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 ) 670 , b ) 650 , c ) 698 , d ) 744 , e ) 700 | b | subtract(815, multiply(divide(subtract(870, 815), subtract(divide(4, const_100), divide(3, const_100))), divide(3, const_100))) | peter invests a sum of money and gets back an amount of $ 815 in 3 years . david invests an equal amount of money and gets an amount of $ 870 in 4 years . if both amounts were invested at the same rate ( simple interest ) what was the sum of money invested ? | "since both peter and david invested the same amount of money at the same rate , they would earn same interest per year . david invested for one year more than peter and hence he got interest amount for one more year . interest earned per year = amount received by david - amount received by peter = 870 - 815 = 55 interest earned for 3 years = 55 * 3 = 165 amount invested = 815 - 165 = 650 answer : b" | a = 870 - 815
b = 4 / 100
c = 3 / 100
d = b - c
e = a / d
f = 3 / 100
g = e * f
h = 815 - g
|
a ) 168 ° , b ) 134 ° , c ) 156 ° , d ) 224 ° , e ) none of these | b | multiply(divide(multiply(add(multiply(multiply(5, const_2), const_10), const_100), const_2), add(add(add(5, 6), 7), 12)), 13) | the ratio of the adjacent angles of a parallelogram is 5 : 13 . also , the ratio of the angles of quadrilateral is 5 : 6 : 7 : 12 . what is the sum of the smaller angle of the parallelogram and the second largest angle of the quadrilateral ? | "the measures of the adjacent angles of a parallelogram add up to be 180 ° given so , 5 x + 13 x = 180 ° or , 18 x = 180 ° or , x = 10 ° hence the angles of the parallelogram are 50 ° and 130 ° further it is given we know sum of all the four angles of a quadrilateral is 360 ° so , 5 y + 6 y + 7 y + 12 y = 360 ° or , 5 y + 6 y + 7 y + 12 y = 360 ° or , 30 y = 360 ° or , y = 12 ° hence the angles of the quadrilateral are 60 ° , 72 , 84 ° and 144 ° will be 50 ° + 84 ° = 134 ° answer : b" | a = 5 * 2
b = a * 10
c = b + 100
d = c * 2
e = 5 + 6
f = e + 7
g = f + 12
h = d / g
i = h * 13
|
a ) 2 / 3 , b ) 5 / 8 , c ) 5 / 9 , d ) 3 / 7 , e ) 4 / 7 | b | divide(add(1, add(divide(1, const_2), add(divide(1, add(1, 3)), divide(3, add(1, 3))))), add(1, 3)) | in a bag containing 3 balls , a white ball was placed and then 1 ball was taken out at random . what is the probability that the extracted ball would turn on to be white , if all possible hypothesis concerning the color of theballs that initiallyin the bag were equally possible ? | "since , all possible hypothesis regarding the colour of the balls are equally likely , therefore these could be 3 white balls , initially in the bag . ∴ required probability = 1 / 4 [ 1 + 3 / 4 + 1 / 2 + 1 / 4 ] = 1 / 4 [ ( 4 + 3 + 2 + 1 ) / 4 ] = 5 / 8 b" | a = 1 / 2
b = 1 + 3
c = 1 / b
d = 1 + 3
e = 3 / d
f = c + e
g = a + f
h = 1 + g
i = 1 + 3
j = h / i
|
a ) 8 , b ) 16 , c ) 10 , d ) 11 , e ) 12 | b | multiply(subtract(13, 5), const_2) | half a number plus 5 is 13 . what is the number ? | "let x be the number . always replace ` ` is ' ' with an equal sign ( 1 / 2 ) x + 5 = 13 ( 1 / 2 ) x = 13 - 5 ( 1 / 2 ) x = 8 x = 16 correct answer is b" | a = 13 - 5
b = a * 2
|
a ) 64 : 25 , b ) 8 : 7 , c ) 5 : 8 , d ) 25 : 64 , e ) 6 : 7 | b | divide(add(const_4, const_4), 7) | a hare and a jackal are running a race . 3 leaps of the hare are equal to 4 leaps of the jackal . for every 6 leaps of the hare , the jackal takes 7 leaps . find the ratio of the speed of the hare to the speed of the jackal . | the hare takes 6 leaps and the jackal takes 7 leaps . 1 hare leap = 4 / 3 jackal leaps thus the hare ' s 6 leaps = 6 * ( 4 / 3 ) = 8 jackal leaps . the ratio of their speeds is 8 : 7 . the answer is b . | a = 4 + 4
b = a / 7
|
a ) 10 , b ) 11 , c ) 12 , d ) 13 , e ) 15 | c | floor(divide(109, 9)) | on dividing 109 by a number , the quotient is 9 and the remainder is 1 . find the divisor . | "d = ( d - r ) / q = ( 109 - 1 ) / 9 = 108 / 9 = 12 c" | a = 109 / 9
b = math.floor(a)
|
a ) $ 800 , b ) $ 900 , c ) $ 1000 , d ) $ 1100 , e ) $ 1200 | d | divide(multiply(multiply(15000, divide(1540, divide(multiply(21000, 8), const_100))), 8), const_100) | a , b and c enter into a partnership by investing $ 15000 , $ 21000 and $ 27000 respectively . at the end of 8 months , b receives $ 1540 as his share . find the share of a . | "the ratio of capital of a , b and c = 15000 : 21000 : 27000 = 5 : 7 : 9 a ' s share = ( 5 / 7 ) * 1540 = $ 1100 the answer is d ." | a = 21000 * 8
b = a / 100
c = 1540 / b
d = 15000 * c
e = d * 8
f = e / 100
|
a ) 27 , b ) 26 , c ) 29 , d ) 25 , e ) 24 | a | multiply(18, const_3) | a is twice as good as workman as b and together they finish a piece of work in 18 days . in how many days will a alone finish the work ? | "a : b = 2 : 1 ab ' s together wrk in 18 days a ; s alone is 1 / 18 * 2 / 3 = 27 days answer : a" | a = 18 * 3
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a ) 4.5 , b ) 5 , c ) 5.6 , d ) 7.5 , e ) 6.5 | d | multiply(divide(15, 12), 6) | when a number is divided by 6 & then multiply by 12 the answer is 15 what is the no . ? | "if $ x $ is the number , x / 6 * 12 = 15 = > 2 x = 15 = > x = 7.5 d" | a = 15 / 12
b = a * 6
|
a ) 12 , b ) 14 , c ) 45 , d ) 10 , e ) 65 | d | inverse(add(divide(divide(const_100, subtract(const_100, 25)), 15), divide(multiply(divide(divide(const_100, subtract(const_100, 25)), 15), 25), const_100))) | by selling 15 pens for a rupee a woman loses 25 % . how many for a rupee should he sell in order to gain 25 % ? | "d 85 % - - - 15 125 % - - - ? 85 / 125 * 15 = 10" | a = 100 - 25
b = 100 / a
c = b / 15
d = 100 - 25
e = 100 / d
f = e / 15
g = f * 25
h = g / 100
i = c + h
j = 1/(i)
|
['a ) 25 : 16', 'b ) 24 : 25', 'c ) 5 : 6', 'd ) 4 : 5', 'e ) 4 : 9'] | b | divide(multiply(2, 3), power(divide(add(2, 3), const_2), const_2)) | if the perimeter of square region z and the perimeter of rectangular region r are equal and the sides of r are in the ratio 2 : 3 then the ratio of the area of r to the area of z | we know perimeter of a square ( pz ) = 4 * side perimeter of a rectangle ( pr ) = 2 ( length + breath ) let us assume 40 to be the perimeter of the square ( since we know each side of a square is equal and the perimeter is divisible by 4 , also take in to account the length and breadth of the rectangle is in the ration 2 k : 3 k = 5 k ; we can assume such a number ) therefore , pz = pr = 40 area of the square = 100 sq . units we know 2 ( length + breadth ) = 40 i . e . length + breadth = 20 ( or 5 k = 20 given that l : b ( or b : l ) = 2 : 3 ) therefore length = 8 , breath = 12 area of the rectangle = 8 * 12 = 96 sq . units question asked = area of the rectangle : area of the square = 96 : 100 = = > 24 : 25 = b | a = 2 * 3
b = 2 + 3
c = b / 2
d = c ** 2
e = a / d
|
a ) 22 , b ) 77 , c ) 78 , d ) 98 , e ) 71 | c | add(add(multiply(divide(60, 9), const_2), 9), add(divide(60, 9), 9)) | the sum of the present ages of two persons a and b is 60 . if the age of a is twice that of b , find the sum of their ages 9 years hence ? | "a + b = 60 , a = 2 b 2 b + b = 60 = > b = 20 then a = 40 . 9 years , their ages will be 49 and 29 . sum of their ages = 49 + 29 = 78 . answer : c" | a = 60 / 9
b = a * 2
c = b + 9
d = 60 / 9
e = d + 9
f = c + e
|
a ) 7 days , b ) 8 days , c ) 9 days , d ) 10 days , e ) 16 days | e | divide(55, divide(add(add(divide(55, 35), divide(55, 20)), add(divide(55, 35), divide(55, 55))), const_2)) | a , band c can do a piece of work in 35 days , 20 days and 55 days respectively , working alone . how soon can the work be done if a is assisted by band c on alternate days ? | "( a + b ) ' s 1 day ' s work = 1 / 35 + 1 / 20 = 11 / 140 ( a + c ) ' s 1 day ' s work = 1 / 35 + 1 / 55 = 18 / 385 work done in 2 day ' s = 11 / 140 + 18 / 385 = 48 / 383 48 / 383 th work done in 2 days work done = 383 / 48 * 2 = 16 days ( approx ) answer : e" | a = 55 / 35
b = 55 / 20
c = a + b
d = 55 / 35
e = 55 / 55
f = d + e
g = c + f
h = g / 2
i = 55 / h
|
a ) $ 13 , b ) $ 15 , c ) $ 20 , d ) $ 28 , e ) $ 48 | c | subtract(divide(add(add(3, 4), 3), subtract(add(add(divide(1, 4), divide(1, 3)), divide(1, 2)), 1)), const_100) | aram ’ s car was fined when he gave joe and peter a ride , so they decided to help aram pay the fine . joe paid $ 3 more than 1 / 4 of the fine and peter paid $ 3 less than 1 / 3 of the fine , leaving pay $ 4 less than 1 / 2 the fine to complete the payment . what fraction of the fine did aram pay ? | as given aram ' payment is [ 1 ] [ / 2 ] * fine - 4 i . e total fine = 2 * ( sam ' s payment + 4 ) now , as everything looks integer , fine should be able to divide by 4 and 3 . because joe ' s contribution has 1 / 4 of the fine and peter ' s contribution has 1 / 3 of the fine so , that leads me to following : options , aram ' s pay , total fine ( f ) , f / 4 , f / 3 a ) , 13 , 34 , no i . e 34 / 4 , no i . e 34 / 3 b ) , 14 , 38 , no , no c ) , 20 , 48 , yes yes , no need to further go on as we have got f / 4 and f / 3 correct , just for confirmation we go further d ) , 28 , 64 , yes , no e ) , 48 , 104 , yes , no correct answer is c | a = 3 + 4
b = a + 3
c = 1 / 4
d = 1 / 3
e = c + d
f = 1 / 2
g = e + f
h = g - 1
i = b / h
j = i - 100
|
a ) 35.2 , b ) 36.1 , c ) 36.2 , d ) 36.4 , e ) none | d | divide(add(multiply(36, 50), subtract(subtract(50, const_2), 23)), 50) | the mean of 50 observations was 36 . it was found later that an observation 43 was wrongly taken as 23 . the corrected new mean is | "solution correct sum = ( 36 x 50 + 43 - 23 ) = 1820 . â ˆ ´ correct mean = 1820 / 50 = 36.4 . answer d" | a = 36 * 50
b = 50 - 2
c = b - 23
d = a + c
e = d / 50
|
a ) a ) 3.12 , b ) b ) 8 , c ) c ) 10 , d ) d ) 2.57 , e ) e ) 24 | d | max(multiply(subtract(add(52, 5), const_1), subtract(divide(5, 36), divide(5, 52))), const_4) | due to construction , the speed limit along an 5 - mile section of highway is reduced from 52 miles per hour to 36 miles per hour . approximately how many minutes more will it take to travel along this section of highway at the new speed limit than it would have taken at the old speed limit ? | "old time in minutes to cross 5 miles stretch = 5 * 60 / 52 = 5 * 15 / 13 = 5.76 new time in minutes to cross 5 miles stretch = 5 * 60 / 36 = 5 * 5 / 3 = 8.33 time difference = 2.57 ans : d" | a = 52 + 5
b = a - 1
c = 5 / 36
d = 5 / 52
e = c - d
f = b * e
g = max(f)
|
a ) 7 : 11 , b ) 7 : 13 , c ) 7 : 17 , d ) 7 : 19 , e ) 7 : 21 | b | divide(7, subtract(16, const_3)) | it is currently 7 : 16 pm . at what time in the morning was it exactly 19,443 minutes earlier ? | converting 19,443 minutes to hours , we get 19,443 / 60 = 324 r 3 ; that is 324 hours and 3 minutes . all of the answers are during the same hour of the morning , thus the hours can be assumed to bring us into the 7 am hour evenly from 7 : 16 pm . thus 324 hours ago was 7 : 16 am . take an extra 3 minutes off , and it was 7 : 13 am . b | a = 16 - 3
b = 7 / a
|
a ) 30.33 , b ) 34.44 , c ) 26.1 , d ) 28.0 , e ) 21.0 | a | subtract(52, divide(subtract(780, multiply(52, 10)), subtract(22, 10))) | 52 is to be divided into two parts such that the sum of 10 times the first and 22 times the second is 780 . the bigger part is : | "explanation : let the two parts be ( 52 - x ) and x . then , 10 ( 52 - x ) + 22 x = 780 = > 12 x = 260 = > x = 21.66 . bigger part = ( 52 - x ) = 30.33 . answer : a ) 30.33" | a = 52 * 10
b = 780 - a
c = 22 - 10
d = b / c
e = 52 - d
|
a ) 3800 , b ) 3300 , c ) 2400 , d ) 4200 , e ) 1400 | c | multiply(divide(subtract(multiply(2, 1600), 1600), subtract(multiply(2, 6), 4)), 12) | the price of 2 sarees and 4 shirts is $ 1600 . with the same money one can buy 1 saree and 6 shirts . if one wants to buy 12 shirts , how much shall he have to pay ? | c $ 2400 let the price of a saree and a shirt be $ x and $ y respectively . then , 2 x + 4 y = 1600 . . . . ( i ) and x + 6 y = 1600 . . . . ( ii ) divide equation ( i ) by 2 , we get the below equation . = > x + 2 y = 800 . - - - ( iii ) now subtract ( iii ) from ( ii ) x + 6 y = 1600 ( - ) x + 2 y = 800 - - - - - - - - - - - - - - - - 4 y = 800 - - - - - - - - - - - - - - - - therefore , y = 200 . now apply value of y in ( iii ) = > x + 2 x 200 = 800 = > x + 400 = 800 therefore x = 400 solving ( i ) and ( ii ) we get x = 400 , y = 200 . cost of 12 shirts = $ ( 12 x 200 ) = $ 2400 . | a = 2 * 1600
b = a - 1600
c = 2 * 6
d = c - 4
e = b / d
f = e * 12
|
a ) a ) 12 , b ) b ) 15 , c ) c ) 16 , d ) d ) 17 , e ) e ) 30 | e | floor(divide(271, 9)) | on dividing 271 by a number , the quotient is 9 and the remainder is 1 . find the divisor ? | "d = ( d - r ) / q = ( 271 - 1 ) / 9 = 270 / 9 = 30 e )" | a = 271 / 9
b = math.floor(a)
|
a ) 4 , b ) 4.5 , c ) 6 , d ) 8 , e ) 6.5 | c | inverse(add(inverse(subtract(12, const_1)), inverse(add(12, 2)))) | working alone at their respective constant rates , a can complete a task in ‘ a ’ days and b in ‘ b ’ days . they take turns in doing the task with each working 2 days at a time . if a starts they finish the task in exactly 12 days . if b starts , they take half a day more . how long does it take to complete the task if they both work together ? | "work done by ab in a day = xy respectively . when a starts : no . of days when a works = 6 no . of days when b works = 6 → 6 x + 6 y = 1 when b starts : no . of days when a works = 6.5 no . of days when b works = 5.5 → 6.5 x + 5.5 y = 1 solving the above two equations for xy x = 1 / 12 y = 1 / 12 → total work done by ab in a day = 1 / 12 + 1 / 12 = 2 / 12 = 1 / 6 → no . of days to complete the work when both work together = 6 answer : c" | a = 12 - 1
b = 1/(a)
c = 12 + 2
d = 1/(c)
e = b + d
f = 1/(e)
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a ) 0 , b ) 5 , c ) 8 , d ) 9 , e ) 4 | e | multiply(subtract(divide(power(3, const_2), 5), floor(divide(power(3, const_2), 5))), 5) | on dividing a number by 5 , we get 3 as remainder . what will the remainder when the square of the this number is divided by 5 ? | "explanation : let the number be x and on dividing x by 5 , we get k as quotient and 3 as remainder . x = 5 k + 3 x ^ 2 = ( 5 k + 3 ) ^ 2 = ( 25 k ^ 2 + 30 k + 9 ) = 5 ( 5 k ^ 2 + 6 k + 1 ) + 4 on dividing x 2 by 5 , we get 4 as remainder . e )" | a = 3 ** 2
b = a / 5
c = 3 ** 2
d = c / 5
e = math.floor(d)
f = b - e
g = f * 5
|
a ) 370 cm 2 , b ) 365 cm 2 , c ) 380 cm 2 , d ) 384 cm 2 , e ) 394 cm 2 | d | add(multiply(multiply(divide(const_1, const_2), 24), sqrt(subtract(multiply(multiply(20, 20), const_4), multiply(24, 24)))), 24) | find the area of a rhombus one side of which measures 20 cm and one diagonal is 24 cm . | "explanation : let other diagonal = 2 x cm . since diagonals of a rhombus bisect each other at right angles , we have : ( 20 ) 2 = ( 12 ) 2 + ( x ) 2 = > x = √ ( 20 ) 2 – ( 12 ) 2 = √ 256 = 16 cm . _ i so , other diagonal = 32 cm . area of rhombus = ( 1 / 2 ) x ( product of diagonals ) = ( 1 / 2 × 24 x 32 ) cm 2 = 384 cm 2 answer : option d" | a = 1 / 2
b = a * 24
c = 20 * 20
d = c * 4
e = 24 * 24
f = d - e
g = math.sqrt(f)
h = b * g
i = h + 24
|
a ) 40 , b ) 42 , c ) 45 , d ) 48 , e ) 49 | c | divide(add(multiply(4, 30), multiply(4, 30)), subtract(add(const_3.0, 4), 4)) | a taxi leaves point a 4 hours after a bus left the same spot . the bus is traveling 30 mph slower than the taxi . find the speed of the taxi , if it overtakes the bus in two hours . | "let the speed of bus be v - 30 , speed of taxi be v the bus travelled a total of 6 hrs and taxi a total of 2 hrs . hence 6 * ( v - 30 ) = 2 v 6 v - 180 = 2 v 4 v = 180 v = 45 mph c" | a = 4 * 30
b = 4 * 30
c = a + b
d = 3 + 0
e = d - 4
f = c / e
|
a ) 1 , b ) 2 , c ) 3 , d ) 4 , e ) 5 | c | divide(15, 5) | m and n are the x and y coordinates , respectively , of a point in the coordinate plane . if the points ( m , n ) and ( m + p , n + 15 ) both lie on the line defined by the equation x = ( y / 5 ) - ( 2 / 5 ) , what is the value of p ? | "x = ( y / 5 ) - ( 2 / 5 ) , and so y = 5 x + 2 . the slope is 5 . ( n + 15 - n ) / ( m + p - m ) = 5 p = 3 the answer is c ." | a = 15 / 5
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a ) 42 , b ) 66 , c ) 98 , d ) 112 , e ) 154 | e | multiply(divide(330, add(7, 8)), 7) | in a certain pet shop , the ratio of dogs to cats to bunnies in stock is 7 : 7 : 8 . if the shop carries 330 dogs and bunnies total in stock , how many dogs are there ? | "let us assume the number of dogs , cats and bunnies to be 7 x , 7 x and 8 x total dogs and bunnies = 15 x . and we are given that 15 x = 330 . hence x = 22 . dogs = 7 x = 7 * 22 = 154 ( option e )" | a = 7 + 8
b = 330 / a
c = b * 7
|
a ) 10 , b ) 30 , c ) 50 , d ) 80 , e ) 100 | a | subtract(subtract(150, 80), 10) | of the 150 employees at company x , 80 are full - time , and 100 have worked at company x for at least a year . there are 10 employees at company x who aren ’ t full - time and haven ’ t worked at company x for at least a year . how many full - time employees of company x have worked at the company for at least a year ? | "full time employee who have not worked for at least one year = a full time employee who have worked for at least one year = b non full time employee who have worked for at least one year = c non full time employee who have not worked for at least one year = d a + b + c + d = 150 a + b = 80 i . e . c + d = 70 b + c = 100 i . e . a + d = 50 d = 20 i . e . c = 70 - 20 = 50 i . e . b = 100 - 50 = 50 i . e . a = 80 - 50 = 30 b = 10 answer : option a" | a = 150 - 80
b = a - 10
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a ) 29 , b ) 58 , c ) 63 , d ) 74 , e ) 77 | e | multiply(add(const_2, const_3), const_2) | if y is the smallest positive integer such that 7,700 multiplied by y is the square of an integer , then y must be | "7700 = 5 * 5 * 2 * 2 * 7 * 11 , so we need one 7 and one 11 to make it a square of a number . so 7 * 11 = 77 ans : e" | a = 2 + 3
b = a * 2
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a ) 18 % , b ) 29.68 % , c ) 19.68 % , d ) 29 % , e ) 68 % | b | multiply(subtract(const_1, divide(multiply(20, const_2), add(add(20, 38), const_1))), const_100) | a bookstore has a shelf that contains biographies which normally sell for $ 20 each and mysteries that normally sell for $ 12 each . during a sale , the biographies and mysteries are discounted at different rates so that a customer saves a total of $ 19 from the normal price by buying 5 discounted biographies and 3 discounted mysteries . if the sum of the discount rates for the two types of books is 38 percent , what is the discount rate on mysteries ? | "let b be the discount on biographies and m be the discount on mysteries so . , b + m = 0.38 - - - - - ( 1 ) and ( 20 * 5 + 12 * 3 ) - ( 20 * 5 * ( 1 - b ) + 12 * 3 * ( 1 - m ) ) = 19 - - > 100 ( 1 - ( 1 - b ) ) + 36 ( 1 - ( 1 - m ) = 19 100 b + 36 m = 19 - - - - - - ( 2 ) solving 12 . , we get m = 0.2968 = 29.68 % b" | a = 20 * 2
b = 20 + 38
c = b + 1
d = a / c
e = 1 - d
f = e * 100
|
a ) 3 , b ) 5 , c ) 6 , d ) 7 , e ) 9 | d | subtract(power(subtract(73, multiply(add(const_3, const_4), const_10)), subtract(320, multiply(floor(divide(320, const_4)), const_4))), multiply(const_2, const_10)) | find the ones digit of 73 ^ 320 | "cyclicity of 3 is 3,9 , 7,1 after 4 multiplication again the cycle repeats . so divide 320 by 4 and we get 87 as quotient and 2 as remainder . so cycle will run for 87 times and then 2 times more . so pick up the 2 nd item from the cycle . hence answer d ." | a = 3 + 4
b = a * 10
c = 73 - b
d = 320 / 4
e = math.floor(d)
f = e * 4
g = 320 - f
h = c ** g
i = 2 * 10
j = h - i
|
a ) 148 , b ) 152 , c ) 156 , d ) 144 , e ) none | c | multiply(39, divide(152, divide(add(add(37, 39), const_2), const_2))) | the ratio of ducks and frogs in a pond is 37 : 39 respectively . the average number of ducks and frogs in the pond is 152 . what is the number of frogs in the pond ? | "solution : ratio of ducks and frogs in pond , = 37 : 39 . average of ducks and frogs in pond , = 152 . so , total number of ducks and frogs in the pond , = 2 * 152 = 304 . therefore , number of frogs , = ( 304 * 39 ) / 76 = 156 . answer : option c" | a = 37 + 39
b = a + 2
c = b / 2
d = 152 / c
e = 39 * d
|
a ) 7 , b ) 8 , c ) 9 , d ) 10 , e ) 11 | a | subtract(subtract(20, const_1), 12) | the “ connection ” between any two positive integers a and b is the ratio of the smallest common multiple of a and b to the product of a and b . for instance , the smallest common multiple of 8 and 12 is 24 , and the product of 8 and 12 is 96 , so the connection between 8 and 12 is 24 / 96 = 1 / 4 the positive integer y is less than 20 and the connection between y and 6 is equal to 1 / 1 . how many possible values of y are there ? | since “ connection ” between y and 6 is 1 / 1 then lcm ( 6 , y ) = 6 y , which means that 6 and y are co - prime ( they do not share any common factor but 1 ) , because if the had any common factor but 1 then lcm ( 6 , y ) would be less than 6 y . so , we should check how many integers less than 20 are co - prime with 6 , which can be rephrased as how many integers less than 20 are not divisible by 2 or 3 ( 6 = 2 * 3 ) . there are ( 18 - 2 ) / 2 + 1 = 9 multiples of 2 in the range from 0 to 20 , not inclusive ; there are ( 18 - 3 ) / 3 + 1 = 6 multiples of 3 in the range from 0 to 20 , not inclusive ; there are 3 multiples of 6 in the range from 0 to 20 , not inclusive ( 6 , 12 , 18 ) - overlap of the above two sets ; total multiples of 2 or 6 in the range from 0 to 20 , not inclusive is 9 + 6 - 3 = 12 ; total integers in the range from 0 to 20 , not inclusive is 19 ; hence , there are total of 19 - 12 = 7 numbers which have no common factor with 6 other than 1 : 1 , 5 , 7 , 11 , 13 , 17 and 19 . answer : a . | a = 20 - 1
b = a - 12
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a ) 0.45 , b ) 0.4 , c ) 0.5 , d ) 0.05 , e ) 0.6 | e | subtract(const_1, subtract(add(0.15, 0.40), 0.15)) | the probability of two events a and b are 0.15 and 0.40 respectively . the probability that both a and b occur is 0.15 . the probability that neither a nor b occur is _________ | "we are apply that formula . . . . . . . . . . . . . . p ( aorb ) = p ( a ) + p ( b ) - p ( a and b ) = . 15 + . 40 - . 15 = . 40 but the probability of neither a nor b = 1 - . 40 = 0.60 answer : e" | a = 0 + 15
b = a - 0
c = 1 - b
|
a ) 0.6 , b ) 0.5 , c ) 0.35 , d ) 0.3 , e ) none of them | d | divide(subtract(power(0.8, 3), power(0.5, 3)), add(add(power(0.8, 2), 0.40), power(0.5, 2))) | ( 0.8 ) ( power 3 ) - ( 0.5 ) ( power 3 ) / ( 0.8 ) ( power 2 ) + 0.40 + ( 0.5 ) ( power 2 ) is : | "given expression = ( 0.8 ) ( power 3 ) - ( 0.5 ) ( power 3 ) / ( 0.8 ) ( power 2 ) + ( 0.8 x 0.5 ) + ( 0.5 ) ( power 2 ) = a ( power 3 ) - b ( power 3 ) / a ( power 2 ) + ab + b ( power 2 ) = ( a - b ) = ( 0.8 - 0.5 ) = 0.30 answer is d ." | a = 0 ** 8
b = 0 ** 5
c = a - b
d = 0 ** 8
e = d + 0
f = 0 ** 5
g = e + f
h = c / g
|
a ) 7 : 1 , b ) 14 : 1 , c ) 21 : 1 , d ) 42 : 1 , e ) 49 : 1 | e | multiply(7, 7) | if the length of the sides of two cubes are in the ratio 7 : 1 , what is the ratio of their total surface area ? | "let x be the length of the small cube ' s side . the total surface area of the small cube is 6 x ^ 2 . the total surface area of the large cube is 6 ( 7 x ) ^ 2 = 294 x ^ 2 . the ratio of surface areas is 49 : 1 . the answer is e ." | a = 7 * 7
|
a ) $ 2160 , b ) $ 2370 , c ) $ 2580 , d ) $ 2790 , e ) $ 2950 | c | add(1,000, divide(110.60, divide(7, const_100))) | when a merchant imported a certain item , she paid a 7 percent import tax on the portion of the total value of the item in excess of $ 1,000 . if the amount of the import tax that the merchant paid was $ 110.60 , what was the total value of the item ? | "let x be the value of the item . 0.07 * ( x - 1000 ) = 110.60 x = 2580 the answer is c ." | a = 7 / 100
b = 110 / 60
c = 1 + 0
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a ) 53 , b ) 108 , c ) 77 , d ) 107 , e ) 109 | d | add(subtract(multiply(multiply(add(const_3, const_3), 3), 3), multiply(add(const_3, const_3), 3)), multiply(subtract(multiply(multiply(add(const_3, const_3), 3), 3), multiply(add(const_3, const_3), 3)), const_2)) | eighteen years ago , a father was 3 times as old as his son . now the father is only twice as old his son . then the sum of the present ages of the son and the father is : | explanation : let the present ages of the father and son be 2 x and x years respectively . then , ( 2 x - 18 ) = 3 ( x - 18 ) = > x = 36 required sum = ( 2 x + x ) = 108 years . answer : option d | a = 3 + 3
b = a * 3
c = b * 3
d = 3 + 3
e = d * 3
f = c - e
g = 3 + 3
h = g * 3
i = h * 3
j = 3 + 3
k = j * 3
l = i - k
m = l * 2
n = f + m
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a ) 16 , b ) 17 , c ) 13 , d ) 14 , e ) 15 | a | subtract(multiply(5, const_3), const_1) | how many pieces can a square pizza be cut into by making 5 linear cuts ? | "1 cut can make 2 pieces a second cut can make 4 pieces by cutting through 2 pieces a third cut can make 7 pieces by cutting through 3 of the pieces a fourth cut can make 11 pieces by cutting through 4 of the pieces a fifth cut can make 16 pieces by cutting through 5 of the pieces a" | a = 5 * 3
b = a - 1
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['a ) 128 m', 'b ) 64 m', 'c ) 64 m ^ 2', 'd ) 128 m ^ 2', 'e ) none'] | b | divide(128, const_2) | the area of a parallelogram is 128 sq m . then the area of a triangle formed by its diagonal is - - - - - - - - - - ? | b * h / 2 = 128 / 2 = 64 m . ans : ( b ) | a = 128 / 2
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a ) 4.1 , b ) 4.5 , c ) 4.6 , d ) 5.4 , e ) 5.5 | c | divide(divide(1080, const_1000), divide(multiply(14, const_60), const_3600)) | a person crosses a 1080 m long street in 14 minutes . what is his speed in km per hour ? | "speed = 1080 / ( 12 x 60 ) m / sec = 1.3 m / sec . converting m / sec to km / hr = 1.3 x ( 18 / 5 ) km / hr = 4.6 km / hr . answer : c" | a = 1080 / 1000
b = 14 * const_60
c = b / 3600
d = a / c
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a ) 2 , b ) 5 / 3 , c ) 3 , d ) 10 / 3 , e ) 14 / 3 | d | divide(add(add(multiply(3, 2), 3), subtract(2, 1)), 3) | if x is a prime number , and x - 1 is the median of the set { x - 1 , 3 x + 3 , 2 x - 4 } , then what is the average ( arithmetic mean ) of the set ? | if x is a prime number , and x - 1 is the median of the set { x - 1 , 3 x + 3 , 2 x - 4 } , then what is the average ( arithmetic mean ) of the set ? a . 2 b . 5 / 3 c . 3 d . 10 / 3 e . 14 / 3 solution : x - 1 is the median of the set , implies if we arrange the set in ascending order , the set would be [ 2 x - 4 , x - 1 , 3 x + 3 ] . this means that : 2 x - 4 < x - 2 < 3 x + 3 solving inequality gives x > - 2 and x < 3 . since x is prime number and the only prime number < 3 is 2 , so put x = 2 . hence set becomes : [ 0 , 1 , 9 ] and avg = 10 / 3 . d | a = 3 * 2
b = a + 3
c = 2 - 1
d = b + c
e = d / 3
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a ) 179 , b ) 208 , c ) 210 , d ) 223 , e ) 229 | a | divide(add(150, subtract(multiply(60, 46), multiply(58, subtract(46, const_2)))), const_2) | the batting average of a particular batsman is 60 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 = 60 * 46 = 2760 runs now excluding the two innings the runs scored = 58 * 44 = 2552 runs hence the runs scored in the two innings = 2760 – 2552 = 208 runs . let the highest score be x , hence the lowest score = x – 150 x + ( x - 150 ) = 208 2 x = 358 x = 179 runs answer a" | a = 60 * 46
b = 46 - 2
c = 58 * b
d = a - c
e = 150 + d
f = e / 2
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a ) 91.5 cm , b ) 92.2 cm , c ) 28.9 cm , d ) 29.2 cm , e ) 98.2 cm | a | 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 135 o . find its perimeter ? | "perimeter of the sector = length of the arc + 2 ( radius ) = ( 135 / 360 * 2 * 22 / 7 * 21 ) + 2 ( 21 ) = 49.5 + 42 = 91.5 cm answer : a" | a = 21 - 3
b = a * 2
c = 4 + 3
d = b / c
e = 2 * d
f = e * 21
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a ) 0.008 , b ) 8 , c ) 0.8 , d ) 80 , e ) 800 | b | multiply(divide(8.008, 1.001), const_100) | 8.008 / 1.001 | "answer is 8 , move the decimal forward three places for both numerator and denominator or just multiply both by a thousand . the result is 8008 / 1001 = 8 answer b" | a = 8 / 8
b = a * 100
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['a ) 36', 'b ) 25', 'c ) 15', 'd ) 28', 'e ) 37'] | a | add(divide(circumface(7), const_2), add(7, 7)) | the radius of a semicircle is 7 . what is the approximate perimeter of the semicircle ? | perimeter of a circle = 2 pi * r perimeter of a semicircle = pi * r + 2 r aprox perimiter = 3.14 * 7 + 2 * 7 = 35.98 approximately 36 answer a | a = circumface / (
b = a + 2
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a ) 16.5 sec , b ) 12 sec , c ) 15 sec , d ) 20 sec , e ) none of these | a | divide(subtract(multiply(12, 15), 15), divide(multiply(12, 15), 18)) | a train consists of 12 boggies , each boggy 15 metres long . the train crosses a telegraph post in 18 seconds . due to some problem , one boggies were detached . the train now crosses a telegraph post in | length of train = 12 ã — 15 = 180 m . then , speed of train = 180 â „ 18 = 10 m / s now , length of train = 11 ã — 15 = 165 m â ˆ ´ required time = 165 â „ 10 = 16.5 sec . answer a | a = 12 * 15
b = a - 15
c = 12 * 15
d = c / 18
e = b / d
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a ) 12 , b ) 12.5 , c ) 13 , d ) 13.5 , e ) 14 | b | divide(divide(const_2, divide(8, const_100)), const_2) | if a sum of money doubles itself in 8 years at simple interest , the ratepercent per annum is | explanation : let sum = x then simple interest = x rate = ( 100 * x ) / ( x * 8 ) = 12.5 option b | a = 8 / 100
b = 2 / a
c = b / 2
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a ) 1.5 , b ) 2.5 , c ) 3.75 , d ) 5 , e ) 7.5 | a | divide(divide(multiply(add(4, 2), 2), 2), 4) | natasha climbs up a hill , and descends along the same way she went up . it takes her 4 hours to reach the top and 2 hours to come back down . if her average speed along the whole journey is 2 kilometers per hour , what was her average speed ( in kilometers per hour ) while climbing to the top ? | "let the distance to the top be x , so the total distance traveled by natasha is 2 x . the total time is 4 + 2 = 6 hours the average speed = total distance / total time taken = 2 x / 6 = x / 3 the average speed of the complete journey is 2 km / hour x / 3 = 2 x = 6 km the average speed while climbing = distance / time = 6 / 4 = 1.5 km / h the answer is a ." | a = 4 + 2
b = a * 2
c = b / 2
d = c / 4
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a ) 55 , b ) 89 , c ) 100 , d ) 109 , e ) 115 | b | subtract(multiply(30, 4), multiply(15, 2)) | the average ( arithmetic mean ) of 4 positive integers is 30 . if the average of 2 of these integers is 15 , what is the greatest possible value that one of the other 2 integers can have ? | "a + b + c + d = 120 a + b = 30 c + d = 90 greatest possible = 89 ( just less than 1 ) answer = b" | a = 30 * 4
b = 15 * 2
c = a - b
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a ) 1 / 4 , b ) 2 / 7 , c ) 5 / 12 , d ) 1 / 2 , e ) 7 / 12 | c | divide(1, 4) | carol spends 1 / 4 of her savings on a stereo and 1 / 3 less than she spent on the stereo for a television . what fraction of her savings did she spend on the stereo and television ? | "total savings = s amount spent on stereo = ( 1 / 4 ) s amount spent on television = ( 1 - 1 / 3 ) ( 1 / 4 ) s = ( 2 / 3 ) * ( 1 / 4 ) * s = ( 1 / 6 ) s ( stereo + tv ) / total savings = s ( 1 / 4 + 1 / 6 ) / s = 5 / 12 answer : c" | a = 1 / 4
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a ) 103 , b ) 102 , c ) 100 , d ) 105 , e ) 106 | c | divide(5000, 50) | jancy had 100 currency notes in all , some of which are of rs 70 denomination and the remaining of rs 50 denomination . the total amount of all these currency notes was rs . 5000 . how much amount ( in rs ) did she have in the denomination of rs 50 ? | let the number of 50 - rupee notes = x then , the number of 70 - rupee notes = ( 100 – x ) 50 x + 70 ( 100 – x ) = 5000 : x = 100 answer : c | a = 5000 / 50
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a ) 3 / 4 , b ) 1 , c ) 4 / 3 , d ) 2 , e ) 7 / 2 | b | divide(4, 4) | a line that passes through ( – 1 , – 4 ) and ( 4 , k ) has a slope = k . what is the value of k ? | "slope = ( y 2 - y 1 ) / ( x 2 - x 1 ) = > k = ( k + 4 ) / ( 4 + 1 ) = > 5 k = k + 4 = > k = 1 ans b it is !" | a = 4 / 4
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a ) 20 % , b ) 25 % , c ) 28.5 % , d ) 30 % , e ) 32 % | b | multiply(divide(subtract(add(add(100, 25), 100), multiply(2, 100)), 100), 100) | a man saves a certain portion of his income during a year and spends the remaining portion on his personal expenses . next year his income increases by 25 % but his savings increase by 100 % . if his total expenditure in 2 years is double his expenditure in 1 st year , what % age of his income in the first year did he save ? | "1 st year income = i 1 st year savings = s 1 st year expense = e 1 2 nd year income = 1.25 i 2 nd year savings = 2 s ( 100 % increase ) 2 nd year expense = e 2 e 1 + e 2 = 2 e 1 e 2 = e 1 that means expenses are same during both years . with increase of 25 % income the savings increased by 100 % . or s = . 25 i or s = 25 % of income b is the answer" | a = 100 + 25
b = a + 100
c = 2 * 100
d = b - c
e = d / 100
f = e * 100
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a ) 24 , b ) 26 , c ) 30 , d ) 46 , e ) 50 | b | subtract(add(24, 30), divide(24, divide(30, const_100))) | the contents of a certain box consist of 24 apples and 30 kiwis . how many kiwis must be added to the box so that exactly 30 % of the pieces of fruit in the box will be apples ? | "apples = ( apples + kiwi + x ) * 0.3 24 = ( 30 + 24 + x ) * 0.3 x = 26 . answer : b ." | a = 24 + 30
b = 30 / 100
c = 24 / b
d = a - c
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a ) 5 days , b ) 10 days , c ) 14 days , d ) 22 days , e ) 40 days | e | add(16, divide(subtract(const_1, divide(16, 80)), add(inverse(80), inverse(48)))) | p and q can complete a work in 80 days and 48 days respectively . p alone started the work and q joined him after 16 days till the completion of the work . how long did the work last ? | explanation : work done by p in 1 day = 1 / 80 work done by q in 1 day = 1 / 48 work done by p in 16 days = 16 ã — ( 1 / 80 ) = 1 / 5 remaining work = 1 â € “ 1 / 5 = 4 / 5 work done by p and q in 1 day = 1 / 80 + 1 / 48 = 1 / 30 number of days p and q take to complete the remaining work = ( 4 / 5 ) / ( 1 / 30 ) = 24 total days = 16 + 24 = 40 answer : option e | a = 16 / 80
b = 1 - a
c = 1/(80)
d = 1/(48)
e = c + d
f = b / e
g = 16 + f
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a ) 4 : 9 , b ) 4 : 3 , c ) 4 : 12 , d ) 4 : 8 , e ) 4 : 5 | b | divide(sqrt(16), sqrt(9)) | two trains , one from howrah to patna and the other from patna to howrah , start simultaneously . after they meet , the trains reach their destinations after 9 hours and 16 hours respectively . the ratio of their speeds is | "let us name the trains as a and b . then , ( a ' s speed ) : ( b ' s speed ) = b : a = 16 : 9 = 4 : 3 . answer : b" | a = math.sqrt(16)
b = math.sqrt(9)
c = a / b
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a ) 3 / 20 , b ) 1 / 3 , c ) 3 / 7 , d ) 2 / 7 , e ) 7 / 2 | c | divide(multiply(3, 3), add(2, multiply(3, 3))) | every student in a room is either a junior or a senior . there is at least one junior and at least one senior in the room . if 2 / 3 of the juniors is equal to 1 / 2 of the seniors , what fraction of the students in the room are juniors ? | "let total number of juniors = j total number of seniors = s ( 2 / 3 ) j = ( 1 / 2 ) s = > s = 4 / 3 j total number of students = j + s = ( 7 / 3 ) j fraction of the students in the room are juniors = j / ( j + s ) = j / [ ( 7 / 3 ) j ] = 3 / 7 answer c" | a = 3 * 3
b = 3 * 3
c = 2 + b
d = a / c
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a ) 40 - 42 , b ) 39 - 41 , c ) 38 - 40 , d ) 37 - 39 , e ) 36 - 37 | c | add(multiply(34.5, divide(15, const_100)), 34.5) | a meal cost $ 34.50 and there was no tax . if the tip was more than 10 pc but less than 15 pc of the price , then the total amount paid should be : | 10 % ( 34.5 ) = 3.45 15 % ( 34.5 ) = 5.175 total amount could have been 34.5 + 3.45 and 34.5 + 5.175 = > could have been between 37.95 and 39.675 = > approximately between 38 and 40 answer is c . | a = 15 / 100
b = 34 * 5
c = b + 34
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a ) 1000 , b ) 1200 , c ) 1500 , d ) 2000 , e ) 1800 | b | multiply(divide(300, subtract(const_100, add(add(add(add(5, 5), divide(multiply(subtract(const_100, add(5, 5)), 25), const_100)), divide(multiply(subtract(const_100, add(5, 5)), 20), const_100)), 24.5))), const_100) | in an examination , questions were asked in 5 sections . out of the total students , 5 % candidates cleared the cut - off in all the sections and 5 % cleared none . of the rest , 25 % cleared only one section and 20 % cleared 4 sections . if 24.5 % of the entire candidates cleared two sections and 300 candidates cleared 3 sections . find out how many candidates appeared at the examination ? | solution : passed in none = 5 % passed in all = 5 % passed in four = 20 % 0 f 90 % = 18 % passed in three = 24.5 % passed in three = ( 100 - 5 - 5 - 22.5 - 24.5 - 18 ) = 25 % . but given 300 students passed in three . hence , 25 % = 300 . so , 100 % = 1200 . 1200 students must have appeared . answer : option b | a = 5 + 5
b = 5 + 5
c = 100 - b
d = c * 25
e = d / 100
f = a + e
g = 5 + 5
h = 100 - g
i = h * 20
j = i / 100
k = f + j
l = k + 24
m = 100 - l
n = 300 / m
o = n * 100
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a ) 5 , b ) 7 , c ) 9 , d ) 14 , e ) 24 | a | divide(multiply(2, add(1, 4)), const_2) | a carpenter worked alone for 1 day on a job that would take him 4 more days to finish . he and another carpenter completed the job in 2 more days . how many days would it have taken the second carpenter to do the complete job working alone ? | "a carpenter worked only 1 day on something that takes him 4 more days . means ; carpenter finishes his work in 5 days . let his buddy finish the same task in x days . respective rates per day : 1 / 5 and 1 / x to complete 1 work : first guy worked for 3 days @ rate = 1 / 5 per day . second one worked for 2 days @ rate = 1 / x per day expression : days * rate = work 3 * 1 / 5 + 2 * 1 / x = 1 3 x + 10 = 5 x 2 x = 10 x = 5 days . ans : a" | a = 1 + 4
b = 2 * a
c = b / 2
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a ) 22 , b ) 24 , c ) 28 , d ) 32 , e ) 44 | b | multiply(2, 12) | g ( x ) is defined as the product of all even integers k such that 0 < k ≤ x . for example , g ( 14 ) = 2 × 4 × 6 × 8 × 10 × 12 × 14 . if g ( z ) is divisible by 4 ^ 11 , what is the smallest possible value for z ? | g ( z ) = 4 ^ 11 = 2 ^ 22 . so we have to find a product with atleast 22 2 ' s in it . in option 1 22 the total no of 2 ' s = [ 22 / 2 ] + [ 22 / 4 ] + [ 22 / 8 ] + [ 22 / 16 ] = 11 + 5 + 2 + 1 = 19 in option 2 24 the total no of 2 ' s = [ 24 / 2 ] + [ 24 / 4 ] + [ 24 / 8 ] + [ 24 / 16 ] = 12 + 6 + 3 + 1 = 22 . hence b | a = 2 * 12
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a ) 19.3 % , b ) 17 % , c ) 36.17 % , d ) 15.5 % , e ) 12.5 % | c | divide(multiply(add(50, 12), 7), 12) | if 12 ounces of a strong vinegar solution are diluted with 50 ounces of water to form a 7 - percent vinegar solution , what was the concentration of the original solution ? | let x be the quantity of non - vinegar in the strong vinegar solution thus vinegar quantity will be 12 - x when 50 ounces of water were added the percentage of vinegar becomes 7 % , thus ( 12 - x ) / 62 = 7 / 100 from this equation x = 7.66 answer ( 12 - 7.66 ) / 12 = 36.17 % answer : c | a = 50 + 12
b = a * 7
c = b / 12
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a ) 16 . , b ) 8 . , c ) 7 . , d ) 2 . , e ) - 2 . | d | subtract(add(5, 3), 6) | if ( a + b ) = 5 , ( b + c ) = 6 and ( c + d ) = 3 , what is the value of ( a + d ) ? | "given a + b = 5 b + c = 6 c + d = 3 ( a + b ) - ( b + c ) + ( c + d ) = ( a + d ) = > 5 - 6 + 3 = 2 . option d . . ." | a = 5 + 3
b = a - 6
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a ) 10.89 , b ) 14.81 , c ) 14.8 , d ) 14.82 , e ) 14.12 | a | divide(multiply(25, 136), 312) | on a map the distance between two mountains is 312 inches . the actual distance between the mountains is 136 km . ram is camped at a location that on the map is 25 inch from the base of the mountain . how many km is he from the base of the mountain ? | "explanation : since 312 inch = 136 km so 1 inch = 136 / 312 km so 25 inch = ( 136 ã — 25 ) / 312 = 10.89 km answer : a" | a = 25 * 136
b = a / 312
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a ) 65 days , b ) 45 days , c ) 8 days , d ) 16 days , e ) 18 days | c | divide(multiply(subtract(31, 27), 400), 200) | a garrison of 400 men had a provision for 31 days . after 27 days 200 persons re - enforcement leave the garrison . find the number of days for which the remaining ration will be sufficient ? | 400 - - - 31 400 - - - 4 200 - - - ? 400 * 4 = 200 * x = > x = 8 days . answer : c | a = 31 - 27
b = a * 400
c = b / 200
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a ) 15456 , b ) 14758 , c ) 15789 , d ) 13250 , e ) 12450 | c | divide(add(15000, subtract(divide(multiply(10, 5000), const_100), divide(multiply(const_4.0, 5000), const_100))), subtract(const_1, divide(5, const_100))) | a salesman commission is 10 % on all sales upto $ 5000 and 5 % on all sales exceeding this . he remits $ 15000 to his parent company after deducting his commission . find the total sales ? | "let his total sales be x total sales - commission = $ 15000 x - [ ( 10 % of 5000 ) + 5 % of ( x - 5000 ) ] = 15000 95 x / 100 = 15000 x = 15789 approximately answer is c" | a = 10 * 5000
b = a / 100
c = 4 * 0
d = c / 100
e = b - d
f = 15000 + e
g = 5 / 100
h = 1 - g
i = f / h
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a ) 30 , b ) 60 , c ) 80 , d ) 100 , e ) 90 | d | add(add(multiply(7.5, 7.5), multiply(2.5, 2.5)), 3.75) | ( 7.5 × 7.5 + 3.75 + 2.5 × 2.5 ) is equal to | "solution given expression = ( 7.5 x 7.5 + 2 × 7.5 × 2.5 + 2.5 × 2.5 ) ² = ( a ² + 2 ab + b ² ) = ( a + b ) ² = ( 7.5 + 2.5 ) ² = 10 ² = 100 . answer d" | a = 7 * 5
b = 2 * 5
c = a + b
d = c + 3
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a ) 40 % , b ) 50 % , c ) 60 % , d ) 30 % , e ) 20 % | c | multiply(divide(divide(40, const_100), subtract(const_1, divide(40, const_100))), const_100) | a merchant has selected two items to be placed on sale , one of which currently sells for 40 percent less than the other . if he wishes to raise the price of the cheaper item so that the two items are equally priced , by what percentage must he raise the price of the less expensive item ? | "expensive item = $ 100 ; cheap item = $ 60 ; we must increase $ 60 to $ 100 , so by $ 40 , which is approximately 60 % increase : ( 100 - 60 ) / 60 = 2 / 3 = ~ 0.66 . answer : c ." | a = 40 / 100
b = 40 / 100
c = 1 - b
d = a / c
e = d * 100
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a ) 2 , b ) 8 , c ) 9 , d ) 6 , e ) 5 | b | add(add(divide(factorial(const_3), factorial(const_2)), divide(factorial(const_4), factorial(const_3))), divide(factorial(divide(50, 10)), factorial(divide(50, 10)))) | a owes b rs . 50 . he agrees to pay b over a number of consecutive days on a monday , paying single note or rs . 10 or rs . 20 on each day . in how many different ways can a repay b . | he can pay by all 10 rupee notes = 1 way 3 ten rupee + 1 twenty rupee = 4 ! 3 ! × 1 ! 4 ! 3 ! × 1 ! = 4 ways 1 ten rupee + 2 twenty rupee notes = 3 ! 2 ! × 1 ! 3 ! 2 ! × 1 ! = 3 ways total ways = 1 + 4 + 3 = 8 answer : b | a = math.factorial(3)
b = math.factorial(2)
c = a / b
d = math.factorial(4)
e = math.factorial(3)
f = d / e
g = c + f
h = 50 / 10
i = math.factorial(h)
j = 50 / 10
k = math.factorial(j)
l = i / k
m = g + l
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a ) 70 % , b ) 90 % , c ) 33.33 % , d ) 50 % , e ) 65 % | c | multiply(divide(1, 9), const_100) | if two positive numbers are in the ratio 1 / 12 : 1 / 9 , then by what percent is the second number more than the first ? | "given ratio = 1 / 12 : 1 / 9 = 9 : 12 let first number be 9 x and the second number be 12 x . the second number is more than first number by 3 x . required percentage = 3 x / 9 x * 100 = 33.33 % . answer : c" | a = 1 / 9
b = a * 100
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a ) 1500,6000 , b ) 4500,3000 , c ) 3500,3000 , d ) 1500,3000 , e ) 2500,3000 | d | divide(multiply(10000, const_1), const_3) | a and b invests rs . 10000 each , a investing for 6 months and b investing for all the 12 months in the year . if the total profit at the end of the year is rs . 4500 , find their shares ? | "the ratio of their profits a : b = 6 : 12 = 1 : 2 share of a in the total profit = 1 / 3 * 4500 = rs . 1500 share of b in the total profit = 2 / 3 * 4500 = rs . 3000 answer : d" | a = 10000 * 1
b = a / 3
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a ) 57 , b ) 67 , c ) 77 , d ) 87 , e ) 97 | a | sqrt(multiply(32.49, const_100)) | a group of students decided to collect as many paise from each member of group as is the number of members . if the total collection amounts to rs . 32.49 , the number of the member is the group is : | "explanation : money collected = ( 32.49 x 100 ) paise = 3249 paise . ∴ number of members = √ ( 3249 ) = 57 . answer : a" | a = 32 * 49
b = math.sqrt(a)
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a ) 120 , b ) 150 , c ) 180 , d ) 210 , e ) 240 | d | multiply(subtract(multiply(subtract(multiply(3, 6), 3), 3), 3), subtract(6, const_1)) | q ' = 3 q - 3 , what is the value of ( 6 ' ) ' ? | ( 6 ' ) ' = ( 3 * 6 - 3 ) ' = 15 ' = 15 * 15 - 15 = 210 answer d | a = 3 * 6
b = a - 3
c = b * 3
d = c - 3
e = 6 - 1
f = d * e
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a ) 18.6 , b ) 9.3 , c ) 6.2 , d ) 3.1 , e ) 1.6 | e | divide(divide(38.3, 3), multiply(4, 2)) | in 1979 approximately 1 / 3 of the 38.3 million airline passengers traveling to or from the united states used kennedy airport . if the number of such passengers that used miami airport was 1 / 2 the number that used kennedy airport and 4 times the number that used logan airport , approximately how many millions of these passengers used logan airport that year ? | number of passengers using kennedy airport = 38 / 3 = ~ 12.67 passengers using miami airport = 12.67 / 2 = ~ 6.3 passengers using logan airport = 6.3 / 4 = ~ 1.58 so e | a = 38 / 3
b = 4 * 2
c = a / b
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a ) 2 , b ) 7 , c ) 5 , d ) 3 , e ) 6 | d | subtract(3, reminder(9, 7)) | when n is divided by 24 , the remainder is 3 . find thee difference between previous remainder and the remainder when 9 n is divided by 7 ? | "let n = 3 ( leaves a remainder of 3 when divided by 24 ) 9 n = 9 ( 3 ) = 27 , which leaves a remainder of 6 when divided by 7 . difference = 6 - 3 = 3 . answer d" | a = 3 - reminder
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a ) rs . 5392 , b ) rs . 6392 , c ) rs . 8392 , d ) rs . 7392 , e ) rs . 9392 | d | add(subtract(6000, divide(multiply(6000, 20), const_100)), divide(multiply(subtract(6000, divide(multiply(6000, 20), const_100)), 20), const_100)) | the initial price of an article is rs . 6000 which increases 40 % increse in its price in the first year , a 20 % decrease in the second year and a 20 % increase in the next year . what is the final price of the article ? | "the initial price of the article is rs . 6000 . in the 1 st year , price of the article = 6000 + 2400 = rs . 8400 . in the 2 nd year , price = 8400 - 20 % of 8400 = 8400 - 1680 = rs . 6720 . in the 3 rd year , price = 6720 + 10 % of 6720 = 6720 + 672 = rs . 7392 required price = = rs . 7392 . answer : d" | a = 6000 * 20
b = a / 100
c = 6000 - b
d = 6000 * 20
e = d / 100
f = 6000 - e
g = f * 20
h = g / 100
i = c + h
|
a ) 76 days , b ) 48 days , c ) 80 days , d ) 31 days , e ) 22 days | c | inverse(subtract(inverse(16), inverse(20))) | a and b can finish a work in 16 days while a alone can do the same work in 20 days . in how many days b alone will complete the work ? | "b = 1 / 16 – 1 / 20 = 1 / 80 = > 80 days answer : c" | a = 1/(16)
b = 1/(20)
c = a - b
d = 1/(c)
|
a ) $ 10800 , b ) $ 18000 , c ) $ 160000 , d ) $ 1800 , e ) none | c | divide(16000, subtract(1, add(add(divide(1, 5), divide(1, 10)), divide(3, 5)))) | a man spend 1 / 5 of his salary on food , 1 / 10 of his salary on house rent and 3 / 5 salary on clothes . he still has $ 16000 left with him . find salary . . | "[ 1 / ( x 1 / y 1 + x 2 / y 2 + x 3 / y 3 ) ] * total amount = balance amount [ 1 - ( 1 / 5 + 1 / 10 + 3 / 5 ) } * total salary = $ 16000 , = [ 1 - 9 / 10 ] * total salary = $ 16000 , total salary = $ 16000 * 10 = $ 160000 , correct answer ( c )" | a = 1 / 5
b = 1 / 10
c = a + b
d = 3 / 5
e = c + d
f = 1 - e
g = 16000 / f
|
a ) 39.25 , b ) 37.25 , c ) 37.26 , d ) 38.21 , e ) 38.32 | b | divide(multiply(subtract(divide(10000, 2), divide(multiply(20, 10000), const_100)), const_100), subtract(10000, divide(multiply(20, 10000), const_100))) | a has 10000 chocolates . 20 % of the chocolates are eaten . what percentage of the uneaten chocolates must be eaten so that 3 / 2 of all chocolates are eaten ? | number of chocolates = 10,000 currently eaten = 10,000 * 20 / 100 = 2000 uneaten chocolates = 10000 - 2000 = 8000 2 / 3 of all chocolates = 5000 the number of chocolates to be eaten to make 2 / 3 of all chocolates eaten = 5000 - 2000 = 3000 so now the question remains - 3000 is how much percentage of uneaten chocolates ( 8000 ) = 3000 * 100 / 8000 = 37.5 answer ( b ) | a = 10000 / 2
b = 20 * 10000
c = b / 100
d = a - c
e = d * 100
f = 20 * 10000
g = f / 100
h = 10000 - g
i = e / h
|
a ) 50 , b ) 52 , c ) 51 , d ) 53 , e ) none of the above | b | divide(add(add(add(multiply(60, 50), multiply(35, 60)), multiply(45, 55)), multiply(42, 45)), add(add(add(60, 35), 45), 42)) | a school has 4 section of chemistry in class x having 60 , 35 , 45 and 42 students . the mean marks obtained in chemistry test are 50 , 60 , 55 and 45 respectively for the 4 sections . determine the overall average of marks per student . | "required average marks = 60 ã — 50 + 35 ã — 60 + 45 ã — 55 + 42 ã — 45 / 60 + 35 + 45 + 42 = 3000 + 2100 + 2475 + 1890 / 182 = 9465 â „ 182 = 52 answer b" | a = 60 * 50
b = 35 * 60
c = a + b
d = 45 * 55
e = c + d
f = 42 * 45
g = e + f
h = 60 + 35
i = h + 45
j = i + 42
k = g / j
|
a ) 101 : 88 , b ) 87 : 100 , c ) 110 : 111 , d ) 85 : 98 , e ) 97 : 84 | e | divide(subtract(const_100, divide(300, const_100)), subtract(subtract(const_100, divide(300, const_100)), add(const_10, divide(300, const_100)))) | the number of employees in obelix menhir co . is a prime number and is less than 300 . the ratio of the number of employees who are graduates and above , to that of employees who are not , can possibly be : - | explanation : going through the given options , option 1 : - 101 : 88 = > 189 . [ divisible by 3 ] option 2 : - 87 : 100 = > 187 . [ divisible by 11 ] option 3 : - 85 : 98 = > 183 . [ divisible by 3 ] option 4 : - 97 : 84 = > 181 . [ prime number ] hence , only 181 is the prime number i . e the total number of employee in the company and the required ratio is 97 : 84 . answer : e | a = 300 / 100
b = 100 - a
c = 300 / 100
d = 100 - c
e = 300 / 100
f = 10 + e
g = d - f
h = b / g
|
a ) rs . 1020.80 , b ) rs . 1025 , c ) rs . 1055 , d ) data inadequate , e ) none of these | c | add(825, divide(multiply(multiply(825, add(divide(multiply(subtract(956, 825), const_100), multiply(825, 3)), 4)), 3), const_100)) | rs . 825 becomes rs . 956 in 3 years at a certain rate of simple interest . if the rate of interest is increased by 4 % , what amount will rs . 825 become in 3 years ? | "solution s . i . = rs . ( 956 - 825 ) = rs . 131 rate = ( 100 x 131 / 825 x 3 ) = 524 / 99 % new rate = ( 524 / 99 + 4 ) % = 920 / 99 % new s . i . = rs . ( 825 x 920 / 99 x 3 / 100 ) rs . 230 . ∴ new amount = rs . ( 825 + 230 ) = rs . 1055 . answer c" | a = 956 - 825
b = a * 100
c = 825 * 3
d = b / c
e = d + 4
f = 825 * e
g = f * 3
h = g / 100
i = 825 + h
|
a ) 22 , b ) 27 , c ) 20 , d ) 21 , e ) 12 | c | divide(multiply(subtract(multiply(540, divide(add(const_100, 15), const_100)), 496.80), const_100), multiply(540, divide(add(const_100, 15), const_100))) | mahesh marks an article 15 % above the cost price of rs . 540 . what must be his discount percentage if he sells it at rs . 496.80 ? | "cp = rs . 540 , mp = 540 + 15 % of 540 = rs . 621 sp = rs . 496.80 , discount = 621 - 496.80 = 124.20 discount % = 124.2 / 621 * 100 = 20 % . answer : c" | a = 100 + 15
b = a / 100
c = 540 * b
d = c - 496
e = d * 100
f = 100 + 15
g = f / 100
h = 540 * g
i = e / h
|
['a ) h / √ 2', 'b ) h / 2', 'c ) h / 4', 'd ) ( h ) ^ 2', 'e ) ( h ) ^ 2 / 4'] | e | power(divide(45, 90), const_2) | a 45 ° - 45 ° - 90 ° right triangle has hypotenuse of length h . what is the area of the triangle t in terms of h ? | if . . . each of the two shorter sides = 3 , then the hypotenuse = h = 3 ( root 2 ) . the area t = ( 1 / 2 ) ( base ) ( height ) = ( 1 / 2 ) ( 3 ) ( 3 ) = 9 / 2 . so we ' re looking for an answer that = 9 / 2 when h = 3 ( root 2 ) . there ' s only one answer that matches . . . e | a = 45 / 90
b = a ** 2
|
a ) 2 , b ) 8 , c ) 9 , d ) 10 , e ) 12 | e | divide(add(divide(18, 3), divide(54, 3)), const_2) | a man swims downstream 54 km and upstream 18 km taking 3 hours each time , what is the speed of the man in still water ? | "54 - - - 3 ds = 18 ? - - - - 1 18 - - - - 3 us = 6 ? - - - - 1 m = ? m = ( 18 + 6 ) / 2 = 12 answer : e" | a = 18 / 3
b = 54 / 3
c = a + b
d = c / 2
|
a ) a ) 46 , b ) b ) 34 , c ) c ) 66 , d ) d ) 76 , e ) e ) 74 | b | subtract(add(multiply(52, 6), multiply(49, 6)), multiply(52, 11)) | the average of 11 results is 52 , if the average of first 6 results is 49 and that of the last 6 is 52 . find the sixth result ? | 1 to 11 = 11 * 52 = 572 1 to 6 = 6 * 49 = 294 6 to 11 = 6 * 52 = 312 6 th = 294 + 312 – 572 = 34 answer : b | a = 52 * 6
b = 49 * 6
c = a + b
d = 52 * 11
e = c - d
|
a ) 20 , b ) 21 , c ) 25 , d ) 26 , e ) 27 | a | sqrt(add(138, multiply(131, const_2))) | sum of the squares of 3 no . is 138 and the sum of their products taken two at a time is 131 . find the sum ? | "( a + b + c ) 2 = a 2 + b 2 + c 2 + 2 ( ab + bc + ca ) = 138 + 2 * 131 a + b + c = √ 400 = 20 a" | a = 131 * 2
b = 138 + a
c = math.sqrt(b)
|
a ) 50 , b ) 100 , c ) 150 , d ) 200 , e ) 250 | d | divide(subtract(multiply(324, 25), multiply(71, const_100)), subtract(25, 20)) | the total of 324 of 20 paise and 25 paise make a sum of rs . 71 . the no of 20 paise coins is : | "explanation : let the number of 20 paise coins be x . then the no of 25 paise coins = ( 324 - x ) . 0.20 × ( x ) + 0.25 ( 324 - x ) = 71 = > x = 200 . answer : d" | a = 324 * 25
b = 71 * 100
c = a - b
d = 25 - 20
e = c / d
|
a ) 8 : 6 , b ) 9 : 3 , c ) 9 : 4 , d ) 9 : 6 , e ) 8 : 2 | c | divide(multiply(subtract(10, const_1), const_2), multiply(const_4, const_2)) | 120 liters of a mixture contains milk and water in the ratio 4 : 2 . if 10 liters of this mixture be replaced by 10 liters of milk , the ratio of milk to water in the new mixture would be ? | quantity of milk in 120 liters if mix = 120 * 4 / 6 = 80 liters quantity of milk in 130 liters of new mix = 80 + 10 = 90 liters quantity of water in it = 130 - 90 = 40 liters ratio of milk and water in new mix = 90 : 40 = 9 : 4 answer is c | a = 10 - 1
b = a * 2
c = 4 * 2
d = b / c
|
a ) 210.6 kg , b ) 221.3 kg , c ) 229.5 kg , d ) 110.8 kg , e ) 114 kg | b | add(multiply(divide(170, add(1, 3)), 3), multiply(divide(250, add(3, 5)), 3)) | 170 kg of an alloy a is mixed with 250 kg of alloy b . if alloy a has lead and tin in the ratio 1 : 3 and alloy b has tin and copper in the ratio 3 : 5 , then the amount of tin in the new alloy is ? | "quantity of tin in 170 kg of a = 170 * 3 / 4 = 127.5 kg quantity of tin in 250 kg of b = 250 * 3 / 8 = 93.75 kg quantity of tin in the new alloy = 127.5 + 93.75 = 221.25 kg answer is b" | a = 1 + 3
b = 170 / a
c = b * 3
d = 3 + 5
e = 250 / d
f = e * 3
g = c + f
|
a ) 88 kmph , b ) 43 kmph , c ) 72 kmph , d ) 16 kmph , e ) 18 kmph | b | divide(divide(360, const_1000), divide(30, const_3600)) | a train 360 m long can cross an electric pole in 30 sec and then find the speed of the train ? | "length = speed * time speed = l / t s = 360 / 30 s = 12 m / sec speed = 12 * 18 / 5 ( to convert m / sec in to kmph multiply by 18 / 5 ) speed = 43 kmph answer : b" | a = 360 / 1000
b = 30 / 3600
c = a / b
|
a ) 2 hours , b ) 3 hours , c ) 4 hours , d ) 5 hours , e ) 6 hours | c | divide(const_1, subtract(divide(const_1, 2), subtract(divide(const_1, 2), divide(const_1, 4)))) | a can do a piece of work in 4 hours ; b and c together can do it in 2 hours , which a and c together can do it in 2 hours . how long will b alone take to do it ? | "a ' s 1 hour work = 1 / 4 ; ( b + c ) ' s 1 hour work = 1 / 2 ; ( a + c ) ' s 1 hour work = 1 / 2 ( a + b + c ) ' s 1 hour work = ( 1 / 4 + 1 / 2 ) = 3 / 4 b ' s 1 hour work = ( 3 / 4 - 1 / 2 ) = 1 / 4 b alone will take 4 hours to do the work . answer : c" | a = 1 / 2
b = 1 / 2
c = 1 / 4
d = b - c
e = a - d
f = 1 / e
|
a ) 91 , b ) 28 , c ) 56 , d ) 89 , e ) none of these | b | gcd(1204, 840) | the maximum number of student amoung them 1204 pens and 840 pencils can be distributed in such a way that each student gets the same number of pens and same number of pencils is : | "solution required number of student = h . c . f of 1204 and 840 = 28 . answer b" | a = math.gcd(1204, 840)
|
a ) 45 , b ) 50 , c ) 55 , d ) 60 , e ) 105 | e | multiply(divide(multiply(divide(1, 12), const_3), divide(divide(1, 30), const_2)), 7) | in the standard formulation of a flavored drink the ratio by volume of flavoring to corn syrup to water is 1 : 12 : 30 . in the sport formulation , the ratio of flavoring to corn syrup is three times as great as in the standard formulation , and the ratio of flavoring to water is half that of the standard formulation . if a large bottle of the sport formulation contains 7 ounces of corn syrup , how many ounces of water does it contain ? | "f : c : w 1 : 12 : 30 sport version : f : c 3 : 12 f : w 1 : 60 or 3 : 180 so c : f : w = 12 : 3 : 180 c / w = 12 / 180 = 3 ounces / x ounces x = 7 * 180 / 12 = 105 ounces of water e" | a = 1 / 12
b = a * 3
c = 1 / 30
d = c / 2
e = b / d
f = e * 7
|
a ) 9 , b ) 15 , c ) 20 , d ) 38 , e ) 56 | a | multiply(power(const_2, 120), factorial(120)) | the product of three consecutive numbers is 120 . then the sum of the smallest two numbers is ? | "product of three numbers = 120 120 = 2 * 2 * 5 * 6 = 4 * 5 * 6 so , the three numbers are 4,5 and 6 . and sum of smallest of these two = 4 + 5 = 9 . answer : option a" | a = 2 ** 120
b = math.factorial(120)
c = a * b
|
a ) 20 , b ) 66 , c ) 18 , d ) 19 , e ) 01 | a | divide(multiply(divide(multiply(12.5, 40), const_100), const_100), 25) | a company pays 12.5 % dividend to its investors . if an investor buys rs . 40 shares and gets 25 % on investment , at what price did the investor buy the shares ? | "explanation : dividend on 1 share = ( 12.5 * 40 ) / 100 = rs . 5 rs . 25 is income on an investment of rs . 100 rs . 5 is income on an investment of rs . ( 5 * 100 ) / 25 = rs . 20 answer : a" | a = 12 * 5
b = a / 100
c = b * 100
d = c / 25
|
a ) 15 , b ) 30 , c ) 31 , d ) 33 , e ) 44 | e | subtract(divide(subtract(subtract(140, 10), const_2), const_2), divide(divide(subtract(subtract(subtract(subtract(140, const_2), multiply(3, const_4)), 3), 3), 3), const_2)) | how many even number in the range between 10 to 140 inclusive are not divisible by 3 | we have to find the number of terms that are divisible by 2 but not by 6 ( as the question asks for the even numbers only which are not divisible by 3 ) for 2 , 10 , 12,14 . . . 140 using ap formula , we can say 140 = 10 + ( n - 1 ) * 2 or n = 66 . for 6 , 12,18 , . . . 138 using ap formula , we can say 138 = 12 + ( n - 1 ) * 6 or n = 22 . hence , only divisible by 2 but not 3 = 66 - 22 = 44 . hence , answer e | a = 140 - 10
b = a - 2
c = b / 2
d = 140 - 2
e = 3 * 4
f = d - e
g = f - 3
h = g - 3
i = h / 3
j = i / 2
k = c - j
|
a ) 760 , b ) 720 , c ) 700 , d ) 786 , e ) 797 | e | subtract(add(multiply(7, 57), multiply(7, 61)), multiply(15, 55)) | the average of 15 result is 55 . average of the first 7 of them is 57 and that of the last 7 is 61 . find the 8 th result ? | "sum of all the 13 results = 15 * 55 = 825 sum of the first 7 of them = 7 * 57 = 399 sum of the last 7 of them = 7 * 61 = 427 so , the 8 th number = 825 + 399 - 427 = 797 . e" | a = 7 * 57
b = 7 * 61
c = a + b
d = 15 * 55
e = c - d
|
a ) 33 , b ) 27 , c ) 25 , d ) 22 , e ) 45 | e | multiply(const_3_6, divide(add(250, 150), 32)) | a train of length 250 m crosses a bridge of length 150 m in 32 seconds . what is the speed of train ? | "sol : ( length of train + length of bridge ) = speed of train x time ( 250 + 150 ) = 32 x speed speed = 400 / 32 = 12.5 m / s = 45 km / h answer = e" | a = 250 + 150
b = a / 32
c = const_3_6 * b
|
a ) 57.75 % , b ) 54.54 % , c ) 63 % , d ) 70 % , e ) none of these | a | multiply(divide(subtract(142, add(multiply(12, const_4), multiply(2, multiply(2, const_3)))), 142), const_100) | a cricketer scored 142 runs which included 12 boundaries and 2 sixes . what percent of his total score did he make by running between the wickets . | "explanation : number of runs made by running = 142 - ( 12 x 4 + 2 x 6 ) = 142 - ( 60 ) = 82 now , we need to calculate 82 is what percent of 142 . = > 82 / 142 * 100 = 57.75 % answer : a" | a = 12 * 4
b = 2 * 3
c = 2 * b
d = a + c
e = 142 - d
f = e / 142
g = f * 100
|
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