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 ) $ 700 , b ) $ 750 , c ) $ 800 , d ) $ 850 , e ) $ 900 | e | divide(multiply(250, multiply(multiply(const_2, const_100), const_100)), divide(multiply(multiply(const_2, const_100), const_100), const_4)) | if $ 5,000 is invested in an account at a simple annual rate of r percent , the interest is $ 250 . when $ 18,000 is invested at the same interest rate , what is the interest from the investment ? | "- > 250 / 5,000 = 5 % and 18,000 * 5 % = 900 . thus , e is the answer ." | a = 2 * 100
b = a * 100
c = 250 * b
d = 2 * 100
e = d * 100
f = e / 4
g = c / f
|
a ) 1 , b ) 2 , c ) 3 , d ) 4 , e ) 5 | c | divide(18, 6) | m and n are the x and y coordinates , respectively , of a point in the coordinate plane . if the points ( m , n ) and ( m + p , n + 18 ) both lie on the line defined by the equation x = ( y / 6 ) - ( 2 / 5 ) , what is the value of p ? | "x = ( y / 6 ) - ( 2 / 5 ) , and so y = 6 x + 12 / 5 . the slope is 6 . ( n + 18 - n ) / ( m + p - m ) = 6 p = 3 the answer is c ." | a = 18 / 6
|
a ) 10 , b ) 12 , c ) 14 , d ) 16 , e ) 18 | c | divide(subtract(30, const_2), const_2) | if w is the product of the integers from 1 to 30 , inclusive , what is the greatest integer k for which 3 ^ k is a factor of w ? | w = 30 ! 8 w = 30 x 29 x 28 x 27 x 26 x 25 x 24 x 24 x 22 x 21 x 20 x 19 x 18 x 17 x 16 x 15 x 14 x 13 x 12 x 11 x 10 x 09 x 8 x 7 x 6 x 5 x 4 x 3 x 2 x 1 out of these 30 , 27 , 24 , 21 , 18 , 15 , 12 , 09 , 06 , 3 are factors of 3 3 x 10 , 3 x 3 x 3 , 3 x 8 , 3 x 3 x 2 , 3 x 5 , 3 x 4 , 3 x 3 x 3 , 3 x 2 , 3 so we have a total of 14 three ' s . . . therefore the maximum value of k can be 14 ( c ) | a = 30 - 2
b = a / 2
|
a ) rs . 7500 , b ) rs . 9000 , c ) rs . 9500 , d ) rs . 10,000 , e ) none | a | multiply(divide(multiply(15000, multiply(const_12, const_2)), add(add(multiply(20000, multiply(const_12, const_2)), multiply(15000, multiply(const_12, const_2))), multiply(20000, subtract(multiply(const_12, const_2), 6)))), 25000) | a and b started a business in partnership investing rs . 20000 and rs . 15000 respectively . after 6 months , c joined them with rs . 20000 . what will be b ' s share in the total profit of rs . 25000 earned at the end of 2 years from the starting of the business ? | solution a : b : c = ( 20000 Γ 24 ) : ( 15000 Γ 24 ) : ( 20000 Γ 18 ) = 4 : 3 : 3 b ' s share = rs . ( 25000 Γ 3 / 10 ) = rs . 7500 . answer a | a = 12 * 2
b = 15000 * a
c = 12 * 2
d = 20000 * c
e = 12 * 2
f = 15000 * e
g = d + f
h = 12 * 2
i = h - 6
j = 20000 * i
k = g + j
l = b / k
m = l * 25000
|
a ) 50 m , b ) 57 m , c ) 80 m , d ) 82 m , e ) 84 m | b | divide(add(divide(5300, 26.50), multiply(const_2, 14)), const_4) | length of a rectangular plot is 14 mtr more than its breadth . if the cost of fencing the plot at 26.50 per meter is rs . 5300 , what is the length of the plot in mtr ? | "let breadth = x metres . then , length = ( x + 14 ) metres . perimeter = 5300 m = 200 m . 26.50 2 [ ( x + 14 ) + x ] = 200 2 x + 14 = 100 2 x = 86 x = 43 . hence , length = x + 14 = 57 m b" | a = 5300 / 26
b = 2 * 14
c = a + b
d = c / 4
|
a ) 1 , b ) 2 , c ) 7 , d ) 4 , e ) 5 | c | divide(log(multiply(9, 9)), log(const_10)) | 9 log 9 ( 7 ) = ? | "exponential and log functions are inverse of each other . hence aloga ( x ) = x , for all x real and positive . and therefore 9 log 9 ( 7 ) = 7 correct answer c" | a = 9 * 9
b = math.log(a)
c = math.log(10)
d = b / c
|
a ) 16 . , b ) 8 . , c ) 7 . , d ) 2 . , e ) - 2 . | d | subtract(add(4, 3), 5) | if ( a + b ) = 4 , ( b + c ) = 5 and ( c + d ) = 3 , what is the value of ( a + d ) ? | "given a + b = 4 b + c = 5 c + d = 3 ( a + b ) - ( b + c ) + ( c + d ) = ( a + d ) = > 4 - 5 + 3 = 2 . option d . . ." | a = 4 + 3
b = a - 5
|
a ) 12.8 , b ) 20 , c ) 10.4 , d ) 11.6 , e ) 15 | c | subtract(subtract(multiply(6500, power(add(const_1, divide(4, const_100)), 2)), 6500), multiply(multiply(6500, divide(4, const_100)), 2)) | indu gave bindu rs . 6500 on compound interest for 2 years at 4 % per annum . how much loss would indu has suffered had she given it to bindu for 2 years at 4 % per annum simple interest ? | "6500 = d ( 100 / 4 ) 2 d = 10.4 answer : c" | a = 4 / 100
b = 1 + a
c = b ** 2
d = 6500 * c
e = d - 6500
f = 4 / 100
g = 6500 * f
h = g * 2
i = e - h
|
a ) 2574 , b ) 2500 , c ) 1365 , d ) 1574 , e ) none of these | a | divide(84942, 33) | if the product of two numbers is 84942 and their h . c . f . is 33 , find their l . c . m . | "explanation : hcf * lcm = 84942 , because we know product of two numbers = product of hcf and lcm lcm = 84942 / 33 = 2574 option a" | a = 84942 / 33
|
a ) 6 only , b ) 6 and 12 both , c ) 12 only , d ) 18 only , e ) 12 and 18 both | b | add(multiply(6, const_100), multiply(2, 6)) | if n is a natural number , then ( 6 n ^ 2 + 6 n ) is always divisible by | "( 6 n ^ 2 + 6 n ) = 6 n ( n + 1 ) , which is divisible by both 6 and 12 , since n ( n + 1 ) is always even . answer b ." | a = 6 * 100
b = 2 * 6
c = a + b
|
a ) 1 : 21 , b ) 1 : 1 , c ) 2 : 1 , d ) 1 : 2 , e ) 2 : 11 | d | divide(divide(add(multiply(3, 3), 2), 3), divide(add(multiply(3, 7), 1), 3)) | express the ratio 3 2 / 3 : 7 1 / 3 in its simplest form . | solution we first convert the mixed numbers 3 2 / 3 and 7 1 / 3 into fractions 3 2 / 3 = 3 * 3 / 3 + 2 / 3 = 11 / 3 7 1 / 3 = 7 * 3 / 3 + 1 / 3 = 22 / 3 the ratio 3 2 / 3 : 7 1 / 3 can be expressed as 11 / 3 Γ· 22 / 3 = 11 / 3 Γ 3 / 22 simplify = 11 / 22 = 1 / 2 the ratio is 1 / 2 or 1 : 2 answer is d | a = 3 * 3
b = a + 2
c = b / 3
d = 3 * 7
e = d + 1
f = e / 3
g = c / f
|
a ) 75 m . , b ) 128 m . , c ) 150 m . , d ) 100 m . , e ) none of the above | b | multiply(64, const_2) | a runs twice as fast as b and gives b a start of 64 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 64 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 64 m is 64 / 1 = 64 m so the total time taken by a and b to reach = 2 * 64 = 128 m answer : b" | a = 64 * 2
|
a ) 212.5 , b ) 213.5 , c ) 214.5 , d ) 215.5 , e ) 216.5 | e | multiply(divide(55, 2.54), divide(25, 2.5)) | on a map , 2.5 inches represent 25 miles . how many miles approximately is the distance if you measured 55 centimeters assuming that 1 - inch is 2.54 centimeters ? | "1 inch = 2.54 cm 2.5 inch = 2.54 * 2.5 cm 6.35 cm = 25 miles 55 cms = 25 / 6.35 * 55 = 216.5 miles answer : e" | a = 55 / 2
b = 25 / 2
c = a * b
|
a ) 36 , b ) 40 , c ) 44 , d ) 50 , e ) 52 | d | add(40, divide(subtract(920, multiply(16, 40)), divide(multiply(16, add(const_100, 75)), const_100))) | a certain bus driver is paid a regular rate of $ 16 per hour for any number of hours that does not exceed 40 hours per week . for any overtime hours worked in excess of 40 hours per week , the bus driver is paid a rate that is 75 % higher than his regular rate . if last week the bus driver earned $ 920 in total compensation , how many total hours did he work that week ? | "for 40 hrs = 40 * 16 = 640 excess = 920 - 640 = 280 for extra hours = . 75 ( 16 ) = 12 + 16 = 28 number of extra hrs = 280 / 28 = 70 / 7 = 10 total hrs = 40 + 10 = 50 answer d 50" | a = 16 * 40
b = 920 - a
c = 100 + 75
d = 16 * c
e = d / 100
f = b / e
g = 40 + f
|
['a ) 0.005', 'b ) 0.002', 'c ) 0.001', 'd ) 0.0005', 'e ) 0.0002'] | b | divide(2, const_1000) | when magnified 1,000 times by an electron microscope , the image of a certain circular piece of tissue has a diameter of 2 centimeter . the actual diameter of the tissue , in centimeters , is | it is very easy if x is the diameter , then the magnified length is 1000 x . ince 1000 x = 2 then x = 2 / 1000 = 0.002 . the answer is b | a = 2 / 1000
|
a ) 40 , b ) 41 , c ) 42 , d ) 45 , e ) 43 | c | subtract(multiply(add(20, 1), add(21, 1)), multiply(20, 21)) | the average age of a class of 20 students is 21 years . the average increased by 1 when the teacher ' s age also included . what is the age of the teacher ? | "total age of all students = 20 Γ£ β 21 total age of all students + age of the teacher = 21 Γ£ β 22 age of the teacher = 21 Γ£ β 22 Γ’ Λ β 21 Γ£ β 20 = 21 ( 22 Γ’ Λ β 20 ) = 21 Γ£ β 2 = 42 answer is c ." | a = 20 + 1
b = 21 + 1
c = a * b
d = 20 * 21
e = c - d
|
a ) 9 , b ) 20 , c ) 55 , d ) 70 , e ) 80 | e | divide(add(multiply(divide(subtract(multiply(71.2, const_10), const_2), const_10), divide(11, divide(const_2, const_10))), 11), divide(11, divide(const_2, const_10))) | when the positive integer k is divided by the positive integer n , the remainder is 11 . if k / n = 71.2 , what is the value of n ? | "here ' s an approach that ' s based on number properties and a bit ofbrute forcemath : we ' re told that k and n are both integers . since k / n = 81.2 , we can say that k = 81.2 ( n ) n has tomultiply outthe . 2 so that k becomes an integer . with the answers that we have to work with , n has to be a multiple of 5 . eliminate a and e . with the remaining answers , we can test the answers and find the one that fits the rest of the info ( k / n = 81.2 and k / n has a remainder of 11 ) answer b : if n = 20 , then k = 1624 ; 1624 / 20 has a remainder of 4 not a match answer c : if n = 55 , then k = 4466 ; 4466 / 55 has a remainder of 11 match . final answer : e" | a = 71 * 2
b = a - 2
c = b / 10
d = 2 / 10
e = 11 / d
f = c * e
g = f + 11
h = 2 / 10
i = 11 / h
j = g / i
|
a ) 1 , b ) 2 , c ) 4 , d ) 8 , e ) 16 | d | multiply(divide(128, power(const_2, 5)), const_2) | if a and b are integers and ( a * b ) ^ 5 = 128 y , y could be : | distribute the exponent . a ^ 5 * b ^ 5 = 128 y find the prime factorization of 96 . this is 2 ^ 5 * 2 ^ 2 . we need 2 ^ 3 ( or some other power of 2 that will give us a multiple of 2 ^ 5 as our second term ) . 2 ^ 3 = 8 the answer is d . | a = 2 ** 5
b = 128 / a
c = b * 2
|
a ) a ) 45 , b ) b ) 33 , c ) c ) 48 , d ) d ) 55 , e ) e ) 61 | c | multiply(subtract(divide(multiply(24, 4), 3), 24), divide(add(subtract(35, 25), 50), subtract(35, 25))) | a train after traveling for 50 km meets with an accident and then proceeds at 3 / 4 of its former speed and arrives at its destination 35 minutes late . had the accident occurred 24 km farther , it would have reached the destination only 25 minutes late . what is the speed t of the train . | "let y be the balance distance to be covered and x be the former speed . a train after traveling for 50 km meets with an accident and then proceeds at 3 / 4 of its former speed and arrives at its destination 35 minutes late so , y / ( 3 x / 4 ) - y / x = 35 / 60 4 y / 3 x - y / x = 7 / 12 y / x ( 4 / 3 - 1 ) = 7 / 12 y / x * 1 / 3 = 7 / 12 y / x = 7 / 4 4 y - 7 x = 0 . . . . . . . . 1 had the accident occurred 24 km farther , it would have reached the destination only 25 minutes late so , ( y - 24 ) / ( 3 x / 4 ) - ( y - 24 ) / x = 25 / 60 4 ( y - 24 ) / 3 x - ( y - 24 ) / x = 5 / 12 ( y - 24 ) / x ( 4 / 3 - 1 ) = 5 / 12 ( y - 24 ) / x * 1 / 3 = 5 / 12 ( y - 24 ) * 12 = 3 x * 5 ( y - 24 ) * 4 = 5 x 4 y - 5 x = 96 . . . . . . . 2 eq 2 - eq 1 2 x = 96 x = 48 = t ans = c" | a = 24 * 4
b = a / 3
c = b - 24
d = 35 - 25
e = d + 50
f = 35 - 25
g = e / f
h = c * g
|
a ) 3 , b ) 7 , c ) 11 , d ) 21 , e ) 25 | c | add(4, subtract(add(12, 5), multiply(const_2, 5))) | jacob is now 12 years younger than michael . if 5 years from now michael will be twice as old as jacob , how old will jacob be in 4 years ? | "jacob = x years , mike = x + 12 years 5 years from now , 2 ( x + 5 ) = x + 17 2 x + 10 = x + 17 x = 7 x + 4 = 11 years answer c" | a = 12 + 5
b = 2 * 5
c = a - b
d = 4 + c
|
a ) 10 , b ) 100 , c ) 1000 , d ) 10000 , e ) none of these | b | multiply(1000, 10) | ( 1000 ) 7 Γ· ( 10 ) 19 = ? | "explanation : = ( 103 ) 7 / ( 10 ) 19 = ( 10 ) 21 / ( 10 ) 19 = 10 ( 2 ) = 100 option b" | a = 1000 * 10
|
a ) rs . 165.50 , b ) rs . 1700 , c ) rs . 175.50 , d ) rs . 180 , e ) none | a | divide(subtract(multiply(143, add(add(1, 1), 2)), add(126, 126)), 2) | tea worth rs . 126 per kg are mixed with a third variety in the ratio 1 : 1 : 2 . if the mixture is worth rs . 143 per kg , the price of the third variety per kg will be | "solution since first second varieties are mixed in equal proportions , so their average price = rs . ( 126 + 135 / 2 ) = rs . 130.50 so , the mixture is formed by mixing two varieties , one at rs . 130.50 per kg and the other at say , rs . x per kg in the ratio 2 : 2 , i . e . , 1 : 1 . we have to find x . x - 143 / 22.50 = 1 = Γ’ β¬ ΒΊ x - 143 = 22.50 = Γ’ β¬ ΒΊ x = 165.50 . hence , price of the third variety = rs . 165.50 per kg . answer a" | a = 1 + 1
b = a + 2
c = 143 * b
d = 126 + 126
e = c - d
f = e / 2
|
a ) 20 , b ) 24 , c ) 28 , d ) 32 , e ) 36 | d | subtract(56, multiply(3, divide(56, 7))) | 3 numbers are in the ratio 3 : 5 : 7 . the largest number is 56 . what is the difference between smallest and largest number ? | the three numbers are 3 x , 5 x , and 7 x . the largest number is 56 = 7 * 8 , so x = 8 . the smallest number is 3 * 8 = 24 . 56 - 24 = 32 the answer is d . | a = 56 / 7
b = 3 * a
c = 56 - b
|
a ) 6 , b ) 7 , c ) 8 , d ) 9 , e ) 10 | a | subtract(multiply(divide(70, subtract(subtract(const_100, 70), 20)), 2), 8) | mixture a is 20 % oil and 80 % material b by weight . if 2 more kilograms of oil are added to the 8 kilograms mixture a , how many kilograms of mixture a must be added to make a 70 % material b in the new mixture ? | old mix has 20 % oil . you add 2 kgs of 100 % oil . you get new mix with 30 % oil ( 70 % material b ) . w 1 / w 2 = ( 100 - 30 ) / ( 30 - 20 ) = 70 / 10 = 7 / 1 for every 7 parts of mix a , you have added 1 part of oil . since you have added 2 kgs oil , mix a must be 7 * 2 = 14 kgs . you already had 8 kgs of mix a , so extra mix a added must be 14 - 8 = 6 kgs . answer ( a ) | a = 100 - 70
b = a - 20
c = 70 / b
d = c * 2
e = d - 8
|
a ) $ 7456.00 , b ) $ 7500.55 , c ) $ 7428.57 , d ) $ 7852.56 , e ) $ 7864.00 | c | divide(add(5000, 200), subtract(const_1, divide(30, const_100))) | kanul spent $ 5000 in buying raw materials , $ 200 in buying machinery and 30 % of the total amount he had as cash with him . what was the total amount ? | "let the total amount be x then , ( 100 - 20 ) % of x = 5000 + 200 70 % of x = 5200 70 x / 100 = 5200 x = $ 52000 / 7 x = $ 7428.57 answer is c" | a = 5000 + 200
b = 30 / 100
c = 1 - b
d = a / c
|
a ) 2354 , b ) 2450 , c ) 2540 , d ) 2650 , e ) 3972 | e | subtract(multiply(add(multiply(const_100, const_100), multiply(multiply(const_100, divide(60, 6)), 3)), multiply(multiply(add(const_1, divide(divide(60, 6), const_100)), add(const_1, divide(divide(60, 6), const_100))), add(const_1, divide(divide(60, 6), const_100)))), add(multiply(const_100, const_100), multiply(multiply(const_100, divide(60, 6)), 3))) | there is 60 % increase in an amount in 6 yrs at si . what will be the ci of rs . 12,000 after 3 years at the same rate ? | "let p = rs . 100 . then , s . i . rs . 60 and t = 6 years . r = 100 x 60 = 10 % p . a . 100 x 6 now , p = rs . 12000 . t = 3 years and r = 10 % p . a . c . i . = rs . 12000 x 1 + 10 3 - 1 100 = rs . 12000 x 331 1000 = 3972 . e" | a = 100 * 100
b = 60 / 6
c = 100 * b
d = c * 3
e = a + d
f = 60 / 6
g = f / 100
h = 1 + g
i = 60 / 6
j = i / 100
k = 1 + j
l = h * k
m = 60 / 6
n = m / 100
o = 1 + n
p = l * o
q = e * p
r = 100 * 100
s = 60 / 6
t = 100 * s
u = t * 3
v = r + u
w = q - v
|
a ) 660 m , b ) 650 m , c ) 570 m , d ) 670 m , e ) 680 m | d | divide(multiply(18, multiply(1.8, const_1000)), 48) | amar takes as much time in running 18 meters as a car takes in covering 48 meters . what will be the distance covered by amar during the time the car covers 1.8 km ? | "distance covered by amar = ( 18 / 48 ) * 1800 = 670 m answer : d" | a = 1 * 8
b = 18 * a
c = b / 48
|
a ) 5 , b ) 7 , c ) 6 , d ) 11 , e ) 12 | a | divide(subtract(7, sqrt(subtract(power(7, 2), multiply(10, 10)))), 2) | find the value of x from the below equation : x ^ 2 β 7 x + 10 = 0 | "here we need to find out a and b such that a + b = - 7 and ab = + 10 a = - 5 and b = - 2 satisfies the above condition . hence x ^ 2 β 7 x + 10 = ( x β 5 ) ( x β 2 ) x 2 β 7 x + 10 = ( x β 5 ) ( x β 2 ) x ^ 2 β 7 x + 10 = 0 β ( x β 5 ) ( x β 2 ) = 0 x 2 β 7 x + 10 = 0 β ( x β 5 ) ( x β 2 ) = 0 step 3 : equate each factor to 0 and solve the equations ( x β 5 ) ( x β 2 ) = 0 β ( x β 5 ) = 0 or ( x β 2 ) = 0 β x = 5 or 2 a" | a = 7 ** 2
b = 10 * 10
c = a - b
d = math.sqrt(c)
e = 7 - d
f = e / 2
|
a ) 9998765 , b ) 9998907 , c ) 9999944 , d ) 9999954 , e ) 9999968 | e | multiply(add(add(add(add(multiply(const_100, const_100), multiply(const_100, const_10)), multiply(const_100, const_3)), multiply(7, const_10)), const_3), 88) | find the largest 7 digit number which is exactly divisible by 88 ? | "largest 7 digit number is 9999999 after doing 9999999 Γ· 88 we get remainder 31 hence largest 7 digit number exactly divisible by 88 = 9999999 - 31 = 9999968 e" | a = 100 * 100
b = 100 * 10
c = a + b
d = 100 * 3
e = c + d
f = 7 * 10
g = e + f
h = g + 3
i = h * 88
|
a ) 600 , b ) 640 , c ) 680 , d ) 720 , e ) 760 | c | multiply(divide(add(15, 53), const_2), divide(add(subtract(53, 15), 2), 2)) | in a theater , the first row has 15 seats and each row has 2 more seats than previous row . if the last row has 53 seats , what is the total number of seats in the theater ? | "the number of seats in the theater is 15 + ( 15 + 2 ) + . . . + ( 15 + 38 ) = 20 ( 15 ) + 2 ( 1 + 2 + . . . + 19 ) = 20 ( 15 ) + 2 ( 19 ) ( 20 ) / 2 = 20 ( 15 + 19 ) = 20 ( 34 ) = 680 the answer is c ." | a = 15 + 53
b = a / 2
c = 53 - 15
d = c + 2
e = d / 2
f = b * e
|
a ) a ) 4 , b ) b ) 8 , c ) c ) 12 , d ) d ) 16 , e ) e ) 20 | b | subtract(20, multiply(multiply(12, 12), 2)) | evaluate : 20 - 12 Γ· 12 Γ 2 = | "according to order of operations , 12 Γ· 12 Γ 2 ( division and multiplication ) is done first from left to right 12 Γ· 12 Γ 2 = 1 Γ 2 = 2 hence 20 - 12 Γ· 12 Γ 2 = 20 - 12 = 8 correct answer is b ) 8" | a = 12 * 12
b = a * 2
c = 20 - b
|
a ) 14 , b ) 20 , c ) 22 , d ) 24 , e ) 15 | e | multiply(add(2, 3), 3) | shipment - - - no . of defective chips / shipment - - - total chips in shipment s 1 - - - - - - - - - - - - - - - - - - - - - - 2 - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - 5,000 s 2 - - - - - - - - - - - - - - - - - - - - - - 4 - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - 12,000 s 3 - - - - - - - - - - - - - - - - - - - - - - 2 - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - 15,000 s 4 - - - - - - - - - - - - - - - - - - - - - - 4 - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - 16,000 a computer chip manufacturer expects the ratio of the number of defective chips to the total number of chips in all future shipments to equal the corresponding ratio for shipments s 1 , s 2 , s 3 , and s 4 combined , as shown in the table above . what β s the expected number of defective chips in a shipment of 60,000 chips ? | for a total of 51000 chips ( adding s 1 , s 2 , s 3 , s 4 ) total number of defective chips is 17 ( ( adding defective chips of s 1 , s 2 , s 3 , s 4 ) so ratio is 12 / 48000 or 1 every 4000 chips . keeping this ratio constant for 60000 chips number of defective chips will be ( 1 / 4000 ) * 60000 = 15 e | a = 2 + 3
b = a * 3
|
a ) 18 , b ) 23 , c ) 28 , d ) 32 , e ) 36 | b | multiply(2, 4) | three numbers are in the ratio of 2 : 3 : 4 and their l . c . m . is 276 . what is their h . c . f . ? | "let the numbers be 2 x , 3 x , and 4 x . lcm of 2 x , 3 x and 4 x is 12 x . 12 x = 276 x = 23 hcf of 2 x , 3 x and 4 x = x = 23 the answer is b ." | a = 2 * 4
|
['a ) 9 1 / 19', 'b ) 9 1 / 15', 'c ) 9 1 / 16', 'd ) 9 1 / 11', 'e ) 9 1 / 10'] | e | subtract(const_100, multiply(divide(const_1, add(const_1, divide(10, const_100))), const_100)) | if the length of a rectangle is increased by 10 % and the area is unchanged , then its corresponding breadth must be decreased by ? | suppose length = 100 m and breadth = x m area = 100 m new length = 110 m and breadth = ( x - y % of x ) then , \ inline 110 \ times ( x - \ frac { y } { 100 } x ) = 100 \ times x \ inline \ rightarrow \ inline 110 \ times ( 1 - \ frac { y } { 100 } ) = 100 \ inline \ rightarrow \ inline 1 - \ frac { y } { 100 } = \ frac { 100 } { 110 } \ inline \ rightarrow \ frac { y } { 100 } = 1 - \ frac { 100 } { 110 } = \ frac { 1 } { 11 } \ inline y = \ frac { 100 } { 11 } = 9 \ frac { 1 } { 10 } % answer : e | a = 10 / 100
b = 1 + a
c = 1 / b
d = c * 100
e = 100 - d
|
a ) 18000 , b ) 22000 , c ) 24000 , d ) 26000 , e ) 32000 | c | multiply(multiply(subtract(1, divide(3, 7)), multiply(multiply(multiply(const_0_25, const_100), const_100), const_10)), add(divide(1, 3), 1)) | in a mayoral election , candidate x received 1 / 3 more votes than candidate y , and candidate y received 1 / 7 fewer votes than z . if z received 21,000 votes how many votes did candidate x received ? | "z = 21 - - > y received 1 / 7 fewer votes than z - - > y = z - 1 / 7 * z = 18 ; x received 1 / 3 more votes than y - - > x = y + 1 / 3 * y = 24 . answer : c ." | a = 3 / 7
b = 1 - a
c = const_0_25 * 100
d = c * 100
e = d * 10
f = b * e
g = 1 / 3
h = g + 1
i = f * h
|
a ) 1 / 16 , b ) 5 / 42 , c ) 1 / 8 , d ) 3 / 16 , e ) 1 / 9 | e | divide(divide(choose(42, const_1), 42), power(const_3, const_2)) | each factor of 200 is inscribed on its own plastic ball , and all of the balls are placed in a jar . if a ball is randomly selected from the jar , what is the probability that the ball is inscribed with a multiple of 42 ? | 200 = 2 * 3 * 5 * 7 , so the # of factors 210 has is ( 1 + 1 ) ( 1 + 1 ) ( 1 + 1 ) ( 1 + 1 ) = 16 ( see below ) ; 42 = 2 * 3 * 7 , so out of 16 factors only two are multiples of 42 : 42 and 210 , itself ; so , the probability is 2 / 16 = 1 / 9 . answer : e . | a = math.comb(42, 1)
b = a / 42
c = 3 ** 2
d = b / c
|
a ) 14 , b ) 16 , c ) 21 , d ) 32 , e ) 48 | e | divide(336, divide(subtract(462, 336), 18)) | a car traveled 462 miles per tankful of gasoline on the highway and 336 miles per tankful of gasoline in the city . if the car traveled 18 fewer miles per gallon in the city than on the highway , how many miles per gallon did the car travel in the city ? | "i treat such problems as work ones . work = rate * time mileage ( m ) = rate ( mpg ) * gallons ( g ) x gallons is a full tank { 462 = rx { 336 = ( r - 18 ) x solve for r , r = 66 66 - 18 = 48 mpg e" | a = 462 - 336
b = a / 18
c = 336 / b
|
['a ) $ 1', 'b ) $ 2', 'c ) $ 4', 'd ) $ 8', 'e ) $ 16'] | e | multiply(4, divide(multiply(multiply(power(const_2, const_4), const_2), const_2), divide(multiply(multiply(const_2, power(const_2, const_2)), const_4), const_2))) | can v and can Π² are both right circular cylinders . the radius of can v is twice the radius of can b , while the height of can v is half the height of can b . if it costs $ 4.00 to fill half of can b with a certain brand of gasoline , how much would it cost to completely fill can v with the same brand of gasoline ? | let x be the radius of b and 2 h be the height of b . therefore , radius of v = 2 x and height = h vol of b = 3.14 * x ^ 2 * 2 h vol of a = 3.14 * 4 x ^ 2 * h cost to fill half of b = $ 4 - - > cost to fill full b = $ 8 - - > 3.14 * x ^ 2 * 2 h = 8 - - > 3.14 * x ^ 2 * h = 4 - - > 4 * ( 3.14 * x ^ 2 * h ) = $ 16 ans e | a = 2 ** 4
b = a * 2
c = b * 2
d = 2 ** 2
e = 2 * d
f = e * 4
g = f / 2
h = c / g
i = 4 * h
|
a ) 112.2 m , b ) 216.3 m , c ) 111.12 m , d ) 213.27 m , e ) 222.40 m | c | divide(multiply(const_1000, subtract(200, 100)), subtract(const_1000, 100)) | x can give y 100 meters start and z 200 meters start in x kilometer race . how much start can y give z in x kilometer race ? | c 111.12 m x runs 1000 m while y runs 900 m and z runs 800 m . the number of meters that z runs when y runs 1000 m , = ( 1000 * 800 ) / 900 = 8000 / 9 = 888.88 m . y can give z = 1000 - 888.88 = 111.12 m . | a = 200 - 100
b = 1000 * a
c = 1000 - 100
d = b / c
|
a ) 30 % , b ) 24 % , c ) 20 % , d ) 19 % , e ) 50 % | b | multiply(divide(subtract(2350, 1890), 1890), const_100) | a sum of money deposited at c . i . amounts to rs . 1890 in 2 years and to rs . 2350 in 3 years . find the rate percent ? | "1890 - - - 460 100 - - - ? = > 24 % answer : b" | a = 2350 - 1890
b = a / 1890
c = b * 100
|
a ) 1 , b ) 2 , c ) 3 , d ) 4 , e ) 5 | e | multiply(subtract(add(const_1, floor(divide(1053, 23))), divide(1053, 23)), 23) | what is the least number that should be added to 1053 , so the sum of the number is divisible by 23 ? | "( 1053 / 23 ) gives a remainder 18 so we need to add 5 . the answer is e ." | a = 1053 / 23
b = math.floor(a)
c = 1 + b
d = 1053 / 23
e = c - d
f = e * 23
|
a ) rs . 7000 , b ) rs . 8000 , c ) rs . 8500 , d ) rs . 9000 , e ) none | d | divide(add(add(add(add(8000, 5000), 16000), 7000), 9000), add(const_4, const_1)) | the salary of a , b , c , d , e is rs . 8000 , rs . 5000 , rs . 16000 , rs . 7000 , rs . 9000 per month respectively , then the average salary of a , b , c , d , and e per month is | "answer average salary = 8000 + 5000 + 16000 + 7000 + 9000 / 5 = rs . 9000 correct option : d" | a = 8000 + 5000
b = a + 16000
c = b + 7000
d = c + 9000
e = 4 + 1
f = d / e
|
a ) 183.62 , b ) 106.07 , c ) 106.04 , d ) 106.03 , e ) 186.01 | a | subtract(multiply(3000, power(add(1, divide(4, const_100)), divide(const_3, 2))), 3000) | find out the c . i on rs . 3000 at 4 % p . a . compound half - yearly for 1 1 / 2 years | "a = 3000 ( 51 / 50 ) 3 = 3183.62 3000 - - - - - - - - - - - 183.62 answer : a" | a = 4 / 100
b = 1 + a
c = 3 / 2
d = b ** c
e = 3000 * d
f = e - 3000
|
a ) 65 , b ) 69 , c ) 30 , d ) 31 , e ) 32 | c | multiply(multiply(subtract(divide(500, multiply(subtract(63, 3), const_0_2778)), const_1), const_10), const_2) | how many seconds will a 500 meter long train take to cross a man walking with a speed of 3 km / hr in the direction of the moving train if the speed of the train is 63 km / hr ? | "let length of tunnel is x meter distance = 800 + x meter time = 1 minute = 60 seconds speed = 78 km / hr = 78 * 5 / 18 m / s = 65 / 3 m / s distance = speed * time 800 + x = ( 65 / 3 ) * 60 800 + x = 20 * 65 = 1300 x = 1300 - 800 = 500 meters answer : option c" | a = 63 - 3
b = a * const_0_2778
c = 500 / b
d = c - 1
e = d * 10
f = e * 2
|
a ) rs . 4935 , b ) rs . 4925 , c ) rs . 5390 , d ) rs . 5290 , e ) rs . 5292 | e | add(divide(492, subtract(power(add(const_1, divide(5, const_100)), const_2), const_1)), 492) | the compound interest earned by sunil on a certain amount at the end of two years at the rate of 5 % p . a . was rs . 492 . find the total amount that sunil got back at the end of two years in the form of principal plus interest earned . | "let the sum be rs . p p { [ 1 + 5 / 100 ] 2 - 1 } = 492 p ( 5 / 100 ) ( 2 + 5 / 100 ) = 492 [ a 2 - b 2 = ( a - b ) ( a + b ) ] p = 492 / ( 0.05 ) ( 2.05 ) = 4800 . amount = 4800 + 492 = rs . 5292 answer : e" | a = 5 / 100
b = 1 + a
c = b ** 2
d = c - 1
e = 492 / d
f = e + 492
|
a ) 4.37 , b ) 5.28 , c ) 6.75 , d ) 8 , e ) 3.1 | b | divide(220, multiply(add(90, 60), const_0_2778)) | a train 220 m long is running with a speed of 90 km / hr . in what time will it pass a bus that is running with a speed of 60 km / hr in the direction opposite to that in which the train is going ? | "speed of train relative to bus = 90 + 60 = 150 km / hr . = 150 * 5 / 18 = 125 / 3 m / sec . time taken to pass the bus = 220 * 3 / 125 = 5.28 sec . answer : b" | a = 90 + 60
b = a * const_0_2778
c = 220 / b
|
a ) 5 , b ) 10 , c ) 12 , d ) 14 , e ) 15 | a | divide(subtract(57, multiply(const_3, 9)), multiply(const_3, const_2)) | a number is doubled and 9 is added . if resultant is trebled , it becomes 57 . what is that number | "explanation : = > 3 ( 2 x + 9 ) = 57 = > 2 x + 9 = 19 = > x = 5 answer : option a" | a = 3 * 9
b = 57 - a
c = 3 * 2
d = b / c
|
a ) 20 miles . , b ) 35 miles . , c ) 45 miles . , d ) 60 miles . , e ) 65 miles . | a | multiply(divide(10, add(5, 10)), 30) | tim and Γ© lan are 30 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 ? | "tim and elan will meet at the same time while their ratio of speed is 2 : 1 respectively . so their individual distance traveled ratio will be same . plugging in the answer choice only answer choice a meet the 2 : 1 ( tim : elan = 20 : 10 ) ratio of maintaining total distance traveled 30 miles so correct answer a" | a = 5 + 10
b = 10 / a
c = b * 30
|
a ) 10 hrs , b ) 12 hrs , c ) 16 hrs , d ) 18 hrs , e ) 20 hrs | c | divide(add(const_1, divide(const_1, const_3)), add(inverse(20), inverse(30))) | 2 pipes can separately fill a tank in 20 hrs and 30 hrs respectively . both the pipes are opened tofill the tank , a leak develops in the tank through which one - third of water supplied by both the pipes goes out . what is the total time taken to fill the tank ? | "1 / 20 + 1 / 30 = 1 / 12 1 + 1 / 3 = 4 / 3 1 - - - 12 4 / 3 - - - ? 4 / 3 * 12 = 16 hrs c" | a = 1 / 3
b = 1 + a
c = 1/(20)
d = 1/(30)
e = c + d
f = b / e
|
a ) 2 , b ) 4 , c ) 8 , d ) 69 , e ) 32 | d | add(subtract(99, add(add(add(15, 10), 5), 3)), 3) | there are 99 people that own pets . 15 people own only dogs , 10 people own only cats , 5 people own only cats and dogs , 3 people own cats , dogs and snakes . how many total snakes are there ? | "lets assign variables to all the areas in venn diagram of three . three different units are dog , cat , snake = total = 99 only dog = d = 15 only cat = c = 10 only snake = s exactly dog and cat = 5 exactly dog and snake = x exactly cat and snake = y all three = 3 so 99 = 15 + 10 + 5 + 3 + x + y + s we need to know total snakes = x + y + s + 3 = 69 answer : d" | a = 15 + 10
b = a + 5
c = b + 3
d = 99 - c
e = d + 3
|
a ) 42 , b ) 70 , c ) 140 , d ) 165 , e ) 252 | e | multiply(multiply(9, 3), 7) | a certain university will select 1 of 7 candidates eligible to fill a position in the mathematics department and 2 of 9 candidates eligible to fill 2 identical positions in the computer science department . if none of the candidates is eligible for a position in both departments , how many different sets of 3 candidates are there to fill the 3 positions ? | "ans : 252 7 c 1 * 9 c 2 answer e )" | a = 9 * 3
b = a * 7
|
a ) 1.2 , b ) 4.8 , c ) 4.4 , d ) 3.21 , e ) none | b | divide(multiply(multiply(8, 12), 5), const_100) | the simple interest on rs . 8 for 12 months at the rate of 5 paise per rupeeper month is | "sol . s . i . = rs . [ 8 * 5 / 100 * 12 ] = rs . 4.8 answer b" | a = 8 * 12
b = a * 5
c = b / 100
|
a ) 23 , b ) 64.8 , c ) 37 , d ) 30 , e ) 28 | b | multiply(divide(360, 24), const_3_6) | an athlete runs 360 metres race in 24 seconds . what is his speed ? | "speed = distance / time = 360 / 20 = 18 m / s = 18 * 18 / 5 = 64.8 km / hr answer : b" | a = 360 / 24
b = a * const_3_6
|
a ) 4 : 9 , b ) 4 : 3 , c ) 4 : 5 , d ) 3 : 1 , e ) 4 : 2 | d | divide(sqrt(36), sqrt(4)) | two trains , one from howrah to patna and the other from patna to howrah , start simultaneously . after they meet , the trains reach their destinations after 4 hours and 36 hours respectively . the ratio of their speeds is ? | "let us name the trains a and b . then , ( a ' s speed ) : ( b ' s speed ) = β b : β a = β 36 : β 4 = 6 : 2 = 3 : 1 answer : d" | a = math.sqrt(36)
b = math.sqrt(4)
c = a / b
|
a ) 40 % , b ) 50 % , c ) 36.67 % , d ) 70 % , e ) 75 % | c | multiply(divide(subtract(120, add(multiply(10, 6), multiply(6, 10))), 120), const_100) | a batsman scored 120 runs which included 10 boundaries and 6 sixes . what % of his total score did he make by running between the wickets | "number of runs made by running = 120 - ( 10 x 4 + 6 x 6 ) = 120 - ( 76 ) = 44 now , we need to calculate 60 is what percent of 120 . = > 44 / 120 * 100 = 36.67 % c" | a = 10 * 6
b = 6 * 10
c = a + b
d = 120 - c
e = d / 120
f = e * 100
|
a ) 50 , b ) 64 , c ) 70 , d ) 80 , e ) 90 | b | add(add(multiply(divide(54, 5), const_2), 5), add(divide(54, 5), 5)) | the sum of the present ages of two persons a and b is 54 . if the age of a is twice that of b , find the sum of their ages 5 years hence ? | "a + b = 54 , a = 2 b 2 b + b = 54 = > b = 18 then a = 36 . 5 years , their ages will be 41 and 23 . sum of their ages = 41 + 23 = 64 . answer : b" | a = 54 / 5
b = a * 2
c = b + 5
d = 54 / 5
e = d + 5
f = c + e
|
a ) 7 % , b ) 9 % , c ) 11 % , d ) 12 % , e ) none of these | b | divide(multiply(6, 18), 12) | if 12 % of x is equal to 6 % of y , then 18 % of x will be equal to how much % of y ? | we have , 12 % of x = 6 % of y = > 2 % of x = 1 % of y = > ( 2 x 9 ) % of x = ( 1 x 9 ) % of y thus , 18 % of x = 9 % of y . answer : b | a = 6 * 18
b = a / 12
|
a ) 2.8 liters . , b ) 2.5 liters . , c ) 8.5 liters . , d ) 2.6 liters . , e ) 10 liters . | e | divide(multiply(divide(subtract(const_100, 90), const_100), 80), divide(subtract(const_100, 20), const_100)) | heinz produces tomato puree by boiling tomato juice . the tomato puree has only 20 % water while the tomato juice has 90 % water . how many liters of tomato puree will be obtained from 80 litres of tomato juice ? | "answer : explanation : in each of the solutions , there is a pure tomato component and some water . so while boiling , water evaporates but tomato not . so we equate tomato part in the both equations . Γ’ β‘ β Γ’ β‘ β 10 % ( 80 ) = 80 % ( x ) Γ’ β‘ β Γ’ β‘ β x = 10 liters . answer : e" | a = 100 - 90
b = a / 100
c = b * 80
d = 100 - 20
e = d / 100
f = c / e
|
a ) 60 , b ) 65 , c ) 55 , d ) 50 , e ) 55 | b | add(divide(const_100, const_2), divide(multiply(30, divide(const_100, const_2)), const_100)) | a technician makes a round - trip to and from a certain service center by the same route . if the technician completes the drive to the center and then completes 30 percent of the drive from the center , what percent of the round - trip has the technician completed ? | round trip means 2 trips i . e . to and fro . he has completed one i . e 50 % completed . then he traveled another 30 % of 50 % i . e 15 % . so he completed 50 + 15 = 65 % of total trip . b | a = 100 / 2
b = 100 / 2
c = 30 * b
d = c / 100
e = a + d
|
a ) a ) 12 , b ) b ) 14 , c ) c ) 16 , d ) d ) 18 , e ) e ) 22 | b | floor(divide(127, 9)) | on dividing 127 by a number , the quotient is 9 and the remainder is 1 . find the divisor . | "d = ( d - r ) / q = ( 127 - 1 ) / 9 = 126 / 9 = 14 b )" | a = 127 / 9
b = math.floor(a)
|
a ) 3 , b ) 4 , c ) 5 , d ) 6 , e ) 9 | a | multiply(6, const_2) | if x + | x | + y = 7 and x + | y | - y = 6 , then x + y = | "if x + | x | + y = 7 and x + | y | - y = 6 , then x + y = can be done in 2.3 mins : there are 4 cases to be tested : 1 ) x is - ve and y is - ve substituting in the equation , we get x - x + y = 7 and x - y - y = 6 solve for x and y we get x = 20 and y = 7 , so x + y = 27 reject 2 ) x is + ve and y is + ve substitute in the equation , we ger x + x + y = 7 and x + y - y = 6 solve for x and y we get x = 6 and y = - 5 , therefore x + y = 1 not on list so reject 3 ) x is - ve and y is + ve substitute , we get x - x = y = 7 and x + y - y = 6 solve fo x and y we get x = 6 and y = 7 , x + y = 13 not on list so reject 4 ) x is + ve and y is - ve substitute , we get x + x = y = 7 and x - y - y = 6 solve for x and y , we get x = 4 and y = - 1 , x + y = 3 , answer choice answer a" | a = 6 * 2
|
a ) 99 , b ) 450 , c ) 350 , d ) 882 , e ) 281 | b | subtract(multiply(divide(300, 18), 45), 300) | a 300 meter long train crosses a platform in 45 seconds while it crosses a signal pole in 18 seconds . what is the length of the platform ? | "speed = [ 300 / 18 ] m / sec = 50 / 3 m / sec . let the length of the platform be x meters . then , x + 300 / 45 = 50 / 3 3 ( x + 300 ) = 2250 Γ¨ x = 450 m . answer : b" | a = 300 / 18
b = a * 45
c = b - 300
|
a ) 24 , b ) 28 , c ) 32 , d ) 56 , e ) 60 | a | multiply(divide(divide(add(add(multiply(multiply(const_4, const_2), const_10), multiply(multiply(const_4, const_2), const_100)), multiply(const_12, const_1000)), 230), 7), 3) | the ratio between the number of sheep and the number of horses at the stewart farm is 3 to 7 , if each horse is fed 230 ounces of horse food per day and the farm needs a total 12,880 ounces of horse food per day , what is the number of sheep in the farm ? | let the number of sheeps and horses be 3 x and 7 x . now total number of horses = total consumption of horse food / consumption per horse = 12880 / 230 = 56 , which is equal to 7 x . = > x = 8 sheeps = 3 x = 3 * 8 = 24 . hence a | a = 4 * 2
b = a * 10
c = 4 * 2
d = c * 100
e = b + d
f = 12 * 1000
g = e + f
h = g / 230
i = h / 7
j = i * 3
|
a ) 23 / 40 , b ) 17 / 30 , c ) 12 / 25 , d ) 7 / 15 , e ) 3 / 10 | c | multiply(divide(subtract(divide(multiply(4, multiply(multiply(4, 4), 5)), 5), divide(multiply(divide(multiply(4, multiply(multiply(4, 4), 5)), 5), 4), 4)), subtract(multiply(multiply(4, 4), 5), divide(multiply(3, multiply(multiply(4, 4), 5)), 4))), const_100) | one day a car rental agency rented 3 / 4 of its cars , including 4 / 5 of its cars with cd players . if 3 / 5 of its cars have cd players , what percent of the cars that were not rented had cd players ? | "the cars with cd players which were not rented is ( 1 / 5 ) ( 3 / 5 ) = 3 / 25 of all the cars . the cars which were not rented is 1 / 4 of all the cars . the percent of non - rented cars which had cd players is ( 3 / 25 ) / ( 1 / 4 ) = 12 / 25 the answer is c ." | a = 4 * 4
b = a * 5
c = 4 * b
d = c / 5
e = 4 * 4
f = e * 5
g = 4 * f
h = g / 5
i = h * 4
j = i / 4
k = d - j
l = 4 * 4
m = l * 5
n = 4 * 4
o = n * 5
p = 3 * o
q = p / 4
r = m - q
s = k / r
t = s * 100
|
a ) 328359 , b ) 557321 , c ) 258327 , d ) 458329 , e ) 4422 | d | power(add(power(add(power(add(power(add(1, 1), 2), 1), 2), 1), 2), 1), 2) | if a ( k ) = ( k + 1 ) ^ 2 , and k = 1 , what is the value of a ( a ( a ( a ( k ) ) ) ) ? | a ( a ( a ( a ( 1 ) ) ) ) = a ( a ( a ( 4 ) ) ) = a ( a ( 25 ) ) = a ( 676 ) = 677 ^ 2 = 458329 thus , the answer is d . | a = 1 + 1
b = a ** 2
c = b + 1
d = c ** 2
e = d + 1
f = e ** 2
g = f + 1
h = g ** 2
|
a ) 80 , b ) 90 , c ) 95 , d ) 100 , e ) 108 | e | multiply(multiply(multiply(const_100, divide(add(const_100, 20), const_100)), divide(subtract(const_100, 25), const_100)), divide(add(const_100, 20), const_100)) | from the beginning to the end of 2007 , the price of a stock rose 20 percent . in 2008 , it dropped 25 percent . in 2009 , it rose 20 percent . what percent of the stock β s 2007 starting price was the price of the stock at the end of 2009 ? | "assume a value at the beginning of 2007 . as this is a % question , assume p = 100 . at the end of 2007 it becmae = 1.2 * 100 = 120 at the end of 2008 it decreased by 25 % = 120 * . 75 = 90 at the end of 2009 it increased by 20 % = 90 * 1.2 = 108 thus ratio = 108 / 100 = 1.08 ( in % terms = 108 % ) . thus e is the correct answer ." | a = 100 + 20
b = a / 100
c = 100 * b
d = 100 - 25
e = d / 100
f = c * e
g = 100 + 20
h = g / 100
i = f * h
|
a ) rs 48 , b ) rs 45 , c ) rs 40 , d ) rs 50 , e ) rs 55 | c | multiply(divide(const_100, 20), 8) | a 8 % stock yields 20 % . the market value of the stock is : | "explanation : for an income of rs . 20 , investment = rs . 100 . for an income of rs 8 , investment = rs . 100 / 20 x 8 = rs 40 market value of rs . 100 stock = rs 40 answer is c" | a = 100 / 20
b = a * 8
|
a ) 12.75 , b ) 50 , c ) 85 , d ) 204 , e ) none | c | power(add(510, const_4), const_4) | the difference between a number and its two fifth is 510 . what is 10 % of that number ? | "solution let the number be x . then , x - 2 / 5 x = 510 . βΉ = βΊ 3 / 5 x = 510 βΉ = βΊ x = [ 510 x 5 / 3 ] = 850 . therefore 10 % of 850 = 85 . answer c" | a = 510 + 4
b = a ** 4
|
a ) 68 % , b ) 70 % , c ) 72 % , d ) 74 % , e ) 77 % | d | divide(add(multiply(70, 65), multiply(95, subtract(100, 70))), 100) | in a certain accounting class of 100 students , 70 % of the students took the final exam on the assigned day while the rest of the students took the exam on a make - up date . if the students on the assigned day had an average score of 65 % , and the students on the make - up date had an average score of 95 % , what was the average score for the entire class ? | "70 % of the class scored 65 % and 30 % of the class scored 95 % . the difference between 65 % and 95 % is 30 % . the average will be 65 % + 0.3 ( 30 % ) = 74 % . the answer is d ." | a = 70 * 65
b = 100 - 70
c = 95 * b
d = a + c
e = d / 100
|
a ) 20 , b ) 40 , c ) 60 , d ) 80 , e ) none of these | a | gcd(1020, 860) | the maximum number of student amoung them 1020 pens and 860 pencils can be distributed in such a way that each student gets the same number of pens and same number of pencils is : | "solution required number of student = h . c . f of 1020 and 860 = 20 . answer a" | a = math.gcd(1020, 860)
|
a ) two , b ) five , c ) six , d ) four , e ) one | a | multiply(2, const_1) | count the numbers between 10 - 99 which yield a remainder of 3 when divided by 9 and also yield a remainder of 2 when divided by 5 ? | answer = a ) two numbers between 10 - 99 giving remainder 3 when divided by 9 = 12 , 21 , 30 , 39 , 48 , 57 , 66 , 75 , 84 , 93 the numbers giving remainder 2 when divided by 5 = 12 , 57 = 2 | a = 2 * 1
|
a ) 10 , b ) 12 , c ) 13 , d ) 14 , e ) 15 | a | subtract(100, add(add(add(subtract(100, 80), subtract(100, 75)), subtract(100, 85)), subtract(100, 70))) | there were totally 100 men . 80 are married . 75 have t . v , 85 have radio , 70 have a . c . how many men have t . v , radio , a . c and also married ? | "100 - ( 100 - 80 ) - ( 100 - 75 ) - ( 100 - 85 ) - ( 100 - 70 ) = 100 - 20 - 25 - 15 - 30 = 100 - 90 = 10 answer : a" | a = 100 - 80
b = 100 - 75
c = a + b
d = 100 - 85
e = c + d
f = 100 - 70
g = e + f
h = 100 - g
|
a ) 1.75 , b ) 1 , c ) 2.25 , d ) 2.4 , e ) 3.2 | a | multiply(5, divide(350, const_1000)) | a boy wanted to calculate his speed on his bike . his starting point was 350 meters from the turning point . he made the round trip 5 times in 30 minutes . what was the boy ' s speed in kilometers per hour ? | the distance between the the starting point and the turning point was 350 meters . the boy made the round trip 5 times so that was 10 times he traveled 350 meters . 10 * 350 = 3500 meters or 3.5 km . 30 minutes is half of an hour . divide the 3500 meters in half to put in the right time interval 3500 / 2 = 1750 meters or 1.75 km correct answer is a | a = 350 / 1000
b = 5 * a
|
a ) 280 , b ) 370 , c ) 460 , d ) 550 , e ) 640 | b | divide(222, divide(60, const_100)) | if it is assumed that 60 percent of those who receive a questionnaire by mail will respond and 222 responses are needed , what is the minimum number of questionnaires that should be mailed ? | "let x be the minimum number of questionnaires to be mailed . 0.6 x = 222 x = 370 the answer is b ." | a = 60 / 100
b = 222 / a
|
a ) 675 , b ) 778 , c ) 654 , d ) 412 , e ) 589 | a | add(multiply(divide(subtract(add(multiply(add(25, 5), 3), 100), multiply(25, 3)), 5), 25), 100) | to deliver an order on time , a company has to make 25 parts a day . after making 25 parts per day for 3 days , the company started to produce 5 more parts per day , and by the last day of work 100 more parts than planned were produced . how many parts the company made ? | "let x be the number of days the company worked . then 25 x is the number of parts they planned to make . at the new production rate they made : 3 β
25 + ( x β 3 ) β
30 = 75 + 30 ( x β 3 ) therefore : 25 x = 75 + 30 ( x β 3 ) β 100 25 x = 75 + 30 x β 90 β 100 190 β 75 = 30 x β 25 115 = 5 x x = 23 so the company worked 23 days and they made 23 β
25 + 100 = 675 pieces . correct answer is a ) 675" | a = 25 + 5
b = a * 3
c = b + 100
d = 25 * 3
e = c - d
f = e / 5
g = f * 25
h = g + 100
|
a ) 442 , b ) 441 , c ) 440 , d ) 320 , e ) 532 | a | multiply(multiply(const_100.0, divide(12, 1092)), 3) | what annual installment will discharge a debt of rs . 1092 due in 3 years at 12 % simple interest ? | "let the installment be rs . x if he keeps the total amount for 3 years , he have to pay the interest of rs . ( 1092 * 12 * 3 ) / 100 . but he paid rs . x after 1 year . so he need not to pay interest for rs . x for remaining 2 years . i . e he need not to pay rs . ( x * 12 * 2 ) / 100 . and he paid another rs . x after 2 years . so he need not to pay interest for for remaining 1 years . i . e . rs . ( x * 12 * 1 ) / 100 . so the total interest he has to pay is rs . ( ( 1092 * 12 * 3 ) / 100 ) - [ ( ( x * 12 * 2 ) / 100 ) + ( ( x * 12 * 1 ) / 100 ) ] . so , the total amount he has to pay is rs . 1092 + rs . ( ( 1092 * 12 * 3 ) / 100 ) - [ ( ( x * 12 * 2 ) / 100 ) + ( ( x * 12 * 1 ) / 100 ) ] . the total amount he paid = 3 x . ( because he paid rs . x for three times ) 1092 + ( ( 1092 * 12 * 3 ) / 100 ) - [ ( ( x * 12 * 2 ) / 100 ) + ( ( x * 12 * 1 ) / 100 ) ] = 3 x by solving , we will get x = 442 . answer : a" | a = 12 / 1092
b = 100 * 0
c = b * 3
|
a ) 50 , b ) 60 , c ) 70 , d ) 80 , e ) 90 | e | add(multiply(divide(100, const_10), multiply(const_2, const_4)), divide(100, const_10)) | how many integerskgreater than 100 and less than 1100 are there such that if the hundreds and the units digits ofkare reversed , the resulting integer is k + 99 ? | numbers will be like 102 = > 201 = 102 + 99 203 = > 302 = 103 + 99 so the hundereth digit and units digit are consecutive where unit digit is bigger than hundred digit . there will be ninet pairs of such numbers for every pair there will 10 numbers like for 12 = > 102 , 112,132 , 142,152 , 162,172 , 182,192 . total = 9 * 10 = 90 hence e . | a = 100 / 10
b = 2 * 4
c = a * b
d = 100 / 10
e = c + d
|
a ) 35 m east , b ) 35 m north , c ) 30 m west , d ) 45 m west , e ) 30 m east | e | add(20, 10) | sandy walked 20 meters towards south . then sandy turned to her left and walked 20 meters . she then turned to her left and walked 20 meters . she then turned to her right and walked 10 meters . what distance is she from the starting point and in which direction ? | "the net distance is 20 + 10 = 30 meters to the east . the answer is e ." | a = 20 + 10
|
a ) 6000 , b ) 7000 , c ) 8000 , d ) 9000 , e ) 10000 | b | subtract(21000, multiply(divide(2, 3), 21000)) | income and expenditure of a person are in the ratio 3 : 2 . if the income of the person is rs . 21000 , then find his savings ? | let the income and the expenditure of the person be rs . 3 x and rs . 2 x respectively . income , 3 x = 21000 = > x = 7000 savings = income - expenditure = 3 x - 2 x = x so , savings = rs . 7000 . answer : b | a = 2 / 3
b = a * 21000
c = 21000 - b
|
a ) 144 , b ) 288 , c ) 12 , d ) 256 , e ) none | a | multiply(factorial(4), factorial(3)) | in how many ways 4 boys and 3 girls can be seated in a row so that they are alternate . | "solution : let the arrangement be , b g b g b g b 4 boys can be seated in 4 ! ways . girl can be seated in 3 ! ways . required number of ways , = 4 ! * 3 ! = 144 . answer : option a" | a = math.factorial(4)
b = math.factorial(3)
c = a * b
|
a ) 1 : 8 , b ) 1 : 6 , c ) 1 : 9 , d ) 16 : 9 , e ) 1 : 2 | d | divide(circle_area(4), circle_area(3)) | the ratio of the radius of two circles is 4 : 3 , and then the ratio of their areas is ? | "r 1 : r 2 = 4 : 3 Ο r 12 : Ο r 22 r 12 : r 22 = 16 : 9 answer : d" | a = circle_area / (
|
a ) 8 , b ) 7 , c ) 2 , d ) 6 , e ) 5 | a | subtract(subtract(28, const_10), const_10) | in a function , every person shakes hand with every other person . if there was a total of 28 handshakes in the function , how many persons were present in the function ? | suppose there are nn persons present in a function and every person shakes hand with every other person . then , total number of handshakes = nc 2 = n ( n β 1 ) / 2 n ( n β 1 ) = 28 Γ 2 n ( n β 1 ) = 56 n = 8 answer a 8 | a = 28 - 10
b = a - 10
|
a ) 121 , b ) 129 , c ) 138 , d ) 152 , e ) 170 | e | add(add(add(multiply(3, 16), multiply(3, 17)), multiply(3, 20)), 11) | 3 * 16 + 3 * 17 + 3 * 20 + 11 = ? | "3 * 16 + 3 * 17 + 3 * 20 + 11 = 48 + 51 + 60 + 11 = 170 the answer is e ." | a = 3 * 16
b = 3 * 17
c = a + b
d = 3 * 20
e = c + d
f = e + 11
|
a ) - 1.5 , b ) - 0.5 , c ) 0.5 , d ) 1.5 , e ) 2.5 | b | add(negate(8), 4) | on the number line , if x is halfway between - 8 and 4 , and if y is halfway between - 4 and 6 , what number is halfway between x and y ? | "x = - 2 and y = 1 . the answer is b ." | a = negate + (
|
a ) 100 , b ) 150 , c ) 160 , d ) 170 , e ) 180 | e | multiply(120, subtract(const_2, const_1)) | a train speeds past a pole in 15 sec and a platform 120 m long in 25 sec , its length is ? | "let the length of the train be x m and its speed be y m / sec . then , x / y = 15 = > y = x / 15 ( x + 120 ) / 25 = x / 15 = > x = 180 m . answer : option e" | a = 2 - 1
b = 120 * a
|
a ) 250 , b ) 276 , c ) 280 , d ) 310 , e ) none | d | divide(add(multiply(add(floor(divide(30, add(const_3, const_4))), const_1), 660), multiply(subtract(30, add(floor(divide(30, add(const_3, const_4))), const_1)), 240)), 30) | a library has an average of 660 visitors on sundays and 240 on other days . the average number of visitors per day in a month of 30 days beginning with a sunday is : | "since the month begins with sunday , to there will be five sundays in the month average required = ( 660 x 5 + 240 x 25 ) / 30 ) = 310 answer : option d" | a = 3 + 4
b = 30 / a
c = math.floor(b)
d = c + 1
e = d * 660
f = 3 + 4
g = 30 / f
h = math.floor(g)
i = h + 1
j = 30 - i
k = j * 240
l = e + k
m = l / 30
|
a ) 43 , 870.34 , b ) 56 , 876.45 , c ) 77 , 789.45 , d ) 78 , 450.70 , e ) 87 , 987.34 | d | add(divide(divide(multiply(subtract(divide(multiply(200000, 28), const_100), 200), const_100), add(43, 28)), const_100), subtract(const_100, const_10)) | mr . riddle invested in fund x and fund y . the total amount she invested , in both funds combined , was $ 200000 . in one year , fund x paid 43 % and fund y paid 28 % . the interest earned in fund y was exactly $ 200 greater than the interest earned in fund x . how much did ms . tom invest in fund x ? | x + y = 200000 0.28 y = 0.43 a + 300 take away decimals first : 28 y + 43 x + 20000 isolate first equation to solve for x ( your goal ) : y = 200000 - y plug in for b : 28 ( 200000 - x ) = 43 x + 30000 5 , 600000 - 28 x = 43 x + 30000 5 , 570000 = 71 x 5 , 570000 / 71 = x 5 , 570000 / 71 = x x = 78 , 450.70 = answer choice d | a = 200000 * 28
b = a / 100
c = b - 200
d = c * 100
e = 43 + 28
f = d / e
g = f / 100
h = 100 - 10
i = g + h
|
a ) 438 , b ) 436 , c ) 430 , d ) 442 , e ) 444 | c | add(subtract(312, divide(114, const_2)), subtract(232, divide(114, const_2))) | at the faculty of aerospace engineering , 312 students study random - processing methods , 232 students study scramjet rocket engines and 114 students study them both . if every student in the faculty has to study one of the two subjects , how many students are there in the faculty of aerospace engineering ? | 312 + 232 - 114 ( since 112 is counted twice ) = 430 c is the answer | a = 114 / 2
b = 312 - a
c = 114 / 2
d = 232 - c
e = b + d
|
a ) 1 / 12 , b ) 5 / 12 , c ) 1 / 6 , d ) 1 / 2 , e ) 1 / 5 | d | divide(add(25, 5), const_60) | two boats are heading towards each other at constant speeds of 5 miles / hr and 25 miles / hr respectively . they begin at a distance 20 miles from each other . how far are they ( in miles ) one minute before they collide ? | "the question asks : how far apart will they be 1 minute = 1 / 60 hours before they collide ? since the combined rate of the boats is 5 + 25 = 30 mph then 1 / 60 hours before they collide they ' ll be rate * time = distance - - > 30 * 1 / 60 = 1 / 2 miles apart . answer : d ." | a = 25 + 5
b = a / const_60
|
a ) 17 , b ) 83 , c ) 12 , d ) 83 , e ) 28 | a | subtract(const_60, multiply(const_60, divide(32, 45))) | excluding stoppages , the speed of a train is 45 kmph and including stoppages it is 32 kmph . of how many minutes does the train stop per hour ? | "explanation : t = 13 / 45 * 60 = 17 answer : option a" | a = 32 / 45
b = const_60 * a
c = const_60 - b
|
a ) 49 , b ) 99 , c ) 24 , d ) 88 , e ) 21 | c | inverse(subtract(inverse(8), inverse(12))) | a and b together can do a piece of work in 8 days . if a alone can do the same work in 12 days , then b alone can do the same work in ? | b = 1 / 8 β 1 / 2 = 1 / 24 = > 24 days answer : c | a = 1/(8)
b = 1/(12)
c = a - b
d = 1/(c)
|
a ) 6 , b ) 7 , c ) 8 , d ) 9 , e ) 12 | e | add(const_4, add(floor(divide(48, add(const_4, const_3))), const_1)) | company z has 48 employees . if the number of employees having birthdays on wednesday is more than the number of employees having birthdays on any other day of the week , each of which have same number of birth - days , what is the minimum number of employees having birthdays on wednesday . | "say the number of people having birthdays on wednesday is x and the number of people having birthdays on each of the other 6 days is y . then x + 6 y = 48 . now , plug options for x . only a and e give an integer value for y . but only for e x > y as needed . answer : e ." | a = 4 + 3
b = 48 / a
c = math.floor(b)
d = c + 1
e = 4 + d
|
a ) 122.6 $ , b ) 128.9 $ , c ) 243.7 $ , d ) 298.85 $ , e ) 312.12 $ | d | add(multiply(multiply(add(65, divide(multiply(65, 120), const_100)), 2), 0.6), multiply(multiply(65, 3), 0.6)) | in a fuel station the service costs $ 2.05 per car , every liter of fuel costs 0.6 $ . assuming that you fill up 3 mini - vans and 2 trucks , how much money will the fuel cost to all the cars owners total , if a mini - van ' s tank is 65 liters and a truck ' s tank is 120 % bigger and they are all empty - ? | "mini van cost = 65 * 0.6 + 2.05 truck capacity = 65 ( 1 + 120 / 100 ) = 143 lts . cost of truck = 143 + 0.6 + 2.05 total cost = 3 * 41.05 ( car ) + 2 * 87.85 ( truck ) = 298.85 answer : d" | a = 65 * 120
b = a / 100
c = 65 + b
d = c * 2
e = d * 0
f = 65 * 3
g = f * 0
h = e + g
|
a ) 32 , b ) 24 , c ) 12 , d ) 7 , e ) 45 | b | divide(18, subtract(const_10, const_1)) | a two - digit number is such that the product of the digits is 8 . when 18 is added to the number , then the digits are reversed . the number is : | "answer : option b let the ten ' s and unit ' s digit be x and 8 / x respectively . then , ( 10 x + 8 / x ) + 18 = 10 * 8 / x + x 9 x 2 + 18 x - 72 = 0 x 2 + 2 x - 8 = 0 ( x + 4 ) ( x - 2 ) = 0 x = 2 so , ten ' s digit = 2 and unit ' s digit = 4 . hence , required number = 24 ." | a = 10 - 1
b = 18 / a
|
a ) 20 % , b ) 21 % , c ) 22 % , d ) 23 % , e ) 24 % | a | multiply(divide(19, 95), const_100) | by selling 95 pens , a trader gains the cost of 19 pens . find his gain percentage ? | "let the cp of each pen be rs . 1 . cp of 95 pens = rs . 95 profit = cost of 19 pens = rs . 19 profit % = 19 / 95 * 100 = 20 % answer : a" | a = 19 / 95
b = a * 100
|
a ) 116 , b ) 85 , c ) 94 , d ) 98 , e ) 108 | a | subtract(add(subtract(193, 59), 23), 41) | there are 193 items that are members of set u . of these items , 41 are members of set b , 59 are not members of either of set a or set b , and 23 are members of both sets a and b . how many of the members of set u are members of set a ? | "you had the answer almost right . the x = 93 refers to only set a . however what ' s being asked is how many members are part of set a . this will include : 1 . only set a 2 . set a and set b so the answer is set a = 93 + set ab = 93 + 23 = 116 a" | a = 193 - 59
b = a + 23
c = b - 41
|
a ) 25 min , b ) 20 min , c ) 10 min , d ) 15 min , e ) 30 min | a | divide(divide(const_1, subtract(divide(const_1, 25), divide(const_1, 50))), const_2) | a machine can filled a tank in 25 minutes and another machine can empty it in 50 minutes . if the tank is already half full and both the taps are opened together , then the tank is filled in how many minutes ? | t = 1 / 2 ( 25 * 50 / 50 - 25 ) = 25 minutes tank is filled in 25 minutes answer is a | a = 1 / 25
b = 1 / 50
c = a - b
d = 1 / c
e = d / 2
|
a ) 180 , b ) 1800 , c ) 18 , d ) 18000 , e ) 1350 | e | divide(multiply(36, 15), divide(multiply(8, 5), const_100)) | a hall 36 m long and 15 m broad is to be paved with stones , each measuring 8 dm by 5 dm . the number of stones required is : | "area of the hall = 3600 * 1500 area of each stone = ( 80 * 50 ) therefore , number of stones = ( 3600 * 1500 / 80 * 50 ) = 1350 answer : e" | a = 36 * 15
b = 8 * 5
c = b / 100
d = a / c
|
a ) 10 % , b ) 25 % , c ) 33 % , d ) 50 % , e ) 67 % | c | multiply(add(const_1, const_10), subtract(subtract(16, 12), const_1)) | at company x , senior sales representatives visit the home office once every 16 days , and junior sales representatives visit the home office once every 12 days . the number of visits that a junior sales representative makes in a 2 - year period is approximately what percent greater than the number of visits that a senior representative makes in the same period ? | each 48 - day period , senior representatives visit the home office 3 times while junior representatives visit 4 times , thus 33 % more . the answer is c . | a = 1 + 10
b = 16 - 12
c = b - 1
d = a * c
|
a ) 2 , b ) 3 , c ) 5 , d ) 6 , e ) 8 | b | add(divide(add(const_1, const_4), divide(divide(divide(60, const_2), const_2), const_3)), const_2) | in n is a positive integer less than 200 , and 14 n / 60 is an integer , then n has how many different positive prime factors q ? | "i like to put the numbers in prime factors so it is easier and faster to visualize . 14 * n / 60 if we write the factors of 14 - - > 2 , 7 , and the factors of 60 - - > 2 , 2 , 3 , 5 , we have ( 2 * 7 * n ) / ( 2 ^ 2 * 3 * 5 ) simplifying 7 * n / ( 2 * 3 * 5 ) the only way the equation above has an integer value is if n has at least the factors 2 , 3 and 5 , so we can simplify again and we have the number 7 . the number could be 2 * 3 * 5 , or 2 * 3 * 5 * 2 , or 2 * 3 * 5 * . . . . . however to be less than 200 we can not add any prime number . 2 * 3 * 5 = 120 if we added the next prime factor 7 , we would have q = 2 * 3 * 5 * 7 = 840 thus , answer b ." | a = 1 + 4
b = 60 / 2
c = b / 2
d = c / 3
e = a / d
f = e + 2
|
Subsets and Splits
No community queries yet
The top public SQL queries from the community will appear here once available.