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a ) 22 kmph , b ) 77 kmph , c ) 54 kmph , d ) 71 kmph , e ) 90 kmph | e | multiply(const_3_6, divide(400, 16)) | a train 400 m in length crosses a telegraph post in 16 seconds . the speed of the train is ? | "s = 400 / 16 * 18 / 5 = 90 kmph answer : e" | a = 400 / 16
b = const_3_6 * a
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a ) 30 days , b ) 35 days , c ) 25 days , d ) 27 days , e ) 29 days | a | divide(const_1, add(inverse(50), inverse(75))) | ravi can do a piece of work in 50 days while prakash can do it in 75 days . in how many days will they finish it together ? | 1 / 50 + 1 / 75 = 5 / 150 30 / 1 = 30 days answer : a | a = 1/(50)
b = 1/(75)
c = a + b
d = 1 / c
|
a ) 2 : 1 , b ) 3 : 2 , c ) 4 : 3 , d ) 5 : 4 , e ) 6 : 5 | b | divide(multiply(3, 5), multiply(5, 2)) | the marks obtained by polly and sandy are in the ratio 3 : 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 = 3 : 5 sandy : willy = 5 : 2 polly : sandy : willy = 3 : 5 : 2 polly : willy = 3 : 2 the answer is b ." | a = 3 * 5
b = 5 * 2
c = a / b
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a ) 9.5 , b ) 10.5 , c ) 11.5 , d ) 12 , e ) 15 | b | add(divide(multiply(10, 30), const_100), divide(multiply(15, 50), const_100)) | add 10 % of 30 and 15 % of 50 . | "10 % of 30 + 15 % of 50 30 * 10 / 100 + 50 * 15 / 100 3 + 7.5 = 10.5 answer b" | a = 10 * 30
b = a / 100
c = 15 * 50
d = c / 100
e = b + d
|
['a ) 98', 'b ) 104', 'c ) 100', 'd ) 61', 'e ) 51'] | a | subtract(volume_cube(5), volume_cube(const_3)) | smita was making a cube with dimensions 5 * 5 * 5 using 1 * 1 * 1 cubes . what is the number of cubes needed to make a hollow cube looking of the same shape ? | cube use to make perfect 5 * 5 * 5 = 125 no need of cube to make hollow = ( 5 - 2 ) * ( 5 - 2 ) * ( 5 - 2 ) = 27 so total cube need to make hollow cube is = 125 - 27 = 98 answer : a | a = volume_cube - (
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a ) 20 , b ) 25 , c ) 40 , d ) 50 , e ) 75 | d | subtract(subtract(multiply(50, add(const_2, const_3)), multiply(40, const_4)), 40) | the average of temperatures at noontime from monday to friday is 50 ; the lowest one is 40 , what is the possible maximum range of the temperatures ? | "there are 5 days so the sum of temperature can be 50 * 5 = 250 lowest is 40 . to find the maximum range we can say the temperature was the lowest for 4 of the 5 days so 4 * 40 = 160 . on the fifth day it is 250 - 160 = 90 range is therefore 90 - 40 = 50 answer d" | a = 2 + 3
b = 50 * a
c = 40 * 4
d = b - c
e = d - 40
|
a ) 166 , b ) 140 , c ) 178 , d ) 177 , e ) 169 | b | add(add(multiply(divide(const_100, 45), 36), multiply(divide(30, 45), 36)), 36) | a sun is divided among x , y and z in such a way that for each rupee x gets , y gets 45 paisa and z gets 30 paisa . if the share of y is rs . 36 , what is the total amount ? | "x : y : z = 100 : 45 : 30 20 : 9 : 6 9 - - - 36 35 - - - ? = > 140 answer : b" | a = 100 / 45
b = a * 36
c = 30 / 45
d = c * 36
e = b + d
f = e + 36
|
a ) 15 % , b ) 17 % , c ) 21 % , d ) 19 % , e ) 25 % | b | multiply(subtract(multiply(subtract(const_1, divide(10, const_100)), const_2), divide(10, const_100)), 10) | a retailer sold 2 houses at each $ 10000 . on 1 st house he got 30 % profit and on 2 nd house he loss 10 % then find his net profit or loss ? | a = 30 % b = - 10 % profit % = 30 - 10 + ( - 300 / 100 ) = 17 % answer is b | a = 10 / 100
b = 1 - a
c = b * 2
d = 10 / 100
e = c - d
f = e * 10
|
a ) 12 , b ) 7 , c ) 0 , d ) 5 , e ) 3 | c | multiply(divide(add(200, const_1), const_3), add(add(reminder(1, 13), reminder(3, 13)), reminder(power(3, const_2), 13))) | the remainder when e = 1 + 3 + 3 ^ 2 + 3 ^ 3 + . . . . . . . . . . + 3 ^ 200 is divided 13 . | e = 1 + 3 + 3 ^ 2 + 3 ^ 3 + . . . . . . . . . . . . . . . . . . + 3 ^ 200 is a geometric progression having common ratio as ' 3 ' and number of terms as ' 201 ' . since sum to n terms in gp = a ( r ^ n - 1 ) / ( r - 1 ) where a = first term and r = common ration hence , 1 * ( 3 ^ 201 - 1 ) / ( 3 - 1 ) rem of ( 3 ^ 201 - 1 ) / 2 divided by 13 3 ^ 201 - 1 / 26 wkt , 3 ^ 3 = 27 = 26 + 1 { ( 26 + 1 ) ^ 67 - 1 } / 26 { 1 - 1 } / 26 = > 0 . c | a = 200 + 1
b = a / 3
c = reminder + (
d = c + reminder
e = b * d
|
a ) $ 6715 , b ) $ 5615 , c ) $ 6415 , d ) $ 6615 , e ) $ 6315 | d | multiply(6000, power(add(const_1, divide(5, const_100)), const_2)) | today joelle opened an interest - bearing savings account and deposited $ 6000 . if the annual interest rate is 5 percent compounded interest , and she neither deposits nor withdraws money for exactly 2 years , how much money will she have in the account ? | interest for 1 st year = 6000 * 5 / 100 = 300 interest for 2 nd year = 6300 * 5 / 100 = 315 total = 6000 + 300 + 315 = 6615 answer : d | a = 5 / 100
b = 1 + a
c = b ** 2
d = 6000 * c
|
a ) 3 hours , b ) 4 hours , c ) 5 hours , d ) 6 hours , e ) 7 hours | e | divide(196, 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 196 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 = 196 km time taken = distance / speed = 196 / 28 = 7 hours answer : option e" | a = 24 + 4
b = 196 / a
|
a ) 200 m , b ) 379 m , c ) 400 m , d ) 589 m , e ) 600 m | c | divide(900, multiply(subtract(78, 1), const_0_2778)) | a train 900 m long is running at a speed of 78 km / hr . if it crosses a tunnel in 1 min , then the length of the tunnel is ? | "speed = 78 * 5 / 18 = 65 / 3 m / sec . time = 1 min = 60 sec . let the length of the train be x meters . then , ( 900 + x ) / 60 = 65 / 3 x = 400 m . answer : c" | a = 78 - 1
b = a * const_0_2778
c = 900 / b
|
a ) rs . 27,000 , b ) rs . 24,000 , c ) rs . 30,000 , d ) rs . 36,000 , e ) none of these | a | multiply(add(multiply(multiply(multiply(const_4, 2), multiply(add(2, const_3), 2)), const_100), multiply(multiply(add(2, const_3), const_100), const_100)), divide(divide(multiply(add(2, const_3), 2), 2), multiply(const_4, const_3))) | jayant opened a shop investing rs . 30,000 . madhu joined him 2 months later , investing rs . 45,000 . they earned a profit of rs . 54,000 after completion of one year . what will be madhu ' s share of profit ? | "30,000 * 12 = 45,000 * 8 1 : 1 madhu ' s share = 1 / 2 * 54,000 i . e . rs . 27,000 answer : a" | a = 4 * 2
b = 2 + 3
c = b * 2
d = a * c
e = d * 100
f = 2 + 3
g = f * 100
h = g * 100
i = e + h
j = 2 + 3
k = j * 2
l = k / 2
m = 4 * 3
n = l / m
o = i * n
|
a ) 15 , b ) 11 , c ) 10 , d ) 8 , e ) 7 | a | divide(multiply(20, 600), add(600, 200)) | 600 men have provisions for 20 days . if 200 more men join them , for how many days will the provisions last now ? | "600 * 20 = 800 * x x = 15 . answer : a" | a = 20 * 600
b = 600 + 200
c = a / b
|
a ) 624 , b ) 450 , c ) 480 , d ) 504 , e ) 555 | a | multiply(52, 12) | the hcf of two numbers is 52 and the other two factors of their lcm are 11 and 12 . what is the largest number . | "explanation : hcf of the two numbers = 52 hcf will be always a factor of lcm 52 is factor of lcm other two factors are 11 & 12 then the numbers are ( 52 * 11 ) and ( 52 x 12 ) = 572 and 624 answer : option a" | a = 52 * 12
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a ) 33 , b ) 47 , c ) 50 , d ) 53 , e ) 62 | a | add(divide(270, 9), 3) | a whale goes on a feeding frenzy that lasts for 9 hours . for the first hour he catches and eats x kilos of plankton . in every hour after the first , it consumes 3 kilos of plankton more than it consumed in the previous hour . if by the end of the frenzy the whale will have consumed a whopping accumulated total 270 kilos of plankton , how many kilos did he consume on the sixth hour ? | "if you list the amount eaten each hour , you ' ll get an equally spaced list , increasing by 3 each hour . in any equally spaced list , the median equals the mean . here , the mean is 270 / 9 = 30 , so the median is also 30 , and that is the amount eaten in the 5 th hour . we need to add 3 to find the total eaten in the next hour , so the answer is 33 . option a" | a = 270 / 9
b = a + 3
|
a ) 24 , b ) 5 , c ) 6 , d ) 8 , e ) 9 | a | divide(divide(multiply(multiply(12, 8), 10), 5), 8) | in a garment industry , 12 men working 8 hours per day complete a piece of work in 10 days . to complete the same work in 8 days , working 5 hours a day , the number of men required is : | "explanation : let the required number of men be x . less days , more men ( indirect proportion ) more working hrs per day , less men ( indirect proportion ) days 8 : 10 working hrs 5 : 8 : : 12 : x = > 8 x 5 x x = 10 x 8 x 12 = > x = 10 x 8 x 12 / ( 8 x 5 ) = > x = 24 answer : a" | a = 12 * 8
b = a * 10
c = b / 5
d = c / 8
|
['a ) 40288.57 cm ^ 2', 'b ) 40388.57 cm ^ 2', 'c ) 40488.57 cm ^ 2', 'd ) 40588.57 cm ^ 2', 'e ) 40688.57 cm ^ 2'] | a | divide(divide(subtract(multiply(const_pi, multiply(multiply(6, 40), multiply(6, 40))), multiply(const_pi, multiply(multiply(40, const_2), multiply(40, const_2)))), 2), 2) | if the radii of umbra and penumbra cast by an object on a wall are of the ratio 2 : 6 , what is the area of the penumbra ring around the umbra of the latter β s radius is 40 cms ? | et the radius of umbra and penumbra are 2 k and 6 k . then as given radius of umbra = 40 cm so 2 k = 40 k = 20 radius of penumbra = 20 * 6 = 120 area of penumbra ring around the umbra = area of penumbra - area of umbra 22 / 7 * [ ( 120 ) ^ 2 - ( 40 ) ^ 2 ] = 40288.57 cm ^ 2 answer : a | a = 6 * 40
b = 6 * 40
c = a * b
d = math.pi * c
e = 40 * 2
f = 40 * 2
g = e * f
h = math.pi * g
i = d - h
j = i / 2
k = j / 2
|
a ) 1098 , b ) 1000 , c ) 1628 , d ) 1098 , e ) 1094 | b | divide(multiply(multiply(6, 5), 4), multiply(multiply(divide(60, const_100), divide(50, const_100)), divide(40, const_100))) | if in a box of dimensions 6 m * 5 m * 4 m smaller boxes of dimensions 60 cm * 50 cm * 40 cm are kept in it , then what will be the maximum number of the small boxes that can be kept in it ? | 6 * 5 * 4 = 60 / 100 * 50 / 100 * 40 / 100 * x 1 = 1 / 10 * 1 / 10 * 1 / 10 * x = > x = 1000 answer : b | a = 6 * 5
b = a * 4
c = 60 / 100
d = 50 / 100
e = c * d
f = 40 / 100
g = e * f
h = b / g
|
a ) 2 min , b ) 4 min , c ) 5 min , d ) 3 min , e ) 3.5 min | d | divide(64816, multiply(7.2, const_1000)) | the area of a square field is 64816 sq m . how long will a lady take to cross the field diagonally at the rate of 7.2 km / hr ? | "area of a square field = 64816 sq m let the side of square = a a ^ 2 = 64816 = > a = 254.59 diagonal = ( 2 ) ^ ( 1 / 2 ) * a = 1.414 * 254.59 = 360 speed of lady = 7.2 km / hour = 7200 m / hour = 120 m / min time taken by lady to cross the field diagonally = 360 / 120 = 3 min answer d" | a = 7 * 2
b = 64816 / a
|
a ) 70 % , b ) 60 % , c ) 30 % , d ) 75 % , e ) none of above | b | multiply(divide(15, 25), const_100) | the ratio 15 : 25 expressed as percent equals to | "explanation : actually it means 15 is what percent of 25 , which can be calculated as , ( 15 / 25 ) * 100 = 15 * 4 = 60 answer : option b" | a = 15 / 25
b = a * 100
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a ) 18 , b ) 16 , c ) 32 , d ) 56 , e ) 58 | b | multiply(divide(divide(12880, 230), 7), 2) | the ratio between the number of sheep and the number of horses at the stewar farm is 2 to 7 . if each of horse is fed 230 ounces of horse food per day and the farm needs a total 12880 ounces of horse food per day . what is number sheep in the form ? ? | "et no of sheep and horses are 2 k and 7 k no of horses = 12880 / 230 = 56 now 7 k = 56 and k = 8 no of sheep = ( 2 * 8 ) = 16 answer : b" | a = 12880 / 230
b = a / 7
c = b * 2
|
a ) 1 : 5 , b ) 1 : 4 , c ) 1 : 3 , d ) 1 : 2 , e ) 2 : 3 | c | multiply(subtract(divide(4, 6), divide(const_1, const_2)), const_2) | the ratio of the arithmetic mean of two numbers to one of the numbers is 4 : 6 . what is the ratio of the smaller number to the larger number ? | for two numbers , the arithmetic mean is the middle of the two numbers . the ratio of the mean to the larger number is 4 : 6 , thus the smaller number must have a ratio of 2 . the ratio of the smaller number to the larger number is 2 : 6 = 1 : 3 . the answer is c . | a = 4 / 6
b = 1 / 2
c = a - b
d = c * 2
|
a ) 10.6 , b ) 10.2 , c ) 10.8 , d ) 10.4 , e ) 10.0 | d | multiply(add(22, 4), divide(24, const_60)) | the speed of a boat in still water in 22 km / hr and the rate of current is 4 km / hr . the distance travelled downstream in 24 minutes is : | "speed downstream = ( 22 + 4 ) = 26 kmph time = 24 minutes = 24 / 60 hour = 2 / 5 hour distance travelled = time Γ speed = 2 / 5 Γ 26 = 10.4 km answer is d ." | a = 22 + 4
b = 24 / const_60
c = a * b
|
a ) 4 , b ) 8 , c ) 10 , d ) 12 , e ) 15 | a | divide(add(45, 25), 25) | the present ratio of students to teachers at a certain school is 45 to 1 . if the student enrollment were to increase by 50 students and the number of teachers were to increase by 5 , the ratio of students to teachers would then be 25 to 1 . what is the present number of teachers ? | "we are given that the ratio of students to teacher is 45 to 1 . we can rewrite this using variable multipliers . students : teachers = 45 x : x we are next given that student enrollment increases by 50 and the number of teachers increases by 5 . with this change the new ratio becomes 25 to 1 . we can put all this into an equation : students / teachers ο 25 / 1 = ( 45 x + 50 ) / ( x + 5 ) if we cross multiply we have : 25 ( x + 5 ) = 45 x + 50 25 x + 125 = 45 x + 50 3.75 = x x ~ 4 since x is the present number of teachers , currently there are 4 teachers . answer a ." | a = 45 + 25
b = a / 25
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a ) 20 , b ) 30 , c ) 40 , d ) 60 , e ) 80 | e | multiply(3, divide(lcm(add(5, 3), add(3, 2)), add(3, 2))) | the number of stamps that kaye and alberto had were in the ration of 5 : 3 respectively . after kaye gave alberto 10 of her stamps , the ration of the number of kaye had to the number of alberto had was 3 : 2 . as a result of the gift , kaye had how many more stamps than alberto ? | "c k 1 = 5 x a 1 = 3 x k 2 = 5 x - 10 a 2 = 3 x + 10 k 2 / a 2 = 3 / 2 ( 5 x - 10 ) / ( 3 x + 10 ) = 3 / 2 2 * ( 5 x - 10 ) = 3 * ( 3 x + 10 ) 10 x - 20 = 9 x + 30 x = 50 k 2 = 5 * 50 - 10 = 240 a 2 = 3 * 50 + 10 = 160 k 2 - a 2 = 80 answer : e" | a = 5 + 3
b = 3 + 2
c = math.lcm(a, b)
d = 3 + 2
e = c / d
f = 3 * e
|
a ) 6.33 , b ) 7.5 , c ) 10 , d ) 15 , e ) 19 | d | subtract(16, divide(16, 16)) | a certain bacteria colony doubles in size every day for 16 days , at which point it reaches the limit of its habitat and can no longer grow . if two bacteria colonies start growing simultaneously , how many days will it take them to reach the habitat β s limit ? | if there is one bacteria colony , then it will reach the limit of its habitat in 16 days . if there are two bacteria colonies , then in order to reach the limit of habitat they would need to double one time less than in case with one colony . thus colonies need to double 15 times . answer : d . similar questions to practice : hope it helps . | a = 16 / 16
b = 16 - a
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a ) 50 cm 2 , b ) 60 cm 2 , c ) 75 cm 2 , d ) 90 cm 2 , e ) 150 cm 2 | c | multiply(multiply(divide(const_1, const_2), add(4, 6)), 15) | find the area of the quadrilateral of one of its diagonals is 15 cm and its off sets 6 cm and 4 cm ? | "1 / 2 * 15 ( 6 + 4 ) = 75 cm 2 answer : c" | a = 1 / 2
b = 4 + 6
c = a * b
d = c * 15
|
a ) 36 , b ) 24 , c ) 72 , d ) 29 , e ) 22 | b | multiply(23, 1) | the average of 10 numbers is 23 . if each number is increased by 1 , what will the new average be ? | "sum of the 10 numbers = 230 if each number is increased by 1 , the total increase = 1 * 10 = 10 the new sum = 230 + 10 = 240 the new average = 240 / 10 = 24 . answer : b" | a = 23 * 1
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a ) 15 , b ) 16 , c ) 18 , d ) 12 , e ) 10 | d | divide(15, add(const_1, divide(25, const_100))) | sakshi can do a piece of work in 15 days . tanya is 25 % more efficient than sakshi . the number of days taken by tanya to do the same piece of work : | "solution ratio of times taken by sakshi and tanya = 125 : 100 = 5 : 4 . suppose tanya taken x days to do the work . 5 : 4 : : 15 : x β x = ( 15 x 4 / 5 ) β x = 12 days . hence , tanya takes 16 days is complete the work . answer d" | a = 25 / 100
b = 1 + a
c = 15 / b
|
a ) 10 hours , b ) 12 hours , c ) 8 hours , d ) 5 hours , e ) 15 hours | a | divide(10, const_1) | an electric pump can fill a tank in 5 hours . because of a leak in the tank , it took 10 hours to fill the tank . if the tank is full , how much time will the leak take to empty it ? | "work done by the leak in 1 hour = 1 / 5 - 1 / 10 = 1 / 10 the leak will empty the tank in 10 hours answer is a" | a = 10 / 1
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a ) 20 , b ) 40 , c ) no = 60 , d ) 80 , e ) 100 | c | subtract(negate(14), multiply(subtract(6, 12), divide(subtract(6, 12), subtract(4, 6)))) | 4 , 6 , 12 , 14 , 28 , 30 , . . ? | "c 60 the given sequence is a combination of two series 4 , 12 , 28 , . . . . and 6 , 14 , 30 , . . . . the pattern is + 8 , + 16 , + 32 . so , the missing number = ( 28 + 32 ) = 60" | a = negate - (
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a ) 48 , b ) 99 , c ) 266 , d ) 28 , e ) 17 | a | multiply(6, multiply(const_2, const_4)) | a cubical block of metal weighs 6 pounds . how much will another cube of the same metal weigh if its sides are twice as long ? | "explanation : if you double the sides of a cube , the ratio of the surface areas of the old and new cubes will be 1 : 4 . the ratio of the volumes of the old and new cubes will be 1 : 8 . weight is proportional to volume . so , if the first weighs 6 pounds , the second weighs 6 x 8 pounds = 48 . answer : a ) 48" | a = 2 * 4
b = 6 * a
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a ) 25 , b ) 33 , c ) 50 , d ) 67 , e ) 75 | b | multiply(divide(multiply(subtract(divide(multiply(multiply(const_4, const_4), const_100), 8), divide(multiply(multiply(const_4, const_4), const_100), add(8, divide(multiply(8, 50), const_100)))), 8), multiply(multiply(const_4, const_4), const_100)), const_100) | working together at their respective constant rates , machine a and machine b can produce 1,600 units in 8 hours . working alone , machine b would complete that same output in 50 % more time . if machine a were to work on its own for an 8 - hour shift , what percent of the 1,600 unit total would it produce ? | 1 / a + 1 / b = 1 / t 1 / a + 1 / 12 = 1 / 8 ( 50 % more of 8 is 12 ) 1 / a = 1 / 24 machine a can produce 1600 units in 24 hrs , so it can produce 1600 * 8 / 24 = 533.33 units is 8 hrs . 533.33 is 33.33 % of 1600 . b is the answer | a = 4 * 4
b = a * 100
c = b / 8
d = 4 * 4
e = d * 100
f = 8 * 50
g = f / 100
h = 8 + g
i = e / h
j = c - i
k = j * 8
l = 4 * 4
m = l * 100
n = k / m
o = n * 100
|
a ) - 12 , b ) 1 , c ) - 2 , d ) - 3 , e ) 4 | a | multiply(subtract(4, const_4), 3) | find the value for x from below equation : x / 3 = - 4 ? | "1 . multiply both sides by 3 : x * 3 / 3 = - 4 / 3 . simplify both sides : x = - 12 a" | a = 4 - 4
b = a * 3
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a ) 16 , b ) 17 , c ) 18 , d ) 19 , e ) 20 | e | inverse(add(divide(const_1, 36), divide(const_1, 45))) | a sum of money is sufficient to pay x ' s wages for 36 days and y ' s wages for 45 days . the same money is sufficient to pay the wages of both for ? | let the total money be $ x x ' s 1 day work = $ x / 36 y ' s 1 day work = $ x / 45 x + y 1 day work = $ x / 20 money is sufficient to pay the wages of both for 20 days answer is e | a = 1 / 36
b = 1 / 45
c = a + b
d = 1/(c)
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a ) 58 m , b ) 60 m , c ) 80 m , d ) 82 m , e ) 84 m | a | divide(add(divide(5300, 26.50), multiply(const_2, 16)), const_4) | length of a rectangular plot is 16 mtr more than its breadth . if the cost of fencing the plot at 26.50 per meter is rs . 5300 , what is the length of the plot in mtr ? | "let breadth = x metres . then , length = ( x + 16 ) metres . perimeter = 5300 m = 200 m . 26.50 2 [ ( x + 16 ) + x ] = 200 2 x + 16 = 100 2 x = 84 x = 42 . hence , length = x + 16 = 58 m a" | a = 5300 / 26
b = 2 * 16
c = a + b
d = c / 4
|
a ) 62.5 , b ) 62.0 , c ) 62.6 , d ) 62.1 , e ) 98.68 | e | divide(multiply(add(75, divide(multiply(75, 25), const_100)), const_100), subtract(const_100, 5)) | at what price must an article costing rs . 75 be marked in order that after deducting 5 % from the list price . it may be sold at a profit of 25 % on the cost price ? | "cp = 75 sp = 75 * ( 125 / 100 ) = 93.75 mp * ( 95 / 100 ) = 93.75 mp = 98.68 answer : e" | a = 75 * 25
b = a / 100
c = 75 + b
d = c * 100
e = 100 - 5
f = d / e
|
a ) 9 , b ) 10 , c ) 11 , d ) 12 , e ) 8 | c | add(8, divide(subtract(155, 20), add(25, 20))) | two stations a and b are 155 km apart on a straight line . one train starts from a at 7 a . m . and travels towards b at 20 kmph . another train starts from b at 8 a . m . and travels towards a at a speed of 25 kmph . at what time will they meet ? | "suppose they meet x hours after 7 a . m . distance covered by a in x hours = 20 x km . distance covered by b in ( x - 1 ) hours = 25 ( x - 1 ) km . therefore 20 x + 25 ( x - 1 ) = 155 45 x = 180 x = 4 . so , they meet at 11 a . m . answer : option c" | a = 155 - 20
b = 25 + 20
c = a / b
d = 8 + c
|
a ) 32 , b ) 36 , c ) 37 , d ) 38 , e ) 39 | a | add(divide(subtract(add(40, 2), 12), 1.5), 12) | each week , harry is paid x dollars per hour for the first 12 hours and 1.5 x dollars for each additional hour worked that week . each week , james is paid x dollars per per hour for the first 40 hours and 2 x dollars for each additional hour worked that week . last week james worked a total of 41 hours if harry and james were paid the same amount last week , how many hours did harry work last week ? | "42 x = 12 x + 1.5 x ( h - 12 ) = = > 42 = 12 + 1.5 ( h - 12 ) = = > h - 12 = 30 / 1.5 = 20 = = > h = 32 answer is a" | a = 40 + 2
b = a - 12
c = b / 1
d = c + 12
|
a ) 20.23 , b ) 20.13 , c ) 30.93 , d ) 19.56 , e ) 10.93 | d | multiply(divide(60, add(add(divide(5, 3), divide(2, 5)), 5)), 2) | a , b and c play a cricket match . the ratio of the runs scored by them in the match is a : b = 5 : 3 and b : c = 5 : 2 . if the total runs scored by all of them are 60 , the runs scored by b are ? | "a : b = 5 : 3 b : c = 5 : 2 a : b : c = 25 : 15 : 6 15 / 46 * 60 = 19.56 answer : d" | a = 5 / 3
b = 2 / 5
c = a + b
d = c + 5
e = 60 / d
f = e * 2
|
a ) 50 kmph , b ) 60 kmph , c ) 70 kmph , d ) 80 kmph , e ) 20 kmph | e | divide(180, multiply(divide(3, 2), 6)) | a van takes 6 hours to cover a distance of 180 km . how much should the speed in kmph be maintained to cover the same direction in 3 / 2 th of the previous time ? | "time = 6 distence = 180 3 / 2 of 6 hours = 6 * 3 / 2 = 9 hours required speed = 180 / 9 = 20 kmph e" | a = 3 / 2
b = a * 6
c = 180 / b
|
a ) s . 20 , b ) s . 48 , c ) s . 42 , d ) s . 20 , e ) s . 60 | a | divide(multiply(subtract(30, 20), const_100), const_2) | find the principal which yields a simple interest of rs . 20 and compound interest of rs . 30 in two years , at the same percent rate per annum ? | "explanation : si in 2 years = rs . 20 , si in 1 year = rs . 10 ci in 2 years = rs . 30 % rate per annum = [ ( ci β si ) / ( si in 1 year ) ] * 100 = [ ( 30 β 20 ) / 20 ] * 100 = 50 % p . a . let the principal be rs . x time = t = 2 years % rate = 50 % p . a . si = ( prt / 100 ) 20 = ( x * 50 * 2 ) / 100 x = rs . 20 answer : a" | a = 30 - 20
b = a * 100
c = b / 2
|
a ) 20 days , b ) 15 days , c ) 30 days , d ) 45 days , e ) 50 days | d | divide(subtract(const_1, multiply(10, divide(const_1, 40))), divide(const_1, 60)) | a can do a work in 40 days , b in 60 days . a alone start the work after 10 days he left work . in how many days b can finish the work ? | 10 days work of a = 10 / 40 = 1 / 4 remaining work = 1 - 1 / 4 = 3 / 4 b can finish 3 / 4 work = 60 * 3 / 4 = 45 days answer is d | a = 1 / 40
b = 10 * a
c = 1 - b
d = 1 / 60
e = c / d
|
a ) 81 , b ) 82 , c ) 84 , d ) 86 , e ) 88 | c | multiply(multiply(7, 3), 4) | a certain university will select 1 of 4 candidates eligible to fill a position in the mathematics department and 2 of 7 candidates eligible to fill 2 identical positions in the computer science department . if none of the candidates is eligible for a position in both departments , how many different sets of 3 candidates are there to fill the 3 positions ? | 1 c 4 * 2 c 7 = 4 * 21 = 84 the answer is ( c ) | a = 7 * 3
b = a * 4
|
a ) 1 / 16 , b ) 5 / 42 , c ) 1 / 8 , d ) 3 / 16 , e ) 1 / 3 | c | divide(divide(choose(15, const_1), 15), power(const_3, const_2)) | each factor of 150 is inscribed on its own plastic ball , and all of the balls are placed in a jar . if a ball is randomly selected from the jar , what is the probability that the ball is inscribed with a multiple of 15 ? | "the # of factors 150 has is 12 so out of 12 factors only four are multiples of 15 : 15 , 30,75 and 150 , itself ; so , the probability is 4 / 12 = 1 / 3 . answer : c ." | a = math.comb(15, 1)
b = a / 15
c = 3 ** 2
d = b / c
|
a ) 0.1503 , b ) 0.001503 , c ) 1.503 , d ) 0.001503 , e ) none of these | a | multiply(divide(15.03, 0.01), const_100) | 15.03 Γ£ β 0.01 = ? | "15.03 Γ£ β 0.01 = ? or , ? = 0.1503 answer a" | a = 15 / 3
b = a * 100
|
a ) 15 , b ) 18 , c ) 22 , d ) 25 , e ) 26 | a | multiply(multiply(divide(const_1, divide(divide(const_1, 4), 15)), subtract(const_1, multiply(multiply(6, 2), divide(divide(const_1, 4), 12)))), divide(const_1, 3)) | 12 men can complete a piece of work in 4 days , while 15 women can complete the same work in 4 days . 6 men start working on the job and after working for 2 days , all of them stopped working . how many women should be put on the job to complete the remaining work , if it is to be completed in 3 days . | one man β s one day β s work = 1 / 48 one woman β s one day β s work = 1 / 60 6 men β s 2 day β s work = ( ( 6 / 48 ) x 2 ) = 1 / 4 remaining work = 3 / 4 now , 1 / 60 work s done in 1 day by 1 woman . thus , 3 / 4 work will be done in 3 days by ( 60 x ( 3 / 4 ) x ( 1 / 3 ) ) = 15 women . answer : a | a = 1 / 4
b = a / 15
c = 1 / b
d = 6 * 2
e = 1 / 4
f = e / 12
g = d * f
h = 1 - g
i = c * h
j = 1 / 3
k = i * j
|
a ) 10 % , b ) 12 % , c ) 15 % , d ) 27 % , e ) 37 % | e | multiply(divide(subtract(divide(50, const_100), divide(20, const_100)), subtract(const_1, divide(20, const_100))), const_100) | mr . kramer , the losing candidate in a two - candidate election , received 942,568 votes , which was exactly 20 percent of all votes cast . approximately what percent of the remaining votes would he need to have received in order to have won at least 50 percent of all the votes cast ? | "lets assume that candidate got 20 % votes and total votes is 100 . candidate won = 20 remaining = 80 to get 50 % , candidate requires 30 votes from 100 which is 30 % and 30 votes from 80 . 30 / 80 = 37.5 % which is approx 37 % . hence the answer is e" | a = 50 / 100
b = 20 / 100
c = a - b
d = 20 / 100
e = 1 - d
f = c / e
g = f * 100
|
a ) 20 , b ) 30 , c ) 40 , d ) 50 , e ) 60 | e | subtract(multiply(subtract(60, divide(60, add(2, 1))), 2), divide(60, add(2, 1))) | in a mixture 60 litres , the ratio of milk and water 2 : 1 . if this ratio is to be 1 : 2 , then the quanity of water to be further added is : | quantity of milk = ( 60 x 2 / 3 ) litres = 40 litres . quantity of water in it = ( 60 - 40 ) litres = 20 litres . new ratio = 1 : 2 let quantity of water to be added further be x litres . then , milk : water = ( 40 / ( 20 + x ) ) now , ( 40 / ( 20 + x ) ) = 1 / 2 = 20 + x = 80 = x = 60 . therefore , quantity of water to be added = 60 litres . answer is e . | a = 2 + 1
b = 60 / a
c = 60 - b
d = c * 2
e = 2 + 1
f = 60 / e
g = d - f
|
a ) 2 hrs , b ) 1 hrs , c ) 10 hrs , d ) 5 hrs , e ) 4 hrs | c | subtract(16, add(3, 3)) | a tap can fill a tank in 16 hours . after half the tank is filled , 3 more similar taps are opened . what is the total time taken to fill the tank completely ? | explanation : a tap can fill a tank in 16 hours . therefore the tap can fill half the tank in 8 hours . one tap take time to fill half of the tank = 8 hrs 4 taps take time to fill half of the tank = 8 / 4 hrs = 2 hrs total time taken to fill the tank = 8 hr + 2 hr = 10 hrs answer : c | a = 3 + 3
b = 16 - a
|
a ) 227 , b ) 358 , c ) 456 , d ) 787 , e ) 191 | c | subtract(subtract(600, divide(multiply(600, 20), const_100)), divide(multiply(subtract(600, divide(multiply(600, 20), const_100)), 5), const_100)) | the sale price sarees listed for rs . 600 after successive discount is 20 % and 5 % is ? | "600 * ( 80 / 100 ) * ( 95 / 100 ) = 456 answer : c" | a = 600 * 20
b = a / 100
c = 600 - b
d = 600 * 20
e = d / 100
f = 600 - e
g = f * 5
h = g / 100
i = c - h
|
a ) 35 , b ) 60 , c ) 70 , d ) 80 , e ) 20 | a | divide(divide(multiply(700, 25), const_100), 5) | a reduction of 25 % in the price of oil enables a house wife to obtain 5 kgs more for rs . 700 , what is the reduced price for kg ? | "a 700 * ( 25 / 100 ) = 175 - - - - 5 ? - - - - 1 = > rs . 35" | a = 700 * 25
b = a / 100
c = b / 5
|
a ) 8 , b ) 12 , c ) 11 , d ) 16 , e ) 18 | b | divide(divide(2, subtract(divide(1, 10), divide(1, 15))), add(const_4, 1)) | robert is travelling on his cycle and has calculated to reach point a at 2 p . m . if he travels at 10 kmph , he will reach there at 12 noon if he travels at 15 kmph . at what speed must he travel to reach a at 1 p . m . ? | "explanation : let the distance travelled by x km . then , = > ( x / 10 ) - ( x / 15 ) = 2 = > 3 x - 2 x = 60 = > x = 60 = > time taken to travel 60 km at 10 km / hr = 60 / 10 = 6 hours . = > so , robert started 6 hours before 2 p . m . i . e . , at 8 a . m . therefore , speed required = ( 60 / 5 ) km / hr = 12 km / hr answer : b" | a = 1 / 10
b = 1 / 15
c = a - b
d = 2 / c
e = 4 + 1
f = d / e
|
a ) 10.5 , b ) 11 , c ) 11.5 , d ) 12 , e ) 12.5 | c | divide(subtract(multiply(10, 16), multiply(add(const_4, const_1), 9)), 10) | the average of 10 consecutive integers is 16 . then , 9 is deducted from the first consecutive number , 8 is deducted from the second , 7 is deducted form the third , and so on until the last number which remains unchanged . what is the new average ? | the total subtracted is ( 9 + 8 + . . . + 1 ) = ( 9 * 10 ) / 2 = 45 on average , each number will be reduced by 45 / 10 = 4.5 therefore , the overall average will be reduced by 4.5 the answer is c . | a = 10 * 16
b = 4 + 1
c = b * 9
d = a - c
e = d / 10
|
a ) 805 , b ) 740 , c ) 670 , d ) 530 , e ) 330 | e | subtract(500, add(multiply(2, 61.5), 47)) | after 2 games , team b had an average of 61.5 points per game . if it got only 47 points in game 3 , how many more points does it need to score to get its total above 500 ? | ( 2 * 61.5 ) + 47 + x > 500 123 + 47 + x > 500 170 + x > 500 = > x > 330 option e | a = 2 * 61
b = a + 47
c = 500 - b
|
a ) 3388 , b ) 27678 , c ) 6000 , d ) 2977 , e ) 27681 | c | divide(divide(multiply(multiply(multiply(multiply(10, const_100), multiply(4, const_100)), multiply(5, const_100)), divide(subtract(10, const_1), 10)), multiply(multiply(25, 15), 8)), 10) | calculate the number of bricks , each measuring 25 cm * 15 cm * 8 cm required to construct a wall of dimensions 10 m * 4 m * 5 m when 10 % of its volume is occupied by mortar ? | 10 * 4 / 100 * 5 * 90 / 100 = 25 / 100 * 15 / 100 * 8 / 100 * x 10 * 20 * 90 = 15 * 2 * x = > x = 6000 answer : c | a = 10 * 100
b = 4 * 100
c = a * b
d = 5 * 100
e = c * d
f = 10 - 1
g = f / 10
h = e * g
i = 25 * 15
j = i * 8
k = h / j
l = k / 10
|
a ) 99 , b ) 48 , c ) 16 , d ) 22 , e ) 60 | b | multiply(add(add(7, 6), 3), 3) | a class has a ratio of 3 : 6 : 7 of children with red hair , blonde hair and black hair respectively . if the class has 9 kids with red hair , how many kids are in the class ? | since there are 9 children with red hair , and the class exists in a 3 : 6 : 7 ratio , there are 3 times more children than the unit ratio . therefore 9 red - haired children 3 * ( 6 ) blonde - haired children 3 * ( 7 ) black - haired children 9 + 18 + 21 = 48 alternatively , a part - to - whole ratio can be set up . since the whole is 3 + 6 + 7 = 16 , 9 / 3 = x / 16 3 * 16 = 48 answer : b ) | a = 7 + 6
b = a + 3
c = b * 3
|
a ) 1 / 720 , b ) 1 / 80 , c ) 1 / 10 , d ) 1 / 9 , e ) 1 / 5 | e | add(divide(const_1, 10), divide(const_1, 10)) | a certain club has 10 members , including parry . one of the 10 members is chosen at random to be the president , one of the remaining 9 members is to be chosen at random to be the secretary , and one of the remaining 8 members is to be chosen at random to be the treasurer . what is the probability that parry will be either the member chose to be secretary or the member chose to be treasurer ? | in order to become a secretary the person can not be chosen as a president . this can be done in 9 / 10 ways . then to become a secretary this person ' s probability is 1 / 9 probability to become a secretary = ( 9 / 10 * 1 / 9 ) = 1 / 10 similar concept for treasurer . probability of not president = 9 / 10 , probability of not secretary = 8 / 9 , probability of treasurer 1 / 8 probability to become a treasurer = ( 9 / 10 * 8 / 9 * 1 / 8 ) = 1 / 10 since , the problem is saying parry secretary or treasurer it is addition : 1 / 10 + 1 / 10 = 1 / 5 = e | a = 1 / 10
b = 1 / 10
c = a + b
|
a ) 1 / 60 , b ) 1 / 15 , c ) 3 / 20 , d ) 3 / 5 , e ) 3 / 4 | c | divide(multiply(subtract(const_1, divide(4, 5)), divide(1, 4)), subtract(const_1, divide(2, 3))) | a millionaire bought a job lot of hats 1 / 4 of which were brown . the millionaire sold 2 / 3 of the hats including 4 / 5 of the brown hats . what fraction of the unsold hats were brown . | if initially he had x hats . x / 4 were brown . sold x * 2 / 3 hats . left with x / 3 hats . sold ( x / 4 ) * ( ( 4 / 5 ) = x / 5 brown hats and left with x / 4 - x / 5 = x / 20 brown hats . fraction of brown hats in total remaining hats = ( x / 20 ) / ( x / 3 ) = 3 / 20 hats answer : c | a = 4 / 5
b = 1 - a
c = 1 / 4
d = b * c
e = 2 / 3
f = 1 - e
g = d / f
|
a ) 36 , b ) 60 , c ) 75 , d ) 80 , e ) 100 | a | divide(divide(3, 5), multiply(divide(divide(divide(3, 4), 5), 30), const_2)) | if three machines working at the same rate can do 3 / 4 of a job in 30 minutes , how many minutes would it take two machines working at the same rate to do 3 / 5 of the job ? | "using the std formula m 1 d 1 h 1 / w 1 = m 2 d 2 h 2 / w 2 substituting the values we have 3 * 1 / 2 * 4 / 3 = 2 * 5 / 3 * x ( converted 30 min into hours = 1 / 2 ) 2 = 10 / 3 * x x = 3 / 5 hour so 36 minutes answer : a" | a = 3 / 5
b = 3 / 4
c = b / 5
d = c / 30
e = d * 2
f = a / e
|
a ) 3025 , b ) 4200 , c ) 3060 , d ) 3210 , e ) none of these | a | subtract(power(add(481, 426), 2), multiply(multiply(4, 481), 426)) | { ( 481 + 426 ) 2 - 4 x 481 x 426 } = ? | "explanation : here , the given statement is like ( a + b ) 2 - 4 ab where a = 481 and b = 426 ( a + b ) 2 - 4 ab = ( a 2 + 2 ab + b 2 ) - 4 ab = a 2 - 2 ab + b 2 = ( a - b ) 2 hence { ( 481 + 426 ) 2 - 4 x 481 x 426 } = ( 481 - 426 ) 2 = 552 = 3025 . answer : option a" | a = 481 + 426
b = a ** 2
c = 4 * 481
d = c * 426
e = b - d
|
a ) 30 , b ) 33 , c ) 50 , d ) 67 , e ) 75 | a | multiply(divide(multiply(subtract(divide(multiply(multiply(const_4, const_4), const_100), 8), divide(multiply(multiply(const_4, const_4), const_100), add(8, divide(multiply(8, 50), const_100)))), 8), multiply(multiply(const_4, const_4), const_100)), const_100) | working together at their respective constant rates , machine a and machine b can produce 900 units in 8 hours . working alone , machine b would complete that same output in 50 % more time . if machine a were to work on its own for an 8 - hour shift , what percent of the 900 unit total would it produce ? | "1 / a + 1 / b = 1 / t 1 / a + 1 / 12 = 1 / 8 ( 50 % more of 8 is 12 ) 1 / a = 1 / 24 machine a can produce 900 units in 24 hrs , so it can produce 900 * 8 / 24 = 300 units is 8 hrs . 300 is 30 % of 900 . a is the answer" | a = 4 * 4
b = a * 100
c = b / 8
d = 4 * 4
e = d * 100
f = 8 * 50
g = f / 100
h = 8 + g
i = e / h
j = c - i
k = j * 8
l = 4 * 4
m = l * 100
n = k / m
o = n * 100
|
a ) 20 , b ) 24 , c ) 26 , d ) 28 , e ) 30 | a | sqrt(add(multiply(131, const_2), 138)) | sum of the squares of 3 no . ' s is 138 and the sum of their products taken two at a time is 131 . find the sum ? | ( a + b + c ) 2 = a 2 + b 2 + c 2 + 2 ( ab + bc + ca ) = 138 + 2 * 131 a + b + c = β 400 = 20 a | a = 131 * 2
b = a + 138
c = math.sqrt(b)
|
a ) 5 : 00 , b ) 5 : 34 , c ) 5 : 42 , d ) 6 : 00 , e ) 6 : 24 | d | subtract(multiply(6, const_10), multiply(multiply(multiply(divide(add(const_60, 15), const_60), divide(subtract(const_60, 20), const_60)), divide(add(const_60, 20), const_60)), subtract(const_60, 10))) | a not - so - good clockmaker has four clocks on display in the window . clock # 1 loses 10 minutes every hour . clock # 2 gains 15 minutes every hour relative to clock # 1 ( i . e . , as clock # 1 moves from 12 : 00 to 1 : 00 , clock # 2 moves from 12 : 00 to 1 : 15 ) . clock # 3 loses 20 minutes every hour relative to clock # 2 . finally , clock # 4 gains 20 minutes every hour relative to clock # 3 . if the clockmaker resets all four clocks to the correct time at 12 noon , what time will clock # 4 display after 6 actual hours ( when it is actually 6 : 00 pm that same day ) ? | "c 1 loses 15 minutes every hour . so after 60 minutes have passed , c 1 displays that 60 - 15 = 45 minutes have passed . c 2 gains 15 minutes for every 60 minutes displayed on c 1 . thus , the time displayed on c 2 is 75 / 60 = 5 / 4 the time displayed on c 1 . so after 60 minutes have passed , c 2 displays the passing of ( 5 / 4 * 45 ) minutes . c 3 loses 20 minutes for every 60 minutes displayed on c 2 . thus , the time displayed on c 3 is 40 / 60 = 2 / 3 the time displayed on c 2 . so after 60 minutes have passed , c 3 displays the passing of ( 2 / 3 * 5 / 4 * 45 ) minutes . c 4 gains 20 minutes for every 60 minutes displayed on c 3 . thus , the time displayed on c 4 is 80 / 60 = 4 / 3 the time displayed on clock 3 . so after 60 minutes have passed , c 4 displays the passing of 4 / 3 * 2 / 3 * 5 / 4 * 45 = 50 minutes . c 4 loses 10 minutes every hour . in 6 hours , c 4 will lose 6 * 10 = 60 minutes = 1 hour . since the correct time after 6 hours will be 6 pm , c 4 will show a time of 6 - 1 = 6 pm . the correct answer is d ." | a = 6 * 10
b = const_60 + 15
c = b / const_60
d = const_60 - 20
e = d / const_60
f = c * e
g = const_60 + 20
h = g / const_60
i = f * h
j = const_60 - 10
k = i * j
l = a - k
|
a ) 3 , b ) 5 , c ) 6 , d ) 8 , e ) 9 | c | add(25, const_1) | if x and y are positive integers and 11 + x + y + xy = 25 , what is the value of x + y ? | "try each answer choices . for a : 11 + 3 + xy = 25 ; xy = 11 ( impossible , 11 prime number . 1 + 11 does n ' t equal 3 ) for b : 11 + 5 + xy = 25 ; xy = 9 ( no combination of xy = 9 and x + y = 5 ) for c : 11 + 6 + xy = 25 ; xy = 8 ( x + y = 6 ; x = 2 , y = 4 or x = 4 , y = 2 ) for d : 11 + 8 + xy = 25 ; xy = 6 ( no combination of xy = 6 and x + y = 8 ) for e : 11 + 9 + xy = 25 ; xy = 5 ( impossible , 5 prime number . 1 + 5 does n ' t equal 9 ) therefore , answer c ." | a = 25 + 1
|
a ) 8 , b ) 12 , c ) 18 , d ) 24 , e ) 36 | e | divide(multiply(add(add(6, 4), 2), divide(6000, 2)), const_1000) | a 6000 liter tank , half - full of water is being filled from a pipe with a flow rate of 1 kiloliter every 2 minutes . at the same time , the tank is losing water from two drains at a rate of 1 kiloliter every 4 minutes and every 6 minutes . how many minutes does it take to fill the tank completely ? | "in : we have : 1,000 / 2 min = 500 litres per minute out : we have : 1,000 / 4 + 1,000 / 6 then do : in - out to figure out the net inflow per minute ( you get 83.3 ) . then divide the total number of litres you need ( 3,000 by that net inflow to get the minutes ) - 36 min . answer e ." | a = 6 + 4
b = a + 2
c = 6000 / 2
d = b * c
e = d / 1000
|
a ) 177 , b ) 150 , c ) 817 , d ) 480 , e ) 616 | d | divide(1040, add(add(divide(4, 2), divide(4, 3)), const_1)) | rs . 1040 is divided amongst a , b , c so that 2 times a ' s share , 3 times b ' s share and 4 times c ' s share are all equal . find a ' s share ? | a + b + c = 1040 2 a = 3 b = 4 c = x a : b : c = 1 / 2 : 1 / 3 : 1 / 4 = 6 : 4 : 3 6 / 13 * 1040 = rs . 480 answer : d | a = 4 / 2
b = 4 / 3
c = a + b
d = c + 1
e = 1040 / d
|
a ) $ 21,000 , b ) $ 18,000 , c ) $ 12,000 , d ) $ 8000 , e ) $ 4000 | c | divide(add(divide(subtract(300, multiply(divide(6, const_100), 1,000)), subtract(divide(8, const_100), divide(6, const_100))), divide(subtract(300, multiply(divide(6, const_100), 1,000)), subtract(divide(8, const_100), divide(6, const_100)))), 1,000) | salesperson a ' s compensation for any week is $ 300 plus 6 percent of the portion of a ' s total sales above $ 1,000 for that week . salesperson b ' s compensation for any week is 8 percent of b ' s total sales for that week . for what amount of total weekly sales would both salespeople earn the same compensation ? | "300 + 0.06 ( x - 1000 ) = 0.08 x 0.02 x = 240 x = $ 12,000 the answer is c ." | a = 6 / 100
b = a * 1
c = 300 - b
d = 8 / 100
e = 6 / 100
f = d - e
g = c / f
h = 6 / 100
i = h * 1
j = 300 - i
k = 8 / 100
l = 6 / 100
m = k - l
n = j / m
o = g + n
p = o / 1
|
a ) 4 , b ) 6 , c ) 8 , d ) 5 , e ) 3 | b | divide(subtract(add(multiply(7, 7), 5), 6), 8) | if the number is decreased by 5 and divided by 7 the result is 7 . what would be the result if 6 is subtracted and divided by 8 ? | explanation : let the number be x . then , ( x - 5 ) / 7 = 7 = > x - 5 = 49 x = 54 . : ( x - 6 ) / 8 = ( 54 - 6 ) / 8 = 6 answer : option b | a = 7 * 7
b = a + 5
c = b - 6
d = c / 8
|
a ) 1 , b ) 2 , c ) 4 , d ) 3 , e ) 5 | d | subtract(divide(40, 5), 5) | ( x + 5 ) is a factor in x ^ 2 - mx - 40 . what is the value of m ? | "i solved the second degree equation and found it like this : x ^ 2 - mx - 40 = 0 ( x - 8 ) ( x + 5 ) = 0 x = 8 or x = - 5 substituting both values for x in the equation we find : x ^ 2 - mx - 40 = > ( - 5 ) ^ 2 - m ( - 5 ) = 40 = > 25 + 5 m = 40 = > 5 m = 40 - 25 = 15 = > m = 3 and with 8 , using a similar process we end up with : ( 8 ) ^ 2 - m ( 8 ) = 40 - 8 m = 40 - 64 = - 24 m = 3 ao , ans d" | a = 40 / 5
b = a - 5
|
a ) 51 , b ) 54 , c ) 57 , d ) 60 , e ) 63 | e | divide(multiply(add(add(3, 7), add(4, 7)), subtract(10, 4)), 4) | in a rectangular coordinate system , what is the area of a quadrilateral whose vertices have the coordinates ( 4 , - 3 ) , ( 4 , 7 ) , ( 10 , 4 ) , ( 10 , - 7 ) ? | "by graphing the points , we can see that this figure is a trapezoid . a trapezoid is any quadrilateral that has one set of parallel sides , and the formula for the area of a trapezoid is : area = ( 1 / 2 ) Γ ( base 1 + base 2 ) Γ ( height ) , where the bases are the parallel sides . we can now determine the area of the quadrilateral : area = 1 / 2 Γ ( 10 + 11 ) Γ 6 = 63 . the answer is e ." | a = 3 + 7
b = 4 + 7
c = a + b
d = 10 - 4
e = c * d
f = e / 4
|
a ) 70 , b ) 75 , c ) 80 , d ) 85 , e ) 90 | c | multiply(divide(add(divide(50, 50), add(divide(const_100, 50), divide(150, 50))), divide(150, 50)), 40) | a sum is divided among b , c and d in such a way that for each rupee b gets , c gets 150 paisa and d gets 50 paisa . if the share of c is rs . 40 , what is the total amount ? | b : c : d = 100 : 150 : 100 20 : 30 : 10 30 - - - 40 60 - - - ? = > 80 answer : c | a = 50 / 50
b = 100 / 50
c = 150 / 50
d = b + c
e = a + d
f = 150 / 50
g = e / f
h = g * 40
|
a ) 74.1 , b ) 56.3 , c ) 67.8 , d ) 79.7 , e ) 19.4 | a | divide(add(multiply(58, 67), multiply(52, 82)), add(58, 52)) | find the average marks of all the students in 2 separate classes , if the average marks of students in the first class of 58 students is 67 and that of another class of 52 students is 82 . | "sum of the marks for the class of 58 students = 58 * 67 = 3886 sum of the marks for the class of 52 students = 52 * 82 = 4264 sum of the marks for the class of 110 students = 3886 + 4262 = 8150 average marks of all the students = 4200 / 80 = 74.1 answer : a" | a = 58 * 67
b = 52 * 82
c = a + b
d = 58 + 52
e = c / d
|
['a ) 675.5 m 2', 'b ) 780.6 m 2', 'c ) 785.8 m 2', 'd ) 850.5 m 2', 'e ) 950.5 m 2'] | d | subtract(multiply(63, 63), multiply(multiply(multiply(multiply(const_4, const_0_25), const_pi), divide(63, const_2)), divide(63, const_2))) | four horses are tethered at four corners of a square plot of side 63 metres so that they just can not reach one another . the area left ungrazed is : | required area = ( 63 * 63 β 4 * 1 / 4 * 22 / 7 * 63 / 2 * 63 / 2 ) = 850.5 m 2 answer : d | a = 63 * 63
b = 4 * const_0_25
c = b * math.pi
d = 63 / 2
e = c * d
f = 63 / 2
g = e * f
h = a - g
|
a ) 10 , b ) 99 , c ) 5 , d ) 22 , e ) 29 | c | multiply(divide(multiply(4, 5), subtract(multiply(7, 4), multiply(3, 4))), 4) | mixture contains alcohol and water in the ratio 4 : 3 . if 7 liters of water is added to the mixture , the ratio becomes 4 : 5 . find the quantity of alcohol in the given mixture . | "let the quantity of alcohol and water be 4 x litres and 3 x litres respectively 28 x = 4 ( 3 x + 5 ) 16 x = 20 x = 1.25 quantity of alcohol = ( 4 x 1.25 ) litres = 5 litres . answer : c" | a = 4 * 5
b = 7 * 4
c = 3 * 4
d = b - c
e = a / d
f = e * 4
|
a ) 13 / 89 , b ) 15 / 87 , c ) 27 / 89 , d ) 27 / 87 , e ) 89 / 27 | c | divide(1, add(3, divide(1, add(3, divide(1, subtract(3, divide(1, 3))))))) | find the value of 1 / ( 3 + 1 / ( 3 + 1 / ( 3 - 1 / 3 ) ) ) ? | "1 / [ 3 + ( 1 / ( 3 + 1 / ( 3 - 1 / 3 ) ) ) ] = > 1 / [ 3 + 1 / ( 3 + 1 / ( 8 / 3 ) ) ] = > 1 / [ 3 + 1 / ( 3 + 3 / 8 ) ] = > 1 / [ 3 + 8 / 27 ] = > 1 / ( 89 / 27 ) = > 27 / 89 c )" | a = 1 / 3
b = 3 - a
c = 1 / b
d = 3 + c
e = 1 / d
f = 3 + e
g = 1 / f
|
a ) 23 , b ) 27 , c ) 20 , d ) 24 , e ) 11 | d | divide(multiply(20, 18), 15) | 18 men can complete a piece of work in 20 days . in how many days can 15 men complete that piece of work ? | "18 * 20 = 15 * x = > x = 24 days answer : d" | a = 20 * 18
b = a / 15
|
a ) 113.2 , b ) 114 , c ) 114.75 , d ) 124 , e ) 124.75 | e | subtract(multiply(divide(9000, 756), 10.5), divide(1, 4)) | the market value of a 10.5 % stock , in which an income of rs . 756 is derived by investing rs . 9000 , brokerage being 1 / 4 % , is : | "face value = rs . 9000 . dividend = 10.5 % . annual income = rs . 756 . brokerage per rs . 100 = rs . 0.25 . dividend is always paid on the face value of a share . face value * dividend / ( market value + brokerage per rs . 100 ) = annual income . = 9000 * 10.5 / 756 = market value of rs . 100 stock + brokerage per rs . 100 . = market value of rs . 100 stock + brokerage per rs . 100 = rs . 125 . = market value of rs . 100 stock = rs . 125 - re . 0.25 . = market value of rs . 100 stock = rs . 124.75 . answer : e" | a = 9000 / 756
b = a * 10
c = 1 / 4
d = b - c
|
a ) 9 : 3 , b ) 3 : 10 , c ) 2 : 3 , d ) 6 : 3 , e ) 10 : 3 | e | divide(multiply(2, 5), 3) | a dog takes 2 leaps for every 3 leaps of a hare . if one leap of the dog is equal to 5 leaps of the hare , the ratio of the speed of the dog to that of the hare is : | "explanation : dog : hare = ( 2 * 5 ) leaps of hare : 3 leaps of hare = 10 : 3 answer : e" | a = 2 * 5
b = a / 3
|
a ) $ 455 , b ) $ 495 , c ) $ 525 , d ) $ 550 , e ) $ 585 | b | subtract(multiply(55, divide(const_100, 10)), 55) | if a 10 percent deposit that has been paid toward the purchase of a certain product is $ 55 , how much more remains to be paid ? | "90 % remains to be paid so the remaining amount is 9 * 55 = $ 495 . the answer is b ." | a = 100 / 10
b = 55 * a
c = b - 55
|
a ) 30 , b ) 40 , c ) 50 , d ) 60 , e ) 39 | b | divide(subtract(multiply(95, const_2), 70), const_3) | if two times of the daughter β s age in years is included to the mother β s age , the total is 70 and if two times of the mother β s age is included to the daughter β s age , the total is 95 . so the mother β s age is , | b 40 let daughter β s age = a and mother β s age = b given : 2 a + b = 70 and , a + 2 b = 95 solving b , we will get b = 40 . | a = 95 * 2
b = a - 70
c = b / 3
|
a ) 120 , b ) 240 , c ) 360 , d ) 480 , e ) 720 | c | divide(factorial(6), const_2) | an auto assembly plant performs 6 functions with each frame that arrives : add axles , add wheels to the axles , install the windshield to the frame , install the instrument panel , install the steering wheel , and install the interior seating . once those 6 tasks are performed , each car goes to a separate building for finishing touches . if these tasks can be arranged along a linear assembly line in any order , except that the axles must be installed before the wheels can be added , how many r ways can the assembly line be arranged ? | c ) 360 short way : there are 6 c ! ways to do the six tasks . half will have wheels before axles and half will have axles before wheels . so we want r = 6 c ! / 2 - > 720 / 2 = 360 | a = math.factorial(6)
b = a / 2
|
a ) 8 : 7 , b ) 7 : 5 , c ) 1 : 3 , d ) 8 : 7 , e ) 9 : 10 | e | divide(add(300, multiply(divide(300, add(subtract(400, const_100), 300)), add(140, multiply(const_2, add(const_3, const_2))))), subtract(400, const_100)) | a and b put in rs . 300 and rs . 400 respectively into a business . a reinvests into the business his share of the first year ' s profit of rs . 140 where as b does not . in what ratio should they divide the second year ' s profit ? | "explanation : 3 : 4 a = 3 / 7 * 140 = 60 360 : 400 39 : 40 answer : e" | a = 400 - 100
b = a + 300
c = 300 / b
d = 3 + 2
e = 2 * d
f = 140 + e
g = c * f
h = 300 + g
i = 400 - 100
j = h / i
|
a ) $ 52000 , b ) $ 61000 , c ) $ 71540 , d ) $ 63100 , e ) $ 56210 | d | divide(multiply(multiply(const_100, const_100), 631), const_100) | the difference between the compound interest and simple interest on a certain sum at 10 % per annum for 2 years is $ 631 . find the sum ? | let the sum be $ x c . i . = x ( 1 + 10 / 100 ) ^ 2 - x = 21 x / 100 s . i . = ( x * 10 * 2 ) / 100 = x / 5 c . i . - s . i . = ( 21 x / 100 ) - ( x / 5 ) = x / 100 x / 100 = 631 x = 63100 answer is d | a = 100 * 100
b = a * 631
c = b / 100
|
a ) 136 , b ) 192 , c ) 198 , d ) 122 , e ) 142 | a | add(lcm(lcm(4, 6), lcm(8, 10)), 16) | what is the smallest number which when diminished by 16 , is divisible 4 , 6 , 8 and 10 ? | required number = ( lcm of 4 , 6,8 , and 10 ) + 16 = 120 + 16 = 136 option a | a = math.lcm(4, 6)
b = math.lcm(8, 10)
c = math.lcm(a, b)
d = c + 16
|
a ) 5 / 21 , b ) 3 / 7 , c ) 4 / 5 , d ) 5 / 7 , e ) 16 / 21 | c | divide(subtract(subtract(subtract(10, 1), 1), 1), subtract(10, 1)) | in a room with 10 people , 6 people have exactly 1 friend in the room and 5 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 ? | ( 6 / 10 ) ( 7 / 9 ) + ( 5 / 10 ) ( 6 / 9 ) if you choose one of the 6 with one other friend , then you have a 7 / 9 chance of not picking their friend 2 nd . if you choose one of the 5 with 2 friends , you have a 6 / 9 chance of not picking one of their friends second . add them up . 42 / 90 + 30 / 90 72 / 90 = 4 / 5 c . 4 / 5 | a = 10 - 1
b = a - 1
c = b - 1
d = 10 - 1
e = c / d
|
a ) 85 , b ) 87 , c ) 89 , d ) 91 , e ) 93 | a | add(divide(subtract(multiply(79, 4), add(6, add(4, 6))), 4), add(4, 6)) | there are 4 people of different heights standing in order of increasing height . the difference is 2 inches between the first person and the second person , and also between the second person and the third person . the difference between the third person and the fourth person is 6 inches and the average height is 79 . how tall is the fourth person ? | "let x be the height of the first person . then the heights are x , x + 2 , x + 4 , and x + 10 . 4 x + 16 = 4 ( 79 ) = 316 x = 75 and the fourth person has a height of 75 + 10 = 85 inches the answer is a ." | a = 79 * 4
b = 4 + 6
c = 6 + b
d = a - c
e = d / 4
f = 4 + 6
g = e + f
|
a ) 5 : 00 , b ) 5 : 34 , c ) 5 : 42 , d ) 6 : 00 , e ) 6 : 24 | b | subtract(multiply(6, const_10), multiply(multiply(multiply(divide(add(const_60, 15), const_60), divide(subtract(const_60, 20), const_60)), divide(add(const_60, 20), const_60)), subtract(const_60, 17))) | a not - so - good clockmaker has four clocks on display in the window . clock # 1 loses 17 minutes every hour . clock # 2 gains 15 minutes every hour relative to clock # 1 ( i . e . , as clock # 1 moves from 12 : 00 to 1 : 00 , clock # 2 moves from 12 : 00 to 1 : 15 ) . clock # 3 loses 20 minutes every hour relative to clock # 2 . finally , clock # 4 gains 20 minutes every hour relative to clock # 3 . if the clockmaker resets all four clocks to the correct time at 12 noon , what time will clock # 4 display after 6 actual hours ( when it is actually 6 : 00 pm that same day ) ? | c 1 loses 15 minutes every hour . so after 60 minutes have passed , c 1 displays that 60 - 15 = 45 minutes have passed . c 2 gains 15 minutes for every 60 minutes displayed on c 1 . thus , the time displayed on c 2 is 75 / 60 = 5 / 4 the time displayed on c 1 . so after 60 minutes have passed , c 2 displays the passing of ( 5 / 4 * 45 ) minutes . c 3 loses 20 minutes for every 60 minutes displayed on c 2 . thus , the time displayed on c 3 is 40 / 60 = 2 / 3 the time displayed on c 2 . so after 60 minutes have passed , c 3 displays the passing of ( 2 / 3 * 5 / 4 * 45 ) minutes . c 4 gains 20 minutes for every 60 minutes displayed on c 3 . thus , the time displayed on c 4 is 80 / 60 = 4 / 3 the time displayed on clock 3 . so after 60 minutes have passed , c 4 displays the passing of 4 / 3 * 2 / 3 * 5 / 4 * 45 = 50 minutes . c 4 loses 10 minutes every hour . in 6 hours , c 4 will lose 6 * 10 = 60 minutes = 1 hour . since the correct time after 6 hours will be 6 pm , c 4 will show a time of 6 - 1 = 5 : 34 pm . the correct answer is b . | a = 6 * 10
b = const_60 + 15
c = b / const_60
d = const_60 - 20
e = d / const_60
f = c * e
g = const_60 + 20
h = g / const_60
i = f * h
j = const_60 - 17
k = i * j
l = a - k
|
a ) 15 , b ) 45 , c ) 75 , d ) 125 , e ) 150 | b | add(add(add(add(add(multiply(multiply(5, 2), 3), multiply(multiply(5, 2), 3)), multiply(multiply(5, 2), 3)), 2), const_4), const_4) | if both 5 ^ 3 and 3 ^ 4 are factors of n x ( 2 ^ 5 ) x ( 12 ^ 2 ) x ( 7 ^ 3 ) x ( 10 ^ 2 ) , what is the smallest possible positive value of n ? | "( 2 ^ 5 ) x ( 12 ^ 2 ) x ( 7 ^ 3 ) x ( 10 ) has two appearances of 3 ( in 12 ^ 2 ) and two appearances of 5 ( in 10 ^ 2 ) . thus n must include at least 3 ^ 2 * 5 = 45 the answer is b ." | a = 5 * 2
b = a * 3
c = 5 * 2
d = c * 3
e = b + d
f = 5 * 2
g = f * 3
h = e + g
i = h + 2
j = i + 4
k = j + 4
|
a ) 960 , b ) 1060 , c ) 1,200 , d ) 945 , e ) none of these | d | multiply(divide(add(812, 448), const_2), add(const_1, divide(50, const_100))) | the profit earned by selling an article for 812 is equal to the loss incurred when the same article is sold for 448 . what should be the sale price of the article for making 50 per cent profit ? | "let the profit or loss be x and 812 β x = 448 + x or , x = 364 β 2 = 182 \ cost price of the article = 812 β x = 448 + x = 630 \ sp of the article = 630 Γ 150 β 100 = 945 answer d" | a = 812 + 448
b = a / 2
c = 50 / 100
d = 1 + c
e = b * d
|
a ) 4 % , b ) 18 % , c ) 36 % , d ) 40 % , e ) 50 % | e | multiply(subtract(divide(subtract(const_1, divide(46, const_100)), multiply(subtract(const_1, divide(40, const_100)), subtract(const_1, divide(40, const_100)))), const_1), const_100) | if w is 40 percent less than e , e is 40 percent less than y , and z is 46 percent less than y , then z is greater than w by what percent of w ? | given w = 0.6 e , e = 0.6 y , z = 0.54 y , substituting , w = 2 / 3 z - - - - > z = 1.5 w and thus z is 50 % greater than w . e is the correct answer . | a = 46 / 100
b = 1 - a
c = 40 / 100
d = 1 - c
e = 40 / 100
f = 1 - e
g = d * f
h = b / g
i = h - 1
j = i * 100
|
a ) 150 , b ) 160 , c ) 140 , d ) 130 , e ) none | d | divide(divide(1540, const_4), const_3) | what is the angle between the minute hand and the hour hand when the time is 1540 hours ? | solution : the total angle made by the minute hand during an hour is 360 o . if it takes 360 o for an hour , it will take ( 40 / 60 ) * 360 = 240 o . the angle between the hour hand the minute hand will therefore , be somewhere between 240 - 90 = 150 o , as the hour hand is between 3 and 4 . the angle made by the hour hand when it moves from say 3 to 4 will be 30 o . that is the hour hand makes 30 o during the course of an hour . the hour hand will therefore , move ( 40 / 60 ) * 30 = 20 o . therefore , the net angle between the hour hand and the minute hand will be 150 - 20 = 130 o . answer d | a = 1540 / 4
b = a / 3
|
a ) 465 , b ) 470 , c ) 452 , d ) 450 , e ) 460 | a | multiply(divide(add(subtract(60, const_3), add(33, const_2)), const_2), add(divide(subtract(subtract(60, const_3), add(33, const_2)), 3), const_1)) | what is the sum of the multiples of 3 from 33 to 60 , inclusive ? | "the formula we want to use in this type of problem is this : average * total numbers = sum first , find the average by taking the sum of the f + l number and divide it by 2 : a = ( f + l ) / 2 second , find the total numbers in our range by dividing our f and l numbers by 7 and add 1 . ( 60 / 3 ) - ( 33 / 3 ) + 1 multiply these together so what we show average * total numbers = sum ( 33 + 60 ) / 2 * ( 60 / 3 ) - ( 33 / 3 ) + 1 = sum 93 / 2 * 10 = 465 a" | a = 60 - 3
b = 33 + 2
c = a + b
d = c / 2
e = 60 - 3
f = 33 + 2
g = e - f
h = g / 3
i = h + 1
j = d * i
|
a ) 25 , b ) 30 , c ) 35 , d ) 15 , e ) 10 | a | subtract(30, 5) | 3 people each took 5 tests . if the ranges of their scores in the 5 practice tests were 15 , 25 and 30 , what is the minimum possible range in scores of the 3 test - takers ? | i simply looked at the 3 different possible scores for each individual test : 15 , 30,25 we have to find the minimum range : 30 - 15 = 15 30 - 25 = 5 25 - 15 = 10 the find the minimum range , you have to make the set of the 5 scores as small as possible . which means that 4 of the 5 scores of each individual person is zero . 5 * 5 = 25 . answer : a | a = 30 - 5
|
a ) 230 m , b ) 250 m , c ) 643 m , d ) 832 m , e ) 270 m | b | multiply(subtract(26, divide(270, multiply(const_0_2778, 72))), multiply(const_0_2778, 72)) | a goods train runs at the speed of 72 kmph and crosses a 270 m long platform in 26 seconds . what is the length of the goods train ? | "speed = ( 72 x 5 / 18 ) m / sec = 20 m / sec . time = 26 sec . let the length of the train be x metres . then , x + 270 / 26 = 20 x + 270 = 520 x = 250 . answer : b" | a = const_0_2778 * 72
b = 270 / a
c = 26 - b
d = const_0_2778 * 72
e = c * d
|
a ) 7000 , b ) 6000 , c ) 5000 , d ) 4000 , e ) 3000 | e | subtract(subtract(divide(18000, divide(subtract(const_100, 50), const_100)), 18000), 15000) | rebecca ' s annual income is $ 15000 and jimmy ' s annual income is $ 18000 . by how much must rebecca ' s annual income increase so that it constitutes 50 % of rebecca and jimmy ' s combined income ? | total rebecca = x + 15000 ; total = x + 15000 + 18000 x + 15000 / x + 33000 = 50 / 100 therefore x = 3000 e | a = 100 - 50
b = a / 100
c = 18000 / b
d = c - 18000
e = d - 15000
|
['a ) 15', 'b ) 22', 'c ) 26', 'd ) 33', 'e ) 44'] | c | subtract(27, const_1) | a cubic object 3 ' ' x 3 ' ' x 3 ' ' is painted blue on all the outside surfaces , including the top and bottom . if the cube is cut into 27 cubes of 1 ' ' x 1 ' ' x 1 ' ' , how many 1 ' ' cubes do have any painted surfaces ? | c 26 only the box in the very centre of the stack will not suffer the strokes of the paint brush , whereas all the other 26 boxes will have at least one side painted . | a = 27 - 1
|
a ) 17 , b ) 25 , c ) 45 , d ) 47 , e ) 50 | c | add(multiply(divide(add(multiply(8, const_3), 76), add(add(5, 8), 7)), 8), 10) | the ratio of ages of aman , bren , and charlie are in the ratio 5 : 8 : 7 respectively . if 8 years ago , the sum of their ages was 76 , what will be the age of charlie 10 years from now ? | "let the present ages of aman , bren , and charlie be 5 x , 8 x and 7 x respectively . 5 x - 8 + 8 x - 8 + 7 x - 8 = 76 x = 5 present age of charlie = 7 * 5 = 35 charlie ' s age 10 years hence = 35 + 10 = 45 answer = c" | a = 8 * 3
b = a + 76
c = 5 + 8
d = c + 7
e = b / d
f = e * 8
g = f + 10
|
a ) 0 , b ) 1 , c ) 2 , d ) 3 , e ) 4 | a | divide(34, 17) | how many of the positive factors of 17 are not factors of 34 ? | "factors of 17 - 1 , 17 factors of 34 - 1 , 2 , 17 , 34 , comparing both , we have three factors of 17 which are not factors of 34 - no factors the answer is a" | a = 34 / 17
|
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