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 |
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a ) 13 : 00 , b ) 11 : 15 , c ) 14 : 00 , d ) 15 : 00 , e ) 15 : 30 | b | add(divide(add(600, 50), subtract(500, divide(600, const_2))), 08) | sari and ken climb up a mountain . at night , they camp together . on the day they are supposed to reach the summit , sari wakes up at 06 : 00 and starts climbing at a constant pace . ken starts climbing only at 08 : 00 , when sari is already 600 meters ahead of him . nevertheless , ken climbs at a constant pace of 500 meters per hour , and reaches the summit before sari . if sari is 50 meters behind ken when he reaches the summit , at what time did ken reach the summit ? | "both sari and ken climb in the same direction . speed of sari = 600 / 2 = 300 meters / hr ( since she covers 700 meters in 2 hrs ) speed of ken = 500 meters / hr at 8 : 00 , distance between ken and sari is 600 meters . ken needs to cover this and another 50 meters . time he will take = total distance to be covered / relative speed = ( 600 + 50 ) / ( 500 - 300 ) = 3.25 hrs starting from 8 : 00 , in 5 hrs , the time will be 11 : 15 answer ( b )" | a = 600 + 50
b = 600 / 2
c = 500 - b
d = a / c
e = d + 8
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a ) 6 % , b ) 25 % , c ) 37 1 / 2 % , d ) 60 % , e ) 75 % | b | multiply(add(divide(const_1, 50), divide(const_1, 5)), const_100) | if x > 0 , x / 50 + x / 5 is what percent of x ? | "just plug and chug . since the question asks for percents , pick 100 . ( but any number will do . ) 100 / 50 + 100 / 5 = 2 + 20 = 25 25 is 25 % of 100 = b" | a = 1 / 50
b = 1 / 5
c = a + b
d = c * 100
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a ) 8 sec , b ) 3 sec , c ) 2 sec , d ) 6 sec , e ) 9 sec | b | divide(60, multiply(72, const_0_2778)) | in what time will a railway train 60 m long moving at the rate of 72 kmph pass a telegraph post on its way ? | "t = 60 / 72 * 18 / 5 = 3 sec answer : b" | a = 72 * const_0_2778
b = 60 / a
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a ) 5 days , b ) 6 days , c ) 7.5 days , d ) 8.5 days , e ) 9 days | e | divide(const_1, add(divide(const_1, 18), divide(const_1, 10))) | a can do a piece of work in 18 days and b alone can do it in 10 days . b works at it for 5 days and then leaves . a alone can finish the remaining work in | "explanation : b ' s 5 days work = 1 / 10 β 5 = 1 / 2 remaining work = 1 β 1 / 2 = 1 / 2 a can finish work = 18 β 1 / 2 = 9 days option e" | a = 1 / 18
b = 1 / 10
c = a + b
d = 1 / c
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a ) 20 : 9 , b ) 2 : 1 , c ) 3 : 2 , d ) 2 : 3 , e ) 5 : 3 | a | divide(multiply(45000, const_12), multiply(36000, add(const_4, const_3))) | x starts a business with rs . 45000 . y joins in the business after 3 months with rs . 36000 . what will be the ratio in which they should share the profit at the end of the year ? | "ratio in which they should share the profit = ratio of the investments multiplied by the time period = 45000 Γ 12 : 36000 Γ 9 = 45 Γ 12 : 36 Γ 9 = 15 Γ 12 : 9 Γ 9 = 20 : 9 answer is a ." | a = 45000 * 12
b = 4 + 3
c = 36000 * b
d = a / c
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a ) 71.11 , b ) 71.12 , c ) 71.1 , d ) 71.17 , e ) 77.42 | e | multiply(320, divide(const_1, add(divide(160, 75), divide(160, 80)))) | a car travels first 160 km at 75 km / hr and the next 160 km at 80 km / hr . what is the average speed for the first 320 km of the tour ? | car travels first 160 km at 75 km / hr time taken to travel first 160 km = distancespeed = 160 / 75 car travels next 160 km at 80 km / hr time taken to travel next 160 km = distancespeed = 160 / 80 total distance traveled = 160 + 160 = 2 Γ 160 total time taken = 160 / 75 + 160 / 80 average speed = total distance traveled / total time taken = 320 / ( 160 / 75 + 160 / 80 ) = 77.42 km / hr answer : e | a = 160 / 75
b = 160 / 80
c = a + b
d = 1 / c
e = 320 * d
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a ) 16 , b ) 0 , c ) 4 , d ) 2 , e ) 1 | b | add(subtract(add(subtract(const_1, const_1), 8), add(subtract(const_1, const_1), 8)), const_1) | if a β 8 = b , then determine the value of | a - b | - | b - a | . | sol . ( b ) | a - b | = | 8 | = 8 = > | b - a | = | - 8 | = 8 = > | a - b | - | b - a | = 8 - 8 = 0 . answer b | a = 1 - 1
b = a + 8
c = 1 - 1
d = c + 8
e = b - d
f = e + 1
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a ) 4500 , b ) 7500 , c ) 5000 , d ) 7200 , e ) none of these | d | divide(18, subtract(power(add(const_1, divide(5, const_100)), 2), add(const_1, multiply(2, divide(5, const_100))))) | the difference between compound interest and simple interest on a certain amount of money at 5 % per annum for 2 years is 18 . find the sum : | "sol . ( d ) let the sum be 100 . therefore , si = 100 Γ 5 Γ 2100 = 10100 Γ 5 Γ 2100 = 10 and ci = 100 ( 1 + 5100 ) 2 β 100100 ( 1 + 5100 ) 2 β 100 β΄ = 100 Γ 21 Γ 2120 Γ 20 β 100 = 414 = 100 Γ 21 Γ 2120 Γ 20 β 100 = 414 difference of ci and si = 41 β 4 - 10 = 1 β 4 if the difference is 1 β 4 , the sum = 100 = > if the difference is 18 , the sum = 400 Γ 18 = 7200 answer d" | a = 5 / 100
b = 1 + a
c = b ** 2
d = 5 / 100
e = 2 * d
f = 1 + e
g = c - f
h = 18 / g
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a ) 10 , b ) 20 , c ) 21 , d ) 25 , e ) 27 | c | multiply(7, divide(multiply(add(7, 9), subtract(9, multiply(divide(5, add(7, 5)), 9))), subtract(multiply(9, 7), multiply(7, 5)))) | a can contains a mixture of liquids a and b is the ratio 7 : 5 . when 9 litres of mixture are drawn off and the can is filled with b , the ratio of a and b becomes 7 : 9 . how many liter w of liquid a was contained by the can initially ? | "as a : b : : 7 : 5 - - - > only option c is a multiple of 7 and hence it is a good place to start . also a : b : : 7 : 5 means that , a = ( 712 ) * total and b = ( 5 / 12 ) * total if a = 21 , b = 15 - - - > remove 9 litres - - - > you remove ( 7 / 12 ) * 9 of a - - - > a remaining = 21 - ( 7 / 12 ) * 9 = 63 / 4 similarly , for b , you remove ( 5 / 12 ) * 9 - - - > b remaining = 15 - ( 5 / 12 ) * 9 = 45 / 4 and then add 9 more litres of b - - - > 9 + 45 / 4 = 81 / 4 thus a / b ( final ratio ) = ( 45 / 4 ) / ( 81 / 4 ) = 7 : 9 , the same as the final ratio mentioned in the question . hence c is the correct answer . a / b = 7 / 9 = ( 7 x - ( 7 / 12 ) * 9 ) / ( 5 x - ( 5 / 12 ) * 9 + 9 ) , where 7 x and 5 x are initial quantities of a and b respectively . thus , 7 / 9 = ( 7 x - ( 7 / 12 ) * 9 ) / ( 5 x - ( 5 / 12 ) * 9 + 9 ) - - - > giving you x = 3 . thus a ( original ) w = 7 * 3 = 21 . c" | a = 7 + 9
b = 7 + 5
c = 5 / b
d = c * 9
e = 9 - d
f = a * e
g = 9 * 7
h = 7 * 5
i = g - h
j = f / i
k = 7 * j
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a ) 38 sec , b ) 35 sec , c ) 43.2 sec , d ) 40 sec , e ) none of these | c | multiply(divide(add(divide(180, const_1000), divide(360, const_1000)), 45), const_3600) | a train 360 m long runs with a speed of 45 km / hr . what time will it take to pass a platform of 180 m long ? | "speed = 45 km / hr = 45 Γ£ β ( 5 / 18 ) m / s = 150 / 12 = 50 / 4 = 25 / 2 m / s total distance = length of the train + length of the platform = 360 + 180 = 540 meter time taken to cross the platform = 540 / ( 25 / 2 ) = 540 Γ£ β 2 / 25 = 43.2 seconds answer : c" | a = 180 / 1000
b = 360 / 1000
c = a + b
d = c / 45
e = d * 3600
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a ) 11 , b ) 5 , c ) 10 , d ) 15 , e ) 20 | a | divide(22, const_2) | if the length of the longest chord of a certain circle is 22 , what is the radius of that certain circle ? | "longest chord of a circle is the diameter of the circle diameter = 2 * radius if diameter of the circle is given as 22 = 2 * 11 so radius of the circle = 11 correct answer - a" | a = 22 / 2
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a ) $ 7,250 , b ) $ 8,500 , c ) $ 8,100 , d ) $ 8,250 , e ) $ 8,600 | c | floor(divide(subtract(subtract(multiply(500, 20.00), multiply(5.00, 100)), multiply(subtract(500, 100), 3.50)), const_1000)) | company c produces toy trucks at a cost of $ 5.00 each for the first 100 trucks and $ 3.50 for each additional truck . if 500 toy trucks were produced by company c and sold for $ 20.00 each , what was company c β s gross profit ? | "cost of 500 trucks : ( 100 * 5 ) + ( 400 * 3.5 ) = 500 + 1400 = $ 1900 revenue : 500 * 20 = $ 10000 profit : 10000 - 1900 = $ 8100 option c is correct" | a = 500 * 20
b = 5 * 0
c = a - b
d = 500 - 100
e = d * 3
f = c - e
g = f / 1000
h = math.floor(g)
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a ) 18.7 , b ) 25.5 , c ) 33.3 , d ) 41.2 , e ) 54.4 | d | multiply(divide(70, add(70, const_100)), const_100) | if y is 70 % greater than x , than x is what % less than y ? | y = 1.7 x x = y / 1.7 = 10 y / 17 x is 7 / 17 less which is about 41.2 % less than y . the answer is d . | a = 70 + 100
b = 70 / a
c = b * 100
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a ) 5 / 21 , b ) 3 / 7 , c ) 4 / 7 , d ) 5 / 7 , e ) 13 / 21 | e | add(multiply(divide(3, 7), divide(subtract(7, 2), subtract(7, const_1))), multiply(divide(3, 7), divide(subtract(7, 3), subtract(7, const_1)))) | in a room filled with 7 people , 3 people have exactly 1 sibling in the room and 4 people have exactly 2 siblings in the room . if two individuals are selected from the room at random , what is the probability that those two individuals are not siblings ? | there are suppose a b c d e f g members in the room 4 people who have exactly one sibling . . . . a b c d . . . . ( a is bs β sssibl β g β ssand β ssviceversa ) β ss ( c β ssis β ssds β sssibl β g β ssand β ssviceversa ) β ss ( c β ssis β ssdssibl β gandviceversa ) ( cisds sibling and viceversa ) ( c is ds sibling and viceversa ) . . . now remaning efg are 4 people who have exactly 2 siblings . . . . ( e has f and g as his / her sibling and so on . . ) there are now 3 different set of siblings ( a and b ) ( c and d ) ; ( efg ) now first selecting 2 people out of 7 is 7 c 2 = 21 first sibling pair - - - - ( a and b ) - - selecting 2 people - - 2 c 2 = 1 second sibling pair ( c and d ) - - selecting 2 people - - 2 c 2 = 1 third sibling pair ( e f g ) - - selecting 2 out of 4 - - 4 c 2 = 6 total = 1 + 1 + 6 = 8 but , a / c to formula p ( success ) - 1 - p ( fail ) here , p ( failure ) is selecting 2 people who are siblings = 8 / 21 ( 21 is 7 c 2 ) = 1 - 8 / 21 = 13 / 21 ans e | a = 3 / 7
b = 7 - 2
c = 7 - 1
d = b / c
e = a * d
f = 3 / 7
g = 7 - 3
h = 7 - 1
i = g / h
j = f * i
k = e + j
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a ) 16000 , b ) 22500 , c ) 15000 , d ) 18000 , e ) 17000 | b | multiply(divide(subtract(27000, divide(multiply(27000, 10), const_100)), add(const_100, 8)), const_100) | mohit sold an article for $ 27000 . had he offered a discount of 10 % on the selling price , he would have earned a profit of 8 % . what is the cost price of the article ? | "let the cp be $ x . had he offered 10 % discount , profit = 8 % profit = 8 / 100 x and hence his sp = x + 8 / 100 x = $ 1.08 x = 27000 - 10 / 100 ( 27000 ) = 27000 - 2700 = $ 22500 = > 1.08 x = 22500 = > x = 22500 b" | a = 27000 * 10
b = a / 100
c = 27000 - b
d = 100 + 8
e = c / d
f = e * 100
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a ) 312 , b ) 490 , c ) 720 , d ) 435 , e ) 679 | a | multiply(divide(add(40, multiply(2, 85)), subtract(260, 85)), 260) | if a farmer wants to plough a farm field on time , he must plough 260 hectares a day . for technical reasons he ploughed only 85 hectares a day , hence he had to plough 2 more days than he planned and he still has 40 hectares left . what is the area of the farm field and how many days the farmer planned to work initially ? | "let x be the number of days in the initial plan . therefore , the whole field is 260 Γ’ βΉ β¦ x hectares . the farmer had to work for x + 2 days , and he ploughed 85 ( x + 2 ) hectares , leaving 40 hectares unploughed . then we have the equation : 260 x = 85 ( x + 2 ) + 40 175 x = 210 x = 1.2 so the farmer planned to have the work done in 6 days , and the area of the farm field is 260 ( 1.2 ) = 312 hectares correct answer a" | a = 2 * 85
b = 40 + a
c = 260 - 85
d = b / c
e = d * 260
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a ) 150 cm , b ) 160 cm , c ) 170 cm , d ) 140 cm , e ) 150 cm | a | divide(159, add(const_1, divide(6, const_100))) | on my sister ' s 15 th birthday , she was 159 cm in height , having grown 6 % since the year before . how tall was she the previous year ? | explanation : given that height on 15 th birthday = 159 cm and growth = 6 % let the previous year height = x then height on 15 th birthday = x Γ ( 100 + 6 ) / 100 = x Γ 106 / 100 β 159 = x Γ 106 / 100 β x = 159 Γ 100 / 106 = 1.5 Γ 100 = 150 cm answer : option a | a = 6 / 100
b = 1 + a
c = 159 / b
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a ) $ 3.15 , b ) $ 4.45 , c ) $ 4.80 , d ) $ 5.50 , e ) $ 5.40 | d | add(multiply(divide(3.6, divide(2, 5)), 0.35), 2.35) | jim Γ’ β¬ β’ s taxi service charges an initial fee of $ 2.35 at the beginning of a trip and an additional charge of $ 0.35 for each 2 / 5 of a mile traveled . what is the total charge for a trip of 3.6 miles ? | "let the fixed charge of jim Γ’ β¬ β’ s taxi service = 2.35 $ and charge per 2 / 5 mile ( . 4 mile ) = . 35 $ total charge for a trip of 3.6 miles = 2.35 + ( 3.6 / . 4 ) * . 35 = 2.35 + 9 * . 35 = 5.5 $ answer d" | a = 2 / 5
b = 3 / 6
c = b * 0
d = c + 2
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a ) $ 180,000 , b ) $ 400,000 , c ) $ 220,000 , d ) $ 402,000 , e ) $ 2 , 200,000 | b | multiply(multiply(400, const_100), const_10) | a special municipal payroll tax charges not tax on a payroll less than $ 200,000 and only 0.2 % on a company β s payroll above $ 200,000 . if belfried industries paid $ 400 in this special municipal payroll tax , then they must have had a payroll of | "answer : b , ( with different approach ) : the 400 paid is 0.2 % of the additional amount above 200,000 . let it be x now 0.2 % of x = 400 therefore x = 200,000 total = 200,000 + x = 400,000" | a = 400 * 100
b = a * 10
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a ) 9.6 , b ) 10.2 , c ) 9.2 , d ) 10.4 , e ) 9.0 | c | multiply(add(20, 3), divide(24, const_60)) | the speed of a boat in still water in 20 km / hr and the rate of current is 3 km / hr . the distance travelled downstream in 24 minutes is : | "speed downstream = ( 20 + 3 ) = 23 kmph time = 24 minutes = 24 / 60 hour = 2 / 5 hour distance travelled = time Γ speed = 2 / 5 Γ 23 = 9.20 km answer is c ." | a = 20 + 3
b = 24 / const_60
c = a * b
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a ) 4 , b ) 3 , c ) 5 , d ) 6 , e ) 2 | c | add(add(const_4, 3), subtract(3, const_4)) | what is the sum of all possible solutions to | x - 3 | ^ 2 + | x - 3 | = 42 ? | "denote | x - 3 | as y : y ^ 2 + y = 42 - - > y = - 7 or y = 6 . discard the first solution since y = | x - 3 | , so it ' s an absolute value and thus can not be negative . y = | x - 3 | = 6 - - > x = 9 or x = - 4 . the sum = 5 . answer : c ." | a = 4 + 3
b = 3 - 4
c = a + b
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a ) 1 , b ) 0 , c ) 3 , d ) 2 , e ) 4 | c | subtract(multiply(multiply(1225, 1227), 1229), subtract(multiply(multiply(1225, 1227), 1229), const_3)) | what is remainder of the division ( 1225 * 1227 * 1229 ) / 12 ? | "remainder will be number / 100 here as the divisor is two digit number = 12 . hence checking for the last two digits = 5 * 7 * 9 = 15 thus remainder = 3 . answer : c" | a = 1225 * 1227
b = a * 1229
c = 1225 * 1227
d = c * 1229
e = d - 3
f = b - e
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a ) 36 , b ) 33 , c ) 30 , d ) 27 , e ) 24 | a | add(add(power(add(add(divide(subtract(subtract(3, const_10), const_2), const_4), const_2), const_2), const_2), power(add(add(add(divide(subtract(subtract(3, const_10), const_2), const_4), const_2), const_2), const_2), const_2)), add(power(divide(subtract(subtract(3, const_10), const_2), const_4), const_2), power(add(divide(subtract(subtract(3, const_10), const_2), const_4), const_2), const_2))) | the sum of three consecutive multiples of 3 is 99 . what is the largest number ? | "let the numbers be 3 x , 3 x + 3 and 3 x + 6 . then , 3 x + ( 3 x + 3 ) + ( 3 x + 6 ) = 99 9 x = 90 x = 10 largest number = 3 x + 6 = 36 answer : a" | a = 3 - 10
b = a - 2
c = b / 4
d = c + 2
e = d + 2
f = e ** 2
g = 3 - 10
h = g - 2
i = h / 4
j = i + 2
k = j + 2
l = k + 2
m = l ** 2
n = f + m
o = 3 - 10
p = o - 2
q = p / 4
r = q ** 2
s = 3 - 10
t = s - 2
u = t / 4
v = u + 2
w = v ** 2
x = r + w
y = n + x
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a ) 8 , b ) 10 , c ) 12 , d ) 14 , e ) 16 | a | add(add(add(2, 2), 2), const_2) | what number should replace the question mark ? 2 , 16 , 4 , 14 , 6 , 12 , - - - ? | answer : a 2 , 16 , 4 , 14 , 6 , 12 , 8 ? there are two alternate sequences : + 2 and - 2 . | a = 2 + 2
b = a + 2
c = b + 2
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a ) 65 days , b ) 45 days , c ) 10 days , d ) 16 days , e ) 18 days | c | divide(multiply(subtract(31, 28), 400), 280) | a garrison of 400 men had a provision for 31 days . after 28 days 280 persons re - enforcement leave the garrison . find the number of days for which the remaining ration will be sufficient ? | "400 - - - 31 400 - - - 3 120 - - - ? 400 * 3 = 120 * x = > x = 10 days . answer : c" | a = 31 - 28
b = a * 400
c = b / 280
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a ) 14 , b ) 17 , c ) 11 , d ) 19 , e ) 26 | e | subtract(divide(62, const_2), 5) | a father said his son , ` ` i was as old as you are at present at the time of your birth . ` ` if the father age is 62 now , the son age 5 years back was | "let the son ' s present age be x years . then , ( 62 - x ) = x x = 31 . son ' s age 5 years back = ( 31 - 5 ) = 26 years answer : e" | a = 62 / 2
b = a - 5
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a ) 20 , b ) 18 , c ) 19 , d ) 15 , e ) 30 | e | multiply(divide(add(add(floor(divide(7, 6)), const_1), floor(divide(49, 6))), const_2), 6) | calculate the average of all the numbers between 7 and 49 which are divisible by 6 . | "explanation : numbers divisible by 6 are 12 , 18,24 , 30,36 , 42,48 , average = ( 12 + 18 + 24 + 30 + 36 + 42 + 48 , ) / 7 = 210 / 7 = 30 answer : e" | a = 7 / 6
b = math.floor(a)
c = b + 1
d = 49 / 6
e = math.floor(d)
f = c + e
g = f / 2
h = g * 6
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a ) $ 1360 , b ) $ 1250 , c ) $ 1260 , d ) $ 1270 , e ) $ 1280 | c | add(add(add(multiply(multiply(const_3, 35), const_2), multiply(const_2, 35)), add(add(multiply(multiply(const_3, 35), const_2), multiply(const_2, 35)), 35)), subtract(multiply(add(add(multiply(multiply(const_3, 35), const_2), multiply(const_2, 35)), 35), const_3), add(multiply(multiply(const_3, 35), const_2), multiply(const_2, 35)))) | a certain sum of money is divided among a , b and c such that a gets one - third of what b and c together get and b gets two - seventh of what a and c together get . if the amount received by a is $ 35 more than that received by b , find the total amount shared by a , b and c . | "a = 1 / 3 ( b + c ) = > c = 3 a - b - - - ( 1 ) b = 2 / 7 ( a + c ) = > c = 3.5 b - a - - ( b ) a - b = $ 35 a = 35 + b ( 1 ) = = = > c = 105 + 3 b - b = 2 b + 105 = = > 2 b - c = - 105 - - - ( 3 ) ( 2 ) = = = > c = 3.5 b - b - 35 = 2.5 b - 35 = = > 2.5 b - c = 35 - - - ( 4 ) from ( 4 ) and ( 3 ) 0.5 b = 140 b = $ 280 a = $ 315 c = 945 - 280 = $ 665 total amount = 315 + 280 + 665 = $ 1260 answer : c" | a = 3 * 35
b = a * 2
c = 2 * 35
d = b + c
e = 3 * 35
f = e * 2
g = 2 * 35
h = f + g
i = h + 35
j = d + i
k = 3 * 35
l = k * 2
m = 2 * 35
n = l + m
o = n + 35
p = o * 3
q = 3 * 35
r = q * 2
s = 2 * 35
t = r + s
u = p - t
v = j + u
|
a ) 45 kmph , b ) 25 kmph , c ) 30 kmph , d ) 50 kmph , e ) none of these | d | divide(divide(subtract(125, multiply(multiply(5, const_0_2778), 5)), 5), const_0_2778) | a train 125 m long passes a man , running at 5 km / hr in the same direction in which the train is going , in 10 seconds . the speed of the train is ? | "explanation : speed of the train relative to man = ( 125 / 10 ) m / sec = ( 25 / 2 ) m / sec . [ ( 25 / 2 ) x ( 18 / 5 ) ] km / hr = 45 km / hr . let the speed of the train be ' s ' km / hr . then , relative speed = ( s - 5 ) km / hr . s - 5 = 45 = > s = 50 km / hr . answer is d" | a = 5 * const_0_2778
b = a * 5
c = 125 - b
d = c / 5
e = d / const_0_2778
|
a ) 25 , b ) 52 , c ) 58 , d ) 60 , e ) 62 | a | divide(multiply(21, const_100), subtract(const_100, 16)) | the number which exceeds 16 % of it by 21 is : | "solution solution let the number be x . x - 16 % of x = 21 x - 16 / 100 x = 21 x - 4 / 25 x = 21 21 / 25 x = 21 x = ( 21 x 25 / 21 ) = 25 answer a" | a = 21 * 100
b = 100 - 16
c = a / b
|
a ) 10 min , b ) 15 min , c ) 12 min , d ) 8 min , e ) 16 min | a | multiply(const_60, divide(subtract(60, 50), 60)) | excluding stoppages , the speed of a bus is 60 kmph and including stoppages , it is 50 kmph . for how many minutes does the bus stop per hour ? | "due to stoppages , it covers 10 km less . time taken to cover 10 km = ( 10 / 60 x 60 ) min = 10 min answer : a" | a = 60 - 50
b = a / 60
c = const_60 * b
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a ) 600 , b ) 700 , c ) 800 , d ) 960 , e ) none of these | d | multiply(4960, divide(6, add(add(add(10, add(4, const_1)), 10), 6))) | if 4 ( p ' s capital ) = 6 ( q ' s capital ) = 10 ( r ' s capital ) , then out of the total profit of rs 4960 , r will receive | "explanation : let p ' s capital = p , q ' s capital = q and r ' s capital = r then 4 p = 6 q = 10 r = > 2 p = 3 q = 5 r = > q = 2 p / 3 r = 2 p / 5 p : q : r = p : 2 p / 3 : 2 p / 5 = 15 : 10 : 6 r ' s share = 4960 * ( 6 / 31 ) = 160 * 6 = 960 . answer : option d" | a = 4 + 1
b = 10 + a
c = b + 10
d = c + 6
e = 6 / d
f = 4960 * e
|
['a ) 8 cm', 'b ) 9 cm', 'c ) 10 cm', 'd ) 11 cm', 'e ) 7 cm'] | a | add(divide(add(subtract(power(13, const_2), power(divide(10, const_2), const_2)), power(divide(10, const_2), const_2)), multiply(sqrt(subtract(power(13, const_2), power(divide(10, const_2), const_2))), const_2)), const_1) | an isosceles triangle those sides are 13 cm , 13 cm , 10 cm long inscribed in a circle . find the radius of the circle | a , b , c are sides of triangle , ad is the median , ad ^ 2 = ab ^ 2 - bd ^ 2 then ad = 12 , the centriod median divides into 2 : 1 ratio so 12 will be divide into 2 : 1 ratio so 8 is radius . so ans is 8 cm answer : a | a = 13 ** 2
b = 10 / 2
c = b ** 2
d = a - c
e = 10 / 2
f = e ** 2
g = d + f
h = 13 ** 2
i = 10 / 2
j = i ** 2
k = h - j
l = math.sqrt(k)
m = l * 2
n = g / m
o = n + 1
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a ) 5 , b ) 6 , c ) 1 , d ) 2 , e ) 3 | b | divide(divide(const_12, const_2), const_1) | n is an 80 - digit positive integer ( in the decimal scale ) . all digits except the 44 th digit ( from the left ) are 2 . if n is divisible by 13 , find the 26 th digit ? | detailed solution to begin with , the question should read ` ` find the 44 th digit ' ' . any number of the form abcabc is a multiple of 1001 . 1001 is 7 * 11 * 13 . so , any number of the form abcabc is a multiple of 13 . so , a number comprising 42 2 ' s would be a multiple of 13 , so would a number comprising 36 2 ' s . so , in effect , we are left with a two digit number 2 a , where a is the 44 th digit . 26 is a multiple of 13 , so the 44 th digit should be 6 . correct answer b | a = 12 / 2
b = a / 1
|
a ) 160 , b ) 170 , c ) 180 , d ) 190 , e ) 240 | e | divide(multiply(60, multiply(40, const_2)), 20) | 60 boys can complete a work in 40 days . how many men need to complete twice the work in 20 days | "one man can complete the work in 40 * 60 = 2400 days = one time work to complete the work twice it will be completed in let m be the no . of worker assign for this therefore the eqn becomes m * 20 = 2 * 2400 m = 240 workers answer : e" | a = 40 * 2
b = 60 * a
c = b / 20
|
a ) 8600 litres , b ) 200 litres , c ) 12800 litres , d ) 11520 litres , e ) 13200 litres | d | multiply(divide(multiply(multiply(8, const_60), add(8, 4)), 4), 8) | an outlet pipe empties a tank which is full in 8 hours . if the inlet pipe is kept open , which lets water in at the rate of 8 litres / min then outlet pipe would take 4 hours longer . find the capacity of the tank . | "let the rate of outlet pipe be x liters / hour ; rate of inlet pipe is 8 litres / min , or 8 * 60 = 480 liters / hour ; net outflow rate when both pipes operate would be x - 480 liters / hour . capacity of the tank = x * 8 hours = ( x - 480 ) * ( 8 + 4 ) hours 8 x = ( x - 480 ) * 12 - - > x = 1440 - - > capacity = 8 x = 11520 liters . answer : d ." | a = 8 * const_60
b = 8 + 4
c = a * b
d = c / 4
e = d * 8
|
a ) 4 : 3 , b ) 4 : 7 , c ) 4 : 9 , d ) 4 : 2 , e ) 4 : 1 | a | divide(subtract(9.2, 8), subtract(8, 7.1)) | how should we mix two rice packets to get one 1 kg packet priced at rs 8 when the prices of 1 kg packets of the two types are rs 7.1 and rs 9.2 . | explanation : no explanation is available for this question ! answer : a | a = 9 - 2
b = 8 - 7
c = a / b
|
a ) 65 : 69 , b ) 56 : 45 , c ) 89 : 71 , d ) 68 : 23 , e ) 63 : 86 | e | divide(multiply(35, 9), multiply(43, 10)) | car a runs at the speed of 35 km / hr & reaches its destination in 9 hr . car b runs at the speed of 43 km / h & reaches its destination in 10 h . what is the respective ratio of distances covered by car a & car b ? | "sol . distance travelled by car a = 35 Γ 9 = 315 km distance travelled by car b = 43 Γ 10 = 430 km ratio = 315 / 430 = 63 : 86 e" | a = 35 * 9
b = 43 * 10
c = a / b
|
a ) 2777 , b ) 2250 , c ) 2789 , d ) 2776 , e ) 2881 | b | divide(multiply(multiply(multiply(const_2, multiply(const_4, add(const_2, const_3))), const_100), subtract(const_1, divide(85, const_100))), add(subtract(const_1, divide(95, const_100)), subtract(const_1, divide(85, const_100)))) | a and b β s salaries together amount to rs . 3,000 . a spends 95 % of his salary and b spends 85 % of his . if now their savings are the same , what is a β s salary ? | "( 5 / 100 ) a = ( 15 / 100 ) b a = 3 b a + b = 3000 4 b = 3000 = > b = 750 a = 2250 answer b" | a = 2 + 3
b = 4 * a
c = 2 * b
d = c * 100
e = 85 / 100
f = 1 - e
g = d * f
h = 95 / 100
i = 1 - h
j = 85 / 100
k = 1 - j
l = i + k
m = g / l
|
a ) 70 kg , b ) 72 kg , c ) 75 kg , d ) 80 kg , e ) none of these | c | subtract(multiply(3, 84), subtract(multiply(const_4, 79), add(subtract(multiply(const_4, 80), multiply(3, 84)), add(subtract(multiply(const_4, 80), multiply(3, 84)), 3)))) | the average weight of 3 packets a , b and c is 84 kg . another packet d added into the group and the average now becomes 80 kg . if another packet e , whose weight is 3 kg more than that of d , replaces a then the average weight of b , c , d and e becomes 79 kg . the weight of a is : | explanation : a + b + c = ( 84 x 3 ) = 252 kg , a + b + c + d = ( 80 x 4 ) = 320 kg . d = ( 320 - 252 ) = 68 , e = ( 68 + 3 ) = 7 l . b + c + d + e = ( 79 x 4 ) = 316 . ( a + b + c + d ) - ( b + c + d + e ) = ( 320 - 316 ) kg = 4 kg . a - e = 4 = a = ( 4 + e ) = 75 kg answer : c | a = 3 * 84
b = 4 * 79
c = 4 * 80
d = 3 * 84
e = c - d
f = 4 * 80
g = 3 * 84
h = f - g
i = h + 3
j = e + i
k = b - j
l = a - k
|
a ) 500 , b ) 450 , c ) 600 , d ) 650 , e ) 750 | a | add(multiply(divide(600, 60), 30), multiply(divide(600, 60), 20)) | the consumption of diesel per hour of a bus varies directly as square of its speed . when the bus is travelling at 60 kmph its consumption is 1 litre per hour . if each litre costs $ 30 and other expenses per hous is $ 20 , then what would be the minimum expenditure required to cover a distance of 600 km ? | 60 kmph consumption is 1 lt / hr so 600 km will take 10 hrs and the consumption is 10 lt for entire distance . 1 lt costs $ 30 so 10 lt costs $ 300 extra expenses for 1 hr - $ 20 10 hrs - $ 200 total expense - $ 300 + $ 200 = $ 500 answer : a | a = 600 / 60
b = a * 30
c = 600 / 60
d = c * 20
e = b + d
|
a ) 15 , b ) 20 , c ) 23 , d ) 24 , e ) 25 | d | add(multiply(10, const_2), multiply(subtract(19.5, multiply(10, const_2)), 10)) | if the average of 10 consecutive integers is 19.5 then the 10 th integer is : - | "the average falls between the 5 th and 6 th integers , integer 5 = 19 , integer 6 = 20 . counting up to the tenth integer we get 24 . answer : d" | a = 10 * 2
b = 10 * 2
c = 19 - 5
d = c * 10
e = a + d
|
a ) 27 , b ) 28 , c ) 25 , d ) 30 , e ) 35 | c | divide(subtract(175, 100), 3) | on the first day of her vacation , louisa traveled 100 miles . on the second day , traveling at the same average speed , she traveled 175 miles . if the 100 - mile trip took 3 hours less than the 175 - mile trip , what was the average speed , in miles per hour ? | "( time ) * ( rate ) = ( distance ) - - > ( rate ) = ( distance ) / ( time ) - - > given : ( rate ) = 100 / t = 175 / ( t + 3 ) - - > 4 / t = 7 / ( t + 3 ) - - > 4 t + 12 = 7 t - - - - > 3 t = 12 . t = 4 - - - - > ( rate ) = 100 / 4 = 25 answer : c" | a = 175 - 100
b = a / 3
|
a ) 28 / 3 , b ) 13 / 6 , c ) 3 , d ) 4 , e ) 14 / 3 | a | subtract(multiply(divide(4, 3), 4), 3) | at a certain paint store forest green is made by mixing 4 parts blue paint with 3 parts yellow paint . verdant green is made by mixing 4 parts yellow paint with 3 parts blue paint . how many liters of yellow paint must be added to 28 liters of forest green to change it to verdant green ? | "28 liter of forset green have 16 liter of blue and 12 liter of yellow suppose we add x liter of yellow to make it a verdant green so the ratio of blue to yellow in verdant green is ΒΎ so the equation is blue / yellow = 16 / ( 12 + x ) = ΒΎ 36 + 3 x = 64 = > x = 28 / 3 answer : a" | a = 4 / 3
b = a * 4
c = b - 3
|
a ) 13.28 % , b ) 14.28 % , c ) 15.28 % , d ) 16.28 % , e ) 25 % | e | subtract(divide(multiply(const_100, 5), subtract(5, 1)), const_100) | in a office work is distribute between p persons . if 1 / 5 members are absent then work increased for each person is ? | "let total % of work is 100 % total person = p 1 / 5 person are absent of total person . so absent person is 1 / 5 p ie p / 5 . left person is , p - p / 5 = 4 p / 5 . p person do the work 100 % 1 person do the work 100 * p % 4 p / 5 person do the work ( 100 * p * 5 ) / 4 p % = 125 % work increased for each person is = ( 125 - 100 ) % = 25 % answer : e" | a = 100 * 5
b = 5 - 1
c = a / b
d = c - 100
|
a ) 21 , b ) 24 , c ) 35 , d ) 62 , e ) 60 | e | multiply(5, const_4) | what could be the range of a set consisting of odd multiples of 5 ? | "range = the difference between the greatest and the smallest numbers in the sequence . our sequence is odd and is a multiple of 7 . every number in that sequence can be represented like this : 5 * ( 2 n + 1 ) where n is any positive integer . range = 5 * ( 2 m + 1 ) - 5 * ( 2 n + 1 ) = 5 * 2 * ( m - n ) = 10 * ( m - n ) . m , n - any positive integers the answer must be divisible by 10 , which is only 60 . the correct answer is e ." | a = 5 * 4
|
a ) 12 , b ) 15 , c ) 20 , d ) 25 , e ) 30 | d | multiply(divide(subtract(75, 60), 60), const_100) | a book is bought for $ 60 and sold for $ 75 . what is the profit in percentage ? | "75 / 60 = 1.25 the answer is d ." | a = 75 - 60
b = a / 60
c = b * 100
|
a ) 15 , b ) 18 , c ) 17 , d ) none of these , e ) can not be determined | a | add(divide(subtract(85, 5), 5), const_1) | how many multiples of 5 are there between 5 to 85 ? | "explanation : as you know , multiples of 5 are integers having 0 or 5 in the digit to the extreme right ( i . e . the units place ) . so the numbers are 10 , 15 , 20 , 25 , 30 , 35 , 40 , 45 , 50 , 55 , 60 , 65 , 70 , 75 , 80 . answer : a" | a = 85 - 5
b = a / 5
c = b + 1
|
a ) 12.75 , b ) 85 , c ) 204 , d ) 250 , e ) none of these | b | divide(multiply(510, add(const_4, const_1)), add(const_1, const_2)) | the difference between a number and its two - fifth is 510 . what is 10 % of that number ? | "explanation : let the number be x . then , x - 2 / 5 x = 510 x = ( 510 * 5 ) / 3 = 850 10 % of 850 = 85 . answer is b" | a = 4 + 1
b = 510 * a
c = 1 + 2
d = b / c
|
a ) 901 , b ) 989 , c ) 990 , d ) 891 , e ) 1,001 | d | subtract(add(const_1000, const_1000), multiply(add(const_1000, const_1000), 0.5)) | in a recent election , james received 0.5 percent of the 2,000 votes cast . to win the election , a candidate needed to receive more than 45 percent of the vote . how many additional votes would james have needed to win the election ? | "james = ( 0.5 / 100 ) * 2000 = 10 votes to win = ( 45 / 100 ) * total votes + 1 = ( 45 / 100 ) * 2000 + 1 = 901 remaining voted needed to win election = 901 - 10 = 891 answer : option d" | a = 1000 + 1000
b = 1000 + 1000
c = b * 0
d = a - c
|
a ) $ 252935 , b ) $ 432935 , c ) $ 122935 , d ) $ 112935 , e ) $ 152935 | d | add(multiply(multiply(subtract(1, divide(23, const_100)), subtract(1, divide(23, 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(23, const_2), const_100), sqrt(const_100))) | the value of a machine depreciates at 23 % 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 = 0.77 * 0.77 * 1 , 50,000 = $ 88,935 sp such that a profit of $ 24,000 is made = 88,935 + 24,000 = $ 1 , 12,935 d" | a = 23 / 100
b = 1 - a
c = 23 / 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 = 23 + 2
p = o * 100
q = math.sqrt(100)
r = p * q
s = n + r
|
a ) 80 kmph , b ) 60 kmph , c ) 86 kmph , d ) 90 kmph , e ) none of these | c | divide(add(240, 450), add(3, 5)) | a train travels 240 km in 3 hours and 450 km in 5 hours . find the average speed of train . | "as we know that speed = distance / time for average speed = total distance / total time taken thus , total distance = 240 + 450 = 690 km thus , total speed = 8 hrs or , average speed = 690 / 8 or , 86 kmph . answer : c" | a = 240 + 450
b = 3 + 5
c = a / b
|
a ) $ 2,040 , b ) $ 2,120 , c ) $ 1,960 , d ) $ 1,920 , e ) $ 1,400 | b | subtract(multiply(const_3, const_1000), add(multiply(divide(add(16, 2), 2), 20), multiply(subtract(divide(add(16, 2), 2), 2), 100))) | a gambler bought $ 3,000 worth of chips at a casino in denominations of $ 20 and $ 100 . that evening , the gambler lost 16 chips , and then cashed in the remainder . if the number of $ 20 chips lost was 2 more or 2 less than the number of $ 100 chips lost , what is the largest amount of money that the gambler could have received back ? | in order to maximize the amount of money that the gambler kept , we should maximize # of $ 20 chips lost and minimize # of $ 100 chips lost , which means that # of $ 20 chips lost must be 2 more than # of $ 100 chips lost . so , if # of $ 20 chips lost is x then # of $ 100 chips lost should be x - 2 . now , given that total # of chips lost is 16 : x + x - 2 = 16 - - > x = 9 : 9 $ 20 chips were lost and 9 - 2 = 7 $ 100 chips were lost . total worth of chips lost is 9 * 20 + 7 * 100 = $ 880 , so the gambler kept $ 3,000 - $ 880 = $ 2,120 . answer : b . | a = 3 * 1000
b = 16 + 2
c = b / 2
d = c * 20
e = 16 + 2
f = e / 2
g = f - 2
h = g * 100
i = d + h
j = a - i
|
a ) 150 meter , b ) 200 meter , c ) 156 meter , d ) 168 meter , e ) 154 meter | b | multiply(divide(multiply(60, const_1000), const_3600), 12) | a train running at the speed of 60 km / hr crosses a pole in 12 seconds . find the length of the train ? | "speed = 60 * ( 5 / 18 ) m / sec = 50 / 3 m / sec length of train ( distance ) = speed * time ( 50 / 3 ) * 12 = 200 meter answer : b" | a = 60 * 1000
b = a / 3600
c = b * 12
|
a ) 18 sec , b ) 12 sec , c ) 17 sec , d ) 21 sec , e ) 16 sec | e | divide(multiply(120, const_2), add(speed(120, 24), speed(120, 12))) | two trains of equal lengths take 12 sec and 24 sec respectively to cross a telegraph post . if the length of each train be 120 m , in what time will they cross other travelling in opposite direction ? | "speed of the first train = 120 / 12 = 10 m / sec . speed of the second train = 120 / 24 = 5 m / sec . relative speed = 10 + 5 = 15 m / sec . required time = ( 120 + 120 ) / 15 = 16 sec . answer : e" | a = 120 * 2
b = speed + (
c = a / b
|
a ) 3 / 5 , b ) 4 / 5 , c ) 17 / 25 , d ) 21 / 25 , e ) 43 / 50 | d | subtract(const_1, multiply(divide(const_2, add(const_2, const_3)), divide(4, add(4, 6)))) | set a contains 4 different positive odd integers and 6 different positive even integers . set b contains two different positive odd integers and 3 different positive even integers . if one integer from set a and one integer from set b are chosen at random , what is the probability that the product of the chosen integers is even ? | there can be an odd product only if both integers are odd . p ( odd product ) = 4 / 10 * 2 / 5 = 4 / 25 p ( even product ) = 1 - 4 / 25 = 21 / 25 the answer is d . | a = 2 + 3
b = 2 / a
c = 4 + 6
d = 4 / c
e = b * d
f = 1 - e
|
a ) 16 days , b ) 55 days , c ) 24 days , d ) 32 days , e ) 22 days | d | divide(multiply(16, const_3), subtract(divide(add(divide(multiply(16, const_3), 12), add(divide(multiply(16, const_3), 16), divide(multiply(16, const_3), 24))), const_2), divide(multiply(16, const_3), 16))) | a and b can do a piece of work in 16 days , b and c in 24 days , c and a in 12 days . how long will c take to do it ? | "2 c = 1 / 24 + 1 / 12 β 1 / 16 = 1 / 16 c = 1 / 32 = > 32 days answer : d" | a = 16 * 3
b = 16 * 3
c = b / 12
d = 16 * 3
e = d / 16
f = 16 * 3
g = f / 24
h = e + g
i = c + h
j = i / 2
k = 16 * 3
l = k / 16
m = j - l
n = a / m
|
a ) 2456789 , b ) 3458009 , c ) 3890043 , d ) 3455048 , e ) 1083875 | e | multiply(1000000, multiply(multiply(add(const_1, divide(15, const_100)), subtract(const_1, divide(35, const_100))), add(const_1, divide(35, const_100)))) | population of a city in 20004 was 1000000 . if in 2005 there isan increment of 15 % , in 2006 there is a decrement of 35 % and in 2007 there is an increment of 45 % , then find the population of city at theend of the year 2007 | "required population = p ( 1 + r 1 / 100 ) ( 1 - r 2 / 100 ) ( 1 + r 3 / 100 ) = p ( 1 + 15 / 100 ) ( 1 - 35 / 100 ) ( 1 + 45 / 100 ) = 1083875 e" | a = 15 / 100
b = 1 + a
c = 35 / 100
d = 1 - c
e = b * d
f = 35 / 100
g = 1 + f
h = e * g
i = 1000000 * h
|
a ) 1 / 4 , b ) 1 / 3 , c ) 1 / 2 , d ) 2 / 3 , e ) 3 / 4 | e | subtract(const_1, divide(5, divide(multiply(const_60, const_1), 3))) | a searchlight on top of the watch - tower makes 3 revolutions per minute . what is the probability that a man appearing near the tower will stay in the dark for at least 5 seconds ? | "3 revolutions per minute = 1 revolution every 20 seconds so no matter what anybody appearing at the tower can not stay in the dark for more than 20 seconds . this will be our total number of possibilities i . e the denominator . p ( man in dark for at least 5 seconds ) = 1 - p ( man in dark for max of 5 seconds ) = 1 - 5 / 20 = 1 - 1 / 4 = 3 / 4 or the other way would be : p ( man in dark for at least 5 seconds ) is like saying he can be in dark for 5 , 6,7 . . . all the way to 20 seconds because that is the max . in this approach it would be 15 / 20 seconds = 3 / 4 ." | a = const_60 * 1
b = a / 3
c = 5 / b
d = 1 - c
|
a ) 4.25 , b ) 5.85 , c ) 6.25 , d ) 7 , e ) 7.5 | b | divide(subtract(282, multiply(10, 4.8)), 40) | in the first 10 overs of a cricket game , the run rate was only 4.8 . what should be the run rate in the remaining 40 overs to reach the target of 282 runs ? | "required run rate = 282 - ( 4.8 x 10 ) = 234 234 / 40 = 5.85 b )" | a = 10 * 4
b = 282 - a
c = b / 40
|
a ) 52 minutes , b ) 45 minutes , c ) 30 minutes , d ) 39 minutes , e ) 44 minutes | d | divide(multiply(4.20, const_60), 5) | the timing of a college is from 12 p . m to 4.20 p . m . six lectures are held in the given duration and a break of 5 minutes after each lecture is given to the students . find the duration of each lecture . | "explanation : total time a student spends in college = 4 hours 20 minutes = 260 minutes as there are 6 lectures , the number of breaks between lectures is 5 . total time of the break = 25 minutes hence , the duration of each lecture is = ( 260 Γ’ β¬ β 25 ) / 6 = 39 minutes answer d" | a = 4 * 20
b = a / 5
|
a ) 6 , b ) 7 , c ) 5 , d ) 8 , e ) 9 | a | divide(subtract(divide(75, 2), divide(45, 2)), const_2) | a man rows his boat 75 km downstream and 45 km upstream , taking 2 1 / 2 hours each time . find the speed of the stream ? | "speed downstream = d / t = 75 / ( 2 1 / 2 ) = 30 kmph speed upstream = d / t = 45 / ( 2 1 / 2 ) = 18 kmph the speed of the stream = ( 30 - 18 ) / 2 = 6 kmph answer : a" | a = 75 / 2
b = 45 / 2
c = a - b
d = c / 2
|
a ) 65 , b ) 66 , c ) 67 , d ) 131 , e ) 132 | d | add(add(const_1, 66), 66) | in the land of oz only one or two - letter words are used . the local language has 66 different letters . the parliament decided to forbid the use of the seventh letter . how many words have the people of oz lost because of the prohibition ? | "the seventh letter ( let it be @ ) can be used with other 65 in two type of arrangents one when seventh letter is in 1 st slot @ _ ( 65 ways ) another , when seventh letter is in 2 nd slot _ @ ( 65 ways ) seventh letter ( @ ) itself lost words = 1 + 65 + 65 = 131 answer : d" | a = 1 + 66
b = a + 66
|
a ) 2 , b ) 2 and 4 , c ) 2 , 4 and 8 , d ) 2 , 4 and 9 , e ) none of these | a | divide(multiply(divide(100, const_2), add(100, 1)), divide(multiply(divide(100, const_2), add(100, 1)), const_2)) | the sum of the first 100 numbers , 1 to 100 is always divisible by | explanation : the sum of the first 100 natural numbers = ( 100 x 101 ) / 2 = ( 50 x 101 ) as 101 is an odd number and 50 is divisible by 2 , so the sum is always divisible by 2 . answer : option a | a = 100 / 2
b = 100 + 1
c = a * b
d = 100 / 2
e = 100 + 1
f = d * e
g = f / 2
h = c / g
|
a ) 354 cm 2 , b ) 380 m 2 , c ) 350 m 2 , d ) 407 m 2 , e ) none of these | b | divide(multiply(power(22, const_2), const_pi), const_4) | a horse is tethered to one corner of a rectangular grassy field 45 m by 25 m with a rope 22 m long . over how much area of the field can it graze ? | "area of the shaded portion = 1 β 4 Γ Ο Γ ( 22 ) 2 = 380 m 2 answer b" | a = 22 ** 2
b = a * math.pi
c = b / 4
|
a ) 17 : 6 , b ) 17 : 0 , c ) 17 : 7 , d ) 17 : 2 , e ) 17 : 3 | c | divide(add(multiply(add(add(const_2, const_3), const_3), multiply(add(const_2, const_3), 2)), add(const_2, const_3)), add(multiply(const_3, multiply(add(const_2, const_3), 2)), add(const_2, const_3))) | p and q started a business investing rs . 85,000 and rs . 35,000 respectively . in what ratio the profit earned after 2 years be divided between p and q respectively ? | p : q = 85000 : 35000 = 17 : 7 . answer : c | a = 2 + 3
b = a + 3
c = 2 + 3
d = c * 2
e = b * d
f = 2 + 3
g = e + f
h = 2 + 3
i = h * 2
j = 3 * i
k = 2 + 3
l = j + k
m = g / l
|
a ) 18 days , b ) 27 days , c ) 26.67 days , d ) 5 days , e ) 12 days | d | add(divide(subtract(const_1, multiply(divide(const_1, 30), 10)), add(divide(const_1, 30), divide(const_1, 30))), 10) | a can complete a project in 30 days and b can complete the same project in 30 days . if a and b start working on the project together and a quits 10 days before the project is completed , in how many days will the project be completed ? | "let x = the number of days taken to complete the project . the amount of work done by a is ( x - 10 ) * ( 1 / 20 ) . the amount of work done by b is ( x ) * ( 1 / 30 ) . ( 1 / 30 ) * ( x - 10 ) + ( 1 / 30 ) * ( x ) = 1 ( x / 30 ) + ( x / 30 ) - ( 10 / 30 ) = 1 2 x / 30 = 1 / 3 x = 5 therefore , the answer is d : 5 ." | a = 1 / 30
b = a * 10
c = 1 - b
d = 1 / 30
e = 1 / 30
f = d + e
g = c / f
h = g + 10
|
a ) 10 mps , b ) 18 mps , c ) 19 mps , d ) 17 mps , e ) 12 mps | e | multiply(const_0_2778, 42) | express a speed of 42 kmph in meters per second ? | "42 * 5 / 18 = 12 mps answer : e" | a = const_0_2778 * 42
|
a ) 8 , b ) 10 , c ) 12 , d ) 14 , e ) 16 | e | divide(subtract(100, multiply(5, const_3)), const_4) | a box has exactly 100 balls , and each ball is either red , blue , or white . if the box has 5 more blue balls than white balls , and thrice as many red balls as blue balls , how many white balls does the box has ? | "x = the number of red balls y = the number of blue balls z = the number of white balls from the first sentence we have equation # 1 : x + y + z = 100 . . . the box has 5 more blue balls than white balls . . . equation # 2 : y = 5 + z . . . thrice as many red balls as blue balls . . . equation # 3 : x = 3 y solve equation # 2 for z : z = y - 5 now , we can replace both x and z with y in equation # 1 3 y + y + ( y - 5 ) = 100 5 y - 5 = 100 5 y = 105 y = 21 there are 21 blue balls . this is 5 more than the number of white balls , so z = 16 . that ' s the answer . just as a check , x = 63 , and 63 + 21 + 16 = 100 . answer = 16 , ( e )" | a = 5 * 3
b = 100 - a
c = b / 4
|
a ) 18 % , b ) 21 % , c ) 20 % , d ) 26.1 % , e ) none of these | d | divide(multiply(subtract(multiply(540, divide(add(const_100, 15), const_100)), 459), const_100), multiply(540, divide(add(const_100, 15), const_100))) | mahesh marks an article 15 % above the cost price of rs . 540 . what must be his discount percentage if he sells it at rs . 459 ? | "cp = rs . 540 , mp = 540 + 15 % of 540 = rs . 621 sp = rs . 459 , discount = 621 - 459 = 162 discount % = 162 / 621 * 100 = 26.1 % answer : d" | a = 100 + 15
b = a / 100
c = 540 * b
d = c - 459
e = d * 100
f = 100 + 15
g = f / 100
h = 540 * g
i = e / h
|
a ) rs . 3600 , b ) rs . 3603 , c ) rs . 3639 , d ) rs . 3000 , e ) rs . 3602 | d | subtract(21000, multiply(divide(6, 7), 21000)) | income and expenditure of a person are in the ratio 7 : 6 . if the income of the person is rs . 21000 , then find his savings ? | "let the income and the expenditure of the person be rs . 7 x and rs . 6 x respectively . income , 7 x = 21000 = > x = 3000 savings = income - expenditure = 7 x - 6 x = x so , savings = rs . 3000 answer : d" | a = 6 / 7
b = a * 21000
c = 21000 - b
|
a ) 8223 , b ) 8123 , c ) 8323 , d ) 8423 , e ) 8523 | b | divide(1760, subtract(const_1, add(add(divide(1, 3), divide(1, 4)), divide(1, 5)))) | a man spends 1 / 3 rd of his salary on food . he spends 1 / 4 th on rent and 1 / 5 th on cloths . if he is left with 1760 , then who much salary does he earn | let total salary be x a / q = > x = ( x / 3 ) + ( x / 4 ) + ( x / 5 ) + 1760 = > x - 47 x / 60 = 1760 = > 13 x / 60 = 1760 = > x = 1760 * 60 / 13 = > x = 8123 ( approax ) answer : b | a = 1 / 3
b = 1 / 4
c = a + b
d = 1 / 5
e = c + d
f = 1 - e
g = 1760 / f
|
a ) 4676 , b ) 4678 , c ) 9984 , d ) 9504 , e ) 9936 | c | multiply(floor(divide(power(const_10, 4), 48)), 48) | what is the largest 4 digit number exactly divisible by 48 ? | largest 4 digit number = 9999 9999 Γ· 48 = 208 , remainder = 15 hence largest 4 digit number exactly divisible by 48 = 9999 - 15 = 9984 answer : c | a = 10 ** 4
b = a / 48
c = math.floor(b)
d = c * 48
|
a ) 8 : 3 , b ) 6 : 5 , c ) 9 : 8 , d ) 9 : 4 , e ) 5 : 4 | d | divide(9, divide(subtract(9, const_1), const_2)) | the ratio of length of two refill is 9 : 8.2 nd refill got damaged when only half length of 2 nd refill get used . the ratio of total no . of pages written by both refill . | the no . of page written is directly proportional to length of refill . so , initial no . of page to be write = 9 x , 8 x . now , 2 nd refill only can write 4 x page so new ratio of page written = 9 x / 4 x = 9 : 4 . answer d | a = 9 - 1
b = a / 2
c = 9 / b
|
a ) 200 , b ) 188 , c ) 204 , d ) 288 , e ) 500 | a | multiply(divide(multiply(45, const_1000), const_3600), 16) | a train running at the speed of 45 km / hr crosses a pole in 16 seconds . find the length of the train . | "speed = 45 * ( 5 / 18 ) m / sec = 25 / 2 m / sec length of train ( distance ) = speed * time ( 25 / 2 ) * 16 = 200 meter answer : a" | a = 45 * 1000
b = a / 3600
c = b * 16
|
a ) 360 liters , b ) 256 , c ) 48 , d ) 525 , e ) 425 | c | multiply(4, multiply(3, 4)) | he number of times a bucket of capacity 4 liters to be used to fill up a tank is less than number of ties another bucket of capacity 3 liters used for the same purpose by 4 . what is the capacity of the tank ? | 4 x = 3 ( x + 4 ) , where x may be number of times 4 l bucket is used . 4 x = 3 ( x + 4 ) - - > 1 . x = 12 , substitute x to lhs or rhs of 1 , which gets you the capacity , and it is 48 . answer : c | a = 3 * 4
b = 4 * a
|
a ) a ) 2500 , b ) b ) 3500 , c ) c ) 5000 , d ) d ) 6000 , e ) e ) 8000 | b | add(multiply(multiply(divide(divide(5000, const_2), 10), const_3), const_4), 500) | a and b start a business , with a investing the total capital of rs . 5000 , on the condition that b pays a interest @ 10 % per annum on his half of the capital . a is a working partner and receives rs . 500 per month from the total profit and any profit remaining is equally shared by both of them . at the end of the year , it was found that the income of a is twice that of b . find the total profit for the year ? | interest received by a from b = 10 % of half of rs . 5000 = 10 % * 2500 = 250 . amount received by a per annum for being a working partner = 500 * 12 = rs . 1000 . let ' p ' be the part of the remaining profit that a receives as his share . total income of a = ( 250 + 1000 + p ) total income of b = only his share from the remaining profit = ' p ' , as a and b share the remaining profit equally . income of a = twice the income of b ( 250 + 1000 + p ) = 2 ( p ) p = 1250 total profit = 2 p + 1000 = 2 * 1250 + 1000 = 3500 answer : b | a = 5000 / 2
b = a / 10
c = b * 3
d = c * 4
e = d + 500
|
a ) 12 % , b ) 20 % , c ) 33 % , d ) 40 % , e ) 50 % | d | multiply(divide(multiply(choose(const_4, const_1), const_2), choose(6, 3)), multiply(multiply(const_5, const_5), const_4)) | two mba admissions committees are to be formed randomly from 6 second year mbas with 3 members each . what is the probability w that jane will be on the same committee as albert ? | "total number of ways to choose 3 member committee - 6 c 3 = ( 6 ! / 3 ! 3 ! ) = 20 no . of ways albert n jane are in same committee : - ( 4 c 1 * 2 ) = 8 probability w = ( 8 / 20 ) * 100 = 40 % . + 1 for me . . : d" | a = math.comb(4, 1)
b = a * 2
c = math.comb(6, 3)
d = b / c
e = 5 * 5
f = e * 4
g = d * f
|
a ) 18 , b ) 750 , c ) 24 , d ) 19 , e ) none of these | c | divide(circle_area(divide(35, multiply(2, const_pi))), 4) | how many plants will be there in a circular bed whose outer edge measure 35 cms , allowing 4 cm 2 for each plant ? | "circumference of circular bed = 35 cm area of circular bed = ( 35 ) 2 Γ’ Β β 4 Γ― β¬ space for each plant = 4 cm 2 Γ’ Λ Β΄ required number of plants = ( 35 ) 2 Γ’ Β β 4 Γ― β¬ Γ£ Β· 4 = 24.36 = 24 ( approx ) answer c" | a = 2 * math.pi
b = 35 / a
c = circle_area / (
|
a ) 3.84 , b ) 1.75 , c ) 3.28 , d ) 2.72 , e ) none of these | c | divide(divide(multiply(4, add(multiply(3, 900), multiply(2, 700))), add(3, 2)), const_1000) | the weights of one liter vegetable ghee packet of two brands β a β and β b β are 900 gm and 700 gm respectively . if they are mixed in the ratio of 3 : 2 by volumes to form a mixture of 4 liters , what is the weight ( in kg ) of the mixture ? | "here ' s how i did it . my notes from reading the problem were : 1 l a = 900 gm 1 l b = 700 gm we are mixing five parts ( 3 parts a plus 2 parts b , 5 parts total ) to get 4 l , so 5 x = 4 - - - > x = 4 / 5 . each part is 4 / 5 of a liter . so if we have 3 parts a , we have 900 * 3 * ( 4 / 5 ) = 2160 if we have 2 parts b , we have 700 * 2 * ( 4 / 5 ) = 1120 2160 + 1120 = 3280 solving for units gives us 3.28 so the answer is c" | a = 3 * 900
b = 2 * 700
c = a + b
d = 4 * c
e = 3 + 2
f = d / e
g = f / 1000
|
a ) a ) 258 , b ) b ) 65 , c ) c ) 58 , d ) d ) 62 , e ) e ) 48 | a | subtract(add(multiply(6, 78), multiply(6, 75)), multiply(11, 60)) | the average of 11 numbers is 60 . out of 11 numbers the average of first 6 no . is 78 , and last 6 numbers is 75 then find 6 th number ? | "6 th number = sum of 1 st 6 no . s + sum of last 6 no . s - sum of 11 no . s answer = 6 * 78 + 6 * 75 - 11 * 60 = 258 answer is a" | a = 6 * 78
b = 6 * 75
c = a + b
d = 11 * 60
e = c - d
|
a ) 100 meter , b ) 150 meter , c ) 200 meter , d ) 250 meter , e ) 300 meter | e | multiply(divide(multiply(120, const_1000), const_3600), 9) | a train running at the speed of 120 km / hr crosses a pole in 9 seconds . find the length of the train . | "speed = 120 * ( 5 / 18 ) m / sec = 200 / 6 m / sec length of train ( distance ) = speed * time ( 200 / 6 ) * 9 = 300 meter answer : e" | a = 120 * 1000
b = a / 3600
c = b * 9
|
a ) 124 % , b ) b ) 120 % , c ) c ) 96 % , d ) d ) 80 % , e ) e ) 64 % | d | multiply(divide(add(const_100, 60), multiply(divide(const_100, subtract(const_100, 50)), const_100)), const_100) | marts income is 60 percent more than tims income and tims income is 50 percent less than juans income . what percentage of juans income is marts income | "m = ( 160 / 100 ) t t = ( 50 / 100 ) j = > m = ( 80 / 100 ) j answer d ." | a = 100 + 60
b = 100 - 50
c = 100 / b
d = c * 100
e = a / d
f = e * 100
|
a ) 12 liters , b ) 62 liters , c ) 41 liters , d ) 50 liters , e ) 34 liters | b | multiply(divide(155, add(3, 2)), 2) | 155 liters of a mixture of milk and water contains in the ratio 3 : 2 . how much water should now be added so that the ratio of milk and water becomes 3 : 4 ? | milk = 3 / 5 * 155 = 93 liters water = 62 liters 93 : ( 62 + p ) = 3 : 4 186 + 3 p = 372 = > p = 62 62 liters of water are to be added for the ratio become 3 : 4 . answer : b | a = 3 + 2
b = 155 / a
c = b * 2
|
a ) 1 / 8 , b ) 1 / 6 , c ) 3 / 8 , d ) 1 / 2 , e ) 3 / 4 | c | divide(const_3, multiply(const_2, const_4)) | the center of a circle lies on the origin of the coordinate plane . if a point ( x , y ) is randomly selected inside of the circle , what is the probability that y > 0 and y > x ? | the line y = x divides the circle into two equal areas . all the points above the line y = x satisfy the condition that y > x . all the points above the x - axis satisfy the condition that y > 0 . the intersection of these two areas is 3 / 8 of the circle . the answer is c . | a = 2 * 4
b = 3 / a
|
a ) 1 , b ) 2 , c ) 3 , d ) 49 , e ) 36 | d | add(power(const_2, const_2), const_2) | how many different positive integers are factors of 196 ? | "15 , 14 Γ 15 , 14 = 7 ^ 2 Γ 2 ^ 2 so total factors = ( 6 + 1 ) ( 6 + 1 ) = 49 answer : d" | a = 2 ** 2
b = a + 2
|
a ) 13 , b ) 15 , c ) 19 , d ) 24 , e ) none of these | c | sqrt(add(power(sqrt(subtract(5, multiply(const_2, 266))), const_2), multiply(const_4, 266))) | the product of two numbers is 266 and their difference is 5 . what is the bigger number ? | "explanation : let the two numbers be a and b , here a > b ab = 266 b = 266 / a - - - - - - - - - - - - - - - - - ( i ) given , a β b = 5 - - - - - - - - - - - ( ii ) substitute from ( i ) in ( ii ) , we get a β 266 / a = 5 a 2 β 5 a + 266 = 0 ( a β 19 ) ( a β 14 ) = 0 therefore , a = 19 or a = 14 hence , bigger number = a = 19 answer : c" | a = 2 * 266
b = 5 - a
c = math.sqrt(b)
d = c ** 2
e = 4 * 266
f = d + e
g = math.sqrt(f)
|
a ) 6 , b ) 4 , c ) 5 , d ) 2 , e ) 3 | a | add(floor(5), const_1) | if 821562 is to be divisible by 5 , what is the least whole number that should be added to it ? | a number is divisible by 8 if the number formed by the last three digits is divisible by 8 . here , 821562 = 562 , the next multiple of 9 is 568 . 6 must be added to 821562 to make it divisible by 8 a | a = math.floor(5)
b = a + 1
|
a ) 80 % , b ) 105 % , c ) 120 % , d ) 124.2 % , e ) 128 % | e | multiply(divide(multiply(16, subtract(const_1, divide(20, const_100))), 10), const_100) | in 2008 , the profits of company n were 10 percent of revenues . in 2009 , the revenues of company n fell by 20 percent , but profits were 16 percent of revenues . the profits in 2009 were what percent of the profits in 2008 ? | "x = profits r = revenue x / r = 0,1 x = 10 r = 100 2009 : r = 80 x / 80 = 0,16 = 16 / 100 x = 80 * 16 / 100 x = 12.8 12.8 / 10 = 1,28 = 128 % , answer e" | a = 20 / 100
b = 1 - a
c = 16 * b
d = c / 10
e = d * 100
|
a ) 75200 , b ) 79970 , c ) 75460 , d ) 24580 , e ) none | b | subtract(multiply(const_10, 8), 8) | the difference between the place values of 8 and 3 in the number 587435 is | sol . = ( place value of 8 ) β ( place value of 3 ) = ( 80000 - 30 ) = 79970 answer b | a = 10 * 8
b = a - 8
|
a ) 25 , b ) 50 , c ) 90 , d ) 140 , e ) it can not be determined from the information given . | c | subtract(multiply(80, const_2), multiply(35, const_2)) | if the average ( arithmetic mean ) of a and b is 35 and the average of b and c is 80 , what is the value of c β a ? | "- ( a + b = 70 ) b + c = 160 c - a = 90 c . 90" | a = 80 * 2
b = 35 * 2
c = a - b
|
a ) 15 , b ) 20 , c ) 42 , d ) 25 , e ) 18 | c | divide(multiply(8, 21), subtract(21, 17)) | a group of men decided to do a work in 17 days , but 8 of them became absent . if the rest of the group did the work in 21 days , find the original number of men ? | "original number of men = 8 * 21 / ( 21 - 17 ) = 42 answer is c" | a = 8 * 21
b = 21 - 17
c = a / b
|
a ) 13.86 , b ) 16.33 , c ) 16.35 , d ) 16.3 , e ) 16.32 | a | divide(add(add(multiply(30, 11.50), multiply(20, 14.25)), multiply(divide(add(multiply(30, 11.50), multiply(20, 14.25)), const_100), 10)), add(30, 20)) | arun purchased 30 kg of wheat at the rate of rs . 11.50 per kg and 20 kg of wheat at the rate of 14.25 per kg . he mixed the two and sold the mixture . approximately what price per kg should be sell the mixture to make 10 % profit ? | "c . p . of 50 kg wheat = ( 30 * 11.50 + 20 * 14.25 ) = rs . 630 . s . p . of 50 kg wheat = 110 % of rs . 630 = 110 / 100 * 630 = rs . 693 s . p . per kg = 693 / 50 = rs . 13.86 answer : a" | a = 30 * 11
b = 20 * 14
c = a + b
d = 30 * 11
e = 20 * 14
f = d + e
g = f / 100
h = g * 10
i = c + h
j = 30 + 20
k = i / j
|
a ) 50 , b ) 70 , c ) 10 , d ) 105 , e ) 140 | c | subtract(subtract(490, divide(subtract(multiply(4, divide(490, 7)), multiply(6, divide(multiply(4, divide(490, 7)), add(6, 1)))), divide(1, 6))), multiply(6, divide(multiply(4, divide(490, 7)), add(6, 1)))) | in an apartment building that has 490 units , 4 out of every 7 units are currently rented , including 1 / 6 of the one - bedroom units . if , of the rented apartments , there is a 6 : 1 ratio of two - bedroom units to one - bedroom units , and the building only consists of two - bedroom and one - bedroom units , how many two - bedroom units are not rented ? | rented = 4 / 7 * 490 = 280 . rented two - bedroom = 6 / 7 * 280 = 240 ; rented one - bedroom = 1 / 7 * 280 = 40 ( or 280 - 240 = 40 ) . rented one - bedroom units are 1 / 5 of all one - bedroom : 1 / 6 * { all one bedroom } = 40 - - > { all one - bedroom } = 240 . { all two - bedroom } = 490 - 240 = 250 . two - bedroom not rented = 250 - 240 = 10 . answer : c . | a = 490 / 7
b = 4 * a
c = 490 / 7
d = 4 * c
e = 6 + 1
f = d / e
g = 6 * f
h = b - g
i = 1 / 6
j = h / i
k = 490 - j
l = 490 / 7
m = 4 * l
n = 6 + 1
o = m / n
p = 6 * o
q = k - p
|
a ) 51 , b ) 52 , c ) 59 , d ) 55 , e ) 56 | b | add(40, divide(subtract(976, multiply(16, 40)), divide(multiply(16, add(const_100, 75)), const_100))) | a certain bus driver is paid a regular rate of $ 16 per hour for any number of hours that does not exceed 40 hours per week . for any overtime hours worked in excess of 40 hours per week , the bus driver is paid a rate that is 75 % higher than his regular rate . if last week the bus driver earned $ 976 in total compensation , how many total hours did he work that week ? | "for 40 hrs = 40 * 16 = 640 excess = 976 - 640 = 336 for extra hours = . 75 ( 16 ) = 12 + 16 = 28 number of extra hrs = 336 / 28 = 12 total hrs = 40 + 12 = 52 answer b" | a = 16 * 40
b = 976 - a
c = 100 + 75
d = 16 * c
e = d / 100
f = b / e
g = 40 + f
|
a ) 4 days , b ) 6 days , c ) 8 days , d ) 10 days , e ) 12 days | b | inverse(add(divide(const_1, 18), multiply(const_2, divide(const_1, 18)))) | a works twice as fast as b . if b can complete a work in 18 days independently , the number of days in which a and b can together finish the work is : | "explanation : ratio of rates of working of a and b = 2 : 1 . so , ratio of times taken = 1 : 2 a ' s 1 day ' s work = 1 / 9 b ' s 1 day ' s work = 1 / 18 ( a + b ) ' s 1 day ' s work = ( 1 / 9 + 1 / 18 ) = 3 / 18 = 1 / 6 so , a and b together can finish the work in 6 days answer is b" | a = 1 / 18
b = 1 / 18
c = 2 * b
d = a + c
e = 1/(d)
|
a ) 75 kg , b ) 55 kg , c ) 45 kg , d ) 85 kg , e ) 25 kg | d | add(multiply(2.5, 8), 65) | he average weight of 8 persons increases by 2.5 kg when a new person comes in place of one of them weighing 65 kg . what might be the weight of the new person ? | explanation : total weight increased = ( 8 x 2.5 ) kg = 20 kg . weight of new person = ( 65 + 20 ) kg = 85 kg . answer : d | a = 2 * 5
b = a + 65
|
a ) 1100 , b ) 800 , c ) 1400 , d ) 1200 , e ) 1600 | e | multiply(8800, divide(const_2, add(add(multiply(const_2, 3), multiply(divide(const_2, 3), 3)), 3))) | a , band c enter into partnership . a invests 3 times as much as b and b invests two - third of what c invests . at the end of the year , the profit earned is rs . 8800 . what is the share of b ? | "let c ' s capital = rs . x . then , b ' s capital = rs . ( 2 / 3 ) x a β s capital = rs . ( 3 x ( 2 / 3 ) . x ) = rs . 2 x . ratio of their capitals = 2 x : ( 2 / 3 ) x : x = 6 : 2 : 3 . hence , b ' s share = rs . ( 8800 x ( 2 / 11 ) ) = rs . 1600 . answer is e" | a = 2 * 3
b = 2 / 3
c = b * 3
d = a + c
e = d + 3
f = 2 / e
g = 8800 * f
|
a ) 25 , b ) 24 , c ) 23 , d ) 22 , e ) 20 | b | divide(100, multiply(4, 1)) | how many numbers are divisible by 4 between 1 and 100 ? | "numbers divisible by 4 till 100 = ( 100 / 4 ) = 25 but we should not consider 100 as we are asked to find the numbers between 1 and 100 which are divisible by 4 . so answer is 24 numbers . answer is option b" | a = 4 * 1
b = 100 / a
|
a ) a ) 10,700 , b ) b ) 10,000 , c ) c ) 10,900 , d ) d ) 11,000 , e ) e ) 11,100 | b | multiply(multiply(const_4, const_2), const_100) | a certain city with a population of 120,000 is to be divided into 11 voting districts , and no district is to have a population that is more than 10 percent greater than the population of any other district what is the minimum possible population that the least populated district could have ? | "let x = number of people in smallest district x * 1.1 = number of people in largest district x will be minimised when the number of people in largest district is maximised 10 * x * 1.1 = 11 x = total number of people in other districts so we have 11 x + x = 120 k x = 10,000 answer : b" | a = 4 * 2
b = a * 100
|
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