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a ) 3 : 7 , b ) 4 : 9 , c ) 10 : 7 , d ) 5 : 7 , e ) 6 : 11 | c | divide(multiply(50, 8), multiply(70, 4)) | car a runs at the speed of 50 km / hr and reaches its destination in 8 hours . car b runs at the speed of 70 km / h and reaches its destination in 4 hours . what is the ratio of distances covered by car a and car b ? | "car a travels 50 Γ 8 = 400 km car b travels 70 Γ 4 = 280 km the ratio is 400 : 280 = 40 : 28 = 10 : 7 the answer is c ." | a = 50 * 8
b = 70 * 4
c = a / b
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a ) 3630 , b ) 9232 , c ) 8828 , d ) 2387 , e ) 3810 | e | multiply(divide(6300, add(add(6300, 4200), 10500)), 12700) | a , b and c invested rs . 6300 , rs . 4200 and rs . 10500 respectively , in a partnership business . find the share of a in profit of rs . 12700 after a year ? | "6300 : 4200 : 10500 3 : 2 : 5 3 / 10 * 12700 = 3810 answer : e" | a = 6300 + 4200
b = a + 10500
c = 6300 / b
d = c * 12700
|
a ) 104 , b ) 108 , c ) 114 , d ) 124 , e ) 142 | c | subtract(25, divide(subtract(25, divide(13, const_2)), const_2)) | sum of two numbers is 25 and their difference is 13 . find their product . | "explanation : friends , this sort of question is quite important in competitive exams , whenever any question come which have relation between sum , product and difference , this formula do the magic : = > ( x + y ) 2 = ( x β y ) 2 + 4 xy < = > ( 25 ) 2 = ( 13 ) 2 + 4 xy < = > 4 xy = ( 25 ) 2 β ( 13 ) 2 < = > xy = 456 / 4 = 114 option c" | a = 13 / 2
b = 25 - a
c = b / 2
d = 25 - c
|
a ) 1 / 4 , b ) 1 / 2 , c ) 1 , d ) 2 , e ) 3 | e | multiply(divide(13, add(13, 13)), 3) | if 13 = 13 w / ( 1 - w ) , then ( 3 w ) 2 = | "13 - 13 w = 13 w 26 w = 13 w = 1 / 2 3 w = 3 / 2 3 w * 2 = 3 / 2 * 2 = 3 answer : e" | a = 13 + 13
b = 13 / a
c = b * 3
|
a ) 8000 , b ) 50000 , c ) 9000 , d ) 9500 , e ) 10000 | b | add(divide(multiply(44000, 12), const_100), 44000) | john ' s bank ' s saving amount is decreased 12 % due to loan payment and current balance is rs . 44000 . find the actual balance before deduction ? | 12 % decreased 88 % balance = 44000 100 % = 44000 / 88 * 100 = 50000 answer : b | a = 44000 * 12
b = a / 100
c = b + 44000
|
a ) 230 , b ) 288 , c ) 267 , d ) 256 , e ) 220 | e | subtract(multiply(multiply(add(120, 80), const_0_2778), 9), 280) | a 280 m long train running at the speed of 120 km / hr crosses another train running in opposite direction at the speed of 80 km / hr in 9 sec . what is the length of the other train ? | "relative speed = 120 + 80 = 200 km / hr . = 200 * 5 / 18 = 500 / 9 m / sec . let the length of the other train be x m . then , ( x + 280 ) / 9 = 500 / 9 = > x = 220 . answer : e" | a = 120 + 80
b = a * const_0_2778
c = b * 9
d = c - 280
|
a ) 10 sec , b ) 20 sec , c ) 15 sec , d ) 16 sec , e ) 14 sec | a | divide(150, add(12, 3)) | an escalator moves towards the top level at the rate of 12 ft . sec and its length is 150 feet . if a person walks on the moving escalator at the rate of 3 feet per second towards the top level , how much time does he take to cover the entire length . | "explanation : time taken to cover the entire length = tot . dist / resultant speed = 150 / ( 12 + 3 ) = 10 sec answer : a" | a = 12 + 3
b = 150 / a
|
a ) 20.5 , b ) 18.4 , c ) 17.5 , d ) 19.9 , e ) 21.5 | d | multiply(divide(subtract(100, 0.5), subtract(subtract(100, 0.5), subtract(100, 3))), 0.5) | in a 100 m sprint race kelly was given a 3 m head start by abel . if abel lost the race to kelly by a distance of 0.50 m , how many meters more would abel need to overtake kelly ? | distance abel ran before completion of race - - - 100 m - 0.50 m = 99.5 m distance gained on kelly over 99.5 m - - - 3 m - 0.50 m = 2.50 m then abel gains 99.5 m / 2.50 m = 1 m on kelly every 39.8 meters . therefore 39.8 divided by half a meter ( i . e the 0.50 m abel was defeated with ) answer : d . 19.9 m | a = 100 - 0
b = 100 - 0
c = 100 - 3
d = b - c
e = a / d
f = e * 0
|
['a ) 40', 'b ) 29', 'c ) 28', 'd ) 27', 'e ) 12'] | a | multiply(divide(divide(divide(12.8, multiply(multiply(5, 8), 5)), divide(const_4, const_10)), divide(const_4, const_10)), const_100) | the volume of a wall , 5 times as high as it is broad and 8 times as long as it is high , is 12.8 cu . meters . find the breadth of the wall . | explanation : let the breadth of the wall be x metres . then , height = 5 x metres and length = 40 x metres . x * 5 x * 40 x = 12.8 = > \ inline \ fn _ jvn x ^ 3 = \ frac { 12.8 } { 200 } = \ frac { 128 } { 2000 } = \ frac { 64 } { 1000 } = > \ inline \ fn _ jvn x = \ frac { 4 } { 10 } m = > \ inline \ fn _ jvn x = \ frac { 4 } { 10 } \ times 100 = 40 cm answer : a ) 40 cm | a = 5 * 8
b = a * 5
c = 12 / 8
d = 4 / 10
e = c / d
f = 4 / 10
g = e / f
h = g * 100
|
a ) 16 % , b ) 32 % , c ) 48 % , d ) 84 % , e ) 92 % | d | subtract(const_100, divide(subtract(const_100, 68), const_2)) | a certain characteristic in a large population has a distribution that is symmetric about the mean m . if 68 % of the distribution lies one standard deviation h of the mean , what percent of the distribution is less than m + h ? | "16 % ________________________________________________ m + h 34 % ________________________________________________ m 34 % ________________________________________________ m - h 16 % since 68 % lies one standard deviation from mean m , = > 50 % of 68 % lies on either side as it is symmetric about m . thus 16 % lie below m - h and 16 % lie above m + h now below m + h = 16 + 34 + 34 = 84 % hence d" | a = 100 - 68
b = a / 2
c = 100 - b
|
a ) 2674 , b ) 1349 , c ) 3343 , d ) 3678 , e ) 3679 | b | add(add(add(multiply(const_100, const_10), multiply(multiply(1, const_3), const_100)), multiply(add(multiply(1, const_3), 1), const_10)), multiply(const_3, multiply(1, const_3))) | what is the 4 digit number whose second digit is thrice the first digit and 3 rd digit is sum of 1 st and 2 nd and last digit is thrice the second digit . | let four digits is a , b , c & d . . . so by given b is 3 a c is 4 a d is 9 a so ans is 1349 answer : b | a = 100 * 10
b = 1 * 3
c = b * 100
d = a + c
e = 1 * 3
f = e + 1
g = f * 10
h = d + g
i = 1 * 3
j = 3 * i
k = h + j
|
a ) 2 , b ) 3 , c ) 4 , d ) 1 , e ) 5 | b | divide(24, 20) | how many of the positive factors of 20 are not factors of 24 | "factors of 20 - 1 , 2 , 4,5 , 10 , 20 factors of 24 - 1 , 2 , 3 , 4 , 6 , 8 , 12 and 24 comparing both , we have 3 factors of 20 which are not factors of 24 - 5 , 10,20 answer : b" | a = 24 / 20
|
a ) 127 , b ) 150 , c ) 177 , d ) 167 , e ) 122 | b | multiply(subtract(divide(525, 3), const_100), 2) | rs . 525 among a , b and c such that b may get 2 / 3 of a and c together get . find the share of c ? | a + b + c = 590 5 a = 6 b = 8 c = x a : b : c = 1 / 5 : 1 / 6 : 1 / 8 = 24 : 20 : 15 15 / 59 * 590 = rs . 150 answer : b | a = 525 / 3
b = a - 100
c = b * 2
|
a ) 65 , b ) 69 , c ) 72 , d ) 75 , e ) none | e | divide(add(add(add(add(81, 68), 82), 69), 90), add(const_1, const_4)) | kamal obtained 81 , 68 , 82 , 69 and 90 marks ( out of 100 ) in english , mathematics , physics , chemistry and biology . what are his average marks ? | "sol . average = 81 + 68 + 82 + 69 + 90 / 5 ) = ( 390 / 5 ) = 78 . answer e" | a = 81 + 68
b = a + 82
c = b + 69
d = c + 90
e = 1 + 4
f = d / e
|
a ) $ 36.30 , b ) $ 38.50 , c ) $ 39.20 , d ) $ 39.50 , e ) $ 40.60 | a | add(multiply(33, divide(40, const_100)), 33) | a farmer spent $ 33 on feed for chickens and goats . he spent 40 % money on chicken feed , which he bought at a 20 % discount off the full price , and spent the rest on goat feed , which he bought at full price . if the farmer had paid full price for both the chicken feed and the goat feed , what amount would he have spent on the chicken feed and goat feed combined ? | "a farmer spent 40 % money on chicken feed , so he spent 0.4 * $ 33 = $ 13.2 on chicken feed , thus he spent the remaining 33 - 13.2 = $ 19.8 on goat feed . now , since he bought chicken feed at a 20 % discount then the original price of it was x * 0.8 = $ 13.2 - - > x = $ 16.5 . therefore if the farmer had paid full price for both the chicken feed and the goat feed , then he would he have spent 16.5 + 19.8 = $ 36.3 . answer : a" | a = 40 / 100
b = 33 * a
c = b + 33
|
a ) 40 / 3 , b ) 40 , c ) 120 , d ) 360 , e ) 420 | d | multiply(multiply(3, 3), 40) | the difference in compound interest earned on a deposit ( compounded annually ) in year 1 and year 2 is $ 40 . had the interest rate been 3 times its present value , the difference c would have been how much ? | case 1 : deposit = $ x ; rate of increase = r . interest yearned in 1 year = xr . deposit in 1 year = x + xr . interest yearned in 2 year = ( x + xr ) r . the difference c = ( x + xr ) r - xr = xr ^ 2 = 40 . case 2 : deposit = $ x ; rate of increase = 3 r . interest yearned in 1 year = x ( 3 r ) . deposit in 1 year = x + 3 xr . interest yearned in 2 year = ( x + 3 xr ) 3 r . the difference = ( x + 3 xr ) 3 r - 3 xr = 9 xr ^ 2 . since from case 1 we know that xr ^ 2 = 40 , then 9 xr ^ 2 = 9 * 40 = 360 . answer : d . | a = 3 * 3
b = a * 40
|
a ) 9 , 10 , b ) 10 , 11 , c ) 11 , 12 , d ) 12 , 13 , e ) none of these | a | divide(subtract(sqrt(add(multiply(subtract(91, const_1), const_4), const_1)), const_1), const_2) | the sum of the squares of two consecutive positive integers exceeds their product by 91 . find the integers ? | let the two consecutive positive integers be x and x + 1 x 2 + ( x + 1 ) 2 - x ( x + 1 ) = 91 x 2 + x - 90 = 0 ( x + 10 ) ( x - 9 ) = 0 = > x = - 10 or 9 . as x is positive x = 9 hence the two consecutive positive integers are 9 and 10 . answer : a | a = 91 - 1
b = a * 4
c = b + 1
d = math.sqrt(c)
e = d - 1
f = e / 2
|
a ) 36 , b ) 2 ^ 4 * 3 , c ) 24 , d ) 38 , e ) 47 | e | subtract(48, const_1) | in a lake , there is a patch of lily pads . every day , the patch doubles in size . it takes 48 days for the patch to cover the entire lake , how many days would it take the patch to cover half of the lake ? | working backward from the day it ' s covered : day 48 : fully covered day 47 : half covered so 47 days answer : e | a = 48 - 1
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a ) rs . 600 , b ) rs . 618 , c ) rs . 525 , d ) rs . 620 , e ) none | b | subtract(add(add(5000, divide(multiply(5000, 12), const_100)), divide(multiply(add(5000, divide(multiply(5000, 12), const_100)), 12), const_100)), 5000) | find compound interest on rs . 5000 at 12 % per annum for 1 year , compounded half - yearly . | "principal = rs . 5000 , rate = 6 % per half year time = 1 year = 2 half - years amount = rs . { 5000 x ( 1 + 6 / 100 ) 2 ] = rs . [ 5000 x 53 / 50 x 53 / 50 ] = rs . 5618 c . i . = rs . ( 5618 β 5000 ) = rs . 618 answer b" | a = 5000 * 12
b = a / 100
c = 5000 + b
d = 5000 * 12
e = d / 100
f = 5000 + e
g = f * 12
h = g / 100
i = c + h
j = i - 5000
|
a ) 16 mtr . , b ) 52 mtr , c ) 70 mtr . , d ) 78 mtr . , e ) 98 mtr . | e | multiply(112, divide(multiply(35, 3), multiply(20, 6))) | if 20 men can build a wall 112 metres long in 6 days , what length of a similar wall can be built by 35 men in 3 days ? | "20 men is 6 days can build 112 metres 35 men in 3 days can build = 112 * ( 35 / 20 ) x ( 3 / 6 ) = 98 meters answer : e ." | a = 35 * 3
b = 20 * 6
c = a / b
d = 112 * c
|
a ) 5 , b ) 4 , c ) 2 , d ) 3 , e ) 1 | c | divide(105, 85) | how many of the positive factors of 85 , 105 and how many common factors are there in numbers ? | "factors of 85 - 1 , 5 , 17 , 85 factors of 105 - 1 , 3 , 5 , 7 , 15 , 21 , 35 , 105 comparing both , we have two common factors of 85 and 105 - 1,5 answer ( c )" | a = 105 / 85
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a ) 4 days , b ) 6 days , c ) 10 days , d ) 16 days , e ) 22 days | d | divide(20, add(const_1, divide(25, const_100))) | sakshi can do a piece of work in 20 days . tanya is 25 % more efficient than sakshi . the number of days taken by tanya to do the same piece of work is ? | ratio of times taken by sakshi and tanya = 125 : 100 = 5 : 4 . suppose tanya takes x days to do the work . 5 : 4 : : 20 : x x = 4 x 20 / 5 x = 16 days . hence , tanya takes 16 days to complete the work . answer : d | a = 25 / 100
b = 1 + a
c = 20 / b
|
a ) 27 / 4 , b ) 27 / 8 , c ) 3 / 4 , d ) 3 / 8 , e ) 1 / 4 | e | divide(multiply(3, 3), multiply(2, multiply(3, 6))) | if a / b = 1 / 3 , b / c = 2 , c / d = 1 / 2 , d / e = 3 and e / f = 1 / 6 , then what is the value of abc / def ? | "say a = 2 . then : a / b = 1 / 3 - - > b = 6 ; b / c = 2 - - > c = 3 ; c / d = 1 / 2 - - > d = 6 ; d / e = 3 - - > e = 2 ; e / f = 1 / 6 - - > f = 12 . abc / def = ( 2 * 6 * 3 ) / ( 6 * 2 * 12 ) = 1 / 4 . answer : e ." | a = 3 * 3
b = 3 * 6
c = 2 * b
d = a / c
|
a ) 3 : 5 , b ) 3 : 8 , c ) 3 : 2 , d ) 9 : 9 , e ) 3 : 1 | c | divide(add(multiply(4000, 6), multiply(multiply(4000, const_2), 6)), multiply(4000, add(6, 6))) | a and b invests rs . 4000 and rs . 4000 respectively in a business . if a doubles his capital after 6 months . in what ratio should a and b divide that year ' s profit ? | "( 4 * 6 + 8 * 6 ) : ( 4 * 12 ) 72 : 48 = > 3 : 2 answer : c" | a = 4000 * 6
b = 4000 * 2
c = b * 6
d = a + c
e = 6 + 6
f = 4000 * e
g = d / f
|
a ) - 5 , b ) - 6 , c ) 4 , d ) 0 , e ) - 4 | b | multiply(7, 7) | find the value of x ? ( 2 / 7 ) ^ - 8 * ( 7 / 2 ) ^ - 2 = ( 2 / 7 ) ^ x | ( 2 / 7 ) ^ - 8 * ( 7 / 2 ) ^ - 2 = ( 2 / 7 ) ^ x now , ( 7 / 2 ) ^ 8 * ( 2 / 7 ) ^ 2 = ( 2 / 7 ) ^ x on solving , ( 7 / 2 ) ^ 6 = ( 2 / 7 ) ^ x ( 2 / 7 ) ^ - 6 = ( 2 / 7 ) ^ x thus , x = - 6 . answer : b | a = 7 * 7
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a ) 384 , b ) 76 , c ) 26 , d ) 28 , e ) 11 | a | divide(multiply(multiply(200, 16), 6), subtract(multiply(5, 16), multiply(5, 6))) | in a maths test , students were asked to find 5 / 16 of a certain number . one of the students by mistake found 5 / 6 th of that number and his answer was 200 more than the correct answer . find the number . | "explanation : let the number be x . 5 * x / 6 = 5 * x / 16 + 200 25 * x / 48 = 200 x = 384 answer : a" | a = 200 * 16
b = a * 6
c = 5 * 16
d = 5 * 6
e = c - d
f = b / e
|
a ) 80 , b ) 100 , c ) 120 , d ) 140 , e ) 160 | b | divide(500, multiply(18, const_0_2778)) | how many seconds does sandy take to cover a distance of 500 meters , if sandy runs at a speed of 18 km / hr ? | "18 km / hr = 18000 m / 3600 s = 5 m / s time = 500 / 5 = 100 seconds the answer is b ." | a = 18 * const_0_2778
b = 500 / a
|
a ) 5 days , b ) 15 days , c ) 28 days , d ) 6 days , e ) 7 days | a | inverse(add(multiply(10, inverse(multiply(24, 10))), add(multiply(inverse(multiply(12, 8)), 10), multiply(inverse(multiply(48, 4)), 4)))) | 8 men can do a piece of work in 12 days . 4 women can do it in 48 days and 10 children can do it in 24 days . in how many days can 10 men , 4 women and 18 children together complete the piece of work ? | "explanation : 1 man β s 1 day β s work = 1 / 8 Γ 12 = 1 / 96 10 men β s 1 day β s work = 1 Γ 10 / 96 = 5 / 48 1 woman β s 1 day β s work = 1 / 192 4 women β s 1 day β s work = 1 / 192 Γ 4 = 1 / 48 1 child β s 1 day β s work = 1 / 240 18 children β s 1 day β s work = 3 / 40 therefore , ( 10 men + 4 women + 18 children ) β s 1 day β s work = 5 / 48 + 1 / 48 + 3 / 40 = 1 / 5 the required no . of days = 5 days answer : option a" | a = 24 * 10
b = 1/(a)
c = 10 * b
d = 12 * 8
e = 1/(d)
f = e * 10
g = 48 * 4
h = 1/(g)
i = h * 4
j = f + i
k = c + j
l = 1/(k)
|
a ) 20 ft , b ) 25 ft , c ) 750 ft , d ) 900 ft , e ) 1050 ft | e | multiply(30, add(divide(multiply(30, divide(const_10, const_2)), const_3), divide(const_10, const_2))) | the circumference of the front wheel of a cart is 30 ft long and that of the back wheel is 35 ft long . what is the distance traveled by the cart , when the front wheel has done five more revolutions than the rear wheel ? | "point to note : both the wheels would have traveled the same distance . now consider , no . of revolutions made by back wheel as x , which implies that the number of revolutions made by the front wheel is ( x + 5 ) . equating the distance traveled by front wheel to back wheel : ( x + 5 ) * 30 = x * 35 . ( formula for calculating the distance traveled by each wheel is : # of revolutions * circumference . ) solving this eqn . gives x = 30 sub x = 30 either in ( x + 5 ) * 30 or in x * 35 to get the distance , which is 1050 . so the correct choice is e ." | a = 10 / 2
b = 30 * a
c = b / 3
d = 10 / 2
e = c + d
f = 30 * e
|
a ) 29 , b ) 27 , c ) 55 , d ) 17 , e ) 10 | c | divide(57, subtract(3, 2)) | a frog can climb up a well at 3 ft per min but due to slipperiness of the well , frog slips down 2 ft before it starts climbing the next minute . if the depth of the well is 57 ft , how much time will the frog take to reach the top ? | explanation : as per given , in 1 min , frog climbs up 3 ft and slips down by 2 ft . so the frog climbs only 1 ft in 1 min so after 54 mins , it would have climbed 54 ft . at the end of 55 mins it climbs up 3 ft to make it 57 ft and come out of the well . once it had reached the destination , it will not slip . so the frog will take only 55 minutes to climb up the well . answer : c | a = 3 - 2
b = 57 / a
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a ) 50 , b ) 100 , c ) 75 , d ) 25 , e ) 10 | e | divide(subtract(multiply(500, divide(5, const_100)), 20), divide(50, const_100)) | if 50 % of x is 20 less than 5 % of 500 , then x is ? | 50 % of x = x / 2 ; 5 % of 500 = 5 / 100 * 500 = 25 given that , x / 2 = 25 - 20 = > x / 2 = 5 = > x = 10 . answer : e | a = 5 / 100
b = 500 * a
c = b - 20
d = 50 / 100
e = c / d
|
['a ) 170', 'b ) 180', 'c ) 190', 'd ) 210', 'e ) 150'] | b | multiply(add(14, multiply(const_2, divide(3, const_10))), add(12, multiply(const_2, divide(3, const_10)))) | a rectangular room 14 m long , 12 m broad is surrounded by a varandah , 3 m wide . find the area of the varandah ? | area of varandah = ( l + b + 2 p ) 2 p = ( 14 + 12 + 6 ) 6 = 180 m ( power ) 2 answer is b . | a = 3 / 10
b = 2 * a
c = 14 + b
d = 3 / 10
e = 2 * d
f = 12 + e
g = c * f
|
a ) 9800 , b ) 9898 , c ) 9702 , d ) 9604 , e ) 9996 | e | multiply(add(const_100, const_2), 98) | calculate the largest 4 digit number which is exactly divisible by 98 ? | largest 4 digit number is 9999 after doing 9999 Γ· 98 we get remainder 3 hence largest 4 digit number exactly divisible by 98 = 9999 - 3 = 9996 e | a = 100 + 2
b = a * 98
|
a ) 120 , b ) 156 , c ) 108 , d ) 105 , e ) 86 | b | add(multiply(20, 6), multiply(multiply(6, 3), const_2)) | 20 6 years ago , a father was 3 times as old as his son . now the father is only twice as old as his son . then the sum of the present ages of the son and the father is : | let the present ages of the father and son be 2 x and x years respectively . then , ( 2 x - 26 ) = 3 ( x - 26 ) x = 52 . required sum = ( 2 x + x ) = 3 x = 156 years . answer is b | a = 20 * 6
b = 6 * 3
c = b * 2
d = a + c
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a ) rs . 17029 , b ) rs . 17028 , c ) rs . 17000 , d ) rs . 17008 , e ) rs . 17067 | c | divide(multiply(const_100, 10200), multiply(divide(multiply(subtract(11730, 10200), const_100), divide(10200, const_2)), const_2)) | the compound and the simple interests on a certain sum at the same rate of interest for two years are rs . 11730 and rs . 10200 respectively . find the sum ? | "the simple interest for the first year is 10200 / 2 is rs . 5100 and compound interest for first year also is rs . 5100 . the compound interest for second year on rs . 5100 for one year so rate of the interest = ( 100 * 1530 ) / ( 5100 * 1 ) = 30 % p . a . so p = ( 100 * 10200 ) / ( 30 * 2 ) = rs . 17000 answer : c" | a = 100 * 10200
b = 11730 - 10200
c = b * 100
d = 10200 / 2
e = c / d
f = e * 2
g = a / f
|
a ) 3 , b ) 6 , c ) 1 / 4 , d ) 1 / 3 , e ) 9 | c | divide(multiply(1, power(3, const_2)), power(6, const_2)) | x varies inversely as square of y . given that y = 3 for x = 1 . the value of x for y = 6 will be equal to : | "explanation : solution : given x = k / y ^ 2 , where k is constant . now , y = 3 and x = 1 gives k = 9 . . ' . x = 9 / y ^ 2 = > x = 9 / 6 ^ 2 = 1 / 4 answer : c" | a = 3 ** 2
b = 1 * a
c = 6 ** 2
d = b / c
|
a ) 22 , b ) 65 , c ) 18 , d ) 36 , e ) 10 | c | multiply(6, 3) | each child has 3 crayons and 12 apples . if there are 6 children , how many crayons are there in total ? | 3 * 6 = 18 . answer is c . | a = 6 * 3
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a ) 5.79 , b ) 5.77 , c ) 5.76 , d ) 5.72 , e ) 5.71 | c | multiply(divide(multiply(add(6, 1.2), subtract(6, 1.2)), add(add(6, 1.2), subtract(6, 1.2))), const_2) | a man can row 6 kmph in still water . when the river is running at 1.2 kmph , it takes him 1 hour to row to a place and black . what is the total distance traveled by the man ? | "m = 6 s = 1.2 ds = 7.2 us = 4.8 x / 7.2 + x / 4.8 = 1 x = 2.88 d = 2.88 * 2 = 5.76 . answer : c" | a = 6 + 1
b = 6 - 1
c = a * b
d = 6 + 1
e = 6 - 1
f = d + e
g = c / f
h = g * 2
|
a ) 50 , b ) 30 , c ) 125 , d ) 25 , e ) none of these | e | divide(18480, 660) | if the l . c . m of two numbers is 660 and their product is 18480 , find the h . c . f of the numbers . | h . c . f = ( product of the numbers ) / ( their l . c . m ) = 18480 / 660 = 28 . answer : e | a = 18480 / 660
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a ) 100 m , b ) 125 m , c ) 130 m , d ) 150 m , e ) 160 m | d | multiply(100, subtract(const_2, const_1)) | a train speeds past a pole in 15 seconds and a platform 100 m long in 25 seconds . its length is : | "explanation : let the length of the train be x meters and its speed be y m / sec . they , x / y = 15 = > y = x / 15 x + 100 / 25 = x / 15 x = 150 m . answer is d" | a = 2 - 1
b = 100 * a
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a ) 5 , b ) 10 , c ) 12 , d ) 15 , e ) 50 | d | add(divide(factorial(5), multiply(factorial(3), factorial(subtract(5, 3)))), divide(factorial(5), multiply(factorial(4), factorial(const_1)))) | show timer statistics to apply for the position of photographer at a local magazine , veronica needs to include 3 or 4 of her pictures in an envelope accompanying her application . if she has pre - selected 5 photos representative of her work , how many choices does she have to provide the photos for the magazine ? | veronica can choose 3 photos in 5 c 3 = 10 ways she can choose 4 photos in 5 c 4 = 5 ways total number of ways = 10 + 5 = 15 answer : d | a = math.factorial(5)
b = math.factorial(3)
c = 5 - 3
d = math.factorial(c)
e = b * d
f = a / e
g = math.factorial(5)
h = math.factorial(4)
i = math.factorial(1)
j = h * i
k = g / j
l = f + k
|
a ) 298 m , b ) 268 m , c ) 500 m , d ) 267 m , e ) 1000 m | e | divide(300, multiply(subtract(78, 1), const_0_2778)) | a train 300 m long is running at a speed of 78 km / hr . if it crosses a tunnel in 1 min , then the length of the tunnel is ? | "speed = 78 * 5 / 18 = 65 / 3 m / sec . time = 1 min = 60 sec . let the length of the train be x meters . then , ( 300 + x ) / 60 = 65 / 3 x = 1000 m . answer : e" | a = 78 - 1
b = a * const_0_2778
c = 300 / b
|
a ) 1 / 16 , b ) 1 / 19 , c ) 1 / 4 , d ) 1 / 3 , e ) none of these | b | divide(choose(const_5, const_2), choose(20, const_2)) | a consignment of 20 picture tubes contains 5 defectives . two tubes are selected one after the other at random . the probability that both are defective assuming that the first tube is not replaced before drawing the second , would be : | probability of drawing a defective tube in first draw is 5 / 20 probability of drawing a defective tube in second draw ( without replacing first tube ) is 4 / 19 . therefore probability of getting both defective is = ( 5 / 20 ) * ( 4 / 19 ) = 1 / 19 answer : b | a = math.comb(5, 2)
b = math.comb(20, 2)
c = a / b
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a ) 40 , b ) 12 , c ) 48 , d ) 51 , e ) 44 | b | add(45, 30) | two goods trains each 125 m long are running in opposite directions on parallel tracks . their speeds are 45 km / hr and 30 km / hr respectively . find the time taken by the slower train to pass the driver of the faster one ? | "relative speed = 45 + 30 = 75 km / hr . 75 * 5 / 18 = 125 / 6 m / sec . distance covered = 125 + 125 = 250 m . required time = 250 * 6 / 125 = 12 sec . answer : option b" | a = 45 + 30
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a ) $ 500 , b ) $ 700 , c ) $ 800 , d ) $ 600 , e ) $ 300 | b | add(600, multiply(500, divide(multiply(2, 10), const_100))) | a sum of $ 500 amounts to 600 in 2 years at simple interest . if the interest rate is increased by 10 % it would amount to how much ? | s . i = 600 - 500 = 100 p = $ 500 t = 2 years r = 100 * 100 / 500 * 2 = 10 % new rate = 10 + 10 = 20 % new s . i . = 500 * 20 * 2 / 100 = $ 200 new amount = 500 + 200 = $ 700 answer is b | a = 2 * 10
b = a / 100
c = 500 * b
d = 600 + c
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a ) 6 , b ) 7 , c ) 8 , d ) 9 , e ) 10 | a | divide(multiply(multiply(divide(778, divide(14, add(const_4, const_1))), 781), divide(775, add(const_4, const_1))), multiply(const_100, const_3)) | if p = 775 Γ 778 Γ 781 , what is the remainder when p is divided by 14 ? | "p 775 / 14 leaves a remainder 5 778 / 14 leaves a remainder 8 781 / 14 leaves a remainder 11 5 * 8 * 11 = 440 so the remainder will be the remainder of 440 / 14 which is 6 ans a" | a = 4 + 1
b = 14 / a
c = 778 / b
d = c * 781
e = 4 + 1
f = 775 / e
g = d * f
h = 100 * 3
i = g / h
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a ) 2 , b ) 8 , c ) 10 , d ) 12 , e ) 15 | a | divide(add(70, 25), 25) | the present ratio of students to teachers at a certain school is 70 to 1 . if the student enrollment were to increase by 50 students and the number of teachers were to increase by 5 , the ratio of students to teachers would then be 25 to 1 . what is the present number of teachers ? | "we are given that the ratio of students to teacher is 70 to 1 . we can rewrite this using variable multipliers . students : teachers = 70 x : x we are next given that student enrollment increases by 50 and the number of teachers increases by 5 . with this change the new ratio becomes 25 to 1 . we can put all this into an equation : students / teachers ο 25 / 1 = ( 70 x + 50 ) / ( x + 5 ) if we cross multiply we have : 25 ( x + 5 ) = 70 x + 50 25 x + 125 = 70 x + 50 1.667 = x x ~ 2 since x is the present number of teachers , currently there are 2 teachers . answer a ." | a = 70 + 25
b = a / 25
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a ) 16 , b ) 25 , c ) 19 , d ) 26 , e ) 17 | b | power(divide(9, subtract(const_2, divide(const_1, add(const_4, const_1)))), const_2) | find out the square of a number which when doubled exceeds its one fifth by 9 ? | a let the number be p , then the square will be p ^ 2 according to question : 2 p = ( p / 5 ) + 9 = > 10 p = p + 45 = > p = 5 p ^ 2 = 5 ^ 2 = 25 . answer : b | a = 4 + 1
b = 1 / a
c = 2 - b
d = 9 / c
e = d ** 2
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a ) 115 , b ) 116 , c ) 117 , d ) 78 , e ) 119 | d | add(add(multiply(divide(const_100, 45), 18), multiply(divide(50, 45), 18)), 18) | a sun is divided among x , y and z in such a way that for each rupee x gets , y gets 45 paisa and z gets 50 paisa . if the share of y is rs . 18 , what is the total amount ? | "x : y : z = 100 : 45 : 50 20 : 9 : 10 9 - - - 18 39 - - - ? = > 78 answer : d" | a = 100 / 45
b = a * 18
c = 50 / 45
d = c * 18
e = b + d
f = e + 18
|
a ) 36 , b ) 25 , c ) 48 , d ) 32 , e ) 56 | b | add(multiply(divide(2, multiply(30, 2)), 2), multiply(divide(2, multiply(30, 2)), 30)) | the l . c . m . of 2 numbers is 30 . the numbers are in the ratio 2 : 3 . find their sum ? | "let the numbers be 2 x and 3 x l . c . m . = 6 x 6 x = 30 x = 5 the numbers are = 10 and 15 required sum = 10 + 15 = 25 answer is b" | a = 30 * 2
b = 2 / a
c = b * 2
d = 30 * 2
e = 2 / d
f = e * 30
g = c + f
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a ) 15 days , b ) 1 / 15 days , c ) 1 day , d ) 30 days , e ) 225 days | a | inverse(inverse(15)) | if 15 horses eat 15 bags of gram in 15 days , in how many days will one horse eat one bag of grain ? | horse * days = bags 15 * 15 = 15 and 1 * days = 1 ( 15 * 15 ) / ( 1 * days ) = 15 / 1 days = 15 answer : a | a = 1/(15)
b = 1/(a)
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a ) 1 , b ) 3 , c ) 5 , d ) 7 , e ) 9 | b | subtract(5, 2) | in a certain game , each player scores either 2 points or 5 points . if n players score 2 points and m players score 5 points , and the total number of points scored is 50 , what is the least possible positive t difference between n and m ? | "we have equation 2 n + 5 m = 50 we have factor 2 in first number and we have factor 5 in second number . lcm ( 2 , 5 ) = 10 so we can try some numbers and we should start from 5 because it will be less list than for 2 2 * 5 = 10 and n should be equal 20 4 * 5 = 20 and n should be equal 15 6 * 5 = 30 and n should be equal 10 8 * 5 = 40 and n should be equal 5 10 * 5 = 50 and n should be equal 0 third variant give us the mininal difference n - m = 10 - 6 = 4 and there is some mistake in my way of thinking because we do n ' t have such answer ) if we change the task and will seek for difference between m and n than minimal result t will be 8 - 5 = 3 and answer b" | a = 5 - 2
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a ) 207 , b ) 209 , c ) 211 , d ) 212 , e ) 213 | d | subtract(multiply(207, const_10), add(add(add(add(add(add(add(add(201, 202), 204), 205), 206), 209), 209), 210), 212)) | if the average of 201 , 202 , 204 , 205 , 206 , 209 , 209 , 210 , 212 and x is 207 , what is the value of x ? | sum of the deviations of the numbers in the set from the mean is always zero 201 , 202 , 204 , 205 , 206 , 209 , 209 , 210 , 212 mean is 207 so the list is - 6 - 5 - 3 - 2 - 1 + 2 + 2 + 3 + 5 . . . this shud total to zero but this is - 5 , hence we need a number that is 5 more than the mean to get a + 5 and make it zero hence the answer is 207 + 5 = 212 d | a = 207 * 10
b = 201 + 202
c = b + 204
d = c + 205
e = d + 206
f = e + 209
g = f + 209
h = g + 210
i = h + 212
j = a - i
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a ) 5 , b ) 7 , c ) 10 , d ) 12 , e ) 14 | c | divide(20, subtract(const_4, const_2)) | in a group of cows and chickens , the number of legs was 20 more than twice the number of heads . the number of cows was : | "let the number of cows be x and their legs be 4 x . let the number of chicken be y and their legs be 2 x . total number of legs = 4 x + 2 y . total number of heads = x + y . the number of legs was 20 more than twice the number of heads . therefore , 2 Γ ( x + y ) + 20 = 4 x + 2 y . or , 2 x + 2 y + 20 = 4 x + 2 y . or , 2 x + 20 = 4 x [ subtracting 2 y from both sides ] . or , 20 = 4 x β 2 x [ subtracting 2 x from both sides ] . or , 20 = 2 x . or , x = 10 [ dividing by 2 on both sides ] . therefore , the number of cows = 10 . correct answer : c ) 10" | a = 4 - 2
b = 20 / a
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a ) 20 , b ) 21 , c ) 22 , d ) 23 , e ) 24 | b | divide(subtract(multiply(44, 4), add(add(add(25, 12), 15), 19)), subtract(44, 39)) | the average marks in mathematics scored by the pupils of a school at the public examination were 39 . if 4 of these pupils who actually scored 25 , 12 , 15 and 19 marks at the examination had not been sent up , the average marks for the school would have been 44 . find the number of pupils sent up for examination from the school ? | 39 x = 25 + 12 + 15 + 19 + ( x β 4 ) 44 39 x = 71 + 44 x - 176 5 x = 105 x = 21 answer : b | a = 44 * 4
b = 25 + 12
c = b + 15
d = c + 19
e = a - d
f = 44 - 39
g = e / f
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a ) $ 40 , b ) $ 35 , c ) $ 50 , d ) $ 65 , e ) $ 15 | b | subtract(50, multiply(divide(30, const_100), 50)) | amanda sees a sale for 30 % off all items , she sees a dress on sale that originally cost $ 50 . how much will it cost amanda to buy the dress after the sale amount of 30 % has been take off ? | final number = original number - 30 % ( original number ) = 50 - 30 % ( 50 ) = 50 - 15 = $ 35 . answer b | a = 30 / 100
b = a * 50
c = 50 - b
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a ) 24 % , b ) 20 % , c ) 30 % , d ) 36 % , e ) 40 % | b | multiply(divide(150, subtract(900, 150)), const_100) | a cricket bat is sold for $ 900 , making a profit of $ 150 . the profit percentage would be | 150 / ( 900 - 150 ) = 150 / 750 = 0.2 = 20 % . answer : b | a = 900 - 150
b = 150 / a
c = b * 100
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a ) 25 % , b ) 28.1 % , c ) 20 % , d ) 12.5 % , e ) 11.1 % | b | multiply(divide(multiply(divide(1, 3), subtract(1, divide(1, 4))), add(multiply(divide(1, 3), subtract(1, divide(1, 4))), subtract(1, divide(1, 3)))), const_100) | of the 3,600 employees of company x , 1 / 3 are clerical . if the clerical staff were to be reduced by 1 / 4 , what percent of the total number of the remaining employees would then be clerical ? | "welcome , just post the question and the choices let ' s see , the way i did it was 1 / 3 are clerical out of 3600 so 1200 are clerical 1200 reduced by 1 / 4 is 1200 * 1 / 4 so it reduced 300 people , so there is 900 clerical people left but since 300 people left , it also reduced from the total of 3600 so there are 3300 people total since 900 clerical left / 3200 people total answer : b" | a = 1 / 3
b = 1 / 4
c = 1 - b
d = a * c
e = 1 / 3
f = 1 / 4
g = 1 - f
h = e * g
i = 1 / 3
j = 1 - i
k = h + j
l = d / k
m = l * 100
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a ) rs . 40000 , b ) rs . 36000 , c ) rs . 50000 , d ) rs . 25000 , e ) none of these | d | divide(5000, subtract(const_1, add(add(divide(60, const_100), divide(10, const_100)), divide(10, const_100)))) | raja spends 60 % of his monthly income on household items , 10 % of his monthly income on buying cloths , 10 % of his monthly income on medicines and saves the remaining amount which is rs . 5000 . find his monthly income . | "savings 20 % - 5000 expenditure 80 % - 20000 total - 25000 answer : d" | a = 60 / 100
b = 10 / 100
c = a + b
d = 10 / 100
e = c + d
f = 1 - e
g = 5000 / f
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a ) 0.22085 , b ) 2.2085 , c ) 0.022085 , d ) 0.0022085 , e ) none of these | c | multiply(divide(22.085, 0.001), const_100) | 22.085 Γ£ β 0.001 = ? | "22.085 Γ£ β 0.001 = ? or , ? = 0.022085 answer c" | a = 22 / 85
b = a * 100
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a ) 1 , b ) 1.0001 , c ) 1.0021 , d ) 1.111 , e ) 1.1111 | e | multiply(divide(0.99999, 0.11111), const_100) | 0.99999 + 0.11111 = ? | "0.99999 + 0.00001 = 1 0.1111 + 1 = 1.1111 correct answer is e ." | a = 0 / 99999
b = a * 100
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a ) a . 480 , b ) b . 300 , c ) c . 600 , d ) d . 410 , e ) e . 240 | a | multiply(0.8, divide(1.08, divide(divide(divide(multiply(multiply(15, 12), 10), const_100), const_100), const_100))) | a certain company has records stored with a record storage firm in 15 - inch by 12 - inch by 10 - inch boxes . the boxes occupy 1.08 million cubic inches of space . if the company pays $ 0.8 per box per month for the record storage , what is the total amount that the company pays each month for record storage ? | "volume per box : 15 x 12 x 10 = 1,800 total volume : 1 , 080,000 number of boxes : total volume / volume per box = 1 , 080,000 / 1,800 = 600 price per month : number of boxes * price per box = 600 * 0.8 = 480 answer : a" | a = 15 * 12
b = a * 10
c = b / 100
d = c / 100
e = d / 100
f = 1 / 8
g = 0 * 8
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a ) 91.5 cm , b ) 92.2 cm , c ) 152.5 cm , d ) 29.2 cm , e ) 98.2 cm | c | multiply(multiply(const_2, divide(multiply(subtract(35, const_3), const_2), add(const_4, const_3))), 35) | the sector of a circle has radius of 35 cm and central angle 135 o . find its perimeter ? | "perimeter of the sector = length of the arc + 2 ( radius ) = ( 135 / 360 * 2 * 22 / 7 * 35 ) + 2 ( 35 ) = 82.5 + 70 = 152.5 cm answer : c" | a = 35 - 3
b = a * 2
c = 4 + 3
d = b / c
e = 2 * d
f = e * 35
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a ) 23 , b ) 19 , c ) 17 , d ) 13 , e ) 11 | d | subtract(divide(26, 2), const_1) | for every even positive integer m , f ( m ) represents the product of all even integers from 2 to m , inclusive . for example , f ( 12 ) = 2 x 4 x 6 x 8 x 10 x 12 . what is the greatest prime factor of f ( 26 ) ? | "f ( 26 ) = 2 * 4 * 6 * 8 * 10 * 12 * 14 * 16 * 18 * 20 * 22 * 24 * 26 the greatest prime factor in this list is 13 . the answer is d ." | a = 26 / 2
b = a - 1
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a ) 0 , b ) 1 , c ) 2 , d ) 3 , e ) 4 | b | subtract(divide(5, const_2), multiply(6, 6)) | what is the remainder when 6 ^ 381 is divided by 5 ? | i also agree that the remainder is ' 1 ' ( using the last digit of the powers of 7 ) . could we have the official answer please ? b | a = 5 / 2
b = 6 * 6
c = a - b
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a ) 82 m , b ) 50 m , c ) 72 m , d ) 150 m , e ) none of these | d | divide(multiply(divide(multiply(subtract(46, 36), const_1000), const_3600), 108), const_2) | two trains of equal length are running on parallel lines in the same directions at 46 km / hr . and 36 km / hr . the faster trains pass the slower train in 108 seconds . the length of each train is : | "explanation : the relative speed of train is 46 - 36 = 10 km / hr = ( 10 x 5 ) / 18 = 25 / 9 m / s 10 Γ 518 = 259 m / s in 108 secs the total distance traveled is 108 x 25 / 9 = 300 m . therefore the length of each train is = 300 / 2 = 150 m . answer d" | a = 46 - 36
b = a * 1000
c = b / 3600
d = c * 108
e = d / 2
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a ) 40 , b ) 50 , c ) 60 , d ) 70 , e ) 80 | a | divide(multiply(30, divide(40, const_100)), subtract(divide(80, const_100), divide(50, const_100))) | a team won 40 percent of its first 30 games in a particular season , and 80 percent of its remaining games . if the team won a total of 50 percent of its games that season , what was the total number of games that the team played ? | 50 % is 10 % - points above 40 % and 30 % - points below 80 % . thus the ratio of ` ` the first 30 games ' ' to ` ` remaining games ' ' is 3 : 1 . so the team played a total of 30 + 10 = 40 games . the answer is a . | a = 40 / 100
b = 30 * a
c = 80 / 100
d = 50 / 100
e = c - d
f = b / e
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a ) 77.14 cm , b ) 47.14 cm , c ) 84.92 cm , d ) 29.85 cm , e ) 23.57 cm | d | divide(circumface(divide(square_edge_by_perimeter(rectangle_perimeter(22, 16)), const_2)), const_2) | the parameter of a square is equal to the perimeter of a rectangle of length 22 cm and breadth 16 cm . find the circumference of a semicircle whose diameter is equal to the side of the square . ( round off your answer to two decimal places ) | "let the side of the square be a cm . parameter of the rectangle = 2 ( 22 + 16 ) = 76 cm parameter of the square = 76 cm i . e . 4 a = 76 a = 19 diameter of the semicircle = 19 cm circimference of the semicircle = 1 / 2 ( β ) ( 19 ) = 1 / 2 ( 22 / 7 ) ( 19 ) = 418 / 14 = 29.85 cm to two decimal places answer : d" | a = square_edge_by_perimeter / (
b = circumface / (
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a ) 1 : 5 , b ) 1 : 6 , c ) 1 : 7 , d ) 1 : 8 , e ) 1 : 9 | e | multiply(subtract(divide(5, 9), divide(const_1, const_2)), const_2) | the ratio of the arithmetic mean of two numbers to one of the numbers is 5 : 9 . what is the ratio of the smaller number to the larger number ? | "for two numbers , the arithmetic mean is the middle of the two numbers . the ratio of the mean to the larger number is 5 : 9 , thus the smaller number must have a ratio of 1 . the ratio of the smaller number to the larger number is 1 : 9 . the answer is e ." | a = 5 / 9
b = 1 / 2
c = a - b
d = c * 2
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a ) 1 , b ) - 2 , c ) 3 , d ) - 3 , e ) 4 | c | subtract(subtract(subtract(100, 10), add(100, 10)), 10) | if | 5 x - 10 | = 100 , then find the sum of the values of x ? | "| 5 x - 10 | = 100 5 x - 10 = 100 or 5 x - 10 = - 100 5 x = 110 or 5 x = - 90 x = 22 or x = - 18 sum = 21 - 18 = 3 answer is c" | a = 100 - 10
b = 100 + 10
c = a - b
d = c - 10
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a ) 31 days , b ) 25 days , c ) 27 days , d ) 29 days , e ) 19 days | c | multiply(add(const_1, const_2), 18) | a is twice as good a workman as b and together they finish a piece of work in 18 days . in how many days will a alone finish the work ? | "if a takes x days to do a work then b takes 2 x days to do the same work . - - > 1 / x + 1 / 2 x = 1 / 18 - - > 3 / 2 x = 1 / 18 - - > x = 27 days . hence , a alone can finish the work in 27 days . answer : c ." | a = 1 + 2
b = a * 18
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a ) 15 minutes . , b ) 30 minutes . , c ) 1 hour and 30 minutes . , d ) 1 hour and 40 minutes , e ) 2 hours | e | multiply(divide(subtract(const_1, add(divide(const_1, 2.5), divide(const_1, 5))), divide(const_1, 5)), const_60) | it takes avery 2.5 hours to build a brick wall while tom can do it in 5 hours . if the two start working together and after an hour avery leaves , how much time will it take tom to complete the wall on his own ? | "avery ' s efficiency is 100 / 2.5 = 40 % tom ' s = 100 / 5 = 20 % they worked together for 1 hour and finished 60 % of the job remaining = 40 % tom will complete 20 % in 60 minutes , 40 % in 120 minutes time taken by tom to finish the remaining on his own = 120 minutes answer : e" | a = 1 / 2
b = 1 / 5
c = a + b
d = 1 - c
e = 1 / 5
f = d / e
g = f * const_60
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a ) 6000 , b ) 4000 , c ) 2000 , d ) 600 , e ) 300 | d | multiply(divide(4000, add(4000, add(multiply(2000, 3), multiply(6000, 3)))), 4000) | a , b and c invests rs . 4000 , rs . 2000 and rs . 6000 in a business . after one year a removed his money ; b and c continued the business for two more year . if the net profit after 3 years be rs . 3300 , then a ' s share in the profit is ? | 4 * 12 : 2 * 36 : 4 * 36 2 : 3 : 6 2 / 11 * 3300 = 600 answer : d | a = 2000 * 3
b = 6000 * 3
c = a + b
d = 4000 + c
e = 4000 / d
f = e * 4000
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a ) 56 , b ) 65 , c ) 75 , d ) 89 , e ) 95 | e | subtract(multiply(82, const_4), subtract(multiply(81, const_4), add(3, subtract(multiply(82, const_4), multiply(80, const_3))))) | the avg weight of a , b & c is 80 kg . if d joins the group , the avg weight of the group becomes 82 kg . if another man e who weights is 3 kg more than d replaces a , then the avgof b , c , d & e becomes 81 kg . what is the weight of a ? | a + b + c = 3 * 80 = 240 a + b + c + d = 4 * 82 = 328 - - - - ( i ) so , d = 88 & e = 88 + 3 = 91 b + c + d + e = 81 * 4 = 324 - - - ( ii ) from eq . ( i ) & ( ii ) a - e = 328 β 324 = 4 a = e + 4 = 91 + 4 = 95 answer : e | a = 82 * 4
b = 81 * 4
c = 82 * 4
d = 80 * 3
e = c - d
f = 3 + e
g = b - f
h = a - g
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a ) 40 , b ) 15 , c ) 25 , d ) 32 , e ) 45 | c | divide(multiply(100, 150), multiply(20, 30)) | rectangular tile each of size 20 cm by 30 cm must be laid horizontally on a rectangular floor of size 100 cm by 150 cm , such that the tiles do not overlap and they are placed with edges jutting against each other on all edges . a tile can be placed in any orientation so long as its edges are parallel to the edges of floor . no tile should overshoot any edge of the floor . the maximum number of tiles that can be accommodated on the floor is : | "area of tile = 20 * 30 = 600 area of floor = 100 * 150 = 15000 no of tiles = 15000 / 600 = 25 so , the no of tile = 25 answer : c" | a = 100 * 150
b = 20 * 30
c = a / b
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a ) 10600 , b ) 11600 , c ) 12600 , d ) 10680 , e ) 20600 | a | divide(subtract(multiply(divide(5, const_100), 5000), 144), subtract(divide(5, const_100), divide(3, const_100))) | 5000 was divided into two parts such a way that when first part was invested at 3 % and the second at 5 % , the whole annual interest from both the investments is rs . 144 , how much was put at 3 % ? | "( x * 3 * 1 ) / 100 + [ ( 5000 - x ) * 5 * 1 ] / 100 = 144 3 x / 100 + ( 25000 β 5 x ) / 100 = 144 = > x = 10600 answer : a" | a = 5 / 100
b = a * 5000
c = b - 144
d = 5 / 100
e = 3 / 100
f = d - e
g = c / f
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a ) 78 % , b ) 250 % , c ) 500 / 3 % , d ) 125 % , e ) 60 % | a | multiply(add(const_1, divide(30, const_100)), subtract(const_100, 40)) | if x is 30 percent more than y and y is 40 percent less than z , then x is what percent of z ? | "z = 100 ; y = 60 so x = 78 x as % of z = 78 / 100 * 100 = > 78 % answer will be ( a )" | a = 30 / 100
b = 1 + a
c = 100 - 40
d = b * c
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a ) 150 , b ) 278 , c ) 179 , d ) 200 , e ) 191 | d | multiply(divide(multiply(60, const_1000), const_3600), 12) | a train running at the speed of 60 km / hr crosses a pole in 12 seconds . find the length of the train . | ": speed = 60 * ( 5 / 18 ) m / sec = 50 / 3 m / sec length of train ( distance ) = speed * time ( 50 / 3 ) * 12 = 200 meter answer : d" | a = 60 * 1000
b = a / 3600
c = b * 12
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a ) 40 % , b ) 50 % , c ) 65 % , d ) 70 % , e ) 80 % | c | multiply(const_100, add(divide(7, 20), divide(3, 10))) | if y > 0 , ( 7 y ) / 20 + ( 3 y ) / 10 is what percent of y ? | "can be reduced to 7 y / 20 + 6 y / 20 = 13 y / 20 = 65 % answer c" | a = 7 / 20
b = 3 / 10
c = a + b
d = 100 * c
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a ) $ 257615 , b ) $ 437615 , c ) $ 127615 , d ) $ 117615 , e ) $ 157615 | d | add(multiply(multiply(subtract(1, divide(21, const_100)), subtract(1, divide(21, const_100))), add(multiply(multiply(const_100, const_100), sqrt(const_100)), multiply(multiply(divide(sqrt(const_100), const_2), const_100), const_100))), multiply(multiply(add(21, const_2), const_100), sqrt(const_100))) | the value of a machine depreciates at 21 % per annum . if its present value is $ 1 , 50,000 , at what price should it be sold after two years such that a profit of $ 24,000 is made ? | "the value of the machine after two years = 0.79 * 0.79 * 1 , 50,000 = $ 96,000 sp such that a profit of $ 24,000 is made = 93,615 + 24,000 = $ 1 , 17,615 d" | a = 21 / 100
b = 1 - a
c = 21 / 100
d = 1 - c
e = b * d
f = 100 * 100
g = math.sqrt(100)
h = f * g
i = math.sqrt(100)
j = i / 2
k = j * 100
l = k * 100
m = h + l
n = e * m
o = 21 + 2
p = o * 100
q = math.sqrt(100)
r = p * q
s = n + r
|
a ) 96.9 , b ) 96.3 , c ) 91.2 , d ) 96.7 , e ) 96.21 | c | add(subtract(100, 9), divide(1, 5)) | what is the cp of rs 100 stock at 9 discount , with 1 / 5 % brokerage ? | "explanation : use the formula , cp = 100 Γ’ β¬ β discount + brokerage % cp = 100 - 9 + 1 / 5 91.2 thus the cp is rs 91.2 . answer : c" | a = 100 - 9
b = 1 / 5
c = a + b
|
a ) 500 , b ) 1,600 / 3 , c ) 1,000 , d ) 1,500 , e ) 2,000 | e | multiply(divide(8000, const_2), divide(const_1, const_2)) | the speeds of three asteroids were compared . asteroids x - 13 and y - 14 were observed for identical durations , while asteroid z - 15 was observed for 2 seconds longer . during its period of observation , asteroid y - 14 traveled three times the distance x - 13 traveled , and therefore y - 14 was found to be faster than x - 13 by 8000 kilometers per second . asteroid z - 15 had an identical speed as that of x - 13 , but because z - 15 was observed for a longer period , it traveled five times the distance x - 13 traveled during x - 13 ' s inspection . asteroid x - 13 traveled how many kilometers during its observation ? | x 13 : ( t , d , s ) y 14 : ( t , 3 d , s + 8000 mi / hour ) z 15 : ( t + 2 seconds , s , 5 d ) d = ? distance = speed * time x 13 : d = s * t x 14 : 3 d = ( s + 8000 ) * t = = = > 3 d = ts + 8000 t z 15 : 5 d = s * ( t + 2 t ) = = = > 5 d = st + 2 st = = = > 5 d - 2 st = st 3 d = 5 d - 2 st + 8000 t - 2 d = - 2 st + 8000 t 2 d = 2 st - 8000 t d = st - 4000 t x 13 : d = s * t st - 4000 t = s * t s - 4000 = s - 2000 = s i got to this point and could n ' t go any further . this seems like a problem where i can set up individual d = r * t formulas and solve but it appears that ' s not the case . for future reference how would i know not to waste my time setting up this problem in the aforementioned way ? thanks ! ! ! the distance of z 15 is equal to five times the distance of x 13 ( we established that x 13 is the baseline and thus , it ' s measurements are d , s , t ) s ( t + 2 ) = 5 ( s * t ) what clues would i have to know to set up the equation in this fashion ? is it because i am better off setting two identical distances together ? st + 2 s = 5 st t + 2 = 5 t 2 = 4 t t = 1 / 2 we are looking for distance ( d = s * t ) so we need to solve for speed now that we have time . speed y 14 - speed x 13 speed = d / t 3 d / t - d / t = 8000 ( remember , t is the same because both asteroids were observed for the same amount of time ) 2 d = 8000 2 = 4000 d = s * t d = 4000 * ( 1 / 2 ) d = 2000 answer : e | a = 8000 / 2
b = 1 / 2
c = a * b
|
a ) 51 , b ) 87 , c ) 35 , d ) 75 , e ) 68 | c | subtract(multiply(8.000001, 77), subtract(multiply(15, 75), multiply(68, 8))) | the average of 15 numbers is 75 . average of the first 8 of them is 68 and that of the last 8.00001 is 77 . find the 8.000001 th number ? | sum of all the 15 numbers = 15 * 75 = 1,125 sum of the first 8 of them = 8 * 68 = 544 sum of the last 8 of them = 8 * 77 = 616 so , the 8 th number = 544 + 616 - 1,125 = 35 . answer : c | a = 8 * 1
b = 15 * 75
c = 68 * 8
d = b - c
e = a - d
|
a ) 110 m , b ) 120 m , c ) 140 m , d ) 160 m , e ) 100 cm | e | divide(12, subtract(divide(12, 10), 5)) | a train covers a distance of 12 km in 10 minutes . if it takes 5 seconds to pass a telegraph post , then the length of the train is | "explanation : speed = 12 / 10 x 60 km / hr = 72 x 5 / 18 m / sec = 20 m / sec . length of the train = ( speed x time ) = ( 20 x 5 ) m = 100 m answer : option e" | a = 12 / 10
b = a - 5
c = 12 / b
|
a ) 4038 , b ) 8076 , c ) 9691.2 , d ) 4845.6 , e ) none of these | a | multiply(divide(divide(48456, divide(divide(multiply(subtract(const_100, 60), 50), const_100), const_100)), multiply(const_3, const_4)), divide(divide(multiply(subtract(const_100, 60), 50), const_100), const_100)) | mr yadav spends 60 % of his monthly salary on consumable items and 50 % of the remaining on clothes and transport . he saves the remaining amount . if his savings at the end of the year were 48456 , how much amount per month would he have spent on clothes and transport ? | β΅ amount , he have spent in 1 month on clothes transport = amount spent on saving per month β΅ amount , spent on clothes and transport = 48456 β 12 = 4038 answer a | a = 100 - 60
b = a * 50
c = b / 100
d = c / 100
e = 48456 / d
f = 3 * 4
g = e / f
h = 100 - 60
i = h * 50
j = i / 100
k = j / 100
l = g * k
|
a ) 500 , b ) 620 , c ) 550 , d ) 340 , e ) 440 | c | add(add(add(add(add(add(const_12, const_2), const_1), add(add(const_12, const_2), add(add(add(add(add(const_2, const_4), const_4), subtract(const_10, const_1)), add(add(const_2, const_4), const_4)), add(const_10, const_2)))), add(add(add(const_12, const_2), const_1), const_1)), 10), add(const_2, const_4)) | what is the sum of all the multiples of 10 between 0 and 100 ? | "the multiples of 10 between 0 and 100 are 10 , 20 , 30 , 40 , 50 , 60 , 70 , 80 , 90 and 100 . if these are all added together , the result is 550 . final answer : c" | a = 12 + 2
b = a + 1
c = 12 + 2
d = 2 + 4
e = d + 4
f = 10 - 1
g = e + f
h = 2 + 4
i = h + 4
j = g + i
k = 10 + 2
l = j + k
m = c + l
n = b + m
o = 12 + 2
p = o + 1
q = p + 1
r = n + q
s = r + 10
t = 2 + 4
u = s + t
|
a ) 50 , b ) 60 , c ) 70 , d ) 80 , e ) 90 | c | divide(336, add(add(divide(7, 5), divide(7, 5)), const_2)) | the length of a rectangular field is 7 / 5 its width . if the perimeter of the field is 336 meters , what is the width of the field ? | "let l be the length and w be the width . l = ( 7 / 5 ) w perimeter : 2 l + 2 w = 336 , 2 ( 7 / 5 ) w + 2 w = 336 solve the above equation to find : w = 70 m and l = 98 m . correct answer c ) 70" | a = 7 / 5
b = 7 / 5
c = a + b
d = c + 2
e = 336 / d
|
a ) 2660 , 1000 , b ) 3660 , 2000 , c ) 3000 , 4160 , d ) 2490 , 4150 , e ) 2475 , 4125 | e | multiply(divide(7.5, 12.5), divide(1650, subtract(const_1, divide(7.5, 12.5)))) | difference of two numbers is 1650 . if 7.5 % of the number is 12.5 % of the other number , find the number ? | "let the numbers be x and y . then , 7.5 % of x = 12.5 % of y x = 125 * y / 75 = 5 * y / 3 . now , x - y = 1650 5 * y / 3 β y = 1650 2 * y / 3 = 1650 y = [ ( 1650 * 3 ) / 2 ] = 2475 . one number = 2475 , second number = 5 * y / 3 = 4125 . answer e ." | a = 7 / 5
b = 7 / 5
c = 1 - b
d = 1650 / c
e = a * d
|
a ) 2 , b ) 4 , c ) 7 , d ) 8 , e ) 11 | e | subtract(subtract(subtract(140, 25), const_4), const_2) | a certain no . when divided by 140 leaves a remainder 25 , what is the remainder if the same no . be divided by 15 ? | "explanation : 140 + 25 = 165 / 15 = 11 ( remainder ) e" | a = 140 - 25
b = a - 4
c = b - 2
|
a ) 12 , b ) 200 , c ) 600 , d ) 800 , e ) 20 | a | divide(multiply(add(multiply(6, const_100), 00), add(multiply(8, const_100), 00)), power(divide(add(multiply(6, const_100), 00), power(const_2, const_4)), const_2)) | a room of 6 m 00 cm long and 8 m 00 cm broad is to be paved with square tiles . find the least number of square tiles required to cover the floor . | "explanation : area of the room = 600 * 800 sq cm size of largest square tile = h . c . f of 600 cm and 800 cm = 200 cm area of 1 tile = 200 * 200 sq cm no . of tiles required = ( 600 * 800 ) / ( 200 * 200 ) = 12 answer : a ) 12" | a = 6 * 100
b = a + 0
c = 8 * 100
d = c + 0
e = b * d
f = 6 * 100
g = f + 0
h = 2 ** 4
i = g / h
j = i ** 2
k = e / j
|
a ) a ) 1500 , b ) b ) 2000 , c ) c ) 2857 , d ) d ) 3000 , e ) e ) 3100 | c | divide(multiply(2, const_100), 0.07) | an inspector rejects 0.07 % of the meters as defective . how many will he examine to reject 2 ? | "let the number of meters to be examined be x then , 0.07 % of x = 2 ( 7 / 100 ) * ( ( 1 / 100 ) * x = 2 x = 2857 answer is c" | a = 2 * 100
b = a / 0
|
a ) rs . 147.50 , b ) rs . 785.50 , c ) rs . 176.50 , d ) rs . 258.50 , e ) none of these | c | divide(subtract(multiply(154, add(add(1, 1), 2)), add(126, 135)), 2) | teas worth rs . 126 per kg and rs . 135 per kg are mixed with a third variety in the ratio 1 : 1 : 2 . if the mixture is worth rs 154 per kg , the price of the third variety per kg will be ? | "explanation : since first and second varieties are mixed in equal proportions . so , their average price = rs . ( 126 + 135 ) / 2 . = > rs . 130.50 . so , the mixture is formed by mixing two varieties , one at rs . 130.50 per kg and the other at say , rs . x per kg in the ratio 2 : 2 , i . e . , 1 : 1 . we have to find x . by the rule of alligation , we have : cost of 1 kg cost of 1 kg of 1 st kind of 2 nd kind ( rs . 130.50 ) ( rs . x ) \ / mean price ( rs . 154 ) / \ x Γ’ Λ β 154 22.50 = > x Γ’ Λ β ( 154 / 22.50 ) = 1 . = > x Γ’ Λ β 154 = 22.50 . = > x = 176.50 rs . answer : c" | a = 1 + 1
b = a + 2
c = 154 * b
d = 126 + 135
e = c - d
f = e / 2
|
a ) 46 , b ) 47 , c ) 48 , d ) 49 , e ) 50 | e | subtract(60, divide(subtract(110, 60), 5)) | mother , her daughter and her grand child weighs 110 kg . daughter and her daughter ( child ) weighs 60 kg . child is 1 / 5 th of her grand mother . what is the age of the daughter ? | "mother + daughter + child = 110 kg daughter + child = 60 kg mother = 110 - 60 = 50 kg child = 1 / 5 th of mother = ( 1 / 5 ) * 50 = 10 kg so now daughter = 110 - ( mother + child ) = 110 - ( 50 + 10 ) = 50 kg answer : e" | a = 110 - 60
b = a / 5
c = 60 - b
|
a ) 5 / 13 , b ) 2 / 15 , c ) 1 / 12 , d ) 3 / 17 , e ) 6 / 19 | c | divide(subtract(10, multiply(const_2, const_3)), 10) | in a simultaneous throw of a pair of dice , find the probability of getting a total more than 10 | "total number of cases = 6 * 6 = 36 favourable cases = [ ( 5,6 ) ( 6,5 ) , ( 6,6 ) ] = 3 so probability = 3 / 36 = 1 / 12 answer is c" | a = 2 * 3
b = 10 - a
c = b / 10
|
a ) 5.5 kg , b ) 6 kg , c ) 30 kg , d ) 36.5 kg , e ) 71 kg | b | subtract(42, divide(subtract(add(multiply(30, subtract(42, add(30, 1))), 42), multiply(subtract(subtract(42, add(30, 1)), const_1), 30)), const_2)) | when a student joe , weighing 42 kg , joins a group of students whose average weight is 30 kg , the average weight goes up by 1 kg . subsequently , if two students , excluding joe , leave the group the average weight comes back to 30 kg . what is the difference between the average weight of the two students who left and the weight of joe ? | after two persons leave the group the average remains the same . that means the weight of the two persons = 42 + 30 = 72 so , the average the two persons = 36 that gives the answer 42 - 36 = 6 answer b | a = 30 + 1
b = 42 - a
c = 30 * b
d = c + 42
e = 30 + 1
f = 42 - e
g = f - 1
h = g * 30
i = d - h
j = i / 2
k = 42 - j
|
a ) 10 , b ) 12 , c ) 16 , d ) 22 , e ) 20 | e | multiply(divide(44, add(5, 6)), 5) | the sum of two numbers is 44 . 5 times one number is equal to 6 times the other . the smaller of the two numbers is | sol . let the numbers be x and ( 44 - x ) . then , 5 x = 6 ( 44 - x ) β 11 x = 264 β x = 24 . so , the numbers are 24 and 20 . answer e | a = 5 + 6
b = 44 / a
c = b * 5
|
a ) 4 , b ) 6 , c ) 10 , d ) 15 , e ) 20 | b | divide(multiply(4, 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 4 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 240 ( which is time in seconds . ) . thus the correct equation is : 60 * x - 20 * x = 240 - - - - > 40 x = 240 - - > x = 6 option b" | a = 4 * 3
b = a / 2
|
a ) 19 , b ) 12 , c ) 11 , d ) 30 , e ) 114 | d | multiply(add(add(const_1, const_2), const_3), 5) | a take twice as much time as b or thrice as much time to finish a piece of work . working together , they can finish the work in 5 days . b can do the work alone in ? | suppose a , b and c take x , x / 2 and x / 3 respectively to finish the work . then , ( 1 / x + 2 / x + 3 / x ) = 1 / 5 6 / x = 1 / 5 = > x = 30 so , b takes 15 hours to finish the work . answer : d | a = 1 + 2
b = a + 3
c = b * 5
|
['a ) 1', 'b ) 2', 'c ) 3', 'd ) 4', 'e ) 5'] | e | divide(multiply(const_1000, const_1000), multiply(200, const_1000)) | the mass of 1 cubic meter of a substance is 200 kg under certain conditions . what is the volume in cubic centimeters of 1 gram of this substance under these conditions ? ( 1 kg = 1,000 grams and 1 cubic meter = 1 , 000,000 cubic centimeters ) | 200 kg - 1 cubic meter ; 200,000 g - 1 cubic meter ; 200,000 g - 1 , 000,000 cubic centimeters ; 1 g - 1 , 000,000 / 200,000 = 10 / 2 = 5 cubic centimeters . answer : e . | a = 1000 * 1000
b = 200 * 1000
c = a / b
|
a ) 244 ft , b ) 88 ft , c ) 122 ft , d ) 178 ft , e ) 66 ft | d | add(multiply(divide(680, 8), const_2), 8) | a rectangular field is to be fenced on three sides leaving a side of 8 feet uncovered . if the area of the field is 680 sq . ft , how many feet of fencing will be required ? | "given that length and area , so we can find the breadth . length x breadth = area 8 x breadth = 680 breadth = 85 feet area to be fenced = 2 b + l = 2 ( 85 ) + 8 = 178 feet answer : d ) 178 ft" | a = 680 / 8
b = a * 2
c = b + 8
|
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