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
values |
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if 17 ^ 4 * 13 ^ 4 * 9 ^ 4 * 8 ^ 4 = x , what is the units digit of x ? | "17 ^ 4 * 13 ^ 4 * 9 ^ 4 * 8 ^ 4 = x units digit of x will be 7 ^ 4 * 3 ^ 4 * 9 ^ 4 * 8 ^ 4 = 1 * 1 * 1 * 6 = 6 therefore , units digit of x = 6 answer c" | a ) 8 , b ) 3 , c ) 6 , d ) 7 , e ) 9 | c | divide(multiply(4, 4), const_10) | multiply(n1,n7)|divide(#0,const_10)| | general |
mark bought a set of 6 flower pots of different sizes at a total cost of $ 8.25 . each pot cost 0.15 more than the next one below it in size . what was the cost , in dollars , of the largest pot ? | "this question can be solved with a handful of different algebra approaches ( as has been shown in the various posts ) . since the question asks for the price of the largest pot , and the answers are prices , we can test the answers . we ' re told that there are 6 pots and that each pot costs 25 cents more than the nex... | a ) $ 1.75 , b ) $ 1.85 , c ) $ 2.00 , d ) $ 2.15 , e ) $ 2.30 | a | add(divide(subtract(8.25, multiply(divide(multiply(subtract(6, const_1), 6), const_2), 0.15)), 6), multiply(subtract(6, const_1), 0.15)) | subtract(n0,const_1)|multiply(n0,#0)|multiply(n2,#0)|divide(#1,const_2)|multiply(n2,#3)|subtract(n1,#4)|divide(#5,n0)|add(#6,#2)| | general |
a pipe can fill a cistern in 66 minutes . find the time in which 1 / 11 part of the cistern will be filled ? | "full cistern filled in = 66 minutes 1 / 11 part filled in = 66 * 1 / 11 = 6 minutes answer is c" | a ) 5 min , b ) 2 min , c ) 6 min , d ) 1 min , e ) 10 min | c | multiply(66, divide(1, 11)) | divide(n1,n2)|multiply(n0,#0)| | physics |
a train crosses a platform of 160 m in 15 sec , same train crosses another platform of length 250 m in 20 sec . then find the length of the train ? | "length of the train be ‘ x ’ x + 160 / 15 = x + 250 / 20 20 x + 3200 = 15 x + 3750 5 x = 550 x = 110 m answer : a" | a ) 110 , b ) 887 , c ) 167 , d ) 197 , e ) 161 | a | subtract(multiply(250, divide(15, divide(15, const_3))), multiply(160, divide(20, divide(15, const_3)))) | divide(n1,const_3)|divide(n1,#0)|divide(n3,#0)|multiply(n2,#1)|multiply(n0,#2)|subtract(#3,#4)| | physics |
if a = 1 , what is the value of – ( a ^ 2 + a ^ 3 + a ^ 4 + a ^ 5 + a ^ 6 ) ? | "if a = - 1 , then putting values in equation = - [ ( 1 ) ^ 2 + ( 1 ) ^ 3 + ( 1 ^ 4 ) + ( 1 ^ 5 ) + ( 1 ^ 6 ) ] = - [ 1 + 1 + 1 + 1 + 1 ] = - 5 answer = e = - 5" | a ) - 14 , b ) - 4 , c ) 0 , d ) 4 , e ) - 5 | e | negate(add(add(add(power(1, 2), power(1, 3)), power(1, 4)), power(1, 5))) | power(n0,n1)|power(n0,n2)|power(n0,n3)|power(n0,n4)|add(#0,#1)|add(#4,#2)|add(#5,#3)|negate(#6)| | general |
a positive number x is multiplied by 2 , and this product is then divided by 5 . if the positive square root of the result of these two operations equals x , what is the value of x ? | "we need to produce an equation from the information given in the problem stem . we are first given that x is multiplied by 2 and then the product is divided by 3 . this gives us : 2 x / 3 next we are given that the positive square root of the result ( which is 2 x / 3 ) is equal to x . this gives us √ ( 2 x / 3 ) = x ... | a ) 9 / 4 , b ) 3 / 2 , c ) 4 / 3 , d ) 2 / 3 , e ) 1 / 2 | b | divide(2, 5) | divide(n0,n1)| | general |
if the number 892 , 132,26 x is divisible by 11 , what must be the value of x ? | "multiplication rule of 11 : ( sum of digits at odd places - sum of digits at even places ) should be divisible by 11 given number : 892 , 132,24 x sum of digits at odd places = 8 + 2 + 3 + 2 + x = 15 + x ( i ) sum of digits at even places = 9 + 1 + 2 + 6 = 18 ( ii ) ( i ) - ( ii ) = 15 + x - 18 = x - 3 hence x should ... | a ) 1 , b ) 2 , c ) 3 , d ) 4 , e ) 5 | c | multiply(multiply(multiply(add(multiply(const_3, const_10), const_1), const_2), const_4), 11) | multiply(const_10,const_3)|add(#0,const_1)|multiply(#1,const_2)|multiply(#2,const_4)|multiply(n2,#3)| | general |
two trains are moving in opposite directions at 60 km / hr and 90 km / hr . their lengths are 1.75 km and 1.10 km respectively . the time taken by the slower train to cross the faster train in seconds is ? | ": relative speed = 60 + 90 = 150 km / hr . = 150 * 5 / 18 = 125 / 3 m / sec . distance covered = 1.75 + 1.10 = 2.75 km = 2750 m . required time = 2750 * 3 / 125 = 66 sec . answer : d" | a ) 48 , b ) 9 , c ) 7 , d ) 66 , e ) 15 | d | subtract(divide(multiply(1.75, const_1000), divide(multiply(60, const_1000), const_3600)), divide(multiply(1.10, const_1000), divide(multiply(90, const_1000), const_3600))) | multiply(n2,const_1000)|multiply(n0,const_1000)|multiply(n3,const_1000)|multiply(n1,const_1000)|divide(#1,const_3600)|divide(#3,const_3600)|divide(#0,#4)|divide(#2,#5)|subtract(#6,#7)| | physics |
a and b put in rs . 300 and rs . 400 respectively into a business . a reinvests into the business his share of the first year ' s profit of rs . 210 where as b does not . in what ratio should they divide the second year ' s profit ? | "3 : 4 a = 3 / 7 * 210 = 90 390 : 400 39 : 40 answer : a" | a ) 39 : 40 , b ) 39 : 45 , c ) 39 : 40 , d ) 39 : 41 , e ) 39 : 42 | a | divide(add(300, multiply(divide(300, add(subtract(400, const_100), 300)), add(210, multiply(const_2, add(const_3, const_2))))), subtract(400, const_100)) | add(const_2,const_3)|subtract(n1,const_100)|add(n0,#1)|multiply(#0,const_2)|add(n2,#3)|divide(n0,#2)|multiply(#4,#5)|add(n0,#6)|divide(#7,#1)| | gain |
tea worth rs . 126 per kg are mixed with a third variety in the ratio 1 : 1 : 2 . if the mixture is worth rs . 153 per kg , the price of the third variety per kg will be | "solution since first second varieties are mixed in equal proportions , so their average price = rs . ( 126 + 135 / 2 ) = rs . 130.50 so , the mixture is formed by mixing two varieties , one at rs . 130.50 per kg and the other at say , rs . x per kg in the ratio 2 : 2 , i . e . , 1 : 1 . we have to find x . x - 153 / 2... | a ) rs . 169.50 , b ) rs . 1700 , c ) rs . 175.50 , d ) rs . 180 , e ) none | c | divide(subtract(multiply(153, add(add(1, 1), 2)), add(126, 126)), 2) | add(n1,n1)|add(n0,n0)|add(n3,#0)|multiply(n4,#2)|subtract(#3,#1)|divide(#4,n3)| | other |
if a and b are integers and ( a * b ) ^ 5 = 144 y , y could be : | "distribute the exponent . a ^ 5 * b ^ 5 = 144 y find the prime factorization of 144 . this is 2 ^ 4 * 3 ^ 2 . we need 2 ^ 1 and 3 ^ 3 i . e 2 * 27 = 54 the answer is e ." | a ) 40 , b ) 48 , c ) 56 , d ) 62 , e ) 54 | e | multiply(divide(144, power(const_2, 5)), const_2) | power(const_2,n0)|divide(n1,#0)|multiply(#1,const_2)| | general |
on dividing 23 by a number , the quotient is 4 and the remainder is 3 . find the divisor . | "d = ( d - r ) / q = ( 23 - 3 ) / 4 = 20 / 4 = 5 d" | a ) 1 , b ) 2 , c ) 4 , d ) 5 , e ) 7 | d | floor(divide(23, 4)) | divide(n0,n1)|floor(#0)| | general |
the diagonal of a rectangle is 41 cm and its area is 20 sq . cm . the perimeter of the rectangle must be : | l 2 + b 2 = 41 . also , lb = 20 . ( l + b ) 2 = ( l 2 + b 2 ) + 2 lb = 41 + 40 = 81 ( l + b ) = 9 . perimeter = 2 ( l + b ) = 18 cm . a ) | ['a ) 18 cm', 'b ) 28 cm', 'c ) 30 cm', 'd ) 38 cm', 'e ) 42 cm'] | a | multiply(const_2, sqrt(add(41, multiply(const_2, 20)))) | multiply(n1,const_2)|add(n0,#0)|sqrt(#1)|multiply(#2,const_2) | geometry |
a boy is travelling from his home to school at 12 km / hr and reached 7 min late . next day he traveled at 16 km / hr and reached 8 min early . distance between home and school ? | "let the distance be x t 1 = x / 12 hr t 2 = x / 16 hr difference in time = 7 + 8 = 15 = 1 / 4 hr x / 12 - x / 16 = 1 / 4 x / 48 = 1 / 4 x = 12 km answer is a" | a ) 12 km , b ) 13 km , c ) 14 km , d ) 15 km , e ) 16 km | a | divide(add(divide(7, const_60), divide(8, const_60)), divide(const_1, 16)) | divide(n1,const_60)|divide(n3,const_60)|divide(const_1,n2)|add(#0,#1)|divide(#3,#2)| | physics |
if a and b get profits of rs . 15,000 and rs . 5,000 respectively at the end of year then ratio of their investments are | "ratio = 15000 / 5000 = 3 : 1 answer : c" | a ) 4 : 1 , b ) 1 : 4 , c ) 3 : 1 , d ) 2 : 3 , e ) 2 : 5 | c | divide(multiply(multiply(multiply(const_3, const_2), const_100), const_100), divide(multiply(multiply(multiply(const_3, const_2), const_100), const_100), multiply(add(const_2, const_3), const_2))) | add(const_2,const_3)|multiply(const_2,const_3)|multiply(#1,const_100)|multiply(#0,const_2)|multiply(#2,const_100)|divide(#4,#3)|divide(#4,#5)| | gain |
after an ice began to melt out from the freezer , in the first hour lost 3 / 4 , in the second hour lost 3 / 4 of its remaining . if after two hours , the volume is 0.4 cubic inches , what is the original volume of the cubic ice , in cubic inches ? | let initial volume of ice be = x ice remaining after 1 hour = x - 0.75 x = 0.25 x ice remaining after 2 hour = ( 1 / 4 ) x - ( 3 / 4 * 1 / 4 * x ) = ( 1 / 16 ) x ( 1 / 16 ) x = 0.4 x = 6.4 alternate solution : try to backsolve . initial volume = 6.4 after one hour - - > ( 1 / 4 ) 6.4 = 1.6 after two hours - - > ( 1 / 4... | ['a ) 2.5', 'b ) 3.0', 'c ) 4.0', 'd ) 6.4', 'e ) 8.0'] | d | divide(divide(0.4, const_0_25), const_0_25) | divide(n4,const_0_25)|divide(#0,const_0_25) | physics |
the speed of a car is 10 km in the first hour and 60 km in the second hour . what is the average speed of the car ? | "s = ( 10 + 60 ) / 2 = 35 kmph answer : c" | a ) 79 kmph , b ) 85 kmph , c ) 35 kmph , d ) 23 kmph , e ) 14 kmph | c | divide(add(10, 60), const_2) | add(n0,n1)|divide(#0,const_2)| | physics |
how many integers between 324,700 and 436,600 have tens digit 1 and units digit 3 ? | "the integers are : 324,713 324,813 etc . . . 436,513 the number of integers is 4366 - 3247 = 1119 the answer is d ." | a ) 327 , b ) 645 , c ) 987 , d ) 1119 , e ) 1339 | d | subtract(436,600, add(add(multiply(const_2, const_100), multiply(add(const_3, const_4), const_10)), const_2)) | add(const_3,const_4)|multiply(const_100,const_2)|multiply(#0,const_10)|add(#1,#2)|add(#3,const_2)|subtract(n1,#4)| | general |
plaudia can choose any two of 4 different candles and any 8 of 9 different flowers for a centerpiece arrangement . given these choices , how many candle + flower groupings can she select ? | plaudia can choose any two of four different candles and any 8 of 9 different flowers for a centerpiece arrangement . 2 candles from 4 candles : 4 c 2 : 6 8 flowers from 9 flowers : 9 c 8 : 9 so total grouping 9 * 6 = 54 . answer is a . | a ) 54 , b ) 72 , c ) 96 , d ) 144 , e ) 432 | a | multiply(subtract(const_10, 4), 9) | subtract(const_10,n0)|multiply(n2,#0) | general |
in town x , 64 percent of the population are employed , and 55 percent of the population are employed males . what percent of the employed people in town x are females ? | "we are asked to find the percentage of females in employed people . total employed people 64 % , out of which 55 are employed males , hence 9 % are employed females . ( employed females ) / ( total employed people ) = 9 / 64 = 14 % answer : a ." | a ) 14 % , b ) 25 % , c ) 32 % , d ) 40 % , e ) 52 % | a | multiply(divide(subtract(64, 55), 64), const_100) | subtract(n0,n1)|divide(#0,n0)|multiply(#1,const_100)| | gain |
for an agricultural experiment , 300 seeds were planted in one plot and 200 were planted in a second plot . if exactly 30 percent of the seeds in the first plot germinated and exactly 35 percent of the seeds in the second plot germinated , what percent of the total number of seeds germinated ? | "in the first plot 30 % of 300 seeds germinated , so 0.3 x 300 = 90 seeds germinated . in the second plot , 35 % of 200 seeds germinated , so 0.35 x 200 = 70 seeds germinated . since 90 + 70 = 260 seeds germinated out of a total of 300 + 200 = 500 seeds , the percent of seeds that germinated is ( 160 / 500 ) x 100 % , ... | a ) 12 % , b ) 26 % , c ) 29 % , d ) 32 % , e ) 60 % | d | multiply(divide(add(multiply(divide(30, const_100), 300), multiply(divide(35, const_100), 200)), add(300, 200)), const_100) | add(n0,n1)|divide(n2,const_100)|divide(n3,const_100)|multiply(n0,#1)|multiply(n1,#2)|add(#3,#4)|divide(#5,#0)|multiply(#6,const_100)| | general |
ifaequals the sum of the even integers from 2 to 20 , inclusive , andbequals the sum of the odd integers from 1 to 19 , inclusive , what is the value of a + b ? | "answer is 10 yes ! there is really a faster way to solve it . sum of consecutive odd or even integers = ( no . of odd or even ints ) * ( first int + last int ) / 2 here a = sum of even ints from 2 to 20 , inclusive number of even ints = 10 , first int + last int = 2 + 20 = 22 a = 10 * 22 / 2 = 110 b = sum of odd ints ... | a ) 110 , b ) 100 , c ) 190 , d ) 200 , e ) 210 | e | subtract(multiply(divide(20, 2), add(divide(20, 2), 1)), multiply(divide(add(19, 1), 2), add(divide(subtract(19, 1), 2), 1))) | add(n2,n3)|divide(n1,n0)|subtract(n3,n2)|add(n2,#1)|divide(#2,n0)|divide(#0,n0)|add(n2,#4)|multiply(#3,#1)|multiply(#6,#5)|subtract(#7,#8)| | general |
a car takes 6 hours to cover a distance of 540 km . how much should the speed in kmph be maintained to cover the same direction in 3 / 2 th of the previous time ? | "time = 6 distance = 540 3 / 2 of 6 hours = 6 * 3 / 2 = 9 hours required speed = 540 / 9 = 60 kmph e )" | a ) 35 kmph , b ) 48 kmph , c ) 52 kmph , d ) 54 kmph , e ) 60 kmph | e | divide(540, divide(multiply(6, 3), 2)) | multiply(n0,n2)|divide(#0,n3)|divide(n1,#1)| | physics |
how long will a boy take to run round a square field of side 50 meters , if he runs at the rate of 12 km / hr ? | "speed = 12 km / hr = 12 * 5 / 18 = 10 / 3 m / sec distance = 50 * 4 = 200 m time taken = 200 * 3 / 10 = 60 sec answer is c" | a ) 52 sec , b ) 45 sec , c ) 60 sec , d ) 25 sec , e ) 39 sec | c | divide(multiply(50, const_4), multiply(12, divide(const_1000, const_3600))) | divide(const_1000,const_3600)|multiply(n0,const_4)|multiply(n1,#0)|divide(#1,#2)| | gain |
a batsman in his 17 th innings makes a score of 85 and their by increasing his average by 3 . what is his average after the 17 th innings ? | "explanation : 16 x + 85 = 17 ( x + 3 ) x = 34 + 3 = 37 answer : d" | a ) 28 , b ) 27 , c ) 12 , d ) 37 , e ) 01 | d | add(subtract(85, multiply(17, 3)), 3) | multiply(n0,n2)|subtract(n1,#0)|add(n2,#1)| | general |
running at the same constant rate , 6 identical machines can produce a total of 270 bottles per minute . at this rate , how many bottles could 5 such machines produce in 4 minutes ? | "6 machines produce 270 bottles per minute ; 1 machine produces 270 / 6 = 45 bottles per minute ; 5 machines produce 45 * 5 = 225 bottles per minute ; in 4 minutes 10 machines produce 225 * 4 = 900 bottles . answer : a ." | a ) 900 , b ) 1,800 , c ) 2,700 , d ) 10,800 , e ) 64,800 | a | multiply(multiply(divide(270, 6), 4), 5) | divide(n1,n0)|multiply(n3,#0)|multiply(n2,#1)| | gain |
when a train travels at a speed of 70 kmph , it reaches the destination on time . when the same train travels at a speed of 50 kmph , it reaches its destination 15 min late . what is the length of journey ? | "let x be the time reached with the speed 70 km / h 50 km / h - - - - > x + 15 distance is equal so 70 ( km / h ) × xhr = 50 ( km / h ) × ( x + 15 ) hr so 70 x = 50 x + 750 so the would be in km and x = 37.5 answer : a" | a ) 37.5 km , b ) 50.5 km , c ) 60 km , d ) 85.5 km , e ) 95 km | a | multiply(const_60, divide(multiply(divide(15, const_60), 50), 50)) | divide(n2,const_60)|multiply(n1,#0)|divide(#1,n1)|multiply(#2,const_60)| | physics |
p , q and r have $ 9000 among themselves . r has two - thirds of the total amount with p and q . find the amount with r ? | "b 3600 let the amount with r be $ r r = 2 / 3 ( total amount with p and q ) r = 2 / 3 ( 9000 - r ) = > 3 r = 18000 - 2 r = > 5 r = 18000 = > r = 3600 ." | a ) 2400 , b ) 3600 , c ) 3998 , d ) 2539 , e ) 1930 | b | divide(multiply(9000, multiply(const_2, const_2)), add(add(multiply(divide(multiply(const_2, const_2), const_3), const_3), multiply(const_1, const_3)), multiply(const_1, const_3))) | multiply(const_2,const_2)|multiply(const_1,const_3)|divide(#0,const_3)|multiply(n0,#0)|multiply(#2,const_3)|add(#4,#1)|add(#5,#1)|divide(#3,#6)| | general |
if two dice are thrown together , the probability of getting an even number on one die and an odd number on the other is - . | "answer : option b the number of exhaustive outcomes is 36 . let e be the event of getting an even number on one die and an odd number on the other . let the event of getting either both even or both odd then = 18 / 36 = 1 / 2 p ( e ) = 1 - 1 / 2 = 1 / 2 ." | a ) 1 / 3 , b ) 1 / 2 , c ) 7 / 8 , d ) 3 / 4 , e ) 7 / 3 | b | divide(const_6, multiply(const_6, const_6)) | multiply(const_6,const_6)|divide(const_6,#0)| | probability |
the sum of the present ages of two persons a and b is 60 . if the age of a is twice that of b , find the sum of their ages 2 years hence ? | "explanation : a + b = 60 , a = 2 b 2 b + b = 60 = > b = 20 then a = 40 . 2 years , their ages will be 42 and 22 . sum of their ages = 42 + 22 = 64 . c )" | a ) 58 , b ) 59 , c ) 64 , d ) 70 , e ) 74 | c | add(add(multiply(divide(60, 2), const_2), 2), add(divide(60, 2), 2)) | divide(n0,n1)|add(#0,n1)|multiply(#0,const_2)|add(#2,n1)|add(#3,#1)| | general |
what is the units digit of the expression 13 ^ 27 + 1 ? | "power of 3 follows repetition pattern of = 3 , 9 , 7 , 1 , 3 , 9 , 7 , 1 . . . . repetition after 4 powers = ( 3 , 9 , 7 , 1 ) 13 ^ 27 = 27 th power = 24 + 3 24 / 4 = 6 ( 6 repetitions ) + 3 = ` ` 7 ' ' 7 + 1 = 8 answer d" | a ) 1 , b ) 3 , c ) 4 , d ) 8 , e ) 9 | d | add(add(const_4, const_3), const_2) | add(const_3,const_4)|add(#0,const_2)| | general |
how long does a train 110 m long traveling at 60 kmph takes to cross a bridge of 340 m in length ? | "d = 110 + 340 = 450 m s = 60 * 5 / 18 = 50 / 3 t = 450 * 3 / 50 = 27.00 sec answer : d" | a ) 18.9 sec , b ) 88.9 sec , c ) 22.9 sec , d ) 27.00 sec , e ) 72.0 sec | d | divide(add(110, 340), multiply(60, const_0_2778)) | add(n0,n2)|multiply(n1,const_0_2778)|divide(#0,#1)| | physics |
the cost price of 30 articles is the same as the selling price of x articles . if the profit is 25 % , what is x ? | let the cost price = y the cost price of 30 articles = 30 y the selling price of x articles = 1.25 y * x 1.25 y * x = 30 y x = 30 / 1.25 = 24 the answer is d . | a ) 18 , b ) 20 , c ) 22 , d ) 24 , e ) 26 | d | divide(multiply(30, const_4), add(const_4, const_1)) | add(const_1,const_4)|multiply(n0,const_4)|divide(#1,#0) | gain |
a is two years older than b who is twice as old as c . if the total of the ages of a , b and c be 42 , the how old is b ? | "explanation : let c ' s age be x years . then , b ' s age = 2 x years . a ' s age = ( 2 x + 2 ) years . ( 2 x + 2 ) + 2 x + x = 42 ⇒ 5 x = 40 ⇒ x = 8 . hence , b ' s age = 2 x = 16 years . answer : a" | a ) 16 , b ) 9 , c ) 8 , d ) 11 , e ) 10 | a | divide(multiply(subtract(42, const_2), const_2), add(const_4, const_1)) | add(const_1,const_4)|subtract(n0,const_2)|multiply(#1,const_2)|divide(#2,#0)| | general |
calculate the ratio between x and y if 15 % of x equal to 50 % of y ? | "explanation : 15 x = 50 y x : y = 15 : 50 = 3 : 10 answer : d" | a ) 3 : 5 , b ) 3 : 7 , c ) 3 : 8 , d ) 3 : 10 , e ) 3 : 9 | d | divide(15, 50) | divide(n0,n1)| | general |
if 0.8 of a number is equal to 0.09 of another number , the ratio of the numbers i | "sol . 0.8 a = 0.09 b â ‡ ” a / b = 0.09 / 0.8 = 9 / 80 = . â ˆ ´ a : b = 9 : 80 . answer c" | a ) 2 : 3 , b ) 3 : 4 , c ) 9 : 80 , d ) 20 : 3 , e ) none | c | divide(multiply(0.09, const_100), multiply(0.8, const_100)) | multiply(n1,const_100)|multiply(n0,const_100)|divide(#0,#1)| | other |
the product of 3 consecutive numbers is 210 . then the sum of the smallest two numbers is ? | product of three numbers = 210 210 = 2 * 3 * 5 * 7 = 5 * 6 * 7 . so , the three numbers are 5 , 6 and 7 . and sum of smallest of these two = 5 + 6 = 11 . answer : option a | a ) 11 , b ) 15 , c ) 20 , d ) 38 , e ) 56 | a | add(add(floor(power(210, const_0_33)), const_1), floor(power(210, const_0_33))) | power(n1,const_0_33)|floor(#0)|add(#1,const_1)|add(#2,#1) | general |
how many integers between 263,205 and 658,805 have tens digit 1 and units digit 3 ? | "there is one number in hundred with 1 in the tens digit and 3 in the units digit : 13 , 113 , 213 , 313 , . . . the difference between 263,205 and is 658,805 - 263205 = 395,600 - one number per each hundred gives 133,900 / 100 = 3,956 answer : c ." | a ) 3,955 , b ) 3,925 , c ) 3,956 , d ) 3,926 , e ) 3,915 | c | subtract(658,805, add(add(multiply(const_2, const_100), multiply(add(const_3, const_4), const_10)), const_2)) | add(const_3,const_4)|multiply(const_100,const_2)|multiply(#0,const_10)|add(#1,#2)|add(#3,const_2)|subtract(n1,#4)| | general |
the diagonals of a rhombus are 11 cm and 20 cm . find its area ? | "1 / 2 * 11 * 20 = 110 answer : e" | a ) 329 , b ) 288 , c ) 150 , d ) 238 , e ) 110 | e | rhombus_area(11, 20) | rhombus_area(n0,n1)| | geometry |
a man can row his boat with the stream at 36 km / h and against the stream in 8 km / h . the man ' s rate is ? | "explanation : ds = 36 us = 8 s = ? s = ( 36 - 8 ) / 2 = 14 kmph answer : d" | a ) 1 kmph , b ) 6 kmph , c ) 7 kmph , d ) 14 kmph , e ) 9 kmph | d | divide(subtract(36, 8), const_2) | subtract(n0,n1)|divide(#0,const_2)| | gain |
ashok secured average of 75 marks in 6 subjects . if the average of marks in 5 subjects is 74 , how many marks did he secure in the 6 th subject ? | "explanation : number of subjects = 6 average of marks in 6 subjects = 75 therefore total marks in 6 subjects = 75 * 6 = 450 now , no . of subjects = 5 total marks in 5 subjects = 74 * 5 = 370 therefore marks in 6 th subject = 450 – 370 = 80 answer : e" | a ) 38 , b ) 27 , c ) 99 , d ) 17 , e ) 80 | e | subtract(multiply(75, 6), multiply(74, 5)) | multiply(n0,n1)|multiply(n2,n3)|subtract(#0,#1)| | general |
a metallic sheet is of rectangular shape with dimensions 48 m x 36 m . from each of its corners , a square is cut off so as to make an open box . if the length of the square is 5 m , the volume of the box ( in m 3 ) is : | "clearly , l = ( 48 - 10 ) m = 38 m , b = ( 36 - 10 ) m = 26 m , h = 8 m . volume of the box = ( 38 x 26 x 5 ) m 3 = 4940 m 3 . answer : option a" | a ) 4940 , b ) 5120 , c ) 6420 , d ) 8960 , e ) 7960 | a | volume_rectangular_prism(subtract(48, multiply(5, const_2)), subtract(36, multiply(5, const_2)), 5) | multiply(n2,const_2)|subtract(n0,#0)|subtract(n1,#0)|volume_rectangular_prism(n2,#1,#2)| | geometry |
a rectangular lawn of dimensions 80 m * 60 m has two roads each 15 m wide running in the middle of the lawn , one parallel to the length and the other parallel to the breadth . what is the cost of traveling the two roads at rs . 3 per sq m ? | "explanation : area = ( l + b – d ) d ( 80 + 60 – 15 ) 15 = > 1875 m 2 1875 * 3 = rs . 5625 answer : option e" | a ) a ) 3820 , b ) b ) 930 , c ) c ) 9309 , d ) d ) 3900 , e ) e ) 5625 | e | multiply(multiply(subtract(add(80, 60), 15), 15), 3) | add(n0,n1)|subtract(#0,n2)|multiply(n2,#1)|multiply(n3,#2)| | geometry |
a rectangular plot measuring 90 metres by 50 metres is to be enclosed by wire fencing . if the poles of the fence are kept 4 metres apart , how many poles will be needed ? | solution perimeter of the plot = 2 ( 90 + 50 ) = 280 m . ∴ number of poles = [ 280 / 4 ] = 70 m answer b | ['a ) 55', 'b ) 70', 'c ) 57', 'd ) 58', 'e ) none of these'] | b | divide(rectangle_perimeter(90, 50), 4) | rectangle_perimeter(n0,n1)|divide(#0,n2) | physics |
if the capacity of a cylindrical tank is 1848 m 3 and the diameter of its base is 14 m , then find the depth of the tank . | let the depth of the tank be h meters . then , ∏ x 72 x h = 1848 h = ( 1848 x ( 7 / 22 ) x ( 1 / 49 ) = 12 m answer is c | ['a ) 10', 'b ) 11', 'c ) 12', 'd ) 13', 'e ) none of them'] | c | divide(1848, circle_area(divide(14, const_2))) | divide(n2,const_2)|circle_area(#0)|divide(n0,#1) | physics |
if a train , travelling at a speed of 160 kmph , crosses a pole in 18 sec , then the length of train is ? | "d = 160 * 5 / 18 * 18 = 800 m answer : a" | a ) 800 , b ) 125 , c ) 288 , d ) 266 , e ) 121 | a | multiply(multiply(160, const_0_2778), 18) | multiply(n0,const_0_2778)|multiply(n1,#0)| | physics |
a tank is 25 m long 12 m wide and 6 m deep . the cost of plastering its walls and bottom at 45 paise per sq m is | "explanation : area to be plastered = [ 2 ( l + b ) ã — h ] + ( l ã — b ) = [ 2 ( 25 + 12 ) ã — 6 ] + ( 25 ã — 12 ) = 744 sq m cost of plastering = 744 ã — ( 45 / 100 ) = rs . 334.80 answer : b" | a ) rs . 234.80 , b ) rs . 334.80 , c ) rs . 434.80 , d ) rs . 534.80 , e ) none of these | b | multiply(divide(45, const_100), add(multiply(25, 12), add(multiply(const_2, multiply(25, 6)), multiply(multiply(12, 6), const_2)))) | divide(n3,const_100)|multiply(n0,n2)|multiply(n1,n2)|multiply(n0,n1)|multiply(#1,const_2)|multiply(#2,const_2)|add(#4,#5)|add(#6,#3)|multiply(#7,#0)| | physics |
a man rows his boat 120 km downstream and 60 km upstream , taking 2 hours each time . find the speed of the stream ? | "speed downstream = d / t = 120 / ( 2 ) = 60 kmph speed upstream = d / t = 60 / ( 2 ) = 30 kmph the speed of the stream = ( 60 - 30 ) / 2 = 15 kmph answer : b" | a ) 6 , b ) 15 , c ) 17 , d ) 18 , e ) 2 | b | divide(subtract(divide(120, 2), divide(60, 2)), const_2) | divide(n0,n2)|divide(n1,n2)|subtract(#0,#1)|divide(#2,const_2)| | physics |
there are 14 slate rocks , 20 pumice rocks , and 10 granite rocks randomly distributed in a certain field . if 2 rocks are chosen at random and without replacement , what is the probability that both rocks will be slate rocks ? | "14 / 44 * 13 / 43 = 91 / 946 the answer is d ." | a ) 1 / 5 , b ) 11 / 22 , c ) 81 / 724 , d ) 91 / 946 , e ) 101 / 987 | d | multiply(divide(14, add(add(14, 20), 10)), divide(subtract(14, const_1), subtract(add(add(14, 20), 10), const_1))) | add(n0,n1)|subtract(n0,const_1)|add(n2,#0)|divide(n0,#2)|subtract(#2,const_1)|divide(#1,#4)|multiply(#3,#5)| | other |
my grandson is about as many days as my son in weeks , and my grandson is as many months as i am in years . my grandson , my son and i together are 150 years . can you tell my son age in years ? | let m be my age in years . if s is my son ' s age in years , then my son is 52 s weeks old . if g is my grandson ' s age in years , then my grandson is 365 g days old . thus , 365 g = 52 s . since my grandson is 12 g months old , 12 g = m . since my grandson , my son and i together are 150 years , g + s + m = 100 . the... | a ) 63 years , b ) 60 years , c ) 50 years , d ) 53 years , e ) 46 years | d | multiply(divide(150, add(const_10, const_10)), const_12) | add(const_10,const_10)|divide(n0,#0)|multiply(#1,const_12)| | physics |
a merchant sells an item at a 20 % discount , but still makes a gross profit of 20 percent of the cost . what percent t of the cost would the gross profit on the item have been if it had been sold without the discount ? | "let the market price of the product is mp . let the original cost price of the product is cp . selling price ( discounted price ) = 100 % of mp - 20 % mp = 80 % of mp . - - - - - - - - - - - - - - - - ( 1 ) profit made by selling at discounted price = 20 % of cp - - - - - - - - - - - - - - ( 2 ) apply the formula : pr... | a ) 20 % , b ) 40 % , c ) 50 % , d ) 60 % , e ) 75 % | c | subtract(const_100, subtract(subtract(const_100, 20), 20)) | subtract(const_100,n0)|subtract(#0,n1)|subtract(const_100,#1)| | gain |
in a race of 1000 m , a can beat by 100 m , in a race of 800 m , b can beat c by 100 m . by how many meters will a beat c in a race of 900 m ? | "when a runs 1000 m , b runs 900 m and when b runs 800 m , c runs 700 m . when b runs 900 m , distance that c runs = ( 900 * 700 ) / 800 = 6300 / 8 = 787.5 m . in a race of 1000 m , a beats c by ( 1000 - 787.5 ) = 212.5 m to c . in a race of 900 m , the number of meters by which a beats c = ( 900 * 212.5 ) / 1000 = 191... | a ) 122.9 m , b ) 127.5 m . , c ) 191.25 m , d ) 222.9 m , e ) 12289 m | c | subtract(1000, divide(multiply(subtract(1000, 100), subtract(800, 100)), 800)) | subtract(n0,n1)|subtract(n2,n1)|multiply(#0,#1)|divide(#2,n2)|subtract(n0,#3)| | physics |
if rs . 7500 are borrowed at c . i at the rate of 4 % per annum , then after 3 years the amount to be paid is ? | a = 7500 ( 26 / 25 ) ^ 3 = 8436 answer : d | a ) 3377 , b ) 2688 , c ) 2688 , d ) 8436 , e ) 1268 | d | multiply(7500, multiply(divide(add(const_100, 4), const_100), divide(add(const_100, 4), const_100))) | add(n1,const_100)|divide(#0,const_100)|multiply(#1,#1)|multiply(n0,#2) | gain |
two friends decide to get together ; so they start riding bikes towards each other . they plan to meet halfway . each is riding at 6 mph . they live 36 miles apart . one of them has a pet carrier pigeon and it starts flying the instant the friends start traveling . the pigeon flies back and forth at 21 mph between the ... | "e 63 it takes 3 hours for the friends to meet ; so the pigeon flies for 3 hours at 21 mph = 63 miles" | a ) 54 , b ) 66 , c ) 68 , d ) 60 , e ) 63 | e | multiply(21, divide(divide(36, 2), 6)) | divide(n1,n3)|divide(#0,n0)|multiply(n2,#1)| | physics |
the sum of number of boys and girls in a school is 400 . if the number of boys is x , then the number of girls becomes x % of the total number of students . the number of boys is ? | "we have x + x % of 400 = 400 x + x / 100 * 400 = 400 5 * x = 400 x = 80 answer is d" | a ) 50 , b ) 40 , c ) 60 , d ) 80 , e ) 70 | d | divide(400, add(divide(400, const_100), const_1)) | divide(n0,const_100)|add(#0,const_1)|divide(n0,#1)| | general |
a customer bought a product at the shop . however , the shopkeeper increased the price of the product by 15 % so that the customer could not buy the required amount of the product . the customer managed to buy only 60 % of the required amount . what is the difference in the amount of money that the customer paid for th... | "let x be the amount of money paid for the first purchase . the second time , the customer paid 0.6 ( 1.15 x ) = 0.69 x . the difference is 31 % . the answer is c ." | a ) 27 % , b ) 29 % , c ) 31 % , d ) 33 % , e ) 35 % | c | multiply(subtract(const_1, multiply(add(divide(15, const_100), const_1), divide(60, const_100))), const_100) | divide(n0,const_100)|divide(n1,const_100)|add(#0,const_1)|multiply(#2,#1)|subtract(const_1,#3)|multiply(#4,const_100)| | general |
the h . c . f . of two numbers is 12 and their l . c . m . is 600 . if one of the number is 30 , find the other ? | "other number = 12 * 600 / 30 = 240 answer is b" | a ) 100 , b ) 240 , c ) 120 , d ) 200 , e ) 150 | b | multiply(12, 30) | multiply(n0,n2)| | physics |
if xy = 1 , x / y = 36 , for positive numbers x and y , y = ? | "very easy question . 2 variables and 2 easy equations . xy = 1 - - - > x = 1 / y - ( i ) x / y = 36 - - - > replacing ( i ) here - - - > 1 / ( y ^ 2 ) = 36 - - - > y ^ 2 = 1 / 36 - - - > y = 1 / 6 or - 1 / 6 the question states that x and y are positive integers . therefore , y = 1 / 6 is the answer . answer e ." | a ) 1 / 2 , b ) 2 , c ) 1 / 3 , d ) 3 , e ) 1 / 6 | e | sqrt(divide(1, 36)) | divide(n0,n1)|sqrt(#0)| | general |
for covering 88 km a wheel revolve 1000 times . what is the radius of wheel ? | distance travel in 1 round = 88000 / 1000 m = 88 m perimeter = 88 m , 2 π r = 882 r = [ 88 * 7 / 22 ] = 28 m hence , diameter = 28 m answer c | a ) 14 m , b ) 24 m , c ) 28 m , d ) 40 m , e ) none of the above | c | multiply(multiply(divide(divide(multiply(divide(88, 1000), add(const_3, const_4)), const_2), add(multiply(const_10, const_2), const_2)), const_1000), const_2) | add(const_3,const_4)|divide(n0,n1)|multiply(const_10,const_2)|add(#2,const_2)|multiply(#0,#1)|divide(#4,const_2)|divide(#5,#3)|multiply(#6,const_1000)|multiply(#7,const_2) | general |
a car covers a distance of 642 km in 6 ½ hours . find its speed ? | 642 / 6 = 107 kmph answer : e | a ) 104 kmph , b ) 176 kmph , c ) 298 kmph , d ) 186 kmph , e ) 107 kmph | e | divide(642, 6) | divide(n0,n1) | physics |
a can complete a certain job in 18 days . b is 80 % more efficient than a . in how many days can b complete the same job ? | "let , total work unit = 180 units a can finish in 18 days = 180 unit work i . e . a can finish in 1 days = 10 unit work i . e . b can finish in 1 days = 10 + ( 80 / 100 ) * 10 = 18 unit work days in which b will complete the work alone = 180 / 18 = 10 days answer : option e" | a ) 6 , b ) 6.25 , c ) 7 , d ) 7.5 , e ) 10 | e | divide(multiply(18, 80), const_100) | multiply(n0,n1)|divide(#0,const_100)| | gain |
i sold a book at a profit of 12 % . had i sold it for $ 18 more , 18 % would have been gained . find the cost price ? | "118 % of cost - 112 % of cost = $ 18 6 % of cost = $ 18 cost = 18 * 100 / 6 = $ 300 answer is c" | a ) $ 100 , b ) $ 250 , c ) $ 300 , d ) $ 200 , e ) $ 180 | c | divide(multiply(divide(multiply(18, 12), subtract(18, 12)), const_100), 12) | multiply(n0,n1)|subtract(n2,n0)|divide(#0,#1)|multiply(#2,const_100)|divide(#3,n0)| | gain |
the speed of a boat in still water is 20 km / hr and the rate of current is 5 km / hr . the distance travelled downstream in 21 minutes is : | "explanation : speed downstream = ( 20 + 5 ) kmph = 25 kmph distance travelled = ( 25 * ( 21 / 60 ) ) km = 8.75 km . answer : c" | a ) 9.75 , b ) 5.75 , c ) 8.75 , d ) 6.75 , e ) 5.15 | c | multiply(divide(21, const_60), add(20, 5)) | add(n0,n1)|divide(n2,const_60)|multiply(#0,#1)| | physics |
if the price of an article went up by 25 % , 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 . 25 % of 100 = 25 . new price = 100 + 25 = rs . 125 required percentage = ( 125 - 100 ) / 125 * 100 = 25 / 125 * 100 = 20 % . answer : a | a ) 20 % , b ) 2 / 8 % , c ) 2 / 1 % , d ) 1 / 3 % , e ) 2 / 7 % | a | multiply(divide(25, add(const_100, 25)), const_100) | add(n0,const_100)|divide(n0,#0)|multiply(#1,const_100) | gain |
a retailer buys 40 pens at the market price of 36 pens from a wholesaler , if he sells these pens giving a discount of 1 % , what is the profit % ? | "let the market price of each pen be rs 1 then , c . p of 40 pens = rs 36 s . p of 40 pens = 99 % of rs 40 = rs 39.60 profit % = ( ( 3.60 * 100 ) / 36 ) % = 10 % answer is b ." | a ) 5 , b ) 10 , c ) 15 , d ) 20 , e ) none of them | b | multiply(const_100, divide(subtract(subtract(1, divide(1, const_100)), divide(36, 40)), divide(36, 40))) | divide(n2,const_100)|divide(n1,n0)|subtract(n2,#0)|subtract(#2,#1)|divide(#3,#1)|multiply(#4,const_100)| | gain |
given a circular wheel of 3.62 m radius how many revolutions will the will make to travel a distance of 1 km ? | 2 * 22 / 7 * 3.62 * x = 22754.3 x = 1000 answer : a | a ) 22754.3 , b ) 22254.2 , c ) 25754.2 , d ) 22753.2 , e ) 22254.2 | a | multiply(multiply(multiply(divide(multiply(add(const_10, const_1), const_2), add(const_3, const_4)), const_2), 3.62), const_1000) | add(const_1,const_10)|add(const_3,const_4)|multiply(#0,const_2)|divide(#2,#1)|multiply(#3,const_2)|multiply(n0,#4)|multiply(#5,const_1000) | physics |
find the least number must be subtracted from 427398 so that remaining no . is divisible by 15 ? | "on dividing 427398 by 15 we get the remainder 3 , so 3 should be subtracted b" | a ) 2 , b ) 3 , c ) 5 , d ) 7 , e ) 8 | b | subtract(427398, multiply(floor(divide(427398, 15)), 15)) | divide(n0,n1)|floor(#0)|multiply(n1,#1)|subtract(n0,#2)| | general |
if 15 students in a class average 73 % on an exam and 10 students average 88 % on the same exam , what is the average in percent for all 25 students ? | "( 15 * 73 + 10 * 88 ) / 25 = 79 % the answer is c ." | a ) 77 % , b ) 78 % , c ) 79 % , d ) 80 % , e ) 81 % | c | divide(add(multiply(15, 73), multiply(10, 88)), 25) | multiply(n0,n1)|multiply(n2,n3)|add(#0,#1)|divide(#2,n4)| | general |
if the price of a tv is first decreased by 10 % and then increased by 30 % , then the net change in the price will be : | "explanation : solution : let the original price be rs . 100 . new final price = 130 % of ( 10 % of 100 ) = rs . 130 / 100 * 90 / 100 * 100 = rs . 117 . . ' . increase = 17 % answer : b" | a ) 4 % increase , b ) 17 % increase , c ) 10 % decrease , d ) 6 % increase , e ) none of these | b | subtract(divide(multiply(subtract(const_100, 10), add(const_100, 30)), const_100), const_100) | add(n1,const_100)|subtract(const_100,n0)|multiply(#0,#1)|divide(#2,const_100)|subtract(#3,const_100)| | general |
if the true discount on a sum due 2 years hence at 14 % per annum be rs . 168 , the sum due is : | solution p . w = 100 xt . d . / r x t = 100 x 168 / 14 x 2 = 600 . ∴ sum = ( p . w . + td . ) = rs . 768 . answer a | a ) rs . 768 , b ) rs . 968 , c ) rs . 1960 , d ) rs . 2400 , e ) none | a | add(divide(multiply(const_100, 168), multiply(14, 2)), 168) | multiply(n2,const_100)|multiply(n0,n1)|divide(#0,#1)|add(n2,#2) | gain |
the difference between the ages of two persons is 30 years . fifteen years ago , the elder one was twice as old as the younger one . the present age of the younger person is ? | "let their ages be x years and ( x + 30 ) years then , ( x + 30 ) - 15 = 2 ( x - 15 ) x + 15 = 2 x - 30 x = 45 answer is c" | a ) 30 yr , b ) 25 yr , c ) 45 yr , d ) 40 yr , e ) 50 yr | c | add(subtract(30, subtract(30, add(const_3, const_2))), multiply(subtract(30, add(const_3, const_2)), const_2)) | add(const_2,const_3)|subtract(n0,#0)|multiply(#1,const_2)|subtract(n0,#1)|add(#2,#3)| | general |
a store owner estimates that the average price of type a products will increase by 20 % next year and that the price of type b products will increase by 11 % next year . this year , the total amount paid for type a products was $ 3500 and the total price paid for type b products was $ 8600 . according to the store owne... | "cost of type a products next year = 1.20 * 3500 = 4200 cost of type b products next year = 1.11 * 8300 = 9546 total 4200 + 9546 = 13746 answer : a" | a ) $ 13,746 , b ) $ 15,325 , c ) $ 16,000 , d ) $ 16,225 , e ) $ 17,155 | a | multiply(divide(const_3, const_4), const_1000) | divide(const_3,const_4)|multiply(#0,const_1000)| | general |
a soft drink company had 6000 small and 10000 big bottles in storage . if 12 % of small 15 % of big bottles have been sold , then the total bottles remaining in storage is | "6000 + 10000 - ( 0.12 * 6000 + 0.15 * 10000 ) = 13780 . answer : b ." | a ) 15360 , b ) 13780 , c ) 15060 , d ) 14930 , e ) 16075 | b | subtract(add(6000, 10000), add(multiply(6000, divide(12, const_100)), multiply(divide(15, const_100), 10000))) | add(n0,n1)|divide(n2,const_100)|divide(n3,const_100)|multiply(n0,#1)|multiply(n1,#2)|add(#3,#4)|subtract(#0,#5)| | general |
in a race where 18 cars are running , the chance that car x will win is 1 / 8 , that y will win is 1 / 12 and that z will win is 1 / 6 . assuming that a dead heat is impossible , find the chance that one of them will win . | "required probability = p ( x ) + p ( y ) + p ( z ) ( all the events are mutually exclusive ) . = 1 / 8 + 1 / 12 + 1 / 6 = 3 / 8 answer : d" | a ) 1 / 140 , b ) 1 / 180 , c ) 3 / 12 , d ) 3 / 8 , e ) 57 / 120 | d | add(add(divide(1, 8), divide(1, 12)), divide(1, 6)) | divide(n1,n2)|divide(n1,n4)|divide(n1,n6)|add(#0,#1)|add(#3,#2)| | general |
the average marks of 25 students in a class is 100 . but a student mark is wrongly noted as 60 instead of 10 then find the correct average marks ? | "correct avg marks = 100 + ( 10 - 60 ) / 25 avg = 100 - 2 = 98 answer is c" | a ) a ) 78 , b ) b ) 82 , c ) c ) 98 , d ) d ) 91 , e ) e ) 85 | c | divide(add(subtract(multiply(100, 25), 60), 10), 25) | multiply(n0,n1)|subtract(#0,n2)|add(n3,#1)|divide(#2,n0)| | general |
what is the sum of the local values of the digits 2 , 3 , 4 , 5 in the number 2345 ? | 2000 + 300 + 40 + 5 = 2345 answer a | a ) 2345 , b ) 4500 , c ) 5000 , d ) 6000 , e ) 7000 | a | add(5, add(add(multiply(2, const_1000), multiply(3, const_100)), multiply(const_10, 4))) | multiply(n0,const_1000)|multiply(n1,const_100)|multiply(n2,const_10)|add(#0,#1)|add(#3,#2)|add(n3,#4) | general |
a started a business with an investment of rs . 70000 and after 6 months b joined him investing rs . 120000 . if the profit at the end of a year is rs . 91000 , then the share of b is ? | "ratio of investments of a and b is ( 70000 * 12 ) : ( 120000 * 6 ) = 7 : 6 total profit = rs . 91000 share of b = 6 / 13 ( 91000 ) = rs . 42000 answer : b" | a ) a ) 44500 , b ) b ) 42000 , c ) c ) 44098 , d ) d ) 43007 , e ) e ) 44098 | b | subtract(91000, multiply(const_60, const_100)) | multiply(const_100,const_60)|subtract(n3,#0)| | gain |
54671 - 14456 - 33490 = ? | "d if we calculate we will get 6725" | a ) 2449 , b ) 5449 , c ) 6749 , d ) 6725 , e ) 6468 | d | subtract(multiply(divide(54671, const_100), 14456), multiply(divide(const_1, const_3), multiply(divide(54671, const_100), 14456))) | divide(n0,const_100)|divide(const_1,const_3)|multiply(n1,#0)|multiply(#1,#2)|subtract(#2,#3)| | general |
if p and q are positive integers each greater than 1 , and 17 ( p + 1 ) = 28 ( q + 1 ) , what is the least possible value of p + q ? | "17 ( p + 1 ) = 29 ( q + 1 ) - - > ( p + 1 ) / ( q + 1 ) = 28 / 17 - - > the least positive value of p + 1 is 28 , so the least value of p is 27 and the least positive value of q + 1 is 17 , so the least value of q is 16 - - > the least value of p + q is 27 + 16 = 43 . answer : d" | a ) 36 , b ) 42 , c ) 44 , d ) 43 , e ) none | d | subtract(add(28, 17), const_2) | add(n1,n3)|subtract(#0,const_2)| | general |
what is the area of square field whose side of length 8 m ? | "8 * 8 = 64 sq m answer : b" | a ) 225 , b ) 64 , c ) 772 , d ) 882 , e ) 21 | b | square_area(8) | square_area(n0)| | geometry |
a thief is spotted by a policeman from a distance of 225 meters . when the policeman starts the chase , the thief also starts running . if the speed of the thief be 8 km / hr and that of the policeman 10 km / hr , how far the thief will have run before he is overtaken ? | "relative speed of the policeman = ( 10 - 8 ) km / hr = 2 km / hr . time taken by police man to cover ( 225 m / 1000 ) x 1 / 2 hr = 9 / 80 hr . in 9 / 80 hrs , the thief covers a distance of 8 x 9 / 80 km = 9 / 10 km = 900 m answer is d ." | a ) 350 m , b ) 200 m , c ) 400 m , d ) 900 m , e ) none of them | d | divide(multiply(225, 8), subtract(10, 8)) | multiply(n0,n1)|subtract(n2,n1)|divide(#0,#1)| | physics |
a , b , c can complete a piece of work in 24,6 and 12 days respectively . working together , they will complete the same work in how many days ? | "( a + b + c ) ' s 1 day work = ( 1 / 24 ) + ( 1 / 6 ) + ( 1 / 12 ) = 7 / 24 a , b , c together will complete the work in 24 / 7 days answer is c" | a ) 13 / 8 , b ) 26 / 11 , c ) 24 / 7 , d ) 12 / 13 , e ) 1 / 2 | c | inverse(add(inverse(12), add(inverse(const_3), inverse(add(add(const_4, const_4), const_1))))) | add(const_4,const_4)|inverse(const_3)|inverse(n1)|add(#0,const_1)|inverse(#3)|add(#1,#4)|add(#5,#2)|inverse(#6)| | physics |
mary works 9 hours per day on monday , wednesday and friday , and 5 hours per day on tuesday and thursday . she does not work on saturday and sunday . she earns $ 407 per week . how much does she earn in dollars per hour ? | so , she works 27 hours in 3 days so , she works 10 hours in 2 days so in a week she works 37 hours ( 27 + 10 ) and earns $ 407 so , hourly wage is 407 / 37 = > 11 hence answer will be ( e ) 11 | a ) 12 , b ) 10 , c ) 9 , d ) 8 , e ) 11 | e | divide(407, add(multiply(9, const_3), multiply(5, const_2))) | multiply(n0,const_3)|multiply(n1,const_2)|add(#0,#1)|divide(n2,#2) | physics |
when a number is divided by 4 & then multiply by 12 the answer is 9 what is the no . ? | if $ x $ is the number , x / 4 * 12 = 9 = > 3 x = 9 = > x = 3 a | a ) 3 , b ) 5 , c ) 5.6 , d ) 5.7 , e ) 6.5 | a | multiply(divide(9, 12), 4) | divide(n2,n1)|multiply(n0,#0) | general |
a toy store ' s revenue in november was 3 / 5 of its revenue in december and its revenue in january was 3 / 4 of its revenue in november , then the store ' s revenue in december was how many times the average ( arithmetic mean ) of its revenues in november and january ? | let dec rev = 100 then nov rev is 3 / 5 ( 100 ) = > 60 therefore jan rev = 3 / 4 ( nov rev ) = 3 / 4 ( 60 ) = > 45 hence dec rev = x * ( nov rev + jan rev ) / 2 100 = x * ( 60 + 45 ) / 2 x = 100 / 52.5 = > 1.90 = 2 ans ) d | a ) 1 / 4 , b ) 1 / 2 , c ) 2 / 3 , d ) 2 , e ) 4 | d | divide(const_1, divide(multiply(add(divide(3, 4), const_1), divide(3, 5)), const_2)) | divide(n0,n3)|divide(n0,n1)|add(#0,const_1)|multiply(#2,#1)|divide(#3,const_2)|divide(const_1,#4) | general |
of the goose eggs laid at a certain pond , 1 / 4 hatched and 4 / 5 of the geese that hatched from those eggs survived the first month . of the geese that survived the first month , 2 / 5 did not survive the first year . if 120 geese survived the first year and if no more than one goose hatched from each egg , how many ... | "let x be the number of eggs that were laid . ( 3 / 5 ) ( 4 / 5 ) ( 1 / 4 ) x = 120 ( 12 / 100 ) x = 120 x = 1000 the answer is e ." | a ) 600 , b ) 700 , c ) 800 , d ) 900 , e ) 1000 | e | divide(divide(divide(120, subtract(const_1, divide(4, 5))), divide(4, 5)), divide(1, 4)) | divide(n4,n5)|divide(n1,n3)|divide(n0,n1)|subtract(const_1,#0)|divide(n6,#3)|divide(#4,#1)|divide(#5,#2)| | general |
in a graduate physics course , 60 percent of the students are male and 30 percent of the students are married . if two - sevenths of the male students are married , what fraction of the female students is single ? | "let assume there are 100 students of which 60 are male and 40 are females if 30 are married then 70 will be single . now its given that two - sevenths of the male students are married that means 2 / 7 of 60 = 17 males are married if 30 is the total number of students who are married and out of that 17 are males then t... | a ) 2 / 7 , b ) 17 / 30 , c ) 1 / 2 , d ) 27 / 40 , e ) 5 / 7 | d | divide(const_10, 30) | divide(const_10,n1)| | gain |
calculate the average of first 18 even numbers is ? | explanation : sum of 10 even numbers = 18 * 19 = 342 average = 342 / 18 = 19 answer : option c | a ) 21 , b ) 29 , c ) 19 , d ) 14 , e ) 10 | c | add(18, const_1) | add(n0,const_1) | general |
an empty fuel tank with a capacity of 200 gallons was filled partially with fuel a and then to capacity with fuel b . fuel a contains 12 % ethanol by volume and fuel b contains 16 % ethanol by volume . if the full fuel tank contains 22 gallons of ethanol , how many gallons of fuel a were added ? | "say there are a gallons of fuel a in the tank , then there would be 200 - a gallons of fuel b . the amount of ethanol in a gallons of fuel a is 0.12 a ; the amount of ethanol in 200 - a gallons of fuel b is 0.16 ( 200 - a ) ; since the total amount of ethanol is 22 gallons then 0.12 a + 0.16 ( 200 - a ) = 22 - - > a =... | a ) 160 , b ) 150 , c ) 250 , d ) 80 , e ) 50 | c | divide(subtract(multiply(200, divide(16, const_100)), 22), subtract(divide(16, const_100), divide(12, const_100))) | divide(n2,const_100)|divide(n1,const_100)|multiply(n0,#0)|subtract(#0,#1)|subtract(#2,n3)|divide(#4,#3)| | gain |
in a class of 36 students 26 play football and play 20 long tennis , if 17 play above , many play neither ? | "26 + 20 - 17 = 29 36 - 29 = 7 play neither answer is a" | a ) 7 , b ) 8 , c ) 10 , d ) 12 , e ) 14 | a | subtract(36, subtract(add(26, 20), 17)) | add(n1,n2)|subtract(#0,n3)|subtract(n0,#1)| | other |
on dividing a number by 56 , we get 29 as remainder . on dividing the same number by 8 , what will be the remainder ? | "formula : ( divisor * quotient ) + remainder = dividend . soln : ( 56 * q ) + 29 = d - - - - - - - ( 1 ) d % 8 = r - - - - - - - - - - - - - ( 2 ) from equation ( 2 ) , ( ( 56 * q ) + 29 ) % 8 = r . = > assume q = 1 . = > ( 56 + 29 ) % 8 = r . = > 85 % 8 = r = > 5 = r . a )" | a ) 5 , b ) 7 , c ) 9 , d ) 11 , e ) 12 | a | multiply(subtract(divide(power(29, const_2), 56), floor(divide(power(29, const_2), 56))), 56) | power(n1,const_2)|divide(#0,n0)|floor(#1)|subtract(#1,#2)|multiply(n0,#3)| | general |
given that p is a positive even integer with a positive units digit , if the units digit of p ^ 3 minus the units digit of p ^ 2 is equal to 0 , what is the units digit of p + 2 ? | "p is a positiveeveninteger with apositive units digit - - > the units digit of p can be 2 , 4 , 6 , or 8 - - > in order the units digit of p ^ 3 - p ^ 2 to be 0 , the units digit of p ^ 3 and p ^ 2 must be the same . thus the units digit of p can be 0 , 1 , 5 or 6 . intersection of values is 6 , thus the units digit o... | a ) 7 , b ) 8 , c ) 9 , d ) 10 , e ) 11 | b | add(add(3, 3), 2) | add(n0,n0)|add(n3,#0)| | general |
dan ' s age after 18 years will be 5 times his age 6 years ago . what is the present age of dan ? | "let dan ' s present age be x . x + 18 = 5 ( x - 6 ) 4 x = 48 x = 12 the answer is c ." | a ) 8 , b ) 10 , c ) 12 , d ) 14 , e ) 16 | c | divide(add(18, multiply(6, 5)), subtract(5, const_1)) | multiply(n1,n2)|subtract(n1,const_1)|add(n0,#0)|divide(#2,#1)| | general |
the end of a blade on an airplane propeller is 10 feet from the center . if the propeller spins at the rate of 1,980 revolutions per second , how many miles will the tip of the blade travel in one minute ? ( 1 mile = 5,280 feet ) | "distance traveled in 1 revolution = 2 π r = 2 π 10 / 5280 revolutions in one second = 1980 revolutions in 60 seconds ( one minute ) = 1980 * 60 total distance traveled = total revolutions * distance traveled in one revolution 1980 * 60 * 2 π 10 / 5280 = 450 π b is the answer" | a ) 200 π , b ) 450 π , c ) 300 π , d ) 480 π , e ) 1,200 π | b | multiply(multiply(multiply(multiply(divide(10, add(multiply(const_2, const_100), multiply(add(const_2, const_3), const_1000))), const_2), divide(add(const_2, multiply(const_2, 10)), add(const_3, const_4))), 1,980), const_60) | add(const_3,const_4)|add(const_2,const_3)|multiply(n0,const_2)|multiply(const_100,const_2)|add(#2,const_2)|multiply(#1,const_1000)|add(#3,#5)|divide(#4,#0)|divide(n0,#6)|multiply(#8,const_2)|multiply(#7,#9)|multiply(n1,#10)|multiply(#11,const_60)| | physics |
a grocery store bought some mangoes at a rate of 5 for a dollar . they were separated into two stacks , one of which was sold at a rate of 3 for a dollar and the other at a rate of 9 for a dollar . what was the ratio of the number of mangoes in the two stacks if the store broke even after having sold all of its mangoes... | "the cost price of a mango = 1 / 5 dollars . the selling price of a mango from the first stack = 1 / 3 dollars - - > the profit from one mango = 1 / 3 - 1 / 5 = 2 / 15 = 4 / 30 dollars . the selling price of a mango from the second stack = 1 / 9 dollars - - > the loss from one mango = 1 / 5 - 1 / 9 = 4 / 45 dollars . t... | a ) 1 : 4 , b ) 1 : 5 , c ) 3 : 2 , d ) 1 : 2 , e ) 2 : 5 | c | divide(subtract(divide(const_1, 3), divide(const_1, 5)), subtract(divide(const_1, 5), divide(const_1, multiply(const_2, const_4)))) | divide(const_1,n1)|divide(const_1,n0)|multiply(const_2,const_4)|divide(const_1,#2)|subtract(#0,#1)|subtract(#1,#3)|divide(#4,#5)| | other |
if 5 % more is gained by selling an article for rs . 420 than by selling it for rs . 380 , the cost of the article is ? | let c . p . be rs . x . then , 5 % of x = 420 - 380 = 40 x / 20 = 40 = > x = 800 answer : e | a ) 500 , b ) 688 , c ) 200 , d ) 121 , e ) 800 | e | divide(subtract(420, 380), divide(5, const_100)) | divide(n0,const_100)|subtract(n1,n2)|divide(#1,#0) | gain |
the area of a square is equal to five times the area of a rectangle of dimensions 36 cm * 20 cm . what is the perimeter of the square ? | "area of the square = s * s = 5 ( 36 * 20 ) = > s = 60 = 60 cm perimeter of the square = 4 * 60 = 240 cm . answer : d" | a ) 289 cm , b ) 800 cm , c ) 829 cm , d ) 240 cm , e ) 289 cm | d | multiply(sqrt(multiply(rectangle_area(36, 20), divide(20, const_2))), const_4) | divide(n1,const_2)|rectangle_area(n0,n1)|multiply(#0,#1)|sqrt(#2)|multiply(#3,const_4)| | geometry |
each of the 39 points is placed either inside or on the surface of a perfect sphere . if 72 % or fewer of the points touch the surface , what is the maximum number of segments which , if connected from those points to form chords , could be the diameter of the sphere ? | "maximum number of points on the surface is 72 % * 39 = 28.08 . . . or 28 since it has to be an integer now note that if two points form a diameter , they can not be part of any other diameter . so in the best case we can pair up the points we have 28 points , so at best we can form 14 pairs ( 28 ) . so , answer is ( d... | a ) 7 , b ) 11 , c ) 13 , d ) 14 , e ) 38 | d | divide(multiply(39, divide(72, const_100)), const_2) | divide(n1,const_100)|multiply(n0,#0)|divide(#1,const_2)| | geometry |
the proportion of water to alcohol in solution a is 5 : 4 and the proportion of water to alcohol in solution b is 6 : 5 . if an equal amount of each solution is mixed together , what is the concentration of alcohol in the new solution ? | "let v be the total volume of the new solution . then a volume of v / 2 was added from each solution a and b . the amount of alcohol added to the new solution was : ( 4 / 9 ) ( v / 2 ) + ( 5 / 11 ) ( v / 2 ) = 2 v / 9 + 5 v / 22 = 89 v / 198 . the concentration of alcohol is 89 / 198 = 44.9 % the answer is e ." | a ) 40.9 % , b ) 41.9 % , c ) 42.9 % , d ) 43.9 % , e ) 44.9 % | e | multiply(divide(add(4, 5), add(add(5, 6), add(4, 5))), const_100) | add(n1,n3)|add(n0,n2)|add(#1,#0)|divide(#0,#2)|multiply(#3,const_100)| | general |
in the coordinate plane , one of the vertices of a square is the point ( - 4 , - 4 ) . if the diagonals of that square intersect at point ( 3 , 2 ) , what is the area of that square ? | "one point ( - 4 - 4 ) , intersection ( 3,2 ) so the distance from the first point - 4 - 3 = - 7 is the midpoint of the square - - > whole side 14 , 14 * 14 = 196 b" | a ) 180 , b ) 196 , c ) 160 , d ) 164 , e ) 172 | b | multiply(multiply(4, 3), multiply(4, 3)) | multiply(n0,n2)|multiply(#0,#0)| | geometry |
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