options stringlengths 37 300 | correct stringclasses 5
values | annotated_formula stringlengths 7 727 | problem stringlengths 5 967 | rationale stringlengths 1 2.74k | program stringlengths 10 646 |
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a ) 15 , b ) 20 , c ) 40 , d ) 44 , e ) 48 | a | divide(rectangle_area(multiply(2, const_100), multiply(25, const_100)), square_area(add(multiply(const_4, const_10), const_1))) | what is the least number of squares tiles required to pave the floor of a room 2 m 25 cm long and 1 m 35 cm broad ? | "length of largest tile = h . c . f . of 225 cm and 135 cm = 45 cm . area of each tile = ( 45 x 45 ) cm 2 . required number of tiles = 225 x 135 / ( 45 ^ 2 ) = 15 . answer : a" | a = 2 * 100
b = 25 * 100
c = rectangle_area / (
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a ) 23 , b ) 24 , c ) 25 , d ) 22 , e ) 27 | d | divide(subtract(multiply(30, 7), multiply(8, 7)), 7) | 7 people average age is 30 . youngest person age is 8 . 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 - 56 = 154 so average age would be 154 / 7 = 22 answer : d" | a = 30 * 7
b = 8 * 7
c = a - b
d = c / 7
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a ) 223 / 26 , b ) 288 / 27 , c ) 309 / 22 , d ) 288 / 26 , e ) 288 / 25 | e | divide(multiply(48, 6), divide(50, const_2)) | 24 people can write 48 book in 6 days working 8 hour a day . then in how many day 192 can be written by 50 people ? | work per day epr hour per person = 48 / ( 6 * 8 * 24 ) / / eq - 1 people = 50 ; let suppose day = p ; per day work for 8 hours acc . to condition work per day epr hour per person = 192 / ( p * 8 * 50 ) / / eq - 2 eq - 1 = = eq - 2 ; p = 288 / 25 answer : e | a = 48 * 6
b = 50 / 2
c = a / b
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a ) 16.12 % , b ) 25 % , c ) 16.56 % , d ) 17.66 % , e ) 18.1 % | b | multiply(divide(subtract(75, 60), 60), const_100) | john makes $ 60 a week from his job . he earns a raise andnow makes $ 75 a week . what is the % increase ? | "increase = ( 15 / 60 ) * 100 = ( 1 / 4 ) * 100 = 25 % . b" | a = 75 - 60
b = a / 60
c = b * 100
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a ) $ 75 , b ) $ 80 , c ) $ 90 , d ) $ 120 , e ) $ 180 | a | divide(5, subtract(1, add(add(divide(1, 10), divide(1, 3)), divide(1, 2)))) | at a supermarket , john spent 1 / 2 of his money on fresh fruits and vegetables , 1 / 3 on meat products , and 1 / 10 on bakery products . if he spent the remaining $ 5 on candy , how much did john spend at the supermarket ? | "let ' s let t = total number of dollars spent at the supermarket . with this variable we can set up an equation and determine t . we are given that john spent 1 / 2 of his money on fresh fruits and vegetables , or ( 1 / 2 ) t , 1 / 3 on meat products , or ( 1 / 3 ) t , and 1 / 10 on bakery products , or ( 1 / 10 ) t . we are also given that he spent the remaining $ 5 on candy . since we know where all his money was allocated , we can sum these values together and set the sum to t . so we have : ( 1 / 2 ) t + ( 1 / 3 ) t + ( 1 / 10 ) t + 5 = t to get rid of the fractions we can multiply the entire equation by 30 , and we obtain : 15 t + 10 t + 3 t + 150 = 30 t 28 t + 150 = 30 t 150 = 2 t t = 75 john spent $ 7 at the supermarket . answer : a" | a = 1 / 10
b = 1 / 3
c = a + b
d = 1 / 2
e = c + d
f = 1 - e
g = 5 / f
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a ) 45 min . , b ) 55 min . , c ) 65 min . , d ) 60 min . , e ) 50 min . | b | multiply(divide(multiply(divide(50, 60), 60), subtract(60, 6)), 60) | a bus started its journey from chennai and reached mumbai in 50 min with its average speed of 60 km / hr . if the average speed of the bus is increased by 6 km / hr , how much time will it take to cover the same distance ? | sol . distance between mumbai and pune = ( 60 * 50 ) / 60 = 50 average speed of the bus is increased by 6 km / hr then the speed of the bus = 66 km / hr required time = 50 * 66 / 60 = 55 min . b | a = 50 / 60
b = a * 60
c = 60 - 6
d = b / c
e = d * 60
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a ) 22 sec , b ) 88 sec , c ) 48 sec , d ) 18 sec , e ) 9 sec | c | add(45, 30) | two goods trains each 500 m long are running in opposite directions on parallel tracks . their speeds are 45 km / hr and 30 km / hr respectively . find the time taken by the slower train to pass the driver of the faster one ? | "relative speed = 45 + 30 = 75 km / hr . 75 * 5 / 18 = 125 / 6 m / sec . distance covered = 500 + 500 = 1000 m . required time = 1000 * 6 / 125 = 48 sec . answer : c" | a = 45 + 30
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a ) 3 : 4 , b ) 3 : 0 , c ) 3 : 1 , d ) 7 : 3 , e ) 3 : 7 | d | divide(add(multiply(4.5, divide(add(4.5, 4.5), subtract(4.5, 3))), 4.5), subtract(multiply(3, divide(add(4.5, 4.5), subtract(4.5, 3))), 4.5)) | the ratio between the present ages of a and b is 4.5 : 3 respectively . the ratio between a ' s age 4 years ago and b ' s age 4 years hence is 1 : 1 . what is the ratio between a ' s age 4 years hence and b ' s age 4 years ago ? | "let the present ages of a and b be 4.5 x and 3 x years respectively . then , ( 4.5 x - 4 ) / ( 3 x + 4 ) = 1 / 1 1.5 x = 8 = > x = 5.33 required ratio = ( 5 x + 4 ) : ( 3 x - 4 ) = 28 : 12 = 7 : 3 . answer : d" | a = 4 + 5
b = 4 - 5
c = a / b
d = 4 * 5
e = d + 4
f = 4 + 5
g = 4 - 5
h = f / g
i = 3 * h
j = i - 4
k = e / j
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a ) 17 kg , b ) 20 kg , c ) 26 kg , d ) 31 kg , e ) 22 kg | e | subtract(add(multiply(40, const_2), multiply(43, const_2)), multiply(46, const_3)) | the average weight of a , b and c is 46 kg . if the average weight of a and b be 40 kg and that of b and c be 43 kg , then the weight of b is : | "let d sum of a , b , c is 3 * 46 = 144 and sum of a and b s 2 * 40 = 80 sum of b and c is 3 * 43 = 86 hence 80 + 86 - 144 = 22 ans = 22 answer : e" | a = 40 * 2
b = 43 * 2
c = a + b
d = 46 * 3
e = c - d
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a ) 25 / 2 , b ) 25 / 4 , c ) 25 / 8 , d ) 25 / 16 , e ) 25 / 32 | a | divide(divide(6400, power(const_2, subtract(divide(add(divide(const_60, const_2), const_60), 10), const_1))), const_2) | the rate of spin of a certain gyroscope doubled every 10 seconds from the moment a particular stopwatch started . if after a minute and a half the gyroscope reached a speed of 6400 meters per second , what was the speed , in meters per second , when the stopwatch was started ? | "let x be the original speed when the stopwatch was started . in 90 seconds , the speed doubled 9 times . 2 ^ 9 * x = 6400 x = ( 2 ^ 8 * 25 ) / 2 ^ 9 = 25 / 2 the answer is a ." | a = const_60 / 2
b = a + const_60
c = b / 10
d = c - 1
e = 2 ** d
f = 6400 / e
g = f / 2
|
a ) 62 , b ) 66 , c ) 70 , d ) 74 , e ) 78 | b | add(add(add(add(add(add(add(6, 5), add(6, const_2)), add(6, const_1)), 6), 5), const_2), const_1) | if two integers x , y ( x > y ) are selected from - 5 to 6 ( inclusive ) , how many cases are there ? | "there are 12 integers from - 5 to 6 inclusive . 12 c 2 = 66 . the answer is b ." | a = 6 + 5
b = 6 + 2
c = a + b
d = 6 + 1
e = c + d
f = e + 6
g = f + 5
h = g + 2
i = h + 1
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a ) 0 , b ) 2 , c ) 4 , d ) 6 , e ) 8 | d | subtract(divide(10, const_2), multiply(2, 2)) | what is the remainder when 2 ^ 20 is divided by 10 ? | "2 ^ 20 = 2 ^ 10 x 2 ^ 10 and 2 ^ 10 = 1024 4 * 4 = 16 , so the unit digit of this multiplication is 6 the remainder is 6 answer : d" | a = 10 / 2
b = 2 * 2
c = a - b
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a ) s . 3944 , b ) s . 3948 , c ) s . 3942 , d ) s . 3808 , e ) s . 3929 | d | multiply(square_perimeter(sqrt(289)), 56) | what will be the cost of building a fence around a square plot with area equal to 289 sq ft , if the price per foot of building the fence is rs . 56 ? | "let the side of the square plot be a ft . a 2 = 289 = > a = 17 length of the fence = perimeter of the plot = 4 a = 68 ft . cost of building the fence = 68 * 56 = rs . 3808 . answer : d" | a = math.sqrt(289)
b = square_perimeter * (
|
a ) 12 , b ) 14 , c ) 18 , d ) 13 , e ) 10 | d | add(subtract(34, add(20, 1)), 1) | the average weight of a group of boys is 20 kg . after a boy of weight 34 kg joins the group , the average weight of the group goes up by 1 kg . find the number of boys in the group originally ? | "let the number off boys in the group originally be x . total weight of the boys = 20 x after the boy weighing 34 kg joins the group , total weight of boys = 20 x + 34 so 20 x + 34 = 21 ( x + 1 ) = > x = 13 . answer : d" | a = 20 + 1
b = 34 - a
c = b + 1
|
a ) 72 , b ) 80 , c ) 25 , d ) 150 , e ) 100 | c | divide(add(12, multiply(12, divide(2, 3))), subtract(const_1, divide(20, const_100))) | in a certain school , 20 % of students are below 8 years of age . the number of students above 8 years of age is 2 / 3 of the number of students of 8 years of age which is 12 . what is the total number of students in the school ? | "explanation : let the number of students be x . then , number of students above 8 years of age = ( 100 - 20 ) % of x = 80 % of x . 80 % of x = 12 + 2 / 3 of 12 80 / 100 x = 20 x = 25 . answer : option c" | a = 2 / 3
b = 12 * a
c = 12 + b
d = 20 / 100
e = 1 - d
f = c / e
|
a ) 40 , b ) 44 , c ) 45 , d ) 46 , e ) 50 | e | add(subtract(74, divide(440, 10)), multiply(10, const_2)) | if w is a positive integer and 10 ^ w – 74 in decimal notation has digits whose sum is 440 , what is the value of w ? | answer e = 50 we have ( 10 ^ w ) - 74 , where the sum of the digits is 440 . 100 - 74 = 26 , so we already have 2 + 6 = 8 from the 440 . because we have a power of 10 , we will have numbers like 100 , 1000 , 10000 and so on . this minus 74 rests 26 and a lot of 9 s . e . g : 1000 - 74 = 9926 . so dividing the 432 / 9 = 48 , that ` s the number of 9 s . and we have 2 other numbers ( 2 and 6 ) wich were 00 before the subtraction . so we have 48 + 2 as an w = 50 | a = 440 / 10
b = 74 - a
c = 10 * 2
d = b + c
|
a ) 22 years , b ) 24 years , c ) 15 years , d ) 16 years , e ) none of these | a | subtract(multiply(30, 30), add(multiply(10, 16), multiply(19, 22))) | the average age of 30 students of a class is 20 years . out of these , the average age of 10 students is 16 years and that of the other 19 students is 22 years . the age of the 30 th student is : | "explanation : age of the 30 th student = [ 30 x 20 - ( 10 x 16 + 19 x 22 ) ] = ( 600 - 278 ) = 22 years . answer : a" | a = 30 * 30
b = 10 * 16
c = 19 * 22
d = b + c
e = a - d
|
a ) 28 km , b ) 20 km , c ) 65 km , d ) 18 km , e ) 16 km | a | multiply(add(4, 3), 4) | roja and pooja start moving in the opposite directions from a pole . they are moving at the speeds of 4 km / hr and 3 km / hr respectively . after 4 hours what will be the distance between them ? | "distance = relative speed * time = ( 4 + 3 ) * 4 = 28 km [ they are travelling in the opposite direction , relative speed = sum of the speeds ] . answer : a" | a = 4 + 3
b = a * 4
|
a ) 440 , b ) 570 , c ) 480 , d ) 520 , e ) 550 | a | divide(176, divide(subtract(70, subtract(const_100, 70)), const_100)) | in an election only two candidates contested . a candidate secured 70 % of the valid votes and won by a majority of 176 votes . find the total number of valid votes ? | "let the total number of valid votes be x . 70 % of x = 70 / 100 * x = 7 x / 10 number of votes secured by the other candidate = x - 7 x / 100 = 3 x / 10 given , 7 x / 10 - 3 x / 10 = 176 = > 4 x / 10 = 176 = > 4 x = 1760 = > x = 440 . answer : a" | a = 100 - 70
b = 70 - a
c = b / 100
d = 176 / c
|
a ) 2.29 , b ) 2.75 , c ) 4.25 , d ) 4.5 , e ) none of these | b | multiply(17, 17) | ( 17 ) 5.25 x ( 17 ) ? = 178 | "solution let ( 17 ) 5.25 * ( 17 ) x = 178 . then , ( 17 ) 5.25 + x = ( 17 ) 8 . ∴ 5.25 + x = 8 ⇔ x = ( 8 - 5.25 ) ⇔ x = 2.75 answer b" | a = 17 * 17
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['a ) 12 % deficit', 'b ) 10 % excess', 'c ) 8 % excess', 'd ) 9 % excess', 'e ) none of these'] | c | subtract(subtract(20, 10), divide(multiply(20, 10), const_100)) | in measuring the sides of a rectangle , one side is taken 20 % in excess and the other 10 % in deficit . find the error per cent in area calculated from the measurement ? | since side 1 x side 2 = area therefore , error % in area = ( x + y + xy / 100 ) % = [ 20 - 10 + ( - 10 x 20 ) / 100 ] % or 8 % i . e . , 8 % excess answer : c | a = 20 - 10
b = 20 * 10
c = b / 100
d = a - c
|
a ) 77 , b ) 18 , c ) 14 , d ) 12 , e ) 17 | b | subtract(54, divide(multiply(54, const_2), const_3)) | the time taken by a man to row his boat upstream is twice the time taken by him to row the same distance downstream . if the speed of the boat in still water is 54 kmph , find the speed of the stream ? | "the ratio of the times taken is 2 : 1 . the ratio of the speed of the boat in still water to the speed of the stream = ( 2 + 1 ) / ( 2 - 1 ) = 3 / 1 = 3 : 1 speed of the stream = 54 / 3 = 18 kmph . answer : b" | a = 54 * 2
b = a / 3
c = 54 - b
|
a ) 6 days , b ) 4 days , c ) 12 days , d ) 3 days , e ) 8 days | d | divide(12, const_4) | some ladies can do a piece of work in 12 days . two times the number of such ladies will do half of that work in : | expl : let x ladies can do the work in 12 days . more ladies , less days ( indirect ) less work , less days ( direct ) ladies 2 x : x work 1 : ½ 2 x : x , 1 : ½ : : 12 : y : . 2 x * 1 * y = x * ½ * 12 or y = 3 hence the required number of days = 3 answer : d | a = 12 / 4
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a ) 24 , b ) 77 , c ) 295 , d ) 267 , e ) 29 | c | multiply(subtract(65, 6), 5) | think of a number , divide it by 5 and add 6 to it . the result is 65 . what is the number thought of ? | "explanation : 65 - 6 = 59 59 x 5 = 295 answer : c" | a = 65 - 6
b = a * 5
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a ) $ 1.56 , b ) $ 1.88 , c ) $ 2.25 , d ) $ 2.50 , e ) $ 3.25 | a | multiply(divide(add(multiply(multiply(32, divide(75, const_100)), multiply(add(const_4, 1), const_2)), 200), add(multiply(subtract(multiply(add(const_4, 1), const_2), 4), 32), multiply(multiply(divide(add(const_100, 25), const_100), 4), 32))), divide(add(const_100, 25), const_100)) | sarah operated her lemonade stand monday through friday over a two week period and made a total profit of 200 dollars . on hot days she sold cups of lemonade for a price that was 25 percent higher than the regular days . each cup she sold had a total cost of 75 cents and sarah did not incur any other costs . if every day she sold exactly 32 cups and 4 of the days were hot , then what was the price of 1 cup on a hot day ? | "6 regular days - - > sales = 6 * 32 * x = 192 x ; 4 hot days - - > sales = 4 * 32 * ( 1.25 x ) = 160 x ; total sales = 192 x + 160 x = 352 x . total cost = 10 * 32 * 0.75 = 240 . profit = 352 x - 240 = 200 - - > x = 1.25 . 1.25 x = ~ 1.56 . answer : a ." | a = 75 / 100
b = 32 * a
c = 4 + 1
d = c * 2
e = b * d
f = e + 200
g = 4 + 1
h = g * 2
i = h - 4
j = i * 32
k = 100 + 25
l = k / 100
m = l * 4
n = m * 32
o = j + n
p = f / o
q = 100 + 25
r = q / 100
s = p * r
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a ) 5100 , b ) 5200 , c ) 5300 , d ) 5400 , e ) 5500 | c | divide(1908, subtract(subtract(const_1, divide(32, const_100)), divide(32, const_100))) | a candidate got 32 % of the votes polled and he lost to his rival by 1908 votes . how many votes were cast ? | "let x be the total number of votes . 0.32 x + 1908 = 0.68 x 0.36 x = 1908 x = 1908 / 0.36 = 5300 the answer is c ." | a = 32 / 100
b = 1 - a
c = 32 / 100
d = b - c
e = 1908 / d
|
a ) 4 : 3 , b ) 1 : 2 , c ) 1 : 3 , d ) 1 : 4 , e ) 2 : 3 | c | divide(sqrt(1), sqrt(9)) | the sub - duplicate ratio of 1 : 9 is | "root ( 1 ) : root ( 9 ) = 1 : 3 answer : c" | a = math.sqrt(1)
b = math.sqrt(9)
c = a / b
|
a ) 3377 , b ) 6886 , c ) 2788 , d ) 2000 , e ) 28712 | d | add(multiply(add(20, const_1), add(20, const_1)), add(20, const_1)) | a sum of money is put out at compound interest for 2 years at 20 % . it would fetch rs . 482 more if the interest were payable half - yearly , then it were pay able yearly . find the sum . | "p ( 11 / 10 ) 4 - p ( 6 / 5 ) 2 = 482 p = 2000 answer : d" | a = 20 + 1
b = 20 + 1
c = a * b
d = 20 + 1
e = c + d
|
a ) 9,99 , 98,623 , b ) 9,99 , 98,704 , c ) 9,99 , 98,800 , d ) 9,99 , 99,521 , e ) 9,99 , 99,932 | b | divide(multiply(add(add(const_100, const_60), const_1), 10), const_100) | what is the value of 10 ^ 8 - 6 ^ 4 ? | "as 10 ^ n will always have last digit as 0 and 6 ^ n will always as last digit 6 . . hence difference of such sum should always be ending with 4 and there is only on option . . answer b" | a = 100 + const_60
b = a + 1
c = b * 10
d = c / 100
|
a ) 160 sq m , b ) 140 sq m , c ) 200 sq m , d ) 400 sq m , e ) 180 sq m | b | divide(square_area(14), const_2) | what is the area of a square field whose diagonal of length 14 m ? | "d 2 / 2 = ( 14 * 14 ) / 2 = 140 answer : b" | a = square_area / (
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a ) 3 hours , b ) 1 hours , c ) 3.20 hours , d ) 2 hours , e ) 1.35 hours | d | inverse(add(inverse(3), inverse(6))) | harish can paint a wall in just 3 hours . however it takes 6 hours for ganpat to complete the same job of painting the wall . how long would it take for both of them to paint the wall , if both work together ? | d two hours . harish who could paint the wall in 3 hours could paint 2 / 3 of the wall in just two hours . ganpat who paint the wall in 6 hours could paint 1 / 3 of the wall in two hours . 2 / 3 + 1 / 3 = 1 . | a = 1/(3)
b = 1/(6)
c = a + b
d = 1/(c)
|
a ) 22 , b ) 24 , c ) 77 , d ) 266 , e ) 182 | b | divide(multiply(subtract(38, 6), 3), 4) | ratio between rahul and deepak is 4 : 3 , after 6 years rahul age will be 38 years . what is deepak present age ? | "present age is 4 x and 3 x , = > 4 x + 6 = 38 = > x = 8 so deepak age is = 3 ( 8 ) = 24 answer : b" | a = 38 - 6
b = a * 3
c = b / 4
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a ) 30 , b ) 35 , c ) 25 , d ) 82 , e ) 22 | a | add(multiply(40, const_100), multiply(multiply(subtract(const_1, multiply(add(divide(const_1, 40), divide(const_1, 120)), const_2)), 40), const_60)) | two pipes a and b can fill a tank in 40 and 120 minutes respectively . if both the pipes are used together , then how long will it take to fill the tank ? | "part filled by a in 1 min . = 1 / 40 part filled by b in 1 min . = 1 / 120 part filled by ( a + b ) in 1 min . = 1 / 40 + 1 / 120 = 1 / 30 . both the pipes can fill the tank in 30 minutes . answer : a" | a = 40 * 100
b = 1 / 40
c = 1 / 120
d = b + c
e = d * 2
f = 1 - e
g = f * 40
h = g * const_60
i = a + h
|
a ) 30000 , b ) 40000 , c ) 50000 , d ) 60000 , e ) 70000 | d | divide(subtract(divide(multiply(70000, 20), const_100), 8000), divide(10, const_100)) | rs 70000 is divided into two parts one part is given to a person with 10 % interest and another part is given to a person with 20 % interest . at the end of first year he gets profit 8000 find money given by 10 % ? | "let first parrt is x and second part is y then x + y = 70000 - - - - - - - - - - eq 1 total profit = profit on x + profit on y 8000 = ( x * 10 * 1 ) / 100 + ( y * 20 * 1 ) / 100 80000 = x + 2 y - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - eq 2 80000 = 70000 + y so y = 10000 then x = 70000 - 10000 = 60000 first part = 60000 answer : d" | a = 70000 * 20
b = a / 100
c = b - 8000
d = 10 / 100
e = c / d
|
a ) 16 , b ) 28 , c ) 32 , d ) 36 , e ) 38 | a | add(multiply(divide(7, add(6, const_1)), 10), multiply(6, divide(7, add(6, const_1)))) | in a two digit number , the digit in the units place is 6 times the digit in 10 ' s place and sum of the digits is equal to 7 . what is the number ? | solution let the ten digits be x . then , units digit = 6 x therefore x + 6 x = 7 ‹ = › 7 x = 7 ‹ = › x = 1 . so , ten ' s digit = 1 , unit ' s digit = 6 . hence , the required number is 16 . answer a | a = 6 + 1
b = 7 / a
c = b * 10
d = 6 + 1
e = 7 / d
f = 6 * e
g = c + f
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a ) 20 , b ) 34 , c ) 38 , d ) 40 , e ) 46 | c | divide(subtract(multiply(const_3, 40), 40), const_3) | shannon and maxine work in the same building and leave work at the same time . shannon lives due north of work and maxine lives due south . the distance between maxine ' s house and shannon ' s house is 40 miles . if they both drive home at the rate 2 r miles per hour , maxine arrives home 40 minutes after shannon . if maxine rider her bike home at the rate of r per hour and shannon still drives at a rate of 2 r miles per hour , shannon arrives home 2 hours before maxine . how far does maxine live from work ? | "nice question + 1 we have that x / 24 - ( 60 - x ) / 2 r = 40 also x / r - ( 60 - x ) / 2 r = 120 so we get that 2 x - 60 = 80 r 3 x - 60 = 240 r get rid of r 120 = 3 x x = 38 hence answer is c" | a = 3 * 40
b = a - 40
c = b / 3
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a ) $ 1100 , b ) $ 520 , c ) $ 780 , d ) $ 1170 , e ) $ 630 | c | multiply(2340, divide(inverse(8), add(inverse(12), add(inverse(6), inverse(8))))) | a , b and c , each working alone can complete a job in 6 , 8 and 12 days respectively . if all three of them work together to complete a job and earn $ 2340 , what will be b ' s share of the earnings ? | explanatory answer a , b and c will share the amount of $ 2340 in the ratio of the amounts of work done by them . as a takes 6 days to complete the job , if a works alone , a will be able to complete 1 / 6 th of the work in a day . similarly , b will complete 1 / 8 th and c will complete 1 / 12 th of the work . so , the ratio of the work done by a : b : c when they work together will be equal to 1 / 6 : 1 / 8 : 1 / 12 multiplying the numerator of all 3 fractions by 24 , the lcm of 6 , 8 and 12 will not change the relative values of the three values . we get 24 / 6 : 24 / 8 : 24 / 12 = 4 : 3 : 2 . i . e . , the ratio in which a : b : c will share $ 2340 will be 4 : 3 : 2 . hence , b ' s share will be 3 * 2340 / 9 = 780 correct choice is ( c ) | a = 1/(8)
b = 1/(12)
c = 1/(6)
d = 1/(8)
e = c + d
f = b + e
g = a / f
h = 2340 * g
|
a ) 384 , b ) 432 , c ) 207 , d ) 269 , e ) 208 | b | divide(add(multiply(divide(44, const_100), 900), add(multiply(divide(64, const_100), 900), multiply(divide(36, const_100), 900))), const_3) | in an exam , amar scored 64 percent , bhavan scored 36 percent and chetan 44 percent . the maximum score awarded in the exam is 900 . find the average mark scored by all the 3 boys ? | average mark scored by all the three boys = [ 64 / 100 ( 900 ) + 36 / 100 ( 900 ) + 44 / 100 ( 900 ) ] / 3 = 432 answer : b | a = 44 / 100
b = a * 900
c = 64 / 100
d = c * 900
e = 36 / 100
f = e * 900
g = d + f
h = b + g
i = h / 3
|
a ) 220 m , b ) 240 m , c ) 260 m , d ) 320 m , e ) 330 m | a | subtract(multiply(multiply(add(120, 80), const_0_2778), 9), 280) | a 280 metres long train running at the speed of 120 kmph crosses another train running in opposite direction at the speed of 80 kmph in 9 seconds . what is the length of the other train ? | "relative speed = ( 120 + 80 ) km / hr = ( 200 x ( 5 / 18 ) ) m / sec = ( 500 / 9 ) m / sec . let the length of the other train be x metres . then , ( x + 280 ) / 9 = 500 / 9 x + 280 = 500 x = 220 . a" | a = 120 + 80
b = a * const_0_2778
c = b * 9
d = c - 280
|
a ) 10 , b ) 15 , c ) 12.5 , d ) 13 , e ) 21 | b | divide(add(add(add(5, const_1), add(add(5, const_1), const_2)), add(subtract(5, 5), subtract(5, const_2))), 5) | find the average of first 5 multiples of 5 ? | "average = ( 5 + 10 + 15 + 20 + 25 ) / 5 = 15 answer is b" | a = 5 + 1
b = 5 + 1
c = b + 2
d = a + c
e = 5 - 5
f = 5 - 2
g = e + f
h = d + g
i = h / 5
|
a ) 229 , b ) 240 , c ) 288 , d ) 480 , e ) 221 | d | multiply(20, multiply(108, const_0_2778)) | a train passes a station platform in 36 sec and a man standing on the platform in 20 sec . if the speed of the train is 108 km / hr . what is the length of the platform ? | "speed = 108 * 5 / 18 = 30 m / sec . length of the train = 30 * 20 = 600 m . let the length of the platform be x m . then , ( x + 600 ) / 36 = 30 = > x = 480 m . answer : d" | a = 108 * const_0_2778
b = 20 * a
|
a ) 1.6 , b ) 1.9 , c ) 2.2 , d ) 2.5 , e ) 2.8 | d | multiply(divide(add(2.9, divide(100, const_1000)), 72), const_60) | a train with a length of 100 meters , is traveling at a speed of 72 km / hr . the train enters a tunnel 2.9 km long . how many minutes does it take the train to pass through the tunnel from the moment the front enters to the moment the rear emerges ? | "72 km / hr = 1.2 km / min the total distance is 3 km . 3 / 1.2 = 2.5 minutes the answer is d ." | a = 100 / 1000
b = 2 + 9
c = b / 72
d = c * const_60
|
a ) 7.2 sec , b ) 9.3 sec , c ) 12 sec , d ) 15.2 sec , e ) 18 sec | a | divide(120, multiply(subtract(68, 8), const_0_2778)) | a train 120 m long is running at a speed of 68 kmph . how long does it take to pass a man who is running at 8 kmph in the same direction as the train ? | speed of the train relative to man = ( 68 - 8 ) kmph = ( 60 * 5 / 18 ) m / sec = ( 50 / 3 ) m / sec time taken by the train to cross the man = time taken by it to cover 120 m at 50 / 3 m / sec = 120 * 3 / 50 sec = 7.2 sec answer : a . | a = 68 - 8
b = a * const_0_2778
c = 120 / b
|
a ) 5 , b ) 6 , c ) 7 , d ) 4 , e ) 2 | e | multiply(7, const_2) | if exactly 7 guests drank only one of the two types of drinks , how many guests drank both types of drinks ? | say x guests drank both drinks . ( 7 - x ) + ( 4 - x ) = 7 - - > x = 2 . answer : e . | a = 7 * 2
|
['a ) 148 / 29', 'b ) 168 / 29', 'c ) 168 / 26', 'd ) 158 / 29', 'e ) 198 / 29'] | b | divide(divide(multiply(triangle_area_three_edges(10, 17, 21), const_2), 21), add(divide(divide(multiply(triangle_area_three_edges(10, 17, 21), const_2), 21), 21), const_1)) | a triangle has sides 10 , 17 , and 21 . a square is inscribed in the triangle . one side of the square lies on the longest side of the triangle . the other two vertices of the square touch the two shorter sides of the triangle . what is the length of the side of the square ? | by heron ' s formula , the area , a , of a triangle with sides a , b , c is given by a = square root [ s ( s − a ) ( s − b ) ( s − c ) ] , where s = ½ ( a + b + c ) is the semi - perimeter of the triangle . then s = ½ ( 10 + 17 + 21 ) = 24 , and a = 84 . a square inscribed in a 10 by 17 by 21 triangle , with perpendicular dropped onto the side of length 21 . now drop a perpendicular of length h onto the side of length 21 . we also have a = ½ × base × perpendicular height . hence a = 21 h / 2 = 84 , from which h = 8 . notice that the triangle above the square is similar to the whole triangle . ( this follows because its base , the top of the square , is parallel to the base of the whole triangle . ) let the square have side of length d . considering the ratio of altitude to base in each triangle , we have 8 / 21 = ( 8 − d ) / d = 8 / d − 1 . therefore the length of the side of the square is 168 / 29 . answer : b | a = triangle_area_three_edges * (
b = a / 2
c = b / 21
|
a ) 281 , b ) 125 , c ) 288 , d ) 266 , e ) 400 | e | multiply(multiply(160, const_0_2778), 9) | if a train , travelling at a speed of 160 kmph , crosses a pole in 9 sec , then the length of train is ? | "d = 160 * 5 / 18 * 9 = 400 m answer : e" | a = 160 * const_0_2778
b = a * 9
|
a ) 5 , b ) 8 , c ) 9.36 , d ) 15 , e ) 24 | c | max(multiply(subtract(add(55, 15), const_1), subtract(divide(15, 35), divide(15, 55))), const_4) | due to construction , the speed limit along an 15 - mile section of highway is reduced from 55 miles per hour to 35 miles per hour . approximately how many minutes more will it take to travel along this section of highway at the new speed limit than it would have taken at the old speed limit ? | "old time in minutes to cross 15 miles stretch = 15 * 60 / 55 = 15 * 12 / 11 = 16.36 new time in minutes to cross 15 miles stretch = 15 * 60 / 35 = 15 * 12 / 7 = 25.71 time difference = 9.36 answer c )" | a = 55 + 15
b = a - 1
c = 15 / 35
d = 15 / 55
e = c - d
f = b * e
g = max(f)
|
a ) 0.045388 , b ) 4.5388 , c ) 453.88 , d ) 473.88 , e ) 0.038904 | e | divide(divide(multiply(3.242, 12), const_100), const_10) | [ ( 3.242 x 12 ) / 100 ] = ? | answer multiplying 3.242 x 12 = 3.8904 now divide 3.8904 by 100 so , 3.8904 ÷ 100 = 0.038904 ∴ shift the decimal two places to the left as 100 correct option : e | a = 3 * 242
b = a / 100
c = b / 10
|
a ) [ 45 ] , b ) [ 30 ] , c ) [ 15 ] , d ) [ 5 ] , e ) [ 50 ] | c | divide(multiply(divide(6, 2), multiply(3, 5)), 3) | for all positive integers m , [ m ] = 3 m when m is odd and [ m ] = ( 1 / 2 ) * m when m is even . what is [ 5 ] * [ 6 ] equivalent to ? | "[ 5 ] * [ 6 ] = 15 * 3 = 45 = [ 15 ] the answer is c ." | a = 6 / 2
b = 3 * 5
c = a * b
d = c / 3
|
a ) 2000 , b ) 10025 , c ) 1025 , d ) 2015 , e ) 2025 | c | subtract(multiply(10000, multiply(add(const_1, divide(const_0_25, const_4)), add(const_1, divide(const_0_25, const_4)))), 10000) | find the c . i . on a sum of rs . 10000 for 9 months at 20 % per annum , interest being compounded quarterly ? | "c . i . = 10000 ( 21 / 20 ) ^ 3 - 10000 = 1025 answer : c" | a = const_0_25 / 4
b = 1 + a
c = const_0_25 / 4
d = 1 + c
e = b * d
f = 10000 * e
g = f - 10000
|
a ) 3.6 , b ) 4.5 , c ) 4.8 , d ) 5.4 , e ) 5.5 | a | divide(divide(720, const_1000), divide(multiply(12, const_60), const_3600)) | a person crosses a 720 m long street in 12 minutes . what is his speed in km per hour ? | "speed = 720 / ( 12 x 60 ) m / sec = 1 m / sec . converting m / sec to km / hr = 1 x ( 18 / 5 ) km / hr = 3.6 km / hr . answer : a" | a = 720 / 1000
b = 12 * const_60
c = b / 3600
d = a / c
|
a ) 70 , b ) 245 , c ) 150 , d ) 200 , e ) 350 | d | divide(28, multiply(divide(subtract(const_100, 30), const_100), divide(20, const_100))) | in a certain boys camp , 20 % of the total boys are from school a and 30 % of those study science . if there are 28 boys in the camp that are from school a but do not study science then what is the total number of boys in the camp ? | "since 30 % of the boys from school a study science , then 70 % of the boys from school a do not study science and since 20 % of the total number of boys are from school a , then 0.2 * 0.7 = 0.14 , or 14 % of the boys in the camp are from school a and do not study science . we are told that this number equals to 28 , so 0.14 * { total } = 28 - - > { total } = 200 . answer : d ." | a = 100 - 30
b = a / 100
c = 20 / 100
d = b * c
e = 28 / d
|
a ) 125 , b ) 187 , c ) 231 , d ) 375 , e ) 460 | c | divide(factorial(22), multiply(factorial(subtract(22, const_2)), factorial(const_2))) | if 22 lines are drawn in a plane such that no two of them are parallel and no three are concurrent , then in how many points do they intersect ? | "if two lines are not parallel , then they will intersect in exactly one point . lines can be extended infinitely on both ends so somewhere they will intersect with each other if they are not parallel . we are also given that no three lines are concurrent . this means that no three lines intersect at the same point . so every pair of two lines we select will have a unique point of intersection which they will not share with any third line . the number of ways to select 2 lines from 22 lines is 22 c 2 = 231 the answer is c ." | a = math.factorial(22)
b = 22 - 2
c = math.factorial(b)
d = math.factorial(2)
e = c * d
f = a / e
|
a ) 4 % , b ) 5 % , c ) 6 % , d ) 8.75 % , e ) none | a | add(multiply(divide(subtract(divide(subtract(subtract(subtract(multiply(multiply(const_10, const_1000), const_10), const_1000), const_1000), multiply(add(2, const_3), const_100)), multiply(add(multiply(add(const_3, const_4), const_10), add(2, const_3)), const_1000)), 1), const_10), const_100), const_4) | the population of a town increased from 1 , 75,000 to 2 , 45,000 in a decade . the average percent increase of population per year is | "solution increase in 10 years = ( 245000 - 175000 ) = 87500 . increase % = ( 70000 / 175000 ã — 100 ) % = 40 % . required average = ( 40 / 10 ) % = 4 % . answer a" | a = 10 * 1000
b = a * 10
c = b - 1000
d = c - 1000
e = 2 + 3
f = e * 100
g = d - f
h = 3 + 4
i = h * 10
j = 2 + 3
k = i + j
l = k * 1000
m = g / l
n = m - 1
o = n / 10
p = o * 100
q = p + 4
|
a ) 190 , b ) 210 , c ) 211 , d ) 230 , e ) 231 | b | subtract(10, add(add(multiply(const_2, const_100), multiply(add(const_3, const_4), const_10)), const_2)) | how many integers between 1 and 10 ^ 20 are such that the sum of their digits is 2 ? | "the integers with a sum of 2 are : 2 , 20 , 200 , . . . , 2 * 10 ^ 19 and there are 20 integers in this list . also , these integers have a sum of 2 : 11 101 , 110 1001 , 1010 , 1100 etc . . . the number of integers in this list is 1 + 2 + . . . + 19 thus , the total number of integers is 1 + 2 + . . . + 19 + 20 = 20 * 21 / 2 = 210 the answer is b ." | a = 2 * 100
b = 3 + 4
c = b * 10
d = a + c
e = d + 2
f = 10 - e
|
a ) 1997 , b ) 1998 , c ) 1999 , d ) 2000 , e ) none of these | b | add(1988, divide(add(divide(40, const_100), subtract(6.3, 4.2)), subtract(divide(40, const_100), divide(15, const_100)))) | the price of commodity x increases by 40 paise every year , while the price of commodity y increases by 15 paise every year . if in 1988 , the price of commodity x was 4.20 and that of y was 6.30 , in which year will commodity x cost 40 paise more than commodity y ? | suppose in ‘ n ’ years the price of comodity x will be more by 40 paise than that of commodity y . ∴ 420 + 40 n – 630 – 15 n = 40 or , 25 n – 210 = 40 or , 25 n = 250 or , n = 250 ⁄ 25 = 10 years . answer b | a = 40 / 100
b = 6 - 3
c = a + b
d = 40 / 100
e = 15 / 100
f = d - e
g = c / f
h = 1988 + g
|
a ) 125 % , b ) 100 % , c ) 25 % , d ) 225 % , e ) 400 % | a | multiply(divide(subtract(divide(3, 4), divide(const_2.0, 3)), divide(4, 3)), const_100) | when tom works alone he chops 4 lb . salad in 3 minutes , and when tammy works alone she chops 3 lb . salad in 4 minutes . they start working together , and after some time finish chopping 65 lb . of salad . of those 80 lb . , the salad quantity chopped by tammy is what percent greater than the quantifying chopped by tom ? . | "tom chops 8 lbs in 12 minutes tammy chopswhen tom works alone he chops 2 lb . salad in 3 minutes , and when tammy works alone she chops 3 lb . salad in 2 minutes . they start working together , and after some time finish chopping 65 lb . of salad . of those 80 lb . , the salad quantity chopped by tammy is what percent greater than the quantifying chopped by tom ? . 9 lbs in 6 minutes so in the same amount of time , tammy chops 125 % more than tom , since 9 is 125 % greater than 4 . so 125 % is the answer . note that the actual time does n ' t matter . if you multiply the time each work by x , you ' ll multiply the work each does by x , and 9 x is still 125 % greater than 4 x . ans : a" | a = 3 / 4
b = 2 / 0
c = a - b
d = 4 / 3
e = c / d
f = e * 100
|
a ) 3.5 , b ) 3.35 , c ) 3.45 , d ) 3.25 , e ) 3.55 | d | multiply(divide(add(multiply(45, divide(5, const_100)), multiply(10, divide(10, const_100))), const_100), const_100) | of the total amount that jill spent on a shopping trip , excluding taxes , she spent 45 percent on clothing , 45 percent on food , and 10 percent on other items . if jill paid a 5 percent tax on the clothing , no tax on the food , and an 10 percent tax on all other items , then the total tax that she paid was what percent of the total amount that she spent , excluding taxes ? | "assume she has $ 200 to spend . tax clothing = 45 % = $ 90 = $ 4.5 food = 45 % = $ 90 = $ 0.00 items = 10 % = $ 20 = $ 2.00 total tax = $ 6.50 % of total amount = 6.5 / 200 * 100 = 3.25 % answer d" | a = 5 / 100
b = 45 * a
c = 10 / 100
d = 10 * c
e = b + d
f = e / 100
g = f * 100
|
a ) 2.9 hrs , b ) 8.9 hrs , c ) 2.9 hrs , d ) 6.7 hrs , e ) 8.6 hrs | d | divide(const_1, subtract(divide(const_1, 4), divide(const_1, 10))) | a cistern can be filled by a tap in 4 hours while it can be emptied by another tap in 10 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 / 10 ) = 3 / 20 the cistern will be filled in 20 / 3 hrs i . e . , 6.7 hrs . answer : d" | a = 1 / 4
b = 1 / 10
c = a - b
d = 1 / c
|
a ) 93.5 , b ) 90 , c ) 6.75 , d ) 6.25 , e ) 2 | d | divide(multiply(25, 25), const_100) | j is 25 % less than p and 20 % less than t . t is t % less than p . what is the value of t ? | usually we can solve every question of this type by choosing appropriate value of the variable and deriving the value of other related variables . let , p = 400 then j = ( 75 / 100 ) * 400 = 300 also j = ( 80 / 100 ) * t i . e . t = 300 * 100 / 80 = 375 and t = [ 1 - ( t / 100 ) ] * p i . e . 100 - t = 100 * t / p = 100 * 375 / 400 = 93.75 i . e . t = 6.25 answer : option d | a = 25 * 25
b = a / 100
|
a ) 12 , b ) 75 , c ) 85 , d ) 54 , e ) 15 | c | divide(add(90, 80), const_2) | the speed of a car is 90 km in the first hour and 80 km in the second hour . what is the average speed of the car ? | "s = ( 90 + 80 ) / 2 = 85 kmph answer : c" | a = 90 + 80
b = a / 2
|
a ) 10 % , b ) 12.5 % , c ) 8 % , d ) 12 % , e ) 6.66 % | c | multiply(subtract(const_1, divide(add(const_1, divide(15, const_100)), add(const_1, divide(25, const_100)))), const_100) | if the price of gasoline increases by 25 % and ron intends to spend only 15 % more on gasoline , by how much % should he reduce the quantity of gasoline that he buys ? | explanatory answer let the price of 1 litre of gasoline be $ x and let ron initially buy ' y ' litres of gasoline . therefore , he would have spent $ xy on gasoline . when the price of gasoline increases by 25 % , the new price per litre of gasoline is 1.25 x . ron intends to increase the amount he spends on gasoline by 15 % . i . e . , he is willing to spend xy + 15 % of xy = 1.15 xy let the new quantity of gasoline that he can get be ' q ' . then , 1.25 x * q = 1.15 xy or q = 1.15 xy / 1.25 x = 0.92 y . as the new quantity that he can buy is 0.92 y , he gets 0.08 y lesser than what he used to get earlier . or a reduction of 8 % . the correct choice is ( c ) | a = 15 / 100
b = 1 + a
c = 25 / 100
d = 1 + c
e = b / d
f = 1 - e
g = f * 100
|
a ) 70 % , b ) 20 % , c ) 10 % , d ) 28 % , e ) 20 % | c | divide(const_100, 10) | in how many years will a sum of money doubles itself at 10 % per annum on simple interest ? | "p = ( p * 10 * r ) / 100 r = 10 % answer : c" | a = 100 / 10
|
a ) rs . 300 , b ) rs . 200 , c ) rs . 250 , d ) rs . 350 , e ) none of these | c | divide(925, add(add(multiply(add(const_1, divide(25, const_100)), add(const_1, divide(20, const_100))), add(const_1, divide(20, const_100))), const_1)) | if x gets 25 % more than y and y gets 20 % more than z , the share of z out of rs . 925 will be : | "z share = z , y = 1.2 z x = 1.25 ã — 1.2 z , x + y + z = 925 ( 1.25 ã — 1.2 + 1.2 + 1 ) z = 925 3.7 z = 925 , z = 250 answer : . c" | a = 25 / 100
b = 1 + a
c = 20 / 100
d = 1 + c
e = b * d
f = 20 / 100
g = 1 + f
h = e + g
i = h + 1
j = 925 / i
|
a ) s : 10123.19 , b ) s : 10123.29 , c ) s : 10123.20 , d ) s : 18221.76 , e ) s : 10123.12 | d | subtract(multiply(multiply(multiply(const_4, const_100), const_100), power(add(const_1, divide(12, const_100)), 3)), multiply(multiply(const_4, const_100), const_100)) | what will be the compound interest on a sum of rs . 45,000 after 3 years at the rate of 12 % p . a . ? | "amount = [ 45000 * ( 1 + 12 / 100 ) 3 ] = 45000 * 28 / 25 * 28 / 25 * 28 / 25 = rs . 63221.76 c . i . = ( 63221.76 - 45000 ) = rs : 18221.76 answer : d" | a = 4 * 100
b = a * 100
c = 12 / 100
d = 1 + c
e = d ** 3
f = b * e
g = 4 * 100
h = g * 100
i = f - h
|
a ) 6 rs , b ) 2 rs , c ) 4 rs , d ) 9 rs , e ) 3 rs | c | divide(60, multiply(const_3, 5)) | 5 men are equal to as many women as are equal to 8 boys . all of them earn rs . 60 only . men â € ™ s wages are ? | "5 m = xw = 8 b 5 m + xw + 8 b - - - - - 60 rs . 5 m + 5 m + 5 m - - - - - 60 rs . 15 m - - - - - - 60 rs . = > 1 m = 4 rs . answer : c" | a = 3 * 5
b = 60 / a
|
a ) 2984 , b ) 2983 , c ) 2982 , d ) 2981 , e ) none of these | a | subtract(3034, divide(1002, 20.04)) | 3034 - ( 1002 / 20.04 ) = ? | option a | a = 1002 / 20
b = 3034 - a
|
a ) 8 , b ) 6 , c ) 3 , d ) 5 , e ) 4 | c | subtract(const_100, multiply(multiply(add(const_1, divide(10, const_100)), subtract(const_1, divide(12, const_100))), const_100)) | the tax on a commodity is diminished by 12 % and its consumption increased by 10 % . the effect on revenue is ? | "100 * 100 = 10000 88 * 110 = 9680 - - - - - - - - - - - 10000 - - - - - - - - - - - 320 100 - - - - - - - - - - - ? = > 3 % decrease answer : c" | a = 10 / 100
b = 1 + a
c = 12 / 100
d = 1 - c
e = b * d
f = e * 100
g = 100 - f
|
a ) 120 % , b ) 140 % , c ) 160 % , d ) 180 % , e ) 200 % | e | multiply(const_100, divide(15, multiply(add(20, const_100), divide(25, const_100)))) | last year a worker saved 15 % of her annual salary . this year , she made 20 % more money than last year and she saved 25 % of her salary . the amount she saved this year was what percent of the amount she saved last year ? | "let x be the worker ' s salary last year . last year , she saved 0.15 x . this year , she saved 0.25 ( 1.2 x ) = 0.3 x 0.3 x / 0.15 x = 200 % the answer is e ." | a = 20 + 100
b = 25 / 100
c = a * b
d = 15 / c
e = 100 * d
|
a ) 11.25 , b ) 11.52 , c ) 1.25 , d ) 9 , e ) 7.2 | d | divide(inverse(divide(inverse(add(const_3, const_2)), add(const_1, 1.25))), 1.25) | one pipe can fill a pool 1.25 times faster than a second pipe . when both pipes are opened , they fill the pool in 4 hours . how long would it take to fill the pool if only the slower pipe is used ? | "say the rate of the slower pipe is r pool / hour , then the rate of the faster pipe would be 1.25 r = 5 r / 4 . since when both pipes are opened , they fill the pool in four hours , then their combined rate is 1 / 4 pool / hour . thus we have that r + 5 r / 4 = 1 / 4 - - > r = 1 / 9 pool / hour - - > time is reciprocal of rate thus it ' s 9 / 1 = 9 hours . answer : d ." | a = 3 + 2
b = 1/(a)
c = 1 + 1
d = b / c
e = 1/(d)
f = e / 1
|
a ) 92 , b ) 43 , c ) 71 , d ) 26 , e ) 31 | d | subtract(add(power(27, 27), 27), multiply(28, floor(divide(add(power(27, 27), 27), 28)))) | what will be the reminder when ( 27 ^ 27 + 27 ) is divided by 28 ? | "( x ^ n + 1 ) will be divisible by ( x + 1 ) only when n is odd ; ( 27 ^ 27 + 1 ) will be divisible by ( 27 + 1 ) ; ( 27 ^ 27 + 1 ) + 26 when divided by 28 will give 26 as remainder . correct option : d" | a = 27 ** 27
b = a + 27
c = 27 ** 27
d = c + 27
e = d / 28
f = math.floor(e)
g = 28 * f
h = b - g
|
a ) 10 % , b ) 15 % , c ) 20 % , d ) 25 % , e ) 50 % | c | subtract(const_100, multiply(const_100, inverse(add(divide(25, const_100), const_1)))) | ram ' s salary is 25 % more than raja ' s . how much percent is raja ' s salary less than ram ' s ? | raja ' s salary less than ram ' s by 25 / ( 100 + 25 ) * 100 % = 20 % answer is c | a = 25 / 100
b = a + 1
c = 1/(b)
d = 100 * c
e = 100 - d
|
a ) 1 , b ) 3 , c ) 5 , d ) 7 , e ) 9 | d | floor(divide(reminder(power(6, reminder(20, add(const_4, const_1))), const_100), const_10)) | what is the tens digit of 6 ^ 20 ? | "the tens digit of 6 in integer power starting from 2 ( 6 ^ 1 has no tens digit ) repeats in a pattern of 5 : { 3 , 1 , 9 , 7 , 5 } : the tens digit of 6 ^ 2 = 36 is 3 . the tens digit of 6 ^ 3 = 216 is 1 . the tens digit of 6 ^ 4 = . . . 96 is 9 . the tens digit of 6 ^ 5 = . . . 76 is 7 . the tens digit of 6 ^ 6 = . . . 56 is 5 . the tens digit of 6 ^ 7 = . . . 36 is 3 again . etc . . . 20 has the form 5 n , so the tens digit of 6 ^ 20 is 7 . the answer is d ." | a = 4 + 1
b = 6 ** reminder
c = reminder / (
d = math.floor(c, 100)
|
['a ) 3.34', 'b ) 3.36', 'c ) 3.34', 'd ) 6.32', 'e ) 3.31'] | a | subtract(circle_area(add(divide(4, const_2), divide(25, const_100))), circle_area(divide(4, const_2))) | a 25 cm wide path is to be made around a circular garden having a diameter of 4 meters . approximate area of the path is square meters is | area of the path = area of the outer circle - area of the inner circle = ∏ { 4 / 2 + 25 / 100 } 2 - ∏ [ 4 / 2 ] 2 = ∏ [ 2.252 - 22 ] = ∏ ( 0.25 ) ( 4.25 ) { ( a 2 - b 2 = ( a - b ) ( a + b ) } = ( 3.14 ) ( 1 / 4 ) ( 17 / 4 ) = 53.38 / 16 = 3.34 sq m . answer : a | a = 4 / 2
b = 25 / 100
c = a + b
d = circle_area - (
|
a ) rs . 30,000 , b ) rs . 50,000 , c ) rs . 60,000 , d ) rs . 80,000 , e ) none | a | divide(multiply(multiply(add(const_1, const_4), const_1000), 1), 3) | x and y invested in a business . they earned some profit which they divided in the ratio of 1 : 3 . if x invested rs . 10,000 . the amount invested by y is | "solution suppose y invested rs . y then , 10000 / y = 1 / 3 â € ¹ = â € º y = ( 10000 ã — 3 / 1 ) . â € ¹ = â € º y = 30000 . answer a" | a = 1 + 4
b = a * 1000
c = b * 1
d = c / 3
|
a ) 40 % , b ) 44 % , c ) 48 % , d ) 52 % , e ) 56 % | c | multiply(divide(subtract(multiply(add(add(multiply(50, const_2), 50), multiply(50, const_2)), divide(4, 5)), 80), add(add(multiply(50, const_2), 50), multiply(50, const_2))), multiply(50, const_2)) | in a certain math department , students are required to enroll in either calculus or trigonometry , each of which is offered in beginner and advanced courses . the number of students enrolled in trigonometry is 50 % greater than the number of students enrolled in calculus , and 80 % of calculus students are enrolled in the beginner course . if 4 / 5 of students are in the beginner courses , and one student is selected at random , what is the probability that a beginner trigonometry student is selected ? | let x be the number of students in calculus . then the number of students in trigonometry is 1.5 x the number of students on beginner calculus is 0.8 x the number of students in beginner trigonometry is 4 / 5 * ( 2.5 x ) - 0.8 x = 1.2 x the percentage in beginner trigonometry is 1.2 x / 2.5 x = 48 % the answer is c . | a = 50 * 2
b = a + 50
c = 50 * 2
d = b + c
e = 4 / 5
f = d * e
g = f - 80
h = 50 * 2
i = h + 50
j = 50 * 2
k = i + j
l = g / k
m = 50 * 2
n = l * m
|
a ) 45 , b ) 20 , c ) 53 , d ) 55 , e ) 60 | b | divide(80, divide(add(multiply(subtract(10, 9), const_60), subtract(10, 10)), const_60)) | a car going at 10 miles per hour set out on an 80 - mile trip at 9 : 00 a . m . exactly 10 minutes later , a second car left from the same place and followed the same route . how fast , in miles per hour , was the second car going if it caught up with the first car at 10 : 30 a . m . ? | "let car a = car that starts at 9 am car b = car that starts at 9 : 10 am time for which car a travels at speed of 10 m per hour = 1.5 hours distance travelled by car a = 10 * 1.5 = 15 miles since car b catches up car a at 10 : 30 , time = 80 mins = 4 / 3 hour speed of car b = 15 / ( 4 / 3 ) = 20 miles per hour answer b" | a = 10 - 9
b = a * const_60
c = 10 - 10
d = b + c
e = d / const_60
f = 80 / e
|
a ) 0.0009 , b ) 0.09 , c ) 0.9 , d ) 9 , e ) 90 | b | divide(0.009, 0.1) | 0.009 / x = 0.1 . find the value of x | x = 0.009 / 0.1 = 0.09 answer : b | a = 0 / 9
|
a ) 10 ^ 5 , b ) 10 ^ 6 , c ) 10 ^ 7 , d ) 10 ^ 8 , e ) 10 ^ 9 | c | divide(0.1, power(0.01, 4)) | the decimal 0.1 is how many times greater than the decimal ( 0.01 ) ^ 4 ? | "0.1 = 10 ^ - 1 ( 0.01 ) ^ 4 = ( 10 ^ - 2 ) ^ 4 = 10 ^ - 8 10 ^ 7 * 10 ^ - 8 = 10 ^ - 1 the answer is c ." | a = 0 ** 1
b = 0 / 1
|
a ) 30 , b ) 35 , c ) 40 , d ) 45 , e ) 50 | c | add(add(50, subtract(60, 50)), 50) | in a group of 70 people , 60 like swimming , 50 like wrestling . how many like both swimming and wrestling ? | "make a venn diagram , and enter your data . let the number of people who like both swimming and wrestling be x 60 - x + x + 50 - x = 70 x = 40 so number who like both swimming and wrestling = 40 answer c" | a = 60 - 50
b = 50 + a
c = b + 50
|
a ) - 4 , b ) 4 , c ) - 1 / 4 , d ) - 1 / 3 , e ) 2 | d | divide(const_1, 3) | in the coordinate plane a slope of the line k is 3 times the y - intercept of the line k . what is the x - intercept of the line k ? | as y = 3 mx + m , from 0 = 3 mx + m we get x = - 1 / 3 . hence , the correct answer choice is d . | a = 1 / 3
|
a ) 7 , b ) 6 , c ) 9 , d ) 10 , e ) 3 | d | subtract(34, multiply(19, const_2)) | a number when divided by 128 gives a remainder 34 , what remainder will be obtained by dividing the same number 19 ? | "explanation : 128 + 34 = 162 / 19 = 10 ( remainder ) answer : d" | a = 19 * 2
b = 34 - a
|
a ) 80 , b ) 95 , c ) 105 , d ) 120 , e ) 210 | d | multiply(divide(120, add(subtract(divide(const_1, const_3), multiply(divide(const_1, const_3), divide(30, const_100))), multiply(subtract(const_1, divide(const_1, const_3)), divide(40, const_100)))), add(multiply(divide(const_1, const_3), divide(30, const_100)), subtract(subtract(const_1, divide(const_1, const_3)), multiply(subtract(const_1, divide(const_1, const_3)), divide(40, const_100))))) | in a certain company , a third of the workers do not have a retirement plan . 30 % of the workers who do not have a retirement plan are women , and 40 % of the workers who do have a retirement plan are men . if 120 of the workers of that company are men , how many of the workers are women ? | "set up equation : x = total number of workers 120 = 0,4 * 2 / 3 * x + 0,7 * 1 / 3 * x 120 = 15 / 30 x x = 240 240 - 120 = 120 answer d" | a = 1 / 3
b = 1 / 3
c = 30 / 100
d = b * c
e = a - d
f = 1 / 3
g = 1 - f
h = 40 / 100
i = g * h
j = e + i
k = 120 / j
l = 1 / 3
m = 30 / 100
n = l * m
o = 1 / 3
p = 1 - o
q = 1 / 3
r = 1 - q
s = 40 / 100
t = r * s
u = p - t
v = n + u
w = k * v
|
a ) - 1 , b ) 1 , c ) 2 , d ) 3 , e ) 4 | c | multiply(3, 2) | find the value of x : x ² − 3 x + 2 | "( x − 1 ) ( x − 2 ) x = 1 or 2 . c" | a = 3 * 2
|
a ) 12 days , b ) 18 days , c ) 21 days , d ) 3 days , e ) 13 days | a | multiply(const_3, 3) | aarti can do a piece of work in 3 days . in how many days will she complete four time of work of same type ? | "we have the important relation , more work , more time ( days ) a piece of work can be done in 3 days . four times of work of same type can be done in 3 x 4 = 12 days answer a" | a = 3 * 3
|
a ) 100 cm 2 , b ) 150 cm 2 , c ) 333 cm 2 , d ) 400 cm 2 , e ) 750 cm 2 | d | multiply(multiply(divide(const_1, const_2), add(9, 11)), 40) | find the area of the quadrilateral of one of its diagonals is 40 cm and its off sets 11 cm and 9 cm ? | "1 / 2 * 40 ( 11 + 9 ) = 400 cm 2 answer : d" | a = 1 / 2
b = 9 + 11
c = a * b
d = c * 40
|
a ) 7 hrs . , b ) 4 hrs . , c ) 5 hrs . , d ) 6 hrs . , e ) 8 hrs . | d | divide(multiply(10, 3), subtract(10, add(divide(multiply(10, 3), 15), 3))) | if 3 taps are opened together a tank is filled in 3 hrs . one of the taps alone can fill it in 10 hrs . and another in 15 hrs . in how many hours will the third tap fill it ? | part filled by three taps together in 1 hr = 1 / 3 part filled by one of the tap in 1 hr = 1 / 10 part filled by another tap in 1 hr = 1 / 15 part filled by third tap in 1 hr = 1 / 3 - ( 1 / 0 + 1 / 15 ) = 5 / 30 therefore the third tap can fill the whole tank in : 30 / 5 hrs . = 6 hrs . answer : d | a = 10 * 3
b = 10 * 3
c = b / 15
d = c + 3
e = 10 - d
f = a / e
|
a ) rs . 2000 , b ) rs . 10,000 , c ) rs . 18,000 , d ) rs . 20,000 , e ) none | c | divide(divide(multiply(5400, const_100), multiply(3, 10)), multiply(divide(const_60, const_4), const_100)) | a man took a loan from a bank at the rate of 10 % p . a . simple interest . after 3 years he had to pay rs . 5400 interest only for the period . the principal amount borrowed by him was : | "solution principal = rs . ( 100 x 5400 / 10 x 3 ) = rs . 18000 . answer c" | a = 5400 * 100
b = 3 * 10
c = a / b
d = const_60 / 4
e = d * 100
f = c / e
|
a ) 6 , b ) 7 , c ) 5 , d ) 8 , e ) 10 | e | add(add(const_4, const_3), const_1) | how many different positive integers exist between 10 ^ 9 and 10 ^ 10 , the sum of whose digits is equal to 2 ? | "so , the numbers should be from 1,000 , 000,000 to 10,000 , 000,000 the following two cases are possible for the sum of the digits to be 2 : 1 . two 1 ' s and the rest are 0 ' s : 1,000 , 000,001 1,000 , 000,010 1,000 , 000,100 1,000 , 001,000 1,000 , 010,000 1,000 , 100,000 1,001 , 000,000 1,010 , 000,000 1,100 , 000,000 9 numbers . 2 . one 2 and the rest are 0 ' s : 2,000 , 000,000 1 number . total = 10 numbers . answer : e" | a = 4 + 3
b = a + 1
|
a ) $ 25 , b ) $ 10 , c ) $ 29.65 , d ) $ 35.95 , e ) $ 45.62 | b | divide(multiply(subtract(const_100, 60), divide(50, const_2)), const_100) | a pair of articles was bought for $ 50 at a discount of 60 % . what must be the marked price of each of the article ? | "s . p . of each of the article = 50 / 2 = $ 25 let m . p = $ x 40 % of x = 25 x = 25 * . 4 = $ 10 answer is b" | a = 100 - 60
b = 50 / 2
c = a * b
d = c / 100
|
a ) 121 , b ) 421 , c ) 143 , d ) 176 , e ) 84 | a | subtract(multiply(add(15, const_1), add(25, 6)), multiply(25, 15)) | the average runs of a cricket player of 15 innings was 25 . how many runs must he make in his next innings so as to increase his average of runs by 6 ? | "explanation : average after 16 innings = 31 required number of runs = ( 31 * 16 ) – ( 25 * 15 ) = 496 – 375 = 121 answer a" | a = 15 + 1
b = 25 + 6
c = a * b
d = 25 * 15
e = c - d
|
a ) 166 , b ) 332 , c ) 400 , d ) 483 , e ) 404 | e | sqrt(subtract(power(multiply(283, 2), 2), power(multiply(202, 2), 2))) | two airplanes take off from one airfield at noon . one flies due east at 202 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 . ? | "e in two hours : the plane flying east will be 404 miles away from airport . the other plane will be 566 miles away from airport . 566 / 404 = ~ 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 404 miles apart . e" | a = 283 * 2
b = a ** 2
c = 202 * 2
d = c ** 2
e = b - d
f = math.sqrt(e)
|
a ) 87801 , b ) 87811 , c ) 87862 , d ) 11803 , e ) 87803 | e | multiply(subtract(subtract(add(11901, 1569), 9211), 1315), divide(const_60, const_2)) | ( x ) + 1315 + 9211 - 1569 = 11901 . calculate the value of x | x + 1315 + 9211 - 1569 = 11901 = x + 1315 + 9211 = 11901 + 1569 = x + 10526 = 13470 = x = 98329 - 10526 = 87803 answer is e | a = 11901 + 1569
b = a - 9211
c = b - 1315
d = const_60 / 2
e = c * d
|
a ) 1600 , b ) 6400 , c ) 16000 , d ) 12000 , e ) 18000 | b | multiply(multiply(8, const_1000), 8) | find the product of the local value and absolute value of 8 in 564823 | "explanation : place value = local value face value = absolute value the place value of 8 in 564823 is 8 x 100 = 800 the face value of 8 in 564823 is nothing but 8 . = > 800 * 8 = 6400 answer : option b" | a = 8 * 1000
b = a * 8
|
a ) 40 , b ) 165 , c ) 120 , d ) 80 , e ) 100 | b | multiply(110, sqrt(divide(9, 4))) | two trains a and b starting from two points and travelling in opposite directions , reach their destinations 9 hours and 4 hours respectively after meeting each other . if the train a travels at 110 kmph , find the rate at which the train b runs . | if two objects a and b start simultaneously from opposite points and , after meeting , reach their destinations in ‘ a ’ and ‘ b ’ hours respectively ( i . e . a takes ‘ a hrs ’ to travel from the meeting point to his destination and b takes ‘ b hrs ’ to travel from the meeting point to his destination ) , then the ratio of their speeds is given by : sa / sb = √ ( b / a ) i . e . ratio of speeds is given by the square root of the inverse ratio of time taken . sa / sb = √ ( 4 / 9 ) = 2 / 3 this gives us that the ratio of the speed of a : speed of b as 2 : 3 . since speed of a is 110 kmph , speed of b must be 110 * ( 3 / 2 ) = 165 kmph answer b | a = 9 / 4
b = math.sqrt(a)
c = 110 * b
|
a ) 1.9 , b ) 1.7 , c ) 1.3 , d ) 1.5 , e ) 1.1 | c | divide(subtract(multiply(3.9, const_2), 3.9), const_3) | a man can row 3.9 km / hr in still water . it takes him twice as long to row upstream as to row downstream . what is the rate of the current ? | "speed of boat in still water ( b ) = 3.9 km / hr . speed of boat with stream ( down stream ) , d = b + u speed of boat against stream ( up stream ) , u = b – u it is given upstream time is twice to that of down stream . ⇒ downstream speed is twice to that of upstream . so b + u = 2 ( b – u ) ⇒ u = b / 3 = 1.3 km / hr . answer : c" | a = 3 * 9
b = a - 3
c = b / 3
|
a ) 115 , b ) 116 , c ) 117 , d ) 118 , e ) 119 | c | add(add(multiply(divide(const_100, 45), 27), multiply(divide(50, 45), 27)), 27) | 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 . 27 , what is the total amount ? | "x : y : z = 100 : 45 : 50 20 : 9 : 10 9 - - - 27 39 - - - ? = > 117 answer : c" | a = 100 / 45
b = a * 27
c = 50 / 45
d = c * 27
e = b + d
f = e + 27
|
a ) − 4 , b ) − 5 , c ) 2 , d ) 46 , e ) 48 | a | add(divide(21, const_10), divide(21, divide(21, const_10))) | if a ( a + 4 ) = 21 and b ( b + 4 ) = 21 , where a ≠ b , then a + b = | "i . e . if a = 3 then b = - 7 or if a = - 7 then b = 3 but in each case a + b = - 7 + 3 = - 4 answer : option a" | a = 21 / 10
b = 21 / 10
c = 21 / b
d = a + c
|
a ) 377 , b ) 267 , c ) 268 , d ) 750 , e ) 271 | d | divide(divide(multiply(90, const_1000), divide(const_60, const_1)), const_2) | the length of a train and that of a platform are equal . if with a speed of 90 k / hr , the train crosses the platform in one minute , then the length of the train ( in meters ) is : | speed = [ 90 * 5 / 18 ] m / sec = 25 m / sec ; time = 1 min . = 60 sec . let the length of the train and that of the platform be x meters . then , 2 x / 60 = 25 è x = 25 * 60 / 2 = 750 . answer : d | a = 90 * 1000
b = const_60 / 1
c = a / b
d = c / 2
|
a ) 160 kg , b ) 175 kg , c ) 180.7 kg , d ) 119.4 kg , e ) 190.8 kg | d | add(multiply(7, 6.2), 76) | the average weight of 7 person ' s increases by 6.2 kg when a new person comes in place of one of them weighing 76 kg . what might be the weight of the new person ? | "total weight increased = ( 7 x 6.2 ) kg = 43.4 kg . weight of new person = ( 76 + 43.4 ) kg = 119.4 kg option d" | a = 7 * 6
b = a + 76
|
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