Datasets:
euclaise
/
mathqa_programs

Dataset card Data Studio Files
xet
Community
1
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 ) 33 , b ) 54 , c ) 81 , d ) 17 , e ) 01
c
multiply(subtract(70, 7), divide(90, 70))
a group of 55 adults and 70 children go for trekking . if there is meal for either 70 adults or 90 children and if 7 adults have their meal , find the total number of children that can be catered with the remaining food .
"explanation : as there is meal for 70 adults and 7 have their meal , the meal left can be catered to 63 adults . now , 70 adults = 90 children 7 adults = 9 children therefore , 63 adults = 81 children hence , the meal can be catered to 81 children . answer : c"
a = 70 - 7 b = 90 / 70 c = a * b
a ) 1 , b ) 4 , c ) 5 , d ) 7 , e ) 2
e
add(const_2, const_3)
if 42 / 330 = 0.1272 , what is the 32 nd digit to the right of the decimal point of the fraction ?
"we are not concerned what 42 / 330 means . . we have to look at the decimal . . 0 . . 1272 means 0.1272727 . . . so leaving first digit to the right of decimal , all odd numbered are 7 and all even numbered are 2 . here 32 is even , so ans is 2 answer is e"
a = 2 + 3
a ) 1 , b ) 6 , c ) 7 , d ) 8 , e ) 2
d
subtract(multiply(const_4, const_3), divide(divide(multiply(36000, multiply(const_4, const_3)), 54000), 2))
a began business with rs . 36000 and was joined afterwards by b with rs . 54000 . when did b join if the profits at the end of the year were divided in the ratio of 2 : 1 ?
"36 * 12 : 54 * x = 2 : 1 x = 4 12 - 4 = 8 answer : d"
a = 4 * 3 b = 4 * 3 c = 36000 * b d = c / 54000 e = d / 2 f = a - e
a ) 5 / 2 , b ) 3 / 4 , c ) 2 / 3 , d ) 1 / 3 , e ) 5 / 4
e
multiply(subtract(16, const_4), 25)
find the value of √ ( 25 / 16 ) .
"√ ( 25 / 16 ) = √ 25 / √ 16 = 5 / 4 answer is e ."
a = 16 - 4 b = a * 25
a ) 2 , b ) 3 , c ) 4 , d ) 1 , e ) 5
d
divide(25, 24)
how many of the positive factors of 24 are not factors of 25
"factors of 24 - 1 , 2 , 3 , 4 , 6 , 8 , 12,24 factors of 25 - 1,5 , 25 comparing both , we have 1 factor of 24 which are not factors of 25 - 1 answer : d"
a = 25 / 24
a ) 12 , b ) 28 , c ) 26 , d ) 19 , e ) 112
a
divide(multiply(4, 3), subtract(4, 3))
a man can do a piece of work in 4 days , but with the help of his daughter , he can do it in 3 days . in what time can his daughter do it alone ?
explanation : daughter ’ s 1 day ’ s work = ( 1 / 3 – 1 / 4 ) = 1 / 12 daughter alone can do the work in 12 / 1 = 12 days answer : a
a = 4 * 3 b = 4 - 3 c = a / b
a ) 50.4 % , b ) 23.8 % , c ) 66.5 % , d ) 34.9 % , e ) 55.3 %
e
multiply(divide(subtract(900, add(add(add(75, 50), add(64, 78)), 135)), 900), const_100)
john had a stock of 900 books in his bookshop . he sold 75 on monday , 50 on tuesday , 64 on wednesday , 78 on thursday and 135 on friday . what percentage of the books were not sold ?
"let n be the total number of books sold . hence n = 75 + 50 + 64 + 78 + 135 = 402 let m be the books not sold m = 900 - n = 900 - 402 = 498 percentage books not sold / total number of books = 498 / 900 = 0.553 = 55.3 % correct answer e"
a = 75 + 50 b = 64 + 78 c = a + b d = c + 135 e = 900 - d f = e / 900 g = f * 100
a ) 647 , b ) 738 , c ) 654 , d ) 847 , e ) 976
b
subtract(825, divide(multiply(subtract(854, 825), 3), 4))
a sum of money at simple interest amounts to rs . 825 in 3 years and to rs . 854 in 4 years . the sum is :
"s . i . for 1 year = rs . ( 854 - 825 ) = rs . 29 . s . i . for 3 years = rs . ( 29 x 3 ) = rs . 87 . principal = rs . ( 825 - 87 ) = rs . 738 answer : b"
a = 854 - 825 b = a * 3 c = b / 4 d = 825 - c
a ) 67 : 19 , b ) 65 : 17 , c ) 61 : 19 , d ) 63 : 15 , e ) 67 : 17
a
divide(subtract(multiply(5, divide(subtract(add(multiply(5, const_3), subtract(multiply(7, 5), 7)), 5), subtract(5, const_3))), subtract(multiply(7, 5), 7)), divide(subtract(add(multiply(5, const_3), subtract(multiply(7, 5), 7)), 5), subtract(5, const_3)))
the age of father 7 years ago was 5 times the age of his son . 5 years hence , father ' s age will be thrice that of his son . the ratio of their present ages is :
let the ages of father and son 7 years ago be 5 x and x years respectively . then , ( 5 x + 7 ) + 5 = 3 [ ( x + 7 ) + 5 ] 5 x + 12 = 3 x + 36 x = 12 . required ratio = ( 5 x + 7 ) : ( x + 7 ) = 67 : 19 . answer : option a
a = 5 * 3 b = 7 * 5 c = b - 7 d = a + c e = d - 5 f = 5 - 3 g = e / f h = 5 * g i = 7 * 5 j = i - 7 k = h - j l = 5 * 3 m = 7 * 5 n = m - 7 o = l + n p = o - 5 q = 5 - 3 r = p / q s = k / r
a ) 5 square inches , b ) 10 square inches , c ) 49 square inches , d ) 25 square inches , e ) 350 square inches
c
subtract(multiply(power(21, const_2), const_3), add(add(power(14, const_2), power(21, const_2)), add(power(14, const_2), power(21, const_2))))
huey ' s hip pizza sells two sizes of square pizzas : a small pizza that measures 14 inches on a side and costs $ 10 , and a large pizza that measures 21 inches on a side and costs $ 20 . if two friends go to huey ' s with $ 30 apiece , how many more square inches of pizza can they buy if they pool their money than if they each purchase pizza alone ?
"in the first case each can buy one pizza of $ 10 and one pizza of $ 20 . in square inches that would be ( 14 * 14 = 196 ) for the small pizza and ( 21 * 21 = 441 ) for the large pizza . in total sq inches that would be ( 196 + 441 ) * 2 = 1274 sq inches . in the second case if they pool their money together they can buy 3 large pizzas . in terms of square inches that would be 3 * 441 = 1323 sq inches . hence , the difference is 49 square inches more ( 1323 - 1274 ) . the correct answer is c"
a = 21 ** 2 b = a * 3 c = 14 ** 2 d = 21 ** 2 e = c + d f = 14 ** 2 g = 21 ** 2 h = f + g i = e + h j = b - i
a ) 432 , b ) 433 , c ) 434 , d ) 435 , e ) 436
a
multiply(6, 72)
if the lcm of a and b is 72 and their hcf is 6 , what is the product of a & b ?
the most important property of lcm ( the least common multiple ) and gcd ( the greatest common divisors ) is : for any positive integers x and y , x βˆ— y = gcd ( x , y ) βˆ— lcm ( x , y ) so ab = 72 * 6 = 432 . answer : a
a = 6 * 72
a ) 22 seconds , b ) 12 seconds , c ) 10 seconds , d ) 8 seconds , e ) 28 seconds
d
subtract(multiply(divide(7, 56), 120), 7)
in a 120 meters race a beats b by 56 m or 7 seconds . a ' s time over the course is :
"b runs 56 m in 7 sec . = > b runs 120 m in 7 / 56 * 120 = 15 seconds since a beats b by 7 seconds , a runs 120 m in ( 15 - 7 ) = 8 seconds hence , a ' s time over the course = 8 seconds answer : d"
a = 7 / 56 b = a * 120 c = b - 7
a ) 1 / 4 , b ) 1 / 3 , c ) 3 / 8 , d ) 7 / 16 , e ) Β½
e
subtract(1, add(multiply(multiply(divide(const_1, const_2), divide(const_1, const_2)), divide(const_1, const_2)), add(multiply(divide(const_1, const_2), divide(const_1, const_2)), multiply(multiply(divide(const_1, const_2), divide(const_1, const_2)), divide(const_1, const_2)))))
if x is a randomly chosen integer between 1 and 20 , inclusive , and y is a randomly chosen integer between 21 and 40 , inclusive , what is the probability that xy is a multiple of 4 ?
there are two ways that i see of getting a multiple of 4 1 ) x and / or y is a multiple of 4 2 ) x is a multiple of 2 and y is a multiple of 2 for 1 we have p ( option 1 ) = ( 5 / 20 ) ( 20 / 20 ) + ( 5 / 20 ) ( 20 / 20 ) - ( 5 / 20 ) ( 5 / 20 ) = 1 / 2 - 1 / 16 since 5 of the numbers in each range are divisible by 4 , and if either x or y is selected to be divisible by 4 the selection of the other variable does not matter . notice the last subtraction term which corrects the fact that we have counted the possibility of both being divisible by 4 twice ( e . g . p ( a union b ) = p ( a ) + p ( b ) - p ( a and b ) ) . notice at this point that the probability of option one is already 7 / 16 , the second largest available option . if we can come up with any other possibilities for getting xy divisble by 4 we must go with option e . of course this is the case because if we select a multiple of 2 that is not a multiple of 4 from both ranges of numbers , the product is divisible by 4 . we can formally compute this as : p ( option 2 ) = ( 5 / 20 ) * ( 5 / 20 ) = 1 / 16 since we do not wish to double count possibilities that fall into category 1 , we only count the probability of selecting even numbers from each set that are not divisible by 4 ( e . g . the other 5 even integers in each set ) . total probability is then p ( option 1 ) + p ( option 2 ) = 1 / 2
a = 1 / 2 b = 1 / 2 c = a * b d = 1 / 2 e = c * d f = 1 / 2 g = 1 / 2 h = f * g i = 1 / 2 j = 1 / 2 k = i * j l = 1 / 2 m = k * l n = h + m o = e + n p = 1 - o
a ) 0 , b ) 1 , c ) 10 , d ) 19 , e ) 20
d
subtract(20, const_1)
the average of 20 number is zero of them at the most , how many may be greater than zero ?
solution average of 20 numbers = 0 . ∴ sum of 20 numbers = ( 0 Γ— 20 ) = 0 . it is quite possible that 19 of these number may be positive and if their sum is a , then 20 th number is ( - a ) . answer d
a = 20 - 1
a ) 1600 , b ) 2777 , c ) 2889 , d ) 27670 , e ) 2771
a
divide(1792, add(divide(multiply(divide(add(multiply(2, 5), 2), 5), 5), const_100), const_1))
find the principle on a certain sum of money at 5 % per annum for 2 2 / 5 years if the amount being rs . 1792 ?
"1792 = p [ 1 + ( 5 * 12 / 5 ) / 100 ] p = 1600 . answer : a"
a = 2 * 5 b = a + 2 c = b / 5 d = c * 5 e = d / 100 f = e + 1 g = 1792 / f
a ) 38,000 , b ) 40,000 , c ) 42,500 , d ) 45,800 , e ) 48,500
d
multiply(multiply(const_100, const_100), const_4)
of the 13 employees in a certain department , 1 has an annual salary of 38,000 , 2 have an annual salary of 45,800 each , 2 have an annual salary of 42,500 each , 3 have an annual salary of 40,000 each and 5 have an annual salary of 48,500 each . what is the median annual salary for the 13 employees ?
"median is just the value in the middle when you arrange all values in the ascending order in this question , the 7 th value would be the median ( since there are 13 employees ) 38 , 40 , 40 , 40 , 42.5 , 42.5 , 45.8 so , answer is d ."
a = 100 * 100 b = a * 4
a ) r = 300 , b ) r = 400 , c ) r = 500 , d ) 600 , e ) none of these
c
divide(200, divide(2, 5))
there are 200 female managers in a certain company . find the total number r of female employees in the company , if 2 / 5 of all the employees are managers and 2 / 5 of all male employees are managers .
"{ managers } = { female managers } + { male managers } ; we are told that the total number of managers in the company is 2 / 5 of all the employees , thus { managers } = 2 / 5 ( m + f ) , where m and f are number of female and male employees , respectively . also , we know that 2 / 5 of all male employees are managers : { male managers } = 2 / 5 * mas well as there are total of 200 female managers : { female managers } = 200 ; thus : 2 / 5 ( m + f ) = 200 + 2 / 5 * m - - > f = 500 . answer : c ."
a = 2 / 5 b = 200 / a
a ) 70 , b ) 245 , c ) 150 , d ) 35 , e ) 350
e
divide(49, multiply(divide(subtract(const_100, 30), const_100), divide(20, const_100)))
in a certain boys camp , 20 % of the total boys are from school a and 30 % of those study science . if there are 49 boys in the camp that are from school a but do not study science then what is the total number of boys in the camp ?
"since 30 % of the boys from school a study science , then 70 % of the boys from school a do not study science and since 20 % of the total number of boys are from school a , then 0.2 * 0.7 = 0.14 , or 14 % of the boys in the camp are from school a and do not study science . we are told that this number equals to 49 , so 0.14 * { total } = 49 - - > { total } = 350 . answer : e ."
a = 100 - 30 b = a / 100 c = 20 / 100 d = b * c e = 49 / d
a ) 2 litres , b ) 1.4 litres , c ) 1 litres , d ) 1.6 litres , e ) none of these
d
multiply(divide(20, const_100), 8)
how many litres of pure acid are there in 8 litres of a 20 % solution ?
"explanation : quantity of pure acid = 8 Γ— ( 20 / 100 ) = 1.6 answer : option d"
a = 20 / 100 b = a * 8
a ) 266 sec , b ) 170 sec , c ) 776 sec , d ) 166 sec , e ) 997 sec
b
divide(add(1200, 500), divide(1200, 120))
a 1200 m long train crosses a tree in 120 sec , how much time will i take to pass a platform 500 m long ?
"l = s * t s = 1200 / 120 s = 10 m / sec . total length ( d ) = 1700 m t = d / s t = 1700 / 10 t = 170 sec answer : b"
a = 1200 + 500 b = 1200 / 120 c = a / b
a ) 130 , b ) 140 , c ) 136 , d ) 156 , e ) 166
a
add(lcm(lcm(6, 7), lcm(9, 18)), 4)
what is the least number which when divided by 6 , 7 , 9 and 18 leaves remainder 4 in each care ?
"explanation : lcm of 6 , 7 , 9 and 18 is 126 required number = 126 + 4 = 130 answer : option a"
a = math.lcm(6, 7) b = math.lcm(9, 18) c = math.lcm(a, b) d = c + 4
a ) 277 , b ) 960 , c ) 277 , d ) 266 , e ) 121
b
multiply(subtract(832, divide(subtract(832, 448), const_2)), add(const_1, divide(50, const_100)))
the profit earned by selling an article for rs . 832 is equal to the loss incurred when the same article is sold for rs . 448 . what should be the sale price for making 50 % profit ?
"let c . p . = rs . x . then , 832 - x = x - 448 2 x = 1280 = > x = 640 required s . p . = 150 % of rs . 640 = 150 / 100 * 640 = rs . 960 . answer : b"
a = 832 - 448 b = a / 2 c = 832 - b d = 50 / 100 e = 1 + d f = c * e
a ) 22636 , b ) 22640 , c ) 22647 , d ) 22533 , e ) 22675
d
divide(multiply(add(multiply(14, const_100), 21), add(multiply(7, const_100), 77)), multiply(subtract(21, add(multiply(const_2, const_4), const_2)), subtract(21, add(multiply(const_2, const_4), const_2))))
a room 14 m 21 cm long and 7 m 77 cm broad is to be paved with square tiles . find the least number of square tiles required to cover the floor .
"explanation : area of the room = ( 1421 x 777 ) cm 2 . size of largest square tile = h . c . f . of 1421 cm and 777 cm = 7 cm . area of 1 tile = ( 7 x 7 ) cm 2 . number of tiles required = ( 1147 Γ— 777 ) / ( 7 Γ— 7 ) = 22533 answer : option d"
a = 14 * 100 b = a + 21 c = 7 * 100 d = c + 77 e = b * d f = 2 * 4 g = f + 2 h = 21 - g i = 2 * 4 j = i + 2 k = 21 - j l = h * k m = e / l
a ) 32 km , b ) 20 km , c ) 65 km , d ) 18 km , e ) 16 km
a
multiply(add(5, 3), 4)
roja and pooja start moving in the opposite directions from a pole . they are moving at the speeds of 5 km / hr and 3 km / hr respectively . after 4 hours what will be the distance between them ?
"distance = relative speed * time = ( 5 + 3 ) * 4 = 32 km [ they are travelling in the opposite direction , relative speed = sum of the speeds ] . answer : a"
a = 5 + 3 b = a * 4
a ) 125 , b ) 52 , c ) 58 , d ) 60 , e ) 62
a
divide(multiply(105, const_100), subtract(const_100, 16))
the number which exceeds 16 % of it by 105 is :
solution solution let the number be x . x - 16 % of x = 105 x - 16 / 100 x = 105 x - 4 / 25 x = 105 21 / 25 x = 105 x = ( 105 x 25 / 21 ) = 125 answer a
a = 105 * 100 b = 100 - 16 c = a / b
a ) 24.5 , b ) 14 , c ) 24.21 , d ) 27 , e ) 25
b
multiply(divide(7, subtract(9, 6)), 6)
sachin is younger than rahul by 7 years . if the ratio of their ages is 6 : 9 , find the age of sachin
"explanation : if rahul age is x , then sachin age is x - 7 , so , 9 x - 63 = 6 x 3 x = 63 x = 21 so sachin age is 21 - 7 = 14 answer : b ) 14"
a = 9 - 6 b = 7 / a c = b * 6
a ) 22 , b ) 23 , c ) 18 , d ) 26 , e ) 28
c
divide(subtract(multiply(44, const_4), 140), const_2)
a man has some hens and cows . if the number of heads be 44 and the number of feet equals 140 , then the number of hens will be :
"let hens be x and cows be y now , feet : x * 2 + y * 4 = 140 heads : x * 1 + y * 1 = 44 implies , 2 x + 4 y = 140 and x + y = 44 solving these two equations , we get x = 18 and y = 26 therefore , hens are 26 . answer : c"
a = 44 * 4 b = a - 140 c = b / 2
a ) 2 / 3 , b ) 2 / 5 , c ) 3 / 4 , d ) 3 / 7 , e ) 5 / 8
b
divide(divide(4, const_2), divide(50, const_10))
in the x - y plane , there are 4 points ( 00 ) , ( 04 ) , ( 54 ) , and ( 50 ) . if these 4 points makes a rectangle , what is the probability that x + y < 4 ?
the line y = - x + 4 intersects the rectangle and these three points of intersection ( 0,0 ) , ( 0,4 ) and ( 4,0 ) form a triangle . the points below the line y = - x + 4 satisfy x + y < 4 . the area of this triangle is ( 1 / 2 ) ( 4 ) ( 4 ) = 8 the area of the rectangle is 20 . p ( x + y < 4 ) = 8 / 20 = 2 / 5 the answer is b .
a = 4 / 2 b = 50 / 10 c = a / b
a ) 0.55 , b ) 0.4 , c ) 0.5 , d ) 0.05 , e ) 0.6
a
subtract(const_1, subtract(add(0.25, 0.35), 0.15))
the probability of two events a and b are 0.25 and 0.35 respectively . the probability that both a and b occur is 0.15 . the probability that neither a nor b occur is _________
"we are apply that formula . . . . . . . . . . . . . . p ( aorb ) = p ( a ) + p ( b ) - p ( a and b ) = . 25 + . 35 - . 15 = . 45 but the probability of neither a nor b = 1 - . 45 = 0.55 answer : a"
a = 0 + 25 b = a - 0 c = 1 - b
a ) 10 , b ) 20 , c ) 30 , d ) 17 , e ) 50
d
multiply(divide(subtract(add(25, add(const_0_25, const_0_25)), 5), add(const_100, subtract(add(25, add(const_0_25, const_0_25)), 5))), const_100)
in a certificate by mistake a candidate gave his height as 25 % more than actual height . in the interview panel , he clarified that his height was 5 feet 8 nches . find the % correction made by the candidate from his stated height to his actual height ?
"his height was = 5 feet 8 inch = 8 + 60 = 68 inch . required % correction = 68 * ( 1.25 - 1 ) = 17 d"
a = const_0_25 + const_0_25 b = 25 + a c = b - 5 d = const_0_25 + const_0_25 e = 25 + d f = e - 5 g = 100 + f h = c / g i = h * 100
a ) 67 , b ) 73 , c ) 53 , d ) 82 , e ) 89
d
multiply(multiply(4, divide(10, add(const_1, 4))), const_10)
the tens digit of a two digit number is 4 times of its unit digit and the sum of the digits be 10 . then the first number is
let the number be 10 x + y according to the question x = 4 y x + y = 10 4 y + y = 10 or y = 2 and x = 8 so required number is = 10 * 8 + 2 = 82 answer : d
a = 1 + 4 b = 10 / a c = 4 * b d = c * 10
a ) 100 m , b ) 120 m , c ) 130 m , d ) 150 m , e ) none of these
c
subtract(180, multiply(divide(3, const_60), const_1000))
a policeman noticed a criminal from a distance of 180 km . the criminal starts running and the policeman chases him . the criminal and the policeman run at the rate of 8 km and 9 km per hour respectively . what is the distance between them after 3 minutes ?
explanation : solution : relative speed = ( 9 - 8 ) = 1 km / hr . distance covered in 3 minutes = ( 1 * 3 / 60 ) km = 1 / 20 km = 50 m . . ' . distance between the criminal and policeman = ( 180 - 50 ) m = 130 m . answer : c
a = 3 / const_60 b = a * 1000 c = 180 - b
a ) 333 $ , b ) 380 $ , c ) 420 $ , d ) 450 $ , e ) 480 $
a
multiply(multiply(0.65, 40), 12)
in a fuel station the service costs $ 1.75 per car , every liter of fuel costs 0.65 $ . assuming that a company owns 12 cars and that every fuel tank contains 40 liters and they are all empty , how much money total will it cost to fuel all cars ?
"total cost = ( 1.75 * 12 ) + ( 0.65 * 12 * 40 ) = 21 + 312 = > 333 hence answer will be ( a ) 333"
a = 0 * 65 b = a * 12
a ) 50 liters , b ) 55 liters , c ) 60 liters , d ) 70 liters , e ) 75 liters
c
subtract(multiply(subtract(60, divide(60, add(2, 1))), 2), divide(60, add(2, 1)))
in a mixture 60 litres , the ratio of milk and water 2 : 1 . if this ratio is to be 1 : 2 , then the quanity of water to be further added is
"quantity of milk = 60 x 2 litres = 40 litres . 3 quantity of water in it = ( 60 - 40 ) litres = 20 litres . new ratio = 1 : 2 let quantity of water to be added further be x litres . then , milk : water = 40 . 20 + x now , 40 = 1 20 + x 2 20 + x = 80 x = 60 . quantity of water to be added = 60 liters . c"
a = 2 + 1 b = 60 / a c = 60 - b d = c * 2 e = 2 + 1 f = 60 / e g = d - f
a ) 210 m , b ) 220 m , c ) 230 m , d ) 240 m , e ) 250 m
b
subtract(multiply(9, multiply(add(120, 80), const_0_2778)), 280)
a 280 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 + 280 / 9 = 500 / 9 x = 220 ans is 220 m answer : b"
a = 120 + 80 b = a * const_0_2778 c = 9 * b d = c - 280
a ) 10 , b ) 20 , c ) 22 , d ) 30 , e ) 15
c
divide(add(add(11, const_4), subtract(34, const_4)), const_2)
find the average of all the numbers between 11 and 34 which are divisible by 5 ?
"average = ( 15 + 20 + 25 + 30 ) 4 = 90 / 4 = 22 answer is c"
a = 11 + 4 b = 34 - 4 c = a + b d = c / 2
a ) 8 days , b ) 10 days , c ) 12 days , d ) 13 days , e ) 14 days
b
add(4, divide(multiply(12, multiply(20, subtract(const_1, divide(4, 20)))), add(20, 12)))
x and y can do a piece of work in 20 days and 12 days respectively . x started the work alone and then after 4 days y joined him till the completion of work . how long did the work last ?
"let the total units of work to be completed be 60 . units of work completed by x in a single day = 60 / 20 = 3 units . units of work completed by y in a single day = 60 / 12 = 5 units . thus , 8 units of work can be done by them in a single day working together . x works for 4 days alone = 12 units of work . remaining work = ( 60 - 12 ) = 48 units . now , x and y works together to complete the remaining work = 48 / 8 = 6 days hence , total days taken to complete the work = ( 6 + 4 ) = 10 days . answer : b"
a = 4 / 20 b = 1 - a c = 20 * b d = 12 * c e = 20 + 12 f = d / e g = 4 + f
a ) 1 / 4 , b ) 1 / 3 , c ) 1 / 2 , d ) 2 / 3 , e ) 3 / 4
a
divide(const_2, 8)
if an integer n is to be selected at random from 1 to 100 , inclusive , what is probability n ( n + 1 ) will be divisible by 8 ?
"because n ( n + 1 ) is always an even product of even * odd or odd * even factors , there is a probability of 1 that that it will be divisible by 2 , and , thus , a probability of 1 / 2 that it will be divisible by 4 and , thus , a probability of 1 / 4 that it will be divisible by 8 1 * 1 / 4 = 1 / 4 answer : a"
a = 2 / 8
a ) 308 , b ) 369 , c ) 434 , d ) 476 , e ) 513
a
multiply(const_pi, multiply(7, 14))
find the curved surface area , if the radius of a cone is 7 m and slant height is 14 m ?
"cone curved surface area = Γ― € rl 22 / 7 Γ£ β€” 7 Γ£ β€” 14 = 22 Γ£ β€” 14 = 308 m ( power 2 ) answer is a ."
a = 7 * 14 b = math.pi * a
a ) 1 , b ) 1 / 2 , c ) 2 , d ) 4 , e ) - 1 / 2
c
divide(16, subtract(9, const_1))
if | x | = 9 x - 16 , then x = ?
substituted option c i . e x = 2 . inequality satisfied . this eliminates all options except answer is c
a = 9 - 1 b = 16 / a
a ) 9680 , b ) 10680 , c ) 2999 , d ) 2774 , e ) 1212
b
add(multiply(multiply(add(divide(2, const_100), divide(divide(subtract(10200, 8000), 3), 8000)), 8000), 3), 8000)
sonika deposited rs . 8000 which amounted to rs . 10200 after 3 years at simple interest . had the interest been 2 % more . she would get how much ?
"( 8000 * 3 * 2 ) / 100 = 480 10200 - - - - - - - - 10680 answer : b"
a = 2 / 100 b = 10200 - 8000 c = b / 3 d = c / 8000 e = a + d f = e * 8000 g = f * 3 h = g + 8000
a ) rs . 3601 , b ) rs . 3618 , c ) rs . 4900 , d ) rs . 3619 , e ) rs . 3610
c
subtract(multiply(add(2400, 100), add(24, const_1)), multiply(2400, 24))
the average monthly salary of 24 employees in an organisation is rs . 2400 . if the manager ' s salary is added , then the average salary increases by rs . 100 . what is the manager ' s monthly salary ?
"manager ' s monthly salary = rs . ( 2500 * 25 - 2400 * 24 ) = rs . 4900 answer : c"
a = 2400 + 100 b = 24 + 1 c = a * b d = 2400 * 24 e = c - d
a ) 3000.33 , b ) 2802.33 , c ) 1833.33 , d ) 2990.33 , e ) 2982.33
c
divide(divide(subtract(multiply(8000, power(add(const_1, divide(20, const_100)), 2)), 8000), 2), multiply(6, divide(16, const_100)))
the s . i . on a certain sum of money for 6 years at 16 % per annum is half the c . i . on rs . 8000 for 2 years at 20 % per annum . the sum placed on s . i . is ?
c . i . = [ 8000 * ( 1 + 20 / 100 ) 2 - 8000 ] = ( 8000 * 6 / 5 * 6 / 5 - 8000 ) = rs . 3520 . sum = ( 3520 * 100 ) / ( 6 * 16 ) = rs . 1833.33 answer : c
a = 20 / 100 b = 1 + a c = b ** 2 d = 8000 * c e = d - 8000 f = e / 2 g = 16 / 100 h = 6 * g i = f / h
['a ) 15 m Γ— 6.67 m', 'b ) 20 m Γ— 5 m', 'c ) 30 m Γ— 3.33 m', 'd ) 40 m Γ— 2.5 m', 'e ) none of these']
b
divide(100, divide(30, const_3))
a farmer wishes to start a 100 square metres rectangular vegetable garden . since he has only 30 m barbed wire , he fences three sides of the garden letting his house compound wall act as the fourth side fencing . the dimension of the garden is :
we have : 2 b + l = 30 β‡’ l = 30 – 2 b . area = 100 m 2 β‡’ l Γ— b = 100 β‡’ b ( 30 – 2 b ) = 100 β‡’ b 2 – 15 b + 50 = 0 β‡’ ( b – 10 ) ( b – 5 ) = 0 β‡’ b = 10 or b = 5 . when b = 10 , l = 10 and when b = 5 , l = 20 . since the garden is rectangular , so its dimension is 20 m Γ— 5 m . answer b
a = 30 / 3 b = 100 / a
a ) 0 , b ) 2 , c ) 4 , d ) 6 , e ) 8
a
multiply(subtract(divide(power(divide(9402, const_3), 2), 9), add(add(add(multiply(multiply(const_1000, subtract(9, 2)), const_100), multiply(multiply(const_1000, subtract(9, 2)), const_10)), multiply(2, const_100)), subtract(const_100, 2))), 9)
if z is a multiple of 9402 , what is the remainder when z ^ 2 is divided by 9 ?
"the sum of the digits is 9 + 4 + 2 = 15 . thus 3 is a factor of 9402 , so 3 is a factor of z . then 3 ^ 3 = 9 is a factor of z ^ 2 . then the remainder when z ^ 2 is divided by 9 is 0 . the answer is a ."
a = 9402 / 3 b = a ** 2 c = b / 9 d = 9 - 2 e = 1000 * d f = e * 100 g = 9 - 2 h = 1000 * g i = h * 10 j = f + i k = 2 * 100 l = j + k m = 100 - 2 n = l + m o = c - n p = o * 9
a ) 12 % , b ) 12.5 % , c ) 13.5 % , d ) 13 % , e ) 14 %
b
multiply(divide(subtract(multiply(divide(12.5, const_100), subtract(subtract(70000, 42000), 12000)), multiply(subtract(multiply(divide(subtract(14, const_1), 14), subtract(70000, 42000)), 12000), divide(12.5, const_100))), multiply(divide(12.5, const_100), subtract(subtract(70000, 42000), 12000))), const_100)
a techno company has 14 machines of equal efficiency in its factory . the annual manufacturing costs are rs 42000 and establishment charges are rs 12000 . the annual output of the company is rs 70000 . the annual output and manufacturing costs are directly proportional to the number of machines . the shareholders get 12.5 profit , which is directly proportional to the annual output of the company . if 7.14 machines remain closed throughout the year , then the percentage decrease in the amount of profit of the shareholders would be :
explanation : the original profit is : - = > 70000 βˆ’ 42000 βˆ’ 12000 = 16000 . if 7.14 % of 14 i . e . one of the machines closed throughout the year , then the change in profit will be : - = > ( 13 / 14 ) Γ— [ 70000 βˆ’ 42000 ] . = > 14000 . thus , decrease in the profit % is : - = > ( 2000 / 16000 ) Γ— 100 . = > 12.5 % . answer : b
a = 12 / 5 b = 70000 - 42000 c = b - 12000 d = a * c e = 14 - 1 f = e / 14 g = 70000 - 42000 h = f * g i = h - 12000 j = 12 / 5 k = i * j l = d - k m = 12 / 5 n = 70000 - 42000 o = n - 12000 p = m * o q = l / p r = q * 100
a ) 35 days , b ) 10 days , c ) 14 days , d ) 22 days , e ) 26 days
a
add(8, divide(subtract(const_1, divide(8, 80)), add(inverse(80), inverse(48))))
p and q can complete a work in 80 days and 48 days respectively . p alone started the work and q joined him after 8 days till the completion of the work . how long did the work last ?
"explanation : work done by p in 1 day = 1 / 80 work done by q in 1 day = 1 / 48 work done by p in 8 days = 8 Γ£ β€” ( 1 / 80 ) = 1 / 10 remaining work = 1 Γ’ € β€œ 1 / 10 = 9 / 10 work done by p and q in 1 day = 1 / 80 + 1 / 48 = 1 / 30 number of days p and q take to complete the remaining work = ( 9 / 10 ) / ( 1 / 30 ) = 27 total days = 8 + 27 = 35 answer : option a"
a = 8 / 80 b = 1 - a c = 1/(80) d = 1/(48) e = c + d f = b / e g = 8 + f
a ) 3 , b ) 8 , c ) 5 , d ) 6 , e ) 7
a
multiply(6, divide(3, 6))
a and b together can complete work in 6 days . a alone starts working and leaves it after working for 3 days completing only half of the work . in how many days it can be completed if the remaining job is undertaken by b ?
explanation : ( a + b ) one day work = 1 / 6 now a does half of the work in 3 days so a can complete the whole work in 6 days a ’ s one day work = 1 / 6 b ’ s one day work = 1 / 6 - 1 / 6 = 1 / 6 b alone can complete the work in 6 days so half of the work in 3 days answer : option a
a = 3 / 6 b = 6 * a
a ) 27 , b ) 30 , c ) 31 , d ) 33 , e ) 46
a
subtract(divide(subtract(subtract(100, 25), const_2), const_2), divide(divide(subtract(subtract(subtract(subtract(100, const_2), multiply(3, const_4)), 3), 3), 3), const_2))
how many even number in the range between 25 to 100 inclusive are not divisible by 3
"we have to find the number of terms that are divisible by 2 but not by 6 ( as the question asks for the even numbers only which are not divisible by 3 ) for 2 , 26 , 28,30 . . . 100 using ap formula , we can say 100 = 10 + ( n - 1 ) * 2 or n = 38 . for 6 , 36 , 42 , . . . 96 using ap formula , we can say 96 = 12 + ( n - 1 ) * 6 or n = 11 . hence , only divisible by 2 but not 3 = 38 - 11 = 27 . hence , answer a"
a = 100 - 25 b = a - 2 c = b / 2 d = 100 - 2 e = 3 * 4 f = d - e g = f - 3 h = g - 3 i = h / 3 j = i / 2 k = c - j
a ) a ) 5 , b ) b ) 9 , c ) c ) 11 , d ) d ) 13 , e ) e ) 15
c
subtract(30, divide(subtract(30, divide(5, const_2)), const_2))
sum of two numbers is 30 . two times of the first exceeds by 5 from the three times of the other . then the numbers will be ?
"explanation : x + y = 30 2 x – 3 y = 5 x = 19 y = 11 c )"
a = 5 / 2 b = 30 - a c = b / 2 d = 30 - c
a ) 26 , b ) 27 , c ) 46 , d ) 57 , e ) none
b
add(add(power(add(add(divide(subtract(subtract(3, const_10), const_2), const_4), const_2), const_2), const_2), power(add(add(add(divide(subtract(subtract(3, const_10), const_2), const_4), const_2), const_2), const_2), const_2)), add(power(divide(subtract(subtract(3, const_10), const_2), const_4), const_2), power(add(divide(subtract(subtract(3, const_10), const_2), const_4), const_2), const_2)))
the sum of three consecutive multiples of 3 is 72 . what is the larget number ?
"sol . let the numbers be 3 x , 3 x + 3 and 3 x + 6 . then , 3 x + ( 3 x + 3 ) + ( 3 x + 6 ) = 72 ⇔ 9 x = 63 ⇔ x = 7 . ∴ largest number = 3 x + 6 = 27 . answer b"
a = 3 - 10 b = a - 2 c = b / 4 d = c + 2 e = d + 2 f = e ** 2 g = 3 - 10 h = g - 2 i = h / 4 j = i + 2 k = j + 2 l = k + 2 m = l ** 2 n = f + m o = 3 - 10 p = o - 2 q = p / 4 r = q ** 2 s = 3 - 10 t = s - 2 u = t / 4 v = u + 2 w = v ** 2 x = r + w y = n + x
a ) 1 , b ) 6 , c ) 3 , d ) 4 , e ) 5
b
divide(log(multiply(9, 9)), log(const_10))
9 log 9 ( 6 ) = ?
"exponential and log functions are inverse of each other . hence aloga ( x ) = x , for all x real and positive . and therefore 9 log 9 ( 6 ) = 6 correct answer b"
a = 9 * 9 b = math.log(a) c = math.log(10) d = b / c
a ) 6 : 1 , b ) 5 : 4 , c ) 20 : 7 , d ) 10 : 9 , e ) 14 : 11
c
divide(subtract(divide(5, add(5, 4)), divide(const_3, add(3, 4))), subtract(divide(3, 5), divide(5, add(5, 4))))
two mixtures of a and b have a and b in the ratio 3 : 2 and 3 : 4 . in what proportion should these two mixtures be mixed to get a new mixture in which the ration of a to b is 5 : 4 ?
i got c but with a different , yet slightly longer approach , which i constantly use for proportion problems such as this one . usually , this approach is very efficient when used with slightly difficult questions , but this problem is quite hard so i had to slightly tweak it . i will show you how i usually use the approach and later show you how to use it with this specific problem . example . 1 liter of solution a contains 45 % alcohol , while 1 liter of solution b contains 20 % alcohol . in what ratio must the two solutions be used to get a solution with 30 % alcohol solution : 1 . 45 / 100 * [ a / ( a + b ) ] + 20 / 100 * [ b / ( a + b ) ] = 30 / 100 2 . multiply 100 to both sides to arrive at 45 a / ( a + b ) + 20 b / ( a + b ) = 30 3 . multiply ( a + b ) to both sides to arrive at 45 a + 20 b = 30 a + 30 b 4 . distribute to arrive at 15 a = 10 b 5 . thus the ratio is a / b = 10 / 15 = 2 / 3 now using this same approach , we tackle gopu 106 ’ s question . it is important to first think of x in the mixture as the alcohol in the problem above ; hence , a mixture of x and y in the ratio of 3 : 2 translates to x is 3 / 5 of the solution . applying this concept to all three equations , we write : 1 . 3 / 5 * [ a / ( a + b ) ] + 3 / 7 * [ b / ( a + b ) ] = 5 / 9 2 . now here is the tweak that must be made to continue with this approach . you must find the common denominator for all three numbers and organize the fractions accordingly . by finding the common denominator of 5 , 79 ( or 315 ) we re - write the equations as follows 3 . 189 / 315 * [ a / ( a + b ) ] + 135 / 315 * [ b / ( a + b ) ] = 175 / 315 4 . multiply 315 to both sides to arrive at 189 a / ( a + b ) + 135 b / ( a + b ) = 175 5 . multiply ( a + b ) to both sides to arrive at 189 a + 135 b = 175 a + 175 b 6 . distribute to arrive at 14 a = 40 b 7 . thus the ratio is a / b = 40 / 14 = 20 / 7 or answer c
a = 5 + 4 b = 5 / a c = 3 + 4 d = 3 / c e = b - d f = 3 / 5 g = 5 + 4 h = 5 / g i = f - h j = e / i
a ) 80 % , b ) 32 % , c ) 48 % , d ) 84 % , e ) 92 %
a
subtract(const_100, divide(subtract(const_100, 60), const_2))
a certain characteristic in a large population has a distribution that is symmetric about the mean m . if 60 percent of the distribution lies within one standard deviation d of the mean , what percent of the distribution is less than m + d ?
"this is easiest to solve with a bell - curve histogram . m here is equal to Β΅ in the gaussian normal distribution and thus m = 50 % of the total population . so , if 60 % is one st . dev , then on either side of m we have 60 / 2 = 30 % . so , 30 % are to the right and left of m ( = 50 % ) . in other words , our value m + d = 50 + 30 = 80 % goingfrom the mean m , to the right of the distributionin the bell shaped histogram . . this means that 80 % of the values are below m + d . like i said , doing it on a bell - curve histogram is much easier to fullygethow this works , or you could apply gmat percentile jargon / theory to it a"
a = 100 - 60 b = a / 2 c = 100 - b
a ) 20 , b ) 18 , c ) 16 , d ) 19 , e ) 17
b
divide(add(multiply(22, 4), multiply(8, 16)), add(8, 4))
8 kilograms of rice costing rs . 16 per kg is mixed with 4 kilograms of rice costing rs . 22 per kg . what is the average price of the mixture ?
p 1 = rs . 16 per kg , p 2 = rs . 22 per kg , q 1 = 8 kg , q 2 = 4 kg now , p = ( p 1 q 1 + p 2 q 2 ) / ( q 1 + q 2 ) average price of the mixture = 8 * 16 + 4 * 22 / 12 = 128 + 88 / 12 = 216 / 12 = 18 answer : b
a = 22 * 4 b = 8 * 16 c = a + b d = 8 + 4 e = c / d
a ) 22 seconds , b ) 65 seconds , c ) 78 seconds , d ) 12 seconds , e ) 14 seconds
e
divide(70, multiply(add(15, 3), const_0_2778))
the speed at which a man can row a boat in still water is 15 kmph . if he rows downstream , where the speed of current is 3 kmph , what time will he take to cover 70 metres ?
"speed of the boat downstream = 15 + 3 = 18 kmph = 18 * 5 / 18 = 5 m / s hence time taken to cover 60 m = 70 / 5 = 14 seconds . answer : e"
a = 15 + 3 b = a * const_0_2778 c = 70 / b
a ) 1 / 50 , b ) 1 / 25 , c ) 1 / 99 , d ) 1 , e ) 2
c
divide(1, 99)
if the numbers 1 to 99 are written on 99 pieces of paper , ( one on each ) and one piece is picked at random , then what is the probability that the number drawn is neither prime nor composite ?
"there are 25 primes , 73 composite numbers from 1 to 100 . the number which is neither prime nor composite is 1 . therefore , required probability = 1 / 99 . answer : c"
a = 1 / 99
a ) 21000 , b ) 21009 , c ) 21029 , d ) 21298 , e ) 21098
a
subtract(subtract(35000, divide(subtract(multiply(divide(add(const_100, 32), const_100), 35000), multiply(divide(add(const_100, 30), const_100), 35000)), subtract(divide(add(const_100, 40), const_100), divide(add(const_100, 30), const_100)))), divide(subtract(multiply(divide(add(const_100, 32), const_100), 35000), multiply(divide(add(const_100, 30), const_100), 35000)), subtract(divide(add(const_100, 40), const_100), divide(add(const_100, 30), const_100))))
a trader purchased two colour televisions for a total of rs . 35000 . he sold one colour television at 30 % profit and the other 40 % profit . find the difference in the cost prices of the two televisions if he made an overall profit of 32 % ?
let the cost prices of the colour television sold at 30 % profit and 40 % profit be rs . x and rs . ( 35000 - x ) respectively . total selling price of televisions = x + 30 / 100 x + ( 35000 - x ) + 40 / 100 ( 35000 - x ) = > 130 / 100 x + 140 / 100 ( 35000 - x ) = 35000 + 32 / 100 ( 35000 ) x = 28000 35000 - x = 7000 difference in the cost prices of televisions = rs . 21000 answer : a
a = 100 + 32 b = a / 100 c = b * 35000 d = 100 + 30 e = d / 100 f = e * 35000 g = c - f h = 100 + 40 i = h / 100 j = 100 + 30 k = j / 100 l = i - k m = g / l n = 35000 - m o = 100 + 32 p = o / 100 q = p * 35000 r = 100 + 30 s = r / 100 t = s * 35000 u = q - t v = 100 + 40 w = v / 100 x = 100 + 30 y = x / 100 z = w - y A = u / z B = n - A
a ) 11.1 % , b ) 12.1 % , c ) 13.1 % , d ) 15.1 % , e ) 16.1 %
e
subtract(const_100, divide(multiply(900, const_100), 775))
an article is bought for rs . 775 and sold for rs . 900 , find the gain percent ?
"775 - - - - 125 100 - - - - ? = > 16.1 % answer : e"
a = 900 * 100 b = a / 775 c = 100 - b
a ) 336 , b ) 346 , c ) 356 , d ) 366 , e ) 504
e
divide(factorial(9), factorial(subtract(9, 3)))
9 ! / ( 9 - 3 ) ! = ?
9 ! / ( 9 - 3 ) ! = 9 ! / 6 ! = 9 * 8 * 7 = 504 . hence , the correct answer is e .
a = math.factorial(9) b = 9 - 3 c = math.factorial(b) d = a / c
a ) 0.3408 , b ) 3.408 , c ) 34.08 , d ) 340.8 , e ) none of these
c
multiply(1.6, 21.3)
if 213 Γ— 16 = 3408 , then 1.6 Γ— 21.3 is equal to :
"solution 1.6 Γ— 21.3 = ( 16 / 10 x 213 / 10 ) = ( 16 x 213 / 100 ) = 3408 / 100 = 34.08 . answer c"
a = 1 * 6
a ) rs . 1287.76 , b ) rs . 1284.76 , c ) rs . 1587.76 , d ) rs . 1266.76 , e ) rs . 1283.76
a
subtract(add(add(divide(multiply(divide(900, multiply(divide(18, const_100), 5)), 18), const_100), divide(900, multiply(divide(18, const_100), 5))), divide(multiply(add(divide(multiply(divide(900, multiply(divide(18, const_100), 5)), 18), const_100), divide(900, multiply(divide(18, const_100), 5))), 18), const_100)), divide(900, multiply(divide(18, const_100), 5)))
if the simple interest on a sum of money for 5 years at 18 % per annum is rs . 900 , what is the compound interest on the same sum at the rate and for the same time ?
"sum = ( 900 * 100 ) / ( 5 * 18 ) = rs . 1 , 000.00 c . i . on rs . rs . 1 , 000.00 for 5 years at 18 % = rs . 2 , 287.76 . = rs . 2 , 287.76 - 1 , 000.00 = rs . 1287.76 answer : a"
a = 18 / 100 b = a * 5 c = 900 / b d = c * 18 e = d / 100 f = 18 / 100 g = f * 5 h = 900 / g i = e + h j = 18 / 100 k = j * 5 l = 900 / k m = l * 18 n = m / 100 o = 18 / 100 p = o * 5 q = 900 / p r = n + q s = r * 18 t = s / 100 u = i + t v = 18 / 100 w = v * 5 x = 900 / w y = u - x
a ) 1200 , b ) 1764 , c ) 1400 , d ) 1354 , e ) none of these
c
add(divide(150, divide(multiply(divide(9, multiply(const_4, const_3)), 16), const_100)), 150)
the true discount on a bill due 9 months hence at 16 % per annum is rs . 150 . the amount of the bill is
explanation : let p . w . be rs . x . then , s . i . on rs . x at 16 % for 9 months = rs . 150 . x Γ£ β€” 16 Γ£ β€” ( 9 / 12 ) Γ£ β€” ( 1 / 100 ) = 150 or x = 1250 . p . w . = rs . 1250 sum due = p . w . + t . d . = rs . ( 1250 150 ) = rs . 1400 answer : c
a = 4 * 3 b = 9 / a c = b * 16 d = c / 100 e = 150 / d f = e + 150
a ) $ 1 , b ) $ 1.81 , c ) $ 1.92 , d ) $ 2.13 , e ) $ 2.15
b
min(divide(add(add(add(add(add(add(1.75, 1.61), 1.79), 2.11), 1.96), 2.09), 1.81), 7), 1.81)
andrew travelling to 7 cities . gasoline prices varied from city to city . $ 1.75 , $ 1.61 , $ 1.79 , $ 2.11 , $ 1.96 , $ 2.09 , $ 1.81 . what is the median gasoline price ?
"ordering the data from least to greatest , we get : $ 1.61 , $ 1.75 , $ 1.79 , $ 1.81 , $ 1.96 , $ 2.09 , $ 2.11 the median gasoline price is $ 1.81 . ( there were 3 states with higher gasoline prices and 3 with lower prices . ) b"
a = 1 + 75 b = a + 1 c = b + 2 d = c + 1 e = d + 2 f = e + 1 g = f / 7 h = min(g)
a ) 59 , b ) 85 , c ) 190 , d ) 215 , e ) 210
a
subtract(divide(add(150, 280), const_2), divide(subtract(subtract(1100, 200), add(150, 280)), 3))
a freight elevator can carry a maximum load of 1100 pounds . sean , who weighs 200 pounds , is in the elevator with two packages weighing 150 pounds and 280 pounds . if he needs to fit 3 more packages in the elevator that weigh as much as possible without exceeding the elevator limit , what is the difference between their average and the average of the two packages already in the elevator ?
the average of existing 2 package is 150 + 280 / 2 = 430 / 2 = 215 remaining allowed weight = 1100 - 200 - 430 = 470 . allowed per package = 470 / 3 = 156 so difference in average of existing and allowable = 215 - 156 = 59 hence a
a = 150 + 280 b = a / 2 c = 1100 - 200 d = 150 + 280 e = c - d f = e / 3 g = b - f
a ) 25 , b ) 18 , c ) 11 , d ) 38 , e ) 0
d
subtract(subtract(75, 15), 22)
power windows : 60 % anti - lock brakes : 25 % cd player : 75 % the table above shows the number of vehicles at bill ' s car dealership that have certain features . no vehicle has all 3 features , but 10 % have power windows and anti - lock brakes , 15 % have anti - lock brakes and a cd player , and 22 % have power windows and a cd player . what percent of the vehicles at bill ' s car dealership have a cd player but no power windows or anti - lock brakes ?
answer : d were looking for the number of cars with a cd player but no other features . we know that 40 % of the cars have a cd player , 15 % have a cd player and anti - lock brakes , while 22 % have a cd player and power windows . since no car has all three features , those account for all of the possibilities except for what were looking for . if a car has a cd player , it must either have anti - lock brakes , power windows , or no other features . since the total of cars with a cd player is 75 % , we can set up the following equation : 75 = 15 + 22 + x x = 38 , choiced .
a = 75 - 15 b = a - 22
a ) 40 sec , b ) 30 sec , c ) 26 sec , d ) 27 sec , e ) 34 sec
b
divide(360, multiply(subtract(60, 140), const_0_2778))
a train 360 m long is running at a speed of 60 km / hr . in what time will it pass a bridge 140 m long ?
"speed = 60 * 5 / 18 = 50 / 3 m / sec total distance covered = 360 + 140 = 500 m required time = 500 * 3 / 50 = 30 sec answer : b"
a = 60 - 140 b = a * const_0_2778 c = 360 / b
a ) 7 / 2 , b ) 19 / 6 , c ) 5 / 3 , d ) 7 , e ) 8
c
divide(log(32), log(8))
the value of ( log 9 27 + log 8 32 ) is :
let log 9 27 = x . then , 9 x = 27 = > ( 32 ) x = 33 = > 2 x = 3 = > x = 3 / 2 let log 8 32 = y . then 8 y = 32 = > ( 23 ) y = 25 = > 3 y = 5 = > y = 5 / 3 answer : c
a = math.log(32) b = math.log(8) c = a / b
a ) 41 - 43 , b ) 39 - 41 , c ) 38 - 40 , d ) 37 - 39 , e ) 36 - 37
a
add(multiply(37.50, divide(15, const_100)), 37.50)
a meal cost $ 37.50 and there was no tax . if the tip was more than 10 pc but less than 15 pc of the price , then the total amount paid should be :
"10 % ( 37.5 ) = 3.75 15 % ( 37.5 ) = 5.625 total amount could have been 37.5 + 3.75 and 37.5 + 5.625 = > could have been between 41.25 and 43.125 = > approximately between 41 and 43 answer is a ."
a = 15 / 100 b = 37 * 50 c = b + 37
a ) 1 , b ) 2 , c ) - 4 , d ) 4 , e ) 5
c
divide(1, 4)
find the slope of the line perpendicular to the line y = ( 1 / 4 ) x - 7
"two lines are perpendicular if the product of their slopes is equal to - 1 . the slope of the given line is equal to 1 / 4 . if m is the slope of the line perpendicular to the given line , then m Γ— ( 1 / 4 ) = - 1 solve for m m = - 4 correct answer c ) - 4"
a = 1 / 4
a ) 60 km , b ) 87 km , c ) 33 km , d ) 67 km , e ) 20 km
c
divide(multiply(multiply(subtract(9, 1), add(9, 1)), 7), add(add(9, 1), subtract(9, 1)))
a person can row at 9 kmph and still water . he takes 7 1 / 2 hours to row from a to b and back . what is the distance between a and b if the speed of the stream is 1 kmph ?
"let the distance between a and b be x km . total time = x / ( 9 + 1 ) + x / ( 9 - 1 ) = 7.5 = > x / 10 + x / 8 = 15 / 2 = > ( 4 x + 5 x ) / 40 = 15 / 2 = > x = 33 km . answer : c"
a = 9 - 1 b = 9 + 1 c = a * b d = c * 7 e = 9 + 1 f = 9 - 1 g = e + f h = d / g
a ) 16 min , b ) 26 min , c ) 34 min , d ) 20 min , e ) 12 min
a
multiply(divide(4, divide(5, 4)), 5)
walking with 4 / 5 of my usual speed , i miss the bus by 4 minutes . what is my usual time ?
"speed ratio = 1 : 4 / 5 = 5 : 4 time ratio = 4 : 5 1 - - - - - - - - 4 4 - - - - - - - - - ? Γ¨ 16 answer : a"
a = 5 / 4 b = 4 / a c = b * 5
a ) s . 520 , b ) s . 480 , c ) s . 420 , d ) s . 100 , e ) s . 200
d
divide(multiply(subtract(22, 20), const_100), const_2)
find the principal which yields a simple interest of rs . 20 and compound interest of rs . 22 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 . 22 % rate per annum = [ ( ci – si ) / ( si in 1 year ) ] * 100 = [ ( 22 – 20 ) / 20 ] * 100 = 10 % p . a . let the principal be rs . x time = t = 2 years % rate = 10 % p . a . si = ( prt / 100 ) 20 = ( x * 10 * 2 ) / 100 x = rs . 100 answer : d
a = 22 - 20 b = a * 100 c = b / 2
a ) 22 , b ) 33 , c ) 44 , d ) 66 , e ) 99
d
multiply(multiply(const_3, 11), const_2)
if x / ( 11 p ) is an odd prime number , where x is a positive integer and p is a prime number , what is the least value of x ?
x / ( 11 p ) = odd prime number x = odd prime number * 11 p least value of x = lowest odd prime number * 11 * lowest value of p = 3 * 11 * 2 = 66 answer d
a = 3 * 11 b = a * 2
a ) rs . 115 , b ) rs . 116 , c ) rs . 117 , d ) rs . 118 , e ) none
c
add(add(12, 9), multiply(8, 12))
9 persons went to a hotel for taking their meals . 8 of them spent rs . 12 each on their meals and the ninth spent rs . 8 more than the average expenditure of all the 9 . what was the total money spent by them .
solution : let the average expenditure of all the nine be x . then , 12 Γ— 8 + ( x + 8 ) = 9 x . therefore , x = 13 . total money spent , = 9 x = rs . ( 9 Γ— 13 ) = rs . 117 answer : option c
a = 12 + 9 b = 8 * 12 c = a + b
a ) 1 : 25 , b ) 1 : 36 , c ) 1 : 52 , d ) 1 : 522 , e ) none
b
divide(const_4, const_100)
a cube of edge 6 cm is cut into cubes each of edge 1 cm . the ratio of the total surface area of one of the small cubes to that of the large cube is equal to :
"sol . required ratio = 6 * 1 * 1 / 6 * 6 * 6 = 1 / 36 = 1 : 36 . answer b"
a = 4 / 100
a ) 4 , b ) 5 , c ) 6 , d ) 8 , e ) 9
a
divide(36, subtract(const_10, const_1))
the difference between a two - digit number and the number obtained by interchanging the positions of its digits is 36 . what is the difference between the two digits of that number ?
"sol . let the ten ’ s digit be x and unit ’ s digit be y , then , ( 10 x + y ) - ( 10 y + x ) = 36 ⇔ 9 ( x - y ) = 36 ⇔ x - y = 4 answer a"
a = 10 - 1 b = 36 / a
a ) 42 , b ) 63 , c ) 77 , d ) 80 , e ) 91
b
subtract(66, 4)
a snail , climbing a 66 feet high wall , climbs up 4 feet on the first day but slides down 2 feet on the second . it climbs 4 feet on the third day and slides down again 2 feet on the fourth day . if this pattern continues , how many days will it take the snail to reach the top of the wall ?
"total transaction in two days = 4 - 2 = 2 feet in 62 days it will climb 62 feet on the 63 rd day , the snail will climb 4 feet , thus reaching the top therefore , total no of days required = 63 b"
a = 66 - 4
a ) 5 , b ) 6 , c ) 4 , d ) 3 , e ) 2
c
divide(divide(360, 15), 6)
if 6 pr = 360 and if 6 cr = 15 , find r ?
solution : npr = ncr οΏ½ r ! 6 pr = 15 * r ! 360 = 15 * r ! r ! = 360 / 15 = 24 r ! = 4 * 3 * 2 * 1 = > r ! = 4 ! therefore , r = 4 . answer : option c
a = 360 / 15 b = a / 6
a ) 16.66 % , b ) 17.66 % , c ) 18.66 % , d ) 19.66 % , e ) 20.66 %
a
multiply(divide(15, 90), const_100)
by selling 90 pens , a trader gains the cost of 15 pens . find his gain percentage ?
"let the cp of each pen be rs . 1 . cp of 90 pens = rs . 90 profit = cost of 15 pens = rs . 15 profit % = 15 / 90 * 100 = 16.66 % answer : a"
a = 15 / 90 b = a * 100
a ) 99 , b ) 18 , c ) 135 , d ) 17 , e ) 12
c
divide(rectangle_area(30, 18), rectangle_area(2, 2))
how many paying stones , each measuring 2 m * 2 m are required to pave a rectangular court yard 30 m long and 18 m board ?
"30 * 18 = 2 * 2 * x = > x = 135 answer : c"
a = rectangle_area / (
a ) 14 sec , b ) 67 sec , c ) 98 sec , d ) 36 sec , e ) 23 sec
a
divide(add(250, 100), multiply(subtract(99, 9), divide(divide(const_10, const_2), divide(subtract(99, 9), const_2))))
a jogger running at 9 km / hr along side a railway track is 250 m ahead of the engine of a 100 m long train running at 99 km / hr in the same direction . in how much time will the train pass the jogger ?
"speed of train relative to jogger = 99 - 9 = 90 km / hr . = 90 * 5 / 18 = 25 m / sec . distance to be covered = 250 + 100 = 350 m . time taken = 350 / 25 = 14 sec . answer : a"
a = 250 + 100 b = 99 - 9 c = 10 / 2 d = 99 - 9 e = d / 2 f = c / e g = b * f h = a / g
a ) 36 , b ) 50 , c ) 88 , d ) 66 , e ) 23
e
divide(divide(subtract(200, multiply(multiply(5, const_0_2778), 5)), 5), const_0_2778)
a train 200 m long passes a man , running at 5 km / hr in the same direction in which the train is going , in 20 seconds . the speed of the train is ?
"speed of the train relative to man = ( 200 / 20 ) m / sec = 5 m / sec . [ 5 * ( 18 / 5 ) ] km / hr = 18 km / hr . let the speed of the train be x km / hr . then , relative speed = ( x - 5 ) km / hr . x - 5 = 18 = = > x = 23 km / hr . answer : e"
a = 5 * const_0_2778 b = a * 5 c = 200 - b d = c / 5 e = d / const_0_2778
a ) 18 , b ) 82 , c ) 16 , d ) 27 , e ) 29
c
multiply(subtract(73, 65), const_2)
a pupil ' s marks were wrongly entered as 73 instead of 65 . due to the average marks for the class got increased by half . the number of pupils in the class is ?
"let there be x pupils in the class . total increase in marks = ( x * 1 / 2 ) = x / 2 x / 2 = ( 73 - 65 ) = > x / 2 = 8 = > x = 16 . answer : c"
a = 73 - 65 b = a * 2
a ) $ 18,714 , b ) $ 18,500 , c ) $ 18,000 , d ) $ 15,850 , e ) $ 12,300
a
divide(subtract(subtract(multiply(multiply(5, 4), multiply(const_4, const_4)), multiply(multiply(5, 5), 5)), multiply(4, 16)), add(const_2, 5))
the average salary of 16 people in the shipping department at a certain firm is $ 20,000 . the salary of 5 of the employees is $ 25,000 each and the salary of 4 of the employees is $ 16,000 each . what is the average salary of the remaining employees ?
total salary . . . 16 * 20 k = 320 k 5 emp @ 25 k = 125 k 4 emp @ 16 k = 64 k remaing 7 emp sal = 320 k - 125 k - 64 k = 131 k average = 131 k / 7 = 18714 ans : a
a = 5 * 4 b = 4 * 4 c = a * b d = 5 * 5 e = d * 5 f = c - e g = 4 * 16 h = f - g i = 2 + 5 j = h / i
a ) $ 120 , b ) $ 100 , c ) $ 91 , d ) $ 77 , e ) $ 69
d
multiply(100, divide(100, add(100, 30)))
a shopkeeper sold an article at $ 100 with 30 % profit . then find its cost price ?
"cost price = selling price * 100 / ( 100 + profit ) c . p . = 100 * 100 / 130 = $ 77 ( approximately ) answer is d"
a = 100 + 30 b = 100 / a c = 100 * b
a ) 1600 , b ) 1800 , c ) 2000 , d ) 2200 , e ) 2400
e
divide(multiply(480, const_100), subtract(const_100, 80))
in an election between two candidates , the first candidate got 80 % of the votes and the second candidate got 480 votes . what was the total number of votes ?
"let v be the total number of votes . 0.2 v = 480 v = 2400 the answer is e ."
a = 480 * 100 b = 100 - 80 c = a / b
a ) 12 % , b ) 12.5 % , c ) 67 % , d ) 25 % , e ) 33 %
d
multiply(divide(divide(subtract(500, 400), 2), subtract(400, multiply(divide(subtract(500, 400), 2), 4))), const_100)
peter invested a certain sum of money in a simple interest bond whose value grew to $ 400 at the end of 4 years and to $ 500 at the end of another 2 years . what was the rate of interest in which he invested his sum ?
"lets assume the principal amount ( initial amount invested ) to be p rate of interest to berand time as t . we need to find r now after a time of 4 years the principal p amounts to $ 400 and after a time of 6 years ( question says after another 6 years so 4 + 2 ) p becomes $ 500 . formulating the above data amount ( a 1 ) at end of 4 years a 1 = p ( 1 + 4 r / 100 ) = 400 amount ( a 2 ) at end of 6 years a 2 = p ( 1 + 2 r / 100 ) = 500 dividing a 2 by a 1 we get ( 1 + 4 r / 100 ) / ( 1 + 2 r / 100 ) = 5 / 4 after cross multiplication we are left with 4 r = 100 which gives r = 25 % option : d"
a = 500 - 400 b = a / 2 c = 500 - 400 d = c / 2 e = d * 4 f = 400 - e g = b / f h = g * 100
['a ) 1299', 'b ) 1305', 'c ) 1311', 'd ) 1317', 'e ) 1323']
e
subtract(divide(divide(multiply(multiply(21, subtract(21, const_1)), subtract(subtract(21, const_1), const_1)), const_3), const_2), add(const_3, const_4))
twenty - one dots are evenly spaced on the circumference of a circle . how many combinations of three dots can we make from these 21 dots that do not form an equilateral triangle ?
the total number of ways we can choose three dots is 21 c 3 = 1330 . we can form seven equilateral triangles from these twenty - one dots . there are 1330 - 7 = 1323 combinations which do not form an equilateral triangle . the answer is e .
a = 21 - 1 b = 21 * a c = 21 - 1 d = c - 1 e = b * d f = e / 3 g = f / 2 h = 3 + 4 i = g - h
a ) 48 , b ) 64 , c ) 70 , d ) 80 , e ) 84
d
subtract(divide(multiply(divide(560, 7), 16), divide(2, add(1, 2))), add(560, multiply(divide(560, 7), 16)))
the ratio of boarders to day scholars at a school is 7 to 16 . however , after a few new students join the initial 560 boarders , the ratio changed to 1 to 2 , respectively . if no boarders became day scholars and vice versa , and no students left the school , how many boarders joined the school ?
let day scholars be ds and boarders be b given b / ds = 7 / 16 - - - > eq 1 initially b = 560 and x be new boarders joined then from eq 1 , we get ds = 16 / 7 * 560 = 1280 now , new ration is 560 + x / ds = 1 / 2 560 + x / 1280 = 1 / 2 = > 1120 + 2 x = 1280 = > 2 x = 160 = > x = 80 ans option d
a = 560 / 7 b = a * 16 c = 1 + 2 d = 2 / c e = b / d f = 560 / 7 g = f * 16 h = 560 + g i = e - h
a ) 151 , b ) 331 , c ) 168 , d ) 691 , e ) 871
c
multiply(18, divide(subtract(add(multiply(15, 2), 3), 6), subtract(18, 15)))
when positive integer n is divided by positive integer p , the quotient is 18 , with a remainder of 6 . when n is divided by ( p + 2 ) , the quotient is 15 and the remainder is 3 . what is the value of n ?
n / p = 18 6 / p = 18 p + 6 n / ( p + 2 ) = 15 2 / ( p + 2 ) = 15 p + 30 + 3 solving these two equations we get p = 9 n = 168 answer is c .
a = 15 * 2 b = a + 3 c = b - 6 d = 18 - 15 e = c / d f = 18 * e
a ) 10 % , b ) 20 % , c ) 60 % , d ) 80 % , e ) 50 %
d
subtract(const_100, subtract(add(25, 35), 40))
in an examination , 25 % of total students failed in hindi , 35 % failed in english and 40 % in both . the percentage of these who passed in both the subjects is :
"pass percentage = 100 - ( 25 + 35 - 40 ) = 100 - 20 = 80 answer : d"
a = 25 + 35 b = a - 40 c = 100 - b
a ) 354354 , b ) 545454 , c ) 465785 , d ) 456573 , e ) 2818075 / 2
e
multiply(1400000, multiply(multiply(add(const_1, divide(15, const_100)), subtract(const_1, divide(35, const_100))), add(const_1, divide(35, const_100))))
population of a city in 20004 was 1400000 . if in 2005 there isan increment of 15 % , in 2006 there is a decrements of 35 % and in 2007 there is an increment of 45 % , then find the population of city at the end of the year 2007
"required population = p ( 1 + r 1 / 100 ) ( 1 - r 2 / 100 ) ( 1 + r 3 / 100 ) = p ( 1 + 15 / 100 ) ( 1 - 35 / 100 ) ( 1 + 45 / 100 ) = 2818075 / 2 e"
a = 15 / 100 b = 1 + a c = 35 / 100 d = 1 - c e = b * d f = 35 / 100 g = 1 + f h = e * g i = 1400000 * h
a ) 26 , b ) 134 , c ) 141 , d ) 190 , e ) 28
c
multiply(multiply(const_pi, 3), 15)
the slant height of a cone is 15 cm and radius of the base is 3 cm , find the curved surface of the cone ?
"Ο€ * 15 * 3 = 141 answer : c"
a = math.pi * 3 b = a * 15
a ) 56 sec , b ) 42 sec , c ) 45 sec , d ) 48 sec , e ) 50 sec
a
divide(360, multiply(subtract(45, 340), const_0_2778))
a train 360 m long is running at a speed of 45 km / hr . in what time will it pass a bridge 340 m long ?
"speed = 45 * 5 / 18 = 25 / 2 m / sec total distance covered = 360 + 340 = 700 m required time = 700 * 2 / 25 = 56 sec answer : a"
a = 45 - 340 b = a * const_0_2778 c = 360 / b
a ) 54 , b ) 55 , c ) 53 , d ) 56 , e ) 57
a
add(55, subtract(55, 56))
if ( 55 ^ 55 + 55 ) is divided by 56 , then the remainder is : ?
solution : ( x ^ n + 1 ) is divisible by ( x + 1 ) , when n is odd . . ' . ( 55 ^ 55 + 1 ) is divisible by ( 55 + 1 ) = 56 . when ( 55 ^ 55 + 1 ) + 54 is divided by 56 , the remainder is 54 . answer a
a = 55 - 56 b = 55 + a
a ) 22 , b ) 24 , c ) 28 , d ) 32 , e ) 44
b
multiply(2, 12)
g ( x ) is defined as the product of all even integers k such that 0 < k ≀ x . for example , g ( 14 ) = 2 Γ— 4 Γ— 6 Γ— 8 Γ— 10 Γ— 12 Γ— 14 . if g ( a ) is divisible by 4 ^ 11 , what is the smallest possible value for a ?
"g ( a ) = 4 ^ 11 = 2 ^ 22 . so we have to find a product with atleast 22 2 ' s in it . in option 1 22 the total no of 2 ' s = [ 22 / 2 ] + [ 22 / 4 ] + [ 22 / 8 ] + [ 22 / 16 ] = 11 + 5 + 2 + 1 = 19 in option 2 24 the total no of 2 ' s = [ 24 / 2 ] + [ 24 / 4 ] + [ 24 / 8 ] + [ 24 / 16 ] = 12 + 6 + 3 + 1 = 22 . hence b"
a = 2 * 12
a ) 115 , b ) 230 , c ) 460 , d ) 575 , e ) 690
b
divide(690, 3)
the least common multiple of positive integer m and 3 - digit integer n is 690 . if n is not divisible by 3 and m is not divisible by 2 , what is the value of n ?
the lcm of n and m is 690 = 2 * 3 * 5 * 23 . m is not divisible by 2 , thus 2 goes to n n is not divisible by 3 , thus 3 goes to m . from above : n must be divisible by 2 and not divisible by 3 : n = 2 * . . . in order n to be a 3 - digit number it must take all other primes too : n = 2 * 5 * 23 = 230 . answer : b .
a = 690 / 3
a ) rs . 50 , b ) rs . 160 , c ) rs . 200 , d ) rs . 225 , e ) rs . 325
c
divide(subtract(350, 340), divide(5, const_100))
if 5 % more is gained by selling an article for rs . 350 than by selling it for rs . 340 , the cost of the article is :
"let c . p . be rs . x . then , 5 % of x = 350 - 340 = 10 x / 20 = 10 = > x = 200 answer : c"
a = 350 - 340 b = 5 / 100 c = a / b
a ) 16 % , b ) 32 % , c ) 48 % , d ) 84 % , e ) 92 %
d
subtract(const_100, divide(subtract(const_100, 68), const_2))
a certain characteristic in a large population has a distribution that is symmetric about the mean m . if 68 percent of the distribution lies within one standard deviation d of the mean , what percent f of the distribution is less than m + d ?
"d the prompt says that 68 % of the population lies between m - d and m + d . thus , 32 % of the population is less than m - d or greater than m + d . since the population is symmetric , half of this 32 % is less than m - d and half is greater than m + d . thus , f = ( 68 + 16 ) % or ( 100 - 16 ) % of the population is less than m + d . d"
a = 100 - 68 b = a / 2 c = 100 - b