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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an woman swims downstream 64 km and upstream 24 km taking 8 hours each time ; what is the speed of the current ? | "64 - - - 8 ds = 8 ? - - - - 1 24 - - - - 8 us = 3 ? - - - - 1 s = ? s = ( 8 - 3 ) / 2 = 2.5 answer : e" | a ) 1.6 , b ) 1.8 , c ) 3.5 , d ) 1.5 , e ) 2.5 | e | divide(subtract(divide(64, 8), divide(24, 8)), const_2) | divide(n0,n2)|divide(n1,n2)|subtract(#0,#1)|divide(#2,const_2)| | physics |
what is 30 % of 210 ? | "x / 210 = 30 / 100 100 x = 210 x 30 x = ( 210 x 30 ) / 100 x = 6300 / 100 = 63 answer = b = 63" | a ) 61 , b ) 63 , c ) 65 , d ) 67 , e ) 69 | b | divide(multiply(30, add(add(multiply(multiply(add(const_3, const_2), const_2), multiply(multiply(const_3, const_4), const_100)), multiply(multiply(add(const_3, const_4), add(const_3, const_2)), multiply(add(const_3, const_2), const_2))), add(const_3, const_3))), const_100) | add(const_2,const_3)|add(const_3,const_4)|add(const_3,const_3)|multiply(const_3,const_4)|multiply(#0,const_2)|multiply(#3,const_100)|multiply(#1,#0)|multiply(#4,#5)|multiply(#6,#4)|add(#7,#8)|add(#9,#2)|multiply(n0,#10)|divide(#11,const_100)| | gain |
company x provides bottled water to its employees in 1 liter bottles , $ 2 each . at least how many 1 l bottles must company x buy monthly , so that new contract with $ 50 fixed monthly rate and $ 1 for 10 l bottle each paid off ? ( assume that no other costs apply ) | let the no . of 1 liter bottles be x , so the no . of 10 l bottles will be x / 10 ( to equate the vol . ) since the total cost will be equal , 2 x = 50 + x / 10 so x = 26.31 or 27 . answer is ( e ) . | a ) 28 , b ) 29 , c ) 30 , d ) 31 , e ) 27 | e | divide(50, subtract(1, divide(const_1, 1))) | divide(const_1,n4)|subtract(n0,#0)|divide(n3,#1)| | general |
the average age of students of a class is 15.8 years . the average age of boys in the class is 16.4 years and that of the girls is 15.3 years , the ratio of the number of boys to the number of girls in the class is | "explanation : let the ratio be k : 1 . then , k * 16.4 + 1 * 15.3 = ( k + 1 ) * 15.8 < = > ( 16.4 - 15.8 ) k = ( 15.8 - 15.3 ) < = > k = 0.5 / 0.6 = 5 / 6 . required ratio = 5 / 6 : 1 = 5 : 6 . answer : e" | a ) 7 : 3 , b ) 2 : 3 , c ) 9 : 3 , d ) 6 : 3 , e ) 5 : 6 | e | divide(subtract(15.8, 15.3), subtract(16.4, 15.8)) | subtract(n0,n2)|subtract(n1,n0)|divide(#0,#1)| | general |
a train can travel 50 % faster than a car . both start from point a at the same time and reach point b 75 kms away from a at the same time . on the way , however , the train lost about 12.5 minutes while stopping at the stations . what is the speed of the car ? | let speed of the car = x kmphthen speed of the train = ( 100 + 50 ) 100 x = 150100 x = 32 x kmphtime taken by the car to travel from a to b = 75 x hourstime taken by the train to travel from a to b = 75 ( 32 x ) + 12.560 hourssince both start from a at the same time and reach point b at the same time 75 x = 75 ( 32 x )... | a ) 22 , b ) 28 , c ) 120 , d ) 287 , e ) 278 | c | divide(multiply(subtract(multiply(12.5, const_2), const_1), multiply(12.5, const_2)), add(const_4, const_1)) | add(const_1,const_4)|multiply(n2,const_2)|subtract(#1,const_1)|multiply(#1,#2)|divide(#3,#0) | physics |
if the sides of a triangle are 39 cm , 32 cm and 10 cm , what is its area ? | "the triangle with sides 39 cm , 32 cm and 10 cm is right angled , where the hypotenuse is 39 cm . area of the triangle = 1 / 2 * 32 * 10 = 160 cm 2 answer : d" | a ) 120 , b ) 772 , c ) 288 , d ) 160 , e ) 2848 | d | divide(multiply(32, 10), const_2) | multiply(n1,n2)|divide(#0,const_2)| | geometry |
the remainder when q = 1 + 3 + 3 ^ 2 + 3 ^ 3 + . . . . . . . . . . + 3 ^ 200 is divided 13 . | q = 1 + 3 + 3 ^ 2 + 3 ^ 3 + . . . . . . . . . . . . . . . . . . + 3 ^ 200 is a geometric progression having common ratio as ' 3 ' and number of terms as ' 201 ' . since sum to n terms in gp = a ( r ^ n - 1 ) / ( r - 1 ) where a = first term and r = common ration hence , 1 * ( 3 ^ 201 - 1 ) / ( 3 - 1 ) rem of ( 3 ^ 201 ... | a ) 12 , b ) 7 , c ) 0 , d ) 5 , e ) 3 | c | power(3, add(200, const_1)) | add(n7,const_1)|power(n1,#0) | general |
p and q invested in a shop . the profits were divided in the ratio of 5 : 2 respectively . if p invested rs . 5600 , the amount invested by q is : | suppose q invested rs . y . then , 5600 / y = 5 / 2 or y = [ 5600 x 2 / 5 ] = 2240 answer e | a ) 3500 , b ) 5600 , c ) 2200 , d ) 2400 , e ) none | e | multiply(divide(5600, 5), 2) | divide(n2,n0)|multiply(n1,#0) | gain |
a car gets 20 kilometers per gallon of gasoline . how many gallons of gasoline would the car need to travel 120 kilometers ? | "each 20 kilometers , 1 gallon is needed . we need to know how many 20 kilometers are there in 120 kilometers ? 120 / 20 = 4.5 * 1 gallon = 6 gallons correct answer e" | a ) 8.5 gallons , b ) 5 gallons , c ) 7 gallons , d ) 5.5 gallons , e ) 6 gallons | e | divide(120, 20) | divide(n1,n0)| | physics |
a cuboidal block 6 cm x 9 cm x 12 cm is cut up into an exact number of equal cubes . the least possible number of equal cubes will be | explanation : volume of block = ( 6 x 9 x 12 ) cm 3 = 648 cm 3 side of largest cube = h . c . f of 6 , 912 = 3 cm volume of the cube = ( 3 x 3 x 3 ) = 27 cm 3 number of cubes = ( 648 / 27 ) = 24 answer : c | a ) 6 , b ) 9 , c ) 24 , d ) 30 , e ) 35 | c | divide(multiply(multiply(6, 9), 12), multiply(multiply(divide(6, const_2), divide(6, const_2)), divide(6, const_2))) | divide(n0,const_2)|multiply(n0,n1)|multiply(n2,#1)|multiply(#0,#0)|multiply(#0,#3)|divide(#2,#4) | general |
in a class of 25 students , 2 students did not borrow any books from the library , 12 students each borrowed 1 book , 4 students each borrowed 2 books , and the rest borrowed at least 3 books . if the average number of books per student was 2 , what is the maximum number of books any single student could have borrowed ... | "the total number of books the students borrowed is 25 * 2 = 50 . the students who borrowed zero , one , or two books borrowed 12 * 1 + 4 * 2 = 20 books . the 7 students who borrowed at least three books borrowed 50 - 20 = 30 books . if 6 of these students borrowed exactly 3 books , then the maximum that one student co... | a ) 8 , b ) 9 , c ) 10 , d ) 11 , e ) 12 | e | subtract(multiply(25, 2), add(multiply(subtract(subtract(25, add(add(multiply(12, 1), 4), 2)), 1), 3), add(multiply(12, 1), multiply(4, 2)))) | multiply(n0,n1)|multiply(n2,n3)|multiply(n1,n4)|add(#1,#2)|add(n4,#1)|add(n1,#4)|subtract(n0,#5)|subtract(#6,n3)|multiply(n6,#7)|add(#3,#8)|subtract(#0,#9)| | general |
the sector of a circle has radius of 21 cm and central angle 180 o . find its perimeter ? | "perimeter of the sector = length of the arc + 2 ( radius ) = ( 135 / 360 * 2 * 22 / 7 * 21 ) + 2 ( 21 ) = 108 cm answer : option c" | a ) 91.5 , b ) 92 , c ) 108 , d ) 94 , e ) 95 | c | multiply(multiply(const_2, divide(multiply(subtract(21, const_3), const_2), add(const_4, const_3))), 21) | add(const_3,const_4)|subtract(n0,const_3)|multiply(#1,const_2)|divide(#2,#0)|multiply(#3,const_2)|multiply(n0,#4)| | physics |
beginning in town a , biker bob rides his bike 30 miles west , 6 miles north , 15 miles east , and then 18 miles north , to town b . how far apart are town a and town b ? ( ignore the curvature of the earth . ) | "using pythagoras we have one side i , e total distance traveled in north direction = 18 + 6 = 24 m other being the base ie distance traveled west - distance traveled eat = 30 - 15 = 15 m now this third side or the distance between town a and town b = 24 ^ 2 + 15 ^ 2 = sq root 801 = 28.3 m answer : d" | a ) 27 miles , b ) 25 miles , c ) 24 miles , d ) 28.3 miles , e ) 23 miles | d | sqrt(add(power(add(6, 18), const_2), power(15, const_2))) | add(n1,n3)|power(n2,const_2)|power(#0,const_2)|add(#2,#1)|sqrt(#3)| | physics |
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 68 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 . = > 68 - x ounces of q is present in the mixture ( as the total = 68 ounces ) given total almond weight = 15 ounces ( 20 x / 100 ) + ( 25 / 100 ) ( 68 - x ) = 15 = > x = 40 = > 68 - 40 = 28 ounces of q is present in the mixture . answer is b ." | a ) 16 , b ) 28 , c ) 32 , d ) 44 , e ) 48 | b | divide(subtract(15, multiply(divide(20, const_100), 68)), 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 |
two trains 200 m and 300 m long run at the speed of 70 kmph and 50 kmph in opposite directions in parallel tracks . the time which they take to cross each other is ? | "relative speed = 70 + 50 = 120 kmph * 5 / 18 = 100 / 3 m / s distance covered in crossing each other = 200 + 300 = 500 m required time = 500 * 3 / 100 = 15 sec answer is b" | a ) 10 sec , b ) 15 sec , c ) 18 sec , d ) 20 sec , e ) 22 sec | b | divide(add(200, 300), multiply(add(70, 50), const_0_2778)) | add(n0,n1)|add(n2,n3)|multiply(#1,const_0_2778)|divide(#0,#2)| | physics |
find the area of a parallelogram with base 12 cm and height 10 cm ? | "area of a parallelogram = base * height = 12 * 10 = 120 cm 2 answer : c" | a ) 297 cm 2 , b ) 384 cm 2 , c ) 120 cm 2 , d ) 267 cm 2 , e ) 186 cm 2 | c | multiply(12, 10) | multiply(n0,n1)| | geometry |
6 lights begin to flash together at 12 : 00 noon and flash respectively at intervals of 3 , 4 , 5 , 6 , 7 , and 8 seconds . not including the flash at 12 : 00 noon , how many more times will all 6 lights flash together before 1 : 00 pm ( i . e . one hour later ) ? | the least common multiple is 2 * 2 * 2 * 3 * 5 * 7 = 840 . 3600 seconds / 840 = 4 + remainder . the answer is a . | a ) 4 , b ) 6 , c ) 8 , d ) 10 , e ) 12 | a | floor(divide(multiply(const_1, const_3600), lcm(lcm(7, 8), lcm(lcm(3, 4), lcm(5, 6))))) | lcm(n7,n8)|lcm(n3,n4)|lcm(n0,n5)|multiply(const_1,const_3600)|lcm(#1,#2)|lcm(#0,#4)|divide(#3,#5)|floor(#6) | physics |
a is the hundreds digit of the 3 digit integer x , b is the tens digit of x , and c is the units digit of x . 2 a = b = 4 c , and a > 0 . what is the difference between the two greatest possible values of x ? tip : dont stop till you have exhausted all answer choices to arrive at the correct one . | ratio of a : b : c = 2 : 4 : 1 two possible greatest single digit values for b are 8 and 4 if b is 8 , then x = 482 if b is 4 , then x = 241 difference = 482 - 241 = 241 b is the answer | a ) 124 , b ) 241 , c ) 394 , d ) 421 , e ) 842 | b | subtract(add(add(multiply(multiply(4, const_1), const_100), multiply(multiply(4, 2), const_10)), divide(multiply(4, const_1), 2)), add(add(multiply(divide(multiply(4, const_1), 2), const_100), multiply(multiply(4, const_1), const_10)), const_1)) | multiply(n2,const_1)|multiply(n1,n2)|divide(#0,n1)|multiply(#0,const_100)|multiply(#1,const_10)|multiply(#0,const_10)|add(#3,#4)|multiply(#2,const_100)|add(#6,#2)|add(#7,#5)|add(#9,const_1)|subtract(#8,#10) | general |
find the slope of the line perpendicular to the line y = ( 1 / 2 ) x - 7 | "two lines are perpendicular if the product of their slopes is equal to - 1 . the slope of the given line is equal to 1 / 2 . if m is the slope of the line perpendicular to the given line , then m × ( 1 / 2 ) = - 1 solve for m m = - 2 correct answer c ) - 2" | a ) 1 , b ) 2 , c ) - 2 , d ) 4 , e ) 5 | c | divide(1, 2) | divide(n0,n1)| | general |
the sale price of an article including the sales tax is rs . 616 . the rate of sales tax is 10 % . if the shopkeeper has made a profit of 25 % , then the cost price of the article is : | "explanation : 110 % of s . p . = 616 s . p . = ( 616 * 100 ) / 110 = rs . 560 c . p = ( 100 * 560 ) / 125 = rs . 448 answer : d" | a ) 500 , b ) 277 , c ) 222 , d ) 448 , e ) 111 | d | divide(multiply(divide(multiply(616, const_100), add(const_100, 10)), add(const_100, 10)), add(const_100, 25)) | add(n1,const_100)|add(n2,const_100)|multiply(n0,const_100)|divide(#2,#0)|multiply(#0,#3)|divide(#4,#1)| | gain |
the triplicate ratio of 1 : 2 is ? | "13 : 23 = 1 : 8 answer : b" | a ) 1 : 7 , b ) 1 : 8 , c ) 1 : 3 , d ) 1 : 1 , e ) 1 : 2 | b | divide(power(1, 2), power(const_3.0, 2)) | power(n0,n1)|power(const_3.0,n1)|divide(#0,#1)| | other |
a number with an interesting property : when i divide it by 2 , the remainder is 1 . when i divide it by 3 , the remainder is 2 . when i divide it by 4 , the remainder is 3 . when i divide it by 5 , the remainder is 4 . when i divide it by 6 , the remainder is 5 . when i divide it by 7 , the remainder is 6 . when i div... | d 2519 the number has to end in 9 . looked brute force for small numbers . 59 and 119 were promising , but no cigar . then looked for agreement among 39 + multiples of 40 , 69 + multiples of 70 and 89 + multiples of 90 smallest one was 2519 . | a ) 2588 , b ) 2474 , c ) 2254 , d ) 2519 , e ) 8725 | d | subtract(multiply(multiply(lcm(lcm(9, 10), 7), 2), 2), const_1) | lcm(n14,n16)|lcm(n10,#0)|multiply(n0,#1)|multiply(n0,#2)|subtract(#3,const_1) | gain |
65 % of a number is 21 less than 4 / 5 th of that number . what is the number ? | let the number be x then , 4 x / 5 - ( 65 % of x ) = 21 4 x / 5 - 65 x / 100 = 21 x = 140 answer is c | a ) 100 , b ) 120 , c ) 140 , d ) 130 , e ) 115 | c | divide(multiply(multiply(21, 5), divide(const_100, 5)), subtract(multiply(divide(const_100, 5), 4), 65)) | divide(const_100,n3)|multiply(n1,n3)|multiply(#0,#1)|multiply(n2,#0)|subtract(#3,n0)|divide(#2,#4) | general |
what is the sum of the integers 45 through 175 inclusive ? | "sum of n consecutive positive integers = n ( n + 1 ) / 2 . . in one case n = 44 and other 175 . . . subtract the sum to get the answer sum of first 175 + ive numbers = 175 * 176 / 2 = 15400 . . sum of first 45 + i ' ve numbers = 45 * 44 / 2 = 990 . . answer = 15400 - 990 = 14410 answer : d" | a ) 12,295 , b ) 13,000 , c ) 14,300 , d ) 14,410 , e ) 28,820 | d | divide(multiply(45, 175), const_4) | multiply(n0,n1)|divide(#0,const_4)| | general |
the ratio of males to females in a class is 2 : 3 . the career preferences of the students in the class are represented in a circle graph . if the area of the graph allocated to each career preference is proportional to the number of students who have that career preference , how many degrees of the circle should be us... | "1 / 2 * 2 / 5 + 1 / 4 * 3 / 5 = 4 / 20 + 3 / 20 = 7 / 20 the number of degrees is 7 / 20 * 360 = 126 degrees the answer is b ." | a ) 120 , b ) 126 , c ) 132 , d ) 138 , e ) 144 | b | divide(multiply(2, const_360), add(2, 3)) | add(n0,n1)|multiply(n0,const_360)|divide(#1,#0)| | geometry |
a vendor sells 60 percent of the apples he had and throws away 40 percent of the remainder . the next day , the vendor sells 50 percent of the remaining apples and throws away the rest . in total , what percent of his apples does the vendor throw away ? | "let x be the original number of apples . on day one , the vendor throws away ( 0.4 ) ( 0.4 ) x = 0.16 x . the remaining apples are ( 0.6 ) ( 0.4 ) x = 0.24 x . on day two , the vendor throws away ( 0.5 ) ( 0.24 ) x = 0.12 x . the vendor throws away a total of 0.16 x + 0.12 x = 0.28 x . the vendor throws away 28 percen... | a ) 24 , b ) 26 , c ) 28 , d ) 30 , e ) 32 | c | add(divide(subtract(subtract(const_100, 60), multiply(divide(40, const_100), subtract(const_100, 60))), const_2), multiply(divide(40, const_100), subtract(const_100, 60))) | divide(n1,const_100)|subtract(const_100,n0)|multiply(#0,#1)|subtract(#1,#2)|divide(#3,const_2)|add(#4,#2)| | general |
a vendor sells 80 percent of the pears he had and throws away 50 percent of the remainder . the next day , the vendor sells 80 percent of the remaining pears and throws away the rest . in total , what percent of his pears does the vendor throw away ? | "let x be the original number of pears . on day one , the vendor throws away ( 0.5 ) ( 0.2 ) x = 0.1 x . the remaining pears are ( 0.5 ) ( 0.2 ) x = 0.1 x . on day two , the vendor throws away ( 0.2 ) ( 0.1 ) x = 0.02 x . the vendor throws away a total of 0.1 x + 0.02 x = 0.12 x . the vendor throws away 12 percent of t... | a ) 6 , b ) 8 , c ) 10 , d ) 12 , e ) 14 | d | multiply(const_100, add(subtract(subtract(subtract(const_1, divide(80, const_100)), multiply(divide(50, const_100), subtract(const_1, divide(80, const_100)))), multiply(subtract(subtract(const_1, divide(80, const_100)), multiply(divide(50, const_100), subtract(const_1, divide(80, const_100)))), divide(80, const_100))),... | divide(n1,const_100)|divide(n0,const_100)|subtract(const_1,#1)|multiply(#0,#2)|subtract(#2,#3)|multiply(#1,#4)|subtract(#4,#5)|add(#3,#6)|multiply(#7,const_100)| | general |
a train 700 m long can cross an electric pole in 20 sec and then find the speed of the train ? | "length = speed * time speed = l / t s = 700 / 20 s = 35 m / sec speed = 35 * 18 / 5 ( to convert m / sec in to kmph multiply by 18 / 5 ) speed = 126 kmph answer : c" | a ) 117 kmph , b ) 178 kmph , c ) 126 kmph , d ) 118 kmph , e ) 119 kmph | c | divide(divide(700, const_1000), divide(20, const_3600)) | divide(n0,const_1000)|divide(n1,const_3600)|divide(#0,#1)| | physics |
the average weight of 8 persons increases by 3.5 kg when a new person comes in place of one of them weighing 65 kg . what might be the weight of the new person ? | "total weight increased = ( 8 x 3.5 ) kg = 28 kg . weight of new person = ( 65 + 28 ) kg = 93 kg . answer : e" | a ) 75 kg , b ) 65 kg , c ) 55 kg , d ) 85 kg , e ) 93 kg | e | add(65, multiply(8, 3.5)) | multiply(n0,n1)|add(n2,#0)| | general |
an industrial loom weaves 1.14 meters of cloth every second . approximately , how much time will it take to weave 52 meters of cloth ? | explanation : given loom weaves 1.14 mts of cloth in one second then 52 mts of cloth can be weaved by loom in , 1.14 - - - - - 1 52.0 - - - - - - ? = > 52 / 1.14 = 45.61 sec answer is d | a ) 29.32 sec , b ) 42.51 sec , c ) 39.25 sec , d ) 45.61 sec , e ) none of these | d | divide(52, 1.14) | divide(n1,n0) | physics |
one pipe can fill a tank three times as fast as another pipe . if together the two pipes can fill tank in 37 min , then the slower pipe alone will be able to fill the tank in ? | "let the slower pipe alone fill the tank in x min . then , faster pipe will fill it in x / 3 min . 1 / x + 3 / x = 1 / 37 4 / x = 1 / 37 = > x = 148 min . answer : d" | a ) 229 , b ) 787 , c ) 144 , d ) 148 , e ) 121 | d | multiply(add(const_1, const_4), 37) | add(const_1,const_4)|multiply(n0,#0)| | physics |
if 10 machines can produce 20 units in 10 hours , how long would it take 40 machines to produce 100 units ? | "10 machines would produce 100 units in 50 hours . increasing the amount of machines by 4 would mean dividing 50 hours by 4 . 50 / 4 = 12.5 answer : e" | a ) 9.5 , b ) 6.4 , c ) 2.3 , d ) 8.9 , e ) 12.5 | e | divide(100, multiply(divide(divide(20, 10), 10), 20)) | divide(n1,n2)|divide(#0,n0)|multiply(n1,#1)|divide(n4,#2)| | physics |
when positive integer n is divided by positive integer j , the remainder is 14 . if n / j = 134.07 , what is value of j ? | "1 ) we know that decimal part of decimal quotient = { remainder / divisor } so 0.07 , the decimal part of the decimal quotient , must equal the remainder , 14 , divided by the divisor j . 0.07 = 14 / j 0.07 * j = 14 j = 14 / 0.07 = 1400 / 7 = 200 so j = 200 , answer = d ." | a ) 22 , b ) 56 , c ) 78 , d ) 200 , e ) 175 | d | divide(14, subtract(134.07, add(const_100, add(multiply(const_4, const_10), const_2)))) | multiply(const_10,const_4)|add(#0,const_2)|add(#1,const_100)|subtract(n1,#2)|divide(n0,#3)| | general |
if m : n is 2 : 9 and n : p is 9 : 4 then m : p is equal to | the two ratios given are having the same number 9 for n in both the ratios . hence - m : n = 2 : 9 n : p = 9 : 4 = > m : p = 2 : 4 = > 1 : 2 answer e | a ) 1 : 4 , b ) 1 : 3 , c ) 5 : 2 , d ) 3 : 2 , e ) 1 : 2 | e | divide(2, 4) | divide(n0,n3) | physics |
the product of two numbers is 1320 and their h . c . f is 6 . the l . c . m of the numbers is | "solution l . c . m = product of numbers / h . c . f ‹ = › 1320 / 6 ‹ = › 220 . answer a" | a ) 220 , b ) 1314 , c ) 1326 , d ) 7920 , e ) none | a | sqrt(add(power(sqrt(subtract(6, multiply(const_2, 1320))), const_2), multiply(const_4, 1320))) | multiply(n0,const_4)|multiply(n0,const_2)|subtract(n1,#1)|sqrt(#2)|power(#3,const_2)|add(#0,#4)|sqrt(#5)| | physics |
a certain taxi company charges $ 2.80 for the first 1 / 5 of a mile plus $ 0.40 for each additional 1 / 5 of a mile . what would this company charge for a taxi ride that was 8 miles long ? | "a certain taxi company charges $ 2.80 for the first 1 / 5 of a mile plus $ 0.40 for each additional 1 / 5 of a mile . what would this company charge for a taxi ride that was 8 miles long ? a . 15.60 b . 16.00 c . 17.50 d . 18.70 e . 19.10 1 / 5 miles = 0.2 miles . the cost of 8 miles long ride would be $ 2.80 for the ... | a ) 15.6 , b ) 16.0 , c ) 18.4 , d ) 18.7 , e ) 19.1 | c | add(2.80, multiply(subtract(divide(8, divide(1, 5)), 1), 0.40)) | divide(n1,n2)|divide(n6,#0)|subtract(#1,n1)|multiply(n3,#2)|add(n0,#3)| | general |
the average of first five multiples of 4 is | "solution average = 4 ( 1 + 2 + 3 + 4 + 5 ) / 5 = 60 / 5 . = 12 answer d" | a ) 3 , b ) 6 , c ) 9 , d ) 12 , e ) 15 | d | add(4, const_1) | add(n0,const_1)| | general |
of the 3 - digit integers greater than 200 , how many have two digits that are equal to each other and the remaining digit different from the other two ? | let the three digit number be represented as x y z . there are 3 cases : case i . [ x = y ] z is not equal to xy : xxzoryyz x can be either 7 , 8 or 9 , so digit at x can be chosen in 3 ways after x is chosen , y can be chosen in 1 way after xy are chosen , z can be chosen in 9 ways thus , possible no of digits = ( 3 w... | a ) 90 , b ) 82 , c ) 80 , d ) 45 , e ) 36 | e | add(divide(subtract(divide(200, const_2), const_1), 3), 3) | divide(n1,const_2)|subtract(#0,const_1)|divide(#1,n0)|add(n0,#2) | general |
a rectangular mat has an area of 120 sq . metres and perimeter of 46 m . the length of its diagonal is : | rectangular area = l × b = 120 and perimeter = 2 ( l + b ) = 46 l + b = 23 ( l - b ) power 2 - 4 lb = ( 23 ) power 2 - 4 × 120 = ( 529 - 480 ) = 49 = l - b = 7 l + b = 23 , l - b = 7 , we get l = 15 , b = 8 diagonal = √ 152 + 82 √ 225 + 64 = > √ 289 = 17 m answer is b . | ['a ) 19 m', 'b ) 17 m', 'c ) 15 m', 'd ) 16 m', 'e ) 20 m'] | b | sqrt(subtract(power(divide(46, const_2), const_2), multiply(120, const_2))) | divide(n1,const_2)|multiply(n0,const_2)|power(#0,const_2)|subtract(#2,#1)|sqrt(#3) | geometry |
a farmer used 1,034 acres of land for beans , wheat , and corn in the ratio of 5 : 2 : 4 , respectively . how many r acres were used for corn ? | "consider 5 x acres of land used for bean consider 2 x acres of land used for wheat consider 4 x acres of land used for corn total given is 1034 acres 11 x = 1034 x = 94 land used for corn r = 4 * 94 = 376 correct option - c" | a ) 188 , b ) 258 , c ) 376 , d ) 470 , e ) 517 | c | multiply(divide(add(multiply(const_1000, const_1), add(multiply(const_10, const_3), 4)), add(add(5, 2), 4)), 4) | add(n1,n2)|multiply(const_10,const_3)|multiply(const_1,const_1000)|add(n3,#1)|add(n3,#0)|add(#3,#2)|divide(#5,#4)|multiply(n3,#6)| | other |
the tax on a commodity is diminished by 40 % but its consumption is increased by 25 % . find the decrease percent in the revenue derived from it ? | "100 * 100 = 10000 60 * 125 = 7500 10000 - - - - - - - 2500 100 - - - - - - - ? = 25 % answer : c" | a ) 18 % , b ) 72 % , c ) 25 % , d ) 12 % , e ) 52 % | c | subtract(const_100, divide(multiply(add(const_100, 25), subtract(const_100, 40)), const_100)) | add(n1,const_100)|subtract(const_100,n0)|multiply(#0,#1)|divide(#2,const_100)|subtract(const_100,#3)| | general |
a lady grows cabbage in her garden that is in the shape of a square . each cabbage takes 1 square foot of area in her garden . this year , she has increased her output by 127 cabbages when compared to last year . the shape of the area used for growing the cabbage has remained a square in both these years . how many cab... | "explanatory answer the shape of the area used for growing cabbage has remained a square in both the years . let the side of the square area used for growing cabbage this year be x ft . therefore , the area of the ground used for cultivation this year = x 2 sq . ft . let the side of the square area used for growing cab... | a ) 4096 , b ) 4098 , c ) 5000 , d ) 5096 , e ) can not be determined | a | power(add(divide(subtract(127, 1), const_2), 1), const_2) | subtract(n1,n0)|divide(#0,const_2)|add(n0,#1)|power(#2,const_2)| | geometry |
a snail , climbing a 20 feet high wall , climbs up 8 feet on the first day but slides down 4 feet on the second . it climbs 8 feet on the third day and slides down again 4 feet on the fourth day . if this pattern continues , how many days will it take the snail to reach the top of the wall ? | "total transaction in two days = 8 - 4 = 4 feet in 7 days it will climb 20 feet thus reaching the top therefore , total no of days required = 7 a" | a ) 7 , b ) 16 , c ) 17 , d ) 20 , e ) 21 | a | subtract(20, 8) | subtract(n0,n1)| | physics |
out of 40 applicants to a law school , 15 majored in political science , 20 had a grade point average higher than 3.0 , and 10 did not major in political science and had a gpa equal to or lower than 3.0 . how many of q applicants majored in political science and had a gpa higher than 3.0 ? | total applicants = 40 political science = 15 and non political science = 40 - 15 = 25 gpa > 3.0 = 20 and gpa < = 3.0 = 20 10 non political science students had gpa < = 3.0 - - > 15 non political science students had gpa > 3.0 gpa > 3.0 in political science = total - ( gpa > 3.0 in non political science ) q = 20 - 15 = ... | a ) 5 , b ) 10 , c ) 15 , d ) 25 , e ) 35 | a | subtract(20, subtract(40, add(10, 15))) | add(n1,n4)|subtract(n0,#0)|subtract(n2,#1) | general |
the sum of the squares of the first 15 positive integers ( 1 ^ 2 + 2 ^ 2 + 3 ^ 2 + . . . + 15 ^ 2 ) is equal to 1270 . what is the sum of the squares of the second 15 positive integers ( 16 ^ 2 + 17 ^ 2 + 18 ^ 2 + . . . + 30 ^ 2 ) ? | "you ' ll never need a formula for the sums of squares on the actual gmat . you do n ' t need to use that formula here , though it ' s not all that straightforward to solve without one . two different approaches : 16 ^ 2 + 17 ^ 2 + 18 ^ 2 + . . . + 30 ^ 2 = ( 15 + 1 ) ^ 2 + ( 15 + 2 ) ^ 2 + ( 15 + 3 ) ^ 2 + . . . + ( 1... | a ) 2480 , b ) 3490 , c ) 6785 , d ) 8245 , e ) 9255 | d | subtract(divide(multiply(multiply(30, add(30, 1)), add(multiply(2, 30), 1)), add(3, 3)), 1270) | add(n17,n1)|add(n5,n5)|multiply(n17,n2)|add(#2,n1)|multiply(n17,#0)|multiply(#3,#4)|divide(#5,#1)|subtract(#6,n9)| | general |
last week vartan spent 15 percent of his wages on recreation . this week , his wages are 10 percent less than last week ʼ s wages and he spent 30 percent of his wages on recreation . the amount he spends on recreation this week is what percent of the amount he spent on recreation last week ? | say vartan ' s wages last week were $ 100 , so he spent 0.15 * 100 = $ 15 on recreation ; this week ' s wages is 0.9 * 100 = $ 90 , so he spends 0.3 * 90 = $ 27 on recreation ; 27 / 15 = 1.8 , hence the amount he spends on recreation this week is 180 % of the amount he spent on recreation last week : 15 * 1.8 = 27 . an... | a ) 100 % , b ) 160 % , c ) 180 % , d ) 200 % , e ) 220 % | c | divide(multiply(30, subtract(const_100, 10)), 15) | subtract(const_100,n1)|multiply(n2,#0)|divide(#1,n0) | general |
the circumferences of two circles are 528 meters and 704 meters . find the difference between the areas of the larger and the smaller circles ? | "let the radii of the smaller and the larger circles be s m and l m respectively . 2 ∏ s = 528 and 2 ∏ l = 704 s = 528 / 2 ∏ and l = 704 / 2 ∏ difference between the areas = ∏ l ^ 2 - ∏ s ^ 2 = ∏ { 264 ^ 2 / ∏ ^ 2 - 352 ^ 2 / ∏ ^ 2 } = 264 ^ 2 / ∏ - 352 ^ 2 / ∏ = ( 264 - 352 ) ( 264 + 352 ) / ∏ = ( 88 ) ( 616 ) / ( 22 ... | a ) 29963 sq m , b ) 28937 sq m , c ) 43162 sq m , d ) 27688 sq m , e ) 17248 sq m | e | subtract(circle_area(divide(704, multiply(const_2, const_pi))), circle_area(divide(528, multiply(const_2, const_pi)))) | multiply(const_2,const_pi)|divide(n1,#0)|divide(n0,#0)|circle_area(#1)|circle_area(#2)|subtract(#3,#4)| | geometry |
if 15 % of a is the same as 25 % of b , then a : b is : | "expl : 15 % of a i = 25 % of b = 15 a / 100 = 25 b / 100 = 5 / 3 = 5 : 3 answer : a" | a ) 5 : 3 , b ) 4 : 3 , c ) 7 : 16 , d ) 12 : 17 , e ) 11 : 13 | a | divide(divide(25, const_100), divide(15, const_100)) | divide(n1,const_100)|divide(n0,const_100)|divide(#0,#1)| | gain |
the ages of old and young total 48 . old is twice as old as young was when old was half as old as young will be when young is 3 times as old was when old was 3 times as old as young . how old is old ? | explanation : from the options itself , we can see that option c old = 30 young = 18 30 + 18 = 48 by reducing this years only by one before 6 years old = 24 ( half of young ) young = 12 ( twice of old ) answer : c | a ) 23 , b ) 27 , c ) 12 , d ) 43 , e ) 37 | c | multiply(add(3, 3), const_2) | add(n1,n1)|multiply(#0,const_2) | general |
the percentage profit earned by selling an article for rs . 1520 is equal to the percentage loss incurred by selling the same article for rs . 1280 . at what price should the article be sold to make 25 % profit ? | "c . p . be rs . x . then , ( 1520 - x ) / x * 100 = ( x - 1280 ) / x * 100 1520 - x = x - 1280 2 x = 2800 = > x = 1400 required s . p . = 125 % of rs . 1400 = 125 / 100 * 1400 = rs . 1750 . answer c" | a ) 3000 , b ) 1230 , c ) 1750 , d ) 5600 , e ) 3400 | c | multiply(divide(add(const_100, 25), const_100), divide(add(1520, 1280), const_2)) | add(n2,const_100)|add(n0,n1)|divide(#0,const_100)|divide(#1,const_2)|multiply(#2,#3)| | gain |
convert 2.5 hectares in ares | "2.5 hectares in ares 1 hectare = 100 ares therefore , 2.5 hectares = 2.5 × 100 ares = 250 ares . answer - e" | a ) 130 ares . , b ) 160 ares . , c ) 180 ares . , d ) 230 ares . , e ) 250 ares . | e | divide(multiply(multiply(multiply(add(const_3, const_2), const_2), multiply(add(const_3, const_2), const_2)), 2.5), multiply(multiply(add(const_3, const_2), const_2), multiply(add(const_3, const_2), const_2))) | add(const_2,const_3)|multiply(#0,const_2)|multiply(#1,#1)|multiply(n0,#2)|divide(#3,#2)| | physics |
if a randomly selected non - negative single digit integer is added to { 2 , 3 , 4 , 9 } . what is the probability that the median of the set will increase but the range still remains the same ? | "we are selecting from non - negative single digit integers , so from { 0 , 1 , 2 , 3 , 4 , 5 , 6 , 7 , 8 , 9 } . these 10 digits represent the total number of outcomes . hence , the total number of outcomes is 10 . we need to find the probability that the median of the set will increase but the range still remains the... | a ) 0.2 , b ) 0.3 , c ) 0.4 , d ) 0.5 , e ) 0.6 | e | divide(const_4, const_10) | divide(const_4,const_10)| | general |
keats library purchases a number of new books , all in the category of biography , and the library does not acquire any other books . with the addition of the new biographies , the biography collection of the library amounts to 32 % of the total number of books in the library . if prior to the purchase , only 20 % of t... | let x be the number of new biographies added to the library . let b be the original number of biographies , so the original number of books was 5 b . 0.32 ( 5 b + x ) = b + x 0.6 b = 0.68 x x = 0.88 b the answer is c . | a ) 60 % , b ) 72 % , c ) 88 % , d ) 95 % , e ) 110 % | c | multiply(divide(divide(subtract(multiply(32, const_100), multiply(20, const_100)), subtract(const_100, 32)), 20), const_100) | multiply(n0,const_100)|multiply(n1,const_100)|subtract(const_100,n0)|subtract(#0,#1)|divide(#3,#2)|divide(#4,n1)|multiply(#5,const_100) | general |
if the volume of two cubes are in the ratio 125 : 1 , the ratio of their edges is : | "explanation : let the edges be a and b of two cubes , then a 3 / b 3 = 125 / 1 = > ( a / b ) 3 = ( 5 / 1 ) 3 a / b = 5 / 1 = > a : b = 5 : 1 option b" | a ) 3 : 1 , b ) 5 : 1 , c ) 3 : 5 , d ) 3 : 7 , e ) none of these | b | cube_edge_by_volume(125) | cube_edge_by_volume(n0)| | geometry |
in a class of 20 students in an examination in mathematics 2 students scored 100 marks each , 3 get zero each and the average of the rest was 40 . what is the average of the whole class ? | total marks obtained by a class of 20 students = 2 * 100 + 3 * 0 + 15 * 40 = 200 + 600 = 800 : average marks of whole class = 800 / 20 = 40 answer is c . | a ) 46 , b ) 49 , c ) 40 , d ) 38 , e ) 35 | c | divide(add(multiply(40, subtract(subtract(20, 2), 3)), multiply(2, 100)), 20) | multiply(n1,n2)|subtract(n0,n1)|subtract(#1,n3)|multiply(n4,#2)|add(#3,#0)|divide(#4,n0) | general |
the cost of an article is decreased by 20 % . if the original cost is $ 40 , find the decrease cost . | "original cost = $ 40 decrease in it = 20 % of $ 40 = 20 / 100 ã — 40 = 800 / 100 = $ 8 therefore , decrease cost = $ 40 - $ 8 = $ 32 answer : b" | a ) 33 , b ) 32 , c ) 68 , d ) 36 , e ) 38 | b | divide(multiply(40, 20), const_100) | multiply(n0,n1)|divide(#0,const_100)| | gain |
two pipes a and b can separately fill a cistern in 24 minutes and 48 minutes respectively . there is a third pipe in the bottom of the cistern to empty it . if all the three pipes are simultaneously opened , then the cistern is full in 12 minutes . in how much time , the third pipe alone can empty the cistern ? | "1 / 12 - ( 1 / 24 + 1 / 48 ) = - 1 / 48 third pipe can empty in 48 minutes answer : b" | a ) 90 min , b ) 48 min , c ) 54 min , d ) 120 min , e ) 130 min | b | inverse(subtract(add(inverse(24), inverse(48)), inverse(12))) | inverse(n0)|inverse(n1)|inverse(n2)|add(#0,#1)|subtract(#3,#2)|inverse(#4)| | physics |
a can do a piece of work in 20 days , and b can do it in 15 days . how long will they take if both work together ? | a + b can do the work in 20 * 15 / 20 + 15 = 20 * 15 / 35 = 8.6 days answer is b | a ) 7 days , b ) 8.6 days , c ) 13 days , d ) 5.6 days , e ) 10 days | b | inverse(add(inverse(15), inverse(20))) | inverse(n1)|inverse(n0)|add(#0,#1)|inverse(#2) | physics |
if the sides of a triangle are 28 cm , 26 cm and 12 cm , what is its area ? | "the triangle with sides 28 cm , 26 cm and 12 cm is right angled , where the hypotenuse is 28 cm . area of the triangle = 1 / 2 * 26 * 12 = 156 cm 2 answer : b" | a ) 120 cm 2 , b ) 156 cm 2 , c ) 178 cm 2 , d ) 189 cm 2 , e ) 176 cm 2 | b | divide(multiply(26, 12), const_2) | multiply(n1,n2)|divide(#0,const_2)| | geometry |
the average weight of 29 students is 28 kg . by the admission of a new student , the average weight is reduced to 27.4 kg . the weight of the new student is | "exp . the total weight of 29 students = 29 * 28 the total weight of 30 students = 30 * 27.4 weight of the new student = ( 30 * 27.4 – 29 * 28 ) = 822 - 812 = 10 answer : d" | a ) 22 kg , b ) 21.6 kg , c ) 22.4 kg , d ) 10 kg , e ) none of these | d | subtract(multiply(add(29, const_1), 27.4), multiply(29, 28)) | add(n0,const_1)|multiply(n0,n1)|multiply(n2,#0)|subtract(#2,#1)| | general |
22 buckets of water fill a tank when the capacity of each bucket is 13.5 litres . how many buckets will be required to fill the same tank if the capacity of each bucket is 9 litres ? | "capacity of the tank = 22 ã — 13.5 = 297 litres when the capacity of each bucket = 9 litres , then the required no . of buckets = 297 â „ 9 = 33 answer a" | a ) 33 , b ) 32 , c ) 60 , d ) data inadequate , e ) none of these | a | divide(multiply(13.5, 22), 9) | multiply(n0,n1)|divide(#0,n2)| | physics |
find a positive number which when increased by 17 is equal to 60 times the reciprocal of the number ? | let the number be x . then , x + 17 = 60 x x 2 + 17 x - 60 = 0 ( x + 20 ) ( x - 3 ) = 0 x = 3 option a | a ) 3 , b ) 5 , c ) 6 , d ) 7 , e ) 9 | a | add(const_2, gcd(60, 17)) | gcd(n0,n1)|add(#0,const_2) | general |
the arithmetic mean and standard deviation of a certain normal distribution are 17.5 and 2.5 , respectively . what value is exactly 2 standard deviations less than the mean ? | "mean = 17.5 two standard deviations is 2.5 + 2.5 = 5.0 there could be two calues for this . mean + two standard deviations = 22.5 mean - two standard deviations = 12.5 answer choice has 12.5 and so e is the answer ." | a ) 10.5 , b ) 11 , c ) 11.5 , d ) 12 , e ) 12.5 | e | subtract(17.5, multiply(2, 2.5)) | multiply(n1,n2)|subtract(n0,#0)| | general |
in a group of 130 people , 90 have an age of more 30 years , and the others have an age of less than 20 years . if a person is selected at random from this group , what is the probability the person ' s age is less than 20 ? | "number of people whose age is less than 20 is given by 130 - 90 = 40 probability p that a person selected at random from the group is less than 20 is given by 40 / 130 = 0.31 correct answer a" | a ) 0.31 , b ) 0.55 , c ) 0.65 , d ) 0.75 , e ) 0.85 | a | divide(subtract(130, 90), 130) | subtract(n0,n1)|divide(#0,n0)| | general |
a store reported total sales of $ 385 million for february of this year . if the total sales for the same month last year was $ 320 million , approximately what was the percent increase e in sales ? | "last year ' s sales = $ 320 million ; this year ' s sales = $ 385 million ; increase e = $ 65 million . now , 20 % of $ 320 million is $ 64 million , which is very close to actual increase of $ 65 million . answer : c ." | a ) 2 % , b ) 17 % , c ) 20 % , d ) 65 % , e ) 83 % | c | multiply(divide(subtract(385, 320), 320), const_100) | subtract(n0,n1)|divide(#0,n1)|multiply(#1,const_100)| | general |
a city with a population of 126,160 is to be divided into 7 voting districts , and no district is to have a population that is more than 10 percent greater than the population of any other district . what is the minimum possible population that the least populated district could have ? | "the minimum possible population occurs when all the other districts have a population that is 10 % greater than the least populated district . let p be the population of the least populated district . then 126,160 = p + 6 ( 1.1 ) p 7.6 p = 126,160 p = 16,600 the answer is b ." | a ) 16,500 , b ) 16,600 , c ) 16,700 , d ) 16,800 , e ) 16,900 | b | divide(divide(subtract(multiply(const_1000, const_100), subtract(subtract(const_3600, const_100), const_1000)), const_1000), add(multiply(add(const_1, divide(10, const_100)), subtract(7, const_1)), const_1)) | divide(n2,const_100)|multiply(const_100,const_1000)|subtract(const_3600,const_100)|subtract(n1,const_1)|add(#0,const_1)|subtract(#2,const_1000)|multiply(#4,#3)|subtract(#1,#5)|add(#6,const_1)|divide(#7,const_1000)|divide(#9,#8)| | general |
find the largest 3 digit number which is exactly divisible by 88 ? | "largest 3 digit number is 999 after doing 999 ÷ 88 we get remainder 31 hence largest 3 digit number exactly divisible by 88 = 999 - 31 = 968 e" | a ) 765 , b ) 907 , c ) 944 , d ) 954 , e ) 968 | e | multiply(add(add(add(add(multiply(const_100, const_100), multiply(const_100, const_10)), multiply(const_100, const_3)), multiply(3, const_10)), const_3), 88) | multiply(const_100,const_100)|multiply(const_10,const_100)|multiply(const_100,const_3)|multiply(n0,const_10)|add(#0,#1)|add(#4,#2)|add(#5,#3)|add(#6,const_3)|multiply(n1,#7)| | general |
if a = { 17 , 27 , 31 , 53 , 61 } , what is the sum of mean and median of the numbers in a ? | mean = ( 17 + 27 + 31 + 53 + 62 ) / 5 = 38 median = 31 sum = 38 + 31 = 69 option a | a ) 69 , b ) 75 , c ) 82 , d ) 91 , e ) 56 | a | add(divide(add(add(add(add(17, 27), 31), 53), 61), add(const_4, const_1)), 31) | add(n0,n1)|add(const_1,const_4)|add(n2,#0)|add(n3,#2)|add(n4,#3)|divide(#4,#1)|add(n2,#5) | general |
the simple interest on a certain sum of money at the rate of 5 % p . a . for 8 years is rs . 840 . at what rate of intrest the same amount of interest can be received on the same sum after 5 years . | "explanation : here firstly we need to calculate the principal amount , then we can calculate the new rate . p = s . i . ∗ 100 / r ∗ t p = 840 ∗ 100 / 5 ∗ 8 p = 2100 required rate = 840 ∗ 100 / 5 ∗ 2100 r = 8 % option d" | a ) 5 % , b ) 6 % , c ) 7 % , d ) 8 % , e ) 9 % | d | multiply(divide(divide(840, 5), divide(divide(840, 8), divide(5, const_100))), const_100) | divide(n2,n0)|divide(n2,n1)|divide(n0,const_100)|divide(#1,#2)|divide(#0,#3)|multiply(#4,const_100)| | gain |
two cards are drawn together from a pack of 52 cards . the probability that one is a club and other is a diamond , is : | explanation : one card drawn . probability of being it club or diamond = ( 13 x 2 ) / 52 . another card drawn . now , probability of being it of other suit ( the suit among the two , not selected in first draw ) = 13 / 51 . hence , probability of event = ( ( 13 x 2 ) / 52 ) x ( 13 / 51 ) = 13 / 102 . answer : d | a ) 29 / 34 , b ) 47 / 100 , c ) 3 / 20 , d ) 13 / 102 , e ) none of these | d | 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 |
a rectangular tiled patio is composed of 160 square tiles . the rectangular patio will be rearranged so that there will be 2 fewer columns of tiles and 4 more rows of tiles . after the change in layout , the patio will still have 160 tiles , and it will still be rectangular . how many rows are in the tile patio before ... | "suppose there are c columns and there are r rows original situation so , number of tiles = c * r = 160 also . reach column has r tiles and each row has c tiles new situation number of tiles in each column is r - 2 and number of tiles in each row is c + 4 so , number of rows = r - 2 and number of columns is c + 4 so , ... | a ) 5 , b ) 6 , c ) 10 , d ) 13 , e ) 28 | c | divide(160, divide(add(negate(4), sqrt(add(power(4, 2), multiply(4, multiply(160, 2))))), 2)) | multiply(n0,n1)|negate(n2)|power(n2,n1)|multiply(n2,#0)|add(#3,#2)|sqrt(#4)|add(#1,#5)|divide(#6,n1)|divide(n0,#7)| | geometry |
8 machines can do a work in 10 days . how many machines are needed to complete the work in 40 days ? | "required number of machines = 8 * 10 / 40 = 2 answer is a" | a ) 2 , b ) 6 , c ) 4 , d ) 7 , e ) 5 | a | divide(multiply(8, 10), 40) | multiply(n0,n1)|divide(#0,n2)| | physics |
a ' s speed is 15 / 13 times that of b . if a and b run a race , what part of the length of the race should a give b as a head start , so that b beats a by 40 % of the length of the race ? | "first calculate the distance , b has covered with his speed in the time , in which a reached 60 % of the race . then add the remaining distance as head start for b to win the race . its best to apply ratios concept here . since a ' s speed is 15 / 13 of b , therefore , b ' s speed is 13 / 15 of a distance covered by b... | a ) 44 % , b ) 48 % , c ) 52 % , d ) 42 % , e ) 46 % | b | multiply(subtract(const_1, multiply(divide(subtract(const_100, 40), const_100), divide(13, 15))), const_100) | divide(n1,n0)|subtract(const_100,n2)|divide(#1,const_100)|multiply(#2,#0)|subtract(const_1,#3)|multiply(#4,const_100)| | general |
what is the radius of the incircle of the triangle whose sides measure 5 , 12 and 13 units ? | explanatory answer in a right angled triangle , the radius of the incircle = s - h , where ' s ' is the semi perimeter of the triangle and ' r ' is the radius of the inscribed circle . the semi perimeter of the triangle = a + b + c / 2 = 5 + 12 + 13 / 2 = 15 therefore , r = 15 - 13 = 2 units . choice a | ['a ) 2 units', 'b ) 12 units', 'c ) 6.5 units', 'd ) 6 units', 'e ) 7.5 units'] | a | divide(multiply(5, 12), add(add(5, 12), 13)) | add(n0,n1)|multiply(n0,n1)|add(n2,#0)|divide(#1,#2) | geometry |
along a yard 414 metres long , 24 trees are palnted at equal distances , one tree being at each end of the yard . what is the distance between two consecutive trees | explanation : 24 trees have 23 gaps between them , required distance ( 414 / 23 ) = 18 option a | a ) 18 , b ) 19 , c ) 10 , d ) 11 , e ) 12 | a | divide(414, subtract(24, const_1)) | subtract(n1,const_1)|divide(n0,#0) | physics |
at the opening of a trading day at a certain stock exchange , the price per share of stock k was $ 8 . if the price per share of stock k was $ 9 at the closing of the day , what was the percent increase in the price per share of stock k for that day ? | "opening = 8 closing = 9 rise in price = 1 so , percent increase = 18 ∗ 10018 ∗ 100 = > 12.50 % thus answer will be ( d )" | a ) 1.4 % , b ) 5.9 % , c ) 11.1 % , d ) 12.5 % , e ) 23.6 % | d | multiply(subtract(divide(9, 8), const_1), const_100) | divide(n1,n0)|subtract(#0,const_1)|multiply(#1,const_100)| | general |
on a certain road 10 % of the motorists exceed the posted speed limit and receive speeding tickets , but 22 % of the motorists who exceed the posted speed limit do not receive speeding tickets . what percent of the motorists on the road exceed the posted speed limit ? | "answer is b . this question is in the og and thus well explained by ets . those who exceed : x so x = 10 % + 0,22 x id est x = 12,8 %" | a ) 10.5 % , b ) 12.8 % , c ) 15 % , d ) 22 % , e ) 30 % | b | divide(const_100, multiply(multiply(divide(10, const_100), divide(22, const_100)), const_100)) | divide(n0,const_100)|divide(n1,const_100)|multiply(#0,#1)|multiply(#2,const_100)|divide(const_100,#3)| | gain |
a fruit seller had some oranges . he sells 40 % oranges and still has 480 oranges . how many oranges he had originally ? | "60 % of oranges = 480 100 % of oranges = ( 480 × 100 ) / 6 = 800 total oranges = 700 answer : a" | a ) 800 , b ) 710 , c ) 720 , d ) 730 , e ) 740 | a | add(480, multiply(480, divide(40, const_100))) | divide(n0,const_100)|multiply(n1,#0)|add(n1,#1)| | gain |
fresh grapes contain 70 % by weight while dried grapes contain 10 % water by weight . what is the weight of dry grapes available from 100 kg of fresh grapes ? | "from the question we know : 100 kg * 70 % = 70 kg of water in the fresh grapes 100 kg - 70 kg of water = 30 kg of non - water mass we are looking for the weight of the dry grapes ( x ) . since the question tells us that 10 % of the weight of the dry graps is water and we know that 30 kg is non - water mass we can set ... | a ) 18.33 kg , b ) 21.50 kg , c ) 25.00 kg , d ) 33.33 kg , e ) 40.39 kg | d | multiply(divide(divide(multiply(subtract(const_100, 70), 100), const_100), subtract(const_100, 10)), const_100) | subtract(const_100,n0)|subtract(const_100,n1)|multiply(n2,#0)|divide(#2,const_100)|divide(#3,#1)|multiply(#4,const_100)| | gain |
a hall 20 m long and 15 m broad is surrounded by a verandah of uniform width of 2.5 m . the cost of flooring the verandah at the rate of rs . 3.50 per sq . meter is | area of verandah = [ ( 25 ã ƒ â € ” 20 ) ? ( 20 ã ƒ â € ” 15 ) ] m â ² = 200 m â ² cost of flooring = rs . ( 200 ã ƒ â € ” 3.50 ) = rs . 700 answer : d | a ) 500 , b ) 600 , c ) 650 , d ) 700 , e ) 750 | d | multiply(subtract(multiply(add(20, multiply(2.5, const_2)), add(15, multiply(2.5, const_2))), multiply(20, 15)), 3.5) | multiply(n2,const_2)|multiply(n0,n1)|add(n0,#0)|add(n1,#0)|multiply(#2,#3)|subtract(#4,#1)|multiply(n3,#5) | physics |
a man rows his boat 90 km downstream and 55 km upstream , taking 3 hours each time . find the speed of the stream ? | "speed downstream = d / t = 90 / ( 3 ) = 30 kmph speed upstream = d / t = 55 / ( 3 ) = 18 kmph the speed of the stream = ( 30 - 18 ) / 2 = 6 kmph answer : b" | a ) 76 kmph , b ) 6 kmph , c ) 14 kmph , d ) 8 kmph , e ) 4 kmph | b | divide(subtract(divide(90, 3), divide(55, 3)), const_2) | divide(n0,n2)|divide(n1,n2)|subtract(#0,#1)|divide(#2,const_2)| | physics |
jim is able to sell a hand - carved statue for $ 540 which was a 35 % profit over his cost . how much did the statue originally cost him ? | 540 = 1.35 * x x = 540 / 1.35 = 400 $ 400 , which is ( a ) . | a ) $ 400.00 , b ) $ 412.40 , c ) $ 455.40 , d ) $ 474.90 , e ) $ 488.20 | a | divide(540, add(divide(35, const_100), const_1)) | divide(n1,const_100)|add(#0,const_1)|divide(n0,#1) | gain |
the speed at which a girl can row a boat in still water is 30 kmph . if she rows downstream , where the speed of current is 6 kmph , what time will he take to cover 240 metres ? | speed of the boat downstream = 30 + 6 = 36 kmph = 36 * 5 / 18 = 10 m / s hence time taken to cover 240 m = 240 / 10 = 24 seconds . answer : e | a ) 21 , b ) 22 , c ) 23 , d ) 20 , e ) 24 | e | divide(multiply(240, const_3_6), add(30, 6)) | add(n0,n1)|multiply(n2,const_3_6)|divide(#1,#0) | physics |
the standard deviation of a normal distribution of data is 2 , and 3 standard deviations below the mean is greater than 45 . what is a possible value for the mean of the distribution ? | "the standard deviation ( { sd } ) = 2 ; 3 standard deviations below the mean is greater than 45 : { mean } - 3 * { sd } > 45 ; { mean } - 6 > 45 ; { mean } > 51 . answer : a ." | a ) 52 , b ) 47 , c ) 48 , d ) 49 , e ) 50 | a | add(add(45, multiply(3, 2)), const_1) | multiply(n0,n1)|add(n2,#0)|add(#1,const_1)| | general |
if 13 = 13 w / ( 1 - w ) , then ( 5 w ) 2 = | "13 - 13 w = 13 w 26 w = 13 w = 1 / 2 5 w = 5 / 2 5 w * 2 = 5 / 2 * 2 = 5 answer : b" | a ) 1 / 4 , b ) 5 , c ) 1 , d ) 2 , e ) 3 | b | multiply(divide(13, add(13, 13)), 5) | add(n0,n0)|divide(n0,#0)|multiply(n3,#1)| | general |
two trains of length 100 m and 280 m are running towards each other on parallel lines at 42 kmph and 30 kmph respectively . in what time will they be clear of each other from the moment they meet ? | "relative speed = ( 42 + 30 ) * 5 / 18 = 4 * 5 = 20 mps . distance covered in passing each other = 100 + 280 = 380 m . the time required = d / s = 380 / 20 = 19 sec . answer : e" | a ) 10 sec , b ) 32 sec , c ) 82 sec , d ) 20 sec , e ) 19 sec | e | divide(add(100, 280), multiply(add(42, 30), const_0_2778)) | add(n0,n1)|add(n2,n3)|multiply(#1,const_0_2778)|divide(#0,#2)| | physics |
the ratio of radius of a circle and the side of a square is 2 : 9 . find the ratio of their areas : | "radius / side = 2 / 9 â ‡ ’ area of circle / area of square = 4 / 81 answer : d" | a ) 2 : 1 , b ) 4 : 7 , c ) 8 : 77 , d ) 4 : 81 , e ) none | d | power(divide(2, 9), 2) | divide(n0,n1)|power(#0,n0)| | geometry |
if 12 men can reap 120 acres of land in 36 days , how many acres of land can 24 men reap in 62 days ? | "12 men 120 acres 36 days 24 men ? 62 days 120 * 24 / 12 * 62 / 36 120 * 2 * 31 / 18 240 * 1.72 = 413.3 answer : b" | a ) 266.5 , b ) 413.3 , c ) 425.8 , d ) 189.3 , e ) 249.6 | b | multiply(120, multiply(divide(24, 12), divide(62, 36))) | divide(n3,n0)|divide(n4,n2)|multiply(#0,#1)|multiply(n1,#2)| | physics |
evaluate : 10010 - 12 * 3 * 2 = ? | "according to order of operations , 12 ? 3 ? 2 ( division and multiplication ) is done first from left to right 12 * * 2 = 4 * 2 = 8 hence 10010 - 12 * 3 * 2 = 10010 - 8 = 10002 correct answer e" | a ) 30002 , b ) 70002 , c ) 50002 , d ) 90002 , e ) 10002 | e | subtract(10010, multiply(multiply(12, 3), 2)) | multiply(n1,n2)|multiply(n3,#0)|subtract(n0,#1)| | general |
a car travels at a speed of 10 miles per hour . how far will it travel in 5 hours ? | "during each hour , the car travels 65 miles . for 5 hours it will travel 10 + 10 + 10 + 10 + 10 = 5 * 10 = 50 miles correct answer a" | a ) 50 miles , b ) 62 miles , c ) 325 miles , d ) 70 miles , e ) 20 miles | a | multiply(10, 5) | multiply(n0,n1)| | physics |
if 42 men do a work in 25 days , in how many days will 50 men do it ? | "42 * 25 = 50 * x x = 21 days answer : e" | a ) 23 , b ) 22 , c ) 19 , d ) 20 , e ) 21 | e | divide(multiply(42, 25), 50) | multiply(n0,n1)|divide(#0,n2)| | physics |
find large number from below question the difference of two numbers is 1345 . on dividing the larger number by the smaller , we get 6 as quotient and the 15 as remainder | let the smaller number be x . then larger number = ( x + 1345 ) . x + 1345 = 6 x + 15 5 x = 1330 x = 266 large number = 266 + 1345 = 1611 e | a ) 1235 , b ) 1345 , c ) 1678 , d ) 1767 , e ) 1611 | e | multiply(divide(subtract(1345, 15), subtract(6, const_1)), 6) | subtract(n0,n2)|subtract(n1,const_1)|divide(#0,#1)|multiply(n1,#2) | general |
company s produces two kinds of stereos : basic and deluxe . of the stereos produced by company s last month , 2 / 3 were basic and the rest were deluxe . if it takes 9 / 5 as many hours to produce a deluxe stereo as it does to produce a basic stereo , then the number of hours it took to produce the deluxe stereos last... | "# of basic stereos was 2 / 3 of total and # of deluxe stereos was 1 / 3 of total , let ' s assume total = 15 , then basic = 10 and deluxe = 5 . now , if time needed to produce one deluxe stereo is 1 unit than time needed to produce one basic stereo would be 9 / 5 units . total time for basic would be 10 * 1 = 10 and t... | a ) 7 / 17 , b ) 14 / 31 , c ) 7 / 15 , d ) 17 / 35 , e ) 9 / 19 | e | divide(add(3, 2), const_10) | add(n0,n1)|divide(#0,const_10)| | general |
a train 250 meters long completely crosses a 300 meters long bridge in 45 seconds . what is the speed of the train is ? | "s = ( 250 + 300 ) / 45 = 550 / 45 * 18 / 5 = 44 answer : d" | a ) 32 , b ) 28 , c ) 29 , d ) 44 , e ) 21 | d | divide(divide(add(250, 300), const_1000), divide(45, const_3600)) | add(n0,n1)|divide(n2,const_3600)|divide(#0,const_1000)|divide(#2,#1)| | physics |
the salaries of a and b together amount to $ 500 . a spends 90 % of his salary and b , 80 % of his . if now , their savings are the same , what is a ' s salary ? | "let a ' s salary is x b ' s salary = 500 - x ( 100 - 90 ) % of x = ( 100 - 80 ) % of ( 500 - x ) x = $ 34 ( approximately ) answer is d" | a ) $ 50 , b ) $ 150 , c ) $ 69 , d ) $ 34 , e ) $ 52 | d | subtract(500, divide(500, add(divide(subtract(const_100, 80), subtract(const_100, 90)), const_1))) | subtract(const_100,n2)|subtract(const_100,n1)|divide(#0,#1)|add(#2,const_1)|divide(n0,#3)|subtract(n0,#4)| | gain |
if there are 10 peanuts in a box and mary puts 8 more peanuts inside , how many peanuts are in the box ? | "10 + 8 = 18 correct answer is e ) 18" | a ) 8 , b ) 9 , c ) 10 , d ) 11 , e ) 18 | e | add(10, 8) | add(n0,n1)| | general |
how many positive integers less than 1960 are such that the product of their digits is 10 . | "1960 is divisible by 10 ( or 2 * 5 ) = 196 196 is divisible by 4 = > 49 49 = 7 * 7 hence 1960 is divisible by 2 , 4,5 , 7 , 7 if there were only one 7 , i . e if the problem were 280 c" | a ) 220 , b ) 230 , c ) 280 , d ) 220 , e ) 240 | c | divide(factorial(subtract(add(const_4, 10), const_1)), multiply(factorial(10), 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 |
the average runs of a cricket player of 10 innings was 32 . how many runs must he makes ih his next innings so as to increase his average of runs by 4 ? | "sol . average after 11 innings = 36 . therefore required numbert of runs = ( 36 × 11 ) – ( 32 × 10 ) = 396 - 320 = 76 . answer d" | a ) 2 , b ) 4 , c ) 70 , d ) 76 , e ) none | d | subtract(multiply(add(10, const_1), add(32, 4)), multiply(32, 10)) | add(n0,const_1)|add(n1,n2)|multiply(n0,n1)|multiply(#0,#1)|subtract(#3,#2)| | general |
an outlet pipe can empty 2 / 3 of a cistern in 16 minutes . in 4 minutes , what part of the cistern will be emptied ? | "4 / 16 * 2 / 3 = 1 / 6 the answer is d ." | a ) 1 / 2 , b ) 1 / 3 , c ) 1 / 4 , d ) 1 / 6 , e ) 1 / 8 | d | divide(multiply(4, divide(2, 3)), 16) | divide(n0,n1)|multiply(n3,#0)|divide(#1,n2)| | physics |
the average monthly income of p and q is rs . 5050 . the average monthly income of q and r is rs . 6250 and the average monthly income of p and r is rs . 5200 . the monthly income of p is ? | "let p , q and r represent their respective monthly incomes . then , we have : p + q = ( 5050 x 2 ) = 10100 . . . . ( i ) q + r = ( 6250 x 2 ) = 12500 . . . . ( ii ) p + r = ( 5200 x 2 ) = 10400 . . . . ( iii ) adding ( i ) , ( ii ) and ( iii ) , we get : 2 ( p + q + r ) = 33000 or p + q + r = 16500 . . . . ( iv ) subt... | a ) rs . 3000 , b ) rs . 4000 , c ) rs . 5000 , d ) rs . 6000 , e ) rs . 7000 | b | subtract(add(5050, 5200), 6250) | add(n0,n2)|subtract(#0,n1)| | general |
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