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 ) rs . 200 , b ) rs . 250 , c ) rs . 300 , d ) rs . 150 , e ) none of these | c | multiply(multiply(3, const_4), divide(100, add(3, 1))) | the cost of 3 pens and 5 pencils is rs . 100 . also the cost of one pen and one pencil is in the ratio of 5 : 1 respectively . what is the cost of one dozen pens ? | "explanation : let the cost of one pen is β 5 x β and pencil is β x β 3 x 5 x + 5 x = rs . 100 15 x + 5 x = rs . 100 x = 100 / 20 = 5 : . cost of 1 pen = 5 x = 5 x 5 = 25 : . cost of 12 pens , i . e . ( one dozen ) = 25 x 12 = rs . 300 answer : option c" | a = 3 * 4
b = 3 + 1
c = 100 / b
d = a * c
|
a ) 4 % , b ) 6 % , c ) 8 % , d ) 10 % , e ) 12 % | b | divide(multiply(subtract(1272, 1200), const_100), 1200) | the compound interest earned on a sum for the second and the third years are $ 1200 and $ 1272 respectively . what is the rate of interest ? | "1272 - 1200 = 72 is the rate of interest on $ 1200 for one year . the rate of interest = ( 100 * 72 ) / ( 1200 ) = 6 % the answer is b ." | a = 1272 - 1200
b = a * 100
c = b / 1200
|
a ) rs . 6184.5 , b ) rs . 4184.5 , c ) rs . 5700 , d ) rs . 4284.5 , e ) rs . 5184.5 | b | divide(multiply(multiply(multiply(24, 4), 4), 170000), multiply(multiply(multiply(10, 26), 6), 10)) | 10 camels cost as much as 24 horses , 26 horses cost as much as 4 oxen and 6 oxen as much as 4 elephants . if the cost of 10 elephants is rs . 170000 , find the cost of a camel ? | "cost of the camel = p 10 camels = 24 horses 26 horses = 4 oxen 6 oxen = 4 elephants 10 elephants = rs . 170000 p = rs . [ ( 24 * 4 * 4 * 170000 ) / ( 10 * 26 * 6 * 10 ) ] p = rs . ( 65280000 / 15600 ) = > p = rs . 4184.5 answer : b" | a = 24 * 4
b = a * 4
c = b * 170000
d = 10 * 26
e = d * 6
f = e * 10
g = c / f
|
a ) 286 m , b ) 899 m , c ) 200 m , d ) 166 m , e ) 258 m | e | subtract(divide(680, const_2), 82) | if the perimeter of a rectangular garden is 680 m , its length when its breadth is 82 m is ? | "2 ( l + 82 ) = 680 = > l = 258 m answer : e" | a = 680 / 2
b = a - 82
|
a ) 0.0002 , b ) 0.0004 , c ) 0.0006 , d ) 0.0008 , e ) 0.001 | c | subtract(multiply(divide(add(add(const_12, const_4), const_2), const_100), divide(add(add(const_12, const_4), const_2), const_100)), 0.0478) | what is the least number . which should be added to 0.0478 to make it a perfect square ? | "0.0478 + 0.0006 = 0.0484 ( 0.22 ) ^ 2 answer : c" | a = 12 + 4
b = a + 2
c = b / 100
d = 12 + 4
e = d + 2
f = e / 100
g = c * f
h = g - 0
|
a ) 33.25 , b ) 32.25 , c ) 65 , d ) 67 , e ) 69 | e | divide(subtract(multiply(7, 56), multiply(4, 46)), 3) | the average score of a cricketer for 7 matches is 56 runs . if the average for first 4 matches is 46 , then average for last 3 matches is | explanation : = ( 56 Γ 7 ) β ( 46 Γ 4 ) / 3 = ( 392 β 184 ) / 3 = 69 answer : option e | a = 7 * 56
b = 4 * 46
c = a - b
d = c / 3
|
a ) 2663 , b ) 1653 , c ) 3693 , d ) 4253 , e ) 2653 | b | add(multiply(multiply(const_3, add(const_10, const_1)), subtract(multiply(add(const_4, const_1), const_10), const_1)), add(multiply(const_3, add(const_10, const_1)), const_3)) | if in a cricket one day match , there were no wide balls , no wides , no extras and no overthrows . what is the maximum number of runs that a batsman can score in an ideal case scenario ? | b 1653 for an ideal case , the batsman will hit a six on each ball . but if he hits six on the last ball of the over , the strike will change in the next over . thus , the best he can do in the last ball is run 3 runs so that he retains the strike even in the next over . thus the total runs that he can score in each over : 6 * 5 + 3 = 33 but this will have to go like it only till the 49 th over . in the last over , he can hit a six in the last ball as well as that will be the last ball of the match . thus runs for the last over will be 6 * 6 = 36 . hence the maximum runs = 33 * 49 + 36 = 1653 | a = 10 + 1
b = 3 * a
c = 4 + 1
d = c * 10
e = d - 1
f = b * e
g = 10 + 1
h = 3 * g
i = h + 3
j = f + i
|
a ) 98 , b ) 99 , c ) 100 , d ) 97 , e ) 95 | b | add(divide(subtract(1001, 11), 10), const_1) | how many multiples of 10 are there between 11 and 1001 ? | "10 * 2 = 20 10 * 100 = 1000 total multiples = ( 100 - 2 ) + 1 = 99 answer b" | a = 1001 - 11
b = a / 10
c = b + 1
|
a ) 17 : 3 , b ) 9 : 1 , c ) 3 : 17 , d ) 5 : 3 , e ) 11 : 2 | b | divide(add(multiply(divide(add(multiply(divide(3, add(3, 2)), subtract(20, 10)), 10), 20), subtract(20, 10)), 10), multiply(divide(multiply(divide(2, add(3, 2)), subtract(20, 10)), 20), subtract(20, 10))) | a 20 litre mixture of milk and water contains milk and water in the ratio 3 : 2 . 10 litres of the mixture is removed and replaced with pure milk and the operation is repeated once more . at the end of the two removal and replacement , what is the ratio of milk and water in the resultant mixture ? | "rom the beginning ; initially milk to water is 12 l : 8 l remove half 6 - 4 add 10 l milk 16 - 4 remove half 8 - 2 add 10 l milk 18 - 2 18 : 2 = 9 : 1 answer : b" | a = 3 + 2
b = 3 / a
c = 20 - 10
d = b * c
e = d + 10
f = e / 20
g = 20 - 10
h = f * g
i = h + 10
j = 3 + 2
k = 2 / j
l = 20 - 10
m = k * l
n = m / 20
o = 20 - 10
p = n * o
q = i / p
|
a ) 3 , b ) 4 , c ) 5 , d ) 6 , e ) 7 | a | divide(subtract(multiply(6, 8), multiply(3, 6)), add(6, 4)) | 6 workers should finish a job in 8 days . after 3 days came 4 workers join them . how many days x do they need to finish the same job ? | "let rate of one worker be r = > ( 6 * r ) * 8 = 1 ( rate * time = work ) = > r = 1 / 48 = > work remaining after 3 days 1 - ( 3 * 6 ) / 48 = 30 / 48 after 4 ppl joined in ( ( 6 + 4 ) * time ) / 48 = 30 / 48 time x = 3 days to finish the task imo a" | a = 6 * 8
b = 3 * 6
c = a - b
d = 6 + 4
e = c / d
|
a ) $ 661.50 , b ) $ 663.50 , c ) $ 665.50 , d ) $ 667.50 , e ) $ 669.50 | a | multiply(600, power(add(const_1, divide(divide(10, const_2), const_100)), const_2)) | if an investor puts $ 600 in a savings account that earns 10 percent annual interest compounded semiannually , how much money will be in the account after one year ? | "1.05 * 1.05 * 600 = $ 661.50 the answer is a ." | a = 10 / 2
b = a / 100
c = 1 + b
d = c ** 2
e = 600 * d
|
a ) 3 , b ) 7 , c ) 21 , d ) 189 , e ) 198 | d | multiply(multiply(multiply(subtract(add(add(subtract(6, 1), add(2, 1)), add(2, 1)), const_10), subtract(add(multiply(2, add(subtract(6, 1), add(2, 1))), 1), const_10)), add(2, 1)), subtract(add(multiply(2, add(add(add(subtract(6, 1), add(2, 1)), add(2, 1)), add(2, 1))), 1), multiply(2, const_10))) | a β sophie germain β prime is any positive prime number p for which 2 p + 1 is also prime . the product of all the possible units digits of sophie germain primes greater than 6 is | in that case , the sophie prime numbers greater than 6 are 7 , 11,23 , 47,59 , . . which yields units digit as 1 , 3,7 and 9 product would be 1 x 3 x 7 x 9 = 189 answer should be d | a = 6 - 1
b = 2 + 1
c = a + b
d = 2 + 1
e = c + d
f = e - 10
g = 6 - 1
h = 2 + 1
i = g + h
j = 2 * i
k = j + 1
l = k - 10
m = f * l
n = 2 + 1
o = m * n
p = 6 - 1
q = 2 + 1
r = p + q
s = 2 + 1
t = r + s
u = 2 + 1
v = t + u
w = 2 * v
x = w + 1
y = 2 * 10
z = x - y
A = o * z
|
a ) 240 , b ) 270 , c ) 295 , d ) 360 , e ) 264 | e | divide(add(1335, 15), subtract(6, const_1)) | the difference of two numbers is 1335 . on dividing the larger number by the smaller , we get 6 as quotient and the 15 as remainder . what is the smaller number ? | "let the smaller number be x . then larger number = ( x + 1335 ) . x + 1335 = 6 x + 15 5 x = 1320 x = 264 smaller number = 264 . answer e" | a = 1335 + 15
b = 6 - 1
c = a / b
|
a ) 180 , b ) 190 , c ) 200 , d ) 210 , e ) 220 | e | divide(multiply(add(30, 36), const_100), 30) | dawson has to secure 30 % marks to clear his exam of class 8 th . he got 30 marks and failed by 36 marks . what is the maximum marks ? | e 220 to pass the exam ravish needs 30 + 36 = 66 marks . = > ( 66 / 30 ) * 100 = 220 | a = 30 + 36
b = a * 100
c = b / 30
|
a ) 10 , b ) 388 , c ) 6.5 , d ) 29 , e ) 22 | c | subtract(const_100, multiply(multiply(divide(subtract(const_100, divide(multiply(const_100, 15), const_100)), const_100), divide(add(const_100, divide(multiply(const_100, 10), const_100)), const_100)), const_100)) | if the price of a book is first decreased by 15 % and then increased by 10 % , then the net change in the price will be : | "explanation : let the original price be rs . 100 . decreased by 15 % = 85 then increased 10 % on rs 85 = 85 + 8.5 = 93.5 net change in price = 100 - 93.5 = 6.5 answer : c" | a = 100 * 15
b = a / 100
c = 100 - b
d = c / 100
e = 100 * 10
f = e / 100
g = 100 + f
h = g / 100
i = d * h
j = i * 100
k = 100 - j
|
a ) 981 , b ) 1000 , c ) 1011 , d ) 10000 , e ) 1110 | d | subtract(power(const_10, 5), multiply(9, power(const_10, 4))) | a warehouse is labeling computer inventory with 5 - digit codes . each code is to consist of 4 digits between the numbers 0 and 9 . the manager is debating whether to allow any 5 - digit codes to have leading zero ' s before a non - zero number is used . if not allowed , then numbers like 0025 cannot be used . the manager wants to measure the magnitude of the number of 5 - digit code possibilities that are lost if he disallows the use of leading zero ' s . how many such codes would be lost ? | the number of codes possible if leading zero is allowed = 10 * 10 * 10 * 10 * 10 = 10 ^ 5 the number of codes possible if leading zero is not allowed = 9 * 10 * 10 * 10 * 10 = 9 * 10 ^ 4 the difference = 10 ^ 5 - 9 * 10 ^ 4 = 10 ^ 4 ( 10 - 9 ) = 10 ^ 4 . answer : d . similar questions to practice : hope it helps . | a = 10 ** 5
b = 10 ** 4
c = 9 * b
d = a - c
|
a ) 30 miles . , b ) 35 miles . , c ) 45 miles . , d ) 60 miles . , e ) 65 miles . | d | multiply(divide(10, add(5, 10)), 90) | tim and Γ© lan are 90 miles away from one another . they are starting to move towards each other simultaneously , tim at a speed of 10 mph and Γ© lan at a speed of 5 mph . if every hour they double their speeds , what is the distance that tim will pass until he meets Γ© lan ? | "in 1 st hour : relative speed of the two will be 10 + 5 = 15 miles / hour . as they move towards each other , the distance between them reduces by 15 miles . distance left between them : 90 - 15 = 75 miles . in 2 nd hour : speed of tim : 20 mph , that of elan : 10 mph . relative speed of the two will be 20 + 10 = 30 miles / hour . as they move towards each other , the distance between them reduces by 30 miles . distance left between them : 75 - 30 = 45 miles . in 3 rd hour : speed of tim : 40 mph , that of elan : 20 mph . relative speed of the two will be 40 + 20 = 60 miles / hour . now in order to cover the remaining distance of 45 miles . they will take : ( 45 / 60 ) hrs . = ( 3 / 4 ) hour . distance travelled by tim ? in first hour : 1 hour * ( 10 miles per hour ) = 10 miles in second hour : 1 hour * ( 20 miles per hour ) = 20 miles in third hour : ( 3 / 4 ) hour * ( 40 miles per hour ) = 30 miles total distance travelled : 10 + 20 + 30 = 60 miles . d is the answer" | a = 5 + 10
b = 10 / a
c = b * 90
|
a ) 81000 , b ) 81007 , c ) 81008 , d ) 79012.3 , e ) 81022 | d | add(add(64000, multiply(divide(1, 9), 64000)), multiply(divide(1, 9), add(64000, multiply(divide(1, 9), 64000)))) | every year an amount increases by 1 / 9 th of itself . how much will it be after two years if its present value is rs . 64000 ? | "64000 * 10 / 9 * 10 / 9 = 79012.3 answer : d" | a = 1 / 9
b = a * 64000
c = 64000 + b
d = 1 / 9
e = 1 / 9
f = e * 64000
g = 64000 + f
h = d * g
i = c + h
|
a ) 2.5 % , b ) 15 % , c ) 25 % , d ) 35 % , e ) 200 % | e | multiply(divide(divide(20, const_100), divide(10, const_100)), const_100) | if c is 20 % of a and 10 % of b , what percent of a is b ? | answer = e 20 a / 100 = 10 b / 100 b = 20 a / 10 = 200 a / 100 = 200 % | a = 20 / 100
b = 10 / 100
c = a / b
d = c * 100
|
a ) 2788 , b ) 4500 , c ) 7282 , d ) 2782 , e ) 2729 | b | multiply(subtract(12, 7), divide(3600, subtract(7, 3))) | an amount of money is to be divided between p , q and r in the ratio of 3 : 7 : 12 . if the difference between the shares of p and q is rs . 3600 , what will be the difference between q and r ' s share ? | "4 - - - 3600 5 - - - ? = > 4500 answer : b" | a = 12 - 7
b = 7 - 3
c = 3600 / b
d = a * c
|
a ) 3 , b ) 6 , c ) 4 , d ) 9 , e ) 1 | c | divide(add(add(1, 2), 5), 2) | the area of a triangle will be when a = 1 m , b = 2 m , c = 5 m , a , b , c being lengths of respective sides ? | s = ( 1 + 2 + 5 ) / 2 = 4 answer : c | a = 1 + 2
b = a + 5
c = b / 2
|
a ) 4.28 days , b ) 4.38 days , c ) 5.28 days , d ) 6.28 days , e ) 2.28 days | a | subtract(add(inverse(add(inverse(10), inverse(25))), 25), const_3) | a and b can do a work in 25 days and 10 days respectively . a starts the work and b joins him after 2 days . in how many days can they complete the remaining work ? | "work done by a in 2 days = 2 / 5 remaining work = 3 / 5 work done by both a and b in one day = 1 / 25 + 1 / 10 = 7 / 50 remaining work = 3 / 5 * 50 / 7 = 4.28 days . answer : a" | a = 1/(10)
b = 1/(25)
c = a + b
d = 1/(c)
e = d + 25
f = e - 3
|
a ) $ 4.95 , b ) $ 4.75 , c ) $ 5.05 , d ) $ 4.50 , e ) $ 5.25 | b | subtract(5, multiply(divide(multiply(add(divide(25, const_100), const_1), 4), const_100), 5)) | a grocery store priced a gallon of milk at 25 % above the wholesale price of $ 4 . if a customer comes in with a coupon for 5 % off milk , how much does that customer pay ? | wholesale cost of milk = $ 4 milk was priced at 25 % above $ 4 = $ 5 % discount given by coupon = 5 % customer paid = . 95 * 5 = $ 4.75 answer b | a = 25 / 100
b = a + 1
c = b * 4
d = c / 100
e = d * 5
f = 5 - e
|
a ) 9 , b ) 10 , c ) 12 , d ) 14 , e ) 15 | a | divide(subtract(81, multiply(const_3, 9)), multiply(const_3, const_2)) | a number is doubled and 9 is added . if resultant is trebled , it becomes 81 . what is that number | "explanation : = > 3 ( 2 x + 9 ) = 81 = > 2 x + 9 = 27 = > x = 9 answer : option a" | a = 3 * 9
b = 81 - a
c = 3 * 2
d = b / c
|
a ) 30 , b ) 25 , c ) 35 , d ) 40 , e ) 45 | d | multiply(divide(divide(multiply(1, add(1, const_1)), const_2), 1), 79) | what is the average of odd numbers from 1 to 79 ? | "sum of n odd natural number is given by , sum of n odd natural numbers = n * n average of n odd natural numbers is given by , = ( n * n ) / n = n . here , n = number of terms = ( 79 + 1 ) / 2 = 40 . so , required average = 40 . answer : d" | a = 1 + 1
b = 1 * a
c = b / 2
d = c / 1
e = d * 79
|
a ) s . 9800 , b ) s . 3800 , c ) s . 9800 , d ) s . 4800 , e ) s . 6880 | d | divide(multiply(multiply(multiply(24, 4), 4), 120000), multiply(multiply(multiply(10, 16), 6), 10)) | 10 camels cost as much as 24 horses , 16 horses cost as much as 4 oxen and 6 oxen as much as 4 elephants . if the cost of 10 elephants is rs . 120000 , find the cost of a camel ? | "cost of the camel = p 10 camels = 24 horses 16 horses = 4 oxen 6 oxen = 4 elephants 10 elephants = rs . 120000 p = rs . [ ( 24 * 4 * 4 * 120000 ) / ( 10 * 16 * 6 * 10 ) ] p = rs . ( 46080000 / 9600 ) = > p = rs . 4800 answer : d" | a = 24 * 4
b = a * 4
c = b * 120000
d = 10 * 16
e = d * 6
f = e * 10
g = c / f
|
a ) 133 , b ) 376 , c ) 377 , d ) 400 , e ) 377 | d | divide(332, divide(83, const_100)) | in an exam ashley secured 332 marks . if she secured 83 % makes , find the maximum marks | let the maximum marks be m . ashley β s marks = 83 % of m ashley secured 332 marks therefore , 83 % of m = 332 β 83 / 100 Γ m = 332 β m = ( 332 Γ 100 ) / 83 β m = 33200 / 83 β m = 400 therefore , ashley got 332 marks out of 400 marks . answer : d | a = 83 / 100
b = 332 / a
|
a ) 20 years , b ) 21 years , c ) 22 years , d ) 24 years , e ) none of these | c | divide(subtract(24, subtract(multiply(const_2, const_2), const_2)), subtract(const_2, const_1)) | a man is 24 years older than his son . in two years , his age will be twice the age of his son . the present age of his son is | "explanation : let the son ' s present age be x years . then , man ' s present age = ( x + 24 ) years = > ( x + 24 ) + 2 = 2 ( x + 2 ) = > x + 26 = 2 x + 4 so , x = 22 option c" | a = 2 * 2
b = a - 2
c = 24 - b
d = 2 - 1
e = c / d
|
a ) z / 2 , b ) 2 z , c ) 11 z / 24 , d ) 3 z / 5 , e ) z / 9 | c | divide(subtract(divide(multiply(3, const_100), const_2), const_2), add(divide(multiply(3, const_100), const_2), const_2)) | if 3 x = 8 y = z , what is x + y , in terms of z ? | "3 x = 8 y = z x = z / 3 and y = z / 8 x + y = z / 3 + z / 8 = 11 z / 24 answer is c" | a = 3 * 100
b = a / 2
c = b - 2
d = 3 * 100
e = d / 2
f = e + 2
g = c / f
|
a ) 68 kmph , b ) 70 kmph , c ) 72 kmph , d ) 60 kmph , e ) 50 kmph | c | multiply(divide(500, subtract(35, 10)), const_3_6) | a train requires 10 seconds to pass a pole while it requires 35 seconds to cross a stationary train which is 500 mtrs long . find the speed of the train . | "in 10 s the train crosses the pole and in 35 sec the train crosses one more stationary train in 25 sec the train travels a distance of 500 mtrs speed = 500 / 25 = 20 m / s = 20 ( 3600 / 1000 ) = 20 * 18 / 5 = 72 kmph answer : c" | a = 35 - 10
b = 500 / a
c = b * const_3_6
|
a ) 20 , b ) 21 , c ) 22 , d ) 23 , e ) 24 | b | add(7, divide(multiply(7, subtract(12000, 8000)), subtract(8000, 6000))) | the average salary of all the workers in a workshop is rs . 8000 . the average salary of 7 technicians is rs . 12000 and the average salary of the rest is rs . 6000 . the total number of workers in the workshop is | "solution let the toatl number of workers be x . then 8000 x = ( 12000 x 7 ) + 6000 ( x - 7 ) 2000 x = 42000 x = 21 . answer b" | a = 12000 - 8000
b = 7 * a
c = 8000 - 6000
d = b / c
e = 7 + d
|
a ) 754 m , b ) 704 m , c ) 1055.04 m , d ) 1058.04 m , e ) 204 m | c | divide(multiply(multiply(multiply(divide(add(multiply(add(const_3, const_4), const_3), const_1), add(const_3, const_4)), 22.6), const_2), 750), const_100) | the radius of a wheel is 22.6 cm . what is the distance covered by the wheel in making 750 resolutions ? | "in one resolution , the distance covered by the wheel is its own circumference . distance covered in 750 resolutions . = 750 * 2 * 22 / 7 * 22.6 = 105504 cm = 1055.04 m answer : c" | a = 3 + 4
b = a * 3
c = b + 1
d = 3 + 4
e = c / d
f = e * 22
g = f * 2
h = g * 750
i = h / 100
|
a ) 320 , b ) 360 , c ) 400 , d ) 408.3 , e ) 440 | b | multiply(multiply(subtract(const_1, divide(20, const_100)), 45), 10) | car z travels 45 miles per gallon of gasoline when driven at a constant rate of 45 miles per hour , but travels 20 percent fewer miles per gallon of gasoline when driven at a constant rate of 60 miles per hour . how many miles does car z travel on 10 gallons of gasoline when driven at a constant rate of 60 miles per hour ? | "the question stem asks us for the distance possible with 10 gallons of fuel at a constant speed of 60 miles per hour . we therefore first calculate the fuel efficiency at that speed . the stem tells us that at 45 miles / hour , the car will run 45 miles / gallon and at 60 miles / hour , that distance decreases by 20 % . we can therefore conclude that the car will travel 36 miles / gallon at a constant speed of 60 miles / gallon . with 10 gallons of fuel , the car can therefore travel 36 miles / gallon * 10 gallons = 360 miles . answer b ." | a = 20 / 100
b = 1 - a
c = b * 45
d = c * 10
|
a ) 3 , b ) 4 , c ) 5 , d ) 6 , e ) 7 | a | divide(3, divide(27, 3)) | evaluate : log 3 27 | "solution log 3 27 = n = 3 n = 27 = 33 n βΉ = βΊ 3 . answer a" | a = 27 / 3
b = 3 / a
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a ) 82.1 sec . , b ) 12.5 sec , c ) 19.1 sec . , d ) 17.1 sec . , e ) 42.1 sec . | b | divide(add(110, 140), multiply(72, const_0_2778)) | how long does a train 110 m long running at the speed of 72 km / hr takes to cross a bridge 140 m length ? | "speed = 72 * 5 / 18 = 20 m / sec total distance covered = 110 + 140 = 250 m . required time = 250 / 20 = 12.5 sec . answer : b" | a = 110 + 140
b = 72 * const_0_2778
c = a / b
|
a ) 35 , b ) 564 , c ) 526 , d ) 48 , e ) 50 | b | subtract(add(multiply(6, 42), multiply(6, 48)), multiply(12, 55)) | the average of 12 result is 55 . average of the first 6 of them is 42 and that of the last 6 is 48 . find the 8 th result ? | "sum of all the 12 results = 12 * 55 = 660 sum of the first 6 of them = 6 * 42 = 252 sum of the last 6 of them = 6 * 48 = 288 so , the 8 th number = 600 + 252 - 288 = 564 . b" | a = 6 * 42
b = 6 * 48
c = a + b
d = 12 * 55
e = c - d
|
a ) 1 , b ) 2.5 , c ) 3 , d ) 4 , e ) 5 | b | divide(multiply(15, 75), 450) | 15 % of 75 is the same as vx % of 450 . find x . | 15 % of 75 = 11.25 450 * x / 100 = 11.25 x = 2.5 answer : b | a = 15 * 75
b = a / 450
|
a ) 10 min , b ) 15 min , c ) 20 min , d ) 25 min , e ) 30 min | d | divide(25, 1) | a fill pipe can fill 1 / 2 of cistern in 25 minutes . in how many minutes , it can fill 1 / 2 of the cistern ? | "required time = 25 * 2 * 1 / 2 = 25 minutes answer is d" | a = 25 / 1
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a ) 259 , b ) 498 , c ) 676 , d ) 823 , e ) 1,002 | a | divide(subtract(520, 2), 2) | a straight line in the xy - plane has a slope of 2 and a y - intercept of 2 . on this line , what is the x - coordinate of the point whose y - coordinate is 520 ? | "slope of 2 and a y - intercept of 2 y - coordinate is 520 y = 2 x + 2 518 = 2 x x = 259 answer : a . 259" | a = 520 - 2
b = a / 2
|
a ) 14 , b ) 16 , c ) 18 , d ) 20 , e ) 22 | b | divide(subtract(multiply(divide(15, const_100), 40), 2), divide(25, const_100)) | if 15 % of 40 is greater than 25 % of a number by 2 , the number is | "explanation : 15 / 100 * 40 - 25 / 100 * x = 2 or x / 4 = 4 so x = 16 option b" | a = 15 / 100
b = a * 40
c = b - 2
d = 25 / 100
e = c / d
|
a ) $ 1.63 , b ) $ 1.71 , c ) $ 1.68 , d ) $ 1.72 , e ) $ 1.76 | b | divide(add(multiply(1325, 1.89), multiply(750, 1.38)), add(1325, 750)) | john purchased 1325 large bottles at $ 1.89 per bottle and 750 small bottles at $ 1.38 per bottle . what was the approximate average price paid per bottle ? | "( 1325 * 1.89 + 750 * 1.38 ) / ( 1325 + 750 ) = ~ 1.71 option ( b )" | a = 1325 * 1
b = 750 * 1
c = a + b
d = 1325 + 750
e = c / d
|
a ) 100555 , b ) 100775 , c ) 100885 , d ) 100995 , e ) 100665 | b | add(subtract(multiply(const_10, multiply(const_100, const_100)), const_100), 35,45) | find the smallest number of six digits exactly divisible by 25 , 35,45 and 15 . | "smallest number of six digits is 100000 . required number must be divisible by l . c . m . of 25,35 , 45,15 i . e 1575 , on dividing 100000 by 1575 , we get 800 as remainder . therefore , required number = 100000 + ( 1575 Γ’ β¬ β 800 ) = 100775 . answer is b ." | a = 100 * 100
b = 10 * a
c = b - 100
d = c + 35
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a ) 60 , b ) 61.2 , c ) 62.5 , d ) 3,780 , e ) 3,672 | d | multiply(1.05, const_3600) | the moon revolves around the earth at a speed of approximately 1.05 kilometers per second . this approximate speed is how many kilometers per hour ? | moon revolves around the earth at a speed of 1.02 kilometers per second . one hour equal to 60 minutes . one minute equals to 60 seconds . so one hour equals to 3600 seconds . so one hour , speed = 1.05 * 3600 = 3780 kilometers per hour . option d is correct | a = 1 * 5
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a ) 12 , b ) 75 , c ) 80 , d ) 54 , e ) 15 | c | divide(add(100, 60), const_2) | the speed of a car is 100 km in the first hour and 60 km in the second hour . what is the average speed of the car ? | "s = ( 100 + 60 ) / 2 = 80 kmph answer : c" | a = 100 + 60
b = a / 2
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a ) 32200 , b ) 41897 , c ) 58799 , d ) 60000 , e ) 62782 | d | add(multiply(add(20, const_1), add(20, const_1)), add(20, const_1)) | a sum of money is put out at compound interest for 2 years at 20 % . it would fetch rs . 1446 more if the interest were payable half - yearly , then it were pay able yearly . find the sum . | "p ( 11 / 10 ) ^ 4 - p ( 6 / 5 ) ^ 2 = 1446 p = 60000 answer : d" | a = 20 + 1
b = 20 + 1
c = a * b
d = 20 + 1
e = c + d
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a ) 1800 , b ) 1420 , c ) 1430 , d ) 1440 , e ) 1540 | a | divide(multiply(subtract(const_100, 10), 2000), const_100) | a man buys a cycle for rs . 2000 and sells it at a loss of 10 % . what is the selling price of the cycle ? | "s . p . = 90 % of rs . 2000 = 90 / 100 x 2000 = rs . 1800 answer : a" | a = 100 - 10
b = a * 2000
c = b / 100
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a ) 75 m . , b ) 80 m . , c ) 150 m . , d ) 142 m . , e ) none of the above | d | multiply(71, const_2) | a runs twice as fast as b and gives b a start of 71 m . how long should the racecourse be so that a and b might reach in the same time ? | "ratio of speeds of a and b is 2 : 1 b is 71 m away from a but we know that a covers 1 meter ( 2 - 1 ) more in every second than b the time taken for a to cover 71 m is 71 / 1 = 71 m so the total time taken by a and b to reach = 2 * 71 = 142 m answer : d" | a = 71 * 2
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a ) 240 , b ) 288 , c ) 277 , d ) 390 , e ) 361 | d | multiply(divide(630, add(add(multiply(3000, 8), multiply(subtract(3000, 1000), subtract(const_12, 8))), add(multiply(4000, 8), multiply(add(4000, 1000), subtract(const_12, 8))))), add(multiply(3000, 8), multiply(subtract(3000, 1000), subtract(const_12, 8)))) | a and b began business with rs . 3000 and rs . 4000 after 8 months , a withdraws rs . 1000 and b advances rs . 1000 more . at the end of the year , their profits amounted to rs . 630 find the share of b . | "( 3 * 8 + 2 * 4 ) : ( 4 * 8 + 5 * 4 ) 8 : 13 13 / 21 * 630 = 390 answer : d" | a = 3000 * 8
b = 3000 - 1000
c = 12 - 8
d = b * c
e = a + d
f = 4000 * 8
g = 4000 + 1000
h = 12 - 8
i = g * h
j = f + i
k = e + j
l = 630 / k
m = 3000 * 8
n = 3000 - 1000
o = 12 - 8
p = n * o
q = m + p
r = l * q
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a ) s . 12.7 , b ) s . 12.4 , c ) s . 12.5 , d ) s . 12.1 , e ) s . 12.9 | c | multiply(divide(divide(multiply(divide(20, const_100), 500), 10), multiply(divide(20, const_100), 500)), const_100) | a reduction of 20 % in the price of salt enables a lady to obtain 10 kgs more for rs . 500 , find the original price per kg ? | "100 * ( 20 / 100 ) = 20 - - - 10 ? - - - 1 = > rs . 2 500 - - - 80 ? - - - 2 = > rs . 12.5 answer : c" | a = 20 / 100
b = a * 500
c = b / 10
d = 20 / 100
e = d * 500
f = c / e
g = f * 100
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a ) 42 , b ) 51 , c ) 62 , d ) 72 , e ) 82 | b | subtract(59, multiply(multiply(12, 3), 2)) | evaluate : 59 - 12 * 3 * 2 = ? | "according to order of operations , 12 ? 3 ? 2 ( division and multiplication ) is done first from left to right 12 * * 2 = 4 * 2 = 8 hence 59 - 12 * 3 * 2 = 59 - 8 = 51 correct answer b" | a = 12 * 3
b = a * 2
c = 59 - b
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a ) 40 years , b ) 41 years , c ) 42 years , d ) 43 years , e ) 44 years | e | subtract(multiply(add(20, 2), add(22, 2)), multiply(20, 22)) | the average age of a class of 20 students is 22 years . the average increased by 2 when the teacher ' s age also included . what is the age of the teacher ? | "if age of the teacher was 22 , average would not have changed . since average increased by 2 , age of the teacher = 19 + 22 Γ 1 = 44 answer : e" | a = 20 + 2
b = 22 + 2
c = a * b
d = 20 * 22
e = c - d
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['a ) 7 cm', 'b ) 7.1 cm', 'c ) 7.2 cm', 'd ) 7.3 cm', 'e ) none of these'] | c | divide(multiply(18, 10), 25) | the length of a rectangle is 18 cm and its breadth is 10 cm . when the length is increased to 25 cm , what will be the breadth of the rectangle if the area remains the same ? | solution let the breadth be b . then , β 25 b = 18 Γ 10 β b = ( 18 x 10 / 25 ) cm = 7.2 cm answer c | a = 18 * 10
b = a / 25
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a ) 38 , b ) 47 , c ) 50 , d ) 53 , e ) 63 | e | add(divide(540, 9), 3) | a whale goes on a feeding frenzy that lasts for 9 hours . for the first hour he catches and eats x kilos of plankton . in every hour after the first , it consumes 3 kilos of plankton more than it consumed in the previous hour . if by the end of the frenzy the whale will have consumed a whopping accumulated total 540 kilos of plankton , how many kilos did he consume on the sixth hour ? | "if you list the amount eaten each hour , you ' ll get an equally spaced list , increasing by 3 each hour . in any equally spaced list , the median equals the mean . here , the mean is 540 / 9 = 60 , so the median is also 60 , and that is the amount eaten in the 5 th hour . we need to add 3 to find the total eaten in the next hour , so the answer is 63 . option e ." | a = 540 / 9
b = a + 3
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a ) 30 , b ) 36 , c ) 42 , d ) 48 , e ) 56 | e | divide(24, subtract(divide(const_1, divide(70, const_100)), const_1)) | walking at 70 % of his usual speed a man takes 24 minutes more to cover a distance . what is his usual time to cover this distance ? | "speed is inversly proprtional to time walking at 70 % of speed meand 7 / 10 s takes 10 / 7 t . it takes 24 minutes extra to cover the distance . then 10 / 7 t = t + 24 10 t = 7 t + 168 3 t = 168 t = 56 . option e is correct" | a = 70 / 100
b = 1 / a
c = b - 1
d = 24 / c
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a ) 220 , b ) 230 , c ) 960 , d ) 780 , e ) 896 | e | multiply(subtract(832, divide(subtract(832, 448), const_2)), add(const_1, divide(40, const_100))) | the profit earned by selling an article for rs . 832 is equal to the loss incurred when the same article is sold for rs . 448 . what should be the sale price for making 40 % profit ? | "c . p . = rs . x . then , 832 - x = x - 448 2 x = 1280 = > x = 640 required s . p . = 140 % of rs . 640 = 140 / 100 * 640 = rs . 896 . e" | a = 832 - 448
b = a / 2
c = 832 - b
d = 40 / 100
e = 1 + d
f = c * e
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a ) s . 150 , b ) s . 200 , c ) s . 250 , d ) s . 350 , e ) s . 400 | e | divide(multiply(880, multiply(add(const_1, const_4), const_2)), multiply(add(multiply(add(const_1, const_4), const_2), const_1), const_2)) | two employees x and y are paid a total of rs . 880 per week by their employer . if x is paid 120 percent of the sum paid to y , how much is y paid per week ? | "let the amount paid to x per week = x and the amount paid to y per week = y then x + y = 880 but x = 120 % of y = 120 y / 100 = 12 y / 10 Γ’ Λ Β΄ 12 y / 10 + y = 880 Γ’ β‘ β y [ 12 / 10 + 1 ] = 880 Γ’ β‘ β 22 y / 10 = 880 Γ’ β‘ β 22 y = 8800 Γ’ β‘ β y = 8800 / 22 = 800 / 2 = rs . 400 e )" | a = 1 + 4
b = a * 2
c = 880 * b
d = 1 + 4
e = d * 2
f = e + 1
g = f * 2
h = c / g
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a ) 500 km , b ) 600 km , c ) 700 km , d ) 900 km , e ) 400 km | d | add(add(divide(multiply(const_100.0, 40), subtract(50, 40)), 1), divide(multiply(1, 40), subtract(50, 40))) | two trains start from p and q respectively and travel towards each other at a speed of 50 km / hr and 40 km / hr respectively . by the time they meet , the first train has travelled 1 ookm more than the second . the distance between p and q is | "explanation : at the time of meeting , let the distance travelled by the second train be x km then , distance covered by the first train is ( x + 100 ) km = > x / 40 = ( x + 100 ) / 50 = > 50 x = 40 x + 4000 = > x = 400 km so , distance between p and q = ( x + x + 100 ) km = 900 km answer : option d" | a = 100 * 0
b = 50 - 40
c = a / b
d = c + 1
e = 1 * 40
f = 50 - 40
g = e / f
h = d + g
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a ) 8 , b ) 9 , c ) 10 , d ) 11 , e ) 12 | e | multiply(divide(3.50, add(0.35, 0.25)), const_2) | a certain store sold pens for $ 0.35 each and pencils for $ 0.25 each . if a customer purchased both pens and pencils from the store for a total of $ 3.50 , what total number of pens and pencils did the customer purchase ? | "answer : algebraically , the question looks like this : 2.5 = 0.35 x + 0.25 y there are six possible numbers of pens that fit that requirement : 1 pen : $ 0.35 2 pens : $ 0.70 3 pens : $ 1.05 4 pens : $ 1.40 5 pens : $ 1.75 balance can fetch 7 pencils for $ 0.25 each . that ' s 5 pens and 7 pencils for a total of 12 pens and pencils . choice ( e ) is correct ." | a = 0 + 35
b = 3 / 50
c = b * 2
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a ) 196 , b ) 284 , c ) 336 , d ) 412 , e ) 572 | c | divide(multiply(42, 32), 4) | dan β s car gets 32 miles per gallon . if gas costs $ 4 / gallon , then how many miles can dan β s car go on $ 42 of gas ? | "42 / 4 = 10.5 gallons 10.5 * 32 = 336 miles the answer is c ." | a = 42 * 32
b = a / 4
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a ) 6 days , b ) 19 days , c ) 24 days , d ) 3 days , e ) 13 days | c | multiply(const_3, 8) | aarti can do a piece of work in 8 days . in how many days will she complete three time of work of same type ? | "we have the important relation , more work , more time ( days ) a piece of work can be done in 8 days . three times of work of same type can be done in 8 x 3 = 24 days answer c" | a = 3 * 8
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a ) 1 / 25 , b ) 28 / 55 , c ) 2 / 35 , d ) 2 / 55 , e ) 1 / 75 | b | divide(divide(factorial(subtract(11, 2)), multiply(factorial(subtract(subtract(11, 2), 3)), factorial(3))), divide(factorial(11), multiply(factorial(subtract(11, 3)), factorial(3)))) | in a box of 11 pencils , a total of 2 are defective . if a customer buys 3 pencils selected at random from the box , what is the probability that neither pencils will be defective ? | "first , there are 9 c 3 ways you can select 3 good pencils from 4 good ones . second , there are 11 c 3 ways you select 3 pencils from 6 ones in the box . then , the probability that neither pen will be defective is : 9 c 3 / 11 c 3 = 84 / 165 = 28 / 55 answer is b" | a = 11 - 2
b = math.factorial(a)
c = 11 - 2
d = c - 3
e = math.factorial(d)
f = math.factorial(3)
g = e * f
h = b / g
i = math.factorial(11)
j = 11 - 3
k = math.factorial(j)
l = math.factorial(3)
m = k * l
n = i / m
o = h / n
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a ) 2177 , b ) 2876 , c ) 4500 , d ) 5300 , e ) 6711 | d | subtract(multiply(8000, const_4), subtract(multiply(8300, 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 . 8300 . 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 * 8300 = 33200 - - - - ( 2 ) ( 2 ) - ( 1 ) i . e . may - jan = 1200 salary of may is rs . 6500 salary of january = rs . 5300 . answer : d | a = 8000 * 4
b = 8300 * 4
c = b - 6500
d = a - c
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a ) 130 , b ) 135 , c ) 140 , d ) 150 , e ) 145 | b | multiply(multiply(divide(1800, 120), divide(9, const_100)), const_100) | by investing rs . 1800 in 9 % stock , syam earns rs . 120 . find out the stock is then quoted at which rate ? | "assume that face value = rs . 100 . dividend per share = rs . 9 ( as it is a 9 % stock ) by investing rs . 1800 , he earns rs . 120 investment needed to earn rs . 9 = 1800 Γ 9 / 120 = rs . 135 ie , stock is then quoted ( then market value ) = rs . 135 answer is b ." | a = 1800 / 120
b = 9 / 100
c = a * b
d = c * 100
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a ) 100 , b ) 170 , c ) 150 , d ) 180 , e ) 199 | b | divide(multiply(multiply(divide(552.50, const_100), 20), const_100), 65) | 65 % of x = 20 % of 552.50 . find the value of x ? | "65 % of x = 20 % of 552.50 then , 65 / 100 * x = 20 / 100 * 5525 / 10 x = 170 answer is b" | a = 552 / 50
b = a * 20
c = b * 100
d = c / 65
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a ) 2134 , b ) 2234 , c ) 2540 , d ) 2560 , e ) 864 | e | divide(multiply(power(12, 3), power(6, 3)), 432) | find the value of y from ( 12 ) ^ 3 x 6 ^ 3 Γ· 432 = y ? | "5184 e" | a = 12 ** 3
b = 6 ** 3
c = a * b
d = c / 432
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a ) 22 , b ) 50 , c ) 42.8 , d ) 288 , e ) 12 | c | divide(divide(subtract(105, multiply(multiply(5, const_0_2778), 5)), 10), const_0_2778) | a train 105 m long passes a man , running at 5 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 = ( 105 / 10 ) m / sec = ( 21 / 2 ) m / sec . [ ( 21 / 2 ) * ( 18 / 5 ) ] km / hr = 37.8 km / hr . let the speed of the train be x km / hr . then , relative speed = ( x - 5 ) km / hr . x - 5 = 37.8 = = > x = 42.8 km / hr . answer : c | a = 5 * const_0_2778
b = a * 5
c = 105 - b
d = c / 10
e = d / const_0_2778
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a ) 19 , b ) 25 , c ) 23 , d ) 20 , e ) 26 | b | add(const_12, multiply(18, const_4)) | find the greatest number that will divide 256 , 193 and 210 leaving 6 , 18 and 100 as remainders respectively | "explanation : answer will be hcf of ( 256 - 6 , 193 - 18 , 210 - 10 ) hcf of ( 250 , 175 , 200 ) = 25 option b" | a = 18 * 4
b = 12 + a
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a ) 100 % , b ) 200 % , c ) 300 % , d ) 600 % , e ) 800 % | c | multiply(const_100, divide(const_2, const_2)) | the area of a circle is increased by 1500 % . by what percent has the radius of the circle increased ? | "the area of the circle is increased by 1500 % , thus the area is increased 16 times . the area of a circle it proportional to the square of the radius ( area = Ο r ^ 2 ) , therefore the radius must increase 4 times ( diameter increase 4 times = area increase 16 times ) , which is increase by 500 % . answer : c" | a = 2 / 2
b = 100 * a
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a ) 1500 , b ) 1600 , c ) 1750 , d ) 1900 , e ) 2000 | e | subtract(multiply(multiply(const_0_25, const_1000), const_10), divide(add(500, multiply(multiply(const_0_25, const_1000), const_10)), 6)) | the total price of a basic computer and printer are $ 2,500 . if the same printer had been purchased with an enhanced computer whose price was $ 500 more than the price of the basic computer , then the price of the printer would have been 1 / 6 of that total . what was the price of the basic computer ? | let the price of basic computer be c and the price of the printer be p : c + p = $ 2,500 . the price of the enhanced computer will be c + 500 and total price for that computer and the printer will be 2,500 + 500 = $ 3,000 . now , we are told that the price of the printer is 1 / 6 of that new total price : p = 1 / 6 * $ 3,000 = $ 500 . plug this value in the first equation : c + 500 = $ 2,500 - - > c = $ 2000 answer : e . | a = const_0_25 * 1000
b = a * 10
c = const_0_25 * 1000
d = c * 10
e = 500 + d
f = e / 6
g = b - f
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a ) $ 18,400 , b ) $ 19,700 , c ) $ 20,200 , d ) $ 20,400 , e ) $ 21,100 | b | multiply(const_2, const_10) | a certain car ' s price decreased by 2.5 % ( from the original price ) each year from 1996 to 2002 , during that time the owner of the car invested in a new carburetor and a new audio system for the car , which increased car ' s price by $ 1,000 . if the price of the car in 1996 was $ 22,000 , what is the car ' s price in 2002 ? | "price in 96 = 22000 price decrease each year = 2.5 / 100 * 22000 = 550 price in 97 = 22000 - 550 price in 98 = 22000 - 2 * 550 price in 99 = 22000 - 3 * 550 price in 00 = 22000 - 4 * 550 price in 01 = 22000 - 5 * 550 price in 02 = 22000 - 6 * 550 = 18700 investment in the car = 1000 net price of the car in 02 = 18700 + 1000 = $ 19700 correct option : b" | a = 2 * 10
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a ) 60 % , b ) 70 % , c ) 80 % , d ) 90 % , e ) 50 % | e | multiply(divide(multiply(divide(20, const_100), multiply(divide(subtract(const_100, 25), const_100), const_100)), multiply(divide(30, const_100), const_100)), const_100) | last week john spent 30 percent of his wages on recreation . this week , his wages are 25 percent less than last week ΚΌ s wages and he spent 20 percent of his wages on recreation . the amount he spends on recreation this week is what percent of the amount he spent on recreation last week ? | say john ' s wages last week were $ 100 , so he spent 0.30 * 100 = $ 30 on recreation ; this week ' s wages is 0.75 * 100 = $ 75 , so he spends 0.2 * 75 = $ 15 on recreation ; 15 / 30 = 0.50 , hence the amount he spends on recreation this week is 50 % of the amount he spent on recreation last week : 30 * 0.5 = 15 answer : e | a = 20 / 100
b = 100 - 25
c = b / 100
d = c * 100
e = a * d
f = 30 / 100
g = f * 100
h = e / g
i = h * 100
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a ) 1 % , b ) 2 % , c ) 1.5 % , d ) 3 % , e ) 4 % | a | divide(subtract(20000, 10000), 10000) | in 2000 there were 10000 women stockbrokers in the united states . by 2005 the number had increased to 20000 . approximately what was the percent increase ? | the percent increase is ( 20000 - 10000 ) / 10000 = 10000 / 10000 = 1 % so the approximate answer is a | a = 20000 - 10000
b = a / 10000
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a ) 1000 , b ) 1500 , c ) 2000 , d ) 2500 , e ) 3500 | c | subtract(2240, multiply(subtract(2600, 2240), 5)) | a sum of money at simple interest amounts to rs . 2240 in 2 years and to rs . 2600 in 5 years . what is the principal amount | "explanation : si for 3 year = 2600 - 2240 = 360 si for 2 year 360 / 3 * 2 = 240 principal = 2240 - 240 = 2000 answer : option c" | a = 2600 - 2240
b = a * 5
c = 2240 - b
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a ) 6 , b ) 10 , c ) 12 , d ) 5 , e ) 8 | d | subtract(add(divide(multiply(9, 8), 3), 6), 25) | if β + β means β Γ β , β - β means β + β , β Γ β means β Γ· β and β Γ· β means β - β then find the value of ; 6 β 9 + 8 Γ 3 Γ· 25 = | d 5 | a = 9 * 8
b = a / 3
c = b + 6
d = c - 25
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a ) 9 : 5 , b ) 2 : 3 , c ) 4 : 7 , d ) 5 : 6 , e ) none of these | a | divide(multiply(3, 3), 5) | a dog takes 3 leaps for every 5 leaps of a hare . if one leap of the dog is equal to 3 leaps of the hare , the ratio of the speed of the dog to that of the hare is : | "explanation : dog : hare = ( 3 * 3 ) leaps of hare : 5 leaps of hare = 9 : 5 . answer : a" | a = 3 * 3
b = a / 5
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a ) 64 , b ) 32 , c ) 8 , d ) 4 , e ) 2 | d | sqrt(power(power(12, divide(const_1, const_2)), const_3)) | what positive number , when squared , is equal to the cube of the positive square root of 12 ? | "let the positive number be x x ^ 2 = ( ( 12 ) ^ ( 1 / 2 ) ) ^ 3 = > x ^ 2 = 4 ^ 3 = 16 = > x = 4 answer d" | a = 1 / 2
b = 12 ** a
c = b ** 3
d = math.sqrt(c)
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a ) 65 mtr . , b ) 28 mtr , c ) 70 mtr . , d ) 78 mtr . , e ) 17 mtr . | b | multiply(112, divide(multiply(10, 3), multiply(20, 6))) | if 20 men can build a wall 112 metres long in 6 days , what length of a similar wall can be built by 10 men in 3 days ? | "20 men is 6 days can build 112 metres 10 men in 3 days can build = 112 * ( 10 / 20 ) x ( 3 / 6 ) = 28 meters answer : b ." | a = 10 * 3
b = 20 * 6
c = a / b
d = 112 * c
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a ) rs . 1200 , b ) rs . 1500 , c ) rs . 1920 , d ) rs . 2000 , e ) none of these | c | multiply(divide(divide(multiply(15, const_2), 10), add(divide(multiply(15, const_2), 10), divide(multiply(15, const_2), 15))), 3200) | a alone can finish a work in 10 days which b alone can finish in 15 days . if they work together and finish it , then out of a total wages of rs . 3200 , a will get : | "explanation : ratio of working days of a : b = 10 : 15 therefore , their wages ratio = reverse ratio = 15 : 10 therefore , a will get 15 units of ratio total ratio = 25 1 unit of ratio = 3000 / 25 = 128 so , a β s amount = 120 Γ 15 = rs . 1920 . answer : option c" | a = 15 * 2
b = a / 10
c = 15 * 2
d = c / 10
e = 15 * 2
f = e / 15
g = d + f
h = b / g
i = h * 3200
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a ) 1 / 8 , b ) 2 / 15 , c ) 3 / 15 , d ) 1 / 4 , e ) 3 / 8 | a | divide(add(3, 3), add(multiply(divide(3, 6), const_60), multiply(divide(3, 10), const_60))) | triathlete dan runs along a 3 - mile stretch of river and then swims back along the same route . if dan runs at a rate of 10 miles per hour and swims at a rate of 6 miles per hour , what is his average rate for the entire trip in miles per minute ? | "dan travels 6 miles round trip . running part : ( 3 / 10 * 60 = 18 minutes ) swimming part : ( 3 / 6 * 60 = 30 minutes ) 6 miles in ( 18 + 30 ) minutes 6 / 48 = 1 / 8 mile per minute answer : 1 / 8 mile per minute a" | a = 3 + 3
b = 3 / 6
c = b * const_60
d = 3 / 10
e = d * const_60
f = c + e
g = a / f
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a ) 114 mph , b ) 110 mph , c ) 88 mph , d ) 100 mph , e ) 99 mph | a | divide(add(140, 88), const_2) | i flew my tiny seaplane to visit my mother . on the flight up , i flew at 140 mph . on the way home , i flew 88 mph . what was my average speed for the trip ? | "( 140 mph + 88 mph ) / 2 = 114 mph correct answer is : a" | a = 140 + 88
b = a / 2
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a ) 3 , b ) 14 , c ) 5 , d ) 6 , e ) 7 | a | divide(sqrt(50), const_2) | how many even integers from 1 to 50 ( both inclusive ) have even number of factors ? | "integers having even number of factors will be perfect squares . even numbers will have even perfect squares . thus , the possible values for the perfect squares are : 4 , 16,36 and the corresponding integers are 2 , 4,6 ( more than 2 ) . thus a is the correct answer ." | a = math.sqrt(50)
b = a / 2
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a ) 31 , b ) 38 , c ) 39 , d ) 40 , e ) 57 | b | add(25, const_1) | the average age of 25 students in a group is 12 years . when teacher ' s age is included to it , the average increases by one . what is the teacher ' s age in years ? | "age of the teacher = ( 26 * 13 - 25 * 12 ) = 38 years . answer : b" | a = 25 + 1
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a ) 15 % , b ) 20 % , c ) 25 % , d ) 29 % , e ) 80 % | d | multiply(divide(subtract(84.95, 59.95), 84.95), const_100) | a $ 84.95 lawn chair was sold for $ 59.95 at a special sale . by approximately what percent was the price decreased ? | "listed selling price of chair = 84.95 $ discounted selling price of chair = 59.95 $ discount = 84.95 - 59.95 = 25 $ % decrease in price of chair = ( 25 / 84.95 ) * 100 % = 29 % approx answer d" | a = 84 - 95
b = a / 84
c = b * 100
|
a ) 58 , b ) 69 , c ) 80 , d ) 91 , e ) 102 | c | divide(multiply(35, add(add(multiply(multiply(add(const_3, const_2), const_2), multiply(multiply(const_3, const_4), const_100)), multiply(multiply(add(const_3, const_4), add(const_3, const_2)), multiply(add(const_3, const_2), const_2))), add(const_3, const_3))), const_100) | what is 35 % of 4 / 13 of 845 ? | "this problem can be solved easily if we just use approximation : 35 % is a little over 1 / 3 , while 4 / 13 is a little less than 4 / 12 , which is 1 / 3 . thus , the answer is about 1 / 3 of 1 / 3 of 845 , or 1 / 9 of 845 . since the first 1 / 3 is a slight underestimate and the second 1 / 3 is a slight overestimate , the errors will partially cancel each other out . our estimate will be relatively accurate . the number 845 is a bit more than 810 , so ( 1 / 9 ) * 845 will be a bit more than 90 . keeping track not only of your current estimate , but also of the degree to which you have overestimated or underestimated , can help you pinpoint the correct answer more confidently . the closest answer is 91 , so this is the answer to choose . the answer is c ." | a = 3 + 2
b = a * 2
c = 3 * 4
d = c * 100
e = b * d
f = 3 + 4
g = 3 + 2
h = f * g
i = 3 + 2
j = i * 2
k = h * j
l = e + k
m = 3 + 3
n = l + m
o = 35 * n
p = o / 100
|
a ) s . 14991 , b ) s . 49930 , c ) s . 14045 , d ) s . 14999 , e ) s . 14578 | c | subtract(multiply(7800, add(4, const_1)), add(3850, add(add(5700, 8550), 6855))) | a grocer has a sale of rs . 5700 , rs . 8550 , rs . 6855 , rs . 3850 for 4 consecutive months . how much sale must he have in the fifth month so that he gets an average sale of rs . 7800 ? | total sale for 4 months = rs . ( 5700 + 8550 + 6855 + 3850 ) = rs . 24955 required sale = rs . [ ( 7800 x 5 ) - 24955 ] = rs . ( 39000 - 24955 ) = rs . 14045 option c | a = 4 + 1
b = 7800 * a
c = 5700 + 8550
d = c + 6855
e = 3850 + d
f = b - e
|
a ) a ) 6 , b ) b ) 14 , c ) c ) 11 , d ) d ) 15 , e ) e ) 23.5 | b | add(20, 5) | a shop produces sarongs . the daily average production is given by 5 n + 20 , where n is the number of workers aside from the owner . in the first k days , 500 units are produced , and then 5 workers are added to the team . after another k days , the cumulative total is 1200 . how many workers were part of the latter production run ? | "the daily average production is given by 5 n + 20 - given in the first k days , 500 units are produced = ( 5 n + 20 ) k = 500 k = 500 / 5 n + 20 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1 5 workers were added = 5 ( n + 5 ) + 20 = 5 n + 45 cumulative is 1200 . . thus for the current period = 1200 - 500 = 700 ( 5 n + 45 ) k = 700 k = 700 / 5 n + 45 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2 equate 1 and 2 500 / 5 n + 20 = 700 / 5 n + 45 500 ( 5 n + 45 ) = 700 ( 5 n + 20 ) 25 n + 225 = 35 n + 140 - 10 n = - 85 n = 8.5 thus n + 5 = 14 hence b" | a = 20 + 5
|
a ) 55 : 23 , b ) 26 : 27 , c ) 52 : 53 , d ) 45 : 53 , e ) 51 : 34 | b | divide(add(const_100, 4), add(const_100, 8)) | the cash difference between the selling prices of an book at a profit of 4 % and 8 % is $ 3 . the ratio of the two selling prices is : | "let c . p . of the book be $ x . then , required ratio = 104 % of x / 108 % of x = 104 / 108 = 26 / 27 = 26 : 27 b" | a = 100 + 4
b = 100 + 8
c = a / b
|
a ) 160 , b ) 152 , c ) 156 , d ) 144 , e ) none | a | multiply(39, divide(152, divide(add(add(35, 39), const_2), const_2))) | the ratio of ducks and frogs in a pond is 35 : 39 respectively . the average number of ducks and frogs in the pond is 152 . what is the number of frogs in the pond ? | solution : ratio of ducks and frogs in pond , = 35 : 39 . average of ducks and frogs in pond , = 152 . so , total number of ducks and frogs in the pond , = 2 * 152 = 304 . therefore , number of frogs , = ( 304 * 39 ) / 74 = 160 . answer : option a | a = 35 + 39
b = a + 2
c = b / 2
d = 152 / c
e = 39 * d
|
a ) 48 , b ) 28 , c ) 36 , d ) 72 , e ) 27 | b | divide(subtract(140, multiply(14, 8)), subtract(15, 14)) | suraj has a certain average of runs for 14 innings . in the 15 th innings he scores 140 runs thereby increasing his average by 8 runs . what is his average after the 15 th innings ? | "to improve his average by 8 runs per innings he has to contribute 14 x 8 = 112 runs for the previous 14 innings . thus , the average after the 15 th innings = 140 - 112 = 28 . answer : b" | a = 14 * 8
b = 140 - a
c = 15 - 14
d = b / c
|
a ) 22 hours , b ) 21 hours , c ) 24 hours , d ) 23 hours , e ) 25 days | c | inverse(subtract(inverse(8), inverse(12))) | a cistern is fille in 8 hours and it takes 12 hours when there is a leak in its bottom . if the cistern is full , in what time shall the leak empty it ? | explanation : work done by 1 hour by filling pipe = 1 / 8 work done by 1 hour by the leak and the filling pipe = 1 / 12 work done by 1 hour by the leak = 1 / 8 β 1 / 12 = 1 / 24 hence the leak can empty it in 24 hours . answer : option c | a = 1/(8)
b = 1/(12)
c = a - b
d = 1/(c)
|
a ) 22 , b ) 87 , c ) 77 , d ) 60 , e ) 85 | d | divide(multiply(12, 80), 16) | if 12 men do a work in 80 days , in how many days will 16 men do it ? | 12 * 80 = 16 * x x = 60 days answer : d | a = 12 * 80
b = a / 16
|
a ) 1 , b ) 5 , c ) 7 , d ) 8 , e ) 9 | a | divide(16, 16) | how many points ( x , y ) lie on the line segment between ( 22 , 12 2 / 3 ) and ( 16 , 17 2 / 3 ) such that x and y are both integers ? | "slope = ( 17 2 / 3 - 12 2 / 3 ) / ( 16 - 22 ) = - 5 / 6 y = mx + b = > 12 2 / 3 = - 22 * 5 / 6 + b = > b = 31 y = - 5 x / 6 + 31 only integer values work , and the only multiples of 6 between 16 and 22 for x values is 18 , thus 1 point . a" | a = 16 / 16
|
a ) 6 , b ) 36 , c ) 48 , d ) 60 , e ) 120 | e | lcm(15, 24) | a certain class of students is being divided into teams . the class can either be divided into 15 teams with an equal number of players on each team or 24 teams with an equal number of players on each team . what is the lowest possible number of students in the class ? | "let total no of students in the class be n so , we are told that n is divisible by both 15 24 so , lets find the least common multiple of 15 24 , ie 120 so our answer is ( e ) 120" | a = math.lcm(15, 24)
|
a ) 4.0 % , b ) 5.0 % , c ) 6.0 % , d ) 7.0 % , e ) 8.0 % | a | multiply(subtract(const_1, multiply(divide(80, const_100), divide(add(const_100, 20), const_100))), const_100) | a department store perceived a run on a lower priced garment and raised the price by 20 % to keep stock available until another shipment arrived . customers fumed and the department store manager allowed them to purchase the garments at 80 % of the new price to save face . what was the difference in the new retail price and the price the customers were allowed to pay instead ? | quantity x rate = price 1 x 1 = 1 0.8 x 1.20 = 0.960 decrease in price = ( 0.040 / 1 ) Γ£ β 100 = 4.0 % answer = option a | a = 80 / 100
b = 100 + 20
c = b / 100
d = a * c
e = 1 - d
f = e * 100
|
a ) 25 % , b ) 19 % , c ) 30 % , d ) 22.5 % , e ) 15 % | b | multiply(subtract(multiply(divide(add(const_100, 40), const_100), divide(subtract(const_100, 15), const_100)), const_1), const_100) | the price of a mobile was increased by 40 percent . the new price was then decreased by 15 percent . a single increase of what percent is equivalent to these two successive changes ? | consider base price - $ 100 25 % increase = 1.40 * 100 = $ 140 then a 15 % decrease on new price = 0.85 * 140 = $ 119 so final price of radio - $ 119 therefore a 19 % increase correct option - b | a = 100 + 40
b = a / 100
c = 100 - 15
d = c / 100
e = b * d
f = e - 1
g = f * 100
|
a ) rs . 80 , b ) rs . 120 , c ) rs . 71.15 , d ) rs . 51.15 , e ) rs . 180 | c | multiply(5, divide(1480, add(add(multiply(16, 3), multiply(9, 4)), multiply(4, 5)))) | a , b and c completed a piece of work , a worked for 16 days , b for 9 days and c for 4 days . their daily wages were in the ratio of 3 : 4 : 5 . find the daily wages of c , if their total earning was rs . 1480 ? | "3 x 4 x 5 x 16 9 4 48 x + 36 x + 20 x = 1480 104 x = 1480 = > x = 14.23 5 x = 71.15 rs . answer : c" | a = 16 * 3
b = 9 * 4
c = a + b
d = 4 * 5
e = c + d
f = 1480 / e
g = 5 * f
|
a ) 650 , b ) 730 , c ) 980 , d ) 1140 , e ) 1480 | d | multiply(subtract(const_1, divide(3, 20)), 20) | there are 20 students in a class . in how many different ways can a committee of 3 students be formed ? | "20 c 3 = 20 * 19 * 18 / 6 = 1140 the answer is d ." | a = 3 / 20
b = 1 - a
c = b * 20
|
a ) 0.092 , b ) 0.92 , c ) 9.2 , d ) 92 , e ) 920 | c | multiply(divide(0.004, 0.03), 69.28) | the closest approximation of w ( 69.28 Γ 0.004 ) / 0.03 is | "w ( 69.28 Γ 0.004 ) / 0.03 1 . 0.004 = 4 Γ 10 ^ ( - 3 ) 2 . 0.03 = 3 Γ 10 ^ ( - 2 ) 3 . ( a Γ b ) / c = a Γ ( b / c ) 4 . 0.004 / 0.03 = 4 Γ 10 ^ ( - 3 ) / ( 3 Γ 10 ^ ( - 2 ) ) = 4 Γ 10 ^ ( - 3 - ( - 2 ) ) / 3 = 4 Γ 10 ^ ( - 1 ) / 3 = ( 4 / 3 ) Γ 10 ^ ( - 1 ) = 1.333 Γ 10 ^ ( - 1 ) therefore , ( 69.28 Γ 0.004 ) / 0.03 = 69.28 Γ ( 0.004 / 0.03 ) = 69.28 Γ 1.33 Γ 10 ^ ( - 1 ) = 69.28 Γ 1.33 / 10 = 6.928 * 1.33 now , 7 Γ 2 = 14 7 Γ 1 = 7 or better : 6.9 Γ 1 = 6.9 6.9 Γ 2 = 13.8 hence , 6.9 < 6.928 Γ 1.33 < 13.8 9.2 is the only answer that satisfies this condition . c" | a = 0 / 4
b = a * 69
|
a ) 6 , b ) 12 , c ) 14 , d ) 18 , e ) 20 | c | divide(112, gcd(72, 112)) | the least number by which 72 must be multiplied in order to produce a multiple of 112 , is : | required number is divisible by 72 as well as by 112 , if it is divisible by their lcm , which is 1008 . now , 1008 when divided by 72 , gives quotient = 14 . therefore , required number = 14 . answer : c | a = math.gcd(72, 112)
b = 112 / a
|
a ) 72 min , b ) 62 min , c ) 70 min , d ) 74 min , e ) 76 min | a | multiply(multiply(const_2, divide(multiply(3, const_60), add(subtract(200, 50), multiply(const_2, 50)))), 50) | if there are 200 questions in a 3 hr examination . among these questions are 50 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 150 x + 50 * 2 x = 180 x = 180 / 250 x = 0.72 time taken for type a problem = 50 * 2 * 0.72 = 72 min answer : a" | a = 3 * const_60
b = 200 - 50
c = 2 * 50
d = b + c
e = a / d
f = 2 * e
g = f * 50
|
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