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a ) 26 , b ) 32 , c ) 33 , d ) 34 , e ) 54 | e | subtract(add(floor(divide(subtract(99, 19), 3)), divide(subtract(99, 19), 2)), floor(divide(subtract(99, 19), multiply(2, 3)))) | if w is the set of all the integers between 19 and 99 , inclusive , that are either multiples of 3 or multiples of 2 or multiples of both , then w contains how many numbers ? | "multiples of 2 from 19 to 99 = multiples of 2 from 1 to 99 - multiples of 2 from 1 to 18 = [ 99 / 2 ] - [ 18 / 2 ] = 49 - 9 = 40 multiples of 3 from 19 to 99 = multiples of 3 from 1 to 99 - multiples of 3 from 1 to 18 = [ 99 / 3 ] - [ 18 / 3 ] = 33 - 6 = 27 multiples of 2 and 3 bothi . e . 6 from 19 to 99 = multiples of 6 from 1 to 99 - multiples of 6 from 1 to 18 = [ 99 / 6 ] - [ 18 / 6 ] = 16 - 3 = 13 these 8 numbers have been counted twice in both the above calculation while calculating multiples of 2 and 3 i . e . total numbers in w = 40 + 27 - 13 = 54 answer option e" | a = 99 - 19
b = a / 3
c = math.floor(b)
d = 99 - 19
e = d / 2
f = c + e
g = 99 - 19
h = 2 * 3
i = g / h
j = math.floor(i)
k = f - j
|
a ) 2550 , b ) 2640 , c ) 3960 , d ) 4280 , e ) 5490 | b | multiply(const_pi, multiply(28, 30)) | find the curved surface area , if the radius of a cone is 28 m and slant height is 30 m ? | "cone curved surface area = ï € rl 22 / 7 ã — 28 ã — 30 = 88 ã — 30 = 2640 m ( power 2 ) answer is b ." | a = 28 * 30
b = math.pi * a
|
a ) 3 , b ) 2 , c ) 1 , d ) 0 , e ) 4 | e | subtract(subtract(16, 5), subtract(12, 5)) | | 16 - 5 | - | 5 - 12 | = ? | "| 16 - 5 | - | 5 - 12 | = | 11 | - | - 7 | = 11 - 7 = 4 correct answer e" | a = 16 - 5
b = 12 - 5
c = a - b
|
a ) 90 seconds , b ) 70 seconds , c ) 135 seconds , d ) 100 seconds , e ) 120 seconds | c | multiply(divide(90, const_2), const_3) | if two - third of a bucket is filled in 90 seconds then the time taken to fill the bucket completely will be . | "2 / 3 filled in 90 seconds 1 / 3 filled in 45 seconds then 2 / 3 + 1 / 3 = 90 + 45 seconds = 135 seconds answer : c" | a = 90 / 2
b = a * 3
|
['a ) 7 meters', 'b ) 9 meters', 'c ) 8 meters', 'd ) 5 meters', 'e ) 3 meters'] | d | multiply(divide(120, multiply(multiply(multiply(const_2, const_3), const_10), 400)), const_1000) | if a tire rotates at 400 revolutions per minute when the car is traveling 120 km / h , what is the circumference of the tire ? | 400 rev / minute = 400 * 60 rev / 60 minutes = 24,000 rev / hour 24,000 * c = 12,0000 m : c is the circumference c = 5 meters correct answer d | a = 2 * 3
b = a * 10
c = b * 400
d = 120 / c
e = d * 1000
|
a ) 4 % , b ) 6 % , c ) 5 % , d ) 50 % , e ) none of these | c | multiply(divide(divide(subtract(add(2, divide(62500, multiply(const_1000, const_100))), add(1, divide(75000, multiply(const_1000, const_100)))), const_10), add(1, divide(75000, multiply(const_1000, const_100)))), const_100) | the population of a town increased from 1 , 75000 to 2 , 62500 in a decade . what is the average percent increase of population per year ? | explanation : increase in the population in 10 years = 2 , 62500 - 1 , 75000 = 87500 % ncrease in the population in 10 years = ( 87500 / 175000 ) × 100 = 8750 / 175 = 50 % average % increase of population per year = 50 % / 10 = 5 % answer : option c | a = 1000 * 100
b = 62500 / a
c = 2 + b
d = 1000 * 100
e = 75000 / d
f = 1 + e
g = c - f
h = g / 10
i = 1000 * 100
j = 75000 / i
k = 1 + j
l = h / k
m = l * 100
|
a ) 50 , b ) 20 , c ) 30 , d ) 10 , e ) 15 | a | divide(multiply(10, 10), 2) | 10 men do a work in 10 days . how many men are needed to finish the work in 2 days ? | "men required to finish the work in 2 days = 10 * 10 / 2 = 50 answer is a" | a = 10 * 10
b = a / 2
|
a ) 14 years , b ) 7 years , c ) 19 years , d ) 21 years , e ) none | b | subtract(add(add(multiply(35, 16), 21), 35), multiply(35, 16)) | the average age of 35 students in a class is 16 years . the average age of 21 students is 14 . what is the average age of remaining 38 students ? | "solution sum of the ages of 14 students = ( 16 x 35 ) - ( 14 x 21 ) = 560 - 294 . = 266 . ∴ required average = 266 / 38 = 7 years . answer b" | a = 35 * 16
b = a + 21
c = b + 35
d = 35 * 16
e = c - d
|
a ) 5.5 , b ) 8.5 , c ) 6.5 , d ) 9.5 , e ) 7.5 | e | divide(multiply(multiply(subtract(8, 2), add(8, 2)), 2), add(add(8, 2), subtract(8, 2))) | a man can row 8 kmph in still water . if the velocity of the current is 2 kmph and it takes him 2 hours to row to a place and come back . how far is the place ? | man ' s rate down stream = 8 + 2 = 10 kmph man ' s rate upstream = 8 - 2 = 6 kmph let the required distance be x km then x / 10 + x / 6 = 2 3 x + 5 x = 60 8 x = 60 x = 7.5 km answer is e . | a = 8 - 2
b = 8 + 2
c = a * b
d = c * 2
e = 8 + 2
f = 8 - 2
g = e + f
h = d / g
|
a ) 601 , b ) 611 , c ) 621 , d ) 631 , e ) 641 | e | add(multiply(divide(add(20, 40), const_2), add(subtract(40, 20), const_1)), add(divide(subtract(40, 20), const_2), const_1)) | if x is equal to the sum of the integers from 20 to 40 , inclusive , and y is the number of even integers from 20 to 40 , inclusive , what is the value of x + y ? | "x = 20 + 21 + . . . + 40 = 21 ( 30 ) = 630 y = 11 x + y = 641 the answer is e ." | a = 20 + 40
b = a / 2
c = 40 - 20
d = c + 1
e = b * d
f = 40 - 20
g = f / 2
h = g + 1
i = e + h
|
a ) 15 , b ) 10 , c ) 20 , d ) 30 , e ) 35 | a | multiply(divide(multiply(subtract(20, 5), divide(5, const_60)), 5), const_60) | the hiker walking at a constant rate of 5 miles per hour is passed by a cyclist traveling in the same direction along the same path at 20 miles per hour . the cyclist stops to wait for the hiker 5 minutes after passing her , while the hiker continues to walk at her constant rate , how many minutes must the cyclist wait until the hiker catches up ? | "after passing the hiker the cyclist travels for 5 minutes at a rate of 20 miles / hour . in those 5 mins the cyclist travels a distance of 5 / 3 miles . in those 5 mins the hiker travels a distance of 5 / 12 miles . so the hiker still has to cover 15 / 12 miles to meet the waiting cyclist . the hiker will need 1 / 4 hours or 15 mins to cover the remaining 15 / 12 miles . so the answer is a ." | a = 20 - 5
b = 5 / const_60
c = a * b
d = c / 5
e = d * const_60
|
a ) 1440 , b ) 2340 , c ) 3240 , d ) 4320 , e ) 5420 | d | divide(multiply(30, power(9, 5)), subtract(add(12, 8), 5)) | a tank with a volume of 30 cubic feet has one inlet pipe and 2 outlet pipes . the inlet pipe fills water into the tank at the rate of 5 cubic inches / min and the 2 outlet pipes empty it out at the rates of 9 cubic inches / min and 8 cubic inches / min respectively . if all 3 pipes are opened when the tank is full , how many minutes does it take to empty the tank ? ( 1 foot = 12 inches ) | "the tank is emptied at this rate : 9 + 8 - 5 = 12 cubic inches / min the tank has a volume of 30 * 12 * 12 * 12 = 51840 cubic inches . the time it takes to empty the tank is 51840 / 12 = 4320 minutes . the answer is d ." | a = 9 ** 5
b = 30 * a
c = 12 + 8
d = c - 5
e = b / d
|
a ) 6.75 sec , b ) 12 sec , c ) 15 sec , d ) 20 sec , e ) none of these | a | divide(subtract(multiply(12, 15), 15), divide(multiply(12, 15), 9)) | a train consists of 12 boggies , each boggy 15 metres long . the train crosses a telegraph post in 9 seconds . due to some problem , three boggies were detached . the train now crosses a telegraph post in | "length of train = 12 ã — 15 = 180 m . then , speed of train = 180 â „ 9 = 20 m / s now , length of train = 9 ã — 15 = 135 m â ˆ ´ required time = 135 â „ 20 = 6.75 sec . answer a" | a = 12 * 15
b = a - 15
c = 12 * 15
d = c / 9
e = b / d
|
a ) 220 meter , b ) 225 meter , c ) 230 meter , d ) 410 meter , e ) none of these | d | subtract(multiply(multiply(add(120, 80), const_0_2778), 9), 90) | a 90 metres long train running at the speed of 120 kmph crosses another train running in opposite direction at the speed of 80 kmph in 9 seconds . what is the length of the other train ? | "explanation : as trains are running in opposite directions so their relative speed will get added so , relative speed = 120 + 80 = 200 kmph = 200 * ( 5 / 18 ) = 500 / 9 m / sec let the length of other train is x meter then x + 90 / 9 = 500 / 9 = > x + 90 = 500 = > x = 410 so the length of the train is 410 meters option d" | a = 120 + 80
b = a * const_0_2778
c = b * 9
d = c - 90
|
a ) 187 km , b ) 480 km , c ) 278 km , d ) 240 km , e ) 671 km | d | divide(1, 2) | with a uniform speed a car covers the distance in 8 hours . had the speed been increased by 2 km / hr , the same distance could have been covered in 7 1 / 2 hours . what is the distance covered ? | "let the distance be x km . then , x / ( 7 1 / 2 ) - x / 8 = 2 2 x / 15 - x / 8 = 2 = > x = 240 km . answer : d" | a = 1 / 2
|
a ) $ 2400 , b ) $ 2412 , c ) $ 2650 , d ) $ 2732 , e ) $ 2800 | b | multiply(2240, divide(add(const_100, 40), add(const_100, 30))) | a store ’ s selling price of $ 2240 for a certain computer would yield a profit of 30 percent of the store ’ s cost for the computer . what selling price would yield a profit of 40 percent of the computer ’ s cost ? | "1.3 x = 2240 x = 2240 / 1.3 so , 1.4 x = 2240 * 1.4 / 1.3 = 2412 answer : - b" | a = 100 + 40
b = 100 + 30
c = a / b
d = 2240 * c
|
a ) 18 square inches , b ) 20 square inches , c ) 24 square inches , d ) 28 square inches , e ) 30 square inches | d | divide(subtract(subtract(220, 24), multiply(175, divide(80, const_100))), const_2) | three table runners have a combined area of 220 square inches . by overlapping the runners to cover 80 % of a table of area 175 square inches , the area that is covered by exactly two layers of runner is 24 square inches . what is the area of the table that is covered with three layers of runner ? | "total = a + b + c - ( sum of exactly 2 - group overlaps ) - 2 * ( all three ) + neither 80 % * 175 = 220 - 24 - 2 * ( all three ) + 0 2 * ( all three ) = 220 - 24 - 140 all three = 28 answer : d" | a = 220 - 24
b = 80 / 100
c = 175 * b
d = a - c
e = d / 2
|
a ) 17 , b ) 18 , c ) 19 , d ) 20 , e ) 21 | d | divide(subtract(multiply(const_4.0, 25), add(add(4, 7), 11)), 4) | the youngest of 4 children has siblings who are 2 , 7 , and 11 years older than she is . if the average ( arithmetic mean ) age of the 4 siblings is 25 , what is the age of the youngest sibling ? | "x + ( x + 2 ) + ( x + 7 ) + ( x + 11 ) = 100 4 x + 20 = 100 4 x = 80 x = 20 the answer is d ." | a = 4 * 0
b = 4 + 7
c = b + 11
d = a - c
e = d / 4
|
a ) 60 , b ) 50 , c ) 46 , d ) 55.56 , e ) 65 | a | multiply(divide(add(multiply(divide(60, const_100), 300), multiply(subtract(const_1, divide(60, const_100)), 240)), add(300, 240)), const_100) | addison high school ’ s senior class has 300 boys and 240 girls . if 60 % of the boys attended the college then 60 % of the total class attended college , what percentage of the girls class attended college ? | "number of boys attending the party : 0.60 * 300 = 180 total students = 300 + 240 = 540 number of students attending the party : 0.60 * 540 = 324 number of girls attending = 324 - 180 = 144 % of girls attending = 144 / 240 = 60 % . a is the correct answer" | a = 60 / 100
b = a * 300
c = 60 / 100
d = 1 - c
e = d * 240
f = b + e
g = 300 + 240
h = f / g
i = h * 100
|
a ) 2 , b ) 4 , c ) 5 , d ) 6 , e ) 7 | a | add(divide(6, 3), divide(3, divide(6, 3))) | if 6 ! / 3 ^ x is an integer , what is the greatest possible value of x ? | "6 - 2 * 3 3 - 1 * 3 hence max of 3 ^ 2 is allowed . imo a ." | a = 6 / 3
b = 6 / 3
c = 3 / b
d = a + c
|
a ) 5 km , b ) 15 km , c ) 8 km , d ) 10 km , e ) 20 km | d | multiply(add(divide(add(multiply(10, divide(10, const_60)), multiply(15, divide(10, const_60))), subtract(15, 10)), divide(10, const_60)), 10) | a person walking at 10 kmph reaches his office 10 minutes late . if he walks at 15 kmph , he reaches there 10 minutes earlier . how far is the office from his house ? | formula = s 1 * s 2 / s 2 - s 1 * t 1 + t 2 / 60 = 10 * 15 / 5 * 20 / 60 = 30 * 20 / 60 = 10 km answer is d | a = 10 / const_60
b = 10 * a
c = 10 / const_60
d = 15 * c
e = b + d
f = 15 - 10
g = e / f
h = 10 / const_60
i = g + h
j = i * 10
|
a ) 1344 , b ) 3360 , c ) 7560 , d ) 50400 , e ) 67200 | c | multiply(multiply(560, divide(18, 8)), 6) | running at the same rate , 8 identical machines can produce 560 paperclips a minute . at this rate , how many paperclips could 18 machines produce in 6 minutes ? | "8 machines produce 560 in 1 min 8 machines produce 560 * 6 in 6 min 18 machine produce 560 * 6 * ( 18 / 8 ) in 6 minutes 560 * 6 * 18 / 8 = 7560 answer is c ." | a = 18 / 8
b = 560 * a
c = b * 6
|
a ) 20 hours , b ) 14 6 / 11 hours , c ) 10 hours , d ) 12 hours , e ) 8 hours | b | divide(const_1, add(divide(const_1, 25), divide(const_1, 35))) | two pipes a and b can fill a tank in 25 hours and 35 hours respectively . if both the pipes are opened simultaneously , how much time will be taken to fill the tank ? | "part filled by a in 1 hour = 1 / 25 part filled by b in 1 hour = 1 / 30 part filled by ( a + b ) in 1 hour = 1 / 25 + 1 / 30 = 11 / 150 both the pipes together fill the tank in 150 / 11 = 14 6 / 11 hours answer is b" | a = 1 / 25
b = 1 / 35
c = a + b
d = 1 / c
|
a ) 3 , b ) 3.5 , c ) 4 , d ) 4.5 , e ) 6 | c | add(divide(subtract(52, divide(subtract(divide(add(14, 2), 2), 2), 2)), add(subtract(divide(add(14, 2), 2), 2), divide(add(14, 2), 2))), divide(const_1, 2)) | tammy climbed a mountain in two days . she spent a total of 14 hours climbing the mountain . on the second day , she walked at an average speed that was half a kilometer per hour faster , but 2 hours less than what she walked on the first day . if the total distance she climbed during the two days is 52 kilometers , how many w kilometers per hour did tammy walk on the second day ? | "ans : c total time = 14 hrs let time traveled during 1 st day = x let time traveled during 2 nd day = x - 2 total time = 14 x + x - 2 = 14 x = 8 speed * time = distance s * 8 + ( s + 0.5 ) ( 8 - 2 ) = 52 solving s = 4.5 now speed for 2 nd day is 0.5 less than the 1 st day which is 4.5 thus speed for 2 nd day = 4 its simple algebra for s * 8 + ( s + 0.5 ) ( 8 - 2 ) = 52 but for some reason im getting 3.5 and not 4.5 . 8 s + 6 s + 3 = 52 14 s = 49 s = 3.5" | a = 14 + 2
b = a / 2
c = b - 2
d = c / 2
e = 52 - d
f = 14 + 2
g = f / 2
h = g - 2
i = 14 + 2
j = i / 2
k = h + j
l = e / k
m = 1 / 2
n = l + m
|
a ) $ 506.00 , b ) $ 726.24 , c ) $ 900.00 , d ) $ 920.24 , e ) $ 926.24 | a | add(multiply(add(multiply(divide(3, const_100), add(multiply(divide(2, const_100), power(const_100, 2)), power(const_100, 2))), add(multiply(divide(2, const_100), power(const_100, 2)), power(const_100, 2))), divide(4, const_100)), multiply(divide(3, const_100), add(multiply(divide(2, const_100), power(const_100, 2)), power(const_100, 2)))) | jolene entered an 14 - month investment contract that guarantees to pay 2 percent interest at the end of 6 months , another 3 percent interest at the end of 12 months , and 4 percent interest at the end of the 18 month contract . if each interest payment is reinvested in the contract , and jolene invested $ 10,000 initially , what will be the total amount of interest paid during the 18 - month contract ? | "if interest were not compounded in every six months ( so if interest were not earned on interest ) then we would have ( 2 + 3 + 4 ) = 9 % simple interest earned on $ 10,000 , which is $ 900 . so , you can rule out a , b and c right away . interest earned after the first time interval : $ 10,000 * 2 % = $ 200 ; interest earned after the second time interval : ( $ 10,000 + $ 200 ) * 3 % = $ 300 + $ 6 = $ 306 ; interest earned after the third time interval : ( $ 10,000 + $ 200 + $ 306 ) * 4 % = $ 400 + $ 8 + ( ~ $ 12 ) = ~ $ 420 ; total : 200 + 306 + ( ~ 420 ) = ~ $ 506.00 answer : a ." | a = 3 / 100
b = 2 / 100
c = 100 ** 2
d = b * c
e = 100 ** 2
f = d + e
g = a * f
h = 2 / 100
i = 100 ** 2
j = h * i
k = 100 ** 2
l = j + k
m = g + l
n = 4 / 100
o = m * n
p = 3 / 100
q = 2 / 100
r = 100 ** 2
s = q * r
t = 100 ** 2
u = s + t
v = p * u
w = o + v
|
['a ) 1.32', 'b ) 3.33', 'c ) 3.58', 'd ) 4.34', 'e ) 5.18'] | b | divide(multiply(triangle_area_three_edges(13, 13, 10), const_2), triangle_perimeter(13, 13, 10)) | an isosceles triangle with sides 13 13 10 and there is a circle inscribed it . find the radius of circle ? | height of triangle = √ ( 13 ^ 2 - ( 10 / 2 ) ^ 2 ) = 12 area of triangle = r * s where , r = radius of inscribed circle & s = ( a + b + c ) / 2 area of triangle = 1 / 2 * b * h = 1 / 2 * 10 * 12 = 60 s = ( 13 + 13 + 10 ) / 2 = 18 r = area of triangle / s = 60 / 18 = 10 / 3 = 3.33 answer : b | a = triangle_area_three_edges * (
b = a / 2
|
a ) 40 , b ) 44 , c ) 48 , d ) 8 , e ) 56 | d | add(add(multiply(5, divide(2, subtract(multiply(divide(3, 5), 5), 1))), divide(2, subtract(multiply(divide(3, 5), 5), 1))), 2) | in a can , there is a mixture of milk and water in the ratio 1 : 5 . if it is filled with an additional 2 litres of milk the can would be full and ratio of milk and water would become 3 : 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 = 1 / 6 ( t - 2 ) after adding milk , quantity of milk in the mixture = 3 / 8 t . 3 t / 8 - 2 = 1 / 6 = ( t - 2 ) 10 t = 96 - 16 = > t = 8 answer : d" | a = 3 / 5
b = a * 5
c = b - 1
d = 2 / c
e = 5 * d
f = 3 / 5
g = f * 5
h = g - 1
i = 2 / h
j = e + i
k = j + 2
|
a ) 22 , b ) 65 , c ) 72 , d ) 78 , e ) 90 | c | multiply(12, 6) | each child has 6 crayons and 12 apples . if there are 12 children , how many crayons are there in total ? | 6 * 12 = 72 . answer is c | a = 12 * 6
|
a ) 27 , b ) 36 , c ) 45 , d ) 72 , e ) 108 | e | subtract(divide(72, subtract(const_1, divide(5, 17))), 72) | in a certain school , the ratio of boys to girls is 5 to 17 . if there are 72 more girls than boys , how many boys are there ? | "the ratio of b to g is 5 : 13 and the other data point is g are more than boys by 72 . . . looking at the ratio we can say that the 12 ( 17 - 5 ) extra parts caused this diff of 72 . so 1 part corresponds to 72 / 8 = 9 and so 5 parts correspond to 12 * 9 = 108 . e" | a = 5 / 17
b = 1 - a
c = 72 / b
d = c - 72
|
a ) 10 kmph , b ) 45 kmph , c ) 60 kmph , d ) 55 kmph , e ) 30 kmph | b | divide(540, divide(multiply(8, 3), 2)) | a car takes 8 hours to cover a distance of 540 km . how much should the speed in kmph be maintained to cover the same direction in 3 / 2 th of the previous time ? | "time = 8 distence = 540 3 / 2 of 8 hours = 8 * 3 / 2 = 12 hours required speed = 540 / 12 = 45 kmph b )" | a = 8 * 3
b = a / 2
c = 540 / b
|
a ) 5 / 21 , b ) 3 / 7 , c ) 4 / 7 , d ) 5 / 7 , e ) 16 / 21 | e | divide(subtract(subtract(subtract(7, 1), 1), 1), subtract(7, 1)) | in a room with 7 people , 4 people have exactly 1 friend in the room and 3 people have exactly 2 friends in the room ( assuming that friendship is a mutual relationship , i . e . if jane is paul ' s friend , paul is jane ' s friend ) . if two individuals are selected from the room at random , what is the probability that those two individuals are not friends ? | "( 4 / 7 ) ( 5 / 6 ) + ( 3 / 7 ) ( 4 / 6 ) if you choose one of the 4 with one other friend , then you have a 5 / 6 chance of not picking their friend 2 nd . if you choose one of the 3 with 2 friends , you have a 4 / 6 chance of not picking one of their friends second . add them up . 20 / 42 + 12 / 42 32 / 42 = 16 / 21 e . 16 / 21" | a = 7 - 1
b = a - 1
c = b - 1
d = 7 - 1
e = c / d
|
a ) 9 , b ) 10 , c ) 13 , d ) 16 , e ) 18 | b | divide(add(divide(20, 2), divide(700, 35)), 3) | assisants are needed to prepare for preparation . each helper can make either 2 large cakes or 35 small cakes / hour . the kitchen is available for 3 hours and 20 large cakes and 700 small cakes are needed . how many helpers are required ? | 20 large cakes will require the equivalent of 10 helpers working for one hour . 700 small cakes will require the equivalent of 20 helpers working for one hour . this means if only one hour were available we would need 30 helpers . but since three hours are available we can use 10 helpers . b | a = 20 / 2
b = 700 / 35
c = a + b
d = c / 3
|
a ) 50 , b ) 100 , c ) 200 , d ) 300 , e ) 400 | d | subtract(divide(1000, const_2), 200) | if the perimeter of a rectangular park is 1000 m , its length when its breadth is 200 m is ? | 2 ( l + 200 ) = 1000 = > l = 300 m answer : d | a = 1000 / 2
b = a - 200
|
a ) 23 : 7 , b ) 22 : 9 , c ) 23 : 8 , d ) 23 : 9 , e ) 23 : 11 | d | divide(divide(add(640, 280), const_2), subtract(640, divide(add(640, 280), const_2))) | a boat running upstream takes 640 min to cover a certain distance , while it takes 280 min to cover the same distance running down stream . what is the ratio between the speed of the boat and speed of water current respectively ? | let the speed of the boat = x and speed of the current = y relative speed in upstream = x - y and in downstream = x + y let the distance ' d ' is to be covered in downstream and upstream , then time taken in upstream = d / ( x - y ) = 640 or d = 640 ( x - y ) - - - - ( i ) and time taken in downstream = d / ( x + y ) = 280 or d = 280 ( x + y ) - - - - ( ii ) comparing ( i ) & ( ii ) , 640 ( x - y ) = 280 ( x + y ) = > 16 ( x - y ) = 7 ( x + y ) = > 16 x - 16 y = 7 x - 7 y = > 9 x = 23 y = > x / y = 23 / 9 answer : d | a = 640 + 280
b = a / 2
c = 640 + 280
d = c / 2
e = 640 - d
f = b / e
|
a ) 162 , b ) 163 , c ) 164 , d ) 165 , e ) 166 | b | subtract(add(add(subtract(add(add(subtract(add(add(subtract(add(add(subtract(add(add(subtract(add(add(1, 1), 8), 2), 1), multiply(8, const_2)), 2), 1), multiply(8, const_3)), 2), 1), multiply(8, const_4)), 2), 1), multiply(8, add(const_3, const_2))), 2), 1), multiply(8, add(const_3, const_3))), 2) | f ( 1 ) = 1 , where f ( x + y ) = f ( x ) + f ( y ) + 8 xy - 2 . then f ( 7 ) = ? | f ( 1 ) = 1 f ( 2 ) = f ( 1 + 1 ) = 1 + 1 + 8 - 2 = 8 f ( 3 ) = f ( 2 + 1 ) = 8 + 1 + ( 8 * 2 * 1 ) - 2 = 23 f ( 6 ) = f ( 3 + 3 ) = 23 + 23 + ( 8 * 3 * 3 ) - 2 = 116 f ( 7 ) = f ( 6 + 1 ) = 116 + 1 + ( 8 * 6 * 1 ) - 2 = 163 thus ans is 163 answer : b | a = 1 + 1
b = a + 8
c = b - 2
d = c + 1
e = 8 * 2
f = d + e
g = f - 2
h = g + 1
i = 8 * 3
j = h + i
k = j - 2
l = k + 1
m = 8 * 4
n = l + m
o = n - 2
p = o + 1
q = 3 + 2
r = 8 * q
s = p + r
t = s - 2
u = t + 1
v = 3 + 3
w = 8 * v
x = u + w
y = x - 2
|
a ) 17 , b ) 25 , c ) 35 , d ) 42 , e ) 47 | e | divide(235, 5) | what is the largest possible value of cc if 5 c + ( d − 12 ) ^ 2 = 235 ? | to maximize c , we should minimize ( d - 12 ) ^ 2 . ( d - 12 ) ^ 2 is a square of a number , thus its smallest possible value is 0 ( for d = 12 ) . in this case we ' d have 5 c + 0 = 235 - - > c = 47 . answer : e . | a = 235 / 5
|
a ) 287 , b ) 132 , c ) 156 , d ) 115.5 , e ) 267 | d | multiply(circumface(divide(35, const_2)), 1.50) | find the cost of fencing around a circular field of diameter 35 m at the rate of rs . 1.50 a meter ? | "2 * 22 / 7 * 17.5 = 110 110 * 1 1 / 2 = rs . 115.5 answer : d" | a = 35 / 2
b = circumface * (
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a ) 945 , b ) 3487 , c ) 28 , d ) 279 , e ) 1287 | a | multiply(divide(divide(subtract(multiply(5, add(const_10, const_1)), multiply(const_1, 5)), subtract(multiply(5, const_2), multiply(const_1, 5))), const_2), add(multiply(5, const_2), multiply(5, add(const_10, const_1)))) | the sum of all two digit numbers divisible by 5 is | "required numbers are 10,15 , 20,25 , . . . , 95 this is an a . p . in which a = 10 , d = 5 and l = 95 . let the number of terms in it be n . then t = 95 so a + ( n - 1 ) d = 95 . 10 + ( n - 1 ) * 5 = 95 , then n = 18 . required sum = n / 2 ( a + l ) = 18 / 2 ( 10 + 95 ) = 945 . answer : a ) 945" | a = 10 + 1
b = 5 * a
c = 1 * 5
d = b - c
e = 5 * 2
f = 1 * 5
g = e - f
h = d / g
i = h / 2
j = 5 * 2
k = 10 + 1
l = 5 * k
m = j + l
n = i * m
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a ) 404 , b ) 452 , c ) 444 , d ) 468 , e ) 422 | e | add(subtract(312, divide(112, const_2)), subtract(222, divide(112, const_2))) | at the faculty of aerospace engineering , 312 students study random - processing methods , 222 students study scramjet rocket engines and 112 students study them both . if every student in the faculty has to study one of the two subjects , how many students are there in the faculty of aerospace engineering ? | "312 + 222 - 112 ( since 112 is counted twice ) = 422 e is the answer" | a = 112 / 2
b = 312 - a
c = 112 / 2
d = 222 - c
e = b + d
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a ) 288 , b ) 266 , c ) 155 , d ) 680 , e ) 441 | d | multiply(subtract(divide(12000, 10000), divide(8000, 10000)), 1700) | a , b and c started a business with capitals of rs . 8000 , rs . 10000 and rs . 12000 respectively . at the end of the year , the profit share of b is rs . 1700 . the difference between the profit shares of a and c is ? | "ratio of investments of a , b and c is 8000 : 10000 : 12000 = 4 : 5 : 6 and also given that , profit share of b is rs . 1700 = > 5 parts out of 15 parts is rs . 1700 now , required difference is 6 - 4 = 2 parts required difference = 2 / 5 ( 1700 ) = rs . 680 answer : d" | a = 12000 / 10000
b = 8000 / 10000
c = a - b
d = c * 1700
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a ) 2 , b ) 4 , c ) 8 , d ) 16 , e ) 32 | b | divide(power(2, divide(power(2, 4), power(2, 2))), power(2, 2)) | if the operation ø is defined for all positive integers x and w by x ø w = ( 2 ^ x ) / ( 2 ^ w ) then ( 4 ø 2 ) ø 2 = ? | "4 ø 2 = 2 ^ 4 / 2 ^ 2 = 4 4 ø 2 = 2 ^ 4 / 2 ^ 2 = 4 the answer is b ." | a = 2 ** 4
b = 2 ** 2
c = a / b
d = 2 ** c
e = 2 ** 2
f = d / e
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a ) 15 , b ) 28 , c ) 37 , d ) 42 , e ) 59 | e | add(add(divide(divide(multiply(multiply(2, 5), 7), const_2), 2), divide(divide(multiply(multiply(2, 5), 7), const_2), 5)), divide(divide(multiply(multiply(2, 5), 7), const_2), 7)) | if x , y , and z are positive integers , and 2 x = 5 y = 7 z , then the least possible value of x + y + z is | "take lcm of 2,5 and 7 = 70 now 2 x = 70 = > x = 35 5 y = 70 = > y = 14 7 z = 70 = > z = 10 35 + 14 + 10 = 59 . option e ." | a = 2 * 5
b = a * 7
c = b / 2
d = c / 2
e = 2 * 5
f = e * 7
g = f / 2
h = g / 5
i = d + h
j = 2 * 5
k = j * 7
l = k / 2
m = l / 7
n = i + m
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a ) 9 , b ) 10 , c ) 11 , d ) 12 , e ) 13 | b | divide(add(16, 16), subtract(6.8, add(divide(16, 8), divide(16, 10)))) | on a trip , a cyclist averaged 8 miles per hour for the first 16 miles and 10 miles per hour for the remaining 16 miles . if the cyclist returned immediately via the same route and took a total of 6.8 hours for the round trip , what was the average speed ( in miles per hour ) for the return trip ? | "the time to go 32 miles was 16 / 8 + 16 / 10 = 2 + 1.6 = 3.6 hours . the average speed for the return trip was 32 miles / 3.2 hours = 10 mph . the answer is b ." | a = 16 + 16
b = 16 / 8
c = 16 / 10
d = b + c
e = 6 - 8
f = a / e
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a ) 1 , b ) 2 , c ) 5 , d ) 7 , e ) 9 | b | divide(log(235), log(add(const_4, const_1))) | for how many unique pairs of nonnegative integers { a , b } is the equation a ^ 2 - b ^ 2 = 235 true ? | "answer b ( a + b ) ( a - b ) = 235 2 cases for ( a + b ) , ( a - b ) 235 , 1 47 , 5 answer b" | a = math.log(235)
b = 4 + 1
c = math.log(b)
d = a / c
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a ) 364724 , b ) 365387 , c ) 365737 , d ) 366757 , e ) none of these | c | multiply(subtract(divide(power(366, const_2), 999), floor(divide(power(366, const_2), 999))), 999) | on dividing a number by 999 , the quotient is 366 and the remainder is 103 . the number is : | "required number = 999 * 366 + 103 = ( 1000 - 1 ) * 366 + 103 = 366000 - 366 + 103 = 365737 . answer : c" | a = 366 ** 2
b = a / 999
c = 366 ** 2
d = c / 999
e = math.floor(d)
f = b - e
g = f * 999
|
a ) 24 , b ) 30 , c ) 35 , d ) 40 , e ) 45 | a | subtract(multiply(divide(20, const_100), 180), multiply(divide(const_1, const_3), multiply(divide(20, const_100), 180))) | one - third less than 20 % of 180 is equal to : | "lots of ways to tackle this . 20 % of 180 = 36 1 / 3 of 36 = 12 so , 1 / 3 , less than 36 is equal to 36 - 12 = 24 answer : a" | a = 20 / 100
b = a * 180
c = 1 / 3
d = 20 / 100
e = d * 180
f = c * e
g = b - f
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a ) 12 h , b ) 10 h , c ) 8 h , d ) 6 h , e ) none of these | c | divide(multiply(const_4, sqrt(576)), 12) | the area of a square field is 576 km 2 . how long will it take for a horse to run around at the speed of 12 km / h ? | "area of field = 576 km 2 . then , each side of field = √ 576 = 24 km distance covered by the horse = perimeter of square field = 24 × 4 = 96 km ∴ time taken by horse = distances / peed = 96 / 12 = 8 h answer c" | a = math.sqrt(576)
b = 4 * a
c = b / 12
|
a ) 7 , b ) 3.75 , c ) 2 , d ) 1 , e ) 8 | b | divide(multiply(6, 10), add(6, 10)) | a and b complete a work in 6 days . a can alone can do it in 10 days . if both can together can do the work in how many days ? | 1 / 6 + 1 / 10 = 8 / 30 = 4 / 15 15 / 4 = 3.75 days answer b | a = 6 * 10
b = 6 + 10
c = a / b
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a ) 670 , b ) 710 , c ) 750 , d ) 790 , e ) 830 | c | divide(subtract(620, multiply(12.5, 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 $ 620 ( before payroll deductions ) ? | total pay = 40 * $ 12.50 + $ 0.16 * x = 620 x = 120 / 0.16 = 750 the answer is c . | a = 12 * 5
b = 620 - a
c = b / 0
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a ) 350 km , b ) 400 km , c ) 800 km , d ) 600 km , e ) 650 km | c | multiply(160, 5) | what is the distance covered by a car if it travels with a speed of 160 kmh for 5 hours ? | distance = time x speed 5 x 160 = 800 answer : c | a = 160 * 5
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a ) 46.2 % , b ) 54.54 % , c ) 63 % , d ) 70 % , e ) none of these | b | multiply(divide(subtract(132, add(multiply(12, const_4), multiply(2, multiply(2, const_3)))), 132), const_100) | a cricketer scored 132 runs which included 12 boundaries and 2 sixes . what percent of his total score did he make by running between the wickets . | "explanation : number of runs made by running = 132 - ( 12 x 4 + 2 x 6 ) = 132 - ( 60 ) = 72 now , we need to calculate 72 is what percent of 132 . = > 72 / 132 * 100 = 54.54 % answer : b" | a = 12 * 4
b = 2 * 3
c = 2 * b
d = a + c
e = 132 - d
f = e / 132
g = f * 100
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a ) 33 days , b ) 44 days , c ) 55 days , d ) 42 days , e ) 32 days | b | add(divide(subtract(const_1, add(multiply(20, divide(const_1, 80)), multiply(20, divide(const_1, 100)))), divide(const_1, 100)), 20) | a can do a piece of work in 80 days and b in 100 days . they work together for 20 days and then b goes away . in how many days will a finish the remaining work ? | "explanation : in this type of questions we first get the finishing of work in 1 minute for both pipes then we will add them to get the result , as : part finished by a in 1 day = 1 / 80 part finished by b in 1 day = 1 / 100 part finished by ( a + b ) in 1 day = 1 / 80 + 1 / 100 = 9 / 400 so ( a + b ) ' s 20 days work = 20 * 9 / 400 = 9 / 20 remaining work = 1 - 9 / 20 = 11 / 20 . number days to finish the remaining work by a = 80 * 11 / 20 = 44 days . option b" | a = 1 / 80
b = 20 * a
c = 1 / 100
d = 20 * c
e = b + d
f = 1 - e
g = 1 / 100
h = f / g
i = h + 20
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a ) 37 / 20 , b ) 1 / 2 , c ) 1 / 4 , d ) 1 , e ) 5 / 6 | a | divide(subtract(add(5, 5), add(divide(5, 6), add(6, 6))), 6) | pumps a , b , and c operate at their respective constant rates . pumps a and b , operating simultaneously , can fill a certain tank in 6 / 5 hours ; pumps a and c , operating simultaneously , can fill the tank in 3 / 2 hours ; and pumps b and c , operating simultaneously , can fill the tank in 1 hours . how many hours does it take pumps a , b , and c , operating simultaneously , to fill the tank . | "a + b = 6 / 5 ; a + c = 3 / 2 , b + c = 1 ; add then 2 * ( a + b + c ) = 6 / 5 + 3 / 2 + 1 = 37 / 10 a + b + c = 37 / 20 hrs a" | a = 5 + 5
b = 5 / 6
c = 6 + 6
d = b + c
e = a - d
f = e / 6
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a ) 2 : 1 , b ) 3 : 2 , c ) 4 : 3 , d ) 5 : 4 , e ) 6 : 5 | a | divide(multiply(4, 5), multiply(5, 2)) | the marks obtained by polly and sandy are in the ratio 4 : 5 and those obtained by sandy and willy are in the ratio of 5 : 2 . the marks obtained by polly and willy are in the ratio of . . . ? | "polly : sandy = 4 : 5 sandy : willy = 5 : 2 polly : sandy : willy = 4 : 5 : 2 polly : willy = 4 : 2 = 2 : 1 the answer is a ." | a = 4 * 5
b = 5 * 2
c = a / b
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a ) 12 , b ) 14 , c ) 16 , d ) 18 , e ) 20 | c | add(multiply(add(3, 3), 3), floor(divide(subtract(120, multiply(divide(multiply(3, subtract(9, 1)), const_2), 9)), 9))) | in a certain supermarket , a triangular display of cans is arranged in 9 rows , numbered 1 through 9 from top to bottom . each successively numbered row contains 3 more cans than the row immediately above it . if there are fewer than 120 cans in the entire display , how many cans are in the six row ? | "let x be the number of cans in row 1 . the total number of cans is x + ( x + 3 ) + . . . + ( x + 24 ) = 9 x + 3 ( 1 + 2 + . . . + 8 ) = 9 x + 3 ( 8 ) ( 9 ) / 2 = 9 x + 108 since the total is less than 120 , x must equal 1 . the number of cans in the 6 th row is 1 + 3 ( 5 ) = 16 the answer is c ." | a = 3 + 3
b = a * 3
c = 9 - 1
d = 3 * c
e = d / 2
f = e * 9
g = 120 - f
h = g / 9
i = math.floor(h)
j = b + i
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a ) 8 , b ) 21 , c ) 14 , d ) 12 , e ) 15 | a | divide(subtract(168, multiply(3, 40)), multiply(3, const_2)) | a man ' s regular pay is $ 3 per hour up to 40 hours . overtime is twice the payment for regular time . if he was paid $ 168 , how many hours overtime did he work ? | "answer is : b at $ 3 per hour up to 40 hours , regular pay = $ 3 x 40 = $ 120 if total pay = $ 168 , overtime pay = $ 168 - $ 120 = $ 48 overtime rate ( twice regular ) = 2 x $ 3 = $ 6 per hour = > number of overtime hours = $ 48 / $ 6 = 8" | a = 3 * 40
b = 168 - a
c = 3 * 2
d = b / c
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a ) 36 , b ) 42 , c ) 28 , d ) 54 , e ) 45 | c | inverse(divide(const_3, multiply(21, const_4))) | if a is thrice as fast as b and together can do a work in 21 days . in how many days a alone can do the work ? | "a ’ s one day ’ s work = 1 / x b ’ s one day ’ s work = 1 / 3 x a + b ’ s one day ’ s work = 1 / x + 1 / 3 x = 1 / 21 = 3 + 1 / 3 x = 4 / 3 x = 1 / 21 x = 21 * 4 / 3 = 28 answer : c" | a = 21 * 4
b = 3 / a
c = 1/(b)
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a ) 83 % , b ) 80 % , c ) 20 % , d ) 17 % , e ) 12 % | d | subtract(20, const_1) | at a local appliance manufacturing facility , the workers received a 20 % hourly pay raise due to extraordinary performance . if one worker decided to reduce the number of hours that he worked so that his overall pay would remain unchanged , by approximately what percent would he reduce the number of hours that he worked ? | "let ' s say he works usually 10 hours and earns 100 per hour . 10 * 100 = 1000 10 * 120 = 1200 ( this are the new earnings after the raise ) to figure out how much he needs to work with the new salary in order to earn the original 1000 : 1000 / 120 = 8.3333 so he can reduce his work by 1.6666 hours . which is > 15 % . answer d ." | a = 20 - 1
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a ) 24 , b ) 27 , c ) 40 , d ) 28 , e ) 20 | a | divide(divide(multiply(1200, 12), const_100), 6) | a reduction of 12 % in the price of oil enables a house wife to obtain 6 kgs more for rs . 1200 , what is the reduced price for kg ? | "1200 * ( 12 / 100 ) = 144 - - - - 6 ? - - - - 1 = > rs . 24 answer : a" | a = 1200 * 12
b = a / 100
c = b / 6
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a ) 8 % , b ) 25 % , c ) 41 % , d ) 59 % , e ) 70 % | c | multiply(subtract(const_1, subtract(const_2, add(divide(8, const_100), const_1))), const_100) | dick and jane each saved $ 3,000 in 1989 . in 1990 dick saved 8 percent more than in 1989 , and together he and jane saved a total of $ 5,000 . approximately what percent less did jane save in 1990 than in 1989 ? | "dick saved = $ 3240 jane saved = $ 1760 ( jane saved $ 1240 less than she did the prior year ) jane saved approximately $ 1240 / $ 3000 ( close to 41 % ) less in 1990 answer : c" | a = 8 / 100
b = a + 1
c = 2 - b
d = 1 - c
e = d * 100
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a ) 20 , b ) 25 , c ) 50 , d ) 45 , e ) 75 | c | subtract(subtract(multiply(40, add(const_2, const_3)), multiply(30, const_4)), 30) | the average of temperatures at noontime from monday to friday is 40 ; the lowest one is 30 , what is the possible maximum range of the temperatures ? | "average = 40 , sum of temperatures = 40 * 5 = 200 as the min temperature is 30 , max would be 200 - 4 * 30 = 80 - - > the range = 80 ( max ) - 30 ( min ) = 50 answer : c" | a = 2 + 3
b = 40 * a
c = 30 * 4
d = b - c
e = d - 30
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['a ) 600 m 2', 'b ) 540 m 2', 'c ) 680 m 2', 'd ) 574 m 2', 'e ) none of these'] | a | multiply(30, 20) | in a parallelogram , the length of one diagonal and the perpendicular dropped on that diagonal are 30 and 20 metres respectively . find its area | in a parallelogram . area = diagonal × length of perpendicular on it . = 30 × 20 = 600 m 2 answer a | a = 30 * 20
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a ) 1 / 13 , b ) 1 / 26 , c ) 2 / 13 , d ) 1 / 52 , e ) 1 / 48 | b | divide(subtract(52, multiply(const_4, const_4)), 52) | a card is drawn from a pack of 52 cards . the probability of getting a queen of spade or an ace of club is : | "explanation : total outcomes = 52 favourable outcomes ( getting a queen of spade or an ace of club ) = 2 probability = 2 / 52 = 1 / 26 . answer : b" | a = 4 * 4
b = 52 - a
c = b / 52
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a ) 1 , b ) 2 , c ) 3 , d ) 4 , e ) 5 | a | subtract(5, const_2) | in a certain game , a large bag is filled with blue , green , purple and red chips worth 1 , 5 , x and 11 points each , respectively . the purple chips are worth more than the green chips , but less than the red chips . a certain number of chips are then selected from the bag . if the product of the point values of the selected chips is 11,000 , how many purple chips were selected ? | "11,000 = 1 * 5 ^ 3 * 8 * 11 the factor of 8 must come from the purple point value , so there is 1 purple chip . the answer is a ." | a = 5 - 2
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a ) 1245 , b ) 1345 , c ) 1455 , d ) 1577 , e ) 1641 | e | multiply(divide(subtract(1370, 15), subtract(6, const_1)), 6) | find large number from below question the difference of two numbers is 1370 . on dividing the larger number by the smaller , we get 6 as quotient and the 15 as remainder | "let the smaller number be x . then larger number = ( x + 1370 ) . x + 1370 = 6 x + 15 5 x = 1355 x = 271 large number = 271 + 1370 = 1641 e" | a = 1370 - 15
b = 6 - 1
c = a / b
d = c * 6
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a ) 70000 , b ) 60000 , c ) 80000 , d ) 90000 , e ) 252000 | e | add(add(multiply(divide(8000, 2000), 36000), multiply(divide(4000, 2000), 36000)), 36000) | a , b and c started a partnership business by investing rs . 8000 , rs . 4000 , rs . 2000 respectively . at the end of the year , the profit were distributed among them . if c ' s share of profit is 36000 , what is the total profit ? | "a : b : c = 8000 : 4000 : 2000 = 4 : 2 : 1 let total profit = p then p ã — 1 / 7 = 36000 p = ( 36000 ã — 7 ) / 1 = 252000 answer is e ." | a = 8000 / 2000
b = a * 36000
c = 4000 / 2000
d = c * 36000
e = b + d
f = e + 36000
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a ) 180 cm 2 , b ) 150 cm 2 , c ) 168 cm 2 , d ) 198 cm 2 , e ) 987 cm 2 | a | multiply(multiply(divide(const_1, const_2), add(6, 9)), 24) | find the area of the quadrilateral of one of its diagonals is 24 cm and its off sets 9 cm and 6 cm ? | "1 / 2 * 24 ( 9 + 6 ) = 180 cm 2 answer : a" | a = 1 / 2
b = 6 + 9
c = a * b
d = c * 24
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a ) rs . 5800 , b ) rs . 3618 , c ) rs . 3600 , d ) rs . 3619 , e ) rs . 3610 | a | subtract(multiply(add(2000, 200), add(18, const_1)), multiply(2000, 18)) | the average monthly salary of 18 employees in an organisation is rs . 2000 . if the manager ' s salary is added , then the average salary increases by rs . 200 . what is the manager ' s monthly salary ? | "manager ' s monthly salary = rs . ( 2200 * 19 - 2000 * 18 ) = rs . 5800 answer : a" | a = 2000 + 200
b = 18 + 1
c = a * b
d = 2000 * 18
e = c - d
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a ) 621 , b ) 276 , c ) 236 , d ) 1600 , e ) 211 | d | multiply(subtract(divide(12000, 10000), divide(8000, 10000)), 4000) | a , b and c started a business with capitals of rs . 8000 , rs . 10000 and rs . 12000 respectively . at the end of the year , the profit share of b is rs . 4000 . the difference between the profit shares of a and c is ? | "explanation : ratio of investments of a , b and c is 8000 : 10000 : 12000 = 4 : 5 : 6 and also given that , profit share of b is rs . 4000 = > 5 parts out of 15 parts is rs . 4000 now , required difference is 6 - 4 = 2 parts required difference = 2 / 5 ( 4000 ) = rs . 1600 answer : d" | a = 12000 / 10000
b = 8000 / 10000
c = a - b
d = c * 4000
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a ) 750 , b ) 700 , c ) 800 , d ) 850 , e ) none of these | c | divide(multiply(const_100, 320), subtract(70, subtract(const_100, 70))) | in an election between the two candidates , the candidates who gets 70 % of votes polled is winned by 320 vote ’ s majority . what is the total number of votes polled ? | "explanation : note : majority ( 40 % ) = difference in votes polled to win ( 70 % ) & defeated candidates ( 30 % ) 40 % = 70 % - 30 % 40 % - - - - - > 320 ( 40 * 8 = 320 ) 100 % - - - - - > 800 ( 100 * 8 = 800 ) answer : option c" | a = 100 * 320
b = 100 - 70
c = 70 - b
d = a / c
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a ) 2 , b ) 3 , c ) 4 , d ) 5 , e ) 6 | c | subtract(multiply(add(multiply(const_4, const_10), const_2), 23), 1100) | what least number must be added to 1100 , so that the sum is completely divisible by 23 ? | "48 * 23 = 1104 1104 - 1100 = 4 answer : c" | a = 4 * 10
b = a + 2
c = b * 23
d = c - 1100
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a ) 380 , b ) 350 , c ) 332 , d ) 310 , e ) none of these | c | multiply(3136, power(add(const_4, const_1), const_4)) | ( 3136 + 6160 ) / 28 | "explanation : as per bodmas rule , first we will solve the equation in bracket then we will go for division = ( 9296 ) / 28 = 332 option c" | a = 4 + 1
b = a ** 4
c = 3136 * b
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a ) 14 sec , b ) 15 sec , c ) 16 sec , d ) 20 sec , e ) none | a | multiply(divide(280, multiply(72, const_1000)), const_3600) | a train 280 m long , running with a speed of 72 km / hr will pass a tree in | "sol . speed = ( 72 x 5 / 18 ) m / sec . = 20 m / sec . time taken = ( 280 x 1 / 20 ) sec = 14 sec answer a" | a = 72 * 1000
b = 280 / a
c = b * 3600
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a ) 5 , b ) 8 , c ) 10 , d ) 4 , e ) 3 | e | divide(add(divide(12, 6), divide(24, 6)), const_2) | a man swims downstream 24 km and upstream 12 km taking 6 hours each time , what is the speed of the man in still water ? | "24 - - - 6 ds = 4 ? - - - - 1 12 - - - - 6 us = 2 ? - - - - 1 m = ? m = ( 4 + 2 ) / 2 = 3 answer : e" | a = 12 / 6
b = 24 / 6
c = a + b
d = c / 2
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a ) 198 , b ) 275 , c ) 299 , d ) 402 , e ) 205 | b | add(55, multiply(subtract(45, const_1), 5)) | s is a set of 45 consecutive multiples of 5 . if the smallest number in s is 55 , then the greatest number in s is | "last term = first term + ( total no . of terms - 1 ) consecutive difference s is a set of 45 consecutive multiples of 5 . if the smallest number in s is 55 , then the greatest number in s is first term = 55 ; total terms = 45 ; difference = 5 55 + ( 44 ) 5 = 275 ans b" | a = 45 - 1
b = a * 5
c = 55 + b
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a ) 10 , b ) 15 , c ) 6 , d ) 5 , e ) 8 | d | subtract(subtract(divide(75, 6), 3), 3) | the sum of ages of 6 children born at the intervals of 3 years each is 75 years . what is the age of the youngest child ? | "let the ages of children be x , ( x + 3 ) , ( x + 6 ) , ( x + 9 ) , ( x + 12 ) , ( x + 15 ) years . then , x + ( x + 3 ) + ( x + 6 ) + ( x + 9 ) + ( x + 12 ) + ( x + 15 ) = 75 6 x = 30 x = 5 . age of the youngest child = x = 4 years . answer : d" | a = 75 / 6
b = a - 3
c = b - 3
|
a ) 315 , b ) 321 , c ) 330 , d ) 415 , e ) 424 | b | add(divide(subtract(312, const_1), multiply(const_2, const_1)), const_1) | the sum of three consecutive integers is 312 . what is the sum of the next three consecutive integers ? | "a + ( a + 1 ) + ( a + 2 ) = 3 a + 3 = 312 ( a + 3 ) + ( a + 4 ) + ( a + 5 ) = ( 3 a + 3 ) + 9 = 312 + 9 = 321 answer : b ." | a = 312 - 1
b = 2 * 1
c = a / b
d = c + 1
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a ) 14 minutes , b ) 15 minutes , c ) 16 minutes , d ) 17 minutes , e ) 18 minutes | e | divide(const_1, subtract(divide(const_1, 9), divide(const_1, 18))) | a can fill tank in 9 minutes , b can empty the tank in 18 minutes . . in what time the tank be filled , if both pipes work simultaneously ? | bigger no = 18 , ratio = 9 : 18 = 1 : 2 time taken to fill the tank = 18 / ( 2 - 1 ) = 18 minutes answer : e | a = 1 / 9
b = 1 / 18
c = a - b
d = 1 / c
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a ) 2 / 20 , b ) 3 / 20 , c ) 4 / 20 , d ) 5 / 20 , e ) none | b | subtract(1, add(multiply(divide(20, const_100), 3), divide(1, 4))) | if you spend 1 / 4 of your salary during the first week of the month and 20 % of it during each of the coming 3 weeks , what part ( fraction ) of your salary will leave unspent at the end of the month ? | solution : 1 / 4 + 3 ( 20 / 100 ) = 1 / 4 + 3 / 5 = 5 + 12 / 20 = 17 / 20 the sallary that will be spent = 17 / 20 unspend will be = 1 − 17 / 20 = 20 − 17 / 20 = 3 / 20 answer b | a = 20 / 100
b = a * 3
c = 1 / 4
d = b + c
e = 1 - d
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a ) 1000 , b ) 1400 , c ) 1320 , d ) 1200 , e ) 1650 | d | 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(24, const_2), const_2), const_3), add(const_4, const_1)), add(const_4, const_1))) | find the l . c . m of 16 , 25 , 24 and 10 . | "explanation : 4 x 5 x 2 x 3 x 5 x 2 = 1200 answer : option d" | a = 3 ** 3
b = 2 ** 3
c = 4 + 1
d = c ** 2
e = b * d
f = a * e
g = 24 / 2
h = g / 2
i = h / 3
j = 4 + 1
k = i / j
l = 4 + 1
m = k / l
n = f * m
|
a ) 8 / 15 , b ) 2 / 5 , c ) 3 / 5 , d ) 11 / 15 , e ) 43 / 91 | e | add(multiply(divide(8, add(6, 8)), divide(subtract(8, const_1), subtract(add(6, 8), const_1))), multiply(divide(6, add(6, 8)), divide(subtract(6, const_1), subtract(add(6, 8), const_1)))) | a bag contains 6 green and 8 white balls . if two balls are drawn simultaneously , the probability that both are of the same colour is | "drawing two balls of same color from six green balls can be done in 6 c ₂ ways . similarly from eight white balls two can be drawn in ⁸ c ₂ ways . p = 6 c ₂ / ¹ ⁵ c ₂ + ⁸ c ₂ / ¹ ⁵ c ₂ = 43 / 91 answer : e" | a = 6 + 8
b = 8 / a
c = 8 - 1
d = 6 + 8
e = d - 1
f = c / e
g = b * f
h = 6 + 8
i = 6 / h
j = 6 - 1
k = 6 + 8
l = k - 1
m = j / l
n = i * m
o = g + n
|
a ) 900 , b ) 750 , c ) 250 , d ) 600 , e ) 800 | a | multiply(multiply(subtract(const_1, divide(20, const_100)), subtract(const_1, divide(55, const_100))), 2000) | in an election between two candidates , one got 55 % of the total valid votes , 20 % of the votes were invalid . if the total number of votes was 2000 , the number of valid votes that the other candidate got , was : | "number of valid votes = 80 % of 2000 = 1600 valid votes polled by other candidate = 45 % of 2000 = ( 45 / 100 ) x 2000 = 900 answer = a" | a = 20 / 100
b = 1 - a
c = 55 / 100
d = 1 - c
e = b * d
f = e * 2000
|
a ) 40 , b ) 45 , c ) 38 , d ) 50 , e ) 70 | e | divide(subtract(14, multiply(14, divide(1, 2))), subtract(divide(1, 2), divide(2, 5))) | a man â € ™ s current age is ( 2 / 5 ) of the age of his father . after 14 years , he will be ( 1 / 2 ) of the age of his father . what is the age of father at now ? | let , father â € ™ s current age is a years . then , man â € ™ s current age = [ ( 2 / 5 ) a ] years . therefore , [ ( 2 / 5 ) a + 14 ] = ( 1 / 2 ) ( a + 14 ) 2 ( 2 a + 70 ) = 5 ( a + 14 ) a = 70 e | a = 1 / 2
b = 14 * a
c = 14 - b
d = 1 / 2
e = 2 / 5
f = d - e
g = c / f
|
a ) 5 , b ) 7 , c ) 8 , d ) 9 , e ) 10 | a | divide(add(5, 10), add(2, 1)) | at a certain committee meeting only associate professors and assistant professors are present . each associate professor has brought 2 pencils and 1 chart to the meeting , while each assistant professor has brought 1 pencil and 2 charts . if a total of 10 pencils and 5 charts have been brought to the meeting , how many people are present ? | "say there are ' a ' associate professors . so we have 2 a pencils and a charts . say there are ' b ' assistant professors . so we have b pencils and 2 b charts . total pencils are 10 so 2 a + b = 10 total charts are 11 so a + 2 b = 5 add both : 3 a + 3 b = 15 so a + b = 5 total number of people = 5 a" | a = 5 + 10
b = 2 + 1
c = a / b
|
a ) seven 7 , b ) 8 , c ) 9 , d ) 15 , e ) 17 | a | subtract(multiply(divide(floor(multiply(add(35, divide(7, 17)), const_2)), const_2), subtract(floor(multiply(add(35, divide(7, 17)), const_2)), 1)), multiply(add(35, divide(7, 17)), subtract(subtract(floor(multiply(add(35, divide(7, 17)), const_2)), 1), 1))) | a set of consecutive positive integers beginning with 1 is written on the blackboard . a student came along and erased one number . the average of the remaining numbers is 35 7 / 17 . what was the number erased ? | "2 ) the average of the new group of numbers is 35 7 / 17 . the denominator of that fraction tells us that the total number of terms in the new list must be a multiple of 17 : 17 , 34 , 51 , 68 , 85 , etc . 3 ) since we ' re removing 1 number from a list of consecutive positive integers , we could quickly limit down the possible number of terms in the original list : if new = 17 , old = 18 , but the numbers from 1 to 18 would not have an average in the mid - 30 s ( it would be much smaller ) . eliminate this option . if new = 34 , old = 35 , but we run into the same problem . the average wo n ' t match here either . eliminate this option . if new = 51 , old = 52 , same problem here . eliminate this option . if new = 68 , old = 69 . . . . here we have a group of numbers that might just be what we ' re looking for . in theold group , the average is themiddle term = 35 , which is really close to the new average once we remove a number . now , we just have to figure out which of the 5 answers was removed ( and that changed the new average to 35 7 / 17 ) . there are a couple of ways to do this math , but i ' m going to use the answers to save myself some steps : sum of 1 to 69 , inclusive = 69 ( 35 ) = 2415 remove answer b . . . . 2415 - 8 = 2407 new average = 2407 / 68 = 35 27 / 68 now , we can compare 27 / 68 to 7 / 17 27 / 68 and 28 / 68 this is really close , but is not the answer . eliminate b . the answer is either a or c . since 28 / 68 is a bigger fraction than 27 / 68 , we need the sum ( of the 68 integers ) to be bigger . . . which means we need to remove a smaller number . final answer : a" | a = 7 / 17
b = 35 + a
c = b * 2
d = math.floor(c)
e = d / 2
f = 7 / 17
g = 35 + f
h = g * 2
i = math.floor(h)
j = i - 1
k = e * j
l = 7 / 17
m = 35 + l
n = 7 / 17
o = 35 + n
p = o * 2
q = math.floor(p)
r = q - 1
s = r - 1
t = m * s
u = k - t
|
a ) 10 , b ) 20 , c ) 30 , d ) 40 , e ) 50 | a | divide(subtract(90, subtract(power(multiply(4, const_2), const_2), power(multiply(1, const_2), const_2))), const_2) | a rectangular photograph is surrounded by a border that is 1 inch wide on each side . the total area of the photograph and the border is m square inches . if the border had been 4 inches wide on each side , the total area would have been ( m + 90 ) square inches . what is the perimeter of the photograph , in inches ? | "let x and y be the width and length of the photograph . ( x + 2 ) ( y + 2 ) = m and so ( 1 ) xy + 2 x + 2 y + 4 = m ( x + 8 ) ( y + 8 ) = m and so ( 2 ) xy + 8 x + 8 y + 64 = m + 90 let ' s subtract equation ( 1 ) from equation ( 2 ) . 6 x + 6 y + 60 = 90 2 x + 2 y = 10 , which is the perimeter of the photograph . the answer is a ." | a = 4 * 2
b = a ** 2
c = 1 * 2
d = c ** 2
e = b - d
f = 90 - e
g = f / 2
|
a ) 22877 , b ) 5655 , c ) 2778 , d ) 6787 , e ) 1771 | b | add(3900, divide(subtract(824.85, subtract(multiply(3900, multiply(divide(9, const_100), const_3)), multiply(3900, multiply(divide(6, const_100), const_3)))), multiply(divide(9, const_100), const_3))) | manoj borrowed rs . 3900 from anwar at 6 % p . a . simple interest for 3 years . he then added some more money to the borrowed sum and lent it to ramu for the same time at 9 % p . a . simple interest . if manoj gains rs . 824.85 by way of interest on the borrowed sum as well as his own amount from the whole transaction , then what is the sum lent by him to ramu ? | let the sum lent by manoj to ramu be rs . p . amount gained by manoj = p . 3.9 / 100 = 3900 . 3.6 / 100 = rs . 824.85 27 p = 70200 = 82485 p = ( 82485 + 70200 ) / 27 = 3055 + 2600 = rs . 5655 answer : b | a = 9 / 100
b = a * 3
c = 3900 * b
d = 6 / 100
e = d * 3
f = 3900 * e
g = c - f
h = 824 - 85
i = 9 / 100
j = i * 3
k = h / j
l = 3900 + k
|
a ) 99 , b ) 277 , c ) 84 , d ) 96 , e ) 22 | c | subtract(divide(multiply(2.10, const_1000), divide(multiply(60, const_1000), const_3600)), divide(multiply(1.4, const_1000), divide(multiply(90, const_1000), const_3600))) | two trains are moving in opposite directions at 60 km / hr and 90 km / hr . their lengths are 2.10 km and 1.4 km respectively . the time taken by the slower train to cross the faster train in seconds is ? | "relative speed = 60 + 90 = 150 km / hr . = 150 * 5 / 18 = 125 / 3 m / sec . distance covered = 2.10 + 1.4 = 3.5 km = 3500 m . required time = 3500 * 3 / 125 = 84 sec . answer : c" | a = 2 * 10
b = 60 * 1000
c = b / 3600
d = a / c
e = 1 * 4
f = 90 * 1000
g = f / 3600
h = e / g
i = d - h
|
a ) 6 , b ) 8.5 , c ) 11.5 , d ) 12 , e ) 12.5 | b | subtract(10.5, multiply(2, 1)) | the arithmetic mean and standard deviation of a certain normal distribution are 10.5 and 1 , respectively . what value is exactly 2 standard deviations less than the mean ? | "mean = 10.5 two standard deviations is 1 + 1 = 2 there could be two calues for this . mean + two standard deviations = 12.5 mean - two standard deviations = 8.5 answer choice has 8.5 and so b is the answer ." | a = 2 * 1
b = 10 - 5
|
['a ) 30', 'b ) 45', 'c ) 60', 'd ) 75', 'e ) 90'] | c | divide(const_180, const_3) | what is the measure of the angle e made by the diagonals of the any adjacent sides of a cube . | c . . 60 degrees all the diagonals are equal . if we take 3 touching sides and connect their diagonals , we form an equilateral triangle . therefore , each angle would be 60 . . c | a = const_180 / 3
|
a ) 643 , b ) 652 , c ) 660 , d ) 564 , e ) 693 | d | multiply(divide(560, add(const_1, divide(20, const_100))), add(const_1, divide(21, const_100))) | if albert ’ s monthly earnings rise by 20 % , he would earn $ 560 . if , instead , his earnings rise by only 21 % , how much ( in $ ) would he earn this month ? | "= 560 / 1.2 ∗ 1.21 = 564 = 564 answer is d" | a = 20 / 100
b = 1 + a
c = 560 / b
d = 21 / 100
e = 1 + d
f = c * e
|
a ) 9 % , b ) 11 % , c ) 13 % , d ) 15 % , e ) 17 % | e | subtract(subtract(30, 10), divide(30, 10)) | a couple who own an appliance store discover that if they advertise a sales discount of 10 % on every item in the store , at the end of one month the number of total items sold increases 30 % . their gross income from sales for one month increases by what percent ? | let p be the original price and let x be the number of items sold originally . the original income is p * x . after the changes , the income is 0.9 p * 1.3 x = 1.17 * ( p * x ) , an increase of 17 % . the answer is e . | a = 30 - 10
b = 30 / 10
c = a - b
|
a ) 24 , b ) 25 , c ) 36 , d ) 39 , e ) 43 | c | add(add(divide(multiply(add(subtract(405, 270), 15), const_2), const_10), divide(divide(multiply(add(subtract(405, 270), 15), const_2), const_10), const_10)), const_1) | 324 ã · ? ã — 15 + 270 = 405 | "explanation : 324 ã · ? ã — 15 = 405 - 270 = 135 ( 324 ã — 15 ) / ? = 135 ? = ( 324 ã — 15 ) / 135 = 36 answer : option c" | a = 405 - 270
b = a + 15
c = b * 2
d = c / 10
e = 405 - 270
f = e + 15
g = f * 2
h = g / 10
i = h / 10
j = d + i
k = j + 1
|
a ) 52 : 56 , b ) 52 : 53 , c ) 52 : 50 , d ) 22 : 56 , e ) 52 : 51 | b | divide(add(const_100, 4), add(const_100, 6)) | the cash difference between the selling prices of an article at a profit of 4 % and 6 % is rs . 3 . the ratio of the two selling prices is ? | "let c . p . of the article be rs . x . then , required ratio = 104 % of x / 106 % of x = 104 / 106 = 52 / 53 = 52 : 53 answer : b" | a = 100 + 4
b = 100 + 6
c = a / b
|
a ) 0.8 , b ) 0.75 , c ) c ) 8 / 9 , d ) 5 / 7 , e ) 6 / 8 | d | divide(subtract(21, 12), 12) | lovely shop runs a sale on goldfish and platys . if one customer buys 13 goldfish and 12 platys for $ 5.60 , and another buys 14 goldfish and 21 platys for $ 8.05 , what is the ratio of the price of goldfish to platys ? | the approach of plugging in the numbers is a fine approach . but , in this case , the answer was the first option and thus , by plugging in a number in the ration given , we arrived at the answer quickly ! however , what if the correct ratio was option d or c ? in any case , the algebraic solution is n ' t that complex if you realize a key concept : the number 7 can be factored out of the second equation ! here ' s the working : lovely shop runs a sale on goldfish and platys . 13 x + 12 y = 5.60 < < < < < equation 1 14 x + 21 y = 8.05 < < < < < equation 2 2 x + 3 y = 1.15 [ factor out 7 from equation 2 ] < < < < < equation 3 8 x + 12 y = 4.6 [ multiply equation 3 with 4 ] 5 x = 1.0 [ equation 1 - equation 2 ] x = 0.2 3 y = 1.15 - 0.4 = 0.75 [ substitute for x in equation 1 ] y = 0.25 x / y = 0.2 / 0.25 = 20 / 25 = 4 / 5 = 0.8 [ answer a ] the whole thing must have taken me about 45 s : d | a = 21 - 12
b = a / 12
|
a ) 1 , b ) 3 , c ) 2 , d ) 4 , e ) 5 | d | subtract(9, 5) | a football team lost 5 yards and then gained 9 . what is the team ' s progress ? | "for lost , use negative . for gain , use positive . progress = - 5 + 9 = 4 yards correct answer is d ) 4" | a = 9 - 5
|
a ) 0.034 , b ) 0.024 , c ) 0.352 , d ) 0.063 , e ) 0.026 | a | divide(divide(add(multiply(3, const_10), const_4), const_100), const_10) | in expressing a length 810472 km as nearly as possible with 3 significant digits , find the percentage error . | error = ( 81.5 – 81.472 ) km = 0.028 . required percentage = [ ( 0.028 / 81.472 ) * 100 ] % = 0.034 % . answer a . | a = 3 * 10
b = a + 4
c = b / 100
d = c / 10
|
a ) $ 280,000 , b ) $ 320,000 , c ) $ 360,000 , d ) $ 450,000 , e ) $ 540,000 | b | divide(const_3600, const_10) | the amount of an investment will double in approximately 70 / p years , where p is the percent interest , compounded annually . if thelma invests $ 40,000 in a long - term cd that pays 5 percent interest , compounded annually , what will be the approximate total value of the investment when thelma is ready to retire 42 years later ? | "the amount of an investment will double in approximately 70 / p years , where p is the percent interest , compounded annually . if thelma invests $ 40,000 in a long - term cd that pays 5 percent interest , compounded annually , what will be the approximate total value of the investment when thelma is ready to retire 42 years later ? the investment gets doubled in 70 / p years . therefore , the investment gets doubled in 70 / 5 = every 14 years . after 42 years , the investment will get doubled 42 / 14 = 3 times . so the amount invested will get doubled thrice . so , 40000 * 2 = 80000 80000 * 2 = 160000 160000 * 2 = 320000 hence , the answer is b ." | a = 3600 / 10
|
a ) 25 , b ) 10 , c ) 50 , d ) 60 , e ) 75 | b | subtract(subtract(subtract(140, 50), 80), divide(subtract(subtract(140, 50), 80), const_3)) | 140 college students were asked in a survey if they preferred windows or mac brand computers . 80 students claimed that they preferred mac to windows brand computers . one third as many of the students who preferred mac to windows , equally preferred both brands . 50 of the students had no preference . how many of the students in the survey preferred windows to mac brand computers ? | "140 = 80 ( mac ) + x ( window ) + 50 ( both ) = > x = 10 answer : b" | a = 140 - 50
b = a - 80
c = 140 - 50
d = c - 80
e = d / 3
f = b - e
|
a ) 10 / 16 , b ) 6 / 16 , c ) 5 / 10 , d ) 6 / 10 , e ) 4 / 10 | c | divide(divide(subtract(15, 5), add(const_1, const_1)), add(divide(subtract(15, 5), add(const_1, const_1)), 5)) | there are 5 more women than there are men on a local co - ed softball team . if there are a total of 15 players on the team , what is the ratio of men to women ? | w = m + 5 w + m = 15 m + 5 + m = 15 2 m = 10 m = 5 w = 10 ratio : 5 : 10 ans : c | a = 15 - 5
b = 1 + 1
c = a / b
d = 15 - 5
e = 1 + 1
f = d / e
g = f + 5
h = c / g
|
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