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a ) 9.5 , b ) 9.4 , c ) 9.3 , d ) 9.2 , e ) 9.1 | c | divide(const_1, subtract(divide(const_1, 4), divide(const_1, 7))) | a cistern can be filled by a tap in 4 hours while it can be emptied by another tap in 7 hours . if both the taps are opened simultaneously , then after how much time will the cistern get filled ? | "net part filled in 1 hour = ( 1 / 4 - 1 / 7 ) = 3 / 28 the cistern will be filled in 28 / 3 hrs i . e . , 9.3 hrs . answer : c" | a = 1 / 4
b = 1 / 7
c = a - b
d = 1 / c
|
a ) 0.1484 , b ) 14.84 , c ) 1.484 , d ) 2.762 , e ) none | c | subtract(multiply(24.2, divide(12, const_100)), multiply(divide(10, const_100), 14.2)) | by how much is 12 % of 24.2 more than 10 % of 14.2 ? | "answer required difference = ( 12 x 24.2 ) / 100 - ( 10 x 14.2 ) / 100 = 2.904 - 1.42 = 1.484 . correct option : c" | a = 12 / 100
b = 24 * 2
c = 10 / 100
d = c * 14
e = b - d
|
a ) 4.56 min , b ) 5.28 min , c ) 5.08 min , d ) 9.28 min , e ) 5.988 min | a | multiply(divide(divide(627, const_1000), add(4.5, 3.75)), const_60) | the jogging track in a sports complex is 627 m in circumference . deepak and his wife start from the same point and walk in opposite directions at 4.5 km / hr and 3.75 km / hr respectively . they will meet for the first time in ? | "clearly , the two will meet when they are 627 m apart . to be ( 4.5 + 3.75 ) = 8.25 km apart , they take 1 hour . to be 627 m apart , they take ( 100 / 825 * 627 / 1000 ) hrs = ( 627 / 8250 * 60 ) min = 4.56 min . answer : a" | a = 627 / 1000
b = 4 + 5
c = a / b
d = c * const_60
|
a ) 76 sec , b ) 72 sec , c ) 98 sec , d ) 36 sec , e ) 23 sec | b | divide(add(240, 120), multiply(subtract(27, 9), divide(divide(const_10, const_2), divide(subtract(27, 9), const_2)))) | a jogger running at 9 km / hr along side a railway track is 240 m ahead of the engine of a 120 m long train running at 27 km / hr in the same direction . in how much time will the train pass the jogger ? | "speed of train relative to jogger = 27 - 9 = 18 km / hr . = 18 * 5 / 18 = 5 m / sec . distance to be covered = 240 + 120 = 360 m . time taken = 360 / 5 = 72 sec . answer : b" | a = 240 + 120
b = 27 - 9
c = 10 / 2
d = 27 - 9
e = d / 2
f = c / e
g = b * f
h = a / g
|
a ) 115 , b ) 116 , c ) 117 , d ) 118 , e ) 156 | e | add(add(multiply(divide(const_100, 45), 36), multiply(divide(50, 45), 36)), 36) | a sun is divided among x , y and z in such a way that for each rupee x gets , y gets 45 paisa and z gets 50 paisa . if the share of y is rs . 36 , what is the total amount ? | "x : y : z = 100 : 45 : 50 20 : 9 : 10 9 - - - 36 39 - - - ? = > 156 answer : e" | a = 100 / 45
b = a * 36
c = 50 / 45
d = c * 36
e = b + d
f = e + 36
|
a ) 2008 , b ) 2004 , c ) 2005 , d ) 2006 , e ) 2007 | a | add(1989, divide(subtract(70, 32), const_2)) | a certain company retirement plan has arule of 70 provision that allows an employee to retire when the employee ' s age plus years of employment with the company total at least 70 . in what year could a female employee hired in 1989 on her 32 nd birthday first be eligible to retire under this provision ? | "she must gain at least 70 points , now she has 32 and every year gives her two more points : one for age and one for additional year of employment , so 32 + 2 * ( # of years ) = 70 - - > ( # of years ) = 19 - - > 1989 + 19 = 2008 . answer : a ." | a = 70 - 32
b = a / 2
c = 1989 + b
|
a ) 5 , b ) 7 , c ) 6 , d ) 4 , e ) 3 | b | divide(subtract(115, multiply(5, 2)), add(10, 5)) | carina has 115 ounces of coffee divided into 5 - and 10 - ounce packages . if she has 2 more 5 - ounce packages than 10 - ounce packages , how many 10 - ounce packages does she have ? | "lets say 5 and 10 ounce packages be x and y respectively . given that , 5 x + 10 y = 115 and x = y + 2 . what is the value of y . substituting the x in first equation , 5 y + 10 + 10 y = 85 - > y = 105 / 15 . = 7 b" | a = 5 * 2
b = 115 - a
c = 10 + 5
d = b / c
|
a ) 3.6 % , b ) 3.9 % , c ) 4.2 % , d ) 4.5 % , e ) 4.8 % | d | multiply(divide(add(const_1, divide(multiply(6.5, 600), const_100)), const_1000), const_100) | by weight , liquid x makes up 1.5 percent of solution p and 6.5 percent of solution q . if 400 grams of solution p are mixed with 600 grams of solution q , then liquid x accounts for what percent of the weight of the resulting solution ? | "the number of grams of liquid x is 1.5 ( 400 ) / 100 + 6.5 ( 600 ) / 100 = 6 + 39 = 45 grams . 45 / 1000 = 4.5 % the answer is d ." | a = 6 * 5
b = a / 100
c = 1 + b
d = c / 1000
e = d * 100
|
a ) 8 % , b ) 62 % , c ) 2 % , d ) 9 % , e ) 22 % | a | multiply(subtract(divide(divide(multiply(subtract(const_100, 40), add(const_100, 80)), const_100), const_100), const_1), const_100) | a trader bought a car at 40 % discount on its original price . he sold it at a 80 % increase on the price he bought it . what percent of profit did he make on the original price ? | "original price = 100 cp = 60 s = 60 * ( 180 / 100 ) = 112 100 - 108 = 8 % answer : a" | a = 100 - 40
b = 100 + 80
c = a * b
d = c / 100
e = d / 100
f = e - 1
g = f * 100
|
a ) - 29 , b ) 50 , c ) - 30 , d ) 30 , e ) 60 | b | add(49, const_1) | what is the sum of 100 consecutive integers from - 49 inclusive , in a increasing order ? | "from - 49 to - 1 - - > 49 nos . zero - - > 1 number from + 1 to + 49 - - > 49 nos . when we add up nos . from - 49 to + 49 sum will be zero . total 99 nos will be added . 100 th number will be 50 . sum of these 100 nos . = 50 . b is the answer ." | a = 49 + 1
|
a ) 6 , b ) 5 , c ) 4 , d ) 3 , e ) 2 | d | multiply(const_12, divide(multiply(40, divide(40, const_100)), 64)) | a reduction of 40 % in the price of bananas would enable a man to obtain 64 more for rs . 40.00001 , what is reduced price per dozen ? | 40 * ( 40 / 100 ) = 16 - - - 64 ? - - - 12 = > rs . 3 answer : d | a = 40 / 100
b = 40 * a
c = b / 64
d = 12 * c
|
a ) 175 seconds , b ) 195 seconds , c ) 155 seconds , d ) 188 seconds , e ) 115 seconds | d | divide(24, 0.128) | an industrial loom weaves 0.128 metres of cloth every second . approximately , how many seconds will it take for the loom to weave 24 metre of cloth ? | "explanation : let the time required by x seconds . then , more cloth means more time ( direct proportion ) so , 0.128 : 1 : : 24 : x = > x = { \ color { blue } \ frac { 24 \ times 1 } { 0.128 } } = > x = 187.5 so time will be approx 188 seconds answer : d" | a = 24 / 0
|
a ) 25 , b ) 30 , c ) 28 , d ) 32 , e ) 42 | d | divide(subtract(multiply(84, 2), subtract(16, 8)), add(2, 3)) | if the average ( arithmetic mean ) of ( 2 a + 16 ) and ( 3 a - 8 ) is 84 , what is the value of a ? | "( ( 2 a + 16 ) + ( 3 a - 8 ) ) / 2 = ( 5 a + 8 ) / 2 = 84 a = 32 the answer is d ." | a = 84 * 2
b = 16 - 8
c = a - b
d = 2 + 3
e = c / d
|
a ) 67 , b ) 55 , c ) 62 , d ) 47 , e ) 57 | e | add(add(multiply(divide(const_100, 50), 30), multiply(divide(40, 50), 30)), 30) | a sun is divided among x , y and z in such a way that for each rupee x gets , y gets 50 paisa and z gets 40 paisa . if the share of x is rs . 30 , what is the total amount ? | "x : y : z = 100 : 50 : 40 10 : 5 : 4 10 - - - 30 19 - - - ? = > 57 answer : e" | a = 100 / 50
b = a * 30
c = 40 / 50
d = c * 30
e = b + d
f = e + 30
|
a ) a ) 29 , b ) b ) 25 , c ) c ) 27 , d ) d ) 31 , e ) e ) 35 | a | sqrt(add(power(sqrt(subtract(289, multiply(const_2, 276))), const_2), multiply(const_4, 276))) | the product of two numbers is 276 and the sum of their squares is 289 . the sum of the number is ? | "let the numbers be x and y . then , xy = 276 and x 2 + y 2 = 289 . ( x + y ) 2 = x 2 + y 2 + 2 xy = 289 + ( 2 x 276 ) = 841 x + y = 29 . option a" | a = 2 * 276
b = 289 - a
c = math.sqrt(b)
d = c ** 2
e = 4 * 276
f = d + e
g = math.sqrt(f)
|
a ) 60 kg , b ) 63 kg , c ) 50 kg , d ) 77 kg , e ) none of these | b | divide(subtract(multiply(27, divide(9.24, add(divide(10, const_100), const_1))), multiply(27, 7)), subtract(9, divide(9.24, add(divide(10, const_100), const_1)))) | how many kilograms of sugar costing rs . 9 per kg must be mixed with 27 kg of sugar costing rs . 7 per kg so that there may be gain of 10 % by selling the mixture at rs . 9.24 per kg ? | "explanation : let the rate of second quality be rs x per kg . step 1 : s . p of 1 kg of mixture = rs . 9.24 gain = 10 % c . p of 1 kg of mixture = [ 100 / ( 100 + 10 ) Γ 9.24 ] . = > rs . 8.40 . thus , the mean price = rs . 8.40 . step 2 : c . p of 1 kg of sugar of 1 st kind = 900 p c . p of 1 kg of sugar of 2 nd kind = 700 p mean price = 840 p by the rule of alligation , we have : c . p . of 1 kg of c . p . of 1 kg of sugar of 1 st sugar of 2 nd kind ( 900 p ) kind ( 700 p ) \ / mean price ( 840 p ) / \ 840 - 700 : 900 - 840 ( 140 ) ( 60 ) = > required ratio = 140 : 60 = 7 : 3 . step 3 : let , x kg of sugar of 1 st kind be mixed with 27 kg of 2 nd kind , then = > 7 : 3 = x : 27 . = > 7 / 3 = x / 27 . = > x = ( 7 / 3 ) x 27 . = > x = 63 . answer : b" | a = 10 / 100
b = a + 1
c = 9 / 24
d = 27 * c
e = 27 * 7
f = d - e
g = 10 / 100
h = g + 1
i = 9 / 24
j = 9 - i
k = f / j
|
a ) 1 / 2 , b ) 2 / 5 , c ) 3 / 5 , d ) 4 / 5 , e ) 5 / 6 | e | divide(multiply(5, 4), add(multiply(5, 4), multiply(1, 4))) | at a loading dock , each worker on the night crew loaded 1 / 4 as many boxes as each worker on the day crew . if the night crew has 4 / 5 as many workers as the day crew , what fraction of all the boxes loaded by the two crews did the day crew load ? | "method : x = no . of boxes loaded by day crew . boxes by night crew = 1 / 4 * 4 / 5 x = 1 / 5 x % loaded by day crew = x / ( x + 1 / 5 x ) = 5 / 6 answer e" | a = 5 * 4
b = 5 * 4
c = 1 * 4
d = b + c
e = a / d
|
a ) 205 , b ) 215 , c ) 225 , d ) 235 , e ) 245 | c | multiply(150, subtract(const_2, const_1)) | a train speeds past a pole in 15 seconds and a platform 150 meters long in 25 seconds . what is the length of the train ( in meters ) ? | "let the length of the train be x meters . the speed of the train is x / 15 . then , x + 150 = 25 * ( x / 15 ) 10 x = 2250 x = 225 meters the answer is c ." | a = 2 - 1
b = 150 * a
|
a ) 48 kmph , b ) 52 kmph , c ) 40 kmph , d ) 63 kmph , e ) 65 kmph | c | divide(420, multiply(divide(3, 2), 7)) | a jeep takes 7 hours to cover a distance of 420 km . how much should the speed in kmph be maintained to cover the same direction in 3 / 2 th of the previous time ? | "time = 7 distance = 420 3 / 2 of 7 hours = 7 * 3 / 2 = 10.5 hours required speed = 420 / 10.5 = 40 kmph c )" | a = 3 / 2
b = a * 7
c = 420 / b
|
a ) 200 Ο , b ) 240 Ο , c ) 300 Ο , d ) 480 Ο , e ) 150 Ο | e | multiply(multiply(multiply(multiply(divide(5, add(multiply(const_2, const_100), multiply(add(const_2, const_3), const_1000))), const_2), divide(add(const_2, multiply(const_2, const_10)), add(const_3, const_4))), 1,320), const_60) | the end of a blade on an airplane propeller is 5 feet from the center . if the propeller spins at the rate of 1,320 revolutions per second , how many miles will the tip of the blade travel in one minute ? ( 1 mile = 5,280 feet ) | "distance traveled in 1 revolution = 2 Ο r = 2 Ο 5 / 5280 revolutions in one second = 1320 revolutions in 60 seconds ( one minute ) = 1320 * 60 total distance traveled = total revolutions * distance traveled in one revolution 1320 * 60 * 2 Ο 5 / 5280 = 150 Ο e is the answer" | a = 2 * 100
b = 2 + 3
c = b * 1000
d = a + c
e = 5 / d
f = e * 2
g = 2 * 10
h = 2 + g
i = 3 + 4
j = h / i
k = f * j
l = k * 1
m = l * const_60
|
a ) 2784 , b ) 2287 , c ) 2977 , d ) 2668 , e ) 1298 | a | multiply(square_perimeter(sqrt(144)), 58) | what will be the cost of building a fence around a square plot with area equal to 144 sq ft , if the price per foot of building the fence is rs . 58 ? | "let the side of the square plot be a ft . a 2 = 144 = > a = 12 length of the fence = perimeter of the plot = 4 a = 48 ft . cost of building the fence = 48 * 58 = rs . 2784 . answer : a" | a = math.sqrt(144)
b = square_perimeter * (
|
a ) 0 , b ) 2 , c ) 4 , d ) 7 , e ) 9 | d | divide(1, 37) | what is the 36 th digit to the right of the decimal point in the decimal expansion of 1 / 37 ? | "1 / 37 = 0.027027 . . . so , we have a repeating cycle of 027 . every third digit ( 3 rd , 6 th , 9 th , . . . ) to the right of the decimal point is 7 , thus 36 th digit to the right of the decimal point is also 7 . answer : d ." | a = 1 / 37
|
a ) β 1 , b ) β 2 , c ) 4 , d ) 46 , e ) 48 | a | subtract(subtract(subtract(subtract(add(add(2, 2), subtract(2, 2)), const_1), const_1), const_1), const_1) | if a ( a - 2 ) = 2 and b ( b - 2 ) = 2 , where a β b , then a + b = | "i . e . if a = - 1 then b = 2 or if a = 2 then b = - 1 but in each case a + b = - 1 + 2 = - 1 answer : option a" | a = 2 + 2
b = 2 - 2
c = a + b
d = c - 1
e = d - 1
f = e - 1
g = f - 1
|
a ) 68 , b ) 54 , c ) 87 , d ) 96 , e ) 15 | a | divide(divide(400, 5280), multiply(4, divide(1, const_3600))) | if an object travels 400 feet in 4 seconds , what is the object ' s approximate speed in miles per hour ? ( note : 1 mile = 5280 feet ) | "1 mile = 5280 feet = > 1 feet = 1 / 5280 miles if the object travels 400 feet in 4 sec then it travels 400 / 4 * 60 * 60 feet in 1 hour ( 1 hr = 60 min * 60 sec ) = 3600 * 100 feet in 1 hour = 360000 feet in 1 hr = 360000 / 5280 miles in 1 hour = 36000 / 528 miles / hr ~ 68 miles / hr answer - a" | a = 400 / 5280
b = 1 / 3600
c = 4 * b
d = a / c
|
a ) 12 , b ) 28 , c ) 160 , d ) 180 , e ) 190 | e | subtract(power(divide(add(10, 19), const_2), const_2), power(subtract(10, divide(add(10, 19), const_2)), const_2)) | if the sum and difference of two numbers are 10 and 19 respectively , then the difference of their square is : | let the numbers be x and y . then , x + y = 10 and x - y = 19 x 2 - y 2 = ( x + y ) ( x - y ) = 10 * 19 = 190 . answer : e | a = 10 + 19
b = a / 2
c = b ** 2
d = 10 + 19
e = d / 2
f = 10 - e
g = f ** 2
h = c - g
|
a ) 95 , b ) 90 , c ) 85 , d ) 80 , e ) 56 | e | divide(14, divide(subtract(const_100, 75), const_100)) | if 14 percent of the students at a certain school went to a camping trip and took more than $ 100 , and 75 percent of the students who went to the camping trip did not take more than $ 100 , what percentage of the students at the school went to the camping trip ? | let x be the number of students in the school . 0.14 x students went to the trip and took more than 100 $ . they compose ( 100 - 75 ) = 25 % of all students who went to the trip . therefore the toal of 0.14 x / 0.25 = 0.56 x students went to the camping which is 56 % . the answer is e | a = 100 - 75
b = a / 100
c = 14 / b
|
a ) 66 , b ) 70 , c ) 72 , d ) 85 , e ) 108 | e | multiply(divide(multiply(120, 3), add(80, 120)), const_60) | cole drove from home to work at an average speed of 80 kmh . he then returned home at an average speed of 120 kmh . if the round trip took a total of 3 hours , how many minutes did it take cole to drive to work ? | "first round distance travelled ( say ) = d speed = 80 k / h time taken , t 2 = d / 80 hr second round distance traveled = d ( same distance ) speed = 120 k / h time taken , t 2 = d / 120 hr total time taken = 3 hrs therefore , 3 = d / 80 + d / 120 lcm of 80 and 120 = 240 3 = d / 80 + d / 120 = > 3 = 3 d / 240 + 2 d / 240 = > d = 240 * 2 / 5 km therefore , t 1 = d / 80 = > t 1 = ( 240 * 3 ) / ( 5 x 80 ) = > t 1 = ( 9 x 60 ) / 5 - - in minutes = > t 1 = 108 minutes . e" | a = 120 * 3
b = 80 + 120
c = a / b
d = c * const_60
|
a ) 126 , b ) 136 , c ) 146 , d ) 208 , e ) 266 | d | divide(multiply(16, 312), 24) | the reciprocal of the hcf and lcm of two are 1 / 16 and 1 / 312 . if one of the number is 24 then other no . is | "reciprocal of the hcf and lcm of two are 1 / 16 and 1 / 312 so , hcf = 16 , lcm = 312 lcm * hcf = product of two numbers = a * b = > b = lcm * hcf / a so , other = 16 * 312 / 24 = 208 answer : d" | a = 16 * 312
b = a / 24
|
a ) 10 , b ) 12 , c ) 14 , d ) 16 , e ) 17 | b | divide(360, add(subtract(31, 2), const_1)) | 360 metres long yard , 31 trees are palnted at equal distances , one tree being at each end of the yard . what is the distance between 2 consecutive trees | "31 trees have 30 gaps between them , required distance ( 360 / 30 ) = 12 b" | a = 31 - 2
b = a + 1
c = 360 / b
|
a ) 11 , b ) 13 , c ) 15 , d ) 17 , e ) 19 | d | add(add(divide(subtract(120, 40), 5), 1), 1) | a test has 120 questions . each question has 5 options , but only 1 option is correct . if test - takers mark the correct option , they are awarded 1 point . however , if an answer is incorrectly marked , the test - taker loses 0.25 points . no points are awarded or deducted if a question is not attempted . a certain group of test - takers attempted different numbers of questions , but each test - taker still received the same net score of 40 . what is the maximum possible number of such test - takers ? | "a correct answers get you 1 point , an incorrect answer gets you minus 1 / 4 point and a skipped question gets you 0 points . since there are 200 total questions , there are a variety of ways to get a total of 40 points . let c be the number of correct answers and let i be the number of incorrect answers . to get 40 points , a test taker must have at least 40 correct answers . then c = > 40 . for every correct question above 40 , the test taker has 4 incorrect answers . then , the i = 4 * ( c - 40 ) . also , i + c < = 120 . thus 5 c < = 280 and so c < = 56 . then 40 < = c < = 56 and c can have 17 possible values . the answer is d ." | a = 120 - 40
b = a / 5
c = b + 1
d = c + 1
|
a ) none , b ) one , c ) two , d ) three , e ) four | b | subtract(15, multiply(3, const_4)) | a = 2 ^ 15 - 625 ^ 3 and a / x is an integer , where x is a positive integer greater than 1 , such that it does not have a factor p such that 1 < p < x , then how many different values for x are possible ? | "this is a tricky worded question and i think the answer is should be d not c . . . here is my reason : the stem says that x is a positive integer such that has no factor grater than 2 and less than x itself . the stem wants to say that x is a prime number . because any prime number has no factor grater than 1 and itself . on the other hand the stem says that x could get how many different number not must get different number ( this is very important issue ) as our friends say , if we simplify numerator more we can obtain : 5 ^ 12 ( 5 ^ 3 - 1 ) = 5 ^ 12 ( 124 ) = 5 ^ 12 ( 31 * 2 * 2 ) divided by x and we are told that this fraction is an integer . so , x could be ( not must be ) 5 , 31 , or 2 ! ! ! so , x could get 1 different values and answer is b . . . ." | a = 3 * 4
b = 15 - a
|
a ) 227 m , b ) 150 m , c ) 187 m , d ) 167 m , e ) 200 m | e | multiply(divide(multiply(80, const_1000), const_3600), 9) | a train running at the speed of 80 km / hr crosses a pole in 9 sec . what is the length of the train ? | "speed = 80 * 5 / 18 = 200 / 9 m / sec length of the train = speed * time = 200 / 9 * 9 = 200 m answer : e" | a = 80 * 1000
b = a / 3600
c = b * 9
|
a ) 7 , b ) 10 , c ) 8 , d ) 2 , e ) 4 | b | divide(180, multiply(add(60, 6), const_0_2778)) | a train 180 m long is running with a speed of 60 km / hr . in what time will it pass a man who is running at 6 km / hr in the direction opposite to that in which the train is going ? | "speed of train relative to man = 60 + 6 = 66 km / hr . = 66 * 5 / 18 = 55 / 3 m / sec . time taken to pass the men = 180 * 3 / 55 = 10 sec . answer b" | a = 60 + 6
b = a * const_0_2778
c = 180 / b
|
a ) 11.3 , b ) 12.4 , c ) 12.7 , d ) 11.2 , e ) 10.3 | d | subtract(divide(multiply(38, 80), const_100), divide(multiply(12, 160), const_100)) | what is the difference between 38 % of 80 nd 12 % of 160 . | "( 38 / 100 ) * 80 Γ’ β¬ β ( 12 / 100 ) * 160 30.4 - 19.2 = 11.2 answer : d" | a = 38 * 80
b = a / 100
c = 12 * 160
d = c / 100
e = b - d
|
a ) 42 , b ) 44 , c ) 46 , d ) 48 , e ) 50 | d | inverse(add(divide(const_1, 80), divide(const_1, 120))) | paul completes a piece of work in 80 days , rose completes the same work in 120 days . if both of them work together , then the number of days required to complete the work is ? | if a can complete a work in x days and b can complete the same work in y days , then , both of them together can complete the work in x y / x + y days . that is , the required no . of days = 80 Γ 120 / 200 = 48 days answer is d | a = 1 / 80
b = 1 / 120
c = a + b
d = 1/(c)
|
a ) 400 , b ) 625 , c ) 1,250 , d ) 2,500 , e ) 10,000 | d | divide(100, divide(2, 100)) | in a certain pond , 100 fish were caught , tagged , and returned to the pond . a few days later , 100 fish were caught again , of which 2 were found to have been tagged . if the percent of tagged fish in the second catch approximates the percent of tagged fish in the pond , what is the approximate number of fish in the pond ? | "total fish = x percentage of second catch = ( 2 / 100 ) * 100 = 2 % so , x * 2 % = 50 x = 2500 ans . d" | a = 2 / 100
b = 100 / a
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a ) s 222 , b ) s 216 , c ) s 258 , d ) s 210 , e ) s 217 | c | add(add(210, divide(500, 100)), multiply(divide(20, 100), add(210, divide(500, 100)))) | the manufacturing cost of a shoe is rs . 210 and the transportation lost is rs . 500 for 100 shoes . what will be the selling price if it is sold at 20 % gains | "explanation : total cost of a watch = 210 + ( 500 / 100 ) = 215 . gain = 20 % = > sp = 1.2 cp = 1.2 x 215 = 258 answer : c" | a = 500 / 100
b = 210 + a
c = 20 / 100
d = 500 / 100
e = 210 + d
f = c * e
g = b + f
|
a ) 31 % . , b ) 44 % . , c ) 49 % . , d ) 29 % . , e ) 51 % . | b | multiply(divide(add(multiply(divide(40, const_100), 2), multiply(divide(60, const_100), 6)), 10), const_100) | a vessel of capacity 2 litre has 40 % of alcohol and another vessel of capacity 6 litre had 60 % alcohol . the total liquid of 8 litre was poured out in a vessel of capacity 10 litre and thus the rest part of the vessel was filled with the water . what is the new concentration of mixture ? | "40 % of 2 litres = 0.8 litres 60 % of 6 litres = 3.6 litres therefore , total quantity of alcohol is 4.4 litres . this mixture is in a 10 litre vessel . hence , the concentration of alcohol in this 10 litre vessel is 44 % b" | a = 40 / 100
b = a * 2
c = 60 / 100
d = c * 6
e = b + d
f = e / 10
g = f * 100
|
a ) 100 , b ) 130 , c ) 150 , d ) 180 , e ) 199 | b | divide(multiply(multiply(divide(422.50, const_100), 20), const_100), 65) | 65 % of x = 20 % of 422.50 . find the value of x ? | "65 % of x = 20 % of 422.50 then , 65 / 100 * x = 20 / 100 * 4225 / 10 x = 845 / 10 * 100 / 65 = 130 answer is b" | a = 422 / 50
b = a * 20
c = b * 100
d = c / 65
|
a ) 20 % , b ) 305 , c ) 50 % , d ) 55 % , e ) 60 % | c | subtract(multiply(divide(240, subtract(240, 80)), const_100), const_100) | on increasing the number of lines in a page by 80 , they become 240 . what is the % of increase in the no . of lines in the page ? | explanation : number of pages increased = 80 now , the number of pages of book = 240 number of pages of the books before increase = 240 β 80 = 160 % increase in the number of pages in the book = 80 160 x 100 % = 50 % c | a = 240 - 80
b = 240 / a
c = b * 100
d = c - 100
|
a ) 40 % , b ) 50 % , c ) 55 % , d ) 65 % , e ) 70 % | b | multiply(divide(add(multiply(3, const_4), multiply(add(3, const_3), 8)), 120), const_100) | a batsman scored 120 runs whichincluded 3 boundaries and 8 sixes . what % of his total score did he make by running between the wickets | number of runs made by running = 110 - ( 3 x 4 + 8 x 6 ) = 120 - ( 60 ) = 60 now , we need to calculate 60 is what percent of 120 . = > 60 / 120 * 100 = 50 % b | a = 3 * 4
b = 3 + 3
c = b * 8
d = a + c
e = d / 120
f = e * 100
|
a ) 3 , b ) 10 , c ) 9 , d ) 8 , e ) 7 | a | subtract(subtract(add(add(add(20, 3), add(20, 8)), add(multiply(4, const_10), 4)), multiply(3, 3)), subtract(86, 3)) | in a group of 86 students , each student is registered for at least one of 3 classes β history , math and english . 20 - 8 students are registered for history , 20 - 3 students are registered for math , and forty - 4 students are registered for english . if only 3 students are registered for all 3 classes , how many students are registered for exactly two classes ? | a u b u c = a + b + c - ab - bc - ac + abc 86 = 28 + 23 + 44 - ab - bc - ac + 3 = > ab + bc + ac = 12 exactly two classes = ab + bc + ac - 3 abc = 12 - 3 * 3 = 3 hence a | a = 20 + 3
b = 20 + 8
c = a + b
d = 4 * 10
e = d + 4
f = c + e
g = 3 * 3
h = f - g
i = 86 - 3
j = h - i
|
a ) 1 kmph , b ) 6 kmph , c ) 7 kmph , d ) 4 kmph , e ) 9 kmph | c | divide(subtract(24, 10), const_2) | a man can row his boat with the stream at 24 km / h and against the stream in 10 km / h . the man ' s rate is ? | "explanation : ds = 24 us = 10 s = ? s = ( 24 - 10 ) / 2 = 7 kmph answer : c" | a = 24 - 10
b = a / 2
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a ) 11 , b ) 12 , c ) 13 , d ) 14 , e ) 15 | b | subtract(100, add(add(add(subtract(100, 82), subtract(100, 75)), subtract(100, 85)), subtract(100, 70))) | there were totally 100 men . 82 are married . 75 have t . v , 85 have radio , 70 have a . c . how many men have t . v , radio , a . c and also married ? | "100 - ( 100 - 82 ) - ( 100 - 75 ) - ( 100 - 85 ) - ( 100 - 70 ) = 100 - 18 - 25 - 15 - 30 = 100 - 88 = 12 answer : b" | a = 100 - 82
b = 100 - 75
c = a + b
d = 100 - 85
e = c + d
f = 100 - 70
g = e + f
h = 100 - g
|
['a ) 38 .', 'b ) 40 .', 'c ) 42 .', 'd ) 44 .', 'e ) 45 .'] | e | subtract(multiply(sqrt(divide(900, 4)), const_4), sqrt(divide(900, 4))) | the roof of an apartment building is rectangular and its length is 4 times longer than its width . if the area of the roof is 900 feet squared , what is the difference between the length and the width of the roof ? | let the width = x x * 4 x = 900 x ^ 2 = 225 x = 15 length = 4 * 15 = 60 difference = 60 - 15 = 45 e is the answer | a = 900 / 4
b = math.sqrt(a)
c = b * 4
d = 900 / 4
e = math.sqrt(d)
f = c - e
|
a ) 24.5 , b ) 63 , c ) 65 , d ) 36 , e ) 24.19 | b | multiply(divide(18, subtract(9, 7)), 7) | sachin is younger than rahul by 18 years . if the ratio of their ages is 7 : 9 , find the age of sachin | "explanation : if rahul age is x , then sachin age is x - 18 , so , 9 x - 162 = 7 x 2 x = 162 x = 81 so sachin age is 81 - 18 = 63 answer : b ) 63" | a = 9 - 7
b = 18 / a
c = b * 7
|
a ) rs . 692.80 , b ) rs . 820 , c ) rs . 1170 , d ) rs . 1385 , e ) none | c | subtract(multiply(735, 4), multiply(590, subtract(4, const_1))) | average monthly income of a family of 4 earning members was rs . 735 . one of the earning members died and therefore , the average income came down to rs 590 . the income of the deceased was ? | "answer income of the deceased = total income of 4 members - total income of remaining 3 members . = 735 x 4 - 590 x 3 rs . = 1170 rs . correct option : c" | a = 735 * 4
b = 4 - 1
c = 590 * b
d = a - c
|
a ) 8 % , b ) 10 % , c ) 9 % , d ) 11 % , e ) 14 % | e | subtract(multiply(divide(subtract(37, 32), subtract(const_100, 32)), const_100), const_10) | two numbers are less than third number by 32 % and 37 % respectively . how much percent is the second number less than by the first | "let the third number is x . then first number = ( 100 - 32 ) % of x = 68 % of x second number is ( 63 x / 100 ) difference = 68 x / 100 - 63 x / 100 = x / 20 so required percentage is , difference is what percent of first number ( x / 20 * 100 / 68 x * 100 ) % = 14 % answer : e" | a = 37 - 32
b = 100 - 32
c = a / b
d = c * 100
e = d - 10
|
a ) 50 % , b ) 25 % , c ) 10 % , d ) 30 % , e ) 18 % | a | multiply(divide(divide(subtract(200, 100), 4), subtract(100, multiply(divide(subtract(200, 100), 4), 2))), const_100) | peter invested a certain sum of money in a simple interest bond whose value grew to $ 100 at the end of 2 years and further to $ 200 at the end of another 4 years . what was the rate of interest in which he invested his sum ? | "explanatory answer initial amount invested = $ x amount at the end of year 2 = $ 100 amount at the end of year 6 ( another 4 years ) = $ 200 therefore , the interest earned for the 4 year period between the 2 rd year and 6 th year = $ 200 - $ 100 = $ 100 as the simple interest earned for a period of 4 years is $ 100 , interest earned per year = $ 25 . therefore , interest earned for 2 years = 2 * 25 = $ 50 . hence , initial amount invested x = amount after 2 years - interest for 2 years = 100 - 50 = $ 50 . rate of interest = ( interest per year / principal ) * 100 = 25 / 50 * 100 = 50 % correct choice is a" | a = 200 - 100
b = a / 4
c = 200 - 100
d = c / 4
e = d * 2
f = 100 - e
g = b / f
h = g * 100
|
a ) 1.5 hours , b ) 1.27 hours , c ) 1.73 hours , d ) 2 hours , e ) 1.69 hours | e | divide(500, add(145, 150)) | two nascar stock cars take off from the starting line at the exact same time , heading in opposite directions . the budweiser car travels at 145 miles per hour , while the stella artois car travels at 150 miles per hour . at this rate , and ignoring other variable , how long will the cars have to drive in order to be 500 miles , in total , from each other ? | the cars travel ( 145 + 150 = 295 ) miles in one hour . 500 miles / 295 miles / hour = 1.69 hours answer is e | a = 145 + 150
b = 500 / a
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a ) 1 / 2 , b ) 2 / 3 , c ) 3 / 5 , d ) 4 / 5 , e ) 5 / 8 | b | divide(multiply(3, 4), add(multiply(3, 4), multiply(3, 2))) | at a loading dock , each worker on the night crew loaded 3 / 4 as many boxes as each worker on the day crew . if the night crew has 2 / 3 as many workers as the day crew , what fraction of all the boxes loaded by the two crews did the day crew load ? | "let x be the number of workers on the day crew . let y be the number of boxes loaded by each member of the day crew . then the number of boxes loaded by the day crew is xy . the number of boxes loaded by the night crew is ( 2 x / 3 ) ( 3 y / 4 ) = xy / 2 the total number of boxes is xy + xy / 2 = 3 xy / 2 the fraction loaded by the day crew is xy / ( 3 xy / 2 ) = 2 / 3 the answer is b ." | a = 3 * 4
b = 3 * 4
c = 3 * 2
d = b + c
e = a / d
|
a ) 784596 , b ) 845796 , c ) 804670 , d ) 784596 , e ) 643736 | e | divide(multiply(multiply(multiply(add(const_3, const_2), const_2), multiply(add(const_3, const_2), const_2)), 400), multiply(multiply(add(const_3, const_2), const_2), multiply(add(const_3, const_2), const_2))) | convert 400 miles into meters ? | "1 mile = 1609.34 meters 400 mile = 400 * 1609.34 = 643736 meters answer is e" | a = 3 + 2
b = a * 2
c = 3 + 2
d = c * 2
e = b * d
f = e * 400
g = 3 + 2
h = g * 2
i = 3 + 2
j = i * 2
k = h * j
l = f / k
|
a ) 20 , b ) 15 , c ) 25 , d ) 30 , e ) 35 | b | divide(add(add(add(multiply(7, const_3), add(7, multiply(7, const_2))), multiply(7, const_4)), multiply(add(const_4, const_1), 7)), 7) | find the average of all numbers between 6 and 36 which are divisible by 7 | explanation : average = ( 7 + 14 + 21 + 28 + 35 ) / 7 = 105 / 7 = 15 option b | a = 7 * 3
b = 7 * 2
c = 7 + b
d = a + c
e = 7 * 4
f = d + e
g = 4 + 1
h = g * 7
i = f + h
j = i / 7
|
a ) 144 , b ) 288 , c ) 36 , d ) 256 , e ) none | c | multiply(factorial(3), factorial(3)) | in how many ways 3 boys and 3 girls can be seated in a row so that they are alternate . | "solution : let the arrangement be , b g b g b g b 34 boys can be seated in 3 ! ways . girl can be seated in 3 ! ways . required number of ways , = 3 ! * 3 ! = 36 . answer : option c" | a = math.factorial(3)
b = math.factorial(3)
c = a * b
|
a ) 256485 , b ) 362880 , c ) 125425 , d ) 0 , e ) 125826 | b | multiply(const_1, subtract(const_1, const_1)) | if you multiply all the numbers on your mobile phone except 0 , what is the answer ? | "we have to multiply 1 to 9 to find the answer . therefore 1 * 2 * 3 * 4 * 5 * 6 * 7 * 8 * 9 = 362880 answer is b" | a = 1 - 1
b = 1 * a
|
a ) 150 , b ) 110 , c ) 100 , d ) 200 , e ) 50 | c | subtract(300, multiply(50, 4)) | a straight line in the xy - plane has slope 4 . on this line the x - coordinate of the point is 50 and y - coordinate is 300 then what is the y intercept of the plane ? | eq of line = y = mx + c m = 4 x = 50 y = 50 * 4 + c , substitute y by 300 as given in question . 300 = 200 + c , c = 100 correct option is c | a = 50 * 4
b = 300 - a
|
a ) 95 , b ) 99 , c ) 26 , d ) 73 , e ) none of the above | b | add(multiply(const_10, add(subtract(13, 5), const_1)), 5) | what two - digit number is less than the sum of the square of its digits by 13 and exceeds their doubled product by 5 ? | let the digits be x and y . the number would be 10 x + y . we are given that 2 xy + 5 = 10 x + y = x ^ 2 y ^ 2 - 13 thus 2 xy + 5 = x ^ 2 + y ^ 2 - 13 x ^ 2 + y ^ 2 - 2 xy = 16 ( x - y ) ^ 2 = 16 ( x - y ) = 4 or - 4 substituting the values of ( x - y ) in the equation 2 xy + 5 = 10 x + y x comes out to be 1 or 9 . . . thus the two numbers can be 15 or 99 thus the answer is b | a = 13 - 5
b = a + 1
c = 10 * b
d = c + 5
|
a ) 6 / Ο , b ) 9 / Ο , c ) 6 , d ) 9 , e ) 12 | d | add(divide(multiply(multiply(4, const_pi), divide(3, const_pi)), 2), multiply(const_pi, divide(3, const_pi))) | the surface area of a sphere is 4 Ο r 2 , where r is the radius of the sphere . if the area of the base of a hemisphere is 3 , what is the surface area q of that hemisphere ? | "given area of the base of a hemisphere is 3 = pi * r ^ 2 thus r = sqrt ( 3 / pi ) . surface area of whole sphere = 4 * pi * r ^ 2 . = 4 * pi * 3 / pi = 12 . since the hemisphere is half of a sphere the surface area of the hemisphere = 12 / 2 = 6 ( curved part , not including the flat rounded base ) . but the total surface area = 6 + area of the base of a hemisphere . = 6 + 3 = 9 . answer is d ! !" | a = 4 * math.pi
b = 3 / math.pi
c = a * b
d = c / 2
e = 3 / math.pi
f = math.pi * e
g = d + f
|
a ) 22277 , b ) 26782 , c ) 22882 , d ) 15000 , e ) 28761 | d | divide(multiply(10000, const_1), const_3) | a and b invests rs . 10000 each , a investing for 8 months and b investing for all the 12 months in the year . if the total profit at the end of the year is rs . 25000 , find their shares ? | "the ratio of their profits a : b = 8 : 12 = 2 : 3 share of a in the total profit = 2 / 5 * 25000 = rs . 10000 share of a in the total profit = 3 / 5 * 25000 = rs . 15000 answer : d" | a = 10000 * 1
b = a / 3
|
a ) 4500 , b ) 2678 , c ) 3600 , d ) 3400 , e ) 2500 | c | divide(multiply(multiply(3, 36), multiply(2, const_1000)), multiply(const_1, const_60)) | a river 3 m deep and 36 m wide is flowing at the rate of 2 kmph the amount of water that runs into the sea per minute is ? | "( 2000 * 3 * 36 ) / 60 = 3600 m 3 answer : c" | a = 3 * 36
b = 2 * 1000
c = a * b
d = 1 * const_60
e = c / d
|
a ) 400 meters , b ) 1111 meters , c ) 1777 meters , d ) 200 meters , e ) none of these | d | multiply(multiply(subtract(100, 64), const_0_2778), 20) | a train traveling at 100 kmph overtakes a motorbike traveling at 64 kmph in 20 seconds . what is the length of the train in meters ? | "train overtakes a bike means that we are talking about total length of the train . ( train ' s head is close to bike when it started and its tail crosses the bike when it overtakes the bike ) relative speed = 100 - 64 = 36 km / h = 36000 m / h time = 20 seconds distance = speed * time 36000 * 20 / 3600 = 200 meters . d is the answer ." | a = 100 - 64
b = a * const_0_2778
c = b * 20
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a ) 23 sec , b ) 30 sec , c ) 49 sec , d ) 54 sec , e ) 59 sec | d | divide(900, multiply(subtract(63, 3), const_0_2778)) | how many seconds will a 900 m long train take to cross a man walking with a speed of 3 km / hr in the direction of the moving train if the speed of the train is 63 km / hr ? | "speed of train relative to man = 63 - 3 = 60 km / hr . = 60 * 5 / 18 = 50 / 3 m / sec . time taken to pass the man = 900 * 3 / 50 = 54 sec . answer : d" | a = 63 - 3
b = a * const_0_2778
c = 900 / b
|
a ) 166 , b ) 332 , c ) 400 , d ) 402 , e ) 566 | d | sqrt(subtract(power(multiply(283, 2), 2), power(multiply(201, 2), 2))) | two airplanes take off from one airfield at noon . one flies due east at 201 miles per hour while the other flies directly northeast at 283 miles per hour . approximately how many miles apart are the airplanes at 2 p . m . ? | "d in two hours : the plane flying east will be 402 miles away from airport . the other plane will be 566 miles away from airport . 566 / 402 = ~ 1.4 = ~ sqrt ( 2 ) this means that planes formed a right isocheles triangle = > sides of such triangles relate as 1 : 1 : sqrt ( 2 ) = > the planes are 402 miles apart . d" | a = 283 * 2
b = a ** 2
c = 201 * 2
d = c ** 2
e = b - d
f = math.sqrt(e)
|
a ) 500 , b ) 620 , c ) 450 , d ) 340 , e ) 780 | e | add(add(add(add(add(add(const_12, const_2), const_1), add(add(const_12, const_2), add(add(add(add(add(const_2, const_4), const_4), subtract(const_10, const_1)), add(add(const_2, const_4), const_4)), add(const_10, const_2)))), add(add(add(const_12, const_2), const_1), const_1)), 10), add(const_2, const_4)) | what is the sum of all the multiples of 10 between 0 and 120 ? | "the multiples of 10 between 0 and 120 are 10 , 20 , 30 , 40 , 50 , 60 , 70 , 80 , 90 , 100 , 110 and 120 . if these are all added together , the result is 780 . final answer : e" | a = 12 + 2
b = a + 1
c = 12 + 2
d = 2 + 4
e = d + 4
f = 10 - 1
g = e + f
h = 2 + 4
i = h + 4
j = g + i
k = 10 + 2
l = j + k
m = c + l
n = b + m
o = 12 + 2
p = o + 1
q = p + 1
r = n + q
s = r + 10
t = 2 + 4
u = s + t
|
a ) 62,000 , b ) 85,500 , c ) 95,500 , d ) 100,500 , e ) 100,000 | c | divide(multiply(multiply(add(const_2, const_3), const_1000), 12), const_2) | if money is invested at r percent interest , compounded annually , the amount of the investment will double in approximately 50 / r years . if luke ' s parents invested $ 11,500 in a long term bond that pays 12 percent interest compounded annually , what will be the approximate total amount of the investment 12 years later , when luke is ready for college ? | answer equals c in 48 years . i thought by 50 th year it would reach 95,500 . options should have been separated more widely for clarity . | a = 2 + 3
b = a * 1000
c = b * 12
d = c / 2
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a ) 20 % , b ) 25 % , c ) 55 % , d ) 65 % , e ) none of above | a | multiply(divide(4, 20), const_100) | the ratio 4 : 20 expressed as percent equals to | explanation : actually it means 4 is what percent of 20 , which can be calculated as , ( 4 / 20 ) * 100 = 4 * 5 = 20 answer : option a | a = 4 / 20
b = a * 100
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a ) zero , b ) $ 81.90 , c ) $ 8190.03 , d ) $ 819.00 , e ) $ 8.19 | d | subtract(subtract(multiply(120000, power(add(const_1, divide(12, multiply(12, const_100))), 12)), 120000), divide(multiply(120000, 12), const_100)) | jack takes a loan of $ 120000 with 12 % annual interest : the interest is paid once , at the end of the year . jill takes a loan of $ 120000 with 12 % annual interest , compounding monthly at the end of each month . at the end of one full year , compared to jack ' s loan interest , approximately how much more does jill have to repay ? | jack ' s interest = $ 120,000 * 0.12 = $ 14400 or $ 1,200 each month . jills β s interest , 12 % / 12 = 1 % each month : for the 1 st month = $ 120,000 * 0.01 = $ 1,200 ; for the 2 nd month = $ 1,200 + 1 % of 1,200 = $ 1,212 , so we would have interest earned on interest ( very small amount ) ; for the 3 rd month = $ 1,212 + 1 % of 1,212 = ~ $ 1,224 ; for the 4 th month = $ 1,224 + 1 % of 1,224 = ~ $ 1,236 ; . . . for the 12 th month = $ 1,320 + 1 % of 1,320 = ~ $ 1,332 . the difference between jack ' s interest and jill β s interest = ~ ( 12 + 24 + . . . + 132 ) = $ 792 . answer : d . | a = 12 * 100
b = 12 / a
c = 1 + b
d = c ** 12
e = 120000 * d
f = e - 120000
g = 120000 * 12
h = g / 100
i = f - h
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a ) 0 , b ) 2 , c ) 2.5 , d ) 7 / 3 , e ) 10 | d | divide(subtract(32, subtract(multiply(4, divide(8, 2)), 12)), 12) | the area of one square is x ^ 2 + 8 x + 16 and the area of another square is 4 x ^ 2 β 12 x + 9 . if the sum of the perimeters of both squares is 32 , what is the value of x ? | "spotting the pattern of equations both are in form of ( x + c ) ^ 2 so a 1 = ( x + 4 ) ^ 2 a 2 = ( 2 x - 3 ) ^ 2 l 1 = x + 5 l 2 = 2 x - 3 p 1 = 4 ( x + 4 ) p 2 = 4 ( 2 x - 3 ) p 1 + p 2 = 32 4 ( x + 4 ) + 4 ( 2 x - 3 ) = 32 . . . . . . . . . . . . . . > x = 7 / 3 answer : d" | a = 8 / 2
b = 4 * a
c = b - 12
d = 32 - c
e = d / 12
|
a ) 700 , b ) 900 , c ) 705 , d ) none of these , e ) 506 | b | divide(const_100.0, divide(05, 45)) | evaluate 45 / . 05 | "explanation : 45 / . 05 = 4500 / 5 = 900 option b" | a = 5 / 45
b = 100 / 0
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a ) 5 , b ) 305 , c ) 365 , d ) 405 , e ) 495 | a | add(add(3, 1), 1) | how many 3 - digit even numbers are possible such that if one of the digits is 5 , the next / succeeding digit to it should be 1 ? | 510 , 512 , 514 , 516 , and 518 , so total 5 . hence option a . | a = 3 + 1
b = a + 1
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a ) 250 , b ) 300 , c ) 350 , d ) 400 , e ) 500 | e | add(multiply(const_2, 150), 200) | brenda and sally run in opposite direction on a circular track , starting at diametrically opposite points . they first meet after brenda has run 150 meters . they next meet after sally has run 200 meters past their first meeting point . each girl runs at a constant speed . what is the length of the track in meters ? | nice problem . + 1 . first timetogetherthey run half of the circumference . second timetogetherthey run full circumference . first time brenda runs 100 meters , thus second time she runs 2 * 150 = 300 meters . since second time ( when they run full circumference ) brenda runs 300 meters and sally runs 200 meters , thus the circumference is 300 + 200 = 500 meters . answer : e . | a = 2 * 150
b = a + 200
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a ) 36 , b ) 72 , c ) 96 , d ) 144 , e ) 180 | c | divide(400, add(divide(100, const_60), divide(150, const_60))) | a metal company ' s old machine makes bolts at a constant rate of 100 bolts per hour . the company ' s new machine makes bolts at a constant rate of 150 bolts per hour . if both machines start at the same time and continue making bolts simultaneously , how many minutes will it take the two machines to make a total of 400 bolts ? | "old machine 100 bolts in 60 mins so , 5 / 3 bolts in 1 min new machine 150 bolts in 60 mins so , 5 / 2 bolts in 1 min together , 5 / 3 + 5 / 2 = 25 / 6 bolts in 1 min so , for 400 bolts 400 * 6 / 25 = 96 mins ans c" | a = 100 / const_60
b = 150 / const_60
c = a + b
d = 400 / c
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a ) 27 , b ) 36 , c ) 29 , d ) 10 , e ) 110 | e | subtract(multiply(divide(subtract(90, 40), subtract(const_12, 9)), const_12), 90) | gopi gives rs . 90 plus one turban as salary to his servant for one year . the servant leaves after 9 months and receives rs . 40 and the turban . find the price of the turban . | "let the price of turban be x . thus , for one year the salary = ( 90 + x ) for 9 months he should earn 3434 ( 90 + x ) . now he gets one turban and rs . 40 . thus , 3434 ( 90 + x ) = 40 + x or 270 + 3 x = 160 + 4 x or x = 110 answer : e" | a = 90 - 40
b = 12 - 9
c = a / b
d = c * 12
e = d - 90
|
a ) 600 m , b ) 300 m , c ) 400 m , d ) 700 m , e ) 500 m | a | divide(multiply(18, multiply(1.6, const_1000)), 48) | amar takes as much time in running 18 meters as a car takes in covering 48 meters . what will be the distance covered by amar during the time the car covers 1.6 km ? | "a 600 m distance covered by amar = 18 / 4.8 ( 1.6 km ) = 3 / 8 ( 1600 ) = 600 m" | a = 1 * 6
b = 18 * a
c = b / 48
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a ) 615 m , b ) 240 m , c ) 168 m , d ) 300 m , e ) 691 m | d | multiply(20, multiply(54, const_0_2778)) | a train passes a station platform in 50 sec and a man standing on the platform in 20 sec . if the speed of the train is 54 km / hr . what is the length of the platform ? | speed = 54 * 5 / 18 = 15 m / sec . length of the train = 15 * 20 = 300 m . let the length of the platform be x m . then , ( x + 300 ) / 50 = 15 = > x = 300 m answer : d | a = 54 * const_0_2778
b = 20 * a
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a ) 1833 , b ) 1853 , c ) 1733 , d ) 1233 , e ) 1832 | a | divide(8247, 6) | the lowest number which should be added to 8247 so that the sum is exactly divisible by 5 , 6 , 7 , 8 and 9 is : | "l . c . m . of 5 , 6 , 7 , 8 and 9 = 2520 . on dividing 8247 by 2520 , the remainder is 687 . number to be added = ( 2520 - 687 ) = 1833 . answer : option ' a '" | a = 8247 / 6
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a ) 3 cm , b ) 9 cm , c ) 12 cm , d ) 15 cm , e ) none of the these | d | add(add(3, 5), 3) | two spherical balls lie on the ground touching . if one of the balls has a radius of 3 cm , and the point of contact is 5 cm above the ground , what is the radius of the other ball ? | "similar triangle properties . . 2 / r + 3 = 3 / r - 3 giving r = 15 . answer : d" | a = 3 + 5
b = a + 3
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a ) 112 , b ) 144 , c ) 136 , d ) 132 , e ) 118 | c | multiply(divide(add(multiply(8, 2), add(8, 10)), 2), 8) | in a class there are a total of 8 rows of desks and each desk can seat one student . there are 10 desks in the first row . in each subsequent row there are 2 more desks than in the previous row . find the maximum number of students seated in the class ? | the 8 rows form an arithmetic progression with first term 10 and last 24 . avg is 17 . so 17 * 8 = 136 answer is c . | a = 8 * 2
b = 8 + 10
c = a + b
d = c / 2
e = d * 8
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a ) 5 , b ) 5 ^ 2 , c ) 30 , d ) 30 ^ 2 , e ) 30 ^ 4 | e | add(const_3, const_4) | what is the smallest positive integer nn such that 6,480 β n β 6,480 β n is a perfect cube ? | "sol : let ' s factorize 6480 and we get 6480 = 3 ^ 4 * 2 ^ 4 * 5 now we need to see for what minimum value of n β n * 6480 = a ^ 3 where a is an integer so from 6480 we already have 2 ^ 4 * 3 ^ 4 * 5 * n β n = ( 2 ^ 2 ) ^ 3 * ( 3 ^ 2 ) ^ 3 * ( 5 ) ^ 3 why cause a is an integer which will need to be have the same factors which are in lhs solving for n β n = ( 2 ^ 6 * 3 ^ 6 * 5 ^ 3 ) / 2 ^ 4 * 3 ^ 4 * 5 and we get n β n = 2 ^ 2 * 3 ^ 2 * 5 ^ 2 or n = 2 ^ 4 * 3 ^ 4 * 5 ^ 4 or 30 ^ 4 ans is e" | a = 3 + 4
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a ) 100 marks , b ) 200 marks , c ) 280 marks , d ) 160 marks , e ) 827 marks | d | add(multiply(divide(add(40, 20), subtract(divide(30, const_100), divide(20, const_100))), divide(20, const_100)), 40) | a candidate who gets 20 % of the marks fails by 40 marks . but another candidate who gets 30 % marks gets 20 marks more than necessary for passing . find the number of marks for passing ? | "20 % - - - - - - - - - - - - 40 30 % - - - - - - - - - - - - 20 - - - - - - - - - - - - - - - - - - - - - - 10 % - - - - - - - - - - - - - 60 20 % - - - - - - - - - - - - - - 120 120 + 40 = 160 marks answer : d" | a = 40 + 20
b = 30 / 100
c = 20 / 100
d = b - c
e = a / d
f = 20 / 100
g = e * f
h = g + 40
|
a ) 7 , b ) 10 , c ) 8 , d ) 9 , e ) 11 | b | inverse(add(divide(6, multiply(10, 8)), divide(10, multiply(3, 12)))) | 10 women can complete a work in 8 days and 10 children take 12 days to complete the work . how many days will 6 women and 3 children together take to complete the work ? | "explanation : 1 women β s 1 day β s work = 1 / 8 / 10 = 1 / 80 1 child β s 1 day β s work = 1 / 12 / 10 = 1 / 120 6 women β s 1 day β s work = 1 / 80 Γ 6 = 3 / 40 3 children β s 1 day β s work = 1 / 120 Γ 3 = 1 / 40 6 women β s + 3 children β s 1 day β s work = 3 / 40 + 1 / 40 = 1 / 10 therefore , they will finish the whole work in 10 days . answer : option b" | a = 10 * 8
b = 6 / a
c = 3 * 12
d = 10 / c
e = b + d
f = 1/(e)
|
a ) 2000 , b ) 2573 , c ) 1600 , d ) 2950 , e ) none of these | c | multiply(200, divide(1600, 200)) | solve for x and check : - 200 x = 1600 | "solution : dividing each side by - 200 , we obtain ( - 200 x / - 200 ) = ( 1600 / - 200 ) therefore : x = - 8 check : - 200 x = 1600 ( - 200 * - 8 ) = 1600 1600 = 1600 answer : c" | a = 1600 / 200
b = 200 * a
|
a ) 50 hours , b ) 40 hours , c ) 17.5 hours , d ) 12 hours , e ) 8 hours | c | divide(140, multiply(divide(divide(20, 10), 5), 20)) | if 5 machines can produce 20 units in 10 hours , how long would it take 20 machines to produce 140 units ? | "here , we ' re told that 5 machines can produce 20 units in 10 hours . . . . that means that each machine works for 10 hours apiece . since there are 5 machines ( and we ' re meant to assume that each machine does the same amount of work ) , then the 5 machines equally created the 20 units . 20 units / 5 machines = 4 units are made by each machine every 10 hours now that we know how long it takes each machine to make 4 units , we can break this down further if we choose to . . . 10 hours / 4 units = 2.5 hours per unit when 1 machine is working . the prompt asks us how long would it take 20 machines to produce 140 units . if 20 machines each work for 2.5 hours , then we ' ll have 20 units . since 140 units is ' 7 times ' 20 , we need ' 7 times ' more time . ( 2.5 hours ) ( 7 times ) = 17.5 hours final answer : [ reveal ] spoiler : c" | a = 20 / 10
b = a / 5
c = b * 20
d = 140 / c
|
a ) 1 % , b ) 1.1 % , c ) 9.1 % , d ) 10 % , e ) 10.8 % | c | divide(multiply(1, const_100), add(1, const_100)) | the annual interest rate earned by an investment increased by 10 percent from last year to this year . if the annual interest rate earned by the investment this year was 1 percent , what was the annual interest rate last year ? | "let i = interest rate i ( this year ) = i ( last year ) + 0.10 i ( last year ) = 1.1 i ( last year ) 1 = 1.1 x i ( last year ) i ( last year ) = 1 / 1.1 = 10 / 11 = 0.90909 . . . or i ( last year ) = 0.91 % answer : c" | a = 1 * 100
b = 1 + 100
c = a / b
|
a ) 400 , b ) 360 , c ) 270 , d ) 180 , e ) 90 | b | subtract(450, divide(450, 5)) | there are 450 birds at a small zoo . the number of birds is 5 times the number of all the other animals combined . how many more birds are there than non bird animals at the zoo ? | total birds = 450 = 5 * other animals ( x ) or x = 90 . so , difference in birds and x = 450 - 90 = 360 . answer is b . | a = 450 / 5
b = 450 - a
|
a ) 15,2 , b ) 15 , 4 , c ) 15 , 8 , d ) 15 , 5 , e ) 15,7 | d | divide(divide(add(20, 10), const_2), const_2) | a man can row downstream at 20 kmph and upstream at 10 kmph . find the speed of the man in still water and the speed of stream respectively ? | "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 = 10 - - - ( 2 ) from ( 1 ) & ( 2 ) 2 x = 30 = > x = 15 , y = 5 . answer : d" | a = 20 + 10
b = a / 2
c = b / 2
|
a ) 29 , b ) 57 , c ) 35 , d ) 20 , e ) 30 | b | multiply(subtract(20, 1), 1) | an seller earns an income of re 1 on the first day of his business . on every subsequent day , he earns an income which is just thrice of that made on the previous day . on the 20 th day of business , he earns an income of : | "2 nd day he earns = 3 ( 2 β 1 ) 3 rd day he earns = 3 ( 3 β 1 ) on 20 th day he earns 3 ( 20 - 1 ) = 57 rupees answer : b" | a = 20 - 1
b = a * 1
|
a ) 2 , b ) 4 , c ) 8 , d ) 12 , e ) 16 | c | divide(16, const_2) | if 16 ^ y = 4 ^ 16 , what is y ? | "16 ^ y = 4 ^ 2 y = 4 ^ 16 2 y = 16 y = 8 the answer is c ." | a = 16 / 2
|
a ) 2 years , b ) 3 years , c ) 4 years , d ) 5 years , e ) 6 years | d | divide(subtract(divide(38, divide(4, 3)), multiply(subtract(4, const_1), 3)), 3) | the sum of the ages of 4 children born at the intervals of 3 years each is 38 years . what is the age of the youngest child ? | "let the ages of the children be x , ( x + 3 ) , ( x + 6 ) , ( x + 9 ) years . then , x + ( x + 3 ) + ( x + 6 ) + ( x + 9 ) = 38 4 x = 20 = > x = 5 . age of youngest child = x = 5 years . answer : d" | a = 4 / 3
b = 38 / a
c = 4 - 1
d = c * 3
e = b - d
f = e / 3
|
a ) 25 % , b ) 45 % , c ) 35 % , d ) 65 % , e ) 55 % | a | subtract(8, 10) | if the selling price of 8 articles is same as the cost price of 10 articles . find the gain or loss percentage ? | "let the c . p of each article be re 1 . then , s . p of 8 articles = c . p of 10 articles = rs . 10 now , c . p of 8 articles = rs . 8 , s . p of 8 articles = rs 10 gain = rs ( 10 - 8 ) = rs 2 . gain % = ( 2 / 8 Γ 100 ) % = 25 % answer : a" | a = 8 - 10
|
a ) 64 , b ) 72 , c ) 85 , d ) 96 , e ) 108 | c | divide(subtract(multiply(subtract(24.85, 0.85), 5), 52), 0.85) | a cricket player whose bowling average was 24.85 runs per wicket , takes 5 wicket for 52 runs in a match . due to this his average decreases by 0.85 . what will be the number of wickets taken by him till the last match ? | "average = total runs / total wickets total runs after last match = 24.85 w + 52 total wickets after last match = w + 5 ( 24.85 w + 52 ) / ( w + 5 ) = 24.85 - 0.85 = 24 w = 80 so total wickets aftr last match = w + 5 = 85 answer : c" | a = 24 - 85
b = a * 5
c = b - 52
d = c / 0
|
a ) 2 / 5 , b ) 3 / 5 , c ) 1 / 6 , d ) 1 , e ) none of these | d | subtract(divide(4, 5), divide(2, 3)) | what is difference between biggest and smallest fraction among 2 / 3 , 3 / 4 , 4 / 5 and 5 / 3 | "explanation : 2 / 3 = . 66 , 3 / 4 = . 75 , 4 / 5 = . 8 and 5 / 3 = 1.66 so biggest is 5 / 3 and smallest is 2 / 3 their difference is 5 / 3 - 2 / 3 = 3 / 3 = 1 option d" | a = 4 / 5
b = 2 / 3
c = a - b
|
a ) 10 , b ) 1000 , c ) 100 , d ) 10000 , e ) 100000 | c | divide(power(10, 222), power(10, 220)) | 10 ^ 222 Γ· 10 ^ 220 = ? | 10 ^ 222 Γ· 10 ^ 220 = 10 ^ ( 222 - 220 ) = 10 ^ 2 = 100 answer is c | a = 10 ** 222
b = 10 ** 220
c = a / b
|
a ) 285 , b ) 337 , c ) 878 , d ) 227 , e ) 291 | a | divide(add(multiply(add(floor(divide(30, add(const_3, const_4))), const_1), 510), multiply(subtract(30, add(floor(divide(30, add(const_3, const_4))), const_1)), 240)), 30) | a library has an average of 510 visitors on sundays and 240 on other days . the average number of visitors per day in a month of 30 days beginning with a sunday is | "explanation : since the month begins with a sunday , so there will be five sundays in the month , required average = ( 510 * 5 + 240 * 25 ) / 30 = 8550 / 30 = 285 . answer : a" | a = 3 + 4
b = 30 / a
c = math.floor(b)
d = c + 1
e = d * 510
f = 3 + 4
g = 30 / f
h = math.floor(g)
i = h + 1
j = 30 - i
k = j * 240
l = e + k
m = l / 30
|
a ) 75 kg , b ) 50 kg , c ) 70 kg , d ) 80 kg , e ) 60 kg | c | add(multiply(8, 2.5), 50) | the average weight of 8 person ' s increases by 2.5 kg when a new person comes in place of one of them weighing 50 kg . what is the weight of the new person ? | "explanation : total increase in weight = 8 Γ£ β 2.5 = 20 if x is the weight of the new person , total increase in weight = x Γ’ Λ β 50 = > 20 = x - 50 = > x = 20 + 50 = 70 answer : option c" | a = 8 * 2
b = a + 50
|
a ) 12.4 , b ) 14 , c ) 16 , d ) 18.6 , e ) none of these | c | add(15, divide(subtract(36, 26), 10)) | the average of 10 numbers is calculated as 15 . it is discovered later on that while calculating the average one number , namely 36 was wrongly read as 26 . the correct average is : | explanation : sum of numbers = ( 10 Γ Γ 15 - 26 + 36 ) = 160 correct average = 160 / 10 = 16 correct option : c | a = 36 - 26
b = a / 10
c = 15 + b
|
a ) s , b ) 3 s , c ) 4 s , d ) - 3 s , e ) - 4 s | b | divide(add(divide(9, gcd(6, 9)), gcd(6, 9)), divide(6, gcd(6, 9))) | if z is not equal to zero , and z = 6 zs β 9 s 2 β β β β β β β β β z = 6 zs β 9 s 2 , then z equals : | squaring on both the sides gives z ^ 2 = 6 zs - 9 s ^ 2 = z ^ 2 - 6 zs + 9 s ^ 2 = z ( z - 3 s ) - 3 s ( z - 3 s ) = ( z - 3 s ) ( z - 3 s ) = z = 3 s answer = b | a = math.gcd(6, 9)
b = 9 / a
c = math.gcd(6, 9)
d = b + c
e = math.gcd(6, 9)
f = 6 / e
g = d / f
|
a ) 18 , b ) 77 , c ) 25 , d ) 55 , e ) 31 | c | add(add(add(multiply(5, const_3), 5), multiply(multiply(5, const_3), add(multiply(5, const_3), 5))), 5) | sushil got thrice as many marks in english as in science . his total marks in english , science and maths are 115 . if the ratio of his marks in english and maths is 5 : 1 , find his marks in science ? | "s : e = 1 : 3 e : m = 5 : 1 - - - - - - - - - - - - s : e : m = 5 : 15 : 3 5 / 23 * 115 = 25 answer : c" | a = 5 * 3
b = a + 5
c = 5 * 3
d = 5 * 3
e = d + 5
f = c * e
g = b + f
h = g + 5
|
a ) 1 / 12 , b ) 7 / 12 , c ) 1 / 6 , d ) 1 / 3 , e ) 1 / 5 | b | divide(add(23, 5), const_60) | two boats are heading towards each other at constant speeds of 5 miles / hr and 23 miles / hr respectively . they begin at a distance 20 miles from each other . how far are they ( in miles ) one minute before they collide ? | "the question asks : how far apart will they be 1 minute = 1 / 60 hours before they collide ? since the combined rate of the boats is 5 + 23 = 25 mph then 1 / 60 hours before they collide they ' ll be rate * time = distance - - > 28 * 1 / 60 = 7 / 15 miles apart . answer : b ." | a = 23 + 5
b = a / const_60
|
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 e 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 . . e = 8 ( 2 + 3 + 4 . . . . + 12 ) = 616 . b" | a = math.floor(2)
b = 2 - a
c = 59 / 32
|
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