . a few days later , 50 fish were caught again , of which 2 were found to have been tagged . if the percent of tagged fish in the second catch approximates the percent of tagged fish in the pond , what is the approximate number of fish in the p... | "this is a rather straight forward ratio problem . 1 . 60 fish tagged 2 . 2 out of the 50 fish caught were tagged thus 2 / 50 2 / 50 = 60 / x thus , x = 1500 think of the analogy : 2 fish is to 50 fish as 50 fish is to . . . ? you ' ve tagged 50 fish and you need to find what that comprises as a percentage of the total... | a ) 400 , b ) 1,500 , c ) 1,250 , d ) 2,500 , e ) 10,000 | b | divide(60, divide(2, 50)) | divide(n2,n1)|divide(n0,#0)| | gain |
how many 4 - digit even numbers are possible such that if one of the digits is 5 , the next / succeeding digit to it should be 7 ? | 5700 , 5710 , . . . . . . 5798 , so total 50 . hence option d . | a ) 5 , b ) 305 , c ) 365 , d ) 50 , e ) 495 | d | add(add(4, 7), 7) | add(n0,n2)|add(n2,#0)| | general |
a man has 1044 candles . after burning , he can make a new candle from 9 stubs left behind . find the maximum number of candles that can be made . | explanation : since , 1 candle is made from 9 stubs . therefore , from 1044 stubs , 1044 / 9 i . e 116 candles will be made . now , from 116 stubs , 116 / 9 i . e 12 candles can be made and 8 stubs will be left . thus , total stubs left is ( 12 + 8 ) = 20 out of which 20 / 9 i . e 2 candles can be made and 2 stubs will... | a ) 116 , b ) 120 , c ) 130 , d ) 140 , e ) none | c | add(divide(1044, 9), floor(divide(divide(1044, 9), 9))) | divide(n0,n1)|divide(#0,n1)|floor(#1)|add(#0,#2) | general |
find the amount on rs . 7000 in 2 years , the rate of interest being 4 % per first year and 5 % for the second year ? | "7000 * 104 / 100 * 105 / 100 = > 7644 answer : d" | a ) 5568 , b ) 6369 , c ) 5460 , d ) 7644 , e ) 6734 | d | divide(multiply(divide(multiply(7000, add(const_100, 4)), const_100), add(const_100, 5)), const_100) | add(n3,const_100)|add(n2,const_100)|multiply(n0,#1)|divide(#2,const_100)|multiply(#0,#3)|divide(#4,const_100)| | gain |
a merchant has 1000 kg of sugar part of which he sells at 8 % profit and the rest at 18 % profit . he gains 14 % on the whole . the quantity sold at 18 % profit is | explanation : by the rule of alligation , we have profit % by selling 1 st part profit % by selling 2 nd part 8 18 net % profit 14 18 - 14 = 4 14 - 8 = 6 = > quantity of part 1 : quantity of part 2 = 4 : 6 = 2 : 3 total quantity is given as 1000 kg so quantity of part 2 ( quantity sold at 18 % profit ) = 1000 Γ 35 = 60... | a ) 300 , b ) 400 , c ) 600 , d ) 500 , e ) none of these | c | multiply(multiply(add(const_2, const_3), const_2), multiply(subtract(add(divide(14, const_100), const_1), add(const_1, divide(8, const_100))), 1000)) | add(const_2,const_3)|divide(n3,const_100)|divide(n1,const_100)|add(#1,const_1)|add(#2,const_1)|multiply(#0,const_2)|subtract(#3,#4)|multiply(n0,#6)|multiply(#5,#7) | gain |
a trader sells 40 metres of cloth for rs . 8200 at a profit of rs . 25 per metre of cloth . how much profit will the trder earn on 40 metres of cloth ? | "explanation : sp of 1 metre cloth = 8200 / 40 = rs . 205 . cp of 1 metre cloth = rs . 205 β 25 = rs . 180 cp on 40 metres = 180 x 40 = rs . 7200 profit earned on 40 metres cloth = rs . 8200 β rs . 7200 = rs . 1000 . answer : option c" | a ) rs . 950 , b ) rs . 1500 , c ) rs . 1000 , d ) rs . 1200 , e ) none of these | c | multiply(25, 40) | multiply(n0,n2)| | gain |
a batsman makes a score of 64 runs in the 16 th innings and thus increased his average by 3 . find his average after the 16 th inning ? | "let the average after the 16 th inning be p . so , the average after the 15 th inning will be ( p - 3 ) hence , 15 ( p - 30 ) + 64 = 16 p = > p = 19 . answer : d" | a ) 23 , b ) 28 , c ) 27 , d ) 19 , e ) 24 | d | add(subtract(64, multiply(16, 3)), 3) | multiply(n1,n2)|subtract(n0,#0)|add(n2,#1)| | general |
when 2 is added to half of one - third of one - fifth of a number , the result is one - fifteenth of the number . find the number ? | "d 60 let the number be 2 + 1 / 2 [ 1 / 3 ( a / 5 ) ] = a / 15 = > 2 = a / 30 = > a = 60" | a ) 40 , b ) 65 , c ) 73 , d ) 60 , e ) 80 | d | divide(2, divide(divide(const_1, multiply(2, add(const_2, 2))), const_2)) | add(const_2,n0)|multiply(#0,n0)|divide(const_1,#1)|divide(#2,const_2)|divide(n0,#3)| | general |
a sells a bicycle to b at a profit of 20 % . b sells it to c at a profit of 25 % . if c pays rs . 225 for it , the cost price of the bicycle for a is | explanation : 125 % of 120 % of a = 225 125 / 100 * 120 / 100 * a = 225 a = 225 * 2 / 3 = 150 . answer : d | a ) 627 , b ) 297 , c ) 127 , d ) 150 , e ) 181 | d | divide(multiply(const_100, divide(multiply(const_100, 225), add(25, const_100))), add(20, const_100)) | add(n1,const_100)|add(n0,const_100)|multiply(n2,const_100)|divide(#2,#0)|multiply(#3,const_100)|divide(#4,#1) | gain |
the perimeter of a triangle is 32 cm and the inradius of the triangle is 3.5 cm . what is the area of the triangle ? | "area of a triangle = r * s where r is the inradius and s is the semi perimeter of the triangle . area of triangle = 3.5 * 32 / 2 = 56 cm 2 answer : c" | a ) 22 , b ) 35 , c ) 56 , d ) 54 , e ) 23 | c | triangle_area(3.5, 32) | triangle_area(n0,n1)| | geometry |
if circles x and y have the same area and circle x has a circumference of 10 Ο , half of the radius of circle y is : | "x be radius of circle x y be radius of circle y given : pi * x ^ 2 = pi * y ^ 2 also , 2 * pi * x = 10 * pi x = 5 thus y = 5 y / 2 = 2.5 ans : d" | a ) 16 , b ) 8 , c ) 4 , d ) 2.5 , e ) 1 | d | divide(divide(10, const_2), const_2) | divide(n0,const_2)|divide(#0,const_2)| | geometry |
how many positive integers , from 2 to 100 , inclusive , are not divisible by even integers greater than 1 ? | "the no will be of the form 2 ^ n to achieve this . since any other form will have odd no in the prime factorization . hence we need to find solution of n for equation - 1 < 2 ^ n < 101 . 2 ^ 6 = 64 , 2 ^ 7 - 128 hence n can take values from 1 to 6 . hence ans - ( b ) 6" | a ) 5 , b ) 6 , c ) 8 , d ) 10 , e ) 50 | b | divide(subtract(multiply(multiply(2, multiply(2, multiply(2, multiply(2, 2)))), 2), const_4), const_10) | multiply(n0,n0)|multiply(n0,#0)|multiply(n0,#1)|multiply(n0,#2)|multiply(n0,#3)|subtract(#4,const_4)|divide(#5,const_10)| | general |
how many terminating zeroes does 200 ! have ? | we must realize that the trailing zeros depend on the number of 10 s ( i . e . 10 s or its multiples ) in the factorial . 10 = 5 * 2 , so we are looking for 5 s and 2 s . by virtue of being the more frequent number , we do n ' t need to count the number of 2 s , because there will always be equal or more 2 s than 5 s i... | a ) 40 , b ) 48 , c ) 49 , d ) 55 , e ) 64 | c | add(divide(200, add(const_4, const_1)), divide(200, multiply(add(const_4, const_1), add(const_4, const_1)))) | add(const_1,const_4)|divide(n0,#0)|multiply(#0,#0)|divide(n0,#2)|add(#1,#3) | other |
two airplanes fly the same exact route from atlanta to chicago . plane a flies 200 mph while plane b flies 300 mph ( assume that simple model versus actual acceleration / deceleration near the airports ) . if plane a took off exactly 40 minutes before plane b , after how many minutes will plane b overtake plane a ? | we can use some form of the equation d = rt [ distance = rate * time ] rate of plane a : 200 mph rate of plane b : 300 mph plane a will be in the sky 40 minutes longer than plane b ( 2 / 3 of an hour ) time of plane a when b overtakes it : t + ( 2 / 3 ) use 2 / 3 of an hour since the rate is in hours time of plane b wh... | a ) 65 , b ) 80 , c ) 90 , d ) 115 , e ) 120 | b | divide(multiply(40, 200), subtract(300, 200)) | multiply(n0,n2)|subtract(n1,n0)|divide(#0,#1) | general |
a certain manufacturer produces items for which the production costs consist of annual fixed costs totaling $ 130,000 and variables costs averaging $ 11 per item . if the manufacturer β s selling price per item is $ 15 , how many items the manufacturer produce and sell to earn an annual profit of $ 150,000 ? | "let the items manufactured or sold bex 130000 + 11 x = 15 x - 150000 4 x = 280000 x = 70000 ans : e" | a ) 2,858 , b ) 8,667 , c ) 21,429 , d ) 35,000 , e ) 70,000 | e | divide(multiply(divide(divide(multiply(add(add(const_3, const_10), 15), power(const_100, const_2)), subtract(15, 11)), multiply(const_4, const_2)), const_3), const_1000) | add(const_10,const_3)|multiply(const_2,const_4)|power(const_100,const_2)|subtract(n2,n1)|add(n2,#0)|multiply(#4,#2)|divide(#5,#3)|divide(#6,#1)|multiply(#7,const_3)|divide(#8,const_1000)| | general |
the area of a parallelogram is 450 sq m and its altitude is twice the corresponding base . then the length of the base is ? | "2 x * x = 450 = > x = 15 answer : e" | a ) 8 , b ) 9 , c ) 7 , d ) 6 , e ) 15 | e | sqrt(divide(450, const_2)) | divide(n0,const_2)|sqrt(#0)| | geometry |
a math teacher has 32 cards , each of which is in the shape of a geometric figure . half of the cards are rectangles , and a third of the cards are rhombuses . if 8 cards are squares , what is the maximum possible number of cards that re circles . | "a square is a special kind of rhombus ( sides are perpendicular ) a square is a special kind of rectangles ( sides with same length ) among the 32 cards with have : 15 rectangles 10 rhombus 8 squares among the 15 rectangles , there could be 8 special ones ( with sides of same length ) that are squares . that lets at l... | a ) 9 , b ) 10 , c ) 11 , d ) 12 , e ) 15 | e | subtract(subtract(divide(divide(multiply(32, 8), const_10), const_2), const_0_25), const_0_25) | multiply(n0,n1)|divide(#0,const_10)|divide(#1,const_2)|subtract(#2,const_0_25)|subtract(#3,const_0_25)| | geometry |
in one hour , a boat goes 15 km along the stream and 5 km against the stream . the speed of the boat in still water ( in km / hr ) is : | "sol . speed in still water = 1 / 2 ( 15 + 5 ) kmph = 10 kmph . answer c" | a ) 2 , b ) 4 , c ) 10 , d ) 12 , e ) 15 | c | divide(add(15, 5), const_2) | add(n0,n1)|divide(#0,const_2)| | physics |
find the least number which when divided by 20 , 2535 and 40 leaves remainders 14 , 1929 and 34 respectively . | here , ( 20 - 14 ) = 6 , ( 25 β 19 ) = 6 , ( 35 - 29 ) = 6 and ( 40 - 34 ) = 6 . therefore , required number = ( l . c . m . of 2025 , 3540 ) β 6 = 1394 . answer is a . | a ) 1394 , b ) 1294 , c ) 1194 , d ) 1094 , e ) none of them | a | subtract(multiply(multiply(const_100, divide(14, const_2)), const_2), multiply(const_2, const_3)) | divide(n3,const_2)|multiply(const_2,const_3)|multiply(#0,const_100)|multiply(#2,const_2)|subtract(#3,#1) | general |
bag a contains red , white and blue marbles such that the red to white marble ratio is 1 : 3 and the white to blue marble ratio is 2 : 3 . bag b contains red and white marbles in the ratio of 1 : 4 . together , the two bags contain 28 white marbles . how many red marbles could be in bag a ? | "# of red marbles in bag a can be either 2 or 6 . no 2 in the choices , so 6 . d . bag a : r : w : b = 2 : 6 : 9 bag b r : w = 1 : 4 6 x + 4 y = 28 i . e 3 x + 2 y = 14 x has to be odd to make an odd sum from the eq . x = 2 , y = 4 or x = 4 , y = 1 so r can be 2 x i . e 4 or 8 . answer c" | a ) 1 , b ) 3 , c ) 4 , d ) 6 , e ) 7 | c | divide(28, add(multiply(3, 2), 4)) | multiply(n1,n2)|add(n5,#0)|divide(n6,#1)| | other |
when average age of 22 members are 0 , how many members greater than 0 ? | average of 22 numbers = 0 . sum of 22 numbers ( 0 x 22 ) = 0 . it is quite possible that 21 of these numbers may be positive and if their sum is a then 22 nd number is ( - a ) answer is 21 ( b ) | a ) 17 , b ) 21 , c ) 22 , d ) 24 , e ) 25 | b | subtract(22, const_1) | subtract(n0,const_1) | general |
if 10 spiders make 5 webs in 5 days , then how many days are needed for 1 spider to make 1 web ? | "explanation : let , 1 spider make 1 web in x days . more spiders , less days ( indirect proportion ) more webs , more days ( direct proportion ) hence we can write as ( spiders ) 10 : 1 ( webs ) 1 : 5 } : : x : 5 Γ’ β‘ β 10 Γ£ β 1 Γ£ β 5 = 1 Γ£ β 5 Γ£ β x Γ’ β‘ β x = 10 answer : option a" | a ) 10 , b ) 9 , c ) 5 , d ) 15 , e ) 20 | a | multiply(1, 10) | multiply(n0,n3)| | physics |
in a certain egg - processing plant , every egg must be inspected , and is either accepted for processing or rejected . for every 96 eggs accepted for processing , 4 eggs are rejected . if , on a particular day , 12 additional eggs were accepted , but the overall number of eggs inspected remained the same , the ratio o... | "straight pluggin in for me . as usual , i started with c and got the answer . lets ' back calculate and see what we get let us consider eggs processed each day to be 400 so initial ratio of eggs processed and rejected is 96 : 4 or 24 : 1 so out of 400 eggs , there will be 384 eggs processed and 16 rejected . now if th... | a ) 100 , b ) 300 , c ) 400 , d ) 3,000 , e ) 4,000 | c | multiply(divide(12, subtract(99, 96)), const_100) | subtract(n3,n0)|divide(n2,#0)|multiply(#1,const_100)| | general |
ram and shakil run a race of 2000 m . first ram gives shakil a start of 200 m and beats him by 1 minute . next ram gives shakil a start of 6 min and is beaten by 1000 meters . find the time in minutes in which ram and shakil can run the race seperately . | let r be velocity of ram and s be the velocity of shakil then ( 2000 / r ) = ( 1800 / s ) - 60 . . . . . . ( 1 ) second case distance shakil travel in 6 min = 360 s again equating time for both 1000 / r = ( 2000 - 360 s ) / s . . . . . . . ( 2 ) solving ( 1 ) & ( 2 ) we get s = 20 / 6 putting value in first we get time... | a ) 12.18 , b ) 10.18 , c ) 10.12 , d ) 8.1 , e ) 8.2 | d | divide(2000, add(200, divide(200, const_4))) | divide(n1,const_4)|add(n1,#0)|divide(n0,#1) | physics |
ravi and kavi start a business by investing Γ’ β ΒΉ 4000 and Γ’ β ΒΉ 72000 , respectively . find the ratio of their profits at the end of year . | "ratio of profit = ratio of investments = 4000 : 72000 = 1 : 18 answer : d" | a ) 2 : 28 , b ) 5 : 28 , c ) 7 : 18 , d ) 1 : 18 , e ) 3 : 18 | d | divide(4000, 72000) | divide(n0,n1)| | gain |
108 . triangle a β s base is 12 % greater than the base of triangle b , and a β s height is 12 % less than the height of triangle b . the area of triangle a is what percent less or more than the area of triangle b ? | "wish the question specified that we are talking about corresponding height . base of a = 13 / 12 * base of b height of a = 11 / 12 * height of b area of a = ( 1 / 2 ) * base of a * height of a = 13 / 12 * 11 / 12 * area of b = 143 / 144 * area of b area of a is 0.7 % less than the area of b . answer ( b )" | a ) 9 % less , b ) 0.7 % less , c ) equal to each other , d ) 1 % more , e ) 9 % more | b | divide(const_100, subtract(multiply(const_100, const_100), multiply(add(const_100, 12), subtract(const_100, 12)))) | add(n1,const_100)|multiply(const_100,const_100)|subtract(const_100,n1)|multiply(#0,#2)|subtract(#1,#3)|divide(const_100,#4)| | geometry |
two airplanes take off from one airfield at noon . one flies due east at 204 miles per hour while the other flies directly northeast at 283 miles per hour . approximately how many miles apart are the airplanes at 2 p . m . ? | b in two hours : the plane flying east will be 408 miles away from airport . the other plane will be 566 miles away from airport . 566 / 408 = ~ 1.4 = ~ sqrt ( 2 ) this means that planes formed a right isocheles triangle = > sides of such triangles relate as 1 : 1 : sqrt ( 2 ) = > the planes are 408 miles apart . b | a ) 166 , b ) 408 , c ) 400 , d ) 483 , e ) 566 | b | sqrt(subtract(power(multiply(283, const_2), const_2), power(multiply(204, 2), const_2))) | multiply(n1,const_2)|multiply(n0,n2)|power(#0,const_2)|power(#1,const_2)|subtract(#2,#3)|sqrt(#4) | physics |
the triplicate ratio of 2 : 5 is ? | "23 : 53 = 8 : 125 answer : b" | a ) 1 : 0 , b ) 8 : 125 , c ) 1 : 7 , d ) 1 : 2 , e ) 1 : 1 | b | divide(power(2, const_3), power(const_3.0, 5)) | power(n0,const_3)|power(const_3.0,n1)|divide(#0,#1)| | other |
how many two - digit numbers that do not contain the digits 1 or 4 are there ? | "we can have 7 digits ( 2 , 3,5 , 6,7 , 8,9 ) for the first place ( tenth ' s place ) . and similarly 8 digits for the unit digit . ( 0,2 , 3,5 , 6,7 , 8,9 ) so in total 7 * 8 = 56 hence b" | a ) 112 , b ) 56 , c ) 64 , d ) 49 , e ) 84 | b | multiply(multiply(multiply(const_10, const_10), subtract(const_10, const_1)), 4) | multiply(const_10,const_10)|subtract(const_10,const_1)|multiply(#0,#1)|multiply(n1,#2)| | general |
marla starts running around a circular track at the same time nick starts walking around the same circular track . marla completes 10 laps around the track per hour and nick completes 5 laps around the track per hour . how many minutes after marla and nick begin moving will marla have completed 4 more laps around the t... | maria ' s rate - 10 laps per hour - - > 10 / 60 laps / min nick ' s rate - 5 laps per hour - - > 5 / 60 laps / min lets set equations : 10 / 60 * t = 4 ( since maria had to run 4 laps before nick would start ) 5 / 60 * t = 0 ( hick has just started and has n ' t run any lap yet ) ( 10 / 60 - 5 / 60 ) * t = 4 - 0 ( sinc... | a ) 5 , b ) 48 , c ) 12 , d ) 15 , e ) 20 | b | divide(4, subtract(divide(10, multiply(add(5, const_1), 10)), divide(5, multiply(add(5, const_1), 10)))) | add(n1,const_1)|multiply(n0,#0)|divide(n0,#1)|divide(n1,#1)|subtract(#2,#3)|divide(n2,#4) | physics |
the average expenditure of a labourer for 6 months was 75 and he fell into debt . in the next 4 months by reducing his monthly expenses to 60 he not only cleared off his debt but also saved 30 . his monthly income i | "income of 6 months = ( 6 Γ 75 ) β debt = 450 β debt income of the man for next 4 months = 4 Γ 60 + debt + 30 = 270 + debt β΄ income of 10 months = 720 average monthly income = 720 Γ· 10 = 72 answer b" | a ) 70 , b ) 72 , c ) 75 , d ) 78 , e ) 80 | b | divide(add(add(multiply(75, 6), multiply(60, 4)), 30), add(6, 4)) | add(n0,n2)|multiply(n0,n1)|multiply(n2,n3)|add(#1,#2)|add(n4,#3)|divide(#4,#0)| | general |
dick and jane each saved $ 2,500 in 1989 . in 1990 dick saved 9 percent more than in 1989 , and together he and jane saved a total of $ 5200 . approximately what percent less did jane save in 1990 than in 1989 ? | "1990 dick saved = $ 2725 jane saved = $ 2475 ( jane saved $ 25 less than she did the prior year ) jane saved approximately $ 25 / $ 2500 ( 1 % ) less in 1990 answer : a" | a ) 1 % , b ) 4 % , c ) 6 % , d ) 8 % , e ) 9 % | a | multiply(subtract(const_1, subtract(const_2, add(divide(9, const_100), const_1))), const_100) | divide(n3,const_100)|add(#0,const_1)|subtract(const_2,#1)|subtract(const_1,#2)|multiply(#3,const_100)| | general |
seed mixture x is 40 % ryegrass and 60 % bluegrass by weight ; seed mixture y is 25 % ryegrass and 75 % fescue . if a mixture of x and y contains 31 % ryegrass , what percent of the weight of the mixture is from mixture x ? | "31 % is 6 % - points above 25 % and 9 % - points below 40 % . thus the ratio of mixture y to mixture x is 3 : 2 . the percent of mixture x is 2 / 5 = 40 % . the answer is c ." | a ) 20 % , b ) 30 % , c ) 40 % , d ) 50 % , e ) 60 % | c | divide(subtract(31, add(25, const_1)), subtract(divide(40, const_100), divide(add(25, const_1), const_100))) | add(n2,const_1)|divide(n0,const_100)|divide(#0,const_100)|subtract(n4,#0)|subtract(#1,#2)|divide(#3,#4)| | gain |
in a tree , 3 / 7 of the birds are parrots while the rest are toucans . if 2 / 3 of the parrots are female and 7 / 10 of the toucans are female , what fraction of the birds in the tree are male ? | "let x be the number of birds in the tree . the fraction of birds that are male parrots is ( 1 / 3 ) ( 3 / 7 ) = 1 / 7 . the fraction of birds that are male toucans is ( 3 / 10 ) ( 4 / 7 ) = 6 / 35 . the total fraction of male birds is 1 / 7 + 6 / 35 = 11 / 35 . the answer is b ." | a ) 3 / 7 , b ) 11 / 35 , c ) 16 / 35 , d ) 31 / 70 , e ) 37 / 70 | b | add(multiply(divide(3, 7), subtract(2, divide(2, 3))), multiply(divide(const_2, 7), subtract(2, divide(3, 10)))) | divide(n0,n1)|divide(n2,n0)|divide(const_2,n1)|divide(n0,n5)|subtract(n2,#1)|subtract(n2,#3)|multiply(#0,#4)|multiply(#2,#5)|add(#6,#7)| | general |
income and expenditure of a person are in the ratio 10 : 8 . if the income of the person is rs . 10000 , then find his savings ? | "let the income and the expenditure of the person be rs . 10 x and rs . 8 x respectively . income , 10 x = 10000 = > x = 1000 savings = income - expenditure = 10 x - 8 x = 2 x = 2 ( 1000 ) so , savings = rs . 2000 . answer : c" | a ) 2003 , b ) 2002 , c ) 2000 , d ) 2001 , e ) 2004 | c | subtract(10000, multiply(divide(8, 10), 10000)) | divide(n1,n0)|multiply(n2,#0)|subtract(n2,#1)| | other |
in a school with 600 students , the average age of the boys is 12 years and that of the girls is 11 years . if the average age of the school is 11 years 9 months , then the number of girls in the school is | "sol . let the number of grils be x . then , number of boys = ( 600 - x ) . then , ( 11 3 / 4 Γ 600 ) β 11 x + 12 ( 600 - x ) β x = 7200 - 7050 β 150 . answer a" | a ) 150 , b ) 200 , c ) 250 , d ) 350 , e ) none | a | subtract(multiply(12, 600), multiply(add(11, divide(9, 12)), 600)) | divide(n4,n1)|multiply(n0,n1)|add(n2,#0)|multiply(n0,#2)|subtract(#1,#3)| | general |
when sold at a 50 % discount , a sweater nets the merchant a 40 % profit on the wholesale cost at which he initially purchased the item . by what % is the sweater marked up from wholesale at its normal retail price ? | we should be careful about what are we measuring % on / what is the base . . let the marked up price = 100 . . selling price = 100 - 50 % of 100 = 50 . . profit = 40 % . . therefore the wholesale purchase cost = x . . . . 1.4 x = 50 or x = 35.71 . . . marked price was 100 so . . . so answer is 64.29 % . . b | a ) 20 % , b ) 64.29 % , c ) 70 % , d ) 80 % , e ) 100 % | b | subtract(const_100, divide(subtract(const_100, 50), add(const_1, divide(40, const_100)))) | divide(n1,const_100)|subtract(const_100,n0)|add(#0,const_1)|divide(#1,#2)|subtract(const_100,#3) | gain |
one fourth of a solution that was 12 % sugar by weight was replaced by a second solution resulting in a solution that was 16 percent sugar by weight . the second solution was what percent sugar by weight ? | "instead of using complex calculations and remembering formulae , why dont u directly get to weighted average . 3 parts of 12 % + 1 part of x ( unknown ) % = 4 parts of 16 % = > x % = 64 % - 36 % = 28 % ans b it is ." | a ) 34 % , b ) 28 % , c ) 22 % , d ) 18 % , e ) 8.5 % | b | multiply(divide(subtract(multiply(const_100, divide(16, const_100)), multiply(subtract(const_100, multiply(divide(const_1, const_4), const_100)), divide(12, const_100))), multiply(divide(const_1, const_4), const_100)), const_100) | divide(n1,const_100)|divide(n0,const_100)|divide(const_1,const_4)|multiply(#0,const_100)|multiply(#2,const_100)|subtract(const_100,#4)|multiply(#1,#5)|subtract(#3,#6)|divide(#7,#4)|multiply(#8,const_100)| | gain |
a palindrome is a number that reads the same forward and backward , such as 144 . how many odd , 4 - digit numbers are palindromes ? | "a palindrome is a number that reads the same forward and backward . examples of four digit palindromes are 1221 , 4334 , 2222 etc you basically get to choose the first two digits and you repeat them in opposite order . say , you choose 45 as your first two digits . the next two digits are 54 and the number is 4554 . a... | a ) 40 , b ) 45 , c ) 50 , d ) 62 , e ) 2500 | d | divide(power(const_10, divide(4, const_2)), const_2) | divide(n1,const_2)|power(const_10,#0)|divide(#1,const_2)| | general |
a diagonal of a polygon is an segment between two non - adjacent vertices of the polygon . how many diagonals does a regular 30 - sided polygon have ? | "there ' s a direct formula for this . number of diagonals in a regular polygon = [ n * ( n - 3 ) ] / 2 , n = number of sides of the regular polygon . here , n = 30 . plugging it in , we get 405 diagonals ! answer ( b ) ." | a ) 875 , b ) 405 , c ) 1425 , d ) 2025 , e ) 2500 | b | divide(multiply(subtract(30, const_3), 30), const_2) | subtract(n0,const_3)|multiply(n0,#0)|divide(#1,const_2)| | geometry |
a batsman makes a score of 66 runs in the 17 th inning and thus increases his averages by 3 . what is his average after 17 th inning ? | "let the average after 17 innings = x total runs scored in 17 innings = 17 x average after 16 innings = ( x - 3 ) total runs scored in 16 innings = 16 ( x - 3 ) total runs scored in 16 innings + 66 = total runs scored in 17 innings = > 16 ( x - 3 ) + 66 = 17 x = > 16 x - 48 + 66 = 17 x = > x = 18 answer is e ." | a ) 25 , b ) 31 , c ) 27 , d ) 29 , e ) 18 | e | add(subtract(66, multiply(17, 3)), 3) | multiply(n1,n2)|subtract(n0,#0)|add(n2,#1)| | general |
find the value of e from ( 14 ) ^ 2 x 5 ^ 3 Γ£ Β· 568 = e . | given exp . = ( 14 ) ^ 2 x 5 ^ 3 Γ£ Β· 568 = e = 196 x 125 Γ£ Β· 568 43.1 b | a ) 13.1 , b ) 43.1 , c ) 40.1 , d ) 23.1 , e ) 45.1 | b | divide(multiply(power(14, 2), power(5, 3)), 568) | power(n0,n1)|power(n2,n3)|multiply(#0,#1)|divide(#2,n4) | general |
a cube of side one meter length is cut into small cubes of side 5 cm each . how many such small cubes can be obtained ? | "along one edge , the number of small cubes that can be cut = 100 / 5 = 20 along each edge 5 cubes can be cut . ( along length , breadth and height ) . total number of small cubes that can be cut = 20 * 20 * 20 = 8000 answer : b" | a ) 2878 , b ) 8000 , c ) 1000 , d ) 2997 , e ) 2701 | b | divide(power(power(5, const_2), const_3), power(5, const_3)) | power(n0,const_2)|power(n0,const_3)|power(#0,const_3)|divide(#2,#1)| | physics |
if 0.75 of a number is equal to 0.025 of another number , the ratio of the numbers is : | "0.75 a = 0.025 b - > a / b = 0.025 / 0.75 = 25 / 750 = 1 / 30 : . a : b = 1 : 30 answer : c" | a ) 2 : 3 , b ) 3 : 4 , c ) 1 : 30 , d ) 20 : 3 , e ) 30 : 7 | c | divide(multiply(0.025, const_100), multiply(0.75, const_100)) | multiply(n1,const_100)|multiply(n0,const_100)|divide(#0,#1)| | other |
a company has two models of computers , model x and model y . operating at a constant rate , a model x computer can complete a certain task in 60 minutes and a model y computer can complete the same task in 30 minutes . if the company used the same number of each model of computer to complete the task in 1 minute , how... | "let ' s say 1 work is processing 60 gb of data . model x : 1 gb per min model y : 2 gb per min working together , 1 x and 1 y = 3 gb per min so , 20 times as many computers would work at 60 gb per min . so no . of x = 20 answer is e" | a ) 18 , b ) 16 , c ) 15 , d ) 17 , e ) 20 | e | divide(multiply(60, 30), add(60, 30)) | add(n0,n1)|multiply(n0,n1)|divide(#1,#0)| | gain |
a certain number of men can do a work in 40 days . if there were 5 men less it could be finished in 10 days more . how many men are there ? | "explanation : x * 40 = ( x - 5 ) * 50 5 x - 4 x = 25 x = 25 men answer : option b" | a ) 15 men , b ) 25 men , c ) 18 men , d ) 12 men , e ) 14 men | b | divide(multiply(add(40, 10), 5), 10) | add(n0,n2)|multiply(n1,#0)|divide(#1,n2)| | physics |
mike earns $ 12 per hour and phil earns $ 6 per hour . approximately how much less , as a percentage , does phil earn than mike per hour ? | "what % less of 12 is 6 let it be x % less , then = 12 ( 1 - x / 100 ) = 6 1 - x / 100 = 6 / 12 x = 100 / 2 x = 50 % ans a" | a ) 50 % , b ) 32.5 % , c ) 37 % , d ) 37.5 % , e ) 40 % | a | multiply(divide(6, 12), const_100) | divide(n1,n0)|multiply(#0,const_100)| | general |
a can do a job in 45 days and b can do it in 30 days . a and b working together will finish four times the amount of work in - - - - - - - days ? | "1 / 45 + 1 / 30 = 5 / 90 = 1 / 18 18 / 1 = 18 * 4 = 72 days answer : e" | a ) 30 days , b ) 45 days , c ) 80 days , d ) 72 1 / 2 days , e ) 72 days | e | add(divide(const_1, 45), divide(const_1, 30)) | divide(const_1,n0)|divide(const_1,n1)|add(#0,#1)| | physics |
a certain bus driver is paid a regular rate of $ 14 per hour for any number of hours that does not exceed 40 hours per week . for any overtime hours worked in excess of 40 hours per week , the bus driver is paid a rate that is 75 % higher than his regular rate . if last week the bus driver earned $ 982 in total compens... | "for 40 hrs = 40 * 14 = 560 excess = 982 - 560 = 422 for extra hours = . 75 ( 14 ) = 10.5 + 14 = 24.5 number of extra hrs = 422 / 24.5 = 17.2 = 17 approx . total hrs = 40 + 17 = 57 answer a" | a ) 57 , b ) 51 , c ) 59 , d ) 55 , e ) 52 | a | add(40, divide(subtract(982, multiply(14, 40)), divide(multiply(14, add(const_100, 75)), const_100))) | add(n3,const_100)|multiply(n0,n1)|multiply(n0,#0)|subtract(n4,#1)|divide(#2,const_100)|divide(#3,#4)|add(n1,#5)| | general |
the least number which when divided by 3 , 4 , 5 and 8 leaves a remainder 3 , but when divided by 9 leaves no remainder , is | "explanation : l . c . m of 3 , 4 , 5 , 8 = 120 therefore required number is of the form 120 k + 3 . least value of k for which ( 120 k + 3 ) is divisible by 9 is k = 2 therefore required number = ( 120 x 2 + 3 ) = 243 answer : b" | a ) 167 , b ) 243 , c ) 252 , d ) 336 , e ) none of these | b | multiply(lcm(lcm(lcm(3, 4), 5), 8), const_2) | lcm(n0,n1)|lcm(n2,#0)|lcm(n3,#1)|multiply(#2,const_2)| | general |
70 + 5 * 12 / ( 180 / 3 ) = ? | "explanation : 70 + 5 * 12 / ( 180 / 3 ) = 70 + 5 * 12 / ( 60 ) = 70 + ( 5 * 12 ) / 60 = 70 + 1 = 71 . answer : b" | a ) 23 , b ) 71 , c ) 27 , d ) 61 , e ) 81 | b | add(70, divide(multiply(5, 12), divide(180, 3))) | divide(n3,n4)|multiply(n1,n2)|divide(#1,#0)|add(n0,#2)| | general |
the length of the rectangular field is double its width . inside the field there is square shaped pond 4 m long . if the area of the pond is 1 / 8 of the area of the field . what is the length of the field ? | "a / 8 = 4 * 4 = > a = 4 * 4 * 8 x * 2 x = 4 * 4 * 8 x = 8 = > 2 x = 16 answer : e" | a ) 54 , b ) 32 , c ) 75 , d ) 28 , e ) 16 | e | sqrt(divide(multiply(square_area(4), 8), inverse(const_2))) | inverse(const_2)|square_area(n0)|multiply(n2,#1)|divide(#2,#0)|sqrt(#3)| | geometry |
14 men can complete a piece of work in 25 days . in how many days can 20 men complete that piece of work ? | "14 * 25 = 20 * x = > x = 17.5 days answer : c" | a ) 27.5 days , b ) 15.5 days , c ) 17.5 days , d ) 16.5 days , e ) 18.5 days | c | divide(multiply(25, 14), 20) | multiply(n0,n1)|divide(#0,n2)| | physics |
a train leaves mumabai at 9 am at a speed of 40 kmph . after one hour , another train leaves mumbai in the same direction as that of the first train at a speed of 50 kmph . when and at what distance from mumbai do the two trains meet ? | "when the second train leaves mumbai the first train covers 40 * 1 = 40 km so , the distance between first train and second train is 40 km at 10.00 am time taken by the trains to meet = distance / relative speed = 40 / ( 50 - 40 ) = 4 hours so , the two trains meet at 2 p . m . the two trains meet 4 * 50 = 200 km away ... | a ) 276 , b ) 271 , c ) 200 , d ) 278 , e ) 268 | c | multiply(divide(multiply(40, const_1), subtract(50, 40)), 50) | multiply(n1,const_1)|subtract(n2,n1)|divide(#0,#1)|multiply(n2,#2)| | physics |
reduce 368 / 598 to the lowest terms | explanation : we can do it easily by in two steps step 1 : we get the hcf of 368 and 598 which is 46 step 2 : divide both by 46 , we will get the answer 8 / 13 answer : option d | a ) 30 / 25 , b ) 28 / 13 , c ) 28 / 29 , d ) 8 / 13 , e ) none of these | d | divide(368, 598) | divide(n0,n1) | general |
a , b and c started a partnership business by investing rs . 5000 , rs . 8000 , rs . 9000 respectively . at the end of the year , the profit were distributed among them . if c ' s share of profit is 36000 , what is the total profit ? | "a : b : c = 5000 : 8000 : 9000 = 5 : 8 : 9 let total profit = p then p Γ£ β 9 / 22 = 36000 p = ( 36000 Γ£ β 22 ) / 9 = 88000 answer is c ." | a ) 70000 , b ) 60000 , c ) 88000 , d ) 90000 , e ) 50000 | c | add(add(multiply(divide(5000, 9000), 36000), multiply(divide(8000, 9000), 36000)), 36000) | divide(n0,n2)|divide(n1,n2)|multiply(n3,#0)|multiply(n3,#1)|add(#2,#3)|add(n3,#4)| | gain |
5 n + 2 > 12 and 7 n - 5 < 37 ; n must be between which numbers ? | "5 n > 10 - - > n > 2 7 n < 42 - - > n < 6 2 < n < 6 answer : e" | a ) 1 and 8 , b ) 2 and 6 , c ) 0 and 9 , d ) 2 and 7 , e ) 2 and 6 | e | add(multiply(2, const_10), divide(add(37, 5), 7)) | add(n4,n5)|multiply(const_10,n1)|divide(#0,n3)|add(#2,#1)| | general |
a man invested rs . 14,400 in rs . 100 shares of a company at 20 % premium . if the company declares 5 % dividend at the end of the year , then how much does he get ? | "solution number of shares = ( 14400 / 120 ) = 120 . face value = rs . ( 100 x 120 ) = rs . 12000 . annual income = rs . ( 5 / 100 x 12000 ) = rs . 600 . answer b" | a ) rs . 500 , b ) rs . 600 , c ) rs . 650 , d ) rs . 720 , e ) none | b | multiply(multiply(100, divide(add(add(multiply(const_10, const_1000), multiply(const_4, const_1000)), multiply(const_4, 100)), add(100, multiply(100, divide(20, 100))))), divide(5, 100)) | divide(n3,n1)|divide(n2,n1)|multiply(const_10,const_1000)|multiply(const_1000,const_4)|multiply(n1,const_4)|add(#2,#3)|multiply(n1,#1)|add(#5,#4)|add(n1,#6)|divide(#7,#8)|multiply(n1,#9)|multiply(#0,#10)| | gain |
two isosceles triangles have equal vertical angles and their areas are in the ratio 16 : 49 . find the ratio of their corresponding heights . | "we are basically given that the triangles are similar . in two similar triangles , the ratio of their areas is the square of the ratio of their sides and also , the square of the ratio of their corresponding heights . therefore , area / area = height ^ 2 / height ^ 2 = 16 / 49 - - > height / height = 4 / 7 . answer : ... | a ) 4 / 5 , b ) 5 / 4 , c ) 3 / 2 , d ) 4 / 7 , e ) 2 / 3 | d | divide(sqrt(16), sqrt(49)) | sqrt(n0)|sqrt(n1)|divide(#0,#1)| | geometry |
two numbers are respectively 50 % and 100 % more than a third number . the percentage that is first of the second is ? | "i ii iii 150 200 100 200 - - - - - - - - - - 150 100 - - - - - - - - - - - ? = > 75 % answer : e" | a ) 65 % , b ) 83 % , c ) 27 % , d ) 42 % , e ) 75 % | e | subtract(const_100, multiply(divide(add(50, const_100), add(100, const_100)), const_100)) | add(n0,const_100)|add(n1,const_100)|divide(#0,#1)|multiply(#2,const_100)|subtract(const_100,#3)| | gain |
if an object travels 300 feet in 5 seconds , what is the object ' s approximate speed in miles per hour ? ( note : 1 mile = 5280 feet ) | "1 mile = 5280 feet = > 1 feet = 1 / 5280 miles if the object travels 300 feet in 5 sec then it travels 300 / 5 * 60 * 60 feet in 1 hour ( 1 hr = 60 min * 60 sec ) = 3600 * 60 feet in 1 hour = 216000 feet in 1 hr = 216000 / 5280 miles in 1 hour = 21600 / 528 miles / hr ~ 41 miles / hr answer - d" | a ) 39 , b ) 38 , c ) 40 , d ) 41 , e ) 34 | d | divide(divide(300, 5280), multiply(5, divide(1, const_3600))) | divide(n0,n3)|divide(n2,const_3600)|multiply(n1,#1)|divide(#0,#2)| | physics |
rs . 875 becomes rs . 956 in 3 years at a certain rate of simple interest . if the rate of interest is increased by 4 % , what amount will rs . 875 become in 3 years ? | "solution s . i . = rs . ( 956 - 875 ) = rs . 81 rate = ( 100 x 81 / 875 x 3 ) = 108 / 35 % new rate = ( 108 / 35 + 4 ) % = 248 / 35 % new s . i . = rs . ( 875 x 248 / 35 x 3 / 100 ) rs . 186 . β΄ new amount = rs . ( 875 + 186 ) = rs . 1061 . answer c" | a ) rs . 1020.80 , b ) rs . 1025 , c ) rs . 1061 , d ) data inadequate , e ) none of these | c | add(875, divide(multiply(multiply(875, add(divide(multiply(subtract(956, 875), const_100), multiply(875, 3)), 4)), 3), const_100)) | multiply(n0,n2)|subtract(n1,n0)|multiply(#1,const_100)|divide(#2,#0)|add(n3,#3)|multiply(n0,#4)|multiply(n2,#5)|divide(#6,const_100)|add(n0,#7)| | gain |
0.003 * ? * 0.0003 = 0.00000027 | "explanation : 0.003 * ? * 0.0003 = 0.00000027 3 / 1000 * ? * 3 / 10000 = 3 / 1000 * 3 / 1000 * 3 / 100 ? = 3 / 10 = 0.3 answer is c" | a ) 9 , b ) 3 , c ) 0.3 , d ) 0.03 , e ) none of these | c | multiply(0.003, 0.0003) | multiply(n0,n1)| | general |
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 . 3400 , 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 = 136 so , a β s amount = 120 Γ 15 = rs . 2040 . answer : option c" | a ) rs . 1200 , b ) rs . 1500 , c ) rs . 2040 , 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))), 3400) | multiply(n1,const_2)|divide(#0,n0)|divide(#0,n1)|add(#1,#2)|divide(#1,#3)|multiply(n2,#4)| | physics |
what will be the reminder when ( 19 ^ 19 + 19 ) is divided by 20 ? | "( x ^ n + 1 ) will be divisible by ( x + 1 ) only when n is odd ; ( 19 ^ 19 + 1 ) will be divisible by ( 19 + 1 ) ; ( 19 ^ 19 + 1 ) + 18 when divided by 20 will give 18 as remainder . correct option : a" | a ) 18 , b ) 40 , c ) 21 , d ) 43 , e ) 28 | a | subtract(add(power(19, 19), 19), multiply(20, floor(divide(add(power(19, 19), 19), 20)))) | power(n0,n0)|add(n0,#0)|divide(#1,n3)|floor(#2)|multiply(n3,#3)|subtract(#1,#4)| | general |
a couple who own an appliance store discover that if they advertise a sales discount of 10 % on every item in the store , at the end of one month the number of total items sold increases 12 % . their gross income from sales for one month increases by what percent ? | "let p be the original price and let x be the number of items sold originally . the original income is p * x . after the changes , the income is 0.9 p * 1.12 x = 1.008 * ( p * x ) , an increase of 0.8 % . the answer is b ." | a ) 0.4 % , b ) 0.8 % , c ) 2 % , d ) 4 % , e ) 8 % | b | subtract(subtract(12, 10), divide(12, 10)) | divide(n1,n0)|subtract(n1,n0)|subtract(#1,#0)| | gain |
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