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a ) 765 , b ) 907 , c ) 944 , d ) 954 , e ) 968 | e | multiply(add(add(add(add(multiply(const_100, const_100), multiply(const_100, const_10)), multiply(const_100, const_3)), multiply(3, const_10)), const_3), 88) | find the largest 3 digit number which is exactly divisible by 88 ? | "largest 3 digit number is 999 after doing 999 Γ· 88 we get remainder 31 hence largest 3 digit number exactly divisible by 88 = 999 - 31 = 968 e" | a = 100 * 100
b = 100 * 10
c = a + b
d = 100 * 3
e = c + d
f = 3 * 10
g = e + f
h = g + 3
i = h * 88
|
a ) 69 , b ) 75 , c ) 82 , d ) 91 , e ) 56 | a | add(divide(add(add(add(add(17, 27), 31), 53), 61), add(const_4, const_1)), 31) | if a = { 17 , 27 , 31 , 53 , 61 } , what is the sum of mean and median of the numbers in a ? | mean = ( 17 + 27 + 31 + 53 + 62 ) / 5 = 38 median = 31 sum = 38 + 31 = 69 option a | a = 17 + 27
b = a + 31
c = b + 53
d = c + 61
e = 4 + 1
f = d / e
g = f + 31
|
a ) 5 % , b ) 6 % , c ) 7 % , d ) 8 % , e ) 9 % | d | multiply(divide(divide(840, 5), divide(divide(840, 8), divide(5, const_100))), const_100) | the simple interest on a certain sum of money at the rate of 5 % p . a . for 8 years is rs . 840 . at what rate of intrest the same amount of interest can be received on the same sum after 5 years . | "explanation : here firstly we need to calculate the principal amount , then we can calculate the new rate . p = s . i . β 100 / r β t p = 840 β 100 / 5 β 8 p = 2100 required rate = 840 β 100 / 5 β 2100 r = 8 % option d" | a = 840 / 5
b = 840 / 8
c = 5 / 100
d = b / c
e = a / d
f = e * 100
|
a ) 29 / 34 , b ) 47 / 100 , c ) 3 / 20 , d ) 13 / 102 , e ) none of these | d | multiply(divide(multiply(divide(52, const_4), divide(52, const_4)), multiply(52, 52)), const_2) | two cards are drawn together from a pack of 52 cards . the probability that one is a club and other is a diamond , is : | explanation : one card drawn . probability of being it club or diamond = ( 13 x 2 ) / 52 . another card drawn . now , probability of being it of other suit ( the suit among the two , not selected in first draw ) = 13 / 51 . hence , probability of event = ( ( 13 x 2 ) / 52 ) x ( 13 / 51 ) = 13 / 102 . answer : d | a = 52 / 4
b = 52 / 4
c = a * b
d = 52 * 52
e = c / d
f = e * 2
|
a ) 5 , b ) 6 , c ) 10 , d ) 13 , e ) 28 | c | divide(160, divide(add(negate(4), sqrt(add(power(4, 2), multiply(4, multiply(160, 2))))), 2)) | a rectangular tiled patio is composed of 160 square tiles . the rectangular patio will be rearranged so that there will be 2 fewer columns of tiles and 4 more rows of tiles . after the change in layout , the patio will still have 160 tiles , and it will still be rectangular . how many rows are in the tile patio before the change in layout ? | "suppose there are c columns and there are r rows original situation so , number of tiles = c * r = 160 also . reach column has r tiles and each row has c tiles new situation number of tiles in each column is r - 2 and number of tiles in each row is c + 4 so , number of rows = r - 2 and number of columns is c + 4 so , number of tiles = ( r - 2 ) * ( c + 4 ) = 160 comparing both of them we get c * r = ( r - 2 ) * ( c + 4 ) = > 4 r - 2 c = 8 c = 2 r - 4 putting it in c * r = 160 ( 2 r - 4 ) * r = 160 2 r ^ 2 - 4 r - 160 = 0 r can not be negative so r = 10 and c = 16 so , answer will be c" | a = negate + (
b = 4 ** 2
c = 160 * 2
d = 4 * c
e = b + d
f = math.sqrt(e)
g = a / f
h = 160 / g
|
a ) 2 , b ) 6 , c ) 4 , d ) 7 , e ) 5 | a | divide(multiply(8, 10), 40) | 8 machines can do a work in 10 days . how many machines are needed to complete the work in 40 days ? | "required number of machines = 8 * 10 / 40 = 2 answer is a" | a = 8 * 10
b = a / 40
|
a ) 44 % , b ) 48 % , c ) 52 % , d ) 42 % , e ) 46 % | b | multiply(subtract(const_1, multiply(divide(subtract(const_100, 40), const_100), divide(13, 15))), const_100) | a ' s speed is 15 / 13 times that of b . if a and b run a race , what part of the length of the race should a give b as a head start , so that b beats a by 40 % of the length of the race ? | "first calculate the distance , b has covered with his speed in the time , in which a reached 60 % of the race . then add the remaining distance as head start for b to win the race . its best to apply ratios concept here . since a ' s speed is 15 / 13 of b , therefore , b ' s speed is 13 / 15 of a distance covered by b = speed x time = ( 13 / 15 ) x ( 0.6 ) = 0.52 % ( which means b would have covered 0.52 of the race length during the time in which a has covered 0.6 of the race length . therefore to win , b needs a headstart of ( 1 - 0.52 = ) 0.48 of the race length . so the answer is b" | a = 100 - 40
b = a / 100
c = 13 / 15
d = b * c
e = 1 - d
f = e * 100
|
['a ) 2 units', 'b ) 12 units', 'c ) 6.5 units', 'd ) 6 units', 'e ) 7.5 units'] | a | divide(multiply(5, 12), add(add(5, 12), 13)) | what is the radius of the incircle of the triangle whose sides measure 5 , 12 and 13 units ? | explanatory answer in a right angled triangle , the radius of the incircle = s - h , where ' s ' is the semi perimeter of the triangle and ' r ' is the radius of the inscribed circle . the semi perimeter of the triangle = a + b + c / 2 = 5 + 12 + 13 / 2 = 15 therefore , r = 15 - 13 = 2 units . choice a | a = 5 * 12
b = 5 + 12
c = b + 13
d = a / c
|
a ) 18 , b ) 19 , c ) 10 , d ) 11 , e ) 12 | a | divide(414, subtract(24, const_1)) | along a yard 414 metres long , 24 trees are palnted at equal distances , one tree being at each end of the yard . what is the distance between two consecutive trees | explanation : 24 trees have 23 gaps between them , required distance ( 414 / 23 ) = 18 option a | a = 24 - 1
b = 414 / a
|
a ) 1.4 % , b ) 5.9 % , c ) 11.1 % , d ) 12.5 % , e ) 23.6 % | d | multiply(subtract(divide(9, 8), const_1), const_100) | at the opening of a trading day at a certain stock exchange , the price per share of stock k was $ 8 . if the price per share of stock k was $ 9 at the closing of the day , what was the percent increase in the price per share of stock k for that day ? | "opening = 8 closing = 9 rise in price = 1 so , percent increase = 18 β 10018 β 100 = > 12.50 % thus answer will be ( d )" | a = 9 / 8
b = a - 1
c = b * 100
|
a ) 10.5 % , b ) 12.8 % , c ) 15 % , d ) 22 % , e ) 30 % | b | divide(const_100, multiply(multiply(divide(10, const_100), divide(22, const_100)), const_100)) | on a certain road 10 % of the motorists exceed the posted speed limit and receive speeding tickets , but 22 % of the motorists who exceed the posted speed limit do not receive speeding tickets . what percent of the motorists on the road exceed the posted speed limit ? | "answer is b . this question is in the og and thus well explained by ets . those who exceed : x so x = 10 % + 0,22 x id est x = 12,8 %" | a = 10 / 100
b = 22 / 100
c = a * b
d = c * 100
e = 100 / d
|
a ) 800 , b ) 710 , c ) 720 , d ) 730 , e ) 740 | a | add(480, multiply(480, divide(40, const_100))) | a fruit seller had some oranges . he sells 40 % oranges and still has 480 oranges . how many oranges he had originally ? | "60 % of oranges = 480 100 % of oranges = ( 480 Γ 100 ) / 6 = 800 total oranges = 700 answer : a" | a = 40 / 100
b = 480 * a
c = 480 + b
|
a ) 18.33 kg , b ) 21.50 kg , c ) 25.00 kg , d ) 33.33 kg , e ) 40.39 kg | d | multiply(divide(divide(multiply(subtract(const_100, 70), 100), const_100), subtract(const_100, 10)), const_100) | fresh grapes contain 70 % by weight while dried grapes contain 10 % water by weight . what is the weight of dry grapes available from 100 kg of fresh grapes ? | "from the question we know : 100 kg * 70 % = 70 kg of water in the fresh grapes 100 kg - 70 kg of water = 30 kg of non - water mass we are looking for the weight of the dry grapes ( x ) . since the question tells us that 10 % of the weight of the dry graps is water and we know that 30 kg is non - water mass we can set up the following equation : x = 1 / 10 ( x ) + 30 kg 9 / 10 ( x ) = 30 kg x = 33.33 kg answer - d" | a = 100 - 70
b = a * 100
c = b / 100
d = 100 - 10
e = c / d
f = e * 100
|
a ) 500 , b ) 600 , c ) 650 , d ) 700 , e ) 750 | d | multiply(subtract(multiply(add(20, multiply(2.5, const_2)), add(15, multiply(2.5, const_2))), multiply(20, 15)), 3.5) | a hall 20 m long and 15 m broad is surrounded by a verandah of uniform width of 2.5 m . the cost of flooring the verandah at the rate of rs . 3.50 per sq . meter is | area of verandah = [ ( 25 Γ£ Ζ Γ’ β¬ β 20 ) ? ( 20 Γ£ Ζ Γ’ β¬ β 15 ) ] m Γ’ Β² = 200 m Γ’ Β² cost of flooring = rs . ( 200 Γ£ Ζ Γ’ β¬ β 3.50 ) = rs . 700 answer : d | a = 2 * 5
b = 20 + a
c = 2 * 5
d = 15 + c
e = b * d
f = 20 * 15
g = e - f
h = g * 3
|
a ) 76 kmph , b ) 6 kmph , c ) 14 kmph , d ) 8 kmph , e ) 4 kmph | b | divide(subtract(divide(90, 3), divide(55, 3)), const_2) | a man rows his boat 90 km downstream and 55 km upstream , taking 3 hours each time . find the speed of the stream ? | "speed downstream = d / t = 90 / ( 3 ) = 30 kmph speed upstream = d / t = 55 / ( 3 ) = 18 kmph the speed of the stream = ( 30 - 18 ) / 2 = 6 kmph answer : b" | a = 90 / 3
b = 55 / 3
c = a - b
d = c / 2
|
a ) $ 400.00 , b ) $ 412.40 , c ) $ 455.40 , d ) $ 474.90 , e ) $ 488.20 | a | divide(540, add(divide(35, const_100), const_1)) | jim is able to sell a hand - carved statue for $ 540 which was a 35 % profit over his cost . how much did the statue originally cost him ? | 540 = 1.35 * x x = 540 / 1.35 = 400 $ 400 , which is ( a ) . | a = 35 / 100
b = a + 1
c = 540 / b
|
a ) 21 , b ) 22 , c ) 23 , d ) 20 , e ) 24 | e | divide(multiply(240, const_3_6), add(30, 6)) | the speed at which a girl can row a boat in still water is 30 kmph . if she rows downstream , where the speed of current is 6 kmph , what time will he take to cover 240 metres ? | speed of the boat downstream = 30 + 6 = 36 kmph = 36 * 5 / 18 = 10 m / s hence time taken to cover 240 m = 240 / 10 = 24 seconds . answer : e | a = 240 * const_3_6
b = 30 + 6
c = a / b
|
a ) 52 , b ) 47 , c ) 48 , d ) 49 , e ) 50 | a | add(add(45, multiply(3, 2)), const_1) | the standard deviation of a normal distribution of data is 2 , and 3 standard deviations below the mean is greater than 45 . what is a possible value for the mean of the distribution ? | "the standard deviation ( { sd } ) = 2 ; 3 standard deviations below the mean is greater than 45 : { mean } - 3 * { sd } > 45 ; { mean } - 6 > 45 ; { mean } > 51 . answer : a ." | a = 3 * 2
b = 45 + a
c = b + 1
|
a ) 1 / 4 , b ) 5 , c ) 1 , d ) 2 , e ) 3 | b | multiply(divide(13, add(13, 13)), 5) | if 13 = 13 w / ( 1 - w ) , then ( 5 w ) 2 = | "13 - 13 w = 13 w 26 w = 13 w = 1 / 2 5 w = 5 / 2 5 w * 2 = 5 / 2 * 2 = 5 answer : b" | a = 13 + 13
b = 13 / a
c = b * 5
|
a ) 10 sec , b ) 32 sec , c ) 82 sec , d ) 20 sec , e ) 19 sec | e | divide(add(100, 280), multiply(add(42, 30), const_0_2778)) | two trains of length 100 m and 280 m are running towards each other on parallel lines at 42 kmph and 30 kmph respectively . in what time will they be clear of each other from the moment they meet ? | "relative speed = ( 42 + 30 ) * 5 / 18 = 4 * 5 = 20 mps . distance covered in passing each other = 100 + 280 = 380 m . the time required = d / s = 380 / 20 = 19 sec . answer : e" | a = 100 + 280
b = 42 + 30
c = b * const_0_2778
d = a / c
|
a ) 2 : 1 , b ) 4 : 7 , c ) 8 : 77 , d ) 4 : 81 , e ) none | d | power(divide(2, 9), 2) | the ratio of radius of a circle and the side of a square is 2 : 9 . find the ratio of their areas : | "radius / side = 2 / 9 Γ’ β‘ β area of circle / area of square = 4 / 81 answer : d" | a = 2 / 9
b = a ** 2
|
a ) 266.5 , b ) 413.3 , c ) 425.8 , d ) 189.3 , e ) 249.6 | b | multiply(120, multiply(divide(24, 12), divide(62, 36))) | if 12 men can reap 120 acres of land in 36 days , how many acres of land can 24 men reap in 62 days ? | "12 men 120 acres 36 days 24 men ? 62 days 120 * 24 / 12 * 62 / 36 120 * 2 * 31 / 18 240 * 1.72 = 413.3 answer : b" | a = 24 / 12
b = 62 / 36
c = a * b
d = 120 * c
|
a ) 30002 , b ) 70002 , c ) 50002 , d ) 90002 , e ) 10002 | e | subtract(10010, multiply(multiply(12, 3), 2)) | evaluate : 10010 - 12 * 3 * 2 = ? | "according to order of operations , 12 ? 3 ? 2 ( division and multiplication ) is done first from left to right 12 * * 2 = 4 * 2 = 8 hence 10010 - 12 * 3 * 2 = 10010 - 8 = 10002 correct answer e" | a = 12 * 3
b = a * 2
c = 10010 - b
|
a ) 50 miles , b ) 62 miles , c ) 325 miles , d ) 70 miles , e ) 20 miles | a | multiply(10, 5) | a car travels at a speed of 10 miles per hour . how far will it travel in 5 hours ? | "during each hour , the car travels 65 miles . for 5 hours it will travel 10 + 10 + 10 + 10 + 10 = 5 * 10 = 50 miles correct answer a" | a = 10 * 5
|
a ) 23 , b ) 22 , c ) 19 , d ) 20 , e ) 21 | e | divide(multiply(42, 25), 50) | if 42 men do a work in 25 days , in how many days will 50 men do it ? | "42 * 25 = 50 * x x = 21 days answer : e" | a = 42 * 25
b = a / 50
|
a ) 1235 , b ) 1345 , c ) 1678 , d ) 1767 , e ) 1611 | e | multiply(divide(subtract(1345, 15), subtract(6, const_1)), 6) | find large number from below question the difference of two numbers is 1345 . on dividing the larger number by the smaller , we get 6 as quotient and the 15 as remainder | let the smaller number be x . then larger number = ( x + 1345 ) . x + 1345 = 6 x + 15 5 x = 1330 x = 266 large number = 266 + 1345 = 1611 e | a = 1345 - 15
b = 6 - 1
c = a / b
d = c * 6
|
a ) 7 / 17 , b ) 14 / 31 , c ) 7 / 15 , d ) 17 / 35 , e ) 9 / 19 | e | divide(add(3, 2), const_10) | company s produces two kinds of stereos : basic and deluxe . of the stereos produced by company s last month , 2 / 3 were basic and the rest were deluxe . if it takes 9 / 5 as many hours to produce a deluxe stereo as it does to produce a basic stereo , then the number of hours it took to produce the deluxe stereos last month was what fraction of the total number of hours it took to produce all the stereos ? | "# of basic stereos was 2 / 3 of total and # of deluxe stereos was 1 / 3 of total , let ' s assume total = 15 , then basic = 10 and deluxe = 5 . now , if time needed to produce one deluxe stereo is 1 unit than time needed to produce one basic stereo would be 9 / 5 units . total time for basic would be 10 * 1 = 10 and total time for deluxe would be 5 * 9 / 5 = 9 - - > total time for both of them would be 10 + 9 = 19 - - > deluxe / total = 9 / 19 . answer : e ." | a = 3 + 2
b = a / 10
|
a ) 32 , b ) 28 , c ) 29 , d ) 44 , e ) 21 | d | divide(divide(add(250, 300), const_1000), divide(45, const_3600)) | a train 250 meters long completely crosses a 300 meters long bridge in 45 seconds . what is the speed of the train is ? | "s = ( 250 + 300 ) / 45 = 550 / 45 * 18 / 5 = 44 answer : d" | a = 250 + 300
b = a / 1000
c = 45 / 3600
d = b / c
|
a ) $ 50 , b ) $ 150 , c ) $ 69 , d ) $ 34 , e ) $ 52 | d | subtract(500, divide(500, add(divide(subtract(const_100, 80), subtract(const_100, 90)), const_1))) | the salaries of a and b together amount to $ 500 . a spends 90 % of his salary and b , 80 % of his . if now , their savings are the same , what is a ' s salary ? | "let a ' s salary is x b ' s salary = 500 - x ( 100 - 90 ) % of x = ( 100 - 80 ) % of ( 500 - x ) x = $ 34 ( approximately ) answer is d" | a = 100 - 80
b = 100 - 90
c = a / b
d = c + 1
e = 500 / d
f = 500 - e
|
a ) 8 , b ) 9 , c ) 10 , d ) 11 , e ) 18 | e | add(10, 8) | if there are 10 peanuts in a box and mary puts 8 more peanuts inside , how many peanuts are in the box ? | "10 + 8 = 18 correct answer is e ) 18" | a = 10 + 8
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a ) 220 , b ) 230 , c ) 280 , d ) 220 , e ) 240 | c | divide(factorial(subtract(add(const_4, 10), const_1)), multiply(factorial(10), factorial(subtract(const_4, const_1)))) | how many positive integers less than 1960 are such that the product of their digits is 10 . | "1960 is divisible by 10 ( or 2 * 5 ) = 196 196 is divisible by 4 = > 49 49 = 7 * 7 hence 1960 is divisible by 2 , 4,5 , 7 , 7 if there were only one 7 , i . e if the problem were 280 c" | a = 4 + 10
b = a - 1
c = math.factorial(b)
d = math.factorial(10)
e = 4 - 1
f = math.factorial(e)
g = d * f
h = c / g
|
a ) 2 , b ) 4 , c ) 70 , d ) 76 , e ) none | d | subtract(multiply(add(10, const_1), add(32, 4)), multiply(32, 10)) | the average runs of a cricket player of 10 innings was 32 . how many runs must he makes ih his next innings so as to increase his average of runs by 4 ? | "sol . average after 11 innings = 36 . therefore required numbert of runs = ( 36 Γ 11 ) β ( 32 Γ 10 ) = 396 - 320 = 76 . answer d" | a = 10 + 1
b = 32 + 4
c = a * b
d = 32 * 10
e = c - d
|
a ) 1 / 2 , b ) 1 / 3 , c ) 1 / 4 , d ) 1 / 6 , e ) 1 / 8 | d | divide(multiply(4, divide(2, 3)), 16) | an outlet pipe can empty 2 / 3 of a cistern in 16 minutes . in 4 minutes , what part of the cistern will be emptied ? | "4 / 16 * 2 / 3 = 1 / 6 the answer is d ." | a = 2 / 3
b = 4 * a
c = b / 16
|
a ) rs . 3000 , b ) rs . 4000 , c ) rs . 5000 , d ) rs . 6000 , e ) rs . 7000 | b | subtract(add(5050, 5200), 6250) | the average monthly income of p and q is rs . 5050 . the average monthly income of q and r is rs . 6250 and the average monthly income of p and r is rs . 5200 . the monthly income of p is ? | "let p , q and r represent their respective monthly incomes . then , we have : p + q = ( 5050 x 2 ) = 10100 . . . . ( i ) q + r = ( 6250 x 2 ) = 12500 . . . . ( ii ) p + r = ( 5200 x 2 ) = 10400 . . . . ( iii ) adding ( i ) , ( ii ) and ( iii ) , we get : 2 ( p + q + r ) = 33000 or p + q + r = 16500 . . . . ( iv ) subtracting ( ii ) from ( iv ) , we get p = 4000 . p ' s monthly income = rs . 4000 . b )" | a = 5050 + 5200
b = a - 6250
|
['a ) 160', 'b ) 240', 'c ) 330', 'd ) 440', 'e ) 560'] | c | add(60, multiply(divide(multiply(60, const_3), const_2), 3)) | a rectangular wall is covered entirely with two kinds of decorative tiles : regular and jumbo . 1 / 3 of the tiles are jumbo tiles , which have a length three times that of regular tiles and have the same ratio of length to width as the regular tiles . if regular tiles cover 60 square feet of the wall , and no tiles overlap , what is the area of the entire wall ? | the number of jumbo tiles = x . the number of regular tiles = 2 x . assume the ratio of the dimensions of a regular tile is a : a - - > area = a ^ 2 . the dimensions of a jumbo tile is 3 a : 3 a - - > area = 9 a ^ 2 . the area of regular tiles = 2 x * a ^ 2 = 60 . the area of jumbo tiles = x * 9 a ^ 2 = 4.5 ( 2 x * a ^ 2 ) = 4.5 * 60 = 270 . total area = 60 + 270 = 330 . answer : c . | a = 60 * 3
b = a / 2
c = b * 3
d = 60 + c
|
a ) 7 , b ) 6 , c ) 5 , d ) 4 , e ) 8 | c | subtract(26, add(7, 10)) | in a group of 26 junior high school students , 7 take french , 10 take spanish , and 4 take both languages . the students taking both french and spanish are not counted with the 7 taking french or the 10 taking spanish . how many students are not taking either french or spanish ? | "c 5 add 7 + 10 + 4 to get 21 . then subtract 21 from the total students β 26 β 21 = 5 . answer is c" | a = 7 + 10
b = 26 - a
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a ) 15 litres , b ) 10 litres , c ) 30 litres , d ) 22 litres , e ) 16 litres | e | multiply(8, const_1) | a mixture contains alcohol and water in the ratio 4 : 3 . if 8 litres of water is added to the mixture , the ratio becomes 4 : 5 . find the quantity of alcohol in the given mixture | "let the quantity of alcohol and water be 4 x litres and 3 x litres respectively 4 x / ( 3 x + 8 ) = 4 / 5 20 x = 4 ( 3 x + 8 ) 8 x = 32 x = 4 quantity of alcohol = ( 4 x 4 ) litres = 16 litres . answer is e ." | a = 8 * 1
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a ) 2012 , b ) 2044 , c ) 2060 , d ) 2072 , e ) 2080 | d | multiply(divide(add(subtract(130, const_3), add(19, const_2)), const_2), add(divide(subtract(subtract(130, const_3), add(19, const_2)), 4), const_1)) | what is the sum of the multiples of 4 between 19 and 130 inclusive ? | "the fastest way in an ap is to find the average and multiply with total integers . . between 19 and 130 , the smallest multiple of 4 is 20 and largest = 128 . . average = ( 20 + 128 ) / 2 = 148 / 2 = 74 . total numbers = ( 128 - 20 ) / 4 + 1 = 27 + 1 = 28 . . sum = 28 * 74 = 2072 ans d" | a = 130 - 3
b = 19 + 2
c = a + b
d = c / 2
e = 130 - 3
f = 19 + 2
g = e - f
h = g / 4
i = h + 1
j = d * i
|
a ) 50 % , b ) 33.33 % , c ) 25 % , d ) 46.66 % , e ) 12.5 % | d | multiply(subtract(divide(3, 4), multiply(divide(3, 4), divide(2, 5))), const_100) | a certain article of clothing was discounted during a special sale to 3 / 4 of its original retail price . when the clothing did n ' t sell , it was discounted even further to 2 / 5 of its original retail price during a second sale . by what percent did the price of this article of clothing decrease from the first sale to the second sale ? | say the original retail price of the item was $ 200 . the price after the first sale = 3 / 4 * $ 200 = $ 150 . the price after the second sale = 2 / 5 * $ 200 = $ 80 . the percent change from the first sale to the second = ( 150 - 80 ) / 150 = 1 / 3 = 46.66 % . answer : d . | a = 3 / 4
b = 3 / 4
c = 2 / 5
d = b * c
e = a - d
f = e * 100
|
a ) 6 , b ) 7 , c ) 8 , d ) 12 , e ) 14 | d | inverse(subtract(divide(const_1, 3), divide(const_1, 4))) | bruce and anne can clean their house in 4 hours working together at their respective constant rates . if anne β s speed were doubled , they could clean their house in 3 hours working at their respective rates . how many hours does it currently take anne to clean the house on her own ? | "lets suppose anne and bruce take a and b hrs working separately so in 1 hour they can together finish 1 / a + 1 / b portion of the work which equals 1 / 4 ( as the work is completed in 4 hours ) after anne doubles her rate of work the portion completed by the both is 1 / a + 2 / b which is equal to 1 / 3 ( as the work is completed in 3 hours ) solving these 2 equations we can find b as 12 answer : d" | a = 1 / 3
b = 1 / 4
c = a - b
d = 1/(c)
|
a ) 9 , b ) 12 , c ) 18 , d ) 28 , e ) 363 | d | multiply(add(3, const_1), add(6, const_1)) | if p and q are prime numbers , how many divisors does the product p ^ 3 * q ^ 6 have ? | ", p ^ 2 , p ^ 3 and q , q ^ 1 , q ^ 2 . . . . q ^ 6 together essentially there are 18 different combinations you can make ( 3 x 6 ) also you have to include the different combinations the p and q exponents can be a divisor . there are 3 different p exponents and 6 different q exponents : p , p ^ 2 , p ^ 3 , q , q ^ 2 , q ^ 3 . . . q ^ 6 also add 1 since ` ` 1 ' ' can also be a divisor . 18 + 3 + 6 + 1 = 28 , answer : d" | a = 3 + 1
b = 6 + 1
c = a * b
|
a ) 5 hours , b ) 6 hours , c ) 8 hours , d ) 10 hours , e ) 12 hours | a | divide(subtract(12, 10), 10) | two men started from the same place walk at the rate of 10 kmph and 12 kmph respectively . what time will they take to be 10 km apart , if they walk in the same direction ? | "to be 2 km apart they take 1 hour to be 10 km apart they take = 1 / 2 * 10 = 5 hours answer is a" | a = 12 - 10
b = a / 10
|
a ) 1,200 , b ) 1,800 , c ) 2,100 , d ) 2,250 , e ) 2,540 | b | divide(540, 540) | the rate of interest on a sum of money is 3 % 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 , occurred on the sum for the total period of 8 years is rs . 540 / - , the sum is | "explanation : i 1 = ( p x 3 x 3 ) / 100 = p / 11 i 2 = ( p x 4 x 4 ) / 100 = 4 p / 25 i 3 = ( p x 1 x 5 ) / 100 = p / 20 p / 11 + 4 p / 25 + p / 20 = 540 30 p / 100 = 540 p = 1800 answer : option b" | a = 540 / 540
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a ) 32,300 , b ) 172,800 , c ) 468,830 , d ) 338,200 , e ) 259,200 | b | multiply(multiply(subtract(6, 2), const_3600), const_12) | in a renowned city , the average birth rate is 6 people every two seconds and the death rate is 2 people every two seconds . estimate the size of the population net increase that occurs in one day . | "every 2 seconds , 4 persons are added ( 6 - 2 ) . every second 2 persons are added . in a day 24 hrs = 24 * 60 minutes = 24 * 60 * 60 = 86400 seconds . 86400 * 2 = 172800 option b is the answer ." | a = 6 - 2
b = a * 3600
c = b * 12
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a ) 12.6 . , b ) 15.8 . , c ) 14.4 . , d ) 16.2 . , e ) 16.4 . | b | subtract(add(multiply(2, 7.2), subtract(8.3, divide(const_4, const_10))), 6.6) | for every x , the action [ x ] is defined : [ x ] is the greatest integer less than or equal to x . what is the value of [ 6.5 ] x [ 2 / 3 ] + [ 2 ] x 7.2 + [ 8.3 ] - 6.6 ? | "[ 6.5 ] x [ 2 / 3 ] + [ 2 ] x 7.2 + [ 8.3 ] - 6.6 = 6 * 0 + 2 * 7.2 + 8 - 6.6 = 0 + 14.4 + 1.4 15.8 answer b" | a = 2 * 7
b = 4 / 10
c = 8 - 3
d = a + c
e = d - 6
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a ) 300 , b ) 320 , c ) 330 , d ) 340 , e ) 290 | a | divide(1, divide(add(multiply(const_3600, divide(1, 300)), 2), const_3600)) | a car traveling at a certain constant speed takes 2 seconds longer to travel 1 kilometer than it would take to travel 1 kilometer at 300 kilometers per hour . at what speed , in kilometers per hour , is the car traveling ? | "answer a 300 * t = 1 km = > t = 1 / 360 km / h v * ( t + 2 / 3600 ) = 1 v ( 1 / 360 + 2 / 3600 ) = 1 = > v = 300 km / h" | a = 1 / 300
b = 3600 * a
c = b + 2
d = c / 3600
e = 1 / d
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a ) 9 , b ) 10 , c ) 11 , d ) 12 , e ) 8 | a | subtract(add(add(13, 15), 6), 25) | in a group of 25 , 13 can speak latin , 15 can speak french , and 6 do n ' t speak either . how many of these speak both latin and french ? | let no . of persons who speeks booth latin and french = x so , no . of persons who speeks only latin = 13 - x no . of persons who speeks only french = 15 - x no . of persons who speeks dont speek any lag = 6 therefore ( 13 - x ) + ( 15 - x ) + x + 6 = 25 34 - x = 25 x = 9 answer : a | a = 13 + 15
b = a + 6
c = b - 25
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a ) 33 , b ) 27 , c ) 66.6 , d ) 22 , e ) 72 | c | multiply(const_3_6, divide(add(250, 120), 20)) | a train of length 250 m crosses a bridge of length 120 m in 20 seconds . what is the speed of train ? | "sol : ( length of train + length of bridge ) = speed of train x time ( 250 + 120 ) = 20 x speed speed = 370 / 20 = 18.5 m / s = 66.6 km / h answer = c" | a = 250 + 120
b = a / 20
c = const_3_6 * b
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a ) 50 , b ) 100 , c ) 150 , d ) 300 , e ) 250 | d | divide(subtract(multiply(150, divide(60, const_100)), multiply(150, divide(55, const_100))), subtract(divide(55, const_100), divide(50, const_100))) | a survey of n people in the town of eros found that 50 % of them preferred brand a . another survey of 150 people in the town of angie found that 60 % preferred brand a . in total , 55 % of all the people surveyed together preferred brand a . what is the total number of people surveyed ? | "it is simply a weighted average question . since the given average of 50 % and 60 % is 55 % ( right in the middle ) , it means the number of people surveyed in eros ( n ) is same as the number of people surveyed in angie . so n = 150 total = 150 + 150 = 300 answer ( d )" | a = 60 / 100
b = 150 * a
c = 55 / 100
d = 150 * c
e = b - d
f = 55 / 100
g = 50 / 100
h = f - g
i = e / h
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a ) 1 : 3 , b ) 1 : 2 , c ) 2 : 3 , d ) 2 : 1 , e ) 3 : 1 | b | divide(subtract(50, 40), subtract(40, 20)) | in what ratio should a 20 % methyl alcohol solution be mixed with a 50 % methyl alcohol solution so that the resultant solution has 40 % methyl alcohol in it ? | two things to remember : - 1 ) the ratio of quantity mixed to get an average is related to the ratio of each qty ' s distance from average . 2 ) more the qty of x from that of y , more closer will be average to x . . qty of 20 % / qty of 50 % = ( 50 β 40 ) / ( 40 β 20 ) = 1 / 2 = 1 : 2 answer : b | a = 50 - 40
b = 40 - 20
c = a / b
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a ) 7 , b ) 11 , c ) 13 , d ) 23 , e ) 38 | c | divide(multiply(42, divide(65, const_100)), const_2) | each of the 42 points is placed either inside or on the surface of a perfect sphere . if 65 % or fewer of the points touch the surface , what is the maximum number of segments which , if connected from those points to form chords , could be the diameter of the sphere ? | "maximum number of points on the surface is 65 % * 42 = 27.3 . . . or 27 since it has to be an integer now note that if two points form a diameter , they can not be part of any other diameter . so in the best case we can pair up the points we have 27 points , so at best we can form 13 pairs ( 26 ) . so , answer is ( c )" | a = 65 / 100
b = 42 * a
c = b / 2
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a ) rupees 1052 , b ) rs 1152 , c ) rs 1252 , d ) rs 1352 , e ) none of these | a | add(800, divide(multiply(multiply(800, add(divide(multiply(subtract(956, 800), const_100), multiply(800, 3)), 4)), 3), const_100)) | rs . 800 becomes rs . 956 in 3 years at a certain rate of simple interest . if the rate of interest is increased by 4 % , what amount will rs . 800 become in 3 years . | explanation : s . i . = 956 - 800 = rs 156 r = 156 β 100 / 800 β 3 r = 6 1 / 2 % new rate = 6 1 / 2 + 4 = 21 / 2 % new s . i . = 800 Γ 21 / 2 Γ 3100 = 252 now amount will be 800 + 252 = 1052 option a | a = 956 - 800
b = a * 100
c = 800 * 3
d = b / c
e = d + 4
f = 800 * e
g = f * 3
h = g / 100
i = 800 + h
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a ) 1 / 2 , b ) 1 / 3 , c ) 1 / 4 , d ) 1 / 5 , e ) 1 / 6 | c | divide(2, 8) | in the rectangular coordinate system , if the line x = 8 y + 5 passes through points ( m , n ) and ( m + 2 , n + p ) , what is the value of p ? | "x = 8 y + 5 , and thus y = x / 8 - 5 / 8 the slope is 1 / 8 . the slope of a line through points ( m , n ) and ( m + 2 , n + p ) is ( n + p - n ) / ( m + 2 - m ) = p / 2 p / 2 = 1 / 8 and thus p = 1 / 4 the answer is c ." | a = 2 / 8
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a ) 20 , b ) 24 , c ) 32 , d ) 16 , e ) 40 | d | divide(add(multiply(5, 10), multiply(5, 10)), subtract(add(5, 5), 5)) | a taxi leaves point a 5 hours after a bus left the same spot . the bus is traveling 10 mph slower than the taxi . find the speed of the taxi , if it overtakes the bus in three hours . | "let the speed of bus be v - 10 , speed of taxi be v the bus travelled a total of 8 hrs and taxi a total of 3 hrs . hence 8 * ( v - 10 ) = 3 v 8 v - 80 = 3 v 5 v = 80 d" | a = 5 * 10
b = 5 * 10
c = a + b
d = 5 + 5
e = d - 5
f = c / e
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a ) 90 , b ) 120 , c ) 240 , d ) 180 , e ) 110 | c | multiply(multiply(const_4.0, 4), const_10) | the length of a rectangular landscape is 12 times its breadth . there is a playground in it whose area is 1200 square mtr & which is 1 / 4 rd of the total landscape . what is the length of the landscape ? | "sol . x * 12 x = 4 * 1200 x = 20 length = 12 * 20 = 240 c" | a = 4 * 0
b = a * 10
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a ) 80 , b ) 91 , c ) 62 , d ) 74 , e ) 82 | e | subtract(100, subtract(82, 62)) | if x , y , and z are positive real numbers such that x + y - z = 62 , y - z + x = 82 , and z - x + y = 100 , then find the x + y + z value ? | "x + y - z = 62 , - - - - - - 1 ) y - z + x = 82 , - - - - - - 2 ) z - x + y = 100 - - - - - - 3 ) from ( 1 ) & ( 2 ) 2 z = 20 z = 10 from ( 2 ) & ( 3 ) 2 y = 182 y = 91 then x = 62 - 81 = - 19 x + y + z = 82 correct option is e" | a = 82 - 62
b = 100 - a
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a ) 24 , b ) 30 , c ) 40 , d ) 45 , e ) 50 | a | divide(400, multiply(const_0_2778, subtract(63, 3))) | how many seconds will a 400 metre long train take to cross a man walking with a speed of 3 km / hr in the direction of the moving train if the speed of the train is 63 km / hr ? | "relative speed of the train = 63 - 3 = 60 kmph = 60 * 5 / 18 = 50 / 3 m / sec t = 400 * 3 / 50 = 24 sec answer : a" | a = 63 - 3
b = const_0_2778 * a
c = 400 / b
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a ) 176 kmph , b ) 108 kmph , c ) 216 kmph , d ) 134 kmph , e ) 161 kmph | c | multiply(divide(240, 4), const_3_6) | a 240 meter long train crosses a man standing on the platform in 4 sec . what is the speed of the train ? | "s = 240 / 4 * 18 / 5 = 216 kmph answer : c" | a = 240 / 4
b = a * const_3_6
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a ) 4300 , b ) 4500 , c ) 8100 , d ) 5230 , e ) 5366 | c | subtract(subtract(10000, multiply(10000, divide(10, const_100))), multiply(subtract(10000, multiply(10000, divide(10, const_100))), divide(10, const_100))) | the population of a town is 10000 . it decreases annually at the rate of 10 % p . a . what will be its population after 2 years ? | "10000 Γ 90 / 100 Γ 90 / 100 = 8100 answer : c" | a = 10 / 100
b = 10000 * a
c = 10000 - b
d = 10 / 100
e = 10000 * d
f = 10000 - e
g = 10 / 100
h = f * g
i = c - h
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a ) 2.5 kmph , b ) 3.5 kmph , c ) 4.5 kmph , d ) 1.5 kmph , e ) 9.5 kmph | a | divide(add(divide(10, 3), divide(10, 6)), const_2) | a boat running downstream covers a distance of 10 km in 3 hr while for covering the same distance upstream , it takes 6 hr . what is the speed of the boat ? | speed upstream = 10 / 3 speed of boat down stream = 10 / 6 speed of boat = 1 / 2 ( 10 / 3 + 10 / 6 ) = 2.5 kmph answer : a | a = 10 / 3
b = 10 / 6
c = a + b
d = c / 2
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a ) 36 , b ) 37 1 / 4 , c ) 42 , d ) 54 , e ) 48 | a | divide(multiply(150, add(25, 15)), subtract(150, 25)) | a fort had provision of food for 150 men for 45 days . after 15 days , 25 men left the fort . the number of days for which the remaining food will last , is : | "after 15 days : 150 men had food for 30 days . suppose 125 men had food for x days . now , less men , more days ( indirect proportion ) therefore 125 : 150 : : 30 : x < = > 125 x x = 150 x 30 = > x = 150 x 30 / 125 = > x = 36 . correct answer is a" | a = 25 + 15
b = 150 * a
c = 150 - 25
d = b / c
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a ) 25650 , b ) 25750 , c ) 26550 , d ) 26750 , e ) 34200 | e | multiply(900, multiply(8, 4.75)) | the length of a room is 8 m and width is 4.75 m . what is the cost of paying the floor by slabs at the rate of rs . 900 per sq . metre . | "area = 8 Γ 4.75 sq . metre . cost for 1 sq . metre . = rs . 900 hence total cost = 8 Γ 4.75 Γ 900 = 8 Γ 4275 = rs . 34200 answer is e ." | a = 8 * 4
b = 900 * a
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a ) a ) 452000 , b ) b ) 200000 , c ) c ) 800000 , d ) d ) 500000 , e ) e ) 652000 | b | multiply(divide(40000, subtract(const_100, add(add(multiply(15, 3), 30), 5))), const_100) | a person distributed 15 % of his income to his 3 children each . he deposited 30 % of his income to his wife ' s account . he donated 5 % of remaining amount to an orphan house . finally he has $ 40000 . find his total income ? | "3 children got = 3 * 15 % = 45 % wife got = 30 % orphan house = 5 % total = 45 + 30 + 5 = 80 % remaining = 100 - 80 = 20 % 20 % = 40000 100 % = 40000 * 100 / 20 = $ 200000 answer is b" | a = 15 * 3
b = a + 30
c = b + 5
d = 100 - c
e = 40000 / d
f = e * 100
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a ) 4 , b ) 5 , c ) 6 , d ) 8 , e ) 9 | e | divide(81, subtract(const_10, const_1)) | the difference between a two - digit number and the number obtained by interchanging the positions of its digits is 81 . what is the difference between the two digits of that number ? | "sol . let the ten β s digit be x and unit β s digit be y , then , ( 10 x + y ) - ( 10 y + x ) = 81 β 9 ( x - y ) = 81 β x - y = 9 answer e" | a = 10 - 1
b = 81 / a
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['a ) 8.5 min', 'b ) 7.5 min', 'c ) 1.5 min', 'd ) 9.5 min', 'e ) 4.5 min'] | b | divide(multiply(subtract(4, const_1), 10), 4) | a and b runs around a circular track . a beats b by one round or 10 minutes . in this race , they had completed 4 rounds . if the race was only of one round , find the a ' s time over the course : | explanation : b runs around the track in 10 min . i . e , speed of b = 10 min per round a beats b by 1 round time taken by a to complete 4 rounds = time taken by b to complete 3 rounds = 30 min a ' s speed = 30 / 4 min per round = 7.5 min per round hence , if the race is only of one round a ' s time over the course = 7 min 30 sec answer : b | a = 4 - 1
b = a * 10
c = b / 4
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a ) 114 , b ) 115 , c ) 116 , d ) 117 , e ) 118 | e | divide(15, 0.127) | a certain industrial loom weaves 0.127 meters of cloth every second . approximately how many seconds will it take for the loom to weave 15 meters of cloth ? | "let the required number of seconds be x more cloth , more time , ( direct proportion ) hence we can write as ( cloth ) 0.127 : 15 : : 1 : x = > 0.127 * x = 15 = > x = 15 / 0.127 = > x = 118 answer : e" | a = 15 / 0
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a ) 52.6 , b ) 52.4 , c ) 52.1 , d ) 57 , e ) 52.9 | d | divide(add(multiply(35, 45), multiply(55, 65)), add(35, 55)) | the average marks of a class of 35 students is 45 and that of another class of 55 students is 65 . find the average marks of all the students ? | "sum of the marks for the class of 35 students = 35 * 45 = 1575 sum of the marks for the class of 55 students = 55 * 65 = 3575 sum of the marks for the class of 90 students = 1575 + 3575 = 5150 average marks of all the students = 5150 / 90 = 57 . answer : d" | a = 35 * 45
b = 55 * 65
c = a + b
d = 35 + 55
e = c / d
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a ) $ 60 , b ) $ 55 , c ) $ 42 , d ) $ 31 , e ) $ 100 | a | divide(100, add(divide(divide(1, 5), divide(3, 10)), const_1)) | a and b together have $ 100 . if 3 / 10 of a ' s amount is equal to 1 / 5 of b ' s amount , how much amount does b have ? | "3 / 10 a = 1 / 5 b a = 1 / 5 * 10 / 3 b a = 2 / 3 b a / b = 2 / 3 a : b = 2 : 3 b ' s share = 100 * 3 / 5 = $ 60 answer is a" | a = 1 / 5
b = 3 / 10
c = a / b
d = c + 1
e = 100 / d
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a ) 65 , b ) 38 , c ) 60 , d ) 28 , e ) 21 | c | subtract(subtract(150, 35), add(30, 30)) | two cars start from the opposite places of a main road , 150 km apart . first car runs for 30 km and takes a right turn and then runs 15 km . it then turns left and then runs for another 25 km and then takes the direction back to reach the main road . in the mean time , due to minor break down the other car has run only 35 km along the main road . what would be the distance between two cars at this point ? | answer : c ) 60 km | a = 150 - 35
b = 30 + 30
c = a - b
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a ) 20 % , b ) 25 % , c ) 18 % , d ) 30 % , e ) none of these | a | multiply(divide(divide(6, const_100), divide(30, const_100)), const_100) | if the given two numbers are respectively 6 % and 30 % of a third number , then what percentage is the first of the second ? | here , l = 6 and m = 30 therefore , first number = l / m x 100 % of second number = 6 / 30 x 100 % of second number = 20 % of second number answer : a | a = 6 / 100
b = 30 / 100
c = a / b
d = c * 100
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a ) 10 , b ) 15 , c ) 20 , d ) 25 , e ) 30 | a | divide(11, add(const_1, divide(10, const_100))) | mary can do a piece of work in 11 days . rosy is 10 % more efficient than mary . the number of days taken by rosy to do the same piece of work is ? | "ratio of times taken by mary and rosy = 110 : 100 = 11 : 10 suppose rosy takes x days to do the work . 11 : 10 : : 11 : x = > x = 10 days . hence , rosy takes 10 days to complete the work . answer : a" | a = 10 / 100
b = 1 + a
c = 11 / b
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a ) 1 / 216 , b ) 1 / 221 , c ) 1 / 212 , d ) 1 / 84 , e ) 1 / 42 | a | divide(const_1, power(subtract(divide(24, const_3), const_1), const_2)) | in a certain game of dice , the player β s score is determined as a sum of three throws of a single die . the player with the highest score wins the round . if more than one player has the highest score , the winnings of the round are divided equally among these players . if jim plays this game against 24 other players , what is the probability of the minimum score that will guarantee jim some monetary payoff ? | "toguaranteethat jim will get some monetary payoff he must score the maximum score of 6 + 6 + 6 = 18 , because if he gets even one less than that so 17 , someone can get 18 and jim will get nothing . p ( 18 ) = 1 / 6 ^ 3 = 1 / 216 . answer : a ." | a = 24 / 3
b = a - 1
c = b ** 2
d = 1 / c
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a ) 5 , b ) 10 , c ) 15 , d ) 20 , e ) 25 | d | multiply(divide(subtract(multiply(24, 5), 25), subtract(multiply(4, 5), 1)), 4) | 24 oz of juice p and 25 oz of juice v are mixed to make smothies m and y . the ratio of p to v in smothie m is 4 is to 1 and that in y is 1 is to 5 . how many ounces of juice p are contained in the smothie m ? | the ratio of p to v in smothie m is 4 is to 1 and that in y is 1 is to 5 . p 1 + p 2 = 24 v 1 + v 2 = 25 p 1 = 4 v 1 p 2 = v 2 / 5 4 v 1 + v 2 / 5 = 24 v 1 + v 2 = 25 4 v 2 - v 2 / 5 = 76 19 v 2 / 5 = 76 = > v 2 = 20 = > v 1 = 5 = > p 1 = 20 answer - d | a = 24 * 5
b = a - 25
c = 4 * 5
d = c - 1
e = b / d
f = e * 4
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a ) 2812 , b ) 8231 , c ) 2734 , d ) 2735 , e ) 4254 | d | divide(10940, 4) | a volume of 10940 l water is in a container of sphere . how many hemisphere of volume 4 l each will be required to transfer all the water into the small hemispheres ? | "a volume of 4 l can be kept in 1 hemisphere therefore , a volume of 10940 l can be kept in ( 10940 / 4 ) hemispheres ans . 2735 answer : d" | a = 10940 / 4
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a ) 87 , b ) 279 , c ) 80 , d ) 278 , e ) 379 | c | multiply(divide(multiply(40, const_1), subtract(80, 40)), 80) | a train leaves mumabai at 9 am at a speed of 40 kmph . after one hour , another train leaves mumbai in the same direction as that of the first train at a speed of 80 kmph . when and at what distance from mumbai do the two trains meet ? | "when the second train leaves mumbai the first train covers 40 * 1 = 40 km so , the distance between first train and second train is 40 km at 10.00 am time taken by the trains to meet = distance / relative speed = 40 / ( 80 - 40 ) = 1 hours so , the two trains meet at 11 a . m . the two trains meet 1 * 80 = 80 km away from mumbai . answer : c" | a = 40 * 1
b = 80 - 40
c = a / b
d = c * 80
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a ) 225 , b ) 325 , c ) 125 , d ) 145 , e ) none | b | subtract(negate(15), multiply(subtract(1, 4), divide(subtract(1, 4), subtract(0, 1)))) | 0 , 1 , 4 , 15 , 64 , ____ | "0 , 1,4 , 15,64 . . . . . 0 * 1 + 1 = 1 1 * 2 + 2 = 4 4 * 3 + 3 = 15 15 * 4 + 4 = 64 64 * 5 + 5 = 325 answer : b" | a = negate - (
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a ) 23 , b ) 32 , c ) 40 , d ) 41 , e ) 44 | c | add(divide(subtract(93, add(add(9, 9), 9)), const_3), add(9, 9)) | virginia , adrienne , and dennis have taught history for a combined total of 93 years . if virginia has taught for 9 more years than adrienne and for 9 fewer years than dennis , for how many years has dennis taught ? | "let number of years taught by virginia = v number of years taught by adrienne = a number of years taught by dennis = d v + a + d = 96 v = a + 9 = > a = v - 9 v = d - 9 = > a = ( d - 9 ) - 9 = d - 18 d - 9 + d - 18 + d = 93 = > 3 d = 93 + 27 = 120 = > d = 40 answer c" | a = 9 + 9
b = a + 9
c = 93 - b
d = c / 3
e = 9 + 9
f = d + e
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a ) 4 % , b ) 7 % , c ) 6 % , d ) 4 % , e ) 3 % | d | divide(divide(const_100, add(const_1, const_4)), 5) | if the simple interest on a certain sum of money for 5 years is one β fifth of the sum , then the rate of interest per annum is | "explanation : let the principal ( p ) be x then , simple interest ( si ) = x / 5 time ( t ) = 5 years rate of interest per annum ( r ) = ( 100 Γ si ) / pt = ( 100 Γ ( x / 5 ) / ( x Γ 5 ) = 20 / 5 = 4 % answer : option d" | a = 1 + 4
b = 100 / a
c = b / 5
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a ) 483224 , b ) 156444 , c ) 229376 , d ) 816416 , e ) 316544 | c | multiply(multiply(multiply(const_10, const_10), subtract(const_10, const_1)), 8) | how many six - digit numbers that do not contain the digits 5 or 8 are there ? | "we can have 7 digits ( 1 , 2,3 , 4,6 , 7,9 ) for the first place ( hundred thousand ' s place ) . and similarly 8 digits for ten thousands , thousands , hundred ' s , tenth ' s and unit digit . ( 0,1 , 2,3 , 4,6 , 7,9 ) so in total 7 * 8 * 8 * 8 * 8 * 8 = 229376 hence c" | a = 10 * 10
b = 10 - 1
c = a * b
d = c * 8
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a ) 761200 , b ) 761400 , c ) 761800 , d ) 17500 , e ) none of them | d | add(multiply(200, 200), multiply(150, 150)) | simplify : 200 x 200 - 150 x 150 | "( 200 ) ^ 2 - ( 150 ) ^ 2 = ( 200 + 150 ) ( 200 - 150 ) = 350 x 50 = 17500 . answer is d" | a = 200 * 200
b = 150 * 150
c = a + b
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a ) 28 , b ) 29 , c ) 30 , d ) 31 , e ) 32 | c | add(subtract(80, multiply(17, 3)), 3) | a batsman in his 17 th innings makes a score of 80 and their by increasing his average by 3 . what is his average after the 17 thinnings ? | "16 x + 80 = 17 ( x + 3 ) x = 27 + 3 = 30 answer : c" | a = 17 * 3
b = 80 - a
c = b + 3
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a ) 5 % , b ) 4 % , c ) 3 % , d ) 2 % , e ) 1 % | e | multiply(subtract(const_1, multiply(add(divide(10, const_100), const_1), divide(90, const_100))), const_100) | a customer bought a product at the shop . however , the shopkeeper increased the price of the product by 10 % so that the customer could not buy the required amount of the product . the customer managed to buy only 90 % of the required amount . what is the difference in the amount of money that the customer paid for the second purchase compared to the first purchase ? | "let x be the amount of money paid for the first purchase . the second time , the customer paid 0.9 ( 1.1 x ) = 0.99 x . the difference is 1 % . the answer is e ." | a = 10 / 100
b = a + 1
c = 90 / 100
d = b * c
e = 1 - d
f = e * 100
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a ) 8 % , b ) 15 % , c ) 25 % , d ) 52 % , e ) 56 % | c | multiply(divide(3, 30), const_100) | a pharmaceutical company received $ 3 million in royalties on the first $ 30 million in sales of and then $ 9 million in royalties on the next $ 120 million in sales . by approximately what percentage did the ratio of royalties to sales decrease from the first $ 30 million in sales to the next $ 120 million in sales ? | "( 9 / 120 ) / ( 3 / 30 ) = 3 / 4 = 75 % it means that 9 / 120 represents only 75 % . therefore a decrease of 25 % . answer c" | a = 3 / 30
b = a * 100
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a ) 2.5 , b ) 2.9 , c ) 2.4 , d ) 2.85 , e ) 2.1 | d | divide(100, multiply(126, const_0_2778)) | in what time will a train 100 m long cross an electric pole , it its speed be 126 km / hr ? | "speed = 126 * 5 / 18 = 35 m / sec time taken = 100 / 35 = 2.85 sec . answer : d" | a = 126 * const_0_2778
b = 100 / a
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a ) 4 , b ) 7 , c ) 12 , d ) 15 , e ) 20 | c | divide(multiply(8, const_3), const_2) | youseff lives x blocks from his office . it takes him 1 minute per block to walk to work and 20 seconds per block to ride his bike to work . it is takes him exactly 8 minutes more to walk to work than to ride his bike to work , then x equals ? | "please follow posting guidelines , link is in my signatures . as for your question , x / 60 = blocks / time / block = block ^ 2 / time . this is not what you want . you are given x blocks and 60 seconds per block . thus you need to put it as 60 * x to give you units of seconds as you are equating this to 480 ( which is time in seconds . ) . thus the correct equation is : 60 * x - 20 * x = 480 - - - - > 40 x = 480 - - > x = 12 . option c" | a = 8 * 3
b = a / 2
|
a ) 24 / 35 , b ) 17 / 35 , c ) 8 / 25 , d ) 6 / 35 , e ) 1 / 25 | d | multiply(add(multiply(divide(const_1, add(const_3, const_4)), divide(const_1, const_5)), multiply(divide(const_2, add(const_3, const_4)), divide(const_1, const_5))), const_2) | there are two names given johnson and jones . if one letter is picked from both simultaneously at random , then find the probability that the letter is same ? | this question can also be solved by using ' brute force ' - you just have to ' map out ' the possibilities . the namejohnsonhas 7 letters and the namejoneshas 5 letters , so there are ( 7 ) ( 5 ) = 35 possible pairings of letters that can occur . of those 35 , the pairings that include the same letter both times are : j and j 1 st o and o 1 st n and n s and s 2 nd o and o 2 nd n and n total : 6 so the probability that the two randomly selected letters will match is 6 / 35 . d | a = 3 + 4
b = 1 / a
c = 1 / 5
d = b * c
e = 3 + 4
f = 2 / e
g = 1 / 5
h = f * g
i = d + h
j = i * 2
|
a ) 0.02 , b ) 0.2 , c ) 0.04 , d ) 0.4 , e ) 4 | c | multiply(0.02, const_2) | the expression ( 11.98 Γ 11.98 + 11.98 Γ x + 0.02 Γ 0.02 ) will be a perfect square for x equal to | "solution given expression = ( 11.98 ) 2 + ( 0.02 ) 2 + 11.98 Γ x . for the given express to be a perfect square , we must have 11.98 Γ x = 2 Γ 11.98 Γ 0.02 or x = 0.04 . answer c" | a = 0 * 2
|
a ) $ 1,850 , b ) $ 2,450 , c ) $ 2,750 , d ) $ 3,150 , e ) $ 3,450 | b | floor(divide(add(divide(101.50, divide(7, const_100)), 1,000), 1,000)) | when a merchant imported a certain item , he paid a 7 percent import tax on the portion of the total value of the item in excess of $ 1,000 . if the amount of the import tax that the merchant paid was $ 101.50 , what was the total value of the item ? | "let x be the value in excess of $ 1,000 . 0.07 x = 101.5 x = $ 1,450 the total value was $ 1,450 + $ 1,000 = $ 2,450 . the answer is b ." | a = 7 / 100
b = 101 / 50
c = b + 1
d = c / 1
e = math.floor(d)
|
a ) 120 sec , b ) 198 sec , c ) 60 sec , d ) 665 sec , e ) 276 sec | c | divide(300, subtract(multiply(54, const_0_2778), multiply(18, const_0_2778))) | a and b go around a circular track of length 300 m on a cycle at speeds of 18 kmph and 54 kmph . after how much time will they meet for the first time at the starting point ? | "time taken to meet for the first time at the starting point = lcm { length of the track / speed of a , length of the track / speed of b } = lcm { 300 / ( 90 * 5 / 18 ) , 300 / ( 54 * 5 / 18 ) } = lcm ( 12 , 20 ) = 60 sec . answer : c" | a = 54 * const_0_2778
b = 18 * const_0_2778
c = a - b
d = 300 / c
|
a ) 5 , b ) 1 , c ) 2 , d ) 3 , e ) 4 | a | subtract(multiply(multiply(multiply(1457, 7863), 4575), 9719), subtract(multiply(multiply(multiply(1457, 7863), 4575), 9719), add(const_4, const_4))) | the unit digit in the product 1457 * 7863 * 4575 * 9719 is ? | "unit digit in the given product = unit digit in 7 * 3 * 5 * 9 = 5 answer is a" | a = 1457 * 7863
b = a * 4575
c = b * 9719
d = 1457 * 7863
e = d * 4575
f = e * 9719
g = 4 + 4
h = f - g
i = c - h
|
a ) 20000 , b ) 10000 , c ) 25000 , d ) 34000 , e ) 40000 | e | multiply(72000, subtract(const_1, divide(multiply(30000, multiply(2, multiply(2, const_3))), add(multiply(45000, subtract(multiply(2, multiply(2, const_3)), 2)), multiply(30000, multiply(2, multiply(2, const_3))))))) | tom opened a shop investing rs . 30000 . jose joined him 2 months later , investing rs . 45000 . they earned a profit of rs . 72000 after completion of one year . what will be jose ' s share of profit ? | "sol = ~ s - so anju β s share = [ 5 / 9 ] x 72000 = 40000 e" | a = 2 * 3
b = 2 * a
c = 30000 * b
d = 2 * 3
e = 2 * d
f = e - 2
g = 45000 * f
h = 2 * 3
i = 2 * h
j = 30000 * i
k = g + j
l = c / k
m = 1 - l
n = 72000 * m
|
a ) 178.89 cm , b ) 178.29 cm , c ) 178.79 cm , d ) 177.89 cm , e ) 178.69 cm | b | floor(divide(add(subtract(multiply(35, 180), 166), 106), 35)) | the average height of 35 boys in a class was calculated as 180 cm . it has later found that the height of one of the boys in the class was wrongly written as 166 cm whereas his actual height was 106 cm . find the actual average height of the boys in the class ( round off your answer to two decimal places ) . | calculated average height of 35 boys = 180 cm . wrong total height of 35 boys = 180 * 35 cm . this was as a result of an actual height of 106 cm being wrongly written as 166 cm . correct total height of 35 boys = 180 * 35 cm - 166 cm + 106 cm = 180 * 35 cm - 166 cm + 106 cm / 35 = 180 cm - 60 / 35 cm = 180 cm - 1.71 cm = 178.29 cm . answer : b | a = 35 * 180
b = a - 166
c = b + 106
d = c / 35
e = math.floor(d)
|
a ) 12 cm , b ) 14 cm , c ) 16 cm , d ) 18 cm , e ) 20 cm | d | divide(const_100, const_3) | the length of a rectangle is twice its breadth . if its length is decreased by 5 cm and breadth is increased by 5 cm , the area of the rectangle is increased by 70 sq . cm . find the length of the rectangle . | "explanation : let breadth = x . then , length = 2 x . then , ( 2 x - 5 ) ( x + 5 ) - 2 x * x = 70 = > 5 x - 25 = 70 = > x = 18 . length of the rectangle = 18 cm . answer : option d" | a = 100 / 3
|
a ) 3 / 2 , b ) 5 / 2 , c ) 7 / 2 , d ) 9 / 2 , e ) 11 / 2 | a | divide(add(2, 1), 2) | a + 2 b = 6 , ab = 4 2 / a + 1 / b = ? | 2 / 2 + 1 / 2 = 1 + 1 / 2 = 3 / 2 answer : a | a = 2 + 1
b = a / 2
|
a ) 15 % , b ) 18 % , c ) 20 % , d ) 25 % , e ) 26 % | c | multiply(divide(subtract(add(add(100, 20), 100), multiply(2, 100)), 100), 100) | a man saves a certain portion of his income during a year and spends the remaining portion on his personal expenses . next year his income increases by 20 % but his savings increase by 100 % . if his total expenditure in 2 years is double his expenditure in 1 st year , what % age of his income in the first year did he save ? | "1 st year income = i 1 st year savings = s 1 st year expense = e 1 2 nd year income = 1.2 i 2 nd year savings = 2 s ( 100 % increase ) 2 nd year expense = e 2 e 1 + e 2 = 2 e 1 e 2 = e 1 that means expenses are same during both years . with increase of 20 % income the savings increased by 100 % . or s = . 2 i or s = 20 % of income c is the answer" | a = 100 + 20
b = a + 100
c = 2 * 100
d = b - c
e = d / 100
f = e * 100
|
a ) 5 / 8 , b ) 3 / 4 , c ) 7 / 8 , d ) 57 / 64 , e ) 15 / 16 | c | subtract(const_1, add(multiply(inverse(power(2, 6)), 6), add(inverse(power(2, 6)), inverse(power(2, 6))))) | a fair 2 sided coin is flipped 6 times . what is the probability that tails will be the result at least twice , but not more than 5 times ? | it would be easier to calculate the probability of opposite event and subtract it from 1 . opposite event : 0 tail , 1 tail , 6 tails . probability of getting no tails : 1 / 2 ^ 6 = 1 / 64 probability of getting 1 tail : 6 c 1 β 1 / 2 ^ 6 = 6 / 64 , we must multiply by 6 c 1 or by 6 as tail can occur for any flip from 6 , hence in 6 ways ; probability of getting 6 tails : 1 / 2 ^ 6 = 1 / 6 ^ 4 p = 1 β ( 1 / 64 + 6 / 64 + 1 / 64 ) = 56 / 64 = 7 / 8 answer : c . | a = 2 ** 6
b = 1/(a)
c = b * 6
d = 2 ** 6
e = 1/(d)
f = 2 ** 6
g = 1/(f)
h = e + g
i = c + h
j = 1 - i
|
a ) rs . 2592 , b ) rs . 2520 , c ) rs . 2880 , d ) rs . 3360 , e ) none of these | a | multiply(8640, divide(add(divide(subtract(90000, add(6000, 3000)), const_3), 6000), 90000)) | a , b and c started a business with a total investment of rs . 90000 . a invests rs . 6000 more than b and b invests rs . 3000 less than c . if the total profit at the end of a year is rs . 8640 , find b ' s share . | "explanation : let c ' s investment = rs . x b ' s investment = rs . ( x - 3000 ) a ' s investment = rs . ( x - 3000 + 6000 ) = rs . ( x + 3000 ) now , ( a + b + c ) ' s investment = rs . 90000 = > x + ( x - 3000 ) + ( x + 3000 ) = 90000 = > 3 x = 90000 = > x = 30000 hence , a ' s investment = rs . 33000 b ' s investment = rs . 27000 c ' s investment = rs . 30000 ratio of the capitals of a , b and c = 33000 : 27000 : 30000 = 11 : 9 : 10 a ' s share = rs . [ ( 9 / 30 ) Γ£ β 8640 ] = rs . 2592 answer : option a" | a = 6000 + 3000
b = 90000 - a
c = b / 3
d = c + 6000
e = d / 90000
f = 8640 * e
|
a ) 5780 litres , b ) 5770 litres , c ) 5760 litres , d ) 5750 litres , e ) 5740 litres | c | divide(multiply(4, multiply(8, const_60)), subtract(divide(multiply(8, const_60), multiply(6, const_60)), const_1)) | a leak in the bottom of a tank can empty the full tank in 6 hours . an inlet pipe fills water at the rate of 4 litres a minute . when the tank is full , the inlet is opened and due to the leak the tank is empty in 8 hours . the capacity of the tank ( in litres ) is | "explanation : work done by the inlet in 1 hour = 1 / 6 β 1 / 8 = 1 / 24 work done by inlet in 1 min = 1 / 24 β 1 / 60 = 1 / 1440 = > volume of 1 / 1440 part = 4 liters volume of whole = ( 1440 * 4 ) litres = 5760 litres . option c" | a = 8 * const_60
b = 4 * a
c = 8 * const_60
d = 6 * const_60
e = c / d
f = e - 1
g = b / f
|
a ) 6 , b ) 7 , c ) 4 , d ) 5 , e ) 8 | c | subtract(subtract(const_100, 95), const_1) | how many digits will be there to the right of the decimal point in the product of 95 and . 02554 ? | product of 95 and . 02554 is 2.4263 . therefore number of digits to right of decimal point is 4 answer is c . | a = 100 - 95
b = a - 1
|
a ) 5 , b ) 6 , c ) 7 , d ) 8 , e ) 9 | e | add(9, const_1) | the average of first 9 odd numbers is ? | "sum of 9 odd no . = 90 average = 81 / 9 = 9 answer : e" | a = 9 + 1
|
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