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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there are 810 male and female participants in a meeting . half the female participants and one - quarterof the male participants are democrats . one - third of all the participants are democrats . how many of the democrats are female ? | "let m be the number of male participants and f be the number of female articipants in the meeting . thetotal number of participants is given as 810 . hence , we have m + f = 810 now , we have that half the female participants and one - quarter of the male participants are democrats . let d equal the number of the demo... | a ) 75 , b ) 100 , c ) 125 , d ) 135 , e ) 225 | d | divide(subtract(multiply(divide(810, const_3), const_4), 810), const_2) | divide(n0,const_3)|multiply(#0,const_4)|subtract(#1,n0)|divide(#2,const_2)| | general |
a square is drawn by joining the mid points of the sides of a given square in the same way and this process continues indefinitely . if a side of the first square is 4 cm , determine the sum of the areas all the square . | solution : side of the first square is 4 cm . side of second square = 2 √ 2 cm . side of third square = 2 cm . and so on . i . e . 4 , 2 √ , 2 , √ 2 , 1 . . . . . . thus , area of these square will be , = 16 , 8 , 4 , 2 , 1 , 1 / 2 . . . . . . . hence , sum of the area of first , second , third square . . . . . . = 16 ... | ['a ) 32 cm 2', 'b ) 16 cm 2', 'c ) 20 cm 2', 'd ) 64 cm 2', 'e ) none of these'] | a | divide(power(4, const_2), divide(const_1, const_2)) | divide(const_1,const_2)|power(n0,const_2)|divide(#1,#0) | geometry |
find the greatest number of 5 digits which is exactly divisible by 12 , 15 and 18 ? | the largest five digit numbers are 13050 , 12960,12080 13050 is not divisible by 12 12960 is divisible by 18 , 12 and 15 answer : b | a ) a ) 13050 , b ) b ) 12960 , c ) c ) 10025 , d ) d ) 11080 , e ) e ) 12080 | b | multiply(const_4, multiply(multiply(12, 15), 18)) | multiply(n1,n2)|multiply(n3,#0)|multiply(#1,const_4) | general |
kim has 4 pairs of shoes ; each pair is a different color . if kim randomly selects 2 shoes without replacement from the 8 shoes , what is the probability that she will select 2 shoes of the same color ? | "total pairs = 8 c 2 = 28 ; same color pairs = 4 c 1 * 1 c 1 = 4 ; prob = 1 / 7 ans a" | a ) 1 / 7 , b ) 1 / 8 , c ) 1 / 9 , d ) 1 / 10 , e ) 1 / 25 | a | divide(4, choose(8, 2)) | choose(n2,n1)|divide(n0,#0)| | probability |
the number 110 can be written as the sum of the squares of 3 different positive integers . what is the sum of these 3 integers ? | "7 ^ 2 + 5 ^ 2 + 6 ^ 2 = 49 + 25 + 36 = 110 7 + 5 + 6 = 18 hence answer is a" | a ) 18 , b ) 16 , c ) 15 , d ) 14 , e ) 13 | a | add(add(add(const_4, 3), add(3, const_2)), 3) | add(n1,const_4)|add(const_2,n1)|add(#0,#1)|add(n1,#2)| | geometry |
in an examination , 20 % of total students failed in hindi , 70 % failed in english and 10 % in both . the percentage of these who passed in both the subjects is : | "pass percentage = 100 - ( 20 + 70 - 10 ) = 100 - 80 = 20 answer : b" | a ) 10 % , b ) 20 % , c ) 30 % , d ) 40 % , e ) 50 % | b | subtract(const_100, subtract(add(20, 70), 10)) | add(n0,n1)|subtract(#0,n2)|subtract(const_100,#1)| | general |
the tax on a commodity is diminished by 25 % and its consumption increased by 13 % . the effect on revenue is ? | "100 * 100 = 10000 75 * 113 = 8475 - - - - - - - - - - - 10000 - - - - - - - - - - - 1525 100 - - - - - - - - - - - ? = > 15 % decrease answer : d" | a ) 18 , b ) 16 , c ) 10 , d ) 15 , e ) 14 | d | subtract(const_100, multiply(multiply(add(const_1, divide(13, const_100)), subtract(const_1, divide(25, 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 |
the famous denali star train starts from anchorge & travels towards fair banksat speed 50 mph . after some time another train glacier discovery train ( at parallel track ) at fair banks and moves towards anchorge at a speed of 70 mph . both the trains denali star & glacier discovery have a length 1 / 6 miles each . aft... | total distance = ( 1 / 6 ) + ( 1 / 6 ) = 1 / 3 miles relative speed = ( 50 + 70 ) mph = 120 mph time taken = ( 1 / 3 ) / ( 120 ) hours = 10 seconds answer : a | a ) 10 sec , b ) 11 sec , c ) 12 sec , d ) 13 sec , e ) 14 sec | a | multiply(divide(add(divide(1, 6), divide(1, 6)), add(50, 70)), const_3600) | add(n0,n1)|divide(n2,n3)|add(#1,#1)|divide(#2,#0)|multiply(#3,const_3600) | physics |
the area of a square is equal to twice the area of a rectangle of dimensions 32 cm * 64 cm . what is the perimeter of the square ? | area of the square = s * s = 2 ( 32 * 64 ) = > s = 64 cm perimeter of the square = 4 * 64 = 256 cm . answer : option b | ['a ) 236', 'b ) 256', 'c ) 456', 'd ) 656', 'e ) 756'] | b | square_perimeter(64) | square_perimeter(n1) | geometry |
how many positive integers between 1 and 300 are there such that they are multiples of 21 ? | "multiples of 21 = 21 , 42,63 , - - - - - 294 number of multiples of 21 = > 21 * 14 = 294 answer is a" | a ) 14 , b ) 16 , c ) 17 , d ) 13 , e ) 15 | a | divide(subtract(300, 1), 21) | subtract(n1,n0)|divide(#0,n2)| | general |
if p , q , and r are distinct positive digits and the product of the two - digit integers pq and pr is 221 , what is the sum of the digits p , q , and r ? | factor out 221 221 = 13 * 17 thus pq = 13 & pr = 17 or vice versa thus p = 1 , q = 3 & r = 7 sum = 1 + 3 + 7 = 11 answer : b | a ) 5 , b ) 11 , c ) 13 , d ) 21 , e ) 23 | b | add(reminder(divide(221, add(const_1, const_12)), const_10), add(const_1, reminder(add(const_1, const_12), const_10))) | add(const_1,const_12)|divide(n0,#0)|reminder(#0,const_10)|add(#2,const_1)|reminder(#1,const_10)|add(#3,#4) | general |
p and q invested in a business . the profit earned was divided in the ratio 3 : 4 . if p invested rs 50000 , the amount invested by q is | "let the amount invested by q = q 50000 : q = 3 : 4 ⇒ 50000 × 4 = 3 q ⇒ q = ( 50000 × 4 ) / 3 = 66666 answer is b ." | a ) 30000 , b ) 66666 , c ) 40000 , d ) 20000 , e ) 60000 | b | multiply(divide(50000, 3), 4) | divide(n2,n0)|multiply(n1,#0)| | gain |
the end of a blade on an airplane propeller is 20 feet from the center . if the propeller spins at the rate of 1,320 revolutions per second , how many miles will the tip of the blade travel in one minute ? ( 1 mile = 5,280 feet ) | "distance traveled in 1 revolution = 2 π r = 2 π 20 / 5280 revolutions in one second = 1320 revolutions in 60 seconds ( one minute ) = 1320 * 60 total distance traveled = total revolutions * distance traveled in one revolution 1320 * 60 * 2 π 20 / 5280 = 600 π c is the answer" | a ) 200 π , b ) 240 π , c ) 600 π , d ) 480 π , e ) 1,200 π | c | multiply(multiply(multiply(multiply(divide(20, add(multiply(const_2, const_100), multiply(add(const_2, const_3), const_1000))), const_2), divide(add(const_2, multiply(const_2, const_10)), add(const_3, const_4))), 1,320), const_60) | add(const_3,const_4)|add(const_2,const_3)|multiply(const_10,const_2)|multiply(const_100,const_2)|add(#2,const_2)|multiply(#1,const_1000)|add(#3,#5)|divide(#4,#0)|divide(n0,#6)|multiply(#8,const_2)|multiply(#7,#9)|multiply(n1,#10)|multiply(#11,const_60)| | physics |
car a runs at the speed of 65 km / hr & reaches its destination in 8 hr . car b runs at the speed of 70 km / h & reaches its destination in 4 h . what is the respective ratio of distances covered by car a & car b ? | "sol . distance travelled by car a = 65 × 8 = 520 km distance travelled by car b = 70 × 4 = 280 km ratio = 520 / 280 = 13 : 7 c" | a ) 10 : 4 , b ) 10 : 7 , c ) 13 : 7 , d ) 14 : 6 , e ) 13 : 9 | c | divide(multiply(65, 8), multiply(70, 4)) | multiply(n0,n1)|multiply(n2,n3)|divide(#0,#1)| | physics |
two cards are drawn together from a pack of 52 cards . the probability that one is a spade and one is a heart , is : | "n ( s ) = ( 52 x 51 ) / ( 2 x 1 ) = 1326 . let e = event of getting 1 spade and 1 heart . n ( e ) = number of ways of choosing 1 spade out of 13 and 1 heart out of 13 = ( 13 x 13 ) = 169 . p ( e ) = n ( e ) / n ( s ) 169 / 1326 = 13 / 102 answer e" | a ) 2 / 109 , b ) 7 / 109 , c ) 8 / 223 , d ) 14 / 263 , e ) 13 / 102 | e | multiply(divide(multiply(divide(52, const_4), divide(52, const_4)), multiply(52, 52)), const_2) | divide(n0,const_4)|multiply(n0,n0)|multiply(#0,#0)|divide(#2,#1)|multiply(#3,const_2)| | probability |
how many positive integers less than 50 are there such that they are multiples of 9 ? | "number of multiples of 9 = > 45 - 9 / 9 + 1 = 5 answer is a" | a ) 5 , b ) 6 , c ) 4 , d ) 8 , e ) 3 | a | divide(factorial(subtract(add(const_4, 9), const_1)), multiply(factorial(9), 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 |
pat , kate and mark charged a total of 189 hours to a certain project . if pat charged twice as much time to the project as kate and 1 / 3 as much times as mark , how many more hours did mark charge to the project than kate . | "let kate charge for x hours , then pat charged for 2 x and mat - for 6 x . so , 2 x + 6 x + x = 189 - total hours charged for , x = 21 . mat charged 6 x - x or 5 x for more hours than kate , or for 105 hours . e is correct" | a ) 18 , b ) 36 , c ) 72 , d ) 90 , e ) 105 | e | multiply(divide(189, add(add(1, const_2), multiply(const_2, 3))), subtract(multiply(const_2, 3), 1)) | add(n1,const_2)|multiply(n2,const_2)|add(#0,#1)|subtract(#1,n1)|divide(n0,#2)|multiply(#4,#3)| | general |
a part of certain sum of money is invested at 9 % per annum and the rest at 18 % per annum , if the interest earned in each case for the same period is equal , then ratio of the sums invested is ? | "18 : 9 = 2 : 1 answer : b" | a ) 4 : 2 , b ) 2 : 1 , c ) 4 : 3 , d ) 4 : 0 , e ) 4 : 9 | b | multiply(divide(18, const_100), 9) | divide(n1,const_100)|multiply(n0,#0)| | gain |
find the amount on rs . 5000 in 2 years , the rate of interest being 10 % per first year and 12 % for the second year ? | 5000 * 110 / 100 * 112 / 100 = > 6160 answer : b | a ) 3377 , b ) 6160 , c ) 5460 , d ) 1976 , e ) 1671 | b | divide(multiply(divide(multiply(5000, add(const_100, 10)), const_100), add(const_100, 12)), const_100) | add(n3,const_100)|add(n2,const_100)|multiply(n0,#1)|divide(#2,const_100)|multiply(#0,#3)|divide(#4,const_100) | gain |
ray writes a two digit number . he sees that the number exceeds 4 times the sum of its digits by 3 . if the number is increased by 18 , the result is the same as the number formed by reversing the digits . find the sum of the digits of the number . | "let the two digit number be xy . 4 ( x + y ) + 3 = 10 x + y - - - ( 1 ) 10 x + y + 18 = 10 y + x - - - ( 2 ) solving 1 st equation , 4 x + 4 y + 3 = 10 x + y 3 y + 3 = 6 x 6 x – 3 y = 3 therefore , 2 x – y = 1 - - - ( 3 ) solving 2 nd equation , xy + 18 = yx ⇒ ⇒ ( 10 x + b ) + 18 = 10 y + x ⇒ ⇒ 18 = 9 y – 9 x ⇒ ⇒ 2 = ... | a ) 2 , b ) 8 , c ) 9 , d ) 3 , e ) 5 | b | divide(subtract(18, 3), 3) | subtract(n2,n1)|divide(#0,n1)| | general |
1000 men have provisions for 15 days . if 200 more men join them , for how many days will the provisions last now ? | "1000 * 15 = 1200 * x x = 12.5 answer b" | a ) 10.5 , b ) 12.5 , c ) 13.5 , d ) 11.5 , e ) 11 | b | divide(multiply(15, 1000), add(1000, 200)) | add(n0,n2)|multiply(n0,n1)|divide(#1,#0)| | physics |
a circle graph shows how the budget of a certain company was spent : 15 percent for transportation , 9 percent for research and development , 5 percent for utilities , 4 percent for equipment , 2 percent for supplies , and the remainder for salaries . if the area of each sector of the graph is proportional to the perce... | "the percent of the budget for salaries is 100 - ( 15 + 9 + 5 + 4 + 2 ) = 65 % 100 % of the circle is 360 degrees . then ( 65 % / 100 % ) * 360 = 234 degrees the answer is d ." | a ) 216 ° , b ) 222 ° , c ) 228 ° , d ) 234 ° , e ) 240 ° | d | subtract(const_360, divide(multiply(add(add(add(add(15, 9), 5), 4), 2), const_360), const_100)) | add(n0,n1)|add(n2,#0)|add(n3,#1)|add(n4,#2)|multiply(#3,const_360)|divide(#4,const_100)|subtract(const_360,#5)| | geometry |
the area of a rectangular field is equal to 750 square meters . its perimeter is equal to 110 meters . find the width of this rectangle . | l * w = 750 : area , l is the length and w is the width . 2 l + 2 w = 110 : perimeter l = 55 - w : solve for l ( 55 - w ) * w = 750 : substitute in the area equation w = 25 and l = 30 correct answer e | ['a ) 5', 'b ) 10', 'c ) 15', 'd ) 20', 'e ) 25'] | e | divide(subtract(divide(110, const_2), sqrt(subtract(multiply(divide(110, const_2), divide(110, const_2)), multiply(const_4, 750)))), const_2) | divide(n1,const_2)|multiply(n0,const_4)|multiply(#0,#0)|subtract(#2,#1)|sqrt(#3)|subtract(#0,#4)|divide(#5,const_2) | geometry |
a number increased by 10 % gives 660 . the number is ? | "formula = total = 100 % , increase = ` ` + ' ' decrease = ` ` - ' ' a number means = 100 % that same number increased by 10 % = 110 % 110 % - - - - - - - > 660 ( 110 ã — 6 = 660 ) 100 % - - - - - - - > 600 ( 100 ã — 6 = 600 ) option ' d '" | a ) 200 , b ) 300 , c ) 500 , d ) 600 , e ) 400 | d | divide(660, add(const_1, divide(10, const_100))) | divide(n0,const_100)|add(#0,const_1)|divide(n1,#1)| | gain |
working simultaneously and independently at an identical constant rate , 10 machines of a certain type can produce a total of x units of product p in 4 days . how many of these machines , working simultaneously and independently at this constant rate , can produce a total of 3 x units of product p in 6 days ? | "the rate of 10 machines is rate = job / time = x / 4 units per day - - > the rate of 1 machine 1 / 10 * ( x / 4 ) = x / 40 units per day ; now , again as { time } * { combined rate } = { job done } then 6 * ( m * x / 40 ) = 3 x - - > m = 20 . answer : d ." | a ) 14 , b ) 15 , c ) 18 , d ) 20 , e ) 22 | d | multiply(multiply(10, 3), divide(4, 6)) | divide(n1,n3)|multiply(n0,n2)|multiply(#0,#1)| | general |
the average of first five prime numbers greater than 3 is ? | "5 + 7 + 11 + 13 + 17 = 53 / 5 = 10.60 answer = b" | a ) 32.2 , b ) 10.6 , c ) 32.3 , d ) 32.8 , e ) 32.4 | b | add(3, const_1) | add(n0,const_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 ) 299 , b ) 266 , c ) 299 , d ) 750 , e ) 261 | 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 age of students of a class is 15.7 years . the average age of boys in the class is 16.4 years and that of the girls is 15.4 years . the ration of the number of boys to the number of girls in the class is ? | "let the ratio be k : 1 . then , k * 16.4 + 1 * 15.4 = ( k + 1 ) * 15.7 = ( 16.4 - 15.7 ) k = ( 15.7 - 15.4 ) = k = 0.3 / 0.7 = 3 / 7 required ratio = 3 / 7 : 1 = 3 : 7 . answer : a" | a ) 3 : 7 , b ) 7 : 3 , c ) 2 : 5 , d ) 2 : 1 , e ) 2 : 4 | a | divide(subtract(15.7, 15.4), subtract(16.4, 15.7)) | subtract(n0,n2)|subtract(n1,n0)|divide(#0,#1)| | general |
in the rectangular coordinate system , points ( 5 , 0 ) and ( – 5 , 0 ) both lie on circle c . what is the maximum possible value of the radius of c ? | "the answer is b it takes 3 distinct points to define a circle . only 2 are given here . the two points essentially identify a single chord of the circle c . since no other information is provided , however , the radius of the circle can essentially be anything . all this information tell us is that the radius isgreate... | a ) 2 , b ) 4 , c ) 8 , d ) 16 , e ) none of the above | b | sqrt(power(5, const_2)) | power(n0,const_2)|sqrt(#0)| | geometry |
train a leaves the station traveling at 30 miles per hour . two hours later train в leaves the same station traveling in the same direction at 42 miles per hour . how many miles from the station was train a overtaken by train b ? | "after two hours , train a is ahead by 60 miles . train b can catch up at a rate of 12 miles per hour . the time to catch up is 60 / 12 = 5 hours . in 5 hours , train a travels another 30 * 5 = 150 miles for a total of 210 miles . the answer is d ." | a ) 150 , b ) 170 , c ) 190 , d ) 210 , e ) 230 | d | multiply(divide(multiply(30, const_2), subtract(42, 30)), 42) | multiply(n0,const_2)|subtract(n1,n0)|divide(#0,#1)|multiply(n1,#2)| | physics |
what sum of money will produce rs . 70 as simple interest in 5 years at 3 1 / 2 percent ? | "70 = ( p * 5 * 7 / 2 ) / 100 p = 400 answer : a" | a ) 400 , b ) 500 , c ) 367 , d ) 368 , e ) 339 | a | divide(70, divide(multiply(5, add(3, divide(1, 2))), const_100)) | divide(n3,n4)|add(n2,#0)|multiply(n1,#1)|divide(#2,const_100)|divide(n0,#3)| | gain |
what is the hcf of 2 / 3 , 5 / 9 and 6 / 5 | "explanation : hcf of fractions = hcf of numerators / lcm of denominators = ( hcf of 2 , 5 , 6 ) / ( lcm of 3 , 9 , 5 ) = 1 / 45 answer : option a" | a ) 1 / 45 , b ) 2 / 45 , c ) 4 / 15 , d ) 8 / 45 , e ) 9 / 45 | a | divide(2, 5) | divide(n0,n5)| | general |
the h . c . f . of two numbers is 38 and the other two factors of their l . c . m . are 12 and 10 . the larger of the two numbers is : | ": explanation : clearly , the numbers are ( 38 x 12 ) and ( 38 x 10 ) . { \ color { blue } \ therefore } larger number = ( 38 x 12 ) = 456 . answer : b ) 456" | a ) 678 , b ) 456 , c ) 234 , d ) 476 , e ) 432 | b | multiply(38, 10) | multiply(n0,n2)| | other |
robert ate 9 chocolates , nickel ate 2 chocolates . how many more chocolates did robert ate than nickel ? | 9 - 2 = 7 . answer is b | a ) a ) 4 , b ) b ) 7 , c ) c ) 9 , d ) d ) 5 , e ) e ) 2 | b | subtract(9, 2) | subtract(n0,n1) | general |
the area of a circle is increased by 800 % . by what percent has the diameter of the circle increased ? | "a diameter of 2 it ' s radius = 1 it ' s area = ( 1 ^ 2 ) pi = 1 pi answer a : if we increase that diameter 100 % , we have . . . . a diameter of 4 it ' s radius = 2 it ' s area = ( 2 ^ 2 ) pi = 4 pi this area has increased ( 4 pi - 1 pi ) / 1 pi = 3 pi / 1 pi = 3 = 300 % answer b : if we increase the diameter 200 % ,... | a ) 100 % , b ) 200 % , c ) 300 % , d ) 600 % , e ) 800 % | e | multiply(const_100, divide(const_2, const_2)) | divide(const_2,const_2)|multiply(#0,const_100)| | geometry |
3 boys are ages 4 , 6 and 7 respectively . 3 girls are ages 5 , 8 and 9 , respectively . if two of the boys and two of the girls are randomly selected and the sum of the selected children ' s ages is q , what is the difference between the probability that q is even and the probability that q is odd ? | age of boys q : 4 , 6 , 7 sum of ages taken 2 at a time : 10 , 13,11 ages of girls : 5 , 8 , 9 sum of ages taken 2 at a time : 13 , 17,14 9 combinations of sum between sets ( 10 , 12,11 ) ( 13 , 17,14 ) = 23 , 27,24 - 16 , 30,17 - 24 , 28,25 prob ( even ) = 5 / 9 prob ( odd ) = 4 / 9 answer = 5 / 9 - 4 / 9 = 1 / 9 | a ) 1 / 9 , b ) 1 / 6 , c ) 2 / 9 , d ) 1 / 4 , e ) 1 / 2 | a | subtract(divide(5, 9), divide(4, 9)) | divide(n5,n7)|divide(n1,n7)|subtract(#0,#1) | general |
if 4 xz + yw = 4 and xw + yz = 8 , what is the value of the expression ( 2 x + y ) ( 2 z + w ) ? | ( 2 x + y ) * ( 2 z + w ) = 4 + 2 ( 8 ) = 20 answer : b | a ) 22 , b ) 20 , c ) 24 , d ) 26 , e ) 28 | b | add(4, multiply(8, 2)) | multiply(n2,n3)|add(n0,#0) | general |
some of 50 % - intensity red paint is replaced with 25 % solution of red paint such that the new paint intensity is 40 % . what fraction of the original paint was replaced ? | "40 % is 15 % - points above 25 % and 10 % - points below 50 % . thus the ratio of 25 % - solution to 50 % - solution is 2 : 3 . 2 / 5 of the original paint was replaced . the answer is c ." | a ) 1 / 30 , b ) 1 / 5 , c ) 2 / 5 , d ) 3 / 4 , e ) 4 / 5 | c | divide(subtract(divide(40, const_100), divide(50, const_100)), subtract(divide(25, const_100), divide(50, const_100))) | divide(n2,const_100)|divide(n0,const_100)|divide(n1,const_100)|subtract(#0,#1)|subtract(#2,#1)|divide(#3,#4)| | gain |
in 10 years , a will be twice as old 5 as b was 10 years ago . if a is now 7 years older than b , the present age of b is | explanation : let b ' s age = x years . then , as age = ( x + 7 ) years . ( x + 7 + 10 ) = 2 ( x — 10 ) hence x = 37 . present age of b = 37 years answer : option b | a ) 35 , b ) 37 , c ) 39 , d ) 41 , e ) 42 | b | add(multiply(const_2, 10), add(7, 10)) | add(n0,n3)|multiply(n0,const_2)|add(#0,#1) | general |
the distance between 2 cities a and b is 1000 km . a train starts from a at 9 a . m . and travels towards b at 100 km / hr . another starts from b at 10 a . m . and travels towards a at 150 km / hr . at what time do they meet ? | "suppose they meet x hrs after 9 a . m . distance moved by first in x hrs + distance moved by second in ( x - 1 ) hrs = 1000 100 x + 150 ( x - 1 ) = 1000 x = 4.60 = 5 hrs they meet at 9 + 5 = 2 p . m . answer is d" | a ) 11 am . , b ) 12 p . m . , c ) 3 pm . , d ) 2 p . m . , e ) 1 p . m . | d | add(divide(add(2, 10), add(9, 10)), 1000) | add(n0,n4)|add(n2,n4)|divide(#0,#1)|add(n1,#2)| | physics |
sachin is younger than rahul by 14 years . if the ratio of their ages is 7 : 9 , find the age of sachin | "explanation : if rahul age is x , then sachin age is x - 14 , so , 9 x - 126 = 7 x 2 x = 126 x = 63 so sachin age is 63 - 14 = 49 answer : e ) 49" | a ) 48 , b ) 24.8 , c ) 24.21 , d ) 24.88 , e ) 49 | e | multiply(divide(14, subtract(9, 7)), 7) | subtract(n2,n1)|divide(n0,#0)|multiply(n1,#1)| | other |
the number x of cars sold each week varies with the price y in dollars according to the equation x = 800000 – 50 y . what would be the total weekly revenue w , in dollars , from the sale of cars priced at $ 15000 ? | number of cars sold = x = 800000 - 50 y y = 15000 x = 800000 - 750000 = 50000 revenue from 50000 cars = 15000 * 50000 = 750000000 e | a ) 50000 , b ) 750000 , c ) 850000 , d ) 7 , 500000 , e ) w = 75000,000 | e | divide(reminder(multiply(subtract(800000, multiply(50, 15000)), 15000), multiply(const_100, multiply(const_1000, const_1000))), const_100) | multiply(n1,n2)|multiply(const_1000,const_1000)|multiply(#1,const_100)|subtract(n0,#0)|multiply(n2,#3)|reminder(#4,#2)|divide(#5,const_100) | general |
two varieties of steel , a and b , have a ratio of iron to chromium as 5 : 1 and 7 : 2 , respectively . steel c is produced by mixing alloys a and b at a ratio of 3 : 2 . what is the ratio of iron to chromium in c ? | in 6 parts of alloy a , 5 parts are iron , and 1 part is chromium . in 9 parts of alloy b , 7 parts are iron and 2 parts are chromium . first , to compare the two alloys , get the same number of parts in total - we can use 18 . so we have : in 18 parts of alloy a , 15 parts are iron and 3 are chromium . in 18 parts of ... | a ) 17 : 73 , b ) 78 : 14 , c ) 45 : 30 , d ) 73 : 17 , e ) 4 : 9 | d | multiply(divide(multiply(divide(5, 1), divide(3, add(3, 2))), multiply(divide(7, 2), divide(2, add(3, 2)))), const_2) | add(n3,n4)|divide(n0,n1)|divide(n2,n3)|divide(n4,#0)|divide(n3,#0)|multiply(#1,#3)|multiply(#2,#4)|divide(#5,#6)|multiply(#7,const_2) | other |
what is the last digit in the product ( 3 ^ 65 x 6 ^ 59 x 7 ^ 71 ) | explanation : unit digit in 34 = 1 unit digit in ( 34 ) 16 = 1 unit digit in 365 = unit digit in [ ( 34 ) 16 x 3 ] = ( 1 x 3 ) = 3 unit digit in 659 = 6 unit digit in 74 unit digit in ( 74 ) 17 is 1 . unit digit in 771 = unit digit in [ ( 74 ) 17 x 73 ] = ( 1 x 3 ) = 3 required digit = unit digit in ( 3 x 6 x 3 ) = 4 .... | a ) 5 , b ) 6 , c ) 4 , d ) 8 , e ) 9 | c | reminder(multiply(multiply(3, 6), reminder(power(7, reminder(71, const_4)), const_10)), const_10) | multiply(n0,n2)|reminder(n5,const_4)|power(n4,#1)|reminder(#2,const_10)|multiply(#0,#3)|reminder(#4,const_10) | general |
there are 11 boys and 10 girls in a class . if three students are selected at random , in how many ways that 3 girl & 2 boys are selected ? | "e = event that 3 girl and 2 boys are selected n ( e ) = we have to select 2 boys from 11 and 3 girl from 10 = 11 c 2 * 10 c 3 = 6600 ans - a" | a ) 6600 , b ) 1300 , c ) 6780 , d ) 1976 , e ) 2448 | a | multiply(choose(11, 2), choose(10, 3)) | choose(n0,n3)|choose(n1,n2)|multiply(#0,#1)| | probability |
two goods trains each 400 m long are running in opposite directions on parallel tracks . their speeds are 45 km / hr and 30 km / hr respectively . find the time taken by the slower train to pass the driver of the faster one ? | "relative speed = 45 + 30 = 75 km / hr . 75 * 5 / 18 = 125 / 6 m / sec . distance covered = 400 + 400 = 800 m . required time = 800 * 6 / 125 = 38.40 sec . answer : a" | a ) 38.4 , b ) 32.6 , c ) 48 , d ) 27.4 , e ) 21 | a | add(45, 30) | add(n1,n2)| | physics |
find the area of trapezium whose parallel sides are 20 cm and 18 cm long , and the distance between them is 13 cm | "area of a trapezium = 1 / 2 ( sum of parallel sides ) * ( perpendicular distance between them ) = 1 / 2 ( 20 + 18 ) * ( 13 ) = 247 cm 2 answer : d" | a ) 178 cm 2 , b ) 179 cm 2 , c ) 285 cm 2 , d ) 247 cm 2 , e ) 197 cm 2 | d | quadrilateral_area(13, 18, 20) | quadrilateral_area(n2,n1,n0)| | physics |
a man purchased 3 blankets @ rs . 200 each , 5 blankets @ rs . 150 each and two blankets at a certain rate which is now slipped off from his memory . but he remembers that the average price of the blankets was rs . 150 . find the unknown rate of two blankets ? | "10 * 150 = 1500 3 * 200 + 5 * 150 = 1350 1350 – 1050 = 300 answer : c" | a ) 100 , b ) 200 , c ) 300 , d ) 400 , e ) 500 | c | subtract(multiply(const_10, 150), add(multiply(3, const_100.0), multiply(5, 150))) | multiply(n3,const_10)|multiply(n0,const_100.0)|multiply(n2,n3)|add(#1,#2)|subtract(#0,#3)| | general |
how many positive integers less than 5,000 are evenly divisible by neither 15 nor 22 ? | "integers less than 5000 divisible by 15 5000 / 15 = 333 . something , so 333 integers less than 5000 divisible by 22 5000 / 22 = 238 . # # , so 238 we have double counted some , so take lcm of 15 and 22 = 105 and divide by 5000 , we get 47 . so all numbers divisible by 15 and 22 = 333 + 238 - 47 = 524 now subtract tha... | a ) 4,514 , b ) 4,475 , c ) 4,521 , d ) 4,428 , e ) 4,349 | a | divide(factorial(subtract(add(const_4, 15), const_1)), multiply(factorial(15), 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 |
a man buys a cycle for rs . 1800 and sells it at a loss of 25 % . what is the selling price of the cycle ? | "s . p . = 75 % of rs . 1800 = rs . 75 / 100 x 1800 = rs . 1350 answer : d" | a ) s . 1090 , b ) s . 1160 , c ) s . 1190 , d ) s . 1350 , e ) s . 1256 | d | divide(multiply(subtract(const_100, 25), 1800), const_100) | subtract(const_100,n1)|multiply(n0,#0)|divide(#1,const_100)| | gain |
a man goes from a to b at a speed of 60 kmph and comes back to a at a speed of 40 kmph . find his average speed for the entire journey ? | "distance from a and b be ' d ' average speed = total distance / total time average speed = ( 2 d ) / [ ( d / 60 ) + ( d / 40 ] = ( 2 d ) / [ 5 d / 120 ) = > 48 kmph . answer : e" | a ) 49 kmph , b ) 58 kmph , c ) 44 kmph , d ) 47 kmph , e ) 48 kmph | e | divide(add(60, 40), const_2) | add(n0,n1)|divide(#0,const_2)| | physics |
5358 x 54 = ? | "5358 x 51 = 5358 x ( 50 + 4 ) = 5358 x 50 + 5358 x 4 = 267900 + 21432 = 289332 . c )" | a ) 272258 , b ) 272358 , c ) 289332 , d ) 274258 , e ) 274358 | c | multiply(divide(5358, 54), const_100) | divide(n0,n1)|multiply(#0,const_100)| | general |
a basketball is dropped from a height of 20 feet . if it bounces back up to a height that is exactly half of its previous height , and it stops bouncing after hitting the ground for the fourth time , then how many total feet will the ball have traveled after 2 full bounces . | initial distance = 40 feet first bounce = 20 feet up + 20 feet down = 40 feet second bouche = 10 feet up + 10 feet down = 20 feet total distance covered = 40 + 40 + 20 = 100 answer is e | a ) 50 , b ) 55 , c ) 60 , d ) 75 , e ) 100 | e | subtract(subtract(subtract(subtract(multiply(add(divide(divide(divide(divide(20, 2), 2), 2), 2), add(add(add(add(add(add(20, divide(20, 2)), divide(20, 2)), divide(divide(20, 2), 2)), divide(divide(20, 2), 2)), divide(divide(divide(20, 2), 2), 2)), divide(divide(divide(20, 2), 2), 2))), 2), divide(divide(divide(20, 2),... | divide(n0,n1)|add(n0,#0)|divide(#0,n1)|add(#1,#0)|divide(#2,n1)|add(#3,#2)|divide(#4,n1)|add(#5,#2)|add(#7,#4)|add(#8,#4)|add(#9,#6)|multiply(n1,#10)|subtract(#11,#4)|subtract(#12,#4)|subtract(#13,#4)|subtract(#14,#4)| | general |
a 200 meter long train crosses a man standing on the platform in 9 sec . what is the speed of the train ? | "s = 200 / 9 * 18 / 5 = 80 kmph answer : e" | a ) 228 , b ) 108 , c ) 1266 , d ) 188 , e ) 80 | e | multiply(divide(200, 9), const_3_6) | divide(n0,n1)|multiply(#0,const_3_6)| | physics |
find the area of a parallelogram with base 32 cm and height 14 cm ? | "area of a parallelogram = base * height = 32 * 14 = 448 cm 2 answer : e" | a ) 498 cm 2 , b ) 384 cm 2 , c ) 430 cm 2 , d ) 128 cm 2 , e ) 448 cm 2 | e | multiply(32, 14) | multiply(n0,n1)| | geometry |
raman mixed 44 kg of butter at rs . 150 per kg with 36 kg butter at the rate of rs . 125 per kg . at what price per kg should he sell the mixture to make a profit of 40 % in the transaction ? | "explanation : cp per kg of mixture = [ 44 ( 150 ) + 36 ( 125 ) ] / ( 44 + 36 ) = rs . 138.75 sp = cp [ ( 100 + profit % ) / 100 ] = 138.75 * [ ( 100 + 40 ) / 100 ] = rs . 194.25 answer : c" | a ) 129.25 , b ) 287.25 , c ) 194.25 , d ) 188.25 , e ) 112.25 | c | add(divide(add(multiply(44, 150), multiply(36, 125)), add(36, 44)), multiply(divide(add(multiply(44, 150), multiply(36, 125)), add(36, 44)), divide(40, const_100))) | add(n0,n2)|divide(n4,const_100)|multiply(n0,n1)|multiply(n2,n3)|add(#2,#3)|divide(#4,#0)|multiply(#5,#1)|add(#5,#6)| | gain |
the average age of 42 students in a group is 16 years . when teacher â € ™ s age is included to it , the average increases by one . what is the teacher â € ™ s age in years ? | "sol . age of the teacher = ( 43 ã — 17 â € “ 42 ã — 16 ) years = 59 years . answer d" | a ) 31 , b ) 56 , c ) 41 , d ) 59 , e ) none | d | add(42, const_1) | add(n0,const_1)| | general |
the l . c . m of two numbers is 2310 and their h . c . f is 30 . if one number is 231 the other is | "the other number = l . c . m * h . c . f / given number = 2310 * 30 / 231 = 300 answer is b ." | a ) 330 , b ) 300 , c ) 270 , d ) 250 , e ) 350 | b | divide(multiply(30, 2310), 231) | multiply(n0,n1)|divide(#0,n2)| | physics |
hcf and lcm two numbers are 12 and 396 respectively . if one of the numbers is 24 , then the other number is ? | "12 * 396 = 24 * x x = 198 answer : b" | a ) 36 , b ) 198 , c ) 132 , d ) 264 , e ) 364 | b | divide(multiply(12, 396), 24) | multiply(n0,n1)|divide(#0,n2)| | physics |
company kw is being sold , and both company a and company b were considering the purchase . the price of company kw is 30 % more than company a has in assets , and this same price is also 100 % more than company b has in assets . if companies a and b were to merge and combine their assets , the price of company kw woul... | let the price of company a ' s assets be 100 price of assets of kw is 30 % more than company a ' s assets which is 130 price of assets of kw is 100 % more than company b ' s assets which means price of company b ' s assets is half the price of kw = 65 a + b = 165 kw = 130 kw / ( a + b ) * 100 = 130 / 165 * 100 = 78.78 ... | a ) 66 % , b ) 79 % , c ) 86 % , d ) 116 % , e ) 150 % | b | multiply(divide(add(100, 30), add(100, divide(add(100, 30), const_2))), const_100) | add(n0,n1)|divide(#0,const_2)|add(n1,#1)|divide(#0,#2)|multiply(#3,const_100) | gain |
two passenger trains start at the same hour in the day from two different stations and move towards each other at the rate of 24 kmph and 21 kmph respectively . when they meet , it is found that one train has traveled 60 km more than the other one . the distance between the two stations is ? | "1 h - - - - - 5 ? - - - - - - 60 12 h rs = 24 + 21 = 45 t = 12 d = 45 * 12 = 540 answer : b" | a ) 288 , b ) 540 , c ) 877 , d ) 278 , e ) 178 | b | add(multiply(divide(60, subtract(21, 24)), 24), multiply(divide(60, subtract(21, 24)), 21)) | subtract(n1,n0)|divide(n2,#0)|multiply(n0,#1)|multiply(n1,#1)|add(#2,#3)| | physics |
what is the sum of the greatest common factor and the lowest common multiple of 48 and 72 ? | "prime factorization of the given numbers 72 = 2 ^ 3 * 3 ^ 2 48 = 2 ^ 4 * 3 greatest common factor = 2 ^ 3 * 3 = 24 lowest common multiple = 2 ^ 4 * 3 ^ 2 = 144 sum = 24 + 144 = 168 answer c" | a ) 192 , b ) 120 , c ) 168 , d ) 160 , e ) 184 | c | divide(multiply(48, 72), const_4) | multiply(n0,n1)|divide(#0,const_4)| | general |
the ration of the number of boys and girls in a college is 2 : 3 . if the percentage is increase in the number of boys and girls be 10 % and 20 % respectively . what will be the new ration ? | "let the number of boys and girls be 2 x and 3 x their increased number is 110 % of 2 x and 120 % of 3 x 2 x * 110 / 100 and 3 x * 120 / 100 11 x / 5 and 18 x / 5 required ratio = 11 x / 5 : 18 x / 5 = 11 : 18 answer is d" | a ) 21 : 22 , b ) 13 : 17 , c ) 15 : 43 , d ) 11 : 18 , e ) 15 : 23 | d | divide(multiply(2, add(divide(10, const_100), const_1)), multiply(3, add(divide(20, const_100), const_1))) | divide(n2,const_100)|divide(n3,const_100)|add(#0,const_1)|add(#1,const_1)|multiply(n0,#2)|multiply(n1,#3)|divide(#4,#5)| | gain |
how many positive integers between 1 and 100 are there such that they are multiples of 13 ? | "multiples of 13 = 13 , 26,39 , - - - - - 91 number of multiples of 13 = > 13 * 7 = 91 answer is d" | a ) 6 , b ) 5 , c ) 4 , d ) 7 , e ) 3 | d | divide(subtract(100, 1), 13) | subtract(n1,n0)|divide(#0,n2)| | general |
in the xy - coordinate plane , the graph of y = - x ^ 2 + 9 intersects line l at ( p , 4 ) 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 ... | a ) 6 , b ) 2 , c ) - 2 , d ) - 6 , e ) - 5.5 | e | divide(subtract(7, 4), subtract(sqrt(add(9, 7)), sqrt(add(4, 9)))) | add(n1,n3)|add(n1,n2)|subtract(n3,n2)|sqrt(#0)|sqrt(#1)|subtract(#3,#4)|divide(#2,#5)| | general |
a chemist mixes one liter of pure water with x liters of a 45 % salt solution , and the resulting mixture is a 15 % salt solution . what is the value of x ? | "concentration of salt in pure solution = 0 concentration of salt in salt solution = 45 % concentration of salt in the mixed solution = 15 % the pure solution and the salt solution is mixed in the ratio of - - > ( 45 - 15 ) / ( 15 - 0 ) = 2 / 1 1 / x = 2 / 1 x = 1 / 2 answer : c" | a ) 1 / 4 , b ) 1 / 3 , c ) 1 / 2 , d ) 1 , e ) 3 | c | divide(15, subtract(45, 15)) | subtract(n0,n1)|divide(n1,#0)| | gain |
the average weight of 5 person ' s increases by 10.0 kg when a new person comes in place of one of them weighing 40 kg . what might be the weight of the new person ? | "explanation : total weight increased = ( 5 x 10.00 ) kg = 50 kg . weight of new person = ( 40 + 50 ) kg = 90 kg . answer : c" | a ) 76 kg , b ) 76.5 kg , c ) 90 kg , d ) data inadequate , e ) none of these | c | add(multiply(5, 10.0), 40) | multiply(n0,n1)|add(n2,#0)| | general |
two trains 140 m and 170 m long run at the speed of 60 km / hr and 40 km / hr respectively in opposite directions on parallel tracks . the time which they take to cross each other is ? | "relative speed = 60 + 40 = 100 km / hr . = 100 * 5 / 18 = 250 / 9 m / sec . distance covered in crossing each other = 140 + 170 = 310 m . required time = 310 * 9 / 250 = 11.16 sec . answer : e" | a ) 10.9 sec , b ) 10.1 sec , c ) 10.6 sec , d ) 10.8 sec , e ) 11.16 sec | e | divide(add(140, 170), multiply(add(60, 40), const_0_2778)) | add(n0,n1)|add(n2,n3)|multiply(#1,const_0_2778)|divide(#0,#2)| | physics |
the units digit of ( 3 ) ^ ( 44 ) + ( 10 ) ^ ( 46 ) is : | "any power of anything ending in zero always has a units digit of 0 . so the first term has a units digit of 0 . the period is 4 . this means , 3 to the power of any multiple of 4 will have a units digit of 1 . 3 ^ 44 has a units digit of 1 . of course 0 + 1 = 1 c" | a ) 2 , b ) 4 , c ) 1 , d ) 8 , e ) 0 | c | add(reminder(multiply(reminder(46, const_4), 10), const_10), reminder(3, const_10)) | reminder(n3,const_4)|reminder(n0,const_10)|multiply(n2,#0)|reminder(#2,const_10)|add(#3,#1)| | general |
if a ( a + 8 ) = 9 and b ( b + 8 ) = 9 , where a ≠b , then a + b = | "a ( a + 8 ) = 9 = > we have a = 1 or - 9 also b ( b + 8 ) = 9 = > b = 1 or - 9 given a ≠b 1 ) when a = 1 , b = - 9 and a + b = - 8 1 ) when a = - 9 , b = 1 and a + b = - 8 answer choice a" | a ) − 8 , b ) − 2 , c ) 2 , d ) 46 , e ) 48 | a | add(divide(9, const_10), divide(9, divide(9, const_10))) | divide(n1,const_10)|divide(n1,#0)|add(#0,#1)| | general |
what is the thousandths digit in the decimal equivalent of 66 / 5000 ? | "66 / 5000 = 66 / ( 5 * 10 ^ 3 ) = ( 66 / 5 ) * 10 ^ - 3 = 13.2 * 10 ^ - 3 = . 0132 thousandths digit = 3 answer c" | a ) 0 , b ) 1 , c ) 3 , d ) 5 , e ) 6 | c | floor(multiply(const_100, divide(66, 5000))) | divide(n0,n1)|multiply(#0,const_100)|floor(#1)| | general |
a contractor undertook to do a piece of work in 6 days . he employed certain number of laboures but 7 of them were absent from the very first day and the rest could finish the work in only 10 days . find the number of men originally employed ? | "let the number of men originally employed be x . 6 x = 10 ( x â € “ 7 ) or x = 17.5 answer c" | a ) 17.2 , b ) 16.5 , c ) 17.5 , d ) 17.9 , e ) 17.3 | c | divide(multiply(10, 7), subtract(10, 6)) | multiply(n1,n2)|subtract(n2,n0)|divide(#0,#1)| | physics |
a container holding 12 ounces of a solution that is 1 part alcohol to 2 parts water is added to a container holding 8 ounces of a solution that is 1 part alcohol to 3 parts water . what is the ratio of alcohol to water in the resulting solution ? | "container 1 has 12 ounces in the ratio 1 : 2 or , x + 2 x = 12 gives x ( alcohol ) = 4 and remaining water = 8 container 2 has 8 ounces in the ratio 1 : 3 or , x + 3 x = 8 gives x ( alcohol ) = 2 and remaining water = 6 mixing both we have alcohol = 4 + 2 and water = 8 + 6 ratio thus alcohol / water = 6 / 14 = 3 / 7 a... | a ) 2 : 5 , b ) 3 : 7 , c ) 3 : 5 , d ) 4 : 7 , e ) 7 : 3 | b | divide(add(multiply(12, divide(1, add(1, 2))), multiply(8, divide(1, add(1, 3)))), subtract(add(12, 8), add(multiply(12, divide(1, add(1, 2))), multiply(8, divide(1, add(1, 3)))))) | add(n1,n2)|add(n1,n5)|add(n0,n3)|divide(n1,#0)|divide(n1,#1)|multiply(n0,#3)|multiply(n3,#4)|add(#5,#6)|subtract(#2,#7)|divide(#7,#8)| | other |
the inside dimensions of a rectangular wooden box are 4 meters by 5 meters by 6 meters . a cylindrical drum is to be placed inside the box so that it stands upright when the closed box rests on one of its six faces . of all such drums that could be used , what is the volume , in cubic meters , of the one that has maxim... | you have three options here . approach is to calculate volumes for all of them and see which one is greatest . 1 . cylinder ' s base rests on a 4 x 5 side of the box . so , height becomes 6 and the maximum possible radius becomes 4 / 2 . so , maximum volume = π . ( 4 / 2 ) ^ 2.6 = π 24 2 . cylinder ' s base rests on a ... | ['a ) 20 π', 'b ) 24 π', 'c ) 25 π', 'd ) 96 π', 'e ) 100 π'] | c | max(max(volume_cylinder(divide(4, const_2), 5), volume_cylinder(divide(4, const_2), 6)), volume_cylinder(divide(5, const_2), 4)) | divide(n0,const_2)|divide(n1,const_2)|volume_cylinder(#0,n1)|volume_cylinder(#0,n2)|volume_cylinder(#1,n0)|max(#2,#3)|max(#5,#4) | geometry |
there are some people in party , 1 / 3 rd left the party . then 2 / 5 th of the remaining left the party , then 2 / 3 rd of the remaining left the party . at last 10 were remaining . how many people were in total ? | "sol : 45 if x persons were there in total , then x × ( 1 – 1 / 3 ) × ( 1 – 2 / 5 ) × ( 1 – 2 / 3 ) = 10 x × 2 / 3 × 3 / 5 × 1 / 3 = 10 x = 75 answer : e" | a ) 45 , b ) 27 , c ) 28 , d ) 26 , e ) 75 | e | divide(10, multiply(multiply(subtract(1, divide(1, 3)), subtract(1, divide(2, 5))), subtract(1, divide(2, 3)))) | divide(n0,n1)|divide(n2,n3)|divide(n2,n1)|subtract(n0,#0)|subtract(n0,#1)|subtract(n0,#2)|multiply(#3,#4)|multiply(#6,#5)|divide(n6,#7)| | general |
the hcf of two numbers is 40 and the other two factors of their lcm are 11 and 12 . what is the largest number . | "explanation : hcf of the two numbers = 40 hcf will be always a factor of lcm 40 is factor of lcm other two factors are 11 & 12 then the numbers are ( 40 * 11 ) and ( 40 x 12 ) = 440 and 480 answer : option d" | a ) 462 , b ) 450 , c ) 488 , d ) 480 , e ) 555 | d | multiply(40, 12) | multiply(n0,n2)| | other |
bert and rebecca were looking at the price of a condominium . the price of the condominium was 110 % more than bert had in savings , and separately , the same price was also 20 % more than rebecca had in savings . what is the ratio of what bert has in savings to what rebecca has in savings . | "suppose bert had 100 so price becomes 210 , this 210 = 1.2 times r ' s saving . . so r ' s saving becomes 175 so required ratio is 100 : 175 = 4 : 7 answer : d" | a ) 1 : 5 , b ) 1 : 4 , c ) 2 : 3 , d ) 4 : 7 , e ) 1 : 2 | d | divide(divide(const_100, add(const_100, 110)), divide(const_100, add(const_100, 20))) | add(n0,const_100)|add(n1,const_100)|divide(const_100,#0)|divide(const_100,#1)|divide(#2,#3)| | gain |
at what rate percent on simple interest will rs . 800 amount to rs . 950 in 5 years ? | "150 = ( 800 * 5 * r ) / 100 r = 3.75 % answer : a" | a ) 3.75 % , b ) 5.93 % , c ) 4.75 % , d ) 5.33 % , e ) 6.33 % | a | multiply(divide(divide(subtract(950, 800), 800), 5), const_100) | subtract(n1,n0)|divide(#0,n0)|divide(#1,n2)|multiply(#2,const_100)| | gain |
in a school with 5 classes , each class has 2 students less than the previous class . how many students are there in the largest class if the total number of students at school is 115 ? | "let x be the number of students in the largest class . then x + ( x - 2 ) + ( x - 4 ) + ( x - 6 ) + ( x - 8 ) = 115 5 x - 20 = 115 5 x = 135 x = 27 the answer is c ." | a ) 25 , b ) 26 , c ) 27 , d ) 28 , e ) 29 | c | add(divide(subtract(115, add(add(add(2, add(2, 2)), add(add(2, 2), 2)), add(add(add(2, 2), 2), 2))), 5), add(add(add(2, 2), 2), 2)) | add(n1,n1)|add(n1,#0)|add(n1,#1)|add(#1,#1)|add(#3,#2)|subtract(n2,#4)|divide(#5,n0)|add(#2,#6)| | general |
the length of 3 ropes are in the ratio 4 : 5 : 6 . if the sum of the weights of the longest and the shortest rope is 100 metres more than the length of the third rope , what is the length of the shortest rope ? | let the lengths of the three ropes be 4 k , 5 k and 6 k respectively . 4 k + 6 k = 5 k + 100 = > 5 k = 100 = > k = 20 therefore the weight of the lightest boy = 4 k = 4 ( 20 ) = 80 m answer : a | a ) 80 m , b ) 180 m , c ) 100 m , d ) 60 m , e ) of these | a | multiply(divide(100, subtract(add(6, 4), 5)), 4) | add(n1,n3)|subtract(#0,n2)|divide(n4,#1)|multiply(n1,#2) | general |
on a ranch , a rancher can place a loop of rope , called a lasso , once in every 2 throws around a cow ’ s neck . what is the probability that the rancher will be able to place a lasso around a cow ’ s neck at least once in 6 attempts ? | "p ( missing all 6 ) = ( 1 / 2 ) ^ 6 = 1 / 64 p ( success on at least one attempt ) = 1 - 1 / 64 = 63 / 64 the answer is e ." | a ) 3 / 4 , b ) 7 / 8 , c ) 15 / 16 , d ) 31 / 32 , e ) 63 / 64 | e | subtract(const_1, power(divide(const_1, 2), 6)) | divide(const_1,n0)|power(#0,n1)|subtract(const_1,#1)| | probability |
two pipes can fill the cistern in 10 hr and 12 hr respectively , while the third empty it in 50 hr . if all pipes are opened simultaneously , then the cistern will be filled in | solution : work done by all the tanks working together in 1 hour . 1 / 10 + 1 / 12 − 1 / 50 = 8 / 49 hence , tank will be filled in 49 / 8 = 6.12 hour option ( a ) | a ) 6.12 hr , b ) 8 hr , c ) 8.5 hr , d ) 10 hr , e ) none of these | a | inverse(subtract(add(divide(const_1, 10), divide(const_1, 12)), divide(const_1, 50))) | divide(const_1,n0)|divide(const_1,n1)|divide(const_1,n2)|add(#0,#1)|subtract(#3,#2)|inverse(#4) | physics |
9 people decided to split the restaurant bill evenly . if the bill was $ 514.16 dollars , how much money did they 1 cent is the smallest unit ? | "this is equivalent to finding the first number that is divisible by 9 that occurs after 51416 . in order to divide the sum in 9 parts , the amount must be divisible by 9 divisibility rule of 9 : the sum of the digits must be divisible by 9 sum of digits of 51416 = 17 and 18 is divisible by 9 . hence , we need to add 1... | a ) $ 514.16 , b ) $ 514.17 , c ) $ 514.18 , d ) $ 514.19 , e ) $ 514.20 | b | add(514.16, divide(const_3, const_100)) | divide(const_3,const_100)|add(n1,#0)| | general |
a certain car increased its average speed by 5 miles per hour in each successive 5 - minute interval after the first interval . if in the first 5 - minute interval its average speed was 26 miles per hour , how many miles did the car travel in the third 5 - minute interval ? | in the third time interval the average speed of the car was 22 + 5 + 5 = 36 miles per hour ; in 5 minutes ( 1 / 12 hour ) at that speed car would travel 36 * 1 / 12 = 3 miles . answer : e . | a ) 1.0 , b ) 1.5 , c ) 2.0 , d ) 2.5 , e ) 3.0 | e | multiply(add(add(26, 5), 5), divide(5, const_60)) | add(n0,n3)|divide(n0,const_60)|add(n0,#0)|multiply(#2,#1) | physics |
there are 13 boys and 10 girls in a class . if three students are selected at random , in how many ways that 1 girl & 2 boys are selected ? | n ( s ) = sample space = 23 c 3 = 1771 e = event that 1 girl and 2 boys are selected n ( e ) = we have to select 2 boys from 13 and 1 girl from 10 = 13 c 2 * 10 c 1 = 780 ans - c | a ) 120 , b ) 480 , c ) 780 , d ) 720 , e ) 800 | c | multiply(choose(13, 2), choose(10, 1)) | choose(n0,n3)|choose(n1,n2)|multiply(#0,#1) | probability |
the grade point average of the entire class is 84 . if the average of one fourth of the class is 96 , what is the average of the rest of the class ? | "let x be the number of students in the class . let p be the average of the rest of the class . 84 x = ( 1 / 4 ) 96 x + ( 3 / 4 ) ( p ) x 336 = 96 + 3 p 3 p = 240 p = 80 . the answer is e ." | a ) 76 , b ) 77 , c ) 78 , d ) 79 , e ) 80 | e | divide(subtract(multiply(84, const_4), 96), subtract(const_4, const_1)) | multiply(n0,const_4)|subtract(const_4,const_1)|subtract(#0,n1)|divide(#2,#1)| | general |
the average weight of 19 students is 15 kg . by the admission of a new student the average weight is reduced to 14.2 kg . the weight of the new student is ? | "answer weight of new student = total weight of all 20 students - total weight of initial 19 students = ( 20 x 14.2 - 19 x 15 ) kg = 1 kg . correct option : a" | a ) 1 kg , b ) 10.8 kg , c ) 11 kg , d ) 14.9 kg , e ) none | a | subtract(multiply(add(19, const_1), 14.2), multiply(19, 15)) | add(n0,const_1)|multiply(n0,n1)|multiply(n2,#0)|subtract(#2,#1)| | general |
if a student loses 6 kilograms , he will weigh twice as much as his sister . together they now weigh 132 kilograms . what is the student ' s present weight in kilograms ? | "let x be the weight of the sister . then the student ' s weight is 2 x + 6 . x + ( 2 x + 6 ) = 132 3 x = 126 x = 42 kg then the student ' s weight is 90 kg . the answer is e ." | a ) 82 , b ) 84 , c ) 86 , d ) 88 , e ) 90 | e | subtract(132, divide(subtract(132, 6), const_3)) | subtract(n1,n0)|divide(#0,const_3)|subtract(n1,#1)| | other |
an ant walks an average of 600 meters in 10 minutes . a beetle walks 25 % less distance at the same time on the average . assuming the beetle walks at her regular rate , what is its speed in km / h ? | "the ant walks an average of 600 meters in 10 minutes 600 meters in 5 / 30 hours the beetle walks 25 % less distance = 600 - 150 = 450 meters in 10 minutes 0.450 km in 5 / 30 = 5 / 30 hours speed = 0.450 * 30 / 5 = 2.7 km / h i guess option d should be 2.7" | a ) 2.215 . , b ) 2.5 . , c ) 2.6 , d ) 2.7 , e ) 3.5 . | d | multiply(divide(divide(600, const_1000), divide(10, const_60)), subtract(const_1, divide(25, const_100))) | divide(n0,const_1000)|divide(n1,const_60)|divide(n2,const_100)|divide(#0,#1)|subtract(const_1,#2)|multiply(#3,#4)| | general |
the width of a rectangular hall is ½ of its length . if the area of the hall is 578 sq . m , what is the difference between its length and breadth ? | "let the length of the hall be x m breadth of the hall = 1 x / 2 m area of the hall = length * breadth 578 = x * 1 x / 2 x ² = 1156 x = 34 difference between the length and breadth of the hall = x - 1 x / 2 = x / 2 = 34 / 2 = 17 m answer : e" | a ) 8 m , b ) 10 m , c ) 12 m , d ) 15 m , e ) 17 m | e | divide(sqrt(divide(578, divide(const_1, const_2))), const_2) | divide(const_1,const_2)|divide(n0,#0)|sqrt(#1)|divide(#2,const_2)| | geometry |
how long does a train 165 meters long running at the rate of 90 kmph take to cross a bridge 660 meters in length ? | "t = ( 660 + 165 ) / 90 * 18 / 5 t = 33 answer : c" | a ) 28 , b ) 27 , c ) 33 , d ) 18 , e ) 12 | c | divide(add(165, 660), multiply(90, const_0_2778)) | add(n0,n2)|multiply(n1,const_0_2778)|divide(#0,#1)| | physics |
a boat having a length 3 m and breadth 2 m is floating on a lake . the boat sinks by 1 cm when a man gets on it . the mass of man is | "solution volume of water displaced = ( 3 x 2 x 0.01 ) m 3 = 0.06 m 3 . mass of man = volume of water displaced × density of water = ( 0.06 × 1000 ) kg = 60 kg . answer b" | a ) 12 kg , b ) 60 kg , c ) 72 kg , d ) 96 kg , e ) none | b | multiply(multiply(multiply(3, 2), divide(1, const_100)), const_1000) | divide(n2,const_100)|multiply(n0,n1)|multiply(#0,#1)|multiply(#2,const_1000)| | physics |
how many different pairs of numbers ( s , t ) such that s = 3 t can be obtained if s and t are selected from the set of number { 0 , 1 , 2 , 3 , 4 , 5 , 6 , 7 , 8 , 9 , 10 , 11 , 12 } | given s = 3 t t can take 0,1 , 2,3 , 4 and so s = 0,3 , 6,9 , 12 4 such pairs can be formed . answer e | a ) 1 , b ) 2 , c ) 3 , d ) 4 , e ) 5 | e | add(divide(12, 3), const_1) | divide(n13,n0)|add(#0,const_1) | probability |
in one alloy there is 10 % chromium while in another alloy it is 8 % . 15 kg of the first alloy was melted together with 35 kg of the second one to form a third alloy . find the percentage of chromium in the new alloy . | "the amount of chromium in the new 15 + 35 = 50 kg alloy is 0.10 * 15 + 0.08 * 35 = 4.3 kg , so the percentage is 4.3 / 50 * 100 = 8.6 % . answer : e ." | a ) 9.0 % , b ) 9.4 % , c ) 9.2 % , d ) 8.8 % , e ) 8.6 % | e | multiply(divide(add(divide(multiply(10, 15), const_100), divide(multiply(8, 35), const_100)), add(15, 35)), const_100) | add(n2,n3)|multiply(n0,n2)|multiply(n1,n3)|divide(#1,const_100)|divide(#2,const_100)|add(#3,#4)|divide(#5,#0)|multiply(#6,const_100)| | gain |
on a certain date , pat invested $ 10,000 at x percent annual interest , compounded annually . if the total value of the investment plus interest at the end of 12 years will be $ 40,000 , in how many years , the total value of the investment plus interest will increase to $ 80,000 ? | "x - interest rate 80.000 = 10.000 ( 1 + x ) ^ year = > 8 = ( 1 + x ) ^ year 40.000 = 10.000 . ( 1 + x ) ^ 12 = > 4 = ( 1 + x ) ^ 12 = > 2 = ( 1 + x ) ^ 6 = > 8 = ( 1 + x ) ^ 18 so , after 18 years , the total value of the investment plus interest will increase to $ 80,000 . answer : c" | a ) 15 , b ) 16 , c ) 18 , d ) 20 , e ) 24 | c | divide(log(divide(multiply(const_3, const_10), add(const_4, const_1))), log(power(divide(multiply(const_2, const_10), add(const_4, const_1)), divide(const_1, 12)))) | add(const_1,const_4)|divide(const_1,n1)|multiply(const_10,const_3)|multiply(const_10,const_2)|divide(#2,#0)|divide(#3,#0)|log(#4)|power(#5,#1)|log(#7)|divide(#6,#8)| | general |
in a river flowing at 2 km / hr , a boat travels 32 km upstream and then returns downstream to the starting point . if its speed in still water be 6 km / hr , find the total journey time . | "explanation : speed of the boat = 6 km / hr speed downstream = ( 6 + 2 ) = 8 km / hr speed upstream = ( 6 - 2 ) = 4 km / hr distance travelled downstream = distance travelled upstream = 32 km total time taken = time taken downstream + time taken upstream = ( 32 / 8 ) + ( 32 / 4 ) = ( 32 / 8 ) + ( 64 / 8 ) = ( 96 / 8 )... | a ) 10 hours , b ) 12 hours , c ) 14 hours , d ) 16 hours , e ) none of these | b | add(divide(32, subtract(6, 2)), divide(32, add(6, 2))) | add(n0,n2)|subtract(n2,n0)|divide(n1,#1)|divide(n1,#0)|add(#2,#3)| | general |
a bowl of nuts is prepared for a party . brand p mixed nuts are 20 % almonds and brand q ' s deluxe nuts are 25 % almonds . if a bowl contains a total of 66 ounces of nuts , representing a mixture of both brands , and 15 ounces of the mixture are almonds , how many ounces of brand q ' s deluxe mixed nuts are used ? | "lets say x ounces of p is mixed with q . = > 66 - x ounces of q is present in the mixture ( as the total = 66 ounces ) given total almond weight = 15 ounces ( 20 x / 100 ) + ( 25 / 100 ) ( 66 - x ) = 15 = > x = 30 = > 66 - 30 = 36 ounces of q is present in the mixture . answer is c ." | a ) 16 , b ) 20 , c ) 36 , d ) 44 , e ) 48 | c | divide(subtract(15, multiply(divide(20, const_100), 66)), subtract(divide(25, const_100), divide(20, const_100))) | divide(n0,const_100)|divide(n1,const_100)|multiply(n2,#0)|subtract(#1,#0)|subtract(n3,#2)|divide(#4,#3)| | general |
what is the units digit of ( 6 ! * 4 ! + 6 ! * 5 ! ) / 6 ? | ( 6 ! * 4 ! + 6 ! * 5 ! ) / 6 = 6 ! ( 4 ! + 5 ! ) / 6 = 720 ( 24 + 120 ) / 6 = ( 720 * 144 ) / 6 = 720 * 24 units digit of the above product will be equal to 0 answer b | a ) 1 , b ) 0 , c ) 2 , d ) 3 , e ) 4 | b | divide(add(multiply(factorial(6), factorial(4)), multiply(factorial(6), factorial(5))), 6) | factorial(n0)|factorial(n1)|factorial(n3)|multiply(#0,#1)|multiply(#0,#2)|add(#3,#4)|divide(#5,n0) | general |
if x and y are sets of integers , x # y denotes the set of integers that belong to set x or set y , but not both . if x consists of 8 integers , y consists of 10 integers , and 6 of the integers are in both x and y , then x # y consists of how many integers ? | "the number of integers that belong to set x only is 8 - 6 = 2 ; the number of integers that belong to set y only is 10 - 6 = 4 ; the number of integers that belong to set x or set y , but not both is 2 + 4 = 6 . answer : a ." | a ) 6 , b ) 16 , c ) 22 , d ) 30 , e ) 174 | a | add(subtract(10, 6), subtract(8, 6)) | subtract(n1,n2)|subtract(n0,n2)|add(#0,#1)| | other |
the mean of 50 observations was 36 . it was found later that an observation 48 was wrongly taken as 21 . the corrected new mean is : | "explanation : correct sum = ( 36 * 50 + 48 - 21 ) = 1827 . correct mean = = 1827 / 50 = 36.54 answer : a ) 36.54" | a ) 36.54 , b ) 36.58 , c ) 36.23 , d ) 36.14 , e ) 36.81 | a | divide(add(multiply(36, 50), subtract(subtract(50, const_2), 21)), 50) | multiply(n0,n1)|subtract(n0,const_2)|subtract(#1,n3)|add(#0,#2)|divide(#3,n0)| | general |
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