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 ) 1 / 2 , b ) 1 / 3 , c ) 1 / 4 , d ) 1 / 13 , e ) 3 / 26 | c | divide(multiply(const_4, const_3), 52) | from a pack of 52 cards , 1 card is drawn at random . what is the probability that a heart is drawn ? | "the total number of cards is 52 . the number of hearts is 13 . p ( heart ) = 13 / 52 = 1 / 4 the answer is c ." | a = 4 * 3
b = a / 52
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a ) 3 , b ) 4 , c ) 2 , d ) 6 , e ) 7 | c | add(const_4, const_4) | if each year the population of the country grows by 50 % , how many years will elapse before the population of the country doubles ? | "till year 2000 , population is 100 . year 2001 : population becomes 150 . . . . . . . . . . . . . 1 year elapsed year 2002 : population > 200 . . . . . . . . . . . . . 2 year elapsed answer : c" | a = 4 + 4
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a ) none , b ) one , c ) two , d ) three , e ) four | a | subtract(subtract(add(multiply(multiply(multiply(7, 3), const_2), const_4), 7), add(multiply(multiply(multiply(7, 3), const_2), const_4), 3)), 1) | for any integer p , * p is equal to the product of all the integers between 1 and p , inclusive . how many prime numbers are there between * 7 + 3 and * 7 + 7 , inclusive ? | "7 ! = 720 * 7 = 5040 # in question = 5043 , 5044 , 5045 , 5046 , 5047 none of these are prime answer - a" | a = 7 * 3
b = a * 2
c = b * 4
d = c + 7
e = 7 * 3
f = e * 2
g = f * 4
h = g + 3
i = d - h
j = i - 1
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a ) 270 , b ) 288 , c ) 266 , d ) 120 , e ) 126 | d | multiply(divide(subtract(72, 36), const_3_6), 12) | two trains are moving in the same direction at 72 kmph and 36 kmph . the faster train crosses a man in the slower train in 12 seconds . find the length of the faster train ? | "relative speed = ( 72 - 36 ) * 5 / 18 = 2 * 5 = 10 mps . distance covered in 12 sec = 27 * 10 = 270 m . the length of the faster train = 120 m . answer : d" | a = 72 - 36
b = a / const_3_6
c = b * 12
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a ) 40 m , b ) 72 m , c ) 80 m , d ) 82 m , e ) 84 m | a | divide(multiply(36, divide(multiply(subtract(44, 36), const_1000), const_3600)), const_2) | two trains of equal are running on parallel lines in the same direction at 44 km / hr and 36 km / hr . the faster train passes the slower train in 36 sec . the length of each train is ? | "let the length of each train be x m . then , distance covered = 2 x m . relative speed = 44 - 36 = 8 km / hr . = 8 * 5 / 18 = 20 / 9 m / sec . 2 x / 36 = 20 / 9 = > x = 40 . answer : a" | a = 44 - 36
b = a * 1000
c = b / 3600
d = 36 * c
e = d / 2
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a ) 25 , b ) 36 , c ) 40 , d ) 50 , e ) 48 | a | divide(original_price_before_loss(20, 50), 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 $ 50 to incur a loss 20 % ? | production cost per article : $ 60 * ( 100 % - 20 % ) / 20 = $ 2.40 required production costs for a loss of 20 % : $ 50 * ( 100 % + 20 % ) = $ 60 number of articles to be sold for $ 60 to incur a 20 % loss : $ 60 / $ 2.40 = 25 thus , solution a is correct . | a = original_price_before_loss / (
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a ) 2 , b ) 5 , c ) 6 , d ) 8 , e ) 9 | c | divide(subtract(subtract(add(add(22, 27), 28), 59), 6), const_2) | each of the 59 members in lourdes school class is required to sign up for a minimum of one and a maximum of 3 academic clubs . the 3 clubs to choose from are the poetry club , the history club , and the writing club . a total of 22 students sign up for the poetry club , 27 students for the history club , and 28 students for the writing club . if 6 students sign up for exactly two clubs , how many students sign up for all 3 clubs ? | each of the 59 members in lourdes school class is required to sign up for a minimum of one and a maximum of three academic clubs . total = g 1 + g 2 + g 3 - ( # in exactly 2 ) - 2 * ( # in 3 sets ) 59 = 22 + 27 + 28 - ( 6 ) - 2 x so , # in 3 sets = 6 = c | a = 22 + 27
b = a + 28
c = b - 59
d = c - 6
e = d / 2
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a ) 565 , b ) 444 , c ) 676 , d ) 456 , e ) 663 | d | add(multiply(divide(60, subtract(22, 16)), 16), multiply(divide(60, subtract(22, 16)), 22)) | two passenger trains start at the same hour in the day from two different stations and move towards each other at the rate of 16 kmph and 22 kmph respectively . when they meet , it is found that one train has traveled 60 km more than the other one . the distance between the two stations is ? | "1 h - - - - - 6 ? - - - - - - 60 12 h rs = 16 + 22 = 38 t = 12 d = 38 * 12 = 456 answer : d" | a = 22 - 16
b = 60 / a
c = b * 16
d = 22 - 16
e = 60 / d
f = e * 22
g = c + f
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a ) 60 sec , b ) 70 sec , c ) 80 sec , d ) 90 sec , e ) 100 sec | d | add(divide(600, subtract(30, 20)), divide(divide(600, subtract(30, 20)), const_2)) | bruce and bhishma are running on a circular track of length 600 m . speed of bruce is 30 m / s and that of bhishma is 20 m / s . they start fro the same point at the same time in the same direction . when will they meet again for the first time ? | actually arjun has to make a lead of 600 m , because when arjun will be 600 m ahead of bhishma , they will be together again as a person when completes the total length it starts retracing he same path and thus arjun and bhishma can be together again . since , they make a difference of 10 m in 1 second . so , he will create 600 m difference in 60 second . d | a = 30 - 20
b = 600 / a
c = 30 - 20
d = 600 / c
e = d / 2
f = b + e
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a ) 19 mins , b ) 29 mins , c ) 39 mins , d ) 38 mins , e ) 37 mins | c | multiply(subtract(divide(10000, 4700), divide(10000, 6800)), const_60) | if a certain computer is capable of printing 4700 monthly credit card bills per hour , while a new model is capable of printing at a rate of 6800 per hour , the old model will take approximately how much longer than the new model to print 10000 bills ? | old model ' s 1 hr work = 4700 new model ' s 1 hr work = 6800 old model ' time taken to print 10000 cards for 1 hr - - 4700 cards x - - 10000 x = 10000 / 4700 = 100 / 47 new model ' time taken to print 10000 cards for 1 hr - - 6800 cards y - - 10000 y = 10000 / 6800 = 100 / 68 x = 10000 / 4700 = 100 / 47 = 2 6 / 47 hr difference is 100 / 47 - 100 / 68 = 6800 - 4700 / ( 47 * 68 ) 2100 / 3196 = 0.66 hrs = > 39 minutes old model ll take 39 minutes longer to print 10000 cards than new model ans : 39 mins answer : c | a = 10000 / 4700
b = 10000 / 6800
c = a - b
d = c * const_60
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a ) 80 , b ) 90 , c ) 110 , d ) 120 , e ) 130 | b | multiply(multiply(const_2, const_4), const_10) | average between two sets of numbers is closer to the set withmore no ' s ? | "if on a test three people answered 90 % of the questions correctly and two people answered 80 % correctly , then the average for the group is not 85 % but rather 3 × 90 + 2 × 805 = 4305 = 86.3 × 90 + 2 × 805 = 4305 = 86 . here , 90 has a weight of 3 = > it occurs 3 times . whereas 80 has a weight of 2 = > it occurs 2 times . so the average is closer to 90 than to 80 as we have just calculated . b" | a = 2 * 4
b = a * 10
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a ) 144 , b ) 119 , c ) 113 , d ) 88 , e ) 31 | d | subtract(119, subtract(add(144, 119), 232)) | in a graduating class of 232 students , 144 took geometry and 119 took biology . what is the difference between the greatest possible number w and the smallest possible number of students that could have taken both geometry and biology ? | "official solution : first of all , notice that since 144 took geometry and 119 took biology , then the number of students who took both geometry and biology can not be greater than 119 . { total } = { geometry } + { biology } - { both } + { neither } ; 232 = 144 + 119 - { both } + { neither } ; { both } = 31 + { neither } . { both } is minimized when { neither } is 0 . in this case { both } = 31 . the greatest possible number w of students that could have taken both geometry and biology , is 119 . thus , the answer is 119 - 31 = 88 . answer : d ." | a = 144 + 119
b = a - 232
c = 119 - b
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a ) 16.8 % , b ) 17.4 % , c ) 17.9 % , d ) 18.5 % , e ) 19.1 % | b | multiply(divide(subtract(653, add(multiply(multiply(const_2, 10), add(multiply(const_2, 10), const_1)), multiply(divide(10, add(divide(25, const_100), const_1)), add(15, const_2)))), add(multiply(multiply(const_2, 10), add(multiply(const_2, 10), const_1)), multiply(divide(10, add(divide(25, const_100), const_1)), add(15, const_2)))), const_100) | rani bought more apples than oranges . she sells apples at ₹ 23 apiece and makes 15 % profit . she sells oranges at ₹ 10 apiece and makes 25 % profit . if she gets ₹ 653 after selling all the apples and oranges , find her profit percentage . | "given : selling price of an apple = 23 - - > cost price = 23 / 1.15 = 20 selling price of an orange = 10 - - > cost price = 10 / 1.25 = 8 a > o 23 * ( a ) + 10 * ( o ) = 653 653 - 23 * ( a ) has to be divisible by 10 - - > units digit has to be 0 values of a can be 1 , 11 , 21 , 31 , . . . . - - > 1 can not be the value between 11 and 21 , if a = 11 , o = 30 - - > not possible if a = 21 , o = 17 - - > possible cost price = 20 * 21 + 8 * 17 = 420 + 136 = 556 profit = 653 - 556 = 97 profit % = ( 97 / 556 ) * 100 = 17.4 % answer : b" | a = 2 * 10
b = 2 * 10
c = b + 1
d = a * c
e = 25 / 100
f = e + 1
g = 10 / f
h = 15 + 2
i = g * h
j = d + i
k = 653 - j
l = 2 * 10
m = 2 * 10
n = m + 1
o = l * n
p = 25 / 100
q = p + 1
r = 10 / q
s = 15 + 2
t = r * s
u = o + t
v = k / u
w = v * 100
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a ) 79 , b ) 89 , c ) 95 , d ) 86.36 , e ) 97.2 | d | floor(divide(add(multiply(6, 100), multiply(5, 70)), add(6, 5))) | a student took 6 courses last year and received an average ( arithmetic mean ) grade of 100 points . the year before , the student took 5 courses and received an average grade of 70 points . to the nearest tenth of a point , what was the student ’ s average grade for the entire two - year period ? | "let the 6 courses that were taken last year be a 1 , a 2 , a 3 , a 4 , a 5 , a 6 a 1 + a 2 + a 3 + a 4 + a 5 + a 6 = 100 * 6 = 600 the year before , the 5 courses be b 1 , b 2 , b 3 , b 4 , b 5 b 1 + b 2 + b 3 + b 4 + b 5 = 70 * 5 = 350 student ' s average = ( 600 + 350 ) / 11 = 86.36 answer d" | a = 6 * 100
b = 5 * 70
c = a + b
d = 6 + 5
e = c / d
f = math.floor(e)
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a ) 1 : 5 , b ) 1 : 4 , c ) 1 : 6 , d ) 1 : 8 , e ) 1 : 7 | b | divide(3, 12) | if m : n is 3 : 9 and n : p is 9 : 12 then m : p is equal to | "the two ratios given are having the same number 9 for n in both the ratios . hence - m : n = 3 : 9 n : p = 9 : 12 = > m : p = 3 : 12 = > 1 : 4 answer b" | a = 3 / 12
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a ) 5 : 7 , b ) 15 : 21 , c ) 21 : 22 , d ) 13 : 17 , e ) 23 : 42 | c | divide(multiply(7, add(divide(20, const_100), const_1)), multiply(8, add(divide(10, const_100), const_1))) | the ration of the number of boys and girls in a college is 7 : 8 . if the percentage is increase in the number of boys and girls be 20 % and 10 % respectively . what will be the new ration ? | let the number of boys and girls be 7 x and 8 x their increased number is 120 % of 7 x and 110 % of 8 x 7 x * 120 / 100 and 8 x * 110 / 100 42 x / 5 and 44 x / 5 required ratio = 42 x / 5 : 44 x / 5 = 21 : 22 answer is c | a = 20 / 100
b = a + 1
c = 7 * b
d = 10 / 100
e = d + 1
f = 8 * e
g = c / f
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a ) 30 days , b ) 35 days , c ) 40 days , d ) 44 days , e ) 45 days | a | divide(subtract(10, add(divide(10, 50), divide(10, 50))), divide(10, 50)) | a and b can do a piece of work in 50 days and 50 days respectively . they work together for 10 days and b leaves . in how many days the whole work is completed ? | "explanation : ( a + b ) ’ s 10 days work = 10 [ 1 / 50 + 1 / 50 ] = 10 [ 1 + 1 / 50 ] = 2 / 5 a complete remaining work in 2 / 5 * 50 = 20 total work = 10 + 20 = 30 days answer : option a" | a = 10 / 50
b = 10 / 50
c = a + b
d = 10 - c
e = 10 / 50
f = d / e
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a ) a ) 2 / 15 , b ) b ) 3 / 15 , c ) c ) 4 / 5 , d ) d ) 2 / 5 , e ) e ) 1 / 5 | e | divide(add(1, 2), multiply(3, 5)) | of all the homes on gotham street , 1 / 3 are termite - ridden , and 2 / 5 of these are collapsing . what fraction of the homes are termite - ridden , but not collapsing ? | let total homes be 15 termite ridden = 1 / 3 ( 15 ) = 5 termite ridden and collapsing = 2 / 5 ( 5 ) = 2 thus homes that are termite ridden , but not collapsing = 5 - 2 = 3 thus required ratio = 3 / 15 = 1 / 5 answer e | a = 1 + 2
b = 3 * 5
c = a / b
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a ) 16 , b ) 80 , c ) 160 , d ) 120 , e ) 240 | d | add(multiply(const_100, 3), const_100) | how many 3 - digit numerals begin with a digit that represents a odd prime and end with a digit that represents a prime number ? | "prime digits 2 , 3,5 and 7 . three digit numbers _ _ _ 1 st place can be filled in 3 ways 2 nd place can be filled in 10 ways 3 rd place can be filled in 4 ways total = 3 * 10 * 4 = 120 ans : d" | a = 100 * 3
b = a + 100
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a ) 116 , b ) 84 , c ) 40 , d ) 28 , e ) 20 | a | divide(multiply(5, 140), add(5, 1)) | a certain mixture of nuts consists of 5 parts almonds to 1 parts walnuts , by weight . what is the number of pounds of almonds in 140 pounds of the mixture ? | "almonds : walnuts = 5 : 1 total mixture has 6 parts in a 140 pound mixture , almonds are 5 / 6 ( total mixture ) = 5 / 6 * 140 = 116 pounds answer ( a )" | a = 5 * 140
b = 5 + 1
c = a / b
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a ) 25 , b ) 30 , c ) 28 , d ) 24 , e ) 35 | e | divide(multiply(subtract(45, 10), 10), subtract(45, subtract(45, 10))) | a certain number of workers can do a work in 45 days . if there were 10 workers more it could be finished in 10 days less . how many workers are there ? | "number of workers = 10 * ( 45 - 10 ) / 10 = 35 answer is e" | a = 45 - 10
b = a * 10
c = 45 - 10
d = 45 - c
e = b / d
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a ) 789 , b ) 769 , c ) 790 , d ) 787 , e ) 788 | a | add(multiply(multiply(power(const_3, 9), power(9, const_4)), add(const_3, const_4)), 9) | the smallest number which when diminished by 9 , is divisible 13 , 26 , 39 , 52 and 65 is : | "required number = ( l . c . m . of 13 , 26 , 39 , 52,65 ) + 9 = 780 + 7 = 789 answer : option a" | a = 3 ** 9
b = 9 ** 4
c = a * b
d = 3 + 4
e = c * d
f = e + 9
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a ) 22000 , b ) 27000 , c ) 16071 , d ) 16789 , e ) none of these | c | divide(multiply(4500, const_100), 28) | sheela deposits rs . 4500 in bank savings account . if this is 28 % of her monthly income . what is her monthly income in ? | "explanation : 28 % of income = rs . 4500 100 % of income = 4500 x 100 / 28 = rs . 16071 answer : c" | a = 4500 * 100
b = a / 28
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a ) 35 yr , b ) 25 yr , c ) 20 yr , d ) 30 yr , e ) 40 yr | a | add(subtract(10, subtract(10, add(const_3, const_2))), multiply(subtract(10, add(const_3, const_2)), const_2)) | the difference between the ages of two persons is 10 years . fifteen years ago , the elder one was twice as old as the younger one . the present age of the elder person is ? | "let their ages be x years and ( x + 10 ) years then , ( x + 10 ) - 15 = 2 ( x - 15 ) x - 5 = 2 x - 30 x = 25 present age of the elder person = 25 + 10 = 35 years answer is a" | a = 3 + 2
b = 10 - a
c = 10 - b
d = 3 + 2
e = 10 - d
f = e * 2
g = c + f
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a ) 768 , b ) 243 , c ) 246 , d ) 288 , e ) 224 | d | add(216, 75) | two tests had the same maximum mark . the pass percentages in the first and the second test were 40 % and 45 % respectively . a candidate scored 216 marks in the second test and failed by 75 marks in that test . find the pass mark in the first test ? | let the maximum mark in each test be m . the candidate failed by 75 marks in the second test . pass mark in the second test = 249 + 75 = 324 45 / 100 m = 324 pass mark in the first test = 40 / 100 m = 40 / 45 * 324 = 288 . answer : d | a = 216 + 75
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a ) 0.52 , b ) 0.26 , c ) 0.48 , d ) 0.64 , e ) 0.46 | b | add(multiply(0.4, 0.5), multiply(0.3, 0.2)) | simplify : 0.4 * 0.5 + 0.3 * 0.2 | "given exp . = 0.4 * 0.5 + ( 0.3 * 0.2 ) = 0.2 + 0.06 = 0.26 answer is b ." | a = 0 * 4
b = 0 * 3
c = a + b
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a ) 4 , b ) 5 , c ) 6 , d ) 7 , e ) 8 | e | subtract(multiply(4.00, 12), 40.00) | elena purchased brand x pens for $ 4.00 apiece and brand y for $ 2.80 apiece . if elena purchased a total of 12 of these pens for $ 40.00 , how many brand x pens did she purchase ? | "4 x + 2.8 y = 42 - - > multiply by 2.5 ( to get the integers ) - - > 10 x + 7 y = 105 - - > only one positive integers solutions x = 8 and y = 5 ( how to solve : 7 y must have the last digit of 5 in order the last digit of the sum to be 5 ) . answer : e ." | a = 4 * 0
b = a - 40
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a ) $ 40000 , b ) $ 15000 , c ) $ 12000 , d ) $ 10000 , e ) $ 9000 | a | multiply(multiply(10000, const_2), const_2) | if money is invested at r percent interest , compounded annually , the amount of investment will double in approximately 70 / r years . if pat ' s parents invested $ 10000 in a long term bond that pays 8 percent interest , compounded annually , what will be the approximate total amount of investment 18 years later , when pat is ready for college ? | since investment doubles in 70 / r years then for r = 8 it ' ll double in 70 / 8 = ~ 9 years ( we are not asked about the exact amount so such an approximation will do ) . thus in 18 years investment will double twice and become ( $ 10,000 * 2 ) * 2 = $ 40,000 ( after 9 years investment will become $ 10,000 * 2 = $ 20,000 and in another 9 years it ' ll become $ 20,000 * 2 = $ 40,000 ) . answer : a . | a = 10000 * 2
b = a * 2
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a ) 72 , b ) 85 , c ) 64 , d ) 51 , e ) 45 | e | add(30, divide(30, const_2)) | in a ratio which is equal to 4 : 6 , if the antecedent is 30 , then the consequent is ? | "we have 4 / 6 = 30 / x 4 x = 30 * 6 x = 45 consequent = 45 answer is e" | a = 30 / 2
b = 30 + a
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a ) 191 , b ) 355 , c ) 737 , d ) 861 , e ) 1,560 | d | divide(multiply(560, const_100), subtract(const_100, 35)) | a side of beef lost 35 percent of its weight in processing . if the side of beef weighed 560 pounds after processing , how many pounds did it weigh before processing ? | "let weight of side of beef before processing = x ( 65 / 100 ) * x = 560 = > x = ( 560 * 100 ) / 65 = 861 answer d" | a = 560 * 100
b = 100 - 35
c = a / b
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a ) 9052 , b ) 8711 , c ) 9279 , d ) 9604 , e ) 9086 | d | subtract(negate(1372), multiply(subtract(98, 686), divide(subtract(98, 686), subtract(49, 98)))) | 49 , 98 , 686 , 1372 . . . . | "49 x 2 = 98 98 x 7 = 684 684 x 2 = 1372 1372 x 7 = 9604 answer = d" | a = negate - (
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a ) 5000 , b ) 2999 , c ) 6000 , d ) 2990 , e ) 2771 | c | divide(multiply(240, const_100), subtract(const_100, add(subtract(const_100, 20), multiply(subtract(const_100, 20), divide(20, const_100))))) | a man saves 20 % of his monthly salary . if an account of dearness of things he is to increase his monthly expenses by 20 % , he is only able to save rs . 240 per month . what is his monthly salary ? | "income = rs . 100 expenditure = rs . 80 savings = rs . 20 present expenditure 80 * ( 20 / 100 ) = rs . 96 present savings = 100 â € “ 96 = rs . 4 100 - - - - - - 4 ? - - - - - - - - - 240 = > 6000 answer : c" | a = 240 * 100
b = 100 - 20
c = 100 - 20
d = 20 / 100
e = c * d
f = b + e
g = 100 - f
h = a / g
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a ) 9 % , b ) 11 % , c ) 15 % , d ) 25 % , e ) 90 % | d | subtract(const_100, multiply(divide(9, 12), const_100)) | a case of 12 rolls of paper towels sells for $ 9 . the cost of one roll sold individually is $ 1 . what is the percent q of savings per roll for the 12 - roll package over the cost of 12 rolls purchased individually ? | cost of 12 paper towels individually = 1 * 12 = 12 cost of a set of 12 paper towels = 9 cost of one roll = 9 / 12 = 3 / 4 = 0.75 savings per roll = 1 - . 75 = 0.25 % of savings is q = . 25 / 1 * 100 = 25 % d is the answer . | a = 9 / 12
b = a * 100
c = 100 - b
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a ) 5 : 8 , b ) 4 : 7 , c ) 3 : 7 , d ) 9 : 5 , e ) 9 : 8 | d | divide(divide(subtract(96, 78), subtract(96, 68)), subtract(const_1, divide(subtract(96, 78), subtract(96, 68)))) | in what ratio mental a at rs . 68 per kg be mixed with another metal at rs . 96 per kg so that cost of alloy ( mixture ) is rs . 78 per kg ? | "( 96 - 78 ) / ( 78 - 68 ) = 18 / 10 = 9 / 5 answer : d" | a = 96 - 78
b = 96 - 68
c = a / b
d = 96 - 78
e = 96 - 68
f = d / e
g = 1 - f
h = c / g
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a ) 3 , b ) 4 , c ) 5 , d ) 6 , e ) 7 | d | divide(multiply(20000, const_12), multiply(4000, const_10)) | arjun started a business with rs . 20000 and is joined afterwards by anoop with rs . 4000 . after how many months did anoop join if the profits at the end of the year are divided equally ? | suppose anoop joined after 3 months . then , 20000 * 12 = 40000 * ( 12 – x ) = > = > x = 6 . answer : d | a = 20000 * 12
b = 4000 * 10
c = a / b
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a ) 1288 , b ) 9600 , c ) 1000 , d ) 10000 , e ) 2887 | b | multiply(add(20, const_100), subtract(const_100, 20)) | the length of a rectangle is increased by 20 % and its breadth is decreased by 20 % . what is the effect on its area ? | "100 * 100 = 10000 120 * 80 = 9600 answer : b" | a = 20 + 100
b = 100 - 20
c = a * b
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a ) 3 , b ) 4 , c ) 5 , d ) 6 , e ) 7 | d | divide(add(const_4, const_2), const_1) | a perfect square is defined as the square of an integer and a perfect cube is defined as the cube of an integer . how many positive integers n are there such that n is less than 100,000 and at the same time n is a perfect square and a perfect cube ? | "if n is a perfect square and a perfect cube , then n = a ^ 6 for some integer a . the numbers are 1 ^ 6 = 1 , 2 ^ 6 = 64 , 3 ^ 6 = 729 , 4 ^ 6 = 4096 , 5 ^ 6 = 15,625 , 6 ^ = 46,656 . the answer is d ." | a = 4 + 2
b = a / 1
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a ) 450 , b ) 550 , c ) 650 , d ) 750 , e ) 465 | e | divide(multiply(90, 95), const_4) | what is the sum of the integers from - 90 to 95 , inclusive ? | "in an arithmetic progression , the nth term is given by tn = a + ( n - 1 ) d here tn = 95 , a = - 90 , d = 1 hence , 95 = - 90 + ( n - 1 ) or n = 186 sum of n terms can be calculated by sn = n / 2 ( a + l ) a = first term , l = last term , n = no . of terms sn = 186 * ( - 90 + 95 ) / 2 sn = 186 * 5 / 2 = 465 answer : e" | a = 90 * 95
b = a / 4
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a ) rs 4462.50 , b ) rs . 8032.50 , c ) rs . 8900 , d ) rs . 8925 , e ) none | d | divide(divide(multiply(4016.25, const_100), 9), 5) | a sum fetched a total simple interest of rs . 4016.25 at the rate of 9 p . c . p . a . in 5 years . what is the sum ? | "solution principal = rs . ( 100 * 4016.25 / 9 * 5 ) = rs . ( 401625 / 45 ) = rs . 8925 answer d" | a = 4016 * 25
b = a / 9
c = b / 5
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a ) 6 , b ) 7 , c ) 8 , d ) 9 , e ) 10 | d | divide(110, divide(add(40, 4), const_3_6)) | a train 110 m long is running with a speed of 40 km / h . in how many seconds will the train pass a man who is running at 4 km / h in the direction opposite to that in which the train is going ? | "the speed of the train relative to the man = 40 + 4 = 44 km / h . 44000 m / h * 1 h / 3600 s = ( 440 / 36 ) m / s ( 110 m ) / ( 440 / 36 m / s ) = ( 110 * 36 ) / 440 = 36 / 4 = 9 seconds the answer is d ." | a = 40 + 4
b = a / const_3_6
c = 110 / b
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a ) 3750 , b ) 3600 , c ) 7400 , d ) 7500 , e ) none of these | a | divide(1200, divide(subtract(add(const_100, 16), subtract(const_100, 16)), const_100)) | vijay sells a cupboard at 16 % below cost price . had he got rs . 1200 more , he would have made a profit of 16 % . what is the cost price of the cupboard ? | "explanation : cost price = 1200 / ( 0.16 + 0.16 ) = 1200 / 0.32 = rs . 3750 answer a" | a = 100 + 16
b = 100 - 16
c = a - b
d = c / 100
e = 1200 / d
|
a ) 2 m , b ) 3 m , c ) 1 m , d ) 2.5 m , e ) 3.5 m | a | divide(multiply(10, 10), subtract(rectangle_area(16.5, 10), rectangle_area(5.5, 10))) | the dimensions of a field are 16.5 m by 10 m . a pit 10 m long , 5.5 m wide and 4 m deep is dug in one corner of the field and the earth removed has been evenly spread over the remaining area of the field . what will be the rise in the height of field as a result of this operation ? | "the volume of the earth removed is 10 * 5.5 * 4 = 220 m ^ 3 . the remaining area of the field is 16.5 * 10 - 10 * 5.5 = 110 m ^ 2 . 220 m ^ 3 of the earth evenly spread over the area of 110 m ^ 2 will rise the height by ( height ) = ( volume ) / ( area ) = 220 / 110 = 2 m . answer : a" | a = 10 * 10
b = rectangle_area - (
c = a / b
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a ) 204.6 cm , b ) 404.6 cm , c ) 224.6 cm , d ) 205.6 cm , e ) 256.6 cm | a | add(add(add(divide(add(multiply(35, 180), subtract(multiply(7, 140), multiply(7, 120))), 35), const_12), const_4), const_4) | a class of 35 students has an average height of 180 cm . 7 students whose average height is 120 cm , left the class and 7 others whose average height is 140 cm , joined . calculate the new average height of the students of the class ( in cm ) is ? | total height = 180 * 35 = 6300 cm . total height of those who left = 120 * 7 = 840 total height of those who joined = 140 * 7 = 980 new average = [ 6300 - 120 + 980 ] / 35 = 7160 / 35 cm . = 204.6 cm answer : a | a = 35 * 180
b = 7 * 140
c = 7 * 120
d = b - c
e = a + d
f = e / 35
g = f + 12
h = g + 4
i = h + 4
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a ) 2.9 kmph , b ) 1.1 kmph , c ) 1.3 kmph , d ) 1.2 kmph , e ) 5.7 kmph | b | multiply(const_3_6, divide(11, 36)) | convert the 11 / 36 m / s into kilometers per hour ? | "11 / 36 m / s = 11 / 36 * 18 / 5 = 11 / 10 = 1.1 kmph . answer : b" | a = 11 / 36
b = const_3_6 * a
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a ) a ) 6 , b ) b ) 10 , c ) c ) 11 , d ) d ) 21 , e ) e ) 23.5 | d | add(20, 5) | a shop produces sarongs . the daily average production is given by 5 n + 20 , where n is the number of workers aside from the owner . in the first k days , 500 units are produced , and then 5 workers are added to the team . after another k days , the cumulative total is 800 . how many workers were part of the latter production run ? | "the daily average production is given by 5 n + 20 - given in the first k days , 500 units are produced = ( 5 n + 20 ) k = 500 k = 500 / 5 n + 20 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1 5 workers were added = 5 ( n + 5 ) + 20 = 5 n + 45 cumulative is 1250 . . thus for the current period = 800 - 500 = 300 ( 5 n + 45 ) k = 300 k = 300 / 5 n + 45 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2 equate 1 and 2 500 / 5 n + 20 = 300 / 5 n + 45 500 ( 5 n + 45 ) = 300 ( 5 n + 20 ) 2500 n + 22500 = 1500 n + 6000 1000 n = - 16500 n = 16 thus n + 5 = 21 hence d" | a = 20 + 5
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a ) 2 , b ) 3 , c ) 4 , d ) 5 , e ) 6 | e | divide(subtract(add(multiply(divide(subtract(multiply(60, 10), multiply(56, 10)), subtract(60, 40)), 40), multiply(subtract(10, divide(subtract(multiply(60, 10), multiply(56, 10)), subtract(60, 40))), 60)), multiply(50, 10)), subtract(60, 50)) | at a certain food stand , the price of each apple is 40 ¢ and the price of each orange is 60 ¢ . mary selects a total of 10 apples and oranges from the food stand , and the average ( arithmetic mean ) price of the 10 pieces of fruit is 56 ¢ . how many oranges must mary put back so that the average price of the pieces of fruit that she keeps is 50 ¢ ? | "if the average price is 56 , then mary has 8 oranges and 2 apples ( a ratio of 4 : 1 ) . for the average price to be 50 , the ratio should be 1 : 1 . mary must put back 6 oranges . the answer is e ." | a = 60 * 10
b = 56 * 10
c = a - b
d = 60 - 40
e = c / d
f = e * 40
g = 60 * 10
h = 56 * 10
i = g - h
j = 60 - 40
k = i / j
l = 10 - k
m = l * 60
n = f + m
o = 50 * 10
p = n - o
q = 60 - 50
r = p / q
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a ) 200 , b ) 210 , c ) 220 , d ) 225 , e ) 230 | b | subtract(divide(multiply(multiply(3500, 12), const_3), const_100), divide(multiply(multiply(3500, 10), 3), const_100)) | if a lends rs . 3500 to b at 10 % per annum and b lends the same sum to c at 12 % per annum then the gain of b in a period of 3 years is ? | ( 3500 * 2 * 3 ) / 100 = > 210 answer : b | a = 3500 * 12
b = a * 3
c = b / 100
d = 3500 * 10
e = d * 3
f = e / 100
g = c - f
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a ) 24 , b ) 55 , c ) 297 , d ) 267 , e ) 29 | b | multiply(subtract(21, 10), 5) | think of a number , divide it by 5 and add 10 to it . the result is 21 . what is the number thought of ? | explanation : 21 - 10 = 11 11 x 5 = 55 answer : b | a = 21 - 10
b = a * 5
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a ) 1.5 % , b ) 0.5 % , c ) 2.5 % , d ) 3.5 % , e ) 4 % | b | multiply(divide(5, 1), const_100) | what percent is 5 gm of 1 kg ? | "1 kg = 1000 gm 5 / 1000 × 100 = 500 / 1000 = 1 / 2 = 0.5 % b )" | a = 5 / 1
b = a * 100
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a ) 5768 , b ) 8925 , c ) 2345 , d ) 8945 , e ) 8723 | d | divide(divide(multiply(4025.25, const_100), 9), 5) | a sum fetched a total simple interest of 4025.25 at the rate of 9 % . p . a . in 5 years . what is the sum ? | "principal = ( 100 x 4025.25 ) / ( 9 x 5 ) = 402525 / 45 = 8945 . answer d" | a = 4025 * 25
b = a / 9
c = b / 5
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a ) 53.33 mph , b ) 56.67 mph , c ) 60 mph , d ) 64 mph , e ) 66.67 mph | e | add(divide(add(multiply(80, 4), multiply(40, 2)), add(4, 2)), subtract(divide(const_100, 4), const_0_33)) | steve traveled the first 2 hours of his journey at 40 mph and the last 4 hours of his journey at 80 mph . what is his average speed of travel for the entire journey ? | "average speed = total distance / total time = ( 40 * 2 + 80 * 4 ) / ( 2 + 4 ) = 400 / 6 = 66.6 answer : e" | a = 80 * 4
b = 40 * 2
c = a + b
d = 4 + 2
e = c / d
f = 100 / 4
g = f - const_0_33
h = e + g
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a ) 21 , b ) 22 , c ) 25 , d ) 19 , e ) 15 | a | multiply(const_100, divide(subtract(power(add(const_100, 10), const_3), power(const_100, const_3)), power(const_100, const_3))) | if each edge of cube increased by 10 % , the percentage increase in | "100 × ( 110 ) / 100 × ( 110 ) / 100 = 121 = > 21 % answer is a ." | a = 100 + 10
b = a ** 3
c = 100 ** 3
d = b - c
e = 100 ** 3
f = d / e
g = 100 * f
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a ) 4.7 , b ) 4.45 , c ) 4.2 , d ) 4.1 , e ) 4.9 | b | divide(subtract(multiply(3.9, 6), add(multiply(3.4, const_2), multiply(3.85, const_2))), const_2) | the average of 6 numbers is 3.9 . the average of two of them is 3.4 , while the average of the other two is 3.85 . what is the average of the remaining two numbers ? | explanation : sum of the remaining two numbers = ( 3.9 * 6 ) - [ ( 3.4 * 2 ) + ( 3.85 * 2 ) ] = 23.40 - ( 6.8 + 7.7 ) = 23.40 - 14.5 = 8.90 . required average = ( 8.9 / 2 ) = 4.45 . answer : b ) 4.45 | a = 3 * 9
b = 3 * 4
c = 3 * 85
d = b + c
e = a - d
f = e / 2
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a ) 22 % , b ) 30 % , c ) 42 6 / 7 % , d ) 70 % , e ) 130 % | c | multiply(divide(divide(30, const_100), subtract(const_1, divide(30, const_100))), const_100) | a merchant has selected two items to be placed on sale , one of which currently sells for 30 percent less than the other . if he wishes to raise the price of the cheaper item so that the two items are equally priced , by what percentage must he raise the price of the less expensive item ? | "expensive item = $ 100 ; cheap item = $ 70 ; we must increase $ 70 to $ 100 , so by $ 30 , which is approximately 40 % increase : ( 100 - 70 ) / 70 = 3 / 7 = ~ 0.42 . answer : c ." | a = 30 / 100
b = 30 / 100
c = 1 - b
d = a / c
e = d * 100
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a ) 4 days , b ) 6 days , c ) 8 days , d ) 18 days , e ) none of these | a | inverse(add(divide(const_1, 12), multiply(const_2, divide(const_1, 12)))) | a works twice as fast as b . if b can complete a work in 12 days independently , the number of days in which a and b can together finish the work is : | solution rates of worktio of raing of a and b = 2 : 1 . so , ratio of times taken = 1 : 2 ∴ a ’ s 1 day ’ s work = 1 / 6 b ’ s 1 day ’ s work = 1 / 12 ( a + b ) ' s 1 day ’ s work = ( 1 / 6 + 1 / 12 ) = 3 / 12 = 1 / 4 so , a and b together can finish the work in 4 days . answer a | a = 1 / 12
b = 1 / 12
c = 2 * b
d = a + c
e = 1/(d)
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a ) 5 : 4 , b ) 5 : 6 , c ) 4 : 3 , d ) 6 : 5 , e ) 17 : 4 | a | divide(add(multiply(add(add(2, const_3), const_3), multiply(add(2, const_3), 2)), add(2, const_3)), add(multiply(const_3, multiply(add(2, const_3), 2)), add(2, const_3))) | p and q started a business investing rs . 25,000 and rs . 20,000 respectively . in what ratio the profit earned after 2 years be divided between p and q respectively ? | "p : q = 25000 : 20000 = 5 : 4 . answer : a" | a = 2 + 3
b = a + 3
c = 2 + 3
d = c * 2
e = b * d
f = 2 + 3
g = e + f
h = 2 + 3
i = h * 2
j = 3 * i
k = 2 + 3
l = j + k
m = g / l
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a ) 33 kg , b ) 31 kg , c ) 32 kg , d ) 36 kg , e ) 37 kg | a | subtract(add(multiply(41, const_2), multiply(43, const_2)), multiply(45, const_3)) | the average weight of a , b and c is 45 kg . if the average weight of a and b be 41 kg and that of b and c be 43 kg , then the weight of b is : | "a 33 kg let a , b , c represent their respective weights . then , we have : a + b + c = ( 45 x 3 ) = 135 . . . . ( i ) a + b = ( 41 x 2 ) = 82 . . . . ( ii ) b + c = ( 43 x 2 ) = 86 . . . . ( iii ) adding ( ii ) and ( iii ) , we get : a + 2 b + c = 168 . . . . ( iv ) subtracting ( i ) from ( iv ) , we get : b = 33 . b ' s weight = 33 kg . a" | a = 41 * 2
b = 43 * 2
c = a + b
d = 45 * 3
e = c - d
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a ) 8 % decrease , b ) 8 % increase , c ) 16 % decrease , d ) 16 % increase , e ) no change | c | subtract(const_100, subtract(add(40, const_100), divide(multiply(add(40, const_100), 40), const_100))) | the salary of a worker is first increased by 40 % and afterwards reduced by 40 % . what is the net change in the worker ' s salary ? | "let x be the original salary . the final salary is 0.6 ( 1.4 x ) = 0.84 x the answer is c ." | a = 40 + 100
b = 40 + 100
c = b * 40
d = c / 100
e = a - d
f = 100 - e
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a ) 2110 , b ) 2220 , c ) 2440 , d ) 2660 , e ) 2880 | e | divide(multiply(20, power(9, 5)), subtract(add(12, 8), 5)) | a tank with a volume of 20 cubic feet has one inlet pipe and 2 outlet pipes . the inlet pipe fills water into the tank at the rate of 5 cubic inches / min and the 2 outlet pipes empty it out at the rates of 9 cubic inches / min and 8 cubic inches / min respectively . if all 3 pipes are opened when the tank is full , how many minutes does it take to empty the tank ? ( 1 foot = 12 inches ) | "the tank is emptied at this rate : 9 + 8 - 5 = 12 cubic inches / min the tank has a volume of 20 * 12 * 12 * 12 = 34560 cubic inches . the time it takes to empty the tank is 34560 / 12 = 2880 minutes . the answer is e ." | a = 9 ** 5
b = 20 * a
c = 12 + 8
d = c - 5
e = b / d
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a ) 1 , b ) 3 , c ) 2 , d ) 0 , e ) 5 | a | subtract(25, reminder(1024, 25)) | what least number should be added to 1024 , so that the sum is completely divisible by 25 ? | "1024 ã · 25 = 40 with remainder = 24 24 + 1 = 25 . hence 1 should be added to 1024 so that the sum will be divisible by 25 answer : option a" | a = 25 - reminder
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a ) 5 , b ) 6 , c ) 7 , d ) 8 , e ) 9 | c | add(const_4, const_3) | a student was asked to find the arithmetic mean of the numbers 3 , 117 , 915 , 138 , 1917 , 2114 and x . he found the mean to be 12 . what should be the remainder in the place of x ? | we have ( 3 + 11 + 7 + 9 + 15 + 13 + 8 + 19 + 17 + 21 + 14 + x ) / 12 = 12 137 + x = 144 x = 7 answer is c | a = 4 + 3
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a ) 750 , b ) 500 , c ) 800 , d ) 850 , e ) none of these | b | divide(multiply(const_100, 200), subtract(70, subtract(const_100, 70))) | in an election between the two candidates , the candidates who gets 70 % of votes polled is winned by 200 vote ’ s majority . what is the total number of votes polled ? | "explanation : note : majority ( 40 % ) = difference in votes polled to win ( 70 % ) & defeated candidates ( 30 % ) 40 % = 70 % - 30 % 40 % - - - - - > 200 ( 40 * 5 = 200 ) 100 % - - - - - > 500 ( 100 * 5 = 500 ) answer : option b" | a = 100 * 200
b = 100 - 70
c = 70 - b
d = a / c
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a ) 4 % , b ) 18 % , c ) 36 % , d ) 40 % , e ) 50 % | e | subtract(divide(const_1, divide(divide(multiply(subtract(const_1, divide(40, const_100)), subtract(const_1, divide(40, const_100))), subtract(const_1, divide(46, const_100))), const_100)), const_100) | if w is 40 percent less than u , u is 40 percent less than y , and z is 46 percent less than y , then z is greater than w by what percent of w ? | given w = 0.6 u , u = 0.6 y , z = 0.54 y , substituting , w = 2 / 3 z - - - - > z = 1.5 w and thus z is 50 % greater than w . e is the correct answer . | a = 40 / 100
b = 1 - a
c = 40 / 100
d = 1 - c
e = b * d
f = 46 / 100
g = 1 - f
h = e / g
i = h / 100
j = 1 / i
k = j - 100
|
a ) 10 % , b ) 17.6 % , c ) 13.7 % , d ) 15.1 % , e ) 12.3 % | b | multiply(divide(multiply(const_100, divide(15, const_100)), subtract(const_100, multiply(const_100, divide(15, const_100)))), const_100) | the salary of a person was reduced by 15 % . by what percent should his reduced salary be raised so as to bring it at par with his original salary ? | "let the original salary be $ 100 new salary = $ 85 increase on 85 = 15 increase on 100 = 15 / 85 * 100 = 17.6 % answer is b" | a = 15 / 100
b = 100 * a
c = 15 / 100
d = 100 * c
e = 100 - d
f = b / e
g = f * 100
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a ) 1 , b ) 2 , c ) 3 , d ) 4 , e ) 5 | c | add(divide(5, const_10), add(add(divide(2007, 2999), divide(8001, 5998)), divide(2001, add(2999, multiply(const_10, const_100))))) | which of the numbers 1 , 2 , 3 , 4 , or 5 is nearest in value to the sum 2007 / 2999 + 8001 / 5998 + 2001 / 3999 = ? | the sum is a little more than ( 2 / 3 ) + ( 8 / 6 ) + ( 2 / 4 ) = 2.5 . correct answer c | a = 5 / 10
b = 2007 / 2999
c = 8001 / 5998
d = b + c
e = 10 * 100
f = 2999 + e
g = 2001 / f
h = d + g
i = a + h
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a ) 108 , b ) 135 , c ) 139 , d ) 147 , e ) 188 | a | add(multiply(divide(subtract(156, 12), const_3), const_2), 12) | if jake loses 12 pounds , he will weigh twice as much as his sister . together they now weigh 156 pounds . what is jake ’ s present weight , in pounds ? | "lets say j is the weight of jack and s is the wt of his sister . if he loses 8 pounds , he s twice as heavy as his sister . j - 12 = 2 * s also , together they weight 156 pounds j + s = 156 solvong the 2 equation , we get j = 108 pounds ! a" | a = 156 - 12
b = a / 3
c = b * 2
d = c + 12
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a ) 100 % , b ) 80 % , c ) 75 % , d ) 66 + 2 / 3 % , e ) 55 % | e | multiply(subtract(divide(add(divide(12, const_100), const_1), multiply(subtract(const_1, divide(20, const_100)), divide(subtract(40, 14), 40))), const_1), const_100) | during a sale of 20 % on everything in a store , a kid is successful in convincing the store manager to give him 40 candies for the discounted price of 14 candies . the store still makes a profit of 12 % on this sale . what is the mark up percentage on each candy ? | hi akhil , i can share the way i solved it . . let ' s say marked price = m so , there ' s a discount of 20 % on m so , new s . p . = 80 % of m now , the child convinces the owner to sell 20 candies for the price of 14 candies . let ' s say each candy after discount is 1 $ . so , s . p . of 20 candies = 20 $ . the child bought it for 14 $ so , he got a discount of 6 / 20 * 100 = 30 % so , the latest s . p . = 70 % of 80 % of m = 0.7 * 0.8 m now , we are given that the shopkeeper still makes a profit of 12 % . so we have , 0.7 * 0.8 * m = 1.12 c . p so , we get , m = 2 c . p . i . e . marked price was kept 100 % above c . p . e | a = 12 / 100
b = a + 1
c = 20 / 100
d = 1 - c
e = 40 - 14
f = e / 40
g = d * f
h = b / g
i = h - 1
j = i * 100
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['a ) 10', 'b ) 15', 'c ) 20', 'd ) 25', 'e ) 30'] | d | subtract(subtract(65, divide(60, const_3)), divide(60, const_3)) | the perimeter of an equilateral triangle is 60 . if one of the sides of the equilateral triangle is the side of an isosceles triangle of perimeter 65 , then how long is the base of isosceles triangle ? | the base of the isosceles triangle is 65 - 20 - 20 = 25 units the answer is d . | a = 60 / 3
b = 65 - a
c = 60 / 3
d = b - c
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a ) 39 , b ) 41 , c ) 40 , d ) 42 , e ) 43 | c | subtract(multiply(add(19, 1), add(20, 1)), multiply(19, 20)) | the average age of a class of 19 students is 20 years . the average increased by 1 when the teacher ' s age also included . what is the age of the teacher ? | total age of all students = 19 ã — 20 total age of all students + age of the teacher = 20 ã — 21 age of the teacher = 20 ã — 21 â ˆ ’ 20 ã — 19 = 20 ( 21 â ˆ ’ 19 ) = 20 ã — 2 = 40 answer is c . | a = 19 + 1
b = 20 + 1
c = a * b
d = 19 * 20
e = c - d
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a ) 0 , b ) - 2 , c ) - 25 , d ) - 49 , e ) - 51 | d | add(add(negate(26), const_1), add(add(negate(26), const_1), const_1)) | the sum of all the integers d such that - 26 < d < 24 is | "easy one - - 25 , - 24 , - 23 , - 22 , . . . . . . - 1,0 , 1 , 2 . . . . , 22 , 23 cancel everyhitng and we ' re left with - - 25 and - 24 d = - 49 . d is the answer ." | a = negate + (
b = a + 1
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a ) 15 , b ) 16 , c ) 28 , d ) 56 , e ) 72 | e | divide(multiply(9, subtract(9, const_1)), const_2) | there are 9 teams in a certain league and each team plays each of the other teams exactly twice . if each game is played by 2 teams , what is the total number of games played ? | "every team plays with 8 teams . . . so total no of matches = 9 x 8 = 72 . now , each match is played twice = > 72 x 2 but 2 teams play a match = > 72 x 2 / 2 = 72 . answer : e" | a = 9 - 1
b = 9 * a
c = b / 2
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a ) 0 , b ) 1 , c ) 2 , d ) 3 , e ) 4 | d | divide(multiply(add(4, 1), const_2), 8) | | x + 1 | – | 4 - x | = | 8 + x | how many solutions will this equation have ? | "you have | x + 3 | - | 4 - x | = | 8 + x | first , look at the three values independently of their absolute value sign , in other words : | x + 3 | - | 4 - x | = | 8 + x | ( x + 3 ) - ( 4 - x ) = ( 8 + x ) now , you ' re looking at x < - 8 , so x is a number less than - 8 . let ' s pretend x = - 10 here to make things a bit easier to understand . when x = - 10 i . ) ( x + 3 ) ( - 10 + 3 ) ( - 7 ) ii . ) ( 4 - x ) ( 4 - [ - 10 ] ) ( double negative , so it becomes positive ) ( 4 + 10 ) ( 14 ) iii . ) ( 8 + x ) ( 8 + - 10 ) ( - 2 ) in other words , when x < - 8 , ( x + 3 ) and ( 8 + x ) are negative . to solve problems like this , we need to check for the sign change . here is how i do it step by step . i . ) | x + 3 | - | 4 - x | = | 8 + x | ii . ) ignore absolute value signs ( for now ) and find the values of x which make ( x + 3 ) , ( 4 - x ) and ( 8 + x ) = to zero as follows : ( x + 3 ) x = - 3 ( - 3 + 3 ) = 0 ( 4 - x ) x = 4 ( 4 - 4 ) = 0 ( 8 + x ) x = - 8 ( 8 + - 8 ) = 3 d" | a = 4 + 1
b = a * 2
c = b / 8
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a ) 10 hours , b ) 12 hours , c ) 8 hours , d ) 5 hours , e ) 15 hours | b | divide(12, const_1) | an electric pump can fill a tank in 6 hours . because of a leak in the tank , it took 12 hours to fill the tank . if the tank is full , how much time will the leak take to empty it ? | "work done by the leak in 1 hour = 1 / 6 - 1 / 12 = 1 / 12 the leak will empty the tank in 12 hours answer is b" | a = 12 / 1
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a ) 4 / 5 , b ) 1 , c ) 2 , d ) 5 , e ) 8 | c | divide(4, 2) | m and n are the x and y coordinates , respectively , of a point in the coordinate plane . if the points ( m , n ) and ( m + p , n + 4 ) both lie on the line defined by the equation x = ( y / 2 ) - ( 2 / 5 ) , what is the value of p ? | "( m , n ) m = x n = y m = n / 2 - 2 / 5 10 m = 5 n - 4 ( multiplied by 10 to simplify equation ) ( m + p , n + 4 ) m + p = x n + 4 = y m + p = ( n + 4 ) / 2 - 2 / 5 multiplied by 10 again 10 m + 10 p = 5 n + 20 - 4 use both equations to solve for p and you get 10 p = 20 p = 2 answer : c" | a = 4 / 2
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a ) 21 , b ) 22 , c ) 23 , d ) 24 , e ) 25 | d | add(10, 14) | lilly has 10 fish and rosy has 14 fish . in total , how many fish do they have in all ? | "10 + 14 = 24 the answer is d ." | a = 10 + 14
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a ) 9 , b ) 12 , c ) 15 , d ) 18 , e ) 21 | c | add(divide(subtract(multiply(add(3, 6), 7), multiply(add(3, 6), 6)), subtract(6, multiply(7, divide(3, 4)))), 3) | 3 years ago , the ratio of ages of p and q was 4 : 3 . 6 years from now , the ratio of their ages will be 7 : 6 . what is p ' s age at present ? | 3 ( p - 3 ) / 4 = ( q - 3 ) . then q = 3 ( p - 3 ) / 4 + 3 6 ( p + 6 ) / 7 = q + 6 6 ( p + 6 ) / 7 = 3 ( p - 3 ) / 4 + 9 24 p + 144 = 21 p + 189 3 p = 45 p = 15 the answer is c . | a = 3 + 6
b = a * 7
c = 3 + 6
d = c * 6
e = b - d
f = 3 / 4
g = 7 * f
h = 6 - g
i = e / h
j = i + 3
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a ) 86 , b ) 165 , c ) 76 , d ) 155 , e ) 26 | d | subtract(add(multiply(90, const_2), multiply(70, const_2)), multiply(55, 3)) | a student scored an average of 55 marks in 3 subjects : physics , chemistry and mathematics . if the average marks in physics and mathematics is 90 and that in physics and chemistry is 70 , what are the marks in physics ? | "given m + p + c = 55 * 3 = 165 - - - ( 1 ) m + p = 90 * 2 = 180 - - - ( 2 ) p + c = 70 * 2 = 140 - - - ( 3 ) where m , p and c are marks obtained by the student in mathematics , physics and chemistry . p = ( 2 ) + ( 3 ) - ( 1 ) = 180 + 140 - 165 = 155 answer : d" | a = 90 * 2
b = 70 * 2
c = a + b
d = 55 * 3
e = c - d
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a ) $ 200 , b ) $ 250 , c ) $ 100 , d ) $ 350 , e ) $ 875 | e | divide(multiply(divide(multiply(divide(60, const_100), 500), divide(30, const_100)), divide(60, const_100)), divide(30, const_100)) | a school has received 30 % of the amount it needs for a new building by receiving a donation of $ 500 each from people already solicited . people already solicited represent 60 % of the people from whom the school will solicit donations . how much average contribution is requited from the remaining targeted people to complete the fund raising exercise ? | "let us suppose there are 100 people . 60 % of them donated $ 30000 ( 500 * 60 ) $ 30000 is 30 % of total amount . so total amount = 30000 * 100 / 60 remaining amount is 70 % of total amount . 70 % of total amount = 30000 * ( 100 / 60 ) * ( 70 / 100 ) = 35000 this amount has to be divided by 40 ( remaining people are 40 ) so per head amount is 35000 / 40 = $ 875 ; answer : e" | a = 60 / 100
b = a * 500
c = 30 / 100
d = b / c
e = 60 / 100
f = d * e
g = 30 / 100
h = f / g
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a ) 150 % , b ) 175 percent , c ) 200 % , d ) 250 % , e ) 350 % | b | multiply(add(add(4, const_1), 2), multiply(add(4, const_1), add(4, const_1))) | if x / y = 4 and y is not ' 0 ' what % of x is 2 x - y | let z be the percentt x * z / 100 = 2 x - y zx = 7 y * 100 z = 700 y / 4 y z = 350 / 2 z = 175 % answer : b | a = 4 + 1
b = a + 2
c = 4 + 1
d = 4 + 1
e = c * d
f = b * e
|
a ) 1000 , b ) 812.88 , c ) 827.78 , d ) 826.82 , e ) 886.08 | e | divide(1120, add(divide(multiply(divide(add(multiply(2, 5), 2), 5), 11), const_100), const_1)) | find the principle on a certain sum of money at 11 % per annum for 2 2 / 5 years if the amount being rs . 1120 ? | "1120 = p [ 1 + ( 5 * 12 / 5 ) / 100 ] p = 886.08 answer : e" | a = 2 * 5
b = a + 2
c = b / 5
d = c * 11
e = d / 100
f = e + 1
g = 1120 / f
|
a ) 0.004 % , b ) 0.04 % , c ) 0.40 % , d ) 3 % , e ) 40 % | d | multiply(divide(multiply(50, 0.006), 10), const_100) | a bowl was filled with 10 ounces of water , and 0.006 ounce of the water evaporated each day during a 50 - day period . what percent of the original amount of water evaporated during this period ? | "total amount of water evaporated each day during a 50 - day period = . 006 * 50 = . 006 * 100 / 2 = . 6 / 2 = . 3 percent of the original amount of water evaporated during this period = ( . 3 / 10 ) * 100 % = 3 % answer d" | a = 50 * 0
b = a / 10
c = b * 100
|
a ) 18,200 , b ) 18,300 , c ) 18,400 , d ) 18,500 , e ) 18,600 | c | divide(divide(subtract(multiply(const_1000, const_100), subtract(subtract(const_3600, const_100), const_1000)), const_1000), add(multiply(add(const_1, divide(10, const_100)), subtract(8, const_1)), const_1)) | a city with a population of 160,080 is to be divided into 8 voting districts , and no district is to have a population that is more than 10 percent greater than the population of any other district . what is the minimum possible population that the least populated district could have ? | "the minimum possible population occurs when all the other districts have a population that is 10 % greater than the least populated district . let p be the population of the least populated district . then 160,080 = p + 7 ( 1.1 ) p 8.7 p = 160,080 p = 18,400 the answer is c ." | a = 1000 * 100
b = 3600 - 100
c = b - 1000
d = a - c
e = d / 1000
f = 10 / 100
g = 1 + f
h = 8 - 1
i = g * h
j = i + 1
k = e / j
|
a ) a ) 497 , b ) b ) 518 , c ) c ) 519 , d ) d ) 521 , e ) e ) 525 | a | add(subtract(414, 18), const_1) | if the average ( arithmetic mean ) of 18 consecutive odd integers is 414 , then the least of these integers is | "a very helpful rule to know in arithmetic is the rule that in evenly spaced sets , average = median . because the average will equal the median in these sets , then we quickly know that the median of this set of consecutive odd integer numbers is 414 . there are 18 numbers in the set , and in a set with an even number of terms the median is just the average of the two most median terms ( here the 9 th and 10 th numbers in the set ) . this means that numbers 9 and 10 in this set are 413 and 415 . because we know that number 9 is 413 , we know that the smallest number is 8 odd numbers below this , which means that it is 8 * 2 = 16 below this ( every odd number is every other number ) . therefore 413 - 16 = 397 , answer choice a" | a = 414 - 18
b = a + 1
|
a ) 25 / 9 , b ) 9 / 5 , c ) 5 / 3 , d ) 7 / 5 , e ) 9 / 25 | d | divide(7, 5) | a number x is multiplied by 7 , and this product is then divided by 5 . if the positive square root of the result of these two operations equals x , what is the value of x if x ≠ 0 ? | sqrt ( 7 x / 5 ) to be perfect square x has to 7 / 5 ans : d | a = 7 / 5
|
a ) 70 , b ) 52 , c ) 51 , d ) 49 , e ) 50 | c | add(add(10, 12), add(16, 13)) | in a games hour 4 different types of players came to the ground ? cricket 10 , hokey 12 , football 16 , softball 13 . how many players are present in the ground ? | total number of players = 10 + 12 + 16 + 13 = 51 answer is c | a = 10 + 12
b = 16 + 13
c = a + b
|
a ) 113 , b ) 213 , c ) 413 , d ) 417 , e ) 414 | c | multiply(56581, 137) | calculate 56581 ÷ ? = 137 | "answer let 56581 / x = 137 then x = 56581 / 137 = 413 option : c" | a = 56581 * 137
|
a ) 9 / 18 , b ) 9 / 16 , c ) 9 / 11 , d ) 9 / 12 , e ) 9 / 13 | b | add(divide(5, subtract(add(5, 12), 1)), divide(multiply(2, 2), subtract(add(5, 12), 1))) | in 2 bags , there are to be put together 5 red and 12 white balls , neither bag being empty . how must the balls be divided so as to give a person who draws 1 ball from either bag - | ( 1 ) put 1 white in first bag , and all the remaining ball in second bag . probability = 12 × 0 + 12 × 51612 × 0 + 12 × 516 = 532532 ( 2 ) to maximize , we put 1 red in the first bag , and all the remaining in the second . probability = 12 × 1 + 12 × 43212 × 1 + 12 × 432 = 9 / 16 answer : b | a = 5 + 12
b = a - 1
c = 5 / b
d = 2 * 2
e = 5 + 12
f = e - 1
g = d / f
h = c + g
|
a ) 33 , b ) 34 , c ) 35 , d ) 36 , e ) 37 | d | add(multiply(divide(subtract(52, multiply(const_2, 6)), add(6, 4)), 6), multiply(const_2, 6)) | sandy leaves her home and walks toward ed ' s house . two hours later , ed leaves his home and walks toward sandy ' s house . the distance between their homes is 52 kilometers , sandy ' s walking speed is 6 km / h , and ed ' s walking speed is 4 km / h . how many kilometers will sandy walk before she meets ed ? | sandy walks 12 km in the first two hours so there are 40 km remaining . when ed starts walking , they complete a total of 10 km per hour . they will meet four hours after ed starts walking . since sandy walks for 6 hours , she walks 36 km . the answer is d . | a = 2 * 6
b = 52 - a
c = 6 + 4
d = b / c
e = d * 6
f = 2 * 6
g = e + f
|
a ) 15.8 sec . , b ) 12.8 sec . , c ) 11.88 sec . , d ) 10.8 sec . , e ) 08.8 sec . | c | divide(add(140, 190), multiply(add(60, 40), const_0_2778)) | two bullet train s 140 m and 190 m long run at the speed of 60 km / hr and 40 km / hr respectively in opposite directions on parallel tracks . the time ( in seconds ) which they take to cross each other , is : | "relative speed = ( 60 + 40 ) km / hr = 100 x 5 / 18 = 250 / 9 m / sec . distance covered in crossing each other = ( 140 + 190 ) m = 330 m . required time = 330 x 9 / 250 = 54 / 5 = 11.88 sec . answer c" | a = 140 + 190
b = 60 + 40
c = b * const_0_2778
d = a / c
|
a ) $ 2.50 , b ) $ 2.75 , c ) $ 3.00 , d ) $ 3.25 , e ) $ 3.50 | c | divide(1.95, add(multiply(const_0_25, const_2), multiply(const_0_33, const_1))) | a customer purchased a package of ground beef at a cost of $ 1.95 per pound . for the same amount of money , the customer could have purchased a piece of steak that weighed 35 percent less than the package of ground beef . what was the cost per pound of the steak ? | "for simplicity , let ' s assume the customer bought 1 pound of ground beef for $ 1.95 . let x be the price per pound for the steak . then 0.65 x = 195 x = 195 / 0.65 = $ 3.00 the answer is c ." | a = const_0_25 * 2
b = const_0_33 * 1
c = a + b
d = 1 / 95
|
a ) 18 , b ) 28 , c ) 32 , d ) 56 , e ) 60 | c | multiply(divide(divide(add(add(multiply(multiply(const_4, const_2), const_10), multiply(multiply(const_4, const_2), const_100)), multiply(const_12, const_1000)), 230), 7), 4) | the ratio between the number of sheep and the number of horses at the stewart farm is 4 to 7 , if each horse is fed 230 ounces of horse food per day and the farm needs a total 12,880 ounces of horse food per day , what is the number of sheep in the farm ? | "actual number of horses = 12,880 / 230 ( hold it since it is a harder calculation ) ratio multiplier = 1288 / ( 23 * 7 ) actual number of sheep = 1288 / ( 23 * 7 ) * 4 . first divide 1288 by 7 to get : 184 * 4 / 23 use the last digit to figure out that 23 will go 8 times into 184 . confirm . 8 * 4 = 32 answer ( c )" | a = 4 * 2
b = a * 10
c = 4 * 2
d = c * 100
e = b + d
f = 12 * 1000
g = e + f
h = g / 230
i = h / 7
j = i * 4
|
a ) a ) 40 , b ) b ) 45 , c ) c ) 36 , d ) d ) 55 , e ) e ) 60 | c | subtract(choose(8, 3), choose(6, 3)) | there are 8 books on a shelf , of which 2 are paperbacks and 6 are hardbacks . how many possible selections of 3 books from this shelf include at least one paperback ? | "approach 1 at - least 1 paper back = total - no paper back 8 c 3 - 6 c 3 = 36 approach 2 at - least 1 paper back = 1 paper back , 3 hard back or 2 paper back 2 hard back = 2 c 1 * 6 c 2 + 2 c 2 * 6 c 1 = 36 answer is c" | a = math.comb(8, 3)
b = math.comb(6, 3)
c = a - b
|
a ) 81 : 49 , b ) 81 : 122 , c ) 81 : 124 , d ) 81 : 126 , e ) 81 : 129 | a | power(divide(729, 343), divide(const_1, const_3)) | the ratio of the volumes of two cubes is 729 : 343 . what is the ratio of their total surface areas ? | "ratio of the sides = 3 √ 729 : 3 √ 343 = 9 : 7 ratio of surface areas = 9 ^ 2 : 7 ^ 2 = 81 : 49 answer : option a" | a = 729 / 343
b = 1 / 3
c = a ** b
|
a ) 150 , b ) 300 , c ) 299 , d ) 322 , e ) 345 | a | multiply(10, 15) | the h . c . f . of two numbers is 10 and the other two factors of their l . c . m . are 11 and 15 . the larger of the two numbers is : | "the numbers are ( 10 x 11 ) and ( 10 x 15 ) . larger number = ( 10 x 15 ) = 150 . answer : a" | a = 10 * 15
|
a ) 22 , b ) 77 , c ) 30 , d ) 99 , e ) 12 | c | divide(add(25, 35), const_2) | a man can row upstream at 25 kmph and downstream at 35 kmph , and then find the speed of the man in still water ? | "us = 25 ds = 35 m = ( 35 + 25 ) / 2 = 30 answer : c" | a = 25 + 35
b = a / 2
|
a ) 29 , b ) 31 , c ) 33 , d ) 35 , e ) 37 | c | subtract(add(multiply(divide(1023, divide(1023, 2)), const_4), const_10), const_1) | the product of two positive integers is 1023 and their difference is 2 . what is the bigger number ? | "let ' s use trial and error to find the two numbers . 32 * 30 = 960 ( too low ) 33 * 31 = 1023 the answer is c ." | a = 1023 / 2
b = 1023 / a
c = b * 4
d = c + 10
e = d - 1
|
a ) 50 m , b ) 72 m , c ) 80 m , d ) 82 m , e ) 92 m | a | divide(multiply(divide(multiply(subtract(46, 36), const_1000), const_3600), 36), const_2) | two trains of equal length are running on parallel lines in the same direction at 46 km / hr and 36 km / hr . the faster train passes the slower train in 36 seconds . the length of each train is : | "let the length of each train be x metres . then , distance covered = 2 x metres . relative speed = ( 46 - 36 ) km / hr = ( 10 x ( 5 / 18 ) m / sec = ( 25 / 9 ) m / sec ( 2 x / 36 ) = ( 25 / 9 ) 2 x = 100 x = 50 . a" | a = 46 - 36
b = a * 1000
c = b / 3600
d = c * 36
e = d / 2
|
a ) 4 , b ) 8 , c ) 12 , d ) 24 , e ) 18 | d | divide(power(12, const_2), 6) | if the area of a square with sides of length 12 centimeters is equal to the area of a rectangle with a width of 6 centimeters , what is the length of the rectangle , in centimeters ? | "let length of rectangle = l 12 ^ 2 = l * 4 = > l = 144 / 6 = 24 answer d" | a = 12 ** 2
b = a / 6
|
a ) 19 , b ) 18 , c ) 16 , d ) 11 , e ) 21 | e | add(20, const_1) | the average of first 20 even numbers is ? | "sum of 20 even numbers = 20 * 21 = 420 average = 420 / 20 = 21 answer : e" | a = 20 + 1
|
a ) 72 , b ) 60 , c ) 32 , d ) 96 , e ) 65 | a | divide(48, subtract(const_1, divide(1, 3))) | the difference between the two numbers is 1 / 3 of the larger number . if the smaller one is 48 the larger one is | explanation : let the larger number be x . then , x - 48 = 1 / 3 x x - x / 3 = 48 = > 2 x / 3 = 48 = > x = 72 answer : option a | a = 1 / 3
b = 1 - a
c = 48 / b
|
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