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 ) 10.5 % , b ) 11.5 % , c ) 12.5 % , d ) 13.5 % , e ) 14.5 % | e | subtract(const_100, multiply(multiply(divide(90, const_100), divide(subtract(const_100, 5), const_100)), const_100)) | a new model car was not selling as expected and the base retail price was lowered by 5 % to increase sales . a customer haggled on the price and negotiated a final agreed price of 90 % of the retail price . how much did the customer save off the retail price ? | quantity x rate = price 1 x 1 = 1 0.9 x 0.95 = 0.855 decrease in price = ( 0.145 / 1 ) ã — 100 = 14.5 % answer = option e | a = 90 / 100
b = 100 - 5
c = b / 100
d = a * c
e = d * 100
f = 100 - e
|
a ) 120 , b ) 125 , c ) 110 , d ) 115 , e ) 121 | a | multiply(divide(add(multiply(1, const_60), 30), 30), 40) | a machine , working at a constant rate , manufactures 40 pencils in 30 minutes . how many pencils does it make in 1 hr 30 min ? | "change 1 hr 30 min to 90 min . for this , we need to set up a simple proportion of pensils per time 40 / 30 = s / 90 the absolutely worst thing you could do at this point in the problem is to cross - multiply . that would be a supremely unstrategic move . instead , cancel before you multiply . for what we can and can ’ t cancel in a proportion , seethis post . we can cancel the factor of 10 in the 40 and 30 . 4 / 3 = s / 90 we can cancel the common factor of 3 in the two denominators . 4 / 1 = s / 30 . now that the fraction is entirely simplified , we can cross - multiply . s = 4 * 30 = 120 the machine would be 120 pencils in 1 hr 30 min . answer : a" | a = 1 * const_60
b = a + 30
c = b / 30
d = c * 40
|
a ) 2 % , b ) 4 % , c ) 6 % , d ) 8 % , e ) 10 % | e | multiply(divide(subtract(1320, add(300, 900)), add(300, 900)), const_100) | sandy buys an old scooter for $ 900 and spends $ 300 on its repairs . if sandy sells the scooter for $ 1320 , what is the gain percent ? | "selling price / total cost = 1320 / 1200 = 1.1 the gain percent is 10 % . the answer is e ." | a = 300 + 900
b = 1320 - a
c = 300 + 900
d = b / c
e = d * 100
|
a ) 150 , b ) 152 , c ) 154 , d ) 156 , e ) 158 | c | divide(divide(divide(770, divide(add(8, 2), const_2)), 2), const_2) | the cross - section of a cannel is a trapezium in shape . if the cannel is 8 m wide at the top and 2 m wide at the bottom and the area of cross - section is 770 sq m , the depth of cannel is ? | "1 / 2 * d ( 8 + 2 ) = 770 d = 154 answer : c" | a = 8 + 2
b = a / 2
c = 770 / b
d = c / 2
e = d / 2
|
a ) 400 meters , b ) 1111 meters , c ) 800 meters , d ) 1822 meters , e ) none of these | c | multiply(multiply(subtract(100, 64), const_0_2778), 80) | a train traveling at 100 kmph overtakes a motorbike traveling at 64 kmph in 80 seconds . what is the length of the train in meters ? | "train overtakes a bike means that we are talking about total length of the train . ( train ' s head is close to bike when it started and its tail crosses the bike when it overtakes the bike ) relative speed = 100 - 64 = 36 km / h = 36000 m / h time = 80 seconds distance = speed * time 36000 * 80 / 3600 = 800 meters . c is the answer ." | a = 100 - 64
b = a * const_0_2778
c = b * 80
|
a ) 127 , b ) 237 , c ) 287 , d ) 450 , e ) 281 | d | divide(multiply(36, 30), divide(multiply(8, 24), 80)) | if 8 men can reap 80 hectares in 24 days , then how many hectares can 36 men reap in 30 days ? | "explanation : let the required no of hectares be x . then more men , more hectares ( direct proportion ) more days , more hectares ( direct proportion ) men \ : 8 : 36 \ \ days \ : 24 : 30 \ end { matrix } \ right \ } : 80 : x \ inline \ fn _ jvn \ therefore \ inline \ fn _ jvn 8 \ times 24 \ times x = 36 \ times 30 \ times 80 \ inline \ fn _ jvn \ leftrightarrow \ inline \ fn _ jvn x = \ frac { 36 \ times 30 \ times 80 } { 8 \ times 24 } \ inline \ fn _ jvn \ leftrightarrow x = 450 answer : d" | a = 36 * 30
b = 8 * 24
c = b / 80
d = a / c
|
a ) 1 / 5 , b ) b . 1 / 4 , c ) 1 / 20 , d ) 3 / 5 , e ) it can not be determined from the given information . | c | subtract(divide(1, 4), divide(1, 4)) | 1 / 4 of all married couples have more than one child . 1 / 5 of all married couples have more than 3 children . what fraction of all married couples have 2 or 3 children ? | "plug in simple numbers . take 100 couples for example . 1 / 4 of 100 couples have more than one child = 25 couples . 1 / 5 of 100 couples have more than 3 kids = 20 couples . this implies that 20 couples are a subset of 25 couples and the complement of 25 couples within those 100 couples , which equals 75 couples have either one or no kids at all . we need to find couples that have 2 or 3 kids , so essentially , it is 25 - 20 = 5 . fraction will be 5 / 100 = 1 / 20 . option c" | a = 1 / 4
b = 1 / 4
c = a - b
|
a ) 92 , b ) 90 , c ) 82 , d ) 78 , e ) 75 | e | divide(multiply(power(power(1000, divide(const_1, const_3)), const_2), add(const_4, const_2)), multiply(const_4, const_2)) | a welder received an order to make a 1 million liter cube - shaped tank . if he has only 4 x 2 meter sheets of metal that can be cut , how many metal sheets will be required for this order r ? ( 1 cubic meter = 1000 liters ) | i get 75 . a cube with 1 million liters cube would be a cube with the dimensions of 100 * 100 * 100 . 4 * 2 covers 8 sq liters so 100 / 8 = 12.5 . r = 12.5 * 6 = 75 . e | a = 1 / 3
b = 1000 ** a
c = b ** 2
d = 4 + 2
e = c * d
f = 4 * 2
g = e / f
|
a ) 8.5 , b ) 8.0 , c ) 9.5 , d ) 7.0 , e ) 8.25 | d | divide(subtract(26, 12), const_2) | a man can row downstream at the rate of 26 kmph and upstream at 12 kmph . find the man ’ s rate in still water and rate of current ? | "rate of still water = 1 / 2 ( down stream + upstream ) = 1 / 2 ( 26 + 12 ) = 19 kmph rate of current = 1 / 2 ( down stream - upstream ) = 1 / 2 ( 26 - 12 ) = 1 / 2 ( 14 ) = 7 kmph answer is d ." | a = 26 - 12
b = a / 2
|
['a ) 18 cm', 'b ) 24 cm', 'c ) 22 cm', 'd ) data inadequate', 'e ) none of these'] | d | multiply(const_2, sqrt(add(const_1, power(12, const_2)))) | the perimeter of a rectangle is equal to the perimeter of aright - angled triangle of height 12 cm . if the base of the triangle is equal to the breadth of the rectangle , what is the length of the rectangle ” | p = 2 ( l + b ) = l + b + h = l + b + 12 . data inadequate . answer d | a = 12 ** 2
b = 1 + a
c = math.sqrt(b)
d = 2 * c
|
a ) 32,300 , b ) 172,800 , c ) 468,830 , d ) 338,200 , e ) 259,200 | e | multiply(multiply(subtract(7, 1), const_3600), const_12) | in a renowned city , the average birth rate is 7 people every two seconds and the death rate is 1 people every two seconds . estimate the size of the population net increase that occurs in one day . | "every 2 seconds , 6 persons are added ( 7 - 1 ) . every second 3 persons are added . in a day 24 hrs = 24 * 60 minutes = 24 * 60 * 60 = 86400 seconds . 86400 * 3 = 259200 option e" | a = 7 - 1
b = a * 3600
c = b * 12
|
a ) 35 , b ) 36 , c ) 37 , d ) 43 , e ) 39 | d | add(divide(subtract(add(40, 2), 30), 1.5), 30) | each week , harry is paid x dollars per hour for the first 30 hours and 1.5 x dollars for each additional hour worked that week . each week , james is paid x dollars per per hour for the first 40 hours and 2 x dollars for each additional hour worked that week . last week james worked a total of 45 hours if harry and james were paid the same amount last week , how many hours did harry work last week ? | "james worked for 41 hours hence he earned 40 * x + 5 * 2 x = 50 x dollars ; we know that harry also earned the same 50 x dollars , out of which he earned 30 x dollars for thefirst 30 hoursplus 20 x additional dollars . since for each additional hour he gets 1.5 x dollars then he worked for 20 x / 1.5 x = 13 additional hours , so harry worked for total of 30 + 13 = 43 hours . answer : d ." | a = 40 + 2
b = a - 30
c = b / 1
d = c + 30
|
a ) 14.1 sec , b ) 12.1 sec , c ) 16.1 sec , d ) 13.1 sec , e ) 12.15 sec | a | divide(add(150, 132), multiply(72, const_0_2778)) | how long does a train 150 m long running at the speed of 72 km / hr takes to cross a bridge 132 m length ? | "speed = 72 * 5 / 18 = 20 m / sec total distance covered = 150 + 132 = 282 m . required time = 282 / 20 = 14.1 sec . answer : a" | a = 150 + 132
b = 72 * const_0_2778
c = a / b
|
a ) 3.84 , b ) 1.75 , c ) 3.44 , d ) 2.72 , e ) none of these | c | divide(divide(multiply(4, add(multiply(3, 900), multiply(2, 800))), add(3, 2)), const_1000) | the weights of one liter vegetable ghee packet of two brands ‘ a ’ and ‘ b ’ are 900 gm and 800 gm respectively . if they are mixed in the ratio of 3 : 2 by volumes to form a mixture of 4 liters , what is the weight ( in kg ) of the mixture ? | "here ' s how i did it . my notes from reading the problem were : 1 l a = 900 gm 1 l b = 800 gm we are mixing five parts ( 3 parts a plus 2 parts b , 5 parts total ) to get 4 l , so 5 x = 4 - - - > x = 4 / 5 . each part is 4 / 5 of a liter . so if we have 3 parts a , we have 900 * 3 * ( 4 / 5 ) = 2160 if we have 2 parts b , we have 800 * 2 * ( 4 / 5 ) = 1280 2160 + 1280 = 3440 solving for units gives us 3.44 so the answer is c" | a = 3 * 900
b = 2 * 800
c = a + b
d = 4 * c
e = 3 + 2
f = d / e
g = f / 1000
|
a ) 8 % , b ) 7 % , c ) 10 % , d ) 3 % , e ) 4 % | d | multiply(divide(subtract(subtract(70, multiply(70, divide(10, const_100))), 61.11), subtract(70, multiply(70, divide(10, const_100)))), const_100) | the list price of an article is rs . 70 . a customer pays rs . 61.11 for it . he was given two successive discounts , one of them being 10 % . the other discount is ? | "explanation : 70 * ( 90 / 100 ) * ( ( 100 - x ) / 100 ) = 61.11 x = 3 % d" | a = 10 / 100
b = 70 * a
c = 70 - b
d = c - 61
e = 10 / 100
f = 70 * e
g = 70 - f
h = d / g
i = h * 100
|
a ) 29 , b ) 23 , c ) 47.119 , d ) 44.586 , e ) 42.686 | e | divide(add(300, 115), divide(multiply(35, const_1000), const_3600)) | a train is 300 meter long is running at a speed of 35 km / hour . in what time will it pass a bridge of 115 meter length ? | "speed = 35 km / hr = 35 * ( 5 / 18 ) m / sec = 175 / 18 m / sec total distance = 300 + 115 = 415 meter time = distance / speed = 415 * ( 18 / 175 ) = 42.686 seconds . answer : e" | a = 300 + 115
b = 35 * 1000
c = b / 3600
d = a / c
|
a ) 15 , b ) 20 , c ) 25 , d ) 30 , e ) 35 | d | subtract(subtract(70, divide(60, const_3)), divide(60, const_3)) | the perimeter of an equilateral triangle is 60 . if one of the sides of the equilateral triangle is the side of an isosceles triangle of perimeter 70 , then how long is the base of isosceles triangle ? | "the base of the isosceles triangle is 70 - 20 - 20 = 30 units the answer is d ." | a = 60 / 3
b = 70 - a
c = 60 / 3
d = b - c
|
['a ) a ) 160 degree', 'b ) b ) 168 degree', 'c ) c ) 191 degree', 'd ) d ) 192 degree', 'e ) e ) 213 degree'] | e | add(multiply(const_360, divide(divide(const_2, add(2, 7)), const_3)), multiply(const_360, multiply(divide(7, add(2, 7)), divide(const_2, const_3)))) | the ratio of male to female in a class is 2 : 7 . the career preference of the students in the class are to be represented in a circle graph . if the area of the graph allocated to each career preference is to be proportional to the number of students who have that career preference , how many degrees of the circle should be used to represent a career that is preferred by one third of the males and two - third of the females in the class ? | here is my approach = > males = > 2 x and females = 7 x = > total = 9 x now 9 x = > 360 therefore 16 x / 3 = > 213 degree . p . s = > 16 x / 3 is nothing but total number of students with the given preference answer e | a = 2 + 7
b = 2 / a
c = b / 3
d = 360 * c
e = 2 + 7
f = 7 / e
g = 2 / 3
h = f * g
i = 360 * h
j = d + i
|
a ) 10 % , b ) 20 % , c ) 40 % , d ) 50 % , e ) 60 % | b | divide(subtract(28, add(25, const_1)), subtract(divide(40, const_100), divide(add(25, const_1), const_100))) | seed mixture x is 40 % ryegrass and 60 % bluegrass by weight ; seed mixture y is 25 % ryegrass and 75 % fescue . if a mixture of x and y contains 28 % ryegrass , what percent of the weight of the mixture is from mixture x ? | "28 % is 3 % - points above 25 % and 12 % - points below 40 % . thus the ratio of mixture y to mixture x is 4 : 1 . the percent of mixture x is 1 / 5 = 20 % . the answer is b ." | a = 25 + 1
b = 28 - a
c = 40 / 100
d = 25 + 1
e = d / 100
f = c - e
g = b / f
|
['a ) 5', 'b ) 6', 'c ) 7', 'd ) 8', 'e ) 9'] | c | divide(divide(divide(divide(divide(divide(2800, const_2), const_2), const_2), const_2), add(const_2, const_3)), add(const_2, const_3)) | which is the smallest number divides 2800 and gives a perfect square . | 7 is the smallest number which divides 2800 and gives a perfect square . as 2800 = 2 * 2 * 2 * 2 * 5 * 5 * 7 and 7 is not in a pair which gives 400 ( a perfect square of 20 ) on dividing 2800 . answer : c | a = 2800 / 2
b = a / 2
c = b / 2
d = c / 2
e = 2 + 3
f = d / e
g = 2 + 3
h = f / g
|
a ) 81 , b ) 101 , c ) 121 , d ) 141 , e ) 161 | c | divide(multiply(divide(const_3600, const_4), const_3), 6) | if a light flashes every 6 seconds , how many times will it flash in 1 / 5 of an hour ? | "in 1 / 5 of an hour there are 12 * 60 = 720 seconds the number of 6 - second intervals = 720 / 6 = 120 after the first flash , there will be 120 more flashes for a total of 121 . the answer is c ." | a = 3600 / 4
b = a * 3
c = b / 6
|
a ) 10 , b ) 7 , c ) 5 , d ) 4 , e ) 2 | b | subtract(divide(divide(800, const_1), const_3), const_3) | in how many ways can the integer 800 be expressed as a product of two different positive integers ? | "800 = ( 2 ^ 5 ) * ( 5 ^ 2 ) since 800 is not a perfect square , no of ways = 7 answer b" | a = 800 / 1
b = a / 3
c = b - 3
|
a ) 80 , b ) 103.5 , c ) 95 , d ) 100 , e ) 108 | b | multiply(multiply(multiply(const_100, divide(add(const_100, 20), const_100)), divide(subtract(const_100, 25), const_100)), divide(add(const_100, 15), const_100)) | from the beginning to the end of 2007 , the price of a stock rose 20 percent . in 2008 , it dropped 25 percent . in 2009 , it rose 15 percent . what percent of the stock â € ™ s 2007 starting price was the price of the stock at the end of 2009 ? | "assume a value at the beginning of 2007 . as this is a % question , assume p = 100 . at the end of 2007 it becmae = 1.2 * 100 = 120 at the end of 2008 it decreased by 25 % = 120 * . 75 = 90 at the end of 2009 it increased by 20 % = 90 * 1.15 = 103.5 thus ratio = 103.5 / 100 = 1.03 ( in % terms = 103 % ) . thus b is the correct answer ." | a = 100 + 20
b = a / 100
c = 100 * b
d = 100 - 25
e = d / 100
f = c * e
g = 100 + 15
h = g / 100
i = f * h
|
['a ) 5', 'b ) 13', 'c ) 14', 'd ) 11', 'e ) 15'] | a | multiply(sqrt(divide(25, add(25, 144))), 13) | d and e are two points respectively on sides ab and ac of triangle abc such that de is parallel to bc . if the ratio of area of triangle ade to that of the trapezium decb is 144 : 25 and de = 13 cm , then find the length of bc . | abc and ade are similar triangles . so ( side of abc / side of ade ) ^ 2 = 25 / 169 side of abc / side of ade = 5 / 13 so the length of bc = 5 answer - a | a = 25 + 144
b = 25 / a
c = math.sqrt(b)
d = c * 13
|
a ) 18 , b ) 28 , c ) 50 , d ) 68 , e ) 70 | c | divide(multiply(divide(multiply(subtract(subtract(300, divide(multiply(70, 300), const_100)), 15), const_100), subtract(const_100, 40)), 40), const_100) | at a particular graduation party with 300 guests , 70 % of the guests brought gifts , and 40 % of the female guests brought gifts . if 15 males did not bring gifts to the party , how many females did bring gifts ? | the correct method total = 300 . . 70 % of 300 = 210 got gifts . . 90 did not get gift , out of which 15 are males , so remaining 90 - 15 = 75 are females . . but 40 % females brought gift , so 60 % did not get it . . so 60 % = 75 , 100 % = 75 * 100 / 60 = 125 . . ans 40 % of 125 = 50 c | a = 70 * 300
b = a / 100
c = 300 - b
d = c - 15
e = d * 100
f = 100 - 40
g = e / f
h = g * 40
i = h / 100
|
a ) 11 , b ) 7 , c ) 8 , d ) 9 , e ) 10 | a | divide(factorial(subtract(add(const_4, 11), const_1)), multiply(factorial(11), factorial(subtract(const_4, const_1)))) | how many positive integers less than 100 have a remainder of 11 when divided by 13 ? | "we have to include 11 also . as 13 * 0 + 11 = 11 if somebody says to divide 11 by 13 , we will be telling we have 0 quotient and remainder as 11 . answer is a" | a = 4 + 11
b = a - 1
c = math.factorial(b)
d = math.factorial(11)
e = 4 - 1
f = math.factorial(e)
g = d * f
h = c / g
|
a ) 182.5 , b ) 184.5 , c ) 167.5 , d ) 172.5 , e ) 187.5 | e | multiply(divide(multiply(add(divide(25, const_100), const_1), 30), 20), const_100) | last year elaine spent 20 % of her annual earnings on rent . this year she earned 25 % more than last year and she spent 30 % of her annual earnings on rent . the amount she spent on rent this year is what percent of the amount spent on rent last year ? | "for this it is easiest to use simple numbers . let ' s assume that elaine ' s annual earnings last year were $ 100 . she would ' ve spent $ 20 of this on rent . this year she earned 25 % more , or $ 125 . she would ' ve spent 30 % of this on rent , or $ 37.5 do $ 37.5 / $ 20 this will give you 187.5 % e is the correct answer ." | a = 25 / 100
b = a + 1
c = b * 30
d = c / 20
e = d * 100
|
a ) 7.9 s , b ) 6 s , c ) 7.5 s , d ) 7.6 s , e ) 7.4 s | b | multiply(divide(divide(150, const_1000), 90), const_3600) | how much time does a train 150 metres long running at 90 km / hr take to pass a pole ? | "explanation : 90 km / hr = 90 * 5 / 18 = 25 m / s speed = distance / time ; v = d / t 25 = 150 / t t = 6 s answer : b" | a = 150 / 1000
b = a / 90
c = b * 3600
|
a ) 12 : 5 , b ) 4 : 3 , c ) 9 : 5 , d ) 9 : 1 , e ) 9 : 2 | a | divide(multiply(4, 3), 5) | a dog takes 4 leaps for every 5 leaps of a hare . if one leap of the dog is equal to 3 leaps of the hare , the ratio of the speed of the dog to that of the hare is : | "explanation : dog : hare = ( 4 * 3 ) leaps of hare : 5 leaps of hare = 12 : 5 . answer : a ) 12 : 5" | a = 4 * 3
b = a / 5
|
a ) 1890 , b ) 2790 , c ) 1990 , d ) 1500 , e ) 1701 | c | multiply(divide(105, 12), const_100) | if an article is sold at 19 % profit instead of 12 % profit , then the profit would be rs . 105 more . what is the cost price ? | "let the cost price of an article be rs . x . ( 19 % of x ) - ( 12 % of x ) = 105 19 x / 100 - 12 x / 100 = 105 = > 7 x = 105 * 100 = > x = 1500 cost price = rs . 1500 answer : c" | a = 105 / 12
b = a * 100
|
a ) 260 , b ) 620 , c ) 500 , d ) 520 , e ) 720 | a | multiply(130, const_2) | on the independence day , bananas were be equally distributed among the children in a school so that each child would get two bananas . on the particular day 130 children were absent and as a result each child got two extra bananas . find the actual number of children in the school ? | "let the number of children in the school be x . since each child gets 2 bananas , total number of bananas = 2 x . 2 x / ( x - 130 ) = 2 + 2 ( extra ) = > 2 x - 260 = x = > x = 260 . answer : a" | a = 130 * 2
|
a ) 287 , b ) 369 , c ) 370 , d ) 371 , e ) 430 | b | add(add(multiply(const_3, const_100), const_60), multiply(const_3, const_3)) | an intern wants to label folders with numbers using digit stickers only . if the intern uses 999 stickers , how many folders have been numbered ? ( the numbers of the folders are consecutive and the number of the first folder is 1 ) . | for the first 9 folders we need 9 stickers . for the next 90 we need 2 stickers each or 180 stickers . for the next 900 folders we need 3 stickers each . the first 99 folders correspond to 189 stickers . subtract 189 from the total number of stickers ( 999 ) . this leaves 810 stickers for folders that use 3 stickers each . so , divide 810 by 3 to get 270 folders . add 99 folders to 270 folders to get the total number of folders , 369 . answer : b | a = 3 * 100
b = a + const_60
c = 3 * 3
d = b + c
|
a ) 191 , b ) 213 , c ) 221 , d ) 223 , e ) 226 | d | divide(multiply(100, 50), subtract(50, divide(multiply(add(divide(multiply(100, 50), 10), 50), 5), 100))) | each month a retailer sells 100 identical items . on each item he makes a profit of $ 50 that constitutes 10 % of the item ' s price to the retailer . if the retailer contemplates giving a 5 % discount on the items he sells , what is the least number of items he will have to sell each month to justify the policy of the discount ? | "for this question , we ' ll need the following formula : sell price = cost + profit we ' re told that the profit on 1 item is $ 20 and that this represents 10 % of the cost : sell price = cost + $ 50 sell price = $ 500 + $ 50 thus , the sell price is $ 550 for each item . selling all 100 items gives the retailer . . . 100 ( $ 50 ) = $ 2,000 of profit if the retailer offers a 5 % discount on the sell price , then the equation changes . . . 5 % ( 550 ) = $ 27.5 discount $ 522.5 = $ 500 + $ 22.5 now , the retailer makes a profit of just $ 22.5 per item sold . to earn $ 2,000 in profit , the retailer must sell . . . . $ 22.5 ( x ) = $ 2,000 x = 2,000 / 22.5 x = 222.222222 items you ' ll notice that this is not among the answer choices . . . . 221 and 223 are . selling 221 items would get us 9 ( 221 ) = $ 1989 which is not enough money . to get back to at least $ 2,000 , we need to sell 223 items . final answer : d" | a = 100 * 50
b = 100 * 50
c = b / 10
d = c + 50
e = d * 5
f = e / 100
g = 50 - f
h = a / g
|
a ) 23.57 , b ) 25.54 , c ) 26.5 , d ) 26.55 , e ) 26.71 | e | floor(divide(circumface(divide(divide(multiply(const_2, add(14, 20)), const_4), const_2)), const_2)) | the perimeter of a square is equal to the perimeter of a rectangle of length 20 cm and breadth 14 cm . find the circumference of a semicircle whose diameter is equal to the side of the square . ( round off your answer to two decimal places ) | "let the side of the square be a cm . perimeter of the rectangle = 2 ( 20 + 14 ) = 68 cm perimeter of the square = 68 cm i . e . 4 a = 68 a = 17 diameter of the semicircle = 17 cm circumference of the semicircle = 1 / 2 ( ∏ ) ( 17 ) = 1 / 2 ( 22 / 7 ) ( 17 ) = 26.71 cm to two decimal places answer : e" | a = 14 + 20
b = 2 * a
c = b / 4
d = c / 2
e = circumface / (
f = math.floor(e, 2)
|
a ) 5 / 2 , b ) 7 / 4 , c ) 8 / 5 , d ) 10 / 3 , e ) 12 / 5 | d | divide(16, divide(const_1, add(divide(const_1, 12), divide(const_1, 8)))) | working alone , printers x , y , and z can do a certain printing job , consisting of a large number of pages , in 16 , 12 , and 8 hours , respectively . what is the ratio of the time it takes printer x to do the job , working alone at its rate , to the time it takes printers y and z to do the job , working together at their individual rates ? | the time it takes printer x is 16 hours . the combined rate of y and z is 1 / 12 + 1 / 8 = 5 / 24 the time it takes y and z is 24 / 5 the ratio of times is 16 / ( 24 / 5 ) = 5 * 16 / 24 = 10 / 3 the answer is d . | a = 1 / 12
b = 1 / 8
c = a + b
d = 1 / c
e = 16 / d
|
a ) 3.496 kg , b ) 3.696 kg , c ) 3.690 kg , d ) 9.696 kg , e ) 3.296 kg | b | divide(multiply(8, multiply(multiply(divide(add(21, const_1), add(const_4, const_3)), subtract(power(const_4, const_2), power(const_3, const_2))), 21)), const_1000) | a hollow iron pipe is 21 cm long and its external diameter is 8 cm . if the thickness of the pipe is 1 cm and iron weighs , then the weight of the pipe is | explanation : external radius = 4 cm , internal radius = 3 cm . volume of iron = = weight of iron = ( 462 x 8 ) gm = 3696 gm = 3.696 kg answer : b ) 3.696 kg | a = 21 + 1
b = 4 + 3
c = a / b
d = 4 ** 2
e = 3 ** 2
f = d - e
g = c * f
h = g * 21
i = 8 * h
j = i / 1000
|
a ) 93.5 , b ) 90 , c ) 6.75 , d ) 6.25 , e ) 2 | d | divide(multiply(25, 25), const_100) | j is 25 % less than p and 20 % less than t . t is r % less than p . what is the value of r ? | "usually we can solve every question of this type by choosing appropriate value of the variable and deriving the value of other related variables . let , p = 400 then j = ( 75 / 100 ) * 400 = 300 also j = ( 80 / 100 ) * t i . e . t = 300 * 100 / 80 = 375 and t = [ 1 - ( r / 100 ) ] * p i . e . 100 - r = 100 * t / p = 100 * 375 / 400 = 93.75 i . e . r = 6.25 answer : option d" | a = 25 * 25
b = a / 100
|
a ) 3 , b ) 4 , c ) 5 , d ) 6 , e ) 9 | e | add(const_1, divide(21, 3)) | if k is the greatest positive integer such that 3 ^ k is a divisor of 21 ! then k = | "21 / 3 = 7 21 / 9 = 2.33 = 2 7 + 2 = 9 k = 9 answer : e" | a = 21 / 3
b = 1 + a
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a ) 1 / 4 , b ) 1 / 3 , c ) 1 / 2 , d ) 1 , e ) 2 | e | divide(20, subtract(30, 20)) | a chemist mixes one liter of pure water with x liters of a 30 % salt solution , and the resulting mixture is a 20 % salt solution . what is the value of x ? | "concentration of salt in pure solution = 0 concentration of salt in salt solution = 30 % concentration of salt in the mixed solution = 20 % the pure solution and the salt solution is mixed in the ratio of - - > ( 30 - 20 ) / ( 20 - 0 ) = 1 / 2 1 / x = 1 / 2 x = 2 answer : e" | a = 30 - 20
b = 20 / a
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a ) a ) 540 , b ) b ) 400 , c ) c ) 700 , d ) d ) 650 , e ) e ) 840 | e | divide(multiply(546, const_100), subtract(const_100, 35)) | in an examination 35 % of the students passed and 546 failed . how many students appeared for the examination ? | "let the number of students appeared be x then , 65 % of x = 546 65 x / 100 = 546 x = 546 * 100 / 65 = 840 answer is e" | a = 546 * 100
b = 100 - 35
c = a / b
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a ) 36 , b ) 42 , c ) 48 , d ) 54 , e ) 60 | c | divide(90, divide(add(divide(90, 30), divide(90, 120)), const_2)) | a person walks from one end to the other of a 90 - meter long moving walkway at a constant rate in 30 seconds , assisted by the walkway . when this person reaches the end , they reverse direction and continue walking with the same speed , but this time it takes 120 seconds because the person is traveling against the direction of the moving walkway . if the walkway were to stop moving , how many seconds would it take this person to walk from one end of the walkway to the other ? | "let v be the speed of the person and let x be the speed of the walkway . 30 ( v + x ) = 90 then 120 ( v + x ) = 360 120 ( v - x ) = 90 when we add the two equations : 240 v = 450 v = 15 / 8 time = 90 / ( 15 / 8 ) = 48 seconds the answer is c ." | a = 90 / 30
b = 90 / 120
c = a + b
d = c / 2
e = 90 / d
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a ) 12 , b ) 15 , c ) 18 , d ) 21 , e ) 24 | d | multiply(divide(subtract(34, 6), 4), 3) | right now , the ratio between the ages of sandy and molly is 4 : 3 . after 6 years , sandy ’ s age will be 34 years . what is molly ' s age right now ? | "now , sandy is 34 - 6 = 28 molly ' s age is ( 3 / 4 ) * 28 = 21 the answer is d ." | a = 34 - 6
b = a / 4
c = b * 3
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a ) 120 , b ) 840 , c ) 350 , d ) 420 , e ) 210 | d | subtract(multiply(divide(420, 30), 60), 420) | a 420 meter long train crosses a platform in 60 seconds while it crosses a signal pole in 30 seconds . what is the length of the platform ? | "speed = [ 420 / 30 ] m / sec = 14 m / sec . let the length of the platform be x meters . then , x + 420 / 60 = 14 x + 420 = 840 è x = 420 m . answer : d" | a = 420 / 30
b = a * 60
c = b - 420
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a ) 180 , b ) 227 , c ) 268 , d ) 198 , e ) 176 | a | multiply(divide(435, add(add(multiply(12, 8), multiply(16, 9)), multiply(18, 6))), multiply(16, 9)) | a , b and c rents a pasture for rs . 435 . a put in 12 horses for 8 months , b 16 horses for 9 months and 18 horses for 6 months . how much should b pay ? | 12 * 8 : 16 * 9 = 18 * 6 8 : 12 : 9 12 / 29 * 435 = 180 answer : a | a = 12 * 8
b = 16 * 9
c = a + b
d = 18 * 6
e = c + d
f = 435 / e
g = 16 * 9
h = f * g
|
a ) 2999 , b ) 2778 , c ) 6000 , d ) 5000 , e ) 6612 | d | multiply(multiply(const_1, const_12), divide(15000, add(add(multiply(const_1, const_12), multiply(subtract(const_12, 6), const_2)), multiply(subtract(const_12, 8), const_3)))) | a , b and c enter into partnership . a invests some money at the beginning , b invests double the amount after 6 months , and c invests thrice the amount after 8 months . if the annual gain be rs . 15000 . a ' s share is ? | "x * 12 : 2 x * 6 : 3 x * 4 1 : 1 : 1 1 / 3 * 15000 = 5000 answer : d" | a = 1 * 12
b = 1 * 12
c = 12 - 6
d = c * 2
e = b + d
f = 12 - 8
g = f * 3
h = e + g
i = 15000 / h
j = a * i
|
a ) 25 , b ) 35 , c ) 45 , d ) 55 , e ) 20 | e | divide(subtract(const_100, 80), divide(80, 80)) | a car traveled 80 % of the way from town x to town y at an average speed of 80 mph . the car traveled at an average speed of v mph for the remaining part of the trip . the average speed for the entire trip was 50 mph . what is v in mph ? | assume total distance = 100 miles time taken for 80 miles = 80 / 80 = 1 hour time taken for the rest of the 20 miles = 20 / v hours . average speed = 50 therefore the total time needed = 2 hours . 2 = 1 + 20 / v hence v = 20 mph answer : e | a = 100 - 80
b = 80 / 80
c = a / b
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a ) 18 , b ) 55 , c ) 63 , d ) 64 , e ) 70 | d | multiply(8, 8) | a guy was asked to specify his age in years . he said , “ take my age 8 years hence , multiply it by 8 and subtract 8 times of my age 8 years ago and get the half of the result and you will know my age . ” what was the age of that guy ? | current age of the guy = a years . then , 8 ( a + 8 ) – 8 ( a – 8 ) = a ( 8 a + 64 ) – ( 8 a – 64 ) = a a = 128 / 2 = 64 d | a = 8 * 8
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a ) 82 , b ) 83 , c ) 84 , d ) 85 , e ) 86 | e | add(add(subtract(subtract(add(74, 75), add(11, 12)), 75), add(11, 12)), 11) | there are 12 students in a class . on the day the test was given , ella was absent . the other 11 students took the test and their average was 74 . the next day , ella took the test , and with this grade included , the new average was 75 . what was ella ' s grade on the test ? | 11 * 74 + ella ' s grade = 12 * 75 ella ' s grade is 12 * 75 - 11 * 74 = 86 . the answer is e . | a = 74 + 75
b = 11 + 12
c = a - b
d = c - 75
e = 11 + 12
f = d + e
g = f + 11
|
a ) 72.5 , b ) 55.5 , c ) 64.2 , d ) 82.5 , e ) 60.5 | c | multiply(divide(divide(multiply(47.50, add(const_100, 25)), const_100), subtract(const_100, 20)), const_100) | at what price must an book costing $ 47.50 be marked in order that after deducting 20 % from the list price . it may be sold at a profit of 25 % on the cost price ? | "c $ 62.50 cp = 47.50 sp = 47.50 * ( 125 / 100 ) = 59.375 mp * ( 80 / 100 ) = 59.375 mp = 74.2" | a = 100 + 25
b = 47 * 50
c = b / 100
d = 100 - 20
e = c / d
f = e * 100
|
['a ) 210.25', 'b ) 75.69', 'c ) 14.5', 'd ) 145', 'e ) 150'] | c | sqrt(divide(126.15, divide(const_3, add(const_1, const_4)))) | three fifth of the square of a certain number is 126.15 . what is the number ? | let the number be x given , 3 / 5 x ^ 2 = 126.5 x ^ 2 = 126.15 * 5 / 3 x ^ 2 = 42.05 * 5 = 210.25 x = 14.5 answer c . | a = 1 + 4
b = 3 / a
c = 126 / 15
d = math.sqrt(c)
|
a ) 5 / 7 , b ) 6 / 7 , c ) 10 / 7 , d ) 12 / 7 , e ) 22 / 7 | b | add(add(const_2, const_2), const_3) | one woman and one man can build a wall together in one hour , but the woman would need the help of two girls in order to complete the same job in the same amount of time . if one man and one girl worked together , it would take them two hours to build the wall . assuming that rates for men , women and girls remain constant , how many hours would it take one woman , one man , and one girl , working together , to build the wall ? | solution : let work done by man , women and girl per hour be m , w , g respectively . then , m + w = 1 / 1 - - > ( 1 ) , w + 2 g = 1 / 1 - - > ( 2 ) and m + g = 1 / 2 - - > ( 3 ) . no . of hours it would take forone woman , one man , and one girl , working together , to build the wall , n = 1 / m + w + g from ( 1 ) and ( 2 ) , m = 2 g and from ( 3 ) g = 1 / 6 , m = 1 / 3 and w = 2 / 3 . so , n = 1 / ( 7 / 6 ) = 6 / 7 option , b | a = 2 + 2
b = a + 3
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a ) 80 , b ) 96 , c ) 108 , d ) 120 , e ) 210 | e | divide(factorial(10), multiply(factorial(divide(const_10, const_2)), factorial(4))) | in a certain circle there are 10 points . what is the number of the triangles connecting 4 points of the 10 points ? | "imo : e here we have to select 4 points out of 10 points . order is not important so the answer will be 10 c 4 = 210" | a = math.factorial(10)
b = 10 / 2
c = math.factorial(b)
d = math.factorial(4)
e = c * d
f = a / e
|
a ) 8 sec , b ) 20 / 7 sec , c ) 33 / 7 sec , d ) 21 / 7 sec , e ) 32 / 7 sec | a | divide(200, multiply(add(54, 36), const_0_2778)) | two trains of length 400 m and 200 m are 200 m apart . they start moving towards each other on parallel tracks , at speeds 54 kmph and 36 kmph . after how much time will the trains meet ? | "they are moving in opposite directions , relative speed is equal to the sum of their speeds . relative speed = ( 54 + 36 ) * 5 / 18 = 25 mps . the time required = d / s = 200 / 25 = 8 sec . answer : a" | a = 54 + 36
b = a * const_0_2778
c = 200 / b
|
a ) 7 : 37 , b ) 6 : 46 , c ) 9 : 49 , d ) 8 : 68 , e ) 6 : 61 | c | divide(const_4, power(7, 3)) | the diagonals of two squares are in the ratio of 3 : 7 then find the ratio of its areas ? | "let the diagonals of the squares be 3 x and 7 x ratio of their areas = 1 / 2 ( 3 x ) ^ 2 : 1 / 2 ( 7 x ) ^ 2 = 9 x ^ 2 : 49 x ^ 2 = 9 : 49 answer is c" | a = 7 ** 3
b = 4 / a
|
a ) 71.25 , b ) 52.9 , c ) 52.1 , d ) 52.3 , e ) 42.5 | a | divide(add(multiply(30, 40), multiply(50, 90)), add(30, 50)) | the average marks of a class of 30 students is 40 and that of another class of 50 students is 90 . find the average marks of all the students ? | "sum of the marks for the class of 30 students = 30 * 40 = 1200 sum of the marks for the class of 50 students = 50 * 90 = 4500 sum of the marks for the class of 80 students = 1200 + 4500 = 5700 average marks of all the students = 5700 / 80 = 71.25 answer : a" | a = 30 * 40
b = 50 * 90
c = a + b
d = 30 + 50
e = c / d
|
a ) 600 , b ) 900 , c ) 1500 , d ) 1600 , e ) 1750 | b | multiply(divide(3, 8), multiply(10, divide(multiply(8, 60), subtract(10, 8)))) | in the storage room of a certain bakery , the ratio of sugar to flour is 3 to 8 , and the ratio of flour to baking soda is 10 to 1 . if there were 60 more pounds of baking soda in the room , the ratio of flour to baking soda would be 8 to 1 . how many pounds of sugar are stored in the room ? | "sugar : flour = 3 : 8 = 15 : 40 ; flour : soda = 10 : 1 = 40 : 4 ; thus we have that sugar : flour : soda = 15 x : 40 x : 4 x . also given that 40 x / ( 4 x + 60 ) = 8 / 1 - - > x = 60 - - > sugar = 15 x = 900 . answer : b ." | a = 3 / 8
b = 8 * 60
c = 10 - 8
d = b / c
e = 10 * d
f = a * e
|
a ) 36 , b ) 38 , c ) 33 , d ) 32 , e ) 38 | a | multiply(18, 2) | jacob is 18 years old . he is 2 times as old as his brother . how old will jacob be when he is twice as old ? | j = 18 ; j = 2 b ; b = 18 / 2 = 9 ; twice as old so b = 9 ( now ) + ( 9 ) = 18 ; jacob is 18 + 18 = 36 answer : a | a = 18 * 2
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a ) 3883203 , b ) 3893103 , c ) 3639403 , d ) 3791203 , e ) none of them | b | multiply(divide(3897, 999), const_100) | 3897 x 999 = ? | "= 3897 x 999 = 3897 x ( 1000 - 1 ) = 3897 x 1000 - 3897 x 1 = 3897000 - 3897 = 3893103 answer is b" | a = 3897 / 999
b = a * 100
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a ) rs . 16003 , b ) rs . 16029 , c ) rs . 11288 , d ) rs . 14100 , e ) rs . 16011 | d | subtract(multiply(add(1500, 600), add(20, const_1)), multiply(1500, 20)) | the average monthly salary of 20 employees in an organisation is rs . 1500 . if the manager ' s salary is added , then the average salary increases by rs . 600 . what is the manager ' s monthly salary ? | "explanation : manager ' s monthly salary rs . ( 2100 * 21 - 1500 * 20 ) = rs . 14100 . answer : d" | a = 1500 + 600
b = 20 + 1
c = a * b
d = 1500 * 20
e = c - d
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a ) 6.75 , b ) 7 , c ) 7.25 , d ) 7.5 , e ) 20 | e | divide(multiply(8, 5), subtract(multiply(const_2, 5), 8)) | noelle walks from point a to point b at an average speed of 5 kilometers per hour . at what speed , in kilometers per hour , must noelle walk from point b to point a so that her average speed for the entire trip is 8 kilometers per hour ? | "let ' s suppose that speed while returning was xkm / h since the distance is same , we can apply the formula of avg speed avg speed = 2 s 1 s 2 / s 1 + s 2 8 = 2 * 5 * x / 5 + x x = 20 e is the answer" | a = 8 * 5
b = 2 * 5
c = b - 8
d = a / c
|
a ) 10 % , b ) 80 % , c ) 30 % , d ) 25 % , e ) 28 % | b | multiply(divide(subtract(90, 50), 50), const_100) | a man buy a book in rs 50 & sale it rs 90 . what is the rate of profit ? ? ? | cp = 50 sp = 90 profit = 90 - 50 = 40 % = 40 / 50 * 100 = 80 % answer : b | a = 90 - 50
b = a / 50
c = b * 100
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a ) 4 sec , b ) 5 sec , c ) 5.55 sec , d ) 6.55 sec , e ) 4.55 sec | c | divide(250, multiply(162, const_0_2778)) | in what time will a train 250 metres long cross an electric pole , if its speed be 162 km / hr ? | "solution speed = ( 162 x 5 / 18 ) m / sec = 45 m / sec time taken = ( 250 / 45 ) sec = 5.55 sec . answer c" | a = 162 * const_0_2778
b = 250 / a
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a ) a - 30 , b ) b - 29 , c ) c - 28 , d ) d - 27 , e ) e - 26 | c | multiply(subtract(3, 2), 30) | an ant is climbing 30 meters tall pole . it climbs 3 meter during the day and slips down 2 meters down during night in sleep , how many days it takes to reach the top of the pole ? | ants overall distance covered is 3 - 2 = 1 meter per day . in 27 days it will climb 27 meters . on 28 th day it will reach the top of the pole . answer : c | a = 3 - 2
b = a * 30
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a ) 1 / 9 , b ) 29 / 180 , c ) 26 / 143 , d ) 3 / 20 , e ) 39 / 121 | d | add(divide(const_1, divide(multiply(50, 4), 10)), divide(const_1, divide(multiply(30, 5), 15))) | if 50 apprentices can finish a job in 4 hours , and 30 journeymen can finish the same job in 5 hours , how much of the job should be completed by 10 apprentices and 15 journeymen in one hour ? | "50 apprentices can finish the job in 4 hours , thus : 10 apprentices can finish the job in 4 * 5 = 20 hours ; in 1 hour 10 apprentices can finish 1 / 20 of the job . 30 journeymen can finish the same job in 4,5 hours , thus : 15 journeymen can finish the job in 5 * 2 = 10 hours ; in 1 hour 15 journeymen can finish 1 / 10 of the job . therefore , in 1 hour 10 apprentices and 15 journeymen can finish 1 / 20 + 1 / 10 = 3 / 20 of the job . answer : d ." | a = 50 * 4
b = a / 10
c = 1 / b
d = 30 * 5
e = d / 15
f = 1 / e
g = c + f
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a ) 7 , b ) 10 , c ) 12 , d ) 14 , e ) 16 | a | multiply(7, divide(multiply(add(7, 3), subtract(3, multiply(divide(5, add(7, 5)), 3))), subtract(multiply(3, 7), multiply(7, 5)))) | a can contains a mixture of liquids a and b is the ratio 7 : 5 . when 3 litres of mixture are drawn off and the can is filled with b , the ratio of a and b becomes 7 : 9 . how many liter of liquid a was contained by the can initially ? | "ci * vi = cf * vf ( 7 / 12 ) * ( v 1 - 3 ) = ( 7 / 16 ) * v 1 ( v 1 - 3 ) / v 1 = 3 / 4 3 accounts for the difference of 1 on ratio scale so initial volume = v 1 = 4 * 3 = 12 litres . 7 / 12 of the initial mixture was liquid a so liquid a was ( 7 / 12 ) * 12 = 7 litres . answer : a" | a = 7 + 3
b = 7 + 5
c = 5 / b
d = c * 3
e = 3 - d
f = a * e
g = 3 * 7
h = 7 * 5
i = g - h
j = f / i
k = 7 * j
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a ) rs . 6000 , b ) rs . 7500 , c ) rs . 8000 , d ) rs . 10,000 , e ) none | b | multiply(1500, add(const_4, const_1)) | rohan spends 40 % of his salary on food , 20 % on house rent , 10 % on entertainment and 10 % on conveyance . if his savings at the end of a month are rs . 1500 . then his monthly salary is | "sol . saving = [ 100 - ( 40 + 20 + 10 + 10 ] % = 20 % . let the monthly salary be rs . x . then , 20 % of x = 1500 ⇔ 20 / 100 x = 1500 ⇔ x = 1500 × 5 = 7500 . answer b" | a = 4 + 1
b = 1500 * a
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a ) 1 , b ) 14 , c ) 9.47 , d ) 21 , e ) 22 | c | multiply(divide(floor(divide(95, const_10)), 95), const_100) | what percentage of numbers from 1 to 95 have squares that end in the digit 0 ? | clearly , the numbers which have 1 or 9 in the unit ' s digit , have squares that end in the digit 1 . such numbers from 1 to 95 are 10 , 20,30 , 40,50 , 60,70 , 80,90 . number of such numbers = 14 . required percentage = ( 9 / 95 * 100 ) = 9.47 % answer : c | a = 95 / 10
b = math.floor(a)
c = b / 95
d = c * 100
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a ) 9 , b ) 36 , c ) 122 , d ) 6 , e ) 10 | e | divide(sqrt(400), 2) | what is the square root of 400 , divided by 2 ? | "square root is a number times itself square root of 400 = 20 , 20 / 2 = 10 ( e ) 1" | a = math.sqrt(400)
b = a / 2
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a ) 2 , b ) 3 , c ) 4 , d ) 5 , e ) 6 | e | add(3, add(1, 2)) | for a recipe for triple berry pie calls for cups of strawberries , raspberries , and blueberries in a ratio of 1 : 2 : 3 . how many total cups of fruit will you need to make the pie ? | the ratio 1 : 2 : 3 represents 1 cup strawberries : 2 cups raspberries : 3 cups blueberries . 1 + 2 + 3 = 6 6 cups will be used for the pie . the answer is e | a = 1 + 2
b = 3 + a
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a ) 7.19 , b ) 7.18 , c ) 7.16 , d ) 7.15 , e ) 7.12 | d | divide(add(121, 165), multiply(add(80, 65), const_0_2778)) | two trains 121 meters and 165 meters in length respectively are running in opposite directions , one at the rate of 80 km and the other at the rate of 65 kmph . in what time will they be completely clear of each other from the moment they meet ? | "t = ( 121 + 165 ) / ( 80 + 65 ) * 18 / 5 t = 7.15 answer : d" | a = 121 + 165
b = 80 + 65
c = b * const_0_2778
d = a / c
|
a ) 75 , b ) 52 , c ) 58 , d ) 60 , e ) 62 | a | divide(multiply(63, const_100), subtract(const_100, 16)) | the number which exceeds 16 % of it by 63 is : | "solution solution let the number be x . x - 16 % of x = 63 x - 16 / 100 x = 63 x - 4 / 25 x = 63 21 / 25 x = 63 x = ( 63 x 25 / 21 ) = 75 answer a" | a = 63 * 100
b = 100 - 16
c = a / b
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a ) 179 , b ) 208 , c ) 210 , d ) 133 , e ) 229 | d | divide(add(150, subtract(multiply(58, 46), multiply(58, subtract(46, const_2)))), const_2) | the batting average of a particular batsman is 58 runs in 46 innings . if the difference in his highest and lowest score is 150 runs and his average excluding these two innings is 58 runs , find his highest score . | "explanation : total runs scored by the batsman = 58 * 46 = 2668 runs now excluding the two innings the runs scored = 58 * 44 = 2552 runs hence the runs scored in the two innings = 2668 – 2552 = 116 runs . let the highest score be x , hence the lowest score = x – 150 x + ( x - 150 ) = 116 2 x = 266 x = 133 runs answer d" | a = 58 * 46
b = 46 - 2
c = 58 * b
d = a - c
e = 150 + d
f = e / 2
|
a ) rs . 4007 , b ) rs . 4000 , c ) rs . 4028 , d ) rs . 4050 , e ) rs . 4032 | b | subtract(add(5050, 5200), 6250) | the average monthly income of p and q is rs . 5050 . the average monthly income of q and r is 6250 and the average monthly income of p and r is rs . 5200 . the monthly income of p is ? | "let p , q and r represent their respective monthly incomes . then , we have : p + q = ( 5050 * 2 ) = 10100 - - - ( i ) q + r = ( 6250 * 2 ) = 12500 - - - ( ii ) p + r = ( 5200 * 2 ) = 10400 - - - ( iii ) adding ( i ) , ( ii ) and ( iii ) , we get : 2 ( p + q + r ) = 33000 = p + q + r = 16500 - - - ( iv ) subtracting ( ii ) from ( iv ) , we get , p = 4000 . p ' s monthly income = rs . 4000 . answer : b" | a = 5050 + 5200
b = a - 6250
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a ) 438 , b ) 629 , c ) 780 , d ) 879 , e ) 924 | b | subtract(387,600, add(add(multiply(const_2, const_100), multiply(add(const_3, const_4), const_10)), const_2)) | how many integers between 324,700 and 387,600 have tens digit 1 and units digit 3 ? | "the integers are : 324,713 324,813 etc . . . 387,513 the number of integers is 3876 - 3247 = 629 the answer is b ." | a = 2 * 100
b = 3 + 4
c = b * 10
d = a + c
e = d + 2
f = 387 - 600
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a ) rs . 7000 , b ) rs . 8400 , c ) rs . 8500 , d ) rs . 9000 , e ) none | b | divide(add(add(add(add(10000, 5000), 11000), 7000), 9000), add(const_4, const_1)) | the salary of a , b , c , d , e is rs . 10000 , rs . 5000 , rs . 11000 , rs . 7000 , rs . 9000 per month respectively , then the average salary of a , b , c , d , and e per month is | "answer average salary = 10000 + 5000 + 11000 + 7000 + 9000 / 5 = rs . 8400 correct option : b" | a = 10000 + 5000
b = a + 11000
c = b + 7000
d = c + 9000
e = 4 + 1
f = d / e
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a ) 648 , b ) 1800 , c ) 2000 , d ) 10800 , e ) 10900 | c | multiply(multiply(divide(300, 6), 4), 10) | running at the same constant rate , 6 identical machines can produce a total of 300 bottles per minute . at this rate , how many bottles could 10 such machines produce in 4 minutes ? | "let the required number of bottles be x . more machines , more bottles ( direct proportion ) more minutes , more bottles ( direct proportion ) machines 6 : 10 : : 300 : x time ( in minutes ) 1 : 4 6 x 1 x x = 10 x 4 x 300 x = ( 10 x 4 x 300 ) / ( 6 ) x = 2000 . answer : c" | a = 300 / 6
b = a * 4
c = b * 10
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a ) 5 % , b ) 6 % , c ) 7 % , d ) 8 % , e ) 9 % | c | sqrt(divide(735, divide(1500, const_100))) | reema took a loan of rs 1500 with simple interest for as many years as the rate of interest . if she paid rs . 735 as interest at the end of the loan period , what was the rate of interest . | explanation : let rate = r % then time = r years . = > 1500 ∗ r ∗ r / 100 = 735 = > r 2 = 49 = > r = 7 % option c | a = 1500 / 100
b = 735 / a
c = math.sqrt(b)
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a ) 6 / 11 , b ) 8 / 11 , c ) 7 / 12 , d ) 9 / 12 , e ) 10 / 11 | d | divide(const_1, add(divide(const_1, 5), divide(const_1, 7))) | a can do a piece of work in 5 days and b can do it in 7 days how long will they both work together to complete the work ? | "explanation : a ’ s one day work = 1 / 5 b ’ s one day work = 1 / 7 ( a + b ) ’ s one day work = 1 / 5 + 1 / 7 = 12 / 35 = > time = 35 / 12 = 2 9 / 12 days answer : option d" | a = 1 / 5
b = 1 / 7
c = a + b
d = 1 / c
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a ) 724533811 , b ) 353654655 , c ) 545463251 , d ) 725117481 , e ) 477899932 | d | multiply(subtract(9999, const_4), 72519) | find the value of m 72519 x 9999 = m ? | "72519 x 9999 = 72519 x ( 10000 - 1 ) = 72519 x 10000 - 72519 x 1 = 725190000 - 72519 = 725117481 d" | a = 9999 - 4
b = a * 72519
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a ) 15 % , b ) 18 % , c ) 20 % , d ) 22 % , e ) 25 % | c | multiply(divide(subtract(1200, add(200, 800)), add(200, 800)), const_100) | sandy buys an old scooter for $ 800 and spends $ 200 on its repairs . if sandy sells the scooter for $ 1200 , what is the gain percent ? | selling price / total cost = 1200 / 1000 = 1.2 the gain percent is 20 % . the answer is c . | a = 200 + 800
b = 1200 - a
c = 200 + 800
d = b / c
e = d * 100
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a ) 6.33 , b ) 7.5 , c ) 20 , d ) 15 , e ) 19 | c | subtract(21, divide(21, 21)) | a certain bacteria colony doubles in size every day for 21 days , at which point it reaches the limit of its habitat and can no longer grow . if two bacteria colonies start growing simultaneously , how many days will it take them to reach the habitat ’ s limit ? | "if there is one bacteria colony , then it will reach the limit of its habitat in 21 days . if there are two bacteria colonies , then in order to reach the limit of habitat they would need to double one time less than in case with one colony . thus colonies need to double 20 times . answer : c . similar questions to practice : hope it helps ." | a = 21 / 21
b = 21 - a
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a ) 6000 , b ) 9000 , c ) 9500 , d ) 12000 , e ) none | b | divide(multiply(multiply(3500, const_12), 3), multiply(subtract(const_12, 5), 2)) | a starts business with rs . 3500 and after 5 months , b joins with a as his partner . after a year , the profit is divided in the ratio 2 : 3 . what is b ’ s contribution in the capital ? | "sol . let b ’ s capital be rs . x . then , 3500 * 12 / 7 x = 2 / 3 ⇔ 14 x = 126000 ⇔ x = 9000 . answer b" | a = 3500 * 12
b = a * 3
c = 12 - 5
d = c * 2
e = b / d
|
a ) 21 , b ) 22 , c ) 23 , d ) 24 , e ) 25 | d | add(divide(100, add(const_4, const_1)), divide(100, multiply(add(const_4, const_1), add(const_4, const_1)))) | how many consecutive zeros are there at the end of 100 ! ( 100 factorial ) ? | number of zeros will be 24 reason 100 / 5 = 20 20 / 5 = 4 so the total number of 5 will be 20 + 4 = 24 answer : d | a = 4 + 1
b = 100 / a
c = 4 + 1
d = 4 + 1
e = c * d
f = 100 / e
g = b + f
|
a ) 20 , b ) 21 , c ) 22 , d ) 23 , e ) 24 | c | multiply(divide(55, subtract(7, 2)), 2) | the ratio of spinsters to cats is 2 to 7 . if there are 55 more cats than spinsters , how many spinsters are there ? | "let 2 x be the number of spinsters . then 7 x is the number of cats . 7 x - 2 x = 55 x = 11 and the number of spinsters is 2 ( 11 ) = 22 . the answer is c ." | a = 7 - 2
b = 55 / a
c = b * 2
|
a ) s . 375 , b ) s . 400 , c ) s . 500 , d ) s . 800 , e ) s . 850 | c | multiply(multiply(subtract(inverse(3), add(inverse(8), inverse(6))), 4000), 3) | a alone can do a piece of work in 6 days and b alone in 8 days . a and b undertook to do it for rs . 4000 . with the help of c , they completed the work in 3 days . how much is to be paid to c ? | "c ' s 1 day ' s work = 1 / 3 - ( 1 / 6 + 1 / 8 ) = 1 / 3 - 7 / 24 = 1 / 24 a ' s wages : b ' s wages : c ' s wages = 1 / 6 : 1 / 8 : 1 / 24 = 4 : 3 : 1 c ' s share ( for 3 days ) = rs . ( 3 * 1 / 24 * 4000 ) = rs . 500 answer = c" | a = 1/(3)
b = 1/(8)
c = 1/(6)
d = b + c
e = a - d
f = e * 4000
g = f * 3
|
a ) 64 , b ) 130 , c ) 140 , d ) 160 , e ) 120 | a | multiply(divide(multiply(8, 20), subtract(13, 8)), const_2) | a sports retailer ordered white and yellow tennis balls in equal number but the dispatch clerk of the tennis ball company erred and dispatched 20 extra yellow balls and making ratio of white balls to yellow balls 8 / 13 . how many tennis balls did the retailer order originally . | "white : yellow = x : ( x + 20 ) = 8 : 13 - - > 13 x = 8 x + 160 - - > x = 32 . the total # of balls originally x + x = 32 + 32 = 64 . answer : a ." | a = 8 * 20
b = 13 - 8
c = a / b
d = c * 2
|
a ) 17 % , b ) 19 % , c ) 24.8 % , d ) 14 % , e ) 16 % | c | multiply(divide(subtract(64900, add(42000, 10000)), add(42000, 10000)), const_100) | ramu bought an old car for rs . 42000 . he spent rs . 10000 on repairs and sold it for rs . 64900 . what is his profit percent ? | "total cp = rs . 42000 + rs . 10000 = rs . 52000 and sp = rs . 64900 profit ( % ) = ( 64900 - 52000 ) / 52000 * 100 = 24.8 % answer : c" | a = 42000 + 10000
b = 64900 - a
c = 42000 + 10000
d = b / c
e = d * 100
|
a ) 26 , b ) 24 , c ) 28 , d ) 30 , e ) 35 | c | divide(add(add(add(multiply(6, const_3), add(6, multiply(6, const_2))), multiply(6, const_4)), multiply(add(const_4, const_1), 6)), 6) | find the average of all numbers between 5 and 45 which are divisible by 6 | "explanation : average = ( 6 + 12 + 18 + 24 + 30 + 36 + 42 / 6 ) = 168 / 6 = 28 option c" | a = 6 * 3
b = 6 * 2
c = 6 + b
d = a + c
e = 6 * 4
f = d + e
g = 4 + 1
h = g * 6
i = f + h
j = i / 6
|
a ) 60 , b ) 15 , c ) 40 , d ) 10 , e ) 25 | b | divide(multiply(5, 60), subtract(60, 40)) | a group of men decided to do a work in 40 days , but 5 of them became absent . if the rest of the group did the work in 60 days , find the original number of men ? | "original number of men = 5 * 60 / ( 60 - 40 ) = 15 answer is b" | a = 5 * 60
b = 60 - 40
c = a / b
|
a ) 30 , b ) 300 , c ) 720 , d ) 1800 , e ) 32400 | e | multiply(multiply(const_3, const_60), const_60) | if an object travels at nine feet per second , how many feet does it travel in one hour ? | "speed = 9 feet per second . 1 hour = 60 x 60 seconds = 3600 seconds . total no of feet traveled in 1 hour = 3600 x 9 = 32400 answer e" | a = 3 * const_60
b = a * const_60
|
a ) 22 , b ) 23 , c ) 24 , d ) 25 , e ) 26 | e | divide(subtract(multiply(4, 30), add(add(4, 6), 7)), 4) | the youngest of 4 children has siblings who are 3 , 6 , and 7 years older than she is . if the average ( arithmetic mean ) age of the 4 siblings is 30 , what is the age of the youngest sibling ? | "x + ( x + 3 ) + ( x + 6 ) + ( x + 7 ) = 120 4 x + 16 = 120 4 x = 104 x = 26 the answer is e ." | a = 4 * 30
b = 4 + 6
c = b + 7
d = a - c
e = d / 4
|
a ) 110 , b ) 112 , c ) 114 , d ) 116 , e ) 118 | b | multiply(2, divide(126, add(2, 16))) | water consists of hydrogen and oxygen , and the approximate ratio , by mass , of hydrogen to oxygen is 2 : 16 . approximately how many grams of oxygen are there in 126 grams of water ? | "( 16 / 18 ) * 126 = 112 grams the answer is b ." | a = 2 + 16
b = 126 / a
c = 2 * b
|
a ) 1 : 3 , b ) 1 : 4 , c ) 1 : 5 , d ) 2 : 5 , e ) 3 : 4 | c | add(subtract(add(1, 4), add(6, 3)), const_1) | in what ratio p : q should the mixture p of milk and water in the ratio of 6 : 1 be mixed with another mixture q of milk and water in the ratio 3 : 4 so that the resultant mixture contains equal quantities of milk and water ? | "( 6 / 7 ) * p + ( 3 / 7 ) * q = ( 1 / 7 ) * p + ( 4 / 7 ) * q 5 p = q p / q = 1 / 5 the answer is c ." | a = 1 + 4
b = 6 + 3
c = a - b
d = c + 1
|
a ) 3 and 5 , b ) 3 and 4 , c ) 4 and 5 , d ) 5 and 10 , e ) 6 and 8 | a | add(15, 8) | sum of two numbers prime to each other is 8 and their l . c . m . is 15 . what are the numbers ? | "as two numbers are prime , only options satisfy ie option a and b and c but option d will not make the product of numbers i . e 15 answer : a" | a = 15 + 8
|
a ) $ 120 , b ) $ 180 , c ) $ 310 , d ) $ 450 , e ) $ 640 | b | multiply(divide(multiply(2.16, multiply(const_1000, const_1000)), multiply(multiply(20, 20), 12)), 0.40) | when greenville state university decided to move its fine arts collection to a new library , it had to package the collection in 20 - inch by 20 - inch by 12 - inch boxes . if the university pays $ 0.40 for every box , and if the university needs 2.16 million cubic inches to package the collection , what is the minimum amount the university must spend on boxes ? | "the volume of each box is 20 * 20 * 12 = 4800 cubic inches . number of boxes = 2 , 160,000 / 4800 = 450 boxes total cost = 450 × $ 0.4 = $ 180 the answer is b ." | a = 1000 * 1000
b = 2 * 16
c = 20 * 20
d = c * 12
e = b / d
f = e * 0
|
a ) 91.5 , b ) 75 , c ) 93 , d ) 94 , e ) 95 | b | multiply(multiply(const_2, divide(multiply(subtract(21, const_3), const_2), add(const_4, const_3))), 21) | the sector of a circle has radius of 21 cm and central angle 90 o . find its perimeter ? | "perimeter of the sector = length of the arc + 2 ( radius ) = ( 90 / 360 * 2 * 22 / 7 * 21 ) + 2 ( 21 ) = 75 cm answer : option b" | a = 21 - 3
b = a * 2
c = 4 + 3
d = b / c
e = 2 * d
f = e * 21
|
a ) 1 / 2 , b ) 5 / 9 , c ) 5 / 1 , d ) 5 / 3 , e ) 7 / 6 | a | divide(divide(subtract(8.75, 7.5), subtract(8.75, 5)), subtract(const_1, divide(subtract(8.75, 7.5), subtract(8.75, 5)))) | in what ratio should a variety of rice costing rs . 5 per kg be mixed with another variety of rice costing rs . 8.75 per kg to obtain a mixture costing rs . 7.50 per kg ? | let us say the ratio of the quantities of cheaper and dearer varieties = x : y by the rule of allegation , x / y = ( 8.75 - 7.50 ) / ( 7.50 - 5 ) = 1 / 2 answer : a | a = 8 - 75
b = 8 - 75
c = a / b
d = 8 - 75
e = 8 - 75
f = d / e
g = 1 - f
h = c / g
|
a ) 91.4 kgs , b ) 88.5 kgs , c ) 86.5 kgs , d ) 67.5 kgs , e ) 88.2 kgs | a | divide(multiply(add(const_1, const_3), 160), 7) | 3 friends a , b , c went for week end party to mcdonald ’ s restaurant and there they measure there weights in some order in 7 rounds . a , b , c , ab , bc , ac , abc . final round measure is 160 kg then find the average weight of all the 7 rounds ? | average weight = [ ( a + b + c + ( a + b ) + ( b + c ) + ( c + a ) + ( a + b + c ) ] / 7 = 4 ( a + b + c ) / 7 = 4 x 160 / 7 = 91.4 kgs answer : a | a = 1 + 3
b = a * 160
c = b / 7
|
a ) 3 % , b ) 5 % , c ) 9 % , d ) 10 % , e ) 12 % | c | multiply(multiply(10, 10), subtract(const_1, divide(add(multiply(9, const_60), 36), add(multiply(10, const_60), 30)))) | bob wants to run a mile in the same time as his sister . if bob ’ s time for a mile is currently 10 minutes 30 seconds and his sister ’ s time is currently 9 minutes 36 seconds , by what percent does bob need to improve his time in order run a mile in the same time as his sister ? | "bob ' s time = 630 secs . his sis ' time = 576 secs . percent increase needed = ( 630 - 576 / 630 ) * 100 = 54 / 640 * 100 = 9 % . ans ( c ) ." | a = 10 * 10
b = 9 * const_60
c = b + 36
d = 10 * const_60
e = d + 30
f = c / e
g = 1 - f
h = a * g
|
a ) a ) 5 , b ) b ) 9 , c ) c ) 11 , d ) d ) 13 , e ) e ) 15 | d | subtract(20, divide(subtract(20, divide(5, const_2)), const_2)) | sum of two numbers is 20 . two times of the first exceeds by 5 from the three times of the other . then the numbers will be ? | "explanation : x + y = 20 2 x – 3 y = 5 x = 13 y = 7 d )" | a = 5 / 2
b = 20 - a
c = b / 2
d = 20 - c
|
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