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 ) rs . 17 , b ) rs . 18 , c ) rs . 12 , d ) rs . 11 , e ) rs . 10 | b | divide(add(1150, 920), add(65, 50)) | rahim bought 65 books for rs . 1150 from one shop and 50 books for rs . 920 from another . what is the average price he paid per book ? | "average price per book = ( 1150 + 920 ) / ( 65 + 50 ) = 2070 / 115 = rs . 18 answer : b" | a = 1150 + 920
b = 65 + 50
c = a / b
|
a ) 100 m sqaure , b ) 103 m sqaure , c ) 152 m sqaure , d ) 164 m sqaure , e ) none of these | b | add(multiply(const_2, add(multiply(add(divide(35, const_100), 1), 6), multiply(add(divide(35, const_100), 1), 10))), multiply(6, 10)) | a cistern 10 m long and 6 m wide contains water up to a breadth of 1 m 35 cm . find the total area of the wet surface . | "explanation : area of the wet surface = 2 [ lb + bh + hl ] - lb = 2 [ bh + hl ] + lb = 2 [ ( 6 * 1.35 + 10 * 1.35 ) ] + 10 * 6 = 103 m square option b" | a = 35 / 100
b = a + 1
c = b * 6
d = 35 / 100
e = d + 1
f = e * 10
g = c + f
h = 2 * g
i = 6 * 10
j = h + i
|
a ) 17,400 , b ) 17,500 , c ) 17,600 , d ) 17,700 , e ) 17,800 | d | divide(divide(subtract(multiply(const_1000, const_100), subtract(subtract(const_3600, const_100), const_1000)), const_1000), add(multiply(add(const_1, divide(10, const_100)), subtract(9, const_1)), const_1)) | a city with a population of 173,460 is to be divided into 9 voting districts , and no district is to have a population that is more than 10 percent greater than the population of any other district . what is the minimum possible population that the least populated district could have ? | "the minimum possible population occurs when all the other districts have a population that is 10 % greater than the least populated district . let p be the population of the least populated district . then 173,460 = p + 8 ( 1.1 ) p 9.8 p = 173,460 p = 17,700 the answer is d ." | a = 1000 * 100
b = 3600 - 100
c = b - 1000
d = a - c
e = d / 1000
f = 10 / 100
g = 1 + f
h = 9 - 1
i = g * h
j = i + 1
k = e / j
|
a ) 2 / 3 , b ) 1 , c ) 4 / 3 , d ) 8 / 5 , e ) 3 | c | divide(80, add(divide(multiply(divide(80, 2), 50), const_100), divide(80, 2))) | car q and car y traveled the same 80 - mile route . if car q took 2 hours and car y traveled at an average speed that was 50 percent faster than the average speed of car q , how many hours did it take car y to travel the route ? | the speed of car q is ( distance ) / ( time ) = 80 / 2 = 40 miles per hour . the speed of car y = 3 / 2 * 40 = 60 miles per hour - - > ( time ) = ( distance ) / ( speed ) = 80 / 60 = 4 / 3 hours . answer : c . or : to cover the same distance at 3 / 2 as fast rate 2 / 3 as much time is needed - - > ( time ) * 2 / 3 = 2 ... | a = 80 / 2
b = a * 50
c = b / 100
d = 80 / 2
e = c + d
f = 80 / e
|
a ) 300 , b ) 350 , c ) 400 , d ) 425 , e ) 275 | a | divide(subtract(1200, multiply(10, 30)), 3) | 3 pig ’ s and 10 hens are brought for rs . 1200 . if the average price of a hen be rs . 30 . what is the average price of a pig . | explanation : average price of a hen = rs . 30 total price of 10 hens = 10 * 30 = rs . 300 but total price of 3 pigs and 10 hens = rs . 1200 total price of 3 pigs is = 1200 - 300 = 900 average price of a pig = 900 / 3 = rs . 300 answer : a | a = 10 * 30
b = 1200 - a
c = b / 3
|
a ) 5 , b ) 8 , c ) 9 , d ) 10 , e ) 20 | d | divide(240, multiply(const_10, const_2)) | how many factors of 240 are also multiples of 3 ? | "take factors of 240 240 = 2 ^ 4 * 3 ^ 1 * 5 ^ 1 total factors of 240 = ( 4 + 1 ) * ( 1 + 1 ) * ( 1 + 1 ) = 5 * 2 * 2 = 20 out of total factors of 20 , half will have 0 as power of 3 and half will have 1 as power of 3 . it is because we have considered only 0 and 1 as power of 3 to compute total factors of 240 . theref... | a = 10 * 2
b = 240 / a
|
a ) 4038 , b ) 8076 , c ) 9691.2 , d ) 4845.6 , e ) 3900 | e | multiply(divide(divide(46800, divide(divide(multiply(subtract(const_100, 60), 50), const_100), const_100)), multiply(const_3, const_4)), divide(divide(multiply(subtract(const_100, 60), 50), const_100), const_100)) | mr yadav spends 60 % of his monthly salary on consumable items and 50 % of the remaining on clothes and transport . he saves the remaining amount . if his savings at the end of the year were 46800 , how much amount per month would he have spent on clothes and transport ? | "∵ amount , he have spent in 1 month on clothes transport = amount spent on saving per month ∵ amount , spent on clothes and transport = 46800 ⁄ 12 = 3900 answer e" | a = 100 - 60
b = a * 50
c = b / 100
d = c / 100
e = 46800 / d
f = 3 * 4
g = e / f
h = 100 - 60
i = h * 50
j = i / 100
k = j / 100
l = g * k
|
a ) 81 , b ) 16 , c ) 33 , d ) 54 , e ) 12 | c | divide(605, multiply(add(60, 6), const_0_2778)) | a train 605 m long is running with a speed of 60 km / hr . in what time will it pass a man who is running at 6 km / hr in the direction opposite to that in which the train is going ? | "speed of train relative to man = 60 + 6 = 66 km / hr . = 66 * 5 / 18 = 55 / 3 m / sec . time taken to pass the men = 605 * 3 / 55 = 33 sec . answer : c" | a = 60 + 6
b = a * const_0_2778
c = 605 / b
|
a ) 3500 , b ) 2800 , c ) 3251 , d ) 4251 , e ) 3508 | a | multiply(6300, subtract(const_1, divide(multiply(3000, multiply(2, multiply(2, const_3))), add(multiply(4500, subtract(multiply(2, multiply(2, const_3)), 2)), multiply(3000, multiply(2, multiply(2, const_3))))))) | tom opened a shop investing rs . 3000 . jose joined him 2 months later , investing rs . 4500 . they earned a profit of rs . 6300 after completion of one year . what will be jose ' s share of profit ? | "sol = ~ s - so anju ’ s share = [ 5 / 9 ] x 6300 = 3500 a" | a = 2 * 3
b = 2 * a
c = 3000 * b
d = 2 * 3
e = 2 * d
f = e - 2
g = 4500 * f
h = 2 * 3
i = 2 * h
j = 3000 * i
k = g + j
l = c / k
m = 1 - l
n = 6300 * m
|
a ) 5 : 7 , b ) 5 : 2 , c ) 5 : 9 , d ) 5 : 3 , e ) 5 : 4 | d | divide(2, 5) | the simple form of the ratio 2 / 3 : 2 / 5 is | "2 / 3 : 2 / 5 = 10 : 6 = 5 : 3 answer : d" | a = 2 / 5
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a ) 299 , b ) 566 , c ) 678 , d ) 700 , e ) 8277 | d | multiply(add(5, 6), const_100) | rs . 1500 is divided into two parts such that if one part is invested at 6 % and the other at 5 % the whole annual interest from both the sum is rs . 83 . how much was lent at 5 % ? | "( x * 5 * 1 ) / 100 + [ ( 1500 - x ) * 6 * 1 ] / 100 = 83 5 x / 100 + 90 – 6 x / 100 = 83 x / 100 = 7 = > x = 700 answer : d" | a = 5 + 6
b = a * 100
|
a ) 25 % , b ) 26 % , c ) 27 % , d ) 28 % , e ) 29 % | d | multiply(subtract(const_1, multiply(divide(subtract(const_100, 20), const_100), divide(subtract(const_100, 10), const_100))), const_100) | a towel , when bleached , was found to have lost 20 % of its length and 10 % of its breadth . the percentage of decrease in area is ? | "explanation : let original length = x and original width = y decrease in area will be = xy − ( 80 x / 100 × 90 y / 100 ) = ( xy − 18 / 25 xy ) = 7 / 25 xy decrease = ( 7 xy / 25 xy × 100 ) % = 28 option d" | a = 100 - 20
b = a / 100
c = 100 - 10
d = c / 100
e = b * d
f = 1 - e
g = f * 100
|
a ) 3 : 2 , b ) 1 : 4 , c ) 3 : 1 , d ) 5 : 2 , e ) 4 : 3 | b | divide(subtract(6, 5), subtract(10, 6)) | gold is 10 times as heavy as water and copper is 5 times as heavy as water . in what ratio should these be mixed to get an alloy 6 times as heavy as water ? | g = 10 w c = 5 w let 1 gm of gold mixed with x gm of copper to get 1 + x gm of the alloy 1 gm gold + x gm copper = x + 1 gm of alloy 10 w + 5 wx = x + 1 * 6 w 10 + 5 x = 6 ( x + 1 ) x = 4 ratio of gold with copper = 1 : 4 = 1 : 4 answer is b | a = 6 - 5
b = 10 - 6
c = a / b
|
a ) 4.13 days , b ) 3.13 days , c ) 513 days , d ) 9.13 days , e ) 2 days | b | divide(multiply(multiply(6, 5), 50), multiply(60, 8)) | calculate how many days it will take for 8 boys to paint a 50 m long wall if 6 boys can paint a 60 m long wall in 5 days , | the length of wall painted by one boy in one day = 60 / 6 * 1 / 5 = 2 m no . of days required to paint 50 m cloth by 8 boys = 50 / 8 * 1 / 2 = 3.13 days . b | a = 6 * 5
b = a * 50
c = 60 * 8
d = b / c
|
a ) 10 , b ) 13 , c ) 12 , d ) 18 , e ) 19 | c | subtract(const_100, subtract(add(const_100, 10), divide(multiply(add(const_100, 10), 20), const_100))) | a fair price shopkeeper takes 10 % profit on his goods . he lost 20 % goods during theft . his loss percent is : | "c 12 % suppose he has 100 items . let c . p . of each item be $ 1 . total cost = $ 100 . number of items left after theft = 80 . s . p . of each item = $ 1.10 total sale = 1.10 * 80 = $ 88 hence , loss % = 12 / 100 * 100 = 12 %" | a = 100 + 10
b = 100 + 10
c = b * 20
d = c / 100
e = a - d
f = 100 - e
|
a ) $ 22.70 , b ) $ 23.20 , c ) $ 24.10 , d ) $ 25.50 , e ) $ 26.90 | c | subtract(subtract(40, 90), multiply(6, const_2)) | a customer went to a shop and paid a total of $ 40 , out of which 90 cents was for sales tax on taxable purchases . if the tax rate was 6 % , then what was the cost of the tax free items ? | "the total cost was $ 40 . the tax was $ 0.90 let the original price of the taxable items = x given that tax rate = 6 % 0.06 x = 0.90 x = $ 15 the cost of the tax free items was $ 40 - $ 15 - $ 0.90 = $ 24.10 the answer is c ." | a = 40 - 90
b = 6 * 2
c = a - b
|
a ) rs . 2007 , b ) rs . 2000 , c ) rs . 2089 , d ) rs . 2067 , e ) rs . 2098 | b | multiply(20, const_100) | albert buys 4 horses and 9 cows for rs . 13,400 . if he sells the horses at 10 % profit and the cows at 20 % profit , then he earns a total profit of rs . 1880 . the cost of a horse is : | "let c . p . of each horse be rs . x and c . p . of each cow be rs . y . then , 4 x + 9 y = 13400 - - ( i ) and , 10 % of 4 x + 20 % of 9 y = 1880 2 / 5 x + 9 / 5 y = 1880 = > 2 x + 9 y = 9400 - - ( ii ) solving ( i ) and ( ii ) , we get : x = 2000 and y = 600 . cost price of each horse = rs . 2000 . answer : b" | a = 20 * 100
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a ) 860 km , b ) 870 km , c ) 960 km , d ) 260 km , e ) 840 km | c | multiply(60, add(divide(60, subtract(64, 60)), 1)) | two motor bikes cover the same distance at the speed of 60 and 64 kmps respectively . find the distance traveled by them if the slower bike takes 1 hour more than the faster bike ? | explanation : 60 ( x + 1 ) = 64 x x = 15 60 x 16 = 960 km answer : c | a = 64 - 60
b = 60 / a
c = b + 1
d = 60 * c
|
a ) 6 , b ) 7 , c ) 8 , d ) 9 , e ) 10 | d | add(add(5, const_2), const_2) | if x < y < z and y - x > 5 , where x is an even integer and y and z are odd integers , what is the least possible value s of z - x ? | "x < y < z to find the least possible value for z - x ; we need to find the values for z and x that can be closest to each other . if x is some even number , then what could be minimum possible odd z . if x is some even number y - x > 5 ; y > x + 5 ; minimum value for y = x + 5 + 2 = x + 7 [ note : x + 5 is as even + o... | a = 5 + 2
b = a + 2
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a ) 1 / 4 , b ) 1 / 5 , c ) 1 / 10 , d ) 2 / 25 , e ) 1 / 20 | d | divide(subtract(5, const_1), multiply(5, const_10)) | if renee earns a raise of between 5 % and 10 % , non - inclusive , by what fraction could her salary have increased ? | 5 % is 5 / 100 = 1 / 20 10 % is 10 / 100 = 1 / 10 the increase must be greater than 1 / 20 and less than 1 / 10 . 1 / 4 , 1 / 5 , and 1 / 10 are not less than 1 / 10 . 1 / 20 is not greater than 1 / 20 . the only option left is 2 / 25 , which is in the specified range . answer : d | a = 5 - 1
b = 5 * 10
c = a / b
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a ) 20 % , b ) 40 % , c ) 50 % , d ) 60 % , e ) 75 % | c | subtract(const_100, subtract(subtract(const_100, 20), 20)) | a merchant sells an item at a 20 % discount , but still makes a gross profit of 20 percent of the cost . what percent r of the cost would the gross profit on the item have been if it had been sold without the discount ? | "let the market price of the product is mp . let the original cost price of the product is cp . selling price ( discounted price ) = 100 % of mp - 20 % mp = 80 % of mp . - - - - - - - - - - - - - - - - ( 1 ) profit made by selling at discounted price = 20 % of cp - - - - - - - - - - - - - - ( 2 ) apply the formula : pr... | a = 100 - 20
b = a - 20
c = 100 - b
|
a ) 9 / 4 , b ) 12 / 5 , c ) 16 / 5 , d ) 30 / 7 , e ) 50 / 3 | d | divide(6, 5) | in δ pqs above , if pq = 5 and ps = 6 , then | "there are two ways to calculate area of pqs . area remains same , so both are equal . 5 * 6 / 2 = pr * 7 / 2 pr = 30 / 7 d" | a = 6 / 5
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a ) 2 / 7 , b ) 1 / 3 , c ) 1 / 2 , d ) 2 / 3 , e ) 5 / 7 | e | divide(const_10, 30) | in a graduate physics course , 70 percent of the students are male and 30 percent of the students are married . if two - sevenths of the male students are married , what fraction of the male students is single ? | "let assume there are 100 students of which 70 are male and 30 are females if 30 are married then 70 will be single . now its given that two - sevenths of the male students are married that means 2 / 7 of 70 = 20 males are married if 30 is the total number of students who are married and out of that 20 are males then t... | a = 10 / 30
|
a ) 2 miles , b ) 6.6 miles , c ) 4.8 miles , d ) 8 miles , e ) 10 miles | b | multiply(divide(const_1, add(divide(const_1, 4), divide(const_1, 20))), const_1_6) | johnny travels a total of one hour to and from school . on the way there he jogs at 4 miles per hour and on the return trip he gets picked up by the bus and returns home at 20 miles per hour . how far is it to the school ? | "answer : b ) 6.6 miles . average speed for round trip = 2 * a * b / ( a + b ) , where a , b are speeds so , average speed was = 2 * 4 * 20 / ( 4 + 20 ) = 6.6 m / hr the distance between schoolhome should be half of that . ie . 6.6 miles answer b" | a = 1 / 4
b = 1 / 20
c = a + b
d = 1 / c
e = d * const_1_6
|
a ) 66 , b ) 72 , c ) 80 , d ) 68 , e ) 73 | a | subtract(90, multiply(sqrt(16), divide(subtract(90, 78), sqrt(4)))) | an engine moves at the speed of 90 kmph without any coaches attached to it . speed of the train reduces at the rate that varies directly as the square root of the number of coaches attached . when 4 coaches are attached speed decreases to 78 kmph . what will be the speed of train when 16 coaches are attached . | "1 . no . of coaches = 4 sqr root = 2 speed decreases by 12 12 = k * 2 k = 6 no . of coaches = 16 swr root = 4 decrease = 4 * 6 = 24 new speed = 90 - 24 = 66 a" | a = math.sqrt(16)
b = 90 - 78
c = math.sqrt(4)
d = b / c
e = a * d
f = 90 - e
|
a ) 6.54 % , b ) 2.54 % , c ) 8 . 2 % , d ) 4.94 % , e ) 1.29 % | e | divide(const_100, 77) | at what rate percent of simple interest will a sum of money double itself in 77 years ? | let sum = x . then , s . i . = x . rate = ( 100 * s . i . ) / ( p * t ) = ( 100 * x ) / ( x * 77 ) = 100 / 77 = 1.29 % answer : e | a = 100 / 77
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a ) 99 , 91,223 , b ) 99 , 92,224 , c ) 99 , 94,265 , d ) 99 , 95,300 , e ) 99 , 96,307 | b | divide(multiply(add(add(const_100, const_60), const_1), 10), const_100) | what is the value of 10 ^ 7 - 6 ^ 5 ? | "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 ) 79 , b ) 99 , c ) 88 , d ) 88 , e ) 76 | e | divide(add(add(add(add(73, 69), 92), 64), 82), divide(const_10, const_2)) | dacid obtained 73 , 69 , 92 , 64 and 82 marks ( out of 100 ) in english , mathematics , physics , chemistry and biology . what are his average marks ? | "average = ( 73 + 69 + 92 + 64 + 82 ) / 5 = 380 / 5 = 76 answer : e" | a = 73 + 69
b = a + 92
c = b + 64
d = c + 82
e = 10 / 2
f = d / e
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a ) 64 % , b ) 66 % , c ) 72 % , d ) 76 % , e ) 80 % | b | multiply(divide(add(const_100, 30), add(add(const_100, 30), const_100)), const_100) | in may mrs lee ' s earnings were 60 percent of the lee family ' s total income . in june mrs lee earned 30 percent more than in may . if the rest of the family ' s income was the same both months , then , in june , mrs lee ' s earnings were approximately what percent of the lee family ' s total income ? | "let in may lee family ' s total income = 100 in may mrs lee ' s income = 60 in may rest of the family ' s income = 40 in june mrs lees income = 60 * 130 / 100 = 78 in june total income = 78 + 40 = 118 % of mrs lee ' s income = 72 / 112 = 66.10 ( b )" | a = 100 + 30
b = 100 + 30
c = b + 100
d = a / c
e = d * 100
|
a ) 24 hours , b ) 5 hours , c ) 10 hours , d ) 15 hours , e ) 20 hours | e | add(15, subtract(15, 10)) | r and s together can plough a field in 10 hours but by himself r requires 15 hours . how long would s take to plough the same field ? | f r and s together can do a piece of work in x days and r alone can do the same work in y days , then s alone can do the same work in x y / y â € “ x days . the no . of hours required by s = 10 ã — 20 / 20 â € “ 10 = 200 / 10 = 20 hours e | a = 15 - 10
b = 15 + a
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a ) 19 , b ) 88 , c ) 267 , d ) 26 , e ) 28 | a | divide(subtract(203, multiply(36, 3)), subtract(8, 3)) | 36 is divided in 2 parts such that 8 times the first part added to 3 times the second part makes 203 . what is the first part ? | explanation : let the first part be x . 8 x + 3 ( 36 – x ) = 203 8 x + 108 – 3 x = 203 5 x + 108 = 203 5 x = 95 x = 19 answer : a | a = 36 * 3
b = 203 - a
c = 8 - 3
d = b / c
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a ) 160 , b ) 168 , c ) 176 , d ) 184 , e ) 192 | e | multiply(add(add(8, subtract(8, 2)), 2), add(subtract(add(8, subtract(8, 2)), divide(8, const_2)), 2)) | roy is now 8 years older than julia and half of that amount older than kelly . if in 2 years , roy will be twice as old as julia , then in 2 years what would be roy ’ s age multiplied by kelly ’ s age ? | "r = j + 8 = k + 4 r + 2 = 2 ( j + 2 ) ( j + 8 ) + 2 = 2 j + 4 j = 6 r = 14 k = 10 in 2 years ( r + 2 ) ( k + 2 ) = 16 * 12 = 192 the answer is e ." | a = 8 - 2
b = 8 + a
c = b + 2
d = 8 - 2
e = 8 + d
f = 8 / 2
g = e - f
h = g + 2
i = c * h
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a ) 13000 , b ) 12500 , c ) 13500 , d ) 14000 , e ) 15000 | b | multiply(power(add(const_2, const_3), 2), 500) | mike works at a science lab that conducts experiments on bacteria . the population of the bacteria multiplies at a constant rate , and his job is to notate the population of a certain group of bacteria each hour . at 1 p . m . on a certain day , he noted that the population was 500 and then he left the lab . he returne... | let the rate be x , then population of the bacteria after each hour can be given as 500,500 x , 500 ( x ^ 2 ) , 500 ( x ^ 3 ) now population at 4 pm = 62,500 thus we have 500 ( x ^ 3 ) = 62,500 = 125 thus x = 5 therefore population at 3 pm = 500 ( 25 ) = 12500 answer : b | a = 2 + 3
b = a ** 2
c = b * 500
|
a ) 12 , b ) 14 , c ) 18 , d ) 24 , e ) 10 | b | add(subtract(35, add(20, 1)), 1) | the average weight of a group of boys is 20 kg . after a boy of weight 35 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 35 kg joins the group , total weight of boys = 20 x + 35 so 20 x + 35 = 21 ( x + 1 ) = > x = 14 . answer : b" | a = 20 + 1
b = 35 - a
c = b + 1
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a ) 7 , b ) 8 , c ) 9 , d ) 10 , e ) 11 | a | divide(subtract(12.80, 0.50), 1.75) | at a certain bowling alley , it costs $ 0.50 to rent bowling shoes for the day and $ 1.75 to bowl 1 game . if a person has $ 12.80 and must rent shoes , what is the greatest number of complete games that person can bowl in one day ? | "after renting bowling shoes the person is left with $ 12.80 - $ 0.5 = $ 12.30 , which is enough for 12.3 / 1.75 = 7.02 - > ~ 7 . answer : a ." | a = 12 - 80
b = a / 1
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a ) 150 , b ) 250 , c ) 300 , d ) 200 , e ) 280 | d | multiply(divide(320, const_3), 5) | the ratio of number of boys and girls in a school is 3 : 5 . if there are 320 students in the school , find the number of girls in the school ? | "let the number of boys and girls be 3 x and 5 x total students = 320 number of girls in the school = 5 * 320 / 8 = 200 answer is d" | a = 320 / 3
b = a * 5
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a ) 10.5 , b ) 11 , c ) 11.5 , d ) 12 , e ) 12.5 | c | subtract(14.5, multiply(2, 1.5)) | the arithmetic mean and standard deviation of a certain normal distribution are 14.5 and 1.5 , respectively . what value is exactly 2 standard deviations less than the mean ? | "the value which isexactlytwo sd less than the mean is : mean - 2 * sd = 14.5 - 2 * 1.5 = 11.5 . answer : c ." | a = 2 * 1
b = 14 - 5
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a ) 230 , b ) 232 , c ) 234 , d ) 236 , e ) 238 | e | subtract(divide(divide(560, 6), const_0_2778), 100) | a girl sitting in a train which is travelling at 100 kmph observes that a goods train travelling in a opposite direction , takes 6 seconds to pass him . if the goods train is 560 m long , find its speed . | "relative speed = ( 560 / 6 ) m / s = ( 560 / 6 ) * ( 18 / 5 ) = 336 kmph speed of goods train = 336 - 100 = 236 kmph answer is e" | a = 560 / 6
b = a / const_0_2778
c = b - 100
|
a ) 20 years , b ) 22 years , c ) 24 years , d ) 25 years , e ) 27 years | d | subtract(multiply(15, 15), add(multiply(4, 14), multiply(9, 16))) | the average age of 15 students of a class is 15 years . out of these , the average age of 4 students is 14 years and that of the other 9 students is 16 years . tee age of the 15 th student is ? | "age of the 15 th student = [ 15 * 15 - ( 14 * 4 + 16 * 9 ) ] = ( 225 - 200 ) = 25 years . answer : d" | a = 15 * 15
b = 4 * 14
c = 9 * 16
d = b + c
e = a - d
|
a ) 2 days , b ) 7 days , c ) 6 days , d ) 9 days , e ) 3 days | c | divide(const_1, add(divide(const_1, 10), divide(const_1, 15))) | a can do a piece of work in 10 days and b alone can do it in 15 days . how much time will both take to finish the work ? | "c 6 days time taken to finish the job = xy / ( x + y ) = 10 x 15 / ( 10 + 15 ) = 150 / 25 = 6 days" | a = 1 / 10
b = 1 / 15
c = a + b
d = 1 / c
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a ) a ) 47 , b ) b ) 45.6 , c ) c ) 44 , d ) d ) 48 , e ) e ) 49.6 | e | divide(add(multiply(45, subtract(25, add(3, 5))), multiply(95, 5)), 25) | in a class of 25 students in an examination in maths 5 students scored 95 marks each , 3 get zero each and the average of the rest was 45 . what is the average of the whole class ? | "explanation : total marks obtained by a class of 25 students = 5 * 95 + 3 * 0 + 17 * 45 = 1240 average marks of whole class = 1240 / 25 = 49.6 answer : option e" | a = 3 + 5
b = 25 - a
c = 45 * b
d = 95 * 5
e = c + d
f = e / 25
|
a ) - 45 , b ) 50 , c ) - 62 , d ) 35 , e ) - 90 | e | subtract(subtract(subtract(190, 10), add(190, 10)), 10) | if | 20 x - 10 | = 190 , then find the product of the values of x ? | "| 20 x - 10 | = 190 20 x - 10 = 190 or 20 x - 10 = - 190 20 x = 200 or 20 x = - 180 x = 10 or x = - 9 product = - 9 * 10 = - 90 answer is e" | a = 190 - 10
b = 190 + 10
c = a - b
d = c - 10
|
a ) 12 , b ) 13 , c ) 16 , d ) 15 , e ) 11 | c | divide(subtract(400, 1), 25) | how many positive integers between 1 and 400 are there such that they are multiples of 25 ? | "multiples of 25 = 25 , 50,75 , - - - - - 400 number of multiples of 25 = > 25 * 16 = 400 answer is c" | a = 400 - 1
b = a / 25
|
a ) 0 , b ) 2 , c ) 12 , d ) 13 , e ) 15 | d | subtract(add(12, const_3), subtract(add(const_10, 4), 12)) | if the range w of the 6 numbers 4 , 314 , 710 and x is 12 , what is the difference between the greatest possible value of x and least possible value of x ? | the range w of a set is the difference between the largest and smallest elements of a set . without x , the difference between the largest and smallest elements of a set is 14 - 3 = 11 < 12 , which means that in order 12 to be the range of the set x must be either the smallest element so that 14 - x = 12 - - - > x = 2 ... | a = 12 + 3
b = 10 + 4
c = b - 12
d = a - c
|
a ) 2748472 , b ) 3748472 , c ) 4748472 , d ) 5748472 , e ) 6748472 | c | add(subtract(subtract(const_1000, const_10), multiply(multiply(const_10, multiply(7, 7)), multiply(const_4, const_2))), const_10) | how many 7 digit number contain number 3 ? | "total 7 digit no . = 9 * 10 * 10 * 10 * 10 * 10 * 10 = 9000000 not containing 3 = 8 * 9 * 9 * 9 * 9 * 9 * 9 = 4251528 total 7 digit number contain 3 = 9000 - 4251528 = 4748472 answer : c" | a = 1000 - 10
b = 7 * 7
c = 10 * b
d = 4 * 2
e = c * d
f = a - e
g = f + 10
|
a ) 500 , b ) 474.6 , c ) 222 , d ) 297 , e ) 111 | b | divide(multiply(divide(multiply(616, const_100), add(const_100, 10)), add(const_100, 10)), add(const_100, 18)) | the sale price of an article including the sales tax is rs . 616 . the rate of sales tax is 10 % . if the shopkeeper has made a profit of 18 % , then the cost price of the article is : | "explanation : 110 % of s . p . = 616 s . p . = ( 616 * 100 ) / 110 = rs . 560 c . p = ( 100 * 560 ) / 118 = rs . 474.6 answer : b" | a = 616 * 100
b = 100 + 10
c = a / b
d = 100 + 10
e = c * d
f = 100 + 18
g = e / f
|
a ) 45 , b ) 49 , c ) 50 , d ) 52 , e ) 56 | b | divide(180, const_10) | how many integers from 10 to 180 , inclusive , are divisible by 3 but not divisible by 7 ? | "we should find # of integers divisible by 3 but not by 3 * 7 = 21 . # of multiples of 21 in the range from 10 to 180 , inclusive is ( 168 - 21 ) / 21 + 1 = 8 ; 57 - 8 = 49 . answer : b ." | a = 180 / 10
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a ) 8 days , b ) 3 days , c ) 7 days , d ) 5 days , e ) 6 days | a | divide(multiply(multiply(12, 10), 8), multiply(8, 15)) | 12 men work 8 hours per day to complete the work in 10 days . to complete the same work in 8 days , working 15 hours a day , the number of men required | "that is , 1 work done = 12 × 8 × 10 then , 12 8 × 10 = ? × 15 × 8 ? ( i . e . no . of men required ) = 12 × 8 × 10 / 15 × 10 = 8 days . a" | a = 12 * 10
b = a * 8
c = 8 * 15
d = b / c
|
a ) 7 , b ) 6 , c ) 5 , d ) 4 , e ) 3 | c | floor(divide(log(divide(220,000, 2.134)), log(10))) | if x is an integer and 2.134 × 10 ^ x is less than 220,000 , what is the greatest possible value for x ? | "x is an integer and 2.134 × 10 x is less than 220,000 , what is the greatest possible value for x ? for 2.134 × 10 x is less than 220,000 to remain true , the greatest number is 213,400 , which makes x = 5 c . 5" | a = 220 / 0
b = math.log(a)
c = math.log(10)
d = b / c
e = math.floor(d)
|
a ) 4586970843 , b ) 4686970743 , c ) 4691100843 , d ) 4586870843 , e ) none | c | multiply(469157, 9999) | calculate 469157 x 9999 = ? | "answer 469157 x 9999 = 469157 x ( 10000 - 1 ) = 4691570000 - 469157 = 4691100843 . option : c" | a = 469157 * 9999
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a ) 12 h , b ) 10 h , c ) 4 h , d ) 6 h , e ) none of these | c | divide(multiply(const_4, sqrt(400)), 20) | the area of a square field is 400 km 2 . how long will it take for a horse to run around at the speed of 20 km / h ? | explanation area of field = 400 km 2 . then , each side of field = √ 400 = 20 km distance covered by the horse = perimeter of square field = 20 × 4 = 80 km ∴ time taken by horse = distances / peed = 80 / 20 = 4 h answer c | a = math.sqrt(400)
b = 4 * a
c = b / 20
|
a ) 516 , b ) 3096 , c ) 6192 , d ) 628 , e ) 4320 | e | multiply(multiply(36, 10), add(const_10, const_2)) | there are 10 dozen mangoes in a box . if there are 36 such boxes , how many mangoes are there in all the boxes together ? | number of mangoes = 12 dozens = 12 × 10 = 120 ∴ number of mangoes in 36 boxes = 36 × 120 = 4320 answer e | a = 36 * 10
b = 10 + 2
c = a * b
|
a ) 3 : 13 , b ) 9 : 13 , c ) 36 : 13 , d ) 13 : 9 , e ) none | b | divide(multiply(3, const_3), 13) | if a and b are in the ratio 3 : 4 , and b and c in the ratio 12 : 13 , then a and c will be in the ratio | solution : ( a / b ) * ( b / c ) = ( 3 / 4 ) * ( 12 / 13 ) ; or , a / b = 36 / 39 = 9 : 13 . answer : option b | a = 3 * 3
b = a / 13
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a ) 196 , b ) 94 , c ) 90 , d ) 100 , e ) 120 | d | subtract(multiply(subtract(33, multiply(const_4, const_4)), 6), const_2) | the sum of a number and the number preceding it is 33 . by how much is two less than 6 times the number ? | two numbers must be 16 and 17 16 + 17 = 33 required number is 17 six times of this number = 6 * 17 = 102 two less than 102 = 102 - 2 = 100 answer : d | a = 4 * 4
b = 33 - a
c = b * 6
d = c - 2
|
a ) 1000 , b ) 1100 , c ) 1210 , d ) 1452 , e ) 1552 | c | multiply(divide(4,18, 100,294), const_100) | 4,18 , 100,294 , ___ | "2 ^ 3 - 2 ^ 2 = 4 ; 3 ^ 3 - 3 ^ 2 = 18 ; 5 ^ 3 - 5 ^ 2 = 100 ; 7 ^ 3 - 7 ^ 2 = 294 ; so , 11 ^ 3 - 11 ^ 2 = 1210 answer : c" | a = 4 / 18
b = a * 100
|
a ) 1 / 6 , b ) 2 / 5 , c ) 3 / 10 , d ) 2 / 9 , e ) 1 / 2 | d | divide(choose(5, 5), choose(add(const_2.0, 5), const_2)) | a bag holds 5 red marbles and 5 green marbles . if you removed two randomly selected marbles from the bag , without replacement , what is the probability that both would be red ? | "probability of selecting first red marble = 5 / 10 probability of selecting second red marble without replacement = 4 / 9 final probability = 5 / 10 * 4 / 9 = 2 / 9 the correct answer is d ." | a = math.comb(5, 5)
b = 2 + 0
c = math.comb(b, 2)
d = a / c
|
a ) 7 , b ) 8 , c ) 9 , d ) 10 , e ) 12 | c | add(floor(divide(24, const_3)), const_1) | what is the smallest integer x for which 27 ^ x > 3 ^ 24 ? | "27 ^ x > 3 ^ 24 converting into the same bases : 27 ^ x > 27 ^ 8 therefore for the equation to hold true , x > 8 or x = 9 option c" | a = 24 / 3
b = math.floor(a)
c = b + 1
|
a ) 14 , b ) 13 , c ) 12 , d ) 16 , e ) 15 | a | divide(divide(add(16, 12), const_2), const_2) | a man can row downstream at 16 kmph and upstream at 12 kmph . find the speed of the man in still water . | "let the speed of the man in still water and speed of stream be x kmph and y kmph respectively . given x + y = 16 - - - ( 1 ) and x - y = 12 - - - ( 2 ) from ( 1 ) & ( 2 ) 2 x = 28 = > x = 14 , y = 2 . answer : a" | a = 16 + 12
b = a / 2
c = b / 2
|
a ) 24 , b ) 28 , c ) 30 , d ) 32 , e ) 44 | e | subtract(divide(multiply(add(const_100, 38), const_100), subtract(const_100, 4)), const_100) | a shopkeeper sold an article offering a discount of 4 % and earned a profit of 38 % . what would have been the percentage of profit earned if no discount was offered ? | "let c . p . be rs . 100 . then , s . p . = rs . 138 let marked price be rs . x . then , 96 / 100 x = 138 x = 13800 / 96 = rs . 143.75 now , s . p . = rs . 143.75 , c . p . = rs . 100 profit % = 44 % . answer : e" | a = 100 + 38
b = a * 100
c = 100 - 4
d = b / c
e = d - 100
|
a ) 10 sec , b ) 32 sec , c ) 82 sec , d ) 24 sec , e ) 89 sec | d | divide(add(160, 320), multiply(add(42, 30), const_0_2778)) | two trains of length 160 m and 320 m are running towards each other on parallel lines at 42 kmph and 30 kmph respectively . in what time will they be clear of each other from the moment they meet ? | "relative speed = ( 42 + 30 ) * 5 / 18 = 4 * 5 = 20 mps . distance covered in passing each other = 160 + 320 = 480 m . the time required = d / s = 480 / 20 = 24 sec . answer : d" | a = 160 + 320
b = 42 + 30
c = b * const_0_2778
d = a / c
|
a ) 0.56 , b ) 0.62 , c ) 0.5 , d ) 0.48 , e ) 0.52 | a | divide(subtract(multiply(multiply(0.76, 0.76), 0.76), 0.008), add(add(multiply(0.76, 0.76), multiply(0.76, 0.2)), 0.04)) | solve ( 0.76 × 0.76 × 0.76 − 0.008 ) / ( 0.76 × 0.76 + 0.76 × 0.2 + 0.04 ) | 0.56 option ' a ' | a = 0 * 76
b = a * 0
c = b - 0
d = 0 * 76
e = 0 * 76
f = d + e
g = f + 0
h = c / g
|
a ) 20 , b ) 30 , c ) 15 , d ) 33 , e ) 31 | b | subtract(multiply(4, const_2), multiply(2, const_2)) | if the average ( arithmetic mean ) of 2 x , 4 x , and 8 x is 140 , what is the value of x ? | "am of 2 x , 4 x and 8 x = 2 x + 4 x + 8 x / 3 = 14 x / 3 given that 14 x / 3 = 140 x = 30 answer is b" | a = 4 * 2
b = 2 * 2
c = a - b
|
a ) 7 / 32 , b ) 11 / 64 , c ) 35 / 128 , d ) 49 / 128 , e ) 65 / 128 | c | multiply(4, power(divide(const_1, const_2), 4)) | if the probability of rain on any given day in chicago during the summer is 50 % , independent of what happens on any other day , what is the probability of having exactly 3 rainy days from july 4 through july 10 inclusive ? | "one possible case is : rainy - rainy - rainy - not rainy - not rainy - not rainy - not rainy . the probability of this case is ( 1 / 2 ) ^ 7 = 1 / 128 the number of possible cases is 7 c 3 = 35 . p ( exactly 3 rainy days ) = 35 * 1 / 128 = 35 / 128 the answer is c ." | a = 1 / 2
b = a ** 4
c = 4 * b
|
a ) 11 , b ) 12 , c ) 13 , d ) 14 , e ) 15 | d | subtract(add(add(add(add(add(const_1, power(const_2, const_2)), power(const_3, const_2)), power(const_4, const_2)), power(add(const_4, const_1), const_2)), power(5, const_2)), add(add(add(const_4, const_3), 5), add(add(add(add(const_2, const_3), add(const_4, const_1)), add(const_4, const_3)), add(add(const_2, const_3),... | what is the positive difference between the sum of the squares of the first 5 positive integers and the sum of the prime numbers between the first square and fourth square ? | "forget conventional ways of solving math questions . in ps , ivy approach is the easiest and quickest way to find the answer . the sum of the squares of the first 4 positive integers = 1 ^ 2 + 2 ^ 2 + 3 ^ 2 + 4 ^ 2 + 5 ^ 2 = 55 the sum of the prime numbers between the first square ( = 1 ) and fourth square ( = 16 ) = ... | a = 2 ** 2
b = 1 + a
c = 3 ** 2
d = b + c
e = 4 ** 2
f = d + e
g = 4 + 1
h = g ** 2
i = f + h
j = 5 ** 2
k = i + j
l = 4 + 3
m = l + 5
n = 2 + 3
o = 4 + 1
p = n + o
q = 4 + 3
r = p + q
s = 2 + 3
t = s + 5
u = r + t
v = m + u
w = k - v
|
a ) 807518799 , b ) 806436469 , c ) 807538799 , d ) 806329359 , e ) 817431046 | d | multiply(subtract(9999, const_4), 80641) | find the value of 80641 x 9999 = m ? | "80641 x 9999 = 80641 x ( 10000 - 1 ) = 80641 x 10000 - 80641 x 1 = 806410000 - 80641 = 806329359 d" | a = 9999 - 4
b = a * 80641
|
a ) 47.095 , b ) 47.752 , c ) 47.932 , d ) 47.95 , e ) none | d | subtract(add(3889, 12.952), 3854.002) | 3889 + 12.952 – ? = 3854.002 | explanation let 3889 + 12.952 – x = 3854.002 . then x = ( 3889 + 12.952 ) – 3854.002 = 3901.952 – 3854.002 = 47.95 . answer d | a = 3889 + 12
b = a - 3854
|
['a ) 5 square inches', 'b ) 10 square inches', 'c ) 15 square inches', 'd ) 25 square inches', 'e ) 34 square inches'] | e | subtract(multiply(power(14, const_2), const_3), add(add(power(9, const_2), power(14, const_2)), add(power(9, const_2), power(14, const_2)))) | huey ' s hip pizza sells two sizes of square pizzas : a small pizza that measures 9 inches on a side and costs $ 10 , and a large pizza that measures 14 inches on a side and costs $ 20 . if two friends go to huey ' s with $ 30 apiece , how many more square inches of pizza can they buy if they pool their money than if t... | in the first case each can buy one pizza of $ 10 and one pizza of $ 20 . in square inches that would be ( 9 * 9 = 81 ) for the small pizza and ( 14 * 14 = 196 ) for the large pizza . in total sq inches that would be ( 81 + 196 ) * 2 = 554 sq inches . in the second case if they pool their money together they can buy 3 l... | a = 14 ** 2
b = a * 3
c = 9 ** 2
d = 14 ** 2
e = c + d
f = 9 ** 2
g = 14 ** 2
h = f + g
i = e + h
j = b - i
|
a ) 3 , b ) 4 , c ) 7 , d ) 18 , e ) 9 | b | power(const_2, const_2) | which number is a factor of 20 ? | a whole number that divides exactly into another whole number is called a factor 20 / 5 = 4 answer b | a = 2 ** 2
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a ) 35.29 kg , b ) 37.25 kg , c ) 42 kg , d ) 38 kg , e ) 29.78 kg | d | divide(add(multiply(24, 40), multiply(16, 35)), add(24, 16)) | there are 2 sections a and b in a class , consisting of 24 and 16 students respectively . if the average weight of section a is 40 kg and that of section b is 35 kg , find the average of the whole class ? | "total weight of 36 + 44 students = 24 * 40 + 16 * 35 = 1520 average weight of the class is = 1520 / 40 = 38 kg answer is d" | a = 24 * 40
b = 16 * 35
c = a + b
d = 24 + 16
e = c / d
|
a ) 1 , b ) 4 , c ) 3 , d ) 2 , e ) 0 | d | multiply(3, 4) | if n divided by 5 has a remainder of 4 , what is the remainder when 3 times n is divided by 5 ? | "as per question = > n = 5 p + 4 for some integer p hence 3 n = > 15 q + 12 but again , 12 can be divided by 5 to get remainder 2 for some integer q hence d" | a = 3 * 4
|
a ) 7 , b ) 6 , c ) 9 , d ) 4 , e ) 2 | d | divide(divide(add(18, 10), const_2), const_2) | a man can row downstream at 18 kmph and upstream at 10 kmph . find the speed of the man in still water and the speed of stream respectively ? | "let the speed of the man in still water and speed of stream be x kmph and y kmph respectively . given x + y = 18 - - - ( 1 ) and x - y = 10 - - - ( 2 ) from ( 1 ) & ( 2 ) 2 x = 28 = > x = 14 , y = 4 . answer : d" | a = 18 + 10
b = a / 2
c = b / 2
|
a ) 48 , b ) 60 , c ) 75 , d ) 80 , e ) 100 | a | divide(divide(3, 5), multiply(divide(divide(divide(3, 4), 5), 30), const_2)) | if four machines working at the same rate can do 3 / 4 of a job in 30 minutes , how many minutes would it take two machines working at the same rate to do 3 / 5 of the job ? | "using the std formula m 1 d 1 h 1 / w 1 = m 2 d 2 h 2 / w 2 substituting the values we have 4 * 1 / 2 * 4 / 3 = 2 * 5 / 3 * x ( converted 30 min into hours = 1 / 2 ) 8 / 3 = 10 / 3 * x x = 4 / 5 hour so 48 minutes answer : a" | a = 3 / 5
b = 3 / 4
c = b / 5
d = c / 30
e = d * 2
f = a / e
|
a ) 150 , b ) 250 , c ) 300 , d ) 370 , e ) 280 | a | multiply(divide(210, const_3), 5) | the ratio of number of boys and girls in a school is 2 : 5 . if there are 210 students in the school , find the number of girls in the school ? | "let the number of boys and girls be 2 x and 5 x total students = 210 number of girls in the school = 5 * 210 / 7 = 150 answer is a" | a = 210 / 3
b = a * 5
|
a ) 420 , b ) 840 , c ) 1260 , d ) 2520 , e ) 5020 | d | lcm(1, 10) | what is the lowest positive integer that is divisible by each of the integers 1 through 10 inclusive ? | "we have to find the lcm of 1 , 2 , 3 , 2 ^ 2 , 5 , 2 * 3 , 7 , 2 ^ 3 , 3 ^ 2 , and 2 * 5 . the lcm is 1 * 2 ^ 3 * 3 ^ 2 * 5 * 7 = 2520 the answer is d ." | a = math.lcm(1, 10)
|
a ) 3 / 2 , b ) 6 / 7 , c ) 1 / 2 , d ) 7 / 2 , e ) 4 / 5 | c | multiply(divide(const_2, const_4), const_100) | there are two positive integers a and b . what is the probability that a + b is odd / | s = adding two numbers is ( even + even ) , ( even + odd ) , ( odd + odd ) , ( odd + even ) n ( s ) = 4 e = ( even + odd ) , ( odd + even ) are the points in the event . n ( e ) = 2 p ( e ) = n ( e ) / n ( s ) = 2 / 4 = 1 / 2 answer is option c | a = 2 / 4
b = a * 100
|
a ) 20 , b ) 30 , c ) 50 , d ) 80 , e ) 100 | d | subtract(subtract(150, 60), 20) | of the 150 employees at company x , 60 are full - time , and 100 have worked at company x for at least a year . there are 20 employees at company x who aren ’ t full - time and haven ’ 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 ... | "full time employee who have not worked for at least one year = a full time employee who have worked for at least one year = b non full time employee who have worked for at least one year = c non full time employee who have not worked for at least one year = d a + b + c + d = 150 a + b = 80 i . e . c + d = 70 b + c = 1... | a = 150 - 60
b = a - 20
|
['a ) 2', 'b ) 5', 'c ) 6', 'd ) 7', 'e ) 14'] | e | add(const_10, const_4) | if f is the smallest positive integer such that 3,150 multiplied by f is the square of an integer , then f must be | solution : this problem is testing us on the rule that when we express a perfect square by its unique prime factors , every prime factor ' s exponent is an even number . let ’ s start by prime factorizing 3,150 . 3,150 = 315 x 10 = 5 x 63 x 10 = 5 x 7 x 3 x 3 x 5 x 2 3,150 = 2 ^ 1 x 3 ^ 2 x 5 ^ 2 x 7 ^ 1 ( notice that ... | a = 10 + 4
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a ) 0.0005 , b ) 0.0016 , c ) 0.0056 , d ) 0.0066 , e ) 0.0006 | e | subtract(multiply(divide(add(add(const_12, const_4), const_2), const_100), divide(add(add(const_12, const_4), const_2), const_100)), 0.0355) | what is the least number . which should be added to 0.0355 to make it a perfect square ? | "0.0355 + 0.0006 = 0.0361 ( 0.19 ) ^ 2 answer : e" | a = 12 + 4
b = a + 2
c = b / 100
d = 12 + 4
e = d + 2
f = e / 100
g = c * f
h = g - 0
|
a ) 6 , b ) 5 , c ) 4 , d ) 3 , e ) 2 | a | add(multiply(5, divide(subtract(multiply(5, 5), 7), subtract(multiply(5, 4), 6))), multiply(subtract(5, multiply(divide(subtract(multiply(5, 5), 7), subtract(multiply(5, 4), 6)), 4)), 3)) | if x + 4 y = 5 and 5 x + 6 y = 7 , then 3 x + 5 y = ? | add the left terms and right terms of the two equation to obtain a new equation x + 5 x + 4 y + 6 y = 5 + 7 group and simplify 6 x + 10 y = 12 divide all terms of the above equation by 2 to obtain a new equation 3 x + 5 y = 6 correct answer a | a = 5 * 5
b = a - 7
c = 5 * 4
d = c - 6
e = b / d
f = 5 * e
g = 5 * 5
h = g - 7
i = 5 * 4
j = i - 6
k = h / j
l = k * 4
m = 5 - l
n = m * 3
o = f + n
|
a ) 5 . , b ) 10 . , c ) 14 . , d ) 15 . , e ) 20 . | a | multiply(divide(add(10, divide(10, const_2)), 6), const_2) | the distance from steve ' s house to work is 10 km . on the way back steve drives twice as fast as he did on the way to work . altogether , steve is spending 6 hours a day on the roads . what is steve ' s speed on the way back from work ? | "time is in the ratio 2 : 1 : : to : fro office therefore , 2 x + 1 x = 6 hrs time take to come back - 2 hrs , distance travelled - 10 km = > speed = 5 kmph a" | a = 10 / 2
b = 10 + a
c = b / 6
d = c * 2
|
a ) 19 , b ) 25 , c ) 35 , d ) 45 , e ) 75 | a | multiply(divide(subtract(multiply(divide(40, 100), 300), multiply(divide(35, 100), multiply(300, divide(60, 100)))), 300), const_100) | 300 first - time customers of a fashion store were surveyed for their shopping experience right after leaving the store . 60 % of the customers in the survey had purchased clothes for less than $ 100 . 40 % of the customers in the survey reported they were overall satisfied with their purchase . 35 % of the customers t... | out of 300 - 180 purchased for less than 100 $ 120 for more out of 300 - 120 responded as satisfied and 180 responded disatisfied out of 180 ( purchased less than 100 $ ) - 35 % = 63 responded as satisfied , so remaining satisfied are 120 - 63 = 57 so 57 is what percentage of 300 - 19 % so the answer should be a | a = 40 / 100
b = a * 300
c = 35 / 100
d = 60 / 100
e = 300 * d
f = c * e
g = b - f
h = g / 300
i = h * 100
|
a ) 1 : 3 , b ) 4 : 3 , c ) 2 : 3 , d ) 4 : 1 , e ) 2 : 4 | d | divide(divide(const_1, multiply(add(30, const_2), const_10)), divide(const_1, multiply(30, const_10))) | a work can be finished in 30 days by 30 women . the same work can be finished in fifteen days by 15 men . the ratio between the capacity of a man and a woman is | "work done by 30 women in 1 day = 1 / 30 work done by 1 woman in 1 day = 1 / ( 30 × 30 ) work done by 15 men in 1 day = 1 / 15 work done by 1 man in 1 day = 1 / ( 15 × 15 ) ratio of the capacity of a man and woman = 1 / ( 15 × 15 ) : 1 / ( 30 × 30 ) ) = 1 / 225 : 1 / 900 = 1 / 1 : 1 / 4 = 4 : 1 option d" | a = 30 + 2
b = a * 10
c = 1 / b
d = 30 * 10
e = 1 / d
f = c / e
|
a ) 1 / 3 , b ) 2 / 3 , c ) 1 / 4 , d ) 1 / 16 , e ) 3 / 5 | d | divide(const_2, 8) | if there is an equal probability of a child being born a boy or a girl , what is the probability that a couple who have 8 children have two children of the same sex and one of the opposite sex ? | "no of ways of selecting a gender - 2 c 1 no of ways of selecting any 2 children out of 8 = 8 c 2 total possible outcomes - 2 ^ 8 ( each child can be either a girl or a boy ) probability = 2 c 1 * 8 c 2 / 2 ^ 8 = 2 * 8 / 2 * 2 * 2 * 2 * 2 * 2 * 2 * 2 = 16 / 256 = 1 / 16 ans = d" | a = 2 / 8
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a ) 277 , b ) 35 , c ) 64 , d ) 72 , e ) none of these | b | divide(multiply(126, const_100), 360) | ? % of 360 = 126 | "? % of 360 = 126 or , ? = 126 × 100 / 360 = 35 answer b" | a = 126 * 100
b = a / 360
|
['a ) 200', 'b ) 225', 'c ) 240', 'd ) 250', 'e ) none'] | d | multiply(const_100, divide(circle_area(5), circumface(5))) | the area of a circle of radius 5 is numerically what percent of its circumference ? | solution required % = [ π x ( 5 ) 2 / 2 π x 5 x 100 ] % ‹ = › 250 % . answer d | a = circle_area / (
b = 100 * a
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a ) 88 , b ) 100 , c ) 198 , d ) 216 , e ) 252 | b | divide(add(multiply(subtract(90, 10), 40), multiply(subtract(90, 70), 40)), 40) | jony walks along sunrise boulevard daily . he starts walking at 07 : 00 from block 10 and walks to block 90 where he turns around and walks back to block 70 , where he stops at 07 : 40 . the blocks along the boulevard are numbered sequentially ( 1 , 2,3 ) , and each block measures 40 meters . what is jony ' s speed in ... | total distance from 10 to 90 = 80 + from 90 to 70 = 20 so the dist is 100 × 40 ( per block dist ) speed = 4000 mts / 40 min = 100 m / min b is the answer | a = 90 - 10
b = a * 40
c = 90 - 70
d = c * 40
e = b + d
f = e / 40
|
a ) 5 , b ) 6 , c ) 7 , d ) 5.5 , e ) 6.5 | b | divide(add(40, 8), subtract(58, 50)) | car a is 40 miles behind car b , which is traveling in the same direction along the same route as car a . car a is traveling at a constant speed of 58 miles per hour and car bis traveling at a constant speed of 50 miles per hour . how many hours will it take for car a to overtake and drive 8 miles ahead of car b ? | "relative speed of car a is 58 - 50 = 8 miles per hour , to catch up 40 miles and drive 8 miles ahead so to drive 48 miles it ' ll need 48 / 8 = 6 hours . answer : b" | a = 40 + 8
b = 58 - 50
c = a / b
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a ) - 6.5 , b ) 2 , c ) - 2 , d ) - 6 , e ) - 10 | a | divide(subtract(8, 5), subtract(sqrt(add(9, 8)), sqrt(add(5, 9)))) | in the xy - coordinate plane , the graph of y = - x ^ 2 + 9 intersects line l at ( p , 5 ) and ( t , - 8 ) . what is the least possible value of the slope of line l ? | "we need to find out the value of p and l to get to the slope . line l and graph y intersect at point ( p , 5 ) . hence , x = p and y = 5 should sactisfy the graph . soliving 5 = - p 2 + 9 p 2 = 4 p = + or - 2 simillarly point ( t , - 8 ) should satisfy the equation . hence x = t and y = - 8 . - 7 = - t 2 + 9 t = + or ... | a = 8 - 5
b = 9 + 8
c = math.sqrt(b)
d = 5 + 9
e = math.sqrt(d)
f = c - e
g = a / f
|
a ) $ 120 , b ) $ 150 , c ) $ 1288 , d ) $ 250 , e ) $ 300 | c | subtract(multiply(1200, power(add(const_1, divide(20, const_100)), 4)), 1200) | find the compound interest on $ 1200 for 4 years at 20 % p . a . if ci is component yearly ? | "a = p ( 1 + r / 100 ) ^ t = 1200 ( 1 + 20 / 100 ) ^ 4 = $ 2488 ci = $ 1288 answer is c" | a = 20 / 100
b = 1 + a
c = b ** 4
d = 1200 * c
e = d - 1200
|
a ) 3 : 2 , b ) 5 : 2 , c ) 2 : 1 , d ) 4 : 3 , e ) 1 : 2 | b | divide(divide(1, 4), divide(10, const_100)) | in a company 10 % of male staff are same in number as 1 / 4 th of the female staff . what is the ratio of male staff to female staff | 10 % of ms = 1 / 4 th of fs - > 10 ms / 100 = 1 / 4 fs - > ms = 5 / 2 fs : . ms / fs = 5 / 2 = ms : fs = 5 : 2 answer : b | a = 1 / 4
b = 10 / 100
c = a / b
|
a ) 40 , b ) 45 , c ) 50 , d ) 75 / 2 , e ) 35 | d | multiply(divide(300, 100), divide(25, const_2)) | the ratio , by volume , of bleach to detergent to water in a certain solution is 2 : 25 : 100 . the solution will be altered so that the ratio of bleach to detergent is tripled while the ratio of detergent to water is halved . if the altered solution will contain 300 liters of water , how many liters of detergent will ... | "b : d : w = 2 : 25 : 100 bnew / dnew = ( 1 / 3 ) * ( 2 / 25 ) = ( 2 / 75 ) dnew / wnew = ( 1 / 2 ) * ( 25 / 100 ) = ( 1 / 8 ) wnew = 300 dnew = wnew / 5 = 300 / 8 = 75 / 2 so , answer will be d" | a = 300 / 100
b = 25 / 2
c = a * b
|
a ) 176 kg , b ) 80 kg , c ) 185 kg , d ) 90 kg , e ) 128 kg | e | add(65, multiply(10, 6.3)) | the average weight of 10 persons increases by 6.3 kg when a new person comes in place of one of them weighing 65 kg . what might be the weight of the new person ? | "solution total weight increased = ( 10 x 6.3 ) kg = 63 kg . weight of new person = ( 65 + 63 ) kg = 128 kg . answer e" | a = 10 * 6
b = 65 + a
|
a ) 50 , b ) 56 , c ) 58 , d ) 62 , e ) 66 | c | subtract(multiply(66, add(1, 2)), multiply(70, 2)) | a charitable association sold an average of 66 raffle tickets per member . among the female members , the average was 70 raffle tickets . the male to female ratio of the association is 1 : 2 . what was the average number t of tickets sold by the male members of the association | "given that , total average t sold is 66 , male / female = 1 / 2 and female average is 70 . average of male members isx . ( 70 * f + x * m ) / ( m + f ) = 66 - > solving this equation after substituting 2 m = f , x = 58 . ans c ." | a = 1 + 2
b = 66 * a
c = 70 * 2
d = b - c
|
a ) 76 kmph , b ) 75 kmph , c ) 87 kmph , d ) 79 kmph , e ) 86 kmph | d | divide(add(98, 60), const_2) | the speed of a car is 98 km in the first hour and 60 km in the second hour . what is the average speed of the car ? | "s = ( 98 + 60 ) / 2 = 79 kmph answer : d" | a = 98 + 60
b = a / 2
|
a ) 6 , b ) 1 , c ) 4 , d ) 3 , e ) 2 | e | subtract(add(subtract(625573, const_1), 3), 625573) | what is the least whole number that should be added to 625573 if it is to be divisible by 3 ? | a number is divisible by 3 if the sum of the digits is divisible by 3 . . here , 6 + 2 + 5 + 5 + 7 + 3 = 28 , the next multiple of 3 is 30 . 2 must be added to 625574 to make it divisible by 3 e | a = 625573 - 1
b = a + 3
c = b - 625573
|
a ) 9 , b ) 8 , c ) 5 , d ) 11 , e ) 15 | d | subtract(2000, multiply(floor(divide(2000, 17)), 17)) | what least number must be subtracted from 2000 to get a number exactly divisible by 17 ? | "on dividing 2000 by 17 , we get 11 as remainder . required number to be subtracted = 11 answer d" | a = 2000 / 17
b = math.floor(a)
c = b * 17
d = 2000 - c
|
a ) 18 , b ) 20 , c ) 22 , d ) 23 , e ) 24 | c | divide(120, multiply(7, 1)) | how many 7 in between 1 to 120 ? | "7 , 17,27 , 37,47 , 57,67 , 70,71 , 72,73 , 74,75 , 76,77 ( two 7 ' s ) , 78,79 , 87,97 , 107,117 22 7 ' s between 1 to 120 answer : c" | a = 7 * 1
b = 120 / a
|
a ) 1992 , b ) 1993 , c ) 1994 , d ) 1995 , e ) 1996 | c | add(1990, multiply(10, multiply(const_2, const_3))) | in 1990 the budgets for projects q and v were $ 620,000 and $ 780,000 , respectively . in each of the next 10 years , the budget for q was increased by $ 30,000 and the budget for v was decreased by $ 10,000 . in which year was the budget for q equal to the budget for v ? | "let the no of years it takes is x . 620 + 30 x = 780 - 10 x - - > 40 x = 160 and x = 4 . thus , it happens in 1994 . c ." | a = 2 * 3
b = 10 * a
c = 1990 + b
|
a ) $ 21 , b ) $ 35 , c ) $ 31 , d ) $ 41 , e ) it can not be determined | b | subtract(add(120, subtract(162.50, 120)), multiply(subtract(162.50, 120), 3)) | at an upscale fast - food restaurant , shin can buy 3 burgers , 7 shakes , and one cola for $ 120 . at the same place it would cost $ 162.50 for 4 burgers , 10 shakes , and one cola . how much would it cost for a meal of one burger , one shake , and one cola ? | "let ' s suppose that the price of a burger is bb , of a shake - ss and that of a cola is cc . we can then construct these equations : 3 b + 7 s + c = 120 4 b + 10 s + c = 162.5 subtracting the first equation from the second gives us b + 3 s = 42.5 now if we subtract the new equation two times from first or 3 times fro... | a = 162 - 50
b = 120 + a
c = 162 - 50
d = c * 3
e = b - d
|
a ) 63 , b ) 44 , c ) 58 , d ) 60 , e ) 62 | a | add(add(18, multiply(add(3, const_1), subtract(divide(18, 3), const_1))), multiply(divide(18, 3), add(3, const_1))) | a hiker walked for 3 days . she walked 18 miles on the first day , walking 3 miles per hour . on the second day she walked for one less hour but she walked one mile per hour , faster than on the first day . on the third day she walked the same number of hours as on the first day at 1 mile faster than 2 nd day for same ... | "first day - 18 miles with 3 miles per hours then total - 6 hours for that day second day - 4 miles per hour and 5 hours - 20 miles third day - 5 miles per hour and 5 hours - 25 miles total 18 + 20 + 25 = 63 answer : option a ." | a = 3 + 1
b = 18 / 3
c = b - 1
d = a * c
e = 18 + d
f = 18 / 3
g = 3 + 1
h = f * g
i = e + h
|
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