options stringlengths 37 300 | correct stringclasses 5 values | annotated_formula stringlengths 7 727 | problem stringlengths 5 967 | rationale stringlengths 1 2.74k | program stringlengths 10 646 |
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a ) β 8 , b ) β 2 , c ) 0 , d ) 2 , e ) 8 | a | subtract(add(power(negate(8), 2), 16), multiply(10, negate(8))) | for what value of x between β 8 and 8 , inclusive , is the value of x ^ 2 β 10 x + 16 the greatest ? | "we can see from the statement that two terms containing x , x ^ 2 will always be positive and - 10 x will be positive if x is - ive . . so the equation will have greatest value if x is - ive , and lower the value of x , greater is the equation . so - 8 will give the greatest value . . ans a" | a = negate ** (
b = a + 2
c = b - 16
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a ) 64 , b ) 32 , c ) 8 , d ) 4 , e ) 2 | c | sqrt(power(power(16, divide(const_1, const_2)), const_3)) | what positive number , when squared , is equal to the cube of the positive square root of 16 ? | "let the positive number be x x ^ 2 = ( ( 16 ) ^ ( 1 / 2 ) ) ^ 3 = > x ^ 2 = 4 ^ 3 = 64 = > x = 8 answer c" | a = 1 / 2
b = 16 ** a
c = b ** 3
d = math.sqrt(c)
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a ) a ) 270 , b ) b ) 280 , c ) c ) 290 , d ) d ) 300 , e ) e ) 310 | d | divide(add(1515, 15), subtract(6, const_1)) | the difference of two numbers is 1515 . on dividing the larger number by the smaller , we get 6 as quotient and the 15 as remainder . what is the smaller number ? | "let the smaller number be x . then larger number = ( x + 1515 ) . x + 1515 = 6 x + 15 5 x = 1500 x = 300 smaller number = 300 . d )" | a = 1515 + 15
b = 6 - 1
c = a / b
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a ) 360 , b ) 383 , c ) 400 , d ) 406 , e ) 412 | b | multiply(divide(subtract(divide(360, subtract(const_1, divide(const_1, 10))), 360), 12), 115) | mr . john used to purchase certain number of mangoes for $ 360 since the price of mangoes is reduced by 10 % he got 12 more mangoes today . find the original price of 115 mangoes . | "mr . john used to purchase certain number of mangoes for $ 360 since the price of mangoes is reduced by 10 % he got 12 more mangoes today . find the original price of 120 mangoes . method 1 : let price per mango = x . let number of mangoes be n . then , nx = 360 . now price = 0.9 x ; number of mangoes = n + 12 . total amount = 0.9 x * ( n + 12 ) = 360 . nx = 0.9 nx + 10.8 x = > 0.1 nx = 10.8 x = > n = 108 = > x = 360 / 108 = 3.33 original price of 115 mangoes = 115 * 3.33 = 383 . answer b" | a = 1 / 10
b = 1 - a
c = 360 / b
d = c - 360
e = d / 12
f = e * 115
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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 r that this product will be a multiple of 45 ? | "there are 3 single digit multiple of 3 , that is , 3 , 6,9 . there are 8 prime nos less than 20 - 2,3 , 5,7 , 11,13 , 17,19 total outcome - 8 * 3 = 24 favourable outcome = 1 ( 9 * 5 ) hence required probability r = 1 / 24 . answer c ." | a = 20 + 4
b = 1 / a
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a ) 3 days , b ) 5 days , c ) 7 days , d ) 1 days , e ) 2 dasy | a | divide(const_1, divide(add(add(inverse(4), inverse(5)), inverse(7)), const_2)) | a and b can do a work in 4 days , b and c in 5 days , c and a in 7 days . if a , b and c work together , they will complete the work in ? | "a + b 1 day work = 1 / 4 b + c 1 day work = 1 / 5 c + a 1 day work = 1 / 7 adding we get 2 ( a + b + c ) = 1 / 4 + 1 / 5 + 1 / 7 = 83 / 140 a + b + c 1 day work = 83 / 280 a , b , c can finish the work in 280 / 83 days = 3 days approximately answer is a" | a = 1/(4)
b = 1/(5)
c = a + b
d = 1/(7)
e = c + d
f = e / 2
g = 1 / f
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a ) 62 , b ) 70 , c ) 72 , d ) 71 , e ) 76 | c | add(choose(9, 2), choose(9, 2)) | there are 9 pairs of socks and 2 socks are worn from that such that the pair of socks worn are not of the same pair . what is the number of pair that can be formed . | "first of all you should remember that there is a difference in left and right sock . now no . of way to select any of the sock = 9 and for second = 8 so total methods = 9 * 8 = 72 answer : c" | a = math.comb(9, 2)
b = math.comb(9, 2)
c = a + b
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a ) 36 , b ) 232 , c ) 216 , d ) 455 , e ) 363 | c | divide(volume_cube(18), volume_cube(divide(3, const_100))) | how many cubes of 3 cm edge can be cut out of a cube of 18 cm edge | "number of cubes = ( 18 x 18 x 18 ) / ( 3 x 3 x 3 ) = 216 answer : c" | a = volume_cube / (
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a ) 180000 , b ) 138800 , c ) 658560 , d ) 116740 , e ) none of them | a | multiply(600, power(223, 600)) | 600 x 223 + 600 x 77 = ? | "= 600 x ( 223 + 77 ) ( by distributive law ) = 600 x 300 = 180000 answer is a" | a = 223 ** 600
b = 600 * a
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a ) 2812 , b ) 8231 , c ) 2734 , d ) 2744 , e ) 4254 | d | divide(10976, 4) | a volume of 10976 l water is in a container of sphere . how many hemisphere of volume 4 l each will be required to transfer all the water into the small hemispheres ? | "a volume of 4 l can be kept in 1 hemisphere therefore , a volume of 10976 l can be kept in ( 10976 / 4 ) hemispheres ans . 2744 answer : d" | a = 10976 / 4
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a ) 50 % , b ) 25 % , c ) 250 % , d ) 75 % , e ) 80 % | a | multiply(divide(divide(multiply(subtract(150, divide(150, const_3)), subtract(subtract(divide(150, const_3), 20), 15)), add(15, subtract(subtract(divide(150, const_3), 20), 15))), subtract(150, divide(150, const_3))), const_100) | at the end of the day , february 14 th , a florist had 150 roses left in his shop , all of which were red , white or pink in color and either long or short - stemmed . a third of the roses were short - stemmed , 20 of which were white and 15 of which were pink . the percentage of pink roses that were short - stemmed equaled the percentage of red roses that were short - stemmed . if none of the long - stemmed roses were white , what percentage of the long - stemmed roses were red ? | "r + w + p = 150 s + l = 150 1 / 3 * 150 = 50 short - stemmed white = 20 short - stemmed pink = 15 = > short - stemmed red = 15 15 / p = 15 / r = > r = p so total long stemmed = 100 and long stemmed red + long stemmed pink = 100 so long stemmed red / long stemmed = ? total white = 20 ( as no long stemmed white ) = > r + r + 20 = 150 = > r = 130 and r = 65 long stemmed r = 65 - 15 = 50 so long stemmed red / r = 50 / 100 = 50 % answer - a" | a = 150 / 3
b = 150 - a
c = 150 / 3
d = c - 20
e = d - 15
f = b * e
g = 150 / 3
h = g - 20
i = h - 15
j = 15 + i
k = f / j
l = 150 / 3
m = 150 - l
n = k / m
o = n * 100
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a ) $ 130.00 , b ) $ 145.60 , c ) $ 163.80 , d ) $ 182.00 , e ) $ 210.00 | b | multiply(260, divide(add(30, 20), const_100)) | a discount electronics store normally sells all merchandise at a discount of 10 percent to 30 percent off the suggested retail price . if , during a special sale , an additional 20 percent were to be deducted from the discount price , what would be the lowest possible price of an item costing $ 260 before any discount ? | "original price : 260 $ max first discount = - 30 % thus : 260 β ( 1 β 30 / 100 ) = 182 second discount on the discounted price = - 20 % thus : 182 β ( 1 β 20 / 100 ) = 145,6 answer b ." | a = 30 + 20
b = a / 100
c = 260 * b
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a ) 420 , b ) 840 , c ) 1260 , d ) 2520 , e ) 27,720 | e | lcm(1, 11) | what is the lowest positive integer that is divisible by each of the integers 1 through 11 inclusive ? | "we have to find the lcm of 1 , 2 , 3 , 2 ^ 2 , 5 , 2 * 3 , 7 , 2 ^ 3 , 3 ^ 2 , 2 * 5 , and 11 . the lcm is 1 * 2 ^ 3 * 3 ^ 2 * 5 * 7 * 11 = 27,720 the answer is e ." | a = math.lcm(1, 11)
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a ) 20 , b ) 16 , c ) 22 , d ) 18 , e ) 24 | d | subtract(multiply(25, 3), multiply(19, 3)) | carl bought 16 gumballs , lewis bought 12 gumballs , and carey bought x gumballs . the average ( arithmetic mean ) number of gumballs the 3 bought is between 19 and 25 , inclusive . what is the difference between the greatest number and the smallest number of gumballs carey could have bought ? | smallest gumballs = ( 19 - 16 ) + ( 19 - 12 ) + 19 = 29 largest gumballs = ( 25 - 16 ) + ( 25 - 12 ) + 25 = 47 difference = 47 - 29 = 18 d | a = 25 * 3
b = 19 * 3
c = a - b
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a ) 10 , b ) 12 , c ) 15 , d ) 18 , e ) 35 | e | multiply(28, inverse(subtract(const_1, divide(8, 40)))) | x can do a piece of work in 40 days . he works at it for 8 days and then y finished it in 28 days . how long will y take to complete the work ? | "work done by x in 8 days = 8 * 1 / 40 = 1 / 5 remaining work = 1 - 1 / 5 = 4 / 5 4 / 5 work is done by y in 28 days whole work will be done by y in 28 * 5 / 4 = 35 days answer is e" | a = 8 / 40
b = 1 - a
c = 1/(b)
d = 28 * c
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a ) 939.6 , b ) 958.9 , c ) 927.78 , d ) 926.82 , e ) 902.1 | b | divide(1120, add(divide(multiply(divide(add(multiply(2, 5), 2), 5), 7), const_100), const_1)) | find the principle on a certain sum of money at 7 % per annum for 2 2 / 5 years if the amount being rs . 1120 ? | "1120 = p [ 1 + ( 7 * 12 / 5 ) / 100 ] p = 958.90 answer : b" | a = 2 * 5
b = a + 2
c = b / 5
d = c * 7
e = d / 100
f = e + 1
g = 1120 / f
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a ) s . 8500 , b ) s . 8700 , c ) s . 7500 , d ) s . 7000 , e ) s . 6500 | b | multiply(multiply(multiply(add(multiply(multiply(multiply(2, 3), const_100), const_100), multiply(multiply(multiply(3, 3), const_100), multiply(add(3, 2), 2))), divide(add(multiply(16, 3), 2), 3)), divide(multiply(3, 3), multiply(2, multiply(2, 3)))), divide(const_1, const_100)) | find the simple interest on rs . 69,600 at 16 2 / 3 % per annum for 9 months . | "p = rs . 69600 , r = 50 / 3 % p . a and t = 9 / 12 years = 3 / 4 years . s . i . = ( p * r * t ) / 100 = rs . ( 69,600 * ( 50 / 3 ) * ( 3 / 4 ) * ( 1 / 100 ) ) = rs . 8700 answer is b ." | a = 2 * 3
b = a * 100
c = b * 100
d = 3 * 3
e = d * 100
f = 3 + 2
g = f * 2
h = e * g
i = c + h
j = 16 * 3
k = j + 2
l = k / 3
m = i * l
n = 3 * 3
o = 2 * 3
p = 2 * o
q = n / p
r = m * q
s = 1 / 100
t = r * s
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a ) 12 , b ) 2 , c ) 4 , d ) 54 , e ) 56 | a | divide(multiply(60, 2), 22) | the lcm and hcf of two numbers are 60 and 2 respectively . find the larger of the two numbers if their sum is 22 . | "there are 2 approaches in solving this . methode 1 . hcf * lcm = the actual number . 60 * 2 = 120 so the answer which we are looking for has to be a factor of 120 . so among the options shortlist the answers by eliminating those numbers which is not divisible by 120 . and then take the highest number as the answer as the question asks abt the highest number . answer is a" | a = 60 * 2
b = a / 22
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a ) 9 , b ) 10 , c ) 12 , d ) 20 , e ) 22 | b | multiply(const_60, divide(subtract(54, 45), 54)) | excluding stoppages , the speed of a bus is 54 kmph and including stoppages , it is 45 kmph . for how many minutes does the bus stop per hour ? | "due to stoppages , it covers 9 km less . time taken to cover 9 km = ( 9 / 54 x 60 ) min = 10 min . answer : b" | a = 54 - 45
b = a / 54
c = const_60 * b
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a ) 25 , b ) 35 , c ) 45 , d ) 55 , e ) 65 | c | divide(add(multiply(15, 2), multiply(15, 1)), add(2, 1)) | the number of stamps that p and q had were in the ratio of 7 : 2 respectively . after p gave q 15 stamps , the ratio of the number of p ' s stamps to the number of q ' s stamps was 2 : 1 . as a result of the gift , p had how many more stamps than q ? | "p started with 7 k stamps and q started with 2 k stamps . ( 7 k - 15 ) / ( 2 k + 15 ) = 2 / 1 3 k = 45 k = 15 p has 7 ( 15 ) - 15 = 90 stamps and q has 2 ( 15 ) + 15 = 45 stamps . the answer is c ." | a = 15 * 2
b = 15 * 1
c = a + b
d = 2 + 1
e = c / d
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a ) 1 , b ) 7 , c ) 8 , d ) 9 , e ) 36 | e | divide(const_1, subtract(subtract(const_0_25, divide(const_1, 12)), divide(const_1, 18))) | if a , b and c together can finish a piece of work in 6 days . a alone in 12 days and b in 18 days , then c alone can do it in ? | "c = 1 / 6 - 1 / 12 β 1 / 18 = 1 / 36 = > 36 days ' answer : e" | a = 1 / 12
b = const_0_25 - a
c = 1 / 18
d = b - c
e = 1 / d
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a ) 35 , b ) 49 , c ) 100 , d ) 105 , e ) 122 | e | add(divide(subtract(multiply(divide(multiply(35, 20), const_100), 33), multiply(divide(multiply(35, 20), const_100), 20)), subtract(multiply(20, divide(40, const_100)), divide(multiply(35, 20), const_100))), 35) | how many liters of a 40 % iodine solution need to be mixed with 35 liters of a 20 % iodine solution to create a 33 % iodine solution ? | solution 1 : assume the iodine solution to be mixed = x lts . iodine = 0.4 x lts , water = 0.6 x lts . solution 2 : 35 liters of a 20 % iodine solution iodine = 7 lts , water = 28 lts . total iodine = 0.4 x + 7 total water = 0.6 x + 28 the resultant is a 35 % idoine solution . hence ( 0.4 x + 7 ) / ( x + 35 ) = 33 / 100 40 x + 700 = 33 x + 1155 7 x = 855 x = 122 lts correct option : e | a = 35 * 20
b = a / 100
c = b * 33
d = 35 * 20
e = d / 100
f = e * 20
g = c - f
h = 40 / 100
i = 20 * h
j = 35 * 20
k = j / 100
l = i - k
m = g / l
n = m + 35
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a ) 36 , b ) 29 , c ) 72 , d ) 29 , e ) 22 | b | multiply(23, 6) | the average of 10 numbers is 23 . if each number is increased by 6 , what will the new average be ? | "sum of the 10 numbers = 230 if each number is increased by 6 , the total increase = 6 * 10 = 60 the new sum = 230 + 60 = 290 the new average = 290 / 10 = 29 . answer : b" | a = 23 * 6
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a ) 45 minutes , b ) 50 minutes , c ) 40 minutes , d ) 55 minutes , e ) 52 minutes | e | subtract(const_60, divide(16, const_2)) | each day a man meets his wife at the train station after work , and then she drives him home . she always arrives exactly on time to pick him up . one day he catches an earlier train and arrives at the station an hour early . he immediately begins walking home along the same route the wife drives . eventually his wife sees him on her way to the station and drives him the rest of the way home . when they arrive home the man notices that they arrived 16 minutes earlier than usual . how much time did the man spend walking ? | "as they arrived 16 minutes earlier than usual , they saved 16 minutes on round trip from home to station ( home - station - home ) - - > 8 minutes in each direction ( home - station ) - - > wife meets husband 8 minutes earlier the usual meeting time - - > husband arrived an hour earlier the usual meeting time , so he must have spent waking the rest of the time before their meeting , which is hour - 8 minutes = 52 minutes . answer : e" | a = 16 / 2
b = const_60 - a
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a ) 174.2 , b ) 212 , c ) 288.1 , d ) 296 , e ) 308 | e | multiply(divide(49, 2.54), divide(24, 1.5)) | on a map , 1.5 inches represent 24 miles . how many miles approximately is the distance if you measured 49 centimeters assuming that 1 - inch is 2.54 centimeters ? | "1.5 inch = 2.54 * 1.5 cm . so , 2.54 * 1.5 represents 24 miles . so for 49 cm . : 49 / ( 2.54 * 1.5 ) = x / 24 - - - > x = 24 * 49 / ( 3.81 ) = 308 answer will be e ." | a = 49 / 2
b = 24 / 1
c = a * b
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a ) 10 , b ) 8 , c ) 7 , d ) 9 , e ) 11 | d | divide(multiply(multiply(9, const_3), const_3), multiply(const_3, const_3)) | if three painters can complete three rooms in three hours , how many painters would it take to do 27 rooms in 9 hours ? | explanation : three painters can complete three rooms in three hours . so 27 rooms can be painted in 9 hrs by 9 painters answer : d ) 9 painters | a = 9 * 3
b = a * 3
c = 3 * 3
d = b / c
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a ) 22 , b ) 27 , c ) 18 , d ) 12 , e ) 81 | a | subtract(multiply(40, divide(80, const_100)), multiply(divide(2, 5), 25)) | how much is 80 % of 40 is greater than 2 / 5 of 25 ? | "( 80 / 100 ) * 40 Γ’ β¬ β ( 2 / 5 ) * 25 32 - 10 = 22 answer : a" | a = 80 / 100
b = 40 * a
c = 2 / 5
d = c * 25
e = b - d
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a ) 4550 , b ) 5000 , c ) 0.5 , d ) 4000 , e ) none | a | subtract(multiply(divide(5000, const_100), 4500), multiply(divide(const_1, const_3), multiply(divide(5000, const_100), 4500))) | 5000 - 4500 Γ· 10.00 = ? | "answer given expression = 5000 - 4500 Γ· 10.00 = 5000 - 450 = 4550 correct option : a" | a = 5000 / 100
b = a * 4500
c = 1 / 3
d = 5000 / 100
e = d * 4500
f = c * e
g = b - f
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a ) 8 , b ) 14 , c ) 15 , d ) 18 , e ) 30 | c | add(multiply(7, const_2), const_1) | the average age of applicants for a new job is 30 , with a standard deviation of 7 . the hiring manager is only willing to accept applications whose age is within one standard deviation of the average age . what is the maximum number of different ages of the applicants ? | "within one standard deviation of the average age means 31 + / - 7 23 - - 30 - - 37 number of dif . ages - 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 total = 15 c" | a = 7 * 2
b = a + 1
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a ) 1.8 , b ) 2.0 , c ) 2.2 , d ) 2.4 , e ) 2.6 | c | inverse(add(divide(const_1, 5), divide(const_1, 4))) | a worker can load one truck in 5 hours . a second worker can load the same truck in 4 hours . if both workers load one truck simultaneously while maintaining their constant rates , approximately how long , in hours , will it take them to fill one truck ? | "the workers fill the truck at a rate of 1 / 5 + 1 / 4 = 9 / 20 of the truck per hour . then the time to fill one truck is 20 / 9 which is about 2.2 hours . the answer is c ." | a = 1 / 5
b = 1 / 4
c = a + b
d = 1/(c)
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a ) 8765 , b ) 8907 , c ) 9944 , d ) 9954 , e ) 9990 | c | multiply(add(add(add(add(multiply(const_100, const_100), multiply(const_100, const_10)), multiply(const_100, const_3)), multiply(4, const_10)), const_3), 88) | find the largest 4 digit number which isexactly divisible by 88 ? | "largest 4 digit number is 9999 after doing 9999 Γ· 88 we get remainder 55 hence largest 4 digit number exactly divisible by 88 = 9999 - 55 = 9944 c" | a = 100 * 100
b = 100 * 10
c = a + b
d = 100 * 3
e = c + d
f = 4 * 10
g = e + f
h = g + 3
i = h * 88
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a ) 6 , b ) 8 , c ) 9 , d ) 11 , e ) 12 | d | divide(divide(divide(600, 4), const_2), const_3) | how many of the positive divisors of 600 are also multiples of 4 not including 600 ? | "600 = 2 ^ 6 * 3 * 5 = ( 4 ) * 2 * 3 * 5 ^ 2 besides ( 4 ) , the exponents of 2 , 3 , and 5 are 1 , 1 , and 2 . there are ( 1 + 1 ) ( 1 + 1 ) ( 2 + 1 ) = 12 ways to make multiples of 4 . we must subtract 1 because one of these multiples is 600 . the answer is d ." | a = 600 / 4
b = a / 2
c = b / 3
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a ) $ 3.85 , b ) $ 4.85 , c ) $ 5.85 , d ) $ 2.85 , e ) $ 1.85 | b | subtract(8.50, add(1.25, add(1.20, 1.20))) | little john had $ 8.50 . he spent $ 1.25 on sweets and gave to his two friends $ 1.20 each . how much money was left ? | "john spent and gave to his two friends a total of 1.25 + 1.20 + 1.20 = $ 3.65 money left 8.50 - 3.65 = $ 4.85 correct answer is b ) $ 4.85" | a = 1 + 20
b = 1 + 25
c = 8 - 50
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a ) 40 , b ) 85 , c ) 60 , d ) 30 , e ) 20 | b | multiply(divide(80, subtract(17, 1)), 17) | the ratio of buses to cars on river road is 1 to 17 . if there are 80 fewer buses than cars on river road , how many cars are on river road ? | b / c = 1 / 17 c - b = 80 . . . . . . . . . > b = c - 80 ( c - 80 ) / c = 1 / 17 testing answers . clearly eliminate acde put c = 85 . . . . . . . . . > ( 85 - 80 ) / 85 = 5 / 85 = 1 / 17 answer : b | a = 17 - 1
b = 80 / a
c = b * 17
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a ) 6 , b ) 8 , c ) 11 , d ) 12 , e ) 14 | a | add(divide(subtract(multiply(60, 4), multiply(55, 4)), subtract(70, 60)), 4) | a car averages 55 mph for the first 4 hours of a trip and averages 70 mph for each additional hour . the average speed for the entire trip was 60 mph . how many hours long is the trip ? | "in 4 hr journey = 55 * 4 = 220 mph let in the next additional hour journey = 70 * x = 70 x mph so , ( 220 + 70 x ) / ( 4 + x ) = 60 = > 220 + 70 x = 240 + 60 x = > 10 x = 20 = > x = 2 h so the trip is ( 4 + x ) = > 6 hours long answer : a" | a = 60 * 4
b = 55 * 4
c = a - b
d = 70 - 60
e = c / d
f = e + 4
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a ) 2 . , b ) 4 . , c ) 5 . , d ) 7 . , e ) 9 . | c | divide(subtract(11, const_1), const_2) | arnold and danny are two twin brothers that are celebrating their birthday . the product of their ages today is smaller by 11 from the product of their ages a year from today . what is their age today ? | "ad = ( a + 1 ) ( d + 1 ) - 11 0 = a + d - 10 a + d = 10 a = d ( as they are twin brothers ) a = d = 5 c is the answer" | a = 11 - 1
b = a / 2
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a ) 4 days , b ) 5 days , c ) 6 days , d ) 7 days , e ) 8 days | b | divide(const_1, multiply(multiply(4, 7), divide(subtract(divide(3, multiply(3, 7)), divide(4, multiply(5, 7))), 4))) | one men and three women working 7 hours a day finish a work in 5 days . four men and four women working 3 hours a day complete the work in 7 days . the number of days in which only 7 men working 4 hours a day will finish the work is ? | "1 m + 3 w - - - - - 35 h 4 m + 4 w - - - - - - - 21 h 7 m - - - - - - - ? d 35 m + 105 w = 84 m + 84 m 21 w = 49 m 4 * 35 = 7 * x = > x = 20 hours 20 / 4 = 5 days answer : b" | a = 4 * 7
b = 3 * 7
c = 3 / b
d = 5 * 7
e = 4 / d
f = c - e
g = f / 4
h = a * g
i = 1 / h
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a ) 5 kg , b ) 2.4 kg , c ) 2.5 kg , d ) 10 kg , e ) none of these | a | multiply(divide(divide(multiply(subtract(const_100, 90), 40), const_100), subtract(const_100, 20)), const_100) | fresh grapes contain 90 % by weight while dried grapes contain 20 % water by weight . what is the weight of dry grapes available from 40 kg of fresh grapes ? | "the weight of non - water in 20 kg of fresh grapes ( which is 100 - 90 = 10 % of whole weight ) will be the same as the weight of non - water in x kg of dried grapes ( which is 100 - 20 = 80 % of whole weight ) , so 40 Γ’ Λ β 0.1 = x Γ’ Λ β 0.8 - - > x = 5 answer : a" | a = 100 - 90
b = a * 40
c = b / 100
d = 100 - 20
e = c / d
f = e * 100
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a ) s . 80 , b ) s . 85 , c ) s . 90 , d ) s . 500 , e ) s . 120 | d | multiply(divide(subtract(multiply(9, 500), multiply(multiply(const_3, const_4), 250)), multiply(multiply(const_3, const_4), const_1)), const_4) | a man engaged a servant on the condition that he would pay him rs . 500 and a uniform after one year service . he served only for 9 months and received uniform and rs . 250 , find the price of the uniform ? | "9 / 12 = 3 / 4 * 500 = 375 250 - - - - - - - - - - - - - 125 1 / 4 - - - - - - - - 125 1 - - - - - - - - - ? = > rs . 500 answer : d" | a = 9 * 500
b = 3 * 4
c = b * 250
d = a - c
e = 3 * 4
f = e * 1
g = d / f
h = g * 4
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a ) 20 , b ) 34 , c ) 38 , d ) 40 , e ) 46 | d | divide(subtract(multiply(const_3, 60), 60), const_3) | shannon and maxine work in the same building and leave work at the same time . shannon lives due north of work and maxine lives due south . the distance between maxine ' s house and shannon ' s house is 60 miles . if they both drive home at the rate 2 r miles per hour , maxine arrives home 40 minutes after shannon . if maxine rider her bike home at the rate of r per hour and shannon still drives at a rate of 2 r miles per hour , shannon arrives home 2 hours before maxine . how far does maxine live from work ? | we have that x / 24 - ( 60 - x ) / 2 r = 40 also x / r - ( 60 - x ) / 2 r = 120 so we get that 2 x - 60 = 80 r 3 x - 60 = 240 r get rid of r 120 = 3 x x = 40 hence answer is d | a = 3 * 60
b = a - 60
c = b / 3
|
a ) 150 , b ) 450 , c ) 750 , d ) 800 , e ) none of them | a | multiply(divide(const_1, add(add(const_4, 3), const_1)), 1200) | a and b undertake to do a piece of work for rs . 1200 . a alone can do it in 6 days while b alone can do it in 8 days . with the help of c , they finish it in 3 days . find the share of c . | "c ' s 1 day ' s work = 1 / 3 - ( 1 / 6 + 1 / 8 ) = 24 a : b : c = ratio of their 1 day ' s work = 1 / 6 : 1 / 8 : 1 / 24 = 4 : 3 : 1 . a β s share = rs . ( 1200 * 4 / 8 ) = rs . 600 , b ' s share = rs . ( 1200 * 3 / 8 ) = rs . 450 c ' s share = rs . [ 1200 - ( 300 + 225 Β» ) = rs . 150 . answer is a" | a = 4 + 3
b = a + 1
c = 1 / b
d = c * 1200
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a ) 36 liters , b ) 40 liters , c ) 45 liters , d ) 54 liters , e ) 60 liters | e | divide(27, subtract(divide(3, 4), divide(30, const_100))) | a big container is 30 % full with water . if 27 liters of water is added , the container becomes 3 / 4 full . what is the capacity of the big container ? | "a big container is 30 % full with water and after 27 liters of water is added , the container becomes 75 % full . hence these 27 liters account for 45 % of the container , which means that the capacity of it is 27 / 0.45 = 60 liters . or : if the capacity of the container is x liters then : 0.3 x + 27 = 0.75 x - - > x = 60 liters . answer : e ." | a = 3 / 4
b = 30 / 100
c = a - b
d = 27 / c
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a ) 22 , b ) 7 , c ) 8 , d ) 6 , e ) 51 | b | subtract(14, const_4) | in a group of cows and hens , the number of legs are 14 more than twice the number of heads . the number of cows is | "let the number of cows be x and the number of hens be y . then , 4 x + 2 y = 2 ( x + y ) + 14 4 x + 2 y = 2 x + 2 y + 14 2 x = 14 x = 7 . answer : b" | a = 14 - 4
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a ) 52.5 , b ) 54.5 , c ) 55.5 , d ) 57.5 , e ) 59.5 | d | divide(add(divide(multiply(500, 15), const_100), 500), multiply(divide(500, const_100), const_2)) | a computer store offers employees a 15 % discount off the retail price . if the store purchased a computer from the manufacturer for $ 500 dollars and marked up the price 15 % to the final retail price , how much would an employee save if he purchased the computer at the employee discount ( 15 % off retail price ) as opposed to the final retail price . | cost price = 500 profit = 15 % = 15 % of 500 = 75 selling price = cp + profit sp = 575 a discount of 10 % to employees means 10 % off on 575 so 10 % of 575 = 57.5 ans d | a = 500 * 15
b = a / 100
c = b + 500
d = 500 / 100
e = d * 2
f = c / e
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a ) 600 , b ) 800 , c ) 1000 , d ) 360 , e ) 1500 | d | multiply(divide(3, subtract(divide(60, 24), const_1)), 60) | working together , printer a and printer b would finish the task in 24 minutes . printer a alone would finish the task in 60 minutes . how many pages does the task contain if printer b prints 3 pages a minute more than printer a ? | answer : d . | a = 60 / 24
b = a - 1
c = 3 / b
d = c * 60
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a ) 22200 , b ) 31897 , c ) 38799 , d ) 40000 , e ) 42782 | d | add(multiply(add(20, const_1), add(20, const_1)), add(20, const_1)) | a sum of money is put out at compound interest for 2 years at 20 % . it would fetch rs . 964 more if the interest were payable half - yearly , then it were pay able yearly . find the sum . | "p ( 11 / 10 ) ^ 4 - p ( 6 / 5 ) ^ 2 = 964 p = 40000 answer : d" | a = 20 + 1
b = 20 + 1
c = a * b
d = 20 + 1
e = c + d
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a ) 7 / 72 , b ) 1 / 6 , c ) 5 / 28 , d ) 15 / 36 , e ) 21 / 36 | c | divide(subtract(8, 3), divide(multiply(8, subtract(8, 1)), const_2)) | a box contains 9 slips that are each labeled with one number : 1 , 3 , 5 , 8 , 13 , 21 , 34 and 55 . two of the slips are drawn at random from the box without replacement . what is the probability that the sum of the numbers on the two slips is equal to one of the numbers left in the box ? | probability = no : of desired outcomes / total no : of outcomes . you are picking two slips out of 8 slips . so total no : of outcomes = 8 c 2 = 28 desired outcome : sum of the numbers on the two slips is equal to one of the numbers left in the box . how many such outcomes are there ? if you look at the numbers closely , you will see that the following pair of numbers will give you the desired outcome . ( 3,5 ) ( 5,8 ) ( 8,13 ) ( 13,21 ) ( 21,34 ) . there are 7 such pairs . if the two numbers which i pick is from any of these 5 pairs , then i get my desired outcome . so no : of desired outcomes = 5 probability = 5 / 28 answer : c | a = 8 - 3
b = 8 - 1
c = 8 * b
d = c / 2
e = a / d
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a ) 18 , b ) 15 , c ) 17 , d ) 19 , e ) 21 | e | add(divide(160, subtract(9, 1)), 1) | the operation is defined for all integers a and b by the equation ab = ( a - 1 ) ( b - 1 ) . if x 9 = 160 , what is the value of x ? | "ab = ( a - 1 ) ( b - 1 ) x 9 = ( x - 1 ) ( 9 - 1 ) = 160 - - > x - 1 = 20 - - > x = 21 answer : e" | a = 9 - 1
b = 160 / a
c = b + 1
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a ) 1.5 % , b ) 0.3 % , c ) 2.5 % , d ) 3.5 % , e ) 4 % | b | multiply(divide(3, 1), const_100) | what percent is 3 gm of 1 kg ? | "1 kg = 1000 gm 3 / 1000 Γ 100 = 300 / 1000 = 3 / 10 = 0.3 % b )" | a = 3 / 1
b = a * 100
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a ) 35 , b ) 55 , c ) 39 , d ) 40 , e ) 60 | a | divide(multiply(30, add(3, divide(const_1, const_2))), 3) | john began driving from home on a trip averaging 30 miles per hour . how many miles per hour must carla drive on average to catch up to him in exactly 3 hours if she leaves 30 minutes after john ? | carla starts 30 minutes later and it takes 3 hr for carla to meet john so john total time travelled = 3 hr + 30 minutes john distance = 30 * ( 3 1 / 2 ) = 105 so carla need to travle 105 to meet john in 3 hrs speed of carla = 105 / 3 = 35 miles per hour answer is a | a = 1 / 2
b = 3 + a
c = 30 * b
d = c / 3
|
a ) 42 , b ) 49 , c ) 56 , d ) 63 , e ) 70 | a | multiply(divide(7, subtract(9, 7)), 12) | sandy is younger than molly by 12 years . if the ratio of their ages is 7 : 9 , how old is sandy ? | let sandy ' s age be 7 x and let molly ' s age be 9 x . 9 x - 7 x = 12 x = 6 sandy is 42 years old . the answer is a . | a = 9 - 7
b = 7 / a
c = b * 12
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a ) 11,550 , b ) 14,550 , c ) 17,550 , d ) 20,550 , e ) 23,550 | c | multiply(divide(add(400, 302), const_2), add(divide(subtract(400, 302), const_2), const_1)) | the sum of the first 50 positive even integers is 2550 . what is the sum of the even integers from 302 to 400 inclusive ? | "2 + 4 + 6 + 8 + . . . + 100 = 2550 302 + 304 + . . . + 400 = 50 ( 300 ) + ( 2 + 4 + . . . + 100 ) = 15,000 + 2550 = 17,550 the answer is c ." | a = 400 + 302
b = a / 2
c = 400 - 302
d = c / 2
e = d + 1
f = b * e
|
a ) 1.2 hr , b ) 2.5 hr , c ) 1.5 hr , d ) 48 min , e ) 58 min | c | divide(multiply(8, 3), 16) | walking at the rate of 8 kmph a man cover certain distance in 3 hrs . . running at a speed of 16 kmph the man will cover the same distance in . | distance = speed * time 8 * 3 = 24 km new speed = 16 kmph therefore time = d / s = 24 / 16 = 1.5 hr . answer : c . | a = 8 * 3
b = a / 16
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a ) 1.9 , b ) 1.7 , c ) 1.2 , d ) 1.5 , e ) 1.1 | e | divide(subtract(multiply(3.3, const_2), 3.3), const_3) | a man can row 3.3 km / hr in still water . it takes him twice as long to row upstream as to row downstream . what is the rate of the current ? | speed of boat in still water ( b ) = 3.3 km / hr . speed of boat with stream ( down stream ) , d = b + u speed of boat against stream ( up stream ) , u = b β u it is given upstream time is twice to that of down stream . β downstream speed is twice to that of upstream . so b + u = 2 ( b β u ) β u = b / 3 = 1.1 km / hr . answer : e | a = 3 * 3
b = a - 3
c = b / 3
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a ) q = 12 , b ) q = 15 , c ) q = 17 , d ) q . 18 , e ) q = 20 | d | multiply(multiply(3, const_2), 3) | two different primes may be said torhymearound an integer if they are the same distance from the integer on the number line . for instance , 3 and 7 rhyme around 5 . what integer q between 1 and 20 , inclusive , has the greatest number of distinct rhyming primes around it ? | "since we are concerned with integers between 1 and 20 , write down the primes till 40 . 2 , 3 , 5 , 7 , 11 , 13 , 17 , 19 , 23 , 29 , 31 , 37 ( you should be very comfortable with the first few primes . . . ) 2 , 3 , 5 , 7 , 11,12 , 13 , 17 , 19 , 23 , 29 , 31 , 37 - three pairs ( 11,13 ) , ( 7,17 ) , ( 5 , 19 ) 2 , 3 , 5 , 7 , 11 , 13 , 15,17 , 19 , 23 , 29 , 31 , 37 - three pairs ( 13 , 17 ) , ( 11 , 19 ) , ( 7 , 23 ) 2 , 3 , 5 , 7 , 11 , 13,17 , 19 , 23 , 29 , 31 , 37 - three pairs ( 11 , 23 ) , ( 5 , 29 ) , ( 3 , 31 ) 2 , 3 , 5 , 7 , 11 , 13 , 17 , 18,19 , 23 , 29 , 31 , 37 - four pairs ( 17 , 19 ) , ( 13 , 23 ) , ( 7 , 29 ) , ( 5 , 31 ) 2 , 3 , 5 , 7 , 11 , 13 , 17 , 19 , 20,23 , 29 , 31 , 37 - definitely can not be more than 4 since there are only 4 primes more than 20 . so must be less than 4 pairs . ignore . answer ( d ) ." | a = 3 * 2
b = a * 3
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['a ) 10', 'b ) 14', 'c ) 18', 'd ) 22', 'e ) 24'] | b | multiply(sqrt(divide(1862, add(add(multiply(3, 3), multiply(2, 2)), multiply(5, 5)))), 2) | if three numbers in the ratio 3 : 2 : 5 be such that the sum of their squares is 1862 , the middle number will be | explanation : let the numbers be 3 x , 2 x and 5 x . then , 9 x + 4 x + 25 x = 1862 β 38 x = 1862 β x = 49 β x = 7 . middle number = 2 x = 14 option b | a = 3 * 3
b = 2 * 2
c = a + b
d = 5 * 5
e = c + d
f = 1862 / e
g = math.sqrt(f)
h = g * 2
|
a ) 80.2 % , b ) 88 % , c ) 87.4 % , d ) 85 % , e ) 83.9 % | a | divide(add(79, 85), const_2) | factory x ' s bulbs work for over 5000 hours in 79 % of cases , whereas factory y ' s bulbs work for over 5000 hours in 85 % of cases . it is known that factory x supplies 80 % of the total bulbs available . what is the chance that a purchased bulb will work for longer than 5000 hours ? | for x , 80 % of 79 % will work . for y , 20 % of 85 % will work . * 20 % is the rest of the bulb supply in the market . so , the probability that a purchased bulb will work is : 0.80 ( 0.79 ) = . 632 0.20 ( 0.85 ) = 0.17 the combined probability then is 63.2 + 17 = 80.2 % ans a | a = 79 + 85
b = a / 2
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a ) 13 km / hr , b ) 59 km / hr , c ) 17 km / hr , d ) 18 km / hr , e ) 12 km / hr | b | divide(divide(subtract(150, multiply(multiply(5, const_0_2778), 5)), 5), const_0_2778) | a train 150 m long passes a man , running at 5 km / hr in the same direction in which the train is going , in 10 seconds . the speed of the train is ? | "speed of the train relative to man = ( 150 / 10 ) m / sec = ( 15 ) m / sec . [ 15 ) * ( 18 / 5 ) ] km / hr = 54 km / hr . let the speed of the train be x km / hr . then , relative speed = ( x - 5 ) km / hr . x - 5 = 54 = = > x = 59 km / hr . answer : b" | a = 5 * const_0_2778
b = a * 5
c = 150 - b
d = c / 5
e = d / const_0_2778
|
a ) 1 / 16 , b ) 5 / 42 , c ) 1 / 8 , d ) 3 / 16 , e ) 1 / 4 | a | divide(divide(choose(40, const_1), 40), power(const_3, const_2)) | each factor of 210 is inscribed on its own plastic ball , and all of the balls are placed in a jar . if a ball is randomly selected from the jar , what is the probability that the ball is inscribed with a multiple of 40 ? | "210 = 2 * 3 * 5 * 7 , so the # of factors 210 has is ( 1 + 1 ) ( 1 + 1 ) ( 1 + 1 ) ( 1 + 1 ) = 16 ( see below ) ; 42 = 2 * 3 * 7 , so out of 16 factors only two are multiples of 42 : 42 and 210 , itself ; so , the probability is 2 / 16 = 1 / 16 . answer : a" | a = math.comb(40, 1)
b = a / 40
c = 3 ** 2
d = b / c
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a ) 100 , b ) 200 , c ) 600 , d ) 800 , e ) 1000 | c | multiply(divide(const_60, 10), 60) | if the population of a certain country increases at the rate of 10 person every 60 seconds , by how many persons does the population increase in 1 hour ? | "answer = 10 * 60 ( 1 hour = 60 minutes ) = 600 answer = c" | a = const_60 / 10
b = a * 60
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a ) 2 , b ) 4 , c ) 6 , d ) 8 , e ) 0 | e | add(reminder(multiply(reminder(46, const_4), 5), const_10), reminder(10, const_10)) | the units digit of ( 10 ) ^ ( 87 ) + ( 5 ) ^ ( 46 ) is : | "any power of anything ending in 5 always has a units digit of 5 . so the first term has a units digit of 5 . done . the second term anything power to 10 unit digit will be zero then 5 + 0 = 5 , e" | a = reminder * (
b = reminder + (
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a ) 0.003 , b ) 0.0005 , c ) 0.25 , d ) 0.02 , e ) none of these | d | multiply(divide(divide(4, const_100), const_4), const_2) | double of quarter of 4 percent written as a decimal is : | explanation : solution : ( 2 ) * ( 1 / 4 ) * 4 % = 2 * ( 1 / 4 * 1 / 100 ) = 0.02 . answer : d | a = 4 / 100
b = a / 4
c = b * 2
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a ) 200 sec , b ) 230 sec , c ) 167 sec , d ) 197 sec , e ) 179 sec | b | divide(add(1200, 1100), divide(1200, 120)) | a 1200 m long train crosses a tree in 120 sec , how much time will i take to pass a platform 1100 m long ? | "l = s * t s = 1200 / 120 s = 10 m / sec . total length ( d ) = 2300 m t = d / s t = 2300 / 10 t = 230 sec answer : b" | a = 1200 + 1100
b = 1200 / 120
c = a / b
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a ) 270 , b ) 279 , c ) 348 , d ) 371 , e ) 372 | c | multiply(divide(841, add(add(multiply(12, 8), multiply(16, 9)), multiply(18, 6))), multiply(16, 9)) | a , b and c rents a pasture for rs . 841 . a put in 12 horses for 8 months , b 16 horses for 9 months and 18 horses for 6 months . how much should b pay ? | "12 * 8 : 16 * 9 = 18 * 6 8 : 12 : 9 12 / 29 * 841 = 348 answer : c" | a = 12 * 8
b = 16 * 9
c = a + b
d = 18 * 6
e = c + d
f = 841 / e
g = 16 * 9
h = f * g
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a ) 8 , b ) 10 , c ) 12 , d ) 14 , e ) 16 | a | add(4, 6) | a one - foot stick is marked in 1 / 4 and 1 / 6 portion . how many total markings will there be , including the end points ? | "lcm of 24 = 12 1 / 4 marking are ( table of 3 ) 0 . . . . . . 3 . . . . . . . . . . . 6 . . . . . . . . . . 9 . . . . . . . . . . 12 ( total = 5 ) 1 / 6 marking are ( table of 2 ) 0 . . . . . . . 2 . . . . . . 4 . . . . . . 6 . . . . . . . . 8 . . . . . . . . . 12 ( ( total = 6 ) overlapping markings are 0 . . . . . . . . 6 . . . . . . . . . 12 ( total = 3 ) total markings = 5 + 6 - 3 = 8 answer = a" | a = 4 + 6
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['a ) 7 : 9', 'b ) 7 : 3', 'c ) 7 : 7', 'd ) 7 : 6', 'e ) 7 : 1'] | b | divide(multiply(divide(divide(924, divide(divide(264, const_pi), const_2)), const_pi), const_2), divide(divide(divide(264, const_pi), const_2), divide(divide(924, divide(divide(264, const_pi), const_2)), const_pi))) | the curved surface area of a cylindrical pillar is 264 sq . m and its volume is 924 cub . m . find the ratio of its diameter to its height | explanation : { \ color { black } \ frac { \ prod r ^ { 2 } h } { 2 \ prod rh } = \ frac { 924 } { 264 } } { \ color { black } \ rightarrow r = \ frac { 924 } { 264 } \ times 2 = 7 m } { \ color { black } and \ ; \ ; 2 \ prod rh = 264 \ rightarrow h = ( 264 \ times \ frac { 7 } { 22 } \ times \ frac { 1 } { 2 } \ times \ frac { 1 } { 7 } ) = 6 m } { \ color { black } \ therefore } required rato = { \ color { black } \ frac { 2 r } { h } } = { \ color { black } \ frac { 14 } { 6 } = 7 : 3 } answer : b ) 7 : 3 | a = 264 / math.pi
b = a / 2
c = 924 / b
d = c / math.pi
e = d * 2
f = 264 / math.pi
g = f / 2
h = 264 / math.pi
i = h / 2
j = 924 / i
k = j / math.pi
l = g / k
m = e / l
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a ) 38 metre , b ) 28 metre , c ) 23 metre , d ) 16 metre , e ) 28 metre | d | subtract(128, multiply(divide(128, 32), 28)) | a can run 128 metre in 28 seconds and b in 32 seconds . by what distance a beat b ? | clearly , a beats b by 4 seconds now find out how much b will run in these 4 seconds speed of b = distance / time taken by b = 128 / 32 = 4 m / s distance covered by b in 4 seconds = speed Γ£ β time = 4 Γ£ β 4 = 16 metre i . e . , a beat b by 16 metre answer is d | a = 128 / 32
b = a * 28
c = 128 - b
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a ) 8 , b ) 7 , c ) 15 , d ) 21 , e ) 25 | a | add(4, subtract(add(13, 9), multiply(const_2, 9))) | jacob is now 13 years younger than michael . if 9 years from now michael will be twice as old as jacob , how old will jacob be in 4 years ? | "jacob = x years , michael = x + 13 years 9 years from now , 2 ( x + 9 ) = x + 22 2 x + 18 = x + 22 x = 4 x + 4 = 8 years answer a" | a = 13 + 9
b = 2 * 9
c = a - b
d = 4 + c
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a ) 93 , b ) 67 , c ) 77 , d ) 87 , e ) 97 | a | sqrt(multiply(86.49, const_100)) | a group of students decided to collect as many paise from each member of group as is the number of members . if the total collection amounts to rs . 86.49 , the number of the member is the group is : | "money collected = ( 86.49 x 100 ) paise = 8649 paise numbers of members = 8649 squareroot = 93 answer a" | a = 86 * 49
b = math.sqrt(a)
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a ) 0.20833 , b ) 0.35424 , c ) 0.10982 , d ) 0.20933 , e ) 0.10984 | a | multiply(divide(divide(480, divide(60, const_100)), add(multiply(multiply(3, const_100), const_1000), multiply(add(multiply(const_4, const_10), const_4), const_1000))), const_100) | lagaan is levied on the 60 percent of the cultivated land . the revenue department collected total rs . 3 , 84,000 through the lagaan from the village of mettur . mettur , a very rich farmer , paid only rs . 480 as lagaan . the percentage of total land of mettur over the total taxable land of the village is : | "total land of sukhiya = \ inline \ frac { 480 x } { 0.6 } = 800 x \ therefore cultivated land of village = 384000 x \ therefore required percentage = \ inline \ frac { 800 x } { 384000 } \ times 100 = 0.20833 a" | a = 60 / 100
b = 480 / a
c = 3 * 100
d = c * 1000
e = 4 * 10
f = e + 4
g = f * 1000
h = d + g
i = b / h
j = i * 100
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a ) 1 / 18 , b ) 1 / 6 , c ) 1 / 3 , d ) 1 / 2 , e ) 7 / 12 | e | subtract(add(12, divide(12, 4)), divide(2, 4)) | in a certain lottery , the probability that a number between 12 and 20 , inclusive , is drawn is 1 / 4 . if the probability that a number 12 or larger is drawn is 2 / 3 , what is the probability that a number less than or equal to 20 is drawn ? | "you can simply use sets concept in this question . the formula total = n ( a ) + n ( b ) - n ( a and b ) is applicable here too . set 1 : number 12 or larger set 2 : number 20 or smaller 1 = p ( set 1 ) + p ( set 2 ) - p ( set 1 and set 2 ) ( combined probability is 1 because every number will be either 12 or moreor 20 or lessor both ) 2 / 3 + p ( set 2 ) - 1 / 4 = 1 p ( set 2 ) = 7 / 12 answer ( e )" | a = 12 / 4
b = 12 + a
c = 2 / 4
d = b - c
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a ) $ 700 , b ) $ 850 , c ) $ 800 , d ) $ 750 , e ) $ 900 | d | divide(multiply(250, multiply(multiply(const_2, const_100), const_100)), divide(multiply(multiply(const_2, const_100), const_100), const_4)) | if $ 5,000 is invested in an account at a simple annual rate of r percent , the interest is $ 250 . when $ 15,000 is invested at the same interest rate , what is the interest from the investment ? | "- > 250 / 5,000 = 5 % and 15,000 * 5 % = 750 . thus , d is the answer ." | a = 2 * 100
b = a * 100
c = 250 * b
d = 2 * 100
e = d * 100
f = e / 4
g = c / f
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a ) 28.32 , b ) 39 , c ) 38.78 , d ) 29 , e ) 39.47 | e | divide(subtract(multiply(65, 40), add(83, 30)), subtract(65, const_2)) | the average of 65 numbers is 40 . if two numbers , 83 and 30 are discarded , then the average of the remaining numbers is nearly | explanation : total sum of 63 numbers = ( 65 * 40 ) - ( 83 + 30 ) = 2600 - 113 = 2487 average = 2487 / 63 = 39.47 answer : e | a = 65 * 40
b = 83 + 30
c = a - b
d = 65 - 2
e = c / d
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a ) 29 , b ) 39 , c ) 19 , d ) 49 , e ) 59 | c | add(const_10, add(subtract(subtract(multiply(17, const_3), multiply(divide(add(multiply(const_3, const_3), const_1), const_2), multiply(const_3, const_3))), const_1), 4)) | a + b + c + d = d + e + f + g = g + h + i = 17 given a = 4 . find value of g and h ? | a + b + c + d = 17 4 + 2 + 6 + 5 = 17 d = 5 d + e + f + g = 17 5 + 3 + 8 + 1 = 17 g = 1 g + h + i = 17 1 + 9 + 7 = 17 g = 1 , h = 9 . answer : c | a = 17 * 3
b = 3 * 3
c = b + 1
d = c / 2
e = 3 * 3
f = d * e
g = a - f
h = g - 1
i = h + 4
j = 10 + i
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a ) $ 2380 , b ) $ 2590 , c ) $ 2760 , d ) $ 2940 , e ) $ 3150 | b | add(1,000, divide(111.30, divide(7, const_100))) | when a merchant imported a certain item , she paid a 7 percent import tax on the portion of the total value of the item in excess of $ 1,000 . if the amount of the import tax that the merchant paid was $ 111.30 , what was the total value of the item ? | "let x be the value of the item . 0.07 * ( x - 1000 ) = 111.30 x = 2590 the answer is b ." | a = 7 / 100
b = 111 / 30
c = 1 + 0
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a ) 25 , b ) 5 , c ) 1 , d ) 23 , e ) 6 | c | divide(divide(divide(divide(divide(13225, const_3), const_3), const_4), const_4), const_4) | find the smallest number which should be multiplied with 13225 to make it a perfect square . | "13225 = 5 * 5 * 23 * 23 required smallest number = 1 1 is the smallest number which should be multiplied with 13225 to make it a perfect square . answer : c" | a = 13225 / 3
b = a / 3
c = b / 4
d = c / 4
e = d / 4
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a ) 72.72 % , b ) 50 % , c ) 57.57 % , d ) 60 % , e ) 65 % | a | multiply(divide(20000, add(add(2500, 5000), 20000)), const_100) | 3 candidates in an election and received 2500 , 5000 and 20000 votes respectively . what % of the total votes did the winningcandidate got in that election ? | "total number of votes polled = ( 2500 + 5000 + 20000 ) = 27500 so , required percentage = 20000 / 27500 * 100 = 72.72 % a" | a = 2500 + 5000
b = a + 20000
c = 20000 / b
d = c * 100
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a ) 36 , b ) 2 ^ 4 * 3 , c ) 24 , d ) 38 , e ) 47 | c | subtract(25, const_1) | in a lake , there is a patch of lily pads . every day , the patch doubles in size . it takes 25 days for the patch to cover the entire lake , how many days would it take the patch to cover half of the lake ? | "so 24 days answer c = 24" | a = 25 - 1
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a ) 1.5 liters , b ) 2.5 liters , c ) 3.5 liters , d ) 4.5 liters , e ) 5.5 liters | b | divide(multiply(divide(50, const_100), 10), const_2) | a 30 % alcohol mixture is added to a 50 % alcohol mixture to form a 10 litre mixture of 45 % alcohol . how much of the 30 % mixture was used ? | solution a = 50 % solution solution b = 30 % solution 30 % - 45 % = 15 % solution b 50 % - 45 % = 5 % solution a so the ratio is 3 : 1 for 30 % : 50 % solutions 3 / 4 * 10 liter = 7.5 for 30 % solution and 2.5 for 50 % solution . . answer : b | a = 50 / 100
b = a * 10
c = b / 2
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a ) $ 160 , b ) $ 176 , c ) $ 282 , d ) $ 302 , e ) $ 286 | d | add(multiply(18.00, add(const_3, const_4)), multiply(11.00, subtract(23, add(const_3, const_4)))) | if the charge of staying in a student youth hostel $ 18.00 / day for the first week , and $ 11.00 / day for each additional week , how much does it cost to stay for 23 days ? | "total number of days of stay = 23 charge of staying in first week = 18 * 7 = 126 $ charge of staying for additional days = ( 23 - 7 ) * 11 = 16 * 11 = 176 $ total charge = 126 + 176 = 302 $ answer d" | a = 3 + 4
b = 18 * 0
c = 3 + 4
d = 23 - c
e = 11 * 0
f = b + e
|
a ) a . 4 , b ) b . 5 / 2 , c ) c . 2 , d ) d . 3 / 2 , e ) e . 5 / 4 | b | divide(add(multiply(3, const_4), 3), const_10) | nails and screws are manufactured at a uniform weight per nail and a uniform weight per screw . if the total weight of one screw and one nail is half that of 6 screws and one nail , then the total weight of 3 screws , and 3 nails is how many times that of 1 screws and 5 nails ? | "let the weight of nail be n and that of screw be s . . so s + w = 1 / 2 * ( 6 s + 1 n ) . . . or 1 n = 4 s . . lets see the weight of 3 s and 3 n = 3 s + 3 * 4 s = 15 s . . and weight of 1 s and 5 n = 1 s + 5 * 1 s = 6 s . . ratio = 15 s / 6 s = 15 / 6 = 5 / 2 b" | a = 3 * 4
b = a + 3
c = b / 10
|
a ) - 4 , b ) - 1 and - 3 , c ) 2 , d ) 4 , e ) can not be determined . | b | divide(power(const_3, const_2), 4) | a number x is multiplied with itself and then added to the product of 4 and x . if the result of these two operations is - 3 , what is the value of x ? | a number x is multiplied with itself - - > x ^ 2 added to the product of 4 and x - - > x ^ 2 + 4 x if the result of these two operations is - 3 - - > x ^ 2 + 4 x = - 3 i . e x ^ 2 + 4 x + 3 = 0 is the quadratic equation which needs to be solved . ( x + 1 ) ( x + 3 ) = 0 hence x = - 1 . x = - 3 imo b | a = 3 ** 2
b = a / 4
|
a ) 35 , b ) 36 , c ) 37 , d ) 38 , e ) 42 | e | add(divide(subtract(add(40, 2), 30), 1.5), 30) | each week , harry is paid x dollars per hour for the first 30 hours and 1.5 x dollars for each additional hour worked that week . each week , james is paid x dollars per hour for the first 40 hours and 2 x dollars for each additional hour worked that week . last week james worked a total of 44 hours . if harry and james were paid the same amount last week , how many hours did harry work last week ? | "amount earned by james = 40 * x + 4 * 2 x = 48 x therefore , amount earned by james = 48 x but we know the amount harry earned assuming working y hours ( y > 30 ) is 30 * x + ( y - 30 ) * 1.5 x [ [ we know y > 30 because in 30 h the most harry could earn is 30 x , but he has earned 48 x ] ] so x * ( 1.5 y - 45 + 30 ) = 48 x or x * ( 1.5 y - 15 ) = 48 x so 1.5 y - 15 = 48 so 1.5 y = 63 so y = 42 answer is e" | a = 40 + 2
b = a - 30
c = b / 1
d = c + 30
|
a ) 100011 , b ) 111111 , c ) 101111 , d ) 1011111 , e ) 211111 | a | add(subtract(multiply(const_10, multiply(const_100, const_100)), const_100), 111) | find the smallest number of 6 digits which is exactly divisible by 111 | "smallest number of 6 digits is 100000 . on dividing 100000 by 111 , we get 100 as remainder . number to be added = ( 111 - 100 ) - 11 . hence , required number = 100011 answer a 100011" | a = 100 * 100
b = 10 * a
c = b - 100
d = c + 111
|
a ) 9 , b ) 20 , c ) 55 , d ) 70 , e ) 81 | b | divide(add(multiply(divide(subtract(multiply(81.2, const_10), const_2), const_10), divide(13, divide(const_2, const_10))), 13), divide(13, divide(const_2, const_10))) | when the positive integer k is divided by the positive integer n , the remainder is 13 . if k / n = 81.2 , what is the value of n ? | "here ' s an approach that ' s based on number properties and a bit ofbrute forcemath : we ' re told that k and n are both integers . since k / n = 81.2 , we can say that k = 81.2 ( n ) n has tomultiply outthe . 2 so that k becomes an integer . with the answers that we have to work with , n has to be a multiple of 5 . eliminate a and e . with the remaining answers , we can test the answers and find the one that fits the rest of the info ( k / n = 81.2 and k / n has a remainder of 11 ) answer b : if n = 20 , then k = 1624 ; 1624 / 20 has a remainder of 4 not a match answer c : if n = 55 , then k = 4466 ; 4466 / 55 has a remainder of 11 match . final answer : b" | a = 81 * 2
b = a - 2
c = b / 10
d = 2 / 10
e = 13 / d
f = c * e
g = f + 13
h = 2 / 10
i = 13 / h
j = g / i
|
a ) 13 , b ) 14 , c ) 15 , d ) 16 , e ) 17 | b | divide(subtract(76, multiply(10, const_2)), add(3, const_1)) | kareem is 3 times as old as his son . after 10 years , the sum of their ages will be 76 years . find their present ages . | sol . let the present age of kareem β s son be x years . then , kareem β s age = 3 x years after 10 years , kareem β s age = 3 x + 10 years and kareem β s son β s age = x + 10 years β΄ ( 3 x + 10 ) + ( x + 10 ) = 76 = > 4 x = 56 = > x = 14 β΄ kareem β s present age = 3 x = 3 Γ 14 = 42 years kareem β s son β s age = x = 14 years . answer b | a = 10 * 2
b = 76 - a
c = 3 + 1
d = b / c
|
a ) 5 , b ) 6 , c ) 3 , d ) 2 , e ) 4 | b | add(divide(subtract(21, const_1), 4), const_1) | what is the greatest value of x such that 4 ^ x is a factor of 21 ! ? | pretty simple , really . if m = 6 , then 4 m = 24 , which is 12 x 2 , both of which are included in 21 ! since 6 is the largest number here , its the answer . answer is b | a = 21 - 1
b = a / 4
c = b + 1
|
a ) 7 , b ) 6 , c ) 5 , d ) 8 , e ) 3 | d | multiply(1, 8) | if 8 spiders make 8 webs in 8 days , then 1 spider will make 1 web in how many days ? | "let the required number days be x . less spiders , more days ( indirect proportion ) less webs , less days ( direct proportion ) spiders 1 : 8 webs 8 : 1 1 x 8 x x = 8 x 1 x 8 = > x = 8 answer is d" | a = 1 * 8
|
a ) 325 , b ) 327 , c ) 323 , d ) 330 , e ) 350 | c | divide(multiply(160, 162), const_4) | what is the sum of the integers from - 160 to 162 , inclusive ? | "in an arithmetic progression , the nth term is given by tn = a + ( n - 1 ) d here tn = 162 , a = - 160 , d = 1 hence , 162 = - 160 + ( n - 1 ) or n = 323 sum of n terms can be calculated by sn = n / 2 ( a + l ) a = first term , l = last term , n = no . of terms sn = 323 * ( - 160 + 162 ) / 2 sn = 323 * 2 / 2 = 323 answer : c" | a = 160 * 162
b = a / 4
|
['a ) they can go on 15 rides and $ 1.50 will be left over .', 'b ) they can go on 3 rides and $ 2.70 will be left over .', 'c ) they can go on 7 rides and $ 0 will be left over .', 'd ) they can go on 6 rides and $ 0 will be left over .', 'e ) they can go on 5 rides and $ 1.50 will be left over .'] | b | subtract(subtract(27, add(multiply(3, 2), 3)), multiply(floor(divide(subtract(27, add(multiply(3, 2), 3)), 1.7)), 1.7)) | chloe has $ 27 dollars to go to the fair . she is bringing 2 other friends along . admission to the fair is $ 3 . each ride costs $ 1.70 . how many rides can she and her friends go on and how much money will be left over ? | admission to the fair is $ 3 per person . if she wants her and her friends to go , the cost will be $ 9 . 27 - 9 is 18 . to go on a ride is $ 5.10 . 5.10 x 3 is 15.30 . 18 - 15.30 is 2.70 . they can go on 3 rides and $ 2.70 will be left over . the correct answer is b . | a = 3 * 2
b = a + 3
c = 27 - b
d = 3 * 2
e = d + 3
f = 27 - e
g = f / 1
h = math.floor(g)
i = h * 1
j = c - i
|
a ) 10 % , b ) 40 % , c ) 30 % , d ) 25 % , e ) 35 % | b | subtract(const_100.0, 30) | if the selling price of 50 articles is equal to the cost price of 30 articles , then the loss or gain percent is : | "let c . p . of each article be re . 1 . then , c . p . of 50 articles = rs . 50 ; s . p . of 50 articles = rs . 30 . loss % = 20 / 50 * 100 = 40 % answer : b" | a = 100 - 0
|
a ) 250 / 3 , b ) 500 / 3 , c ) 400 , d ) 480 , e ) 600 | e | multiply(divide(subtract(25, 10), subtract(30, 25)), 200) | solution x is 10 percent alcohol by volume , and solution y is 30 percent alcohol by volume . how many milliliters of solution y must be added to 200 milliliters of solution x to create a solution that is 25 percent alcohol by volume ? | we know that x is 10 % , y is 30 % and w . avg = 25 % . what does this mean with respect to w . avg technique ? w . avg is 1 portion away from y and 3 portion away from x so for every 1 portion of x we will have to add 3 portions of y . if x = 200 then y = 600 answer : e | a = 25 - 10
b = 30 - 25
c = a / b
d = c * 200
|
a ) s . 10123.77 , b ) s . 10123.21 , c ) s . 10123.20 , d ) s . 10123.28 , e ) s . 10528.12 | e | subtract(multiply(multiply(multiply(const_4, const_100), const_100), power(add(const_1, divide(12, const_100)), 3)), multiply(multiply(const_4, const_100), const_100)) | what will be the compound interest on a sum of rs . 26,000 after 3 years at the rate of 12 % p . a . ? | "amount = [ 26000 * ( 1 + 12 / 100 ) 3 ] = 26000 * 28 / 25 * 28 / 25 * 28 / 25 = rs . 36528.12 c . i . = ( 36528.12 - 26000 ) = rs . 10528.12 answer : e" | a = 4 * 100
b = a * 100
c = 12 / 100
d = 1 + c
e = d ** 3
f = b * e
g = 4 * 100
h = g * 100
i = f - h
|
a ) 50 yards , b ) 60 yards , c ) 70 yards , d ) 80 yards , e ) 90 yards | e | subtract(150, 60) | the star running back on our football team got most of his total yardage running . the rest was catching passes . he caught passes for 60 yards . his total yardage was 150 yards . the running back for the other team got 200 yards . how many yards did the star running back on our football team get running ? | . the other team is extra information . 150 β 60 = 90 he got 90 yards running . correct answer e | a = 150 - 60
|
a ) 28 : 15 , b ) 6 : 7 , c ) 7 : 6 , d ) 5 : 21 , e ) 21 : 5 | e | divide(divide(subtract(35, 56), subtract(30, 56)), subtract(const_1, divide(subtract(35, 56), subtract(30, 56)))) | in what ratio must rice of rs . 30 per kg be mixed with rice of rs . 56 per kg so that cost of mixture is rs . 35 per kg ? | "( 35 - 56 ) / ( 30 - 35 ) = 6 / 2 = 21 : 5 answer : e" | a = 35 - 56
b = 30 - 56
c = a / b
d = 35 - 56
e = 30 - 56
f = d / e
g = 1 - f
h = c / g
|
a ) 1305 , b ) 1375 , c ) 1345 , d ) 1415 , e ) 1455 | c | add(add(multiply(multiply(25, divide(4, 5)), subtract(subtract(subtract(60, multiply(60, divide(30, const_100))), multiply(60, divide(30, const_100))), 10)), multiply(25, 10)), add(multiply(25, multiply(60, divide(30, const_100))), multiply(25, multiply(60, divide(30, const_100))))) | an automobile parts supplier charges $ 25 per package of gaskets . when a customer orders more than 10 packages of gaskets , the supplier charges 4 / 5 the price for each package in excess of 10 . during a certain week , the supplier sold 60 packages of gaskets . if 30 percent of the gaskets went to company x , 15 percent to company y , and the rest to company z , what was the total amount , in dollars , that the parts supplier received in payment for the gaskets ? | "$ 25 per packet of gasket in case a customer orders less than 10 in case a customer orders > 10 price per gasket = 25 * 4 / 5 = 20 a certain week the supplier sold 60 gasket 1 . he sold 30 % of the gaskets to x = 18 gaskets = 25 * 10 + 20 * 8 = 250 + 160 = 410 2 . he sold 15 % of the gaskets to y = 9 gaskets = 25 * 9 = 225 3 . he sold remaining 55 % to z = 33 gaskets = 25 * 10 = 250 + 20 * 23 = 710 thus , total money earned 410 + 225 + 710 = 1345 answer is c" | a = 4 / 5
b = 25 * a
c = 30 / 100
d = 60 * c
e = 60 - d
f = 30 / 100
g = 60 * f
h = e - g
i = h - 10
j = b * i
k = 25 * 10
l = j + k
m = 30 / 100
n = 60 * m
o = 25 * n
p = 30 / 100
q = 60 * p
r = 25 * q
s = o + r
t = l + s
|
a ) 65 , b ) 38 , c ) 20 , d ) 28 , e ) 21 | d | subtract(subtract(113, 35), add(25, 25)) | two cars start from the opposite places of a main road , 113 km apart . first car runs for 25 km and takes a right turn and then runs 15 km . it then turns left and then runs for another 25 km and then takes the direction back to reach the main road . in the mean time , due to minor break down the other car has run only 35 km along the main road . what would be the distance between two cars at this point ? | answer : d ) 28 km | a = 113 - 35
b = 25 + 25
c = a - b
|
a ) 233 , b ) 299 , c ) 333 , d ) 199 , e ) 339 | d | divide(subtract(800, 4), 4) | a straight line in the xy - plane has a slope of 4 and a y - intercept of 4 . on this line , what is the x - coordinate of the point whose y - coordinate is 800 ? | eq of line = y = mx + c m = 4 , c = 4 y = 4 x + 4 , substitute y by 800 as given in question . 800 = 4 x + 4 , x = 199 . correct option is d | a = 800 - 4
b = a / 4
|
a ) 30 % , b ) 40 % , c ) 50 % , d ) 66.67 % , e ) 70 % | d | multiply(divide(10, subtract(25, 10)), const_100) | a shop owner sells 25 mtr of cloth and gains sp of 10 mtrs . find the gain % ? | "here , selling price of 10 m cloth is obtained as profit . profit of 10 m cloth = ( s . p . of 25 m cloth ) β ( c . p . of 25 m cloth ) selling price of 15 m cloth = selling price of 25 m of cloth let cost of each metre be rs . 100 . therefore , cost price of 15 m cloth = rs . 1500 and s . p . of 15 m cloth = rs . rs . 2500 profit % = 10 / 15 Γ 100 = 66.67 % profit of 66.67 % was made by the merchant . d" | a = 25 - 10
b = 10 / a
c = b * 100
|
a ) 130 , b ) 300 , c ) 200 , d ) 30 , e ) 75 | b | multiply(2, 150) | what number has a 150 : 1 ratio to the number 2 ? | "150 : 1 = x : 2 x = 150 * 2 x = 300 answer : b" | a = 2 * 150
|
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