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 ) 20 , b ) 21 , c ) 22 , d ) 23 , e ) 24 | e | add(divide(16, const_2), subtract(32, 16)) | jane started baby - sitting when she was 16 years old . whenever she baby - sat for a child , that child was no more than half her age at the time . jane is currently 32 years old , and she stopped baby - sitting 10 years ago . what is the current age of the oldest person for whom jane could have baby - sat ? | "check two extreme cases : jane = 16 , child = 8 , years ago = 32 - 16 = 16 - - > child ' s age now = 8 + 16 = 24 ; jane = 22 , child = 11 , years ago = 32 - 22 = 10 - - > child ' s age now = 11 + 10 = 21 . answer : e ." | a = 16 / 2
b = 32 - 16
c = a + b
|
['a ) 1', 'b ) 2', 'c ) 3', 'd ) 4', 'e ) 5'] | c | sqrt(divide(divide(divide(volume_cylinder(6, 15), const_10), add(multiply(multiply(divide(const_1, const_3), const_pi), const_4), multiply(divide(const_2, const_3), const_pi))), const_3)) | a cylindrical container of radius 6 cm and height 15 cm is filled with ice - cream . the whole icecream has to be distributed to 10 children in equal cones with hemispherical tops . if the height of the conical portion is four times the radius of its base , find the radius of the ice - cream cone . | volume of given ice cream = ( pie ) * 6 ² * 15 = 540 ( pie ) dividing to 10 children volume given to a child = 54 ( pie ) let r be radius and h be the height of the cone , then ( pie ) r ² h / 3 + 2 ( pie ) r ³ / 3 = 54 ( pie ) also given h = 4 r 4 r ³ / 3 + 2 r ³ / 3 = 54 2 r ³ = 54 r ³ = 27 r = 3 answer : c | a = volume_cylinder / (
b = a / 10
c = 1 / 3
d = c * math.pi
e = d * 4
f = 2 / 3
g = f * math.pi
h = e + g
i = b / h
j = math.sqrt(i)
|
a ) 1 , b ) 2 , c ) 3 , d ) 4 , e ) 5 | c | subtract(subtract(divide(multiply(add(3, 3), 9), 15), const_0_33), const_0_33) | if 1 = 3 2 = 6 3 = 9 4 = 12 5 = 15 then 9 = ? hint : its a logic riddle not a mathematical riddle | c 3 as stated 3 = 9 = > 9 = 3 answer is c | a = 3 + 3
b = a * 9
c = b / 15
d = c - const_0_33
e = d - const_0_33
|
a ) 150 , b ) 750 , c ) 1,250 , d ) 1,700 , e ) 2,500 | d | multiply(50, 5) | in a forest 170 deer were caught , tagged with electronic markers , then released . a week later , 50 deer were captured in the same forest . of these 50 deer , it was found that 5 had been tagged with the electronic markers . if the percentage of tagged deer in the second sample approximates the percentage of tagged deer in the forest , and if no deer had either left or entered the forest over the preceding week , what is the approximate number of deer in the forest ? | "the percentage of tagged deer in the second sample = 5 / 50 * 100 = 10 % . so , 170 tagged deers comprise 10 % of total # of deers - - > total # of deers = 170 * 10 = 1,700 . answer : d ." | a = 50 * 5
|
a ) 60 , b ) 80 , c ) 70 , d ) 45 , e ) 50 | c | divide(add(290, 100), add(5, 2)) | in an examination , a student scores 5 marks for every correct answer and loses 2 marks for every wrong answer . if he attempts all 100 questions and secures 290 marks , the number of questions he attempts correctly , is : | "let the number of correct answers be x . number of incorrect answers = ( 100 â € “ x ) . 5 x â € “ 2 ( 100 â € “ x ) = 290 or 7 x = 490 or x = 70 . answer : c" | a = 290 + 100
b = 5 + 2
c = a / b
|
a ) 15 , b ) 30 , c ) 25 , d ) 10 , e ) 15 | a | divide(divide(300, 2), 10) | there are 3 boats b 1 , b 2 and b 3 working together they carry 60 people in each trip . one day an early morning b 1 carried 50 people in few trips alone . when it stopped carrying the passengers b 2 and b 3 started carrying the people together . it took a total of 10 trips to carry 300 people by b 1 , b 2 and b 3 . it is known that each day on an average 300 people cross the river using only one of the 3 boats b 1 , b 2 and b 3 . how many trips it would take to b 1 , to carry 150 passengers alone ? | explanation : combined efficiency of all the three boats = 60 passenger / trip now , consider option ( a ) 15 trips and 150 passengers means efficiency of b 1 = 10 p / t which means in carrying 50 passengers b 1 must has taken 5 trips . so the rest trips equal to 5 ( 10 - 5 = 5 ) in which b 2 and b 3 together carried remaining 250 ( 300 - 50 = 250 ) passengers . therefore the efficiency of b 2 and b 3 = 250 / 5 = 50 p / t since , the combined efficiency of b 1 , b 2 and b 3 is 60 . which is same as given in the first statement hence option ( a ) is correct answer : a | a = 300 / 2
b = a / 10
|
a ) 22 , b ) 30 , c ) 34 , d ) 38 , e ) 27 | c | add(divide(add(21, 23), const_2), multiply(const_1, 12)) | the average age of 12 men is increased by years when two of them whose ages are 21 years and 23 years are replaced by two new men . the average age of the two new men is | "total age increased = ( 12 * 2 ) years = 24 years . sum of ages of two new men = ( 21 + 23 + 24 ) years = 68 years average age of two new men = ( 68 / 2 ) years = 34 years . answer : c" | a = 21 + 23
b = a / 2
c = 1 * 12
d = b + c
|
a ) 3 hours , b ) 4 hours , c ) 5 hours , d ) 6 hours , e ) 7 hours | b | divide(112, add(24, 4)) | a boat can travel with a speed of 24 km / hr in still water . if the speed of the stream is 4 km / hr , find the time taken by the boat to go 112 km downstream | "explanation : speed of the boat in still water = 24 km / hr speed of the stream = 4 km / hr speed downstream = ( 22 + 5 ) = 28 km / hr distance travelled downstream = 112 km time taken = distance / speed = 112 / 28 = 4 hours answer : option b" | a = 24 + 4
b = 112 / a
|
a ) 37800 , b ) 37600 , c ) 37200 , d ) 37500 , e ) none of them | a | multiply(multiply(power(const_3, const_3), multiply(power(const_2, const_3), power(add(const_4, const_1), const_2))), divide(divide(divide(divide(divide(2100, const_2), const_2), const_3), add(const_4, const_1)), add(const_4, const_1))) | find the l . c . m . of 72 , 108 and 2100 . | 72 = 2 ^ 3 x 3 ^ 2 , 108 = 3 ^ 3 x 2 ^ 2 , 2100 = 2 ^ 2 x 5 ^ 2 x 3 x 7 . l . c . m . = 2 ^ 3 x 3 ^ 3 x 5 ^ 2 x 7 = 37800 . answer is a . | a = 3 ** 3
b = 2 ** 3
c = 4 + 1
d = c ** 2
e = b * d
f = a * e
g = 2100 / 2
h = g / 2
i = h / 3
j = 4 + 1
k = i / j
l = 4 + 1
m = k / l
n = f * m
|
a ) 17 , b ) 19 , c ) 25 , d ) 31 , e ) 46 | b | divide(add(subtract(multiply(16, 10), 25), 55), 10) | the average of 10 numbers is calculated as 16 . it is discovered later on that while calculating the average , the number 55 was incorrectly read as 25 , and this incorrect number was used in the calculation . what is the correct average ? | "the total sum of the numbers should be increased by 30 . then the average will increase by 30 / 10 = 3 . the correct average is 19 . the answer is b ." | a = 16 * 10
b = a - 25
c = b + 55
d = c / 10
|
a ) 2 , b ) 4 , c ) 9 , d ) 5 , e ) 7 | c | subtract(subtract(multiply(3, 16), add(subtract(21, 16), 3)), 16) | the average of 1 st 3 of 4 numbers is 16 and of the last 3 are 15 . if the sum of the first and the last number is 21 . what is the last numbers ? | "a + b + c = 48 b + c + d = 45 a + d = 21 a – d = 3 a + d = 21 2 d = 18 d = 9 answer c" | a = 3 * 16
b = 21 - 16
c = b + 3
d = a - c
e = d - 16
|
a ) 48.74 inches , b ) 58.74 inches , c ) 78.74 inches , d ) 28.74 inches , e ) 88.74 inches | c | multiply(divide(2000, const_10), const_0_3937) | how many inches are in 2000 millimeters ? ( round your answer to the nearest hundredth of of an inch ) . | one inch is the same as 25.4 mm . let x inches be the same as 1000 mm x = 1 inch * 2000 mm / 25.4 mm = 78.74 inches correct answer c | a = 2000 / 10
b = a * const_0_3937
|
a ) 2.3 , b ) 2.6 , c ) 6.9 , d ) 4.5 , e ) 4.6 | c | divide(subtract(multiply(6, 6.40), add(multiply(2, 6.2), multiply(2, 6.1))), 2) | the average of 6 no . ' s is 6.40 . the average of 2 of them is 6.2 , while the average of the other 2 is 6.1 . what is the average of the remaining 2 no ' s ? | "sum of the remaining two numbers = ( 6.4 * 6 ) - [ ( 6.2 * 2 ) + ( 6.1 * 2 ) ] = 38.40 - ( 12.4 + 12.2 ) = 38.40 - 24.6 = 13.80 required average = ( 13.8 / 2 ) = 6.9 answer : c" | a = 6 * 6
b = 2 * 6
c = 2 * 6
d = b + c
e = a - d
f = e / 2
|
a ) 69 miles , b ) 88 miles , c ) 96 miles , d ) 91 miles , e ) 112 miles | d | divide(multiply(multiply(subtract(20, 6), add(20, 6)), 10), add(add(20, 6), subtract(20, 6))) | the current in a river is 6 mph . a boat can travel 20 mph in still water . how far up the river can the boat travel if the round trip is to take 10 hours ? | "upstream speed = 20 - 6 = 14 mph downstream speed = 20 + 6 = 26 mph d / 14 + d / 26 = 10 hours solving for d we get d = 91 answer : d" | a = 20 - 6
b = 20 + 6
c = a * b
d = c * 10
e = 20 + 6
f = 20 - 6
g = e + f
h = d / g
|
a ) 2 , b ) 4 , c ) 5 , d ) 6 , e ) 7 | b | divide(add(add(6, 5), 5), 4) | a school has 6 maths 5 physics and 5 chemistry teachers each teacher can teach 4 subjects max what is he minimum number of teachers required | "total subjects = 6 + 5 + 5 = 16 max subjects by 1 teacher = 4 so , min of teachers required = 16 / 4 = 4 answer : b" | a = 6 + 5
b = a + 5
c = b / 4
|
a ) 36 min , b ) 37 min , c ) 38 min , d ) 39 min , e ) 32 min | b | add(divide(subtract(60, 6), divide(3, const_2)), const_1) | if a monkey can climb 6 meter & slip down 3 meter in alternate minutes , than how much time can be taken by the monkey to climb up to a height of 60 meter . | in 1 st min it climb 6 m and in 2 nd min it slips 3 m , so it final displacement is 3 m in 2 min . now for 51 m it has taken 34 min . in 35 min it goes to 57 m . in 36 min come down to 54 m . in 37 min it goes to 60 answer : b | a = 60 - 6
b = 3 / 2
c = a / b
d = c + 1
|
a ) 174.2 , b ) 212 , c ) 288.1 , d ) 290 , e ) 282.4 | d | multiply(divide(46, 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 46 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 46 cm . : 46 / ( 2.54 * 1.5 ) = x / 24 - - - > x = 24 * 46 / ( 3.81 ) = 290 answer will be d ." | a = 46 / 2
b = 24 / 1
c = a * b
|
a ) 104 , b ) 106 , c ) 102 , d ) 192 , e ) 122 | a | subtract(104.25, divide(1, 4)) | the cash realised on selling a 14 % stock is rs . 104.25 , brokerage being 1 / 4 % is ? | "cash realised = rs . ( 104.25 - 0.25 ) = rs . 104 . answer : a" | a = 1 / 4
b = 104 - 25
|
a ) 22 , b ) 33 , c ) 77 , d ) 28 , e ) 22 | b | floor(divide(200, multiply(2, 3))) | how many numbers up to 200 are divisible by 2 and 3 both ? | explanation : 200 / 6 = 33 2 / 6 = > 33 numbers answer : b | a = 2 * 3
b = 200 / a
c = math.floor(b)
|
a ) a ) 7 , b ) b ) 27 , c ) c ) 23 , d ) d ) 19 , e ) e ) 21 | c | add(15, 8) | a dog has 15 bones , then he finds 8 more bones . now how many bones does he have . | 15 + 8 = 23 . answer is c . | a = 15 + 8
|
a ) $ 752 , b ) $ 755 , c ) $ 765 , d ) $ 773 , e ) $ 775 | c | divide(add(multiply(add(750, 20), 20), multiply(6, 750)), 26) | last year manfred received 26 paychecks . each of his first 6 paychecks was $ 750 ; each of his remaining paychecks was $ 20 more than each of his first 6 paychecks . to the nearest dollar , what was the average ( arithmetic mean ) amount of his pay checks for the year ? | = ( 750 * 6 + 770 * 20 ) / 26 = 765 answer is c . posted from my mobile device | a = 750 + 20
b = a * 20
c = 6 * 750
d = b + c
e = d / 26
|
a ) 5 % , b ) 7 % , c ) 6.25 % , d ) 2 % , e ) 4 % | c | divide(multiply(const_100, 200), multiply(800, 4)) | what is the rate percent when the simple interest on rs . 800 amount to rs . 200 in 4 years ? | "200 = ( 800 * 4 * r ) / 100 r = 6.25 % answer : c" | a = 100 * 200
b = 800 * 4
c = a / b
|
['a ) 1 : 6', 'b ) 1 : 4', 'c ) 1 : 0', 'd ) 1 : 2', 'e ) 1 : 1'] | b | multiply(const_2, divide(volume_cube(const_1), volume_cube(2))) | the sides of a cube are in the ratio 1 : 2 the ratio of their volume is ? | 1 : 4 answer : b | a = volume_cube / (
b = 2 * a
|
a ) 12 1 / 2 days , b ) 10 1 / 2 days , c ) 12 1 / 3 days , d ) 10 1 / 3 days , e ) 11 1 / 2 days | a | divide(const_1, subtract(divide(const_1, 12), divide(const_1, 15))) | a can do a piece of work in 12 days . he worked for 15 days and then b completed the remaining work in 10 days . both of them together will finish it in . | explanation : 15 / 25 + 10 / x = 1 = > x = 25 1 / 25 + 1 / 25 = 2 / 25 25 / 2 = 12 1 / 2 days answer a | a = 1 / 12
b = 1 / 15
c = a - b
d = 1 / c
|
a ) 150 , b ) 872 , c ) 3.33 , d ) 3.21 , e ) 30.1 | c | multiply(divide(multiply(6, const_1000), const_3600), 2) | a train running at the speed of 6 km / hr crosses a pole in 2 seconds . find the length of the train ? | "speed = 6 * ( 5 / 18 ) m / sec = 5 / 3 m / sec length of train ( distance ) = speed * time 5 / 3 ) * 2 = 3.33 meter answer : c" | a = 6 * 1000
b = a / 3600
c = b * 2
|
a ) 6 / 11 , b ) 8 / 13 , c ) 11 / 16 , d ) 12 / 17 , e ) 14 / 19 | a | divide(divide(subtract(multiply(7, add(1, 5)), 12), subtract(12, 7)), add(divide(subtract(multiply(7, add(1, 5)), 12), subtract(12, 7)), 5)) | the denominator of a fraction is 5 greater than the numerator . if the numerator and the denominator are increased by 1 , the resulting fraction is equal to 7 â „ 12 . what is the value of the original fraction ? | let the numerator be x . then the denominator is x + 5 . x + 1 / x + 6 = 7 / 12 . 12 x + 12 = 7 x + 42 . 5 x = 30 . x = 6 . the original fraction is 6 / 11 . the answer is a . | a = 1 + 5
b = 7 * a
c = b - 12
d = 12 - 7
e = c / d
f = 1 + 5
g = 7 * f
h = g - 12
i = 12 - 7
j = h / i
k = j + 5
l = e / k
|
a ) 108 , b ) 119 , c ) 120 , d ) 135 , e ) 143 | c | multiply(divide(subtract(22, divide(44, 22)), divide(44, 22)), add(divide(subtract(22, divide(44, 22)), divide(44, 22)), divide(44, 22))) | if the sum of two positive integers is 22 and the difference of their squares is 44 , what is the product of the two integers ? | "let the 2 positive numbers x and y x + y = 22 - - 1 x ^ 2 - y ^ 2 = 44 = > ( x + y ) ( x - y ) = 44 - - 2 using equation 1 in 2 , we get = > x - y = 2 - - 3 solving equation 1 and 3 , we get x = 12 y = 10 product = 12 * 10 = 120 answer c" | a = 44 / 22
b = 22 - a
c = 44 / 22
d = b / c
e = 44 / 22
f = 22 - e
g = 44 / 22
h = f / g
i = 44 / 22
j = h + i
k = d * j
|
a ) 800 , b ) 900 , c ) 1000 , d ) 1100 , e ) 1200 | c | divide(subtract(660, multiply(12.50, 40)), 0.16) | suppose you work for a manufacturing plant that pays you $ 12.50 an hour plus $ 0.16 for each widget you make . how many widgets must you produce in a 40 hour week to earn $ 660 ( before payroll deductions ) ? | "total pay = 40 * $ 12.50 + $ 0.16 * x = 660 x = 160 / 0.16 = 1000 the answer is c ." | a = 12 * 50
b = 660 - a
c = b / 0
|
a ) 18 kmph , b ) 22 kmph , c ) 16 kmph , d ) 20 kmph , e ) none of these | b | divide(add(10, 10), divide(57, const_60)) | in a stream running at 3 kmph , a motor boat goes 10 km upstream and back again to the starting point in 57 minutes . find the speed of motor boat in still water ? | "explanation : let the speed of motor boat instill water be x kmph then , speed in downstream = ( x + 3 ) km and . speed in upstream = ( x - 3 ) kmph time taken to row 10 km & back = ( 10 / x + 3,10 / x - 3 ) 10 / x + 3 + 10 / x - 3 = 57 / 60 11 x 2 - 240 x - 44 = 0 ( x - 22 ) ( 11 x + 2 ) = 0 x = 22 or x = - 2 / 11 then x = 22 kmph answer : option b" | a = 10 + 10
b = 57 / const_60
c = a / b
|
a ) 40 , b ) 44 , c ) 48 , d ) 22 , e ) none of these | d | add(add(multiply(5, divide(4, subtract(multiply(divide(6, 5), 5), 4))), divide(4, subtract(multiply(divide(6, 5), 5), 4))), 4) | in a can , there is a mixture of milk and water in the ratio 4 : 5 . if it is filled with an additional 4 litres of milk the can would be full and ratio of milk and water would become 6 : 5 . find the capacity of the can ? | "let the capacity of the can be t litres . quantity of milk in the mixture before adding milk = 4 / 9 ( t - 4 ) after adding milk , quantity of milk in the mixture = 6 / 11 t . 6 t / 11 - 4 = 4 / 9 ( t - 4 ) 10 t = 396 - 176 = > t = 22 . answer : d" | a = 6 / 5
b = a * 5
c = b - 4
d = 4 / c
e = 5 * d
f = 6 / 5
g = f * 5
h = g - 4
i = 4 / h
j = e + i
k = j + 4
|
['a ) 21', 'b ) 22', 'c ) 23', 'd ) 24', 'e ) 20'] | b | multiply(const_2, subtract(18, 7)) | people were sitting in a circle . 7 th one is direct opposite to 18 th one . . then how many were there in that group ? | 7 th one is exactly opposite represents half of circle . with 7 and 18 total members are 12 for remaing half = 12 * 2 - 2 ( 7 th and 18 th ( already taken ) ) = 22 answer : b | a = 18 - 7
b = 2 * a
|
a ) 3 , b ) 6 , c ) 9 , d ) 11 , e ) 14 | c | divide(add(221, 43), 17) | a no . when divided by 221 gives a remainder 43 , what remainder will be obtainedby dividingthe same no . 17 ? | "221 + 43 = 264 / 17 = 9 ( remainder ) c" | a = 221 + 43
b = a / 17
|
a ) 8 , b ) 5 , c ) 6 , d ) 7 , e ) 10 | a | subtract(8, const_1) | there are 25 balls which are red , blue or green . if 8 balls are blue and the sum of green balls and blue balls is less than 17 , at most how many green balls are there ? | "r + g + b = 25 b = 8 b + g < 17 = > 8 + g < 17 = > g < 9 = > at most 8 green balls answer : a" | a = 8 - 1
|
a ) 1 / 9 , b ) 1 / 6 , c ) 1 / 3 , d ) 4 / 9 , e ) 11 / 15 | e | divide(11, add(11, 4)) | a waitress ' s income consists of her salary and tips . during one week , her tips were 11 / 4 of her salary . what fraction of her income for the week came from tips ? | "her tips were 11 / 4 of her salary . let ' s say her salary = $ 4 this mean her tips = ( 11 / 4 ) ( $ 4 ) = $ 11 so , her total income = $ 4 + $ 11 = $ 15 what fraction of her income for the week came from tips $ 11 / $ 15 = 11 / 15 = e" | a = 11 + 4
b = 11 / a
|
a ) 30 , b ) 40 , c ) 50 , d ) 60 , e ) 70 | c | subtract(subtract(multiply(5, divide(add(multiply(15, 17), multiply(15, 7)), subtract(multiply(5, 7), 17))), 15), add(divide(add(multiply(15, 17), multiply(15, 7)), subtract(multiply(5, 7), 17)), 15)) | on a certain farm the ratio of horses to cows is 5 : 1 . if the farm were to sell 15 horses and buy 15 cows , the ratio of horses to cows would then be 17 : 7 . after the transaction , how many more horses than cows would the farm own ? | originally , there were 5 k horses and k cows . 7 ( 5 k - 15 ) = 17 ( k + 15 ) 35 k - 17 k = 255 + 105 18 k = 360 k = 20 the difference between horses and cows is ( 5 k - 15 ) - ( k + 15 ) = 4 k - 30 = 50 the answer is c . | a = 15 * 17
b = 15 * 7
c = a + b
d = 5 * 7
e = d - 17
f = c / e
g = 5 * f
h = g - 15
i = 15 * 17
j = 15 * 7
k = i + j
l = 5 * 7
m = l - 17
n = k / m
o = n + 15
p = h - o
|
a ) a ) 36 , b ) b ) 38 , c ) c ) 90 , d ) d ) 88 , e ) e ) 37 | a | divide(add(130, 50), 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 50 questions and secures 130 marks , the no of questions he attempts correctly is : | "let the number of correct answers be x . number of incorrect answers = ( 60 – x ) . 4 x – ( 50 – x ) = 130 = > 5 x = 180 = > x = 36 answer : a" | a = 130 + 50
b = 4 + 1
c = a / b
|
a ) 1240 , b ) 1120 , c ) 1190 , d ) 1634 , e ) none of these | d | add(684, divide(multiply(684, const_100), multiply(12, 6))) | the banker ' s gain on a sum due 6 years hence at 12 % per annum is rs . 684 . what is the banker ' s discount ? | "explanation : td = ( bg × 100 ) / tr = ( 684 × 100 ) / ( 6 × 12 ) = rs . 950 bg = bd – td = > 684 = bd - 950 = > bd = 684 + 950 = 1634 answer : option d" | a = 684 * 100
b = 12 * 6
c = a / b
d = 684 + c
|
a ) $ 153 , b ) $ 698 , c ) $ 330 , d ) $ 549 , e ) $ 675 | c | subtract(780, divide(multiply(subtract(930, 780), const_2.0), 4)) | a sum of money at simple interest amounts to $ 780 in 3 years and to $ 930 in 4 years . the sum is : | "c $ 330 s . i . for 1 year = $ ( 930 - 780 ) = $ 150 . s . i . for 3 years = $ ( 150 x 3 ) = $ 450 . principal = $ ( 780 - 450 ) = $ 330 ." | a = 930 - 780
b = a * 2
c = b / 4
d = 780 - c
|
a ) 6.6 , b ) 7.8 , c ) 8.9 , d ) 9.4 , e ) 10.2 | a | divide(multiply(15, add(add(multiply(multiply(add(const_3, const_2), const_2), multiply(multiply(const_3, const_4), const_100)), multiply(multiply(add(const_3, const_4), add(const_3, const_2)), multiply(add(const_3, const_2), const_2))), add(const_3, const_3))), const_100) | what is 15 percent of 44 ? | "( 15 / 100 ) * 44 = 6.6 the answer is a ." | a = 3 + 2
b = a * 2
c = 3 * 4
d = c * 100
e = b * d
f = 3 + 4
g = 3 + 2
h = f * g
i = 3 + 2
j = i * 2
k = h * j
l = e + k
m = 3 + 3
n = l + m
o = 15 * n
p = o / 100
|
a ) 2010 , b ) 2011 , c ) 2012 , d ) 2013 , e ) 2014 | d | add(2001, divide(add(divide(90, const_100), subtract(7.30, 5.20)), subtract(divide(45, const_100), subtract(7.30, 5.20)))) | the price of commodity x increases by 45 cents every year , while the price of commodity y increases by 20 cents every year . in 2001 , the price of commodity x was $ 5.20 and the price of commodity y was $ 7.30 . in which year will the price of commodity x be 90 cents more than the price of commodity y ? | "the price of commodity x increases 25 cents each year relative to commodity y . the price difference is $ 2.10 and commodity x needs to be 90 cents more than commodity y . $ 3.00 / 25 cents = 12 years the answer is 2001 + 12 years = 2013 . the answer is d ." | a = 90 / 100
b = 7 - 30
c = a + b
d = 45 / 100
e = 7 - 30
f = d - e
g = c / f
h = 2001 + g
|
a ) 6 , b ) 7 , c ) 8 , d ) 9 , e ) 10 | d | subtract(negate(9), multiply(subtract(1, 3), divide(subtract(1, 3), subtract(1, 1)))) | 1 , 1 , 3 , 9 , 5 , 25 , 7 , 49 _ , 81 | "as seq is 1 , 1 , 3 , 9 , 5 , 25 , 7 , 49 9 , 81 two series are there . 1 , 3,5 , 7,9 . . . odd numbers and 1 , 9,25 , 49,81 . . square of odd numbers answer : d" | a = negate - (
|
a ) 20 % , b ) 32 % , c ) 23.365 % , d ) 22.109 % , e ) 2 % | b | subtract(multiply(add(20, const_100), divide(add(10, const_100), const_100)), const_100) | a shopkeeper cheats both his supplier and customer by using faulty weights . when he buys from the supplier , he takes 20 % more than the indicated weight . when he sells to his customer , he gives the customer a weight such that 10 % of that is added to the weight , the weight claimed by the trader is obtained . if he charges the cost price of the weight that he claims , find his profit percentage . | anyways , one can infer that he ' steals ' 20 % from suppliers and then charges 10 % extra to customers so basically 1.2 * 1.1 = 1.32 given that 1 is start point , we get 32 % more hence answer is b | a = 20 + 100
b = 10 + 100
c = b / 100
d = a * c
e = d - 100
|
a ) 18 , b ) 19 , c ) 45 , d ) 21 , e ) 22 | c | divide(multiply(165, 3), add(multiply(4, const_2), 3)) | of 3 numbers , the third is 4 times the second and the second is two times the first . if their average is 165 , the smallest of the 3 numbers is : | explanation : let first number be x . so , 2 nd no . = 2 x & 3 rd no . = 8 x . so , x + 2 x + 8 x = 165 × 3 = 495 11 x = 495 / 11 x = 495 / 11 hence , smallest number x = 45 answer : c | a = 165 * 3
b = 4 * 2
c = b + 3
d = a / c
|
a ) 8 , b ) 8.5 , c ) 9 , d ) 9.5 , e ) 10 | e | divide(550, add(30, add(10, 20))) | a train travels at the rate of 10 miles / hr for the first hour of a trip , at 20 miles / hr for the second hour , at 30 miles / hr for the third hour and so on . how many hours will it take the train to complete a 550 - mile journey ? assume that the train makes no intermediate stops . | "a train travels at the rate of 10 miles / hr for the first hour of a trip , at 20 miles / hr for the second hour , at 30 miles / hr for the third hour and so on . how many hours will it take the train to complete a 550 - mile journey ? assume that the train makes no intermediate stops . i think the easiest way to solve this problem would be simply to count the number of miles it travels per hour ( and in total ) hour miles / hour total miles 1 10 10 2 20 30 3 30 60 4 40 100 5 50 150 6 60 210 7 70 280 8 80 360 9 90 450 10 100 550 it takes a total of nine hours to cover the 550 mile distance . answer : e" | a = 10 + 20
b = 30 + a
c = 550 / b
|
a ) 1 , b ) 10 , c ) 19 , d ) 5 , e ) 2 | d | subtract(multiply(divide(20, 2), add(divide(20, 2), 1)), multiply(divide(add(19, 1), 2), add(divide(subtract(19, 1), 2), 1))) | ifaequals the sum of the even integers from 2 to 20 , inclusive , andbequals the sum of the odd integers from 1 to 19 , inclusive , what is the value of ( a - b ) / 2 ? | "answer is 10 yes ! there is really a faster way to solve it . sum of consecutive odd or even integers = ( no . of odd or even ints ) * ( first int + last int ) / 2 here a = sum of even ints from 2 to 20 , inclusive number of even ints = 10 , first int + last int = 2 + 20 = 22 a = 10 * 22 / 2 = 110 b = sum of odd ints from 1 to 19 , inclusive number of odd ints = 10 , first int + last int = 1 + 19 = 20 a = 10 * 20 / 2 = 100 ( a - b ) / 2 = 110 - 100 = 10 / 2 = 5 ans : d" | a = 20 / 2
b = 20 / 2
c = b + 1
d = a * c
e = 19 + 1
f = e / 2
g = 19 - 1
h = g / 2
i = h + 1
j = f * i
k = d - j
|
a ) 68.8 , b ) 73.6 , c ) 75.2 , d ) 76.16 , e ) 81.6 | d | multiply(add(add(7.5, 7.8), 8.5), 3.2) | in a certain diving competition , 5 judges score each dive on a scale from 1 to 10 . the point value of the dive is obtained by dropping the highest score and the lowest score and multiplying the sum of the remaining scores by the degree of difficulty . if a dive with a degree of difficulty of 3.2 received scores of 7.5 , 7.8 , 9.0 , 6.0 , and 8.5 , what was the point value of the dive ? | degree of difficulty of dive = 3.2 scores are 6.0 , 7.5 , 8.0 , 8.5 and 9.0 we can drop 6.0 and 9.0 sum of the remaining scores = ( 7.5 + 7.8 + 8.5 ) = 23.8 point of value of the dive = 23.8 * 3.2 = 76.16 answer d | a = 7 + 5
b = a + 8
c = b * 3
|
a ) 160 , b ) 220 , c ) 336 , d ) 360 , e ) 420 | c | add(divide(multiply(divide(55, const_100), 28), subtract(divide(60, const_100), divide(55, const_100))), 28) | the workforce of company x is 60 % female . the company hired 28 additional male workers , and as a result , the percent of female workers dropped to 55 % . how many employees did the company have after hiring the additional male workers ? | "let ' s xx be total quantity of employees 0.6 x = females before adding men 0.55 ( x + 28 ) = females after adding men as quantity of women does n ' t change we can make an equation : 0.6 x = 0.55 ( x + 28 ) 0.05 x = 15.4 x = 308 - this is quantity of employees before adding 28 men so after adding it will be 336 answer is c" | a = 55 / 100
b = a * 28
c = 60 / 100
d = 55 / 100
e = c - d
f = b / e
g = f + 28
|
a ) 9.9 , b ) 8.8 , c ) 5.6 , d ) 11.9 , e ) 11.11 | e | add(25, const_1) | the average of first 25 prime numbers is ? | "explanation : average = ( 2 + 3 + 5 + 7 + 11 + 13 + 17 + 19 + 23 / 9 = 100 / 9 = 11.11 ( approx ) answer is e" | a = 25 + 1
|
a ) 23 days , b ) 26 days , c ) 17 days , d ) 29 days , e ) 20 days | c | divide(multiply(22, 14), 18) | 14 men can complete a piece of work in 22 days . in how many days can 18 men complete that piece of work ? | "c 17 days 14 * 22 = 18 * x = > x = 17 days" | a = 22 * 14
b = a / 18
|
a ) $ 8080 , b ) $ 8130 , c ) $ 8260 , d ) $ 8320 , e ) $ 8470 | e | multiply(7000, power(add(const_1, divide(10, const_100)), const_2)) | what amount does an investor receive if the investor invests $ 7000 at 10 % p . a . compound interest for two years , compounding done annually ? | "a = ( 1 + r / 100 ) ^ n * p ( 1.1 ) ^ 2 * 7000 = 1.21 * 7000 = 8470 the answer is e ." | a = 10 / 100
b = 1 + a
c = b ** 2
d = 7000 * c
|
a ) 11 , b ) 15 , c ) 13 , d ) 14 , e ) can not be determined | b | divide(150, divide(150, 15)) | 15 times a number gives 150 . the number is | explanation : let the number be ' n ' 15 × n = 150 ⇒ n = 10 correct option : b | a = 150 / 15
b = 150 / a
|
a ) 30 , b ) 16 , c ) 18 , d ) 20 , e ) 24 | a | divide(log(divide(multiply(const_3, const_10), add(const_4, const_1))), log(power(divide(multiply(const_2, const_10), add(const_4, const_1)), divide(const_1, 12)))) | on a certain date , pat invested $ 10,000 at x percent annual interest , compounded annually . if the total value of the investment plus interest at the end of 12 years will be $ 40,000 , in how many years , the total value of the investment plus interest will increase to $ 320,000 ? | "if i were to choose during the test , would go for 18 or 20 . probably 18 cuz it wont take too long to get the value doubled . . . . i found a method : rule of 72 . given an x % return , it takes 10,000 to quadralope 12 years . so according to the rule : 72 / x is the no of years 10 , 000.00 took to double 20 , 000.00 . again , 20 , 000.00 took to double 40 , 000.00 same ( 72 / x ) no of years . 72 / x + 72 / x = 12 x = 12 % ( though rate here is not very much required ) . again , 40 , 000.00 takes the same ( 72 / x ) no of years to double 320 , 000.00 . 72 / x = 6 years . so altogather : 10,000 - 20,000 = 6 years 20,000 - 40,000 = 6 years 40,000 - 80,000 = 6 years 80,000 - 160,000 = 6 years 160,000 - 320,000 = 6 years total 30 years ." | a = 3 * 10
b = 4 + 1
c = a / b
d = math.log(c)
e = 2 * 10
f = 4 + 1
g = e / f
h = 1 / 12
i = g ** h
j = math.log(i)
k = d / j
|
a ) 10 , b ) 13 , c ) 14 , d ) 15 , e ) 26 | e | divide(multiply(multiply(39, 12), 10), multiply(30, 6)) | 39 persons can repair a road in 12 days , working 10 hours a day . in how many days will 30 persons , working 6 hours a day , complete the work ? | "let the required number of days be x . less persons , more days ( indirect proportion ) more working hours per day , less days ( indirect proportion ) persons 30 : 39 : : 12 : x working hours / day 6 : 10 30 x 6 x x = 39 x 10 x 12 x = ( 39 x 10 x 12 ) / ( 30 x 6 ) x = 26 . answer : e" | a = 39 * 12
b = a * 10
c = 30 * 6
d = b / c
|
a ) 26.3 , b ) 35.6 , c ) 32.9 , d ) 25.5 , e ) 28.8 | e | subtract(subtract(subtract(multiply(0.6, const_4), const_4), const_4), const_1) | find the lcm of 0.6 , 9.6 and 0.36 . | "the given numbers are equivalent to 0.60 , 9.60 and 0.36 . now , find the lcm of 60,960 and 36 . which is equal to 2880 . the required lcm = 28.80 . answer is e" | a = 0 * 6
b = a - 4
c = b - 4
d = c - 1
|
a ) 4 % , b ) 4.1 % , c ) 4.2 % , d ) 4.3 % , e ) 4.5 % | a | divide(multiply(const_100, 320), multiply(4000, 2)) | what is the rate percent when the simple interest on rs . 4000 amount to rs . 320 in 2 years ? | "interest for 1 year = 320 / 2 = 160 interest on rs 4000 p / a = 160 interest rate = 160 / 4000 * 100 = 4 % answer : a" | a = 100 * 320
b = 4000 * 2
c = a / b
|
a ) 30 , b ) 35 , c ) 40 , d ) 70 , e ) 45 | b | divide(add(30, 40), const_2) | man can row upstream at 30 kmph and downstream at 40 kmph , and then find the speed of the man in still water ? | "us = 30 ds = 40 m = ( 30 + 40 ) / 2 = 35 answer : b" | a = 30 + 40
b = a / 2
|
a ) 4 : 9 , b ) 4 : 3 , c ) 4 : 12 , d ) 4 : 8 , e ) 5 : 2 | e | divide(sqrt(10), sqrt(4)) | two trains , one from howrah to patna and the other from patna to howrah , start simultaneously . after they meet , the trains reach their destinations after 4 hours and 10 hours respectively . the ratio of their speeds is | let us name the trains as a and b . then , ( a ' s speed ) : ( b ' s speed ) = b : a = 10 : 4 = 5 : 2 . answer : e | a = math.sqrt(10)
b = math.sqrt(4)
c = a / b
|
a ) 74 , b ) 76 , c ) 78 , d ) 80 , e ) 82 | a | divide(add(multiply(58, 3), multiply(98, 2)), add(2, 3)) | for a certain exam , a score of 58 was 2 standard deviations below mean and a score of 98 was 3 standard deviations above mean . what was the mean score q for the exam ? | a score of 58 was 2 standard deviations below the mean - - > 58 = mean - 2 d a score of 98 was 3 standard deviations above the mean - - > 98 = mean + 3 d solving above for mean q = 74 . answer : a . | a = 58 * 3
b = 98 * 2
c = a + b
d = 2 + 3
e = c / d
|
a ) 46 , b ) 54 , c ) 48 , d ) 49 , e ) 50 | b | add(add(47, multiply(3, 2)), const_1) | the standard deviation of a normal distribution of data is 2 , and 3 standard deviations below the mean is greater than 47 . what is a possible value for the mean of the distribution ? | the standard deviation ( { sd } ) = 2 ; 3 standard deviations below the mean is greater than 47 : { mean } - 3 * { sd } > 47 ; { mean } - 6 > 47 ; { mean } > 53 . answer : b . | a = 3 * 2
b = 47 + a
c = b + 1
|
a ) 11 , b ) 24 , c ) 99 , d ) 77 , e ) 18 | b | multiply(divide(subtract(1120, 900), 900), const_100) | a cycle is bought for rs . 900 and sold for rs . 1120 , find the gain percent ? | "900 - - - - 220 100 - - - - ? = > 24 % answer : b" | a = 1120 - 900
b = a / 900
c = b * 100
|
a ) 0.002 , b ) 0.02 , c ) 0.2 , d ) 20 , e ) 2 | e | multiply(divide(6.006, 3.003), const_100) | 6.006 / 3.003 | "answer is 2 , move the decimal forward three places for both numerator and denominator or just multiply both by a thousand . the result is 6006 / 3003 = 2 answer e" | a = 6 / 6
b = a * 100
|
a ) 2 : 9 , b ) 2 : 5 , c ) 3 : 6 , d ) 2 : 0 , e ) 1 : 3 | e | divide(subtract(4, 3), 3) | the ratio between the sale price and the cost price of an article is 4 : 3 . what is the ratio between the profit and the cost price of that article ? | "let c . p . = rs . 3 x and s . p . = rs . 4 x . then , gain = rs . x required ratio = x : 3 x = 1 : 3 answer : e" | a = 4 - 3
b = a / 3
|
a ) 50.5 , b ) 88.5 , c ) 37.5 , d ) 55.5 , e ) 23.5 | c | divide(multiply(divide(multiply(subtract(46, 36), const_1000), const_3600), 27), const_2) | two trains of equal length are running on parallel lines in the same direction at 46 km / hr and 36 km / hr . the faster train passes the slower train in 27 seconds . what is the length of each train ( in meters ) ? | "the relative speed = 46 - 36 = 10 km / hr = 10 * 5 / 18 = 25 / 9 m / s in 27 seconds , the relative difference in distance traveled is 27 * 25 / 9 = 75 meters this distance is twice the length of each train . the length of each train is 75 / 2 = 37.5 meters the answer is c ." | a = 46 - 36
b = a * 1000
c = b / 3600
d = c * 27
e = d / 2
|
a ) 125 , b ) 189 , c ) 297 , d ) 267 , e ) 298 | a | multiply(25, 5) | the average of 35 numbers is 25 . if each number is multiplied by 5 , find the new average ? | "sum of the 35 numbers = 35 * 25 = 875 if each number is multiplied by 5 , the sum also gets multiplied by 5 and the average also gets multiplied by 5 . thus , the new average = 25 * 5 = 125 . answer : a" | a = 25 * 5
|
a ) 20 , b ) 27 , c ) 25 , d ) 99 , e ) 21 | a | divide(subtract(multiply(90, 5), multiply(5, 50)), subtract(90, 80)) | the average mark of the students of a class in a particular exam is 80 . if 5 students whose average mark in that exam is 50 are excluded , the average mark of the remaining will be 90 . find the number of students who wrote the exam . | "let the number of students who wrote the exam be x . total marks of students = 80 x . total marks of ( x - 5 ) students = 90 ( x - 5 ) 80 x - ( 5 * 50 ) = 90 ( x - 5 ) 200 = 10 x = > x = 20 . answer : a" | a = 90 * 5
b = 5 * 50
c = a - b
d = 90 - 80
e = c / d
|
a ) 0 , b ) 1 , c ) 4 , d ) 6 , e ) 8 | a | divide(log(5), log(power(5, 11))) | if n = 5 ^ 11 – 5 , what is the units digit of n ? | always divide the power ( incase 11 ) by 4 and use the remainder as the new power . the question now becomes 5 ^ 3 - 5 . now 5 ^ 3 has last digit 5 . we subtract 5 from 5 = 0 is the answer . option a | a = math.log(5)
b = 5 ** 11
c = math.log(b)
d = a / c
|
a ) rs . 3000 , b ) rs . 1600 , c ) rs . 2400 , d ) rs . 4000 , e ) none of these | b | multiply(divide(4000, add(const_1, divide(const_2, const_3))), divide(const_2, const_3)) | p , q and r have rs . 4000 among themselves . r has two - thirds of the total amount with p and q . find the amount with r ? | "let the amount with r be rs . r r = 2 / 3 ( total amount with p and q ) r = 2 / 3 ( 4000 - r ) = > 3 r = 8000 - 2 r = > 5 r = 8000 = > r = 1600 . answer : b" | a = 2 / 3
b = 1 + a
c = 4000 / b
d = 2 / 3
e = c * d
|
a ) 310 , b ) 312 , c ) 320 , d ) 325 , e ) 345 | c | add(8500, divide(1600, 1)) | 8500 + ( 1600 ÷ ? ) of 1 / 5 = 8501 | "explanation : = > 8500 + ( 1600 / ? ) × 1 / 5 = 8501 = > 1600 / ? × 1 / 5 = 8501 - 8500 = 1 = > ? = 1600 / 5 = 320 answer : option c" | a = 1600 / 1
b = 8500 + a
|
a ) 2 , b ) 0.02 , c ) 0.2 , d ) 20 , e ) 2 | a | multiply(divide(8.008, 4.004), const_100) | 8.008 / 4.004 | "answer is 2 , move the decimal forward three places for both numerator and denominator or just multiply both by a thousand . the result is 8008 / 4004 = 2 answer a" | a = 8 / 8
b = a * 100
|
a ) 250 , b ) 255 , c ) 254 , d ) 253 , e ) 251 | e | subtract(power(const_2, multiply(4, const_2)), const_1) | two friends a , b decided to share a lot of apples . each of them had half of the total plus half an apple in order . after each of them took their share 4 time , no apples were left . how many apples were there ? | whenever the rate of reduction is ' half of the total and half of it ' , the answer is always ( 2 ^ n ) - 1 , where ' n ' is the number of times the process is repeated . here , the process is repeated 8 times . so answer is ( 2 ^ 8 ) - 1 = 255 answer : e | a = 4 * 2
b = 2 ** a
c = b - 1
|
a ) $ 410 , b ) $ 500 , c ) $ 650 , d ) $ 710 , e ) $ 900 | e | multiply(10000, divide(9, const_100)) | find the simple interest on $ 10000 at 9 % per annum for 12 months ? | p = $ 10000 r = 9 % t = 12 / 12 years = 1 year s . i . = p * r * t / 100 = 10000 * 9 * 1 / 100 = $ 900 answer is e | a = 9 / 100
b = 10000 * a
|
a ) 2 , b ) 4 , c ) 6 , d ) 8 , e ) 10 | a | subtract(15, reminder(3, 5)) | when positive integer n is divided by 3 , the remainder is 1 . when n is divided by 5 , the remainder is 3 . what is the smallest positive integer k such that k + n is a multiple of 15 ? | "n = 3 p + 1 = 5 q + 3 n + 2 = 3 p + 3 = 5 q + 5 n + 2 is a multiple of 3 and 5 , so it is a multiple of 15 . the answer is a ." | a = 15 - reminder
|
a ) 47 , b ) 49 , c ) 42 , d ) 41 , e ) 40 | a | add(divide(subtract(256, 64), 4), const_1) | how many multiples of 4 are there between 64 and 256 ? | "it should be mentioned whether 64 and 256 are inclusive . if 64 and 256 are inclusive , then the answer is ( 256 - 64 ) / 4 + 1 = 49 . if 64 and 256 are not inclusive , then the answer is ( 252 - 68 ) / 4 + 1 = 47 . since oa is a , then we have not inclusive case ." | a = 256 - 64
b = a / 4
c = b + 1
|
a ) 2 , b ) 5 , c ) 10 , d ) 12 , e ) 15 | c | divide(divide(14, const_2), divide(70, const_100)) | at the end of the month , a certain ocean desalination plant ’ s reservoir contained 14 million gallons of water . this amount is twice the normal level . if this amount represents 70 % of the reservoir ’ s total capacity , how many million gallons short of total capacity is the normal level ? | the q talks of total capacity , normal level , present level , shortage etc . . so it is all about not going wrong in these terms 14 mg = 70 % of total . . total = 14 / . 7 = 20 mg . . normal level = 1 / 2 of 20 = 10 mg . . shortage of normal level = 20 - 10 = 10 mg . . c | a = 14 / 2
b = 70 / 100
c = a / b
|
a ) 14.3 % , b ) 16.67 % , c ) 33 % , d ) 28.6 % , e ) 49.67 % | b | multiply(multiply(divide(divide(divide(2, 7), 2), add(divide(divide(2, 7), 2), subtract(const_1, divide(2, 7)))), const_100), const_3) | of the 14,210 employees of the anvil factory , 2 / 7 are journeymen . if half of the journeymen were laid off , what percentage of the total remaining employees would be journeymen ? | "the number of employees is not reqd . . . 1 / 2 of 2 / 7 are shunted out so 1 / 7 left . . . there are another 5 equal parts ( 5 / 7 ) . . so total remaining parts = 6 , in which 1 part is journeyman . . . . therefore % = 1 / 6 * 100 = 16.67 % answer : b" | a = 2 / 7
b = a / 2
c = 2 / 7
d = c / 2
e = 2 / 7
f = 1 - e
g = d + f
h = b / g
i = h * 100
j = i * 3
|
a ) 20 , b ) - 6 , c ) 30 , d ) 32 , e ) 48 | b | multiply(multiply(multiply(negate(2), 1), 1), subtract(negate(2), 1)) | what is the value of x ^ 2 yz − xyz ^ 2 , if x = 1 , y = 1 , and z = 3 ? | "1 * 1 * 3 - ( 1 * 1 * 9 ) = 3 - 9 = - 6 ans : b" | a = negate * (
b = a * 1
c = b * 1
|
a ) 8 ° , b ) 10 ° , c ) 18 ° , d ) 54 ° , e ) 52 ° | d | divide(multiply(subtract(const_100, add(add(add(add(14, 24), 10), 29), 8)), divide(const_3600, const_10)), const_100) | a circle graph shows how the megatech corporation allocates its research and development budget : 14 % microphotonics ; 24 % home electronics ; 10 % food additives ; 29 % genetically modified microorganisms ; 8 % industrial lubricants ; and the remainder for basic astrophysics . if the arc of each sector of the graph is proportional to the percentage of the budget it represents , how many degrees of the circle are used to represent basic astrophysics research ? | "14 % microphotonics ; 24 % home electronics ; 10 % food additives ; 29 % genetically modified microorganisms ; 8 % industrial lubricants ; 100 - ( 14 + 24 + 10 + 29 + 8 ) = 15 % basic astrophysics . 15 % of 360 ° is 54 ° . answer : d ." | a = 14 + 24
b = a + 10
c = b + 29
d = c + 8
e = 100 - d
f = 3600 / 10
g = e * f
h = g / 100
|
a ) 121 , b ) 11 , c ) 9 , d ) 10 , e ) 3 | a | power(add(7, const_4), const_4) | the difference between a 7 digit number and the number formed by reversing its digit is not a multiple of | "another approach is to test a number . let ' s say the original number is 1231231 so , the reversed number is 1321321 the difference = 1321321 - 1231231 = 90090 no check the answer choices 90090 is a multiple of 3,9 , 10,11 121 is not a multiple of 90090 answer : a" | a = 7 + 4
b = a ** 4
|
a ) 120 sec , b ) 125 sec , c ) 135 sec , d ) 140 sec , e ) none of these | c | divide(add(150, 120), subtract(divide(120, 10), divide(150, 15))) | two trains of length 120 meter and 150 meter crosses a stationary man in 10 and 15 seconds respectively . in what time they will cross each other when they are moving in same direction . | explanation : 120 = a * 10 , a = 12 m / sec ( speed of first train ) 150 = b * 15 , b = 10 m / sec ( speed of second train ) 270 = ( 2 ) * t , t = 135 seconds answer – c | a = 150 + 120
b = 120 / 10
c = 150 / 15
d = b - c
e = a / d
|
a ) 11 minutes , b ) 10 minutes , c ) 12 minutes , d ) 13 minutes , e ) 14 minutes | b | divide(const_1, subtract(add(divide(const_1, divide(45, const_3)), divide(const_1, add(const_3, divide(45, const_3)))), divide(const_1, 45))) | a , b two inlet pipes takes 15,18 minutes to fill the tank and c an oulet pipe takes 45 minutes to empty the tank respectively . in what time the tank be filled if all of them are operated together ? | lcm = 90 no of days = [ 90 / ( 90 / 15 + 90 / 18 - 90 / 45 ) = [ 90 / ( 6 + 5 - 2 ) ] = [ 90 / 9 ] = 10 minutes answer : b | a = 45 / 3
b = 1 / a
c = 45 / 3
d = 3 + c
e = 1 / d
f = b + e
g = 1 / 45
h = f - g
i = 1 / h
|
a ) 24 . , b ) 32 . , c ) 36 . , d ) 42 . , e ) 44 . | d | inverse(add(inverse(63), divide(inverse(63), const_2))) | a car traveled from san diego to san francisco at an average speed of 63 miles per hour . if the journey back took twice as long , what was the average speed of the trip ? | "let the time taken be = x one way distance = 63 x total distance traveled = 2 * 63 x = 126 x total time taken = x + 2 x = 3 x average speed = 126 x / 3 x = 42 answer : d" | a = 1/(63)
b = 1/(63)
c = b / 2
d = a + c
e = 1/(d)
|
a ) 1.4 % , b ) 5.9 % , c ) 11.1 % , d ) 12.5 % , e ) 33.33 % | e | multiply(subtract(divide(8, 6), const_1), const_100) | at the opening of a trading day at a certain stock exchange , the price per share of stock k was $ 6 . if the price per share of stock k was $ 8 at the closing of the day , what was the percent increase in the price per share of stock k for that day ? | "opening = 6 closing = 8 rise in price = 2 so , percent increase = 2 / 6 * 100 = 33.33 % answer : e" | a = 8 / 6
b = a - 1
c = b * 100
|
a ) 377 , b ) 197 , c ) 187 , d ) 100 , e ) 822 | d | multiply(divide(subtract(multiply(9, 900), multiply(multiply(const_3, const_4), 650)), multiply(multiply(const_3, const_4), const_1)), const_4) | a man engaged a servant on the condition that he would pay him rs . 900 and a uniform after one year service . he served only for 9 months and received uniform and rs . 650 , find the price of the uniform ? | "explanation : 9 / 12 = 3 / 4 * 900 = 675 650 - - - - - - - - - - - - - 25 1 / 4 - - - - - - - - 25 1 - - - - - - - - - ? = > rs . 100 answer : d" | a = 9 * 900
b = 3 * 4
c = b * 650
d = a - c
e = 3 * 4
f = e * 1
g = d / f
h = g * 4
|
a ) 55 , b ) 45 , c ) 35 , d ) 25 , e ) 15 | c | divide(multiply(29.75, const_100), 85) | how many pieces of 85 cm length can be cut from a rod of 29.75 meters long ? | "number of pieces = 2975 / 85 = 35 the answer is c ." | a = 29 * 75
b = a / 85
|
a ) 2 , b ) 5 , c ) 11 , d ) 13 , e ) 19 | d | subtract(divide(30, 2), const_1) | for every even positive integer m , f ( m ) represents the product of all even integers from 2 to m , inclusive . for example , f ( 12 ) = 2 x 4 x 6 x 8 x 10 x 12 . what is the greatest prime factor of f ( 30 ) ? | "f ( 30 ) = 2 * 4 * 6 * 8 * 10 * 12 * 14 * 16 * 18 * 20 * 22 * 24 * 26 * 28 * 30 the greatest prime factor in this list is 13 . the answer is d ." | a = 30 / 2
b = a - 1
|
a ) 13.28 % , b ) 14 % , c ) 15 % , d ) 16 % , e ) 50 % | e | subtract(divide(multiply(const_100, 3), subtract(3, 1)), const_100) | in a office work is distribute between p persons . if 1 / 3 members are absent then work increased for each person is ? | "let total % of work is 100 % total person = p 1 / 3 person are absent of total person . so absent person is 1 / 3 p ie p / 3 . left person is , p - p / 3 = 2 p / 3 . p person do the work 100 % 1 person do the work 100 * p % 2 p / 3 person do the work ( 100 * p * 3 ) / 2 p % = 150 % work increased for each person is = ( 150 - 100 ) % = 50 % answer : e" | a = 100 * 3
b = 3 - 1
c = a / b
d = c - 100
|
a ) 960 cm 2 , b ) 240 cm 2 , c ) 480 cm 2 , d ) 45 cm 2 , e ) 250 cm 2 | a | multiply(60, 16) | find the area of a parallelogram with base 60 cm and height 16 cm ? | "area of a parallelogram = base * height = 60 * 16 = 960 cm 2 answer : a" | a = 60 * 16
|
['a ) 11647', 'b ) 11550', 'c ) 12654', 'd ) 26537', 'e ) 32651'] | b | multiply(divide(multiply(20, add(20, const_1)), const_2), divide(multiply(add(10, const_1), 10), const_2)) | in front of you lies a figure made up with 20 x 10 square blocks . will you be able to find out the number of unique squares and rectangles that are being formed inside this figure by combining two or more distinct squares ? | b 11550 there is just one way to find it out and that is the generic computing method . the number of squares and rectangles formed = ( summation of column numbers ) = ( 1 + 2 + 3 + . . . . . . + 19 + 20 ) * ( 1 + 2 + 3 + . . . . . . + 9 + 10 ) = 210 * 55 = 11550 | a = 20 + 1
b = 20 * a
c = b / 2
d = 10 + 1
e = d * 10
f = e / 2
g = c * f
|
a ) s . 4580 , b ) s . 4570 , c ) s . 4500 , d ) s . 4550 , e ) s . 4900 | e | subtract(multiply(8000, const_4), subtract(multiply(8400, const_4), 6500)) | the average salary of a person for the months of january , february , march and april is rs . 8000 and that for the months february , march , april and may is rs . 8400 . if his salary for the month of may is rs . 6500 , find his salary for the month of january ? | "sum of the salaries of the person for the months of january , february , march and april = 4 * 8000 = 32000 - - - - ( 1 ) sum of the salaries of the person for the months of february , march , april and may = 4 * 8400 = 33600 - - - - ( 2 ) ( 2 ) - ( 1 ) i . e . may - jan = 1600 salary of may is rs . 6500 salary of january = rs . 4900 answer : e" | a = 8000 * 4
b = 8400 * 4
c = b - 6500
d = a - c
|
a ) 181 , b ) 211 , c ) 241 , d ) 271 , e ) 301 | c | divide(multiply(divide(const_3600, const_4), const_3), 6) | if a light flashes every 6 seconds , how many times will it flash in 2 / 5 of an hour ? | "in 2 / 5 of an hour there are 24 * 60 = 1440 seconds the number of 6 - second intervals = 1440 / 6 = 240 after the first flash , there will be 240 more flashes for a total of 241 . the answer is c ." | a = 3600 / 4
b = a * 3
c = b / 6
|
['a ) 32 cm', 'b ) 48 cm', 'c ) 50 cm', 'd ) 28 cm', 'e ) 30 cm'] | b | multiply(divide(104, add(add(divide(1, 4), divide(1, 3)), divide(1, 2))), divide(1, 2)) | the sides of the triangle are in the ratio 1 / 2 : 1 / 3 : 1 / 4 and its perimeter is 104 cm . the length of the longest side is ? | ratio of sides = 1 / 2 : 1 / 3 : 1 / 4 = 6 : 4 : 3 largest side = 104 * 6 / 13 = 48 cm answer is b | a = 1 / 4
b = 1 / 3
c = a + b
d = 1 / 2
e = c + d
f = 104 / e
g = 1 / 2
h = f * g
|
a ) 5 , b ) 7 , c ) 8 , d ) 10 , e ) 11 | b | add(divide(25, 5), const_2) | on a race track a maximum of 5 horses can race together at a time . there are a total of 25 horses . there is no way of timing the races . what is the minimum number w of races we need to conduct to get the top 3 fastest horses ? | "w = 7 is the correct answer . good solution buneul . b" | a = 25 / 5
b = a + 2
|
a ) 19.33 , b ) 54.55 , c ) 89.33 , d ) 97.21 , e ) 68.77 | c | divide(multiply(66.67, add(add(multiply(multiply(add(const_3, const_2), const_2), multiply(multiply(const_3, const_4), const_100)), multiply(multiply(add(const_3, const_4), add(const_3, const_2)), multiply(add(const_3, const_2), const_2))), add(const_3, const_3))), const_100) | what is 66.67 % of 804 of 0.1666 ? | "66.67 % = 2 / 3 2 / 3 of 804 = 2 / 3 * 804 = 536 536 of 0.166 = 536 * ( 1 / 6 ) = 89.33 . . . . ans - c" | a = 3 + 2
b = a * 2
c = 3 * 4
d = c * 100
e = b * d
f = 3 + 4
g = 3 + 2
h = f * g
i = 3 + 2
j = i * 2
k = h * j
l = e + k
m = 3 + 3
n = l + m
o = 66 * 67
p = o / 100
|
a ) rs . 700 , b ) rs . 780 , c ) rs . 750 , d ) rs . 1000 , e ) none of these | c | divide(870, add(divide(multiply(divide(add(multiply(3, 5), 3), 5), 5), const_100), const_1)) | find the principle on a certain sum of money at 5 % per annum for 3 1 / 5 years if the amount being rs . 870 ? | "explanation : 870 = p [ 1 + ( 5 * 16 / 5 ) / 100 ] p = 750 answer : option c" | a = 3 * 5
b = a + 3
c = b / 5
d = c * 5
e = d / 100
f = e + 1
g = 870 / f
|
a ) 21 , b ) 22 , c ) 29 , d ) 26 , e ) 28 | c | add(divide(subtract(60, 1), 2), const_1) | how many multiples of 2 are there between 1 and 60 , exclusive ? | "29 multiples of 2 between 1 and 60 exclusive . from 2 * 1 upto 2 * 29 , ( 1,2 , 3,4 , . . . , 29 ) . hence , 29 multiples ! correct option is c" | a = 60 - 1
b = a / 2
c = b + 1
|
a ) 2880 , b ) 2870 , c ) 2860 , d ) 2850 , e ) 2840 | a | divide(multiply(480, 576), subtract(576, 480)) | the bankers discount of a certain sum of money is rs . 576 and the true discount on the same sum for the same time is rs . 480 . the sum due is : | "sum = ( b . d * t . d ) / ( b . d - t . d ) ( 576 * 480 ) / 576 - 480 ; 2880 answer : a" | a = 480 * 576
b = 576 - 480
c = a / b
|
a ) $ 25 , b ) $ 17.50 , c ) $ 29.65 , d ) $ 35.95 , e ) $ 45.62 | b | divide(multiply(subtract(const_100, 30), divide(50, const_2)), const_100) | a pair of articles was bought for $ 50 at a discount of 30 % . what must be the marked price of each of the article ? | "s . p . of each of the article = 50 / 2 = $ 25 let m . p = $ x 70 % of x = 25 x = 25 * . 7 = $ 17.50 answer is b" | a = 100 - 30
b = 50 / 2
c = a * b
d = c / 100
|
a ) 45 , b ) 35 , c ) 55 , d ) 30 , e ) 40 | e | subtract(185, add(add(50, 100), divide(add(50, subtract(150, add(add(100, 110), 50))), 2))) | in an intercollegiate competition that lasted for 3 days , 185 students took part on day 1 , 150 on day 2 and 200 on day 3 . if 100 took part on day 1 and day 2 and 110 took part on day 2 and day 3 and 50 took part on all three days , how many students took part only on day 3 ? | "day 1 & 2 = 100 ; only day 1 & 2 ( 100 - 50 ) = 50 , day 2 & 3 = 110 ; only day 2 & 3 ( 110 - 50 ) = 60 , only day 3 = 200 - ( 50 + 60 + 50 ) = 40 answer : e" | a = 50 + 100
b = 100 + 110
c = b + 50
d = 150 - c
e = 50 + d
f = e / 2
g = a + f
h = 185 - g
|
a ) 10.78 mps , b ) 12.78 mps , c ) 97.78 mps , d ) 17.78 mps , e ) 18.78 mps | b | multiply(const_0_2778, 46) | express a speed of 46 kmph in meters per second ? | "46 * 5 / 18 = 12.78 mps answer : b" | a = const_0_2778 * 46
|
a ) 23 , b ) 25 , c ) 89 , d ) 30 , e ) 32 | c | add(80, add(subtract(8, 5), add(subtract(6, 5), subtract(7, 2)))) | simplify 80 â ˆ ’ [ 5 â ˆ ’ ( 6 + 2 ( 7 â ˆ ’ 8 â ˆ ’ 5 â ¯ â ¯ â ¯ â ¯ â ¯ â ¯ â ¯ â ¯ â ¯ â ¯ â ¯ ) ) ] | explanation : = 80 â ˆ ’ [ 5 â ˆ ’ ( 6 + 2 ( 7 â ˆ ’ 8 + 5 ) ) ] ( please check due to overline , sign has been changed ) = 80 â ˆ ’ [ 5 â ˆ ’ ( 6 + 2 ã — 4 ) ) ] = 80 â ˆ ’ [ â ˆ ’ 9 ] = 80 + 9 = 89 option c | a = 8 - 5
b = 6 - 5
c = 7 - 2
d = b + c
e = a + d
f = 80 + e
|
Subsets and Splits
No community queries yet
The top public SQL queries from the community will appear here once available.