Problem stringlengths 5 967 | Rationale stringlengths 1 2.74k | options stringlengths 37 300 | correct stringclasses 5
values | annotated_formula stringlengths 7 6.48k | linear_formula stringlengths 8 925 | category stringclasses 6
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a person travels equal distances with speed of 3 km / hr , 4 km / hr and 5 km / hr and takes a total of 47 minutes . find the total distane | explanation : let the distance be 3 x km , then , x / 3 + x / 4 + x / 5 = 47 / 60 47 x / 60 = 47 / 60 x = 1 so total distance = 3 * 1 = 3 km option a | a ) 3 km , b ) 4 km , c ) 6 km , d ) 9 km , e ) 5 km | a | multiply(const_3, divide(divide(47, const_60), add(add(divide(const_1, 3), divide(const_1, 4)), divide(const_1, 5)))) | divide(n3,const_60)|divide(const_1,n0)|divide(const_1,n1)|divide(const_1,n2)|add(#1,#2)|add(#4,#3)|divide(#0,#5)|multiply(#6,const_3) | physics |
a train running at the speed of 60 km / hr crosses a pole in 3 seconds . find the length of the train . | ": speed = 60 * ( 5 / 18 ) m / sec = 50 / 3 m / sec length of train ( distance ) = speed * time ( 50 / 3 ) * 3 = 50 meter answer : b" | a ) 60 , b ) 50 , c ) 75 , d ) 100 , e ) 120 | b | multiply(divide(multiply(60, const_1000), const_3600), 3) | multiply(n0,const_1000)|divide(#0,const_3600)|multiply(n1,#1)| | physics |
sam invested $ 15000 @ 10 % per annum for one year . if the interest is compounded half yearly , then the amount received by sam at the end of the year will be ? | p = $ 15000 r = 10 % p . a . = 5 % t = 2 half years amount = 15000 * ( 1 + 5 / 100 ) ^ 2 = 15000 * 21 / 20 * 21 / 20 = $ 16537.50 answer is a | a ) $ 16537.50 , b ) $ 24512.56 , c ) $ 18475.89 , d ) $ 15478.56 , e ) $ 12456.35 | a | multiply(multiply(add(divide(add(const_4, const_1), const_100), const_1), 15000), add(divide(add(const_4, const_1), const_100), const_1)) | add(const_1,const_4)|divide(#0,const_100)|add(#1,const_1)|multiply(n0,#2)|multiply(#2,#3) | gain |
two cars start at the same time from opposite ends of a highway that is 45 miles long . one car is riding at 14 mph and the second car is riding at 16 mph . how long after they begin will they meet ? | "as cars are moving in opposite directions their speeds will be added . so their relative speeds : 16 + 14 = 30 mph total distance to be covered = 45 miles . time taken would be : 45 miles / 30 mph = 1.5 hours e is the answer ." | a ) 1 , b ) 1.2 , c ) 1.25 , d ) 1.35 , e ) 1.5 | e | divide(45, add(14, 16)) | add(n1,n2)|divide(n0,#0)| | physics |
a discount electronics store normally sells all merchandise at a discount of 10 percent to 30 percent off the suggested retail price . if , during a special sale , an additional 20 percent were to be deducted from the discount price , what would be the lowest possible price of an item costing $ 300 before any discount ... | "since the question is essentially just about multiplication , you can do the various mathstepsin a variety of ways ( depending on whichever method you find easiest ) . we ' re told that the first discount is 10 % to 30 % , inclusive . we ' re told that the next discount is 20 % off of the discounted price . . . . we '... | a ) $ 230.00 , b ) $ 145.60 , c ) $ 168.80 , d ) $ 182.00 , e ) $ 210.00 | c | multiply(300, divide(add(30, 20), const_100)) | add(n1,n2)|divide(#0,const_100)|multiply(n3,#1)| | gain |
tim and é lan are 180 miles away from one another . they are starting to move towards each other simultaneously , tim at a speed of 10 mph and é lan at a speed of 5 mph . if every hour they double their speeds , what is the distance that tim will pass until he meets é lan ? | "tim and elan will meet at the same time while their ratio of speed is 2 : 1 respectively . so their individual distance traveled ratio will be same . plugging in the answer choice only answer choice d meet the 2 : 1 ( tim : elan = 120 : 60 ) ratio of maintaining total distance traveled 180 miles socorrect answer d" | a ) 130 miles . , b ) 135 miles . , c ) 145 miles . , d ) 120 miles . , e ) 165 miles . | d | multiply(divide(10, add(5, 10)), 180) | add(n1,n2)|divide(n1,#0)|multiply(n0,#1)| | physics |
for any integer k > 1 , the term “ length of an integer ” refers to the number of positive prime factors , not necessarily distinct , whose product is equal to k . for example , if k = 24 , the length of k is equal to 4 , since 24 = 2 × 2 × 2 × 3 . if x and y are positive integers such that x > 1 , y > 1 , and x + 3 y ... | "we know that : x > 1 , y > 1 , and x + 3 y < 960 , and it is given that length means no of factors . for any value of x and y , the max no of factors can be obtained only if factor is smallest no all factors are equal . hence , lets start with smallest no 2 . 2 ^ 1 = 2 2 ^ 2 = 4 2 ^ 3 = 8 2 ^ 4 = 16 2 ^ 5 = 32 2 ^ 6 =... | a ) 10 , b ) 16 , c ) 12 , d ) 14 , e ) 19 | b | add(add(4, 3), add(add(4, 4), 1)) | add(n2,n7)|add(n2,n2)|add(n0,#1)|add(#0,#2)| | general |
a man rows his boat 90 km downstream and 45 km upstream , taking 2 1 / 2 hours each time . find the speed of the stream ? | "speed downstream = d / t = 90 / ( 2 1 / 2 ) = 36 kmph speed upstream = d / t = 45 / ( 2 1 / 2 ) = 18 kmph the speed of the stream = ( 36 - 18 ) / 2 = 9 kmph answer : a" | a ) 9 kmph , b ) 6 kmph , c ) 5 kmph , d ) 8 kmph , e ) 7 kmph | a | divide(subtract(divide(90, 2), divide(45, 2)), const_2) | divide(n0,n2)|divide(n1,n2)|subtract(#0,#1)|divide(#2,const_2)| | physics |
what is the smallest positive integer k such that the product of 1575 x k is a perfect square ? | "perfect square , is just an integer that can be written as the square of some other integer . for example 16 = 4 ^ 2 , is a perfect square . now , 1575 = 3 ^ 2 * 5 ^ 2 * 7 , so if k = 7 then 1575 k = ( 3 * 5 * 7 ) ^ 2 , which is a perfect square ( basically the least positive value of k must complete only the power of... | a ) 7 , b ) 9 , c ) 15 , d ) 25 , e ) 63 | a | add(const_3, const_4) | add(const_3,const_4)| | general |
in a public show 75 % of the seats were filled . if there were 600 seats in the hall , how many seats were vacant ? | "75 % of 600 = 75 / 100 × 600 = 450 therefore , the number of vacant seats = 600 - 450 = 150 . answer : e" | a ) 100 , b ) 110 , c ) 120 , d ) 140 , e ) 150 | e | divide(multiply(600, subtract(const_100, 75)), const_100) | subtract(const_100,n0)|multiply(n1,#0)|divide(#1,const_100)| | gain |
the true discount on a bill due 9 months hence at 16 % per annum is rs . 180 . the amount of the bill is | "explanation : let p . w . be rs . x . then , s . i . on rs . x at 16 % for 9 months = rs . 180 . x ã — 16 ã — ( 9 / 12 ) ã — ( 1 / 100 ) = 180 or x = 1500 . p . w . = rs . 1500 . sum due = p . w . + t . d . = rs . ( 1500 180 ) = rs . 1680 . answer : c" | a ) 1200 , b ) 1764 , c ) 1680 , d ) 1354 , e ) none of these | c | add(divide(180, divide(multiply(divide(9, multiply(const_4, const_3)), 16), const_100)), 180) | multiply(const_3,const_4)|divide(n0,#0)|multiply(n1,#1)|divide(#2,const_100)|divide(n2,#3)|add(n2,#4)| | gain |
a project scheduled to be carried out over a single fiscal year has a budget of $ 12600 , divided into 12 equal monthly allocations . at the end of the sixth month of that fiscal year , the total amount actually spent on the project was $ 6580 . by how much was the project over its budget ? | difficulty level : 600 each month ' s budget = 12600 / 12 = 1050 budget for 6 months = 6 * 1050 = 6300 actual amount spent = 6580 amount spent over the budget = 6580 - 6300 = 280 answer ( c ) , regards , | a ) $ 380 , b ) $ 540 , c ) $ 280 , d ) $ 1380 , e ) $ 1430 | c | subtract(6580, multiply(divide(12600, 12), add(const_2, const_4))) | add(const_2,const_4)|divide(n0,n1)|multiply(#0,#1)|subtract(n2,#2) | general |
a man rows his boat 85 km downstream and 25 km upstream , taking 2 1 / 2 hours each time . find the speed of the stream ? | "speed downstream = d / t = 85 / ( 2 1 / 2 ) = 34 kmph speed upstream = d / t = 25 / ( 2 1 / 2 ) = 10 kmph the speed of the stream = ( 34 - 10 ) / 2 = 12 kmph answer : b" | a ) 6 , b ) 12 , c ) 5 , d ) 8 , e ) 9 | b | divide(subtract(divide(85, 2), divide(25, 2)), const_2) | divide(n0,n2)|divide(n1,n2)|subtract(#0,#1)|divide(#2,const_2)| | physics |
a cement mixture is composed of 3 elements . by weight , 1 / 4 of the mixture is sand , 2 / 5 of the mixture is water , and the remaining 14 pounds of the mixture is gravel . what is the weight of the entire mixture in pounds ? | let the total weight be x . sand content = ( 1 / 4 ) x water content = ( 2 / 5 ) x gravel = x - ( 1 / 4 ) x - ( 2 / 5 ) x = ( 7 / 20 ) x = 14 x = 40 then answer will be d = 40 | a ) 9 , b ) 14 , c ) 28 , d ) 40 , e ) 60 | d | divide(14, subtract(const_1, add(divide(1, 4), divide(2, 5)))) | divide(n1,n2)|divide(n3,n4)|add(#0,#1)|subtract(const_1,#2)|divide(n5,#3) | general |
the length of a rectangle is 2 times its width . if the width of the rectangle is 5 inches , what is the rectangle ' s area , in square inches ? | if the width is 5 in and the length is 2 times the width , then the length is 2 * 5 = 10 in the area is given by 5 * 10 = 50 square inches correct answer a | ['a ) 50 square inches', 'b ) 75 square inches', 'c ) 68 square inches', 'd ) 89 square inches', 'e ) 92 square inches'] | a | rectangle_area(5, multiply(2, 5)) | multiply(n0,n1)|rectangle_area(n1,#0) | geometry |
if a is the smallest positive integer such that 5880 multiplied by a is the square of an integer , then ya must be | 5880 = 2 * 2 * 7 * 7 * 5 * 6 , so we need one 5 and one 6 to make it a square of a number . so 5 * 6 = 30 ans : a | a ) 30 , b ) 15 , c ) 12 , d ) 23 , e ) 17 | a | divide(5880, multiply(power(add(const_3, const_4), const_2), power(const_2, const_2))) | add(const_3,const_4)|power(const_2,const_2)|power(#0,const_2)|multiply(#2,#1)|divide(n0,#3) | geometry |
working simultaneously and independently at an identical constant rate , 6 machines of a certain type can produce a total of x units of product p in 10 days . how many of these machines , working simultaneously and independently at this constant rate , can produce a total of 5 x units of product p in 6 days ? | "6 machines do x units in 10 days we have x / 10 = > rate of the 6 machines we know that we need to have 5 x units in 6 days therefore , we need to get to 5 x / 6 rate of the machines . rate of one machine is x / 10 * 1 / 6 = x / 60 . now , we need to know how many machines need to work simultaneously , to get 5 x done... | a ) 50 , b ) 18 , c ) 16 , d ) 12 , e ) 8 | a | multiply(multiply(6, 5), divide(10, 6)) | divide(n1,n3)|multiply(n0,n2)|multiply(#0,#1)| | general |
if the difference between the compound interest & simple interest of a certain sum of money is rs . 72 at 12 % p . a . per two year . find the sum ( in rs . ) ? | 200 / 17 c . i = a - p a = p ( 1 + 12 / 100 ) ^ 2 s . i = p * 2 * 12 / 100 c . i - s . i = 72 given substitute and solve p = 200 / 17 answer : b | a ) 200 / 15 , b ) 200 / 17 , c ) 220 / 17 , d ) 240 / 17 , e ) 260 / 17 | b | divide(multiply(72, multiply(const_100, const_100)), power(12, const_2)) | multiply(const_100,const_100)|power(n1,const_2)|multiply(n0,#0)|divide(#2,#1) | gain |
each year for 4 years , a farmer increased the number of trees in a certain orchard by 1 / 4 of the number of trees in the orchard of the preceding year . if all of the trees thrived and there were 3125 trees in the orchard at the end of 4 year period , how many trees were in the orchard at the beginning of the 4 year ... | "trees increase by 1 / 4 the number of trees in preceding year . hence , correct answer must be divisible by 4 . based on divisibility rules , if last 2 digits are divisible by 4 then the number is divisible by 4 . thus , we can eliminate a , b , c , e the answer to be d again , trees increase by 1 / 4 the number of tr... | a ) 1287 , b ) 1285 , c ) 1283 , d ) 1280 , e ) 1281 | d | divide(3125, power(add(divide(1, 4), 1), 4)) | divide(n1,n0)|add(n1,#0)|power(#1,n0)|divide(n3,#2)| | general |
if n is a natural number , then 6 n ^ 2 + 6 n is always divisible by ? | "6 n ^ 2 + 6 n = 6 n ( n + 1 ) , which is always divisible by 6 and 12 both , since n ( n + 1 ) is always even . answer is b" | a ) 6 only , b ) 6 and 12 , c ) 12 only , d ) 18 only , e ) 20 only | b | add(multiply(6, const_100), multiply(2, 6)) | multiply(n0,const_100)|multiply(n0,n1)|add(#0,#1)| | general |
how many seconds will a train 100 meters long take to cross a bridge 120 meters long if the speed of the train is 36 kmph ? | "explanation : d = 100 + 120 = 220 s = 36 * 5 / 18 = 10 mps t = 250 / 10 = 22 sec answer : option a" | a ) 22 sec , b ) 27 sec , c ) 25 sec , d ) 24 sec , e ) 11 sec | a | divide(add(120, 100), multiply(36, const_0_2778)) | add(n0,n1)|multiply(n2,const_0_2778)|divide(#0,#1)| | physics |
a certain fruit stand sold apples for $ 0.70 each and bananas for $ 0.60 each . if a customer purchased both apples and bananas from the stand for a total of $ 6.30 , what total number of apples and bananas did the customer purchase ? | "let ' s start with 1 apple for $ 0.70 . let ' s subtract $ 0.70 from $ 6.30 until we get a multiple of $ 0.60 . $ 6.30 , $ 5.60 , $ 4.90 , $ 4.20 = 7 * $ 0.60 the customer purchased 7 bananas and 3 apples . the answer is c ." | a ) 8 , b ) 9 , c ) 10 , d ) 11 , e ) 12 | c | add(divide(subtract(6.30, multiply(0.70, const_2)), 0.60), const_2) | multiply(n0,const_2)|subtract(n2,#0)|divide(#1,n1)|add(#2,const_2)| | other |
a man ' s speed with the current is 15 km / hr and the speed of the current is 2.5 km / hr . the man ' s speed against the current is : | "explanation : man ' s speed with the current = 15 km / hr = > speed of the man + speed of the current = 15 km / hr speed of the current is 2.5 km / hr hence , speed of the man = 15 - 2.5 = 12.5 km / hr man ' s speed against the current = speed of the man - speed of the current = 12.5 - 2.5 = 10 km / hr answer is b" | a ) 8.5 km / hr , b ) 10 km / hr , c ) 12.5 km / hr , d ) 9 km / hr , e ) none of these | b | subtract(subtract(15, 2.5), 2.5) | subtract(n0,n1)|subtract(#0,n1)| | gain |
a shopkeeper sold an article at $ 1200 and gained a 20 % profit . what was the cost price ? | "let x be the cost price . 1.2 x = 1200 x = 1200 / 1.2 = 1000 the answer is a ." | a ) $ 1000 , b ) $ 1020 , c ) $ 1040 , d ) $ 1060 , e ) $ 1080 | a | multiply(const_100.0, divide(const_100, add(1200, 20))) | add(n1,const_100)|divide(const_100,#0)|multiply(n0,#1)| | gain |
at company x , senior sales representatives visit the home office once every 20 days , and junior sales representatives visit the home office once every 15 days . the number of visits that a junior sales representative makes in a 2 - year period is approximately what percent greater than the number of visits that a sen... | "each 60 - day period , senior representatives visit the home office 3 times while junior representatives visit 4 times , thus 33 % more . the answer is c ." | a ) 10 % , b ) 25 % , c ) 33 % , d ) 50 % , e ) 67 % | c | multiply(add(const_1, const_10), subtract(subtract(20, 15), const_1)) | add(const_1,const_10)|subtract(n0,n1)|subtract(#1,const_1)|multiply(#0,#2)| | general |
daniel went to a shop and bought things worth rs . 25 , out of which 30 paise went on sales tax on taxable purchases . if the tax rate was 5 % , then what was the cost of the tax free items ? | "total cost of the items he purchased = rs . 25 given that out of this rs . 25 , 30 paise is given as tax = > total tax incurred = 30 paise = rs . 30 / 100 let the cost of the tax free items = x given that tax rate = 5 % ∴ ( 25 − 30 / 100 − x ) 5 / 100 = 30 / 100 ⇒ 5 ( 25 − 0.3 − x ) = 30 ⇒ ( 25 − 0.3 − x ) = 6 ⇒ x = 2... | a ) a ) 18.7 , b ) b ) 20 , c ) c ) 21.3 , d ) d ) 21.5 , e ) e ) 22 | a | subtract(subtract(25, divide(30, const_100)), divide(30, 5)) | divide(n1,const_100)|divide(n1,n2)|subtract(n0,#0)|subtract(#2,#1)| | gain |
a car gets 40 kilometers per gallon of gasoline . how many gallons of gasoline would the car need to travel 190 kilometers ? | "each 40 kilometers , 1 gallon is needed . we need to know how many 40 kilometers are there in 190 kilometers ? 190 ã · 40 = 4.75 ã — 1 gallon = 4.75 gallons correct answer is d ) 4.75 gallons" | a ) 3.5 gallons , b ) 2.7 gallons , c ) 5.7 gallons , d ) 4.75 gallons , e ) 7.5 gallons | d | divide(190, 40) | divide(n1,n0)| | physics |
in the faculty of reverse - engineering , 300 second year students study numeric methods , 600 second year students study automatic control of airborne vehicles and 200 second year students study them both . how many students are there in the faculty if the second year students are approximately 70 % of the total ? | "total number of students studying both are 300 + 600 - 200 = 700 ( subtracting the 200 since they were included in the both the other numbers already ) . so 70 % of total is 700 , so 100 % is 1000 answer is e" | a ) 750 , b ) 800 , c ) 900 , d ) 950 , e ) 1000 | e | add(300, 600) | add(n0,n1)| | general |
a cistern can be filled by a tap in 3 hours while it can be emptied by another tap in 8 hours . if both the taps are opened simultaneously , then after how much time will the cistern get filled ? | explanation : part filled by one tap in 1 hour = 1 / 4 part emptied by second tap 1 hour = 1 / 8 net part filled by both these taps in 1 hour = 1 / 3 - 1 / 8 = 5 / 24 i . e , the cistern gets filled in 24 / 5 hours = 4.8 hrs answer : option a | a ) 4.8 hr , b ) 2.4 hr , c ) 3.6 hr , d ) 1.8 hr , e ) 2.0 hr | a | divide(const_1, subtract(divide(const_1, 3), divide(const_1, 8))) | divide(const_1,n0)|divide(const_1,n1)|subtract(#0,#1)|divide(const_1,#2) | physics |
in a graduating class of 233 students , 144 took geometry and 119 took biology . what is the difference between the greatest possible number and the smallest possible number of students that could have taken both geometry and biology ? | "greatest possible number taken both should be 144 ( as it is maximum for one ) smallest possible number taken both should be given by total - neither = a + b - both both = a + b + neither - total ( neither must be 0 to minimize the both ) so 144 + 119 + 0 - 233 = 30 greatest - smallest is 144 - 30 = 114 so answer must... | a ) 114 , b ) 119 , c ) 113 , d ) 88 , e ) 31 | a | subtract(119, subtract(add(144, 119), 233)) | add(n1,n2)|subtract(#0,n0)|subtract(n2,#1)| | other |
on a map , 1.5 inches represent 24 miles . how many miles approximately is the distance if you measured 48 centimeters assuming that 1 - inch is 2.54 centimeters ? | "1.5 inch = 2.54 * 1.5 cm . so , 2.54 * 1.5 represents 24 miles . so for 48 cm . : 48 / ( 2.54 * 1.5 ) = x / 24 - - - > x = 24 * 47 / ( 3.81 ) = 302 answer will be e ." | a ) 174.2 , b ) 212 , c ) 288.1 , d ) 296 , e ) 302 | e | multiply(divide(48, 2.54), divide(24, 1.5)) | divide(n2,n4)|divide(n1,n0)|multiply(#0,#1)| | physics |
if | 4 x + 6 | = 42 , what is the sum of all the possible values of x ? | "there will be two cases 4 x + 6 = 42 or 4 x + 6 = - 42 = > x = 9 or x = - 12 sum of both the values will be - 12 + 9 = - 3 answer is b" | a ) 2 , b ) - 3 , c ) 4 , d ) - 5 , e ) 6 | b | divide(subtract(42, 6), 4) | subtract(n2,n1)|divide(#0,n0)| | general |
a pipe can empty 1 / 4 th of a cistern in 12 mins . in 8 mins , what part of the cistern will be empty ? | "1 / 4 - - - - 12 ? - - - - - 8 = = > 1 / 6 d" | a ) 2 / 3 , b ) 3 / 7 , c ) 4 / 9 , d ) 1 / 6 , e ) 4 / 7 | d | divide(multiply(divide(1, 4), 8), 12) | divide(n0,n1)|multiply(n3,#0)|divide(#1,n2)| | physics |
4 mat - weavers can weave 4 mats in 4 days . at the same rate , how many mats would be woven by 16 mat - weavers in 16 days ? | "let the required number of bottles be x . more weavers , more mats ( direct proportion ) more days , more mats ( direct proportion ) wavers 4 : 16 : : 4 : x days 4 : 16 4 * 4 * x = 16 * 16 * 4 x = ( 16 * 16 * 4 ) / ( 4 x 4 ) x = 64 . answer is e ." | a ) 25 , b ) 19 , c ) 39 , d ) 61 , e ) 64 | e | multiply(multiply(4, divide(16, 4)), divide(16, 4)) | divide(n3,n0)|multiply(n0,#0)|multiply(#0,#1)| | gain |
suppose 8 monkeys take 8 minutes to eat 8 banana ' s . how many minutes would it take 3 monkeys to eat 3 banana ' s ? | there are equal no . of monkeys and equal no . of bananas and they take equal time and the time is 8 mint to eat a banana so each monkey take 8 mints to eat a banana ' s so 3 monkeys will take 8 mints to eat 3 banana ' s . answer : d | a ) 5 min , b ) 6 min , c ) 7 min , d ) 8 min , e ) 9 min | d | divide(3, divide(3, 8)) | divide(n3,n0)|divide(n3,#0) | physics |
if x + | x | + y = 4 and x + | y | - y = 3 what is x + y = ? | "if x < 0 and y < 0 , then we ' ll have x - x + y = 7 and x - y - y = 6 . from the first equation y = 7 , so we can discard this case since y is not less than 0 . if x > = 0 and y < 0 , then we ' ll have x + x + y = 7 and x - y - y = 6 . solving gives x = 4 > 0 and y = - 1 < 0 - - > x + y = 3 . since in ps questions on... | a ) 1 , b ) - 1 , c ) 3 , d ) 5 , e ) 13 | c | multiply(3, const_2) | multiply(n1,const_2)| | general |
there were 35 students in a hostel . if the number of students increases by 7 , the expenses of the mess increase by rs . 42 per day while the average expenditure per head diminishes by re 1 . find the original expenditure of the mess . | let d be the average daily expenditure original expenditure = 35 × d new expenditure = 35 × d + 42 new average expenditure will be : ( 35 × d + 42 ) / 42 = d - 1 on solving , we get d = 12 therefore original expenditure = 35 × 12 = 420 answer : c | a ) rs . 480 , b ) rs . 520 , c ) rs . 420 , d ) rs . 460 , e ) none of these | c | multiply(divide(add(42, 42), subtract(42, 35)), 35) | add(n2,n2)|subtract(n2,n0)|divide(#0,#1)|multiply(n0,#2) | general |
two numbers are in the ratio of 1 : 2 . if 8 be added to both , their ratio changes to 3 : 5 . the greater number is | "let the ratio be x : y , given x / y = 1 / 2 , ( x + 8 ) / ( y + 8 ) = 3 / 5 = > x = 16 and y = 32 answer : d" | a ) 20 , b ) 24 , c ) 28 , d ) 32 , e ) 36 | d | multiply(1, 8) | multiply(n0,n2)| | other |
if xerox paper costs 5 cents a sheet and a buyer gets 10 % discount on all xerox paper one buys after the first 200 papers and 20 % discount after first 1000 papers , how much will it cost to buy 2500 sheets of xerox paper ? | "30 sec approach - solve it using approximation 2500 sheet at full price , 5 cent = 125 2500 sheet at max discount price , 4 cent = 100 your ans got to be between these two . ans b it is ." | a ) $ 125 , b ) $ 106 , c ) $ 135 , d ) $ 90 , e ) $ 100 | b | multiply(subtract(1000, 200), multiply(subtract(const_1, divide(const_1, 10)), divide(const_1, 10))) | divide(const_1,n1)|subtract(n4,n2)|subtract(const_1,#0)|multiply(#0,#2)|multiply(#3,#1)| | gain |
a and b can do a work in 10 days and 15 days respectively . a starts the work and b joins him after 3 days . in how many days can they complete the remaining work ? | "work done by a in 3 days = 3 / 10 remaining work = 7 / 10 work done by both a and b in one day = 1 / 10 + 1 / 15 = 5 / 30 = 1 / 6 remaining work = 7 / 10 * 6 / 1 = 21 / 5 = 4 1 / 5 days . answer : d" | a ) 6 days , b ) 2 days , c ) 8 days , d ) 4 1 / 5 days , e ) 9 days | d | subtract(add(inverse(add(inverse(15), inverse(10))), 10), const_3) | inverse(n1)|inverse(n0)|add(#0,#1)|inverse(#2)|add(n0,#3)|subtract(#4,const_3)| | physics |
how many integers between 265,205 and 758,805 have tens digit 1 and units digit 3 ? | "there is one number in hundred with 1 in the tens digit and 3 in the units digit : 13 , 113 , 213 , 313 , . . . the difference between 265,205 and 758,805 is 758,805 - 265,205 = 493,600 - one number per each hundred gives 133,900 / 100 = 4,936 numbers . answer : e ." | a ) 4,629 , b ) 4,639 , c ) 3,946 , d ) 4,926 , e ) 4,936 | e | subtract(758,805, add(add(multiply(const_2, const_100), multiply(add(const_3, const_4), const_10)), const_2)) | add(const_3,const_4)|multiply(const_100,const_2)|multiply(#0,const_10)|add(#1,#2)|add(#3,const_2)|subtract(n1,#4)| | general |
the shopkeeper increased the price of a product by 25 % so that customer finds it difficult to purchase the required amount . but somehow the customer managed to purchase only 60 % of the required amount . what is the net difference in the expenditure on that product ? | "quantity x rate = price 1 x 1 = 1 0.6 x 1.25 = 0.755 decrease in price = ( 0.125 / 1 ) × 100 = 12.5 % a )" | a ) a ) 12.5 % , b ) b ) 13 % , c ) c ) 13.15 % , d ) d ) 14 % , e ) e ) 15 % | a | divide(multiply(subtract(multiply(const_100, const_100), multiply(add(const_100, 25), 60)), const_100), multiply(const_100, const_100)) | add(n0,const_100)|multiply(const_100,const_100)|multiply(n1,#0)|subtract(#1,#2)|multiply(#3,const_100)|divide(#4,#1)| | general |
a batsman scored 160 runs which included 3 boundaries and 8 sixes . what percent of his total score did he make by running between the wickets . | "explanation : number of runs made by running = 160 - ( 3 x 4 + 8 x 6 ) = 160 - ( 60 ) = 100 now , we need to calculate 100 is what percent of 160 . = > 100 / 160 * 100 = 68 option d" | a ) 64 % , b ) 66 % , c ) 60 % , d ) 68 % , e ) 70 % | d | multiply(divide(subtract(160, add(multiply(3, 8), multiply(8, 3))), 160), const_100) | multiply(n1,n2)|multiply(n1,n2)|add(#0,#1)|subtract(n0,#2)|divide(#3,n0)|multiply(#4,const_100)| | general |
a train covers a distance of 6 km in 10 min . if it takes 12 sec to pass a telegraph post , then the length of the train is ? | "speed = ( 6 / 10 * 60 ) km / hr = ( 36 * 5 / 18 ) m / sec = 10 m / sec . length of the train = 10 * 12 = 120 m . answer : a" | a ) 120 m , b ) 180 m , c ) 240 m , d ) 220 m , e ) 280 m | a | divide(6, subtract(divide(6, 10), 12)) | divide(n0,n1)|subtract(#0,n2)|divide(n0,#1)| | physics |
60 % of the employees of a company are men . 60 % of the men in the company speak french and 50 % of the employees of the company speak french . what is % of the women in the company who do not speak french ? | "no of employees = 100 ( say ) men = 60 women = 40 men speaking french = 0.6 * 60 = 36 employees speaking french = 0.5 * 100 = 50 therefore women speaking french = 50 - 36 = 14 and women not speaking french = 40 - 14 = 26 % of women not speaking french = 26 / 40 * 100 = 65 % answer e" | a ) 4 % , b ) 10 % , c ) 96 % , d ) 90 % , e ) 65 % | e | multiply(divide(subtract(divide(subtract(const_100, 60), const_100), subtract(divide(50, const_100), multiply(divide(60, const_100), divide(60, const_100)))), divide(subtract(const_100, 60), const_100)), const_100) | divide(n2,const_100)|divide(n0,const_100)|divide(n1,const_100)|subtract(const_100,n0)|divide(#3,const_100)|multiply(#1,#2)|subtract(#0,#5)|subtract(#4,#6)|divide(#7,#4)|multiply(#8,const_100)| | gain |
an error 2 % in excess is made while measuring the side ofa square . the % of error in the calculated area of the square is ? | 100 cm is read as 102 cm . a 1 = ( 100 x 100 ) cm 2 and a 2 ( 102 x 102 ) cm 2 . ( a 2 - a 1 ) = [ ( 102 ) 2 - ( 100 ) 2 ] = ( 102 + 100 ) x ( 102 - 100 ) = 404 cm 2 . percentage error = 404 x 100 % = 4.04 % 100 x 100 b | ['a ) 4 %', 'b ) 4.04 %', 'c ) 4.23 %', 'd ) 4.56 %', 'e ) 5 %'] | b | divide(multiply(subtract(square_area(add(const_100, 2)), square_area(const_100)), const_100), square_area(const_100)) | add(n0,const_100)|square_area(const_100)|square_area(#0)|subtract(#2,#1)|multiply(#3,const_100)|divide(#4,#1) | geometry |
in a group of donkeys and pigs , the numbers of legs are 16 more than twice the number of heads . the number of donkeys is | let the number of donkeys be x and the number of pigs be y . then , 4 x + 2 y = 2 ( x + y ) = 16 or 2 x + ( 2 x + 2 y ) = ( 2 x + 2 y ) + 16 or 2 x = 16 or x = 8 . answer : b | a ) 6 , b ) 8 , c ) 11 , d ) 12 , e ) 13 | b | multiply(divide(16, add(const_2, const_2)), const_2) | add(const_2,const_2)|divide(n0,#0)|multiply(#1,const_2) | general |
a furniture dealer purchased a desk for $ 240 and then set the selling price equal to the purchase price plus a markup that was 40 % of the selling price . if the dealer sold the desk at the selling price , what was the amount of the dealer ' s gross profit from the purchase and the sale of the desk ? | "purchase price = 240 selling price = x 240 + 0.4 * x = x 0.6 * x = 240 x = 400 profit = 400 - 240 = 160 answer : c" | a ) $ 40 , b ) $ 60 , c ) $ 160 , d ) $ 90 , e ) $ 100 | c | divide(multiply(subtract(divide(240, subtract(const_1, divide(40, const_100))), 240), const_100), 240) | divide(n1,const_100)|subtract(const_1,#0)|divide(n0,#1)|subtract(#2,n0)|multiply(#3,const_100)|divide(#4,n0)| | gain |
a , b and c enter into a partnership . a invests 3 times as much as b invests and 2 / 3 of what c invests . at the end of the year , the profit earned is rs . 12375 . what is the share of c ? | "explanation : let the investment of c be rs . x . the inverstment of b = rs . ( 2 x / 3 ) the inverstment of a = rs . ( 3 × ( 2 / 3 ) x ) = rs . ( 2 x ) ratio of capitals of a , b and c = 2 x : 2 x / 3 : x = 6 : 2 : 3 c ' s share = rs . [ ( 3 / 11 ) × 12375 ] = rs . 3375 answer : option b" | a ) rs . 2250 , b ) rs . 3375 , c ) rs . 6750 , d ) rs . 5625 , e ) none of these | b | multiply(12375, inverse(add(add(divide(2, 3), multiply(divide(2, 3), 3)), const_1))) | divide(n1,n0)|multiply(n0,#0)|add(#0,#1)|add(#2,const_1)|inverse(#3)|multiply(n3,#4)| | gain |
a shopkeeper has 280 kg of apples . he sells 40 % of these at 20 % profit and remaining 60 % at 20 % profit . find his % profit on total . | "if the total quantity was 100 then 40 x 20 % + 60 x 20 % = 20 this profit will remain same for any total quantity unless the % of products remains the same . hence ' b ' is the answer" | a ) 24 % , b ) 20 % , c ) 26 % , d ) 28 % , e ) 35 % | b | divide(multiply(subtract(add(multiply(divide(multiply(280, 40), const_100), divide(add(const_100, 20), const_100)), multiply(divide(multiply(280, 60), const_100), divide(add(const_100, 20), const_100))), 280), const_100), 280) | add(n2,const_100)|add(n4,const_100)|multiply(n0,n1)|multiply(n0,n3)|divide(#2,const_100)|divide(#0,const_100)|divide(#3,const_100)|divide(#1,const_100)|multiply(#4,#5)|multiply(#6,#7)|add(#8,#9)|subtract(#10,n0)|multiply(#11,const_100)|divide(#12,n0)| | gain |
how many positive integers less than 30 have no common prime factor with 30 ? | "30 = 2 * 3 * 5 . so , the number must be less than 30 and not have primes 2 , 3 , or 5 . this means that the number could be : 1 , 7 , 11 , 13 , 17 , 19 , 23 , or 29 . total of 8 numbers . answer : d ." | a ) 5 , b ) 6 , c ) 7 , d ) 8 , e ) 9 | d | divide(factorial(subtract(add(const_4, 30), const_1)), multiply(factorial(30), factorial(subtract(const_4, const_1)))) | add(n1,const_4)|factorial(n1)|subtract(const_4,const_1)|factorial(#2)|subtract(#0,const_1)|factorial(#4)|multiply(#1,#3)|divide(#5,#6)| | other |
a 300 m long train crosses a platform in 38 sec while it crosses a signal pole in 18 sec . what is the length of the platform ? | "speed = 300 / 18 = 50 / 3 m / sec . let the length of the platform be x meters . then , ( x + 300 ) / 38 = 50 / 3 3 x + 900 = 1900 = > x = 333 m . answer : e" | a ) 288 , b ) 350 , c ) 889 , d ) 276 , e ) 333 | e | subtract(multiply(speed(300, 18), 38), 300) | speed(n0,n2)|multiply(n1,#0)|subtract(#1,n0)| | physics |
two trains of length 120 m and 280 m are running towards each other on parallel lines at 42 kmph and 36 kmph respectively . in what time will they be clear of each other from the moment they meet ? | "relative speed = ( 42 + 36 ) * 5 / 18 = 21.7 mps . distance covered in passing each other = 120 + 280 = 400 m . the time required = d / s = 400 / 21.7 = 18.4 sec . answer : a" | a ) 18.4 sec , b ) 77 sec , c ) 76 sec , d ) 20 sec , e ) 66 sec | a | divide(add(120, 280), multiply(add(42, 36), const_0_2778)) | add(n0,n1)|add(n2,n3)|multiply(#1,const_0_2778)|divide(#0,#2)| | physics |
tough and tricky questions : mixture problems . a drink holding 6 ounces of an alcoholic drink that is 1 part rum to 2 parts coke is added to a jug holding 35 ounces of an alcoholic drink that is 1 part rum to 3 parts coke . what is the ratio of rum to coke in the resulting mixture ? | answer = b ) 5 : 15 . . . . . . . . . . . . . . . . . . . . . . . . rum . . . . . . . . . . . . . . . . . . . . . coke . . . . . . . . . . . . . . . . . total first drink . . . . . . . . . . . 2 . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4 . . . . . . . . . . . . . . . . . . . . . 6 ( ratio 1 : 2 ) second ... | a ) 2 : 5 , b ) 5 : 15 , c ) 3 : 5 , d ) 4 : 7 , e ) 14 : 5 | b | divide(add(divide(6, const_3), divide(35, multiply(2, divide(6, const_3)))), multiply(add(divide(6, const_3), divide(35, multiply(2, divide(6, const_3)))), const_3)) | divide(n0,const_3)|multiply(n2,#0)|divide(n3,#1)|add(#0,#2)|multiply(#3,const_3)|divide(#3,#4) | general |
excluding stoppages , the speed of a bus is 70 km / hr and including stoppages , it is 40 km / hr . for how many minutes does the bus stop per hour ? | "due to stoppages , it covers 30 km less . time taken to cover 30 km = 30 / 70 * 60 = 25 min . answer : e" | a ) 118 min , b ) 10 min , c ) 18 min , d ) 16 min , e ) 25 min | e | multiply(const_60, divide(subtract(70, 40), 70)) | subtract(n0,n1)|divide(#0,n0)|multiply(#1,const_60)| | physics |
49 x 49 x 49 x 49 = 7 ? | "solution 49 * 49 * 49 * 49 = ( 72 * 72 * 72 * 72 ) = 7 ( 2 + 2 + 2 + 2 ) = 78 so , the correct answer is 8 answer c" | a ) 4 , b ) 7 , c ) 8 , d ) 16 , e ) 20 | c | divide(subtract(multiply(49, 49), multiply(49, 49)), 7) | multiply(n0,n1)|multiply(n2,n3)|subtract(#0,#1)|divide(#2,n4)| | general |
calculate 12351 ÷ ? = 69 | "answer let 12351 ÷ ? = 69 then x = 12351 / 69 = 179 . option : a" | a ) 179 , b ) 119 , c ) 129 , d ) 173 , e ) 156 | a | multiply(12351, 69) | multiply(n0,n1)| | general |
the ratio between the length and the breadth of a rectangular park is 3 : 2 . if a man cycling along the boundary of the park at the speed of 12 km / hr completes one round in 8 minutes , then the area of the park ( in sq . m ) is ? | "let length = 3 x metres and breadth = 2 x metres . then , 2 ( 3 x + 2 x ) = 1600 or x = 160 . length = 480 m and breadth = 320 m . area = ( 480 x 320 ) m ^ 2 = 153600 m ^ 2 . answer : a" | a ) 153600 , b ) 153629 , c ) 151299 , d ) 153129 , e ) 152109 | a | rectangle_area(divide(divide(multiply(multiply(divide(12, multiply(const_10, multiply(const_3, const_2))), 8), const_1000), add(3, 2)), const_2), multiply(divide(divide(multiply(multiply(divide(12, multiply(const_10, multiply(const_3, const_2))), 8), const_1000), add(3, 2)), const_2), 2)) | add(n0,n1)|multiply(const_2,const_3)|multiply(#1,const_10)|divide(n2,#2)|multiply(n3,#3)|multiply(#4,const_1000)|divide(#5,#0)|divide(#6,const_2)|multiply(n1,#7)|rectangle_area(#7,#8)| | physics |
if 9 engines consume 24 metric tonnes of coal , when each is working 8 hoursa day , how much coal will be required for 8 engines , each running 13 hours a day , it is being given that 3 engines of former type consume as much as 4 engines of latter type ? | explanation : let 3 engines of former type consume 1 unit in 1 hour . then , 4 engines of latter type consume 1 unit in 1 hour . therefore 1 engine of former type consumes ( 1 / 3 ) unit in 1 hour . 1 engine of latter type consumes ( 1 / 4 ) unit in 1 hour . let the required consumption of coal be x units . less engine... | a ) 23 , b ) 25 , c ) 26 , d ) 28 , e ) 30 | c | multiply(divide(multiply(const_0_25, 24), divide(9, 3)), 13) | divide(n0,n5)|multiply(n1,const_0_25)|divide(#1,#0)|multiply(n4,#2) | physics |
it takes 40 identical printing presses 15 hours to print 500,000 papers . how many hours would it take 30 of these printing presses to print 500,000 papers ? | 40 printing presses can do 1 / 15 of the job each hour . 30 printing presses can do 3 / 4 * 1 / 15 = 1 / 20 of the job each hour . the answer is c . | a ) 16 , b ) 18 , c ) 20 , d ) 21 , e ) 24 | c | divide(multiply(divide(const_1000, const_2), const_1000), multiply(divide(divide(multiply(divide(const_1000, const_2), const_1000), 40), 15), 30)) | divide(const_1000,const_2)|multiply(#0,const_1000)|divide(#1,n0)|divide(#2,n1)|multiply(n3,#3)|divide(#1,#4) | physics |
a rectangular field is to be fenced on three sides leaving a side of 20 feet uncovered . if the area of the field is 400 sq . ft , how many feet of fencing will be required ? | "explanation : we are given with length and area , so we can find the breadth . as length * breadth = area = > 20 * breadth = 400 = > breadth = 20 feet area to be fenced = 2 b + l = 2 * 20 + 20 = 60 feet answer : option a" | a ) 60 feet , b ) 50 feet , c ) 40 feet , d ) 30 feet , e ) 5 feet | a | add(multiply(divide(400, 20), const_2), 20) | divide(n1,n0)|multiply(#0,const_2)|add(n0,#1)| | geometry |
one night 20 percent of the female officers on a police force were on duty . if 100 officers were on duty that night and half of these were female officers , how many female officers were on the police force ? | "let total number of female officers in the police force = f total number of officers on duty on that night = 100 number of female officers on duty on that night = 100 / 2 = 50 ( 20 / 100 ) * f = 50 = > f = 250 answer c" | a ) 90 , b ) 180 , c ) 250 , d ) 500 , e ) 1,000 | c | divide(divide(100, const_2), divide(20, const_100)) | divide(n1,const_2)|divide(n0,const_100)|divide(#0,#1)| | gain |
if a certain number of people can dig earth 100 m deep , 25 m long , and 30 m broad in 12 days , then find the number of days the same number of people will require to dig earth 75 m deep , 20 m long and 50 m broad . | explanation : more number of days means – more length , more depth and more width . hence , it ’ s a direct proportion . ( 100 * 25 * 30 ) : ( 75 * 20 * 50 ) : : 12 : x 75000 : 75000 : : 12 : x x = 12 answer a | a ) 12 , b ) 18 , c ) 6 , d ) 1 , e ) 17 | a | divide(multiply(multiply(multiply(100, 25), 30), 12), multiply(multiply(75, 20), 50)) | multiply(n0,n1)|multiply(n4,n5)|multiply(n2,#0)|multiply(n6,#1)|multiply(n3,#2)|divide(#4,#3) | physics |
a mixture of 125 gallons of wine and water contains 20 % water . how much water must be added to the mixture in order to increase the percentage of water to 25 % of the new mixture ? | "in 125 gallons of the solution there are 0.2 ∗ 125 = 250 gallons of water . we want to add ww gallons of water to 125 gallons of solution so that 25 + w gallons of water to be 25 % of new solution : 25 + w = 0.25 ( 125 + w ) - - > w = 25 / 3 ≈ 8.33 answer : e ." | a ) 10 glas , b ) 8.5 gals , c ) 8 gals , d ) 6.66 gals , e ) 8.33 gals | e | divide(subtract(multiply(divide(multiply(divide(20, const_100), 125), const_100), 125), 25), subtract(const_1, divide(multiply(divide(20, const_100), 125), const_100))) | divide(n1,const_100)|multiply(n0,#0)|divide(#1,const_100)|multiply(n0,#2)|subtract(const_1,#2)|subtract(#3,n2)|divide(#5,#4)| | general |
the average ( arithmetic mean ) of 16 students first quiz scores in a difficult english class is 61.5 . when one student dropped the class , the average of the remaining scores increased to 64.0 . what is the quiz score of the student who dropped the class ? | "total score of 16 students is 16 * 61.50 = 984 total score of 15 students is 15 * 64 = 960 so , the score of the person who left is 24 ( 984 - 960 ) answer will be ( b )" | a ) a ) 10 , b ) b ) 24 , c ) c ) 40 , d ) d ) 55 , e ) e ) 70 | b | subtract(multiply(16, 61.5), multiply(subtract(16, const_1), 64.0)) | multiply(n0,n1)|subtract(n0,const_1)|multiply(n2,#1)|subtract(#0,#2)| | general |
the ratio , by volume , of soap to alcohol to water in a certain solution is 4 : 40 : 90 . the solution will be altered so that the ratio of soap to alcohol is doubled while the ratio of soap to water is halved . if the altered solution will contain 200 cubic centimeters of alcohol , how many cubic centimeters of water... | soap : alcohol initial ratio soap : alcohol : water - - > 4 : 40 : 90 initial soap : alcohol = 4 / 40 = 4 : 40 after doubled soap : alcohol = 2 * 4 / 40 = 8 : 40 initial soap : water = 4 / 90 = 4 : 90 after halved soap : water : 1 / 2 * 4 / 90 = 2 / 90 = 2 : 90 after soap : alcohol : water - - > 8 : 40 : 360 - - > 1 : ... | ['a ) 1300', 'b ) 1500', 'c ) 1600', 'd ) 1700', 'e ) 1800'] | e | divide(divide(divide(divide(divide(volume_rectangular_prism(200, 90, 40), const_3), const_2), const_4), 4), const_4) | volume_rectangular_prism(n1,n2,n3)|divide(#0,const_3)|divide(#1,const_2)|divide(#2,const_4)|divide(#3,n0)|divide(#4,const_4) | geometry |
there is enough provisions for 2400 women in an army camp for 100 days . if there were 1200 men less , how long will the provision last ? | "we have , m 1 d 1 = m 2 d 2 2400 * 100 = 1200 * d 2 d 2 = 2400 * 100 / 1200 = 200 days . answer : d" | a ) 170 , b ) 180 , c ) 190 , d ) 200 , e ) 210 | d | divide(multiply(2400, 100), 1200) | multiply(n0,n1)|divide(#0,n2)| | physics |
a thief steals a car at 2.30 pm and drives it at 60 kmph . the theft is discovered at 3 pm and the owner sets off in another car at 75 kmph when will he overtake the thief | let the thief is overtaken x hrs after 2.30 pm distance covered by the thief in x hrs = distance covered by the owner in x - 1 / 2 hr 60 x = 75 ( x - ½ ) x = 5 / 2 hr thief is overtaken at 2.30 pm + 2 ½ hr = 5 pm answer is d . | a ) 3.3 pm , b ) 5.5 pm , c ) 4.9 pm , d ) 5 pm , e ) 5.2 pm | d | add(3, divide(divide(60, const_2), subtract(75, 60))) | divide(n1,const_2)|subtract(n3,n1)|divide(#0,#1)|add(n2,#2) | physics |
two cars a and b start from boston and new york respectively simultaneously and travel towards each other at constant speeds along the same route . after meeting at a point between boston and new york the two cars a and b proceed to their respective destinations of new york and boston . car a reaches new york 100 minut... | "both cars leave at the same time both cars travel at constant speed stealing a useful piece of information from paragkan : if two objects a and b start from opposite points and , after having met en route , reach their respective destinations in a and b mins ( or any other measure of time ) respectively , then the rat... | a ) 1 hour , b ) 1 hour 10 minutes , c ) 2 hours 30 minutes , d ) 1 hour 40 minutes , e ) 2 hours 40 minutes | e | add(multiply(sqrt(divide(100, 90)), 90), 100) | divide(n0,n1)|sqrt(#0)|multiply(n1,#1)|add(n0,#2)| | physics |
if a person walks at 15 km / hr instead of 10 km / hr , he would have walked 20 km more . the actual distance traveled by him is : | "let the actual distance travelled be x km . x / 10 = ( x + 20 ) / 15 15 x = 10 x + 200 5 x = 200 x = 40 km . answer : d" | a ) 50 km , b ) 56 km , c ) 60 km , d ) 40 km , e ) 80 km | d | multiply(10, divide(20, subtract(15, 10))) | subtract(n0,n1)|divide(n2,#0)|multiply(n1,#1)| | general |
nina has exactly enough money to purchase 6 widgets . if the cost of each widget were reduced by $ 1.5 , then nina would have exactly enough money to purchase 8 widgets . how much money does nina have ? | "d its is . let price = x ( x - 1.5 ) 8 = 6 x x = 6 hence total money = 6 * 6 = 36" | a ) $ 22 , b ) $ 24 , c ) $ 30 , d ) $ 36 , e ) $ 40 | d | multiply(divide(multiply(1.5, 8), subtract(8, 6)), 6) | multiply(n1,n2)|subtract(n2,n0)|divide(#0,#1)|multiply(n0,#2)| | general |
there are 40 doors marked with numbers 1 to 40 . there are 40 individuals marked 1 to 40 . an operation on a door is defined as changing the status of the door from open to closed or vice versa . all the doors are closed to start with . one at a time , one randomly chosen individual goes and operates the doors . the in... | if a door is closed at the start , it requires an odd number of people to operate to be open at the end . only the perfect squares have an odd number of factors . the doors which are open at the end are : 1 , 4 , 9 , 16 , 25 , 36 for a total of 6 doors . the answer is d . | a ) 3 , b ) 4 , c ) 5 , d ) 6 , e ) 7 | d | subtract(10, const_4) | subtract(n8,const_4) | general |
for any number y , y * is defined as the greatest positive even integer less than or equal to y . what is the value of 7.2 – 7.2 * ? | since y * is defined as the greatest positive even integer less than or equal to y , then 7.2 * = 6 ( the greatest positive even integer less than or equal to 7.2 is 6 ) . hence , 7.2 – 7.2 * = 7.2 - 6 = 1.2 answer : a . | a ) 1.2 , b ) 0.2 , c ) 1.8 , d ) 2.2 , e ) 4.0 | a | subtract(7.2, subtract(floor(7.2), const_1)) | floor(n0)|subtract(#0,const_1)|subtract(n0,#1) | general |
p and q started a business investing rs 85000 and rs 10000 resp . in what ratio the profit earned after 2 years be divided between p and q respectively . | "explanation : in this type of question as time frame for both investors is equal then just get the ratio of their investments . p : q = 85000 : 10000 = 85 : 10 = 17 : 2 option d" | a ) 17 : 5 , b ) 17 : 3 , c ) 17 : 6 , d ) 17 : 2 , e ) 17 : 8 | d | divide(85000, 10000) | divide(n0,n1)| | gain |
a bag contains 5 white marbles and 5 black marbles . if each of 5 girls and 5 boys randomly selects and keeps a marble , what is the probability that all of the girls select the same colored marble ? | "consider a case when all girls select black marbles . . . so girl 1 will chose 1 black marble out of 10 . . and black marble will be chosen from 5 black marbles girl 1 will have a probability of picking black marble 5 / 10 girl 2 will have to pick a black marble out of 4 and total remaining no . of marbles 9 = 4 / 9 s... | a ) 1 / 126 , b ) 1 / 120 , c ) 1 / 24 , d ) 4 / 25 , e ) 1 / 2 | a | divide(const_2, choose(add(5, 5), 5)) | add(n0,n0)|choose(#0,n0)|divide(const_2,#1)| | probability |
if the radius of a circle that centers at the origin is 5 , how many m points on the circle have integer coordinates ? | "i understand this might not be required but i used the equation of a circle . since the origin is at 0 , x ^ 2 + y ^ 2 = 5 ^ 2 . x , y could be + / - ( 0,5 or 5,0 ) - 4 possibilities . x , y could be + / - ( 3,4 or 4,3 ) - 8 possibilities . ans : m = c" | a ) 4 , b ) 8 , c ) 12 , d ) 15 , e ) 20 | c | subtract(subtract(multiply(5, 5), add(5, const_4)), power(const_2, const_2)) | add(n0,const_4)|multiply(n0,n0)|power(const_2,const_2)|subtract(#1,#0)|subtract(#3,#2)| | physics |
the angle between the minute hand and the hour hand of a clock when the time is 8.30 , is | "angle between hands of a clock when the minute hand is behind the hour hand , the angle between the two hands at m minutes past h ' o clock = 30 ( h − m / 5 ) + m / 2 degree when the minute hand is ahead of the hour hand , the angle between the two hands at m minutes past h ' o clock = 30 ( m / 5 − h ) − m / 2 degree ... | a ) 35 ° , b ) 65 ° , c ) 45 ° , d ) 75 ° , e ) 95 ° | d | divide(multiply(subtract(multiply(divide(multiply(const_3, const_4), subtract(multiply(const_3, const_4), const_1)), multiply(add(const_4, const_1), subtract(multiply(const_3, const_4), const_1))), divide(const_60, const_2)), subtract(multiply(const_3, const_4), const_1)), const_2) | add(const_1,const_4)|divide(const_60,const_2)|multiply(const_3,const_4)|subtract(#2,const_1)|divide(#2,#3)|multiply(#0,#3)|multiply(#4,#5)|subtract(#6,#1)|multiply(#7,#3)|divide(#8,const_2)| | physics |
ravi and sunil are partners in a business . ravi invests rs . 10,000 for 8 months and sunil invested rs . 12000 for 10 months then after one year ratio of their profits will be | "= ( 15000 * 8 ) : ( 8000 * 10 ) = 80000 : 120000 = 2 : 3 answer : b" | a ) 1 : 3 , b ) 2 : 3 , c ) 3 : 2 , d ) 3 : 1 , e ) 3 : 4 | b | divide(multiply(multiply(multiply(multiply(8, const_3), multiply(10, const_2)), const_100), 8), multiply(12000, 10)) | multiply(const_3,n1)|multiply(n3,const_2)|multiply(n2,n3)|multiply(#0,#1)|multiply(#3,const_100)|multiply(n1,#4)|divide(#5,#2)| | gain |
dan can do a job alone in 15 hours . annie , working alone , can do the same job in just 10 hours . if dan works alone for 3 hours and then stops , how many hours will it take annie , working alone , to complete the job ? | "dan can complete 1 / 15 of the job per hour . in 3 hours , dan completes 3 ( 1 / 15 ) = 1 / 5 of the job . annie can complete 1 / 10 of the job per hour . to complete the job , annie will take 4 / 5 / 1 / 10 = 8 hours . the answer is d ." | a ) 2 , b ) 4 , c ) 6 , d ) 8 , e ) 10 | d | multiply(subtract(const_1, divide(3, 15)), 10) | divide(n2,n0)|subtract(const_1,#0)|multiply(n1,#1)| | physics |
alan buys 4 tvs , a 26 inch , a 28 inch , a 30 inch , and a 32 inch , for his new house . each tv costs $ 200 more than the size below it . alan spends a total of $ 3,800 . how much would he have spent if he had bought only the 28 inch and 30 inch tvs ? | assume the cost of the least sized ( 26 inch ) tv = x cost of 28 inches tv = x + 200 cost of 30 inches tv = x + 400 cost of 32 inches tv = x + 600 total cost = 4 x + 1200 = 3800 therefore x = 2600 / 4 = 650 price of 28 inch + 30 inch = 850 + 1050 = 1900 option c | a ) 850 , b ) 1,100 , c ) 1,900 , d ) 2,550 , e ) 3,400 | c | divide(add(add(divide(subtract(add(200, const_3600), multiply(add(add(const_1, const_2), const_3), 200)), const_4), 200), add(add(divide(subtract(add(200, const_3600), multiply(add(add(const_1, const_2), const_3), 200)), const_4), 200), 200)), subtract(multiply(const_2, const_10), const_1)) | add(n5,const_3600)|add(const_1,const_2)|multiply(const_10,const_2)|add(#1,const_3)|subtract(#2,const_1)|multiply(n5,#3)|subtract(#0,#5)|divide(#6,const_4)|add(n5,#7)|add(n5,#8)|add(#8,#9)|divide(#10,#4) | general |
1 , 3,5 , 7,9 , . 50 find term of sequnce | "this is an arithmetic progression , and we can write down a = 1 a = 1 , d = 2 d = 2 , n = 50 n = 50 . we now use the formula , so that sn = 12 n ( 2 a + ( n − 1 ) l ) sn = 12 n ( 2 a + ( n − 1 ) l ) s 50 = 12 × 50 × ( 2 × 1 + ( 50 − 1 ) × 2 ) s 50 = 12 × 50 × ( 2 × 1 + ( 50 − 1 ) × 2 ) = 25 × ( 2 + 49 × 2 ) = 25 × ( 2... | a ) 1234 , b ) 1367 , c ) 1789 , d ) 1568 , e ) 2500 | e | subtract(negate(50), multiply(subtract(3,5, 7,9), divide(subtract(3,5, 7,9), subtract(1, 3,5)))) | negate(n3)|subtract(n1,n2)|subtract(n0,n1)|divide(#1,#2)|multiply(#3,#1)|subtract(#0,#4)| | general |
how many multiples of 3 are there between 81 and 358 ? | "3 * 27 = 81 3 * 119 = 357 total multiples of 3 = ( 119 - 27 ) + 1 = 93 answer is c ." | a ) 94 , b ) 95 , c ) 93 , d ) 97 , e ) 98 | c | add(divide(subtract(358, 81), 3), const_1) | subtract(n2,n1)|divide(#0,n0)|add(#1,const_1)| | general |
a trader bought a car at 10 % discount on its original price . he sold it at a 40 % increase on the price he bought it . what percent of profit did he make on the original price ? | "original price = 100 cp = 80 s = 80 * ( 140 / 100 ) = 126 100 - 126 = 26 % answer : d" | a ) 118 , b ) 110 , c ) 112 , d ) 126 , e ) 115 | d | multiply(subtract(divide(divide(multiply(subtract(const_100, 10), add(const_100, 40)), const_100), const_100), const_1), const_100) | add(n1,const_100)|subtract(const_100,n0)|multiply(#0,#1)|divide(#2,const_100)|divide(#3,const_100)|subtract(#4,const_1)|multiply(#5,const_100)| | gain |
according to the formula f = 9 / 5 ( c ) + 32 , if the temperature in degrees farenheit ( f ) increases by 30 , by how much does the temperature in degrees celsius ( c ) increase ? | "you can plug in values . c = 5 / 9 * ( f - 32 ) f = 32 - - > c = 0 ; f = 32 + 30 = 62 - - > c = 5 / 9 * 30 = 16.66 . increase = 16.66 degrees . answer : b ." | a ) 9 , b ) 16.66 , c ) 47 , d ) 48 3 / 5 , e ) 59 | b | divide(multiply(30, 5), 9) | multiply(n1,n3)|divide(#0,n0)| | general |
a train covers a distance of 18 km in 10 min . if it takes 9 sec to pass a telegraph post , then the length of the train is ? | "speed = ( 18 / 10 * 60 ) km / hr = ( 108 * 5 / 18 ) m / sec = 30 m / sec . length of the train = 30 * 9 = 270 m . answer : c" | a ) 120 m , b ) 180 m , c ) 270 m , d ) 220 m , e ) 280 m | c | divide(18, subtract(divide(18, 10), 9)) | divide(n0,n1)|subtract(#0,n2)|divide(n0,#1)| | physics |
the smallest value of n , for which n + 1 is not a prime number , is | "( 1 + 1 ) = 2 . ( 2 + 1 ) = 3 . ( 3 + 1 ) = 4 . ( 4 + 1 ) = 5 . which is not prime , n = 3 . answer : d" | a ) 5 , b ) 6 , c ) 8 , d ) 3 , e ) 10 | d | add(1, 1) | add(n0,n0)| | general |
a number divided by 24 leaves remainder 56 what is the remainder when same number divided by 7 | "add 24 + 56 = 80 now 80 divided by 7 so we get 3 as reaminder answer : d" | a ) 9 , b ) 7 , c ) 5 , d ) 3 , e ) 1 | d | divide(subtract(56, 7), subtract(56, 7)) | subtract(n1,n2)|divide(#0,#0)| | general |
a shop owner professes to sell his articles at certain cost price but he uses false weights with which he cheats by 14 % while buying and by 20 % while selling . what is his percentage profit ? | the owner buys 100 kg but actually gets 114 kg ; the owner sells 100 kg but actually gives 80 kg ; profit : ( 114 - 80 ) / 80 * 100 = ~ 42.5 % answer : d . | a ) 10.22 % , b ) 20.22 % , c ) 21.22 % , d ) 42.5 % , e ) ca n ' t be calculated | d | divide(multiply(subtract(add(const_100, 14), subtract(const_100, 20)), const_100), subtract(const_100, 20)) | add(n0,const_100)|subtract(const_100,n1)|subtract(#0,#1)|multiply(#2,const_100)|divide(#3,#1) | gain |
9548 + 7314 = 3362 + ( ? ) | 9548 16862 = 3362 + x + 7314 x = 16862 - 3362 - - - - - = 13500 16862 - - - - - c ) | a ) 8300 , b ) 8400 , c ) 13500 , d ) 8700 , e ) 9000 | c | subtract(add(9548, 7314), 3362) | add(n0,n1)|subtract(#0,n2) | general |
john had a stock of 620 books in his bookshop . he sold 50 on monday , 82 on tuesday , 60 on wednesday , 48 on thursday and 40 on friday . what percentage of the books were not sold ? | "let n be the total number of books sold . hence n = 50 + 82 + 64 + 78 + 135 = 280 let m be the books not sold m = 620 - n = 620 - 280 = 340 percentage books not sold / total number of books = 340 / 620 = 0.55 = 54.84 % correct answer a" | a ) 54.84 % , b ) 36.5 % , c ) 46.5 % , d ) 56.5 % , e ) 66.5 % | a | multiply(divide(subtract(620, add(add(add(50, 82), add(60, 48)), 40)), 620), const_100) | add(n1,n2)|add(n3,n4)|add(#0,#1)|add(n5,#2)|subtract(n0,#3)|divide(#4,n0)|multiply(#5,const_100)| | gain |
a sum of money is distributed among w , x , y , z in the proportion of 1 : 6 : 2 : 4 . if x gets $ 1500 more than y , what is the w ' s share ? | let the shares of w , x , y , z are 1 a , 6 a , 2 a , 4 a 6 a - 2 a = 1500 4 a = $ 1500 , a = 375 w ' s share = 1 a = $ 375 answer is c | a ) $ 1500 , b ) $ 150 , c ) $ 375 , d ) $ 200 , e ) $ 450 | c | divide(divide(1500, subtract(divide(6, 2), const_1)), 2) | divide(n1,n2)|subtract(#0,const_1)|divide(n4,#1)|divide(#2,n2) | general |
a student has to obtain 33 % of the total marks to pass . he got 125 marks and failed by 40 marks . the maximum marks are : | "explanation : given that the student got 125 marks and still he failed by 40 marks = > the minimum pass mark = 125 + 40 = 165 given that minimum pass mark = 33 % of the total mark answer : a ) 500" | a ) 500 , b ) 387 , c ) 298 , d ) 267 , e ) 191 | a | divide(add(125, 40), divide(33, const_100)) | add(n1,n2)|divide(n0,const_100)|divide(#0,#1)| | general |
what is 15 percent of 58 ? | "( 15 / 100 ) * 58 = 8.7 the answer is c ." | a ) 6.9 , b ) 7.4 , c ) 8.7 , d ) 9.3 , e ) 10.4 | c | divide(multiply(15, 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) | add(const_2,const_3)|add(const_3,const_4)|add(const_3,const_3)|multiply(const_3,const_4)|multiply(#0,const_2)|multiply(#3,const_100)|multiply(#1,#0)|multiply(#4,#5)|multiply(#6,#4)|add(#7,#8)|add(#9,#2)|multiply(n0,#10)|divide(#11,const_100)| | gain |
if equation | x / 4 | + | y / 4 | = 2.5 encloses a certain region on the coordinate plane , what is the area of this region ? | the equation can be reduced to intercept form as | x / 10 | + | y / 10 | = 1 , so these are lines in four quadrants with x and y intercept as 10 , so it is a rhombus with diagonals of 20 each and hence area = 1 / 2 * d 1 * d 2 = 1 / 2 * 20 * 20 = 200 . answer e . | a ) 20 , b ) 50 , c ) 100 , d ) 400 , e ) 200 | e | add(power(multiply(const_2.0, 4), 4), power(multiply(4, 4), 4)) | multiply(const_2.0,const_2.0)|power(#0,n0)|add(#1,#1)| | general |
if 60 honey bees make 60 grams of honey in 60 days , then 1 honey bee will make 1 gram of honey in how many days ? | "explanation : let the required number days be x . less honey bees , more days ( indirect proportion ) less honey , less days ( direct proportion ) honey bees 1 : 60 : : 60 : x honey 60 : 1 = > 1 x 60 x x = 60 x 1 x 60 = > x = 60 . answer : e" | a ) 1 , b ) 3.5 , c ) 20 , d ) 49 , e ) 60 | e | divide(1, divide(1, 60)) | divide(n3,n0)|divide(n3,#0)| | physics |
what is the probability of randomly selecting one of the shortest diagonals from all the diagonals of a regular 7 - sided polygon ) ? | "from any vertex , there are two vertices on sides , which do not make a diagonal but a side . so the remaining n - 3 vertices make diagonals . there are 2 of these diagonals which are the shortest . the probability of choosing one of the shortest diagonals is 2 / 4 = 1 / 2 . the answer is a ." | a ) 1 / 2 , b ) 1 / 3 , c ) 1 / 4 , d ) 1 / 6 , e ) 1 / 7 | a | divide(7, divide(multiply(7, subtract(7, const_3)), const_2)) | subtract(n0,const_3)|multiply(n0,#0)|divide(#1,const_2)|divide(n0,#2)| | probability |
find the remainder when 3 ^ 27 is divided by 5 ? | "solution : 3 ^ 27 = ( ( 3 ^ 4 ) ^ 6 ) * ( 3 ^ 3 ) = ( 81 ^ 6 ) * 27 then unit digit of ( 81 ^ 6 ) is 1 so on multiplying with 27 , unit digit in the result will be 7 . now , 7 when divided by 5 gives 2 as remainder . answer b" | a ) 3 , b ) 2 , c ) 4 , d ) 1 , e ) 5 | b | reminder(power(3, 27), 5) | power(n0,n1)|reminder(#0,n2)| | general |
a is twice as good a work man as b and together they finish a piece of work in 16 days . the number of days taken by a alone to finish the work is : | "solution ( a ’ s 1 day ’ s work ) : ( b ’ s 1 day ’ s work ) = 2 : 1 . ( a + b ) ' s 1 day ’ s work = 1 / 16 divide 1 / 14 in the ratio 2 : 1 . ∴ a ’ s 1 day ’ s work = ( 1 / 14 x 2 / 3 ) = 1 / 24 hence , a alone can finish the work in 24 days . answer b" | a ) 36 days , b ) 24 days , c ) 46 days , d ) 16 days , e ) 14 days | b | inverse(divide(inverse(16), add(const_2, const_1))) | add(const_1,const_2)|inverse(n0)|divide(#1,#0)|inverse(#2)| | physics |
a cube of edge 15 cm is imersed completely in a rectangular vessel containing water . if the dimensions of the base of vessel are 20 cm * 15 cm , find the rise in waer level . | sol . increase in volume = volume of the cube = ( 15 * 15 * 15 ) cm ³ . ∴ rise in water level = [ volume / area ] = [ 15 * 15 * 15 / 20 * 15 ] cm = 11.25 cm . answer a | ['a ) 11.25 cm .', 'b ) 12.25 cm', 'c ) 13.25 cm', 'd ) 14.25 cm', 'e ) none of these'] | a | divide(volume_cube(15), multiply(15, 20)) | multiply(n0,n1)|volume_cube(n0)|divide(#1,#0) | geometry |
a volunteer organization is recruiting new members . in the fall they manage to increase their number by 7 % . by the spring however membership falls by 19 % . what is the total change in percentage from fall to spring ? | "( 100 % + 7 % ) * ( 100 % - 19 % ) = 1.07 * . 81 = 0.8667 1 - 0.8667 = 13.33 % lost = - 13.33 % the answer is d the organization has lost 13.33 % of its total volunteers from fall to spring ." | a ) 16.16 % , b ) 15.55 % , c ) 14.44 % , d ) 13.33 % , e ) 12.22 % | d | subtract(const_100, multiply(multiply(add(const_1, divide(7, const_100)), subtract(const_1, divide(19, const_100))), const_100)) | divide(n0,const_100)|divide(n1,const_100)|add(#0,const_1)|subtract(const_1,#1)|multiply(#2,#3)|multiply(#4,const_100)|subtract(const_100,#5)| | general |
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