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 ) 30 , b ) 40 , c ) 10 , d ) 20 , e ) 50 | b | divide(add(subtract(divide(rectangle_area(const_360, const_1000), const_10), multiply(const_1000, multiply(const_3, const_2))), add(multiply(const_3, const_1000), multiply(15, const_10))), divide(add(subtract(divide(rectangle_area(const_360, const_1000), const_10), multiply(const_1000, multiply(const_3, const_2))), add(multiply(const_3, const_1000), multiply(15, const_10))), const_10)) | a rectangular block 6 cm by 12 cm by 15 cm is cut up into an exact number of equal cubes . find the least possible number of cubes . | "explanation : volume of the block = ( 6 x 12 x 15 ) cm 3 = 1080 cm 3 side of the largest cube = h . c . f . of 6 cm , 12 cm , 15 cm = 3 cm . volume of this cube = ( 3 x 3 x 3 ) cm 3 = 27 cm 3 number of cubes = ( 1080 / 27 ) = 40 . answer : b" | a = rectangle_area / (
b = a - 10
c = 3 * 2
d = 1000 * c
e = b + d
f = 3 * 1000
g = 15 * 10
h = f + g
i = e / h
|
a ) 40 sec , b ) 60 sec , c ) 45 sec , d ) 48 sec , e ) 50 sec | b | divide(360, multiply(subtract(45, 390), const_0_2778)) | a train 360 m long is running at a speed of 45 km / hr . in what time will it pass a bridge 390 m long ? | "speed = 45 * 5 / 18 = 25 / 2 m / sec total distance covered = 360 + 140 = 750 m required time = 750 * 2 / 25 = 40 sec answer : b" | a = 45 - 390
b = a * const_0_2778
c = 360 / b
|
a ) 300 , b ) 150 , c ) 100 , d ) 80 , e ) 50 | a | divide(subtract(multiply(200, divide(16, const_100)), 20), subtract(divide(16, const_100), divide(12, const_100))) | an empty fuel tank with a capacity of 200 gallons was filled partially with fuel a and then to capacity with fuel b . fuel a contains 12 % ethanol by volume and fuel b contains 16 % ethanol by volume . if the full fuel tank contains 20 gallons of ethanol , how many gallons of fuel a were added ? | "say there are a gallons of fuel a in the tank , then there would be 200 - a gallons of fuel b . the amount of ethanol in a gallons of fuel a is 0.12 a ; the amount of ethanol in 200 - a gallons of fuel b is 0.16 ( 200 - a ) ; since the total amount of ethanol is 20 gallons then 0.12 a + 0.16 ( 200 - a ) = 20 - - > a = 300 . answer : a ." | a = 16 / 100
b = 200 * a
c = b - 20
d = 16 / 100
e = 12 / 100
f = d - e
g = c / f
|
['a ) 40', 'b ) 45', 'c ) 50', 'd ) 55', 'e ) 60'] | c | subtract(80, divide(subtract(multiply(30, 80), 2100), subtract(30, 20))) | the circus sells two kinds of tickets : lower seats for $ 30 and upper seats for $ 20 . on a certain night , the circus sells 80 tickets and gets $ 2100 in revenue from the sales . how many tickets for lower seats did they sell ? | let l be the number of lower seat tickets . let u be the number of upper seat tickets . l + u = 80 and u = 80 - l . 30 l + 20 u = 2100 . 30 l + 20 ( 80 - l ) = 2100 10 l + 1600 = 2100 . 10 l = 500 . l = 50 . the answer is c . | a = 30 * 80
b = a - 2100
c = 30 - 20
d = b / c
e = 80 - d
|
a ) 76 kg , b ) 77 kg , c ) 76.5 kg , d ) data inadequate , e ) none of these | b | add(65, multiply(8, 1.5)) | the average weight of 8 persons increases by 1.5 kg . if a person weighing 65 kg is replaced by a new person , what could be the weight of the new person ? | "total weight increases = 8 × 1.5 = 12 kg so the weight of new person = 65 + 12 = 77 kg answer b" | a = 8 * 1
b = 65 + a
|
a ) 18 , b ) 27 , c ) 28 , d ) 12 , e ) 25 | a | divide(150, multiply(add(26, 4), const_0_2778)) | the speed at which a man can row a boat in still water is 26 kmph . if he rows downstream , where the speed of current is 4 kmph , what time will he take to cover 150 metres ? | speed of the boat downstream = 26 + 4 = 30 kmph = 30 * 5 / 18 = 8.33 m / s hence time taken to cover 150 m = 150 / 8.33 = 18 seconds . answer : a | a = 26 + 4
b = a * const_0_2778
c = 150 / b
|
a ) 73.41 , b ) 74.31 , c ) 72.43 , d ) 73.43 , e ) 73.82 | e | divide(subtract(multiply(35, 72), subtract(99, 35)), 35) | a mathematics teacher tabulated the marks secured by 35 students of 8 th class . the average of their marks was 72 . if the marks secured by reema was written as 35 instead of 99 then find the correct average marks up to two decimal places . | "total marks = 35 x 72 = 2520 corrected total marks = 2520 - 35 + 99 = 2584 correct average = 2584 / 35 = 73.82 answer : e" | a = 35 * 72
b = 99 - 35
c = a - b
d = c / 35
|
a ) $ 220 , b ) $ 240 , c ) $ 260 , d ) $ 300 , e ) $ 320 | c | divide(572, add(divide(120, const_100), const_1)) | two employees m and n are paid a total of $ 572 per week by their employer . if m is paid 120 percent of the salary paid to n , how much is n paid per week ? | "1.2 n + n = 572 2.2 n = 572 n = 260 the answer is c ." | a = 120 / 100
b = a + 1
c = 572 / b
|
a ) s . 300 , b ) s . 360 , c ) s . 389 , d ) s . 368 , e ) s . 372 | e | divide(527, add(add(multiply(divide(2, 3), divide(1, 4)), divide(1, 4)), const_1)) | if rs . 527 be divided among a , b , c in such a way that a gets 2 / 3 of what b gets and b gets 1 / 4 of what c gets , then their shares are respectively ? | ( a = 2 / 3 b and b = 1 / 4 c ) = a / b = 2 / 3 and b / c = 1 / 4 a : b = 2 : 3 and b : c = 1 : 4 = 3 : 12 a : b : c = 2 : 3 : 12 a ; s share = 527 * 2 / 17 = rs . 62 b ' s share = 527 * 3 / 17 = rs . 93 c ' s share = 527 * 12 / 17 = rs . 372 . answer : e | a = 2 / 3
b = 1 / 4
c = a * b
d = 1 / 4
e = c + d
f = e + 1
g = 527 / f
|
a ) 1 % , b ) 1.1 % , c ) 8.18 % , d ) 10 % , e ) 10.8 % | c | divide(multiply(9, const_100), add(9, const_100)) | the annual interest rate earned by an investment increased by 10 percent from last year to this year . if the annual interest rate earned by the investment this year was 9 percent , what was the annual interest rate last year ? | "9 = 1.1 * x x = 8.18 % answer c )" | a = 9 * 100
b = 9 + 100
c = a / b
|
a ) 1 / 68 , b ) 62 / 89 , c ) 12 / 68 , d ) 11 / 84 , e ) 5 / 6 | e | divide(divide(subtract(8.75, 7.50), subtract(8.75, 6)), subtract(const_1, divide(subtract(8.75, 7.50), subtract(8.75, 6)))) | in what ratio should a variety of rice costing rs . 6 per kg be mixed with another variety of rice costing rs . 8.75 per kg to obtain a mixture costing rs . 7.50 per kg ? | "explanation : let us say the ratio of the quantities of cheaper and dearer varieties = x : y by the rule of allegation , x / y = ( 87.5 - 7.50 ) / ( 7.50 - 6 ) = 5 / 6 answer : e" | a = 8 - 75
b = 8 - 75
c = a / b
d = 8 - 75
e = 8 - 75
f = d / e
g = 1 - f
h = c / g
|
a ) 35 , b ) 36 , c ) 40 , d ) 42 , e ) 45 | c | divide(60, add(divide(30, 60), divide(subtract(60, 30), 30))) | a driver goes on a trip of 60 kilometers , the first 30 kilometers at 60 kilometers per hour and the remaining distance at 30 kilometers per hour . what is the average speed of the entire trip in kilometers per hour ? | "the time for the first part of the trip was 30 / 60 = 1 / 2 hours . the time for the second part of the trip was 30 / 30 = 1 hour . the total time for the trip was 3 / 2 hours . the average speed for the trip was 60 / ( 3 / 2 ) = 40 kph the answer is c ." | a = 30 / 60
b = 60 - 30
c = b / 30
d = a + c
e = 60 / d
|
a ) 2 , b ) 3 , c ) 4 , d ) 5 , e ) 6 | b | subtract(multiply(3, 7), 10) | a certain meter records voltage between 0 and 10 volts inclusive . if the average value of 3 recordings on the meter was 7 volts , what was the smallest possible recording in volts ? | "if average of 3 is 7 so sum of 3 should be 21 3 recording can be from 0 - 10 inclusive to find one smallest other two should be highest so , lets assume three var are a , b , c say a is smallest and give b and c greatest readings say 9 and 9 so a has to be 3 b" | a = 3 * 7
b = a - 10
|
a ) 13 √ 4 , b ) 13 √ 2 , c ) 23 √ 2 , d ) 13.9 √ 2 , e ) 13 √ 9 | d | sqrt(multiply(add(power(divide(48, const_4), const_2), power(divide(28, const_4), const_2)), const_2)) | the perimeter of one square is 48 cm and that of another is 28 cm . find the perimeter and the diagonal of a square which is equal in area to these two combined ? | "4 a = 48 4 a = 28 a = 12 a = 7 a 2 = 144 a 2 = 49 combined area = a 2 = 193 = > a = 13.9 d = 13.9 √ 2 answer : d" | a = 48 / 4
b = a ** 2
c = 28 / 4
d = c ** 2
e = b + d
f = e * 2
g = math.sqrt(f)
|
a ) 15.8 sec . , b ) 12.8 sec . , c ) 11.8 sec . , d ) 10.8 sec . , e ) 08.8 sec . | d | divide(add(140, 160), multiply(add(60, 40), const_0_2778)) | two bullet train s 140 m and 160 m long run at the speed of 60 km / hr and 40 km / hr respectively in opposite directions on parallel tracks . the time ( in seconds ) which they take to cross each other , is : | "d 10.8 sec . relative speed = ( 60 + 40 ) km / hr = 100 x 5 / 18 = 250 / 9 m / sec . distance covered in crossing each other = ( 140 + 160 ) m = 300 m . required time = 300 x 9 / 250 = 54 / 5 = 10.8 sec ." | a = 140 + 160
b = 60 + 40
c = b * const_0_2778
d = a / c
|
a ) 13 , b ) 10 , c ) 11 , d ) 12 , e ) 14 | b | divide(add(18, subtract(power(18, const_2), multiply(80, const_4))), const_2) | 18 times a positive integer is more than its square by 80 , then the positive integer is | "explanation : let the number be x . then , 18 x = x 2 + 80 = > x 2 - 18 x + 80 = 0 = > ( x - 10 ) ( x - 8 ) = 0 = > x = 10 or 8 answer : b" | a = 18 ** 2
b = 80 * 4
c = a - b
d = 18 + c
e = d / 2
|
a ) - 16 , b ) - 12 , c ) - 8 , d ) - 3 , e ) 11 | d | add(3, 15) | if | x + 3 | = 15 , what is the sum of all the possible values of x ? | "there will be two cases x + 3 = 15 or x + 3 = - 15 = > x = 12 or x = - 18 sum of both the values will be - 18 + 15 = - 3 answer : d" | a = 3 + 15
|
['a ) 46', 'b ) 64', 'c ) 164', 'd ) 146', 'e ) 56'] | c | add(add(multiply(4, add(add(12, 12), const_4)), multiply(add(const_4, const_2), add(5, 3))), 4) | a box has 12 shapes ; 3 circles , 5 squares , and 4 triangles . how many groups of 3 shapes can be made with at least 1 triangle ? | the problem asks for a combination , since order does n ' t matter . now , selecting r items from a set of n gives the combination formula : ncr = n ! / r ! ( n - r ) ! n = 12 r = 3 so , total groups is 12 c 3 = 12 ! / ( 3 ! ( 12 - 3 ) ! ) = 220 , and n = 12 - 4 = 8 r = 3 for groups without a triangle is 8 c 3 = 8 ! / ( 3 ! ( 8 - 3 ) ! ) = 56 , so , groups with at least 1 triangle = 220 - 56 = 164 answer : c | a = 12 + 12
b = a + 4
c = 4 * b
d = 4 + 2
e = 5 + 3
f = d * e
g = c + f
h = g + 4
|
a ) 35 , b ) 22.5 , c ) 27 , d ) 16 2 / 3 , e ) 15 | b | subtract(50, multiply(divide(50, const_100), 5)) | how many liters of water must be evaporated from 50 liters of a 3 - percent sugar solution to get a 5 - percent solution ? | "how many liters of water must be evaporated from 50 liters of a 3 - percent sugar solution to get a 5 - percent solution ? 3 % of a 50 liter solution is 1.5 l . so you are trying to determine how many liters must a solution be for the 1.5 l to represent 5 % of the solution . set up an inequality and solve for x : 1.5 / x = 1 / 5 x = 7.5 since you need a 15 l solution , you must evaporate 22.5 of the original 50 l solution to get a 5 % solution . answer is b ." | a = 50 / 100
b = a * 5
c = 50 - b
|
a ) a ) 36 , b ) b ) 30 , c ) c ) 40 , d ) d ) 42 , e ) e ) 44 | b | subtract(540, multiply(85, 6)) | we bought 85 hats at the store . blue hats cost $ 6 and green hats cost $ 7 . the total price was $ 540 . how many green hats did we buy ? | "let b be the number of blue hats and let g be the number of green hats . b + g = 85 . b = 85 - g . 6 b + 7 g = 540 . 6 ( 85 - g ) + 7 g = 540 . 510 - 6 g + 7 g = 540 . g = 540 - 510 = 30 . the answer is b ." | a = 85 * 6
b = 540 - a
|
a ) 185 , b ) 160 , c ) 170 , d ) 145 , e ) 225 | c | add(multiply(divide(10, 20), divide(80, 40)), 10) | if â € œ * â € is called â € œ + â € , â € œ / â € is called â € œ * â € , â € œ - â € is called â € œ / â € , â € œ + â € is called â € œ - â € . 80 / 40 â € “ 10 * 20 + 10 = ? | "explanation : given : 80 / 40 â € “ 10 * 20 + 10 = ? substituting the coded symbols for mathematical operations , we get , 80 * 40 / 20 + 20 â € “ 10 = ? 80 * 2 + 20 â € “ 10 = ? 160 + 20 â € “ 10 = ? 180 â € “ 10 = 170 answer : c" | a = 10 / 20
b = 80 / 40
c = a * b
d = c + 10
|
a ) rs . 15250 , b ) rs . 13375 , c ) rs . 16750 , d ) rs . 18000 , e ) none of these | d | multiply(66000, inverse(add(add(divide(2, 3), multiply(divide(2, 3), 3)), const_1))) | a , b and c enter into a partnership . a invests 3 times as much as b invests and 2 / 3 of what c invests . at the end of the year , the profit earned is rs . 66000 . what is the share of c ? | "explanation : let the investment of c be rs . x . the inverstment of b = rs . ( 2 x / 3 ) the inverstment of a = rs . ( 3 × ( 2 / 3 ) x ) = rs . ( 2 x ) ratio of capitals of a , b and c = 2 x : 2 x / 3 : x = 6 : 2 : 3 c ' s share = rs . [ ( 3 / 11 ) × 66000 ] = rs . 18000 answer : option d" | a = 2 / 3
b = 2 / 3
c = b * 3
d = a + c
e = d + 1
f = 1/(e)
g = 66000 * f
|
a ) 400 , b ) 475 , c ) 550 , d ) 625 , e ) 600 | e | multiply(divide(600, 11), 8) | a factory has three types of machines , each of which works at its own constant rate . if 7 machine as and 11 machine bs can produce 525 widgets per hour , and if 8 machine as and 22 machine cs can produce 600 widgets per hour , how many widgets could one machine a , one machine b , and one machine c produce in one 8 - hour day ? | "let machine a produce a widgets per hour . b produce b widgets per hour and c produce c widgets per hour . 7 a + 11 b = 525 - - - ( 1 ) 8 a + 22 c = 600 - - - ( 2 ) dividing ( 2 ) by 2 4 a + 11 c = 300 . . . . . ( 3 ) adding ( 1 ) ( 3 ) 11 a + 11 b + 11 c = 825 a + b + c = 75 per hour so for eight hrs = 75 * 8 = 600 = answer = e" | a = 600 / 11
b = a * 8
|
a ) 18 , b ) 22 , c ) 28 , d ) 26 , e ) 98 | a | divide(multiply(60, 70), const_100) | find number which is 70 % less than 60 . | "explanation : 70 % less is 30 % of the given number therefore , 30 % of 60 is 18 . answer : a" | a = 60 * 70
b = a / 100
|
a ) $ 100 , b ) $ 200 , c ) $ 300 , d ) $ 500 , e ) $ 600 | c | multiply(divide(4, add(add(2, const_3.0), 3)), 1000) | a person want to give his money of $ 1000 to his 4 children a , b , c , d in the ratio 2 : 4 : 3 : 1 . what is the a + d share ? | "a ' s share = 1000 * 2 / 10 = $ 200 d ' s share = 1000 * 1 / 10 = $ 100 a + d = $ 300 answer is c" | a = 2 + 3
b = a + 3
c = 4 / b
d = c * 1000
|
a ) 720 , b ) 120 , c ) 300 , d ) 30 , e ) 333 | e | subtract(subtract(subtract(divide(divide(divide(factorial(14), factorial(subtract(14, 3))), factorial(3)), const_2), 14), 14), const_10) | mariah has decided to hire three workers . to determine whom she will hire , she has selected a group of 14 candidates . she plans to have one working interview with 3 of the 14 candidates every day to see how well they work together . how many days will it take her to have working interviews with all the different combinations of job candidates ? | "333 . answer e" | a = math.factorial(14)
b = 14 - 3
c = math.factorial(b)
d = a / c
e = math.factorial(3)
f = d / e
g = f / 2
h = g - 14
i = h - 14
j = i - 10
|
a ) 23 , b ) 38 , c ) 37 , d ) 36 , e ) 28 | d | multiply(divide(200, 20), const_3_6) | an athlete runs 200 metres race in 20 seconds . what is his speed ? | "speed = distance / time = 200 / 20 = 10 m / s = 10 * 18 / 5 = 36 km / hr answer : d" | a = 200 / 20
b = a * const_3_6
|
a ) 2 , b ) 3 , c ) 4 , d ) 5 , e ) 6 | a | subtract(multiply(3, const_4.0), 10) | a certain meter records voltage between 0 and 10 volts inclusive . if the average value of 3 recordings on the meter was 6 volts , what was the smallest possible recording in volts ? | "if average of 3 is 6 so sum of 3 should be 18 3 recording can be from 0 - 10 inclusive to find one smallest other two should be highest so , lets assume three var are a , b , c say a is smallest and give b and c greatest readings say 8 and 8 so a has to be 2 a" | a = 3 * 4
b = a - 10
|
a ) 5 , b ) 6 , c ) 8 , d ) 9 , e ) 4 | e | subtract(subtract(add(multiply(287, 287), multiply(269, 269)), multiply(multiply(2, 287), 269)), add(multiply(const_10, const_2), multiply(const_3, const_100))) | find the last digit in the product 287 x 287 + 269 x 269 - 2 x 287 x 269 ? | explanation : given exp . = a 2 + b 2 - 2 ab , where a = 287 and b = 269 = ( a - b ) 2 = ( 287 - 269 ) 2 = ( 182 ) = 324 answer is e | a = 287 * 287
b = 269 * 269
c = a + b
d = 2 * 287
e = d * 269
f = c - e
g = 10 * 2
h = 3 * 100
i = g + h
j = f - i
|
a ) 1 / 3 , b ) 1 / 4 , c ) 3 / 8 , d ) 2 / 9 , e ) 5 / 18 | d | add(multiply(subtract(const_1, divide(5, 6)), subtract(const_1, divide(2, 3))), multiply(subtract(const_1, divide(3, 4)), divide(2, 3))) | at a monthly meeting , 2 / 3 of the attendees were males and 3 / 4 of the male attendees arrived on time . if 5 / 6 of the female attendees arrived on time , what fraction of the attendees at the monthly meeting did not arrive on time ? | "males who did not arrive on time are 1 / 4 * 2 / 3 = 1 / 6 of the attendees . females who did not arrive on time are 1 / 6 * 1 / 3 = 1 / 18 of the attendees . the fraction of all attendees who did not arrive on time is 1 / 6 + 1 / 18 = 2 / 9 the answer is d ." | a = 5 / 6
b = 1 - a
c = 2 / 3
d = 1 - c
e = b * d
f = 3 / 4
g = 1 - f
h = 2 / 3
i = g * h
j = e + i
|
a ) 3 , b ) 18 , c ) 10 , d ) 99 , e ) 38 | a | divide(subtract(divide(36, divide(4, 4)), multiply(subtract(4, const_1), 4)), 4) | the sum of the ages of 4 children born at the intervals of 4 years each is 36 years . what is the age of the youngest child ? | "let x = the youngest child . each of the other four children will then be x + 4 , x + 8 , x + 12 we know that the sum of their ages is 36 . so , x + ( x + 4 ) + ( x + 8 ) + ( x + 12 ) = 36 therefore the youngest child is 3 years old answer : a" | a = 4 / 4
b = 36 / a
c = 4 - 1
d = c * 4
e = b - d
f = e / 4
|
a ) 50.12 , b ) 52.12 , c ) 51.12 , d ) 53.12 , e ) none of the above | b | divide(add(add(add(multiply(50, 50), multiply(35, 60)), multiply(45, 55)), multiply(42, 45)), add(add(add(50, 35), 45), 42)) | a school has 4 section of chemistry in class x having 50 , 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 = 50 ã — 50 + 35 ã — 60 + 45 ã — 55 + 42 ã — 45 / 50 + 35 + 45 + 42 = 2500 + 2100 + 2475 + 1890 / 172 = 8965 â „ 172 = 52.12 answer b" | a = 50 * 50
b = 35 * 60
c = a + b
d = 45 * 55
e = c + d
f = 42 * 45
g = e + f
h = 50 + 35
i = h + 45
j = i + 42
k = g / j
|
a ) 60 , b ) 62 , c ) 64 , d ) 66 , e ) 68 | a | add(add(add(add(10, const_1), add(add(10, const_1), const_2)), add(add(add(10, const_1), const_2), 4)), add(add(add(add(10, const_1), const_2), 4), const_2)) | what is the least integer that is a sum of 4 different prime numbers each greater than 10 ? | the sum of the four smallest primes greater than 10 is 11 + 13 + 17 + 19 = 60 . the answer is a . | a = 10 + 1
b = 10 + 1
c = b + 2
d = a + c
e = 10 + 1
f = e + 2
g = f + 4
h = d + g
i = 10 + 1
j = i + 2
k = j + 4
l = k + 2
m = h + l
|
a ) 2 , b ) 3 , c ) 5 , d ) 7 , e ) 11 | c | add(const_3, const_4) | what is the smallest positive integer that can be multiplied by 605 to make it a perfect square ? | "605 is multiplied by 5 gives 3025 its a square root of 55 answer is 5 - option c" | a = 3 + 4
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a ) 8 , b ) 10 , c ) 12 , d ) 15 , e ) 9 | a | add(5, 1) | set a consists of the integers from 1 to 12 , inclusive , while set b consists of the integers from 5 to 15 , inclusive . how many distinct integers do belong to the both sets at the same time ? | "a = { 1,2 , 3,4 , 5,6 , 7 , 8 , 9 , 10 , 11 , 12 } b = { 5,6 , 7 , 8 , 9 , 10 , 11 , 12 , 13 , 14 , 15 } common elements = { 5,6 , 7 , 8 , 9 , 10 , 11 , 12 } = 8 elements answer : option a ." | a = 5 + 1
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a ) 2388 , b ) 3722 , c ) 2400 , d ) 2872 , e ) 1231 | c | multiply(divide(6000, add(const_1, divide(const_2, const_3))), divide(const_2, const_3)) | p , q and r have rs . 6000 among themselves . r has two - thirds of the total amount with p and q . find the amount with r ? | explanation : let the amount with r be rs . r r = 2 / 3 ( total amount with p and q ) r = 2 / 3 ( 6000 - r ) = > 3 r = 12000 - 2 r = > 5 r = 12000 = > r = 2400 . answer : c | a = 2 / 3
b = 1 + a
c = 6000 / b
d = 2 / 3
e = c * d
|
a ) 5 , b ) 6 , c ) 7 , d ) 8 , e ) 9 | e | add(divide(subtract(32, add(4, multiply(const_2, 4))), 4), 4) | on sunday , bill ran 4 more miles than he ran on saturday . julia did not run on saturday , but she ran twice the number of miles on sunday that bill ran on sunday . if bill and julia ran a total of 32 miles on saturday and sunday , how many miles did bill run on sunday ? | "let bill run x on saturday , so he will run x + 4 on sunday . . julia will run 2 * ( x + 4 ) on sunday . . totai = x + x + 4 + 2 x + 8 = 32 . . 4 x + 12 = 32 . . x = 5 . . ans = x + 4 = 5 + 4 = 9 answer e" | a = 2 * 4
b = 4 + a
c = 32 - b
d = c / 4
e = d + 4
|
a ) 5000 , b ) 2999 , c ) 2878 , d ) 2990 , e ) 6500 | e | divide(multiply(260, const_100), subtract(const_100, add(subtract(const_100, 20), multiply(subtract(const_100, 20), divide(20, const_100))))) | a man saves 20 % of his monthly salary . if an account of dearness of things he is to increase his monthly expenses by 20 % , he is only able to save rs . 260 per month . what is his monthly salary ? | "income = rs . 100 expenditure = rs . 80 savings = rs . 20 present expenditure 80 * ( 20 / 100 ) = rs . 96 present savings = 100 â € “ 96 = rs . 4 100 - - - - - - 4 ? - - - - - - - - - 260 = > 6500 answer : e" | a = 260 * 100
b = 100 - 20
c = 100 - 20
d = 20 / 100
e = c * d
f = b + e
g = 100 - f
h = a / g
|
a ) 176 kmph , b ) 144 kmph , c ) 176 kmph , d ) 134 kmph , e ) 161 kmph | b | multiply(divide(160, 4), const_3_6) | a 160 meter long train crosses a man standing on the platform in 4 sec . what is the speed of the train ? | "s = 160 / 4 * 18 / 5 = 144 kmph answer : b" | a = 160 / 4
b = a * const_3_6
|
a ) 7 : 5 , b ) 17 : 3 , c ) 4 : 1 , d ) 17 : 7 , e ) 5 : 8 | c | divide(60000, 15000) | p and q started a business investing rs 60000 and rs 15000 resp . in what ratio the profit earned after 2 years be divided between p and q respectively . | explanation : in this type of question as time frame for both investors is equal then just get the ratio of their investments . p : q = 60000 : 15000 = 60 : 15 = 4 : 1 option c | a = 60000 / 15000
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a ) 200 , b ) 100 , c ) 50 , d ) 70 , e ) 25 | b | divide(multiply(50, 15), divide(60, const_2)) | 20 people can write 50 book in 15 days working 8 hour a day . then in how many day 200 can be written by 60 people ? | "work per day epr hour per person = 50 / ( 15 * 8 * 20 ) / / eq - 1 people = 60 ; let suppose day = p ; per day work for 8 hours acc . to condition work per day epr hour per person = 200 / ( p * 8 * 60 ) / / eq - 2 eq - 1 = = eq - 2 ; p = 100 answer : b" | a = 50 * 15
b = 60 / 2
c = a / b
|
a ) 3 , b ) 2 , c ) 4 , d ) 5 , e ) 6 | a | subtract(multiply(add(multiply(const_4, const_10), const_2), 23), 1078) | what least number must be added to 1078 , so that the sum is completely divisible by 23 ? | "47 * 23 = 1081 1081 - 1078 = 3 answer : a" | a = 4 * 10
b = a + 2
c = b * 23
d = c - 1078
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a ) 55.1 , b ) 51.1 , c ) 53.1 , d ) 52.1 , e ) 59.1 | e | divide(add(multiply(55, 60), multiply(48, 58)), add(55, 48)) | find the average marks of all the students in 2 separate classes , if the average marks of students in the first class of 55 students is 60 and that of another class of 48 students is 58 | sum of the marks for the class of 55 students = 55 * 60 = 3300 sum of the marks for the class of 48 students = 48 * 58 = 2784 sum of the marks for the class of 103 students = 3300 + 2784 = 6084 average marks of all the students = 6084 / 103 = 59.1 answer : e | a = 55 * 60
b = 48 * 58
c = a + b
d = 55 + 48
e = c / d
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a ) 10 seconds , b ) 12 seconds , c ) 14 seconds , d ) 16 seconds , e ) none of these | b | multiply(const_3600, divide(divide(100, const_1000), 30)) | a train is 100 meter long and is running at the speed of 30 km per hour . find the time it will take to pass a man standing at a crossing . | explanation : as we need to get answer in seconds , so never forget to convert speed into meter per second . speed = 30 km / hr = 30 * 5 / 18 m / sec = 25 / 3 m / sec distance = length of train = 100 meter required time = distancespeed = 100253 = 100 ∗ 325 = 12 sec answer : b | a = 100 / 1000
b = a / 30
c = 3600 * b
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a ) 2 / 6 , b ) 1 / 6 , c ) 1 / 8 , d ) 1 / 3 , e ) 1 / 2 | b | divide(const_1, multiply(3, const_2)) | on a game show , a contestant is given 3 keys , each of which opens exactly one of the 3 identical boxes . the contestant assigns each key to one of the boxes and wins the amount of money contained in any box that is opened by the key assigned to it . the first box contains $ 4 , the second $ 400 , and the third $ 4000 . what is the probability that a contestant will win more than $ 4000 ? | let ' s call the boxes that contain $ 4 , $ 400 , and $ 4000 , respectively , box a , box b , box c . these are opened , respectively , by key a , key b , and key c . we want to know the probability of winning more than $ 4000 . notice that if the distribution of keys is : box a = key b box b = key a box c = key c then the contestant wins exactly $ 4000 , not more than $ 4000 . the only configuration that leads to winning more than $ 1000 is : box a = key a box b = key b box c = key c i . e . , getting all three keys correct . that ' s the only way to be more than $ 4000 . so , really , the question can be rephrased : what is the probability of guessing the order of keys so that each key matches the correct box ? well , for a set of three items , the number of possible permutations is 3 ! = 3 * 2 * 1 = 6 . of those 6 possible permutations , only one of them leads to all three keys being paired with the right box . so , the answer is probability = 1 / 6 answer : b | a = 3 * 2
b = 1 / a
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a ) 8 , b ) 10 , c ) 12 , d ) 14 , e ) 16 | c | divide(divide(divide(1701, const_3), const_3), add(1, const_4)) | for any integer k greater than 1 , the symbol k * denotes the product of all integers between 1 and k , inclusive . if k * is a multiple of 1701 what is the least possible value of k ? | "1701 = 3 * 3 * 3 * 3 * 3 * 7 thus k must include numbers at least up to the number 12 so that there are at least five appearances of 3 ( that is : 3 , 6 , 9 = 3 * 3 , and 12 ) . the answer is c ." | a = 1701 / 3
b = a / 3
c = 1 + 4
d = b / c
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a ) 14 , b ) 16 , c ) 21 , d ) 22 , e ) 32 | e | divide(336, divide(subtract(462, 336), 12)) | a car traveled 462 miles per tankful of gasoline on the highway and 336 miles per tankful of gasoline in the city . if the car traveled 12 fewer miles per gallon in the city than on the highway , how many miles per gallon did the car travel in the city ? | "i treat such problems as work ones . work = rate * time mileage ( m ) = rate ( mpg ) * gallons ( g ) x gallons is a full tank { 462 = rx { 336 = ( r - 12 ) x solve for r , r = 44 44 - 12 = 32 mpg e" | a = 462 - 336
b = a / 12
c = 336 / b
|
a ) 244.85 , b ) 224.95 , c ) 244.85 , d ) 224.95 , e ) 244.95 | e | divide(multiply(10, sqrt(subtract(power(multiply(const_2, 25), const_2), power(10, const_2)))), const_2) | the side of a rhombus is 25 m and length of one of its diagonals is 10 m . the area of the rhombus is ? | "area of the rhombus = 1 / 2 * p * √ 4 ( a ) 2 - ( p ) 2 a = 25 ; p = 10 a = 1 / 2 * 10 * √ 4 ( 25 ) 2 - ( 10 ) 2 = 1 / 2 * 10 * √ 2500 - 100 = 1 / 2 * 10 * √ 2400 a = 244.95 answer : e" | a = 2 * 25
b = a ** 2
c = 10 ** 2
d = b - c
e = math.sqrt(d)
f = 10 * e
g = f / 2
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a ) 12 , b ) 15 , c ) 18 , d ) 20 , e ) 25 | d | divide(add(negate(multiply(3, 5)), sqrt(subtract(power(multiply(3, 5), const_2), multiply(multiply(const_4, 3), negate(multiply(300, 5)))))), multiply(3, const_2)) | two dogsled teams raced across a 300 - mile course in wyoming . team a finished the course in 3 fewer hours than did team b . if team a ’ s average speed was 5 miles per hour greater than that of team b , what was team b ’ s average speed , in miles per hour ? | speed of b = s speed of a = s + 5 time taken by b = 300 / s time taken by a = 300 / ( s + 5 ) difference in their time is 3 hrs . 300 / s - 3 = 300 / ( s + 5 ) plug in the values from the option no need of choosing 12 & 18 it wo n ' t satisfy a ) if we take 15 then t ( b ) = 300 / 15 = 20 t ( a ) = 300 / 20 = 15 . hence from this option difference comes out to be 5 which is not correct b ) if we take 20 then t ( b ) = 300 / 20 = 15 t ( a ) = 300 / 25 = 12 difference in timing is 3 hence option d is correct | a = 3 * 5
b = negate + (
c = 3 * 5
d = c ** 2
e = 4 * 3
f = 300 * 5
g = e * negate
h = d - g
i = math.sqrt(h)
j = b / i
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a ) 10 % , b ) 9 % , c ) 11.11 % , d ) 42.85 % , e ) none of these | d | multiply(divide(add(multiply(const_2, const_100), divide(const_100, const_2)), 700), const_100) | a shopkeeper forced to sell at cost price , uses a 700 grams weight for a kilogram . what is his gain percent ? | "shopkeeper sells 700 g instead of 1000 g . so , his gain = 1000 - 700 = 300 g . thus , % gain = ( 300 * 100 ) / 700 = 42.85 % . answer : option d" | a = 2 * 100
b = 100 / 2
c = a + b
d = c / 700
e = d * 100
|
a ) 5 , b ) 7 , c ) 15 / 2 , d ) 21 / 4 , e ) 18 / 5 | d | divide(multiply(7, 3), subtract(7, 3)) | a man can do a piece of work in 7 days , but with the help of his son , he can do it in 3 days . in what time can the son do it alone ? | "son ' s 1 day ' s work = ( 1 / 3 ) - ( 1 / 7 ) = 4 / 21 the son alone can do the work in 21 / 4 days answer is d" | a = 7 * 3
b = 7 - 3
c = a / b
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a ) 1 , b ) 4 , c ) 81 , d ) log 80 243 , e ) e 81 | b | divide(log(81), log(3)) | what is log 3 ( 4 ) log 4 ( 5 ) . . . log 80 ( 81 ) ? | recall the change of base formula logb ( a ) = ln ( a ) ln ( b ) : ( alternatively , we can substitute logc for ln on the right , as long as the base is the same on the top and on the bottom . ) using this , we can rewrite the entire expression using natural logarithms , as ln ( 4 ) ln ( 3 ) ln ( 5 ) ln ( 4 ) ln ( 81 ) ln ( 80 ) = ln ( 81 ) ln ( 3 ) = log 3 ( 81 ) = log 3 ( 34 ) = 4 : correct answer b | a = math.log(81)
b = math.log(3)
c = a / b
|
a ) 7534 , b ) 255.4 , c ) 756.4 , d ) 755.4 , e ) 752.4 | d | divide(0.0005521, 0.0000110) | 0.0005521 / 0.0000110 x 15.05 = ? | "explanation : ? = 0.0005521 / 0.0000110 x 15.05 = 755.4 answer : option d" | a = 0 / 5521
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a ) 1 , b ) 3 , c ) 9 , d ) 6 , e ) 7 | c | subtract(multiply(subtract(16, 1), add(6, 1)), multiply(16, 6)) | the average of 6 observations is 16 . a new observation is included and the new average is decreased by 1 . the seventh observation is ? | "let seventh observation = x . then , according to the question we have = > ( 96 + x ) / 7 = 15 = > x = 9 hence , the seventh observation is 9 . answer : c" | a = 16 - 1
b = 6 + 1
c = a * b
d = 16 * 6
e = c - d
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a ) 16 , b ) 24 , c ) c is elder than a , d ) data inadequate , e ) none | a | multiply(16, const_1) | the total age of a and b is 16 years more than the total age of b and c . c is how many years younger than a ? | "solution [ ( a + b ) - ( b + c ) ] = 16 â € ¹ = â € º a - c = 16 . answer a" | a = 16 * 1
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a ) 23 % , b ) 28 % , c ) 33 % , d ) 38 % , e ) 43 % | c | multiply(divide(divide(subtract(multiply(25, const_100), multiply(20, const_100)), subtract(const_100, 25)), 20), const_100) | keats library purchases a number of new books , all in the category of biography , and the library does not acquire any other books . with the addition of the new biographies , the biography collection of the library amounts to 25 % of the total number of books in the library . if prior to the purchase , only 20 % of the books in keats library were biographies , by what percent has the number of biographies in the library increased ? | "let x be the number of new biographies added to the library . let b be the original number of biographies , so the original number of books was 5 b . 0.25 ( 5 b + x ) = b + x 0.25 b = 0.75 x x = 0.33 b the answer is c ." | a = 25 * 100
b = 20 * 100
c = a - b
d = 100 - 25
e = c / d
f = e / 20
g = f * 100
|
a ) 186 m , b ) 176 m , c ) 100 m , d ) 600 m , e ) 765 m | d | multiply(divide(multiply(60, const_1000), const_3600), 36) | a train running at the speed of 60 km / hr crosses a pole in 36 seconds . what is the length of the train ? | "speed = ( 60 * 5 / 18 ) m / sec = ( 50 / 3 ) m / sec length of the train = ( speed x time ) = ( 50 / 3 * 36 ) m = 600 m . answer : d" | a = 60 * 1000
b = a / 3600
c = b * 36
|
a ) 11 , b ) 15 , c ) 20 , d ) 38 , e ) 56 | b | multiply(power(const_2, 504), factorial(const_4)) | the product of three consecutive numbers is 504 . then the sum of the smallest two numbers is ? | "product of three numbers = 504 504 = 7 * 8 * 9 . so , the three numbers are 7 , 8 and 9 . and sum of smallest of these two = 7 + 8 = 15 . answer : option b" | a = 2 ** 504
b = math.factorial(4)
c = a * b
|
a ) 18 , b ) 20 , c ) 22 , d ) 24 , e ) 26 | b | divide(log(multiply(power(2, 20), power(3, 20))), log(6)) | if 3 ^ 20 x 2 ^ 20 = 6 ^ n what is the value of n ? | "3 ^ 20 * 2 ^ 20 = 6 ^ n 6 ^ 20 = 6 ^ n therefore n = 20 b" | a = 2 ** 20
b = 3 ** 20
c = a * b
d = math.log(c)
e = math.log(6)
f = d / e
|
a ) 0 days , b ) 10 days , c ) 20 days , d ) 30 days , e ) 40 days | a | divide(subtract(multiply(20, 40), multiply(40, 20)), 40) | 20 men can do a work in 40 days . when should 20 men leave the work so that the entire work is completed in 40 days after they leave the work ? | "total work to be done = 20 * 40 = 800 let 20 men leave the work after ' p ' days , so that the remaining work is completed in 40 days after they leave the work . 40 p + ( 20 * 40 ) = 800 40 p = 0 = > p = 0 days answer : a" | a = 20 * 40
b = 40 * 20
c = a - b
d = c / 40
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a ) 45645433 , b ) 46457457 , c ) 4675 , d ) 3000 , e ) 3432 | d | divide(divide(multiply(800, const_3), const_2), divide(subtract(700, 300), add(700, 300))) | mary and harry enter into a partnership by investing $ 700 and $ 300 respectively . at the end of one year , they divided their profits such that a third of the profit is divided equally for the efforts they have put into the business and the remaining amount of profit is divided in the ratio of the investments they made in the business . if mary received $ 800 more than harry did , what was the profit made by their business in that year ? | say the profit was $ x . mary ' s share = x / 6 ( half of the third ) + ( x - x / 3 ) * 0.7 harry share = x / 6 ( half of the third ) + ( x - x / 3 ) * 0.3 thus ( x - x / 3 ) * 0.7 - ( x - x / 3 ) * 0.3 = 800 - - > x = 3000 . answer is d | a = 800 * 3
b = a / 2
c = 700 - 300
d = 700 + 300
e = c / d
f = b / e
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a ) a ) 2 , b ) b ) 4 , c ) c ) 7 , d ) d ) 5 , e ) e ) 8 | b | subtract(divide(multiply(divide(const_1, const_2), 45), subtract(50, 45)), divide(const_1, const_2)) | a thief goes away with a santro car at a speed of 45 kmph . the theft has been discovered after half an hour and the owner sets off in a bike at 50 kmph when will the owner over take the thief from the start ? | explanation : | - - - - - - - - - - - 20 - - - - - - - - - - - - - - - - - - - - | 50 45 d = 20 rs = 50 – 45 = 5 t = 20 / 5 = 4 hours answer : option b | a = 1 / 2
b = a * 45
c = 50 - 45
d = b / c
e = 1 / 2
f = d - e
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a ) 814 , b ) 800 , c ) 100 , d ) 200 , e ) 456 | a | divide(rectangle_area(multiply(15, const_100), multiply(17, const_100)), square_area(add(multiply(const_4, const_10), const_1))) | what is the least number of squares tiles required to pave the floor of a room 15 m 17 cm long and 9 m 2 cm broad ? | "length of largest tile = h . c . f . of 1517 cm and 902 cm = 41 cm . area of each tile = ( 41 x 41 ) cm 2 . required number of tiles = ( 1517 x 902 ) / 41 x 41 = 814 . answer : a" | a = 15 * 100
b = 17 * 100
c = rectangle_area / (
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a ) 87 , b ) 16 , c ) 10 , d ) 84 , e ) 17 | d | subtract(multiply(add(10, const_1), add(4, 40)), multiply(10, 40)) | the average of runs of a cricket player of 10 innings was 40 . how many runs must he make in his next innings so as to increase his average of runs by 4 ? | "average after 11 innings = 44 required number of runs = ( 44 * 11 ) - ( 40 * 10 ) = 484 - 400 = 84 . answer : d" | a = 10 + 1
b = 4 + 40
c = a * b
d = 10 * 40
e = c - d
|
a ) 1235 , b ) 1346 , c ) 1378 , d ) 1634.4 , e ) 1489 | d | multiply(divide(subtract(1365, 18), subtract(6, const_1)), 6) | find large number from below question the difference of two numbers is 1365 . on dividing the larger number by the smaller , we get 6 as quotient and the 18 as remainder | "let the smaller number be x . then larger number = ( x + 1365 ) . x + 1365 = 6 x + 18 5 x = 1347 x = 269.4 number = 269.4 + 1365 = 1634.4 d" | a = 1365 - 18
b = 6 - 1
c = a / b
d = c * 6
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a ) 16 , b ) 17 , c ) 18 , d ) 19 , e ) 14 | e | sqrt(divide(588, const_3)) | the length of a rectangular plot is thrice its breadth . if the area of the rectangular plot is 588 sq m , then what is the breadth of the rectangular plot ? | "let the breadth of the plot be b m . length of the plot = 3 b m ( 3 b ) ( b ) = 588 3 b 2 = 588 b 2 = 196 b = 14 m . answer : option e" | a = 588 / 3
b = math.sqrt(a)
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a ) $ 30 , b ) $ 31 , c ) $ 32 , d ) $ 33 , e ) $ 34 | a | divide(add(multiply(divide(20, const_100), 20), 20), divide(subtract(const_100, 20), const_100)) | a distributor sells a product through an online store , which take a commission of 20 % of the price set by the distributor . the distributor obtains the product from a producer at the price of $ 20 per item . what is the price that the buyer observers online if the distributor wants to maintain a 20 % profit on the cost of the item ? | "let x be the price that buyers see online . the distributor wants to receive 1.2 ( original price ) which should be 80 % of x . 1.2 ( 20 ) = 0.8 x x = 1.2 ( 20 ) / 0.8 = 1.5 ( 20 ) = $ 30 the answer is a ." | a = 20 / 100
b = a * 20
c = b + 20
d = 100 - 20
e = d / 100
f = c / e
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a ) 120 , b ) 250 , c ) 110 , d ) 220 , e ) none | e | add(200, divide(add(multiply(30, 20), 200), 20)) | 21 friends went to a hotel and decided to pay the bill amount equally . but 20 of them could pay rs . 30 each as a result 21 st has to pay rs . 200 extra than his share . find the amount paid by him . | "explanation : average amount paid by 20 persons = rs . 30 increase in average due to rs . 200 paid extra by the 21 st men = rs . 200 / 20 = rs . 10 therefore , average expenditure of 21 friends = rs . 30 + rs . 10 = rs . 40 therefore , amount paid by the 21 st men = rs . 40 + rs . 200 = rs . 240 correct option : e" | a = 30 * 20
b = a + 200
c = b / 20
d = 200 + c
|
a ) 48 , b ) 50 , c ) 28 , d ) 26 , e ) 29 | a | divide(divide(subtract(125, multiply(multiply(3, const_0_2778), 3)), 3), const_0_2778) | a train 125 m long passes a man , running at 3 km / hr in the same direction in which the train is going , in 10 seconds . the speed of the train is ? | "speed of the train relative to man = ( 125 / 10 ) m / sec = ( 25 / 2 ) m / sec . [ ( 25 / 2 ) * ( 18 / 5 ) ] km / hr = 45 km / hr . let the speed of the train be x km / hr . then , relative speed = ( x - 3 ) km / hr . x - 3 = 45 = = > x = 48 km / hr answer : a" | a = 3 * const_0_2778
b = a * 3
c = 125 - b
d = c / 3
e = d / const_0_2778
|
a ) 1 minutes , b ) 10 minute , c ) 100 minutes , d ) 10000 minutes , e ) 1000 minutes | b | multiply(divide(10, 10), 10) | if 10 lions can kill 10 deers in 10 minutes how long will it take 100 lions to kill 100 deers ? | "we can try the logic of time and work , our work is to kill the deers so 10 ( lions ) * 10 ( min ) / 10 ( deers ) = 100 ( lions ) * x ( min ) / 100 ( deers ) hence answer is x = 10 answer : b" | a = 10 / 10
b = a * 10
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a ) 20 cs , b ) cs / 2 , c ) 36 cs , d ) ( 2 cs ) / 12 , e ) ( 24 c ) / s | c | multiply(3, const_12) | a certain school implemented a reading program for its students , with the goal of getting each student to read 3 books per month year - round . if the school has c classes made up of s students in each class , how many books will the entire student body read in one year ? | "ans : c solution : simple multiplication s students , c classes , 3 books / month = 36 books a year total number of books = 36 cs" | a = 3 * 12
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a ) 110.22 meters , b ) 111.22 meters , c ) 222.22 meters , d ) 333.22 meters , e ) none of these | c | subtract(multiply(const_100, const_10), divide(multiply(multiply(const_100, const_10), subtract(multiply(const_100, const_10), 300)), subtract(multiply(const_100, const_10), 100))) | a can give b 100 meters start and c 300 meters start in a kilometer race . how much start can b give c in a kilometer race ? | "explanation : a runs 1000 meters while b runs 900 meters and c runs 700 meters . therefore , b runs 900 meters while c runs 700 meters . so , the number of meters that c runs when b runs 1000 meters = ( 1000 x 700 ) / 900 = 777.77 meters thus , b can give c ( 1000 - 777.77 ) = 222.22 meters start answer : c" | a = 100 * 10
b = 100 * 10
c = 100 * 10
d = c - 300
e = b * d
f = 100 * 10
g = f - 100
h = e / g
i = a - h
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a ) rs . 300 , b ) rs . 200 , c ) rs . 240 , d ) rs . 350 , e ) none of these | b | divide(740, add(add(multiply(add(const_1, divide(25, const_100)), add(const_1, divide(20, const_100))), add(const_1, divide(20, const_100))), const_1)) | if x gets 25 % more than y and y gets 20 % more than z , the share of z out of rs . 740 will be : | "z share = z , y = 1.2 z x = 1.25 × 1.2 z , x + y + z = 740 ( 1.25 × 1.2 + 1.2 + 1 ) z = 74 3.7 z = 740 , z = 200 answer : . b ." | a = 25 / 100
b = 1 + a
c = 20 / 100
d = 1 + c
e = b * d
f = 20 / 100
g = 1 + f
h = e + g
i = h + 1
j = 740 / i
|
a ) rs . 660 , b ) rs . 760 , c ) rs . 600 , d ) rs . 960 , e ) none of these | c | multiply(750, subtract(const_1, divide(20, const_100))) | a man buys an item at rs . 750 and sells it at the loss of 20 percent . then what is the selling price of that item | "explanation : here always remember , when ever x % loss , it means s . p . = ( 100 - x ) % of c . p when ever x % profit , it means s . p . = ( 100 + x ) % of c . p so here will be ( 100 - x ) % of c . p . = 80 % of 750 = 80 / 100 * 750 = 600 option c" | a = 20 / 100
b = 1 - a
c = 750 * b
|
a ) 1 / 3 , b ) 2 / 3 , c ) 3 / 3 , d ) 3 / 1 , e ) 3 / 2 | c | divide(subtract(subtract(subtract(10, 2), 3), 2), 3) | harry started a 4 - mile hike with a full 10 - cup canteen of water and finished the hike in 2 hours with 2 cup of water remaining in the canteen . if the canteen leaked at the rate of 1 cup per hour and harry drank 3 cups of water during the last mile , how many cups did he drink per mile during the first 3 miles of the hike ? | "no of cups leaked during the trip = 2 cups . no of cups harry drank = 6 cups . no of cups harry drank during the first 3 miles = 3 . drink / mile = 3 / 3 answer : c" | a = 10 - 2
b = a - 3
c = b - 2
d = c / 3
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a ) $ 6000 , b ) $ 8000 , c ) $ 10,000 , d ) $ 12,000 , e ) $ 14,000 | c | divide(add(divide(subtract(260, multiply(divide(6, const_100), 1,000)), subtract(divide(8, const_100), divide(6, const_100))), divide(subtract(260, multiply(divide(6, const_100), 1,000)), subtract(divide(8, const_100), divide(6, const_100)))), 1,000) | salesperson a ' s compensation for any week is $ 260 plus 6 percent of the portion of a ' s total sales above $ 1,000 for that week . salesperson b ' s compensation for any week is 8 percent of b ' s total sales for that week . for what amount of total weekly sales would both salespeople earn the same compensation ? | "260 + 0.06 ( x - 1000 ) = 0.08 x 0.02 x = 200 x = $ 10,000 the answer is c ." | a = 6 / 100
b = a * 1
c = 260 - b
d = 8 / 100
e = 6 / 100
f = d - e
g = c / f
h = 6 / 100
i = h * 1
j = 260 - i
k = 8 / 100
l = 6 / 100
m = k - l
n = j / m
o = g + n
p = o / 1
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a ) 20,000 , b ) 21,200 , c ) 1,800 , d ) 20,500 , e ) none of these | c | divide(add(multiply(subtract(const_12, 6), 1000), multiply(multiply(const_2, multiply(const_100, const_100)), 6)), multiply(const_100, const_10)) | a , b and c start a business each investing 2,000 . after 6 months a withdrew 1000 , b withdrew 1000 and c invests 1000 more . at the end of the year , a total profit of 6600 was recorded . find the share of b . | "ratio of the capitals of a , b and c = 2000 ã — 6 + 1000 ã — 6 : 2000 ã — 6 + 1000 ã — 6 : 2000 ã — 6 + 3000 ã — 6 = 18000 : 18000 : 30000 = 18 : 18 : 30 b â € ™ s share = ( 6600 ã — 18 â „ 66 ) = 1800 answer c" | a = 12 - 6
b = a * 1000
c = 100 * 100
d = 2 * c
e = d * 6
f = b + e
g = 100 * 10
h = f / g
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a ) 0 . , b ) 6 . , c ) 8 . , d ) 10 . , e ) 12 . | a | min(divide(96, const_3), divide(96, 6)) | in the third grade of windblown school there are 96 students , one third of them failed the math test and 1 / 6 failed that literature test . at least how many students failed both tests ? | "total = 96 failed in math = 96 / 3 = 32 failed in literature = 108 / 6 = 16 the least failed in both can be 0 while max can be 16 answer a" | a = 96 / 3
b = 96 / 6
c = min(a)
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a ) 3 , b ) 4 , c ) 5 , d ) 6 , e ) 7 | d | divide(24, const_4) | rajat , vikas and abhishek are submitting questions in the ratio 7 : 3 : 2 . if total number of questions submitted by them is 24 . find the number of questions submitted by vikas . | explanation : number of questions submitted by vikas = ( 24 * 3 ) / ( 7 + 3 + 2 ) = 6 answer : d | a = 24 / 4
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a ) 7 , b ) 8 , c ) 6 , d ) 3 , e ) 1 | e | divide(subtract(37, 5), subtract(37, 5)) | a number divided by 64 leaves remainder 37 what is the remainder when same number divided by 5 | add 64 + 37 = 101 now 101 divided by 5 so we get 1 as reaminder answer : e | a = 37 - 5
b = 37 - 5
c = a / b
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a ) 45 , b ) 36 , c ) 40 , d ) 50 , e ) 48 | c | divide(original_price_before_loss(20, 80), divide(original_price_before_gain(20, 60), 20)) | a man sold 20 articles for $ 60 and gained 20 % . how many articles should he sell for $ 80 to incur a loss 20 % ? | "production cost per article : $ 60 * ( 100 % - 20 % ) / 20 = $ 2.40 required production costs for a loss of 20 % : $ 80 * ( 100 % + 20 % ) = $ 96 number of articles to be sold for $ 96 to incur a 20 % loss : $ 96 / $ 2.40 = 40 thus , solution c is correct ." | a = original_price_before_loss / (
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a ) s . 800 , b ) s . 400 , c ) s . 600 , d ) s . 900 , e ) s . 1100 | e | divide(1100, const_3) | divide rs . 1100 among a , b and c so that a receives 1 / 3 as much as b and c together and b receives 2 / 3 as a and c together . a ' s share is ? | "a + b + c = 1100 a = 1 / 3 ( b + c ) ; b = 2 / 3 ( a + c ) a / ( b + c ) = 1 / 3 a = 1 / 4 * 1100 = > 275 answer : e" | a = 1100 / 3
|
a ) 35 , b ) 91 , c ) 120 , d ) 126 , e ) 150 | b | subtract(choose(9, 5), choose(subtract(9, 2), 2)) | a meeting has to be conducted with 5 managers . find the number of ways in which the managers be selected from among 9 managers , if 2 managers will not attend the meeting together ? | "we can either choose all 5 people from 7 manager who have no problems or choose 4 from the 7 and 1 from the 2 managers who have a problem sitting together so 7 c 5 + ( 7 c 4 * 2 c 1 ) this is 21 + 70 91 ans : b" | a = math.comb(9, 5)
b = 9 - 2
c = math.comb(b, 2)
d = a - c
|
a ) 1.0 , b ) 1.5 , c ) 2.0 , d ) 2.5 , e ) 3.0 | a | multiply(add(add(2, 5), 5), divide(5, const_60)) | a certain car increased its average speed by 5 miles per hour in each successive 5 - minute interval after the first interval . if in the first 5 - minute interval its average speed was 2 miles per hour , how many miles did the car travel in the third 5 - minute interval ? | "in the third time interval the average speed of the car was 2 + 5 + 5 = 12 miles per hour ; in 5 minutes ( 1 / 12 hour ) at that speed car would travel 12 * 1 / 12 = 1 miles . answer : a ." | a = 2 + 5
b = a + 5
c = 5 / const_60
d = b * c
|
a ) 10 sec , b ) 16 sec , c ) 12 sec , d ) 67 sec , e ) 13 sec | c | multiply(multiply(multiply(const_0_2778, subtract(60, 40)), 60), inverse(multiply(const_0_2778, add(60, 40)))) | two trains of equal length , running with the speeds of 60 and 40 kmph , take 60 seconds to cross each other while they are running in the same direction . what time will they take to cross each other if they are running in opposite directions ? | "rs = 60 - 40 = 20 * 5 / 18 = 100 / 18 t = 50 d = 60 * 100 / 18 = 1000 / 3 rs = 60 + 40 = 100 * 5 / 18 t = 1000 / 3 * 18 / 500 = 12 sec answer : c" | a = 60 - 40
b = const_0_2778 * a
c = b * 60
d = 60 + 40
e = const_0_2778 * d
f = 1/(e)
g = c * f
|
a ) 3 / 7 , b ) 3 / 5 , c ) 3 / 2 , d ) 3 / 7 , e ) 3 / 8 | b | divide(circle_area(divide(0.60, const_2)), const_2) | what will be the vulgar fraction of 0.60 | "explanation : 0.6 = 60 / 100 = 3 / 5 option b" | a = 0 / 60
b = circle_area / (
|
a ) 128 , b ) 142 , c ) 143 , d ) 141 , e ) 129 | e | divide(subtract(subtract(multiply(const_100, const_10), const_1), add(multiply(add(const_10, const_4), 7), 5)), 7) | how many 3 digit positive integers w exist that when divided by 7 leave a remainder of 5 ? | "minimum three digit number is 100 and maximum three digit number is 999 . the first three digit number that leaves remainder 5 when divided by 7 is 103 . 14 * 7 = 98 + 5 = 103 the second three digit number that leaves remainder 5 when divided by 7 is 110 . 15 * 7 = 105 + 5 = 110 the third three digit number that leaves remainder 5 when divided by 7 is 117 and so on the last three digit number that leaves remainder 5 when divided by 7 is 999 142 * 7 = 994 + 5 = 999 therefore , we identify the sequence 103 , 110,117 . . . . . 999 use the formula of last term last term = first term + ( n - 1 ) * common difference you will get the answer 129 that is definitely e ." | a = 100 * 10
b = a - 1
c = 10 + 4
d = c * 7
e = d + 5
f = b - e
g = f / 7
|
a ) 50 , b ) 100 , c ) 10 , d ) 1000 , e ) 110 | b | divide(0.02, divide(0.2, const_100)) | find the missing figures : 0.2 % of ? = 0.02 | "let 0.2 % of x = 0.02 . then , 0.2 * x / 100 = 0.02 x = [ ( 0.02 * 100 ) / 0.2 ] = 100 . answer is b ." | a = 0 / 2
b = 0 / 2
|
a ) 194852714 , b ) 183941604 , c ) 183851704 , d ) 183951714 , e ) 183768244 | b | multiply(subtract(9999, const_4), 18396) | find the value of 18396 x 9999 = m ? | "18396 x 9999 = 18396 x ( 10000 - 1 ) = 18396 x 10000 - 18396 x 1 = 183960000 - 18396 = 183941604 b" | a = 9999 - 4
b = a * 18396
|
a ) 20 , b ) 30 , c ) 40 , d ) 50 , e ) none of these | e | add(multiply(sqrt(divide(subtract(100, 48), const_2)), const_100), sqrt(subtract(100, divide(subtract(100, 48), const_2)))) | the sum of the squares of three numbers is 100 , while the sum of their products taken two at a time is 48 . their sum is : | "x ^ + y ^ 2 + z ^ 2 = 100 xy + yz + zx = 48 as we know . . ( x + y + z ) ^ 2 = x ^ 2 + y ^ 2 + z ^ 2 + 2 ( xy + yz + zx ) so ( x + y + z ) ^ 2 = 100 + ( 2 * 48 ) ( x + y + z ) ^ 2 = 196 so x + y + z = 14 answer : e" | a = 100 - 48
b = a / 2
c = math.sqrt(b)
d = c * 100
e = 100 - 48
f = e / 2
g = 100 - f
h = math.sqrt(g)
i = d + h
|
a ) 89 , b ) 49 , c ) 68 , d ) 268 , e ) 55 | d | add(13, multiply(15, 17)) | if , 1 * 3 * 5 = 16 3 * 5 * 7 = 38 then find , 13 * 15 * 17 = ? | d 268 ( 17 * 15 ) + 13 = 268 | a = 15 * 17
b = 13 + a
|
a ) 53 m , b ) 46 m , c ) 94 m , d ) 49 m , e ) 64 m | d | multiply(66, divide(multiply(86, 8), multiply(20, 6))) | if 20 men can build a wall 66 metres long in 6 days , what length of a similar can be built by 86 men in 8 days ? | "let the required length be x metres more men , more length built ( direct proportion ) less days , less length built ( direct proportion ) men 20 : 35 days 6 : 3 : : 56 : x therefore ( 20 x 6 x x ) = ( 35 x 3 x 56 ) x = ( 35 x 3 x 56 ) / 120 = 49 hence , the required length is 49 m . answer is d" | a = 86 * 8
b = 20 * 6
c = a / b
d = 66 * c
|
a ) 2288 , b ) 2052 , c ) 2778 , d ) 2719 , e ) 1711 | b | add(multiply(subtract(multiply(const_4, const_4), const_2), const_2), 2024) | the calendar of the year 2024 can be used again in the year ? | "explanation : given year 2024 when divided by 4 , leaves a remainder 0 . note : when remainder is 0 , 28 is added to the given year to get the result . so , 2024 + 28 = 2052 answer : b ) 2052" | a = 4 * 4
b = a - 2
c = b * 2
d = c + 2024
|
a ) 20 km , b ) 30 km , c ) 40 km , d ) 50 km , e ) 60 km | b | multiply(60, divide(multiply(5, 6), const_60)) | the pinedale bus line travels at an average speed of 60 km / h , and has stops every 5 minutes along its route . yahya wants to go from his house to the pinedale mall , which is 6 stops away . how far away , in kilometers , is pinedale mall away from yahya ' s house ? | number of stops in an hour : 60 / 5 = 12 distance between stops : 60 / 12 = 5 km distance between yahya ' s house and pinedale mall : 5 x 6 = 30 km imo , correct answer is ` ` b . ' ' | a = 5 * 6
b = a / const_60
c = 60 * b
|
['a ) 50 sq units', 'b ) 60 sq units', 'c ) 70 sq units', 'd ) 80 sq units', 'e ) 90 sq units'] | e | subtract(multiply(subtract(divide(42, const_2), const_10), const_10), multiply(subtract(divide(42, const_2), const_1), const_1)) | a sixth grade teacher asked her students to draw rectangles with positive integer length and a perimeter of 42 . the difference between the largest and smallest possible ares of the rectangles that the students could have come up with is ? | sum of length and breadth will be 21 units . 42 / 2 = 21 area will be max when lxb = 11 x 10 = 110 sq units area will be min when lxb = 20 x 1 = 20 sq units . . the difference between the largest and smallest possible ares of the rectangles that the students could have come up with = 110 - 20 = 90 sq units answer : e | a = 42 / 2
b = a - 10
c = b * 10
d = 42 / 2
e = d - 1
f = e * 1
g = c - f
|
a ) rs . 4000 , b ) rs . 6000 , c ) rs . 7500 , d ) rs . 6600 , e ) none | b | multiply(divide(4, add(add(3, 4), 4)), 16500) | 3 partners a , b , c starts a business . twice a ' s capital is equal to thrice b ' s capital and b ' s capital is 4 times c ' s capital . out of a total profit of rs . 16500 at the end of the year , b ' s share is | solution let c = x . then , b = 4 x and 2 a = 3 x 4 x = 12 x or a = 6 x . ∴ a : b : c = 6 x : 4 x : x = 6 : 4 : 1 . so b ' s capital = rs ( 16500 x 4 / 11 ) = rs . 6000 . answer b | a = 3 + 4
b = a + 4
c = 4 / b
d = c * 16500
|
a ) $ 37.00 , b ) $ 40.00 , c ) $ 45.50 , d ) $ 48.00 , e ) $ 52.50 | e | divide(subtract(multiply(6, 50), add(40, 50)), subtract(6, 2)) | a retailer sells 6 shirts . the first 2 he sells for $ 40 and $ 50 . if the retailer wishes to sell the 6 shirts for an overall average price of over $ 50 , what must be the minimum average price of the remaining 4 shirts ? | "first 2 shirts are sold for $ 40 and $ 50 = $ 90 . to get average price of $ 50 , total sale should be 6 * $ 50 = $ 300 so remaining 4 shirts to be sold for $ 300 - $ 90 = $ 210 answer should be 210 / 4 = $ 52.50 that is e" | a = 6 * 50
b = 40 + 50
c = a - b
d = 6 - 2
e = c / d
|
a ) s . 3300 , b ) s . 4570 , c ) s . 4500 , d ) s . 4550 , e ) s . 2500 | a | subtract(multiply(8000, const_4), subtract(multiply(8800, const_4), 6500)) | the average salary of a person for the months of january , february , march and april is rs . 8000 and that for the months february , march , april and may is rs . 8800 . if his salary for the month of may is rs . 6500 , find his salary for the month of january ? | "sum of the salaries of the person for the months of january , february , march and april = 4 * 8000 = 32000 - - - - ( 1 ) sum of the salaries of the person for the months of february , march , april and may = 4 * 8800 = 35200 - - - - ( 2 ) ( 2 ) - ( 1 ) i . e . may - jan = 3200 salary of may is rs . 6500 salary of january = rs . 3300 answer : a" | a = 8000 * 4
b = 8800 * 4
c = b - 6500
d = a - c
|
a ) 9,000 , b ) 1,500 , c ) 1,750 , d ) 1,200 , e ) 1,300 | b | divide(divide(50, subtract(divide(subtract(divide(const_60, 20), inverse(2)), const_60), inverse(multiply(const_60, 2)))), const_1000) | with both inlets open , a water tank will be filled with water in 20 minutes . the first inlet alone would fill the tank in 2 hours . if in every minutes the second inlet admits 50 cubic meters of water than the first , what is the capacity of the tank ? | "the work done by inlet a and b together in 1 min = 1 / 20 the work done by inlet a ( first inlet ) in 1 min = 1 / 120 the work done by inlet b ( second inlet ) in 1 min = ( 1 / 20 ) - ( 1 / 120 ) = 1 / 24 difference of work done by b and a = b - a = 50 cubic meter i . e . ( 1 / 24 ) - ( 1 / 120 ) = 50 cubic meter i . e . 1500 cubic meter answer : option b" | a = const_60 / 20
b = 1/(2)
c = a - b
d = c / const_60
e = const_60 * 2
f = 1/(e)
g = d - f
h = 50 / g
i = h / 1000
|
['a ) 1 : 28', 'b ) 1 : 27', 'c ) 1 : 29', 'd ) 1 : 39', 'e ) 1 : 22'] | b | divide(1, power(3, const_3)) | the ratio between the radii of two spheres is 1 : 3 . find the ratio between their volumes ? | explanation : r 1 : r 2 = 1 : 3 r 13 : r 23 = 1 : 27 answer : option b | a = 3 ** 3
b = 1 / a
|
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