options stringlengths 37 300 | correct stringclasses 5
values | annotated_formula stringlengths 7 727 | problem stringlengths 5 967 | rationale stringlengths 1 2.74k | program stringlengths 10 646 |
|---|---|---|---|---|---|
a ) 264 , b ) 209 , c ) 57 , d ) 171 , e ) 181 | a | multiply(multiply(8, subtract(11, 8)), 11) | there are 456 doctors and nurses in a hospital . if the ratio of the doctors to the nurses is 8 : 11 , then how many nurses are there in the hospital ? | given , the ratio of the doctors to the nurses is 8 : 11 number of nurses = 11 / 19 x 456 = 264 answer : a | a = 11 - 8
b = 8 * a
c = b * 11
|
a ) s . 8500 , b ) s . 8000 , c ) s . 7500 , d ) s . 8800 , e ) s . 6500 | d | multiply(multiply(multiply(add(multiply(multiply(multiply(2, 3), const_100), const_100), multiply(multiply(multiply(3, 3), const_100), multiply(add(3, 2), 2))), divide(add(multiply(16, 3), 2), 3)), divide(multiply(3, 3), multiply(2, multiply(2, 3)))), divide(const_1, const_100)) | find the simple interest on rs . 70,400 at 16 2 / 3 % per annum for 9 months . | "p = rs . 70400 , r = 50 / 3 % p . a and t = 9 / 12 years = 3 / 4 years . s . i . = ( p * r * t ) / 100 = rs . ( 70,400 * ( 50 / 3 ) * ( 3 / 4 ) * ( 1 / 100 ) ) = rs . 8800 answer is d ." | a = 2 * 3
b = a * 100
c = b * 100
d = 3 * 3
e = d * 100
f = 3 + 2
g = f * 2
h = e * g
i = c + h
j = 16 * 3
k = j + 2
l = k / 3
m = i * l
n = 3 * 3
o = 2 * 3
p = 2 * o
q = n / p
r = m * q
s = 1 / 100
t = r * s
|
a ) rs . 1386 , b ) rs . 1764 , c ) rs . 1275 , d ) rs . 2268 , e ) none of these | c | add(divide(153, divide(multiply(divide(9, multiply(const_4, const_3)), 16), const_100)), 153) | the true discount on a bill due 9 months hence at 16 % per annum is rs . 153 . the amount of the bill is | "solution 32.5 let p . w . be rs . x . then , s . i . on rs . x at 16 % for 9 months = rs . 153 . ∴ x 16 x 9 / 12 x 1 / 100 } = 153 or x = 1275 . ∴ p . w . = rs . 1275 . answer c" | a = 4 * 3
b = 9 / a
c = b * 16
d = c / 100
e = 153 / d
f = e + 153
|
['a ) 2', 'b ) 7', 'c ) 6', 'd ) 889', 'e ) 1'] | b | add(multiply(const_3, const_2), const_1) | a voltage will influence current only if the circuit is : | answer : b | a = 3 * 2
b = a + 1
|
a ) s . 1000 , b ) s . 1009 , c ) s . 1166.7 , d ) s . 1006 , e ) s . 1002 | c | divide(multiply(140, const_100), subtract(add(const_100, 2), subtract(const_100, 10))) | a watch was sold at a loss of 10 % . if it was sold for rs . 140 more , there would have been a gain of 2 % . what is the cost price ? | "explanation : 90 % 102 % - - - - - - - - 12 % - - - - 140 100 % - - - - ? = > rs . 1166.7 answer : c" | a = 140 * 100
b = 100 + 2
c = 100 - 10
d = b - c
e = a / d
|
a ) 4 . , b ) 6 . , c ) 7 . , d ) 9 . , e ) 5 . | d | add(multiply(15, divide(const_1, 2)), const_1) | in the junior basketball league there are 15 teams , 2 / 3 of them are bad and ½ are rich . what ca n ' t be the number of teams that are rich and bad ? | "total teams = 16 bad teams = ( 2 / 3 ) * 15 = 10 rich teams = 8 so maximum value that the both rich and bad can take will be 8 . so e = 9 can not be that value . ans d ." | a = 1 / 2
b = 15 * a
c = b + 1
|
a ) 22 , b ) 28 , c ) 77 , d ) 99 , e ) 378 | e | multiply(divide(multiply(4, 9), 2), add(17, 4)) | a movie buff buys movies on dvd and on blu - ray in a ratio of 17 : 4 . if she returns 4 blu - ray movies , that ratio would change to 9 : 2 . if she buys movies on no other medium , what was the original number of movies purchased ? | if u can just keep an eye on the options 99 is the only multiple of 9 in options given . . so you can mark it wid in seconds . now coming to the process m ( d ) = 17 x and b ( d ) = 4 x now from the next line the new eqn becomes 17 x / ( 4 x - 4 ) = 9 / 2 solving it 34 x = 36 x - 36 x = 18 which means m ( d ) = 306 and b ( d ) = 72 so total initially is m ( d ) + b ( d ) = 378 e | a = 4 * 9
b = a / 2
c = 17 + 4
d = b * c
|
a ) 80 % , b ) 105 % , c ) 120 % , d ) 124.2 % , e ) 138 % | c | multiply(divide(multiply(subtract(const_1, divide(20, const_100)), divide(15, const_100)), divide(10, const_100)), const_100) | in 1998 the profits of company n were 10 percent of revenues . in 1999 , the revenues of company n fell by 20 percent , but profits were 15 percent of revenues . the profits in 1999 were what percent of the profits in 1998 ? | "let ' s test : 1998 revenues = $ 100 profits = $ 10 next we ' re told that , in 1999 , revenue fell by 20 % , but profits were 15 % of revenues . using the numbers from 1998 ( above ) , we end up with . . . 1999 revenues = $ 80 profits = $ 12 we ' re asked to compare the profits in 1999 to the profits in 1998 ( as a percentage ) : $ 12 / $ 10 = 1.2 = 120 % answer : c" | a = 20 / 100
b = 1 - a
c = 15 / 100
d = b * c
e = 10 / 100
f = d / e
g = f * 100
|
a ) 1 : 5 , b ) 3 : 6 , c ) 1 : 2 , d ) 2 : 9 , e ) 1 : 3 | a | divide(multiply(multiply(add(const_1, const_4), divide(const_1, const_2)), const_2), multiply(add(multiply(add(const_1, const_4), divide(const_1, const_2)), const_1), const_2)) | a cube has two of its faces painted half red and half white . the other faces are completely painted white . what is the ratio between the red painted areas and the white painted areas of the cube ? | "let x be the area of each face of the cube . the area painted red is 2 ( x / 2 ) = x the area painted white is 2 ( x / 2 ) + 4 x = 5 x the ratio of red to white is x : 5 x which is 1 : 5 . the answer is a ." | a = 1 + 4
b = 1 / 2
c = a * b
d = c * 2
e = 1 + 4
f = 1 / 2
g = e * f
h = g + 1
i = h * 2
j = d / i
|
a ) 1 kmph , b ) 2 kmph , c ) 1.5 kmph , d ) 12 kmph , e ) 15 kmph | a | divide(subtract(divide(24, 4), divide(16, 4)), const_2) | if athul rows 16 km upstream and 24 km down steam taking 4 hours each , then the speed of the stream | speed upstream = 16 / 4 = 4 kmph speed down stream = 24 / 4 = 6 kmph speed of stream = ½ ( 6 - 4 ) = 1 kmph answer : a | a = 24 / 4
b = 16 / 4
c = a - b
d = c / 2
|
a ) 59 , b ) 55 , c ) 61 , d ) 45 , e ) 36 | e | subtract(add(add(multiply(30, 5), 1), 30), multiply(30, 5)) | the average age of 30 students in a class is 5 years . if teacher ' s age is also included then average increases 1 year then find the teacher ' s age ? | "total age of 50 students = 30 * 5 = 150 total age of 51 persons = 31 * 6 = 186 age of teacher = 186 - 150 = 36 years answer is e" | a = 30 * 5
b = a + 1
c = b + 30
d = 30 * 5
e = c - d
|
a ) rs . 2250 , b ) rs . 3000 , c ) rs . 6750 , d ) rs . 5625 , e ) none of these | b | multiply(11000, inverse(add(add(divide(2, 3), multiply(divide(2, 3), 3)), const_1))) | 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 . 11000 . 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 × ( 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 ) × 11000 ] = rs . 3000 answer : option b" | a = 2 / 3
b = 2 / 3
c = b * 3
d = a + c
e = d + 1
f = 1/(e)
g = 11000 * f
|
a ) 30 % , b ) 50 % , c ) 60 % , d ) 80 % , e ) 90 % | b | multiply(divide(subtract(120, add(multiply(3, 8), multiply(8, 3))), 120), const_100) | a batsman scored 120 runs which included 3 boundaries and 8 sixes . what % of his total score did he make by running between the wickets | "number of runs made by running = 110 - ( 3 x 4 + 8 x 6 ) = 120 - ( 60 ) = 60 now , we need to calculate 60 is what percent of 120 . = > 60 / 120 * 100 = 50 % b" | a = 3 * 8
b = 8 * 3
c = a + b
d = 120 - c
e = d / 120
f = e * 100
|
a ) 49 , b ) 249 , c ) 349 , d ) 449 , e ) none of these | d | inverse(add(inverse(8), inverse(10))) | worker a takes 8 hours to do a job . worker b takes 10 hours to do a job . how long should it take both a and b , working together to do same job | explanation : in this type of questions , first we need to calculate 1 hours work , then their collective work as a ' s 1 hour work is 1 / 8 b ' s 1 hour work is 1 / 10 ( a + b ) ' s 1 hour work = 1 / 8 + 1 / 10 = 9 / 40 so both will finish the work in 40 / 9 hours = 449 answer : d | a = 1/(8)
b = 1/(10)
c = a + b
d = 1/(c)
|
a ) 10.09 , b ) 1.06 , c ) 10.06 , d ) 100.9 , e ) 100.6 | d | divide(4.036, 0.04) | 4.036 divided by 0.04 gives : | "= 4.036 / 0.04 = 403.6 / 4 = 100.9 answer is d ." | a = 4 / 36
|
a ) - 11 , b ) - 12 , c ) - 13 , d ) - 14 , e ) - 15 | c | add(7, add(4, 2)) | find the constant k so that : - x 2 - ( k + 7 ) x - 8 = - ( x - 2 ) ( x - 4 ) | "- x 2 - ( k + 7 ) x - 8 = - ( x - 2 ) ( x - 4 ) : given - x 2 - ( k + 7 ) x - 8 = - x 2 + 6 x - 8 - ( k + 7 ) = 6 : two polynomials are equal if their corresponding coefficients are equal . k = - 13 : solve the above for k correct answer c" | a = 4 + 2
b = 7 + a
|
a ) 40 , b ) 50 , c ) 60 , d ) 80 , e ) 85 | d | divide(360, 4.5) | a car takes 4.5 hours to travel from a to b , which is 360 miles apart . what is the average speed of the car ? | average speed = 360 / 4.5 hours = 360 / 270 = 1.33 miles per min = 80 miles per hour answer : d | a = 360 / 4
|
a ) 4 , b ) 5 , c ) 6 , d ) 7 , e ) 8 | b | subtract(7, const_1) | there are 20 balls which are red , blue or green . if 7 balls are green and the sum of red balls and green balls is less than 13 , at most how many red balls are there ? | "explanation : given : red + green + blue = 20 green = 7 solution : r + g < 13 therefore r + 7 < 13 so r < 6 ans : at most 5 red balls b . 5 hence ( b ) is correct . answer : b" | a = 7 - 1
|
a ) 23777 , b ) 25000 , c ) 29977 , d ) 26777 , e ) 19871 | b | divide(subtract(multiply(100000, divide(add(9, divide(1, 2)), const_100)), multiply(100000, divide(9, const_100))), subtract(divide(11, const_100), divide(9, const_100))) | an amount of rs . 100000 is invested in two types of shares . the first yields an interest of 9 % p . a and the second , 11 % p . a . if the total interest at the end of one year is 9 1 / 2 % , then the amount invested at 11 % was ? | "let the sum invested at 9 % be rs . x and that invested at 11 % be rs . ( 100000 - x ) . then , ( x * 9 * 1 ) / 100 + [ ( 100000 - x ) * 11 * 1 ] / 100 = ( 100000 * 19 / 2 * 1 / 100 ) ( 9 x + 1100000 - 11 x ) = 950000 x = 75000 sum invested at 9 % = rs . 75000 sum invested at 11 % = rs . ( 100000 - 75000 ) = rs . 25000 . answer : b" | a = 1 / 2
b = 9 + a
c = b / 100
d = 100000 * c
e = 9 / 100
f = 100000 * e
g = d - f
h = 11 / 100
i = 9 / 100
j = h - i
k = g / j
|
a ) 72.5 , b ) 64.5 , c ) 62.5 , d ) 82.5 , e ) 60.5 | b | multiply(divide(divide(multiply(47.5, add(const_100, 25)), const_100), subtract(const_100, 8)), const_100) | at what price must an book costing $ 47.50 be marked in order that after deducting 8 % from the list price . it may be sold at a profit of 25 % on the cost price ? | c $ 62.50 cp = 47.50 sp = 47.50 * ( 125 / 100 ) = 59.375 mp * ( 92 / 100 ) = 59.375 mp = 64.5 b | a = 100 + 25
b = 47 * 5
c = b / 100
d = 100 - 8
e = c / d
f = e * 100
|
a ) 76 % , b ) 60 % , c ) 50 % , d ) 40 % , e ) 24 % | d | subtract(const_100, multiply(divide(36, subtract(const_100, 40)), const_100)) | exactly 36 % of the numbers in set s are even multiples of 3 . if 40 % of the even integers in set s are not multiples of 3 , what percent of the numbers in set s are not even integers ? | let s be the total number of elements in set s . we know that 0.36 * s is the number of even multiples of three . let n be the number of even numbers . we know 0.40 * n are even numbers not multiple of three . this also means , 0.60 * n are even numbers that are multiples of three . therefore : 0.06 * n = 0.36 * s n = ( 0.36 / 0.60 ) * s = ( 36 / 60 ) * s = ( 6 / 10 ) * s = 0.60 * s = 60 % of s so even numbers are 60 % of s . that means , 40 % of s are not even integers . answer = d | a = 100 - 40
b = 36 / a
c = b * 100
d = 100 - c
|
a ) 18 , b ) 56 , c ) 12 , d ) 17 , e ) 14 | a | subtract(48, multiply(10, 3)) | the average age of a group of 10 persons was decreased by 3 years when one person , whose age was 48 years , was replaced by a new person . find the age of the new person ? | "initial average age of the 10 persons be p . age of the new person q . sum of the ages of the initial 10 persons = 10 p new average = ( p - 3 ) 10 ( p - 3 ) = 10 p - 48 + q = > q = 18 answer : a" | a = 10 * 3
b = 48 - a
|
a ) 3.00 , b ) 5.00 , c ) 4.50 , d ) 5.50 , e ) 6.75 | c | multiply(45, divide(10, 100)) | a glucose solution contains 10 grams of glucose per 100 cubic centimeters of solution . if 45 cubic centimeters of the solution were poured into an empty container , how many grams of glucose would be in the container ? | "we are given that a glucose solution contains 10 grams of glucose per 100 cubic centimeters of solution . since we are dealing with a solution , we know that the grams of glucose is proportional to the number of cubic centimeters of solution . thus , to determine how many grams of glucose would be in the container when we have 45 cubic centimeters of solution , we can set up a proportion . we can say : “ 10 grams of glucose is to 100 cubic centimeters of solution as x grams of glucose is to 45 cubic centimeters of solution . ” let ’ s now set up the proportion and solve for x . 10 / 100 = x / 45 when we cross multiply we obtain : ( 10 ) ( 45 ) = 100 x 450 = 100 x 4.50 = x there are 4.5 grams of glucose in the solution in the container . the answer is c ." | a = 10 / 100
b = 45 * a
|
a ) 20 , b ) 21 , c ) 22 , d ) 23 , e ) 24 | b | subtract(multiply(7, 4), 7) | find the two digit number , such that the ratio is 7 / 4 of original number to the number formed by reversing the digits . | here ratio is 7 / 4 . so lets check one by one . for 1 : 7 * 1 / 4 * 1 = 7 / 4 for 2 : 7 * 2 / 4 * 2 = 14 / 8 , which does n ' t satisfy the conditions . for 3 : 7 * 3 / 4 * 3 = 21 / 12 answer : b | a = 7 * 4
b = a - 7
|
a ) rs . 6289 , b ) rs . 6298 , c ) rs . 6681 , d ) rs . 6725 , e ) rs . 6708 | c | divide(7350, add(const_1, divide(10, const_100))) | the owner of a furniture shop charges his customer 10 % more than the cost price . if a customer paid rs . 7350 for a computer table , then what was the cost price of the computer table ? | "cp = sp * ( 100 / ( 100 + profit % ) ) = 7350 ( 100 / 110 ) = rs . 6681 . answer : c" | a = 10 / 100
b = 1 + a
c = 7350 / b
|
a ) 87 days , b ) 20 days , c ) 16 days , d ) 19 days , e ) 36 days | b | divide(multiply(4, 5), divide(subtract(multiply(4, 5), multiply(add(divide(multiply(4, 5), 4), divide(multiply(4, 5), 5)), 2)), 2)) | a can do a piece of work in 4 days . b can do it in 5 days . with the assistance of c they completed the work in 2 days . find in how many days can c alone do it ? | "c = 1 / 2 - 1 / 4 - 1 / 5 = 1 / 20 = > 20 days answer : b" | a = 4 * 5
b = 4 * 5
c = 4 * 5
d = c / 4
e = 4 * 5
f = e / 5
g = d + f
h = g * 2
i = b - h
j = i / 2
k = a / j
|
a ) 10 , b ) 12 , c ) 14 , d ) 8 , e ) 9 | c | subtract(10154, multiply(floor(divide(10154, 30)), 30)) | what least no . must be subtracted from 10154 so that remaining no . is divisible by 30 ? | "explanation : on dividing 10154 by 30 we get the remainder 14 , so 14 should be subtracted option c" | a = 10154 / 30
b = math.floor(a)
c = b * 30
d = 10154 - c
|
a ) 100 , b ) 120 , c ) 200 , d ) 220 , e ) 264 | e | add(240, divide(multiply(240, 10), const_100)) | the present population of a town is 240 . population increase rate is 10 % p . a . find the population of town after 1 years ? | "p = 240 r = 10 % required population of town = p * ( 1 + r / 100 ) ^ t = 240 * ( 1 + 10 / 100 ) = 240 * ( 11 / 10 ) = 264 answer is e" | a = 240 * 10
b = a / 100
c = 240 + b
|
a ) 25 , b ) 31 , c ) 26 , d ) 29 , e ) 39 | c | add(subtract(74, multiply(17, 3)), 3) | a batsman makes a score of 74 runs in the 17 th inning and thus increases his averages by 3 . what is his average after 17 th inning ? | "let the average after 17 innings = x total runs scored in 17 innings = 17 x average after 16 innings = ( x - 3 ) total runs scored in 16 innings = 16 ( x - 3 ) total runs scored in 16 innings + 74 = total runs scored in 17 innings = > 16 ( x - 3 ) + 74 = 17 x = > 16 x - 48 + 74 = 17 x = > x = 26 answer is c" | a = 17 * 3
b = 74 - a
c = b + 3
|
a ) 122 , b ) 210 , c ) 216 , d ) 217 , e ) 225 | c | subtract(multiply(factorial(5), const_2), factorial(4)) | a 5 - digit number divisible by 3 is to be formed using numerical 0 , 1 , 2 , 3 , 4 and 5 without repetition . the total number q of ways this can be done is : | we should determine which 5 digits from given 6 , would form the 5 digit number divisible by 3 . we have six digits : 0 , 1 , 2 , 3 , 4 , 5 . their sum = 15 . for a number to be divisible by 3 the sum of the digits must be divisible by 3 . as the sum of the six given numbers is 15 ( divisible by 3 ) only 5 digits good to form our 5 digit number would be 15 - 0 = { 1 , 2 , 3 , 4 , 5 } and 15 - 3 = { 0 , 1 , 2 , 4 , 5 } . meaning that no other 5 from given six will total the number divisible by 3 . second step : we have two set of numbers : 1 , 2 , 3 , 4 , 5 and 0 , 1 , 2 , 4 , 5 . how many 5 digit numbers can be formed using these two sets : 1 , 2 , 3 , 4 , 5 - - > 5 ! as any combination of these digits would give us 5 digit number divisible by 3 . 5 ! = 120 . 0 , 1 , 2 , 4 , 5 - - > here we can not use 0 as the first digit , otherwise number wo n ' t be any more 5 digit and become 4 digit . so , desired # would be total combinations 5 ! , minus combinations with 0 as the first digit ( combination of 4 ) 4 ! - - > 5 ! - 4 ! = 4 ! ( 5 - 1 ) = 4 ! * 4 = 96 120 + 96 = 216 = q answer : c . | a = math.factorial(5)
b = a * 2
c = math.factorial(4)
d = b - c
|
a ) 168.58 cm , b ) 158.58 cm , c ) 179.29 cm , d ) 168.58 cm , e ) 178.58 cm | e | floor(divide(add(subtract(multiply(35, 180), 156), 106), 35)) | the average height of 35 boys in a class was calculated as 180 cm . it has later found that the height of one of the boys in the class was wrongly written as 156 cm whereas his actual height was 106 cm . find the actual average height of the boys in the class ( round off your answer to two decimal places ) . | "calculated average height of 35 boys = 180 cm . wrong total height of 35 boys = 180 * 35 cm . this was as a result of an actual height of 106 cm being wrongly written as 156 cm . correct total height of 35 boys = 180 * 35 cm - 156 cm + 106 cm = 180 * 35 cm - 156 cm + 106 cm / 35 = 180 cm - 50 / 35 cm = 180 cm - 1.42 cm = 178.58 cm . answer : e" | a = 35 * 180
b = a - 156
c = b + 106
d = c / 35
e = math.floor(d)
|
a ) 45 , b ) 47 , c ) 48 , d ) 49 , e ) 53 | d | subtract(100, 51) | in a group of 100 cars , 37 cars do not have air conditioning . if at least 51 cars have racing stripes , what is the greatest number of cars that could have air conditioning but not racing stripes ? | "lets assume ac = 63 ( includesonly ac carsandcars with ac and racing stripes ) lets assume rs ( racing stripes ) > = 51 ( includescars with ac and racing stripesandonly racing stripes ) . now since we want to maximize ( only ac ) we have to see to it thatcars with ac and racing stripesis minimal ( assume 0 ) but since rs > = 51 . . we have to assign atleast 14 tocars with ac and racing stripes . hence ac = 63 - 14 = 49 . the answer is d" | a = 100 - 51
|
a ) 20 % , b ) 25 % , c ) 40 % , d ) 60 % , e ) 75 % | b | multiply(subtract(const_1, divide(divide(60, multiply(multiply(const_2, const_5), multiply(const_2, const_5))), subtract(const_1, divide(20, multiply(multiply(const_2, const_5), multiply(const_2, const_5)))))), multiply(multiply(const_2, const_5), multiply(const_2, const_5))) | the maitre ' d at an expensive manhattan restaurant has noticed that 60 % of the couples order dessert and coffee . however , 20 % of the couples who order dessert do n ' t order coffee . what is the probability r that the next couple the maitre ' d seats will not order dessert ? | could you use a venn diagram and just go with the number 100 . 60 people order dessert and coffee . . . which is the union of d and c . r = 2 / 10 of d are n ' t in d u c = so 8 / 10 of d are in duc which means = 60 = 8 / 10 d . so d in total = 75 , and 15 d ' s are n ' t in d union c . which means 25 people are in c only + neither . b 25 % | a = 2 * 5
b = 2 * 5
c = a * b
d = 60 / c
e = 2 * 5
f = 2 * 5
g = e * f
h = 20 / g
i = 1 - h
j = d / i
k = 1 - j
l = 2 * 5
m = 2 * 5
n = l * m
o = k * n
|
a ) $ 1.63 , b ) $ 1.64 , c ) $ 1.68 , d ) $ 1.73 , e ) $ 1.76 | d | divide(add(multiply(1365, 1.89), multiply(720, 1.42)), add(1365, 720)) | john purchased 1365 large bottles at $ 1.89 per bottle and 720 small bottles at $ 1.42 per bottle . what was the approximate average price paid per bottle ? | ( 1365 * 1.89 + 720 * 1.42 ) / ( 1365 + 720 ) = ~ 1.73 option ( d ) | a = 1365 * 1
b = 720 * 1
c = a + b
d = 1365 + 720
e = c / d
|
a ) 32 kmph , b ) 96 kmph , c ) 34 kmph , d ) 43 kmph , e ) 40 kmph | b | divide(divide(add(100, 300), const_1000), divide(15, const_3600)) | a train 100 meters long completely crosses a 300 meters long bridge in 15 seconds . what is the speed of the train is ? | "s = ( 100 + 300 ) / 45 = 400 / 15 * 18 / 5 = 96 answer : b" | a = 100 + 300
b = a / 1000
c = 15 / 3600
d = b / c
|
a ) 52000 , b ) 32400 , c ) 22400 , d ) 22800 , e ) 24200 | c | subtract(divide(multiply(4, add(4, const_1)), const_2), divide(multiply(subtract(500, const_1), 500), const_2)) | ( 500 + 200 + 100 ) ã — 4 ã — ( 3 + 4 ) = ? | "( 500 + 200 + 100 ) ã — 4 ã — ( 3 + 4 ) = ? or , ? = 800 ã — 4 ã — 7 = 22400 answer c" | a = 4 + 1
b = 4 * a
c = b / 2
d = 500 - 1
e = d * 500
f = e / 2
g = c - f
|
a ) 20 % , b ) 80 % , c ) 100 % , d ) 180 % , e ) 200 % | c | multiply(divide(10, subtract(subtract(const_100, 80), 10)), const_100) | jane makes toy bears . when she works with an assistant , she makes 80 percent more bears per week and works 10 percent fewer hours each week . having an assistant increases jane ’ s output of toy bears per hour by what percent e ? | "c . let ' s assume just jane 40 bears per 40 / hrs a week , so that is 1 bear / hr . with an assistant she makes 72 bears per 36 hours a week or 2 bears / hr ( [ 40 bears * 1.8 ] / [ 40 hrs * . 90 ] ) . e = [ ( 2 - 1 ) / 1 ] * 100 % = 100 % . c" | a = 100 - 80
b = a - 10
c = 10 / b
d = c * 100
|
a ) 16.8 , b ) 26 , c ) 14 , d ) 12 , e ) 91 | a | divide(1056, multiply(multiply(const_2, divide(add(add(multiply(const_3, const_100), multiply(const_1, const_10)), const_4), const_100)), 10)) | if the wheel is 10 cm then the number of revolutions to cover a distance of 1056 cm is ? | "2 * 22 / 7 * 10 * x = 1056 = > x = 16.8 answer : a" | a = 3 * 100
b = 1 * 10
c = a + b
d = c + 4
e = d / 100
f = 2 * e
g = f * 10
h = 1056 / g
|
a ) 22 , b ) 28 , c ) 98 , d ) 24.5 , e ) 13 | d | divide(126, add(const_2, const_pi)) | the perimeter of a semi circle is 126 cm then the radius is ? | "36 / 7 r = 126 = > r = 24.5 answer : d" | a = 2 + math.pi
b = 126 / a
|
a ) 1 hr , b ) 2 hrs , c ) 3 hrs , d ) 5 hrs , e ) 6 hrs | b | divide(50, add(10, 15)) | two cyclist start from the same places in opposite directions . one is going towards north at 10 kmph and the other is going towards south 15 kmph . what time will they take to be 50 km apart ? | "to be ( 10 + 15 ) km apart , they take 1 hour to be 50 km apart , they take 1 / 25 * 50 = 2 hrs answer is b" | a = 10 + 15
b = 50 / a
|
a ) none , b ) two , c ) four , d ) five , e ) seven | d | subtract(const_4, const_1) | r is the set of positive even integers less than 101 , and s is the set of the squares of the integers in r . how many elements does the intersection of r and s contain ? | "r is the set of positive even integers less than 101 , and s is the set of the squares of the integers in r . how many elements does the intersection of r and s contain ? r = 2,4 , 6,8 , 10,12 . . . s = 4,16 , 36,64 . . . numbers : 4 , 16 , 36 , 64 , and 100 are even integers ( less than 101 ) that are in both sets . solution : five answer : d" | a = 4 - 1
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a ) 48 min , b ) 36 min , c ) 25 min , d ) 30 min , e ) 50 min | a | divide(16, 1) | a fill pipe can fill 1 / 4 of cistern in 16 minutes in how many minutes , it can fill 3 / 4 of the cistern ? | "1 / 4 of the cistern can fill in 16 min 3 / 4 of the cistern can fill in = 16 * 4 * 3 / 4 = 48 min answer is a" | a = 16 / 1
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a ) 17 , b ) 19 , c ) 20 , d ) 21 , e ) 22 | b | add(subtract(subtract(const_1000, const_10), multiply(multiply(const_10, multiply(78, 78)), multiply(const_4, const_2))), const_10) | how many three digit numbers e are divisible by 78 or 91 ? | "the answer will be 19 . explanation : 78 = 2 * 3 * 13 now multiples of 78 , 156 . . . . 780 , now 1000 - 780 = 220 only two more muktiples of 78 can exists . so total number of 3 digit multiples of 78 are 9 + 2 = 11 91 = 13 * 7 - - total number of three digit multiples - - 9 no remember we have a common multiples as well - - 13 * 7 * 6 = 91 * 6 = 546 so total number of multiples e - - 11 + 9 - 1 = 19 . hence answer is 19 . b" | a = 1000 - 10
b = 78 * 78
c = 10 * b
d = 4 * 2
e = c * d
f = a - e
g = f + 10
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a ) 4.65 % increase , b ) 5.65 % increase , c ) 6.65 % increase , d ) 6.65 % decrease , e ) 7.65 % increase | c | multiply(divide(subtract(multiply(add(const_100, 35), subtract(const_100, 21)), multiply(const_100, const_100)), multiply(const_100, const_100)), const_100) | calculate the effect changes in dimension of a rectangle will have on its area , if length is increased by 35 % and its breadth is decreased by 21 % ? | "let l and b be 100 each 100 * 100 = 10000 l increase by 35 % = 135 b decrease by 21 % = 79 135 * 79 = 10665 6.65 % increase answer : c" | a = 100 + 35
b = 100 - 21
c = a * b
d = 100 * 100
e = c - d
f = 100 * 100
g = e / f
h = g * 100
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a ) 9 % , b ) 56.6 % , c ) 11 % , d ) 12 % , e ) 15 % | b | multiply(subtract(divide(18, const_100), divide(subtract(3, multiply(divide(18, const_100), multiply(18, 2))), subtract(multiply(15, 3), multiply(18, 2)))), const_100) | fox jeans regularly sell for $ 15 a pair and pony jeans regularly sell for $ 18 a pair . during a sale these regular unit prices are discounted at different rates so that a total of $ 3 is saved by purchasing 5 pairs of jeans : 3 pairs of fox jeans and 2 pairs of pony jeans . if the sum of the two discounts rates is 18 percent , what is the discount rate on pony jeans ? | "you know that fox jeans costs $ 15 , and pony jeans costs $ 18 , you also know that 3 pairs of fox jeans and 2 pairs of pony jeans were purchased . so 3 ( 15 ) = 45 - fox 2 ( 18 ) = 36 - pony the total discount discount is $ 3 and you are asked to find the percent discount of pony jeans , so 45 ( 18 - x ) / 100 + 36 ( x ) / 100 = 3 or 45 * 18 - 45 * x + 36 * x = 3 * 100 or 9 x = - 3 * 100 + 45 * 18 x = 510 / 9 = 56.6 % b" | a = 18 / 100
b = 18 / 100
c = 18 * 2
d = b * c
e = 3 - d
f = 15 * 3
g = 18 * 2
h = f - g
i = e / h
j = a - i
k = j * 100
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a ) 240 , b ) 288 , c ) 277 , d ) 455 , e ) 361 | d | multiply(divide(630, add(add(multiply(2000, 8), multiply(subtract(2000, 1000), subtract(const_12, 8))), add(multiply(4000, 8), multiply(add(4000, 1000), subtract(const_12, 8))))), add(multiply(2000, 8), multiply(subtract(2000, 1000), subtract(const_12, 8)))) | a and b began business with rs . 2000 and rs . 4000 after 8 months , a withdraws rs . 1000 and b advances rs . 1000 more . at the end of the year , their profits amounted to rs . 630 find the share of b . | "( 2 * 8 + 1 * 4 ) : ( 4 * 8 + 5 * 4 ) 5 : 13 13 / 18 * 630 = 455 answer : d" | a = 2000 * 8
b = 2000 - 1000
c = 12 - 8
d = b * c
e = a + d
f = 4000 * 8
g = 4000 + 1000
h = 12 - 8
i = g * h
j = f + i
k = e + j
l = 630 / k
m = 2000 * 8
n = 2000 - 1000
o = 12 - 8
p = n * o
q = m + p
r = l * q
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a ) 65 , b ) 67 , c ) 100 , d ) 95 , e ) 33 | d | add(divide(39.9, divide(subtract(70, 28), const_100)), 28) | if 70 % of a number exceeds 28 % of it by 39.9 , then find the number ? | "use the elimination method to find the correct option . of all the options only 95 fits 70 % of 95 = 66.5 28 % of 95 = 26.6 66.5 - 26.6 = 39.9 required number is 95 . answer : d" | a = 70 - 28
b = a / 100
c = 39 / 9
d = c + 28
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a ) 200 , b ) 202 , c ) 204 , d ) 206 , e ) 208 | d | divide(subtract(multiply(7.68, const_1000), multiply(35, 190)), 5) | 40 onions on a scale weigh 7.68 kg . when 5 onions are removed from the scale , the average weight of the 35 onions is 190 grams . what is the average weight ( in grams ) of the 5 onions which were removed ? | "35 * 190 = 6650 . the other 5 onions weigh a total of 1030 grams . the average weight is 1030 / 5 = 206 grams . the answer is d ." | a = 7 * 68
b = 35 * 190
c = a - b
d = c / 5
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a ) 6819.59775 , b ) 6981.59775 , c ) 6918.59775 , d ) 7113.81775 , e ) 6891.59775 | d | subtract(6402.5, multiply(multiply(640.25, 64.025), 6.4025)) | evaluate : 6402.5 + 640.25 + 64.025 + 6.4025 + 0.64025 | "6402.5 640.25 64.025 6.4025 + 0.64025 - - - - - - - - - - - - - - - 7113.81775 answer is d ." | a = 640 * 25
b = a * 6
c = 6402 - 5
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a ) 14 , b ) 19 , c ) 27 , d ) 18 , e ) 15 | a | divide(add(16, multiply(const_3, 4)), const_2) | mudit ' s age 16 years hence will be thrice his age 4 years ago . find mudit ' s present age ? | explanation : let mudit ' s present age be ' m ' years . m + 16 = 3 ( m - 4 ) = > 2 m = 28 = > m = 14 years . answer : a | a = 3 * 4
b = 16 + a
c = b / 2
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a ) 2 % , b ) 4 % , c ) 6 % , d ) 8 % , e ) 10 % | e | divide(multiply(subtract(1540, 1400), const_100), 1400) | the compound interest earned on a sum for the second and the third years are $ 1400 and $ 1540 respectively . what is the rate of interest ? | 1540 - 1400 = 140 is the rate of interest on $ 1400 for one year . the rate of interest = ( 100 * 140 ) / ( 1400 ) = 10 % the answer is e . | a = 1540 - 1400
b = a * 100
c = b / 1400
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a ) 220 , b ) 200 , c ) 140 , d ) 80 , e ) 60 | a | add(multiply(7, subtract(multiply(5, 10), multiply(3, 10))), multiply(4, subtract(multiply(5, 10), multiply(3, 10)))) | on a certain planet where people usually tend to live for more than a century , the ratio of present ages of father and son is 7 : 4 . 10 years later , the ratio of their ages will be 5 : 3 . what is the sum of their present ages ? | the ratio of present ages of father and son is 7 : 4 , so suppose the present ages are 7 x and 4 x , respectively . 10 years later , the ages of father and son , respectively , shall be 7 x + 10 and 4 x + 10 . the ratio of ages 10 years later is given to be 5 : 3 . this means ( 7 x + 10 ) / ( 4 x + 10 ) = 5 / 3 solving we get x = 20 . so present age of father is 7 x = 140 and present age of son is 4 x = 80 . the sum of present ages is thus 220 . ans a | a = 5 * 10
b = 3 * 10
c = a - b
d = 7 * c
e = 5 * 10
f = 3 * 10
g = e - f
h = 4 * g
i = d + h
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a ) 66.66 , b ) 62.5 , c ) 58 , d ) 60 , e ) 62 | a | subtract(multiply(divide(const_1, subtract(const_1, divide(40, const_100))), const_100), const_100) | a certain candy manufacturer reduced the weight of candy bar m by 40 percent buy left the price unchanged . what was the resulting percent increase in the price per ounce of candy bar m ? | "assume 1 oz candy cost $ 1 before . now price remain same $ 1 but weight of candy reduces to 0.6 oz new price of candy = 1 / 0.6 = 1.6666 price increase 66.66 % a" | a = 40 / 100
b = 1 - a
c = 1 / b
d = c * 100
e = d - 100
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a ) 19 % , b ) 10 % , c ) 21 % , d ) 16 % , e ) none | d | add(15, multiply(subtract(15, 10), divide(1, 5))) | weights of two friends ram and shyam are in the ratio 1 : 5 . if ram ' s weight is increased by 10 % and total weight of ram and shyam become 82.8 kg , with an increases of 15 % . by what percent did the weight of shyam has to be increased ? | solution : given ratio of ram and shayam ' s weight = 1 : 5 hence , ( x - 15 ) / ( 15 - 10 ) = 1 / 5 or , x = 16 % . answer : option d | a = 15 - 10
b = 1 / 5
c = a * b
d = 15 + c
|
a ) 9.5 , b ) 8.6 , c ) 9.3 , d ) 9.6 , e ) 9.8 | d | add(20, const_1) | the average of first 20 prime numbers is ? | "explanation : average = ( 2 + 3 + 5 + 7 + 11 + 13 + 17 + 19 / 8 = 77 / 8 = 9.6 ( approx ) answer is d" | a = 20 + 1
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a ) 20 , b ) 24 , c ) 30 , d ) 32 , e ) 48 | e | multiply(multiply(multiply(negate(2), 1), 4), subtract(negate(2), 1)) | what is the value of x ^ 2 yz − xyz ^ 2 , if x = − 2 , y = 1 , and z = 4 ? | "4 * 1 * 4 - ( - 2 * 1 * 16 ) = 16 + 32 = 48 ans : e" | a = negate * (
b = a * 1
c = b * 4
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a ) 6 , b ) 3 , c ) 4 , d ) 9 , e ) 3.6 | e | add(divide(multiply(add(35, const_1), 60), const_1000), 1.5) | a train of 35 carriages , each of 60 meters length , when an engine also of 60 meters length is running at a speed of 60 kmph . in what time will the train cross a bridge 1.5 km long ? | "d = 35 * 60 + 1500 = 3600 m t = 3600 / 60 * 18 / 5 = 216 sec = 3.6 mins answer : e" | a = 35 + 1
b = a * 60
c = b / 1000
d = c + 1
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a ) 90 % , b ) 99 % , c ) 100 % , d ) 101 % , e ) 110 % | b | multiply(10, 10) | 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 = 10 * 10
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a ) 28,22 , b ) 28,17 , c ) 27,17 , d ) 26,17 , e ) 28,19 | b | add(subtract(multiply(divide(const_10, const_2), 9), divide(add(11, multiply(divide(const_10, const_2), 9)), const_2)), divide(const_10, const_2)) | the difference of two numbers is 11 and one - fifth of their sum is 9 . find the numbers | if two numbers are x , y x - y = 11 - - ( i ) & ( x + y ) / 5 = 9 - - ( ii ) solving ( i ) & ( ii ) , x = 28 , y = 17 answer : b | a = 10 / 2
b = a * 9
c = 10 / 2
d = c * 9
e = 11 + d
f = e / 2
g = b - f
h = 10 / 2
i = g + h
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a ) $ 10800 , b ) $ 18000 , c ) $ 180000 , d ) $ 1800 , e ) none | c | divide(18000, subtract(1, add(add(divide(1, 5), divide(1, 10)), divide(3, 5)))) | a man spend 1 / 5 of his salary on food , 1 / 10 of his salary on house rent and 3 / 5 salary on clothes . he still has $ 18000 left with him . find salary . . | "[ 1 / ( x 1 / y 1 + x 2 / y 2 + x 3 / y 3 ) ] * total amount = balance amount [ 1 - ( 1 / 5 + 1 / 10 + 3 / 5 ) } * total salary = $ 18000 , = [ 1 - 9 / 10 ] * total salary = $ 18000 , total salary = $ 18000 * 10 = $ 180000 , correct answer ( c )" | a = 1 / 5
b = 1 / 10
c = a + b
d = 3 / 5
e = c + d
f = 1 - e
g = 18000 / f
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a ) $ 36 , b ) $ 44 , c ) $ 52 , d ) $ 60 , e ) $ 68 | b | multiply(multiply(inverse(subtract(1, divide(1, 4))), add(divide(9, const_2), 12)), const_2) | bert left the house with n dollars . he spent 1 / 4 of this at the hardware store , then $ 9 at the dry cleaners , and then half of what was left at the grocery store . when he got home , he had $ 12 left in his pocket . what was the value of n ? | "started to test answer b if he had 44 , then he spent 11 at hardware store now he was left with 33 $ he spent 9 dollars on cleaning , thus he remained with 24 $ he then spent 1 / 2 of 24 , or 12 , and was left with 12 . hence , the only option that can be right is b" | a = 1 / 4
b = 1 - a
c = 1/(b)
d = 9 / 2
e = d + 12
f = c * e
g = f * 2
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a ) 22 , b ) 28 , c ) 17 , d ) 12 , e ) 40 | e | divide(3520, multiply(multiply(const_2, divide(add(add(multiply(const_3, const_100), multiply(const_1, const_10)), const_4), const_100)), 14)) | if the wheel is 14 cm then the number of revolutions to cover a distance of 3520 cm is ? | "2 * 22 / 7 * 14 * x = 3520 = > x = 40 answer : e" | a = 3 * 100
b = 1 * 10
c = a + b
d = c + 4
e = d / 100
f = 2 * e
g = f * 14
h = 3520 / g
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a ) s . 10123.22 , b ) s . 5823.20 , c ) s . 9123.20 , d ) s . 7256.20 , e ) s . 6073.92 | e | subtract(multiply(multiply(multiply(const_4, const_100), const_100), power(add(const_1, divide(12, const_100)), 3)), multiply(multiply(const_4, const_100), const_100)) | what will be the compound interest on a sum of rs . 15000 after 3 years at the rate of 12 % p . a . ? | "explanation : amount = [ 15000 * ( 1 + 12 / 100 ) 3 ] = 15000 * 28 / 25 * 28 / 25 * 28 / 25 = rs . 35123.20 c . i . = ( 35123.20 - 15000 ) = rs . 6073.92 answer : e" | a = 4 * 100
b = a * 100
c = 12 / 100
d = 1 + c
e = d ** 3
f = b * e
g = 4 * 100
h = g * 100
i = f - h
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a ) 1 / 4 , b ) 1 / 2 , c ) 1 , d ) 2 , e ) 3 | c | multiply(divide(13, add(13, 13)), 2) | if 13 = 13 w / ( 1 - w ) , then ( w ) 2 = | 13 - 13 w = 13 w 26 w = 13 w = 1 / 2 2 w * 2 = 1 / 2 * 2 = 1 answer : c | a = 13 + 13
b = 13 / a
c = b * 2
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a ) 170 , b ) 179 , c ) 37 , d ) 67 , e ) 32 | b | divide(subtract(15968, 37), 89) | on dividing 15968 by a certain number , the quotient is 89 and the remainder is 37 . find the divisor . | "divisor = ( dividend – remainder / quotient ) = ( 15968 - 37 ) / 89 = 179 answer b 179" | a = 15968 - 37
b = a / 89
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a ) 20 , b ) 22 , c ) 28 , d ) 27 , e ) 32 | c | subtract(negate(40), multiply(subtract(52, 46), divide(subtract(52, 46), subtract(58, 52)))) | 58 , 52 , 46 , 40 , 34 , . . . ? | "each number is 6 less than the previous number . 58 - 6 = 52 52 - 6 = 46 46 - 6 = 40 40 - 6 = 34 34 - 6 = 28 answer : c" | a = negate - (
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a ) 1 / 4 , b ) 1 / 8 , c ) 1 / 2 , d ) 1 / 16 , e ) 1 / 32 | d | inverse(power(2, 4)) | if a coin is flipped , the probability that the coin will land tails is 1 / 2 . if the coin is flipped 4 times , what is the probability that it will land tails up on the first 2 flips and not on the last 2 flips ? | ( 1 / 2 ) * ( 1 / 2 ) * ( 1 / 2 ) * ( 1 / 2 ) = 1 / 16 answer : d | a = 2 ** 4
b = 1/(a)
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a ) rs . 575 , b ) rs . 595 , c ) rs . 590 , d ) rs . 580 , e ) rs . 585 | d | subtract(multiply(subtract(540, 460), 6), subtract(540, 460)) | if rs . 460 amount to rs . 540 in 4 years , what will it amount to in 6 years at the same rate % per annum ? | "80 = ( 460 * 4 * r ) / 100 r = 4.34 % i = ( 460 * 6 * 4.34 ) / 100 = 120 460 + 120 = 580 answer : d" | a = 540 - 460
b = a * 6
c = 540 - 460
d = b - c
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a ) 227 , b ) 500 , c ) 550 , d ) 525 , e ) 171 | d | multiply(15, multiply(54, const_0_2778)) | a train passes a station platform in 50 sec and a man standing on the platform in 15 sec . if the speed of the train is 54 km / hr . what is the length of the platform ? | "speed = 54 * 5 / 18 = 15 m / sec . length of the train = 15 * 15 = 225 m . let the length of the platform be x m . then , ( x + 225 ) / 50 = 15 = > x = 525 m . answer : d" | a = 54 * const_0_2778
b = 15 * a
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a ) 22.25 , b ) 99 , c ) 26.25 , d ) 66 , e ) 887 | c | divide(multiply(70, 3), add(divide(70, 70), divide(multiply(2, 70), 20))) | a trained covered x km at 70 kmph and another 2 x km at 20 kmph . find the average speed of the train in covering the entire 3 x km . | "total time taken = x / 70 + 2 x / 20 hours = 4 x / 35 hours average speed = 3 x / ( 4 x / 35 ) = 26.25 kmph answer : c" | a = 70 * 3
b = 70 / 70
c = 2 * 70
d = c / 20
e = b + d
f = a / e
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a ) 120 , b ) 180 , c ) 240 , d ) 280 , e ) 320 | d | multiply(divide(420, const_3), const_2) | a salesman sold twice as much pears in the afternoon than in the morning . if he sold $ 420 kilograms of pears that day , how many kilograms did he sell in the afternoon ? | "3 x = 420 x = 140 therefore , the salesman sold 140 kg in the morning and 2 ⋅ 140 = 280 kg in the afternoon . so answer is d ." | a = 420 / 3
b = a * 2
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a ) rs . 15000 , b ) rs . 15550 , c ) rs . 15600 , d ) rs . 20625 , e ) none of these | d | multiply(1000, multiply(5.5, 3.75)) | the length of a room is 5.5 m and width is 3.75 m . find the cost of paving the floor by slabs at the rate of rs . 1000 per sq . metre . | solution area of the floor = ( 5.5 × 3.75 ) m 2 = 20.625 m 2 ∴ cost of paving = rs . ( 1000 × 20.625 ) = 20625 . answer d | a = 5 * 5
b = 1000 * a
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a ) 9 / 12 , b ) 10 / 12 , c ) 11 / 12 , d ) 1 , e ) 2 | c | subtract(const_1, multiply(subtract(const_1, divide(2, 3)), subtract(const_1, divide(3, 4)))) | the probability that a can solve the problem is 2 / 3 and b can solve it is 3 / 4 . if both of them attemp the problem , then what is the probability that the problem get solved ? | explanation : the event is defined as : a solves the problem and b does not solve the problem or a does n ' t solve the problem and b solves the problem or a solves the problem and b solves the problem numerically this is equivalent to : ( 2 / 3 ) x ( 1 / 4 ) + ( 1 / 3 ) x ( 3 / 4 ) + ( 2 / 3 ) x ( 3 / 4 ) = 11 / 12 answer : c | a = 2 / 3
b = 1 - a
c = 3 / 4
d = 1 - c
e = b * d
f = 1 - e
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a ) 41.25 days , b ) 30.25 days , c ) 60.25 days , d ) 71.25 days , e ) 51.25 days | a | divide(multiply(11, 15), subtract(15, 11)) | a and b finish the job in 15 days . while a , b and c can finish it in 11 days . c alone will finish the job in | explanation : 11 = ( 15 * x ) / ( 15 + x ) 165 + 11 x = 15 x 4 x = 165 x = 41.25 answer : option a | a = 11 * 15
b = 15 - 11
c = a / b
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a ) 5.2 , b ) 7.4 , c ) 13.7 , d ) 21.2 , e ) 28.7 | c | divide(383.6, 28) | on a map , 1 inch represents 28 miles . how many x xxinches would be necessary to represent a distance of 383.6 miles ? | "x inches necessary to represent a distance of 383.6 miles = 383.6 / 28 = 13.7 answer c" | a = 383 / 6
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a ) 68 , b ) 66 , c ) 565 , d ) 16 , e ) 35 | d | add(add(subtract(13, const_1), const_1), const_3) | tim has 13 10 - dollar bills , 11 5 - dollar bills , and 17 one - dollar bills . if tim needs to pay exactly $ 128 , what is the least number of bills he will need to use ? | 128 is the total sum of money . as we have 13 10 dollar bills so the closest we can get to 128 using the 10 dollar bills is by using 12 * 10 dollar bils 128 - 120 = 8 so now we need to get 8 from either 5 dollar bills or 1 dollar bills 8 - 5 ( use only 1 5 dollar bill ) = 3 so we can get 3 from using 3 $ 1 bills . hence 16 ( 12 + 1 + 3 ) answer is d | a = 13 - 1
b = a + 1
c = b + 3
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a ) 20 m , b ) 26 m , c ) 11 m , d ) 10 m , e ) 15 m | b | multiply(divide(subtract(25, 20), 25), 130) | if in a race of 130 m , a covers the distance in 20 seconds and b in 25 seconds , then a beats b by : | "explanation : the difference in the timing of a and b is 5 seconds . hence , a beats b by 5 seconds . the distance covered by b in 5 seconds = ( 130 * 5 ) / 25 = 26 m hence , a beats b by 26 m . answer b" | a = 25 - 20
b = a / 25
c = b * 130
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a ) 19 , b ) 19.5 , c ) 20 , d ) 20.5 , e ) 21 | d | divide(subtract(multiply(10, 25), multiply(add(const_4, const_1), 9)), 10) | the average of 10 consecutive integers is 25 . then , 9 is deducted from the first consecutive number , 8 is deducted from the second , 7 is deducted form the third , and so on until the last number which remains unchanged . what is the new average ? | "the total subtracted is ( 9 + 8 + . . . + 1 ) = ( 9 * 10 ) / 2 = 45 on average , each number will be reduced by 45 / 10 = 4.5 therefore , the overall average will be reduced by 4.5 the answer is d ." | a = 10 * 25
b = 4 + 1
c = b * 9
d = a - c
e = d / 10
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a ) 1.12 , b ) 1.65 , c ) 1.2 , d ) 1.3 , e ) none of these | b | divide(multiply(0.75, 11), 5) | if 0.75 : x : : 5 : 11 , then x is equal to : | "explanation : ( x * 5 ) = ( 0.75 * 11 ) x = 8.25 / 5 = 1.65 answer : b" | a = 0 * 75
b = a / 5
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a ) 41 % , b ) 42 % , c ) 43 % , d ) 43.75 % , e ) 44 % | d | divide(subtract(const_100, divide(subtract(const_100, multiply(divide(3, 4), const_100)), const_2)), const_2) | an empty fuel tank is filled with brand z gasoline . when the tank is 3 / 4 empty , it is filled with brand x gasoline . when the tank is half empty again , it is filled with brand z gasoline . when the tank is half empty again , it is filled with brand x gasoline . at this time , what percent of the gasoline in the tank is brand z ? | "work with fraction of brand z in the tank . 1 st step : brand z is 1 2 nd step : brand z is 1 / 2 3 rd step : brand z is ( 3 / 4 ) * ( 1 / 2 ) + 1 / 2 = 7 / 8 4 th step : brand z is ( 1 / 2 ) * ( 7 / 8 ) = 7 / 16 = 43.75 % answer ( d )" | a = 3 / 4
b = a * 100
c = 100 - b
d = c / 2
e = 100 - d
f = e / 2
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a ) 24 , b ) 25 , c ) 28 , d ) 32 , e ) 36 | e | divide(multiply(9, subtract(9, const_1)), const_2) | there are 9 teams in a certain league and each team plays each of the other teams exactly once . what is the total number of games played ? | "9 c 2 = 36 the answer is e" | a = 9 - 1
b = 9 * a
c = b / 2
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a ) 13 , b ) 13.5 , c ) 11.5 , d ) 12.5 , e ) 12 | d | divide(multiply(divide(multiply(500, 8), const_100), 4), multiply(divide(8, const_100), 160)) | in how many years rs 160 will produce the same interest at 8 % as rs . 500 produce in 4 years at 8 % | "explanation : clue : firstly we need to calculate the si with prinical 500 , time 4 years and rate 8 % , it will be rs . 160 then we can get the time as time = ( 100 * 160 ) / ( 160 * 8 ) = 12.5 option d" | a = 500 * 8
b = a / 100
c = b * 4
d = 8 / 100
e = d * 160
f = c / e
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a ) 14 , b ) 13 , c ) 9 , d ) 7 , e ) 2 | e | add(subtract(add(7, 9), subtract(15, 1)), subtract(9, 7)) | of 15 applicants for a job , 7 had at least 4 years ' experience , 9 had degrees , and 1 had less than 4 years ' experience and did not have a degree . how many of the applicants had at least 4 years ' experience and a degree ? | "15 - 1 = 14 14 - 7 - 9 = - 2 then 2 are in the intersection between 4 years experience and degree . answer : e" | a = 7 + 9
b = 15 - 1
c = a - b
d = 9 - 7
e = c + d
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a ) rs . 3225 , b ) rs . 2580 , c ) rs . 8000 , d ) rs . 1290 , e ) none | a | divide(multiply(divide(divide(add(divide(multiply(8000, 15), const_100), divide(multiply(add(8000, divide(multiply(8000, 15), const_100)), 15), const_100)), 2), 5), const_100), 8) | simple interest on a certain sum of money for 5 years at 8 % per annum is half the compound interest on rs . 8000 for 2 years at 15 % per annum . the sum placed on simple interest is | solution c . i . = rs [ 8000 x ( 1 + 15 / 100 ) â ² - 8000 ] rs . ( 8000 x 115 / 100 x 115 / 100 - 8000 ) = rs . 2580 . sum = rs . [ 1290 x 100 / 5 x 8 ] = rs . 3225 . answer a | a = 8000 * 15
b = a / 100
c = 8000 * 15
d = c / 100
e = 8000 + d
f = e * 15
g = f / 100
h = b + g
i = h / 2
j = i / 5
k = j * 100
l = k / 8
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a ) $ 120 , b ) $ 125 , c ) $ 130 , d ) $ 135 , e ) $ 140 | a | subtract(multiply(2.75, 80), multiply(1.25, 80)) | for each color copy , print shop x charges $ 1.25 and print shop y charges $ 2.75 . how much greater is the charge for 80 color copies at print shop y than at print shop x ? | "the difference in the two prices is $ 2.75 - $ 1.25 = $ 1.50 for each color copy . each color copy will cost an extra $ 1.50 at print shop y . 80 * $ 1.50 = $ 120 the answer is a ." | a = 2 * 75
b = 1 * 25
c = a - b
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a ) 1 / 5 , b ) 1 / 3 , c ) 1 / 2 , d ) 2 / 3 , e ) 4 / 5 | c | divide(const_4, add(0,0, const_10)) | in the xy - plane , a triangle has vertices ( 0,0 ) , ( 4,0 ) and ( 4,8 ) . if a point ( a , b ) is selected at random from the triangular region , what is the probability that a - b > 0 ? | "the area of the right triangle is ( 1 / 2 ) * 4 * 8 = 16 . only the points ( a , b ) below the line y = x satisfy a - b > 0 . the part of the triangle which is below the line y = x has an area of ( 1 / 2 ) ( 4 ) ( 4 ) = 8 . p ( a - b > 0 ) = 8 / 16 = 1 / 2 the answer is c ." | a = 0 + 0
b = 4 / a
|
a ) 2 km , b ) 2.6 km , c ) 3.6 km , d ) 4 km , e ) none | c | multiply(divide(12, const_60), add(15, 3)) | the speed of a boat in still water is 15 km / hr and the rate of current is 3 km / hr the distance travelled downstream in 12 minutes is : | sol . speed downstream = ( 15 + 3 ) kmph = 18 kmph . distance travelled = [ 18 * 12 / 60 ] km = 3.6 km . answer c | a = 12 / const_60
b = 15 + 3
c = a * b
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['a ) 12 b ^ 2', 'b ) 10 b ^ 2', 'c ) 6 b ^ 2', 'd ) 4 b ^ 2', 'e ) 2 b ^ 2'] | b | multiply(add(multiply(multiply(const_1, const_2), const_2), multiply(const_1, const_1)), const_2) | two cubes each with side b are joined to form a cuboids . what is the surface area of this cuboids ? | l = 2 b ; b = b ; h = b ; surface area = 2 * ( lb + bh + hl ) = 2 * ( 2 b ^ 2 + b ^ 2 + 2 b ^ 2 ) = 10 b ^ 2 answer : b | a = 1 * 2
b = a * 2
c = 1 * 1
d = b + c
e = d * 2
|
a ) 25.0 % , b ) 8.3 % , c ) 3.1 % , d ) 2.7 % , e ) 1.0 % | c | multiply(divide(const_1, add(multiply(7, 4), 4)), const_100) | bob is attempting to feed his pets . he feeds his pets as follows : - no dog receives more food than any other dog - no cat receives any less food than any other cat - the cats , as a group , receive exactly as much food as a single dog if bob has 7 dogs and 4 cats , approximately what percentage of the food bag does a cat receive ? | c . each cat gets 1 / 4 as much food as a dog . a dog gets 1 / 8 of the food . 1 / 4 of 1 / 8 is 1 / 32 , which is about 3.1 % , so the answer is c . | a = 7 * 4
b = a + 4
c = 1 / b
d = c * 100
|
a ) 6 , b ) 2 , c ) - 2 , d ) 1 , e ) - 10 | d | divide(subtract(7, 5), subtract(sqrt(add(9, 7)), sqrt(add(5, 9)))) | in the xy - coordinate plane , the graph of y = - x ^ 2 + 9 intersects line l at ( p , 5 ) and ( t , 7 ) . what is the least possible value of the slope of line l ? | "we need to find out the value of p and l to get to the slope . line l and graph y intersect at point ( p , 5 ) . hence , x = p and y = 5 should sactisfy the graph . soliving 5 = - p 2 + 9 p 2 = 4 p = + or - 2 simillarly point ( t , 7 ) should satisfy the equation . hence x = t and y = 7 . - 7 = - t 2 + 9 t = + or - 4 considering p = - 2 and t = 4 , the least slope is ( 7 - 5 ) / ( 4 - 2 ) = 1 imo option d is correct answer ." | a = 7 - 5
b = 9 + 7
c = math.sqrt(b)
d = 5 + 9
e = math.sqrt(d)
f = c - e
g = a / f
|
a ) 724 , b ) 804 , c ) 69 , d ) 844 , e ) none | c | divide(multiply(add(multiply(1, const_100), 50), multiply(11, const_100)), power(50, const_2)) | what is the least number of square tiles required to pave the floor of a room 11 m 50 cm long and 1 m 50 cm broad ? | "solution length of largest tile = h . c . f . of 1150 cm & 150 cm = 50 cm . area of each tile = ( 50 x 50 ) cm 2 ∴ required number of tiles = [ 1150 x 150 / 50 x 50 ] = 69 . answer c" | a = 1 * 100
b = a + 50
c = 11 * 100
d = b * c
e = 50 ** 2
f = d / e
|
a ) 62 , b ) 45 , c ) 28 , d ) 20 , e ) 25 | d | divide(subtract(multiply(36, 5), add(add(multiply(const_2, 5), const_3), 107)), const_3) | in a cricket match , 5 batsmen a , b , c , d and e scored an average of 36 runs . d scored 5 more than e ; e scored 8 fewer than a ; b scored as many as d and e combined ; and b and c scored 107 between them . how many runs did e score ? | explanation : total runs scored = ( 36 x 5 ) = 180 . let the runs scored by e be x . then , runs scored by d = x + 5 ; runs scored by a = x + 8 ; runs scored by b = x + x + 5 = 2 x + 5 ; runs scored by c = ( 107 - b ) = 107 - ( 2 x + 5 ) = 102 - 2 x . so , total runs = ( x + 8 ) + ( 2 x + 5 ) + ( 102 - 2 x ) + ( x + 5 ) + x = 3 x + 120 . therefore 3 x + 120 = 180 3 x = 60 x = 20 . answer : option d | a = 36 * 5
b = 2 * 5
c = b + 3
d = c + 107
e = a - d
f = e / 3
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a ) 25600.24 , b ) 25600.27 , c ) 25600.28 , d ) 25600.21 , e ) 25600.29 | a | divide(21093, power(subtract(1, divide(add(divide(1, 4), 6), const_100)), const_3)) | a property decreases in value every year at the rate of 6 1 / 4 % of its value at the beginning of the year its value at the end of 3 years was rs . 21093 . find its value at the beginning of the first year ? | 6 1 / 4 % = 1 / 16 x * 15 / 16 * 15 / 16 * 15 / 16 = 21093 x = 25600.24 answer : a | a = 1 / 4
b = a + 6
c = b / 100
d = 1 - c
e = d ** 3
f = 21093 / e
|
a ) 4.49 seconds , b ) 8.89 seconds , c ) 26 seconds , d ) 12 seconds , e ) 16 seconds | a | divide(40, multiply(add(24, 8), const_0_2778)) | the speed at which a man can row a boat in still water is 24 kmph . if he rows downstream , where the speed of current is 8 kmph , what time will he take to cover 40 metres ? | "speed of the boat downstream = 24 + 8 = 32 kmph = 32 * 5 / 18 = 8.89 m / s hence time taken to cover 40 m = 40 / 8.89 = 4.49 seconds . answer : a" | a = 24 + 8
b = a * const_0_2778
c = 40 / b
|
a ) 12 kg , b ) 13 kg , c ) 22 kg , d ) 14 kg , e ) 15 kg | a | multiply(divide(60, const_100), multiply(5, const_4)) | a manufacturing company has 15 % cobalt , 25 % led and 60 % of copper . if 5 kg of led is used in a mixture how much copper we need to use | 25 % = 5 kg 100 % = 20 kg 60 % of 20 kg = 12 kg 12 kg answer : a | a = 60 / 100
b = 5 * 4
c = a * b
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a ) 12 / 5 , b ) 5 / 12 , c ) 6 / 4 , d ) 2 / 3 , e ) 4 / 7 | b | divide(add(add(7, const_4), const_4), multiply(add(const_4, const_2), add(const_4, const_2))) | in a simultaneous throw of pair of dice . find the probability of getting the total more than 7 ? | "here n ( s ) = ( 6 * 6 ) = 36 let e = event of getting a total more than 7 = { ( 2,6 ) , ( 3,5 ) , ( 3,6 ) , ( 4,4 ) , ( 4,5 ) , ( 4,6 ) , ( 5,3 ) , ( 5,4 ) , ( 5,5 ) , ( 5,6 ) , ( 6,2 ) , ( 6,3 ) , ( 6,4 ) , ( 6,5 ) , ( 6,6 ) } p ( e ) = n ( e ) / n ( s ) = 15 / 36 = 5 / 12 option b" | a = 7 + 4
b = a + 4
c = 4 + 2
d = 4 + 2
e = c * d
f = b / e
|
a ) 1008 , b ) 1055 , c ) 1022 , d ) 1032 , e ) 1048 | b | add(multiply(multiply(power(const_3, const_2.0), power(const_2.0, const_4)), add(const_3, const_4)), 15) | the smallest number which when diminished by 15 , is divisible 10 , 15 , 25 , 30 and 35 is : | "required number = ( l . c . m . of 10 , 15 , 25 , 30 , 35 ) + 15 = 1050 + 15 = 1055 answer : option b" | a = 3 ** 2
b = 2 ** 0
c = a * b
d = 3 + 4
e = c * d
f = e + 15
|
a ) 37 , b ) 35 , c ) 34 , d ) 36 , e ) 38 | b | sqrt(85) | the length of the longest tape in cm which can be used to measure exactly , the length 7 m ; 3 m 85 cm ; and 12 m 95 cm is : | "the three lengths in cm are 700 , 385 & 1295 . hcf of 700 , 385 & 1295 is 35 . hence , the answer is 35 cm . answer : b" | a = math.sqrt(85)
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a ) 20 , b ) 22 , c ) 24 , d ) 26 , e ) 28 | a | subtract(add(divide(const_1, 12), divide(const_1, 15)), divide(const_1, 10)) | two pipes can fill a tank in 12 minutes and 15 minutes . an outlet pipe can empty the tank in 10 minutes . if all the pipes are opened when the tank is empty , then how many minutes will it take to fill the tank ? | "let v be the volume of the tank . the rate per minute at which the tank is filled is : v / 12 + v / 15 - v / 10 = 3 v / 60 = v / 20 per minute the tank will be filled in 20 minutes . the answer is a ." | a = 1 / 12
b = 1 / 15
c = a + b
d = 1 / 10
e = c - d
|
a ) 198 , b ) 200 , c ) 204 , d ) 240 , e ) 210 | d | add(192, multiply(192, divide(25, const_100))) | an article with cost price of 192 is sold at 25 % profit . what is the selling price ? | "sp = 1.25 * 192 = 240 answer : d" | a = 25 / 100
b = 192 * a
c = 192 + b
|
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