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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 * 2 / 3 = 4 / 3 hours . answer : c . | 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 β 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 . therefore factors that are multiple of 3 = 20 / 2 = 10 ans - d" | 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 ? | "β΅ amount , he have spent in 1 month on clothes transport = amount spent on saving per month β΅ amount , spent on clothes and transport = 46800 β 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 β 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
|
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 β 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 β ( 80 x / 100 Γ 90 y / 100 ) = ( xy β 18 / 25 xy ) = 7 / 25 xy decrease = ( 7 xy / 25 xy Γ 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
|
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 + odd = odd and nearest odd greater than x + 5 is x + 5 + 2 ] minimum value for z = y + 2 = x + 7 + 2 = x + 9 [ note : z = y + 2 because both z and y are odd . difference between two odd numbers is 2 ] s = z - x = x + 9 - x = 9 ans : d" | a = 5 + 2
b = a + 2
|
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
|
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 : profit r = selling price - original cost price = > 20 % of cp = 80 % of mp - 100 % cp = > mp = 120 cp / 80 = 3 / 2 ( cp ) now if product is sold without any discount , then , profit = selling price ( without discount ) - original cost price = market price - original cost price = mp - cp = 3 / 2 cp - cp = 1 / 2 cp = 50 % of cp thus , answer should bec ." | 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 Ξ΄ 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
|
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 the remaining 10 will be females who are married . total females = 70 married males = 20 then single males = 70 - 20 = 50 we need to find the fraction of male students who are single i . e single male students / total male student = 50 / 70 = 5 / 7 [ e ]" | 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
|
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
|
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 Γ’ β¬ β x days . the no . of hours required by s = 10 Γ£ β 20 / 20 Γ’ β¬ β 10 = 200 / 10 = 20 hours e | a = 15 - 10
b = 15 + a
|
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 β x ) = 203 8 x + 108 β 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
|
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 β s age multiplied by kelly β 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
|
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 returned in time to take a reading at 4 p . m . , by which point the population had grown to 62,500 . now he has to fill in the missing data for 2 p . m . and 3 p . m . what was the population at 3 p . m . ? | 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
|
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
|
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
|
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
|
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
|
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
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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
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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 or x must the largest element so that x - 3 = 12 - - > x = 15 . the the difference between the greatest possible value of x and least possible value of x is 15 - 2 = 13 . answer : d . | 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 Γ 8 Γ 10 then , 12 8 Γ 10 = ? Γ 15 Γ 8 ? ( i . e . no . of men required ) = 12 Γ 8 Γ 10 / 15 Γ 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 Γ 10 ^ x is less than 220,000 , what is the greatest possible value for x ? | "x is an integer and 2.134 Γ 10 x is less than 220,000 , what is the greatest possible value for x ? for 2.134 Γ 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 = β 400 = 20 km distance covered by the horse = perimeter of square field = 20 Γ 4 = 80 km β΄ 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 Γ 10 = 120 β΄ number of mangoes in 36 boxes = 36 Γ 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
|
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 Γ 0.76 Γ 0.76 β 0.008 ) / ( 0.76 Γ 0.76 + 0.76 Γ 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), 5)))) | 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 ) = 2 + 3 + 5 + 7 + 11 + 13 = 41 . so the difference between 41 and 55 is 14 . so the answer is ( d ) ." | 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 β ? = 3854.002 | explanation let 3889 + 12.952 β x = 3854.002 . then x = ( 3889 + 12.952 ) β 3854.002 = 3901.952 β 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 they each purchase pizza alone ? | 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 large pizzas . in terms of square inches that would be 3 * 196 = 588 sq inches . hence , the difference is 34 square inches more ( 588 - 554 ) . the correct answer is e | 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 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 ? | "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 = 100 i . e . a + d = 50 d = 20 i . e . c = 70 - 20 = 50 i . e . b = 100 - 50 = 50 i . e . a = 80 - 50 = 30 b = 80 answer : option d" | 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 β 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 the exponents of both 2 and 7 are not even numbers . this tells us that 3,150 itself is not a perfect square . ) we also are given that 3,150 multiplied by f is the square of an integer . we can write this as : 2 ^ 1 x 3 ^ 2 x 5 ^ 2 x 7 ^ 1 x f = square of an integer according to our rule , we need all unique prime factors ' exponents to be even numbers . thus , we need one more 2 and one more 7 . therefore , y = 7 x 2 = 14 answer is e . | 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 that had purchased clothes for less than $ 100 reported they were overall satisfied with their purchase . what percent of the customers surveyed purchased clothes for at least $ 100 and reported that they were not overall satisfied with their purchase ? | 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
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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 Γ 30 ) work done by 15 men in 1 day = 1 / 15 work done by 1 man in 1 day = 1 / ( 15 Γ 15 ) ratio of the capacity of a man and woman = 1 / ( 15 Γ 15 ) : 1 / ( 30 Γ 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 Γ 100 / 360 = 35 answer b" | a = 126 * 100
b = a / 360
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['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 % = [ Ο x ( 5 ) 2 / 2 Ο x 5 x 100 ] % βΉ = βΊ 250 % . answer d | a = circle_area / (
b = 100 * a
|
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 meters per minute ? | total distance from 10 to 90 = 80 + from 90 to 70 = 20 so the dist is 100 Γ 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
|
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 - 4 considering p = - 2 and t = 4 , the least slope is ( - 8 - 5 ) / ( 4 - 2 ) = - 6.5 imo option a is correct answer ." | 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 it contain ? | "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 from second we will get b + s + c = 35 . in any case , there is no necessity to know each item ' s price , just the sum . answer : b" | 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 hours . how many miles in total did she walk ? | "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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