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 ) 90 , b ) 110 , c ) 120 , d ) 140 , e ) 220 | d | subtract(divide(subtract(multiply(12, 360), add(add(multiply(const_3, const_1000), multiply(const_3, const_100)), multiply(const_2, const_10))), subtract(12, 8)), subtract(360, divide(subtract(multiply(12, 360), add(add(multiply(const_3, const_1000), multiply(const_3, const_100)), multiply(const_2, const_10))), subtract(12, 8)))) | a theater charges $ 12 for seats in the orchestra and $ 8 for seats in the balcony . on a certain night , a total of 360 tickets were sold for a total cost of $ 3,320 . how many more tickets were sold that night for seats in the balcony than for seats in the orchestra ? | "orchestra seats - a balcony seats - b a + b = 360 and 12 a + 8 b = 3320 solving equations simultaneously ( multiply equation 1 with 8 and subtract from second equation ) 4 a = 3320 - 8 * 360 = 3320 - 2880 = 440 i . e . a = 110 and b = 360 - 110 = 250 more seats in balcony than orchestra = b - a = 250 - 110 = 140 answer : option d" | a = 12 * 360
b = 3 * 1000
c = 3 * 100
d = b + c
e = 2 * 10
f = d + e
g = a - f
h = 12 - 8
i = g / h
j = 12 * 360
k = 3 * 1000
l = 3 * 100
m = k + l
n = 2 * 10
o = m + n
p = j - o
q = 12 - 8
r = p / q
s = 360 - r
t = i - s
|
a ) 120 metres , b ) 180 metres , c ) 324 metres , d ) 90 metres , e ) 100 meters | d | multiply(divide(multiply(36, const_1000), const_3600), 9) | a train running at the speed of 36 km / hr crosses a pole in 9 seconds . what is the length of the train ? | "speed = 36 x 5 / 18 m / sec = 30 / 3 m / sec . length of the train = ( speed x time ) . length of the train = 30 / 3 x 9 m = 90 m . answer : d" | a = 36 * 1000
b = a / 3600
c = b * 9
|
a ) 4 , b ) 18 , c ) 29 , d ) 8 , e ) 116 | d | divide(add(power(56, 2), power(56, 2)), power(28, 2)) | tough and tricky questions : arithmetic . ( 56 ^ 2 + 56 ^ 2 ) / 28 ^ 2 = | ans is 8 my approach was : ( 56 ^ 2 + 56 ^ 2 ) / 28 ^ 2 = 56 ( 56 + 56 ) / 28 * 28 = 56 * 112 / 28 * 28 = 2 * 4 = 8 d | a = 56 ** 2
b = 56 ** 2
c = a + b
d = 28 ** 2
e = c / d
|
a ) 90 , b ) 110 , c ) 140 , d ) 130 , e ) 120 | a | multiply(divide(480, multiply(const_4, const_4)), const_3) | a boy has rs . 480 in the denominations of one - rupee notes , 5 - rupee notes and 10 - rupee notes . the number of notes of each denomination is equal . what is the total number of notes that he has ? | let number of notes of each denomination be x . then x + 5 x + 10 x = 480 16 x = 480 x = 30 . hence , total number of notes = 3 x = 90 . a | a = 4 * 4
b = 480 / a
c = b * 3
|
a ) 12 , b ) 13 , c ) 14 , d ) 15 , e ) 16 | d | subtract(divide(multiply(130, const_2), subtract(11, 6)), 37) | glen and hannah drive at constant speeds toward each other on a highway . glen drives at a constant speed of 37 km per hour . at a certain time they pass by each other , and then keep driving away from each other , maintaining their constant speeds . if glen is 130 km away from hannah at 6 am , and also 130 km away from hannah at 11 am , then how fast is hannah driving ( in kilometers per hour ) ? | glen and hannah complete 260 km / 5 hours = 52 km / hour hannah ' s speed is 52 - 37 = 15 km / hour the answer is d . | a = 130 * 2
b = 11 - 6
c = a / b
d = c - 37
|
a ) 20 % , b ) 24 % , c ) 30 % , d ) 32 % , e ) 79 % | b | divide(multiply(subtract(add(add(const_100, 10), multiply(add(const_100, 10), divide(20, const_100))), const_100), const_100), add(add(const_100, 10), multiply(add(const_100, 10), divide(20, const_100)))) | the output of a factory was increased by 10 % 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 ? | "let initial output is o then after 10 % increase it will be 1.1 o and after 20 % increase on this new output the latest output will be 1.1 o * 1.20 = 1.32 o now we have to decrease the output by some percentage so that the new output is same as the starting output ( o ) so , 1.32 o * ( 1 - x / 100 ) = o = > x = 24.24 % so , answer will be b" | a = 100 + 10
b = 100 + 10
c = 20 / 100
d = b * c
e = a + d
f = e - 100
g = f * 100
h = 100 + 10
i = 100 + 10
j = 20 / 100
k = i * j
l = h + k
m = g / l
|
a ) 7 , b ) 9 , c ) 13 , d ) 14 , e ) 18 | c | add(const_2, divide(divide(subtract(233, 79), const_2), add(const_3, const_4))) | integer d is the product of the integers a , b , and c and 1 < a < b < c . if the remainder when 233 is divided by d is 79 , what is the value of a + c ? | 79 is the remainder , therefore the number ( a * b * c ) or multiple of the number we need is 79 less than 233 hence 233 - 79 = 154 therefore , 154 is either a number we want or it is a multiple of the number we want but the largest factor of 154 is 77 because 77 * 2 = 154 but 77 is less than 79 hence it can not be a number thus 154 is the number we want a * b * c = 154 now factorize 154 = 1 * 2 * 7 * 11 all are prime also we know 1 < a < b < c thus a = 2 and c = 11 therefore a + c = 2 + 11 = 13 answer : c | a = 233 - 79
b = a / 2
c = 3 + 4
d = b / c
e = 2 + d
|
a ) 10 , b ) 12 , c ) 14 , d ) 16 , e ) 18 | b | divide(subtract(add(multiply(10, divide(subtract(multiply(20, 20), multiply(20, 18)), subtract(20, 10))), multiply(20, subtract(20, divide(subtract(multiply(20, 20), multiply(20, 18)), subtract(20, 10))))), multiply(20, 15)), subtract(20, 15)) | each of the 20 boxes in a certain shipment weighs either 10 pounds or 20 pounds , and the average ( arithmetic mean ) weight of the boxes in the shipment is 18 pounds . if the average weight of the boxes in the shipment is to be reduced to 15 pounds by removing some of the 20 - pound boxes , how many 20 - pound boxes must be removed ? | "if the average of 10 - pound and 20 - pound boxes is 18 , the ratio of 10 - pound boxes : 20 - pound boxes is 1 : 4 . so out of 20 boxes , 4 are 10 - pound boxes and 16 are 20 - pound boxes . if the average of 10 and 20 - pound boxes is to be 15 , the ratio of 10 - pound boxes : 20 - pound boxes should be 1 : 1 . the number of 10 pound boxes remains the same so we still have 4 of them . to get a ratio of 1 : 1 , the number of 20 - pound boxes must be 4 . we need to remove 12 of the 20 - pound boxes . the answer is b ." | a = 20 * 20
b = 20 * 18
c = a - b
d = 20 - 10
e = c / d
f = 10 * e
g = 20 * 20
h = 20 * 18
i = g - h
j = 20 - 10
k = i / j
l = 20 - k
m = 20 * l
n = f + m
o = 20 * 15
p = n - o
q = 20 - 15
r = p / q
|
a ) 120 , b ) 56 , c ) 68 , d ) 87 , e ) 92 | a | multiply(add(const_10, const_2), divide(3600, rectangle_area(18, 20))) | the monthly rent of a shop of dimension 18 feet Γ 20 feet is rs . 3600 . what is the annual rent per square foot of the shop ? | "sol . monthly rent per square feet = 3600 / ( 18 * 20 ) = 10 & annual rent per square feet = 12 * 10 = 120 answer : a" | a = 10 + 2
b = 3600 / rectangle_area
c = a * b
|
a ) 50 kmph , b ) 60 kmph , c ) 70 kmph , d ) 42 kmph , e ) 90 kmph | d | divide(378, multiply(divide(3, 2), 6)) | a van takes 6 hours to cover a distance of 378 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 = 378 3 / 2 of 6 hours = 6 * 3 / 2 = 9 hours required speed = 378 / 9 = 42 kmph d" | a = 3 / 2
b = a * 6
c = 378 / b
|
a ) $ 40,000 , b ) $ 52,000 , c ) $ 64,000 , d ) $ 66,667 , e ) $ 80,000 | b | add(multiply(multiply(const_4, const_10), const_1000), divide(subtract(multiply(multiply(const_4, const_2), const_1000), multiply(divide(14, const_100), multiply(multiply(const_4, const_10), const_1000))), divide(20, const_100))) | country x taxes each of its citizens an amount equal to 14 percent of the first $ 40,000 of income , plus 20 percent of all income in excess of $ 40,000 . if a citizen of country x is taxed a total of $ 8,000 , what is her income ? | "equation is correct , so math must be a problem . 0.14 * 40,000 + 0.2 * ( x - 40,000 ) = 8,000 - - > 5,600 + 0.2 x - 8,000 = 8,000 - - > 0.2 x = 10,400 - - > x = 52,000 . answer : b ." | a = 4 * 10
b = a * 1000
c = 4 * 2
d = c * 1000
e = 14 / 100
f = 4 * 10
g = f * 1000
h = e * g
i = d - h
j = 20 / 100
k = i / j
l = b + k
|
a ) 60 % , b ) 23 % , c ) 25 % , d ) 56 % , e ) 73 % | c | subtract(divide(multiply(add(const_100, 18.75), const_100), subtract(const_100, 5)), const_100) | a shopkeeper sold an article offering a discount of 5 % and earned a profit of 18.75 % . what would have been the percentage of profit earned if no discount was offered ? | "let c . p . be rs . 100 . then , s . p . = rs . 118.75 let marked price be rs . x . then , 95 / 100 x = 118.75 x = 11875 / 95 = rs . 125 now , s . p . = rs . 125 , c . p . = rs . 100 profit % = 25 % . answer : c" | a = 100 + 18
b = a * 100
c = 100 - 5
d = b / c
e = d - 100
|
a ) 2 % , b ) 1 % , c ) 7 % , d ) 5 % , e ) 8 % | c | multiply(divide(divide(const_3, const_4.0), add(multiply(const_3, const_4), add(const_0_25, const_0_25))), const_100) | a sum of 15,600 amounts to 21,200 in 5 years at the rate of simple interest . what is the rate of interest ? | "c 7 % s . i . = ( 21200 - 15600 ) = 5600 . rate = ( 100 x 5600 ) / ( 15600 x 5 ) % = 7 %" | a = 3 / 4
b = 3 * 4
c = const_0_25 + const_0_25
d = b + c
e = a / d
f = e * 100
|
a ) 120 , b ) 75 , c ) 100 , d ) 150 , e ) 175 | a | divide(subtract(multiply(16, const_2), 8), subtract(subtract(subtract(1, divide(20, const_100)), multiply(subtract(1, divide(20, const_100)), divide(1, const_4))), multiply(const_2, divide(20, const_100)))) | a tank holds x gallons of a saltwater solution that is 20 % salt by volume . one fourth of the water is evaporated , leaving all of the salt . when 8 gallons of water and 16 gallons of salt are added , the resulting mixture is 33 1 / 3 % salt by volume . what is the value of x ? | "nope , 150 . i can only get it by following pr ' s backsolving explanation . i hate that . original mixture has 20 % salt and 80 % water . total = x out of which salt = 0.2 x and water = 0.8 x now , 1 / 4 water evaporates and all salt remains . so what remains is 0.2 x salt and 0.6 x water . now 16 gallons salt is added and 8 gallons of water is added . so salt now becomes - > ( 0.2 x + 16 ) and water - - > ( 0.6 x + 8 ) amount of salt is 33.33 % of total . so amount of water is 66.66 % . so salt is half of the volume of water . so ( 0.2 x + 16 ) = ( 0.6 x + 8 ) / 2 = > 0.2 x = 24 solving , x = 120 answer : a" | a = 16 * 2
b = a - 8
c = 20 / 100
d = 1 - c
e = 20 / 100
f = 1 - e
g = 1 / 4
h = f * g
i = d - h
j = 20 / 100
k = 2 * j
l = i - k
m = b / l
|
a ) 5 : 1 , b ) 5 : 5 , c ) 5 : 8 , d ) 5 : 4 , e ) 5 : 2 | b | divide(divide(subtract(multiply(450, const_100), multiply(6000, 7)), subtract(8, 7)), divide(subtract(multiply(450, const_100), multiply(6000, 7)), subtract(8, 7))) | rs . 6000 is lent out in two parts . one part is lent at 7 % p . a simple interest and the other is lent at 8 % p . a simple interest . the total interest at the end of one year was rs . 450 . find the ratio of the amounts lent at the lower rate and higher rate of interest ? | "let the amount lent at 7 % be rs . x amount lent at 8 % is rs . ( 6000 - x ) total interest for one year on the two sums lent = 7 / 100 x + 8 / 100 ( 6000 - x ) = 480 - x / 100 = > 480 - 1 / 100 x = 450 = > x = 3000 amount lent at 10 % = 3000 required ratio = 3000 : 3000 = 5 : 5 answer : b" | a = 450 * 100
b = 6000 * 7
c = a - b
d = 8 - 7
e = c / d
f = 450 * 100
g = 6000 * 7
h = f - g
i = 8 - 7
j = h / i
k = e / j
|
a ) $ 600 , b ) $ 650 , c ) $ 700 , d ) $ 750 , e ) $ 800 | c | divide(subtract(multiply(multiply(const_100, multiply(add(const_2, const_3), const_2)), add(divide(6, const_100), const_1)), add(add(multiply(const_100, multiply(add(const_2, const_3), const_2)), multiply(multiply(add(const_2, const_3), const_2), const_4)), multiply(const_2, const_3))), subtract(add(divide(6, const_100), const_1), add(divide(4, const_100), const_1))) | we invested a total of $ 1,000 . we invested one part of the money at 4 % and the rest of the money at 6 % . the total investment with interest at the end of the year was $ 1,046 . how much money did we invest at 4 % ? | let x be the money invested at 4 % . 1.04 x + 1.06 ( 1000 - x ) = 1046 . 0.02 x = 1060 - 1046 . 0.02 x = 14 . 2 x = 1400 . x = 700 . the answer is c . | a = 2 + 3
b = a * 2
c = 100 * b
d = 6 / 100
e = d + 1
f = c * e
g = 2 + 3
h = g * 2
i = 100 * h
j = 2 + 3
k = j * 2
l = k * 4
m = i + l
n = 2 * 3
o = m + n
p = f - o
q = 6 / 100
r = q + 1
s = 4 / 100
t = s + 1
u = r - t
v = p / u
|
a ) 52 mph , b ) 56.67 mph , c ) 53.33 mph , d ) 64 mph , e ) 66.67 mph | a | add(divide(add(multiply(60, 3), multiply(40, 2)), add(3, 2)), subtract(divide(const_100, 3), const_0_33)) | steve traveled the first 2 hours of his journey at 40 mph and the last 3 hours of his journey at 60 mph . what is his average speed of travel for the entire journey ? | "answer average speed of travel = total distance travelled / total time taken total distance traveled by steve = distance covered in the first 2 hours + distance covered in the next 3 hours . distance covered in the first 2 hours = speed * time = 40 * 2 = 80 miles . distance covered in the next 3 hours = speed * time = 60 * 3 = 180 miles . therefore , total distance covered = 80 + 180 = 260 miles . total time taken = 2 + 3 = 5 hours . hence , average speed = total distance travelled / total time taken = 260 / 5 = 52 miles per hour . choice a" | a = 60 * 3
b = 40 * 2
c = a + b
d = 3 + 2
e = c / d
f = 100 / 3
g = f - const_0_33
h = e + g
|
['a ) 4 cm', 'b ) 8 cm', 'c ) 16 cm', 'd ) 24 cm', 'e ) none of the above'] | a | sqrt(divide(add(8, sqrt(add(multiply(const_4, 128), power(8, const_2)))), const_2)) | the volume of a box with a square base is 128 cubic cm . the area of the base of the box is 8 inches more than the height . what is the length of the base of the box ? | l x w x h = 64 l x w - 8 = h l = w solving the 3 equations we get area of base = 16 square cm and height = 8 cm , which means length = 4 cm and width = 4 cm correct answer : a | a = 4 * 128
b = 8 ** 2
c = a + b
d = math.sqrt(c)
e = 8 + d
f = e / 2
g = math.sqrt(f)
|
a ) 0.94 , b ) 9.4 , c ) 0.094 , d ) 94 , e ) none | a | divide(subtract(const_100, 6), const_100) | subtracting 6 % of a from a is equivalent to multiplying a by how much ? | "answer let a - 6 % of a = ab . β ( 94 x a ) / 100 = ab β΄ b = 0.94 correct option : a" | a = 100 - 6
b = a / 100
|
a ) 146 , b ) 148 , c ) 150 , d ) 152 , e ) none of these | c | add(subtract(subtract(const_1000, const_10), multiply(multiply(const_10, multiply(3, 3)), multiply(const_4, const_2))), const_10) | how many 3 digit numbers are completely divisible 6 ? | "explanation : 100 / 6 = 16 , remainder = 4 . hence 2 more should be added to 100 to get the minimum 3 digit number divisible by 6 . = > minimum 3 digit number divisible by 6 = 100 + 2 = 102 999 / 6 = 166 , remainder = 3 . hence 3 should be decreased from 999 to get the maximum 3 digit number divisible by 6 . = > maximum 3 digit number divisible by 6 = 999 - 3 = 996 hence , the 3 digit numbers divisible by 6 are 102 , 108 , 114 , . . . 996 this is arithmetic progression with a = 102 , d = 6 , l = 996 number of terms = ( l β a ) / d + 1 = ( 996 β 102 ) / 6 + 1 = 894 / 6 + 1 = 149 + 1 = 150 . answer : option c" | a = 1000 - 10
b = 3 * 3
c = 10 * b
d = 4 * 2
e = c * d
f = a - e
g = f + 10
|
a ) 52 , b ) 35 , c ) 42 , d ) 41 , e ) 43 | e | subtract(divide(100, const_2), multiply(7, 7)) | what is the remainder if 7 ^ 3 is divided by 100 ? | "7 * 7 * 7 / 100 = 343 / 100 remainder 43 answer : e" | a = 100 / 2
b = 7 * 7
c = a - b
|
a ) 20 m , b ) 28 m , c ) 22.5 m , d ) 9 m , e ) 12 m | b | multiply(divide(140, 45), subtract(45, 36)) | in 140 m race , a covers the distance in 36 seconds and b in 45 seconds . in this race a beats b by : | "distance covered by b in 9 sec . = 140 / 45 x 9 m = 28 m . a beats b by 28 metres . answer : option b" | a = 140 / 45
b = 45 - 36
c = a * b
|
a ) 22 , b ) 24 , c ) 44 , d ) 48 , e ) 52 | c | divide(subtract(multiply(const_2, multiply(const_2, const_12)), const_4), const_2) | how many times in a day , the hands of a clock are straight ? | "in 12 hours , the hands coincide or are in opposite direction 22 times . in 24 hours , the hands coincide or are in opposite direction 44 times a day . answer : option c" | a = 2 * 12
b = 2 * a
c = b - 4
d = c / 2
|
['a ) 500 ( Ο β 3 )', 'b ) 500 ( Ο β 2.5 )', 'c ) 500 ( Ο β 2 )', 'd ) 550 ( Ο β 1.5 )', 'e ) 500 ( Ο β 1 )'] | d | divide(volume_cylinder(5, 22), const_2) | a right circular cylinder has a height of 22 and a radius of 5 . a rectangular solid with a height of 15 and a square base , is placed in the cylinder such that each of the corners of the solid is tangent to the cylinder wall . liquid is then poured into the cylinder such that it reaches the rim . what is the volume of the liquid ? | [ quote = bunuel ] a right circular cylinder has a height of 20 and a radius of 5 . a rectangular solid with a height of 15 and a square base , is placed in the cylinder such that each of the corners of the solid is tangent to the cylinder wall . liquid is then poured into the cylinder such that it reaches the rim . what is the volume of the liquid ? the square base has sides of sqrt ( 50 ) due to the 45 - 45 - 90 triangle 22 * 25 * pi - 15 * sqrt ( 50 ) ^ 2 = 550 ( Ο β 1.5 ) d . 500 ( Ο β 1.5 ) | a = volume_cylinder / (
|
a ) 90 , b ) 120 , c ) 160 , d ) 360 , e ) 560 | e | divide(140, subtract(const_1, divide(3, 4))) | the visitors of a modern art museum who watched a certain picasso painting were asked to fill in a short questionnaire indicating whether they had enjoyed looking at the picture and whether they felt they had understood it . according to the results of the survey , all 140 visitors who did not enjoy the painting also did not feel they had understood the painting , and the number of visitors who enjoyed the painting was equal to the number of visitors who felt they had understood the painting . if 3 / 4 of the visitors who answered the questionnaire both enjoyed the painting and felt they had understood the painting , then how many visitors answered the questionnaire ? | "if we exclude those cases and take the question at face value , then it seems straightforward . group # 1 = ( did n ' t like , did n ' t understand ) = 120 group # 2 = ( likeunderstood ) = 3 / 4 ( 1 / 4 ) n = 560 n = 480 answer = ( e )" | a = 3 / 4
b = 1 - a
c = 140 / b
|
a ) 870 , b ) 600 , c ) 287 , d ) 771 , e ) 191 | a | subtract(multiply(430, 9), subtract(multiply(430, 9), 870)) | the average monthly salary of 8 workers and one supervisor in a factory was 430 . @ sswhen @ ssthe @ sssupervisor @ cc @ sswhose @ sssalary @ sswas @ ss 430 . @ sswhen @ ssthe @ sssupervisor @ cc @ sswhose @ sssalary @ sswas @ ss 430 . whenthesupervisor , whosesalarywas 430 . when the supervisor , whose salary was 870 per month , retired , a new person was appointed and then the average salary of 9 people was $ $ 430 per month . the salary of the new supervisor is : | "explanation : total salary of 8 workers and supervisor together = 9 Γ£ β 430 = 3870 now total salary of 8 workers = 3870 Γ’ Λ β 870 = 3000 total salary of 9 workers including the new supervisor = 9 Γ£ β 430 = 3870 salary of the new supervisor = 3870 Γ’ Λ β 3000 = 870 answer : a" | a = 430 * 9
b = 430 * 9
c = b - 870
d = a - c
|
a ) 12 , b ) 15 , c ) 18 , d ) 21 , e ) 24 | a | multiply(factorial(3), factorial(2)) | 3 men and 2 women are lined up in a row . what is the number of cases where they stand with each other in turn ? ( the number of cases in which men ( or women ) do not stand next to each other ) | the list should be wmwmw . hence , from women 2 ! and men 3 ! , we get ( 2 ! ) ( 3 ! ) = 12 . therefore , the correct answer is a . | a = math.factorial(3)
b = math.factorial(2)
c = a * b
|
a ) 7 , b ) 8 , c ) 9 , d ) 10 , e ) 11 | c | floor(divide(40, divide(10, const_2))) | what is the greatest integer m for which the number 40 ! / 10 ^ m is an integer ? | "10 ^ m = 2 ^ m * 5 ^ m . let ' s figure out how many 5 ' s are in the prime factorization of 40 ! the multiples of 5 are : 5 , 10 , 15 , 20 , 5 * 5 , 30 , 35 , 40 . thus 5 ^ 9 will divide 40 ! but 5 ^ 10 will not . clearly 2 ^ 9 will divide 40 ! so m = 9 is the largest possible integer . the answer is c ." | a = 10 / 2
b = 40 / a
c = math.floor(b)
|
a ) 20 cm , b ) 25 cm , c ) 35 cm , d ) 50 cm , e ) none of these | b | divide(multiply(4, 50), multiply(40, 20)) | 50 men took a dip in a water tank 40 m long and 20 m broad on a religious day . if the average displacement of water by a man is 4 m 3 , then the rise in the water level in the tank will be : | "explanation : total volume of water displaced = ( 4 x 50 ) m 3 = 200 m 3 rise in water level = 200 / 40 Γ 20 = 0.25 m = 25 cm answer : b" | a = 4 * 50
b = 40 * 20
c = a / b
|
a ) 10.5 , b ) 0.5 , c ) 25.5 , d ) 30 , e ) 60 | b | subtract(subtract(divide(const_60, const_2), 25), divide(const_60, divide(20, 1.5))) | darcy lives 1.5 miles from work . she can walk to work at a constant rate of 3 miles per hour , or she can ride the train to work at a constant rate of 20 miles per hour . if she rides the train , there is an additional x minutes spent walking to the nearest train station , waiting for the train , and walking from the final train station to her work . if it takes darcy a total of 25 more minutes to commute to work by walking than it takes her to commute to work by riding the train , what is the value of x ? | "the time it takes darcy to walk to work is ( 1.5 / 3 ) * 60 = 30 minutes the time it takes darcy to take the train is ( 1.5 / 20 ) * 60 + x = 4.5 + x minutes it takes 15 minutes longer to walk , so 30 = 4.5 + x + 25 x = 0.5 minutes answer : b" | a = const_60 / 2
b = a - 25
c = 20 / 1
d = const_60 / c
e = b - d
|
a ) 26 , b ) 24 , c ) 21 , d ) 28 , e ) 20 | c | divide(log(multiply(power(8, 7), power(5, 21))), log(const_10)) | if 5 ^ 21 x 8 ^ 7 = 10 ^ n what is the value of n ? | 5 ^ 21 * ( 2 ^ 3 ) ^ 7 = 10 ^ n or 5 ^ 21 * 2 ^ 21 = 10 ^ n or 10 ^ 21 = 10 ^ n n = 21 c | a = 8 ** 7
b = 5 ** 21
c = a * b
d = math.log(c)
e = math.log(10)
f = d / e
|
a ) $ 10,000 , b ) $ 11,200 , c ) $ 12,000 , d ) $ 12,800 , e ) $ 26,400 | e | multiply(floor(multiply(divide(subtract(add(multiply(divide(20, const_100), 60000), multiply(add(divide(40, const_100), 1), subtract(60000, multiply(divide(20, const_100), 60000)))), add(multiply(multiply(2, 20000), divide(20, const_100)), multiply(subtract(multiply(2, 20000), multiply(multiply(2, 20000), divide(20, const_100))), add(divide(40, const_100), 1)))), 60000), const_10)), const_3) | at a certain supplier , a machine of type a costs $ 20000 and a machine of type b costs $ 60000 . each machine can be purchased by making a 20 percent down payment and repaying the remainder of the cost and the finance charges over a period of time . if the finance charges are equal to 40 percent of the remainder of the cost , how much less would 2 machines of type a cost than 1 machine of type b under this arrangement ? | total cost of 2 machines of type a = 20 % of ( cost of 2 machine a ) + remainder + 40 % remainder = 20 % of 40000 + ( 40000 - 20 % of 40000 ) + 40 % of ( 40000 - 20 % of 40000 ) = 52800 total cost of 1 machine of type b = 20 % of ( cost of 1 machine b ) + remainder + 40 % remainder = 20 % of 60000 + ( 60000 - 20 % of 60000 ) + 40 % of ( 60000 - 20 % of 60000 ) = 79200 diff = 79200 - 52800 = 26400 hence , e . | a = 20 / 100
b = a * 60000
c = 40 / 100
d = c + 1
e = 20 / 100
f = e * 60000
g = 60000 - f
h = d * g
i = b + h
j = 2 * 20000
k = 20 / 100
l = j * k
m = 2 * 20000
n = 2 * 20000
o = 20 / 100
p = n * o
q = m - p
r = 40 / 100
s = r + 1
t = q * s
u = l + t
v = i - u
w = v / 60000
x = w * 10
y = math.floor(x)
z = y * 3
|
a ) 2378 , b ) 277 , c ) 208 , d ) 270 , e ) 350 | e | divide(subtract(11600, multiply(16, divide(subtract(multiply(11600, const_2), 11600), subtract(multiply(37, const_2), 16)))), 24) | the wages of 24 men and 16 women amounts to rs . 11600 per day . half the number of men and 37 women earn the same amount per day . what is the daily wage of a man ? | let the wage of a man is m and woman be w . 24 m + 16 w = 11600 12 m + 37 w = 11600 solving we get m = 350 answer : e | a = 11600 * 2
b = a - 11600
c = 37 * 2
d = c - 16
e = b / d
f = 16 * e
g = 11600 - f
h = g / 24
|
a ) 2.5 % , b ) 20 % , c ) 28.3 % , d ) 45.2 % , e ) 73.6 % | b | multiply(divide(multiply(divide(30, const_100), subtract(1, divide(1, 2))), subtract(1, divide(1, 3))), const_100) | a library branch originally contained 18360 volumes , 30 % of which were fiction novels . 1 / 3 of the volumes were transferred to another location and 1 / 2 of the volumes transferred were fiction novels . what percent of the remaining collection was fiction novels ? | "fiction novels = 5,508 transferred to another location = 6,120 transferred fiction novels = 3,060 non transferred fiction novels = 2,448 percent of the remaining collection was fiction novels = 2,448 / ( 18360 - 6120 ) * 100 = > 20 % hence answer will be ( b )" | a = 30 / 100
b = 1 / 2
c = 1 - b
d = a * c
e = 1 / 3
f = 1 - e
g = d / f
h = g * 100
|
['a ) 2998', 'b ) 2799', 'c ) 1782', 'd ) 1485', 'e ) 2780'] | c | multiply(volume_cylinder(divide(3, const_2), 14), 18) | find the expenditure on digging a well 14 m deep and of 3 m diameter at rs . 18 per cubic meter ? | 22 / 7 * 14 * 3 / 2 * 3 / 2 = 99 m 2 99 * 18 = 1782 answer : c | a = 3 / 2
b = volume_cylinder * (
|
a ) 3616.5 , b ) 3613.5 , c ) 3313.5 , d ) 3616.5 , e ) 3113.5 | b | divide(multiply(multiply(multiply(const_3, const_100), const_100), multiply(5, divide(8.5, multiply(8.5, const_3)))), const_100) | what is the compound interest on rs . 7000 at 8.5 % p . a . compounded half - yearly for 5 years . | "compound interest : a = p ( 1 + r / n ) nt a = 10 , 613.50 c . i . > > 10 , 613.50 - 7000 > > rs . 3613.5 answer : b" | a = 3 * 100
b = a * 100
c = 8 * 5
d = 8 / 5
e = 5 * d
f = b * e
g = f / 100
|
a ) 5 : 3 , b ) 8 : 11 , c ) 8 : 7 , d ) 6 : 5 , e ) 3 : 11 | c | divide(add(multiply(4, divide(14, add(4, 3))), 8), add(multiply(3, divide(14, add(4, 3))), 8)) | the ratio of the ages of mini and minakshi is 4 : 3 . the sum of their ages is 14 years . the ratio of their ages after 8 years will be | "let mini β s age = 4 x and minakshi β s age = 3 x then 4 x + 3 x = 14 x = 2 mini β s age = 8 years and minakshi β s age = 6 years ratio of their ages after 8 years = ( 8 + 8 ) : ( 6 + 8 ) = 16 : 14 = 8 : 7 answer : c" | a = 4 + 3
b = 14 / a
c = 4 * b
d = c + 8
e = 4 + 3
f = 14 / e
g = 3 * f
h = g + 8
i = d / h
|
a ) 400 , b ) 365 , c ) 385 , d ) 315 , e ) 355 | d | divide(subtract(700700, multiply(divide(700700, const_2), divide(const_1, add(const_1, const_4)))), const_2) | a large field of 700700 hectares is divided into two parts . the difference of the areas of the two parts is one - fifth of the average of the two areas . what is the area of the smaller part in hectares ? | "explanation : average of the two areas = 700 / 2 = 350 one - fifth of the average of the two areas = 350 / 5 = 70 β difference of the two areas = 70 = 70 let area of the smaller part = x hectares . then , area of the larger part = x + 70 hectares . x + ( x + 70 ) = 700 β 2 x = 630 β x = 315 answer : option d" | a = 700700 / 2
b = 1 + 4
c = 1 / b
d = a * c
e = 700700 - d
f = e / 2
|
a ) 15 % , b ) 25 % , c ) 35 % , d ) 40 % , e ) 55 % | e | add(multiply(divide(12, 48), const_100), 30) | the purchase price of an article is $ 48 . in order to include 30 % of cost for overhead and to provide $ 12 of net profit , the markup should be | "cost price of article = 48 $ % of overhead cost = 30 net profit = 12 $ we need to calculate % markup net profit as % of cost price = ( 12 / 48 ) * 100 = 25 % total markup should be = 25 + 30 = 55 % answer e" | a = 12 / 48
b = a * 100
c = b + 30
|
a ) 201 , b ) 159 , c ) 179 , d ) 189 , e ) 209 | e | divide(multiply(subtract(23, const_1), subtract(subtract(23, const_1), const_3)), const_2) | how many diagonals does a polygon with 23 sides have , if one of its vertices does not connect to any diagonal ? | "if i calculate it using the formulae , # diagonals = n ( n - 3 ) / 2 each vertex sends of n - 3 diagonals n = 23 - 1 then 22 * ( 22 - 3 ) / 2 = 209 correct option : e" | a = 23 - 1
b = 23 - 1
c = b - 3
d = a * c
e = d / 2
|
a ) $ 220 , b ) $ 230 , c ) $ 240 , d ) $ 200 , e ) $ 250 | d | subtract(260, multiply(divide(subtract(350, 260), 3), 2)) | mary invested a certain sum of money in a bank that paid simple interest . the amount grew to $ 260 at the end of 2 years . she waited for another 3 years and got a final amount of $ 350 . what was the principal amount that she invested at the beginning ? | what shall be the rate of interest . ? does that is not required for the calculation ? not really ! keep in mind that the interest earned each year will be the same in simple interest . at the end of 2 years , amount = $ 260 at the end of 5 years , amount = $ 350 this means she earned an interest of $ 90 in 3 years . or $ 30 in each year . we know that the interest earned each year will be the same . therefore she must have earned $ 60 in 2 years . hence principal amount = $ 260 - $ 60 = $ 200 option d | a = 350 - 260
b = a / 3
c = b * 2
d = 260 - c
|
a ) 2.8 liters . , b ) 2.5 liters . , c ) 8.5 liters . , d ) 6.25 liters . , e ) 2.1 liters . | d | divide(multiply(divide(subtract(const_100, 90), const_100), 50), divide(subtract(const_100, 20), const_100)) | heinz produces tomato puree by boiling tomato juice . the tomato puree has only 20 % water while the tomato juice has 90 % water . how many liters of tomato puree will be obtained from 50 litres of tomato juice ? | "answer : explanation : in each of the solutions , there is a pure tomato component and some water . so while boiling , water evaporates but tomato not . so we equate tomato part in the both equations . Γ’ β‘ β Γ’ β‘ β 10 % ( 50 ) = 80 % ( x ) Γ’ β‘ β Γ’ β‘ β x = 6.25 liters . answer : d" | a = 100 - 90
b = a / 100
c = b * 50
d = 100 - 20
e = d / 100
f = c / e
|
a ) 44 , b ) 50 , c ) 28 , d ) 27 , e ) 18 | a | add(22, 24) | laxmi and prasanna set on a journey . laxmi moves northwards at a speed of 22 kmph and prasanna moves southward at a speed of 24 kmph . how far will be prasanna from laxmi after 60 minutes ? | "explanation : we know 60 min = 1 hr total northward laxmi ' s distance = 22 kmph x 1 hr = 22 km total southward prasanna ' s distance = 24 kmph x 1 hr = 24 km total distance between prasanna and laxmi is = 22 + 24 = 44 km . answer : a" | a = 22 + 24
|
a ) 2 : 1 , b ) 3 : 1 , c ) 5 : 2 , d ) 1 : 1 , e ) none of these | a | divide(1, subtract(divide(60, 40), 1)) | a good train and a passenger train are running on parallel tracks in the same direction . the driver of the goods train observes that the passenger train coming from behind overtakes and crosses his train completely in 60 sec . whereas a passenger on the passenger train marks that he crosses the goods train in 40 sec . if the speeds of the trains be in the ratio 1 : 2 . find the ratio of their lengths . | explanation : let , the speeds of the two trains be s and 2 s m / s respectively . also , suppose that the lengths of the two trains are p and q metres respectively . then , = > ( p + q ) / ( 2 s β s ) = 60 . - - - - - - - - - - - - - - ( 1 ) and , = > p / ( 2 s β s ) = 40 . - - - - - - - - - - - - - - ( 2 ) on dividing ( 1 ) by ( 2 ) , we get : - = > ( p + q ) / p = 60 / 40 . the required ratio is p : q = 2 : 1 . answer : a | a = 60 / 40
b = a - 1
c = 1 / b
|
a ) 380 , b ) 350 , c ) 333 , d ) 310 , e ) none of these | c | multiply(3164, power(add(const_4, const_1), const_4)) | ( 3164 + 6160 ) / 28 | "explanation : as per bodmas rule , first we will solve the equation in bracket then we will go for division = ( 9324 ) / 28 = 333 option c" | a = 4 + 1
b = a ** 4
c = 3164 * b
|
a ) $ 200 , b ) $ 220 , c ) $ 285 , d ) $ 300 , e ) $ 360 | c | subtract(subtract(600, divide(multiply(600, 2), 5)), multiply(subtract(subtract(15, divide(multiply(15, 3), 5)), const_1), 15)) | a prize of $ 600 is to be distributed among 15 winners , each of whom must be awarded at least $ 15 . if 2 / 5 of the prize will be distributed to 3 / 5 of the winners , what is the greatest possible individual award ? | "total value of the prize = $ 600 number of people = 15 2 / 5 of 600 ( = $ 240 ) should be distributed among 3 / 5 of 15 ( = 9 people ) with each getting $ 15 each . remaining money = 600 - 240 = $ 360 . now in order to ' maximize ' 1 prize , we need to minimise the others and we have been given that each should get $ 15 . thus , minimising the remaining 5 people ( = 15 - 9 - 1 . ' - 1 ' to exclude 1 that needs to be maximised ) = 5 * 15 = 75 . thus the maximum award can be = 360 - 75 = $ 285 , hence c is the correct answer ." | a = 600 * 2
b = a / 5
c = 600 - b
d = 15 * 3
e = d / 5
f = 15 - e
g = f - 1
h = g * 15
i = c - h
|
a ) 675 , rs . 2025 , b ) 575 , rs . 1350 , c ) 1350 , rs . 675 , d ) 1450 , rs . 775 , e ) 1550 , rs . 875 | a | multiply(subtract(rectangle_area(add(75, multiply(2.5, const_2)), add(55, multiply(2.5, 3))), rectangle_area(75, 55)), 3) | a rectangular grass field is 75 m * 55 m , it has a path of 2.5 m wide all round it on the outside . find the area of the path and the cost of constructing it at rs . 3 per sq m ? | "area = ( l + b + 2 d ) 2 d = ( 75 + 55 + 2.5 * 2 ) 2 * 2.5 = > 675 675 * 3 = rs . 2025 answer : a" | a = 2 * 5
b = 75 + a
c = 2 * 5
d = 55 + c
e = rectangle_area - (
f = e * rectangle_area
|
a ) 10 ΒΊ , b ) 75 ΒΊ , c ) 180 ΒΊ , d ) 270 ΒΊ , e ) 360 ΒΊ | b | divide(multiply(subtract(multiply(divide(multiply(const_3, const_4), subtract(multiply(const_3, const_4), const_1)), multiply(add(const_4, const_1), subtract(multiply(const_3, const_4), const_1))), divide(const_60, const_2)), subtract(multiply(const_3, const_4), const_1)), const_2) | the angle between the minute hand and the hour hand of a clock when the time is 8.30 , is : | "angle traced by hour hand in 17 / 2 hrs = ( 360 / 12 x 17 / 2 ) ΒΊ = 255 . angle traced by min . hand in 30 min . = ( 360 / 60 x 30 ) ΒΊ = 180 . required angle = ( 255 - 180 ) ΒΊ = 75 ΒΊ . answer b" | a = 3 * 4
b = 3 * 4
c = b - 1
d = a / c
e = 4 + 1
f = 3 * 4
g = f - 1
h = e * g
i = d * h
j = const_60 / 2
k = i - j
l = 3 * 4
m = l - 1
n = k * m
o = n / 2
|
a ) 3775 , b ) 3665 , c ) 3456 , d ) 3459 , e ) 3569 | a | multiply(divide(add(100, 51), const_2), subtract(51, const_1)) | evaluate ( 51 + 52 + 53 + . . . + 100 ) | sn = ( 1 + 2 + 3 + . . . + 50 + 51 + 52 + . . . + 100 ) - ( 1 + 2 + 3 + . . . + 50 ) = ( 50 x 101 ) - ( 25 x 51 ) = ( 5050 - 1275 ) = 3775 . option a | a = 100 + 51
b = a / 2
c = 51 - 1
d = b * c
|
a ) 18 , b ) 16 , c ) 12 , d ) 10 , e ) 6 | e | subtract(90, add(add(38, const_1), 45)) | 90 people are attending a newspaper conference . 45 of them are writers and more than 38 are editors . of the people at the conference , x are both writers and editors and 2 x are neither . what is the largest possible number of people who are both writers and editors ? | { total } = { writers } + { editors } - { both } + { neither } . { total } = 90 ; { writers } = 45 ; { editors } > 38 ; { both } = x ; { neither } = 2 x ; 90 = 45 + { editors } - x + 2 x - - > x = 45 - { editors } . we want to maximize x , thus we should minimize { editors } , minimum possible value of { editors } is 39 , thus x = { both } = 45 - 39 = 6 . answer : e . | a = 38 + 1
b = a + 45
c = 90 - b
|
a ) 23 , b ) 37 , c ) 28 , d ) 44 , e ) 81 | d | subtract(const_100, subtract(add(34, 44), 22)) | in an examination , 34 % of total students failed in hindi , 44 % failed in english and 22 % in both . the percentage of these who passed in both the subjects is : | explanation : formula n ( a Γ’ Λ Βͺ b ) = n ( a ) + n ( b ) Γ’ Λ β n ( a Γ’ Λ Β© b ) fail in hindi or english = 34 + 44 Γ’ β¬ β 22 = 56 therefore students who passed = 100 Γ’ β¬ β 56 = 44 . answer : d | a = 34 + 44
b = a - 22
c = 100 - b
|
a ) 50 , b ) 45 , c ) 40 , d ) 35 , e ) 30 | e | add(divide(multiply(3, 12), 3), divide(multiply(3, 12), subtract(5, 3))) | nicky and cristina are running a 500 meter race . since cristina is faster than nicky , she gives him a 12 second head start . if cristina runs at a pace of 5 meters per second and nicky runs at a pace of only 3 meters per second , how many seconds will nicky have run before cristina catches up to him ? | "the distance traveled by both of them is the same at the time of overtaking . 3 ( t + 12 ) = 5 t t = 18 . cristina will catch up nicky in 18 seconds . so in 18 seconds cristina would cover = 18 * 5 = 90 meter . now time taken my nicky to cover 90 meter = 90 / 3 = 30 seconds . e" | a = 3 * 12
b = a / 3
c = 3 * 12
d = 5 - 3
e = c / d
f = b + e
|
a ) 1992 , b ) 1993 , c ) 1994 , d ) 1995 , e ) 1996 | e | add(1990, multiply(10, multiply(const_2, const_3))) | in 1990 the budgets for projects q and v were $ 540,000 and $ 780,000 , respectively . in each of the next 10 years , the budget for q was increased by $ 30,000 and the budget for v was decreased by $ 10,000 . in which year was the budget for q equal to the budget for v ? | let the no of years it takes is x . 540 + 30 x = 780 - 10 x - - > 40 x = 240 and x = 6 . thus , it happens in 1996 . e . | a = 2 * 3
b = 10 * a
c = 1990 + b
|
a ) 320 , b ) 340 , c ) 360 , d ) 380 , e ) 400 | d | multiply(divide(multiply(divide(multiply(divide(320, 5), 5), 2), 19), 24), 5) | there is a train and car . the ratio between the speed of a train & a car is 24 : 19 respectively . also , a bus covered a distance of 320 km in 5 hours . the speed of the bus is 2 / 3 rd of the speed of the train . how many kilometers will the car cover in 5 hours ? | "the speed of the bus is 320 / 5 = 64 km / hr the speed of the train is ( 64 * 3 ) / 2 = 96 km / hr the speed of the car is 96 / 24 * 19 = 76 km / hr the distance covered by the car in 5 hours is 76 Γ 5 = 380 km the answer is d ." | a = 320 / 5
b = a * 5
c = b / 2
d = c * 19
e = d / 24
f = e * 5
|
a ) 4 , b ) 7 , c ) 10 , d ) 15 , e ) 18 | e | divide(multiply(12, const_3), const_2) | youseff lives x blocks from his office . it takes him 1 minute per block to walk to work and 20 seconds per block to ride his bike to work . it is takes him exactly 12 minutes more to walk to work than to ride his bike to work , then x equals ? | "please follow posting guidelines , link is in my signatures . as for your question , x / 60 = blocks / time / block = block ^ 2 / time . this is not what you want . you are given x blocks and 60 seconds per block . thus you need to put it as 60 * x to give you units of seconds as you are equating this to 720 ( which is time in seconds . ) . thus the correct equation is : 60 * x - 20 * x = 7200 - - - - > 40 x = 720 - - > x = 18 . option e" | a = 12 * 3
b = a / 2
|
a ) 1 , b ) 2 , c ) 3 , d ) 4 , e ) 5 | c | subtract(4, 2) | buffalo gives 4 liter milk , cow gives ( 1 / 2 ) liter milk and goat gives 1 / 4 liter milk . you have to get 20 liter milk by 20 animals . what is the number of buffalos ? | "assume number of respective animals are x , y , z . x + y + z = 20 - - - ( 1 ) as the total number of animal has to be 20 amt of milk will be 4 x + ( 1 / 2 ) y + ( 1 / 4 ) z = 20 - - - ( 2 ) solving equation 1 and 2 we get 15 x + y = 60 - - - - ( 3 ) since buffalo gives 4 litre and total milk is 20 , x < 5 but from eq 3 , x can not be more than 4 ; further if x = 1 or 2 ; y > 20 . . . not possible , since total animal is 20 thus , x = 3 , y = 15 , z = 2 ans : 3 buffalos answer : c" | a = 4 - 2
|
a ) 288 , b ) 560 , c ) 155 , d ) 600 , e ) 441 | b | multiply(subtract(divide(12000, 10000), divide(8000, 10000)), 1400) | a , b and c started a business with capitals of rs . 8000 , rs . 10000 and rs . 12000 respectively . at the end of the year , the profit share of b is rs . 1400 . the difference between the profit shares of a and c is ? | "ratio of investments of a , b and c is 8000 : 10000 : 12000 = 4 : 5 : 6 and also given that , profit share of b is rs . 1400 = > 5 parts out of 15 parts is rs . 1400 now , required difference is 6 - 4 = 2 parts required difference = 2 / 5 ( 1400 ) = rs . 560 answer : b" | a = 12000 / 10000
b = 8000 / 10000
c = a - b
d = c * 1400
|
a ) 7 / 15 , b ) 1 / 5 , c ) 4 / 15 , d ) 1 / 3 , e ) 2 / 5 | a | subtract(divide(1, 2), divide(const_1, multiply(3, const_10))) | of all the students in a certain dormitory , 1 / 2 are first - year students and the rest are second - year students . if 4 / 5 of the first - year students have not declared a major and if the fraction of second - year students who have declared a major is 1 / 3 times the fraction of first - year students who have declared a major , what fraction of all the students in the dormitory are second - year students who have not declared a major ? | tot students = x 1 st year student = x / 2 - - - - > non majaor = 4 / 5 ( x / 2 ) - - - - - > maj = 1 / 5 ( x / 2 ) 2 nd year student = x / 2 - - - - > maj = 1 / 3 ( 1 / 5 ( x / 2 ) ) = 1 / 30 ( x ) - - - > non major = x / 2 - 1 / 30 ( x ) = 7 / 15 ( x ) hence 7 / 15 a | a = 1 / 2
b = 3 * 10
c = 1 / b
d = a - c
|
a ) 42 , b ) 56 , c ) 76 , d ) 84 , e ) 85 | a | add(multiply(multiply(3, const_4.0), const_100), multiply(4, 72)) | two numbers are in the ratio 3 : 4 . if their l . c . m . is 72 . what is sum of the numbers ? | "explanation : let the numbers be 3 x and 4 x lcm of 3 x and 4 x = 12 x ( since lcm of 3 and 4 is 12 . hence lcm of 3 x and 4 x is 12 x ) given that lcm of 3 x and 4 x is 72 = > 12 x = 72 = > x = 72 / 12 = 6 sum of the numbers = 3 x + 4 x = 7 x = 7 x 6 = 42 answer : option a" | a = 3 * 4
b = a * 100
c = 4 * 72
d = b + c
|
a ) 9 % , b ) 10 % , c ) 11 % , d ) 12 % , e ) 13 % | c | multiply(divide(subtract(9990, 9000), 9000), const_100) | a sum of money deposited at c . i . amounts to rs . 9000 in 10 years and to rs . 9990 in 11 years . find the rate percent ? | "9000 - - - 990 100 - - - ? = > 11 % answer : c" | a = 9990 - 9000
b = a / 9000
c = b * 100
|
a ) 125 , b ) 280 , c ) 384 , d ) 480 , e ) 500 | d | divide(multiply(multiply(250, 16), 6), subtract(multiply(5, 16), multiply(5, 6))) | in a maths test , students were asked to find 5 / 16 of a certain number . one of the students by mistake found 5 / 6 th of that number and his answer was 250 more than the correct answer . find the number . | "explanation : let the number be x . 5 * x / 6 = 5 * x / 16 + 250 25 * x / 48 = 250 x = 480 answer d" | a = 250 * 16
b = a * 6
c = 5 * 16
d = 5 * 6
e = c - d
f = b / e
|
a ) 132 , b ) 39 , c ) 42 , d ) 65 , e ) 156 | a | multiply(multiply(multiply(power(2, 2), 3), divide(11, 2)), 2) | if 2 ^ 5 , 3 ^ 3 , and 11 ^ 2 are all factors of the product of 936 and w where w is a positive integer , what is the smallest possible value of w ? | "here 156 has three two ' s two three ' s and one 11 rest of them must be in w so w = 11 * 3 * 4 = 132 smash a" | a = 2 ** 2
b = a * 3
c = 11 / 2
d = b * c
e = d * 2
|
a ) 132 Β° f , b ) 140 Β° f , c ) 148 Β° f , d ) 156 Β° f , e ) 164 Β° f | b | add(multiply(divide(subtract(212, 32), 100), 60), 32) | water boils at 212 Β° f or 100 Β° c and ice melts at 32 Β° f or 0 Β° c . if the temperature of a pot of water is 60 Β° c , what is the temperature of the pot of water in Β° f ? | "let f and c denote the temperature in fahrenheit and celsius respectively . ( f - 32 ) / ( 212 - 32 ) = ( c - 0 ) / ( 100 - 0 ) f = 9 c / 5 + 32 f = 9 ( 60 ) / 5 + 32 = 140 Β° f the answer is b ." | a = 212 - 32
b = a / 100
c = b * 60
d = c + 32
|
a ) 2 , b ) 4 , c ) 8 , d ) 256 , e ) 32 | d | power(4, multiply(const_4, 1)) | xy = 1 then what is ( 4 ^ ( x + y ) ^ 2 ) / ( 4 ^ ( x - y ) ^ 2 ) | "( x + y ) ^ 2 - ( x - y ) ^ 2 ( x + y + x - y ) ( x + y - x + y ) ( 2 x ) ( 2 y ) 4 xy 4 4 ^ 4 = 256 answer d" | a = 4 * 1
b = 4 ** a
|
a ) 150 cm , b ) 140 cm , c ) 142 cm , d ) 148 cm , e ) 146 cm | b | subtract(154, divide(multiply(154, 10), const_100)) | on my sister ' s birthday , she was 154 cm in height , having grown 10 % since the year before . how tall was she the previous year ? | "let the previous year ' s height be x . 1.1 x = 154 x = 140 the answer is b ." | a = 154 * 10
b = a / 100
c = 154 - b
|
a ) 3 / 5 , b ) 5 / 9 , c ) 1 / 24 , d ) 4 / 9 , e ) 7 / 15 | e | divide(subtract(divide(16, const_2), const_1), subtract(16, const_1)) | an empty wooden vessel weighs 16 % of its total weight when filled with paint . if the weight of a partially filled vessel is one half that of a completely filled vessel , what fraction of the vessel is filled . | "an empty wooden vessel weighs 16 % of its total weight when filled with paint : vessel = 0.16 ( vessel + paint ) ; 16 v = v + p ( so the weight of completely filled vessel is 16 v ) p = 15 v ( so the weight of the paint when the vessels is completely filled is 15 v ) . the weight of a partially filled vessel is one half that of a completely filled vessel : v + p ' = 1 / 2 * 16 v ; p ' = 7 v ( so the weight of the paint when the vessels is partially filled is 7 v ) . what fraction of the vessel is filled ? so , we need to find the ratio of the weight of the paint when the vessel iscompletely filledto the weight of the paint when the vessel ispartially filled : p ' / p = 7 v / 15 v = 7 / 15 . answer : e ." | a = 16 / 2
b = a - 1
c = 16 - 1
d = b / c
|
a ) 5 seconds , b ) 4.5 seconds , c ) 3 seconds , d ) 2.3 seconds , e ) none of these | d | divide(120, multiply(184, const_0_2778)) | in what time will a train 120 meters long cross an electric pole , if its speed is 184 km / hr | "explanation : first convert speed into m / sec speed = 184 * ( 5 / 18 ) = 51 m / sec time = distance / speed = 120 / 51 = 2.3 seconds option d" | a = 184 * const_0_2778
b = 120 / a
|
a ) 4 / 99 , b ) 2 / 25 , c ) 8 / 99 , d ) 49 / 100 , e ) 86 / 99 | e | divide(subtract(add(multiply(divide(const_100, 4), const_2), multiply(divide(const_100, 5), const_2)), 4), subtract(const_100, 1)) | if x is a positive integer with fewer than 3 digits , what is the probability r that x * ( x + 1 ) is a multiple of either 4 or 5 ? | "interesting question ! also one that we should be able to answer very quickly be keeping an eye on our best friends , the answer choices . we know that x belongs to the set { 1 , 2 , 3 , . . . , 99 } . we want to know the probability r that x ( x + 1 ) is a multiple of either 4 or 5 . when will this happen ? if either x or ( x + 1 ) is a multiple of 4 or 5 . since 4 * 5 is 20 , let ' s look at the first 20 numbers to get a rough idea of how often this happens . out of the numbers from 1 to 20 : 4 , 5 , 6 , 8 , 9 , 10 , 11 , 12 , 13 , 15 , 16 , 17 , 20 so , 14 out of the first 20 numbers match our criteria . since : probability = ( # of desired outcomes ) / ( total # of possibilities ) , we guesstimate the answer to be 14 / 20 . since ( e ) is the only answer greater than 1 / 2 , we go with ( e ) ." | a = 100 / 4
b = a * 2
c = 100 / 5
d = c * 2
e = b + d
f = e - 4
g = 100 - 1
h = f / g
|
a ) 432 , b ) 288 , c ) 376 , d ) 397 , e ) 592 | a | multiply(multiply(divide(96, add(multiply(const_3, const_2), multiply(const_1, const_2))), const_3), divide(96, add(multiply(const_3, const_2), multiply(const_1, const_2)))) | the length of rectangle is thrice its breadth and its perimeter is 96 m , find the area of the rectangle ? | "2 ( 3 x + x ) = 96 l = 36 b = 12 lb = 36 * 12 = 432 answer : a" | a = 3 * 2
b = 1 * 2
c = a + b
d = 96 / c
e = d * 3
f = 3 * 2
g = 1 * 2
h = f + g
i = 96 / h
j = e * i
|
a ) 2 , b ) 3 , c ) 4 , d ) 5 , e ) 6 | b | add(subtract(12, 11), 2) | there are 12 pieces of radioactive metal a that look identical . 11 of the pieces give the same radiation count when measured , the 12 th piece is a counterfeit and gives a different radiation level , which may be more or less than the other 11 . we are given a radiation scale , which can take 2 sets of samples and compare their added up radiation levels to tell us if the sums are the same or if different , which set has the higher level of radiation . what is the minimum number of comparisons we need on this scale to identify the counterfeit sample and to also determine whether it has more or less radiation than the other samples ? | first of all if you are down to just 3 pieces and you know that if the offending piece is less or more active , then it takes exactly 1 measurement to find out the offending piece . so you know you have to reduce the problem to three . now when you are down to either a or b after measurement 1 , you need the next measurement to ( a ) reduce the problem set to 3 and ( b ) to know whether anser is more or less . now you can not compare a group of 4 to 4 , as in the best case it will only reduce the problem to 4 elements which is not good enough . if you have to choose a set of 3 to compare , you can not pick any 3 on the same side from the same set ( a or b ) because if you do this , a quick check will show you that in every choice there is a case where you can only get down to 4 elements . eg . if you weighed { 1 , 23 } v / s { 5 , 910 } and they were equal you ' re problem would only reduce to { 46 , 78 } the easiest way to solve this then is to compare 3 to 3 , and make sure each side has elements from both ab such that whatever the measurement outcome in the worst case the problem reduces to 3 elements only . which is why the sets { 1 , 59 } and { 2 , 67 } or { a , b , c } { a , b , b } . the extra element from c is just taken to make the problem symmetric so to say , we have 8 elements and we make it 9 , to compose 3 sets of 3 each . = b | a = 12 - 11
b = a + 2
|
a ) 2 / 3 , b ) 5 / 8 , c ) 5 / 4 , d ) 5 / 6 , e ) 6 / 7 | b | divide(add(1, add(divide(1, const_2), add(divide(1, add(1, 5)), divide(5, add(1, 5))))), add(1, 5)) | in a bag containing 5 balls , a white ball was placed and then 1 ball was taken out at random . what is the probability that the extracted ball would turn on to be white , if all possible hypothesis concerning the color of the balls that initiallyin the bag wereequally possible ? | "since , all possible hypothesis regarding the colour of the balls are equally likely , therefore these could be 3 white balls , initially in the bag . β΄ required probability = 1 / 4 [ 1 + 3 / 4 + 1 / 2 + 1 / 4 ] = 1 / 4 [ ( 4 + 3 + 2 + 1 ) / 4 ] = 5 / 8 b" | a = 1 / 2
b = 1 + 5
c = 1 / b
d = 1 + 5
e = 5 / d
f = c + e
g = a + f
h = 1 + g
i = 1 + 5
j = h / i
|
a ) 20 cs , b ) cs / 2 , c ) 60 cs , d ) ( 2 cs ) / 12 , e ) ( 24 c ) / s | c | multiply(5, const_12) | a certain school implemented a reading program for its students , with the goal of getting each student to read 5 books per month year - round . if the school has c classes made up of s students in each class , how many books will the entire student body read in one year ? | "ans : c solution : simple multiplication s students , c classes , 5 books / month = 60 books a year total number of books = 60 cs" | a = 5 * 12
|
a ) 14 , b ) 17 , c ) 11 , d ) 19 , e ) 99 | b | subtract(divide(44, const_2), 5) | a father said his son , ` ` i was as old as you are at present at the time of your birth . ` ` if the father age is 44 now , the son age 5 years back was | "let the son ' s present age be x years . then , ( 44 - x ) = x x = 22 . son ' s age 5 years back = ( 22 - 5 ) = 17 years answer : b" | a = 44 / 2
b = a - 5
|
a ) 16 , b ) 32 , c ) 64 , d ) 96 , e ) 128 | c | multiply(16, multiply(16, divide(4, multiply(4, 4)))) | 4 weavers can weave 4 mats in 4 days . at the same rate , how many mats would be woven by 16 weavers in 16 days ? | "1 weaver can weave 1 mat in 4 days . 16 weavers can weave 16 mats in 4 days . 16 weavers can weave 64 mats in 16 days . the answer is c ." | a = 4 * 4
b = 4 / a
c = 16 * b
d = 16 * c
|
a ) 16 , b ) 20 , c ) 32 , d ) 40 , e ) 48 | d | divide(subtract(15, multiply(divide(20, const_100), 65)), subtract(divide(25, const_100), divide(20, const_100))) | a bowl of nuts is prepared for a party . brand p mixed nuts are 20 % almonds and brand q ' s deluxe nuts are 25 % almonds . if a bowl contains a total of 65 ounces of nuts , representing a mixture of both brands , and 15 ounces of the mixture are almonds , how many ounces of brand q ' s deluxe mixed nuts are used ? | lets say x ounces of p is mixed with q . = > 65 - x ounces of q is present in the mixture ( as the total = 65 ounces ) given total almond weight = 15 ounces ( 20 x / 100 ) + ( 25 / 100 ) ( 65 - x ) = 15 = > x = 25 = > 65 - 25 = 40 ounces of q is present in the mixture . answer is d . | a = 20 / 100
b = a * 65
c = 15 - b
d = 25 / 100
e = 20 / 100
f = d - e
g = c / f
|
a ) 1 / 3 , b ) ΒΌ , c ) 9 / 25 , d ) 5 / 16 , e ) 0 | d | divide(add(3, const_2), multiply(const_4, const_4)) | if a number n is chosen at random from the set of two - digit integers whose digits are both prime numbers , what is the probability q that n is divisible by 3 ? | "prime digits are : 2 , 3 , 5 , 7 total number of 2 digit # s with both digits prime are : 4 * 4 = 16 out of these numbers divisible by 3 = 33 , 27 , 57 , 72 and 75 . i had to find the numbers manually using the 4 numbers above . = > prob = 5 / 16 . ans d . took me 3 : 20 mins ." | a = 3 + 2
b = 4 * 4
c = a / b
|
a ) 37 min , b ) 55 min , c ) 47 min , d ) 67 min , e ) 45 min | e | divide(add(330, multiply(multiply(const_0_2778, 36), 12)), multiply(const_0_2778, 36)) | a train running at a speed of 36 kmph crosses an electric pole in 12 seconds . in how much time will it cross a 330 m long platform ? | "e 45 min let the length of the train be x m . when a train crosses an electric pole , the distance covered is its own length . so , x = 12 * 36 * 5 / 18 m = 120 m . time taken to cross the platform = ( 120 + 330 ) / 36 * 5 / 18 = 45 min ." | a = const_0_2778 * 36
b = a * 12
c = 330 + b
d = const_0_2778 * 36
e = c / d
|
a ) 3 , b ) 4 , c ) 5 , d ) 6 , e ) 7 | d | subtract(subtract(10, 3), const_1) | if 3 < x < 6 < y < 10 , then what is the greatest possible positive integer difference of x and y ? | "3 < x < 6 < y < 10 ; 3 < x y < 10 3 + y < x + 10 y - x < 7 . positive integer difference is 6 ( for example y = 9.5 and x = 3.5 ) answer : d ." | a = 10 - 3
b = a - 1
|
a ) 8 , b ) 10 , c ) 12 , d ) 14 , e ) - 2 | e | subtract(6, 8) | what is 10 - 8 + 6 - 4 + . . . + ( - 14 ) ? | the expression considers all even numbers between 10 and - 14 with alternate addition and subtraction of the numbers . the numbers to be used are : 10 , 8 , 6 , 4 , 2 , 0 , - 2 , - 4 , - 6 , - 8 , - 10 , - 12 , and - 14 now , the first term is positive and the next term is subtracted . so , the required expression becomes , 10 - 8 + 6 - 4 + 2 - 0 + ( - 2 ) - ( - 4 ) + ( - 6 ) - ( - 8 ) + ( - 10 ) - ( - 12 ) + ( - 14 ) = 10 - 8 + 6 - 4 + 2 - 0 - 2 + 4 - 6 + 8 - 10 + 12 - 14 = 42 - 44 = - 2 hence the correct answer choice is e . | a = 6 - 8
|
a ) rs . 16003 , b ) rs . 16029 , c ) rs . 22500 , d ) rs . 16108 , e ) rs . 16011 | c | subtract(multiply(add(1500, 1000), add(20, const_1)), multiply(1500, 20)) | the average monthly salary of 20 employees in an organisation is rs . 1500 . if the manager ' s salary is added , then the average salary increases by rs . 1000 . what is the manager ' s monthly salary ? | "explanation : manager ' s monthly salary rs . ( 2500 * 21 - 1500 * 20 ) = rs . 22500 . answer : c" | a = 1500 + 1000
b = 20 + 1
c = a * b
d = 1500 * 20
e = c - d
|
a ) 15 mile , b ) 14 mile , c ) 11 mile , d ) 10 mile , e ) 12.5 mile | e | multiply(divide(subtract(30, multiply(24, divide(add(1, 5), const_60))), add(5, add(1, 5))), 5) | stacy and heather are 30 miles apart and walk towards each other along the same route . stacy walks at constant rate that is 1 mile per hour fast than heather ' s constant rate of 5 miles / hour . if heather starts her journey 24 minutes after stacy , how far from the original destination has heather walked when the two meet ? . | "ss - stacy ' s speed = 6 m / hr sh - heather ' s speed = 5 m / hr in 24 minutes stacy will cover = ( 24 / 60 ) * 6 = 2.4 miles now since both are walking in opposite directions , add their speeds - 6 + 5 = 11 m / hr and distance to cover is 30 - 2.4 = 17.6 time taken = distance / speed = 27.6 / 11 = 2.5 hrs heather will cover = 5 * 2.5 = 12.5 miles . answer e" | a = 1 + 5
b = a / const_60
c = 24 * b
d = 30 - c
e = 1 + 5
f = 5 + e
g = d / f
h = g * 5
|
a ) 672 m , b ) 6738 m , c ) 634 m , d ) 671 m , e ) 636 m | a | subtract(224, multiply(8, speed(224, 32))) | for a race a distance of 224 meters can be covered by p in 8 seconds and q in 32 seconds . by what distance does p defeat q eventually ? | "explanation : this is a simple speed time problem . given conditions : = > speed of p = 224 / 8 = 28 m / s = > speed of q = 224 / 32 = 7 m / s = > difference in time taken = 24 seconds therefore , distance covered by p in that time = 28 m / s x 24 seconds = 672 metres answer : a" | a = 8 * speed
b = 224 - a
|
a ) 1 / 3 , b ) 1 / 4 , c ) 2 / 3 , d ) 2 / 5 , e ) 3 / 7 | a | divide(const_2, add(4, const_2)) | in a single throw of a die , what is the probability of getting a number greater than 4 ? | s = { 1,2 , 3,4 , 5,6 } e = { 5,6 } probability = 2 / 6 = 1 / 3 answer is a | a = 4 + 2
b = 2 / a
|
a ) 10 , b ) 12 , c ) 16 , d ) 20 , e ) 4 | e | divide(subtract(multiply(12, subtract(40, 6)), multiply(12, 32)), 6) | the average age of an adult class is 40 years . 12 new students with an avg age of 32 years join the class . therefore decreasing the average by 6 year . find what was the original average age of the class ? | let original strength = y then , 40 y + 12 x 32 = ( y + 12 ) x 34 Γ’ β‘ β 40 y + 384 = 34 y + 408 Γ’ β‘ β 6 y = 24 Γ’ Λ Β΄ y = 4 e | a = 40 - 6
b = 12 * a
c = 12 * 32
d = b - c
e = d / 6
|
a ) 10 kmph , b ) 20 kmph , c ) 15 kmph , d ) 30 kmph , e ) 25 kmph | b | divide(180, divide(multiply(6, 3), 2)) | a car takes 6 hours to cover a distance of 180 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 distance = 280 3 / 2 of 6 hours = 6 * 3 / 2 = 9 hours required speed = 180 / 9 = 20 kmph b )" | a = 6 * 3
b = a / 2
c = 180 / b
|
a ) 1 and 8 , b ) 2 and 6 , c ) 2 and 9 , d ) 2 and 7 , e ) 2 and 9 | c | add(multiply(const_2, const_10), divide(add(44, 19), 7)) | 5 n + 2 > 12 and 7 n - 19 < 44 ; n must be between which numbers ? | 5 n + 2 > 12 5 n > 10 n > 2 7 n - 19 < 44 7 n < 63 n < 9 so n must be between 2 and 9 2 < n < 9 correct answer c | a = 2 * 10
b = 44 + 19
c = b / 7
d = a + c
|
a ) β 2 % , b ) 2 % , c ) 22 % , d ) 25 % , e ) can not be determined | d | multiply(const_100, subtract(divide(add(const_1, divide(const_10, const_100)), subtract(const_1, divide(const_12, const_100))), const_1)) | the price of a consumer good increased by pp % during 20122012 and decreased by 1212 % during 20132013 . if no other change took place in the price of the good and the price of the good at the end of 20132013 was 1010 % higher than the price at the beginning of 20122012 , what was the value of pp ? | as per question = > price was simple 10 percent greater hence x [ 1 + 10 / 100 ] must be the final price . equating the two we get = > x [ 110 / 100 ] = x [ 1 + p / 100 ] [ 88 / 100 ] = > 44 p + 4400 = 5500 = > 44 p = 1100 = > p = 1100 / 44 = > 100 / 4 = > 25 . so p must be 25 answer : d | a = 10 / 100
b = 1 + a
c = 12 / 100
d = 1 - c
e = b / d
f = e - 1
g = 100 * f
|
a ) 4 , b ) 6 , c ) 12 , d ) 24 , e ) 30 | d | multiply(divide(multiply(2, 6), subtract(4, 2)), 4) | two integers are in the ratio of 1 to 4 . if 6 is added to the smaller number , the ratio becomes 1 to 2 . find the larger integer . | "assume the integers to be x and y , where x < y given x / y = 1 / 4 - ( i ) or y = 4 x and x + 6 / y = 1 / 2 - ( ii ) or y = 2 x + 12 substituting the value of y from ( i ) , 4 x = 2 x + 12 x = 6 hence y = 4 * 6 = 24 answer d" | a = 2 * 6
b = 4 - 2
c = a / b
d = c * 4
|
a ) 4 , b ) 5 , c ) 6 , d ) 8 , e ) 9 | b | divide(divide(multiply(multiply(12, 8), 10), 16), 12) | in a garment industry , 12 men working 8 hours per day complete a piece of work in 10 days . to complete the same work in 12 days , working 16 hours a day , the number of men required is : | "explanation : let the required number of men be x . less days , more men ( indirect proportion ) more working hrs per day , less men ( indirect proportion ) days 8 : 10 working hrs 16 : 8 : : 12 : x = > 12 x 16 x x = 10 x 8 x 12 = > x = 10 x 8 x 12 / ( 12 x 16 ) = > x = 5 answer : b" | a = 12 * 8
b = a * 10
c = b / 16
d = c / 12
|
['a ) 952 cm ^ 2', 'b ) 957 cm ^ 2', 'c ) 954 cm ^ 2', 'd ) 958 cm ^ 2', 'e ) none of them'] | b | add(multiply(multiply(power(divide(7, const_2), const_2), const_pi), const_2), multiply(multiply(7, const_pi), 40)) | find the volume , curved surface area and the total surface area of a cylinder with diameter of base 7 cm and height 40 cm . | volume = β r 2 h = ( ( 22 / 7 ) x ( 7 / 2 ) x ( 7 / 2 ) x 40 ) = 1540 cm ^ 3 . . curved surface area = 2 β rh = ( 2 x ( 22 / 7 ) x ( 7 / 2 ) x 40 ) = 880 cm ^ 2 . total surface area = 2 β rh + 2 β r 2 = 2 β r ( h + r ) = ( 2 x ( 22 / 7 ) x ( 7 / 2 ) x ( 40 + 3.5 ) ) cm 2 = 957 cm ^ 2 answer is b | a = 7 / 2
b = a ** 2
c = b * math.pi
d = c * 2
e = 7 * math.pi
f = e * 40
g = d + f
|
a ) 4,514 , b ) 4,475 , c ) 4,521 , d ) 4,428 , e ) 4,349 | c | divide(factorial(subtract(add(const_4, 15), const_1)), multiply(factorial(15), factorial(subtract(const_4, const_1)))) | how many positive integers less than 5,000 are evenly divisible by neither 15 nor 23 ? | "integers less than 5000 divisible by 15 5000 / 15 = 333 . something , so 333 integers less than 5000 divisible by 23 5000 / 23 = 238 . # # , so 238 we have double counted some , so take lcm of 15 and 23 = 105 and divide by 5000 , we get 47 . so all numbers divisible by 15 and 23 = 333 + 238 - 47 = 524 now subtract that from 4999 . 4999 - 524 = 4521 answer c ." | a = 4 + 15
b = a - 1
c = math.factorial(b)
d = math.factorial(15)
e = 4 - 1
f = math.factorial(e)
g = d * f
h = c / g
|
a ) 1000 , b ) 6000 , c ) 5000 , d ) 8000 , e ) 1900 | a | multiply(multiply(divide(50, multiply(20, 4)), const_100), multiply(20, 4)) | find the sum the difference between the compound and s . i . on a certain sum of money for 4 years at 20 % per annum is rs . 50 of money ? | "p = 50 ( 100 / 20 ) 4 = > p = 1000 answer : a" | a = 20 * 4
b = 50 / a
c = b * 100
d = 20 * 4
e = c * d
|
a ) rs . 27,000 , b ) rs . 26,000 , c ) rs . 30,000 , d ) rs . 36,000 , e ) none of these | b | multiply(add(multiply(multiply(multiply(const_4, 2), multiply(add(2, const_3), 2)), const_100), multiply(multiply(add(2, const_3), const_100), const_100)), divide(divide(multiply(add(2, const_3), 2), 2), multiply(const_4, const_3))) | jayant opened a shop investing rs . 30,000 . madhu joined him 2 months later , investing rs . 45,000 . they earned a profit of rs . 52,000 after completion of one year . what will be madhu ' s share of profit ? | "30,000 * 12 = 45,000 * 8 1 : 1 madhu ' s share = 1 / 2 * 52,000 i . e . rs . 26,000 answer : b" | a = 4 * 2
b = 2 + 3
c = b * 2
d = a * c
e = d * 100
f = 2 + 3
g = f * 100
h = g * 100
i = e + h
j = 2 + 3
k = j * 2
l = k / 2
m = 4 * 3
n = l / m
o = i * n
|
a ) $ 28,300 , b ) $ 30,400 , c ) $ 31,300 , d ) $ 31,200 , e ) $ 35,100 | d | multiply(divide(234, divide(9, multiply(const_3, const_4))), const_100) | an investment yields an interest payment of $ 234 each month . if the simple annual interest rate is 9 % , what is the amount of the investment ? | let the principal amount = p simple annual interest = 9 % simple monthly interest = ( 9 / 12 ) = ( 3 / 4 ) % ( 3 / 4 ) * ( p / 100 ) = 234 = > p = ( 234 * 4 * 10 ^ 2 ) / 3 = 78 * 4 * 10 ^ 2 = 312 * 10 ^ 2 = 31200 answer d | a = 3 * 4
b = 9 / a
c = 234 / b
d = c * 100
|
a ) 0.8 , b ) 1.25 , c ) 8.0 , d ) 2.5 , e ) 80.0 | d | subtract(divide(power(const_100, const_3), multiply(400, 1,000)), const_2) | the mass of 1 cubic meter of a substance is 400 kilograms under certain conditions . what is the volume , in cubic centimeters , of 1 gram of this substance under these conditions ? ( 1 kilogram = 1,000 grams and 1 cubic meter = 1 , 000,000 cubic centimeters ) | "density is mass divided by volume . so density of the given substance will be mass / volume = 400 kg / 1 m ^ 3 = 400 kg / m ^ 3 or 1 g / 2.5 cm ^ 3 = 0.4 g / cm ^ 3 . next , ask yourself if 400,000 g is equivalent to 1 , 000,000 cubic centimeters then 1 g is equivalent to how many cubic centimeters ? - - > 1 g - 1 , 000,000 / 400,000 = 10 / 4 = 2.5 cubic centimeters . answer is d" | a = 100 ** 3
b = 400 * 1
c = a / b
d = c - 2
|
a ) 10 , b ) 12 , c ) 14 , d ) 16 , e ) 18 | d | divide(80, add(const_4, divide(const_2, const_2))) | a certain number of horses and an equal number of men are going somewhere . half of the owners are on their horses ' back while the remaining ones are walking along leading their horses . if the number of legs walking on the ground is 80 , how many horses are there ? | "legs 16 * 4 = 64 now half on their horses so remaining on the walk so 8 men 8 men has 16 legs so , 16 + 64 = 80 legs walking answer : d" | a = 2 / 2
b = 4 + a
c = 80 / b
|
a ) 33 , b ) 34 , c ) 26 , d ) 28 , e ) 19 | b | subtract(54, divide(subtract(780, multiply(54, 10)), subtract(22, 10))) | 54 is to be divided into two parts such that the sum of 10 times the first and 22 times the second is 780 . the bigger part is : | "explanation : let the two parts be ( 54 - x ) and x . then , 10 ( 54 - x ) + 22 x = 780 = > 12 x = 240 = > x = 20 . bigger part = ( 54 - x ) = 34 . answer : b ) 34" | a = 54 * 10
b = 780 - a
c = 22 - 10
d = b / c
e = 54 - d
|
a ) 50 hrs , b ) 30 hrs , c ) 70 hrs , d ) 80 hrs , e ) 90 hrs | b | divide(const_1, subtract(divide(const_1, 10), divide(const_1, 15))) | a cistern is filled by pipe a in 10 hours and the full cistern can be leaked out by an exhaust pipe b in 15 hours . if both the pipes are opened , in what time the cistern is full ? | "time taken to full the cistern = ( 1 / 10 - 1 / 15 ) hrs = 1 / 30 = 30 hrs answer : b" | a = 1 / 10
b = 1 / 15
c = a - b
d = 1 / c
|
a ) 65 , b ) 68 , c ) 72 , d ) 70 , e ) 58 | d | multiply(subtract(45, 10), const_2) | all the students of class are told to sit in circle shape . here the boy at the 10 th position is exactly opposite to 45 th boy . total number of boys in the class ? | "as half the circle shape consist of 45 - 10 = 35 boys , so total number of boys in full circle = 2 * 35 = 70 answer : d" | a = 45 - 10
b = a * 2
|
a ) 10 , b ) 20 , c ) 60 , d ) 120 , e ) 600 | b | divide(factorial(multiply(divide(2000, const_1000), divide(3000, const_1000))), multiply(factorial(divide(3000, const_1000)), factorial(divide(3000, const_1000)))) | how many integers between 2000 and 3000 that have distinct digits and increase from left to right ? | since the numbers must be distinct and increasing from left to right , the only arrangements we could come - up with are : 245 _ - - > 4 246 _ - - > 3 247 _ - - > 2 248 _ - - > 1 256 _ - - > 3 257 _ - - > 2 258 _ - - > 1 267 _ - - > 2 268 _ - - > 1 number of integers = 20 b | a = 2000 / 1000
b = 3000 / 1000
c = a * b
d = math.factorial(c)
e = 3000 / 1000
f = math.factorial(e)
g = 3000 / 1000
h = math.factorial(g)
i = f * h
j = d / i
|
Subsets and Splits
No community queries yet
The top public SQL queries from the community will appear here once available.