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 ) 9 : 6 , b ) 9 : 8 , c ) 9 : 14 , d ) 9 : 9 , e ) 9 : 5 | c | divide(add(multiply(3000, 6), multiply(multiply(3000, const_2), 6)), multiply(7000, add(6, 6))) | a and b invests rs . 3000 and rs . 7000 respectively in a business . if a doubles his capital after 6 months . in what ratio should a and b divide that year ' s profit ? | "( 3 * 6 + 6 * 6 ) : ( 7 * 12 ) 54 : 84 = > 9 : 14 . answer : c" | a = 3000 * 6
b = 3000 * 2
c = b * 6
d = a + c
e = 6 + 6
f = 7000 * e
g = d / f
|
['a ) 64', 'b ) 72', 'c ) 86', 'd ) 98', 'e ) 102'] | e | subtract(power(add(const_1, add(const_1, const_3)), const_3), 23) | cubes with each side one inch long are glued together to form a larger cube . the larger cube ' s face is painted with red color and the entire assembly is taken apart . 23 small cubes are found with no paints on them . how many of unit cubes have at least one face that is painted red ? | use the options . the options which after getting added to 23 shows a cube of a number could be right . here 64 + 23 = 87 72 + 23 = 95 86 + 23 = 109 98 + 23 = 121 102 + 23 = 125 - - - ( 5 * 5 * 5 ) so we have 102 as the answer ! e | a = 1 + 3
b = 1 + a
c = b ** 3
d = c - 23
|
a ) 0 % , b ) 20 % increase , c ) 20 % decrease , d ) 6 % decrease , e ) insufficient data | d | subtract(const_100, divide(multiply(add(const_100, 25), subtract(const_100, 25)), const_100)) | what is the % change in the area of a rectangle when its length increases by 25 % and its width decreases by 25 % ? | "( 125 / 10 ) * ( 75 / 10 ) = 9375 / 100 ~ 94 of original area 0.94 is a 6 % decrease from 100 / 100 - > d" | a = 100 + 25
b = 100 - 25
c = a * b
d = c / 100
e = 100 - d
|
a ) 1 , b ) 3 , c ) 4 , d ) 6 , e ) 14 | e | divide(70, add(multiply(3, 2), 4)) | bag a contains red , white and blue marbles such that the red to white marble ratio is 1 : 3 and the white to blue marble ratio is 2 : 3 . bag b contains red and white marbles in the ratio of 1 : 4 . together , the two bags contain 70 white marbles . how many red marbles could be in bag a ? | "6 is the answer . bag a - r : w : b = 2 : 6 : 9 let w in bag a be 6 k bab b - r : w = 1 : 4 let w in bag b be 4 k w = 70 = 6 k + 4 k = > k = 7 total red ' s in bag a will be 2 k = 14 e" | a = 3 * 2
b = a + 4
c = 70 / b
|
a ) 18 , b ) 20 , c ) 22 , d ) 24 , e ) 26 | d | divide(add(multiply(2, 30), 60), add(2, 3)) | sandy gets 3 marks for each correct sum and loses 2 marks for each incorrect sum . sandy attempts 30 sums and obtains 60 marks . how many sums did sandy get correct ? | "let x be the correct sums and ( 30 - x ) be the incorrect sums . 3 x - 2 ( 30 - x ) = 60 5 x = 120 x = 24 the answer is d ." | a = 2 * 30
b = a + 60
c = 2 + 3
d = b / c
|
a ) 30 % , b ) 35 % , c ) 40 % , d ) 45 % , e ) 50 % | a | multiply(divide(add(multiply(40, divide(20, const_100)), multiply(80, divide(35, const_100))), add(40, 80)), const_100) | a car dealership has 40 cars on the lot , 20 % of which are silver . if the dealership receives a new shipment of 80 cars , 35 % of which are not silver , what percentage of total number of cars are silver ? | "the number of silver cars is 0.2 * 40 + 0.35 * 80 = 36 the percentage of cars which are silver is 36 / 120 = 30 % the answer is a ." | a = 20 / 100
b = 40 * a
c = 35 / 100
d = 80 * c
e = b + d
f = 40 + 80
g = e / f
h = g * 100
|
a ) $ 0.50 , b ) $ 0.75 , c ) $ 1.25 , d ) $ 1.50 , e ) $ 1.75 | b | add(multiply(0.25, subtract(5, 1)), 0.25) | at a certain company , each employee has a salary grade s that is at least 1 and at most 5 . each employee receives an hourly wage p , in dollars , determined by the formula p = 9.50 + 0.25 ( s β 1 ) . an employee with a salary grade of 5 receives how many more dollars per hour than an employee with a salary grade of 2 ? | "salary grade of 5 is p ( 5 ) = 9.50 + 0.25 ( 5 β 1 ) = 9.50 + 0.25 * 4 ; salary grade of 2 is p ( 2 ) = 9.50 + 0.25 ( 2 β 1 ) = 9.50 + 0.25 ; p ( 5 ) - p ( 2 ) = 9.50 + 0.25 * 4 - 9.50 - 0.25 = 0.75 . answer : b ." | a = 5 - 1
b = 0 * 25
c = b + 0
|
a ) β 29 , b ) β 12 , c ) - 8 , d ) 29 , e ) 168 | c | divide(8, const_1) | what is the product of all the possible values of x if x ^ 2 - 2 x - 8 ? | "explanation : = > y = x ^ 2 - 2 x - 8 = > y = ( x + 2 ) ( x - 4 ) = > x = - 2 , y = 4 product x = ( - 2 ) ( 4 ) = - 8 answer option - 8 answer : c" | a = 8 / 1
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a ) 5 , b ) 8 , c ) 16 , d ) 18 , e ) 20 | c | add(subtract(multiply(30, 2), add(add(10, multiply(8, 2)), multiply(5, 4))), 4) | in a class of 30 students , 5 students did not borrow any books from the library , 10 students each borrowed 1 book , 8 students each borrowed 2 books , and the rest of the students each borrowed at least 4 books . if the average ( arithmetic mean ) number of books borrowed per student was 2 , what is the maximum number of books that any single student could have borrowed ? | "the average number of books per student was 2 means that total of 2 * 30 = 60 books were borrowed ; 5 + 12 + 8 = 25 students borrowed total of 5 * 0 + 12 * 1 + 8 * 2 = 28 books ; so 60 - 28 = 32 books are left to distribute among 30 - 25 = 5 students , these 5 arethe rest who borrowed at least 4 books ; tomaximizethe number of books one student from above 5 could have borrowed we shouldminimizethe number of books other 4 students from 5 could have borrowed . minimum these 4 students could have borrowed is 4 books per student , so total number of books they could have borrowed is 4 * 4 = 16 books . so the 5 th student could have borrowed is 32 - 16 = 16 books . answer : c ." | a = 30 * 2
b = 8 * 2
c = 10 + b
d = 5 * 4
e = c + d
f = a - e
g = f + 4
|
a ) 33.33 % , b ) 24.33 % , c ) 15.33 % , d ) 16.33 % , e ) 17.33 % | a | subtract(divide(multiply(const_100, 4), subtract(4, 1)), const_100) | in a office work is distribute between p persons . if 1 / 4 members are absent then work increased for each person is ? | "let total % of work is 100 % total person = p 1 / 4 person are absent of total person . so absent person is 1 / 4 p ie p / 4 . left person is , p - p / 4 = 3 p / 4 . p person do the work 100 % 1 person do the work 100 * p % 3 p / 4 person do the work ( 100 * p * 4 ) / 3 p % = 133.33 % work increased for each person is = ( 133.33 - 100 ) % = 33.33 % answer : a" | a = 100 * 4
b = 4 - 1
c = a / b
d = c - 100
|
a ) $ 200 , b ) $ 400 , c ) $ 600 , d ) $ 800 , e ) $ 1,200 | a | subtract(add(multiply(const_100, const_10), 800), divide(add(800, add(multiply(const_100, const_10), 800)), const_2)) | john has $ 1,200 at the beginning of his trip , after spending money , he still has exactly $ 800 less than he spent on the trip . how much money does john still have ? | "suppose total money spent = x not spend ( money he still has ) = x - 800 x + x - 800 = 1200 x = 1000 money not spend = 1000 - 800 = 200 answer : a" | a = 100 * 10
b = a + 800
c = 100 * 10
d = c + 800
e = 800 + d
f = e / 2
g = b - f
|
a ) 5 days , b ) 4 days , c ) 3 days , d ) 2 days , e ) 7 days | c | divide(const_1, subtract(subtract(const_0_25, divide(const_1, 12)), divide(const_1, 18))) | if a , b and c together can finish a piece of work in 2 days . a alone in 12 days and b in 18 days , then c alone can do it in ? | "c = 1 / 2 - 1 / 12 β 1 / 18 = 1 / 3 = > 3 days answer : c" | a = 1 / 12
b = const_0_25 - a
c = 1 / 18
d = b - c
e = 1 / d
|
a ) 2110 , b ) 3210 , c ) 4320 , d ) 5430 , e ) 6540 | c | divide(multiply(30, power(9, 3)), subtract(add(9, 6), 3)) | a tank with a volume of 30 cubic feet has one inlet pipe and 2 outlet pipes . the inlet pipe fills water into the tank at the rate of 3 cubic inches / min and the 2 outlet pipes empty it out at the rates of 9 cubic inches / min and 6 cubic inches / min respectively . if all 3 pipes are opened when the tank is full , how many minutes does it take to empty the tank ? ( 1 foot = 12 inches ) | "the tank is emptied at this rate : 9 + 6 - 3 = 12 cubic inches / min the tank has a volume of 30 * 12 * 12 * 12 = 51840 cubic inches . the time it takes to empty the tank is 51840 / 12 = 4320 minutes . the answer is c ." | a = 9 ** 3
b = 30 * a
c = 9 + 6
d = c - 3
e = b / d
|
a ) 33 , b ) 84 , c ) 40 , d ) 99 , e ) 11 | b | inverse(subtract(divide(const_1, 12), divide(const_1, add(12, const_2)))) | a cistern is normally filled in 12 hours but takes two hours longer to fill because of a leak in its bottom . if the cistern is full , the leak will empty it in ? | "1 / 12 - 1 / x = 1 / 14 x = 84 answer : b" | a = 1 / 12
b = 12 + 2
c = 1 / b
d = a - c
e = 1/(d)
|
a ) 400 , b ) 350 , c ) 700 , d ) 750 , e ) 800 | c | divide(28, subtract(154.04, add(const_100, add(multiply(const_4, const_10), const_2)))) | when positive integer n is divided by positive integer j , the remainder is 28 . if n / j = 154.04 , what is value of j ? | "when a number is divided by another number , we can represent it as : dividend = quotient * divisor + remainder so , dividend / divisor = quotient + remainder / divisor given that n / j = 154.04 here 154 is the quotient . given that remainder = 28 so , 154.04 = 154 + 28 / j so , j = 700 answer : c" | a = 4 * 10
b = a + 2
c = 100 + b
d = 154 - 4
e = 28 / d
|
a ) 80 litres , b ) 90 litres , c ) 120 litres , d ) 170 litres , e ) none of these | b | multiply(1200, multiply(800, divide(45, multiply(800, 600)))) | 45 litres of diesel is required to travel 600 km using a 800 cc engine . if the volume of diesel required to cover a distance varies directly as the capacity of the engine , then how many litres of diesel is required to travel 800 km using 1200 cc engine ? | "explanatory answer to cover a distance of 800 kms using a 800 cc engine , the amount of diesel required = 800 / 600 * 45 = 60 litres . however , the vehicle uses a 1200 cc engine and the question states that the amount of diesel required varies directly as the engine capacity . i . e . , for instance , if the capacity of engine doubles , the diesel requirement will double too . therefore , with a 1200 cc engine , quantity of diesel required = 1200 / 800 * 60 = 90 litres . answer b" | a = 800 * 600
b = 45 / a
c = 800 * b
d = 1200 * c
|
a ) 16.5 , b ) 17.5 , c ) 18.5 , d ) 19.5 , e ) 20.5 | e | add(divide(subtract(21, 20), const_2), 20) | christine and siri have 21 rs between them . christine has 20 rs more than siri . how much does christine have . | let siri has x rs , then christine has 20 + x rs given x + ( 20 + x ) = 21 = > 2 x = 1 or x = 0.5 so christine and siri has 20.5 and 0.5 rs respectively . answer : e | a = 21 - 20
b = a / 2
c = b + 20
|
a ) 37 , b ) 38 , c ) 39 , d ) 34 , e ) 41 | d | add(add(multiply(const_2, 10), 10), 4) | in 10 years , a will be twice as old as b was 10 years ago . if a is now 4 years older than b the present age of b is | let present age of a be a and b be b a + 10 = 2 * ( b - 10 ) = > 2 b - a = 30 . . . . . . ( i ) a = b + 4 = > 2 b - b - 4 = 30 b = 34 so the present age of b is 34 years answer : d | a = 2 * 10
b = a + 10
c = b + 4
|
a ) 347.4 , b ) 984.4 , c ) 877.4 , d ) 637.4 , e ) 667.4 | b | add(add(multiply(4, multiply(divide(24, 10), divide(multiply(23, 6), 2))), multiply(3, divide(multiply(23, 6), 2))), multiply(5, 23)) | the cost of 10 kg of mangos is equal to the cost of 24 kg of rice . the cost of 6 kg of flour equals the cost of 2 kg of rice . the cost of each kg of flour is $ 23 . find the total cost of 4 kg of mangos , 3 kg of rice and 5 kg of flour ? | let the costs of each kg of mangos and each kg of rice be $ a and $ r respectively . 10 a = 24 r and 6 * 23 = 2 r a = 12 / 5 r and r = 69 a = 165.6 required total cost = 4 * 165.6 + 3 * 69 + 5 * 23 = 662.4 + 207 + 115 = $ 984.40 b | a = 24 / 10
b = 23 * 6
c = b / 2
d = a * c
e = 4 * d
f = 23 * 6
g = f / 2
h = 3 * g
i = e + h
j = 5 * 23
k = i + j
|
a ) 171 , b ) 281 , c ) 391 , d ) 491 , e ) 611 | e | add(multiply(divide(add(50, 60), const_2), add(subtract(60, 50), const_1)), add(divide(subtract(60, 50), const_2), const_1)) | if x is equal to the sum of the integers from 50 to 60 , inclusive , and y is the number of even integers from 50 to 60 , inclusive , what is the value of x + y ? | "sum s = n / 2 { 2 a + ( n - 1 ) d } = 11 / 2 { 2 * 50 + ( 11 - 1 ) * 1 } = 11 * 55 = 605 = x number of even number = ( 60 - 50 ) / 2 + 1 = 6 = y x + y = 605 + 6 = 611 e" | a = 50 + 60
b = a / 2
c = 60 - 50
d = c + 1
e = b * d
f = 60 - 50
g = f / 2
h = g + 1
i = e + h
|
a ) 12.5 % , b ) 85.7 % , c ) 80 % , d ) 11 % , e ) 1 % | b | multiply(divide(subtract(7, const_1), 7), const_100) | a number x is 7 times another number y . the percentage that y is less than x is | "say y = 1 and x = 7 . then y = 1 is less than x = 7 by ( 7 - 1 ) / 7 * 100 = 6 / 7 * 100 = 85.7 % . answer : b ." | a = 7 - 1
b = a / 7
c = b * 100
|
a ) 15 , b ) 20 , c ) 12 , d ) 14 , e ) 7 | c | subtract(divide(multiply(48, subtract(40, 25)), const_12), 48) | an alloy weighing 48 ounces is 25 % gold . how many ounces of pure gold must be added to create an alloy that is 40 % gold ? | an alloy of 48 oz which is 25 % gold means there is 12 oz of gold . to get to an alloy that is 40 % gold , let ' s use this expression : ( 12 + x ) / ( 48 + x ) = 0.40 with x representing the amount of pure gold that must be added to get to 40 % . the expression we are using represents the new total weight of pure gold over the new total weight of the alloy and this fraction should represent 40 % or 0.4 . you will see that 12 is the correct answer , as 24 / 60 = 0.4 choose c | a = 40 - 25
b = 48 * a
c = b / 12
d = c - 48
|
a ) s . 5000 , b ) s . 7000 , c ) s . 14000 , d ) s . 17000 , e ) s . 27000 | b | divide(420, divide(multiply(subtract(15, 12), const_2), const_100)) | a certain sum is invested at simple interest at 15 % p . a . for two years instead of investing at 12 % p . a . for the same time period . therefore the interest received is more by rs . 420 . find the sum ? | "let the sum be rs . x . ( x * 15 * 2 ) / 100 - ( x * 12 * 2 ) / 100 = 420 = > 30 x / 100 - 24 x / 100 = 420 = > 6 x / 100 = 420 = > x = 7000 . answer : b" | a = 15 - 12
b = a * 2
c = b / 100
d = 420 / c
|
a ) 343 , b ) 393 , c ) 363 , d ) 391 , e ) 409 | d | multiply(divide(subtract(multiply(7, 20), multiply(const_3, 7)), 7), add(subtract(7, const_1), divide(subtract(multiply(7, 20), multiply(const_3, 7)), 7))) | if the average ( arithmetic mean ) of 7 consecutive integers is 20 , then the product of the greatest and least integer is | n = smallest number n + 6 = largest number ( n + n + 6 ) / 2 = 20 = > ( 2 n + 6 ) / 2 = 20 = > n + 3 = 20 = > n = 17 so product of n and n + 6 = ( 17 ) ( 23 ) = 391 answer - d | a = 7 * 20
b = 3 * 7
c = a - b
d = c / 7
e = 7 - 1
f = 7 * 20
g = 3 * 7
h = f - g
i = h / 7
j = e + i
k = d * j
|
a ) 270 m , b ) 255 m , c ) 235 m , d ) 220 m , e ) 240 m | b | subtract(multiply(multiply(45, const_0_2778), 30), 120) | a train , 120 meters long travels at a speed of 45 km / hr crosses a bridge in 30 seconds . the length of the bridge is | "explanation : assume the length of the bridge = x meter total distance covered = 120 + x meter total time taken = 30 s speed = total distance covered / total time taken = ( 120 + x ) / 30 m / s = > 45 Γ£ β ( 10 / 36 ) = ( 120 + x ) / 30 = > 45 Γ£ β 10 Γ£ β 30 / 36 = 120 + x = > 45 Γ£ β 10 Γ£ β 10 / 12 = 120 + x = > 15 Γ£ β 10 Γ£ β 10 / 4 = 120 + x = > 15 Γ£ β 25 = 120 + x = 375 = > x = 375 - 120 = 255 answer : option b" | a = 45 * const_0_2778
b = a * 30
c = b - 120
|
a ) 3 : 2 , b ) 2 : 1 , c ) 1 : 2 , d ) 4 : 5 , e ) 2 : 3 | e | multiply(divide(2, 3), multiply(divide(2, 3), divide(6, 3))) | find the compound ratio of ( 2 : 3 ) , ( 6 : 11 ) and ( 11 : 6 ) is | "required ratio = 2 / 3 * 6 / 11 * 11 / 6 = 2 / 1 = 2 : 3 answer is e" | a = 2 / 3
b = 2 / 3
c = 6 / 3
d = b * c
e = a * d
|
a ) 17 , b ) 16 , c ) 15 , d ) 14 , e ) 13 | b | add(add(add(const_4, const_3), add(const_3, const_2)), 4) | the number 90 can be written as the sum of the squares of 4 different positive integers . what is the sum of these 4 integers ? | "4 ^ 2 + 3 ^ 2 + 8 ^ 2 + 1 ^ 2 = 90 - - > 1 + 3 + 8 + 4 = 16 . b" | a = 4 + 3
b = 3 + 2
c = a + b
d = c + 4
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a ) 3576 s , b ) 3589 s , c ) 3596 s , d ) 3598 s , e ) 3576 s | e | subtract(divide(multiply(subtract(divide(add(36, 24), subtract(36, 24)), const_1), multiply(multiply(const_10, const_3), const_60)), const_2), 24) | a train with 120 wagons crosses john who is going in the same direction , in 36 seconds . it travels for half an hour from the time it starts overtaking the john ( he is riding on the horse ) before it starts overtaking the mike ( who is also riding on his horse ) coming from the opposite direction in 24 seconds . in how much time ( in secs ) after the train has crossed the mike do the john meets to mike ? | "let the length of the train be l metres and speeds of the train arun and sriram be r , a and s respectively , then - - - - - - - - - - ( i ) and - - - - - - - - - ( ii ) from eq . ( i ) and ( ii ) 3 ( r - a ) = 2 ( r + k ) r = 3 a + 2 k in 30 minutes ( i . e 1800 seconds ) , the train covers 1800 r ( distance ) but the arun also covers 1800 a ( distance ) in the same time . therefore distance between arun and sriram , when the train has just crossed sriram = 1800 ( r - a ) - 24 ( a + k ) time required = = ( 3600 - 24 ) = 3576 s e" | a = 36 + 24
b = 36 - 24
c = a / b
d = c - 1
e = 10 * 3
f = e * const_60
g = d * f
h = g / 2
i = h - 24
|
a ) 10.9 sec , b ) 5.76 sec , c ) 10.6 sec , d ) 10.8 sec , e ) 20.8 sec | b | divide(add(100, 60), multiply(add(60, 40), const_0_2778)) | two trains 100 m and 60 m long run at the speed of 60 km / hr and 40 km / hr respectively in opposite directions on parallel tracks . the time which they take to cross each other is ? | "relative speed = 60 + 40 = 100 km / hr . = 100 * 5 / 18 = 250 / 9 m / sec . distance covered in crossing each other = 100 + 60 = 160 m . required time = 160 * 9 / 250 = 5.76 = 5.76 sec . answer : b" | a = 100 + 60
b = 60 + 40
c = b * const_0_2778
d = a / c
|
a ) $ 11.73 , b ) $ 12.00 , c ) $ 13.80 , d ) $ 14.00 , e ) $ 15.87 | b | multiply(multiply(divide(207.00, add(const_100, 15)), const_100), divide(const_1, 15)) | the price of lunch for 15 people was $ 207.00 , including a 15 percent gratuity for service . what was the average price per person , excluding the gratuity ? | "if the amount spent excluding the gratuity was x , then x ( 1.15 ) = 207 = > x = 207 / 1.15 = 180 then the average price per person = 180 / 15 = 12 option ( b )" | a = 100 + 15
b = 207 / 0
c = b * 100
d = 1 / 15
e = c * d
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a ) 16 , b ) 18 , c ) 20 , d ) 22 , e ) 24 | e | divide(0.12, divide(0.50, const_100)) | find the missing figures : 0.50 % of ? = 0.12 | "let 0.50 % of x = 0.12 . then , 0.50 * x / 100 = 0.12 x = [ ( 0.12 * 100 ) / 0.50 ] = 24 . answer is e ." | a = 0 / 50
b = 0 / 12
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a ) 2 , b ) 8 , c ) 9 , d ) 11 , e ) 15 | b | divide(add(220, 43), 17) | a no . when divided by 220 gives a remainder 43 , what remainder will beobtained by dividingthe same no . 17 ? | "220 + 43 = 263 / 17 = 8 ( remainder ) b" | a = 220 + 43
b = a / 17
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a ) 14 , b ) 20 , c ) 22 , d ) 24 , e ) 25 | b | divide(subtract(16000, subtract(subtract(subtract(60000, 18000), 16000), 12000)), const_100) | shipment - - - - - - no . of defective chips / shipment - - - total chips in shipment s 2 - - - - - - - - - - - - - - 5 - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - 12000 s 3 - - - - - - - - - - - - - - 6 - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - 18000 s 4 - - - - - - - - - - - - - - 4 - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - 16000 a computer chip manufacturer expects the ratio of the number of defective chips to be total number of chips in all future shipments equal to the corresponding ratio for shipmemts s 1 , s 2 , s 3 and s 4 comined as shown in the table above . what is the expected number q of defective chips in a shipment of 60000 chips ? | i agree with your solution = 20 . but the question is : there are different combination to get 60000 chips . for example : 1 * s 3 + 2 * s 4 + 2 * s 2 . in this way , we ship 60000 chips with only 6 + 4 * 2 + 2 * 2 = 18 defective chips , better than the average of 20 . the question is to find the expected number q of defective chips , i guess it assume the minimum # , therefore it might not be 20 . | a = 60000 - 18000
b = a - 16000
c = b - 12000
d = 16000 - c
e = d / 100
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a ) a ) 1 , b ) b ) 8 , c ) c ) 17 / 5 , d ) d ) 18 / 5 , e ) e ) 4 | b | divide(add(divide(subtract(multiply(7, 2), 5), subtract(multiply(2, 2), const_1)), subtract(7, multiply(2, divide(subtract(multiply(7, 2), 5), subtract(multiply(2, 2), const_1))))), 8) | if 2 x + y = 7 and x + 2 y = 5 , then 8 xy / 3 = ? | "2 * ( x + 2 y = 5 ) equals 2 x + 4 y = 10 2 x + 4 y = 10 - 2 x + y = 7 = 3 y = 3 therefore y = 1 plug and solve . . . 2 x + 1 = 7 2 x = 6 x = 3 ( 8 * 3 * 1 ) / 3 = 24 / 3 = 8 b" | a = 7 * 2
b = a - 5
c = 2 * 2
d = c - 1
e = b / d
f = 7 * 2
g = f - 5
h = 2 * 2
i = h - 1
j = g / i
k = 2 * j
l = 7 - k
m = e + l
n = m / 8
|
a ) 40 , b ) 45 , c ) 50 , d ) 55 , e ) 60 | a | gcd(640, 520) | the maximum number of students among them 640 pens and 520 pencils can be distributed in such a way that each student get the same number of pens and same number of pencils ? | "number of pens = 640 number of pencils = 520 required number of students = h . c . f . of 640 and 520 = 40 answer is a" | a = math.gcd(640, 520)
|
a ) 9 , b ) 10 , c ) 11 , d ) 12 , e ) 13 | d | divide(80, add(3, 4)) | a certain airline ' s fleet consisted of 80 type a planes at the beginning of 1980 . at the end of each year , starting with 1980 , the airline retired 3 of the type a planes and acquired 4 new type b plans . how many years did it take before the number of type a planes left in the airline ' s fleet was less than 50 percent of the fleet ? | "let x be the number of years . 4 x > 80 - 3 x 7 x > 80 x > 11 + 3 / 7 the answer is d ." | a = 3 + 4
b = 80 / a
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a ) 1 / 50 , b ) 1 / 25 , c ) 1 / 96 , d ) 1 , e ) 2 | c | divide(1, 96) | if the numbers 1 to 96 are written on 96 pieces of paper , ( one on each ) and one piece is picked at random , then what is the probability that the number drawn is neither prime nor composite ? | there are 25 primes , 70 composite numbers from 1 to 96 . the number which is neither prime nor composite is 1 . therefore , required probability = 1 / 96 . answer : c | a = 1 / 96
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a ) 32 meters east , b ) 32 meters north , c ) 32 meters west , d ) 42 meters east , e ) 52 meters west | a | add(25, 12) | sandy walked 25 meters towards south . then sandy turned to her left and walked 20 meters . she then turned to her left and walked 25 meters . she then turned to her right and walked 12 meters . what distance is she from the starting point and in which direction ? | "the net distance is 20 + 12 = 32 meters to the east . the answer is a ." | a = 25 + 12
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a ) 35 , b ) 40 , c ) 42 , d ) 45 , e ) 48 | d | subtract(50, multiply(divide(50, const_100), 10)) | how many liters of water must be evaporated from 50 liters of a 1 - percent sugar solution to get a 10 - percent solution ? | 1 % of a 50 liter solution is 0.5 l which is 10 % of the solution at the end . the solution at the end must be 5 l . we need to evaporate 45 liters . the answer is d . | a = 50 / 100
b = a * 10
c = 50 - b
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a ) 129 , b ) 287 , c ) 190 , d ) 188 , e ) 112 | c | add(divide(add(multiply(27, 150), multiply(36, 125)), add(36, 27)), multiply(divide(add(multiply(27, 150), multiply(36, 125)), add(36, 27)), divide(40, const_100))) | raman mixed 27 kg of butter at rs . 150 per kg with 36 kg butter at the rate of rs . 125 per kg . at what price per kg should he sell the mixture to make a profit of 40 % in the transaction ? | "explanation : cp per kg of mixture = [ 27 ( 150 ) + 36 ( 125 ) ] / ( 27 + 36 ) = rs . 135.71 sp = cp [ ( 100 + profit % ) / 100 ] = 135.71 * [ ( 100 + 40 ) / 100 ] = rs . 190 . answer : c" | a = 27 * 150
b = 36 * 125
c = a + b
d = 36 + 27
e = c / d
f = 27 * 150
g = 36 * 125
h = f + g
i = 36 + 27
j = h / i
k = 40 / 100
l = j * k
m = e + l
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a ) 22.35 , b ) 33.25 , c ) 22.25 , d ) 51.35 , e ) 56.25 | e | divide(subtract(multiply(50, 56), add(45, 55)), subtract(50, const_2)) | the average of 50 numbers id 56 . if two numbers , namely 45 and 55 are discarded , the average of the remaining numbers is : | explanation : total of 50 numbers = ( 50 Γ 56 ) = 2800 total of 48 numbers = ( 2800 - ( 45 + 55 ) ] = 2700 required average = 2700 / 48 = 56.25 answer : e | a = 50 * 56
b = 45 + 55
c = a - b
d = 50 - 2
e = c / d
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a ) 0 , b ) 1 , c ) 2 , d ) 3 , e ) 9 | e | subtract(12, 14) | a set consists of 12 numbers , all are even or multiple of 5 . if 7 numbers are even and 14 numbers are multiple of 5 , how many numbers is multiple of 10 ? | "{ total } = { even } + { multiple of 5 } - { both } + { nether } . since { neither } = 0 ( allare even or multiple of 5 ) then : 12 = 7 + 14 - { both } + 0 ; { both } = 9 ( so 1 number is both even and multiple of 5 , so it must be a multiple of 10 ) . answer : e ." | a = 12 - 14
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a ) 17 , b ) 19 , c ) 21 , d ) 13 , e ) 18 | a | sqrt(add(149, multiply(70, const_2))) | sum of the squares of 3 no . is 149 and the sum of their products taken two at a time is 70 . find the sum ? | "( a + b + c ) 2 = a 2 + b 2 + c 2 + 2 ( ab + bc + ca ) = 149 + 2 * 70 a + b + c = Γ’ Λ Ε‘ 289 = 17 answer a" | a = 70 * 2
b = 149 + a
c = math.sqrt(b)
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a ) 0 , b ) 1 , c ) 2 , d ) 3 , e ) 4 | c | divide(add(2, const_1), 5) | if x and y are positive integers and x = 5 y + 2 , what is the remainder when x is divided by 5 ? | "this question asks what is . . . ( the answer ) , so we know that the answer will be consistent . as such , we can test values to quickly get the solution . we ' re told that x and y are positive integers and x = 5 y + 2 . we ' re asked for the remainder when x is divided by 5 . if . . . . y = 1 x = 7 7 / 5 = 1 remainder 2 final answer : c" | a = 2 + 1
b = a / 5
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a ) 106 , b ) 49 , c ) 68 , d ) 38 , e ) 55 | a | add(multiply(9, 11), 7) | if , 1 * 3 * 5 = 16 3 * 5 * 7 = 38 5 * 7 * 9 = 68 then find , 7 * 9 * 11 = ? | a 106 ( 11 * 9 ) + 7 = 106 | a = 9 * 11
b = a + 7
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a ) 25 days , b ) 88 days , c ) 21 days , d ) 11 days , e ) 13 days | c | multiply(divide(multiply(7, add(const_2, const_1)), const_2), const_2) | a is twice as good a workman as b and they took 7 days together to do the work b alone can do it in ? | "wc = 2 : 1 2 x + x = 1 / 7 x = 1 / 21 = > 21 days answer : c" | a = 2 + 1
b = 7 * a
c = b / 2
d = c * 2
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a ) 1 , b ) 2 , c ) 3 , d ) 4 , e ) 5 | d | subtract(4, 2) | points a , b , c , and d , in that order , lie on a line . if ab = 2 cm , ac = 4 cm , and bd = 6 cm , what is cd , in centimeters ? | "putting a value to each point , lets use the following : a - 0 b - 2 ( ab = 2 ) c - 4 ( ac = 4 ) d - 8 ( bd = 6 ) cd is 8 - 4 = 4 . ans d" | a = 4 - 2
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a ) s . 650 , b ) s . 690 , c ) s . 698 , d ) s . 728 , e ) s . 760 | d | subtract(815, divide(multiply(subtract(844, 815), 3), 4)) | a sum of money at simple interest amounts to rs . 815 in 3 years and to rs . 844 in 4 years . the sum is : | "s . i . for 1 year = rs . ( 844 - 815 ) = rs . 29 . s . i . for 3 years = rs . ( 29 x 3 ) = rs . 87 . principal = rs . ( 815 - 87 ) = rs . 728 . answer : option d" | a = 844 - 815
b = a * 3
c = b / 4
d = 815 - c
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a ) 191 , b ) 400 , c ) 737 , d ) 840 , e ) 1,560 | b | divide(multiply(240, const_100), subtract(const_100, 40)) | a side of beef lost 40 percent of its weight in processing . if the side of beef weighed 240 pounds after processing , how many pounds did it weigh before processing ? | "let weight of side of beef before processing = x ( 60 / 100 ) * x = 240 = > x = ( 240 * 100 ) / 60 = 400 answer b" | a = 240 * 100
b = 100 - 40
c = a / b
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a ) 18 , b ) 20 , c ) 17 , d ) 24 , e ) 21 | d | divide(multiply(18, 36), 27) | 36 men can complete a piece of work in 18 days . in how days will 27 men can complete the work ? | "let the required number of days be x less men , more days ( indirect proportion ) 27 : 36 : : 18 : x 27 x = 36 * 18 x = ( 36 * 18 ) / 27 x = 24 days option is d" | a = 18 * 36
b = a / 27
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a ) 17 % , b ) 62 % , c ) 12 % , d ) 19 % , e ) 71 % | e | multiply(subtract(divide(divide(multiply(subtract(const_100, 50), add(const_100, 80)), const_100), const_100), const_1), const_100) | a trader bought a car at 50 % discount on its original price . he sold it at a 80 % increase on the price he bought it . what percent of profit did he make on the original price ? | "original price = 100 cp = 95 s = 95 * ( 180 / 100 ) = 112 100 - 171 = 71 % answer : e" | a = 100 - 50
b = 100 + 80
c = a * b
d = c / 100
e = d / 100
f = e - 1
g = f * 100
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a ) 4.8 , b ) 1.2 , c ) 4.4 , d ) 4.6 , e ) 4.7 | a | inverse(subtract(divide(const_1, 3), divide(const_1, 8))) | renu can do a piece of work in 8 days , but with the help of her friend suma , she can do it in 3 days . in what time suma can do it alone ? | "renu Γ’ β¬ β’ s one day Γ’ β¬ β’ s work = 1 / 8 suma Γ’ β¬ β’ s one day Γ’ β¬ β’ s work = 1 / 3 - 1 / 8 = 5 / 24 suma can do it alone in 4.8 days . answer : a" | a = 1 / 3
b = 1 / 8
c = a - b
d = 1/(c)
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a ) 800 , b ) 550 , c ) 555 , d ) 600 , e ) 605 | a | divide(multiply(35, 65), const_4) | what is the sum of the odd integers from 35 to 65 , inclusive ? | "the mean is 50 . sum = mean ( # of elements ) there are 16 odd numbers between 35 - 65 inclusive . 16 * 50 = 800 a" | a = 35 * 65
b = a / 4
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a ) 20 min , b ) 8 min , c ) 9 min , d ) 10 min , e ) 11 min | a | multiply(const_60, divide(subtract(54, 36), 54)) | excluding stoppages , the speed of a bus is 54 kmph and including stoppages , it is 36 kmph . for how many minutes does the bus stop per hour ? | "due to stoppages , it covers 18 km less . time taken to cover 18 km = ( 18 / 54 ) x 60 = 20 min answer : a" | a = 54 - 36
b = a / 54
c = const_60 * b
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a ) 2.7 , b ) 2.9 , c ) 3.1 , d ) 3.3 , e ) 3.5 | a | inverse(add(divide(const_1, 6), divide(const_1, 5))) | a worker can load one truck in 6 hours . a second worker can load the same truck in 5 hours . if both workers load one truck simultaneously while maintaining their constant rates , approximately how long , in hours , will it take them to fill one truck ? | "the workers fill the truck at a rate of 1 / 6 + 1 / 5 = 11 / 30 of the truck per hour . then the time to fill one truck is 30 / 11 which is about 2.7 hours . the answer is a ." | a = 1 / 6
b = 1 / 5
c = a + b
d = 1/(c)
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a ) 1.1 , b ) - 11 , c ) 11 / 11 , d ) 11 / 1 , e ) 1 / 11 | d | divide(divide(divide(divide(divide(divide(11, const_4), const_3), const_4), const_3), const_2), const_2) | which is equal to 11 ? | "11 / 1 is equal to 11 . answer : d" | a = 11 / 4
b = a / 3
c = b / 4
d = c / 3
e = d / 2
f = e / 2
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a ) 50 kmph , b ) 51 kmph , c ) 41 kmph , d ) 61 kmph , e ) 21 kmph | b | multiply(divide(30, const_60), 102) | the speed of a train is 102 kmph . what is the distance covered by it in 30 minutes ? | "102 * 30 / 60 = 51 kmph answer : b" | a = 30 / const_60
b = a * 102
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['a ) 24', 'b ) 28', 'c ) 36', 'd ) 48', 'e ) none of these'] | a | multiply(add(8, 4), const_2) | a rectangular of certain dimensions is chopped off from one corner of a larger rectangle as shown . ab = 8 cm and bc = 4 cm . the perimeter of the figure abcpqra ( in cm ) is : | solution required perimeter = ( ab + bc + cp + pq + qr + ra ) = ab + bc + ( cp + qr ) + ( pq + ra ) = ab + bc + ab + bc = 2 ( ab + bc ) = [ 2 ( 8 + 4 ) ] cm = 24 cm . answer a | a = 8 + 4
b = a * 2
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a ) 10 , b ) 6 , c ) 4 , d ) 7 , e ) 5 | e | divide(multiply(20, 10), 40) | 20 machines can do a work in 10 days . how many machines are needed to complete the work in 40 days ? | "required number of machines = 20 * 10 / 40 = 5 answer is e" | a = 20 * 10
b = a / 40
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a ) $ 880 , b ) $ 990 , c ) $ 1,000 , d ) $ 1,100 , e ) $ 1,210 | b | subtract(multiply(110, divide(const_100, 10)), 110) | if a 10 percent deposit that has been paid toward the purchase of a certain product is $ 110 , how much more remains to be paid ? | "10 / 100 p = 110 > > p = 110 * 100 / 10 = 1100 1100 - 110 = 990 answer : b" | a = 100 / 10
b = 110 * a
c = b - 110
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['a ) 1 : 9', 'b ) 1 : 3', 'c ) 3 : 1', 'd ) 1 : 10', 'e ) can not be determined'] | a | multiply(divide(const_1, 3), divide(const_1, 3)) | in a trapezium abcd , ab is parallel to dc , ab = 3 * dc , and the diagonals of the trapezium intersect at o . the ratio of the area of β ocd to the area of β oab is : | triangles are similar , so ratio will be 1 : 9 a | a = 1 / 3
b = 1 / 3
c = a * b
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a ) 0 , b ) 50 , c ) 450 , d ) 495 , e ) 694 | e | subtract(multiply(multiply(add(const_3, const_4), const_1000), divide(1, 10)), multiply(divide(divide(1, 10), const_100), multiply(add(const_3, const_4), const_1000))) | when 1 / 10 percent of 6,000 is subtracted from 1 / 10 of 6,000 , the difference is | "we can break this problem into two parts : 1 ) what is 1 / 10 percent of 6,000 ? 2 ) what is 1 / 10 of 6,000 ? to calculate 1 / 10 percent of 6,000 we must first remember to divide 1 / 10 by 100 . so we have : ( 1 / 10 ) / ( 100 ) to divide a number by 100 means to multiply it by 1 / 100 , so we have : 1 / 10 x 1 / 100 = 1 / 1,000 thus , 1 / 10 percent of 6,000 = 1 / 1,000 x 6,000 = 6 . now let ' s concentrate on part 2 . we need to calculate 1 / 10 of 5,000 . to do this we simply multiply 1 / 10 by 6,000 . 1 / 10 x 6,000 = 600 the answer to part 1 is 6 , and the answer to part 2 is 600 . their difference is 600 β 6 = 694 . answer e ." | a = 3 + 4
b = a * 1000
c = 1 / 10
d = b * c
e = 1 / 10
f = e / 100
g = 3 + 4
h = g * 1000
i = f * h
j = d - i
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a ) 17.14286 min , b ) 62 min , c ) 70.1346 min , d ) 74 min , e ) 76.1346 min | a | multiply(multiply(const_2, divide(multiply(3, const_60), add(subtract(200, 10), multiply(const_2, 10)))), 10) | if there are 200 questions in a 3 hr examination . among these questions are 10 type a problems , which requires twice as much as time be spent than the rest of the type b problems . how many minutes should be spent on type a problems ? | x = time for type b prolems 2 x = time for type a problem total time = 3 hrs = 180 min 190 x + 10 * 2 x = 180 x = 180 / 210 x = 0.72 time taken for type a problem = 10 * 2 * 0.857143 = 17.14286 min answer : a | a = 3 * const_60
b = 200 - 10
c = 2 * 10
d = b + c
e = a / d
f = 2 * e
g = f * 10
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a ) 1 / 2 , b ) 2 / 5 , c ) 3 / 4 , d ) 1 / 4 , e ) 1 / 5 | b | divide(4, const_10) | working individually , jane can bake cakes for 4 hours and roy can bake cakes in 5 hours . if jane and roy work together but independently at the task for 2 hours , at which point roy leaves , how many remaining hours will it take jane to complete the task alone ? | in first 2 hrs roy will finish 2 / 5 = 2 / 5 of work and jane will finish 2 / 4 work so total 2 / 5 + 1 / 2 = 9 / 10 work is finished and 1 - 9 / 10 = 1 / 10 work remaining . now jane will take ( 1 / 10 ) * 4 = 2 / 5 hrs to finish it . so answer is b . | a = 4 / 10
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a ) 0.54 , b ) 0.77 , c ) 0.82 , d ) 0.97 , e ) 1.62 | b | divide(2.22, const_3) | the arithmetic mean of set x is zero . if y = { - 2.22 ; - 1.96 ; - 1.68 ; 1.62 ; 1.94 ; 2.16 } is the subset of x consisting of all those elements in x which are more than two but less than 3 standard deviations away from the arithmetic mean of x , what could be equal to the standard deviation of x ? | this is a fairly straightforward question that can be solved quickly by just applying the options we are provided with mean , m = 0 and y = { - 2.22 ; - 1.96 ; - 1.68 ; 1.62 ; 1.94 ; 2.16 } is the subset of s y consists of all those elements in x that are more than 2 but less than 3 xds away from the arithmetic mean of x if an element is 1 xd away from the mean , we can write it as either m + xd or m - xd similarly , if an element is 2 xds away from the mean , we can write it as either m + 2 * xd or m - 2 * xd so , if these elements lie within 2 and 3 xds of mean , m = 0 we can find which one of these values of xd satisfies each value within y only xd = 0.77 does answer : b | a = 2 / 22
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a ) 1,705 , b ) 1,500 , c ) 1,240 , d ) 1,120 , e ) 1,100 | a | divide(85, 85) | what is the sum of odd integers from 25 to 85 , inclusive ? | number of odd integers = ( 85 - 25 ) / 2 + 1 = 60 / 2 + 1 = 31 sum of odd integers = ( 25 + 85 ) / 2 * 31 = 1705 answer a | a = 85 / 85
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a ) $ 150 , b ) $ 248.75 , c ) $ 200 , d ) $ 171.6 , e ) $ 190 | b | floor(multiply(25, 9.95)) | carrie likes to buy t - shirts at the local clothing store . they cost $ 9.95 each . one day , she bought 25 t - shirts . how much money did she spend ? | $ 9.95 * 25 = $ 248.75 . answer is b . | a = 25 * 9
b = math.floor(a)
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a ) - 500 , b ) - 1000 , c ) - 999 , d ) - 1001 , e ) 500500 | a | divide(1000, 2) | evaluate the sum 1 - 2 + 3 - 4 + 5 - 6 + . . . + 997 - 998 + 999 - 1000 | one groups the first two terms and each successive two terms to obtain a sum of 500 expressions each of which is - 1 . the answer is - 500 correct answer a | a = 1000 / 2
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a ) ( 40,0 ) , b ) ( 30,0 ) , c ) ( 0,40 ) , d ) ( 40,30 ) , e ) ( 0,30 ) | a | multiply(negate(divide(subtract(negate(39), multiply(negate(12), divide(3, 4))), divide(3, 4))), const_10) | a line has a slope of 3 / 4 and intersects the point t ( - 12 , - 39 ) . at which point does this line intersect the x - axis ? | "assume that the equation of the line is y = mx + c , where m and c are the slope and y - intercept . you are also given that the line crosses the point ( - 12 , - 39 ) , this means that this point will also lie on the line above . thus you get - 39 = m * ( - 12 ) + c , with m = 3 / 4 as the slope is given to be 3 / 4 . after substituting the above values , you get c = - 30 . thus the equation of the line is y = 0.75 * x - 30 and the point where it will intersect the x - axis will be with y coordinate = 0 . put y = 0 in the above equation of the line and you will get , x = 40 . thus , the point t of intersection is ( 40,0 ) . a is the correct answer ." | a = negate - (
b = negate * (
c = 3 / 4
d = a / b
e = negate * (
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a ) rs . 6725 , b ) rs . 6727 , c ) rs . 6908 , d ) rs . 6725 , e ) rs . 6947.5 | e | divide(8337, add(const_1, divide(20, const_100))) | the owner of a furniture shop charges his customer 20 % more than the cost price . if a customer paid rs . 8337 for a computer table , then what was the cost price of the computer table ? | "cp = sp * ( 100 / ( 100 + profit % ) ) = 8337 ( 100 / 120 ) = rs . 6947.5 answer : e" | a = 20 / 100
b = 1 + a
c = 8337 / b
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a ) 3.75 hours , b ) 5.45 hours , c ) 7.10 hours , d ) 4.75 hours , e ) 3.33 hours | a | divide(subtract(4, const_1), subtract(multiply(divide(const_1, 4), 4), divide(const_1, 5))) | two drivers began their journey with the same amount of petrol in their cars at the same time . the only difference is that the first driver οΏ½ s car could drive 4 hours in that amount of petrol and the second one could drive 5 hours . however , they only drove for some time and found that the amount of petrol that was left in one of the cars was 4 times the petrol left in the other one . for how long had they driven at this point of time ? | a 3.75 hours while you can solve it as you like , a simple mathematical equation can be used to find out . let m be the amount of petrol initially . let n be the time for which they drove . according to the question , the amount of petrol used by first car in n hours = mn / 4 the amount of petrol used by second car in n hours = mn / 5 hence , the amount of petrol left in the first car = ( m - mn / 4 ) the amount of petrol left in the second car = ( m - mn / 5 ) as per the details given in the question , we can form the below equation : m - mn / 5 = 4 ( m - mn / 4 ) n = 15 / 4 or 3.75 hours . hence , both the drivers have driven the car for 3.75 hours at that particular time . | a = 4 - 1
b = 1 / 4
c = b * 4
d = 1 / 5
e = c - d
f = a / e
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a ) 30 , 19 , b ) 21 , 37 , c ) 15 , 34 , d ) 15 , 27 , e ) none of these | d | sqrt(divide(5, add(power(5, 9), add(power(5, const_2), power(9, 9))))) | the ratio of two numbers is 5 : 9 . if each number is decreased by 5 , the ratio becomes 5 : 11 . find the numbers . | "explanation : let the two numbers be 5 x and 9 x . ( 5 x - 5 ) / ( 9 x - 5 ) = 5 : 11 ( 5 x - 5 ) * 11 = ( 9 x - 5 ) * 5 55 x β 55 = 45 x β 25 10 x = 30 x = 3 therefore , the numbers are 15 and 27 . answer : d" | a = 5 ** 9
b = 5 ** 2
c = 9 ** 9
d = b + c
e = a + d
f = 5 / e
g = math.sqrt(f)
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a ) 80 , b ) 190 , c ) 132 , d ) 170 , e ) 60 | c | multiply(add(11, const_1), 11) | tim came second in math . when his mother asked him how much he had scored , he answered that he got the sum of the first 11 even numbers . his mother immediately worked out the answer . how much had he scored in math ? | "c 132 sum = ( n x n ) + n hence , 11 x 11 = 121 + 11 = 132" | a = 11 + 1
b = a * 11
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a ) 142.3 cms , b ) 162.3 cms , c ) 160.5 cms , d ) 142.8 cms , e ) 159.8 cms | e | divide(add(multiply(161, 12), multiply(158, const_10)), 20) | the average height of 12 girls out of a class of 20 is 161 cm . and that of the remaining girls is 158 cm . the average height of the whole class is : | "explanation : average height of the whole class = ( 12 Γ 161 + 8 Γ 158 / 20 ) = 159.8 cms answer e" | a = 161 * 12
b = 158 * 10
c = a + b
d = c / 20
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a ) 550.87 , b ) 506.07 , c ) 506.04 , d ) 506.03 , e ) 306.01 | a | subtract(multiply(9000, power(add(1, divide(4, const_100)), divide(const_3, 2))), 9000) | find out the c . i on rs . 9000 at 4 % p . a . compound half - yearly for 1 1 / 2 years | "a = 9000 ( 51 / 50 ) 3 = 9550.87 9000 - - - - - - - - - - - 550.87 answer : a" | a = 4 / 100
b = 1 + a
c = 3 / 2
d = b ** c
e = 9000 * d
f = e - 9000
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a ) 432 , b ) 212 , c ) 252 , d ) 992 , e ) 147 | e | multiply(multiply(divide(56, add(multiply(const_3, const_2), multiply(const_1, const_2))), const_3), divide(56, add(multiply(const_3, const_2), multiply(const_1, const_2)))) | the length of rectangle is thrice its breadth and its perimeter is 56 m , find the area of the rectangle ? | "2 ( 3 x + x ) = 56 l = 21 b = 7 lb = 21 * 7 = 147 answer : e" | a = 3 * 2
b = 1 * 2
c = a + b
d = 56 / c
e = d * 3
f = 3 * 2
g = 1 * 2
h = f + g
i = 56 / h
j = e * i
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a ) 3 hr . , b ) 5 hr . , c ) 2 hr . , d ) 7 hr . , e ) 1 hr . | e | divide(600, divide(multiply(200, 30), const_10)) | in a flight of 600 km , an aircraft was slowed down due to bad weather . its average speed for the trip was reduced by 200 km / hr and the time of flight increased by 30 minutes . the duration of the flight is | "e 1 hr . let the duration of the flight be x hours . then , 600 / x - 600 / 0.5 x = 200 - - > 600 / x - 1200 / ( 2 x + 1 ) = 200 - - > x ( 2 x + 1 ) = 3 - - > 2 x ^ 2 + x - 3 = 0 - - > ( 2 x + 3 ) ( x - 1 ) = 0 - - > x = 1 hr ." | a = 200 * 30
b = a / 10
c = 600 / b
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a ) $ 3,750 , b ) $ 19,800 , c ) $ 8,100 , d ) $ 15,000 , e ) $ 22,500 | b | multiply(2200, power(const_3, divide(28, divide(112, 8)))) | money invested at x % , compounded annually , triples in value in approximately every 112 / x years . if $ 2200 is invested at a rate of 8 % , compounded annually , what will be its approximate worth in 28 years ? | "x = 8 % 112 / x years = 112 / 8 = 14 years now , money triples every 14 years therefore , in 14 yrs , if $ 2200 triples to $ 6600 , in 28 years , it will again triple to $ 6600 * 3 = $ 19,800 answer b" | a = 112 / 8
b = 28 / a
c = 3 ** b
d = 2200 * c
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a ) 140 , b ) 150 , c ) 155 , d ) 160 , e ) 175 | d | divide(add(multiply(12, 150), 600), add(12, 3)) | a rocket soars for 12 seconds at 150 meters per second . it then plummets 600 meters in 3 seconds . what is the average speed of the rocket in meters per second ? | when soaring , the rocket travels a total distance of 1800 m ( 150 m / s for 12 s ) while plummeting the rockets travel distance of 600 m in 3 s the total distance traveled is 2400 m ( 1800 + 600 ) the time taken to travel this distance is 15 s ( 12 + 3 ) average speed = 2400 / 15 = 160 ans : ( option d ) | a = 12 * 150
b = a + 600
c = 12 + 3
d = b / c
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a ) none , b ) one , c ) two , d ) three , e ) four | a | add(1, 1) | for any integer n greater than 1 , # n denotes the product of all the integers from 1 to n , inclusive . how many prime numbers w are there between # 6 + 2 and # 6 + 6 , inclusive ? | none is the answer . a . because for every k 6 ! + k : : k , because 6 ! : : k , since k is between 2 and 6 . a | a = 1 + 1
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a ) 2 , b ) 7 , c ) 9 , d ) 11 , e ) 12 | e | divide(add(224, 43), 17) | a no . when divided by 224 gives a remainder 43 , what remainder will beobtained by dividingthe same no . 17 ? | "224 + 43 = 267 / 17 = 12 ( remainder ) e" | a = 224 + 43
b = a / 17
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a ) 2 . , b ) 4 . , c ) x = 5 . , d ) x = 6 . , e ) 8 . | c | divide(multiply(3, 20), 12) | 20 beavers , working together in a constant pace , can build a dam in 3 hours . how many hours x will it take 12 beavers that work at the same pace , to build the same dam ? | "c . 5 hrs if there were 10 beavers it qould have taken double x = 6 hrs . . so closest to that option is 5 ." | a = 3 * 20
b = a / 12
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a ) 22 , b ) 77 , c ) 88 , d ) 12 , e ) 62 | d | sqrt(add(power(sqrt(subtract(26, multiply(const_2, 168))), const_2), multiply(const_4, 168))) | the product of two numbers is 168 and the sum of these two numbers is 26 . what is the largest of these two numbers ? | "let the numbers be x and ( 26 β x ) . therefore , x ( 26 β x ) = 168 x 2 β 26 x + 168 = 0 ( x β 14 ) ( x β 12 ) = 0 x = 14 or x = 12 so , the numbers are 14 and 12 . answer : d" | a = 2 * 168
b = 26 - a
c = math.sqrt(b)
d = c ** 2
e = 4 * 168
f = d + e
g = math.sqrt(f)
|
a ) $ 9.40 , b ) $ 10.00 , c ) $ 13.20 , d ) $ 17.80 , e ) $ 22.10 | b | add(divide(40, const_100), add(add(const_4, const_3), add(divide(40, const_100), divide(divide(20, const_4), const_100)))) | a small company is planning to rent either computer a or computer b to print customer mailing lists . both computer a and computer b must be rented on an hourly basis . the rental fee is based only on the amount of time the computer is turned on . it will cost 40 percent more per hour to rent computer a than to rent computer b . computer b would , however , require 20 hours more than computer a to do the job . if either computer a , or computer b were rented the total cost to rent the computer would be $ 700.00 . what would be the approximate hourly charge to rent computer b ? | "pa = price of a pb = price of b ta = time for a to complete the job tb = time for b to complete the job given pa = 1.4 pb ta + 20 = tb pa * ta = pb * tb = 700 1.4 pb * ( tb - 20 ) = pb * tb 1.4 pb tb - pb tb = 1.4 pb * 20 0.4 pbtb = 28 pb tb = 28 / 0.4 = 70 pb = 700 / 70 = 10 b" | a = 40 / 100
b = 4 + 3
c = 40 / 100
d = 20 / 4
e = d / 100
f = c + e
g = b + f
h = a + g
|
a ) 37 / 66 , b ) 8 / 41 , c ) 9 / 348 , d ) 1 / 8 , e ) 41 / 91 | a | divide(subtract(330, add(subtract(54, divide(multiply(12.5, 104), multiply(const_1, const_100))), 104)), 330) | in a survey of 330 employees , 104 of them are uninsured , 54 work part time , and 12.5 percent of employees who are uninsured work part time . if a person is to be randomly selected from those surveyed , what is the probability that the person will neither work part time nor be uninsured ? | "- - - - - - - - - ui - - - - - - - - - - - - - - - - nui - - - - - - - total pt - - - - ( 12.5 / 100 ) * 104 = 13 - - - - - - - - - - - - - 54 npt - - - 104 - 13 - - - - - - - - - - - - - - x - - - - - - - - 276 total - - 104 - - - - - - - - - - - - - - - - - - - - - - - - - - - - 330 we have to find not part time and not uninsured . in other words not part time and insured = x / 330 = ( 276 - 104 + 13 ) / 330 = 37 / 66 answer is a ." | a = 12 * 5
b = 1 * 100
c = a / b
d = 54 - c
e = d + 104
f = 330 - e
g = f / 330
|
a ) 22 % , b ) 25 % , c ) 77 % , d ) 99 % , e ) 11.6 % | e | subtract(multiply(divide(const_100, 896), multiply(const_100, multiply(add(const_3, const_2), const_2))), const_100) | a dishonest dealer professes to sell goods at the cost price but uses a weight of 896 grams per kg , what is his percent ? | "896 - - - 104 100 - - - ? = > 11.6 % answer : e" | a = 100 / 896
b = 3 + 2
c = b * 2
d = 100 * c
e = a * d
f = e - 100
|
a ) 3 / 14 , b ) 2 / 7 , c ) 1 / 3 , d ) 1 / 2 , e ) 12 / 21 | a | divide(3, multiply(7, 2)) | jack has two dice , one has 6 equally probable sides , labeled 1 , 2 , 3 , 4 , 5 , 6 , and the other has 7 equally probable sides , labeled 1 , 2 , 3 , 4 , 5 , 6 , 7 . if jack rolls both dice what is the probability that both of the numbers will be even ? | probability that the number on first die is even = 3 / 6 [ because 3 out of 6 faces are even ] probability that the number on second die is even = 3 / 7 [ because 3 out of 7 faces are even ] probability that both dice result in odd numbers = ( 3 / 6 ) * ( 3 / 7 ) = 3 / 14 answer : option a | a = 7 * 2
b = 3 / a
|
a ) s . 10123.22 , b ) s . 10823.20 , c ) s . 10123.20 , d ) s . 10123.29 , e ) s . 10528.13 | e | subtract(multiply(multiply(multiply(const_4, const_100), const_100), power(add(const_1, divide(12, const_100)), 3)), multiply(multiply(const_4, const_100), const_100)) | what will be the compound interest on a sum of rs . 26000 after 3 years at the rate of 12 % p . a . ? | "explanation : amount = [ 26000 * ( 1 + 12 / 100 ) 3 ] = 26000 * 28 / 25 * 28 / 25 * 28 / 25 = rs . 35123.20 c . i . = ( 36528.13 - 26000 ) = rs . 10528.13 answer : e" | a = 4 * 100
b = a * 100
c = 12 / 100
d = 1 + c
e = d ** 3
f = b * e
g = 4 * 100
h = g * 100
i = f - h
|
a ) 345466 , b ) 465767 , c ) 565676 , d ) 645469 , e ) 859622 | e | multiply(456, subtract(add(floor(divide(859622, 456)), const_1), divide(859622, 456))) | which number need to add to 859622 to get a no . exactly divisible by 456 ? | "dividend = quotient * divisor + reminder 859622 / 456 gives quotient = 1885 and reminder = 62 . so , the next number divisible by 456 is 456 places infront of 456 * 1885 which means 456 β 62 = 394 should be added to 859622 . e" | a = 859622 / 456
b = math.floor(a)
c = b + 1
d = 859622 / 456
e = c - d
f = 456 * e
|
a ) 9 % , b ) 11 % , c ) 15 % , d ) 25 % , e ) 90 % | d | subtract(const_100, multiply(divide(9, 12), const_100)) | a case of 12 rolls of paper towels sells for $ 9 . the cost of one roll sold individually is $ 1 . what is the percent r of savings per roll for the 12 - roll package over the cost of 12 rolls purchased individually ? | "cost of 12 paper towels individually = 1 * 12 = 12 cost of a set of 12 paper towels = 9 cost of one roll = 9 / 12 = 3 / 4 = 0.75 savings per roll = 1 - . 75 = 0.25 % of savings is r = . 25 / 1 * 100 = 25 % d is the answer ." | a = 9 / 12
b = a * 100
c = 100 - b
|
a ) 50 hrs , b ) 60 hrs , c ) 70 hrs , d ) 80 hrs , e ) 90 hrs | b | divide(const_1, subtract(divide(const_1, 10), divide(const_1, 12))) | a cistern is filled by pipe a in 10 hours and the full cistern can be leaked out by an exhaust pipe b in 12 hours . if both the pipes are opened , in what time the cistern is full ? | "time taken to full the cistern = ( 1 / 10 - 1 / 12 ) hrs = 1 / 60 = 60 hrs answer : b" | a = 1 / 10
b = 1 / 12
c = a - b
d = 1 / c
|
a ) 8 , b ) 9 , c ) 10 , d ) 11 , e ) 12 | a | subtract(multiply(20, 2), add(multiply(subtract(subtract(20, add(add(multiply(8, 1), 3), 2)), 1), 3), add(multiply(8, 1), multiply(3, 2)))) | in a class of 20 students , 2 students did not borrow any books from the library , 8 students each borrowed 1 book , 3 students each borrowed 2 books , and the rest borrowed at least 3 books . if the average number of books per student was 2 , what is the maximum number of books any single student could have borrowed ? | "the total number of books the students borrowed is 20 * 2 = 40 . the students who borrowed zero , one , or two books borrowed 8 * 1 + 3 * 2 = 14 books . the 7 students who borrowed at least three books borrowed 40 - 14 = 26 books . if 6 of these students borrowed exactly 3 books , then the maximum that one student could have borrowed is 26 - 18 = 8 books . the answer is a ." | a = 20 * 2
b = 8 * 1
c = b + 3
d = c + 2
e = 20 - d
f = e - 1
g = f * 3
h = 8 * 1
i = 3 * 2
j = h + i
k = g + j
l = a - k
|
a ) 9.25 , b ) 11.75 , c ) 13.25 , d ) 15.75 , e ) 17.25 | a | add(divide(multiply(divide(1, 4), subtract(17, 8)), divide(30, const_100)), multiply(subtract(15, 8), divide(1, 4))) | a manufacturer produces a certain men ' s athletic shoe in integer sizes from 8 to 17 . for this particular shoe , each unit increase in size corresponds to a 1 / 4 - inch increase in the length of the shoe . if the largest size of this shoe is 30 % longer than the smallest size , how long , in inches , is the shoe in size 15 ? | "let x be the length of the size 8 shoe . then 0.3 x = 9 / 4 x = 90 / 12 = 15 / 2 inches the size 15 shoe has a length of 15 / 2 + 7 / 4 = 37 / 4 = 9.25 inches the answer is a ." | a = 1 / 4
b = 17 - 8
c = a * b
d = 30 / 100
e = c / d
f = 15 - 8
g = 1 / 4
h = f * g
i = e + h
|
a ) 6.5 , b ) 12.5 , c ) 28 , d ) 30.5 , e ) 40 | d | divide(triangle_area_three_edges(11, 60, 61), divide(triangle_perimeter(11, 60, 61), const_2)) | what is the measure of the radius of the circle that circumscribes a triangle whose sides measure 11 , 60 and 61 ? | "some of pyhtagron triplets we need to keep it in mind . like { ( 2 , 3,5 ) , ( 5 , 12,13 ) , ( 7 , 24,25 ) , ( 11 , 60,61 ) . so now we know the triangle is an right angle triangle . the circle circumscribes the triangle . the circumraduis of the circle that circumscribes the right angle triangle = hypotanse / 2 = 61 / 2 = 30.5 ans . d" | a = triangle_area_three_edges / (
|
a ) 17.5 % , b ) 16 % , c ) 15.4 % , d ) 12.6 % , e ) 13.5 % | a | divide(multiply(subtract(subtract(const_100, 50), 15), subtract(const_100, 50)), const_100) | a man spends 50 % of his income on food , 15 % on children education and 50 % of the remaining on house rent . what percent of his income he is left with ? | "let the total income be x then , income left = ( 100 - 50 ) % of x - [ 100 - ( 50 + 15 ) ] % of x = 50 % of 35 % of x = 17.5 % of x answer is a" | a = 100 - 50
b = a - 15
c = 100 - 50
d = b * c
e = d / 100
|
a ) 104 , b ) 277 , c ) 298 , d ) 269 , e ) 208 | e | divide(624, 3) | a car covers a distance of 624 km in 3 hours . find its speed ? | 624 / 3 = 208 kmph answer : e | a = 624 / 3
|
a ) 20 days . , b ) 15 days . , c ) 22 days . , d ) 30 days . , e ) 18 days . | a | multiply(10, divide(50, 25)) | it was calculated that 50 men could complete a piece of work in 10 days . when work was scheduled to commence , it was found necessary to send 25 men to another project . how much longer will it take to complete the work ? | "one day work = 1 / 10 one man β s one day work = 1 / ( 10 * 50 ) now : no . of workers = 25 one day work = 25 * 1 / ( 10 * 50 ) the total no . of days required to complete the work = ( 50 * 10 ) / 25 = 20 answer : a" | a = 50 / 25
b = 10 * a
|
a ) 4 : 3 , b ) 15 : 6 , c ) 45 : 16 , d ) 15 : 4 , e ) 27 : 5 | e | divide(add(8, 3), const_2) | at a supermarket , the ratio of the number of shampoos to the number of soaps is 8 to 5 , and the ratio of the number of drinks to the number of shampoos is 3 to 4 . if the ratio of the soaps to the number of cereals is 9 to 2 , what is the ratio of the number of drinks to the number of cereals ? | sh : soaps = 8 : 5 drinks : sh = 3 : 4 = > 6 : 8 drinks : sh : soaps = 6 : 8 : 5 soaps : cer = 9 : 2 = > 9 * 5 : 2 * 5 = 45 : 10 now , drinks : sh : soaps = > 6 * 9 : 8 * 9 : 5 * 9 = 54 : 72 : 45 drinks : sh : soaps : cer = 54 : 72 : 45 : 10 drinks : cer = 54 : 10 = > 27 : 5 e | a = 8 + 3
b = a / 2
|
a ) 2 , b ) 625 , c ) 8 , d ) 16 , e ) 32 | b | power(5, multiply(const_4, 1)) | xy = 1 then what is ( 5 ^ ( x + y ) ^ 2 ) / ( 5 ^ ( x - y ) ^ 2 ) | "( x + y ) ^ 2 - ( x - y ) ^ 2 ( x + y + x - y ) ( x + y - x + y ) ( 2 x ) ( 2 y ) 4 xy 4 5 ^ 4 = 625 answer b" | a = 4 * 1
b = 5 ** a
|
a ) 3 : 2 , b ) 1 : 2 , c ) 3 : 1 , d ) 5 : 2 , e ) 4 : 3 | a | divide(subtract(8, 5), subtract(10, 8)) | gold is 10 times as heavy as water and copper is 5 times as heavy as water . in what ratio should these be mixed to get an alloy 8 times as heavy as water ? | "g = 10 w c = 5 w let 1 gm of gold mixed with x gm of copper to get 1 + x gm of the alloy 1 gm gold + x gm copper = x + 1 gm of alloy 10 w + 5 wx = x + 1 * 8 w 10 + 5 x = 8 ( x + 1 ) x = 2 / 3 ratio of gold with copper = 1 : 2 / 3 = 3 : 2 answer is a" | a = 8 - 5
b = 10 - 8
c = a / b
|
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