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 average of 10 consecutive integers is 25 . then , 9 is deducted from the first consecutive number , 8 is deducted from the second , 7 is deducted form the third , and so on until the last number which remains unchanged . what is the new average ? | "the total subtracted is ( 9 + 8 + . . . + 1 ) = ( 9 * 10 ) / 2 = 45 on average , each number will be reduced by 45 / 10 = 4.5 therefore , the overall average will be reduced by 4.5 the answer is d ." | a ) 19 , b ) 19.5 , c ) 20 , d ) 20.5 , e ) 21 | d | divide(subtract(multiply(10, 25), multiply(add(const_4, const_1), 9)), 10) | add(const_1,const_4)|multiply(n0,n1)|multiply(n2,#0)|subtract(#1,#2)|divide(#3,n0)| | general |
if 0.75 : x : : 5 : 11 , then x is equal to : | "explanation : ( x * 5 ) = ( 0.75 * 11 ) x = 8.25 / 5 = 1.65 answer : b" | a ) 1.12 , b ) 1.65 , c ) 1.2 , d ) 1.3 , e ) none of these | b | divide(multiply(0.75, 11), 5) | multiply(n0,n2)|divide(#0,n1)| | general |
an empty fuel tank is filled with brand z gasoline . when the tank is 3 / 4 empty , it is filled with brand x gasoline . when the tank is half empty again , it is filled with brand z gasoline . when the tank is half empty again , it is filled with brand x gasoline . at this time , what percent of the gasoline in the ta... | "work with fraction of brand z in the tank . 1 st step : brand z is 1 2 nd step : brand z is 1 / 2 3 rd step : brand z is ( 3 / 4 ) * ( 1 / 2 ) + 1 / 2 = 7 / 8 4 th step : brand z is ( 1 / 2 ) * ( 7 / 8 ) = 7 / 16 = 43.75 % answer ( d )" | a ) 41 % , b ) 42 % , c ) 43 % , d ) 43.75 % , e ) 44 % | d | divide(subtract(const_100, divide(subtract(const_100, multiply(divide(3, 4), const_100)), const_2)), const_2) | divide(n0,n1)|multiply(#0,const_100)|subtract(const_100,#1)|divide(#2,const_2)|subtract(const_100,#3)|divide(#4,const_2)| | gain |
there are 9 teams in a certain league and each team plays each of the other teams exactly once . what is the total number of games played ? | "9 c 2 = 36 the answer is e" | a ) 24 , b ) 25 , c ) 28 , d ) 32 , e ) 36 | e | divide(multiply(9, subtract(9, const_1)), const_2) | subtract(n0,const_1)|multiply(n0,#0)|divide(#1,const_2)| | general |
what is the value of â ˆ š 25 % ? | "explanation : br > â ˆ š 25 % = â ˆ š 25 / â ˆ š 100 = 5 / 10 = 50 / 100 = 50 % correct answer is b ) 50 %" | a ) 10 % , b ) 50 % , c ) 25 % , d ) 45 % , e ) 68 % | b | circle_area(divide(25, multiply(const_2, const_pi))) | multiply(const_2,const_pi)|divide(n0,#0)|circle_area(#1)| | gain |
in how many years rs 160 will produce the same interest at 8 % as rs . 500 produce in 4 years at 8 % | "explanation : clue : firstly we need to calculate the si with prinical 500 , time 4 years and rate 8 % , it will be rs . 160 then we can get the time as time = ( 100 * 160 ) / ( 160 * 8 ) = 12.5 option d" | a ) 13 , b ) 13.5 , c ) 11.5 , d ) 12.5 , e ) 12 | d | divide(multiply(divide(multiply(500, 8), const_100), 4), multiply(divide(8, const_100), 160)) | divide(n1,const_100)|multiply(n2,n4)|divide(#1,const_100)|multiply(n0,#0)|multiply(n3,#2)|divide(#4,#3)| | gain |
of 15 applicants for a job , 7 had at least 4 years ' experience , 9 had degrees , and 1 had less than 4 years ' experience and did not have a degree . how many of the applicants had at least 4 years ' experience and a degree ? | "15 - 1 = 14 14 - 7 - 9 = - 2 then 2 are in the intersection between 4 years experience and degree . answer : e" | a ) 14 , b ) 13 , c ) 9 , d ) 7 , e ) 2 | e | add(subtract(add(7, 9), subtract(15, 1)), subtract(9, 7)) | add(n1,n3)|subtract(n0,n4)|subtract(n3,n1)|subtract(#0,#1)|add(#3,#2)| | general |
simple interest on a certain sum of money for 5 years at 8 % per annum is half the compound interest on rs . 8000 for 2 years at 15 % per annum . the sum placed on simple interest is | solution c . i . = rs [ 8000 x ( 1 + 15 / 100 ) â ² - 8000 ] rs . ( 8000 x 115 / 100 x 115 / 100 - 8000 ) = rs . 2580 . sum = rs . [ 1290 x 100 / 5 x 8 ] = rs . 3225 . answer a | a ) rs . 3225 , b ) rs . 2580 , c ) rs . 8000 , d ) rs . 1290 , e ) none | a | divide(multiply(divide(divide(add(divide(multiply(8000, 15), const_100), divide(multiply(add(8000, divide(multiply(8000, 15), const_100)), 15), const_100)), 2), 5), const_100), 8) | multiply(n2,n4)|divide(#0,const_100)|add(n2,#1)|multiply(n4,#2)|divide(#3,const_100)|add(#1,#4)|divide(#5,n3)|divide(#6,n0)|multiply(#7,const_100)|divide(#8,n1) | gain |
for each color copy , print shop x charges $ 1.25 and print shop y charges $ 2.75 . how much greater is the charge for 80 color copies at print shop y than at print shop x ? | "the difference in the two prices is $ 2.75 - $ 1.25 = $ 1.50 for each color copy . each color copy will cost an extra $ 1.50 at print shop y . 80 * $ 1.50 = $ 120 the answer is a ." | a ) $ 120 , b ) $ 125 , c ) $ 130 , d ) $ 135 , e ) $ 140 | a | subtract(multiply(2.75, 80), multiply(1.25, 80)) | multiply(n1,n2)|multiply(n0,n2)|subtract(#0,#1)| | general |
in the xy - plane , a triangle has vertices ( 0,0 ) , ( 4,0 ) and ( 4,8 ) . if a point ( a , b ) is selected at random from the triangular region , what is the probability that a - b > 0 ? | "the area of the right triangle is ( 1 / 2 ) * 4 * 8 = 16 . only the points ( a , b ) below the line y = x satisfy a - b > 0 . the part of the triangle which is below the line y = x has an area of ( 1 / 2 ) ( 4 ) ( 4 ) = 8 . p ( a - b > 0 ) = 8 / 16 = 1 / 2 the answer is c ." | a ) 1 / 5 , b ) 1 / 3 , c ) 1 / 2 , d ) 2 / 3 , e ) 4 / 5 | c | divide(const_4, add(0,0, const_10)) | add(n0,const_10)|divide(const_4,#0)| | geometry |
three cubes of metal whose edges are 9 , 12 and 24 cm respectively , are melted and one new cube is made . find the edge of the new cube ? | 93 + 123 + 243 = a 3 = > a = 25.3 answer : e | ['a ) 28', 'b ) 77', 'c ) 66', 'd ) 18', 'e ) 25.3'] | e | power(add(add(volume_cube(9), volume_cube(12)), volume_cube(24)), const_0_33) | volume_cube(n0)|volume_cube(n1)|volume_cube(n2)|add(#0,#1)|add(#3,#2)|power(#4,const_0_33) | physics |
the speed of a boat in still water is 15 km / hr and the rate of current is 3 km / hr the distance travelled downstream in 12 minutes is : | sol . speed downstream = ( 15 + 3 ) kmph = 18 kmph . distance travelled = [ 18 * 12 / 60 ] km = 3.6 km . answer c | a ) 2 km , b ) 2.6 km , c ) 3.6 km , d ) 4 km , e ) none | c | multiply(divide(12, const_60), add(15, 3)) | add(n0,n1)|divide(n2,const_60)|multiply(#0,#1) | physics |
two cubes each with side b are joined to form a cuboids . what is the surface area of this cuboids ? | l = 2 b ; b = b ; h = b ; surface area = 2 * ( lb + bh + hl ) = 2 * ( 2 b ^ 2 + b ^ 2 + 2 b ^ 2 ) = 10 b ^ 2 answer : b | ['a ) 12 b ^ 2', 'b ) 10 b ^ 2', 'c ) 6 b ^ 2', 'd ) 4 b ^ 2', 'e ) 2 b ^ 2'] | b | multiply(add(multiply(multiply(const_1, const_2), const_2), multiply(const_1, const_1)), const_2) | multiply(const_1,const_2)|multiply(const_1,const_1)|multiply(#0,const_2)|add(#2,#1)|multiply(#3,const_2) | geometry |
bob is attempting to feed his pets . he feeds his pets as follows : - no dog receives more food than any other dog - no cat receives any less food than any other cat - the cats , as a group , receive exactly as much food as a single dog if bob has 7 dogs and 4 cats , approximately what percentage of the food bag does a... | c . each cat gets 1 / 4 as much food as a dog . a dog gets 1 / 8 of the food . 1 / 4 of 1 / 8 is 1 / 32 , which is about 3.1 % , so the answer is c . | a ) 25.0 % , b ) 8.3 % , c ) 3.1 % , d ) 2.7 % , e ) 1.0 % | c | multiply(divide(const_1, add(multiply(7, 4), 4)), const_100) | multiply(n0,n1)|add(n1,#0)|divide(const_1,#1)|multiply(#2,const_100) | general |
in the xy - coordinate plane , the graph of y = - x ^ 2 + 9 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 + 9 p 2 = 4 p = + or - 2 simillarly point ( t , 7 ) should satisfy the equation . hence x = t and y = 7 . - 7 = - t 2 + 9 t = + or - 4 ... | a ) 6 , b ) 2 , c ) - 2 , d ) 1 , e ) - 10 | d | divide(subtract(7, 5), subtract(sqrt(add(9, 7)), sqrt(add(5, 9)))) | add(n1,n3)|add(n1,n2)|subtract(n3,n2)|sqrt(#0)|sqrt(#1)|subtract(#3,#4)|divide(#2,#5)| | general |
what is the least number of square tiles required to pave the floor of a room 11 m 50 cm long and 1 m 50 cm broad ? | "solution length of largest tile = h . c . f . of 1150 cm & 150 cm = 50 cm . area of each tile = ( 50 x 50 ) cm 2 ∴ required number of tiles = [ 1150 x 150 / 50 x 50 ] = 69 . answer c" | a ) 724 , b ) 804 , c ) 69 , d ) 844 , e ) none | c | divide(multiply(add(multiply(1, const_100), 50), multiply(11, const_100)), power(50, const_2)) | multiply(n2,const_100)|multiply(n0,const_100)|power(n3,const_2)|add(n3,#0)|multiply(#3,#1)|divide(#4,#2)| | physics |
in a cricket match , 5 batsmen a , b , c , d and e scored an average of 36 runs . d scored 5 more than e ; e scored 8 fewer than a ; b scored as many as d and e combined ; and b and c scored 107 between them . how many runs did e score ? | explanation : total runs scored = ( 36 x 5 ) = 180 . let the runs scored by e be x . then , runs scored by d = x + 5 ; runs scored by a = x + 8 ; runs scored by b = x + x + 5 = 2 x + 5 ; runs scored by c = ( 107 - b ) = 107 - ( 2 x + 5 ) = 102 - 2 x . so , total runs = ( x + 8 ) + ( 2 x + 5 ) + ( 102 - 2 x ) + ( x + 5 ... | a ) 62 , b ) 45 , c ) 28 , d ) 20 , e ) 25 | d | divide(subtract(multiply(36, 5), add(add(multiply(const_2, 5), const_3), 107)), const_3) | multiply(n0,n1)|multiply(n0,const_2)|add(#1,const_3)|add(n4,#2)|subtract(#0,#3)|divide(#4,const_3) | general |
a property decreases in value every year at the rate of 6 1 / 4 % of its value at the beginning of the year its value at the end of 3 years was rs . 21093 . find its value at the beginning of the first year ? | 6 1 / 4 % = 1 / 16 x * 15 / 16 * 15 / 16 * 15 / 16 = 21093 x = 25600.24 answer : a | a ) 25600.24 , b ) 25600.27 , c ) 25600.28 , d ) 25600.21 , e ) 25600.29 | a | divide(21093, power(subtract(1, divide(add(divide(1, 4), 6), const_100)), const_3)) | divide(n1,n2)|add(n0,#0)|divide(#1,const_100)|subtract(n1,#2)|power(#3,const_3)|divide(n4,#4) | gain |
the speed at which a man can row a boat in still water is 24 kmph . if he rows downstream , where the speed of current is 8 kmph , what time will he take to cover 40 metres ? | "speed of the boat downstream = 24 + 8 = 32 kmph = 32 * 5 / 18 = 8.89 m / s hence time taken to cover 40 m = 40 / 8.89 = 4.49 seconds . answer : a" | a ) 4.49 seconds , b ) 8.89 seconds , c ) 26 seconds , d ) 12 seconds , e ) 16 seconds | a | divide(40, multiply(add(24, 8), const_0_2778)) | add(n0,n1)|multiply(#0,const_0_2778)|divide(n2,#1)| | physics |
a manufacturing company has 15 % cobalt , 25 % led and 60 % of copper . if 5 kg of led is used in a mixture how much copper we need to use | 25 % = 5 kg 100 % = 20 kg 60 % of 20 kg = 12 kg 12 kg answer : a | a ) 12 kg , b ) 13 kg , c ) 22 kg , d ) 14 kg , e ) 15 kg | a | multiply(divide(60, const_100), multiply(5, const_4)) | divide(n2,const_100)|multiply(n3,const_4)|multiply(#0,#1) | general |
in a simultaneous throw of pair of dice . find the probability of getting the total more than 7 ? | "here n ( s ) = ( 6 * 6 ) = 36 let e = event of getting a total more than 7 = { ( 2,6 ) , ( 3,5 ) , ( 3,6 ) , ( 4,4 ) , ( 4,5 ) , ( 4,6 ) , ( 5,3 ) , ( 5,4 ) , ( 5,5 ) , ( 5,6 ) , ( 6,2 ) , ( 6,3 ) , ( 6,4 ) , ( 6,5 ) , ( 6,6 ) } p ( e ) = n ( e ) / n ( s ) = 15 / 36 = 5 / 12 option b" | a ) 12 / 5 , b ) 5 / 12 , c ) 6 / 4 , d ) 2 / 3 , e ) 4 / 7 | b | divide(add(add(7, const_4), const_4), multiply(add(const_4, const_2), add(const_4, const_2))) | add(n0,const_4)|add(const_2,const_4)|add(#0,const_4)|multiply(#1,#1)|divide(#2,#3)| | general |
the smallest number which when diminished by 15 , is divisible 10 , 15 , 25 , 30 and 35 is : | "required number = ( l . c . m . of 10 , 15 , 25 , 30 , 35 ) + 15 = 1050 + 15 = 1055 answer : option b" | a ) 1008 , b ) 1055 , c ) 1022 , d ) 1032 , e ) 1048 | b | add(multiply(multiply(power(const_3, const_2.0), power(const_2.0, const_4)), add(const_3, const_4)), 15) | add(const_3,const_4)|power(const_3,const_2.0)|power(const_2.0,const_4)|multiply(#1,#2)|multiply(#0,#3)|add(#4,n0)| | general |
the length of the longest tape in cm which can be used to measure exactly , the length 7 m ; 3 m 85 cm ; and 12 m 95 cm is : | "the three lengths in cm are 700 , 385 & 1295 . hcf of 700 , 385 & 1295 is 35 . hence , the answer is 35 cm . answer : b" | a ) 37 , b ) 35 , c ) 34 , d ) 36 , e ) 38 | b | sqrt(85) | sqrt(n2)| | physics |
a rectangular box measures internally 1.6 m long , 1 m broad and 60 cm deep . the number of cubical blocks each of edge 20 cm that can be packed inside the box is | sol . number of blocks = 160 x 100 x 60 20 x 20 x 20 ] = 120 . answer c | ['a ) 30', 'b ) 60', 'c ) 120', 'd ) 150', 'e ) none'] | c | volume_rectangular_prism(divide(1.6, divide(20, const_100)), divide(1, divide(20, const_100)), divide(divide(60, const_100), divide(20, const_100))) | divide(n3,const_100)|divide(n2,const_100)|divide(n0,#0)|divide(n1,#0)|divide(#1,#0)|volume_rectangular_prism(#2,#3,#4) | physics |
two pipes can fill a tank in 12 minutes and 15 minutes . an outlet pipe can empty the tank in 10 minutes . if all the pipes are opened when the tank is empty , then how many minutes will it take to fill the tank ? | "let v be the volume of the tank . the rate per minute at which the tank is filled is : v / 12 + v / 15 - v / 10 = 3 v / 60 = v / 20 per minute the tank will be filled in 20 minutes . the answer is a ." | a ) 20 , b ) 22 , c ) 24 , d ) 26 , e ) 28 | a | subtract(add(divide(const_1, 12), divide(const_1, 15)), divide(const_1, 10)) | divide(const_1,n0)|divide(const_1,n1)|divide(const_1,n2)|add(#0,#1)|subtract(#3,#2)| | physics |
an article with cost price of 192 is sold at 25 % profit . what is the selling price ? | "sp = 1.25 * 192 = 240 answer : d" | a ) 198 , b ) 200 , c ) 204 , d ) 240 , e ) 210 | d | add(192, multiply(192, divide(25, const_100))) | divide(n1,const_100)|multiply(n0,#0)|add(n0,#1)| | gain |
two pipes a and b can fill a tank in 50 hour and 75 hours respectively . if both the pipes are opened simultaneously , how much time will be taken to fill the tank ? | part filled by a in 1 hour = 1 / 50 part filled by b in 1 hour = 1 / 75 part filled by a + b in 1 hour = 1 / 50 + 1 / 75 = 1 / 30 both the pipes will together will fill the tank in 30 hours . answer is d | a ) 15 hrs , b ) 20 hrs , c ) 25 hrs , d ) 30 hrs , e ) 32 hrs | d | divide(const_1, add(divide(const_1, 50), divide(const_1, 75))) | divide(const_1,n0)|divide(const_1,n1)|add(#0,#1)|divide(const_1,#2) | physics |
how many integers from 0 to 50 inclusive have a remainder of 3 when divided by 7 ? | "the numbers should be of the form 7 c + 3 . the minimum is 3 when c = 0 . the maximum is 45 when c = 6 . there are 7 such numbers . the answer is c ." | a ) 5 , b ) 6 , c ) 7 , d ) 8 , e ) 9 | c | divide(50, const_10) | divide(n1,const_10)| | general |
addison high school ’ s senior class has 160 boys and 200 girls . if 75 % of the boys attended the college and 40 % of the girls did not attend the college , what percentage of the total class attended college ? | number of boys attending the college : 0.75 * 160 = 120 number of girls not attending the college : 0.40 * 200 = 80 number of girls attending the college : 200 - 80 = 120 total attending = 120 + 120 = 240 total people = 160 + 200 = 360 % of people attending = 240 / 360 = 66.67 % . c is the correct answer | a ) 35 , b ) 64 , c ) 66.67 , d ) 60 , e ) 84 | c | multiply(divide(add(multiply(divide(75, const_100), 160), multiply(subtract(const_1, divide(40, const_100)), 200)), add(160, 200)), const_100) | add(n0,n1)|divide(n2,const_100)|divide(n3,const_100)|multiply(n0,#1)|subtract(const_1,#2)|multiply(n1,#4)|add(#3,#5)|divide(#6,#0)|multiply(#7,const_100) | general |
if x dollars is invested at 10 percent for one year and y dollars is invested at 8 percent for one year , the annual income from the 10 percent investment will exceed the annual income from the 8 percent investment by $ 92 . if $ 2,000 is the total amount invested , how much is invested at 8 percent ? | 0.1 x = 0.08 ( 2000 - x ) + 92 0.18 x = 252 x = 1400 then the amount invested at 8 % is $ 2000 - $ 1400 = $ 600 the answer is a . | a ) $ 600 , b ) $ 700 , c ) $ 800 , d ) $ 900 , e ) $ 1000 | a | subtract(multiply(multiply(const_100, const_10), const_2), divide(add(multiply(multiply(10, 8), const_2), 92), add(divide(10, const_100), divide(8, const_100)))) | divide(n0,const_100)|divide(n1,const_100)|multiply(const_10,const_100)|multiply(n0,n1)|add(#0,#1)|multiply(#2,const_2)|multiply(#3,const_2)|add(n4,#6)|divide(#7,#4)|subtract(#5,#8) | general |
there is a 30 % increase in the price of an article in the first year , a 20 % decrease in the second year and a 10 % increase in the next year . if the final price of the article is rs . 3432 , then what was the price of the article initially ? | "let the price of the article , four years age be rs . 100 in the 1 st year , price of the article = 100 + 30 = rs . 130 . in the 2 nd year , price = 130 - 20 % of 130 = 130 - 26 = rs . 104 . in the 3 rd year , price = 104 + 10 % of 104 = 104 + 10.4 = rs . 114.40 . but present price of the article is rs . 3432 for 114.... | a ) rs . 2000 , b ) rs . 4000 , c ) rs . 3000 , d ) rs . 5000 , e ) rs . 6000 | c | divide(3432, add(const_1, divide(add(subtract(30, 20), 10), const_100))) | subtract(n0,n1)|add(n2,#0)|divide(#1,const_100)|add(#2,const_1)|divide(n3,#3)| | general |
the balance of a trader weighs 10 % less than it should . still the trader marks up his goods to get an overall profit of 40 % . what is the mark up on the cost price ? | "the most natural way to deal with ' weights ' questions is by assuming values . say the trader ' s balance shows 100 gms . it is actually 90 gms because it weighs 10 % less . say , the cost price is $ 90 ( $ 1 / gm ) . since he gets a profit of 40 % , the selling price must be 90 + ( 40 / 100 ) * 90 = $ 126 since the ... | a ) 40 % , b ) 8 % , c ) 25 % , d ) 16.66 % , e ) 26 % | e | subtract(add(multiply(multiply(const_4, 10), divide(40, const_100)), multiply(const_4, 10)), const_100) | divide(n1,const_100)|multiply(n0,const_4)|multiply(#0,#1)|add(#2,#1)|subtract(#3,const_100)| | gain |
jaclyn buys $ 60 000 worth of debentures in a company . she earns 9.5 % p . a . simple interest , paid to her quarterly ( that is , every 3 months ) . if the agreed period of the debenture was 18 months : calculate the amount of interest jaclyn will earn for each quarter | "explanation : i = ( p x r x t ) / 100 = 60000 * 9.5 / 100 * ( 18 / 12 ) ^ 1 / 6 = 1425 answer : d" | a ) 1187.5 , b ) 1234 , c ) 1289 , d ) 1425 , e ) none of these | d | divide(divide(multiply(multiply(const_100, const_100), 9.5), const_100), const_4) | multiply(const_100,const_100)|multiply(n2,#0)|divide(#1,const_100)|divide(#2,const_4)| | gain |
danny obtained 76 , 65 , 82 , 67 and 75 marks ( out of 100 ) in english , mathematics , physics , chemistry and biology . what are his average marks ? | average = ( 76 + 65 + 82 + 67 + 75 ) / 5 = 365 / 5 = 73 . answer : e | a ) a ) 87 , b ) b ) 99 , c ) c ) 68 , d ) d ) 82 , e ) e ) 73 | e | divide(add(add(add(add(76, 65), 82), 67), 75), divide(const_10, const_2)) | add(n0,n1)|divide(const_10,const_2)|add(n2,#0)|add(n3,#2)|add(n4,#3)|divide(#4,#1) | general |
in the land of oz only one or two - letter words are used . the local language has 69 different letters . the parliament decided to forbid the use of the seventh letter . how many words have the people of oz lost because of the prohibition ? | "the answer to the question is indeed e . the problem with above solutions is that they do not consider words like aa , bb , . . . the number of 1 letter words ( x ) that can be made from 69 letters is 69 ; the number of 2 letter words ( xx ) that can be made from 69 letters is 69 * 69 , since each x can take 69 values... | a ) 65 , b ) 66 , c ) 67 , d ) 131 , e ) 138 | e | add(add(const_1, 69), 69) | add(n0,const_1)|add(n0,#0)| | general |
a candidate got 35 % of the votes and lost to the rival by 2400 votes . how many votes were cast ? | "35 % - - - - - - - - - - - l 65 % - - - - - - - - - - - w - - - - - - - - - - - - - - - - - - 30 % = 2400 10 % = 800 100 % of the votes = 8000 answer : b" | a ) 7700 , b ) 8000 , c ) 8300 , d ) 8600 , e ) 8900 | b | divide(2400, subtract(subtract(const_1, divide(35, const_100)), divide(35, const_100))) | divide(n0,const_100)|subtract(const_1,#0)|subtract(#1,#0)|divide(n1,#2)| | gain |
a bag contains 9 green and 8 white balls . if two balls are drawn simultaneously , the probability that both are of the same colour is | "drawing two balls of same color from nine green balls can be done in 9 c ₂ ways . similarly from eight white balls two can be drawn in ⁸ c ₂ ways . p = 9 c ₂ / ¹ ⁵ c ₂ + ⁸ c ₂ / ¹ ⁵ c ₂ = 8 / 17 answer : a" | a ) 8 / 17 , b ) 2 / 5 , c ) 3 / 5 , d ) 11 / 15 , e ) 7 / 15 | a | add(multiply(divide(8, add(9, 8)), divide(subtract(8, const_1), subtract(add(9, 8), const_1))), multiply(divide(9, add(9, 8)), divide(subtract(9, const_1), subtract(add(9, 8), const_1)))) | 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(#4,#7)|add(#8,#9)| | other |
speed of a boat in standing water is 8 kmph and speed of the stream is 0.5 kmph . a man can rows to a place at a distance of 105 km and comes back to the starting point . the total time taken by him is ? | "speed upstream = 7.5 kmph speed downstream = 8.5 kmph total time taken = 105 / 7.5 + 105 / 8.5 = 26.35 hours answer is b" | a ) 27.35 hours , b ) 26.35 hours , c ) 25.35 hours , d ) 24.35 hours , e ) 23.35 hours | b | add(multiply(add(add(8, 0.5), subtract(8, 0.5)), 105), multiply(subtract(add(divide(105, add(8, 0.5)), divide(105, subtract(8, 0.5))), add(add(8, 0.5), subtract(8, 0.5))), const_60)) | add(n0,n1)|subtract(n0,n1)|add(#0,#1)|divide(n2,#0)|divide(n2,#1)|add(#3,#4)|multiply(n2,#2)|subtract(#5,#2)|multiply(#7,const_60)|add(#6,#8)| | physics |
a , b and c started a business with a total investment of rs . 90000 . a invests rs . 6000 more than b and b invests rs . 3000 less than c . if the total profit at the end of a year is rs . 8640 , find a ' s share . | "explanation : let c ' s investment = rs . x b ' s investment = rs . ( x - 3000 ) a ' s investment = rs . ( x - 3000 + 6000 ) = rs . ( x + 3000 ) now , ( a + b + c ) ' s investment = rs . 90000 = > x + ( x - 3000 ) + ( x + 3000 ) = 90000 = > 3 x = 90000 = > x = 30000 hence , a ' s investment = rs . 33000 b ' s investme... | a ) rs . 3168 , b ) rs . 2520 , c ) rs . 2880 , d ) rs . 3360 , e ) none of these | a | multiply(8640, divide(add(divide(subtract(90000, add(6000, 3000)), const_3), 6000), 90000)) | add(n1,n2)|subtract(n0,#0)|divide(#1,const_3)|add(n1,#2)|divide(#3,n0)|multiply(n3,#4)| | general |
on a certain day , orangeade was made by mixing a certain amount of orange juice with an equal amount of water . on the next day , orangeade was made by mixing the same amount of orange juice with twice the amount of water . on both days , all the orangeade that was made was sold . if the revenue from selling the orang... | on the first day 1 unit of orange juice and 1 unit of water was used to make 2 units of orangeade ; on the second day 1 unit of orange juice and 2 units of water was used to make 3 units of orangeade ; so , the ratio of the amount of orangeade made on the first day to the amount of orangeade made on the second day is 2... | a ) $ 0.15 , b ) $ 0.26 , c ) $ 0.30 , d ) $ 0.40 , e ) $ 0.45 | b | divide(multiply(add(const_1, const_1), 0.4), add(const_1, const_2)) | add(const_1,const_1)|add(const_1,const_2)|multiply(n0,#0)|divide(#2,#1) | general |
the largest four - digit number which is a perfect cube , is | solution clearly , 9261 is a perfect cube satisfying the given property . answer b | ['a ) 8000', 'b ) 9261', 'c ) 9999', 'd ) none of these', 'e ) can not be determined'] | b | power(add(multiply(const_10, const_2), const_1), const_3) | multiply(const_10,const_2)|add(#0,const_1)|power(#1,const_3) | geometry |
a number when divided by 6 leaves a remainder 3 . when the square of the same number is divided by 6 , the remainder is | "sol . let x = 6 q + 3 . then x 2 = ( 6 q + 3 ) 2 = 36 q 2 + 36 q + 9 = 6 ( 6 q 2 + 6 q + 1 ) + 3 . so , when x 2 is divided by 6 , remainder = 3 . answer d" | a ) 0 , b ) 1 , c ) 2 , d ) 3 , e ) 4 | d | divide(multiply(6, 3), 6) | multiply(n0,n1)|divide(#0,n0)| | geometry |
the contents of a certain box consist of 40 apples and 30 oranges . how many oranges must be added to the box so that exactly 40 % of the pieces of fruit in the box will be apples ? | "apple = ( apple + orange + x ) * 0.4 40 = ( 30 + 40 + x ) * 0.4 x = 30 answer : e" | a ) 40 , b ) 50 , c ) 60 , d ) 70 , e ) 30 | e | subtract(add(40, 30), divide(40, divide(40, const_100))) | add(n0,n1)|divide(n2,const_100)|divide(n0,#1)|subtract(#0,#2)| | general |
two numbers are in the ratio 5 : 8 and thier h . c . f is 4 . find the numbers . | explanation : let two numbers 5 x , 8 x their h . c . f = 4 therefore h . c . f is 4 i . e , x in this case . therefore the numbers are 5 x 4 = 20 8 x 4 = 32 answer : option b | a ) 25 and 40 , b ) 20 and 32 , c ) 30 and 48 , d ) 15 and 24 , e ) 16 and 25 | b | add(multiply(multiply(5, 4), const_100), multiply(8, 4)) | multiply(n0,n2)|multiply(n1,n2)|multiply(#0,const_100)|add(#2,#1) | other |
a ' s speed is 20 / 13 times that of b . if a and b run a race , what part of the length of the race should a give b as a head start , so that b beats a by 60 % of the length of the race ? | "first calculate the distance , b has covered with his speed in the time , in which a reached 40 % of the race . then add the remaining distance as head start for b to win the race . its best to apply ratios concept here . since a ' s speed is 20 / 13 of b , therefore , b ' s speed is 13 / 20 of a distance covered by b... | a ) 44 % , b ) 48 % , c ) 52 % , d ) 42 % , e ) 74 % | e | multiply(subtract(const_1, multiply(divide(subtract(const_100, 60), const_100), divide(13, 20))), const_100) | divide(n1,n0)|subtract(const_100,n2)|divide(#1,const_100)|multiply(#2,#0)|subtract(const_1,#3)|multiply(#4,const_100)| | general |
a contract is to be completed in 46 days sad 117 men were said to work 8 hours a day . after 33 days , ( 4 / 7 ) of the work is completed . how many additional men may be employed so that the work may be completed in time , each man now working 9 hours a day ? | remaining work = ( 1 - ( 4 / 7 ) = ( 3 / 7 ) remaining period = ( 46 - 33 ) days = 13 days let the total men working at it be x . less work , less men ( direct proportion ) less days , more men ( indirect proportion ) more hours per day , less men ( indirect proportion ) work ( 4 / 7 ) : ( 3 / 7 ) days 13 : 33 } : : 11... | a ) 91 , b ) 87 , c ) 85 , d ) 81 , e ) none of them | d | subtract(divide(multiply(multiply(multiply(subtract(const_1, divide(4, 7)), 33), 8), 117), multiply(multiply(divide(4, 7), subtract(46, 33)), 9)), 117) | divide(n4,n5)|subtract(n0,n3)|multiply(#0,#1)|subtract(const_1,#0)|multiply(n3,#3)|multiply(n6,#2)|multiply(n2,#4)|multiply(n1,#6)|divide(#7,#5)|subtract(#8,n1) | physics |
a person borrows 1000000 for 7 years at 3 % p . a . simple interest . he immediately lends it to another person at 5 % p . a . for 7 years . find his gain in the transaction per year . | "gain in 7 years = [ ( 1000000 ã — 5 ã — 7 / 100 ) â ˆ ’ ( 1000000 ã — 7 ã — 3 / 100 ) ] = ( 350000 â € “ 210000 ) = 140000 â ˆ ´ gain in 1 year = ( 140000 / 7 ) = 20000 answer c" | a ) 20068 , b ) 20050 , c ) 20000 , d ) 25000 , e ) 28000 | c | multiply(1000000, divide(const_1, const_100)) | divide(const_1,const_100)|multiply(n0,#0)| | gain |
12 : 21 : : 31 : ? | "ans 13 reverse of 31 answer : a" | a ) 13 , b ) 14 , c ) 30 , d ) 31 , e ) 40 | a | multiply(31, divide(12, 21)) | divide(n0,n1)|multiply(n2,#0)| | general |
if ( n + 2 ) ! / n ! = 72 , n = ? | ( n + 2 ) ! / n ! = 72 rewrite as : [ ( n + 2 ) ( n + 1 ) ( n ) ( n - 1 ) ( n - 2 ) . . . . ( 3 ) ( 2 ) ( 1 ) ] / [ ( n ) ( n - 1 ) ( n - 2 ) . . . . ( 3 ) ( 2 ) ( 1 ) ] = 132 cancel out terms : ( n + 2 ) ( n + 1 ) = 72 from here , we might just test the answer choices . since ( 9 ) ( 8 ) = 72 , we can see that n = 7 a | a ) 7 , b ) 9 , c ) 10 , d ) 11 , e ) 12 | a | subtract(add(const_4, const_4), const_1) | add(const_4,const_4)|subtract(#0,const_1) | general |
a room 5 m 55 cm long and 3 m 74 cm broad is to be paved with square tiles . find the least number of square tiles required to cover the floor | "let x and y be the sides of the rectangle . then , correct area = xy . area of the room = ( 544 x 374 ) cm 2 . size of largest square tile = h . c . f . of 544 cm and 374 cm = 34 cm . area of 1 tile = ( 34 x 34 ) cm 2 . number of tiles required = ( 544 * 374 ) / ( 34 * 34 ) = 176 answer a 176" | a ) 176 , b ) 170 , c ) 160 , d ) 179 , e ) 174 | a | divide(multiply(add(multiply(5, const_100), 55), add(multiply(3, const_100), 74)), multiply(subtract(55, add(multiply(const_2, const_4), const_2)), subtract(55, add(multiply(const_2, const_4), const_2)))) | multiply(n0,const_100)|multiply(n2,const_100)|multiply(const_2,const_4)|add(n1,#0)|add(n3,#1)|add(#2,const_2)|multiply(#3,#4)|subtract(n1,#5)|multiply(#7,#7)|divide(#6,#8)| | physics |
the ages of two persons differ by 16 years . if 6 years agao , the elder one be 3 times as old as the younger one , find their present ages . | "sol . let the age of the younger person be x years . then , age of the elder perosn = ( x + 16 ) years . ∴ 3 ( x - 6 ) = ( x + 16 - 6 ) ⇔ 3 x - 18 = x + 10 ⇔ 2 x = 28 ⇔ x = 14 hence , their present ages are 14 years and 30 years . answer b" | a ) 20 years . , b ) 30 years . , c ) 40 years . , d ) 50 years . , e ) 60 years . | b | subtract(add(divide(multiply(16, 6), subtract(6, const_1)), 6), 16) | multiply(n0,n1)|subtract(n1,const_1)|divide(#0,#1)|add(n1,#2)|subtract(#3,n0)| | general |
working at a constant rate , p can finish a job in 4 hours . q , also working at a constant rate , can finish the same job in 15 hours . if they work together for 3 hours , how many more minutes will it take p to finish the job , working alone at his constant rate ? | "each hour they complete 1 / 4 + 1 / 15 = 19 / 60 of the job . in 3 hours , they complete 3 ( 19 / 60 ) = 19 / 20 of the job . the time for p to finish is ( 1 / 20 ) / ( 1 / 4 ) = ( 1 / 5 ) hour = 12 minutes the answer is d ." | a ) 5 , b ) 8 , c ) 10 , d ) 12 , e ) 15 | d | multiply(divide(subtract(const_1, multiply(add(inverse(4), inverse(15)), 3)), inverse(4)), const_60) | inverse(n0)|inverse(n1)|add(#0,#1)|multiply(#2,n2)|subtract(const_1,#3)|divide(#4,#0)|multiply(#5,const_60)| | physics |
a man walking at the speed of 4 kmph crosses a square field diagonally in 3 minutes . the area of the field is : | length of the diagonal = distance covered in 3 min . at 4 km / hr . = ( 4000 / 60 * 3 ) = 200 m . therefore , area of the field = 1 / 2 * diagonal 2 = ½ * 200 * 200 = 20000 m 2 answer : b | ['a ) 18000 m 2', 'b ) 20000 m 2', 'c ) 19000 m 2', 'd ) 25000 m 2', 'e ) 15000 m 2'] | b | square_area(divide(multiply(multiply(4, const_1000), divide(3, multiply(multiply(const_2, const_3), const_10))), sqrt(const_2))) | multiply(const_2,const_3)|multiply(n0,const_1000)|sqrt(const_2)|multiply(#0,const_10)|divide(n1,#3)|multiply(#4,#1)|divide(#5,#2)|square_area(#6) | physics |
how many positive integers less than 256 are multiple of 4 but not multiples of 6 ? | from here , it might be easier to go up in bounds of 60 , so we know that 61 - 120 gives 10 more numbers . 121 - 180 and 181 - 240 as well . this brings us up to 240 with 40 numbers . a cursory glance at the answer choices should confirm that it must be 42 , as all the other choices are very far away . the numbers 244 ... | a ) 20 , b ) 31 , c ) 42 , d ) 43 , e ) 64 | d | divide(factorial(subtract(add(const_4, 4), const_1)), multiply(factorial(4), factorial(subtract(const_4, const_1)))) | add(n1,const_4)|factorial(n1)|subtract(const_4,const_1)|factorial(#2)|subtract(#0,const_1)|factorial(#4)|multiply(#1,#3)|divide(#5,#6)| | general |
jill invests $ 10000 in an account that pays an annual rate of 3.96 % , compounding semi - annually . approximately how much e does she have in her account after two years ? | "ps . i guess one can use simple interest to solve cause the answer choices are quite spread between you can easily arrive at something near 8 % hence b the answer" | a ) $ 10079.44 , b ) e = $ 10815.83 , c ) $ 12652.61 , d ) $ 14232.14 , e ) $ 20598.11 | b | multiply(10000, power(add(const_1, divide(divide(3.96, const_100), const_2)), const_4)) | divide(n1,const_100)|divide(#0,const_2)|add(#1,const_1)|power(#2,const_4)|multiply(n0,#3)| | gain |
two numbers are respectively 19 % and 40 % more than a third number . the percentage that is first of the second is ? | i ii iii 119 140 100 140 - - - - - - - - - - 119 100 - - - - - - - - - - - ? = > 15 % answer : a | a ) 15 % , b ) 20 % , c ) 25 % , d ) 30 % , e ) 35 % | a | subtract(const_100, multiply(divide(add(19, const_100), add(40, const_100)), const_100)) | add(n0,const_100)|add(n1,const_100)|divide(#0,#1)|multiply(#2,const_100)|subtract(const_100,#3) | gain |
the c . p of 5 books is equal to the s . p of 20 books . find his gain % or loss % ? | "explanation : 5 cp = 20 sp 20 - - - 15 cp loss 100 - - - ? = > 75 % loss answer : e" | a ) 35 % , b ) 20 % , c ) 5 % , d ) 25 % , e ) 75 % | e | multiply(subtract(const_1, divide(5, 20)), const_100) | divide(n0,n1)|subtract(const_1,#0)|multiply(#1,const_100)| | gain |
for all positive integers m and v , the expression m θ v represents the remainder when m is divided by v . what is the value of ( ( 90 θ 33 ) θ 17 ) - ( 99 θ ( 33 θ 17 ) ) ? | "( ( 90 θ 33 ) θ 17 ) the remainder of 90 divided by 33 is 24 ; the remainder of 24 divided by 17 is 7 ; ( 99 θ ( 33 θ 17 ) ) the remainder of 33 divided by 17 is 16 ; 7 - 3 = 4 . answer : d ." | a ) 10 , b ) 8 , c ) 6 , d ) 4 , e ) 2 | d | subtract(reminder(reminder(90, 33), 17), reminder(99, reminder(33, 17))) | reminder(n0,n1)|reminder(n1,n2)|reminder(#0,n2)|reminder(n3,#1)|subtract(#2,#3)| | general |
convert the 10 / 36 m / s into kilometers per hour ? | "10 / 36 m / s = 10 / 36 * 18 / 5 = 1 / 1 = 1 kmph . answer : d" | a ) 2.9 kmph , b ) 9.9 kmph , c ) 1.3 kmph , d ) 1 kmph , e ) 5.7 kmph | d | multiply(const_3_6, divide(10, 36)) | divide(n0,n1)|multiply(#0,const_3_6)| | physics |
what is the remainder when 9 ^ 1 + 9 ^ 2 + 9 ^ 3 + . . . + 9 ^ 9 is divided by 3 ? | "notice that in the brackets we have the sum of 8 odd multiples of 3 , which yields remainder of 0 upon division by 3 . answer : a" | a ) 0 , b ) 3 , c ) 2 , d ) 5 , e ) none of the above | a | divide(subtract(power(9, 1), power(9, 1)), 9) | power(n0,n1)|power(n2,n1)|subtract(#0,#1)|divide(#2,n4)| | general |
if the price of gasoline increases by 25 % and a driver intends to spend only 10 % more on gasoline , by how much percent should the driver reduce the quantity of gasoline that he buys ? | "let x be the amount of gasoline the driver buys originally . let y be the new amount of gasoline the driver should buy . let p be the original price per liter . ( 1.25 * p ) y = 1.1 ( p * x ) y = ( 1.1 / 1.25 ) * x = 0.88 x which is a reduction of 12 % . the answer is d ." | a ) 5 % , b ) 8 % , c ) 10 % , d ) 12 % , e ) 15 % | d | multiply(divide(subtract(add(25, const_100), add(10, const_100)), add(25, const_100)), const_100) | add(n0,const_100)|add(n1,const_100)|subtract(#0,#1)|divide(#2,#0)|multiply(#3,const_100)| | general |
a square was given . inside the square there are white tiles and black tiles . black tiles was among the diagonal of the square and dimensions of both white as well as black tiles is 1 cm x 1 cm . if there are 81 black tiles in the square . then find the no of white tiles in it . | in a square , number of squares on the diagonal is equal to the tiles on a single row . if there are even number of square on a side , then total squares on the diagonal is 2 n - 1 , otherwise 2 n . as the total tiles on the diagonal are given as 81 , then number of tiles on a side = 2 n - 1 = 81 so n = 41 . so number ... | ['a ) 2399', 'b ) 1600', 'c ) 2679', 'd ) 1697', 'e ) 1711'] | b | subtract(power(divide(add(81, 1), const_2), const_2), 81) | add(n0,n2)|divide(#0,const_2)|power(#1,const_2)|subtract(#2,n2) | geometry |
if each edge of cube increased by 15 % , the percentage increase in | "100 × ( 115 ) / 100 × ( 115 ) / 100 = 132.25 = > 32.25 % answer is b ." | a ) 30.25 , b ) 32.25 , c ) 38.25 , d ) 37.25 , e ) 41.25 | b | multiply(const_100, divide(subtract(power(add(const_100, 15), const_3), power(const_100, const_3)), power(const_100, const_3))) | add(n0,const_100)|power(const_100,const_3)|power(#0,const_3)|subtract(#2,#1)|divide(#3,#1)|multiply(#4,const_100)| | gain |
the average age of 16 students of a class is 16 years . out of these , the average age of 5 students is 14 years and that of the other 9 students is 16 years , the age of the 12 th student is | "explanation : age of the 16 th student = [ 16 * 16 - ( 14 * 5 + 16 * 9 ) ] = ( 256 - 214 ) = 42 years . answer : e" | a ) 11 , b ) 66 , c ) 18 , d ) 16 , e ) 42 | e | subtract(multiply(16, 16), add(multiply(5, 14), multiply(9, 16))) | multiply(n0,n0)|multiply(n2,n3)|multiply(n4,n5)|add(#1,#2)|subtract(#0,#3)| | general |
the difference between the place value and the face value of 8 in the numeral 856973 is | "( place value of 8 ) - ( face value of 8 ) = ( 800000 - 8 ) = 799992 answer : option e" | a ) 973 , b ) 6973 , c ) 5994 , d ) 6084 , e ) none of these | e | subtract(multiply(const_10, 8), 8) | multiply(n0,const_10)|subtract(#0,n0)| | general |
the average speed of a car is 1 4 / 5 times the avg speed of a bike . a tractor covers 575 km in 23 hrs . how much distance will the car cover in 6 hrs if the speed of the bike is twice speed of the tractor ? | "sol . average speed of a tractor = 25 km / h the speed of a bus in an hour = 25 × 2 = 50 km the speed of a car in an hour = 9 / 5 * 50 = 90 km so , the distance covered by car in 6 h is 90 × 6 = 540 km ans . ( d )" | a ) 400 km , b ) 500 km , c ) 360 km , d ) 540 km , e ) 600 km | d | multiply(multiply(add(1, divide(4, 5)), multiply(const_2, divide(575, 23))), 6) | divide(n1,n2)|divide(n3,n4)|add(n0,#0)|multiply(#1,const_2)|multiply(#2,#3)|multiply(n5,#4)| | general |
33 men can complete a piece of work in 18 days . in how many days will 27 men complete the same work ? | "explanation : less men , means more days { indirect proportion } let the number of days be x then , 27 : 33 : : 18 : x [ please pay attention , we have written 27 : 33 rather than 33 : 27 , in indirect proportion , if you get it then chain rule is clear to you : ) ] { \ color { blue } x = \ frac { 33 \ times 18 } { 27... | a ) 22 , b ) 77 , c ) 36 , d ) 25 , e ) 13 | a | divide(multiply(18, 33), 27) | multiply(n0,n1)|divide(#0,n2)| | physics |
john makes $ 65 a week from his job . he earns a raise andnow makes $ 70 a week . what is the % increase ? | "increase = ( 5 / 65 ) * 100 = ( 1 / 13 ) * 100 = 7.69 % . a" | a ) 7.69 % , b ) 16.66 % , c ) 17.8 % , d ) 19 % , e ) 21 % | a | multiply(divide(subtract(70, 65), 65), const_100) | subtract(n1,n0)|divide(#0,n0)|multiply(#1,const_100)| | gain |
an error 38 % in excess is made while measuring the side of a square . now what is the percentage of error in the calculated area of the square ? | "percentage error in calculated area = ( 38 + 38 + ( 38 ã — 38 ) / 100 ) % = 90.44 % answer : e" | a ) 6.64 % , b ) 16.64 % , c ) 15.64 % , d ) 26.64 % , e ) 90.44 % | e | divide(multiply(subtract(square_area(add(const_100, 38)), square_area(const_100)), const_100), square_area(const_100)) | add(n0,const_100)|square_area(const_100)|square_area(#0)|subtract(#2,#1)|multiply(#3,const_100)|divide(#4,#1)| | gain |
if n is a positive integer , what is the remainder when ( 7 ^ ( 4 n + 3 ) ) ( 3 ^ n ) is divided by 10 ? | "this one took me bout 3 1 / 2 min . just testin numbers and what not . first notice that n is positive . save time by noticing thati worked out one solution where n = 0 only to find that thats not an option : p . 1 - 7 stands for ^ 1 thru 7 1 : 7 * 1 = 7 2 : 7 * 7 = 9 3 : 7 * 9 = 3 4 : 7 * 3 = 1 5 : 7 * 1 = 7 6 : 7 * ... | a ) 1 , b ) 2 , c ) 4 , d ) 6 , e ) 8 | b | reminder(multiply(multiply(const_3, const_3), 3), 10) | multiply(const_3,const_3)|multiply(n3,#0)|reminder(#1,n4)| | general |
in kaya ' s teacher ' s desk there are 6 pink highlighters , 2 yellow highlighters , and 4 blue highlighters . how many highlighters are there in all ? | "add the numbers of highlighters . 6 + 2 + 4 = 12 . answer is b ." | a ) 11 , b ) 12 , c ) 77 , d ) 33 , e ) 88 | b | add(add(6, 2), 4) | add(n0,n1)|add(n2,#0)| | general |
if a speaks the truth 20 % of the times , b speaks the truth 80 % of the times . what is the probability that at least one will tell the truth | "probability of a speaks truth p ( a ) = 2 / 10 ; false = 8 / 10 probability of b speaks truth p ( b ) = 8 / 10 ; false = 2 / 10 . for given qtn ans = 1 - ( neither of them tell truth ) . because a & b are independent events = 1 - [ ( 8 / 10 ) * ( 2 / 10 ) ] = 1 - 16 / 100 = 1 - 0.16 = 0.84 answer : b" | a ) 0.82 , b ) 0.84 , c ) 1.5 , d ) 1.2 , e ) 1.4 | b | multiply(divide(20, multiply(multiply(const_4, const_5), const_5)), divide(80, multiply(multiply(const_4, const_5), const_5))) | multiply(const_4,const_5)|multiply(#0,const_5)|divide(n0,#1)|divide(n1,#1)|multiply(#2,#3)| | gain |
a man can do a piece of work in 15 days , but with the help of his son , he can do it in 6 days . in what time can the son do it alone ? | "son ' s 1 day ' s work = ( 1 / 6 ) - ( 1 / 15 ) = 3 / 30 = 1 / 10 the son alone can do the work in 10 days answer is e" | a ) 13 , b ) 9 , c ) 15 , d ) 8 , e ) 10 | e | divide(multiply(15, 6), subtract(15, 6)) | multiply(n0,n1)|subtract(n0,n1)|divide(#0,#1)| | physics |
how many of the positive factors of 45 , 16 and how many common factors are there in numbers ? | "factors of 45 - 1 , 3 , 9 , 15 and 45 factors of 16 - 1 , 2 , 4 , 8 and 16 comparing both , we have three common factors of 45,16 - 1 answer ( a )" | a ) 1 , b ) 2 , c ) 3 , d ) 4 , e ) 5 | a | divide(16, 45) | divide(n1,n0)| | other |
what is the unit ’ s digit of 7 ^ 5 | "7 ^ 1 = 7 7 ^ 2 = 49 7 ^ 3 = 343 7 ^ 4 = 1 ( last digit ) 7 ^ 5 = 7 ( last digit ) and the cycle repeats after every 4 powers therefore , last digit of 7 ^ 5 = 7 answer d" | a ) 1 , b ) 3 , c ) 5 , d ) 7 , e ) 9 | d | reminder(power(7, const_2), const_10) | power(n0,const_2)|reminder(#0,const_10)| | general |
if y is the smallest positive integer such that 8820 multiplied by y is the square of an integer , then y must be | "8820 = 2 ^ 2 * 3 ^ 2 * 5 * 7 ^ 2 to be perfect square , we need to multiply by at least 5 . the answer is b ." | a ) 2 , b ) 5 , c ) 6 , d ) 7 , e ) 14 | b | multiply(add(const_2, const_3), const_2) | add(const_2,const_3)|multiply(#0,const_2)| | geometry |
if 20 typists can type 60 letters in 20 minutes , then how many letters will 30 typists working at the same rate complete in 1 hour ? | "no . of letters typing by 20 typists in 20 minutes = 60 no . of letters typing by 20 typists in 60 minutes = 60 * 3 = 180 no . of letters typing by 30 typists in 60 minutes = 180 / 20 * 30 = 270 answer : c" | a ) 240 , b ) 200 , c ) 270 , d ) 300 , e ) 310 | c | multiply(divide(multiply(60, const_3), 20), 30) | multiply(n1,const_3)|divide(#0,n0)|multiply(n3,#1)| | physics |
a batsman makes a score of 85 runs in the 17 th inning and thus increases his averages by 3 . find his average after 17 th inning ? | "let the average after 17 th inning = x then average after 16 th inning = ( x - 3 ) therefore 16 ( x - 3 ) + 85 = 17 x therefore x = 37 answer : c" | a ) 19 , b ) 29 , c ) 37 , d ) 49 , e ) 59 | c | add(subtract(85, multiply(17, 3)), 3) | multiply(n1,n2)|subtract(n0,#0)|add(n2,#1)| | general |
of the families in city x in 1998 , 20 percent owned a personal computer . the number of families in city x owning a computer in 2002 was 40 percent greater than it was in 1998 , and the total number of families in city x was 8 percent greater in 2002 than it was in 1998 . what percent of the families in city x owned a... | "say a 100 families existed in 1998 then the number of families owning a computer in 1998 - 20 number of families owning computer in 2002 = 20 * 140 / 100 = 28 number of families in 2002 = 108 the percentage = 28 / 108 * 100 = 25.92 % . answer : e" | a ) 49.66 % , b ) 52.66 % , c ) 28.22 % , d ) 33.33 % , e ) 25.92 % | e | multiply(const_100, divide(divide(multiply(add(40, const_100), 20), const_100), add(const_100, 8))) | add(n3,const_100)|add(n5,const_100)|multiply(n1,#0)|divide(#2,const_100)|divide(#3,#1)|multiply(#4,const_100)| | general |
three unbiased coins are tossed . what is the probability of getting 3 heads ? | "let , h - - > head , t - - > tail here s = { ttt , tth , tht , htt , thh , hth , hht , hhh } let e = event of getting 3 heads then e = { hhh } p ( e ) = n ( e ) / n ( s ) = 1 / 8" | a ) 3 / 4 , b ) 1 / 4 , c ) 3 / 8 , d ) 7 / 8 , e ) 1 / 8 | e | negate_prob(divide(const_1, power(const_2, const_3))) | power(const_2,const_3)|divide(const_1,#0)|negate_prob(#1)| | probability |
three numbers are in the ratio of 2 : 3 : 4 and their l . c . m . is 240 . their h . c . f . is : | "explanation : let the number be 2 x , 3 x and 4 x lcm of 2 x , 3 x and 4 x = 12 x 12 x = 240 = > x = 240 / 12 = 20 hcf of 2 x , 3 x and 4 x = x = 20 answer : option b" | a ) 30 , b ) 20 , c ) 40 , d ) 60 , e ) 70 | b | multiply(2, 4) | multiply(n0,n2)| | other |
if the radius of a circle is increased by 10 % then the area is increased by : | "initially a = pi * r 2 now r 2 = 110 / 100 r then area = pi * ( 110 r / 100 ) 2 area = 121 r / 100 that is area increases from 100 to 121 = increase in area = 21 % answer : e" | a ) 44 % , b ) 120 % , c ) 144 % , d ) 40 % , e ) 21 % | e | multiply(subtract(divide(const_100, const_100), power(subtract(divide(const_100, const_100), divide(10, const_100)), const_2)), const_100) | divide(const_100,const_100)|divide(n0,const_100)|subtract(#0,#1)|power(#2,const_2)|subtract(#0,#3)|multiply(#4,const_100)| | geometry |
217 x 217 + 183 x 183 = ? | "= ( 217 ) ^ 2 + ( 183 ) ^ 2 = ( 200 + 17 ) ^ 2 + ( 200 - 17 ) ^ 2 = 2 [ ( 200 ) ^ 2 + ( 17 ) ^ 2 ] = 2 [ 40000 + 289 ] = 2 x 40289 = 80578 answer is b" | a ) 79698 , b ) 80578 , c ) 80698 , d ) 81268 , e ) none of them | b | multiply(217, power(217, 183)) | power(n1,n2)|multiply(n0,#0)| | general |
kim has 6 pairs of shoes ; each pair is a different color . if kim randomly selects 2 shoes without replacement from the 12 shoes , what is the probability that she will select 2 shoes of the same color ? | total pairs = 12 c 2 = 66 ; same color pairs = 6 c 1 * 1 c 1 = 6 ; prob = 1 / 11 ans d | a ) 2 / 5 , b ) 1 / 5 , c ) 1 / 9 , d ) 1 / 11 , e ) 1 / 25 | d | divide(6, choose(12, 2)) | choose(n2,n1)|divide(n0,#0) | probability |
in a kilometer race , a beats b by 40 meters or 10 seconds . what time does a take to complete the race ? | "time taken by b run 1000 meters = ( 1000 * 10 ) / 40 = 250 sec . time taken by a = 250 - 10 = 240 sec . answer : e" | a ) 290 sec , b ) 190 sec , c ) 199 sec , d ) 192 sec , e ) 240 sec | e | subtract(divide(multiply(const_1, const_1000), divide(40, 10)), 10) | divide(n0,n1)|multiply(const_1,const_1000)|divide(#1,#0)|subtract(#2,n1)| | physics |
a crow leaves its nest , and flies back and forth from its nest to a nearby ditch to gather worms . the distance between the nest and the ditch is 100 meters . in one and a half hours , the crow manages to bring worms to its nest 15 times . what is the speed of the crow in kilometers per hour ? | the distance between the nest and the ditch is 100 meters . 15 times mean = a crow leaves its nest , and flies back ( going and coming back ) i . e . 2 times we get total 30 rounds . so the distance is 30 * 100 = 3000 . d = st 3000 / 1.5 = t , i think we can take 3000 meters as 3 km , then only we get t = 2 . ( 1000 me... | a ) 1 , b ) 2 , c ) 3 , d ) 5 , e ) 7 | d | divide(divide(multiply(100, multiply(15, const_2)), const_1000), divide(15, const_10)) | divide(n1,const_10)|multiply(n1,const_2)|multiply(n0,#1)|divide(#2,const_1000)|divide(#3,#0) | physics |
s is a set of 100 consecutive multiples of 8 . if the smallest number in s is 108 , then the greatest number in s is | "last term = first term + ( total no . of terms - 1 ) consecutive difference s is a set of 100 consecutive multiples of 8 . if the smallest number in s is 108 , then the greatest number in s is first term = 102 ; total terms = 100 ; difference = 3 108 + ( 99 ) 8 = 900 ans e" | a ) 998 , b ) 297 , c ) 999 , d ) 902 , e ) 900 | e | add(108, multiply(subtract(100, const_1), 8)) | subtract(n0,const_1)|multiply(n1,#0)|add(n2,#1)| | general |
the batting average of a particular batsman is 62 runs in 46 innings . if the difference in his highest and lowest score is 150 runs and his average excluding these two innings is 58 runs , find his highest score . | "explanation : total runs scored by the batsman = 62 * 46 = 2852 runs now excluding the two innings the runs scored = 58 * 44 = 2552 runs hence the runs scored in the two innings = 2852 – 2552 = 300 runs . let the highest score be x , hence the lowest score = x – 150 x + ( x - 150 ) = 300 2 x = 450 x = 225 runs answer ... | a ) 179 , b ) 208 , c ) 210 , d ) 225 , e ) 229 | d | divide(add(150, subtract(multiply(62, 46), multiply(58, subtract(46, const_2)))), const_2) | multiply(n0,n1)|subtract(n1,const_2)|multiply(n3,#1)|subtract(#0,#2)|add(n2,#3)|divide(#4,const_2)| | general |
what is the greatest value of positive integer x such that 2 ^ x is a factor of 100 ^ 50 ? | put in simple words , we need to find the highest power of 2 in 100 ^ 50 100 = 2 ^ 2 * 5 ^ 2 therefore 100 ^ 50 = ( 2 ^ 2 * 5 ^ 2 ) ^ 50 = 2 ^ 100 * 5 ^ 100 answer : d | a ) 70 , b ) 80 , c ) 90 , d ) 100 , e ) 120 | d | subtract(100, subtract(power(2, divide(100, 50)), const_1)) | divide(n1,n2)|power(n0,#0)|subtract(#1,const_1)|subtract(n1,#2) | general |
a towel , when bleached , lost 10 % of its length and 20 % of its breadth . what is the percentage decrease in area ? | "percentage change in area = ( − 10 − 20 + ( 10 × 20 ) / 100 ) % = − 28 % i . e . , area is decreased by 28 % answer : a" | a ) 28 % , b ) 30 % , c ) 44 % , d ) 54 % , e ) 64 % | a | divide(subtract(multiply(const_100, const_100), multiply(subtract(const_100, 10), subtract(const_100, 20))), const_100) | multiply(const_100,const_100)|subtract(const_100,n0)|subtract(const_100,n1)|multiply(#1,#2)|subtract(#0,#3)|divide(#4,const_100)| | gain |
if 5 ^ 5 × 5 ^ 7 = ( 5 ^ 4 ) ^ x , then what is the value of x ? | 5 ^ 5 × 5 ^ 7 = ( 5 ^ 4 ) ^ x 5 ^ ( 5 + 7 ) = 5 ^ ( 12 ) with multiplying numbers with exponents it ' s the same as adding the exponents 5 ^ 12 = ( 5 ^ 4 ) ^ x since base is same , we can equate powers : when raising an exponent to an exponent , it ' s the same as multiplying them , therefore : 12 = 4 x x = 3 ans . b )... | a ) 2 , b ) 3 , c ) 4 , d ) 5 , e ) 6 | b | divide(log(multiply(power(5, 5), power(5, 7))), log(power(5, 4))) | power(n0,n0)|power(n0,n3)|power(n0,n5)|log(#2)|multiply(#0,#1)|log(#4)|divide(#5,#3) | general |
in a group of cows and hens , the number of legs are 16 more than twice the number of heads . the number of cows is | "explanation : let the number of cows be x and the number of hens be y . then , 4 x + 2 y = 2 ( x + y ) + 16 4 x + 2 y = 2 x + 2 y + 16 2 x = 16 x = 8 . answer : b" | a ) 5 , b ) 8 , c ) 10 , d ) 12 , e ) 14 | b | subtract(16, const_4) | subtract(n0,const_4)| | general |
on rainy mornings , mo drinks exactly n cups of hot chocolate ( assume that n is an integer ) . on mornings that are not rainy , mo drinks exactly 2 cups of tea . last week mo drank a total of 16 cups of tea and hot chocolate together . if during that week mo drank 4 more tea cups than hot chocolate cups , then how man... | "t = the number of cups of tea c = the number of cups of hot chocolate t + c = 16 t - c = 4 - > t = 10 . c = 6 . mo drinks 5 cups of tea a day then number of days that are not rainy = 10 / 2 = 5 so number of rainy days = 7 - 5 = 2 e is the answer ." | a ) 6 , b ) 3 , c ) 4 , d ) 5 , e ) 2 | e | subtract(add(const_4, 2), divide(divide(add(16, 4), const_2), 2)) | add(n0,const_4)|add(n1,n2)|divide(#1,const_2)|divide(#2,n0)|subtract(#0,#3)| | general |
if the price of a tv is first decreased by 20 % and then increased by 30 % , then the net change in the price will be : | explanation : solution : let the original price be rs . 100 . new final price = 130 % of ( 80 % of 100 ) = rs . 130 / 100 * 80 / 100 * 100 = rs . 104 . . ' . increase = 4 % answer : a | a ) 4 % increase , b ) 10 % increase , c ) 10 % decrease , d ) 6 % increase , e ) none of these | a | subtract(divide(multiply(subtract(const_100, 20), add(const_100, 30)), const_100), const_100) | add(n1,const_100)|subtract(const_100,n0)|multiply(#0,#1)|divide(#2,const_100)|subtract(#3,const_100) | general |
equal amount of water were poured into two empty jars of different capacities , which made one jar 1 / 5 full and other jar 1 / 4 full . if the water in the jar with lesser capacity is then poured into the jar with greater capacity , what fraction of the larger jar will be filled with water ? | "same amount of water made bigger jar 1 / 5 full , then the same amount of water ( stored for a while in smaller jar ) were added to bigger jar , so bigger jar is 1 / 5 + 1 / 5 = 2 / 5 full . answer : b ." | a ) 1 / 7 , b ) 2 / 5 , c ) 1 / 2 , d ) 7 / 12 , e ) 2 / 3 | b | divide(const_2, 5) | divide(const_2,n1)| | general |
if the average ( arithmetic mean ) of x + 3 , x + 4 , and x + 5 is 0 , then x = | "( x + 3 + x + 4 + x + 5 ) / 3 = 0 = > 3 x + 12 = 0 = > x = - 4 answer a" | a ) – 4 , b ) – 3 , c ) – 2 , d ) – 1 , e ) 0 | a | subtract(multiply(4, const_2), multiply(3, const_2)) | multiply(n1,const_2)|multiply(n0,const_2)|subtract(#0,#1)| | general |
dan can do a job alone in 15 hours . annie , working alone , can do the same job in just 10 hours . if dan works alone for 6 hours and then stops , how many hours will it take annie , working alone , to complete the job ? | "dan can complete 1 / 15 of the job per hour . in 6 hours , dan completes 6 ( 1 / 15 ) = 2 / 5 of the job . annie can complete 1 / 10 of the job per hour . to complete the job , annie will take 3 / 5 / 1 / 10 = 6 hours . the answer is c ." | a ) 2 , b ) 4 , c ) 6 , d ) 8 , e ) 10 | c | multiply(subtract(const_1, divide(6, 15)), 10) | divide(n2,n0)|subtract(const_1,#0)|multiply(n1,#1)| | physics |
when processing flower - nectar into honey bees ' extract , a considerable amount of water gets reduced . how much flower - nectar must be processed to yield 1 kg of honey , if nectar contains 50 % water , and the honey obtained from this nectar contains 20 % water ? | "explanation : flower - nectar contains 50 % of non - water part . in honey this non - water part constitutes 80 % ( 100 - 20 ) . therefore 0.5 x amount of flower - nectar = 0.80 x amount of honey = 0.80 x 1 kg therefore amount of flower - nectar needed = ( 0.80 / 0.51 ) kg = 1.6 kgs answer : d" | a ) 1.2 kg , b ) 1.5 kg , c ) 1.7 kg , d ) 1.6 kg , e ) none of these | d | divide(divide(subtract(const_100, 20), const_100), divide(50, const_100)) | divide(n1,const_100)|subtract(const_100,n2)|divide(#1,const_100)|divide(#2,#0)| | general |
in a throw of dice what is the probability of ge æ « ing number greater than 2 | "explanation : number greater than 2 is 3 , 4 , 5 & 6 , so only 4 number total cases of dice = [ 1,2 , 3,4 , 5,6 ] so probability = 4 / 6 = 2 / 3 answer : a" | a ) 2 / 3 , b ) 1 / 3 , c ) 1 / 5 , d ) 1 / 6 , e ) none of these | a | divide(subtract(const_6, 2), const_6) | subtract(const_6,n0)|divide(#0,const_6)| | probability |
find the area of a cuboid of length 8 cm , breadth 10 cm . and height 12 cm . | "area of a cuboid = lxbxh = 8 cm x 10 cm x 12 cm = 960 cm cube answer : e" | a ) 900 cm cube , b ) 910 cm cube , c ) 920 cm cube , d ) 930 cm cube , e ) 960 cm cube | e | multiply(multiply(8, 10), 12) | multiply(n0,n1)|multiply(n2,#0)| | physics |
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