options stringlengths 37 300 | correct stringclasses 5
values | annotated_formula stringlengths 7 727 | problem stringlengths 5 967 | rationale stringlengths 1 2.74k | program stringlengths 10 646 |
|---|---|---|---|---|---|
a ) 16 sec , b ) 12 sec , c ) 17 sec , d ) 15 sec , e ) 23 sec | d | divide(multiply(120, const_2), add(speed(120, 20), speed(120, 12))) | two trains of equal lengths take 12 sec and 20 sec respectively to cross a telegraph post . if the length of each train be 120 m , in what time will they cross other travelling in opposite direction ? | "speed of the first train = 120 / 12 = 10 m / sec . speed of the second train = 120 / 20 = 6 m / sec . relative speed = 10 + 6 = 16 m / sec . required time = ( 120 + 120 ) / 16 = 15 sec . answer : d" | a = 120 * 2
b = speed + (
c = a / b
|
a ) 2 % , b ) 4 % , c ) 5 % , d ) 8 % , e ) 12 % | d | subtract(subtract(20, 10), divide(20, 10)) | a couple who own an appliance store discover that if they advertise a sales discount of 10 % on every item in the store , at the end of one month the number of total items sold increases 20 % . their gross income from sales for one month increases by what percent ? | "let list price of an item = 100 discount on each item = 10 % discounted price of an item = . 9 * 100 = 90 if they advertise a sales discount of 10 % on every item in the store , at the end of one month the number of total items sold increases 20 % originally if 10 items were sold in a month , with the new discount 12 items will be sold original revenue = no of items * price of an item = 10 * 100 = 1000 new revenue = 12 * 90 = 1080 increase in gross income = 1080 - 1000 = 80 % increase in gross revenue = 80 / 1000 * 100 % = 8 % answer d" | a = 20 - 10
b = 20 / 10
c = a - b
|
a ) 2.6 , b ) 8.6 , c ) 7.6 , d ) 3.6 , e ) 1.6 | d | multiply(const_12, divide(multiply(48, divide(48, const_100)), 64)) | a reduction of 48 % in the price of bananas would enable a man to obtain 64 more for rs . 40 , what is reduced price per dozen ? | "explanation : 40 * ( 48 / 100 ) = 19.2 - - - 64 ? - - - 12 = > rs . 3.6 answer : d" | a = 48 / 100
b = 48 * a
c = b / 64
d = 12 * c
|
a ) 3000 , b ) 2802 , c ) 1750 , d ) 2990 , e ) 2982 | c | divide(divide(subtract(multiply(4000, power(add(const_1, divide(10, const_100)), 2)), 4000), 2), multiply(3, divide(8, const_100))) | the s . i . on a certain sum of money for 3 years at 8 % per annum is half the c . i . on rs . 4000 for 2 years at 10 % per annum . the sum placed on s . i . is ? | "c . i . = [ 4000 * ( 1 + 10 / 100 ) 2 - 4000 ] = ( 4000 * 11 / 10 * 11 / 10 - 4000 ) = rs . 840 . sum = ( 420 * 100 ) / ( 3 * 8 ) = rs . 1750 answer : c" | a = 10 / 100
b = 1 + a
c = b ** 2
d = 4000 * c
e = d - 4000
f = e / 2
g = 8 / 100
h = 3 * g
i = f / h
|
a ) $ 80,000 , b ) $ 130,000 , c ) $ 240,000 , d ) $ 290,000 , e ) $ 220,000 | e | divide(subtract(subtract(multiply(multiply(add(70, 20), const_1000), const_100), multiply(20, multiply(add(70, 20), const_1000))), multiply(multiply(multiply(divide(20, const_2), 20), const_1000), 70)), 20) | in plutarch enterprises , 70 % of the employees are marketers , 20 % are engineers , and the rest are managers . marketers make an average salary of $ 60,000 a year , and engineers make an average of $ 80,000 . what is the average salary for managers if the average for all employees is also $ 80,000 ? | "for sake of ease , let ' s say there are 10 employees : 7 marketers , 2 engineers , and 1 manager . average company salary * number of employees = total company salary > > > $ 80,000 * 10 = $ 800,000 subtract the combined salaries for the marketers ( 7 * $ 60,000 ) and the engineers ( 2 * $ 80,000 ) > > > $ 800,000 - $ 420,000 - $ 160,000 = $ 220,000 . the correct answer is e ." | a = 70 + 20
b = a * 1000
c = b * 100
d = 70 + 20
e = d * 1000
f = 20 * e
g = c - f
h = 20 / 2
i = h * 20
j = i * 1000
k = j * 70
l = g - k
m = l / 20
|
a ) 542 , b ) 882 , c ) 772 , d ) 662 , e ) 521 | a | add(460, multiply(460, divide(18, const_100))) | a person buys an article at rs . 460 . at what price should he sell the article so as to make a profit of 18 % ? | "cost price = rs . 460 profit = 18 % of 460 = rs . 82 selling price = cost price + profit = 460 + 82 = 542 answer : a" | a = 18 / 100
b = 460 * a
c = 460 + b
|
a ) 1 / 4 , b ) 3 / 8 , c ) 1 / 2 , d ) 55 / 86 , e ) 3 / 4 | d | divide(add(divide(86, 2), divide(86, 8)), 86) | if an integer n is to be chosen at random from the integers 1 to 86 , inclusive , what is the probability that n ( n + 1 ) ( n + 2 ) will be divisible by 8 ? | "n ( n + 1 ) ( n + 2 ) will be divisible by 8 when n is a multiple of 2 or when ( n + 1 ) is a multiple of 8 . thus when n is even , this whole expression will be divisible by 8 . from 1 to 86 , there are 43 even integers . now when ( n + 1 ) is multiple by 8 , we have 12 such values for ( n + 1 ) probability that n ( n + 1 ) ( n + 2 ) will be divisible by 8 = ( 43 + 12 ) / 96 = 55 / 86 ans isd" | a = 86 / 2
b = 86 / 8
c = a + b
d = c / 86
|
a ) $ 200 , b ) $ 177.78 , c ) $ 100 , d ) $ 277.78 , e ) $ 377.78 | d | divide(multiply(divide(multiply(divide(40, const_100), 600), divide(60, const_100)), divide(40, const_100)), divide(60, const_100)) | a school has received 60 % of the amount it needs for a new building by receiving a donation of $ 600 each from people already solicited . people already solicited represent 40 % 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 the amount school needs = x let total people school plans to solicit = t school has received 60 % of x = > ( 3 / 5 ) x people already solicited = 40 % of t = > ( 2 / 5 ) t now , as per the information given in the question : ( 3 / 5 ) x = $ 400 . ( 2 / 5 ) . t - - - - - - - - - - - - - - - - - - - - - - - - - - - 1 remaning amount is 40 % i . e ( 2 / 5 ) x - - - - - - because school has already received 60 % and the remaining people are 60 % i . e ( 3 / 5 ) . t - - - - - because 40 % of the people are already solicited so , average contribution required from the remaining targeted people is ( 2 / 5 ) x = ( amount required ) . ( 3 / 5 ) . t - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - 2 divide eqn 1 by eqn 2 amount required = $ 277.78 d | a = 40 / 100
b = a * 600
c = 60 / 100
d = b / c
e = 40 / 100
f = d * e
g = 60 / 100
h = f / g
|
a ) 13 , b ) 10 , c ) 9 , d ) 7 , e ) 6 | c | log(power(2, const_10)) | the “ length of integer x ” refers to the number of prime factors , not necessarily distinct , that x has . ( if x = 60 , the length of x would be 4 because 60 = 2 × 2 × 3 × 5 . ) what is the greatest possible length of integer z if z < 1,000 ? | "to maximize the length of z , we should minimize its prime base . the smallest prime is 2 and since 2 ^ 9 = 512 < 1,000 , then the greatest possible length of integer z is 9 . the answer is c ." | a = 2 ** 10
b = math.log(a)
|
a ) 209 , b ) 200 , c ) 210 , d ) 225 , e ) 221 | a | divide(factorial(add(6, 4)), multiply(factorial(4), factorial(6))) | in a group of 6 boys and 4 girls are to be selected . in how many different ways can they be selected such that at least one boy should be there ? | no of boys = 6 no of girls = 4 therefore , no of ways at least 1 boy can be selected = 6 c 4 * 4 c 0 + 6 c 3 * 4 c 1 + 6 c 2 * 4 c 2 + 6 c 1 * 4 c 3 = 15 + 80 + 90 + 24 = 209 ans - a | a = 6 + 4
b = math.factorial(a)
c = math.factorial(4)
d = math.factorial(6)
e = c * d
f = b / e
|
a ) 10 , b ) 11 , c ) 12 , d ) 13 , e ) 14 | e | divide(subtract(multiply(4, 17), add(add(4, 4), 7)), 4) | the youngest of 4 children has siblings who are 1 , 4 , and 7 years older than she is . if the average ( arithmetic mean ) age of the 4 siblings is 17 , what is the age of the youngest sibling ? | "x + ( x + 1 ) + ( x + 4 ) + ( x + 8 ) = 68 4 x + 12 = 68 4 x = 56 x = 14 the answer is e ." | a = 4 * 17
b = 4 + 4
c = b + 7
d = a - c
e = d / 4
|
a ) 95 , b ) 95.42 , c ) 95.45 , d ) 94 , e ) 93 | b | divide(multiply(const_4, 167), 7) | a , b , c weighted separately 1 st like a , b , c , then a & b , then b & c , then c & a at last a & b & c , the last weight was 167 , then what will be the average weight of the 7 reading ? | the sum of 7 weights = weights of [ a + b + c + ( a + b ) + ( b + c ) + ( c + a ) + ( a + b + c ) ] = weights of [ 4 * ( a + b + c ) = 4 * 167 = 668 therefore the average weight of 7 readings is 668 / 7 = 95.428 answer : b | a = 4 * 167
b = a / 7
|
a ) 2 km , b ) 2.4 km , c ) 2.5 km , d ) 3.4 km , e ) none | b | divide(multiply(subtract(5, 1), const_3), 5) | a man can row at 5 kmph in still water . if the velocity of current is 1 kmph and it takes him 1 hour to row to a plce and come back , how far is the place ? | "sol . speed downstream = ( 5 + 1 ) kmph = 6 kmph ; speed upstream = ( 5 - 1 ) kmph = 4 kmph . let the required distance be x km . then , x / 6 + x / 4 = 1 ⇔ 2 x + 3 x = 12 ⇔ 5 x = 12 ⇔ x = 2.4 km . answer b" | a = 5 - 1
b = a * 3
c = b / 5
|
a ) 4 / 19 , b ) 7 / 19 , c ) 12 / 19 , d ) 24 / 77 , e ) none | d | subtract(const_1, divide(choose(subtract(24, 4), const_2), choose(24, const_2))) | a box contains 24 electric bulbs , out of which 4 are defective . two bulbs are chosen at random from this box . the probability that at least one of these is defective is | "solution p ( none is defective ) = 20 c 2 / 24 c 2 = 53 / 77 . p ( at least one is defective ) = ( 1 - 53 / 77 ) = 24 / 77 . answer d" | a = 24 - 4
b = math.comb(a, 2)
c = math.comb(24, 2)
d = b / c
e = 1 - d
|
a ) 3 / 4 , b ) 2 / 3 , c ) 4 / 7 , d ) 1 / 2 , e ) 1 / 7 | e | divide(subtract(subtract(const_100, 30), subtract(90, 30)), subtract(const_100, 30)) | in a sample of christ college students , 30 percent are third year students and 90 percent are not second - year students . what fraction of those students who are not third - year students are second - year students ? | = 10 / 70 . = 1 / 7 my answer is e too | a = 100 - 30
b = 90 - 30
c = a - b
d = 100 - 30
e = c / d
|
a ) 4 hours , b ) 12 hours , c ) 8 hours , d ) 5 hours , e ) 15 hours | a | divide(12, const_1) | an electric pump can fill a tank in 3 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 / 3 - 1 / 12 = 1 / 4 the leak will empty the tank in 4 hours answer is a" | a = 12 / 1
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a ) 120 , b ) 128 , c ) 136 , d ) 144 , e ) 152 | c | subtract(800, add(add(multiply(divide(45, const_100), 800), multiply(divide(23, const_100), 800)), multiply(divide(15, const_100), 800))) | in a school of 800 students , 45 % wear blue shirts , 23 % wear red shirts , 15 % wear green shirts , and the remaining students wear other colors . how many students wear other colors ( not blue , not red , not green ) ? | "45 + 23 + 15 = 83 % 100 – 83 = 17 % 800 * 17 / 100 = 136 the answer is c ." | a = 45 / 100
b = a * 800
c = 23 / 100
d = c * 800
e = b + d
f = 15 / 100
g = f * 800
h = e + g
i = 800 - h
|
a ) 7 / 45 , b ) 2 / 45 , c ) 4 / 15 , d ) 1 / 30 , e ) 9 / 45 | d | divide(2, 5) | what is the hcf of 2 / 3 , 4 / 12 and 6 / 5 | "explanation : hcf of fractions = hcf of numerators / lcm of denominators = ( hcf of 2 , 4 , 6 ) / ( lcm of 3 , 12 , 5 ) = 2 / 60 = 1 / 30 answer : option d" | a = 2 / 5
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a ) 127 % , b ) 120 % , c ) 121 % , d ) 135 % , e ) 140 % | a | multiply(power(add(1, divide(divide(10, const_100), const_2)), add(const_2, const_3)), const_100) | michelle deposited a certain sum of money in a savings account on july 1 st , 2007 . she earns an 10 % interest compounded semiannually . the sum of money in the account on december 31 st , 2009 is approximately what percent of the initial deposit ? | "since michelle earns 10 % interest compounded semiannually , then she earns 5 % interest every 6 months . now , the simple interest earned in 5 periods ( 30 months = 5 * 6 months ) would be 5 % * 5 = 25 % . but , since the interest is compounded every 6 months , then there would be interest earned on interest ( very small amount ) thus the actual interest earned would be a little bit more than 25 % , only answer choice a fits . answer : a" | a = 10 / 100
b = a / 2
c = 1 + b
d = 2 + 3
e = c ** d
f = e * 100
|
a ) 20 % , b ) 30 % , c ) 40 % , d ) 50 % , e ) 60 % | b | subtract(100, 70) | john want to buy a $ 100 trouser at the store , but he think it â € ™ s too expensive . finally , it goes on sale for $ 70 . what is the percent decrease ? | "the is always the difference between our starting and ending points . in this case , it â € ™ s 100 â € “ 70 = 30 . the â € œ original â € is our starting point ; in this case , it â € ™ s 100 . ( 30 / 100 ) * 100 = ( 0.3 ) * 100 = 30 % . b" | a = 100 - 70
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a ) 23 , b ) 19 , c ) 22 , d ) 20 , e ) 14 | e | add(7, divide(multiply(7, subtract(10000, 8000)), subtract(8000, 6000))) | the average salary of all the workers in a workshop is rs . 8000 . the average salary of 7 technicians is rs . 10000 and the average salary of the rest is rs . 6000 . the total number of workers in the workshop is : | "let the total number of workers be x . then , 8000 x = ( 10000 * 7 ) + 6000 ( x - 7 ) = 2000 x = 28000 = x = 14 . answer : e" | a = 10000 - 8000
b = 7 * a
c = 8000 - 6000
d = b / c
e = 7 + d
|
['a ) 76', 'b ) 65', 'c ) 59.5', 'd ) 49', 'e ) 38.5'] | e | multiply(subtract(8.5, multiply(1.5, const_2)), subtract(10, multiply(1.5, const_2))) | a picture is copied onto a sheet of paper 8.5 inches by 10 inches . a 1.5 inch margin is left all around . what area in square inches does the picture cover ? | area covered by picture = ( 8.5 - 3 ) * ( 10 - 3 ) = 5.5 * 7 = 38.5 answer e | a = 1 * 5
b = 8 - 5
c = 1 * 5
d = 10 - c
e = b * d
|
a ) 142 m sqaure , b ) 122 m sqaure , c ) 135 m sqaure , d ) 188 m sqaure , e ) none of these | b | add(multiply(const_2, add(multiply(add(divide(25, const_100), 2), 6), multiply(add(divide(25, const_100), 2), 9))), multiply(6, 9)) | a cistern 9 m long and 6 m wide contains water up to a breadth of 2 m 25 cm . find the total area of the wet surface . | "explanation : area of the wet surface = 2 [ lb + bh + hl ] - lb = 2 [ bh + hl ] + lb = 2 [ ( 6 * 2.25 + 9 * 2.25 ) ] + 9 * 6 = 122 m square option b" | a = 25 / 100
b = a + 2
c = b * 6
d = 25 / 100
e = d + 2
f = e * 9
g = c + f
h = 2 * g
i = 6 * 9
j = h + i
|
['a ) 18', 'b ) 20', 'c ) 24', 'd ) 26', 'e ) 40'] | a | divide(add(add(multiply(const_2, 4), multiply(const_2, 3)), sqrt(add(multiply(const_4, subtract(120, multiply(multiply(const_2, 3), multiply(const_2, 4)))), power(add(multiply(const_2, 4), multiply(const_2, 3)), const_2)))), const_2) | a tailor trims 4 feet from opposite edges of a square piece of cloth , and 3 feet from the other two edges . if 120 square feet of cloth remain , what was the length of a side of the original piece of cloth ? | let the original side of the square be x . ( x - 8 ) * ( x - 6 ) = 120 = 10 * 12 x = 18 the answer is a . | a = 2 * 4
b = 2 * 3
c = a + b
d = 2 * 3
e = 2 * 4
f = d * e
g = 120 - f
h = 4 * g
i = 2 * 4
j = 2 * 3
k = i + j
l = k ** 2
m = h + l
n = math.sqrt(m)
o = c + n
p = o / 2
|
a ) 55 % , b ) 60 % , c ) 65 % , d ) 70 % , e ) 75 % | d | multiply(const_100, divide(subtract(subtract(const_100, 54), subtract(60, multiply(60, divide(70, const_100)))), subtract(const_100, 60))) | in a company , 54 percent of the employees are men . if 60 percent of the employees are unionized and 70 percent of these are men , what percent of the non - union employees are women ? | the percent of employees who are unionized and men is 0.7 * 0.6 = 42 % the percent of employees who are unionized and women is 60 - 42 = 18 % 46 % of all employees are women , so non - union women are 46 % - 18 % = 28 % 40 % of all employees are non - union . the percent of non - union employees who are women is 28 % / 40 % = 70 % the answer is d . | a = 100 - 54
b = 70 / 100
c = 60 * b
d = 60 - c
e = a - d
f = 100 - 60
g = e / f
h = 100 * g
|
a ) 5 days , b ) 7 days , c ) 8 days , d ) 9 days , e ) 11 days | b | inverse(add(multiply(divide(const_1, multiply(10, 7)), 5), multiply(divide(const_1, multiply(10, 14)), 10))) | 10 women can complete a work in 7 days & 10 children take 14 days to complete the work . how many days will 5 women and 10 children take to complete the work ? | 1 woman ' s 1 day ' s work = 1 / 70 1 child ' s 1 day ' s work = 1 / 140 5 women and 10 children 1 day work = ( 5 / 70 + 10 / 140 ) = 1 / 7 so 5 women and 10 children will finish the work in 7 days . b ) | a = 10 * 7
b = 1 / a
c = b * 5
d = 10 * 14
e = 1 / d
f = e * 10
g = c + f
h = 1/(g)
|
a ) 1 and 8 , b ) 2 and 6 , c ) 0 and 9 , d ) 1 and 7 , e ) 2 and 9 | d | add(multiply(7, const_10), divide(add(44, 5), 7)) | 5 n + 7 > 12 and 7 n - 5 < 44 ; n must be between which numbers ? | "5 n > 5 - - > n > 1 7 n < 49 - - > n < 7 1 < n < 7 answer : d" | a = 7 * 10
b = 44 + 5
c = b / 7
d = a + c
|
a ) $ 200 , b ) $ 360 , c ) $ 400 , d ) $ 500 , e ) $ 600 | b | multiply(multiply(4, 3), divide(30, subtract(4, 3))) | the total cost of a vacation was divided among 3 people . if the total cost of the vacation had been divided equally among 4 people , the cost per person would have been $ 30 less . what was the total cost cost of the vacation ? | "c for cost . p price per person . c = 3 * p c = 4 * p - 120 substituting the value of p from the first equation onto the second we get p = 120 . plugging in the value of p in the first equation , we get c = 360 . which leads us to answer choice b" | a = 4 * 3
b = 4 - 3
c = 30 / b
d = a * c
|
a ) 2 % , b ) 5 % , c ) 14 % , d ) 60 % , e ) 73 % | d | floor(multiply(subtract(divide(9, 60), divide(7, 75)), const_100)) | a survey was sent to 75 customers , 7 of whom responded . then the survey was redesigned and sent to another 60 customers , 9 of whom responded . by approximately what percent did the response rate increase from the original survey to the redesigned survey ? | "rate of first survey = 7 / 75 rate of second survey = 9 / 60 % response rate increase ( 9 / 60 - 7 / 75 ) / ( 7 / 75 ) = 60 % answer is d = 60 %" | a = 9 / 60
b = 7 / 75
c = a - b
d = c * 100
e = math.floor(d)
|
a ) 128 % , b ) b . 120 % , c ) 96 % , d ) 80 % , e ) 64 % | a | add(subtract(const_100, 20), multiply(subtract(const_100, 20), divide(60, const_100))) | mary ' s income is 60 % more than tim ' s income and tim ' s income is 20 % less than juan ' s income . what % of juan ' s income is mary ' s income . | even i got 96 % j = 100 t = 100 * 0.8 = 80 m = 80 * 1.6 = 128 if mary ' s income is x percent of j m = j * x / 100 x = m * 100 / j = 128 * 100 / 100 = 128 ans : a | a = 100 - 20
b = 100 - 20
c = 60 / 100
d = b * c
e = a + d
|
a ) 11 , b ) 50 , c ) 88 , d ) 65 , e ) 54 | e | divide(divide(subtract(125, multiply(multiply(9, const_0_2778), 9)), 9), const_0_2778) | a train 125 m long passes a man , running at 9 km / hr in the same direction in which the train is going , in 10 seconds . the speed of the train is ? | "speed of the train relative to man = ( 125 / 10 ) m / sec = ( 25 / 2 ) m / sec . [ ( 25 / 2 ) * ( 18 / 5 ) ] km / hr = 45 km / hr . let the speed of the train be x km / hr . then , relative speed = ( x - 9 ) km / hr . x - 9 = 45 = = > x = 54 km / hr . answer : e" | a = 9 * const_0_2778
b = a * 9
c = 125 - b
d = c / 9
e = d / const_0_2778
|
a ) 60 mph , b ) 56.67 mph , c ) 53.33 mph , d ) 64 mph , e ) 66.67 mph | d | add(divide(add(multiply(80, 3), multiply(40, 2)), add(3, 2)), subtract(divide(const_100, 3), const_0_33)) | steve traveled the first 2 hours of his journey at 40 mph and the last 3 hours of his journey at 80 mph . what is his average speed of travel for the entire journey ? | "answer average speed of travel = total distance travelled / total time taken total distance traveled by steve = distance covered in the first 2 hours + distance covered in the next 3 hours . distance covered in the first 2 hours = speed * time = 40 * 2 = 80 miles . distance covered in the next 3 hours = speed * time = 80 * 3 = 240 miles . therefore , total distance covered = 80 + 240 = 320 miles . total time taken = 2 + 3 = 5 hours . hence , average speed = total distance travelled / total time taken = 320 / 5 = 64 miles per hour . choice d" | a = 80 * 3
b = 40 * 2
c = a + b
d = 3 + 2
e = c / d
f = 100 / 3
g = f - const_0_33
h = e + g
|
a ) 24000 , b ) 23000 , c ) 22000 , d ) 25000 , e ) 26000 | d | divide(divide(divide(multiply(multiply(multiply(25, const_100), multiply(2, const_100)), multiply(0.75, const_100)), 20), 10), 7.5) | a brick measures 20 cm * 10 cm * 7.5 cm how many bricks will be required for a wall 25 m * 2 m * 0.75 m ? | "25 * 2 * 0.75 = 20 / 100 * 10 / 100 * 7.5 / 100 * x 25 = 1 / 100 * x = > x = 25000 answer : d" | a = 25 * 100
b = 2 * 100
c = a * b
d = 0 * 75
e = c * d
f = e / 20
g = f / 10
h = g / 7
|
a ) $ 1500 , b ) $ 4000 , c ) $ 2000 , d ) $ 2500 , e ) $ 3200 | b | multiply(divide(5, 2), 1600) | the ratio of the incomes of a and b is 5 : 4 and the ratio of their expenditure is 3 : 2 . if at the end of the year , each saves $ 1600 then the income of a is ? | let the income of a and b be $ 5 x and $ 4 x let their expenditures be $ 3 y and $ 2 y 5 x - 3 y = 1600 - - - - - - - 1 ) 4 x - 2 y = 1600 - - - - - - - 2 ) from 1 ) and 2 ) x = 800 a ' s income = 5 x = 5 * 800 = $ 4000 answer is b | a = 5 / 2
b = a * 1600
|
a ) 9800000 , b ) 1000000 , c ) 7500000 , d ) 1200000 , e ) none of these | b | divide(multiply(multiply(multiply(const_4.0, const_100), multiply(7, const_100)), multiply(8, const_100)), multiply(multiply(8, 7), 7)) | a wooden box of dimensions 8 m x 7 m x 6 m is to carry rectangularboxes of dimensions 8 cm x 7 cm x 6 cm . the maximum number ofboxes that can be carried in the wooden box , is | explanation : number = ( 800 * 700 * 600 ) / 8 * 7 * 6 = 1000000 answer : b | a = 4 * 0
b = 7 * 100
c = a * b
d = 8 * 100
e = c * d
f = 8 * 7
g = f * 7
h = e / g
|
a ) 565 , b ) 444 , c ) 676 , d ) 420 , e ) 663 | d | add(multiply(divide(60, subtract(19, 16)), 16), multiply(divide(60, subtract(19, 16)), 19)) | 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 19 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 - - - - - 3 ? - - - - - - 60 12 h rs = 16 + 19 = 35 t = 12 d = 35 * 12 = 420 answer : d" | a = 19 - 16
b = 60 / a
c = b * 16
d = 19 - 16
e = 60 / d
f = e * 19
g = c + f
|
a ) 8 hours , b ) 6 hours , c ) 4 hours , d ) 15 hours , e ) 6 hours | d | divide(60, subtract(12, subtract(divide(60, 3), 12))) | the speed of the boat in still water in 12 kmph . it can travel downstream through 60 kms in 3 hrs . in what time would it cover the same distance upstream ? | "still water = 12 km / hr downstream = 60 / 3 = 20 km / hr upstream = > > still water = ( u + v / 2 ) = > > 12 = u + 20 / 2 = 4 km / hr so time taken in upstream = 60 / 4 = 15 hrs answer : d" | a = 60 / 3
b = a - 12
c = 12 - b
d = 60 / c
|
a ) 40 min , b ) 62 min , c ) 70 min , d ) 74 min , e ) 76 min | a | multiply(multiply(const_2, divide(multiply(3, const_60), add(subtract(200, 25), multiply(const_2, 25)))), 25) | if there are 200 questions in a 3 hr examination . among these questions are 25 type a problems , which requires twice as much as time be spent than the rest of the type b problems . how many minutes should be spent on type a problems ? | "x = time for type b prolems 2 x = time for type a problem total time = 3 hrs = 180 min 175 x + 25 * 2 x = 180 x = 180 / 225 x = 0.8 time taken for type a problem = 25 * 2 * 0.8 = 40 min answer : a" | a = 3 * const_60
b = 200 - 25
c = 2 * 25
d = b + c
e = a / d
f = 2 * e
g = f * 25
|
a ) 2 , b ) 4 , c ) 16 , d ) 38 , e ) 40 | d | divide(subtract(power(20, const_2), power(18, const_2)), const_2) | the size of a television screen is given as the length of the screen ' s diagonal . if the screens were flat , then the area of a square 20 - inch screen would be how many square inches greater than the area of a square 18 - inch screen ? | "pythogoras will help here ! let the sides be x and diagonal be d then d ^ 2 = 2 x ^ 2 and area = x ^ 2 now plug in the given diagonal values to find x values and then subtract the areas ans will be 20 ^ 2 / 2 - 18 ^ 2 / 2 = 76 / 2 = 38 ans d ." | a = 20 ** 2
b = 18 ** 2
c = a - b
d = c / 2
|
a ) 56 , b ) 46 , c ) 64 , d ) 65 , e ) 10 | b | add(1, multiply(multiply(subtract(8, 3), 3), 3)) | in a group of 8 engineers , 3 engineers have a phd degree ; the others have only an ms or bs degree . a team of 3 engineers is to be chosen which must have at least 1 engineer with a phd , how many different teams can be chosen ? | the problem asks for a combination , since order does n ' t matter . now , selecting r items from a set of n gives the combination formula : ncr = n ! / r ! ( n - r ) ! n = 8 r = 3 so , total teams is 8 c 3 = 8 ! / ( 3 ! ( 8 - 3 ) ! ) = 56 , and n = 8 - 3 = 5 r = 3 for teams without a phd is 5 c 3 = 5 ! / ( 3 ! ( 5 - 3 ) ! ) = 10 , so , teams with at least 1 phd = 56 - 10 = 46 answer : b | a = 8 - 3
b = a * 3
c = b * 3
d = 1 + c
|
a ) 45 , b ) 50 , c ) 55 , d ) 60 , e ) 63 | d | multiply(divide(multiply(divide(1, 12), const_3), divide(divide(1, 30), const_2)), 4) | in the standard formulation of a flavored drink the ratio by volume of flavoring to corn syrup to water is 1 : 12 : 30 . in the sport formulation , the ratio of flavoring to corn syrup is three times as great as in the standard formulation , and the ratio of flavoring to water is half that of the standard formulation . if a large bottle of the sport formulation contains 4 ounces of corn syrup , how many ounces of water does it contain ? | "f : c : w 1 : 12 : 30 sport version : f : c 3 : 12 f : w 1 : 60 or 3 : 180 so c : f : w = 12 : 3 : 180 c / w = 12 / 180 = 3 ounces / x ounces x = 4 * 180 / 12 = 60 ounces of water d" | a = 1 / 12
b = a * 3
c = 1 / 30
d = c / 2
e = b / d
f = e * 4
|
a ) 10000 , b ) 5000 , c ) 15000 , d ) 20000 , e ) 6000 | b | divide(1000, divide(20, const_100)) | in an election between two candidates , the winner has a margin of 20 % of the votes polled . if 1000 people change their mind and vote for the loser , the loser would have won by a margin of 20 % of the votes polled . find the total number of votes polled in the election ? | winner - looser 60 % - 40 % if 1000 people change their mind and vote for the loser : winner - looser 40 % - 60 % thus 1,000 people compose 20 % of all voters , which means that the total number of votes is 5,000 . answer : b | a = 20 / 100
b = 1000 / a
|
a ) 243 , b ) 265 , c ) 375 , d ) 390 , e ) 400 | c | multiply(divide(600, 5), 3) | there are 600 students in a school . the ratio of boys and girls in this school is 3 : 5 . find the total of girls & boys are there in this school ? | in order to obtain a ratio of boys to girls equal to 3 : 5 , the number of boys has to be written as 3 x and the number of girls as 5 x where x is a common factor to the number of girls and the number of boys . the total number of boys and girls is 600 . hence 3 x + 5 x = 600 solve for x 8 x = 600 x = 75 number of boys 3 x = 3 × 75 = 225 number of girls 5 x = 5 × 75 = 375 c | a = 600 / 5
b = a * 3
|
a ) 1 / 16 , b ) 1 / 4 , c ) 1 , d ) 4 , e ) 8 | a | divide(multiply(divide(2, 8), const_100), multiply(divide(8, 2), const_100)) | 8 is 2 % of a , and 2 is 8 % of b . c equals b / a . what is the value of c ? | "2 a / 100 = 8 a = 400 8 b / 100 = 2 b = 25 c = b / a = 25 / 400 = 1 / 16 the answer is a ." | a = 2 / 8
b = a * 100
c = 8 / 2
d = c * 100
e = b / d
|
a ) $ 120 , b ) $ 135 , c ) $ 150 , d ) $ 165 , e ) $ 252 | e | add(multiply(divide(subtract(400, 30), 5), 3), 30) | a certain psychologist charges $ 30 more for the first hour of therapy than for each additional hour . if the total charge to a patient who receives 5 hours of therapy is $ 400 , what is the total charge to a patient who receives only 3 hours of therapy ? | "let the charge for first hour = x + 30 then charge for each additional hour = x x + 30 + 4 x = 400 = > 5 x = 370 = > x = 74 total charge for patient for 3 hours of therapy = x + 30 + 2 x = 3 x + 30 = 252 $ answer e" | a = 400 - 30
b = a / 5
c = b * 3
d = c + 30
|
a ) 1 hr , b ) 2 hr , c ) 1.5 hr , d ) 2.5 hr , e ) 3 hr | b | divide(add(30, multiply(3, 30)), const_60) | in a race , the speeds of a and b are in the ratio 3 : 4 . a takes 30 minutes more than b to reach the destination . the time taken by a to reach the destination is ? | explanation : ratio of speeds = 3 : 4 distance remaining constant , the ratio of time taken = 4 : 3 a takes 0.5 hours more than bhence time taken by a = 4 × 0.5 = 2 hour answer : b | a = 3 * 30
b = 30 + a
c = b / const_60
|
a ) s . 486 , b ) s . 455 , c ) s . 640 , d ) s . 480 , e ) s . 489 | c | divide(multiply(800, const_100), add(const_100, 25)) | by selling an article at rs . 800 , a profit of 25 % is made . find its cost price ? | sp = 800 cp = ( sp ) * [ 100 / ( 100 + p ) ] = 800 * [ 100 / ( 100 + 25 ) ] = 800 * [ 100 / 125 ] = rs . 640 answer : c | a = 800 * 100
b = 100 + 25
c = a / b
|
a ) $ 1010 , b ) $ 1025 , c ) $ 1125 , d ) $ 1100 , e ) $ 1000 | e | divide(add(500, 400), subtract(const_1, divide(10, const_100))) | kanul spent $ 500 in buying raw materials , $ 400 in buying machinery and 10 % of the total amount he had as cash with him . what was the total amount ? | "let the total amount be x then , ( 100 - 10 ) % of x = 500 + 400 90 % of x = 900 90 x / 100 = 9000 / 9 x = $ 1000 answer is e" | a = 500 + 400
b = 10 / 100
c = 1 - b
d = a / c
|
a ) 2 , b ) 4 , c ) 6 , d ) 8 , e ) 10 | b | divide(subtract(44, subtract(multiply(4, divide(8, 2)), 8)), 8) | the area of one square is x ^ 2 + 8 x + 16 and the area of another square is 4 x ^ 2 − 20 x + 25 . if the sum of the perimeters of both squares is 44 , what is the value of x ? | "the areas are ( x + 4 ) ^ 2 and ( 2 x - 5 ) ^ 2 . the lengths of the sides are x + 4 and 2 x - 5 . if we add the two perimeters : 4 ( x + 4 ) + 4 ( 2 x - 5 ) = 44 12 x = 48 x = 4 the answer is b ." | a = 8 / 2
b = 4 * a
c = b - 8
d = 44 - c
e = d / 8
|
a ) 1 / 32 , b ) 3 / 32 , c ) 1 / 4 , d ) 5 / 16 , e ) 1 / 2 | d | divide(const_2, choose(add(const_3, const_3), const_3)) | what is the probability of getting exactly three heads on five flips of a fair coin ? | "5 flips of a fair coin to get = hhhtt = no . of ways this can be achieved = 5 ! / 3 ! x 2 ! = 10 probability to get any of the above 10 arrangements ( hhhtt ) = ( 1 / 2 ) ^ 5 = 1 / 32 total probability = 1 / 32 x 10 = 5 / 16 answer : d" | a = 3 + 3
b = math.comb(a, 3)
c = 2 / b
|
a ) rs . 3000 , b ) rs . 607.39 , c ) rs . 1214.78 , d ) rs . 1518.48 , e ) none | d | divide(multiply(divide(divide(add(divide(multiply(3000, 12), const_100), divide(multiply(add(3000, divide(multiply(3000, 12), const_100)), 12), const_100)), 3), 4), const_100), 10) | simple interest on a certain sum of money for 4 years at 10 % per annum is half the compound interest on rs . 3000 for 3 years at 12 % per annum . the sum placed on simple interest is | "solution c . i . = rs [ 3000 x ( 1 + 12 / 100 ) 3 - 3000 ] rs . ( 3000 x 112 / 100 x 112 / 100 x 112 / 100 - 3000 ) = rs . 1214.78 . sum = rs . [ 607.39 x 100 / 4 x 10 ] = rs . 1518.48 . answer d" | a = 3000 * 12
b = a / 100
c = 3000 * 12
d = c / 100
e = 3000 + d
f = e * 12
g = f / 100
h = b + g
i = h / 3
j = i / 4
k = j * 100
l = k / 10
|
a ) 1 / 18 , b ) 1 / 6 , c ) 1 / 2 , d ) 1 / 3 , e ) 5 / 6 | c | subtract(add(14, divide(14, 3)), divide(2, 6)) | in a certain lottery , the probability that a number between 14 and 20 , inclusive , is drawn is 1 / 6 . if the probability that a number 14 or larger is drawn is 2 / 3 , what is the probability that a number less than or equal to 20 is drawn ? | "you can simply use sets concept in this question . the formula total = n ( a ) + n ( b ) - n ( a and b ) is applicable here too . set 1 : number 14 or larger set 2 : number 20 or smaller 1 = p ( set 1 ) + p ( set 2 ) - p ( set 1 and set 2 ) ( combined probability is 1 because every number will be either 14 or moreor 20 or lessor both ) 2 / 3 + p ( set 2 ) - 1 / 6 = 1 p ( set 2 ) = 1 / 2 answer ( c )" | a = 14 / 3
b = 14 + a
c = 2 / 6
d = b - c
|
a ) 0 , b ) 5 , c ) 6 , d ) 10 , e ) 12 | c | add(divide(50, const_10), const_1) | x is the product of each integer from 1 to 50 , inclusive and y = 100 ^ k , where k is an integer . what is the greatest value of k for which y is a factor of x ? | x = 1 * 2 * 3 * 4 * 5 * . . . . . . * 49 * 50 = 50 ! y = 100 ^ k 100 = 10 ^ 2 = 2 ^ 2 * 5 ^ 2 on prime factorization of x , power of 5 will always be greater than power of 2 concept : in every factorial value ( x ! ) , on prime factorization , the power of bigger prime number > the power of smaller prime number also , power of any prime number in any factorial can be calculated by following understanding power of prime x in n ! = [ n / x ] + [ n / x ^ 2 ] + [ n / x ^ 3 ] + [ n / x ^ 4 ] + . . . and so on where , [ x ] = greatest integer less than or equal to x and the explanation of terms is as follows [ n / x ] = no . of integers that are multiple of x from 1 to n [ n / x ^ 2 ] = no . of integers that are multiple of x ^ 2 from 1 to n whose first power has been counted in previous step and second is being counted at this step [ n / x ^ 3 ] = no . of integers that are multiple of x ^ 3 from 1 to n whose first two powers have been counted in previous two step and third power is counted at this step and so on . . . . . where [ n / x ] is greatest integer value of ( n / x ) less than or equal to ( n / x ) i . e . [ 100 / 3 ] = [ 33.33 ] = 33 i . e . [ 100 / 9 ] = [ 11.11 ] = 11 etc . power of 5 in x = [ 50 / 5 ] + [ 50 / 5 ^ 2 ] + [ 50 / 5 ^ 3 ] . . . = 10 + 2 + 0 = 12 power of 2 in x = [ 50 / 2 ] + [ 50 / 2 ^ 2 ] + [ 50 / 2 ^ 3 ] . . . = 25 + 12 + 6 + 3 + 1 = 47 i . e . x = 50 ! = 2 ^ 47 * 5 ^ 12 * . . . = 10 ^ 12 * . . . = 100 ^ 6 * . . . i . e . x is divisible by a 6 th power of 100 at the most i . e . y = 100 ^ k = 100 ^ 6 i . e . k = 6 answer : option c | a = 50 / 10
b = a + 1
|
a ) 16.6 , b ) 44 , c ) 27 , d ) 50 , e ) 56 | a | add(add(multiply(7, divide(8, subtract(multiply(divide(4, 7), 7), 3))), divide(8, subtract(multiply(divide(4, 7), 7), 3))), 8) | in a can , there is a mixture of milk and water in the ratio 3 : 7 . if it is filled with an additional 8 litres of milk the can would be full and ratio of milk and water would become 4 : 7 . find the capacity of the can ? | "let the capacity of the can be t litres . quantity of milk in the mixture before adding milk = 3 / 10 ( t - 8 ) after adding milk , quantity of milk in the mixture = 4 / 11 t . 4 t / 11 - 8 = 3 / 10 ( t - 8 ) 37 t = 880 - 264 = > t = 16.6 . answer : a" | a = 4 / 7
b = a * 7
c = b - 3
d = 8 / c
e = 7 * d
f = 4 / 7
g = f * 7
h = g - 3
i = 8 / h
j = e + i
k = j + 8
|
a ) 20.5 , b ) 21.5 , c ) 22.5 , d ) 23.5 , e ) 24.5 | c | divide(add(38, 7), const_2) | dawson and henry are in a relay race . dawson runs the first leg of the course in 38 seconds . henry runs the second leg of the course in 7 seconds . what was the average time they took to run a leg of the course ? | ( 38 seconds + 7 seconds ) / 2 = 22.5 seconds correct option is : c | a = 38 + 7
b = a / 2
|
a ) 20 % , b ) 40 % , c ) 60 % , d ) 80 % , e ) 100 % | b | multiply(subtract(const_1, divide(1080, multiply(1500, divide(add(const_100, 20), const_100)))), const_100) | if 1500 is increased by 20 % , and then reduced by y % , yielding 1080 , what is y ? | soln : - 1,500 * 1.2 = 1,800 . 1,800 ( 1 - y / 100 ) = 1,080 - - > ( 1 - y / 100 ) = 0.6 - - > y = 40 % . answer : b | a = 100 + 20
b = a / 100
c = 1500 * b
d = 1080 / c
e = 1 - d
f = e * 100
|
a ) 36 , b ) 42 , c ) 28 , d ) 32 , e ) 45 | d | inverse(divide(const_3, multiply(24, const_4))) | if a is thrice as fast as b and together can do a work in 24 days . in how many days a alone can do the work ? | "a ’ s one day ’ s work = 1 / x b ’ s one day ’ s work = 1 / 3 x a + b ’ s one day ’ s work = 1 / x + 1 / 3 x = 1 / 24 = 3 + 1 / 3 x = 4 / 3 x = 1 / 24 x = 24 * 4 / 3 = 32 answer : d" | a = 24 * 4
b = 3 / a
c = 1/(b)
|
a ) 5 ½ ft , b ) 7 ½ ft , c ) 8 ½ ft , d ) 9 ½ ft , e ) 10 ft | b | subtract(divide(12, const_2), divide(const_1, const_2)) | the figure above shows the dimensions of a semicircular cross section of a one - way tunnel . the single traffic lane is 12 feet wide and is equidistant from the sides of the tunnel . if vehicles must clear the top of the tunnel by at least ½ foot when they are inside the traffic lane , what should be the limit u on the height of vehicles that are allowed to use the tunnel ? | "let ' s label the midpoint of the circle o . since the base of the semi - circle is 20 , we know that the diameter is 20 and , accordingly , the radius is 10 . we also know that the traffic lane is 12 feet long and there ' s an equal amount of space on either side , so the traffic lane extends 6 feet on either side of o . let ' s call the leftmost point on the base of the traffic lane a . so , the distance oa is 6 . now draw a line straight up from a to the top of the tunnel . let ' s label the point at which the line intersects the circle b . the answer to the question will , therefore , be the height ab - . 5 feet ( we need to leave . 5 feet of clearance ) . here ' s the key to solving the question : if we draw a line from o to b , that line is a radius of the circle and , therefore , has length 10 . we now have right triangle oab ( the right angle is at point a ) , with leg oa = 6 and hypotenuse ob = 10 . we can now solve for leg ab = 8 ( either by applying the pythagorean theorum or by applying the 3 / 4 / 5 special right triangle ratio ) . finally : ab = 8 , so the correct answer u is 8 - . 5 = 7.5 . . . choose ( b ) ! from a strategic guessing point of view , as soon as we realize that the height of the tunnel is 10 in the middle , we should quickly eliminate ( d ) and ( e ) as too big ; worse case you have a 1 / 3 shot at picking up the points . b" | a = 12 / 2
b = 1 / 2
c = a - b
|
a ) 1000 , b ) 1200 , c ) 1340 , d ) 1450 , e ) 1600 | e | multiply(multiply(4, const_1000), 4) | find the product of the local value and absolute value of 4 in 20468 ? | "local value of 4 = 4 x 100 = 400 place value of 4 = 4 there fore = 4 x 400 = 1600 e" | a = 4 * 1000
b = a * 4
|
a ) 32 , b ) 28 , c ) 49 , d ) 64 , e ) 21 | d | divide(divide(add(500, 300), const_1000), divide(45, const_3600)) | a train 500 meters long completely crosses a 300 meters long bridge in 45 seconds . what is the speed of the train is ? | "s = ( 500 + 300 ) / 45 = 800 / 45 * 18 / 5 = 64 answer : d" | a = 500 + 300
b = a / 1000
c = 45 / 3600
d = b / c
|
a ) 9 : 6 , b ) 9 : 8 , c ) 9 : 0 , d ) 9 : 2 , e ) 9 : 1 | b | divide(add(multiply(3000, 6), multiply(multiply(3000, const_2), 6)), multiply(4000, add(6, 6))) | a and b invests rs . 3000 and rs . 4000 respectively in a business . if a doubles his capital after 6 months . in what ratio should a and b divide that year ' s profit ? | "( 3 * 6 + 6 * 6 ) : ( 4 * 12 ) 54 : 48 = > 9 : 8 answer : b" | a = 3000 * 6
b = 3000 * 2
c = b * 6
d = a + c
e = 6 + 6
f = 4000 * e
g = d / f
|
a ) 35 , b ) 40 , c ) 45 , d ) 50 , e ) 55 | a | divide(subtract(100, 30), const_2) | we have boxes colored either red or blue . in each blue box there is a fixed number of blueberries . in each red box there is a fixed number of strawberries . if we dispose of one blue box for one additional red box , the total number of berries would increase by 30 , and the difference between the total number of strawberries and the total number of blueberries would increase by 100 . each blue box contains how many blueberries ? | "let x be the number of blueberries in each blue box . then there are x + 30 strawberries in each red box . x + ( x + 30 ) = 100 x = 35 the answer is a ." | a = 100 - 30
b = a / 2
|
a ) 124 , b ) 97 , c ) 117 , d ) 229 , e ) 362 | e | add(117, multiply(7, add(34, const_1))) | there are an average of 117 jellybeans in each of 34 bags of various sizes . after adding another bag , the average number of jellybeans per bag increases by 7 . how many jellybeans are in the new bag ? | total jellybeans in 34 bags = 34 * 117 = 3978 total jellybeans in 35 bags = 35 * 124 = 4340 number of jellybeans in new bag = 3978 - 4340 = 362 jellybeans answer is e | a = 34 + 1
b = 7 * a
c = 117 + b
|
a ) 450 , b ) 300 , c ) 800 , d ) 610 , e ) 175 | c | divide(add(175, 89), divide(33, const_100)) | a student has to obtain 33 % of the total marks to pass . he got 175 marks and failed by 89 marks . the maximum marks are ? | "let the maximum marks be x then , 33 % of x = 175 + 89 33 x / 100 = 264 x = 800 answer is c" | a = 175 + 89
b = 33 / 100
c = a / b
|
a ) 2 , b ) 1.5 , c ) 3.5 , d ) 2.5 , e ) 3 | c | divide(subtract(divide(divide(multiply(90, 7), const_10), divide(3, 2)), 10), divide(90, const_10)) | a solution of 90 litres contains milk and water in the ratio 7 : x . if 10 litres of water is added to the solution , the ratio becomes 3 : 2 , find the value of x ? | exp . total new quantity = original sol + water = 90 + 10 = 100 new ratio = 3 : 2 , new quantity of milk = 3 / 5 * 100 = 60 lit , new quantity of water = 2 / 5 * 100 = 40 lit water present initially = ( 40 - 10 ) = 30 lit ratio = 60 / 30 = 7 / x there for x = 3.5 answer : c | a = 90 * 7
b = a / 10
c = 3 / 2
d = b / c
e = d - 10
f = 90 / 10
g = e / f
|
a ) 600 , b ) 700 , c ) 800 , d ) 850 , e ) none of these | a | divide(multiply(const_100, 240), subtract(70, subtract(const_100, 70))) | in an election between the two candidates , the candidates who gets 70 % of votes polled is winned by 240 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 % - - - - - > 280 ( 40 * 6 = 240 ) 100 % - - - - - > 600 ( 100 * 6 = 600 ) answer : option a" | a = 100 * 240
b = 100 - 70
c = 70 - b
d = a / c
|
a ) 5 kmph , b ) 6 kmph , c ) 6.25 kmph , d ) 7 kmph , e ) 7.8 kmph | d | divide(subtract(42, divide(42, 2)), add(1, 2)) | in covering a distance of 42 km , abhay takes 2 hours more than sameer . if abhay doubles his speed , then he would take 1 hour less than sameer . abhay ' s speed is : | "let abhay ' s speed be x km / hr . then , 42 / x - 42 / 2 x = 3 6 x = 42 x = 7 km / hr . answer : option d" | a = 42 / 2
b = 42 - a
c = 1 + 2
d = b / c
|
a ) 81 , b ) 84 , c ) 85 , d ) 88 , e ) 90 | b | divide(subtract(multiply(96, const_2), multiply(96, divide(60, const_100))), add(divide(60, const_100), const_1)) | a store sells chairs and tables . if the price of 2 chairs and 1 table is 60 % of the price of 1 chair and 2 tables , and the price of 1 table and 1 chair is $ 96 , what is the price , in dollars , of 1 table ? ( assume that every chair has the same price and every table has the same price . ) | let c = chair ; t = table 2 c + 1 t = 0.6 ( 1 c + 2 t ) or c ( 2 - 0.6 ) = 1.2 t - 1 t or 1.4 c = 0.2 t therefore c = 0.2 / 1.4 t = 1 / 7 t ic + 1 t = 96 or 1 / 7 t + 1 t = 96 therefore t = 96 * 7 / 8 = 84 b | a = 96 * 2
b = 60 / 100
c = 96 * b
d = a - c
e = 60 / 100
f = e + 1
g = d / f
|
a ) 1 / 10 , b ) 3 / 20 , c ) 1 / 5 , d ) 1 / 4 , e ) 9 / 20 | e | subtract(add(divide(const_2, add(add(const_2, add(const_2, 4)), 12)), divide(add(const_2, add(const_2, 4)), add(add(const_2, add(const_2, 4)), 12))), multiply(divide(const_2, add(add(const_2, add(const_2, 4)), 12)), divide(add(const_2, add(const_2, 4)), add(add(const_2, add(const_2, 4)), 12)))) | maths , physics and chemistry books are stored on a library shelf that can accommodate 25 books . currently , 20 % of the shelf spots remain empty . there are twice as many maths books as physics books and the number of physics books is 4 greater than that of chemistry books . among all the books , 12 books are soft cover and the remaining are hard - cover . if there are a total of 7 hard - cover books among the maths and physics books . what is the probability t , that a book selected at random is either a hard cover book or a chemistry book ? | "first phase of this problem requires you to determine how many mathematics and chemistry books are even on the shelf . to do so , you have the equations : m + p + c = 20 ( since 4 / 5 of the 25 spots are full of books ) m = 2 p p = 4 + c from that , you can use substitution to get everything down to one variable . c = p - 4 m = 2 p p = p then ( p - 4 ) + 2 p + p = 20 , so 4 p = 24 and p = 6 . that means that there are 12 math , 6 physics , and 2 chemistry books on the shelf . with those numbers , you also know that there are 8 total hardcovers , 1 of which is chemistry . so if your goal is to get either a hardcover or a chemistry , there are 9 ways towin - either one of the 7 hardcovers that are n ' t chemistry or the two chemistry books . so out of the 20 total , t = 9 provide the desired outcome , making the answer e ." | a = 2 + 4
b = 2 + a
c = b + 12
d = 2 / c
e = 2 + 4
f = 2 + e
g = 2 + 4
h = 2 + g
i = h + 12
j = f / i
k = d + j
l = 2 + 4
m = 2 + l
n = m + 12
o = 2 / n
p = 2 + 4
q = 2 + p
r = 2 + 4
s = 2 + r
t = s + 12
u = q / t
v = o * u
w = k - v
|
a ) 45823 , b ) 53836 , c ) 44824 , d ) 66524 , e ) none of them | b | subtract(86, multiply(multiply(863, 86), 237)) | evaluate : 86 x 863 - 86 x 237 | "86 x 863 + 986 x 237 = 986 x ( 863 - 237 ) = 86 x 626 = 53836 . answer is b" | a = 863 * 86
b = a * 237
c = 86 - b
|
a ) 960 km , b ) 224 km , c ) 230 km , d ) 232 km , e ) 234 km | a | multiply(const_2, divide(multiply(multiply(20, 30), 40), add(20, 30))) | a man complete a journey in 40 hours . he travels first half of the journey at the rate of 20 km / hr and second half at the rate of 30 km / hr . find the total journey in km . | "0.5 x / 20 + 0.5 x / 30 = 40 - - > x / 20 + x / 30 = 80 - - > 5 x = 60 x 80 - - > x = ( 60 x 80 ) / 5 = 960 km . answer : a" | a = 20 * 30
b = a * 40
c = 20 + 30
d = b / c
e = 2 * d
|
a ) 475 , b ) 350 , c ) 450 , d ) 550 , e ) 650 | a | multiply(19, 25) | a large box contains 19 small boxes and each small box contains 25 chocolate bars . how many chocolate bars are in the large box ? | "the number of chocolate bars is equal to 19 * 25 = 475 correct answer a" | a = 19 * 25
|
a ) 2 / 5 , b ) 3 / 5 , c ) 2 / 3 , d ) 1 / 2 , e ) 3 / 7 | e | divide(choose(6, 4), choose(10, 6)) | a store has 10 bottles of juice , including 6 bottles of apple juice . in the evening , 6 bottles of juice are sold one by one . what is the probability of selling 4 bottles of apple juice among the 6 bottles ? assume that every bottle has an equal chance of being bought . | "the total number of ways to sell 6 bottles from 10 is 10 c 6 = 210 . the number of ways to sell 4 bottles of apple juice is 6 c 4 * 4 c 2 = 15 * 6 = 90 p ( selling 4 bottles of apple juice ) = 90 / 210 = 9 / 21 = 3 / 7 the answer is e ." | a = math.comb(6, 4)
b = math.comb(10, 6)
c = a / b
|
a ) 8 , b ) 12 , c ) 15 , d ) 16 , e ) 18 | a | subtract(50, add(add(subtract(18, 2), subtract(26, 2)), 2)) | in a class of 50 students , 18 take chorus , 26 take band , and 2 take both chorus and band . how many students in the class are not enrolled in either chorus or band ? | 16 + 2 + 24 + x = 50 42 + x = 50 x = 8 students a ) | a = 18 - 2
b = 26 - 2
c = a + b
d = c + 2
e = 50 - d
|
a ) 28.5 , b ) 27.675 , c ) 30 , d ) data inadequate , e ) 42 | e | subtract(divide(multiply(add(const_100, 34.9), const_100), subtract(const_100, 5)), const_100) | a shopkeeper sold an article offering a discount of 5 % and earned a profit of 34.9 % . what would have been the percentage of profit earned if no discount had been offered ? | "giving no discount to customer implies selling the product on printed price . suppose the cost price of the article is 100 . then printed price = 100 ã — ( 100 + 34.9 ) / ( 100 â ˆ ’ 5 ) = 142 hence , required % profit = 142 â € “ 100 = 42 % answer e" | a = 100 + 34
b = a * 100
c = 100 - 5
d = b / c
e = d - 100
|
a ) 52 sec , b ) 45 sec , c ) 60 sec , d ) 25 sec , e ) 36 sec | e | divide(multiply(30, const_4), multiply(12, divide(const_1000, const_3600))) | how long will a boy take to run round a square field of side 30 meters , if he runs at the rate of 12 km / hr ? | "speed = 12 km / hr = 12 * 5 / 18 = 10 / 3 m / sec distance = 30 * 4 = 120 m time taken = 120 * 3 / 10 = 36 sec answer is e" | a = 30 * 4
b = 1000 / 3600
c = 12 * b
d = a / c
|
a ) 60 , b ) 90 , c ) 120 , d ) 150 , e ) 240 | c | add(divide(multiply(280, 2), 5), multiply(2, 5)) | the number of timeshare condos available at sunset beach is 2 / 5 the number of timeshare condos available at playa del mar . if the total number of timeshare condos available at the two beaches combined is 280 , what is the difference between the number of condos available at sunset beach and the number of condos available at playa del mar ? | "let x be the number of timeshare condos available at playa del mar . then number of timeshare condos available at sunset beach = 3 / 5 x we know , x + 2 / 5 x = 280 hence , x = 200 . so , number of timeshare condos available at playa del mar = 200 the difference between the number of condos available at sunset beach and the number of condos available at playa del mar = x - 2 / 5 x = 2 / 5 x = 3 / 5 ( 200 ) = 120 the correct answer is c" | a = 280 * 2
b = a / 5
c = 2 * 5
d = b + c
|
a ) 70 % , b ) 15 % , c ) 20 % , d ) 25 % , e ) 30 % | a | multiply(divide(350, 500), const_100) | the price of a coat in a certain store is $ 500 . if the price of the coat is to be reduced by $ 350 , by what percent is the price to be reduced ? | "price of a coat in a certain store = $ 500 the price of the coat is to be reduced by $ 350 % change = ( final value - initial value ) * 100 / initial value % reduction = ( reduction in price ) * 100 / initial value i . e . % reduction = ( 350 ) * 100 / 500 = 70 % answer : option a" | a = 350 / 500
b = a * 100
|
a ) 18 : 11 , b ) 8 : 1 , c ) 11 : 18 , d ) 1 : 8 , e ) 1 : 1 | a | divide(multiply(const_2, multiply(divide(33, subtract(multiply(divide(add(const_100, 20), const_100), 5), divide(9, 7))), divide(9, 7))), add(divide(33, subtract(multiply(divide(add(const_100, 20), const_100), 5), divide(9, 7))), 4)) | ratio of two numbers x and y is 7 : 9 . if x is increased by 20 % and y is increased by 33 then the new ratio becomes 1 : 5 . what is the ratio 2 y : ( x + 4 ) | let : x = 7 n y = 9 n 8.4 n / ( 9 n + 33 ) = 1 / 5 42 n = 9 n + 33 n = 1 so , x = 7 ; y = 9 2 y / ( x + 4 ) = 18 / 11 answer = a | a = 100 + 20
b = a / 100
c = b * 5
d = 9 / 7
e = c - d
f = 33 / e
g = 9 / 7
h = f * g
i = 2 * h
j = 100 + 20
k = j / 100
l = k * 5
m = 9 / 7
n = l - m
o = 33 / n
p = o + 4
q = i / p
|
a ) 54 , b ) 59 , c ) 63 , d ) 64 , e ) 92 | d | add(multiply(subtract(const_1, divide(const_1, const_3)), subtract(70, 52)), 52) | of 70 players on a football team , 52 are throwers . the rest of the team is divided so one third are left - handed and the rest are right handed . assuming that all throwers are right handed , how many right - handed players are there total ? | "total = 70 thrower = 52 rest = 70 - 52 = 18 left handed = 18 / 3 = 6 right handed = 12 if all thrower are right handed then total right handed is 52 + 12 = 64 so d . 64 is the right answer" | a = 1 / 3
b = 1 - a
c = 70 - 52
d = b * c
e = d + 52
|
a ) 65 , b ) 86 , c ) 30 , d ) 78 , e ) 25 | e | divide(add(15, 35), const_2) | a man can row upstream at 15 kmph and downstream at 35 kmph , and then find the speed of the man in still water ? | "us = 15 ds = 35 m = ( 35 + 15 ) / 2 = 25 answer : e" | a = 15 + 35
b = a / 2
|
a ) rs . 45,000 , b ) rs . 50,000 , c ) rs . 60,000 , d ) rs . 80,000 , e ) none | c | divide(multiply(multiply(add(const_1, const_4), const_1000), 2), 3) | x and y invested in a business . they earned some profit which they divided in the ratio of 2 : 3 . if x invested rs . 40,000 . the amount invested by y is | "solution suppose y invested rs . y then , 40000 / y = 2 / 3 ‹ = › y = ( 40000 × 3 / 2 ) . ‹ = › y = 60000 . answer c" | a = 1 + 4
b = a * 1000
c = b * 2
d = c / 3
|
a ) 2574 , b ) 2500 , c ) 1500 , d ) 1574 , e ) none of these | c | divide(82500, 55) | if the product of two numbers is 82500 and their h . c . f . is 55 , find their l . c . m . | "explanation : hcf * lcm = 82500 , because we know product of two numbers = product of hcf and lcm lcm = 82500 / 55 = 1500 option c" | a = 82500 / 55
|
['a ) 24 sq units', 'b ) 48 sq units', 'c ) 24 under root 3', 'd ) 48 under root 3', 'e ) such a triangle does not exist'] | a | subtract(triangle_area(add(multiply(8, divide(sqrt(const_3), const_2)), 12), divide(8, const_2)), triangle_area(divide(8, const_2), multiply(8, divide(sqrt(const_3), const_2)))) | what is the area of an obtuse angled triangle whose two sides are 8 and 12 and the angle included between the two sides is 150 o ? | explanatory answer if two sides of a triangle and the included angle ' y ' is known , then the area of the triangle = 1 / 2 * ( product of sides ) * sin y substituting the values in the formula , we get 1 / 2 * 8 * 12 * sin 150 o = 1 / 2 * 8 * 12 * 1 / 2 . note : sin 150 o = 1 / 2 area = 24 sq units . choice a | a = math.sqrt(3)
b = a / 2
c = 8 * b
d = c + 12
e = 8 / 2
f = triangle_area - (
|
a ) 8 mph , b ) 5.25 mph , c ) 3.5 mph , d ) 4.5 mph , e ) 0.5 mph | d | divide(add(1, 8), const_2) | tabby is training for a triathlon . she swims at a speed of 1 mile per hour . she runs at a speed of 8 miles per hour . she wants to figure out her average speed for these two events . what is the correct answer for her ? | "( 1 mph + 8 mph ) / 2 = 4.5 mph correct option is : d" | a = 1 + 8
b = a / 2
|
a ) 15 , b ) 16 , c ) 15.5 , d ) insufficient data , e ) 17 | d | subtract(16, 15) | the average age of boys in a class is 16 years and that of the girls is 15 years . what is the average age for the whole class ? | explanation : we do not have the number of boys and girls . hence we can not find out the answer . answer : option d | a = 16 - 15
|
a ) 11 , b ) 12 , c ) 13 , d ) 14 , e ) 15 | d | subtract(100, add(add(add(subtract(100, 84), subtract(100, 75)), subtract(100, 85)), subtract(100, 70))) | there were totally 100 men . 84 are married . 75 have t . v , 85 have radio , 70 have a . c . how many men have t . v , radio , a . c and also married ? | "100 - ( 100 - 84 ) - ( 100 - 75 ) - ( 100 - 85 ) - ( 100 - 70 ) = 100 - 16 - 25 - 15 - 30 = 100 - 86 = 14 answer : d" | a = 100 - 84
b = 100 - 75
c = a + b
d = 100 - 85
e = c + d
f = 100 - 70
g = e + f
h = 100 - g
|
a ) 22 , b ) 88 , c ) 24 , d ) 20 , e ) 99 | c | subtract(multiply(const_1, const_60), multiply(divide(35, 60), const_60)) | excluding stoppages , the average speed of a bus is 60 km / hr and including stoppages , the average speed of the bus is 35 km / hr . for how many minutes does the bus stop per hour ? | "in 1 hr , the bus covers 60 km without stoppages and 35 km with stoppages . stoppage time = time take to travel ( 60 - 35 ) km i . e 25 km at 60 km / hr . stoppage time = 25 / 60 hrs = 24 min . answer : c" | a = 1 * const_60
b = 35 / 60
c = b * const_60
d = a - c
|
a ) 5 , b ) 4 , c ) 3 , d ) 2 , e ) 1 | e | divide(divide(80, 8), const_10) | one paper sheet can be used to create 8 paper boats or 6 paper planes . if total exact 80 paper toys are to be made , minimum how many sheets need to be used to make paper boats ? | let us say x sheets are used to create paper boats and y sheets are used to create paper planes . since total 80 paper toys are to be created , 8 x + 6 y = 80 . = > 8 x = 80 - 6 y = > x = 10 - ( 6 / 8 ) y = > x = 10 - ( 3 / 4 ) y to minimize x , we need to maximize y . and y needs to be multiple of 4 . y = 4 , x = 10 - ( 3 / 4 ) 4 = 7 y = 8 , x = 10 - ( 3 / 4 ) 8 = 4 y = 12 , x = 10 - ( 3 / 4 ) 12 = 1 beyond this , x will become negative for greater values of y . so , minimum 1 sheet is required to be used for creating paper boats . answer : e | a = 80 / 8
b = a / 10
|
a ) 100 , b ) 240 , c ) 120 , d ) 200 , e ) 150 | c | multiply(12, 60) | the h . c . f . of two numbers is 12 and their l . c . m . is 600 . if one of the number is 60 , find the other ? | "other number = 12 * 600 / 60 = 120 answer is c" | a = 12 * 60
|
a ) a ) 2572816 , b ) b ) 3624216 , c ) c ) 3624316 , d ) d ) 3625116 , e ) e ) 3625216 | a | multiply(divide(1604, 1604), const_100) | 1604 x 1604 = ? | "1604 x 1604 = ( 1604 ) 2 = ( 1600 + 4 ) 2 = ( 1600 ) 2 + ( 4 ) 2 + ( 2 x 1600 x 4 ) = 2560000 + 16 + 12800 . = 2572816 . a )" | a = 1604 / 1604
b = a * 100
|
['a ) 0.586', 'b ) 0.687', 'c ) 0.784', 'd ) 0.874', 'e ) 0.974'] | a | subtract(const_2, 1.414) | if abc is a quarter circle and a smaller circle is inscribed in it ; if radius of quarter circle is 1.414 units . find the radius of smaller circle | if r be radius of smaller circle then distance b / n centres = √ 2 * r so , r + r √ 2 = r here , r = 1.414 = √ 2 = > r = ( √ 2 - 1 ) r r = 0 . 586 unit answer : a | a = 2 - 1
|
a ) 4 kg , b ) 8 kg , c ) 7 kg , d ) 3 kg , e ) 9 kg | b | divide(16, const_2) | a bag of potatoes weighs 16 kg divided by a quarter of its weight . how much does the bag weigh ? | b 8 kg 8 kg : 16 / 2 | a = 16 / 2
|
a ) 8 % , b ) 12 % , c ) 10 % , d ) 14 % , e ) 21 | a | multiply(divide(subtract(1590, 1450), 1590), const_100) | the cost price of a radio is rs . 1590 and it was sold for rs . 1450 , find the loss % ? | "1590 - - - - 140 100 - - - - ? = > 8 % answer : a" | a = 1590 - 1450
b = a / 1590
c = b * 100
|
a ) 81 , b ) 100 , c ) 120 , d ) 135 , e ) 160 | c | divide(multiply(add(90, divide(multiply(90, 20), const_100)), const_100), multiply(multiply(const_3, const_3), 10)) | a retailer bought a machine at a wholesale price of $ 90 and later on sold it after a 10 % discount of the retail price . if the retailer made a profit equivalent to 20 % of the whole price , what is the retail price of the machine ? | "my solution : wholesale price = 90 retail price , be = x he provides 10 % discount on retail price = x - 10 x / 100 this retail price = 20 % profit on wholesale price x - 10 x / 100 = 90 + 1 / 5 ( 90 ) x = 120 ; answer : c" | a = 90 * 20
b = a / 100
c = 90 + b
d = c * 100
e = 3 * 3
f = e * 10
g = d / f
|
a ) 21 , b ) 22 , c ) 23 , d ) 24 , e ) 25 | d | subtract(multiply(add(floor(divide(1056, 27)), const_1), 27), 1056) | what is the least number should be added to 1056 , so the sum of the number is completely divisible by 27 ? | "( 1056 / 27 ) gives remainder 3 24 + 3 = 27 , so we need to add 24 d" | a = 1056 / 27
b = math.floor(a)
c = b + 1
d = c * 27
e = d - 1056
|
a ) 620 , b ) 630 , c ) 616 , d ) 600 , e ) 650 | c | multiply(divide(880, const_100), subtract(const_100, 30)) | if 30 % of the 880 students at a certain college are enrolled in biology classes , how many students at the college are not enrolled in a biology class ? | "we know 37.5 % people study biology , therefore the no of people not studying = 100 - 30 = 70 % > therefore the people not studying biology out of a total 880 people are = 70 % of 880 > ( 70 / 100 ) * 880 = 616 people c" | a = 880 / 100
b = 100 - 30
c = a * b
|
a ) 6 , b ) 9 , c ) 8 , d ) 2 , e ) 3 | b | subtract(add(multiply(10, 5), multiply(9, 5)), multiply(9, 9)) | the average of 9 observations was 9 , that of the 1 st of 5 being 10 and that of the last 5 being 8 . what was the 5 th observation ? | "1 to 9 = 9 * 9 = 81 1 to 5 = 5 * 10 = 50 5 to 9 = 5 * 8 = 40 5 th = 50 + 40 = 90 – 81 = 9 answer : b" | a = 10 * 5
b = 9 * 5
c = a + b
d = 9 * 9
e = c - d
|
a ) 33 % , b ) 34 % , c ) 35 % , d ) 36 % , e ) 37 % | b | multiply(add(multiply(divide(30, const_100), divide(20, const_100)), multiply(divide(subtract(const_100, 30), 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 30 % of all the employees are men . what % of all the employee went to the picnic ? | "total men in company 30 % means total women in company 70 % ( assume total people in company 100 % ) no of men employees attended picnic = 30 x ( 20 / 100 ) = 6 no of women employees attended picnic = 70 x ( 40 / 100 ) = 28 total percentage of employees attended the picnic = 6 + 28 = 34 % answer : b" | a = 30 / 100
b = 20 / 100
c = a * b
d = 100 - 30
e = d / 100
f = 40 / 100
g = e * f
h = c + g
i = h * 100
|
a ) 35 % , b ) 30 % , c ) 45 % , d ) 50 % , e ) 55 % | b | multiply(divide(30, 100), const_100) | 30 is what percent of 100 ? | 30 / 100 × 100 = 30 % answer : b | a = 30 / 100
b = a * 100
|
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