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 ) 66 , b ) 87 , c ) 75 , d ) 88 , e ) 90 | c | multiply(multiply(2, 2), add(2, const_4)) | find a two digit number , given that the sum of the digits is 12 and the difference of the digits is 2 . ? | "using elimination method find which of the options fit the description of the number . . . from the option only 75 meets this description sum of digits - - - 7 + 5 = 12 difference of digits - - - 7 - 5 = 2 answer c ." | a = 2 * 2
b = 2 + 4
c = a * b
|
a ) 6 , b ) 8 , c ) 10 , d ) 12 , e ) 14 | d | divide(24, subtract(3, const_1)) | lisa and robert have taken the same number of photos on their school trip . lisa has taken 3 times as many photos as claire and robert has taken 24 more photos than claire . how many photos has claire taken ? | l = r l = 3 c r = c + 24 3 c = c + 24 c = 12 the answer is d . | a = 3 - 1
b = 24 / a
|
a ) 49 , b ) 56 , c ) 60 , d ) 43 , e ) 70 | d | divide(multiply(25, add(add(multiply(multiply(add(const_3, const_2), const_2), multiply(multiply(const_3, const_4), const_100)), multiply(multiply(add(const_3, const_4), add(const_3, const_2)), multiply(add(const_3, const_2), const_2))), add(const_3, const_3))), const_100) | what is 25 % of 4 / 12 of 520 ? | 25 % = 25 / 100 = 1 / 4 of 4 / 12 = 1 / 4 * 4 / 12 = 1 / 12 of 520 = 1 / 12 * 520 = 43.33 = = 43 ans - d | a = 3 + 2
b = a * 2
c = 3 * 4
d = c * 100
e = b * d
f = 3 + 4
g = 3 + 2
h = f * g
i = 3 + 2
j = i * 2
k = h * j
l = e + k
m = 3 + 3
n = l + m
o = 25 * n
p = o / 100
|
a ) 10 , b ) 16 , c ) 13 , d ) 14 , e ) 19 | d | divide(multiply(multiply(42, 12), 5), multiply(30, 6)) | 42 persons can repair a road in 12 days , working 5 hours a day . in how many days will 30 persons , working 6 hours a day , complete the work ? | "let the required number of days be x . less persons , more days ( indirect proportion ) more working hours per day , less days ( indirect proportion ) persons 30 : 42 : : 12 : x working hours / day 6 : 5 30 x 6 x x = 42 x 5 x 12 x = ( 42 x 5 x 12 ) / ( 30 x 6 ) x = 14 answer d" | a = 42 * 12
b = a * 5
c = 30 * 6
d = b / c
|
a ) 27 % , b ) 26 % , c ) 29 % , d ) 30 % , e ) 60 % | a | multiply(divide(add(multiply(divide(25, const_100), 300), multiply(divide(30, const_100), 200)), add(300, 200)), const_100) | for an agricultural experiment , 300 seeds were planted in one plot and 200 were planted in a second plot . if exactly 25 percent of the seeds in the first plot germinated and exactly 30 percent of the seeds in the second plot germinated , what percent of the total number of seeds germinated ? | "in the first plot 25 % of 300 seeds germinated , so 0.25 x 300 = 75 seeds germinated . in the second plot , 30 % of 200 seeds germinated , so 0.30 x 200 = 60 seeds germinated . since 75 + 60 = 135 seeds germinated out of a total of 300 + 200 = 500 seeds , the percent of seeds that germinated is ( 135 / 500 ) x 100 % , or 27 % . answer : a ." | a = 25 / 100
b = a * 300
c = 30 / 100
d = c * 200
e = b + d
f = 300 + 200
g = e / f
h = g * 100
|
a ) 14 : 5 , b ) 15 : 5 , c ) 16 : 5 , d ) 19 : 6 , e ) none of these | d | divide(add(multiply(3, 4), 7), add(2, 4)) | if x : y = 1 : 4 , then find the value of ( 7 x + 3 y ) : ( 2 x + y ) | explanation : let x = 1 k and y = 4 k , so = 7 ( k ) + 3 ( 4 k ) / 2 ( k ) + 1 ( 4 k ) = 19 k / 6 k = 19 : 6 option d | a = 3 * 4
b = a + 7
c = 2 + 4
d = b / c
|
a ) 12,28 , b ) 14,30 , c ) 16,32 , d ) 18,42 , e ) 19,34 | d | subtract(divide(subtract(add(multiply(6, 6), 24), 6), subtract(6, const_1)), const_1) | ages of two persons differ by 24 years . if 6 year ago , the elder one be 3 times as old the younger one , find their present age | "explanation : let the age of younger person is x , then elder person age is ( x + 24 ) = > 3 ( x - 6 ) = ( x + 24 - 6 ) [ 6 years before ] = > 3 x - 18 = x + 18 = > x = 18 . so other person age is x + 24 = 42 answer : option d" | a = 6 * 6
b = a + 24
c = b - 6
d = 6 - 1
e = c / d
f = e - 1
|
a ) 20 % , b ) 30 % , c ) 40 % , d ) 50 % , e ) 45 % | e | subtract(100, 55) | john want to buy a $ 100 trouser at the store , but he think it Γ’ β¬ β’ s too expensive . finally , it goes on sale for $ 55 . what is the percent decrease ? | "the is always the difference between our starting and ending points . in this case , it Γ’ β¬ β’ s 100 Γ’ β¬ β 55 = 45 . the Γ’ β¬ Ε original Γ’ β¬ Β is our starting point ; in this case , it Γ’ β¬ β’ s 100 . ( 45 / 100 ) * 100 = ( 0.45 ) * 100 = 45 % . e" | a = 100 - 55
|
a ) 10 , b ) 8 , c ) 32 , d ) 14 , e ) 16 | c | divide(800, subtract(26, const_1)) | in a garden , 26 trees are planted at equal distances along a yard 800 metres long , one tree being at each end of the yard . what is the distance between two consecutive trees ? | "26 trees have 25 gaps between them . length of each gap = 800 / 25 = 32 i . e . , distance between two consecutive trees = 32 answer is c ." | a = 26 - 1
b = 800 / a
|
a ) 19 , b ) 26 , c ) 20 , d ) 32 , e ) 21 | b | subtract(multiply(28, const_3), multiply(29, const_2)) | the average age of a , b and c is 28 years . if the average age of a and c is 29 years , what is the age of b in years ? | "age of b = age of ( a + b + c ) Γ’ β¬ β age of ( a + c ) = 28 Γ£ β 3 Γ’ β¬ β 29 Γ£ β 2 = 84 Γ’ β¬ β 58 = 26 years b" | a = 28 * 3
b = 29 * 2
c = a - b
|
a ) 986000 , b ) 968000 , c ) 978000 , d ) 987000 , e ) none of them | a | subtract(986, multiply(multiply(237, 986), 863)) | evaluate : 986 x 237 + 986 x 863 | "986 x 137 + 986 x 863 = 986 x ( 137 + 863 ) = 986 x 1000 = 986000 . answer is a ." | a = 237 * 986
b = a * 863
c = 986 - b
|
a ) 70 , b ) 50 , c ) 60 , d ) 80 , e ) 65 | a | divide(divide(multiply(700, 30), const_100), 3) | a reduction of 30 % in the price of oil enables a house wife to obtain 3 kgs more for rs . 700 , what is the reduced price for kg ? | "700 * ( 30 / 100 ) = 210 - - - - 3 ? - - - - 1 = > rs . 70 answer : a" | a = 700 * 30
b = a / 100
c = b / 3
|
a ) 24 , b ) 120 , c ) 625 , d ) 720 , e ) 1024 | c | power(5, 4) | a multiple choice test consists of 4 questions , and each question has 5 answer choices . in how many f ways can the test be completed if every question is unanswered ? | "5 choices for each of the 4 questions , thus total f of 5 * 5 * 5 * 5 = 5 ^ 4 = 625 ways to answer all of them . answer : c ." | a = 5 ** 4
|
a ) 17 : 7 , b ) 18 : 8 , c ) 19 : 9 , d ) 19 : 5 , e ) 19 : 4 | a | divide(add(divide(multiply(62.5, 8), const_100), divide(multiply(87.5, 4), const_100)), add(subtract(8, divide(multiply(62.5, 8), const_100)), subtract(4, divide(multiply(87.5, 4), const_100)))) | two vessels p and q contain 62.5 % and 87.5 % of alcohol respectively . if 8 litres from vessel p is mixed with 4 litres from vessel q , the ratio of alcohol and water in the resulting mixture is ? | "quantity of alcohol in vessel p = 62.5 / 100 * 8 = 5 litres quantity of alcohol in vessel q = 87.5 / 100 * 4 = 7 / 2 litres quantity of alcohol in the mixture formed = 5 + 7 / 2 = 17 / 2 = 8.5 litres as 12 litres of mixture is formed , ratio of alcohol and water in the mixture formed = 8.5 : 3.5 = 17 : 7 . answer : a" | a = 62 * 5
b = a / 100
c = 87 * 5
d = c / 100
e = b + d
f = 62 * 5
g = f / 100
h = 8 - g
i = 87 * 5
j = i / 100
k = 4 - j
l = h + k
m = e / l
|
a ) 720 , b ) 120 , c ) 300 , d ) 30 , e ) 333 | b | subtract(subtract(subtract(divide(divide(divide(factorial(10), factorial(subtract(10, 3))), factorial(3)), const_2), 10), 10), const_10) | mariah has decided to hire three workers . to determine whom she will hire , she has selected a group of 10 candidates . she plans to have one working interview with 3 of the 10 candidates every day to see how well they work together . how many days will it take her to have working interviews with all the different combinations of job candidates ? | "there are c 310 = 120 different groups of three possible out of 10 candidates and since each of these groups is interviewed every day then total of 120 days are needed . answer : b" | a = math.factorial(10)
b = 10 - 3
c = math.factorial(b)
d = a / c
e = math.factorial(3)
f = d / e
g = f / 2
h = g - 10
i = h - 10
j = i - 10
|
a ) 22 , b ) 15 , c ) 77 , d ) 21 , e ) 182 | d | divide(multiply(subtract(34, 6), 3), 4) | ratio between rahul and deepak is 4 : 3 , after 6 years rahul age will be 34 years . what is deepak present age ? | "present age is 4 x and 3 x , = > 4 x + 6 = 34 = > x = 7 so deepak age is = 3 ( 7 ) = 21 answer : d" | a = 34 - 6
b = a * 3
c = b / 4
|
a ) 81 , b ) 100 , c ) 120 , d ) 132 , e ) 160 | d | divide(multiply(add(99, divide(multiply(99, 20), const_100)), const_100), multiply(multiply(const_3, const_3), 10)) | a retailer bought a machine at a wholesale price of $ 99 and later on sold it after a 10 % discount of the retail price . if the retailer made a profit equivalent to 20 % of the whole price , what is the retail price of the machine ? | "my solution : wholesale price = 99 retail price , be = x he provides 10 % discount on retail price = x - 10 x / 100 this retail price = 20 % profit on wholesale price x - 10 x / 100 = 99 + 1 / 5 ( 99 ) x = 132 ; answer : d" | a = 99 * 20
b = a / 100
c = 99 + b
d = c * 100
e = 3 * 3
f = e * 10
g = d / f
|
a ) 2 % , b ) 17 % , c ) 50 % , d ) 65 % , e ) 83 % | c | multiply(divide(subtract(480, 320), 320), const_100) | a store reported total sales of $ 480 million for february of this year . if the total sales for the same month last year was $ 320 million , approximately what was the percent increase in sales ? | "new value β old value ) / old value x 100 we are given : february sales this year = 480 million february sales last year = 320 million we need to determine the percent increase between sales from last year to sales this year . thus , the new value = 480 million and the old value = 320 million . let β s plug them into our percent change formula . ( new value β old value ) / old value x 100 [ ( 480 β 320 ) / 320 ] x 100 = 50 % . the answer is c ." | a = 480 - 320
b = a / 320
c = b * 100
|
a ) 24 % , b ) 25 % , c ) 34 % , d ) 36 % , e ) 40 % | c | multiply(divide(215, subtract(850, 215)), const_100) | a cricket bat is sold for $ 850 , making a profit of $ 215 . the profit percentage would be | "215 / ( 850 - 215 ) = 215 / 635 = 43 / 127 = 34 % . answer : c ." | a = 850 - 215
b = 215 / a
c = b * 100
|
a ) 16.39 , b ) 16.33 , c ) 16.35 , d ) 17.01 , e ) 16.32 | d | divide(add(add(multiply(30, 11.5), multiply(20, 14.25)), multiply(divide(add(multiply(30, 11.5), multiply(20, 14.25)), const_100), 35)), add(30, 20)) | arun purchased 30 kg of wheat at the rate of rs . 11.50 per kg and 20 kg of wheat at the rate of 14.25 per kg . he mixed the two and sold the mixture . approximately what price per kg should be sell the mixture to make 35 % profit ? | c . p . of 50 kg wheat = ( 30 * 11.50 + 20 * 14.25 ) = rs . 630 . s . p . of 50 kg wheat = 135 % of rs . 630 = 135 / 100 * 630 = rs . 850.50 . s . p . per kg = 850.50 / 50 = rs . 16.38 = 16.30 . answer : d | a = 30 * 11
b = 20 * 14
c = a + b
d = 30 * 11
e = 20 * 14
f = d + e
g = f / 100
h = g * 35
i = c + h
j = 30 + 20
k = i / j
|
a ) 80 , b ) 82 , c ) 84 , d ) 86 , e ) 88 | e | divide(multiply(multiply(multiply(const_12, const_2), 2), subtract(multiply(const_12, const_4), const_4)), multiply(const_12, const_2)) | how many times are the hands of a clock at right angle in 2 days ? | "in 1 day , they are at right angles 44 times . in 2 days , they are at right angles 88 times . answer : option e" | a = 12 * 2
b = a * 2
c = 12 * 4
d = c - 4
e = b * d
f = 12 * 2
g = e / f
|
a ) 9.28 , b ) 8.48 , c ) 8.78 , d ) 8.98 , e ) 9.18 | c | divide(add(8, 10), add(divide(8, 10), divide(10, 8))) | a cyclist rides a bicycle 8 km at an average speed of 10 km / hr and again travels 10 km at an average speed of 8 km / hr . what is the average speed for the entire trip ? | "distance = 18 km time = 8 / 10 + 10 / 8 = ( 64 + 100 ) / 80 = 164 / 80 = 41 / 20 hours average speed = ( 18 * 20 ) / 41 = 8.78 km / h the answer is c ." | a = 8 + 10
b = 8 / 10
c = 10 / 8
d = b + c
e = a / d
|
a ) 456578972 , b ) 436567874 , c ) 725087484 , d ) 725117481 , e ) 357889964 | c | multiply(subtract(9999, const_4), 72516) | find the value of 72516 x 9999 = m ? | "72516 x 9999 = 72516 x ( 10000 - 1 ) = 72516 x 10000 - 72516 x 1 = 725160000 - 72516 = 725087484 c" | a = 9999 - 4
b = a * 72516
|
a ) 0 , b ) 0 , c ) 2 , d ) 3 , e ) 4 | b | floor(divide(reminder(power(7, reminder(1001, add(const_4, const_1))), const_100), const_10)) | what is the tens digit of 7 ^ 1001 ? | "7 ^ 1 = 7 7 ^ 2 = 49 7 ^ 3 = 343 7 ^ 4 = 2401 7 ^ 5 = 16807 7 ^ 6 = 117649 we should see this as pattern recognition . we have a cycle of 4 . ( we can multiply the last 2 digits only as we care about ten ' s digit ) 0 , 4 , 4 , 0 . 1001 = 4 * 250 + 1 the ten ' s digit will be 1 . answer b" | a = 4 + 1
b = 7 ** reminder
c = reminder / (
d = math.floor(c, 100)
|
a ) s . 345 , b ) s . 350 , c ) s . 352 , d ) s . 362 , e ) s . 368 | c | multiply(subtract(multiply(12000, divide(subtract(multiply(3, 4), 3), multiply(3, 4))), multiply(9000, divide(subtract(multiply(3, 4), 4), multiply(3, 4)))), divide(3872, add(add(18000, multiply(12000, divide(subtract(multiply(3, 4), 3), multiply(3, 4)))), multiply(9000, divide(subtract(multiply(3, 4), 4), multiply(3, 4)))))) | suresh started a business , investing rs . 18000 . after 3 months and 4 months respectively , rohan and sudhir joined him with capitals of 12000 and 9000 . at the end of the year the total profit was rs . 3872 . what is the difference between rohan β s and sudhir β s share in the profit ? | "suresh : rohan : sudhir ratio of their investments = 18000 Γ 12 : 12000 Γ 9 : 9000 Γ 8 = 6 : 3 : 2 the difference between rohan β s and sudhir β s share = 1 share : . i . e . = rs . 3872 Γ 1 / 11 = rs . 352 . c" | a = 3 * 4
b = a - 3
c = 3 * 4
d = b / c
e = 12000 * d
f = 3 * 4
g = f - 4
h = 3 * 4
i = g / h
j = 9000 * i
k = e - j
l = 3 * 4
m = l - 3
n = 3 * 4
o = m / n
p = 12000 * o
q = 18000 + p
r = 3 * 4
s = r - 4
t = 3 * 4
u = s / t
v = 9000 * u
w = q + v
x = 3872 / w
y = k * x
|
a ) 10000 , b ) 10100.5 , c ) 20000 , d ) 15000 , e ) 19000 | b | divide(subtract(multiply(multiply(multiply(4, const_100), const_100), power(add(1, divide(divide(4, const_100), 4)), 4)), multiply(multiply(4, const_100), const_100)), subtract(power(add(1, divide(divide(2, const_100), 2)), 2), 1)) | john invests $ x at the semi - annual constant compounded rate of 2 percent and also does $ 5,000 at the quarterly constant compounded rate of 4 percent . if the interests are the same after 1 year , what is the value of x ? ? | "a = p ( 1 + r / n ) ^ nt a = total amount accrued p = principal deposited r = rate of interest in decimal form n = number of times per year , interest compounded t = time in number of years . . x ( 1 + 0.02 / 2 ) ^ 2 - x = 5,000 ( 1 + 0.04 / 4 ) ^ 4 - 5,000 [ when the principal is subtracted from the total amount accrued , the resulting difference is the interest portion and question states interests are equal ) = > x [ ( 1.01 ) ^ 2 - 1 ] = 5,000 [ ( 1.01 ) ^ 4 - 1 ] = > x [ ( 1.01 ) ^ 2 - 1 ] = 5,000 [ ( 1.01 ) ^ 2 + 1 ] [ ( 1.01 ) ^ 2 - 1 ] - - > using a ^ 2 - b ^ 2 = a + b x a - b formula and cancel common expression on both sides = > x = 5,000 ( 1.0201 + 1 ) = 10 , 100.5 hence answer is b ." | a = 4 * 100
b = a * 100
c = 4 / 100
d = c / 4
e = 1 + d
f = e ** 4
g = b * f
h = 4 * 100
i = h * 100
j = g - i
k = 2 / 100
l = k / 2
m = 1 + l
n = m ** 2
o = n - 1
p = j / o
|
a ) 11 , b ) 12 , c ) 13 , d ) 14 , e ) 15 | e | subtract(subtract(98, 82), divide(98, add(const_1, const_10))) | what number is obtained by adding the units digits of 734 ^ 98 and 347 ^ 82 ? | "the units digit of 734 ^ 98 is 6 because 4 raised to the power of an even integer ends in 6 . the units digit of 347 ^ 82 is 9 because powers of 7 end in 7 , 9 , 3 , or 1 cyclically . since 82 is in the form 4 n + 2 , the units digit is 9 . then 6 + 9 = 15 . the answer is e ." | a = 98 - 82
b = 1 + 10
c = 98 / b
d = a - c
|
a ) 4 , b ) 3 , c ) 14 , d ) 20 , e ) 28 | a | subtract(60, add(20, 36)) | in a certain alphabet , 20 letters contain a dot and a straight line . 36 letters contain a straight line but do not contain a dot . if that alphabet has 60 letters , all of which contain either a dot or a straight line or both , how many letters contain a dot but do not contain a straight line ? | "we are told that all of the letters contain either a dot or a straight line or both , which implies that there are no letters without a dot and a line ( no line / no dot box = 0 ) . first we find the total # of letters with lines : 20 + 36 = 56 ; next , we find the total # of letters without line : 60 - 56 = 4 ; finally , we find the # of letters that contain a dot but do not contain a straight line : 4 - 0 = 4 ." | a = 20 + 36
b = 60 - a
|
a ) $ 115,000 , b ) $ 165,000 , c ) $ 215,000 , d ) $ 240,000 , e ) $ 365,000 | b | add(multiply(25, 10), 10) | an auction house charges a commission of 25 % on the first $ 50,000 of the sale price of an item , plus 10 % on the amount of of the sale price in excess of $ 50,000 . what was the price of a painting for which the house charged a total commission of $ 24,000 ? | "say the price of the house was $ x , then 0.25 * 50,000 + 0.1 * ( x - 50,000 ) = 24,000 - - > x = $ 165,000 ( 25 % of $ 50,000 plus 10 % of the amount in excess of $ 50,000 , which is x - 50,000 , should equal to total commission of $ 24,000 ) . answer : b ." | a = 25 * 10
b = a + 10
|
a ) 41 - 44 , b ) 39 - 41 , c ) 38 - 40 , d ) 37 - 39 , e ) 36 - 37 | a | add(divide(multiply(35.50, 15), const_100), 35.50) | a meal cost $ 35.50 adn there was no tax . if the tip was more than 15 pc but less than 25 pc of the price , then the total amount paid should be : | "15 % ( 35.5 ) = 5.325 25 % ( 35.5 ) = 8.875 total amount could have been 35.5 + 5.325 and 35.5 + 8.875 = > could have been between 40.825 and 44.375 = > approximately between 41 and 44 answer is a ." | a = 35 * 50
b = a / 100
c = b + 35
|
a ) 167 , b ) 175 , c ) 183 , d ) 191 , e ) 199 | d | add(multiply(95, const_2), 1) | the sum of the even numbers between 1 and n is 95 * 96 , where n is an odd number , then n = ? | "let n - 1 = 2 a . 2 + 4 + . . . + 2 a = 2 * ( 1 + 2 + . . . + a ) = 2 * ( a ) ( a + 1 ) / 2 = ( a ) ( a + 1 ) = 95 * 96 then a = 95 and n = 191 . the answer is d ." | a = 95 * 2
b = a + 1
|
a ) 3 , b ) 8 , c ) 10 , d ) 12 , e ) 15 | a | divide(add(50, 25), 25) | the present ratio of students to teachers at a certain school is 50 to 1 . if the student enrollment were to increase by 50 students and the number of teachers were to increase by 5 , the ratio of students to teachers would then be 25 to 1 . what is the present number of teachers ? | we are given that the ratio of students to teacher is 50 to 1 . we can rewrite this using variable multipliers . students : teachers = 50 x : x we are next given that student enrollment increases by 50 and the number of teachers increases by 5 . with this change the new ratio becomes 25 to 1 . we can put all this into an equation : students / teachers ο 25 / 1 = ( 30 x + 50 ) / ( x + 5 ) if we cross multiply we have : 25 ( x + 5 ) = 50 x + 50 25 x + 125 = 50 x + 50 3 = x since x is the present number of teachers , currently there are 3 teachers . answer a . | a = 50 + 25
b = a / 25
|
a ) 5 / 3 , b ) 3 / 9 , c ) 3 / 2 , d ) 3 / 5 , e ) 3 / 4 | e | divide(const_3, 4) | julie decided to save a certain amount of her monthly salary each month and her salary was unchanged from month to month . if julie ' s savings by the end of the year from these monthly savings were 4 times the amount she spent per month , what should be the fraction of her salary that she spent each month ? | let julie ' s monthly savings = s julie ' s monthly pay = p julie ' s monthly expenditure = p - s julie ' s savings by the end of the year from these monthly savings were four times the amount she spent per month 12 s = 4 * ( p - s ) = > 3 s = p - s = > p = 4 s julie ' s monthly expenditure = p - s = 4 s - s = 3 s fraction of her salary that julie spent each month = 3 s / 4 s = p - s / p = 3 / 4 . answer is e | a = 3 / 4
|
a ) 1 , b ) 2 , c ) 3 , d ) 4 , e ) 5 | c | power(2, 2) | the function f ( n ) is defined as the product of all the consecutive positive integers between 2 and n ^ 2 , inclusive , whereas the function g ( n ) is defined as the product of the squares of all the consecutive positive integers between 1 and n , inclusive . the exponent on 2 in the prime factorization of f ( 3 ) / g ( 3 ) is | "f ( 3 ) / g ( 3 ) = product ( 1 to 3 ^ 2 ) / 1.2 ^ 2.3 ^ 2 = 1 . 2.3 . 4.5 . 6.7 . 8.9 / 1 . 4.9 = 1 . 2.3 . ( 2 ^ 2 ) . 5 . ( 2.3 ) . 7 . ( 2 ^ 3 ) . 9 / 1 . ( 2 ^ 2 ) . 9 = 1 . ( 2 ^ 7 ) . 3.5 . 7.9 / 1 . ( 2 ^ 2 ) . 9 loof for 2 ^ 7 / 2 ^ 2 = 2 ^ 5 - - - - exponent 3 answer : c" | a = 2 ** 2
|
a ) 6209 , b ) 6200 , c ) 6799 , d ) 6199 , e ) 6685 | b | divide(multiply(const_100, 441), 7) | calculate the amount that an investor needs to be invest to earn $ 441 in interest in 12 months if the investor plans to invest x dollars in a savings account that pays interest at an annual rate of 7 % compounded semi - annually ? | the approach is substitution , our interest requirement is $ 441 after 12 months , 2 compounding period . calculate the compound interest on each option and find out the one that yields $ 441 in 12 months 6200 yielded $ 441 using the formula a = p ( 1 + r / n ) nt hence answer is b | a = 100 * 441
b = a / 7
|
a ) 65 , b ) 62 , c ) 61 , d ) 56 , e ) 58 | d | divide(factorial(subtract(add(const_4, 5), const_1)), multiply(factorial(5), factorial(subtract(const_4, const_1)))) | how many positive integers less than 6,000 are there in which the sum of the digits equals 5 ? | "basically , the question asks how many 4 digit numbers ( including those in the form 0 xxx , 00 xx , and 000 x ) have digits which add up to 5 . think about the question this way : we know that there is a total of 5 to be spread among the 4 digits , we just have to determine the number of ways it can be spread . let x represent a sum of 1 , and | represent a seperator between two digits . as a result , we will have 5 x ' s ( digits add up to the 5 ) , and 3 | ' s ( 3 digit seperators ) . so , for example : xx | x | x | x = 2111 | | xxx | xx = 0032 etc . there are 8 c 3 ways to determine where to place the separators . hence , the answer is 8 c 3 = 56 . d" | a = 4 + 5
b = a - 1
c = math.factorial(b)
d = math.factorial(5)
e = 4 - 1
f = math.factorial(e)
g = d * f
h = c / g
|
a ) 4.5 , b ) 8.9 , c ) 10.5 , d ) 4.5 , e ) 5.6 | c | divide(add(multiply(8, 8), multiply(5, 4)), 8) | the average ( arithmetic mean ) of 8 numbers is 8 . if 4 is added from each of 5 of the numbers , what is the new average ? | sum of 8 numbers = 8 * 8 = 64 if 4 is added from each of five of the numbers , we added 4 * 5 = 20 from the total sum sum of 8 number after adding 4 from each of five of the numbers = 64 + 20 = 84 new average = 84 / 8 = 10.5 answer c | a = 8 * 8
b = 5 * 4
c = a + b
d = c / 8
|
a ) 5 , b ) 6 , c ) 7 , d ) 11 , e ) 12 | a | subtract(subtract(subtract(10, 3), const_1), const_1) | list k consists of 10 consecutive integers . if - 3 is the least integer in list k , what is the range of the positive integers in list k ? | "answer = a = 5 if least = - 3 , then largest = 6 range = 6 - 1 = 5" | a = 10 - 3
b = a - 1
c = b - 1
|
a ) 24 % , b ) 25 % , c ) 26 % , d ) 28 % , e ) 35 % | c | divide(multiply(subtract(add(multiply(divide(multiply(280, 40), const_100), divide(add(const_100, 20), const_100)), multiply(divide(multiply(280, 60), const_100), divide(add(const_100, 30), const_100))), 280), const_100), 280) | a shopkeeper has 280 kg of apples . he sells 40 % of these at 20 % profit and remaining 60 % at 30 % profit . find his % profit on total . | "total number of apples = 280 let the cost price be x selling price at 20 % profit = 1.2 x selling price at 30 % profit = 1.3 x profit % = ( ( sp - cp ) / cp ) * 100 profit % = ( ( 2 / 5 ) * 280 * 1.2 x + ( 3 / 5 ) * 280 * 1.3 x - 280 x ) / 280 x * 100 = ( 1 / 5 * ( 2.4 + 3.9 ) - 1 ) * 100 = ( 6.3 - 5 ) * 20 = 26 % answer : c" | a = 280 * 40
b = a / 100
c = 100 + 20
d = c / 100
e = b * d
f = 280 * 60
g = f / 100
h = 100 + 30
i = h / 100
j = g * i
k = e + j
l = k - 280
m = l * 100
n = m / 280
|
a ) 3 , b ) 4 , c ) 5 , d ) 6 , e ) 7 | b | multiply(subtract(3, 2), divide(288, add(multiply(3, 14), multiply(2, 15)))) | we run a business that rents out canoes and kayaks . a canoe rental costs $ 14 per day , and a kayak rental costs $ 15 dollars per day . one day , our business rents out 3 canoes for every 2 kayaks and receives a total of $ 288 in revenue . how many more canoes than kayaks were rented out ? | let x be the number of canoes . then 2 x / 3 is the number of kayaks . 14 x + ( 2 x / 3 ) * 15 = 288 14 x + 10 x = 288 24 x = 288 x = 12 ( canoes ) 2 x / 3 = 8 ( kayaks ) there were 12 - 8 = 4 more canoes rented out . the answer is b . | a = 3 - 2
b = 3 * 14
c = 2 * 15
d = b + c
e = 288 / d
f = a * e
|
a ) 25 , b ) 31 , c ) 15 , d ) 26 , e ) 23 | b | subtract(subtract(45, 5), const_10) | positive integers from 1 to 45 , inclusive are placed in 5 groups of 9 each . what is the highest possible average of the medians of these 5 groups ? | answer we need to maximize the median in each group in order to maximize the average of all the medians . the highest possible median is 41 as there should be 4 numbers higher than the median in a group of 9 . so , if we have a group that has a , b , c , d , 41 , 42 , 43 , 44 , 45 , the median will be 41 . in this set , it is essential not to expend any more high values on a , b , c , or d as these do not affect the median . the median of a group that comprises 1 , 2 , 3 , 4 , 41 , 42 , 43 , 44 , 45 will be 41 . the next group can be 5 , 6 , 7 , 8 , 36 , 37 , 38 , 39 , 40 . the median will be 36 . extrapolating the findings in the two sets listed above , to maximize medians in all the 5 groups , the medians of the 5 groups will have to be 21 , 26 , 31 , 36 and 41 . the average of the highest possible medians will be the average of these 5 numbers = 31 . answer b | a = 45 - 5
b = a - 10
|
a ) 22 % , b ) 24 % , c ) 25 % , d ) 27 % , e ) 28 % | e | multiply(divide(add(multiply(divide(20, const_100), 500), multiply(divide(40, const_100), subtract(800, 500))), 800), const_100) | for each of her sales , a saleswoman receives a commission equal to 20 percent of the first $ 500 of the total amount of the sale , plus 40 percent of the total amount in excess of $ 500 . if the total amount of one of her sales was $ 800 , the saleswoman β s commission was approximately what percent of the total amount of the sale ? | "total sales = 800 comission = ( 20 / 100 ) * 500 + ( 40 / 100 ) * 300 = 100 + 120 = 220 % comission = ( 220 / 800 ) * 100 = 27.5 ~ 28 % answer is e" | a = 20 / 100
b = a * 500
c = 40 / 100
d = 800 - 500
e = c * d
f = b + e
g = f / 800
h = g * 100
|
a ) 180 , b ) 185 , c ) 190 , d ) 195 , e ) 200 | b | divide(divide(multiply(add(multiply(5000, const_2), multiply(subtract(const_12, const_1), 100)), const_12), const_2), add(add(multiply(const_3, 100), multiply(multiply(const_2, const_3), const_10)), add(const_2, const_3))) | a salt manufacturing company produced a total of 5000 tonnes of salt in january of a particular year . starting from february its production increased by 100 tonnes every month over the previous months until the end of the year . find its ave 66 rage daily production for that year ? | "total production of salt by the company in that year = 5000 + 5100 + 5200 + . . . . + 6100 = 66600 . average monthly production of salt for that year = 66600 / 365 = 185 answer : b" | a = 5000 * 2
b = 12 - 1
c = b * 100
d = a + c
e = d * 12
f = e / 2
g = 3 * 100
h = 2 * 3
i = h * 10
j = g + i
k = 2 + 3
l = j + k
m = f / l
|
a ) 85.5 , b ) 86.5 , c ) 87.5 , d ) 88.5 , e ) 89.5 | c | add(multiply(9, 2.5), 65) | the average weight of 9 person ' s increases by 2.5 kg when a new person comes in place of one of them weighing 65 kg . what is the weight of the new person ? | "total increase in weight = 9 Γ 2.5 = 22.5 if x is the weight of the new person , total increase in weight = x β 65 = > 22.5 = x - 65 = > x = 22.5 + 65 = 87.5 answer : c" | a = 9 * 2
b = a + 65
|
a ) 53 , b ) 44 , c ) 58 , d ) 60 , e ) 62 | a | add(add(18, multiply(5, const_4)), multiply(5, 3)) | a hiker walked for 3 days . she walked 18 miles on the first day , walking 3 miles per hour . on the second day she walked for one less hour but she walked one mile per hour , faster than on the first day . on the third day she walked at 5 miles per hour for 3 hours . how many miles in total did she walk ? | "first day - 18 miles with 3 miles per hours then total - 6 hours for that day second day - 4 miles per hour and 5 hours - 20 miles third day - 5 miles per hour and 3 hours - 15 miles total 18 + 20 + 15 = 53 answer : option a ." | a = 5 * 4
b = 18 + a
c = 5 * 3
d = b + c
|
a ) 2 / 15 , b ) 8 / 15 , c ) 3 / 11 , d ) 5 / 12 , e ) 6 / 13 | d | subtract(const_1, multiply(5, add(divide(const_1, 15), divide(const_1, 20)))) | a can do a job in 15 days and b in 20 days . if they work on it together for 5 days , then the fraction of the work that is left is ? | "a ' s 1 day work = 1 / 15 b ' s 1 day work = 1 / 20 a + b 1 day work = 1 / 15 + 1 / 20 = 7 / 60 a + b 5 days work = 7 / 60 * 5 = 7 / 12 remaining work = 1 - 7 / 12 = 5 / 12 answer is d" | a = 1 / 15
b = 1 / 20
c = a + b
d = 5 * c
e = 1 - d
|
a ) 90 , b ) 91 , c ) 94 , d ) none , e ) 95 | c | add(add(divide(380, add(const_4, const_1)), divide(subtract(380, add(const_4, const_1)), power(add(const_4, const_1), const_2))), divide(subtract(380, add(const_4, const_1)), power(add(const_4, const_1), const_3))) | how many zeros are the end of 380 ! ? | 380 ! has 380 / 5 + 380 / 5 ^ 2 + 380 / 5 ^ 3 = 76 + 15 + 3 = 94 trailing zeros ( take only the quotient into account ) . answer : c . | a = 4 + 1
b = 380 / a
c = 4 + 1
d = 380 - c
e = 4 + 1
f = e ** 2
g = d / f
h = b + g
i = 4 + 1
j = 380 - i
k = 4 + 1
l = k ** 3
m = j / l
n = h + m
|
a ) 107 , b ) 109 , c ) 111 , d ) 113 , e ) 115 | e | add(1, lcm(38, 3)) | find the least number which when divided by 38 and 3 leaves a remainder of 1 in each case . | "the least number which when divided by different divisors leaving the same remainder in each case = lcm ( different divisors ) + remainder left in each case . hence the required least number = lcm ( 38 , 3 ) + 1 = 115 . answer : e" | a = math.lcm(38, 3)
b = 1 + a
|
a ) 13 , b ) 15 , c ) 17 , d ) 19 , e ) 21 | d | add(multiply(add(3, 3), 3), floor(divide(subtract(150, multiply(divide(multiply(3, subtract(10, 1)), const_2), 10)), 10))) | in a certain supermarket , a triangular display of cans is arranged in 10 rows , numbered 1 through 10 from top to bottom . each successively numbered row contains 3 more cans than the row immediately above it . if there are fewer than 150 cans in the entire display , how many cans are in the seventh row ? | let x be the number of cans in row 1 . the total number of cans is x + ( x + 3 ) + . . . + ( x + 27 ) = 10 x + 3 ( 1 + 2 + . . . + 9 ) = 10 x + 3 ( 9 ) ( 10 ) / 2 = 10 x + 135 since the total is less than 150 , x must equal 1 . the number of cans in the 7 th row is 1 + 3 ( 6 ) = 19 the answer is d . | a = 3 + 3
b = a * 3
c = 10 - 1
d = 3 * c
e = d / 2
f = e * 10
g = 150 - f
h = g / 10
i = math.floor(h)
j = b + i
|
a ) 96 , b ) 108 , c ) 120 , d ) 132 , e ) 144 | d | multiply(subtract(multiply(subtract(multiply(3, 5), 3), 3), 3), subtract(5, const_1)) | q ' = 3 q - 3 , what is the value of ( 5 ' ) ' ? | "( 5 ' ) ' = ( 3 * 5 - 3 ) ' = 12 ' = 12 * 12 - 12 = 132 answer d" | a = 3 * 5
b = a - 3
c = b * 3
d = c - 3
e = 5 - 1
f = d * e
|
a ) 32 , b ) 28 , c ) 29 , d ) 54 , e ) 16 | e | divide(multiply(multiply(subtract(9, 1), add(9, 1)), 3), add(add(9, 1), subtract(9, 1))) | a person can row at 9 kmph and still water . he takes 3 1 / 2 hours to row from a to b and back . what is the distance between a and b if the speed of the stream is 1 kmph ? | "let the distance between a and b be x km . total time = x / ( 9 + 1 ) + x / ( 9 - 1 ) = 3.5 = > x / 10 + x / 8 = 7 / 2 = > ( 4 x + 5 x ) / 40 = 7 / 2 = > x = 16 km . answer : e" | a = 9 - 1
b = 9 + 1
c = a * b
d = c * 3
e = 9 + 1
f = 9 - 1
g = e + f
h = d / g
|
a ) 330 , b ) 300 , c ) 270 , d ) 250 , e ) 350 | a | divide(multiply(30, 2310), 210) | the l . c . m of two numbers is 2310 and their h . c . f is 30 . if one number is 210 the other is | "the other number = l . c . m * h . c . f / given number = 2310 * 30 / 210 = 330 answer is a ." | a = 30 * 2310
b = a / 210
|
a ) 1439 , b ) 1440 , c ) 1459 , d ) 1449 , e ) 1500 | a | add(lcm(lcm(lcm(12, 18), lcm(24, 32)), 40), 1) | the smallest number when increased by ` ` 1 ` ` is exactly divisible by 12 , 18 , 24 , 32 and 40 is : | lcm = 1440 1440 - 1 = 1439 answer : a | a = math.lcm(12, 18)
b = math.lcm(24, 32)
c = math.lcm(a, b)
d = math.lcm(c, 40)
e = d + 1
|
a ) 7 , b ) 8 , c ) 15 , d ) 16 , e ) 17 | e | power(divide(10, 5), const_4) | if ( 2 ^ 17 ) ( 25 ^ s ) = 5 ( 10 ^ m ) what is the value of m ? | "given 2 ^ 17 * 25 ^ s = 5 * 10 ^ 2 = > 2 ^ 17 * 5 ^ ( 2 s ) = 2 ^ m * 5 ^ ( m + 1 ) ans e on comparing the power of 2 = > m = 17" | a = 10 / 5
b = a ** 4
|
a ) 30 , b ) 36 , c ) 42 , d ) 45 , e ) 48 | d | subtract(subtract(multiply(divide(720, 25), const_2), const_4), const_10) | of the 720 students at a certain university , 25 percent are seniors . if x seniors were to graduate early and leave the university and no additional students entered or left the university , what value of x would reduce the number of seniors at the university to 20 percent ? | sol : 180 - x = 20 % of ( 720 - x ) 4 x = 180 x = 45 answer : d | a = 720 / 25
b = a * 2
c = b - 4
d = c - 10
|
a ) 1 / 2 , b ) 1 , c ) 3 , d ) 4 , e ) 5 | a | divide(const_1, 2) | from given equation find the value of x : 2 x Β² + 9 x β 5 | that quadratic is factored as follows : 2 x Β² + 9 x β 5 = ( 2 x β 1 ) ( x + 5 ) . lesson 17 . now , it is easy to see that the second factor will be 0 when x = β 5 . as for the value of x that will make 2 x β 1 = 0 , we must solve that little equation . ( lesson 9 . ) we have : 2 x = 1 x = 1 2 the solutions are : x = 1 / 2 or β 5 a | a = 1 / 2
|
a ) 2 / 15 , b ) 8 / 15 , c ) 13 / 20 , d ) 1 / 12 , e ) 6 / 13 | c | subtract(const_1, multiply(3, add(divide(const_1, 15), divide(const_1, 20)))) | a can do a job in 15 days and b in 20 days . if they work on it together for 3 days , then the fraction of the work that is left is ? | a ' s 1 day work = 1 / 15 b ' s 1 day work = 1 / 20 a + b 1 day work = 1 / 15 + 1 / 20 = 7 / 60 a + b 3 days work = 7 / 60 * 3 = 7 / 20 remaining work = 1 - 7 / 20 = 13 / 20 answer is c | a = 1 / 15
b = 1 / 20
c = a + b
d = 3 * c
e = 1 - d
|
a ) 10 , b ) 15 , c ) 21 , d ) 18 , e ) 19 | c | divide(subtract(500, 100), 19) | how many positive integers between 100 and 500 are there such that they are multiples of 19 ? | multiples of 19 = 114 , 133,152 , - - - - - , 494 number of multiples of 19 = > 494 - 114 / 19 + 1 = 21 answer is c | a = 500 - 100
b = a / 19
|
a ) 35.5 , b ) 36.5 , c ) 37.5 , d ) 38.5 , e ) 39.5 | c | divide(multiply(100, const_3), subtract(multiply(const_3, const_3), const_1)) | p says to q ` ` i am thrice as old as you were when i was as old as you are ' ' . if the sum of their present age is 100 years , then the present age of q ? | let the present age of q be x the age of q years back ( to be defined ) is x / 3 as the age of p will be x by then . years passed = x - x / 3 = 2 x / 3 ( subtracting age of q ) so p acurrent age is x + 2 x / 3 and q current age is x adding them to 100 we get x = 37.5 answer : c | a = 100 * 3
b = 3 * 3
c = b - 1
d = a / c
|
a ) 20 , b ) 40 , c ) 60 , d ) 80 , e ) 100 | a | multiply(divide(subtract(divide(12, multiply(subtract(15, 12), divide(40, const_60))), multiply(subtract(15, 12), divide(40, const_60))), subtract(15, 12)), const_60) | annie and sam set out together on bicycles traveling at 15 and 12 km per hour respectively . after 40 minutes , annie stops to fix a flat tire . if it takes annie 15 minutes to fix the flat tire and sam continues to ride during this time , how many minutes will it take annie to catch up with sam assuming that annie resumes riding at 15 km per hour ? | "annie gains 3 km per hour ( or 1 km every 20 minutes ) on sam . after 40 minutes annie is 2 km ahead . in the next 15 minutes , sam rides 3 km so sam will be 1 km ahead . it will take annie 20 minutes to catch sam . the answer is a ." | a = 15 - 12
b = 40 / const_60
c = a * b
d = 12 / c
e = 15 - 12
f = 40 / const_60
g = e * f
h = d - g
i = 15 - 12
j = h / i
k = j * const_60
|
a ) 0.25 % , b ) 4 % , c ) 25 % , d ) 40 % , e ) 60 % | e | multiply(divide(75, 45), const_100) | what percent of 75 is 45 ? | "75 * x / 100 = 45 x = 4 * 45 / 3 x = 60 ans : e" | a = 75 / 45
b = a * 100
|
a ) 102 , b ) 80 , c ) 75 , d ) 70 , e ) 65 | a | divide(add(multiply(divide(30, add(30, 40)), 50), multiply(divide(40, add(30, 40)), 90)), divide(add(30, 40), const_60)) | a car was driving at 50 km / h for 30 minutes , and then at 90 km / h for another 40 minutes . what was its average speed ? | "driving at 50 km / h for 30 minutes , distance covered = 50 * 1 / 2 = 25 km driving at 90 km / h for 40 minutes , distance covered = 90 * 2 / 3 = 60 km average speed = total distance / total time = 85 / 5 / 6 = 102 km / h answer : a" | a = 30 + 40
b = 30 / a
c = b * 50
d = 30 + 40
e = 40 / d
f = e * 90
g = c + f
h = 30 + 40
i = h / const_60
j = g / i
|
a ) 15 % , b ) 17 % , c ) 24 % , d ) 30 % , e ) 33 % | e | multiply(divide(subtract(divide(40, const_100), multiply(divide(25, const_100), divide(40, const_100))), subtract(const_1, multiply(divide(25, const_100), divide(40, const_100)))), const_100) | in february wilson β s earnings were 40 percent of his family β s total income . in march wilson earned 25 percent less than in february . if the rest of his family β s income was the same in both months , then , in march , wilson β s earnings were approximately what percent w of his family β s total income ? | "lets suppose the total family income in feb = 100 x wilson ' s earning in feb = 40 % of 100 x = 40 x earnings of remaining family in feb = 100 x - 40 x = 60 x wilson ' s earning in march = 75 % of wilson ' s feb earnings = 75 % of 40 x = 30 x earnings of remaining family in march = earnings of remaining family in feb = 60 x thus wilson ' s earning as % of total family income in march w = 30 x / ( 30 + 60 ) x = 30 x / 90 x = 33.33 % thus answer is e" | a = 40 / 100
b = 25 / 100
c = 40 / 100
d = b * c
e = a - d
f = 25 / 100
g = 40 / 100
h = f * g
i = 1 - h
j = e / i
k = j * 100
|
a ) 1 / 130 , b ) 1 / 5 , c ) 3 / 13 , d ) 10 / 13 , e ) 30 / 31 | e | divide(6, add(divide(20, const_100), 6)) | a committee is reviewing a total of 20 x black - and - white films and 6 y color films for a festival . if the committee selects y / x % of the black - and - white films and all of the color films , what fraction r of the selected films are in color ? | "it ' s y / xpercentnot y / x . if x = 20 and y = 10 . then : 20 x = 400 black - and - white films ; 6 y = 60 color films . y / x % = 10 / 20 % = 0.5 % of the black - and - white films , so 2 black - and - white films and all 60 color films , thus total of 62 films were selected . color films thus compose r = 60 / 62 = 30 / 31 of the selected films . answer : e ." | a = 20 / 100
b = a + 6
c = 6 / b
|
a ) 10 , b ) 148 , c ) 12 , d ) 13 , e ) 189 | b | divide(multiply(1221, add(1221, 1234)), 123) | 1234 + 123 + 12 + x = 1221 . find the value of x . | "x = 1234 + 123 + 12 - 1221 x = 148 correct answer : b" | a = 1221 + 1234
b = 1221 * a
c = b / 123
|
a ) 20 m , b ) 25 m , c ) 22 m , d ) 9 m , e ) 12 m | c | multiply(divide(110, 45), subtract(45, 36)) | in 110 m race , a covers the distance in 36 seconds and b in 45 seconds . in this race a beats b by : | "distance covered by b in 9 sec . = 110 / 45 x 9 m = 22 m . a beats b by 20 metres . answer : option c" | a = 110 / 45
b = 45 - 36
c = a * b
|
a ) 13.0 , b ) 13.3 , c ) 13.6 , d ) 13.9 , e ) 14.2 | a | multiply(sqrt(divide(16.9, 10.0)), 10.0) | at 1 : 00 pm , there were 10.0 grams of bacteria . the bacteria increased to x grams at 4 : 00 pm , and 16.9 grams at 7 : 00 pm . if the amount of bacteria present increased by the same fraction during each of the 3 - hour periods , how many grams of bacteria were present at 4 : 00 pm ? | "let x be the factor by which the bacteria increases every three hours . at 4 : 00 pm , the amount of bacteria was 10 x and at 7 : 00 pm it was 10 x ^ 2 . 10 x ^ 2 = 16.9 x ^ 2 = 1.69 x = 1.3 at 4 : 00 pm , the amount of bacteria was 10 ( 1.3 ) = 13 grams . the answer is a ." | a = 16 / 9
b = math.sqrt(a)
c = b * 10
|
a ) 82 % , b ) 6.5 % , c ) 0.82 % , d ) 65 % , e ) 0.0065 % | d | multiply(divide(65, 100), const_100) | a certain tax rate is $ 65 per $ 100.00 . what is the rate , expressed as a percent ? | here in question it is asking $ 65 is what percent of $ 100 . suppose $ 65 is x % of 100 means 100 * ( x / 100 ) = 65 hence x = 65 so answer is d | a = 65 / 100
b = a * 100
|
a ) 457 km , b ) 444 km , c ) 547 km , d ) 645 km , e ) 576 km | e | add(multiply(divide(60, subtract(21, 27)), 27), multiply(divide(60, subtract(21, 27)), 21)) | two passenger trains start at the same hour in the day from two different stations and move towards each other at the rate of 27 kmph and 21 kmph respectively . when they meet , it is found that one train has traveled 60 km more than the other one . the distance between the two stations is ? | "1 h - - - - - 5 ? - - - - - - 60 12 h rs = 27 + 21 = 48 t = 12 d = 48 * 12 = 576 answer : e" | a = 21 - 27
b = 60 / a
c = b * 27
d = 21 - 27
e = 60 / d
f = e * 21
g = c + f
|
a ) 2000 , b ) 2200 , c ) 2300 , d ) 2450 , e ) 2500 | c | multiply(divide(add(subtract(145, const_3), add(53, const_2)), const_2), add(divide(subtract(subtract(145, const_3), add(53, const_2)), 4), const_1)) | what is the sum of the multiples of 4 between 53 and 145 inclusive ? | "the fastest way in an ap is to find the average and multiply with total integers . . between 53 and 145 , the smallest multiple of 4 is 56 and largest = 144 . . average = ( 56 + 144 ) / 2 = 100 . . total numbers = ( 144 - 56 ) / 4 + 1 = 22 + 1 = 23 . . sum = 23 * 100 = 2300 ans c" | a = 145 - 3
b = 53 + 2
c = a + b
d = c / 2
e = 145 - 3
f = 53 + 2
g = e - f
h = g / 4
i = h + 1
j = d * i
|
a ) 5 % , b ) 6 % , c ) 7 % , d ) 8 % , e ) 9 % | a | multiply(divide(divide(1680, 4), divide(divide(1680, 5), divide(4, const_100))), const_100) | the simple interest on a certain sum of money at the rate of 4 % p . a . for 5 years is rs . 1680 . at what rate of interest the same amount of interest can be received on the same sum after 4 years ? | s . i . = 1680 , r = 4 % t = 5 years principal = ( 100 * 1680 ) / ( 5 * 4 ) = 8400 so p = 8400 rate = ( 100 * 1680 ) / ( 8400 * 4 ) = 5 % answer : a | a = 1680 / 4
b = 1680 / 5
c = 4 / 100
d = b / c
e = a / d
f = e * 100
|
a ) 8 , b ) 12.5 , c ) 16 , d ) 24 , e ) 36 | b | divide(multiply(25, 2), const_4) | if ( 1 / 2 ) ^ 25 ( 1 / 81 ) ^ k = 1 / 18 ^ 25 , then k = | "i ' m going to focus on denominator only . . ( 2 ^ 25 ) . ( ( 3 ^ 4 ) ^ k = 18 ^ 25 ( 2 ^ 25 ) . ( ( 3 ^ 4 k ) = ( 2 . 3 ^ 2 ) ^ 25 ( 2 ^ 25 ) . ( ( 3 ^ 4 k ) = ( 2 ^ 25 ) . ( 3 ^ 2 ) ^ 25 hence 4 k = 50 k = 12,5 answer b i hope it ' s quite clear" | a = 25 * 2
b = a / 4
|
a ) 100 , b ) 150 , c ) 140 , d ) 120 , e ) 110 | b | subtract(subtract(add(multiply(400, divide(3, 4)), add(multiply(400, divide(1, 2)), multiply(400, divide(5, 8)))), multiply(2, multiply(400, divide(1, 8)))), 400) | a high school has 400 students 1 / 2 attend the airthmetic club , 5 / 8 attend the biology club and 3 / 4 attend the chemistry club . 1 / 4 attend all 3 clubs . if every student attends at least one club how many students attend exactly 2 clubs . | "a - club has 200 members ( 1 / 2 of 400 ) b - club has 250 members ( 5 / 8 of 400 ) c - club has 300 members ( 3 / 4 of 400 ) we can create an equation to solve this : 200 + 250 + 300 = n + x + 2 y where n is the number of students , x is the number of students in two clubs , and y is the number of students in three clubs . the question provides y for us ( 100 ) . 750 = 400 + x + 200 x = 150 b" | a = 3 / 4
b = 400 * a
c = 1 / 2
d = 400 * c
e = 5 / 8
f = 400 * e
g = d + f
h = b + g
i = 1 / 8
j = 400 * i
k = 2 * j
l = h - k
m = l - 400
|
a ) 18 , b ) 12 , c ) 14 , d ) 16 , e ) 17 | a | divide(180, add(subtract(11, 2), const_1)) | 180 metres long yard , 11 trees are palnted at equal distances , one tree being at each end of the yard . what is the distance between 2 consecutive trees | "11 trees have 10 gaps between them , required distance ( 180 / 10 ) = 18 a" | a = 11 - 2
b = a + 1
c = 180 / b
|
a ) 600,400 , b ) 500,500 , c ) 300,700 , d ) 800,200 , e ) 550,450 | a | divide(multiply(4, 6), add(4, 6)) | a can do a work in 4 days . b can do the same work in 6 days . both a & b together will finish the work and they got $ 1000 from that work . find their shares ? | "ratio of their works a : b = 4 : 6 ratio of their wages a : b = 3 : 2 a ' s share = ( 3 / 5 ) 1000 = 600 b ' s share = ( 2 / 5 ) 1000 = 400 correct option is a" | a = 4 * 6
b = 4 + 6
c = a / b
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a ) 400 , b ) 428.5 , c ) 480 , d ) 500 , e ) 600 | b | divide(300, divide(70, const_100)) | if it is assumed that 70 percent of those who receive a questionnaire by mail will respond and 300 responses are needed , what is the minimum number of questionnaires that should be mailed ? | "minimum no of mail to be sent for getting 300 responses at 70 % = 300 / 0.7 = 428.5 option b" | a = 70 / 100
b = 300 / a
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a ) 4 / 5 , b ) 4 / 7 , c ) 5 , d ) 5 / 7 , e ) none | a | divide(const_4, subtract(9, const_4)) | the difference between a positive proper fraction and its reciprocal is 9 / 20 . the fraction is ? | "let the required fraction be x . then 1 - x = 9 x 20 1 - x 2 = 9 x 20 20 - 20 x 2 = 9 x 20 x 2 + 9 x - 20 = 0 20 x 2 + 25 x - 16 x - 20 = 0 5 x ( 4 x + 5 ) - 4 ( 4 x + 5 ) = 0 ( 4 x + 5 ) ( 5 x - 4 ) = 0 x = 4 / 5 option a" | a = 9 - 4
b = 4 / a
|
a ) 1.7 , b ) 0.5 , c ) 1.1 , d ) 1.5 , e ) 1.9 | b | divide(1.5, const_3) | a man can swim in still water at 1.5 km / h , but takes twice as long to swim upstream than downstream . the speed of the stream is ? | "m = 1.5 s = x ds = 1.5 + x us = 1.5 + x 1.5 + x = ( 1.5 - x ) 2 1.5 + x = 3 - 2 x 3 x = 1.5 x = 0.5 answer : b" | a = 1 / 5
|
a ) 5 , b ) 4 , c ) not possible , d ) 3 , e ) 6 | c | add(add(const_3, const_2), 1) | by using the numbers 1 , 2 , 3 , 7 and 9 only once , how many 5 digit numbers can be made that are divisible by 25 ? | a number to be divisible by 25 must end with 00 , 25 , 50 , or 75 . so , not possible . answer : c . | a = 3 + 2
b = a + 1
|
a ) 0.003 , b ) 0.0005 , c ) 0.025 , d ) 0.005 , e ) none of these | c | multiply(divide(divide(5, const_100), 5), const_2) | double of quarter of 5 percent written as a decimal is : | "explanation : solution : ( 2 ) * ( 1 / 4 ) * 5 % = 0.025 answer : c" | a = 5 / 100
b = a / 5
c = b * 2
|
a ) $ 500 , b ) $ 504 , c ) $ 505 , d ) $ 560 , e ) $ 600 | e | add(divide(multiply(multiply(25, const_1000), subtract(const_1, divide(10, const_100))), 60), multiply(divide(divide(12, const_100), 12), multiply(multiply(25, const_1000), subtract(const_1, divide(10, const_100))))) | a car is purchased on hire - purchase . the cash price is $ 25 000 and the terms are a deposit of 10 % of the price , then the balance to be paid off over 60 equal monthly installments . interest is charged at 12 % p . a . what is the monthly installment ? | "explanation : cash price = $ 25 000 deposit = 10 % Γ£ β $ 25 000 = $ 2500 loan amount = $ 25000 Γ’ Λ β $ 2500 number of payments = 60 = $ 22500 i = p * r * t / 100 i = 13500 total amount = 22500 + 13500 = $ 36000 regular payment = total amount / number of payments = 600 answer : e" | a = 25 * 1000
b = 10 / 100
c = 1 - b
d = a * c
e = d / 60
f = 12 / 100
g = f / 12
h = 25 * 1000
i = 10 / 100
j = 1 - i
k = h * j
l = g * k
m = e + l
|
a ) $ 19,250 , b ) $ 18,500 , c ) $ 18,000 , d ) $ 19,100 , e ) $ 12,300 | d | divide(subtract(subtract(multiply(multiply(5, 4), multiply(4, 4)), multiply(multiply(5, 5), 5)), multiply(4, 15)), add(const_2, 5)) | the average salary of 15 people in the shipping department at a certain firm is $ 20,000 . the salary of 5 of the employees is $ 25,000 each and the salary of 4 of the employees is $ 15,000 each . what is the average salary of the remaining employees ? | "total salary . . . 15 * 20 k = 300 k 5 emp @ 25 k = 125 k 4 emp @ 15 k = 60 k remaing 6 emp sal = 300 k - 125 k - 60 k = 115 k average = 115 k / 6 = 19100 ans : d" | a = 5 * 4
b = 4 * 4
c = a * b
d = 5 * 5
e = d * 5
f = c - e
g = 4 * 15
h = f - g
i = 2 + 5
j = h / i
|
a ) 2 / 7 , b ) 5 / 21 , c ) 4 / 21 , d ) 3 / 7 , e ) 1 / 2 | b | multiply(divide(5, add(5, 2)), divide(2, subtract(add(5, 2), const_1))) | a jar contains 5 black and 2 white balls . if you pick two balls at the same time , what ' s the probability that one ball is black and one is white ? | p ( 1 st black , 2 nd white ) = 5 / 7 * 2 / 6 = 10 / 42 ; p ( 1 st white , 2 nd black ) = 2 / 7 * 5 / 6 = 10 / 42 . p = 10 / 42 + 10 / 42 = 20 / 84 = 10 / 42 = 5 / 21 answer : b . | a = 5 + 2
b = 5 / a
c = 5 + 2
d = c - 1
e = 2 / d
f = b * e
|
a ) 10 / 16 , b ) 6 / 16 , c ) 4 / 16 , d ) 7 / 9 , e ) 4 / 10 | d | divide(divide(subtract(16, 2), add(const_1, const_1)), add(divide(subtract(16, 2), add(const_1, const_1)), 2)) | there are 2 more women than there are men on a local co - ed softball team . if there are a total of 16 players on the team , what is the ratio of men to women ? | "w = m + 2 w + m = 16 m + 2 + m = 16 2 m = 14 m = 7 w = 9 ratio : 7 : 9 ans : d" | a = 16 - 2
b = 1 + 1
c = a / b
d = 16 - 2
e = 1 + 1
f = d / e
g = f + 2
h = c / g
|
a ) $ 16.32 , b ) $ 18.00 , c ) $ 21.60 , d ) $ 34 , e ) $ 28.80 | d | multiply(divide(subtract(const_100, 15), const_100), multiply(0.40, 100)) | the regular price per can of a certain brand of soda is $ 0.40 . if the regular price per can is discounted 15 percent when the soda is purchased in 24 - can cases , what is the price of 100 cans of this brand of soda purchased in 24 - can cases ? | "the discounted price of one can of soda is ( 0.85 ) ( $ 0.40 ) , or $ 0.34 . therefore , the price of 72 cans of soda at the discounted price would be ( 100 ) ( $ 0.34 ) = 34 answer : d ." | a = 100 - 15
b = a / 100
c = 0 * 40
d = b * c
|
a ) 50 , b ) 8 , c ) 60 , d ) 42 , e ) 32 | c | add(add(25, 32), const_2) | a lady has fine gloves and hats in her closet - 18 blue , 32 red , and 25 yellow . the lights are out and it is totally dark . in spite of the darkness , she can make out the difference between a hat and a glove . she takes out an item out of the closet only if she is sure that if it is a glove . how many gloves must she take out to make sure she has a pair of each color ? | in the first case the lady takes up 32 red colour then he takes up 24 y ( cause he should take up pair ) then he takes up 1 y + 1 blue then 2 blue ( cause to make sure he has a pair of each color in hand ) 32 + 24 + 2 + 2 = 60 answer : c | a = 25 + 32
b = a + 2
|
a ) 110 : 100 , b ) 105 : 100 , c ) 95 : 100 , d ) 85 : 100 , e ) 120 : 100 | b | divide(51.27, 48.73) | in a 2000 census , 51.27 % of the population are male , and 48.73 % are female . what is the ratio of men to every 100 women ? | the ratio of a to b , where a = the percent of males , and b = the percent of females . a = 51.27 b = 48.73 a : b = 51.27 : 48.73 so , 100 ( 5,127 / 4,873 ) : 100 giving a male to female ratio of 105.20 men for every 100 women . answer is b | a = 51 / 27
|
a ) 32.8 , b ) 32.4 , c ) 72 , d ) 32.2 , e ) 32.9 | c | add(divide(circumface(14), const_2), multiply(14, const_2)) | the radius of a semi circle is 14 cm then its perimeter is ? | "diameter = 28 cm 1 / 2 * 22 / 7 * 28 + 28 = 72 answer : c" | a = circumface / (
b = a + 2
|
a ) 435 hectares . , b ) 425 hectares . , c ) 445 hectares . , d ) 415 hectares . , e ) 405 hectares . | e | divide(subtract(900, multiply(divide(900, const_2), divide(const_1, add(const_1, const_4)))), const_2) | a large field of 900 hectares is divided into two parts . the difference of the areas of the two parts is one - fifth of the average of the two areas . what is the area of the smaller part in hectares ? | "let the areas of the parts be x hectares and ( 900 - x ) hectares . difference of the areas of the two parts = x - ( 900 - x ) = 2 x - 900 one - fifth of the average of the two areas = 1 / 5 [ x + ( 900 β x ) ] / 2 = 1 / 5 Γ ( 900 / 2 ) = 450 / 5 = 90 given that difference of the areas of the two parts = one - fifth of the average of the two areas = > 2 x - 900 = 90 = > 2 x = 990 β x = 990 / 2 = 495 hence , area of smaller part = ( 900 - x ) = ( 900 β 495 ) = 405 hectares . answer is e ." | a = 900 / 2
b = 1 + 4
c = 1 / b
d = a * c
e = 900 - d
f = e / 2
|
a ) 615 m , b ) 420 m , c ) 168 m , d ) 197 m , e ) 691 m | b | multiply(20, multiply(54, const_0_2778)) | a train passes a station platform in 60 sec and a man standing on the platform in 20 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 * 20 = 300 m . let the length of the platform be x m . then , ( x + 300 ) / 36 = 15 = > x = 420 m answer : b" | a = 54 * const_0_2778
b = 20 * a
|
a ) 10 days , b ) 11 days , c ) 9 days , d ) 8 days , e ) 12 days | a | divide(multiply(multiply(15, 10), 8), multiply(8, 15)) | 15 men work 8 hours per day to complete the work in 10 days . to complete the same work in 8 days , working 15 hours a day , the number of men required ? | "that is , 1 work done = 15 Γ 8 Γ 10 then , 12 8 Γ 10 = ? Γ 15 Γ 8 ? ( i . e . no . of men required ) = 15 Γ 8 Γ 10 / 15 Γ 8 = 10 days a )" | a = 15 * 10
b = a * 8
c = 8 * 15
d = b / c
|
a ) 1140 toys , b ) 2375 toys , c ) 3375 toys , d ) 4375 toys , e ) 5375 toys | a | divide(4560, 4) | a factory produces 4560 toys per week . if the workers at this factory work 4 days a week and if these workers make the same number of toys everyday , how many toys are produced each day ? | "to find the number of toys produced every day , we divide the total number of toys produced in one week ( of 4 days ) by 4 . 4560 / 4 = 1140 toys correct answer a" | a = 4560 / 4
|
a ) 33 , b ) 38 , c ) 70 , d ) 123 , e ) 12 | b | add(add(power(add(add(divide(subtract(subtract(140, const_10), const_2), const_4), const_2), const_2), const_2), power(add(add(add(divide(subtract(subtract(140, const_10), const_2), const_4), const_2), const_2), const_2), const_2)), add(power(divide(subtract(subtract(140, const_10), const_2), const_4), const_2), power(add(divide(subtract(subtract(140, const_10), const_2), const_4), const_2), const_2))) | the sum of four consecutive even numbers is 140 . what would be the largest number ? | "let the four consecutive even numbers be 2 ( x - 2 ) , 2 ( x - 1 ) , 2 x , 2 ( x + 1 ) their sum = 8 x - 4 = 140 = > x = 18 smallest number is : 2 ( x + 1 ) = 38 . answer : b" | a = 140 - 10
b = a - 2
c = b / 4
d = c + 2
e = d + 2
f = e ** 2
g = 140 - 10
h = g - 2
i = h / 4
j = i + 2
k = j + 2
l = k + 2
m = l ** 2
n = f + m
o = 140 - 10
p = o - 2
q = p / 4
r = q ** 2
s = 140 - 10
t = s - 2
u = t / 4
v = u + 2
w = v ** 2
x = r + w
y = n + x
|
a ) 25 , b ) 31 , c ) 27 , d ) 29 , e ) 34 | e | add(subtract(82, multiply(17, 3)), 3) | a batsman makes a score of 82 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 + 82 = total runs scored in 17 innings = > 16 ( x - 3 ) + 82 = 17 x = > 16 x - 48 + 82 = 17 x = > x = 34 answer is e ." | a = 17 * 3
b = 82 - a
c = b + 3
|
a ) 15 , b ) 20 , c ) 30 , d ) 35 , e ) 45 | e | divide(subtract(const_100, add(35, 20)), const_3) | a polling company surveyed a certain country , and it found that 35 % of that country β s registered voters had an unfavorable impression of both of that state β s major political parties and that 20 % had a favorable impression only of party a . if one registered voter has a favorable impression of both parties for every two registered voters who have a favorable impression only of party b , then what percentage of the country β s registered voters have a favorable impression of both parties ( assuming that respondents to the poll were given a choice between favorable and unfavorable impressions only ) ? | "assume the total pool of registered voters = 100 , so 35 of the country β s registered voters had an unfavorable impression of both of that state β s major political parties and 20 had a favorable impression only of party a let x = # of voters with a favorable impression of both parties let 2 x = # of voters with a favorable impression only of party b so unfavorable a and favorable b # of voters = 2 x - x = x 35 + x = number of unfavorable a 20 + 35 + x = 100 55 + x = 100 x = 45 answer : e" | a = 35 + 20
b = 100 - a
c = b / 3
|
a ) 42.8 % , b ) 20 % , c ) 25 % , d ) 30 % , e ) 35 % | a | multiply(divide(subtract(const_1, divide(subtract(const_100, 30), const_100)), divide(subtract(const_100, 30), const_100)), const_100) | the length of a rectangle is reduced by 30 % . by what % would the width have to be increased to maintain the original area ? | "sol . required change = ( 30 * 100 ) / ( 100 - 30 ) = 42.8 % a" | a = 100 - 30
b = a / 100
c = 1 - b
d = 100 - 30
e = d / 100
f = c / e
g = f * 100
|
a ) 5 , b ) 4 , c ) 1 , d ) 2 , e ) 10 | c | subtract(divide(100, const_2), multiply(7, 7)) | what is the remainder if 7 ^ 4 is divided by 100 ? | "7 * 7 * 7 * 7 / 100 = 2401 / 100 = 24 reminder 1 answer : c" | a = 100 / 2
b = 7 * 7
c = a - b
|
a ) 8 , b ) 12 , c ) 16 , d ) 18 , e ) 24 | a | divide(subtract(80, const_10), const_10) | how many positive factors do 120 and 80 have in common ? | "the number of common factors will be same as number of factors of the highest common factor ( hcf ) hcf of 120 and 80 is 40 number of factors of 40 = 8 answer : a" | a = 80 - 10
b = a / 10
|
a ) 234 , b ) 267 , c ) 324 , d ) 356 , e ) 577 | c | subtract(power(add(422, 404), const_2), multiply(multiply(4, 422), 404)) | find value of x : ( 422 + 404 ) ^ 2 β ( 4 Γ 422 Γ 404 ) = x . | given equation is in the form ( a + b ) 2 β 4 ab where a = 422 and b = 404 hence answer = ( a + b ) 2 β 4 ab = ( a β b ) 2 = ( 422 β 404 ) 2 = 182 = 324 c | a = 422 + 404
b = a ** 2
c = 4 * 422
d = c * 404
e = b - d
|
a ) s . 6000 , b ) s . 9000 , c ) s . 10800 , d ) s . 9357 , e ) s . 9980 | d | divide(multiply(6400, const_100), subtract(subtract(subtract(const_100, 20), divide(multiply(subtract(const_100, 20), 10), const_100)), divide(multiply(subtract(subtract(const_100, 20), divide(multiply(subtract(const_100, 20), 10), const_100)), 5), const_100))) | after successive discounts of 20 % , 10 % and 5 % a certain good is sold for rs . 6400 . find the actual price of the good . | "let actual price was 100 . after three successive discount this will become , 100 = = 20 % discount = > 80 = = 10 % discount = > 72 = = 5 % discount = 68.4 now compare , 68.4 = 6400 1 = 6400 / 68.4 100 = ( 6400 * 100 ) / 68.4 = rs . 9357 . answer : option d" | a = 6400 * 100
b = 100 - 20
c = 100 - 20
d = c * 10
e = d / 100
f = b - e
g = 100 - 20
h = 100 - 20
i = h * 10
j = i / 100
k = g - j
l = k * 5
m = l / 100
n = f - m
o = a / n
|
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