Problem stringlengths 5 967 | Rationale stringlengths 1 2.74k | options stringlengths 37 300 | correct stringclasses 5
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a train with a length of 100 meters , is traveling at a speed of 72 km / hr . the train enters a tunnel 2.9 km long . how many minutes does it take the train to pass through the tunnel from the moment the front enters to the moment the rear emerges ? | "72 km / hr = 1.2 km / min the total distance is 3 km . 3 / 1.2 = 2.5 minutes the answer is d ." | a ) 1.6 , b ) 1.9 , c ) 2.2 , d ) 2.5 , e ) 2.8 | d | multiply(divide(add(2.9, divide(100, const_1000)), 72), const_60) | divide(n0,const_1000)|add(n2,#0)|divide(#1,n1)|multiply(#2,const_60)| | physics |
a train 120 m long is running at a speed of 68 kmph . how long does it take to pass a man who is running at 8 kmph in the same direction as the train ? | speed of the train relative to man = ( 68 - 8 ) kmph = ( 60 * 5 / 18 ) m / sec = ( 50 / 3 ) m / sec time taken by the train to cross the man = time taken by it to cover 120 m at 50 / 3 m / sec = 120 * 3 / 50 sec = 7.2 sec answer : a . | a ) 7.2 sec , b ) 9.3 sec , c ) 12 sec , d ) 15.2 sec , e ) 18 sec | a | divide(120, multiply(subtract(68, 8), const_0_2778)) | subtract(n1,n2)|multiply(#0,const_0_2778)|divide(n0,#1) | physics |
if exactly 7 guests drank only one of the two types of drinks , how many guests drank both types of drinks ? | say x guests drank both drinks . ( 7 - x ) + ( 4 - x ) = 7 - - > x = 2 . answer : e . | a ) 5 , b ) 6 , c ) 7 , d ) 4 , e ) 2 | e | multiply(7, const_2) | multiply(n0,const_2) | other |
a triangle has sides 10 , 17 , and 21 . a square is inscribed in the triangle . one side of the square lies on the longest side of the triangle . the other two vertices of the square touch the two shorter sides of the triangle . what is the length of the side of the square ? | by heron ' s formula , the area , a , of a triangle with sides a , b , c is given by a = square root [ s ( s − a ) ( s − b ) ( s − c ) ] , where s = ½ ( a + b + c ) is the semi - perimeter of the triangle . then s = ½ ( 10 + 17 + 21 ) = 24 , and a = 84 . a square inscribed in a 10 by 17 by 21 triangle , with perpendicu... | ['a ) 148 / 29', 'b ) 168 / 29', 'c ) 168 / 26', 'd ) 158 / 29', 'e ) 198 / 29'] | b | divide(divide(multiply(triangle_area_three_edges(10, 17, 21), const_2), 21), add(divide(divide(multiply(triangle_area_three_edges(10, 17, 21), const_2), 21), 21), const_1)) | triangle_area_three_edges(n0,n1,n2)|multiply(#0,const_2)|divide(#1,n2)|divide(#2,n2)|add(#3,const_1)|divide(#2,#4) | geometry |
if a train , travelling at a speed of 160 kmph , crosses a pole in 9 sec , then the length of train is ? | "d = 160 * 5 / 18 * 9 = 400 m answer : e" | a ) 281 , b ) 125 , c ) 288 , d ) 266 , e ) 400 | e | multiply(multiply(160, const_0_2778), 9) | multiply(n0,const_0_2778)|multiply(n1,#0)| | physics |
when a certain number x is divided by 63 , the remainder is 11 . what is the remainder when x is divided by 9 ? | "when a certain number x is divided by 63 , the remainder is 11 . what is the remainder when x is divided by 9 ? putting a value say x = 11 we get remainder as 11 when divided by 63 . when 11 divided by 9 we get 2 as remainder . a is the answer ." | a ) 2 , b ) 3 , c ) 4 , d ) 5 , e ) 6 | a | reminder(11, 9) | reminder(n1,n2)| | general |
due to construction , the speed limit along an 15 - mile section of highway is reduced from 55 miles per hour to 35 miles per hour . approximately how many minutes more will it take to travel along this section of highway at the new speed limit than it would have taken at the old speed limit ? | "old time in minutes to cross 15 miles stretch = 15 * 60 / 55 = 15 * 12 / 11 = 16.36 new time in minutes to cross 15 miles stretch = 15 * 60 / 35 = 15 * 12 / 7 = 25.71 time difference = 9.36 answer c )" | a ) 5 , b ) 8 , c ) 9.36 , d ) 15 , e ) 24 | c | max(multiply(subtract(add(55, 15), const_1), subtract(divide(15, 35), divide(15, 55))), const_4) | add(n0,n1)|divide(n0,n2)|divide(n0,n1)|subtract(#0,const_1)|subtract(#1,#2)|multiply(#3,#4)|max(#5,const_4)| | physics |
[ ( 3.242 x 12 ) / 100 ] = ? | answer multiplying 3.242 x 12 = 3.8904 now divide 3.8904 by 100 so , 3.8904 ÷ 100 = 0.038904 ∴ shift the decimal two places to the left as 100 correct option : e | a ) 0.045388 , b ) 4.5388 , c ) 453.88 , d ) 473.88 , e ) 0.038904 | e | divide(divide(multiply(3.242, 12), const_100), const_10) | multiply(n0,n1)|divide(#0,const_100)|divide(#1,const_10) | general |
for all positive integers m , [ m ] = 3 m when m is odd and [ m ] = ( 1 / 2 ) * m when m is even . what is [ 5 ] * [ 6 ] equivalent to ? | "[ 5 ] * [ 6 ] = 15 * 3 = 45 = [ 15 ] the answer is c ." | a ) [ 45 ] , b ) [ 30 ] , c ) [ 15 ] , d ) [ 5 ] , e ) [ 50 ] | c | divide(multiply(divide(6, 2), multiply(3, 5)), 3) | divide(n4,n2)|multiply(n0,n3)|multiply(#0,#1)|divide(#2,n0)| | general |
find the c . i . on a sum of rs . 10000 for 9 months at 20 % per annum , interest being compounded quarterly ? | "c . i . = 10000 ( 21 / 20 ) ^ 3 - 10000 = 1025 answer : c" | a ) 2000 , b ) 10025 , c ) 1025 , d ) 2015 , e ) 2025 | c | subtract(multiply(10000, multiply(add(const_1, divide(const_0_25, const_4)), add(const_1, divide(const_0_25, const_4)))), 10000) | divide(const_0_25,const_4)|add(#0,const_1)|multiply(#1,#1)|multiply(n0,#2)|subtract(#3,n0)| | gain |
a person crosses a 720 m long street in 12 minutes . what is his speed in km per hour ? | "speed = 720 / ( 12 x 60 ) m / sec = 1 m / sec . converting m / sec to km / hr = 1 x ( 18 / 5 ) km / hr = 3.6 km / hr . answer : a" | a ) 3.6 , b ) 4.5 , c ) 4.8 , d ) 5.4 , e ) 5.5 | a | divide(divide(720, const_1000), divide(multiply(12, const_60), const_3600)) | divide(n0,const_1000)|multiply(n1,const_60)|divide(#1,const_3600)|divide(#0,#2)| | physics |
in a certain boys camp , 20 % of the total boys are from school a and 30 % of those study science . if there are 28 boys in the camp that are from school a but do not study science then what is the total number of boys in the camp ? | "since 30 % of the boys from school a study science , then 70 % of the boys from school a do not study science and since 20 % of the total number of boys are from school a , then 0.2 * 0.7 = 0.14 , or 14 % of the boys in the camp are from school a and do not study science . we are told that this number equals to 28 , s... | a ) 70 , b ) 245 , c ) 150 , d ) 200 , e ) 350 | d | divide(28, multiply(divide(subtract(const_100, 30), const_100), divide(20, const_100))) | divide(n0,const_100)|subtract(const_100,n1)|divide(#1,const_100)|multiply(#2,#0)|divide(n2,#3)| | gain |
if 22 lines are drawn in a plane such that no two of them are parallel and no three are concurrent , then in how many points do they intersect ? | "if two lines are not parallel , then they will intersect in exactly one point . lines can be extended infinitely on both ends so somewhere they will intersect with each other if they are not parallel . we are also given that no three lines are concurrent . this means that no three lines intersect at the same point . s... | a ) 125 , b ) 187 , c ) 231 , d ) 375 , e ) 460 | c | divide(factorial(22), multiply(factorial(subtract(22, const_2)), factorial(const_2))) | factorial(n0)|factorial(const_2)|subtract(n0,const_2)|factorial(#2)|multiply(#3,#1)|divide(#0,#4)| | physics |
the population of a town increased from 1 , 75,000 to 2 , 45,000 in a decade . the average percent increase of population per year is | "solution increase in 10 years = ( 245000 - 175000 ) = 87500 . increase % = ( 70000 / 175000 ã — 100 ) % = 40 % . required average = ( 40 / 10 ) % = 4 % . answer a" | a ) 4 % , b ) 5 % , c ) 6 % , d ) 8.75 % , e ) none | a | add(multiply(divide(subtract(divide(subtract(subtract(subtract(multiply(multiply(const_10, const_1000), const_10), const_1000), const_1000), multiply(add(2, const_3), const_100)), multiply(add(multiply(add(const_3, const_4), const_10), add(2, const_3)), const_1000)), 1), const_10), const_100), const_4) | add(n2,const_3)|add(const_3,const_4)|multiply(const_10,const_1000)|multiply(#2,const_10)|multiply(#0,const_100)|multiply(#1,const_10)|add(#0,#5)|subtract(#3,const_1000)|multiply(#6,const_1000)|subtract(#7,const_1000)|subtract(#9,#4)|divide(#10,#8)|subtract(#11,n0)|divide(#12,const_10)|multiply(#13,const_100)|add(#14,co... | general |
how many integers between 1 and 10 ^ 20 are such that the sum of their digits is 2 ? | "the integers with a sum of 2 are : 2 , 20 , 200 , . . . , 2 * 10 ^ 19 and there are 20 integers in this list . also , these integers have a sum of 2 : 11 101 , 110 1001 , 1010 , 1100 etc . . . the number of integers in this list is 1 + 2 + . . . + 19 thus , the total number of integers is 1 + 2 + . . . + 19 + 20 = 20 ... | a ) 190 , b ) 210 , c ) 211 , d ) 230 , e ) 231 | b | subtract(10, add(add(multiply(const_2, const_100), multiply(add(const_3, const_4), const_10)), const_2)) | add(const_3,const_4)|multiply(const_100,const_2)|multiply(#0,const_10)|add(#1,#2)|add(#3,const_2)|subtract(n1,#4)| | general |
the price of commodity x increases by 40 paise every year , while the price of commodity y increases by 15 paise every year . if in 1988 , the price of commodity x was 4.20 and that of y was 6.30 , in which year will commodity x cost 40 paise more than commodity y ? | suppose in ‘ n ’ years the price of comodity x will be more by 40 paise than that of commodity y . ∴ 420 + 40 n – 630 – 15 n = 40 or , 25 n – 210 = 40 or , 25 n = 250 or , n = 250 ⁄ 25 = 10 years . answer b | a ) 1997 , b ) 1998 , c ) 1999 , d ) 2000 , e ) none of these | b | add(1988, divide(add(divide(40, const_100), subtract(6.3, 4.2)), subtract(divide(40, const_100), divide(15, const_100)))) | divide(n0,const_100)|divide(n1,const_100)|subtract(n4,n3)|add(#0,#2)|subtract(#0,#1)|divide(#3,#4)|add(n2,#5) | general |
when tom works alone he chops 4 lb . salad in 3 minutes , and when tammy works alone she chops 3 lb . salad in 4 minutes . they start working together , and after some time finish chopping 65 lb . of salad . of those 80 lb . , the salad quantity chopped by tammy is what percent greater than the quantifying chopped by t... | "tom chops 8 lbs in 12 minutes tammy chopswhen tom works alone he chops 2 lb . salad in 3 minutes , and when tammy works alone she chops 3 lb . salad in 2 minutes . they start working together , and after some time finish chopping 65 lb . of salad . of those 80 lb . , the salad quantity chopped by tammy is what percent... | a ) 125 % , b ) 100 % , c ) 25 % , d ) 225 % , e ) 400 % | a | multiply(divide(subtract(divide(3, 4), divide(const_2.0, 3)), divide(4, 3)), const_100) | divide(n1,const_2.0)|divide(const_2.0,n1)|subtract(#0,#1)|divide(#2,#1)|multiply(#3,const_100)| | physics |
of the total amount that jill spent on a shopping trip , excluding taxes , she spent 45 percent on clothing , 45 percent on food , and 10 percent on other items . if jill paid a 5 percent tax on the clothing , no tax on the food , and an 10 percent tax on all other items , then the total tax that she paid was what perc... | "assume she has $ 200 to spend . tax clothing = 45 % = $ 90 = $ 4.5 food = 45 % = $ 90 = $ 0.00 items = 10 % = $ 20 = $ 2.00 total tax = $ 6.50 % of total amount = 6.5 / 200 * 100 = 3.25 % answer d" | a ) 3.5 , b ) 3.35 , c ) 3.45 , d ) 3.25 , e ) 3.55 | d | multiply(divide(add(multiply(45, divide(5, const_100)), multiply(10, divide(10, const_100))), const_100), const_100) | divide(n3,const_100)|divide(n4,const_100)|multiply(n0,#0)|multiply(n2,#1)|add(#2,#3)|divide(#4,const_100)|multiply(#5,const_100)| | general |
a cistern can be filled by a tap in 4 hours while it can be emptied by another tap in 10 hours . if both the taps are opened simultaneously , then after how much time will the cistern get filled ? | "net part filled in 1 hour = ( 1 / 4 - 1 / 10 ) = 3 / 20 the cistern will be filled in 20 / 3 hrs i . e . , 6.7 hrs . answer : d" | a ) 2.9 hrs , b ) 8.9 hrs , c ) 2.9 hrs , d ) 6.7 hrs , e ) 8.6 hrs | d | divide(const_1, subtract(divide(const_1, 4), divide(const_1, 10))) | divide(const_1,n0)|divide(const_1,n1)|subtract(#0,#1)|divide(const_1,#2)| | physics |
j is 25 % less than p and 20 % less than t . t is t % less than p . what is the value of t ? | usually we can solve every question of this type by choosing appropriate value of the variable and deriving the value of other related variables . let , p = 400 then j = ( 75 / 100 ) * 400 = 300 also j = ( 80 / 100 ) * t i . e . t = 300 * 100 / 80 = 375 and t = [ 1 - ( t / 100 ) ] * p i . e . 100 - t = 100 * t / p = 10... | a ) 93.5 , b ) 90 , c ) 6.75 , d ) 6.25 , e ) 2 | d | divide(multiply(25, 25), const_100) | multiply(n0,n0)|divide(#0,const_100) | gain |
the speed of a car is 90 km in the first hour and 80 km in the second hour . what is the average speed of the car ? | "s = ( 90 + 80 ) / 2 = 85 kmph answer : c" | a ) 12 , b ) 75 , c ) 85 , d ) 54 , e ) 15 | c | divide(add(90, 80), const_2) | add(n0,n1)|divide(#0,const_2)| | physics |
if the price of gasoline increases by 25 % and ron intends to spend only 15 % more on gasoline , by how much % should he reduce the quantity of gasoline that he buys ? | explanatory answer let the price of 1 litre of gasoline be $ x and let ron initially buy ' y ' litres of gasoline . therefore , he would have spent $ xy on gasoline . when the price of gasoline increases by 25 % , the new price per litre of gasoline is 1.25 x . ron intends to increase the amount he spends on gasoline b... | a ) 10 % , b ) 12.5 % , c ) 8 % , d ) 12 % , e ) 6.66 % | c | multiply(subtract(const_1, divide(add(const_1, divide(15, const_100)), add(const_1, divide(25, const_100)))), const_100) | divide(n1,const_100)|divide(n0,const_100)|add(#0,const_1)|add(#1,const_1)|divide(#2,#3)|subtract(const_1,#4)|multiply(#5,const_100) | general |
in how many years will a sum of money doubles itself at 10 % per annum on simple interest ? | "p = ( p * 10 * r ) / 100 r = 10 % answer : c" | a ) 70 % , b ) 20 % , c ) 10 % , d ) 28 % , e ) 20 % | c | divide(const_100, 10) | divide(const_100,n0)| | gain |
if x gets 25 % more than y and y gets 20 % more than z , the share of z out of rs . 925 will be : | "z share = z , y = 1.2 z x = 1.25 ã — 1.2 z , x + y + z = 925 ( 1.25 ã — 1.2 + 1.2 + 1 ) z = 925 3.7 z = 925 , z = 250 answer : . c" | a ) rs . 300 , b ) rs . 200 , c ) rs . 250 , d ) rs . 350 , e ) none of these | c | divide(925, add(add(multiply(add(const_1, divide(25, const_100)), add(const_1, divide(20, const_100))), add(const_1, divide(20, const_100))), const_1)) | divide(n1,const_100)|divide(n0,const_100)|add(#0,const_1)|add(#1,const_1)|multiply(#3,#2)|add(#2,#4)|add(#5,const_1)|divide(n2,#6)| | general |
what will be the compound interest on a sum of rs . 45,000 after 3 years at the rate of 12 % p . a . ? | "amount = [ 45000 * ( 1 + 12 / 100 ) 3 ] = 45000 * 28 / 25 * 28 / 25 * 28 / 25 = rs . 63221.76 c . i . = ( 63221.76 - 45000 ) = rs : 18221.76 answer : d" | a ) s : 10123.19 , b ) s : 10123.29 , c ) s : 10123.20 , d ) s : 18221.76 , e ) s : 10123.12 | d | subtract(multiply(multiply(multiply(const_4, const_100), const_100), power(add(const_1, divide(12, const_100)), 3)), multiply(multiply(const_4, const_100), const_100)) | divide(n2,const_100)|multiply(const_100,const_4)|add(#0,const_1)|multiply(#1,const_100)|power(#2,n1)|multiply(#3,#4)|subtract(#5,#3)| | gain |
5 men are equal to as many women as are equal to 8 boys . all of them earn rs . 60 only . men â € ™ s wages are ? | "5 m = xw = 8 b 5 m + xw + 8 b - - - - - 60 rs . 5 m + 5 m + 5 m - - - - - 60 rs . 15 m - - - - - - 60 rs . = > 1 m = 4 rs . answer : c" | a ) 6 rs , b ) 2 rs , c ) 4 rs , d ) 9 rs , e ) 3 rs | c | divide(60, multiply(const_3, 5)) | multiply(n0,const_3)|divide(n2,#0)| | general |
3034 - ( 1002 / 20.04 ) = ? | option a | a ) 2984 , b ) 2983 , c ) 2982 , d ) 2981 , e ) none of these | a | subtract(3034, divide(1002, 20.04)) | divide(n1,n2)|subtract(n0,#0)| | general |
the tax on a commodity is diminished by 12 % and its consumption increased by 10 % . the effect on revenue is ? | "100 * 100 = 10000 88 * 110 = 9680 - - - - - - - - - - - 10000 - - - - - - - - - - - 320 100 - - - - - - - - - - - ? = > 3 % decrease answer : c" | a ) 8 , b ) 6 , c ) 3 , d ) 5 , e ) 4 | c | subtract(const_100, multiply(multiply(add(const_1, divide(10, const_100)), subtract(const_1, divide(12, const_100))), const_100)) | divide(n1,const_100)|divide(n0,const_100)|add(#0,const_1)|subtract(const_1,#1)|multiply(#2,#3)|multiply(#4,const_100)|subtract(const_100,#5)| | general |
last year a worker saved 15 % of her annual salary . this year , she made 20 % more money than last year and she saved 25 % of her salary . the amount she saved this year was what percent of the amount she saved last year ? | "let x be the worker ' s salary last year . last year , she saved 0.15 x . this year , she saved 0.25 ( 1.2 x ) = 0.3 x 0.3 x / 0.15 x = 200 % the answer is e ." | a ) 120 % , b ) 140 % , c ) 160 % , d ) 180 % , e ) 200 % | e | multiply(const_100, divide(15, multiply(add(20, const_100), divide(25, const_100)))) | add(n1,const_100)|divide(n2,const_100)|multiply(#0,#1)|divide(n0,#2)|multiply(#3,const_100)| | gain |
one pipe can fill a pool 1.25 times faster than a second pipe . when both pipes are opened , they fill the pool in 4 hours . how long would it take to fill the pool if only the slower pipe is used ? | "say the rate of the slower pipe is r pool / hour , then the rate of the faster pipe would be 1.25 r = 5 r / 4 . since when both pipes are opened , they fill the pool in four hours , then their combined rate is 1 / 4 pool / hour . thus we have that r + 5 r / 4 = 1 / 4 - - > r = 1 / 9 pool / hour - - > time is reciproca... | a ) 11.25 , b ) 11.52 , c ) 1.25 , d ) 9 , e ) 7.2 | d | divide(inverse(divide(inverse(add(const_3, const_2)), add(const_1, 1.25))), 1.25) | add(const_2,const_3)|add(n0,const_1)|inverse(#0)|divide(#2,#1)|inverse(#3)|divide(#4,n0)| | physics |
what will be the reminder when ( 27 ^ 27 + 27 ) is divided by 28 ? | "( x ^ n + 1 ) will be divisible by ( x + 1 ) only when n is odd ; ( 27 ^ 27 + 1 ) will be divisible by ( 27 + 1 ) ; ( 27 ^ 27 + 1 ) + 26 when divided by 28 will give 26 as remainder . correct option : d" | a ) 92 , b ) 43 , c ) 71 , d ) 26 , e ) 31 | d | subtract(add(power(27, 27), 27), multiply(28, floor(divide(add(power(27, 27), 27), 28)))) | power(n0,n0)|add(n0,#0)|divide(#1,n3)|floor(#2)|multiply(n3,#3)|subtract(#1,#4)| | general |
ram ' s salary is 25 % more than raja ' s . how much percent is raja ' s salary less than ram ' s ? | raja ' s salary less than ram ' s by 25 / ( 100 + 25 ) * 100 % = 20 % answer is c | a ) 10 % , b ) 15 % , c ) 20 % , d ) 25 % , e ) 50 % | c | subtract(const_100, multiply(const_100, inverse(add(divide(25, const_100), const_1)))) | divide(n0,const_100)|add(#0,const_1)|inverse(#1)|multiply(#2,const_100)|subtract(const_100,#3) | general |
what is the tens digit of 6 ^ 20 ? | "the tens digit of 6 in integer power starting from 2 ( 6 ^ 1 has no tens digit ) repeats in a pattern of 5 : { 3 , 1 , 9 , 7 , 5 } : the tens digit of 6 ^ 2 = 36 is 3 . the tens digit of 6 ^ 3 = 216 is 1 . the tens digit of 6 ^ 4 = . . . 96 is 9 . the tens digit of 6 ^ 5 = . . . 76 is 7 . the tens digit of 6 ^ 6 = . .... | a ) 1 , b ) 3 , c ) 5 , d ) 7 , e ) 9 | d | floor(divide(reminder(power(6, reminder(20, add(const_4, const_1))), const_100), const_10)) | add(const_1,const_4)|reminder(n1,#0)|power(n0,#1)|reminder(#2,const_100)|divide(#3,const_10)|floor(#4)| | general |
a 25 cm wide path is to be made around a circular garden having a diameter of 4 meters . approximate area of the path is square meters is | area of the path = area of the outer circle - area of the inner circle = ∏ { 4 / 2 + 25 / 100 } 2 - ∏ [ 4 / 2 ] 2 = ∏ [ 2.252 - 22 ] = ∏ ( 0.25 ) ( 4.25 ) { ( a 2 - b 2 = ( a - b ) ( a + b ) } = ( 3.14 ) ( 1 / 4 ) ( 17 / 4 ) = 53.38 / 16 = 3.34 sq m . answer : a | ['a ) 3.34', 'b ) 3.36', 'c ) 3.34', 'd ) 6.32', 'e ) 3.31'] | a | subtract(circle_area(add(divide(4, const_2), divide(25, const_100))), circle_area(divide(4, const_2))) | divide(n1,const_2)|divide(n0,const_100)|add(#0,#1)|circle_area(#0)|circle_area(#2)|subtract(#4,#3) | geometry |
x and y invested in a business . they earned some profit which they divided in the ratio of 1 : 3 . if x invested rs . 10,000 . the amount invested by y is | "solution suppose y invested rs . y then , 10000 / y = 1 / 3 â € ¹ = â € º y = ( 10000 ã — 3 / 1 ) . â € ¹ = â € º y = 30000 . answer a" | a ) rs . 30,000 , b ) rs . 50,000 , c ) rs . 60,000 , d ) rs . 80,000 , e ) none | a | divide(multiply(multiply(add(const_1, const_4), const_1000), 1), 3) | add(const_1,const_4)|multiply(#0,const_1000)|multiply(n0,#1)|divide(#2,n1)| | gain |
in a certain math department , students are required to enroll in either calculus or trigonometry , each of which is offered in beginner and advanced courses . the number of students enrolled in trigonometry is 50 % greater than the number of students enrolled in calculus , and 80 % of calculus students are enrolled in... | let x be the number of students in calculus . then the number of students in trigonometry is 1.5 x the number of students on beginner calculus is 0.8 x the number of students in beginner trigonometry is 4 / 5 * ( 2.5 x ) - 0.8 x = 1.2 x the percentage in beginner trigonometry is 1.2 x / 2.5 x = 48 % the answer is c . | a ) 40 % , b ) 44 % , c ) 48 % , d ) 52 % , e ) 56 % | c | multiply(divide(subtract(multiply(add(add(multiply(50, const_2), 50), multiply(50, const_2)), divide(4, 5)), 80), add(add(multiply(50, const_2), 50), multiply(50, const_2))), multiply(50, const_2)) | divide(n2,n3)|multiply(n0,const_2)|add(n0,#1)|add(#2,#1)|multiply(#3,#0)|subtract(#4,n1)|divide(#5,#3)|multiply(#6,#1) | gain |
a car going at 10 miles per hour set out on an 80 - mile trip at 9 : 00 a . m . exactly 10 minutes later , a second car left from the same place and followed the same route . how fast , in miles per hour , was the second car going if it caught up with the first car at 10 : 30 a . m . ? | "let car a = car that starts at 9 am car b = car that starts at 9 : 10 am time for which car a travels at speed of 10 m per hour = 1.5 hours distance travelled by car a = 10 * 1.5 = 15 miles since car b catches up car a at 10 : 30 , time = 80 mins = 4 / 3 hour speed of car b = 15 / ( 4 / 3 ) = 20 miles per hour answer ... | a ) 45 , b ) 20 , c ) 53 , d ) 55 , e ) 60 | b | divide(80, divide(add(multiply(subtract(10, 9), const_60), subtract(10, 10)), const_60)) | subtract(n4,n2)|subtract(n0,n4)|multiply(#0,const_60)|add(#2,#1)|divide(#3,const_60)|divide(n1,#4)| | physics |
0.009 / x = 0.1 . find the value of x | x = 0.009 / 0.1 = 0.09 answer : b | a ) 0.0009 , b ) 0.09 , c ) 0.9 , d ) 9 , e ) 90 | b | divide(0.009, 0.1) | divide(n0,n1) | general |
the decimal 0.1 is how many times greater than the decimal ( 0.01 ) ^ 4 ? | "0.1 = 10 ^ - 1 ( 0.01 ) ^ 4 = ( 10 ^ - 2 ) ^ 4 = 10 ^ - 8 10 ^ 7 * 10 ^ - 8 = 10 ^ - 1 the answer is c ." | a ) 10 ^ 5 , b ) 10 ^ 6 , c ) 10 ^ 7 , d ) 10 ^ 8 , e ) 10 ^ 9 | c | divide(0.1, power(0.01, 4)) | power(n1,n2)|divide(n0,#0)| | general |
in a group of 70 people , 60 like swimming , 50 like wrestling . how many like both swimming and wrestling ? | "make a venn diagram , and enter your data . let the number of people who like both swimming and wrestling be x 60 - x + x + 50 - x = 70 x = 40 so number who like both swimming and wrestling = 40 answer c" | a ) 30 , b ) 35 , c ) 40 , d ) 45 , e ) 50 | c | add(add(50, subtract(60, 50)), 50) | subtract(n1,n2)|add(n2,#0)|add(n2,#1)| | other |
in the coordinate plane a slope of the line k is 3 times the y - intercept of the line k . what is the x - intercept of the line k ? | as y = 3 mx + m , from 0 = 3 mx + m we get x = - 1 / 3 . hence , the correct answer choice is d . | a ) - 4 , b ) 4 , c ) - 1 / 4 , d ) - 1 / 3 , e ) 2 | d | divide(const_1, 3) | divide(const_1,n0) | general |
a number when divided by 128 gives a remainder 34 , what remainder will be obtained by dividing the same number 19 ? | "explanation : 128 + 34 = 162 / 19 = 10 ( remainder ) answer : d" | a ) 7 , b ) 6 , c ) 9 , d ) 10 , e ) 3 | d | subtract(34, multiply(19, const_2)) | multiply(n2,const_2)|subtract(n1,#0)| | general |
in a certain company , a third of the workers do not have a retirement plan . 30 % of the workers who do not have a retirement plan are women , and 40 % of the workers who do have a retirement plan are men . if 120 of the workers of that company are men , how many of the workers are women ? | "set up equation : x = total number of workers 120 = 0,4 * 2 / 3 * x + 0,7 * 1 / 3 * x 120 = 15 / 30 x x = 240 240 - 120 = 120 answer d" | a ) 80 , b ) 95 , c ) 105 , d ) 120 , e ) 210 | d | multiply(divide(120, add(subtract(divide(const_1, const_3), multiply(divide(const_1, const_3), divide(30, const_100))), multiply(subtract(const_1, divide(const_1, const_3)), divide(40, const_100)))), add(multiply(divide(const_1, const_3), divide(30, const_100)), subtract(subtract(const_1, divide(const_1, const_3)), mul... | divide(const_1,const_3)|divide(n0,const_100)|divide(n1,const_100)|multiply(#0,#1)|subtract(const_1,#0)|multiply(#2,#4)|subtract(#0,#3)|add(#5,#6)|subtract(#4,#5)|add(#3,#8)|divide(n2,#7)|multiply(#9,#10)| | gain |
find the value of x : x ² − 3 x + 2 | "( x − 1 ) ( x − 2 ) x = 1 or 2 . c" | a ) - 1 , b ) 1 , c ) 2 , d ) 3 , e ) 4 | c | multiply(3, 2) | multiply(n0,n1)| | general |
aarti can do a piece of work in 3 days . in how many days will she complete four time of work of same type ? | "we have the important relation , more work , more time ( days ) a piece of work can be done in 3 days . four times of work of same type can be done in 3 x 4 = 12 days answer a" | a ) 12 days , b ) 18 days , c ) 21 days , d ) 3 days , e ) 13 days | a | multiply(const_3, 3) | multiply(n0,const_3)| | physics |
find the area of the quadrilateral of one of its diagonals is 40 cm and its off sets 11 cm and 9 cm ? | "1 / 2 * 40 ( 11 + 9 ) = 400 cm 2 answer : d" | a ) 100 cm 2 , b ) 150 cm 2 , c ) 333 cm 2 , d ) 400 cm 2 , e ) 750 cm 2 | d | multiply(multiply(divide(const_1, const_2), add(9, 11)), 40) | add(n1,n2)|divide(const_1,const_2)|multiply(#0,#1)|multiply(n0,#2)| | geometry |
if 3 taps are opened together a tank is filled in 3 hrs . one of the taps alone can fill it in 10 hrs . and another in 15 hrs . in how many hours will the third tap fill it ? | part filled by three taps together in 1 hr = 1 / 3 part filled by one of the tap in 1 hr = 1 / 10 part filled by another tap in 1 hr = 1 / 15 part filled by third tap in 1 hr = 1 / 3 - ( 1 / 0 + 1 / 15 ) = 5 / 30 therefore the third tap can fill the whole tank in : 30 / 5 hrs . = 6 hrs . answer : d | a ) 7 hrs . , b ) 4 hrs . , c ) 5 hrs . , d ) 6 hrs . , e ) 8 hrs . | d | divide(multiply(10, 3), subtract(10, add(divide(multiply(10, 3), 15), 3))) | multiply(n0,n2)|divide(#0,n3)|add(n0,#1)|subtract(n2,#2)|divide(#0,#3) | physics |
a man took a loan from a bank at the rate of 10 % p . a . simple interest . after 3 years he had to pay rs . 5400 interest only for the period . the principal amount borrowed by him was : | "solution principal = rs . ( 100 x 5400 / 10 x 3 ) = rs . 18000 . answer c" | a ) rs . 2000 , b ) rs . 10,000 , c ) rs . 18,000 , d ) rs . 20,000 , e ) none | c | divide(divide(multiply(5400, const_100), multiply(3, 10)), multiply(divide(const_60, const_4), const_100)) | divide(const_60,const_4)|multiply(n2,const_100)|multiply(n0,n1)|divide(#1,#2)|multiply(#0,const_100)|divide(#3,#4)| | gain |
how many different positive integers exist between 10 ^ 9 and 10 ^ 10 , the sum of whose digits is equal to 2 ? | "so , the numbers should be from 1,000 , 000,000 to 10,000 , 000,000 the following two cases are possible for the sum of the digits to be 2 : 1 . two 1 ' s and the rest are 0 ' s : 1,000 , 000,001 1,000 , 000,010 1,000 , 000,100 1,000 , 001,000 1,000 , 010,000 1,000 , 100,000 1,001 , 000,000 1,010 , 000,000 1,100 , 000... | a ) 6 , b ) 7 , c ) 5 , d ) 8 , e ) 10 | e | add(add(const_4, const_3), const_1) | add(const_3,const_4)|add(#0,const_1)| | general |
a pair of articles was bought for $ 50 at a discount of 60 % . what must be the marked price of each of the article ? | "s . p . of each of the article = 50 / 2 = $ 25 let m . p = $ x 40 % of x = 25 x = 25 * . 4 = $ 10 answer is b" | a ) $ 25 , b ) $ 10 , c ) $ 29.65 , d ) $ 35.95 , e ) $ 45.62 | b | divide(multiply(subtract(const_100, 60), divide(50, const_2)), const_100) | divide(n0,const_2)|subtract(const_100,n1)|multiply(#0,#1)|divide(#2,const_100)| | gain |
the average runs of a cricket player of 15 innings was 25 . how many runs must he make in his next innings so as to increase his average of runs by 6 ? | "explanation : average after 16 innings = 31 required number of runs = ( 31 * 16 ) – ( 25 * 15 ) = 496 – 375 = 121 answer a" | a ) 121 , b ) 421 , c ) 143 , d ) 176 , e ) 84 | a | subtract(multiply(add(15, const_1), add(25, 6)), multiply(25, 15)) | add(n0,const_1)|add(n1,n2)|multiply(n0,n1)|multiply(#0,#1)|subtract(#3,#2)| | general |
two airplanes take off from one airfield at noon . one flies due east at 202 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 . ? | "e in two hours : the plane flying east will be 404 miles away from airport . the other plane will be 566 miles away from airport . 566 / 404 = ~ 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 404 miles apart . e" | a ) 166 , b ) 332 , c ) 400 , d ) 483 , e ) 404 | e | sqrt(subtract(power(multiply(283, 2), 2), power(multiply(202, 2), 2))) | multiply(n1,n2)|multiply(n0,n2)|power(#0,n2)|power(#1,n2)|subtract(#2,#3)|sqrt(#4)| | physics |
( x ) + 1315 + 9211 - 1569 = 11901 . calculate the value of x | x + 1315 + 9211 - 1569 = 11901 = x + 1315 + 9211 = 11901 + 1569 = x + 10526 = 13470 = x = 98329 - 10526 = 87803 answer is e | a ) 87801 , b ) 87811 , c ) 87862 , d ) 11803 , e ) 87803 | e | multiply(subtract(subtract(add(11901, 1569), 9211), 1315), divide(const_60, const_2)) | add(n2,n3)|divide(const_60,const_2)|subtract(#0,n1)|subtract(#2,n0)|multiply(#1,#3) | general |
find the product of the local value and absolute value of 8 in 564823 | "explanation : place value = local value face value = absolute value the place value of 8 in 564823 is 8 x 100 = 800 the face value of 8 in 564823 is nothing but 8 . = > 800 * 8 = 6400 answer : option b" | a ) 1600 , b ) 6400 , c ) 16000 , d ) 12000 , e ) 18000 | b | multiply(multiply(8, const_1000), 8) | multiply(n0,const_1000)|multiply(n0,#0)| | general |
two trains a and b starting from two points and travelling in opposite directions , reach their destinations 9 hours and 4 hours respectively after meeting each other . if the train a travels at 110 kmph , find the rate at which the train b runs . | if two objects a and b start simultaneously from opposite points and , after meeting , reach their destinations in ‘ a ’ and ‘ b ’ hours respectively ( i . e . a takes ‘ a hrs ’ to travel from the meeting point to his destination and b takes ‘ b hrs ’ to travel from the meeting point to his destination ) , then the rat... | a ) 40 , b ) 165 , c ) 120 , d ) 80 , e ) 100 | b | multiply(110, sqrt(divide(9, 4))) | divide(n0,n1)|sqrt(#0)|multiply(n2,#1) | physics |
a man can row 3.9 km / hr in still water . it takes him twice as long to row upstream as to row downstream . what is the rate of the current ? | "speed of boat in still water ( b ) = 3.9 km / hr . speed of boat with stream ( down stream ) , d = b + u speed of boat against stream ( up stream ) , u = b – u it is given upstream time is twice to that of down stream . ⇒ downstream speed is twice to that of upstream . so b + u = 2 ( b – u ) ⇒ u = b / 3 = 1.3 km / hr ... | a ) 1.9 , b ) 1.7 , c ) 1.3 , d ) 1.5 , e ) 1.1 | c | divide(subtract(multiply(3.9, const_2), 3.9), const_3) | multiply(n0,const_2)|subtract(#0,n0)|divide(#1,const_3)| | general |
a sun is divided among x , y and z in such a way that for each rupee x gets , y gets 45 paisa and z gets 50 paisa . if the share of y is rs . 27 , what is the total amount ? | "x : y : z = 100 : 45 : 50 20 : 9 : 10 9 - - - 27 39 - - - ? = > 117 answer : c" | a ) 115 , b ) 116 , c ) 117 , d ) 118 , e ) 119 | c | add(add(multiply(divide(const_100, 45), 27), multiply(divide(50, 45), 27)), 27) | divide(const_100,n0)|divide(n1,n0)|multiply(n2,#0)|multiply(n2,#1)|add(#2,#3)|add(n2,#4)| | general |
if a ( a + 4 ) = 21 and b ( b + 4 ) = 21 , where a ≠ b , then a + b = | "i . e . if a = 3 then b = - 7 or if a = - 7 then b = 3 but in each case a + b = - 7 + 3 = - 4 answer : option a" | a ) − 4 , b ) − 5 , c ) 2 , d ) 46 , e ) 48 | a | add(divide(21, const_10), divide(21, divide(21, const_10))) | divide(n1,const_10)|divide(n1,#0)|add(#0,#1)| | general |
the length of a train and that of a platform are equal . if with a speed of 90 k / hr , the train crosses the platform in one minute , then the length of the train ( in meters ) is : | speed = [ 90 * 5 / 18 ] m / sec = 25 m / sec ; time = 1 min . = 60 sec . let the length of the train and that of the platform be x meters . then , 2 x / 60 = 25 è x = 25 * 60 / 2 = 750 . answer : d | a ) 377 , b ) 267 , c ) 268 , d ) 750 , e ) 271 | d | divide(divide(multiply(90, const_1000), divide(const_60, const_1)), const_2) | divide(const_60,const_1)|multiply(n0,const_1000)|divide(#1,#0)|divide(#2,const_2) | physics |
the average weight of 7 person ' s increases by 6.2 kg when a new person comes in place of one of them weighing 76 kg . what might be the weight of the new person ? | "total weight increased = ( 7 x 6.2 ) kg = 43.4 kg . weight of new person = ( 76 + 43.4 ) kg = 119.4 kg option d" | a ) 160 kg , b ) 175 kg , c ) 180.7 kg , d ) 119.4 kg , e ) 190.8 kg | d | add(multiply(7, 6.2), 76) | multiply(n0,n1)|add(n2,#0)| | general |
the cost of 3 pens and 5 pencils is rs . 100 . also the cost of one pen and one pencil is in the ratio of 5 : 1 respectively . what is the cost of one dozen pens ? | "explanation : let the cost of one pen is ‘ 5 x ’ and pencil is ‘ x ’ 3 x 5 x + 5 x = rs . 100 15 x + 5 x = rs . 100 x = 100 / 20 = 5 : . cost of 1 pen = 5 x = 5 x 5 = 25 : . cost of 12 pens , i . e . ( one dozen ) = 25 x 12 = rs . 300 answer : option c" | a ) rs . 200 , b ) rs . 250 , c ) rs . 300 , d ) rs . 150 , e ) none of these | c | multiply(multiply(3, const_4), divide(100, add(3, 1))) | add(n0,n4)|multiply(n0,const_4)|divide(n2,#0)|multiply(#2,#1)| | other |
the compound interest earned on a sum for the second and the third years are $ 1200 and $ 1272 respectively . what is the rate of interest ? | "1272 - 1200 = 72 is the rate of interest on $ 1200 for one year . the rate of interest = ( 100 * 72 ) / ( 1200 ) = 6 % the answer is b ." | a ) 4 % , b ) 6 % , c ) 8 % , d ) 10 % , e ) 12 % | b | divide(multiply(subtract(1272, 1200), const_100), 1200) | subtract(n1,n0)|multiply(#0,const_100)|divide(#1,n0)| | gain |
10 camels cost as much as 24 horses , 26 horses cost as much as 4 oxen and 6 oxen as much as 4 elephants . if the cost of 10 elephants is rs . 170000 , find the cost of a camel ? | "cost of the camel = p 10 camels = 24 horses 26 horses = 4 oxen 6 oxen = 4 elephants 10 elephants = rs . 170000 p = rs . [ ( 24 * 4 * 4 * 170000 ) / ( 10 * 26 * 6 * 10 ) ] p = rs . ( 65280000 / 15600 ) = > p = rs . 4184.5 answer : b" | a ) rs . 6184.5 , b ) rs . 4184.5 , c ) rs . 5700 , d ) rs . 4284.5 , e ) rs . 5184.5 | b | divide(multiply(multiply(multiply(24, 4), 4), 170000), multiply(multiply(multiply(10, 26), 6), 10)) | multiply(n1,n3)|multiply(n0,n2)|multiply(n3,#0)|multiply(n4,#1)|multiply(n7,#2)|multiply(n0,#3)|divide(#4,#5)| | general |
if the perimeter of a rectangular garden is 680 m , its length when its breadth is 82 m is ? | "2 ( l + 82 ) = 680 = > l = 258 m answer : e" | a ) 286 m , b ) 899 m , c ) 200 m , d ) 166 m , e ) 258 m | e | subtract(divide(680, const_2), 82) | divide(n0,const_2)|subtract(#0,n1)| | physics |
what is the least number . which should be added to 0.0478 to make it a perfect square ? | "0.0478 + 0.0006 = 0.0484 ( 0.22 ) ^ 2 answer : c" | a ) 0.0002 , b ) 0.0004 , c ) 0.0006 , d ) 0.0008 , e ) 0.001 | c | subtract(multiply(divide(add(add(const_12, const_4), const_2), const_100), divide(add(add(const_12, const_4), const_2), const_100)), 0.0478) | add(const_12,const_4)|add(#0,const_2)|divide(#1,const_100)|multiply(#2,#2)|subtract(#3,n0)| | general |
the average score of a cricketer for 7 matches is 56 runs . if the average for first 4 matches is 46 , then average for last 3 matches is | explanation : = ( 56 × 7 ) − ( 46 × 4 ) / 3 = ( 392 − 184 ) / 3 = 69 answer : option e | a ) 33.25 , b ) 32.25 , c ) 65 , d ) 67 , e ) 69 | e | divide(subtract(multiply(7, 56), multiply(4, 46)), 3) | multiply(n0,n1)|multiply(n2,n3)|subtract(#0,#1)|divide(#2,n4) | general |
if in a cricket one day match , there were no wide balls , no wides , no extras and no overthrows . what is the maximum number of runs that a batsman can score in an ideal case scenario ? | b 1653 for an ideal case , the batsman will hit a six on each ball . but if he hits six on the last ball of the over , the strike will change in the next over . thus , the best he can do in the last ball is run 3 runs so that he retains the strike even in the next over . thus the total runs that he can score in each ov... | a ) 2663 , b ) 1653 , c ) 3693 , d ) 4253 , e ) 2653 | b | add(multiply(multiply(const_3, add(const_10, const_1)), subtract(multiply(add(const_4, const_1), const_10), const_1)), add(multiply(const_3, add(const_10, const_1)), const_3)) | add(const_1,const_10)|add(const_1,const_4)|multiply(#0,const_3)|multiply(#1,const_10)|add(#2,const_3)|subtract(#3,const_1)|multiply(#2,#5)|add(#4,#6) | physics |
how many multiples of 10 are there between 11 and 1001 ? | "10 * 2 = 20 10 * 100 = 1000 total multiples = ( 100 - 2 ) + 1 = 99 answer b" | a ) 98 , b ) 99 , c ) 100 , d ) 97 , e ) 95 | b | add(divide(subtract(1001, 11), 10), const_1) | subtract(n2,n1)|divide(#0,n0)|add(#1,const_1)| | general |
a 20 litre mixture of milk and water contains milk and water in the ratio 3 : 2 . 10 litres of the mixture is removed and replaced with pure milk and the operation is repeated once more . at the end of the two removal and replacement , what is the ratio of milk and water in the resultant mixture ? | "rom the beginning ; initially milk to water is 12 l : 8 l remove half 6 - 4 add 10 l milk 16 - 4 remove half 8 - 2 add 10 l milk 18 - 2 18 : 2 = 9 : 1 answer : b" | a ) 17 : 3 , b ) 9 : 1 , c ) 3 : 17 , d ) 5 : 3 , e ) 11 : 2 | b | divide(add(multiply(divide(add(multiply(divide(3, add(3, 2)), subtract(20, 10)), 10), 20), subtract(20, 10)), 10), multiply(divide(multiply(divide(2, add(3, 2)), subtract(20, 10)), 20), subtract(20, 10))) | add(n1,n2)|subtract(n0,n3)|divide(n1,#0)|divide(n2,#0)|multiply(#2,#1)|multiply(#3,#1)|add(n3,#4)|divide(#5,n0)|divide(#6,n0)|multiply(#7,#1)|multiply(#8,#1)|add(n3,#10)|divide(#11,#9)| | general |
6 workers should finish a job in 8 days . after 3 days came 4 workers join them . how many days x do they need to finish the same job ? | "let rate of one worker be r = > ( 6 * r ) * 8 = 1 ( rate * time = work ) = > r = 1 / 48 = > work remaining after 3 days 1 - ( 3 * 6 ) / 48 = 30 / 48 after 4 ppl joined in ( ( 6 + 4 ) * time ) / 48 = 30 / 48 time x = 3 days to finish the task imo a" | a ) 3 , b ) 4 , c ) 5 , d ) 6 , e ) 7 | a | divide(subtract(multiply(6, 8), multiply(3, 6)), add(6, 4)) | add(n0,n3)|multiply(n0,n1)|multiply(n0,n2)|subtract(#1,#2)|divide(#3,#0)| | physics |
if an investor puts $ 600 in a savings account that earns 10 percent annual interest compounded semiannually , how much money will be in the account after one year ? | "1.05 * 1.05 * 600 = $ 661.50 the answer is a ." | a ) $ 661.50 , b ) $ 663.50 , c ) $ 665.50 , d ) $ 667.50 , e ) $ 669.50 | a | multiply(600, power(add(const_1, divide(divide(10, const_2), const_100)), const_2)) | divide(n1,const_2)|divide(#0,const_100)|add(#1,const_1)|power(#2,const_2)|multiply(n0,#3)| | gain |
a “ sophie germain ” prime is any positive prime number p for which 2 p + 1 is also prime . the product of all the possible units digits of sophie germain primes greater than 6 is | in that case , the sophie prime numbers greater than 6 are 7 , 11,23 , 47,59 , . . which yields units digit as 1 , 3,7 and 9 product would be 1 x 3 x 7 x 9 = 189 answer should be d | a ) 3 , b ) 7 , c ) 21 , d ) 189 , e ) 198 | d | multiply(multiply(multiply(subtract(add(add(subtract(6, 1), add(2, 1)), add(2, 1)), const_10), subtract(add(multiply(2, add(subtract(6, 1), add(2, 1))), 1), const_10)), add(2, 1)), subtract(add(multiply(2, add(add(add(subtract(6, 1), add(2, 1)), add(2, 1)), add(2, 1))), 1), multiply(2, const_10))) | add(n0,n1)|multiply(n0,const_10)|subtract(n2,n1)|add(#0,#2)|add(#3,#0)|multiply(n0,#3)|add(n1,#5)|add(#4,#0)|subtract(#4,const_10)|multiply(n0,#7)|subtract(#6,const_10)|add(n1,#9)|multiply(#8,#10)|multiply(#0,#12)|subtract(#11,#1)|multiply(#13,#14) | general |
the difference of two numbers is 1335 . on dividing the larger number by the smaller , we get 6 as quotient and the 15 as remainder . what is the smaller number ? | "let the smaller number be x . then larger number = ( x + 1335 ) . x + 1335 = 6 x + 15 5 x = 1320 x = 264 smaller number = 264 . answer e" | a ) 240 , b ) 270 , c ) 295 , d ) 360 , e ) 264 | e | divide(add(1335, 15), subtract(6, const_1)) | add(n0,n2)|subtract(n1,const_1)|divide(#0,#1)| | general |
dawson has to secure 30 % marks to clear his exam of class 8 th . he got 30 marks and failed by 36 marks . what is the maximum marks ? | e 220 to pass the exam ravish needs 30 + 36 = 66 marks . = > ( 66 / 30 ) * 100 = 220 | a ) 180 , b ) 190 , c ) 200 , d ) 210 , e ) 220 | e | divide(multiply(add(30, 36), const_100), 30) | add(n0,n3)|multiply(#0,const_100)|divide(#1,n0) | gain |
if the price of a book is first decreased by 15 % and then increased by 10 % , then the net change in the price will be : | "explanation : let the original price be rs . 100 . decreased by 15 % = 85 then increased 10 % on rs 85 = 85 + 8.5 = 93.5 net change in price = 100 - 93.5 = 6.5 answer : c" | a ) 10 , b ) 388 , c ) 6.5 , d ) 29 , e ) 22 | c | subtract(const_100, multiply(multiply(divide(subtract(const_100, divide(multiply(const_100, 15), const_100)), const_100), divide(add(const_100, divide(multiply(const_100, 10), const_100)), const_100)), const_100)) | multiply(n0,const_100)|multiply(n1,const_100)|divide(#0,const_100)|divide(#1,const_100)|add(#3,const_100)|subtract(const_100,#2)|divide(#5,const_100)|divide(#4,const_100)|multiply(#6,#7)|multiply(#8,const_100)|subtract(const_100,#9)| | general |
a warehouse is labeling computer inventory with 5 - digit codes . each code is to consist of 4 digits between the numbers 0 and 9 . the manager is debating whether to allow any 5 - digit codes to have leading zero ' s before a non - zero number is used . if not allowed , then numbers like 0025 cannot be used . the mana... | the number of codes possible if leading zero is allowed = 10 * 10 * 10 * 10 * 10 = 10 ^ 5 the number of codes possible if leading zero is not allowed = 9 * 10 * 10 * 10 * 10 = 9 * 10 ^ 4 the difference = 10 ^ 5 - 9 * 10 ^ 4 = 10 ^ 4 ( 10 - 9 ) = 10 ^ 4 . answer : d . similar questions to practice : hope it helps . | a ) 981 , b ) 1000 , c ) 1011 , d ) 10000 , e ) 1110 | d | subtract(power(const_10, 5), multiply(9, power(const_10, 4))) | power(const_10,n0)|power(const_10,n1)|multiply(n3,#1)|subtract(#0,#2) | general |
tim and é lan are 90 miles away from one another . they are starting to move towards each other simultaneously , tim at a speed of 10 mph and é lan at a speed of 5 mph . if every hour they double their speeds , what is the distance that tim will pass until he meets é lan ? | "in 1 st hour : relative speed of the two will be 10 + 5 = 15 miles / hour . as they move towards each other , the distance between them reduces by 15 miles . distance left between them : 90 - 15 = 75 miles . in 2 nd hour : speed of tim : 20 mph , that of elan : 10 mph . relative speed of the two will be 20 + 10 = 30 m... | a ) 30 miles . , b ) 35 miles . , c ) 45 miles . , d ) 60 miles . , e ) 65 miles . | d | multiply(divide(10, add(5, 10)), 90) | add(n1,n2)|divide(n1,#0)|multiply(n0,#1)| | physics |
every year an amount increases by 1 / 9 th of itself . how much will it be after two years if its present value is rs . 64000 ? | "64000 * 10 / 9 * 10 / 9 = 79012.3 answer : d" | a ) 81000 , b ) 81007 , c ) 81008 , d ) 79012.3 , e ) 81022 | d | add(add(64000, multiply(divide(1, 9), 64000)), multiply(divide(1, 9), add(64000, multiply(divide(1, 9), 64000)))) | divide(n0,n1)|multiply(n2,#0)|add(n2,#1)|multiply(#2,#0)|add(#2,#3)| | general |
if c is 20 % of a and 10 % of b , what percent of a is b ? | answer = e 20 a / 100 = 10 b / 100 b = 20 a / 10 = 200 a / 100 = 200 % | a ) 2.5 % , b ) 15 % , c ) 25 % , d ) 35 % , e ) 200 % | e | multiply(divide(divide(20, const_100), divide(10, const_100)), const_100) | divide(n0,const_100)|divide(n1,const_100)|divide(#0,#1)|multiply(#2,const_100) | gain |
an amount of money is to be divided between p , q and r in the ratio of 3 : 7 : 12 . if the difference between the shares of p and q is rs . 3600 , what will be the difference between q and r ' s share ? | "4 - - - 3600 5 - - - ? = > 4500 answer : b" | a ) 2788 , b ) 4500 , c ) 7282 , d ) 2782 , e ) 2729 | b | multiply(subtract(12, 7), divide(3600, subtract(7, 3))) | subtract(n1,n0)|subtract(n2,n1)|divide(n3,#0)|multiply(#2,#1)| | general |
the area of a triangle will be when a = 1 m , b = 2 m , c = 5 m , a , b , c being lengths of respective sides ? | s = ( 1 + 2 + 5 ) / 2 = 4 answer : c | a ) 3 , b ) 6 , c ) 4 , d ) 9 , e ) 1 | c | divide(add(add(1, 2), 5), 2) | add(n0,n1)|add(n2,#0)|divide(#1,n1)| | geometry |
a and b can do a work in 25 days and 10 days respectively . a starts the work and b joins him after 2 days . in how many days can they complete the remaining work ? | "work done by a in 2 days = 2 / 5 remaining work = 3 / 5 work done by both a and b in one day = 1 / 25 + 1 / 10 = 7 / 50 remaining work = 3 / 5 * 50 / 7 = 4.28 days . answer : a" | a ) 4.28 days , b ) 4.38 days , c ) 5.28 days , d ) 6.28 days , e ) 2.28 days | a | subtract(add(inverse(add(inverse(10), inverse(25))), 25), const_3) | inverse(n1)|inverse(n0)|add(#0,#1)|inverse(#2)|add(n0,#3)|subtract(#4,const_3)| | physics |
a grocery store priced a gallon of milk at 25 % above the wholesale price of $ 4 . if a customer comes in with a coupon for 5 % off milk , how much does that customer pay ? | wholesale cost of milk = $ 4 milk was priced at 25 % above $ 4 = $ 5 % discount given by coupon = 5 % customer paid = . 95 * 5 = $ 4.75 answer b | a ) $ 4.95 , b ) $ 4.75 , c ) $ 5.05 , d ) $ 4.50 , e ) $ 5.25 | b | subtract(5, multiply(divide(multiply(add(divide(25, const_100), const_1), 4), const_100), 5)) | divide(n0,const_100)|add(#0,const_1)|multiply(n1,#1)|divide(#2,const_100)|multiply(n2,#3)|subtract(n2,#4) | gain |
a number is doubled and 9 is added . if resultant is trebled , it becomes 81 . what is that number | "explanation : = > 3 ( 2 x + 9 ) = 81 = > 2 x + 9 = 27 = > x = 9 answer : option a" | a ) 9 , b ) 10 , c ) 12 , d ) 14 , e ) 15 | a | divide(subtract(81, multiply(const_3, 9)), multiply(const_3, const_2)) | multiply(n0,const_3)|multiply(const_2,const_3)|subtract(n1,#0)|divide(#2,#1)| | general |
what is the average of odd numbers from 1 to 79 ? | "sum of n odd natural number is given by , sum of n odd natural numbers = n * n average of n odd natural numbers is given by , = ( n * n ) / n = n . here , n = number of terms = ( 79 + 1 ) / 2 = 40 . so , required average = 40 . answer : d" | a ) 30 , b ) 25 , c ) 35 , d ) 40 , e ) 45 | d | multiply(divide(divide(multiply(1, add(1, const_1)), const_2), 1), 79) | add(n0,const_1)|multiply(n0,#0)|divide(#1,const_2)|divide(#2,n0)|multiply(n1,#3)| | general |
10 camels cost as much as 24 horses , 16 horses cost as much as 4 oxen and 6 oxen as much as 4 elephants . if the cost of 10 elephants is rs . 120000 , find the cost of a camel ? | "cost of the camel = p 10 camels = 24 horses 16 horses = 4 oxen 6 oxen = 4 elephants 10 elephants = rs . 120000 p = rs . [ ( 24 * 4 * 4 * 120000 ) / ( 10 * 16 * 6 * 10 ) ] p = rs . ( 46080000 / 9600 ) = > p = rs . 4800 answer : d" | a ) s . 9800 , b ) s . 3800 , c ) s . 9800 , d ) s . 4800 , e ) s . 6880 | d | divide(multiply(multiply(multiply(24, 4), 4), 120000), multiply(multiply(multiply(10, 16), 6), 10)) | multiply(n1,n3)|multiply(n0,n2)|multiply(n3,#0)|multiply(n4,#1)|multiply(n7,#2)|multiply(n0,#3)|divide(#4,#5)| | general |
in an exam ashley secured 332 marks . if she secured 83 % makes , find the maximum marks | let the maximum marks be m . ashley ’ s marks = 83 % of m ashley secured 332 marks therefore , 83 % of m = 332 ⇒ 83 / 100 × m = 332 ⇒ m = ( 332 × 100 ) / 83 ⇒ m = 33200 / 83 ⇒ m = 400 therefore , ashley got 332 marks out of 400 marks . answer : d | a ) 133 , b ) 376 , c ) 377 , d ) 400 , e ) 377 | d | divide(332, divide(83, const_100)) | divide(n1,const_100)|divide(n0,#0) | gain |
a man is 24 years older than his son . in two years , his age will be twice the age of his son . the present age of his son is | "explanation : let the son ' s present age be x years . then , man ' s present age = ( x + 24 ) years = > ( x + 24 ) + 2 = 2 ( x + 2 ) = > x + 26 = 2 x + 4 so , x = 22 option c" | a ) 20 years , b ) 21 years , c ) 22 years , d ) 24 years , e ) none of these | c | divide(subtract(24, subtract(multiply(const_2, const_2), const_2)), subtract(const_2, const_1)) | multiply(const_2,const_2)|subtract(const_2,const_1)|subtract(#0,const_2)|subtract(n0,#2)|divide(#3,#1)| | general |
a number when divided by 100 leaves 11 as a remainder . what will be the remainder if the number is divided by 11 ? | "let 100 + 11 = 111 111 divided evenly by eleven leaves a remainder of one . correct answer is b ." | a ) 11 , b ) 1 , c ) 2 , d ) 3 , e ) 21 | b | reminder(add(100, 11), 11) | add(n0,n1)|reminder(#0,n2)| | general |
if 3 x = 8 y = z , what is x + y , in terms of z ? | "3 x = 8 y = z x = z / 3 and y = z / 8 x + y = z / 3 + z / 8 = 11 z / 24 answer is c" | a ) z / 2 , b ) 2 z , c ) 11 z / 24 , d ) 3 z / 5 , e ) z / 9 | c | divide(subtract(divide(multiply(3, const_100), const_2), const_2), add(divide(multiply(3, const_100), const_2), const_2)) | multiply(n0,const_100)|divide(#0,const_2)|add(#1,const_2)|subtract(#1,const_2)|divide(#3,#2)| | general |
a train requires 10 seconds to pass a pole while it requires 35 seconds to cross a stationary train which is 500 mtrs long . find the speed of the train . | "in 10 s the train crosses the pole and in 35 sec the train crosses one more stationary train in 25 sec the train travels a distance of 500 mtrs speed = 500 / 25 = 20 m / s = 20 ( 3600 / 1000 ) = 20 * 18 / 5 = 72 kmph answer : c" | a ) 68 kmph , b ) 70 kmph , c ) 72 kmph , d ) 60 kmph , e ) 50 kmph | c | multiply(divide(500, subtract(35, 10)), const_3_6) | subtract(n1,n0)|divide(n2,#0)|multiply(#1,const_3_6)| | physics |
the average salary of all the workers in a workshop is rs . 8000 . the average salary of 7 technicians is rs . 12000 and the average salary of the rest is rs . 6000 . the total number of workers in the workshop is | "solution let the toatl number of workers be x . then 8000 x = ( 12000 x 7 ) + 6000 ( x - 7 ) 2000 x = 42000 x = 21 . answer b" | a ) 20 , b ) 21 , c ) 22 , d ) 23 , e ) 24 | b | add(7, divide(multiply(7, subtract(12000, 8000)), subtract(8000, 6000))) | subtract(n2,n0)|subtract(n0,n3)|multiply(n1,#0)|divide(#2,#1)|add(n1,#3)| | general |
the radius of a wheel is 22.6 cm . what is the distance covered by the wheel in making 750 resolutions ? | "in one resolution , the distance covered by the wheel is its own circumference . distance covered in 750 resolutions . = 750 * 2 * 22 / 7 * 22.6 = 105504 cm = 1055.04 m answer : c" | a ) 754 m , b ) 704 m , c ) 1055.04 m , d ) 1058.04 m , e ) 204 m | c | divide(multiply(multiply(multiply(divide(add(multiply(add(const_3, const_4), const_3), const_1), add(const_3, const_4)), 22.6), const_2), 750), const_100) | add(const_3,const_4)|multiply(#0,const_3)|add(#1,const_1)|divide(#2,#0)|multiply(n0,#3)|multiply(#4,const_2)|multiply(n1,#5)|divide(#6,const_100)| | physics |
car z travels 45 miles per gallon of gasoline when driven at a constant rate of 45 miles per hour , but travels 20 percent fewer miles per gallon of gasoline when driven at a constant rate of 60 miles per hour . how many miles does car z travel on 10 gallons of gasoline when driven at a constant rate of 60 miles per ho... | "the question stem asks us for the distance possible with 10 gallons of fuel at a constant speed of 60 miles per hour . we therefore first calculate the fuel efficiency at that speed . the stem tells us that at 45 miles / hour , the car will run 45 miles / gallon and at 60 miles / hour , that distance decreases by 20 %... | a ) 320 , b ) 360 , c ) 400 , d ) 408.3 , e ) 440 | b | multiply(multiply(subtract(const_1, divide(20, const_100)), 45), 10) | divide(n2,const_100)|subtract(const_1,#0)|multiply(n0,#1)|multiply(n4,#2)| | gain |
evaluate : log 3 27 | "solution log 3 27 = n = 3 n = 27 = 33 n ‹ = › 3 . answer a" | a ) 3 , b ) 4 , c ) 5 , d ) 6 , e ) 7 | a | divide(3, divide(27, 3)) | divide(n1,n0)|divide(n0,#0)| | other |
how long does a train 110 m long running at the speed of 72 km / hr takes to cross a bridge 140 m length ? | "speed = 72 * 5 / 18 = 20 m / sec total distance covered = 110 + 140 = 250 m . required time = 250 / 20 = 12.5 sec . answer : b" | a ) 82.1 sec . , b ) 12.5 sec , c ) 19.1 sec . , d ) 17.1 sec . , e ) 42.1 sec . | b | divide(add(110, 140), multiply(72, const_0_2778)) | add(n0,n2)|multiply(n1,const_0_2778)|divide(#0,#1)| | physics |
the average of 12 result is 55 . average of the first 6 of them is 42 and that of the last 6 is 48 . find the 8 th result ? | "sum of all the 12 results = 12 * 55 = 660 sum of the first 6 of them = 6 * 42 = 252 sum of the last 6 of them = 6 * 48 = 288 so , the 8 th number = 600 + 252 - 288 = 564 . b" | a ) 35 , b ) 564 , c ) 526 , d ) 48 , e ) 50 | b | subtract(add(multiply(6, 42), multiply(6, 48)), multiply(12, 55)) | multiply(n2,n3)|multiply(n2,n5)|multiply(n0,n1)|add(#0,#1)|subtract(#3,#2)| | general |
15 % of 75 is the same as vx % of 450 . find x . | 15 % of 75 = 11.25 450 * x / 100 = 11.25 x = 2.5 answer : b | a ) 1 , b ) 2.5 , c ) 3 , d ) 4 , e ) 5 | b | divide(multiply(15, 75), 450) | multiply(n0,n1)|divide(#0,n2) | gain |
a fill pipe can fill 1 / 2 of cistern in 25 minutes . in how many minutes , it can fill 1 / 2 of the cistern ? | "required time = 25 * 2 * 1 / 2 = 25 minutes answer is d" | a ) 10 min , b ) 15 min , c ) 20 min , d ) 25 min , e ) 30 min | d | divide(25, 1) | divide(n2,n0)| | physics |
a straight line in the xy - plane has a slope of 2 and a y - intercept of 2 . on this line , what is the x - coordinate of the point whose y - coordinate is 520 ? | "slope of 2 and a y - intercept of 2 y - coordinate is 520 y = 2 x + 2 518 = 2 x x = 259 answer : a . 259" | a ) 259 , b ) 498 , c ) 676 , d ) 823 , e ) 1,002 | a | divide(subtract(520, 2), 2) | subtract(n2,n0)|divide(#0,n0)| | general |
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