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 ) 114 , b ) 120 , c ) 134 , d ) 144 , e ) none of these | d | divide(multiply(378, multiply(4, 2)), add(multiply(10, 2), 1)) | p , q , r enter into a partnership & their share are in the ratio 1 / 2 : 1 / 3 : 1 / 4 , after two months , p withdraws half of the capitals & after 10 months , a profit of rs 378 is divided among them . what is q ' s share ? | explanation : the ratio of their initial investment = 1 / 2 : 1 / 3 : 1 / 4 = 6 : 4 : 3 let ' s take the initial investment of p , q and r as 6 x , 4 x and 3 x respectively a : b : c = ( 6 x * 2 + 3 x * 10 ) : 4 x * 12 : 3 x * 12 = ( 12 + 30 ) : 4 * 12 : 3 * 12 = ( 4 + 10 ) : 4 * 4 : 12 = 14 : 16 : 12 = 7 : 8 : 6 b ' s share = 378 * ( 8 / 21 ) = 18 * 8 = 144 . answer : option d | a = 4 * 2
b = 378 * a
c = 10 * 2
d = c + 1
e = b / d
|
a ) − 48 , b ) − 5 , c ) 2 , d ) 46 , e ) 48 | b | add(divide(36, const_10), divide(36, divide(36, const_10))) | if a ( a + 5 ) = 36 and b ( b + 5 ) = 36 , where a ≠ b , then a + b = | "i . e . if a = 4 then b = - 9 or if a = - 9 then b = 4 but in each case a + b = - 9 + 4 = - 5 answer : option b" | a = 36 / 10
b = 36 / 10
c = 36 / b
d = a + c
|
a ) 22 , b ) 38 , c ) 62 , d ) 122 , e ) 672 | d | multiply(2, divide(divide(2013, add(const_10, const_1)), const_3)) | a computer is programmed to multiply consecutive even integers 2 * 4 * 6 * 8 * … * n until the product is divisible by 2013 , what is the value of n ? | "same logic but with application of factorials . 2 * 4 * 6 * 8 * . . . * n = 2 ( 1 * 2 * 3 * 4 * . . . n / 2 ) = 2 * ( n / 2 ) ! now we have : 2 * ( n / 2 ) ! / 2013 = 2 * ( n / 2 ) ! / ( 3 × 11 × 61 ) for the expression 2 * ( n / 2 ) ! to be divisible by 2013 min value of n / 2 should be 61 . so n / 2 = 61 = > n = 122 answer d ." | a = 10 + 1
b = 2013 / a
c = b / 3
d = 2 * c
|
a ) 242 , b ) 312 , c ) 424 , d ) 525 , e ) 622 | a | add(add(add(const_10, const_10), const_2), add(multiply(multiply(132, divide(const_1, 3)), const_2), 132)) | billy worked for 3 straight hours on his homework questions . if he solved 132 questions in the third hour , which was twice as many as he solved in the second hour , and 3 times as many questions as he solved in the first hour , how many questions did he solve total ? | total questions solved = questions solved in 3 rd hr + in 2 nd hr + in 1 st hr 132 + 132 / 2 + 132 / 3 = 242 a is the answer | a = 10 + 10
b = a + 2
c = 1 / 3
d = 132 * c
e = d * 2
f = e + 132
g = b + f
|
a ) 0 and 3 , b ) 3 and 6 , c ) 6 and 9 , d ) 9 and 12 , e ) 12 and 15 | b | add(add(multiply(4, 4), const_10), const_10) | if e = x ^ 4 + y ^ 4 = 100 , then the greatest possible value of x is between | my attempt : if e = x ^ 4 + y ^ 4 = 100 , then the greatest possible value of x would be when y is minimum . let y ^ 4 be 0 . now x ^ 4 = 100 . x should be definitely greater than 3 but less than 4 . the only option that fits this range is b hence answer is - - b ) 3 and 6 . | a = 4 * 4
b = a + 10
c = b + 10
|
a ) rs . 660 , b ) rs . 675 , c ) rs . 860 , d ) rs . 960 , e ) none of these | b | multiply(750, subtract(const_1, divide(10, const_100))) | a man buys an item at rs . 750 and sells it at the loss of 10 percent . then what is the selling price of that item | "explanation : here always remember , when ever x % loss , it means s . p . = ( 100 - x ) % of c . p when ever x % profit , it means s . p . = ( 100 + x ) % of c . p so here will be ( 100 - x ) % of c . p . = 90 % of 750 = 90 / 100 * 750 = 675 option b" | a = 10 / 100
b = 1 - a
c = 750 * b
|
a ) 32 , b ) 83 , c ) 45 , d ) 53 , e ) 64 | a | subtract(multiply(8, 61), subtract(multiply(13, 60), multiply(55, 7))) | the average of 13 numbers is 60 . average of the first 7 of them is 55 and that of the last 7 is 61 . find the 8 th number ? | "sum of all the 13 numbers = 13 * 60 = 780 sum of the first 7 of them = 7 * 55 = 385 sum of the last 7 of them = 7 * 61 = 427 so , the 8 th number = 427 + 385 - 780 = 32 . answer : a" | a = 8 * 61
b = 13 * 60
c = 55 * 7
d = b - c
e = a - d
|
a ) 0 , b ) 1 , c ) 2 , d ) 3 , e ) 4 | c | subtract(add(1, multiply(15, 6)), multiply(15, 6)) | how many two - digit numbers yield a remainder of 1 when divided by both 6 and 15 ? | "easier to start with numbers that are of the form 15 p + 1 - - - > 16,31 , 46,61 , 76,91 . out of these 2 ( 31,61 ) are also of the form 6 q + 1 . thus 2 is the answer . c is the correct answer ." | a = 15 * 6
b = 1 + a
c = 15 * 6
d = b - c
|
a ) 2 / 7 , b ) 9 / 4 , c ) 16 / 25 , d ) 7 / 1 , e ) 16 / 9 | e | divide(power(2, add(2, 2)), power(3, 2)) | for what value of â € œ k â € will the equation ( 2 kx 2 + 3 kx + 2 ) = 0 have equal roots ? | for a 2 nd degree equation ax 2 + bx _ c = 0 has equal roots the condition is b 2 - 4 ac = 0 in the given equation ( 3 k ) ^ 2 - 4 * 2 k * 2 = 0 by solving this equation we get k = 0 , k = 16 / 9 answer : e | a = 2 + 2
b = 2 ** a
c = 3 ** 2
d = b / c
|
a ) 80 , b ) 63 , c ) 120 , d ) 1008 , e ) 1260 | d | multiply(divide(multiply(9, subtract(9, const_1)), const_2), divide(multiply(8, subtract(8, const_1)), const_2)) | if 9 parallel lines in a plane is intersected by a family of another 8 parallel lines , how many parallelograms are there in the network thus formed ? | "parallelogram can formed by 2 horizontal and 2 vertical lines for horizontal 9 c 2 for vertical 8 c 2 total parallelogram is 9 c 2 * 8 c 2 = 36 * 28 = 1008 answer : d" | a = 9 - 1
b = 9 * a
c = b / 2
d = 8 - 1
e = 8 * d
f = e / 2
g = c * f
|
a ) 60 , b ) 50 , c ) 66 , d ) 54 , e ) none | a | subtract(divide(add(divide(multiply(20, const_100), const_100), const_100), subtract(const_1, divide(25, const_100))), const_100) | entry fee in an examination was re . 1 . later , this was reduced by 25 % which increased the sale by 20 % . the percentage increase in the number of visitors is | sol . let the total original sale be rs . 100 . then , original number of visitors = 100 . new number of visitors = 120 / 0.75 = 160 . ∴ increase % = 60 % . answer a | a = 20 * 100
b = a / 100
c = b + 100
d = 25 / 100
e = 1 - d
f = c / e
g = f - 100
|
a ) 2 , b ) 4 8 / 19 , c ) 5 , d ) 67 , e ) 9 | b | inverse(add(inverse(28), add(inverse(7), inverse(21)))) | a , b and c can do a piece of work in 7 days , 21 days and 28 days respectively . how long will they taken , if all the three work together ? | "1 / 7 + 1 / 21 + 1 / 28 = 19 / 84 = > 84 / 19 = > 4 8 / 19 days answer : b" | a = 1/(28)
b = 1/(7)
c = 1/(21)
d = b + c
e = a + d
f = 1/(e)
|
a ) 81 , b ) 100 , c ) 120 , d ) 135 , e ) 168 | e | divide(multiply(add(126, divide(multiply(126, 20), const_100)), const_100), multiply(multiply(const_3, const_3), 10)) | a retailer bought a machine at a wholesale price of $ 126 and later on sold it after a 10 % discount of the retail price . if the retailer made a profit equivalent to 20 % of the whole price , what is the retail price of the machine ? | my solution : wholesale price = 126 retail price , be = x he provides 10 % discount on retail price = x - 10 x / 100 this retail price = 20 % profit on wholesale price x - 10 x / 100 = 126 + 1 / 5 ( 126 ) x = 168 ; answer : e | a = 126 * 20
b = a / 100
c = 126 + b
d = c * 100
e = 3 * 3
f = e * 10
g = d / f
|
a ) 18.20 , b ) 18.33 , c ) 18 , d ) 18.3 , e ) none of these | b | multiply(110, divide(subtract(add(10, 110), const_100), add(10, 110))) | if rs . 10 be allowed as true discount on a bill of rs . 110 at the end of a certain time , then the discount allowed on the same sum due at the end of double the time is | "explanation : present worth = amount - truediscount = 110 - 10 = rs . 100 si on rs . 100 for a certain time = rs . 10 si on rs . 100 for doube the time = rs . 20 truediscount on rs . 120 = 120 - 100 = rs . 20 truediscount on rs . 110 = ( 110 x 20 / 120 ) = rs . 18.33 answer : b" | a = 10 + 110
b = a - 100
c = 10 + 110
d = b / c
e = 110 * d
|
a ) 76 , b ) 77 , c ) 78 , d ) 79 , e ) 80 | d | divide(subtract(multiply(85, const_4), 97), subtract(const_4, const_1)) | the grade point average of the entire class is 85 . if the average of one third of the class is 97 , what is the average of the rest of the class ? | "let x be the number of students in the class . let p be the average of the rest of the class . 85 x = ( 1 / 3 ) 97 x + ( 2 / 3 ) ( p ) x 255 = 97 + 2 p 2 p = 158 p = 79 . the answer is d ." | a = 85 * 4
b = a - 97
c = 4 - 1
d = b / c
|
a ) 4.4 m , b ) 220 m , c ) 440 m , d ) 560 , e ) 650 | c | multiply(multiply(add(const_3, subtract(const_0_25, divide(add(add(const_1000, const_60), add(multiply(const_2, const_10), const_4)), multiply(const_1000, const_10)))), 0.7), 200) | the diameter of a wheel is 0.7 m . how far will it travel in 200 revolutions | diameter ( d ) = 0.7 m circumference of the wheel , = 2 * pi * r = d * pi = 0.7 * 3.14 = 2.198 m . in one revolution , wheel will covered = 2.198 m , so , in 200 revolution , it will cover = 200 * 2.198 = 4.396 m = 440 m ( approx ) . answer : option c | a = 1000 + const_60
b = 2 * 10
c = b + 4
d = a + c
e = 1000 * 10
f = d / e
g = const_0_25 - f
h = 3 + g
i = h * 0
j = i * 200
|
a ) 45 % , b ) 65 % , c ) 85 % , d ) 115 % , e ) 125 % | d | multiply(divide(divide(subtract(multiply(35, const_100), multiply(20, const_100)), subtract(const_100, 35)), 20), const_100) | keats library purchases a number of new books , all in the category of biography , and the library does not acquire any other books . with the addition of the new biographies , the biography collection of the library amounts to 35 % of the total number of books in the library . if prior to the purchase , only 20 % of the books in keats library were biographies , by what percent has the number of biographies in the library increased ? | "let x be the number of new biographies added to the library . let b be the original number of biographies , so the original number of books was 5 b . 0.35 ( 5 b + x ) = b + x 0.75 b = 0.65 x x = 1.15 b the answer is d ." | a = 35 * 100
b = 20 * 100
c = a - b
d = 100 - 35
e = c / d
f = e / 20
g = f * 100
|
a ) 1 / 6 , b ) 1 / 12 , c ) 1 / 18 , d ) 1 / 2 , e ) 1 / 4 | a | multiply(multiply(multiply(divide(const_1, 6), divide(const_1, 6)), divide(const_1, 6)), divide(const_1, 6)) | two 6 faced dice are thrown together . the probability that both show the same number on them is ? | "it all 3 numbers have to be same basically we want doublets . 11 , 22 , 33 , 44 , 55 and 66 . those are six in number . further the three dice can fall in 6 * 6 = 36 ways . hence the probability is 6 / 36 = 1 / 6 answer : a" | a = 1 / 6
b = 1 / 6
c = a * b
d = 1 / 6
e = c * d
f = 1 / 6
g = e * f
|
a ) 56.33 , b ) 34.89 , c ) 23.67 , d ) 54.65 , e ) 43.98 | a | divide(add(add(multiply(40, 45), multiply(50, 55)), multiply(60, 65)), add(add(40, 50), 60)) | if the average marks of 3 batches of 40 , 50 and 60 students respectively is 45 , 55 , 65 , what is the average marks of all the students ? | average marks of batch 1 = 45 students in batch 1 = 40 total marks of batch 1 = 40 * 45 average marks of batch 2 = 55 students in batch 2 = 50 total marks of batch 2 = 50 × 55 average marks of batch 3 = 65 students in batch 3 = 60 total marks of batch 3 = 65 × 60 total students = 40 + 60 + 50 = 150 average marks of all the students = ( ( 40 * 45 ) + ( 50 × 55 ) + ( 65 × 60 ) ) / 150 = 169 / 3 = 56.333 answer is a . | a = 40 * 45
b = 50 * 55
c = a + b
d = 60 * 65
e = c + d
f = 40 + 50
g = f + 60
h = e / g
|
a ) 2 : 3 , b ) 5 : 8 , c ) 5 : 6 , d ) 3 : 4 , e ) 1 : 2 | e | divide(divide(subtract(12, 10), subtract(16, 10)), subtract(const_1, divide(subtract(12, 10), subtract(16, 10)))) | in what ratio must rice of rs . 16 per kg be mixed with rice of rs . 10 per kg so that cost of mixture is rs . 12 per kg ? | "( 12 - 10 ) / ( 16 - 12 ) = 2 / 4 = 1 : 2 answer : e" | a = 12 - 10
b = 16 - 10
c = a / b
d = 12 - 10
e = 16 - 10
f = d / e
g = 1 - f
h = c / g
|
a ) 52 , b ) 49 , c ) 91 , d ) 88 , e ) 66 | c | add(multiply(add(15, subtract(15, 3)), 3), 10) | in a bus left side are 15 seats available , 3 few seats in right side because in rear exit door . each seat hold 3 people . in addition , there is a seat back can sit 10 people all together . how many people can sit in a bus ? | right side = 15 seat left side = 15 - 3 ( 3 few seat in right side ) = 12 seat total = 15 + 12 = 27 people can seat in 27 seat = 27 * 3 = 81 people can seat in last seat = 10 total people can seat = 81 + 10 = 91 answer : c | a = 15 - 3
b = 15 + a
c = b * 3
d = c + 10
|
a ) 3 , b ) 4 , c ) 5 , d ) 6 , e ) 7 | e | subtract(subtract(12, 4), const_1) | if 4 < x < 8 < y < 12 , then what is the greatest possible positive integer difference of x and y ? | "let x = 4.1 and y = 11.1 greatest possible difference = 11.1 - 4.1 = 7 answer e" | a = 12 - 4
b = a - 1
|
a ) 22 , b ) 10 , c ) 25 , d ) 99 , e ) 21 | b | divide(subtract(multiply(90, 5), multiply(5, 70)), subtract(90, 80)) | the average mark of the students of a class in a particular exam is 80 . if 5 students whose average mark in that exam is 70 are excluded , the average mark of the remaining will be 90 . find the number of students who wrote the exam . | "let the number of students who wrote the exam be x . total marks of students = 80 x . total marks of ( x - 5 ) students = 90 ( x - 5 ) 80 x - ( 5 * 70 ) = 90 ( x - 5 ) 100 = 10 x = > x = 10 . answer : b" | a = 90 * 5
b = 5 * 70
c = a - b
d = 90 - 80
e = c / d
|
a ) 1500 , b ) 1400 , c ) 1700 , d ) 1800 , e ) 1600 | a | divide(multiply(power(add(divide(5, const_100), const_1), 4), 6450), 4) | what annual payment will discharge a debt of rs . 6450 due in 4 years at 5 % per annum ? | "explanation : he amount needs to be repaid in 4 years = rs . 6450 suppose rs . x is paid annually to repay this debt then , amount paid after 1 st year = rs . x interest for this rs . x for the remaining 3 years = x × 5 × 3 / 100 = 15 x / 100 then , amount paid after 2 nd year = rs . x interest for this rs . x for the remaining 2 years = x × 5 × 2 / 100 = 10 x / 100 amount paid after 3 rd year = rs . x interest for this rs . x for the remaining 1 year = x × 5 × 1 / 100 = 5 x / 100 amount paid after 4 th year = rs . x and this closes the entire debt = > x + 15 x / 100 + x + 10 x / 100 + x + 5 x / 100 + x = 6450 = > 4 x + 30 x / 100 = 6450 = > 4 x + 3 x / 10 = 6450 = > 40 x + 3 x = 64500 = > 43 x = 64500 = > x = 1500 answer : option a" | a = 5 / 100
b = a + 1
c = b ** 4
d = c * 6450
e = d / 4
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a ) $ 320 , b ) $ 800 , c ) $ 320 , d ) $ 600 , e ) $ 850 | b | divide(multiply(subtract(multiply(multiply(add(const_3, const_4), const_4), const_1000), multiply(multiply(add(const_4, const_1), const_4), const_1000)), 8), const_100) | in township k each property is taxed at 8 percent of its assessed value . if the assessed value of a property in township k is increased from $ 18,000 to $ 28,000 , by how much will the property tax increase ? | "increase in house value = $ 28,000 - $ 18,000 = $ 10000 so , tax increase = 8 % of $ 10000 = $ 800 answer : b" | a = 3 + 4
b = a * 4
c = b * 1000
d = 4 + 1
e = d * 4
f = e * 1000
g = c - f
h = g * 8
i = h / 100
|
a ) 25 % , b ) 33.3 % , c ) 40 % , d ) 60 % , e ) 66.6 % | a | divide(multiply(subtract(multiply(8, 8), multiply(6, 8)), const_100), multiply(8, 8)) | at a special sale , 8 tickets can be purchased for the price of 6 tickets . if 8 tickets are purchased at the sale , the amount saved will be what percent of the original price of the 8 tickets ? | "let the price of a ticket be rs . 100 , so 6 tickets cost 600 & 8 tickets cost 800 8 tickets purchased at price of 6 tickets ie . , for 600 , so amount saved s rs . 200 , % of 6 tickets = ( 200 / 800 ) * 100 = 25 % answer : a" | a = 8 * 8
b = 6 * 8
c = a - b
d = c * 100
e = 8 * 8
f = d / e
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a ) 21 , b ) 22 , c ) 23 , d ) 26 , e ) 28 | d | divide(subtract(add(add(29, 3), 29), subtract(11, 2)), subtract(11, subtract(11, 2))) | the cricket team of 11 members is 29 yrs old & the wicket keeper is 3 yrs older . if the ages ofthese 2 are excluded , the average age of theremaining players is 1 year less than the average age of the whole team . what is the average age of the team ? | "let the average age of the whole team be x years . 11 x - ( 29 + 32 ) = 9 ( x - 1 ) = > 11 x - 9 x = 52 = > 2 x = 52 = > x = 26 . so , average age of the team is 26 years . d" | a = 29 + 3
b = a + 29
c = 11 - 2
d = b - c
e = 11 - 2
f = 11 - e
g = d / f
|
a ) 90 , b ) 100 , c ) 110 , d ) 70 , e ) none of these | e | divide(2560, multiply(power(const_2, const_4), const_2)) | what is the greater of the two numbers whose product is 2560 , given that the sum of the two numbers exceeds their difference by 64 ? | let the greater and the smaller number be g and s respectively . gs = 2560 g + s exceeds g - s by 64 i . e . , g + s - ( g - s ) = 64 i . e . , 2 s = 64 = > s = 32 . g = 2560 / s = 80 . answer : e | a = 2 ** 4
b = a * 2
c = 2560 / b
|
a ) 76 hrs , b ) 99 hrs , c ) 55 hrs , d ) 90 hrs , e ) 10 hrs | e | inverse(subtract(divide(const_1, 5), divide(const_1, const_10))) | a cistern which could be filled in 5 hours takes one hour more to be filled owing to a leak in its bottom . if the cistern is full in what time will the leak empty it ? | "1 / 5 - 1 / x = 1 / 10 = > 10 hrs answer : e" | a = 1 / 5
b = 1 / 10
c = a - b
d = 1/(c)
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a ) a ) 12 , b ) b ) 18 , c ) c ) 24 , d ) d ) 30 , e ) e ) 36 | c | subtract(const_60, multiply(const_60, divide(36, 60))) | excluding stoppages , the speed of a train is 60 kmph and including stoppages it is 36 kmph . of how many minutes does the train stop per hour ? | "explanation : t = 24 / 60 * 60 = 24 answer : option c" | a = 36 / 60
b = const_60 * a
c = const_60 - b
|
a ) - 5 , b ) - 4 , c ) 4 , d ) 3 , e ) 2 | e | subtract(3, 1) | find value for x from below equation ? x + 1 = 3 | "1 . subtract 1 from both sides : x + 1 - 1 = 3 - 1 2 . simplify both sides : x = 2 e" | a = 3 - 1
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a ) 35 ° , b ) 65 ° , c ) 45 ° , d ) 195 ° , 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) | the angle between the minute hand and the hour hand of a clock when the time is 12.30 , 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 here h = 12 , m = 30 and minute hand is behind the hour hand . hence the angle = 30 ( h − m / 5 ) + m / 2 = 30 ( 12 − 30 / 5 ) + 30 / 2 = 30 ( 12 − 6 ) + 15 = 30 × 6 + 15 = 195 ° answer is d ." | a = 3 * 4
b = 3 * 4
c = b - 1
d = a / c
e = 4 + 1
f = 3 * 4
g = f - 1
h = e * g
i = d * h
j = const_60 / 2
k = i - j
l = 3 * 4
m = l - 1
n = k * m
o = n / 2
|
a ) 20 , b ) 21 , c ) 22 , d ) 23 , e ) 24 | d | add(divide(18, const_2), subtract(32, 18)) | jane started baby - sitting when she was 18 years old . whenever she baby - sat for a child , that child was no more than half her age at the time . jane is currently 32 years old , and she stopped baby - sitting 12 years ago . what is the current age of the oldest person for whom jane could have baby - sat ? | "check two extreme cases : jane = 18 , child = 9 , years ago = 32 - 18 = 14 - - > child ' s age now = 9 + 14 = 23 ; jane = 20 , child = 10 , years ago = 32 - 20 = 12 - - > child ' s age now = 10 + 12 = 22 . answer : d ." | a = 18 / 2
b = 32 - 18
c = a + b
|
a ) 16 % , b ) 33.33 % , c ) 33.69 % , d ) 33.98 % , e ) 37 % | b | multiply(divide(subtract(40, 30), 30), const_100) | john makes $ 30 a week from his job . he earns a raise andnow makes $ 40 a week . what is the % increase ? | "increase = ( 10 / 30 ) * 100 = ( 1 / 3 ) * 100 = 33.33 % . b" | a = 40 - 30
b = a / 30
c = b * 100
|
a ) - 3 , b ) - 1 , c ) - 1 / 3 , d ) 0 , e ) 1 | e | divide(add(divide(subtract(8, 4), 2), 2), divide(add(2, 6), 2)) | line m lies in the xy - plane . the y - intercept of line m is - 2 , and line m passes through the midpoint of the line segment whose endpoints are ( 2 , 8 ) and ( 6 , - 4 ) . what is the slope of line m ? | "the midpoint of ( 2,8 ) and ( 6 , - 4 ) is ( 4,2 ) . the slope of a line through ( 0 , - 2 ) and ( 4,2 ) is ( 2 - ( - 2 ) ) / ( 4 - 0 ) = 4 / 4 = 1 the answer is e ." | a = 8 - 4
b = a / 2
c = b + 2
d = 2 + 6
e = d / 2
f = c / e
|
a ) 20 , b ) 25 , c ) b = 30 , d ) b = 35 , e ) b = 40 | d | divide(subtract(const_100, multiply(const_10, const_3)), multiply(divide(subtract(39.4, 19.4), 1000), const_100)) | the annual birth and death rate in a country per 1000 are 39.4 and 19.4 respectively . the number of years b in which the population would be doubled assuming there is no emigration or immigration is | "suppose the population of the country in current year is 1000 . so annual increase is 1000 + 39.4 - 19.4 = 1020 hence every year there is an increase of 2 % . 2000 = 1000 ( 1 + ( 2 / 100 ) ) ^ n n = 35 answer is d ." | a = 10 * 3
b = 100 - a
c = 39 - 4
d = c / 1000
e = d * 100
f = b / e
|
a ) 12 , b ) 38 , c ) 60 , d ) 88 , e ) 72 | a | divide(multiply(36, 5), 15) | if 36 men can do a piece of work in 5 hours , in how many hours will 15 men do it ? | "explanation : let the required no of hours be x . then less men , more hours ( indirect proportion ) \ inline \ fn _ jvn \ therefore 15 : 36 : : 5 : x \ inline \ fn _ jvn \ leftrightarrow ( 15 x x ) = ( 36 x 5 ) \ inline \ fn _ jvn \ leftrightarrow \ inline \ fn _ jvn x = \ frac { 36 \ times 5 } { 15 } = 12 hence , 15 men can do it in 12 hours . answer : a ) 12" | a = 36 * 5
b = a / 15
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a ) 13333 , b ) 26758 , c ) 45896 , d ) 14639 , e ) 89576 | a | multiply(multiply(multiply(multiply(add(1, 4), 1), const_100), const_100), divide(1, add(add(2, 4), 1))) | in business , a and c invested amounts in the ratio 4 : 2 , whereas the ratio between amounts invested by a and b was 4 : 1 , if rs 20,000 was their profit , how much amount did a receive . | "explanation : a : b = 4 : 1 = 4 : 1 = > a : c = 4 : 2 = 4 : 2 = > a : b : c = 4 : 1 : 2 a share = ( 4 / 6 ) * 20000 = 13333 option a" | a = 1 + 4
b = a * 1
c = b * 100
d = c * 100
e = 2 + 4
f = e + 1
g = 1 / f
h = d * g
|
a ) 15 , b ) 20 , c ) 23 , d ) 29 , e ) 25 | d | add(multiply(10, const_2), multiply(subtract(24.5, multiply(10, const_2)), 10)) | if the average of 10 consecutive integers is 24.5 then the 10 th integer is : - | "the average falls between the 5 th and 6 th integers , integer 5 = 24 , integer 6 = 25 . counting up to the tenth integer we get 29 . answer : d" | a = 10 * 2
b = 10 * 2
c = 24 - 5
d = c * 10
e = a + d
|
a ) 1 / 18 , b ) 1 / 12 , c ) 1 / 4 , d ) 1 / 8 , e ) 1 / 6 | e | divide(const_1, factorial(const_3)) | the letters o , l and t are rearranged to form the word ' lot ' . find its probability . | explanation : there are total 3 letters . the probability that o gets the first position is 1 / 3 . likewise , probability for l and t are 1 / 2 and 1 / 1 . hence , the required probability is : - = > ( 1 / 3 ) x ( 1 / 2 ) x 1 . = > 1 / 6 . answer : e | a = math.factorial(3)
b = 1 / a
|
a ) a ) 23 , b ) b ) 25 , c ) c ) 27 , d ) d ) 31 , e ) e ) 35 | b | sqrt(add(power(sqrt(subtract(289, multiply(const_2, 168))), const_2), multiply(const_4, 168))) | the product of two numbers is 168 and the sum of their squares is 289 . the sum of the number is ? | "let the numbers be x and y . then , xy = 168 and x 2 + y 2 = 289 . ( x + y ) 2 = x 2 + y 2 + 2 xy = 289 + ( 2 x 168 ) = 625 x + y = 25 . option b" | a = 2 * 168
b = 289 - a
c = math.sqrt(b)
d = c ** 2
e = 4 * 168
f = d + e
g = math.sqrt(f)
|
a ) 55 , b ) 60 , c ) 73 , d ) 82 , e ) 88 | e | subtract(100, divide(subtract(100, 64), const_3)) | a teacher grades students ’ tests by subtracting twice the number of incorrect responses from the number of correct responses . if student a answers each of the 100 questions on her test and receives a score of 64 , how many questions did student a answer correctly ? | "let the number of correct responses be x then the number of incorrect responses = 100 - x according to question x - 2 ( 100 - x ) = 64 ( subtracting twice of incorrect from correct ) 3 x = 264 x = 88 answer : e" | a = 100 - 64
b = a / 3
c = 100 - b
|
a ) 40 , b ) 50 , c ) 10 ^ 4 , d ) 10 ^ 5 , e ) 10 ^ 6 | d | power(const_10, subtract(7, 2)) | on the richter scale , which measures the total amount of energy released during an earthquake , a reading of x - 1 indicates one - tenth the released energy as is indicated by a reading of x . on that scale , the frequency corresponding to a reading of 7 is how many times as great as the frequency corresponding to a reading of 2 ? | "if richter scale reading goes from x - 1 to x it will be 10 if richter scale reading goes from 2 to 3 it will be 10 if richter scale reading goes from 3 to 4 it will be 10 if richter scale reading goes from 4 to 5 it will be 10 if richter scale reading goes from 5 to 6 it will be 10 similarly if richter scale reading goes from 6 to 7 it will be 10 so it will from 2 to 7 i . e 3 , 4,5 , 6,7 = 10 * 10 * 10 * 10 * 10 = 10 ^ 5 answer is d" | a = 7 - 2
b = 10 ** a
|
a ) 340 , b ) 330 , c ) 320 , d ) 310 , e ) 890 | e | divide(add(add(multiply(100, 8000), divide(multiply(100, 8000), 10)), 10000), 1000) | a wildlife preserve is being planned for 8000 rhinoceroses . the preserve is to contain a total of 10000 acres of watering area , plus 100 acres of grazing area for each rhinoceros . if the number of rhinoceroses is expected to increase by 10 percent , how many 1000 acres should the preserve have in order to provide for the increased population ? | number of rhinos = 8000 watering area = 10,000 acres number of rhino to increase by 10 percent , then number of rhino = 8800 grazing area for a rhino = 100 total grazing area for 8800 rhinos = 8800 * 100 = 8 , 80,000 total area required for the wildlife preserve = 8 , 80,000 + 10,000 = 8 , 90,000 = 890 * 1000 answer e | a = 100 * 8000
b = 100 * 8000
c = b / 10
d = a + c
e = d + 10000
f = e / 1000
|
a ) 1 : 2 , b ) 1 : 3 , c ) 2 : 3 , d ) 3 : 2 , e ) 4 : 7 | e | divide(subtract(multiply(add(const_1, divide(1.2, const_100)), const_1000), multiply(subtract(const_1, divide(6, const_100)), const_1000)), subtract(multiply(add(const_1, divide(14, const_100)), const_1000), multiply(add(const_1, divide(1.2, const_100)), const_1000))) | at a certain organisation , the number of male members went up by 14 % in the year 2001 from year 2000 , and the number of females members went down by 6 % in the same time period . if the total membership at the organisation went up by 1.2 % from the year 2000 to 2001 , what was the ratio of male members to female members in the year 2000 ? | "men increase by 14 % = = > 1.14 m = males in 2001 women decrease by 6 % = = > 0.94 f = women in 2001 total employees increase by 1.2 % = = > 1.012 * ( m + f ) = total number of employees in 2001 obviously ( males in 2001 ) + ( females in 2001 ) = total number of employees in 2001 1.14 m + 0.94 f = 1.012 * ( m + f ) 1.14 m + 0.94 f = 1.012 m + 1.012 f 1.14 m - 1.012 m = 1.012 f - 0.94 f 0.128 m = 0.072 f m / f = ( 0.072 ) / ( 0.128 ) = 72 / 128 = 4 / 7 answer = ( e )" | a = 1 / 2
b = 1 + a
c = b * 1000
d = 6 / 100
e = 1 - d
f = e * 1000
g = c - f
h = 14 / 100
i = 1 + h
j = i * 1000
k = 1 / 2
l = 1 + k
m = l * 1000
n = j - m
o = g / n
|
a ) 16 , b ) 18 , c ) 20 , d ) 24 , e ) 30 | c | multiply(4, divide(50, add(4, 6))) | maxwell leaves his home and walks toward brad ' s house at the same time that brad leaves his home and runs toward maxwell ' s house . if the distance between their homes is 50 kilometers , maxwell ' s walking speed is 4 km / h , and brad ' s running speed is 6 km / h . what is the distance traveled by maxwell when they meet in the middle ? | "distance between maxwell ' s and brad ' s home = 50 km maxwell ' s speed = 4 km / h brad ' s speed = 6 km / h relative speed of approach = 10 km / h time = 50 / 10 = 5 hours distance traveled by maxwell when they meet in the middle = 4 * 5 = 20 km answer c" | a = 4 + 6
b = 50 / a
c = 4 * b
|
a ) $ 112.00 , b ) $ 64.80 , c ) $ 90.00 , d ) $ 100.80 , e ) $ 154.80 | a | divide(multiply(subtract(1720, 600), 10), const_100) | a tourist does not have to pay tax on the first $ 600 of goods he purchases in country b , but does have to pay a 10 percent tax on the portion of the total value that is in excess of $ 600 . what tax must be paid by a tourist if he buys goods with a total value of $ 1720 ? | "correct answer : a the tourist must pay tax on $ 1720 - $ 600 = $ 1120 . thus , the amount of tax he has to pay is 0.1 ( $ 1120 ) = $ 112 . the correct answer is a ." | a = 1720 - 600
b = a * 10
c = b / 100
|
a ) 78 years , b ) 22 years , c ) 88 years , d ) 35 years , e ) 65 years | d | divide(subtract(37, subtract(multiply(const_2, const_2), const_2)), subtract(const_2, const_1)) | a man is 37 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 | "let ' s son age is x , then father age is x + 37 . = > 2 ( x + 2 ) = ( x + 37 + 2 ) = > 2 x + 4 = x + 39 = > x = 35 years answer : d" | a = 2 * 2
b = a - 2
c = 37 - b
d = 2 - 1
e = c / d
|
a ) 345 , b ) 340 , c ) 343 , d ) 349 , e ) 338 | a | add(multiply(subtract(10, const_1), subtract(40, const_2)), 10) | what is the sum of the digits of integer k , if k = ( 10 ^ 40 - 79 ) | "there are 41 digits in 10 ^ 40 when we subtract 79 from it , there will be 40 digits left . 10 ^ 40 can be written as 9999999 . . . . ( 40 times ) + 1 so , 10 ^ 40 - 79 = 9999999 . . . . ( 40 times ) + 1 - 79 = 9999999 . . . . ( 40 times ) - 78 consider the last 2 digits , 99 - 78 = 21 the last 2 digits will be 21 . and our number would be 99999 . . . . . . 99921 with 2 less 9 s after subtraction . number of 9 s left are 38 and the last two digits are 21 the sum of the digits will be ( 38 * 9 ) + 2 + 1 = 345 answer : - a" | a = 10 - 1
b = 40 - 2
c = a * b
d = c + 10
|
a ) 5760 , b ) 7667 , c ) 4445 , d ) 4566 , e ) 3344 | c | multiply(divide(multiply(multiply(multiply(const_4, const_2), multiply(const_4, const_2)), const_3), divide(const_100, const_2)), subtract(1500, multiply(const_4, const_100))) | by investing in 16 % stock at 64 , one earns rs . 1500 . the investment made is : | "explanation : to earn rs . , investment = rs . 64 . to earn rs . 1500 , investment = = rs . 5760 . answer : c ) 5760" | a = 4 * 2
b = 4 * 2
c = a * b
d = c * 3
e = 100 / 2
f = d / e
g = 4 * 100
h = 1500 - g
i = f * h
|
a ) 960 , b ) 1000 , c ) 1020 , d ) 1080 , e ) 1075 | a | divide(multiply(80, 600), 50) | if 80 percent of 600 is 50 percent of x , then x = ? | 0.8 * 600 = 0.5 * x x = 8 / 5 * 600 = 960 | a = 80 * 600
b = a / 50
|
a ) 4 , b ) 52 , c ) 16 , d ) 24 , e ) 64 | d | sqrt(add(560, multiply(8, 2))) | if a 2 + b 2 + c 2 = 560 and ab + bc + ca = 8 , then a + b + c is | "by formula , ( a + b + c ) ^ 2 = a ^ 2 + b ^ 2 + c ^ 2 + 2 ( ab + bc + ca ) , since , a ^ 2 + b ^ 2 + c ^ 2 = 560 and ab + bc + ca = 8 , ( a + b + c ) ^ 2 = 560 + 2 ( 8 ) = 576 = 24 ^ 2 therefore : a + b + c = 24 answer : d" | a = 8 * 2
b = 560 + a
c = math.sqrt(b)
|
a ) 1 kmph , b ) 5 kmph , c ) 5.5 kmph , d ) 4 kmph , e ) 7 kmph | c | divide(add(6.5, subtract(6.5, 2.5)), const_2) | a man whose speed is 6.5 kmph in still water rows to a certain upstream point and back to the starting point in a river which flows at 2.5 kmph , find his average speed for the total journey ? | "explanation : speed of man = 6.5 kmph speed of stream = 2.5 kmph speed in downstream = 9 kmph speed in upstream = 4 kmph average speed = ( 2 x 9 x 4 ) / 13 = 5.5 kmph . answer : c" | a = 6 - 5
b = 6 + 5
c = b / 2
|
a ) 200 , b ) 768 , c ) 150 , d ) 280 , e ) 279 | c | multiply(divide(multiply(subtract(add(multiply(divide(const_100, subtract(const_100, 15)), 1000), multiply(divide(const_100, add(const_100, 15)), 1000)), add(1000, 1000)), const_100), add(multiply(divide(const_100, subtract(const_100, 15)), 1000), multiply(divide(const_100, add(const_100, 15)), 1000))), const_100) | a shopkeeper buys two articles for rs . 1000 each and then sells them , making 15 % profit on the first article and 15 % loss on second article . find the net profit or loss percent ? | "profit on first article = 150 % of 1000 = 150 . this is equal to the loss he makes on the second article . that , is he makes neither profit nor loss . answer : c" | a = 100 - 15
b = 100 / a
c = b * 1000
d = 100 + 15
e = 100 / d
f = e * 1000
g = c + f
h = 1000 + 1000
i = g - h
j = i * 100
k = 100 - 15
l = 100 / k
m = l * 1000
n = 100 + 15
o = 100 / n
p = o * 1000
q = m + p
r = j / q
s = r * 100
|
a ) 7 , b ) 6 , c ) 5 , d ) 8 , e ) 9 | d | divide(multiply(8, add(divide(8, const_2), const_2)), 6) | a and b can do a work in 8 days , b and c can do the same work in 12 days . a , b and c together can finish it in 6 days . a and c together will do it in how many days ? | "let the total units of work be 48 . units of work completed by a and b in one day = 6 . units of work completed by b and c in one day = 4 units of work completed by a , b , and c in one day = 8 . from the above information we get the work completed by a and c in one day is 6 units . therefore , the number of days taken by a and c to complete the whole work = 48 / 6 = 8 days . answer : d" | a = 8 / 2
b = a + 2
c = 8 * b
d = c / 6
|
a ) a . 40 , b ) b . 100 , c ) c . 400 , d ) d . 1,000 , e ) e . 10,000 | c | divide(const_180, const_1000) | the volume of a sphere with radius r is ( 4 / 3 ) * pi * r ^ 3 and the surface area is 4 * pi * r ^ 3 . if a sperical balloon has a volume of 288 pi cubic centimeters , what is hte surface area of the balloon in square centimeters ? | "the surface area is 4 . pi . r ^ 2 ( its area remember not volume ) as 4 / 3 . pi . r ^ 3 = 288 pi r = 6 so area = 4 . pi . r ^ 2 = 144 . pi = 324 x 3.14 = 400 ( approx ) c" | a = const_180 / 1000
|
a ) 21 % , b ) 30 % , c ) 25 % , d ) 40 % , e ) 50 % | a | multiply(subtract(multiply(divide(add(const_100, 10), const_100), divide(add(const_100, 10), const_100)), const_1), const_100) | the percentage increase in the area of rectangle , if each of its side is increased by 10 % is ? | "let original length = x meters original breadth = y meters original area = xy m ^ 2 new length = 110 x / 100 new breadth = 110 y / 100 = 11 y / 10 new area = 11 x / 10 * 11 y / 10 = 121 xy / 100 m ^ 2 increase percent = 21 xy / 100 * 1 / xy * 100 = 21 % answer is a" | a = 100 + 10
b = a / 100
c = 100 + 10
d = c / 100
e = b * d
f = e - 1
g = f * 100
|
a ) 20 min , b ) 25 min , c ) 30 min , d ) 35 min , e ) 40 min | d | divide(35, 1) | a fill pipe can fill 1 / 2 of cistern in 35 minutes . in how many minutes , it can fill 1 / 2 of the cistern ? | "required time = 35 * 2 * 1 / 2 = 20 minutes answer is d" | a = 35 / 1
|
a ) 675 , rs . 1350 , b ) 575 , rs . 675 , c ) 1350 , rs . 675 , d ) 1450 , rs . 775 , e ) 1550 , rs . 875 | b | multiply(subtract(rectangle_area(add(75, multiply(2.5, const_2)), add(55, multiply(2.5, const_2))), rectangle_area(75, 55)), 1) | a rectangular grass field is 75 m * 55 m , it has a path of 2.5 m wide all round it on the outside . find the area of the path and the cost of constructing it at rs . 1 per sq m ? | "area = ( l + b + 2 d ) 2 d = ( 75 + 55 + 2.5 * 2 ) 2 * 2.5 = > 675 675 * 1 = rs . 675 answer : b" | a = 2 * 5
b = 75 + a
c = 2 * 5
d = 55 + c
e = rectangle_area - (
f = e * rectangle_area
|
['a ) 33', 'b ) 45', 'c ) 66', 'd ) 77', 'e ) 21'] | b | divide(volume_cylinder(divide(6, const_2), 5), const_pi) | the diameter of a cylindrical tin is 6 cm and height is 5 cm . find the volume of the cylinder ? | r = 3 h = 5 π * 3 * 3 * 5 = 45 π cc answer : b | a = 6 / 2
b = volume_cylinder / (
|
a ) 3.75 days , b ) 4.73 days , c ) 5.75 days , d ) 6.75 days , e ) 7.73 days | b | divide(multiply(10, 9), add(9, 10)) | b completes a work in 9 days . a alone can do it in 10 days . if both work together , the work can be completed in how many days ? | "1 / 9 + 1 / 10 = 19 / 90 90 / 19 = 4.73 days answer : b" | a = 10 * 9
b = 9 + 10
c = a / b
|
a ) − 48 , b ) − 2 , c ) − 7 , d ) 46 , e ) 48 | c | add(divide(18, const_10), divide(18, divide(18, const_10))) | if a ( a + 7 ) = 18 and b ( b + 7 ) = 18 , where a ≠ b , then a + b = | "a ( a + 7 ) = 18 = > we have a = 2 or - 9 also b ( b + 7 ) = 18 = > b = 2 or - 9 given a ≠ b 1 ) when a = 2 , b = - 9 and a + b = - 7 1 ) when a = - 9 , b = 2 and a + b = - 7 answer choice c" | a = 18 / 10
b = 18 / 10
c = 18 / b
d = a + c
|
a ) 20 , b ) 24 , c ) 28 , d ) 36 , e ) 40 | a | subtract(divide(add(40, 0), const_2), divide(add(80, 0), const_2)) | the average ( arithmetic mean ) of the even integers from 0 to 80 inclusive is how much greater than the average ( arithmetic mean ) of the even integers from 0 to 40 inclusive ? | "the sum of even numbers from 0 to n is 2 + 4 + . . . + n = 2 ( 1 + 2 + . . . + n / 2 ) = 2 ( n / 2 ) ( n / 2 + 1 ) / 2 = ( n / 2 ) ( n / 2 + 1 ) the average is ( n / 2 ) ( n / 2 + 1 ) / ( n / 2 + 1 ) = n / 2 the average of the even numbers from 0 to 80 is 80 / 2 = 40 the average of the even numbers from 0 to 40 is 40 / 2 = 20 the answer is a ." | a = 40 + 0
b = a / 2
c = 80 + 0
d = c / 2
e = b - d
|
a ) 33 : 25 , b ) 9 : 5 , c ) 47 : 25 , d ) 22 : 25 , e ) 97 : 50 | b | divide(add(80, const_100), const_100) | the number of sailors on a ship is 80 % more than the number of officers . the ratio of sailors to officers would be | "sailor = 1.80 * officer sailor / officer = 1.80 / 1 = 180 / 100 = 9 / 5 answer will be b" | a = 80 + 100
b = a / 100
|
a ) 23,500 , b ) 24,500 , c ) 25,500 , d ) 26,500 , e ) 27,500 | a | floor(divide(divide(subtract(550, multiply(1,000, divide(8, const_100))), subtract(divide(10, const_100), divide(8, const_100))), 1,000)) | angelo and isabella are both salespersons . in any given week , angelo makes $ 550 in base salary plus 8 percent of the portion of his sales above $ 1,000 for that week . isabella makes 10 percent of her total sales for any given week . for what amount of weekly sales would angelo and isabella earn the same amount of money ? | "let the weekly sales of both = x 550 + ( x − 1000 ) 8 / 100 = 10 / 100 x x = 23500 answer : a" | a = 8 / 100
b = 1 * 0
c = 550 - b
d = 10 / 100
e = 8 / 100
f = d - e
g = c / f
h = g / 1
i = math.floor(h)
|
a ) 182 , b ) 208 , c ) 224 , d ) 254 , e ) 244 | e | divide(195, divide(subtract(const_100, 20), const_100)) | a small college reduced its faculty by approximately 20 percent to 195 professors . what was the original number of faculty members ? | "if x is the original number of faculty members , then after 20 % reduction in faculty members number is . 80 x but we are given . 80 x = 195 x = 244 so the original number of faculty members is 244 correct answer - e" | a = 100 - 20
b = a / 100
c = 195 / b
|
a ) 10.5 , b ) 11 , c ) 11.5 , d ) 12 , e ) 12.5 | e | subtract(15.5, multiply(2, 1.5)) | the arithmetic mean and standard deviation of a certain normal distribution are 15.5 and 1.5 , respectively . what value is exactly 2 standard deviations less than the mean ? | "the value which isexactlytwo sd less than the mean is : mean - 2 * sd = 15.5 - 2 * 1.5 = 12.5 . answer : e ." | a = 2 * 1
b = 15 - 5
|
a ) 31 , b ) 33 , c ) 35 , d ) 38 , e ) 39 | c | divide(subtract(565, 5), gcd(180, 565)) | the greatest number that divides 180 , 565 and 705 leaving remainder 5 , 5 and 5 respectively is : | "explanation : 180 - 5 = 175 , 565 - 5 = 560 and 705 - 5 = 700 highest number that divides 175 , 560 and 700 is hcf of numbers . hcf of 175 , 560 and 700 = 35 answer : c" | a = 565 - 5
b = math.gcd(180, 565)
c = a / b
|
a ) 6 , b ) 7 , c ) 4 , d ) 9 , e ) 10 | c | divide(factorial(subtract(add(const_4, 4), const_1)), multiply(factorial(4), factorial(subtract(const_4, const_1)))) | how many positive integers less than 100 have a remainder of 4 when divided by 13 ? | "we have to include 4 also . as 13 * 0 + 4 = 2 if somebody says to divide 4 by 13 , we will be telling we have 0 quotient and remainder as 4 . answer is c" | a = 4 + 4
b = a - 1
c = math.factorial(b)
d = math.factorial(4)
e = 4 - 1
f = math.factorial(e)
g = d * f
h = c / g
|
a ) 3 . , b ) 4 . , c ) 5 . , d ) 7 . , e ) 8 . | d | divide(divide(divide(120, 4), const_2), const_3) | how many of the positive divisors f of 120 are also multiples of 4 not including 120 ? | "4 , 8,12 , 20,24 , 40,60 . ( 7 ) is the answer other way : factors of 120 = 2 ^ 3 * 3 * 5 separate 2 ^ 2 ( which means 4 ) now , calculate the number of other factors . f = 2 * 3 * 5 = total positive factors are 2 * 2 * 2 = 8 this 8 factors include 120 so subtract 1 from 8 ans is 7 = d" | a = 120 / 4
b = a / 2
c = b / 3
|
['a ) 4 cm ^ 2', 'b ) 6 cm ^ 2', 'c ) 5 cm ^ 2', 'd ) 7 cm ^ 2', 'e ) 8 cm ^ 2'] | b | sqrt(multiply(6, 6)) | a circle is inscribed in a triangle of side 6 cm . and a square is inscribed in the circle . what is the area of square ? | the radius of the incircle in a equilateral triangle is a / 2 ( sqrt ) 3 where a = side of triangle = 6 cm so the diameter of the circle would be the diagonal of square so diagonal = 2 * a / 2 ( sqrt ) 3 = a / ( sqrt ) 3 so area of square = 1 / 2 * ( diagonal ) ^ 2 = 6 * 6 / 2 * 3 = 6 cm ^ 2 answer : b | a = 6 * 6
b = math.sqrt(a)
|
a ) 4.85 , b ) 7.85 , c ) 6.85 , d ) 5.85 , e ) 6.62 | e | divide(add(111, 165), multiply(add(60, 90), const_0_2778)) | two trains 111 meters and 165 meters in length respectively are running in opposite directions , one at the rate of 60 km and the other at the rate of 90 kmph . in what time will they be completely clear of each other from the moment they meet ? | "t = ( 111 + 165 ) / ( 60 + 90 ) * 18 / 5 t = 6.62 answer : e" | a = 111 + 165
b = 60 + 90
c = b * const_0_2778
d = a / c
|
a ) a ) 70 , b ) b ) 77 , c ) c ) 79 , d ) d ) 84 , e ) e ) 82 | d | multiply(divide(divide(multiply(20, add(20, const_1)), const_2), 20), 7) | what is the average of first 20 multiples of 7 ? | "required average = 7 ( 1 + 2 + . . . . + 20 ) / 20 ( 8 / 20 ) x ( ( 20 x 21 ) / 2 ) ( because sum of first 21 natural numbers ) = 84 d" | a = 20 + 1
b = 20 * a
c = b / 2
d = c / 20
e = d * 7
|
a ) 1184 , b ) 1723 , c ) 1129 , d ) 2613 , e ) 1372 | a | divide(148, subtract(inverse(const_2), subtract(const_1, divide(62.5, const_100)))) | all the milk in container a which was filled to its brim was poured into two containers b and c . the quantity of milk in container b was 62.5 % less than the capacity of container a . if 148 liters was now transferred from c to b , then both the containers would have equal quantities of milk . what was the initial quantity of milk in container a ? | "a 1184 b has 62.5 % or ( 5 / 8 ) of the milk in a . therefore , let the quantity of milk in container a ( initially ) be 8 k . quantity of milk in b = 8 k - 5 k = 3 k . quantity of milk in container c = 8 k - 3 k = 5 k container : a b c quantity of milk : 8 k 3 k 5 k it is given that if 148 liters was transferred from container c to container b , then both the containers would have equal quantities of milk . 5 k - 148 = 3 k + 148 = > 2 k = 296 = > k = 148 the initial quantity of milk in a = 8 k = 8 * 148 = 1184 liters ." | a = 1/(2)
b = 62 / 5
c = 1 - b
d = a - c
e = 148 / d
|
a ) 5336 , b ) 4306 , c ) 4236 , d ) 4336 , e ) none | a | multiply(multiply(5, 46), divide(multiply(5, 46), 10)) | in a division sum , the divisor is 10 times the quotient and 5 times the remainder . if the remainder is 46 , the dividend is | sol . divisor = ( 5 × 46 ) = 230 . also , 10 × q = 230 ⇒ q = 23 . and , r = 46 . dividend = ( 230 × 23 + 46 ) = 5336 . answer a | a = 5 * 46
b = 5 * 46
c = b / 10
d = a * c
|
a ) 8.8 % , b ) 9 % , c ) 9.2 % , d ) 9.1 % , e ) 8.4 % | d | multiply(divide(add(divide(multiply(12, 15), const_100), divide(multiply(8, 40), const_100)), add(15, 40)), const_100) | in one alloy there is 12 % chromium while in another alloy it is 8 % . 15 kg of the first alloy was melted together with 40 kg of the second one to form a third alloy . find the percentage of chromium in the new alloy . | "the amount of chromium in the new 15 + 40 = 55 kg alloy is 0.12 * 15 + 0.08 * 40 = 5 kg , so the percentage is 5 / 55 * 100 = 9.1 % . answer : d ." | a = 12 * 15
b = a / 100
c = 8 * 40
d = c / 100
e = b + d
f = 15 + 40
g = e / f
h = g * 100
|
a ) 2299 , b ) 2699 , c ) 2670 , d ) 6725 , e ) 5000 | e | divide(5750, add(const_1, divide(15, const_100))) | the owner of a furniture shop charges his customer 15 % more than the cost price . if a customer paid rs . 5750 for a computer table , then what was the cost price of the computer table ? | "cp = sp * ( 100 / ( 100 + profit % ) ) = 5750 ( 100 / 115 ) = rs . 5000 answer : e" | a = 15 / 100
b = 1 + a
c = 5750 / b
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a ) 5 , b ) 4 , c ) 12 , d ) 32 , e ) 35 | a | subtract(36, reminder(3, 7)) | when positive integer n is divided by 5 , the remainder is 1 . when n is divided by 7 , the remainder is 3 . what is the smallest positive integer k such that k + n is a multiple of 36 . | "first , let us say i have a number n which is divisible by 5 and by 7 . we all agree that it will be divisible by 35 , the lcm of 5 and 7 . now , if i have a number n which when divided by 5 gives a remainder 1 and when divided by 7 gives a remainder 1 , we can say the number is of the form n = 5 a + 1 e . g . 5 + 1 , 10 + 1 , 15 + 1 , 20 + 1 , 25 + 1 , 30 + 1 , 35 + 1 etc and n = 7 b + 1 e . g . 7 + 1 , 14 + 1 , 21 + 1 , 28 + 1 , 35 + 1 etc so when it is divided by the lcm , 35 , it will give 1 as remainder ( as is apparent above ) next , if i have a number n which when divided by 5 gives a remainder 1 and when divided by 7 gives a remainder 3 , we can say the number is of the form n = 5 a + 1 and n = 7 b + 3 now , the only thing you should try to understand here is that when n is divided by 5 and if i say the remainder is 1 , it is the same as saying the remainder is - 4 . e . g . when 6 is divided by 5 , remainder is 1 because it is 1 more than a multiple of 5 . i can also say it is 4 less than the next multiple of 5 , ca n ' t i ? 6 is one more than 5 , but 4 less than 10 . therefore , we can say n = 5 x - 4 and n = 7 y - 4 ( a remainder of 3 when divided by 7 is the same as getting a remainder of - 4 ) now this question is exactly like the question above . so when you divide n by 36 , remainder will be - 4 i . e . n will be 4 less than a multiple of 36 . so you must add 5 to n to make it a multiple of 36 a" | a = 36 - reminder
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a ) 8 , b ) 12 , c ) 20 , d ) 32 , e ) 36 | e | multiply(subtract(const_10, const_1), 4) | on june 1 a bicycle dealer noted that the number of bicycles in stock had decreased by 4 for each of the past 5 months . if the stock continues to decrease at the same rate for the rest of the year , how many fewer bicycles will be in stock on october 1 than were in stock on january 1 ? | jan 1 = c feb 1 = c - 4 march 1 = c - 8 april 1 = c - 12 may 1 = c - 16 june 1 = c - 20 july 1 = c - 24 aug 1 = c - 28 sept 1 = c - 32 oct 1 = c - 36 difference between stock on october 1 than were in stock on january 1 will be - c - ( c - 36 ) = 36 hence answer will be ( e ) | a = 10 - 1
b = a * 4
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a ) 10 , b ) 50 , c ) 30 , d ) 15 , e ) 25 | e | divide(5, subtract(105.20, add(const_100, add(multiply(const_4, const_10), const_2)))) | when positive integer n is divided by positive integer j , the remainder is 5 . if n / j = 105.20 , what is value of j ? | "when a number is divided by another number , we can represent it as : dividend = quotient * divisor + remainder so , dividend / divisor = quotient + remainder / divisor given that n / j = 105.20 here 105 is the quotient . given that remainder = 5 so , 105.20 = 105 + 5 / j so , j = 25 answer : e" | a = 4 * 10
b = a + 2
c = 100 + b
d = 105 - 20
e = 5 / d
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a ) 18 seconds , b ) 8.4 seconds , c ) 5 seconds , d ) 9.1 seconds , e ) 9.5 econds | d | divide(45.5, multiply(add(9.5, 8.5), const_0_2778)) | the speed at which a man can row a boat in still water is 9.5 kmph . if he rows downstream , where the speed of current is 8.5 kmph , what time will he take to cover 45.5 metres ? | "speed of the boat downstream = 9.5 + 8.5 = 18 kmph = 18 * 5 / 18 = 5 m / s hence time taken to cover 45.5 m = 45.5 / 5 = 9.1 seconds . answer : d" | a = 9 + 5
b = a * const_0_2778
c = 45 / 5
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a ) 8 sec , b ) 7 sec , c ) 2 sec , d ) 6.5 sec , e ) 9 sec | d | divide(65, multiply(36, const_0_2778)) | in what time will a railway train 65 m long moving at the rate of 36 kmph pass a telegraph post on its way ? | "t = 65 / 36 * 18 / 5 = 6.5 sec answer : d" | a = 36 * const_0_2778
b = 65 / a
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a ) 54 , b ) 59 , c ) 63 , d ) 71 , e ) 62 | e | add(multiply(subtract(const_1, divide(const_1, const_3)), subtract(70, 46)), 46) | of 70 players on a football team , 46 are throwers . the rest of the team is divided so one third are left - handed and the rest are right handed . assuming that all throwers are right handed , how many right - handed players are there total ? | "total = 70 thrower = 46 rest = 70 - 46 = 24 left handed = 24 / 3 = 8 right handed = 16 if all thrower are right handed then total right handed is 46 + 16 = 62 so e . 62 is the right answer" | a = 1 / 3
b = 1 - a
c = 70 - 46
d = b * c
e = d + 46
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a ) 2 , b ) 5 , c ) 4 , d ) 3.61 , e ) 4.3 | d | multiply(divide(106, add(add(multiply(add(const_3, const_4), const_100), multiply(multiply(const_2, const_3), const_10)), const_1000)), const_60) | how many minutes will it take for a car that is traveling at a constant rate of 9 miles per hour to travel a distance of 106 yards ? ( 1 mile = 1,160 yards ) | speed = 9 miles / hr = 4.4 yard / s = 29.33 yards / min distance = 106 yards time = distance / speed = 106 / 29.33 = 3.61 minutes ans - d | a = 3 + 4
b = a * 100
c = 2 * 3
d = c * 10
e = b + d
f = e + 1000
g = 106 / f
h = g * const_60
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a ) 40 , b ) 44 , c ) 48 , d ) 50 , e ) 56 | b | add(add(multiply(5, divide(8, subtract(multiply(divide(6, 5), 5), 4))), divide(8, subtract(multiply(divide(6, 5), 5), 4))), 8) | in a can , there is a mixture of milk and water in the ratio 4 : 5 . if it is filled with an additional 8 litres of milk the can would be full and ratio of milk and water would become 6 : 5 . find the capacity of the can ? | "let the capacity of the can be t litres . quantity of milk in the mixture before adding milk = 4 / 9 ( t - 8 ) after adding milk , quantity of milk in the mixture = 6 / 11 t . 6 t / 11 - 8 = 4 / 9 ( t - 8 ) 10 t = 792 - 352 = > t = 44 . answer : b" | a = 6 / 5
b = a * 5
c = b - 4
d = 8 / c
e = 5 * d
f = 6 / 5
g = f * 5
h = g - 4
i = 8 / h
j = e + i
k = j + 8
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a ) 122.9 m , b ) 170 m . , c ) 122.2 m , d ) 222.9 m , e ) 12289 m | b | subtract(1000, divide(multiply(subtract(1000, 100), subtract(800, 100)), 800)) | in a race of 1000 m , a can beat by 100 m , in a race of 800 m , b can beat c by 100 m . by how many meters will a beat c in a race of 800 m ? | "when a runs 1000 m , b runs 900 m and when b runs 800 m , c runs 700 m . when b runs 900 m , distance that c runs = ( 900 * 700 ) / 800 = 6300 / 8 = 787.5 m . in a race of 1000 m , a beats c by ( 1000 - 787.5 ) = 212.5 m to c . in a race of 800 m , the number of meters by which a beats c = ( 800 * 212.5 ) / 1000 = 170 m . answer : b" | a = 1000 - 100
b = 800 - 100
c = a * b
d = c / 800
e = 1000 - d
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a ) 10 , b ) 11 , c ) 13 , d ) 15 , e ) 16 | c | subtract(18, 5) | 5 ^ 100 is divided by 18 then what is the remainder ? | here n = 18 = 2 × 32 ϕ ( 18 ) = 18 ( 1 − 12 ) ( 1 − 13 ) = 6 so 56 when divided by 18 , remainder is 1 . so we can write the given expression 5100 = ( 56 ) 16 × 54 = ( 1 ) 16 × 54 = 52 × 52 = 7 × 7 = 49 now 49 when divided by 18 , remainder is 13 . c | a = 18 - 5
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a ) 16 . , b ) 8 . , c ) 7 . , d ) 2 . , e ) - 2 . | d | subtract(add(4, 5), 7) | if ( a + b ) = 4 , ( b + c ) = 7 and ( c + d ) = 5 , what is the value of ( a + d ) ? | "given a + b = 4 b + c = 7 c + d = 5 ( a + b ) - ( b + c ) + ( c + d ) = ( a + d ) = > 4 - 7 + 5 = 2 . option d . . ." | a = 4 + 5
b = a - 7
|
a ) 600 , b ) 470 , c ) 750 , d ) 650 , e ) 560 | b | subtract(subtract(multiply(45, 8), multiply(22, 20)), multiply(22, 15)) | the average of 45 results is 8 . 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 8 * 45 - 15 * 22 + 20 * 22 = 360 - 330 + 440 = 470 answer is b" | a = 45 * 8
b = 22 * 20
c = a - b
d = 22 * 15
e = c - d
|
a ) 4200 , b ) 5400 , c ) 6600 , d ) 7800 , e ) 8900 | c | multiply(330, 20) | a man started driving at a constant speed , from the site of a blast , the moment he heard the blast . he heard a second blast after a time of 30 mins and 20 seconds . if the second blast occurred exactly 30 mins after the first , how many meters was he from the site when he heard the second blast ? ( speed of sound = 330 m / s ) | "the distance the sound traveled to the man is 20 * 330 = 6600 meters the answer is c ." | a = 330 * 20
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a ) 180 , b ) 245 , c ) 240 , d ) 200 , e ) 190 | b | divide(multiply(70, multiply(const_3, 5)), 5) | in the set of positive integers from 1 to 70 , what is the sum of all the odd multiples of 5 ? | "reduce 1 - 70 5 - 15 - 25 - 35 - 45 + 55 + 65 are valid multiples 0 f 5 . add them - - > 245 b" | a = 3 * 5
b = 70 * a
c = b / 5
|
['a ) 12 cm', 'b ) 10 cm', 'c ) 11 cm', 'd ) 15 cm', 'e ) 20 cm'] | a | divide(multiply(square_area(6), const_2), 6) | if the area of a triangle with base 6 cm is equal to the area of a square with side 6 cm , then the altitude of the triangle is | area of square = 6 ^ 2 = 36 area of triangle = 36 1 / 2 bh = 36 h = 12 cm answer a | a = square_area * (
b = a / 2
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a ) 298 m , b ) 279 m , c ) 500 m , d ) 289 m , e ) 700 m | e | divide(600, multiply(subtract(78, 1), const_0_2778)) | a train 600 m long is running at a speed of 78 km / hr . if it crosses a tunnel in 1 min , then the length of the tunnel is ? | "speed = 78 * 5 / 18 = 65 / 3 m / sec . time = 1 min = 60 sec . let the length of the train be x meters . then , ( 600 + x ) / 60 = 65 / 3 x = 700 m . answer : e" | a = 78 - 1
b = a * const_0_2778
c = 600 / b
|
a ) 2 hr , b ) 1 hr , c ) 3 hr , d ) 1.5 hr , e ) 2.5 hr | a | divide(multiply(multiply(divide(multiply(20, 3), 4), 4), const_2), const_60) | walking at 3 / 4 of his usual place , a man reaches his office 20 minute late . find his usual time ? | let the original speed be s and time be t if new speed = s ã — 3 / 4 , then new time would be t ã — 4 / 3 ( d = st = constant ) . given , 3 t / 4 â ˆ ’ t = 20 t / 3 â ‡ ’ t = 60 minutes = 1 hour answer : a | a = 20 * 3
b = a / 4
c = b * 4
d = c * 2
e = d / const_60
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a ) $ 1 , b ) $ 1.86 , c ) $ 1.92 , d ) $ 2.13 , e ) $ 2.15 | b | min(divide(add(add(add(add(add(add(1.75, 1.61), 1.79), 2.11), 1.96), 2.09), 1.86), 7), 1.86) | andrew travelling to 7 cities . gasoline prices varied from city to city . $ 1.75 , $ 1.61 , $ 1.79 , $ 2.11 , $ 1.96 , $ 2.09 , $ 1.86 . what is the median gasoline price ? | "ordering the data from least to greatest , we get : $ 1.61 , $ 1.75 , $ 1.79 , $ 1.86 , $ 1.96 , $ 2.09 , $ 2.11 the median gasoline price is $ 1.86 . ( there were 3 states with higher gasoline prices and 3 with lower prices . ) b" | a = 1 + 75
b = a + 1
c = b + 2
d = c + 1
e = d + 2
f = e + 1
g = f / 7
h = min(g)
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a ) 12 , b ) 13 , c ) 14 , d ) 15 , e ) 16 | d | add(10, const_4) | how many odd numbers between 10 and 1,100 are the squares of integers ? | "the square of an odd number is an odd number : 10 < odd < 1,000 10 < odd ^ 2 < 1,000 3 . something < odd < 31 . something ( by taking the square root ) . so , that odd number could be any odd number from 5 to 31 , inclusive : 5 , 7 , 9 , 11 , 13 , 15 , 17 , 19 , 21 , 23 , 25 , 27 , 29 , and 31 . 15 numbers . answer : d ." | a = 10 + 4
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a ) 12 hr , b ) 11 hr , c ) 13 hr , d ) 14 hr , e ) 16 hr | a | divide(multiply(6, 12), subtract(12, 6)) | a pump can fill a tank with water in 6 hours . because of a leak , it took 12 hours to fill the tank . the leak can drain all the water in ? | work done by the leak in 1 hour = 1 / 6 - 1 / 12 = 1 / 12 leak will empty the tank in 12 hours answer is a | a = 6 * 12
b = 12 - 6
c = a / b
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a ) 52 , b ) 166 , c ) 84 , d ) 122 , e ) 168 | b | divide(multiply(multiply(8, 14), 10), const_4) | if 10 x = 8 y = 14 z , then what is a possible sum of positive integers x , y , and z ? | "10 x = 8 y = 14 z 5 x = 4 y = 7 z 5 ( 4 * 7 ) = 4 ( 5 * 7 ) = 7 ( 5 * 4 ) addition = 28 + 20 + 35 = 83 answer would be multiple of 83 which is 166 answer : b" | a = 8 * 14
b = a * 10
c = b / 4
|
a ) 3 , b ) 5 , c ) 6 , d ) 8 , e ) 9 | c | add(23, const_1) | if x and y are positive integers and 9 + x + y + xy = 23 , what is the value of x + y ? | "try each answer choices . for a : 9 + 3 + xy = 23 ; xy = 11 ( impossible , 11 prime number . 1 + 11 does n ' t equal 3 ) for b : 9 + 5 + xy = 23 ; xy = 9 ( no combination of xy = 9 and x + y = 5 ) for c : 9 + 6 + xy = 23 ; xy = 8 ( x + y = 6 ; x = 2 , y = 4 or x = 4 , y = 2 ) for d : 9 + 8 + xy = 23 ; xy = 6 ( no combination of xy = 6 and x + y = 8 ) for e : 9 + 9 + xy = 23 ; xy = 5 ( impossible , 5 prime number . 1 + 5 does n ' t equal 9 ) therefore , answer c ." | a = 23 + 1
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a ) 2 : 5 , b ) 3 : 7 , c ) 3 : 5 , d ) 4 : 7 , e ) 7 : 16 | e | divide(add(multiply(15, divide(1, add(1, 2))), multiply(8, divide(1, add(1, 3)))), subtract(add(15, 8), add(multiply(15, divide(1, add(1, 2))), multiply(8, divide(1, add(1, 3)))))) | a container holding 15 ounces of a solution that is 1 part alcohol to 2 parts water is added to a container holding 8 ounces of a solution that is 1 part alcohol to 3 parts water . what is the ratio of alcohol to water in the resulting solution ? | container 1 has 15 ounces in the ratio 1 : 2 or , x + 2 x = 15 gives x ( alcohol ) = 5 and remaining water = 10 container 2 has 8 ounces in the ratio 1 : 3 or , x + 3 x = 8 gives x ( alcohol ) = 2 and remaining water = 6 mixing both we have alcohol = 5 + 2 and water = 10 + 6 ratio thus alcohol / water = 7 / 16 answer e | a = 1 + 2
b = 1 / a
c = 15 * b
d = 1 + 3
e = 1 / d
f = 8 * e
g = c + f
h = 15 + 8
i = 1 + 2
j = 1 / i
k = 15 * j
l = 1 + 3
m = 1 / l
n = 8 * m
o = k + n
p = h - o
q = g / p
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