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 ) 19', 'b ) 750', 'c ) 24', 'd ) 120', 'e ) none of these'] | a | divide(circle_area(divide(31, multiply(2, const_pi))), const_4) | how many plants will be there in a circular bed whose outer edge measure 31 cms , allowing 4 cm 2 for each plant ? | circumference of circular bed = 31 cm area of circular bed = ( 31 ) 2 â „ 4 ï € space for each plant = 4 cm 2 â ˆ ´ required number of plants = ( 31 ) 2 â „ 4 ï € ã · 4 = 19.11 = 19 ( approx ) answer a | a = 2 * math.pi
b = 31 / a
c = circle_area / (
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a ) 0.06 , b ) 0.12 , c ) 0.21 , d ) 0.29 , e ) 0.28 | e | multiply(multiply(subtract(const_1, 0.3), subtract(const_1, 0.2)), subtract(const_1, 0.5)) | 3 cannons are firing at a target . if their individual probabilities to hit the target are 0.5 , 0.2 , and 0.3 respectively , what is the probability that none of the cannons will hit the target after one round of fire ? prob . that all the cannons will hit the target = . 06 prob . that none of the cannons will hit = 1 - . 06 = . 94 | the probability that eachdoesn ' thit is : 0.5 , 0.8 , and 0.7 . when we have multiple independent events , we multiply the probabilities : . 5 * . 8 * . 7 = 0.28 . option : e | a = 1 - 0
b = 1 - 0
c = a * b
d = 1 - 0
e = c * d
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a ) $ 20000 , b ) $ 15000 , c ) $ 12000 , d ) $ 10000 , e ) $ 9000 | a | multiply(multiply(5000, const_2), const_2) | if money is invested at r percent interest , compounded annually , the amount of investment will double in approximately 70 / r years . if pat ' s parents invested $ 5000 in a long term bond that pays 4 percent interest , compounded annually , what will be the approximate total amount of investment 36 years later , when pat is ready for college ? | "since investment doubles in 70 / r years then for r = 4 it ' ll double in 70 / 4 = ~ 18 years ( we are not asked about the exact amount so such an approximation will do ) . thus in 36 years investment will double twice and become ( $ 5,000 * 2 ) * 2 = $ 20,000 ( after 18 years investment will become $ 5,000 * 2 = $ 10,000 and in another 18 years it ' ll become $ 10,000 * 2 = $ 20,000 ) . answer : a ." | a = 5000 * 2
b = a * 2
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a ) 511 , b ) 419 , c ) 616 , d ) 636 , e ) 686 | c | multiply(multiply(multiply(divide(2, 5), divide(3, 7)), divide(7, 11)), 64) | 2 / 5 of 3 / 7 of 23 / 11 of ? = 64 × 3 | "explanation : 2 / 5 & time ; 3 / 7 & time ; 20 / 11 & time ; x = 192 x = 192 & time ; 11 & time ; 7 & time ; 5 / 2 & time ; 3 & time ; 20 = 616 answer : option c" | a = 2 / 5
b = 3 / 7
c = a * b
d = 7 / 11
e = c * d
f = e * 64
|
a ) 17 % , b ) 19 % , c ) 18 % , d ) 20 % , e ) 22 % | d | divide(multiply(22, const_100), add(22, const_100)) | the annual interest rate earned by an investment increased by 11 percent from last year to this year . if the annual interest rate earned by the investment this year was 22 percent , what was the annual interest rate last year ? | "let i = interest rate i ( this year ) = i ( last year ) + 0.11 i ( last year ) = 1.11 i ( last year ) 22 = 1.11 x i ( last year ) i ( last year ) = 22 / 1.1 = 220 / 11 = 20 % answer : d" | a = 22 * 100
b = 22 + 100
c = a / b
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a ) 1 / 32 , b ) 1 / 28 , c ) 1 / 24 , d ) 1 / 16 , e ) 1 / 14 | c | divide(const_1, add(20, const_4)) | if a randomly selected positive single digit multiple of 3 is multiplied by a randomly selected prime number less than 20 , what is the probability that this product will be a multiple of 45 ? | there are 3 single digit multiple of 3 , that is , 3 , 69 . there are 8 prime nos less than 20 - 23 , 57 , 1113 , 1719 total outcome - 8 * 3 = 24 favourable outcome = 1 ( 9 * 5 ) hence required probability = 1 / 24 . answer c . | a = 20 + 4
b = 1 / a
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a ) 1 / 10 , b ) 1 / 5 , c ) 2 / 5 , d ) 3 / 5 , e ) 3 / 10 | b | divide(const_2, 10) | if p ^ 2 – 13 p + 40 = h , and p is a positive integer between 1 and 10 , inclusive , what is the probability that h < 0 ? | p 2 – 13 p + 40 = h so ( p – 8 ) ( p – 5 ) = h for q to be negative , the expressions ( p – 8 ) and ( p – 5 ) must have opposite signs . which integers on the number line will yield opposite signs for the expressions ( p – 8 ) and ( p – 5 ) ? those integers in the range 5 < p < 8 ( notice 5 and 8 are not included because they would both yield a value of zero and zero is a nonnegative integer ) . that means that there are only two integer values for p , 6 and 7 , that would yield a negative q . with a total of 10 possible p values , only 2 yield a negative q , so the probability is 2 / 10 or 1 / 5 . the correct answer is b . | a = 2 / 10
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a ) 3008 , b ) 2002 , c ) 1008 , d ) 2016 , e ) 3000 | b | divide(4004, const_2) | abcd is a square where ab = â ˆ š 4004 . let x be a point on ab and y be a point on cd such that ax = cy . compute the area of trapezoid axyd . | "note that trapezoids axy d and bxy c are congruent , so the area of axy d is always 4004 / 2 = 2002 . correct answer b" | a = 4004 / 2
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a ) 5 / 12 , b ) 12 / 5 , c ) 1 / 3 , d ) 144 / 25 , e ) 146 / 25 | c | sqrt(divide(multiply(16, const_3), multiply(216, const_2))) | two - third of a positive number and 16 / 216 of its reciprocal are equal . the number is : | let the number be x . then , 2 / 3 x = 16 / 216 * 1 / x x 2 = 16 / 216 * 3 / 2 = 16 / 144 = 1 / 9 x = 1 / 3 answer : c | a = 16 * 3
b = 216 * 2
c = a / b
d = math.sqrt(c)
|
a ) 6 , b ) 7 , c ) 8 , d ) 13 , e ) 10 | d | add(add(9, const_2), const_2) | if x < y < z and y - x > 9 , where x is an even integer and y and z are odd integers , what is the least possible value of z - x ? | we have : 1 ) x < y < z 2 ) y - x > 9 3 ) x = 2 k ( x is an even number ) 4 ) y = 2 n + 1 ( y is an odd number ) 5 ) z = 2 p + 1 ( z is an odd number ) 6 ) z - x = ? least value z - x = 2 p + 1 - 2 k = 2 p - 2 k + 1 = 2 ( p - k ) + 1 - that means that z - x must be an odd number . we can eliminate answer choices a , c and e we are asked to find the least value , so we have to pick the least numbers since y is odd and x is even , y - x must be odd . since y - x > 9 , the least value for y - x must be 11 , the least value for x must be 2 , and , thus , the least possible value for y must be 13 ( y - 2 = 11 , y = 13 ) 2 < 13 < z , since z is odd , the least possible value for z is 15 z - x = 15 - 2 = 13 answer d | a = 9 + 2
b = a + 2
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a ) 3 , b ) 5 , c ) 7 , d ) 11 , e ) 17 | c | divide(1365, multiply(multiply(add(const_2, const_3), add(add(const_2, const_3), const_2)), add(const_10, const_3))) | in a certain business school class , p students are accounting majors , q students are finance majors , r students are marketing majors , and s students are strategy majors . if pqrs = 1365 , and if 1 < p < q < r < s , how many students in the class are marketing majors ? | "pqrs = 1365 = 3 * 5 * 7 * 13 since 1 < p < q < r < s , the number of students who are marketing majors is r = 7 . the answer is c ." | a = 2 + 3
b = 2 + 3
c = b + 2
d = a * c
e = 10 + 3
f = d * e
g = 1365 / f
|
a ) − 100 , b ) 0.007 , c ) 0.25 , d ) 4 , e ) 8 | b | divide(divide(2, 2), const_1000) | if x = 8 and y = − 2 , what is the value of ( x − 2 y ) ^ y ? | "quickly we can spot that answer is neither integer nor negative . eliminate a , de by inversing and squaring 0.007 answer : b" | a = 2 / 2
b = a / 1000
|
a ) 83.33 , b ) 882 , c ) 772 , d ) 252 , e ) 121 | a | multiply(divide(multiply(60, const_1000), const_3600), 5) | a train running at the speed of 60 km / hr crosses a pole in 5 seconds . find the length of the train . | "speed = 60 * ( 5 / 18 ) m / sec = 50 / 3 m / sec length of train ( distance ) = speed * time ( 50 / 3 ) * 5 = 83.33 meter . answer : a" | a = 60 * 1000
b = a / 3600
c = b * 5
|
a ) 16.7 % , b ) 16.2 % , c ) 16.5 % , d ) 15.5 % , e ) 15.3 % | e | multiply(divide(subtract(add(multiply(10, 12), 18), multiply(10, 12)), 118), const_100) | on a purchase of $ 118 , a store offered a payment plan consisting of a $ 18 down payment and 12 monthly payments of $ 10 each . what percent of the purchase price , to the nearest tenth of a percent , did the customer pay in interest by using this plan ? | "12 * 10 + 18 = 138 ( 18 / 118 ) * 100 = 15.3 answer : e" | a = 10 * 12
b = a + 18
c = 10 * 12
d = b - c
e = d / 118
f = e * 100
|
a ) 1 , b ) 3.5 , c ) 20 , d ) 49 , e ) 70 | e | divide(1, divide(1, 70)) | if 70 honey bees make 70 grams of honey in 70 days , then 1 honey bee will make 1 gram of honey in how many days ? | "explanation : let the required number days be x . less honey bees , more days ( indirect proportion ) less honey , less days ( direct proportion ) honey bees 1 : 70 : : 70 : x honey 70 : 1 = > 1 x 70 x x = 70 x 1 x 70 = > x = 70 . answer : e" | a = 1 / 70
b = 1 / a
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a ) 15 , b ) 17 , c ) 19 , d ) 21 , e ) 26 | a | add(add(multiply(const_2, const_4), 3), const_4) | 3 times the first of 3 consecutive odd integers is 3 more than twice the third . the third integer is ? | let the three integers be x , x + 2 and x + 4 . then , 3 x = 2 ( x + 4 ) + 3 x = 11 . third integer = x + 4 = 15 . option a | a = 2 * 4
b = a + 3
c = b + 4
|
a ) 76 km / hr , b ) 66 km / hr , c ) 48 km / hr , d ) 67 km / hr , e ) 22 km / hr | c | divide(multiply(multiply(divide(add(100, 100), multiply(10, add(const_1, const_2))), const_2), const_3600), const_1000) | two trains , each 100 m long , moving in opposite directions , cross other in 10 sec . if one is moving twice as fast the other , then the speed of the faster train is ? | let the speed of the slower train be x m / sec . then , speed of the train = 2 x m / sec . relative speed = ( x + 2 x ) = 3 x m / sec . ( 100 + 100 ) / 10 = 3 x = > x = 20 / 3 . so , speed of the faster train = 40 / 3 = 40 / 3 * 18 / 5 = 48 km / hr . answer : c | a = 100 + 100
b = 1 + 2
c = 10 * b
d = a / c
e = d * 2
f = e * 3600
g = f / 1000
|
a ) 18 , b ) 28 , c ) 48 , d ) 38 , e ) 54 | e | divide(multiply(subtract(69, 9), 9), add(9, const_1)) | if a certain number is divided by 9 , the quotient , dividend , and divisor , added together , will amount to 69 . what is the number ? | let x = the number sought . then x / 9 + x + 9 = 69 . and x = 54 . e | a = 69 - 9
b = a * 9
c = 9 + 1
d = b / c
|
a ) s . 532 , b ) s . 600 , c ) s . 822 , d ) s . 812 , e ) s . 810 | b | multiply(2400, divide(1, 3)) | a , b , c and d enter into partnership . a subscribes 1 / 3 of the capital b 1 / 4 , c 1 / 5 and d the rest . how much share did b get in a profit of rs . 2400 ? | "let the total amount in the partnership be ' x ' . then a ' s share = x / 3 b ' s share = x / 4 c ' s share = x / 5 d ' s share = x - ( x / 3 + x / 4 + x / 5 ) = 13 x / 60 a : b : c : d = x / 3 : x / 4 : x / 5 : 13 x / 60 = 20 : 15 : 12 : 13 b ' s share in the profit of rs . 2400 = 15 ( 2400 / 60 ) = rs . 600 . answer : b" | a = 1 / 3
b = 2400 * a
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a ) rs . 49.17 , b ) rs . 51.03 , c ) rs . 54.17 , d ) rs . 55.33 , e ) none of the above | d | divide(add(multiply(10, 50), multiply(5, 66)), add(10, 5)) | if 10 litres of an oil of rs . 50 per litres be mixed with 5 litres of another oil of rs . 66 per litre then what is the rate of mixed oil per litre ? | "50 * 10 = 500 66 * 5 = 330 830 / 15 = 55.33 answer : d" | a = 10 * 50
b = 5 * 66
c = a + b
d = 10 + 5
e = c / d
|
a ) 20 % , b ) 10 % , c ) 25 % , d ) 18 % , e ) 30 % | b | subtract(const_100, multiply(multiply(divide(subtract(const_100, divide(multiply(const_100, 25), const_100)), const_100), divide(add(const_100, divide(multiply(const_100, 20), const_100)), const_100)), const_100)) | if the price of a book is first decreased by 25 % and then increased by 20 % , then the decrease in the price will be ? | "let the original price be $ 100 new final price = 120 % of ( 75 % of $ 100 ) = 120 / 100 * 75 / 100 * 100 = $ 90 decrease is 10 % answer is b" | a = 100 * 25
b = a / 100
c = 100 - b
d = c / 100
e = 100 * 20
f = e / 100
g = 100 + f
h = g / 100
i = d * h
j = i * 100
k = 100 - j
|
a ) 75 kg , b ) 85 kg , c ) 95 kg , d ) 65 kg , e ) 55 kg | e | add(multiply(2.5, 8), 35) | the average weight of 8 people increases by 2.5 kg when a new person comes in place of one of them weighing 35 kg . what is the weight of the new person ? | "the total weight increase = ( 8 x 2.5 ) kg = 20 kg weight of new person = ( 35 + 20 ) kg = 55 kg the answer is e ." | a = 2 * 5
b = a + 35
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a ) 1 , b ) 2 , c ) 3 , d ) 4 , e ) 5 | b | subtract(divide(multiply(const_3, multiply(8, 2)), add(8, 4)), 2) | from an island , it is possible to reach the mainland by either ferry p or ferry q . ferry p travels for 2 hours at 8 kilometers per hour , while ferry q takes a route that is three times longer . if ferry p is slower than ferry q by 4 kilometers per hour , how many hours longer is the journey of ferry q compared with the journey of ferry p ? | the distance traveled by ferry p is 16 km . then the distance traveled by ferry q is 48 km . ferry q travels at a speed of 12 kph . the time of the journey for ferry q is 48 / 12 = 4 hours , which is 2 hours more than ferry p . the answer is b . | a = 8 * 2
b = 3 * a
c = 8 + 4
d = b / c
e = d - 2
|
a ) 50 kmph , b ) 60 kmph , c ) 65 kmph , d ) 36 kmph , e ) none | d | divide(324, divide(multiply(6, 3), 2)) | a car takes 6 hours to cover a distance of 324 km . how much should the speed in kmph be maintained to cover the same direction in 3 / 2 th of the previous time ? | "time = 6 distence = 324 3 / 2 of 6 hours = 6 * 3 / 2 = 9 hours required speed = 324 / 9 = 36 kmph d )" | a = 6 * 3
b = a / 2
c = 324 / b
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a ) 86,400 , b ) 172,800 , c ) 468,830 , d ) 338,200 , e ) 259,200 | a | multiply(multiply(subtract(4, 2), const_3600), const_12) | in a renowned city , the average birth rate is 4 people every two seconds and the death rate is 2 people every two seconds . estimate the size of the population net increase that occurs in one day . | "every 2 seconds , 2 persons are added ( 4 - 2 ) . every second 1 person are added . in a day 24 hrs = 24 * 60 minutes = 24 * 60 * 60 = 86400 seconds . 86400 * 1 = 86400 option a" | a = 4 - 2
b = a * 3600
c = b * 12
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a ) 53.33 mph , b ) 88 mph , c ) 60 mph , d ) 64 mph , e ) 66.67 mph | b | add(divide(add(multiply(80, 3), multiply(40, 5)), add(3, 5)), subtract(divide(const_100, const_3), const_0_33)) | steve traveled the first 5 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 ? | average speed = total distance / total time = ( 40 * 5 + 80 * 3 ) / ( 5 + 3 ) = 440 / 5 = 88 answer : b | a = 80 * 3
b = 40 * 5
c = a + b
d = 3 + 5
e = c / d
f = 100 / 3
g = f - const_0_33
h = e + g
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a ) 16 , b ) 5 , c ) 4 , d ) 2 , e ) 1 | b | divide(divide(20, const_2), const_2) | if circles x and y have the same area and circle x has a circumference of 20 π , half of the radius of circle y is : | "x b é radius of circle x ý b é radius of circle y given : pi * x ^ 2 = pi * ý ^ 2 also , 2 * pi * x = 20 * pi x = 10 thus ý = 10 y / 2 = 5 ans : b" | a = 20 / 2
b = a / 2
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a ) 1 : 1 , b ) 2 : 3 , c ) 3 : 2 , d ) 3 : 1 , e ) 3 : 4 | a | divide(multiply(multiply(multiply(multiply(const_4, const_3), multiply(10, const_2)), const_100), 8), multiply(16000, 10)) | ravi and sunil are partners in a business . ravi invests rs . 20,000 for 8 months and sunil invested rs . 16000 for 10 months then after one year ratio of their profits will be | "= ( 15000 * 8 ) : ( 8000 * 10 ) = 160000 : 160000 = 1 : 1 answer : a" | a = 4 * 3
b = 10 * 2
c = a * b
d = c * 100
e = d * 8
f = 16000 * 10
g = e / f
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['a ) 75 %', 'b ) 80 %', 'c ) 100 %', 'd ) 120 %', 'e ) 125 %'] | b | multiply(divide(volume_cylinder(divide(8, const_2), 10), volume_cylinder(divide(10, const_2), 8)), const_100) | tanks c and b are each in the shape of a right circular cylinder . the interior of tank c has a height of 10 meters and a circumference of 8 meters , and the interior of tank b has a height of 8 meters and a circumference of 10 meters . the capacity of tank c is what percent of the capacity of tank b ? | b . for c , r = 8 / 2 pi . its capacity = ( 4 pi ) ^ 2 * 10 = 160 pi for b , r = 10 / pi . its capacity = ( 5 pi ) ^ 2 * 8 = 200 pi c / b = 160 pi / 200 pi = 0.8 | a = 8 / 2
b = volume_cylinder / (
c = 10 / 2
d = b * volume_cylinder
|
a ) 7 , b ) 9 , c ) 8 , d ) 11 , e ) 10 | c | divide(multiply(subtract(22, const_2), const_2), add(const_4, const_1)) | a is two years older than b who is twice as old as c . if the total of the ages of a , b and c be 22 , the how old is b ? | "explanation : let c ' s age be x years . then , b ' s age = 2 x years . a ' s age = ( 2 x + 2 ) years . ( 2 x + 2 ) + 2 x + x = 22 ⇒ 5 x = 20 ⇒ x = 4 . hence , b ' s age = 2 x = 8 years . answer : c" | a = 22 - 2
b = a * 2
c = 4 + 1
d = b / c
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a ) 69 . , b ) 73 . , c ) 53 , d ) 91 , e ) 108 | c | subtract(subtract(420, 312), subtract(250, divide(multiply(250, 78), const_100))) | in the fifth grade at parkway elementary school there are 420 students . 312 students are boys and 250 students are playing soccer . 78 % of the students that play soccer are boys . how many girl student are in parkway that is not playing soccer ? | "total students = 420 boys = 312 , girls = 108 total playing soccer = 250 78 % of 250 = 195 are boys who play soccer . girls who play soccer = 55 . total girls who do not play soccer = 108 - 55 = 53 . correct option : c" | a = 420 - 312
b = 250 * 78
c = b / 100
d = 250 - c
e = a - d
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a ) 700 , b ) 300 , c ) 250 , d ) 550 , e ) 800 | e | multiply(divide(1500, add(multiply(divide(subtract(12, 6), subtract(3, 1)), 3), 6)), 8) | the price of 3 pants and 6 t - shirts is rs . 1500 . with the same money one can buy 1 pant and 12 t - shirts . if one wants to buy 8 t - shirts , how much shall she have to pay ? | let the price of a pant and a t - shirt be rs . x and rs . y respectively . then , 3 x + 6 y = 1500 . . . . ( i ) and x + 12 y = 1500 . . . . ( ii ) divide equation ( i ) by 3 , we get the below equation . = x + 2 y = 500 . - - - ( iii ) now subtract ( iii ) from ( ii ) x + 12 y = 1500 ( - ) x + 2 y = 500 - - - - - - - - - - - - - - - - 10 y = 1000 - - - - - - - - - - - - - - - - cost of 8 t - shirts = 8 * 100 = 800 answer : e | a = 12 - 6
b = 3 - 1
c = a / b
d = c * 3
e = d + 6
f = 1500 / e
g = f * 8
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a ) rs . 260 , b ) rs . 270 , c ) rs . 280 , d ) rs . 290 , e ) rs . 300 | d | add(add(divide(subtract(770, add(add(40, 40), 30)), const_3), 30), 40) | rs . 770 were divided among a , b , c in such a way that a had rs . 40 more than b and c had rs 30 more than a . how much was b â € ™ s share ? | let b gets rs x . then we can say a gets rs ( x + 40 ) and c gets rs ( x + 70 ) . x + 40 + x + x + 70 = 770 3 x = 660 x = 220 . b â € ™ s share = rs ( 220 + 70 ) = rs . 290 d | a = 40 + 40
b = a + 30
c = 770 - b
d = c / 3
e = d + 30
f = e + 40
|
a ) 14 , b ) 15 , c ) 16 , d ) 17 , e ) 18 | a | add(divide(subtract(39, 3), 3), const_2) | what is the greatest of 3 consecutive integers whose sum is 39 ? | "39 / 3 = 13 the three numbers are 12 , 13 , and 14 . the answer is a ." | a = 39 - 3
b = a / 3
c = b + 2
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a ) 3 , b ) 3 1 / 3 , c ) 3 1 / 2 , d ) 4 , e ) 1 1 / 4 | e | divide(add(subtract(15, 9), subtract(9, subtract(multiply(divide(40, const_60), 15), multiply(divide(40, const_60), 9)))), subtract(multiply(divide(40, const_60), 15), multiply(divide(40, const_60), 9))) | john and jacob set out together on bicycle traveling at 15 and 9 miles per hour , respectively . after 40 minutes , john stops to fix a flat tire . if it takes john one hour to fix the flat tire and jacob continues to ride during this time , how many hours will it take john to catch up to jacob assuming he resumes his ride at 15 miles per hour ? ( consider john ' s deceleration / acceleration before / after the flat to be negligible ) | john ' s speed - 15 miles / hr jacob ' s speed - 9 miles / hr after 40 min ( i . e 2 / 3 hr ) , distance covered by john = 15 x 2 / 3 = 10 miles . jacob continues to ride for a total of 1 hour and 40 min ( until john ' s bike is repaired ) . distance covered in 1 hour 40 min ( i . e 5 / 3 hr ) = 9 x 5 / 3 = 15 miles . now , when john starts riding back , the distance between them is 7.5 miles . jacob and john are moving in the same direction . for john to catch jacob , the effective relative speed will be 15 - 9 = 6 miles / hr . thus , to cover 7.5 miles at 6 miles / hr , john will take 7.5 / 6 = 1.25 hours answer e | a = 15 - 9
b = 40 / const_60
c = b * 15
d = 40 / const_60
e = d * 9
f = c - e
g = 9 - f
h = a + g
i = 40 / const_60
j = i * 15
k = 40 / const_60
l = k * 9
m = j - l
n = h / m
|
a ) 65 , b ) 75 , c ) 85 , d ) 95 , e ) 45 | c | multiply(23, 19) | 23 × 19 × ? ÷ 19 ÷ 17 = 115 | "explanation : 23 × 19 × ? ÷ 19 ÷ 17 = 115 = > ( 23 × 19 × ? ) / ( 19 × 17 ) = 115 = > ? = ( 115 × 19 × 17 ) / ( 23 × 19 ) = 85 answer : option c" | a = 23 * 19
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a ) 15,000 , b ) 15,500 , c ) 16,000 , d ) 16,500 , e ) 17,000 | a | divide(divide(97500, add(multiply(divide(add(const_100, const_10), const_100), subtract(6, const_1)), const_1)), const_1000) | a city with a population of 97500 is to be divided into 6 voting districts , and no district is to have a population that is more than 10 percent greater than the population of any other district . what is the minimum possible population that the least populated district could have ? | the minimum possible population occurs when all the other districts have a population that is 10 % greater than the least populated district . let p be the population of the least populated district . then 97,500 = p + 5 ( 1.1 ) p 6.5 p = 97,5000 p = 15,000 the answer is a . | a = 100 + 10
b = a / 100
c = 6 - 1
d = b * c
e = d + 1
f = 97500 / e
g = f / 1000
|
a ) 1 hr , b ) 30 min , c ) 15 min , d ) 17 min , e ) 10 min | d | multiply(add(add(divide(10, 10), divide(10, 20)), divide(10, 40)), 10) | pipe a can fill a tank in 10 hr , pipe b in 20 hr and pipe c in 40 hr . if all the pipes are open in how many hours will tank be filled ? | part filled by ( a + b + c ) in 1 hour = 1 / 10 + 1 / 20 + 1 / 40 = 7 / 40 all the 3 pipes together will fill the tank in 17 min approximately answer is d | a = 10 / 10
b = 10 / 20
c = a + b
d = 10 / 40
e = c + d
f = e * 10
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a ) 1 / 4 , b ) 2 / 5 , c ) 1 / 2 , d ) 3 / 5 , e ) 2 / 3 | e | divide(divide(subtract(4, multiply(divide(0.375, 5), 40)), subtract(divide(1.375, 10), divide(0.375, 5))), subtract(40, divide(subtract(4, multiply(divide(0.375, 5), 40)), subtract(divide(1.375, 10), divide(0.375, 5))))) | a 40 kg metal bar made of alloy of tin and silver lost 4 kg of its weight in the water . 10 kg of tin loses 1.375 kg in the water ; 5 kg of silver loses 0.375 kg . what is the ratio of tin to silver in the bar ? | you can simply use this formula to avoid confusion : w 1 / w 2 = ( a 2 - aavg ) / ( avg - a 1 ) here is how you will find the values of a 1 an a 2 . we have an overall loss ( average loss ) . the average loss is 4 kg when 40 kg alloy is immersed . this is a loss of ( 4 / 40 ) * 100 = 10 % . this is aavg the loss of tin is 1.375 kg for every 10 kg . this means it loses ( 1.375 / 10 ) * 100 = 13.75 % of its weight in water . this is a 1 . the loss of silver is . 375 kg for every 5 kg . this means it loses ( . 375 / 5 ) * 100 = 7.5 % of its weight in water . this is a 2 . weight of tin / weight of silver = ( silver ' s loss - avg loss ) / ( avg loss - tin ' s loss ) x / y = ( 7.5 - 10 ) / ( 10 - 13.75 ) = 2 / 3 e | a = 0 / 375
b = a * 40
c = 4 - b
d = 1 / 375
e = 0 / 375
f = d - e
g = c / f
h = 0 / 375
i = h * 40
j = 4 - i
k = 1 / 375
l = 0 / 375
m = k - l
n = j / m
o = 40 - n
p = g / o
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a ) 7 , b ) 8 , c ) 9 , d ) 10 , e ) 11 | d | subtract(add(add(17, 19), 2), 28) | in a sports club with 28 members , 17 play badminton and 19 play tennis and 2 do not play either . how many members play both badminton and tennis ? | "let x play both badminton and tennis so 17 - x play only badminton and 19 - x play only tennis . 2 play none and there are total 30 students . hence , ( 17 - x ) + ( 19 - x ) + x + 2 = 28 38 - 2 x + x = 28 38 - x = 28 x = 10 so 8 members play both badminton and tennis . d" | a = 17 + 19
b = a + 2
c = b - 28
|
a ) 15 , b ) 16 , c ) 11 , d ) 18 , e ) 13 | b | divide(subtract(74000, 42000), add(800, 1200)) | village x has a population of 74000 , which is decreasing at the rate of 1200 per year . village y has a population of 42000 , which is increasing at the rate of 800 per year . in how many years will the population of the two villages be equal ? | "let the population of two villages be equal after p years then , 74000 - 1200 p = 42000 + 800 p 2000 p = 32000 p = 16 answer is b ." | a = 74000 - 42000
b = 800 + 1200
c = a / b
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a ) 32.7 , b ) 32.4 , c ) 33.4 , d ) 32.8 , e ) 32.1 | c | add(divide(circumface(6.5), const_2), multiply(6.5, const_2)) | the radius of a semi circle is 6.5 cm then its perimeter is ? | "36 / 7 r = 6.5 = 33.4 answer : c" | a = circumface / (
b = a + 2
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a ) 12.5 % , b ) 66.6 % , c ) 80 % , d ) 11 % , e ) 1 % | b | multiply(divide(subtract(3, const_1), 3), const_100) | a number x is 3 times another number y . the percentage that y is less than x is | "say y = 1 and x = 3 . then y = 1 is less than x = 3 by ( 3 - 1 ) / 3 * 100 = 2 / 3 * 100 = 66.6 % . answer : b ." | a = 3 - 1
b = a / 3
c = b * 100
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a ) 12 sec , b ) 13 sec , c ) 15 sec , d ) 16 sec , e ) 17 sec | d | divide(100, multiply(add(54, 36), const_0_2778)) | two trains of length 100 m and 200 m are 100 m apart . they start moving towards each other on parallel tracks , at speeds 54 kmph and 36 kmph . in how much time will the trains cross each other ? | "d relative speed = ( 54 + 36 ) * 5 / 18 = 5 * 5 = 25 mps . the time required = d / s = ( 100 + 100 + 200 ) / 25 = 400 / 25 = 16 sec ." | a = 54 + 36
b = a * const_0_2778
c = 100 / b
|
a ) 35 kmh , b ) 40 kmh , c ) 50 kmh , d ) 60 kmh , e ) 65 kmh | b | divide(subtract(sqrt(add(multiply(multiply(const_2, multiply(60, 20)), const_4), power(20, const_2))), 20), const_2) | if a car had traveled 20 kmh faster than it actually did , the trip would have lasted 30 minutes less . if the car went exactly 60 km , at what speed did it travel ? | "time = distance / speed difference in time = 1 / 2 hrs 60 / x - 60 / ( x + 20 ) = 1 / 2 substitute the value of x from the options . - - > x = 40 - - > 60 / 40 - 60 / 60 = 3 / 2 - 1 = 1 / 2 answer : b" | a = 60 * 20
b = 2 * a
c = b * 4
d = 20 ** 2
e = c + d
f = math.sqrt(e)
g = f - 20
h = g / 2
|
a ) 450 , b ) 300 , c ) 500 , d ) 610 , e ) 175 | b | divide(add(59, 40), divide(33, const_100)) | a student has to obtain 33 % of the total marks to pass . he got 59 marks and failed by 40 marks . the maximum marks are ? | "let the maximum marks be x then , 33 % of x = 59 + 40 33 x / 100 = 99 x = 300 answer is b" | a = 59 + 40
b = 33 / 100
c = a / b
|
a ) 144 : 121 , b ) 141 : 127 , c ) 144 : 191 , d ) 181 : 161 , e ) 81 : 182 | a | power(divide(1728, 1331), divide(const_1, const_3)) | the ratio of the volumes of two cubes is 1728 : 1331 . what is the ratio of their total surface areas ? | "explanation : ratio of the sides = ³ √ 1728 : ³ √ 1331 = 12 : 11 ratio of surface areas = 12 ^ 2 : 11 ^ 2 = 144 : 121 answer : option a" | a = 1728 / 1331
b = 1 / 3
c = a ** b
|
a ) s . 9621 , b ) s . 6921 , c ) s . 9724 , d ) s . 6261 , e ) s . 6361 | c | multiply(8000, power(add(const_1, divide(5, const_100)), 4)) | the amount of principal rs . 8000 at compound interest at the ratio of 5 % p . a . for 4 years is | "c . i = p ( 1 + r / 100 ) ^ n = 8000 ( 1 + 5 / 100 ) ^ 4 = rs 9724 answer : c" | a = 5 / 100
b = 1 + a
c = b ** 4
d = 8000 * c
|
a ) 225 , b ) 235 , c ) 215 , d ) 255 , e ) 245 | e | add(multiply(multiply(4, 6), 10), divide(subtract(525, multiply(multiply(multiply(4, 6), 10), 2.1)), 4.2)) | there are 6 working days in a regular week and for each day , the working hours are 10 . a man earns rs . 2.10 per hour for regular work and rs . 4.20 per hour for overtime . if he earns rs . 525 in 4 weeks , how many hours did he work ? | egular working hours in 4 weeks = ( 4 × 6 × 10 ) = 240 hours amount earned by working in these regular working hours = 240 × 2.10 = rs . 504 additional amount he earned = 525 - 504 = rs . 21 hours he worked overtime = 21 / 4.2 = 210 / 42 = 5 hours total hours he worked = 240 + 5 = 245 hours answer is e . | a = 4 * 6
b = a * 10
c = 4 * 6
d = c * 10
e = d * 2
f = 525 - e
g = f / 4
h = b + g
|
a ) 2 , b ) 3 , c ) 4 , d ) 5 , e ) 6 | c | max(multiply(subtract(add(55, 7), const_1), subtract(divide(7, 35), divide(7, 55))), const_4) | due to construction , the speed limit along an 7 - mile section of highway is reduced from 55 miles per hour to 35 miles per hour . approximately how many minutes more will it take to travel along this section of highway at the new speed limit than it would have taken at the old speed limit ? | "7 / 35 - 7 / 55 = 7 / 5 * ( 11 - 7 ) / 77 = 7 / 5 * 4 / 77 * 60 min = 7 * 12 * 4 / 77 = 336 / 77 ~ 4.3 answer - c" | a = 55 + 7
b = a - 1
c = 7 / 35
d = 7 / 55
e = c - d
f = b * e
g = max(f)
|
a ) 20 % , b ) 16 % , c ) 15 % , d ) 12 % , e ) 10 % | b | multiply(divide(subtract(add(25, const_100), add(5, const_100)), add(25, const_100)), const_100) | if the price of gasoline increases by 25 % and a driver intends to spend only 5 % more on gasoline , by how much percent should the driver reduce the quantity of gasoline that he buys ? | "let x be the amount of gasoline the driver buys originally . let y be the new amount of gasoline the driver should buy . let p be the original price per liter . ( 1.25 * p ) y = 1.05 ( p * x ) y = ( 1.05 / 1.25 ) x = 0.84 x which is a reduction of 16 % . the answer is b ." | a = 25 + 100
b = 5 + 100
c = a - b
d = 25 + 100
e = c / d
f = e * 100
|
a ) 4000 , b ) 345 , c ) 5673 , d ) 6530 , e ) 8154 | e | multiply(subtract(subtract(multiply(multiply(const_2, 12), add(15, 25)), multiply(6, 3)), multiply(3, multiply(4, 3))), 9) | the dimensions of a room are 25 feet * 15 feet * 12 feet . what is the cost of white washing the four walls of the room at rs . 9 per square feet if there is one door of dimensions 6 feet * 3 feet and three windows of dimensions 4 feet * 3 feet each ? | "area of the four walls = 2 h ( l + b ) since there are doors and windows , area of the walls = 2 * 12 ( 15 + 25 ) - ( 6 * 3 ) - 3 ( 4 * 3 ) = 906 sq . ft . total cost = 906 * 9 = rs . 8154 answer : option e" | a = 2 * 12
b = 15 + 25
c = a * b
d = 6 * 3
e = c - d
f = 4 * 3
g = 3 * f
h = e - g
i = h * 9
|
a ) 12 , b ) 5 , c ) 14 , d ) 11 , e ) 15 | b | multiply(sqrt(divide(144, add(144, 25))), 13) | d and e are two points respectively on sides ab and ac of triangle abc such that de is parallel to bc . if the ratio of area of triangle ade to that of the trapezium decb is 25 : 144 and bc = 13 cm , then find the length of de . | "abc and ade are similar triangles . so ( side of abc / side of ade ) ^ 2 = 169 / 25 side of abc / side of ade = 13 / 5 so the length of de = 5 answer - b" | a = 144 + 25
b = 144 / a
c = math.sqrt(b)
d = c * 13
|
a ) 800 , b ) 820 , c ) 650 , d ) 870 , e ) 900 | c | multiply(add(24, const_2), subtract(add(24, const_2), const_1)) | there are 24 stations between ernakulam and chennai . how many second class tickets have to be printed , so that a passenger can travel from one station to any other station ? | "the total number of stations = 26 from 26 stations we have to choose any two stations and the direction of travel ( ernakulam to chennai is different from chennai to ernakulam ) in 26 p 2 ways . 26 p 2 = 26 * 25 = 650 answer : c" | a = 24 + 2
b = 24 + 2
c = b - 1
d = a * c
|
a ) 1187.5 , b ) 1234 , c ) 1289 , d ) 1345 , e ) none of these | a | divide(divide(multiply(multiply(const_100, const_100), 9.5), const_100), const_4) | jaclyn buys $ 50 000 worth of debentures in a company . she earns 9.5 % p . a . simple interest , paid to her quarterly ( that is , every 3 months ) . if the agreed period of the debenture was 18 months : calculate the amount of interest jaclyn will earn for each quarter | "explanation : i = ( p x r x t ) / 100 answer : a" | a = 100 * 100
b = a * 9
c = b / 100
d = c / 4
|
a ) 18 , b ) 12 , c ) 20 , d ) 30 , e ) 15 | a | inverse(add(divide(const_1, 30), divide(const_1, 45))) | x does a work in 30 days . y does the same work in 45 days . in how many days they together will do the same work ? | "x ' s 1 day ' s work = 1 / 30 y ' s 1 day ' s work = 1 / 45 ( x + y ) ' s 1 day ' s work = ( 1 / 30 + 1 / 45 ) = 1 / 18 both together will finish the work in 18 days . correct option is a" | a = 1 / 30
b = 1 / 45
c = a + b
d = 1/(c)
|
a ) 50 , b ) 33 , c ) 16 , d ) 35 , e ) 14 | c | divide(3, 2) | how many positive integer solutions does the equation 2 x + 3 y = 100 have ? | "formula : ( constant ) / ( lcm of two nos ) = 100 / ( 2 * 3 ) = 15.5 answer : c" | a = 3 / 2
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a ) 7 , b ) 8 , c ) 11 , d ) 6 , e ) 9 | e | divide(add(multiply(7, 3), 7), 21) | if 3 : 7 : : x : 21 , then find the value of x | explanation : treat 3 : 7 as 3 / 7 and x : 21 as x / 21 , treat : : as = so we get 3 / 7 = x / 21 = > 7 x = 63 = > x = 9 option e | a = 7 * 3
b = a + 7
c = b / 21
|
a ) 30 , b ) 40 , c ) 86 , d ) 89 , e ) 95 | c | divide(36.5, 0.42) | how many pieces of 0.42 meteres can be cut from a rod 36.5 meteres long | "explanation : we need so simple divide 36.5 / 0.42 , = ( 3650 / 42 ) = 86 option c" | a = 36 / 5
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a ) 5.07 % , b ) 6 % , c ) 5.7 % , d ) 6.50 % , e ) 6.25 % | e | multiply(divide(subtract(const_100, 94), 94), const_100) | if the cost price is 94 % of sp then what is the profit % | "sol . sp = rs 100 : then cp = rs 94 : profit = rs 6 . profit = { ( 6 / 94 ) * 100 } % = 6.25 % answer is e ." | a = 100 - 94
b = a / 94
c = b * 100
|
a ) 1 . , b ) 3 . , c ) 5 . , d ) 7 . , e ) 9 . | a | power(add(multiply(1, 2), 2), 2) | if a is a positive integer , and if the units digit of a ^ 2 is 1 and the units digit of ( a + 1 ) ^ 2 is 0 , what is the units digit of ( a + 2 ) ^ 2 ? | "if the units digit of a ^ 2 is 1 , then the units digit of a is either 1 or 9 . if the units digit of ( a + 1 ) ^ 2 is 0 , then the units digit of a + 1 is 0 . to satisfy both conditions , the units digit of a must be 9 . then a + 2 has the units digit of 1 , thus the units digit of ( a + 2 ) ^ 2 will be 1 . the answer is a ." | a = 1 * 2
b = a + 2
c = b ** 2
|
['a ) 2.5 ft', 'b ) 1 ft', 'c ) 0.5 ft', 'd ) 0.25 ft', 'e ) 0.025 ft'] | e | divide(volume_cylinder(5, 10), circle_area(100)) | a full stationary oil tank that is a right circular cylinder has a radius of 100 feet and a height of 25 feet . oil is pumped from the stationary tank to an oil truck that has a tank that is a right circular cylinder until the truck ' s tank is completely filled . if the truck ' s tank has a radius of 5 feet and a height of 10 feet , how far did the oil level drop in the stationary tank ? | in order to fill up the tank , same volume is shifted from one tank to the other . pi r 1 ^ 2 h 1 = pi r 2 ^ 2 h 2 5 * 5 * 10 = 100 * 100 * h 2 h 2 = . 025 ft e is the answer | a = volume_cylinder / (
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a ) 8 % , b ) 10 % , c ) 2 % , d ) 5 % , e ) 6 % | b | add(floor(multiply(divide(10, add(10, 100)), const_100)), const_1) | a person saved $ 10 in buying an item on sale . if he spent $ 100 for the item , approximately how much percent he saved in the transaction ? | actual price = 100 + 10 = $ 110 saving = 10 / 110 * 100 = 100 / 11 = 10 % approximately answer is b | a = 10 + 100
b = 10 / a
c = b * 100
d = math.floor(c)
e = d + 1
|
a ) 2 , b ) 7 , c ) 10 , d ) 16 , e ) 22 | b | subtract(add(42, 65), 100) | two family reunions are happening at the same hotel , the oates reunion and the yellow reunion . all 100 guests at the hotel attend at least one of the reunions . if 42 people attend the oates reunion and 65 people attend the yellow reunion , how many people attend both reunions ? | no of people in oates reunion = 42 no of people in yellow reunion = 65 attending both = x all guests attend at least one . therefore , 100 = 42 + 65 - ( both ) both = 7 answer b | a = 42 + 65
b = a - 100
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a ) 140 , b ) 146 , c ) 152 , d ) 162 , e ) 170 | b | multiply(multiply(multiply(200, subtract(1, divide(1, 10))), subtract(1, divide(1, 10))), subtract(1, divide(1, 10))) | in a certain animal population , for each of the first 3 months of life , the probability that an animal will die during that month is 1 / 10 . for a group of 200 newborn members of the population , approximately how many would be expected to survive the first 3 months of life ? | "number of newborns that can die in first month = 1 / 10 * 200 = 20 survived = 180 number of newborns that can die in second month = 1 / 10 * 180 = 18 survived = 162 number of newborns that can die in third month = 1 / 10 * 162 = 16 survived = 146 answer : b" | a = 1 / 10
b = 1 - a
c = 200 * b
d = 1 / 10
e = 1 - d
f = c * e
g = 1 / 10
h = 1 - g
i = f * h
|
a ) 35 , 49 , b ) 40 , 56 , c ) 45 , 63 , d ) 50 , 70 , e ) 55 , 77 | a | subtract(multiply(multiply(subtract(multiply(3, 7), multiply(2, 7)), 5), divide(7, 5)), multiply(subtract(multiply(3, 7), multiply(2, 7)), 5)) | two numbers are in the ratio of 5 : 7 . if 7 is subtracted from each number , they are in the ratio of 2 : 3 . what are the two numbers ? | "( 5 x - 7 ) : ( 7 x - 7 ) = 2 y : 3 y 5 x - 7 = 2 y 7 x - 7 = 3 y 49 - 35 = 15 y - 14 y 14 = y 5 x - 7 = 28 x = 7 the two numbers are 5 x and 7 x which are 35 and 49 . the answer is a ." | a = 3 * 7
b = 2 * 7
c = a - b
d = c * 5
e = 7 / 5
f = d * e
g = 3 * 7
h = 2 * 7
i = g - h
j = i * 5
k = f - j
|
a ) 875 , b ) 740 , c ) 1425 , d ) 2025 , e ) 2500 | b | divide(multiply(subtract(40, const_3), 40), const_2) | a diagonal of a polygon is an segment between two non - adjacent vertices of the polygon . how many diagonals does a regular 40 - sided polygon have ? | "there ' s a direct formula for this . number of diagonals in a regular polygon = [ n * ( n - 3 ) ] / 2 , n = number of sides of the regular polygon . here , n = 40 . plugging it in , we get 740 diagonals ! answer ( b ) ." | a = 40 - 3
b = a * 40
c = b / 2
|
a ) 4 kmph , b ) 6 kmph , c ) 6.25 kmph , d ) 7.5 kmph , e ) 7.8 kmph | a | divide(subtract(24, divide(24, 2)), add(1, 2)) | in covering a distance of 24 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 , 24 / x - 24 / 2 x = 3 6 x = 24 x = 4 km / hr . answer : option a" | a = 24 / 2
b = 24 - a
c = 1 + 2
d = b / c
|
a ) 7 , b ) 6 , c ) 15 , d ) 4 , e ) 3 | c | divide(subtract(multiply(20, 30), multiply(subtract(20, 5), 35)), 5) | 20 workers can finish a work in 30 days . after how many days should 5 workers leave the job so the work is completed in 35 days ? | work done by 15 workers in 35 days = 15 * 35 = 525 men days . now , work remaining = 600 – 525 = 75 men days and we have 5 workers . so 5 workers will work for 75 / 5 = 15 days . answer : c | a = 20 * 30
b = 20 - 5
c = b * 35
d = a - c
e = d / 5
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a ) 0.006 % , b ) 0.06 % , c ) 6 % , d ) 2 % , e ) 60 % | c | multiply(divide(multiply(0.02, 30), 10), const_100) | a glass was filled with 10 ounces of water , and 0.02 ounce of the water evaporated each day during a 30 - day period . what percent of the original amount of water evaporated during this period ? | "in 30 days 30 * 0.02 = 0.6 ounces of water evaporated , which is 0.6 / 10 â ˆ — 100 = 6 of the original amount of water . answer : c ." | a = 0 * 2
b = a / 10
c = b * 100
|
a ) 77 kmph , b ) 55 kmph , c ) 57 kmph , d ) 58 kmph , e ) 76 kmph | c | multiply(const_3_6, divide(284, 18)) | a train 284 m in length crosses a telegraph post in 18 seconds . the speed of the train is ? | "s = 284 / 18 * 18 / 5 = 57 kmph answer : c" | a = 284 / 18
b = const_3_6 * a
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a ) 72 , b ) 84 , c ) 88 , d ) 96 , e ) 108 | d | multiply(add(divide(subtract(add(8, 2), multiply(3, 2)), subtract(3, const_1)), add(8, 2)), subtract(add(divide(subtract(add(8, 2), multiply(3, 2)), subtract(3, const_1)), add(8, 2)), divide(8, const_2))) | roy is now 8 years older than julia and half of that amount older than kelly . if in 2 years , roy will be 3 times as old as julia , then in 2 years what would be roy ’ s age multiplied by kelly ’ s age ? | r = j + 8 = k + 4 r + 2 = 3 ( j + 2 ) ( j + 8 ) + 2 = 3 j + 6 j = 2 r = 10 k = 6 in 2 years ( r + 2 ) ( k + 2 ) = 12 * 8 = 96 the answer is d . | a = 8 + 2
b = 3 * 2
c = a - b
d = 3 - 1
e = c / d
f = 8 + 2
g = e + f
h = 8 + 2
i = 3 * 2
j = h - i
k = 3 - 1
l = j / k
m = 8 + 2
n = l + m
o = 8 / 2
p = n - o
q = g * p
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a ) 2 liters , b ) 3 liters , c ) 8 liters , d ) 5 liters , e ) 7 liters | c | divide(subtract(multiply(divide(25, const_100), 120), multiply(divide(20, const_100), 120)), subtract(const_1, divide(25, const_100))) | a mixture of 120 liters of wine and water contains 20 % water . how much more water should be added so that water becomes 25 % of the new mixture ? | number of liters of water in 120 liters of the mixture = 20 % of 120 = 20 / 100 * 120 = 24 liters . p liters of water added to the mixture to make water 25 % of the new mixture . total amount of water becomes ( 24 + p ) and total volume of mixture is ( 120 + p ) . ( 24 + p ) = 25 / 100 * ( 120 + p ) 96 + 4 p = 120 + p p = 8 liters . answer : c | a = 25 / 100
b = a * 120
c = 20 / 100
d = c * 120
e = b - d
f = 25 / 100
g = 1 - f
h = e / g
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a ) 18 , b ) 18.8 , c ) 19 , d ) 17.8 , e ) 17.9 | b | divide(subtract(100, divide(multiply(6, 100), const_100)), 5) | 5 friends visited a fruit stall , and everyone decided to contribute equally to the total bill of $ 100 . if one of the friends had a coupon for 6 % off the total bill , and if each friend still contributed equally after the coupon was applied to the bill , how much did each friend pay ? | at the non - discounted price , each friend would pay $ 20 , as $ 100 divided by 5 friends is $ 20 per friend . but if the bill is 6 % off , then each friend would pay 6 % less . 6 % of $ 20 is $ 1.2 , so each friend saves $ 1.2 and pays the remaining 18.8 correct option : option b | a = 6 * 100
b = a / 100
c = 100 - b
d = c / 5
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a ) 1 / 5 , b ) 1 / 3 , c ) 1 / 2 , d ) 2 / 3 , e ) 3 / 4 | e | subtract(1, divide(divide(5, 15), add(divide(5, 15), 1))) | a certain country is divided into 5 provinces . each province consists entirely of progressives and traditionalists . if each province contains the same number of traditionalists and the number of traditionalists in any given province is 1 / 15 the total number of progressives in the entire country , what fraction of the country is traditionalist ? | "let p be the number of progressives in the country as a whole . in each province , the number of traditionalists is p / 15 the total number of traditionalists is 5 p / 15 = p / 3 . the total population is p + p / 3 = 4 p / 3 p / ( 4 p / 3 ) = 3 / 4 the answer is e ." | a = 5 / 15
b = 5 / 15
c = b + 1
d = a / c
e = 1 - d
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a ) 5 % , b ) 7 % , c ) 9 % , d ) 11 % , e ) 25 % | e | multiply(divide(subtract(22050, 17640), 17640), const_100) | an amount at compound interest sums to rs . 17640 / - in 2 years and to rs . 22050 / - in 3 years at the same rate of interest . find the rate percentage ? | "explanation : the difference of two successive amounts must be the simple interest in 1 year on the lower amount of money . s . i = 22050 / - - 17640 / - = rs . 4410 / - rate of interest = ( 4410 / 17640 ) × ( 100 / 1 ) = > 25 % answer : option e" | a = 22050 - 17640
b = a / 17640
c = b * 100
|
a ) 294 , b ) 306 , c ) 318 , d ) 338 , e ) 362 | a | subtract(456, reminder(859722, 456)) | which number can we add to 859722 to get a number exactly divisible by 456 ? | "859722 / 456 = 1885 with a remainder of 162 . we need to add 456 - 162 = 294 the answer is a ." | a = 456 - reminder
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a ) 240 , b ) 600 , c ) 320 , d ) 150 , e ) 250 | b | multiply(multiply(power(const_3, const_3), multiply(power(const_2, const_3), power(add(const_4, const_1), const_2))), divide(divide(divide(divide(divide(24, const_2), const_2), const_3), add(const_4, const_1)), add(const_4, const_1))) | find the l . c . m of 12 , 25 , 24 and 20 . | "explanation : 6 x 2 x 2 x 5 x 5 = 600 answer : option b" | a = 3 ** 3
b = 2 ** 3
c = 4 + 1
d = c ** 2
e = b * d
f = a * e
g = 24 / 2
h = g / 2
i = h / 3
j = 4 + 1
k = i / j
l = 4 + 1
m = k / l
n = f * m
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a ) 16 min , b ) 26 min , c ) 34 min , d ) 20 min , e ) 12 min | e | multiply(divide(3, divide(5, 4)), 5) | walking with 4 / 5 of my usual speed , i miss the bus by 3 minutes . what is my usual time ? | speed ratio = 1 : 4 / 5 = 5 : 4 time ratio = 4 : 51 - - - - - - - - 3 4 - - - - - - - - - ? è 12 answer : e | a = 5 / 4
b = 3 / a
c = b * 5
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a ) 1 / 6 , b ) 1 / 8 , c ) 1 / 9 , d ) 1 / 12 , e ) none of these | c | divide(const_2, choose(add(const_3, const_3), const_3)) | what is the probability of getting a sum 9 from two throws of a dice ? | "in two throws of a dice , n ( s ) = ( 6 x 6 ) = 36 . let e = event of getting a sum = { ( 3 , 6 ) , ( 4 , 5 ) , ( 5 , 4 ) , ( 6 , 3 ) } . p ( e ) = n ( e ) / n ( s ) = 4 / 36 = 1 / 9 . answer : c" | a = 3 + 3
b = math.comb(a, 3)
c = 2 / b
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a ) 150 . , b ) 145 . , c ) 100 . , d ) 30 . , e ) 25 . | c | add(multiply(divide(subtract(const_100, 5), const_100), 100), multiply(divide(5, const_100), 100)) | in the hillside summer camp there are 100 children . 90 % of the children are boys and the rest are girls . the camp administrator decided to make the number of girls only 5 % of the total number of children in the camp . how many more boys must she bring to make that happen ? | "given there are 100 students , 90 % of 50 = 90 boys and remaining 10 girls . now here 90 % are boys and 10 % are girls . now question is asking about how many boys do we need to add , to make the girls percentage to 5 or 5 % . . if we add 1000 to existing 90 then the count will be 190 and the girls number will be 10 as it . now boys are 95 % and girls are 5 % . ( out of 200 students = 190 boys + 10 girls ) . imo option c is correct ." | a = 100 - 5
b = a / 100
c = b * 100
d = 5 / 100
e = d * 100
f = c + e
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a ) 0 , b ) 5 , c ) 375 , d ) 1905 , e ) 965 | d | divide(multiply(185, 195), const_4) | what is the sum of the integers from - 185 to 195 inclusive ? | "sum / n = average . sum = ( average ) ( n ) average = a + b / 2 = - 185 + 195 / 2 = 5 number of items ( n ) = b - a + 1 = 195 - ( - 185 ) + 1 = 195 + 186 = 381 . sum = average * n = 5 * 381 = 1905 . answer is d" | a = 185 * 195
b = a / 4
|
a ) 1 / 60 , b ) 1 / 45 , c ) 2 / 45 , d ) 3 / 22 , e ) 6 / 22 | e | divide(multiply(choose(5, 2), choose(3, 1)), choose(add(add(5, 4), 3), 3)) | a jar contains only red , yellow , and orange marbles . if there are 3 red , 5 yellow , and 4 orange marbles , and 2 marbles are chosen from the jar at random without replacing any of them , what is the probability that 2 yellow , 1 red , and no orange marbles will be chosen ? | "i started by finding the 2 probabilities , without calculation , like this : p ( yyr ) p ( yry ) p ( ryy ) i calculated the first one and ended in 1 / 22 . i looked at the answer choices at this point and saw answer d : 3 / 22 . this helped me realise that for the 6 possible orderings the probabbility is the same . so , it should be ( 1 / 22 ) * ( 6 ) , which indeed is 6 / 22 . e" | a = math.comb(5, 2)
b = math.comb(3, 1)
c = a * b
d = 5 + 4
e = d + 3
f = math.comb(e, 3)
g = c / f
|
a ) 1 , b ) 7 , c ) 3 , d ) 9 , e ) 2 | c | divide(subtract(subtract(multiply(24, add(4, 2)), add(multiply(24, 4), multiply(10, 4))), 2), const_2) | 10 years ago , the average age of a family of 4 members was 24 years . two children having been born ( with age diference of 2 years ) , the present average age of the family is the same . the present age of the youngest child is ? | total age of 4 members , 10 years ago = ( 24 x 4 ) years = 96 years . total age of 4 members now = [ 96 + ( 10 x 4 ) ] years = 136 years . total age of 6 members now = ( 24 x 6 ) years = 144 years . sum of the ages of 2 children = ( 144 - 136 ) years = 8 years . let the age of the younger child be years . then , age of the elder child = years . so , age of younger child = 3 years . answer : c | a = 4 + 2
b = 24 * a
c = 24 * 4
d = 10 * 4
e = c + d
f = b - e
g = f - 2
h = g / 2
|
a ) 1 / 4 , b ) 1 / 3 , c ) 1 / 2 , d ) 2 / 3 , e ) 7 / 9 | e | inverse(add(divide(subtract(35, 25), subtract(70, 35)), const_1)) | a certain quantity of 70 % solution is replaced with 25 % solution such that the new concentration is 35 % . what is the fraction of the solution that was replaced ? | let ' s say that the total original mixture a is 100 ml the original mixture a thus has 70 ml of alcohol out of 100 ml of solution you want to replace some of that original mixture a with another mixture b that contains 25 ml of alcohol per 100 ml . thus , the difference between 70 ml and 25 ml is 45 ml per 100 ml of mixture . this means that every time you replace 100 ml of the original mixture a by 100 ml of mixture b , the original alcohol concentration will decrease by 45 % . the question says that the new mixture , let ' s call it c , must be 35 % alcohol , a decrease of only 35 % . therefore , 35 out of 45 is 7 / 9 and e is the answer . | a = 35 - 25
b = 70 - 35
c = a / b
d = c + 1
e = 1/(d)
|
a ) 7 , b ) 6 , c ) 78 , d ) 9 , e ) 1 | a | add(2, 5) | f 1 ( x ) f 1 ( x ) = 2 x - 1 and fn ( x ) = f 1 ( fn − 1 ( x ) ) fn ( x ) = f 1 ( fn − 1 ( x ) ) for n ≥ ≥ 2 . find f 5 ( 2 ) | answer : a | a = 2 + 5
|
a ) 2 , b ) 4 , c ) 6 , d ) 8 , e ) 10 | e | subtract(subtract(22, subtract(10, 4)), 6) | of 60 children , 30 are happy , 10 are sad , and 20 are neither happy nor sad . there are 22 boys and 38 girls . if there are 6 happy boys and 4 sad girls , how many boys are neither happy nor sad ? | "venn diagrams are useful for multiple values of a single variable e . g . state of mind - happy / sad / neither . when you have two or more variables such as here where you have gender - boy / girl too , it becomes unwieldy . in this case , either use the table or logic . table method is shown above ; here is how you will use logic : there are 6 happy boys . there are 4 sad girls but total 10 sad children . so rest 6 sad children must be sad boys . we have 6 happy boys and 6 sad boys . total we have 22 boys . so 22 - 6 - 6 = 10 boys must be neither happy nor sad . answer ( e )" | a = 10 - 4
b = 22 - a
c = b - 6
|
a ) 38 / 6 , b ) 38 / 5 , c ) 77 / 81 , d ) 38 / 7 , e ) 38 / 10 | c | multiply(divide(divide(multiply(divide(add(subtract(const_100, 10), multiply(subtract(const_100, 10), divide(subtract(const_100, 10), const_100))), const_2), subtract(const_100, 10)), const_100), multiply(subtract(const_100, 10), divide(subtract(const_100, 10), const_100))), const_10) | real - estate salesman z is selling a house at a 10 percent discount from its retail price . real - estate salesman x vows to match this price , and then offers an additional 10 percent discount . real - estate salesman y decides to average the prices of salesmen z and x , then offer an additional 10 percent discount . salesman y ' s final price is what fraction of salesman x ' s final price ? | "let the retail price be = x selling price of z = 0.90 x selling price of x = 0.90 * 0.90 x = 0.81 x selling price of y = ( ( 0.90 x + 0.81 x ) / 2 ) * 0.90 = 0.86 x * 0.90 = 0.77 x 0.77 x = k * 0.81 x k = 0.77 / 0.81 = 77 / 81 answer : c" | a = 100 - 10
b = 100 - 10
c = 100 - 10
d = c / 100
e = b * d
f = a + e
g = f / 2
h = 100 - 10
i = g * h
j = i / 100
k = 100 - 10
l = 100 - 10
m = l / 100
n = k * m
o = j / n
p = o * 10
|
a ) 7 km , b ) 3 km , c ) 4 km , d ) 5 km , e ) 6 km | a | divide(multiply(42, divide(multiply(10, const_1000), const_60)), const_1000) | find the distance covered by a man walking for 42 min at a speed of 10 km / hr ? | "distance = 10 * 42 / 60 = 7 km answer is a" | a = 10 * 1000
b = a / const_60
c = 42 * b
d = c / 1000
|
a ) 20 % , b ) 27 % , c ) 30 % , d ) 32 % , e ) 79 % | b | divide(multiply(subtract(add(add(const_100, 15), multiply(add(const_100, 15), divide(20, const_100))), const_100), const_100), add(add(const_100, 15), multiply(add(const_100, 15), divide(20, const_100)))) | the output of a factory was increased by 15 % to keep up with rising demand . to handle the holiday rush , this new output was increased by 20 % . by approximately what percent would the output now have to be decreased in order to restore the original output ? | the original output increases by 15 % and then 20 % . total % change = a + b + ab / 100 total % change = 15 + 20 + 15 * 20 / 100 = 38 % now , you want to change it to 0 , so , 0 = 38 + x + 38 x / 100 x = - 38 ( 100 ) / 138 = 27 % approximately answer is b | a = 100 + 15
b = 100 + 15
c = 20 / 100
d = b * c
e = a + d
f = e - 100
g = f * 100
h = 100 + 15
i = 100 + 15
j = 20 / 100
k = i * j
l = h + k
m = g / l
|
a ) 10 , b ) 20 , c ) 30 % , d ) 40 , e ) 50 | c | subtract(60, 30) | in a sample of associates at a law firm , 30 percent are second - year associates , and 60 percent are not first - year associates . what percentage of the associates at the law firm have been there for more than two years ? | let the sample be 100 . given 60 % are not first year i . e . there could be second year ( x ) and more than 2 years ( y ) possibility . 60 % = ( x ) + y given 30 % = second year ( x ) club both , we get y = 30 % . answer : option c is correct answer . | a = 60 - 30
|
a ) 40 , b ) 45 , c ) 50 , d ) 55 , e ) 60 | c | subtract(80, subtract(subtract(160, 90), 40)) | in a company of 160 employees , 90 are females . a total of 80 employees have advanced degrees and the rest have a college degree only . if 40 employees are males with college degree only , how many employees are females with advanced degrees ? | "the number of males is 160 - 90 = 70 . the number of males with advanced degrees is 70 - 40 = 30 . the number of females with advanced degrees is 80 - 30 = 50 . the answer is c ." | a = 160 - 90
b = a - 40
c = 80 - b
|
a ) 10000 , b ) 20000 , c ) 15000 , d ) 61000 , e ) 26000 | b | multiply(multiply(multiply(multiply(add(1, 3), 1), const_100), const_100), divide(const_3.0, add(add(2, const_2.0), 1))) | in business , a and c invested amounts in the ratio 3 : 2 , whereas the ratio between amounts invested by a and b was 3 : 1 , if rs 60000 was their profit , how much amount did c receive . | "explanation : a : b = 3 : 1 = 3 : 1 = > a : c = 3 : 2 = 3 : 2 = > a : b : c = 3 : 1 : 2 c share = ( 2 / 6 ) * 60000 = 20000 option b" | a = 1 + 3
b = a * 1
c = b * 100
d = c * 100
e = 2 + 2
f = e + 1
g = 3 / 0
h = d * g
|
a ) 40 , b ) 45 , c ) 55 , d ) 45 , e ) 50 | d | subtract(add(add(20, 40), 60), add(add(multiply(5, const_3), 10), 50)) | the average ( arithmetic mean ) of 20 , 40 , and 60 is 5 more than the average of 10 , 50 , and what number ? | "a 1 = 120 / 3 = 40 a 2 = a 1 - 5 = 35 sum of second list = 35 * 3 = 105 therefore the number = 105 - 60 = 45 d" | a = 20 + 40
b = a + 60
c = 5 * 3
d = c + 10
e = d + 50
f = b - e
|
a ) 24 , b ) 26 , c ) 30 , d ) 49 , e ) 40 | d | floor(49) | an old man distributed all the gold coins he had to his two sons into two different numbers such that the difference between the squares of the two numbers is 49 times the difference between the two numbers . how many coins did the old man have ? | "let the number of coins one son got be x and the number of coins another got be y . total = x + y . x ^ 2 - y ^ 2 = 49 ( x - y ) - - > x + y = 49 . answer : d ." | a = math.floor(49)
|
a ) 512 , b ) 300 , c ) 294 , d ) 264 , e ) 385 | b | multiply(multiply(multiply(1, const_10), const_10), const_3) | how many times will the digit 4 be written when listing the integers from 1 to 1000 ? | "many approaches are possible . for example : consider numbers from 0 to 999 written as follows : 1 . 000 2 . 001 3 . 002 4 . 003 . . . . . . . . . 1000 . 999 we have 1000 numbers . we used 3 digits per number , hence used total of 3 * 1000 = 3000 digits . now , why should any digit have preferences over another ? we used each of 10 digits equal # of times , thus we used each digit ( including 4 ) 3000 / 10 = 300 times . answer : b ." | a = 1 * 10
b = a * 10
c = b * 3
|
a ) 5 minutes , b ) 11 minutes , c ) 15 minutes , d ) 18 minutes , e ) 20 minutes | b | divide(20, add(divide(20, 24), add(divide(20, 50), divide(20, 30)))) | jonathan can type a 20 page document in 50 minutes , susan can type it in 30 minutes , and jack can type it in 24 minutes . working together , how much time will it take them to type the same document ? | you may set up common equation like this : job / a + job / b + job / c = job / x memorize this universal formula , you will need it definitely for gmat . and find x from this equation in this specific case , the equation will look like this : 20 / 50 + 20 / 30 + 20 / 24 = 20 / x if you solve this equation , you get the same answer b ( 11 ) | a = 20 / 24
b = 20 / 50
c = 20 / 30
d = b + c
e = a + d
f = 20 / e
|
a ) 8 , b ) 10 , c ) 6 , d ) 2 , e ) 4 | e | divide(6000, multiply(150, 10)) | in a certain village , 150 litres of water are required per household per month . at this rate , if there are 10 households in the village , how long ( in months ) will 6000 litres of water last ? | "i find it much easier to understand with real numbers , so choose ( almost ) any numbers to replace m , n and p : in a certain village , m 150 litres of water are required per household per month . at this rate , if there aren 10 households in the village , how long ( in months ) willp 2000 litres of water last ? water required is 150 * 10 = 1500 ( m * n ) water available is 6000 ( p ) it will last 4 months ( p / m * n ) ans : e" | a = 150 * 10
b = 6000 / a
|
a ) 10 and 3 , b ) 7 and 10 , c ) 11 and 8 , d ) 3 and 10 , e ) 10 and 10 | c | subtract(add(divide(20, 2), 1), 3) | one side of a rectangle is 3 cm shorter than the other side . if we increase the length of each side by 1 cm , then the area of the rectangle will increase by 20 cm 2 . find the lengths of all sides . | "let x be the length of the longer side x > 3 , then the other side ' s length is x − 3 cm . then the area is s 1 = x ( x - 3 ) cm 2 . after we increase the lengths of the sides they will become ( x + 1 ) and ( x − 3 + 1 ) = ( x − 2 ) cm long . hence the area of the new rectangle will be a 2 = ( x + 1 ) ⋅ ( x − 2 ) cm 2 , which is 20 cm 2 more than the first area . therefore a 1 + 20 = a 2 x ( x − 3 ) + 20 = ( x + 1 ) ( x − 2 ) x 2 − 3 x + 20 = x 2 + x − 2 x − 2 2 x = 22 x = 11 . so , the sides of the rectangle are 11 cm and ( 11 − 3 ) = 8 cm long . so answer is c ." | a = 20 / 2
b = a + 1
c = b - 3
|
a ) 1217 , b ) 1219 , c ) 1210 , d ) 565.7 , e ) 1312 | d | multiply(subtract(power(18, const_2), power(const_10, const_2)), divide(add(multiply(12, const_2), const_2), add(const_4, const_3))) | a rope of which a calf is tied is increased from 12 m to 18 m , how much additional grassy ground shall it graze ? | "π ( 182 – 122 ) = 565.7 answer : d" | a = 18 ** 2
b = 10 ** 2
c = a - b
d = 12 * 2
e = d + 2
f = 4 + 3
g = e / f
h = c * g
|
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