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 ) 3 / 2 , b ) 3 / 4 , c ) 4 / 5 , d ) 1 / 2 , e ) 1 / 5 | c | divide(multiply(8, 5), add(multiply(8, 5), multiply(5, 2))) | a call center has two teams . each member of team a was able to process 2 / 5 calls as compared to each member of team b . if team a has 5 / 8 as many number of call center agents as team b , what fraction of the total calls was processed by team b ? | let team b has 8 agents , so team a has 5 agents let each agent of team b picked up 5 calls , so total calls by team b = 40 so , each agent in team a picked up 2 calls , so total calls for team a = 10 fraction for team b = 40 / ( 40 + 10 ) = 4 / 5 = answer = c | a = 8 * 5
b = 8 * 5
c = 5 * 2
d = b + c
e = a / d
|
a ) 18 , b ) 11.2 , c ) 14 , d ) 12 , e ) 91 | b | divide(1056, multiply(multiply(const_2, divide(add(add(multiply(const_3, const_100), multiply(const_1, const_10)), const_4), const_100)), 15)) | if the wheel is 15 cm then the number of revolutions to cover a distance of 1056 cm is ? | "2 * 22 / 7 * 15 * x = 1056 = > x = 11.2 answer : b" | a = 3 * 100
b = 1 * 10
c = a + b
d = c + 4
e = d / 100
f = 2 * e
g = f * 15
h = 1056 / g
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a ) rs . 120 , b ) rs . 210 , c ) rs . 240 , d ) rs . 300 , e ) none | b | subtract(subtract(multiply(divide(1190, const_10), const_2), const_12), const_12) | a sum of rs . 1190 has been divided among a , b and c such that a gets of what b gets and b gets of what c gets . b β s share is : | explanation let c β s share = rs . x then , b β s share = rs . x / 4 , a β s share = rs . ( 2 / 3 x x / 4 ) = rs . x / 6 = x / 6 + x / 4 + x = 1190 = > 17 x / 12 = 1190 = > 1190 x 12 / 17 = rs . 840 hence , b β s share = rs . ( 840 / 4 ) = rs . 210 . answer b | a = 1190 / 10
b = a * 2
c = b - 12
d = c - 12
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a ) 1680 , b ) 1600 , c ) 1200 , d ) 1500 , e ) 600 | c | divide(1680, add(const_1, divide(40, const_100))) | a number increased by 40 % gives 1680 . the number is ? | "formula = total = 100 % , increase = ` ` + ' ' decrease = ` ` - ' ' a number means = 100 % that same number increased by 40 % = 140 % 140 % - - - - - - - > 1680 ( 140 Γ£ β 12 = 1680 ) 100 % - - - - - - - > 1200 ( 100 Γ£ β 12 = 1200 ) option ' c '" | a = 40 / 100
b = 1 + a
c = 1680 / b
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a ) 6 , b ) 24 , c ) 72 , d ) 96 , e ) 120 | e | factorial(5) | if each digit in the set a = { 1 , 2 , 3 , 4 , 5 } is used exactly once , in how many ways can the 5 digits be arranged ? | "use the slot method for all the possible arrangements : we have 5 options for the 1 st slot , 4 for the 2 nd , 3 for 3 rd , 2 for 4 th and 1 for the 5 th , giving the total number of arrangements = 5 * 4 * 3 * 2 * 1 = 120 , e is the correct answer ." | a = math.factorial(5)
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a ) 20 hours , b ) 15 hours , c ) 10 hours , d ) 12 hours , e ) 7 4 / 9 hours | e | divide(const_1, add(divide(const_1, 12), divide(const_1, 15))) | two pipes a and b can fill a tank in 12 hours and 15 hours respectively . if both the pipes are opened simultaneously , how much time will be taken to fill the tank ? | "part filled by a in 1 hour = 1 / 12 part filled by b in 1 hour = 1 / 15 part filled by ( a + b ) in 1 hour = 1 / 12 + 1 / 15 = 9 / 60 both the pipes together fill the tank in 60 / 9 = 7 4 / 9 hours answer is e" | a = 1 / 12
b = 1 / 15
c = a + b
d = 1 / c
|
['a ) 150 cm', 'b ) 200 cm', 'c ) 162 cm', 'd ) 220 cm', 'e ) 280 cm'] | e | divide(multiply(7, 720), add(7, add(5, 6))) | the sides of the triangle are in the ratio 5 : 6 : 7 and its perimeter is 720 cm . the length of the longest side is ? | ratio of sides = 5 : 6 : 7 largest side = 720 * 7 / 18 = 280 cm answer is e | a = 7 * 720
b = 5 + 6
c = 7 + b
d = a / c
|
a ) 18 , b ) 12 , c ) 24 , d ) 36 , e ) 42 | b | multiply(const_3, divide(60, const_10)) | jackie has two solutions that are 2 percent sulfuric acid and 12 percent sulfuric acid by volume , respectively . if these solutions are mixed in appropriate quantities to produce 60 liters of a solution that is 10 percent sulfuric acid , approximately how many liters of the 2 percent solution will be required ? | "let a = amount of 2 % acid and b = amount of 12 % acid . now , the equation translates to , 0.02 a + . 12 b = . 1 ( a + b ) but a + b = 60 therefore . 02 a + . 12 b = . 1 ( 60 ) = > 2 a + 12 b = 600 but b = 60 - a therefore 2 a + 12 ( 60 - a ) = 600 = > 10 a = 120 hence a = 12 . answer : b" | a = 60 / 10
b = 3 * a
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a ) 5500 , b ) 2999 , c ) 2878 , d ) 2990 , e ) 2771 | a | divide(multiply(220, 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 . 220 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 Γ’ β¬ β 96 = rs . 4 100 - - - - - - 4 ? - - - - - - - - - 220 = > 5500 answer : a" | a = 220 * 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 : 10', 'b ) 17 : 19', 'c ) 23 : 27', 'd ) 13 : 17', 'e ) 16 : 19'] | e | sqrt(divide(1536, 2166)) | triangle atriangle b are similar triangles with areas 1536 units square and 2166 units square respectively . the ratio of there corresponding height would be | let x be the height of triangle a and y be the height of triangle of b . since triangles are similar , ratio of area of a and b is in the ratio of x ^ 2 / y ^ 2 therefore , ( x ^ 2 / y ^ 2 ) = 1536 / 2166 ( x ^ 2 / y ^ 2 ) = ( 16 * 16 * 6 ) / ( 19 * 19 * 6 ) ( x ^ 2 / y ^ 2 ) = 17 ^ 2 / 19 ^ 2 x / y = 16 / 19 ans = e | a = 1536 / 2166
b = math.sqrt(a)
|
a ) 20 % , b ) 25 % , c ) 30 % , d ) 35 % , e ) 72 % | c | multiply(divide(108, divide(const_3600, const_10)), const_100) | the megatek corporation is displaying its distribution of employees by department in a circle graph . the size of each sector of the graph representing a department is proportional to the percentage of total employees in that department . if the section of the circle graph representing the manufacturing department takes up 108 Β° of the circle , what percentage of megatek employees are in manufacturing ? | "answer : c 108 Β° divided by 360 Β° equals 0.3 , therefore the sector is equal to 30 % of the total" | a = 3600 / 10
b = 108 / a
c = b * 100
|
a ) $ 2.50 , b ) $ 3.00 , c ) $ 3.50 , d ) $ 4.00 , e ) $ 5.00 | d | add(2.50, multiply(0.50, subtract(add(divide(subtract(18, 2), 4), const_1), 2))) | the toll t , in dollars , for a truck using a certain bridge is given by the formula t = 2.50 + 0.50 ( x β 2 ) , where x is the number of axles on the truck . what is the toll for an 18 - wheel truck that has 2 wheels on its front axle and 4 wheels on each of its other axles ? | "number of wheels in truck = 18 number of wheels on its front axle = 2 number of wheels remaining = 16 number of axles remaining axles = 16 / 4 = 4 total number of axles = 5 t = 2.50 + 0.50 ( x β 2 ) = 2.50 + . 5 * 3 = 2.5 + 1.5 = 4 $ answer d" | a = 18 - 2
b = a / 4
c = b + 1
d = c - 2
e = 0 * 50
f = 2 + 50
|
a ) 80 % , b ) 105 % , c ) 120 % , d ) 124.2 % , e ) 160 % | e | multiply(divide(multiply(20, subtract(const_1, divide(20, const_100))), 10), const_100) | in 2008 , the profits of company n were 10 percent of revenues . in 2009 , the revenues of company n fell by 20 percent , but profits were 20 percent of revenues . the profits in 2009 were what percent of the profits in 2008 ? | "x = profits r = revenue x / r = 0,1 x = 10 r = 100 2009 : r = 80 x / 80 = 0,2 = 15 / 100 x = 80 * 20 / 100 x = 16 16 / 10 = 1,6 = 160 % , answer e" | a = 20 / 100
b = 1 - a
c = 20 * b
d = c / 10
e = d * 100
|
a ) 5 , b ) 15 , c ) 10 , d ) 20 , e ) 25 | c | multiply(const_60, divide(multiply(divide(10, const_60), subtract(12, 6)), 6)) | a man walking at a constant rate of 6 miles per hour is passed by a woman traveling in the same direction along the same path at a constant rate of 12 miles per hour . the woman stops to wait for the man 10 minutes after passing him , while the man continues to walk at his constant rate . how many minutes must the woman wait until the man catches up ? | "when the woman passes the man , they are aligned ( m and w ) . they are moving in the same direction . after 5 minutes , the woman ( w ) will be ahead the man ( m ) : m - - - - - - m - - - - - - - - - - - - - - - w w in the 5 minutes , after passing the man , the woman walks the distance mw = ww , which is 10 * 12 / 60 = 2 miles and the man walks the distance mm , which is 10 * 6 / 60 = 1 mile . the difference of 2 - 1 = 1 miles ( mw ) will be covered by the man in ( 1 ) / 6 = 1 / 6 of an hour , which is 10 minutes . answer c ." | a = 10 / const_60
b = 12 - 6
c = a * b
d = c / 6
e = const_60 * d
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a ) 7 , b ) 11 , c ) 9 , d ) 2 , e ) 3 | b | subtract(32, multiply(12, const_2)) | a number when divided by 219 gives a remainder 32 , what remainder will be obtained by dividing the same number 12 ? | "explanation : 219 + 32 = 251 / 12 = 9 ( remainder ) answer : b" | a = 12 * 2
b = 32 - a
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a ) s . 2.04 , b ) s . 2.08 , c ) s . 2.02 , d ) s . 2.86 , e ) s . 2.42 | d | subtract(multiply(7000, multiply(multiply(add(1, divide(2, const_100)), add(1, divide(2, const_100))), add(1, divide(2, const_100)))), multiply(7000, multiply(add(1, divide(2, const_100)), add(1, divide(4, const_100))))) | what is the difference between the c . i . on rs . 7000 for 1 1 / 2 years at 4 % per annum compounded yearly and half - yearly ? | "c . i . when interest is compounded yearly = [ 7000 * ( 1 + 4 / 100 ) * ( 1 + ( 1 / 2 * 4 ) / 100 ] = 7000 * 26 / 25 * 51 / 50 = rs . 7425.6 c . i . when interest is compounded half - yearly = [ 7000 * ( 1 + 2 / 100 ) 2 ] = ( 7000 * 51 / 50 * 51 / 50 * 51 / 50 ) = rs . 7428.46 difference = ( 7428.46 - 7425.6 ) = rs . 2.86 . answer : d" | a = 2 / 100
b = 1 + a
c = 2 / 100
d = 1 + c
e = b * d
f = 2 / 100
g = 1 + f
h = e * g
i = 7000 * h
j = 2 / 100
k = 1 + j
l = 4 / 100
m = 1 + l
n = k * m
o = 7000 * n
p = i - o
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a ) 124 % , b ) 102 % , c ) 96 % , d ) 80 % , e ) 64 % | b | multiply(multiply(subtract(const_1, divide(40, const_100)), add(const_1, divide(70, const_100))), const_100) | mary ' s income is 70 percent more than tim ' s income , and tim ' s income is 40 percent less than juan ' s income . what percent of juan ' s income is mary ' s income ? | juan ' s income = 100 ( assume ) ; tim ' s income = 60 ( 40 percent less than juan ' s income ) ; mary ' s income = 102 ( 70 percent more than tim ' s income ) . thus , mary ' s income ( 102 ) is 102 % of juan ' s income ( 100 ) . answer : b . | a = 40 / 100
b = 1 - a
c = 70 / 100
d = 1 + c
e = b * d
f = e * 100
|
a ) $ 360 , b ) $ 380 , c ) $ 400 , d ) $ 420 , e ) $ 440 | c | subtract(480, multiply(subtract(680, 480), divide(2, 5))) | a sum of money lent out at s . i . amounts to a total of $ 480 after 2 years and to $ 680 after a further period of 5 years . what was the initial sum of money that was invested ? | "s . i for 5 years = $ 680 - $ 480 = $ 200 the s . i . is $ 40 / year s . i . for 2 years = $ 80 principal = $ 480 - $ 80 = $ 400 the answer is c ." | a = 680 - 480
b = 2 / 5
c = a * b
d = 480 - c
|
a ) 2.05 , b ) 2.50025 , c ) 2.501 , d ) 2.5025 , e ) 2.5 | e | multiply(divide(5.005, 2.002), const_100) | 5.005 / 2.002 = | "5.005 / 2.002 = 5005 / 2002 = 5 ( 1001 ) / 2 ( 1001 ) = 5 / 2 = 2.5 the answer is e ." | a = 5 / 5
b = a * 100
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a ) 0.125 % , b ) 1.25 % , c ) 56 % , d ) 125 % , e ) 0.152 % | c | multiply(divide(divide(multiply(148, 5.17), add(add(9.18, 5.17), 2.05)), 148), const_100) | irin , ingrid and nell bake chocolate chip cookies in the ratio of 9.18 : 5.17 : 2.05 . if altogether they baked a batch of 148 cookies , what percent of the cookies did irin bake ? | "9.18 x + 5.17 x + 2.05 x = 16.4 x = 150 cookies x = 150 / 16.4 = 9.14 ( approx ) so , irin baked 9.14 * 9.18 cookies or 84 cookies ( approx ) % share = 84 / 150 = 56 approx hence , answer is c ." | a = 148 * 5
b = 9 + 18
c = b + 2
d = a / c
e = d / 148
f = e * 100
|
a ) 11 , b ) 25 , c ) 27 , d ) 22 , e ) 91 | d | divide(subtract(24, subtract(multiply(const_2, const_2), const_2)), subtract(const_2, const_1)) | a man is 24 years older than his son . in two years , his age will be twice the age of his son . the present age of the son is ? | "let the son ' s present age be x years . then , man ' s present age = ( x + 24 ) years . ( x + 24 ) + 2 = 2 ( x + 2 ) x + 26 = 2 x + 4 = > x = 22 . answer : d" | a = 2 * 2
b = a - 2
c = 24 - b
d = 2 - 1
e = c / d
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a ) 0.275 d , b ) 0.225 d , c ) 0.265 d , d ) 0.245 d , e ) 0.205 d | b | subtract(divide(subtract(const_100, 55), const_100), multiply(divide(subtract(const_100, 55), const_100), divide(50, const_100))) | a dress on sale in a shop is marked at $ d . during the discount sale its price is reduced by 55 % . staff are allowed a further 50 % reduction on the discounted price . if a staff member buys the dress what will she have to pay in terms of d ? | "effective discount = a + b + ab / 100 = - 55 - 50 + ( - 55 ) ( - 50 ) / 100 = 77.5 sale price = d * ( 1 - 77.5 / 100 ) sale price = . 225 * d answer ( b )" | a = 100 - 55
b = a / 100
c = 100 - 55
d = c / 100
e = 50 / 100
f = d * e
g = b - f
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a ) 18,104 , b ) 18,303 , c ) 18 , 356.732 , d ) 19,502 , e ) 18,909 | d | add(100, const_1) | if 100 < x < 190 and 10 < y < 100 , then the product xy can not be equal to : | "correct answer : ( d ) determine the range of xy by multiplying the two extremes of each individual range together . the smallest value of xy must be greater than 100 * 10 . the largest value must be less than 190 * 100 . this means that 1000 < xy < 19,000 . ( d ) is outside of this range , so it is not a possible product of xy ." | a = 100 + 1
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a ) 25 % loss , b ) 25 % profit , c ) 20 % loss , d ) 20 % profit , e ) 50 % loss | e | multiply(const_100, divide(subtract(const_100, divide(multiply(const_100, 40), 20)), divide(multiply(const_100, 40), 20))) | if the cost price of 20 articles is equal to the selling price of 40 articles , what is the % profit or loss made by the merchant ? | "let the cost price of 1 article be $ 1 . therefore , cost price of 20 articles = 20 * 1 = $ 20 the selling price of 40 articles = cost price of 40 articles = $ 40 . now , we know the selling price of 40 articles . let us find the cost price of 40 articles . cost price of 40 articles = 40 * 1 = $ 40 . therefore , profit made on sale of 40 articles = selling price of 40 articles - cost price of 40 articles = 20 - 40 = - $ 20 . as the profit is in the negative , the merchant has made a loss of $ 20 . therefore , % loss = loss / cp * 100 % loss = - 20 / 40 * 100 = 50 % loss . e" | a = 100 * 40
b = a / 20
c = 100 - b
d = 100 * 40
e = d / 20
f = c / e
g = 100 * f
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a ) 6 , b ) 7 , c ) 8 , d ) 9 , e ) 10 | d | add(add(5, const_2), const_2) | if x < y < z and y - x > 5 , where x is an even integer and y and z are odd integers , what is the least possible value of z - x ? | "we want to minimize z β xz β x , so we need to maximize xx . say z = 11 = oddz = 11 = odd , then max value of yy will be 9 ( as yy is also odd ) . now , since y β 5 > xy β 5 > x - - > 9 β 5 > x 9 β 5 > x - - > 4 > x 4 > x , then max value of xx is 2 ( as xx is even ) . hence , the least possible value of z β xz β x is 11 - 2 = 9 . answer : d ." | a = 5 + 2
b = a + 2
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a ) 70 , b ) 75 , c ) 80 , d ) 85 , e ) 60 | b | divide(add(add(add(add(65, 67), 76), 82), 85), add(const_4, const_1)) | reeya obtained 65 , 67 , 76 , 82 and 85 out of 100 in different subjects , what will be the average | "explanation : ( 65 + 67 + 76 + 82 + 855 ) = 75 answer : option b" | a = 65 + 67
b = a + 76
c = b + 82
d = c + 85
e = 4 + 1
f = d / e
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a ) 56 , b ) 40 , c ) 44 , d ) 48 , e ) 52 | a | add(40, divide(subtract(982, multiply(15, 40)), divide(multiply(15, add(const_100, 75)), const_100))) | a certain bus driver is paid a regular rate of $ 15 per hour for any number of hours that does not exceed 40 hours per week . for any overtime hours worked in excess of 40 hours per week , the bus driver is paid a rate that is 75 % higher than his regular rate . if last week the bus driver earned $ 982 in total compensation , how many total hours did he work that week ? | "for 40 hrs = 40 * 15 = 600 excess = 982 - 600 = 382 for extra hours = . 75 ( 15 ) = 11.25 + 16 = 27.25 number of extra hrs = 382 / 27.25 = 14 total hrs = 40 + 14 = 56 answer a 56" | a = 15 * 40
b = 982 - a
c = 100 + 75
d = 15 * c
e = d / 100
f = b / e
g = 40 + f
|
a ) 28 , b ) 27 , c ) 55 , d ) 59 , e ) 12 | d | divide(add(165, 720), multiply(54, const_0_2778)) | how long does a train 165 meters long running at the rate of 54 kmph take to cross a bridge 720 meters in length ? | "t = ( 720 + 165 ) / 54 * 18 / 5 t = 59 answer : d" | a = 165 + 720
b = 54 * const_0_2778
c = a / b
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a ) 125 , b ) 173 , c ) 225 , d ) 250 , e ) 500 | b | sqrt(divide(multiply(90, const_100), divide(30, const_100))) | 90 students represent x percent of the boys at jones elementary school . if the boys at jones elementary make up 30 % of the total school population of x students , what is x ? | "90 = x / 100 * 30 / 100 * x = > x ^ 2 = 9 * 10000 / 3 = > x = 173 b" | a = 90 * 100
b = 30 / 100
c = a / b
d = math.sqrt(c)
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a ) $ 40,000 , b ) $ 41,667 , c ) $ 42,000 , d ) $ 46 , 460.17 , e ) $ 60,000 | d | add(divide(divide(divide(multiply(add(divide(8, const_100), const_1), multiply(subtract(add(const_1000, const_60), const_10), const_1000)), add(multiply(subtract(21, const_1), add(divide(8, const_100), const_1)), 1)), add(divide(8, const_100), const_1)), const_100), add(multiply(const_100, const_2), const_3)) | last year , company x paid out a total of $ 1 , 050,000 in salaries to its 21 employees . if no employee earned a salary that is more than 8 % greater than any other employee , what is the lowest possible salary that any one employee earned ? | employee 1 earned $ x ( say ) employee 2 will not earn more than $ 1.08 x therfore , to minimize the salary of any one employee , we need to maximize the salaries of the other 20 employees ( 1.08 x * 20 ) + x = 1 , 050,000 solving for x = $ 46 , 460.17 answer d | a = 8 / 100
b = a + 1
c = 1000 + const_60
d = c - 10
e = d * 1000
f = b * e
g = 21 - 1
h = 8 / 100
i = h + 1
j = g * i
k = j + 1
l = f / k
m = 8 / 100
n = m + 1
o = l / n
p = o / 100
q = 100 * 2
r = q + 3
s = p + r
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a ) 9 % , b ) 15 % , c ) 29 % , d ) 50 % , e ) 100 % | c | multiply(divide(divide(40, const_100), divide(add(40, const_100), const_100)), const_100) | a part - time employee β s hourly wage was increased by 40 % . she decided to decrease the number of hours worked per week so that her total income did not change . by approximately what percent should the number of hours worked be decreased ? | "let ' s plug in somenicenumbers and see what ' s needed . let ' s say the employee used to make $ 1 / hour and worked 100 hours / week so , the total weekly income was $ 100 / week after the 40 % wage increase , the employee makes $ 1.40 / hour we want the employee ' s income to remain at $ 100 / week . so , we want ( $ 1.40 / hour ) ( new # of hours ) = $ 100 divide both sides by 1.40 to get : new # of hours = 100 / 1.40 β 71 hours so , the number of hours decreases from 100 hours to ( approximately ) 71 hours . this represents a 29 % decrease ( approximately ) . answer : c" | a = 40 / 100
b = 40 + 100
c = b / 100
d = a / c
e = d * 100
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a ) 1 kmph , b ) 2 kmph , c ) 3 kmph , d ) 2.5 kmph , e ) 1.85 kmph | e | divide(subtract(multiply(divide(1, 9), const_60), multiply(divide(1, 20), const_60)), const_2) | a boat moves upstream at the rate of 1 km in 20 minutes and down stream 1 km in 9 minutes . then the speed of the current is : | "rate upstream = ( 1 / 20 * 60 ) = 3 kmph rate dowm stream = 1 / 9 * 60 = 6.7 kmph rate of the current = Β½ ( 6.7 - 3 ) = 1.85 kmph answer : e" | a = 1 / 9
b = a * const_60
c = 1 / 20
d = c * const_60
e = b - d
f = e / 2
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a ) 10 % , b ) 20 % , c ) 30 % , d ) 32 % , e ) none | b | subtract(10, 12) | if the selling price of 10 articles is same as the cost price of 12 articles . find the gain or loss percentage ? | "let the c . p of each article be re 1 . then , s . p of 10 articles = c . p of 12 articles = rs . 12 now , c . p of 10 articles = rs . 10 , s . p of 10 articles = rs 12 gain = rs ( 12 - 10 ) = rs 2 . gain % = ( 2 / 10 Γ 100 ) % = 20 % b )" | a = 10 - 12
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a ) 2 , b ) 8.5 , c ) 9.5 , d ) 6.5 , e ) 11.5 | e | divide(add(divide(18, 3), divide(51, 3)), const_2) | a man swims downstream 51 km and upstream 18 km taking 3 hours each time , what is the speed of the man in still water ? | "51 - - - 3 ds = 17 ? - - - - 1 18 - - - - 3 us = 6 ? - - - - 1 m = ? m = ( 17 + 6 ) / 2 = 11.5 answer : e" | a = 18 / 3
b = 51 / 3
c = a + b
d = c / 2
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a ) 18 , b ) 20 , c ) 22 , d ) 24 , e ) 26 | d | divide(subtract(sqrt(add(multiply(multiply(312, 2), const_4), power(2, 2))), 2), 2) | a jar of 312 marbles is divided equally among a group of marble - players today . if 2 people joined the group in the future , each person would receive 1 marble less . how many people are there in the group today ? | "312 = 24 * 13 = 26 * 12 there are 24 people in the group today . the answer is d ." | a = 312 * 2
b = a * 4
c = 2 ** 2
d = b + c
e = math.sqrt(d)
f = e - 2
g = f / 2
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a ) s . 27.5 , b ) s . 34 , c ) s . 26 , d ) s . 25.28 , e ) s . 25.32 | e | multiply(422, divide(3, const_100)) | find the simple interest on rs . 422 for 3 months at 2 paisa per month ? | "explanation : i = ( 422 * 3 * 2 ) / 100 = 25.32 answer : option e" | a = 3 / 100
b = 422 * a
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a ) 32.75 , b ) 32.45 , c ) 22.45 , d ) 34.45 , e ) 32.15 | d | add(divide(circumface(6.7), const_2), multiply(6.7, const_2)) | the radius of a semi circle is 6.7 cm then its perimeter is ? | "36 / 7 r = 6.7 = 34.45 answer : d" | a = circumface / (
b = a + 2
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a ) 21 , b ) 26 , c ) 28 , d ) 30 , e ) 32 | a | multiply(divide(7, 8), add(19, const_4)) | john was 19 years old when he married betty . they just celebrated their fifth wedding anniversary , and betty ' s age is now 7 / 8 of john ' s . how old is betty ? | "assume betty ' s age on marriage = x years . john ' s age on marriage = 19 john ' s age after 5 years = 24 years . betty ' s age after 5 years = x + 5 given : x + 5 = 7 / 8 ( 24 ) = 21 therefore betty ' s current age = 21 option a" | a = 7 / 8
b = 19 + 4
c = a * b
|
a ) 140 sec , b ) 160 sec , c ) 176 sec , d ) 150 sec , e ) 170 sec | a | multiply(divide(70, multiply(63, const_1000)), const_3600) | a train 70 m long , running with a speed of 63 km / hr will pass a tree in ? | "speed = 63 * 5 / 18 = 35 / 2 m / sec time taken = 70 * 2 / 35 = 140 sec answer : a" | a = 63 * 1000
b = 70 / a
c = b * 3600
|
a ) 8 , b ) 10 , c ) 12 , d ) 14 , e ) 16 | c | divide(subtract(100, multiply(10, const_3)), const_4) | a box has exactly 100 balls , and each ball is either red , blue , or white . if the box has 10 more blue balls than white balls , and thrice as many red balls as blue balls , how many white balls does the box has ? | "x = the number of red balls y = the number of blue balls z = the number of white balls from the first sentence we have equation # 1 : x + y + z = 100 . . . the box has 10 more blue balls than white balls . . . equation # 2 : y = 10 + z . . . thrice as many red balls as blue balls . . . equation # 3 : x = 3 y solve equation # 2 for z : z = y - 10 now , we can replace both x and z with y in equation # 1 3 y + y + ( y - 10 ) = 100 5 y - 10 = 100 5 y = 110 y = 22 there are 22 blue balls . this is 10 more than the number of white balls , so z = 12 . that ' s the answer . just as a check , x = 66 , and 66 + 22 + 12 = 100 . answer = 12 , ( c )" | a = 10 * 3
b = 100 - a
c = b / 4
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a ) $ 110 , b ) $ 120 , c ) $ 180 , d ) $ 220 , e ) $ 210 | e | divide(multiply(21, const_100), subtract(multiply(add(12, 1), 10), add(const_100, 20))) | right now , al and eliot have bank accounts , and al has more money than eliot . the difference between their two accounts is 1 / 12 of the sum of their two accounts . if al β s account were to increase by 10 % and eliot β s account were to increase by 20 % , then al would have exactly $ 21 more than eliot in his account . how much money does eliot have in his account right now ? | "lets assume al have amount a in his bank account and eliot ' s bank account got e amount . we can form an equation from the first condition . a - e = 1 / 12 * ( a + e ) = = > 11 a = 13 e - - - - - - - - - - - - ( 1 ) second condition gives two different amounts , al ' s amount = 1.1 a and eliot ' s amount = 1.2 e 1.1 a = 21 + 1.2 e = = > 11 a = 210 + 12 e - - - - - - - ( 2 ) substituting ( 1 ) in ( 2 ) : 13 e = 210 + 12 e = = > e = 210 e" | a = 21 * 100
b = 12 + 1
c = b * 10
d = 100 + 20
e = c - d
f = a / e
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a ) $ 320 , b ) $ 400 , c ) $ 600 , d ) $ 800 , e ) $ 1000 | a | subtract(divide(negate(subtract(multiply(20, 8), multiply(30, 8))), divide(35, const_100)), subtract(multiply(20, 8), multiply(30, 8))) | at a florist shop on a certain day , all corsages sold for either $ 20 or $ 30 . if 8 of the corsages that sold for $ 30 had instead sold for $ 20 , then the store ' s revenue from corsages that day would have been reduced by 35 percent . what was the store ' s actual revenue from corsages that day ? | i am doing it elaborately , hope it will help you . let , no . of corsages @ $ 20 = x , no . of corsages @ $ 30 = y and revenue = r so , 20 x + 30 y = r . . . . . . . . . ( 1 ) now , given the situation , 20 ( x + 8 ) + 30 ( y - 8 ) = r - . 25 r = > 20 x + 160 + 30 y - 240 = . 75 r = > 20 x + 30 y = . 75 r + 80 . . . . . . . . . . . . ( 2 ) so , r = . 75 r + 80 = > r = 320 the answer is a . | a = 20 * 8
b = 30 * 8
c = a - b
d = negate / (
e = 35 / 100
f = d - e
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a ) 30 , b ) 60 , c ) 40 , d ) 45 , e ) 50 | b | subtract(add(40, 30), 10) | in a group of houses , 40 had dogs , 30 had cats and 10 houses having both dogs and cats . what is the number of houses ? | make a venn diagram , and enter your data . let the number of houses be x 30 + 10 + 20 = x x = 60 so number of houses were = 60 answer b | a = 40 + 30
b = a - 10
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a ) 6 , b ) 6.25 , c ) 7.25 , d ) 7.5 , e ) 8 | b | divide(subtract(282, multiply(3.2, 10)), 40) | cricket match is conducted in us . the run rate of a cricket game was only 3.2 in first 10 over . what should be the run rate in the remaining 40 overs to reach the target of 282 runs ? | "required run rate = 282 - ( 3.2 x 10 ) = 250 = 6.25 40 40 b" | a = 3 * 2
b = 282 - a
c = b / 40
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a ) 16 % , b ) 23.07 % , c ) 25 % , d ) 30 % , e ) 35 % | b | multiply(subtract(const_1, divide(const_100, add(const_100, 30))), const_100) | the product of x and y is a constant . if the value of x is increased by 30 % , by what percentage must the value of y be decreased ? | "x * y = constt . let x = y = 100 in beginning i . e . x * y = 100 * 100 = 10000 x ( 100 ) - - - becomes - - - > 1.3 x ( 130 ) i . e . 130 * new ' y ' = 10000 i . e . new ' y ' = 10000 / 130 = 76.92 i . e . y decreases from 100 to 76.92 i . e . decrease of 23.07 % b" | a = 100 + 30
b = 100 / a
c = 1 - b
d = c * 100
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a ) rs . 150 , b ) rs . 190 , c ) rs . 200 , d ) rs . 250 , e ) rs . 300 | d | divide(550, add(divide(120, const_100), const_1)) | two employees a and b are paid a total of rs . 550 per week by their employer . if a is paid 120 percent of the sum paid to b , how much is b paid per week ? | "let the amount paid to a per week = x and the amount paid to b per week = y then x + y = 550 but x = 120 % of y = 120 y / 100 = 12 y / 10 β΄ 12 y / 10 + y = 550 β y [ 12 / 10 + 1 ] = 550 β 22 y / 10 = 550 β 22 y = 5500 β y = 5500 / 22 = 500 / 2 = rs . 250 d )" | a = 120 / 100
b = a + 1
c = 550 / b
|
a ) 7 : 3 , b ) 1 : 3 , c ) 9 : 3 , d ) 6 : 3 , e ) 2 : 5 | b | divide(subtract(15.8, 15.6), subtract(16.4, 15.8)) | the average age of students of a class is 15.8 years . the average age of boys in the class is 16.4 years and that of the girls is 15.6 years , the ratio of the number of boys to the number of girls in the class is | "explanation : let the ratio be k : 1 . then , k * 16.4 + 1 * 15.6 = ( k + 1 ) * 15.8 < = > ( 16.4 - 15.8 ) k = ( 15.8 - 15.6 ) < = > k = 0.2 / 0.6 = 1 / 3 . required ratio = 1 / 3 : 1 = 1 : 3 . answer : b" | a = 15 - 8
b = 16 - 4
c = a / b
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a ) 3 / 8 , b ) 5 / 8 , c ) 7 / 16 , d ) 9 / 16 , e ) 23 / 32 | b | multiply(multiply(divide(1, 7), divide(1, 8)), subtract(1, divide(1, 6))) | jack , jill , and sandy each have one try to make a basket from half court . if their individual probabilities of making the basket are 1 / 6 , 1 / 7 , and 1 / 8 respectively , what is the probability that all three will miss ? | "the probability that all three will miss is 5 / 6 * 6 / 7 * 7 / 8 = 5 / 8 . the answer is b ." | a = 1 / 7
b = 1 / 8
c = a * b
d = 1 / 6
e = 1 - d
f = c * e
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a ) 1888 , b ) 2999 , c ) 3159 , d ) 2777 , e ) 2991 | c | divide(subtract(multiply(subtract(const_1, divide(20, const_100)), 9720), divide(multiply(15, 9720), const_100)), const_2) | in an election between two candidates a and b , the number of valid votes received by a exceeds those received by b by 15 % of the total number of votes polled . if 20 % of the votes polled were invalid and a total of 9720 votes were polled , then how many valid votes did b get ? | "let the total number of votes polled in the election be 100 k . number of valid votes = 100 k - 20 % ( 100 k ) = 80 k let the number of votes polled in favour of a and b be a and b respectively . a - b = 15 % ( 100 k ) = > a = b + 15 k = > a + b = b + 15 k + b now , 2 b + 15 k = 80 k and hence b = 32.5 k it is given that 100 k = 9720 32.5 k = 32.5 k / 100 k * 9720 = 3159 the number of valid votes polled in favour of b is 3159 . answer : c" | a = 20 / 100
b = 1 - a
c = b * 9720
d = 15 * 9720
e = d / 100
f = c - e
g = f / 2
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a ) 3 , b ) 6 , c ) 9 , d ) 12 , e ) 15 | c | divide(multiply(const_12, log(2)), log(2)) | if 2 ^ ( 2 w ) = 8 ^ ( w β 3 ) , what is the value of w ? | "2 ^ ( 2 w ) = 8 ^ ( w β 3 ) 2 ^ ( 2 w ) = 2 ^ ( 3 * ( w β 3 ) ) 2 ^ ( 2 w ) = 2 ^ ( 3 w - 9 ) let ' s equate the exponents as the bases are equal . 2 w = 3 w - 9 w = 9 the answer is c ." | a = math.log(2)
b = 12 * a
c = math.log(2)
d = b / c
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a ) 25 , b ) 46 , c ) 88 , d ) 13 , e ) 12 | a | divide(add(multiply(30, 20), multiply(15, 10)), add(20, 10)) | the average runs scored by a batsman in 20 matches is 30 . in the next 10 matches the batsman scored an average of 15 runs . find his average in all the 30 matches ? | "total score of the batsman in 20 matches = 600 . total score of the batsman in the next 10 matches = 150 . total score of the batsman in the 30 matches = 750 . average score of the batsman = 750 / 30 = 25 . answer : a" | a = 30 * 20
b = 15 * 10
c = a + b
d = 20 + 10
e = c / d
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a ) 5 , b ) 6 , c ) 7 , d ) 8 , e ) 9 | c | multiply(divide(subtract(subtract(41, const_1), multiply(subtract(14, const_1), const_2)), subtract(multiply(subtract(14, const_1), const_4), multiply(subtract(14, const_1), const_2))), subtract(14, const_1)) | a luxury liner , queen marry ii , is transporting several cats as well as the crew ( sailors , a cook , and one - legged captain ) to a nearby port . altogether , these passengers have 14 heads and 41 legs . how many cats does the ship host ? | "sa ' s + co + ca + cats = 14 . sa ' s + 1 + 1 + cats = 14 or sa ' s + cats = 12 . sa ' s ( 2 ) + 2 + 1 + cats * 4 = 41 sa ' s * 2 + cats * 4 = 38 or sa ' s + cats * 2 = 19 or 12 - cats + cat * 2 = 19 then cats = 7 c" | a = 41 - 1
b = 14 - 1
c = b * 2
d = a - c
e = 14 - 1
f = e * 4
g = 14 - 1
h = g * 2
i = f - h
j = d / i
k = 14 - 1
l = j * k
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a ) 9 m , b ) 9.5 m , c ) 10.5 m , d ) 12 m , e ) none | b | multiply(cosine(divide(multiply(60, divide(add(19, const_3), add(const_4, const_3))), multiply(const_60, const_3))), 19) | a ladder learning against a wall makes an angle of 60 Β° with the ground . if the length of the ladder is 19 m , find the distance of the foot of the ladder from the wall . | let ab be the wall and bc be the ladder . then , < abc = 60 Β° and , bc = 19 m . ; ac = x metres ac / bc = cos 60 Β° = x / 19 = 1 / 2 x = 19 / 2 = 9.5 m . answer b | a = 19 + 3
b = 4 + 3
c = a / b
d = 60 * c
e = const_60 * 3
f = d / e
g = cosine * (
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a ) 352 , b ) 435 , c ) 224 , d ) 646 , e ) 572 | e | multiply(12.5, 5.2) | 12.5 * 5.2 * 8.8 = ? | "e 572 ? = 12.5 * 5.2 * 8.8 = 572" | a = 12 * 5
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a ) t = 21 , b ) t = 22 , c ) t = 23 , d ) 24 , e ) 27 | c | subtract(power(5, 2), 2) | if x + ( 1 / x ) = 5 , what is the value of t = x ^ 2 + ( 1 / x ) ^ 2 ? | "squaring on both sides , x ^ 2 + ( 1 / x ) ^ 2 + 2 ( x ) ( 1 / x ) = 5 ^ 2 x ^ 2 + ( 1 / x ) ^ 2 = 23 answer : c" | a = 5 ** 2
b = a - 2
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a ) $ 20,000 , b ) $ 25,000 , c ) $ 35,000 , d ) $ 40,000 , e ) $ 50,000 | e | multiply(divide(multiply(2.1, multiply(10, 7)), 420), const_100) | a certain company had a total annual expenditure of 2.1 β 10 ^ 7 on employee salaries last year . if the company employed 420 people , what was the average employee salary ? | "given : total annual expenditure of 2.1 β 10 ^ 7 on employee salaries total employees = 420 observe that 420 * 5 = 2100 therefore try to bring the numerator in terms of 2100 average salary = ( 2100 * 10 ^ 4 ) / 420 = 5 * 10 ^ 4 = 50,000 option e" | a = 10 * 7
b = 2 * 1
c = b / 420
d = c * 100
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a ) a ) 12 , b ) b ) 15 , c ) c ) 16 , d ) d ) 17 , e ) e ) 20 | e | floor(divide(181, 9)) | on dividing 181 by a number , the quotient is 9 and the remainder is 1 . find the divisor ? | "d = ( d - r ) / q = ( 181 - 1 ) / 9 = 180 / 9 = 20 e )" | a = 181 / 9
b = math.floor(a)
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a ) rs . 75 , b ) rs . 36 , c ) rs . 54 , d ) rs . 50 , e ) none | a | divide(9, divide(12, const_100)) | the banker ' s gain on a bill due due 1 year hence at 12 % per annum is rs . 9 . the true discount is | "solution t . d = [ b . g x 100 / r x t ] = rs . ( 9 x 100 / 12 x 1 ) = rs . 75 . answer a" | a = 12 / 100
b = 9 / a
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a ) $ 37.80 , b ) $ 38.50 , c ) $ 39.20 , d ) $ 39.60 , e ) $ 40.60 | d | add(multiply(36, divide(40, const_100)), 36) | a farmer spent $ 36 on feed for chickens and goats . he spent 40 % money on chicken feed , which he bought at a 20 % discount off the full price , and spent the rest on goat feed , which he bought at full price . if the farmer had paid full price for both the chicken feed and the goat feed , what amount would he have spent on the chicken feed and goat feed combined ? | "a farmer spent 40 % money on chicken feed , so he spent 0.4 * $ 36 = $ 14.4 on chicken feed , thus he spent the remaining 36 - 14.4 = $ 21.6 on goat feed . now , since he bought chicken feed at a 20 % discount then the original price of it was x * 0.8 = $ 14.4 - - > x = $ 18 therefore if the farmer had paid full price for both the chicken feed and the goat feed , then he would he have spent 18 + 21.6 = $ 39.6 . answer : d ." | a = 40 / 100
b = 36 * a
c = b + 36
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a ) 45 days , b ) 55 days , c ) 35 days , d ) 25 days , e ) 40 days | a | divide(multiply(subtract(31, 28), 300), 280) | a garrison of 300 men had a provision for 31 days . after 28 days 280 persons re - enforcement leave the garrison . find the number of days for which the remaining ration will be sufficient ? | "400 - - - 31 400 - - - 3 120 - - - ? 300 * 3 = 20 * x = > x = 45 days answer : a" | a = 31 - 28
b = a * 300
c = b / 280
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a ) 1 kmph , b ) 4 kmph , c ) 3 kmph , d ) 2 kmph , e ) 1.9 kmph | d | divide(subtract(10, 6), const_2) | what is the speed of the stream if a canoe rows upstream at 6 km / hr and downstream at 10 km / hr | sol . speed of stream = 1 / 2 ( 10 - 6 ) kmph = 2 kmph . answer d | a = 10 - 6
b = a / 2
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a ) 12.5 % , b ) 30 % , c ) 35 % , d ) 37.5 % , e ) 40 % | a | subtract(multiply(divide(subtract(const_100, 25), const_100), multiply(add(const_100, 20), divide(add(const_100, 25), const_100))), const_100) | a particular store purchased a stock of turtleneck sweaters and marked up its cost by 20 % . during the new year season , it further marked up its prices by 25 % of the original retail price . in february , the store then offered a discount of 25 % . what was its profit on the items sold in february ? | "assume the total price = 100 x price after 20 % markup = 120 x price after 25 % further markup = 1.25 * 120 x = 150 x price after the discount = 0.75 * 150 x = 112.5 x hence total profit = 12.5 % option a" | a = 100 - 25
b = a / 100
c = 100 + 20
d = 100 + 25
e = d / 100
f = c * e
g = b * f
h = g - 100
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a ) 21 , b ) 28 , c ) 90 , d ) 100 , e ) 11 | d | subtract(divide(8925, 85), 5) | a trader sells 85 meters of cloth for rs . 8925 at the profit of rs . 5 per metre of cloth . what is the cost price of one metre of cloth ? | "explanation : sp of 1 m of cloth = 8925 / 85 = rs . 105 cp of 1 m of cloth = sp of 1 m of cloth - profit on 1 m of cloth = rs . 105 - rs . 5 = rs . 100 . answer : d" | a = 8925 / 85
b = a - 5
|
['a ) 34 %', 'b ) 45 %', 'c ) 50 %', 'd ) 60 %', 'e ) 67 %'] | c | multiply(divide(subtract(multiply(const_3, divide(const_1, const_2)), const_1), const_1), const_100) | the length of a rectangle is halved , while its breadth is tripled . wa tis the % changein area ? | let original length = x and original breadth = y . original area = xy . new length = x . 2 new breadth = 3 y . new area = x x 3 y = 3 xy . 2 2 increase % = 1 xy x 1 x 100 % = 50 % . 2 xy c | a = 1 / 2
b = 3 * a
c = b - 1
d = c / 1
e = d * 100
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a ) 36 , b ) 42 , c ) 48 , d ) 56 , e ) 60 | b | divide(divide(84, const_2), const_2) | if k ^ 3 is divisible by 84 , what is the least possible value of integer k ? | "k ^ 3 = 84 * x = 2 ^ 2 * 3 * 7 * x the factors of k must at minimum include 2 * 3 * 7 = 42 . the answer is b ." | a = 84 / 2
b = a / 2
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a ) 450 km , b ) 500 km , c ) 360 km , d ) 550 km , e ) 600 km | a | multiply(multiply(add(1, divide(4, 5)), multiply(const_2, divide(575, 23))), 5) | the average speed of a car is 1 4 / 5 times the avg speed of a bike . a tractor covers 575 km in 23 hrs . how much distance will the car cover in 5 hrs if the speed of the bike is twice speed of the tractor ? | "sol . average speed of a tractor = 25 km / h the speed of a bike in an hour = 25 Γ 2 = 50 km the speed of a car in an hour = 9 / 5 * 50 = 90 km so , the distance covered by car in 5 h is 90 Γ 5 = 450 km ans . ( a )" | a = 4 / 5
b = 1 + a
c = 575 / 23
d = 2 * c
e = b * d
f = e * 5
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a ) 30 , b ) 35 , c ) 40 , d ) 45 , e ) 50 | a | divide(650, multiply(subtract(117, 325), const_0_2778)) | a train 650 m long is running at a speed of 117 km / hr . in what time will it pass a bridge 325 m long ? | "speed = 117 * 5 / 18 = 65 / 2 m / sec total distance covered = 650 + 325 = 975 m required time = 975 * 2 / 65 = 30 sec answer : a" | a = 117 - 325
b = a * const_0_2778
c = 650 / b
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a ) 11 , b ) 17 , c ) 18 , d ) 101 , e ) 1322 | a | sqrt(divide(363, const_3)) | the length of a rectangular plot is thrice its breadth . if the area of the rectangular plot is 363 sq m , then what is the breadth of the rectangular plot ? | "let the breadth of the plot be b m . length of the plot = 3 b m ( 3 b ) ( b ) = 363 3 b 2 = 363 b 2 = 121 = 11 ( b > 0 ) b = 11 m . answer : a" | a = 363 / 3
b = math.sqrt(a)
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a ) 30 , b ) 10 , c ) 18 , d ) 28 , e ) 32 | c | subtract(subtract(add(40, 12), 22), 12) | in a class of 40 students , 12 enrolled for both english and german . 22 enrolled for german . if the students of the class enrolled for at least one of the two subjects , then how many students enrolled for only english and not german ? | "answer let a be the set of students who have enrolled for english and b be the set of students who have enrolled for german . then , ( a u b ) is the set of students who have enrolled at least one of the two subjects . as the students of the class have enrolled for at least one of the two subjects , a u b = 40 we know a u b = a + b - ( a n b ) i . e , 40 = a + 22 - 12 or a = 30 which is the set of students who have enrolled for english and includes those who have enrolled for both the subjects . however , we need to find out the number of students who have enrolled for only english = students enrolled for english - students enrolled for both german and english = 30 - 12 = 18 . choice is ( c )" | a = 40 + 12
b = a - 22
c = b - 12
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a ) 2 , b ) 50 , c ) 92 , d ) 96 , e ) 98 | b | multiply(divide(subtract(100, 99), subtract(100, 98)), 100) | each of the cucumbers in 100 pounds of cucumbers is composed of 99 % water , by weight . after some of the water evaporates , the cucumbers are now 98 % water by weight . what is the new weight of the cucumbers , in pounds ? | "out of 100 pounds 99 % or 99 pounds is water and 1 pound is non - water . after some water evaporates the cucumbers become 98 % water and 2 % of non - water , so now 1 pound of non - water composes 2 % of cucucmbers , which means that the new weight of cucumbers is 1 / 0.02 = 50 pounds . answer : b ." | a = 100 - 99
b = 100 - 98
c = a / b
d = c * 100
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a ) 18 , b ) 91 , c ) 16.5 , d ) 17 , e ) 12 | c | subtract(divide(multiply(60, 50), const_100), divide(multiply(45, 30), const_100)) | how much 60 % of 50 is greater than 45 % of 30 ? | "( 60 / 100 ) * 50 β ( 45 / 100 ) * 30 30 - 13.5 = 16.5 answer : c" | a = 60 * 50
b = a / 100
c = 45 * 30
d = c / 100
e = b - d
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a ) 9 , b ) 15 , c ) 12 , d ) 11 , e ) 10 | a | divide(multiply(120, 150), multiply(50, 40)) | rectangular tile each of size 50 cm by 40 cm must be laid horizontally on a rectangular floor of size 120 cm by 150 cm , such that the tiles do not overlap and they are placed with edges jutting against each other on all edges . a tile can be placed in any orientation so long as its edges are parallel to the edges of floor . no tile should overshoot any edge of the floor . the maximum number of tiles that can be accommodated on the floor is : | "area of tile = 50 * 40 = 2000 area of floor = 120 * 150 = 18000 no of tiles = 18000 / 2000 = 9 so , the no of tile = 9 answer : a" | a = 120 * 150
b = 50 * 40
c = a / b
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a ) 12 , b ) 18 , c ) 20 , d ) 24 , e ) 30 | b | divide(multiply(60, add(multiply(4, const_2), 4)), 40) | it takes printer a 4 more minutes more than printer b to print 40 pages . working together , the two printers can print 50 pages in 6 minutes . how long will it take printer a to print 60 pages ? | "if it takes 4 more minutes for a to print 40 pages than it takes b , it takes 5 more minutes for a to print 50 pages than it takes b . thus if b is the number of minutes than b takes to print 50 pages , we can write : 1 / b + 1 / ( b + 5 ) = 1 / 6 ( since in 1 minute , they print 1 / 6 th of the 50 page job ) 6 ( 2 b + 5 ) = b ( b + 5 ) b ^ 2 - 7 b - 30 = 0 ( b - 10 ) ( b + 3 ) = 0 b = 10 thus it takes a 15 minutes to print 50 pages and 15 * 60 / 50 = 18 minutes to print 60 pages ( answer b )" | a = 4 * 2
b = a + 4
c = 60 * b
d = c / 40
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a ) 28 , b ) 266 , c ) 990 , d ) 20 , e ) 24 | e | divide(add(200, 280), multiply(add(42, 30), const_0_2778)) | two trains of length 200 m and 280 m are running towards each other on parallel lines at 42 kmph and 30 kmph respectively . in what time will they be clear of each other from the moment they meet ? | "relative speed = ( 42 + 30 ) * 5 / 18 = 4 * 5 = 20 mps . distance covered in passing each other = 200 + 280 = 480 m . the time required = d / s = 480 / 20 = 24 sec . answer : e" | a = 200 + 280
b = 42 + 30
c = b * const_0_2778
d = a / c
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a ) 360 , b ) 350 , c ) 340 , d ) 370 , e ) 380 | a | multiply(divide(subtract(multiply(22, 5), 80), subtract(24, 22)), 24) | a woodworker normally makes a certain number of parts in 24 days . but he was able to increase his productivity by 5 parts per day , and so he not only finished the job in only 22 days but also he made 80 extra parts . how many parts does the woodworker normally makes per day and how many pieces does he make in 24 days ? | let x be the number of parts the woodworker normally makes daily . in 24 days he makes 24 β
x pieces . his new daily production rate is x + 5 pieces and in 22 days he made 22 β
( x + 5 ) parts . this is 80 more than 24 β
x . therefore the equation is : 24 β
x + 80 = 22 ( x + 5 ) 30 = 2 x x = 15 normally he makes 15 parts a day and in 24 days he makes 15 β
24 = 360 parts . so answer is a . | a = 22 * 5
b = a - 80
c = 24 - 22
d = b / c
e = d * 24
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a ) 28 , b ) 12 , c ) - 28 , d ) 12 , e ) 0 | c | add(20, 8) | julian owes his classmate jenny 20 dollars . if he borrows 8 dollars how much will he owe her ? | add how he has already borrowed and how much he is going to borrow - 20 + - 8 correct answer c ) - 28 | a = 20 + 8
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a ) $ 6000 , b ) $ 5500 , c ) $ 5000 , d ) $ 5700 , e ) $ 5800 | c | divide(subtract(2500, multiply(multiply(multiply(const_3, multiply(const_2, const_3)), const_1000), divide(10, const_100))), subtract(divide(20, const_100), divide(10, const_100))) | country c imposes a two - tiered tax on imported cars : the first tier imposes a tax of 20 % of the car ' s price up to a certain price level . if the car ' s price is higher than the first tier ' s level , the tax on the portion of the price that exceeds this value is 10 % . if ron imported a $ 20,000 imported car and ended up paying $ 2500 in taxes , what is the first tier ' s price level ? | "let t be the tier price , p be total price = 20000 per the given conditions : 0.20 t + 0.10 ( p - t ) = 2500 0.20 t + 0.10 * 20000 - 0.10 t = 2500 0.10 t + 2000 = 2500 0.10 t = 2500 - 2000 = 500 t = 500 / 0.10 = 5000 answer c" | a = 2 * 3
b = 3 * a
c = b * 1000
d = 10 / 100
e = c * d
f = 2500 - e
g = 20 / 100
h = 10 / 100
i = g - h
j = f / i
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a ) 2 , b ) 9 , c ) 6 , d ) 4 , e ) 0 | a | subtract(subtract(divide(90, 5), 8), 8) | the sum of ages of 5 children born at the intervals of 8 years each is 90 years . what is the age of the youngest child ? | "let the ages of children be x , ( x + 8 ) , ( x + 16 ) , ( x + 24 ) and ( x + 32 ) years . then , x + ( x + 8 ) + ( x + 16 ) + ( x + 24 ) + ( x + 32 ) = 90 5 x = 10 x = 2 . age of the youngest child = x = 2 years . answer : a" | a = 90 / 5
b = a - 8
c = b - 8
|
a ) 57 , b ) 59 , c ) 61 , d ) 63 , e ) 65 | c | add(power(divide(subtract(9, sqrt(subtract(power(9, 2), multiply(const_4, 9)))), 2), 2), power(divide(add(9, sqrt(subtract(power(9, 2), multiply(const_4, 9)))), 2), 2)) | if a and b are the roots of the equation x 2 - 9 x + 20 = 0 , find the value of a 2 + b 2 + ab ? | "a 2 + b 2 + ab = a 2 + b 2 + 2 ab - ab i . e . , ( a + b ) 2 - ab from x 2 - 9 x + 20 = 0 , we have a + b = 9 and ab = 20 . hence the value of required expression ( 9 ) 2 - 20 = 61 . option c" | a = 9 ** 2
b = 4 * 9
c = a - b
d = math.sqrt(c)
e = 9 - d
f = e / 2
g = f ** 2
h = 9 ** 2
i = 4 * 9
j = h - i
k = math.sqrt(j)
l = 9 + k
m = l / 2
n = m ** 2
o = g + n
|
a ) 8 , b ) 9 , c ) 10 , d ) 11 , e ) 12 | d | subtract(reminder(reminder(96, 33), 17), reminder(94, reminder(33, 17))) | for all positive integers m and v , the expression m ΞΈ v represents the remainder when m is divided by v . what is the value of ( ( 96 ΞΈ 33 ) ΞΈ 17 ) - ( 94 ΞΈ ( 33 ΞΈ 17 ) ) ? | "( ( 96 ΞΈ 33 ) ΞΈ 17 ) the remainder of 96 divided by 33 is 30 ; the remainder of 30 divided by 17 is 13 ; ( 97 ΞΈ ( 33 ΞΈ 17 ) ) the remainder of 33 divided by 17 is 16 ; the remainder of 94 divided by 16 is 1 . 13 - 2 = 11 . answer : d ." | a = reminder - (
|
a ) 13.89 % , b ) 15.5 % , c ) 14 % , d ) 14.25 % , e ) 14.95 % | a | divide(multiply(multiply(40, 60), divide(12, const_100)), multiply(40, subtract(60, 6))) | a man bought 40 shares of rs . 60 at 6 discount , the rate of dividend being 12 1 / 2 % the rate of interest obtained is | "explanation : face value of a share = rs . 60 he bought each share at rs . 60 - rs . 6 = rs . 54 number of shares = 40 dividend = 12 1 / 2 % = 25 / 2 % dividend per share = 60 Γ 25 / 2 Γ 100 = rs . 7.5 total dividend = ( 40 Γ 7.5 ) ie , he got a dividend of ( 40 Γ 7.5 ) for an investment of rs . ( 40 Γ 54 ) interest obtained = 40 Γ 7.5 Γ 100 / 40 Γ 54 = 13.89 % answer : option a" | a = 40 * 60
b = 12 / 100
c = a * b
d = 60 - 6
e = 40 * d
f = c / e
|
a ) 2 , b ) 5 , c ) 6 , d ) 8 , e ) 10 | c | subtract(multiply(multiply(multiply(224, 607), 214), 863), subtract(multiply(multiply(multiply(224, 607), 214), 863), add(const_4, const_4))) | the unit digit in the product ( 224 * 607 * 214 * 863 ) is : | "explanation : unit digit in the given product = unit digit in ( 4 * 7 * 4 * 3 ) = 6 answer : c" | a = 224 * 607
b = a * 214
c = b * 863
d = 224 * 607
e = d * 214
f = e * 863
g = 4 + 4
h = f - g
i = c - h
|
a ) 49 % , b ) 18 % , c ) 17 % , d ) 13 % , e ) 16 % | a | divide(multiply(add(70, const_100), subtract(const_100, 70)), const_100) | a number is increased by 70 % and then decreased by 70 % . find the net increase or decrease per cent . | let the number be 100 . increase in the number = 70 % = 70 % of 100 = ( 70 / 100 Γ£ β 100 ) = 70 therefore , increased number = 100 + 70 = 170 this number is decreased by 70 % therefore , decrease in number = 70 % of 170 = ( 70 / 100 Γ£ β 170 ) = 11900 / 100 = 119 therefore , new number = 170 - 119 = 51 thus , net decreases = 100 - 51 = 49 hence , net percentage decrease = ( 49 / 100 Γ£ β 100 ) % = ( 4900 / 100 ) % = 49 % answer : a | a = 70 + 100
b = 100 - 70
c = a * b
d = c / 100
|
a ) 40 sec , b ) 35 sec , c ) 33 sec , d ) 42 sec , e ) 41 sec | a | divide(400, subtract(divide(50, const_3_6), divide(divide(14, const_2), const_3_6))) | a train which has 400 m long , is running 50 kmph . in what time will it cross a person moving at 14 kmph in same direction ? | "time taken to cross a moving person = length of train / relative speed time taken = 400 / ( ( 50 - 14 ) ( 5 / 18 ) = 400 / 36 * ( 5 / 18 ) = 400 / 10 = 40 sec answer : a" | a = 50 / const_3_6
b = 14 / 2
c = b / const_3_6
d = a - c
e = 400 / d
|
a ) 240 , b ) 369 , c ) 469 , d ) 520 , e ) 599 | c | subtract(add(multiply(divide(factorial(multiply(4, const_2)), factorial(subtract(multiply(4, const_2), const_2))), 4), multiply(add(4, const_1), divide(factorial(multiply(4, const_2)), factorial(subtract(multiply(4, const_2), const_2))))), multiply(add(4, const_1), add(const_3, const_4))) | how many even 4 - digit integers greater than 8000 are there such that all their digits are different ? | case 1 : numbers starting with 8 followed by even number the hundreds digit can take 4 numbers ( 0,2 , 4,6 ) so 4 ways the unit digit can take remaining 3 even numbers left after using 2 , to be even so 3 ways . the tens digit can take remaining 7 numbers left after using 3 numbers so 7 ways total = 4 * 3 * 7 = 84 case 2 : numbers starting with 8 followed by odd number the hundreds digit can take 5 numbers ( 1 , 3,5 , 7,9 ) so 5 ways the unit digit can take remaining 4 even numbers ( 0,2 , 4,6 ) to be even so 4 ways . the tens digit can take remaining 7 numbers left after using 3 numbers so 7 ways total = 5 * 4 * 7 = 140 case 3 : numbers starting with 9 followed by even number the hundreds digit can take 5 numbers ( 0 , 2,4 , 6,8 ) so 5 ways the unit digit can take remaining 4 even numbers left after using 1 , to be even so 3 ways . the tens digit can take remaining 7 numbers left after using 3 numbers so 7 ways total = 5 * 3 * 7 = 105 case 4 : numbers starting with 9 followed by odd number the hundreds digit can take 4 numbers ( 1,3 , 5,7 ) so 4 ways the unit digit can take 5 numbers ( 0 , 2,4 , 6,8 ) to be even so 5 ways . the tens digit can take remaining 7 numbers left after using 3 numbers so 7 ways total = 4 * 5 * 7 = 140 hence 84 + 140 + 105 + 140 = 469 , correct answer is c | a = 4 * 2
b = math.factorial(a)
c = 4 * 2
d = c - 2
e = math.factorial(d)
f = b / e
g = f * 4
h = 4 + 1
i = 4 * 2
j = math.factorial(i)
k = 4 * 2
l = k - 2
m = math.factorial(l)
n = j / m
o = h * n
p = g + o
q = 4 + 1
r = 3 + 4
s = q * r
t = p - s
|
a ) $ 5,330 , b ) $ 3,360 , c ) $ 1,350 , d ) $ 600 , e ) $ 150 | d | multiply(60, divide(multiply(360, 1210), divide(multiply(360, 1210), const_10))) | a certain farmer pays $ 60 per acre per month to rent farmland . how much does the farmer pay per month to rent a rectangular plot of farmland that is 360 feet by 1210 feet ? ( 43,560 square feet = 1 acre ) | basically the question an error . 1 acre = 43,560 square feet and if it is then the answer is 600 ( d ) | a = 360 * 1210
b = 360 * 1210
c = b / 10
d = a / c
e = 60 * d
|
a ) $ 6715 , b ) $ 5615 , c ) $ 6415 , d ) $ 6615 , e ) $ 6315 | d | add(add(multiply(multiply(multiply(const_3, const_2), const_100), const_10), divide(multiply(multiply(multiply(multiply(const_3, const_2), const_100), const_10), 5), const_100)), divide(multiply(add(multiply(multiply(multiply(const_3, const_2), const_100), const_10), divide(multiply(multiply(multiply(multiply(const_3, const_2), const_100), const_10), 5), const_100)), 5), const_100)) | today joelle opened an interest - bearing savings account and deposited $ 6,000 . if the annual interest rate is 5 percent compounded interest , and she neither deposits nor withdraws money for exactly 2 years , how much money will she have in the account ? | interest for 1 st year = 6000 * 5 / 100 = 300 interest for 2 nd year = 6300 * 5 / 100 = 315 total = 6000 + 300 + 315 = 6615 answer : d | a = 3 * 2
b = a * 100
c = b * 10
d = 3 * 2
e = d * 100
f = e * 10
g = f * 5
h = g / 100
i = c + h
j = 3 * 2
k = j * 100
l = k * 10
m = 3 * 2
n = m * 100
o = n * 10
p = o * 5
q = p / 100
r = l + q
s = r * 5
t = s / 100
u = i + t
|
a ) 25 , b ) 50 , c ) 75 , d ) 100 , e ) 4 | b | subtract(100, multiply(2, 25)) | what is x if x + 2 y = 100 and y = 25 ? | "substitute y by 25 in x + 2 y = 100 x + 2 ( 25 ) = 100 x + 50 = 100 if we substitute x by 50 in x + 50 = 100 , we have 50 + 50 = 100 . hence x = 50 correct answer b" | a = 2 * 25
b = 100 - a
|
a ) 5 days , b ) 7 days , c ) 12 days , d ) 4.5 days , e ) 10 days | d | inverse(multiply(inverse(3), subtract(const_1, multiply(5, inverse(15))))) | a can do a piece of work in 15 days . a does the work for 5 days only and leaves the job . b does the remaining work in 3 days . in how many days b alone can do the work ? | "explanation : a β s 5 day work = 5 * 1 / 15 = 1 / 3 remaining work = 1 - 1 / 3 = 2 / 3 b completes 2 / 3 work in 6 days b alone can do in x days 2 / 3 * x = 3 x = 4.5 days answer : option d" | a = 1/(3)
b = 1/(15)
c = 5 * b
d = 1 - c
e = a * d
f = 1/(e)
|
a ) 600 m , b ) 700 m , c ) 800 m , d ) 1000 m , e ) 1200 m | c | multiply(divide(multiply(50, 8), subtract(multiply(8, 5), multiply(4, 5))), multiply(8, 5)) | a tiger is 50 of its own leaps behind a deer . the tiger takes 5 leaps per minutes to deer β s 4 . if the tiger and the deer cover 8 m and 5 m per leap respectively , what distance will the tiger have to run before it catches the deer ? | tiger takes 5 leaps / m and deer takes 4 leaps / m thus , tiger runs at the speed of 5 x 8 = 40 m / min and deer runs at the speed of 4 x 5 = 20 m / min tiger is 50 leaps behind , which is 50 x 8 = 400 m relative speed of tiger and deer is = 20 m / min thus , 400 / 20 = 20 m thus , the tiger will catch deer = 20 x 40 = 800 m answer : c | a = 50 * 8
b = 8 * 5
c = 4 * 5
d = b - c
e = a / d
f = 8 * 5
g = e * f
|
a ) 22 , b ) 37 , c ) 32 , d ) 27 , e ) 18 | c | divide(add(25, 39), const_2) | a man can row upstream at 25 kmph and downstream at 39 kmph , and then find the speed of the man in still water ? | "us = 25 ds = 39 m = ( 39 + 25 ) / 2 = 32 answer : c" | a = 25 + 39
b = a / 2
|
a ) 2 / 9 , b ) 4 / 5 , c ) 1 / 6 , d ) 6 / 6 , e ) 3 / 9 | c | divide(add(3, 120), multiply(3, 120)) | if the sum of two numbers is 60 and the h . c . f and l . c . m of these numbers are 3 and 120 respectively , then the sum of the reciprocal of the numbers is equal to : | "let the numbers be a and b . then , a + b = 60 and ab = 3 * 120 = 360 . required sum = 1 / a + 1 / b = ( a + b ) / ab = 60 / 360 = 1 / 6 . answer : c" | a = 3 + 120
b = 3 * 120
c = a / b
|
a ) 1 / 4 , b ) 4 / 15 , c ) 2 / 13 , d ) 4 / 11 , e ) 4 / 5 | c | divide(2, add(subtract(const_12, const_1), 2)) | last year department store x had a sales total for december that was 2 times the average ( arithmetic mean ) of the monthly sales totals for january through november . the sales total for december was what fraction of the sales total for the year ? | "let avg for 11 mos . = 10 therefore , dec = 20 year total = 11 * 10 + 20 = 130 answer = 20 / 130 = 2 / 13 = c" | a = 12 - 1
b = a + 2
c = 2 / b
|
a ) 9 , b ) 12 , c ) 17 , d ) 21 , e ) 24 | c | add(multiply(divide(24, subtract(24, 18)), const_2), const_1) | a worker earns $ 24 on the first day and spends $ 18 on the second day . the worker earns $ 24 on the third day and spends $ 18 on the fourth day . if this pattern continues , on which day will the worker first reach a net total of $ 72 ? | "every two days , the net total is $ 6 . after 16 days , the worker will have $ 48 . on day 17 , the worker will receive $ 24 for a net total of $ 72 . the answer is c ." | a = 24 - 18
b = 24 / a
c = b * 2
d = c + 1
|
a ) 1 / 3 , b ) 2 / 3 , c ) 4 / 3 , d ) 5 / 3 , e ) 7 / 3 | e | divide(subtract(divide(2, 3), divide(1, 5)), divide(1, 5)) | in a class of students , 2 / 3 of the number of girls is equal to 1 / 5 of the total number of students . what is the ratio of boys to girls in the class ? | ( 2 / 3 ) g = ( 1 / 5 ) ( b + g ) 10 g = 3 b + 3 g 7 g = 3 b b / g = 7 / 3 . the answer is e . | a = 2 / 3
b = 1 / 5
c = a - b
d = 1 / 5
e = c / d
|
a ) 6 , b ) 7 , c ) 8 , d ) 9 , e ) 10 | b | add(add(const_1, const_2), 4) | jaya two digits age when appended with the two digits age of amitabh the 4 digits number formed is a perfect square . what is the the sum of the 4 digits no . ? | min no could be = 7 ( 1024 ) let jaya has two digits number = 10 and amitabh has two digits number = 24 after appending = 1024 which is perfect square of 32 and sum of four digits number 1024 = 7 answer : b | a = 1 + 2
b = a + 4
|
a ) 2 litres , b ) 5 litres , c ) 7 litres , d ) 11 litres , e ) none of these | b | divide(subtract(multiply(25, 3.6), multiply(25, 3)), 3) | pure milk costs 3.60 per litre . a milkman adds water to 25 litres of pure milk and sells the mixture at 3 per litre . how many litres of water does he add ? | in mixture , quantity of pure milk / quantity of water = 3 β 0 / 3.6 β 3 = 3 / 0.6 = 5 / 1 since in every 5 litres of milk , he adds 1 litre of water . β΄ in every 25 litres of milk , he adds 5 litres of water . answer b | a = 25 * 3
b = 25 * 3
c = a - b
d = c / 3
|
a ) 2 , b ) 3 , c ) 4 , d ) 6 , e ) 60 / 7 | e | divide(12000, subtract(divide(6000, 2.5), subtract(divide(3000, 1), divide(6000, 3)))) | machine a can process 6000 envelopes in 3 hours . machines b and c working together but independently can process the same number of envelopes in 2.5 hours . if machines a and c working together but independently process 3000 envelopes in 1 hour , then how many hours would it take machine b to process 12000 envelopes . | for 1 hour - machine a rate - 2000 envelopes machine b + c rate - 2400 envelopes since a + c = 3000 envelopes a ' s rate is 2000 envelopes as above , c has a rate of 1000 envelopes per hour . which makes machine b ' s rate as 1400 envelopes per hour . thus , it will take 8 hours to manufacture 12000 envelopes . answer : - e | a = 6000 / 2
b = 3000 / 1
c = 6000 / 3
d = b - c
e = a - d
f = 12000 / e
|
a ) 5 / 4 , b ) 2 / 3 , c ) 1 / 6 , d ) 5 / 2 , e ) 1 / 8 | c | subtract(divide(add(multiply(const_10, 2), 2), 3), divide(add(const_10, 3), 4)) | what is 2 4 / 3 - 3 1 / 4 divided by 2 / 3 - 1 / 6 ? | "2 4 / 3 - 3 1 / 4 = 10 / 3 - 13 / 4 = ( 40 - 39 ) / 12 = 1 / 12 2 / 3 - 1 / 6 = ( 4 - 1 ) / 6 = 3 / 6 = 1 / 2 so 1 / 12 / 1 / 2 = 2 / 12 = 1 / 6 answer - c" | a = 10 * 2
b = a + 2
c = b / 3
d = 10 + 3
e = d / 4
f = c - e
|
a ) 45 , b ) 65 , c ) 75 , d ) 78 , e ) 90 | d | subtract(multiply(80, const_4), subtract(multiply(79, const_4), add(const_3.0, subtract(multiply(80, const_4), multiply(84, 6))))) | the avg weight of a , b & c is 84 kg . if d joins the group , the avg weight of the group becomes 80 kg . if another man e who weights is 6 kg more than d replaces a , then the avgof b , c , d & e becomes 79 kg . what is the weight of a ? | "a + b + c = 3 * 84 = 252 a + b + c + d = 4 * 80 = 320 - - - - ( i ) so , d = 68 & e = 68 + 6 = 74 b + c + d + e = 79 * 4 = 316 - - - ( ii ) from eq . ( i ) & ( ii ) a - e = 320 β 316 = 4 a = e + 4 = 74 + 4 = 78 d" | a = 80 * 4
b = 79 * 4
c = 80 * 4
d = 84 * 6
e = c - d
f = 3 + 0
g = b - f
h = a - g
|
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