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a ) 4857 , b ) 4184 , c ) 4012 , d ) 4891 , e ) 5291
d
subtract(multiply(6500, add(5, const_1)), add(add(add(6855, add(6535, 6927)), 7230), 6562))
a grocery shop has a sale of rs . 6535 , rs . 6927 , rs . 6855 , rs . 7230 and rs . 6562 for 5 consecutive months . how much sale must he have in the sixth month so that he gets an average sale of rs . 6500 ?
"let the sale in the sixth month = x then ( 6535 + 6927 + 6855 + 7230 + 6562 + x ) / 6 = 6500 = > 6535 + 6927 + 6855 + 7230 + 6562 + x = 6 Γ— 6500 = > 34109 + x = 39000 = > x = 39000 βˆ’ 34109 = 4891 answer : d"
a = 5 + 1 b = 6500 * a c = 6535 + 6927 d = 6855 + c e = d + 7230 f = e + 6562 g = b - f
a ) 110 , b ) 111 , c ) 271 , d ) 300 , e ) 304
d
multiply(multiply(multiply(1, const_10), const_10), const_3)
how many times will the digit 7 be written when listing the integers from 1 to 1000 ?
another approach : in the range 0 - 100 : 7 as units digit - 10 times ( 7 , 17 , 27 , . . . , 97 ) ; 7 as tens digit - 10 time ( 71 , 72 , 73 , . . . , 79 ) ; so in first one hundred numbers 7 is written 10 + 10 = 20 times . in 10 hundreds 7 as units or tens digit will be written 10 * 20 = 200 times . plus 100 times when 7 is written as hundreds digit ( 700 , 701 , 702 , . . . , 799 ) . total 200 + 100 = 300 . answer : d .
a = 1 * 10 b = a * 10 c = b * 3
a ) 24 , b ) 26 , c ) 28 , d ) 30 , e ) 32
d
divide(rectangle_area(8, 15), 4)
carol and jordan draw rectangles of equal area . if carol ' s rectangle measures 8 inches by 15 inches and jordan ' s rectangle is 4 inches long , how wide is jordan ' s rectangle , in inches ?
"area of first rectangle is 8 * 15 = 120 hence area of second would be 4 x = 120 x x = 30 answer is d"
a = rectangle_area / (
a ) 2177 , b ) 2876 , c ) 4700 , d ) 2981 , e ) 2711
c
subtract(multiply(8000, const_4), subtract(multiply(8450, const_4), 6500))
the average salary of a person for the months of january , february , march and april is rs . 8000 and that for the months february , march , april and may is rs . 8450 . if his salary for the month of may is rs . 6500 , find his salary for the month of january ?
"sum of the salaries of the person for the months of january , february , march and april = 4 * 8000 = 32000 - - - - ( 1 ) sum of the salaries of the person for the months of february , march , april and may = 4 * 8450 = 33800 - - - - ( 2 ) ( 2 ) - ( 1 ) i . e . may - jan = 1800 salary of may is rs . 6500 salary of january = rs . 4700 . answer : c"
a = 8000 * 4 b = 8450 * 4 c = b - 6500 d = a - c
a ) 60 , b ) 80 , c ) 100 , d ) 120 , e ) 140
c
divide(divide(multiply(multiply(45, divide(4, 5)), divide(4, 5)), subtract(const_1, multiply(divide(4, 5), divide(4, 5)))), divide(4, 5))
a student was asked to find 4 / 5 of a number . but the student divided the number by 4 / 5 , thus the student got 45 more than the correct answer . find the number .
"let the number be x . ( 5 / 4 ) * x = ( 4 / 5 ) * x + 45 25 x = 16 x + 900 9 x = 900 x = 100 the answer is c ."
a = 4 / 5 b = 45 * a c = 4 / 5 d = b * c e = 4 / 5 f = 4 / 5 g = e * f h = 1 - g i = d / h j = 4 / 5 k = i / j
a ) 1 / 12 , b ) 1 / 3 , c ) 1 / 4 , d ) 1 / 2 , e ) 1 / 6
a
multiply(divide(const_1, const_2), multiply(divide(const_1, const_3), divide(const_1, const_2)))
a quiz has one mcq question with a , b and c as options . and two questions with true / false answers . what is the probability of giving all 3 answers correct ?
probability of getting mcq answer correct = 1 / 3 probability of getting one true / false answer correct = 1 / 2 probability of getting all 3 answers correct = 1 / 3 * 1 / 2 * 1 / 2 = 1 / 12 answer : a
a = 1 / 2 b = 1 / 3 c = 1 / 2 d = b * c e = a * d
a ) 1045 , b ) 4860 , c ) 5000 , d ) 5500 , e ) 5550
a
add(100, const_1)
what is the sum of natural numbers between 90 and 100
"a = first number l = last number sn = n / 2 [ a + l ] between 90 and 100 numbers = 11 = > 100 - 90 = 10 + 1 = 11 sn = 11 / 2 Γ£ β€” 190 = 11 Γ£ β€” 95 = 1045 answer : a"
a = 100 + 1
a ) 30 m , b ) 28 m , c ) 26 m , d ) 24 m , e ) 22 m
d
add(add(multiply(subtract(12, const_1), 2), divide(10, 2)), divide(10, 2))
in a garden , there are 10 rows and 12 columns of mango trees . the distance between the two trees is 2 metres and a distance of one metre is left from all sides of the boundary of the garden . what is the length of the garden ?
"explanation : between the 12 mango trees , there are 11 gaps and each gap has 2 meter length also , 1 meter is left from all sides of the boundary of the garden . hence , length of the garden = ( 11 Γ— 2 ) + 1 + 1 = 24 meter answer : d"
a = 12 - 1 b = a * 2 c = 10 / 2 d = b + c e = 10 / 2 f = d + e
a ) 0.125 % , b ) 0.36 % , c ) 0.8 % , d ) 1.25 % , e ) 2.0 %
b
multiply(9, divide(4, const_100))
in the manufacture of a certain product , 9 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 = 9 % 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 ) * ( 9 / 100 ) * 100 % = ( 36 / 10000 ) * 100 % = ( 36 / 100 ) % = . 36 % answer b
a = 4 / 100 b = 9 * a
a ) 53 , b ) 54 , c ) 55 , d ) 56 , e ) 57
d
add(subtract(70, 52), 38)
70 students are required to paint a picture . 52 use green color and some children use red , 38 students use both the colors . how many students use red color ?
70 = red + green - both red = 70 - green + both = 70 - 52 + 38 = 56 answer : d
a = 70 - 52 b = a + 38
a ) 17 % , b ) 16 % , c ) 19 % , d ) 78 % , e ) 28 %
a
multiply(divide(subtract(1500, 1245), 1500), const_100)
the cost price of a radio is rs . 1500 and it was sold for rs . 1245 , find the loss % ?
"1500 - - - - 255 100 - - - - ? = > 17 % answer : a"
a = 1500 - 1245 b = a / 1500 c = b * 100
a ) 560 , b ) 449 , c ) 600 , d ) 740 , e ) 460
c
add(500, multiply(500, divide(20, const_100)))
a person buys an article at $ 500 . at what price should he sell the article so as to make a profit of 20 % ?
"c 600 cost price = $ 500 profit = 20 % of 500 = $ 100 selling price = cost price + profit = 500 + 100 = 600"
a = 20 / 100 b = 500 * a c = 500 + b
a ) 16 % , b ) 17 % , c ) 18 % , d ) 54 % , e ) 13 %
d
multiply(divide(subtract(68400, add(36400, 8000)), add(36400, 8000)), const_100)
ramu bought an old car for rs . 36400 . he spent rs . 8000 on repairs and sold it for rs . 68400 . what is his profit percent ?
"total cp = rs . 36400 + rs . 8000 = rs . 44400 and sp = rs . 68400 profit ( % ) = ( 68400 - 44400 ) / 44400 * 100 = 54 % answer : d"
a = 36400 + 8000 b = 68400 - a c = 36400 + 8000 d = b / c e = d * 100
a ) a ) 657 , b ) b ) 200 , c ) c ) 438 , d ) d ) 290 , e ) e ) 427
b
divide(multiply(480, add(const_100, 19)), add(subtract(const_100, 15), add(const_100, 19)))
i bought two books ; for rs . 480 . i sold one at a loss of 15 % and other at a gain of 19 % and then i found each book was sold at the same price . find the cost of the book sold at a gain ?
"x = price of a book sold at loss x * ( 85 / 100 ) = ( 480 - x ) 119 / 100 x = 280 price of book sold at gain = 480 - 280 = 200 answer : b"
a = 100 + 19 b = 480 * a c = 100 - 15 d = 100 + 19 e = c + d f = b / e
a ) 35 m , b ) 45 m , c ) 30 m , d ) 15 m , e ) 25 m
d
sqrt(subtract(power(25, const_2), power(20, const_2)))
a ladder 25 m long reaches a window of building 20 m above the ground . then the distance of the foot of the ladder from the building is ?
let it forms a right angled triangle and follow c ^ 2 = a ^ 2 + b ^ 2 then 25 ^ 2 = 20 ^ 2 + b ^ 2 b ^ 2 = 625 - 400 b = 15 answer : d
a = 25 ** 2 b = 20 ** 2 c = a - b d = math.sqrt(c)
a ) 125 , b ) 56 , c ) 78 , d ) 112 , e ) 175
a
divide(10, subtract(134.08, add(const_100, add(multiply(const_4, const_10), const_2))))
when positive integer n is divided by positive integer j , the remainder is 10 . if n / j = 134.08 , what is value of j ?
"when a number is divided by another number , we can represent it as : dividend = quotient * divisor + remainder so , dividend / divisor = quotient + remainder / divisor given that n / j = 134.08 here 134 is the quotient . given that remainder = 10 so , 134.08 = 134 + 10 / j so , j = 125 ans a"
a = 4 * 10 b = a + 2 c = 100 + b d = 134 - 8 e = 10 / d
a ) 600 , b ) 200 , c ) 400 , d ) 800 , e ) 1000
a
divide(18, subtract(204.03, add(const_100, add(multiply(const_4, const_10), const_2))))
when positive integer n is divided by positive integer j , the remainder is 18 . if n / j = 204.03 , what is value of j ?
"when a number is divided by another number , we can represent it as : dividend = quotient * divisor + remainder so , dividend / divisor = quotient + remainder / divisor given that n / j = 204.03 here 204 is the quotient . given that remainder = 18 so , 204.03 = 204 + 18 / j so , j = 600 answer : a"
a = 4 * 10 b = a + 2 c = 100 + b d = 204 - 3 e = 18 / d
a ) 125 , b ) 126 , c ) 127 , d ) 128 , e ) 129
d
add(power(4, const_3), power(8, const_2))
if 2 + 7 = 57 ; 3 + 6 = 63 ; 5 + 9 = 206 then 4 + 8 = ?
2 ^ 3 + 7 ^ 2 = 57 3 ^ 3 + 6 ^ 2 = 63 5 ^ 3 + 9 ^ 2 = 206 and 4 ^ 3 + 8 ^ 2 = 128 answer : d
a = 4 ** 3 b = 8 ** 2 c = a + b
a ) 30 % , b ) 35 % , c ) 40 % , d ) 45 % , e ) 50 %
e
multiply(divide(subtract(divide(multiply(6, subtract(const_100, 30)), const_100), multiply(0.4, 6)), subtract(6, multiply(0.4, 6))), const_100)
a tank is filled to one quarter of its capacity with a mixture consisting of water and sodium chloride . the proportion of sodium chloride in the tank is 30 % by volume and the capacity of the tank is 24 gallons . if the water evaporates from the tank at the rate of 0.4 gallons per hour , and the amount of sodium chloride stays the same , what will be the concentration of water in the mixture in 6 hours ?
"the number of gallons in the tank is ( 1 / 4 ) 24 = 6 gallons the amount of sodium chloride is 0.3 ( 6 ) = 1.8 gallons at the start , the amount of water is 0.7 ( 6 ) = 4.2 gallons after 6 hours , the amount of water is 4.2 - 0.4 ( 6 ) = 1.8 gallons the concentration of water is 1.8 / ( 1.8 + 1.8 ) = 1.8 / 3.6 = 50 % the answer is e ."
a = 100 - 30 b = 6 * a c = b / 100 d = 0 * 4 e = c - d f = 0 * 4 g = 6 - f h = e / g i = h * 100
a ) 8 , b ) 9 , c ) 6 , d ) 14 , e ) 10
e
multiply(divide(const_1, multiply(add(const_100, 30), divide(const_1, subtract(const_100, 35)))), 20)
by selling 20 pencils for a rupee a man loses 35 % . how many for a rupee should he sell in order to gain 30 % ?
65 % - - - 20 130 % - - - ? 65 / 130 * 20 = 10 answer : e
a = 100 + 30 b = 100 - 35 c = 1 / b d = a * c e = 1 / d f = e * 20
a ) a ) 3254416 , b ) b ) 3624216 , c ) c ) 3624316 , d ) d ) 3625116 , e ) e ) 3625216
a
multiply(divide(1804, 1804), const_100)
1804 x 1804 = ?
"1804 x 1804 = ( 1804 ) 2 = ( 1800 + 4 ) 2 = ( 1800 ) 2 + ( 4 ) 2 + ( 2 x 1800 x 4 ) = 3240000 + 16 + 14400 . = 3254416 . a )"
a = 1804 / 1804 b = a * 100
a ) rs . 1550 , b ) rs . 1430 , c ) rs . 1450 , d ) rs . 1400 , e ) rs . 1500
e
multiply(divide(90, 30), 500)
in order to obtain an income of rs . 500 from 30 % stock at rs . 90 , one must make an investment of
"explanation : market value = rs . 90 required income = rs . 500 . here face value is not given . take face value as rs . 100 if it is not given in the question to obtain rs . 30 ( ie , 30 % of the face value 100 ) , investment = rs . 90 to obtain rs . 15000 , investment = 90 / 30 Γ£ β€” 500 = rs . 1500 answer : option e"
a = 90 / 30 b = a * 500
a ) 6 , b ) 12 , c ) 24 , d ) 36 , e ) 48
b
multiply(sqrt(divide(72, 2)), 2)
if n is a positive integer and n ^ 2 is divisible by 72 , then the largest positive integer q that must divide n is ?
"q : if n is a positive integer and n ^ 2 is divisible by 72 , then the largest positive integer q that must divide n is : a 6 , b 12 , c 24 , d 36 , e 48 n ^ 2 is divisible by 72 , but it must also be greater than 72 . if n is an integer , then n ^ 2 must be a perfect square . the factorization of 72 is ( 8 ) ( 9 ) , so if it is multiplied by 2 , it will be ( 2 ) ( 8 ) ( 9 ) = ( 16 ) ( 9 ) = 144 , a perfect square . so n ^ 2 must be at least 144 or a multiple of 144 , which means that n must be 12 or a multiple of 12 . b"
a = 72 / 2 b = math.sqrt(a) c = b * 2
a ) . 25 , b ) . 235 , c ) 4 , d ) 3 , e ) 2
c
divide(subtract(power(add(2.502, 0.064), const_2), power(subtract(2.502, 0.064), const_2)), multiply(2.502, 0.064))
the value of ( 2.502 + 0.064 ) ^ 2 - ( 2.502 - 0.064 ) ^ 2 / 2.502 * 0.064
"( 2.502 + 0.064 ) 2 - ( 2.502 - 0.064 ) 2 / 2.502 * 0.064 = ( a + b ) 2 - ( a - b ) / ab = 4 ab / ab = 4 answer : c"
a = 2 + 502 b = a ** 2 c = 2 - 502 d = c ** 2 e = b - d f = 2 * 502 g = e / f
a ) 50 , b ) 78 , c ) 267 , d ) 20 , e ) 27
d
subtract(add(200, 320), 500)
a , b and c have rs . 500 between them , a and c together have rs . 200 and b and c rs . 320 . how much does c have ?
a + b + c = 500 a + c = 200 b + c = 320 - - - - - - - - - - - - - - a + b + 2 c = 520 a + b + c = 500 - - - - - - - - - - - - - - - - c = 20 answer : d
a = 200 + 320 b = a - 500
a ) - 8 , b ) 10 , c ) 12 , d ) - 4 , e ) 16
d
subtract(6, 8)
what is 10 - 8 + 6 - 4 + . . . + ( - 18 ) ?
"the expression considers all even numbers between 10 and - 18 with alternate addition and subtraction of the numbers . the numbers to be used are : 10 , 8 , 6 , 4 , 2 , 0 , - 2 , - 4 , - 6 , - 8 , - 10 , - 12 , - 14 , - 16 , and - 18 . now , the first term is positive and the next term is subtracted . so , the required expression becomes , 10 - 8 + 6 - 4 + 2 - 0 + ( - 2 ) - ( - 4 ) + ( - 6 ) - ( - 8 ) + ( - 10 ) - ( - 12 ) + ( - 14 ) - ( - 16 ) + ( - 18 ) = 10 - 8 + 6 - 4 + 2 - 0 - 2 + 4 - 6 + 8 - 10 + 12 - 14 + 16 - 18 = 58 - 62 = - 4 hence the correct answer choice is d ."
a = 6 - 8
a ) 11 : 5 , b ) 11 : 8 , c ) 13 : 7 , d ) 15 : 7 , e ) 2 : 1
e
divide(multiply(70, 10), multiply(35, 10))
car a runs at the speed of 70 km / hr & reaches its destination in 10 hr . car b runs at the speed of 35 km / h & reaches its destination in 10 h . what is the respective ratio of distances covered by car a & car b ?
"sol . distance travelled by car a = 70 Γ— 10 = 700 km distance travelled by car b = 35 Γ— 10 = 350 km ratio = 700 / 350 = 2 : 1 answer : e"
a = 70 * 10 b = 35 * 10 c = a / b
a ) 105 , b ) 120 , c ) 140 , d ) 147 , e ) 98
e
multiply(35, divide(multiply(divide(72, const_60), 35), subtract(50, 35)))
car x began traveling at an average speed of 35 miles per hour . after 72 minutes , car y began traveling at an average speed of 50 miles per hour . when both cars had traveled the same distance , both cars stopped . how many miles did car x travel from the time car y began traveling until both cars stopped ?
"car y began travelling after 72 minutes or 1.2 hours . let t be the time for which car y travelled before it stopped . both cars stop when they have travelled the same distance . so , 35 ( t + 1.2 ) = 50 t t = 2.8 distance travelled by car x from the time car y began traveling until both cars stopped is 35 x 2.8 = 98 miles answer : - e"
a = 72 / const_60 b = a * 35 c = 50 - 35 d = b / c e = 35 * d
a ) 10 , b ) 6 , c ) 77 , d ) 55 , e ) 21
b
add(inverse(subtract(divide(const_1, 4), divide(const_1, 14))), divide(const_2, add(const_2, const_3)))
a and b together can do a work in 4 days . if a alone can do it in 14 days . in how many days can b alone do it ?
1 / 4 – 1 / 12 = 1 / 6 = > 6 answer : b
a = 1 / 4 b = 1 / 14 c = a - b d = 1/(c) e = 2 + 3 f = 2 / e g = d + f
a ) 2 / 3 , b ) 29 / 60 , c ) 17 / 30 , d ) 19 / 30 , e ) 11 / 15
a
subtract(1, add(multiply(inverse(3), inverse(const_2)), inverse(6)))
sally has a gold credit card with a certain spending limit , and a platinum card with twice the spending limit of the gold card . currently , she has a balance on her gold card that is 1 / 3 of the spending limit on that card , and she has a balance on her platinum card that is 1 / 6 of the spending limit on that card . if sally transfers the entire balance on her gold card to her platinum card , what portion of her limit on the platinum card will remain unspent ?
"let s assume the platinum card spending limit = x gold card spending limit will be = x / 2 balance on gold card is = x / 2 * 1 / 3 = x / 6 platinum card unspent limit is = x - 1 / 6 x = 5 / 6 x so if gold card balance is transferred then the rest unspent will be 5 / 6 x - x / 6 = 2 / 3 x so the ans is a"
a = 1/(3) b = 1/(2) c = a * b d = 1/(6) e = c + d f = 1 - e
a ) 3 : 2 , b ) 2 : 1 , c ) 1 : 2 , d ) 4 : 5 , e ) 2 : 3
b
multiply(divide(2, 3), multiply(divide(2, 3), divide(6, 3)))
find the compound ratio of ( 2 : 3 ) , ( 6 : 11 ) and ( 11 : 2 ) is
"required ratio = 2 / 3 * 6 / 11 * 11 / 2 = 2 / 1 = 2 : 1 answer is b"
a = 2 / 3 b = 2 / 3 c = 6 / 3 d = b * c e = a * d
a ) 5 : 00 , b ) 7 : 30 , c ) 6 : 00 , d ) 6 : 30 , e ) 7 : 00
b
divide(add(5, const_2), 30)
city a and city b are 140 miles apart . train c departs city a , heading towards city b , at 4 : 00 and travels at 40 miles per hour . train d departs city b , heading towards city a , at 5 : 30 and travels at 20 miles per hour . the trains travel on parallel tracks . at what time do the two trains meet ?
train c has traveled 20 mi in the half hour before train d has started its journey . 140 - 20 = 120 40 + 20 = 60 mph 120 mi / 60 mph = 2 hrs 5 : 30 pm + 2 hrs = 7 : 30 pm answer : b . 7 : 30
a = 5 + 2 b = a / 30
a ) 2.8 , b ) 1 , c ) 3.5 , d ) 14 , e ) 1.6
a
divide(const_1, multiply(divide(const_1, multiply(7, const_2)), add(const_2, 3)))
you cook a pot of beef stew for you and your roommate . normally it lasts 7 days between the two of you . if 3 guests are now staying with you and your roommate , how many days will the beef stew now last if everyone eats it ?
first set up a ratio with all the relevant values and also the unknown variable : 2 * 7 = 5 * x now solve by multiplying 2 by 7 and then dividing that product by 5 : 2 * 7 = 14 14 / 5 = x x = 2.8 the value of x represents the amount of days a bot of beef stew will last for 5 total people . answer : a
a = 7 * 2 b = 1 / a c = 2 + 3 d = b * c e = 1 / d
a ) 1 , b ) 4 , c ) 5 , d ) 6 , e ) 7
a
divide(10, subtract(12, 2))
a person can swim in still water at 12 km / h . if the speed of water 2 km / h , how many hours will the man take to swim back against the current for 10 km ?
"m = 12 s = 2 us = 12 - 2 = 10 d = 10 t = 10 / 10 = 1 answer : a"
a = 12 - 2 b = 10 / a
a ) 135 , b ) 278 , c ) 255 , d ) 150 , e ) 287
a
multiply(divide(multiply(54, const_1000), const_3600), 9)
a train running at the speed of 54 km / hr crosses a pole in 9 seconds . what is the length of the train ?
"speed = ( 54 * 5 / 18 ) m / sec = ( 15 ) m / sec length of the train = ( speed x time ) = ( 15 * 9 ) m = 135 m . answer : a"
a = 54 * 1000 b = a / 3600 c = b * 9
a ) 31 kg , b ) 32 kg , c ) 33 kg , d ) 34 kg , e ) none of these
c
subtract(add(multiply(40, const_2), multiply(44, const_2)), multiply(45, const_3))
the average weight of a , b and c is 45 kg . if the average weight of a and b be 40 kg and that of b and c be 44 kg , then the weight of b is :
"explanation : let a , b , c represent their respective weights . then , we have : a + b + c = ( 45 x 3 ) = 135 … . ( i ) a + b = ( 40 x 2 ) = 80 … . ( ii ) b + c = ( 44 x 2 ) = 88 … . ( iii ) adding ( ii ) and ( iii ) , we get : a + 2 b + c = 168 … . ( iv ) subtracting ( i ) from ( iv ) , we get : b = 33 . b ’ s weight = 33 kg . answer : c"
a = 40 * 2 b = 44 * 2 c = a + b d = 45 * 3 e = c - d
a ) 1 : 2 , b ) 1 : 4 , c ) 1 : 8 , d ) 9 : 16 , e ) 1 : 13
d
divide(power(3, const_2), power(4, const_2))
the duplicate ratio of 3 : 4 is ?
"3 ^ 2 : 4 ^ 2 = 9 : 16 answer : d"
a = 3 ** 2 b = 4 ** 2 c = a / b
a ) 5.55 , b ) 6.25 , c ) 6.75 , d ) 7.02 , e ) 6.48
e
divide(subtract(275, multiply(15, 3.2)), 35)
in the first 15 overs of a cricket game , the run rate was only 3.2 . what should be the run rate in the remaining 35 overs to reach the target of 275 runs ?
required run rate = 275 - ( 3.2 x 15 ) / 35 = 227 / 35 = 6.48 option e
a = 15 * 3 b = 275 - a c = b / 35
a ) 13 , b ) 14 , c ) 15 , d ) 16 , e ) 17
a
divide(subtract(multiply(1800, const_2), 1000), 200)
a car salesman earns a base salary of $ 1000 per month plus a commission of $ 200 for each car he sells . if the car salesman earned $ 1800 in january , how many cars does he need to sell in february in order to double his january earnings ?
1000 + 200 x = 3600 x = 13 cars . the answer is a .
a = 1800 * 2 b = a - 1000 c = b / 200
a ) 7 , b ) 7.67 , c ) 8.75 , d ) 8.98 , e ) 8.28
a
multiply(divide(const_1, const_2), multiply(4, 3.5))
the area of a sector of a circle of radius 4 cm formed by an arc of length 3.5 cm is ?
"( 4 * 3.5 ) / 2 = 7 answer : a"
a = 1 / 2 b = 4 * 3 c = a * b
a ) 1 , b ) 3 , c ) 4 , d ) 6 , e ) 9
d
inverse(subtract(inverse(4), inverse(12)))
working alone , a can complete a certain kind of job in 12 hours . a and d , working together at their respective rates , can complete one of these jobs in 4 hours . in how many hours can d , working alone , complete one of these jobs ?
"let total time taken by s to complete the job = s total time taken by a to complete the job = 12 work done by a in an hour 1 / a = 1 / 12 working together a and d can complete the job in 4 hours 1 / a + 1 / d = 1 / 4 = > 1 / d = 1 / 4 - 1 / 12 = 1 / 4 - 1 / 12 = 1 / 6 = > d = 6 hours answer d"
a = 1/(4) b = 1/(12) c = a - b d = 1/(c)
a ) 1 / 22 , b ) 1 / 19 , c ) 3 / 19 , d ) 2 / 19 , e ) 3 / 22
e
divide(subtract(22, 19), 22)
a ’ s speed is 22 / 19 times that of b . if a and b run a race , what part of the length of the race should a give b as a head start , so that the race ends in a dead heat ?
"we have the ratio of a ’ s speed and b ’ s speed . this means , we know how much distance a covers compared with b in the same time . this is what the beginning of the race will look like : ( start ) a _________ b ______________________________ if a covers 22 meters , b covers 19 meters in that time . so if the race is 22 meters long , when a reaches the finish line , b would be 3 meters behind him . if we want the race to end in a dead heat , we want b to be at the finish line too at the same time . this means b should get a head start of 3 meters so that he doesn ’ t need to cover that . in that case , the time required by a ( to cover 22 meters ) would be the same as the time required by b ( to cover 19 meters ) to reach the finish line . so b should get a head start of 3 / 22 th of the race . answer ( e )"
a = 22 - 19 b = a / 22
a ) 94 kg , b ) 95 kg , c ) 96 kg , d ) 97 kg , e ) 98 kg
a
add(multiply(8, 3), 70)
the average weight of 8 girls increases by 3 kg when a new girl comes in place of one of them weighing 70 kg . what might be the weight of the new girl ?
total weight increased = 8 x 3 kg = 24 kg . weight of new person = 70 + 24 kg = 94 kg answer : a
a = 8 * 3 b = a + 70
a ) 92 , b ) 95 , c ) 98 , d ) 101 , e ) 104
b
add(divide(multiply(35, 9), 5), 32)
if t = 5 / 9 * ( k - 32 ) , and if t = 35 , then what is the value of k ?
"k - 32 = 9 t / 5 k = 9 t / 5 + 32 k = 9 ( 35 ) / 5 + 32 = 95 the answer is b ."
a = 35 * 9 b = a / 5 c = b + 32
a ) 1 , b ) 2 , c ) 3 , d ) 4 , e ) 5
a
subtract(divide(multiply(const_3, multiply(6, 3)), add(6, const_4.0)), 3)
from an island , it is possible to reach the mainland by either ferry p or ferry q . ferry p travels for 3 hours at 6 kilometers per hour , while ferry q takes a route that is two times longer . if ferry p is slower than ferry q by 3 kilometers per hour , how many hours longer is the journey of ferry q compared with the journey of ferry p ?
"the distance traveled by ferry p is 18 km . then the distance traveled by ferry q is 36 km . ferry q travels at a speed of 9 kph . the time of the journey for ferry q is 36 / 9 = 4 hours , which is 1 hour more than ferry p . the answer is a ."
a = 6 * 3 b = 3 * a c = 6 + 4 d = b / c e = d - 3
a ) 16.089 , b ) 16.35 , c ) 16.89 , d ) 16.944 , e ) none of them
d
multiply(divide(48.95, 32.006), const_100)
48.95 - 32.006 = ?
"48.950 - 32.006 - - - - - - - - - 16.944 answer is d"
a = 48 / 95 b = a * 100
a ) 6.06 % , b ) 8.16 % , c ) 6.08 % , d ) 6.09 % , e ) 6.19 %
b
add(add(divide(8, const_2), divide(8, const_2)), divide(multiply(divide(8, const_2), divide(8, const_2)), const_100))
the effective annual rate of interest corresponding to a nominal rate of 8 % per annum payable half - yearly is ?
amount of rs . 100 for 1 year when compounded half - yearly = [ 100 * ( 1 + 4 / 100 ) 2 ] = rs . 108.16 effective rate = ( 108.16 - 100 ) = 8.16 % answer : b
a = 8 / 2 b = 8 / 2 c = a + b d = 8 / 2 e = 8 / 2 f = d * e g = f / 100 h = c + g
a ) 10 , b ) 12 , c ) 14 , d ) 16 , e ) 18
a
divide(50, add(const_4, divide(const_2, const_2)))
a certain number of horses and an equal number of men are going somewhere . half of the owners are on their horses ' back while the remaining ones are walking along leading their horses . if the number of legs walking on the ground is 50 , how many horses are there ?
"legs 10 * 4 = 40 now half on their horses so remaining on the walk so 5 men 5 men has 10 legs so , 10 + 40 = 50 legs walking answer : a"
a = 2 / 2 b = 4 + a c = 50 / b
a ) 15060000 , b ) 0.001506 , c ) 1.506 e - 06 , d ) 0.1506 , e ) none of these
c
multiply(divide(15.06, 0.0000001), const_100)
15.06 * 0.0000001 = ?
"explanation : clearly after decimal 9 digits should be there . option c"
a = 15 / 6 b = a * 100
a ) 36 kmh , b ) 40 kmh , c ) 42 kmh , d ) 44 kmh , e ) 50 kmh
c
divide(const_3, add(add(divide(const_1, 80), divide(const_1, 24)), divide(const_1, 60)))
if a car went the first third of the distance at 80 kmh , the second third at 24 kmh , and the last third at 60 kmh , what was the average speed of the car for the entire trip ?
"assume d / 3 = 240 ( this number is convenient because it is divisible by 80 , 24 and 60 ) so : 240 = 80 * t 1 = 3 hrs 240 = 24 * t 2 = 10 hrs 240 = 60 * t 3 = 4 hrs t = t 1 + t 2 + t 3 = 17 hrs d = rt ( 240 * 3 ) = r * 17 r = 42.35 answer : c"
a = 1 / 80 b = 1 / 24 c = a + b d = 1 / 60 e = c + d f = 3 / e
a ) 1992 , b ) 1993 , c ) 1994 , d ) 1995 , e ) 1996
e
add(1990, const_4)
in 1990 the budgets for projects q and v were $ 540000 and $ 780000 , respectively . in each of the next 10 years , the budget for q was increased by $ 30000 and the budget for v was decreased by $ 10000 . in which year was the budget for q equal to the budget for v ?
let the no of years it takes is x . 540 + 30 x = 780 - 10 x - - > 40 x = 240 and x = 6 . thus , it happens in 1996 . e .
a = 1990 + 4
a ) s . 40 , b ) s . 12 , c ) s . 96 , d ) s . 50 , e ) s . 13
a
add(divide(15000, 500), 10)
a shopkeeper sells 500 metres of cloth for rs . 15000 at a loss of rs . 10 per metre . find his cost price for one metre of cloth ?
"sp per metre = 15000 / 500 = rs . 30 loss per metre = rs . 10 cp per metre = 30 + 10 = rs . 40 answer : a"
a = 15000 / 500 b = a + 10
a ) 19 , b ) 20 , c ) 21 , d ) 22 , e ) 23
a
subtract(subtract(multiply(3, 10), 10), 1)
in a colony the municipality decided to number the houses and the shops . they want to number the shops with the numbers containing 3 between 1 - 100 . the houses are 10 - 20 floored and at the back of the houses 8 feet is left for fire exit . how many shops can be numbered ?
3 numbers containing 3 - - - > 3 , 1323 , . . . = 10 numbers - - - - > 30 , 3132 . . . = 10 numbers total = 20 - 1 = 19 answer : a
a = 3 * 10 b = a - 10 c = b - 1
a ) 24 , b ) 12 , c ) 8 , d ) 2 , e ) 4
e
subtract(196, multiply(64, 3))
a sporting goods store sold 64 frisbees in one week , some for $ 3 and the rest for $ 4 each . if receipts from frisbee sales for the week totaled $ 196 what is the fewest number of $ 4 frisbees that could have been sold ?
"in this question however , because we are told that exactly 64 frisbees have been sold and revenue was exactly $ 204 , there is only one possible solution for the number of $ 3 and $ 4 frisbees sold . to solve , we have 2 equations and 2 unknowns let x = number of $ 3 frisbees sold let y = number of $ 4 frisbees sold x + y = 64 3 x + 4 y = 196 x = 64 - y 3 ( 64 - y ) + 4 y = 196 192 - 3 y + 4 y = 196 y = 4 answer : e"
a = 64 * 3 b = 196 - a
a ) 2 , b ) 3 / 2 , c ) 4 , d ) 5 , e ) can not be determined from the information given .
b
multiply(subtract(multiply(divide(const_1, const_2), 3), const_1), 3)
the average ( arithmetic man ) of 3 integers a , b , and c is exactly half the median . if a < b < c and a = 0 , what is the value of c / b ?
the average of three integers a , b , and c is exactly twice the median - - > ( a + b + c ) / 3 = 1 b / 2 - - > since a = 0 , then ( 0 + b + c ) / 3 = 1 b / 2 - - > c = 3 b / 2 - - > c / b = 3 / 2 . answer : b .
a = 1 / 2 b = a * 3 c = b - 1 d = c * 3
a ) 5 / 21 , b ) 3 / 7 , c ) 4 / 7 , d ) 5 / 7 , e ) 49 / 45
e
divide(subtract(subtract(subtract(10, 1), 1), 1), subtract(10, 1))
in a room with 10 people , 7 people have exactly 1 friend in the room and 6 people have exactly 2 friends in the room ( assuming that friendship is a mutual relationship , i . e . if jane is paul ' s friend , paul is jane ' s friend ) . if two individuals are selected from the room at random , what is the probability that those two individuals are not friends ?
"( 7 / 10 ) ( 8 / 9 ) + ( 6 / 10 ) ( 7 / 9 ) if you choose one of the 7 with one other friend , then you have a 8 / 9 chance of not picking their friend 2 nd . if you choose one of the 6 with 2 friends , you have a 7 / 9 chance of not picking one of their friends second . add them up . 56 / 90 + 42 / 90 98 / 90 = 49 / 45 e . 49 / 45"
a = 10 - 1 b = a - 1 c = b - 1 d = 10 - 1 e = c / d
a ) 124 , b ) 256 , c ) 368 , d ) 424 , e ) 512
c
divide(multiply(46, 32), 4)
dan ’ s car gets 32 miles per gallon . if gas costs $ 4 / gallon , then how many miles can dan ’ s car go on $ 46 of gas ?
"46 / 4 = 11.5 gallons 11.5 * 32 = 368 miles the answer is c ."
a = 46 * 32 b = a / 4
a ) 6 / 16 , b ) 14 / 31 , c ) 7 / 15 , d ) 17 / 35 , e ) 1 / 2
a
divide(add(3, 2), const_10)
company s produces two kinds of stereos : basic and deluxe . of the stereos produced by company s last month , 2 / 3 were basic and the rest were deluxe . if it takes 6 / 5 as many hours to produce a deluxe stereo as it does to produce a basic stereo , then the number of hours it took to produce the deluxe stereos last month was what fraction of the total number of hours it took to produce all the stereos ?
"# of basic stereos was 2 / 3 of total and # of deluxe stereos was 1 / 3 of total , let ' s assume total = 15 , then basic = 10 and deluxe = 5 . now , if time needed to produce one deluxe stereo is 1 unit than time needed to produce one basic stereo would be 6 / 5 units . total time for basic would be 10 * 1 = 10 and total time for deluxe would be 5 * 6 / 5 = 6 - - > total time for both of them would be 10 + 6 = 16 - - > deluxe / total = 6 / 16 . answer : a ."
a = 3 + 2 b = a / 10
a ) 48 , b ) 54 , c ) 60 , d ) 66 , e ) 72
e
add(multiply(negate(12), power(subtract(add(3, 2), 3), 2)), 120)
an object thrown directly upward is at a height of h feet after t seconds , where h = - 12 ( t - 3 ) ^ 2 + 120 . at what height , in feet , is the object 2 seconds after it reaches its maximum height ?
"we see that h will be a maximum h = 120 when t - 3 = 0 , that is when t = 3 . at t = 5 , h = - 12 * ( 5 - 3 ) ^ 2 + 120 = - 12 ( 4 ) + 120 = 72 the answer is e ."
a = negate * ( b = 3 + 2 c = b - 3 d = c ** 2 e = a + d
a ) 46 , b ) 10 , c ) 56 , d ) - 46 , e ) 28
b
subtract(add(23, 5), subtract(23, 5))
if | x - 5 | = 23 what is the sum of all the values of x .
"there will be two cases x - 5 = 23 and x - 5 = - 23 solve for x = > x = 23 + 5 = > x = 28 or x = - 23 + 5 = > x = - 18 the sum of both values will be 28 + - 18 = 10 answer is b"
a = 23 + 5 b = 23 - 5 c = a - b
a ) 22 , b ) 25 , c ) 15 , d ) 35 , e ) 30
b
multiply(8, divide(lcm(add(6, 4), add(8, 6)), add(8, 6)))
the number of stamps that kaye and alberto had were in the ration of 6 : 4 respectively . after kaye gave alberto 5 of her stamps , the ration of the number of kaye had to the number of alberto had was 8 : 6 . as a result of the gift , kaye had how many more stamps than alberto ?
"k 1 = 6 x a 1 = 4 x k 2 = 6 x - 5 a 2 = 4 x + 5 k 2 / a 2 = 8 / 6 ( 6 x - 5 ) / ( 4 x + 5 ) = 8 / 6 6 * ( 6 x - 5 ) = 8 * ( 4 x + 5 ) 36 x - 30 = 32 x + 40 4 x = 70 x = 17.5 k 2 = 6 * 17.5 - 5 = 100 a 2 = 4 * 17.5 + 5 = 75 k 2 - a 2 = 100 - 75 = 25 answer : b"
a = 6 + 4 b = 8 + 6 c = math.lcm(a, b) d = 8 + 6 e = c / d f = 8 * e
a ) 76 , b ) 1936 , c ) 1304 , d ) 1296 , e ) none
b
power(multiply(4, 11), const_2)
find √ ? / 11 = 4 ?
answer let √ n / 11 = 4 then √ n = 11 x 4 = 44 ∴ n = 44 x 44 = 1936 . correct option : b
a = 4 * 11 b = a ** 2
a ) $ 015 , b ) $ 0.20 , c ) $ 0.30 , d ) $ 0.40 , e ) $ 0.45
d
divide(multiply(add(const_1, const_1), 0.60), add(const_1, const_2))
on a certain day , orangeade was made by mixing a certain amount of orange juice with an equal amount of water . on the next day , orangeade was made by mixing the same amount of orange juice with twice the amount of water . on both days , all the orangeade that was made was sold . if the revenue from selling the orangeade was the same for both days and if the orangeade was sold at $ 0.60 per glass on the first day , what was the price w per glass on the second day ?
"on the first day 1 unit of orange juice and 1 unit of water was used to make 2 units of orangeade ; on the second day 1 unit of orange juice and 2 units of water was used to make 3 units of orangeade ; so , the ratio of the amount of orangeade made on the first day to the amount of orangeade made on the second day is 2 to 3 . naturally the ratio of the # of glasses of orangeade made on the first day to the # of glasses of orangeade made on the second day is 2 to 3 . we are told thatthe revenue from selling the orangeade was the same for both daysso the revenue from 2 glasses on the first day equals to the revenue from 3 glasses on the second day . say the price of the glass of the orangeade on the second day was $ x then 2 * 0.6 = 3 * x - - > x = $ 0.4 . answer : d ."
a = 1 + 1 b = a * 0 c = 1 + 2 d = b / c
a ) 1 / 8 , b ) 1 / 4 , c ) 2 / 4 , d ) 1 / 4 , e ) 3 / 4
e
multiply(subtract(const_1, multiply(add(divide(const_1, 12), divide(const_1, 13)), 3)), 13)
two pipes p and q can fill a cistern in 12 and 13 minutes respectively . both are opened together , but at the end of 3 minutes the first is turned off . how much longer will the cistern take to fill ?
"3 / 12 + x / 13 = 1 x = 7 3 / 4 answer : e"
a = 1 / 12 b = 1 / 13 c = a + b d = c * 3 e = 1 - d f = e * 13
a ) 198 mph , b ) 110 mph , c ) 88 mph , d ) 100 mph , e ) 109 mph
e
divide(add(130, 88), const_2)
i flew my tiny seaplane to visit my mother . on the flight up , i flew at 130 mph . on the way home , i flew 88 mph . what was my average speed for the trip ?
"( 130 mph + 88 mph ) / 2 = 109 mph correct answer is : e"
a = 130 + 88 b = a / 2
a ) 27.5 % , b ) 30 % , c ) 35 % , d ) 37.5 % , e ) 40 %
c
subtract(multiply(divide(subtract(const_100, 10), const_100), multiply(add(const_100, 20), divide(add(const_100, 25), const_100))), const_100)
a particular store purchased a stock of turtleneck sweaters and marked up its cost by 20 % . during the new year season , it further marked up its prices by 25 % of the original retail price . in february , the store then offered a discount of 10 % . what was its profit on the items sold in february ?
"assume the total price = 100 x price after 20 % markup = 120 x price after 25 % further markup = 1.25 * 120 x = 150 x price after the discount = 0.9 * 150 x = 135 x hence total profit = 35 % option c"
a = 100 - 10 b = a / 100 c = 100 + 20 d = 100 + 25 e = d / 100 f = c * e g = b * f h = g - 100
a ) 10 % decrease , b ) 20 % decrease , c ) 36 % decrease , d ) 40 % decrease , e ) 50 % decrease
c
subtract(const_100, multiply(power(divide(20, const_100), const_2), const_100))
if the radius of a circle is decreased 20 % , what happens to the area ?
"area of square = pi * radius ^ 2 new radius = 4 / 5 * old radius so new area = ( 4 / 5 ) ^ 2 old area = > 16 / 25 of old area = > 64 % old area answer : c"
a = 20 / 100 b = a ** 2 c = b * 100 d = 100 - c
a ) 22 , b ) 24 , c ) 26 , d ) 20 , e ) 28
b
divide(subtract(multiply(44, const_4), 128), const_2)
a man has some hens and cows . if the number of heads be 44 and the number of feet equals 128 , then the number of hens will be
"explanation : let number of hens = h and number of cows = c number of heads = 44 = > h + c = 44 - - - ( equation 1 ) number of feet = 128 = > 2 h + 4 c = 128 = > h + 2 c = 64 - - - ( equation 2 ) ( equation 2 ) - ( equation 1 ) gives 2 c - c = 64 - 44 = > c = 20 substituting the value of c in equation 1 , we get h + 22 = 44 = > h = 44 - 20 = 24 i . e . , number of hens = 24 answer : b"
a = 44 * 4 b = a - 128 c = b / 2
a ) 40 sec , b ) 47 sec , c ) 33 sec , d ) 35 sec , e ) none of these
c
subtract(multiply(divide(7, 35), 200), 7)
in a 200 meters race a beats b by 35 m or 7 seconds . a ’ s time over the course is :
"solution b runs 35 m in 7 sec . ∴ b covers 200 m in ( 7 / 35 x 200 ) = 40 sec b ’ s time over the course = 40 sec . ∴ a ’ s time over the course = ( 40 - 7 ) sec = 33 sec answer c"
a = 7 / 35 b = a * 200 c = b - 7
a ) 13 1 / 3 , b ) 878 , c ) 30 , d ) 88 , e ) 11
a
inverse(subtract(5, divide(5, 8)))
a and b can do a piece of work in 8 days . with the help of c they finish the work in 5 days . c alone can do that piece of work in ?
"c = 1 / 5 Γ’ € β€œ 1 / 8 = 3 / 40 = > 40 / 3 = 13 1 / 3 days answer : a"
a = 5 / 8 b = 5 - a c = 1/(b)
a ) 2436 , b ) 2801 , c ) - 2801 , d ) - 2071 , e ) none of these
d
add(multiply(negate(84), 29), 365)
- 84 * 29 + 365 = ?
"= > - 84 * ( 30 - 1 ) + 365 ; = > - ( 84 * 30 ) + 84 + 365 ; = > - 2520 + 449 = - 2071 . correct option : d"
a = negate * ( b = a + 29
a ) 3 : 4 , b ) 3 : 5 , c ) 4 : 3 , d ) 1 : 3 , e ) 1 : 5
b
divide(300, 500)
if shares of two persons in profits are rs . 300 and rs . 500 then ratio of their capitals is
"total profit = 1000 ratio = 300 / 500 = 3 : 5 answer : b"
a = 300 / 500
a ) 1 / 25 , b ) 12 / 49 , c ) 1 / 4 , d ) 24 / 49 , e ) 1 / 2
d
multiply(multiply(divide(subtract(divide(50, 2), const_1), subtract(50, const_1)), divide(divide(50, 2), 50)), 2)
there are 2 available positions and 50 candidates , one half of whom are democrats and another half are republicans . if it was decided that the positions would be filled at random , then what is the probability r that the both positions will be taken by members of just one party ?
"r probability of one party having both spots : ( 1 / 2 ) * ( 24 / 49 ) = 12 / 49 ( 1 / 2 ) or ( 25 / 50 ) because it does not matter which party or which person gets the first spot . ( 24 / 49 ) because after one person from a particular party is chosen , there are 24 members of the same party left out of 49 total candidates . since this result can happen for both parties , ( 12 / 49 ) + ( 12 / 49 ) = ( 24 / 49 ) answer : d"
a = 50 / 2 b = a - 1 c = 50 - 1 d = b / c e = 50 / 2 f = e / 50 g = d * f h = g * 2
a ) 12 sec , b ) 9 sec , c ) 16 sec , d ) 20 sec , e ) none
b
multiply(divide(270, multiply(108, const_1000)), const_3600)
a train 270 m long , running with a speed of 108 km / hr will pass a tree in
"sol . speed = ( 108 x 5 / 18 ) m / sec . = 30 m / sec . time taken = ( 270 x 1 / 30 ) sec = 9 sec answer b"
a = 108 * 1000 b = 270 / a c = b * 3600
a ) 2 : 5 , b ) 3 : 7 , c ) 3 : 5 , d ) 4 : 7 , e ) 7 : 17
e
divide(add(multiply(12, divide(1, add(1, 2))), multiply(12, divide(1, add(1, 3)))), subtract(add(12, 12), add(multiply(12, divide(1, add(1, 2))), multiply(12, divide(1, add(1, 3))))))
a container holding 12 ounces of a solution that is 1 part alcohol to 2 parts water is added to a container holding 12 ounces of a solution that is 1 part alcohol to 3 parts water . what is the ratio of alcohol to water in the resulting solution ?
"container 1 has 12 ounces in the ratio 1 : 2 or , x + 2 x = 12 gives x ( alcohol ) = 4 and remaining water = 8 container 2 has 12 ounces in the ratio 1 : 3 or , x + 3 x = 12 gives x ( alcohol ) = 3 and remaining water = 9 mixing both we have alcohol = 4 + 3 and water = 8 + 9 ratio thus alcohol / water = 7 / 17 answer e"
a = 1 + 2 b = 1 / a c = 12 * b d = 1 + 3 e = 1 / d f = 12 * e g = c + f h = 12 + 12 i = 1 + 2 j = 1 / i k = 12 * j l = 1 + 3 m = 1 / l n = 12 * m o = k + n p = h - o q = g / p
a ) 6 : 8 , b ) 9 : 8 , c ) 7 : 9 , d ) 9 : 5 , e ) 1 : 1
e
divide(add(multiply(3000, 6), multiply(multiply(3000, const_2), 6)), multiply(4500, add(6, 6)))
a and b invests rs . 3000 and rs . 4500 respectively in a business . if a doubles his capital after 6 months . in what ratio should a and b divide that year ' s profit ?
"( 3 * 6 + 6 * 6 ) : ( 4.5 * 12 ) 54 : 54 = > 1 : 1 answer : e"
a = 3000 * 6 b = 3000 * 2 c = b * 6 d = a + c e = 6 + 6 f = 4500 * e g = d / f
a ) 0.5 l , b ) l , c ) 1.5 l , d ) 2 l , e ) 3
b
subtract(2, const_1)
a train of length l is traveling at a constant velocity and passes a pole in t seconds . if the same train travelling at the same velocity passes a platform in 2 t seconds , then what is the length of the platform ?
"the train passes a pole in t seconds , so velocity v = l / t ( l + p ) / v = 2 t ( l + p ) / ( l / t ) = 2 t p = l the answer is b ."
a = 2 - 1
a ) 70 , b ) 80 , c ) 90 , d ) 100 , e ) 110
c
subtract(multiply(86, 5), multiply(4, 85))
on a test average ( arithmetic mean ) test score for 4 students is 85 . what must be 5 th student ' s score that average score for 5 students to be 86 ?
( 4 * 85 + x ) / 5 = 86 x = ( 5 * 86 ) - ( 4 * 85 ) x = 430 - 340 total score required 430 - 340 = 90 correct answer is c
a = 86 * 5 b = 4 * 85 c = a - b
a ) 210 m , b ) 300 m , c ) 230 m , d ) 240 m , e ) 250 m
b
subtract(multiply(9, multiply(add(120, 80), const_0_2778)), 200)
a 200 meter long train running at the speed of 120 kmph crosses another train running in the opposite direction at the speed of 80 kmph in 9 seconds . what is the lenght of other train .
"relative speeds = ( 120 + 80 ) km / hr = 200 km / hr = ( 200 * 5 / 18 ) m / s = ( 500 / 9 ) m / s let length of train be xm x + 200 / 9 = 500 / 9 x = 300 ans is 300 m answer : b"
a = 120 + 80 b = a * const_0_2778 c = 9 * b d = c - 200
a ) 8 , b ) 9 , c ) 9.5 , d ) 10 , e ) 11
d
divide(460, add(multiply(10, const_3), multiply(8, const_2)))
sharmila works 10 hours per day on monday , wednesday and friday , and 8 hours per day on tuesday and thursday . she does not work on saturday and sunday . she earns $ 460 per week . how much does she earn in dollars per hour ?
so , she works 30 hours in 3 days so , she works 16 hours in 2 days so in a week she works 46 hours ( 30 + 16 ) and earns $ 460 so , hourly wage is 460 / 46 = > 10 hence answer will be ( d ) 10
a = 10 * 3 b = 8 * 2 c = a + b d = 460 / c
a ) 0.86 , b ) 0.68 , c ) 0.96 , d ) 0.69 , e ) 0.76
e
divide(subtract(power(0.86, 3), power(0.1, 3)), add(add(power(0.86, 2), 0.086), power(0.1, 2)))
( 0.86 ) ( power 3 ) - ( 0.1 ) ( power 3 ) / ( 0.86 ) ( power 2 ) + 0.086 + ( 0.1 ) ( power 2 ) is :
given expression = ( 0.86 ) ( power 3 ) - ( 0.1 ) ( power 3 ) / ( 0.86 ) ( power 2 ) + ( 0.86 x 0.1 ) + ( 0.1 ) ( power 2 ) = a ( power 3 ) - b ( power 3 ) / a ( power 2 ) + ab + b ( power 2 ) = ( a - b ) = ( 0.86 - 0.1 ) = 0.76 answer is e
a = 0 ** 86 b = 0 ** 1 c = a - b d = 0 ** 86 e = d + 0 f = 0 ** 1 g = e + f h = c / g
a ) 5 : 2 , b ) 5 : 9 , c ) 5 : 3 , d ) 5 : 1 , e ) 5 : 8
a
divide(divide(divide(20, 2), const_2), 2)
find the ratio in which wheat of inferior quality ( rs . 14 / kg ) be mixed with wheat of superior quality ( rs . 28 / kg ) so that the shopkeeper gains rs . 2 by selling the resulting mixture at rs . 20 / kg .
explanation : let the resulting mixture be 1 kg , and x kg be the amount of wheat of inferior quality . therefore , ( 1 - x ) kg is the amount of wheat of superior quality . as the shopkeeper gains rs . 2 , the cost of the mixture is rs . 18 14 * x + 28 * ( 1 - x ) = 18 14 x - 28 x + 28 = 18 14 x = 10 x = 5 / 7 ( 1 – x ) = 2 / 7 x : ( 1 - x ) = 5 / 7 : 2 / 7 = 5 : 2 answer : a
a = 20 / 2 b = a / 2 c = b / 2
a ) 5 , b ) 4 , c ) 2 , d ) 6 , e ) 7
c
divide(subtract(subtract(subtract(75, multiply(75, divide(3, 7))), multiply(subtract(75, multiply(75, divide(3, 7))), divide(3, 7))), 3), subtract(subtract(subtract(75, multiply(75, divide(3, 7))), multiply(subtract(75, multiply(75, divide(3, 7))), divide(3, 7))), 3))
george baked a total of 75 pizzas for 7 straight days , beginning on saturday . he baked 3 / 5 of the pizzas the first day , and 3 / 5 of the remaining pizzas the second day . if each successive day he baked fewer pizzas than the previous day , what is the maximum number of pizzas he could have baked on wednesday ?
"3 / 5 of the 75 pizzas cooked on saturday = 45 pizzas 3 / 5 of the remaining pizzas on sunday = 18 pizzas we ' re left with ( 75 - 45 - 18 ) = 12 pizzas for the remaining 5 days . the prompt tells us that each day has fewer pizzas than the day before it , so we ca n ' t have duplicate numbers . m t w th f 5 4 2 1 0 = 12 w = 2 c"
a = 3 / 7 b = 75 * a c = 75 - b d = 3 / 7 e = 75 * d f = 75 - e g = 3 / 7 h = f * g i = c - h j = i - 3 k = 3 / 7 l = 75 * k m = 75 - l n = 3 / 7 o = 75 * n p = 75 - o q = 3 / 7 r = p * q s = m - r t = s - 3 u = j / t
a ) 100 , b ) 250 , c ) 300 , d ) 400 , e ) 500
b
subtract(800, add(add(150, subtract(500, 150)), subtract(400, subtract(500, 150))))
- - - - - - - - - - - - - - - - yes - - - - - - - - - no - - - - unsure subject m - - - - 500 - - - - - - - - 200 - - - - - 100 subject r - - - - 400 - - - - - - - - 100 - - - - - 300 a total of 800 students were asked whether they found two subjects , m and r , interesting . each answer was either yes or no or unsure , and the numbers of students who gave these answers are listed in the table above . if 150 students answered yes only for subject m , how many of the students did not answer yes for either subject ?
"since 150 students answered yes only for subject m , then the remaining 350 students who answered yes for subject m , also answered yes for subject r . so , 350 students answered yes for both subjects . if 350 students answered yes for both subjects , then 400 - 350 = 50 students answered yes only for subject r . so , we have that : 200 students answered yes only for subject m ; 50 students answered yes only for subject r ; 300 students answered yes for both subjects ; therefore 800 - ( 200 + 50 + 300 ) = 250 students did not answer yes for either subject . answer : b ."
a = 500 - 150 b = 150 + a c = 500 - 150 d = 400 - c e = b + d f = 800 - e
a ) 12 / 55 , b ) 3 / 5 , c ) 3 / 11 , d ) 1 / 4 , e ) 7 / 16
a
divide(multiply(choose(4, const_2), choose(add(6, 2), const_1)), choose(add(add(6, 2), 4), 6))
a bag contains 6 red , 2 yellow and 4 green balls . 3 balls are drawn randomly . what is the probability that the balls drawn contain balls of different colours ?
"total number of balls = 6 + 2 + 4 = 12 n ( s ) = 12 c 3 = 12 * 11 * 10 / 3 * 2 = 220 n ( e ) = 6 c 1 * 2 c 1 * 4 c 1 = 48 probability = 48 / 220 = 12 / 55 answer is a"
a = math.comb(4, 2) b = 6 + 2 c = math.comb(b, 1) d = a * c e = 6 + 2 f = e + 4 g = math.comb(f, 6) h = d / g
a ) 30 % , b ) 34 % , c ) 35 % , d ) 36 % , e ) 37 %
a
multiply(add(multiply(divide(50, const_100), divide(20, const_100)), multiply(divide(subtract(const_100, 50), const_100), divide(40, const_100))), const_100)
in a certain company 20 % of the men and 40 % of the women attended the annual company picnic . if 50 % of all the employees are men . what % of all the employee went to the picnic ?
"total men in company 50 % means total women in company 50 % ( assume total people in company 100 % ) no of men employees attended picnic = 50 x ( 20 / 100 ) = 10 no of women employees attend picnic = 50 x ( 40 / 100 ) = 20 total percentage of employees attend the picnic = 10 + 20 = 30 % answer : a"
a = 50 / 100 b = 20 / 100 c = a * b d = 100 - 50 e = d / 100 f = 40 / 100 g = e * f h = c + g i = h * 100
a ) 118 , b ) 88 , c ) 90 , d ) 42 , e ) 22
a
subtract(divide(10000, 80), 7)
a trader sells 80 meters of cloth for rs . 10000 at the profit of rs . 7 per metre of cloth . what is the cost price of one metre of cloth ?
"sp of 1 m of cloth = 8925 / 85 = rs . 125 cp of 1 m of cloth = sp of 1 m of cloth - profit on 1 m of cloth = rs . 125 - rs . 7 = rs . 118 answer : a"
a = 10000 / 80 b = a - 7
a ) 5 , b ) 8 , c ) 10 , d ) 12 , e ) 15
b
divide(add(35, 25), 25)
the present ratio of students to teachers at a certain school is 35 to 1 . if the student enrollment were to increase by 50 students and the number of teachers were to increase by 5 , the ratio of students to teachers would then be 25 to 1 . what is the present number of teachers ?
"we are given that the ratio of students to teacher is 35 to 1 . we can rewrite this using variable multipliers . students : teachers = 35 x : x we are next given that student enrollment increases by 50 and the number of teachers increases by 5 . with this change the new ratio becomes 25 to 1 . we can put all this into an equation : students / teachers οƒ  25 / 1 = ( 35 x + 50 ) / ( x + 5 ) if we cross multiply we have : 25 ( x + 5 ) = 35 x + 50 25 x + 125 = 35 x + 50 7.5 = x x ~ 8 since x is the present number of teachers , currently there are 8 teachers . answer b ."
a = 35 + 25 b = a / 25
a ) 3 , b ) 2 , c ) 4 , d ) 1 , e ) 0
b
divide(reminder(multiply(653, const_100), 80), const_10)
if x and y are the two digits of the number 653 xy . such that this number is divisible by 80 , then x + y is equal to
since 653 xy is divisible by 5 as well as 2 . so y = 0 now 653 x 0 must be divisible by 8 so 3 x 0 must be 8 this happens when x = 2 hence , x + y = 2 + 0 = 2 answer b 2
a = 653 * 100 b = reminder / (
a ) 1361 , b ) 1362 , c ) 1363 , d ) 1364 , e ) 1785
e
divide(multiply(61, add(61, 22)), 23)
22 + 23 + 24 + . . . 61 + 62 + 63 = ?
"sum = 22 + 23 + 24 + . . . 61 + 62 + 63 sum of n consecutive positive integers starting from 1 is given as n ( n + 1 ) / 2 sum of first 63 positive integers = 63 * 64 / 2 sum of first 21 positive integers = 21 * 22 / 2 sum = 22 + 23 + 24 + . . . 61 + 62 + 63 = 63 * 64 / 2 - 21 * 22 / 2 = 1376 answer : e"
a = 61 + 22 b = 61 * a c = b / 23
a ) 24 , b ) 25 , c ) 26 , d ) 27 , e ) 28
a
divide(10, subtract(const_1, multiply(12, divide(const_1, 24))))
matt and peter can do together a piece of work in 24 days . after they have worked together for 12 days matt stops and peter completes the remaining work in 10 days . in how many days peter complete the work separately .
"together they complete the job in 24 days means they complete 12 / 24 of the job after 12 days . peter completes the remaining ( 12 / 24 ) of the job in 10 days which means that the whole job ( 1 ) can be completed in x days . x = 12 / ( 12 / 24 ) = 24 a"
a = 1 / 24 b = 12 * a c = 1 - b d = 10 / c
a ) 42 sec , b ) 44 sec , c ) 46 sec , d ) 40 sec , e ) 50 sec
d
divide(600, subtract(divide(60, const_3_6), divide(divide(6, const_2), const_3_6)))
a train which has 600 m long , is running 60 kmph . in what time will it cross a person moving at 6 kmph in same direction ?
"time taken to cross a moving person = length of train / relative speed time taken = 600 / ( ( 60 - 6 ) ( 5 / 18 ) = 600 / 54 * ( 5 / 18 ) = 600 / 15 = 40 sec answer : d"
a = 60 / const_3_6 b = 6 / 2 c = b / const_3_6 d = a - c e = 600 / d
a ) 1 / 8 , b ) 1 / 4 , c ) 1 / 3 , d ) 3 / 8 , e ) 1 / 2
b
divide(const_1, power(const_2, 3))
when tossed , a certain coin has equal probability of landing on either side . if the coin is tossed 3 times , what is the probability that it will land on the same side each time ?
"winning scenario is if we ' ll have either three tails ( ttt ) or three heads ( hhh ) : 1 / 2 * 1 / 2 * 1 / 2 + 1 / 2 * 1 / 2 * 1 / 2 = 1 / 4 . answer : b ."
a = 2 ** 3 b = 1 / a
a ) 3 pm , b ) 2 pm , c ) 1 : 00 pm , d ) 11 am , e ) 12 am
c
floor(add(11, divide(subtract(const_1, multiply(subtract(11, 9), add(divide(const_1, 12), add(divide(const_1, 8), divide(const_1, 10))))), add(divide(const_1, 10), divide(const_1, 12)))))
machine p can print one lakh books in 8 hours . machine q can print the same number of books in 10 hours while machine r can print the same in 12 hours . all the machines started printing at 9 a . m . machine p is stopped at 11 a . m . and the remaining two machines complete work . approximately at what time will the printing of one lakh books be completed ?
explanation : work done by p in 1 hour = 1 / 8 work done by q in 1 hour = 1 / 10 work done by r in 1 hour = 1 / 12 work done by p , q and r in 1 hour = 1 / 8 + 1 / 10 + 1 / 12 = 37 / 120 work done by q and r in 1 hour = 1 / 10 + 1 / 12 = 22 / 120 = 11 / 60 from 9 am to 11 am , all the machines were operating . ie , they all operated for 2 hours and work completed = 2 Γ— ( 37 / 120 ) = 37 / 60 pending work = 1 - 37 / 60 = 23 / 60 hours taken by q an r to complete the pending work = ( 23 / 60 ) / ( 11 / 60 ) = 23 / 11 which is approximately equal to 2 hence the work will be completed approximately 2 hours after 11 am ; ie around 1 pm answer : option c
a = 11 - 9 b = 1 / 12 c = 1 / 8 d = 1 / 10 e = c + d f = b + e g = a * f h = 1 - g i = 1 / 10 j = 1 / 12 k = i + j l = h / k m = 11 + l n = math.floor(m)
a ) 33 , b ) 77 , c ) 36 , d ) 18 , e ) 16
c
add(multiply(multiply(4, 6), const_100), multiply(5, 6))
three numbers are in the ratio 4 : 5 : 6 and their average is 30 . the largest number is :
"explanation : let the numbers be 4 x , 5 x and 6 x . therefore , ( 4 x + 5 x + 6 x ) / 3 = 30 15 x = 90 x = 6 largest number = 6 x = 36 . answer : c"
a = 4 * 6 b = a * 100 c = 5 * 6 d = b + c
a ) 6 , b ) 7 , c ) 8 , d ) 9 , e ) 10
e
add(floor(divide(subtract(subtract(45, 10), subtract(20, 5)), const_2)), 1)
the scoring system in a certain football competition goes as follows : 3 points for victory , 1 point for a draw , and 0 points for defeat . each team plays 20 matches . if a team scored 10 points after 5 games , what is the least number of the remaining matches it has to win to reach the 45 - point mark by the end of the tournament ?
"to get 45 points as end of season we need another 35 points or more from remaining 15 matches : option a = 6 * 3 + 9 * 1 = 27 option b = 7 * 3 + 8 * 1 = 29 option c = 8 * 3 + 7 * 1 = 31 option d = 9 * 3 + 6 * 1 = 33 option e = 10 * 3 + 5 * 1 = 35 hence option e - 10"
a = 45 - 10 b = 20 - 5 c = a - b d = c / 2 e = math.floor(d) f = e + 1
a ) $ 11.73 , b ) $ 12.60 , c ) $ 13.80 , d ) $ 14.00 , e ) $ 15.87
b
multiply(multiply(divide(208.00, add(const_100, 15)), const_100), divide(const_1, 15))
the price of lunch for 15 people was $ 208.00 , including a 15 percent gratuity for service . what was the average price per person , excluding the gratuity ?
"take the initial price before the gratuity is 100 the gratuity is calculated on the final price , so as we assumed the final bill before adding gratuity is 100 so gratuity is 15 % of 100 is 15 so the total price of meals is 115 so the given amount i . e 208 is for 115 then we have to calculate for 100 for 115 208 for 100 x so by cross multiplication we get 115 x = 100 * 208 = > x = 100 * 208 / 110 by simplifying we get x as 189 which is the price of lunch before gratuity so the gratuity is 19 so as the question ask the average price person excluding gratuity is 189 / 15 = 12.6 so our answer is b )"
a = 100 + 15 b = 208 / 0 c = b * 100 d = 1 / 15 e = c * d
a ) 20 % , b ) 18 % , c ) 36 % , d ) 64 % , e ) none of these
a
multiply(subtract(const_1, sqrt(divide(subtract(const_100, 36), const_100))), const_100)
if the area of a circle decreases by 36 % , then the radius of a circle decreases by
"if area of a circle decreased by x % then the radius of a circle decreases by ( 100 βˆ’ 10 √ 100 βˆ’ x ) % = ( 100 βˆ’ 10 √ 100 βˆ’ 36 ) % = ( 100 βˆ’ 10 √ 64 ) % = 100 - 80 = 20 % answer a"
a = 100 - 36 b = a / 100 c = math.sqrt(b) d = 1 - c e = d * 100
a ) 10 , b ) 20 , c ) 30 , d ) 40 , e ) 50
c
multiply(sqrt(divide(450, 2)), 2)
if n is a positive integer and n ^ 2 is divisible by 450 , then what is the largest positive integer that must divide n ?
"450 = 2 * 3 ^ 2 * 5 ^ 2 if 450 divides n ^ 2 , then n must be divisible by 2 * 3 * 5 = 30 the answer is c ."
a = 450 / 2 b = math.sqrt(a) c = b * 2
a ) 20 % , b ) 25 % , c ) 18 % , d ) 50 % , e ) none of these
d
multiply(divide(divide(7, const_100), divide(14, const_100)), const_100)
if the given two numbers are respectively 7 % and 14 % of a third number , then what percentage is the first of the second ?
"here , l = 7 and m = 14 therefore , first number = l / m x 100 % of second number = 7 / 14 x 100 % of second number = 50 % of second number answer : d"
a = 7 / 100 b = 14 / 100 c = a / b d = c * 100