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
|
|---|---|---|---|---|---|
a ) a ) 4 , b ) b ) 5 , c ) c ) 6 , d ) d ) 3 , e ) e ) 7 | d | sqrt(51) | from below option 51 is divisible by which one ? | "51 / 3 = 17 d" | a = math.sqrt(51)
|
a ) 16 , b ) 12 , c ) 18 , d ) 22 , e ) 08 | d | multiply(2, 11) | each child has 2 pencils and 13 skittles . if there are 11 children , how many pencils are there in total ? | 2 * 11 = 22 . answer is d . | a = 2 * 11
|
a ) 104 , b ) 55 , c ) 66 , d ) 98 , e ) 100 | d | divide(390, 4) | . a car covers a distance of 390 km in 4 hours . find its speed ? | 390 / 4 = 98 kmph answer : d | a = 390 / 4
|
a ) 61376 , b ) 54411 , c ) 612314 , d ) 64170 , e ) 64171 | a | multiply(add(add(const_100, const_4), subtract(multiply(const_100, const_10), 8)), divide(add(divide(subtract(subtract(multiply(const_100, const_10), 8), add(const_100, const_4)), 8), const_1), const_2)) | find the sum of all 3 digit natural numbers , which are divisible by 8 | "the three digit natural numbers divisible by 8 are 104 , 112 , 120 , β¦ . 992 . let sndenote their sum . that is , sn = 104 112 120 128 , 992 g + + + + + . now , the sequence 104 , 112 , 120 , g , 992 forms an a . p . a = 104 , d = 8 , l = 992 n = l - a / d n = 112 s 112 = n / 2 ( a + l ) = 61376 answer a 61376" | a = 100 + 4
b = 100 * 10
c = b - 8
d = a + c
e = 100 * 10
f = e - 8
g = 100 + 4
h = f - g
i = h / 8
j = i + 1
k = j / 2
l = d * k
|
a ) 55 , b ) 82 , c ) 73 , d ) 75 , e ) 85 | d | divide(add(multiply(25, subtract(25, 20)), multiply(25, subtract(25, 30))), add(25, 25)) | a man buys 25 lts of liquid which contains 20 % of the liquid and the rest is water . he then mixes it with 25 lts of another mixture with 30 % of liquid . what is the % of water in the new mixture ? | "20 % in 25 lts is 5 . so water = 25 - 5 = 20 lts . 30 % of 25 lts = 7.5 . so water in 2 nd mixture = 25 - 7.5 = 17.5 lts . now total quantity = 25 + 25 = 50 lts . total water in it will be 20 + 17.5 = 37.5 lts . % of water = ( 100 * 37.5 ) / 50 = 75 . answer : d" | a = 25 - 20
b = 25 * a
c = 25 - 30
d = 25 * c
e = b + d
f = 25 + 25
g = e / f
|
a ) 235 miles . , b ) 245 miles . , c ) 255 miles . , d ) 265 miles . , e ) 275 miles . | c | add(multiply(45, 2), multiply(3, 55)) | john left home and drove at the rate of 45 mph for 2 hours . he stopped for lunch then drove for another 3 hours at the rate of 55 mph to reach his destination . how many miles did john drive ? | "the total distance d traveled by john is given by d = 45 * 2 + 3 * 55 = 255 miles . answer c" | a = 45 * 2
b = 3 * 55
c = a + b
|
a ) 8 , b ) 10 , c ) 15 , d ) 6 , e ) 19 | d | subtract(30, divide(add(multiply(7.50, 30), 555), add(7.50, 25))) | a contractor is engaged for 30 days on the condition thathe receives rs . 25 for each day he works & is fined rs . 7.50 for each day is absent . he gets rs . 555 in all . for how many days was he absent ? | "30 * 25 = 750 455 - - - - - - - - - - - 195 25 + 7.50 = 32.5 195 / 32.5 = 6 d" | a = 7 * 50
b = a + 555
c = 7 + 50
d = b / c
e = 30 - d
|
a ) 23 days , b ) 37 days , c ) 37 Β½ days , d ) 40 days , e ) 41 days | c | inverse(subtract(inverse(multiply(3, 5)), inverse(multiply(divide(5, 4), 20)))) | a does 4 / 5 th of a work in 20 days . he then calls in b and they together finish the remaining work in 3 days . how long b alone would take to do the whole work ? | explanation : a can finish the whole work in 20 Γ 5 / 4 days = 25 days a and b together finish the whole work in 5 Γ 3 days = 15 days therefore , b can finish the whole work in 25 b / 25 + b = 15 25 b = 15 ( 25 + b ) = 375 + 15 b 10 b = 375 and b = 375 / 10 = 37 Β½ days . answer : option c | a = 3 * 5
b = 1/(a)
c = 5 / 4
d = c * 20
e = 1/(d)
f = b - e
g = 1/(f)
|
a ) 11 , b ) 17 , c ) 10 , d ) 17 , e ) 18 | e | subtract(add(inverse(add(inverse(36), inverse(12))), 12), const_3) | a and b can do a work in 12 days and 36 days respectively . if they work on alternate days beginning with b , in how many days will the work be completed ? | the work done in the first two days = 1 / 12 + 1 / 36 = 1 / 9 so , 9 such two days are required to finish the work . i . e . , 18 days are required to finish the work . answer : e | a = 1/(36)
b = 1/(12)
c = a + b
d = 1/(c)
e = d + 12
f = e - 3
|
a ) 270 m , b ) 245 m , c ) 235 m , d ) 220 m , e ) 240 m | c | subtract(multiply(multiply(45, const_0_2778), 30), 140) | a train , 140 meters long travels at a speed of 45 km / hr crosses a bridge in 30 seconds . the length of the bridge is | "explanation : assume the length of the bridge = x meter total distance covered = 140 + x meter total time taken = 30 s speed = total distance covered / total time taken = ( 140 + x ) / 30 m / s = > 45 Γ£ β ( 10 / 36 ) = ( 140 + x ) / 30 = > 45 Γ£ β 10 Γ£ β 30 / 36 = 140 + x = > 45 Γ£ β 10 Γ£ β 10 / 12 = 140 + x = > 15 Γ£ β 10 Γ£ β 10 / 4 = 140 + x = > 15 Γ£ β 25 = 140 + x = 375 = > x = 375 - 140 = 235 answer : option c" | a = 45 * const_0_2778
b = a * 30
c = b - 140
|
a ) 2 , b ) 4 , c ) 6 , d ) 8 , e ) 10 | c | divide(add(multiply(18, 12), 18), 9) | if 12 : 18 : : x : 9 , then find the value of x | explanation : treat 12 : 18 as 12 / 18 and x : 9 as x / 9 , treat : : as = so we get 12 / 18 = x / 9 = > 18 x = 108 = > x = 6 option c | a = 18 * 12
b = a + 18
c = b / 9
|
a ) 30 days , b ) 65 days , c ) 86 days , d ) 45 days , e ) 17 days | a | divide(10, subtract(const_1, divide(add(10, 10), 30))) | a can do a piece of work in 30 days ; b can do the same in 30 days . a started alone but left the work after 10 days , then b worked at it for 10 days . c finished the remaining work in 10 days . c alone can do the whole work in ? | 10 / 30 + 10 / 30 + 10 / x = 1 x = 30 days answer : a | a = 10 + 10
b = a / 30
c = 1 - b
d = 10 / c
|
a ) 22 , b ) 27 , c ) 236 , d ) 90 , e ) 81 | d | add(subtract(subtract(divide(subtract(300, 40), add(2, 2)), 10), 10), subtract(subtract(divide(subtract(300, 40), add(2, 2)), 10), 10)) | the distance between towns a and b is 300 km . one train departs from town a and another train departs from town b , both leaving at the same moment of time and heading towards each other . we know that one of them is 10 km / hr faster than the other . find the speeds of both trains if 2 hours after their departure the distance between them is 40 km . | let the speed of the slower train be xx km / hr . then the speed of the faster train is ( x + 10 ) ( x + 10 ) km / hr . in 2 hours they cover 2 x 2 x km and 2 ( x + 10 ) 2 ( x + 10 ) km , respectively . therefore if they did n ' t meet yet , the whole distance from a to b is 2 x + 2 ( x + 10 ) + 40 = 4 x + 602 x + 2 ( x + 10 ) + 40 = 4 x + 60 km . however , if they already met and continued to move , the distance would be 2 x + 2 ( x + 10 ) β 40 = 4 x β 202 x + 2 ( x + 10 ) β 40 = 4 x β 20 km . so we get the following equations : 4 x + 60 = 3004 x + 60 = 300 4 x = 2404 x = 240 x = 60 x = 60 or 4 x β 20 = 3004 x β 20 = 300 4 x = 3204 x = 320 x = 80 x = 80 hence the speed of the slower train is 60 km / hr or 80 km / hr and the speed of the faster train is 70 km / hr or 90 km / hr . answer : d | a = 300 - 40
b = 2 + 2
c = a / b
d = c - 10
e = d - 10
f = 300 - 40
g = 2 + 2
h = f / g
i = h - 10
j = i - 10
k = e + j
|
a ) 12 , b ) 15 , c ) 10 , d ) 20 , e ) 14 | a | divide(subtract(sqrt(add(multiply(multiply(16, 2), const_4), power(16, const_2))), 16), const_2) | tom read a book containing 480 pages by reading the same number of pages each day . if he would have finished the book 2 days earlier by reading 16 pages a day more , how many days did tom spend reading the book ? | "actually u can set up 2 equation p - - stands for the pages d - - stands for the days 1 ) p * d = 480 ( we want to find the days , sop = 480 / d ) 2 ) ( p + 16 ) ( d - 2 ) = 480 = > pd - 2 p + 16 d - 32 = 480 as the 1 ) stated u can put 1 ) into 2 ) = > 480 - 2 p + 16 d - 32 = 480 = > 16 d - 2 p = 32 put the bold one into it = > 16 d - 2 ( 480 / d ) = 80 the we get the final equation 16 d ^ 2 - 960 = 80 d ( divide 16 ) = > d ^ 2 - 5 d - 60 = 0 ( d - 12 ) ( d + 5 ) = 0 so d = 12 days ans : a" | a = 16 * 2
b = a * 4
c = 16 ** 2
d = b + c
e = math.sqrt(d)
f = e - 16
g = f / 2
|
a ) 13000 , b ) 7000 , c ) 10000 , d ) 5000 , e ) none of these | d | divide(subtract(75000, 50000), add(const_2, const_3)) | a textile manufacturing firm employees 70 looms . it makes fabrics for a branded company . the aggregate sales value of the output of the 70 looms is rs 00000 and the monthly manufacturing expenses is rs 1 , 50000 . assume that each loom contributes equally to the sales and manufacturing expenses are evenly spread over the number of looms . monthly establishment charges are rs 75000 . if one loom breaks down and remains idle for one month , the decrease in profit is : | explanation : profit = 5 , 00,000 Γ’ Λ β ( 1 , 50,000 + 75,000 ) = rs . 2 , 75,000 . since , such loom contributes equally to sales and manufacturing expenses . but the monthly charges are fixed at rs 75,000 . if one loan breaks down sales and expenses will decrease . new profit : - = > 500000 Γ£ β ( 69 / 70 ) Γ’ Λ β 150000 Γ£ β ( 69 / 70 ) Γ’ Λ β 75000 . = > rs 2 , 70,000 . decrease in profit = > 2 , 75,000 Γ’ Λ β 2 , 70,000 = > rs . 5,000 . answer : d | a = 75000 - 50000
b = 2 + 3
c = a / b
|
a ) rs . 11.81 , b ) rs . 12 , c ) rs . 18.94 , d ) rs . 12.31 , e ) none | c | divide(multiply(14, add(const_100, 15)), subtract(const_100, 15)) | a fruit seller sells mangoes at the rate of rs . 14 per kg and thereby loses 15 % . at what price per kg , he should have sold them to make a profit of 15 % ? | "solution 85 : 14 = 115 : x x = ( 14 Γ£ β 115 / 85 ) = rs . 18.94 hence , s . p per kg = rs . 18.94 answer c" | a = 100 + 15
b = 14 * a
c = 100 - 15
d = b / c
|
a ) 1 , b ) 2 , c ) 3 , d ) 9 , e ) 27 | a | subtract(3, 2) | for a 3 - digit number xyz , where x , y , and z are the digits of the number , f ( xyz ) = 5 ^ x 2 ^ y 3 ^ z . if f ( abc ) = 3 * f ( def ) , what is the value of abc - def ? | since f ( abc ) = 3 * f ( def ) , i would assume that f = c - 1 from the function above . the answer should be ( a ) | a = 3 - 2
|
a ) 33 , b ) 44 , c ) 55 , d ) 77 , e ) 22 | c | divide(add(165, 660), multiply(54, const_0_2778)) | how long does a train 165 meters long running at the rate of 54 kmph take to cross a bridge 660 meters in length | "t = ( 660 + 165 ) / 54 * 18 / 5 t = 55 answer : c" | a = 165 + 660
b = 54 * const_0_2778
c = a / b
|
a ) 6 hours , b ) 5 hours , c ) 7 hours , d ) 8 hours , e ) none | c | divide(add(392, 70), add(divide(392, 8), 18)) | a truck covers a distance of 392 km at a certain speed in 8 hours . how much time would a car take at an average speed which is 18 kmph more than that of the speed of the truck to cover a distance which is 70 km more than that travelled by the truck ? | "explanation : speed of the truck = distance / time = 392 / 8 = 49 kmph now , speed of car = ( speed of truck + 18 ) kmph = ( 48 + 18 ) = 66 kmph distance travelled by car = 392 + 70 = 462 km time taken by car = distance / speed = 462 / 66 = 7 hours . answer β c" | a = 392 + 70
b = 392 / 8
c = b + 18
d = a / c
|
a ) 145646 , b ) 236578 , c ) 645353 , d ) 456546 , e ) 220070 | e | add(multiply(multiply(add(555, 445), 2), subtract(555, 445)), 70) | a no . when divided by the sum of 555 and 445 gives 2 times their difference as quotient & 70 as remainder . find the no . is ? | "( 555 + 445 ) * 2 * 110 + 70 = 220000 + 70 = 220070 e" | a = 555 + 445
b = a * 2
c = 555 - 445
d = b * c
e = d + 70
|
a ) 600 , b ) 480 , c ) 750 , d ) 650 , e ) 560 | e | subtract(subtract(multiply(45, 10), multiply(22, 20)), multiply(22, 15)) | the average of 45 results is 10 . the average of first 22 of them is 15 and that of last 22 is 20 . find the 23 result ? | "23 th result = sum of 45 results - sum of 44 results 10 * 45 - 15 * 22 + 20 * 22 = 450 - 330 + 440 = 560 answer is e" | a = 45 * 10
b = 22 * 20
c = a - b
d = 22 * 15
e = c - d
|
a ) $ 880 , b ) $ 990 , c ) $ 1,000 , d ) $ 1,100 , e ) $ 1,260 | e | subtract(multiply(140, divide(const_100, 10)), 140) | if a 10 percent deposit that has been paid toward the purchase of a certain product is $ 140 , how much more remains to be paid ? | "10 / 100 p = 140 > > p = 140 * 100 / 10 = 1400 1400 - 140 = 1260 answer : e" | a = 100 / 10
b = 140 * a
c = b - 140
|
a ) - $ 56 , b ) - $ 6 , c ) $ 0 , d ) $ 6 , e ) $ 8 | e | subtract(multiply(add(4, 4), 21), multiply(add(7, 21), divide(const_1, add(divide(const_1, add(4, 4)), divide(const_1, 20))))) | a professional janitor can clean a certain high school in ( 4 + 4 ) hours , working at a constant rate . a student sentenced to detention can clean that same high school in 20 hours , also working at a constant rate . if the student is paid $ 7 total per hour and the janitor is paid $ 21 per hour , how much more would it cost the school to pay the janitor to do the job himself than it would to pay the student and the janitor to do the job together ? | a professional janitor can clean a certain high school in ( 4 + 4 ) or 8 hours so ( applying rule # 1 ) , the janitor can clean 1 / 8 of the school in one hour a student sentenced to detention can clean that same high school in 20 hours so ( applying rule # 1 ) , the student can clean 1 / 20 of the school in one hour so , combined , the student and janitor can clean ( 1 / 8 + 1 / 20 ) of the school in one hour 1 / 8 + 1 / 20 = 5 / 40 + 2 / 40 = 7 / 40 so , in one hour they can clean 7 / 40 of the school . applying rule # 2 , it will takethem 40 / 7 hoursto clean the entire school . the janitor earns $ 21 / hour and the student earns $ 7 / hour , so their combined rate is $ 28 / hour . theircombined wages = ( pay rate ) ( time ) = ( $ 28 / hour ) ( 40 / 7 hours ) = $ 160 working alone , the janitor takes 8 hours and earns $ 21 / hour so , working alone , the janitor ' s earnings = ( pay rate ) ( time ) = ( $ 21 / hour ) ( 8 hours ) = $ 168 $ 168 - $ 160 = $ 8 , so the answer is e | a = 4 + 4
b = a * 21
c = 7 + 21
d = 4 + 4
e = 1 / d
f = 1 / 20
g = e + f
h = 1 / g
i = c * h
j = b - i
|
a ) 1 / 2 , b ) 1 / 10 , c ) 1 / 18 , d ) 1 / 16 , e ) 1 / 11 | a | divide(add(multiply(divide(subtract(8, 2), subtract(6, 3)), 2), 3), add(multiply(6, divide(subtract(8, 2), subtract(6, 3))), 2)) | 3 men and 8 women complete a task in same time as 6 men and 2 women do . how much fraction of work will be finished in same time if 2 men and 3 women will do that task . | "3 m + 8 w = 6 m + 2 w 3 m = 6 w 1 m = 2 w therefore 3 m + 8 w = 14 w 2 m + 3 w = 7 w answer is 7 / 14 = 1 / 2 answer : a" | a = 8 - 2
b = 6 - 3
c = a / b
d = c * 2
e = d + 3
f = 8 - 2
g = 6 - 3
h = f / g
i = 6 * h
j = i + 2
k = e / j
|
a ) 105 , b ) 67 , c ) 80 , d ) 60 , e ) 100 | b | multiply(60, add(const_1, divide(12, const_100))) | a light has a rating of 60 watts , it is replaced with a new light that has 12 % higher wattage . how many watts does the new light have ? | "final number = initial number + 12 % ( original number ) = 60 + 12 % ( 60 ) = 60 + 7 = 67 answer b" | a = 12 / 100
b = 1 + a
c = 60 * b
|
a ) 50 / 9 , b ) 50 / 13 , c ) 50 / 17 , d ) 25 / 9 , e ) 5 5 / 1 | d | multiply(divide(subtract(divide(7, 6), const_1), 6), const_100) | a sum of money becomes 7 / 6 of itself in 6 years at a certain rate of simple interest . the rate per annum is ? | "let sum = x . then , amount = 7 x / 6 s . i . = 7 x / 6 - x = x / 6 ; time = 6 years . rate = ( 100 * x ) / ( x * 6 * 6 ) = 25 / 9 % . answer : d" | a = 7 / 6
b = a - 1
c = b / 6
d = c * 100
|
a ) 50 , b ) 45 , c ) 150 , d ) 450 , e ) 500 | a | divide(0.15, divide(0.3, const_100)) | find the missing figures : 0.3 % of ? = 0.15 | let 0.3 % of x = 0.15 . then , 0.30 * x / 100 = 0.15 x = [ ( 0.15 * 100 ) / 0.3 ] = 50 . answer is a . | a = 0 / 3
b = 0 / 15
|
a ) rs . 435 , b ) rs . 350 , c ) rs . 275 , d ) rs . 425 , e ) none of these | a | divide(subtract(multiply(30, 350), multiply(15, 265)), 15) | the mean daily profit made by a shopkeeper in a month of 30 days was rs . 350 . if the mean profit for the first fifteen days was rs . 265 , then the mean profit for the last 15 days would be | average would be : 350 = ( 265 + x ) / 2 on solving , x = 435 . answer : a | a = 30 * 350
b = 15 * 265
c = a - b
d = c / 15
|
a ) 26 , b ) 28 , c ) 30 , d ) 32 , e ) 34 | c | add(divide(subtract(multiply(multiply(5, 2), 3), multiply(5, 2)), 2), subtract(multiply(multiply(5, 2), 3), multiply(5, 2))) | the ratio of the number of females to males at a party was 1 : 2 but when 5 females and 5 males left , the ratio became 1 : 3 . how many people were at the party originally ? | "the total number of people are x females + 2 x males . 3 * ( x - 5 ) = 2 x - 5 x = 10 there were 3 x = 30 people at the party originally . the answer is c ." | a = 5 * 2
b = a * 3
c = 5 * 2
d = b - c
e = d / 2
f = 5 * 2
g = f * 3
h = 5 * 2
i = g - h
j = e + i
|
a ) 427 , b ) 859 , c ) 869 , d ) 856 , e ) none of these | d | subtract(lcm(lcm(lcm(24, 32), 36), 54), 8) | the least number which when increased by 8 each divisible by each one of 24 , 32 , 36 and 54 is : | "solution required number = ( l . c . m . of 24 , 32 , 36 , 54 ) - 8 = 864 - 8 = 856 . answer d" | a = math.lcm(24, 32)
b = math.lcm(a, 36)
c = math.lcm(b, 54)
d = c - 8
|
['a ) 15840', 'b ) 9280', 'c ) 2667', 'd ) 8766', 'e ) 66711'] | b | multiply(square_perimeter(square_edge_by_area(53824)), 10) | find the length of the wire required to go 10 times round a square field containing 53824 m 2 . | a 2 = 53824 = > a = 232 4 a = 928 928 * 10 = 9280 answer : b | a = square_perimeter * (
|
a ) rs . 500 , b ) rs . 600 , c ) rs . 650 , d ) rs . 720 , e ) none | e | multiply(multiply(100, divide(add(add(multiply(const_10, const_1000), multiply(const_4, const_1000)), multiply(const_4, 100)), add(100, multiply(100, divide(20, 100))))), divide(4, 100)) | a man invested rs . 14,400 in rs . 100 shares of a company at 20 % premium . if the company declares 4 % dividend at the end of the year , then how much does he get ? | "solution number of shares = ( 14400 / 120 ) = 120 . face value = rs . ( 100 x 120 ) = rs . 12000 . annual income = rs . ( 4 / 100 x 12000 ) = rs . 480 . answer e" | a = 10 * 1000
b = 4 * 1000
c = a + b
d = 4 * 100
e = c + d
f = 20 / 100
g = 100 * f
h = 100 + g
i = e / h
j = 100 * i
k = 4 / 100
l = j * k
|
a ) 24 , b ) 25 , c ) 26 , d ) 27 , e ) 28 | d | subtract(multiply(add(add(add(add(const_10, const_10), add(const_10, const_10)), add(const_10, const_10)), add(const_4, const_3)), add(add(add(add(const_10, const_10), add(const_10, const_10)), add(const_10, const_10)), add(const_4, const_3))), 4329) | what should be added to 4329 so that it may become a perfect square ? | "66 x 66 = 4356 4356 - 4329 = 27 if added to 27 get perfect square answer = d" | a = 10 + 10
b = 10 + 10
c = a + b
d = 10 + 10
e = c + d
f = 4 + 3
g = e + f
h = 10 + 10
i = 10 + 10
j = h + i
k = 10 + 10
l = j + k
m = 4 + 3
n = l + m
o = g * n
p = o - 4329
|
a ) 2220000 , b ) 2850000 , c ) 2121000 , d ) 1855000 , e ) none of these | e | divide(2664000, multiply(divide(add(20, const_100), const_100), divide(add(20, const_100), const_100))) | the income of a company increases 20 % per annum . if its income is 2664000 in the year 1999 what was its income in the year 1997 ? | let income in 1997 = x according to the question , income in 1998 = x + x β 5 = 6 x β 5 income in 1999 = 6 x β 5 + 6 x β 25 = 36 x β 25 but given , income in 1999 = 2664000 β΄ 36 x β 25 = 2664000 β x = 1850000 answer e | a = 20 + 100
b = a / 100
c = 20 + 100
d = c / 100
e = b * d
f = 2664000 / e
|
a ) 11 , b ) 17 , c ) 13 , d ) 20 , e ) none of these | d | subtract(add(11, 18), 9) | when 242 is divided by a certain divisor the remainder obtained is 11 . when 698 is divided by the same divisor the remainder obtained is 18 . however , when the sum of the two numbers 242 and 698 is divided by the divisor , the remainder obtained is 9 . what is the value of the divisor ? | "let that divisor be x since remainder is 11 or 18 it means divisor is greater than 18 . now 242 - 11 = 231 = kx ( k is an integer and 234 is divisble by x ) similarly 698 - 18 = 680 = lx ( l is an integer and 689 is divisible by x ) adding both 698 and 242 = ( 231 + 680 ) + 11 + 18 = x ( k + l ) + 29 when we divide this number by x then remainder will be equal to remainder of ( 29 divided by x ) = 9 hence x = 29 - 9 = 20 hence d" | a = 11 + 18
b = a - 9
|
a ) 220 , b ) 420 , c ) 250 , d ) 700 , e ) 500 | e | divide(multiply(divide(multiply(1100, divide(10, divide(20, 10))), add(const_1, divide(const_1, 10))), 10), multiply(10, 10)) | anil brought a scooter for a certain sum of money . he spent 10 % of the cost on repairs and sold the scooter for a profit of rs . 1100 . how much did he spend on repairs if he made a profit of 20 % ? | e c . p . be rs . x . then , 20 % of x = 1100 20 / 100 * x = 1100 = > x = 5500 c . p . = rs . 5500 , expenditure on repairs = 10 % actual price = rs . ( 100 * 5500 ) / 110 = rs . 5000 expenditures on repairs = ( 5500 - 5000 ) = rs . 500 . | a = 20 / 10
b = 10 / a
c = 1100 * b
d = 1 / 10
e = 1 + d
f = c / e
g = f * 10
h = 10 * 10
i = g / h
|
a ) 3 days , b ) 8 days , c ) 9 days , d ) 11 days , e ) 7 days | e | divide(divide(multiply(7, 7), divide(7, 9)), 9) | if 7 persons can do 7 times of a particular work in 7 days , then , 9 persons can do 9 times of that work in ? | that is , 1 person can do one time of the work in 7 days . therefore , 9 persons can do 9 times work in the same 7 days itself . option ' e ' | a = 7 * 7
b = 7 / 9
c = a / b
d = c / 9
|
a ) $ 4 , b ) $ 0.1 , c ) $ 1 , d ) $ 3 , e ) $ 1.65 | b | multiply(divide(add(multiply(divide(add(100, 100), 100), divide(3, const_2)), multiply(divide(3, const_2), divide(subtract(100, 90), 100))), 3.15), divide(subtract(100, 90), 100)) | the cost per pound of green tea and coffee were the same in june . in july , the price of coffee shot up by 100 % and that of green tea dropped by 90 % . if in july , a mixture containing equal quantities of green tea and coffee costs $ 3.15 for 3 lbs , how much did a pound of green tea cost in july ? | "lets assume price of coffee in june = 100 x price of green tea in june = 100 x price of coffee in july = 200 x ( because of 100 % increase in price ) price of green tea in july = 10 x ( because of 90 % decrease in price ) price of 1.5 pound of coffee 1.5 pound of green tea in july will be = 300 x + 15 x = 315 x as per question 315 x = 3.15 $ x = 0.01 s so the price of green tea in july = 10 x = 10 x 0.01 = 0.1 $ / pound answer b" | a = 100 + 100
b = a / 100
c = 3 / 2
d = b * c
e = 3 / 2
f = 100 - 90
g = f / 100
h = e * g
i = d + h
j = i / 3
k = 100 - 90
l = k / 100
m = j * l
|
a ) 1 / 25 , b ) 1 / 6 , c ) 1 / 5 , d ) 1 / 3 , e ) 6 | d | divide(divide(4, 6), 6) | if xy > 0 , 1 / x + 1 / y = 4 , and 1 / xy = 6 , then ( x + y ) / 2 = ? | "( 1 / x + 1 / y ) = 4 canbe solved as { ( x + y ) / xy } = 6 . substituting for 1 / xy = 6 , we get x + y = 4 / 6 = = > ( x + y ) / 2 = 4 / ( 6 * 2 ) = 1 / 3 . d" | a = 4 / 6
b = a / 6
|
a ) 0.12356 , b ) 1.2356 , c ) 12.356 , d ) 0.012356 , e ) 0.0012356 | b | divide(multiply(0.01, add(add(multiply(multiply(add(const_3, const_2), const_2), multiply(multiply(const_3, const_4), const_100)), multiply(multiply(add(const_3, const_4), add(const_3, const_2)), multiply(add(const_3, const_2), const_2))), add(const_3, const_3))), const_100) | what is 0.01 percent of 12,356 ? | "soln : - 0.01 % of 12,356 = 0 . 011000.01100 x 12,356 = 1100 β 1001100 β 100 x 12,356 = 12,356100 β 10012,356100 β 100 = 1.2356 answer : b" | a = 3 + 2
b = a * 2
c = 3 * 4
d = c * 100
e = b * d
f = 3 + 4
g = 3 + 2
h = f * g
i = 3 + 2
j = i * 2
k = h * j
l = e + k
m = 3 + 3
n = l + m
o = 0 * 1
p = o / 100
|
a ) 1 , b ) 2 , c ) 3 , d ) 4 , e ) 5 | a | subtract(subtract(multiply(2500, power(add(const_1, divide(4, const_100)), 2)), 2500), multiply(multiply(2500, divide(4, const_100)), 2)) | indu gave bindu rs . 2500 on compound interest for 2 years at 4 % per annum . how much loss would indu has suffered had she given it to bindu for 2 years at 2 % per annum simple interest ? | "2500 = d ( 100 / 2 ) 2 d = 1 answer : a" | a = 4 / 100
b = 1 + a
c = b ** 2
d = 2500 * c
e = d - 2500
f = 4 / 100
g = 2500 * f
h = g * 2
i = e - h
|
a ) 9 % , b ) 11 % , c ) 15 % , d ) 25 % , e ) 90 % | d | subtract(const_100, multiply(divide(9, 12), const_100)) | a case of 12 rolls of paper towels sells for $ 9 . the cost of one roll sold individually is $ 1 . what is the percent e of savings per roll for the 12 - roll package over the cost of 12 rolls purchased individually ? | "cost of 12 paper towels individually = 1 * 12 = 12 cost of a set of 12 paper towels = 9 cost of one roll = 9 / 12 = 3 / 4 = 0.75 savings per roll = 1 - . 75 = 0.25 % of savings is e = . 25 / 1 * 100 = 25 % d is the answer ." | a = 9 / 12
b = a * 100
c = 100 - b
|
a ) 12 , b ) 9 , c ) 3 , d ) 7.5 , e ) 2.5 | b | divide(multiply(12, 3), const_4.0) | for what values of k will the pair of equations 3 ( 3 x + 4 y ) = 36 and kx + 12 y = 30 does not have a unique solution ? | "we have 2 equations 1 . 3 ( 3 x + 4 y ) = 36 - - > 3 x + 4 y = 12 - - > 9 x + 12 y = 36 2 . kx + 12 y = 30 substract 1 - 2 , we get ( 9 - k ) x = 6 i . e . x = 6 / ( 9 - k ) then , by looking at options , we get some value of x except for b . when we put k = 9 , x becomes 6 / 0 and hence answer is b" | a = 12 * 3
b = a / 4
|
a ) a . 10 , b ) b . 12 , c ) c . 14 , d ) d . 26 , e ) e . 24 | d | add(add(add(add(add(add(add(add(add(add(add(add(const_1, add(3, 5)), const_1), const_1), const_1), const_1), 5), const_1), const_1), const_1), const_1), const_1), const_1) | working at constant rate , pump x pumped out half of the water in a flooded basement in 5 hours . the pump y was started and the two pumps , working independently at their respective constant rates , pumped out rest of the water in 3 hours . how many hours would it have taken pump y , operating alone at its own constant rate , to pump out all of the water that was pumped out of the basement ? | "rate of x = 1 / 8 rate of x + y = 1 / 6 rate of y = 1 / 6 - 1 / 8 = 1 / 24 26 hours d" | a = 3 + 5
b = 1 + a
c = b + 1
d = c + 1
e = d + 1
f = e + 1
g = f + 5
h = g + 1
i = h + 1
j = i + 1
k = j + 1
l = k + 1
m = l + 1
|
a ) 3003 , b ) 3027 , c ) 3024 , d ) 3021 , e ) 3018 | a | add(3000, 3) | there 3 kinds of books in the library physics , chemistry and biology . ratio of physics to chemistry is 3 to 2 ; ratio of chemistry to biology is 4 to 3 , and the total of the books is more than 3000 . which one of following can be the total r of the book ? | "first , you have to find the common ratio for all 3 books . you have : p : c : b 3 : 2 - - > multiply by 2 ( gives you row 3 ) 4 : 6 6 : 4 : 3 hence : p : c : b : t ( total ) r 6 : 4 : 3 : 13 - - - - > this means , the total number must be a multiple of 13 . answer a is correct since 299 is divisible by 13 , hence is 2990 and so is 3003 ( 2990 + 13 ) ." | a = 3000 + 3
|
a ) 5 , b ) 3 , c ) 4 , d ) 2 , e ) 6 | d | subtract(add(const_4, const_3), divide(divide(add(36, 14), const_2), 5)) | on rainy mornings , mo drinks exactly n cups of hot chocolate ( assume that n is an integer ) . on mornings that are not rainy , mo drinks exactly 5 cups of tea . last week mo drank a total of 36 cups of tea and hot chocolate together . if during that week mo drank 14 more tea cups than hot chocolate cups , then how many rainy days were there last week ? | "t = the number of cups of tea c = the number of cups of hot chocolate t + c = 36 t - c = 14 - > t = 25 . c = 11 . mo drinks 5 cups of tea a day then number of days that are not rainy = 25 / 5 = 5 so number of rainy days = 7 - 5 = 2 d is the answer ." | a = 4 + 3
b = 36 + 14
c = b / 2
d = c / 5
e = a - d
|
a ) 33.5 kg , b ) 37.25 kg , c ) 42.45 kg , d ) 55.12 kg , e ) 29.78 kg | a | divide(add(multiply(16, 20), multiply(14, 25)), add(16, 14)) | there are 2 sections a and b in a class , consisting of 16 and 14 students respectively . if the average weight of section a is 20 kg and that of section b is 25 kg , find the average of the whole class ? | "total weight of 36 + 44 students = 16 * 20 + 14 * 25 = 670 average weight of the class is = 670 / 20 = 33.5 kg answer is a" | a = 16 * 20
b = 14 * 25
c = a + b
d = 16 + 14
e = c / d
|
a ) 10 , b ) 20 , c ) 25 , d ) 30 , e ) 37.5 | b | divide(subtract(add(40, 60), 75), subtract(divide(add(40, 60), 40), divide(75, 60))) | a car traveled 75 % of the way from town a to town b at an average speed of 60 miles per hour . the car travels at an average speed of s miles per hour for the remaining part of the trip . the average speed for the entire trip was 40 miles per hour . what is s ? | total distance = 100 miles ( easier to work with % ) 75 % of the distance = 75 miles 25 % of the distance = 25 miles 1 st part of the trip β 75 / 60 = 1.25 2 nd part of the trip β 25 / s = t total trip β ( 75 + 25 ) / 40 = 1.25 + t Β» 100 / 40 = 1.25 + t Β» 2.5 = 1.25 + t Β» t = 1.25 back to 2 nd part of the trip formula : 25 / s = 1.25 Β» s = 20 ans b | a = 40 + 60
b = a - 75
c = 40 + 60
d = c / 40
e = 75 / 60
f = d - e
g = b / f
|
a ) 70.9 , b ) 75 , c ) 48 , d ) 65 , e ) 67.5 | a | multiply(const_100, divide(subtract(const_100, subtract(subtract(const_100, 35), multiply(subtract(const_100, 35), divide(10, const_100)))), subtract(subtract(const_100, 35), multiply(subtract(const_100, 35), divide(10, const_100))))) | the price of a jacket is reduced by 35 % . during a special sale the price of the jacket is reduced another 10 % . by approximately what percent must the price of the jacket now be increased in order to restore it to its original amount ? | "1 ) let the price of jacket initially be $ 100 . 2 ) then it is decreased by 35 % , therefore bringing down the price to $ 65 . 3 ) again it is further discounted by 10 % , therefore bringing down the price to $ 58.5 4 ) now 58.5 has to be added byx % in order to equal the original price . 58.5 + ( x % ) 58.5 = 100 . solving this eq for x , we get x = 70.9 ans is a ." | a = 100 - 35
b = 100 - 35
c = 10 / 100
d = b * c
e = a - d
f = 100 - e
g = 100 - 35
h = 100 - 35
i = 10 / 100
j = h * i
k = g - j
l = f / k
m = 100 * l
|
a ) 10 , b ) 25 , c ) 40 , d ) 65 , e ) 90 | a | divide(add(multiply(13, 5), multiply(13, 4)), add(5, 4)) | the number of stamps that p and q had were in the ratio of 7 : 3 respectively . after p gave q 13 stamps , the ratio of the number of p ' s stamps to the number of q ' s stamps was 5 : 4 . as a result of the gift , p had how many more stamps than q ? | "p started with 7 k stamps and q started with 3 k stamps . ( 7 k - 13 ) / ( 3 k + 13 ) = 5 / 4 28 k - 15 k = 117 k = 9 p has 7 ( 9 ) - 13 = 50 stamps and q has 3 ( 9 ) + 13 = 40 stamps . the answer is a ." | a = 13 * 5
b = 13 * 4
c = a + b
d = 5 + 4
e = c / d
|
a ) 14 , b ) 25 , c ) 63 , d ) 84 , e ) 252 | b | add(10, const_1) | if x and y are positive integers and 25 x = 10 y what is the least possible value of xy ? | "25 x = 10 y = > x / y = 2 / 5 = > 5 x = 2 y 5 ( 3 ) = 2 ( 3 ) = > x * y = 9 but it is not given 5 ( 5 ) = 2 ( 5 ) = > x * y = 25 b" | a = 10 + 1
|
a ) 1 , b ) 2 , c ) 3 , d ) 4 , e ) 5 | c | subtract(multiply(divide(27, add(const_1, divide(80, 100))), const_2), 27) | a retailer bought a shirt at wholesale and marked it up 80 % to its initial price of $ 27 . by how many more dollars does he need to increase the price to achieve a 100 % markup ? | "let x be the wholesale price . then 1.8 x = 27 and x = 27 / 1.8 = 15 . to achieve a 100 % markup , the price needs to be $ 30 . the retailer needs to increase the price by $ 3 more . the answer is c ." | a = 80 / 100
b = 1 + a
c = 27 / b
d = c * 2
e = d - 27
|
a ) 196 : 121 , b ) 81 : 127 , c ) 181 : 196 , d ) 81 : 161 , e ) 81 : 182 | a | power(divide(2744, 1331), divide(const_1, const_3)) | the ratio of the volumes of two cubes is 2744 : 1331 . what is the ratio of their total surface areas ? | "explanation : ratio of the sides = Β³ β 2744 : Β³ β 1331 = 14 : 11 ratio of surface areas = 14 ^ 2 : 11 ^ 2 = 196 : 121 answer : option a" | a = 2744 / 1331
b = 1 / 3
c = a ** b
|
a ) 224 , b ) 242 , c ) 252 , d ) 262 , e ) 282 | a | divide(28, divide(350, 28)) | evaluate 28 % of 350 + 45 % of 280 | "explanation : = ( 28 / 100 ) * 350 + ( 45 / 100 ) * 280 = 98 + 126 = 224 answer : option a" | a = 350 / 28
b = 28 / a
|
a ) 12000 , 20000 , b ) 9600 , 22400 , c ) 12000 , 20007 , d ) 12000 , 20006 , e ) 12000 , 20001 | b | multiply(subtract(7, const_2), divide(32000, add(3, subtract(7, const_2)))) | divide rs . 32000 in the ratio 3 : 7 ? | 3 / 10 * 32000 = 9600 7 / 8 * 32000 = 22400 answer : b | a = 7 - 2
b = 7 - 2
c = 3 + b
d = 32000 / c
e = a * d
|
a ) 2 second , b ) 4 second , c ) 6 second , d ) 8 second , e ) 10 second | c | divide(110, multiply(add(60, 6), const_0_2778)) | a train is running with a speed of 60 kmph and its length is 110 metres . calculate the time by which it will pass a man running opposite with speed of 6 kmph | explanation : from the given question , we will first calculate the speed of train relative to man , = > ( 60 + 6 ) = 66 km / hr ( we added 6 because man is running opposite ) convert it in metre / second = 66 Γ£ β 5 / 18 = 55 / 3 m / sec time it will take to pass man = 110 Γ£ β 3 / 55 = 6 seconds answer is c | a = 60 + 6
b = a * const_0_2778
c = 110 / b
|
a ) 71 , b ) 44 , c ) 54 , d ) 16 , e ) 18 | c | subtract(66, multiply(4, const_3)) | the total of the ages of amar , akbar and anthony is 66 years . what was the total of their ages 4 years ago ? | explanation : required sum = ( 66 - 3 x 4 ) years = ( 66 - 12 ) years = 54 years . answer : c | a = 4 * 3
b = 66 - a
|
a ) 2 / 5 , b ) 3 / 5 , c ) 8 / 15 , d ) 1 / 2 , e ) 7 / 5 | b | subtract(const_1, add(add(divide(const_3, multiply(add(const_2, const_3), 3)), divide(const_2, multiply(add(const_2, const_3), 3))), divide(const_1, multiply(add(const_2, const_3), 3)))) | kim finds a 3 - meter tree branch and marks it off in thirds and fifths . she then breaks the branch along all the markings and removes one piece of every distinct length . what fraction of the original branch remains ? | 3 pieces of 1 / 5 length and two piece each of 1 / 15 and 2 / 15 lengths . removing one piece each from pieces of each kind of lengths the all that will remain will be 2 pieces of 1 / 5 i . e 2 / 5 , 1 piece of 1 / 15 , and 1 piece of 2 / 15 which gives us 2 / 5 + 1 / 15 + 2 / 15 - - - - - > 3 / 5 answer is b | a = 2 + 3
b = a * 3
c = 3 / b
d = 2 + 3
e = d * 3
f = 2 / e
g = c + f
h = 2 + 3
i = h * 3
j = 1 / i
k = g + j
l = 1 - k
|
a ) 1 / 5 , b ) 1 / 3 , c ) 1 / 2 , d ) 2 / 3 , e ) 3 / 5 | e | subtract(1, divide(divide(8, 12), add(divide(8, 12), 1))) | a certain country is divided into 8 provinces . each province consists entirely of progressives and traditionalists . if each province contains the same number of traditionalists and the number of traditionalists in any given province is 1 / 12 the total number of progressives in the entire country , what fraction of the country is traditionalist ? | "let p be the number of progressives in the country as a whole . in each province , the number of traditionalists is p / 12 the total number of traditionalists is 8 p / 12 = 2 p / 3 . the total population is p + 2 p / 3 = 5 p / 3 p / ( 5 p / 3 ) = 3 / 5 the answer is e ." | a = 8 / 12
b = 8 / 12
c = b + 1
d = a / c
e = 1 - d
|
a ) 59 , b ) 49 , c ) 58 , d ) 113 , e ) 131 | b | subtract(multiply(const_10, 6), const_10) | n and m are each 3 - digit integers . each of the numbers 1 , 2 , 3 , 45 and 6 is a digit of either n or m . what is the smallest possible positive difference between n and m ? | you have 6 digits : 12 , 3 , 4 , 5 , 6 each digit needs to be used to make two 3 digit numbers . this means that we will use each of the digits only once and in only one of the numbers . the numbers need to be as close to each other as possible . the numbers can not be equal so the greater number needs to be as small as possible and the smaller number needs to be as large as possible to be close to each other . the first digit ( hundreds digit ) of both numbers should be consecutive integers now let ' s think about the next digit ( the tens digit ) . to minimize the difference between the numbers , the tens digit of the greater number should be as small as possible and the tens digit of the smaller number should be as large as possible . so let ' s not use 1 and 6 in the hundreds places and reserve them for the tens places now what are the options ? try and make a pair with ( 2 * * and 3 * * ) . make the 2 * * number as large as possible and make the 3 * * number as small as possible . 265 and 314 ( difference is 49 ) or try and make a pair with ( 4 * * and 5 * * ) . make the 4 * * number as large as possible and make the 5 * * number as small as possible . we get 463 and 512 ( difference is 49 ) b | a = 10 * 6
b = a - 10
|
a ) 9,000 , b ) 4,500 , c ) 1,750 , d ) 1,000 , e ) 2,000 | b | divide(divide(50, subtract(divide(subtract(divide(const_60, 36), inverse(2)), const_60), inverse(multiply(const_60, 2)))), const_1000) | with both inlets open , a water tank will be filled with water in 36 minutes . the first inlet alone would fill the tank in 2 hours . if in every minutes the second inlet admits 50 cubic meters of water than the first , what is the capacity of the tank ? | "the work done by inlet a and b together in 1 min = 1 / 36 the work done by inlet a ( first inlet ) in 1 min = 1 / 120 the work done by inlet b ( second inlet ) in 1 min = ( 1 / 36 ) - ( 1 / 120 ) = 1 / 51 difference of work done by b and a = b - a = 50 cubic meter i . e . ( 1 / 51 ) - ( 1 / 120 ) = 50 cubic meter i . e . 4500 cubic meter answer : option b" | a = const_60 / 36
b = 1/(2)
c = a - b
d = c / const_60
e = const_60 * 2
f = 1/(e)
g = d - f
h = 50 / g
i = h / 1000
|
a ) 625 deg , b ) 420 deg , c ) 145 deg , d ) 150 deg , e ) 300 deg | b | subtract(multiply(40, multiply(const_3, const_2)), 16) | what is the angle between the hands of a clock when time is 16 : 40 ? | "angle between two hands = 40 h - 11 / 2 m = 40 * 16 - 40 * 11 / 2 = 640 - 220 = 420 deg answer : b" | a = 3 * 2
b = 40 * a
c = b - 16
|
a ) - 3 , b ) - 2 , c ) - 1 , d ) 1 , e ) 2 | a | divide(add(2, 8), add(8, 8)) | solve the equation for x : 8 ( x + y + 2 ) = 8 y - 8 | "a - 3 8 ( x + y + 2 ) = 8 y - 8 8 x + 8 y + 16 = 8 y - 8 8 x + 16 = - 8 8 x = - 24 = > x = - 3" | a = 2 + 8
b = 8 + 8
c = a / b
|
a ) 80 % , b ) 105 % , c ) 115 % , d ) 120 % , e ) 140 % | b | multiply(divide(multiply(subtract(const_1, divide(add(divide(multiply(add(8, subtract(8, const_1)), subtract(const_100, 8)), const_100), 10), 25)), const_100), multiply(subtract(const_1, divide(add(10, divide(multiply(add(8, const_4), subtract(const_100, 8)), const_100)), 20)), const_100)), const_100) | during a special promotion , a certain filling station is offering a 8 percent discount on gas purchased after the first 10 gallons . if kim purchased 20 gallons of gas , and isabella purchased 25 gallons of gas , then isabella β s total per - gallon discount is what percent of kim β s total per - gallon discount ? | "kim purchased 20 gallons of gas . she paid for 4 + 0.9 * 16 = 18.4 gallons , so the overall discount she got was 1.6 / 20 = 8 % . isabella purchased 25 gallons of gas . she paid for 4 + 0.9 * 21 = 22.9 gallons , so the overall discount she got was 2.1 / 25 = 8.4 % . 8.4 / 8 * 100 = 105 % . answer : b ." | a = 8 - 1
b = 8 + a
c = 100 - 8
d = b * c
e = d / 100
f = e + 10
g = f / 25
h = 1 - g
i = h * 100
j = 8 + 4
k = 100 - 8
l = j * k
m = l / 100
n = 10 + m
o = n / 20
p = 1 - o
q = p * 100
r = i / q
s = r * 100
|
a ) rs . 6 , b ) rs . 6.5 , c ) rs . 8 , d ) rs . 5 , e ) rs . 2 | a | divide(90, multiply(const_3, 5)) | 5 men are equal to as many women as are equal to 8 boys . all of them earn rs . 90 only . men Γ’ β¬ β’ s wages are ? | "answer : option a 5 m = xw = 8 b 5 m + xw + 8 b - - - - - 90 rs . 5 m + 5 m + 5 m - - - - - 90 rs . 15 m - - - - - - 90 rs . = > 1 m = 6 rs ." | a = 3 * 5
b = 90 / a
|
a ) 9 , b ) 8 , c ) 11 , d ) 8.5 , e ) 6 | c | divide(33, 3) | stacy has a 33 page history paper due in 3 days . how many pages per day would she have to write to finish on time ? | "33 / 3 = 11 answer : c" | a = 33 / 3
|
a ) 3 , b ) 4 , c ) 5 , d ) none , e ) 6 | b | add(2, 2) | the smallest value of n , for which 2 n + 1 is not a prime number , is | "solution ( 2 Γ 1 + 1 ) = 3 . ( 2 Γ 2 + 1 ) = 5 . ( 2 Γ 3 + 1 ) = 7 . ( 2 Γ 4 + 1 ) = 9 . which is not prime , n = 4 . answer b" | a = 2 + 2
|
a ) 150 , b ) 188 , c ) 250 , d ) 288 , e ) 300 | c | multiply(divide(multiply(50, const_1000), const_3600), 18) | a train running at the speed of 50 km / hr crosses a pole in 18 seconds . find the length of the train . | "speed = 50 * ( 5 / 18 ) m / sec = 125 / 9 m / sec length of train ( distance ) = speed * time ( 125 / 9 ) * 18 = 250 meter answer : c" | a = 50 * 1000
b = a / 3600
c = b * 18
|
a ) 4000 , b ) 8877 , c ) 2877 , d ) 2678 , e ) 1011 | a | divide(7000, power(add(subtract(divide(9261, 7000), const_1), const_1), 2)) | what sum of money put at c . i amounts in 2 years to rs . 7000 and in 3 years to rs . 9261 ? | "7000 - - - - 2261 100 - - - - ? = > 32.3 % x * 1323 / 100 * 1323 / 100 = 7000 x * 1.75 = 7000 x = 7000 / 1.75 = > 3999.25 answer : a" | a = 9261 / 7000
b = a - 1
c = b + 1
d = c ** 2
e = 7000 / d
|
a ) $ 245 , b ) $ 255 , c ) $ 265 , d ) $ 275 , e ) $ 285 | c | divide(583, add(divide(120, const_100), const_1)) | two employees m and n are paid a total of $ 583 per week by their employer . if m is paid 120 percent of the salary paid to n , how much is n paid per week ? | "1.2 n + n = 583 2.2 n = 583 n = 265 the answer is c ." | a = 120 / 100
b = a + 1
c = 583 / b
|
a ) 2 , b ) 4 , c ) 6 , d ) 14 , e ) 16 | b | divide(subtract(217, 1), 54) | if remainder is 1 , quotient is 54 and dividend is 217 then what is divisor ? | "we know dividend = divisor * quotient + remainder = = = > 217 = divisor * 54 + 1 = = = = = > 216 / 54 = divisor = = = > divisor = 4 ans - b" | a = 217 - 1
b = a / 54
|
a ) 22 , b ) 35 , c ) 97 , d ) 32 , e ) 25 | c | subtract(negate(25), multiply(subtract(7, 13), divide(subtract(7, 13), subtract(4, 7)))) | 4 , 7 , 13 , 25 , 49 , ( . . . ) | "explanation : 4 4 Γ 2 - 1 = 7 7 Γ 2 - 1 = 13 13 Γ 2 - 1 = 25 25 Γ 2 - 1 = 49 49 Γ 2 - 1 = 97 answer : option c" | a = negate - (
|
a ) 1 , b ) 5 , c ) 7 , d ) 9 , e ) 13 | c | subtract(1387, multiply(subtract(const_100, multiply(const_2, const_4)), 15)) | find the least number that must be subtracted from 1387 so that the remaining number is divisible by 15 . | on dividing 1387 by 15 we get the remainder 7 , so 7 should be subtracted . the answer is c . | a = 2 * 4
b = 100 - a
c = b * 15
d = 1387 - c
|
a ) 20 % , b ) 24 % , c ) 30 % , d ) 32 % , e ) 79 % | a | divide(multiply(subtract(add(add(const_100, 5), multiply(add(const_100, 5), divide(20, const_100))), const_100), const_100), add(add(const_100, 5), multiply(add(const_100, 5), divide(20, const_100)))) | the output of a factory was increased by 5 % to keep up with rising demand . to handle the holiday rush , this new output was increased by 20 % . by approximately what percent would the output now have to be decreased in order to restore the original output ? | "the original output increases by 5 % and then 20 % . total % change = a + b + ab / 100 total % change = 5 + 20 + 5 * 20 / 100 = 26 % now , you want to change it to 0 , so , 0 = 26 + x + 26 x / 100 x = - 26 ( 100 ) / 126 = 20 % approximately answer is a" | a = 100 + 5
b = 100 + 5
c = 20 / 100
d = b * c
e = a + d
f = e - 100
g = f * 100
h = 100 + 5
i = 100 + 5
j = 20 / 100
k = i * j
l = h + k
m = g / l
|
a ) 6 , b ) 12 , c ) 24 , d ) 36 , e ) 48 | b | multiply(sqrt(divide(62, 2)), 2) | if n is a positive integer and n ^ 2 is divisible by 62 , then the largest positive integer that must divide n is | "the question asks aboutthe largest positive integer that must divide n , not could divide n . since the least value of n for which n ^ 2 is a multiple of 72 is 12 then the largest positive integer that must divide n is 12 . complete solution of this question is given above . please ask if anything remains unclear . i spent a few hours on this one alone and i ' m still not clear . i chose 12 at first , but then changed to 48 . i ' m not a native speaker , so here is how i interpreted this question : the largest positive integer that must divide n = the largest positive factor of n . since n is a variable ( i . e . n is moving ) , so is its largest factor . please correct if i ' m wrong here . i know that if n = 12 , n ^ 2 = 144 = 2 * 72 ( satisfy the condition ) . when n = 12 , the largest factor of n is n itself , which is 12 . check : 12 is the largest positive number that must divide 12 - - > true however if n = 48 , n ^ 2 = 48 * 48 = 32 * 72 ( satisfy the condition too ) . when n = 48 , the largest factor of n is n itself , which is 48 . check : 48 is the largest positive number that must divide 48 - - > true so , i also notice that the keyword ismust , notcould . the question is , why is 48 notmust divide 48 , but instead onlycould divide 48 ? i ' m not clear right here . why is 12 must divide 12 ? what ' s the difference between them ? only restriction we have on positive integer n is that n ^ 2 is divisible by 72 . the least value of n for which n ^ 2 is divisible by 72 is 12 , thus nmustbe divisible by 12 ( n is in any case divisible by 12 ) . for all other values of n , for which n ^ 2 is divisible by 72 , n will still be divisible by 12 . this means that n is always divisible by 12 if n ^ 2 is divisible by 72 . now , ask yourself : if n = 12 , is n divisible by 48 ? no . so , n is not always divisible by 48 . b" | a = 62 / 2
b = math.sqrt(a)
c = b * 2
|
a ) 4 : 1 , b ) 10 : 1 , c ) 3 : 2 , d ) 2 : 3 , e ) 2 : 5 | b | divide(multiply(multiply(multiply(const_3, const_2), const_100), const_100), divide(multiply(multiply(multiply(const_3, const_2), const_100), const_100), multiply(add(const_2, const_3), const_2))) | if a and b get profits of rs . 60,000 and rs . 6,000 respectively at the end of year then ratio of their investments are | ratio = 60000 / 6000 = 10 : 1 answer : b | a = 3 * 2
b = a * 100
c = b * 100
d = 3 * 2
e = d * 100
f = e * 100
g = 2 + 3
h = g * 2
i = f / h
j = c / i
|
a ) 1 , b ) 2 , c ) 4 , d ) 7 , e ) 9 | c | floor(multiply(divide(add(multiply(divide(99, 9), 2), multiply(divide(99, 11), 3)), 99), const_10)) | what is the 99 th digit after the decimal point in the decimal expansion of 2 / 9 + 3 / 11 ? | 2 / 9 = 0.22222 . . . . = = > 99 th digit is 2 3 / 11 = 0.27272727 . . . . = = > every odd digit is 2 . so , 99 th digit will be 2 . 2 + 2 = 4 answer : c | a = 99 / 9
b = a * 2
c = 99 / 11
d = c * 3
e = b + d
f = e / 99
g = f * 10
h = math.floor(g)
|
a ) 17 , b ) 25 , c ) 27 , d ) 35 , e ) 50 | e | add(multiply(divide(add(multiply(8, const_3), 76), add(add(5, 8), 7)), 8), 10) | the ratio of ages of aman , bren , and charlie are in the ratio 5 : 8 : 7 respectively . if 8 years ago , the sum of their ages was 76 , what will be the age of bren 10 years from now ? | let the present ages of aman , bren , and charlie be 5 x , 8 x and 7 x respectively . 5 x - 8 + 8 x - 8 + 7 x - 8 = 76 x = 5 present age of bren = 8 * 5 = 40 bren ' s age 10 years hence = 40 + 10 = 50 answer = e | a = 8 * 3
b = a + 76
c = 5 + 8
d = c + 7
e = b / d
f = e * 8
g = f + 10
|
['a ) 12300', 'b ) 14500', 'c ) 15400', 'd ) 16700', 'e ) 18200'] | c | divide(circle_area(140), const_4) | a trainer is standing in one corner of a square ground of side 25 m . his voice can be heard upto 140 m . find the area of the ground in which his voice can be heard ? | area covered by goat = pi * r ^ 2 / 4 ( here we divide by 4 because the trainer is standing in a corner of the ground and only in 1 / 4 part , the voice can be heard ) where r = 14 m = length reaching the voice so area = ( 22 / 7 ) * 140 * 140 / 4 = 15400 sq m answer : c | a = circle_area / (
|
a ) 1 / 2 , b ) 7 / 13 , c ) 4 / 13 , d ) 8 / 29 , e ) 6 / 33 | b | divide(7, subtract(multiply(7, 2), const_1)) | machine m , n , o working simultaneously machine m can produce x units in 3 / 4 of the time it takes machine n to produce the same amount of units . machine n can produce x units in 2 / 7 the time it takes machine o to produce that amount of units . if all 3 machines are working simultaneously , what fraction of the total output is produced by machine n ? | now ultimately the speed of every machine is given with respect to mach o . so lets assume the speed of o , say 12 hrs to make x units ( assuming 6 because we can see we will need to divide by 3 and 4 mach o makes x units in 12 hrs so , mach n = 2 / 7 of o = 2 / 7 * 12 = 24 / 7 hrs to make x units and mach m = 3 / 4 of n = 3 / 4 * 24 / 7 = 1 / 6 hrs to make x units no they are running simultaneously . lets see how much each mach makes in 1 hr mach o = x / 12 units mach n = 7 / 24 units mach m = x / 6 units in 1 hr , together they make - x / 12 + 7 / 24 + x / 6 = 13 / 24 so what ratio of this has mach n made ? ( 7 / 24 ) / ( 13 / 24 ) = 7 / 13 ans : b = 7 / 13 | a = 7 * 2
b = a - 1
c = 7 / b
|
a ) 12 / 28 , b ) 48 / 19 , c ) 16 / 48 , d ) 18 / 48 , e ) 12 / 64 | b | divide(const_1, divide(add(add(inverse(2), inverse(6)), inverse(8)), 2)) | a and b can do a work in 2 days , b and c in 6 days and c and a in 8 days . in how many days will the work be completed , if all three of them work together ? | "one day work of a and b = 1 / 2 one day work of b and c = 1 / 6 one day work of c and a = 1 / 8 2 ( a + b + c ) = 1 / 2 + 1 / 6 + 1 / 8 2 ( a + b + c ) = 19 / 24 ( a + b + c ) = 19 / 48 number of days required = 48 / 19 days . answer : b" | a = 1/(2)
b = 1/(6)
c = a + b
d = 1/(8)
e = c + d
f = e / 2
g = 1 / f
|
a ) 21 years , b ) 22 years , c ) 23 years , d ) 12 years , e ) 16 years | e | divide(subtract(18, subtract(multiply(const_2, const_2), const_2)), subtract(const_2, const_1)) | a man is 18 years older than his son . in two years , his age will be twice the age of his son . the present age of this son is | "explanation : let ' s son age is x , then father age is x + 18 . = > 2 ( x + 2 ) = ( x + 18 + 2 ) = > 2 x + 4 = x + 20 = > x = 16 years option e" | a = 2 * 2
b = a - 2
c = 18 - b
d = 2 - 1
e = c / d
|
a ) 950 , b ) 940 , c ) 980 , d ) 960 , e ) 990 | e | divide(multiply(0.0088, 4.5), multiply(multiply(0.05, 0.1), 0.008)) | ( 0.0088 ) ( 4.5 ) / ( 0.05 ) ( 0.1 ) ( 0.008 ) = | "( 0.0088 ) ( 4.5 ) / ( 0.05 ) ( 0.1 ) ( 0.008 ) = 0.0088 * 450 / 5 * ( 0.1 ) ( 0.008 ) = 0.088 * 90 / 1 * 0.008 = 88 * 90 / 8 = 11 * 90 = 990 answer : e" | a = 0 * 88
b = 0 * 5
c = b * 0
d = a / c
|
a ) s . 1000 , b ) s . 1009 , c ) s . 1007 , d ) s . 1006 , e ) s . 1500 | e | divide(multiply(210, const_100), subtract(add(const_100, const_4), subtract(const_100, 10))) | a watch was sold at a loss of 10 % . if it was sold for rs . 210 more , there would have been a gain of 4 % . what is the cost price ? | explanation : 90 % 104 % - - - - - - - - 14 % - - - - 210 100 % - - - - ? = > rs . 1500 answer : e | a = 210 * 100
b = 100 + 4
c = 100 - 10
d = b - c
e = a / d
|
a ) 1865113 , b ) 1775123 , c ) 1765013 , d ) 1675123 , e ) none of them | c | multiply(4300631, power(add(const_4, const_1), const_4)) | ( 4300631 ) - ? = 2535618 | "let 4300631 - x = 2535618 then x = 4300631 - 2535618 = 1765013 answer is c" | a = 4 + 1
b = a ** 4
c = 4300631 * b
|
a ) 10.6 , b ) 10.9 , c ) 10.4 , d ) 27 , e ) 10.1 | d | divide(add(250, 500), multiply(add(60, 40), const_0_2778)) | two trains 250 m and 500 m long run at the speed of 60 km / hr and 40 km / hr respectively in opposite directions on parallel tracks . the time which they take to cross each other is ? | "relative speed = 60 + 40 = 100 km / hr . = 100 * 5 / 18 = 250 / 9 m / sec . distance covered in crossing each other = 250 + 500 = 750 m . required time = 750 * 9 / 250 = 27 sec answer : d" | a = 250 + 500
b = 60 + 40
c = b * const_0_2778
d = a / c
|
a ) 12 , b ) 24 , c ) 20 , d ) 30 , e ) 36 | a | multiply(add(3, 1), add(1, 2)) | in a rectangular coordinate system , what is the area of a rectangle whose vertices have the coordinates ( - 3 , 1 ) , ( 1 , 1 ) , ( 1 , - 2 ) and ( - 3 , - 2 ) ? | "length of side 1 = 3 + 1 = 4 length of side 2 = 2 + 1 = 3 area of rectangle = 4 * 3 = 12 a is the answer" | a = 3 + 1
b = 1 + 2
c = a * b
|
a ) 45 % , b ) 125 % , c ) 145 % , d ) 158 % , e ) 225 % | d | divide(subtract(22,947, 8,902), 8,902) | in 1970 there were 8,902 women stockbrokers in the united states . by 1978 the number had increased to 22,947 . approximately what was the percent increase ? | "the percent increase is ( 22947 - 8902 ) / 8902 = 14045 / 8902 = 1.58 so the approximate answer is d" | a = 22 - 947
b = a / 8
|
a ) 4 , b ) 5 , c ) 6 , d ) 2 , e ) 8 | d | divide(subtract(const_1, add(multiply(divide(const_1, 4), const_2), multiply(divide(const_1, 8), const_2))), divide(const_1, 8)) | a can finish a piece of work in 4 days . b can do it in 8 days . they work together for two days and then a goes away . in how many days will b finish the work ? | "2 / 4 + ( 2 + x ) / 8 = 1 = > x = 2 days answer : d" | a = 1 / 4
b = a * 2
c = 1 / 8
d = c * 2
e = b + d
f = 1 - e
g = 1 / 8
h = f / g
|
a ) 80 / 6 , b ) 80 / 7 , c ) 80 / 9 , d ) 80 / 2 , e ) 80 / 1 | c | divide(100, multiply(add(90, 72), const_0_2778)) | two trains of length 100 m and 200 m are 100 m apart . they start moving towards each other on parallel tracks , at speeds 90 kmph and 72 kmph . in how much time will the trains cross each other ? | "relative speed = ( 90 + 72 ) * 5 / 18 = 45 mps . the time required = d / s = ( 100 + 100 + 200 ) / 45 = 400 / 45 = 80 / 9 sec . answer : c" | a = 90 + 72
b = a * const_0_2778
c = 100 / b
|
a ) 32 , b ) 43 , c ) 44 , d ) 45 , e ) 46 | a | add(subtract(65, multiply(12, 3)), 3) | a batsman in his 12 th innings makes a score of 65 and thereby increases his average by 3 runs . what is his average after the 12 th innings if he had never been β not out β ? | "let β x β be the average score after 12 th innings β 12 x = 11 Γ ( x β 3 ) + 65 β΄ x = 32 answer a" | a = 12 * 3
b = 65 - a
c = b + 3
|
a ) 14 , b ) 13 , c ) 9 , d ) 6 , e ) 5 | d | add(subtract(add(13, 18), subtract(30, 3)), subtract(18, 13)) | of 30 applicants for a job , 13 had at least 4 years ' experience , 18 had degrees , and 3 had less than 4 years ' experience and did not have a degree . how many of the applicants had at least 4 years ' experience and a degree ? | "d . 6 30 - 3 = 27 27 - 13 - 18 = - 6 then 6 are in the intersection between 4 years experience and degree . answer d" | a = 13 + 18
b = 30 - 3
c = a - b
d = 18 - 13
e = c + d
|
a ) 65 seconds , b ) 46 seconds , c ) 25 seconds , d ) 97 seconds , e ) 26 seconds | c | divide(add(360, 140), divide(multiply(72, const_1000), const_3600)) | a train is 360 meter long is running at a speed of 72 km / hour . in what time will it pass a bridge of 140 meter length ? | "speed = 72 km / hr = 72 * ( 5 / 18 ) m / sec = 20 m / sec total distance = 360 + 140 = 500 meter time = distance / speed = 500 * ( 1 / 20 ) = 25 seconds answer : c" | a = 360 + 140
b = 72 * 1000
c = b / 3600
d = a / c
|
a ) 42 , b ) 10 , c ) 40 , d ) 65 , e ) 15 | e | multiply(multiply(5, 3), 3) | a certain university will select 2 of 3 candidates eligible to fill a position in the mathematics department and 4 of 5 candidates eligible to fill 2 identical positions in the computer science department . if none of the candidates is eligible for a position in both departments , how many different sets of 3 candidates are there to fill the 3 positions ? | "3 c 2 * 5 c 4 = 3 * 5 = 15 the answer is ( e )" | a = 5 * 3
b = a * 3
|
a ) 5 days , b ) 7 days , c ) 8 days , d ) 9 days , e ) 6 days | b | multiply(1, 9) | if 9 spiders make 9 webs in 9 days , then how many days are needed for 1 spider to make 1 web ? | "let , 1 spider make 1 web in a days . more spiders , less days ( indirect proportion ) more webs , more days ( direct proportion ) hence we can write as ( spiders ) 9 : 1 } : : a : 9 ( webs ) 1 : 9 β 7 Γ 1 Γ 7 = 1 Γ 7 Γ a β a = 7 answer : b" | a = 1 * 9
|
a ) 23 sec , b ) 15 sec , c ) 12 sec , d ) 11 sec , e ) 16 sec | c | divide(220, divide(multiply(add(59, 7), const_1000), const_3600)) | a bullet train 220 m long is running with a speed of 59 kmph . in what time will it pass a man who is running at 7 kmph in the direction opposite to that in which the bullet train is going ? | "c 12 sec speed of the bullet train relative to man = ( 59 + 7 ) kmph = 66 * 5 / 18 m / sec = 55 / 3 m / sec . time taken by the bullet train to cross the man = time taken by it to cover 220 m at ( 55 / 3 ) m / sec = ( 220 * 3 / 55 ) sec = 12 sec" | a = 59 + 7
b = a * 1000
c = b / 3600
d = 220 / c
|
a ) 1,108 , b ) 2,100 , c ) 2,108 , d ) 2,160 , e ) 2,256 | d | multiply(divide(208, 22.95), 250) | at the wholesale store you can buy an 8 - pack of hot dogs for $ 1.55 , a 20 - pack for $ 3.05 , and a 250 - pack for $ 22.95 . what is the greatest number of hot dogs you can buy at this store with $ 208 ? | "i think it should be d . i can buy 8 250 - pack for rs 22.95 * 8 = $ 183.60 now , i can buy 8 20 - pack for 3.05 * 8 = $ 24.40 now , i am left with only $ 1.15 . i can not but anything with this . hence total hotdogs = 250 * 8 + 20 * 5 = 2160" | a = 208 / 22
b = a * 250
|
a ) 3 / 1 , b ) 9 / 1 , c ) 3 / 3 , d ) 3 / 5 , e ) 5 / 2 | b | divide(subtract(27, 26), subtract(26, 17)) | two trains running in opposite directions cross a man standing on the platform in 27 seconds and 17 seconds respectively and they cross each other in 26 seconds . the ratio of their speeds is ? | "let the speeds of the two trains be x m / sec and y m / sec respectively . then , length of the first train = 27 x meters , and length of the second train = 17 y meters . ( 27 x + 17 y ) / ( x + y ) = 26 = = > 27 x + 17 y = 26 x + 26 y = = > 1 x = 9 y = = > x / y = 9 / 1 . answer : b" | a = 27 - 26
b = 26 - 17
c = a / b
|
a ) 2.66 cm , b ) 2.5 cm , c ) 3 cm , d ) 3.5 cm , e ) 4 cm | b | multiply(power(subtract(subtract(power(divide(3, 2), const_3), power(divide(add(1, divide(1, 2)), 2), const_3)), 1), inverse(3)), const_2) | the spherical ball of lead 3 cm in diameter is melted and recast into 3 spherical balls . the diameters of two of these are 1 Β½ cm and 2 cm respectively . the diameter of third ball is ? | 4 / 3 Ο * 3 * 3 * 3 = 4 / 3 Ο [ ( 3 / 2 ) 3 + 23 + r 3 ] r = 1.25 d = 2.5 answer : b | a = 3 / 2
b = a ** 3
c = 1 / 2
d = 1 + c
e = d / 2
f = e ** 3
g = b - f
h = g - 1
i = 1/(3)
j = h ** i
k = j * 2
|
a ) 10 , b ) 12 , c ) 14 , d ) 16 , e ) 17 | c | divide(196, subtract(15, const_1)) | find the number which when multiplied by 15 is increased by 196 | "explanation : let the number be x . then , 15 x = x + 196 = βΊ 14 x = 196 = βΊ x = 14 . answer : option c" | a = 15 - 1
b = 196 / a
|
Subsets and Splits
No community queries yet
The top public SQL queries from the community will appear here once available.