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a ) 11 , b ) 15 , c ) 20 , d ) 38 , e ) 56 | a | add(add(floor(power(210, const_0_33)), const_1), floor(power(210, const_0_33))) | the product of 3 consecutive numbers is 210 . then the sum of the smallest two numbers is ? | product of three numbers = 210 210 = 2 * 3 * 5 * 7 = 5 * 6 * 7 . so , the three numbers are 5 , 6 and 7 . and sum of smallest of these two = 5 + 6 = 11 . answer : option a | a = 210 ** const_0_33
b = math.floor(a)
c = b + 1
d = 210 ** const_0_33
e = math.floor(d)
f = c + e
|
a ) 3,955 , b ) 3,925 , c ) 3,956 , d ) 3,926 , e ) 3,915 | c | subtract(658,805, add(add(multiply(const_2, const_100), multiply(add(const_3, const_4), const_10)), const_2)) | how many integers between 263,205 and 658,805 have tens digit 1 and units digit 3 ? | "there is one number in hundred with 1 in the tens digit and 3 in the units digit : 13 , 113 , 213 , 313 , . . . the difference between 263,205 and is 658,805 - 263205 = 395,600 - one number per each hundred gives 133,900 / 100 = 3,956 answer : c ." | a = 2 * 100
b = 3 + 4
c = b * 10
d = a + c
e = d + 2
f = 658 - 805
|
a ) 1 kmph , b ) 6 kmph , c ) 7 kmph , d ) 14 kmph , e ) 9 kmph | d | divide(subtract(36, 8), const_2) | a man can row his boat with the stream at 36 km / h and against the stream in 8 km / h . the man ' s rate is ? | "explanation : ds = 36 us = 8 s = ? s = ( 36 - 8 ) / 2 = 14 kmph answer : d" | a = 36 - 8
b = a / 2
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a ) 38 , b ) 27 , c ) 99 , d ) 17 , e ) 80 | e | subtract(multiply(75, 6), multiply(74, 5)) | ashok secured average of 75 marks in 6 subjects . if the average of marks in 5 subjects is 74 , how many marks did he secure in the 6 th subject ? | "explanation : number of subjects = 6 average of marks in 6 subjects = 75 therefore total marks in 6 subjects = 75 * 6 = 450 now , no . of subjects = 5 total marks in 5 subjects = 74 * 5 = 370 therefore marks in 6 th subject = 450 – 370 = 80 answer : e" | a = 75 * 6
b = 74 * 5
c = a - b
|
a ) a ) 3820 , b ) b ) 930 , c ) c ) 9309 , d ) d ) 3900 , e ) e ) 5625 | e | multiply(multiply(subtract(add(80, 60), 15), 15), 3) | a rectangular lawn of dimensions 80 m * 60 m has two roads each 15 m wide running in the middle of the lawn , one parallel to the length and the other parallel to the breadth . what is the cost of traveling the two roads at rs . 3 per sq m ? | "explanation : area = ( l + b – d ) d ( 80 + 60 – 15 ) 15 = > 1875 m 2 1875 * 3 = rs . 5625 answer : option e" | a = 80 + 60
b = a - 15
c = b * 15
d = c * 3
|
['a ) 55', 'b ) 70', 'c ) 57', 'd ) 58', 'e ) none of these'] | b | divide(rectangle_perimeter(90, 50), 4) | a rectangular plot measuring 90 metres by 50 metres is to be enclosed by wire fencing . if the poles of the fence are kept 4 metres apart , how many poles will be needed ? | solution perimeter of the plot = 2 ( 90 + 50 ) = 280 m . ∴ number of poles = [ 280 / 4 ] = 70 m answer b | a = rectangle_perimeter / (
|
['a ) 10', 'b ) 11', 'c ) 12', 'd ) 13', 'e ) none of them'] | c | divide(1848, circle_area(divide(14, const_2))) | if the capacity of a cylindrical tank is 1848 m 3 and the diameter of its base is 14 m , then find the depth of the tank . | let the depth of the tank be h meters . then , ∏ x 72 x h = 1848 h = ( 1848 x ( 7 / 22 ) x ( 1 / 49 ) = 12 m answer is c | a = 14 / 2
b = 1848 / circle_area
|
a ) 800 , b ) 125 , c ) 288 , d ) 266 , e ) 121 | a | multiply(multiply(160, const_0_2778), 18) | if a train , travelling at a speed of 160 kmph , crosses a pole in 18 sec , then the length of train is ? | "d = 160 * 5 / 18 * 18 = 800 m answer : a" | a = 160 * const_0_2778
b = a * 18
|
a ) rs . 234.80 , b ) rs . 334.80 , c ) rs . 434.80 , d ) rs . 534.80 , e ) none of these | b | multiply(divide(45, const_100), add(multiply(25, 12), add(multiply(const_2, multiply(25, 6)), multiply(multiply(12, 6), const_2)))) | a tank is 25 m long 12 m wide and 6 m deep . the cost of plastering its walls and bottom at 45 paise per sq m is | "explanation : area to be plastered = [ 2 ( l + b ) ã — h ] + ( l ã — b ) = [ 2 ( 25 + 12 ) ã — 6 ] + ( 25 ã — 12 ) = 744 sq m cost of plastering = 744 ã — ( 45 / 100 ) = rs . 334.80 answer : b" | a = 45 / 100
b = 25 * 12
c = 25 * 6
d = 2 * c
e = 12 * 6
f = e * 2
g = d + f
h = b + g
i = a * h
|
a ) 6 , b ) 15 , c ) 17 , d ) 18 , e ) 2 | b | divide(subtract(divide(120, 2), divide(60, 2)), const_2) | a man rows his boat 120 km downstream and 60 km upstream , taking 2 hours each time . find the speed of the stream ? | "speed downstream = d / t = 120 / ( 2 ) = 60 kmph speed upstream = d / t = 60 / ( 2 ) = 30 kmph the speed of the stream = ( 60 - 30 ) / 2 = 15 kmph answer : b" | a = 120 / 2
b = 60 / 2
c = a - b
d = c / 2
|
a ) 1 / 5 , b ) 11 / 22 , c ) 81 / 724 , d ) 91 / 946 , e ) 101 / 987 | d | multiply(divide(14, add(add(14, 20), 10)), divide(subtract(14, const_1), subtract(add(add(14, 20), 10), const_1))) | there are 14 slate rocks , 20 pumice rocks , and 10 granite rocks randomly distributed in a certain field . if 2 rocks are chosen at random and without replacement , what is the probability that both rocks will be slate rocks ? | "14 / 44 * 13 / 43 = 91 / 946 the answer is d ." | a = 14 + 20
b = a + 10
c = 14 / b
d = 14 - 1
e = 14 + 20
f = e + 10
g = f - 1
h = d / g
i = c * h
|
a ) 63 years , b ) 60 years , c ) 50 years , d ) 53 years , e ) 46 years | d | multiply(divide(150, add(const_10, const_10)), const_12) | my grandson is about as many days as my son in weeks , and my grandson is as many months as i am in years . my grandson , my son and i together are 150 years . can you tell my son age in years ? | let m be my age in years . if s is my son ' s age in years , then my son is 52 s weeks old . if g is my grandson ' s age in years , then my grandson is 365 g days old . thus , 365 g = 52 s . since my grandson is 12 g months old , 12 g = m . since my grandson , my son and i together are 150 years , g + s + m = 100 . the above system of 3 equations in 3 unknowns ( g , s and m ) can be solved as follows : 52 s / 365 + s + 12 x ( 52 s / 365 ) = 150 or 52 s + 365 s + 365 x ( 624 s / 365 ) = 365 x 150 or s = 54,750 / 1,041 = 53 years answer = d | a = 10 + 10
b = 150 / a
c = b * 12
|
a ) 20 % , b ) 40 % , c ) 50 % , d ) 60 % , e ) 75 % | c | subtract(const_100, subtract(subtract(const_100, 20), 20)) | a merchant sells an item at a 20 % discount , but still makes a gross profit of 20 percent of the cost . what percent t of the cost would the gross profit on the item have been if it had been sold without the discount ? | "let the market price of the product is mp . let the original cost price of the product is cp . selling price ( discounted price ) = 100 % of mp - 20 % mp = 80 % of mp . - - - - - - - - - - - - - - - - ( 1 ) profit made by selling at discounted price = 20 % of cp - - - - - - - - - - - - - - ( 2 ) apply the formula : profit t = selling price - original cost price = > 20 % of cp = 80 % of mp - 100 % cp = > mp = 120 cp / 80 = 3 / 2 ( cp ) now if product is sold without any discount , then , profit = selling price ( without discount ) - original cost price = market price - original cost price = mp - cp = 3 / 2 cp - cp = 1 / 2 cp = 50 % of cp thus , answer should bec ." | a = 100 - 20
b = a - 20
c = 100 - b
|
a ) 122.9 m , b ) 127.5 m . , c ) 191.25 m , d ) 222.9 m , e ) 12289 m | c | subtract(1000, divide(multiply(subtract(1000, 100), subtract(800, 100)), 800)) | in a race of 1000 m , a can beat by 100 m , in a race of 800 m , b can beat c by 100 m . by how many meters will a beat c in a race of 900 m ? | "when a runs 1000 m , b runs 900 m and when b runs 800 m , c runs 700 m . when b runs 900 m , distance that c runs = ( 900 * 700 ) / 800 = 6300 / 8 = 787.5 m . in a race of 1000 m , a beats c by ( 1000 - 787.5 ) = 212.5 m to c . in a race of 900 m , the number of meters by which a beats c = ( 900 * 212.5 ) / 1000 = 191.25 m . answer : c" | a = 1000 - 100
b = 800 - 100
c = a * b
d = c / 800
e = 1000 - d
|
a ) 3377 , b ) 2688 , c ) 2688 , d ) 8436 , e ) 1268 | d | multiply(7500, multiply(divide(add(const_100, 4), const_100), divide(add(const_100, 4), const_100))) | if rs . 7500 are borrowed at c . i at the rate of 4 % per annum , then after 3 years the amount to be paid is ? | a = 7500 ( 26 / 25 ) ^ 3 = 8436 answer : d | a = 100 + 4
b = a / 100
c = 100 + 4
d = c / 100
e = b * d
f = 7500 * e
|
a ) 54 , b ) 66 , c ) 68 , d ) 60 , e ) 63 | e | multiply(21, divide(divide(36, 2), 6)) | two friends decide to get together ; so they start riding bikes towards each other . they plan to meet halfway . each is riding at 6 mph . they live 36 miles apart . one of them has a pet carrier pigeon and it starts flying the instant the friends start traveling . the pigeon flies back and forth at 21 mph between the 2 friends until the friends meet . how many miles does the pigeon travel ? | "e 63 it takes 3 hours for the friends to meet ; so the pigeon flies for 3 hours at 21 mph = 63 miles" | a = 36 / 2
b = a / 6
c = 21 * b
|
a ) 50 , b ) 40 , c ) 60 , d ) 80 , e ) 70 | d | divide(400, add(divide(400, const_100), const_1)) | the sum of number of boys and girls in a school is 400 . if the number of boys is x , then the number of girls becomes x % of the total number of students . the number of boys is ? | "we have x + x % of 400 = 400 x + x / 100 * 400 = 400 5 * x = 400 x = 80 answer is d" | a = 400 / 100
b = a + 1
c = 400 / b
|
a ) 27 % , b ) 29 % , c ) 31 % , d ) 33 % , e ) 35 % | c | multiply(subtract(const_1, multiply(add(divide(15, const_100), const_1), divide(60, const_100))), const_100) | a customer bought a product at the shop . however , the shopkeeper increased the price of the product by 15 % so that the customer could not buy the required amount of the product . the customer managed to buy only 60 % of the required amount . what is the difference in the amount of money that the customer paid for the second purchase compared to the first purchase ? | "let x be the amount of money paid for the first purchase . the second time , the customer paid 0.6 ( 1.15 x ) = 0.69 x . the difference is 31 % . the answer is c ." | a = 15 / 100
b = a + 1
c = 60 / 100
d = b * c
e = 1 - d
f = e * 100
|
a ) 100 , b ) 240 , c ) 120 , d ) 200 , e ) 150 | b | multiply(12, 30) | the h . c . f . of two numbers is 12 and their l . c . m . is 600 . if one of the number is 30 , find the other ? | "other number = 12 * 600 / 30 = 240 answer is b" | a = 12 * 30
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a ) 1 / 2 , b ) 2 , c ) 1 / 3 , d ) 3 , e ) 1 / 6 | e | sqrt(divide(1, 36)) | if xy = 1 , x / y = 36 , for positive numbers x and y , y = ? | "very easy question . 2 variables and 2 easy equations . xy = 1 - - - > x = 1 / y - ( i ) x / y = 36 - - - > replacing ( i ) here - - - > 1 / ( y ^ 2 ) = 36 - - - > y ^ 2 = 1 / 36 - - - > y = 1 / 6 or - 1 / 6 the question states that x and y are positive integers . therefore , y = 1 / 6 is the answer . answer e ." | a = 1 / 36
b = math.sqrt(a)
|
a ) 14 m , b ) 24 m , c ) 28 m , d ) 40 m , e ) none of the above | c | multiply(multiply(divide(divide(multiply(divide(88, 1000), add(const_3, const_4)), const_2), add(multiply(const_10, const_2), const_2)), const_1000), const_2) | for covering 88 km a wheel revolve 1000 times . what is the radius of wheel ? | distance travel in 1 round = 88000 / 1000 m = 88 m perimeter = 88 m , 2 π r = 882 r = [ 88 * 7 / 22 ] = 28 m hence , diameter = 28 m answer c | a = 88 / 1000
b = 3 + 4
c = a * b
d = c / 2
e = 10 * 2
f = e + 2
g = d / f
h = g * 1000
i = h * 2
|
a ) 104 kmph , b ) 176 kmph , c ) 298 kmph , d ) 186 kmph , e ) 107 kmph | e | divide(642, 6) | a car covers a distance of 642 km in 6 ½ hours . find its speed ? | 642 / 6 = 107 kmph answer : e | a = 642 / 6
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a ) 6 , b ) 6.25 , c ) 7 , d ) 7.5 , e ) 10 | e | divide(multiply(18, 80), const_100) | a can complete a certain job in 18 days . b is 80 % more efficient than a . in how many days can b complete the same job ? | "let , total work unit = 180 units a can finish in 18 days = 180 unit work i . e . a can finish in 1 days = 10 unit work i . e . b can finish in 1 days = 10 + ( 80 / 100 ) * 10 = 18 unit work days in which b will complete the work alone = 180 / 18 = 10 days answer : option e" | a = 18 * 80
b = a / 100
|
a ) $ 100 , b ) $ 250 , c ) $ 300 , d ) $ 200 , e ) $ 180 | c | divide(multiply(divide(multiply(18, 12), subtract(18, 12)), const_100), 12) | i sold a book at a profit of 12 % . had i sold it for $ 18 more , 18 % would have been gained . find the cost price ? | "118 % of cost - 112 % of cost = $ 18 6 % of cost = $ 18 cost = 18 * 100 / 6 = $ 300 answer is c" | a = 18 * 12
b = 18 - 12
c = a / b
d = c * 100
e = d / 12
|
a ) 9.75 , b ) 5.75 , c ) 8.75 , d ) 6.75 , e ) 5.15 | c | multiply(divide(21, const_60), add(20, 5)) | the speed of a boat in still water is 20 km / hr and the rate of current is 5 km / hr . the distance travelled downstream in 21 minutes is : | "explanation : speed downstream = ( 20 + 5 ) kmph = 25 kmph distance travelled = ( 25 * ( 21 / 60 ) ) km = 8.75 km . answer : c" | a = 21 / const_60
b = 20 + 5
c = a * b
|
a ) 20 % , b ) 2 / 8 % , c ) 2 / 1 % , d ) 1 / 3 % , e ) 2 / 7 % | a | multiply(divide(25, add(const_100, 25)), const_100) | if the price of an article went up by 25 % , then by what percent should it be brought down to bring it back to its original price ? | let the price of the article be rs . 100 . 25 % of 100 = 25 . new price = 100 + 25 = rs . 125 required percentage = ( 125 - 100 ) / 125 * 100 = 25 / 125 * 100 = 20 % . answer : a | a = 100 + 25
b = 25 / a
c = b * 100
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a ) 5 , b ) 10 , c ) 15 , d ) 20 , e ) none of them | b | multiply(const_100, divide(subtract(subtract(1, divide(1, const_100)), divide(36, 40)), divide(36, 40))) | a retailer buys 40 pens at the market price of 36 pens from a wholesaler , if he sells these pens giving a discount of 1 % , what is the profit % ? | "let the market price of each pen be rs 1 then , c . p of 40 pens = rs 36 s . p of 40 pens = 99 % of rs 40 = rs 39.60 profit % = ( ( 3.60 * 100 ) / 36 ) % = 10 % answer is b ." | a = 1 / 100
b = 1 - a
c = 36 / 40
d = b - c
e = 36 / 40
f = d / e
g = 100 * f
|
a ) 22754.3 , b ) 22254.2 , c ) 25754.2 , d ) 22753.2 , e ) 22254.2 | a | multiply(multiply(multiply(divide(multiply(add(const_10, const_1), const_2), add(const_3, const_4)), const_2), 3.62), const_1000) | given a circular wheel of 3.62 m radius how many revolutions will the will make to travel a distance of 1 km ? | 2 * 22 / 7 * 3.62 * x = 22754.3 x = 1000 answer : a | a = 10 + 1
b = a * 2
c = 3 + 4
d = b / c
e = d * 2
f = e * 3
g = f * 1000
|
a ) 2 , b ) 3 , c ) 5 , d ) 7 , e ) 8 | b | subtract(427398, multiply(floor(divide(427398, 15)), 15)) | find the least number must be subtracted from 427398 so that remaining no . is divisible by 15 ? | "on dividing 427398 by 15 we get the remainder 3 , so 3 should be subtracted b" | a = 427398 / 15
b = math.floor(a)
c = b * 15
d = 427398 - c
|
a ) 77 % , b ) 78 % , c ) 79 % , d ) 80 % , e ) 81 % | c | divide(add(multiply(15, 73), multiply(10, 88)), 25) | if 15 students in a class average 73 % on an exam and 10 students average 88 % on the same exam , what is the average in percent for all 25 students ? | "( 15 * 73 + 10 * 88 ) / 25 = 79 % the answer is c ." | a = 15 * 73
b = 10 * 88
c = a + b
d = c / 25
|
a ) 4 % increase , b ) 17 % increase , c ) 10 % decrease , d ) 6 % increase , e ) none of these | b | subtract(divide(multiply(subtract(const_100, 10), add(const_100, 30)), const_100), const_100) | if the price of a tv is first decreased by 10 % and then increased by 30 % , then the net change in the price will be : | "explanation : solution : let the original price be rs . 100 . new final price = 130 % of ( 10 % of 100 ) = rs . 130 / 100 * 90 / 100 * 100 = rs . 117 . . ' . increase = 17 % answer : b" | a = 100 - 10
b = 100 + 30
c = a * b
d = c / 100
e = d - 100
|
a ) rs . 768 , b ) rs . 968 , c ) rs . 1960 , d ) rs . 2400 , e ) none | a | add(divide(multiply(const_100, 168), multiply(14, 2)), 168) | if the true discount on a sum due 2 years hence at 14 % per annum be rs . 168 , the sum due is : | solution p . w = 100 xt . d . / r x t = 100 x 168 / 14 x 2 = 600 . ∴ sum = ( p . w . + td . ) = rs . 768 . answer a | a = 100 * 168
b = 14 * 2
c = a / b
d = c + 168
|
a ) 30 yr , b ) 25 yr , c ) 45 yr , d ) 40 yr , e ) 50 yr | c | add(subtract(30, subtract(30, add(const_3, const_2))), multiply(subtract(30, add(const_3, const_2)), const_2)) | the difference between the ages of two persons is 30 years . fifteen years ago , the elder one was twice as old as the younger one . the present age of the younger person is ? | "let their ages be x years and ( x + 30 ) years then , ( x + 30 ) - 15 = 2 ( x - 15 ) x + 15 = 2 x - 30 x = 45 answer is c" | a = 3 + 2
b = 30 - a
c = 30 - b
d = 3 + 2
e = 30 - d
f = e * 2
g = c + f
|
a ) $ 13,746 , b ) $ 15,325 , c ) $ 16,000 , d ) $ 16,225 , e ) $ 17,155 | a | multiply(divide(const_3, const_4), const_1000) | a store owner estimates that the average price of type a products will increase by 20 % next year and that the price of type b products will increase by 11 % next year . this year , the total amount paid for type a products was $ 3500 and the total price paid for type b products was $ 8600 . according to the store owner ' s estimate , and assuming the number of products purchased next year remains the same as that of this year , how much will be spent for both products next year ? | "cost of type a products next year = 1.20 * 3500 = 4200 cost of type b products next year = 1.11 * 8300 = 9546 total 4200 + 9546 = 13746 answer : a" | a = 3 / 4
b = a * 1000
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a ) 15360 , b ) 13780 , c ) 15060 , d ) 14930 , e ) 16075 | b | subtract(add(6000, 10000), add(multiply(6000, divide(12, const_100)), multiply(divide(15, const_100), 10000))) | a soft drink company had 6000 small and 10000 big bottles in storage . if 12 % of small 15 % of big bottles have been sold , then the total bottles remaining in storage is | "6000 + 10000 - ( 0.12 * 6000 + 0.15 * 10000 ) = 13780 . answer : b ." | a = 6000 + 10000
b = 12 / 100
c = 6000 * b
d = 15 / 100
e = d * 10000
f = c + e
g = a - f
|
a ) 1 / 140 , b ) 1 / 180 , c ) 3 / 12 , d ) 3 / 8 , e ) 57 / 120 | d | add(add(divide(1, 8), divide(1, 12)), divide(1, 6)) | in a race where 18 cars are running , the chance that car x will win is 1 / 8 , that y will win is 1 / 12 and that z will win is 1 / 6 . assuming that a dead heat is impossible , find the chance that one of them will win . | "required probability = p ( x ) + p ( y ) + p ( z ) ( all the events are mutually exclusive ) . = 1 / 8 + 1 / 12 + 1 / 6 = 3 / 8 answer : d" | a = 1 / 8
b = 1 / 12
c = a + b
d = 1 / 6
e = c + d
|
a ) a ) 78 , b ) b ) 82 , c ) c ) 98 , d ) d ) 91 , e ) e ) 85 | c | divide(add(subtract(multiply(100, 25), 60), 10), 25) | the average marks of 25 students in a class is 100 . but a student mark is wrongly noted as 60 instead of 10 then find the correct average marks ? | "correct avg marks = 100 + ( 10 - 60 ) / 25 avg = 100 - 2 = 98 answer is c" | a = 100 * 25
b = a - 60
c = b + 10
d = c / 25
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a ) 2345 , b ) 4500 , c ) 5000 , d ) 6000 , e ) 7000 | a | add(5, add(add(multiply(2, const_1000), multiply(3, const_100)), multiply(const_10, 4))) | what is the sum of the local values of the digits 2 , 3 , 4 , 5 in the number 2345 ? | 2000 + 300 + 40 + 5 = 2345 answer a | a = 2 * 1000
b = 3 * 100
c = a + b
d = 10 * 4
e = c + d
f = 5 + e
|
a ) a ) 44500 , b ) b ) 42000 , c ) c ) 44098 , d ) d ) 43007 , e ) e ) 44098 | b | subtract(91000, multiply(const_60, const_100)) | a started a business with an investment of rs . 70000 and after 6 months b joined him investing rs . 120000 . if the profit at the end of a year is rs . 91000 , then the share of b is ? | "ratio of investments of a and b is ( 70000 * 12 ) : ( 120000 * 6 ) = 7 : 6 total profit = rs . 91000 share of b = 6 / 13 ( 91000 ) = rs . 42000 answer : b" | a = const_60 * 100
b = 91000 - a
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a ) 2449 , b ) 5449 , c ) 6749 , d ) 6725 , e ) 6468 | d | subtract(multiply(divide(54671, const_100), 14456), multiply(divide(const_1, const_3), multiply(divide(54671, const_100), 14456))) | 54671 - 14456 - 33490 = ? | "d if we calculate we will get 6725" | a = 54671 / 100
b = a * 14456
c = 1 / 3
d = 54671 / 100
e = d * 14456
f = c * e
g = b - f
|
a ) 36 , b ) 42 , c ) 44 , d ) 43 , e ) none | d | subtract(add(28, 17), const_2) | if p and q are positive integers each greater than 1 , and 17 ( p + 1 ) = 28 ( q + 1 ) , what is the least possible value of p + q ? | "17 ( p + 1 ) = 29 ( q + 1 ) - - > ( p + 1 ) / ( q + 1 ) = 28 / 17 - - > the least positive value of p + 1 is 28 , so the least value of p is 27 and the least positive value of q + 1 is 17 , so the least value of q is 16 - - > the least value of p + q is 27 + 16 = 43 . answer : d" | a = 28 + 17
b = a - 2
|
a ) 350 m , b ) 200 m , c ) 400 m , d ) 900 m , e ) none of them | d | divide(multiply(225, 8), subtract(10, 8)) | a thief is spotted by a policeman from a distance of 225 meters . when the policeman starts the chase , the thief also starts running . if the speed of the thief be 8 km / hr and that of the policeman 10 km / hr , how far the thief will have run before he is overtaken ? | "relative speed of the policeman = ( 10 - 8 ) km / hr = 2 km / hr . time taken by police man to cover ( 225 m / 1000 ) x 1 / 2 hr = 9 / 80 hr . in 9 / 80 hrs , the thief covers a distance of 8 x 9 / 80 km = 9 / 10 km = 900 m answer is d ." | a = 225 * 8
b = 10 - 8
c = a / b
|
a ) 13 / 8 , b ) 26 / 11 , c ) 24 / 7 , d ) 12 / 13 , e ) 1 / 2 | c | inverse(add(inverse(12), add(inverse(const_3), inverse(add(add(const_4, const_4), const_1))))) | a , b , c can complete a piece of work in 24,6 and 12 days respectively . working together , they will complete the same work in how many days ? | "( a + b + c ) ' s 1 day work = ( 1 / 24 ) + ( 1 / 6 ) + ( 1 / 12 ) = 7 / 24 a , b , c together will complete the work in 24 / 7 days answer is c" | a = 1/(12)
b = 1/(3)
c = 4 + 4
d = c + 1
e = 1/(d)
f = b + e
g = a + f
h = 1/(g)
|
a ) 12 , b ) 10 , c ) 9 , d ) 8 , e ) 11 | e | divide(407, add(multiply(9, const_3), multiply(5, const_2))) | mary works 9 hours per day on monday , wednesday and friday , and 5 hours per day on tuesday and thursday . she does not work on saturday and sunday . she earns $ 407 per week . how much does she earn in dollars per hour ? | so , she works 27 hours in 3 days so , she works 10 hours in 2 days so in a week she works 37 hours ( 27 + 10 ) and earns $ 407 so , hourly wage is 407 / 37 = > 11 hence answer will be ( e ) 11 | a = 9 * 3
b = 5 * 2
c = a + b
d = 407 / c
|
a ) 3 , b ) 5 , c ) 5.6 , d ) 5.7 , e ) 6.5 | a | multiply(divide(9, 12), 4) | when a number is divided by 4 & then multiply by 12 the answer is 9 what is the no . ? | if $ x $ is the number , x / 4 * 12 = 9 = > 3 x = 9 = > x = 3 a | a = 9 / 12
b = a * 4
|
a ) 1 / 4 , b ) 1 / 2 , c ) 2 / 3 , d ) 2 , e ) 4 | d | divide(const_1, divide(multiply(add(divide(3, 4), const_1), divide(3, 5)), const_2)) | a toy store ' s revenue in november was 3 / 5 of its revenue in december and its revenue in january was 3 / 4 of its revenue in november , then the store ' s revenue in december was how many times the average ( arithmetic mean ) of its revenues in november and january ? | let dec rev = 100 then nov rev is 3 / 5 ( 100 ) = > 60 therefore jan rev = 3 / 4 ( nov rev ) = 3 / 4 ( 60 ) = > 45 hence dec rev = x * ( nov rev + jan rev ) / 2 100 = x * ( 60 + 45 ) / 2 x = 100 / 52.5 = > 1.90 = 2 ans ) d | a = 3 / 4
b = a + 1
c = 3 / 5
d = b * c
e = d / 2
f = 1 / e
|
a ) 600 , b ) 700 , c ) 800 , d ) 900 , e ) 1000 | e | divide(divide(divide(120, subtract(const_1, divide(4, 5))), divide(4, 5)), divide(1, 4)) | of the goose eggs laid at a certain pond , 1 / 4 hatched and 4 / 5 of the geese that hatched from those eggs survived the first month . of the geese that survived the first month , 2 / 5 did not survive the first year . if 120 geese survived the first year and if no more than one goose hatched from each egg , how many goose eggs were laid at the pond ? | "let x be the number of eggs that were laid . ( 3 / 5 ) ( 4 / 5 ) ( 1 / 4 ) x = 120 ( 12 / 100 ) x = 120 x = 1000 the answer is e ." | a = 4 / 5
b = 1 - a
c = 120 / b
d = 4 / 5
e = c / d
f = 1 / 4
g = e / f
|
a ) 2 / 7 , b ) 17 / 30 , c ) 1 / 2 , d ) 27 / 40 , e ) 5 / 7 | d | divide(const_10, 30) | in a graduate physics course , 60 percent of the students are male and 30 percent of the students are married . if two - sevenths of the male students are married , what fraction of the female students is single ? | "let assume there are 100 students of which 60 are male and 40 are females if 30 are married then 70 will be single . now its given that two - sevenths of the male students are married that means 2 / 7 of 60 = 17 males are married if 30 is the total number of students who are married and out of that 17 are males then the remaining 13 will be females who are married . total females = 40 married females = 13 then single females = 40 - 13 = 27 we need to find the fraction of female students who are single i . e single female students / total female student = 27 / 40 [ d ]" | a = 10 / 30
|
a ) 21 , b ) 29 , c ) 19 , d ) 14 , e ) 10 | c | add(18, const_1) | calculate the average of first 18 even numbers is ? | explanation : sum of 10 even numbers = 18 * 19 = 342 average = 342 / 18 = 19 answer : option c | a = 18 + 1
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a ) 160 , b ) 150 , c ) 250 , d ) 80 , e ) 50 | c | divide(subtract(multiply(200, divide(16, const_100)), 22), subtract(divide(16, const_100), divide(12, const_100))) | an empty fuel tank with a capacity of 200 gallons was filled partially with fuel a and then to capacity with fuel b . fuel a contains 12 % ethanol by volume and fuel b contains 16 % ethanol by volume . if the full fuel tank contains 22 gallons of ethanol , how many gallons of fuel a were added ? | "say there are a gallons of fuel a in the tank , then there would be 200 - a gallons of fuel b . the amount of ethanol in a gallons of fuel a is 0.12 a ; the amount of ethanol in 200 - a gallons of fuel b is 0.16 ( 200 - a ) ; since the total amount of ethanol is 22 gallons then 0.12 a + 0.16 ( 200 - a ) = 22 - - > a = 250 . answer : c ." | a = 16 / 100
b = 200 * a
c = b - 22
d = 16 / 100
e = 12 / 100
f = d - e
g = c / f
|
a ) 7 , b ) 8 , c ) 10 , d ) 12 , e ) 14 | a | subtract(36, subtract(add(26, 20), 17)) | in a class of 36 students 26 play football and play 20 long tennis , if 17 play above , many play neither ? | "26 + 20 - 17 = 29 36 - 29 = 7 play neither answer is a" | a = 26 + 20
b = a - 17
c = 36 - b
|
a ) 5 , b ) 7 , c ) 9 , d ) 11 , e ) 12 | a | multiply(subtract(divide(power(29, const_2), 56), floor(divide(power(29, const_2), 56))), 56) | on dividing a number by 56 , we get 29 as remainder . on dividing the same number by 8 , what will be the remainder ? | "formula : ( divisor * quotient ) + remainder = dividend . soln : ( 56 * q ) + 29 = d - - - - - - - ( 1 ) d % 8 = r - - - - - - - - - - - - - ( 2 ) from equation ( 2 ) , ( ( 56 * q ) + 29 ) % 8 = r . = > assume q = 1 . = > ( 56 + 29 ) % 8 = r . = > 85 % 8 = r = > 5 = r . a )" | a = 29 ** 2
b = a / 56
c = 29 ** 2
d = c / 56
e = math.floor(d)
f = b - e
g = f * 56
|
a ) 7 , b ) 8 , c ) 9 , d ) 10 , e ) 11 | b | add(add(3, 3), 2) | given that p is a positive even integer with a positive units digit , if the units digit of p ^ 3 minus the units digit of p ^ 2 is equal to 0 , what is the units digit of p + 2 ? | "p is a positiveeveninteger with apositive units digit - - > the units digit of p can be 2 , 4 , 6 , or 8 - - > in order the units digit of p ^ 3 - p ^ 2 to be 0 , the units digit of p ^ 3 and p ^ 2 must be the same . thus the units digit of p can be 0 , 1 , 5 or 6 . intersection of values is 6 , thus the units digit of p + 2 is 6 + 2 = 8 . answer : b ." | a = 3 + 3
b = a + 2
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a ) 8 , b ) 10 , c ) 12 , d ) 14 , e ) 16 | c | divide(add(18, multiply(6, 5)), subtract(5, const_1)) | dan ' s age after 18 years will be 5 times his age 6 years ago . what is the present age of dan ? | "let dan ' s present age be x . x + 18 = 5 ( x - 6 ) 4 x = 48 x = 12 the answer is c ." | a = 6 * 5
b = 18 + a
c = 5 - 1
d = b / c
|
a ) 200 π , b ) 450 π , c ) 300 π , d ) 480 π , e ) 1,200 π | b | multiply(multiply(multiply(multiply(divide(10, add(multiply(const_2, const_100), multiply(add(const_2, const_3), const_1000))), const_2), divide(add(const_2, multiply(const_2, 10)), add(const_3, const_4))), 1,980), const_60) | the end of a blade on an airplane propeller is 10 feet from the center . if the propeller spins at the rate of 1,980 revolutions per second , how many miles will the tip of the blade travel in one minute ? ( 1 mile = 5,280 feet ) | "distance traveled in 1 revolution = 2 π r = 2 π 10 / 5280 revolutions in one second = 1980 revolutions in 60 seconds ( one minute ) = 1980 * 60 total distance traveled = total revolutions * distance traveled in one revolution 1980 * 60 * 2 π 10 / 5280 = 450 π b is the answer" | a = 2 * 100
b = 2 + 3
c = b * 1000
d = a + c
e = 10 / d
f = e * 2
g = 2 * 10
h = 2 + g
i = 3 + 4
j = h / i
k = f * j
l = k * 1
m = l * const_60
|
a ) 1 : 4 , b ) 1 : 5 , c ) 3 : 2 , d ) 1 : 2 , e ) 2 : 5 | c | divide(subtract(divide(const_1, 3), divide(const_1, 5)), subtract(divide(const_1, 5), divide(const_1, multiply(const_2, const_4)))) | a grocery store bought some mangoes at a rate of 5 for a dollar . they were separated into two stacks , one of which was sold at a rate of 3 for a dollar and the other at a rate of 9 for a dollar . what was the ratio of the number of mangoes in the two stacks if the store broke even after having sold all of its mangoes ? | "the cost price of a mango = 1 / 5 dollars . the selling price of a mango from the first stack = 1 / 3 dollars - - > the profit from one mango = 1 / 3 - 1 / 5 = 2 / 15 = 4 / 30 dollars . the selling price of a mango from the second stack = 1 / 9 dollars - - > the loss from one mango = 1 / 5 - 1 / 9 = 4 / 45 dollars . the profit from one mango from the first stack is 4 times the loss from one mango from the second stack . the ratio is 4 / 30 * 45 / 4 = 3 : 2 c" | a = 1 / 3
b = 1 / 5
c = a - b
d = 1 / 5
e = 2 * 4
f = 1 / e
g = d - f
h = c / g
|
a ) 500 , b ) 688 , c ) 200 , d ) 121 , e ) 800 | e | divide(subtract(420, 380), divide(5, const_100)) | if 5 % more is gained by selling an article for rs . 420 than by selling it for rs . 380 , the cost of the article is ? | let c . p . be rs . x . then , 5 % of x = 420 - 380 = 40 x / 20 = 40 = > x = 800 answer : e | a = 420 - 380
b = 5 / 100
c = a / b
|
a ) 7 , b ) 11 , c ) 13 , d ) 14 , e ) 38 | d | divide(multiply(39, divide(72, const_100)), const_2) | each of the 39 points is placed either inside or on the surface of a perfect sphere . if 72 % or fewer of the points touch the surface , what is the maximum number of segments which , if connected from those points to form chords , could be the diameter of the sphere ? | "maximum number of points on the surface is 72 % * 39 = 28.08 . . . or 28 since it has to be an integer now note that if two points form a diameter , they can not be part of any other diameter . so in the best case we can pair up the points we have 28 points , so at best we can form 14 pairs ( 28 ) . so , answer is ( d )" | a = 72 / 100
b = 39 * a
c = b / 2
|
a ) 40.9 % , b ) 41.9 % , c ) 42.9 % , d ) 43.9 % , e ) 44.9 % | e | multiply(divide(add(4, 5), add(add(5, 6), add(4, 5))), const_100) | the proportion of water to alcohol in solution a is 5 : 4 and the proportion of water to alcohol in solution b is 6 : 5 . if an equal amount of each solution is mixed together , what is the concentration of alcohol in the new solution ? | "let v be the total volume of the new solution . then a volume of v / 2 was added from each solution a and b . the amount of alcohol added to the new solution was : ( 4 / 9 ) ( v / 2 ) + ( 5 / 11 ) ( v / 2 ) = 2 v / 9 + 5 v / 22 = 89 v / 198 . the concentration of alcohol is 89 / 198 = 44.9 % the answer is e ." | a = 4 + 5
b = 5 + 6
c = 4 + 5
d = b + c
e = a / d
f = e * 100
|
a ) 180 , b ) 196 , c ) 160 , d ) 164 , e ) 172 | b | multiply(multiply(4, 3), multiply(4, 3)) | in the coordinate plane , one of the vertices of a square is the point ( - 4 , - 4 ) . if the diagonals of that square intersect at point ( 3 , 2 ) , what is the area of that square ? | "one point ( - 4 - 4 ) , intersection ( 3,2 ) so the distance from the first point - 4 - 3 = - 7 is the midpoint of the square - - > whole side 14 , 14 * 14 = 196 b" | a = 4 * 3
b = 4 * 3
c = a * b
|
a ) 100 m , b ) 80 m , c ) 130 m , d ) 150 m , e ) none of these | b | subtract(180, multiply(divide(3, const_60), const_1000)) | a policeman noticed a criminal from a distance of 180 km . the criminal starts running and the policeman chases him . the criminal and the policeman run at the rate of 8 km and 10 km per hour respectively . what is the distance between them after 3 minutes ? | "explanation : solution : relative speed = ( 10 - 8 ) = 2 km / hr . distance covered in 3 minutes = ( 2 * 3 / 60 ) km = 1 / 10 km = 100 m . . ' . distance between the criminal and policeman = ( 180 - 100 ) m = 80 m . answer : b" | a = 3 / const_60
b = a * 1000
c = 180 - b
|
a ) 121 , b ) 123 , c ) 119 , d ) 125 , e ) 127 | a | divide(multiply(22, 55), 10) | if 22 men do a work in 55 days , in how many days will 10 men do it ? | "22 * 55 = 10 * x x = 121 days answer : a" | a = 22 * 55
b = a / 10
|
a ) 85 , b ) 94 , c ) 82 , d ) 72 , e ) none | c | add(multiply(divide(add(10, 6), const_2), 10), subtract(10, divide(add(10, 6), const_2))) | the sum of digits of a two digit number is 10 , the difference between the digits is 6 . find the number | description : = > x + y = 10 , x - y = 6 adding these 2 x = 16 = > x = 8 , y = 2 . thus the number is 82 answer c | a = 10 + 6
b = a / 2
c = b * 10
d = 10 + 6
e = d / 2
f = 10 - e
g = c + f
|
a ) $ 350 , b ) $ 400 , c ) $ 365 , d ) $ 385 , e ) $ 375 | c | divide(subtract(multiply(7, 400), add(add(add(add(406, 413), 420), 436), 395)), const_2) | tough and tricky questions : word problems . a salesman ' s income consists of commission and base salary . his weekly income totals over the past 5 weeks have been $ 406 , $ 413 , $ 420 , $ 436 and $ 395 . what must his average ( arithmetic mean ) income over the next two weeks be to decrease his average weekly income to $ 400 over the 7 - week period ? | official solution : ( c ) first , we need to add up the wages over the past 5 weeks : $ 406 + $ 413 + $ 420 + $ 436 + $ 395 = $ 2070 . to average $ 400 over 7 weeks , the salesman would need to earn : $ 400 × 7 = $ 2800 . subtract $ 2070 from $ 2800 to determine how much he would need to earn , in total , over the next 2 weeks to average $ 400 for the 7 weeks : $ 2800 – $ 2070 = $ 730 . dividing $ 730 by 2 will give us the amount he needs to earn on average over the next 2 weeks : $ 730 / 2 = $ 365 . the correct answer is choice ( c ) . | a = 7 * 400
b = 406 + 413
c = b + 420
d = c + 436
e = d + 395
f = a - e
g = f / 2
|
a ) 258 , b ) 318 , c ) 322 , d ) 324 , e ) 330 | a | add(multiply(6, const_4), multiply(divide(40, 10), const_60)) | a man walks at a rate of 10 mph . after every ten miles , he rests for 6 minutes . how much time does he take to walk 40 miles ? | "to cover 40 miles the man needs ( time ) = ( distance ) / ( rate ) = 40 / 10 = 4 hours = 240 minutes . he will also rest 3 times ( after 10 , 20 , and 30 miles ) , so total resting time = 3 * 6 = 18 minutes . total time = 240 + 18 = 258 minutes . answer : a ." | a = 6 * 4
b = 40 / 10
c = b * const_60
d = a + c
|
a ) 2 / 7 , b ) 3 / 5 , c ) 3 / 11 , d ) 18 / 85 , e ) 7 / 16 | d | divide(multiply(choose(const_4.0, const_2), choose(add(const_3.0, 6), const_1)), choose(add(add(3, 6), 8), 3)) | a bag contains 3 red , 6 yellow and 8 green balls . 3 balls are drawn randomly . what is the probability that the balls drawn contain balls of different colours ? | "total number of balls = 3 + 6 + 8 = 17 n ( s ) = 17 c 3 = 680 n ( e ) = 3 c 1 * 6 c 1 * 8 c 1 = 144 probability = 144 / 680 = 18 / 85 answer is d" | a = math.comb(4, 0)
b = 3 + 0
c = math.comb(b, 1)
d = a * c
e = 3 + 6
f = e + 8
g = math.comb(f, 3)
h = d / g
|
a ) 2 , b ) 2.4 , c ) 2.7 , d ) 3 , e ) 3.5 | d | divide(240, add(divide(240, 12), divide(240, 4))) | while working alone at their constant rates , computer x can process 240 files in 12 hours , and computer y can process 240 files in 4 hours . if all files processed by these computers are the same size , how many hours would it take the two computers , working at the same time at their respective constant rates , to process a total of 240 files ? | "both computers together process files at a rate of 240 / 12 + 240 / 4 = 20 + 60 = 80 files per hour . the time required to process 240 files is 240 / 80 = 3 hours the answer is d ." | a = 240 / 12
b = 240 / 4
c = a + b
d = 240 / c
|
a ) 573 , b ) 608 , c ) 613 , d ) 616 , e ) 621 | a | add(multiply(subtract(72, const_1), 8), 5) | let s be the set of all positive integers that , when divided by 8 , have a remainder of 5 . what is the 72 th number in this set ? | the set s = { 5 , 13 , 21 , 29 , . . . . . . . . . . . . . . . . . . . . . } 1 st number = 8 * 0 + 5 = 5 2 nd number = 8 * 1 + 5 = 13 3 rd number = 8 * 2 + 5 = 21 72 th number = 8 * ( 72 - 1 ) + 5 = 573 answer = a | a = 72 - 1
b = a * 8
c = b + 5
|
a ) 4 , b ) 6 , c ) 8 , d ) 12 , e ) 16 | c | multiply(multiply(add(const_1, const_1), add(const_1, const_1)), add(const_1, const_1)) | how many positive integers will divide evenly into 190 ? | the question is asking how many factors 190 has . 190 = 2 * 5 * 19 the number of factors is 2 ^ 3 = 8 the answer is c . | a = 1 + 1
b = 1 + 1
c = a * b
d = 1 + 1
e = c * d
|
a ) 150 , b ) 500 / 3 , c ) 400 , d ) 480 , e ) 600 | a | multiply(divide(subtract(25, 10), subtract(30, 25)), 50) | solution x is 10 percent alcohol by volume , and solution y is 30 percent alcohol by volume . how many milliliters of solution y must be added to 50 milliliters of solution x to create a solution that is 25 percent alcohol by volume ? | "we know that x is 10 % , y is 30 % and w . avg = 25 % . what does this mean with respect to w . avg technique ? w . avg is 1 portion away from y and 3 portion away from x so for every 1 portion of x we will have to add 3 portions of y . if x = 50 then y = 150 answer : a" | a = 25 - 10
b = 30 - 25
c = a / b
d = c * 50
|
a ) 14 , b ) 18 , c ) 22 , d ) 26 , e ) 30 | e | add(divide(400, 10), subtract(subtract(10, 4), const_1)) | in a certain quiz that consists of 10 questions , each question after the first is worth 4 points more than the preceding question . if the 10 questions on the quiz are worth a total of 400 points , how many points is the third question worth ? | "x x + 4 x + 8 x + 12 x + 16 x + 20 x + 24 x + 28 x + 32 x + 36 10 x + 180 = 400 10 x = 220 x = 22 3 rd question = x + 8 = 22 + 8 = 30 answer e" | a = 400 / 10
b = 10 - 4
c = b - 1
d = a + c
|
a ) 8 % , b ) 15 % , c ) 45 % , d ) 52 % , e ) 66 % | e | multiply(divide(5, 40), const_100) | a pharmaceutical company received $ 5 million in royalties on the first $ 40 million in sales of and then $ 9 million in royalties on the next $ 210 million in sales . by approximately what percentage did the ratio of royalties to sales decrease from the first $ 40 million in sales to the next $ 210 million in sales ? | "( 9 / 210 ) / ( 5 / 40 ) = 12 / 35 = 34 % it means that 9 / 210 represents only 34 % . therefore a decrease of 66 % . answer e" | a = 5 / 40
b = a * 100
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a ) 20 % , b ) 80 % , c ) 100 % , d ) 180 % , e ) 200 % | d | multiply(divide(20, subtract(subtract(const_100, 80), 20)), const_100) | jane makes toy bears . when she works with an assistant , she makes 80 percent more bears per week and works 20 percent fewer hours each week . having an assistant increases jane ’ s output of toy bears per hour by what percent ? | "we can use fractional equivalents here to solve the problem 80 % = 4 / 5 ; this means that in 1 st case if she prepares 5 bears , in 2 nd case she prepares 9 bears 10 % = 1 / 10 ; this means that in 1 st case if she needs 10 hours , in 2 nd case she needs 9 hours now we come to productivity based on above fractional values the productivity in 1 st case is 0.5 bears / hour and in the 2 nd case it is 1 bear / hour hence the productivity is double with the assistant i . e . the increase in productivity is 180 % d" | a = 100 - 80
b = a - 20
c = 20 / b
d = c * 100
|
a ) 60 m 2 , b ) 64 m 2 , c ) 68 m 2 , d ) 66 m 2 , e ) none of these | d | multiply(5, multiply(multiply(multiply(3, divide(22, 7)), divide(1.4, 3)), 3)) | the diameter of a garden roller is 1.4 m and it is 3 m long . how much area will it cover in 5 revolutions ? ( use ï € = 22 â „ 7 ) | "required area covered in 5 revolutions = 5 ã — 2 ï € rh = 5 ã — 2 ã — 22 â „ 7 ã — 0.7 ã — 3 = 66 m 2 answer d" | a = 22 / 7
b = 3 * a
c = 1 / 4
d = b * c
e = d * 3
f = 5 * e
|
a ) 1 / 20 , b ) 1 / 50 , c ) 1 / 75 , d ) 2 / 25 , e ) none of these | d | divide(circle_area(divide(8, const_2)), const_2) | what will be the fraction of 8 % | "explanation : 8 * 1 / 100 = 2 / 25 . option d" | a = 8 / 2
b = circle_area / (
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a ) 14 , b ) 20 , c ) 22 , d ) 24 , e ) 15 | e | divide(60000, divide(add(add(add(5000, 12000), 15000), 16000), add(add(add(2, 4), 2), 4))) | shipment - - - no . of defective chips / shipment - - - total chips in shipment s 1 - - - - - - - - - - - - - - - - - - - - - - 2 - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - 5000 s 2 - - - - - - - - - - - - - - - - - - - - - - 4 - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - 12000 s 3 - - - - - - - - - - - - - - - - - - - - - - 2 - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - 15000 s 4 - - - - - - - - - - - - - - - - - - - - - - 4 - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - 16000 a computer chip manufacturer expects the ratio of the number of defective chips to the total number of chips in all future shipments to equal the corresponding ratio for shipments s 1 , s 2 , s 3 , and s 4 combined , as shown in the table above . what ’ s the expected number of defective chips in a shipment of 60000 chips ? | for a total of 51000 chips ( adding s 1 , s 2 , s 3 , s 4 ) total number of defective chips is 17 ( ( adding defective chips of s 1 , s 2 , s 3 , s 4 ) so ratio is 12 / 48000 or 1 every 4000 chips . keeping this ratio constant for 60000 chips number of defective chips will be ( 1 / 4000 ) * 60000 = 15 e | a = 5000 + 12000
b = a + 15000
c = b + 16000
d = 2 + 4
e = d + 2
f = e + 4
g = c / f
h = 60000 / g
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a ) 392 , b ) 229 , c ) 753 , d ) 493 , e ) 540 | c | multiply(multiply(multiply(divide(3, 4), divide(1, 2)), divide(2, 5)), 5020) | 3 / 4 of 1 / 2 of 2 / 5 of 5020 = ? | "c 753 ? = 5020 * ( 2 / 5 ) * ( 1 / 2 ) * ( 3 / 4 ) = 753" | a = 3 / 4
b = 1 / 2
c = a * b
d = 2 / 5
e = c * d
f = e * 5020
|
a ) 19 , b ) 27 , c ) 29 , d ) 48 , e ) 28 | d | divide(add(360, multiply(multiply(const_0_2778, 36), 12)), multiply(const_0_2778, 36)) | a train running at a speed of 36 kmph crosses an electric pole in 12 seconds . in how much time will it cross a 360 m long platform ? | "let the length of the train be x m . when a train crosses an electric pole , the distance covered is its own length . so , x = 12 * 36 * 5 / 18 m = 120 m . time taken to cross the platform = ( 120 + 360 ) / 36 * 5 / 18 = 48 min . answer : d" | a = const_0_2778 * 36
b = a * 12
c = 360 + b
d = const_0_2778 * 36
e = c / d
|
a ) 120 , b ) 240 , c ) 360 , d ) 480 , e ) 720 | c | multiply(factorial(const_4), divide(divide(factorial(6), factorial(const_4)), factorial(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 t 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 t = 6 c ! / 2 - > 720 / 2 = 360 | a = math.factorial(4)
b = math.factorial(6)
c = math.factorial(4)
d = b / c
e = math.factorial(2)
f = d / e
g = a * f
|
a ) 1 , 4,6 , b ) 12 , 24,36 , c ) 10 , 20 , 30 , d ) 12 , 24 , 36 , e ) 9 , 36,54 . | e | add(multiply(multiply(1, 6), const_100), multiply(4, 6)) | three numbers are in the ratio 1 : 4 : 6 and their h . c . f is 9 . the numbers are : | "let the required numbers be x , 4 x and 6 x . then , their h . c . f = x . so , x = 9 . the numbers are 9 , 36,54 . answer : e" | a = 1 * 6
b = a * 100
c = 4 * 6
d = b + c
|
a ) 52.5 , b ) 52.9 , c ) 52.1 , d ) 48.75 , e ) 42.5 | d | divide(add(multiply(30, 30), multiply(50, 60)), add(30, 50)) | the average marks of a class of 30 students is 30 and that of another class of 50 students is 60 . find the average marks of all the students ? | "sum of the marks for the class of 30 students = 30 * 30 = 900 sum of the marks for the class of 50 students = 50 * 60 = 3000 sum of the marks for the class of 80 students = 900 + 3000 = 3900 average marks of all the students = 3900 / 80 = 48.75 answer : d" | a = 30 * 30
b = 50 * 60
c = a + b
d = 30 + 50
e = c / d
|
a ) 25 / 323 , b ) 21 / 969 , c ) 28 / 989 , d ) 74 / 879 , e ) 23 / 589 | a | divide(multiply(choose(6, 2), choose(5, 1)), choose(add(add(5, 6), 8), 3)) | an urn contains 5 red , 6 blue and 8 green balls . 3 balls are randomly selected from the urn , find the probability that the drawn ball are 2 blue and 1 red ? | "sample space = no . of ways 3 balls were drawn from urn = 19 c 3 = 969 no . ways 2 blue balls and 1 red were drawn from bag = 6 c 2 * 5 c 1 = 75 probability = 75 / 969 = 25 / 323 ans - a" | a = math.comb(6, 2)
b = math.comb(5, 1)
c = a * b
d = 5 + 6
e = d + 8
f = math.comb(e, 3)
g = c / f
|
a ) $ 1.63 , b ) $ 1.64 , c ) $ 1.68 , d ) $ 1.70 , e ) $ 1.76 | d | divide(add(multiply(1300, 1.89), multiply(750, 1.38)), add(1300, 750)) | john purchased 1300 large bottles at $ 1.89 per bottle and 750 small bottles at $ 1.38 per bottle . what was the approximate average price paid per bottle ? | "( 1300 * 1.89 + 750 * 1.38 ) / ( 1300 + 750 ) = ~ 1.70 option ( d )" | a = 1300 * 1
b = 750 * 1
c = a + b
d = 1300 + 750
e = c / d
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a ) 41.4 , b ) 34.1 , c ) 13.4 , d ) 12.4 , e ) 10.8 | a | add(inverse(subtract(divide(const_1, 12.5), divide(const_1, 30))), inverse(subtract(divide(const_1, 7.5), divide(const_1, 12)))) | two consultants can type up a report in 12.5 hours and edit it in 7.5 hours . if mary needs 30 hours to type the report and jim needs 12 hours to edit it alone , how many hours q will it take if jim types the report and mary edits it immediately after he is done ? | break down the problem into two pieces : typing and editing . mary needs 30 hours to type the report - - > mary ' s typing rate = 1 / 30 ( rate reciprocal of time ) ( point 1 in theory below ) ; mary and jim can type up a report in 12.5 and - - > 1 / 30 + 1 / x = 1 / 12.5 = 2 / 25 ( where x is the time needed for jim to type the report alone ) ( point 23 in theory below ) - - > x = 150 / 7 ; jim needs 12 hours to edit the report - - > jim ' s editing rate = 1 / 12 ; mary and jim can edit a report in 7.5 and - - > 1 / y + 1 / 12 = 1 / 7.5 = 2 / 15 ( where y is the time needed for mary to edit the report alone ) - - > y = 20 ; how many q hours will it take if jim types the report and mary edits it immediately after he is done - - > x + y = 150 / 7 + 20 = ~ 41.4 answer : a . | a = 1 / 12
b = 1 / 30
c = a - b
d = 1/(c)
e = 1 / 7
f = 1 / 12
g = e - f
h = 1/(g)
i = d + h
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a ) 4 3 / 6 , b ) 4 3 / 9 , c ) 4 3 / 8 , d ) 4 3 / 4 , e ) 4 3 / 1 | c | add(multiply(add(3, divide(const_1, const_2)), subtract(add(5, divide(const_3, 4)), add(4, divide(const_1, const_2)))), const_2) | two men a and b start from place x walking at 4 ½ kmph and 5 ¾ kmph respectively . how many km apart they are at the end of 3 ½ hours if they are walking in the same direction ? | "rs = 5 3 / 4 - 4 1 / 2 = 1 1 / 4 t = 3 1 / 2 h . d = 5 / 4 * 7 / 2 = 35 / 8 = 4 3 / 8 km answer : c" | a = 1 / 2
b = 3 + a
c = 3 / 4
d = 5 + c
e = 1 / 2
f = 4 + e
g = d - f
h = b * g
i = h + 2
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a ) 198 , b ) 288 , c ) 432 , d ) 396 , e ) 484 | c | multiply(multiply(multiply(power(2, const_2.0), 3), divide(12, 2)), 2) | if 2 ^ 4 , 3 ^ 3 , and 12 ^ 3 are factors of the product of 1,452 and w , where w is a positive integer , what is the smallest possible value of w ? | i will go with c ( pending elements to match is 2 ^ 2 * 3 ^ 2 * 12 ^ 1 = 432 | a = 2 ** 2
b = a * 3
c = 12 / 2
d = b * c
e = d * 2
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a ) - 10 , b ) - 2 , c ) 8 , d ) 13 , e ) 17 | d | subtract(reminder(reminder(94, 33), 17), reminder(96, reminder(33, 17))) | for all positive integers m and v , the expression m θ v represents the remainder when m is divided by v . what is the value of ( ( 94 θ 33 ) θ 17 ) - ( 96 θ ( 33 θ 17 ) ) ? | "( ( 94 θ 33 ) θ 17 ) the remainder of 98 divided by 33 is 28 ; the remainder of 28 divided by 17 is 11 ; ( 96 θ ( 33 θ 17 ) ) the remainder of 33 divided by 17 is 16 ; the remainder of 96 divided by 16 is 0 . 11 - 0 = 11 . answer : d ." | a = reminder - (
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a ) 50 , b ) 100 , c ) 490 , d ) 500 , e ) 75 | e | divide(735, 9.8) | a sports equipment store sold ping pong rackets for a total of $ 735 . if the average ( arithmetic mean ) price of a pair of rackets is $ 9.8 , how many pairs were sold ? | "average price for a pair of rackets = $ 9.8 total cost = $ 9.8 * x = $ 735 x = 75 pairs were sold . answer : e" | a = 735 / 9
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a ) 45 , b ) 67 , c ) 70 , d ) 77 , e ) 98 | b | add(add(add(divide(lcm(lcm(lcm(3, 1), 1), 3), 3), divide(lcm(lcm(lcm(3, 1), 1), 3), 1)), divide(lcm(lcm(lcm(3, 1), 1), 3), 1)), divide(lcm(lcm(lcm(3, 1), 1), 3), 3)) | john distributes his pencil among his 4 friends rose , mary , ranjan , and rohit in the ratio 1 / 3 : 1 / 3 : 1 / 4 : 1 / 5 . what is the minimum no . of pencils that the person should have ? | "rakesh : rahul : ranjan : rohit = 1 / 3 : 1 / 3 : 1 / 4 : 1 / 5 step 1 : at first we need to do is lcm of 3 , 3,4 and 5 is 60 . step 2 : then pencil are distributed in ratio among friends , rakesh = ( 1 / 3 x 60 ) = 20 . rahul = ( 1 / 3 x 60 ) = 20 . ranjan = ( 1 / 4 x 60 ) = 15 . rohit = ( 1 / 5 x 60 ) = 12 . step 3 : total number of pencils are ( 20 x + 20 x + 15 x + 12 x ) = 67 x . for minimum number of pencils x = 1 . the person should have atleast 67 pencils . b )" | a = math.lcm(3, 1)
b = math.lcm(a, 1)
c = math.lcm(b, 3)
d = c / 3
e = math.lcm(3, 1)
f = math.lcm(e, 1)
g = math.lcm(f, 3)
h = g / 1
i = d + h
j = math.lcm(3, 1)
k = math.lcm(j, 1)
l = math.lcm(k, 3)
m = l / 1
n = i + m
o = math.lcm(3, 1)
p = math.lcm(o, 1)
q = math.lcm(p, 3)
r = q / 3
s = n + r
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a ) 6500 , b ) 3250 , c ) 1100 , d ) 1300 , e ) 1200 | d | divide(3250, const_3) | divide rs . 3250 among a , b and c so that a receives 1 / 5 as much as b and c together and b receives 2 / 3 as a and c together . b ' s share is ? | "a + b + c = 3250 a = 1 / 5 ( b + c ) ; b = 2 / 3 ( a + c ) b / ( a + c ) = 2 / 3 b = 1 / 5 * 6500 = > 1300 answer : d" | a = 3250 / 3
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a ) 150 , b ) 185 , c ) 190 , d ) 210 , e ) 220 | b | subtract(200, 15) | the mean of 50 observations is 200 . but later he found that there is decrements of 15 from each observations . what is the the updated mean is ? | "185 answer is b" | a = 200 - 15
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a ) 150 m , b ) 899 m , c ) 200 m , d ) 166 m , e ) 187 m | a | subtract(divide(600, const_2), 150) | if the perimeter of a rectangular garden is 600 m , its length when its breadth is 150 m is ? | "2 ( l + 150 ) = 600 = > l = 150 m answer : a" | a = 600 / 2
b = a - 150
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a ) 4.31 , b ) 6.75 , c ) 7.92 , d ) 5.5 , e ) 6.5 | d | divide(110, multiply(add(60, 12), const_0_2778)) | a train 110 m long is running with a speed of 60 km / hr . in what time will it pass a trolley that is running with a speed of 12 km / hr in the direction opposite to that in which the train is going ? | speed of train relative to trolley = 60 + 12 = 72 km / hr . = 72 * 5 / 18 = 20 m / sec . time taken to pass the trolley = 110 * 1 / 20 = 5.5 sec . answer : d | a = 60 + 12
b = a * const_0_2778
c = 110 / b
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a ) 75 , b ) 80 , c ) 85 , d ) 90 , e ) 95 | c | divide(subtract(divide(const_3600, const_10), multiply(10, const_2)), const_4) | in a certain parallelogram the degree measure of one angle exceeds that of the other by 10 what is the degree measure of the smaller angle ? | "in a parallelogram opposite angles are equal and the angles at each side are supplementary to each other ( supplementary angles are two angles that add up to 180 ° ) . given : x + ( x + 10 ) = 180 - - > x = 85 . answer : c ." | a = 3600 / 10
b = 10 * 2
c = a - b
d = c / 4
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a ) 20 , b ) 21 , c ) 22 , d ) 23 , e ) 24 | c | divide(add(multiply(10, 18), subtract(66, 26)), 10) | the average of 10 numbers was calculated as 18 . it is discovered later on that while calculating the average , one number , namely 66 , was incorrectly read as 26 . what is the correct average ? | "10 * 18 - 26 + 66 = 220 220 / 10 = 22 the answer is c ." | a = 10 * 18
b = 66 - 26
c = a + b
d = c / 10
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a ) 22 , b ) 65 , c ) 12 , d ) 36 , e ) 50 | e | multiply(10, 5) | each child has 5 crayons and 14 apples . if there are 10 children , how many crayons are there in total ? | 5 * 10 = 50 . answer is e . | a = 10 * 5
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a ) 4 , b ) 5 , c ) 6 , d ) 7 , e ) 3 | e | divide(1050, lcm(lcm(10, 25), 35)) | how many integers between 1 and 1050 are divisible by 10 , 25 , and 35 ? | prime factorization of given numbers 10 = 2 * 5 25 = 5 ^ 2 35 = 5 * 7 lcm of the given numbers = 2 * 5 ^ 2 * 7 = 50 * 7 = 350 therefore , number of integers = 1050 / 350 = 3 answer : option e | a = math.lcm(10, 25)
b = math.lcm(a, 35)
c = 1050 / b
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a ) - 12 number , b ) - 14 , c ) 2 , d ) 8 , e ) 6 | a | subtract(negate(30), multiply(subtract(64, 48), divide(subtract(64, 48), subtract(78, 64)))) | 78 , 64 , 48 , 30 10 , ( . . . ) | "explanation : 78 - 14 = 64 64 - 16 = 48 48 - 18 = 30 30 - 20 = 10 10 - 22 = - 12 answer : option a" | a = negate - (
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a ) 39 % , b ) 20 % , c ) 23 % , d ) 74 % , e ) 60 % | e | multiply(divide(subtract(1440, 900), 900), const_100) | a cycle is bought for rs . 900 and sold for rs . 1440 , find the gain percent ? | "900 - - - - 540 100 - - - - ? = > 60 % answer : e" | a = 1440 - 900
b = a / 900
c = b * 100
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a ) 52 / 7 , b ) 8 , c ) 14 / 2 , d ) 6 , e ) 43 / 6 | a | add(const_1, inverse(divide(divide(const_1, 7), subtract(const_1, divide(const_1, 8))))) | a and b can finish a work in 7 days and 8 days respectively . if both do work one day and leave one day . and a start the work then in how much days work will finish ? | ( a + b ) work in 2 days = [ 1 / 7 + 1 / 8 ] = 15 / 56 ( a + b ) work in 6 days = 45 / 56 now , the turn of a , work of a in 1 day = 1 / 7 . till now completed work = [ 45 / 56 + 1 / 7 ] = 53 / 56 , remaining work = [ 1 - 53 / 56 ] = 3 / 56 now , turn of b , 1 / 8 work b do in 1 day so , 3 / 56 part of work b do = [ 8 * 3 / 56 ] = 3 / 7 days total time taken = 6 + 1 + 3 / 7 = 52 / 7 days answer a | a = 1 / 7
b = 1 / 8
c = 1 - b
d = a / c
e = 1/(d)
f = 1 + e
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