options stringlengths 37 300 | correct stringclasses 5
values | annotated_formula stringlengths 7 727 | problem stringlengths 5 967 | rationale stringlengths 1 2.74k | program stringlengths 10 646 |
|---|---|---|---|---|---|
a ) a ) 51 , b ) b ) 50 , c ) c ) 45 , d ) d ) 40 , e ) e ) 36 | c | divide(multiply(30, 3), subtract(3, const_1)) | if the number is divided by 3 , it reduced by 30 . the number is | "explanation : let the number be x . then , x - ( x / 3 ) = 30 = > 2 x / 3 = 30 = > x = 45 answer : option c" | a = 30 * 3
b = 3 - 1
c = a / b
|
a ) 30 , b ) 90 , c ) 148 , d ) 172 , e ) 188 | c | subtract(400, subtract(add(multiply(divide(400, const_100), 51), multiply(divide(400, const_100), 46)), multiply(divide(400, const_100), 34))) | among 400 students , 51 % study sociology , 46 % study mathematics and 40 % study biology . if 34 % of students study both mathematics and sociology , what is the largest possible number of students who study biology but do not study either mathematics or sociology ? | i would just like to add a bit of explanation after the step where you calculate that the number of students studying both m and s = 136 using your analysis : we see that the total number of students who study either maths or sociology = 184 + 204 - 136 = 252 so , in the image we know that the number of students in the zone with the black boundary = 252 let ' s assume the number of students who studyonlybiology to beb ( this is the number that we have to maximize ) and , let ' s assume the number of students who study none of the three subjects , that is the number of students in the white space = w since the total number of students = 400 , we can write : 252 + b + w = 400 or , b + w = 400 - 252 = 148 that is , b = 148 - w so , the maximum value ofbwill happen forw = 0 this is how we get , the maximum value ofb = 148 c | a = 400 / 100
b = a * 51
c = 400 / 100
d = c * 46
e = b + d
f = 400 / 100
g = f * 34
h = e - g
i = 400 - h
|
a ) 0 . , b ) 6 . , c ) 8 . , d ) 10 . , e ) 12 . | a | min(divide(102, const_3), divide(102, 6)) | in the third grade of windblown school there are 102 students , one third of them failed the math test and 1 / 6 failed that literature test . at least how many students failed both tests ? | "total = 102 failed in math = 102 / 3 = 34 failed in literature = 108 / 6 = 17 the least failed in both can be 0 while max can be 17 answer a" | a = 102 / 3
b = 102 / 6
c = min(a)
|
a ) s . 3096 , b ) s . 4076 , c ) s . 4085 , d ) s . 4096 , e ) s . 5096 | e | divide(5913, power(add(1, divide(add(6, divide(1, 4)), const_100)), 3)) | the principal that amounts to rs . 5913 in 3 years at 6 1 / 4 % per annum c . i . compounded annually , is ? | "principal = [ 4913 / ( 1 + 25 / ( 4 * 100 ) ) 3 ] = 5913 * 16 / 17 * 16 / 17 * 16 / 17 = rs . 5096 . answer : e" | a = 1 / 4
b = 6 + a
c = b / 100
d = 1 + c
e = d ** 3
f = 5913 / e
|
a ) 1.01 , b ) 1.1 , c ) 0.11 , d ) 11 , e ) none | d | divide(multiply(1100, 2.55), 2805) | if 2805 / 2.55 = 1100 , then 280.5 / 25.5 is equal to ? | "answer given expression 280.5 / 25.5 = 2805 / 255 = 2805 / ( 2.55 x 100 ) = 1100 / 100 = 11 correct option : d" | a = 1100 * 2
b = a / 2805
|
a ) 1.075 days , b ) 0.185 days , c ) 0.065 days , d ) 0.045 days , e ) 0.0775 days | d | inverse(add(inverse(50), inverse(40))) | a and b complete a work in 50 days . a alone can do it in 40 days . if both together can do the work in how many days ? | "1 / 50 + 1 / 40 = 0.045 days answer : d" | a = 1/(50)
b = 1/(40)
c = a + b
d = 1/(c)
|
a ) 1642 , b ) 1640 , c ) 1632 , d ) 2789 , e ) 1704 | e | subtract(2769, divide(multiply(multiply(3, 5), 2769), add(multiply(3, 5), multiply(8, 3)))) | a sum of rs . 2769 is lent into two parts so that the interest on the first part for 8 years at 3 % per annum may be equal to the interest on the second part for 3 years at 5 % per annum . find the second sum ? | "( x * 8 * 3 ) / 100 = ( ( 2769 - x ) * 3 * 5 ) / 100 24 x / 100 = 41535 / 100 - 15 x / 100 39 x = 41535 = > x = 1065 second sum = 2769 β 1065 = 1704 answer : e" | a = 3 * 5
b = a * 2769
c = 3 * 5
d = 8 * 3
e = c + d
f = b / e
g = 2769 - f
|
a ) 0.099 , b ) 0.0099 , c ) 9.9 , d ) 990 , e ) 99 | d | divide(multiply(0.0077, 3.6), multiply(multiply(0.04, 0.1), 0.007)) | ( 0.0077 ) ( 3.6 ) / ( 0.04 ) ( 0.1 ) ( 0.007 ) = | "( 0.0077 ) ( 3.6 ) / ( 0.04 ) ( 0.1 ) ( 0.007 ) = 0.0077 * 360 / 4 * ( 0.1 ) ( 0.007 ) = 0.077 * 90 / 1 * 0.007 = 77 * 90 / 7 = 11 * 90 = 990 answer : d" | a = 0 * 77
b = 0 * 4
c = b * 0
d = a / c
|
a ) 22 % , b ) 23 % , c ) 24 % , d ) 25 % , e ) 34 % | d | divide(22, divide(subtract(const_100, 12), const_100)) | in a certain candy store , 22 % of the customers are caught sampling the candy and are charged a small fine , but 12 % of the customers who sample the candy are not caught . what is the total percent of all customers who sample candy ? | "since 12 % of the customers who sample the candy are not caught , then 88 % of the customers who sample the candy are caught : { % of customers who sample candy } * 0.88 = 0.22 ; { % of customers who sample candy } = 0.25 . answer : d ." | a = 100 - 12
b = a / 100
c = 22 / b
|
a ) 16 % , b ) 12 % , c ) 74 % , d ) 10 % , e ) 8 % | e | multiply(divide(1760, add(multiply(5000, 2), multiply(3000, 4))), const_100) | a lent rs . 5000 to b for 2 years and rs . 3000 to c for 4 years on simple interest at the same rate of interest and received rs . 1760 in all from both of them as interest . the rate of interest per annum is ? | "let the rate be r % p . a . then , ( 5000 * r * 2 ) / 100 + ( 3000 * r * 4 ) / 100 = 1760 100 r + 120 r = 1760 r = 8 % answer : e" | a = 5000 * 2
b = 3000 * 4
c = a + b
d = 1760 / c
e = d * 100
|
a ) 22 , b ) 33 , c ) 77 , d ) 99 , e ) 27 | b | subtract(multiply(20, 3), subtract(multiply(15, 3), 18)) | out of 4 numbers , the average of first 3 is 20 and that of the last 3 is 15 . if the last number is 18 , the first number is : | explanation : let the numbers be a , b , c , d given , a + b + c = 60 , b + c + d = 45 now , d = 18 thus , b + c + 18 = 45 β b + c = 27 putting the value of b + c in a + b + c = 60 a + 27 = 60 β a = 33 answer : b | a = 20 * 3
b = 15 * 3
c = b - 18
d = a - c
|
a ) rs . 6152 , b ) rs . 6159 , c ) rs . 6156 , d ) rs . 6150 , e ) rs . 6196 | d | multiply(subtract(add(add(add(divide(multiply(divide(16, 15), 1125), divide(15, 12)), divide(multiply(divide(18, 18), 1125), divide(15, 12))), divide(multiply(divide(12, 20), 1125), divide(15, 12))), 1125), multiply(4, const_100)), const_2) | 4 car rental agencies a , b , c and d rented a plot for parking their cars during the night . a parked 15 cars for 12 days , b parked 12 cars for 20 days , c parked 18 cars for 18 days and d parked 16 cars for 15 days . if a paid rs . 1125 as rent for parking his cars , what is the total rent paid by all the 4 agencies ? | the ratio in which the four agencies will be paying the rents = 15 * 12 : 12 * 20 : 18 * 18 : 16 * 15 = 180 : 240 : 324 : 240 = 45 : 60 : 81 : 60 let us consider the four amounts to be 45 k , 60 k , 81 k and 60 k respectively . the total rent paid by the four agencies = 45 k + 60 k + 81 k + 60 k = 246 k it is given that a paid rs . 1125 45 k = 1125 = > k = 25 246 k = 246 ( 25 ) = rs . 6150 thus the total rent paid by all the four agencies is rs . 6150 . answer : d | a = 16 / 15
b = a * 1125
c = 15 / 12
d = b / c
e = 18 / 18
f = e * 1125
g = 15 / 12
h = f / g
i = d + h
j = 12 / 20
k = j * 1125
l = 15 / 12
m = k / l
n = i + m
o = n + 1125
p = 4 * 100
q = o - p
r = q * 2
|
a ) 1 / 6 , b ) 2 / 6 , c ) 3 / 6 , d ) 4 / 6 , e ) 5 / 6 | d | divide(subtract(subtract(subtract(9, 2), 2), 1), 6) | harry started a 7 - mile hike with a full 9 - cup canteen of water and finished the hike in 2 hours with 3 cup of water remaining in the canteen . if the canteen leaked at the rate of 1 cup per hour and harry drank 2 cups of water during the last mile , how many cups did he drink per mile during the first 6 miles of the hike ? | "no of cups leaked during the trip = 2 cups . no of cups harry drank = 6 cups . no of cups harry drank during the first 6 miles = 4 . drink / mile = 4 / 6 answer d" | a = 9 - 2
b = a - 2
c = b - 1
d = c / 6
|
a ) 1240 , b ) 1120 , c ) 1190 , d ) 2150 , e ) none of these | d | add(900, divide(multiply(900, const_100), multiply(12, 6))) | the banker ' s gain on a sum due 6 years hence at 12 % per annum is rs . 900 . what is the banker ' s discount ? | "explanation : td = ( bg Γ 100 ) / tr = ( 900 Γ 100 ) / ( 6 Γ 12 ) = rs . 1250 bg = bd β td = > 900 = bd - 1250 = > bd = 2150 answer : option d" | a = 900 * 100
b = 12 * 6
c = a / b
d = 900 + c
|
a ) 10 , b ) 12 , c ) 16 , d ) 17 , e ) 19 | c | subtract(17, 8) | 8 ^ 100 is divisible by 17 . find the remainder for this ? | "this is an extremely difficult problem to solve with out fermat ' s little theorem . by applying fermat ' s little theorem , we know that 816 when divided by 17 , the remainder is 1 . so divide 100 by 16 and find the remainder . remainder = 4 therefore , 100 = ( 16 Γ 6 ) + 4 now this problem can be written as 810017 = 816 Γ 6 + 417 = ( 816 ) 6 Γ 8417 now this problem simply boils down to ( 1 ) 6 Γ 8417 = 8417 84 = 82 Γ 82 , we need to find the remainder when 64 Γ 64 is divisible by 17 . or 13 Γ 13 = 169 . when 169 is divided by 17 , remainder is 16 . c" | a = 17 - 8
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a ) 140 , b ) 120 , c ) 130 , d ) 110 , e ) 150 | e | multiply(const_100, divide(add(const_100, 42.5), subtract(const_100, 5))) | a shopkeeper sold an book offering a discount of 5 % and earned a profit of 42.5 % . what would have been the percentage of profit earned if no discount was offered ? | "let c . p . be $ 100 . then , s . p . = $ 142.50 let marked price be $ x . then , 95 / 100 x = 142.5 x = 14250 / 95 = $ 150 now , s . p . = $ 150 , c . p . = $ 100 profit % = 50 % . e" | a = 100 + 42
b = 100 - 5
c = a / b
d = 100 * c
|
a ) 12 , b ) 15 , c ) 20 , d ) 25 , e ) 30 | e | divide(multiply(subtract(1, divide(1, 4)), const_10), subtract(1, subtract(1, divide(1, 4)))) | a driver would have reduced the time it took to drive from home to the store by 1 / 4 if the average speed had been increased by 10 miles per hour . what was the actual average speed , in miles per hour , when the driver drove from home to the store ? | since the distance remains the same ( we ' re just changing the rate and time ) , any increase in rate or time is met with a decrease in the other term . decreasing the time by 1 / 4 would give us : d = ( r ) ( t ) = ( 3 t / 4 ) ( x * r ) x = 4 / 3 since ( 3 t / 4 ) ( 4 r / 3 ) = ( r ) ( t ) = d 4 r / 3 = r + 10 r / 3 = 10 r = 30 the answer is e . | a = 1 / 4
b = 1 - a
c = b * 10
d = 1 / 4
e = 1 - d
f = 1 - e
g = c / f
|
a ) 14 days , b ) 16 days , c ) 20 days , d ) 11 days , e ) 19 days | c | add(divide(const_1, 15), divide(const_1, 30)) | a can do a job in 15 days and b can do it in 30 days . a and b working together will finish twice the amount of work in - - - - - - - days ? | "c 1 / 15 + 1 / 30 = 1 / 10 10 * 2 = 20 days" | a = 1 / 15
b = 1 / 30
c = a + b
|
a ) $ 60 , b ) $ 80 , c ) $ 90 , d ) $ 120 , e ) $ 150 | e | divide(10, subtract(1, add(add(divide(1, 10), divide(1, 3)), divide(1, 2)))) | at a supermarket , john spent 1 / 2 of his money on fresh fruits and vegetables , 1 / 3 on meat products , and 1 / 10 on bakery products . if he spent the remaining $ 10 on candy , how much did john spend at the supermarket ? | "let ' s let t = total number of dollars spent at the supermarket . with this variable we can set up an equation and determine t . we are given that john spent 1 / 2 of his money on fresh fruits and vegetables , or ( 1 / 2 ) t , 1 / 3 on meat products , or ( 1 / 3 ) t , and 1 / 10 on bakery products , or ( 1 / 10 ) t . we are also given that he spent the remaining $ 10 on candy . since we know where all his money was allocated , we can sum these values together and set the sum to t . so we have : ( 1 / 2 ) t + ( 1 / 3 ) t + ( 1 / 10 ) t + 10 = t to get rid of the fractions we can multiply the entire equation by 30 , and we obtain : 15 t + 10 t + 3 t + 300 = 30 t 28 t + 300 = 30 t 300 = 2 t t = 150 john spent $ 90 at the supermarket . answer : e" | a = 1 / 10
b = 1 / 3
c = a + b
d = 1 / 2
e = c + d
f = 1 - e
g = 10 / f
|
a ) 100 , b ) 60 , c ) 120 , d ) 200 , e ) 150 | b | multiply(12, 120) | the h . c . f . of two numbers is 12 and their l . c . m . is 600 . if one of the number is 120 , find the other ? | "other number = 12 * 600 / 120 = 60 answer is b" | a = 12 * 120
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a ) 6.62 , b ) 66.2 , c ) 662 , d ) 0.662 , e ) 0.0662 | a | multiply(divide(66.2, 1000), const_100) | 66.2 is what percent of 1000 ? | "we assume that 1000 is 100 % assume ' x ' is value we looking for here , 1000 = 100 % and x % = 66.2 therefore , 100 / x = 1000 / 66.2 100 / x = 15.105 x = 6.62 a" | a = 66 / 2
b = a * 100
|
a ) 50 , b ) 28 , c ) 160 , d ) 180 , e ) 18 | a | subtract(power(divide(add(5, 10), const_2), const_2), power(subtract(5, divide(add(5, 10), const_2)), const_2)) | if the sum and difference of two numbers are 5 and 10 respectively , then the difference of their square is : | "let the numbers be x and y . then , x + y = 5 and x - y = 10 x 2 - y 2 = ( x + y ) ( x - y ) = 5 * 10 = 50 . answer : a" | a = 5 + 10
b = a / 2
c = b ** 2
d = 5 + 10
e = d / 2
f = 5 - e
g = f ** 2
h = c - g
|
a ) 9632 , b ) 7896 , c ) 8741 , d ) 1683 , e ) 8523 | d | multiply(multiply(multiply(add(const_2, const_3), add(3, 3)), add(3, const_4)), add(const_4, const_4)) | find the least number which when divided by 56 , 78 leaves a remainder 3 but when divided by 9 leaves no remainder | l . c . m of 5,6 , 7,8 = 840 required number is of the form of 840 k + 3 least value of k for which ( 840 k + 3 ) is divided by 9 is k = 2 required number = ( 840 * 2 + 3 ) = 1683 answer ( d ) | a = 2 + 3
b = 3 + 3
c = a * b
d = 3 + 4
e = c * d
f = 4 + 4
g = e * f
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a ) a ) 70 , b ) b ) 77 , c ) c ) 79 , d ) d ) 81 , e ) e ) 88 | e | multiply(divide(divide(multiply(21, add(21, const_1)), const_2), 21), 8) | what is the average of first 21 multiples of 8 ? | "required average = 8 ( 1 + 2 + . . . . + 21 ) / 21 ( 8 / 21 ) x ( ( 21 x 22 ) / 2 ) ( because sum of first 21 natural numbers ) = 88 e" | a = 21 + 1
b = 21 * a
c = b / 2
d = c / 21
e = d * 8
|
a ) $ 100 , b ) $ 75 , c ) $ 20 , d ) $ 120 , e ) $ 60 | e | multiply(100, subtract(const_1, divide(divide(200, const_2), add(150, divide(200, const_2))))) | a invested $ 150 in a business after 6 months b invested $ 200 in the business . end of the year if they got $ 100 as profit . find a shares ? | "a : b = 150 * 12 : 200 * 6 a : b = 3 : 2 a ' s share = 100 * 3 / 5 = $ 60 answer is e" | a = 200 / 2
b = 200 / 2
c = 150 + b
d = a / c
e = 1 - d
f = 100 * e
|
a ) 0.2 , b ) 1.2 , c ) 1.8 , d ) 2.2 , e ) 4.0 | b | subtract(9.2, subtract(floor(9.2), const_1)) | for any number y , y * is defined as the greatest positive even integer less than or equal to y . what is the value of 9.2 β 9.2 * ? | "since y * is defined as the greatest positive even integer less than or equal to y , then 9.2 * = 8 ( the greatest positive even integer less than or equal to 9.2 is 4 ) . hence , 9.2 β 9.2 * = 9.2 - 8 = 1.2 answer : b ." | a = math.floor(9, 2)
b = a - 1
c = 9 - 2
|
a ) 399 , b ) 272 , c ) 932.4 , d ) 277 , e ) 311 | c | multiply(multiply(subtract(multiply(sqrt(3136), const_4), multiply(const_2, 1)), 1.40), 3) | the area of a square field 3136 sq m , if the length of cost of drawing barbed wire 3 m around the field at the rate of rs . 1.40 per meter . two gates of 1 m width each are to be left for entrance . what is the total cost ? | "answer : option c explanation : a 2 = 3136 = > a = 56 56 * 4 * 3 = 672 Γ’ β¬ β 6 = 666 * 1.4 = 932.4 answer : c" | a = math.sqrt(3136)
b = a * 4
c = 2 * 1
d = b - c
e = d * 1
f = e * 3
|
a ) $ 100 , b ) $ 150 , c ) $ 160 , d ) $ 200 , e ) $ 250 | c | multiply(divide(400, add(divide(2, 3), const_1)), divide(2, 3)) | $ 400 is divided amongst a , b and c so that a may get 2 / 3 as much as b and c together , b may get 6 / 9 as much as a and c together , then the share of a is | "a : ( b + c ) = 2 : 3 a ' s share = 400 * 2 / 5 = $ 160 answer is c" | a = 2 / 3
b = a + 1
c = 400 / b
d = 2 / 3
e = c * d
|
a ) 5 , b ) 10 , c ) 8 , d ) 15 , e ) 20 | b | add(multiply(multiply(2, 2), 2), 2) | 2 . the value of x + x ( xx ) when x = 2 is : | x + x ( xx ) put the value of x = 2 in the above expression we get , 2 + 2 ( 22 ) = 2 + 2 ( 2 Γ 2 ) = 2 + 2 ( 4 ) = 2 + 8 = 10 answer : b | a = 2 * 2
b = a * 2
c = b + 2
|
a ) no answer , b ) 25 , c ) 1 , d ) 0 , e ) - 2 | a | divide(subtract(5, sqrt(subtract(power(5, 2), multiply(25, 25)))), 2) | find the value of x from the below equation ? : x ^ 2 + 5 x + 25 = 0 | "a = 1 , b = 5 , c = 25 x 1,2 = ( - 5 Γ’ Β± Γ’ Λ Ε‘ ( 5 ^ 2 - 4 Γ£ β 1 Γ£ β 25 ) ) / ( 2 Γ£ β 1 ) = ( - 5 Γ’ Β± Γ’ Λ Ε‘ ( 25 - 100 ) ) / 2 = ( - 5 Γ’ Β± Γ’ Λ Ε‘ - 75 ) / 2 no real roots a" | a = 5 ** 2
b = 25 * 25
c = a - b
d = math.sqrt(c)
e = 5 - d
f = e / 2
|
a ) 8 % , b ) 9 % , c ) 11 % , d ) 12.5 % , e ) 20 % | e | multiply(divide(subtract(multiply(divide(2, 15), 75), 18), multiply(divide(2, 15), 75)), const_100) | a doctor prescribed 18 cubic centimeters of a certain drug to a patient whose body weight was 75 pounds . if the typical dosage is 2 cubic centimeters per 15 pounds of the body weight , by what percent was the prescribed dosage greater than the typical dosage ? | "typical dosage is dose : weight : : 2 : 15 . now if weight is 75 : ( 75 / 15 ) ) then typical dosage would be 2 * 5 = 10 cc . dosage = 18 cc . dosage is greater by 2 cc . % dosage is greater : ( 2 / 10 ) * 100 = 20 % e is the answer ." | a = 2 / 15
b = a * 75
c = b - 18
d = 2 / 15
e = d * 75
f = c / e
g = f * 100
|
a ) 70 % , b ) 90 % , c ) 60 % , d ) 40 % , e ) 65 % | d | multiply(divide(1, 5), const_100) | if two positive numbers are in the ratio 1 / 7 : 1 / 5 , then by what percent is the second number more than the first ? | "given ratio = 1 / 7 : 1 / 5 = 5 : 7 let first number be 5 x and the second number be 7 x . the second number is more than first number by 2 x . required percentage = 2 x / 5 x * 100 = 40 % . answer : d" | a = 1 / 5
b = a * 100
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a ) 556 , b ) 612.5 , c ) 756.8 , d ) 610 , e ) 820 | d | multiply(divide(30.5, 2000), const_100) | 30.5 % of 2000 | 1 % of 2000 = 20 30 % of 2000 = 30 * 20 = 600 0.5 % of 2000 = 0.5 * 2000 = 10 total = 610 answer : d | a = 30 / 5
b = a * 100
|
a ) β 1 , b ) 0 , c ) 1 , d ) 25 , e ) 29 | d | power(subtract(power(subtract(negate(1), 1), 2), negate(1)), 2) | if the operation x is defined by a x b = ( b - a ) ^ 2 / a ^ 2 for all numbers a and b , and a β 0 , then β 1 x ( 1 x β 1 ) = | on solving the inner bracket we get 4 . then we get equation as - 1 x 4 where b = 4 and a = - 1 answer d | a = negate - (
b = a ** 1
c = b - 2
d = c ** negate
|
a ) 16 , b ) 3 / β 2 , c ) 8 , d ) 2 β 2 , e ) ( β 2 ) / 3 | b | sqrt(divide(multiply(3, 3), const_2)) | the two lines y = x and x = - 3 intersect on the coordinate plane . if z represents the area of the figure formed by the intersecting lines and the x - axis , what is the side length of a cube whose surface area is equal to 6 z ? | 800 score official solution : the first step to solving this problem is to actually graph the two lines . the lines intersect at the point ( - 3 , - 3 ) and form a right triangle whose base length and height are both equal to 4 . as you know , the area of a triangle is equal to one half the product of its base length and height : a = ( 1 / 2 ) bh = ( 1 / 2 ) ( 3 Γ 3 ) = 9 / 2 ; so z = 9 / 2 . the next step requires us to find the length of a side of a cube that has a face area equal to 9 / 2 . as you know the 6 faces of a cube are squares . so , we can reduce the problem to finding the length of the side of a square that has an area of 9 / 2 . since the area of a square is equal to s Β² , where s is the length of one of its side , we can write and solve the equation s Β² = 9 / 2 . clearly s = β 9 / 2 = 3 / β 2 , oranswer choice ( b ) . | a = 3 * 3
b = a / 2
c = math.sqrt(b)
|
a ) 0.96 , b ) 9.4 , c ) 0.094 , d ) 94 , e ) none | a | divide(subtract(const_100, 4), const_100) | subtracting 4 % of a from a is equivalent to multiplying a by how much ? | "answer let a - 4 % of a = ab . β ( 96 x a ) / 100 = ab β΄ b = 0.96 correct option : a" | a = 100 - 4
b = a / 100
|
a ) 1 / 252 , b ) 1 / 10 , c ) 1 / 8 , d ) 1 / 2 , e ) 5 / 9 | a | multiply(multiply(multiply(divide(add(const_4, const_1), add(9, const_1)), divide(subtract(add(5, const_1), const_1), subtract(add(9, const_1), const_1))), divide(subtract(subtract(add(5, const_1), const_1), const_1), subtract(subtract(add(9, const_1), const_1), const_1))), divide(subtract(subtract(subtract(add(5, const_1), const_1), const_1), const_1), subtract(subtract(subtract(add(9, const_1), const_1), const_1), const_1))) | each of the integers from 0 to 9 , inclusive , is written on a separate slip of blank paper and the ten slips are dropped into a hat . if 5 of the slips are the drawn , without replacement , what is the probability that all 5 have a even number written on it ? | "key is that there is no replacement , so each successive choice will become more skewed towards picking a neg ( i . e . the pool of positives decreases , while the pool of negatives stay the same ) p ( + on 1 st pick ) = 5 / 10 p ( + on 2 nd pick ) = 4 / 9 p ( + on 3 rd pick ) = 3 / 8 p ( + on 4 rd pick ) = 2 / 7 p ( + on 5 rd pick ) = 1 / 6 5 / 10 * 4 / 9 * 3 / 8 * 2 / 7 * 1 / 6 = 1 / 252 a" | a = 4 + 1
b = 9 + 1
c = a / b
d = 5 + 1
e = d - 1
f = 9 + 1
g = f - 1
h = e / g
i = c * h
j = 5 + 1
k = j - 1
l = k - 1
m = 9 + 1
n = m - 1
o = n - 1
p = l / o
q = i * p
r = 5 + 1
s = r - 1
t = s - 1
u = t - 1
v = 9 + 1
w = v - 1
x = w - 1
y = x - 1
z = u / y
A = q * z
|
a ) 16 , b ) 1 , c ) 8 , d ) 12 , e ) 24 | b | multiply(divide(32, power(const_2, 5)), const_2) | if a and b are integers and ( a * b ) ^ 5 = 32 y , y could be : | "distribute the exponent . a ^ 5 * b ^ 5 = 32 y find the prime factorization of 32 . this is 2 ^ 5 * 1 . we need 1 the answer is b ." | a = 2 ** 5
b = 32 / a
c = b * 2
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a ) 45 days . , b ) 40 days . , c ) 38 days . , d ) 35 days . , e ) 30 days . | b | multiply(20, divide(100, 50)) | it was calculated that 100 men could complete a piece of work in 20 days . when work was scheduled to commence , it was found necessary to send 50 men to another project . how much longer will it take to complete the work ? | one day work = 1 / 20 one man β s one day work = 1 / ( 20 * 100 ) now : no . of workers = 50 one day work = 50 * 1 / ( 20 * 100 ) the total no . of days required to complete the work = ( 100 * 20 ) / 50 = 40 answer : b | a = 100 / 50
b = 20 * a
|
a ) 36 , b ) 20 , c ) 28 , d ) 54 , e ) 45 | b | inverse(divide(const_3, multiply(15, const_4))) | if a is thrice as fast as b and together can do a work in 15 days . in how many days a alone can do the work ? | "a β s one day β s work = 1 / x b β s one day β s work = 1 / 3 x a + b β s one day β s work = 1 / x + 1 / 3 x = 1 / 15 = 3 + 1 / 3 x = 4 / 3 x = 1 / 15 x = 15 * 4 / 3 = 20 answer : b" | a = 15 * 4
b = 3 / a
c = 1/(b)
|
a ) - 33 , b ) 33 , c ) 35 , d ) 36 , e ) - 35 | a | divide(subtract(multiply(5, 10), 16), 2) | average of 5 numbers is - 10 , and the sum of 3 of the numbers is 16 , wat is the average of the other 2 numbers ? | let the five numbers be a , b , c , d , e . then their average is ( a + b + c + d + e 5 ) = 10 . ( a + b + c + d + e 5 ) = 10 . now three of the numbers have a sum of 16 , say , a + b + c = 16 a + b + c = 16 . so substitute 16 for a + b + ca + b + c in the average above : ( 16 + d + e 5 ) = 10 . ( 16 + d + e 5 ) = 10 . solving this equation for d + ed + e gives d + e = β 66 d + e = β 66 . finally , dividing by 2 ( to form the average ) gives ( d + e 2 ) = β 33 . ( d + e 2 ) = β 33 . hence , the answer is a : - 33 | a = 5 * 10
b = a - 16
c = b / 2
|
a ) 2 , b ) - 2 , c ) 4 , d ) - 3 , e ) 6 | d | divide(subtract(50, 6), 4) | if | 4 x + 6 | = 50 , what is the sum of all the possible values of x ? | "there will be two cases 4 x + 6 = 50 or 4 x + 6 = - 50 = > x = 11 or x = - 14 sum of both the values will be - 14 + 11 = - 3 answer is d" | a = 50 - 6
b = a / 4
|
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 e 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 e = 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 ) 18 , b ) 99 , c ) 27 , d ) 26 , e ) 21 | e | multiply(divide(subtract(2300, 1800), 2300), const_100) | the cost price of a radio is rs . 2300 and it was sold for rs . 1800 , find the loss % ? | "2300 - - - - 500 100 - - - - ? = > 21 % answer : e" | a = 2300 - 1800
b = a / 2300
c = b * 100
|
a ) 30 , b ) 60 , c ) 40 , d ) 20 , e ) 50 | b | multiply(30, divide(20, subtract(30, 20))) | a and b can finish a work 30 days if they work together . they worked together for 20 days and then b left . a finished the remaining work in another 20 days . in how many days a alone can finish the work ? | amount of work done by a and b in 1 day = 1 / 30 amount of work done by a and b in 20 days = 20 Γ ( 1 / 30 ) = 20 / 30 = 2 / 3 remaining work β 1 β 2 / 3 = 1 / 3 a completes 1 / 3 work in 20 days amount of work a can do in 1 day = ( 1 / 3 ) / 20 = 1 / 60 = > a can complete the work in 60 days answer is b . | a = 30 - 20
b = 20 / a
c = 30 * b
|
a ) 4 , b ) 0.4 , c ) 0.04 , d ) 40 , e ) none of these | c | multiply(divide(0.0004, 0.01), const_100) | 0.0004 ? = 0.01 | "explanation : required answer = 0.0004 / 0.01 = 0.04 / 1 = 0.04 . answer : option c" | a = 0 / 4
b = a * 100
|
a ) 430 , b ) 516 , c ) 660 , d ) 860 , e ) none | d | divide(multiply(860, 50), 50) | 860 % of 50 + 50 % of 860 = ? | "solution given expression = [ 860 / 100 x 50 + 50 / 100 x 860 ] = 430 + 430 = 860 . answer d" | a = 860 * 50
b = a / 50
|
a ) 36 , b ) 2 ^ 4 * 3 , c ) 24 , d ) 38 , e ) 47 | d | subtract(39, const_1) | in a lake , there is a patch of lily pads . every day , the patch doubles in size . it takes 39 days for the patch to cover the entire lake , how many days would it take the patch to cover half of the lake ? | "working backward from the day it ' s covered : day 39 : fully covered day 38 : half covered so 38 days answer : d" | a = 39 - 1
|
a ) 1000 , b ) 1100 , c ) 1200 , d ) 1300 , e ) 1600 | e | multiply(multiply(4, const_1000), 4) | find the product of the localvalue and absolutevalue of 4 in 20468 ? | "local value of 4 = 4 x 100 = 400 place value of 4 = 4 there fore = 4 x 400 = 1600 e" | a = 4 * 1000
b = a * 4
|
a ) 21 , b ) 20 , c ) 19 , d ) 16 , e ) 15 | e | subtract(20, add(floor(divide(9, const_2)), const_1)) | marcella has 20 pairs of shoes . if she loses 9 individual shoes , what is the greatest number of matching pairs she could have left ? | "marcella has 20 pairs of shoes and loses 9 shoes . to minimize the loss of identical pairs of shoes we want marcella to lose as many identical pairs as possible . this would yield 4 identical pairs and 1 additional shoe ( destroying 5 pairs of shoes ) . the 20 pairs of shoes minus the 5 ' destroyed ' pairs yields 15 pairs that still fulfill the requirements . answer : e" | a = 9 / 2
b = math.floor(a)
c = b + 1
d = 20 - c
|
a ) 50 % , b ) 67 % , c ) 85 % , d ) 91 % , e ) 97 % | e | multiply(subtract(1, power(divide(divide(10, const_2), 10), const_2)), const_100) | a miniature roulette wheel is divided into 10 equal sectors , each bearing a distinct integer from 1 to 10 , inclusive . each time the wheel is spun , a ball randomly determines the winning sector by settling in that sector . if the wheel is spun five times , approximately what is the probability that the product of the five winning sectors β integers will be even ? | "the only way to have an odd product is if all 5 integers are odd . p ( odd product ) = 1 / 2 * 1 / 2 * 1 / 2 * 1 / 2 * 1 / 2 = 1 / 32 p ( even product ) = 1 - 1 / 32 = 31 / 32 which is about 97 % the answer is e ." | a = 10 / 2
b = a / 10
c = b ** 2
d = 1 - c
e = d * 100
|
a ) 15 , b ) 37 , c ) 49 , d ) 110 , e ) 180 | c | multiply(divide(7, 35), 245) | on a map , 7 centimeters represents 35 kilometers . two cities 245 kilometers apart would be separated on the map by how many centimeters ? | 1 centimeter represents 5 kilometers ( 35 / 7 ) x = 245 / 5 = 49 answer : c | a = 7 / 35
b = a * 245
|
a ) 0.09 , b ) 0.65 , c ) 0.54 , d ) 0.85 , e ) 0.91 | b | divide(add(add(22, 10), 7), 60) | a certain bag contains 60 balls β 22 white , 10 green , 7 yellow , 15 red , and 6 purple . if a ball is to be chosen at random , what is the probability that the ball will be neither red nor purple ? | "according to the stem the ball can be white , green or yellow , so the probability is ( white + green + yellow ) / ( total ) = ( 22 + 10 + 17 ) / 60 = 39 / 60 = 0.65 . answer : b ." | a = 22 + 10
b = a + 7
c = b / 60
|
a ) 1 , b ) 2 , c ) 3 , d ) 6 , e ) 5 | d | subtract(5, 2) | points a , b , c , and d , in that order , lie on a line . if ab = 2 cm , ac = 5 cm , and bd = 9 cm , what is cd , in centimeters ? | "putting a value to each point , lets use the following : a - 0 b - 2 ( ab = 2 ) c - 5 ( ac = 5 ) d - 11 ( bd = 9 ) cd is 11 - 5 = 6 . ans d" | a = 5 - 2
|
a ) 4 , b ) 8 , c ) 2 , d ) 16 , e ) 18 | c | divide(power(4, const_2), 8) | if the area of a square with sides of length 4 centimeters is equal to the area of a rectangle with a width of 8 centimeters , what is the length of the rectangle , in centimeters ? | "let length of rectangle = l 4 ^ 2 = l * 8 = > l = 16 / 8 = 2 answer c" | a = 4 ** 2
b = a / 8
|
a ) 9.5 , b ) 8.0 , c ) 7.2 , d ) 5.0 , e ) 4.0 | c | subtract(multiply(9.2, 6), multiply(6.8, 6)) | the average of 6 numbers is 6.8 . if one of the numbers is multiplied by a factor of 3 , the average of the numbers increases to 9.2 . what number is multiplied by 3 ? | "the average of 6 numbers is 6.8 the sum of 6 numbers will be 6.8 x 6 = 40.8 the average of 6 number after one of the number is multiplied by 3 is 9.2 the sum of the numbers will now be 9.2 x 6 = 55.2 so the sum has increased by 55.2 - 40.8 = 14.4 let the number multiplied by 3 be n then , 3 n = n + 14.4 or 2 n = 14.4 or n = 7.2 answer : - c" | a = 9 * 2
b = 6 * 8
c = a - b
|
a ) 10.6 , b ) 10.9 , c ) 36 , d ) 10.8 , e ) 10.1 | c | divide(add(200, 800), multiply(add(60, 40), const_0_2778)) | two trains 200 m and 800 m long run at the speed of 60 km / hr and 40 km / hr respectively in opposite directions on parallel tracks . the time which they take to cross each other is ? | "relative speed = 60 + 40 = 100 km / hr . = 100 * 5 / 18 = 250 / 9 m / sec . distance covered in crossing each other = 200 + 800 = 1000 m . required time = 1000 * 9 / 250 = 36 sec . answer : c" | a = 200 + 800
b = 60 + 40
c = b * const_0_2778
d = a / c
|
a ) 3 , b ) 4 , c ) 5 , d ) 6 , e ) 7 | a | subtract(subtract(7, 3), const_1) | if 3 < x < 6 < y < 7 , then what is the greatest possible positive integer difference of x and y ? | "3 < x < 6 < y < 7 ; 3 < x y < 7 3 + y < x + 7 y - x < 4 . positive integer difference is 3 ( for example y = 6.5 and x = 3.5 ) answer : a ." | a = 7 - 3
b = a - 1
|
a ) 11 , b ) 12 , c ) 13 , d ) 14 , e ) 15 | b | divide(96, const_10) | the ratio between the perimeter and the width of a rectangle is 5 : 1 . if the area of the rectangle is 96 square centimeters , what is the length of the rectangle in centimeters ? | "perimeter = 2 ( w + l ) = 5 w 3 w = 2 l w = 2 l / 3 wl = 96 2 l ^ 2 / 3 = 96 l ^ 2 = 144 l = 12 cm the answer is b ." | a = 96 / 10
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a ) 50 , b ) 51 , c ) 52 , d ) 53 , e ) 54 | d | subtract(90, subtract(add(55, 33), 51)) | in a group of 90 people , 55 have visited iceland and 33 have visited norway . if 51 people have visited both iceland and norway , how many people have visited neither country ? | "this is an example of a standard overlapping sets question . it has no ' twists ' to it , so you ' ll likely find using the overlapping sets formula to be a fairly easy approach . if you ' re not familiar with it , then here is the formula : 90 = 55 + 33 - 51 + ( # in neither group ) = 53 the prompt gives you all of the numbers you need to get to the correct answer . just plug in and solve . d" | a = 55 + 33
b = a - 51
c = 90 - b
|
a ) a ) 182 , b ) b ) 227 , c ) c ) 220 , d ) d ) 224 , e ) e ) 302 | b | divide(195, divide(subtract(const_100, 14), const_100)) | a small company reduced its faculty by approximately 14 percent to 195 employees . what was the original number of employees ? | "if x is the original number of employees , then after 14 % reduction in employees number is . 86 x but we are given . 86 x = 195 x = 227 so the original number of employees is 227 correct answer - b" | a = 100 - 14
b = a / 100
c = 195 / b
|
a ) 1 , b ) 1 / 40 , c ) 40 , d ) 80 , e ) 1 / 80 | c | multiply(divide(40, 40), 40) | in a dairy farm , 40 cows eat 40 bags of husk in 40 days . in how many days one cow will eat one bag of husk ? | "assume that in x days , one cow will eat one bag of husk . more cows , less days ( indirect proportion ) more bags , more days ( direct proportion ) hence we can write as cowsbags 40 : 11 : 40 β« β β¬ : : x : 40 β 40 Γ 1 Γ 40 = 1 Γ 40 Γ x β x = 40 answer : c" | a = 40 / 40
b = a * 40
|
a ) 22 , b ) 30 , c ) 99 , d ) 26 , e ) 27 | d | add(divide(add(21, 23), const_2), multiply(const_1, 4)) | the average age of 4 men is increased by years when two of them whose ages are 21 years and 23 years are replaced by two new men . the average age of the two new men is | total age increased = ( 8 * 2 ) years = 8 years . sum of ages of two new men = ( 21 + 23 + 8 ) years = 52 years average age of two new men = ( 52 / 2 ) years = 26 years . answer : d | a = 21 + 23
b = a / 2
c = 1 * 4
d = b + c
|
a ) 28 % , b ) 25 % , c ) 30 % , d ) 27 % , e ) 35 % | d | multiply(add(multiply(6, divide(25, const_100)), multiply(4, divide(30, const_100))), add(6, 4)) | one type of liquid contains 25 % of kerosene , the other contains 30 % of kerosene . p can is filled with 6 parts of the first liquid and 4 parts of the second liquid . find the percentage of the kerosene in the new mixture . | let p be filled by 60 lts of 1 st liquid and 40 lts . of 2 nd liquid . amount of kerosene = ( 25 * 60 / 100 ) + ( 30 * 40 / 100 ) = 27 lts . % of kerosene = 27 % answer : d | a = 25 / 100
b = 6 * a
c = 30 / 100
d = 4 * c
e = b + d
f = 6 + 4
g = e * f
|
a ) rs 1600 , b ) rs 1800 , c ) rs 2000 , d ) rs 2200 , e ) none of these | c | add(800, multiply(divide(subtract(multiply(800, 5), multiply(800, 4)), subtract(multiply(3, 4), multiply(2, 5))), 3)) | the ratio of income of anand to that of balu is 5 : 4 and the expenditure of anand to that of balu is 3 : 2 . if at the end of the year , each saves rs , 800 , the income of anand is : | explanation : let the income of anand and balu be 5 x and 4 x and . the expenditures of anand and balu be 3 y and 2 y . then , 5 x 3 y = 800 and 4 x 2 y = 800 . on solving we get : x = 400 . as income = 5 x = rs . 2000 . answer : c | a = 800 * 5
b = 800 * 4
c = a - b
d = 3 * 4
e = 2 * 5
f = d - e
g = c / f
h = g * 3
i = 800 + h
|
a ) 16 , b ) 24 , c ) 48 , d ) 96 , e ) 78 | c | multiply(multiply(subtract(3, const_1), multiply(subtract(3, const_1), subtract(3, const_1))), add(multiply(subtract(3, const_1), subtract(3, const_1)), subtract(3, const_1))) | the largest natural number by which the product of 3 consecutive even natural numbers is always divisible , is | solution required number = ( 2 Γ 4 Γ 6 ) = 48 . answer c | a = 3 - 1
b = 3 - 1
c = 3 - 1
d = b * c
e = a * d
f = 3 - 1
g = 3 - 1
h = f * g
i = 3 - 1
j = h + i
k = e * j
|
a ) 2099 , b ) 6086 , c ) 2976 , d ) 2000 , e ) 3862 | d | add(multiply(add(20, const_1), add(20, const_1)), add(20, const_1)) | a sum of money is put out at compound interest for 2 years at 20 % . it would fetch rs . 482 more if the interest were payable half - yearly , then it were pay able yearly . find the sum ? | "p ( 11 / 10 ) 4 - p ( 6 / 5 ) 2 = 482 p = 2000 answer : d" | a = 20 + 1
b = 20 + 1
c = a * b
d = 20 + 1
e = c + d
|
a ) 4 , b ) 6 , c ) 8 , d ) 12 , e ) 15 | e | multiply(multiply(5, const_2), divide(const_3, const_2)) | a group of hikers is planning a trip that will take them up a mountain using one route and back down using another route . they plan to travel down the mountain at a rate of one and a half times the rate they will use on the way up , but the time each route will take is the same . if they will go up the mountain at a rate of 5 miles per day and it will take them two days , how many miles long is the route down the mountain ? | "on the way down , the rate is 1.5 * 5 = 7.5 miles per day . the distance of the route down the mountain is 2 * 7.5 = 15 miles . the answer is e ." | a = 5 * 2
b = 3 / 2
c = a * b
|
a ) 6 , b ) 2 , c ) - 4 , d ) - 6 , e ) - 10 | c | divide(subtract(7, 5), subtract(sqrt(add(4, 7)), sqrt(add(5, 4)))) | in the xy - coordinate plane , the graph of y = - x ^ 2 + 4 intersects line l at ( p , 5 ) and ( t , - 7 ) . what is the least possible value of the slope of line l ? | "we need to find out the value of p and l to get to the slope . line l and graph y intersect at point ( p , 5 ) . hence , x = p and y = 5 should sactisfy the graph . soliving 5 = - p 2 + 4 p 2 = 1 p = + or - 1 simillarly point ( t , - 7 ) should satisfy the equation . hence x = t and y = - 7 . - 7 = - t 2 + 9 t = + or - 4 considering p = - 2 and t = 4 , the least slope is ( - 7 - 5 ) / ( 4 - 1 ) = - 4 imo option c is correct answer ." | a = 7 - 5
b = 4 + 7
c = math.sqrt(b)
d = 5 + 4
e = math.sqrt(d)
f = c - e
g = a / f
|
a ) $ 258375 , b ) $ 438375 , c ) $ 128375 , d ) $ 108375 , e ) $ 158375 | d | add(multiply(multiply(subtract(1, divide(25, const_100)), subtract(1, divide(25, const_100))), add(multiply(multiply(const_100, const_100), sqrt(const_100)), multiply(multiply(divide(sqrt(const_100), const_2), const_100), const_100))), multiply(multiply(add(25, const_2), const_100), sqrt(const_100))) | the value of a machine depreciates at 25 % per annum . if its present value is $ 1 , 50,000 , at what price should it be sold after two years such that a profit of $ 24,000 is made ? | "the value of the machine after two years = 0.75 * 0.75 * 1 , 50,000 = $ 84,375 sp such that a profit of $ 24,000 is made = 84,375 + 24,000 = $ 1 , 08,375 d" | a = 25 / 100
b = 1 - a
c = 25 / 100
d = 1 - c
e = b * d
f = 100 * 100
g = math.sqrt(100)
h = f * g
i = math.sqrt(100)
j = i / 2
k = j * 100
l = k * 100
m = h + l
n = e * m
o = 25 + 2
p = o * 100
q = math.sqrt(100)
r = p * q
s = n + r
|
a ) 9 : 10 , b ) 18 : 19 , c ) 23 : 27 , d ) 13 : 17 , e ) 15 : 23 | b | sqrt(divide(2268, 2527)) | triangle atriangle b are similar triangles with areas 2268 units square and 2527 units square respectively . the ratio of there corresponding height would be | "let x be the height of triangle a and y be the height of triangle of b . since triangles are similar , ratio of area of a and b is in the ratio of x ^ 2 / y ^ 2 therefore , ( x ^ 2 / y ^ 2 ) = 2268 / 2527 ( x ^ 2 / y ^ 2 ) = ( 18 * 18 * 7 ) / ( 19 * 19 * 7 ) ( x ^ 2 / y ^ 2 ) = 18 ^ 2 / 19 ^ 2 x / y = 18 / 19 ans = b" | a = 2268 / 2527
b = math.sqrt(a)
|
a ) 80 litres , b ) 90 litres , c ) 120 litres , d ) 170 litres , e ) none of these | a | multiply(1200, multiply(800, divide(40, multiply(800, 600)))) | 40 litres of diesel is required to travel 600 km using a 800 cc engine . if the volume of diesel required to cover a distance varies directly as the capacity of the engine , then how many litres of diesel is required to travel 800 km using 1200 cc engine ? | "explanatory answer to cover a distance of 800 kms using a 800 cc engine , the amount of diesel required = 800 / 600 * 40 = 53.33 litres . however , the vehicle uses a 1200 cc engine and the question states that the amount of diesel required varies directly as the engine capacity . i . e . , for instance , if the capacity of engine doubles , the diesel requirement will double too . therefore , with a 1200 cc engine , quantity of diesel required = 1200 / 800 * 53.33 = 80 litres . answer a" | a = 800 * 600
b = 40 / a
c = 800 * b
d = 1200 * c
|
a ) 120 sec , b ) 176 sec , c ) 178 sec , d ) 180 sec , e ) 189 sec | d | divide(600, subtract(multiply(60, const_0_2778), multiply(24, const_0_2778))) | a and b go around a circular track of length 600 m on a cycle at speeds of 24 kmph and 60 kmph . after how much time will they meet for the first time at the starting point ? | "time taken to meet for the first time at the starting point = lcm { length of the track / speed of a , length of the track / speed of b } = lcm { 600 / ( 24 * 5 / 18 ) , 600 / ( 60 * 5 / 18 ) } = 180 sec . answer : d" | a = 60 * const_0_2778
b = 24 * const_0_2778
c = a - b
d = 600 / c
|
a ) 20 years , b ) 21 years , c ) 22 years , d ) 23 years , e ) 25 years | e | divide(subtract(add(28, add(28, 3)), multiply(3, 3)), const_2) | the captain of a cricket team of 11 members is 28 years old and the wicket keeper is 3 years older . if the ages of these two are excluded , the average age of the remaining players is one year less than the average age of the whole team . what is the average age of the team ? | "explanation : let the average age of the whole team by x years . 11 x Γ’ β¬ β ( 28 + 31 ) = 9 ( x - 1 ) 11 x Γ’ β¬ β 9 x = 50 2 x = 50 x = 25 so , average age of the team is 25 years . answer e" | a = 28 + 3
b = 28 + a
c = 3 * 3
d = b - c
e = d / 2
|
a ) 287,000 , b ) 290,000 , c ) 284,000 , d ) 285,000 , e ) 286,000 | c | multiply(150, divide(subtract(2007, 2003), const_2)) | during 2003 , a company produced an average of 2,000 products per month . how many products will the company need to produce from 2004 through 2007 in order to increase its monthly average for the period from 2003 through 2007 by 150 % over its 2003 average ? | company produced 12 * 2000 = 24,000 products in 2003 . if company produces x products from 2004 to 2007 , then total amount of product produced in 4 years ( 2003 through 2007 ) is x + 24,000 . the gives the average of ( x + 24,000 ) / 4 . this average needs to be 300 % higher than that in 2003 . in math terms , 36,000 + 150 % ( 36,000 ) = 80,000 . so : ( x + 36,000 ) / 4 = 80,000 x + 36,000 = 320,000 x = 284,000 the answer is c . | a = 2007 - 2003
b = a / 2
c = 150 * b
|
a ) 280 , b ) 285 , c ) 290 , d ) 295 , e ) 275 | b | divide(add(multiply(add(floor(divide(30, add(const_3, const_4))), const_1), 510), multiply(subtract(30, add(floor(divide(30, add(const_3, const_4))), const_1)), 240)), 30) | a library has an average of 510 visitors on sundays and 240 on other day . the average number of visitors in a month of 30 days starting with sunday is | "explanation : as the month begin with sunday , so there will be five sundays in the month . so result will be : = ( 510 Γ 5 + 240 Γ 25 / 30 ) = ( 8550 / 30 ) = 285 option b" | a = 3 + 4
b = 30 / a
c = math.floor(b)
d = c + 1
e = d * 510
f = 3 + 4
g = 30 / f
h = math.floor(g)
i = h + 1
j = 30 - i
k = j * 240
l = e + k
m = l / 30
|
a ) 100 , b ) 150 , c ) 160 , d ) 225 , e ) 180 | d | multiply(150, subtract(const_2, const_1)) | a train speeds past a pole in 15 sec and a platform 150 m long in 25 sec , its length is ? | "let the length of the train be x m and its speed be y m / sec . then , x / y = 15 = > y = x / 15 ( x + 150 ) / 25 = x / 15 = > x = 225 m . answer : option d" | a = 2 - 1
b = 150 * a
|
a ) 50.78 kg , b ) 49.32 kg , c ) 61.67 kg , d ) 69.15 kg , e ) 54.28 kg | e | divide(add(multiply(40, 50), multiply(30, 60)), add(40, 30)) | there are 2 sections a and b in a class , consisting of 40 and 30 students respectively . if the average weight of section a is 50 kg and that of section b is 60 kg , find the average of the whole class ? | "total weight of 40 + 30 students = 40 * 50 + 30 * 60 = 2000 + 1800 average weight of the class is = 3800 / 70 = 54.28 kg answer is e" | a = 40 * 50
b = 30 * 60
c = a + b
d = 40 + 30
e = c / d
|
a ) $ 2.00 , b ) $ 2.25 , c ) $ 2.50 , d ) $ 2.75 , e ) $ 3.00 | d | divide(1.76, add(multiply(const_0_25, const_2), multiply(const_0_33, const_1))) | a customer purchased a package of ground beef at a cost of $ 1.76 per pound . for the same amount of money , the customer could have purchased a piece of steak that weighed 36 percent less than the package of ground beef . what was the cost per pound of the steak ? | "for simplicity , let ' s assume the customer bought 1 pound of ground beef for $ 1.76 . let x be the price per pound for the steak . then 0.64 x = 176 x = 176 / 0.64 = $ 2.75 the answer is d ." | a = const_0_25 * 2
b = const_0_33 * 1
c = a + b
d = 1 / 76
|
a ) 7 sec , b ) 5 sec , c ) 15 sec , d ) 6 sec , e ) 12 sec | c | multiply(const_3600, divide(divide(275, const_1000), add(60, 6))) | a train 275 meters long is running with a speed of 60 kmph . in what time will it pass a man who is running at 6 kmph in the direction opposite to that in which the train is going ? | "speed of train relative to man = ( 60 + 6 ) km / hr = 66 km / hr [ 66 * 5 / 18 ] m / sec = [ 55 / 3 ] m / sec . time taken to pass the man = [ 275 * 3 / 55 ] sec = 15 sec answer : c" | a = 275 / 1000
b = 60 + 6
c = a / b
d = 3600 * c
|
a ) 0.20833 , b ) 0.14544 , c ) 0.25632 , d ) 0.2139 , e ) 0.63435 | d | multiply(divide(divide(480, divide(60, const_100)), add(multiply(multiply(3, const_100), const_1000), multiply(add(multiply(const_4, const_10), const_4), const_1000))), const_100) | lagaan is levied on the 60 percent of the cultivated land . the revenue department collected total rs . 3 , 74,000 through the lagaan from the village of mutter . mutter , a very rich farmer , paid only rs . 480 as lagaan . the percentage of total land of mutter over the total taxable land of the village is : | "total land of sukhiya = \ inline \ frac { 480 x } { 0.6 } = 800 x \ therefore cultivated land of village = 374000 x \ therefore required percentage = \ inline \ frac { 800 x } { 374000 } \ times 100 = 0.21390 d" | a = 60 / 100
b = 480 / a
c = 3 * 100
d = c * 1000
e = 4 * 10
f = e + 4
g = f * 1000
h = d + g
i = b / h
j = i * 100
|
a ) 10 , b ) 5 , c ) 15 , d ) 20 , e ) 7 | b | subtract(divide(add(56, 14), const_2), divide(add(44, 16), const_2)) | the average ( arithmetic mean ) of the even integers from 16 to 44 inclusive is how much greater than the average ( arithmetic mean ) of the even integers from 14 to 56 inclusive ? | "so , according to a mean of a set of even numbers from 16 to 44 = ( 16 + 44 ) / 2 = 30 and mean of a set of even numbers from 14 to 56 = ( 14 + 56 ) / 2 = 35 difference = 35 - 30 = 5 answer : b ." | a = 56 + 14
b = a / 2
c = 44 + 16
d = c / 2
e = b - d
|
a ) 4 / 15 , b ) 4 / 10 , c ) 4 / 18 , d ) 4 / 13 , e ) 4 / 11 | d | add(divide(const_3, const_52), divide(divide(const_52, const_4), const_52)) | if a card is drawn from a well shuffled pack of cards , the probability of drawing a spade or a king is ? | "p ( s α΄ k ) = p ( s ) + p ( k ) - p ( s β© k ) , where s denotes spade and k denotes king . p ( s α΄ k ) = 13 / 52 + 4 / 52 - 1 / 52 = 4 / 13 answer : d" | a = 3 / const_52
b = const_52 / 4
c = b / const_52
d = a + c
|
a ) s . 130 , b ) s . 140 , c ) s . 150 , d ) s . 228 , e ) s . 282 | d | divide(570, add(divide(150, const_100), const_1)) | two employees a and b are paid a total of rs . 570 per week by their employer . if a is paid 150 percent of the sum paid to b , how much is b paid per week ? | "let the amount paid to a per week = x and the amount paid to b per week = y then x + y = 570 but x = 150 % of y = 150 y / 100 = 15 y / 10 β΄ 15 y / 10 + y = 570 β y [ 15 / 10 + 1 ] = 570 β 25 y / 10 = 570 β 25 y = 5700 β y = 5700 / 25 = rs . 228 d )" | a = 150 / 100
b = a + 1
c = 570 / b
|
a ) 12 , b ) 11 , c ) 10 , d ) 9 , e ) 8 | a | inverse(add(inverse(18), inverse(36))) | a company has two models of computers , model m and model n . operating at a constant rate , a model m computer can complete a certain task in 36 minutes and a model n computer can complete the same task in 18 minutes . if the company used the same number of each model of computer to complete the task in 1 minute , how many model m computers were used ? | "let ' s say 1 work is processing 36 gb of data . model m : 1 gb per min model n : 2 gb per min working together , 1 m and 1 n = 3 gb per min so , 12 times as many computers would work at 36 gb per min . so no . of m = 12 answer is a" | a = 1/(18)
b = 1/(36)
c = a + b
d = 1/(c)
|
a ) $ 14 , b ) $ 16 , c ) $ 20 , d ) $ 24 , e ) $ 25 | a | subtract(70, multiply(divide(const_1, add(divide(25, 70), const_1)), 70)) | a retailer sold an appliance for $ 70 . if the retailer ' s gross profit on the appliance was 25 percent of the retailer ' s cost for the appliance , how many dollars was the retailer ' s gross profit ? | "gross profit = selling price - cost = 70 - x 70 - x = 1 / 4 x x = 56 gp = 70 - 56 = 14" | a = 25 / 70
b = a + 1
c = 1 / b
d = c * 70
e = 70 - d
|
a ) 24 , b ) 12 , c ) 20 , d ) 4 , e ) 2 | c | subtract(200, multiply(60, 3)) | a sporting goods store sold 60 frisbees in one week , some for $ 3 and the rest for $ 4 each . if receipts from frisbee sales for the week totaled $ 200 , what is the fewest number of $ 4 frisbees that could have been sold ? | the wording of the question is slightly misleading . when a question asks what is thefewestthatcouldhave been sold , it is suggesting that there could be many different quantities sold . in this question however , because we are told that exactly 64 frisbees have been sold and revenue was exactly $ 204 , there is only one possible solution for the number of $ 3 and $ 4 frisbees sold . to solve , we have 2 equations and 2 unknowns let x = number of $ 3 frisbees sold let y = number of $ 4 frisbees sold x + y = 60 3 x + 4 y = 200 x = 60 - y 3 ( 60 - y ) + 4 y = 200 180 - 3 y + 4 y = 200 y = 20 answer : c | a = 60 * 3
b = 200 - a
|
a ) 7 , b ) 23 , c ) 12 , d ) 17 , e ) 25 | b | add(16, 7) | there are 16 bees in the hive , then 7 more fly . how many bees are there in all ? | 16 + 7 = 23 . answer is b . | a = 16 + 7
|
a ) $ 2.10 , b ) $ 37.10 , c ) $ 37.16 , d ) $ 38.10 , e ) $ 38.80 | b | add(multiply(divide(6, const_100), 35), 35) | laura took out a charge account at the general store and agreed to pay 6 % simple annual interest . if she charges $ 35 on her account in january , how much will she owe a year later , assuming she does not make any additional charges or payments ? | "principal that is amount taken by laura at year beginning = 35 $ rate of interest = 6 % interest = ( 6 / 100 ) * 35 = 2.10 $ total amount that laura owes a year later = 35 + 2.1 = 37.1 $ answer b" | a = 6 / 100
b = a * 35
c = b + 35
|
a ) 2,000 , b ) 10,500 , c ) 11,750 , d ) 12,000 , e ) 13,000 | a | divide(divide(50, subtract(divide(subtract(divide(const_60, 24), inverse(const_2)), const_60), inverse(multiply(const_60, const_2)))), const_1000) | with both inlets open , a water tank will be filled with water in 24 minutes . the first inlet alone would fill the tank in 2 hours . if in every minutes the second inlet admits 50 cubic meters of water than the first , what is the capacity of the tank ? | the work done by inlet a and b together in 1 min = 1 / 24 the work done by inlet a ( first inlet ) in 1 min = 1 / 120 the work done by inlet b ( second inlet ) in 1 min = ( 1 / 24 ) - ( 1 / 120 ) = 1 / 30 difference of work done by b and a = b - a = 50 cubic meter i . e . ( 1 / 30 ) - ( 1 / 120 ) = 50 cubic meter i . e . tank = 40 * 50 = 2000 cubic meter answer : option a | a = const_60 / 24
b = 1/(2)
c = a - b
d = c / const_60
e = const_60 * 2
f = 1/(e)
g = d - f
h = 50 / g
i = h / 1000
|
a ) 1 day , b ) 2 days , c ) 3 days , d ) 4 days , e ) 5 days | a | divide(36, multiply(divide(48, multiply(4, 2)), 6)) | if 4 men can colour 48 m long cloth in 2 days , then 6 men can colour 36 m long cloth in ? | "the length of cloth painted by one man in one day = 48 / 4 Γ 2 = 6 m no . of days required to paint 36 m cloth by 6 men = 36 / 6 Γ 6 = 1 day . option ' a '" | a = 4 * 2
b = 48 / a
c = b * 6
d = 36 / c
|
a ) 100 cm 2 , b ) 150 cm 2 , c ) 300 cm 2 , d ) 368 cm 2 , e ) 400 cm 2 | d | add(multiply(multiply(divide(const_1, const_2), 23), sqrt(subtract(multiply(multiply(20, 20), const_4), multiply(23, 23)))), 23) | find the area of a rhombus one side of which measures 20 cm and one diagonal is 23 cm . | "explanation : let other diagonal = 2 x cm . since diagonals of a rhombus bisect each other at right angles , we have : ( 20 ) 2 = ( 12 ) 2 + ( x ) 2 = > x = β ( 20 ) 2 β ( 12 ) 2 = β 256 = 16 cm . _ i so , other diagonal = 32 cm . area of rhombus = ( 1 / 2 ) x ( product of diagonals ) = ( 1 / 2 Γ 23 x 32 ) cm 2 = 368 cm 2 answer : option d" | a = 1 / 2
b = a * 23
c = 20 * 20
d = c * 4
e = 23 * 23
f = d - e
g = math.sqrt(f)
h = b * g
i = h + 23
|
a ) rs . 473 , b ) rs . 374 , c ) rs . 495 , d ) rs . 945 , e ) none of the above | d | subtract(multiply(6000, power(add(1, divide(divide(10, 2), const_100)), multiply(add(1, divide(1, 2)), 2))), 6000) | compound interest of rs . 6000 at 10 % per annum for 1 1 / 2 years will be ( interest compounded half yearly ) . | "10 % interest per annum will be 5 % interest half yearly for 3 terms ( 1 1 / 2 years ) so compound interest = 6000 [ 1 + ( 5 / 100 ) ] ^ 3 - 6000 = 6000 [ ( 21 / 20 ) ^ 3 - 1 ] = 6000 ( 9261 - 8000 ) / 8000 = 6 * 1261 / 8 = 945 answer : d" | a = 10 / 2
b = a / 100
c = 1 + b
d = 1 / 2
e = 1 + d
f = e * 2
g = c ** f
h = 6000 * g
i = h - 6000
|
a ) 9 : 2 , b ) 3 : 2 , c ) 18 : 20 , d ) 18 : 4 , e ) 13 : 4 | e | divide(add(multiply(multiply(const_3, const_3), add(multiply(const_3, const_3), const_1)), 2), multiply(2, add(multiply(const_3, const_3), const_1))) | a and b started a business investing rs . 75,000 and rs 20,000 respectively . in what ratio the profit earned after 2 years be divided between a and b respectively ? | "a : b = 75000 : 20000 = 75 : 20 = 13 : 4 answer : e" | a = 3 * 3
b = 3 * 3
c = b + 1
d = a * c
e = d + 2
f = 3 * 3
g = f + 1
h = 2 * g
i = e / h
|
a ) 112.50 , b ) 125 , c ) 150 , d ) 167.5 , e ) none of these | a | multiply(5000, divide(const_1, const_100)) | a person borrows 5000 for 2 years at 4 % p . a . simple interest . he immediately lends it to another person at 61 β 4 % p . a . for 2 years . find his gain in the transaction per year . | "gain in 2 years = [ ( 5000 Γ 25 / 4 Γ 2 / 100 ) β ( 5000 Γ 4 Γ 2 / 100 ) ] = ( 625 β 400 ) = 225 . β΄ gain in 1 year = ( 225 β 2 ) = 112.50 answer a" | a = 1 / 100
b = 5000 * a
|
a ) 714.28 , b ) 712 , c ) 714.88 , d ) 714.25 , e ) 714 | a | divide(multiply(500, const_100), subtract(const_100, 30)) | a side of beef lost 30 percent of its weight in processing . if the side of beef weighed 500 pounds after processing , how many pounds did it weigh before processing ? | "let weight of side of beef before processing = x ( 70 / 100 ) * x = 500 = > x = ( 500 * 100 ) / 70 = 714.28 answer a" | a = 500 * 100
b = 100 - 30
c = a / b
|
a ) 4 % increase , b ) 12 % increase , c ) 10 % decrease , d ) 6 % increase , e ) none of these | b | subtract(divide(multiply(subtract(const_100, 20), add(const_100, 40)), const_100), const_100) | if the price of a tv is first decreased by 20 % and then increased by 40 % , then the net change in the price will be : | "explanation : solution : let the original price be rs . 100 . new final price = 140 % of ( 80 % of 100 ) = rs . 140 / 100 * 80 / 100 * 100 = rs . 112 . . ' . increase = 12 % answer : b" | a = 100 - 20
b = 100 + 40
c = a * b
d = c / 100
e = d - 100
|
a ) 4 , b ) 7 , c ) 8 , d ) 12 , e ) it can not be determined from the information given . | d | add(subtract(add(const_3, const_4), const_1), add(subtract(add(const_3, const_2), 3), subtract(add(const_2, const_4), subtract(add(const_3, const_2), 3)))) | two sets of 3 consecutive positive odd integers have exactly one integer in common . the sum of the integers in the set with greater numbers is how much greater than the sum of the integers in the other set ? | a = ( 1 , 3,5 ) , sum of this = 9 b = ( 5 , 7,9 ) , sum of this = 21 , the differenct between 21 - 9 = 12 hence , 12 is the answer i . e . d | a = 3 + 4
b = a - 1
c = 3 + 2
d = c - 3
e = 2 + 4
f = 3 + 2
g = f - 3
h = e - g
i = d + h
j = b + i
|
a ) 8 hrs , b ) 10 hrs , c ) 12 hrs , d ) 16 hrs , e ) 6 hrs | d | multiply(divide(1, 5), 20) | a train running at 1 / 5 of its own speed reached a place in 20 hours . how much time could be saved if the train would have run at its own speed ? | "time taken if run its own speed = 1 / 5 * 20 = 4 hrs time saved = 20 - 4 = 16 hrs answer : d" | a = 1 / 5
b = a * 20
|
a ) 400 , b ) 1,500 , c ) 1,250 , d ) 2,500 , e ) 10,000 | b | divide(60, divide(2, 50)) | in a certain pond , 60 fish were caught , tagged , and returned to the pond . a few days later , 50 fish were caught again , of which 2 were found to have been tagged . if the percent of tagged fish in the second catch approximates the percent of tagged fish in the pond , what is the approximate number of fish in the pond ? | "this is a rather straight forward ratio problem . 1 . 60 fish tagged 2 . 2 out of the 50 fish caught were tagged thus 2 / 50 2 / 50 = 60 / x thus , x = 1500 think of the analogy : 2 fish is to 50 fish as 50 fish is to . . . ? you ' ve tagged 50 fish and you need to find what that comprises as a percentage of the total fish population - we have that information with the ratio of the second catch . b" | a = 2 / 50
b = 60 / a
|
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