options stringlengths 37 300 | correct stringclasses 5 values | annotated_formula stringlengths 7 727 | problem stringlengths 5 967 | rationale stringlengths 1 2.74k | program stringlengths 10 646 |
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a ) – 19 , b ) – 11 , c ) – 4 , d ) 4 , e ) 16 | a | add(negate(multiply(add(negate(3), 5), const_2)), multiply(negate(3), 5)) | if x * y = xy – 2 ( x + y ) for all integers x and y , then 5 * ( – 3 ) = | 5 * ( - 3 ) = 5 * ( - 3 ) - 2 ( 5 + ( - 3 ) ) = - 15 - 4 = - 19 option ( a ) | a = negate + (
b = a * 5
c = negate + (
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a ) 50 , b ) 52 , c ) 53 , d ) 55 , e ) 56 | c | subtract(98, subtract(add(floor(divide(98, add(const_4, const_1))), floor(divide(98, const_4))), floor(divide(98, add(const_10, add(const_4, const_1)))))) | two assembly line inspectors , lauren and steven , inspect widgets as they come off the assembly line . if lauren inspects every fourth widget , starting with the fourth , and steven inspects every third , starting with the third , how many of the 98 widgets produced in the first hour of operation are not inspected by either inspector ? | "widgets inspected by lauren : ( ( 96 - 4 ) / 4 ) + 1 = 23 + 1 = 24 widgets inspected by steven : ( ( 96 - 3 ) / 3 ) + 1 = 31 + 1 = 32 widgets inspected by both : ( ( 96 / 12 ) + 1 = 9 total : 24 + 32 - 9 = 45 hence , widgets not inspected : 98 - 45 = 53 option c" | a = 4 + 1
b = 98 / a
c = math.floor(b)
d = 98 / 4
e = math.floor(d)
f = c + e
g = 4 + 1
h = 10 + g
i = 98 / h
j = math.floor(i)
k = f - j
l = 98 - k
|
a ) 2.04 % , b ) 6.12 % , c ) 8 % , d ) 8.25 % , e ) 9 % | e | multiply(2.2, const_4) | on the first of the year , james invested x dollars at proudstar bank in an account that yields 2.2 % in interest every quarter year . at the end of the year , during which he made no additional deposits or withdrawals , he had y dollars in the account . if james had invested the same amount in an account which pays interest on a yearly basis , what must the interest rate be for james to have y dollars at the end of the year ? | "if the interest were compounded annually instead of quarterly then in one year the interest would be 2.2 * 4 = 8.8 % . now , since the interest is compounded quarterly then there would be interest earned on interest ( very small amount ) thus the actual interest would be a little bit more than 8.8 % answer : e ." | a = 2 * 2
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a ) 36 , b ) 35 , c ) 34 , d ) 38 , e ) 40 | c | multiply(add(subtract(9, const_1), 9), divide(22, add(subtract(6, const_1), 6))) | a certain clock marks every hour by striking a number of times equal to the hour , and the time require for a stroke is exactly equal to the time interval between strokes . at 6 : 00 the time lapse between the beginning of the first stoke and the end of the last stroke is 22 seconds . at 9 : 00 , how many seconds elapse between the beginning of the first stroke and the end of the last stroke ? | at 6 ' o clock , there would be 6 strikes . first strike , then a short interval , the second strike , then a short interval and so on till the 6 th strike . so there would be in all 5 intervals between 6 strikes . similarly , between 9 strikes , there would be 8 intervals . according to the question , the time spent in the strike and the interval is same . at 6 ' o clock , the 6 strikes and the 5 intervals together take 22 sec so each strike and each interval takes 2 secs . at 12 ' o clock , the 9 strikes and 8 intervals will take 2 * ( 9 + 8 ) = 34 secs c | a = 9 - 1
b = a + 9
c = 6 - 1
d = c + 6
e = 22 / d
f = b * e
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a ) $ 645.56 , b ) $ 4121.60 , c ) $ 954.26 , d ) $ 745.69 , e ) $ 1020.45 | b | subtract(multiply(power(add(divide(divide(4, const_100), 2), const_1), 4), 50000), 50000) | find the compound interest on $ 50000 in 2 years at 4 % per annum , the interest being compounded half - yearly ? | principle = $ 50000 rate = 2 % half yearly = 4 half years amount = 50000 * ( 1 + 2 / 100 ) ^ 4 = 50000 * 51 / 50 * 51 / 50 * 51 / 50 * 51 / 50 = $ 54121.60 c . i . = 54121.60 - 50000 = $ 4121.60 answer is b | a = 4 / 100
b = a / 2
c = b + 1
d = c ** 4
e = d * 50000
f = e - 50000
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a ) 4 hours , b ) 2 hours , c ) 3 hours , d ) 2 hours 24 minutes , e ) none | c | divide(subtract(multiply(60, 5), 240), subtract(60, 40)) | i travel the first part of my journey at 40 kmph and the second part at 60 kmph and cover the total distance of 240 km to my destination in 5 hours . how long did the first part of my journey last ? | "explanatory answer the total time of journey = 5 hours . let ' x ' hours be the time that i traveled at 40 kmph therefore , 5 - x hours would be time that i traveled at 60 kmph . hence , i would have covered x * 40 + ( 5 - x ) 60 kms in the 5 hours = 240 kms . solving , for x in the equation 40 x + ( 5 - x ) * 60 = 240 , we get 40 x + 300 - 60 x = 240 = > 20 x = 60 or x = 3 hours . answer c" | a = 60 * 5
b = a - 240
c = 60 - 40
d = b / c
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a ) 81 , b ) 63 , c ) 54 , d ) 36 , e ) 90 | d | divide(divide(factorial(divide(add(sqrt(add(multiply(const_4, multiply(36, const_2)), const_1)), const_1), const_2)), factorial(7)), const_2) | the set s has 36 different subsets each of which contains exactly two elements . how many subsets of s could contain exactly 7 elements each ? | nc 2 = 36 = > n * ( n - 1 ) / 2 = 36 by middle term factor and n can not be negative = > n = 9 nc 7 = 9 c 7 = 9 ! / 7 ! * ( 9 - 7 ) ! = 9 * 8 * 7 ! / 7 ! * 2 = 36 so , answer is d . | a = 36 * 2
b = 4 * a
c = b + 1
d = math.sqrt(c)
e = d + 1
f = e / 2
g = math.factorial(f)
h = math.factorial(7)
i = g / h
j = i / 2
|
a ) 17.5 litres , b ) 16.67 litres , c ) 17.67 litres , d ) 16.5 litres , e ) 16 litres | b | multiply(100, subtract(const_1, sqrt(divide(25, 36)))) | a 100 - litre mixture of milk and water contains 36 litres of milk . ' x ' litres of this mixture is removed and replaced with an equal quantum of water . if the process is repeated once , then the concentration of the milk stands reduced at 25 % . what is the value of x ? | "in 100 l mixture we have 36 l milk and 64 l water . hence m and w are in the ratio of 9 : 16 . x l of solution has been removed . hence we have 36 - 9 / 25 x of milk and 64 - 16 / 25 x of water . for calculation simplicity multiply and divide 9 / 25 x by 4 we get 36 / 100 x since this procedure is repeated 2 times . this was my equation 25 = 36 * ( 36 - 36 / 100 x ) ^ 2 ) / ( 36 ) ^ 2 by solving it . we get x = 100 / 6 or 50 / 3 or 16.67 answer : b" | a = 25 / 36
b = math.sqrt(a)
c = 1 - b
d = 100 * c
|
a ) 2.25 , b ) 3.25 , c ) 4.25 , d ) 14 , e ) 6.25 | d | subtract(power(4, 2), 2) | x + ( 1 / x ) = 4 find x ^ 2 + ( 1 / x ^ 2 ) | "squaring on both sides ( x + 1 / x ) ^ 2 = 4 ^ 2 x ^ 2 + 1 / x ^ 2 = 16 - 2 x ^ 2 + 1 / x ^ 2 = 14 answer : d" | a = 4 ** 2
b = a - 2
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a ) $ 972 , b ) $ 810 , c ) $ 915 , d ) $ 715 , e ) $ 795 | a | multiply(1200, power(subtract(const_1, divide(10, const_100)), 2)) | a present value of a machine is $ 1200 . its value depletion rate is 10 % per annum then find the machine value after 2 years ? | "p = $ 1200 r = 10 % t = 2 years machine value after 2 years = p [ ( 1 - r / 100 ) ^ t ] = 1200 * 9 / 10 * 9 / 10 = $ 972 answer is a" | a = 10 / 100
b = 1 - a
c = b ** 2
d = 1200 * c
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a ) 3 cm , b ) 4 cm , c ) 6 cm , d ) 8 cm , e ) none | d | sqrt(divide(multiply(multiply(const_pi, multiply(16, divide(16, const_2))), const_2), multiply(const_pi, const_4))) | the surface area of a sphere is same as the curved surface area of a right circular cylinder whose height and diameter are 16 cm each . the radius of the sphere is | "solution 4 î r 2 = 2 î 8 x 16 â ‡ ’ r 2 = ( 8 x 16 / 2 ) â ‡ ’ 64 â ‡ ’ r = 8 cm . answer d" | a = 16 / 2
b = 16 * a
c = math.pi * b
d = c * 2
e = math.pi * 4
f = d / e
g = math.sqrt(f)
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a ) 15 , b ) 16 , c ) 18 , d ) 19 , e ) 17 | e | add(const_10, add(4, const_3)) | for what value of k will the two equations 2 x + 4 = 4 ( x - 2 ) and - x + k = 2 x - 1 have the same solution ? | solve the first equation 2 x + 4 = 4 ( x - 2 ) to obtain . x = 6 substitute x by 6 ( same solution ) in the second equation and solve for k . - 6 + k = 2 ( 6 ) - 1 solve for k . k = 17 correct answer e | a = 4 + 3
b = 10 + a
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a ) 7297 , b ) 6425 , c ) 2871 , d ) 6725 , e ) 2981 | b | divide(7967, add(const_1, divide(24, const_100))) | the owner of a furniture shop charges his customer 24 % more than the cost price . if a customer paid rs . 7967 for a computer table , then what was the cost price of the computer table ? | "explanation : cp = sp * ( 100 / ( 100 + profit % ) ) = 7967 ( 100 / 124 ) = rs . 6425 . answer : b" | a = 24 / 100
b = 1 + a
c = 7967 / b
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a ) 16 , b ) 18 , c ) 17 , d ) 12 , e ) 12 | d | subtract(const_100, subtract(add(const_100, 10), divide(multiply(add(const_100, 10), 20), const_100))) | a fair price shopkeeper takes 10 % profit on his goods . he lost 20 % goods during theft . his loss percent is ? | "suppose he has 100 items . let c . p . of each item be re . 1 . total cost = rs . 100 . number of items left after theft = 80 . s . p . of each item = rs . 1.10 total sale = 1.10 * 80 = rs . 88 hence , loss % = 12 / 100 * 100 = 12 % answer : d" | a = 100 + 10
b = 100 + 10
c = b * 20
d = c / 100
e = a - d
f = 100 - e
|
a ) 12 kg , b ) 60 kg , c ) 72 kg , d ) 80 kg , e ) none of these | d | multiply(multiply(multiply(4, 2), divide(1, const_100)), const_1000) | a boat having a length 4 m and breadth 2 m is floating on a lake . the boat sinks by 1 cm when a man gets on it . the mass of the man is : | "explanation : volume of water displaced = ( 4 x 2 x 0.01 ) m 3 = 0.08 m 3 . ∴ mass of man = volume of water displaced x density of water = ( 0.08 x 1000 ) kg = 80 kg . answer : d" | a = 4 * 2
b = 1 / 100
c = a * b
d = c * 1000
|
a ) 1 , b ) 2 , c ) 3 , d ) 4 , e ) 5 | d | subtract(divide(add(divide(210, 6), divide(7, 4)), 6), const_2) | for an international mathematics olympiad , country d will send 6 delegates in total — two will be supervisors and 4 will be contestants . there are 210 ways in which the 6 delegates can be chosen and there are 7 candidates competing for the 4 contestants ’ places available . how many candidates are competing for the two supervisors ’ slots available ? | option a wrong as 2 supervisor need to be selected . going by statement : 4 contestant chosen from 7 - - 7 c 4 = = 35 and total ways = 210 consider 2 supervisor will be chosen from : x people . so as per question : 35 * x = 210 x = 6 so by using options , option a already decided wrong . option b 2 c 2 will be 1 not 6 . option c 3 c 2 will be 3 not 6 option d 4 c 2 will be 6 correct answer option e 5 c 2 will be 10 not 6 . ans : d | a = 210 / 6
b = 7 / 4
c = a + b
d = c / 6
e = d - 2
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a ) 15 % , b ) 16 % , c ) 25 % , d ) 30 % , e ) 40 % | b | divide(divide(multiply(40, 80), multiply(80, 80)), add(divide(multiply(40, 80), multiply(80, 80)), 8)) | a certain car can travel 40 minutes on a gallon of gasoline at 80 miles per hour . if the car had started with a full tank and had 8 gallons of gasoline left in its tank at the end , then what percent of the tank was used to travel 80 miles at 80 mph ? | "total time for travelling 80 miles @ 60 mph = 80 / 80 = 1 hour = 60 minutes . given , the car uses 1 gallon for every 40 minutes of driving @ 80 mph . thus in 60 minutes it will use = 1.5 gallons . thus , full tank = 1.5 + 8 = 9.5 gallons - - - > 1.5 / 9.5 = 16 % of the fuel used . b is the correct answer ." | a = 40 * 80
b = 80 * 80
c = a / b
d = 40 * 80
e = 80 * 80
f = d / e
g = f + 8
h = c / g
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['a ) 50', 'b ) 60', 'c ) 70', 'd ) 80', 'e ) 90'] | c | divide(700, divide(add(12, 8), const_2)) | the cross - section of a water channel is a trapezium in shape . if the channel is 12 meters wide at the top and 8 meters wide at the bottom and the area of cross - section is 700 square meters , what is the depth of the channel ( in meters ) ? | 1 / 2 * d * ( 12 + 8 ) = 700 d = 70 the answer is c . | a = 12 + 8
b = a / 2
c = 700 / b
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a ) 6 , b ) 6.5 , c ) 7.25 , d ) 7.5 , e ) 8 | b | divide(subtract(292, multiply(3.2, 10)), 40) | 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 292 runs ? | "required run rate = 262 - ( 3.2 x 10 ) = 250 = 6.5 40 40 b" | a = 3 * 2
b = 292 - a
c = b / 40
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a ) 17523 , b ) 2788 , c ) 1750 , d ) 2787 , e ) 29899 | c | divide(multiply(divide(divide(add(divide(multiply(4000, 10), const_100), divide(multiply(add(4000, divide(multiply(4000, 10), const_100)), 10), const_100)), 2), 3), const_100), 8) | simple interest on a certain sum of money for 3 years at 8 % per annum is half the compound interest on rs . 4000 for 2 years at 10 % per annum . the sum placed on simple interest is : | "explanation : c . i . = rs . [ 4000 * ( 1 + 10 / 100 ) ^ 2 - 4000 ] = rs . 840 sum = rs . ( 420 * 100 ) / 3 * 8 = rs . 1750 \ answer : c ) rs , 1750" | a = 4000 * 10
b = a / 100
c = 4000 * 10
d = c / 100
e = 4000 + d
f = e * 10
g = f / 100
h = b + g
i = h / 2
j = i / 3
k = j * 100
l = k / 8
|
a ) 26 , b ) 35 , c ) 39 , d ) 51 , e ) 55 | d | multiply(subtract(20, const_4), const_3) | the product z of two prime numbers is between 20 and 56 . if one of the prime numbers is greater than 2 but less than 6 and the other is greater than 14 but less than 30 then what is z ? | "the smallest possible product is 51 which is 3 * 17 . all other products are too big . the answer is d ." | a = 20 - 4
b = a * 3
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a ) 574580 , b ) 574200 , c ) 584250 , d ) 576460 , e ) none of these | b | multiply(divide(5800, 99), const_100) | 5800 * 99 | "explanation : 5800 * ( 100 - 1 ) = 580000 - 5800 = 574200 option b" | a = 5800 / 99
b = a * 100
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a ) 8 , b ) 9 , c ) 10 , d ) 12 , e ) 15 | c | divide(add(115, 85), multiply(const_2, const_10)) | 3 numbers are such that the second is as much lesser than thr third as the first is lesser than the second . if the product of the two smaller numbers is 85 and product of two larger number is 115 then find the middle number | explanation : its a bit funny sum . if you see the first sentence actually it saying that the numbers are in a . p . now as the first two numbers product is 85 so if you take factor 85 you get 5 and 17 . . now as per option 17 can not be the middle number so from intuition and seeing the option only 10 could be the middle number where 8.5 is the first number and 11.5 is the last answer : c | a = 115 + 85
b = 2 * 10
c = a / b
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a ) 280 , b ) 250 , c ) 260 , d ) 270 , e ) 300 | b | divide(200, subtract(const_1, divide(20, const_100))) | after decreasing 20 % in the price of an article costs rs . 200 . find the actual cost of an article ? | "cp * ( 80 / 100 ) = 200 cp = 2.5 * 100 = > cp = 250 answer : b" | a = 20 / 100
b = 1 - a
c = 200 / b
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a ) 298 , b ) 237 , c ) 306 , d ) 876 , e ) 291 | c | subtract(subtract(480, divide(multiply(480, 15), const_100)), divide(multiply(subtract(480, divide(multiply(480, 15), const_100)), 25), const_100)) | the sale price sarees listed for rs . 480 after successive discount is 15 % and 25 % is ? | "480 * ( 85 / 100 ) * ( 75 / 100 ) = 306 answer : c" | a = 480 * 15
b = a / 100
c = 480 - b
d = 480 * 15
e = d / 100
f = 480 - e
g = f * 25
h = g / 100
i = c - h
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a ) – 7 , b ) 7 , c ) 10 , d ) 12 , e ) 16 | e | multiply(7, 4) | the sum of all solutions for x in the equation x ^ 2 – 8 x + 21 = | x – 4 | + 7 is equal to : | "x ^ 2 - 8 x + 14 = | x - 4 | rhs can be - ve or + ve x ^ 2 - 9 x + 18 = 0 x ^ 2 - 7 x + 10 = 0 x = 6,5 , 3,2 we test all 4 values in original equation , all ok . thus , sum = 6 + 5 + 3 + 2 = 16 ans ( e )" | a = 7 * 4
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a ) rs . 40 , b ) rs . 70 , c ) rs . 90 , d ) rs . 100 , e ) rs . 120 | b | subtract(add(divide(600, 5), divide(910, 7)), divide(1620, 9)) | p , q and r together earn rs . 1620 in 9 days . p and r can earn rs . 600 in 5 days . q and r in 7 days can earn rs . 910 . how much amount does r can earn per day ? | "explanation : amount earned by p , q and r in 1 day = 1620 / 9 = 180 - - - ( 1 ) amount earned by p and r in 1 day = 600 / 5 = 120 - - - ( 2 ) amount earned by q and r in 1 day = 910 / 7 = 130 - - - ( 3 ) ( 2 ) + ( 3 ) - ( 1 ) = > amount earned by p , q and 2 r in 1 day - amount earned by p , q and r in 1 day = 120 + 130 - 180 = 70 = > amount earned by r in 1 day = 70 answer : option b" | a = 600 / 5
b = 910 / 7
c = a + b
d = 1620 / 9
e = c - d
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a ) 351 , b ) 425 , c ) 748 , d ) 854 , e ) 945 | e | multiply(multiply(15, 3), 9) | a certain university will select 1 of 9 candidates eligible to fill a position in the mathematics department and 2 of 15 candidates eligible to fill 2 identical positions in the computer science department . if none of the candidates is eligible for a position in both departments , how many different sets of 3 candidates are there to fill the 3 positions ? | "1 c 9 * 2 c 15 = 9 * 105 = 945 the answer is ( e )" | a = 15 * 3
b = a * 9
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a ) 54 mins , b ) 55 mins , c ) 56 mins , d ) 57 mins , e ) 58 mins | a | add(45, multiply(divide(subtract(multiply(6, divide(45, const_60)), multiply(4, divide(45, const_60))), add(4, 6)), const_60)) | two motor cycles a & b are started from one point at 4 kmph & 6 kmph ; after 45 min b starts returning , at what time they will meet ? | distance covered by a in 45 mins = 3 kms . distance covered by b in 45 mins = 4.5 kms . distance between and b after 45 mins = 1.5 kms . relative speed when b is returning = 4 + 6 = 10 kmph distance between a and b is covered in 1.5 / 10 = 0.15 hrs = 9 mins so they will meet again after 54 mins from starting time . answer : a | a = 45 / const_60
b = 6 * a
c = 45 / const_60
d = 4 * c
e = b - d
f = 4 + 6
g = e / f
h = g * const_60
i = 45 + h
|
a ) 20 , b ) 120 , c ) 360 , d ) 6000 , e ) 820 | d | divide(multiply(120, 1000), 20) | if 20 % of a number = 1000 , then 120 % of that number will be ? | "let the number x . then , 20 % of x = 1000 x = ( 1000 * 100 ) / 20 = 5000 120 % of x = ( 120 / 100 * 5000 ) = 6000 . answer : d" | a = 120 * 1000
b = a / 20
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a ) 80 , b ) 90 , c ) 100 , d ) 120 , e ) 140 | c | divide(subtract(multiply(120, 35), multiply(120, 15)), subtract(39, 15)) | the average of marks obtained by 120 candidates was 35 . if the avg of marks of passed candidates was 39 & that of failed candidates was 39 and that of failed candidates was 15 , the no . of candidates who passed the examination is ? | "let the number of candidate who passed = y then , 39 y + 15 ( 120 - y ) = 120 x 35 ⇒ 24 y = 4200 - 1800 ∴ y = 2400 / 24 = 100 c" | a = 120 * 35
b = 120 * 15
c = a - b
d = 39 - 15
e = c / d
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a ) 288 , b ) 262 , c ) 72 , d ) 205 , e ) 164 | e | multiply(divide(32, 40), add(add(const_100, 65), 40)) | a certain sum of money is divided among a , b and c so that for each rs . a has , b has 65 paisa and c 40 paisa . if c ' s share is rs . 32 , find the sum of money ? | "a : b : c = 100 : 65 : 40 = 20 : 13 : 8 8 - - - - 32 41 - - - - ? = > rs . 164 answer : e" | a = 32 / 40
b = 100 + 65
c = b + 40
d = a * c
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a ) 40 % , b ) 33 8 / 3 % , c ) 33 1 / 3 % , d ) 33 2 / 3 % , e ) 33 1 / 2 % | a | subtract(const_100, divide(multiply(945, const_100), 675)) | an article is bought for rs . 675 and sold for rs . 945 , find the gain percent ? | "675 - - - - 270 100 - - - - ? = > 40 % answer : a" | a = 945 * 100
b = a / 675
c = 100 - b
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a ) 2.04 % , b ) 6.12 % , c ) 8 % , d ) 8.25 % , e ) 10 % | b | multiply(1.5, const_4) | on the first of the year , james invested x dollars at proudstar bank in an account that yields 1.5 % in interest every quarter year . at the end of the year , during which he made no additional deposits or withdrawals , he had y dollars in the account . if james had invested the same amount in an account which pays interest on a yearly basis , what must the interest rate be for james to have y dollars at the end of the year ? | "if the interest were compounded annually instead of quarterly then in one year the interest would be 1.5 * 4 = 6 % . now , since the interest is compounded quarterly then there would be interest earned on interest ( very small amount ) thus the actual interest would be a little bit more than 6 % . answer : b ." | a = 1 * 5
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a ) 181 / 3 , b ) 182 / 3 , c ) 183 / 3 , d ) 184 / 3 , e ) 182 / 5 | b | add(add(multiply(divide(63, add(multiply(const_2, subtract(13, 5)), subtract(add(6, 15), const_2))), subtract(13, 5)), multiply(subtract(23, const_1), divide(63, add(multiply(const_2, subtract(13, 5)), subtract(add(6, 15), const_2))))), const_4) | a 1 , a 5 , a 13 ( terms of ap ) are in gp and a 6 + a 15 = 63 . find a 23 | a 1 a 5 = a 1 + 4 d a 13 = a 1 + 12 d now a 5 / a 1 = a 13 / a 5 - - - - > > a 1 = 4 d . . . . . . . . ( 1 ) given a 6 + a 15 = 63 - - - - > a 1 + 5 d + a 1 + 14 d = 63 - - - - > 2 a 1 + 19 d = 63 - - - - > 8 d + 19 d = 63 from ( 1 ) - - - - > d = 63 / 27 = 7 / 3 so a = 4 d = 4 * 7 / 3 = 28 / 3 from ( 1 ) a 23 = a 1 + 22 d = 28 / 3 + 22 * 7 / 3 = 182 / 3 answer : b | a = 13 - 5
b = 2 * a
c = 6 + 15
d = c - 2
e = b + d
f = 63 / e
g = 13 - 5
h = f * g
i = 23 - 1
j = 13 - 5
k = 2 * j
l = 6 + 15
m = l - 2
n = k + m
o = 63 / n
p = i * o
q = h + p
r = q + 4
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a ) 5 / 24 , b ) 7 / 12 , c ) 7 / 24 , d ) 8 / 24 , e ) 9 / 24 | b | subtract(divide(3, 4), divide(3, 18)) | if 3 / p = 4 & 3 / q = 18 then p - q = ? | "p = 3 / 4 , q = 3 / 18 = > q = 1 / 6 therefore p - q = ( 3 / 4 ) - ( 1 / 6 ) = 7 / 12 answer : b" | a = 3 / 4
b = 3 / 18
c = a - b
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a ) 11 , b ) 13 , c ) 15 , d ) data inadequate , e ) none | d | subtract(63, multiply(29, const_2)) | a number when divided by 899 gives a remainder 63 . if the same number is divided by 29 , the remainder will be | "sol . number = ( 31 x q ) + 29 . given data is inadequate . answer d" | a = 29 * 2
b = 63 - a
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a ) 11 , b ) 17 , c ) 18 , d ) 101 , e ) 30 | e | sqrt(divide(2700, const_3)) | the length of a rectangular plot is thrice its breadth . if the area of the rectangular plot is 2700 sq m , then what is the breadth of the rectangular plot ? | "let the breadth of the plot be b m . length of the plot = 3 b m ( 3 b ) ( b ) = 2700 3 b 2 = 2700 b 2 = 900 = 30 ( b > 0 ) b = 30 m . answer : e" | a = 2700 / 3
b = math.sqrt(a)
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a ) 334 , b ) 160 , c ) 387 , d ) 278 , e ) 112 | b | divide(multiply(sqrt(16), 120), sqrt(9)) | two trains a and b start simultaneously in the opposite direction from two points p and q and arrive at their destinations 16 and 9 hours respectively after their meeting each other . at what speed does the second train b travel if the first train travels at 120 km / h | answer : b ) 160 km / h | a = math.sqrt(16)
b = a * 120
c = math.sqrt(9)
d = b / c
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a ) 12 days , b ) 10 days , c ) 9 days , d ) 8 days , e ) 11 days | a | inverse(subtract(inverse(3), add(inverse(6), inverse(12)))) | a man , a woman and a boy can together complete a piece of work in 3 days . if a man alone can do it in 6 days and a boy alone in 12 days , how long will a woman take to complete the work ? | "explanation : ( 1 man + 1 woman + 1 boy ) ’ s 1 day ’ s work = 1 / 3 1 man ’ s 1 day work = 1 / 6 1 boy ’ s 1 day ’ s work = 1 / 1 ( 1 man + 1 boy ) ‘ s 1 day ’ s work = 1 / 6 + 1 / 12 = 1 / 4 therefore , 1 woman ’ s 1 day ’ s work = 1 / 3 – 1 / 4 = 1 / 12 therefore , the woman alone can finish the work in 12 days . answer : option a" | a = 1/(3)
b = 1/(6)
c = 1/(12)
d = b + c
e = a - d
f = 1/(e)
|
a ) 3.1 feet , b ) 3.2 feet , c ) 3.3 feet , d ) 3.4 feet , e ) 3.83 feet | e | divide(add(add(multiply(add(2, 6), 2), 6), 12), add(2, 6)) | carmen made a sculpture from small pieces of wood . the sculpture is 2 feet 6 inches tall . carmen places her sculpture on a base that is 12 inches tall . how tall are the sculpture andbase together ? | "we know 1 feet = 12 inch then 2 feet = 24 inch 24 + 10 = 34 then 34 + 12 = 44 46 / 12 = 3.83 feet answer : e" | a = 2 + 6
b = a * 2
c = b + 6
d = c + 12
e = 2 + 6
f = d / e
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a ) rs . 960 , b ) rs . 250 , c ) rs . 300 , d ) rs . 150 , e ) none of these | a | multiply(multiply(3, const_4), divide(340, add(3, 1))) | the cost of 3 pens and 5 pencils is rs . 340 . also the cost of one pen and one pencil is in the ratio of 4 : 1 respectively . what is the cost of one dozen pens ? | "explanation : let the cost of one pen is ‘ 4 x ’ and pencil is ‘ x ’ 3 x 4 x + 5 x = rs . 340 12 x + 5 x = rs . 340 x = 340 / 17 = 20 : . cost of 1 pen = 4 x = 4 x 20 = 80 : . cost of 12 pens , i . e . ( one dozen ) = 80 x 12 = rs . 960 answer : option a" | a = 3 * 4
b = 3 + 1
c = 340 / b
d = a * c
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a ) $ 8250 , b ) $ 8350 , c ) $ 8650 , d ) $ 8450 , e ) $ 8500 | a | multiply(400, multiply(5.5, 3.75)) | the length of a room is 5.5 m and width is 3.75 m . what is the cost of paying the floor by slabs at the rate of $ 400 per sq . metre . | "area = 5.5 × 3.75 sq . metre . cost for 1 sq . metre . = $ 400 hence , total cost = 5.5 × 3.75 × 400 = $ 8250 a" | a = 5 * 5
b = 400 * a
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a ) 45 , b ) 56 , c ) 60.4 , d ) 85 , e ) 90 | c | add(multiply(power(2, multiply(divide(60, 10), subtract(const_1, 2))), 120), 60) | the temperature of a certain cup of coffee 10 minutes after it was poured was 120 degrees fahrenheit . if the temperature f of the coffee t minutes after it was poured can be determined by the formula f = 120 ( 2 ^ - at ) + 60 , where f is in degrees fahrenheit and a is a constant . then the temperature of the coffee 80 minutes after it was poured was how many degrees fahrenheit ? | "answer : b the temperature of coffee 10 minutes after it was poured ( 120 f ) will help in solving the constant “ a ” . 120 = 120 ( 2 ^ 10 a ) + 60 2 ^ - 1 = 2 ^ 10 a a = - 1 / 10 the temperature of coffee 80 minutes after it was poured is : f = 120 ( 2 ^ - 80 / 10 ) + 60 f = 120 * 1 / 256 + 60 f = 15 / 32 + 60 f = 1935 / 32 = 60.4 c" | a = 60 / 10
b = 1 - 2
c = a * b
d = 2 ** c
e = d * 120
f = e + 60
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a ) 35 , b ) 50 , c ) 100 , d ) 120 , e ) 150 | a | subtract(subtract(multiply(11, 20), multiply(5, 22)), multiply(5, 15)) | the average of 11 results is 20 . the average of first 5 of them is 15 and that of last 5 is 22 . find the 6 th result ? | "6 th result = sum of 11 results - sum of 10 results = 11 * 20 - 5 * 15 - 5 * 22 = 220 - 75 - 110 = 35 answer is a" | a = 11 * 20
b = 5 * 22
c = a - b
d = 5 * 15
e = c - d
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a ) 0 , b ) 2 / 15 , c ) 2 / 5 , d ) 9 / 20 , e ) 5 / 6 | e | divide(5, 6) | 1 / 3 + 1 / 2 - 5 / 6 + 1 / 5 + 1 / 4 - 9 / 20 - 5 / 6 = | 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 5 / 6 . answer : e | a = 5 / 6
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a ) 11 , b ) 15 , c ) 20 , d ) 38 , e ) 56 | a | multiply(power(const_2, 210), factorial(210)) | the product of three consecutive numbers is 210 . then the sum of the smallest two numbers is ? | "product of three numbers = 210 210 = 2 * 3 * 5 * 7 = 5 * 6 * 7 . so , the three numbers are 5 , 6 and 7 . and sum of smallest of these two = 5 + 6 = 11 . answer : option a" | a = 2 ** 210
b = math.factorial(210)
c = a * b
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a ) 14 , b ) 52 , c ) 54 , d ) 180 , e ) 240 | b | add(add(divide(240, 40), divide(240, 40)), 40) | frank the fencemaker needs to fence in a rectangular yard . he fences in the entire yard , except for one full side of the yard , which equals 40 feet . the yard has an area of 240 square feet . how many feet offence does frank use ? | "area = length x breadth 240 = 40 x breadth so , breadth = 6 units fencing required is - breadth + breadth + length 6 + 6 + 40 = > 52 feet answer must be ( b ) 52" | a = 240 / 40
b = 240 / 40
c = a + b
d = c + 40
|
a ) 250 , b ) 277 , c ) 278 , d ) 200 , e ) 288 | a | divide(subtract(multiply(334, 25), multiply(71, const_100)), subtract(25, 20)) | the total of 334 of 20 paise and 25 paise make a sum of rs . 71 . the no of 20 paise coins is | "explanation : let the number of 20 paise coins be x . then the no of 25 paise coins = ( 334 - x ) . 0.20 * ( x ) + 0.25 ( 334 - x ) = 71 = > x = 250 . . answer : a ) 250" | a = 334 * 25
b = 71 * 100
c = a - b
d = 25 - 20
e = c / d
|
a ) 1 , b ) 2 , c ) 3 , d ) 4 , e ) 5 | a | power(2, 2) | the function f ( n ) is defined as the product of all the consecutive positive integers between 1 and n ^ 2 , inclusive , whereas the function g ( n ) is defined as the product of the squares of all the consecutive positive integers between 1 and n , inclusive . the exponent on 3 in the prime factorization of f ( 3 ) / g ( 3 ) is | "f ( 3 ) / g ( 3 ) = product ( 1 to 3 ^ 2 ) / 1.2 ^ 2.3 ^ 2 = 1 . 2.3 . 4.5 . 6.7 . 8.9 / 1 . 4.9 = 1 . 2.3 . ( 2 ^ 2 ) . 5 . ( 2.3 ) . 7 . ( 2 ^ 3 ) . 9 / 1 . ( 2 ^ 2 ) . 9 = 1 . ( 2 ^ 7 ) . 3.5 . 7.9 / 1 . ( 2 ^ 2 ) . 9 loof for 2 ^ 7 / 2 ^ 2 = 2 ^ 5 - - - - exponent 1 answer : a" | a = 2 ** 2
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a ) rs 18221.76 , b ) rs 18123.30 , c ) rs 18123.40 , d ) rs 18123.50 , e ) none of these | a | subtract(multiply(multiply(multiply(const_4, const_100), const_100), power(add(const_1, divide(12, const_100)), 3)), multiply(multiply(const_4, const_100), const_100)) | what will be the compound interest on rs . 45000 after 3 years at the rate of 12 % per annum | "explanation : ( 45000 × ( 1 + 12 / 100 ) 3 ) = > 45000 × 28 / 25 × 28 / 25 × 28 / 25 = > 63221.76 so compound interest will be 63221.76 - 45000 = rs 18221.76 option a" | a = 4 * 100
b = a * 100
c = 12 / 100
d = 1 + c
e = d ** 3
f = b * e
g = 4 * 100
h = g * 100
i = f - h
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a ) 160 , b ) 161 , c ) 162 , d ) 163 , e ) 169 | e | add(floor(divide(337, 2)), const_1) | the guests at a football banquet consumed a total of 337 pounds of food . if no individual guest consumed more than 2 pounds of food , what is the minimum number of guests that could have attended the banquet ? | "to minimize one quantity maximize other . 168 * 2 ( max possible amount of food a guest could consume ) = 336 pounds , so there must be more than 168 guests , next integer is 169 . answer : e ." | a = 337 / 2
b = math.floor(a)
c = b + 1
|
a ) 2430 , b ) 2700 , c ) 3300 , d ) 4860 , e ) 5400 | b | multiply(divide(factorial(divide(10, 2)), const_2), divide(factorial(10), multiply(factorial(subtract(10, 2)), factorial(2)))) | a plant manager must assign 10 new workers to one of five shifts . she needs a first , second , and third shift , and two alternate shifts . each of the shifts will receive 2 new workers . how many different ways can she assign the new workers ? | "my take selecting team of 2 out of 10 to assign to the shifts = 10 c 2 = 45 ways . now 2 out of 10 means total of 5 group possible . so putting them in shifts = counting methode : first , second , third , alt , alt = 5 * 4 * 3 * 2 * 1 = 120 here alt and alt are the same : so 120 / 2 = 60 ways . total ways of selecting = ( selecting 2 out of 10 ) * arranging those teams in shifts = 45 * 60 = 2700 ans : b" | a = 10 / 2
b = math.factorial(a)
c = b / 2
d = math.factorial(10)
e = 10 - 2
f = math.factorial(e)
g = math.factorial(2)
h = f * g
i = d / h
j = c * i
|
a ) 1 / 18 , b ) 1 / 14 , c ) 5 / 18 , d ) 1 / 15 , e ) 1 / 16 | c | divide(choose(5, const_2), choose(add(5, 4), const_2)) | there are 5 red shoes & 4 green shoes . if two of red shoes are drawn what is the probability of getting red shoes | taking 2 red shoe the probability is 5 c 2 from 9 shoes probability of taking 2 red shoes is 5 c 2 / 9 c 2 = 5 / 18 answer : c | a = math.comb(5, 2)
b = 5 + 4
c = math.comb(b, 2)
d = a / c
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a ) 50 , b ) 100 , c ) 150 , d ) 200 , e ) 250 | e | divide(200, divide(4, divide(const_10, const_2))) | to fill a tank , 200 buckets of water is required . how many buckets of water will be required to fill the same tank if the capacity of the bucket is reduced to 4 - fifths of its present ? | let the capacity of 1 bucket = x . then , the capacity of tank = 200 x . new capacity of bucket = 4 / 5 x therefore , required number of buckets = ( 200 x ) / ( 4 x / 5 ) = ( 200 x ) x 5 / 4 x = 1000 / 4 = 250 answer is e . | a = 10 / 2
b = 4 / a
c = 200 / b
|
a ) 1 / 6 , b ) 1 / 8 , c ) 1 / 9 , d ) 1 / 12 , e ) 1 / 13 | c | divide(const_2, choose(add(const_3, const_3), const_3)) | what is the probability of getting a sum 9 from two throw of a dice ? | "in two throws of die , n ( s ) = 36 let e = event of getting a sum 9 = { ( 3,6 ) , ( 4,5 ) , ( 5,4 ) , ( 6,3 ) } p ( e ) = 4 / 36 = 1 / 9 answer c 1 / 9" | a = 3 + 3
b = math.comb(a, 3)
c = 2 / b
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a ) 24 , b ) 26 , c ) 28 , d ) 30 , e ) 32 | c | multiply(divide(7, 9), add(31, const_4)) | john was 31 years old when he married betty . they just celebrated their fifth wedding anniversary , and betty ' s age is now 7 / 9 of john ' s . how old is betty ? | "assume betty ' s age on marriage = x years . john ' s age on marriage = 31 john ' s age after 5 years = 36 years . betty ' s age after 5 years = x + 5 given : x + 5 = 7 / 9 ( 36 ) = 28 therefore betty ' s current age = 28 option c" | a = 7 / 9
b = 31 + 4
c = a * b
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a ) a ) 1040 , b ) b ) 1145 , c ) c ) 1055 , d ) d ) 1060 , e ) e ) 1075 | b | add(multiply(8, 70), multiply(9, 65)) | tom purchased 8 kg of apples at the rate of 70 per kg and 9 kg of mangoes at the rate of 65 per kg . how much amount did he pay to the shopkeeper ? | "cost of 8 kg apples = 70 × 8 = 560 . cost of 9 kg of mangoes = 65 × 9 = 585 . total cost he has to pay = 560 + 585 = 1145 . b )" | a = 8 * 70
b = 9 * 65
c = a + b
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a ) 2 km , b ) 4 km , c ) 5 km , d ) 2.25 km , e ) none of these | d | divide(divide(90, const_60), add(inverse(add(4, 2)), inverse(subtract(4, 2)))) | a man can row 4 kmph is still water . if the river is running at 2 kmph it takes 90 min to row to a place and back . how far is the place | explanation : speed in still water = 4 kmph speed of the stream = 2 kmph speed upstream = ( 4 - 2 ) = 2 kmph speed downstream = ( 4 + 2 ) = 6 kmph total time = 90 minutes = 90 ⁄ 60 hour = 3 ⁄ 2 hour let l be the distance . then ( l / 6 ) + ( l / 2 ) = 32 = > l + 3 l = 9 = > 4 l = 9 = > l = 9 ⁄ 4 = 2.25 km . answer : option d | a = 90 / const_60
b = 4 + 2
c = 1/(b)
d = 4 - 2
e = 1/(d)
f = c + e
g = a / f
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a ) 1 , b ) 2 , c ) 5 , d ) 6 , e ) 8 | a | subtract(multiply(add(floor(divide(929, 30)), const_1), 30), 929) | what is the least number should be added to 929 , so the sum of the number is completely divisible by 30 ? | "( 929 / 30 ) gives remainder 29 29 + 1 = 30 , so we need to add 1 answer : a" | a = 929 / 30
b = math.floor(a)
c = b + 1
d = c * 30
e = d - 929
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a ) 288 , b ) 378 , c ) 342 , d ) 662 , e ) 262 | b | subtract(subtract(495, divide(multiply(495, 15), const_100)), divide(multiply(subtract(495, divide(multiply(495, 15), const_100)), 10), const_100)) | the sale price sarees listed for rs . 495 after successive discount is 15 % and 10 % is ? | "495 * ( 85 / 100 ) * ( 90 / 100 ) = 378 answer : b" | a = 495 * 15
b = a / 100
c = 495 - b
d = 495 * 15
e = d / 100
f = 495 - e
g = f * 10
h = g / 100
i = c - h
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a ) 5 , b ) 7 , c ) 9 , d ) 8 , e ) 12 | d | add(divide(subtract(multiply(floor(divide(119, 11)), 11), multiply(add(floor(divide(29, 11)), const_1), 11)), 11), const_1) | how many numbers from 29 to 119 are exactly divisible by 11 ? | "29 / 11 = 2 and 119 / 11 = 10 = = > 10 - 2 = 8 numbers answer : d" | a = 119 / 11
b = math.floor(a)
c = b * 11
d = 29 / 11
e = math.floor(d)
f = e + 1
g = f * 11
h = c - g
i = h / 11
j = i + 1
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a ) 58 , b ) 70 , c ) 63 , d ) 65 , e ) 72 | c | divide(multiply(subtract(multiply(divide(add(6, 2), 2), 4), 2), add(multiply(divide(add(6, 2), 2), subtract(4, 2)), const_1)), const_2) | the points a ( 0 , 0 ) , b ( 0 , 4 a - 2 ) and c ( 2 a + 1 , 2 a + 6 ) form a triangle . if angle abc = 90 , what is the area of triangle abc ? | 1 / 2 bh = 1 / 2 ( 2 a + 1 ) ( 2 a + 6 ) now 4 a - 2 = 2 a + 6 2 a = 8 . a = 4 therefore , a ( 0,0 ) ; b ( 0,14 ) ; c ( 9,14 ) 1 / 2 * 9 * 14 = 63 answer : c | a = 6 + 2
b = a / 2
c = b * 4
d = c - 2
e = 6 + 2
f = e / 2
g = 4 - 2
h = f * g
i = h + 1
j = d * i
k = j / 2
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a ) 1050 , b ) 1220 , c ) 2500 , d ) 1060 , e ) 1110 | c | divide(900, subtract(const_1, divide(multiply(8, 8), const_100))) | a fellow borrowed a certain sum of money at 8 % per annum at simple interest and in 8 years the interest amounted to rs . 900 less than the sum lent . what was the sum lent ? | "p - 900 = ( p * 8 * 8 ) / 100 p = 2500 answer : c" | a = 8 * 8
b = a / 100
c = 1 - b
d = 900 / c
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a ) 6 days , b ) 18 days , c ) 21 days , d ) 30 days , e ) 13 days | d | multiply(const_3, 3) | aarti can do a piece of work in 3 days . in how many days will she complete 10 time of work of same type ? | "we have the important relation , more work , more time ( days ) a piece of work can be done in 3 days . 10 times of work of same type can be done in 3 x 10 = 30 days answer d" | a = 3 * 3
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a ) 457 km , b ) 444 km , c ) 547 km , d ) 600 km , e ) 453 km | d | add(multiply(divide(60, subtract(21, 29)), 29), multiply(divide(60, subtract(21, 29)), 21)) | two passenger trains start at the same hour in the day from two different stations and move towards each other at the rate of 29 kmph and 21 kmph respectively . when they meet , it is found that one train has traveled 60 km more than the other one . the distance between the two stations is ? | "1 h - - - - - 5 ? - - - - - - 60 12 h rs = 29 + 21 = 50 t = 12 d = 50 * 12 = 600 answer : d" | a = 21 - 29
b = 60 / a
c = b * 29
d = 21 - 29
e = 60 / d
f = e * 21
g = c + f
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a ) 20 % , b ) 40 % , c ) 50 % , d ) 53.85 % , e ) 100 % | d | subtract(const_100, divide(subtract(const_100, 40), add(const_1, divide(30, const_100)))) | when sold at a 40 % discount , a sweater nets the merchant a 30 % profit on the wholesale cost at which he initially purchased the item . by what % is the sweater marked up from wholesale at its normal retail price ? | "we should be careful about what are we measuring % on / what is the base . . let the marked up price = 100 . . selling price = 100 - 40 % of 100 = 60 . . profit = 30 % . . therefore the wholesale purchase cost = x . . . . 1.3 x = 60 or x = 46.15 . . . marked price was 100 so . . . so answer is 53.85 % . . d" | a = 100 - 40
b = 30 / 100
c = 1 + b
d = a / c
e = 100 - d
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a ) 5 , b ) 9 , c ) 10 , d ) 20 , e ) 30 | d | multiply(subtract(9, 10), 10) | what is the greatest positive integer x such that 3 ^ x is a factor of 9 ^ 10 ? | "9 ^ 10 = ( 3 ^ 2 ) ^ 10 = 3 ^ 20 the answer is d . 20" | a = 9 - 10
b = a * 10
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a ) 1360 , b ) 1450 , c ) 1600 , d ) 947 , e ) none | d | divide(multiply(85, 78), subtract(85, 78)) | the simple interest and the true discount on a certain sum for a given time and at a given rate are rs . 85 and rs . 78 respectively . the sum is : | sol . sum = s . i . * t . d . / ( s . i ) - ( t . d . ) = 85 * 78 / ( 85 - 78 ) = rs . 947 . answer d | a = 85 * 78
b = 85 - 78
c = a / b
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a ) 22 , b ) 60 , c ) 99 , d ) 88 , e ) 11 | b | multiply(divide(subtract(1080, 675), 675), const_100) | a cycle is bought for rs . 675 and sold for rs . 1080 , find the gain percent ? | "675 - - - - 180 100 - - - - ? = > 60 % answer : b" | a = 1080 - 675
b = a / 675
c = b * 100
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a ) 4 hours , b ) 5 hours , c ) 6 hours , d ) 7 hours , e ) none of these | c | sqrt(multiply(add(4, divide(1, 2)), 8)) | two workers a and b are engaged to do a work . a working alone takes 8 hours more to complete the job than if both worked together . if b worked alone , he would need 4 1 / 2 hours more to complete the job than they both working together . what time would they take to do the work together ? | let a and be together take x hours to complete the work . then , a alone takes ( x + 8 ) hrs and b alone takes ( x + 9 / 2 ) hrs to complete the work . then , 1 / ( x + 8 ) + 1 / ( x + 9 / 2 ) = 1 / x , 1 / ( x + 8 ) + 2 / ( 2 x + 9 ) = 1 / x x ( 4 x + 25 ) = ( x + 8 ) ( 2 x + 9 ) 2 x ^ 2 = 72 , x ^ 2 = 36 , x = 6 correct option : c | a = 1 / 2
b = 4 + a
c = b * 8
d = math.sqrt(c)
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a ) 28 , b ) 40 , c ) 68 , d ) 88 , e ) 78 | a | add(multiply(divide(80, 20), const_2), 20) | a rectangular field is to be fenced on three sides leaving a side of 20 feet uncovered . if the area of the field is 80 sq . feet , how many feet of fencing will be required ? | "we have : l = 20 ft and lb = 80 sq . ft . so , b = 4 ft . length of fencing = ( l + 2 b ) = ( 20 + 8 ) ft = 28 ft . answer : a" | a = 80 / 20
b = a * 2
c = b + 20
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a ) 4 , b ) 5 , c ) 6 , d ) 10 , e ) 15 | a | divide(const_1, add(divide(const_1, 6), divide(const_1, 12))) | a can do a work in 6 days . b can do in 12 days . if both a & b are working together in how many days they can finish the work ? | 1 day work of a = 1 / 6 1 day work of b = 1 / 12 1 day work of a & b = 1 / 6 + 1 / 12 = 1 / 4 a & b finish the work in 4 days answer is a | a = 1 / 6
b = 1 / 12
c = a + b
d = 1 / c
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a ) 8 , b ) 10 , c ) 12 , d ) 12 , e ) 16 | a | lcm(2, 3) | if 2 and 3 are positive integers , then 2 * 3 + 2 is | answer : a | a = math.lcm(2, 3)
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a ) 230 m , b ) 240 m , c ) 260 m , d ) 320 m , e ) 330 m | c | subtract(multiply(multiply(add(120, 80), const_0_2778), 9), 240) | a 240 metres long train running at the speed of 120 kmph crosses another train running in opposite direction at the speed of 80 kmph in 9 seconds . what is the length of the other train ? | "relative speed = ( 120 + 80 ) km / hr = ( 200 x ( 5 / 18 ) ) m / sec = ( 500 / 9 ) m / sec . let the length of the other train be x metres . then , ( x + 240 ) / 9 = 500 / 9 x + 240 = 500 x = 260 . c" | a = 120 + 80
b = a * const_0_2778
c = b * 9
d = c - 240
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a ) 10 % , b ) 25 % , c ) 64 % , d ) 42 % , e ) 17 % | b | multiply(divide(subtract(600, add(add(add(add(25, 70), 100), 110), 145)), 600), const_100) | kavi had a stock of 600 bags in his bookshop . he sold 25 on monday , 70 on tuesday , 100 on wednesday , 110 on thursday and 145 on friday . what percentage of the bags were not sold ? | let n be the total number of bags sold . hence n = 25 + 70 + 100 + 110 + 145 = 450 let m be the bags not sold m = 600 - n = 600 - 450 = 150 percentage bags not sold / total number of bags = 150 / 600 = 0.25 = 25 % correct answer b | a = 25 + 70
b = a + 100
c = b + 110
d = c + 145
e = 600 - d
f = e / 600
g = f * 100
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a ) 6 , b ) 7 , c ) 5 , d ) 8 , e ) 9 | d | divide(subtract(divide(85, 2), divide(45, 2)), const_2) | a man rows his boat 85 km downstream and 45 km upstream , taking 2 1 / 2 hours each time . find the speed of the stream ? | "speed downstream = d / t = 85 / ( 2 1 / 2 ) = 34 kmph speed upstream = d / t = 45 / ( 2 1 / 2 ) = 18 kmph the speed of the stream = ( 34 - 18 ) / 2 = 8 kmph answer : d" | a = 85 / 2
b = 45 / 2
c = a - b
d = c / 2
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a ) 1 / 7 , b ) 3 , c ) 12 / 3 , d ) 6 / 6 , e ) 1 | a | divide(multiply(1, const_1), subtract(multiply(const_2, const_4), 1)) | if f ( x ) = 1 / x and x is a natural number , what can not be the answer for f ( f ( x ) ) ? | answer a is impossible because the invers of 1 / x is x , and the only way to have something other than a natural number is to input something other than a natural number . with the specification that only natural numbers may be used 1 / 7 is not a possibility for f ( f ( x ) ) | a = 1 * 1
b = 2 * 4
c = b - 1
d = a / c
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a ) 0 , b ) 1 , c ) 2 , d ) 3 , e ) 4 | e | subtract(add(const_2, const_3), const_2) | the product of the squares of two positive integers is 900 . how many pairs of positive integers satisfy this condition ? | "ans : e - 4 pairs ( x ˆ 2 ) ( y ˆ 2 ) = 900 [ square root both sides ] xy = 30 20 = 1 x 30 , 3 x 10 , 6 x 5 , 5 x 6 , 10 x 3 , 30 x 1 , 15 x 2 , 2 x 15 cancel the repeats this leaves us with exactly 4 options . hence , e" | a = 2 + 3
b = a - 2
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a ) 2 , b ) 0 , c ) 1 , d ) 3 , e ) 4 | c | divide(divide(divide(lcm(23, 57), 57), const_4), const_4) | what is the least value of x . so that 23 x 57 is divisible by 3 . | "explanation : the sum of the digits of the number is divisible by 3 , then the number is divisible by 3 . 2 + 3 + x + 5 + 7 = 17 + x least value of x may be 1 therefore 17 + 1 = 18 is divisible by 3 . answer : option c" | a = math.lcm(23, 57)
b = a / 57
c = b / 4
d = c / 4
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a ) 1 : 2 , b ) 2 : 1 , c ) 2 : 3 , d ) 3 : 2 , e ) none of these | d | divide(divide(const_1, multiply(10, 4)), divide(const_1, multiply(4, const_10))) | six women can do a work in 10 days . ten men can complete the same work in 4 days . what is the ratio between the capacity of a man and a woman ? | "explanation : ( 6 ã — 10 ) women can complete the work in 1 day . â ˆ ´ 1 woman ' s 1 day ' s work = 1 / 60 ( 10 ã — 4 ) men can complete the work in 1 day . â ˆ ´ 1 man ' s 1 day ' s work = 1 / 40 so , required ratio = 1 / 60 : 1 / 40 = 3 : 2 answer : d" | a = 10 * 4
b = 1 / a
c = 4 * 10
d = 1 / c
e = b / d
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a ) 32 , b ) 42 , c ) 52 , d ) 62 , e ) 72 | c | subtract(subtract(subtract(add(multiply(5, 28), 232), multiply(5, 28)), multiply(5, 28)), multiply(const_4, const_10)) | if 28 less than 5 times a certain number is 232 . what is the number ? | 5 x − 28 subtraction is built backwards , multiply the unknown by 5 5 x − 28 = 232 is translates to equals + 28 + 28 add 28 to both sides 5 x = 260 the variable ismultiplied by 5 5 5 divide both sides by 5 x = 52 the number is 52 . correct answer c | a = 5 * 28
b = a + 232
c = 5 * 28
d = b - c
e = 5 * 28
f = d - e
g = 4 * 10
h = f - g
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a ) 1217 , b ) 1348.2 , c ) 1210 , d ) 1212 , e ) 1312 | b | multiply(subtract(power(23, const_2), power(const_10, const_2)), divide(add(multiply(const_10, const_2), const_2), add(const_4, const_3))) | a rope of which a calf is tied is increased from 10 m to 23 m , how much additional grassy ground shall it graze ? | π ( 232 – 102 ) = 1348.2 answer : b | a = 23 ** 2
b = 10 ** 2
c = a - b
d = 10 * 2
e = d + 2
f = 4 + 3
g = e / f
h = c * g
|
a ) 50 kg , b ) 60 kg , c ) 70 kg , d ) 80 kg , e ) none of these | b | multiply(multiply(multiply(3, 2), divide(1, const_100)), const_1000) | a boat having a length 3 m and breadth 2 m is floating on a lake . the boat sinks by 1 cm when a man gets into it . the mass of the man is : | explanation : in this type of question , first we will calculate the volume of water displaces then will multiply with the density of water . volume of water displaced = 3 * 2 * 0.01 = 0.06 m cube mass of man = volume of water displaced * density of water = 0.06 * 1000 = 60 kg option b | a = 3 * 2
b = 1 / 100
c = a * b
d = c * 1000
|
a ) 60 % , b ) 80 % , c ) 100 % , d ) 120 % , e ) 125 % | a | multiply(multiply(power(divide(6, 10), const_2), divide(10, 6)), const_100) | 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 6 meters , and the interior of tank b has a height of 6 meters and a circumference of 10 meters . the capacity of tank a is what percent of the capacity of tank b ? | "the radius of tank a is 6 / ( 2 * pi ) . the capacity of tank a is 10 * pi * 36 / ( 4 * pi ^ 2 ) = 180 / ( 2 * pi ) the radius of tank b is 10 / ( 2 * pi ) . the capacity of tank b is 6 * pi * 100 / ( 4 * pi ^ 2 ) = 300 / ( 2 * pi ) tank a / tank b = 180 / 300 = 6 / 10 = 60 % the answer is a ." | a = 6 / 10
b = a ** 2
c = 10 / 6
d = b * c
e = d * 100
|
a ) 12 % , b ) 14 % , c ) 16 % , d ) 20 % , e ) 23 % | e | subtract(const_100, divide(multiply(add(const_100, 10), subtract(const_100, 30)), const_100)) | the tax on a commodity is diminished by 30 % but its consumption is increased by 10 % . find the decrease percent in the revenue derived from it ? | "explanation : 100 * 100 = 10000 70 * 110 = 7700 10000 - - - - - - - 2300 100 - - - - - - - ? = 23 % e )" | a = 100 + 10
b = 100 - 30
c = a * b
d = c / 100
e = 100 - d
|
a ) 3.6 sec , b ) 18 sec , c ) 36 sec , d ) 39 sec , e ) none of these | d | multiply(multiply(divide(divide(add(270, 120), const_1000), subtract(45, 9)), const_60), const_60) | a jogger running at 9 kmph along side a railway track is 270 metres ahead of the engine of a 120 metre long train running at 45 kmph in the same direction . in how much time will the train pass the jogger ? | "speed of train relative to jogger = ( 45 – 9 ) km / h = 36 km / h = ( 36 × 5 ⁄ 18 ) m / sec = 10 m / sec distance to be covered = ( 270 + 120 ) m = 390 m . ∴ time taken = ( 390 ⁄ 10 ) sec = 39 sec . answer d" | a = 270 + 120
b = a / 1000
c = 45 - 9
d = b / c
e = d * const_60
f = e * const_60
|
a ) 20 , b ) 87 , c ) 266 , d ) 288 , e ) 11 | a | subtract(divide(multiply(add(39, const_1), add(add(39, const_1), const_1)), const_2), 800) | some consecutive natural numbers , starting with 1 , are written on the board . now , one of the numbers was erased and the average of the remaining numbers is 800 / 39 . find the number which was erased . | we know that average of n consecutive numbes average = n × ( n + 1 ) 2 n = ( n + 1 ) 2 n × ( n + 1 ) 2 n = ( n + 1 ) 2 if the given n is sufficiently large , the average does not change much even though we exclude one or two numbers from it . so the approximate number of observations is almost double to the average ( remember : the average of consecutive numbers almost lies in the middle ) the approximate average is 800 / 39 = approx 20 . so the initial numbers may be nearer to 40 . in this question it is actually 40 as from the denominator of the new average 800 / 39 . the initial numbers are 40 . sum of 40 consecutive numbers = 40 × ( 40 + 1 ) 2 = 82040 × ( 40 + 1 ) 2 = 820 sum of 39 numbers = average x numbenumber of observations = 80039 × 3980039 × 39 = 800 so the number excluded = 820 - 800 = 20 answer : a | a = 39 + 1
b = 39 + 1
c = b + 1
d = a * c
e = d / 2
f = e - 800
|
a ) 50 cm 2 , b ) 100 cm 2 , c ) 150 cm 2 , d ) 200 cm 2 , e ) 250 cm 2 | a | multiply(multiply(divide(const_1, const_2), add(3, 7)), 10) | find the area of the quadrilateral of one of its diagonals is 10 cm and its off sets 7 cm and 3 cm ? | "1 / 2 * 10 ( 7 + 3 ) = 50 cm 2 answer : a" | a = 1 / 2
b = 3 + 7
c = a * b
d = c * 10
|
a ) 30 , b ) 35 , c ) 40 , d ) 45 , e ) 50 | d | divide(rectangle_area(15, 24), 8) | carol and jordan draw rectangles of equal area . if carol ' s rectangle measures 15 inches by 24 inches and jordan ' s rectangle is 8 inches long , how wide is jordan ' s rectangle , in inches ? | "area of first rectangle is 15 * 24 = 360 hence area of second would be 8 x = 360 x x = 45 answer is d" | a = rectangle_area / (
|
a ) 2327 , b ) 2757 , c ) 3147 , d ) 3587 , e ) 3997 | c | add(lcm(lcm(18, 70), lcm(25, 21)), 3) | what is the smallest number which when increased by 3 is divisible by 18 , 70 , 25 and 21 ? | "when increased by 3 , the number must include at least 2 * 3 ^ 2 * 5 ^ 2 * 7 = 3150 the answer is c ." | a = math.lcm(18, 70)
b = math.lcm(25, 21)
c = math.lcm(a, b)
d = c + 3
|
a ) 2 m , b ) 3 m , c ) 4 m , d ) 5 m , e ) 1.5 m | c | divide(multiply(10, 10), subtract(rectangle_area(10, 10), rectangle_area(5, 10))) | the dimensions of a field are 10 m by 10 m . a pit 10 m long , 5 m wide and 4 m deep is dug in one corner of the field and the earth removed has been evenly spread over the remaining area of the field . what will be the rise in the height of field as a result of this operation ? | "the volume of the earth removed is 10 * 5 * 4 = 200 m ^ 3 . the remaining area of the field is 10 * 10 - 10 * 5 = 50 m ^ 2 . 200 m ^ 3 of the earth evenly spread over the area of 50 m ^ 2 will rise the height by ( height ) = ( volume ) / ( area ) = 200 / 50 = 4 m . answer : c" | a = 10 * 10
b = rectangle_area - (
c = a / b
|
a ) 25.5 % , b ) 21.5 % , c ) 17.5 % , d ) 13.5 % , e ) 9.5 % | c | add(subtract(subtract(const_100, 60), multiply(divide(3, 4), subtract(const_100, 60))), subtract(60, multiply(divide(7, 8), 60))) | in a survey of parents , exactly 7 / 8 of the mothers and 3 / 4 of the fathers held full - time jobs . if 60 percent of the parents surveyed were women , what percent of the parents did not hold full - time jobs ? | "fathers without full - time jobs are 1 / 4 * 2 / 5 = 2 / 20 of all the parents surveyed . mothers without full - time jobs are 1 / 8 * 3 / 5 = 3 / 40 of all the parents surveyed . the percent of parents without full - time jobs is 2 / 20 + 3 / 40 = 7 / 40 = 17.5 % the answer is c ." | a = 100 - 60
b = 3 / 4
c = 100 - 60
d = b * c
e = a - d
f = 7 / 8
g = f * 60
h = 60 - g
i = e + h
|
a ) 6 hours , b ) 5 hours , c ) 5.5 hours , d ) 8 hours , e ) none | c | divide(add(296, 6.5), add(divide(296, 8), 18)) | a truck covers a distance of 296 km at a certain speed in 8 hours . how much time would a car take at an average speed which is 18 kmph more than that of the speed of the truck to cover a distance which is 6.5 km more than that travelled by the truck ? | explanation : speed of the truck = distance / time = 296 / 8 = 37 kmph now , speed of car = ( speed of truck + 18 ) kmph = ( 37 + 18 ) = 55 kmph distance travelled by car = 296 + 6.5 = 302.5 km time taken by car = distance / speed = 302.5 / 55 = 5.5 hours . answer – c | a = 296 + 6
b = 296 / 8
c = b + 18
d = a / c
|
a ) 9 / 100 , b ) 2 / 19 , c ) 1 / 8 , d ) 3 / 20 , e ) 3 / 10 | e | divide(choose(12, 3), choose(add(12, 12), 3)) | a bag contains 12 red jellybeans and 12 blue jellybeans . if 3 jellybeans are removed one at a time , at random and are not replaced , what is the probability that all 3 jellybeans removed from the bag are blue ? | "method - 1 10 red jellybeans and 10 blue jellybeans total outcomes = no . of ways to choose 3 jelly bean at random out of a total 20 jellybeans = 20 c 3 = 1140 favourable outcomes = no . of ways to choose 3 jelly bean such that they are all blue out of 10 blue = 10 c 3 = 120 probability = favourable outcomes / total outcomes = 10 c 3 / 20 c 3 probability = 120 / 1140 = 2 / 19 answer : option b method - 2 probability of first jelly bean to be blue = 10 / 20 [ total 10 blue out of total 20 jellybeans ] probability of second jelly bean to be blue = 9 / 19 [ total 9 blue remaining out of total 19 jellybeans remaining ] probability of third jelly bean to be blue = 8 / 18 [ total 8 blue remaining out of total 18 jellybeans remaining ] required probability = ( 10 / 20 ) * ( 9 / 19 ) * ( 8 / 18 ) = 3 / 10 answer : option e" | a = math.comb(12, 3)
b = 12 + 12
c = math.comb(b, 3)
d = a / c
|
a ) 7500 , b ) 20000 , c ) 2775 , d ) 5496 , e ) 6851 | b | divide(16000, subtract(subtract(const_1, divide(10, const_100)), divide(10, const_100))) | a candidate got 10 % of the votes polled and he lost to his rival by 16000 votes . how many votes were cast ? | "10 % - - - - - - - - - - - l 90 % - - - - - - - - - - - w - - - - - - - - - - - - - - - - - - 80 % - - - - - - - - - - 16000 100 % - - - - - - - - - ? = > 20000 answer : b" | a = 10 / 100
b = 1 - a
c = 10 / 100
d = b - c
e = 16000 / d
|
a ) 3.4 km , b ) 2.9 km , c ) 2.4 km , d ) 2.6 km , e ) 2.8 km | c | multiply(divide(multiply(8, 9), subtract(9, 8)), divide(add(14, 16), const_60)) | if i walk at 8 km / h , i miss the bus by 14 minutes . if i walk at 9 km / h , i reach 16 minutes before the arrival of the bus . how far i walk to reach the bus stand ? | d = product of speed difference of time / difference of speed d = 8 x 9 / 60 [ 14 â ˆ ’ ( â ˆ ’ 16 ) / 9 - 8 ] [ here , â € “ ve sign indicates before the schedule time ] â ‡ ’ d = 2.4 km answer c | a = 8 * 9
b = 9 - 8
c = a / b
d = 14 + 16
e = d / const_60
f = c * e
|
['a ) l = 2 , w = 9', 'b ) l = 5 , w = 8', 'c ) l = 6 , w = 4', 'd ) l = 1 , w = 7', 'e ) l = 2 , w = 3'] | c | add(divide(subtract(divide(20, const_2), 2), const_2), 2) | the length of a rectangle is 2 cm more than the width of the rectangle . the perimeter of the rectangle is 20 cm . find the length and the width of the rectangle . | let length l = x , width w = x − 2 and perimeter = p ∴ p = 2 l + 2 w = 2 x + 2 ( x − 2 ) 20 = 2 x + 2 x − 4 4 x = 24 x = 6 l = 6 cm and w = l − 2 = 4 cm answer is c . | a = 20 / 2
b = a - 2
c = b / 2
d = c + 2
|
a ) 24 , b ) 55 , c ) 77 , d ) 99 , e ) 01 | a | multiply(divide(subtract(10, const_1), subtract(4, const_1)), 8) | a certain sum becomes 4 times itself at simple interest in 8 years . in how many years does it become 10 times itself ? | let the sum be rs . x , then it becomes rs . 4 x in eight years rs . 3 x is the interest on x for eight years . r = ( 100 * 3 x ) / ( x * 8 ) = 300 / 8 % if the sum becomes ten times itself , then interest is 9 x . the required time period = ( 100 * 9 x ) / ( x * 300 / 8 ) = ( 100 * 9 x * 8 ) / ( x * 300 ) = 24 years . answer : a | a = 10 - 1
b = 4 - 1
c = a / b
d = c * 8
|
a ) 290 / 289 , b ) 26 / 25 , c ) 290 / 90 , d ) 290 / 19 , e ) none of these | b | add(power(divide(const_1, const_1), const_2), power(divide(const_1, 5), const_2)) | product of two natural numbers is 5 . then , the sum of reciprocals of their squares is | "explanation : if the numbers are a , b , then ab = 5 , as 17 is a prime number , so a = 1 , b = 5 . 1 / a 2 + 1 / b 2 = 1 / 1 ( 2 ) + 1 / 5 ( 2 ) = 26 / 25 option b" | a = 1 / 1
b = a ** 2
c = 1 / 5
d = c ** 2
e = b + d
|
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