options stringlengths 37 300 | correct stringclasses 5 values | annotated_formula stringlengths 7 727 | problem stringlengths 5 967 | rationale stringlengths 1 2.74k | program stringlengths 10 646 |
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a ) β 2 , b ) β 1 , c ) 0 , d ) 1 , e ) 2 | a | add(0.5, negate(3)) | what is the least integer greater than β 3 + 0.5 ? | "this question is just about doing careful arithmetic and remembering what makes a numberbiggerorsmallercompared to another number . first , let ' s take care of the arithmetic : ( - 3 ) + ( 0.5 ) = - 2.5 on a number line , since we ' re adding + . 5 to a number , the total moves to the right ( so we ' re moving from - 3 to - 2.5 ) . next , the question asks for the least integer that is greater than - 2.5 again , we can use a number line . numbers become greater as you move to the right . the first integer to the right of - 2.5 is - 2 . final answer : a" | a = 0 + 5
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a ) $ 60 , b ) $ 70 , c ) $ 75 , d ) $ 96 , e ) can not be determined | b | divide(add(50, 90), 2) | if greg buys 3 shirts , 4 trousers and 2 ties , the total cost is $ 90 . if greg buys 7 shirts , 2 trousers and 2 ties , the total cost is $ 50 . how much will it cost him to buy 3 trousers , 5 shirts and 2 ties ? | "solution : 3 x + 4 y + 2 z = 90 7 x + 2 y + 2 z = 50 adding both the equations = 10 x + 6 y + 4 z = 140 5 x + 3 y + 2 z = 70 ans b" | a = 50 + 90
b = a / 2
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a ) 22 % , b ) 23.9 % , c ) 24 % , d ) 25 % , e ) 34 % | b | divide(22, divide(subtract(const_100, 8), const_100)) | in a certain candy store , 22 % of the customers are caught sampling the candy and are charged a small fine , but 8 % of the customers who sample the candy are not caught . what is the total percent of all customers who sample candy ? | "since 8 % of the customers who sample the candyare notcaught , then 88 % of the customers who sample the candyarecaught : { % of customers who sample candy } * 0.92 = 0.22 ; { % of customers who sample candy } = 0.239 . answer : b ." | a = 100 - 8
b = a / 100
c = 22 / b
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a ) 0 , b ) 1 , c ) 2 , d ) 3 , e ) 4 | a | divide(multiply(add(4, 3), const_2), 8) | | x + 3 | β | 4 - x | = | 8 + x | how many u solutions will this equation have ? | "| x | = x when x > = 0 ( x is either positive or 0 ) | x | = - x when x < 0 ( note here that you can put the equal to sign here as well x < = 0 because if x = 0 , | 0 | = 0 = - 0 ( all are the same ) so the ' = ' sign can be put with x > 0 or with x < 0 . we usually put it with ' x > 0 ' for consistency . a" | a = 4 + 3
b = a * 2
c = b / 8
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a ) 12 days , b ) 12 1 / 4 days , c ) 14 days , d ) 24 1 / 2 days , e ) none of these | a | divide(const_1, add(divide(const_1, 21), divide(const_1, 28))) | a sum of money is sufficient to pay a ' s wages for 21 days and b ' s wages for 28 days . the same money is sufficient to pay the wages of both for ? | let total money be rs . x a ' s 1 day ' s wages = rs . x / 21 , b ' s 1 day ' s wages = rs . x / 28 ( a + b ) ' s 1 day ' s wages = rs . ( x / 21 + x / 28 ) = rs . x / 12 ; money is sufficient to pay the wages of both for 12 days . correct option : a | a = 1 / 21
b = 1 / 28
c = a + b
d = 1 / c
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a ) 50 , b ) 80 , c ) 120 , d ) 150 , e ) 100 | a | divide(add(24, multiply(24, 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 24 . 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 = 24 + 2 / 3 of 24 80 / 100 x = 40 x = 50 . answer : option a" | a = 2 / 3
b = 24 * a
c = 24 + b
d = 20 / 100
e = 1 - d
f = c / e
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a ) 28 , b ) 32 , c ) 40 , d ) 53 , e ) 54 | c | add(multiply(divide(48, multiply(2, 3)), 3), multiply(divide(48, multiply(2, 3)), 2)) | the l . c . m . of two numbers is 48 . the numbers are in the ratio 2 : 3 . then sum of the number is : | "let the numbers be 2 x and 3 x . then , their l . c . m . = 6 x . so , 6 x = 48 or x = 8 . the numbers are 16 and 24 . hence , required sum = ( 16 + 24 ) = 40 . answer : option c" | a = 2 * 3
b = 48 / a
c = b * 3
d = 2 * 3
e = 48 / d
f = e * 2
g = c + f
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a ) r = 21 , b ) r = 22 , c ) r = 23 , d ) 24 , e ) 27 | c | subtract(power(5, 2), 2) | if x + ( 1 / x ) = 5 , what is the value of r = x ^ 2 + ( 1 / x ) ^ 2 ? | "squaring on both sides , x ^ 2 + ( 1 / x ) ^ 2 + 2 ( x ) ( 1 / x ) = 5 ^ 2 x ^ 2 + ( 1 / x ) ^ 2 = 23 answer : c" | a = 5 ** 2
b = a - 2
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a ) 21 , b ) 22 , c ) 27 , d ) 26 , e ) 28 | c | add(divide(subtract(56, 1), 2), const_1) | how many multiples of 2 are there between 1 and 56 , exclusive ? | "27 multiples of 2 between 1 and 56 exclusive . from 2 * 1 upto 2 * 27 , ( 1,2 , 3,4 , . . . , 27 ) . hence , 27 multiples ! correct option is c" | a = 56 - 1
b = a / 2
c = b + 1
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a ) 100 , b ) 110 , c ) 120 , d ) 130 , e ) 140 | c | divide(subtract(multiply(39, 120), multiply(120, 15)), subtract(39, 15)) | the average of marks obtained by 120 boys was 39 . if the average of marks of passed boys was 39 and that of failed boys was 15 , the number of boys who passed the examination is ? | "let the number of boys who passed = x . then , 39 x x + 15 x ( 120 - x ) = 120 x 39 24 x = 4680 - 1800 = > x = 2880 / 24 x = 120 . hence , the number of boys passed = 120 . answer : c" | a = 39 * 120
b = 120 * 15
c = a - b
d = 39 - 15
e = c / d
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['a ) rs . 4233', 'b ) rs . 4350', 'c ) rs . 4457', 'd ) rs . 4670', 'e ) rs . 4756'] | c | multiply(circumface(sqrt(divide(multiply(multiply(const_1000, const_10), 17.56), const_pi))), const_3) | the area of a circular place is 17.56 hectares . find the cost of fencing it at the rate of rs . 3 / meter approximately . | area = ( 17.56 x 10000 ) m 2 = 175600 m 2 . Ο r 2 = 175600 β ( r ) 2 = ( 175600 x ( 7 / 22 ) ) β r = 236.37 m . circumference = 2 Ο r = ( 2 x ( 22 / 7 ) x 236.37 ) m = 1485.78 m . cost of fencing = rs . ( 1485.78 x 3 ) = rs . 4457 . c | a = 1000 * 10
b = a * 17
c = b / math.pi
d = math.sqrt(c)
e = circumface * (
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a ) 1 / 221 , b ) 8 / 221 , c ) 4 / 589 , d ) 4 / 587 , e ) 7 / 654 | a | divide(multiply(divide(52, const_4), divide(52, const_4)), choose(52, const_2)) | from a pack of 52 cards , two cards are drawn at random together at random what is the probability of both the cards being kings ? | "let s be the sample space then n ( s ) = 52 c 2 = ( 52 * 51 ) / ( 2 * 1 ) = 1326 let e be the event of getting 2 kings out of 4 n ( e ) = 4 c 2 = ( 4 * 3 ) / ( 2 * 1 ) = 6 p ( e ) = n ( e ) / n ( s ) = 6 / 1326 = 1 / 221 answer ( a )" | a = 52 / 4
b = 52 / 4
c = a * b
d = math.comb(52, 2)
e = c / d
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a ) 40 % , b ) 45 % , c ) 50 % , d ) 60 % , e ) 67 % | c | multiply(divide(subtract(multiply(const_3, divide(const_1, const_2)), const_1), const_1), const_100) | the length of a rectangle is halved , while its breadth is tripled . wat isthe % change in area ? | "let original length = x and original breadth = y . original area = xy . new length = x . 2 new breadth = 3 y . new area = x x 3 y = 3 xy . 2 2 increase % = 1 xy x 1 x 100 % = 50 % . 2 xy c" | a = 1 / 2
b = 3 * a
c = b - 1
d = c / 1
e = d * 100
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a ) 43 , b ) 53 , c ) 63 , d ) 65 , e ) 74 | e | subtract(negate(10), multiply(subtract(7,12, 18,25), divide(subtract(7,12, 18,25), subtract(3, 7,12)))) | 3 , 7,12 , 18,25 . . . . . . . . . . . . . . 10 th terms | "3 + 4 = 7 7 + 5 = 12 12 + 6 = 18 18 + 7 = 25 25 + 8 = 33 33 + 9 = 42 42 + 10 = 52 52 + 11 = 63 63 + 11 = 74 answer : e" | a = negate - (
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a ) 127 days , b ) 150 days , c ) 177 days , d ) 187 days , e ) 225 days | b | multiply(divide(multiply(const_1, 100), subtract(multiply(const_1, 100), multiply(10, 5))), multiply(15, 5)) | 10 men and 15 women together can complete a work in 5 days . it takes 100 days for one man alone to complete the same work . how many days will be required for one woman alone to complete the same work ? | "1 man ' s 1 day work = 1 / 100 ( 10 men + 15 women ) ' s 1 day work = 1 / 5 15 women ' s 1 day work = ( 1 / 5 - 10 / 100 ) = 1 / 10 1 woman ' s 1 day work = 1 / 150 1 woman alone can complete the work in 150 days . answer : b" | a = 1 * 100
b = 1 * 100
c = 10 * 5
d = b - c
e = a / d
f = 15 * 5
g = e * f
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a ) 1 : 3 , b ) 4 : 1 , c ) 2 : 3 , d ) 3 : 2 , e ) 3 : 4 | b | divide(subtract(27, 25), subtract(25, 17)) | two trains running in opposite directions cross a man standing on the platform in 27 seconds and 17 seconds respectively . if they cross each other in 25 seconds , what is the ratio of their speeds ? | "let the speed of the trains be x and y respectively length of train 1 = 27 x length of train 2 = 17 y relative speed = x + y time taken to cross each other = 25 s = ( 27 x + 17 y ) / ( x + y ) = 25 = ( 27 x + 17 y ) / = 25 ( x + y ) = 2 x = 8 y = x / y = 8 / 2 = 4 / 1 i . e 4 : 1 answer : b" | a = 27 - 25
b = 25 - 17
c = a / b
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a ) s . 48 , b ) s . 50 , c ) s . 55 , d ) s . 60 , e ) s . 70 | a | divide(720, subtract(20, 5)) | on selling 20 balls at rs . 720 , there is a loss equal to the cost price of 5 balls . the cost price of a ball is : | "( c . p . of 20 balls ) - ( s . p . of 20 balls ) = ( c . p . of 5 balls ) c . p . of 15 balls = s . p . of 20 balls = rs . 720 . c . p . of 1 ball = rs . 720 / 15 = rs . 48 . answer : option a" | a = 20 - 5
b = 720 / a
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a ) 5.2 , b ) 5.5 , c ) 7 , d ) 9 , e ) 10 | c | add(13, const_1) | the average of first 13 natural numbers is ? | "sum of 13 natural no . = 182 / 2 = 91 average = 91 / 13 = 7 answer : c" | a = 13 + 1
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a ) 20000 , b ) 24000 , c ) 34000 , d ) 35000 , e ) 30000 | e | multiply(54000, subtract(const_1, divide(multiply(30000, multiply(const_2, multiply(const_2, const_3))), add(multiply(45000, subtract(multiply(const_2, multiply(const_2, const_3)), 2)), multiply(30000, multiply(const_2, multiply(const_2, const_3))))))) | tom opened a shop investing rs . 30000 . jose joined him 2 months later , investing rs . 45000 . they earned a profit of rs . 54000 after completion of one year . what will be jose ' s share of profit ? | sol = ~ s - so anju β s share = [ 5 / 9 ] x 54000 = 30000 e | a = 2 * 3
b = 2 * a
c = 30000 * b
d = 2 * 3
e = 2 * d
f = e - 2
g = 45000 * f
h = 2 * 3
i = 2 * h
j = 30000 * i
k = g + j
l = c / k
m = 1 - l
n = 54000 * m
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a ) 75 kgs , b ) 64 kgs , c ) 72 kgs , d ) 65 kgs , e ) 70 kgs | a | add(divide(multiply(25, subtract(const_100, 25)), const_100), 75) | fresh grapes contain 75 % water by weight and raisins obtained by drying fresh grapes contain 25 % water by weight . how many kgs of fresh grapes are needed to get 25 kgs of raisins ? | "the weight of non - water in 20 kg of dried grapes ( which is 100 - 25 = 75 % of whole weight ) will be the same as the weight of non - water in x kg of fresh grapes ( which is 100 - 75 = 25 % of whole weight ) , so 25 * 0.75 = x * 0.25 - - > x = 75 . answer : a ." | a = 100 - 25
b = 25 * a
c = b / 100
d = c + 75
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a ) 81000 , b ) 22887 , c ) 26877 , d ) 26771 , e ) 17121 | a | add(add(64000, multiply(divide(1, 8), 64000)), multiply(divide(1, 8), add(64000, multiply(divide(1, 8), 64000)))) | every year an amount increases by 1 / 8 th of itself . how much will it be after two years if its present value is rs . 64000 ? | 64000 * 9 / 8 * 9 / 8 = 81000 answer : a | a = 1 / 8
b = a * 64000
c = 64000 + b
d = 1 / 8
e = 1 / 8
f = e * 64000
g = 64000 + f
h = d * g
i = c + h
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a ) 298 m , b ) 188 m , c ) 120 m , d ) 160 m , e ) 189 m | d | divide(12, subtract(divide(12, 10), 5)) | a train covers a distance of 12 km in 10 min . if it takes 5 sec to pass a telegraph post , then the length of the train is ? | "speed = ( 12 / 10 * 60 ) km / hr = ( 72 * 5 / 18 ) m / sec = 20 m / sec . length of the train = 20 * 8 = 160 m . answer : d" | a = 12 / 10
b = a - 5
c = 12 / b
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a ) 334 , b ) 500 , c ) 376 , d ) 288 , e ) 300 | e | divide(285, divide(95, const_100)) | victor gets 95 % marks in examinations . if these are 285 marks , find the maximum marks . | let the maximum marks be m then 95 % of m = 285 β 95 / 100 Γ m = 285 β m = ( 285 Γ 100 ) / 95 β m = 28500 / 95 β m = 300 therefore , maximum marks in the examinations are 300 . answer : e | a = 95 / 100
b = 285 / a
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a ) 1 / 6 , b ) 1 / 2 , c ) 5 / 8 , d ) 5 / 6 , e ) 15 / 16 | c | divide(subtract(0.75, 0.5), subtract(0.9, 0.5)) | last month , john rejected 0.5 % of the products that he inspected and jane rejected 0.9 percent of the products that she inspected . if total of 0.75 percent of the products produced last month were rejected , what fraction of the products did jane inspect ? | "x - fraction of products jane inspected ( 1 - x ) - fraction of products john inspected 0.9 ( x ) + 0.5 ( 1 - x ) = 0.75 0.4 x = 0.75 - 0.5 x = 0.25 / 0.4 x = 5 / 8 therefore the answer is c : 5 / 8 ." | a = 0 - 75
b = 0 - 9
c = a / b
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a ) 3 , b ) 4 , c ) 5 , d ) 6 , e ) 7 | e | add(divide(lcm(375, 150), 375), divide(lcm(375, 150), 150)) | a company wants to spend equal amounts of money for the purchase of two types of computer printers costing $ 375 and $ 150 per unit , respectively . what is the fewest number of computer printers that the company can purchase ? | the smallest amount that the company can spend is the lcm of 375 and 150 , which is 750 for each , which is total 1500 . the number of 1 st type of computers which costing $ 375 = 750 / 375 = 2 . the number of 2 nd type of computers which costing $ 150 = 750 / 150 = 5 . total = 2 + 5 = 7 answer is e . | a = math.lcm(375, 150)
b = a / 375
c = math.lcm(375, 150)
d = c / 150
e = b + d
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a ) 50 , b ) 60 , c ) 70 , d ) 80 , e ) 100 | e | add(multiply(divide(100, const_10), multiply(const_2, const_4)), divide(100, const_10)) | how many integerskgreater than 100 and less than 1200 are there such that if the hundreds and the units digits ofkare reversed , the resulting integer is k + 99 ? | "numbers will be like 102 = > 201 = 102 + 99 203 = > 302 = 103 + 99 so the hundereth digit and units digit are consecutive where unit digit is bigger than hundred digit . there will be ten pairs of such numbers for every pair there will 10 numbers like for 12 = > 102 , 112,132 , 142,152 , 162,172 , 182,192 . total = 10 * 10 = 100 hence e" | a = 100 / 10
b = 2 * 4
c = a * b
d = 100 / 10
e = c + d
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a ) 21 / 31 , b ) 9 / 4 , c ) 5 / 9 , d ) 7 / 5 , e ) 9 / 7 | a | add(subtract(1, divide(2, 3)), subtract(divide(2, 3), divide(1, 5))) | a batch of cookies was divided amomg 3 tins : 2 / 3 of all the cookies were placed in either the blue or the green tin , and the rest were placed in the red tin . if 1 / 5 of all the cookies were placed in the blue tin , what fraction of the cookies that were placed in the other tins were placed in the green tin | "this will help reduce the number of variables you have to deal with : g + b = 2 / 3 r = 1 / 3 b = 1 / 5 we can solve for g which is 7 / 10 what fraction ( let it equal x ) of the cookies that were placed in the other tins were placed in the green tin ? so . . x * ( g + r ) = g x * ( 7 / 10 + 1 / 3 ) = 7 / 10 x = 21 / 31 answer : a" | a = 2 / 3
b = 1 - a
c = 2 / 3
d = 1 / 5
e = c - d
f = b + e
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a ) 360 , b ) 252 , c ) 280 , d ) 300 , e ) 420 | a | divide(multiply(multiply(30, 42), 60), multiply(multiply(7, 6), 5)) | a grocer is storing soap boxes in cartons that measure 30 inches by 42 inches by 60 inches . if the measurement of each soap box is 7 inches by 6 inches by 5 inches , then what is the maximum number of soap boxes that can be placed in each carton ? | "however the process of dividing the volume of box by the volume of a soap seems flawed but it does work in this case due to the numbers dimensions of the box = 30 * 42 * 60 dimensions of the soap = 5 * 6 * 7 we get = 6 x 6 x 10 = 360 so the question is why this particular arrangement , in order to maximize number of soaps we need to minimize the space wasted and this is the only config where we dont waste any space so we can expect the maximum number the answer is ( a )" | a = 30 * 42
b = a * 60
c = 7 * 6
d = c * 5
e = b / d
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a ) 4 . , b ) 8 . , c ) 12 . , d ) 16 , e ) 64 | d | power(subtract(5, divide(add(13, 5), const_2)), const_2) | if ( a - b - c + d = 13 ) and ( a + b - c - d = 5 ) , what is the value of ( b - d ) ^ 2 ? | a - b - c + d = 13 - - equation 1 a + b - c - d = 5 - - equation 2 adding 1 and 2 , we get 2 a - 2 c = 18 = > a - c = 9 - - equation 3 using equation 3 in 2 , we get b - d = 5 - 9 = - 4 = > ( b - d ) ^ 2 = 16 answer d | a = 13 + 5
b = a / 2
c = 5 - b
d = c ** 2
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a ) 37 kmph , b ) 34 kmph , c ) 32 kmph , d ) 38 kmph , e ) 76 kmph | c | multiply(const_3_6, divide(225, 15)) | a train 225 m in length crosses a telegraph post in 15 seconds . the speed of the train is ? | "s = 225 / 25 * 18 / 5 = 32 kmph answer : c" | a = 225 / 15
b = const_3_6 * a
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a ) 800 , b ) 500 , c ) 900 , d ) 1600 , e ) 2400 | e | multiply(divide(multiply(multiply(3.6, 0.48), 2.50), multiply(multiply(0.12, 0.09), 0.5)), 3) | find the value of 3 x [ ( 3.6 x 0.48 x 2.50 ) / ( 0.12 x 0.09 x 0.5 ) ] | "answer 3 x [ ( 3.6 x 0.48 x 2.50 ) / ( 0.12 x 0.09 x 0.5 ) ] = 3 x [ ( 36 x 48 x 250 ) / ( 12 x 9 x 5 ) ] = 3 x 4 x 4 x 50 = 2400 correct option : e" | a = 3 * 6
b = a * 2
c = 0 * 12
d = c * 0
e = b / d
f = e * 3
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a ) 9 , b ) 8 , c ) 10 , d ) 12 , e ) 15 | a | divide(28, divide(add(3.5, 2.8), const_2)) | sara bought both german chocolate and swiss chocolate for some cakes she was baking . the swiss chocolate cost $ 3.5 per pound , and german chocolate cost $ 2.8 per pound . if the total the she spent on chocolate was $ 28 and both types of chocolate were purchased in whole number of pounds , how many total pounds of chocolate she purchased ? | "if there were all the expensive ones , 3.5 . . . . there would be 28 / 3.5 or 8 of them but since 2.8 $ ones are also there , answer has to be > 8 . . if all were 2.8 $ ones , there will be 28 / 2.8 or 10 . . . so only 9 is left answer a . ." | a = 3 + 5
b = a / 2
c = 28 / b
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a ) 127 , b ) 225 , c ) 287 , d ) 450 , e ) 281 | b | divide(multiply(36, 30), divide(multiply(16, 24), 80)) | if 16 men can reap 80 hectares in 24 days , then how many hectares can 36 men reap in 30 days ? | "explanation : let the required no of hectares be x . then men - - - hectares - - - days 16 - - - - - - - - - 80 - - - - - - - - - 24 36 - - - - - - - - - x - - - - - - - - - 30 more men , more hectares ( direct proportion ) more days , more hectares ( direct proportion ) x = 36 / 16 * 30 / 24 * 80 x = 225 answer : b" | a = 36 * 30
b = 16 * 24
c = b / 80
d = a / c
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a ) 21 , b ) 22 , c ) 23 , d ) 24 , e ) 25 | b | add(add(add(divide(15, 3), 1), 1), 15) | a shop sells 1 chocolate at the rate rupee 1 each . you can exchange 3 warppers for one chocolate . if you have rs . 15 , what is the maximum number of chocolates you can get ? | rs . 15 = 15 chocolates 15 wrappers = 5 chocolates 5 wrappers = 1 chocolate ( u have 2 wrappers with you . . ) count this wrapper with already existing wrappers = 15 + 5 + 1 + 1 = 22 chocolates . . answer : b | a = 15 / 3
b = a + 1
c = b + 1
d = c + 15
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a ) 42 , b ) 46 , c ) 50 , d ) 54 , e ) 58 | b | multiply(add(divide(add(subtract(multiply(6, 5), 6), multiply(9, const_2)), subtract(9, 6)), 9), const_2) | jim is now twice as old as fred , who is 9 years older than sam . 6 years ago , jim was 5 times as old as sam . how old is jim now ? | j = 2 f = 2 ( s + 9 ) = 2 s + 18 j - 6 = 5 * ( s - 6 ) ( 2 s + 18 ) - 6 = 5 s - 30 s = 14 and so j = 46 the answer is b . | a = 6 * 5
b = a - 6
c = 9 * 2
d = b + c
e = 9 - 6
f = d / e
g = f + 9
h = g * 2
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a ) 228 , b ) 278 , c ) 300 , d ) 500 , e ) 821 | c | divide(204, subtract(const_1, divide(multiply(4, 8), const_100))) | a person lent a certain sum of money at 4 % per annum at simple interest and in 8 years the interest amounted to rs . 204 less than the sum lent . what was the sum lent ? | p - 204 = ( p * 4 * 8 ) / 100 p = 300 answer : c | a = 4 * 8
b = a / 100
c = 1 - b
d = 204 / c
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a ) 36 , b ) 40 , c ) 48 , d ) 32 , e ) 56 | b | add(multiply(divide(2, multiply(48, 2)), 2), multiply(divide(2, multiply(48, 2)), 48)) | the l . c . m . of 2 numbers is 48 . the numbers are in the ratio 2 : 3 . find their sum ? | "let the numbers be 2 x and 3 x l . c . m . = 6 x 6 x = 48 x = 8 the numbers are = 16 and 24 required sum = 16 + 24 = 40 answer is b" | a = 48 * 2
b = 2 / a
c = b * 2
d = 48 * 2
e = 2 / d
f = e * 48
g = c + f
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a ) 20.8 , b ) 30.4 , c ) 31.8 , d ) 39.1 , e ) 33.8 | d | multiply(divide(subtract(69.0, 42.0), 69.0), const_100) | the credit card and a global payment processing companies have been suffering losses for some time now . a well known company recently announced its quarterly results . according to the results , the revenue fell to $ 42.0 billion from $ 69.0 billion , a year ago . by what percent did the revenue fall ? | "$ 69 - $ 42 = 27 $ ( 27 / 69 ) * 100 = 39.13 % answer : d" | a = 69 - 0
b = a / 69
c = b * 100
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a ) 5 , b ) 7 , c ) 8 , d ) 9 , e ) 6 | c | divide(factorial(subtract(add(const_4, 2), const_1)), multiply(factorial(2), factorial(subtract(const_4, const_1)))) | how many positive integers less than 100 have a remainder of 2 when divided by 13 ? | "take the multiples of 13 and add 2 0 x 13 + 2 = 2 . . . . 13 x 7 + 2 = 93 there are 14 numbers answer c" | a = 4 + 2
b = a - 1
c = math.factorial(b)
d = math.factorial(2)
e = 4 - 1
f = math.factorial(e)
g = d * f
h = c / g
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a ) 105600 yards , b ) 35200 yards , c ) 39520 yards , d ) 42560 yards , e ) 41520 yards | a | divide(multiply(multiply(multiply(add(const_3, const_2), const_2), multiply(add(const_3, const_2), const_2)), 60), multiply(multiply(add(const_3, const_2), const_2), multiply(add(const_3, const_2), const_2))) | convert 60 miles into yards ? | "1 mile = 1760 yards 60 miles = 60 * 1760 = 105600 yards answer is a" | a = 3 + 2
b = a * 2
c = 3 + 2
d = c * 2
e = b * d
f = e * 60
g = 3 + 2
h = g * 2
i = 3 + 2
j = i * 2
k = h * j
l = f / k
|
a ) $ 4.50 , b ) $ 10.2 , c ) $ 5.30 , d ) $ 5.50 , e ) $ 5.60 | b | add(3.2, multiply(divide(subtract(10.8, 3.2), add(5, subtract(sqrt(const_2), 1))), 5)) | the price of a bushel of corn is currently $ 3.20 , and the price of a peck of wheat is $ 10.80 . the price of corn is increasing at a constant rate of 5 x cents per day while the price of wheat is decreasing at a constant rate of x ( 2 ^ 1 / 2 ) - x cents per day . what is the approximate price when a bushel of corn costs the same amount as a peck of wheat ? | i tried using time / rate approach : - initial price difference = 10.80 - 3.20 = 7.60 price of corn increasing by 5 x price of wheat decreasing by x ( 1.4 ) - x = . 4 x since both the quantities are moving towards reducing the price gap hence : - relative increase = 5 x + . 4 x let t be the time by which gap is filled so , 7.6 = t ( 5.4 x ) - > t = ( 7.6 ) / 5.4 x final price = 3.20 + 5 x * t - > 3.20 + 5 * 7.6 / 5.4 = 10.2 answer b . | a = 10 - 8
b = math.sqrt(2)
c = b - 1
d = 5 + c
e = a / d
f = e * 5
g = 3 + 2
|
a ) 3 , b ) 16 , c ) 75 , d ) 24 , e ) 26 | e | subtract(add(subtract(300, 200), const_1), add(add(add(divide(subtract(300, 200), 2), const_1), add(floor(subtract(add(divide(subtract(300, 200), 3), const_1), add(multiply(5, 3), 2))), const_1)), subtract(add(divide(subtract(300, 200), 5), const_1), multiply(5, 3)))) | how many positive integers e between 200 and 300 ( both inclusive ) are not divisible by 2 , 3 or 5 ? | 1 ) i figured there are 101 integers ( 300 - 200 + 1 = 101 ) . since the set begins with an even and ends with an even , there are 51 evens . 2 ) question says integers are not divisible by 2 , leaving all of the odds ( 101 - 51 = 50 integers ) . 3 ) question says integers are not divisible by 5 , removing all the integers ending in 5 ( already took out those ending in 0 ) . take out 10 integers ( 2 ? 5 , ? = 0 to 9 ) , leaving us with 40 integers . 4 ) now the painstaking part . we have to remove the remaining numbers that are multiples of 3 . those are 201 , 207 , 213 , 219 , 231 , 237 , 243 , 249 , 261 , 267 , 273 , 279 , 291 , and 297 . . . a total of 14 numbers . 26 numbers left ! 6 ) answer choice e . | a = 300 - 200
b = a + 1
c = 300 - 200
d = c / 2
e = d + 1
f = 300 - 200
g = f / 3
h = g + 1
i = 5 * 3
j = i + 2
k = h - j
l = math.floor(k)
m = l + 1
n = e + m
o = 300 - 200
p = o / 5
q = p + 1
r = 5 * 3
s = q - r
t = n + s
u = b - t
|
a ) 280 , b ) 284 , c ) 292 , d ) 320 , e ) 322 | e | multiply(23, 14) | the hcf of two numbers is 23 and the other two factors of their lcm are 13 and 14 . what is the largest number ? | hcf of the two numbers = 23 since hcf will be always a factor of lcm , 23 is a factor of the lcm . given that other two factors in the lcm are 13 and 14 . hence factors of the lcm are 23 , 13 , 14 so , numbers can be taken as ( 23 Γ 13 ) and ( 23 Γ 14 ) = 299 and 322 hence , largest number = 322 e ) | a = 23 * 14
|
a ) 1 / 2 , b ) 2 / 3 , c ) 1 / 5 , d ) 1 / 6 , e ) none of these | b | divide(subtract(const_6, 2), const_6) | in a throw of dice what is the probability of ge Γ¦ Β« ng number greater than 2 | explanation : number greater than 2 is 3 , 4 , 5 & 6 , so only 4 number total cases of dice = [ 1,2 , 3,4 , 5,6 ] so probability = 4 / 6 = 2 / 3 answer : b | a = 6 - 2
b = a / 6
|
a ) 15 , b ) 20 , c ) 10 , d ) 5 , e ) 8 | d | multiply(subtract(multiply(add(const_1, divide(50, const_100)), subtract(const_1, divide(30, const_100))), const_1), const_100) | a retailer purchases shirts from a wholesaler and then sells the shirts in her store at a retail price that is 50 percent greater than the wholesale price . if the retailer decreases the retail price by 30 percent this will have the same effect as increasing the wholesale price by what percent ? | answer : d = 5 . assume rs . 100 to be the price at which the retailer buys from wholesaler . 50 % increase makes retail price = 150 . now 30 % decrease - > ( 1 - 30 / 100 ) * 150 = 105 . now compared to the wholesale price of 100 , 5 % increase is what will have the same effect as increasing the wholesale price . | a = 50 / 100
b = 1 + a
c = 30 / 100
d = 1 - c
e = b * d
f = e - 1
g = f * 100
|
a ) 9 , b ) 11 , c ) 13 , d ) 15 , e ) 17 | e | floor(divide(172, 10)) | on dividing 172 by a number , the quotient is 10 and the remainder is 2 . find the divisor . | "d = ( d - r ) / q = ( 172 - 2 ) / 10 = 170 / 10 = 17 e" | a = 172 / 10
b = math.floor(a)
|
a ) 25 % , b ) 33 1 / 3 % , c ) 83 1 / 3 % , d ) 33 4 / 3 % , e ) 73 1 / 3 % | a | divide(multiply(const_100, const_100), multiply(4, const_100)) | in what time a sum of money double itself at 4 % per annum simple interest ? | "p = ( p * 4 * r ) / 100 r = 25 % answer : a" | a = 100 * 100
b = 4 * 100
c = a / b
|
a ) 80 , b ) 90 , c ) 100 , d ) 110 , e ) 120 | e | divide(48, subtract(divide(3, 4), divide(35, const_100))) | a big container is 35 % full with water . if 48 liters of water is added , the container becomes 3 / 4 full . what is the capacity of the big container in liters ? | "48 liters is 40 % of the capacity c . 48 = 0.4 c c = 48 / 0.4 = 120 liters . the answer is e ." | a = 3 / 4
b = 35 / 100
c = a - b
d = 48 / c
|
a ) 2 / 3 , b ) 3 / 4 , c ) 7 / 19 , d ) 8 / 21 , e ) 10 / 21 | d | divide(8, add(add(8, 7), 6)) | in a box , there are 8 orange , 7 black and 6 white balls . if one ball is picked up randomly . what is the probability that it is neither orange nor white ? | total balls are 21 and neither black nor white . so favor case will be only 8 c 1 and total case 21 c 1 probability will be 8 c 1 / 21 c 1 = 8 / 21 answer : d | a = 8 + 7
b = a + 6
c = 8 / b
|
a ) 48 , b ) 64 , c ) 36 , d ) 72 , e ) 27 | c | divide(subtract(96, multiply(12, 5)), subtract(13, 12)) | suraj has a certain average of runs for 12 innings . in the 13 th innings he scores 96 runs thereby increasing his average by 5 runs . what is his average after the 13 th innings ? | "to improve his average by 5 runs per innings he has to contribute 12 x 5 = 60 runs for the previous 12 innings . thus , the average after the 13 th innings = 96 - 60 = 36 . answer : c" | a = 12 * 5
b = 96 - a
c = 13 - 12
d = b / c
|
a ) 50 , b ) 60 , c ) 70 , d ) 80 , e ) 90 | b | divide(multiply(150, const_100), add(150, const_100)) | total number of boys and girls in a school is 150 . if the number of boys is x , then girls become x % of the total number of students . the number of boys is | explanation : clearly , x % of 150 = 150 - x [ as x is number of boys ] = > x + x / 100 β 150 = 150 = > 5 / 2 x = 150 = > x = 60 option b | a = 150 * 100
b = 150 + 100
c = a / b
|
a ) 2 years , b ) 5 years , c ) 7 years , d ) 8 years , e ) 10 years | e | divide(subtract(60000, 20000), multiply(20000, divide(20, const_100))) | $ 20000 will become $ 60000 at 20 % p . a . find the time ? | "si = simple interest = a - p = 60000 - 20000 = $ 40000 r = 100 si / pt = 100 * 40000 / 20000 * 20 = 10 years answer is e" | a = 60000 - 20000
b = 20 / 100
c = 20000 * b
d = a / c
|
a ) 9 : 8 , b ) 8 : 9 , c ) 6 : 2 , d ) 2 : 3 , e ) 1 : 2 | b | divide(divide(multiply(const_4, 3), multiply(3, 3)), divide(multiply(3, const_4), multiply(2, const_4))) | a certain car dealership sells economy cars , luxury cars , and sport utility vehicles . the ratio of economy to luxury cars is 6 : 2 . the ratio of economy cars to sport utility vehicles is 5 : 3 . what is the ratio of luxury cars to sport utility vehicles ? | the ratio of economy to luxury cars is 6 : 2 - - > e : l = 6 : 2 = 30 : 10 . the ratio of economy cars to sport utility vehicles is 5 : 3 - - > e : s = 5 : 3 = 30 : 18 . thus , l : s = 10 : 18 = 5 : 6 . answer : b . | a = 4 * 3
b = 3 * 3
c = a / b
d = 3 * 4
e = 2 * 4
f = d / e
g = c / f
|
a ) - 2 , b ) 10 , c ) - 1 , d ) 3 , e ) 0 | c | divide(subtract(negate(const_1), 3), subtract(subtract(power(negate(const_1), const_2), multiply(8, negate(const_1))), 20)) | what is the smallest integer that satisfies the inequality ( ( x - 3 ) / ( x ^ 2 - 8 x - 20 ) ) > 0 ? | let us factorize the denominator and rewrite the expression as ( x Γ’ Λ β 3 ) / ( ( x Γ’ Λ β 10 ) ( x + 2 ) ) > 0 equate each of the terms of the expression to zero to identify the values of x that are relevant to test whether the inequality holds good . the values that are relevant to us are x = 3 , x = 10 and x = - 2 . let us arrange these values in ascending order : - 2 , 3 and 10 . the quickest way to solve inequalities questions after arriving at these values is verifying whether the inequality holds good at the following intervals . interval 1 : x < - 2 . pick a value in that range and check whether the inequality holds good . let us take x = - 10 . when x = - 10 , ( x Γ’ Λ β 3 ) / ( ( x Γ’ Λ β 10 ) ( x + 2 ) ) < 0 ; the inequality does not hold good in this interval . interval 2 : - 2 < x < 3 . let us take x = 0 . when x = 0 , ( x Γ’ Λ β 3 ) / ( ( x Γ’ Λ β 10 ) ( x + 2 ) ) = ( 0 Γ’ Λ β 3 ) / ( ( 0 Γ’ Λ β 10 ) ( 0 + 2 ) ) > 0 ; the inequality holds good in this interval . the least integer value that x can take if x > - 2 is x = - 1 . answer : c . | a = negate - (
b = a / 3
|
a ) 35 , b ) 30 , c ) 25 , d ) 20 , e ) 15 | e | subtract(add(add(20, 40), 60), add(add(multiply(5, const_3), 10), 80)) | the average ( arithmetic mean ) of 20 , 40 , and 60 is 5 more than the average of 10 , 80 , and what number ? | a 1 = 120 / 3 = 40 a 2 = a 1 - 5 = 35 sum of second list = 35 * 3 = 105 therefore the number = 105 - 90 = 15 e | a = 20 + 40
b = a + 60
c = 5 * 3
d = c + 10
e = d + 80
f = b - e
|
a ) 25 , b ) 42 , c ) 45 , d ) 49 , e ) 54 | a | divide(power(105, 3), multiply(multiply(21, 49), 45)) | if a = 105 and a ^ 3 = 21 * 49 * 45 * b , what is the value of b ? | "first step will be to break down all the numbers into their prime factors . 105 = 3 * 5 * 7 21 = 7 * 3 49 = 7 * 7 45 = 3 * 3 * 5 so , ( 105 ) ^ 3 = 3 * 7 * 7 * 7 * 3 * 3 * 5 * b therefore ( 3 * 5 * 7 ) ^ 3 = 3 ^ 3 * 5 * 7 ^ 3 * b therefore , b = 3 ^ 3 * 5 ^ 3 * 7 ^ 3 / 3 ^ 3 * 5 * 7 ^ 3 b = 5 ^ 2 = 25 correct answer a ." | a = 105 ** 3
b = 21 * 49
c = b * 45
d = a / c
|
a ) 54.0 , b ) 54.9 , c ) 92.5 , d ) 57.0 , e ) 63.0 | c | add(add(multiply(divide(subtract(const_100, 20), const_100), 30), 30), multiply(divide(add(const_100, 25), const_100), 30)) | james took a 3 - hour bike ride . in the second hour he traveled 30 miles , which was 20 percent farther than he traveled the first hour . if he traveled 25 percent farther in the third hour than he did in the second hour , how many miles did jose travel during the entire ride ? | let the distance travelled in the first hour be x . thus , 1.2 x = 30 , x = 25 . now , the distance travelled in the 3 rd hour = 30 + 1 / 4 Γ’ Λ β 30 = 37.5 the only option ending with a 0.5 in the decimal place is c . answer : c | a = 100 - 20
b = a / 100
c = b * 30
d = c + 30
e = 100 + 25
f = e / 100
g = f * 30
h = d + g
|
a ) 1 , b ) 5 , c ) 3 , d ) 2 , e ) 4 | e | subtract(3, divide(subtract(44, 20), add(10, 20))) | calculate the value of n from the below equation : n ^ 3 Γ’ Λ β 10 n + 20 = 44 | "use elimination method to find the correct option . you find that of all the options when 4 is the correct value for n answer : e" | a = 44 - 20
b = 10 + 20
c = a / b
d = 3 - c
|
a ) a ) 5.61 , b ) b ) 3.42 , c ) c ) 10 , d ) d ) 15 , e ) e ) 24 | b | max(multiply(subtract(add(45, 9), const_1), subtract(divide(9, 35), divide(9, 45))), const_4) | due to construction , the speed limit along an 9 - mile section of highway is reduced from 45 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 9 miles stretch = 9 * 60 / 45 = 9 * 4 / 3 = 12 new time in minutes to cross 9 miles stretch = 9 * 60 / 35 = 9 * 12 / 7 = 15.42 time difference = 3.42 ans : b" | a = 45 + 9
b = a - 1
c = 9 / 35
d = 9 / 45
e = c - d
f = b * e
g = max(f)
|
a ) 25 / 9 , b ) 9 / 5 , c ) 5 / 3 , d ) 3 / 5 , e ) 9 / 25 | c | divide(5, 3) | a number x is multiplied by 5 , and this product is then divided by 3 . if the positive square root of the result of these two operations equals x , what is the value of x if x β 0 ? | "sqrt ( 5 x / 3 ) to be perfect square x has to 5 / 3 ans : c" | a = 5 / 3
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a ) $ 4720 , b ) $ 4840 , c ) $ 4920 , d ) $ 5080 , e ) $ 5160 | b | multiply(4000, power(add(const_1, divide(10, const_100)), const_2)) | what amount does an investor receive if the investor invests $ 4000 at 10 % p . a . compound interest for two years , compounding done annually ? | "a = ( 1 + r / 100 ) ^ n * p ( 1.1 ) ^ 2 * 5000 = 1.21 * 5000 = 4840 the answer is b ." | a = 10 / 100
b = 1 + a
c = b ** 2
d = 4000 * c
|
a ) $ 14,755 , b ) $ 15,430 , c ) $ 16,000 , d ) $ 16,225 , e ) $ 17,155 | b | multiply(divide(const_3, const_4), const_1000) | a store owner estimates that the average price of type a products will increase by 40 % next year and that the price of type b products will increase by 10 % next year . this year , the total amount paid for type a products was $ 4500 and the total price paid for type b products was $ 8300 . according to the store owner ' s estimate , and assuming the number of products purchased next year remains the same as that of this year , how much will be spent for both products next year ? | "cost of type a products next year = 1.40 * 4500 = 6300 cost of type b products next year = 1.1 * 8300 = 9130 total 6300 + 9130 = 15430 option b" | a = 3 / 4
b = a * 1000
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a ) 3 , b ) 9 , c ) 12 , d ) 18 , e ) 24 | e | multiply(divide(9, 3), add(add(4, 3), 1)) | the school soccer team looks at their record and finds that they win , lose , and tie games in a ratio of 4 : 3 : 1 . how many games did they play if they lost 9 games ? | the ratio is 4 wins : 3 losses : 1 tie . think of ratio as ` ` parts . ' ' divide 9 ( total losses ) by 3 ( ratio losses ) to find 1 ` ` part ' ' of the ratio . 9 / 3 = 3 this means the team tied 3 games . multiply 3 ( 1 ` ` part ' ' of ratio ) by 4 ( ratio wins ) to find total wins . 3 * 4 = 12 . this mean the team won 12 games . add up wins , losses , and ties . 12 + 9 + 3 = 24 the answer is e | a = 9 / 3
b = 4 + 3
c = b + 1
d = a * c
|
a ) 20 , b ) 24 , c ) 36 , d ) 40 , e ) 54 | e | add(multiply(divide(9, subtract(divide(9, 10), divide(3, 4))), divide(3, 4)), 9) | if 9 gallons of gasoline are added to a tank that is already filled to 3 / 4 of its capacity , the tank is then filled to 9 / 10 of its capacity . how many gallons does the tank hold ? | "let the capacity of the tank = c ( 3 / 4 ) c + 9 = ( 9 / 10 ) c = > ( 9 / 10 ) c - ( 3 / 4 ) c = 9 = > ( 3 / 20 ) c = 9 = > c = ( 9 * 20 ) / 3 = 60 number of gallons of gasoline that the tank currently holds = 3 / 4 * c + 9 = 45 + 9 = 54 answer e" | a = 9 / 10
b = 3 / 4
c = a - b
d = 9 / c
e = 3 / 4
f = d * e
g = f + 9
|
a ) 18 , b ) 36 , c ) 72 , d ) 0 , e ) 1 | a | multiply(divide(99, add(const_10, const_1)), const_2) | what is the remainder when 121212 . . . . . . . ( 300 ) digits is divided by 99 . | 12 / 99 = 12 1212 / 99 = 24 121212 / 99 = 36 . . . 121212 . . . ( 300 ) / 99 = ( 12 * 300 ) % 99 = 36 if they say 300 digits then ( 12 * 150 ) % 99 = 18 answer : a | a = 10 + 1
b = 99 / a
c = b * 2
|
a ) 22 , b ) 99 , c ) 27 , d ) 36 , e ) 20 | d | multiply(4, 9) | walking 9 / 8 of his usual rate , a boy reaches his school 4 min early . find his usual time to reach the school ? | "speed ratio = 1 : 9 / 8 = 8 : 9 time ratio = 9 : 8 1 - - - - - - - - 9 4 - - - - - - - - - ? 36 m . answer : d" | a = 4 * 9
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a ) 47 , b ) 47.03 , c ) 48.03 , d ) 48 , e ) 49 | c | subtract(multiply(55, 30), multiply(48, 30)) | the average of 30 numbers is 48 . if three numbers namely 45 , 55 and 43 are discarded , the average of remaining numbers is ? | "explanation : total of 30 numbers = 30 * 48 = 1440 average o 27 numbers = 1440 - ( 45 + 55 + 43 ) / 27 = 48.03 answer : option c" | a = 55 * 30
b = 48 * 30
c = a - b
|
a ) 5 kmph , b ) 7 kmph , c ) 9 kmph , d ) 8 kmph , e ) 10 kmph | e | divide(subtract(divide(95, 2), divide(45, 2)), const_2) | a man rows his boat 95 km downstream and 45 km upstream , taking 2 1 / 2 hours each time . find the speed of the stream ? | "speed downstream = d / t = 95 / ( 2 1 / 2 ) = 38 kmph speed upstream = d / t = 45 / ( 2 1 / 2 ) = 18 kmph the speed of the stream = ( 38 - 18 ) / 2 = 10 kmph answer : e" | a = 95 / 2
b = 45 / 2
c = a - b
d = c / 2
|
a ) $ 35,000 , b ) $ 45,000 , c ) $ 55,000 , d ) $ 60,000 , e ) $ 75,000 | d | divide(add(multiply(multiply(multiply(const_3, const_3), const_10), multiply(const_100, const_10)), multiply(multiply(const_3, const_10), multiply(const_100, const_10))), add(75, 15)) | a company has 15 managers and 75 associates . the 15 managers have an average salary of $ 210,000 . the 75 associates have an average salary of $ 30,000 . what is the average salary for the company ? | "another method is to get ratios say 30000 = a and we know the # of people are in 1 : 5 ratio average = ( 7 a * 1 + a * 5 ) / 6 = 12 a / 6 = 2 a = 60000 answer is d . $ 60,000" | a = 3 * 3
b = a * 10
c = 100 * 10
d = b * c
e = 3 * 10
f = 100 * 10
g = e * f
h = d + g
i = 75 + 15
j = h / i
|
a ) 19 , b ) 20 , c ) 21 , d ) 22 , e ) 23 | a | divide(subtract(multiply(18, 18), 1), 17) | a batsman scored 1 runs in his 18 th innings and that makes his average 18 . find his average upto the 17 th innings ? | avg = sum of the value / no . of the value 18 = ( sum of the first 17 innings score + 1 ) / 18 ( 18 * 18 - 1 ) = sum of the first 17 innings score sum of the first 17 innings score = 323 ave upto 17 innings = 323 / 17 = 19 answer : a | a = 18 * 18
b = a - 1
c = b / 17
|
a ) 4 , b ) 4.5 , c ) 6 , d ) 6.25 , e ) 7.2 | e | divide(multiply(9, const_60), add(15, const_60)) | if jack walked 9 miles in 1 hour and 15 minutes , what was his rate of walking in miles per hour ? | "distance walked in 1 hour and 15 mins = 9 miles speed per hour = distance / time = 9 / ( 5 / 4 ) = 7.2 miles per hour answer e" | a = 9 * const_60
b = 15 + const_60
c = a / b
|
a ) 90 , b ) 129 , c ) 120 , d ) 160 , e ) 200 | b | divide(add(110, 148), 2) | a student chose a number , multiplied it by 2 , then subtracted 148 from the result and got 110 . what was the number he chose ? | "solution : let x be the number he chose , then 2 * x * 148 = 110 2 x = 258 x = 129 correct answer b" | a = 110 + 148
b = a / 2
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a ) 27 , b ) 3 , c ) 6 , d ) 8 , e ) 12 | b | divide(divide(multiply(multiply(8, 12), 3), 12), 8) | a crate measures 3 feet by 8 feet by 12 feet on the inside . a stone pillar in the shape of a right circular cylinder must fit into the crate for shipping so that it rests upright when the crate sits on at least one of its six sides . what is the radius , in feet , of the pillar with the largest volume that could still fit in the crate ? | "we can find the radius of all the three cases of cylinders . the only crux to find the answer faster is that : voulme is pi * r ^ 2 * h . the volume is a function of r ^ 2 . so r has to be the highest to find the largest volume . so r = 3 for the surface 8 * 12 face . volume = 27 pi answer b" | a = 8 * 12
b = a * 3
c = b / 12
d = c / 8
|
a ) 3 , b ) 4 , c ) 5 , d ) 6 , e ) 7 | e | subtract(add(3.4, 10.7), 1.6) | if [ x ] is the greatest integer less than or equal to x , what is the value of [ 1.6 ] + [ - 3.4 ] + [ 10.7 ] ? | "you are asked what the closest lesser integer value to [ x ] is . [ 1.6 ] = 1.0 [ - 3.4 ] = - 4.0 [ 10.7 ] = 10.0 therefore , answer is : 1.0 - 4.0 + 10.0 = 7.0 option e ." | a = 3 + 4
b = a - 1
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a ) 12 , b ) 14 , c ) 16 , d ) 18 , e ) 20 | e | multiply(divide(75, subtract(multiply(137, 2), 61)), 2) | a firm is comprised of partners and associates in a ratio of 2 : 61 . if 75 more associates were hired , the ratio of partners to associates would be 4 : 137 . how many partners are currently in the firm ? | "the ratio 2 : 61 = 4 : 122 so the ratio changed from 4 : 122 to 4 : 137 . 137 - 122 = 15 which is 1 / 5 of the increase in 75 associates . the ratio changed from 20 : 610 to 20 : 685 . thus the number of partners is 20 . the answer is e ." | a = 137 * 2
b = a - 61
c = 75 / b
d = c * 2
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a ) 20 , b ) 25 , c ) 27 , d ) 22 , e ) 17 | b | subtract(37, multiply(multiply(18, 3), 2)) | evaluate : 37 - 18 Γ· 3 Γ 2 = | "according to order of operations , 18 Γ· 3 Γ 2 ( division and multiplication ) is done first from left to right 18 Γ· 3 Γ 2 = 6 Γ 2 = 12 hence 37 - 18 Γ· 3 Γ 2 = 37 - 12 = 25 correct answer b ) 25" | a = 18 * 3
b = a * 2
c = 37 - b
|
a ) 271 , b ) 266 , c ) 400 , d ) 277 , e ) 232 | c | subtract(multiply(25, multiply(90, const_0_2778)), 225) | a train 225 m long running at 90 kmph crosses a platform in 25 sec . what is the length of the platform ? | "length of the platform = 90 * 5 / 18 * 25 = 625 β 225 = 400 answer : c" | a = 90 * const_0_2778
b = 25 * a
c = b - 225
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a ) 22 , b ) 389 , c ) 38 , d ) 45 , e ) 01 | d | multiply(divide(add(add(add(const_10, 20), multiply(divide(41, const_100), const_1000)), const_1000), multiply(add(200, 120), const_10)), const_100) | in an election only two candidates contested 20 % of the voters did not vote and 120 votes were declared as invalid . the winner got 200 votes more than his opponent thus he secured 41 % votes of the total voters on the voter list . percentage votes of the defeated candidate out of the total votes casted is : | let there be x voters and k votes goes to loser then 0.8 x - 120 = k + ( k + 200 ) k + 200 = 0.41 x \ inline \ rightarrow k = 1440 and ( k + 200 ) = 1640 therefore \ inline \ frac { 1440 } { 3200 } \ times 100 = 45 % answer : d ) 45 % | a = 10 + 20
b = 41 / 100
c = b * 1000
d = a + c
e = d + 1000
f = 200 + 120
g = f * 10
h = e / g
i = h * 100
|
a ) 800 , b ) 700 , c ) 360 , d ) 370 , e ) 380 | b | subtract(multiply(speed(600, 18), 39), 600) | a 600 m long train crosses a platform in 39 sec while it crosses a signal pole in 18 sec . what is the length of the platform ? | "speed = 600 / 18 = 100 / 3 m / sec . let the length of the platform be x meters . then , ( x + 300 ) / 39 = 100 / 3 = > x = 1300 m . l = 1300 - 600 = 700 m answer : option b" | a = speed * (
b = a - 39
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a ) 11 , b ) 12 , c ) 13.2 , d ) none , e ) can not be determined | a | divide(187, divide(187, 17)) | 17 times a number gives 187 . the number is | "explanation : let the number be ' n ' 17 Γ n = 187 β n = 11 correct option : a" | a = 187 / 17
b = 187 / a
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a ) 8.5 , b ) 6.6 , c ) 7.6 , d ) 6.4 , e ) 5.7 | b | divide(0.0008154, 0.00205) | 0.0008154 / 0.00205 x 16.5 = ? | "explanation : ? = 0.0008154 / 0.00205 x 16.5 = 6.6 answer : option b" | a = 0 / 8154
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a ) 3 , b ) 3.5 , c ) 4 , d ) 4.5 , e ) 6 | c | add(divide(subtract(52, divide(subtract(divide(add(14, 2), 2), 2), 2)), add(subtract(divide(add(14, 2), 2), 2), divide(add(14, 2), 2))), divide(const_1, 2)) | tammy climbed a mountain in two days . she spent a total of 14 hours climbing the mountain . on the second day , she walked at an average speed that was half a kilometer per hour faster , but 2 hours less than what she walked on the first day . if the total distance she climbed during the two days is 52 kilometers , how many r kilometers per hour did tammy walk on the second day ? | "ans : c total time = 14 hrs let time traveled during 1 st day = x let time traveled during 2 nd day = x - 2 total time = 14 x + x - 2 = 14 x = 8 speed * time = distance s * 8 + ( s + 0.5 ) ( 8 - 2 ) = 52 solving s = 4.5 now speed for 2 nd day is 0.5 less than the 1 st day which is 4.5 thus speed for 2 nd day = 4 its simple algebra for s * 8 + ( s + 0.5 ) ( 8 - 2 ) = 52 but for some reason im getting 3.5 and not 4.5 . 8 s + 6 s + 3 = 52 14 s = 49 s = 3.5" | a = 14 + 2
b = a / 2
c = b - 2
d = c / 2
e = 52 - d
f = 14 + 2
g = f / 2
h = g - 2
i = 14 + 2
j = i / 2
k = h + j
l = e / k
m = 1 / 2
n = l + m
|
a ) 32 kmph , b ) 58 kmph , c ) 62 kmph , d ) 65 kmph , e ) 75 kmph | a | subtract(multiply(divide(280, 9), const_3_6), 80) | a man sitting in a train which is traveling at 80 kmph observes that a goods train , traveling in opposite direction , takes 9 seconds to pass him . if the goods train is 280 m long , find its speed . ? | "relative speed = 280 / 9 m / sec = ( ( 280 / 9 ) * ( 18 / 5 ) ) kmph = 112 kmph . speed of goods train = ( 112 - 80 ) kmph = 32 kmph . answer : a" | a = 280 / 9
b = a * const_3_6
c = b - 80
|
a ) 80 , b ) 100 , c ) 110 , d ) 120 , e ) 140 | d | multiply(multiply(3, const_10), const_4) | howmany 3 digit numbers are there tens digit place is more than hundreds digit place and units place less than hundreds digit place ? | we have numbers { 01 , 23 , 45 , 67 , 89 } now we would decide the cases on behalf of value of hundred place digit . and accordingly choose unit ' s digit and ten ' s digit . value of hundred place digit must be choosen in such a manner that there should exist a greater digit for ten ' s place and a smaller digit for unit place . thus we can have values of hundred ' s digit from 12 , 34 , 56 , 78 only now starting from hundred ' s digit = 8 : hundred ' s digit possible values for ten ' s digit possible values for unit digit total ways 8 9 01 , 23 , 45 , 67 1 * 1 * 8 = 8 7 89 0 , 12 , 34 , 56 1 * 2 * 7 = 14 6 7 , 89 01 , 23 , 45 1 * 3 * 6 = 18 5 67 , 89 0 , 12 , 34 1 * 4 * 5 = 20 4 5 , 67 , 89 01 , 23 1 * 5 * 4 = 20 3 45 , 67 , 89 0 . 1.2 1 * 6 * 3 = 18 2 3 , 45 , 67 , 89 01 1 * 7 * 2 = 14 1 23 , 45 , 67 , 89 0 1 * 8 * 1 = 8 so , total no of cases = 8 + 14 + 18 + 20 + 20 + 18 + 14 + 8 = 120 hence total no of possible 3 digit numbers = 120 answer : d | a = 3 * 10
b = a * 4
|
a ) none , b ) one , c ) two , d ) three , e ) four | e | subtract(15, multiply(3, const_4)) | a = 7 ^ 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 4 different values and answer is e . . . ." | a = 3 * 4
b = 15 - a
|
a ) 70 , b ) 76 , c ) 78 , d ) 80 , e ) 88 | b | subtract(multiply(add(32, 4), add(10, const_1)), multiply(10, 32)) | average of 10 matches is 32 , how many runs one should should score to increase his average by 4 runs . | "explanation : average after 11 innings should be 36 so , required score = ( 11 * 36 ) - ( 10 * 32 ) = 396 - 320 = 76 answer : option b" | a = 32 + 4
b = 10 + 1
c = a * b
d = 10 * 32
e = c - d
|
a ) 120 , b ) 140 , c ) 160 , d ) 180 , e ) 200 | c | divide(subtract(300, 244), subtract(const_1, divide(65, const_100))) | a particular library has 300 books in a special collection , all of which were in the library at the beginning of the month . these book are occasionally loaned out through an inter - library program . if , by the end of the month , 65 percent of books that were loaned out are returned and there are 244 books in the special collection at that time , how many books of the special collection were loaned out during that month ? | "the total number of books is 300 . let x be the number of books which were loaned out . 65 % of books that were loaned out are returned . 35 % of books that were loaned out are not returned . now , there are 244 books , thus the number of un - returned books is 300 - 244 = 56 books . 0.35 x = 56 x = 160 the answer is c ." | a = 300 - 244
b = 65 / 100
c = 1 - b
d = a / c
|
a ) 57 , b ) 2 ^ 4 * 3 , c ) 24 , d ) 38 , e ) 47 | a | subtract(58, const_1) | in a lake , there is a patch of lily pads . every day , the patch doubles in size . it takes 58 days for the patch to cover the entire lake , how many days would it take the patch to cover half of the lake ? | "so 57 days answer a = 57" | a = 58 - 1
|
a ) 1 / 15 , b ) 1 / 12 , c ) 1 / 9 , d ) 1 / 6 , e ) 1 / 3 | e | divide(divide(factorial(6), multiply(factorial(2), factorial(1))), factorial(6)) | joshua and jose work at an auto repair center with 1 other workers . for a survey on health care insurance , 2 of the 6 workers will be randomly chosen to be interviewed . what is the probability that joshua and jose will both be chosen ? | "two methods 1 ) probability of chosing josh first = 1 / 3 probability of chosing jose second = 1 / 2 total = 1 / 6 probability of chosing jose first = 1 / 3 probability of chosing josh second = 1 / 2 total = 1 / 6 final = 1 / 6 + 1 / 6 = 1 / 3 e" | a = math.factorial(6)
b = math.factorial(2)
c = math.factorial(1)
d = b * c
e = a / d
f = math.factorial(6)
g = e / f
|
a ) 47 , b ) 25 , c ) 37 , d ) 33 , e ) 29 | b | add(subtract(multiply(sqrt(169), const_2), multiply(const_2, 1)), 1) | what is the value of n if the sum of the consecutive odd intergers e from 1 to n equals 169 ? | "before you tackle this question you must first understand that the question is comprised of two key parts , 1 st is finding out how manytermsis in that sequence and 2 nd whatactual number valuethat term is . in an arithmetic progression , in this case consecutive odd integers 1 , 3 , 5 , . . . . , there are two set of rules . rule # 1 ( arithmetic sequence ) : xn = a + d ( n - 1 ) identifies what the actual # in the sequence would be . each number in the sequence has a term such as 1 ( is the first term ) , 3 ( is the second term ) and so on . so if i were to ask you to find out what the 10 th term is of that sequence you would use that formula to find that value . a = 1 ( first term ) d = 2 ( the common difference ) remember in the sequence 1 , 3 , 5 , 7 the common difference is always 2 * on a side note we use n - 1 because we do n ' t have d in the first term , therefore if we were solving for the first term we would get 0 as n - 1 and 0 times d would give us 0 , leaving only the first term . this works regardless what your first term is in any sequence . but remember the question askswhat is thevalueof n if the sum of the consecutive odd integers from 1 to n equals 169 ? which means we first need a consecutive sequence that sums up to 169 and than find what the value of the n is , in this case it would be the last number in that sequence . in order to find that we first need to knowhow many terms ( how many of the n there is ) in order to be able to plug n in this formula given we know what the sum is . for that to happen we need to use rule # 2 . rule # 2 ( summing an arithmetic series ) : 169 = n / 2 ( 2 a + ( n - 1 ) d ) . given the question gives us what the sum is ( 169 in this case ) we would simply use this formula to solve for n . once we solve for n ( 13 in this case ) we can simply plug n into the first formula ( rule 1 ) and find the value . it feels very confusing and difficult at first , but once you identify the steps all you need to do is plug and play . we have the sum ( 169 ) of a sequence , the number of terms in that sequence is ( unknown ) . rule # 2 tells us how many numbers there are in that sequence and rule # 1 gives us what that last term is ." | a = math.sqrt(169)
b = a * 2
c = 2 * 1
d = b - c
e = d + 1
|
a ) 7858 , b ) 8301 , c ) 14667 , d ) 63840 , e ) 146,667 | e | divide(multiply(subtract(3.25, 2.85), divide(55000, multiply(const_10, const_100))), subtract(2.8, 2.65)) | joseph completes work worth $ 3.25 in hour and his cost to company per hour is $ 2.85 . ray completes worth worth of $ 2.80 in an hour and his cost to company is $ 2.65 per hour . if joseph works for 55000 hours , how many must ray work so that company makes at least as much in total gross profit as in case of joseph ? | joseph : profit / hour = 3.25 - 2.85 = 0.4 : no of hours = 55,000 : gross profit = 55,000 * 0.4 = 22,000 ray : profit / hour = 0.15 : gross profit = 22,000 : no of hours = 22,000 / 0.15 = 220,000 / 1.5 ( only closes is 146,667 ) answer e | a = 3 - 25
b = 10 * 100
c = 55000 / b
d = a * c
e = 2 - 8
f = d / e
|
a ) 95 , b ) 50 , c ) 12 , d ) 13 , e ) 67 | a | divide(divide(subtract(250, multiply(multiply(5, const_0_2778), 5)), 5), const_0_2778) | a train 250 m long passes a man , running at 5 km / hr in the same direction in which the train is going , in 10 seconds . the speed of the train is : | "speed of the train relative to man = ( 250 / 10 ) m / sec = ( 25 ) m / sec . [ ( 25 ) * ( 18 / 5 ) ] km / hr = 90 km / hr . let the speed of the train be x km / hr . then , relative speed = ( x - 5 ) km / hr . x - 5 = 90 = = > x = 95 km / hr . answer : option a" | a = 5 * const_0_2778
b = a * 5
c = 250 - b
d = c / 5
e = d / const_0_2778
|
a ) 20 , b ) 30 , c ) 10 , d ) 80 , e ) 100 | c | subtract(subtract(200, 50), 10) | of the 200 employees at company x , 50 are full - time , and 150 have worked at company x for at least a year . there are 10 employees at company x who aren β t full - time and haven β t worked at company x for at least a year . how many full - time employees of company x have worked at the company for at least a year ? | "200 employees 50 are full - time 150 have worked at company x for at least a year 10 employees at company x who aren β t full - time and haven β t worked at company x for at least a year . how many full - time employees of company x have worked at the company for at least a year ? 200 - 50 = 150 employees not full time 150 - 10 = 140 employees not full time who worked over a year 150 employees have worked at company x for at least a year - 140 employees not full time who worked over a year = 10 full - time employees of company x have worked at the company for at least a year ans c" | a = 200 - 50
b = a - 10
|
a ) 26630 , b ) 26640 , c ) 36644 , d ) 31944 , e ) 26844 | d | multiply(24000, add(const_1, divide(multiply(3, 10), const_100))) | population is 24000 . population increases by 10 % every year , then the population after 3 years is ? | "population after 1 st year = 24000 * 10 / 100 = 2400 = = = > 24000 + 2400 = 26400 population after 2 nd year = 26400 * 10 / 100 = 2640 = = = > 26400 + 2640 = 29040 population after 3 rd year = 29040 * 10 / 100 = 2904 = = = > 29040 + 2904 = 31944 answer : d" | a = 3 * 10
b = a / 100
c = 1 + b
d = 24000 * c
|
a ) 278 , b ) 900 , c ) 278 , d ) 450 , e ) 772 | b | subtract(multiply(const_10, 150), add(multiply(3, 100), multiply(2, 150))) | a man purchased 3 blankets @ rs . 100 each , 2 blankets @ rs . 150 each and two blankets at a certain rate which is now slipped off from his memory . but he remembers that the average price of the blankets was rs . 150 . find the unknown rate of two blankets ? | "10 * 150 = 1500 3 * 100 + 2 * 150 = 600 1500 β 600 = 900 answer : b" | a = 10 * 150
b = 3 * 100
c = 2 * 150
d = b + c
e = a - d
|
a ) 170 , b ) 172 , c ) 174 , d ) 176 , e ) 178 | c | add(4, lcm(34, 5)) | find the least number which when divided by 34 and 5 leaves a remainder of 4 in each case . | "the least number which when divided by different divisors leaving the same remainder in each case = lcm ( different divisors ) + remainder left in each case . hence the required least number = lcm ( 31 , 5 ) + 4 = 174 . answer : c" | a = math.lcm(34, 5)
b = 4 + a
|
a ) 1 / pi , b ) sqrt ( 2 / pi ) , c ) sqrt ( 4.5 / pi ) , d ) 2 / sqrt ( pi ) , e ) pi / 2 | c | sqrt(divide(divide(square_area(3), 2), const_pi)) | an artist wishes to paint a circular region on a square poster that is 3 feet on a side . if the area of the circular region is to be 1 / 2 the area of the poster , what must be the radius of the circular region in feet ? | "area of the poster is 3 x 3 = 9 1 / 2 the area = 4.5 pi * r ^ 2 = 4.5 r ^ 2 = 4.5 / pi r = sqrt ( 4.5 / pi ) answer ( c )" | a = square_area / (
b = a / 2
c = math.sqrt(b)
|
a ) 10.9 sec , b ) 14.8 sec , c ) 10.6 sec , d ) 10.8 sec , e ) 9.27 sec | e | divide(add(120, 150), multiply(add(60, 40), const_0_2778)) | two trains 120 m and 150 m long run at the speed of 60 km / hr and 40 km / hr respectively in opposite directions on parallel tracks . the time which they take to cross each other is ? | "relative speed = 60 + 40 = 100 km / hr . = 100 * 5 / 18 = 250 / 9 m / sec . distance covered in crossing each other = 120 + 150 = 270 m . required time = 270 * 9 / 250 = 9.27 = 9.27 sec . answer : e" | a = 120 + 150
b = 60 + 40
c = b * const_0_2778
d = a / c
|
a ) 96 kmph , b ) 94 kmph , c ) 92 kmph , d ) 86 kmph , e ) 72 kmph | e | multiply(divide(160, 8), const_3_6) | a 160 meter long train crosses a man standing on the platform in 8 sec . what is the speed of the train ? | "s = 160 / 8 * 18 / 5 = 72 kmph answer : e" | a = 160 / 8
b = a * const_3_6
|
a ) 2 . , b ) 4 . , c ) 5 . , d ) 6 , e ) 18 . | e | divide(multiply(12, 18), 12) | 18 beavers , working together in a constant pace , can build a dam in 12 hours . how many hours will it take 12 beavers that work at the same pace , to build the same dam ? | "total work = 18 * 12 = 216 beaver hours 12 beaver * x = 216 beaver hours x = 216 / 12 = 18 answer : e" | a = 12 * 18
b = a / 12
|
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