Problem stringlengths 5 967 | Rationale stringlengths 1 2.74k | options stringlengths 37 164 | correct stringclasses 5 values | annotated_formula stringlengths 7 1.65k | linear_formula stringlengths 8 925 | category stringclasses 6 values | answer stringclasses 5 values |
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points a , b , c , and d , in that order , lie on a line . if ab = 2 cm , ac = 5 cm , and bd = 6 cm , what is cd , in centimeters ? | putting a value to each point , lets use the following : a - 0 b - 2 ( ab = 2 ) c - 5 ( ac = 5 ) d - 8 ( bd = 6 ) cd is 8 - 5 = 3 . ans c | a ) 1 , b ) 2 , c ) 3 , d ) 4 , e ) 5 | c | subtract(5, 2) | subtract(n1,n0) | physics | C |
a train 100 meters long completely crosses a 300 meters long bridge in 12 seconds . what is the speed of the train is ? | "s = ( 100 + 300 ) / 12 = 400 / 12 * 18 / 5 = 120 answer : c" | a ) 32 kmph , b ) 76 kmph , c ) 120 kmph , d ) 43 kmph , e ) 40 kmph | c | divide(divide(add(100, 300), const_1000), divide(12, const_3600)) | add(n0,n1)|divide(n2,const_3600)|divide(#0,const_1000)|divide(#2,#1)| | physics | C |
express a speed of 72 kmph in meters per second ? | "72 * 5 / 18 = 20 mps answer : c" | a ) 10 mps , b ) 76 mps , c ) 20 mps , d ) 97 mps , e ) 16 mps | c | multiply(const_0_2778, 72) | multiply(n0,const_0_2778)| | physics | C |
what will be the difference between simple and compound interest at 6 % per annum on a sum of rs . 1000 after 4 years ? | "s . i . = ( 1000 * 6 * 4 ) / 100 = rs . 240 c . i . = [ 1000 * ( 1 + 6 / 100 ) 4 - 1000 ] = rs . 262.5 difference = ( 262.5 - 240 ) = rs . 22.5 answer : e" | a ) 24.19 , b ) 24.12 , c ) 22.1 , d ) 24.1 , e ) 22.5 | e | subtract(subtract(multiply(1000, power(add(divide(6, const_100), const_1), 4)), 1000), multiply(multiply(1000, divide(6, const_100)), 4)) | divide(n0,const_100)|add(#0,const_1)|multiply(n1,#0)|multiply(n2,#2)|power(#1,n2)|multiply(n1,#4)|subtract(#5,n1)|subtract(#6,#3)| | gain | E |
7 machines at a certain factory operate at the same constant rate . if 6 of these machines , operating simultaneously , take 42 hours to fill a certain production order , how many fewer hours does it take all 7 machines , operating simultaneously , to fill the same production order ? | the total work is 6 * 42 = 252 machine hours the time required for seven machines is 252 / 7 = 36 hours , thus 6 fewer hours . the answer is d . | a ) 3 , b ) 4 , c ) 5 , d ) 6 , e ) 7 | d | subtract(42, divide(multiply(6, 42), 7)) | multiply(n1,n2)|divide(#0,n0)|subtract(n2,#1) | other | D |
the average ( arithmetic mean ) of 14 , 32 , and 53 is 3 more than the average of 21 , 47 , and x . what is x ? | "the average of 14 , 32 , and 53 is 33 . the average of 21 , 47 and x is 30 . then 21 + 47 + x = 90 . x = 22 . the answer is c ." | a ) 20 , b ) 21 , c ) 22 , d ) 23 , e ) 24 | c | subtract(add(add(14, 32), 53), add(add(multiply(3, const_3), 21), 47)) | add(n0,n1)|multiply(n3,const_3)|add(n2,#0)|add(n4,#1)|add(n5,#3)|subtract(#2,#4)| | general | C |
the present ratio of students to teachers at a certain school is 45 to 1 . if the student enrollment were to increase by 50 students and the number of teachers were to increase by 5 , the ratio of students to teachers would then be 25 to 1 . what is the present number of teachers ? | "we are given that the ratio of students to teacher is 45 to 1 . we can rewrite this using variable multipliers . students : teachers = 45 x : x we are next given that student enrollment increases by 50 and the number of teachers increases by 5 . with this change the new ratio becomes 25 to 1 . we can put all this into an equation : students / teachers ο 25 / 1 = ( 45 x + 50 ) / ( x + 5 ) if we cross multiply we have : 25 ( x + 5 ) = 45 x + 50 25 x + 125 = 45 x + 50 3.75 = x x ~ 4 since x is the present number of teachers , currently there are 4 teachers . answer a ." | a ) 4 , b ) 8 , c ) 10 , d ) 12 , e ) 15 | a | divide(add(45, 25), 25) | add(n0,n4)|divide(#0,n4)| | other | A |
an isosceles triangle those sides are 13 cm , 13 cm , 10 cm long inscribed in a circle . find the radius of the circle | a , b , c are sides of triangle , ad is the median , ad ^ 2 = ab ^ 2 - bd ^ 2 then ad = 12 , the centriod median divides into 2 : 1 ratio so 12 will be divide into 2 : 1 ratio so 8 is radius . so ans is 8 cm answer : a | ['a ) 8 cm', 'b ) 9 cm', 'c ) 10 cm', 'd ) 11 cm', 'e ) 7 cm'] | a | add(divide(add(subtract(power(13, const_2), power(divide(10, const_2), const_2)), power(divide(10, const_2), const_2)), multiply(sqrt(subtract(power(13, const_2), power(divide(10, const_2), const_2))), const_2)), const_1) | divide(n2,const_2)|power(n0,const_2)|power(#0,const_2)|subtract(#1,#2)|add(#2,#3)|sqrt(#3)|multiply(#5,const_2)|divide(#4,#6)|add(#7,const_1) | geometry | A |
if a certain number is divided by 12 , the quotient , dividend , and divisor , added together , will amount to 64 . what is the number ? | "let x = the number sought . then x / 12 + x + 12 = 64 . and x - 624 / 13 = 48 ." | a ) 18 , b ) 28 , c ) 48 , d ) 38 , e ) 59 | a | divide(multiply(subtract(64, 12), 12), add(12, const_1)) | add(n0,const_1)|subtract(n1,n0)|multiply(n0,#1)|divide(#2,#0)| | general | A |
a man complete a journey in 10 hours . he travels first half of the journey at the rate of 20 km / hr and second half at the rate of 20 km / hr . find the total journey in km . | "0.5 x / 20 + 0.5 x / 20 = 10 - - > x / 20 + x / 20 = 20 - - > 2 x = 20 x 20 - - > x = ( 20 x 20 ) / 2 = 200 km . answer : e ." | a ) 220 km , b ) 224 km , c ) 230 km , d ) 232 km , e ) 200 km | e | multiply(const_2, divide(multiply(multiply(20, 20), 10), add(20, 20))) | add(n1,n2)|multiply(n1,n2)|multiply(n0,#1)|divide(#2,#0)|multiply(#3,const_2)| | physics | E |
two trains travel in opposite directions at 36 kmph and 45 kmph and a man sitting in slower train passes the faster train in 10 seconds . the length of the faster train is | "solution relative speed = ( 36 + 45 ) km / hr = ( 81 x 5 / 18 ) m / sec = ( 45 / 2 ) m / sec length of the train = ( 45 / 2 x 10 ) m = 225 m . answer a" | a ) 225 m , b ) 100 m , c ) 120 m , d ) 180 m , e ) none | a | multiply(multiply(add(36, 45), const_0_2778), 10) | add(n0,n1)|multiply(#0,const_0_2778)|multiply(n2,#1)| | physics | A |
if y is the smallest positive integer such that 6300 multiplied by y is the square of an integer , then y must be | "6300 = 2 ^ 2 * 3 ^ 2 * 5 ^ 2 * 7 to be perfect square , we need to multiply by at least 7 . the answer is b ." | a ) 5 , b ) 7 , c ) 15 , d ) 35 , e ) 45 | b | multiply(add(const_2, const_3), const_2) | add(const_2,const_3)|multiply(#0,const_2)| | geometry | B |
61 + 5 * 12 / ( 180 / 3 ) = ? | "61 + 5 * 12 / ( 180 / 3 ) = 61 + 5 * 12 / ( 60 ) = 61 + ( 5 * 12 ) / 60 = 61 + 1 = 62 . answer : d" | a ) 22 , b ) 77 , c ) 29 , d ) 62 , e ) 21 | d | add(61, divide(multiply(5, 12), divide(180, 3))) | divide(n3,n4)|multiply(n1,n2)|divide(#1,#0)|add(n0,#2)| | general | D |
the number 130 can be written as the sum of the squares of 2 different positive integers . what is the sum of these 2 integers ? | "11 ^ 2 + 3 ^ 2 = 130 - - > 11 + 3 = 14 . d" | a ) 17 , b ) 16 , c ) 15 , d ) 14 , e ) 13 | d | add(add(add(const_4, const_3), add(const_3, const_2)), 2) | add(const_3,const_4)|add(const_2,const_3)|add(#0,#1)|add(n1,#2)| | geometry | D |
what is the rate percent when the simple interest on rs . 1200 amount to rs . 400 in 4 years ? | "interest for 4 yrs = 400 interest for 1 yr = 100 interest rate = 100 / 1200 x 100 = 8.33 % answer : a" | a ) 8.33 % , b ) 6 % , c ) 2 % , d ) 95 % , e ) 1 % | a | divide(multiply(const_100, 400), multiply(1200, 4)) | multiply(n1,const_100)|multiply(n0,n2)|divide(#0,#1)| | gain | A |
how many trucks are there if each truck carrying 70 packages and total of 490 packages ? | sol . total packages 490 each truck carries 70 packages = 490 / 70 = 7 answer : a | a ) a ) 7 , b ) b ) 6 , c ) c ) 9 , d ) d ) 11 , e ) e ) none of the above | a | divide(490, 70) | divide(n1,n0) | general | A |
if the speed of a man is 51 km per hour , then what is the distance traveled by him in 30 seconds ? | "the distance traveled in 30 sec = 51 * ( 5 / 18 ) * 30 = 425 m answer : d" | a ) 275 m , b ) 360 m , c ) 375 m , d ) 425 m , e ) 440 m | d | multiply(multiply(51, const_0_2778), 30) | multiply(n0,const_0_2778)|multiply(n1,#0)| | physics | D |
4.2 x 4.2 - 1.9 x 1.9 / 2.3 x 6.1 is equal to : | "given expression = ( a 2 - b 2 ) / ( a + b ) ( a - b ) = ( a 2 - b 2 ) / ( a 2 - b 2 ) = 1 . answer is b ." | a ) 2 , b ) 1 , c ) 3 , d ) 4 , e ) 8 | b | divide(subtract(multiply(4.2, 4.2), multiply(1.9, 1.9)), 2.3) | multiply(n0,n1)|multiply(n2,n3)|subtract(#0,#1)|divide(#2,n4)| | general | B |
let f ( x ) = x ^ 2 + bx + c . if f ( 1 ) = 0 and f ( - 7 ) = 0 , then f ( x ) crosses the y - axis at what y - coordinate ? | "when x = 1 and when x = - 7 , the expression f ( x ) = x Β² + bx + c equals 0 . then f ( x ) = ( x - 1 ) ( x + 7 ) f ( 0 ) = - 7 the answer is a ." | a ) - 7 , b ) - 1 , c ) 0 , d ) 1 , e ) 7 | a | negate(divide(subtract(power(7, 2), 1), add(7, 1))) | add(n3,n1)|power(n3,n0)|subtract(#1,n1)|divide(#2,#0)|negate(#3)| | general | A |
if the numbers 1 to 96 are written on 96 pieces of paper , ( one on each ) and one piece is picked at random , then what is the probability that the number drawn is neither prime nor composite ? | there are 25 primes , 70 composite numbers from 1 to 96 . the number which is neither prime nor composite is 1 . therefore , required probability = 1 / 96 . answer : c | a ) 1 / 50 , b ) 1 / 25 , c ) 1 / 96 , d ) 1 , e ) 2 | c | divide(1, 96) | divide(n0,n1) | other | C |
a shopkeeper sells 20 % of his stock at 25 % profit ans sells the remaining at a loss of 5 % . he incurred an overall profit of rs . 400 . find the total worth of the stock ? | "let the total worth of the stock be rs . x . the sp of 20 % of the stock = 1 / 5 * x * 5 / 4 = 5 x / 20 the sp of 80 % of the stock = 4 / 5 * x * 0.95 = 152 x / 200 total sp = 50 x / 200 + 152 x / 200 = 202 x / 200 overall profit = 152 x / 200 - x = 2 x / 200 x / 50 = 400 = > x = 40000 answer : e" | a ) 25500 , b ) 25000 , c ) 40500 , d ) 20000 , e ) 40000 | e | divide(400, subtract(multiply(divide(5, const_100), divide(subtract(const_100, 20), const_100)), multiply(divide(25, const_100), divide(20, const_100)))) | divide(n2,const_100)|divide(n1,const_100)|divide(n0,const_100)|subtract(const_100,n0)|divide(#3,const_100)|multiply(#1,#2)|multiply(#0,#4)|subtract(#6,#5)|divide(n3,#7)| | gain | E |
the product of two successive numbers is 462 . which is the smaller of the two numbers ? | a 21 from the given alternatives , 21 Γ 22 = 462 β΄ smaller number = 21 | a ) 21 , b ) 23 , c ) 26 , d ) 27 , e ) 29 | a | sqrt(462) | sqrt(n0) | general | A |
the number x of cars sold each week varies with the price y in dollars according to the equation x = 800000 β 50 y . what would be the total weekly revenue , in dollars , from the sale of cars priced at $ 12000 ? | number of cars sold = x = 800000 - 50 y y = 12000 x = 800000 - 600000 = 200000 revenue from 50000 cars = 12000 * 200000 = 2400000000 answer e | a ) 50,000 , b ) 750,000 , c ) 850,000 , d ) 7 , 500,000 , e ) 2,400 , 000,000 | e | add(divide(divide(divide(multiply(subtract(800000, multiply(50, 12000)), 12000), const_1000), const_1000), const_3), 50) | multiply(n1,n2)|subtract(n0,#0)|multiply(n2,#1)|divide(#2,const_1000)|divide(#3,const_1000)|divide(#4,const_3)|add(n1,#5) | general | E |
the least number which must be subtracted from 820 to make it exactly divisible by 9 is : | "on dividing 820 by 9 , we get remainder = 1 therefore , required number to be subtracted = 1 answer : c" | a ) a ) 2 , b ) b ) 3 , c ) c ) 1 , d ) d ) 5 , e ) e ) 6 | c | subtract(820, multiply(add(multiply(add(const_4, const_1), const_10), add(const_4, const_2)), 9)) | add(const_2,const_4)|add(const_1,const_4)|multiply(#1,const_10)|add(#0,#2)|multiply(n1,#3)|subtract(n0,#4)| | general | C |
in the fifth grade at parkway elementary school there are 420 students . 312 students are boys and 250 students are playing soccer . 86 % of the students that play soccer are boys . how many girl student are in parkway that is not playing soccer ? | total students = 420 boys = 312 , girls = 108 total playing soccer = 250 86 % of 250 = 215 are boys who play soccer . girls who play soccer = 35 . total girls who do not play soccer = 108 - 35 = 73 . correct option : b | a ) 69 . , b ) 73 . , c ) 81 , d ) 91 , e ) 108 | b | subtract(subtract(420, 312), subtract(250, divide(multiply(250, 86), const_100))) | multiply(n2,n3)|subtract(n0,n1)|divide(#0,const_100)|subtract(n2,#2)|subtract(#1,#3) | gain | B |
a man ' s speed with the current is 21 km / hr and the speed of the current is 2.5 km / hr . the man ' s speed against the current is ? | "man ' s speed with the current = 21 km / hr = > speed of the man + speed of the current = 21 km / hr speed of the current is 2.5 km / hr hence , speed of the man = 21 - 2.5 = 18.5 km / hr man ' s speed against the current = speed of the man - speed of the current = 18.5 - 2.5 = 16 km / hr answer is c ." | a ) 10 , b ) 20 , c ) 16 , d ) 30 , e ) 40 | c | subtract(subtract(21, 2.5), 2.5) | subtract(n0,n1)|subtract(#0,n1)| | gain | C |
a works twice as fast as b . if b can complete a work in 12 days independently , the number of days in which a and b can together finish the work is : | solution rates of worktio of raing of a and b = 2 : 1 . so , ratio of times taken = 1 : 2 β΄ a β s 1 day β s work = 1 / 6 b β s 1 day β s work = 1 / 12 ( a + b ) ' s 1 day β s work = ( 1 / 6 + 1 / 12 ) = 3 / 12 = 1 / 4 so , a and b together can finish the work in 4 days . answer a | a ) 4 days , b ) 6 days , c ) 8 days , d ) 18 days , e ) none of these | a | inverse(add(divide(const_1, 12), multiply(const_2, divide(const_1, 12)))) | divide(const_1,n0)|multiply(#0,const_2)|add(#0,#1)|inverse(#2) | physics | A |
on july 1 of last year , total employees at company e was decreased by 10 percent . without any change in the salaries of the remaining employees , the average ( arithmetic mean ) employee salary was 10 percent more after the decrease in the number of employees than before the decrease . the total of the combined salaries of all the employees at company e after july 1 last year was what percent q of thatbeforejuly 1 last year ? | "the total number of employees = n the average salary = x total salary to all emplyoees = xn after the total number of employees = n - 0.1 n = 0.9 n the average salary = x + 10 % of x = 1.1 x total salary to all emplyoees = 0.9 n ( 1.1 x ) total salary after as a % of total salary before q = [ 0.9 n ( 1.1 x ) ] / xn = 0.99 or 99 % = b" | a ) 90 % , b ) 99 % , c ) 100 % , d ) 101 % , e ) 110 % | b | multiply(10, 10) | multiply(n1,n1)| | general | B |
the simple form of the ratio 8 / 5 : 2 / 3 is ? | "8 / 5 : 2 / 3 = 12 : 5 answer : a" | a ) 12 : 5 , b ) 12 : 7 , c ) 7 : 12 , d ) 5 : 7 , e ) 4 : 7 | a | divide(2, 3) | divide(n2,n3)| | other | A |
rs . 1500 is divided into two parts such that if one part is invested at 6 % and the other at 5 % the whole annual interest from both the sum is rs . 81 . how much was lent at 5 % ? | "( x * 5 * 1 ) / 100 + [ ( 1500 - x ) * 6 * 1 ] / 100 = 81 5 x / 100 + 90 β 6 x / 100 = 81 x / 100 = 9 = > x = 900 answer : d" | a ) 299 , b ) 666 , c ) 778 , d ) 900 , e ) 977 | d | multiply(add(5, 6), const_100) | add(n1,n2)|multiply(#0,const_100)| | gain | D |
find the area of trapezium whose parallel sides are 20 cm and 18 cm long , and the distance between them is 12 cm | "area of a trapezium = 1 / 2 ( sum of parallel sides ) * ( perpendicular distance between them ) = 1 / 2 ( 20 + 18 ) * ( 12 ) = 228 cm 2 answer : e" | a ) 178 cm 2 , b ) 179 cm 2 , c ) 285 cm 2 , d ) 167 cm 2 , e ) 228 cm 2 | e | quadrilateral_area(12, 18, 20) | quadrilateral_area(n2,n1,n0)| | physics | E |
evaluate : ( 1 - 1 / 10 ) ( 1 - 1 / 11 ) ( 1 - 1 / 12 ) . . . ( 1 - 1 / 99 ) ( 1 - 1 / 100 ) | ( 1 - 1 / 10 ) ( 1 - 1 / 11 ) ( 1 - 1 / 12 ) . . . ( 1 - 1 / 99 ) ( 1 - 1 / 100 ) = ( 9 / 10 ) ( 10 / 11 ) ( 11 / 12 ) . . . ( 98 / 99 ) ( 99 / 100 ) = 9 / 100 : simplify correct answer e | a ) 5 / 100 , b ) 6 / 100 , c ) 7 / 100 , d ) 8 / 100 , e ) 9 / 100 | e | divide(subtract(10, 1), 100) | subtract(n2,n0)|divide(#0,n14) | general | E |
if n is a positive integer and the product of all integers from 1 to n , inclusive , is a multiple of 210 , what is the least possible value of n ? | "210 = 2 * 3 * 5 * 7 , so n must be at least 7 . the answer is a ." | a ) 7 , b ) 9 , c ) 11 , d ) 13 , e ) 14 | a | divide(divide(divide(divide(210, const_2), const_3), const_4), divide(const_10, const_2)) | divide(n1,const_2)|divide(const_10,const_2)|divide(#0,const_3)|divide(#2,const_4)|divide(#3,#1)| | general | A |
a and b β s salaries together amount to rs . 3,000 . a spends 95 % of his salary and b spends 85 % of his . if now their savings are the same , what is a β s salary ? | "( 5 / 100 ) a = ( 15 / 100 ) b a = 3 b a + b = 3000 4 b = 3000 = > b = 750 a = 2250 answer b" | a ) 2777 , b ) 2250 , c ) 2789 , d ) 2776 , e ) 2881 | b | divide(multiply(multiply(multiply(const_2, multiply(const_4, add(const_2, const_3))), const_100), subtract(const_1, divide(85, const_100))), add(subtract(const_1, divide(95, const_100)), subtract(const_1, divide(85, const_100)))) | add(const_2,const_3)|divide(n2,const_100)|divide(n1,const_100)|multiply(#0,const_4)|subtract(const_1,#1)|subtract(const_1,#2)|add(#5,#4)|multiply(#3,const_2)|multiply(#7,const_100)|multiply(#8,#4)|divide(#9,#6)| | gain | B |
108 . triangle a β s base is 9 % greater than the base of triangle b , and a β s height is 9 % less than the height of triangle b . the area of triangle a is what percent less or more than the area of triangle b ? | "wish the question specified that we are talking about corresponding height . base of a = 10 / 9 * base of b height of a = 8 / 9 * height of b area of a = ( 1 / 2 ) * base of a * height of a = 10 / 9 * 8 / 9 * area of b = 80 / 81 * area of b area of a is 1.3 % more than the area of b . answer ( d )" | a ) 9 % less , b ) 1 % less , c ) equal to each other , d ) 1.3 % more , e ) 9 % more | d | divide(const_100, subtract(multiply(const_100, const_100), multiply(add(const_100, 9), subtract(const_100, 9)))) | add(n1,const_100)|multiply(const_100,const_100)|subtract(const_100,n1)|multiply(#0,#2)|subtract(#1,#3)|divide(const_100,#4)| | geometry | D |
total 85 matches are conducted in knockout match type . how many players will be participated in that tournament ? | 84 players answer : e | a ) 85 , b ) 81 , c ) 53 , d ) 82 , e ) 84 | e | add(85, const_1) | add(n0,const_1) | general | E |
the least number , which when divided by 12 , 31 , 20 and 54 leaves in each case a remainder of 8 is : | "required number = ( l . c . m . of 12 , 31 , 20 , 54 ) + 8 = 16740 + 8 = 16748 . answer : c" | a ) 448 , b ) 488 , c ) 16748 , d ) 548 , e ) 560 | c | multiply(54, const_10) | multiply(n3,const_10)| | general | C |
it takes joey the postman 1 hours to run a 4 mile long route every day . he delivers packages and then returns to the post office along the same path . if the average speed of the round trip is 5 mile / hour , what is the speed with which joey returns ? | "let his speed for one half of the journey be 4 miles an hour let the other half be x miles an hour now , avg speed = 5 mile an hour 2 * 4 * x / 4 + x = 5 8 x = 5 x + 20 = > 3 x = 20 = > x = 6 2 / 3 a" | a ) 6 2 / 3 , b ) 12 , c ) 13 , d ) 14 , e ) 15 | a | divide(4, subtract(divide(multiply(const_2, 4), 5), 1)) | multiply(n1,const_2)|divide(#0,n2)|subtract(#1,n0)|divide(n1,#2)| | physics | A |
if n is a positive integer and the product of all the integers from 1 to n , inclusive , is a multiple of 170 , what is the least possible value of n ? | 17 , since n needs to be the largest of the prime factors . therefore , the answer is d . | a ) 14 , b ) 13 , c ) 15 , d ) 17 , e ) 16 | d | divide(170, const_2) | divide(n1,const_2)| | general | D |
a train running at the speed of 56 km / hr crosses a pole in 9 sec . what is the length of the train ? | "speed = 56 * 5 / 18 = 140 / 9 m / sec length of the train = speed * time = 140 / 9 * 9 = 140 m answer : a" | a ) 140 m , b ) 786 m , c ) 566 m , d ) 546 m , e ) 445 m | a | multiply(divide(multiply(56, const_1000), const_3600), 9) | multiply(n0,const_1000)|divide(#0,const_3600)|multiply(n1,#1)| | physics | A |
a women purchased 3 towels @ rs . 100 each , 5 towels @ rs . 150 each and two towels at a certain rate which is now slipped off from his memory . but she remembers that the average price of the towels was rs . 170 . find the unknown rate of two towels ? | "10 * 170 = 1700 3 * 100 + 5 * 150 = 1050 1700 β 1050 = 650 b" | a ) a ) 400 , b ) b ) 650 , c ) c ) 500 , d ) d ) 550 , e ) e ) 600 | b | subtract(subtract(multiply(add(add(3, 5), const_2), 170), multiply(5, 150)), multiply(3, 100)) | add(n0,n2)|multiply(n2,n3)|multiply(n0,n1)|add(#0,const_2)|multiply(n4,#3)|subtract(#4,#1)|subtract(#5,#2)| | general | B |
how many shots of 1 cm radius can be prepared from a sphere of 4 cm radius ? | "4 / 3 Γ― β¬ * 4 * 4 * 4 = 4 / 3 Γ― β¬ * 1 * 1 * 1 * x x = 64 answer : d" | a ) 33 , b ) 88 , c ) 27 , d ) 64 , e ) 99 | d | multiply(power(4, const_2), 4) | power(n1,const_2)|multiply(n1,#0)| | physics | D |
in a single throw of a die , what is the probability of getting a number greater than 4 ? | s = { 1,2 , 3,4 , 5,6 } e = { 5,6 } probability = 2 / 6 = 1 / 3 answer is a | a ) 1 / 3 , b ) 1 / 4 , c ) 2 / 3 , d ) 2 / 5 , e ) 3 / 7 | a | divide(const_2, add(4, const_2)) | add(n0,const_2)|divide(const_2,#0) | probability | A |
5 / 8 th of a job is completed in 10 days . if a person works at the same pace , how many days will he take to complete the job ? | explanation : solution : it is given that 5 / 8 th of the work is completed in 10 days . = > remaining work = 3 / 8 th of total applying unitary method : total work will be completed in 10 * 8 / 5 days = > it takes 16 days to complete total work = > hence , remaining work days = 16 - 10 = 6 days answer : c | a ) 8 , b ) 9 , c ) 6 , d ) 2 , e ) 1 | c | multiply(multiply(subtract(const_1, divide(5, 8)), 10), divide(8, 5)) | divide(n1,n0)|divide(n0,n1)|subtract(const_1,#1)|multiply(n2,#2)|multiply(#0,#3) | physics | C |
if a - b = 3 and a ^ 2 + b ^ 2 = 35 , find the value of ab . | "2 ab = ( a ^ 2 + b ^ 2 ) - ( a - b ) ^ 2 = 35 - 9 = 26 ab = 13 . answer is b ." | a ) 12 , b ) 13 , c ) 10 , d ) 18 , e ) 13 | b | multiply(multiply(add(3, divide(subtract(sqrt(35), 3), 2)), divide(subtract(sqrt(35), 3), 2)), 2) | sqrt(n3)|subtract(#0,n0)|divide(#1,n1)|add(n0,#2)|multiply(#3,#2)|multiply(n1,#4)| | general | B |
if 30 % of 100 is greater than 50 % of a number by 10 , then the number is ? | "let the number be x 30 % of 100 - 50 % of x = 10 50 / 100 x = ( 30 / 100 * 100 ) - 10 x / 2 = 20 x = 40 answer is a" | a ) 40 , b ) 30 , c ) 60 , d ) 80 , e ) 50 | a | divide(subtract(multiply(divide(30, const_100), 100), 10), divide(50, const_100)) | divide(n0,const_100)|divide(n2,const_100)|multiply(n1,#0)|subtract(#2,n3)|divide(#3,#1)| | gain | A |
a survey of employers found that during 1993 employment costs rose 3.5 percent , where employment costs consist of salary costs and fringe benefit costs . if salary costs rose 3 percent and fringe - benefit costs rose 5.5 percent during 1993 , then fringe - benefit costs represented what percent of employment costs at the beginning of 1993 ? | "the amount by which employment costs rose is equal to 0.035 ( salary costs + fringe benefit costs ) ; on the other hand the amount by which employment costs rose is equal to 0.03 * salary costs + 0.055 * fringe benefit costs ; so , 35 ( s + f ) = 30 s + 55 f - - > s = 4 f - - > f / s = 1 / 4 - - > f / ( s + f ) = 1 / ( 1 + 4 ) = 1 / 5 = 0.2 . answer : b ." | a ) 16.5 % , b ) 20 % , c ) 35 % , d ) 55 % , e ) 65 % | b | multiply(divide(subtract(3.5, 3), subtract(5.5, 3)), const_100) | subtract(n1,n2)|subtract(n3,n2)|divide(#0,#1)|multiply(#2,const_100)| | gain | B |
amar takes as much time in running 18 meters as a car takes in covering 48 meters . what will be the distance covered by amar during the time the car covers 1.6 km ? | "a 600 m distance covered by amar = 18 / 4.8 ( 1.6 km ) = 3 / 8 ( 1600 ) = 600 m" | a ) 600 m , b ) 300 m , c ) 400 m , d ) 700 m , e ) 500 m | a | divide(multiply(18, multiply(1.6, const_1000)), 48) | multiply(n2,const_1000)|multiply(n0,#0)|divide(#1,n1)| | physics | A |
a doctor prescribed 20 cubic centimeters of a certain drug to a patient whose body weight was 120 pounds . if the typical dosage is 2 cubic centimeters per 15 pounds of the body weight , by what percent was the prescribed dosage greater than the typical dosage ? | "typical dosage per 15 pound of the body weight = 2 c . c typical dosage per 120 pound of the body weight = 2 * ( 120 / 15 ) = 2 * 8 = 16 c . c dosage prescribed by doctor for 120 pound patient = 20 c . c % prescribed dosage greater than the typical dosage = ( 20 - 16 / 16 ) * 100 % = ( 2 / 16 ) * 100 % = 25 % answer a" | a ) 25 % , b ) 9 % , c ) 11 % , d ) 12.5 % , e ) 14.8 % | a | multiply(divide(subtract(multiply(divide(2, 15), 120), 20), multiply(divide(2, 15), 120)), const_100) | divide(n2,n3)|multiply(n1,#0)|subtract(#1,n0)|divide(#2,#1)|multiply(#3,const_100)| | gain | A |
donovan and michael are racing around a circular 500 - meter track . if donovan runs each lap in 45 seconds and michael runs each lap in 40 seconds , how many laps will michael have to complete in order to pass donovan , assuming they start at the same time ? | "one way of approaching this question is by relative speed method 1 . speed / rate of donovan = distance / time = > 500 / 45 = > 100 / 9 2 . speed / rate of michael = distance / time = > 500 / 40 = > 50 / 4 relative speed between them = 50 / 4 - 100 / 9 = > 25 / 36 ( we subtract the rates if moving in the same direction and add the rates if moving in the opposite direction ) in order to pass donovan - distance to be covered = 500 , relative rate = 50 / 36 total time taken by micheal to surpass donovan = distance / rate = > 500 * 36 / 50 = > 3600 / 10 = > 360 no . of laps taken by michael = total time / michael ' s rate = > 360 / 40 = > 9 hence correct answer is 9 laps . a" | a ) 9 , b ) 10 , c ) 11 , d ) 12 , e ) 13 | a | divide(divide(500, subtract(divide(500, 40), divide(500, 45))), 40) | divide(n0,n2)|divide(n0,n1)|subtract(#0,#1)|divide(n0,#2)|divide(#3,n2)| | physics | A |
if the price of an article went up by 15 % , then by what percent should it be brought down to bring it back to its original price ? | "let the price of the article be rs . 100 . 15 % of 100 = 15 new price = 100 + 15 = rs . 115 required percentage = ( 115 - 100 ) / 115 * 100 = 20 / 115 * 100 = 17.39 % . answer : a" | a ) 17.39 % , b ) 2 / 8 % , c ) 2 / 1 % , d ) 1 / 3 % , e ) 2 / 7 % | a | multiply(divide(15, add(const_100, 15)), const_100) | add(n0,const_100)|divide(n0,#0)|multiply(#1,const_100)| | gain | A |
for any integer n greater than 1 , n * denotes the product of all the integers from 1 to n , inclusive . how many prime numbers r are there between 6 * + 2 and 6 * + 6 , inclusive ? | "given that n * denotes the product of all the integers from 1 to n , inclusive so , 6 * + 2 = 6 ! + 2 and 6 * + 6 = 6 ! + 6 . now , notice that we can factor out 2 our of 6 ! + 2 so it can not be a prime number , we can factor out 3 our of 6 ! + 3 so it can not be a prime number , we can factor out 4 our of 6 ! + 4 so it can not be a prime number , . . . the same way for all numbers between 6 * + 2 = 6 ! + 2 and 6 * + 6 = 6 ! + 6 , inclusive . which means that there are no primes r in this range . answer : a ." | a ) none , b ) one , c ) two , d ) three , e ) four | a | divide(add(factorial(6), 6), add(factorial(6), 6)) | factorial(n2)|add(n2,#0)|divide(#1,#1)| | general | A |
a man can row downstream at the rate of 12 km / hr and upstream at 6 km / hr . find man ' s rate in still water and the rate of current ? | "explanation : rate of still water = 1 / 2 ( 12 + 6 ) = 9 km / hr rate of current = 1 / 2 ( 12 - 6 ) = 3 km / hr answer : option c" | a ) 9,6 , b ) 6,3 , c ) 9,3 , d ) 6,6 , e ) none of these | c | divide(subtract(12, 6), const_2) | subtract(n0,n1)|divide(#0,const_2)| | gain | C |
the sale price sarees listed for rs . 540 after successive discount is 5 % and 2 % is ? | "540 * ( 95 / 100 ) * ( 98 / 100 ) = 502 answer : c" | a ) 288 , b ) 500 , c ) 502 , d ) 540 , e ) 262 | c | subtract(subtract(540, divide(multiply(540, 5), const_100)), divide(multiply(subtract(540, divide(multiply(540, 5), const_100)), 2), const_100)) | multiply(n0,n1)|divide(#0,const_100)|subtract(n0,#1)|multiply(n2,#2)|divide(#3,const_100)|subtract(#2,#4)| | gain | C |
company z has 48 employees . if the number of employees having birthdays on wednesday is more than the number of employees having birthdays on any other day of the week , each of which have same number of birth - days , what is the minimum number of employees having birthdays on wednesday . | "say the number of people having birthdays on wednesday is x and the number of people having birthdays on each of the other 6 days is y . then x + 6 y = 48 . now , plug options for x . only a and e give an integer value for y . but only for e x > y as needed . answer : e ." | a ) 6 , b ) 7 , c ) 8 , d ) 9 , e ) 12 | e | add(const_4, add(floor(divide(48, add(const_4, const_3))), const_1)) | add(const_3,const_4)|divide(n0,#0)|floor(#1)|add(#2,const_1)|add(#3,const_4)| | general | E |
natasha climbs up a hill , and descends along the same way she went up . it takes her 4 hours to reach the top and 2 hours to come back down . if her average speed along the whole journey is 4 kilometers per hour , what was her average speed ( in kilometers per hour ) while climbing to the top ? | "let the distance to the top be x , so the total distance traveled by natasha is 2 x . the total time is 4 + 2 = 6 hours the average speed = total distance / total time taken = 2 x / 6 = x / 3 the average speed of the complete journey is 4 km / hour x / 3 = 4 x = 12 km the average speed while climbing = distance / time = 12 / 4 = 3 km / h the answer is c ." | a ) 2 , b ) 2.5 , c ) 3 , d ) 3.25 , e ) 3.5 | c | divide(divide(multiply(add(4, 2), 4), 2), 4) | add(n0,n1)|multiply(n2,#0)|divide(#1,n1)|divide(#2,n0)| | physics | C |
if 4 people are selected from a group of 7 married couples , what is the probability that none of them would be married to each other ? | "if we are to select 4 people from 7 couples without any restriction , how many ways can we make the selection ? 14 ! / 4 ! 10 ! = 1001 if we are to select 4 people from 7 couples with restriction that no married couple can both make it to the group , only a representative ? 7 ! / 4 ! 3 ! = 35 but we know that to select a person from each couple , take 2 possibilities 35 * 2 * 2 * 2 * 2 = 560 probability = desired / all possibilities = 560 / 1001 = 47 / 84 answer : e" | a ) 1 / 33 , b ) 2 / 33 , c ) 1 / 3 , d ) 16 / 33 , e ) 47 / 84 | e | multiply(multiply(multiply(divide(multiply(7, const_2), multiply(7, const_2)), divide(multiply(4, 4), subtract(multiply(7, const_2), const_1))), divide(subtract(multiply(4, 4), const_2), multiply(4, 4))), divide(subtract(subtract(multiply(4, 4), const_2), const_2), subtract(multiply(4, 4), const_1))) | multiply(n0,n0)|multiply(n1,const_2)|divide(#1,#1)|subtract(#0,const_2)|subtract(#0,const_1)|subtract(#1,const_1)|divide(#3,#0)|divide(#0,#5)|subtract(#3,const_2)|divide(#8,#4)|multiply(#2,#7)|multiply(#6,#10)|multiply(#9,#11)| | probability | E |
a pump can fill a tank with water in 2 hours . because of a leak , it took hours to fill the tank . the leak can drain all the water of the tank in : | explanation : work done by the leak in 1 hour = { \ color { black } \ left ( \ frac { 1 } { 2 } - \ frac { 3 } { 7 } \ right ) = \ frac { 1 } { 14 } } { \ color { black } \ therefore } leak will empty the tank in 14 hrs answer : d ) 14 hours | a ) 33 , b ) 37 , c ) 76 , d ) 14 , e ) 29 | d | inverse(subtract(divide(const_1, const_2), inverse(add(2, divide(const_1, const_3))))) | divide(const_1,const_2)|divide(const_1,const_3)|add(n0,#1)|inverse(#2)|subtract(#0,#3)|inverse(#4) | physics | D |
length of a rectangular plot is 20 mtr more than its breadth . if the cost of fencing the plot at 26.50 per meter is rs . 5300 , what is the length of the plot in mtr ? | "let breadth = x metres . then , length = ( x + 18 ) metres . perimeter = 5300 m = 200 m . 26.50 2 [ ( x + 18 ) + x ] = 200 2 x + 18 = 100 2 x = 82 x = 41 . hence , length = x + 18 = 59 m a" | a ) 59 m , b ) 60 m , c ) 80 m , d ) 82 m , e ) 84 m | a | divide(add(divide(5300, 26.50), multiply(const_2, 20)), const_4) | divide(n2,n1)|multiply(n0,const_2)|add(#0,#1)|divide(#2,const_4)| | physics | A |
a cube of side 3 meter length is cut into small cubes of side 15 cm each . how many such small cubes can be obtained ? | "along one edge , the number of small cubes that can be cut = 300 / 15 = 20 along each edge 20 cubes can be cut . ( along length , breadth and height ) . total number of small cubes that can be cut = 20 * 20 * 20 = 8000 answer : e" | a ) 8078 , b ) 1040 , c ) 1058 , d ) 1021 , e ) 8000 | e | divide(power(power(3, const_2), const_3), power(3, const_3)) | power(n0,const_2)|power(n0,const_3)|power(#0,const_3)|divide(#2,#1)| | physics | E |
a , b and c enter into a partnership . a invests 3 times as much as b invests and 2 / 3 of what c invests . at the end of the year , the profit earned is rs . 55000 . what is the share of c ? | explanation : let the investment of c be rs . x . the inverstment of b = rs . ( 2 x / 3 ) the inverstment of a = rs . ( 3 Γ ( 2 / 3 ) x ) = rs . ( 2 x ) ratio of capitals of a , b and c = 2 x : 2 x / 3 : x = 6 : 2 : 3 c ' s share = rs . [ ( 3 / 11 ) Γ 55000 ] = rs . 15000 answer : option d | a ) rs . 12250 , b ) rs . 13375 , c ) rs . 16750 , d ) rs . 15000 , e ) none of these | d | multiply(55000, inverse(add(add(divide(2, 3), multiply(divide(2, 3), 3)), const_1))) | divide(n1,n0)|multiply(n0,#0)|add(#0,#1)|add(#2,const_1)|inverse(#3)|multiply(n3,#4) | gain | D |
lovely shop runs a sale on goldfish and platys . if one customer buys 13 goldfish and 12 platys for $ 5.60 , and another buys 14 goldfish and 21 platys for $ 8.05 , what is the ratio of the price of goldfish to platys ? | the approach of plugging in the numbers is a fine approach . but , in this case , the answer was the first option and thus , by plugging in a number in the ration given , we arrived at the answer quickly ! however , what if the correct ratio was option d or c ? in any case , the algebraic solution is n ' t that complex if you realize a key concept : the number 7 can be factored out of the second equation ! here ' s the working : lovely shop runs a sale on goldfish and platys . 13 x + 12 y = 5.60 < < < < < equation 1 14 x + 21 y = 8.05 < < < < < equation 2 2 x + 3 y = 1.15 [ factor out 7 from equation 2 ] < < < < < equation 3 8 x + 12 y = 4.6 [ multiply equation 3 with 4 ] 5 x = 1.0 [ equation 1 - equation 2 ] x = 0.2 3 y = 1.15 - 0.4 = 0.75 [ substitute for x in equation 1 ] y = 0.25 x / y = 0.2 / 0.25 = 20 / 25 = 4 / 5 = 0.8 [ answer a ] the whole thing must have taken me about 45 s : d | a ) 0.8 , b ) 0.75 , c ) c ) 8 / 9 , d ) 5 / 7 , e ) 6 / 8 | d | divide(subtract(21, 12), 12) | subtract(n4,n1)|divide(#0,n1) | other | D |
6 ) a marketing firm determined that , of 300 households surveyed , 80 used neither brand a nor brand b soap . 60 used only brand a soap and for every household that used both brands of soap , 3 used only brand b soap . how many of the 200 household surveyed used both brands of soap ? | "220 = at least one of soap a or b both brands = x brand b = 3 x = > 60 + x + 3 x = 220 = > 4 x = 160 = > x = 40 answer - d" | a ) a ) 15 , b ) b ) 20 , c ) c ) 30 , d ) d ) 40 , e ) e ) 45 | d | divide(subtract(subtract(300, 80), 60), const_4) | subtract(n1,n2)|subtract(#0,n3)|divide(#1,const_4)| | other | D |
how many different ways can 2 students be seated in a row of 3 desks , so that there is always at least one empty desk between the students ? | "1 ways to seat the students : with two empty seats between 1 empty w / one student on the left most 1 empty . . . . right most two students can be interchanged 1 x 2 = 2 a" | a ) 2 , b ) 3 , c ) 4 , d ) 6 , e ) 12 | a | permutation(subtract(3, const_1), 2) | subtract(n1,const_1)|permutation(#0,n0)| | probability | A |
a small college reduced its faculty by approximately 25 percent to 195 professors . what was the original number of faculty members ? | "if x is the original number of faculty members , then after 25 % reduction in faculty members number is . 75 x but we are given . 75 x = 195 x = 260 so the original number of faculty members is 260 correct answer - b" | a ) 182 , b ) 260 , c ) 224 , d ) 254 , e ) 302 | b | divide(195, divide(subtract(const_100, 25), const_100)) | subtract(const_100,n0)|divide(#0,const_100)|divide(n1,#1)| | gain | B |
a class of 35 students has an average height of 180 cm . 7 students whose average height is 120 cm , left the class and 7 others whose average height is 140 cm , joined . calculate the new average height of the students of the class ( in cm ) is ? | total height = 180 * 35 = 6300 cm . total height of those who left = 120 * 7 = 840 total height of those who joined = 140 * 7 = 980 new average = [ 6300 - 120 + 980 ] / 35 = 7160 / 35 cm . = 204.6 cm answer : a | a ) 204.6 cm , b ) 404.6 cm , c ) 224.6 cm , d ) 205.6 cm , e ) 256.6 cm | a | add(add(add(divide(add(multiply(35, 180), subtract(multiply(7, 140), multiply(7, 120))), 35), const_12), const_4), const_4) | multiply(n0,n1)|multiply(n2,n5)|multiply(n2,n3)|subtract(#1,#2)|add(#0,#3)|divide(#4,n0)|add(#5,const_12)|add(#6,const_4)|add(#7,const_4) | general | A |
if the average of t , b , c , 14 and 15 is 12 . what is the average value of t , b , c and 29 | t + b + c + 14 + 15 = 12 * 5 = 60 = > t + b + c = 60 - 29 = 31 t + b + c + 29 = 31 + 29 = 60 average = 60 / 4 = 15 answer d | a ) 12 , b ) 13 , c ) 14 , d ) 15 , e ) 16 | d | divide(multiply(12, add(const_2, const_3)), const_4) | add(const_2,const_3)|multiply(n2,#0)|divide(#1,const_4) | general | D |
find the least number of complete years in which a sum of money put out at 33.3 % compound interest will be more than double of itself ? | "3 years answer : a" | a ) 3 , b ) 4 , c ) 5 , d ) 6 , e ) 7 | a | floor(add(divide(log(const_2), log(add(const_1, divide(33.3, const_100)))), const_1)) | divide(n0,const_100)|log(const_2)|add(#0,const_1)|log(#2)|divide(#1,#3)|add(#4,const_1)|floor(#5)| | general | A |
the speed at which a man can row a boat in still water is 18 kmph . if he rows downstream , where the speed of current is 3 kmph , what time will he take to cover 60 metres ? | "speed of the boat downstream = 18 + 3 = 21 kmph = 21 * 5 / 18 = 35 / 9 m / s hence time taken to cover 60 m = 60 * 9 / 35 = 15.4 seconds . answer : b" | a ) 16 seconds , b ) 15.4 seconds , c ) 26 seconds , d ) 12 seconds , e ) 18 seconds | b | divide(60, multiply(add(18, 3), const_0_2778)) | add(n0,n1)|multiply(#0,const_0_2778)|divide(n2,#1)| | physics | B |
in a coconut grove , ( x + 1 ) trees yield 60 nuts per year , x trees yield 120 nuts per year and ( x β 1 ) trees yield 180 nuts per year . if the average yield per year per tree be 100 , find x . | "( x + 1 ) Γ 60 + x Γ 120 + ( x β 1 ) Γ 180 / ( x + 1 ) + x + ( x β 1 ) = 100 β 360 x β 120 / 3 x = 100 β 60 x = 120 β x = 2 answer a" | a ) 2 , b ) 4 , c ) 5 , d ) 6 , e ) none of the above | a | divide(subtract(multiply(180, 1), multiply(60, 1)), subtract(add(add(60, 120), 180), multiply(100, const_3))) | add(n1,n2)|multiply(n0,n4)|multiply(n0,n1)|multiply(n5,const_3)|add(n4,#0)|subtract(#1,#2)|subtract(#4,#3)|divide(#5,#6)| | general | A |
a group of hikers is planning a trip that will take them up a mountain using one route and back down using another route . they plan to travel down the mountain at a rate of one and a half times the rate they will use on the way up , but the time each route will take is the same . if they will go up the mountain at a rate of 4 miles per day and it will take them two days , how many miles long is the route down the mountain ? | "d = s * t given condition is for up the mountain hikers takes 4 miles per day and total 2 days . . . miles = 8 miles . . . for down speed they take = 3 / 2 ( 4 ) = 6 miles per day and for the same 2 days . . . miles = 12 miles . . answer : option d is correct answer" | a ) 4 , b ) 6 , c ) 8 , d ) 12 , e ) 16 | d | multiply(multiply(4, const_2), divide(const_3, const_2)) | divide(const_3,const_2)|multiply(n0,const_2)|multiply(#0,#1)| | physics | D |
the ratio of pens to pencils is 5 to 6 . there are 9 more pencils than pens . how many pencils are there ? | let the number of pens be 5 x and the number of pencils be 6 x . 6 x - 5 x = 9 x = 9 the number of pencils is 54 . the answer is d . | a ) 36 , b ) 42 , c ) 48 , d ) 54 , e ) 60 | d | add(multiply(9, 5), 9) | multiply(n0,n2)|add(n2,#0) | other | D |
what will be the result of 12 ^ 4 + 3 + 2 ^ 9 + 3 - 8 / 2 ? | "the multiplication ( e . g . 12 ^ 4 ) and division ( e . g . 8 / 2 ) is done first and then the sum ( e . g . 48 + 3 ) and subtraction ( e . g . 4 - 4 ) , and the answer is 69 , option d ." | a ) 58 , b ) 14 , c ) 145 , d ) 69 , e ) 74 | d | subtract(multiply(add(divide(subtract(multiply(12, 4), 3), 2), 9), 3), 12) | multiply(n0,n1)|subtract(#0,n2)|divide(#1,n3)|add(n4,#2)|multiply(n5,#3)|subtract(#4,n0)| | general | D |
if a tire rotates at 400 revolutions per minute when the car is traveling 144 km / h , what is the circumference of the tire ? | "400 rev / minute = 400 * 60 rev / 60 minutes = 24,000 rev / hour 24,000 * c = 144,000 m : c is the circumference c = 6 meters correct answer b" | a ) 2 meters , b ) 6 meters , c ) 5 meters , d ) 3 meters , e ) 7 meters | b | multiply(divide(144, multiply(multiply(multiply(const_2, const_3), const_10), 400)), const_1000) | multiply(const_2,const_3)|multiply(#0,const_10)|multiply(n0,#1)|divide(n1,#2)|multiply(#3,const_1000)| | physics | B |
the area of a square garden is q square feet and the perimeter is p feet . if q = 2 p + 33 , what is the perimeter of the garden in feet ? | "let x be the length of one side of the square garden . x ^ 2 = 8 x + 33 x ^ 2 - 8 x - 33 = 0 ( x - 11 ) ( x + 3 ) = 0 x = 11 , - 3 p = 4 ( 11 ) = 44 the answer is e ." | a ) 28 , b ) 32 , c ) 36 , d ) 40 , e ) 44 | e | multiply(2, 33) | multiply(n1,n0)| | geometry | E |
a certain kind of fuel is 1515 high octane and 4545 regular octane . if high octane costs 3 times as much as regular octane , what fraction of the cost of the fuel is due to the high octane ? | let x be the amount of fuel high octane = x / 5 regular = 4 x / 5 price of regular = y price of high octante = 3 y cost of high octane in the fuel = 3 y * x / 5 total cost = 3 y * x / 5 + 4 x / 5 * y % = cost of high octante / total cost = 3 / 7 answer is c | a ) 1 / 5 , b ) 2 / 3 , c ) 3 / 7 , d ) 3 / 5 , e ) 4 / 5 | c | divide(3, add(const_3, const_4)) | add(const_3,const_4)|divide(n2,#0) | general | C |
x and y start a business with rs . 2000 and rs . 12000 respectively . hoe should they share their profits at the end of one year ? | "they should share the profits in the ratio of their investments . the ratio of the investments made by x and y = 2000 : 12000 = > 1 : 6 . answer : a" | a ) 1 : 6 , b ) 6 : 1 , c ) 1 : 5 , d ) 5 : 1 , e ) 1 : 4 | a | divide(2000, 12000) | divide(n0,n1)| | gain | A |
in a certain town , the ratio of ny yankees fans to ny mets fans is 2 : 1 , and the ratio of ny mets fans to boston red sox fans is 3 : 7 . if there are 320 baseball fans in the town , each of whom is a fan of exactly one of those three teams , how many ny mets fans are there in this town ? | "the ratio of yankees : mets : red sox = 6 : 3 : 7 the mets fans are 3 / 16 of the population . ( 3 / 16 ) * 320 = 60 the answer is d ." | a ) 48 , b ) 52 , c ) 56 , d ) 60 , e ) 64 | d | multiply(divide(320, add(add(multiply(2, 1), 3), 7)), 3) | multiply(n0,n1)|add(n2,#0)|add(n3,#1)|divide(n4,#2)|multiply(n2,#3)| | other | D |
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 | A |
what is the probability q of randomly selecting an arrangement of the letters of the wordmediterraneanin which the first letter is e and the last letter is r ? | why do you calculate the total combinations and everything , it is much easier to calculate as follows : 3 / 13 * 1 * 1 * 1 * 1 . . . . . * 2 / 12 = 6 / 156 = 1 / 26 this is because 3 / 13 = probability that first letter will be e and 2 / 12 = probability that thefirstletter will be r . between everything else cuts down to 1 * 1 * 1 . . . e . g . probability q of the second letter is 11 / 11 = 1 , > we do not care what ' s in between . c | a ) 1 / 13 , b ) 1 / 20 , c ) 1 / 26 , d ) 1 / 50 , e ) 1 / 100 | c | divide(divide(factorial(subtract(multiply(const_3, const_4), const_1)), power(factorial(const_2), const_3)), divide(factorial(add(subtract(multiply(const_3, const_4), const_1), const_2)), multiply(factorial(const_3), power(factorial(const_2), const_3)))) | factorial(const_2)|factorial(const_3)|multiply(const_3,const_4)|power(#0,const_3)|subtract(#2,const_1)|add(#4,const_2)|factorial(#4)|multiply(#1,#3)|divide(#6,#3)|factorial(#5)|divide(#9,#7)|divide(#8,#10) | probability | C |
if an amount of rs 42,900 is distributed equally amongst 22 persons , how much amount would each person get ? | required amount = 42900 / 22 = rs 1950 answer a | a ) rs 1950 , b ) rs 2000 , c ) rs 745 , d ) rs 765 , e ) none | a | divide(subtract(multiply(add(multiply(const_4, const_10), const_3), const_1000), const_100), 22) | multiply(const_10,const_4)|add(#0,const_3)|multiply(#1,const_1000)|subtract(#2,const_100)|divide(#3,n1) | general | A |
there are 408 boys and 312 girls in a school which are to be divided into equal sections of either boys or girls alone . find the total number of sections thus formed . | "explanation : hcf ( 408 , 312 ) = 24 the number of boys or girls that can be placed in a section = 24 . thus the total number of sections is given by 408 / 24 + 312 / 24 = 17 + 13 = 30 answer : d" | a ) 31 , b ) 32 , c ) 35 , d ) 30 , e ) 45 | d | divide(add(408, 312), multiply(multiply(multiply(const_2, const_2), const_2), const_3)) | add(n0,n1)|multiply(const_2,const_2)|multiply(#1,const_2)|multiply(#2,const_3)|divide(#0,#3)| | general | D |
the β s - number β of a number x is defined as the ones digit of 2 ^ x . antony rolls a die with 6 sides labeled with the integers from 1 to 6 , each of which has an equal probability of landing face - up . he then takes 3 ^ s , where s is the s - number of the result of his die roll , and plots 3 ^ s on a number line as the point a . finally , he repeats this entire process , this time plotting the result as the point b . what is the probability that the distance between a and b is greater than the value of b ? | if you calculate 3 ^ s for 1 st roll , all 6 results will be 9 , 81 , 6561 , 729 , 9 , 81 . this result is the same for 2 nd roll . 9 , 81 , 6561 , 729 , 9 , 81 . about distance : if the first result is 9 and the second is also 9 , the distance is 9 - 9 = 0 which is smaller than 9 . if the first result is 9 and the second is 81 , the distance is 81 - 9 = 72 which is also smaller than b which has the value of 81 . if the first result is 81 and the second is 9 , the distance will be greater than b . distance 81 - 9 = 72 > 9 . on the first roll , the probability of getting result 9 is 2 / 6 . in this case no other alternative values for second roll which would make the distance greater than b . so probability is 0 . so next estimations are : probability of getting 81 on the first roll ( 2 / 6 ) * probability of getting 9 on the second roll ( 2 / 6 ) = 1 / 9 probability of getting 729 on the first roll ( 1 / 6 ) * probability of getting 9 , 81 on the second roll ( 4 / 6 ) = 1 / 9 probability of getting 6561 on the first roll ( 1 / 6 ) * probability of getting 9 , 81 , 729 on the first roll ( 5 / 6 ) = 5 / 36 all together : 1 / 9 + 1 / 9 + 5 / 36 = 13 / 36 = b | a ) 3 / 8 , b ) 13 / 36 , c ) 17 / 36 , d ) 19 / 36 , e ) 23 / 36 | b | divide(add(add(add(6, 1), 3), 3), multiply(6, 6)) | add(n1,n2)|multiply(n1,n1)|add(n4,#0)|add(n4,#2)|divide(#3,#1) | general | B |
the perimeter of one square is 40 cm and that of another is 20 cm . find the perimeter and the diagonal of a square which is equal in area to these two combined ? | "4 a = 40 4 a = 20 a = 10 a = 5 a 2 = 100 a 2 = 25 combined area = a 2 = 125 = > a = 11.2 d = 11.2 β 2 answer : a" | a ) 11.2 β 2 , b ) 13 β 2 , c ) 23 β 2 , d ) 12 β 4 , e ) 13 β 9 | a | sqrt(multiply(add(power(divide(40, const_4), const_2), power(divide(20, const_4), const_2)), const_2)) | divide(n0,const_4)|divide(n1,const_4)|power(#0,const_2)|power(#1,const_2)|add(#2,#3)|multiply(#4,const_2)|sqrt(#5)| | geometry | A |
the average age of the boys in a class is 16 years and that of the girls is 15 years . the average age for the whole class is | clearly , to find the average , we ought to know the number of boys , girls or students in the class , neither of which has been given . so , the data provided is inadequate answer : d | a ) 27 , b ) 29 , c ) 11 , d ) data inadequate , e ) 28 | d | divide(add(16, 15), const_2) | add(n0,n1)|divide(#0,const_2) | general | D |
tanks m and b are each in the shape of a right circular cylinder . the interior of tank m has a height of 10 meters and a circumference of 8 meters , and the interior of tank b has a height of 8 meters and a circumference of 10 meters . the capacity of tank m is what percent of the capacity of tank b ? | b . for m , r = 8 / 2 pi . its capacity = ( 4 pi ) ^ 2 * 10 = 160 pi for b , r = 10 / pi . its capacity = ( 5 pi ) ^ 2 * 8 = 200 pi m / b = 160 pi / 200 pi = 0.8 | a ) 75 % , b ) 80 % , c ) 100 % , d ) 120 % , e ) 125 % | b | multiply(divide(multiply(multiply(divide(8, const_2), divide(8, const_2)), 10), multiply(multiply(divide(10, const_2), divide(10, const_2)), 8)), const_100) | divide(n1,const_2)|divide(n0,const_2)|multiply(#0,#0)|multiply(#1,#1)|multiply(n0,#2)|multiply(n1,#3)|divide(#4,#5)|multiply(#6,const_100) | physics | B |
50 liters of a mixture contains milk and water in the ratio 3 : 2 . if 5 liters of this mixture be replaced by 5 liters of milk , the ratio of milk to water in the new mixture would be ? | "quantity of milk in 50 liters if mix = 50 * 3 / 5 = 30 liters quantity of milk in 55 liters of new mix = 30 + 5 = 35 liters quantity of water in it = 55 - 35 = 20 liters ratio of milk and water in new mix = 35 : 20 = 7 : 4 answer is d" | a ) 6 : 5 , b ) 6 : 2 , c ) 6 : 4 , d ) 7 : 4 , e ) 7 : 2 | d | divide(multiply(subtract(5, const_1), 2), multiply(3, 2)) | multiply(n2,n1)|subtract(n3,const_1)|multiply(#1,n2)|divide(#2,#0)| | other | D |
how many words , with or without meaning , can be formed using all letters of the word red using each letter exactly once ? | the word red has exactly 3 letters which are all different . therefore the number of words that can be formed = number of permutations of 3 letters taken all at a time . = p ( 3 , 3 ) = 3 ! = 3 x 2 Γ 1 = 6 answer : a | a ) 6 , b ) 7 , c ) 8 , d ) 9 , e ) 10 | a | factorial(const_3) | factorial(const_3) | general | A |
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 | B |
at a certain organisation , the number of male members went up by 12 % in the year 2001 from year 2000 , and the number of females members went down by 4 % in the same time period . if the total membership at the organisation went up by 1.2 % from the year 2000 to 2001 , what was the ratio of male members to female members in the year 2000 ? | "men increase by 12 % = = > 1.12 m = males in 2001 women decrease by 4 % = = > 0.96 f = women in 2001 total employees increase by 1.2 % = = > 1.012 * ( m + f ) = total number of employees in 2001 obviously ( males in 2001 ) + ( females in 2001 ) = total number of employees in 2001 1.12 m + 0.96 f = 1.012 * ( m + f ) 1.12 m + 0.96 f = 1.012 m + 1.012 f 1.12 m - 1.012 m = 1.012 f - 0.96 f 0.108 m = 0.052 f m / f = ( 0.052 ) / ( 0.108 ) = 52 / 108 = 1 / 2 answer = ( a )" | a ) 1 : 2 , b ) 1 : 3 , c ) 2 : 3 , d ) 3 : 2 , e ) 2 : 1 | a | divide(subtract(multiply(add(const_1, divide(1.2, const_100)), const_1000), multiply(subtract(const_1, divide(4, const_100)), const_1000)), subtract(multiply(add(const_1, divide(12, const_100)), const_1000), multiply(add(const_1, divide(1.2, const_100)), const_1000))) | divide(n4,const_100)|divide(n3,const_100)|divide(n0,const_100)|add(#0,const_1)|add(#2,const_1)|subtract(const_1,#1)|multiply(#3,const_1000)|multiply(#5,const_1000)|multiply(#4,const_1000)|subtract(#6,#7)|subtract(#8,#6)|divide(#9,#10)| | other | A |
if 60 percent of 500 is 50 percent of x , then x = ? | "0.6 * 500 = 0.5 * x x = 6 / 5 * 500 = 600" | a ) 600 , b ) 620 , c ) 650 , d ) 700 , e ) 720 | a | divide(multiply(60, 500), 50) | multiply(n0,n1)|divide(#0,n2)| | general | A |
a total of $ 20,000 was invested in two certificates of deposit at simple annual interest rates of 6 percent and 9 percent , respectively . if the total interest on the two certificates was $ 1,440 at the end of one year , what fractional part of the $ 20.000 was invested at the higher rate ? | x * 6 / 100 * 1 + ( 20000 - x ) * 9 / 100 * 1 = 1440 6 x - 8 x = 144000 - 180000 = > - 2 x = - 18000 = > x = 9000 so 11000 / 20000 = 11 / 20 answer - a | a ) 11 / 20 , b ) 2 / 5 , c ) 1 / 2 , d ) 3 / 5 , e ) 3 / 4 | a | divide(subtract(add(divide(divide(add(add(multiply(const_100, multiply(const_2, add(const_1, const_4))), multiply(const_4, const_100)), multiply(const_4, multiply(const_2, add(const_1, const_4)))), 20), 9), 9), 6), 20) | add(const_1,const_4)|multiply(const_100,const_4)|multiply(#0,const_2)|multiply(#2,const_100)|multiply(#2,const_4)|add(#3,#1)|add(#5,#4)|divide(#6,n4)|divide(#7,n2)|add(n2,#8)|subtract(#9,n1)|divide(#10,n4) | gain | A |
what is the area of a triangle with the following vertices l ( 2 , 3 ) , m ( 5 , 1 ) , and n ( 3 , 5 ) ? | there is a direct formula to calculate the are of a triangle based on coordinates of its vertices and one could use it to solve this problem . though if you make a diagram minimum simple calculations will be needed : answer : b . | ['a ) 3', 'b ) 4', 'c ) 5', 'd ) 6', 'e ) 7'] | b | triangle_area_three_edges(sqrt(add(power(subtract(5, 2), const_2), power(subtract(1, 3), const_2))), sqrt(add(power(subtract(2, 3), const_2), power(subtract(3, 5), const_2))), sqrt(add(power(subtract(1, 5), const_2), power(subtract(3, 5), const_2)))) | subtract(n2,n0)|subtract(n3,n1)|subtract(n0,n1)|subtract(n1,n2)|subtract(n3,n2)|power(#0,const_2)|power(#1,const_2)|power(#2,const_2)|power(#3,const_2)|power(#4,const_2)|add(#5,#6)|add(#7,#8)|add(#9,#8)|sqrt(#10)|sqrt(#11)|sqrt(#12)|triangle_area_three_edges(#13,#14,#15) | geometry | B |
7 / 10 of the population of the country of venezia lives in montague province , while the rest lives in capulet province . in the upcoming election , 80 % of montague residents support romeo , while 70 % of capulet residents support juliet ; each resident of venezia supports exactly one of these two candidates . rounded if necessary to the nearest percent , the probability that a juliet supporter chosen at random resides in capulet is | "total population = 100 ( assume ) . 7 / 10 * 100 = 70 people from montague . 3 / 10 * 100 = 30 people from capulet . 0.2 * 70 = 14 people from montague support juliet . 0.7 * 30 = 21 people from capulet support juliet . the probability that a juliet supporter chosen at random resides in capulet is 21 / ( 14 + 21 ) = ~ 60 . answer : c" | a ) 28 % , b ) 41 % , c ) 60 % , d ) 72 % , e ) 78 % | c | multiply(divide(multiply(divide(70, const_100), divide(subtract(10, 7), 10)), add(multiply(divide(subtract(const_100, 80), const_100), divide(7, 10)), multiply(divide(70, const_100), divide(subtract(10, 7), 10)))), const_100) | divide(n3,const_100)|divide(n0,n1)|subtract(n1,n0)|subtract(const_100,n2)|divide(#2,n1)|divide(#3,const_100)|multiply(#0,#4)|multiply(#5,#1)|add(#7,#6)|divide(#6,#8)|multiply(#9,const_100)| | gain | C |
what is the smallest integer that is multiple of 3 , 5,9 | "it is the lcm of 3 , 5 and 9 which is 45 . the answer is b ." | a ) a ) 70 , b ) b ) 45 , c ) c ) 200 , d ) d ) 280 , e ) e ) 140 | b | add(const_3, const_4) | add(const_3,const_4)| | general | B |
if a ( k ) = ( 2 k + 1 ) ^ k , and k = 0 , what is the value of a ( a ( a ( k ) ) ) ? | a ( a ( a ( 0 ) ) ) = a ( a ( 1 ) ) = a ( 3 ) = ( 6 + 1 ) ^ 3 = 343 . thus , the answer is a . | a ) 343 , b ) 199 , c ) 121 , d ) 69 , e ) 249 | a | power(add(multiply(2, power(add(multiply(2, power(add(1, multiply(2, 0)), 0)), 1), 1)), 1), power(add(multiply(2, power(add(1, multiply(2, 0)), 0)), 1), 1)) | multiply(n0,n2)|add(n1,#0)|power(#1,n2)|multiply(n0,#2)|add(n1,#3)|power(#4,n1)|multiply(n0,#5)|add(n1,#6)|power(#7,#5) | general | A |
by travelling at 50 kmph , a person reaches his destination on time . he covered two - third the total distance in one - third of the total time . what speed should he maintain for the remaining distance to reach his destination on time ? | "let the time taken to reach the destination be 3 x hours . total distance = 50 * 3 x = 150 x km he covered 2 / 3 * 150 x = 100 x km in 1 / 3 * 3 x = x hours so , the remaining 50 x km , he has to cover in 2 x hours . required speed = 50 x / 2 x = 25 kmph . answer : c" | a ) 23 kmph , b ) 24 kmph , c ) 25 kmph , d ) 26 kmph , e ) 27 kmph | c | divide(subtract(multiply(50, const_3), divide(multiply(multiply(50, const_3), const_2), const_3)), subtract(const_3, const_1)) | multiply(n0,const_3)|subtract(const_3,const_1)|multiply(#0,const_2)|divide(#2,const_3)|subtract(#0,#3)|divide(#4,#1)| | physics | C |
a , b and c rent a pasture . if a puts 10 oxen for 7 months , b puts 12 oxen for 5 months and c puts 15 oxen for 3 months for grazing and the rent of the pasture is rs . 280 , then how much amount should c pay as his share of rent ? | a : b : c = 10 Γ 7 : 12 Γ 5 : 15 Γ 3 = 2 Γ 7 : 12 Γ 1 : 3 Γ 3 = 14 : 12 : 9 amount that c should pay = 280 Γ 9 / 35 = 8 Γ 9 = 72 answer is d | a ) 35 , b ) 45 , c ) 25 , d ) 72 , e ) 55 | d | multiply(280, divide(multiply(15, 3), add(add(multiply(10, 7), multiply(12, 5)), multiply(15, 3)))) | multiply(n4,n5)|multiply(n0,n1)|multiply(n2,n3)|add(#1,#2)|add(#3,#0)|divide(#0,#4)|multiply(n6,#5) | general | D |
real - estate salesman z is selling a house at a 30 percent discount from its retail price . real - estate salesman x vows to match this price , and then offers an additional 15 percent discount . real - estate salesman y decides to average the prices of salesmen z and x , then offer an additional 40 percent discount . salesman y ' s final price is what fraction of salesman x ' s final price ? | let the retail price be = x selling price of z = 0.7 x selling price of x = 0.85 * 0.7 x = 0.60 x selling price of y = ( ( 0.7 x + 0.6 x ) / 2 ) * 0.60 = 0.65 x * 0.60 = 0.39 x 0.39 x = k * 0.60 x k = 0.39 / 0.6 = 39 / 6 answer : a | a ) 39 / 6 , b ) 62 / 11 , c ) 11 / 61 , d ) 21 / 61 , e ) 20 / 61 | a | multiply(divide(divide(multiply(divide(add(subtract(const_100, 30), multiply(subtract(const_100, 30), divide(subtract(const_100, 15), const_100))), const_2), subtract(const_100, 40)), const_100), multiply(subtract(const_100, 30), divide(subtract(const_100, 15), const_100))), const_10) | subtract(const_100,n1)|subtract(const_100,n0)|subtract(const_100,n2)|divide(#0,const_100)|multiply(#3,#1)|add(#4,#1)|divide(#5,const_2)|multiply(#6,#2)|divide(#7,const_100)|divide(#8,#4)|multiply(#9,const_10) | general | A |
at a speed of 40 miles per hour , a certain car uses 1 gallon of gasoline every 40 miles . if the car starts with a full 12 gallon tank of gasoline and travels for 5 hours at 40 miles per hour , the amount of gasoline used would be what fraction of a full tank ? | "gas used = ( 5 hours ) * ( 40 miles / hour ) * ( 1 gallon / 40 miles ) = 5 gallons portion used = ( 5 ) / 12 = 5 / 12 ans b" | a ) 3 / 25 , b ) 5 / 12 , c ) 7 / 12 , d ) 2 / 3 , e ) 25 / 36 | b | divide(divide(multiply(40, 5), 40), 12) | multiply(n0,n4)|divide(#0,n2)|divide(#1,n3)| | physics | B |
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