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 ) 11 sec , b ) 70 sec , c ) 21 sec , d ) 20 sec , e ) 19 sec | a | divide(add(100, 120), multiply(add(42, 30), const_0_2778)) | two trains of length 100 m and 120 m are running towards each other on parallel lines at 42 kmph and 30 kmph respectively . in what time will they be clear of each other from the moment they meet ? | "relative speed = ( 42 + 30 ) * 5 / 18 = 4 * 5 = 20 mps . distance covered in passing each other = 100 + 120 = 220 m . the time required = d / s = 220 / 20 = 11 sec . answer : a" | a = 100 + 120
b = 42 + 30
c = b * const_0_2778
d = a / c
|
a ) 150 , b ) 250 , c ) 400 , d ) 300 , e ) 100 | e | divide(add(300, 100), add(3, 2)) | in an examination , a student scores 3 marks for every correct answer and loses 2 marks for every wrong answer . if he attempts all 100 questions and secures 300 marks , the number of questions he attempts correctly , is : | "let the number of correct answers be x . number of incorrect answers = ( 100 â € “ x ) . 3 x â € “ 2 ( 100 â € “ x ) = 300 or 5 x = 500 or x = 100 . answer : e" | a = 300 + 100
b = 3 + 2
c = a / b
|
a ) 0 % , b ) 20 % increase , c ) 20 % decrease , d ) 9 % decrease , e ) insufficient data | d | subtract(const_100, divide(multiply(add(const_100, 30), subtract(const_100, 30)), const_100)) | what is the % change in the area of a rectangle when its length increases by 30 % and its width decreases by 30 % ? | "( 13 / 10 ) * ( 7 / 10 ) = 91 / 100 of original area 91 / 100 is a 9 % decrease from 100 / 100 - > d" | a = 100 + 30
b = 100 - 30
c = a * b
d = c / 100
e = 100 - d
|
['a ) rs . 3944', 'b ) rs . 3942', 'c ) rs . 3987', 'd ) rs . 3929', 'e ) rs . 3938'] | a | multiply(square_perimeter(sqrt(289)), 58) | what will be the cost of building a fence around a square plot with area equal to 289 sq ft , if the price per foot of building the fence is rs . 58 ? | let the side of the square plot be a ft . a 2 = 289 = > a = 17 length of the fence = perimeter of the plot = 4 a = 68 ft . cost of building the fence = 68 * 58 = rs . 3944 . answer : a | a = math.sqrt(289)
b = square_perimeter * (
|
a ) 1 , b ) 8 , c ) 3 , d ) 4 , e ) 5 | b | subtract(4, 3) | points a , b , c , and d , in that order , lie on a line . if ab = 3 cm , ac = 4 cm , and bd = 9 cm , what is cd , in centimeters ? | "putting a value to each point , lets use the following : a - 0 b - 3 ( ab = 3 ) c - 4 ( ac = 4 ) d - 12 ( bd = 9 ) cd is 12 - 4 = 8 . ans b" | a = 4 - 3
|
a ) 8 kmph , b ) 10 kmph , c ) 12 kmph , d ) 14 kmph , e ) 16 kmph | c | divide(divide(2, subtract(divide(1, 10), divide(1, 15))), subtract(divide(divide(2, subtract(divide(1, 10), divide(1, 15))), 10), 1)) | robert is travelling on his cycle andhas calculated to reach point a at 2 pm . if he travels at 10 kmph , he will reach there at 12 pm if he travels at 15 kmph . at what speed musthe travel to reach a at 1 pm ? | "let the distance travelled by x km . then , x - x = 2 10 15 3 x - 2 x = 60 x = 60 km . time taken to travel 60 km at 10 km / hr = 60 hrs = 6 hrs . 10 so , robert started 6 hours before 2 p . m . i . e . , at 8 a . m . required speed = 60 kmph . = 12 kmph . 5 c" | a = 1 / 10
b = 1 / 15
c = a - b
d = 2 / c
e = 1 / 10
f = 1 / 15
g = e - f
h = 2 / g
i = h / 10
j = i - 1
k = d / j
|
a ) 724533811 , b ) 353654655 , c ) 545463251 , d ) 725117481 , e ) 325157481 | e | multiply(subtract(9999, const_4), 32519) | find the value of m 32519 x 9999 = m ? | "32519 x 9999 = 32519 x ( 10000 - 1 ) = 32519 x 10000 - 32519 x 1 = 325190000 - 32519 = 325157481 e" | a = 9999 - 4
b = a * 32519
|
a ) 11 , b ) 10 , c ) 9 , d ) 8 , e ) 7 | e | divide(252, add(multiply(8, const_3), multiply(6, const_2))) | sheila works 8 hours per day on monday , wednesday and friday , and 6 hours per day on tuesday and thursday . she does not work on saturday and sunday . she earns $ 252 per week . how much does she earn in dollars per hour ? | "let sheila earn x dollars per hour so , on monday , wednesday and friday , she earns 8 x each and , on tuesday and thursday , she earns 6 x each in total , over the week she should earn , 3 ( 8 x ) + 2 ( 6 x ) = 36 x she earns $ 252 per week 36 x = 252 x = 7 correct option : e" | a = 8 * 3
b = 6 * 2
c = a + b
d = 252 / c
|
a ) 88 m , b ) 60 m , c ) 80 m , d ) 65 m , e ) 75 m | a | add(multiply(4, divide(divide(66, 4), subtract(4, const_1))), 66) | a can run 4 times as fast as b and gives b a start of 66 m . how long should the race course be so that a and b might reach in the same time ? | speed of a : speed of b = 4 : 1 means in a race of 4 m a gains 3 m . then in a race of 66 m he gains 66 * ( 4 / 3 ) i . e 88 m answer : a | a = 66 / 4
b = 4 - 1
c = a / b
d = 4 * c
e = d + 66
|
a ) 9 , b ) 18 , c ) 17 , d ) none of these , e ) 19 | e | add(divide(subtract(105, 5), 5), const_1) | how many multiples of 5 are there between 5 to 105 ? | "explanation : as you know , multiples of 5 are integers having 0 or 5 in the digit to the extreme right ( i . e . the units place ) . so the numbers are 10 , 15 , 20 , 25 , 30 , 35 , 40 , 45 , 50 , 55 , 60 , 65 , 70 , 75 , 80 , 85 , 90 , 95 , 100 . answer : e" | a = 105 - 5
b = a / 5
c = b + 1
|
a ) 120 sec , b ) 360 sec , c ) 178 sec , d ) 187 sec , e ) 189 sec | b | divide(600, subtract(multiply(54, const_0_2778), multiply(30, const_0_2778))) | a and b go around a circular track of length 600 m on a cycle at speeds of 30 kmph and 54 kmph . after how much time will they meet for the first time at the starting point ? | "time taken to meet for the first time at the starting point = lcm { length of the track / speed of a , length of the track / speed of b } = lcm { 600 / ( 30 * 5 / 18 ) , 600 / ( 54 * 5 / 18 ) } = 360 sec . answer : b" | a = 54 * const_0_2778
b = 30 * const_0_2778
c = a - b
d = 600 / c
|
['a ) a ) 160 degree', 'b ) b ) 168 degree', 'c ) c ) 191 degree', 'd ) d ) 72 degree', 'e ) e ) 204 degree'] | d | multiply(divide(const_1, 3), multiply(divide(3, add(2, 3)), const_360)) | the ratio of male to female in a class is 2 : 3 . 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 of the males and one of the females in the class ? | let the common ratio be x . . so m = 2 x and f = 3 x and total = 5 x 1 of m = 2 x and 1 of f = 3 x . . total preferring that carrer = 5 x now 5 x = 360 , so x = 360 / 5 = 72 . . so x * 8 / 3 = 72 * 8 / 3 = 192 d | a = 1 / 3
b = 2 + 3
c = 3 / b
d = c * 360
e = a * d
|
a ) 228 , b ) 480 , c ) 512 , d ) 288 , e ) 276 | c | subtract(divide(multiply(800, const_100), add(25, const_100)), divide(multiply(divide(multiply(800, const_100), add(25, const_100)), 20), const_100)) | by selling an article at rs . 800 , a shopkeeper makes a profit of 25 % . at what price should he sell the article so as to make a loss of 20 % ? | sp = 800 profit = 25 % cp = ( sp ) * [ 100 / ( 100 + p ) ] = 800 * [ 100 / 125 ] = 640 loss = 25 % = 25 % of 640 = rs . 128 sp = cp - loss = 640 - 128 = rs . 512 answer : c | a = 800 * 100
b = 25 + 100
c = a / b
d = 800 * 100
e = 25 + 100
f = d / e
g = f * 20
h = g / 100
i = c - h
|
a ) 20 , b ) 200 , c ) 300 , d ) 400 , e ) 100 | e | subtract(divide(divide(9540, 26.50), const_2), multiply(const_2, 20)) | the length of a rectangular plot is 20 metres more than its breadth . if the cost of fencing the plot @ rs . 26.50 per metre is rs . 9540 , what is the length of the plot in metres ? | "let length of plot = l meters , then breadth = l - 20 meters and perimeter = 2 [ l + l - 20 ] = [ 4 l - 40 ] meters [ 4 l - 40 ] * 26.50 = 9540 [ 4 l - 40 ] = 9540 / 26.50 = 360 4 l = 400 l = 400 / 4 = 100 meters . answer : e" | a = 9540 / 26
b = a / 2
c = 2 * 20
d = b - c
|
a ) 12 , b ) 13 , c ) 14 , d ) 15 , e ) 16 | b | add(divide(subtract(36, 3), 3), const_2) | what is the greatest of 3 consecutive integers whose sum is 36 ? | "36 / 3 = 12 the three numbers are 11 , 12 , and 13 . the answer is b ." | a = 36 - 3
b = a / 3
c = b + 2
|
a ) 3 , b ) 11 , c ) 13 , d ) 23 , e ) 38 | a | divide(multiply(43, divide(16, const_100)), const_2) | each of the 43 points is placed either inside or on the surface of a perfect sphere . if 16 % or fewer of the points touch the surface , what is the maximum number of segments which , if connected from those points to form chords , could be the diameter of the sphere ? | "maximum number of points on the surface is 16 % * 43 = 6.88 . . . or 6 since it has to be an integer now note that if two points form a diameter , they can not be part of any other diameter . so in the best case we can pair up the points we have 6 points , so at best we can form 3 pairs ( 6 ) . so , answer is ( a )" | a = 16 / 100
b = 43 * a
c = b / 2
|
a ) 48 , b ) 59 , c ) 60 , d ) 56 , e ) 50 | a | add(add(add(12, 12), 12), 12) | xavier starts from p towards q at a speed of 60 kmph and after every 12 mins increases his speed by 10 kmph . if the distance between p and q is 60 km , then how much time does he take to cover the distance ? | "first 12 min = 60 * 12 / 60 = 12 km 2 nd 12 min = 70 * 12 / 60 = 14 km 3 rd 12 min = 80 * 12 / 60 = 16 km 4 th 12 min = 90 * 12 / 60 = 18 km total time 12.4 = 48 min a" | a = 12 + 12
b = a + 12
c = b + 12
|
a ) 28000 , b ) 30000 , c ) 42000 , d ) 34000 , e ) none of these | c | divide(6000, divide(multiply(const_1, const_1), add(multiply(const_3, const_2), multiply(const_1, const_1)))) | a & b started a partnership business . a ' s investment was thrice the investment of b and the period of his investment was two times the period of investments of b . if b received rs 6000 as profit , what is their total profit ? | explanation : suppose b ' s investment = x . then a ' s investment = 3 x suppose bs period of investment = y , then a ' s period of investment = 2 y a : b = 3 x * 2 y : xy = 6 : 1 total profit * 1 / 7 = 6000 = > total profit = 6000 * 7 = 42000 . answer : option c | a = 1 * 1
b = 3 * 2
c = 1 * 1
d = b + c
e = a / d
f = 6000 / e
|
a ) 21.5 , b ) 10.6 , c ) 8 , d ) 9.5 , e ) none of these | a | divide(subtract(multiply(5, 11), multiply(3, 4)), 2) | the average of 5 quantities is 11 . the average of 3 of them is 4 . what is the average of remaining 2 numbers ? | "answer : a ( 5 x 11 - 3 x 4 ) / 2 = 21.5" | a = 5 * 11
b = 3 * 4
c = a - b
d = c / 2
|
a ) 50 , b ) 288 , c ) 270 , d ) 129 , e ) 281 | a | divide(45, divide(45, const_100)) | 10 % of a number is added to 45 , the result is the same number . find the number ? | "( 10 / 100 ) * x + 45 = x 9 x = 450 x = 50 answer : a" | a = 45 / 100
b = 45 / a
|
a ) 10 , b ) 20 , c ) 15 , d ) 26 , e ) 18 | b | divide(80, 4) | if 4 x + y + z = 80 , 2 x - y - z = 4003 x + y - z = 20 for integers of x , y and z , find x = ? | 4 x + y + z = 80 - - - - - - - - - - 1 ) 2 x - y - z = 40 - - - - - - - - - - - - - 2 ) 3 x + y - z = 20 - - - - - - - - - - - - 3 ) from 1 ) and 2 ) 6 x = 120 x = 20 answer is b | a = 80 / 4
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a ) 1.29 hrs , b ) 2.29 hrs , c ) 4.29 hrs , d ) 5.29 hrs , e ) 3.29 hrs | c | add(divide(reminder(subtract(multiply(15, const_60), multiply(multiply(15, divide(5, 7)), const_60)), const_60), const_100), floor(divide(subtract(multiply(15, const_60), multiply(multiply(15, divide(5, 7)), const_60)), const_60))) | calculate how much time could be saved if the train would run at its usual speed , given that it ran at 5 / 7 of its own speed and got to its destination in 15 hours ? | "new speed = 5 / 7 of usual speed new time = 5 / 7 of usual time 5 / 7 of usual time = 15 hrs usual time = 15 * 5 / 7 = 10.71 hrs time saved = 15 - 10.71 = 4.29 hrs answer is c" | a = 15 * const_60
b = 5 / 7
c = 15 * b
d = c * const_60
e = a - d
f = reminder / (
g = f + 100
|
a ) 35.5 , b ) 37.5 , c ) 38.5 , d ) 40.5 , e ) none of these | c | divide(add(add(add(10, const_1), add(add(10, const_1), const_2)), add(subtract(7, 10), subtract(7, const_2))), 10) | find the average of first 10 multiples of 7 | "explanation : = 7 ( 1 + 2 + 3 + . . . + 10 ) 10 = 7 ( 10 ( 10 + 1 ) ) 10 × 2 = 7 ( 110 ) 10 × 2 = 38.5 answer : option c" | a = 10 + 1
b = 10 + 1
c = b + 2
d = a + c
e = 7 - 10
f = 7 - 2
g = e + f
h = d + g
i = h / 10
|
a ) 36 , b ) 86 , c ) 27 , d ) 86 , e ) 27 | a | add(multiply(3, divide(9, multiply(3, 5))), multiply(5, divide(9, multiply(3, 5)))) | two numbers are in the ratio 3 : 5 . if 9 be subtracted from each , they are in the ratio of 9 : 17 . the first number ? | "( 3 x - 9 ) : ( 5 x - 9 ) = 9 : 17 x = 12 = > 3 x = 36 answer : a" | a = 3 * 5
b = 9 / a
c = 3 * b
d = 3 * 5
e = 9 / d
f = 5 * e
g = c + f
|
a ) 16 % , b ) 16.66 % , c ) 18 % , d ) 21 % , e ) 33.33 % | e | multiply(divide(subtract(40, 30), 30), const_100) | john makes $ 30 a week from his job . he earns a raise and now makes $ 40 a week . what is the % increase ? | "increase = ( 10 / 30 ) * 100 = 33.33 % . e" | a = 40 - 30
b = a / 30
c = b * 100
|
a ) 1 : 2 , b ) 3 : 4 , c ) 2 : 5 , d ) 3 : 7 , e ) 1 : 3 | b | divide(550, multiply(divide(33, 45), const_1000)) | a truck covers a distance of 550 metres in 1 minute whereas a bus covers a distance of 33 kms in 45 minutes . the ratio of their speeds is ? | ratio of speeds = ( 550 / 60 ) 18 / 5 : 60 * 33 / 45 = 33 : 44 = 3 : 4 answer is b | a = 33 / 45
b = a * 1000
c = 550 / b
|
a ) 40 , b ) 60 , c ) 80 , d ) 100 , e ) 120 | c | divide(subtract(150, 122), subtract(const_1, divide(65, const_100))) | a particular library has 150 books in a special collection , all of which were in the library at the beginning of the month . these book are occasionally loaned out through an inter - library program . if , by the end of the month , 65 percent of books that were loaned out are returned and there are 122 books in the special collection at that time , how many books of the special collection were loaned out during that month ? | "the total number of books is 150 . let x be the number of books which were loaned out . 65 % of books that were loaned out are returned . 35 % of books that were loaned out are not returned . now , there are 122 books , thus the number of un - returned books is 150 - 122 = 28 books . 0.35 x = 28 x = 80 the answer is c ." | a = 150 - 122
b = 65 / 100
c = 1 - b
d = a / c
|
a ) 26 , b ) 43 , c ) 44 , d ) 45 , e ) 46 | a | add(subtract(48, multiply(12, 2)), 2) | a batsman in his 12 th innings makes a score of 48 and thereby increases his average by 2 runs . what is his average after the 12 th innings if he had never been ‘ not out ’ ? | "let ‘ x ’ be the average score after 12 th innings ⇒ 12 x = 11 × ( x – 2 ) + 48 ∴ x = 26 answer a" | a = 12 * 2
b = 48 - a
c = b + 2
|
a ) 4 , b ) 5 , c ) 3.5 , d ) 7 , e ) 8 | c | subtract(subtract(multiply(3, 7), add(subtract(13, 7), const_3.0)), 7) | the average of 1 st 3 of 4 numbers is 7 and of the last 3 are 5 . if the sum of the first and the last number is 13 . what is the last numbers ? | "a + b + c = 21 b + c + d = 15 a + d = 13 a – d = 6 a + d = 13 2 d = 7 d = 3.5 answer : c" | a = 3 * 7
b = 13 - 7
c = b + 3
d = a - c
e = d - 7
|
a ) 143.5 , b ) 154 , c ) 160 , d ) 148.75 , e ) 142.25 | a | subtract(divide(power(24, const_2), const_2), divide(power(17, const_2), const_2)) | the size of a flat - screen television is given as the length of the screen ’ s diagonal . how many square inches greater is the screen of a square 24 - inch flat - screen television than a square 17 - inch flat - screen television ? | "if we take a square with side length x and draw a diagonal , we get two isosceles right triangles . if we focus on one such right triangle , we see that the legs have length x . square 24 - inch flat - screen television the diagonal ( hypotenuse ) = 24 so , we can apply the pythagorean theorem to get x ² + x ² = 24 ² simplify : 2 x ² = 24 ² divide both sides by 2 to get : x ² = 24 ² / 2 since the area of the square = x ² , we can see that the area of this square is 24 ² / 2 square 17 - inch flat - screen television the diagonal ( hypotenuse ) = 17 so , we can apply the pythagorean theorem to get x ² + x ² = 17 ² simplify : 2 x ² = 17 ² divide both sides by 2 to get : x ² = 17 ² / 2 since the area of the square = x ² , we can see that the area of this square is 17 ² / 2 difference in areas = 24 ² / 2 - 17 ² / 2 = 288 - 144.5 i . e = 143.5 a" | a = 24 ** 2
b = a / 2
c = 17 ** 2
d = c / 2
e = b - d
|
a ) 76 kg , b ) 76.5 kg , c ) 85 kg , d ) 90 kg , e ) none | c | add(65, multiply(8, 2.5)) | the average weight of 8 persons increases by 2.5 kg when a new person comes in place of one of them weighing 65 kg . what might be the weight of the new person ? | "sol . total weight increased = ( 8 × 2.5 ) kg = 20 kg . weight of new person = ( 65 + 20 ) kg = 85 kg . answer c" | a = 8 * 2
b = 65 + a
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a ) 3 / 8 , b ) 1 / 2 , c ) 11 / 16 , d ) 5 / 8 , e ) 3 / 4 | d | divide(add(const_1, divide(subtract(multiply(divide(6, 8), const_2), const_1), divide(const_1, subtract(multiply(divide(3, 4), const_2), const_1)))), const_2) | a , b , and c are integers and a < b < c . s is the set of all integers from a to b , inclusive . q is the set of all integers from b to c , inclusive . the median of set s is ( 3 / 4 ) * b . the median of set q is ( 6 / 8 ) * c . if r is the set of all integers from a to c , inclusive , what fraction of c is the median of set r ? | "the answer isc : 11 / 16 . the key to this problem is remembering that the median for a consecutive set of numbers is equivalent to its mean . for example , the mean and median of a set consisting of x , x + 1 , x + 2 , . . . , y will always be ( x + y ) / 2 . for set s , consisting of numbers ( a , a + 1 , . . . , b ) , the median is given to be 3 / 4 * b : ( a + b ) / 2 = ( 3 / 4 ) * b a = b / 2 for set q , consisting of numbers ( b , b + 1 , . . . , c ) , the median is given to be 6 / 8 * c : ( b + c ) / 2 = ( 6 / 8 ) * c b = ( 1 / 2 ) * c for set r , consisting of numbers ( a , a + 1 , . . . c ) , the median needs to be found : a = b / 2 = ( 1 / 2 * c ) / 2 = ( 1 / 4 ) * c median = ( a + c ) / 2 = ( 1 / 4 * c + c ) / 2 = ( 5 / 4 ) * c / 2 = ( 5 / 8 ) * c ( answer d )" | a = 6 / 8
b = a * 2
c = b - 1
d = 3 / 4
e = d * 2
f = e - 1
g = 1 / f
h = c / g
i = 1 + h
j = i / 2
|
a ) 48 , b ) 52 , c ) 56 , d ) 60 , e ) 64 | c | add(subtract(divide(75, 5), const_2), add(divide(75, 3), divide(75, 5))) | line q has the equation 5 y – 3 x = 75 . if line s is perpendicular to q , has an integer for its y - intercept , and intersects q in the second quadrant , then how many possible line s ’ s exist ? ( note : intersections on one of the axes do not count . ) | "5 y - 3 x = 75 and so y = 3 x / 5 + 15 when x = 0 , then y = 15 . when y = 0 , then x = - 25 the slope is 3 / 5 , so the slope of line s is - 5 / 3 . through the point ( - 25 , 0 ) , 0 = - 5 ( - 25 ) / 3 + c the y - intercept is c = - 125 / 3 < - 41 . thus the perpendicular line s can have y - intercepts from - 41 up to 14 . the number of possible lines is 41 + 14 + 1 = 56 the answer is c ." | a = 75 / 5
b = a - 2
c = 75 / 3
d = 75 / 5
e = c + d
f = b + e
|
a ) $ 1100 , b ) $ 1240 , c ) $ 1360 , d ) $ 1480 , e ) $ 1550 | b | multiply(surface_rectangular_prism(3, 5, 2), 20) | a rectangular tank needs to be coated with insulation . the tank has dimensions of 3 feet , 5 feet , and 2 feet . each square foot of insulation costs $ 20 . how much will it cost to cover the surface of the tank with insulation ? | "the total surface area is 2 ( 2 * 3 + 3 * 5 + 2 * 5 ) = 62 square feet the total cost is 62 * $ 20 = $ 1240 the answer is b ." | a = surface_rectangular_prism * (
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['a ) 2', 'b ) 4', 'c ) 5', 'd ) 6', 'e ) 78'] | b | subtract(divide(divide(divide(528, 7), const_2), const_pi), divide(divide(divide(352, 7), const_2), const_pi)) | two concentric circles form a ring . the inner and outer circumference of the ring are 352 / 7 m and 528 / 7 m respectively . find the width of the ring . | explanation : let the inner and outer radii be r and r meters then , \ inline { \ color { black } 2 \ prod r } = 352 / 7 = > r = ( 352 / 7 ) * ( 7 / 22 ) * ( 1 / 2 ) = 8 m \ inline { \ color { black } 2 \ prod r } = 528 / 7 = > r = ( 528 / 7 ) * ( 7 / 22 ) * ( 1 / 2 ) = 12 m width of the ring = r - r = 12 - 8 = 4 m answer : b ) 4 | a = 528 / 7
b = a / 2
c = b / math.pi
d = 352 / 7
e = d / 2
f = e / math.pi
g = c - f
|
a ) 4 % , b ) 5 % , c ) 6 % , d ) 7 % , e ) 12 % | e | floor(multiply(const_100, divide(subtract(2564.36, 2300), 2300))) | a shopkeeper sold an article for rs 2564.36 . approximately what was his profit percent if the cost price of the article was rs 2300 | "explanation : gain % = ( 264.36 * 100 / 2300 ) = 11.49 % = 12 % approx option e" | a = 2564 - 36
b = a / 2300
c = 100 * b
d = math.floor(c)
|
a ) 50 , b ) 50.5 , c ) 56 , d ) 60 , e ) 63 | b | multiply(divide(add(3000, divide(multiply(6000, 0.5), const_100)), 6000), const_100) | in a recent election , geoff received 0.5 percent of the 6000 votes cast . to win the election , a candidate needed to receive more than x % of the vote . if geoff needed exactly 3000 more votes to win the election , what is the value of x ? | word problems are tricky in somehow more than other problem because you have the additional step to translate . breaking the problem : geoff how many votes he receives ? ? 30 votes he needs 3571 more votes so : 30 + 3000 = 3030 now what ' s the problem wants ? ? a x % . . . . . . . . 3030 is what % of total votes 6000 . . . . . . . . translating : 3030 = x / 100 * 6000 - - - x = 50.5 % . . . . . . . . . . b | a = 6000 * 0
b = a / 100
c = 3000 + b
d = c / 6000
e = d * 100
|
a ) 1 / 2 , b ) 5 / 8 , c ) 9 / 14 , d ) 16 / 21 , e ) 9 / 10 | c | divide(add(multiply(divide(24, const_2), add(const_2, const_3)), multiply(24, const_2)), multiply(24, add(const_3, const_4))) | a certain electric - company plan offers customers reduced rates for electricity used between 8 p . m . and 8 a . m . weekdays and 24 hours a day saturdays and sundays . under this plan , the reduced rates w apply to what fraction of a week ? | "number of hours between 8 pm to 8 am = 12 number of hours with reduced rates = ( 12 * 5 ) + ( 24 * 2 ) hours with reduced rates w / total number of hours in a week = ( 12 * 5 ) + ( 24 * 2 ) / ( 24 * 7 ) = 108 / ( 24 * 7 ) = 9 / 14 answer : c" | a = 24 / 2
b = 2 + 3
c = a * b
d = 24 * 2
e = c + d
f = 3 + 4
g = 24 * f
h = e / g
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a ) 42 , b ) 58 , c ) 44 , d ) 45 , e ) 46 | b | add(add(power(add(add(divide(subtract(subtract(115, const_10), const_2), const_4), const_2), const_2), const_2), power(add(add(add(divide(subtract(subtract(115, const_10), const_2), const_4), const_2), const_2), const_2), const_2)), add(power(divide(subtract(subtract(115, const_10), const_2), const_4), const_2), power(add(divide(subtract(subtract(115, const_10), const_2), const_4), const_2), const_2))) | the sum of two consecutive number is 115 . which is the larger number ? | "let consecutive number be x , x + 1 therefore sum of the consecutive number is x + x + 1 = 115 2 x + 1 = 115 2 x = 114 x = 57 therefore larger number is x + 1 = 58 answer : b" | a = 115 - 10
b = a - 2
c = b / 4
d = c + 2
e = d + 2
f = e ** 2
g = 115 - 10
h = g - 2
i = h / 4
j = i + 2
k = j + 2
l = k + 2
m = l ** 2
n = f + m
o = 115 - 10
p = o - 2
q = p / 4
r = q ** 2
s = 115 - 10
t = s - 2
u = t / 4
v = u + 2
w = v ** 2
x = r + w
y = n + x
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a ) 3 / 8 , b ) 2 / 8 , c ) 6 / 7 , d ) 3 / 5 , e ) 2 / 5 | a | multiply(divide(3, 30), divide(const_1, multiply(3, add(divide(const_1, 30), add(divide(const_1, 20), divide(const_1, 20)))))) | three pipes a , b and c can fill a tank from empty to full in 20 minutes , 20 minutes and 30 minutes respectively . when the tank is empty , all the three pipes are opened . a , b and c discharge chemical solutions p , q and r respectively . what is the proportion of solution p in the liquid in the tank after 3 minutes ? | "part filled by ( a + b + c ) in 3 minutes = 3 ( 1 / 20 + 1 / 20 + 1 / 30 ) = 6 / 15 part filled by a in 3 minutes = 3 / 20 required ratio = 3 / 20 * 15 / 6 = 3 / 8 answer : a" | a = 3 / 30
b = 1 / 30
c = 1 / 20
d = 1 / 20
e = c + d
f = b + e
g = 3 * f
h = 1 / g
i = a * h
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a ) s . 5000 , b ) s . 5100 , c ) s . 5800 , d ) s . 6000 , e ) s . 3600 | e | divide(36, multiply(divide(10, const_100), divide(10, const_100))) | if difference between compound interest and simple interest on a sum at 10 % p . a . for 2 years is rs . 36 then sum is | "p ( r / 100 ) ^ 2 = c . i - s . i p ( 10 / 100 ) ^ 2 = 36 3600 answer : e" | a = 10 / 100
b = 10 / 100
c = a * b
d = 36 / c
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a ) 148 , b ) 152 , c ) 156 , d ) 169 , e ) none | d | multiply(39, divide(152, divide(add(add(31, 39), const_2), const_2))) | the ratio of ducks and frogs in a pond is 31 : 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 , = 31 : 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 ) / 70 = 169 . answer : option d" | a = 31 + 39
b = a + 2
c = b / 2
d = 152 / c
e = 39 * d
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a ) 3 , b ) 4 , c ) 6 , d ) 7 , e ) 8 | a | divide(add(negate(17), sqrt(subtract(power(17, const_2), multiply(const_4, multiply(const_1, negate(60)))))), const_2) | x is a positive number which when increased by 17 is equal to 60 times the reciprocal of the number . find value of x ? | let the number be x . then , x + 17 = 60 x x 2 + 17 x - 60 = 0 ( x + 20 ) ( x - 3 ) = 0 x = 3 a | a = negate + (
b = 17 ** 2
c = 1 * negate
d = 4 * c
e = b - d
f = math.sqrt(e)
g = a / f
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a ) rs . 1200 , b ) rs . 1300 , c ) rs . 1500 , d ) rs . 3000 , e ) none of these | d | multiply(multiply(8, const_1000), divide(subtract(6, 5), subtract(8, 6))) | a man invests rs . 6,000 at the rate of 5 % per annum . how much more should he invest at the rate of 8 % , so that he can earn a total of 6 % per annum ? | "explanation : interest on rs . 6000 at 5 % per annum = ( 6000 × 5 × 1 ) / 100 = rs . 300 let his additional investment at 8 % = x interest on rs . x at 8 % per annum = ( x × 8 × 1 ) / 100 = 2 x / 25 . to earn 6 % per annum for the total , interest = ( 6000 + x ) × 6 × 1 / 100 . = > 300 + 2 x / 25 = ( 6000 + x ) × 6 × 1 / 100 . = > 30000 + 8 x = ( 6000 + x ) × 6 . = > 30000 + 8 x = 36000 + 6 x . = > 2 x = 6000 . = > x = 3000 . answer : d" | a = 8 * 1000
b = 6 - 5
c = 8 - 6
d = b / c
e = a * d
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a ) 715 , b ) 825 , c ) 330 , d ) 582 , e ) 517 | e | divide(multiply(2310, 47), 210) | the lcm of two numbers is 2310 and hcf is 47 . if one of the numbers is 210 . then what is the other number ? | first number * second number = lcm * hcf other number = 2310 * 47 / 210 = 11 * 47 = 517 answer : e | a = 2310 * 47
b = a / 210
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a ) 26 , b ) 37.5 , c ) 42 , d ) 19 , e ) 50 | d | multiply(subtract(multiply(add(const_1, divide(70, const_100)), subtract(const_1, divide(30, const_100))), const_1), const_100) | a retailer purchases shirts from a wholesaler and then sells the shirts in her store at a retail price that is 70 percent greater than the wholesale price . if the retailer decreases the retail price by 30 percent this will have the same effect as increasing the wholesale price by what percent ? | "answer : d = 19 . assume rs . 100 to be the price at which the retailer buys from wholesaler . 70 % increase makes retail price = 170 . now 30 % decrease - > ( 1 - 30 / 100 ) * 170 = 119 . now compared to the wholesale price of 100 , 19 % increase is what will have the same effect as increasing the wholesale price ." | a = 70 / 100
b = 1 + a
c = 30 / 100
d = 1 - c
e = b * d
f = e - 1
g = f * 100
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a ) 5 , b ) 2 , c ) 55 , d ) 95 , e ) 125 | b | divide(subtract(220, divide(220, add(divide(10, const_100), const_1))), 10) | machine p and machine q are each used to manufacture 220 sprockets . it takes machine p 10 hours longer to produce 220 sprockets than machine q . machine q produces 10 % more sprockets per hour than machine a . how many sprockets per hour does machine a produce ? | "p makes x sprockets per hour . then q makes 1.1 x sprockets per hour . 220 / x = 220 / 1.1 x + 10 1.1 ( 220 ) = 220 + 11 x 11 x = 22 x = 2 the answer is b ." | a = 10 / 100
b = a + 1
c = 220 / b
d = 220 - c
e = d / 10
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['a ) 3 √ 9', 'b ) 3 √ 8', 'c ) 3 √ 7', 'd ) 3 √ 5', 'e ) 3 √ 1'] | b | divide(triangle_area_three_edges(const_1, const_1, const_1), square_area(sqrt(divide(power(const_2, const_2), add(const_1, const_1))))) | the diagonal of a square is twice the side of equilateral triangle then the ratio of area of the triangle to the area of square is ? | let the side of equilateral triangle = 1 unit . we know that area of an equilateral triangle = 3 – √ 4 a 234 a 2 as side = 1 unit area of the equilateral triangle = 3 – √ 434 now diagonal of the square = 2 ( side of the equilateral triangle ) = 2 we know that area of the square = 12 d 212 d 2 where d = diagonal so area of the square = 12 ( 22 ) = 212 ( 22 ) = 2 ratio of the areas of equilateral triangle and square = 3 – √ 434 : 2 ⇒ ⇒ 3 √ 8 answer : b | a = triangle_area_three_edges / (
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a ) 14 , b ) 25 , c ) 35 , d ) 45 , e ) 55 | a | divide(subtract(multiply(90, 5), multiply(5, 20)), subtract(90, 65)) | the average mark of the students of a class in a particular exam is 65 . if 5 students whose average mark in that exam is 20 are excluded , the average mark of the remaining will be 90 . find the number of students who wrote the exam . | "let the number of students who wrote the exam be x . total marks of students = 75 x . total marks of ( x - 5 ) students = 90 ( x - 5 ) 65 x - ( 5 * 20 ) = 90 ( x - 5 ) 350 = 25 x = > x = 14 answer : a" | a = 90 * 5
b = 5 * 20
c = a - b
d = 90 - 65
e = c / d
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a ) 1250 , b ) 1800 , c ) 2024 , d ) 3256 , e ) 2864 | c | multiply(subtract(const_1, divide(3, 24)), 24) | there are 24 students in a class . in how many different ways can a committee of 3 students be formed ? | "24 c 3 = 24 * 23 * 22 / 6 = 2024 the answer is c ." | a = 3 / 24
b = 1 - a
c = b * 24
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a ) 5 , b ) 9 , c ) 25 , d ) 30 , e ) 45 | c | multiply(5, divide(35, add(2, 5))) | fred and sam are standing 35 miles apart and they start walking in a straight line toward each other at the same time . if fred walks at a constant speed of 2 miles per hour and sam walks at a constant speed of 5 miles per hour , how many miles has sam walked when they meet ? | relative distance = 35 miles relative speed = 2 + 5 = 7 miles per hour time taken = 35 / 7 = 5 hours distance travelled by sam = 5 * 5 = 25 miles = c | a = 2 + 5
b = 35 / a
c = 5 * b
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a ) 10.03 % , b ) 11.03 % , c ) 12.03 % , d ) 13.03 % , e ) 11 % | b | divide(multiply(subtract(72.5, add(11.5, add(multiply(2, 12.45), multiply(2, 14.05)))), const_100), 72.5) | a camera lens filter kit containing 5 filters sells for $ 72.50 . if the filters are purchased individually , 2 of them are priced at $ 12.45 each , 2 at $ 14.05 each , 1 at $ 11.50 . the amount saved by purchasing the kit is what percent of the total price of the 5 filters purchased individually ? | cost of kit = $ 72.50 if filters are purchased individually - $ 12.45 * 2 + $ 14.05 * 2 + $ 11.50 = $ 64.50 amount saved = $ 72.50 - $ 64.50 = $ 8 required % age = ( $ 8 / $ 72.50 ) * 100 = 11.03 % so , the correct answer is b . | a = 2 * 12
b = 2 * 14
c = a + b
d = 11 + 5
e = 72 - 5
f = e * 100
g = f / 72
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a ) 4 / 7 , b ) 1 / 3 , c ) 5 / 3 , d ) 12 / 7 , e ) 17 / 20 | a | divide(multiply(2, 2), add(3, multiply(2, 2))) | 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 1 / 2 of the juniors is equal to 2 / 3 of the seniors , what fraction of the students in the room are juniors ? | let total number of juniors = j total number of seniors = s ( 1 / 2 ) j = ( 2 / 3 ) s = > s = 3 / 4 j total number of students = j + s = ( 7 / 4 ) j fraction of the students in the room are juniors = j / ( j + s ) = j / [ ( 7 / 4 ) j ] = 4 / 7 answer a | a = 2 * 2
b = 2 * 2
c = 3 + b
d = a / c
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a ) 6 , b ) 8 , c ) 14 , d ) 20 , e ) 28 | a | subtract(40, add(10, 24)) | in a certain alphabet , 10 letters contain a dot and a straight line . 24 letters contain a straight line but do not contain a dot . if that alphabet has 40 letters , all of which contain either a dot or a straight line or both , how many letters contain a dot but do not contain a straight line ? | we are told that all of the letters contain either a dot or a straight line or both , which implies that there are no letters without a dot and a line ( no line / no dot box = 0 ) . first we find the total # of letters with lines : 10 + 24 = 34 ; next , we find the total # of letters without line : 40 - 34 = 6 ; finally , we find the # of letters that contain a dot but do not contain a straight line : 6 - 0 = 6 . a | a = 10 + 24
b = 40 - a
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a ) $ 77.40 , b ) $ 80.00 , c ) $ 83.64 , d ) $ 87.40 , e ) $ 88.00 | e | multiply(divide(add(10, const_100), add(15, const_100)), 92) | a certain manufacturer increased its gross profit on a product from 10 percent of the cost of the product to 15 percent of the cost by changing the selling price . if the new selling price was $ 92.00 and the cost of the product remained the same , what was the old selling price ? | given that { cost of the product } * 1.15 = $ 92 - - > { cost of the product } = $ 80 . the old price was $ 80 * 1.1 = $ 88 . answer : e . | a = 10 + 100
b = 15 + 100
c = a / b
d = c * 92
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a ) 50 % , b ) 53 % , c ) 54 % , d ) 55 % , e ) 47 % | e | add(multiply(60, divide(65, const_100)), multiply(subtract(const_100, 60), divide(20, const_100))) | in a certain city , 60 percent of the registered voters are democrats and the rest are republicans . in a mayoral race , if 65 percent of the registered voters who are democrats and 20 percent of the registered voters who are republicans are expected to vote for candidate a , what percent of the registered voters are expected to vote for candidate a ? | "say there are total of 100 registered voters in that city . thus 60 are democrats and 40 are republicans . 60 * 0.65 = 39 democrats are expected to vote for candidate a ; 40 * 0.20 = 8 republicans are expected to vote for candidate a . thus total of 39 + 8 = 47 registered voters are expected to vote for candidate a , which is 47 % of the total number of registered voters . answer : e ." | a = 65 / 100
b = 60 * a
c = 100 - 60
d = 20 / 100
e = c * d
f = b + e
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a ) 14439 , b ) 44398 , c ) 44400 , d ) 44402 , e ) 15005 | b | add(lcm(lcm(lcm(12, 30), lcm(48, 74)), 100), 2) | the smallest number when increased by ` ` 2 ` ` is exactly divisible by 12 , 30 , 48 , 74 and 100 is : | "lcm = 44400 44400 - 2 = 44398 answer : b" | a = math.lcm(12, 30)
b = math.lcm(48, 74)
c = math.lcm(a, b)
d = math.lcm(c, 100)
e = d + 2
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a ) 4 , b ) 5 , c ) 6 , d ) 7 , e ) 8 | e | divide(1400, lcm(25, 35)) | how many integers between 1 and 1400 are divisible by 25 , and 35 ? | lcm of the given numbers = 175 therefore , number of integers = 1400 / 175 = 8 answer is option e | a = math.lcm(25, 35)
b = 1400 / a
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['a ) 55', 'b ) 86', 'c ) 25', 'd ) 72', 'e ) 35'] | d | add(multiply(const_pi, 14), multiply(const_2, 14)) | a semicircular cubicle has a radius of 14 . what is the approximate perimeter of the cubicle ? | perimeter of a circle = 2 pi * r perimeter of a semicircle = pi * r + 2 r aprox perimiter = 3.14 * 14 + 2 * 14 = 71.96 approximately 72 answer d | a = math.pi * 14
b = 2 * 14
c = a + b
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a ) 22.5 days , b ) 21.5 days , c ) 23.5 days , d ) 24.5 days , e ) 25.5 days | a | inverse(add(inverse(multiply(const_3, divide(60, subtract(const_3, const_1)))), inverse(divide(60, subtract(const_3, const_1))))) | a works thrice as much as b . if a takes 60 days less than b to do a work then find the number of days it would take to complete the work if both work together ? | if a finishes the job in x days b finishes it in 3 x days 3 x = x + 60 thus x = 30 in one hour 1 / t = 1 / 30 + 1 / 90 = 4 / 90 where t is the number of days they finish the job together . t = 90 / 4 = 22.5 days answer : a | a = 3 - 1
b = 60 / a
c = 3 * b
d = 1/(c)
e = 3 - 1
f = 60 / e
g = 1/(f)
h = d + g
i = 1/(h)
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a ) 81 , b ) 100 , c ) 120 , d ) 135 , e ) 160 | c | divide(multiply(add(90, divide(multiply(90, 20), const_100)), const_100), multiply(multiply(const_3, const_3), 10)) | a retailer bought a machine at a wholesale price of $ 90 and later on sold it after a 10 % discount of the retail price . if the retailer made a profit equivalent to 20 % of the whole price , what is the retail price t of the machine ? | "since the wholesale price was $ 90 and the profit was 20 % of the wholesale price ( [ . 2 ] [ 90 ] = $ 18 ) , the retail price would have to be above $ 108 , but not that much greater than that . let ' s start by testing answer c : $ 120 . . . . if . . . . . retail price t = $ 120 10 % discount off = $ 120 - ( . 1 ) ( 120 ) = 120 - 12 = 108 20 % profit on wholesale price = 90 + ( . 2 ) ( 90 ) = 90 + 18 = 108 these two numbers match , so this must be the answer ! final answer : [ reveal ] spoiler : c" | a = 90 * 20
b = a / 100
c = 90 + b
d = c * 100
e = 3 * 3
f = e * 10
g = d / f
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a ) 80 , b ) 95 , c ) 105 , d ) 112 , e ) 210 | d | multiply(divide(128, add(subtract(divide(const_1, const_3), multiply(divide(const_1, const_3), divide(20, const_100))), multiply(subtract(const_1, divide(const_1, const_3)), divide(40, const_100)))), add(multiply(divide(const_1, const_3), divide(20, const_100)), subtract(subtract(const_1, divide(const_1, const_3)), multiply(subtract(const_1, divide(const_1, const_3)), divide(40, const_100))))) | in a certain company , a third of the workers do not have a retirement plan . 20 % of the workers who do not have a retirement plan are women , and 40 % of the workers who do have a retirement plan are men . if 128 of the workers of that company are men , how many of the workers are women ? | "set up equation : x = total number of workers 128 = 0,4 * 2 / 3 * x + 0,8 * 1 / 3 * x 128 = 16 / 30 x x = 240 240 - 128 = 112 answer d" | a = 1 / 3
b = 1 / 3
c = 20 / 100
d = b * c
e = a - d
f = 1 / 3
g = 1 - f
h = 40 / 100
i = g * h
j = e + i
k = 128 / j
l = 1 / 3
m = 20 / 100
n = l * m
o = 1 / 3
p = 1 - o
q = 1 / 3
r = 1 - q
s = 40 / 100
t = r * s
u = p - t
v = n + u
w = k * v
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a ) 520 , b ) 650.2 , c ) 685.5 , d ) 703.1 , e ) 712.2 | d | divide(divide(multiply(multiply(multiply(10, 10), 5), 90), const_100), divide(multiply(0.8, 80), const_100)) | calculate the number of urns of water needed to fill a tank of dimensions of 10 x 10 x 5 m to 90 % full , if the volume of the urn is 0.8 m 3 and can only be filled to 80 % . | volume of the tank that needed to be filled = 10 x 10 x 5 x 0.9 = 450 volume of the urn that can be filled = 0.8 x 0.8 = 0.64 no . of urns of water needed to fill the tank = 450 / 0.64 = 703.1 answer : d | a = 10 * 10
b = a * 5
c = b * 90
d = c / 100
e = 0 * 8
f = e / 100
g = d / f
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a ) 36 , b ) 39 , c ) 42 , d ) 44 , e ) none of the above | d | add(subtract(92, multiply(17, 3)), 3) | a batsman makes a score of 92 runs in the 17 th inning and thus increases his average by 3 . find his average after 17 th inning . | "let the average after 17 th inning = x . then , average after 16 th inning = ( x – 3 ) . ∴ 16 ( x – 3 ) + 92 = 17 x or x = ( 92 – 48 ) = 44 . answer d" | a = 17 * 3
b = 92 - a
c = b + 3
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a ) 15 mins , b ) 20 mins , c ) 25 mins , d ) 30 mins , e ) 35 mins | d | divide(multiply(multiply(40, const_3), const_2), add(const_4, const_4)) | a tank can be filled by two pipes a and b in 60 minutes and 40 minutes respectively . how many minutes will it take to fill the tank from empty state if b is used for the first half time and then a and b fill it together for the other half . | explanation : let the total time be x mins . part filled in first half means in x / 2 = 1 / 40 part filled in second half means in x / 2 = 1 / 60 + 1 / 40 = 1 / 24 total = x / 2 x 1 / 40 + x / 2 x 1 / 24 = 1 x = 30 mins answer is d | a = 40 * 3
b = a * 2
c = 4 + 4
d = b / c
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a ) 1595 , b ) 1569 , c ) 1265 , d ) 1555 , e ) 1565 | e | add(multiply(65, 24), 5) | in a division sum , the quotient is 65 , the divisor 24 and the remainder 5 , find the dividend ? | "explanation : 65 * 24 + 5 = 1565 answer : e" | a = 65 * 24
b = a + 5
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a ) s . 3096 , b ) s . 4076 , c ) s . 4085 , d ) s . 4096 , e ) s . 5096 | a | divide(3913, power(add(1, divide(add(6, divide(1, 4)), const_100)), 3)) | the principal that amounts to rs . 3913 in 3 years at 6 1 / 4 % per annum c . i . compounded annually , is ? | "principal = [ 4913 / ( 1 + 25 / ( 4 * 100 ) ) 3 ] = 3913 * 16 / 17 * 16 / 17 * 16 / 17 = rs . 3096 . answer : a" | a = 1 / 4
b = 6 + a
c = b / 100
d = 1 + c
e = d ** 3
f = 3913 / e
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a ) 50 , b ) 60 , c ) 70 , d ) 80 , e ) 90 | a | divide(240, add(add(divide(7, 5), divide(7, 5)), const_2)) | the length of a rectangular field is 7 / 5 its width . if the perimeter of the field is 240 meters , what is the width of the field ? | "let l be the length and w be the width . l = ( 7 / 5 ) w perimeter : 2 l + 2 w = 240 , 2 ( 7 / 5 ) w + 2 w = 240 solve the above equation to find : w = 50 m and l = 70 m . correct answer a ) 50" | a = 7 / 5
b = 7 / 5
c = a + b
d = c + 2
e = 240 / d
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a ) 58 , b ) 59 , c ) 62 , d ) 70 , e ) 80 | e | add(add(multiply(divide(60, 10), const_2), 10), add(divide(60, 10), 10)) | 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 10 years hence ? | "explanation : a + b = 60 , a = 2 b 2 b + b = 60 = > b = 20 then a = 40 . 10 years , their ages will be 50 and 30 . sum of their ages = 50 + 30 = 80 . e )" | a = 60 / 10
b = a * 2
c = b + 10
d = 60 / 10
e = d + 10
f = c + e
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a ) 6 , b ) 7 , c ) 8 , d ) 9 , e ) 10 | a | divide(add(8, 10), add(2, 1)) | at a certain committee meeting only associate professors and assistant professors are present . each associate professor has brought 2 pencils and 1 chart to the meeting , while each assistant professor has brought 1 pencil and 2 charts . if a total of 10 pencils and 8 charts have been brought to the meeting , how many people are present ? | "say there are ' a ' associate professors . so we have 2 a pencils and a charts . say there are ' b ' assistant professors . so we have b pencils and 2 b charts . total pencils are 10 so 2 a + b = 10 total charts are 11 so a + 2 b = 8 add both : 3 a + 3 b = 18 so a + b = 6 total number of people = 6 a" | a = 8 + 10
b = 2 + 1
c = a / b
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a ) 10 min . , b ) 41 min . , c ) 17 min . , d ) 23 min . , e ) 25 min . | d | divide(multiply(14, divide(3, add(2, 3))), multiply(divide(3, add(2, 3)), divide(3, add(2, 3)))) | it is the new year and mandy has made a resolution to lose weight this year . she plans to exercise and do yoga . for exercise she plans to workout at the gym and ride her bicycle in the ratio of 2 : 3 everyday . she will also do yoga in the ratio , yoga : exercise = 2 : 3 . if she rides her bike for 14 minutes , how much time will she spend doing yoga ? ( rounded to minutes ) | "the ratio is 2 : 3 = gym : ride , so ( 14 ) ( 3 / 2 ) = 21 minutes at the gym , and 21 + 14 = 35 minutes exercise , so ( 2 / 3 ) ( 35 ) = 23 minutes yoga . answer : d" | a = 2 + 3
b = 3 / a
c = 14 * b
d = 2 + 3
e = 3 / d
f = 2 + 3
g = 3 / f
h = e * g
i = c / h
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a ) $ 3.15 , b ) $ 4.45 , c ) $ 4.80 , d ) $ 5.05 , e ) $ 5.85 | e | add(multiply(divide(3.6, divide(2, 5)), 0.4), 2.25) | jim â € ™ s taxi service charges an initial fee of $ 2.25 at the beginning of a trip and an additional charge of $ 0.4 for each 2 / 5 of a mile traveled . what is the total charge for a trip of 3.6 miles ? | "let the fixed charge of jim â € ™ s taxi service = 2.25 $ and charge per 2 / 5 mile ( . 4 mile ) = . 4 $ total charge for a trip of 3.6 miles = 2.25 + ( 3.6 / . 4 ) * . 4 = 2.25 + 9 * . 4 = 5.85 $ answer e" | a = 2 / 5
b = 3 / 6
c = b * 0
d = c + 2
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a ) 0.33 , b ) 0.44 , c ) 0.55 , d ) 44 , e ) 55 | d | multiply(divide(divide(8, divide(const_1, const_2)), const_12), multiply(0.33, 100)) | in a market , a dozen eggs cost as much as a pound of rice , and a half - liter of kerosene costs as much as 8 eggs . if the cost of each pound of rice is $ 0.33 , then how many w cents does a liter of kerosene cost ? [ one dollar has 100 cents . ] | "main thing to remember is answer is asked in cents , however when we calculate , it comes up as 0.44 $ just multiply by 100 , answer w = 44 . d" | a = 1 / 2
b = 8 / a
c = b / 12
d = 0 * 33
e = c * d
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a ) 66.8 kmph , b ) 65.8 kmph , c ) 54.8 kmph , d ) 64.8 kmph , e ) 44.8 kmph | d | multiply(divide(360, subtract(25, 5)), const_3_6) | a train requires 5 seconds to pass a pole while it requires 25 seconds to cross a stationary train which is 360 mtrs long . find the speed of the train . | "in 5 s the train crosses the pole and in 25 sec the train crosses one more stationary train in 20 sec the train travels a distance of 360 mtrs speed = 360 / 20 = 18 m / s = 18 ( 3600 / 1000 ) = 18 * 18 / 5 = 64.8 kmph answer : d" | a = 25 - 5
b = 360 / a
c = b * const_3_6
|
a ) 2233 , b ) 2222 , c ) 2211 , d ) 6382 , e ) 23,17 | d | add(multiply(divide(const_100, const_2), const_10), power(18, const_3)) | a two digit number is 18 less than the square of the sum of its digits . how many such numbers are there ? | option 2 take n = 10 a + b . given that , ( 10 a + b ) + 18 = k 2 = ( a + b ) 2 given number = k 2 - 18 = ( 10 a + b ) that means , when we add 18 to the given number it should be a perfect square . so k 2 takes the following values . 1 , 4 , 9 , 16 , 25 , 36 , 49 , 64 , 81 , 100 , 121 , . . . . 1 to 16 are ruled out as if we subtract 18 from them , the resulting number is a single digit number . now 25 - 18 = 7 36 - 18 = 18 49 - 18 = 31 64 - 18 = 46 81 - 18 = 63 100 - 18 = 82 121 - 18 = 103 now 63 , 82 satisfies . answer : d | a = 100 / 2
b = a * 10
c = 18 ** 3
d = b + c
|
a ) 90 , b ) 95 , c ) 70 , d ) 85 , e ) 50 | e | divide(divide(multiply(add(10, 40,50), add(divide(subtract(40,50, 10), 20,30), const_1)), const_2), add(divide(subtract(40,50, 10), 10), const_1)) | what is the average ( arithmetic mean ) of 10 , 20,30 , 40,50 , 60,70 , 80,90 ? | "so addition of all term - 10 , 20 , 30 , . . . . . . . 90 so average = ( 10 + 20 + 30 + 40 + 50 + 60 + 70 + 80 + 90 ) / 9 = ( 450 ) / 9 = 50 hence , the correct answer is e ." | a = 10 + 40
b = 40 - 50
c = b / 20
d = c + 1
e = a * d
f = e / 2
g = 40 - 50
h = g / 10
i = h + 1
j = f / i
|
a ) 22 , b ) 20 , c ) 21 , d ) 23 , e ) 24 | b | multiply(6, const_2) | if the average of 6 x and 8 y is greater than 100 , and x is twice y , what is the least integer value of x ? | "substitution can be used in the following way : always start with the equation : x = 2 y . it is more straight forward to manage as compared to the inequality . substitute y = x / 2 , not the other way because you need to find the minimum value of x . so you can get rid of y . now go on to the inequality . so 8 y = 8 x / 2 = 4 x now average of 6 x and 4 x is greater than 100 . average of 6 x and 4 x is 5 x . so , 5 x > 100 x > 20 answer : b" | a = 6 * 2
|
a ) 7 : 9 , b ) 2 : 5 , c ) 3 : 5 , d ) 1 : 5 , e ) 4 : 5 | a | divide(power(343, const_0_33), power(729, const_0_33)) | two cubes of their volumes in the ratio 343 : 729 . the ratio of their surface area is : | "the ratio of their surface area is 343 : 729 7 : 9 answer is a ." | a = 343 ** const_0_33
b = 729 ** const_0_33
c = a / b
|
a ) 11 1 / 2 min , b ) 11 1 / 4 min , c ) 11 2 / 4 min , d ) 11 1 / 5 min , e ) 10 1 / 2 min | e | multiply(subtract(const_1, multiply(add(divide(const_1, 12), divide(const_1, 14)), 3)), 14) | two pipes p and q can fill a cistern in 12 and 14 minutes respectively . both are opened together , but at the end of 3 minutes the first is turned off . how much longer will the cistern take to fill ? | "3 / 12 + x / 14 = 1 x = 10 1 / 2 answer : e" | a = 1 / 12
b = 1 / 14
c = a + b
d = c * 3
e = 1 - d
f = e * 14
|
a ) 18 days , b ) 27 days , c ) 26.67 days , d ) 16 days , e ) 15 days | e | add(divide(subtract(const_1, multiply(divide(const_1, 20), 10)), add(divide(const_1, 20), divide(const_1, 20))), 10) | a can complete a project in 20 days and b can complete the same project in 20 days . if a and b start working on the project together and a quits 10 days before the project is completed , in how many days will the project be completed ? | "let x = the number of days taken to complete the project . the amount of work done by a is ( x - 10 ) * ( 1 / 20 ) . the amount of work done by b is ( x ) * ( 1 / 30 ) . ( 1 / 20 ) * ( x - 10 ) + ( 1 / 20 ) * ( x ) = 1 ( x / 20 ) + ( x / 20 ) - ( 10 / 20 ) = 1 x / 10 = 3 / 2 x = 15 therefore , the answer is e : 15 ." | a = 1 / 20
b = a * 10
c = 1 - b
d = 1 / 20
e = 1 / 20
f = d + e
g = c / f
h = g + 10
|
a ) 2 , b ) 3 , c ) 1 1 / 3 , d ) 4 , e ) 4 4 / 5 | c | add(5, divide(multiply(4, 5), add(4, 11))) | a work crew of 4 men takes 5 days to complete one - half of a job . if 11 men are then added to the crew and the men continue to work at the same rate , how many days will it take the enlarged crew to do the rest of the job ? | "suppose 1 man can do work in x days . . so 4 men will do in . . 4 / x = 1 / 5 * 1 / 2 as half job is done x = 40 now 11 more are added then 15 / 40 = 1 / 2 * 1 / d for remaining half job d = 1 1 / 3 number of days c" | a = 4 * 5
b = 4 + 11
c = a / b
d = 5 + c
|
a ) 67 kmph , b ) 50 kmph , c ) 55 kmph , d ) 60 kmph , e ) 70 kmph | a | subtract(divide(120, multiply(6, const_0_2778)), 5) | a train 120 meters long takes 6 seconds to cross a man walking at 5 kmph in the direction opposite to that of the train . find the speed of the train . | "explanation : let the speed of the train be x kmph . speed of the train relative to man = ( x + 5 ) kmph = ( x + 5 ) × 5 / 18 m / sec . therefore 120 / ( ( x + 5 ) × 5 / 18 ) = 6 < = > 30 ( x + 5 ) = 2160 < = > x = 67 speed of the train is 67 kmph . answer : option a" | a = 6 * const_0_2778
b = 120 / a
c = b - 5
|
a ) 182 , b ) 208 , c ) 220 , d ) 224 , e ) 302 | d | divide(195, divide(subtract(const_100, 13), const_100)) | a small company reduced its faculty by approximately 13 percent to 195 employees . what was the original number of employees ? | "if x is the original number of employees , then after 13 % reduction in employees number is . 87 x but we are given . 87 x = 195 x = 224 so the original number of employees is 224 correct answer - d" | a = 100 - 13
b = a / 100
c = 195 / b
|
a ) 26630 , b ) 29282 , c ) 36620 , d ) 26620 , e ) 26820 | b | multiply(22000, add(const_1, divide(multiply(3, 10), const_100))) | population is 22000 . population increases by 10 % every year , then the population after 3 years is ? | "population after 1 st year = 22000 * 10 / 100 = 2200 = = = > 22000 + 2200 = 24000 population after 2 nd year = 24000 * 10 / 100 = 2400 = = = > 24000 + 2400 = 26620 population after 3 rd year = 26620 * 10 / 100 = 2662 = = = > 26620 + 2662 = 29282 answer : b" | a = 3 * 10
b = a / 100
c = 1 + b
d = 22000 * c
|
a ) 54 kmph , b ) 108 kmph , c ) 176 kmph , d ) 134 kmph , e ) 161 kmph | a | multiply(divide(180, 12), const_3_6) | a 180 meter long train crosses a man standing on the platform in 12 sec . what is the speed of the train ? | "s = 180 / 12 * 18 / 5 = 54 kmph answer : a" | a = 180 / 12
b = a * const_3_6
|
a ) 2339 , b ) 2229 , c ) 4429 , d ) 5529 , e ) 2669 | b | add(divide(multiply(multiply(26.3, 12), 20), const_3), 125) | ( 26.3 * 12 * 20 ) / 3 + 125 = ? | b 2229 ? = [ ( 26.3 * 12 * 20 ) / 3 ] + 125 ? = 2104 + 125 = 2229 | a = 26 * 3
b = a * 20
c = b / 3
d = c + 125
|
a ) 35 , b ) 40 , c ) 25 , d ) 18 , e ) 20 | e | divide(multiply(250, 180), multiply(45, 50)) | rectangular tile each of size 45 cm by 50 cm must be laid horizontally on a rectangular floor of size 250 cm by 180 cm , such that the tiles do not overlap and they are placed with edges jutting against each other on all edges . a tile can be placed in any orientation so long as its edges are parallel to the edges of floor . no tile should overshoot any edge of the floor . the maximum number of tiles that can be accommodated on the floor is : | "area of tile = 45 * 50 = 2250 area of floor = 250 * 180 = 45000 no of tiles = 45000 / 2250 = 20 so , the no of tile = 20 answer : e" | a = 250 * 180
b = 45 * 50
c = a / b
|
a ) $ 1.00 , b ) $ 16 , c ) $ 5.00 , d ) $ 6.00 , e ) $ 20.00 | b | subtract(30.00, multiply(subtract(15, 1.00), 1.00)) | if 15 people contributed a total of $ 30.00 toward a gift and each of them contributed at least $ 1.00 , then the maximum possible amount any one person could have contributed is | "b for me 14 people with 1 $ each - > maximum = 16" | a = 15 - 1
b = a * 1
c = 30 - 0
|
['a ) 30 cm', 'b ) 10 cm', 'c ) 50 cm', 'd ) 90 cm', 'e ) 70 cm'] | b | divide(const_1000, 100) | a water tank , having the shape of a rectangular prism of base 100 square centimeters , is being filled at the rate of 1 liter per minute . find the rate at which the height of the water in the water tank increases . express your answer in centimeters per minute . | the volume of a rectangular prism is given by area of base ? height the rate at which the tank is being filled is 1 liter per minute which we need to convert 1 liter = 1 dm 3 but 1 dm = 10 cm . hence 1 liter = ( 10 cm ) 3 = 1000 cm 3 the height h of the water is related to the volume by volume = area of base * h in one minute the volume increases by 1 liter or 1000 cm 3 and the height increases by h = volume / area of base = 1000 cm 3 / 100 cm 2 = 10 cm correct answer b | a = 1000 / 100
|
a ) 4 , b ) 5 , c ) 6 , d ) 8 , e ) 9 | d | divide(72, subtract(const_10, const_1)) | the difference between a two - digit number and the number obtained by interchanging the positions of its digits is 72 . what is the difference between the two digits of that number ? | "sol . let the ten ’ s digit be x and unit ’ s digit be y , then , ( 10 x + y ) - ( 10 y + x ) = 72 ⇔ 9 ( x - y ) = 72 ⇔ x - y = 8 answer d" | a = 10 - 1
b = 72 / a
|
a ) 4 hrs , b ) 6 hrs , c ) 8 hrs , d ) 10 hrs , e ) 16 hrs | c | multiply(subtract(const_1, divide(6, 8)), inverse(multiply(inverse(8), inverse(4)))) | machine a can put caps on a set of bottles in 8 hours at a constant rate . if machine a does the job for 6 hours and machine b does the rest of the job , which works at 1 / 4 constant rate of machine a . how long will it take for machine b alone to do the rest of the job ? | machine a will do 6 / 8 in 6 hrs , so a does 3 / 4 of the work . . therefore , b will do the remaining 1 / 4 th work alone . . as the speed of b is 1 / 4 rate of a , b will do the 1 / 4 th work in same time that a takes to complete full job . . . ans 8 c | a = 6 / 8
b = 1 - a
c = 1/(8)
d = 1/(4)
e = c * d
f = 1/(e)
g = b * f
|
a ) 1 / 8 , b ) 1 / 16 , c ) 1 / 15 , d ) 1 / 32 , e ) 1 / 31 | b | divide(subtract(16, 15), 16) | a ’ s speed is 16 / 15 times that of b . if a and b run a race , what part of the length of the race should a give b as a head start , so that the race ends in a dead heat ? | "let x be the fraction of the distance that b runs . let v be the speed at which b runs . the time should be the same for both runners . time = d / ( 16 v / 15 ) = xd / v ( 15 / 16 ) * d / v = x * d / v x = 15 / 16 b should have a head start of 1 / 16 of the full distance . the answer is b ." | a = 16 - 15
b = a / 16
|
a ) 3 , b ) 4 , c ) 5 , d ) 6 , e ) 7 | a | divide(30, multiply(multiply(divide(divide(divide(75, 6), 8), 5), 8), 4)) | 6 women can do 75 unit of work in 8 days by working 5 hr / day in how many days 4 women do 30 units of work by working 8 hr / day | ( 6 x 8 x 5 ) / 75 = ( ax 4 x 8 ) / 30 a = 3 answer : a | a = 75 / 6
b = a / 8
c = b / 5
d = c * 8
e = d * 4
f = 30 / e
|
a ) 640 , b ) 278 , c ) 800 , d ) 267 , e ) 121 | a | divide(multiply(add(divide(multiply(650, 20), const_100), 190), const_100), 50) | 50 % of a number is more than 20 % of 650 by 190 . find the number ? | "( 50 / 100 ) * x â € “ ( 20 / 100 ) * 650 = 190 1 / 2 x = 320 x = 640 answer : a" | a = 650 * 20
b = a / 100
c = b + 190
d = c * 100
e = d / 50
|
a ) 0 , b ) 3 , c ) 8 , d ) 6 , e ) 4 | c | add(add(const_4, const_3), const_2) | what is the units digit of the expression 14 ^ 7 − 16 ^ 4 ? | i think answer on this one should be c too . since we know that 14 ^ 7 > 16 ^ 4 , as will said one should always check if the number is positive . | a = 4 + 3
b = a + 2
|
a ) a ) 1780 , b ) b ) 1250 , c ) c ) 10780 , d ) d ) 10500 , e ) e ) 8600 | a | subtract(add(8280, 18500), 25000) | when xyz co discontinued an agency arrangement with john , management held his commission of rs . 25000 / - for one month . earlier john had availed an advance agency fees of 8280 / - from company . but robert forgot that . after one month john asked his fees and accountant gives rs . 18500 / - to him . what is the incentive amount given to john ? | total fees = rs . 25000 / - advance fees = 8280 / - balance fees = 25000 - 8280 = 16720 paid amount = 18500 / - incentive amount = 18500 - 16720 = 1780 / - answer is a | a = 8280 + 18500
b = a - 25000
|
a ) 2 , b ) - 2 , c ) 6 , d ) - 3 , e ) 5 | a | subtract(subtract(subtract(150, 5), add(150, 5)), 5) | if | 5 x - 5 | = 150 , what is the sum of all the possible values of x ? | "there will be two cases 5 x - 5 = 150 or 5 x - 5 = - 150 = > x = 31 or x = - 29 sum of both the values will be - 29 + 31 = 2 answer is a" | a = 150 - 5
b = 150 + 5
c = a - b
d = c - 5
|
a ) 5 , b ) 6 , c ) 8 , d ) 10 , e ) 12 | e | add(divide(100, const_10), 1) | x is the product of each integer from 1 to 100 , inclusive and y = 100 ^ k , where k is an integer . what is the greatest value of k for which y is a factor of x ? | "the number of trailing zeros in the decimal representation of n ! , the factorial of a non - negative integer n , can be determined with this formula : n 5 + n 52 + n 53 + . . . + n 5 k , where k must be chosen such that 5 k ≤ n x = 1 * 2 * 3 . . . . * 100 = 100 ! no . of trailing zeros in 100 ! = 100 / 5 + 100 / 5 ^ 2 = 20 + 4 = 24 100 ^ k = 10 ^ 2 k → k = 24 / 2 = 12 e" | a = 100 / 10
b = a + 1
|
a ) 1 / 49 , b ) 4 / 7 , c ) 3 / 7 , d ) 16 / 49 , e ) 40 / 49 | b | subtract(const_1, sqrt(divide(9, 49))) | jean drew a gumball at random from a jar of pink and blue gumballs . since the gumball she selected was blue and she wanted a pink one , she replaced it and drew another . the second gumball also happened to be blue and she replaced it as well . if the probability of her drawing the two blue gumballs was 9 / 49 , what is the probability that the next one she draws will be pink ? | the probability of drawing a pink gumball both times is the same . the probability that she drew two blue gumballs = 9 / 49 = ( 3 / 7 ) * ( 3 / 7 ) therefore probability that the next one she draws is pink = 4 / 7 option ( b ) | a = 9 / 49
b = math.sqrt(a)
c = 1 - b
|
a ) rs . 750 , b ) rs . 700 , c ) rs . 820 , d ) rs . 940 , e ) rs . 900 | a | multiply(divide(divide(60, 2), 4), const_100) | a sum was put a simple interest at a certain rate for 2 years . had it been put at 4 % higher rate , it would have fetched rs . 60 more . the sum is : | "explanation : let the sub be rs . x and the initial rate be r % . then x × ( r + 4 ) × 2 / 100 − x × r × 2 / 100 = 60 ⇒ x × 4 × 2 / 100 = 60 ⇒ x × 2 / 100 = 15 ⇒ 2 x = 1500 ⇒ x = 750 answer : option a" | a = 60 / 2
b = a / 4
c = b * 100
|
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