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 . 300 , b ) rs . 200 , c ) rs . 100 , d ) rs . 350 , e ) none of these | c | divide(370, add(add(multiply(add(const_1, divide(25, const_100)), add(const_1, divide(20, const_100))), add(const_1, divide(20, const_100))), const_1)) | if x gets 25 % more than y and y gets 20 % more than z , the share of z out of rs . 370 will be : | z share = z , y = 1.2 z x = 1.25 Γ£ β 1.2 z , x + y + z = 740 ( 1.25 Γ£ β 1.2 + 1.2 + 1 ) z = 37 3.7 z = 370 , z = 100 answer : . c | a = 25 / 100
b = 1 + a
c = 20 / 100
d = 1 + c
e = b * d
f = 20 / 100
g = 1 + f
h = e + g
i = h + 1
j = 370 / i
|
a ) 14 , b ) 19 , c ) 11 , d ) 18 , e ) 13 | a | divide(subtract(70000, 42000), add(800, 1200)) | village x has a population of 70000 , which is decreasing at the rate of 1200 per year . village y has a population of 42000 , which is increasing at the rate of 800 per year . in how many years will the population of the two villages be equal ? | "let the population of two villages be equal after p years then , 70000 - 1200 p = 42000 + 800 p 2000 p = 28000 p = 14 answer is a ." | a = 70000 - 42000
b = 800 + 1200
c = a / b
|
a ) 2 , b ) 4 , c ) 8 , d ) 10 , e ) 12 | c | divide(const_2, add(add(divide(const_1, 11), divide(const_1, 20)), add(divide(const_1, 11), divide(const_1, 55)))) | a , b , c can do a piece of work in 11 days , 20 days and 55 days respectively , working alone . how soon can the work be done if a is assisted by b and c on alternate days ? | a + b 1 day work = 1 / 11 + 1 / 20 = 31 / 220 a + c 1 day work = 1 / 11 + 1 / 55 = 6 / 55 work done in 2 days = 31 / 220 + 6 / 55 = 55 / 220 = 1 / 4 1 / 4 work is done by a in 2 days whole work will be done in 2 * 4 = 8 days answer is c | a = 1 / 11
b = 1 / 20
c = a + b
d = 1 / 11
e = 1 / 55
f = d + e
g = c + f
h = 2 / g
|
a ) 15 , b ) 20 , c ) 30 , d ) 25 , e ) 18 | a | divide(multiply(3, 10), subtract(10, 8)) | a group of men decided to do a work in 8 days , but 3 of them became absent . if the rest of the group did the work in 10 days , find the original number of men ? | "original number of men = 3 * 10 / ( 10 - 8 ) = 15 answer is a" | a = 3 * 10
b = 10 - 8
c = a / b
|
a ) 9 , b ) 18 , c ) 27 , d ) 36 , e ) 45 | c | subtract(power(2, const_4), const_4) | x , y , and z are different prime numbers . the product x ^ 2 * y ^ 2 * z ^ 2 is divisible by how many different positive numbers ? | "the exponents of x ^ 2 * y ^ 2 * z ^ 2 are 2 , 2 , and 2 . the number of factors is ( 2 + 1 ) ( 2 + 1 ) ( 2 + 1 ) = 27 the answer is c ." | a = 2 ** 4
b = a - 4
|
a ) 3 , b ) 4 , c ) 6 , d ) 9 , e ) 12 | d | divide(const_1, multiply(36, add(divide(const_1, 36), divide(const_1, const_2.0)))) | a company has two types of machines , type r and type s . operating at a constant rate a machine of r does a certain job in 36 hours and a machine of type s does the job in 36 hours . if the company used the same number of each type of machine to do job in 12 hours , how many machine r were used ? | "yes there is a typo in the question , i got the same ques on my gmat prep last week , and the questions goes as : a company has two types of machines , type r and type s . operating at a constant rate a machine of r does a certain job in 36 hours and a machine of type s does the job in 36 hours . if the company used the same number of each type of machine to do job in 2 hours , how many machine r were used ? so for a job to be done in 2 hours r = 1 / 2 r _ a ( rate of machine r ) = 1 / 36 r _ s ( rate of machine s ) = 1 / 36 lets say x machines are used to attain the desired rate , thus x / 36 + x / 36 = 1 / 2 ( desired r = 1 / 2 i . e . to complete the job in 2 hours ) ( x + x ) / 36 = 1 / 2 2 x / 36 = 1 / 2 x = 9 . qa = 9 ( answer d )" | a = 1 / 36
b = 1 / 2
c = a + b
d = 36 * c
e = 1 / d
|
a ) 545 , b ) 685 , c ) 865 , d ) 1730 , e ) 534 | d | divide(multiply(173, 240), 24) | ? x 24 = 173 x 240 | "let y x 24 = 173 x 240 then y = ( 173 x 240 ) / 24 = 173 x 10 = 1730 answer : d" | a = 173 * 240
b = a / 24
|
a ) 25 , b ) 35 , c ) 10 , d ) 1 , e ) 15 | d | divide(divide(20, 10), 1) | in how many no . between 10 and 20 exactly two of the digits is 1 ? | "it ' s simple can be solved by elimination of answer choices . option a and b are too large , not possible . even ce are large to have correct choice . ans : d" | a = 20 / 10
b = a / 1
|
a ) rs . 25561.80 , b ) rs . 37500.80 , c ) rs . 50561.80 , d ) rs . 60000 , e ) none of these | c | divide(multiply(divide(multiply(54000, const_100), add(const_100, 20)), const_100), subtract(const_100, 11)) | a man sells a car to his friend at 11 % loss . if the friend sells it for rs . 54000 and gains 20 % , the original c . p . of the car was : | explanation : s . p = rs . 54,000 . gain earned = 20 % c . p = rs . [ 100 / 120 Γ£ β 54000 ] = rs . 45000 this is the price the first person sold to the second at at loss of 11 % . now s . p = rs . 45000 and loss = 11 % c . p . rs . [ 100 / 89 Γ£ β 45000 ] = rs . 50561.80 . correct option : c | a = 54000 * 100
b = 100 + 20
c = a / b
d = c * 100
e = 100 - 11
f = d / e
|
a ) 1.9532 , b ) 1.0025 , c ) 1.5693 , d ) 1.0266 , e ) none | a | multiply(divide(268, const_100), divide(74, const_100)) | given that 268 x 74 = 19532 , find the value of 2.68 x . 74 . | "solution sum of decimals places = ( 2 + 2 ) = 4 . therefore , = 2.68 Γ . 74 = 1.9532 answer a" | a = 268 / 100
b = 74 / 100
c = a * b
|
a ) { 1,5 } , b ) { 1,3 , 5,6 } , c ) { 2,6 } , d ) { 8,9 } , e ) { 4,12 } | b | add(divide(add(add(add(add(1, 3), 5), 3), 5), add(const_4, const_1)), 5) | if a = { 1 , 3 , 5 } , b = { 3 , 5 , 6 } . find a βͺ b | "a = { 1 , 3,5 } b = { 3 , 5,6 } therefore , correct answer : a βͺ b = { 1 , 3 , 5 , 6 } b" | a = 1 + 3
b = a + 5
c = b + 3
d = c + 5
e = 4 + 1
f = d / e
g = f + 5
|
a ) 28 , b ) 30 , c ) 40 , d ) 50 , e ) 64 | e | divide(multiply(1, 48), 3) | the l . c . m of two numbers is 48 . the numbers are in the ratio 1 : 3 . the sum of numbers is : | "let the numbers be 1 x and 3 x . then , their l . c . m = 3 x . so , 3 x = 48 or x = 16 . the numbers are 16 and 48 . hence , required sum = ( 16 + 48 ) = 64 . answer : e" | a = 1 * 48
b = a / 3
|
a ) 45 % , b ) 48 % , c ) 54 % , d ) 58 % , e ) 65 % | c | multiply(divide(245, divide(add(115, 245), divide(80, const_100))), const_100) | after a storm deposits 115 billion gallons of water into the city reservoir , the reservoir is 80 % full . if the original contents of the reservoir totaled 245 billion gallons , the reservoir was approximately what percentage full before the storm ? | "when the storm deposited 115 billion gallons , volume of water in the reservoir = 245 + 115 = 360 billion gallons if this is only 80 % of the capacity of the reservoir , the total capacity of the reservoir = 360 / 0.8 = 450 billion gallons therefore percentage of reservoir that was full before the storm = ( 245 / 450 ) * 100 = 54.4 % option c" | a = 115 + 245
b = 80 / 100
c = a / b
d = 245 / c
e = d * 100
|
a ) 6 , b ) 18 , c ) 10 , d ) 99 , e ) 38 | a | divide(subtract(divide(50, divide(5, const_2)), multiply(subtract(5, const_1), 2)), const_2) | the sum of the ages of 5 children born at the intervals of 2 years each is 50 years . what is the age of the youngest child ? | let x = the youngest child . each of the other four children will then be x + 2 , x + 4 , x + 6 , x + 8 . we know that the sum of their ages is 50 . so , x + ( x + 2 ) + ( x + 4 ) + ( x + 6 ) + ( x + 8 ) = 50 therefore the youngest child is 6 years old answer : a | a = 5 / 2
b = 50 / a
c = 5 - 1
d = c * 2
e = b - d
f = e / 2
|
a ) 1200 , b ) 2400 , c ) 1400 , d ) 2500 , e ) none of these | d | divide(multiply(2, const_100), 0.08) | an inspector rejects 0.08 % of the meters as defective , how many meters he examine to reject 2 meteres | explanation : it means that 0.08 % of x = 2 = > ( 8 / 100 Γ 100 Γ x ) = 2 = > x = 2 Γ 100 Γ 100 / 8 = > x = 2500 option d | a = 2 * 100
b = a / 0
|
a ) 1 / 7 , b ) 4 / 7 , c ) 3 / 7 , d ) 16 / 49 , e ) 40 / 49 | a | subtract(const_1, sqrt(divide(36, 49))) | jean drew a gumball at random from a jar of pink and blue gumballs . since the gumball she selected was blue and she wanted a pink one , she replaced it and drew another . the second gumball also happened to be blue and she replaced it as well . if the probability of her drawing the two blue gumballs was 36 / 49 , what is the probability that the next one she draws will be pink ? | "the probability of drawing a pink gumball both times is the same . the probability that she drew two blue gumballs = 36 / 49 = ( 6 / 7 ) * ( 6 / 7 ) therefore probability that the next one she draws is pink = 1 / 7 option ( a )" | a = 36 / 49
b = math.sqrt(a)
c = 1 - b
|
a ) 31.25 , b ) 37.5 , c ) 50.0 , d ) 52.5 , e ) 69.0 | e | multiply(subtract(multiply(add(divide(30, const_100), const_1), add(divide(30, const_100), const_1)), const_1), const_100) | increasing the original price of a certain item by 30 percent and then increasing the new price by 30 percent is equivalent to increasing the original price by what percent ? | "we ' re told that the original price of an item is increased by 30 % and then that price is increased by 30 % . . . . if . . . . starting value = $ 100 + 30 % = 100 + . 30 ( 100 ) = 130 + 30 % = 130 + . 30 ( 130 ) = 130 + 39 = 169 the question asks how the final price relates to the original price . this is essentially about percentage change , which means we should use the percentage change formula : percentage change = ( new - old ) / old = difference / original doing either calculation will yield the same result : 59 / 100 = 69 % final answer : e" | a = 30 / 100
b = a + 1
c = 30 / 100
d = c + 1
e = b * d
f = e - 1
g = f * 100
|
a ) 28 , b ) 36 , c ) 40 , d ) 56 , e ) 64 | c | subtract(subtract(add(const_10, multiply(15, 2)), const_0_25), const_0_25) | the area of a square garden is a square feet and the perimeter is p feet . if a = 2 p + 15 , what is the perimeter of the garden , in feet ? | "perimeter of square = p side of square = p / 4 area of square = ( p ^ 2 ) / 16 = a given that a = 2 p + 15 ( p ^ 2 ) / 16 = 2 p + 15 p ^ 2 = 32 p + 240 p ^ 2 - 32 p - 240 = 0 p ^ 2 - 40 p + 6 p - 240 = 0 p ( p - 40 ) + 6 ( p + 40 ) = 0 ( p - 40 ) ( p + 6 ) = 0 p = 40 or - 6 discarding negative value , p = 40 answer is c" | a = 15 * 2
b = 10 + a
c = b - const_0_25
d = c - const_0_25
|
a ) 237 , b ) 287 , c ) 197 , d ) 740 , e ) 720 | d | multiply(370, const_2) | on the independence day , bananas were be equally distributed among the children in a school so that each child would get two bananas . on the particular day 370 children were absent and as a result each child got two extra bananas . find the actual number of children in the school ? | explanation : let the number of children in the school be x . since each child gets 2 bananas , total number of bananas = 2 x . 2 x / ( x - 370 ) = 2 + 2 ( extra ) = > 2 x - 740 = x = > x = 740 . answer : d | a = 370 * 2
|
a ) 338 m , b ) 778 m , c ) 200 m , d ) 260 m , e ) 971 m | d | subtract(divide(900, const_2), 190) | if the perimeter of a rectangular garden is 900 m , its length when its breadth is 190 m is ? | "2 ( l + 190 ) = 900 = > l = 260 m answer : d" | a = 900 / 2
b = a - 190
|
a ) 36 , b ) 42 , c ) 48 , d ) 54 , e ) 60 | e | subtract(add(divide(const_1, 20), divide(const_1, 15)), divide(const_1, 10)) | two pipes can fill a tank in 20 minutes and 15 minutes . an outlet pipe can empty the tank in 10 minutes . if all the pipes are opened when the tank is empty , then how many minutes will it take to fill the tank ? | "let v be the volume of the tank . the rate per minute at which the tank is filled is : v / 20 + v / 15 - v / 10 = v / 60 per minute the tank will be filled in 60 minutes . the answer is e ." | a = 1 / 20
b = 1 / 15
c = a + b
d = 1 / 10
e = c - d
|
a ) 250 , b ) 300 , c ) 350 , d ) 400 , e ) 500 | d | add(multiply(const_2, 100), 200) | brenda and sally run in opposite direction on a circular track , starting at diametrically opposite points . they first meet after brenda has run 100 meters . they next meet after sally has run 200 meters past their first meeting point . each girl runs at a constant speed . what is the length of the track in meters ? | "nice problem . + 1 . first timetogetherthey run half of the circumference . second timetogetherthey run full circumference . first time brenda runs 100 meters , thus second time she runs 2 * 100 = 200 meters . since second time ( when they run full circumference ) brenda runs 200 meters and sally runs 200 meters , thus the circumference is 200 + 200 = 400 meters . answer : d ." | a = 2 * 100
b = a + 200
|
a ) 9 , b ) 11 , c ) 13 , d ) 15 , e ) 17 | c | add(11, 2) | if ( 2 to the x ) - ( 2 to the ( x - 2 ) ) = 3 ( 2 to the 11 ) , what is the value of x ? | "( 2 to the power x ) - ( 2 to the power ( x - 2 ) ) = 3 ( 2 to the power 11 ) 2 ^ x - 2 ^ ( x - 2 ) = 3 . 2 ^ 11 hence x = 13 . answer is c" | a = 11 + 2
|
a ) 5 , b ) 4 , c ) 1 / 2 , d ) 1 / 5 , e ) 1 / 8 | d | divide(const_1, divide(add(subtract(9, 1), const_2), const_2)) | the vertex of a rectangle are ( 1 , 0 ) , ( 9 , 0 ) , ( 1 , 2 ) and ( 9 , 2 ) respectively . if line l passes through the origin and divided the rectangle into two identical quadrilaterals , what is the slope of line l ? | if line l divides the rectangle into two identical quadrilaterals , then it must pass through the center ( 5 , 1 ) . the slope of a line passing through ( 0,0 ) and ( 5 , 1 ) is 1 / 5 . the answer is d . | a = 9 - 1
b = a + 2
c = b / 2
d = 1 / c
|
a ) 8 / 5 , b ) 5 / 2 , c ) 3 , d ) 33 / 25 , e ) 4 | d | divide(multiply(subtract(const_12, const_1), subtract(inverse(subtract(const_12, const_3)), inverse(const_10))), subtract(inverse(add(const_1, const_4)), inverse(subtract(const_12, const_3)))) | for each month of a given year except december , a worker earned the same monthly salary and donated one - tenth of that salary to charity . in december , the worker earned n times his usual monthly salary and donated one - third of his earnings to charity . if the worker ' s charitable contributions totaled one - eighth of his earnings for the entire year , what is the value of n ? | "let monthly salary for each of the 11 months except december was x , then 11 x * 1 / 10 + nx * 1 / 3 = 1 / 8 ( 11 x + nx ) ; 11 / 10 + n / 3 = 1 / 8 ( 11 + n ) 33 + 10 n / 30 = 11 + n / 8 = > 264 + 80 n = 330 + 30 n = > 50 n = 66 n = 66 / 50 = 33 / 25 answer : d ." | a = 12 - 1
b = 12 - 3
c = 1/(b)
d = 1/(10)
e = c - d
f = a * e
g = 1 + 4
h = 1/(g)
i = 12 - 3
j = 1/(i)
k = h - j
l = f / k
|
a ) 4 , b ) 3 , c ) 2 , d ) 0 , e ) 1 | d | divide(add(multiply(factorial(5), factorial(4)), multiply(factorial(5), factorial(5))), 5) | what is the units digit of ( 5 ! * 4 ! + 6 ! * 5 ! ) / 3 ? | "( 5 ! * 4 ! + 6 ! * 5 ! ) / 3 = 5 ! ( 4 ! + 6 ! ) / 3 = 120 ( 24 + 720 ) / 3 = ( 120 * 744 ) / 3 = 120 * 248 units digit of the above product will be equal to 0 answer d" | a = math.factorial(5)
b = math.factorial(4)
c = a * b
d = math.factorial(5)
e = math.factorial(5)
f = d * e
g = c + f
h = g / 5
|
a ) 1 : 1 , b ) 2 : 3 , c ) 5 : 2 , d ) 4 : 3 , e ) 7 : 9 | d | divide(add(multiply(4, 5), multiply(3, divide(add(5, 3), const_2))), add(multiply(4, 3), multiply(3, divide(add(5, 3), const_2)))) | two alloys a and b are composed of two basic elements . the ratios of the compositions of the two basic elements in the two alloys are 5 : 3 and 1 : 1 , respectively . a new alloy x is formed by mixing the two alloys a and b in the ratio 4 : 3 . what is the ratio of the composition of the two basic elements in alloy x ? | mixture a has a total of 5 + 3 = 8 parts . if in the final mixture this represents 4 parts , then the total number of parts in mixture b should be ( 8 / 4 ) * 3 = 6 . so , we should take of mixture b a quantity with 3 and 3 parts , respectively . this will give us in the final mixture ( 5 + 3 ) : ( 3 + 3 ) , which means 4 : 3 answer d . | a = 4 * 5
b = 5 + 3
c = b / 2
d = 3 * c
e = a + d
f = 4 * 3
g = 5 + 3
h = g / 2
i = 3 * h
j = f + i
k = e / j
|
a ) 77.7 , b ) 91.0 , c ) 88.0 , d ) 70.9 , e ) 71.2 | a | add(70, multiply(divide(11, const_100), 70)) | if x is 11 percent greater than 70 , then x = | "11 % of 70 = ( 70 * 0.11 ) = 7.7 11 % greater than 70 = 70 + 7.7 = 77.7 answer is clearly a ." | a = 11 / 100
b = a * 70
c = 70 + b
|
a ) 55 % , b ) 60 % , c ) 65 % , d ) 75 % , e ) none . | a | subtract(const_100, subtract(add(subtract(const_100, 63), subtract(const_100, 65)), 27)) | if an examination 63 % of the candidates in english , 65 % passed in mathematics , and 27 % failed in both subjects . what is the pass percentage ? | fail in english = 100 - 63 = 37 % fail in maths = 100 - 65 = 35 % so pass % = 100 - ( 37 + 35 - 27 ) = 55 % answer : a | a = 100 - 63
b = 100 - 65
c = a + b
d = c - 27
e = 100 - d
|
a ) 250 meter , b ) 876 meter , c ) 167 meter , d ) 719 meter , e ) 169 meter | a | multiply(divide(multiply(60, const_1000), const_3600), 15) | a train running at the speed of 60 km / hr crosses a pole in 15 seconds . find the length of the train ? | "speed = 60 * ( 5 / 18 ) m / sec = 50 / 3 m / sec length of train ( distance ) = speed * time ( 50 / 3 ) * 15 = 250 meter answer : a" | a = 60 * 1000
b = a / 3600
c = b * 15
|
a ) 1 / 2 , b ) 7 / 8 , c ) 1 / 4 , d ) 1 / 8 , e ) 1 / 16 | a | add(add(add(add(divide(1, 2), divide(divide(1, 2), 2)), divide(divide(divide(1, 2), 2), 2)), divide(divide(divide(divide(1, 2), 2), 2), 2)), divide(divide(divide(divide(divide(1, 2), 2), 2), 2), 2)) | if 1 / 2 of the air in a tank is removed with each stroke of a vacuum pump , what fraction of the original amount of air has been removed after 1 strokes ? | "left after 1 st stroke = 1 / 2 so removed = 1 - 1 / 2 = 1 / 2" | a = 1 / 2
b = 1 / 2
c = b / 2
d = a + c
e = 1 / 2
f = e / 2
g = f / 2
h = d + g
i = 1 / 2
j = i / 2
k = j / 2
l = k / 2
m = h + l
n = 1 / 2
o = n / 2
p = o / 2
q = p / 2
r = q / 2
s = m + r
|
a ) 5 , b ) 9 , c ) 20 , d ) 30 , e ) 45 | c | multiply(4, divide(40, add(4, 4))) | fred and sam are standing 40 miles apart and they start walking in a straight line toward each other at the same time . if fred walks at a constant speed of 4 miles per hour and sam walks at a constant speed of 4 miles per hour , how many miles has sam walked when they meet ? | "relative distance = 40 miles relative speed = 4 + 4 = 8 miles per hour time taken = 40 / 8 = 5 hours distance travelled by sam = 4 * 5 = 20 miles = c" | a = 4 + 4
b = 40 / a
c = 4 * b
|
a ) 20 , b ) 16 , c ) 15 , d ) 8 , e ) 10 | a | add(5, const_1) | the average of first seven multiples of 5 is : | "explanation : ( 5 ( 1 + 2 + 3 + 4 + 5 + 6 + 7 ) / 7 = 5 x 28 / 7 = 20 answer : a" | a = 5 + 1
|
a ) 40 , b ) 120 , c ) 50 , d ) data inadequate , e ) none of these | e | subtract(divide(divide(5,300, 26.50), const_2), multiply(const_2, 20)) | the length of a rectangular plot is 20 metres more than its breadth . if the cost of fencing the plot at the rate of 26.50 per metre is 5,300 , what is the length of the plot ( in metres ) ? | "perimeter of the rectangular plot = [ ( b + 20 ) + b ] Γ 2 = 5300 / 26.5 = 200 = 200 β΄ ( 2 b + 20 ) 2 = 200 β b = 40 β l = 40 + 20 = 60 m answer e" | a = 5 / 300
b = a / 2
c = 2 * 20
d = b - c
|
a ) 2998 , b ) 1900 , c ) 2788 , d ) 2662 , e ) 1122 | b | divide(multiply(subtract(7400, divide(multiply(5, 7400), const_100)), multiply(10000, 3)), add(add(multiply(6500, 6), multiply(8400, 5)), multiply(10000, 3))) | a , b and c are entered into a partnership . a invested rs . 6500 for 6 months , b invested rs . 8400 for 5 months and c invested for rs . 10000 for 3 months . a is a working partner and gets 5 % of the total profit for the same . find the share of c in a total profit of rs . 7400 ? | 65 * 6 : 84 * 5 : 100 * 3 26 : 28 : 20 c share = 74000 * 95 / 100 = 7030 * 20 / 74 = > 1900 answer : b | a = 5 * 7400
b = a / 100
c = 7400 - b
d = 10000 * 3
e = c * d
f = 6500 * 6
g = 8400 * 5
h = f + g
i = 10000 * 3
j = h + i
k = e / j
|
a ) 66 , b ) 53 , c ) 55 , d ) 60 , e ) 61 | d | divide(multiply(25, 96), 40) | if 25 men do a work in 96 days , in how many days will 40 men do it ? | "25 * 96 = 40 * x x = 60 days answer : d" | a = 25 * 96
b = a / 40
|
a ) 53 , b ) 54 , c ) 55 , d ) 56 , e ) 57 | d | add(choose(8, 2), choose(8, 2)) | there are 8 pairs of socks and 2 socks are worn from that such that the pair of socks worn are not of the same pair . what is the number of pair that can be formed . | "first of all you should remember that there is a difference in left and right sock . now no . of way to select any of the sock = 8 and for second = 7 so total methods = 8 * 7 = 56 answer : d" | a = math.comb(8, 2)
b = math.comb(8, 2)
c = a + b
|
a ) 0.28 , b ) 0.26 , c ) 1 , d ) 0 , e ) 2 | a | multiply(0.14, const_2) | the expression ( 12.86 Γ 12.86 + 12.86 Γ p + 0.14 Γ 0.14 ) will be a perfect square for p equal to | explanation : 12.86 Γ 12.86 + 12.86 Γ p + 0.14 Γ 0.14 = ( 12.86 ) 2 + 12.86 Γ p + ( 0.14 ) 2 this can be written as ( 12.86 + 0.14 ) 2 = 132 , if 12.86 Γ p = 2 Γ 12.86 Γ 0.14 i . e . , if p = 2 Γ 0.14 = 0.28 hence , p = 0.28 . answer : option a | a = 0 * 14
|
a ) $ 460 , b ) $ 510 , c ) $ 440 , d ) $ 500 , e ) $ 550 | c | subtract(add(divide(multiply(add(divide(multiply(400, const_100), multiply(20, 2)), divide(multiply(divide(multiply(400, const_100), multiply(20, 2)), 20), const_100)), 20), const_100), add(divide(multiply(400, const_100), multiply(20, 2)), divide(multiply(divide(multiply(400, const_100), multiply(20, 2)), 20), const_100))), divide(multiply(400, const_100), multiply(20, 2))) | if the sample interest on a sum of money 20 % per annum for 2 years is $ 400 , find the compound interest on the same sum for the same period at the same rate ? | "rate = 20 % time = 2 years s . i . = $ 400 principal = 100 * 400 / 20 * 2 = $ 1000 amount = 1000 ( 1 + 20 / 100 ) ^ 2 = $ 1440 c . i . = 1440 - 1000 = $ 440 answer is c" | a = 400 * 100
b = 20 * 2
c = a / b
d = 400 * 100
e = 20 * 2
f = d / e
g = f * 20
h = g / 100
i = c + h
j = i * 20
k = j / 100
l = 400 * 100
m = 20 * 2
n = l / m
o = 400 * 100
p = 20 * 2
q = o / p
r = q * 20
s = r / 100
t = n + s
u = k + t
v = 400 * 100
w = 20 * 2
x = v / w
y = u - x
|
a ) 36 , b ) 40 , c ) 99 , d ) 13 , e ) 12 | b | multiply(subtract(83, 63), const_2) | a pupil ' s marks were wrongly entered as 83 instead of 63 . due to that the average marks for the class got increased by half . the number of pupils in the class is | "let there be x pupils in the class . total increase in marks = ( x * 1 / 2 ) = x / 2 . x / 2 = ( 83 - 63 ) = > x / 2 = 20 = > x = 40 . answer : b" | a = 83 - 63
b = a * 2
|
a ) 26 km , b ) 32 km , c ) 30 km , d ) 28 km , e ) none | b | subtract(multiply(16, 7), 80) | a man travelled a distance of 80 km in 7 hours partly on foot at the rate of 8 km per hour and partly on bicycle at 16 km per hour . find the distance travelled on foot . | total time = 7 hrs let the distance travelled by foot @ 8 kmph be x kms ? distance travlled by bicycle @ 16 kmph be ( 80 - x ) kms atq . 7 hr = x / 8 + ( 80 - x ) / 16 ? 7 = ( 2 x + 80 - x ) / 16 ? x = 32 kms answer : b . | a = 16 * 7
b = a - 80
|
a ) 1400 , b ) 1079 , c ) 1200 , d ) 1023 , e ) 1523 | b | divide(820, subtract(const_1, divide(24, const_100))) | after decreasing 24 % in the price of an article costs rs . 820 . find the actual cost of an article ? | "cp * ( 76 / 100 ) = 820 cp = 10.78 * 100 = > cp = 1079 answer : b" | a = 24 / 100
b = 1 - a
c = 820 / b
|
a ) - 9 , b ) - 5 , c ) - 3 , d ) - 1 , e ) 5 | b | subtract(divide(add(const_1, sqrt(add(const_1, multiply(const_4, 2)))), const_2), 4) | if x ^ 2 β x = 2 , then one possible value of x β 4 = | x ^ 2 β x = 2 i . e . x ^ 2 β x - 2 = 0 i . e . x ^ 2 + x - 2 x - 2 = 0 i . e . ( x - 2 ) ( x + 1 ) = 0 i . e . x = 2 or - 1 i . e . x - 4 = 2 - 4 or - 1 - 4 i . e . x - 4 = - 2 or - 5 answer : option b | a = 4 * 2
b = 1 + a
c = math.sqrt(b)
d = 1 + c
e = d / 2
f = e - 4
|
a ) 16.16 % , b ) 15.15 % , c ) 14.14 % , d ) 13.33 % , e ) 12.52 % | e | subtract(const_100, multiply(multiply(add(const_1, divide(8, const_100)), subtract(const_1, divide(19, const_100))), const_100)) | a volunteer organization is recruiting new members . in the fall they manage to increase their number by 8 % . by the spring however membership falls by 19 % . what is the total change in percentage from fall to spring ? | ( 100 % + 8 % ) * ( 100 % - 19 % ) = 1.08 * . 81 = 0.8748 1 - 0.8748 = 12.52 % lost = - 12.52 % the answer is e the organization has lost 12.52 % of its total volunteers from fall to spring . | a = 8 / 100
b = 1 + a
c = 19 / 100
d = 1 - c
e = b * d
f = e * 100
g = 100 - f
|
a ) 75 , b ) 25 , c ) 26 , d ) 23 , e ) 22 | a | multiply(multiply(const_0_2778, 54), subtract(25, 20)) | a train passes a platform in 25 seconds . the same train passes a man standing on the platform in 20 seconds . if the speed of the train is 54 km / hr , the length of the platform is | speed of the train = 54 km / hr = ( 54 Γ 10 ) / 36 m / s = 15 m / s length of the train = speed Γ time taken to cross the man = 15 Γ 20 = 300 m let the length of the platform = l time taken to cross the platform = ( 300 + l ) / 15 = > ( 300 + l ) / 15 = 20 = > 300 + l = 15 Γ 25 = 375 = > l = 375 - 300 = 75 meter answer is a . | a = const_0_2778 * 54
b = 25 - 20
c = a * b
|
a ) 8 , b ) 13 , c ) 28 , d ) 6 , e ) 2 | b | multiply(divide(5, 8), const_100) | 5 + 8 | b | a = 5 / 8
b = a * 100
|
a ) 652 , b ) 827 , c ) 939 , d ) 1045 , e ) 1136 | c | subtract(418,600, add(add(multiply(const_2, const_100), multiply(add(const_3, const_4), const_10)), const_2)) | how many integers between 324,700 and 418,600 have tens digit 1 and units digit 3 ? | "the integers are : 324,713 324,813 etc . . . 418,513 the number of integers is 4186 - 3247 = 939 the answer is c ." | a = 2 * 100
b = 3 + 4
c = b * 10
d = a + c
e = d + 2
f = 418 - 600
|
a ) 12500 , b ) 20000 , c ) 20289 , d ) 20027 , e ) 20026 | a | divide(250, subtract(multiply(divide(5, const_100), divide(subtract(const_100, 20), const_100)), multiply(divide(10, const_100), divide(20, const_100)))) | a shopkeeper sells 20 % of his stock at 10 % profit ans sells the remaining at a loss of 5 % . he incurred an overall loss of rs . 250 . find the total worth of the stock ? | "let the total worth of the stock be rs . x . the sp of 20 % of the stock = 1 / 5 * x * 1.1 = 11 x / 50 the sp of 80 % of the stock = 4 / 5 * x * 0.95 = 19 x / 25 = 38 x / 50 total sp = 11 x / 50 + 38 x / 50 = 49 x / 50 overall loss = x - 49 x / 50 = x / 50 x / 50 = 250 = > x = 12500 answer : a" | a = 5 / 100
b = 100 - 20
c = b / 100
d = a * c
e = 10 / 100
f = 20 / 100
g = e * f
h = d - g
i = 250 / h
|
a ) 10 cm , b ) 25 / 2 cm , c ) 35 / 3 cm , d ) 15 cm , e ) none of the these | c | add(add(5, 7), 5) | two spherical balls lie on the ground touching . if one of the balls has a radius of 5 cm , and the point of contact is 7 cm above the ground , what is the radius of the other ball ? | "similar triangle properties . . 2 / r + 5 = 5 / r - 5 giving r = 35 / 3 . answer : c" | a = 5 + 7
b = a + 5
|
a ) $ 115 , b ) $ 120 , c ) $ 125 , d ) $ 130 , e ) $ 135 | b | divide(multiply(divide(multiply(158.40, const_100), add(const_100, 20)), const_100), add(const_100, 10)) | a couple spent $ 158.40 in total while dining out and paid this amount using a credit card . the $ 158.40 figure included a 20 percent tip which was paid on top of the price which already included a sales tax of 10 percent on top of the price of the food . what was the actual price of the food before tax and tip ? | "let the price of the meal be x . after a 10 % sales tax addition , the price is 1.1 * x after a 20 % tip on this amount , the total is 1.2 * 1.1 * x = 1.32 x 1.32 x = 158.40 x = 120 the correct answer is b ." | a = 158 * 40
b = 100 + 20
c = a / b
d = c * 100
e = 100 + 10
f = d / e
|
['a ) a 125', 'b ) b 120', 'c ) c 70', 'd ) d 40', 'e ) e 10'] | a | add(cube_edge_by_volume(120), 120) | we have a rectangular metallic piece of paper that covers exactly the area of a cube . the length of the piece of paper is 120 inches and the width is 108 inches . what is the volume of the cube in cubic feet is 1 feet is 12 inches ? | l = 120 / 12 = 10 ft w = 108 / 12 = 9 ft area of paper = 90 area of cube = 10 * side ^ 2 side of cube = 5 v of cube = 125 | a = cube_edge_by_volume + (
|
a ) 1 / 2 , b ) 1 , c ) 2 , d ) 5 / 2 , e ) 4 | b | divide(subtract(add(3, 2), 3), 2) | in the xy - coordinate system , if ( m , n ) and ( m + 2 , n + k ) are two points on the line with the equation x = 2 y + 3 , then k = | "since ( m , n ) and ( m + 2 , n + k ) are two points on the line with the equation x = 2 y + 5 they should satisfy m = 2 n + 3 and m + 2 = 2 * ( n + k ) + 3 . by 1 st equation we have m - 2 n = 3 and by 2 nd equation m - 2 n = 2 k + 1 - - - > 3 = 2 k + 1 - - - > k = 1 . the answer is , therefore , ( b ) ." | a = 3 + 2
b = a - 3
c = b / 2
|
a ) 96 , b ) 75 , c ) 48 , d ) 25 , e ) 12 | d | divide(5, subtract(96.2, floor(96.2))) | when positive integer x is divided by positive integer y , the remainder is 5 . if x / y = 96.2 , what is the value of y ? | "guys , one more simple funda . 5 / 2 = 2.5 now . 5 x 2 = 1 is the remainder 25 / 4 = 6.25 now . 25 x 4 = 1 is the remainder 32 / 5 = 6.4 now . 4 x 5 = 2 is the remainder given x / y = 96.2 and remainder is 5 so . 2 x y = 5 hence y = 25 ans d" | a = math.floor(96, 2)
b = 96 - 2
c = 5 / b
|
a ) 5729 , b ) 5040 , c ) 2889 , d ) 2870 , e ) 2799 | b | divide(multiply(3.5, multiply(8, const_60)), subtract(divide(multiply(8, const_60), multiply(6, const_60)), const_1)) | a leak in the bottom of a tank can empty the full tank in 6 hours . an inlet pipe fills water at the rate of 3.5 liters per minute . when the tank is full in inlet is opened and due to the leak the tank is empties in 8 hours . the capacity of the tank is ? | "1 / x - 1 / 6 = - 1 / 8 x = 24 hrs 24 * 60 * 3.5 = 5040 . answer : b" | a = 8 * const_60
b = 3 * 5
c = 8 * const_60
d = 6 * const_60
e = c / d
f = e - 1
g = b / f
|
a ) a ) 90 , b ) b ) 32 , c ) c ) 64 , d ) d ) 70 , e ) of these | c | divide(2048, multiply(power(const_2, const_4), const_2)) | what is the greater of the two numbers whose product is 2048 , given that the sum of the two numbers exceeds their difference by 64 ? | "let the greater and the smaller number be g and s respectively . gs = 2048 g + s exceeds g - s by 64 i . e . , g + s - ( g - s ) = 64 i . e . , 2 s = 64 = > s = 32 . g = 2048 / s = 64 . answer : c" | a = 2 ** 4
b = a * 2
c = 2048 / b
|
a ) 1 / 2 , b ) 1 / 6 , c ) 1 / 7 , d ) 1 / 8 , e ) none of these | a | add(divide(const_1, 6), multiply(divide(const_1, 6), const_2)) | a can finish a work in 6 days and b can do same work in half the time taken by a . then working together , what part of same work they can finish in a day ? | "explanation : please note in this question , we need to answer part of work for a day rather than complete work . it was worth mentioning here because many do mistake at this point in hurry to solve the question so lets solve now , a ' s 1 day work = 1 / 6 b ' s 1 day work = 1 / 3 [ because b take half the time than a ] ( a + b ) ' s one day work = ( 1 / 6 + 1 / 3 ) = 1 / 2 so in one day 1 / 2 work will be done answer : a" | a = 1 / 6
b = 1 / 6
c = b * 2
d = a + c
|
a ) 16 , b ) 18 , c ) 20 , d ) 22 , e ) 24 | e | add(divide(subtract(multiply(50, 4), multiply(35, 4)), subtract(53, 50)), 4) | a car averages 35 miles per hour for the first 4 hours of a trip and averages 53 miles per hour for each additional hour of travel time . if the average speed for the entire trip is 50 miles per hour , how many hours long is the trip ? | "let t be the total time of the trip . 35 * 4 + 53 ( t - 4 ) = 50 t 3 t = 212 - 140 t = 24 the answer is e ." | a = 50 * 4
b = 35 * 4
c = a - b
d = 53 - 50
e = c / d
f = e + 4
|
a ) 6 , b ) 7 , c ) 8 , d ) 9 , e ) 10 | d | divide(multiply(multiply(8, 9), 24), multiply(12, 16)) | 8 men , working 9 hours a day can complete a work in 24 days . how many hours a day must 12 men work to complete the same work in 16 days ? | "the number of hours required to complete the work is 8 * 9 * 24 = 1728 12 Γ 16 Γ ( x ) = 1728 x = 9 the answer is d ." | a = 8 * 9
b = a * 24
c = 12 * 16
d = b / c
|
a ) 8 kmph , b ) 6 kmph , c ) 7 kmph , d ) 5 kmph , e ) 4 kmph | a | divide(24, add(const_1, const_2)) | the speed of a boat in still water is 24 kmph . what is the speed of the stream if the boat can cover 64 km downstream or 32 km upstream in the same time ? | "x = the speed of the stream ( 24 + x ) / ( 24 - x ) = 2 / 1 24 + x = 48 - 2 x 3 x = 24 x = 8 km / hour if the speed of the stream is 8 km / hour , then the ' downstream ' speed of the boat is 24 + 8 = 32 km / hour and the ' upstream ' speed of the boat is 24 - 8 = 16 km / hour . in that way , if the boat traveled for 2 hours , it would travel 2 x 32 = 64 km downstream and 2 x 16 = 32 km / hour upstream . answer : a" | a = 1 + 2
b = 24 / a
|
a ) 74 , b ) m = 75 , c ) m = 175 , d ) m = 680 , e ) 690 | b | add(multiply(34, const_2), divide(subtract(multiply(34, const_2), multiply(8, 5)), subtract(5, const_1))) | a number when divided by 5 gives a number which is 8 more than the remainder obtained on dividing the same number by 34 . such a least possible number m is | "i solved this question by plugging in numbers from the answer choices . a . ) 74 starting with answer choice a , i immediately eliminated it because 74 is not even divisible by 5 . b . ) 75 i divide 75 / 5 and get 15 as an answer . i divide 75 / 34 and get a remainder of 7 . 15 - 7 = 8 so i know the correct answer isb" | a = 34 * 2
b = 34 * 2
c = 8 * 5
d = b - c
e = 5 - 1
f = d / e
g = a + f
|
a ) 8 Β° , b ) 10 Β° , c ) 18 Β° , d ) 36 Β° , e ) 72 Β° | e | divide(multiply(subtract(const_100, add(add(add(add(14, 24), 15), 19), 8)), divide(const_3600, const_10)), const_100) | a circle graph shows how the megatech corporation allocates its research and development budget : 14 % microphotonics ; 24 % home electronics ; 15 % food additives ; 19 % genetically modified microorganisms ; 8 % industrial lubricants ; and the remainder for basic astrophysics . if the arc of each sector of the graph is proportional to the percentage of the budget it represents , how many degrees of the circle are used to represent basic astrophysics research ? | "14 % microphotonics ; 24 % home electronics ; 15 % food additives ; 19 % genetically modified microorganisms ; 8 % industrial lubricants ; 100 - ( 14 + 24 + 15 + 19 + 8 ) = 20 % basic astrophysics . 10 % of 360 Β° is 72 Β° . answer : e ." | a = 14 + 24
b = a + 15
c = b + 19
d = c + 8
e = 100 - d
f = 3600 / 10
g = e * f
h = g / 100
|
a ) 1 , b ) 3 , c ) 4 , d ) 6 , e ) 8 | e | divide(36, add(multiply(3, 2), 4)) | bag a contains red , white and blue marbles such that the red to white marble ratio is 1 : 3 and the white to blue marble ratio is 2 : 3 . bag b contains red and white marbles in the ratio of 1 : 4 . together , the two bags contain 36 white marbles . how many red marbles could be in bag a ? | "6 is the answer . bag a - r : w : b = 2 : 6 : 9 let w in bag a be 6 k bab b - r : w = 1 : 4 let w in bag b be 4 p w = 36 = 6 k + 4 p = > k = 4 , p = 3 total red ' s in bag a will be 2 k = 8 e" | a = 3 * 2
b = a + 4
c = 36 / b
|
a ) 2 pm , b ) 9 pm , c ) 3 pm , d ) 4 pm , e ) 6 pm | d | subtract(multiply(1, const_12), divide(multiply(1, const_12), add(divide(1, 2), const_1))) | when asked what the time is , a person answered that the amount of time left is 1 / 2 of the time already completed . what is the time . | "a day has 24 hrs . assume x hours have passed . remaining time is ( 24 - x ) 24 β x = 1 / 2 x β x = 16 time is 4 pm answer : d" | a = 1 * 12
b = 1 * 12
c = 1 / 2
d = c + 1
e = b / d
f = a - e
|
a ) 10 hrs , b ) 15 hrs , c ) 20 hrs , d ) 25 hrs , e ) 18 hrs | a | divide(500, 50) | ajay can ride 50 km in 1 hour . in how many hours he can ride 500 km ? | "1 hour he ride 50 km he ride 500 km in = 500 / 50 * 1 = 10 hours answer is a" | a = 500 / 50
|
a ) 32,300 , b ) 172,800 , c ) 345,600 , d ) 338,200 , e ) 259,200 | c | multiply(multiply(subtract(10, 2), const_3600), const_12) | in a renowned city , the average birth rate is 10 people every two seconds and the death rate is 2 people every two seconds . estimate the size of the population net increase that occurs in one day . | "every 2 seconds , 8 persons are added ( 10 - 2 ) . every second 4 persons are added . in a day 24 hrs = 24 * 60 minutes = 24 * 60 * 60 = 86400 seconds . 86400 * 4 = 345600 option c" | a = 10 - 2
b = a * 3600
c = b * 12
|
a ) $ 270 , b ) $ 280 , c ) $ 290 , d ) $ 300 , e ) $ 310 | b | divide(616, add(divide(120, const_100), const_1)) | two employees m and n are paid a total of $ 616 per week by their employer . if m is paid 120 percent of the salary paid to n , how much is n paid per week ? | "1.2 n + n = 616 2.2 n = 616 n = 280 the answer is b ." | a = 120 / 100
b = a + 1
c = 616 / b
|
a ) 16 % , b ) 25 % , c ) 32 % , d ) 40 % , e ) 52 % | d | multiply(divide(subtract(64, 38), 64), const_100) | in town x , 64 percent of the population are employed , and 38 percent of the population are employed males . what percent of the employed people in town x are females ? | "we are asked to find the percentage of females in employed people . total employed people 64 % , out of which 38 are employed males , hence 26 % are employed females . ( employed females ) / ( total employed people ) = 26 / 64 = 40 % answer : d ." | a = 64 - 38
b = a / 64
c = b * 100
|
a ) 550 , b ) 744 , c ) 255 , d ) 199 , e ) 231 | a | divide(multiply(divide(100, divide(subtract(55, subtract(const_100, 55)), const_100)), 55), const_100) | there were two candidates in an election . winner candidate received 55 % of votes and won the election by 100 votes . find the number of votes casted to the winning candidate ? | "w = 55 % l = 45 % 55 % - 45 % = 10 % 10 % - - - - - - - - 100 55 % - - - - - - - - ? = > 550 answer : a" | a = 100 - 55
b = 55 - a
c = b / 100
d = 100 / c
e = d * 55
f = e / 100
|
a ) 25 % , b ) 28 % , c ) 33 % , d ) 40 % , e ) 50 % | a | divide(multiply(50, multiply(5, 5)), 50) | 5 friends adi , brian , close , derek and eli appeared in two aptitude tests . in the first aptitude test , derek score 50 % less than the average score of the 5 people . in the second aptitude test , derek score 50 % more than what he scored on the first aptitude test . if the score of his friend in the second aptitude test were same as their score in the first test , by approximately what percentage was derek ' s score less than the average score of the 5 people in the second aptitude ? | average score in first test be x so derby score = 0.5 x ( first test ) derby score = 1.5 ( 0.5 x ) ( second test ) = . 75 x so less than x by . 25 answer a 25 % | a = 5 * 5
b = 50 * a
c = b / 50
|
a ) 10 sec , b ) 8 sec , c ) 9 sec , d ) 7 sec , e ) 12 sec | b | divide(160, add(12, 8)) | an escalator moves towards the top level at the rate of 12 ft . sec and its length is 160 feet . if a person walks on the moving escalator at the rate of 8 feet per second towards the top level , how much time does he take to cover the entire length . | "explanation : time taken to cover the entire length = tot . dist / resultant speed = 160 / ( 12 + 8 ) = 8 sec answer : b" | a = 12 + 8
b = 160 / a
|
a ) s . 5000 , b ) s . 5100 , c ) s . 5800 , d ) s . 6000 , e ) s . 4100 | e | divide(41, multiply(divide(10, const_100), divide(10, const_100))) | if difference between compound interest and simple interest on a sum at 10 % p . a . for 2 years is rs . 41 then sum is | "p ( r / 100 ) ^ 2 = c . i - s . i p ( 10 / 100 ) ^ 2 = 41 4100 answer : e" | a = 10 / 100
b = 10 / 100
c = a * b
d = 41 / c
|
a ) 7580 , b ) 7960 , c ) 8290 , d ) 2160 , e ) none | d | divide(multiply(6, multiply(12, const_60)), subtract(divide(multiply(12, const_60), multiply(4, const_60)), const_1)) | a leak in the bottom of a tank can empty the full tank in 4 hours . an inlet pipe fills water at the rate of 6 litres a minute . when the tank is full , the inlet is opened and due to the leak , the tank is empty in 12 hours . how many litres does the cistern hold ? | solution work done by the inlet in 1 hour = ( 1 / 4 - 1 / 12 ) = 1 / 6 . work done by the inlet in 1 min . = ( 1 / 6 Γ 1 / 60 ) = 0.002778 volume of 0.002778 part = 6 litres . therefore , volume of whole = ( . 002778 Γ 6 ) βΉ = βΊ 2160 litres . answer d | a = 12 * const_60
b = 6 * a
c = 12 * const_60
d = 4 * const_60
e = c / d
f = e - 1
g = b / f
|
a ) 38 metre , b ) 28 metre , c ) 23 metre , d ) 15 metre , e ) 28 metre | b | subtract(224, multiply(divide(224, 32), 28)) | a can run 224 metre in 28 seconds and b in 32 seconds . by what distance a beat b ? | "clearly , a beats b by 4 seconds now find out how much b will run in these 4 seconds speed of b = distance / time taken by b = 224 / 32 = 28 / 4 = 7 m / sdistance covered by b in 4 seconds = speed Γ time = 7 Γ 4 = 28 metre i . e . , a beat b by 28 metre answer is b" | a = 224 / 32
b = a * 28
c = 224 - b
|
a ) 0.125 % , b ) 0.32 % , c ) 0.8 % , d ) 1.25 % , e ) 2.0 % | b | multiply(8, divide(4, const_100)) | in the manufacture of a certain product , 8 percent of the units produced are defective and 4 percent of the defective units are shipped for sale . what percent of the units produced are defective units that are shipped for sale ? | "percent of defective produced = 8 % percent of the defective units that are shipped for sale = 4 % percent of units produced are defective units that are shipped for sale = ( 4 / 100 ) * ( 8 / 100 ) * 100 % = ( 32 / 10000 ) * 100 % = ( 32 / 100 ) % = . 32 % answer b" | a = 4 / 100
b = 8 * a
|
a ) $ 100 , b ) $ 150 , c ) $ 125 , d ) $ 280 , e ) $ 250 | d | multiply(divide(700, add(divide(const_2, const_3), const_1)), divide(const_2, const_3)) | $ 700 is divided amongst a , b and c so that a may get 2 / 3 as much as b and c together , b may get 6 / 9 as much as a and c together , then the share of a is | a : ( b + c ) = 2 : 3 a ' s share = 700 * 2 / 5 = $ 280 answer is d | a = 2 / 3
b = a + 1
c = 700 / b
d = 2 / 3
e = c * d
|
a ) 27 , b ) 33 , c ) 96 , d ) 81 , e ) 162 | c | multiply(multiply(multiply(multiply(const_2, const_2), const_2), const_3), 12) | the number of boxes in a warehouse can be divided evenly into 12 equal shipments by boat or 32 equal shipments by truck . what is the smallest number of boxes that could be in the warehouse ? | "answer is the lcm of 12 and 32 = 96 answer c" | a = 2 * 2
b = a * 2
c = b * 3
d = c * 12
|
a ) 4 % decrease , b ) 8 % decrease , c ) 4 % increase , d ) 8 % increase , e ) no change | a | subtract(const_100, subtract(add(20, const_100), divide(multiply(add(20, const_100), 20), const_100))) | the salary of a worker is first increased by 20 % and afterwards reduced by 20 % . what is the net change in the worker ' s salary ? | "let x be the original salary . the final salary is 0.8 ( 1.2 x ) = 0.96 x the answer is a ." | a = 20 + 100
b = 20 + 100
c = b * 20
d = c / 100
e = a - d
f = 100 - e
|
a ) 148 mins , b ) 140 mins , c ) 136 mins , d ) 132 minw , e ) none of these | a | multiply(add(const_1, const_4), 37) | one pipe can fill a tank three times as fast as another pipe . if together the two pipes can fill the tank in 37 minutes , then the slower pipe alone will be able to fill the tank in | "explanation : let the slower pipe alone fill the tank in x minutes then faster will fill in x / 3 minutes . part filled by slower pipe in 1 minute = 1 / x part filled by faster pipe in 1 minute = 3 / x part filled by both in 1 minute = 1 / x + 3 / x = 1 / 37 = > 4 / x = 1 / 37 x = 37 β 4 = 148 mins option a" | a = 1 + 4
b = a * 37
|
a ) 11 , b ) 14 , c ) 15 , d ) 18 , e ) 30 | a | add(multiply(5, const_2), const_1) | the average age of applicants for a new job is 31 , with a standard deviation of 5 . the hiring manager is only willing to accept applications whose age is within one standard deviation of the average age . what is the maximum number of different ages of the applicants ? | within one standard deviation of the average age means 31 + / - 7 26 - - 31 - - 36 number of dif . ages - 26 27 28 29 30 31 32 33 34 35 36 total = 11 a | a = 5 * 2
b = a + 1
|
a ) 2000 , b ) 2600 , c ) 2200 , d ) 2300 , e ) 2400 | b | multiply(divide(78, 3), const_100) | a sum was put at simple interest at certain rate for 3 years . had it been put at 1 % higher rate it would have fetched rs . 78 more . the sum is : a . rs . 2,400 b . rs . 2,100 c . rs . 2,200 d . rs . 2,480 | "1 percent for 3 years = 78 1 percent for 1 year = 26 = > 100 percent = 2600 answer : b" | a = 78 / 3
b = a * 100
|
a ) 86 , b ) 87 , c ) 88 , d ) 89 , e ) 90 | c | add(multiply(16, 6), 2) | what is the dividend . divisor 16 , the quotient is 6 and the remainder is 2 | "c = d * q + r c = 16 * 6 + 2 c = 86 + 2 c = 88" | a = 16 * 6
b = a + 2
|
a ) 8 / 17 , b ) 7 / 15 , c ) 3 / 15 , d ) 8 / 15 , e ) 1 / 15 | e | subtract(const_1, multiply(add(divide(const_1, 15), divide(const_1, 20)), 8)) | a can do a work in 15 days and b in 20 days . if they work on it together for 8 days , then the fraction of the work that is left is | "person ( a ) ( b ) ( a + b ) time - ( 15 ) ( 20 ) ( - ) rate - ( 20 ) ( 15 ) ( 35 ) work - ( 300 ) ( 300 ) ( 300 ) therefore a + b requires ( 300 / 35 ) days to complete entire work for 1 st 4 days they work 35 * 8 = 280 remaining work is 300 - 280 = 20 remaining fraction of work is = 20 / 300 = 1 / 15 answer e" | a = 1 / 15
b = 1 / 20
c = a + b
d = c * 8
e = 1 - d
|
a ) 200 , b ) 240 , c ) 160 , d ) 250 , e ) 310 | b | add(multiply(divide(5, 7), 140), 140) | in a college the ratio of the numbers of boys to the girls is 5 : 7 . if there are 140 girls , the total number of students in the college is ? | "let the number of boys and girls be 5 x and 7 x then , 7 x = 140 x = 20 total number of students = 12 x = 12 * 20 = 240 answer is b" | a = 5 / 7
b = a * 140
c = b + 140
|
a ) 624545037 , b ) 627745452 , c ) 624545077 , d ) 624545066 , e ) 625454211 | d | multiply(divide(62467, 9998), const_100) | 62467 Γ 9998 = ? | "d 624545066 62467 Γ 9998 = 62467 Γ ( 10000 - 2 ) = 62467 Γ 10000 - 62467 Γ 2 = 624670000 - 124934 = 624545066" | a = 62467 / 9998
b = a * 100
|
a ) 240 km , b ) 120 km , c ) 360 km , d ) 180 km , e ) none of these | c | divide(38, add(divide(const_1, add(14, 4)), divide(const_1, multiply(subtract(14, 4), const_2)))) | a boat takes 38 hours for travelling downstream from point a to point b and coming back to point c midway between a and b . if the velocity of the stream is 4 kmph and the speed of the boat in still water is 14 kmph , what is the distance between a and b ? | "explanation : velocity of the stream = 4 kmph speed of the boat in still water is 14 kmph speed downstream = ( 14 + 4 ) = 18 kmph speed upstream = ( 14 - 4 ) = 10 kmph let the distance between a and b be x km time taken to travel downstream from a to b + time taken to travel upstream from b to c ( mid of a and b ) = 38 hours β x / 18 + ( x / 2 ) 10 = 38 β x / 18 + x / 20 = 38 β 19 x / 180 = 38 β x / 180 = 2 β x = 360 i . e . , the distance between a and b = 360 km . answer : option c" | a = 14 + 4
b = 1 / a
c = 14 - 4
d = c * 2
e = 1 / d
f = b + e
g = 38 / f
|
a ) 40 minutes , b ) 1 hour 7 min , c ) 1 hour 15 min , d ) 1 hour 30 min , e ) 1 hour 10 min | b | divide(5, divide(add(multiply(divide(1, 10), const_60), divide(3, 3)), const_2)) | a boatman goes 3 km against the current of the stream in 1 hour and goes 1 km along the current in 10 minutes . how long will it take to go 5 km in stationary water ? | "speed ( upstream ) = 3 / 1 = 3 kmhr speed ( downstream ) = 1 / ( 10 / 60 ) = 6 kmhr speed in still water = 1 / 2 ( 3 + 6 ) = 4.5 kmhr time taken in stationary = 5 / 4.5 = 1 hrs 7 min answer : b" | a = 1 / 10
b = a * const_60
c = 3 / 3
d = b + c
e = d / 2
f = 5 / e
|
a ) 30 , b ) 32 , c ) 45 , d ) data inadequate , e ) none of these | c | divide(multiply(13.5, 30), 9) | 30 buckets of water fill a tank when the capacity of each bucket is 13.5 litres . how many buckets will be required to fill the same tank if the capacity of each bucket is 9 litres ? | "capacity of the tank = 30 Γ£ β 13.5 = 405 litres when the capacity of each bucket = 9 litres , then the required no . of buckets = 405 Γ’ Β β 9 = 45 answer c" | a = 13 * 5
b = a / 9
|
a ) 1.96 % , b ) 2.56 % , c ) 3.12 % , d ) 4.65 % , e ) 5.12 % | a | multiply(divide(subtract(add(multiply(divide(const_100, add(const_100, 14)), 675958), multiply(divide(const_100, subtract(const_100, 14)), 675958)), add(675958, 675958)), add(multiply(divide(const_100, add(const_100, 14)), 675958), multiply(divide(const_100, subtract(const_100, 14)), 675958))), const_100) | a man two flats for $ 675958 each . on one he gains 14 % while on the other he loses 14 % . how much does he gain or lose in the whole transaction ? | "in such a case there is always a loss loss % = ( 14 / 10 ) ^ 2 = 49 / 25 = 1.96 % answer is a" | a = 100 + 14
b = 100 / a
c = b * 675958
d = 100 - 14
e = 100 / d
f = e * 675958
g = c + f
h = 675958 + 675958
i = g - h
j = 100 + 14
k = 100 / j
l = k * 675958
m = 100 - 14
n = 100 / m
o = n * 675958
p = l + o
q = i / p
r = q * 100
|
a ) 15 sec , b ) 18 sec , c ) 12 sec , d ) 10 sec , e ) 11 sec | b | divide(250, multiply(80, const_0_2778)) | two trains each 250 m in length are running on the same parallel lines in opposite directions with the speed of 80 kmph and 20 kmph respectively . in what time will they cross each other completely ? | "explanation : d = 250 m + 250 m = 500 m rs = 80 + 20 = 100 * 5 / 18 = 250 / 9 t = 500 * 9 / 250 = 18 sec answer : option b" | a = 80 * const_0_2778
b = 250 / a
|
a ) 0 , b ) 1 , c ) 2 , d ) 3 , e ) 4 | a | subtract(const_1, const_1) | how many pairs ( r , r + 1 ) have one or more prime factors common , where r is an integer and 2 β€ r β€ 9 ? | r and r + 1 are consecutive integers . two consecutive integers are co - prime , which means that they do n ' t share any common factor but 1 . for example 20 and 21 are consecutive integers , thus only common factor they share is 1 . answer : a . | a = 1 - 1
|
a ) 81 mins , b ) 64 mins , c ) 52 mins , d ) 40 mins , e ) none | d | divide(multiply(multiply(divide(multiply(10, 4), 5), 5), const_2), const_60) | walking at 4 / 5 of his usual speed a man is 10 mins too late . find his usual time . | "speed = s , time = t , 4 / 5 s * ( t + 10 ) = st t = 40 answer : d" | a = 10 * 4
b = a / 5
c = b * 5
d = c * 2
e = d / const_60
|
a ) $ 360 , b ) $ 380 , c ) $ 430 , d ) $ 290 , e ) $ 520 | b | divide(subtract(multiply(400, 9), add(add(add(add(add(add(406, 413), 420), 436), 395), 410), 360)), const_2) | tough and tricky questions : word problems . a salesman ' s income consists of commission and base salary . his weekly income totals over the past 7 weeks have been $ 406 , $ 413 , $ 420 , $ 436 , $ 395 , $ 410 , $ 360 . what must his average ( arithmetic mean ) income over the next two weeks be to decrease his average weekly income to $ 400 over the 9 - week period ? | official solution : ( b ) first , we need to add up the wages over the past 7 weeks : $ 406 + $ 413 + $ 420 + $ 436 + $ 395 + $ 410 + $ 360 = $ 2840 . to average $ 400 over 9 weeks , the salesman would need to earn : $ 400 Γ 9 = $ 3600 . subtract $ 2840 from $ 3600 to determine how much he would need to earn , in total , over the next 2 weeks to average $ 400 for the 9 weeks : $ 3600 β $ 2840 = $ 760 . dividing $ 760 by 2 will give us the amount he needs to earn on average over the next 2 weeks : $ 760 / 2 = $ 380 . the correct answer is choice ( b ) . | a = 400 * 9
b = 406 + 413
c = b + 420
d = c + 436
e = d + 395
f = e + 410
g = f + 360
h = a - g
i = h / 2
|
a ) 14 , b ) 13 , c ) 11 , d ) 9 , e ) 7 | d | divide(const_1, add(add(divide(const_1, multiply(12, const_2)), divide(const_1, multiply(15, const_2))), add(divide(divide(const_1, multiply(12, const_2)), const_2), divide(divide(const_1, multiply(15, const_2)), const_2)))) | a furniture manufacturer has two machines , but only one can be used at a time . machine q is utilized during the first shift and machine b during the second shift , while both work half of the third shift . if machine q can do the job in 12 days working two shifts and machine b can do the job in 15 days working two shifts , how many days will it take to do the job with the current work schedule ? | machine q finish the job in 2 * 12 shifts = 24 shifts machine b finish the job in 2 * 15 shifts = 30 shifts lets assume total work require 120 shifts therefore , rate of q = 5 shifts / day rate of b = 4 shifts / day rate of ( q + b ) = 9 shifts / day according to current schedule work complete in a day = 5 + 4 + ( 9 / 2 ) = 13.5 shifts / day therefore , time required to finish 120 shifts = ( 120 / 13.5 ) = 8.88 . . days ~ 9 days = d | a = 12 * 2
b = 1 / a
c = 15 * 2
d = 1 / c
e = b + d
f = 12 * 2
g = 1 / f
h = g / 2
i = 15 * 2
j = 1 / i
k = j / 2
l = h + k
m = e + l
n = 1 / m
|
a ) 3 , b ) 4 , c ) 9 / 5 , d ) 9 / 8 , e ) 12 | c | divide(const_1, multiply(const_2.0, add(divide(const_1, 36), divide(const_1, 4)))) | a company has two types of machines , type r and type s . operating at a constant rate a machine of r does a certain job in 36 hours and a machine of type s does the job in 4 hours . if the company used the same number of each type of machine to do job in 12 hours , how many machine r were used ? | "yes there is a typo in the question , i got the same ques on my gmat prep last week , and the questions goes as : a company has two types of machines , type r and type s . operating at a constant rate a machine of r does a certain job in 36 hours and a machine of type s does the job in 4 hours . if the company used the same number of each type of machine to do job in 2 hours , how many machine r were used ? so for a job to be done in 2 hours r = 1 / 2 r _ a ( rate of machine r ) = 1 / 36 r _ s ( rate of machine s ) = 1 / 4 lets say x machines are used to attain the desired rate , thus x / 36 + x / 4 = 1 / 2 ( desired r = 1 / 2 i . e . to complete the job in 2 hours ) ( x + 9 x ) / 36 = 1 / 2 10 x / 36 = 1 / 2 x = 9 / 5 . qa = 9 / 5 ( answer c )" | a = 1 / 36
b = 1 / 4
c = a + b
d = 2 * 0
e = 1 / d
|
a ) 33.3 % , b ) 32.5 % , c ) 37 % , d ) 37.5 % , e ) 40 % | a | multiply(divide(10, 15), const_100) | mike earns $ 15 per hour and phil earns $ 10 per hour . approximately how much less , as a percentage , does phil earn than mike per hour ? | "what % less of 15 is 10 let it be x % less , then = 15 ( 1 - x / 100 ) = 10 1 - x / 100 = 10 / 15 x = 100 / 3 x = 33.3 % ans a" | a = 10 / 15
b = a * 100
|
a ) 55 , b ) 60 , c ) 93 , d ) 82 , e ) 91 | c | subtract(100, divide(subtract(100, 79), const_3)) | a teacher grades students β tests by subtracting twice the number of incorrect responses from the number of correct responses . if student a answers each of the 100 questions on her test and receives a score of 79 , how many questions did student a answer correctly ? | "let the number of correct responses be x then the number of incorrect responses = 100 - x according to question x - 2 ( 100 - x ) = 79 ( subtracting twice of incorrect from correct ) 3 x = 279 x = 93 answer : c" | a = 100 - 79
b = a / 3
c = 100 - b
|
a ) 4.0 , b ) 4.2 , c ) 4.4 , d ) 4.6 , e ) 4.8 | d | power(add(power(4, 3), add(2, power(3, 3))), const_0_33) | the edges of three metal cubes are 2 cm , 3 cm , and 4 cm respectively . a new cube is made by melting these three cubes together . what is the edge of the new cube ( in centimeters ) ? | "the total volume is 2 ^ 3 + 3 ^ 3 + 4 ^ 3 = 99 the edge of the new cube is the cube root of 99 which is about 4.6 cm . the answer is d ." | a = 4 ** 3
b = 3 ** 3
c = 2 + b
d = a + c
e = d ** const_0_33
|
a ) 28 , b ) 20 , c ) 27 , d ) 29 , e ) 21 | b | divide(add(1000, 800), add(50, 40)) | rahim bought 50 books for rs . 1000 from one shop and 40 books for rs . 800 from another . what is the average price he paid per book ? | "average price per book = ( 1000 + 800 ) / ( 50 + 40 ) = 1800 / 90 = rs . 20 answer : b" | a = 1000 + 800
b = 50 + 40
c = a / b
|
a ) 1400 , b ) 2500 , c ) 3000 , d ) 1500 , e ) 2100 | e | multiply(subtract(divide(add(multiply(add(100, 40), 10), 400), 40), 10), add(100, 40)) | in a hostel there were 100 students . to accommodate 40 more students the average is decreased by rupees 10 . but total expenditure decreased by rs . 400 . find the total expenditure of the hostel now ? | "100 x - 400 = 140 ( x β 10 ) x = 25 100 * 25 - 400 = 2100 answer : e" | a = 100 + 40
b = a * 10
c = b + 400
d = c / 40
e = d - 10
f = 100 + 40
g = e * f
|
a ) $ 2678 , b ) $ 2464 , c ) $ 2650 , d ) $ 2732 , e ) $ 2800 | a | multiply(2500, divide(add(const_100, 50), add(const_100, 40))) | a store β s selling price of $ 2500 for a certain computer would yield a profit of 40 percent of the store β s cost for the computer . what selling price would yield a profit of 50 percent of the computer β s cost ? | 1.4 x = 2500 x = 2500 / 1.4 so , 1.5 x = 2500 * 1.5 / 1.4 = 2478 answer : - a | a = 100 + 50
b = 100 + 40
c = a / b
d = 2500 * c
|
Subsets and Splits
No community queries yet
The top public SQL queries from the community will appear here once available.