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 ) 32.5 , b ) 36.5 , c ) 40.5 , d ) 44.5 , e ) 48.5 | c | divide(multiply(9, 9), const_2) | the two lines y = x and x = - 9 intersect on the coordinate plane . what is the value of the area of the figure formed by the intersecting lines and the x - axis ? | the point of intersection is ( - 9 , - 9 ) . the triangle has a base of length 9 and a height of 9 . area = ( 1 / 2 ) * base * height = ( 1 / 2 ) * 9 * 9 = 40.5 the answer is c . | a = 9 * 9
b = a / 2
|
a ) 1 , b ) 2 , c ) 3 , d ) 4 , e ) 5 | d | divide(subtract(add(multiply(divide(subtract(multiply(60, 10), multiply(54, 10)), subtract(60, 40)), 40), multiply(subtract(10, divide(subtract(multiply(60, 10), multiply(54, 10)), subtract(60, 40))), 60)), multiply(50, 10)), subtract(60, 50)) | at a certain food stand , the price of each apple is 40 ¢ and the price of each orange is 60 ¢ . mary selects a total of 10 apples and oranges from the food stand , and the average ( arithmetic mean ) price of the 10 pieces of fruit is 54 ¢ . how many oranges must mary put back so that the average price of the pieces of fruit that she keeps is 50 ¢ ? | "if the average price is 54 , then mary has 7 oranges and 3 apples ( a ratio of 7 : 3 ) . for the average price to be 50 , the ratio should be 1 : 1 . mary must put back 4 oranges . the answer is d ." | a = 60 * 10
b = 54 * 10
c = a - b
d = 60 - 40
e = c / d
f = e * 40
g = 60 * 10
h = 54 * 10
i = g - h
j = 60 - 40
k = i / j
l = 10 - k
m = l * 60
n = f + m
o = 50 * 10
p = n - o
q = 60 - 50
r = p / q
|
a ) 2 / 7 , b ) 3 / 11 , c ) 4 / 15 , d ) 5 / 21 , e ) 6 / 25 | d | divide(choose(5, 4), choose(10, 5)) | a store has 10 bottles of juice , including 5 bottles of apple juice . in the evening , 6 bottles of juice are sold one by one . what is the probability of selling 4 bottles of apple juice among the 6 bottles ? assume that every bottle has an equal chance of being bought . | "the total number of ways to sell 6 bottles from 10 is 10 c 6 = 210 . the number of ways to sell 4 bottles of apple juice is 5 c 4 * 5 c 2 = 5 * 10 = 50 p ( selling 4 bottles of apple juice ) = 50 / 210 = 5 / 21 the answer is d ." | a = math.comb(5, 4)
b = math.comb(10, 5)
c = a / b
|
a ) 63 % , b ) 75 % , c ) 77 % , d ) 79 % , e ) 81 % | a | subtract(multiply(const_3, 75), add(71, 81)) | a student got 71 % in math and 81 % in history . to get an overall average of 75 % , how much should the student get in the third subject ? | "71 + 81 + x = 3 * 75 x = 73 the answer is a ." | a = 3 * 75
b = 71 + 81
c = a - b
|
a ) 1 , b ) 2 , c ) 3 , d ) 4 , e ) 5 | c | divide(9, 3) | m and n are the x and y coordinates , respectively , of a point in the coordinate plane . if the points ( m , n ) and ( m + p , n + 9 ) both lie on the line defined by the equation x = ( y / 3 ) - ( 2 / 5 ) , what is the value of p ? | "x = ( y / 3 ) - ( 2 / 5 ) , and so y = 3 x + 6 / 5 . the slope is 3 . ( n + 9 - n ) / ( m + p - m ) = 3 p = 3 the answer is c ." | a = 9 / 3
|
a ) 1 / 4 , b ) 1 / 8 , c ) 1 / 6 , d ) 1 / 2 , e ) 1 | d | divide(add(multiply(5, const_0_33), multiply(5, divide(const_2, const_3))), add(5, 5)) | at an international conference , “ red ” world countries and “ blue ” world countries are the only participants . the ratio of “ red ” world participants to “ blue ” world participants is 5 : 5 . if one - third of “ red ” world participants are left - handed and two - thirds of “ blue ” world participants are left - handed , then what is the fraction of the participants who are left - handed ? | "red : blue = 5 : 5 let red = 5 x and blue = 5 x 1 / 3 of red are left handed = > 1 / 3 * 5 x = 5 x / 3 red left handed 2 / 3 of blue are left handed = > 2 / 3 * 5 x = 10 x / 3 blue left handed fraction of participants who are left handed = total left handed / total participants = ( red left handed + blue left handed ) / total participants = ( 5 x / 3 + 10 x / 3 ) / ( 5 x + 5 x ) = ( 15 x / 30 x ) = 1 / 2 answer : d" | a = 5 * const_0_33
b = 2 / 3
c = 5 * b
d = a + c
e = 5 + 5
f = d / e
|
a ) 1 , b ) 2 , c ) 4 , d ) 5 , e ) 7 | b | subtract(divide(8, const_2), multiply(42, 42)) | what is the remainder when 42 * 49 is divided by 8 ? | "we can make use of the rule : remainder of { ( a * b ) / n } } = remainder of ( a / n ) * remainder of ( b / n ) here remainder of { 42 * 49 ) / 8 } } = remainder of ( 42 / 8 ) * remainder of ( 49 / 8 ) = 2 * 1 = 2 answer : b" | a = 8 / 2
b = 42 * 42
c = a - b
|
a ) 76 , b ) 68 , c ) 78 , d ) 66 , e ) 67 | c | divide(multiply(12, add(12, 1)), 2) | 1 + 2 + 3 + . . . + 12 = ? | 1 + 2 + 3 + ⋯ + n = ∑ n = n ( n + 1 ) / 2 1 + 2 + 3 + ⋯ + 12 = n ( n + 1 ) / 2 = ( 12 ( 12 + 1 ) ) / 2 = ( 12 × 13 ) / 2 = 6 × 13 = 78 answer is c | a = 12 + 1
b = 12 * a
c = b / 2
|
a ) 12076 , b ) 12070 , c ) 12077 , d ) 12075 , e ) 12074 | d | divide(multiply(add(multiply(6, const_100), 44), add(multiply(3, const_100), 00)), power(divide(add(multiply(6, const_100), 44), power(const_2, const_4)), const_2)) | a room of 6 m 44 cm long and 3 m 00 cm broad is to be paved with square tiles . find the least number of square tiles required to cover the floor . | "explanation : area of the room = 644 * 300 sq cm size of largest square tile = h . c . f of 644 cm and 300 cm = 4 cm area of 1 tile = 4 * 4 sq cm no . of tiles required = ( 644 * 300 ) / ( 4 * 4 ) = 12075 answer : d ) 12075" | a = 6 * 100
b = a + 44
c = 3 * 100
d = c + 0
e = b * d
f = 6 * 100
g = f + 44
h = 2 ** 4
i = g / h
j = i ** 2
k = e / j
|
a ) 23 , b ) 24 , c ) 25 , d ) 26 , e ) 27 | b | divide(subtract(multiply(30, 7), multiply(6, 7)), 7) | 7 people average age is 30 . youngest person age is 6 . find average of the people when youngest was born . | "average age of people = 30 so have total age = 210 before 7 years we have to deduct each person age by seven years 210 - 42 = 161 so average age would be 161 / 7 = 24 answer : b" | a = 30 * 7
b = 6 * 7
c = a - b
d = c / 7
|
a ) 320 , b ) 240 , c ) 369 , d ) 489 , e ) 125 | a | multiply(40, multiply(divide(16, 4), divide(30, 15))) | if 4 men can reap 40 acres of land in 15 days , how many acres of land can 16 men reap in 30 days ? | "4 men 40 acres 15 days 16 men ? 30 days 40 * 16 / 4 * 30 / 15 40 * 4 * 2 40 * 8 = 320 answer : a" | a = 16 / 4
b = 30 / 15
c = a * b
d = 40 * c
|
a ) 90 % , b ) 87.5 % , c ) 66.7 % , d ) 50 % , e ) 36 % | a | multiply(divide(subtract(subtract(const_100, 46), subtract(60, multiply(divide(70, const_100), 60))), subtract(const_100, 60)), const_100) | in the excel manufacturing company , 46 percent of the employees are men . if 60 percent of the employees are unionized and 70 percent of these are men , what percent of the non - union employees are women ? | lets say there are total 100 employees then men = 46 & women = 54 unionized emp = 60 & non unionized emp = 40 men unionized = . 7 * 60 = 42 women unionized = 18 women non - unionized = 54 - 18 = 36 percentage of women non unionized = 36 / 40 * 100 = 90 % the answer is option a | a = 100 - 46
b = 70 / 100
c = b * 60
d = 60 - c
e = a - d
f = 100 - 60
g = e / f
h = g * 100
|
a ) 2.5 % , b ) 17.67 % , c ) 28.3 % , d ) 45.2 % , e ) 32.5 % | e | multiply(divide(multiply(divide(30, const_100), subtract(1, divide(1, 4))), 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 / 4 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 = 1,530 non transferred fiction novels = 3,978 percent of the remaining collection was fiction novels = 3978 / ( 18360 - 6120 ) * 100 = > 32.5 % hence answer will be ( e )" | a = 30 / 100
b = 1 / 4
c = 1 - b
d = a * c
e = 1 / 3
f = 1 - e
g = d / f
h = g * 100
|
a ) 18 , b ) 92 , c ) 72 , d ) 26 , e ) 19 | c | multiply(add(6, 6), 6) | a man was asked to state his age in years . his reply was , ` ` take my age 6 years hence , multiply it by 6 and subtract 6 times my age 6 years ago and you will know how old i am . ' ' what was the age of the person ? | explanation : let the present age of person be x years . then , 6 ( x + 6 ) - 6 ( x - 6 ) = x < = > ( 6 x + 36 ) - ( 6 x - 36 ) = x < = > x = 72 . . answer : c | a = 6 + 6
b = a * 6
|
a ) 20 , b ) 50 , c ) 65 , d ) 70 , e ) 140 | d | subtract(multiply(60, const_2), multiply(40, const_2)) | if the average ( arithmetic mean ) of x and y is 40 , and z – x = 60 , what is the average of y and z ? | "x + y / 2 = 40 = > x + y = 80 x = z - 60 . . . sub this value z - 60 + y = 80 = > z + y = 140 = > z + y / 2 = 70 . answer : d" | a = 60 * 2
b = 40 * 2
c = a - b
|
a ) s . 57.8 , b ) s . 56.4 , c ) s . 29.3 , d ) s . 33 , e ) s . 74 | b | multiply(235, divide(3, const_100)) | find the simple interest on rs . 235 for 3 months at 8 paisa per month ? | "explanation : i = ( 235 * 3 * 8 ) / 100 = 56.4 answer : option b" | a = 3 / 100
b = 235 * a
|
a ) 720 , b ) 266 , c ) 155 , d ) 600 , e ) 441 | a | multiply(subtract(divide(12000, 10000), divide(8000, 10000)), 1800) | 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 . 1800 . 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 . 1800 = > 5 parts out of 15 parts is rs . 1800 now , required difference is 6 - 4 = 2 parts required difference = 2 / 5 ( 1800 ) = rs . 720 answer : a" | a = 12000 / 10000
b = 8000 / 10000
c = a - b
d = c * 1800
|
a ) 255 , b ) 285 , c ) 315 , d ) 345 , e ) 375 | a | subtract(multiply(divide(multiply(45, const_1000), const_3600), 30), 120) | the length of a bridge in meters , which a train 120 - meters long and traveling at 45 km / hr can cross in 30 seconds is ? | "45 km / h = 45000 m / 3600 s = 12.5 m / s in 30 seconds , the train can go 30 ( 12.5 ) = 375 meters let x be the length of the bridge . x + 120 = 375 meters x = 255 meters the answer is a ." | a = 45 * 1000
b = a / 3600
c = b * 30
d = c - 120
|
a ) 54 , b ) 59 , c ) 63 , d ) 71 , e ) 92 | c | add(multiply(subtract(const_1, divide(const_1, const_3)), subtract(70, 49)), 49) | of 70 players on a football team , 49 are throwers . the rest of the team is divided so one third are left - handed and the rest are right handed . assuming that all throwers are right handed , how many right - handed players are there total ? | total = 70 thrower = 49 rest = 70 - 49 = 21 left handed = 21 / 3 = 7 right handed = 14 if all thrower are right handed then total right handed is 49 + 14 = 63 so c . 63 is the right answer | a = 1 / 3
b = 1 - a
c = 70 - 49
d = b * c
e = d + 49
|
a ) 3 / 14 , b ) 2 / 7 , c ) 1 / 3 , d ) 1 / 42 , e ) 12 / 21 | d | multiply(divide(const_1, 6), divide(const_1, 7)) | jack has two dice , one has 6 equally probable sides , labeled 1 , 2 , 3 , 4 , 5 , 6 , and the other has 7 equally probable sides , labeled 1 , 2 , 3 , 4 , 5 , 6 , 7 . if jack rolls both dice what is the probability that both of the numbers will be 3 ? | method - 1 probability that the number on first die is 3 = 1 / 6 probability that the number on second die is 3 = 1 / 7 probability that both dice result in odd numbers = ( 1 / 6 ) * ( 1 / 7 ) = 1 / 42 answer : option d | a = 1 / 6
b = 1 / 7
c = a * b
|
a ) 7 , b ) 8 , c ) 9 , d ) 10 , e ) 11 | b | divide(subtract(12, 12), subtract(12, 12)) | a number divided by 44 leaves remainder 12 what is the remainder when same number divided by 12 | "add 44 + 12 = 56 now 56 divided by 12 so we get 8 as reaminder answer : b" | a = 12 - 12
b = 12 - 12
c = a / b
|
a ) 2 , b ) 2.5 , c ) 3 , d ) 3.5 , e ) 4 | c | add(divide(divide(16, const_2), 4), divide(divide(16, const_2), 8)) | a person walks at a speed of 4 km / hr and runs at a speed of 8 km / hr . how many hours will the person require to cover a distance of 16 km , if the person completes half of the distance by walking and the other half by running ? | time = 8 / 4 + 8 / 8 = 3 hours the answer is c . | a = 16 / 2
b = a / 4
c = 16 / 2
d = c / 8
e = b + d
|
a ) 5 , b ) 8 , c ) 7 , d ) 10 , e ) 11 | a | subtract(subtract(subtract(subtract(255, const_100), const_100), multiply(const_10, const_4)), const_10) | change the below lagarithm log 255 = 1 / 2 to exponential form | solution : log 255 = 251 / 2 = 5 answer is a | a = 255 - 100
b = a - 100
c = 10 * 4
d = b - c
e = d - 10
|
a ) 10 cm , b ) 40 cm , c ) 20 cm , d ) 30 cm , e ) 50 cm | b | divide(const_100, const_3) | the length of a rectangle is twice its breadth . if its length is decreased by 5 cm and breadth is increased by 5 cm , the area of the rectangle is increased by 75 sq . cm . find the length of the rectangle ? | "let the breadth of the given rectangle be x then length is 2 x . thus area of the given rect is \ fn _ jvn { \ color { black } 2 x ^ { 2 } } after dec 5 cm from length and inc 5 cm breadth , new lenght becomes 2 x - 5 and breadth is x + 5 . thus new area = ( 2 x - 5 ) ( x + 5 ) = \ fn _ jvn { \ color { black } 2 x ^ { 2 } + 5 x - 25 } since new area is 75 units greater than original area thus \ fn _ jvn { \ color { black } 2 x ^ { 2 } + 75 = 2 x ^ { 2 } + 5 x - 25 } 5 x = 75 + 25 5 x = 100 therefore x = 20 hence length of the rectangle is 40 cm . answer : b" | a = 100 / 3
|
a ) 8 , b ) 9 , c ) 12 , d ) 89 , e ) 81 | c | multiply(divide(const_1, multiply(add(const_100, 20), divide(const_1, subtract(const_100, 10)))), 16) | by selling 16 pencils for a rupee a man loses 10 % . how many for a rupee should he sell in order to gain 20 % ? | "90 % - - - 16 120 % - - - ? 90 / 120 * 16 = 12 answer : c" | a = 100 + 20
b = 100 - 10
c = 1 / b
d = a * c
e = 1 / d
f = e * 16
|
a ) 1 ⁄ 3 , b ) 4 ⁄ 3 , c ) 90 , d ) 120 , e ) 180 | c | divide(multiply(1, const_60), divide(2, 3)) | if it takes a machine 2 ⁄ 3 minute to produce one item , how many items will it produce in 1 hours ? | "1 item takes 2 / 3 min so it takes 60 min to produce x 2 x / 3 = 60 the x = 90 answer : c" | a = 1 * const_60
b = 2 / 3
c = a / b
|
a ) 17 , b ) 39 , c ) 42 , d ) 45 , e ) none of the above | a | add(subtract(65, multiply(17, 3)), 3) | a batsman makes a score of 65 runs in the 17 th inning and thus increases his average by 3 . find his average after 17 th inning . | let the average after 17 th inning = x . then , average after 16 th inning = ( x – 3 ) . ∴ 16 ( x – 3 ) + 65 = 17 x or x = ( 65 – 48 ) = 17 . answer a | a = 17 * 3
b = 65 - a
c = b + 3
|
a ) 87 % , b ) 88 % , c ) 78 % , d ) 75 % , e ) 73 % | c | subtract(const_100, divide(multiply(1200, const_100), 675)) | an article is bought for rs . 675 and sold for rs . 1200 , find the gain percent ? | "675 - - - - 525 100 - - - - ? = > = 78 % answer : c" | a = 1200 * 100
b = a / 675
c = 100 - b
|
a ) 09 pm , b ) 07 pm , c ) 11 pm , d ) 05 am , e ) 03 am | b | add(divide(add(1410, 75), add(60, 75)), 8) | the distance between two cities a and b is 1410 km . a train starts from a at 8 a . m . and travel towards b at 60 km / hr . another train starts from b at 9 a . m and travels towards a at 75 km / hr . at what time do they meet ? | "explanation : suppose they meet x hrs after 8 a . m then , [ distance moved by first in x hrs ] + [ distance moved by second in ( x - 1 ) hrs ] = 1410 . therefore , 60 x + 75 ( x - 1 ) = 1410 . = > x = 3 . so , they meet at ( 8 + 11 ) i . e , 7 pm . answer : b" | a = 1410 + 75
b = 60 + 75
c = a / b
d = c + 8
|
a ) 2000 , b ) 2200 , c ) 2309 , d ) 2350 , e ) 3250 | d | multiply(4230, subtract(const_1, divide(multiply(3000, multiply(2, multiply(2, const_3))), add(multiply(4000, subtract(multiply(2, multiply(2, const_3)), 2)), multiply(3000, multiply(2, multiply(2, const_3))))))) | tom opened a shop investing rs . 3000 . jose joined him 2 months later , investing rs . 4000 . they earned a profit of rs . 4230 after completion of one year . what will be jose ' s share of profit ? | "sol = ~ s - so anju ’ s share = [ 5 / 9 ] x 4230 = 2350 d" | a = 2 * 3
b = 2 * a
c = 3000 * b
d = 2 * 3
e = 2 * d
f = e - 2
g = 4000 * f
h = 2 * 3
i = 2 * h
j = 3000 * i
k = g + j
l = c / k
m = 1 - l
n = 4230 * m
|
a ) 12 , b ) 16 , c ) 18 , d ) 24 , e ) 26 | c | divide(54, add(const_1, const_2)) | a student got twice as many sums wrong as he got right . if he attempted 54 sums in all , how many did he solve correctly ? | "explanation : suppose the boy got x sums right and 2 x sums wrong . then , x + 2 x = 54 3 x = 54 x = 18 . answer : c" | a = 1 + 2
b = 54 / a
|
a ) 1 / 12 , b ) 4 / 17 , c ) 2 / 3 , d ) 1 1 / 9 , e ) 2 1 / 9 | b | subtract(add(divide(4, 17), divide(const_3, const_10)), divide(9, 30)) | the instructions state that deepthi needs 4 / 17 square yards of one type of material and 3 / 10 square yards of another type of material for a project . she buys exactly that amount . after finishing the project , however , she has 9 / 30 square yards left that she did not use . what is the total amount of square yards of material deepthi used ? | total bought = 4 / 17 + 3 / 10 left part 9 / 30 - - - > 3 / 10 so used part 4 / 17 + 3 / 10 - 3 / 10 = 4 / 17 answer : b | a = 4 / 17
b = 3 / 10
c = a + b
d = 9 / 30
e = c - d
|
a ) a ) 10,700 , b ) b ) 10,800 , c ) c ) 9,000 , d ) d ) 11,000 , e ) e ) 11,100 | c | multiply(multiply(const_4, const_2), const_100) | a certain city with a population of 108,000 is to be divided into 11 voting districts , and no district is to have a population that is more than 10 percent greater than the population of any other district what is the minimum possible population that the least populated district could have ? | "let x = number of people in smallest district x * 1.1 = number of people in largest district x will be minimised when the number of people in largest district is maximised 10 * x * 1.1 = 11 x = total number of people in other districts so we have 11 x + x = 108 k x = 9,000 answer : c" | a = 4 * 2
b = a * 100
|
a ) 391 , b ) 408 , c ) 410 , d ) 423 , e ) 445 | b | divide(470, add(const_1, divide(15, const_100))) | company p had 15 percent more employees in december than it had in january . if company p had 470 employees in december , how many employees did it have in january ? | "d = number of employees in december j = number of employees in january j x 1.15 = d j x 1.15 = 470 j = 470 / 1.15 j = 47,000 / 115 = 408 thus b is the correct answer ." | a = 15 / 100
b = 1 + a
c = 470 / b
|
a ) 2 , b ) 4 , c ) 2 8 / 11 , d ) 67 , e ) 9 | c | inverse(add(inverse(15), add(inverse(5), inverse(10)))) | a , b and c can do a piece of work in 5 days , 10 days and 15 days respectively . how long will they taken , if all the three work together ? | "1 / 5 + 1 / 10 + 1 / 15 = 11 / 30 = > 30 / 11 = > 2 8 / 11 days answer : c" | a = 1/(15)
b = 1/(5)
c = 1/(10)
d = b + c
e = a + d
f = 1/(e)
|
a ) 550 , b ) 3000 , c ) 250 , d ) 300 , e ) 400 | b | divide(multiply(6800, 60), 136) | find the annual income derived by investing $ 6800 in 60 % stock at 136 . | "by investing $ 136 , income obtained = $ 60 . by investing $ 6800 , income obtained = $ [ ( 60 / 136 ) * 6800 ] = $ 3000 . answer b ." | a = 6800 * 60
b = a / 136
|
a ) 560 , b ) 616 , c ) 672 , d ) 728 , e ) 784 | b | divide(59.32, subtract(2, floor(2))) | when positive integer x is divided by positive integer y , the result is 59.32 . what is the sum w of all possible 2 - digit remainders for x / y ? | "ans b 616 . . . remainders = . 32 = 32 / 100 = 8 / 25 = 16 / 50 and so on . . so two digit remainders are 16 + 24 + 32 + . . . . + 96 . . w = 8 ( 2 + 3 + 4 . . . . + 12 ) = 616" | a = math.floor(2)
b = 2 - a
c = 59 / 32
|
a ) 87 , b ) 89 , c ) 90 , d ) 93 , e ) 95 | b | subtract(multiply(4, add(81, 2)), multiply(81, 3)) | jerry ’ s average ( arithmetic mean ) score on the first 3 of 4 tests is 81 . if jerry wants to raise his average by 2 points , what score must he earn on the fourth test ? | "total score on 3 tests = 81 * 3 = 243 jerry wants the average to be = 83 hence total score on 4 tests should be = 83 * 4 = 332 score required on the fourth test = 332 - 243 = 89 option b" | a = 81 + 2
b = 4 * a
c = 81 * 3
d = b - c
|
['a ) 34', 'b ) 59', 'c ) 68', 'd ) 88', 'e ) 92'] | b | add(multiply(divide(390, 20), const_2), 20) | a rectangular field is to be fenced on three sides leaving a side of 20 feet uncovered . if the area of the field is 390 sq . feet , how many feet of fencing will be required ? | given that length and area , so we can find the breadth . length x breadth = area 20 x breadth = 390 breadth = 19.5 feet area to be fenced = 2 b + l = 2 ( 19.5 ) + 20 = 59 feet answer : b | a = 390 / 20
b = a * 2
c = b + 20
|
a ) 20 % , b ) 25 % , c ) 50 % , d ) 60 % , e ) 27 % | a | divide(multiply(2,000, const_100), add(add(multiply(const_2, const_1000), const_100), 2,000)) | in a certain apartment building , there are one - bedroom and two - bedroom apartments . the rental prices of the apartment depend on a number of factors , but on average , two - bedroom apartments have higher rental prices than do one - bedroom apartments . let m be the average rental price for all apartments in the building . if m is $ 2,000 higher than the average rental price for all one - bedroom apartments , and if the average rental price for all two - bedroom apartments is $ 8,000 higher that m , then what percentage of apartments in the building are two - bedroom apartments ? | "ratio of 2 bedroom apartment : 1 bedroom apartment = 2000 : 8000 - - - - - > 1 : 4 let total number of apartments be x no . of 2 bedroom apartment = ( 1 / 5 ) * x percentage of apartments in the building are two - bedroom apartments - - - - > ( 1 / 5 ) * 100 - - - > 20 % answer : a" | a = 2 * 0
b = 2 * 1000
c = b + 100
d = c + 2
e = a / d
|
['a ) 2', 'b ) 3', 'c ) 5', 'd ) 7', 'e ) 11'] | d | add(const_3, const_4) | what is the smallest positive integer that can be multiplied by 1,008 to make it a perfect square ? | 1008 = 2 ^ 4 x 3 ^ 2 x 7 . therefore the smallest integer to multiplied to 1008 to make it a perfect square is 7 . answer d . | a = 3 + 4
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a ) 3 , b ) 4 , c ) 5 , d ) 6 , e ) 15 | e | divide(660, multiply(const_10, const_2)) | how many factors of 660 are odd numbers greater than 1 ? | when factorized , 660 has 5 prime factors . of these prime factors 4 are odd and 1 is even . hence total number of odd factors is 2 * 2 * 2 * 2 ( 16 ) , which includes 1 . the total number of odd factors greater than 1 are 15 . ( option e ) | a = 10 * 2
b = 660 / a
|
a ) 3 , b ) 4 , c ) 5 , d ) 6 , e ) 7 | d | add(const_3, const_3) | a palindrome is a number that reads the same front - to - back as it does back - to - front ( e . g . 202 , 575 , 1991 , etc . ) p is the smallest integer greater than 300 that is both a prime and a palindrome . what is the sum of the digits of p ? | given that p is smallest integer greater than 300 - assume there is a 3 - digit that satisfies the above conditions . let the number be xyx ; question asks us the values of 2 x + y we can straight away cross out options a ) and d ) - sum of digits 3 or 6 implies it is divisible by 3 - - - > we know that p is a prime number coming to option b ) 2 x + y = 4 - - > only x = 2 and y = 0 satisfy this equation ( x > 2 will never give sum of digits = 4 ) ; but 202 is divisible by 2 ; we know that p is a prime number similarly option c ) 2 x + y = 5 - - > only x = 2 and y = 1 satisfy this equation ( x > 2 will never give sum of digits = 5 ) ; but 212 is divisible by 2 ; we know that p is a prime number therefore answer option should be e - - - > can be verified by taking 2 x + y = 7 - - - > x = 3 and y = 1 ; gives 313 d | a = 3 + 3
|
['a ) 4.00', 'b ) 6.00', 'c ) 5.50', 'd ) 6.50', 'e ) 6.75'] | b | divide(multiply(60, 10), 100) | a sucrose solution contains 10 grams of sucrose per 100 cubic centimeters of solution . if 60 cubic centimeters of the solution were poured into an empty container , how many grams of sucrose would be in the container ? | we are given that a sucrose solution contains 10 grams of sucrose per 100 cubic centimeters of solution . since we are dealing with a solution , we know that the grams of sucrose is proportional to the number of cubic centimeters of solution . thus , to determine how many grams of sucrose would be in the container when we have 60 cubic centimeters of solution , we can set up a proportion . we can say : “ 10 grams of sucrose is to 100 cubic centimeters of solution as x grams of sucrose is to 60 cubic centimeters of solution . ” let ’ s now set up the proportion and solve for x . 10 / 100 = x / 60 when we cross multiply we obtain : ( 10 ) ( 60 ) = 100 x 600 = 100 x 6.00 = x there are 6.00 grams of sucrose in the solution in the container . the answer is b . | a = 60 * 10
b = a / 100
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a ) 2 , b ) 4 , c ) 6 , d ) 8 , e ) 12 | c | divide(divide(multiply(multiply(8, 12), 6), 12), 8) | a crate measures 6 feet by 8 feet by 12 feet on the inside . a stone pillar in the shape of a right circular cylinder must fit into the crate for shipping so that it rests upright when the crate sits on at least one of its six sides . what is the radius , in feet , of the pillar with the largest volume that could still fit in the crate ? | "we can find the radius of all the three cases of cylinders . the only crux to find the answer faster is that : voulme is pi * r ^ 2 * h . the volume is a function of r ^ 2 . so r has to be the highest to find the largest volume . so r = 6 for the surface 8 * 12 face . volume = 216 pi answer c" | a = 8 * 12
b = a * 6
c = b / 12
d = c / 8
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a ) 13 , 3 , b ) 12 , 6 , c ) 16 , 4 , d ) 14 , 4 , e ) none of these | c | divide(divide(add(20, 12), const_2), const_2) | a man can row downstream at 20 kmph and upstream at 12 kmph . find the speed of the man in still water and the speed of stream respectively ? | "explanation : let the speed of the man in still water and speed of stream be x kmph and y kmph respectively . given x + y = 20 - - - ( 1 ) and x - y = 12 - - - ( 2 ) from ( 1 ) & ( 2 ) 2 x = 32 = > x = 16 , y = 4 . answer : option c" | a = 20 + 12
b = a / 2
c = b / 2
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a ) 240 sec , b ) 165 sec , c ) 186 sec , d ) 167 sec , e ) 168 sec | a | divide(1200, subtract(multiply(54, const_0_2778), multiply(36, const_0_2778))) | a and b go around a circular track of length 1200 m on a cycle at speeds of 36 kmph and 54 kmph . after how much time will they meet for the first time at the starting point ? | "time taken to meet for the first time at the starting point = lcm { length of the track / speed of a , length of the track / speed of b } = lcm { 1200 / ( 36 * 5 / 18 ) , 1200 / ( 54 * 5 / 18 ) } = lcm ( 120 , 80 ) = 240 sec . answer : a" | a = 54 * const_0_2778
b = 36 * const_0_2778
c = a - b
d = 1200 / c
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a ) 4 , b ) 7 , c ) 8 , d ) 5 , e ) 3 | e | divide(subtract(add(multiply(7, 7), 5), 24), 10) | if the number is decreased by 5 and divided by 7 the result is 7 . what would be the result if 24 is subtracted and divided by 10 ? | "explanation : let the number be x . then , ( x - 5 ) / 7 = 7 = > x - 5 = 49 x = 54 . : ( x - 24 ) / 10 = ( 54 - 24 ) / 10 = 3 answer : option e" | a = 7 * 7
b = a + 5
c = b - 24
d = c / 10
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a ) 19 , b ) 18 , c ) 9.12 , d ) 20 , e ) none | c | divide(subtract(sqrt(add(multiply(multiply(16, 5), const_4), power(16, const_2))), 16), const_2) | tom read a book containing 560 pages by reading the same number of pages each day . if he would have finished the book 5 days earlier by reading 16 pages a day more , how many days did tom spend reading the book ? | "actually u can set up 2 equation p - - stands for the pages d - - stands for the days 1 ) p * d = 560 ( we want to find the days , sop = 560 / d ) 2 ) ( p + 16 ) ( d - 5 ) = 560 = > pd - 5 p + 16 d - 80 = 560 as the 1 ) stated u can put 1 ) into 2 ) = > 560 - 5 p + 16 d - 80 = 560 = > 16 d - 5 p = 80 put the bold one into it = > 16 d - 5 ( 560 / d ) = 80 the we get the final equation 16 d ^ 2 - 2800 = 80 d ( divide 16 ) = > d ^ 2 - 5 d - 175 = 0 ( d - 15 ) ( d + 10 ) = 0 so d = 9.12 days ans : c" | a = 16 * 5
b = a * 4
c = 16 ** 2
d = b + c
e = math.sqrt(d)
f = e - 16
g = f / 2
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a ) 46 , b ) 47 , c ) 58 , d ) 44 , e ) 54 | d | divide(407, 9.25) | john bought 9.25 m of cloth for $ 407 . find the cost price per metre . | "cloth bought by john = 9.25 m cost of 9.25 m = $ 407 cost per metre = 407 ÷ 9.25 the cost of the cloth per metre = $ 44 answers : d" | a = 407 / 9
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a ) 20 , b ) 21 , c ) 22 , d ) 23 , e ) 24 | c | subtract(multiply(6, const_3), const_1) | how many pieces can a square pizza be cut into by making 6 linear cuts ? | "1 cut makes 2 pieces a second cut can make 4 pieces by cutting through both pieces a third cut can make 7 pieces by cutting through 3 of the pieces a fourth cut can make 11 pieces by cutting through 4 of the pieces a fifth cut can make 16 pieces by cutting through 5 of the pieces a sixth cut can make 22 pieces by cutting through 6 of the pieces c" | a = 6 * 3
b = a - 1
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a ) 6 , b ) 9 , c ) 24 , d ) 32 , e ) 64 | c | subtract(add(power(3, const_2), power(const_4, const_2)), const_2) | on the xy - plane , each point on circle c has coordinates such that x 0 and y 0 . if the center of circle c is the point ( 3 ; 9 ) , what is the maximum possible area of c ? | i think that theintentof this question is that the circle will have no co - ordinates that fit the following 2 forms : ( 0 , y ) or ( x , 0 ) . in real basic terms , this circle will neither touch ( nor cross ) the x - axis nor will it touch ( nor cross ) the y - axis . with a center at ( 3,8 ) , the longest possible radius would almost touch the x - axis , so the radius would be a little less than 3 . if the radius was 3 exactly , then the area would be 9 pi , which is approximately 28.26 since the circle ca n ' t touch the x - axis , the area has to be less than approximately 28.26 as it ' s written , none of the answers is clearly the correct one . c | a = 3 ** 2
b = 4 ** 2
c = a + b
d = c - 2
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a ) 14 % , b ) 25 % , c ) 48 % , d ) 34 % , e ) 50 % | c | multiply(divide(add(multiply(divide(30, const_100), 40), 10), add(40, 10)), const_100) | if 10 gallons of grape juice are added to 40 gallons of a mixture , which contains 30 percent grape juice then what percent of the resulting mixture is grape juice ? | "official solution : if we start with 40 gallons of a mixture that is 10 % grape juice , then we have : 40 × 0.30 = 12 gallons of grape juice . 40 × 0.70 = 28 gallons of other components . if we add 10 gallons of grape juice , we will end up with 24 gallons of grape juice and 36 gallons of other components , and we will have a total of 50 gallons of the mixture . so 24 / 50 of the new mixture is grape juice . now we convert this to a percent : percent grape juice = 24 / 50 = 48 / 100 = 48 % . the correct answer is choice ( c )" | a = 30 / 100
b = a * 40
c = b + 10
d = 40 + 10
e = c / d
f = e * 100
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a ) 5 : 2 , b ) 2 : 3 , c ) 1 : 2 , d ) 3 : 7 , e ) 2 : 5 | c | divide(divide(10, const_100), divide(1, 5)) | in a school 10 % of the boys are same in number as 1 / 5 th of the girls . what is the ratio of boys to the girls in the school ? | 10 % of b = 1 / 5 g 10 b / 100 = g / 5 b = g / 2 b / g = 1 / 2 b : g = 1 : 2 answer is c | a = 10 / 100
b = 1 / 5
c = a / b
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a ) 70 , b ) 80 , c ) 85 , d ) 90 , e ) 95 | a | divide(subtract(divide(const_3600, const_10), multiply(40, const_2)), const_4) | in a certain parallelogram the degree measure of one angle exceeds that of the other by 40 what is the degree measure of the smaller angle ? | "in a parallelogram opposite angles are equal and the angles at each side are supplementary to each other ( supplementary angles are two angles that add up to 180 ° ) . given : x + ( x + 40 ) = 180 - - > x = 70 . answer : a ." | a = 3600 / 10
b = 40 * 2
c = a - b
d = c / 4
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['a ) 100', 'b ) 104', 'c ) 108', 'd ) 112', 'e ) none of these'] | b | add(multiply(subtract(27, const_2), const_4), const_4) | a farmer built a fence around his square plot . he used 27 fence poles on each side of the square . how many poles did he need altogether ? | one side = 2 corners ( 1 nd 1 each ) plus 25 . . . 1 25 1 25 25 1 25 1 der 4 ans = 25 * 4 + 4 = 104 answer : b | a = 27 - 2
b = a * 4
c = b + 4
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a ) 8 , b ) 10 , c ) 12 , d ) 14 , e ) 16 | a | add(divide(add(power(4, 4), sqrt(add(power(power(4, 4), const_2), power(subtract(208, power(4, 4)), const_2)))), multiply(power(4, const_2), const_2)), subtract(divide(add(power(4, 4), sqrt(add(power(power(4, 4), const_2), power(subtract(208, power(4, 4)), const_2)))), multiply(power(4, const_2), const_2)), 4)) | if one positive integer is greater than another positive integer by 4 , and the difference of their cubes is 208 , what is their sum ? | "1 ^ 3 = 1 2 ^ 3 = 8 3 ^ 3 = 27 4 ^ 3 = 64 5 ^ 3 = 125 6 ^ 3 = 216 the two numbers are 2 and 6 . the answer is a ." | a = 4 ** 4
b = 4 ** 4
c = b ** 2
d = 4 ** 4
e = 208 - d
f = e ** 2
g = c + f
h = math.sqrt(g)
i = a + h
j = 4 ** 2
k = j * 2
l = i / k
m = 4 ** 4
n = 4 ** 4
o = n ** 2
p = 4 ** 4
q = 208 - p
r = q ** 2
s = o + r
t = math.sqrt(s)
u = m + t
v = 4 ** 2
w = v * 2
x = u / w
y = x - 4
z = l + y
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a ) 0 , b ) 1 , c ) 2 , d ) 3 , e ) 4 | a | multiply(divide(add(add(add(add(add(2, 1), 3), 4), 5), 6), add(const_2, const_4)), divide(add(3, 4), const_2)) | if s = { 1 , 2 , 3 , 4 , 5 , 6 , 7 } , how much less is the mean of the numbers in s than the median of the numbers in s ? | "mean = ( 1 + 2 + 3 + 4 + 5 + 6 + 7 ) / 7 = 4 median = 4 difference = 4 - 4 = 0 option a" | a = 2 + 1
b = a + 3
c = b + 4
d = c + 5
e = d + 6
f = 2 + 4
g = e / f
h = 3 + 4
i = h / 2
j = g * i
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a ) 16.2 , b ) 4 , c ) 5 , d ) 500 , e ) 8 | e | divide(multiply(divide(40, const_100), 900), 45) | 45 x ? = 40 % of 900 | "answer let 45 x a = ( 40 x 900 ) / 100 ∴ a = ( 40 x 9 ) / 45 = 8 correct option : e" | a = 40 / 100
b = a * 900
c = b / 45
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a ) 22 , b ) 77 , c ) 18 , d ) 77 , e ) 65 | c | multiply(divide(subtract(64900, add(42000, 13000)), add(42000, 13000)), const_100) | ramu bought an old car for rs . 42000 . he spent rs . 13000 on repairs and sold it for rs . 64900 . what is his profit percent ? | "total cp = rs . 42000 + rs . 13000 = rs . 55000 and sp = rs . 64900 profit ( % ) = ( 64900 - 55000 ) / 55000 * 100 = 18 % answer : c" | a = 42000 + 13000
b = 64900 - a
c = 42000 + 13000
d = b / c
e = d * 100
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a ) 2.6 , b ) 5.6 , c ) 6.3 , d ) 8 , e ) 10 | a | divide(add(multiply(6, 3), 8), 10) | a honey bee flies for 10 seconds , from a daisy to a rose . it immediately continues to a poppy , flying for 6 additional seconds . the distance the bee passed , flying from the daisy to the rose is 8 meters longer than the distance it passed flying from the rose to the poppy . the bee flies to the poppy at 3 meters per second faster than her speed flying to the rose . the bee flies how many meters per second from the daisy to the rose ? | let the speed be ' s ' and let the distance between rose and poppy be ' x ' the problem boils down to : rose to poppy : s + 3 = x / 6 - - - - - - - 1 daisy to rose : s = ( x + 8 ) / 10 - - - - - - 2 so from 1 we can re write x as x = 6 s + 18 substitute the value of x in 2 gives us s = 2.6 m / s a | a = 6 * 3
b = a + 8
c = b / 10
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a ) 5 / 9 , b ) 8 / 15 , c ) 23 / 30 , d ) 43 / 60 , e ) 53 / 90 | b | divide(add(multiply(multiply(5, 2), const_2.0), 3), multiply(2, multiply(3, 5))) | of the female students at barkely university , 5 / 6 are on the honor roll . of the male students , 2 / 3 are on the honor roll . if 3 / 5 of the students are female , what is ratio of male to female students on honor roll ? | "let the total students be 30 given 3 / 5 of the students are females = 18 then males = 2 / 5 = 12 5 / 6 of the females are on honor roll = 15 males on the honor roll = 2 / 3 = 8 ratio of m : f students on honor roll = 8 / 15 b" | a = 5 * 2
b = a * 2
c = b + 3
d = 3 * 5
e = 2 * d
f = c / e
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a ) 20 m , b ) 16 m , c ) 11 m , d ) 10 m , e ) 18 m | e | multiply(divide(subtract(25, 20), 25), 90) | if in a race of 90 m , a covers the distance in 20 seconds and b in 25 seconds , then a beats b by : | explanation : the difference in the timing of a and b is 5 seconds . hence , a beats b by 5 seconds . the distance covered by b in 5 seconds = ( 90 * 5 ) / 25 = 18 m hence , a beats b by 18 m . answer e | a = 25 - 20
b = a / 25
c = b * 90
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a ) 5 , b ) 8 , c ) 7 , d ) 6 , e ) 9 | d | subtract(899830, multiply(floor(divide(899830, 16)), 16)) | find the least number must be subtracted from 899830 so that remaining no . is divisible by 16 ? | "on dividing 899830 by 16 we get the remainder 6 , so 6 should be subtracted d" | a = 899830 / 16
b = math.floor(a)
c = b * 16
d = 899830 - c
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a ) 130 , b ) 125 , c ) 140 , d ) 154 , e ) 145 | a | subtract(multiply(multiply(multiply(const_60, const_3), const_2), divide(40, const_60)), multiply(divide(divide(multiply(multiply(const_60, const_3), const_2), const_3), const_4), add(divide(40, const_60), 3))) | at 3 : 40 , the hour hand and the minute hand of a clock form an angle of : | angle traced by hour hand in 12 hrs . = 360 â ° . abgle traced it in 11 / 3 hrs = ( 360 / 12 * 11 / 3 ) = 110 angle traced by min hand in 60 min = 360 angle traced by it in 40 min = ( 360 / 60 * 40 ) = 240 req ( 240 - 110 ) = 130 d | a = const_60 * 3
b = a * 2
c = 40 / const_60
d = b * c
e = const_60 * 3
f = e * 2
g = f / 3
h = g / 4
i = 40 / const_60
j = i + 3
k = h * j
l = d - k
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a ) 200 , b ) 182 , c ) 216 , d ) 127 , e ) 400 | c | multiply(divide(60, const_1000), const_3600) | express 60 mps in kmph ? | "60 * 18 / 5 = 216 kmph answer : c" | a = 60 / 1000
b = a * 3600
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a ) 3 / 14 , b ) 2 / 7 , c ) 9 / 14 , d ) 5 / 7 , e ) 11 / 14 | b | divide(divide(factorial(add(4, 1)), multiply(factorial(subtract(add(4, 1), 3)), factorial(3))), divide(factorial(add(4, 3)), multiply(factorial(4), factorial(3)))) | set a : { 1 , 3 , 4 , 6 , 9 , 12 , 15 } if 3 numbers are randomly selected from set a without replacement , what is the probability that the sum of the 3 numbers is divisible by 3 ? | probability of an event = ( # of favorable outcomes ) / ( total # of outcomes ) . there are 5 numbers you can choose from ( 3 , 6 , 9 , 12 , 15 ) so that the sum to be divisible by 3 and there are total of 7 numbers so p = c 35 c 37 = 27 p = c 53 c 73 = 27 . answer : b . | a = 4 + 1
b = math.factorial(a)
c = 4 + 1
d = c - 3
e = math.factorial(d)
f = math.factorial(3)
g = e * f
h = b / g
i = 4 + 3
j = math.factorial(i)
k = math.factorial(4)
l = math.factorial(3)
m = k * l
n = j / m
o = h / n
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a ) 21 . , b ) 12 . , c ) 35 . , d ) 49 . , e ) 52 . | b | add(const_3, const_2) | in a rectangular axis system , what is the area of a parallelogram with the coordinates : ( 1,3 ) , ( 5,3 ) , ( 2,6 ) , ( 6,6 ) ? | "delta x will give us the dimension of one side of the parallelogram = 5 - 1 = 4 unit delta y will give us the dimension of the other side of parallelogram = 6 - 3 = 3 unit area of parallelogram = 4 * 3 = 12 answer is b" | a = 3 + 2
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a ) 960 , b ) 840 , c ) 1020 , d ) 760 , e ) 920 | c | add(270, divide(multiply(270, const_100), multiply(12, 3))) | the banker ' s gain on a sum due 3 years hence at 12 % per year is rs . 270 . the banker ' s discount is : | "t . d . = ( b . g . x 100 / r x t ) = rs . ( 270 x 100 / 12 x 3 ) = rs . 750 . therefore , b . d . = rs . ( 750 + 270 ) = rs . 1020 . answer is c" | a = 270 * 100
b = 12 * 3
c = a / b
d = 270 + c
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a ) 3.5 kmph . , b ) 2.5 kmph . , c ) 1 kmph . , d ) 1.5 kmph . , e ) 1.9 kmph . | c | multiply(const_3_6, divide(12, 43)) | convert the 12 / 43 m / s into kilometers per hour ? | "12 / 43 m / s = 12 / 43 * 18 / 5 = 1 ( 1 / 250 ) = 1 kmph . answer : c" | a = 12 / 43
b = const_3_6 * a
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a ) 2288 , b ) 2779 , c ) 2779 , d ) 5200 , e ) 2781 | d | multiply(multiply(subtract(add(80, 60), 10), 10), 4) | a rectangular lawn of dimensions 80 m * 60 m has two roads each 10 m wide running in the middle of the lawn , one parallel to the length and the other parallel to the breadth . what is the cost of traveling the two roads at rs . 4 per sq m ? | "area = ( l + b â € “ d ) d ( 80 + 60 â € “ 10 ) 10 = > 1300 m 2 1300 * 4 = rs . 5200 answer : d" | a = 80 + 60
b = a - 10
c = b * 10
d = c * 4
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['a ) h / √ 2', 'b ) h / 2', 'c ) h / 4', 'd ) ( h ) ^ 2', 'e ) ( h ) ^ 2 / 4'] | e | inverse(sine(45)) | a 45 ° - 45 ° - 90 ° right triangle has hypotenuse of length h . what is the area of the triangle r in terms of h ? | if . . . each of the two shorter sides = 3 , then the hypotenuse = h = 3 ( root 2 ) . the area r = ( 1 / 2 ) ( base ) ( height ) = ( 1 / 2 ) ( 3 ) ( 3 ) = 9 / 2 . so we ' re looking for an answer that = 9 / 2 when h = 3 ( root 2 ) . there ' s only one answer that matches . . . e | a = 1/(sine)
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a ) - 4212 , b ) 4601 , c ) - 4801 , d ) - 3471 , e ) none of these | a | add(multiply(negate(88), 49), 100) | - 88 * 49 + 100 = ? | "= > - 88 * ( 50 - 1 ) + 100 ; = > - ( 88 * 50 ) + 88 + 100 ; = > - 4400 + 188 = - 4212 . correct option : a" | a = negate * (
b = a + 49
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a ) 31 , b ) 61 , c ) 50 , d ) 99 , e ) 101 | b | floor(add(const_1, multiply(divide(log(2), log(const_10)), 200))) | how many digits 2 ^ 200 has ? | "2 ^ 10 = 1.024 * 10 ^ 3 = > 2 ^ 100 = ( 1.024 ) ^ 10 * 10 ^ 60 therefore 61 digits would be my best guess b" | a = math.log(2)
b = math.log(10)
c = a / b
d = c * 200
e = 1 + d
f = math.floor(e)
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a ) 25 , b ) 45 , c ) 55 , d ) 70 , e ) 30 | b | subtract(175, add(add(20, 80), divide(add(20, subtract(210, add(add(80, 70), 20))), 2))) | in an intercollegiate competition that lasted for 3 days , 175 students took part on day 1 , 210 on day 2 and 150 on day 3 . if 80 took part on day 1 and day 2 and 70 took part on day 2 and day 3 and 20 took part on all three days , how many students took part only on day 1 ? | day 1 & 2 = 80 ; only day 1 & 2 ( 80 - 20 ) = 60 , day 2 & 3 = 70 ; only day 2 & 3 ( 70 - 20 ) = 50 , only day 1 = 175 - ( 60 + 50 + 20 ) = 45 answer : b | a = 20 + 80
b = 80 + 70
c = b + 20
d = 210 - c
e = 20 + d
f = e / 2
g = a + f
h = 175 - g
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a ) 388 , b ) 1500 , c ) 277 , d ) 500 , e ) 271 | b | multiply(add(5, 6), const_100) | rs . 1500 is divided into two parts such that if one part is invested at 6 % and the other at 5 % the whole annual interest from both the sum is rs . 75 . how much was lent at 5 % ? | "( x * 5 * 1 ) / 100 + [ ( 1500 - x ) * 6 * 1 ] / 100 = 75 5 x / 100 + 90 – 6 x / 100 = 75 x / 100 = 15 = > x = 1500 . answer : b" | a = 5 + 6
b = a * 100
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a ) $ 153 , b ) $ 848 , c ) $ 398 , d ) $ 549 , e ) $ 675 | b | subtract(980, divide(multiply(subtract(1024, 980), 3), 4)) | a sum of money at simple interest amounts to $ 980 in 3 years and to $ 1024 in 4 years . the sum is : | "b $ 848 s . i . for 1 year = $ ( 1024 - 980 ) = $ 44 . s . i . for 3 years = $ ( 44 x 3 ) = $ 132 . principal = $ ( 980 - 132 ) = $ 848 ." | a = 1024 - 980
b = a * 3
c = b / 4
d = 980 - c
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a ) 8 , b ) 2 , c ) 7 , d ) 5 , e ) 1 | d | subtract(const_4, const_3) | the perimeter of a rectangular yard is completely surrounded by a fence that measures 18 meters . what is the length of the yard if the area of the yard is 20 meters squared ? | "perimeter of rectangular yard = 2 ( l + b ) = 18 - - > l + b = 9 area = l * b = 20 b = 9 - l l ( 9 - l ) = 20 29 l - l ^ 2 = 20 l ^ 2 - 9 l + 20 = 0 upon simplifying we get l = 5 or 4 . only 5 is there in the answer choice . answer : d" | a = 4 - 3
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a ) 12.5 % , b ) 13 % , c ) 16 % , d ) 15 % , e ) 19 % | d | divide(multiply(subtract(multiply(const_100, const_100), multiply(add(const_100, 25), 68)), const_100), multiply(const_100, const_100)) | the shopkeeper increased the price of a product by 25 % so that customer finds it difficult to purchase the required amount . but somehow the customer managed to purchase only 68 % of the required amount . what is the net difference in the expenditure on that product ? | "quantity x rate = price 1 x 1 = 1 0.68 x 1.25 = 0.85 decrease in price = ( 0.15 / 1 ) × 100 = 15 % d )" | a = 100 * 100
b = 100 + 25
c = b * 68
d = a - c
e = d * 100
f = 100 * 100
g = e / f
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a ) 80.0 , b ) 80.9 , c ) 77.4 , d ) 81.1 , e ) 81.9 | c | multiply(multiply(divide(subtract(const_100, 10), const_100), divide(subtract(const_100, 14), const_100)), const_100) | a store reduced the price of all items in the store by 10 % on the first day and by another 14 % on the second day . the price of items on the second day was what percent of the price before the first reduction took place ? | "consider price of the all items as $ 100 after a initial reduction of 10 % price becomes = 0.9 * 100 = $ 90 after the final reduction of 14 % price becomes = 0.86 * 90 = $ 77.4 price of all items on second day is 77.4 % of price on first day correct answer option c" | a = 100 - 10
b = a / 100
c = 100 - 14
d = c / 100
e = b * d
f = e * 100
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a ) 20 % , b ) 15 % , c ) 30 % , d ) 25 % , e ) none of these | d | multiply(divide(subtract(25, 15), add(25, 15)), const_100) | if 25 % of ( x - y ) = 15 % of ( x + y ) , then what percent of x is y ? | "explanation : solution : 25 % of ( x - y ) = 15 % of ( x + y ) 25 ( x - y ) / 100 = 15 ( x + y ) / 100 5 ( x - y ) = 3 ( x + y ) x = 4 y . . ' . required percentage = ( y * 100 / x ) % = y * 100 / 4 y = 25 % answer : d" | a = 25 - 15
b = 25 + 15
c = a / b
d = c * 100
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a ) 65 kg , b ) 70 kg , c ) 75 kg , d ) 80 kg , e ) 85 kg | c | subtract(multiply(80, const_4), subtract(multiply(79, const_4), add(subtract(multiply(80, const_4), multiply(84, 3)), 3))) | the average weight of 3 men a , b and c is 84 kg . the average weight becomes 80 kg when d joins them . if e whose weight is 3 kg more than d joins the group replacing a , then the average weight of b , c , d and e becomes 79 kg . the weight of a is : | wt of abc = 84 * 3 = 252 kg wt of abcd = 80 * 4 = 320 kg wt of d = 68 kg wt of e = 71 kg wt of abcde = 320 + 71 = 391 kg wt of bcde = 79 * 4 = 316 kg wt of a = 391 - 316 = 75 kg answer : c | a = 80 * 4
b = 79 * 4
c = 80 * 4
d = 84 * 3
e = c - d
f = e + 3
g = b - f
h = a - g
|
a ) 1 : 4 , b ) 3 : 4 , c ) 1 : 2 , d ) 3 : 1 , e ) 3 : 2 | b | divide(add(multiply(divide(2, 3), divide(subtract(multiply(6, 3), 6), subtract(multiply(multiply(const_1, const_3), divide(2, 3)), 1))), 4), add(divide(subtract(multiply(6, 3), 6), subtract(multiply(multiply(const_1, const_3), divide(2, 3)), 1)), 4)) | the ratio of the present ages of two friends is 2 : 3 and 6 years back , the ratio was 1 : 3 . what will be the ratio of their ages after 4 years ? | let the ages be 2 x , 3 x 6 yrs back , so 2 x - 6 / 3 x - 6 = 1 / 3 x = 4 after 4 yrs 2 x + 43 x + 4 2 ( 4 ) + 43 ( 4 ) + 4 12 : 16 3 : 4 answer : b | a = 2 / 3
b = 6 * 3
c = b - 6
d = 1 * 3
e = 2 / 3
f = d * e
g = f - 1
h = c / g
i = a * h
j = i + 4
k = 6 * 3
l = k - 6
m = 1 * 3
n = 2 / 3
o = m * n
p = o - 1
q = l / p
r = q + 4
s = j / r
|
a ) 46.75 , b ) 40 , c ) 33.66 , d ) 66.66 , e ) 66.86 | a | multiply(multiply(subtract(const_1, divide(45, const_100)), divide(85, const_100)), const_100) | if 85 % of the population of an ant colony is red , and of these 45 % are females , then what % of the total ant population are male red ants ? | 55 % are males percentage of male red ants is 55 * . 85 = 46.75 answer : a | a = 45 / 100
b = 1 - a
c = 85 / 100
d = b * c
e = d * 100
|
a ) $ 20000 , b ) $ 15000 , c ) $ 12000 , d ) $ 10000 , e ) $ 9000 | c | divide(multiply(multiply(add(const_2, const_3), const_1000), 6), const_2) | if money is invested at r percent interest , compounded annually , the amount of the investment will double in approximately 54 / r years . if pat ' s parents invested $ 6,000 in a long - term bond that pays 6 percent interest , compounded annually , what will be the approximate total amount of the investment 18 years later , when pat is ready for college ? | "since investment doubles in 54 / r years , then for r = 6 it ' ll double in 54 / 6 = ~ 9 years ( we are not asked about the exact amount so such an approximation will do ) . thus after 18 years investment will become $ 6,000 * 2 = $ 12,000 . answer : c ." | a = 2 + 3
b = a * 1000
c = b * 6
d = c / 2
|
a ) 28 sec , b ) 32 sec , c ) 36 sec , d ) 38 sec , e ) 40 sec | a | multiply(const_3600, divide(divide(210, const_1000), add(25, 2))) | a train 210 metres long is moving at a speed of 25 kmph . it will cross a man coming from the opposite direction at a speed of 2 km per hour in : | "relative speed = ( 25 + 2 ) km / hr = 27 km / hr = ( 27 × 5 / 18 ) m / sec = 15 / 2 m / sec . time taken by the train to pass the man = ( 210 × 2 / 15 ) sec = 28 sec answer : a" | a = 210 / 1000
b = 25 + 2
c = a / b
d = 3600 * c
|
a ) 100 , b ) 350 , c ) 400 , d ) 450 , e ) 500 | a | subtract(divide(multiply(multiply(3, 500), 40), const_100), add(divide(multiply(500, 55), const_100), divide(multiply(500, 45), const_100))) | in 3 annual examinations , of which the aggregate marks of each was 500 , a student secured average marks 45 % and 55 % in the first and the second yearly examinations respectively . to secure 40 % average total marks , it is necessary for him in third yearly examination to secure marks : | total marks : 1500 for three exams 40 % of 1500 = 600 first exam marks = 45 % of 500 = 225 second exam marks = 55 % of 500 = 275 let x be the third exam marks 225 + 275 + x = 600 x = 100 answer : a | a = 3 * 500
b = a * 40
c = b / 100
d = 500 * 55
e = d / 100
f = 500 * 45
g = f / 100
h = e + g
i = c - h
|
a ) 22 , b ) 23 , c ) 24 , d ) 26 , e ) 28 | a | divide(subtract(multiply(46, const_4), 140), const_2) | a man has some hens and cows . if the number of heads be 46 and the number of feet equals 140 , then the number of hens will be : | "let hens be x and cows be y now , feet : x * 2 + y * 4 = 140 heads : x * 1 + y * 1 = 46 implies , 2 x + 4 y = 140 and x + y = 46 solving these two equations , we get x = 22 and y = 24 therefore , hens are 22 . answer : a" | a = 46 * 4
b = a - 140
c = b / 2
|
a ) 484 , b ) 726 , c ) 1,100 , d ) 1,320 , e ) 1,560 | e | multiply(divide(subtract(subtract(multiply(20, const_100), const_10), const_10), add(add(const_1, divide(20, const_100)), const_1)), add(const_1, divide(20, const_100))) | yesterday ' s closing prices of 2,860 different stocks listed on a certain stock exchange were all different from today ' s closing prices . the number of stocks that closed at a higher price today than yesterday was 20 percent greater than the number that closed at a lower price . how many of the stocks closed at a higher price today than yesterday ? | "lets consider the below - the number of stocks that closed at a higher price = h the number of stocks that closed at a lower price = l we understand from first statement - > h + l = 2860 - - - - ( 1 ) we understand from second statement - > h = ( 120 / 100 ) l = > h = 1.2 l - - - - ( 2 ) solve eq ( 1 ) ( 2 ) to get h = 1560 . e is my answer ." | a = 20 * 100
b = a - 10
c = b - 10
d = 20 / 100
e = 1 + d
f = e + 1
g = c / f
h = 20 / 100
i = 1 + h
j = g * i
|
['a ) 4 ⁄ 5', 'b ) 3 ⁄ 5', 'c ) 5 ⁄ 6', 'd ) 6 ⁄ 7', 'e ) none of these'] | a | multiply(divide(circle_area(divide(10, const_2)), square_area(10)), const_100) | from a square piece of a paper having each side equal to 10 cm , the largest possible circle is being cut out . the ratio of the area of the circle to the area of the original square is nearly : | area of the square = ( 10 ) 2 = 100 cm 2 area of the circle = 22 ⁄ 7 × ( 5 ) 2 = 22 × 25 / 7 required ratio = 22 × 25 / 7 × 100 = 22 / 28 = 11 / 14 = 0.785 ≈ 0.8 = 4 ⁄ 5 answer a | a = 10 / 2
b = circle_area / (
c = b * square_area
|
a ) 381 , b ) 382 , c ) 383 , d ) 384 , e ) 385 | c | divide(add(add(const_2, 47), multiply(add(20, add(const_2, const_60)), const_60)), 13) | light glows for every 13 seconds . how many times did it between 1 : 57 : 58 and 3 : 20 : 47 am | the diff in sec between 1 : 57 : 58 and 3 : 20 : 47 is 4969 sec , 4969 / 13 = 382 . so total 383 times light ll glow answer : c | a = 2 + 47
b = 2 + const_60
c = 20 + b
d = c * const_60
e = a + d
f = e / 13
|
a ) 2 : 5 , b ) 2 : 3 , c ) 1 : 8 , d ) 4 : 1 , e ) 3 : 2 | d | divide(add(2, 2), const_1) | if 7 men and 2 boys working together , can do 6 times as much work per hour as a man and a boy together . find the ratio of the work done by a man and that of a boy for a given time ? | 7 m + 2 b = 6 ( 1 m + 1 b ) 7 m + 2 b = 6 m + 6 b 1 m = 4 b the required ratio of work done by a man and a boy = 4 : 1 answer : d | a = 2 + 2
b = a / 1
|
a ) 68 , b ) 62.4 , c ) 86 , d ) 72.8 , e ) 108 | b | add(52, multiply(divide(20, const_100), 52)) | if x is 20 percent greater than 52 , then x = | "x is 20 % greater than 52 means x is 1.2 times 52 ( in other words 52 + 20 / 100 * 52 = 1.2 * 52 ) therefore , x = 1.2 * 88 = 62.4 answer : b" | a = 20 / 100
b = a * 52
c = 52 + b
|
a ) 10 am , b ) 12 am , c ) 10.30 am , d ) 12.30 am , e ) 1 pm | e | add(divide(add(245, 25), add(20, 25)), 7) | two stations p and q are 245 km apart on a straight track . one train starts from p at 7 a . m . and travels towards q at 20 kmph . another train starts from q at 8 a . m . and travels towards p at a speed of 25 kmph . at what time will they meet ? | "assume both trains meet after x hours after 7 am distance covered by train starting from p in x hours = 20 x km distance covered by train starting from q in ( x - 1 ) hours = 25 ( x - 1 ) total distance = 245 = > 20 x + 25 ( x - 1 ) = 245 = > 45 x = 270 = > x = 6 means , they meet after 6 hours after 7 am , ie , they meet at 1 pm answer is e ." | a = 245 + 25
b = 20 + 25
c = a / b
d = c + 7
|
a ) 130 , b ) 132 , c ) 135 , d ) 136 , e ) 138 | c | divide(1215, subtract(43, 34)) | a girl was asked to multiply a certain number by 43 . she multiplied it by 34 and got his answer less than the correct one by 1215 . find the number to be multiplied . | let the required number be x . then , 43 x – 34 x = 1215 or 9 x = 1215 or x = 135 . required number = 135 answer : c | a = 43 - 34
b = 1215 / a
|
a ) 18.75 , b ) 19.75 , c ) 20.75 , d ) 21.75 , e ) 22.75 | a | subtract(multiply(divide(const_100, 64), divide(multiply(60, 84), const_100)), 60) | there is 60 lit of milk and water in which milk forms 84 % . howmuch water must be added to this solution to make it solution in which milk forms 64 % | "60 * 84 / 100 = 50.40 lit milk that is 9.60 lit water let x lit water will be added then ( 60 + x ) * 64 / 100 = 50.40 so x = 18.75 answer : a" | a = 100 / 64
b = 60 * 84
c = b / 100
d = a * c
e = d - 60
|
a ) 55 , b ) 65 , c ) 50 , d ) 75 , e ) 85 | c | subtract(power(10, 2), multiply(2, 25)) | if x + y = - 10 , and x = 25 / y , what is the value of x ^ 2 + y ^ 2 ? | "x ^ 2 + y ^ 2 should make you think of these formulas : ( x + y ) ( x + y ) = x ^ 2 + y ^ 2 + 2 xy we already know ( x + y ) = - 10 and x * y = 25 ( x + y ) ( x + y ) = ( - 10 ) ( - 10 ) = x ^ 2 + y ^ 2 + 2 * ( 25 ) x ^ 2 + y ^ 2 = 100 - 50 = 50 answer : c" | a = 10 ** 2
b = 2 * 25
c = a - b
|
a ) 1 / 5 , b ) 29 / 35 , c ) 5 / 6 , d ) 35 / 29 , e ) 35 / 12 | d | divide(const_1, add(divide(2, 5), divide(3, 7))) | a type r machine can complete a job in 5 hours and a type b machine can complete the job in 7 hours . how many hours will it take 2 type r machines and 3 type b machines working together and independently to complete the job ? | now d should be the answer . r need 5 hours to complete and b needs 7 hours to compete so 2 r + 3 b will complete 2 / 5 + 3 / 7 or 29 / 35 portion of the job in 1 hour so the whole job will take 35 / 29 hours . . . . = d | a = 2 / 5
b = 3 / 7
c = a + b
d = 1 / c
|
a ) 86.6 km , b ) 46.6 km , c ) 35.2 km , d ) 35.6 km , e ) 26.6 km | c | multiply(add(42, 6), divide(44, const_60)) | the speed of a boat in still water in 42 km / hr and the rate of current is 6 km / hr . the distance travelled downstream in 44 minutes is : | "speed downstream = ( 42 + 6 ) = 48 kmph time = 44 minutes = 44 / 60 hour = 11 / 15 hour distance travelled = time × speed = 11 / 15 × 48 = 35.2 km answer : c" | a = 42 + 6
b = 44 / const_60
c = a * b
|
a ) 7 , b ) 13 , c ) 16 , d ) 21 , e ) 23 | e | add(subtract(100, multiply(12, 7)), 7) | a basketball team composed of 12 players scored 100 points in a particular contest . if none of the individual players scored fewer than 7 points , what is the greatest number of points l that an individual player might have scored ? | "general rule for such kind of problems : to maximize one quantity , minimize the others ; to minimize one quantity , maximize the others . thus to maximize the number of points of one particular player minimize the number of points of all other 11 players . minimum number of points for a player is 7 , so the minimum number of points of 11 players is 7 * 11 = 77 . therefore , the maximum number of points l for 12 th player is 100 - 77 = 23 . answer : e ." | a = 12 * 7
b = 100 - a
c = b + 7
|
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