regisss/math_qa huggingface dataset:
Problem: a bowl of nuts is prepared for a party . brand p mixed nuts are 20 % almonds and brand q ' s deluxe nuts are 25 % almonds . if a bowl contains a total of 66 ounces of nuts , representing a mixture of both brands , and 15 ounces of the mixture are almonds , how many ounces of brand q ' s deluxe mixed nuts are used ?
Options: a ) 16 , b ) 20 , c ) 36 , d ) 44 , e ) 48
Rationale: "lets say x ounces of p is mixed with q . = > 66 - x ounces of q is present in the mixture ( as the total = 66 ounces ) given total almond weight = 15 ounces ( 20 x / 100 ) + ( 25 / 100 ) ( 66 - x ) = 15 = > x = 30 = > 66 - 30 = 36 ounces of q is present in the mixture . answer is c ."
Answer: c
Problem: what is the units digit of ( 6 ! * 4 ! + 6 ! * 5 ! ) / 6 ?
Options: a ) 1 , b ) 0 , c ) 2 , d ) 3 , e ) 4
Rationale: ( 6 ! * 4 ! + 6 ! * 5 ! ) / 6 = 6 ! ( 4 ! + 5 ! ) / 6 = 720 ( 24 + 120 ) / 6 = ( 720 * 144 ) / 6 = 720 * 24 units digit of the above product will be equal to 0 answer b
Answer: b
Problem: if x and y are sets of integers , x # y denotes the set of integers that belong to set x or set y , but not both . if x consists of 8 integers , y consists of 10 integers , and 6 of the integers are in both x and y , then x # y consists of how many integers ?
Options: a ) 6 , b ) 16 , c ) 22 , d ) 30 , e ) 174
Rationale: "the number of integers that belong to set x only is 8 - 6 = 2 ; the number of integers that belong to set y only is 10 - 6 = 4 ; the number of integers that belong to set x or set y , but not both is 2 + 4 = 6 . answer : a ."
Answer: a
Problem: the mean of 50 observations was 36 . it was found later that an observation 48 was wrongly taken as 21 . the corrected new mean is :
Options: a ) 36.54 , b ) 36.58 , c ) 36.23 , d ) 36.14 , e ) 36.81
Rationale: "explanation : correct sum = ( 36 * 50 + 48 - 21 ) = 1827 . correct mean = = 1827 / 50 = 36.54 answer : a ) 36.54"
Answer: a
Problem: if an object travels at ten feet per second , how many feet does it travel in two hours ?
Options: a ) 60000 , b ) 30000 , c ) 18000 , d ) 72000 , e ) 22200
Rationale: "if an object travels at 10 feet per second it covers 10 x 60 feet in one minute , and 10 x 60 x 2 x 60 feet in two hour . answer = 72000 answer : d"
Answer: d
Problem: 3 medical experts , working together at the same constant rate , can write an anatomy textbook in 24 days . how many additional experts , working together at this same constant rate , are needed to write the textbook in 18 days ?
Options: a ) 1 , b ) 2 , c ) 3 , d ) 4 , e ) 5
Rationale: each expert can write 1 / 72 of the book per day . to complete the book in 18 days , we need 72 / 18 = 4 experts , thus 1 more expert is needed . the answer is a .
Answer: a
Problem: the value of ( 34.31 * 0.473 * 1.567 ) / ( 0.0673 * 23.25 * 7.57 ) is close to
Options: a ) 2 , b ) 1.15 , c ) 2.05 , d ) 2.15 , e ) 2.35
Rationale: "( 34.31 * 0.473 * 1.567 ) / ( 0.0673 * 23.25 * 7.57 ) = 25.4303 / 11.845 = 2.15 answer : d"
Answer: d
Problem: the cost of carpeting a room 15 meters long with a carpet 75 cm wide at 30 paise per meter is rs . 36 . the breadth of the room is ?
Options: a ) 6 meters , b ) 7 meters , c ) 8 meters , d ) 9 meters , e ) 10 meters
Rationale: "length of carpet = total cost / rate = 3600 / 30 = 120 m area of carpet = ( 120 x 75 ) / 100 m 2 = 90 m 2 ∴ area of the room = 90 m 2 breadth of the room = area / length = 90 / 15 m = 6 m answer : a"
Answer: a
Problem: an aeroplane covers a certain distance of 590 kmph in 8 hours . to cover the same distance in 2 3 / 4 hours , it must travel at a speed of
Options: a ) 1780 , b ) 1716 , c ) 1890 , d ) 1980 , e ) 1450
Rationale: speed of aeroplane = 590 kmph distance travelled in 8 hours = 590 * 8 = 4720 km speed of aeroplane to acver 4720 km in 11 / 4 = 4720 * 4 / 11 = 1716 km answer b .
Answer: b
Problem: a work can be completed by 12 boys in 24 days and 12 girls in 12 days . in how many days would the 6 boys and 6 girls working together complete the work ?
Options: a ) 12 , b ) 18 , c ) 16 , d ) 20 , e ) 24
Rationale: in 1 day , 12 boys does 1 / 24 of the total work . = > in 1 day , 1 boy does 1 / ( 24 * 12 ) of the total work = > in 1 days , 6 boys do 6 / ( 24 * 12 ) = 1 / 48 of the total work in 1 day , 12 girls do 1 / 12 of the total work . = > in 1 day , 1 girl does 1 / ( 12 * 12 ) of the total work = > in 1 day , 6 girls do 6 / ( 12 * 12 ) = 1 / 24 of the total work in 1 day , 6 boys and 6 girls do 1 / 48 + 1 / 24 of the total work = > in 1 days , 6 boys and 6 girls do 3 / 48 = 1 / 16 of the total work so , 6 boys and 6 girls complete all work in 16 days . hence , the answer is c
Answer: c
Problem: walking at 75 % of his usual speed a man takes 24 minutes more to cover a distance . what is his usual time to cover this distance ?
Options: a ) 30 , b ) 36 , c ) 42 , d ) 48 , e ) 72
Rationale: "speed is inversly proprtional to time walking at 75 % of speed meand 3 / 4 s takes 4 / 3 t . it takes 24 minutes extra to cover the distance . then 4 / 3 t = t + 24 4 t = 3 t + 72 t = 72 option e is correct"
Answer: e
Problem: anusha , banu and esha run a running race of 100 meters . anusha is the fastest followed by banu and then esha . anusha , banu and esha maintain constant speeds during the entire race . when anusha reached the goal post , banu was 10 m behind . when banu reached the goal post esha was 10 m behind . how far was behind anusha when the latter reached the goal post .
Options: a ) 22 , b ) 88 , c ) 81 , d ) 66 , e ) 22
Rationale: by that time anusha covered 100 m , bhanu covered 90 m . so ratio of their speeds = 10 : 9 by that time bhanu reached 100 m , esha covered 90 m . so ratio of their speeds = 10 : 9 ratio of the speed of all the three = 100 : 90 : 81 by that time anusha covered 100 m , esha covers only 81 . answer : c
Answer: c
Problem: the output of a factory was increased by 10 % to keep up with rising demand . to handle the holiday rush , this new output was increased by 30 % . by approximately what percent would the output now have to be decreased in order to restore the original output ?
Options: a ) 20 % , b ) 24 % , c ) 30 % , d ) 32 % , e ) 79 %
Rationale: "the original output increases by 10 % and then 30 % . total % change = a + b + ab / 100 total % change = 10 + 30 + 10 * 30 / 100 = 43 % now , you want to change it to 0 , so , 0 = 43 + x + 43 x / 100 x = - 43 ( 100 ) / 143 = 30 % approximately answer is c"
Answer: c
Problem: if the cost price is 89 % of selling price then what is the profit percentage .
Options: a ) 8.35 , b ) 9.35 , c ) 10.35 , d ) 11.35 , e ) 12.35
Rationale: "selling price = rs 100 : then cost price = rs 89 : profit = rs 11 . profit = { ( 11 / 89 ) * 100 } % = 12.35 % answer is e ."
Answer: e
Problem: the rate of interest on a sum of money is 2 % p . a . for the first 3 years , 4 % p . a . for the next 4 years , and 5 % for the period beyond 7 years . if the s . i , occured on the sum for the total period of 8 years is rs . 540 / - , the sum is
Options: a ) 2,200 , b ) 2,000 , c ) 2,100 , d ) 2,250 , e ) 2,560
Rationale: "explanation : i 1 = ( p x 3 x 2 ) / 100 = 3 p / 50 i 2 = ( p x 4 x 4 ) / 100 = 4 p / 25 i 3 = ( p x 1 x 5 ) / 100 = p / 20 3 p / 50 + 4 p / 25 + p / 20 = 540 the l . c . m of 50 , 25 , 20 = 100 ( 6 p + 16 p + 5 p ) / 100 = 540 27 p / 100 = 540 27 p = 54000 p = 54000 / 27 p = 2000 answer : option b"
Answer: b
Problem: at what rate percent per annum will the simple interest on a sum of money be 7 / 5 of the amount in 10 years ?
Options: a ) 4 % , b ) 7 % , c ) 9 % , d ) 3 % , e ) 14 %
Rationale: "let sum = x . then , s . i . = 7 x / 5 , time = 10 years . rate = ( 100 * 7 x ) / ( x * 5 * 10 ) = 14 % answer : e"
Answer: e
Problem: two trains are moving in opposite directions at 60 km / hr and 90 km / hr . their lengths are 2.9 km and 1.6 km respectively . the time taken by the slower train to cross the faster train in seconds is ?
Options: a ) 99 , b ) 277 , c ) 48 , d ) 96 , e ) 108
Rationale: "relative speed = 60 + 90 = 150 km / hr . = 150 * 5 / 18 = 125 / 3 m / sec . distance covered = 2.9 + 1.6 = 4.5 km = 4500 m . required time = 4500 * 3 / 125 = 108 sec . answer : e"
Answer: e
Problem: the edge of a cube is 3 a cm . find its surface ?
Options: a ) 24 a 8 , b ) 24 a 4 , c ) 24 a 1 , d ) 54 a 2 , e ) 24 a 7
Rationale: "6 a 2 = 6 * 3 a * 3 a = 54 a 2 answer : d"
Answer: d
Problem: how many integers between 324,700 and 458,600 have a 2 in the tens digit and a 1 in the units digit ?
Options: a ) 1339 , b ) 2300 , c ) 4200 , d ) 1340 , e ) 2414
Rationale: "there is one number in hundred with 2 in th tens digit and 1 in the units digit : 21 , 121 , 221 , 321 , . . . the difference between 324,700 and 458,600 is 458,600 - 324,700 = 133,900 - one number per each hundred gives 133,900 / 100 = 1,339 numbers . answer : 1,339 a"
Answer: a
Problem: in a graduating class of 242 students , 144 took geometry and 119 took biology . what is the difference between the greatest possible number and the smallest possible number of students that could have taken both geometry and biology ?
Options: a ) 144 , b ) 119 , c ) 113 , d ) 88 , e ) 123
Rationale: "greatest possible number taken both should be 144 ( as it is maximum for one ) smallest possible number taken both should be given by total - neither = a + b - both both = a + b + neither - total ( neither must be 0 to minimize the both ) so 144 + 119 + 0 - 242 = 21 greatest - smallest is 144 - 21 = 123 so answer must be e . 123"
Answer: e
Problem: the average mark of the students of a class in a particular exam is 80 . if 5 students whose average mark in that exam is 20 are excluded , the average mark of the remaining will be 95 . find the number of students who wrote the exam .
Options: a ) 15 , b ) 45 , c ) 35 , d ) 25 , e ) 55
Rationale: let the number of students who wrote the exam be x . total marks of students = 80 x . total marks of ( x - 5 ) students = 95 ( x - 5 ) 80 x - ( 5 * 20 ) = 95 ( x - 5 ) 375 = 15 x = > x = 25 answer : d
Answer: d
Problem: real - estate salesman z is selling a house at a 30 percent discount from its retail price . real - estate salesman x vows to match this price , and then offers an additional 30 percent discount . real - estate salesman y decides to average the prices of salesmen z and x , then offer an additional 30 percent discount . salesman y ' s final price is what fraction of salesman x ' s final price ?
Options: a ) 12 / 13 , b ) 11 / 15 , c ) 11 / 17 , d ) 11 / 14 , e ) 11 / 7
Rationale: "let the retail price be = x selling price of z = 0.70 x selling price of x = 0.70 * 0.80 x = 0.56 x selling price of y = ( ( 0.70 x + 0.56 x ) / 2 ) * 0.70 = 0.63 x * 0.70 = 0.44 x 0.44 x = k * 0.56 x k = 0.44 / 0.56 = 44 / 56 = 11 / 14 answer : d"
Answer: d
Problem: a student has to obtain 33 % of the total marks to pass . he got 92 marks and failed by 40 marks . the maximum marks are ?
Options: a ) 400 , b ) 300 , c ) 500 , d ) 610 , e ) 175
Rationale: "let the maximum marks be x then , 33 % of x = 92 + 40 33 x / 100 = 132 x = 400 answer is a"
Answer: a
Problem: p works 25 % more efficiently than q and q works 50 % more efficiently than r . to complete a certain project , p alone takes 50 days less than q alone . if , in this project p alone works for 60 days and then q alone works for 150 days , in how many days can r alone complete the remaining work ?
Options: a ) 50 days , b ) 75 days , c ) 120 days , d ) 150 days , e ) 80 days
Rationale: "p works 25 % more efficiently than q : something that takes q 5 days , takes p 4 days q works 50 % more efficiently than r : something that takes r 7.5 days , takes q 5 days p alone takes 50 days less than q : for every 4 days p works , q has to work an extra day . hence p alone can do it in 200 days and q alone in 250 days and hence r alone in 400 days p works for 60 days - - > 60 / 200 work done = > 30 % q works for 150 days - - > 150 / 250 work done = > 60 % 30 % work left . . . r alone will take 30 % * 400 = 120 days answer is ( c )"
Answer: c
Problem: what is the sum of the multiples of 6 from 48 to 72 , inclusive ?
Options: a ) 500 , b ) 600 , c ) 700 , d ) 300 , e ) 400
Rationale: "the formula we want to use in this type of problem is this : average * total numbers = sum first , find the average by taking the sum of the f + l number and divide it by 2 : a = ( f + l ) / 2 second , find the total numbers in our range by dividing our f and l numbers by 7 and add 1 . ( 72 / 6 ) - ( 48 / 6 ) + 1 multiply these together so what we show average * total numbers = sum ( 48 + 72 ) / 2 * ( 72 / 6 ) - ( 48 / 6 ) + 1 = sum 60 * 5 = 300 d"
Answer: d
Problem: if d = 1 / ( 2 ^ 3 * 5 ^ 10 ) is expressed as a terminating decimal , how many nonzero digits will d have ?
Options: a ) one , b ) two , c ) three , d ) seven , e ) ten
Rationale: another way to do it is : we know x ^ a * y ^ a = ( x * y ) ^ a given = 1 / ( 2 ^ 3 * 5 ^ 10 ) = multiply and divide by 2 ^ 7 = 2 ^ 7 / ( 2 ^ 3 * 2 ^ 7 * 5 ^ 10 ) = 2 ^ 7 / 10 ^ 10 = > non zero digits are 128 = > ans c
Answer: c
Problem: 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 . 3300 . what is the share of b ?
Options: a ) 1100 , b ) 800 , c ) 1400 , d ) 600 , e ) none of them
Rationale: "let c ' s capital = rs . x . then , b ' s capital = rs . ( 2 / 3 ) x a ’ 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 . ( 3300 x ( 2 / 11 ) ) = rs . 600 answer is d ."
Answer: d
Problem: tanks a and b are each in the shape of a right circular cylinder . the interior of tank a has a height of 10 meters and a circumference of 11 meters , and the interior of tank b has a height of 11 meters and a circumference of 10 meters . the capacity of tank a is what percent of the capacity of tank b ?
Options: a ) 75 % , b ) 80 % , c ) 100 % , d ) 110 % , e ) 125 %
Rationale: "the radius of tank a is 11 / ( 2 * pi ) . the capacity of tank a is 10 * pi * 121 / ( 4 * pi ^ 2 ) = 605 / ( 2 * pi ) the radius of tank b is 10 / ( 2 * pi ) . the capacity of tank b is 11 * pi * 100 / ( 4 * pi ^ 2 ) = 550 / ( 2 * pi ) tank a / tank b = 605 / 550 = 11 / 10 = 110 % the answer is d ."
Answer: d
Problem: the cost price of a radio is rs . 1500 and it was sold for rs . 1335 , find the loss % ?
Options: a ) 18 , b ) 16 , c ) 26 , d ) 17 , e ) 11
Rationale: "explanation : 1500 - - - - 165 100 - - - - ? = > 11 % answer : e"
Answer: e
Problem: each of the products produced yesterday was checked by worker x or worker y . 0.5 % of the products checked by worker x are defective and 0.8 % of the products checked by worker y are defective . if the total defective rate of all the products checked by worker x and worker y is 0.6 % , what fraction of the products was checked by worker y ?
Options: a ) 1 / 5 , b ) 1 / 4 , c ) 1 / 3 , d ) 2 / 5 , e ) 3 / 7
Rationale: "x : 0.5 % is 0.1 % - points from 0.6 % . y : 0.8 % is 0.2 % - points from 0.6 % . therefore the ratio of products checked by y : x is 1 : 2 . thus , worker y checked 1 / 3 of the products . the answer is c ."
Answer: c
Problem: in 1998 the profits of company n were 10 percent of revenues . in 1999 , the revenues of company n fell by 30 percent , but profits were 15 percent of revenues . the profits in 1999 were what percent of the profits in 1998 ?
Options: a ) 80 % , b ) 105 % , c ) 120 % , d ) 124.2 % , e ) 138 %
Rationale: "0,105 r = x / 100 * 0.1 r answer b"
Answer: b
Problem: a shopkeeper sells 300 metres of cloth for rs . 18000 at a loss of rs . 5 per metre . find his cost price for one metre of cloth ?
Options: a ) 12 , b ) 27 , c ) 29 , d ) 50 , e ) 65
Rationale: sp per metre = 18000 / 300 = rs . 60 loss per metre = rs . 5 cp per metre = 60 + 5 = rs . 65 . answer : e
Answer: e
Problem: ? % of 360 = 144
Options: a ) 277 , b ) 36 , c ) 64 , d ) 40 , e ) none of these
Rationale: "? % of 360 = 144 or , ? = 144 × 100 / 360 = 40 answer d"
Answer: d
Problem: linda spent 5 / 6 of her savings on furniture and the rest on a tv . if the tv cost her $ 500 , what were her original savings ?
Options: a ) $ 9000 , b ) $ 3000 , c ) $ 6000 , d ) $ 7000 , e ) $ 8000
Rationale: "if linda spent 5 / 6 of her savings on furniture , the rest 6 / 6 - 5 / 6 = 1 / 6 on a tv but the tv cost her $ 500 . so 1 / 6 of her savings is $ 500 . so her original savings are 6 times $ 500 = $ 3000 correct answer b"
Answer: b
Problem: how many bricks , each measuring 40 cm x 11.25 cm x 6 cm , will be needed to build a wall of 8 m x 6 m x 22.5 cm ?
Options: a ) 1400 , b ) 2400 , c ) 4000 , d ) 7000 , e ) 3400
Rationale: "number of bricks = volume of the wall / volume of 1 brick = ( 800 x 600 x 22.5 ) / ( 40 x 11.25 x 6 ) = 4000 answer : c"
Answer: c
Problem: find the simple interest on rs . 64,000 at 16 2 / 3 % per annum for 9 months .
Options: a ) s . 8500 , b ) s . 8000 , c ) s . 7500 , d ) s . 7000 , e ) s . 6500
Rationale: "p = rs . 64000 , r = 50 / 3 % p . a and t = 9 / 12 years = 3 / 4 years . s . i . = ( p * r * t ) / 100 = rs . ( 64,000 * ( 50 / 3 ) * ( 3 / 4 ) * ( 1 / 100 ) ) = rs . 8000 answer is b ."
Answer: b
Problem: three unbiased coins are tossed . what is the probability of getting at most 2 heads ?
Options: a ) 7 / 8 , b ) 5 / 8 , c ) 3 / 8 , d ) 7 / 5 , e ) 3 / 5
Rationale: "here s = { ttt , tth , tht , htt , thh , hth , hht , hhh } let e = event of getting at least two heads = { ttt , tth , tht , htt , thh , hth , hht } p ( e ) = n ( e ) / n ( s ) = 7 / 8 . answer a ."
Answer: a
Problem: the average height of 15 girls out of a class of 60 is 138 cm . and that of the remaining girls is 142 cm . the average height of the whole class is :
Options: a ) 132 cms , b ) 141 cms , c ) 142 cms , d ) 152 cms , e ) 161 cms
Rationale: "explanation : average height of the whole class = ( 15 × 138 + 45 × 142 / 60 ) = 141 cms answer b"
Answer: b
Problem: a cistern can be filled by a tap in 2 hours while it can be emptied by another tap in 9 hours . if both the taps are opened simultaneously , then after how much time will the cistern get filled ?
Options: a ) 2.57 hrs , b ) 5 hrs , c ) 6.57 hrs , d ) 7.2 hrs , e ) 9.27 hrs
Rationale: "net part filled in 1 hour = 1 / 2 - 1 / 9 = 7 / 18 therefore the cistern will be filled in 18 / 7 hours or 2.57 hours . answer : a"
Answer: a
Problem: the edges of a cuboid are 6 cm , 5 cm and 6 cm . find the volume of the cuboid ?
Options: a ) 120 , b ) 160 , c ) 210 , d ) 368 , e ) 180
Rationale: "6 * 5 * 6 = 180 answer : e"
Answer: e
Problem: a began business with rs . 45000 and was joined afterwards by b with rs . 27000 . when did b join if the profits at the end of the year were divided in the ratio of 2 : 1 ?
Options: a ) 1 , b ) 6 , c ) 7 , d ) 8 , e ) 2
Rationale: "45 * 12 : 27 * x = 2 : 1 x = 10 12 - 10 = 2 answer : e"
Answer: e
Problem: 1 / 3 + 1 / 2 - 5 / 6 + 1 / 5 + 1 / 4 - 9 / 20 - 9 / 20 =
Options: a ) 0 , b ) 2 / 15 , c ) 2 / 5 , d ) 9 / 20 , e ) 5 / 6
Rationale: "we need to determine the result of 1 / 3 + 1 / 2 - 5 / 6 + 1 / 5 + 1 / 4 - 9 / 20 let ’ s add the given fractions in two groups . in the group of the first three fractions , notice that 1 / 3 and 1 / 2 share a common denominator of 6 with 5 / 6 . 1 / 2 + 1 / 3 = 3 / 6 + 2 / 6 = 5 / 6 thus , 5 / 6 – 5 / 6 = 0 looking at the 2 nd group of the fractions ( 1 / 5 , 1 / 4 , and 9 / 20 ) , notice that 1 / 5 and 1 / 4 share a common denominator of 20 with 9 / 20 . 1 / 5 + 1 / 4 = 4 / 20 + 5 / 20 = 9 / 20 thus , 9 / 20 – 9 / 20 = 0 . thus , the result of 1 / 3 + 1 / 2 – 5 / 6 + 1 / 5 + 1 / 4 – 9 / 20 is 9 / 20 . answer : d"
Answer: d
Problem: a certain car dealership sells economy cars , luxury cars , and sport utility vehicles . the ratio of economy to luxury cars is 4 : 3 . the ratio of economy cars to sport utility vehicles is 6 : 5 . what is the ratio of luxury cars to sport utility vehicles ?
Options: a ) 9 : 10 , b ) 8 : 9 , c ) 3 : 2 , d ) 2 : 3 , e ) 1 : 2
Rationale: "the ratio of economy to luxury cars is 4 : 3 - - > e : l = 4 : 3 = 24 : 18 . the ratio of economy cars to sport utility vehicles is 6 : 5 - - > e : s = 6 : 5 = 24 : 20 . thus , l : s = 18 : 20 = 9 : 10 . answer : a ."
Answer: a
Problem: which digits should come in place of @ and # if the number 62684 @ # is divisible by both 8 and 5 ?
Options: a ) 4,0 , b ) 0,4 , c ) 4,4 , d ) 0,0 , e ) none
Rationale: explanation : since the given number is divisible by 5 , so 0 or 5 must come in place of # . but , a number ending with 5 is never divisible by 8 . so , 0 will replace # . now , the number formed by the last three digits is 4 @ 0 , which becomes divisible by 8 , if @ is replaced by 4 . hence , digits in place of @ and # are 4 and 0 respectively . answer : a
Answer: a
Problem: a invested $ 300 in a business after 6 months b invested $ 200 in the business . end of the year if they got $ 100 as profit . find b ' s shares ?
Options: a ) $ 100 , b ) $ 75 , c ) $ 25 , d ) $ 120 , e ) $ 50
Rationale: "a : b = 300 * 12 : 200 * 6 a : b = 1 : 1 b ' s share = 100 * 1 / 4 = $ 25 answer is c"
Answer: c
Problem: ashok secured average of 76 marks in 6 subjects . if the average of marks in 5 subjects is 74 , how many marks did he secure in the 6 th subject ?
Options: a ) 38 , b ) 27 , c ) 99 , d ) 17 , e ) 86
Rationale: "explanation : number of subjects = 6 average of marks in 6 subjects = 76 therefore total marks in 6 subjects = 76 * 6 = 456 now , no . of subjects = 5 total marks in 5 subjects = 74 * 5 = 370 therefore marks in 6 th subject = 456 – 370 = 86 answer : e"
Answer: e
Problem: a man has a certain number of small boxes to pack into parcles . if he packs 3 , 4 , 5 or 6 in a parcel , he is left with one over ; if he packs 7 in a parcle , none is left over . what is the number of boxes , he may have to pack ?
Options: a ) 106 , b ) 301 , c ) 309 , d ) 400 , e ) 450
Rationale: explanation : clearly , the required number would be such that it leaves a remainder of 1 when divided by 3 , 4 , 5 , or 6 and no remainder when divided by 7 . thus , the number must be of the form ( l . c . m of 3 , 4 , 5 , 6 ) x + 1 i . e . , ( 60 x + 1 ) and a multiple of 7 . clearly , for x = 5 , the number is a multiple of 7 . so the number is 301 . answer : b ) 301
Answer: b
Problem: find b and c so that the parabola with equation y = 4 x 2 - bx - c has a vertex at ( 2 , 4 ) ?
Options: a ) - 10 , b ) - 20 , c ) - 30 , d ) - 40 , e ) - 50
Rationale: h = b / 8 = 2 : formula for x coordinate of vertex b = 16 : solve for b y = 4 for x = 2 : the vertex point is a solution to the equation of the parabola 4 ( 2 ) 2 - 16 ( 2 ) - c = 4 c = - 20 : solve for c correct answer b
Answer: b
Problem: pipe a can fill a tank in 8 minutes and pipe b cam empty it in 24 minutes . if both the pipes are opened together after how many minutes should pipe b be closed , so that the tank is filled in 30 minutes ?
Options: a ) 18 , b ) 27 , c ) 98 , d ) 27 , e ) 66
Rationale: let the pipe b be closed after x minutes . 30 / 8 - x / 24 = 1 = > x / 24 = 30 / 8 - 1 = 11 / 4 = > x = 11 / 4 * 24 = 66 . answer : e
Answer: e
Problem: a call center has two teams . each member of team a was able to process 2 / 5 calls as compared to each member of team b . if team a has 5 / 8 as many number of call center agents as team b , what fraction of the total calls was processed by team b ?
Options: a ) 3 / 2 , b ) 3 / 4 , c ) 4 / 5 , d ) 1 / 2 , e ) 1 / 5
Rationale: let team b has 8 agents , so team a has 5 agents let each agent of team b picked up 5 calls , so total calls by team b = 40 so , each agent in team a picked up 2 calls , so total calls for team a = 10 fraction for team b = 40 / ( 40 + 10 ) = 4 / 5 = answer = c
Answer: c
Problem: if the wheel is 15 cm then the number of revolutions to cover a distance of 1056 cm is ?
Options: a ) 18 , b ) 11.2 , c ) 14 , d ) 12 , e ) 91
Rationale: "2 * 22 / 7 * 15 * x = 1056 = > x = 11.2 answer : b"
Answer: b
Problem: a sum of rs . 1190 has been divided among a , b and c such that a gets of what b gets and b gets of what c gets . b ’ s share is :
Options: a ) rs . 120 , b ) rs . 210 , c ) rs . 240 , d ) rs . 300 , e ) none
Rationale: explanation let c ’ s share = rs . x then , b ’ s share = rs . x / 4 , a ’ s share = rs . ( 2 / 3 x x / 4 ) = rs . x / 6 = x / 6 + x / 4 + x = 1190 = > 17 x / 12 = 1190 = > 1190 x 12 / 17 = rs . 840 hence , b ’ s share = rs . ( 840 / 4 ) = rs . 210 . answer b
Answer: b
Problem: a number increased by 40 % gives 1680 . the number is ?
Options: a ) 1680 , b ) 1600 , c ) 1200 , d ) 1500 , e ) 600
Rationale: "formula = total = 100 % , increase = ` ` + ' ' decrease = ` ` - ' ' a number means = 100 % that same number increased by 40 % = 140 % 140 % - - - - - - - > 1680 ( 140 ã — 12 = 1680 ) 100 % - - - - - - - > 1200 ( 100 ã — 12 = 1200 ) option ' c '"
Answer: c
Problem: if each digit in the set a = { 1 , 2 , 3 , 4 , 5 } is used exactly once , in how many ways can the 5 digits be arranged ?
Options: a ) 6 , b ) 24 , c ) 72 , d ) 96 , e ) 120
Rationale: "use the slot method for all the possible arrangements : we have 5 options for the 1 st slot , 4 for the 2 nd , 3 for 3 rd , 2 for 4 th and 1 for the 5 th , giving the total number of arrangements = 5 * 4 * 3 * 2 * 1 = 120 , e is the correct answer ."
Answer: e
Problem: two pipes a and b can fill a tank in 12 hours and 15 hours respectively . if both the pipes are opened simultaneously , how much time will be taken to fill the tank ?
Options: a ) 20 hours , b ) 15 hours , c ) 10 hours , d ) 12 hours , e ) 7 4 / 9 hours
Rationale: "part filled by a in 1 hour = 1 / 12 part filled by b in 1 hour = 1 / 15 part filled by ( a + b ) in 1 hour = 1 / 12 + 1 / 15 = 9 / 60 both the pipes together fill the tank in 60 / 9 = 7 4 / 9 hours answer is e"
Answer: e
Problem: the sides of the triangle are in the ratio 5 : 6 : 7 and its perimeter is 720 cm . the length of the longest side is ?
Options: a ) 150 cm
b ) 200 cm
c ) 162 cm
d ) 220 cm
e ) 280 cm
Rationale: ratio of sides = 5 : 6 : 7 largest side = 720 * 7 / 18 = 280 cm answer is e
Answer: e
Problem: jackie has two solutions that are 2 percent sulfuric acid and 12 percent sulfuric acid by volume , respectively . if these solutions are mixed in appropriate quantities to produce 60 liters of a solution that is 10 percent sulfuric acid , approximately how many liters of the 2 percent solution will be required ?
Options: a ) 18 , b ) 12 , c ) 24 , d ) 36 , e ) 42
Rationale: "let a = amount of 2 % acid and b = amount of 12 % acid . now , the equation translates to , 0.02 a + . 12 b = . 1 ( a + b ) but a + b = 60 therefore . 02 a + . 12 b = . 1 ( 60 ) = > 2 a + 12 b = 600 but b = 60 - a therefore 2 a + 12 ( 60 - a ) = 600 = > 10 a = 120 hence a = 12 . answer : b"
Answer: b
Problem: a man saves 20 % of his monthly salary . if an account of dearness of things he is to increase his monthly expenses by 20 % , he is only able to save rs . 220 per month . what is his monthly salary ?
Options: a ) 5500 , b ) 2999 , c ) 2878 , d ) 2990 , e ) 2771
Rationale: "income = rs . 100 expenditure = rs . 80 savings = rs . 20 present expenditure 80 * ( 20 / 100 ) = rs . 96 present savings = 100 â € “ 96 = rs . 4 100 - - - - - - 4 ? - - - - - - - - - 220 = > 5500 answer : a"
Answer: a
Problem: triangle atriangle b are similar triangles with areas 1536 units square and 2166 units square respectively . the ratio of there corresponding height would be
Options: a ) 9 : 10
b ) 17 : 19
c ) 23 : 27
d ) 13 : 17
e ) 16 : 19
Rationale: 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 ) = 1536 / 2166 ( x ^ 2 / y ^ 2 ) = ( 16 * 16 * 6 ) / ( 19 * 19 * 6 ) ( x ^ 2 / y ^ 2 ) = 17 ^ 2 / 19 ^ 2 x / y = 16 / 19 ans = e
Answer: e
Problem: the megatek corporation is displaying its distribution of employees by department in a circle graph . the size of each sector of the graph representing a department is proportional to the percentage of total employees in that department . if the section of the circle graph representing the manufacturing department takes up 108 ° of the circle , what percentage of megatek employees are in manufacturing ?
Options: a ) 20 % , b ) 25 % , c ) 30 % , d ) 35 % , e ) 72 %
Rationale: "answer : c 108 ° divided by 360 ° equals 0.3 , therefore the sector is equal to 30 % of the total"
Answer: c
Problem: the toll t , in dollars , for a truck using a certain bridge is given by the formula t = 2.50 + 0.50 ( x − 2 ) , where x is the number of axles on the truck . what is the toll for an 18 - wheel truck that has 2 wheels on its front axle and 4 wheels on each of its other axles ?
Options: a ) $ 2.50 , b ) $ 3.00 , c ) $ 3.50 , d ) $ 4.00 , e ) $ 5.00
Rationale: "number of wheels in truck = 18 number of wheels on its front axle = 2 number of wheels remaining = 16 number of axles remaining axles = 16 / 4 = 4 total number of axles = 5 t = 2.50 + 0.50 ( x − 2 ) = 2.50 + . 5 * 3 = 2.5 + 1.5 = 4 $ answer d"
Answer: d
Problem: 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 20 percent of revenues . the profits in 2009 were what percent of the profits in 2008 ?
Options: a ) 80 % , b ) 105 % , c ) 120 % , d ) 124.2 % , e ) 160 %
Rationale: "x = profits r = revenue x / r = 0,1 x = 10 r = 100 2009 : r = 80 x / 80 = 0,2 = 15 / 100 x = 80 * 20 / 100 x = 16 16 / 10 = 1,6 = 160 % , answer e"
Answer: e
Problem: a man walking at a constant rate of 6 miles per hour is passed by a woman traveling in the same direction along the same path at a constant rate of 12 miles per hour . the woman stops to wait for the man 10 minutes after passing him , while the man continues to walk at his constant rate . how many minutes must the woman wait until the man catches up ?
Options: a ) 5 , b ) 15 , c ) 10 , d ) 20 , e ) 25
Rationale: "when the woman passes the man , they are aligned ( m and w ) . they are moving in the same direction . after 5 minutes , the woman ( w ) will be ahead the man ( m ) : m - - - - - - m - - - - - - - - - - - - - - - w w in the 5 minutes , after passing the man , the woman walks the distance mw = ww , which is 10 * 12 / 60 = 2 miles and the man walks the distance mm , which is 10 * 6 / 60 = 1 mile . the difference of 2 - 1 = 1 miles ( mw ) will be covered by the man in ( 1 ) / 6 = 1 / 6 of an hour , which is 10 minutes . answer c ."
Answer: c
Problem: a number when divided by 219 gives a remainder 32 , what remainder will be obtained by dividing the same number 12 ?
Options: a ) 7 , b ) 11 , c ) 9 , d ) 2 , e ) 3
Rationale: "explanation : 219 + 32 = 251 / 12 = 9 ( remainder ) answer : b"
Answer: b
Problem: what is the difference between the c . i . on rs . 7000 for 1 1 / 2 years at 4 % per annum compounded yearly and half - yearly ?
Options: a ) s . 2.04 , b ) s . 2.08 , c ) s . 2.02 , d ) s . 2.86 , e ) s . 2.42
Rationale: "c . i . when interest is compounded yearly = [ 7000 * ( 1 + 4 / 100 ) * ( 1 + ( 1 / 2 * 4 ) / 100 ] = 7000 * 26 / 25 * 51 / 50 = rs . 7425.6 c . i . when interest is compounded half - yearly = [ 7000 * ( 1 + 2 / 100 ) 2 ] = ( 7000 * 51 / 50 * 51 / 50 * 51 / 50 ) = rs . 7428.46 difference = ( 7428.46 - 7425.6 ) = rs . 2.86 . answer : d"
Answer: d
Problem: mary ' s income is 70 percent more than tim ' s income , and tim ' s income is 40 percent less than juan ' s income . what percent of juan ' s income is mary ' s income ?
Options: a ) 124 % , b ) 102 % , c ) 96 % , d ) 80 % , e ) 64 %
Rationale: juan ' s income = 100 ( assume ) ; tim ' s income = 60 ( 40 percent less than juan ' s income ) ; mary ' s income = 102 ( 70 percent more than tim ' s income ) . thus , mary ' s income ( 102 ) is 102 % of juan ' s income ( 100 ) . answer : b .
Answer: b
Problem: a sum of money lent out at s . i . amounts to a total of $ 480 after 2 years and to $ 680 after a further period of 5 years . what was the initial sum of money that was invested ?
Options: a ) $ 360 , b ) $ 380 , c ) $ 400 , d ) $ 420 , e ) $ 440
Rationale: "s . i for 5 years = $ 680 - $ 480 = $ 200 the s . i . is $ 40 / year s . i . for 2 years = $ 80 principal = $ 480 - $ 80 = $ 400 the answer is c ."
Answer: c
Problem: 5.005 / 2.002 =
Options: a ) 2.05 , b ) 2.50025 , c ) 2.501 , d ) 2.5025 , e ) 2.5
Rationale: "5.005 / 2.002 = 5005 / 2002 = 5 ( 1001 ) / 2 ( 1001 ) = 5 / 2 = 2.5 the answer is e ."
Answer: e
Problem: irin , ingrid and nell bake chocolate chip cookies in the ratio of 9.18 : 5.17 : 2.05 . if altogether they baked a batch of 148 cookies , what percent of the cookies did irin bake ?
Options: a ) 0.125 % , b ) 1.25 % , c ) 56 % , d ) 125 % , e ) 0.152 %
Rationale: "9.18 x + 5.17 x + 2.05 x = 16.4 x = 150 cookies x = 150 / 16.4 = 9.14 ( approx ) so , irin baked 9.14 * 9.18 cookies or 84 cookies ( approx ) % share = 84 / 150 = 56 approx hence , answer is c ."
Answer: c
Problem: a man is 24 years older than his son . in two years , his age will be twice the age of his son . the present age of the son is ?
Options: a ) 11 , b ) 25 , c ) 27 , d ) 22 , e ) 91
Rationale: "let the son ' s present age be x years . then , man ' s present age = ( x + 24 ) years . ( x + 24 ) + 2 = 2 ( x + 2 ) x + 26 = 2 x + 4 = > x = 22 . answer : d"
Answer: d
Problem: a dress on sale in a shop is marked at $ d . during the discount sale its price is reduced by 55 % . staff are allowed a further 50 % reduction on the discounted price . if a staff member buys the dress what will she have to pay in terms of d ?
Options: a ) 0.275 d , b ) 0.225 d , c ) 0.265 d , d ) 0.245 d , e ) 0.205 d
Rationale: "effective discount = a + b + ab / 100 = - 55 - 50 + ( - 55 ) ( - 50 ) / 100 = 77.5 sale price = d * ( 1 - 77.5 / 100 ) sale price = . 225 * d answer ( b )"
Answer: b
Problem: if 100 < x < 190 and 10 < y < 100 , then the product xy can not be equal to :
Options: a ) 18,104 , b ) 18,303 , c ) 18 , 356.732 , d ) 19,502 , e ) 18,909
Rationale: "correct answer : ( d ) determine the range of xy by multiplying the two extremes of each individual range together . the smallest value of xy must be greater than 100 * 10 . the largest value must be less than 190 * 100 . this means that 1000 < xy < 19,000 . ( d ) is outside of this range , so it is not a possible product of xy ."
Answer: d
Problem: if the cost price of 20 articles is equal to the selling price of 40 articles , what is the % profit or loss made by the merchant ?
Options: a ) 25 % loss , b ) 25 % profit , c ) 20 % loss , d ) 20 % profit , e ) 50 % loss
Rationale: "let the cost price of 1 article be $ 1 . therefore , cost price of 20 articles = 20 * 1 = $ 20 the selling price of 40 articles = cost price of 40 articles = $ 40 . now , we know the selling price of 40 articles . let us find the cost price of 40 articles . cost price of 40 articles = 40 * 1 = $ 40 . therefore , profit made on sale of 40 articles = selling price of 40 articles - cost price of 40 articles = 20 - 40 = - $ 20 . as the profit is in the negative , the merchant has made a loss of $ 20 . therefore , % loss = loss / cp * 100 % loss = - 20 / 40 * 100 = 50 % loss . e"
Answer: e
Problem: if x < y < z and y - x > 5 , where x is an even integer and y and z are odd integers , what is the least possible value of z - x ?
Options: a ) 6 , b ) 7 , c ) 8 , d ) 9 , e ) 10
Rationale: "we want to minimize z − xz − x , so we need to maximize xx . say z = 11 = oddz = 11 = odd , then max value of yy will be 9 ( as yy is also odd ) . now , since y − 5 > xy − 5 > x - - > 9 − 5 > x 9 − 5 > x - - > 4 > x 4 > x , then max value of xx is 2 ( as xx is even ) . hence , the least possible value of z − xz − x is 11 - 2 = 9 . answer : d ."
Answer: d
Problem: reeya obtained 65 , 67 , 76 , 82 and 85 out of 100 in different subjects , what will be the average
Options: a ) 70 , b ) 75 , c ) 80 , d ) 85 , e ) 60
Rationale: "explanation : ( 65 + 67 + 76 + 82 + 855 ) = 75 answer : option b"
Answer: b
Problem: if the sides of a square are multiplied by sqrt ( 2 ) , the area of the original square is how many times as large as the area of the resultant square ?
Options: a ) 2 % , b ) 4 % , c ) 50 % , d ) 100 % , e ) 200 %
Rationale: "let x be the original length of one side . then the original area is x ^ 2 . the new square has sides of length sqrt ( 2 ) * x , so the area is 2 x ^ 2 . the area of the original square is 1 / 2 = 50 % times the area of the new square . the answer is c ."
Answer: c
Problem: a certain bus driver is paid a regular rate of $ 15 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 $ 982 in total compensation , how many total hours did he work that week ?
Options: a ) 56 , b ) 40 , c ) 44 , d ) 48 , e ) 52
Rationale: "for 40 hrs = 40 * 15 = 600 excess = 982 - 600 = 382 for extra hours = . 75 ( 15 ) = 11.25 + 16 = 27.25 number of extra hrs = 382 / 27.25 = 14 total hrs = 40 + 14 = 56 answer a 56"
Answer: a
Problem: how long does a train 165 meters long running at the rate of 54 kmph take to cross a bridge 720 meters in length ?
Options: a ) 28 , b ) 27 , c ) 55 , d ) 59 , e ) 12
Rationale: "t = ( 720 + 165 ) / 54 * 18 / 5 t = 59 answer : d"
Answer: d
Problem: 90 students represent x percent of the boys at jones elementary school . if the boys at jones elementary make up 30 % of the total school population of x students , what is x ?
Options: a ) 125 , b ) 173 , c ) 225 , d ) 250 , e ) 500
Rationale: "90 = x / 100 * 30 / 100 * x = > x ^ 2 = 9 * 10000 / 3 = > x = 173 b"
Answer: b
Problem: last year , company x paid out a total of $ 1 , 050,000 in salaries to its 21 employees . if no employee earned a salary that is more than 8 % greater than any other employee , what is the lowest possible salary that any one employee earned ?
Options: a ) $ 40,000 , b ) $ 41,667 , c ) $ 42,000 , d ) $ 46 , 460.17 , e ) $ 60,000
Rationale: employee 1 earned $ x ( say ) employee 2 will not earn more than $ 1.08 x therfore , to minimize the salary of any one employee , we need to maximize the salaries of the other 20 employees ( 1.08 x * 20 ) + x = 1 , 050,000 solving for x = $ 46 , 460.17 answer d
Answer: d
Problem: a part - time employee ’ s hourly wage was increased by 40 % . she decided to decrease the number of hours worked per week so that her total income did not change . by approximately what percent should the number of hours worked be decreased ?
Options: a ) 9 % , b ) 15 % , c ) 29 % , d ) 50 % , e ) 100 %
Rationale: "let ' s plug in somenicenumbers and see what ' s needed . let ' s say the employee used to make $ 1 / hour and worked 100 hours / week so , the total weekly income was $ 100 / week after the 40 % wage increase , the employee makes $ 1.40 / hour we want the employee ' s income to remain at $ 100 / week . so , we want ( $ 1.40 / hour ) ( new # of hours ) = $ 100 divide both sides by 1.40 to get : new # of hours = 100 / 1.40 ≈ 71 hours so , the number of hours decreases from 100 hours to ( approximately ) 71 hours . this represents a 29 % decrease ( approximately ) . answer : c"
Answer: c
Problem: a boat moves upstream at the rate of 1 km in 20 minutes and down stream 1 km in 9 minutes . then the speed of the current is :
Options: a ) 1 kmph , b ) 2 kmph , c ) 3 kmph , d ) 2.5 kmph , e ) 1.85 kmph
Rationale: "rate upstream = ( 1 / 20 * 60 ) = 3 kmph rate dowm stream = 1 / 9 * 60 = 6.7 kmph rate of the current = ½ ( 6.7 - 3 ) = 1.85 kmph answer : e"
Answer: e
Problem: if the selling price of 10 articles is same as the cost price of 12 articles . find the gain or loss percentage ?
Options: a ) 10 % , b ) 20 % , c ) 30 % , d ) 32 % , e ) none
Rationale: "let the c . p of each article be re 1 . then , s . p of 10 articles = c . p of 12 articles = rs . 12 now , c . p of 10 articles = rs . 10 , s . p of 10 articles = rs 12 gain = rs ( 12 - 10 ) = rs 2 . gain % = ( 2 / 10 × 100 ) % = 20 % b )"
Answer: b
Problem: a man swims downstream 51 km and upstream 18 km taking 3 hours each time , what is the speed of the man in still water ?
Options: a ) 2 , b ) 8.5 , c ) 9.5 , d ) 6.5 , e ) 11.5
Rationale: "51 - - - 3 ds = 17 ? - - - - 1 18 - - - - 3 us = 6 ? - - - - 1 m = ? m = ( 17 + 6 ) / 2 = 11.5 answer : e"
Answer: e
Problem: a jar of 312 marbles is divided equally among a group of marble - players today . if 2 people joined the group in the future , each person would receive 1 marble less . how many people are there in the group today ?
Options: a ) 18 , b ) 20 , c ) 22 , d ) 24 , e ) 26
Rationale: "312 = 24 * 13 = 26 * 12 there are 24 people in the group today . the answer is d ."
Answer: d
Problem: find the simple interest on rs . 422 for 3 months at 2 paisa per month ?
Options: a ) s . 27.5 , b ) s . 34 , c ) s . 26 , d ) s . 25.28 , e ) s . 25.32
Rationale: "explanation : i = ( 422 * 3 * 2 ) / 100 = 25.32 answer : option e"
Answer: e
Problem: amit and ian paint a wall in alternating shifts . first amit paints alone , then ian paints alone , then amit paints alone , etc . during each of his shifts , amit paints 1 / 2 of the remaining unpainted area of the wall , while ian paints 1 / 3 of the remaining unpainted area of the wall during each of his shifts . if amit goes first , what fraction of the wall ' s area will remain unpainted after amit has completed his 6 th shift ?
Options: a ) 1 / 27
b ) 1 / 54
c ) 1 / 81
d ) 1 / 162
e ) 1 / 486
Rationale: fraction of area unpainted after first shift of amit = 1 - ( 1 / 2 ) = 1 / 2 fraction of area unpainted after first shift of ian = ( 1 / 2 ) - ( 1 / 3 ) ( 1 / 2 ) = ( 2 / 3 ) ( 1 / 2 ) fraction of area unpainted after second shift of amit = ( 1 / 2 ) ( 2 / 3 ) ( 1 / 2 ) fraction of area unpainted after second shift of ian = ( 2 / 3 ) ( 1 / 2 ) ( 2 / 3 ) ( 1 / 2 ) fraction of area unpainted after third shift of amit = ( 1 / 2 ) ( 2 / 3 ) ( 1 / 2 ) ( 2 / 3 ) ( 1 / 2 ) fraction of area unpainted after third shift of ian = ( 2 / 3 ) ( 1 / 2 ) ( 2 / 3 ) ( 1 / 2 ) ( 2 / 3 ) ( 1 / 2 ) fraction of area unpainted after forth shift of amit = ( 1 / 2 ) ( 2 / 3 ) ( 1 / 2 ) ( 2 / 3 ) ( 1 / 2 ) ( 2 / 3 ) ( 1 / 2 ) fraction of area unpainted after forth shift of ian = ( 2 / 3 ) ( 1 / 2 ) ( 2 / 3 ) ( 1 / 2 ) ( 2 / 3 ) ( 1 / 2 ) ( 2 / 3 ) ( 1 / 2 ) fraction of area unpainted after fifth shift of amit = ( 1 / 2 ) ( 2 / 3 ) ( 1 / 2 ) ( 2 / 3 ) ( 1 / 2 ) ( 2 / 3 ) ( 1 / 2 ) ( 2 / 3 ) ( 1 / 2 ) fraction of area unpainted after fifth shift of ian = ( 2 / 3 ) ( 1 / 2 ) ( 2 / 3 ) ( 1 / 2 ) ( 2 / 3 ) ( 1 / 2 ) ( 2 / 3 ) ( 1 / 2 ) ( 2 / 3 ) ( 1 / 2 ) fraction of area unpainted after sixth shift of amit = ( 1 / 2 ) ( 2 / 3 ) ( 1 / 2 ) ( 2 / 3 ) ( 1 / 2 ) ( 2 / 3 ) ( 1 / 2 ) ( 2 / 3 ) ( 1 / 2 ) ( 2 / 3 ) ( 1 / 2 ) = 486 answer : option e
Answer: e
Problem: the radius of a semi circle is 6.7 cm then its perimeter is ?
Options: a ) 32.75 , b ) 32.45 , c ) 22.45 , d ) 34.45 , e ) 32.15
Rationale: "36 / 7 r = 6.7 = 34.45 answer : d"
Answer: d
Problem: john was 19 years old when he married betty . they just celebrated their fifth wedding anniversary , and betty ' s age is now 7 / 8 of john ' s . how old is betty ?
Options: a ) 21 , b ) 26 , c ) 28 , d ) 30 , e ) 32
Rationale: "assume betty ' s age on marriage = x years . john ' s age on marriage = 19 john ' s age after 5 years = 24 years . betty ' s age after 5 years = x + 5 given : x + 5 = 7 / 8 ( 24 ) = 21 therefore betty ' s current age = 21 option a"
Answer: a
Problem: a train 70 m long , running with a speed of 63 km / hr will pass a tree in ?
Options: a ) 140 sec , b ) 160 sec , c ) 176 sec , d ) 150 sec , e ) 170 sec
Rationale: "speed = 63 * 5 / 18 = 35 / 2 m / sec time taken = 70 * 2 / 35 = 140 sec answer : a"
Answer: a
Problem: a box has exactly 100 balls , and each ball is either red , blue , or white . if the box has 10 more blue balls than white balls , and thrice as many red balls as blue balls , how many white balls does the box has ?
Options: a ) 8 , b ) 10 , c ) 12 , d ) 14 , e ) 16
Rationale: "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 10 more blue balls than white balls . . . equation # 2 : y = 10 + z . . . thrice as many red balls as blue balls . . . equation # 3 : x = 3 y solve equation # 2 for z : z = y - 10 now , we can replace both x and z with y in equation # 1 3 y + y + ( y - 10 ) = 100 5 y - 10 = 100 5 y = 110 y = 22 there are 22 blue balls . this is 10 more than the number of white balls , so z = 12 . that ' s the answer . just as a check , x = 66 , and 66 + 22 + 12 = 100 . answer = 12 , ( c )"
Answer: c
Problem: right now , al and eliot have bank accounts , and al has more money than eliot . the difference between their two accounts is 1 / 12 of the sum of their two accounts . if al ’ s account were to increase by 10 % and eliot ’ s account were to increase by 20 % , then al would have exactly $ 21 more than eliot in his account . how much money does eliot have in his account right now ?
Options: a ) $ 110 , b ) $ 120 , c ) $ 180 , d ) $ 220 , e ) $ 210
Rationale: "lets assume al have amount a in his bank account and eliot ' s bank account got e amount . we can form an equation from the first condition . a - e = 1 / 12 * ( a + e ) = = > 11 a = 13 e - - - - - - - - - - - - ( 1 ) second condition gives two different amounts , al ' s amount = 1.1 a and eliot ' s amount = 1.2 e 1.1 a = 21 + 1.2 e = = > 11 a = 210 + 12 e - - - - - - - ( 2 ) substituting ( 1 ) in ( 2 ) : 13 e = 210 + 12 e = = > e = 210 e"
Answer: e
Problem: at a florist shop on a certain day , all corsages sold for either $ 20 or $ 30 . if 8 of the corsages that sold for $ 30 had instead sold for $ 20 , then the store ' s revenue from corsages that day would have been reduced by 35 percent . what was the store ' s actual revenue from corsages that day ?
Options: a ) $ 320 , b ) $ 400 , c ) $ 600 , d ) $ 800 , e ) $ 1000
Rationale: i am doing it elaborately , hope it will help you . let , no . of corsages @ $ 20 = x , no . of corsages @ $ 30 = y and revenue = r so , 20 x + 30 y = r . . . . . . . . . ( 1 ) now , given the situation , 20 ( x + 8 ) + 30 ( y - 8 ) = r - . 25 r = > 20 x + 160 + 30 y - 240 = . 75 r = > 20 x + 30 y = . 75 r + 80 . . . . . . . . . . . . ( 2 ) so , r = . 75 r + 80 = > r = 320 the answer is a .
Answer: a
Problem: in a group of houses , 40 had dogs , 30 had cats and 10 houses having both dogs and cats . what is the number of houses ?
Options: a ) 30 , b ) 60 , c ) 40 , d ) 45 , e ) 50
Rationale: make a venn diagram , and enter your data . let the number of houses be x 30 + 10 + 20 = x x = 60 so number of houses were = 60 answer b
Answer: b
Problem: cricket match is conducted in us . the run rate of a cricket game was only 3.2 in first 10 over . what should be the run rate in the remaining 40 overs to reach the target of 282 runs ?
Options: a ) 6 , b ) 6.25 , c ) 7.25 , d ) 7.5 , e ) 8
Rationale: "required run rate = 282 - ( 3.2 x 10 ) = 250 = 6.25 40 40 b"
Answer: b
Problem: the product of x and y is a constant . if the value of x is increased by 30 % , by what percentage must the value of y be decreased ?
Options: a ) 16 % , b ) 23.07 % , c ) 25 % , d ) 30 % , e ) 35 %
Rationale: "x * y = constt . let x = y = 100 in beginning i . e . x * y = 100 * 100 = 10000 x ( 100 ) - - - becomes - - - > 1.3 x ( 130 ) i . e . 130 * new ' y ' = 10000 i . e . new ' y ' = 10000 / 130 = 76.92 i . e . y decreases from 100 to 76.92 i . e . decrease of 23.07 % b"
Answer: b
Problem: two employees a and b are paid a total of rs . 550 per week by their employer . if a is paid 120 percent of the sum paid to b , how much is b paid per week ?
Options: a ) rs . 150 , b ) rs . 190 , c ) rs . 200 , d ) rs . 250 , e ) rs . 300
Rationale: "let the amount paid to a per week = x and the amount paid to b per week = y then x + y = 550 but x = 120 % of y = 120 y / 100 = 12 y / 10 ∴ 12 y / 10 + y = 550 ⇒ y [ 12 / 10 + 1 ] = 550 ⇒ 22 y / 10 = 550 ⇒ 22 y = 5500 ⇒ y = 5500 / 22 = 500 / 2 = rs . 250 d )"
Answer: d
Problem: the average age of students of a class is 15.8 years . the average age of boys in the class is 16.4 years and that of the girls is 15.6 years , the ratio of the number of boys to the number of girls in the class is
Options: a ) 7 : 3 , b ) 1 : 3 , c ) 9 : 3 , d ) 6 : 3 , e ) 2 : 5
Rationale: "explanation : let the ratio be k : 1 . then , k * 16.4 + 1 * 15.6 = ( k + 1 ) * 15.8 < = > ( 16.4 - 15.8 ) k = ( 15.8 - 15.6 ) < = > k = 0.2 / 0.6 = 1 / 3 . required ratio = 1 / 3 : 1 = 1 : 3 . answer : b"
Answer: b