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 ) 10 , b ) 7 , c ) 11 , d ) 5 , e ) 3 | c | subtract(add(add(8, 15), 12), add(4, 20)) | the sum of 4 th and 20 th term of a . p . is equal to the sum of 8 th , 15 th and 12 th term . find the term which is 0 | t 4 + t 20 = t 8 + t 15 + t 12 = > a + 3 d + a + 19 d = a + 7 d + a + 14 d + a + 11 d = > a + 10 d = 0 = > t 11 = 0 i . e . 11 th term is zero . answer : c | a = 8 + 15
b = a + 12
c = 4 + 20
d = b - c
|
a ) 20 , b ) 30 , c ) 35 , d ) 40 , e ) 55 | e | divide(subtract(75, 64), subtract(const_1, divide(80, const_100))) | a particular library has 75 books in a special collection , all of which were in the library at the beginning of the month . these book are occasionally loaned out through an inter - library program . if , by the end of the month , 80 percent of books that were loaned out are returned and there are 64 books in the special collection at that time , how many books of the special collection were loaned out during that month ? | "there are 11 books less ( 75 - 64 ) which represents 20 % of the loaned books ( 100 - 80 ) so total loaned out books = 55 answer e" | a = 75 - 64
b = 80 / 100
c = 1 - b
d = a / c
|
a ) 2.4 , b ) 2.7 , c ) 3.0 , d ) 3.3 , e ) 3.6 | e | subtract(6, multiply(divide(multiply(6, 20), const_100), const_2)) | a 6 - liter solution is 20 % alcohol . how many liters of pure alcohol must be added to produce a solution that is 50 % alcohol ? | "let x be the amount of pure alcohol required . 0.2 ( 6 ) + x = 0.5 ( x + 6 ) 0.5 x = 3 - 1.2 x = 3.6 liters the answer is e ." | a = 6 * 20
b = a / 100
c = b * 2
d = 6 - c
|
a ) 21 , b ) 22 , c ) 25 , d ) 26 , e ) 28 | c | add(divide(subtract(52, 1), 2), const_1) | how many multiples of 2 are there between 1 and 52 , exclusive ? | "25 multiples of 2 between 1 and 52 exclusive . from 2 * 1 upto 2 * 25 , ( 1,2 , 3,4 , . . . , 25 ) . hence , 25 multiples ! correct option is c" | a = 52 - 1
b = a / 2
c = b + 1
|
a ) 25650 , b ) 25750 , c ) 26550 , d ) 26750 , e ) 25725 | a | multiply(900, multiply(6, 4.75)) | the length of a room is 6 m and width is 4.75 m . what is the cost of paying the floor by slabs at the rate of rs . 900 per sq . metre . | "area = 6 × 4.75 sq . metre . cost for 1 sq . metre . = rs . 900 hence total cost = 6 × 4.75 × 900 = 6 × 4275 = rs . 25650 answer is a ." | a = 6 * 4
b = 900 * a
|
a ) a . 4 , b ) b . 5 / 2 , c ) c . 2 , d ) d . 3 / 2 , e ) e . 5 / 4 | d | divide(add(multiply(3, const_4), 3), const_10) | nails and screws are manufactured at a uniform weight per nail and a uniform weight per screw . if the total weight of one screw and one nail is half that of 6 screws and one nail , then the total weight of 3 screws , and 3 nails is how many times that of 2 screws and 4 nails ? | "let the weight of nail be n and that of screw be s . . so s + w = 1 / 2 * ( 6 s + 1 n ) . . . or 1 n = 4 s . . lets see the weight of 3 s and 3 n = 3 s + 3 * 4 s = 15 s . . and weight of 2 s and 4 n = 2 s + 4 * 2 s = 10 s . . ratio = 15 s / 10 s = 15 / 10 = 3 / 2 d" | a = 3 * 4
b = a + 3
c = b / 10
|
a ) 12 , b ) 16 , c ) 17 , d ) 19 , e ) 21 | b | add(14, const_2) | a box contains 10 tablets of medicine a and 14 tablets of medicine b . what is the least number of tablets that should be taken from the box to ensure that at least two tablets of each kind are among the extracted . | the worst case scenario will be if we remove all 14 tablets of medicine b first . the next 2 tablets we remove have to be of medicine a , so to guarantee that at least two tablets of each kind will be taken we should remove minimum of 14 + 2 = 16 tablets . answer : b . | a = 14 + 2
|
a ) 10 , b ) 99 , c ) 77 , d ) 55 , e ) 21 | a | add(inverse(subtract(divide(const_1, 6), divide(const_1, 15))), divide(const_2, add(const_2, const_3))) | a and b together can do a work in 6 days . if a alone can do it in 15 days . in how many days can b alone do it ? | "1 / 6 – 1 / 15 = 1 / 10 = > 10 answer : a" | a = 1 / 6
b = 1 / 15
c = a - b
d = 1/(c)
e = 2 + 3
f = 2 / e
g = d + f
|
a ) 25 , b ) 24 , c ) 27 , d ) 28 , e ) 29 | b | add(divide(add(27, 7), const_2), 7) | the sum of present age of abe and the age before 7 years is 27 . find the present age of abe . what will be his age after 7 years ? | "present age = x before 7 yrs , y = x - 7 after 7 yrs , z = x + 7 by the qn , x + ( x - 7 ) = 27 2 x - 7 = 27 2 x = 27 + 7 x = 34 / 2 x = 17 z = x + 7 = 17 + 7 = 24 answer : b" | a = 27 + 7
b = a / 2
c = b + 7
|
a ) rs . 3300 , b ) rs . 3100 , c ) rs . 3500 , d ) rs . 4200 , e ) rs . 3400 | a | multiply(divide(6600, add(add(6600, 4400), 11000)), 12100) | a , b and c invested rs . 6600 , rs . 4400 and rs . 11000 respectively , in a partnership business . find the share of a in profit of rs . 12100 after a year ? | "6600 : 4400 : 11000 3 : 2 : 5 3 / 10 * 11000 = rs . 3300 answera" | a = 6600 + 4400
b = a + 11000
c = 6600 / b
d = c * 12100
|
a ) 23 miles , b ) 25 miles , c ) 22 miles , d ) 24 miles , e ) 26 miles | d | subtract(divide(add(20, 30), const_2), const_1) | a man drive his car to the office at 20 miles / hr . after reaching office , he realize that its a new year holiday so he went back home at a speed of 30 miles / hr . discounting the time spent in the stoppage what was his average speed of his journey ? | d 24 miles / hr ( not 25 miles / hr which might be guessed by many ) d : distance traveled t 1 : time of going to office t 2 : time of returning back y : average speed d = 20 * t 1 t 1 = d / 20 d = 30 * t 2 t 2 = d / 30 2 d = y ( t 2 + t 1 ) 2 d = y ( d / 20 + d / 30 ) 2 d = y ( 3 d / 60 + 2 d / 60 ) 2 d = y ( 5 d / 60 ) y = 120 d / 5 d = > y = 24 | a = 20 + 30
b = a / 2
c = b - 1
|
a ) 11 , b ) 15 , c ) 16 , d ) 28 , e ) 19 | d | multiply(4, 7) | walking 7 / 6 of his usual rate , a boy reaches his school 4 min early . find his usual time to reach the school ? | "speed ratio = 1 : 7 / 6 = 6 : 7 time ratio = 7 : 6 1 - - - - - - - - 7 4 - - - - - - - - - ? è 28 m answer : d" | a = 4 * 7
|
a ) 1000 , b ) 800 , c ) 2889 , d ) 2777 , e ) 2991 | b | divide(896, add(divide(multiply(divide(add(multiply(2, 5), 2), 5), 5), const_100), const_1)) | find the principle on a certain sum of money at 5 % per annum for 2 2 / 5 years if the amount being rs . 896 ? | "896 = p [ 1 + ( 5 * 12 / 5 ) / 100 ] p = 800 answer : b" | a = 2 * 5
b = a + 2
c = b / 5
d = c * 5
e = d / 100
f = e + 1
g = 896 / f
|
a ) 99 , b ) 98 , c ) 97 , d ) 96 , e ) 95 | d | add(multiply(const_2, const_3), subtract(const_100, const_10)) | a number is said to be prime saturated if the product of all the different positive prime factors of g is less than the square root of g . what is the greatest two digit prime saturated integer ? | "g = 96 = 3 * 32 = 3 * 2 ^ 5 answer is d ." | a = 2 * 3
b = 100 - 10
c = a + b
|
a ) 8 % , b ) 1.6 % , c ) 2 % , d ) 5 % , e ) 6 % | b | add(floor(multiply(divide(10, add(10, 600)), 600)), const_1) | a person saved $ 10 in buying an item on sale . if he spent $ 600 for the item , approximately how much percent he saved in the transaction ? | "actual price = 600 + 10 = $ 610 saving = 10 / 610 * 100 = 100 / 61 = 1.6 % approximately answer is b" | a = 10 + 600
b = 10 / a
c = b * 600
d = math.floor(c)
e = d + 1
|
a ) 27.65 % , b ) 33 1 / 2 % , c ) 33 1 / 3 % , d ) 33 5 / 3 % , e ) 35 1 / 3 % | a | subtract(const_100, divide(multiply(900, const_100), 705)) | an article is bought for rs . 705 and sold for rs . 900 , find the gain percent ? | "705 - - - - 195 100 - - - - ? = > 27.65 % answer : a" | a = 900 * 100
b = a / 705
c = 100 - b
|
a ) 25 % , b ) 28 % , c ) 21 % , d ) 30 % , e ) 32 % | c | multiply(subtract(power(add(const_1, divide(10, const_100)), const_2), const_1), const_100) | what will be the percentage increase in the area of the cube ' s surface if each of the cube ' s edges grows by 10 % ? | "the question is very easy . my logic is the following : a surface = 6 * a ^ 2 after 10 % increase a surface = 6 * ( ( 1.1 a ) ^ 2 ) = 6 * 1.21 * a ^ 2 the increase in the surface area = ( 6 * 1.21 * a ^ 2 - 6 * a ^ 2 ) / 6 * a ^ 2 = ( 6 * a ^ 2 ( 1.21 - 1 ) ) / ( 6 * a ^ 2 ) = 1.21 - 1 = 0.21 = 21 % answer : c" | a = 10 / 100
b = 1 + a
c = b ** 2
d = c - 1
e = d * 100
|
a ) 34 , b ) 677 , c ) 197 , d ) 177 , e ) 191 | a | subtract(subtract(subtract(500, 1), const_4), const_1) | how many positive integers less than 500 can be formed using the numbers 1 , 2,3 , and 5 for digits , each digit being used only once . | "single digit numbers = 4 double digit numbers = 4 x 3 = 12 three digit numbers = 3 x 3 x 2 x 1 = 18 total = 34 answer : a" | a = 500 - 1
b = a - 4
c = b - 1
|
a ) 34 , b ) 38 , c ) 36 , d ) 40 , e ) 42 | e | multiply(add(subtract(11, const_1), 11), divide(22, add(subtract(6, const_1), 6))) | a certain clock marks every hour by striking a number of times equal to the hour , and the time require for a stroke is exactly equal to the time interval between strokes . at 6 : 00 the time lapse between the beginning of the first stoke and the end of the last stroke is 22 seconds . at 11 : 00 , how many seconds elapse between the beginning of the first stroke and the end of the last stroke ? | "at 6 ' o clock , there would be 6 strikes . first strike , then a short interval , the second strike , then a short interval and so on till the 6 th strike . so there would be in all 5 intervals between 6 strikes . similarly , between 11 strikes , there would be 10 intervals . according to the question , the time spent in the strike and the interval is same . at 6 ' o clock , the 6 strikes and the 5 intervals together take 22 sec so each strike and each interval takes 2 secs . at 11 ' o clock , the 11 strikes and 10 intervals will take 2 * ( 11 + 10 ) = 42 secs e" | a = 11 - 1
b = a + 11
c = 6 - 1
d = c + 6
e = 22 / d
f = b * e
|
['a ) 11', 'b ) 12', 'c ) 13', 'd ) 14', 'e ) none'] | a | divide(add(divide(34, const_2), 5), const_2) | the length of a rectangle is 5 more than the width . what are the dimensions of the rectangle if the perimeter is 34 ? | pretend width = 1 , then length = 6 ( 1 + 5 ) 2 × 1 + 2 × 7 = 2 + 14 = 16 . notice that 16 is far from a perimeter of 34 try much bigger number . how about if we . . . pretend width = 4 , then length = 9 ( 4 + 5 ) 2 × 4 + 2 × 9 = 6 + 18 = 24 . we are getting closer to a perimeter of 34 pretend width = 5 , then length = 10 ( 5 + 5 ) 2 × 5 + 2 × 10 = 10 + 30 = 30 . pretend width = 7 , then length = 12 ( 7 + 5 ) 2 × 7 + 2 × 12 = 14 + 24 = 38 . this is higher than a perimeter of 34 . so width should be higher than 5 and smaller than 7 . may be a width of 6 will work . pretend width = 6 , then length = 11 ( 6 + 5 ) 2 × 6 + 2 × 11 = 12 + 22 = 34 . use of algebra : let width = x let length = x + 5 p = 2 × l + 2 × w 34 = 2 × ( x + 5 ) + 2 × x 34 = 2 x + 10 + 2 x 34 = 4 x + 10 34 - 10 = 4 x + 10 - 10 24 = 4 x 24 / 4 = 4 x / 4 6 = x therefore , width = 6 and length = x + 5 = 6 + 5 = 11 answer a | a = 34 / 2
b = a + 5
c = b / 2
|
a ) 11 , b ) 12 , c ) 13 , d ) 14 , e ) 15 | c | divide(subtract(54, 2), 4) | if there are only 2 wheelers and 4 wheelers parked in a school located at the heart of the city , find the number of 4 wheelers parked there if the total number of wheels is 54 ? | four wheeler = 13 * 4 = 52 ( max ) 2 wheel = 1 so no of 4 wheeler = 13 answer : c | a = 54 - 2
b = a / 4
|
a ) 250 , b ) 200 , c ) 300 , d ) 240 , e ) 180 | e | subtract(multiply(30, divide(multiply(36, const_1000), const_3600)), multiply(divide(multiply(36, const_1000), const_3600), 12)) | a train passes a train standing in platform in 30 sec and a man standing on the platform in 12 sec . if the speed of the train is 36 km / hr . what is the length of the platform ? | speed = 36 * 5 / 18 = 10 m / sec . length of the train = 10 * 12 = 120 m . let the length of the platform be x m . then , ( x + 120 ) / 30 = 10 = > x = 180 m . answer : e | a = 36 * 1000
b = a / 3600
c = 30 * b
d = 36 * 1000
e = d / 3600
f = e * 12
g = c - f
|
a ) 5 , b ) 7 , c ) 9 , d ) 1 , e ) 0 | c | floor(divide(divide(subtract(multiply(factorial(4), factorial(4)), multiply(factorial(4), factorial(2))), 4), multiply(const_100, const_10))) | what is the tenth digit of ( 4 ! * 3 ! + 4 ! * 2 ! ) / 3 ? | "( 4 ! * 3 ! + 4 ! * 2 ! ) / 2 = 4 ! ( 3 ! + 2 ! ) / 2 = 24 ( 6 + 2 ) / 2 = 96 units digit of the above product will be equal to 9 answer c" | a = math.factorial(4)
b = math.factorial(4)
c = a * b
d = math.factorial(4)
e = math.factorial(2)
f = d * e
g = c - f
h = g / 4
i = 100 * 10
j = h / i
k = math.floor(j)
|
a ) 70,000 , b ) 60,000 , c ) 128,000 , d ) 90,000 , e ) 10,000 | c | multiply(1, const_60) | bullock likes to keep a spare tyre in his car every time . on a certain day , he travels 1 , 60,000 km and just to make the most of all the tyres , he changes the tyres between his journey such that each tyre runs the same distance . what is the distance traveled by each tyre ? | the distance traveled by each tyre : 4 / 5 * 1 , 60 , 000 km = 128,000 km . c | a = 1 * const_60
|
a ) 52 , b ) 18 , c ) 42 , d ) 102 , e ) 36 | c | subtract(60, multiply(multiply(9, 3), 6)) | evaluate : 60 - 9 ÷ 3 × 6 = | "according to order of operations , 9 ÷ 3 × 6 ( division and multiplication ) is done first from left to right 9 ÷ 3 × 6 = 3 × 6 = 18 hence 60 - 9 ÷ 3 × 6 = 60 - 18 = 42 correct answer c ) 42" | a = 9 * 3
b = a * 6
c = 60 - b
|
a ) 23.75 , b ) 22 , c ) 20 , d ) 19.2 , e ) none of these | d | add(divide(multiply(divide(add(multiply(80, 15), multiply(20, 20)), add(80, 20)), 20), const_100), divide(add(multiply(80, 15), multiply(20, 20)), add(80, 20))) | a trader mixes 80 kg of tea at 15 per kg with 20 kg of tea at cost price of 20 per kg . in order to earn a profit of 20 % , what should be the sale price of the mixed tea ? | "c . p . of mixture = 80 × 15 + 20 × 20 / 80 + 20 = 16 ∴ s . p . = ( 100 + 20 ) / 100 × 16 = 19.2 answer d" | a = 80 * 15
b = 20 * 20
c = a + b
d = 80 + 20
e = c / d
f = e * 20
g = f / 100
h = 80 * 15
i = 20 * 20
j = h + i
k = 80 + 20
l = j / k
m = g + l
|
a ) 60 , b ) 80 , c ) 40 , d ) 120 , e ) 160 | b | divide(30, subtract(add(const_1, divide(12.5, const_100)), subtract(const_1, divide(25, const_100)))) | the difference between the value of a number increased by 12.5 % and the value of the original number decreased by 25 % is 30 . what is the original number w ? | "( 1 + 1 / 8 ) x - ( 1 - 1 / 4 ) x = 30 ( 9 / 8 ) x - ( 3 / 4 ) x = 30 x = 80 = w answer : b" | a = 12 / 5
b = 1 + a
c = 25 / 100
d = 1 - c
e = b - d
f = 30 / e
|
a ) 3 , b ) 6 , c ) 7.2 , d ) 7.8 , e ) 9 | a | multiply(divide(20, const_100), 20) | uncle bruce is baking chocolate chip cookies . he has 36 ounces of dough ( with no chocolate ) and 12 ounces of chocolate . how many ounces of chocolate are left over if he uses all the dough but only wants the cookies to consist of 20 % chocolate ? | "answer is a . x / x + 36 = 1 / 5 x = 9 12 - 9 = 3" | a = 20 / 100
b = a * 20
|
a ) 6 / 13 , b ) 5 / 9 , c ) 1 / 24 , d ) 4 / 9 , e ) 2 / 5 | a | divide(subtract(divide(14, const_2), const_1), subtract(14, const_1)) | an empty wooden vessel weighs 14 % of its total weight when filled with paint . if the weight of a partially filled vessel is one half that of a completely filled vessel , what fraction of the vessel is filled . | "an empty wooden vessel weighs 14 % of its total weight when filled with paint : vessel = 0.14 ( vessel + paint ) ; 14 v = v + p ( so the weight of completely filled vessel is 14 v ) p = 13 v ( so the weight of the paint when the vessels is completely filled is 13 v ) . the weight of a partially filled vessel is one half that of a completely filled vessel : v + p ' = 1 / 2 * 14 v ; p ' = 6 v ( so the weight of the paint when the vessels is partially filled is 6 v ) . what fraction of the vessel is filled ? so , we need to find the ratio of the weight of the paint when the vessel iscompletely filledto the weight of the paint when the vessel ispartially filled : p ' / p = 6 v / 13 v = 6 / 13 . answer : a ." | a = 14 / 2
b = a - 1
c = 14 - 1
d = b / c
|
a ) 53 , b ) 47 , c ) 61 , d ) 59 , e ) 45 | c | multiply(43, 23) | 23 , 29 , 31 , 37 , 41 , 43 , 47 , 53 , 59 , ( . . . ) | "explanation : all are prime numbers in their order , starting from 23 hence , next number is 61 answer : c" | a = 43 * 23
|
a ) 9.6 , b ) 11 , c ) 11.5 , d ) 12 , e ) 12.5 | a | subtract(12, multiply(2, 1.2)) | the arithmetic mean and standard deviation of a certain normal distribution are 12 and 1.2 , respectively . what value is exactly 2 standard deviations less than the mean ? | "mean = 12 two standard deviations is 1.2 + 1.2 = 2.4 there could be two calues for this . mean + two standard deviations = 14.4 mean - two standard deviations = 9.6 answer choice has 9.6 and so a is the answer ." | a = 2 * 1
b = 12 - a
|
a ) 1 , b ) 9 / 4 , c ) 16 / 25 , d ) 7 / 1 , e ) 7 / 2 | a | divide(power(2, add(2, 2)), power(4, 2)) | for what value of â € œ k â € will the equation ( 2 kx 2 + 4 kx + 2 ) = 0 have equal roots ? | "for a 2 nd degree equation ax 2 + bx _ c = 0 has equal roots the condition is b 2 - 4 ac = 0 in the given equation ( 4 k ) ^ 2 - 4 * 2 k * 2 = 0 by solving this equation we get k = 0 , k = 1 answer : a" | a = 2 + 2
b = 2 ** a
c = 4 ** 2
d = b / c
|
a ) 4 , b ) 12 , c ) 14 , d ) 21 , e ) 28 | c | divide(multiply(7, const_4), 2) | a is the average ( arithmetic mean ) of the first 7 positive multiples of 7 and b is the median of the first 3 positive multiples of positive integer n . if the value of a ^ 2 – b ^ 2 is zero , what is the value of n ? | "if a ^ 2 - b ^ 2 = 0 , then let ' s assume that a = b . a must equal the 4 th positive multiple of 4 , thus a = 28 , which also equals b . b is the second positive multiple of n , thus n = 28 / 2 = 14 . the answer is c ." | a = 7 * 4
b = a / 2
|
a ) 4 , b ) 6 , c ) 12 , d ) 18 , e ) none of these | c | gcd(36, 84) | find the highest common factor of 36 and 84 ? | 36 = 2 ^ 2 * 3 ^ 2 ; 84 = 2 ^ 2 * 3 * 7 h . c . f = 2 ^ 2 * 3 = 12 . correct option : c | a = math.gcd(36, 84)
|
['a ) 87', 'b ) 16', 'c ) 17', 'd ) 60', 'e ) 18'] | d | divide(multiply(30, 6), const_3) | the area of a base of a cone is 30 cm 2 . if the height of the cone is 6 cm , find its volume ? | π r 2 = 30 h = 6 1 / 3 * 30 * 6 = 60 answer : d | a = 30 * 6
b = a / 3
|
a ) 12 : 7 , b ) 15 : 11 , c ) 25 : 16 , d ) 30 : 23 , e ) 32 : 25 | e | divide(rectangle_area(3, 2), rectangle_area(divide(divide(rectangle_perimeter(3, 2), 2), add(3, 1)), multiply(divide(divide(rectangle_perimeter(3, 2), 2), add(3, 1)), 3))) | an order was placed for a carpet whose length and width were in the ratio of 3 : 2 . subsequently , the dimensions of the carpet were altered such that its length and width were in the ratio 3 : 1 but were was no change in its perimeter . what is the ratio of the areas of the carpets ? | "let the length and width of one carpet be 3 x and 2 x . let the length and width of the other carpet be 3 y and y . 2 ( 3 x + 2 x ) = 2 ( 3 y + y ) 5 x = 4 y ( 5 / 4 ) * x = y the ratio of the areas of the carpet in both cases : = 3 x * 2 x : 3 y * y = 6 x ^ 2 : 3 y ^ 2 = 6 x ^ 2 : 3 * ( 25 / 16 ) * x ^ 2 = 96 : 75 = 32 : 25 the answer is e ." | a = rectangle_area / (
|
a ) 5 , b ) 9 , c ) 16 , d ) 20 , e ) 30 | c | multiply(subtract(9, 8), 8) | what is the greatest positive integer x such that 3 ^ x is a factor of 9 ^ 8 ? | "what is the greatest positive integer x such that 3 ^ x is a factor of 9 ^ 8 ? 9 ^ 8 = ( 3 ^ 2 ) ^ 8 = 3 ^ 16 c . 16" | a = 9 - 8
b = a * 8
|
a ) 15 , b ) 20 , c ) 25 , d ) 30 , e ) 35 | c | divide(add(add(11, const_4), subtract(36, const_4)), const_2) | find the average of all the numbers between 11 and 36 which are divisible by 5 | "explanation : average = ( 15 + 20 + 25 + 30 + 35 ) / 5 = 125 / 5 = 25 answer c" | a = 11 + 4
b = 36 - 4
c = a + b
d = c / 2
|
a ) rs . 11900 , b ) rs . 8400 , c ) rs . 14700 , d ) rs . 13600 , e ) rs . 14600 | c | subtract(floor(divide(multiply(divide(add(divide(subtract(subtract(multiply(const_10, 5000), 5000), add(4000, 5000)), const_3), add(4000, 5000)), multiply(const_10, 5000)), multiply(add(const_3, const_4), 5000)), const_1000)), const_1) | a , b , c subscribe rs . 50000 for a business . a subscribes rs . 4000 more than b and b rs . 5000 more than c . out of a total profit of rs . 35000 a receives : | "let c = x . then b = x + 5000 and a + x + 5000 + 4000 = x + 9000 therefore , x + x + 5000 + x + 9000 = 50000 û x = 12000 . therefore , a : b : c = 21000 : 17000 : 12000 = 21 : 17 : 12 therefore , a ’ s share = rs ( 35000 * 21 / 50 ) = rs . 14700 answer : c" | a = 10 * 5000
b = a - 5000
c = 4000 + 5000
d = b - c
e = d / 3
f = 4000 + 5000
g = e + f
h = 10 * 5000
i = g / h
j = 3 + 4
k = j * 5000
l = i * k
m = l / 1000
n = math.floor(m)
o = n - 1
|
a ) 6 miles , b ) 8,4 miles , c ) 9 miles , d ) 9,6 miles , e ) 18 miles | e | multiply(divide(28, add(3, 2)), 2) | one hour before john started walking from p to q , a distance of 28 miles , ann had started walking along the same road from q to p . ann walked at a constant speed of 3 miles per hour and john at 2 miles per hour . how many miles had ann walked when they met ? | "ann walks from q to p at a speed of 3 miles / hr for one hour . she covers 3 miles in 1 hour and now distance between john and ann is 28 - 3 = 25 miles . ann walks at 3 mph and john at 2 mph so their relative speed is 3 + 2 = 5 mph . they have to cover 25 miles so it will take them 25 / 5 = 5 hours to meet . in 5 hrs , ann would have covered 5 hrs * 3 miles per hour = 15 miles . adding this to the 3 miles she covered before john , ann covered a total of 3 + 15 = 18 miles . answer ( e )" | a = 3 + 2
b = 28 / a
c = b * 2
|
a ) 156 , b ) 162 , c ) 318 , d ) 324 , e ) 325 | c | add(multiply(floor(divide(680, 50)), multiply(const_12, const_2)), floor(divide(subtract(680, multiply(floor(divide(680, 50)), 50)), 4.5))) | roses can be purchased individually for $ 4.50 , one dozen for $ 36 , or two dozen for $ 50 . what is the greatest number of roses that can be purchased for $ 680 ? | buy as many $ 50 deals as possible . we can by 650 / 50 = 13 two dozen roses , thus total of 13 * 24 = 312 roses . we are left with 680 - 650 = $ 30 . we can buy 30 / 4.5 = ~ 6 roses for that amount . total = 312 + 6 = 318 . answer : c . | a = 680 / 50
b = math.floor(a)
c = 12 * 2
d = b * c
e = 680 / 50
f = math.floor(e)
g = f * 50
h = 680 - g
i = h / 4
j = math.floor(i)
k = d + j
|
a ) 12100 , b ) 14400 , c ) 14500 , d ) 14600 , e ) 14700 | a | add(10000, multiply(divide(multiply(10000, 10), const_100), 2)) | the population of a town is 10000 . it increases annually at the rate of 10 % p . a . what will be its population after 2 years ? | "formula : 10000 × 110 / 100 × 110 / 100 = 12100 answer : a" | a = 10000 * 10
b = a / 100
c = b * 2
d = 10000 + c
|
a ) 2200 , b ) 2550 , c ) 5050 , d ) 6275 , e ) 11325 | a | multiply(subtract(108, 8), add(divide(subtract(50, 8), const_2), const_1)) | set a contains all the even numbers between 8 and 50 inclusive . set b contains all the even numbers between 108 and 150 inclusive . what is the difference between the sum of elements of set b and the sum of the elements of set a ? | "set a contains 8 , 10 , 12 . . . 50 set b contains 108 , 110 , 112 . . . 150 number of terms in each set = 22 difference between corresponding terms in set a and b = 100 difference between sum of set b and set a = 100 * 22 = 2200 answer a" | a = 108 - 8
b = 50 - 8
c = b / 2
d = c + 1
e = a * d
|
a ) 16 % , b ) 46.66 % , c ) 44.44 % , d ) 36.98 % , e ) 17 % | c | multiply(divide(subtract(75, 45), 45), const_100) | john makes $ 45 a week from his job . he earns a raise andnow makes $ 75 a week . what is the % increase ? | "increase = ( 20 / 45 ) * 100 = ( 4 / 9 ) * 100 = 44.44 % . c" | a = 75 - 45
b = a / 45
c = b * 100
|
a ) 18 hours , b ) 23 hours , c ) 28 hours , d ) 30 hours , e ) 40 hours | d | divide(multiply(20, 12), subtract(20, 12)) | pipe a can fill a tank in 12 hours . due to a leak at the bottom it takes 20 hours to fill the tank . in what time the leak alone can empty the full tank ? | "let leak can empty the full tank in x hours . 1 / 12 - 1 / x = 1 / 20 = > 1 / 12 – 1 / 20 = 1 / x 1 / x = 5 - 3 / 60 = 1 / 30 x = 30 hours answer : d" | a = 20 * 12
b = 20 - 12
c = a / b
|
a ) 72 , b ) 94 , c ) 86 , d ) 74 , e ) 110 | a | divide(30, 240) | find 30 % of 240 | "we know that r % of m is equal to r / 100 × m . so , we have 30 % of 240 30 / 100 × 240 = 72 answer : a" | a = 30 / 240
|
a ) 84 m , b ) 88 m , c ) 14 m , d ) 137 m , e ) none | c | divide(11, multiply(power(divide(1, const_2), const_2), const_pi)) | 11 cubic centimetres of silver is drawn into a wire 1 mm in diameter . the length of the wire in metres will be : | "sol . let the length of the wire b h . radius = 1 / 2 mm = 1 / 20 cm . then , 22 / 7 * 1 / 20 * 1 / 20 * h = 11 ⇔ = [ 11 * 20 * 20 * 7 / 22 ] = 1400 cm = 14 m . answer c" | a = 1 / 2
b = a ** 2
c = b * math.pi
d = 11 / c
|
a ) 337 , b ) 500 , c ) 700 , d ) 288 , e ) 211 | c | divide(70, divide(multiply(4, add(2, divide(1, 2))), const_100)) | what sum of money will produce rs . 70 as simple interest in 4 years at 2 1 / 2 percent ? | "70 = ( p * 4 * 5 / 2 ) / 100 p = 700 answer : c" | a = 1 / 2
b = 2 + a
c = 4 * b
d = c / 100
e = 70 / d
|
a ) 0 , b ) 1 , c ) 2 , d ) 3 , e ) 4 | e | divide(16, add(add(2, 2), 1)) | if 2 / z = 2 / ( z + 1 ) + 2 / ( z + 16 ) which of these integers could be the value of z ? | "solving for z algebraically in this problem would not be easy . instead , we can follow the hint in the question ( “ which of these integers … ” ) and test each answer choice : a . 2 / 0 = 2 / 1 + 2 / 16 incorrect ( division by zero ) a . 2 / 1 = 2 / 2 + 2 / 17 incorrect a . 2 / 2 = 2 / 3 + 2 / 18 incorrect a . 2 / 3 = 2 / 4 + 2 / 19 incorrect a . 2 / 4 = 2 / 5 + 2 / 20 correct the correct answer is e , because it contains the only value that makes the equation work . notice how quickly this strategy worked in this case" | a = 2 + 2
b = a + 1
c = 16 / b
|
a ) 6 hours , b ) 5 hours , c ) 7 hours , d ) 8 hours , e ) none | d | divide(add(360, 14), add(divide(360, 8), 18)) | a truck covers a distance of 360 km at a certain speed in 8 hours . how much time would a car take at an average speed which is 18 kmph more than that of the speed of the truck to cover a distance which is 14 km more than that travelled by the truck ? | "explanation : speed of the truck = distance / time = 360 / 8 = 45 kmph now , speed of car = ( speed of truck + 18 ) kmph = ( 45 + 18 ) = 63 kmph distance travelled by car = 360 + 81 = 441 km time taken by car = distance / speed = 441 / 63 = 8 hours . answer – d" | a = 360 + 14
b = 360 / 8
c = b + 18
d = a / c
|
a ) 31 , b ) 35 , c ) 39 , d ) 43 , e ) 47 | c | add(add(add(add(divide(3.6, divide(8, const_60)), 3), 3), 3), 3) | the average speed of a car decreased by 3 miles per hour every successive 8 - minutes interval . if the car traveled 3.6 miles in the fifth 8 - minute interval , what was the average speed of the car , in miles per hour , in the first 8 minute interval ? | ( 3.6 miles / 8 minutes ) * 60 minutes / hour = 27 mph let x be the original speed . x - 4 ( 3 ) = 27 x = 39 mph the answer is c . | a = 8 / const_60
b = 3 / 6
c = b + 3
d = c + 3
e = d + 3
f = e + 3
|
['a ) 33', 'b ) 60', 'c ) 88', 'd ) 27', 'e ) 26'] | b | divide(subtract(power(multiply(68, divide(15, subtract(68, multiply(const_4, const_2)))), const_2), power(multiply(divide(15, subtract(68, multiply(const_4, const_2))), 52), const_2)), const_2) | a took 15 seconds to cross a rectangular field diagonally walking at the rate of 52 m / min and b took the same time to cross the same field along its sides walking at the rate of 68 m / min . the area of the field is ? | explanation : length of the diagonal = 52 * ( 15 / 60 ) = 13 m sum of length and breadth = 68 * ( 15 / 60 ) = 17 m \ inline { \ color { black } \ sqrt { l ^ { 2 } + b ^ { 2 } } = 13 \ ; or \ ; l + b = 17 } area = lb = 1 / 2 [ ( 2 lb ) ] = 1 / 2 [ ( l + b ) ² - ( l ² + b ² ) ] = 1 / 2 [ 17 ² - 169 ] = 1 / 2 * 120 = 60 sq meter answer : b ) 60 | a = 4 * 2
b = 68 - a
c = 15 / b
d = 68 * c
e = d ** 2
f = 4 * 2
g = 68 - f
h = 15 / g
i = h * 52
j = i ** 2
k = e - j
l = k / 2
|
['a ) 28.32', 'b ) 36.57', 'c ) 44.57', 'd ) 48.93', 'e ) 54.24'] | b | add(subtract(square_perimeter(8), 8), divide(circumface(divide(8, const_2)), const_2)) | a geometric shape is obtained by combining a square of side 8 cm and a semicircle of diameter 8 cm . what is the perimeter of the shape obtained ? | circumference of the semicircle = ï € xd / 2 = ( 22 / 7 ) x 4 = 12.57 perimeter of the square plot = 3 x 8 = 24 perimeter of the shape = 36.57 cm answer : b | a = square_perimeter - (
b = a + 8
|
a ) $ 602.5 , b ) $ 625.8 , c ) $ 710.6 , d ) $ 745 , e ) $ 790.3 | a | divide(divide(subtract(multiply(100, power(add(const_1, divide(5, const_100)), 2)), 100), 2), multiply(2, divide(10, const_100))) | the simple interest on a certain sum of money for 2 years at 10 % per annum is half the compound interest on $ 100 for 2 years at 5 % per annum . the sum placed on simple interest is ? | "c . i . = 100 * ( 1 + 5 / 100 ) ^ 2 - 100 = 100 * 21 / 20 * 21 / 20 - 100 = $ 120.5 sum = 120.5 * 100 / 2 * 10 = $ 602.5 answer is a" | a = 5 / 100
b = 1 + a
c = b ** 2
d = 100 * c
e = d - 100
f = e / 2
g = 10 / 100
h = 2 * g
i = f / h
|
a ) $ 2100 , b ) $ 2000 , c ) $ 2050 , d ) $ 2070 , e ) $ 2500 | b | multiply(subtract(40, multiply(15, const_2)), 500) | redo ’ s manufacturing costs for sets of horseshoes include a $ 10,500 initial outlay , and $ 15 per set . they can sell the sets $ 40 . if profit is revenue from sales minus manufacturing costs , and the company producessells 500 sets of horseshoes , what was their profit ? | "total manufacturing cost = 10500 + 500 * 15 = 18000 total selling cost = 500 * 40 = 20000 profit = 20000 - 18000 = 2000 answer : b" | a = 15 * 2
b = 40 - a
c = b * 500
|
a ) 9 , b ) 8 , c ) 0 , d ) 2 , e ) 1 | b | divide(multiply(lcm(const_100, 8), gcd(const_100, 8)), const_100) | find the highest value of ' a ' so that 365 a 16 is divisible by 8 . | explanation : given , number is divisible by 8 only if ' a 16 ' is divisible by 8 . . : highest value of a is ' 8 ' . answer : option b | a = math.lcm(100, 8)
b = math.gcd(100, 8)
c = a * b
d = c / 100
|
a ) 127 , b ) 240 , c ) 287 , d ) 450 , e ) 281 | b | divide(multiply(36, 30), divide(multiply(15, 24), 80)) | if 15 men can reap 80 hectares in 24 days , then how many hectares can 36 men reap in 30 days ? | explanation : let the required no of hectares be x . then men - - - hectares - - - days 15 - - - - - - - - - 80 - - - - - - - - - 24 36 - - - - - - - - - x - - - - - - - - - 30 more men , more hectares ( direct proportion ) more days , more hectares ( direct proportion ) x = 36 / 15 * 30 / 24 * 80 x = 240 answer : b | a = 36 * 30
b = 15 * 24
c = b / 80
d = a / c
|
a ) 13.28 % , b ) 14.28 % , c ) 15.28 % , d ) 16.28 % , e ) 17.28 % | b | subtract(divide(multiply(const_100, 8), subtract(8, 1)), const_100) | in a office work is distribute between p persons . if 1 / 8 members are absent then work increased for each person is ? | "let total % of work is 100 % total person = p 1 / 8 person are absent of total person . so absent person is 1 / 8 p ie p / 8 . left person is , p - p / 8 = 7 p / 8 . p person do the work 100 % 1 person do the work 100 * p % 7 p / 8 person do the work ( 100 * p * 8 ) / 7 p % = 114.28 % work increased for each person is = ( 114.28 - 100 ) % = 14.28 % answer : b" | a = 100 * 8
b = 8 - 1
c = a / b
d = c - 100
|
a ) 1 / 2 , b ) 1 / 4 , c ) 3 / 4 , d ) 2 / 3 , e ) none of these | b | divide(multiply(65, 250), multiply(1000, 65)) | if a * b * c = 65 , b * c * d = 65 , c * d * e = 1000 and d * e * f = 250 the ( a * f ) / ( c * d ) = ? | "explanation : a â ˆ — b â ˆ — c / b â ˆ — c â ˆ — d = 65 / 65 = > a / d = 1 d â ˆ — e â ˆ — f / c â ˆ — d â ˆ — e = 250 / 1000 = > f / c = 1 / 4 a / d * f / c = 1 * 1 / 4 = 1 / 4 answer : b" | a = 65 * 250
b = 1000 * 65
c = a / b
|
a ) rs . 1090 , b ) rs . 1160 , c ) rs . 1120 , d ) rs . 1202 , e ) none | c | divide(multiply(subtract(const_100, 20), 1400), const_100) | a man buys a cycle for rs . 1400 and sells it at a loss of 20 % . what is the selling price of the cycle ? | "solution s . p = 80 % of rs . 1400 = rs . ( 80 / 100 × 1400 ) rs . 1120 . answer c" | a = 100 - 20
b = a * 1400
c = b / 100
|
a ) 8.8 % , b ) 9 % , c ) 9.2 % , d ) 8.6 % , e ) 8.4 % | c | multiply(divide(add(divide(multiply(12, 15), const_100), divide(multiply(8, 35), const_100)), add(15, 35)), const_100) | in one alloy there is 12 % chromium while in another alloy it is 8 % . 15 kg of the first alloy was melted together with 35 kg of the second one to form a third alloy . find the percentage of chromium in the new alloy . | the amount of chromium in the new 15 + 35 = 50 kg alloy is 0.12 * 15 + 0.08 * 35 = 4.6 kg , so the percentage is 4.6 / 50 * 100 = 9.2 % . answer : c . | a = 12 * 15
b = a / 100
c = 8 * 35
d = c / 100
e = b + d
f = 15 + 35
g = e / f
h = g * 100
|
a ) 92 , b ) 35 , c ) 64 , d ) 46 , e ) 355 | a | multiply(divide(add(const_100, 20), add(add(const_100, 20), const_100)), const_100) | in may mrs lee ' s earnings were 90 percent of the lee family ' s total income . in june mrs lee earned 20 percent more than in may . if the rest of the family ' s income was the same both months , then , in june , mrs lee ' s earnings were approximately what percent of the lee family ' s total income ? | "lets say the family income is 100 in may , lee earned 90 family income is 10 in june , lee earned 20 % more than may , so it is ( 90 + 20 * 90 / 100 = 108 ) family income is same 10 in june lee ' s income percent is 108 * 100 / 118 ~ 92 ans is a" | a = 100 + 20
b = 100 + 20
c = b + 100
d = a / c
e = d * 100
|
a ) 277 , b ) 36 , c ) 64 , d ) 72 , e ) 46 | e | divide(multiply(165.6, const_100), 360) | ? % of 360 = 165.6 | "? % of 360 = 165.6 or , ? = 165.6 × 100 / 360 = 46 answer e" | a = 165 * 6
b = a / 360
|
a ) $ 492 , b ) $ 512 , c ) $ 675 , d ) $ 745 , e ) $ 1020 | c | multiply(6000, divide(15, const_100)) | find the simple interest on $ 6000 at 15 % per annum for 9 months ? | "p = $ 6000 r = 15 % t = 9 / 12 years = 3 / 4 years s . i . = p * r * t / 100 = 6000 * 15 * 3 / 400 = $ 675 answer is c" | a = 15 / 100
b = 6000 * a
|
a ) 1,349 , b ) 1,353 , c ) 10,040 , d ) 1,354 , e ) 10,400 | a | subtract(458,705, add(add(multiply(const_2, const_100), multiply(add(const_3, const_4), const_10)), const_2)) | how many integers between 324,805 and 458,705 have tens digit 1 and units digit 3 ? | "there is one number in hundred with 1 in the tens digit and 3 in the units digit : 13 , 113 , 213 , 313 , . . . the difference between 324,805 and 458,705 is 458,705 - 324,805 = 134,900 - one number per each hundred gives 134,900 / 100 = 1,349 numbers . answer : a ." | a = 2 * 100
b = 3 + 4
c = b * 10
d = a + c
e = d + 2
f = 458 - 705
|
a ) 25.5 liters , b ) 29.16 liters , c ) 30.98 liters , d ) 42.15 liters , e ) 48.32 liters | b | subtract(subtract(subtract(40, 4), 4), 4) | a container contains 40 liters of milk , from this container 4 liters of milk was taken out and replaced by water . this process was repeated further 2 times . how much milk is now contained by the container ? | "amount of milk left after 3 operations = 40 ( 1 - 4 / 40 ) ^ 3 = 40 * 9 / 10 * 9 / 10 * 9 / 10 = 29.16 liters answer is b" | a = 40 - 4
b = a - 4
c = b - 4
|
a ) 30 % , b ) 40 % , c ) 50 % , d ) 19 % , e ) 29 % | c | divide(const_100, add(const_1, 25)) | solve the quickfire maths brain teaser â ˆ š 25 % = ? | "â ˆ š 25 % = > â ˆ š 25 / â ˆ š 100 = > 5 / 10 = > 50 / 100 = > 50 % c" | a = 1 + 25
b = 100 / a
|
a ) 5 / 6 , b ) 1 / 5 , c ) 1 / 6 , d ) 1 / 9 , e ) 1 / 24 | d | divide(24, add(24, multiply(24, multiply(divide(1, 3), 24)))) | a satellite is composed of 24 modular units , each of which is equipped with a set of sensors , some of which have been upgraded . each unit contains the same number of non - upgraded sensors . if the number of non - upgraded sensors on one unit is 1 / 3 the total number of upgraded sensors on the entire satellite , what fraction of the sensors on the satellite have been upgraded ? | let x be the number of upgraded sensors on the satellite . the number of non - upgraded sensors per unit is x / 3 . the number of non - upgraded sensors on the whole satellite is 24 ( x / 3 ) = 8 x . the fraction of sensors which have been upgraded is x / ( x + 8 x ) = x / 9 x = 1 / 9 the answer is d . | a = 1 / 3
b = a * 24
c = 24 * b
d = 24 + c
e = 24 / d
|
a ) 12 , b ) 20 , c ) 24 , d ) 30 , e ) 36 | b | divide(const_1, subtract(divide(const_3, 36), divide(const_2, 60))) | tom drives from town a to town b , driving at a constant speed of 60 miles per hour . from town b tom immediately continues to town c . the distance between a and b is twice the distance between b and c . if the average speed of the whole journey was 36 mph , then what is tom ' s speed driving from b to c in miles per hour ? | let ' s assume that it takes 4 hours to go from point a to b . then the distance between them becomes 240 which makes distance between b and c 120 . ( 240 + 120 ) / ( 4 + x ) gives us the average speed which is 36 . you find x = 6 . so the question simplifies itself to 120 / 6 = 20 hence the answer is b | a = 3 / 36
b = 2 / 60
c = a - b
d = 1 / c
|
a ) 42 , b ) 44 , c ) 46 , d ) 48 , e ) 50 | c | subtract(add(add(22, 33), 53), add(add(multiply(8, const_3), 30), 56)) | the average ( arithmetic mean ) of 22 , 33 , and 53 is 8 less than the average of 30 , 56 , and x . what is x ? | "the average of 22 , 33 , and 53 is 108 / 3 = 36 . the average of 30 , 56 and x is 44 . then 30 + 56 + x = 132 . x = 46 . the answer is c ." | a = 22 + 33
b = a + 53
c = 8 * 3
d = c + 30
e = d + 56
f = b - e
|
a ) 100 , b ) 150 , c ) 200 , d ) 250 , e ) 300 | b | add(multiply(const_100, 3), const_100) | how many 3 - digit numbers are completely divisible 6 ? | "3 - digit number divisible by 6 are : 102 , 108 , 114 , . . . , 996 this is an a . p . in which a = 102 , d = 6 and l = 996 let the number of terms be n . then tn = 996 . a + ( n - 1 ) d = 996 102 + ( n - 1 ) x 6 = 996 6 x ( n - 1 ) = 894 ( n - 1 ) = 149 n = 150 number of terms = 150 . b )" | a = 100 * 3
b = a + 100
|
a ) 61 , b ) 66 , c ) 76 , d ) 86 , e ) 83 | b | subtract(negate(31), multiply(subtract(6, 16), divide(subtract(6, 16), subtract(1, 6)))) | 1 , 6 , 16 , 31 , 51 , ( . . . . ) | "explanation : the pattern is 5 , 10 , 15 , 20 , 25 , etc hence 25 = 66 answer : b" | a = negate - (
|
a ) 100 , b ) 110 , c ) 120 , d ) 140 , e ) 240 | e | divide(multiply(600, subtract(const_100, 60)), const_100) | in a public show 60 % of the seats were filled . if there were 600 seats in the hall , how many seats were vacant ? | "75 % of 600 = 60 / 100 × 600 = 360 therefore , the number of vacant seats = 600 - 360 = 240 . answer : e" | a = 100 - 60
b = 600 * a
c = b / 100
|
a ) 0.0016 , b ) 0.0625 , c ) 0.16 , d ) 0.25 , e ) 0.01 | e | power(divide(1, 10), 2) | what is the decimal equivalent of ( 1 / 10 ) ^ 2 ? | "( 1 / 10 ) ² = ( 1 / 10 ) ( 1 / 10 ) = 1 / 100 approach # 1 : use long division to divide 100 into 1 to get 1 / 100 = 0.01 e" | a = 1 / 10
b = a ** 2
|
a ) 1 / 16 , b ) 1 / 8 , c ) 3 / 16 , d ) 3 / 8 , e ) 13 / 16 | b | divide(divide(const_1, const_3), multiply(multiply(multiply(divide(const_1, const_3), 2), 2), 2)) | for a bake sale , simon baked 2 n more pies than theresa . theresa baked half as many pies as roger , who baked 1313 n pies . no other pies were baked for the sale . what fraction of the total pies for sale did roger bake ? | s : simon t : theresa r : roger let theresa baked a . as per question , we have s = a + 2 n r = 2 a = n / 3 or a = n / 6 now , total = a + ( a + 2 n ) + 2 a = 8 n / 3 fraction , ( n / 3 ) / ( 8 n / 3 ) = 1 / 8 answer : b | a = 1 / 3
b = 1 / 3
c = b * 2
d = c * 2
e = d * 2
f = a / e
|
a ) 140 , b ) 175 , c ) 200 , d ) 245 , e ) 280 | c | subtract(30, subtract(50, 30)) | sarah is driving to the airport . after driving at 50 miles per hour for one hour , she realizes that if she continues at that same average rate she will be an hour late for her flight . she then travels 100 miles per hour for the rest of the trip , and arrives 30 minutes before her flight departs . how many miles did she drive in all ? | "after driving at 50 miles per hourfor one hour , this distance left to cover is d - 50 . say this distance is x miles . now , we know that the difference in time between covering this distance at 50 miles per hour and 100 miles per hour is 1 + 1 / 2 = 3 / 2 hours . so , we have that x / 50 - x / 100 = 3 / 2 - - > 2 x / 100 - x / 100 = 3 / 2 - - > x / 100 = 3 / 2 - - > x = 150 total distance = x + 50 = 200 miles . answer : c ." | a = 50 - 30
b = 30 - a
|
a ) 10 s , b ) 15 s , c ) 4 s , d ) 8 s , e ) 12 s | b | divide(add(225, 150), add(divide(multiply(54, const_1000), const_3600), divide(multiply(36, const_1000), const_3600))) | two trains a and b are 225 m and 150 m long and are moving at one another at 54 km / hr and 36 km / hr respectively . arun is sitting on coach b 1 of train a . calculate the time taken by arun to completely cross train b . | "detailed solution speed of a = 54 ∗ 1000 / 60 ∗ 60 = 15 m / s speed of b = 36 ∗ 1000 / 60 ∗ 60 = 10 m / s relative speed = s 1 + s 2 = 15 + 10 m / s = 25 m / s the length that needs to be crossed = length of train b = 150 m . therefore time taken = 150 / 25 = 6 s . what is the time taken for trains to completely cross each other ? the length that needs to be crossed = 225 + 150 = 375 m . time taken = 375 / 25 = 15 s . correct answer b ." | a = 225 + 150
b = 54 * 1000
c = b / 3600
d = 36 * 1000
e = d / 3600
f = c + e
g = a / f
|
a ) 10,000 , b ) 11,025 , c ) 14,400 , d ) 12,696 , e ) can not be determined | a | power(add(divide(199, const_2), add(const_0_25, const_0_25)), const_2) | a gardener grows cabbages in her garden that is in the shape of a square . each cabbage takes 1 square feet of area in her garden . this year , she has increased her output by 199 cabbages as compared to last year . the shape of the area used for growing the cabbages has remained a square in both these years . how many cabbages did she produce this year ? | "let the side for growing cabbages this year be x ft . thus the area is x ^ 2 . let the side for growing cabbages last year be y ft . thus , the area was y ^ 2 . the area would have increased by 199 sq ft as each cabbage takes 1 sq ft space . x ^ 2 - y ^ 2 = 199 ( x + y ) ( x - y ) = 199 199 is a prime number and thus it will be ( 100 + 99 ) * ( 100 - 99 ) . thus x = 100 and y = 99 x ^ 2 = 100 ^ 2 = 10,000 the answer is a ." | a = 199 / 2
b = const_0_25 + const_0_25
c = a + b
d = c ** 2
|
a ) 33 , b ) 17 , c ) 49 , d ) 16 , e ) 23 | b | subtract(subtract(multiply(4, 7), 7), 4) | if x , y , and z are positive integers and x = 4 y = 7 z , then the least possible value of x - y - z is | x - y - z = x - x / 4 - x / 7 = ( 28 - 7 - 4 ) x / 28 = 17 x / 28 17 is not divisible by 28 ( it ' s a prime # ) , so for least value , x = 28 answer - b | a = 4 * 7
b = a - 7
c = b - 4
|
a ) 38.07 , b ) 40 , c ) 42 , d ) 39 , e ) 40.07 | a | multiply(divide(subtract(80, 8), add(8, 9)), 9) | one hour after yolanda started walking from x to y , a distance of 80 miles , bob started walking along the same road from y to x . if yolanda â s walking rate was 8 miles per hour and bob â s was 9 miles per hour , how many miles had bob walked when they met ? | "let t be the number of hours that bob had walked when he met yolanda . then , when they met , bob had walked 4 t miles and yolanda had walked 8 ( t + 1 ) miles . these distances must sum to 80 miles , so 9 t + 8 ( t + 1 ) = 80 , which may be solved for t as follows 9 t + 8 ( t + 1 ) = 80 9 t + 8 t + 8 = 80 17 t = 72 t = 4.23 ( hours ) therefore , bob had walked 9 t = 9 ( 4.23 ) = 38.07 miles when they met . the best answer is a ." | a = 80 - 8
b = 8 + 9
c = a / b
d = c * 9
|
a ) 2 hours , b ) 3 hours , c ) 3 hours 48 minutes , d ) 5 hours , e ) none | c | divide(68, add(13, 5)) | a boat can travel with a speed of 13 km / hr in still water . if the speed of the stream is 5 km / hr . find the time taken by the boat to go 68 km downstream ? | "solution speed downstream = ( 13 + 5 ) km / hr = 18 km / hr . time taken to travel 68 km downstream = ( 68 / 18 ) hrs = 3 hrs 48 minutes . answer c" | a = 13 + 5
b = 68 / a
|
a ) 15 kmph , b ) 11 kmph , c ) 88 kmph , d ) 19 kmph , e ) 12 kmph | d | multiply(divide(12, const_60), 94) | the speed of a train is 94 kmph . what is the distance covered by it in 12 minutes ? | "94 * 12 / 60 = 18.8 kmph answer : d" | a = 12 / const_60
b = a * 94
|
a ) 50 hours , b ) 40 hours , c ) 22.5 hours , d ) 12 hours , e ) 8 hours | c | divide(180, multiply(divide(divide(20, 10), 5), 20)) | if 5 machines can produce 20 units in 10 hours , how long would it take 20 machines to produce 180 units ? | "here , we ' re told that 5 machines can produce 20 units in 10 hours . . . . that means that each machine works for 10 hours apiece . since there are 5 machines ( and we ' re meant to assume that each machine does the same amount of work ) , then the 5 machines equally created the 20 units . 20 units / 5 machines = 4 units are made by each machine every 10 hours now that we know how long it takes each machine to make 4 units , we can break this down further if we choose to . . . 10 hours / 4 units = 2.5 hours per unit when 1 machine is working . the prompt asks us how long would it take 20 machines to produce 180 units . if 20 machines each work for 2.5 hours , then we ' ll have 20 units . since 180 units is ' 9 times ' 20 , we need ' 9 times ' more time . ( 2.5 hours ) ( 9 times ) = 22.5 hours final answer : [ reveal ] spoiler : c" | a = 20 / 10
b = a / 5
c = b * 20
d = 180 / c
|
a ) $ 28 , b ) $ 32 , c ) $ 34 , d ) $ 35 , e ) $ 40 | c | divide(51, add(const_1, divide(50, const_100))) | a worker ' s daily wage is increased by 50 % and the new wage is $ 51 per day . what was the worker ' s daily wage before the increase ? | "let x be the daily wage before the increase . 1.5 x = $ 51 x = $ 34 the answer is c ." | a = 50 / 100
b = 1 + a
c = 51 / b
|
a ) 18 % , b ) 30 % , c ) 36 % , d ) 48 % , e ) 54 % | c | multiply(divide(subtract(divide(60, const_100), multiply(divide(72, const_100), divide(2, const_3))), divide(1, const_3)), const_100) | in an election , candidate douglas won 60 percent of the total vote in counties x and y . he won 72 percent of the vote in county x . if the ratio of people who voted in county x to county y is 2 : 1 , what percent of the vote did candidate douglas win in county y ? | "given voters in ratio 2 : 1 let x has 200 votersy has 100 voters for x 72 % voted means 72 * 200 = 144 votes combined for xy has 300 voters and voted 60 % so total votes = 180 balance votes = 180 - 144 = 36 as y has 100 voters so 36 votes means 36 % of votes required ans c" | a = 60 / 100
b = 72 / 100
c = 2 / 3
d = b * c
e = a - d
f = 1 / 3
g = e / f
h = g * 100
|
a ) 23 , b ) 41 , c ) 48 , d ) 62 , e ) 86 | d | divide(50, divide(1600, multiply(const_2, const_1000))) | a ferry can transport 50 tons of vehicles . automobiles range in weight from 1600 to 3200 pounds . what is the greatest number of automobiles that can be loaded onto the ferry ? | to get maximum vehicles we must take into consideration the minimum weight i . e 1600 pounds here since , 1 ton = 2000 pounds 50 tons will be 100,000 pounds from the answer choices : let max number of vehicles be 62 total weight will be = 62 * 1600 = 99200 pounds , which is lesser than the maximum weight allowed . ans : d | a = 2 * 1000
b = 1600 / a
c = 50 / b
|
a ) 7.9 s , b ) 2.5 s , c ) 7.5 s , d ) 7.6 s , e ) 10 s | e | multiply(divide(divide(150, const_1000), 54), const_3600) | how much time does a train 150 metres long running at 54 km / hr take to pass a pole ? | "explanation : 54 km / hr = 54 * 5 / 18 = 15 m / s speed = distance / time ; v = d / t 15 = 150 / t t = 10 s answer : e" | a = 150 / 1000
b = a / 54
c = b * 3600
|
a ) 123 , b ) 106 , c ) 100 , d ) 156 , e ) 104 | e | subtract(104.25, divide(1, 4)) | the cash realised on selling a 14 % stock is rs . 104.25 , brokerage being 1 / 4 % is | "explanation : cash realised = rs . ( 104.25 - 0.25 ) = rs . 104 . answer : e" | a = 1 / 4
b = 104 - 25
|
a ) 12 , b ) 9 , c ) 10 , d ) 89 , e ) 81 | a | multiply(divide(const_1, multiply(add(const_100, 18), divide(const_1, subtract(const_100, 12)))), 12) | by selling 12 pencils for a rupee a man loses 12 % . how many for a rupee should he sell in order to gain 18 % ? | "88 % - - - 12 112 % - - - ? 88 / 112 * 12 = 9 answer : a" | a = 100 + 18
b = 100 - 12
c = 1 / b
d = a * c
e = 1 / d
f = e * 12
|
a ) 5,9 , b ) 14,18 , c ) 16,20 , d ) 18,22 , e ) 19,23 | a | subtract(divide(subtract(add(multiply(3, 3), 4), 3), subtract(3, const_1)), const_1) | ages of two persons differ by 4 years . if 3 years ago , the elder one was 3 times as old the younger one , find their present age | "explanation : let the age of the younger person be x , then the elder person ' s age is ( x + 4 ) = > 3 ( x - 3 ) = ( x + 4 - 3 ) [ 3 years before ] = > 3 x - 9 = x + 1 = > x = 5 . so the elder person ' s age is x + 4 = 9 answer : option a" | a = 3 * 3
b = a + 4
c = b - 3
d = 3 - 1
e = c / d
f = e - 1
|
a ) 190 , b ) 284.6 , c ) 300 , d ) 280 , e ) 312 | d | multiply(12.2, 23) | a type of extra - large suv averages 12.2 miles per gallon ( mpg ) on the highway , but only 7.6 mpg in the city . what is the maximum distance , in miles , that this suv could be driven on 23 gallons of gasoline ? | "so 12.2 * 23 = 280 . . imo option d is correct answer . ." | a = 12 * 2
|
a ) 10 days , b ) 30 days , c ) 20 days , d ) 80 days , e ) 40 days | b | inverse(add(divide(9, multiply(12, 54)), divide(21, multiply(20, 54)))) | if 12 men or 20 women can do a piece of work in 54 days , then in how many days can 9 men and 21 women together do the work ? | "b 30 days given that 12 m = 20 w = > 3 m = 5 w 9 men + 21 women = 15 women + 21 women = 36 women 20 women can do the work in 54 days . so , 36 women can do it in ( 20 * 54 ) / 36 = 30 days ." | a = 12 * 54
b = 9 / a
c = 20 * 54
d = 21 / c
e = b + d
f = 1/(e)
|
a ) 45 , b ) 35 , c ) 40 , d ) 50 , e ) 48 | b | divide(original_price_before_loss(20, 70), divide(original_price_before_gain(20, 60), 20)) | a man sold 20 articles for $ 60 and gained 20 % . how many articles should he sell for $ 70 to incur a loss 20 % ? | "production cost per article : $ 60 * ( 100 % - 20 % ) / 20 = $ 2.40 required production costs for a loss of 20 % : $ 70 * ( 100 % + 20 % ) = $ 84 number of articles to be sold for $ 84 to incur a 20 % loss : $ 84 / $ 2.40 = 35 thus , solution b is correct ." | a = original_price_before_loss / (
|
a ) 20 % , b ) 30 % , c ) 40 % , d ) 50 % , e ) 60 % | a | multiply(divide(subtract(divide(60, const_100), divide(55, const_100)), add(subtract(divide(60, const_100), divide(55, const_100)), subtract(divide(80, const_100), divide(60, const_100)))), const_100) | solution p is 20 percent lemonade and 80 percent carbonated water by volume ; solution q is 45 percent lemonade and 55 percent carbonated water by volume . if a mixture of pq contains 60 percent carbonated water , what percent of the volume of the mixture is p ? | "60 % is 20 % - points below 80 % and 5 % - points above 55 % . so the ratio of solution p to solution q is 1 : 4 . mixture p is 1 / 5 = 20 % of the volume of mixture pq . the answer is a ." | a = 60 / 100
b = 55 / 100
c = a - b
d = 60 / 100
e = 55 / 100
f = d - e
g = 80 / 100
h = 60 / 100
i = g - h
j = f + i
k = c / j
l = k * 100
|
a ) 5 , b ) 6 , c ) 7 , d ) 8 , e ) 9 | a | divide(50, const_10) | how many integers from 0 to 50 inclusive have a remainder of 3 when divided by 11 ? | "the numbers should be of the form 11 c + 3 . the minimum is 3 when c = 0 . the maximum is 47 when c = 4 . there are 5 such numbers . the answer is a ." | a = 50 / 10
|
a ) 3 , b ) 5 , c ) 7 , d ) 12 , e ) 15 | a | add(const_2, gcd(60, 17)) | find a positive number which when increased by 17 is equal to 60 times the reciprocal of the number . | "sol . let the number be x . then , x + 17 = 60 / x ⇔ x 2 + 17 x - 60 = 0 ⇔ ( x + 20 ) ( x - 3 ) = 0 ⇔ x = 3 . answer a" | a = math.gcd(60, 17)
b = 2 + a
|
a ) 220880 , b ) 145778 , c ) 220081 , d ) 220080 , e ) 220030 | d | add(multiply(multiply(add(555, 445), 2), subtract(555, 445)), 80) | a no . when divided by the sum of 555 and 445 gives 2 times their difference as quotient & 80 as remainder . find the no . is ? | ( 555 + 445 ) * 2 * 110 + 80 = 220000 + 80 = 220080 d | a = 555 + 445
b = a * 2
c = 555 - 445
d = b * c
e = d + 80
|
a ) 9525 , b ) 5565 , c ) 7545 , d ) 6555 , e ) 8535 | c | multiply(multiply(120, 30), const_2) | the sum of two number is 120 and their difference is 30 . find the numbers . | let the two number be x and y . x + y = 120 . . . . . . . ( 1 ) x - y = 30 . . . . . . . . . ( 2 ) = > 2 x = 150 = > x = 75 . x - y = 30 = > y = 45 . the two numbers are 7545 ans : c | a = 120 * 30
b = a * 2
|
a ) 50 , b ) 45 , c ) 52.63 , d ) 40 , e ) 35 | c | divide(50, multiply(subtract(const_1, divide(20, const_100)), add(divide(10, const_100), const_1))) | the prices of tea and coffee per kg were the same in june . in july the price of coffee shot up by 10 % and that of tea dropped by 20 % . if in july , a mixture containing equal quantities of tea and coffee costs 50 / kg . how much did a kg of coffee cost in june ? | "let the price of tea and coffee be x per kg in june . price of tea in july = 1.1 x price of coffee in july = 0.8 x . in july the price of 1 / 2 kg ( 500 gm ) of tea and 1 / 2 kg ( 500 gm ) of coffee ( equal quantities ) = 50 1.1 x ( 1 / 2 ) + 0.8 x ( 1 / 2 ) = 50 = > x = 52.63 c" | a = 20 / 100
b = 1 - a
c = 10 / 100
d = c + 1
e = b * d
f = 50 / e
|
a ) 1 : 20 , b ) 1 : 10 , c ) 1 : 120 , d ) 23 : 4 , e ) 23 : 120 | e | divide(subtract(33, divide(36, divide(add(const_100, 30), const_100))), divide(36, divide(add(const_100, 30), const_100))) | a dishonest person wants to make a profit on the selling of milk . he would like to mix water ( costing nothing ) with milk costing 33 $ per litre so as to make a profit of 30 % on cost when he sells the resulting milk and water mixture for 36 $ . in what ratio should he mix the water and milk ? | "first of all , let ' s consider 1 liter of the stuff he is going to sell - - - naive customers think it ' s pure milk , but we know it ' s some milk - water mixture . he is going to sell this liter of milk - water for $ 36 . this $ 36 should be a 30 % increase over cost . here , we need to think about percentage increases as multipliers . using multipliers ( cost ) * 1.30 = $ 36 cost = 36 / 1.3 = 360 / 12 = $ 27.69 if he wants a 30 % increase over cost on the sale of one liter of his milk - water , the cost has to be $ 27.69 . well , a liter of milk costs $ 33 , so if he is going to use just $ 30 of milk in his mixture , that ' s 27.69 / 33 = 120 / 143 of a liter . if milk is 120 / 143 of the liter , then water is 23 / 143 of the liter , and the ratio of water to milk is 23 : 120 . answer choice ( e )" | a = 100 + 30
b = a / 100
c = 36 / b
d = 33 - c
e = 100 + 30
f = e / 100
g = 36 / f
h = d / g
|
Subsets and Splits
No community queries yet
The top public SQL queries from the community will appear here once available.