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 |
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a ) 8 ° , b ) 40 ° , c ) 18 ° , d ) 36 ° , e ) 52 ° | b | divide(multiply(subtract(const_100, add(add(add(add(13, 24), 15), 29), 8)), divide(const_3600, const_10)), const_100) | a circle graph shows how the megatech corporation allocates its research and development budget : 13 % microphotonics ; 24 % home electronics ; 15 % food additives ; 29 % genetically modified microorganisms ; 8 % industrial lubricants ; and the remainder for basic astrophysics . if the arc of each sector of the graph is proportional to the percentage of the budget it represents , how many degrees of the circle are used to represent basic astrophysics research ? | "here all percentage when summed we need to get 100 % . as per data 13 + 24 + 15 + 29 + 8 = 89 % . so remaining 11 % is the balance for the astrophysics . since this is a circle all percentage must be equal to 360 degrees . 100 % - - - - 360 degrees then 11 % will be 40 degrees . . imo option b ." | a = 13 + 24
b = a + 15
c = b + 29
d = c + 8
e = 100 - d
f = 3600 / 10
g = e * f
h = g / 100
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a ) 52.112 , b ) 53.156 , c ) 54 , d ) 89 , e ) 88 | b | subtract(negate(23.63), multiply(subtract(10.5, 15.75), divide(subtract(10.5, 15.75), subtract(7, 10.5)))) | 7 , 10.5 , 15.75 , 23.63 , 35.43 , ( . . . ) | "7 ( 7 ã — 3 ) ã · 2 = 10.5 ( 10.5 ã — 3 ) ã · 2 = 15.75 ( 15.75 ã — 3 ) ã · 2 = 23.63 ( 23.63 ã — 3 ) ã · 2 = 35.43 ( 35.43 ã — 3 ) ã · 2 = 53.156 answer is b" | a = negate - (
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a ) $ 3.15 , b ) $ 4.5 , c ) $ 4.80 , d ) $ 5.05 , e ) $ 5.40 | b | add(2.25, multiply(0.25, divide(3.6, divide(2, 5)))) | jim ’ s taxi service charges an initial fee of $ 2.25 at the beginning of a trip and an additional charge of $ 0.25 for each 2 / 5 of a mile traveled . what is the total charge for a trip of 3.6 miles ? | "let the fixed charge of jim ’ s taxi service = 2.25 $ and charge per 2 / 5 mile ( . 4 mile ) = . 25 $ total charge for a trip of 3.6 miles = 2.25 + ( 3.6 / . 4 ) * . 25 = 2.25 + 9 * . 25 = 4.5 $ answer b" | a = 2 / 5
b = 3 / 6
c = 0 * 25
d = 2 + 25
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a ) a ) 60 , b ) b ) 150 , c ) c ) 130 , d ) d ) 90 , e ) e ) 210 | a | divide(20, divide(divide(2, 3), const_2)) | payal finished 2 / 3 of a book . she calculated that she finished 20 more pages than she has yet to read . how long is her book ? | let x be the total number of pages in the book , then she finished 2 / 3 * x pages . then she has x − 2 / 3 * x = 1 / 3 * x pages left . 2 / 3 * x − 1 / 3 * x = 20 1 / 3 * x = 20 x = 60 so the book is 270 pages long . answer is a . | a = 2 / 3
b = a / 2
c = 20 / b
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a ) 2 % decrease , b ) 3.5 % increase , c ) 9 % decrease , d ) 3 % decrease , e ) 2 % increase | b | subtract(const_100, multiply(multiply(add(const_1, divide(15, const_100)), subtract(const_1, divide(10, const_100))), const_100)) | the tax on a commodity is diminished by 10 % and its consumption increased by 15 % . the effect on revenue is ? | "100 * 100 = 10000 90 * 115 = 10350 - - - - - - - - - - - 10000 - - - - - - - - - - - 350 100 - - - - - - - - - - - ? = > 3.5 % decrease answer : b" | a = 15 / 100
b = 1 + a
c = 10 / 100
d = 1 - c
e = b * d
f = e * 100
g = 100 - f
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a ) 21 st , b ) 22 nd , c ) 23 rd , d ) 35 th , e ) none of these | d | subtract(multiply(2, 19), 1) | a monkey ascends a greased pole 19 metres high . he ascends 2 metres in first minute and slips down 1 metre in the alternate minute . in which minute , he reaches the top ? | "in 2 minutes , he ascends = 1 metre â ˆ ´ 17 metres , he ascends in 34 minutes . â ˆ ´ he reaches the top in 35 th minute . answer d" | a = 2 * 19
b = a - 1
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a ) 20 cm , b ) 25 cm , c ) 26 cm , d ) 100 / 3 cm , e ) 23 cm | d | divide(const_100, const_3) | the length of a rectangle is twice its breadth . if its length is decreased by 5 cm and breadth is increased by 4 cm , the area of the rectangle is increased by 75 sq . cm . find the length of the rectangle . | explanation : let breadth = x . then , length = 2 x . then , ( 2 x - 5 ) ( x + 4 ) - 2 x * x = 75 = > 3 x - 25 = 75 = > x = 100 / 3 . length of the rectangle = 100 / 3 cm . answer : option d | a = 100 / 3
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a ) 24 , b ) 25 , c ) 26 , d ) 27 , e ) 28 | c | add(divide(multiply(multiply(multiply(divide(2, 3), add(1, 1)), 3), const_12), 2), 2) | there is a 1 km long wire placed on some number of poles which are in equal distance . if the number of poles is reduced by 1 then the distance of wire between each poles increases 1 ( 2 / 3 ) . how many poles are there initially . | 26 answer : c | a = 2 / 3
b = 1 + 1
c = a * b
d = c * 3
e = d * 12
f = e / 2
g = f + 2
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a ) rs . 80 , b ) rs . 96 , c ) rs . 106 , d ) rs . 108 , e ) rs . 118 | b | multiply(multiply(divide(1620, 135), divide(8, const_100)), const_100) | by investing rs . 1620 in 8 % stock , michael earns rs . 135 . the stock is then quoted at : | michel earns rs 135 by investing rs 1620 to earn rs 8 how much he have to invest ? = ( 8 * 1620 ) / 135 = rs 96 answer : b | a = 1620 / 135
b = 8 / 100
c = a * b
d = c * 100
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a ) rs . 1748 , b ) rs . 1948 , c ) rs . 1848 , d ) rs . 2048 , e ) rs . 2148 | c | add(subtract(2000, divide(multiply(2000, 20), const_100)), divide(multiply(subtract(2000, divide(multiply(2000, 20), const_100)), 10), const_100)) | the initial price of an article is rs . 2000 which increases 30 % increse in its price in the first year , a 20 % decrease in the second year and a 10 % increase in the next year . what is the final price of the article ? | the initial price of the article , four years age is rs . 2000 in the 1 st year , price of the article = 2000 + 600 = rs . 2600 . in the 2 nd year , price = 2600 - 20 % of 2600 = 2200 - 520 = rs . 1680 . in the 3 rd year , price = 1680 + 10 % of 1680 = 1680 + 168 = rs . 1848 . required price = rs . 1848 answer : c | a = 2000 * 20
b = a / 100
c = 2000 - b
d = 2000 * 20
e = d / 100
f = 2000 - e
g = f * 10
h = g / 100
i = c + h
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a ) 6.25 % , b ) 1.25 % , c ) 4 % , d ) 5 % , e ) 3.25 % | b | multiply(divide(divide(subtract(900, 750), 750), 16), const_100) | at what rate percent on simple interest will rs . 750 amount to rs . 900 in 16 years ? | "150 = ( 750 * 16 * r ) / 100 r = 1.25 % answer : b" | a = 900 - 750
b = a / 750
c = b / 16
d = c * 100
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a ) 11 , b ) 12 , c ) 13 , d ) 14 , e ) 15 | d | divide(subtract(58, 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 58 ? | "four wheeler = 14 * 4 = 56 ( max ) 2 wheel = 1 so no of 4 wheeler = 14 answer : d" | a = 58 - 2
b = a / 4
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a ) 0.125 , b ) 0.375 , c ) 0.325 , d ) 0.5 , e ) 0.666 | b | multiply(power(divide(const_1, const_2), const_3), 3) | if a coin has an equal probability of landing heads up or tails up each time it is flipped , what is the probability that the coin will land tails up exactly once in 3 consecutive flips ? | total number of ways in which h or t can appear in 3 tosses of coin is = 2 * 2 * 2 = 8 ways for 2 t and 1 th thus probability is = p ( htt ) + p ( tth ) + p ( tht ) = 1 / 8 + 1 / 8 + 1 / 8 = 3 / 8 = . 375 answer : b | a = 1 / 2
b = a ** 3
c = b * 3
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a ) 50 min , b ) 60 min , c ) 90 min , d ) 80 min , e ) 120 min | e | multiply(const_10, multiply(const_1, 2)) | a pipe takes a hours to fill the tank . but because of a leakage it took 2 times of its original time . find the time taken by the leakage to empty the tank | "pipe a can do a work 60 min . lets leakage time is x ; then 1 / 60 - 1 / x = 1 / 120 x = 120 min answer : e" | a = 1 * 2
b = 10 * a
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a ) 15 , b ) 9 , c ) 10 , d ) 11 , e ) 35 | b | divide(320, multiply(5, 7)) | if the ratio of two number is 5 : 7 and lcm of the number is 320 then what is the number . | "product of two no = lcm * hcf 5 x * 7 x = 320 * x x = 9 answer : b" | a = 5 * 7
b = 320 / a
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a ) rs . 2000 , b ) rs . 1750 , c ) rs . 2010 , d ) rs . 2005 , e ) none of these | b | divide(2030, add(divide(multiply(divide(add(multiply(3, 5), 3), 5), 5), const_100), const_1)) | find the principle on a certain sum of money at 5 % per annum for 3 1 / 5 years if the amount being rs . 2030 ? | "explanation : 2030 = p [ 1 + ( 5 * 16 / 5 ) / 100 ] p = 1750 answer : option b" | a = 3 * 5
b = a + 3
c = b / 5
d = c * 5
e = d / 100
f = e + 1
g = 2030 / f
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a ) 2 , b ) 3 , c ) 4 , d ) 5 , e ) 60 / 7 | d | divide(7000, subtract(divide(6000, 2.5), subtract(divide(3000, 1), divide(6000, 3)))) | machine a can process 6000 envelopes in 3 hours . machines b and c working together but independently can process the same number of envelopes in 2.5 hours . if machines a and c working together but independently process 3000 envelopes in 1 hour , then how many hours would it take machine b to process 7000 envelopes . | "you can either take the amount of work done as the same as karishma has done or take the work done by each in the same time . i will do the latter 1 . work done in 1 hr by a is 2000 envelopes 2 . work done in 1 hr by a and c is 3000 envelopes 3 . so work done in 1 hr by c is 1000 envelopes 4 . work done in 1 hr by b and c is 2400 envelopes 5 . so work done in 1 hr by b is 1400 envelopes 6 . so to process 7000 envelopes b will take 7000 / 1400 hrs = 5 hrs so the answer is choice d" | a = 6000 / 2
b = 3000 / 1
c = 6000 / 3
d = b - c
e = a - d
f = 7000 / e
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['a ) 1 : 4', 'b ) 8 : 1', 'c ) 4 : 1', 'd ) 3 : 1', 'e ) 1 : 3'] | b | divide(divide(const_1, divide(25, const_100)), divide(const_1, const_2)) | abcd is a square aegf is a rectangle . . such that the rectangle shares 25 % of the area of the suare also ae lies on the line ab and ag lies on segment of ad . if the square shares half the area of the rectangle what is the ratio ae : ag ? | a - - - - - - - - - - - - b - - - - - - - - e | | 50 % | g - - - - - - - - - - - - | - - - - - - - - f | | | 75 % | d - - - - - - - - - - - c remaining square 75 % and remaining rectangle 50 % rectangle share 25 % area of square so , ab x ag = ( ad x ab ) / 4 ad = 4 ag - - - - - - - - - - - - - - - - - - ( 1 ) square share half the area of rectangle so , ag x ab = ( ae x ag ) / 2 ab = ae / 2 - - - - - - - - - - - - - - - - - - ( 2 ) in square all sides are equal i . e . ab = bc = cd = ad so , equation ( 1 ) and ( 2 ) both are equal ae / 2 = 4 ag ae / ag = 8 ae : ag = 8 : 1 answer : b | a = 25 / 100
b = 1 / a
c = 1 / 2
d = b / c
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a ) 420 , b ) 400 , c ) 430 , d ) 410 , e ) 450 | b | divide(subtract(1500, multiply(10, 70)), 2) | 2 cow ’ s and 10 goats are brought for rs . 1500 . if the average price of a goat be rs . 70 . what is the average price of a cow . | "explanation : average price of a goat = rs . 70 total price of 8 goats = 10 * 70 = rs . 700 but total price of 2 cows and 8 goats = rs . 1500 total price of 2 cows is = 1500 - 700 = 800 average price of a cow = 800 / 2 = rs . 400 answer : b" | a = 10 * 70
b = 1500 - a
c = b / 2
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a ) 6 , b ) 7 , c ) 8 , d ) 9 , e ) 10 | d | add(add(5, const_2), const_2) | if x < y < z and y - x > 5 , where x is an even integer and y and z are odd integers , what is the least possible value f of z - x ? | "x < y < z to find the least possible value for z - x ; we need to find the values for z and x that can be closest to each other . if x is some even number , then what could be minimum possible odd z . if x is some even number y - x > 5 ; y > x + 5 ; minimum value for y = x + 5 + 2 = x + 7 [ note : x + 5 is as even + odd = odd and nearest odd greater than x + 5 is x + 5 + 2 ] minimum value for z = y + 2 = x + 7 + 2 = x + 9 [ note : z = y + 2 because both z and y are odd . difference between two odd numbers is 2 ] f = z - x = x + 9 - x = 9 ans : d" | a = 5 + 2
b = a + 2
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a ) 160 , b ) 168 , c ) 191 , d ) 192 , e ) 204 | b | divide(multiply(2, const_360), add(2, 3)) | the ratio of males to females in a class is 2 : 3 . the career preferences of the students in the class are represented in a circle graph . if the area of the graph allocated to each career preference is proportional to the number of students who have that career preference , how many degrees of the circle should be used to represent a career that is preferred by two - thirds of the males and one - third of the females in the class ? | "2 / 3 * 2 / 5 + 1 / 3 * 3 / 5 = 4 / 15 + 3 / 15 = 7 / 15 the number of degrees is 7 / 15 * 360 = 168 degrees the answer is b ." | a = 2 * 360
b = 2 + 3
c = a / b
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a ) $ 752 , b ) $ 796 , c ) $ 765 , d ) $ 773 , e ) $ 775 | b | divide(add(multiply(add(750, 60), 60), multiply(6, 750)), 26) | last year manfred received 26 paychecks . each of his first 6 paychecks was $ 750 ; each of his remaining paychecks was $ 60 more than each of his first 6 paychecks . to the nearest dollar , what was the average ( arithmetic mean ) amount of his pay checks for the year ? | "= ( 750 * 6 + 810 * 20 ) / 26 = 796 answer is b . posted from my mobile device" | a = 750 + 60
b = a * 60
c = 6 * 750
d = b + c
e = d / 26
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a ) 180 , b ) 185 , c ) 190 , d ) 160 , e ) 165 | a | divide(multiply(60, multiply(const_3, 5)), 5) | in the set of positive integers from 1 to 60 , what is the sum of all the odd multiples of 5 ? | 1 - 60 5 - 15 - 25 - 35 - 45 are valid multiples of 5 . add them 5 + 15 + 25 + 35 + 45 + 55 = 180 a | a = 3 * 5
b = 60 * a
c = b / 5
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a ) 65 seconds , b ) 33.33 seconds , c ) 40 seconds , d ) 97 seconds , e ) 26 seconds | b | divide(add(360, 140), divide(multiply(54, const_1000), const_3600)) | a train is 360 meter long is running at a speed of 54 km / hour . in what time will it pass a bridge of 140 meter length ? | "speed = 54 km / hr = 54 * ( 5 / 18 ) m / sec = 15 m / sec total distance = 360 + 140 = 500 meter time = distance / speed = 500 * ( 1 / 15 ) = 33.33 seconds answer : b" | a = 360 + 140
b = 54 * 1000
c = b / 3600
d = a / c
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a ) 18 , b ) 28 , c ) 48 , d ) 38 , e ) 59 | a | divide(multiply(subtract(64, 12), 12), add(12, const_1)) | if a certain number is divided by 12 , the quotient , dividend , and divisor , added together , will amount to 64 . what is the number ? | "let x = the number sought . then x / 12 + x + 12 = 64 . and x - 624 / 13 = 48 ." | a = 64 - 12
b = a * 12
c = 12 + 1
d = b / c
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a ) 20,20 , b ) 20,10 , c ) 25,15 , d ) 35,11 , e ) none of these | d | subtract(add(divide(multiply(24, 5), subtract(5, const_1)), 5), 24) | the ages of two persons differ by 24 years . if 5 years ago , the elder one be 5 times as old as the younger one , their present ages ( in years ) are respectively | "explanation : let their ages be x and ( x + 24 ) years . 5 ( x - 5 ) = ( x + 24 - 5 ) or 4 x = 44 or x = 11 . their present ages are 35 years and 11 years option d" | a = 24 * 5
b = 5 - 1
c = a / b
d = c + 5
e = d - 24
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a ) 0 , b ) 1 , c ) 2 , d ) 4 , e ) 5 | a | divide(5, 5) | what is the remainder when the number t = 14 ^ 2 * 15 ^ 8 is divided by 5 ? | "14 ^ 2 has units digit 6 15 ^ 8 has units digit 5 thus t = 14 ^ 2 * 15 ^ 8 has units digit 0 and will be divisible by 5 . the remainder will be zero answer : ( a )" | a = 5 / 5
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a ) 187888 , b ) 276889 , c ) 267777 , d ) 504000 , e ) 297112 | d | multiply(multiply(multiply(15000, add(const_1, divide(12, const_100))), divide(5, 2)), 12) | the monthly incomes of a and b are in the ratio 5 : 2 . b ' s monthly income is 12 % more than c ' s monthly income . if c ' s monthly income is rs . 15000 , then find the annual income of a ? | "b ' s monthly income = 15000 * 112 / 100 = rs . 16800 b ' s monthly income = 2 parts - - - - > rs . 16800 a ' s monthly income = 5 parts = 5 / 2 * 16800 = rs . 42000 a ' s annual income = rs . 42000 * 12 = rs . 504000 answer : d" | a = 12 / 100
b = 1 + a
c = 15000 * b
d = 5 / 2
e = c * d
f = e * 12
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a ) $ 500 , b ) $ 600 , c ) $ 700 , d ) $ 800 , e ) $ 900 | d | multiply(divide(72, 9), const_100) | if an article is sold at 18 % profit instead of 9 % profit , then the profit would be $ 72 more . what is the cost price ? | "9 % * cost price = $ 72 1 % * cost price = $ 72 / 9 = $ 8 the cost price is $ 800 . the answer is d ." | a = 72 / 9
b = a * 100
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a ) 173 % , b ) 516 % , c ) 461 % , d ) 350 % , e ) 290 % | d | multiply(divide(subtract(3, divide(1, 3)), divide(1, 3)), const_100) | by approximately what percent is x greater than 1 / 3 if ( 2 / 3 ) ( x ) = 1 ? | "what percent is x greater than 1 / 3 if ( 2 / 3 ) ( x ) = 1 ? = > x = 3 / 2 % change = [ ( 3 / 2 - 1 / 3 ) / ( 1 / 3 ) ] * 100 = 350 ans d , 350 %" | a = 1 / 3
b = 3 - a
c = 1 / 3
d = b / c
e = d * 100
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a ) 106 , b ) 107 , c ) 108 , d ) 109 , e ) 110 | c | divide(divide(1500, const_1000), divide(50, const_3600)) | a train 1500 m long can cross an electric pole in 50 sec and then find the speed of the train ? | "length = speed * time speed = l / t s = 1500 / 50 s = 30 m / sec speed = 30 * 18 / 5 ( to convert m / sec in to kmph multiply by 18 / 5 ) speed = 108 kmph answer : c" | a = 1500 / 1000
b = 50 / 3600
c = a / b
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a ) 33 % , b ) 34 % , c ) 35 % , d ) 36 % , e ) 37 % | a | multiply(add(multiply(divide(35, const_100), divide(20, const_100)), multiply(divide(subtract(const_100, 35), const_100), divide(40, const_100))), const_100) | in a certain company 20 % of the men and 40 % of the women attended the annual company picnic . if 35 % of all the employees are men . what % of all the employee went to the picnic ? | total men in company 35 % means total women in company 65 % ( assume total people in company 100 % ) no of men emplyess attended picnic = 35 x ( 20 / 100 ) = 7 no of women empolyees attened picnin = 65 x ( 40 / 100 ) = 26 total percentage of empolyess attened the picnic = 7 + 26 = 33 % answer : a | a = 35 / 100
b = 20 / 100
c = a * b
d = 100 - 35
e = d / 100
f = 40 / 100
g = e * f
h = c + g
i = h * 100
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a ) 34 , b ) 40 , c ) 68 , d ) 88 , e ) 26 | e | add(multiply(divide(60, 20), const_2), 20) | a rectangular field is to be fenced on three sides leaving a side of 20 feet uncovered . if the area of the field is 60 sq . feet , how many feet of fencing will be required ? | "we have : l = 20 ft and lb = 60 sq . ft . so , b = 3 ft . length of fencing = ( l + 2 b ) = ( 20 + 6 ) ft = 26 ft . answer : e" | a = 60 / 20
b = a * 2
c = b + 20
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a ) 1 / 4 , b ) 3 / 8 , c ) 1 / 2 , d ) 5 / 8 , e ) 3 / 4 | d | divide(add(divide(96, 8), divide(96, 2)), 96) | if an integer w is to be chosen at random from the integers 1 to 96 , inclusive , what is the probability that w ( w + 1 ) ( w + 2 ) will be divisible by 8 ? | for w total numbers 8 * 12 there are 12 numbers divisible by 8 - > 3 * 12 ( if 8 is an example - ( 6 , 78 ) , ( 7 , 89 ) , ( 8 , 910 ) ) and 12 numbers divisible by 4 but not divisible by 8 - > 2 * 12 ( if 4 is an example ( 2 , 34 ) and ( 4 , 56 ) ) the answer 5 / 8 - > d | a = 96 / 8
b = 96 / 2
c = a + b
d = c / 96
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a ) 1 / 3 , b ) 2 / 5 , c ) 3 / 10 , d ) 3 / 7 , e ) 1 / 7 | a | divide(divide(24, 3), 24) | tickets numbered from 1 to 24 are mixed and then a ticket is selected randomly . what is the probability that the selected ticket bears a number which is a multiple of 3 ? | here , s = [ 1 , 2 , 3 , 4 , … . , 19 , 20 , 21 , 22 , 23 , 24 ] let e = event of getting a multiple of 3 = [ 3 , 6 , 9 , 12 , 15 , 18 , 21 , 24 ] p ( e ) = n ( e ) / n ( s ) = 8 / 24 = 1 / 3 the answer is a . | a = 24 / 3
b = a / 24
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a ) 15 , b ) 16 , c ) 18 , d ) 20 , e ) none of these | a | add(subtract(multiply(divide(const_10, const_2), 12), divide(add(20, multiply(divide(const_10, const_2), 12)), const_2)), divide(const_10, const_2)) | the difference of two numbers is 20 % of the larger number . if the smaller number is 12 , the larger one is : | "let the larger number be x . then , x - 12 = 20 % of x < > x - ( x / 5 ) = 12 4 x / 5 = 12 < = > x = ( 12 ∗ 5 ) / 5 = 15 . answer : a" | a = 10 / 2
b = a * 12
c = 10 / 2
d = c * 12
e = 20 + d
f = e / 2
g = b - f
h = 10 / 2
i = g + h
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a ) 36 , b ) 45 , c ) 54 , d ) 63 , e ) 72 | d | multiply(divide(7, subtract(9, 7)), 14) | sandy is younger than molly by 14 years . if their ages are in the respective ratio of 7 : 9 , how old is molly ? | "s = m - 14 s / m = 7 / 9 9 s = 7 m 9 ( m - 14 ) = 7 m m = 63 the answer is d ." | a = 9 - 7
b = 7 / a
c = b * 14
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a ) $ 150 , b ) $ 70 , c ) $ 200 , d ) $ 171.6 , e ) $ 190 | b | floor(multiply(18, 8.75)) | carrie likes to buy t - shirts at the local clothing store . they cost $ 8.75 each . one day , she bought 18 t - shirts . how much money did she spend ? | $ 8.75 * 18 = $ 70 . answer is b . | a = 18 * 8
b = math.floor(a)
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a ) 8 % , b ) 15 % , c ) 46 % , d ) 52 % , e ) 56 % | c | multiply(divide(3, 20), const_100) | a pharmaceutical company received $ 3 million in royalties on the first $ 20 million in sales of and then $ 8 million in royalties on the next $ 108 million in sales . by approximately what percentage did the ratio of royalties to sales decrease from the first $ 20 million in sales to the next $ 108 million in sales ? | "( 8 / 108 ) / ( 3 / 20 ) = 30 / 54 = 49,3 % it means that 8 / 108 represents only 49,3 % . therefore a decrease of 46 % . answer c" | a = 3 / 20
b = a * 100
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a ) 33 , b ) 54 , c ) 18 , d ) 17 , e ) 72 | e | multiply(subtract(70, 14), divide(90, 70)) | a group of 55 adults and 70 children go for trekking . if there is meal for either 70 adults or 90 children and if 14 adults have their meal , find the total number of children that can be catered with the remaining food . | "explanation : as there is meal for 70 adults and 14 have their meal , the meal left can be catered to 56 adults . now , 70 adults = 90 children 7 adults = 9 children therefore , 56 adults = 72 children hence , the meal can be catered to 72 children . answer : e" | a = 70 - 14
b = 90 / 70
c = a * b
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a ) 15 / 2 , b ) 9 / 4 , c ) 5 / 11 , d ) 7 / 5 , e ) 9 / 7 | c | add(subtract(1, divide(2, 3)), subtract(divide(2, 3), divide(1, 4))) | a batch of cookies was divided amomg 2 tins : 2 / 3 of all the cookies were placed in either the blue or the green tin , and the rest were placed in the red tin . if 1 / 4 of all the cookies were placed in the blue tin , what fraction of the cookies that were placed in the other tins were placed in the green tin | "this will help reduce the number of variables you have to deal with : g + b = 2 / 3 r = 1 / 2 b = 1 / 4 we can solve for g which is 5 / 12 what fraction ( let it equal x ) of the cookies that were placed in the other tins were placed in the green tin ? so . . x * ( g + r ) = g x * ( 5 / 12 + 1 / 2 ) = 5 / 12 x = 5 / 11 answer : c . 5 / 11" | a = 2 / 3
b = 1 - a
c = 2 / 3
d = 1 / 4
e = c - d
f = b + e
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a ) 10 % , b ) 20 % , c ) 25 % , d ) 36 % , e ) 50 % | d | multiply(divide(subtract(multiply(const_100, const_100), multiply(subtract(const_100, 60), add(const_100, 60))), multiply(const_100, const_100)), const_100) | robert ' s salary was decreased by 60 % and subsequently increased by 60 % . how much percentage does he lose ? | let original salary be $ 100 salary after decreasing 60 % = 100 - 100 x 60 / 100 = $ 40 salary after increasing 60 % on $ 40 = 40 + 40 x 60 / 100 = $ 64 percentage of loss = 100 - 64 = 36 % answer : d | a = 100 * 100
b = 100 - 60
c = 100 + 60
d = b * c
e = a - d
f = 100 * 100
g = e / f
h = g * 100
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a ) 2000 , b ) 2400 , c ) 2800 , d ) 3200 , e ) 3600 | a | divide(640, subtract(subtract(const_1, divide(34, const_100)), divide(34, const_100))) | a candidate got 34 % of the votes polled and he lost to his rival by 640 votes . how many votes were cast ? | "let x be the total number of votes . 0.34 x + 640 = 0.66 x 0.32 x = 640 x = 640 / 0.32 = 2000 the answer is a ." | a = 34 / 100
b = 1 - a
c = 34 / 100
d = b - c
e = 640 / d
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a ) 44 , b ) 5 , c ) 10 , d ) 15 , e ) 20 | a | power(add(sqrt(11), sqrt(11)), const_2) | if x ¤ y = ( x + y ) ^ 2 - ( x - y ) ^ 2 . then √ 11 ¤ √ 11 = | x = √ 11 and y also = √ 11 applying the function ( √ 11 + √ 11 ) ^ 2 - ( √ 11 - √ 11 ) ^ 2 = ( 2 √ 11 ) ^ 2 - 0 = 4 x 11 = 44 . note : alternative approach is the entire function is represented as x ^ 2 - y ^ 2 = ( x + y ) ( x - y ) which can be simplified as ( x + y + x - y ) ( x + y - ( x - y ) ) = ( 2 x ) ( 2 y ) = 4 xy . substituting x = √ 11 and y = √ 11 you get the answer 44 . answer a | a = math.sqrt(11)
b = math.sqrt(11)
c = a + b
d = c ** 2
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a ) 80 , b ) 160 , c ) 250 , d ) 350 , e ) 480 | c | add(divide(200, 4), 200) | of the people who responded to a market survey , 200 preferred brand x and the rest preferred brand y . if the respondents indicated a preference for brand x over brand y by ratio of 4 to 1 , how many people responded to the survey ? | "ratio = 4 : 1 = > 4 x respondents preferred brand x and x preferred brand y since , no . of respondents who preferred brand x = 200 = > 4 x = 200 = > x = 50 hence total no . of respondents = 200 + 50 = 250 hence c is the answer ." | a = 200 / 4
b = a + 200
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a ) 272258 , b ) 272358 , c ) 294690 , d ) 274258 , e ) 274358 | c | multiply(divide(5358, 55), const_100) | 5358 x 55 = ? | "5358 x 51 = 5358 x ( 50 + 5 ) = 5358 x 50 + 5358 x 5 = 267900 + 26790 = 294690 . c )" | a = 5358 / 55
b = a * 100
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a ) 1.5 , b ) 3.0 , c ) 3.9 , d ) 4.5 , e ) 6.0 | d | subtract(multiply(9.2, 5), multiply(7.4, 5)) | the average of 5 numbers is 7.4 . if one of the numbers is multiplied by a factor of 3 , the average of the numbers increases to 9.2 . what number is multiplied by 3 ? | "the average of 5 numbers is 7.4 the sum of 5 numbers will be 7.4 x 5 = 37 the average of 5 number after one of the number is multiplied by 3 is 9.2 the sum of the numbers will now be 9.2 x 5 = 46 so the sum has increased by 46 - 37 = 9 let the number multiplied by 3 be n then , 3 n = n + 9 or 2 n = 9 or n = 4.5 answer : - d" | a = 9 * 2
b = 7 * 4
c = a - b
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a ) 3800 , b ) 4500 , c ) 5200 , d ) 3400 , e ) 4200 | e | multiply(120, 35) | the average marks obtained by 120 candidates in a certain examination is 35 . find the total marks . | "following the above formula , we have the total marks = 120 * 35 = 4200 answer is e" | a = 120 * 35
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a ) a ) 35 , b ) b ) 34 , c ) c ) 50 , d ) d ) 67 , e ) e ) 100 | a | divide(divide(multiply(280, 6), 12), const_4) | according to the directions on the can of frozen orange juice concentrate , 1 can of concentrate is to be mixed with 3 cans of water to make orange juice . how many 12 ounces cans of the concentrate are required to prepare 280 6 ounces servings of orange juice ? | "its a . total juice rquired = 280 * 6 = 1680 ounce 12 ounce concentate makes = 12 * 4 = 48 ounce juice total cans required = 1680 / 48 = 35 . answer a" | a = 280 * 6
b = a / 12
c = b / 4
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a ) 5 % , b ) 10 % , c ) 15 % , d ) 20 % , e ) it can not be determined | a | subtract(add(60, 45), const_100) | a box contains either blue or red flags . the total number of flags in the box is an even number . a group of children are asked to pick up two flags each . if all the flags are used up in the process such that 60 % of the children have blue flags , and 45 % have red flags , what percentage of children have flags of both the colors ? | "solution : let the total number of flags be 100 ( even number ) let the total number of ' blue ' flags alone be ' a ' let the total number of ' red ' flags alone be ' b ' let the total number of ' both ' flags be ' c ' we have given , total number of blue flags = 60 % = 60 = a + c total number of red flags = 45 % = 45 = b + c total number of flags = a + b + c = 100 ( since all the flag have been utilized ) so , substituting for c in the third equation , we have , 60 - c + c + 45 - c = 100 c = 5 option a ." | a = 60 + 45
b = a - 100
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a ) 300 , b ) 200 , c ) 100 , d ) 400 , e ) 500 | c | subtract(divide(1200, const_2), 500) | if the perimeter of a rectangular playground is 1200 m , its length when its breadth is 500 m is ? | 2 ( l + 500 ) = 1200 = > l = 100 m answer : c | a = 1200 / 2
b = a - 500
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a ) 8 % , b ) 9 % , c ) 11 % , d ) 12.5 % , e ) 13.03 % | e | multiply(divide(subtract(multiply(divide(2, 15), 115), 18), multiply(divide(2, 15), 115)), const_100) | a doctor prescribed 18 cubic centimeters of a certain drug to a patient whose body weight was 115 pounds . if the typical dosage is 2 cubic centimeters per 15 pounds of the body weight , by what percent was the prescribed dosage greater than the typical dosage ? | "typical dosage is dose : weight : : 2 : 15 . now if weight is 115 : ( 115 / 15 ) ) then typical dosage would be 2 * 7.67 = 15.33 cc . dosage = 18 cc . dosage is greater by 2 cc . % dosage is greater : ( 2 / 15.33 ) * 100 = 13.03 % e is the answer ." | a = 2 / 15
b = a * 115
c = b - 18
d = 2 / 15
e = d * 115
f = c / e
g = f * 100
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a ) 190 , b ) 255 , c ) 200 , d ) 205 , e ) 210 | b | divide(divide(multiply(add(10, 500), add(divide(subtract(500, 10), 10), const_1)), const_2), add(divide(subtract(500, 10), 10), const_1)) | what is the average ( arithmetic mean ) of all multiples of 10 from 10 to 500 inclusive ? | this question can be solved with the average formula and ' bunching . ' we ' re asked for the average of all of the multiples of 10 from 10 to 500 , inclusive . to start , we can figure out the total number of terms rather easily : 1 ( 10 ) = 10 2 ( 10 ) = 20 . . . 50 ( 10 ) = 500 so we know that there are 50 total numbers . we can now figure out the sum of those numbers with ' bunching ' : 10 + 500 = 510 20 + 490 = 510 30 + 480 = 510 etc . since there are 50 total terms , this pattern will create 25 ' pairs ' of 510 . thus , since the average = ( sum of terms ) / ( number of terms ) , we have . . . ( 25 ) ( 510 ) / ( 50 ) = 255 final answer : b | a = 10 + 500
b = 500 - 10
c = b / 10
d = c + 1
e = a * d
f = e / 2
g = 500 - 10
h = g / 10
i = h + 1
j = f / i
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a ) 4 , b ) 7 , c ) 8 , d ) 10 , e ) none of these | d | subtract(divide(64, 64), const_1) | 64 ã — 64 ã — 64 ã — 64 x 64 = 8 ^ ? | "64 ã — 64 ã — 64 ã — 64 x 64 = 8 ^ ? or , 8 ( 2 ) ã — 8 ( 2 ) ã — 8 ( 2 ) ã — 8 ( 2 ) x 8 ( 2 ) = 8 ? or 7 ( 10 ) = 8 ? or , ? = 10 answer d" | a = 64 / 64
b = a - 1
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a ) 3 , b ) 6 , c ) 8 , d ) 9 , e ) 7 | e | divide(14, subtract(4, 2)) | a person can swim in still water at 4 km / h . if the speed of water 2 km / h , how many hours will the man take to swim back against the current for 14 km ? | m = 4 s = 2 us = 4 - 2 = 2 d = 14 t = 14 / 2 = 7 answer : e | a = 4 - 2
b = 14 / a
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a ) 1500 , b ) 2677 , c ) 1997 , d ) 2677 , e ) 1971 | a | multiply(multiply(divide(15, multiply(10, 2)), const_100), multiply(10, 2)) | find the sum the difference between the compound and s . i . on a certain sum of money for 2 years at 10 % per annum is rs . 15 of money ? | p = 15 ( 100 / 10 ) 2 = > p = 1500 answer : a | a = 10 * 2
b = 15 / a
c = b * 100
d = 10 * 2
e = c * d
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a ) 1 / 4 , b ) 1 / 3 , c ) 3 / 8 , d ) 2 / 3 , e ) 1 / 2 | b | divide(add(divide(const_1, const_2), divide(const_2, const_4)), add(add(subtract(const_2, divide(const_2, const_4)), subtract(const_1, divide(const_1, const_2))), add(divide(const_1, const_2), divide(const_2, const_4)))) | one bottle is half - full of oil and another bottle with twice the capacity is one quarter full of oil . if water is added so that both the bottles are full and the contents of both are then poured into a third bottle that is empty and large enough to hold the contents of both , what fractions of the contents in the third bottle is oil ? | let assume that 1 st bottle capacity = 1 litre ; therefore , it contain 1 / 2 litre oil and 1 / 2 litre water . ( i . e ) water = 1 / 2 litre oil = 1 / 2 litre 2 nd bottle is twice the capacity i . e . 2 litre capacity it means , 1 / 2 litre filled with oil and 1 ( 1 / 2 ) litre filled with water . i . e . oil = 1 / 2 liter water = 1 ( 1 / 2 ) litre in total the third bottle will contain , 1 litre oil and 2 litre water . i . e . , oil = 1 litre water = 2 liter therefore , oil proportion is 1 / 3 answer : b | a = 1 / 2
b = 2 / 4
c = a + b
d = 2 / 4
e = 2 - d
f = 1 / 2
g = 1 - f
h = e + g
i = 1 / 2
j = 2 / 4
k = i + j
l = h + k
m = c / l
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a ) 1 / 190 , b ) 1 / 23 , c ) 1 / 19 , d ) 1 / 10 , e ) 1 / 9 | b | divide(const_1, subtract(24, const_1)) | a box contains 12 pairs of shoes ( 24 shoes in total ) . if two shoes are selected at random , what it is the probability that they are matching shoes ? | "the problem with your solution is that we do n ' t choose 1 shoe from 24 , but rather choose the needed one after we just took one and need the second to be the pair of it . so , the probability would simply be : 1 / 1 * 1 / 23 ( as after taking one at random there are 23 shoes left and only one is the pair of the first one ) = 1 / 23 answer : b" | a = 24 - 1
b = 1 / a
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a ) 26 , b ) 39 , c ) 120 , d ) 65 , e ) 156 | c | multiply(multiply(multiply(power(2, 2), 3), divide(10, 2)), 2) | if 2 ^ 5 , 3 ^ 3 , and 10 ^ 2 are all factors of the product of 936 and w where w is a positive integer , what is the smallest possible value of w ? | "here 156 has three two ' s two three ' s and one 10 rest of them must be in w so w = 10 * 3 * 4 = 120 smash c" | a = 2 ** 2
b = a * 3
c = 10 / 2
d = b * c
e = d * 2
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a ) 15.7 sec , b ) 15.1 sec , c ) 15.5 sec , d ) 17.1 sec , e ) 16.7 sec | c | divide(add(130, 150), multiply(65, const_0_2778)) | how long does a train 130 m long running at the speed of 65 km / hr takes to cross a bridge 150 m length ? | "speed = 65 * 5 / 18 = 18.1 m / sec total distance covered = 130 + 150 = 280 m . required time = 280 / 18.1 ' = 15.5 sec . answer : c" | a = 130 + 150
b = 65 * const_0_2778
c = a / b
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a ) 36 days . , b ) 17 days . , c ) 18 days . , d ) 19 days . , e ) 20 days . | a | subtract(60, multiply(divide(60, 10), 4)) | arun and tarun can do a work in 10 days . after 4 days tarun went to his village . how many days are required to complete the remaining work by arun alone . arun can do the work alone in 60 days . | "they together completed 4 / 10 work in 4 days . balance 6 / 10 work will be completed by arun alone in 60 * 6 / 10 = 36 days . answer : a" | a = 60 / 10
b = a * 4
c = 60 - b
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a ) 17 , b ) 19 , c ) 21 , d ) 23 , e ) 25 | d | add(divide(subtract(95, add(add(add(2, add(2, 2)), add(add(2, 2), 2)), add(add(add(2, 2), 2), 2))), 5), add(add(add(2, 2), 2), 2)) | in a school with 5 classes , each class has 2 students less then the previous class . how many students are there in the largest class if the total number of students at school is 95 ? | "total classes = 5 total students = 95 . average = 19 if the classes had average number of students : 19 19 19 19 19 given case = 23 21 19 17 15 hence number of students in the largest class = 23 correct option : d" | a = 2 + 2
b = 2 + a
c = 2 + 2
d = c + 2
e = b + d
f = 2 + 2
g = f + 2
h = g + 2
i = e + h
j = 95 - i
k = j / 5
l = 2 + 2
m = l + 2
n = m + 2
o = k + n
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a ) 12 , b ) 14 , c ) 16 , d ) 18 , e ) 20 | c | add(add(choose(6, const_1), choose(6, const_1)), choose(const_4, const_1)) | jane and thomas are among the 6 people from which a committee of 3 people is to be selected . how many different possible committees of 3 people can be selected from these 6 people if at least one of either jane or thomas is to be selected ? | the total number of ways to choose 3 people from 6 is 6 c 3 = 20 . the number of committees without jane or thomas is 4 c 3 = 4 . there are 20 - 4 = 16 possible committees which include jane and / or thomas . the answer is c . | a = math.comb(6, 1)
b = math.comb(6, 1)
c = a + b
d = math.comb(4, 1)
e = c + d
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a ) 50 , b ) 200 , c ) 380 , d ) 398 , e ) 400 | d | multiply(inverse(10), multiply(multiply(const_100, 10), add(const_4, const_4))) | when 1 / 20 % of 4,000 is subtracted from 1 / 10 of 4,000 , the difference is | "( 1 / 10 ) * 4000 - ( 1 / 20 * 100 ) * 4000 = 400 - 2 = 398 answer d" | a = 1/(10)
b = 100 * 10
c = 4 + 4
d = b * c
e = a * d
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a ) 85 , b ) 86 , c ) 88 , d ) 90 , e ) 92 | c | add(add(60, 4), add(4, 1)) | the average weight of a class is x pounds . when a new student weighing 60 pounds joins the class , the average decreases by 1 pound . in a few months the student ’ s weight increases to 110 pounds and the average weight of the class becomes x + 4 pounds . none of the other students ’ weights changed . what is the value of x ? | "when the student weighs 80 pounds the average weight is x - 1 pounds ; when the student weighs 110 pounds the average weight is x + 4 pounds . so , the increase in total weight of 110 - 80 = 30 pounds corresponds to the increase in average weight of ( x + 4 ) - ( x - 1 ) = 5 pounds , which means that there are 30 / 5 = 6 students ( including the new one ) . so , initially there were 5 student . total weight = 5 x + 80 = 6 ( x - 1 ) - - > x = 88 pounds . answer : c ." | a = 60 + 4
b = 4 + 1
c = a + b
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a ) 250 m , b ) 112 m , c ) 117 m , d ) 125 m , e ) 123 m | a | multiply(multiply(180, const_0_2778), 5) | if a train , travelling at a speed of 180 kmph , crosses a pole in 5 sec , then the length of train is ? | "a a = 180 * 5 / 18 * 5 = 250 m" | a = 180 * const_0_2778
b = a * 5
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a ) 3 , b ) 4 , c ) 10 , d ) 6 , e ) 7 | c | subtract(subtract(17, 6), const_1) | if 6 < x < 10 < y < 17 , then what is the greatest possible positive integer difference of x and y ? | "let x = 6.1 and y = 16.1 greatest possible difference = 16.1 - 6.1 = 10 answer c" | a = 17 - 6
b = a - 1
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['a ) 4 cm', 'b ) 6 cm', 'c ) 8 cm', 'd ) 10 cm', 'e ) 12 cm'] | c | multiply(divide(divide(divide(divide(multiply(divide(volume_cylinder(divide(16, const_2), 12), const_pi), const_3), const_4), 9), const_4), const_4), const_2) | 9 spheres of the same size are made from melting a solid cylinder of 16 cm diameter and 12 cm height . what is the diameter of each sphere ? | volume of cylinder = pi * r ^ 2 * h volume of a sphere = 4 * pi * r ^ 3 / 3 9 * 4 * pi * r ^ 3 / 3 = pi * r ^ 2 * h r ^ 3 = r ^ 2 * h / 12 = 64 cm ^ 3 r = 4 cm d = 8 cm the answer is c . | a = 16 / 2
b = volume_cylinder / (
c = b * math.pi
d = c / 3
e = d / 4
f = e / 9
g = f / 4
h = g * 4
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a ) 2 % , b ) 6 % , c ) 14 % , d ) 28 % , e ) 63 % | b | floor(multiply(subtract(divide(9, 63), divide(7, 90)), const_100)) | a survey was sent to 90 customers , 7 of whom responded . then the survey was redesigned and sent to another 63 customers , 9 of whom responded . by approximately what percent did the response rate increase from the original survey to the redesigned survey ? | "case 1 : ( 7 / 90 ) = x / 100 x = 8 % case 2 : ( 9 / 63 ) = y / 100 y = 14 % so percent increase is = ( y - x ) = ( 14 - 8 ) % = 6 % answer is b" | a = 9 / 63
b = 7 / 90
c = a - b
d = c * 100
e = math.floor(d)
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a ) 41 km , b ) 76 km , c ) 25 km , d ) 15 km , e ) 30 km | a | divide(add(add(29, multiply(2, 10)), 29), 2) | a car started running at a speed of 29 km / hr and the speed of the car was increased by 2 km / hr at the end of every hour . find the total distance covered by the car in the first 10 hours of the journey . | "a 41 km the total distance covered by the car in the first 10 hours = 32 + 34 + 36 + 38 + 40 + 42 + 44 + 46 + 48 + 50 = sum of 10 terms in ap whose first term is 32 and last term is 50 = 10 / 2 [ 32 + 50 ] = 410 km ." | a = 2 * 10
b = 29 + a
c = b + 29
d = c / 2
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a ) 50 , b ) 99 , c ) 88 , d ) 77 , e ) 23 | d | divide(50, const_2) | q is as much younger than r as he is older than t . if the sum of the ages of r and t is 50 years . what is definitely the difference between r and q ' s age ? | "r - q = r - t q = t . also r + t = 50 ; r + q = 50 so , ( r - q ) can not be determined . answer : d" | a = 50 / 2
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a ) 30 min , b ) 3 min , c ) 1 hr , d ) 1 min , e ) 2 min | d | divide(60, 60) | a deer passed a certain tree at a constant speed of 50 miles per hour while being chased by a cheetah . then , 2 minutes later , the cheetah passed the same tree at a constant speed of 60 miles per hour . if both animals maintained their speeds , how long after the cheetah passed the tree did the cheetah catch up with the deer ? | when the cheetah is at the tree , the deer is 50 / 30 miles ahead on the highway . ( the distance covered in 2 min ) every hour , cheetah runs 10 miles more than the deer . how many hours will it takes it to cover 50 / 30 miles more ? the answer is ( 50 / 30 ) / 10 = 1 / 60 = 1 min answer d | a = 60 / 60
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a ) 199 , b ) 249 , c ) 233 , d ) 215 , e ) 229 | e | divide(add(add(multiply(multiply(const_4, const_4), const_1000), multiply(1, const_100)), multiply(add(150, 1), 250)), 250) | a computer manufacturer produces a certain electronic component at a cost of $ 150 per component . shipping costs for delivering the components are $ 1 per unit . further , the manufacturer has costs of $ 19,500 a month related to the electronic component regardless of how many it produces . if the manufacturer produces and sells 250 components a month , what is the lowest price it can sell them for such that the costs do n ' t exceed the revenues ? | "$ 19500 is a fixed cost each component is $ 151 ( $ 150 to produce , $ 1 to ship ) manufacturer will be producing and selling 250 components so therefore the equation to find price would be 250 * p = 19500 + ( 250 * 150 ) + ( 250 * 1 ) p = ( 19500 + 37500 + 250 ) / 250 p = 229 answer : e" | a = 4 * 4
b = a * 1000
c = 1 * 100
d = b + c
e = 150 + 1
f = e * 250
g = d + f
h = g / 250
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['a ) 4 tiles', 'b ) 5 tiles', 'c ) 6 tiles', 'd ) 7 tiles', 'e ) 8 tiles'] | c | divide(24, const_4) | the minimum number of tiles of size 16 by 24 required to form a square by placing them adjacent to one another is | lcm of 16,24 = 48 48 * 48 is the minimum size of square made with 16 by 24 tiles number of tiles required = area of square / area of one tile = 48 * 48 / ( 16 * 24 ) = 6 tiles answer : c | a = 24 / 4
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a ) $ 13 , b ) $ 15 , c ) $ 17 , d ) $ 19 , e ) $ 21 | d | divide(add(1380, 900), add(65, 55)) | sandy bought 65 books for $ 1380 from one shop and 55 books for $ 900 from another shop . what is the average price that sandy paid per book ? | average price per book = ( 1380 + 900 ) / ( 65 + 55 ) = 2280 / 120 = $ 19 the answer is d . | a = 1380 + 900
b = 65 + 55
c = a / b
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a ) 3.45 , b ) 4.5 , c ) 2.25 , d ) 3.21 , e ) none | a | divide(multiply(multiply(23, 3), 5), const_100) | the simple interest on rs . 23 for 3 months at the rate of 5 paise per rupeeper month is | sol . s . i . = rs . [ 23 * 5 / 100 * 3 ] = rs . 3.45 answer a | a = 23 * 3
b = a * 5
c = b / 100
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a ) 1 / 5 , b ) 1 / 3 , c ) 2 / 5 , d ) 9 / 17 , e ) 9 / 23 | d | divide(subtract(60, 15), subtract(const_100, 15)) | the weight of a glass of jar is 15 % of the weight of the jar filled with coffee beans . after some of the beans have been removed , the weight of the jar and the remaining beans is 60 % of the original total weight . what fraction part of the beans remain in the jar ? | "let weight of jar filled with beans = 100 g weight of jar = 15 g weight of coffee beans = 85 g weight of jar and remaining beans = 60 g weight of remaining beans = 45 g fraction remaining = 45 / 85 = 9 / 17 answer is d ." | a = 60 - 15
b = 100 - 15
c = a / b
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a ) 100 , b ) 10 , c ) 11 , d ) 12 , e ) 14 | a | add(divide(subtract(multiply(floor(divide(1000, 9)), 9), multiply(add(floor(divide(10, 9)), const_1), 9)), 9), const_1) | how many numbers from 10 to 1000 are exactly divisible by 9 ? | "10 / 9 = 1 and 1000 / 9 = 111 = = > 111 - 1 = 100 . therefore 100 answer : a" | a = 1000 / 9
b = math.floor(a)
c = b * 9
d = 10 / 9
e = math.floor(d)
f = e + 1
g = f * 9
h = c - g
i = h / 9
j = i + 1
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a ) 20 , b ) 36 , c ) 48 , d ) 38 , e ) 59 | d | divide(subtract(multiply(add(25, 5), 48), multiply(25, 50)), 5) | in a factory , an average of 50 tv ' s are produced per day for the fist 25 days of the months . a few workers fell ill for the next 5 days reducing the daily avg for the month to 48 sets / day . the average production per day for day last 5 days is ? | "production during these 5 days = total production in a month - production in first 25 days . = 30 x 48 - 25 x 50 = 190 ∴ average for last 5 days = 190 / 5 = 38 d" | a = 25 + 5
b = a * 48
c = 25 * 50
d = b - c
e = d / 5
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a ) 300 , b ) 600 , c ) 550 , d ) 1000 , e ) 900 | c | divide(add(200, 20), divide(40, const_100)) | pradeep has to obtain 40 % of the total marks to pass . he got 200 marks and failed by 20 marks . the maximum marks are | "explanation : let their maximum marks be x . then , 40 % of x = 200 + 20 = > 40 / 100 x = 220 x = ( 22000 / 40 ) x = 550 . answer : c" | a = 200 + 20
b = 40 / 100
c = a / b
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a ) 25 , b ) 8 , c ) 27 , d ) 29 , e ) 39 | b | add(subtract(56, multiply(17, 3)), 3) | a batsman makes a score of 56 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 + 56 = total runs scored in 17 innings = > 16 ( x - 3 ) + 56 = 17 x = > 16 x - 48 + 56 = 17 x = > x = 8 answer is b ." | a = 17 * 3
b = 56 - a
c = b + 3
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a ) 1 , b ) 4 / 3 , c ) 17 / 5 , d ) 11 / 9 , e ) 4 | d | divide(add(divide(subtract(multiply(6, 2), 5), subtract(multiply(2, 2), const_1)), subtract(6, multiply(2, divide(subtract(multiply(6, 2), 5), subtract(multiply(2, 2), const_1))))), 3) | if 2 x + y = 6 and x + 2 y = 5 , then ( x + y ) / 3 = | we have two equations : 2 x + y = 6 x + 2 y = 5 notice that something nice happens when we add them . we get : 3 x + 3 y = 11 divide both sides by 3 to get : x + y = 11 / 3 so , ( x + y ) / 3 = 11 / 9 answer : d | a = 6 * 2
b = a - 5
c = 2 * 2
d = c - 1
e = b / d
f = 6 * 2
g = f - 5
h = 2 * 2
i = h - 1
j = g / i
k = 2 * j
l = 6 - k
m = e + l
n = m / 3
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a ) 295 , b ) 324 , c ) 385 , d ) 391 , e ) 399 | c | power(12, negate(22)) | what is the value of ( 12 + 22 + 32 + 42 + - - - - - + 102 ) | "explanation : ( 12 + 22 + … . . + n 2 ) = ( 1 / 6 ) n ( n + 1 ) ( 2 n + 1 ) here , n = 10 therefore , ( 12 + 22 + … . . + 102 ) = ( 1 / 6 ) 10 ( 10 + 1 ) ( 2 x 10 + 1 ) = ( 1 / 6 ) x 10 x 11 x 21 = 385 answer : c" | a = 12 ** negate
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a ) 0 , b ) 1 , c ) 2 , d ) 3 , e ) 4 | e | subtract(subtract(multiply(15, const_2), 1), multiply(5, 5)) | if m and n are positive integers and m = 15 n - 1 , what is the remainder when m is divided by 5 ? | this question asks what is . . . ( the answer ) , so we know that the answer will be consistent . as such , we can test values to quickly get the solution . we ' re told that m and n are positive integers and m = 15 n - 1 . we ' re asked for the remainder when m is divided by 5 . if . . . . n = 1 m = 14 14 / 5 = 2 remainder 4 final answer : e | a = 15 * 2
b = a - 1
c = 5 * 5
d = b - c
|
a ) 4.6 % , b ) 3.6 % , c ) 6.4 % , d ) 7.8 % , e ) 8.9 % | c | divide(subtract(subtract(500, multiply(divide(4000, const_100), 5)), multiply(divide(3500, const_100), 4)), divide(subtract(subtract(multiply(const_100, const_100), 4000), 3500), const_100)) | a man has $ 10,000 to invest . he invests $ 4000 at 5 % and $ 3500 at 4 % . in order to have a yearly income of $ 500 , he must invest the remainder at : | "c 6.4 % income from $ 4000 at 5 % in one year = $ 4000 of 5 % . = $ 4000 × 5 / 100 . = $ 4000 × 0.05 . = $ 200 . income from $ 3500 at 4 % in one year = $ 3500 of 4 % . = $ 3500 × 4 / 100 . = $ 3500 × 0.04 . = $ 140 . total income from 4000 at 5 % and 3500 at 4 % = $ 200 + $ 140 = $ 340 . remaining income amount in order to have a yearly income of $ 500 = $ 500 - $ 340 . = $ 160 . total invested amount = $ 4000 + $ 3500 = $ 7500 . remaining invest amount = $ 10000 - $ 7500 = $ 2500 . we know that , interest = principal × rate × time interest = $ 160 , principal = $ 2500 , rate = r [ we need to find the value of r ] , time = 1 year . 160 = 2500 × r × 1 . 160 = 2500 r 160 / 2500 = 2500 r / 2500 [ divide both sides by 2500 ] 0.064 = r r = 0.064 change it to a percent by moving the decimal to the right two places r = 6.4 % therefore , he invested the remaining amount $ 2500 at 6.4 % in order to get $ 500 income every year ." | a = 4000 / 100
b = a * 5
c = 500 - b
d = 3500 / 100
e = d * 4
f = c - e
g = 100 * 100
h = g - 4000
i = h - 3500
j = i / 100
k = f / j
|
a ) 350 , b ) 400 , c ) 500 , d ) 550 , e ) 590 | c | divide(multiply(multiply(subtract(3.25, 3), const_1000), const_100), 50) | workers decided to raise rs . 3 lacs by equal contribution from each . had they contributed rs . 50 eachextra , the contribution would have been rs . 3.25 lacs . how many workers were they ? | "n * 50 = ( 325000 - 300000 ) = 25000 n = 25000 / 50 = 500 c" | a = 3 - 25
b = a * 1000
c = b * 100
d = c / 50
|
a ) 1 / 2 , b ) 1 , c ) 2 , d ) 5 / 2 , e ) 4 | b | multiply(2, divide(1, 2)) | in the xy - coordinate system , if ( m , n ) and ( m 1 2 , n 1 k ) are two points on the line with the equation x 5 2 y 1 5 , then k 5 | step 1 : analyze the question for any question involving the equation of a line , a good place to start is the slope - intercept form of the line , y = mx 1 b . remember that if you have two points on a line , you can derive the entire equation , and if you have an equation of the line , you can calculate any points on that line . step 2 : state the task we are solving for k , which is the amount by which the y - coordinate increases when the x - coordinate increases by 2 . step 3 : approach strategically the slope of a line is the ratio between the change in y and the change in x . in other words , every time the x - coordinate increases by 1 , the y - coordinate increases by the amount of the slope . the equation of the line in the question stem is defined as x = 2 y + 5 . we must isolate y to have slope - intercept form : so the slope of this line is 1 / 2 . this means that for every change of + 1 in the x direction , there is a change of + 1 / 2 in the y direction . then we know that , because there is an increase in 2 units in the x direction when moving from m to m + 2 , there must be a change of 1 unit in the y direction when moving from n to n + k . so k = 1 . since there are variables that eventually cancel ( m and n are not part of the answers ) , we can pick numbers . let ’ s say that you choose the y - coordinate of the point ( m , n ) to be 0 to allow for easier calculations . using the equation we ’ re given to relate x - and y - coordinates , we can calculate the x - coordinate : so ( m , n ) is the point ( 5 , 0 ) . now we ’ ll plug our values of m and n into the next point : ( m + 2 , n + k ) . that yields ( 7 , k ) . all we have to do is plug an x - coordinate of 7 into the equation to solve for k , the y - coordinate : answer is b | a = 1 / 2
b = 2 * a
|
a ) - 4 , b ) - 1 / 4 , c ) 0 , d ) 1 / 4 , e ) 4 | d | divide(const_1, 4) | if 625 ^ ( - x ) + 25 ^ ( - 2 x ) + 5 ^ ( - 4 x ) = 11 , what is the value of x ? | we ' re told that 625 ^ ( - x ) + 25 ^ ( - 2 x ) + 5 ^ ( - 4 x ) = 15 . we ' re asked for the value of x . since each of the calculated terms must be positive ( regardless of what the exponent is ) , we can use thebasesto our advantage . . . . . with answer a , we ' d have 625 ^ 4 , which is much bigger than 15 ( and we ' d be adding to that big number ) . eliminate a . with answer e , we ' d have 625 ^ ( - 4 ) , which would create a tiny fraction ( and we ' d add some other fractions to it , so the total would be much too small ) . eliminate e . with answer d , we ' d have 625 ^ ( - 1 / 4 ) , which will also be a fraction ( just not as tiny as the one in answer e ) , but the total would still be too small . eliminate d . with answer c , anything to the ' 0 power ' is 1 , so we ' d have 1 + 1 + 1 = 3 . this is not 15 . eliminate c . d | a = 1 / 4
|
a ) 76 , b ) 5776 , c ) 304 , d ) 2704 , e ) none | d | power(multiply(4, 13), const_2) | find √ ? / 13 = 4 ? | "answer let √ n / 13 = 4 then √ n = 13 x 4 = 52 ∴ n = 52 x 52 = 2704 . correct option : d" | a = 4 * 13
b = a ** 2
|
a ) 23 , b ) 38 , c ) 37 , d ) 30 , e ) 28.8 | e | multiply(divide(200, 25), const_3_6) | an athlete runs 200 metres race in 25 seconds . what is his speed ? | "speed = distance / time = 200 / 25 = 8 m / s = 8 * 18 / 5 = 28.8 km / hr answer : e" | a = 200 / 25
b = a * const_3_6
|
a ) 29 , b ) 34 , c ) 44 , d ) 54 , e ) 64 | a | add(subtract(125, const_100), const_4) | how many odd prime numbers are there less than 125 ? | odd prime number less than 125 : 3 , 5 , 7 , 11 , 13 , 17 , 19 , 23 , 29 , 31 , 37 , 41 , 43 , 47 , 53 , 59 , 61 , 67 , 71 , 73 , 79 , 83 , 89 , 97 , 101 , 103 , 107 , 109 , 113 there is 29 the odd prime number answer is a | a = 125 - 100
b = a + 4
|
a ) 15.94 , b ) 12.41 , c ) 16.1 , d ) 32.92 , e ) 32.3 | a | add(divide(circumface(3.1), const_2), multiply(3.1, const_2)) | the radius of a semi circle is 3.1 cm then its perimeter is ? | "36 / 7 r = 6.3 = 15.94 answer : a" | a = circumface / (
b = a + 2
|
a ) 5 days , b ) 8 days , c ) 9 days , d ) 10 days , e ) 11 days | a | divide(55, divide(add(add(divide(55, 11), divide(55, 5)), add(divide(55, 11), divide(55, 55))), const_2)) | a , band c can do a piece of work in 11 days , 5 days and 55 days respectively , working alone . how soon can the work be done if a is assisted by band c on alternate days ? | ( a + b ) ' s 1 day ' s work = 1 / 11 + 1 / 5 = 16 / 55 ( a + c ) ' s 1 day ' s work = 1 / 11 + 1 / 55 = 6 / 55 work done in 2 day ' s = 16 / 55 + 6 / 55 = 2 / 5 2 / 5 th work done in 2 days work done = 5 / 2 * 2 = 5 days answer : a | a = 55 / 11
b = 55 / 5
c = a + b
d = 55 / 11
e = 55 / 55
f = d + e
g = c + f
h = g / 2
i = 55 / h
|
a ) 237 , b ) 270 , c ) 177 , d ) 166 , e ) 111 | c | floor(divide(8200, add(20, divide(2.5, const_100)))) | find the number of shares that can be bought for rs . 8200 if the market value is rs . 20 each with brokerage being 2.5 % . | "explanation : cost of each share = ( 20 + 2.5 % of 20 ) = rs . 20.5 therefore , number of shares = 8200 / 20.5 = 400 answer : c" | a = 2 / 5
b = 20 + a
c = 8200 / b
d = math.floor(c)
|
a ) 5.6 sec , b ) 46.67 sec , c ) 10.8 sec , d ) 12.6 sec , e ) 15 sec | b | divide(add(300, 400), multiply(add(36, 18), const_0_2778)) | two trains 300 m and 400 m long run at the speed of 36 kmph and 18 kmph in opposite directions in parallel tracks . the time which they take to cross each other is ? | "relative speed = 36 + 18 = 54 kmph * 5 / 18 = 15 m / s distance covered in crossing each other = 300 + 400 = 700 m required time = 700 * 1 / 15 = 46.67 sec answer is b" | a = 300 + 400
b = 36 + 18
c = b * const_0_2778
d = a / c
|
a ) - 1 , b ) 2 , c ) 1 , d ) - 2 , e ) 0 | d | subtract(add(1, 1), 1) | if 1 / ( x + 2 ) + 1 / ( x - 2 ) = 1 / ( x + 2 ) , what is the value of x ? | "if we solve the question , we get x = - 2 . option : d" | a = 1 + 1
b = a - 1
|
a ) $ 120 , b ) $ 150 , c ) $ 240 , d ) $ 250 , e ) $ 300 | c | subtract(multiply(1200, power(add(const_1, divide(20, const_100)), 1)), 1200) | find the compound interest on $ 1200 for 1 year at 20 % p . a . if ci is component yearly ? | "a = p ( 1 + r / 100 ) ^ t = 1200 ( 1 + 20 / 100 ) ^ 1 = 1200 * 6 / 5 = $ 1440 ci = a - p = 1440 - 1200 = $ 240 answer is c" | a = 20 / 100
b = 1 + a
c = b ** 1
d = 1200 * c
e = d - 1200
|
a ) 4 : 3 , b ) 5 : 4 , c ) 4 : 7 , d ) 3 : 4 , e ) 4 : 5 | b | divide(5, 4) | if the sides of a cube are in the ratio 5 : 4 . what is the ratio of their diagonals ? | "explanation : diagonal of a cube = a √ 3 where a is side a 1 : a 2 = 5 : 4 d 1 : d 2 = 5 : 4 where √ 3 cancelled both side answer : b" | a = 5 / 4
|
a ) 3 % , b ) 5 % , c ) 8 % , d ) 10 % , e ) 13 % | e | multiply(multiply(10, 10), subtract(const_1, divide(add(multiply(9, const_60), 17), add(multiply(10, const_60), 40)))) | bob wants to run a mile in the same time as his sister . if bob ’ s time for a mile is currently 10 minutes 40 seconds and his sister ’ s time is currently 9 minutes 17 seconds , by what percent does bob need to improve his time in order run a mile in the same time as his sister ? | bob ' s time = 640 secs . his sis ' time = 557 secs . percent increase needed = ( 640 - 557 / 640 ) * 100 = 83 / 640 * 100 = 13 % . ans ( e ) . | a = 10 * 10
b = 9 * const_60
c = b + 17
d = 10 * const_60
e = d + 40
f = c / e
g = 1 - f
h = a * g
|
a ) 7 : 00 , b ) 8 : 00 , c ) 9 : 00 , d ) 10 : 00 , e ) 11 : 00 | c | divide(add(70, multiply(70, divide(const_1, const_2))), subtract(80, 70)) | a train sets off at 9 : 00 am at the speed of 70 km / h . another train starts at 10 : 30 am in the same direction at the rate of 80 km / h . at what time will the second train catch the first train ? | in one hour and thirty minutes the first train travels 105 km . the second train catches the first train at a rate of 80 km / h - 70 km / h = 10 km / h . the second train will catch the first train in 105 / 10 = 10.5 hours , so at 9 : 00 pm . the answer is c . | a = 1 / 2
b = 70 * a
c = 70 + b
d = 80 - 70
e = c / d
|
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