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 ) 140 sq . units , b ) 120 sq . units , c ) 187 sq . units , d ) 607 sq . units , e ) 815 sq . units | b | multiply(10, multiply(const_2, divide(sqrt(900), divide(10, const_2)))) | the length of a rectangle is two - fifths of the radius of a circle . the radius of the circle is equal to the side of the square , whose area is 900 sq . units . what is the area ( in sq . units ) of the rectangle if the rectangle if the breadth is 10 units ? | "given that the area of the square = 900 sq . units = > side of square = √ 900 = 30 units the radius of the circle = side of the square = 30 units length of the rectangle = 2 / 5 * 30 = 12 units given that breadth = 10 units area of the rectangle = lb = 12 * 10 = 120 sq . units answer : b" | a = math.sqrt(900)
b = 10 / 2
c = a / b
d = 2 * c
e = 10 * d
|
a ) 1350 , b ) 1200 , c ) 9676 , d ) 1679 , e ) 2691 | b | multiply(subtract(rectangle_area(add(60, multiply(2.5, 2)), add(55, multiply(2.5, 2))), rectangle_area(60, 55)), 2) | a rectangular grass field is 60 m * 55 m , it has a path of 2.5 m wide all round it on the outside . find the area of the path and the cost of constructing it at rs . 2 per sq m ? | "area = ( l + b + 2 d ) 2 d = ( 60 + 55 + 2.5 * 2 ) 2 * 2.5 = > 600 600 * 2 = rs . 1200 answer : b" | a = 2 * 5
b = 60 + a
c = 2 * 5
d = 55 + c
e = rectangle_area - (
f = e * rectangle_area
|
a ) 30 min , b ) 35 min , c ) 45 min , d ) 40 min , e ) 55 min | d | multiply(add(const_3, 5), 5) | a clock shows the time as 9 a . m . if the minute hand gains 5 minutes every hour , how many minutes will the clock gain by 5 p . m . ? | "there are 8 hours in between 9 a . m . to 5 p . m . 8 * 5 = 40 minutes . answer : d" | a = 3 + 5
b = a * 5
|
a ) 24 , b ) 27 , c ) 33 , d ) 45 , e ) 54 | d | multiply(divide(factorial(divide(6, 2)), const_2), divide(factorial(6), multiply(factorial(subtract(6, 2)), factorial(2)))) | a plant manager must assign 6 new workers to one of five shifts . she needs a first , second , and third shift , and two alternate shifts . each of the shifts will receive 2 new workers . how many different ways can she assign the new workers ? | "whatever : my take selecting team of 2 out of 10 to assign to the shifts = 6 c 2 = 15 ways . now 2 out of 10 means total of 3 group possible . so putting them in shifts = counting methode : first , second , third , alt , alt = 3 * 2 * 1 = 6 here alt and alt are the same : so 6 / 2 = 3 ways . total ways of selecting = ( selecting 2 out of 6 ) * arranging those teams in shifts = 15 * 3 = 45 ans : d" | a = 6 / 2
b = math.factorial(a)
c = b / 2
d = math.factorial(6)
e = 6 - 2
f = math.factorial(e)
g = math.factorial(2)
h = f * g
i = d / h
j = c * i
|
a ) 3 , b ) 9 , c ) 10 , d ) 15 , e ) 17 | c | add(4, const_1) | the average of first four multiples of 4 is : | "solution average = 4 ( 1 + 2 + 3 + 4 ) / 4 = 40 / 4 = 10 answer c" | a = 4 + 1
|
a ) $ 92.00 , b ) $ 88.00 , c ) $ 87.04 , d ) $ 80.96 , e ) $ 60.00 | e | multiply(subtract(10, divide(multiply(15, 28), const_100)), const_10) | an item is being sold for $ 10 each . however , if a customer will “ buy at least 3 ” they have a promo discount of 15 % . also , if a customer will “ buy at least 10 ” items they will deduct an additional 28 % to their “ buy at least 3 ” promo price . if sam buys 10 pcs of that item how much should he pay ? | without any discount sam should pay 10 * 10 = $ 100 . now , the overall discount would be slightly less than 43 % , thus he must pay slightly more than $ 57 . only answer choice e fits . answer : e . | a = 15 * 28
b = a / 100
c = 10 - b
d = c * 10
|
a ) 1920 , b ) 2201 , c ) 3240 , d ) 1130 , e ) 3840 | d | divide(multiply(45, 125), const_4) | what is the sum of the greatest common factor and the lowest common multiple of 45 and 125 ? | "prime factorization of the given numbers 45 = 3 ^ 2 125 = 5 ^ 3 greatest common factor = 5 lowest common multiple = 3 ^ 2 * 5 ^ 2 * 5 = 1125 sum = 5 + 1125 = 1130 answer d" | a = 45 * 125
b = a / 4
|
a ) 15 / 64 , b ) 30 / 64 , c ) 1 / 2 , d ) 1 / 4 , e ) 52 / 64 | a | divide(choose(6, 4), inverse(multiply(power(divide(const_1, const_2), 4), power(divide(const_1, const_2), const_2)))) | the chance of rain on any given day in tel - aviv is 50 % . what is the probability e that it rained in tel - aviv on exactly 4 out of 6 randomly chosen days ? | "chances of rain on exactly 4 days and not rain on 2 days = ( 1 / 2 ) ^ 4 * ( 1 / 2 ) ^ 2 = 1 / 64 chosing 4 days out of 6 = 6 ! / ( 4 ! * 2 ! ) = 15 chances of rain on exactly 4 days out of 6 days e = 15 / 64 ans a it is ." | a = math.comb(6, 4)
b = 1 / 2
c = b ** 4
d = 1 / 2
e = d ** 2
f = c * e
g = 1/(f)
h = a / g
|
a ) 39 , b ) 40 , c ) 41 , d ) 42 , e ) 59 | e | subtract(multiply(add(22, 10), const_2), 19) | you collect pens . suppose you start out with 10 . mike gives you another 22 pens . since her father makes pens , cindy decides to double your pens . since you ' re nice , you give sharon 19 pens . how many pens do you have at the end ? | "solution start with 10 pens . mike gives you 22 pens : 10 + 22 = 39 pens . cindy doubles the number of pens you have : 39 ã — 2 = 78 pens . sharon takes 19 pens from you : 78 - 19 = 59 pens . so you have 59 at the end . correct answer : e" | a = 22 + 10
b = a * 2
c = b - 19
|
a ) two , b ) three , c ) four , d ) five , e ) six | d | floor(subtract(divide(subtract(17, 5), 3), divide(subtract(1, 5), 3))) | how many integer values are there for x such that 1 < 3 x + 5 < 17 ? | the inequality given is 1 < 3 x + 5 < 17 it can further reduced to - 4 < 3 x < 12 finally - 4 / 3 < x < 4 so can only take 5 integer values i . e . - 1 , 0,1 , 2,3 correct answer - d | a = 17 - 5
b = a / 3
c = 1 - 5
d = c / 3
e = b - d
f = math.floor(e)
|
a ) 5 / 11 , b ) 6 / 11 , c ) 7 / 11 , d ) 9 / 11 , e ) 13 / 6 | d | divide(subtract(85, 40), subtract(85, 30)) | a portion of the 85 % solution of chemicals was replaced with an equal amount of 30 % solution of chemicals . as a result , 40 % solution of chemicals resulted . what part of the original solution was replaced ? | "this is a weighted average question . say x % of the solution was replaced - - > equate the amount of chemicals : 0.85 ( 1 - x ) + 0.3 * x = 0.4 - - > x = 9 / 11 . answer : d ." | a = 85 - 40
b = 85 - 30
c = a / b
|
a ) 10 , b ) 388 , c ) 4 , d ) 29 , e ) 22 | c | subtract(const_100, multiply(multiply(divide(subtract(const_100, divide(multiply(const_100, 20), const_100)), const_100), divide(add(const_100, divide(multiply(const_100, 20), const_100)), const_100)), const_100)) | if the price of a book is first decreased by 20 % and then increased by 20 % , then the net change in the price will be : | "explanation : let the original price be rs . 100 . decreased by 20 % = 80 then increased 20 % on rs 80 = 80 + 16 = 96 net change in price = 100 - 96 = 4 answer : c" | a = 100 * 20
b = a / 100
c = 100 - b
d = c / 100
e = 100 * 20
f = e / 100
g = 100 + f
h = g / 100
i = d * h
j = i * 100
k = 100 - j
|
a ) 50 , b ) 82.5 , c ) 60 , d ) 75 , e ) 80 | b | subtract(subtract(subtract(subtract(multiply(add(divide(divide(divide(divide(30, 3), 3), 3), 3), add(add(add(add(add(add(30, divide(30, 3)), divide(30, 3)), divide(divide(30, 3), 3)), divide(divide(30, 3), 3)), divide(divide(divide(30, 3), 3), 3)), divide(divide(divide(30, 3), 3), 3))), 3), divide(divide(divide(30, 3), 3), 3)), divide(divide(divide(30, 3), 3), 3)), divide(divide(divide(30, 3), 3), 3)), divide(divide(divide(30, 3), 3), 3)) | a basketball is dropped from a height of 30 feet . if it bounces back up to a height that is exactly half of its previous height , and it stops bouncing after hitting the ground for the fourth time , then how many total feet will the ball have traveled after 3 full bounces . | "initial distance = 30 feet first bounce = 15 feet up + 15 feet down = 30 feet second bouche = 7.5 feet up + 7.5 feet down = 15 feet third bounce = 375 feet up and 3.75 feet down = 7.5 feet total distance covered = 30 + 30 + 15 + 7.5 = 82.5 answer is b" | a = 30 / 3
b = a / 3
c = b / 3
d = c / 3
e = 30 / 3
f = 30 + e
g = 30 / 3
h = f + g
i = 30 / 3
j = i / 3
k = h + j
l = 30 / 3
m = l / 3
n = k + m
o = 30 / 3
p = o / 3
q = p / 3
r = n + q
s = 30 / 3
t = s / 3
u = t / 3
v = r + u
w = d + v
x = w * 3
y = 30 / 3
z = y / 3
A = z / 3
B = x - A
C = 30 / 3
D = C / 3
E = D / 3
F = B - E
G = 30 / 3
H = G / 3
I = H / 3
J = F - I
K = 30 / 3
L = K / 3
M = L / 3
N = J - M
|
a ) 20 sec , b ) 15 sec , c ) 30 sec , d ) 50 sec , e ) 40 min | e | divide(600, add(8, 7)) | two cyclist start on a circular track from a given point but in opposite direction with speeds of 7 m / s and 8 m / s . if the circumference of the circle is 600 meters , after what time will they meet at the starting point ? | "they meet every 600 / 7 + 8 = 40 sec answer is e" | a = 8 + 7
b = 600 / a
|
a ) 20 , b ) 25 , c ) 10 , d ) 30 , e ) 15 | a | divide(add(10, multiply(3, 10)), subtract(3, const_1)) | praveen age after 10 years will be 3 times his age 3 years back . what is the present age of chris ? | chris present age = x after 10 years = x + 10 3 years back = x - 3 x + 10 = 3 ( x - 3 ) x = 20 answer is a | a = 3 * 10
b = 10 + a
c = 3 - 1
d = b / c
|
a ) 27 , b ) 25 , c ) 24 , d ) 21 , e ) 22 | e | add(divide(subtract(52, 10), 2), const_1) | how many multiples of 2 are there between 10 and 52 , 10 and 52 inclusive ? | "2 multiples are . . . 10,12 , 14,16 , 18,20 , 22,24 , 26,28 , 30,32 , 34,36 , 38,40 , 42,44 , 46,48 , 50,52 , . . . , the answer is = 22 answer is e" | a = 52 - 10
b = a / 2
c = b + 1
|
a ) 154 , b ) 220 , c ) 129 , d ) 131 , e ) 224 | d | add(const_100, add(reminder(129, const_100), const_2)) | what is the next number 12 13 15 17 111 113 117 119 123 129 ? | 131 . these are the first 10 prime numbers ( 2 , 3 , 5 . . . ) prefixed with a 1 answer : d | a = reminder + (
b = 100 + a
|
a ) 1 / 6 , b ) 1 / 7 , c ) 1 / 8 , d ) 1 / 4 , e ) 1 / 3 | d | divide(const_4.0, add(subtract(const_10, 2), 2)) | david works at a resort from the beginning of march to the end of september . during the month of august this past year , he made 2 times the average ( arithmetic mean ) of his monthly totals in tips for the other months . his total tips for august were what fraction of his total tips for all of the months he worked ? | "first notice the number of months for which he worked - march to sept i . e . 7 months avg of monthly totals in tips for months other than august = x tips in august = 2 x total tips for all months = 6 * x + 2 x = 8 x tips for august / total tips for all months = 2 x / 8 x = 1 / 4 answer : d" | a = 10 - 2
b = a + 2
c = 4 / 0
|
a ) 12 , b ) 14 , c ) 16 , d ) 10 , e ) 20 | d | multiply(10, const_2) | if the average of 10 x and 8 y is greater than 70 , and x is twice y , what is the least integer value of x ? | "substitution can be used in the following way : always start with the equation : x = 2 y . it is more straight forward to manage as compared to the inequality . substitute y = x / 2 , not the other way because you need to find the minimum value of x . so you can get rid of y . now go on to the inequality . so 8 y = 8 x / 2 = 4 x now average of 10 x and 4 x is greater than 70 . average of 10 x and 4 x is 7 x . so , 7 x > 70 x > 10 answer : d" | a = 10 * 2
|
a ) 1.3 kg , b ) 1.5 kg , c ) 1.7 kg , d ) 1.9 kg , e ) none of these | a | divide(divide(subtract(const_100, 35), const_100), divide(50, const_100)) | when processing flower - nectar into honey bees ' extract , a considerable amount of water gets reduced . how much flower - nectar must be processed to yield 1 kg of honey , if nectar contains 50 % water , and the honey obtained from this nectar contains 35 % water ? | explanation : flower - nectar contains 50 % of non - water part . in honey this non - water part constitutes 65 % ( 100 - 35 ) . therefore 0.5 x amount of flower - nectar = 0.65 x amount of honey = 0.65 x 1 kg therefore amount of flower - nectar needed = ( 0.65 / 0.51 ) kg = 1.3 kgs answer : a | a = 100 - 35
b = a / 100
c = 50 / 100
d = b / c
|
a ) 22200 , b ) 31897 , c ) 48799 , d ) 50000 , e ) 52782 | d | add(multiply(add(20, const_1), add(20, const_1)), add(20, const_1)) | a sum of money is put out at compound interest for 2 years at 20 % . it would fetch rs . 1205 more if the interest were payable half - yearly , then it were pay able yearly . find the sum . | "p ( 11 / 10 ) ^ 4 - p ( 6 / 5 ) ^ 2 = 1205 p = 50000 answer : d" | a = 20 + 1
b = 20 + 1
c = a * b
d = 20 + 1
e = c + d
|
a ) $ 0.00 , b ) $ 4.50 , c ) $ 3.40 , d ) $ 5.00 , e ) $ 6.80 | b | subtract(multiply(61.2, divide(add(const_100, 25), const_100)), divide(61.2, divide(subtract(const_100, 15), const_100))) | cindy has her eye on a sundress but thinks it is too expensive . it goes on sale for 15 % less than the original price . before cindy can buy the dress , however , the store raises the new price by 25 % . if the dress cost $ 61.2 after it went on sale for 15 % off , what is the difference between the original price and the final price ? | "0.85 * { original price } = $ 61.2 - - > { original price } = $ 72 . { final price } = $ 61.2 * 1.25 = $ 76.5 . the difference = $ 76.5 - $ 72 = $ 4.5 answer : b ." | a = 100 + 25
b = a / 100
c = 61 * 2
d = 100 - 15
e = d / 100
f = 61 / 2
g = c - f
|
a ) 1008 , b ) 1015 , c ) 1020 , d ) 1032 , e ) 1048 | c | add(multiply(multiply(power(const_3, const_2.0), power(const_2.0, const_4)), add(const_3, const_4)), 12) | the smallest number which when diminished by 12 , is divisible 12 , 24 , 36 , 48 and 56 is : | "required number = ( l . c . m . of 12 , 24 , 36 , 48,56 ) + 12 = 1008 + 12 = 1020 answer : option c" | a = 3 ** 2
b = 2 ** 0
c = a * b
d = 3 + 4
e = c * d
f = e + 12
|
a ) 20 , b ) 99 , c ) 28 , d ) 27 , e ) 12 | a | divide(subtract(multiply(100, 87.5), multiply(100, 84)), subtract(87.5, multiply(add(const_4, const_3), const_10))) | there are 100 students in 3 sections a , b and c of a class . the average marks of all the 3 sections was 84 . the average of b and c was 87.5 and the average marks of students in a section was ? | numbers of students in section a = x numbers of students in section b and c = ( 100 – x ) x 70 + ( 100 – x ) 87.5 = 84 100 70 x + 87.5 100 – 87.5 x = 8400 8750 – 17.5 x = 8400 17.5 x = 8750 – 8400 = > x = 20 . answer : a | a = 100 * 87
b = 100 * 84
c = a - b
d = 4 + 3
e = d * 10
f = 87 - 5
g = c / f
|
a ) 17 % , b ) 62 % , c ) 52 % , d ) 19 % , e ) 22 % | c | multiply(subtract(divide(divide(multiply(subtract(const_100, 5), add(const_100, 60)), const_100), const_100), const_1), const_100) | a trader bought a car at 5 % discount on its original price . he sold it at a 60 % increase on the price he bought it . what percent of profit did he make on the original price ? | original price = 100 cp = 95 s = 95 * ( 160 / 100 ) = 152 100 - 152 = 52 % answer : c | a = 100 - 5
b = 100 + 60
c = a * b
d = c / 100
e = d / 100
f = e - 1
g = f * 100
|
a ) rs . 200 , b ) rs . 350 , c ) rs . 275 , d ) rs . 425 , e ) none of these | d | divide(subtract(multiply(30, 350), multiply(15, 275)), 15) | the mean daily profit made by a shopkeeper in a month of 30 days was rs . 350 . if the mean profit for the first fifteen days was rs . 275 , then the mean profit for the last 15 days would be | "average would be : 350 = ( 275 + x ) / 2 on solving , x = 425 . answer : d" | a = 30 * 350
b = 15 * 275
c = a - b
d = c / 15
|
a ) 8 , b ) 16 , c ) 11 , d ) 12 , e ) 21 | b | divide(100, multiply(add(18, 72), const_0_2778)) | two trains of length 100 m and 200 m are 100 m apart . they start moving towards each other on parallel tracks , at speeds 18 kmph and 72 kmph . in how much time will the trains cross each other ? | "relative speed = ( 18 + 72 ) * 5 / 18 = 25 mps . the time required = d / s = ( 100 + 100 + 200 ) / 25 = 400 / 25 = 16 sec . answer : b" | a = 18 + 72
b = a * const_0_2778
c = 100 / b
|
a ) 11 : 3 , b ) 11 : 2 , c ) 3 : 2 , d ) 7 : 2 , e ) 9 : 5 | b | divide(add(power(2, 3), 3), add(add(1, 3), 3)) | if x : y = 1 : 3 , find the value of ( 2 x + 3 y ) : ( 5 x – y ) | "explanation : given : x / y = 1 / 3 ( 2 x + 3 y ) / ( 3 x – y ) = ( 2 * 1 + 3 * 3 ) : ( 5 * 1 – 3 ) = 11 : 2 answer : b" | a = 2 ** 3
b = a + 3
c = 1 + 3
d = c + 3
e = b / d
|
a ) 4 / 3 , b ) 1 / 3 , c ) 1 / 12 , d ) - 1 / 12 , e ) 1 / 24 | e | divide(power(divide(power(negate(1), 3), 3), 3), negate(3)) | if x # y is defined to equal x ^ 3 / y for all x and y , then ( - 1 # 2 ) # 3 = | "( - 1 ) ^ 3 / 2 = 1 / 2 ( 1 / 2 ) ^ 3 / 3 = 1 / 24 so e is my answer" | a = negate ** (
b = a / 3
c = b ** 3
d = c / 3
|
a ) 30 % , b ) 55 % , c ) 90 % , d ) 100 % , e ) none of these | b | multiply(const_100, divide(subtract(add(multiply(const_100, divide(150, const_100)), const_100), multiply(const_100, add(const_1, divide(12, const_100)))), add(multiply(const_100, divide(150, const_100)), const_100))) | in a restaurant , the profit is 150 % of the cost . if the cost increases by 12 % but the selling price remains constant , approximately what percentage of the selling price is the profit ? | "explanation : let c . p . = rs . 100 . then , profit = rs . 150 , s . p . = rs . 250 . new c . p . = 112 % of rs . 100 = rs . 112 new s . p . = rs . 250 . profit = rs . ( 250 - 112 ) = rs . 138 . required percentage = ( 138 / 250 * 100 ) % = 55 % appox answer : b" | a = 150 / 100
b = 100 * a
c = b + 100
d = 12 / 100
e = 1 + d
f = 100 * e
g = c - f
h = 150 / 100
i = 100 * h
j = i + 100
k = g / j
l = 100 * k
|
a ) 10 , b ) 20 , c ) 21 , d ) 25 , e ) 27 | c | multiply(7, divide(multiply(add(7, 9), subtract(9, multiply(divide(5, add(7, 5)), 9))), subtract(multiply(9, 7), multiply(7, 5)))) | a can contains a mixture of liquids a and b is the ratio 7 : 5 . when 9 litres of mixture are drawn off and the can is filled with b , the ratio of a and b becomes 7 : 9 . how many liter e of liquid a was contained by the can initially ? | "as a : b : : 7 : 5 - - - > only option c is a multiple of 7 and hence it is a good place to start . also a : b : : 7 : 5 means that , a = ( 712 ) * total and b = ( 5 / 12 ) * total if a = 21 , b = 15 - - - > remove 9 litres - - - > you remove ( 7 / 12 ) * 9 of a - - - > a remaining = 21 - ( 7 / 12 ) * 9 = 63 / 4 similarly , for b , you remove ( 5 / 12 ) * 9 - - - > b remaining = 15 - ( 5 / 12 ) * 9 = 45 / 4 and then add 9 more litres of b - - - > 9 + 45 / 4 = 81 / 4 thus a / b ( final ratio ) = ( 45 / 4 ) / ( 81 / 4 ) = 7 : 9 , the same as the final ratio mentioned in the question . hence c is the correct answer . a / b = 7 / 9 = ( 7 x - ( 7 / 12 ) * 9 ) / ( 5 x - ( 5 / 12 ) * 9 + 9 ) , where 7 x and 5 x are initial quantities of a and b respectively . thus , 7 / 9 = ( 7 x - ( 7 / 12 ) * 9 ) / ( 5 x - ( 5 / 12 ) * 9 + 9 ) - - - > giving you x = 3 . thus a ( original ) e = 7 * 3 = 21 . c" | a = 7 + 9
b = 7 + 5
c = 5 / b
d = c * 9
e = 9 - d
f = a * e
g = 9 * 7
h = 7 * 5
i = g - h
j = f / i
k = 7 * j
|
a ) 18 , b ) 77 , c ) 66 , d ) 55 , e ) 14 | e | subtract(inverse(subtract(multiply(divide(const_1, 8), subtract(const_1, multiply(2, divide(const_1, 14)))), divide(const_1, 14))), add(14, 8)) | a can do a piece of work in 14 days . when he had worked for 2 days b joins him . if the complete work was finished in 8 days . in how many days b alone can finish the work ? | "8 / 14 + 6 / x = 1 x = 14 days answer : e" | a = 1 / 8
b = 1 / 14
c = 2 * b
d = 1 - c
e = a * d
f = 1 / 14
g = e - f
h = 1/(g)
i = 14 + 8
j = h - i
|
a ) 66 , b ) 52 , c ) 58 , d ) 55 , e ) 56 | e | multiply(8, subtract(8, const_1)) | 8 chess players take part in a tournament . every player plays twice with each of his opponents . how many games are to be played ? | "though 2 * ( 8 c 2 ) is the correct approcah to do this , but for people like me who find perm , comb n prob a nightmare , an easy approach can be used . the first guy has to play 2 matches with the rest of 7 , so he ' ll play 14 matches . similarly , second guy has to play with the rest of 6 as his 2 games with the first guy are already played . so he plays 12 matches . this continues like this and the total matches are 14 + 12 + 10 . . . + 2 14 + 12 + . . . + 2 = 2 ( 7 + 6 + . . . + 1 ) = 2 ( ( 7 * 8 ) / 2 ) = 7 * 8 = 56 . answer : e" | a = 8 - 1
b = 8 * a
|
a ) 182 , b ) 208 , c ) 227 , d ) 254 , e ) 302 | c | divide(195, divide(subtract(const_100, 14), const_100)) | a small college reduced its faculty by approximately 14 percent to 195 professors . what was the original number of faculty members ? | if x is the original number of faculty members , then after 14 % reduction in faculty members number is . 86 x but we are given . 86 x = 195 x = 227 so the original number of faculty members is 227 correct answer - c | a = 100 - 14
b = a / 100
c = 195 / b
|
a ) 140 , b ) 150 , c ) 180 , d ) 192 , e ) 250 | d | divide(multiply(divide(16, multiply(multiply(divide(const_1, const_4), divide(const_1, const_3)), divide(const_2, add(const_2, const_3)))), 40), const_100) | one fourth of one third of two fifth of a number is 16 . what will be 40 % of that number | "explanation : ( 1 / 4 ) * ( 1 / 3 ) * ( 2 / 5 ) * x = 16 then x = 16 * 30 = 480 40 % of 480 = 192 answer : option d" | a = 1 / 4
b = 1 / 3
c = a * b
d = 2 + 3
e = 2 / d
f = c * e
g = 16 / f
h = g * 40
i = h / 100
|
a ) 40 , b ) 50 , c ) 60 , d ) 70 , e ) 80 | a | divide(2400, multiply(multiply(3, 4), 5)) | hree number are in the ratio of 3 : 4 : 5 and their l . c . m . is 2400 . their h . c . f . is : | let the numbers be 3 x , 4 x and 5 x . then , their l . c . m . = 60 x . so , 60 x = 2400 or x = 40 . the numbers are ( 3 x 40 ) , ( 4 x 40 ) and ( 5 x 40 ) . hence , required h . c . f . = 40 . answer : a | a = 3 * 4
b = a * 5
c = 2400 / b
|
a ) 140 , b ) 99 , c ) 100 , d ) 72 , e ) 12 | c | multiply(10, multiply(const_2, divide(sqrt(625), divide(10, const_2)))) | the length of a rectangle is two - fifths of the radius of a circle . the radius of the circle is equal to the side of the square , whose area is 625 sq . units . what is the area ( in sq . units ) of the rectangle if the rectangle if the breadth is 10 units ? | "given that the area of the square = 625 sq . units = > side of square = √ 625 = 25 units the radius of the circle = side of the square = 25 units length of the rectangle = 2 / 5 * 25 = 10 units given that breadth = 10 units area of the rectangle = lb = 10 * 10 = 100 sq . units answer : c" | a = math.sqrt(625)
b = 10 / 2
c = a / b
d = 2 * c
e = 10 * d
|
a ) 200 , b ) 225 , c ) 50 , d ) 115 , e ) 150 | b | divide(add(250, 200), const_2) | if x + y = 250 , x - y = 200 , for integers of x and y , y = ? | "x + y = 250 x - y = 200 2 x = 50 x = 25 y = 225 answer is b" | a = 250 + 200
b = a / 2
|
a ) 102 , b ) 103 , c ) 104 , d ) 105 , e ) 106 | d | divide(multiply(42, add(const_4, const_1)), const_2) | to fill a tank , 42 buckets of water is required . how many buckets of water will be required to fill the same tank if the capacity of the bucket is reduced to two - fifths of its present ? | "let the capacity of 1 bucket = x . then , the capacity of tank = 42 x . new capacity of bucket = 2 / 5 x therefore , required number of buckets = ( 42 x ) / ( 2 x / 5 ) = ( 42 x ) x 5 / 2 x = 210 / 2 = 105 answer is d ." | a = 4 + 1
b = 42 * a
c = b / 2
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a ) 84 , b ) 85 , c ) 86 , d ) 87 , e ) 88 | d | divide(870, 10) | a factory produces 1 defective bulb out of 10 bulbs a yr . if it produces 870 bulbs a yr , how many defective bulbs are produced ? | 10 out of 1 is defective 20 out of 2 is defective 100 out of 10 is defective and so on 800 out of 80 is defective 70 out of 7 is defective 80 + 7 = 87 answer : d | a = 870 / 10
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a ) s . 40 , b ) s . 48 , c ) s . 49 , d ) s . 20 , e ) s . 60 | a | divide(multiply(subtract(25, 20), const_100), const_2) | find the principal which yields a simple interest of rs . 20 and compound interest of rs . 25 in two years , at the same percent rate per annum ? | "explanation : si in 2 years = rs . 20 , si in 1 year = rs . 10 ci in 2 years = rs . 25 % rate per annum = [ ( ci – si ) / ( si in 1 year ) ] * 100 = [ ( 25 – 20 ) / 20 ] * 100 = 25 % p . a . let the principal be rs . x time = t = 2 years % rate = 25 % p . a . si = ( prt / 100 ) 20 = ( x * 25 * 2 ) / 100 x = rs . 40 answer : a" | a = 25 - 20
b = a * 100
c = b / 2
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a ) $ 20 , b ) $ 43 , c ) $ 50 , d ) $ 70 , e ) $ 40 | c | add(divide(18000, 400), 5) | a shopkeeper sells 400 metres of cloth for $ 18000 at a loss of rs . 5 per metre . find his cost price for one metre of cloth ? | "c $ 50 sp per metre = 18000 / 400 = $ 45 loss per metre = $ 5 cp per metre = 45 + 5 = $ 50" | a = 18000 / 400
b = a + 5
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a ) 3 , b ) 2 , c ) 6 , d ) 1 , e ) 5 | c | multiply(36, 36) | 36 * 36 * 36 = 6 ^ ? | "6 ^ 2 * 6 ^ 2 * 6 ^ 2 = 6 ^ ( 2 + 2 + 2 ) = 6 ^ 6 answer : 6 answer : c" | a = 36 * 36
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a ) 130 , b ) 140 , c ) 136 , d ) 184 , e ) 196 | d | add(lcm(lcm(5, 9), lcm(12, 18)), 4) | what is the least number which when divided by 5 , 9 , 12 and 18 leaves remainder 4 in each care ? | "explanation : lcm of 5 , 9 , 12 and 18 is 180 required number = 180 + 4 = 184 answer : option d" | a = math.lcm(5, 9)
b = math.lcm(12, 18)
c = math.lcm(a, b)
d = c + 4
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a ) rs 460 , b ) rs 484 , c ) rs 550 , d ) rs 664 , e ) none of these | b | add(divide(1210, 5), divide(1210, 5)) | abhinav and bhupathi together have rs . 1210 . if 4 / 15 of abhinav ' s amount is equal to 2 / 5 of bhupathi ' s amount , how much amount does bhupathi have ? | explanation : let a = abhinav , b = bhupathi . 4 / 15 a = 2 / 5 b = > a = ( 2 / 515 / 4 ) b = > a = 3 / 2 b = > a / b = 3 / 2 = > a : b = 3 : 2 b ' s share = rs . ( 12102 / 5 ) = rs . 484 . answer : b | a = 1210 / 5
b = 1210 / 5
c = a + b
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a ) 2660 , 1000 , b ) 3660 , 2000 , c ) 3000 , 4160 , d ) 2430 , 4050 , e ) 4660 , 3000 | d | multiply(divide(7.5, 12.5), divide(1620, subtract(const_1, divide(7.5, 12.5)))) | difference of two numbers is 1620 . if 7.5 % of the number is 12.5 % of the other number , find the number ? | "let the numbers be x and y . then , 7.5 % of x = 12.5 % of y x = 125 * y / 75 = 5 * y / 3 . now , x - y = 1620 5 * y / 3 – y = 1620 2 * y / 3 = 1620 y = [ ( 1620 * 3 ) / 2 ] = 2430 . one number = 2430 , second number = 5 * y / 3 = 4050 . answer d ." | a = 7 / 5
b = 7 / 5
c = 1 - b
d = 1620 / c
e = a * d
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a ) 72 % , b ) 70 % , c ) 52 % , d ) 50 % , e ) 28 % | c | subtract(60, multiply(divide(60, const_100), 10)) | a shirt goes on sale for 60 % of its original price . one week later , the sale price is marked down 10 % . the final price is what percent of the original price ? | "just assume original price is 100 . sale price = 60 then it is marked down by 10 % = 60 - 6 = 52 . hence it is 52 % od the original price . hence answer is c ." | a = 60 / 100
b = a * 10
c = 60 - b
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a ) 2 : 3 , b ) 4 : 3 , c ) 6 : 7 , d ) 9 : 16 , e ) none | b | divide(sqrt(16), sqrt(9)) | 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 9 hours and 16 hours respectively the ratio of their speeds is | "solution let us name the trains as a and b . = ( a ' s speed ) : ( b ' s speed ) = √ b : √ a = √ 16 : √ 9 = 4 : 3 answer b" | a = math.sqrt(16)
b = math.sqrt(9)
c = a / b
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a ) 64 , b ) 128 , c ) 152 , d ) 155 , e ) 256 | d | subtract(volume_cube(add(cube_edge_by_volume(61), const_2)), 61) | 61 small identical cubes are used to form a large cube . how many more cubes are needed to add one top layer of small cube all over the surface of the large cube ? | "61 small cube will make a large cube with 4 cubes in each line i . e . adding one layer will require one cube at each end and hence new cube will have 6 cubes in each line . total number of small cubes in new cube = 6 ^ 3 = 216 extra cube required = 216 - 61 = 155 hence , d is the answer ." | a = cube_edge_by_volume + (
b = volume_cube - (
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a ) 1 hr , b ) 2 hr , c ) 3 hr , d ) 4 hr , e ) 5 hr | c | multiply(add(add(divide(5, 5), divide(5, 10)), divide(5, 30)), 5) | pipe a can fill a tank in 5 hr , pipe b in 10 hr and pipe c in 30 hr . if all the pipes are open in how many hours will tank be filled ? | "part filled by ( a + b + c ) in 1 hour = 1 / 5 + 1 / 10 + 1 / 30 = 1 / 3 all the 3 pipes together will fill the tank in 3 hours answer is c" | a = 5 / 5
b = 5 / 10
c = a + b
d = 5 / 30
e = c + d
f = e * 5
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a ) 50 min , b ) 60 min , c ) 90 min , d ) 80 min , e ) 75 min | e | multiply(const_10, multiply(const_1, 5)) | a pipe takes a hours to fill the tank . but because of a leakage it took 5 times of its original time . find the time taken by the leakage to empty the tank | "pipe a can do a work 60 min . lets leakage time is x ; then 1 / 60 - 1 / x = 1 / 300 x = 75 min answer : e" | a = 1 * 5
b = 10 * a
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a ) 10.6 kg , b ) 10.8 kg , c ) 11 kg , d ) 14.9 kg , e ) none | e | subtract(multiply(add(19, const_1), 14.7), multiply(19, 15)) | the average weight of 19 students is 15 kg . by the admission of a new student the average weight is reduced to 14.7 kg . the weight of the new student is ? | "answer weight of new student = total weight of all 20 students - total weight of initial 19 students = ( 20 x 14.7 - 19 x 15 ) kg = 9 kg . correct option : e" | a = 19 + 1
b = a * 14
c = 19 * 15
d = b - c
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a ) 20 m east , b ) 40 m north , c ) 30 m west , d ) 40 m east , e ) 20 m west | d | add(20, 20) | 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 20 meters . what distance is she from the starting point and in which direction ? | "the net distance is 20 + 20 = 40 meters to the east . the answer is d ." | a = 20 + 20
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['a ) 32', 'b ) 8', 'c ) 16', 'd ) 24', 'e ) 0'] | c | multiply(divide(32, const_4), const_2) | the following questions are based on the information given . choose the appropriate answer that suits the question based on the information . a cube is cut in two equal parts along a plane parallel to one of its faces . one piece is then colored red on the two larger faces and green on remaining . while the other is colored green on two smaller adjacent faces and red on the remaining . each is then cut into 32 cubes of same size and mixed up . 21 . how many cubes have only one colored face each ? | 16 = 2 ^ 3 each cube will have 2 ^ 3 cubes painted 1 side . answer : c | a = 32 / 4
b = a * 2
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a ) 3277 , b ) 2977 , c ) 22500 , d ) 6077 , e ) 17112 | c | multiply(multiply(4, multiply(const_100, const_100)), power(divide(3, 4), 2)) | the value of a scooter depreciates in such a way that its value of the end of each year is 3 / 4 of its value of the beginning of the same year . if the initial value of the scooter is rs . 40,000 , what is the value at the end of 2 years ? | explanation : 40,000 * ( 3 / 4 ) ^ 2 answer : c | a = 100 * 100
b = 4 * a
c = 3 / 4
d = c ** 2
e = b * d
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a ) 1 / 130 , b ) 1 / 5 , c ) 3 / 13 , d ) 10 / 13 , e ) 30 / 31 | e | divide(6, add(divide(20, const_100), 6)) | a committee is reviewing a total of 20 x black - and - white films and 6 y color films for a festival . if the committee selects y / x % of the black - and - white films and all of the color films , what fraction e of the selected films are in color ? | "it ' s y / xpercentnot y / x . if x = 20 and y = 10 . then : 20 x = 400 black - and - white films ; 6 y = 60 color films . y / x % = 10 / 20 % = 0.5 % of the black - and - white films , so 2 black - and - white films and all 60 color films , thus total of 62 films were selected . color films thus compose e = 60 / 62 = 30 / 31 of the selected films . answer : e ." | a = 20 / 100
b = a + 6
c = 6 / b
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a ) 9800000 , b ) 1000000 , c ) 7500000 , d ) 2000000 , e ) none of these | d | divide(multiply(multiply(multiply(4, const_100), multiply(const_2.0, const_100)), multiply(8, const_100)), multiply(multiply(const_4.0, 7), 7)) | a wooden box of dimensions 8 m x 7 m x 6 m is to carry rectangularboxes of dimensions 4 cm x 7 cm x 6 cm . the maximum number ofboxes that can be carried in the wooden box , is | explanation : number = ( 800 * 700 * 600 ) / 8 * 7 * 6 = 2000000 answer : d | a = 4 * 100
b = 2 * 0
c = a * b
d = 8 * 100
e = c * d
f = 4 * 0
g = f * 7
h = e / g
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a ) 1 / 4 , b ) 1 / 3 , c ) 1 / 2 , d ) 2 / 3 , e ) 3 / 4 | c | subtract(const_1, divide(10, divide(multiply(const_60, const_1), 3))) | a searchlight on top of the watch - tower makes 3 revolutions per minute . what is the probability that a man appearing near the tower will stay in the dark for at least 10 seconds ? | "3 revolutions per minute = 1 revolution every 20 seconds so no matter what anybody appearing at the tower can not stay in the dark for more than 20 seconds . this will be our total number of possibilities i . e the denominator . p ( man in dark for at least 5 seconds ) = 1 - p ( man in dark for max of 10 seconds ) = 1 - 10 / 20 = 1 - 1 / 2 = 1 / 2 or the other way would be : p ( man in dark for at least 5 seconds ) is like saying he can be in dark for 5 , 6,7 . . . all the way to 20 seconds because that is the max . in this approach it would be 10 / 20 seconds = 1 / 2 . answer is c" | a = const_60 * 1
b = a / 3
c = 10 / b
d = 1 - c
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a ) 40 , b ) 56 , c ) 60 , d ) 70 , e ) 80 | b | divide(add(negate(10), sqrt(subtract(power(negate(10), const_2), multiply(const_4, negate(divide(multiply(750, 10), const_2)))))), const_2) | the time it took car p to travel 750 miles was 2 hours less than the time it took car r to travel the same distance . if car p ’ s average speed was 10 miles per hour greater than that of car r , what was car r ’ s average speed , in miles per hour ? | let speed of car r be = x then speed of car p = x + 10 a / q , ( 750 / x ) - ( 750 / ( x + 10 ) ) = 2 solving for x = 56 miles \ hr . b | a = negate + (
b = negate ** (
c = b - 2
d = 750 * 10
e = d / 2
f = 4 * negate
g = math.sqrt(c)
h = a / g
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a ) 500 , b ) 400 , c ) 200 , d ) 300 , e ) 100 | c | divide(subtract(350, 340), divide(5, const_100)) | if 5 % more is gained by selling an book for $ 350 than by selling it for $ 340 , the cost of the book is : | "c 200 let c . p . be $ x . then , 5 % of x = 350 - 340 = 10 x / 20 = 10 = > x = 200" | a = 350 - 340
b = 5 / 100
c = a / b
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a ) 7 , b ) 8 , c ) 9 , d ) 10 , e ) 11 | b | divide(112, subtract(13, const_1)) | find the number which when added to itself 13 times , gives 112 . | "let the number be x . then , x + 13 x = 112 14 x = 112 x = 8 . answer : b" | a = 13 - 1
b = 112 / a
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a ) 1 / 3 , b ) 1 / 4 , c ) 1 / 2 , d ) 1 / 5 , e ) 1 / 6 | a | divide(choose(6, const_2), choose(add(6, 4), const_2)) | there are 6 red shoes & 4 green shoes . if two of red shoes are drawn what is the probability of getting red shoes | "taking 2 red shoe the probablity is 6 c 2 from 10 shoes probablity of taking 2 red shoe is 6 c 2 / 10 c 2 = 1 / 3 answer : a" | a = math.comb(6, 2)
b = 6 + 4
c = math.comb(b, 2)
d = a / c
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a ) 39 , b ) 35 , c ) 42 , d ) 40 , e ) 41.5 | d | add(subtract(200, multiply(17, 10)), 10) | a batsman makes a score of 200 runs in the 17 th inning and thus increases his averages by 10 what is his average after 17 th inning ? | "let the average after 16 th inning = x then total run after 16 th inning = 16 x then total run after 17 th inning = 16 x + 200 then average run after 17 th inning = ( 16 x + 200 ) / 17 ( 16 x + 200 ) / 17 = x + 10 = > 16 x + 200 = 17 x + 170 x = 30 ; average after 17 th inning = 30 + 10 = 40 answer : d" | a = 17 * 10
b = 200 - a
c = b + 10
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a ) 3 , b ) 4 , c ) 6 , d ) 8 , e ) 12 | d | add(add(const_3, 2), const_3) | if w and y are both odd prime numbers and w < y , how many distinct positive integer factors does 2 wy have ? | since 2 wy prime factors are w ^ 1 * y ^ 1 * 2 ^ 1 , its total number or factors must be ( 1 + 1 ) ( 1 + 1 ) ( 1 + 1 ) = 2 ^ 3 = 8 . thus , i think d would be the correct answer . d | a = 3 + 2
b = a + 3
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a ) 25 % , b ) 35 % , c ) 45 % , d ) 70 % , e ) 80 % | b | add(const_10, divide(add(25, 25), const_2)) | in goshawk - eurasian nature reserve 30 percent of the birds are hawks , and 40 percent of the non - hawks are paddyfield - warblers . if there are 25 percent as many kingfishers as paddyfield - warblers in the reserve , then what percent of the birds r in the nature reserve are not hawks , paddyfield - warblers , or kingfishers ? | 1 . we are given the following percentages : 30 ( 70 ) , 40 ( 60 ) , 25 ( 75 ) . there are two threads from here . first starts at 30 % and finishes there . second one starts at 70 , then 40 , and then 25 . we need a value that is divisible by 7 , 2 , and 5 at least once . lets pick a number now , say 700 . so say if non hawks are 700 ( this is 70 % of the total , so total = 1000 ) , then paddy warbs are 2 / 5 x 700 = 1400 / 5 = 280 . kingfishers , therefore , are 280 / 4 = 70 . lets add them up . 300 hawks + 280 peddy warbs + 70 kingsifhers = 650 . so all others are 1000 - 650 = 350 or 35 % of total birds . the main job here to to identify the smart number to start the question with . this can be time consuming , but once identified , this question can be solved fairly quickly . 2 . another method : if x is total - - > non hawks = 0.7 x - - > warbs = 0.4 ( 0.7 x ) - - > kfs = 0.25 ( 0.4 ( 0.7 x ) ) . our job is to find out r : ( 0.3 x + 0.28 x + 0.07 x ) / x . or 0.65 x / x = 0.65 . we need to find 1 - 0.65 = 0.35 or 35 % . b | a = 25 + 25
b = a / 2
c = 10 + b
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a ) 4 , b ) 3 , c ) 2 , d ) 5 , e ) none of the above | e | divide(subtract(power(8, 1), power(8, 1)), 8) | what is the remainder when 8 ^ 1 + 8 ^ 2 + 8 ^ 3 + . . . + 8 ^ 9 is divided by 4 ? | "notice that in the brackets we have the sum of 9 even multiples of 4 , hence the sum in the brackets will be even multiple of 4 ( the sum of 9 even numbers is even ) . which yields remainder of 0 upon division by 4 . answer : e" | a = 8 ** 1
b = 8 ** 1
c = a - b
d = c / 8
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a ) rs . 55.50 , b ) rs . 67.50 , c ) rs . 86.50 , d ) rs . 91.00 , e ) none of these | d | divide(multiply(rectangle_perimeter(multiply(3, sqrt(divide(8112, multiply(3, 4)))), multiply(4, sqrt(divide(8112, multiply(3, 4))))), 25), const_100) | the sides of a rectangular field are in the ratio 3 : 4 . if the area of the field is 8112 sq . m , the cost of fencing the field @ 25 paise per metre is | "solution let length = ( 3 x ) metres and breadth = ( 4 x ) metres . then , 3 x × 4 x = 7600 ⇔ 12 x 2 = 8112 ⇔ x 2 = 676 ⇔ x = 26 so , length = 78 m and breadth = 104 m . perimeter = [ 2 ( 78 + 104 ) ] m = 364 m . ∴ cost of fencing = rs . ( 0.25 × 364 ) = rs . 91 . answer d" | a = 3 * 4
b = 8112 / a
c = math.sqrt(b)
d = 3 * c
e = 3 * 4
f = 8112 / e
g = math.sqrt(f)
h = 4 * g
i = rectangle_perimeter * (
j = i / 25
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a ) 90 , b ) 100 , c ) 200 , d ) 300 , e ) 400 | d | divide(90, multiply(divide(6, const_100), 5)) | a sum was put at simple interest at a certain rate for 5 years had it been put at 6 % higher rate , it would have fetched 90 more . find the sum . | "difference in s . i . = p × t / 100 ( r 1 − r 2 ) ⇒ 90 = p × 5 x 6 / 100 ( ∵ r 1 - r 2 = 2 ) ⇒ p = 90 × 100 / 5 x 6 = 300 answer d" | a = 6 / 100
b = a * 5
c = 90 / b
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a ) 18 days , b ) 27 days , c ) 26.67 days , d ) 15 days , e ) 12 days | d | add(divide(subtract(const_1, multiply(divide(const_1, 30), 5)), add(divide(const_1, 20), divide(const_1, 30))), 5) | a can complete a project in 20 days and b can complete the same project in 30 days . if a and b start working on the project together and a quits 5 days before the project is completed , in how many days will the project be completed ? | "let x = the number of days taken to complete the project . the amount of work done by a is ( x - 10 ) * ( 1 / 20 ) . the amount of work done by b is ( x ) * ( 1 / 30 ) . ( 1 / 20 ) * ( x - 5 ) + ( 1 / 30 ) * ( x ) = 1 ( x / 20 ) + ( x / 30 ) - ( 5 / 20 ) = 1 5 x / 60 = 5 / 4 x = 60 / 4 x = 15 therefore , the answer is d : 15 ." | a = 1 / 30
b = a * 5
c = 1 - b
d = 1 / 20
e = 1 / 30
f = d + e
g = c / f
h = g + 5
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a ) 2 : 5 , b ) 3 : 7 , c ) 10 : 9 , d ) 7 : 3 , e ) none of these | c | divide(divide(const_1, const_4), divide(30, const_100)) | if 30 % of a number is equal to one - third of another number , what is the ratio of first number to the second number ? | "explanation : let 30 % of a = 1 / 3 b then , 30 a / 100 = 1 b / 3 = > 3 a / 10 = b / 3 = > a : b = 10 : 9 answer : c" | a = 1 / 4
b = 30 / 100
c = a / b
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a ) 81000 , b ) 33888 , c ) 77678 , d ) 200988 , e ) 40500 | e | add(add(32000, multiply(divide(1, 8), 32000)), multiply(divide(1, 8), add(32000, multiply(divide(1, 8), 32000)))) | every year an amount increases by 1 / 8 th of itself . how much will it be after two years if its present value is rs . 32000 ? | "32000 * 9 / 8 * 9 / 8 = 40500 answer : e" | a = 1 / 8
b = a * 32000
c = 32000 + b
d = 1 / 8
e = 1 / 8
f = e * 32000
g = 32000 + f
h = d * g
i = c + h
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a ) 75 , b ) 100 , c ) 125 , d ) 150 , e ) none of these | b | subtract(add(175, 325), subtract(450, 50)) | out of 450 students of a school , 325 play football , 175 play cricket and 50 neither play football nor cricket . how many students play both football and cricket ? | "explanation : students who play cricket , n ( a ) = 325 students who play football , n ( b ) = 175 total students who play either or both games , = n ( a ∪ b ) = 450 − 50 = 400 required number , n ( a ∩ b ) = n ( a ) + n ( b ) − n ( a ∪ b ) = 325 + 175 − 400 = 100 option b" | a = 175 + 325
b = 450 - 50
c = a - b
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a ) 8 , b ) 10 , c ) 14 , d ) 13 , e ) 16 | d | divide(subtract(99, multiply(const_3, 7)), multiply(const_3, const_2)) | a number is doubled and 7 is added . if resultant is trebled , it becomes 99 . what is that number | "explanation : = > 3 ( 2 x + 7 ) = 99 = > 2 x + 7 = 33 = > x = 13 option d" | a = 3 * 7
b = 99 - a
c = 3 * 2
d = b / c
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a ) 1 / 10 , b ) 1 / 6 , c ) 3 / 10 , d ) 1 / 5 , e ) 1 / 2 | d | divide(choose(3, 3), choose(add(const_2.0, 3), const_2)) | a bag holds 3 red marbles and 3 green marbles . if you removed two randomly selected marbles from the bag , without replacement , what is the probability that both would be green ? | "given : 3 r and 3 g marbles required : probability that 2 marbles removed without replacement are both red initially we have to pick one red from a total of 3 red and 3 green marbles after one green has been picked , we need to pick 1 green from a total of 2 greenand 3 red marbles . p ( both green ) = ( 3 / 6 ) * ( 2 / 5 ) = 1 / 5 option d" | a = math.comb(3, 3)
b = 2 + 0
c = math.comb(b, 2)
d = a / c
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a ) 20 m , b ) 50 m , c ) 45 m , d ) 60 m , e ) none of these | b | multiply(divide(200, 60), subtract(60, 45)) | in a 200 m race , a covers the distance in 45 seconds and b in 60 second . in this race a beats b by : | "solution distance covered by b in 15 sec . = ( 200 / 60 x 15 ) m = 50 m . ∴ a beats b by 50 metres . answer b" | a = 200 / 60
b = 60 - 45
c = a * b
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a ) 15 , b ) 14 , c ) 13 , d ) 12 , e ) 11 | b | subtract(divide(power(negate(8), 2), 4), 2) | find the value of a / b + b / a , if a and b are the roots of the quadratic equation x 2 + 8 x + 4 = 0 ? | "a / b + b / a = ( a 2 ] / ab a + b = + b 2 ) / ab = ( a 2 + b 2 + a + b ) / ab = [ ( a + b ) 2 - 2 ab - 8 / 1 = - 8 ab = 4 / 1 = 4 hence a / b + b / a = [ ( - 8 ) 2 - 2 ( 4 ) ] / 4 = 56 / 4 = 14 . answer : b" | a = negate ** (
b = a / 2
c = b - 4
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a ) 6 ^ 5 , b ) 5 ^ 6 , c ) ( 10 / 3 ) ^ 5 , d ) 6 ^ 3 , e ) 15 ^ 3 | c | divide(power(20, 10), power(20, 5)) | 20 ^ 10 / 120 ^ 5 = ? | "20 ^ 10 / 120 ^ 5 = ? a . 6 ^ 5 b . 5 ^ 6 c . ( 10 / 3 ) ^ 5 d . 6 ^ 3 e . 15 ^ 3 - > 20 ^ 10 / 120 ^ 5 = ( 20 ^ 10 ) / ( 6 ^ 5 ) ( 20 ^ 5 ) = ( 20 ^ 5 ) / ( 6 ^ 5 ) = ( 2 ^ 5 ) ( 10 ^ 5 ) / ( 2 ^ 5 ) ( 3 ^ 5 ) = ( 10 / 3 ) ^ 5 . thus , c is the answer ." | a = 20 ** 10
b = 20 ** 5
c = a / b
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a ) 77 , b ) 57 , c ) 34 , d ) 65 , e ) 21 | b | add(multiply(5, 6), multiply(3, 9)) | darren bought 5 packs of white t - shirts and 3 packs of blue t - shirts for his basketball team . the white t - shirts come in packs of 6 , and the blue t - shirts come in packs of 9 . how many t - shirts did darren buy in all ? | step 1 : find the number of white t - shirts . 5 × 6 = 30 step 2 : find the number of blue t - shirts . 3 × 9 = 27 step 3 : find the total number of t - shirts . 30 + 27 = 57 darren bought 57 t - shirts . answer is b . | a = 5 * 6
b = 3 * 9
c = a + b
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a ) 1.5 , b ) 2.25 , c ) 2.75 , d ) 2.5 , e ) 3.5 | b | divide(add(10, 8), subtract(58, 50)) | car a is 10 miles behind car b , which is traveling in the same direction along the same route as car a . car a is traveling at a constant speed of 58 miles per hour and car bis traveling at a constant speed of 50 miles per hour . how many hours will it take for car a to overtake and drive 8 miles ahead of car b ? | "relative speed of car a is 58 - 50 = 8 miles per hour , to catch up 10 miles and drive 8 miles ahead so to drive 18 miles it ' ll need 18 / 8 = 2.25 hours . answer : b" | a = 10 + 8
b = 58 - 50
c = a / b
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a ) 20 m , b ) 125 m , c ) 120 m , d ) 160 m , e ) none of these | d | multiply(divide(400, 100), subtract(100, 60)) | in a 400 m race , a covers the distance in 60 seconds and b in 100 second . in this race a beats b by : | "solution distance covered by b in 40 sec . = ( 400 / 100 x 40 ) m = 160 m . ∴ a beats b by 160 metres . answer d" | a = 400 / 100
b = 100 - 60
c = a * b
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a ) 39.55 $ , b ) 40.63 $ , c ) 38.63 $ , d ) 37.88 $ , e ) 35.15 $ | c | divide(50, add(divide(add(7, 20), const_100), const_1)) | a business executive and his client are charging their dinner tab on the executive ' s expense account . the company will only allow them to spend a total of 50 $ for the meal . assuming that they will pay 7 % in sales tax for the meal and leave a 20 % tip , what is the most their food can cost ? | let x is the cost of the food 1.07 x is the gross bill after including sales tax 1.20 * 1.07 x = 50 x = 37.88 hence , the correct option is c | a = 7 + 20
b = a / 100
c = b + 1
d = 50 / c
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a ) 88 , b ) 99 , c ) 76 , d ) 55 , e ) 66 | b | multiply(11, 77) | the h . c . f . of two numbers is 11 and their l . c . m . is 693 . if one of the number is 77 , find the other ? | "other number = 11 * 693 / 77 = 99 answer is b" | a = 11 * 77
|
a ) 80 , b ) 70 , c ) 85 , d ) 90 , e ) 95 | d | add(multiply(const_100, const_100), divide(multiply(195, 9), add(add(divide(multiply(9, 12), 6), 9), 12))) | p , q and r can do a work in 6 , 9 and 12 days respectively . they completed the work and got rs . 195 . what is the share of p ? | "the ratio of their working rates = 1 / 6 : 1 / 9 : 1 / 12 = 6 : 4 : 3 . since , they work together , the share of p = 6 / 13 * 195 = rs . 90 answer : d" | a = 100 * 100
b = 195 * 9
c = 9 * 12
d = c / 6
e = d + 9
f = e + 12
g = b / f
h = a + g
|
a ) 2 , b ) 3 , c ) 4 , d ) 5 , e ) 6 | a | divide(220, divide(multiply(multiply(10, 220), divide(add(const_100, 10), const_100)), subtract(multiply(220, divide(add(const_100, 10), const_100)), 220))) | machine a and machine b are each used to manufacture 220 sprockets . it takes machine a 10 hours longer to produce 220 sprockets than machine b . machine b produces 10 percent more sprockets per hour than machine a . how many sprockets per hour does machine a produces ? | "machine b : takes x hours to produce 220 sprockets machine a : takes ( x + 10 ) hours to produce 220 sprockets machine b : in 1 hour , b makes 220 / x sprockets machine a : in 1 hour , a makes 220 / ( x + 10 ) sprockets equating : 1.1 ( 220 / ( x + 10 ) ) = 220 / x 242 / ( x + 10 ) = 220 / x 242 x = 220 x + 2200 22 x = 2200 x = 100 a makes 220 / ( 110 ) = 2 sprockets per hour answer : a" | a = 10 * 220
b = 100 + 10
c = b / 100
d = a * c
e = 100 + 10
f = e / 100
g = 220 * f
h = g - 220
i = d / h
j = 220 / i
|
a ) 10.22 % , b ) 20.22 % , c ) 21.22 % , d ) 30 % , e ) ca n ' t be calculated | d | divide(multiply(subtract(add(const_100, 30), subtract(const_100, 10)), const_100), subtract(const_100, 10)) | a shop owner professes to sell his articles at certain cost price but he uses false weights with which he cheats by 30 % while buying and by 10 % while selling . what is his percentage profit ? | "the owner buys 100 kg but actually gets 130 kg ; the owner sells 100 kg but actually gives 90 kg ; profit : ( 130 - 90 ) / 90 * 100 = 30 % answer : d ." | a = 100 + 30
b = 100 - 10
c = a - b
d = c * 100
e = 100 - 10
f = d / e
|
a ) 24 % , b ) 25 % , c ) 30 % , d ) 50 % , e ) 40 % | d | multiply(divide(300, subtract(900, 300)), const_100) | a cricket bat is sold for $ 900 , making a profit of $ 300 . the profit percentage would be | "300 / ( 900 - 300 ) = 50 % . answer : d" | a = 900 - 300
b = 300 / a
c = b * 100
|
a ) 9 , b ) 12 , c ) 11 , d ) 10 , e ) 3 | a | add(divide(15, const_2), 1) | each of the integers from 1 to 15 is written on the a seperate index card and placed in a box . if the cards are drawn from the box at random without replecement , how many cards must be drawn to ensure that the product of all the integers drawn is even ? | "out of the 15 integers : 8 are odd and 7 are even . if we need to make sure that the product of all the integers withdrawn is even then we need to make sure that we have at least one even number . in the worst case : 1 . we will end up picking odd numbers one by one , so we will pick all 8 odd numbers first 2 . 9 th number will be the first even number so we need to withdraw at least 9 numbers to make sure that we get one even number and the product of all the integers picked is even . so , answer will be 9 . ( a )" | a = 15 / 2
b = a + 1
|
a ) 40 , b ) 42 , c ) 44 , d ) 46 , e ) 48 | a | divide(add(130, 150), 7) | a student chose a number , multiplied it by 7 , then subtracted 150 from the result and got 130 . what was the number he chose ? | "let x be the number he chose , then 7 ⋅ x − 150 = 130 7 x = 280 x = 40 correct answer a" | a = 130 + 150
b = a / 7
|
a ) rs . 692.80 , b ) rs . 820 , c ) rs . 990 , d ) rs . 1385 , e ) none | e | subtract(multiply(735, 4), multiply(550, subtract(4, const_1))) | average monthly income of a family of 4 earning members was rs . 735 . one of the earning members died and therefore , the average income came down to rs 550 . the income of the deceased was ? | "answer income of the deceased = total income of 4 members - total income of remaining 3 members . = 735 x 4 - 550 x 3 rs . = 1290 rs . correct option : e" | a = 735 * 4
b = 4 - 1
c = 550 * b
d = a - c
|
a ) 5 / 100 , b ) 6 / 100 , c ) 7 / 100 , d ) 8 / 100 , e ) 9 / 100 | e | divide(subtract(10, 1), 100) | evaluate : ( 1 - 1 / 10 ) ( 1 - 1 / 11 ) ( 1 - 1 / 12 ) . . . ( 1 - 1 / 99 ) ( 1 - 1 / 100 ) | ( 1 - 1 / 10 ) ( 1 - 1 / 11 ) ( 1 - 1 / 12 ) . . . ( 1 - 1 / 99 ) ( 1 - 1 / 100 ) = ( 9 / 10 ) ( 10 / 11 ) ( 11 / 12 ) . . . ( 98 / 99 ) ( 99 / 100 ) = 9 / 100 : simplify correct answer e | a = 10 - 1
b = a / 100
|
a ) 7 , b ) 8 , c ) 9 , d ) 10 , e ) 11 | a | subtract(divide(subtract(39, 15), add(2, 1)), 1) | when a person aged 39 is added to a group of n people , the average age increases by 2 . when a person aged 15 is added instead , the average age decreases by 1 . what is the value of w ? | "a simple and elegant solution . as addition of 39 , shifts mean by 2 , and addition of 15 , shifts mean by 1 to the other side , we have the mean lying between 3915 , and in a ratio of 2 : 1 39 - 15 = 24 24 divide by 3 is 8 . meaning mean of the n terms is 15 + 8 = 39 - 16 = 23 now , from first statement , when a person aged 39 is added to a group of n people , the average age increases by 2 . w * 23 + 39 = 25 * ( w + 1 ) w = 7 ans . ( a )" | a = 39 - 15
b = 2 + 1
c = a / b
d = c - 1
|
a ) 25 , b ) 40 , c ) 50 , d ) 60 , e ) 75 | d | subtract(subtract(subtract(210, 60), 90), divide(subtract(subtract(210, 60), 90), const_3)) | 210 college students were asked in a survey if they preferred windows or mac brand computers . 90 students claimed that they preferred mac to windows brand computers . one third as many of the students who preferred mac to windows , equally preferred both brands . 60 of the students had no preference . how many of the students in the survey preferred windows to mac brand computers ? | "210 = 90 ( mac ) + x ( window ) + 60 ( both ) = > x = 60 answer : d" | a = 210 - 60
b = a - 90
c = 210 - 60
d = c - 90
e = d / 3
f = b - e
|
a ) 25 , b ) 40 , c ) 50 , d ) 60 , e ) 75 | b | subtract(210, add(add(add(90, 60), const_10), const_10)) | 210 school students were asked in a survey if they preferred windows or mac brand computers . 60 students claimed that they preferred mac to windows brand computers . one third as many of the students who preferred mac to windows , equally preferred both brands . 90 of the students had no preference . how many of the students in the survey preferred windows to mac brand computers ? | we are told that 60 students claimed that they preferred mac to windows , which means that 60 preferred mac but not windows , so # of students who preferred mac ( p ( a ) as you wrote ) , does not equal to 60 , it equals to 60 + 20 ( 20 is # of students who equally preferred both brands ) . also we are asked to find # of the students who preferred windows to mac , so if you denote x as those who prefer windows then you should calculate x - 20 . so , if we use your formula it should be : 210 = { mac } + { windows } - { both } + { neither } = ( 60 + 20 ) + x - 20 + 90 - - > x = 60 ( # of student who prefer windows ) - - > # of the students who preferred windows to mac is x - 20 = 40 . | a = 90 + 60
b = a + 10
c = b + 10
d = 210 - c
|
a ) 21.12 hours , b ) 20.12 hours , c ) 19.12 hours , d ) 18.12 hours , e ) 17.12 hours | b | add(multiply(add(add(11, 2.5), subtract(11, 2.5)), 105), multiply(subtract(add(divide(105, add(11, 2.5)), divide(105, subtract(11, 2.5))), add(add(11, 2.5), subtract(11, 2.5))), const_60)) | speed of a boat in standing water is 11 kmph and speed of the stream is 2.5 kmph . a man can rows to a place at a distance of 105 km and comes back to the starting point . the total time taken by him is ? | "speed upstream = 8.5 kmph speed downstream = 13.5 kmph total time taken = 105 / 8.5 + 105 / 13.5 = 20.12 hours answer is b" | a = 11 + 2
b = 11 - 2
c = a + b
d = c * 105
e = 11 + 2
f = 105 / e
g = 11 - 2
h = 105 / g
i = f + h
j = 11 + 2
k = 11 - 2
l = j + k
m = i - l
n = m * const_60
o = d + n
|
a ) 42 , b ) 43 , c ) 44 , d ) 45 , e ) 46 | c | divide(650, multiply(subtract(81, 340), const_0_2778)) | a train 650 m long is running at a speed of 81 km / hr . in what time will it pass a bridge 340 m long ? | "speed = 81 * 5 / 18 = 45 / 2 m / sec total distance covered = 650 + 340 = 990 m required time = 990 * 2 / 45 = 44 sec answer : c" | a = 81 - 340
b = a * const_0_2778
c = 650 / b
|
a ) 8 / 25 , b ) 11 / 24 , c ) 12 / 25 , d ) 12 / 25 , e ) 12 / 23 | d | multiply(divide(4, 5), divide(3, 5)) | what is the probability that carol and bernie will get selected for a job they both applied for in a company , given that their chances of getting selected is 4 / 5 and 3 / 5 respectively ? | explanation : p ( carol ) = 4 / 5 p ( bernie ) = 3 / 5 e = { carol and bernie both get selected } p ( e ) = p ( carol ) * p ( bernie ) = 4 / 5 * 3 / 5 = 12 / 25 answer : d | a = 4 / 5
b = 3 / 5
c = a * b
|
a ) 1 : 2 , b ) 2 : 3 , c ) 3 : 4 , d ) 4 : 5 , e ) 5 : 6 | c | divide(add(1, divide(7, 2)), add(divide(7, 2), divide(5, 2))) | the weight of every type a widget is the same , the weight of every type b widget is the same , and the weight of every type c widget is the same . if the weight of 7 type a widgets is equal to the weight of 2 type b widgets , and the weight of 5 type b widgets is equal to the weight of 7 type c widgets . what is the ratio of the total weight of 1 type a widget and 1 type b widget , to the total weight of 1 type b widget and 1 type c widget ? | "5 b = 7 c and so b = 7 c / 5 7 a = 2 b and so a = 2 b / 7 = 2 c / 5 a + b = 2 c / 5 + 7 c / 5 = 9 c / 5 b + c = 7 c / 5 + c = 12 c / 5 the ratio of a + b : b + c = 9 : 12 = 3 : 4 the answer is c ." | a = 7 / 2
b = 1 + a
c = 7 / 2
d = 5 / 2
e = c + d
f = b / e
|
a ) 0.9 , b ) 0.75 , c ) 0.6 , d ) 0.8 , e ) 0.5 | e | divide(subtract(100, add(47, 3)), 100) | a certain bag contains 100 balls â € ” 10 white , 30 green , 10 yellow , 47 red , and 3 purple . if a ball is to be chosen at random , what is the probability that the ball will be neither red nor purple ? | "according to the stem the ball can be white , green or yellow , so the probability is ( white + green + yellow ) / ( total ) = ( 10 + 30 + 10 ) / 100 = 50 / 100 = 0.5 . answer is e" | a = 47 + 3
b = 100 - a
c = b / 100
|
a ) 5600 , b ) 6000 , c ) 242 , d ) 7200 , e ) 8600 | c | divide(multiply(multiply(multiply(8, const_100), multiply(1, const_100)), 5), multiply(multiply(25, 11), 6)) | how many bricks , each measuring 25 cm x 11 cm x 6 cm , will be needed to build a wall of 8 m x 1 m x 5 cm ? | "number of bricks = volume of the wall / volume of 1 brick = ( 800 x 100 x 5 ) / ( 25 x 11 x 6 ) = 242 . answer : option c" | a = 8 * 100
b = 1 * 100
c = a * b
d = c * 5
e = 25 * 11
f = e * 6
g = d / f
|
a ) 37500 , b ) 38000 , c ) 38500 , d ) 39000 , e ) 39500 | a | multiply(divide(divide(800, const_2), add(divide(60, const_100), const_1)), subtract(divide(800, const_2), divide(divide(800, const_2), add(divide(60, const_100), const_1)))) | the breadth of a rectangular field is 60 % of its length . if the perimeter of the field is 800 m , find out the area of the field . | "given that breadth of the rectangular field is 60 % of its length . ⇒ b = 60 l 100 = 3 l 5 perimeter of the field = 800 m ⇒ 2 ( l + b ) = 800 ⇒ 2 ( l + 3 l 5 ) = 800 ⇒ l + 3 l 5 = 400 ⇒ 8 l 5 = 400 ⇒ l 5 = 50 ⇒ l = 5 × 50 = 250 m b = 3 l 5 = 3 × 2505 = 3 × 50 = 150 m area = lb = 250 × 150 = 37500 m 2 answer is a ." | a = 800 / 2
b = 60 / 100
c = b + 1
d = a / c
e = 800 / 2
f = 800 / 2
g = 60 / 100
h = g + 1
i = f / h
j = e - i
k = d * j
|
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