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 ) 12 , b ) 8 , c ) 36 , d ) 5 , e ) 7 | e | divide(add(divide(30, 5), divide(40, 5)), const_2) | a man swims downstream 40 km and upstream 30 km taking 5 hours each time , what is the speed of the man in still water ? | "40 - - - 5 ds = 8 ? - - - - 1 30 - - - - 5 us = 6 ? - - - - 1 m = ? m = ( 8 + 6 ) / 2 = 7 answer : e" | a = 30 / 5
b = 40 / 5
c = a + b
d = c / 2
|
a ) 36 , b ) 53 , c ) 88 , d ) 66 , e ) 22 | b | divide(divide(subtract(125, multiply(multiply(8, const_0_2778), 8)), 8), const_0_2778) | a train 125 m long passes a man , running at 8 km / hr in the same direction in which the train is going , in 10 seconds . the speed of the train is ? | "speed of the train relative to man = ( 125 / 10 ) m / sec = ( 25 / 2 ) m / sec . [ ( 25 / 2 ) * ( 18 / 5 ) ] km / hr = 45 km / hr . let the speed of the train be x km / hr . then , relative speed = ( x - 8 ) km / hr . x - 8 = 45 = = > x = 53 km / hr . answer : b" | a = 8 * const_0_2778
b = a * 8
c = 125 - b
d = c / 8
e = d / const_0_2778
|
a ) 13 β 5 , b ) 13 β 2 , c ) 13 β 9 , d ) 12 β 9 , e ) 13 β 1 | b | divide(24, divide(24, const_100)) | 85 % of a number is added to 24 , the result is the same number . find the number ? | "4 a = 48 4 a = 20 a = 12 a = 5 a 2 = 144 a 2 = 25 combined area = a 2 = 169 = > a = 13 d = 13 β 2 answer : b" | a = 24 / 100
b = 24 / a
|
a ) 20 , 10 , b ) 10 , 20 , c ) 20 , 30 , d ) 30 , 20 , e ) 10 , 10 | b | sqrt(divide(10, add(power(2, 1), add(power(2, 1), power(1, 1))))) | the ratio of 2 numbers is 1 : 2 and their h . c . f . and l . c . m are 10 and 20 . find the numbers . | "let the numbers be x and 2 x their h . c . f . = 10 l . c . m = 20 ie ) 2 x ^ 2 = 200 x ^ 2 = 100 , x = 10 t the numbers are x and 2 x , ie ) 10 , 20 respectively . answer is b" | a = 2 ** 1
b = 2 ** 1
c = 1 ** 1
d = b + c
e = a + d
f = 10 / e
g = math.sqrt(f)
|
a ) 68 , b ) 70.4 , c ) 86 , d ) 105.6 , e ) 108 | d | add(88, multiply(divide(20, const_100), 88)) | if x is 20 percent greater than 88 , then x = | "x is 20 % greater than 88 means x is 1.2 times 88 ( in other words 88 + 20 / 100 * 88 = 1.2 * 88 ) therefore , x = 1.2 * 88 = 105.6 answer : d" | a = 20 / 100
b = a * 88
c = 88 + b
|
a ) 12 , b ) 13 , c ) 14 , d ) 17 , e ) 23 | c | subtract(subtract(36, const_2), add(add(15, const_4), const_1)) | set a consists of all the prime numbers between 15 and 36 . what is the range of set a ? | "the range of a set of data is the difference between the highest and lowest values in the set in this set , highest number = 31 lowest number = 17 range = highest - lowest = 31 - 17 = 14 option c" | a = 36 - 2
b = 15 + 4
c = b + 1
d = a - c
|
a ) 3 , b ) 5 , c ) 10 , d ) 9 , e ) 15 | e | divide(add(21, 9), const_2) | in one hour , a boat goes 21 km along the stream and 9 km against the stream . the speed of the boat in still water ( in km / hr ) is : | "solution speed in still water = 1 / 2 ( 21 + 9 ) kmph . = 15 kmph . answer e" | a = 21 + 9
b = a / 2
|
a ) 515 . , b ) 545 . , c ) 618 . , d ) 644 . , e ) 736 . | e | add(226, 423) | in the faculty of reverse - engineering , 226 second year students study numeric methods , 423 second year students study automatic control of airborne vehicles and 134 second year students study them both . how many students are there in the faculty if the second year students are approximately 70 % of the total ? | "answer is e : 736 solution : total number of students studying both are 423 + 226 - 134 = 515 ( subtracting the 134 since they were included in the both the other numbers already ) . so 70 % of total is 515 , so 100 % is approx . 736 ." | a = 226 + 423
|
a ) 48 , b ) 42 , c ) 45 , d ) 98 , e ) 96 | e | divide(multiply(24, divide(add(6, 12), 12)), 12) | machines x and y produce bottles at their respective constant rates . machine x produces k bottles in 6 hours and machine y produces k bottles in 12 hours . how many hours does it take machines x and y , working simultaneously , to produce 24 k bottles ? | "x rate = k / 6 y rate = k / 12 k / 6 + k / 12 = 24 k / t solving t = 96 answer e" | a = 6 + 12
b = a / 12
c = 24 * b
d = c / 12
|
a ) - 1 , b ) - 1 / 2 , c ) 0 , d ) 1 / 2 , e ) 1 | e | add(negate(4), 4) | on the number line , if x is halfway between - 4 and 4 , and if y is halfway between - 2 and 6 , what number is halfway between x and y ? | "x = 0 and y = 2 . the answer is e ." | a = negate + (
|
a ) 2988 , b ) 2776 , c ) 4079 , d ) 2871 , e ) 881 | c | floor(divide(3294, multiply(divide(subtract(const_100, 5), const_100), divide(subtract(const_100, 15), const_100)))) | 5 % people of a village in sri lanka died by bombardment , 15 % of the remainder left the village on account of fear . if now the population is reduced to 3294 , how much was it in the beginning ? | x * ( 95 / 100 ) * ( 85 / 100 ) = 3294 x = 4079 answer : c | a = 100 - 5
b = a / 100
c = 100 - 15
d = c / 100
e = b * d
f = 3294 / e
g = math.floor(f)
|
a ) 48 % , b ) 80 % , c ) 100 % , d ) 120 % , e ) 125 % | a | multiply(multiply(power(divide(8, 10), const_2), divide(6, 8)), const_100) | tanks a and b are each in the shape of a right circular cylinder . the interior of tank a has a height of 6 meters and a circumference of 8 meters , and the interior of tank b has a height of 8 meters and a circumference of 10 meters . the capacity of tank a is what percent of the capacity of tank b ? | "for a , r = 8 / 2 pi . its capacity = ( 4 pi ) ^ 2 * 6 = 96 pi for b , r = 10 / pi . its capacity = ( 5 pi ) ^ 2 * 8 = 200 pi a / b = 96 pi / 200 pi = 0.48 a" | a = 8 / 10
b = a ** 2
c = 6 / 8
d = b * c
e = d * 100
|
a ) 1 : 2 , b ) 2 : 5 , c ) 1 : 3 , d ) 2 : 3 , e ) 3 : 5 | b | divide(subtract(const_100, 2), add(const_100, 12)) | the number of singles that a baseball player hit increased 12 percent from her first season to her second season , and the number of doubles that she hit in the same time period decreased by 2 percent . if the total number of singles and doubles that she hit increased 2 percent from her first season to her second season , what is the ratio of the number of singles she hit in her first season to the number of doubles she hit in her first season ? | "soln : - 1.12 s + 0.98 d = 1.02 [ s + d ] 0.10 s = 0.04 d s / d = 2 / 5 answer : b" | a = 100 - 2
b = 100 + 12
c = a / b
|
a ) 237 , b ) 287 , c ) 197 , d ) 287 , e ) 700 | e | multiply(350, const_2) | on the independence day , bananas were be equally distributed among the children in a school so that each child would get two bananas . on the particular day 350 children were absent and as a result each child got two extra bananas . find the actual number of children in the school ? | "explanation : let the number of children in the school be x . since each child gets 2 bananas , total number of bananas = 2 x . 2 x / ( x - 350 ) = 2 + 2 ( extra ) = > 2 x - 700 = x = > x = 700 . answer : e" | a = 350 * 2
|
a ) 10 , b ) 11 , c ) 1001 , d ) 1011 , e ) 10001 | c | add(divide(100100000, 10010000), const_1000) | every digit of a number written in binary is either 0 or 1 . to translate a number from binary , multiply the nth digit ( reading from right to left ) by 2 ^ ( n - 1 ) what is the largest prime number ( written in binary ) that is a factor of both 10010000 and 100100000 ? | binary divison can provide a quick answer if you are comfortable with it . as option e is the biggest binary number we try with it first : 100010000 / 1001 = 10000 1000100000 / 1001 = 100000 so answer is option is c | a = 100100000 / 10010000
b = a + 1000
|
a ) 27 , b ) 20 , c ) 40 , d ) 73 , e ) 72 | c | inverse(subtract(divide(const_1, 8), divide(const_1, add(8, const_2)))) | a cistern is normally filled in 8 hours but takes two hours longer to fill because of a leak in its bottom . if the cistern is full , the leak will empty it in ? | "1 / 8 - 1 / x = 1 / 10 x = 40 answer : c" | a = 1 / 8
b = 8 + 2
c = 1 / b
d = a - c
e = 1/(d)
|
a ) 1 km , b ) 3 km , c ) 4 km , d ) 5 km , e ) 6 km | c | divide(multiply(24, divide(multiply(10, const_1000), const_60)), const_1000) | find the distance covered by a man walking for 24 min at a speed of 10 km / hr ? | "distance = 10 * 24 / 60 = 4 km answer is c" | a = 10 * 1000
b = a / const_60
c = 24 * b
d = c / 1000
|
a ) 69.44 , b ) 62.0 , c ) 62.6 , d ) 62.1 , e ) 62.2 | a | divide(multiply(add(50, divide(multiply(50, 25), const_100)), const_100), subtract(const_100, 10)) | at what price must an article costing rs . 50 be marked in order that after deducting 10 % from the list price . it may be sold at a profit of 25 % on the cost price ? | "cp = 50 sp = 50 * ( 125 / 100 ) = 62.50 mp * ( 90 / 100 ) = 62.50 mp = 69.44 answer : a" | a = 50 * 25
b = a / 100
c = 50 + b
d = c * 100
e = 100 - 10
f = d / e
|
a ) 45 , b ) 54 , c ) 63 , d ) 72 , e ) 81 | d | multiply(add(add(subtract(multiply(2, const_10), 2), 1), multiply(2, const_10)), 3) | every letter in the alphabet has a number value that is equal to its place in the alphabet . thus , the letter a has a value of 1 , the letter b has a value of 2 , the letter c has a value of 3 , etc . . . the number value of a word is obtained by adding up the value of the letters in the word and then multiplying that sum by the length of the word . what is the number value of the word ` ` cat ' ' ? | "` ` cat ' ' = ( 3 + 1 + 20 ) * 3 = 72 . the answer is d ." | a = 2 * 10
b = a - 2
c = b + 1
d = 2 * 10
e = c + d
f = e * 3
|
a ) 10 % , b ) 5 % , c ) 12 % , d ) 6 % , e ) 4 % | e | multiply(divide(subtract(add(multiply(divide(const_100, add(const_100, 20)), 512456), multiply(divide(const_100, subtract(const_100, 20)), 512456)), add(512456, 512456)), add(multiply(divide(const_100, add(const_100, 20)), 512456), multiply(divide(const_100, subtract(const_100, 20)), 512456))), const_100) | a man two flats for $ 512456 each . on one he gains 20 % while on the other he loses 20 % . how much does he gain or lose in the whole transaction ? | in such a case there is always a loss loss % = ( 20 / 10 ) ^ 2 = 4 % answer is e | a = 100 + 20
b = 100 / a
c = b * 512456
d = 100 - 20
e = 100 / d
f = e * 512456
g = c + f
h = 512456 + 512456
i = g - h
j = 100 + 20
k = 100 / j
l = k * 512456
m = 100 - 20
n = 100 / m
o = n * 512456
p = l + o
q = i / p
r = q * 100
|
a ) 80 kmph , b ) 60 kmph , c ) 70 kmph , d ) 90 kmph , e ) none of these | c | divide(add(225, 370), add(3.5, 5)) | a train travels 225 km in 3.5 hours and 370 km in 5 hours . find the average speed of train . | "as we know that speed = distance / time for average speed = total distance / total time taken thus , total distance = 225 + 370 = 595 km thus , total speed = 8.5 hrs or , average speed = 595 / 8.5 or , 70 kmph . answer : c" | a = 225 + 370
b = 3 + 5
c = a / b
|
a ) 51 : 52 , b ) 52 : 54 , c ) 53 : 54 , d ) 54 : 55 , e ) none of these | b | divide(add(const_100, 4), add(const_100, 8)) | the cash difference between the selling prices of an article at a profit of 4 % and 8 % is rs 3 . the ratio of two selling prices is | "explanation : let the cost price of article is rs . x required ratio = ( 104 % of x ) / ( 108 % of x ) = 104 / 108 = 52 / 54 = 52 : 54 . answer : b" | a = 100 + 4
b = 100 + 8
c = a / b
|
a ) - 3 , b ) 4 , c ) 3 , d ) - 5 , e ) 6 | a | divide(subtract(36, 6), 4) | if | 4 x + 6 | = 36 , then find the sum of the values of x ? | "| 4 x + 6 | = 36 4 x + 6 = 36 or 4 x + 6 = - 36 4 x = 30 or 4 x = - 42 x = 7.5 or x = - 10.5 sum = 7.5 - 10.5 = - 3 answer is a" | a = 36 - 6
b = a / 4
|
a ) 86.6 km , b ) 46.6 km , c ) 36.6 km , d ) 35.6 km , e ) 26.6 km | c | multiply(add(42, 8), divide(44, const_60)) | the speed of a boat in still water in 42 km / hr and the rate of current is 8 km / hr . the distance travelled downstream in 44 minutes is : | "speed downstream = ( 42 + 8 ) = 50 kmph time = 44 minutes = 44 / 60 hour = 11 / 15 hour distance travelled = time Γ speed = 11 / 15 Γ 50 = 36.6 km answer : c" | a = 42 + 8
b = 44 / const_60
c = a * b
|
a ) 15 % , b ) 35 % , c ) 25 % , d ) 50 % , e ) 30 % | b | subtract(divide(subtract(const_100, 10), divide(2, 3)), const_100) | an article is sold at a certain price . by selling it at 2 / 3 of that price one loses 10 % . find the gain percent at original price . | "let the original s . p be rs . x . then new s . p = rs . 2 / 3 x , loss = 10 % so c . p = rs . [ 100 / 90 * 2 / 3 x ] = 20 x / 27 . now c . p = rs . 20 x / 27 , s . p = rs . x , gain = rs . [ x - 20 x / 27 ] = rs . 7 x / 27 . gain % = [ 7 x / 27 * 27 / 20 x * 100 ] % = 35 % answer : b" | a = 100 - 10
b = 2 / 3
c = a / b
d = c - 100
|
a ) 100 , b ) 150 , c ) 120 , d ) 520 , e ) none | e | multiply(120, divide(520, 5)) | 120 + 520 Γ· 5.00 = ? | answer given expression = 120 + 520 Γ· 5.00 = 120 + 104 = 224 correct option : e | a = 520 / 5
b = 120 * a
|
a ) 1 / 9 , b ) 29 / 180 , c ) 26 / 143 , d ) 2 / 15 , e ) 39 / 121 | d | add(divide(const_1, divide(multiply(50, 4), const_10)), divide(const_1, divide(multiply(30, 6), 15))) | if 50 apprentices can finish a job in 4 hours , and 30 journeymen can finish the same job in 6 hours , how much of the job should be completed by 10 apprentices and 15 journeymen in one hour ? | 50 apprentices can finish the job in 4 hours , thus : 10 apprentices can finish the job in 4 * 5 = 20 hours ; in 1 hour 10 apprentices can finish 1 / 20 of the job . 30 journeymen can finish the same job in 4,5 hours , thus : 15 journeymen can finish the job in 6 * 2 = 12 hours ; in 1 hour 15 journeymen can finish 1 / 12 of the job . therefore , in 1 hour 10 apprentices and 15 journeymen can finish 1 / 20 + 1 / 12 = 2 / 15 of the job . answer : d . | a = 50 * 4
b = a / 10
c = 1 / b
d = 30 * 6
e = d / 15
f = 1 / e
g = c + f
|
a ) s . 59 , b ) s . 58 , c ) s . 09 , d ) s . 70 , e ) s . 13 | d | add(divide(36000, 600), 10) | a shopkeeper sells 600 metres of cloth for rs . 36000 at a loss of rs . 10 per metre . find his cost price for one metre of cloth ? | "sp per metre = 36000 / 600 = rs . 60 loss per metre = rs . 10 cp per metre = 60 + 10 = rs . 70 answer : d" | a = 36000 / 600
b = a + 10
|
a ) 50 hours , b ) 33 hours , c ) 72 hours , d ) 82 hours , e ) 92 hours | c | multiply(8, 9) | how long will take a leak at the bottom of a tank to empty it if it will take 8 hours to fill it without the leak , but due to the leak it will take 9 hours to be filled ? | part filled without leak in 1 hour = 1 / 8 part filled with leak in 1 hour = 1 / 9 work done by leak in 1 hour = 1 / 8 Γ’ Λ β 1 / 9 = 72 hours answer : c | a = 8 * 9
|
a ) 15 % , b ) 14.25 % , c ) 9.23 % , d ) 10.5 % , e ) 11.5 % | c | divide(multiply(const_100, subtract(subtract(340, divide(325, 2)), divide(325, 2))), divide(325, 2)) | on a sum of money , the simple interest for 2 years is rs . 325 , while the compound interest is rs . 340 , the rate of interest being the same in both the cases . the rate of interest is | "explanation : the difference between compound interest and simple interest on rs . p for 2 years at r % per annum = ( r Γ£ β si ) / ( 2 Γ£ β 100 ) difference between the compound interest and simple interest = 340 - 325 = 15 ( r Γ£ β si ) / ( 2 Γ£ β 100 ) = 15 ( r Γ£ β 325 ) / ( 2 Γ£ β 100 ) = 15 r = 9.23 % answer : option c" | a = 325 / 2
b = 340 - a
c = 325 / 2
d = b - c
e = 100 * d
f = 325 / 2
g = e / f
|
a ) s . 7 , b ) s . 9 , c ) s . 10 , d ) s . 11 , e ) s . 19 | c | multiply(divide(divide(multiply(divide(20, const_100), 400), 10), multiply(divide(20, const_100), 400)), const_100) | a reduction of 20 % in the price of salt enables a lady to obtain 10 kgs more for rs . 400 , find the original price per kg ? | 100 * ( 20 / 100 ) = 20 - - - 10 ? - - - 1 = > rs . 2 400 - - - 80 ? - - - 2 = > rs . 10 answer : c | a = 20 / 100
b = a * 400
c = b / 10
d = 20 / 100
e = d * 400
f = c / e
g = f * 100
|
a ) 2 , b ) 5 , c ) 4 , d ) 24 , e ) 35 | c | subtract(7, divide(add(multiply(7, const_2), 1), subtract(7, const_2))) | the product of a and b is equal to 1 more than twice the sum of a and b . if b = 7 , what is the value of b - a ? | "ab = 1 + 2 ( a + b ) 7 a = 1 + 2 a + 14 5 a = 15 a = 3 b - a = 7 - 3 = 4 c is the answer" | a = 7 * 2
b = a + 1
c = 7 - 2
d = b / c
e = 7 - d
|
a ) 15 cm , b ) 25 cm , c ) 35 cm , d ) 42 cm , e ) 45 cm | c | multiply(add(const_4, const_1), 7) | he greatest possible length which can be used to measure exactly the lengths 7 m , 3 m 85 cm , 12 m 95 cm is : | required length = h . c . f . of 700 cm , 385 cm and 1295 cm = 35 cm . answer : option c | a = 4 + 1
b = a * 7
|
a ) 1 / 2 , b ) 1 / 3 , c ) 1 / 4 , d ) 1 / 6 , e ) 1 / 12 | c | divide(multiply(divide(6, 12), const_100), multiply(divide(12, 6), const_100)) | 12 is 6 % of a , and 6 is 12 % of b . c equals b / a . what is the value of c ? | 6 a / 100 = 12 a = 200 12 b / 100 = 6 b = 50 c = b / a = 50 / 200 = 1 / 4 the answer is c . | a = 6 / 12
b = a * 100
c = 12 / 6
d = c * 100
e = b / d
|
a ) 3.5 , b ) 2.75 , c ) 4.25 , d ) 4.5 , e ) none of these | a | multiply(17, 17) | ( 17 ) 4.5 x ( 17 ) ? = 178 | "solution let ( 17 ) 4.5 * ( 17 ) x = 178 . then , ( 17 ) 4.5 + x = ( 17 ) 8 . β΄ 4.5 + x = 8 β x = ( 8 - 4.5 ) β x = 3.5 answer a" | a = 17 * 17
|
a ) 5 / 2 , b ) 8 / 3 , c ) 9 / 4 , d ) 10 / 3 , e ) 12 / 5 | c | divide(15, divide(const_1, add(divide(const_1, 10), divide(const_1, 20)))) | working alone , printers x , y , and z can do a certain printing job , consisting of a large number of pages , in 15 , 10 , and 20 hours , respectively . what is the ratio of the time it takes printer x to do the job , working alone at its rate , to the time it takes printers y and z to do the job , working together at their individual rates ? | "the time it takes printer x is 15 hours . the combined rate of y and z is 1 / 10 + 1 / 20 = 3 / 20 the time it takes y and z is 20 / 3 the ratio of times is 15 / ( 20 / 3 ) = 3 * 15 / 20 = 9 / 4 the answer is c ." | a = 1 / 10
b = 1 / 20
c = a + b
d = 1 / c
e = 15 / d
|
a ) 82 , b ) 17 , c ) 22 , d ) 82 , e ) 18 | c | subtract(const_60, multiply(const_60, divide(22, 35))) | excluding stoppages , the speed of a train is 35 kmph and including stoppages it is 22 kmph . of how many minutes does the train stop per hour ? | "explanation : t = 13 / 35 * 60 = 22 answer : option c" | a = 22 / 35
b = const_60 * a
c = const_60 - b
|
a ) 20 , b ) 25 , c ) 30 , d ) 35 , e ) 40 | b | add(add(choose(7, const_1), choose(7, const_1)), choose(const_4, const_1)) | jane and thomas are among the 7 people from which a committee of 3 people is to be selected . how many different possible committees of 3 people can be selected from these 7 people if at least one of either jane or thomas is to be selected ? | "the total number of ways to choose 3 people from 7 is 7 c 3 = 35 . the number of committees without jane or thomas is 5 c 3 = 10 . there are 35 - 10 = 25 possible committees which include jane and / or thomas . the answer is b ." | a = math.comb(7, 1)
b = math.comb(7, 1)
c = a + b
d = math.comb(4, 1)
e = c + d
|
a ) 1,200 , b ) 2,000 , c ) 2,100 , d ) 2,250 , e ) 2,540 | a | divide(540, 540) | the rate of interest on a sum of money is 8 % p . a . for the first 3 years , 4 % p . a . for the next 4 years , and 5 % for the period beyond 7 years . if the s . i , occured on the sum for the total period of 8 years is rs . 540 / - , the sum is | explanation : i 1 = ( p x 3 x 8 ) / 100 = 6 p / 25 i 2 = ( p x 4 x 4 ) / 100 = 4 p / 25 i 3 = ( p x 1 x 5 ) / 100 = p / 20 6 p / 25 + 4 p / 25 + p / 20 = 540 9 p / 20 = 540 p = 1200 answer : option a | a = 540 / 540
|
a ) 665 , b ) 127 , c ) 272 , d ) 287 , e ) 227 | a | subtract(add(multiply(84, 8), 85), 92) | a team of 8 persons joins in a shooting competition . the best marksman scored 85 points . if he had scored 92 points , the average score for the team would have been 84 . the number of points , the team scored was | "let the total score be x . ( x + 92 - 85 ) / 8 = 84 . so , x + 7 = 672 = > x = 665 . answer : a" | a = 84 * 8
b = a + 85
c = b - 92
|
a ) $ 58 , b ) $ 60.60 , c ) $ 70.60 , d ) $ 40.60 , e ) $ 50.60 | a | add(50.50, divide(multiply(50.50, 15), const_100)) | if tim had lunch at $ 50.50 and he gave 15 % tip , how much did he spend ? | "the tip is 20 % of what he paid for lunch . hence tip = 15 % of 50.50 = ( 15 / 100 ) * 50.50 = $ 7.575 total spent 50.50 + 7.575 = $ 58 correct answer a" | a = 50 * 50
b = a / 100
c = 50 + 50
|
a ) 1 / 50 , b ) 1 / 25 , c ) 1 / 98 , d ) 1 , e ) 2 | c | divide(1, 98) | if the numbers 1 to 98 are written on 98 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 , 72 composite numbers from 1 to 98 . the number which is neither prime nor composite is 1 . therefore , required probability = 1 / 98 . answer : c" | a = 1 / 98
|
a ) 16 , b ) 18 , c ) 20 , d ) 22 , e ) 24 | d | multiply(sqrt(subtract(multiply(65, const_3.0), multiply(3, 3))), 2) | if ( x + 3 ) ^ 2 / ( 3 x + 65 ) = 2 , then the difference between the two possible values of x is : | "( x + 3 ) ^ 2 / ( 3 x + 65 ) = 2 ( x + 3 ) ^ 2 = 2 ( 3 x + 65 ) x ^ 2 + 6 x + 9 = 6 x + 130 x ^ 2 - 121 = 0 ( x - 11 ) ( x + 11 ) = 0 x = 11 or x = - 11 the answer is d ." | a = 65 * 3
b = 3 * 3
c = a - b
d = math.sqrt(c)
e = d * 2
|
a ) 11 , b ) 12 , c ) 13 , d ) 14 , e ) 17 | e | divide(subtract(70, 2), 4) | if there are only 2 wheelers and 4 wheelers parked in a school located at the heart of the city , find the number of 4 wheelers parked there if the total number of wheels is 70 ? | "four wheeler = 17 * 4 = 68 ( max ) 2 wheel = 1 so no of 4 wheeler = 17 answer : e" | a = 70 - 2
b = a / 4
|
a ) 10.7 % , b ) 11.7 % , c ) 12.7 % , d ) 13.7 % , e ) 14.7 % | a | multiply(divide(subtract(add(multiply(10, 12), 12), multiply(10, 12)), 112), const_100) | on a purchase of $ 112 , a store offered a payment plan consisting of a $ 12 down payment and 12 monthly payments of $ 10 each . what percent of the purchase price , to the nearest tenth of a percent , did the customer pay in interest by using this plan ? | "12 * 10 + 12 = 132 ( 12 / 112 ) * 100 = 10.7 answer : a" | a = 10 * 12
b = a + 12
c = 10 * 12
d = b - c
e = d / 112
f = e * 100
|
a ) 46 , b ) 47 , c ) 58 , d ) 56 , e ) 54 | a | divide(425.5, 9.25) | john bought 9.25 m of cloth for $ 425.50 . find the cost price per metre . | cloth bought by john = 9.25 m cost of 9.25 m = $ 425.50 cost per metre = 425.50 Γ· 9.25 the cost of the cloth per metre = $ 46 answers : a | a = 425 / 5
|
a ) 30 , b ) 25 , c ) 15 , d ) 10 , e ) 5 | d | divide(30, const_3) | a dog is tied to a tree by a long nylon cord . if the dog runs from the due north side of the tree to the due south side of the tree with the cord extended to its full length at all items , and the dog ran approximately 30 feet , what was the approximate length of the nylon cord e , in feet ? | "because the cord was extended to its full length at all items , the dog ran along a semi - circular path , from north to south . the circumference of a full circle is 2 * pi * r , but since we only care about the length of half the circle , the semi - circle path is pi * r . e = pi * r = 30 . round pi = 3 , then r = 10 . chord is about 10 feet long . d" | a = 30 / 3
|
a ) 20 min , b ) 35 min , c ) 50 min , d ) 70 min , e ) none | c | divide(multiply(40, 10), 8) | a can go round a circular path 8 times in 40 minutes . if the diameter of the circle is increased to 10 times the original diameter , then the time required by a to go round the new path once , travelling at the same speed as before is | solution let original diameter be d metres . then , its circumference = ( Ο d ) metres . time taken to cover ( 8 Ο d ) m = 40 min . new diameter = ( 10 d ) m . then , its circumference = ( Ο x 10 d ) m . time taken to go round it once βΉ = βΊ [ 40 / 8 Ο d x 10 Ο d ] m βΉ = βΊ 50 min . answer c | a = 40 * 10
b = a / 8
|
a ) rs 400 , b ) rs 300 , c ) rs 3109 , d ) rs 250 , e ) rs 400 | c | subtract(add(add(8000, divide(multiply(8000, 15), const_100)), divide(multiply(add(8000, divide(multiply(8000, 15), const_100)), 15), const_100)), 8000) | find compound interest on rs . 8000 at 15 % per annum for 2 years 4 months , compounded annually . | "time = 2 years 4 months = 2 ( 4 / 12 ) years = 2 ( 1 / 3 ) years . amount = rs ' . [ 8000 x ( 1 + Β¬ ( 15 / 100 ) ) 2 x ( 1 + ( ( 1 / 3 ) * 15 ) / 100 ) ] = rs . [ 8000 * ( 23 / 20 ) * ( 23 / 20 ) * ( 21 / 20 ) ] = rs . 11109 . . : . c . i . = rs . ( 11109 - 8000 ) = rs . 3109 . answer c rs 3109" | a = 8000 * 15
b = a / 100
c = 8000 + b
d = 8000 * 15
e = d / 100
f = 8000 + e
g = f * 15
h = g / 100
i = c + h
j = i - 8000
|
a ) 8 , b ) 7 , c ) 6 , d ) 9 , e ) 10 | d | add(divide(75, const_10), 1) | x is the product of each integer from 1 to 75 , inclusive and y = 100 ^ k , where k is an integer . what is the greatest value of k for which y is a factor of x ? | "the number of trailing zeros in the decimal representation of n ! , the factorial of a non - negative integer n , can be determined with this formula : n 5 + n 52 + n 53 + . . . + n 5 k , where k must be chosen such that 5 k β€ n x = 1 * 2 * 3 . . . . * 75 = 50 ! no . of trailing zeros in 75 ! = 75 / 5 + 75 / 5 ^ 2 = 15 + 3 = 18 100 ^ k = 10 ^ 2 k β k = 18 / 2 = 9 d" | a = 75 / 10
b = a + 1
|
a ) $ 4096 , b ) $ 5904 , c ) $ 7892 , d ) $ 10052 , e ) $ 20736 | d | divide(20000, const_2) | a will was written such that an estate would be divided among 3 guys with the remainder donated to charity . one stipulation in the will is that no one from the guys can receive an amount within 30 % of another ' s amount . if one of the guys received $ 20000 , what is the smallest possible range between the highest and lowest amounts ( rounded to the nearest dollar ) among the 3 guys ? | 1 st - 20000 . 2 nd - 0.7 * 20000 = 14000 3 rd - 0.7 * 14000 = 9800 range = 20000 - 9800 = 10200 . answer : d . | a = 20000 / 2
|
a ) 96 , b ) 94 , c ) 86 , d ) 176 , e ) 110 | d | divide(40, 440) | find 40 % of 440 | "we know that r % of m is equal to r / 100 Γ m . so , we have 40 % of 440 40 / 100 Γ 440 = 176 answer : d" | a = 40 / 440
|
a ) 3 , b ) 5 , c ) 10 , d ) 13 , e ) 15 | c | add(subtract(multiply(20, 2), add(add(8, multiply(5, 2)), multiply(2, 3))), 3) | in a class of 20 students , 2 students did not borrow any books from the library , 8 students each borrowed 1 book , 5 students each borrowed 2 books , and the rest of the students each borrowed at least 3 books . if the average ( arithmetic mean ) number of books borrowed per student was 2 , what is the maximum number of books that any single student could have borrowed ? | "the average number of books per student was 2 means that total of 2 * 20 = 40 books were borrowed ; 2 + 8 + 5 = 15 students borrowed total of 2 * 0 + 8 * 1 + 5 * 2 = 18 books ; so 40 - 18 = 22 books are left to distribute among 20 - 15 = 5 students , these 5 arethe rest who borrowed at least 3 books ; tomaximizethe number of books one student from above 5 could have borrowed we shouldminimizethe number of books other 4 students from 5 could have borrowed . minimum these 4 students could have borrowed is 3 books per student , so total number of books they could have borrowed is 4 * 3 = 12 books . so the 5 th student could have borrowed is 22 - 12 = 10 books . answer : c ." | a = 20 * 2
b = 5 * 2
c = 8 + b
d = 2 * 3
e = c + d
f = a - e
g = f + 3
|
a ) 5 % , b ) 6 % , c ) 8 % , d ) 10 % , e ) none of these | b | multiply(divide(subtract(27160, add(10000, 12000)), add(multiply(10000, add(3, const_2)), multiply(12000, 3))), const_100) | mukesh borrowed a sum of rs . 10000 at simple interest from a bank . after two years he takes rs . 12000 more as loan and discharge all the debt at the end of next 3 years depositing rs . 27160 to the bank . what is the rate of interest per annum ? | si = 27160 - ( principal amount ) = 27160 - 22000 = 5160 . now , ( 10000 * 5 * r / 100 ) + ( 12000 * 3 * r / 100 ) = 5160 . 500 r + 360 r = 5160 . . . . . rate = 6 % . answer : b | a = 10000 + 12000
b = 27160 - a
c = 3 + 2
d = 10000 * c
e = 12000 * 3
f = d + e
g = b / f
h = g * 100
|
a ) 4 , b ) 5 , c ) 6 , d ) 7 , e ) 8 | c | subtract(7, divide(subtract(13, 7), add(4, const_2))) | a player holds 13 cards of 4 suits , of which 7 are black and 6 are red . there are twice as many diamonds as spades and twice as many hearts as diamonds . how many clubs does he hold | explanation : clearly , the black cards are either clubs or spades while the red cards are either diamonds or hearts . let the number of spades be x . then , number of clubs = ( 7 - x ) . number of diamonds = 2 x number of spades = 2 x ; number of hearts = 2 x number of diamonds = 4 x . total number of cards = x + 2 x + 4 x + 7 - x = 6 x + 7 . therefore 6 x + 7 = 13 6 x = 6 x - 1 . hence , number of clubs = ( 7 - x ) = 6 . answer : c | a = 13 - 7
b = 4 + 2
c = a / b
d = 7 - c
|
a ) 2.6 , b ) 2.8 , c ) 3.0 , d ) 3.2 , e ) 3.4 | e | inverse(add(divide(const_1, 6), divide(const_1, 8))) | a worker can load one truck in 6 hours . a second worker can load the same truck in 8 hours . if both workers load one truck simultaneously while maintaining their constant rates , approximately how long , in hours , will it take them to fill one truck ? | "the workers fill the truck at a rate of 1 / 6 + 1 / 8 = 14 / 48 = 7 / 24 of the truck per hour . then the time to fill one truck is 24 / 7 which is about 3.4 hours . the answer is e ." | a = 1 / 6
b = 1 / 8
c = a + b
d = 1/(c)
|
a ) 11 , b ) 15 , c ) 18 , d ) 17 , e ) 19 | b | divide(const_1, subtract(const_1, add(add(divide(const_1, add(const_1, const_4)), divide(const_1, const_3)), multiply(subtract(divide(const_1, const_3), divide(const_1, add(const_1, const_4))), 3)))) | here is a simple arithmetical puzzle in his own flowery , poetical language . if one - fifth of a hive of bees flew to the badamba flower , one - third flew to the slandbara , 3 times the difference of these two numbers flew to an arbour , and one bee continued to fly about , attracted on each side by the fragrant ketaki and malati , what was the total number of bees ? | b 15 the number of bees was 15 | a = 1 + 4
b = 1 / a
c = 1 / 3
d = b + c
e = 1 / 3
f = 1 + 4
g = 1 / f
h = e - g
i = h * 3
j = d + i
k = 1 - j
l = 1 / k
|
a ) 37 , b ) 41 , c ) 3.5 e - 05 , d ) 35 , e ) 3.5 e - 07 | b | multiply(divide(multiply(multiply(multiply(10, 10), subtract(multiply(10, 10), const_1)), divide(35, subtract(multiply(10, 10), const_1))), const_1000), 10) | if the digits 35 in the decimal 0.00035 repeat indefinitely , what is the value of ( 10 ^ 5 - 10 ^ 3 ) ( 0.00035 ) ? | "99 * 0.35 = 34.65 approx . 35 answer : b" | a = 10 * 10
b = 10 * 10
c = b - 1
d = a * c
e = 10 * 10
f = e - 1
g = 35 / f
h = d * g
i = h / 1000
j = i * 10
|
a ) 222 , b ) 117 , c ) 278 , d ) 767 , e ) 298 | b | multiply(117, const_1) | product of two co - prime numbers is 117 . their l . c . m should be | h . c . f of co - prime numbers is 1 . so , l . c . m = \ inline \ fn _ jvn \ frac { 117 } { 1 } = 117 answer : b | a = 117 * 1
|
a ) 1280 , b ) 1270 , c ) 1260 , d ) 1250 , e ) none | a | divide(subtract(multiply(125, 7392), multiply(462, 720)), 462) | if to a certain number , 720 be added , and the sum be divided by 125 ; the quotient will be equal to 7392 divided by 462 . what is that number ? | solution let x = the number required . a = 720 d = 7392 b = 125 h = 462 then by the conditions of the problem ( x + a ) / b = d / h therefore x = ( bd - ah ) / h restoring the numbers , x = [ ( 125.7392 ) - ( 720.462 ) ] / 462 = 1280 . answer a | a = 125 * 7392
b = 462 * 720
c = a - b
d = c / 462
|
a ) 4 , b ) 5 , c ) 7 , d ) 8 , e ) 2 | d | inverse(add(divide(6, multiply(10, 9)), divide(10, multiply(10, 12)))) | 10 women can complete a work in 9 days and 10 children take 12 days to complete the work . how many days will 6 women and 7 children take to complete the work ? | "1 women ' s 1 day work = 1 / 90 1 child ' s 1 day work = 1 / 120 ( 6 women + 7 children ) ' s 1 day work = ( 6 / 90 + 7 / 120 ) = 1 / 8 6 women and 7 children will complete the work in 8 days . d" | a = 10 * 9
b = 6 / a
c = 10 * 12
d = 10 / c
e = b + d
f = 1/(e)
|
a ) 64 , b ) 72 , c ) 121 , d ) 984 , e ) 108 | c | divide(subtract(multiply(subtract(22.5, 0.5), 5), 52), 0.5) | a cricket player whose bowling average was 22.5 runs per wicket , takes 5 wicket for 52 runs in a match . due to this his average decreases by 0.5 . what will be the number of wickets taken by him till the last match ? | average = total runs / total wickets total runs after last match = 22.5 w + 52 total wickets after last match = w + 5 ( 22.5 w + 52 ) / ( w + 5 ) = 22.5 - 0.5 = 22 w = 116 so total wickets aftr last match = w + 5 = 121 answer : c | a = 22 - 5
b = a * 5
c = b - 52
d = c / 0
|
a ) 200 , b ) 250 , c ) 300 , d ) 500 , e ) none of these | d | divide(divide(multiply(const_100, multiply(const_1000, 66)), multiply(const_60, const_1)), multiply(multiply(const_2, 35), add(const_3, divide(add(const_2, multiply(const_3, const_4)), power(add(const_2, multiply(const_4, const_2)), const_2))))) | the radius of the wheel of a bus is 35 cms and the speed of the bus is 66 km / h , then the r . p . m . ( revolutions per minutes ) of the wheel is | radius of the wheel of bus = 35 cm . then , circumference of wheel = 2 Γ― β¬ r = 70 Γ― β¬ = 220 cm distance covered by bus in 1 minute = 66 Γ’ Β β 60 Γ£ β 1000 Γ£ β 100 cms distance covered by one revolution of wheel = circumference of wheel = 220 cm Γ’ Λ Β΄ revolutions per minute = 6600000 / 60 Γ£ β 220 = 500 answer d | a = 1000 * 66
b = 100 * a
c = const_60 * 1
d = b / c
e = 2 * 35
f = 3 * 4
g = 2 + f
h = 4 * 2
i = 2 + h
j = i ** 2
k = g / j
l = 3 + k
m = e * l
n = d / m
|
a ) 14 , b ) 13 , c ) 11 , d ) 9 , e ) 7 | d | inverse(add(add(add(divide(const_1, multiply(12, const_2)), divide(divide(const_1, multiply(12, const_2)), const_2)), divide(const_1, multiply(15, const_2))), divide(divide(const_1, multiply(15, const_2)), const_2))) | a furniture manufacturer has two machines , but only one can be used at a time . machine e is utilized during the first shift and machine b during the second shift , while both work half of the third shift . if machine e can do the job in 12 days working two shifts and machine b can do the job in 15 days working two shifts , how many days will it take to do the job with the current work schedule ? | machine e finish the job in 2 * 12 shifts = 24 shifts machine b finish the job in 2 * 15 shifts = 30 shifts lets assume total work require 120 shifts therefore , rate of e = 5 shifts / day rate of b = 4 shifts / day rate of ( e + b ) = 9 shifts / day according to current schedule work complete in a day = 5 + 4 + ( 9 / 2 ) = 13.5 shifts / day therefore , time required to finish 120 shifts = ( 120 / 13.5 ) = 8.88 . . days ~ 9 days = d | a = 12 * 2
b = 1 / a
c = 12 * 2
d = 1 / c
e = d / 2
f = b + e
g = 15 * 2
h = 1 / g
i = f + h
j = 15 * 2
k = 1 / j
l = k / 2
m = i + l
n = 1/(m)
|
a ) 7 sec , b ) 6 sec , c ) 8 sec , d ) 4.4 sec , e ) 2 sec | d | divide(110, multiply(add(84, 6), const_0_2778)) | a train 110 m long is running with a speed of 84 km / hr . in what time will it pass a man who is running at 6 km / hr in the direction opposite to that in which the train is going ? | "speed of train relative to man = 84 + 6 = 90 km / hr . = 90 * 5 / 18 = 25 m / sec . time taken to pass the men = 110 / 25 = 4.4 sec . answer : d" | a = 84 + 6
b = a * const_0_2778
c = 110 / b
|
a ) 500 , b ) 450 , c ) 552 , d ) 600 , e ) 400 | c | divide(add(multiply(25, 500), multiply(add(500, 100), subtract(52, 25))), 52) | a certain debt will be paid in 52 installments from january 1 to december 31 of a certain year . each of the first 25 payments is to be $ 500 ; each of the remaining payments is to be $ 100 more than each of the first 25 payments . what is the average ( arithmetic mean ) payment that will be made on the debt for the year ? | "total = 500 ( 25 ) + 600 ( 27 ) number of installments = 52 average = total / number of installments = 552 approximately answer : c" | a = 25 * 500
b = 500 + 100
c = 52 - 25
d = b * c
e = a + d
f = e / 52
|
a ) 90 , b ) 120 , c ) 225 , d ) 180 , e ) 200 | c | divide(multiply(multiply(6, const_4), multiply(6, 6)), power(factorial(2), 2)) | there are 6 fictions and 6 non - fictions . how many cases are there such that 2 fictions and 2 non - fictions are selected from them ? | "number of ways of selecting 2 fiction books = 6 c 2 number of ways of selecting 2 non fiction books = 6 c 2 6 c 2 * 6 c 2 = 15 * 15 = 225 answer : c" | a = 6 * 4
b = 6 * 6
c = a * b
d = math.factorial(2)
e = d ** 2
f = c / e
|
a ) 20 cs , b ) cs / 2 , c ) 48 cs , d ) ( 2 cs ) / 12 , e ) ( 24 c ) / s | c | multiply(4, const_12) | a certain school implemented a reading program for its students , with the goal of getting each student to read 4 books per month year - round . if the school has c classes made up of s students in each class , how many books will the entire student body read in one year ? | "ans : c solution : simple multiplication s students , c classes , 4 books / month = 48 books a year total number of books = 48 cs" | a = 4 * 12
|
a ) 39 , b ) 24 , c ) 28 , d ) 26 , e ) 30 | b | add(divide(340, 10), subtract(subtract(10, 4), const_1)) | in a certain quiz that consists of 10 questions , each question after the first is worth 4 points more than the preceding question . if the 10 questions on the quiz are worth a total of 340 points , how many points is the third question worth ? | "x x + 4 x + 8 x + 12 x + 16 x + 20 x + 24 x + 28 x + 32 x + 36 10 x + 180 = 340 10 x = 160 x = 16 3 rd question = x + 8 = 16 + 8 = 24 answer b" | a = 340 / 10
b = 10 - 4
c = b - 1
d = a + c
|
a ) 26 , b ) 25 , c ) 24 , d ) 22 , e ) 23 | a | add(divide(subtract(subtract(200, const_4), add(43, const_1)), 6), const_1) | how many natural numbers are there between 43 and 200 which are exactly divisible by 6 ? | "43 / 6 = 7 , remainder = 1 . hence 5 more should be added to 43 to get the minimum number divisible by 6 between 43 and 200 . = > minimum number divisible by 6 between 43 and 200 = 43 + 5 = 48 200 / 6 = 33 , remainder = 2 . hence 2 should be decreased from 200 to get the maximum number divisible by 6 between 43 and 200 . = > maximum number divisible by 6 between 43 and 200 = 200 - 2 = 198 hence , natural numbers numbers divisible by 6 between 43 and 200 are 48 , 54 , 60 , . . . 198 this is arithmetic progression with a = 48 , d = 6 , l = 198 number of terms = ( l β a ) / d + 1 = ( 198 β 48 ) / 6 + 1 = 150 / 6 + 1 = 25 + 1 = 26 answer is a" | a = 200 - 4
b = 43 + 1
c = a - b
d = c / 6
e = d + 1
|
a ) 110 , b ) 1,100 , c ) 9,900 , d ) 10,000 , e ) 12,100 | c | multiply(multiply(subtract(const_10, const_1), multiply(add(6, 4), subtract(16, 5))), const_10) | right triangle pqr is to be constructed in the xy - plane so that the right angle is at p and pr is parallel to the x - axis . the x - and y - coordinates of p , q , and r are to be integers that satisfy the inequalities - 4 < = x < = 5 and 6 < = y < = 16 . how many different triangles with these properties could be constructed ? | "p can take a total of 10 * 11 co - ordinates . for a given p , r can take a total of 9 co - ordinates and q can take a total of 10 co - ordinates . hence , total = 110 * 9 * 10 = 9900 ans : c" | a = 10 - 1
b = 6 + 4
c = 16 - 5
d = b * c
e = a * d
f = e * 10
|
a ) 35 , b ) 38 , c ) 40 , d ) 42 , e ) 44 | e | divide(add(160, 60), add(4, 1)) | in an examination , a student scores 4 marks for every correct answer and loses 1 mark for every wrong answer . if he attempts all 60 questions and secures 160 marks , the no of questions he attempts correctly is : | "explanation : let the number of correct answers be x . number of incorrect answers = ( 60 β x ) . 4 x β ( 60 β x ) = 160 = > 5 x = 220 = > x = 44 answer : e" | a = 160 + 60
b = 4 + 1
c = a / b
|
a ) 4.5 , b ) 7 , c ) 8 , d ) 11 , e ) 12 | d | multiply(multiply(18, 2), divide(1, 2)) | in the coordinate plane , points ( x , 1 ) and ( 18 , y ) are on line k . if line k passes through the origin and has slope 1 / 2 , then x + y = | "line k passes through the origin and has slope 1 / 2 means that its equation is y = 1 / 2 * x . thus : ( x , 1 ) = ( 2 , 1 ) and ( 18 , y ) = ( 18,9 ) - - > x + y = 2 + 9 = 11 . answer : d ." | a = 18 * 2
b = 1 / 2
c = a * b
|
a ) $ 1200 , b ) $ 1000 , c ) $ 500 , d ) $ 1100 , e ) $ 1500 | a | multiply(1200, divide(5, const_100)) | find the simple interest on $ 1200 for 5 years at 20 % per annum ? | "si = ptr / 100 = 1200 * 5 * 20 / 100 = $ 1200 answer is a" | a = 5 / 100
b = 1200 * a
|
a ) 39 , b ) 18 , c ) 42 , d ) 40.5 , e ) 41.5 | b | add(subtract(50, multiply(17, 2)), 2) | a batsman makes a score of 50 runs in the 17 th inning and thus increases his averages by 2 . what is his average after 17 th inning ? | "let the average after 16 th inning = x then total run after 16 th inning = 16 x then total run after 17 th inning = 16 x + 50 then average run after 17 th inning = ( 16 x + 50 ) / 17 ( 16 x + 50 ) / 17 = x + 2 = > 16 x + 50 = 17 x + 34 = > x = 16 x = 16 ; average after 17 th inning = 16 + 2 = 18 answer : b" | a = 17 * 2
b = 50 - a
c = b + 2
|
a ) 10000 , b ) 11000 , c ) 12000 , d ) 13000 , e ) 14000 | a | subtract(const_100, 90) | to produce an annual income of rs . 800 from a 8 % stock at 90 , what is the amount of stock needed ? | "since face value is not given , take it as rs . 100 . as it is an 8 % stock , income ( dividend ) per stock = rs . 8 ie , for an income of rs . 8 , amount of stock needed = rs . 100 for an income of rs . 800 , amount of stock needed = 100 Γ 800 / 8 = 10000 answer is a ." | a = 100 - 90
|
a ) a ) 14 % , b ) b ) 16 % , c ) c ) 18 % , d ) d ) 21.6 % , e ) of these | d | multiply(divide(subtract(64900, add(40000, 13000)), add(40000, 13000)), const_100) | ramu bought an old car for rs . 40000 . he spent rs . 13000 on repairs and sold it for rs . 64900 . what is his profit percent ? | "total cp = rs . 40000 + rs . 13000 = rs . 53000 and sp = rs . 64900 profit ( % ) = ( 64900 - 53000 ) / 55000 * 100 = 21.6 % answer : d" | a = 40000 + 13000
b = 64900 - a
c = 40000 + 13000
d = b / c
e = d * 100
|
a ) 174.2 , b ) 277 , c ) 288.1 , d ) 296 , e ) 282.4 | b | multiply(divide(44, 2.54), divide(24, 1.5)) | on a map , 1.5 inches represent 24 miles . how many miles approximately is the distance if you measured 44 centimeters assuming that 1 - inch is 2.54 centimeters ? | "1.5 inch = 2.54 * 1.5 cm . so , 2.54 * 1.5 represents 24 miles . so for 44 cm . : 44 / ( 2.54 * 1.5 ) = x / 24 - - - > x = 24 * 44 / ( 3.81 ) = 277 answer will be b ." | a = 44 / 2
b = 24 / 1
c = a * b
|
a ) 0.001 , b ) 0.01 , c ) 0.1 , d ) 1 , e ) 0 | c | sqrt(divide(0.12, 12)) | if 0.12 Γ· x 2 = 12 , than x = ? | explanation : 0.12 Γ· x 2 = 12 = > 0.12 / x 2 = 12 = > x 2 = 0.12 / 12 = 0.01 = > x = 0.1 answer : option c | a = 0 / 12
b = math.sqrt(a)
|
a ) 334 km . , b ) 216 km . , c ) 314 km . , d ) 224 km . , e ) 672 km . | e | multiply(multiply(divide(multiply(30, 24), add(24, 21)), 21), const_2) | a woman complete a journey in 30 hours . she travels first half of the journey at the rate of 21 km / hr and second half at the rate of 24 km / hr . find the total journey in km . | "0.5 x / 21 + 0.5 x / 24 = 30 - - > x / 21 + x / 24 = 60 - - > x = 672 km . e" | a = 30 * 24
b = 24 + 21
c = a / b
d = c * 21
e = d * 2
|
a ) 6 , b ) 8 , c ) 10 , d ) 12 , e ) 14 | a | subtract(35, subtract(add(26, 20), 17)) | in a class of 35 students 26 play football and play 20 long tennis , if 17 play above , many play neither ? | "26 + 20 - 17 = 29 35 - 29 = 6 play neither answer is a" | a = 26 + 20
b = a - 17
c = 35 - b
|
a ) 480 , b ) 488 , c ) 882 , d ) 269 , e ) 298 | a | multiply(24, 20) | a cistern has a leak which would empty the cistern in 20 minutes . a tap is turned on which admits 4 liters a minute into the cistern , and it is emptied in 24 minutes . how many liters does the cistern hold ? | 1 / x - 1 / 20 = - 1 / 24 x = 120 120 * 4 = 480 answer : a | a = 24 * 20
|
a ) 3 , b ) 4 , c ) 5 , d ) 6 , e ) 7 | a | add(divide(subtract(multiply(21, 6), multiply(7, 16)), subtract(21, 16)), divide(const_2, const_10)) | the least whole number which when subtracted from both the terms of the ratio 6 : 7 to give a ratio less than 16 : 21 , is ? | let x be subtracted . then , 6 - x 7 - x < 16 21 21 ( 6 - x ) < 16 ( 7 - x ) 5 x > 14 x > 2.8 least such whole number is 3 . answer : a | a = 21 * 6
b = 7 * 16
c = a - b
d = 21 - 16
e = c / d
f = 2 / 10
g = e + f
|
a ) 12.5 % , b ) 20 % , c ) 25 % , d ) 50 % , e ) 100 % | c | multiply(divide(subtract(divide(multiply(const_10, const_4), multiply(divide(subtract(const_100, 20), const_100), const_10)), const_4), const_4), const_100) | a part - time employee whose hourly wage was decreased by 20 percent decided to increase the number of hours worked per week so that the employee ' s total income did not change . by what percent should the number of hours worked be increased ? | "assume hours worked : 10 hours hourly wage : 10 $ weekly wage : 100 $ after decrease of 20 % in hourly wage would become 8 $ hours worked would have to be 12.5 hours in order to maintain weekly wage of 100 $ % increase in number of hours worked = ( 12.5 - 10 ) / 10 = 0.25 * 100 = 25 % answer is c" | a = 10 * 4
b = 100 - 20
c = b / 100
d = c * 10
e = a / d
f = e - 4
g = f / 4
h = g * 100
|
a ) 40 days , b ) 20 days , c ) 8 days , d ) 60 days , e ) 40 days | c | inverse(subtract(4, divide(4, 8))) | a and b can do a piece of work in 8 days . with the help of c they finish the work in 4 days . c alone can do that piece of work in ? | "c 8 days c = 1 / 4 β 1 / 8 = 1 / 8 = > 8 days" | a = 4 / 8
b = 4 - a
c = 1/(b)
|
a ) 0.75 , b ) 0.8 , c ) 1 , d ) 1.33 , e ) 1.75 | d | inverse(divide(60, add(60, 20))) | patrick purchased 60 pencils and sold them at a loss equal to the selling price of 20 pencils . the cost of 60 pencils is how many times the selling price of 60 pencils ? | "say the cost price of 60 pencils was $ 60 ( $ 1 per pencil ) and the selling price of 1 pencil was p . selling at a loss : 60 - 60 p = 20 p - - > p = 3 / 4 . ( cost price ) / ( selling price ) = 1 / ( 3 / 4 ) = 4 / 3 = 1.33 . answer : d ." | a = 60 + 20
b = 60 / a
c = 1/(b)
|
a ) 09 am , b ) 07 am , c ) 01 pm , d ) 05 pm , e ) 03 pm | c | add(divide(add(600, 75), add(60, 75)), 8) | the distance between two cities a and b is 600 km . a train starts from a at 8 a . m . and travel towards b at 60 km / hr . another train starts from b at 9 a . m and travels towards a at 75 km / hr . at what time do they meet ? | "explanation : suppose they meet x hrs after 8 a . m then , [ distance moved by first in x hrs ] + [ distance moved by second in ( x - 1 ) hrs ] = 600 . therefore , 60 x + 75 ( x - 1 ) = 600 . = > x = 5 . so , they meet at ( 8 + 5 ) i . e , 1 p . m . answer : c )" | a = 600 + 75
b = 60 + 75
c = a / b
d = c + 8
|
a ) 2 / 125 , b ) 3 / 125 , c ) c ) 2 / 25 , d ) 3 / 25 , e ) 1 / 25 | e | divide(subtract(divide(multiply(800, 1), 5), 128), 800) | in a group of 800 people , 1 / 5 play at least one instrument , 128 play two or more . what is the probability that one student play exactly one instrument ? | "p ( playing 2 or more instruments ) = 128 / 800 = 4 / 25 . then , the probability of playing exactly one instrument is given by : p ( playing 1 or more instruments ) - p ( playing 2 or more instruments ) = 1 / 5 - 4 / 25 = 1 / 25 . answer e ." | a = 800 * 1
b = a / 5
c = b - 128
d = c / 800
|
a ) 36 , b ) 52 , c ) 66 , d ) 96 , e ) none | d | sqrt(multiply(576, 16)) | the present worth of a sum due sometime hence is rs . 576 and the banker β s gain is rs . 16 . the true discount is : | "sol . t . d . = β p . w . * b . g . = β 576 * 16 = 96 . answer d" | a = 576 * 16
b = math.sqrt(a)
|
a ) 62 : 25 , b ) 66 : 52 , c ) 68 : 15 , d ) 65 : 41 , e ) 23 : 59 | d | divide(multiply(52, 10), multiply(46, 8)) | car a runs at the speed of 52 km / hr & reaches its destination in 10 hr . car b runs at the speed of 46 km / h & reaches its destination in 8 h . what is the respective ratio of distances covered by car a & car b ? | "sol . distance travelled by car a = 52 Γ 10 = 520 km distance travelled by car b = 46 Γ 8 = 368 km ratio = 520 / 368 = 65 : 41 d" | a = 52 * 10
b = 46 * 8
c = a / b
|
a ) 45 , b ) 48 , c ) 50 , d ) 52 , e ) 60 | e | divide(60, subtract(divide(60, 40), divide(30, const_60))) | the distance from city a to city b is 60 miles . while driving from city a to city b , bob drives at a constant speed of 40 miles per hour . alice leaves city a 30 minutes after bob . what is the minimum constant speed in miles per hour that alice must exceed in order to arrive in city b before bob ? | "the time it takes bob to drive to city b is 60 / 40 = 1.5 hours . alice needs to take less than 1 hour for the trip . alice needs to exceed a constant speed of 60 / 1 = 60 miles per hour . the answer is e ." | a = 60 / 40
b = 30 / const_60
c = a - b
d = 60 / c
|
a ) 28 , b ) 18 , c ) 24 , d ) 21 , e ) 16 | b | divide(multiply(multiply(18, 100), 42), multiply(140, 30)) | if 18 persons can build a wall 140 m long in 42 days , the number of days that 30 persons will take to complete a similar wall 100 m long , is ? | "explanation : ( length 140 : 100 ) : ( persons 30 : 18 ) : : 42 : x 140 x 30 x x = 100 x 18 x 42 or x = 18 answer : b" | a = 18 * 100
b = a * 42
c = 140 * 30
d = b / c
|
a ) 8 % , b ) 15 % , c ) 45 % , d ) 52 % , e ) 76 % | e | multiply(divide(subtract(multiply(divide(7, 20), const_100), multiply(divide(9, 108), const_100)), multiply(divide(7, 20), const_100)), const_100) | a pharmaceutical company received $ 7 million in royalties on the first $ 20 million in sales of the generic equivalent of one of its products and then $ 9 million in royalties on the next $ 108 million in sales . by approximately what percent did the ratio of royalties to sales decrease from the first $ 20 million in sales to the next $ 108 million in sales ? | "solution : this is a percent decrease problem . we will use the formula : percent change = ( new β old ) / old x 100 to calculate the final answer . we first set up the ratios of royalties to sales . the first ratio will be for the first 20 million in sales , and the second ratio will be for the next 108 million in sales . because all of the sales are in millions , we do not have to express all the trailing zeros in our ratios . first 20 million royalties / sales = 7 / 20 next 108 million royalties / sales = 9 / 108 = 1 / 12 because each ratio is not an easy number to use , we can simplify each one by multiplying each by the lcm of the two denominators , which is 60 . keep in mind that we are able to do this only because our answer choices are expressed in percents . first 20 million royalties / sales = ( 7 / 20 ) x 60 = 21 next 108 million royalties / sales = 9 / 108 = ( 1 / 12 ) x 60 = 5 we can plug 9 and 5 into our percent change formula : ( new β old ) / old x 100 [ ( 5 β 21 ) / 21 ] x 100 - 1600 / 21 x 100 at this point we can stop and consider the answer choices . since we know that 1600 / 21 is just a bit less than Β½ , we know that - 1600 / 21 x 100 is about a 76 % decrease . answer e ." | a = 7 / 20
b = a * 100
c = 9 / 108
d = c * 100
e = b - d
f = 7 / 20
g = f * 100
h = e / g
i = h * 100
|
a ) 14 , b ) 35 , c ) 39 , d ) 31 , e ) 12 | d | add(add(power(add(add(divide(subtract(subtract(93, const_10), const_2), const_4), const_2), const_2), const_2), power(add(add(add(divide(subtract(subtract(93, const_10), const_2), const_4), const_2), const_2), const_2), const_2)), add(power(divide(subtract(subtract(93, const_10), const_2), const_4), const_2), power(add(divide(subtract(subtract(93, const_10), const_2), const_4), const_2), const_2))) | the sum of three consecutive even numbers is 93 . find the middle number of the three ? | "middle value = 93 / 3 = 31 ans d" | a = 93 - 10
b = a - 2
c = b / 4
d = c + 2
e = d + 2
f = e ** 2
g = 93 - 10
h = g - 2
i = h / 4
j = i + 2
k = j + 2
l = k + 2
m = l ** 2
n = f + m
o = 93 - 10
p = o - 2
q = p / 4
r = q ** 2
s = 93 - 10
t = s - 2
u = t / 4
v = u + 2
w = v ** 2
x = r + w
y = n + x
|
a ) 4 cm , b ) 6 cm , c ) 8 cm , d ) 10 cm , e ) 12 cm | a | multiply(divide(divide(divide(divide(multiply(divide(volume_cylinder(divide(16, const_2), 2), const_pi), const_3), const_4), 12), const_4), const_4), const_2) | 12 spheres of the same size are made from melting a solid cylinder of 16 cm diameter and 2 cm height . find the diameter of each sphere . | "explanation : in this type of question , just equate the two volumes to get the answer as , volume of cylinder = Ο r 2 h volume of sphere = 43 Ο r 3 = > 12 β 4 / 3 Ο r 3 = Ο r 2 h = > 12 β 4 / 3 Ο r 3 = Ο β 8 β 8 β 2 = > r 3 = 8 β 8 β 2 β 3 / 12 β 4 = > r 3 = 8 = > r = 2 cm = > diameter = 2 β 2 = 4 cm option a" | a = 16 / 2
b = volume_cylinder / (
c = b * math.pi
d = c / 3
e = d / 4
f = e / 12
g = f / 4
h = g * 4
|
a ) 0.0123 , b ) 0.0625 , c ) 0.16 , d ) 0.25 , e ) 0.5 | a | power(divide(1, 9), 2) | what is the decimal equivalent of ( 1 / 9 ) ^ 2 ? | "( 1 / 9 ) Β² = ( 1 / 9 ) ( 1 / 9 ) = 1 / 81 approach # 1 : use long division to divide 81 into 1 to get 1 / 81 = 0.0123 a" | a = 1 / 9
b = a ** 2
|
a ) 4242 , b ) 4155 , c ) 1123 , d ) 11023 , e ) none of them | b | multiply(divide(7589, 3434), const_100) | 7589 - ? = 3434 | "let 7589 - x = 3434 then x = 7589 - 3434 = 4155 answer is b" | a = 7589 / 3434
b = a * 100
|
a ) 6.25 , b ) 5.5 , c ) 7.4 , d ) 8.3 , e ) 6.2 | d | divide(subtract(282, multiply(10, 3.2)), 30) | in the first 10 overs of a cricket game , the run rate was only 3.2 . what should be the run rate in the remaining 30 overs to reach the target of 282 runs ? | "explanation : runs scored in the first 10 overs = 10 Γ 3.2 = 32 total runs = 282 remaining runs to be scored = 282 - 32 = 250 remaining overs = 30 run rate needed = 250 / 30 = 8.3 answer : option d" | a = 10 * 3
b = 282 - a
c = b / 30
|
a ) - 19 , b ) 18 , c ) - 37 , d ) 50 , e ) 37 | c | subtract(19, multiply(14, 4)) | with a slope of 4 , with a straight line in the xy - plane which has a point with x - coordinate of the point 14 and y - coordinate 19 , find the y - intercept . | eq of line = y = mx + c m = 4 x = 14 y = 19 c = y - mx substitute given : c = 19 - ( 4 ) ( 14 ) c = - 37 correct option is c | a = 14 * 4
b = 19 - a
|
a ) 1 , b ) 4 , c ) 9 , d ) 13 , e ) 16 | e | multiply(4, 4) | elena β s bread recipe calls for 3 ounces of butter for each 4 cups of flour used . she needs to make 4 times the original recipe . if 12 ounces of butter is used , then how many cups of flour are needed ? | "number of cups flour needed for 3 ounces of butter = 4 number of cups flour needed for 1 ounce of butter = 4 / 3 number of cups flour needed for 12 ounces of butter = 4 / 3 * 12 = 16 answer e" | a = 4 * 4
|
Subsets and Splits
No community queries yet
The top public SQL queries from the community will appear here once available.